THEORY OF STRUCTURES
Copyright @ 1965 by McGraw-Hili, Inc. AU Rights Reserved. Copyright
1945 by McGra\v-Hill, Inc. All Rights Reserved. Printed in the United
States of America. l"his book, or parts thereof, may not be reproduced in
any form without pcrn...ission of the publishers.
Library of Congress Catalog Card NU11tber 64-7941
64868
II

Preface
This second edition of "Theory of Structures," like the first, is intended
primarily as a textbook for undergraduate and first-year graduate courses in
structural analysis for civil engineers. To serve this purpose, every effort
has been made to maintain a close connection bet\veen the methods that arc
developed for the analysis of various types of structures and the fundan1cntal
principles of mechanics on \vhich they are based. It is only through a sound
understanding of these principles that the engineer can successfully adapt his
methods of analysis to the ever-changing problems that will confront him in
this modern era.
The book ITIay be roughly divided into two parts: the first part dealing
\vith statically determinate structures, and the second part dealing \vith sta-
tically jndeterminate structures. On this plan, the first four chaptcrs deal
successively \\,ith a rcvic\v of statics (primarily graphic statics), statically
determinate plane trusses, influencc lines for beams and tr,usses, and statically
determinate space trusses. F ollo\\ring this, Chapters 5 and 6 treat the funda-
mental theorems relating to elastic systems and their applications to the cal-
culation of deflcctions of beams and trusses. In turn, there are chapters dealing
\vith the analysis of staticaJly indeterminate trusses, arches, and frames.
The final chapters arc devoted to an introduction to matrix methods in struc-
tural analysis, the analysis of sti tfened suspension bridges, and an introduction
to the dynamics of structures.
The first seven chapters in this second edition of "Theory of Structures"
remain essentially the same as in the first edition. Chapters 8 and 9, dealing
\vith arches and frames, have been completely re\\'ritten. The present treat-
ment of arches has been simplified b}' basing ir on the theorem of least v/ork
and using the concept of elastic center. Several articles on the analysis of
portal-type frames, using the elastic-ccnter concept, havc also been added.
In rewriting the chapter on the uses of slope-deflection equations in the analysis
of continuous beams and frames, \\'e have extended the treatment to include
systems \\rjth nonprismatic members and have included many exan1plcs of
this kind.
y

vi
PREFACE
Since the first appearance of this book (1945), tVlO ne\\' aspects of structural
analysis have beconle very inlportant, namely, the use of matrix methods of
fornlulating problcnls and the analysis of structurcs under dynamic loading.
"l'hesc are both very extcnsive subjects, and a number of complcte books on
each arc no\v available. Chapters 10 and 12 here arc intended only as intro-
ductions to these topics, but \ve hope that they \vill encourage the readcr to
continue his studies in these directions.
In the preparation of the first edition of this book, the senior author's
Russian book "Theory of Structures" (Leningrad, 1926) \vas extensively
used. Ackno\vledgmcnt is also due to Otto l\1ohr's "Abhandlugen aus dcm
Gebiete der technischen /lechanik" and to 1-1. MiiIlcr-Breslau's "Die graph-
ische Statik cler Baukonstruktionen." 'rhc authors also \vish to give special
thanks to Mr. P. Rabcevich of e\v York City for the use of a number of
examples and problems appearing in (:hapters 8 and 9 and to '1iss Rose
Marie Stanlpfcl and 1is Martha lcc Young for their careful typing of the
nc\\!' portions of the manuscript.
s. P. Ti11tOshmko
D. H. Young

1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.1
3.2
Contents
Preface v
chapter 1 ELEMENTS OF PLANE STATICS 1
CONCURRENT FORCES IN A J>LAF: 1
THREE FORCES 11' EQUILIBIUUIvI 5
EQGATIONS OF EQUILIBRIUM 10
INTER N AL FORCES 14
FUNICULAR POLYGON 19
APPLICATIONS OF THE FUNICULAR POLYGON 24
I"UNICULAR CURVES FOR DISTRIBUTED FORCE 2 9
FLEXIBLE SUSPESION CABLES 33
GRAPHICAL CONSTRUCTION OF BENDING-MOMENT DIAGRAMS
PIUNCIPLE OF VIRTUAL DISPJ.ACEIENTS 4)
39
chapter 2 STATICALLY DETERMINATE PLANE TRUSSES S2
S[lPLE TRUSSES 52
REAcrIONS 57
rvlETHOD OF JOINTS 62
rvIAX\\"ELL DIAGRAMS 65
1\.tETHOD OF SECTIONS 70
COMPOUND TRUSSES 77
GENERAL THEORY 01.' PLANE TRUSSRS 85
COivfPLRX TRUSSES: IIENNEBERG'S METHOD
METHOD OF VIRTUAL ()ISPLACE1.ENTS 98
92
chapter 3 INFLUENCE LINES IDS
tOVING LOADS AND INFLUENCE LINES
INFLUENCE LINES FOR BEAM REACrIONS
105
112
vII

vIII CONTENTS
3.3
3.4
3.5
3.6
3.7
3.8
4.1
4.2
4.3
4.4
4.5
4.6
4.7
5.1
5.2
5.3
5.4
5.5
5.6
5.7
6.1
6.2
6.3
6.4
6.5
7.1
7.2
INFLUENCE LINES FOR SHEARING FORCE 119
INFLl;ENCE LINES FOR RENDING i\10l\1ENT 127
GIRDERS \VITH FLOOR BEAMS 134
INFLUENCE LINES FOR THREE-HINGED ARCH RIBS
INFLGENCE LINES FOR SII\.lPLE TRUSSES 147
INFLUENCE LINES FOR CO!vIPOl;ND TRUSSES 1)4
140
chapter 4 STATICALLY DETERMINATE SPACE
STRUCTURES 161
COCUH.RF.NT FORCES IN SPACE /6/
SIlv1PLE SPACE THCSSES: l\-1ETHOD OF jOl:\'"TS
STATICALLY DETERl\1INATE CONSTRAIT OF A
RIGID BODY IN SPACE 176
COlvlPOUND SPACE TRUSSES: lvlETHOn OF SECTIOi':S
169
/c3
GEN£l{Al.. THEORY OF 5TATJCALf,Y I)ETERIINATE
SPACE TRUSSES 188
ANALYSIS OF COI\{PLEX SPACE TRUSSES 195
HENNEBRRG'S lvtETHOD 205
chapter 5 GENERAL THEOREMS RELATING TO ELASTIC
SYSTEMS 215
STRAIN El'ERGY IN TENSION, TORSION, AND BENDING
PRINCIPLE OF SUPERPOSITION 2/9
STRAIN ENERGY IN GENERALIZED FOR1\.l 223
CASTIGLIANO'S FIRST THEOREl\.1 229
CASTIGLIANO'S SECOND THEORErv1 234
TIJEORErvl OF LEJ'.ST \VORI( 241
THE RECIPROCAL THF..OREi\t 247
21$
chapter 6 DEFLECTION OF PIN-JOINTED TRUSSES 257
APPLICATIONS OF CASTIGLIANO'S THEOREl\1 257
MAX\"ELL-1\.tOHR METHOD OF CALCULATIG DEFLECrlONS
GRAPHICAL DETERrvllNATION OF TRUSS DEFLECrIONS 267
TvlETHOD OF FICTITIOVS LOADS 276
ALTERNATIVE J\.fETHOD OF FICTITIOl:S LOADS 285
263
chapter 7 STATICALLY INDETERMINATE PIN-JOINTED
TRUSSES 294
GENERAL CONSIDERATIONS 294
TRUSSES \VITH ONE REDUl'DANT ELE1\'fENT
297

CONTENTS
Ix
7.3 TRUSSES 'VITH SEVERAL REDU='DANT ME!vIBERS 303
7.4 ASSEMBLY AND THERMAL STRESSES IN STATICALLY
IDETERlINATE TRUSSES 310
7.5
7.6
INFLUENCE LINES FOR STATICALLY INDETERMINATE TRUSSES
STATICALLY INDETERMIJ';ATE SPACE STRUCTURES 325
316
chapter 8 ARCHES AND FRAMES 332
JNTH.ODUCrION 332
SYITvIETIUCAL T\\'O-IIINGED ARCHES 335
 41
SY11\1F:TRICAI. HINGELESS ARCHES .J
NUlvlERICAL CAI.CL:LATION OF REDUDANT ELEMENTS
FUNICULAR CURVE AS THE CENTER LINE OF AN ARCH
UNSYMj\JIETRICAL ARCHES 37/
FRAMES \\'lTH01JT lUNGES 38/
FRA1\1ES \VITH HINGES 390
EFFECTS OF TE1\'IPERATURE CHANGES AND
SUPPORT SETTLErvtENT ,94
H.I0 RINGS 398
H.l
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9. I
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
10. ]
10.2
10.3
10.4
10.5
10.6
351
358
chapter 9 CONTINUOUS BEAMS AND FRAMES 402
SLOPE-UEFLECTION EQUATIONS 402
BEAMS \VITII FIXED ENDS 408
CONTINUOUS BEAMS 412
BEAMS OF VARIABLE CROSS SECTION 421
CONTll'VOUS BEA1S OF VAlUABLE Cl{OSS SECTION
SIMPLE FRAMES 'VITH PRISIvlATIC ME1\.IRERS 441
CONTINUOU') }RA1vlES \\TflI PRIS!\.1ATIC METvtBERS
MOMRNT-DISTRIB{;TION '1ETHOD 460
ANALYSIS OF BUILDING FRAMES 469
FRAMES ""ITH NONPRISlvlATIC 1EMBERS 475
43/
451
chapter 10 MATRIX METHODS IN STRUCTURAL
ANAL YSIS 480
FORCE AND DEFOR"A TION METIIODS
ELF:lvIENTS OF MATRIX AI.GEBRA
480
484
APPLICATION OF l\1ATRIX IETHODS TO PLANE TRUSSES
49/
I\fATRIX ANALYSIS OF CONTINUOUS BEArvlS
lvlATRIX TRATf\.1ENT OF ARCHES AND FRAMES
IvIATRIX ANALYSIS OF CONTINUOUS FRAMES
500
510
516

x
CONTENTS
chapter 11 SUSPENSION BRIDGES 523
11.1 PARABOLIC FUNICULAR CURVE 5'23
11.2 DEFLECflONS Of' UNSTIFFENED SUSPENSION BRIDGES )25
11.3 FUNDAMENTAL EQUATIONS FUR STIFFENED
SUSPENSION BRIDGES ) 33
11.4 ANALYSIS OF STIFFENING TRUSSES 538
11.5 APPLICATION OF TRIGONOMETRIC SERIES IN
CALCULATING DEFLECTIONS 543
11.6 THREE-SPAN SUSPENSION BRIDGES 'VITH SIMPLY SUPPORTED
STIFFENING TRUSSES 547
11.7 THREE-SPAN SUSPENSION BRIDGE \VITH CONTINUOUS
STIFFENING TRUSS 549
lIFFENING TRUSS OF VARIABLE CHUSS SECTION
11.8
12.1
12.2
12.3
12.4
] 2.5
12.6
12.7
12.8
12.9
12.10
556
chapter 12 STRUCTURAL DYNAMICS 562
FREE VIBRATIONS: ONE DEGREE OF FREEDOM
RAYLIGH'S METHOD 568
FORCED VIBRATIONS: STEADY STAT
GENERAL CASE OJ.' A DISTURBING FORCE
NUMERICAL INTEGRATION 588
GRAPHICAL INTEGRATION $94
STATICAL AND nYNAMIC STRESSES IN RAILS
LATERAL VIBRATIONS OF PRISMATIC BEAMS
VIBRATION OF BIUDGES 613
STRUCTURES SUBJECTED TO EARTHQUAKES
562
575
581
601
608
617
Name index
Subject index
625
627

Chapter 1
Elements of plane statics
1.1 CONCURRENT FORCES IN A PLANE
The theory of strUC/IITrS j s based to a large extent upon the principles of statics
\"ith \vhich the reader is assumed to be familiar. Ho\\'cvcr, \.ve shall revie\v
here some pares of statics that arc n10st useful in the analysis of engineering
structures. '/Ve begin \vith the principle of the parallclogranl of forces as
fo1l0\\'5: 17.1)0 forteJ PI and P2, aJ represented by the vectors ITA and OB in
Fig. 1.1 a, are equivalel1t i11 actio11 to a single resultallt force R obtained as the
diagonal 0(' of the parallelogrt1111forl11ed on the given vectors as show11. The same
resultant force can be obtained also from the triangle of forces shown in Fig.
I. I b. lhis foIlo\vs from the fact that the triangle A BC in Fig. 1.1 b is identical
\virh the triangle (JAC in Fig. l.Ia.
If several forces in a plane act at a single point 0 (J."ig. 1.2a), they can always
be reduced to one resultant force \vhich also acts through that point. This
resultant force {.:an be found by successive applications of the parallelogram of
1

2
ELEMENTS OF PLANE STATICS
o
A
B
\
,
\
,
\
FIG. 1.1
(a)
c
o
A
B
R
(b)
c
P3
/
/
/
/ /
/ / I /
/ / P2 , "
/ / // R"
/ ./ I /
j_____l"/
R
FIG.1.2 (a)
;'.:
B
8 2
o
FI G. 1.3
(a)
(b)
P 3
x
JV
8 2
(b)
forces as illustrated in Fig. ] .2a or as the closing side of the polygon of forces
constructed as shown in Fig. 1.2b.
If the polygon of forces is closed, the resultant force vanishes and the given
forces arc in equilib1'itnl1. "hus, if sevcral concurrcnt forccs in a plane arc
kno\vn to be in equilibrium, their frce vectors rnust build a cloJcd polygon oj
forces.
The above graphical condition of equilibrilun is vcry useful in the analysis
of structures. Let us consider, for example, the vcry simple casc of a \veight
W supportcd by strings as sho\vn in Fig. 1.311. Isolating .thc ring () as 'J .free
body, \\'c conclude that it is in equilibrium under the action of three forces 
Sit and S2' the lines of action of \vhich coincide \vith the strings and the mag-
nitudes of \vhich represent the axial forces, or tel1si01ls, in these strings. If thc

ARTICLE 1.1
3
magnitude of the \veight HI is kno\Jln, the magnitudes of the tensions S. and S2
are found from the closcd triangle of forces as sho\vn in Fig. 1.3 b, froIn \vhich
\ve obtain S1 = f' csc ex and S2 = J1! cot a.
From the fact that sevcral concurrent, coplanar forces in equilibrium build
a closed polygon, it follo\vs that the algebraic sums of projections of the forces
on any system of orthogonal axes x and)' in their plane of action must be zero.
lhus \VC arrive at the fanliliar equatiOl/S of equilibrilllll
};X j = 0
}; l:. = 0
(1.1)
\\There LY i and ):. denote, respectively, the projections of any force Pi on the
axes x and y, and the sumnlations arc understood to include all forces in the
systcn1. 11hese analytical conditions of equilibrium are equivalent to the
graphical condition of a closed polygon of forces, but they are sometimes more
convenient to use. Applying the equations of equilibrium to the system shO\\fn
in Fig. 1.311, for example, ,ve obtain
S1 COS a - S2 = 0
S1 sin ex - I,V = 0
fronl \vhich \ve find, as before, 51 = lJ' csc a and S2 = Jtv cot a.
'1 he foregoing graphical and analytical conditions of equilibrium are par-
ticularly useful in the analysis of pin-connected trusses. Consider, for
example, the truss loaded as sho\\'n in Fig. I .411. The analysis of such a truss
entails finding the axial forces induced in the various bars by the action of the
external load P. 'rhe \vork can be greatly simplified in this case by noting
that some of the bars are ill(ll:th.y" i .c., unstressed . For instance, if \VC isolate
the hingc /1 as a free body (Fig. 1.4b), \\'e see at once that the bars 1 and 2 of
the truss are inactive, since t\\'O forces can be in equilibrium only if they are
collinear in action, and the axes of these t\VO bars arc not collinear. I laving
concluded that the bar 2 is inactive, \ve consider the equilibrium of the hinge B
(Fig. 1.4c), \vhere \VC find the possibility of three forces in equilibrium, t\VO of
\\'hich are collinear in action. Then from the first of Eqs. (1.1) it follo\\'5
E 1
2A
(b)
8 3
A 4 B
D B
FIG. 1.4 (a) p (c) p

4
ELEMENTS OF PLANE STATICS
that the force :n bar 4 must be zero, and \ve conclude that this bar also is
inactive. finally, then, only the bars sho\vn in the figure by heavy lincs carry
axial forces di:fcrenr from zero. Considering, further, the equilibrium of
hinge B and using the second of Eqs. (1.1), \\'e conclude that the bar 3 carries
a tension equal to the load P. "The axial forces in the bars 5 and 6 can be found
by considering the conditions of equilibrium for the hinge C, and this completes
the analysis of the truss.
PROBLEMS
1 Find the tensile force 8 1 in the Inenlber A C and the conlpressive force 52 in the
nlcmber AB of the simple truss sho\\'o in Fig. 1.5.
A1/s, SI = P csc a, S = P cot a.
2 To produce biaxial c0J11pression of a concrete cube AI, the system of hinged bars
sho\vn in Fig. 1.6 is used. Find the compressive forces exerted on the faces of the
cube j f the frame has the fOCln of a square and the inclined bars lie along its diagonals.
Ans. S = V2 P.
3 Idcnti fy, by inspection, the inactive nlembcrs of the truss sho\vn in Fig. 1.7, (a)
\vhen there is a vertical load P e1t F, (b) \vhen the same load is at fJ.
4 Ho\v is the acrion of the simple truss sho\vn in Fig. 1.8 affected by changing the
direction of the diagonal from AD as sho\vn to BC?
D
c
c
p
A P
B 2
P A B
FI G. 1.5 FIG. 1.6
D
C
P
FIG. 1.7
FIG. 1.8

ARTICLE 1.2
5
FIG. 1.9
(a)
p
(b)
5 Distinguish by heavy lines the active members of the t\VO trusses supported and
loaded as sho\vn in Fig. 1.9.
1.2 THREE FORCES IN EQUILIBRIUM
Three nonparallel forces in a planc can be in equilibrium only if their lines of
action meet in one point. To prove this statelnent, \ve refer to Fig. 1.10,
\vhere P and Q arc any t\VO forces that intersect at a point O. l"hen a third
force S can hold thesc t\VO forces in equilibrium on]y if it is equal, opposite,
and collinear \vith their resultant R, \vhich, of course, acts through point O.
Hence, the force S also must act through point 0 as sho\vn.
l"he foregoing theOre11"l of three forces is very useful in the determination of the
reactions induced at the points of support of a body or structure under the action
of given forces. Consider, for example, the cranc sho\vn in Fig. 1.11, the mast
of \vhich is supported in bearings at A and B so that the crane can be rotated
around the vertical axis. Under the action of a vertical load P, the crane exerts
pressures on its bearings at A and B, and these actions of the crane on its supports
inducc equal and opposite reactiolls on the cranc as sho\vn. l"hus tbe vcrtical
force P together \\,ith the reactions Ra and Rb arc three forces in equilibrium.
Neglecting friction in the bearing at B, the reaction Rb must be a horizontal
force, and hence the kno\vn lines of action of t\VO of the forces (P and R b )
determinc the point of concurrence 0 of the system. The third force ROo, then,
must also pass through point 0; and so, by the theorem of three forces, its line
of action AO is established. Kno\ving the magnitude of the force P, the mag-
FIG. 1.10
s

6 ELEMENTS OF PLANE STATICS
 12'---1
Rb B 0
---------71-
I,
I 1
I
I I
I
C
P
16'
I' RIJ
I
I
I
I
I
I P
I
A I
FIG. 1.11 (a) (b)
nitudes of the reactions Ra and Rb are found from the closed triangle of forces
sho\\'n in F'ig. 1. lIb. Since this triangle is similar to triangle BA() in Fig.
l.lla, \ve obtain Ra = iP and Rb = iP.
As a second example of the application of the theorem of three forces in
calculating reactions, let us find the axial forces induced in the hinged bars 1, 2,
3, \vhich support a horizontal beam A B under the action of an applied force P,
as sho\vn in :Fig. l.I2a. Replacing the supporting bars by the reactions SI,
S2, S3, \vhich they exert on the beam by virtue of the axial forces induced in
them, \ve find ourselves \vith a system of four forces (P, Sl, 8 2 , 8 3 ) in a plane
that are in equilibrium. To reduce this system to the case of three forces in
equilibrium, Vie imagine that Sl and S2 are replaced temporarily by their
0 a
- -'11 C
--- - I
R -- I '
 -- / I
I I
I I 8 1
I
I
I 8 3
I
A B
8 8
b
(a) (b)
FIG. 1.12

ARTICLE 1.2
7
a
p
c
(a)
FIG. 1.13
b
resultant R, as yet unkno\vn in direction but evidently acting through point ])
obtained as the jntersection of the kno\vn lines of action of Sl and 52. No\\!,
instead of four forces, \ve have the three forces (P, Sa, and R) that are in
equilibrium. The poine of concurrence of this system of thrce forces is
evidently point 0, obtained as the intersection of the kno\\'n lines of action of P
and S31 Thus, the line of action ])0 of the third force R is established as
sho\vn, and the closed triangle of forces ((be (Fig. I. I 2b) can be constructed.
From this construction, .the magnitudes of S3 and R arc determined. Finally,
kno\ving the forcc R, as reprcsented by the vector ca , \ve find the magnitudes
of 52 and 51 by resolving R into the t\VO components td and da having the
kno\vn directions of S2 and S1, respectively; and, if all constructions have been
made to scale, the thrce rcactions are determined. F con1 the directions of the
arro\\'s on the vectors in Fig. 1.12b, \\'e conclude that the bars I and 3 are
under compression \vhile the bar 2 carries tension.
A common type of engineering structure is the three-hinged arch sho\vn in
:Fig. 1.1 3((. l"he reactions at the points of support A and B of such a structurc
under the action of a load P acting as sho\\'n can be found by using the theorem
of three forces. By virtue of the hinge at the crown C, \VC conclude that the
reaction Rb must act along the line BC, \\,hich intersects the kno\vn line of
action of the force P at point O. Thus 0 is the point of concurrence of the
system, and the line of action A 0 of Ru is established. The magnitudes of the
reactions are now found from the closed triangle of forces shawn in Fig. 1.13b.
If, in addition to the load P, there is a load Q on the rib C'B, thc same proce-
dure can be used. We find first the reactions at A and B due ta the load P
alone, as above. Then repeating the same procedure, \VC find the reactions due
to the load Q alone. Thus, \VC shall have t\VO reactions at A and t\\'o at B.
lhc resultants of these t\VO forces at each point of support are the cesired
reactions due to the simultaneous action of P and Q. *
As a last example, let us consider the COll1POU'lld beal11 consisting of two
* A Inorc expedient method of handling several forces on the arch is discusstd on page 27.

8
ELEMENTS OF PLANE STATICS
d
F
.........
I '\ ........ 8 1
I '\ .......
I '\
'\
AI '\ B
c
"
'\..
8 1 , 8 4 a
"-
,
'\.. P
\ b
',\ ,
'\..\ I

FIG. 1.14 (a) n
e
portions AC and BC hinged together at C and supported by four hinged bars
as sho\\'o in Fig. I.l4a. 10 find the axial forces Sl, S2, S3, and S, induced
in the four supporting bars by the action of a load P, \\'C begin \,"ith a con-
sideration of the cquilibrium of the beam CB, which is actcd upon by only three
forces. Thesc forces arc the t\VO reactions Sa and 5'" togethcr \vith a force at
the hinge C that represcnts the action of the beam AC on the beam CB. The
kno\\'n lines of action of S3 and 54 dctcrnline the point of concurrence ]J of
these three forces and hence the line of action ('D of the force on the hinge C.
The equal and opposite reaction exerted by the beam CB on the bcam AC has
the same line of action. Hence, thc beam AC' is in the same condition as if it
had, in addition to the supporting bars 1 and 2, a third supporting bar along the
line CD. With this conclusion in mind, \\'c see that, in considering the
equilibrium of the beam AC', we may procecd in exactly the same manncr as
\ve did for the beam in Fig. 1.12. Thus, thc line of action FE of the rcsultant
of Sl and S2 is established, and the closed triangle of forces abc in Fig. t .14b
can b constructed. The remainder of the solution reduces simply to the
resolution of the vectors en and he into their respective components 51, 51 2 and
S4, S3, as shown in Fig. t.14b. Bars 1 and 3 are in compression while 2 and 4
. .
are In tensIon.
PROBLEMS
Using the theorem of three forces, find the reactions induced at the points of sup-
port A and B of the sin1ple beam supported and loaded as sho\\'n in Fig. 1.15.
Ans. Ra = 0.882P, Rb = 0.667 P.
2 Find, graphically, the reactions at A and R for the beam shown in Fig. 1.16.
Ails. Ra = 2.24P, Rb = 2.82P.

ARTIC LE 1.2
9
C
A B C
I
2
A
I D
P
FIG. 1.15 FIG. 1.16
B
T
7.2'
1 A
A o 1. 5.4' -I
FIG. 1.17 FIG. 1.18
3 Find, graphicaJly, rhc reactions at A and B for the hearT) sho\vn in Fig. 1.17.
AlIs. R.a = P, lb = I .41 P.
4 .\ priSIl1atic har A B of \"eight H' rests on a horizontal Hoor at .4 and against a
vertical \\7a11 at B and is kept frolll falling by a string ()J), as sho,vn in Fig. 1.18.
Neglecting friction at the points of support, find, graphically, the reactions at A
and R and the tension in the string OD.
S Find, graphically, the reactions Ra and Rb .induced at the points of support of the
con1pound structure loaded as sho\vn in Fig. t .19.
Hint The reaction at B nl\.1Sr be a horizontal force.
FIG. 1.19
.
\0 '
50'

10 ELEMENTS OF PLANE STATICS
p1
1
II --A
- - - ./
-- --(----
- b a.......
.......

P4
A
FIG. 1.20
E
(a)
(b)
1.3 EQUATIONS OF EQUILIBRIUM
In general, a system of coplanar forces, the lines of action of \\,'hich do t10t
intersect in one point, ITIay rcduce to (I) 11 resultant force, (2) a resultant couple,
or (3) a state of cquilibri'l4111.
If the givcn forces PI, P2, P 3 , /)4 arc slIch that their free vectors build an
ullclosed polygon as sho\vn in Fig. 1.20b, the systenl \vill ah\'ays rcduce to a
resultant force. The rnagnitude and direction of this resultant arc givcn by
the cloJil1g side AE of the polygon of forces, and its line of action is found by
the construction illustrated in Fig. ] .20a.
If the given forces arc such that their free vectors build a closed polygon
(Fig. 1.21h), the resultant force vanishes, but thcrc is still the possibility of a
,'.
\
A
\
\
r.
\
\
\
\ d
 )
P5/
/
/
;:
B
E
D
(a)
(b)
FIG. 1.21

ARTICLE 1.3
u
resultant couple. In this case, \VC arbitrarily divide the forces into t\VO groups
PI, P2, P3 and P4, J'r" P6, having equal and opposite resultants Rl and R 2 as
represented by thc vectors 7f[j and [)/1 in Fig. 1.21 b. l-hen, proceeding \vith
each of these groups in the same manner as in the preceding case, \ve find the
lines of action be and ef of their resultants as sho\vn in Fig. 1.2Ia, and these t\VO
equal and opposite forces rcpresent a resultant couple.
In the particular case, \vhcrc the lines of action of the forces R 1 and R'}. in
Fig. 1.21a are found to coincide, the resultant couple vanishcs, and the system
of forces is in cquilibrium. Thus, the conditions of equilibrium for any sys-
tem of forces in a plane arc a closed polygon of forces and a vanishing resultant
couple. From the condition of a closed polygon of forces, \ve conclude that
the algebraic sums of projections of the given forces on any pair of orthogonal
axes must bc zero. From the condition of a vanishing resultant couple, \\'e
concludc, with the help of Varignon's thcorcrTI,l that the algebraic sum of
moments of the given forces \vith respect to any center in their plane must be
zero. These conditions arc expressed analytically by the folIo\ving equation...
of equi/ihrilnl1:
..Xi = 0
Yi = 0
};J\;1 i = 0
(1.2(1)
whcre "'\'i and  arc orthogonal projcctions of any force Pi, and I\J i is the
ITIOment of the same force \vith respect to a choscn ccntcr. "I'he sunlmations
arc understood to include all forces of the system.
It can easily bc sho\vn that the abovc conditions of equilibriwn can also bc
expressed by the three momcnt equations 2
2;AJ A = 0
};M n = 0
2;,\;[ c = 0
(1.2b)
in \\'hich A, B, and ( are three arbitrary ccnters that form a triangle in the
plane of action of the forces.
Equations (1.2) find a \vide application in thc theory of structures, par-
ticularly in the determination of reactions and in the analysis of trusses. As
a first cxanlplc, let us consider the dctermination of the reactions induccd at
the points of support of the simple crane ABC' loaded as sho\vn in Fig. 1.22.
If \\,'C neglect friction in the guide at B, the supports of the crane can be rcplaced
by reactions R b , )(a, and 1, as sho\vn, \vhere Xa and  arc horizontal and
vertical components of thc unkno\vn rcaction at A. The system of forces as
shown is in equilibrium; and, if point A is uscd as a moment center, Eqs. (1.2a)
1 Th morncnt of the resultant \\,irh respect to any center in the plane of action of the forces
is equal to the algebraic sUin of the rnO(l1cnts of the given forces \virh respect to the sanlC
center. See S. 'rirnoshenko and I). H. Young, "Engineering ivlechanics," 4th ed., p. 4L
McGra\v-HiH Book C:ompany, New York, }956.
2 See ibid., p. 109.

12
ELEMENTS OF PLANE STATICS
Rb
B
(j
E'
A C
,X a
l, Q
FIG. 1.22 FIG. 1.23
become
Xa - Rb = 0
l:,-P-Q=O
Rbc - Pa - Qb = 0
From these equations, \\le find Xa = Rb = Pal c + Qb/ c and
}Ta=P+Q
i\S a second exanlple, let us consider the detcrmination of the axial forces
induced in the bars I, 2, and 3 of the sjnlpJe tfUSS loaded and supported as
sho\\!n in Fig. 1.23. Considering only the shadcd portion HC'D of the truss
and replacing the bars 1, 2, and 3 by the reactions that they exert upon this
portion, \ve obtain the free-body diagran1 as sht)\vn. Here \ve have t1ve forces
in cquilibriunl, and the algebraic Slun of their moments \vi th respect to any'
ccntcr in the plane of the truss ITIUst be zero. l)pon choosing, successively,
the points A, D, and C as moment centcrs, Eqs. (1.2b) becolne
S}(l - Qa - 2Qa = 0
S 3tl (
V3 - Qa = 0
Q'1 - Sv 1 = 0
and \VC find 51 = 3Q (tension), Sa = V3 Q (coll1pression), and 8 2 = Q
(compression). It should be noted that the nlonlcnt centers are chosen so as
to give us only one unkno\vn in each equation.

ARTICLE 1.3
13
PROBLEMS
1\ prisrnaric bar Af: of negligiblc \veight is hinged to a vertical \vall at A and sup-
portcd by a strut RD. as ShOVlO in Fig. 1.24. Find rhe reactions induced at A and
lJ through the action of a sphere of \vcight Q tied by a string Cf'that is parallel to
AI. l-hc radius of the sphere is such that the points Band C arc at the sante level.
/11/J. Ru = O.65Q, Rb = 1.52Q.
2 Find the axial forces induced in the supporting bars I, 2, and 3 of the sirnp)e truss
sho\vn in Fig. .1.25.
Ans. Sl = +S.33P, 8 2 = +2.01P, 53 = -7.12P.
3 Evaluate the horizontal and vertical componenrs of the reactions induced at the
supports A and B of the con1pound structure sho\\'" in Fig. 1.26.
Ans. X rs = 0.167 P, X b = 0.833]>, Y a = 0.125P, Y b = 1.125P.
4- Dctcnnine rhe tension induced in the tic bar AR of the plane structure 5upported
and loaded as sho\vn in Fig. 1.27.
Ans. S = P12.
5 In Fig. 1.24, ho\v \vill the reactions at A and B be changed j f the inclined string
CF is replaced by a hodzontal srring CB?
p
p
p
p
D
T
12'

FIG. 1.24
FIG. 1.25
p
c
p
p

I

FIG. 1.26
FIG. 1.27

14 ELEMENTS OF PLANE STATICS
1.4 INTERNAL FORCES
In previous articles, vie have discussed the dctcrrnination of external reactions
induced at the supports of a rigid body constrained and loaded in one plane.
V\'c shaH now consider the internal forces, or stresses, that are set up in such a
.
constrained body under the action of applied external forces. Consider, for
example, the rigid body supported and loadcd in one plane as sho\vn in Fig.
1.28a. Under the action of the applied loads, reactions Ra and Rb arc indDced
at the supports, and these reactive forces are evaluated in the usual manner
by considering the conditions of equilibrium of the entire body. l'hus, the
system of forces extcrnaJ to the body as a \\'hoJe is completcJy defined. No\v
imagine that an arbitrary plane section 11111 dividcs the body into t\VO parts /
and K as sho\vn. It is evident that forces of internal constraint Inust exist
between these t\VO parts of the body to hold them togethcr. Such intcrnal
forces, of course, al\\-'ays occur in pairs of equal and opposite forces at each
point of the body ad do not enter into OtIr consideration of the equilibrium of
the body as a \\,hole. To bring them undcr consideration, \\'e isolate one
portion ]( as a free body (Fig. 1.28b). 1hcn the action of the rcmoved portion
I is represented by the forces that its various particlcs exert on those of the
free body K, and in this \vay the required internal forces are brought under
\ ...
consideration.
Although the actual distribution of these forces over the section may be
complicated, it is evident that they must be staticaHy equivalent to the system I
of forces acting externally on part f and can l\vays be represented by a
resultant force R applied at the centroid of the cross section together \\,ith a
couple of moment .t\J. The force R in turn can be resolved into rectangular
components ]\1 and V as sho\\"n in f"ig. 1.28&. The three quantities 1\7,  and
M on the section 11111 are called the 1101111(11 force, the shearing force, and the
bending 111(J111el1t, respectively. They are usually considcred positive when
P3
P3
P3
N
(a)
FIG. 1.,28
(c)
(6)

ARTICLE 1.4
15
m
ml
w per unit length
n I III
I n t-- x ---,
(a) r
I "-1
(a) I
J
N x V x r
(b)
(b) wx
,-
p
t
I
I
-,
Pcosa
L
: Shearing force diagram
I
T
PI COB ex
.L
Shearing force diagram
Bending moment diagram
Bending moment diagram
(c)
(c)
FIG. 1.29
FIG. 1.30
directed as sho\\'n in the figure. In general, their magnitudes \viH depend on
the position and orientation of the chosen section, but in any case they can be
evaluated by the three equations of equilibrium applicable to the free body K.
The simplest and most important class of structures for \vhich \\.'c shall ",ish
to investigate the internal forces on various sections are bemns submitted to
transverse loads. Such beams are usually prismatic in form and are con-
strained and loaded in a plane of symmetry. Under these conditions, \ve can
best define the gcncral state of stress in tcrms of the internal forces on cross
sections thar are nornlal to the axis of the beam. Consider, for example, the
cantilever beam AB, loaded as sho\vn in Fig. 1.290. Let 1nn be any normal
section defined by its distance .t from the free end. Then considering the
equilibrium of the free body to the right of this section (Fig. I.29b) and using
Eqs. (1.2a), we find
N z = +P sin a
 = +p CDS ex
M z = - Px COS Q

16 ELEMENTS OF PLANE STATICS
f'rom these expressions, \ve see that the normal force N z and the shearing force
V % are indepcndent of the position of the section along the beam, \vhi Ie thc
bending moment M z is proportional to the distance..\'. '[he variations in
normal force, shearing force, and bending moment along the bcanl can be
represented graphically by the diagrams sho\vn in Fig. 1.29c, \vhich accordingly
are calJed 110m/lll-jorct. .rhenring-jorce, and !Jc1Jdi11.g-111011/Cll1 diagr(l111.r.
As a second example, considcr the same cantilever beam uniformly loadcd
along its length as sho\vn in Fig. 1. 30a. In this case the exprcssions' for
normal force, shearing force, and bending moment on any cross section, defined
by its distance.\" from the free end of the hearn, becollle
N = 0
:r;
V:r; = +wx
wx 2
M = - --
:t 2
l'he corresponding diagrams are shown in Fig. 1.30c.
In the case of curved bean1s, we may find it convenient to proceed in a
slightly different manner. Consider, for example, the cantilevcr beam that
has a circular axis of radius R and is loaded as shown in Fig. 1.31a. In this
case we can most easily define the position of an arbitrary normal cross section
by the angular coordinate cp mcasured as sho\vn in the figure. l'hen, applying
Eqs. (1.2,1) to the free body in Fig. 1.3 I b, \\'C find
N9' = -p sin  V9' = +1) cas cp 1\{9' = -PR sjn 
As a last example, let us consider the simply supported beam \vith overhang
as shown in Fig. 1.32a. Undcr the action of a load uniformly distributed
over the span I, the reaction at A is wl/ 2, and the bending moment at any
section distance x to the right of this support is
wi x wx
M = - r - wx- = - ( /- X )
2: 2. 2 2
Thus the bending-moment diagram is a parabola with maximum ordinate wl 2 /8
at mid-span as shown. The frce overhang is \vithout internal forces if \\'c
negJect the ,,'eight of the beam itself.
If the same beam carries a conccntrated load j> at thc free end as shown in
.Fig. 1.3 2b, the rcaction at A is /)a/ I, and the bending moment at any section
FIG. 1.31 (a)
p
Mf
N A-\ p
'P \ I
\I
(i
\ ,

(b) 0
m
B

ARTICLE 1.4
17
w per unit length
A
I
wi
"2
I
I
I
wi
"2
I
I
I
C
I
I
a
I
I
J
I
,
(a)
p
A
X 2
(b)
C
I
I a
1
f
PII.(I+a) I
I
Pa
I
.
W per unit length P
C
I
I I

I. I
, .., -' . I
FIG. 1.32 (c)
bct\veen A and B is
Pa
ill = - - ' 1
:I: I
\\There :t't defincs the location of the section as sho\vn. Likewise, for any
section distance 1'2 to the left of the free end C of the beam, the bending
momcnt IS
iVl:1: = - PX2
Thus, in this case, the bending-moment diagram is made up of t\VO straight
lincs having the con1mon maximum 'ordinate - Pa at the support B as sho\vn.
If the beam carries both the distributed load and the concentrated load
simultaneously, we obtain the corresponding bending-moment diagrarn, as
shown in Fig. 1.32c, simply by superposing the diagrams of Fig. 1.32a and
1.32b. l'he same procedure can be used in the construction of shcaring-
force and normal-force diagrams.

18
ELEMENTS OF PLANE STATICS
p
p
I
4
p
I
a-l

A
B
(a)
(a)
f- 4 -+--- j ---j
 A
(b) I f- i ---+- 
(b)
p -L
 a-i   B
I i-+---4 
(c) P (c)
FIG. 1.33 FIG. 1.34
PI
FI G. 1.35
- - - - - - - - 1- - - - - - - -
I
2

I
2

PROBLEMS
I Construct shearing-force and bending-momcnt diagrams for each of the cantilever
beams sho\vn in Fig. 1.33.
2 Construct normal-force, shearing-force, and bending-n1ot11cnt diagrams for cach
of the healTIS shown in Figs. 1.15 to 1.17.
3 Construct shcaring-force and bcnding-mon1enr diagrams for each of the sin1ply
supported bean1s sho\vn jn Fig. 1.34.
4 (:onsrruct normal-force, shearing-force, and bcnding-mon1cnt diagrams tor the bar
BC of the structure shown in Fig. 1.27.

ARTICLE 1.5
19
5 Prove that the vertical ordinates in Fig. 1.35 represent the bending nlomcnts for
corresponding points on the semicircular axis of the three-hinged arch loaded as
"ho\vn. Evaluate the factor by \vhich each ordinate must be nlultiplied in order to
give the corresponding bending mon1cnt.
Au!. P /3.
1.5 FUNICULAR POLYGON
We shall no\v develop a general graphical treatment for coplanar forces that
has some practical advantages ovcr the analytical trcatn1cnt discussed in Art.
I .3. V\'e hegin \vith the very simple case of t\VO forccs P and Q as sho\vn in
Fig. 1.36a. As \ve already kno\v, the closing side AC' of the polygon of
forces (F;g. 1.36b) gives both the magnitude and the direction of the rcsultant
R, but the position of its line of action remains to be determined. In general,
this can be accomplished as follo\vs: In Fig. 1.3611, \ve take the arbitrary
point 0, called a pole, and join it by lincs I, 2, and 3 \vith the apices of the
polygon of forces. lhese, lines arc called TiT)'S and, like the other lincs in this
figure, may be regarded as free force vectors. laking, for cxan1ple, the
directions as indicated hy the arro\vs inside the triangle AOB, \ve may consider
the force P as thc resultant of the forces 1 and 2. In the same manner, the
force Q can be considered as the resultant of the forces 2 and 3, the directions
of \vhich are indicated by the arro\vs inside the trianglc BOC. Referring no\\"
to Fig. 1. 36a, it is evident that the action of the forces [J and Q \vill not be
changed if each of them is replaced by its t\\'o componcnts indicated in Fig.
1 .3 6b. These replacements \vi 11 be Inade in the foJIo\ving manncr: Bcginning
at any point tl in the plane of action of the forces, \ve dra\\' the line ab parallel
to the ray AO. From the point of intersection b of this line \vith the line of
action of the force P, \ve dra\v the linc be parallel to the ray BO; and, fron1 the
point of intersection cof this line \vith the line of action of the force Q, \ve dra\v
A
a
"
"
"
1
2
d
/'
./
./
3
B
"
"
"

"-
"-
R "-
.... 2 "
r - -/ 0
./
/'
A/
/'
/'
e
FIG. 1.36 (a)
R
(b)
c

20 ELEMENTS or PLANE STATICS
/g
-b
/'f
// I
X // /
/ I
/./e P4 I
./ I
I
I
I
I
I
I
h
(a)
a,......l Y; ,___ j __-
, c' d
P ,
1 P '
2........ P
...... 3
......
"-
,
......
,
,
FIG. 1.37
A
,
......
........
,
B ",1
--- ........
- - 2 '
P ----'
2 ---=-=-3_-- -::.   0
C ----  4 ///1
R /5/',
ttfP,It" 6 J
7 0 '
./
I
I
F
E
(b)
the line cd parallel to the ray CO. The polygon abed, obtained in this \\'ay, is
called a funicular polygon for the forces P and Q. The apices of this polygon
are on the lines of action of the given forces, and its sides are parallel to the
rays of the polygon of torces. We assume no\\' that at point b the forces 1
and 2, replacing the force P, and at point c the forces 2 and 3, replacing the
force Q, arc applicd as sho\vn in Fig. 1.36a. 'T'hus, the given system of forces
P and Q is replaccd by a systcnl of four forces applied at points hand e. Since
the pair of forces 2 acting along the linc be arc equal and opposite, they may be
removed from the system and there remain only the forces 1 and 3, \vhich are
equivalent in action to the givcn forces P and Q. The magnitude and direction
of the resultant of these forces are given by the vector AC in Fig. 1.36b, and a
point on its line of action is given by the point of intersection e of thc forces 1
and 3 acting along the first and last sides of the funicular polygon in Fig. 1.3 6a.
If \'le imaginc a \veightlcss string going along the sides ab, be, and ed of the
funicular polygon \vith its ends fixed at a and d, this string will be in equilibrium
under the action of the forces P and Q. "fhc tensile forces in the portions ab
and be of the string \vill be numcricaIly equal to the forces 1 and 2, and their
actions on the knot at b \vill balance the force P. In the same \vay the tensile
forces in the portions he and cd of the string balance the force Q. This relation
between the constructed polygon abed and the configuration of equilibrium of a
string submitted to the action of the given forces explains the origin of the name
funicular polygol1, i.e., string polygon.
lhe graphical constructions discussed above are perfectly general and can be
used also in the case of several coplanar forces PI, . . . , Pr, acting as sho\vn
in Fig. 1.37a.. Again, \ve begin \vith the construction of the polygon of forces

ARTICLE 1.5
21
ABCDEF (Fig. 1.37 b). Choosing an arbitrary polc 0, dra\ving the rays 1,
2, 3, 4, 5, 6, and constructing, in the planc of action of the forces, the lines ab,
be, . . . ,1.'1" parallel to these rays, \ve obtain the funicular polygon abedefg
as sho\vn. .A,t the apiccs of this polygon, each of the given forces Ph . . . ,
Po is replaced by its t\\'o con1ponents as represented by rays in Fig. 1. 3 7 b.
The forces acting along the sides be, ed, de, and ef are pairs of equal and opposite
forces in equilibriun1 and may be removed from the system. 1.here remain
only the forces 1 and 6 acting at points band f, \vhich are equivalcnt to the
given forces Ph . . . , 1-\. 'fhe magnitude and direction of the resultant of
these forces are given by the closing side Af ' of the polygon of forces (Fig.
1.37 b), and a point on its line of action by the intersection h 'of the first and last
sides of the funicular polygon (Fig. 1.3 7a) .
If the polygon of forces is closed, the possibility of a resultant force vanishes.
In such a :ase the first and last rays coincidc; hence the flrst and last sides of
the funicular polygon become parallel or coincide. In the first case, the t\VO
equal and oppositc forces acting along these sides represent a resultant couple.
In the second casc, they balance each other, and the given systcm of forces is
in equiIibriunL l"hese t\VO cases are illustrated in Fig. 1.38. l-ohree given
forces Ph P2, and I): have such rnagnitudes and directions that thcir frce vectors
build a closed polygon (Fig. 1.38c). l-fence, the rays 1 and 4, directcd to thc
beginning of the first vector A B and to the end of the last vector CA , coincide
in one ra)' OA.
If the given forces have the lines of action sho\\rn in Fig. 1.3 8a, the first side
ab and the last sjd de of the funicular polygon are parallel but do not coincide.
Hence, the unbalanced forces 1 and 4 acting along these sides f'epresent a
couple that is the resultant of the given system of forces.
If the given forces PI, . . . , P3 have the lines of action sho\vn in Fig. 1.3 8b,
the first and last sides of the funicular polygon coincide, and the equal but
opposite forces 1 and 4 acting along these coincident sides balance each other.
Hence, in this case, the given system of forces is in equilibrium.
We see that, by using the polygon of forces together \\rith the funicular
P3__e Pa 
a-- b Pz
p.
PI (a) (b)
FIG. 1.38
A __.!t__AO
--- /-r
.... I
/2 I
y- /3
I
I
I
I
I
(c) C

22 ELEMENTS OF PLANE STATICS
polygon, the three possibilities for a systcm of forces in a plane (see pagc 10)
can be investigated entirely by graphical methods. If the polygon of forces
is unclosed, the givcn systcm reduces to a resultant force. If the polygon of
forccs is closed but the first and last sides of thc funicular polygon, which are
parallel, do not coincide, the system reduces to a resultant couple. If the
polygon of forces is closed and the first and last sides of the funicular polygon
coincide, i.c., the funicular polygon is also closed, the system of forces is in
equilibrium. '"rhus, the graphical conditions of equilibrium arc a closed
polygon of forces and a closcd funicular polygon.
As an example, let us dctcrmine graphically the reactions at the supports A
and B of the beam /1C loaded as sho\vn in Fig. 1.39a. Beginning \\rith the load
P as represented by the free vector AB in Fig. 1.39b and selecting a pole 0,
\ve construct the rays 1 and 2, after \vhich the corrcsponding sides ab and be
of the funicular polygon can be dra\\'n as sho\\'n in Fig. 1.39a. Then the
closing side ac of this polygon deternlines the direction of thc corresponding ray
3 in Fig. 1.39b, and the rcactions Rb and Ru are graphically determincd by the
vectors BC and CA . respectively. As \vould be expected, the reaction Ra is
directed do\\'n\vard.
.A.s a second exanlple, \ve take thc case of a roof truss ABC supported and
loadcd as sho\\,'n in Fig. 1.40a. Hcre again, to find the reactions at the sup-
ports A and B, \ve begin \vith the polygon of forces ABCDEF, sclcct a pole
0, and dra\v the rays 1 to 6 as sho\\ln in Fig. 1.40b. No\v, in this case, \vhen
\ve come to construct the funicular polygon in Fig. 1.40a, \ve note that the
hinge A is the only kno\vn point on the line of action of Ra. Hence, \\'c must
start the funicular polygon at this point. Then, since the load P also acts
through point A, that side of the funicular polygon corresponding to the ray 1
vanishes. Other\visc, the funicular polygon is constructed in the usual man-
ner, and \VC obtain thc closing side A b as sho\vn. After this, \ve rcturn to
Fig. 1.40b, \vhere \\'C obtain the apex G of the polygon of forces by the
intersection of the fay 7 (parallel to the closing side of the funicular polygon)
p C
,
,
B ,
,
C Ra "
,3
A ,
,
1 ,
,
,
Rb ,
,
0
b
FIG. 1.39 c B (b)

ARTICLE 1.5
5P
FIG. 1.40
23
A
B '-
"-
,'-
"
:-..... ,"-
R ...........
a ,1
..... "'3............. 2
 "'
:-..., ...... 
, ......
G 4""" 
...... " ,
- ........7'''
-- .........' 
-- 5 ::-.::: ,
Rb -   
--O
-6--
---
F
(b)
and the known vertical direction of Rb. The required reactions are no\v
complctely detcrmined by the vectors F(; aRd (;A as sho\vn.
PROBLEMS
1 Determine graphically the n1agnitudcs of the reactions lu and Rb of the simply
supported beanl loaded as sho\vn in Fig. ].340.
Ans. Ra = 1.25P, Rb = O.75P.
2 Using the funicular polygon, detern\ine the rn,1gnitudes of the reactions at the sup-
ports A and B of the structure sho\vn in Fig. 1.27.
Al1S. Ru = Rb = P.
3 Using the funicular polygon, determine the axial forces in the bars 1, 2, and 3 of
the simple truss loaded as sho\vn in Fig. ] .25.
4 Dctcrnline graphically the reactions at the supports A and B of the sirnple truss
sh()\vn in Fig. 1.40t7 if the hinge A is on roUers \vhilc the hinge B is fixed. .Assume
AR = 40 ft, LCAR = LCBA = 15°, P = 1,000 lb.
Ans. Ra = 9,315 Ib, Rb = 6,465 lb.
5 Dcrern1ine graphically the axial forces in the bars 1, 2, and 3 \vhich support the
beam A B loaded as sho\vn in Fig. 1.41. .AsSlll11e P = 1,000 lb.
Aus. SI = -1,538 lb, S2 = + 1,414Ib, S3 = -1,392 lb.
p
FIG. 1.41

24
ELEMENTS OF PLANE STATICS
1.6 APPLICATIONS OF THE FUNICULAR POLYGON
In the preceding article, \ve used the funicular polygon simply as a tool in the
graphical composition of forces in a plane and in the evaluation of reactions.
\\"c have seen, ho\vcver, that it possesses a certain physical significance in
itself, namely: the configuration of equilibriuITl of a string under the action of a
given system of forces. l'his concept can be gcneralized to thc extent that \\rc
imaginc a string capable of sustaining compression as \vell as tension. 'rhus,
for example, the funicular polygon {lbcdef sho\\'n in Fig. 1.42a nlay be rcgarded
as a systcnl of hinged bars supported at II and f so as to form an arch that is in
equilibrium under the action of the applied forccs. Sometimes \ve need to
construct such a funicular polygon to satisfy certain specified conditions, such
as passing through t\VO or three given points, for cxarnple. This general
problem call be grcatly simplified by using the follo\ving theorem:
TI-IEORE'f r, for a given S)'ste1J1 o.f forces in (1 plane, any 1'7..1.)0 funicular
polygon.r are drawn .for t'U)O different poles () and 0', correspoudil1S!.
sides 0.( tht'se polygons 'lviII'll/tel iu pointJ that lie (JII a Jtl'aip;ht line
parallel to tlu: line O()' joining the two jJolt's.
rhis theorem can be proved as follo\vs: Let fb F 2 , F' 3 , . . . (Fig. 1.43a)
be any systcrn of forces in a plane, and let ABC'])E . . . (Fig. I.43b) bc
the corresponding polygon of forces, \vith 0 and 0' any t\VO arbitrary poles
and I, 2, 3, 4, . . . and }/, 2', 3 / , 4', . . . the corresponding rays. lhen,
beginning at any t\\lO points ,7 and a ' in the plane of action of the forces, t\VO
funicular polygons abcde . . . and a' h' c' d't' . . . corresponding to the poles
o and 0' can be constructed in the usual nlanncr. No\\' it foJIo\vs from the
discussion of Art. 1.5 that at point b, in Fig. 1.43a, \ve can rcplace the force FI
by its components 1 and 2 as represented by the vectors A() and ()B in Fig.
1.43b. In the same \vay, the forces 2' and I " represented by the vectors BO '
and 0' A (Fig. 1.43b), \,rhen applied at point b' (Fig. 1.43a), arc equivalent to
the equilibrant of the force Fl. Hence, the forces I , 2, 1', and 2', directed as
(b)
P3
c
./
./
./
./
,,/ .---.-
/ -
o-:::. P2
,=" -----
" '
" "
" '
...... '
"
"-
"
PI
FI G. 1.42

ARTIC LE 1.6
25
('
A
Fl "
"
\ \,.1
\ 
,,----\ " 0
',2 \:::
" ..........1 \
3' '(' \
.......... '/ / X,I' 
.... ..... ...... /' ,.,/ 2' , 
C 4/ /" \,
- - / 5/ 3 ,' \ \
/ - -/- - ", \
/ I 4' ::: j 0 ,
/ !--- ./
/ -- "
- / 5' /
I ".
I "
F4 1./ /
E
Fl
(b)
FIG. 1.43
shown in Fig. 1.4341, are four forces in equilibrium. .From this \ve conclude
that if at point 111 in the planc of action of the forces \\'c replacc the forces 1 I
and 1 by their resultant as represented by the vector 0 ' 0 (Fig. 1.43/1) and at
point 11 \ve rcplace the forces 2 and 2' by their rcsultant as rcpresented by the
vector 00' (Fig. 1.43b), these t\VO equal and opposite forces are in equilibrium
and consequently must be collinear in action. That is, the points of inter-
section 111 and 11 of the corresponding first and second sidcs of the t\VO funicular
polygons lie on a straight line parallel to the line 00' joining the t\VO poles.
Applying the same reasoning in connection \vith the force F'1.' we conclude
that the intersections 11 and 0 of the corresponding second and third sides of the
t\VO funicular polygons also lie on a straight line parallel to 00 ' , etc. Hence
the points 111, 11, 0, p, q, . . . aU lie on one straight line paraJlel to 00 ' , and
the theorem is proved.
FUNICULAR POLYGON THROUGH THREE POINTS By the aid of the
foregoing theorcm, it is possible to construct, for any system of forces in a
plane, a funicular polygon three sides of \vhich \vill pass through three given
points in the plane of action of the forces. We begin \vith the case of t\VO
copJanar forces P and Q as sho\vn in Fig. I.44a and for \vhich \ve wish to
construct a funicular polygon the three sides of \vhich \vill pass, respectively,
through the given points 111, 11, and p. We first construct the polygon of forces
ABC as sho\vn in Fig. 1.44h. Then" ihstcad of choosing an arbitrary pole,
\VC fifst construct two sides ab and be of a funicular polygon so that these sides
do pass through the given points 111 and 11, respectively, as sho\\rn in Fig. 1.44a.
The position of the pole 0 corresponding to the funicular polygon that \ve have
thus started and that we shall call the trial funicular polygon can now be found

26 ELEMENTS OF PLANE STATICS
a' A
a
d
'-
\ "
P l' \ "" 1
'0' '-
2' --  '-
--- I '- 0
B -------
I 2 //
3'/ ,/
I //3
1//
(6) c
FI G. 1.44
as the intersection of the rays I and 2 (Fig. 1.44b) dra\vn through the apices A
and B parallel, respectively, to the sides ab and be of the trial funicular polygon.
Having the pole 0, \ve can dra\v the ray 3, and thcn the corresponding sidc cd
of the trial funicular polygon can be constructed parallel to this ray as sho\vn
in Fig. 1.44a. In general, this last sidc of the trial funicular polygon \viB not
pass through the given point p. Ho\vever, fronl the trial funicular polygon
together \vith the above :heorcm, \ve can easily deternlinc a funicular polygon
for the given forces I and Q" the three sides of \vhich \vill pass through the
given points 111, 11, and p, respectively, and \vhich \ve shall accordingly refer to
as the true fUllicular pol)'gu/l. To accomplish this, \VC note that the first t\yo
sides of the trial funicular polygon pass through the given points 111 and 11,
respectively, and that the first t\\'o sides of the true funicular polygon ITIUst also
pass through these poin:s. Hence, the straight line 1111/ (Fig. l.44a) is the
line on v...hich the intersections of corresponding sides of the trial and true
funicular polygons must all lie. Prolonging the third side cd of the trial
funicular polygon to its intersection q \vith the line 11111, \VC obtain a point
through \vhich the third side of the true funicular polygon must pass. Remem-
bering that the third side of the true funicular polygon must also pass through
point p, \ve have the position of this side c'd' as sho\vn in Fig. 1.44a. No\\'
through the apex C of the polygon of forces \ve dra\v the ray 3 ' parallel to
e ' d ' , and through point 0 \ve dra\v a line parallel to the straight line 11111q.
The intersection of these t\\lO lines detcrmines the pole 0' for the true funicular
polygon. The rays 1 I and 2 ' can nO\'l be dra\vn and the funicular polygon
a' b'e' d' constucted as sho\vn in Fig. l.44a. The three sides of this funicular
polygon pass, respectively, through the three given points 111, 11, and p, as
desi red.
If there are several forces in a planc for \vhich \ve \vish to construct a
funicular polygon passing through three given points 111, 11, and p, \ve first
replace the forces bet\veen 111 and 11 by their resultant P and the forces bct\veen
11 and p by their resultant Q and then proceed as illustrated in Fig. 1.44.
The funicular polygon through three given points is of practical value in
connection \\lith the dctermination of the reactions at the supports of a three-

ARTICLE 1.6
27
(a)
Rb
\
\
\
d'
D
II
/ I
/ I
RIll' 11
afl 10
I ,f', 2
I //, ,
I / I Re "-
O,v_--+-' E
r 2 ' ---
\ 'R'
c
 \ 3' ,
\ '3
Rb' I
\ ,
\ I
d
FIG. 1.45
/ a
I
I
a'
F
(b)
hinged arch as sho\\'n in Fig. 1.45a. \\'e first construct, for the applied forces
P and Q, a funicular polygon '1' 17 ' c' d ' , the three sides of \vhich pass, respectively,
through the centers of the hinges A, C, and B. This construction is carried
out as explained above and is completely illustrated in Fig. 1.45. Obviously,
the rays I', 2', and 3 I, \vhen considered as vectors acting as sho\vn in Fig. 1.45 b,
represent thc reactions la, Rc = R, and lb, since, \vhen applicd at points A,
C, and B, respectively, they fulfill all conditions of equilibrium for the arch as a
\vhole as \vell as for either portion of it.
PROBLEMS
A string ACB of length / hangs bct\vccn t\yO vertical \\'alls as sho\vn in Fig. 1.46.
Along this string a 5111all pulley C, fron) \vhich is suspended a load P, can roll \yirh-
out friction. In the particular case \vhere I = 2a = 4b, \vhat configuration of
equilibriun1 \vill the SYStCl11 assume? Make the solutio entirely by graphical
constructIons.
Ans. .t' = 0.3550.
a
A
FIG, 1.46

28
ELEMENTS OF PLANE STATICS
2 lOhe hoisting cable of a crane is carried over a series of pulleys a, b, c, d, and e, as
sho\vn in Fig. 1.47". Using the cable as a funicular polygon for the system of
forces that the pulleys exert on the joints of the crane, sho\v that these forces are
conlpJctely defined by the polygon ABCDEF in Fig. 1.47 b.
3 Determine the proper shape of the top chord nbcdt'fgh of the plane truss sho\vn in
Fig. 1.48 so that, for the given system of loads P, all the diagonal \veb mernbcrs \vill
be inactive. '''''hat arc the corresponding forces in the chord rnel11bers if I = 70 ft
and k = 18 ft?
4 Construct a funicular polygon through the three points A, C, B to determine the
reactions at the supports of the three-hinged arch \vith scn1icircular rib as shown
in Fig. 1.49.
Al1S. Ra = 6880 lb, Rb = 8,220 lb.
5 l\ flexible \veightless string is fastened at A, passes over a pulley at D, and carries
a load P at its free end, as sho\vn in Fig. 1.50. i\ stiff bar BC \vith small rollers
at its ends carries weights P and Q as sho\vn and can roll freely along the string.
Neglecting friction, determine the configuration of equilibrium of the system.
Numerical data are given as follo\vs: a = 18ft, b = 4 ft, C = 8 ft, P = 2Q. ""hat
con1pressivc force S is induced in the bar BC? Make the solution entirely by
graphical constructions.
Alls. S = O.6Q.
e
B
p
a
I
I
I
I
I
I I /
1/// _---
0r\---
I P
I
r
t
I

- E
A
,
I
I
/
/
/
,;
FIG. 1.47
(a)
F
(b)
d e
FIG. 1.48
p
p
p
p
p
p

ARTICLE 1.7
29
B
a
2 tn
1 tn
A
1
b
t
p
40'
401--
FIG. 1.49
FIG. 1.50
1.7 FUNICULAR CURVES FOR DISTRIBUTED FORCE
In prcvious discussions, we have dealt entirely \vith concentrated forces.
Sometimes \\le need to consider the equilibrium of structures under the action
of distributed _force. Consider, for example, the prismatic beam AB (Fig.
1.51.7), the \veight of \vhich is uniformly distributed along its length. Such a
distribution of force is completely defined by its intensity q, that is, the \veight
per unit length of the beam. It is represented graphically by the rectangle
AabB, which is called a load diagram. In a more general case, we may have
nonuniformly distributed load along the beam. Here also the distribution of
force can be completely defincd by a load diagram like that sho'n in }'ig. 1.51 h,
where the intensity of load at each point is indicated by the corresponding
ordinate.
Any given load diagram can al\vays be approximated by a series of trape-
zoidal elements as sho\vn in Fig. 1.51 b. Then, if q is the average inten-
sity over the length x, the corresponding element must represent a force
p = q X ilx that acts through the ccntroid of that element. In this \vay, we
" 'I J ! .1 . I' '; b
. I q.,
,  : I
A ' B
(a)
FIG. 1.51 (b)

30 ELEMENTS OF PLANE STATICS
A
\
\ H
\
m II PI \
B \
, \
, \
"1 I IN " \
I I J p'!. " " \
I J I J I " \
I PtJ P'2 I P;1 1 p I S. '
C I
I I I I -- --O
a I I I I I //
I , I I I S /{
I I I I PJ j// /
//
I I I '/ I
I I I / /
/
I I I D /
I P-4! /
I I
I I !
I /
I E
FI G. 1.52 (a) c i d (6)
obtain a series of parallel forces the magnitudes of \vhich are represented by
the areas of the corresponding trapezoids and the moments of \\'hich, \\'ith
respect to any point, arc identical with the corresponding statical moments of
those trapezoids. r"rom this it follo,"'s that the resultant force is represented
by the total area of the load diagram and acts through its centroid.
""'e have already seen that any funicular polygon for a system of coplanar
forces may be interpreted as a possible configuration of equilibrium of a string
fixed at its ends and submitted to thc action of the givcn forces. In thc case of
continuously distributed force in a plane, as discussed above, the corrcsponding
configuration of equilibrium of a string becomes a sn100th curve called a
funicular curve. Physically, \ve see the manifestation of such a curve in the
case of a heavy flexible chain that is fixed at its t\\lO ends and othcr\vise
unsupported in the gravIty field. '[his particular curve is called a catenary.
I-icr MnfnN (Fig. 1.52a) be the load diagram foc a given distribution of
vertical force in a plane, To construct a cocresponding funicular curve, \\'e
begin by dividing the load diagram into sevcral parts, each small enough so
that, \vithout scrious error, it may be considered as a trapezoid. l'hen it
foIlo\vs that each trapezoid represents a vertical force equal in magnitude to
the area of that trapezoid and acting through its centroid. We may, therefore,
replace the continuous distribution of forcc by several vertical forces Ph
P2, . . . as sho\vn and construct a funicular polygon abcdef for these forces
in the usual manner. This done, \VC inscri be a smooth curve tangent to the
sides of the funicular pclygon at the points g, h, i, j, and k, and this is the
required funicular curve.
f'rom the foregoing discussion it follows that between any funicular curve
(Fig.l.52a) and the corresponding diagram of forces (Fig.l.52b) from\vhich
it is dcrived therc exist the following relationships: (1) For every tangent

ARTICLE 1.7
31
to the string, there is a corresponding parallel ray in the diagram of forces, thc
length of \vhich represents the tensile force in the string at the point of tangency.
(2) lhe shortest ray, caBed the pole distance H, represents the tension at the
vertex of the funicular curve and, Ii ke\visc, the horizontal projection of the
tension at any other point. (3) For any t\VO points on the funicular curve,
rays parallel to the corresponding tangcnts cut from the load line 1f1! that
portion of the total load \vhich acts bet\vecn these t\\'O points.
As an example of application of the foregoing discussion, let us consider the
case of a symmetrical t\vo-hinged arch rib ACB subjected to the action of an
earth-fill load as sho\vn in Fig. 1.53a. In such a case, it \vill be desirable to
rnakc the axis of the rib a funicular curve for the load diagram AA' B' R so that
the ri b \vi II be subjected to pure compression \vithout bending. 1 Ho\vcver,
since the load diagram depends upon the shape of the axis of the rib, \vhereas
the proper shape of this axis, in turn, depends upon the load diagram, it is
evident that a graphical solution of the problem must be made by trial and
crror.
\\, begin by noting that because of symmetry, the left portion AC' of the
arch must have the same shape as the right portion.BC. Hence, only one-half
of the arch need be considered. If the arch were to carry a load uniformly
distributed ovcr the span, the proper shape of its axis \\Tould be a parabola.
It is, therefore, convenicnt to hegin \vith AA'("C (Fig. 1.53h) as a trail load
diagram whcre the points A, D, Ii, r', and C lie on a parabola \vith vertical
axis and vertex at C.
Assuming earth fill to weigh 100 Ibjft 3 and considering a strip of the arch of
unit \vidth in the direction nornlal to the plane of the figure, \ve find that the
loads represented by the four trapezoidal portions of the load diagram \viII be
proportional to the areas of these trapezoids. rhe actual numerical values
are sho\vn at the top of each trapezoid. 1hese loads PI, . . . , Poi act through
the ccntroids of thc areas of the trapezoids by which they are represented.
Constructing the polygon of forces (Fig. 1.53g) for these loads and choosing
an arbitrary pole (J, \\'e may construct a funicular polygon abcdef (Fig. 1.53c)
in the usual manner, and \ve note that the line of action of the resultant of the
forces Pt,. . . , Pot is detcrnlined by the point of intersection g of the first
and last sides of this funicular polygon. No\v it is evident that the first and
last sides of any other funicular polygon for these same forces must also
intersect on thc line of action of their resultant. Since \ve are seeking a
funicular curve which passes through point (' \vith a horizontal tangent and
\vhich passes through point A, \ve may no\v draw the first and last sides ag
and gf (Fig. 1.53d) of the corresponding funicular polygon and thus determine
the pole 0 1 (Fig. 1.5 3g). Having this pole, \ve nlay dra\v the remaining sides
be, cd, and de of the funicular polygon. The inscribed curve (not shown)
tangent to the sides of this funicular polygon at points A, D, E, F, and C
1 Shortening of the rib due to axial corn pression is neglected in this discussion.

32
ELEMENTS OF PLANE STATICS
represents the first approximation to the desired funicular curve. Comparing
this, as sho\vn by the dashed curve abcdef in Fig. 1.53b, \vith the assumed
parabola AIJEFC, we find considerable discrepancy, \vhich indicates that a
second approximation must be made.
To do this, \\'e now use the diagram AA'C'C of "'ig. 1.53d as a load diagram
A'
B'
(a) f
___________________1__________________
1/2 1 1/2
A' D' E' C'
13,200 8,400 5,200
- -- f
IC
I '"
I /
(6) /
1 /
I /
I / PI
.-
I /
I "
f .-
.-
A /
.-
"
/ Pz
/
"
/
/
(c) /
E' /
.- P3
4.600 "
/
a /
f /" Pol (g)
IC L_ -___
I 0 0 1
(d) I
I
I /
I Scale: 1" .::20,000 lb .-
I ;'
f' .- Pi
e ' I /
.-
I "
I "
I .-
.-
I .-
"
f' "
"
/ Pi
"
(e) /
"
"
"
" P 3
"
/
/
Scale: 1" = 20' / P4
"
(f) " (h)
L._____
0' Oi
a'
FIG. 1.53

ARTIC LE 1.8
33
and, as before, find the forces l), . . . , I) as indicated. l"he magnitude
of each load, proportional to the area of the trapezoid by \vhich it is repre-
sented, is sho\vn at the top of that trapezoid. Thc polygon of forces (Fig.
1.53h) for these loads is no\\' constructed, an arbitrary pole 0 ' chosen, and the
funicular polygon a ' b' c ' d ' e'J' (Fig. 1.5 3e) dra\vn. 'fhe intersection J{' of the
first and last sides of this funicular polygon determines the line of action of
the resultant of the loads p, . . . , P. l'hcn, as before, we dctcrnline the
pole O (Fig. 1.5311) corrcsponding to the funicular polygon \\'hich passes
through point A and horizontally through point C. 'I'his funicular polygon
a'b'e'd'e'f' (Fig. 1.53/) agrees so closely \vith abed,! of Fig. 1.53d that \VC
may consider the inscribed curvc ADEP'C (Fig. 1.53f) as the desircd funicular
curve. It is interesting to note that in this particular case the obtained curve
is not greatly differcnt fronl thc arc of a circle.
1.8 FLEXIBLE SUSPENSION CABLES
In engineering structures, \ve sometinlcs encounter flcxible cables or chains
suspended bet\veen supports at the ends and subjected to some l<ind of dis-
tributcd vertical load. The cquilibrinnl shape assunlcd by such a cable is,
of course, a funicular curve for the loading to \\'hich the cablc is subjected.
In dealing \vith this problem, it is dcsirable to develop the equilibrilull shape
of the cable analytically rather than graphically.
To do this, \VC refer to Fig. I .54a, \vhere AII11111!\T represents a given distribu-
tion of vcrtical load and abed a funicular curve corrcsponding to the pole 0,
as sho\vn in Fig. 1.54b. On this funicular curve, \VC choose t\VO adjacent
points band c as defined by the coordinates x and x + d:x. The corresponding
slopes can be reprcsented by the exprcssions
d . 'Y d)' d ( dY )
dx and d' + dx d dx
II A
--....
.......
.......
--....
n "
.......
"
m E ------o
-::;;
- /
-- ./'
M, B - ./' /
I ./' /
-
C - /
I. x -I /
I I /
a I /
/
I / H
I
(a) (b) D
FIG. 1.54

34 ELEMENTS OF PLANE STATICS
respectively. "rhen, from the relationships discussed in the prccedipg article,
we have
dy  BE
.-":- = tan BOE = --"
dx OE
d)' d 2 v .. CE
dx + d:t:'2 dx = tan COE = OE
(0)
(b)
Subtracting (a) from (b), \ve obtain
d 2 y . _ BC _ q dx
-dx-- --
dx 2 O H
\vhich reduces to
d 2 y
H dx 2 = q
(1.3)
This is the differential equation of a family of funicular curves corresponding
to the arbitrary pole distance H. For any particular distribution of load as
defincd by the function q, the corrcsponding family of curves can be found by
integration of this differential cquation. T\\'o particular cascs of load dis-
tribution \vi]) no\v be considered in further detail.
LOAD Ul'"IFORMLY DISTRIBUTED AWNG THE SPAN Referring to
Fig. 1.5 Sa, lct A('B be a flexible cable supported at A and B and acted upon by a
vertical load (not sho\vn) of unifonn intensity qo \vith respect to the horizontal
span /. With the lo\vest point C on the curve as origin, let rectangular
coordinate axes x and y be taken as shown. letfl andf2 be the)' coordinates
of the points of support /1 and B, and a and b the x coordinatcs of these points.
Then \vith q = qo = const, the solution of Eq. (1.3) becomes
qox2
)' = ---
2H
(1.4)
and \ve see that the curve of equilibrium assumed by the cable is a parllbola
\vith vertical axis.
From the equilibrium conditions of any portion CD of the cable, \ve have,
for the tension S at point D,
(c)
S = vi H2 + (qo.'t) 2
l'his sho\vs that the tension in the cable is a Ininin1um at the JO\\'cst point C\
\\.here it is cqual to H, and that it increases to\vard the ends, being a maXin111nl
at the highest support.

ARTICLE 1.8
35
g
f 1
- I
1-
qox
1
x
II
(a)
FIG. 1.55
(b)
For the tensions at the ends A and B of the cable, respectiveI y, \ve have
from Eq. (c)
Sa = v'H2 + qo2tt 2 Sb = v'H2 + qo2b 2 (d)
To determine the distances iT and h, locating the low point C \vith reference
to the supports A and B, \ve use Eq. (1.4) successively for the portions AC
and BC and note that \vhen x = - a, y = /1, while \vhen .t. = b, Y = /2.
Thus, \VC obtain
q ()tl2
/1 = --
2H
q o b 2
/2 = - .--
2H
(c)
Subtracting the first of these expressions from the second and uSIng the
notation /2 - /1 = h, \ve obtain
2hl-I = qo(b 2 - a 2 )
--[hen, since a + b = I, \VC have
J hH
a=----
2 qol
b = _ + hH
2 qol
(I)
It \\Till be noted that \vhen h = 0, that is, when the supports are on the same
level, \VC obtain a = b = 1/2.
Substituting the above expression for b into the second of Eqs. (e), \ve
obtain the follo\"ing equation for calculating H:
H = ;2 ((2 -  :t Vfd 2 ) (g)
In E{J. (g)  the minus sign should be used for all cases \vhcre the vertex of the
parabola lies bet\veen the supports, \vhereas the plus sign should be used \vhen
the vertex ljes to the same side of both supports, as represented by the dashed
curve AB in Fig. 1.55. In the particular case \vhere the supports are on the

36 ELEMENTS OF PLANE STATICS
san1C level, /1 = 12 = .f, a == b = 1/2, and proceeding as before, \\'C obtain
H = q o/
81
(g')
In most practical probleols the span I, the sags /1 and /2, and the intensity
qo of the uniforolIy distrbuted load \vill be given so that fronl E<I. (g) or (/)
the rninimuol tension H in the cable may be found, aftcr \\lhich any of the
quantities previously defined in terms of H can be calculatcd \vithout difficulty.
T'he length of the portion C'B of the cable is
( B ( b I ( dY ) 2
[ = J c ds = J 0 \j 1 + dt d:r
Noting froo1 Fig. 1.55 b that dy/d.\" = qo."l/ H and substituting this into the
expression for L, \VC obtain
L = Job I + ( jr)2 dx
\vhich, after integration, beconlcs
I, = b 11 + ( 9b ) 2 + !- sinh-1 9
2 '\j Ii 2qo H
(11)
This can also be used for calculating the length of the portion AC if the distance
b is replaced by a.
LOAD lJNIFORM.LY nISTHIHUTED ALOG TilE CABLE In Fig. 1.56a,
let ACB represcnt the configuration of equilibriulll of a flexible cable or chain
hanging freely under the action of its o\vn \veight wliforn1ly distributed along
the length of the curve itself. As before, lct the lo\vest point C on thc curve
be taken as the origin of coordinate axes x and )', and let qo denote no\\' the
\veight per unit lcngth of the cable. ...t\lso, let s bc thc distance from C to [J
measured along the curvc. 'l'hen, the intensity of load q \vith respect to x
becomes q = £10 ds/d:t:, and Eg. (1.3) takes the fOrIl1
d 2 y _  ( d Y ) 2
H - - q o 1 + -
dx 2 dx
the solution of \vhich is
V = H ( cosh tjo.t - 1 ) (1.5)
. £10 H
This is the equation of a cart"nary \vhich can be constructed by using numerical
tables for cosh (qo.t/H).

ARTICLE 1.8
37
y
I
.1
qoB
c
x
H
qoB
(a)
(b)
FIG. 1.56
The length s of the portion CD of the curve may be determined by integrating
the differential relat ionship
ds = ydx 2 + dy2 = I + (  y dx (i)
\vhere, from the equilibrium conditions of thc portion CD of the cable (Fig.
I. 56b) , \ve have
d y q oS
d = H
(j)
Substitution of (j) into (i) and subsequent integration give
H . h qox
s = - sIn ---
qo H
Also, from the conditions of equilibrium of the segment C/J of the cable
(Fig. 1.5 6b), \\'e have
S = yH2 + (qoS)2
(k)
(I)
Substitution of expression (k) into expression (I) gives
S = H cosh qo.'t'
H
which, \vith the use of 1!:q. (1.5), finally becoInes
S '= H + qoY
(111 )
Again, \ve see that the tension S in the cable is a nlinimum at the Io\vcst point
C, \",here it is equal to H, and that it increases to\vard the ends, being a max i-
mum at the highest support.

38 ELEMENTS OF PLANE STATICS
For the tcnsions at the ends A and B, respectively, \ve have, from Eq. (111),
So. = H + qo/!
56 = 1-/ + qO/2
(11)
To determine the value of H, \\'e usc Eq. (]. 5)
A C and BC of the cable, obtaining
11 = H ( cOSh qo - I )
qo H
successively for the portions
/2 = H ( cosh q oh - 1 )
qo H
(0)
\vhich may also be writtcn
a = H cosh-1 ( 1 ifl + 1 ) h = FI cosh- 1 ( 9.0! + 1 ) (p)
qo H qo If
Adding these last two equations and remembering that (l + b = I, \ve obtain
r: = cosh- 1 (q;{l + I) + cosh- 1 (q;f + I) (q)
In most practical problems the span I, the sags /1 and /2, and the \veight qo
per unit length of cable \viIl be given so that from Eq. (q) the minimum tcnsion
H in the cable can be found, aftcr which any of thc quantities, previously
defined in terms of H, may be calculated wjthout difficulty.
PROBLEMS
lhe cable sho\vn in Fig. 1.57 carries a. total vertical load Q = HX),OOO Ib, uni-
formly distributed with respect to the horizontal span /. Dctermine the J11aximum
tensile force in the cable if I = 100 ft,11 = 12 ft, h = 10 ft.
Ans. Sin"" = 94,700 lb.
Z rhc cable shown in Fig. 1.57 carries a total vertical load Q = 100,000 Ib, uni-
formly distributed with respect to the horizontal span I. I f the lT1aximun1 allov/able
tension for the cable is 120,000 lb, \vhat length L of cable should be used? 'rhc
tollo\ving numerical data are given: 1 = 100 ft, h = 10 ft.
Al1S. L = 104 ft.
t
h
r-- L
I'
I
I
B
c
c
I
D
FIO. 1.57
F I O. 1.58

ARTICLE 1.9
39
h
x
FIG. 1.59 A B
3 }\ flexihle chain 1 no ft long and \veighing 5 Ih per tC)ot of length is freely suspended
at its ends fronl t\\'o supports 50 ft apart and having the saIne elevation. Find the
sag f at the Iniddle of the span.
Aus. f = 39.8 ft.
4 rhe flexible cable shO\VI1 in Fig. 1.57 hangs freely under its o\\'n \vcight, and the
to)l()\ving numerical data are given: I = 50 fr,11 = 10ft," = 20 ft, and £10 = 10 Ib
per f()ot of cable. Find the rnaxinlum tension.
Aus. Sb = 496 lb.
5 [)etermine the shortest overall length L of a fiexi ble chain of uni fort11 \veight pCI.
llnit lcngth that can hang in equilibriun1 as sh()\vn in Fig. J.SH. 1\:cglcct friction,
nd aSS\1mc that the radius of the pulley at B is very sOlall.
Al1J. 1- mln = I. 14/ + 0.801 = 1 .94/.
(-, "'lith reference to the coordinate axes sho\\'n in Fig. 1.59, develop the equation of
rhe curve A()B that \vill be a funicular curve for the load diagram defined by itself
:tnd the line )' = - II. Lct ",:)/1 be the intensity of load at O.
AJ1J. Y = Iz(cosh V wi H .\' - J).
1.9 GRAPHICAL CONSTRUCTION OF BENDING-MOMENT
DIAGRAMS
In previous articles, \ve have seen that the funicular polygon for a system of
forces in a plane has a variety of practical applications. As a further applica-
tion, \VC shall no\\' sho\v ho\v it can bc used in the construction of bending-
moment diagrams for transversely loaded beams.
Referring to Fig. 1.60, Jet P be a given force for \vhich ab and be arc t\\'o
c
a
A
p
.....
.....
.....
"'-
"'-
.....
"'-
"'-
.....
"'-
---------;0
'/
'/
'/
,/
'/
,/
,/
/
/
'/
II
p
B
FIG, 1.60
(a)
(b)

40
ELEMENTS OF PLANE STATICS
f
-
d
/'
A
h "
d'- - -  "
.......
Ie
l
B
"'-
.........
"
PI '....
"
"'-
---.. "'-
p.) ..... - "'-
- - .....
c - ---- - t l O
/1
/'1
P3 /'/ I
/' I
/' /
/' I
D / I
P. I II
I
I
H--
a
c
- - -  - c"
P l
......
'" J
'/
/'
P2 / /' G
/
/
/
e'
F
I
I
I
I
r
"'-- K /'
.... ......./
/- /
-;---
I ............'-..
/ ""-'-..
;/......... P3
"
,
...... I
a
P.i
'-
'-.. b'
(a)
FI G. 1.01
E
(b)
adjacent sides of a funicular polygon corrcsponding to the pole (). No\\r
through any point C: (Fig. 1.60a) let a line be dra\vn parallel to P so that it
intcrsects the t\VO sides of the funicular polygon in points d and e as sho\vn.
'fhcn the intercept h of this line bct\veen the t\VO adjacent sides of the funicular
polygon, \vhen multiplied by the pole distallfc II (Fig. I .60b), represents the
moment of the force P \vith respect to point L'. That is,
I-/h = PI/ (a)
\vherc Hand, P arc Incasurcd to the scale of force (Fig. 1 .60b) and h and I
to the scale of length (Fig. 1.60L1). This follo\\'s at once from the fact that
triangle bde is sinlilar to triangle OJ:4B, and hence \\lC may \vrite
I:h = H:P
\\'hich is equiv"alent to expression (a).
The above conclusion can be used also in the more general case of several
parallel forces PI, P2, . . . in one plane as sho\vn in Fig. 1.61. Assume,
for example, that corresponding to the pole 0 the funicular polygon abede!
has been constructed in the usual n1anner. No\v through any arbitrary
momcnt centcr (J' \ve dra\v a line parallel to the lines of action of thc given
forces. Then the intercept of this line bet\\'cen any t\VO sides of the funicular
polygon, \vhen nllJltiplied by the pole distance H, represents the algebraic sum
of moments, \vith respect to point C, of those forces included bct\veen the same
t\1/0 sides of the funicular polygon. rhus, for example, to obtain the algebraic
sum of moments of all the forces \vith respecr to point C, \ve take the product
of the intercept F(; and the pole distance H. Again, to obtain the algebraic
slim of nloments of the forces PI and /)2 \vith respect to point C, \ve take the
intercept G J.. multiplied by H; for the monlcnt of P" \ve take the intercept

ARTICLE 1.9
41
/./j X H, etc. rrhe rnoment is positive or negative accordingly as the intercept
is to the right or to the lcft of the intersection of the t\VO lines that detcrmine it.
AU this follo\vs dircctly from Eq. (a) since the intersection of any t\\'o sides of
the funicular polygon detern1incs the line of action of the resultant of those
forces included bet\veen these t\VO sides and since the moment of the resultant
is always equal to the algebraic sum of the moments of its components.
No\v, rcferring to Fig. 1.62, let us again considcr a system of parallel forces
Ph P2, P3, and P", for \vhich abcdef is a funicular polygon corre,sponding to the
pole 0, If, parallel to the lines of action of the forces, ,ve construct all
intercepts bet\veen the first sidc aa' and the remainder of the funicular polygon
as sho\vn, we obtain a 111UI11e11t diagrff/ll. Any ordinate of this diagram, \\lhen
multiplied by the pole distance H, represents the algebraic sum of moments of
those forces to the left of that ordinate \\lith respect to a point any\vhere on
the ordinate. l'his statement foJ!o\\'s directly from the discussion in reference
to Fig. 1.61. No\v, in the case of a transversely loaded beam, we know that
this idea of the sum of moments of all forces to one side of a point is of particular
value because it defines the bcndill?, 111rmtCllt in the beam at that point. l'hus,
the diagram sho\vn in Fig. 1 .62 can be considered as a bending-moment
diagram for a cantilever beaIn AB built in at B and submitted to the action of
the forces Pl, . . . , P 4 as sho\vn.
As a second example, let us considcr the simply supportcd beam loaded
transversely as sho\vn in Fig. 1.63a. In this case, the closed funicular polygon
abcde, \vhich has been constructed in the usual manner for the purpose of
determining the reactions Ro and R b , may be used directly as a bcnding-
moment diagram for the beam. It is necessary only to multiply each ordinate
of the diagram by the pole distance H to have the corresponding bending
moment. This foI10\\'5 from the fact that any ordinatc of the diagram is
simply the intercept bet\veen those t\VO sides of the funicular polygon \\lhich
intersect on the line of action of the resultant of all forces to one sidc of the
.
Intercept.
a
f
PI
.......
'"
.......
'-
a'
.......
.......
p'!. - - - - - ..............
---_::-::.O
-"/
././/
P3 ./ /' / /
/' /
/' /
/' /
/
P4 / / H
/
PI P2
P3
P-4
(b)
A
FIG. 1.62
(a)

42
ELEMENTS OF PLANE STATICS
r l r 2 r 3 ,
,
,
A B ,
  '1
,
,
Ra ,
,
"
,
e - --2- -- -- 0
_-,/'i
5......- //
- //
p...... 3/
2 ". /
/ 4
// /
Rb ",,- /
P3/ /
;I'
a /
!  H
2 IC (a) (6)
FIG. 1.63
p
,
\
Ra Q\
\
\ \
Q' \
\ \
, " \
'" \ \
Rb Q \\
\
----O
: e
(6) (a) f
FIG. 1.64
In the case of a beam under distributed load, \ve divide the load diagram into
several finite portions, as sho\vn in Fig. 1.64, and then construct the closed
funicular polygon abcdef and determine the reactions Ia and Rb in the usual
manner. 'J'hen, to have a bending-moment diagram for the beam, it is neces-
sary only to inscribe the smooth curve apqr corresponding to the distributed
load and take the ordinates as sho\vn. We see that in this case thcrc \\,ill be a
change in the sign of bending moment as we cross the section 11111. As in the
previous cases, the ordinates in t"ig. 1.64a must be measurcd to the scale of
length and thcn nlultipJicd by the pole distance H, measured to the scale of
force, before \ve obtain the true bending moment.

ARTICLE 1.9
43
PI
p"
P3
A
B
c
D
Ra
"
"
,
,
,
....., '6
....... '1 '
....... ' 2 ..... "-
R -, "-
b -........, "-
-
_ -- -3- - - - ::>O
5 _--/
- ",
-- /
...4 ..
Rd", / /
// (b)
(
(a)
FI G. 1.65
In the prcceding examples, \ve have seen that a funicular polygon con-
structed for the purpose of determining the reactions at the supports of a beam
can also be used as a bending-molnent diagram. Sometimes the situation can
be reversed and the fact that such a funicular polygon is a bending-moment
diagram can help us in the determination of reactions. Consider, for example,
the compound beam sho\\'n in Fig. 1.65a, for \vhich it is desircd to deternline
graphically the reactions at the supports A, B, and D. We begin by con-
structing in the usual manner the funicular polygon ahClfe for the given loads
/)1, P2, and P3 as sho\vn. At this stage of the construction thc rays 5 and 6 and
the corresponding sides ef and fa of the closed funicular polygon for the entire
system are missing, but the fact that there can be no bending 1l10mcnt in the
beam at the hinge C enables us to find thern. For, by this condition, \ve con-
clude that that side of the closed funicular polygon \vhich is comInon to Rd
and Rb must intcrsect the established side cd on the vertical through the hinge
C, that is, at point q. Thus the sidc ef can be dra\vn, and thence the closing
side fa. The corresponding rays 5 and 6 determine the reactions as sho\vn.
PROBLEMS
For the beanl sho\vn in Fig. 1.66, construct a closed funicular polygon such that
the ordinates of the diagram f()r bcnding Ill0l11cnt thus obtained \vill he to the scale
1 in. = 1,000 ft-Ib.
400'300& 500 t 250. 350'
2' 5' 3' 5'
A B
FI G. 1.66 3' -+ S'--/- 4'

.
ELEMENTS OF PLANE STATICS
2 Prove that the shaded portion of the funicular diagram sho\vn in Fig. 1.67 repre-
sents the bending-moment diagranl for the bearn AR, \vhile the unshaded portion is
the bending-momentdiagra,n for CIJ.
3 Construct the bending-nlornent diagrarn for the gitder All sho\vn in Fig. 1.68.
The magnitu.dcs of the loads arc in kips (I kip = 1,000 lb).
4 Deterrnine, by means of 'J funicular polygon, the reactions at the supports A, R.
E, and F of the compound beam loaded as sho\vn in Fig. 1.69.
5 Referring to the three-hinged arch loaded as sho\vn in Fig. 1.70, assume that a
pole 0 has been so chosen that the corresponding funicular polygon ahcde passes
P
A C 0 B
.......
......
Ra ......
.......
.-J __::,,0
P I
I
Rb I
I
I
I
1---0
FIG. 1.67 (a) (b)
15 30 30 30 30 19.5 19.5 19.5 19.5
A
B
12'
12'
12'
12'
FIG. 1.68
A
p
F
FIG. 1.69

ARTICLE 1.10
45
P..
PI
P:1
/
I p]
I
I
I ...
o  ==_ P'2
\
\ P 3
\
\
FIG. 1.70 a
(a)
('
(b)
through the hinges, A, C, and B of the arch as sho\vn. rhen prove that any ordi-
nate of the shaded diagranl, \vhcn 11luitipIicd by the pole distance f/, represents the
cDrrcsponding bending n10nlcnt in the arch rib.
1.10 PRINCIPLE OF VIRTUAL DISPLACEMENTS
The various analytical and graphical tnethods of solving pronlems of statics so
far considercd have neen based on the principle of the parallelognllll of forceJ.
V\l e shall no\v consider another general principle of statics called the principle of
virtual displaceWle1l1s. In certain cases, methods of solution bascd on the latter
principle have a decided advantage over any of those previously considered.
"rhe principle is pcrfectly general, but here \VC shall limit our discussion of its
developrT1cnt and application to coplanar systcrns.
"'Te begin \vith the case of a single particle P under the action of a ,systenl of
concurrent forces F h F 2, . . . , Frt in one plane as sho\\rn in Fig. 1.7 I .
Inlagine no\v that this particle is given a small arbitrary displacement ds as
sho\vn. l'hen any force F,. of the systcm is said to produce 'lL'ork on this dis-
placement. rrhis \vork is defined as the product of the displacement and the
projection of the force on the direction of the displacement, i.e., by the
FIG. 1.71

46
ELEMENTS OF PLANE STATICS
FIG. 1.72
expressIon
tis Pi COS exi
(a)
We see, by this definition, that the \vork ,viII be positive or negative accordingly
as the sign of the projection of the force agrees or disagrees \vith that of the
displaceInent. "[he net "V)ork of all the forces, i.e., the algebraic sum of
expressions (a), may be \vritten as follo\v5:
i-n ,i=n
L (ds}i cas at) = ds 2: F j cas cx)' = ds R cos a
i=l i-I
(b)
\vhere R cas a is the projection of thc resultant R on thc direction of the
displaccment. From expression (b), \ve conclude that the nct \vork of thc
systeln of forces on any displacement ds of the particle is equal to the \vork of
their resultant on the same displacement. Thus, if the nct \vork of the forces
on each of t\VO orthogonal projections dx and dy of the displacement ds is
zero, \\'e conclude that the rcsultant force vanishes and the systcm is in
cquilibrium. Conversely, if a particle is knowl1 to be il1 equilibrill1n, the algebraic
SU111 of the works of all for&es acting upon it l1Il/St be zero for any arbitrary S111all
displacnnent of the pa1"ticle. rhis is the principle of virtual displace1ncllts, or
virtual work as it is sometimes called.
Now let us consider the case of a partially constrained particle, for example,
a small bead that can slide along a smooth wire, as 8ho\\'0 in Fig. 1.72. In
such a case, we distinguish bct\\'cen t\VO kinds of 1rccs acting on thc particle:
active forces, such as Ft, F 2 , . . . , Fn, and reactive forces, such as N, exerted
on the bead by the \\,ire. As a condition of equilibrium, \ve conclude that the
algebraic sun1 of the \\fork of all forces (both active and reactive) on any small
displacement of the particle must be zero. Suppose, ho\vever, that \ve agree
to restrict such a displacement to one that is consistcnt \\lith the constraint,
i.e., in the direction of the tangent to the \vire at P. * Such an imaginary small
displacen1ent of the particle is called a virtual displace111cl1t, i.e., a possible
displacement, and it \viII be denoted by the symbol os. If the ,",ire is \vithout
* Since the displacelllent is infinitesinlal, an e1enlent of the axis of the \vire of length ds
{nay be considered as coincident \vith the tangent at P.

ARTIC LE 1.10
47
friction (an ideal cOll:;traint), the reactive force ]\' is normal to this virtual
displacemcnt s and does not produce \vork. Thus, the \vork of all the forces
is identical \vith the \vork of the active forces alonc, and \ve conclude that, for
any ,virtual displact1llcl1t of all ideally constrained particle, the algebraic sunl of the
work.f of the active forces alone 111l1Jt be zero. Reactive forces need not be con-
sidered at all.
We come finally to the general case of a systen1 of particles. It is in
application to such systems that the principle of virtual displaccments is of
special value. As a specific exalnple of a system of particles, let us consider
t\VO blocks A and B that rest on inclined planes and are connected by a flexible
but incxtensiblc string overrunning a pulley, as sho\vn in Fig. 1.73. Neglecting
friction in the pulley and on the inclincd planes, we call this an ideal Jyste11J.
Here again we distinguish bct\\leen active forces such as the \veights P and Q
and reactive forces such as "Ala, l\T", and S. Further, \ve notc that the reactive
forces arc of t\VO kinds, those external to the systcm as a \\'hole, like J\Ta and
N b , and those internal to the system, like the tensile forces S exerted on the
particles by the string. A virtual displacement of the system, i.e., a small
change in position consistent \vith the constraints (both external and internal),
\vill be obtained by giving to each particle an imaginary displacemcnt ox along
its supporting plane as sho\vn. By virtuc of the inextensibility of the string,
these displacenlcnts must be c<}ual. Then, if the system is in equilibrium,
each particle is in equilibrium, and the algebraic sum of the \\'orks of all forces
on such a displacement of the system as a \vhole must vanish. Ho\vcver, the
forces lV a and }-lb, being normal to the inclined planes along \vhich the particles
move; do not produce \vork and need not be considered. Further, the forces
S, being equal in magnitude, produce \vorks of oppositc sign, \vhich cancel,
and they also can be disrcgarded. 1 'hus again, for a1JY virtual displacelllCllt of
an ideal J)'ste1n of particles i1/ equi!ibrilnn, the algebraic .fll1lt of he works of the
activt' forces alonc 11111st he zero. i\pplying this conclusion to d1 case in hand,
\\'e \vrlte
- P sin a ox + Q sin {3 x = 0
\vhich gives, as a condition of equilibriulTI of the system,
p sin {3
Q - si n a
(c)
FIG. 1.73

'48 ELEMENTS OF PLANE STATICS
Q
- --- p
be
--
---
FIG. 1.74
-a=+- a
"'[0 get the above result by the usual equations of equilibrium, we consider
each particle separately and \\rritc
- P sin a + S = 0
Q sin {j - S = 0
"hen, upon eliminating S bet\\'een these two equations, \\'c obtain condition
(c). For a more complicated systcm of several particles, this latter method
\vill require the \vriting of a number of simultaneous equations of equilibrium,
the solution of \vhich can become very complicated. In such cases, the
method based on the principle of virtual displaccments often proves to be more
practicable.
The principle can be applied also to any ideally connected system of rigid
bodies. Consider, for example, the system of levers arranged as sho\vn in
Fig. 1.74. To find the relation bet\\'een P and Q for equilibrium, wc note that
a virtual displacement of the system can be defined by a small angle of rotation
08 of the horizontal bars. The corrcsponding displacemcnts of the gravity
centers of the weights P and Q are indcpendent of the positions of these bodies
on the pans and are always equal. Hence, by the principle of virtual dis-
placements, the weights P and Q must be equal for equilibriwn.
Sometimes the principle of virtual displacements can be used to advantage
in finding reactions. Consider, for example, the system of connected beams
supported and loaded as sho\vn in Fig. 1.7 Sa. To find the reaction at B, \\'C
imagine that the constraint there is replaced by a vertical force Rb as sho\\'n
in Fig. 1.7 5 b. In this vlay \VC obtain a nonrigid system for \vhich we have the
problem of finding the relation bct\veen the forces P and Rb for equilibrium.
A virtual displacement of the system can be completely defined by a small
vertical displacement ox of the hinge C as indicated in thc figure. The cor-
responding equation of virtual work becomes
x a
P - 8x - Rb 8x = 0
a a + b

ARTICLE 1.10
49
and \ve find Rb = Px (a + b) / a 2 . The reactions at the other supports can be
found in a similar manner. rhe advantage of the present method lies in the
fact that only one reaction need be considered at a time and no consideration
need be given to the internal forces of the systcm. That is, \ve do not have
to take the systen1 apart and consider separate free-body diagrams for cach
of the beams AC, C'/), and Dr'.
As another example, consider the simple frame structure sho\\'n in Fig.
I .7 6a. To find the horizontal component of .the reaction at B, \\'e remove the
p
x
b
a
b
A
B
c
D E
(a)
a
p
A
---
J
FIG. 1.75 (b)
Rb
c
(a)  I ..{
0 68b b
()
/41
/ I
/ /
/ /
/ I
(c) / I
/
/ I
/ I
C / /
/ I
/ P
I
I
I 68
rJ
D
FIG. 1.76 (6) H

50 ELEMENTS OF PLANE STATICS
corresponding constraint at this support and replace it by a force H as sho\vn
in Fig. 1.7 6b. l'he systenl is no\\" fil0vablc, and its configuration can be
defined by one coordinate 8. A virtual displacement of the systcm \vill be
rcprescnted by a small change of) in this angle. 1.o\s a result of such rotation of
the bar AC, point C moves pcrpendicularly to ,,4C, and point B moves hori-
zontally in compliance \vith the rcrnaining constraints of the system. lhjs
means that thc bar CB rotates about its illstanti11leOl/S cenftr 0, obtained as the
intersection of a vertical Jine through B and the prolongation of AC. O\ving
to such rotation of (.'B, te point ]) moves in the direction pcrpendicular to
()]J, and thc virtual displacernents of points C, B, and l) bccornc
oSc = a 08
DB
OSb = 0(' t1 06
(d)
OD
OSd = ()C' a 08
Thus, the cquation of vir:ual \vork is
- H g {( MJ + P cos (P, OSd)ZP {( 00 + Q cos 0 {( 00 = 0
and \ve find
OC 0 [J
H = ()B Q cos 8 + OB P cas (P, OSd)
(e)
The ratios OCIOB and OD/OB and the angle (P, OSd) bet\veen the direction of
the force P and the direction of the displacement lJs d can bc found fronl the
figure if it is dra\vn to scale.
If from a pole 0 (Fig. 1.76c) \\'e construct vectors ob , od , and OC , representing
to a chosen scale the displacenlents OSb, OSd, and oSc, respectively, \ve find that
the figure obdc, so obtaintd, is geometrically silnilar to the figure OBDC', since
corresponding sides are mutually pcrpendicular and their lengths are in a
constant ratio. This relationship bet\veen the t\VO figures holds in all cases
of displacement of a body in a plane, and such a figure as obdc is called a
displace1llellt diaj{rtl111. Vlith the aid of such diagrallls, a set of cOfi1patible
virtual displacements for various points of a system \vith one degree of freedom
can readily be found, after \vhich the equation of virtual \\fork can be \vritten
\\'irhour difficulty.

ARTICLE 1.10
51
p
A
r a
b -r- b -t- a ---i
B
FIG. 1.77
c
1
2
p
FIG. 1.78 P
PROBLEMS
Referring to the beam In Fig. 1.41 and using the nlethod of virtual displacements,
find the tensile force 52 in the inclined bar due to the loads P acting as shown.
Ans. S2 = 1.41P, tension.
2 A conlpound beam ACE is supportcd and loaded as sho\vn in Fig. 1.77. Using
the principle of virtual displacclllcnts, find the axial forces induced in the vertical
bars I and 2.
Ans. S1 = -P, S2 = +Pbl(a + b).
3 Calculate the reaction at the support E of thc compound beam shown in Fig. I .69
by thc method of virtual \vork.
Ans. Rc = 13P14.
4 Using the principle of virtual displacements, find the horizontal and vertical con1-
ponents of the reactions induced at the supports A and B of thc pin-connected franlc
loaded as sho\vn in Fig. 1.78.
Ails. Xu = -X b = .y3 P, la = }'b = 2P.
5 For the sen1icircular three-hinged arch sho\vn in Fig. 1.35, construct a displacc-
anent diagran1 and find the horizontal thrust H due to the load P acting as sho\vn.
Ans. 1-/ = P/3.

Chapter 2
Statically determinate plane
trusses
2.1 SIMPLE TRUSSES
The plane truss is one of the most important of all structural forms. In
general, it may be defined as a system of bars all lying in one planc and joined
together at their ends so as to form a rigid frame\vork. Consider, for example,
the two simple frames shown in "'ig. 2.1. 'I'he rectangular frame, consisting
of four bars pinned together at their ends as sho\vn in Fig. 2.la, obviously is
not rigid but can be collapsed as indicated by dashed lincs. The same con-
clusion holds for any othcr frame composed of more than four bars that arc
pinned together in the form of a polygon. On the othcr hand, three bars
pinned together at their ends in the form of a triallf{/e (Fig. 2.1 b) constitute a
rigid frame that cannot be coJIapscd. That is, neglecting possible small
changes in thc lengths of the bars, the relative positions of the pins A, B, and C
cannot be changed. 10]8, the triangular frame alone behaves like a rigid
body and may be considered as the simplest form of truss.
52

ARTICLE 2.1
53
C D
 . 'D
C'----::::: = = -:.:. -. =-:9D'
I' I'
" '1
" "
,II ,'
, , ,
,',' II'"
, ,
c
A.
.B
FIG. 2.1
(a)
(b)
D
E
D
A
FIG. 2.2
(a)
(b)
A
B
E
G
c'
C D F
1\7\
,. YL - V "
FIG. 2.3
(a)
(b)
Beginning \\'ith a rigid triangle A B(' (Fig. 2.2a) and attaching to this the
bars 4D and BD, which are pinned together at D, \\'C obtain the rigid frame
ABCD. This foHows directly from the fact that the rigid triangle ABC
together with the bars AD and BJJ are arranged in triangular form. In the
same way, the rigid truss ABCDE shown in Fig. 2.2h is obtained by adding to
the rigid portion ABCD the t\VO bars DE and ("'E, \vhich arc pinned together
at E.
Since the procedure above may be continued indefinitely, we conclude that
(1 rigid pla11c truss can always be fanned by begillni1lg with three bars pinned
together at thcir ends ;11 the fOr111 of a triangle and then adding thereto two new bars
for ca(h new pili. Any system of bars assembled in accordance with this rule
is called a simple truss. Figure 2.3a, for example, shows a truss of this kind.

S4 STATICALL Y DETIRMINATE PLANE TRUSSES
lhc trianglc ,,4 BC' has been taken as a rigid starting unit and the rcmainder of
the joints D, E, }', and G established in alphabetical order, each by the addition
of £\VO bars to the existing Systenl.
'"- .
In Fig. 2.3b, \ve have a simple truss fornlcd in accordance \vith a slight
variation of the foregoing rule. In this case, instead of starring \vhh a triangle,
\VC begin directly \vith a rigid foundation (say a vertical \vall) and establish
the joints of the truss in alphabetical order, each by the addition of t\VO bars
to the existing systenl. 1'his gilnplc truss in Fig. 2.3h differs froIl1 that in
Fig. 2.3a sinlply by the fact rh2t it must be considered an integral part of its
foundation, \vhcrcas thc rigidity of thc truss in Fig. 2.3a is quite independent
of its attachment to any foundation.
It can easily be sho\vn that, tJr each of the t\\lO variations of a simple truss
as sho\vn in Fig. 2.3, there must exist a definite rclationship bet\veen the
nurnber of bars, or 111t:'111bers, '111 and the number of pins, or joints, ). In the
case of the truss in Fig. 2.3a, \VC note that, exclusive of the starting triangle
ABC, \vhich contains three bars and three joints, therc arc t\VO bars for each
joint. Hence, \VC may \vritc
'IJ1 - 3 = 2 (j - 3)
from \vhich
111 = 2j - 3
(2.1a)
In the second case (Fig. 2.3 b) the question of the starting triangle is eliminated,
and \ve have simply
111 = 2j
(2. 1 b)
In this latter case the points of attachment to the foundation arc not to be
countcd as joints.
In arranging the bars of a trl1S, the actual fonn in any situation \vill depend
largely on the structural and architectural functions that are to be served.
There arc nlany standardized fcrn1s of trusses for various types of structures,
and sevcral exalnples typical of roof and bridge construction arc sho\vn in
Fig. 2.4. For each of these examples, thc readcr should verify the fact that
Eq. (2.1a) regarding the relationship bet\veen number of bars 'In and number
of joints.i is satisfied. .A.t thosc intersections \vherc no connections are sho\\'n
it is understood that the bars pass each other freely; such intcrsections .should
not be counted as joints. \"Idcd or riveted joints as sho\\'o in (a) and (b) arc
the most cOlnmon forms of connection in nl0dern steel structures, although
pinncd joints as sho\vn in (c) and (d) arc still to be found.
EssentiaJIy the function of every truss is SiOlply that of a large beanl to carry
loads across an open span. In Fig. 2.5 \VC see a very silnplc Jo\v-rruss bridge,
\vhich \vill serve to illustrate this fundamental structural action of it truss.

ART. CLI: 2.1
55
(a)
(b)
.
FI G. 2.4
(c)
(d)
FIG. 2.5
Herc \ve have t\\'o identical parallel trusses supported at their ends on piers
and carrying bct\veen thcln a floor system to \vhich active loads can be applied.
This floor system consists of croSSbea11lJ, supported bct\vccn corresponding
joints of the trusses, and str;l1f.{CrS, \vhich in turn are supported on the cross-
beams. It is evident from such construction that all loads applied to the floor
of the bridge \vill be transmitted to the two trusses only at their joints.
I..oads applied to the floor of the bridge as \\feU as the \"eight of the bridge
itself produce bending of the trusses principally in their o\vn planes. Such
plane bending of a truss is illustrated in Fig. 2.6b. The distortion of the truss
as a \vhole results from slight changes in the lengths of the various members, and
in each there is induced a corresponding tension or c0l11pression, j .e., an axial
force. 11hc stress (force per unit area) corresponding to this axial force in any
bar is called the lJri1nar)' stress in that bar.
In the case of a truss \\'irh riveted or \\'cldcd connections (Fig. 2.6c), bending
of the truss as a, ,vhole also induces some bending of the individual members

56 STATICALLY DETERMINATE PLANE TRUSSES
(:  B
(a)
(b)
FI G. 2.6
FIG. 2.7
because of the rigidity of the joints. Such bending of the bars of a truss
superimpose:; additional bending stresses, \\'hich arc called secondary stresses;
these must be investigated as a scparate problem.- Hovlcver, if thc bars are
carefully arranged so that their center lines nleet in one point at each joint, \ve
shall find that the presence of secondary stresses due to the rigidity of the
joints usually does not greatly affect the magnitudes of primary stresses.
Thus, in calculating the lattcr, the rigidity of the joints can be ignorcd and
pinned joints asslr/ned. . In this \\lay \\'e disposc temporarily of the necessity to
distinguish bet\vecn trusses \vith riveted joints and those \vith pinned joints.
Even in the case of a truss that actually does have pinned joints, there \vill,
of course, be some bending of the bars duc to thcir o\\'n \\'eights. This
bending, however, is usually slight. It is common practice to ignore it and
to replace the distributed weight of each bar by two equal concentrated forccs
on the joints at its ends; i.e., the \\'cight of thc truss is assumed to be con-
centrated in its joints.
"'inally, then, for purposes of analysis, we rcplace an actual physical truss
(Fig. 2.7«) by a corresponding idealized truss (Fig. 2.7 b) consisting of a
system of weightless bars all lying in one planc and joined together at thcir
cnds by frictionless hinges to \\rhich cxternal forces acting only in the planc of
the truss can be applied. Under such idealized conditions, ea.h bar is under
simple tension or compression without bending, and consequently the two equal
but opposite forces that it exerts on thc pins at its cnds arc collinear \vith the
axis of the bar. Thus, at each joint of the truss, \ve have in equilibrium a
system of concurrent coplanar forces \vith known lines of action, and a
dctertl1ination of thc magnitudes of these internal forces constitutes the analysis
of the truss. Ho\vever, before we can consider general methods of analysis of

ARTICLE 2.2
57
plane trusses, \ve must first consider the qucstion of st:pport of a truss in its
plane and the related problenl of detcrnlination of reactions induced at sllch
supports.
2.2 REACTIONS
Any sinlplc truss of the kind sho\vn in Fig. 2.3b, being fonned as an integral
part of the foundation, is, of course, complctely constrained in its o\vn plane
and can hold in equilibrium any system of external loads applicd at its joints.
Ho\vever, to con}plte thc constraint in one plane of a simple truss like that in
Fig. 2.3t1, \ve have yet to attach it to a foundation, either by additional bars or
by other suitable supports. The nlost connnon proccdure \vill be to anchor one
point 1 to the foundation by means of a fixed hinge as sho\vn in Fig. 2.6c.
Such a constraint elinlinates all possibility of translation of he truss; thcrc
re1l1ains only the possibility of rotation around point A. During such rotation,
another point B could move only in a direction normal to the line AB. I-fence,
to complcte the constraint of the truss in its 0\\-'11 plane it is necessary not to
fix the hinge B conlplctely but only to support it on rollers that can move frcely
in one direction. If this one degree of frcedonl of the .inge B is incompatiblc
\virh rotation about A, it follo\vs that such rollers 1 cOIT1pIcte the constraint of
the truss in its plane.
Practicaltncans of attaining desired degrees of constraint in the support of
actual trusses are sho\\'n in Fig. 2.8. Such dctai Is depend very la rgel y on the
size, \'leight, and span of the truss; the sketches sho\vn here are intended to
give only a general idea of rheir design.
.Another gcneral nlcthod of supporting a truss in its o\vn plane is to attach it
to the foundation by means of additional bars likc its o\\'n nlcmbers. rhe
simple truss ..4B("' sho\vn in Fig. 2.9, for exaolple, can he attached to the
1 "'e assurne that these rollcrs arc on a special track that prevents upward as \1lel] as do\\'n-
\'lard motion.
FIG. 2.8
fa)
(b)

58 STATICALLY DETERMINATE PLANE TRUSSES
FI G. 2.9
I
I
I /
1/
v
G
/
/
/
foundation in this manner by the three bars I, 2, 3. In general, complete
constraint of a truss in one plane can al\vays be attained with three bars so
arranged that their axes neither are parallel nor intersect in one point. This
may be proved as follo\\'s: Imagine first that the truss in Fig. 2.9 is attached
to the foundation by two bars AD and BE that arc not parallel. Then points
A and B of the truss can move, respectively, in directions normal to the axes
of these bars. The result will be equivalent to a rotation of the truss around
point G, \\,here the axes of the bars AD and BE intersect. A third bar CF,
which constrains point C to move in only one direction not compatible \vith
rotation around point G (\vhich means simply that the axis of the bar CF must
not pass through point (), obviously completes the constraint of the truss in
its o\\'n plane.
In the special case where the truss is supported by three parallel bars (Fig.
2.100), it is self-evident that there is freedom for lateral movement and the
truss is not completely constrained. In the same way, a truss supported by
three bars the axes of \vhich intersect in one point (Fig. 2.10b) is somc\vhat free
to rotate about this point and is incompletely constrained. Thus, in general,
three bars, the axes of \vhich neither are parallel nor intersect in one point,
FIG. 2.10
(a)
\11
\V
o
(b)

ART' CLE 2.2
59

Ef
(b)
FIG. 2.11
are both necessary and sufficient for thc completc constraint of a rigid truss
in one plane.
Any supports of a rigid plane truss in excess of those both nccessary and
sufficicnt for complete constraint in its o\vn plane are called redundaut fupports.
For example, the truss supported by four hinged bars as sho\\'n in Fig. 2.1la
is, of course, anlply constrained in its plane; but since anyone of these bars
may be rcmoved \vithout destroying thc complete constraint of the truss, any
one of them may be considercd as a rcdundant support. Similarly, it is
evident that anyone of the rollers by \vhich the truss in Fig. 2.11 b is supported
may be considercd as a rcdundant support; but not so with thc bar DE, since
\vithout this the truss \\,ill be free to move horizontally.
When a simple truss, complctely constrained in its o\vn plane, is sublnitted
to the action of a system of forces also in this plane, reactions arc developed at
the points of support, and active forces togcther \vith reactive forces constitutc
a coplanar systenl of forces in cquilibrium. For such a system of forces, \VC
have thrcc conditions of cquilibrium as reprcscnted by Eqs. (1.2) (page 11),
and hence \\'e can determine not morc than threc un kno\vn elemcnts pertaining
to the system of reactions. This, ho\vever, happens to be cxactly the number
of degrees of constraint required for completely fixing the truss in its plane.
Thue;, in thc case of a simple truss supportcd in a manncr both necessary and
sufficient for conlplctc constraint, \ve conclude that the reactions induced at
thc supports by any system of applied forces can be completely determined
by equations of statics. For this reason, such a system of supports is said to
be statically deteT111il1ate. Methods of determining such rcactions have already
been discussed in Arts. 1.2, 1.3, and 1.5.
If the system of supports of a rigid planc truss involves morc than three
degrces of constraint, i.e., if there are redundant supports as in Fig. 2.11, the
three equations of statics \vill be insufficient for determining the unkno\vns, and
the system is said to be statically indeter111il1ate. In such a case, the \vay in \vhich
the loads transmitted through the bars of the truss arc divided among the
several supports depends on the elastic properties of the bars, and the reactions
can be determined only by considering these properties. 1
1 This problem is con1plerely discussed in Chap. 7.

60 STATICALLY DETERMINATE PLANE TRUSSES
In special cases like those sho\vn in Fig. 2.10, a systcrn of supports in one
plane, although involving only three degrees of constraint, \vin also prove to be
statically indeterminate. Takc, for example, the case sho\vn in Fig. 2.10b,
and suppose that the resultant of all active forces applied to the truss is a
force \vhich passes through thc point o. Snce the thrce reactions induced in
thc supporting bars must also intersect in this point, \VC have a system of
concurrent forces in a plane for \vhich thcrc arc only t\VO independcnt con-
ditions of equilibriunl fEqs. (1.1), page 3]. Thus, the problem is statically
indeterminate. On the other hand, if the resultant of alJ active forces applied
to the truss is a couple, or a force that does not pass through point 0, the
situation \vill be quite different. In this case, if we decide to take point 0 as a
moment center, \ve see at once that it \vill bc necessary to have infinite forces
developed in some of the supporting bars since, \\,ith zero-mon1cnt arms, the
reactions exerted on the truss by these bars can in no other \\lay develop a
finite moment \vith respect to point 0 by \vhich to balance the resultant
moment of thc applied forces. In such a casc, \vhat actually happens is that
the supporting bars defornl (elongatc or contract) sufficiently to allo\v thc truss
to rotate into a nc\\' position such that thc axes of the threc supporting bars no
longcr intersect in one point. -rhus, the final configuration of equilibrium of
the systcnl, and the corresponding forces developed in the three supporting
bars depend on the elastic properties of the system, and \VC have again a
statically indeterminate problem.
PROBLEMS
Find the reactions at A and B for each of the trusses supported and loaded as sho\vn
in Fig. 2.12.
Al1S. (a) Ra = 5P, X b = 2.5 V3 P, }Tb = I.5P.
2 Find the reactions at A and B for each of the trusses supportcd and loaded as sho\vn
in Fig. 2.13.
A1/s. (a) Rb = 5P/V3, Xu = 2P, Y a = P/V3.
A
c
c
p
p
p p
(0)
p
p
p p
(b)
FIG. 2.12

ARTICLE 2.2
61
p
p
A
p
a+a+a+a
p
p
aaa+a
FIG. 2.13
(a)
(b)
FIG. 2.14
40
p
p
FIG. 2.15
3 Find the reactions at A and JJ of the silnple roof truss sho\vn in Fig. 2.14, ('1) \vhcn
the truss is loaded as sho\vn by the solid vectors and (b) \vhen loaded similarly as
sho\vn by dashed vectors. The loads are given in kips.
A?lS. (a) Rb = 184.7 kips, Xa = 160 kips, Yo = 92.4 ,kips.
4 Find the axial force induced in each of the' supporting bars I, 2, 3 of the truss
sho\vn in Fig. 2.15, (a) \vhen arranged as sho\\'n in Fig. 2.15a and (b) \vhen
arranged as sho\vn in Fig. 2.15b.
Al1S. (a) S. = - 2.45P, 8 2 = - 3.87 P, S3 = +4.46P.
" \.Vhat axial forces \vill be induced in the supporting bars of the truss in Fig. 2.9
by the action of a horizontal force P applied at joint K? ASSUlllC AI = AI) =
12 ft, BC = 10 ft.
Alls. 8 1 = 0, 8 2 :- 1.56P, Sa = -I.20P.

62 STATICALLY DETERMINATE PLANE TRUSSES
2.3 METHOD OF JOINTS
c shall nO\\l consider in detail the analysis of a simple truss ABC supported
and loaded as sho\vn in Fig. 2.16a. O\ving to the applied loads P, vertical
reactions each equal to -p are induced at the supports A and B, and the truss
as a whole is in equilibriun1 under the action of the balanccd system of vertical
forces sho\vn. l\.S was pointed out in Art. 2.1, these external loads induce
axial forces in the various bars of the truss, and cach bar in turn exerts on the
hinges at its ends t\VO equal but oppositely directed forces having for their
common linc of action the axis of the bar. lhe rnagnitude of either of these
forces reprcsents the axial force in the bar; and \vhcther jt is directed 3\\'ay
from the hinge or to\vard jt detcrmines, accordingly, \\'hcther the bar is in
tcnsion or in compression. To avoid confusion in dealing \\lith these internal
forces, \ve shall number each bar of the truss as shov.'n and then denote by
Si the magnitude of axial forcc in any bar i.
We begin the nal¥-sis of the truss in this case by consideration of thc
equilibriun1 of thc hingc .4. raking this hinge as a free body as indicated by
the circle around joint i4, \VC fin acting upon it the external reaction jP,
together \\lith the internal reactions SI and 52 exerted, respectively, by the
bars I and 2. We shall not kno\v at once \vhether these last t\VO forces should
bc directed a\\'ay fron1 the hinge or to\vard it, but only that thcir lines of action
are rcpresented by the axes of the bars exerting them. From this fact, ho\v-
ever, a closed triangle of forces (Fig. 2.16") can be constructed and the
p
FIG. 2.16
p
8 6
(6)
(d)

ARTICLE 2.1
61
(;
x
a
a
a
FIG. 2.17
p
p
p
nlagnitudes of the forces 5'1 and .)'). scaled or calculated frool this. A.lso, from
the directions of the arro\vs, \vhich must folto\v tail to head around the closed
triangle of forces, \\'C see that the bar 1 is in tcnsion \\'hile the bar 2 is in
coolprcssion. l'\rrO\Vs indicating such action can no\\' be placed at 44 as sho\\'n.
\," arc no\v in a position to consider the equilibrium of the hinge F. \IVC
first replace the bar 2 by its kno\vn reaction S2 directed to\vard the hinge I:.
There rCfilain then but t\VO unkno\\'n forces at 1/ representing the reactions on
this hinge of the bars 3 and 4. i\gain, \VC do not kno\v \vhich \vay thcse last
t\VO forces should be directed; but, kno\ving their lines of action, \Vt can
construct the polygon of forces as sho\vn in Fig. 2.16(. Fron} this closed
polygon, the magnitudes of axial force in the bars J and 4 can be found as
before. In this case, \ve see from the arro\vs on the polygon of forces that
both bars are in conlprcssion, and arro\vs indicating such action can no\\' be
inserted in Fig. 2. I 6a.
Proceeding next to a consideration of the hinge D, the reactions exerted on
this hinge by the bars I and 3 arc already kno\vn, and again there remain only
t\VO bars (5 and 6) for \vhich the axial forces are llnkno\vn. The closed
polygon of forces for this hinge, froln \vhich Su and 5'6 can be found, is sho\vn
in Fjg. 2.16d. This cornpletcs the analysis of the truss in this particular case,
for, D\ving to symmetry, it is evident that each bar on the right \vill carry the
same axial force as the corresponding bar on the left.
If desired, the t\VO equations of cquili briuOl [Eqs. (1.1), page 3] can be
applied successively to the joints of a sinlple truss instead of constructing the
various polygons of forces as \vas done above. -rhis alternative \vill usually
be p:-efcrablc in cases \vhere the angles bet\vcen the bars of the truss are sllch
that the projections of the forces arc rcadily calculated. By \vay of illusrrating
this procedure, let us consider the simple truss supported and loaded as sho\\>'n
in Fig. 2.] 7. In applying equations of equilibriunl instead of the graphical

64 STATICALLY DETERMINATE PLANE TRUSSES
procedure above, it is convenient to assume at the outset that all bars arc in
tension as indicated by the reactions directed a\\'ay from the hinges in the
figure. Then, autor,natically, results \vith the plus sign \\Till indicatc tension,
\vhereas those with the minus sign \vi II indicate cOlnpression. For exanlple,
\ve imagine the hinge /1 as an isolatcd free body acted upon by the thrce
forces P, Sl, and 52 directed as sho\vn. Then, equating to zero the algebraic
sums of projections of these forces on thc orthogonal axes .'t' and y, \ve obtain
- y 51 - S2 = 0
2
- I> + iSl = 0
from \vhich Sl = + 2P and S2 = - y3 P. Thus, the bar 1 carries a tensile
force equal to 2P, \vhile the bar 2 carries a compressive force equal to y3 ]:1.
In using these values in subsequent calculations, \ve should consider the sign
Jas an integral part of the result.
Upon proceeding to the hinge B and projecting all forces on the same
orthogonal axes, we see that the equations of equilibriurTI \vill beconle
+S2 - S6 = 0
-p + S3 = 0
Upon substituting for S its previously deterrnined value - y} P, \VC obtain
from these equations S3 = +P and S6 = - y3 P (tension and compression,
respectively) . For the hinge C' it \vill be most convenient to project all forces
onto the orthogonal axes x' and y' directed as shown, For such axes, the
equations of equilibrium become
y} y}
- ---2 53 - . -2 - S6 = 0
-Sl - -!Sa + .i S 6 + S4 = 0
Using S3 = +P in the first of these, \\'e find S5 = - P, after which the second
equation gives 8 4 = + 31. It is left as an exercisc for the readcr to consider
the hinges D and' E and complete the analysis of the truss.
'[he foregoing procedure in the analysis of a simple truss is called the
1ncthod of joints. It ah\:ays can be applied either graphicaJly or analytically
to any truss the bars of which arc assembled in accordance \vith the rule given
on page 53, i.e., to any si1nple trllS.C rhis follo\vs from the fact that in any
such truss there nlust exist at least onc joint (the last one added in accordance
\vith the rule) at \vhich only two bars nlcet. Hence the axial forces in these
t\vO bars can be detcrmined from the t\VO conditions of equilibrium existing
for that joint and the two bars replaced by the reactions that they exert on t\VO
othcr hinges of the truss. l"hen, again, there must be one joint (the next to

ARTICLE 2.4
65
the last one established in accordance \vith the rule) \vhere only t\va bars \vith
unkno\vn forces \vi II be encountered, and these forces can be determined.
'rht:s, by considering the joints of the truss one by one in the reverse order
from \\'hich they \vere established, \VC shall find at each joint only t\VO bars
\.vith unkno\vn forces, and the analysis proceeds \vithout difficulty. It must
be pointed out, of course, that it \vill often be necessary to dcrcrnlinc the
external rcactions at the points of support as a separatc problem before the
analysis of the truss proper can be commenced.
PROBLEMS
Dctcrnlinc the axial, force in each bar of rhe sirnple truss supported and loaded as
sho\vn in Fig. 2.12&1.
2 Jv1akc a cOInplete analysis of eac of the sirnple trusses supported and loaded as
sho\\'11 in Fig. 2.13.
3 Referring to the roof truss sho\vn in Fig. 2.1-t, make a c0l11plcre anal ysis tor each
of the indicated conditions of loading.
c
B E
H
K
N
p
o
L
p
FIG. 2.18
FIG. 2.19
4 i\'lake a complete analysis, by the method of joints, of the silnple truss sho\vn in
Fig. 2.18. ,All inclined bars are at 45° \vith the horizontal.
5 Make a complete analysis, by the method of joints, of the K truss sho\\'n in Fig.
2.) 9. The inclined bars are at 45° \vith the horizontal, and the panel distances are
uni torm.
2.4 MAXWELL DIAGRAMS
Referring to Fig. 2.16, in \vhich the analysis of a truss has been n1ade by con-
structing a separate force polygon for each hinge, \VC note that each axial-force
vector appears in t\VO diffcrent force polygons. To avoid this duplication of

66 STATICALLY DETERMINATE PLANE TRUSSES
vectors, the separate polygons of forces, under certain conditions, can be
superimposed to forn1 one composite diagram called a i\1a:rwe/l diag1'tl111 for
the truss . For exarnplc, the three polygons of forces shown in Fig. 2.16,
when superimposed, make the composite diagram sho\vn in Fig. 2.20b. Such
superposition is desirable since it reduces the amount of nccessary construc-
tion and also makes a n10re compact record of the final results. Ho\vevcr, in
ordcr to avoid duplication of any vector, the constructions nlust be carried out
in a definite manner, \vhich \VC shall no\\' consider. .
We begin \vith the siIT1plcst case of a triangular frarne ABC, \\,hich i's in
equilibrium undcr the action of three external forces P, Q, and R acting in the
plane of the triangle as sho\vn in Fig. 2.21a. These three forces, being in
equilibriunl, must intcrsect in onc point D, and their frce vectors must build a
closed triangle (the as sho\vn in Fig. 2.21b. 'I'his done, a closed triangle of
forces for each hinge can be superimposed directly on this triangle as sho\vn.
For example, the vectors t1b , bd , and dtl dirccted in accordance \vith the arro\vs
insidc thc abd rcpresent the forces that are in equilibrium at thc hinge A.
Similarly, Lhcd and L.cad \vith vectors directed in accordance \vith the inside
arro\vs arc closed triangles of forces for the hinges Band C, respectivel y .
(a)
a
j
b
e
f
c
FIG. 2.20
(6)
d

ARTICLE 2.4
67
b
b
a
a
c
c
(b)
(c)
FIG. 2.21
"htJS, in Fig. 2.21 b, all triangles of forces are compactly superimposed, and
no vcctor is duplicatcd. "!'his \vill be possiblc only if each polygon of forces
has its vectors assembled in the san1C order by \vhich the forces that they
represcnt are encountered in going around the corrcsponding hinge consistently
in onc direction (either clock\vise or counterclock\vise). We may note no\v
that the basic triangle abc assembles the external forces in thc ordcr PQR
obtained by going around thc truss in a clock\\,'isc direction. "rhus, in the
construction of thc rcmaining force polygons that are superimposed on the
abc, \\'C must go around each hinge in a clock\\'ise direction and take the
forces in the order in \vhich they are so encountered. Only by this consistent
procedure can \ve avoid duplication of some vectors. For cxample, if \ve go
clock\visc around the hinges Rand C but counterclockwise around the hinge A,
\\'e obtain the composite diagram sho\vn in Fig. 2.21c, in \vhich there is
duplication of the vcctors bd and da .
A study of the diagrams in Fig. 2.21a and /J sho\vs that they bear a definite
relationship to each other. In ordcr to define this relationship bet\vecn the
t\VO figures, let us visualize each onc as the plane projection of a four-faced
polyhedron. 1'he polyhedron in Fig. 2.2Ia, for example, has the faces ACD,
ABD, B(1), and ABC\ \vhich \VC designate, respectively, by the lo\vercasc
letters a, b, c, d, and the vcrtices A, B, C, D. Sinlilarly, the polyhedron in
Fig. 2.21 b has the faces abd, hcd, acd, and abc, designated by A, B, C, 1), and the
vertices a, h, c, d. l.hus, for each face of thc polyhedron in Fig. 2.21a, there is
a corresponding vertex to the one in Itig. 2.21 b, for each vcrtex of thc first, a
corresponding face to the second, and their edges are rl1utually parallel and
equal in number. T\vo such polyhedrons are said to be reciprocal, and the t\VO
plane figures representing their projections on a common plane are called
reciprocal figures. It follo\vs at once from such reciprocity that if forces repre-
sented in n1agnitl1de by the lines of one such plane figure are made to act
between the extremities of the corresponding lines of the reciprocal figure,
the points of the reciprocal figure \vill all be in equilibrium under the action
of these forces. This £0110\\'5 from thc fact that the forces meeting in any

D a
It.
/ 1\
/ 1\
/ I \
/ I \
/ I \
/ I \
P / p \ p f b
/ \
/ b c
h
d g c
68 STATICALLY DETERMINATE PLANE TRUSSES
FI G. 2.22
(b)
d
one point of thc second figure are proportional to thc sidcs of a closed polygon
in the first figure. This observation \vas made in 1864 by Clcrk Maxwell in
discussing the significance of reciprocal figures to problems of statics.!
Because Max\\'ell \vas the first to point out the reciprocity bet\vcen diagrams
Ii ke those in Fig. 2.21, the composite vector diagram is called a Alaxwell
diagrt1111 for the truss. For the truss in Fig. 2.20a, the complete Max\vell
diagram is sho\\'n in Fig. 2.20c. In a n10re complex case of this kind it is
practically impossible to visualize the corresponding polyhedrons, but this is
of no consequence so long as the t\VO diagrams fulfill the requirements of
reciprocal figures.
In the construction and use of Max\vell diagrams for the analysis of trusses,
a system of notations called BfYW's notation is convenient. In this systcm the
spaces bet\vcen the lines of action of the forces acting on the joints of the truss
arc given a lo\vercase Jetter. Then, any force is corrcspondingly designated
by the letters of the t\\'o spaces separated by its line of action. Consider, for
example, the sin1ple truss supported and loaded as sho\\'n in Fig. 2.22a. In
accordance \'lith Bo\v's system, \ve letter the spaces bet\veen the five external
forces a, b, c, d, e and those bet\veen the bars of the truss f, K, h, as sho\vn.
Thus, reading clockwise around the truss, we refer to the first load P on the
left as ab , the second as he , the external reaction at (' as dC, etc. l.like\vise, the
reaction of the vertical bar on the hinge C is hd ; that of the horizontal bar on
the hinge A is 7i]:, etc.
We arc no\v ready to consider the construction of a Maxwell diagram (Fig.
2.22h), \vhich begins \vith a closed polygon abcde for the balanced system of
forces external to the truss. With this basic polygon, the reciprocal diagram
1 See Clerk Max\vcll, On Reciprocal Figures and Diagrams of Forces, Phil. lvlag., vol. 26,
p. 250, 1864.

ARTICLE 2.4
69
is completed simply by dra\ving through each of its vertices lines parallel to
all those which bound the corresponding space in the truss diagram. For
example, through points band c \ve dra\v horizontal and inclined lines parallel,
respectively, to AB and AC of thc truss. Since thesc t\\'O lines also help to
bound space f, their intersection determines the vertex f of the reciprocal
diagram. Then, through points f and c, we draw vertical and horizontal lines,
respectively, and their interscction determines the vcrtex g of the diagram.
Finally, through points g and e, \ve dra\v the inclined lincs \\,hose intersection
h on the line ad determines thc final vertex, and the diagram is complctcd.
l"he readcr \vill do \vell to identify in this diagram the separate polygon of
forces for each hinge of the truss.
Bo\v's notation is particularly advantageous \vhen \ve come to decide
v.'hcrher a given bar of the truss is in tension or in compression. L,et us
consider, for example, the force in the bar BC that is represented in magnitude
by the length of the line hd in the Max\\lcll diagram. Going around the hinge
B in a clock\vise direction, the reaction that this bar exerts on the hinge \vill
be read as dh. No\v, in the Max\vell diagram, the vector dJi is directed
up\\1ard, indicating pressure on the hinge B, and \ve conclude that the bar BC
is in compression. If, instead, \VC choose to consider the reaction of this
same bar on the hingc C, reading cJock\\lise around C, \ve have hd instead of
dh, and in the 1\1ax\\'cIl diagram the vector lid is directed do\vnward, indicating
pressurc on (,' and consequently compression, as before.
PROBLEMS
1 Construct a Maxwell diagranl for the truss sho\vn in Fig. 2.23, and determine from
it the axial force in each bar.
2 Construct a l\1ax\vcll diagram for the truss sh()\vn in Fig. 2.18, and determine from
it the axial force in each bar.
3 Construct a l\'lax\vell diagram for the truss sho\vn in Fig. 2.24, and deternline frolll
it the axial force in each bar.
4 Construct a Nlaxwell diagranl for the truss in Fig. 2.19, and determine from it the
axial force in each bar.
5 Construct a 1\1ax\vell diagralll for the truss sho\vn in Fig. 2.25, and .deterrnine from
it the axial force in each bar.
p
p
p
A
B
FIG. 2.23
tp
Ip

70 STATICALLY DETERMINATE PLANE TRUSSES
c
A
Ip
iP
B
FI G. 2.24
p
p
P
p
p
p
A
B
FI G. 2.25
Sp
Bp
2.5 METHOD OF SECTIONS
\\'e shall no\\' consider another method of analysis of trusses, the use of \vhich
makes it possible to determine the axial force in son1e chosen bar \vithout
going through successive considerations of the equilibri un1 of all hinges as \vas
done in the two preceding articles. Referring to the truss shown in Fig. 2.26a,
let us assumc that it is required to dctcrmine the axial forces in the bars 10, 11,
and 12, due to the loading sho\vn. Instcad of considering the equilibrium of
the hinges A, H, B, I, C, and J in succession, as \vould be necessary by the
method of joints, \ve imagine that a section 111n cuts the truss into t\VO parts and
then consider the conditions of equilibriun1 of the part to the left of this section
(Fig. 2.26b). Acting upon this free body \ve have the vcrtical reaction at
A, three vertical loads P, and the three unkno\vn forces 5 10 , Su, and S12,
representing the axial forces in the bars that \vere cut by the section 11ln.
The directions of these forces must, of coursc, coincide vlith the axes of the
bars so that only their magnitudes remain unkno\\'n. Thus we obtain
altogether a system of coplanar forces in equilibrium, and Eqs. (1.2) (page 11)
can be employed to determine the rnagnitudes of the three unkno\vn forces.
For example, equating to zero the algebraic sum of mon1cnts of all forces with
rcspect to point 1), \\'e obtain
-Stoh + Pa + P2a + P3a - iP3a = 0
from \vhich S10 = - 9Pa12h. The negative sign, of course, indicates com-

ART. CLE 2.5
71
prcssion instead of tension as assumed. In a similar manner, by taking point
J as a mon1ent center, \\'e find S12 = +4Pa/h. Su can be evaluatcd most
readily by equating to zero the algebraic sum of projections of all forces on a
vertical axis. Thus, \\'c obtain
h
i lJ - 3P - Si1 --==----=: = 0
V a 2 + h 2
from \vhich 8 11 = + P va 2 + h 2 1211.
The foregoing procedure in the analysis of trusses is called the 111ethod of
sectiol1S. It consists, essentially, in the isolation of a portion of the truss by a
section so chosen as to cause those internal forces which we \vish to evaluate
to appear as external forces on the isolated frec body. By this procedure, \ve
usually obtain the general case of a coplanar system of forccs in equilibrium,
and Eqs. (1.2) can al\vays be uscd in evaluating the unkno\vns as \vas donc
above. "rhe success or failure of the ITIcthod rests entirely upon the choice of
section. In general, a section should cut only. three bars, since only three
unkno\vns can bc dctermined from three equations of equilibrium. I-Io\vever,
there are occasional exceptions to this rulc, some of \vhich are illustrated belo\\'.
p p p p p p P
22
P N
h 1 25
lA
aa n
(a)
p p p
m
H I J
A
8 12
(6)
FI G. 2.26
jp
n

72 STATICALL Y DETERMINATE PLANE TRUSSES
p
p
p
p
p
III m'
--r-
h
2
+
h
2'
-LA
p+ a n
n'
(a)
p
tA S2
(c) ER: 3 ---x
p
p
p
p
m
m'
8 1
8 1
8 4
8 4
Ip
n
p
(b)
(d)
FI G. 2.27
Sornetimes, in order to obtain the desired results, it may be necessary to
make n10re than one section or to use the method of sections in conjunction
\vith the nlethod of joints. Suppose, for exarnple, that it is rcquired to find
the axial force in each of the bars 1, 2, 3, 4 of the simple K truss sho\vn in
Fig. 2.27a. To accomplish this, \VC considcr t\VO sections and one joint as
indicated by the free-body diagrams (b), (c), and (d). Bcginning \\lith Fig.
2.27 b and using points D and C successively as l11on1cnt centers, \VC obtain
SI = -4Pa/h and 8 4 = +4Pa/h. --rhen, considering the equilibriu1l1 of the
hinge E (Fig. 2.27c) and projecting all forces onto a horizontal x axis, \VC
conclude that the forces in the bars 2 and 3 n1ust be of equal tl1agnitllde but
opposite sign, that is, 52 = -8 3 . l(eeping this condition in mind and proceed-
ing to Fig. 2.27d, \ve have only to project all forces onto a vertical axis to find
S3 = -S2 = if csc a.
It is son1ctimcs advantageous to be ahle to clnploy the method of sections in
a pureJy graphical manner; this can be done \vith vcry little trouble. By \vay

ARTICLE 2.5
73
of illustration, \ve take the truss sho\vn in Fig. 2.28a and assume that \ve \vish
to evaluate graphically the axial forces in the bars I, 2, and 3. rhcn as a
first stcp \VC considcr the cquilibrium of the entire truss and make the polygon
of forces (Fig. 2.28b) and the corresponding funicular polygon (Fig. 2.28c)
from \vhich \"c find thc reactions at the supports A and B as sho\vn. This
done, \VC isolate that portion of the truss to the left of the section 1Jt1J and
consider its conditions of cquilibriuln. It \viJl be remembered from the dis-
cussion of Art. 1.9 that the resultant of all external forces to the left of section
1/11/ is a vertical force R acting through the intersection q of thc sidcs 3 and 9 of
the funicular polygon as sho\vn. This follo\vs from the fact that the rays 9
and 3 in Fig. 2.28b are components of R a'nd must intersect on its line of action.
Since the internal f<;>rces SI, S2, and S3 must hold the force R in equilibrium,
,ve scc that our problem is simply onc of resolution of a given force R' (the
equilibrant of R) into three componcnts having specified Jines of action (the
axes of the bars 1, 2, and 3). This is a completely determinate problem and
may be carried out graphically as follo\vs: In Fig. 2.28a, ,ve.extend the line
of action of 51 to its intersection (' \vith the resultant force R. At this point
the cquili brant of R can be resolved into t\\'o components acting along the lines
CD and CE, respectively, and having the magnitudes Sl and Q as sho\vn in
Fig. 2.28d. lhen, in turn, the force Q can be transmitted to point E and there
resolved into componcnts S2 and S3 coinciding \vith the axes of the bars 2 and 3.
FinaUy, ignoring the force Q, \ve have in Fig. 2.28d thc closed polygon of
forces for that portion of the truss to the left of 1nn.
Sometimes the analysis of a truss can bc rnade in a vcry simple manner by
P3 P5
P 7
(c)
FIG. 2.28
(b)
PI "
"-
,
R  1
,.....
.....0 '2'
'3. ..... "
-- ," 
-4
--9--===O
__5-;'
-- ", //
.....6.7/
..... /" /
Rb/ 8
/

s - _...Q.. _ _ R'
8 3
(d)

74
STATICALLY DETERMINATE PLANE TRUSSES
considering it as a beam and using conventional bending-moment and shearing-
force diagran1s as discussed in Art. 1.4. Consider, for example, the simple
truss \vith parallel top and bottom chords loaded as sho\vn in Fig. 2.29a.
1aking a vertical section 11111 and \vriting equations of equilibrium for that
portion of the truss to the left of this section, \\'e obtain
Ra4a - 1\2 (1
h
S1 =
«(1)
8 2 = + a. - 
sIn a
(b)
S3 = + Ra 3a - Pl 
h
(c)
An examination of these equations no\\' sho\vs that the numerators represent,
respectively, the bending moment at point E, the shearing force at the section
1111/, and the bending moment at point D, \vhen \\re consider the truss as a beam.
The corresponding axial forces Sl, S2, and S3 arc seen to be proportional to
these quantities. Since similar conclusions may be rcached for any section
other than 11117, \ve conclude that a bending-moment diagram and a shearing-
force diagran1 arc all that is nceded for a cornplcte analysis of the truss.
P 3
,
\.
\.
,
\.
Ra 1
\.
"
\.
--2-:::=.:::O
__- 15 ///
3.... /
P ,; /
2 // 4
R i //
/
/
/
PI
Ra
a
PI
a a
.
a
a
a
a
R.
,
(a)
r-H
(b)
(c)
2
'2
(d)
FI G. 2.29

ARTICLE: 2.5
75
Pl \
R
\
p \
I \
(a) Pi P'l R 2 I ' \
,, \\
I I I
I I I " ' \
I I I "\
I I I " \
I I P2 R 2 " " ' \
I I -----O
I
I -
I P 3 - /"
..-""" /"
(c) ,.".. ./
-- ./
p.. /'
/'
./
I H
(+) (+) (b)
(d) (-)
FIG. 2.30
In such cases it is expedient to \\'ork graphically. First \VC determine the
reactions as sho\\'n in Fig. 2.29/7, and thereby \VC obtain the desircd bcnding-
n10lnent diagram as represented by thc closed funicular polygon in Fig. 2.29c.
A shearing-force diagranl (Fig. 2.29d) can ahvays bc constructed \vithout
difficulty. As \vas sho\vn in Art. 1.9, the bcnding moment for any point on
thc truss is obtained silnply by multiplying the corrcsponding ordinate of the
closed funicular polygon by the pole distance H. lhus, for example, the
ordinate (1 (Fig. 2.29c) when m.ultiplicd by H (Fig. 2.29b) gives the bending
momcnt \vith respect to point E, that is, the numerator of expression (a).
We conclude then that the force H \vhen multiplied by thc factor c11 h gives
directly the axial force SI. Similarly, the force H multiplied by c3/h gives the
axial force S3, etc. "'e obtain the force 8 2 simply by multiplying the ordinate
/2 of the shearing-force diagram by the factor csc a.
l"he bending-moment and the shearing-force diagrams for a truss arc somc-
times helpful in distinguishing bct\veen tension and compression members.
In general, for trusses \vith parallel chords, it may be ohscrved that for positive
bending moment, top chord n1embers \vill be in compression and bottom chord
members in tension, as the extreme fibcrs of a beam. For negative bcnding
moment, these conditions are simply revcrsed. For positive shear, \veb mem-
bers sloping do\vn to the right \vill be in tension, and those sloping up to the
right \vill be in olnpression. For negative shear, these conditions also are
reversed. An example, illustrating the foregoing rClnarks, is sho\vn in Fig.
2.30, Here compression members are sho\vn by heavy lines and tension
members by fine lines. "rith the aid of the bending-moment and shearing-

76 STATICALLY DETERMINATE PLANE TRUSSES
(b)
(a)
FIG. 2.31 (c)
force diagrams sho\\'n belo\\' the truss, the reader should vcri fy these results
for himself.
Another example is shown in Fig. 2.31. For a uniforn1 loading of this
truss, \\'c see from the shearing-force diagranl (Fig. 2.31b) that for the
arrangement of \vcb mcnlbcrs sho\\'n in Fig. 2.3 la, \VC shall have tension in cach
diagonal and compression in each vertical, \vhereas for the arrangement sho,vn
in Fig. 2.31c, these conditions \vill be reversed. For a steel truss, the arrange-
ment in Fig. 2.31 a is better because the short verticals can carry con1pression
more efficiently than the longer diagonals, \vhich may have a tendency to
buckle. ()n the other hand, for a \vooden truss, \vith respect to \vhich the
question of buckling is not likely to be significant, the arrangement of \veb
members in Fig. 2.31 c is ideal, especially if steel rods arc used for the verticals.
PROBLEMS
1 Referring to Fig. 2.26 and using the method of sections, find the axial forces in
the bars 2, 5, and 8. l-\.sSUl11e P = 10,000 Ib, (f = 9 ft, h = 12 ft.
Ans. S2 = -1.875P, S6 = -2.5P, 58 = + 1.875P.
2 Using the Incthod of sections, conlpute the axial fi)rces in the bars 1, 2, and 3 of
the truss sho\vn in Fig. 2.32.
Al1S. 51 = +2.25P, S2 = -O.75P, S3 = -2.25P.
r- a -ra-r-a-t-a-ra-t- a-r- a ----1
C
1 450
FI G. 2.32
p

ARTICLE 2.6
77
c
A
FIG. 2.33
p
p
p
p
p
p
p
2
A  /Ju/1 D

FIG, 2.34 B P P P P P C
3 Construct hcnding-rnoment and shearing-force diagran1s tor the truss in Fig. 2.32,
and distinguish bct\\'cen tension and compression Illenlbers accordingly.
4- Referring to Fig. 2.33, prove that the axial force in the nth vcrtical from the free
end is 8 11 = - [('/1 - 1) j2]P.
5 "'lith the aid of bcnding-nl0n1cnt and shearing-force diagrams, distinguish bct\veen
'tension and cOll1pression members in the bridge loaded as sho\vn in Fig. 2.34.
2.6 COMPOUND TRUSSES
In preceding articles \\'e have considered only simple trusses formed in accord-
ance \vith the rule given on page 53. Another kind of plane tfUSS, called a
c01npolilld truss, can be forllled by interconnecting t\VO or more simple trusses
in accordance \vith the requiremcnts for complete constraint of a rigid body
in one plane (see page 58). "rhe trusses in Fig. 2.35, for example, arc of this
kind. In Fig. 2.35a t\\'o simple trusses AJJC and CEB (shaded in the figure)
arc hinged togcther at (' and othcr\visc interconnccted by the bar DE. Like-
\vise, in Fig. 2.35 b, the same t\VO simple trusses arc interconnected by three
bars so arranged that their axes ncither are parallel nor intersect in onc point.
Although sllch interconnection of t\\'o sin1ple trusses al\\rays makes a rigid
and statically detcnninate system, it frequently happens that such a truss
cannot be conlp]ete]y analyzed by the method of joints alone.
Consider, for example, th compound truss sho\vn in Fig. 2.3 6a, \\'hich
consists of two shaded simple trusses hinged together at C and also connected
by a bar DE. A a first step in the analysis of this truss, \\'C detcrminc the

78 STATICALLY DETERMINATE PLANE TRUSSES
c
(a)
B
A
D
E
C D
FI G. 2.35 (6)
F
A
B
E
p
p
(a)
p
B
8 5
(b)
FIG. 2.36
P(4 co. a -see a +1>
P(sec a +1>
external reactions at A and B as sho\vn in :Fig. 2.36b. This done, \ve can find
the axial forces in the bar,) I, 2, 3, 4 without difficulty by the tncthod of joints,
but beyond this no progress can be made by this method because there is no
joint \vhere \ve ecounter less than three unkno\vn internal forces. 'rhus the
shaded portion D]i'CGE (Fig. 2.36b) does not lend itself to analysis by the,

ARTICLE 2.6
79
mcthod of joints, and to proceed further \VC must resort to the method of
sections. 1\1aking a section 1111/ as sho\vn and considcring the equilibrium of
that portion of the truss to the right of this section, \VC take (1 as a momnt
center and \vrite
(P sec a + .g.P) 3a cos a - l(l cos a - 2Pa cos a - Soh = 0
fronl \vhich S6 = +(3Palh)(1 + - cas a). As soon as S5 is kno\vn, the
analysis may be complcted by the method of joints \vithollt further difficulty.
If \VC undertake to construct a Max \vell diagram for this truss loaded as
sho\"n in Fig. 2.37(1, \ve shall, of course, encounter the same difficulty. As
soon as the constructions indicated by heavy lines in Fig. 2.37 c have been
completed, we shall be unable to finish the reciprocal figure in the usual
manner. Consequently, to proceed further, \ve introduce the section 11111 as
d
g
.... R
....... .......
A
(a) a h
/
/
/
/
/
/
/
/
/
(b) /
/'
/'
Y
a
b
\
c'
\
d' \
\ \
, , \
" \ ,
, '\
'\
. '&
I 
-----O
(c) .... /1/
....... /11
....... //1/
....... ....
.... ....... / /1
.... e III
I /
3 fl/
I
g
FI G. 2.37 4 h

80 STATICALLY DETERMINATE: PLANE TRUSSES
shown and complctc the polygon of forces efghiSe for that portion of the truss
to thc right of this scction. 1"hcsc constructions, indicatcd by dashed lines,
arc carricd out as explained in connection with Fig. 2.28 (see page 73) and
scrve to establish the apex 5 of the reciprocal diagram. Point 5 having been
established, the remainder of the diagram as indicated by finc lines can be
constructed in thc usual !nanner \vithout further difficulty.
In Fig. 2.38a \ve have a compound truss for \vhich the method of joints fails
at the very beginning becausc there is no joint to \vhich less than three bars are
attached. It also appears at first glance that thcre is no possibility of using the
method of sections, since we cannot makc any section that cuts only thrce bars
that do not nlcct in one point. Ho\vevcr, if we note that the truss consists
essentially of t\\'O triangles ARf and BCD, \vhich arc interconnected by three
bars that ncither are parallel nor intersect in one point, \ve see at once ho\v to
proceed \vith the analysis. After the external reactions at A and B have becn
found as sho\vn, \ve isolate the triangle BCD as shown in Fig. 2.38b. In this
\vay \\'e obtain a statically deternlinate system of coplanar forces in equilibrium
and by using Eqs. (1.2) (page 11) can find the three unkno\\'n forces SJ, S2, and
S3 \vithout difficulty. As soon as these forces are known, the remainder of
the analysis can be made by the method of joints.
Another method of forming a compound tfUSS, different from those con-
p
I
P(l- a)
I
(a)
Po
I
p
B
FI G. 2.38
(b)
Po
I

ARTICLE 2.6
81
c
FI G. 2.39 A
G
F
H
B
sidered above, is illustrated in Fig. 2.39. Here \ve have several simple trusses
(shaded in the figure) that together \vith single bars are arranged in accordance
\vith the rule for assembling the bars of a simple truss. It is obvious that any
such arrangclTlent of bars and simple trusses must constitlte a rigid and
statically determinate system. Hov.rever, such compound trusses usually
require special methods of analysis. We have already seen that compound
trusses like those in Fig. 2.35 cannot bc complctely analyzed by the method
of joints alone but that it is necessary to employ also the mcthod of sections.
Similarly, for a compound truss like that sho\vn in Fig. 2.39, it may happen
that the method of joints and the method of sections, even \"hen used in con-
junction, ",ill be inadequate for a completc analysis. A general method of
analysis applicable to all such cases \vill no\\' be considered.
Referring to f'ig. 2.39, \ve note that the simple truss clements in this systcm
really serve t\\'o functions. They act as single bars in the main truss, and at
the same time they function as secondary tn/sses to transfer their 0\\-'11 loads to
the )ints at thcir ends. By separating these t\VO functions, \ve always can
make a complcte analysis of the system \vithout difficulty. In Fig. 2.40a,
for example, suppose that AIJE is a portion of a truss that has \vithin it a
secondary truss {,'E as sho\vn. In the analysis of this system, then, \ve first
replace the secondary truss by a fictitious bar CE and the system of loads Ph
P2, . . . acting upon it by a statically equivalent system consisting of tv.'o
parallel forces R 1 and R 2 at C and E as shown in Fig. 2.40b. l These substitu-
tions will not affect any of the bars outside the secondary truss, and we may
no\v proceed with the analysis of the main truss by any of the methods already
discussed, thereby obtaining the axial force S in the fictitious bar. Now, from
a study of Fig. 2.40b, it is evident that the forces sho\vn at the hinges C and E
must represent completely the action of the secondary truss on the remainder
of the system. Hence, by reversing thesc forces, \ve obtain the reactions for
the secondary truss as sho\vn in :Fig. 2.40c and can no\v make a complete
analysis of this system. If thcrc arc other secondary trusses within the main
truss, they maybe hand led in the same manncr.
1 Any coplanar system of forces can ahvays be resolved into two parallel components
applied at t\VO given points in their plane of action.

82 STATICALLY DETERMINATE PLANE TRUSSES
s
s
FIG. 2.40 (c)
R i
PI P 2
A specific application of the foregoing procedure is illustrated in Fig. 2.41,
in \vhich VlC have a roof truss \vith secondary trusses for top chord members.
We begin thc analysis by replacing the actual systcm of loads (Fig. 2.41a) by
the statically equivalent system sho\vn in Fig. 2.41 b. This leaves the second-
ary trusses free from all loads so that they function simply as single bars, and
\ve obtain a simple truss as sho\vn. A Maxwell diagram for this system, from
\vhich the axial forces in all major members can be scaled, is sho\vn in :Fig.
2.41c. Thus we arc rcady to consider the secondary trusses. Taking AF
as a typical example, \ve see that the external forces can be broken down into
t\VO balanced systems, (1) the loads P together \vith their vertical reactions
at A and F and (2) the equal, opposite, and collincar forces Sl found from the
Max\vell diagram in Fig. 2.41c. Undcr the action of this latter system, only
the top chord members \vill be active, and \ve already have the corresponding
axial force in this chord from Fig. 2.41c. Hence, in the remaining analysis of
the secondary truss, \\'e ignore the forces Sl and make a Max\vell diagram for
the vertical loading only, as sho\vn in Fig. 2.41e. To obtain the total force in

ARTICLE 2.6
83
any bar of the C0I11pound truss (Fig. 2.41a), \ve simply superinlposc the results
frolll the t\\'o separate J\/lax\\rell diagranls.
111e foregoing n1cthod of analysis of a cotnpollnd truss is sometinlcs useful
even in cases \\,here it is not strictly necessary. (:onsider, for cxanlple, the
truss sho\vn in Fig. 2.42a. i\lthough not nccessary in this case, the notion of
secondary-truss action can be used to advantage. Proceeding on this basis,
\ve regard ,4B(J" loaded as sho\vn in Fig. 2.42b, as the basic truss. Super-
p
(a)
4P
b 8 1 /
" ;Jp
" 
e h'
1
c e'
c'
2
(c) d l' d'
a
p
Sl /
/
/
(e)
FIG. 2.41

84 STATICALL Y DETERMINATE PLANE TRUSSES
c
A
p p p p p p p
(a)
E
A
p
p
c
4P
(b)
E
D
A
p p p
(c)
FIG. 2.42
1
AG
;  ;
2" P "2
D
(d)
(e)
imposed upon this arc the t\VO secondary trusses ADE and BI'V, loadcd as
sho\vn in Fig. 2.42c. Each of these secondary trusses again can be regarded
as consisting of a main truss ADE (Fig. 2.42d) that has \virhin it t\VO smallcr
trusses as sho\vn in Fig. 2.42e. FinaJly, then, by making a siolplc analysis by
inspection of each of the trusses in (/7), (d), and (e) and superilllposing the
results, \ve obtain a complete analysis of the conlpollnd truss in (a).
PROBLEMS
,1ake a complete analysis of the compound truss sho\vn in Fig. 2.43. The tri-
angles ABC and DEF arc equilateral.
c
p
p
p
p
B
 --1 A
p
FI G. 2.43 FIG. 2.44

ARTICLE 2.7
85
FI G. 2.45 P
ppppppppp
(a)
A a+a+a+a-1B
p p p p p p p p p
FI G. 2.46
(b)
A
2 Ilake complete analysis of the cotl1pound truss sho\vn in Fig. 2.44. ABC/JEFGH
is a portion of a regular duodccagon.
3 Make a complete analysis of the compound truss sho\vn in Fig. 2.45. A.ssume
P = 10 kips.
4 Make a completc analysis of the compound truss sho\vn in Fig. 2.46a, and conl-
pare \vith that in Fig. 2.46b.
5 Make a cornplete analysis of the compound trllS sho\vn in Fig. 2.46b, and com-
pare \vith that in Fig. 2.46c1.
2.7 GENERAL THEORY OF PLANE TRUSSES
\\e return no\V to the general problem of assembling a system of bars in one
plane so as to form a rigid truss. In Art. 2.1 \ve have already seen that this
can be donc in t\VO \\Tays. In one case \ve begin \vith three bars pinned
together at their ends in the form of a triangle and attach each joint thercafter
by means of two additional bars. By this procedure we obtain a simple truss

86 STATICALLY DETERMINATE PLANE TRUSSES
(Fig. 2.3a) for \vhich there \vill always exist, bct\veen the number of members
111 and the number of joints j, the relationship
111 = 2) - 3
(2.1 a)
The rigidity of such a truss is, of course, entirely indepcndent of any attach-
ment to a foundation. In the othcr case, we begin directly with a foundation
and establish each joint by means of two intersecting bars as sho\vn in Fig. 2.3b.
In this \yay, \ve obtain a simple truss the rigidity of \\-.hich depcnds on its
interconnection \\rith points of thc foundation and for \vhich, instead of Eq.
(2.] a), \ve have
111 = 2;
(2.1 b)
As we have already seen in Art. 2.2 any simple truss of the first kind requircs,
for the completion of its constraint in onc plane, three additional bars or their
equivalent, \vhcrcas for any simple truss of the second kind the constraint is
alrcady complcte. Thus, in either case \ve come finally to the same con-
clusion: namcly, for the complete constraint of j pins in one plane \ve must
intcrconncct thenl bet\veen themselves and the foundation by 1n = 2j bars Of
equivalent constraints.
A great variety of plane trusses satisfying the foregoing general rcquirement
of rigidity can be obtained by variously rearranging the bars of a simple truss
A
B
c
D
D
E
F
G
H
FI G. 2.47
(c)

ARTICLE 2.7
87
c
/)
(a)
FIG. 2.48
(b)
in such a \vay that neither the total nunlbcr of bars nor the total number of
joints is changed. Consider, for example, thc simple truss sho\vn in Fig.
2.47a. Regarding this as a systcm of two sin1ple trusses that are intercon-
nected by three bars that neither are parallel nor intcrsect in one point, we
conclude that \ve can substitute for the bar CF a bar BH and obtain the rigid
truss sho\vn in Fig. 2.47 b. By this substitution \ve change neither the numbcr
of bars nor the number of joints, but \VC no\\' have a compound truss instead
of a simple truss.
As a second example, considcr the simple truss sho\vn in Fig. 2.48a. Replac-
ing thc bar CG' by a bar AD, \ve obtain the truss sho\vn in Fig. 2.48b. lhis
arrangen1cnt of bars and external constraints still satisfies the relationship
111 = 2j but other\vise fails to fulfill either the definition of a sinlplc truss or
that of a cOInpound truss. Such a system is called a ct»l1plcx truss.
It must not be concluded from thc foregoing discussion that \ve can indis-
crirninatcly interconnect 2j bars \vith.i joints and expect to obtain a rigid
system. l'hat is, the condition 111 = 2) is not alone a complete criterion of
rigidity. Consider again, for exaIl1ple, the silnple truss sho\\'n in Fig. 2.47a.
If \VC renlOVC the diagonal f("1 from the middlc panel, \\le destroy the rigidity
of the systcm and introduce the possibility of relative translation bet\veen the
t\\'O rigid shadcd portions. A bar BI I as sho\vn in Fig. 2.47 h prevents such
distortion and is thcrcfore a legitinlate substitute for the bar F(. In fact, as
already noted, \VC no\\' have a cOlnpound tfUSS. ()n the other hand, if \ve
replace the bar }(; by a second diagonal ('I-T in the end panel, as shown in
Fig. 2.47 c, we do n.ot restore the rigidity of the truss; there is still the same

88 STATICALLY DETERMINATE PLANE TRUSSES
freedom for relative translation bct\vcen the t\VO shaded portions. Thus, not-
\vithstanding thc fact that 111 = 2j, the system in Fig. 2.47c is not a rigid one.
Accordingly, \ve conclude that \\'e must rnodify our criterion of rigidity and
say that 2j bars arc necessary and, when properly arrallKed, sufficient for the
rigid interconnection between themsclves and the foundation of) joints in one
plane.
\\/C shall no\\' consider a further significance of the relationship 111 = 2).
In Fig. 2.49, \ve take any completcly constrained plane truss (simple, com-
pound, or complex) comprised of 111 bars and j joints and load it in its O\Vn
planc and at the joints only as sho\vn. Then, to make a complete analysis of
the truss, \ve must determine thc axial force in each of the 111 bars. Replacing
each bar by the two equal but opposite reactions that it exerts on the pins at
its ends, \\'e obtain j systems of concurrent coplanar forces for cach of which
therc cxist t\VO conditions of equilibrium [Eqs. (1.1), page 3]. I-Ience, \\'c
have altogether 2j simultaneous equations involving 111 unkno\vn axial forces;
and we see that, if 111 = 2j, there arc exactly as many unkno\\'ns as there arc
equations of statics. Thus, in all but exceptional cases, to be considered later,
these equations give a definite solution to the problem. For this reason any
completely constrained plane truss that satisfies the condition 111 = i is said
to be statically dete r111i '/late. That is, the axial forces in the bars can be found
from equations of statics alone; it is unnecessary to take account of the elastic
deformations throughout the system. Consider, for exanlplc, the system
shown in Fig. 2.50a, for \\rhich 111 = 2j. Under the action of a load P applied
as sho\vn, \ve see that the bars AC' and AD arc inactive, \vhilc BC and BD
carry, respectively, tension and compression the magnitudes of which can be
found from statical conidcrations of the hinge B. O\ving to these internal
forces, BC \vill be slightly elongated, \vhilc BD \vill be shortened; consequently,
the joint B moves slightly do\vn\\'ard and to the right, \\,hilc the joint A
remains stationary. Thus, the distance AB is greater after loading than before;
but this small elastic distortion of the system does not affect the internal forces,
and we need take no account of it.
P3
FIG. 2.49
Ps

ARTICLE 2.7 89
p p
B P
B P
A
II
B
A
A
D
c
D
c
D
c
m=2j
m>2j
m<2j
FIG 2.50
(a)
(b)
(c)
If 111 > j, there \\,ill, of course, be more unkno\vn axial forces than there
are :ndcpendent equations of statics, and these equations fail to yield a unique
solution. Accordingly, the truss is said to be statically i'11deter111il1ate. Such a
case is sho\vn in "'ig. 2.50b, \vhich is identical \vith Fig. 2.50a except for the
extra bar AB. In this case, o\ving to the presence of the bar AB, the joint
B cannot move relative to the joint A, as before, \vithout stretching the bar
AB and thus inducing some movement of A, also. 'rhus, part of the load j:J
,\Till be transn1ittcd to the joint A, and the bars AC and AD also become active.
The \vay in \vhich j) is di vidcd bet\veen B and A in this case depends on the
relative rigidities of the bars. For example if AB is very flexible in comparison
\vith the other four bars, most of the load \vill be carricd at B; in the extreme
case \vhcrc A B has no rigidity, all the load is carried at B as in Fig. 2. 50a.
On the othcr hand, if AB is relatively vcry rigid, it will rcmain practically
constant in length, and A and B \vill have to movc about the same amount.
Accordingly, the load \vill be about equally dividcd bct\vccn thcse t\VO joints.
Thus, \ve see that, \\rhen a truss is statically indeterminate (111 > 2j), the
distribution of internal forces depends on the clastic deformations throughout
the system, and, to make an analysis, these deformations must bc taken into
acccunt. 1
If 111 < 2j (Fig. 2.50c), the system is not rigid and can be in equilibrium
only undcr certain conditions of external loading. '[hat is, since there are
more equations of statics than unkno\vn axial forces, it follo\\'s that these
equations \vill determine the unkno\vns and, in addition, impose certain limita-
tions on the system of external loads. In Fig. 2.50c, for example, \ve can have
equilibrium f for vertical loads as sho\vn, but arty lateral loads \vill cause the
frame to col lapse.
In follo\ving the various rules of formation of plane trusses, already dis-
cussed, there is ah\rays the possibility of accidcntally obtaining a so-called
critical fonn, i.e., a ccrtain configuration of the truss that is nonrigid, \vhereas
1 ¥.uious rnerhods of analysis of statically indeterrninate trusses \vill be discussed in Chap. 7.

90 STATICALLY DETERMINATE PLANE TRUSSES
adjacent configurations are rigid. Sometimes these critical forms are self-
evident; sometimes they are not. "Ve consider first the obvious example of a
critical form of the simple truss sho\vn in Fig. 2.51. The configurations (a)
and (c) of this truss are complctcly rigid, but bet\veen them lies thc possibility
of the critical form (b). Herc the bars CE and DE are collinear, and appre-
ciable movement of the joint E relative to the other joints can result from the
most minute changes in the lengths of the bars Of from slight play in the hinges.
In short, such a truss is not com'pletcly rigid. In the same \vay, the conlound
truss shown in Fig. 2.52a has a critical form \vhen the bars 1, 2, and 3 are
parallel as sho\vn in Fig. 2.52b or intersect in one point as sho\vn in Fig. 2.52c.
In each of these latter cases, there is a limited freedom for relative lateral
translation bet\vccn the t\VO shaded portions, and \ve must rcgard thcnl as
nonrigid forms. Such critical forms in the case of a compound tfUSS arc, of
course, complctely similar to the incompletely rigid systems of support illus-
trated in f"ig. 2.10.
Another peculiarity of the critical form is that it is ahvays statically
indeternlinate not\vithstanding the fact that the condition 111 = j is satisfied.
Consider, for example, the critical form of simple truss sho\vn in Fig. 2.51 b.
Loaded as sho\\'n in Fig. 2.53a, th,is system behaves as the truss in Fig. 2.50b,
and \\FC conclude accordingly that it is statically indeterminate. That is,
the portion of the load 1 J that is transmitted to Cl) (\vhich functions as one
bar in this case) depends on the relative elastic deformations of the bars.
When loaded as sho\vn in Fig. 2.53b, we see, by making a free body of the
hinge E, that the bars CE and DE must carry infinitc tensions in ordcr to
balance the vertical component of the external force P; and this, of coursc, is
E
FIG. 2.51
(a)
FIG. 2.52
(a)
c
E
D
(b)
(b)
c
D
(c)
(c)

ARTICLE 2.7
91
physically impossible. .Li\ctually, upon application of the load 1>, all bars of
the systcn1 \vill deform slightly, allo\ving the hinge E to assume an appreciably
lo\\rer position \"hereby CE and DE become sufficiently inclined to balance the
load P \vith finite tensions. Ho\vever, these elongations and consequently
the final configuration of equilibriurn of the system depend on the clastic
deformations of the bars, and these deformations must be taken into account
in the analysis of the truss. Thus, again, the truss is statically indeterminate.
In the more general case of a complex truss, it is not al\vays possible to
detect a critical form by inspection. The cOlnplex truss in Fig. 2.54h, for
example, has a critical form, \vhcreas the onc in Fig. 2.54a has not. In such
cases a general nlethod of detecting a critical form of truss can be based on a
consideration of the defer/11illallt of the system of j equations of equilibrium
for its j joints. If this determinant is different from zero, thc equations yield
a unique solution: namely, there is one and only one set of values for the axial
forces that can satisfy the conditions of equilibrium at each joint, and the truss
is rigid and statically determinate. On the other hand, if the dcterminant
happens to be zero, the equations of cquilibriuln fail to yield a unique solution,
and this \vill ahvays be an indication that the truss has a critical form. 'rhis
suggests, thcn, a convenient mcthod of testing for critical form, generally
I<no\\'n as the zero-load test. With no loads on the truss, \ve see at once that
one possible solution satisfying the conditions of equilibrium at each joint \\rill
be obtained by assulning all bars to be inactive, i.e., \\rith zero axial force.
Hence, if under the sanlC condition of loading \\'e can find another set of values
different from zero that also satisfy the conditions of equilibrium at each joint,
\VC shall kno\\' that the truss has a critical form.
p
c
E
D P
A
FIG. 2.53
(a)
(b)
FI G. 2.54
(a)
(b)

92
STATICALL Y DETERMINATE PLANE TRUSSES
Consider, for example, the truss in Fig. 2.51 b. 'Vith no external forces at
the joints, \ve aSSUOlC a tcnsion S in the bar CE and an equal tension in the
bar lJE. 'fhcn, \vith the same tcnsion S in each of the bars AC, BI), and AB
and \vith a con1pression y2 S in each of the diagonal bars, \VC can satisfy the
conditions of equilibrium at all joints. 'rhus, under zcro load, \ve may have
forccs in the bars different from zero, and this indicates that the truss has a
critical form. Since the determinant rnentioned above depends only on the
configuration of the truss and not at all on ho\v it is loaded, \VC conclude that a
truss of critical form will al\vays bc statically indetcrminatc regardless of how
it n1ay be loaded.
PROBLEMS
'The plane truss in Fig. 2.55 has one redundant rncrnbcr; that is, 1JJ = 2j - 3 + I.
I f BE is removed, \VC shall have a sin1plc truss. "Vhat kind of truss shall \ve have
if AB is removed: CD, nD, etc., for each bar jn turn?
c
E
A
F
B
D
FI G. 2.55
2 Using the zero-load test, prove that the complex truss in Fig. 2.48b is statically
determinate.
3 Test each of the c0l11pound trusses in Fig. 2.52 for critical tC}rnl by the rnethod of
zero loads.
4 Apply the zero-load rest to each of the conlplcx trusses sho\vn in Fig. 2.54, and
prove that (b) has a critical form \vhile (a) has not.
2.8 COMPLEX TRUSSES: HENNEBERG'S METHOD
In previous articles, \VC have considered the analysis of simple and compound
trusses by the method of joints and the n1erhod of sections. These are very
useful methods of analysis for trusses and are applicable in the majority of
practical cases. Ho\vever, in the case of a complex truss, it usually happens
that these elementary methods of analysis are not directly applicable. In
such cases, of coursc, \ve can aI\\rays proceed \\,ith the solution of 2j simul-
taneous equations of equilibriurn for the j joints of the system, but this \vill

ARTICLE 2.8
93
p
p
p
E
(a)
FI G. 2.56
(b)
(c)
(d)
usually prove highly irnpracticablc. The first \vorkable method of analysis
for cornplcx trusses \vas developed by l-Jenncberg 1 and \vill no\\' be explained.
...A.s a specific exanlplc, let us consider the conlplcx truss supported and
loaded as sho\vn in Fig. 2.5 6a. \\, notc at once that, the reactions at L4 and B
having been found, no further progrcss \vith the analysis can he rnadc by either
the method of joints or the rnethod of sections. \\'e observe, ho\vever, that
by substituting for the bar "ID a bar BE, \VC obtain a simple truss as sho\vn
in Fig. 2.56b. A complete analysis of this simple truss under any given con-
dition of loading can be made by the method of joints. Let us assume, then,
that this sirnple truss, corresponding to the given complex truss, has been
completcly analyzcd f()r t\VO particular conditions of loading, as follo\vs: (1)
the same loading as that on thc given con1plex truss (Fig. 2.56b) and (2) t\\'o
equal and opposite unit forces acting bet\\'een A and D (Fig. 2.56c). Let S:
denote the axial force in any bar of the simple truss due to the loading of the
first case (Fig. 2.56b) and .< the corresponding axial force due to the loading
of the second case (Fig. 2.5 6c). In the second case, if \VC have forces of
magnitude \.. instead of unit forces, it is obvious that the axial force in any
bar \vill be simply .r: )( instead of <. Finally, then, by superposition of thcse
t\VO cases, \VC conclude that for the combined loading sho\\'n in Fig. 2.5 6d the
axial force in any bar of the truss \vill be
5'. - s/ + / A'
· I - i Sj
(a)
and that in the particular case of the substituted bar BE, for \vhich \ve shall use
the sl1bseript a, it \vill be
Sa = S + s:..\
(b)
No\v, if \ve choose }{ of such Inagnitude that Sa froIn Eg. (b) becomes zero,
the substituted bar BE beC0l11CS inactive and rnay be rernoved, and the truss
in Fig. 2.5 6d is then identical \vith the given truss (Fig. 2.5 6a) except that the
1 Sce L. Hcnneberg, 'Statik cler Starrcn Systclnc," Bcrgstrasscr, DaCiTIstadr, GCrfl1any,
1886.

94 STATICALLY DETERMINATE PLANE TRUSSES
1
1
A
E
A
(a)
I
(b)
(e)
I
I
FI G. 2.57
(c)
(d)
action of the bar AI) on the rest of the systcln is replaced by the forces '(.
Hence, \ve conclude that that value of X \vhich makes Sa equal to zero in
Eq. (b) reprcsents the true axial force in the bar AD. Procecding in this
manner, \\'e \\'flte
s + s:X = 0
(c)
from which
x=
S'
a
s'
a
(d)
With the value of X from Eq. (d), the force in any other bar of the given
complex truss can be fcund by using Eq. (a).
A similar procedure may be uscd in the analysis of a complex truss like
that in I.'ig. 2.57 a, \vhere, to arrive at a simple truss, it is necessary to replace
t\VO bars AH and BG by fictitious bars (;C and [-ID, respectively, as sho\vn in
.Fig. 2.57 b. Considering three conditions of loading of this simple truss as
sho'W'n in (b), (c), and (d), \ve can find the corrcsponding axial forces S, s,
and S' in each bar of the truss v.,ithout difficulty by the method of joints.
Denoting, then, by X and Y the unknov.rn forces in the bars AH and BC;,
respectively, of the given complex truss and using the idca of superposition,

ARTICLE 2.8
95
\\'c find for th axial force in any bar of the fictitious truss loaded as sho\\'n
in Fig. 2.57e the follo\ving:
Si = S; + <LY + s'Y (t')
For the fictitious bars GC' and HI), denoted, respectively, by the subscripts a
and b, \\,re have
Sa = S + s:X + s Y
Sb = S + sX + S' y
Setting these values of Sa and Sb equal to zero, as bcfore, in order to realize by
the superimposed systcnls the case of the given complex truss, \ve obtain
(f)
J" S' - S"S'
x= b a a b
r's" - s's"
'ba ab
' s ' , S '
y _ Sb a - Sa b
- S' s"=- S' S"
a b b a
(g)
As soon as the values of J\ and Y have becn found from Eqs. (g), the axial force
in any bar of the complex truss is found from Eq. (e).
In the case of a complex truss of critical form \\lC shall find that the denomina-
tor of expression (d) or of exprcssions (g) becomes zero. This, of course,
indicates that the system is statically indeternlinate.
As a specific application of the I-Ienncberg nlethod, let us consider no\\' a
complete analysis of the complex truss in Fig. 2.5Ra. Assuming P = 1,000 lb,
a = 5 ft, h = 9 ft, \ve begin \vith the corresponding simple truss and make a
complete analysis (1) for the loading sho\vn in Fig. 2.58b and (2) for the unit
forces acting along the line AD in Fig. 2.58c. lhesc analyses can readily be
made by the method of joints, and the results arc recorded in colunlns 2 and 3,
respectively, of Table 2. I .
E E E
C c
T-
h 1
l A A
p p P
2 2
P
(a) (6) (c)
FI G. 2.58

96 STATICALLY DETERMINATE PLANE TRUSSES
TABLE 2.1
Bar S , /x s,
S.
I . .
(I) (2) (3) (4) (5)
1 -500 -0.574 -546 - 1,046
2 0 - O. 749 -714 -714
3 +4-55 +0.522 +498 +953
4 -391 -0.488 -427 -818
5 0 - O. 749 -714 -714
6 + 1 ,000 -0. 191 -182 +818
7 0 -0.858 -818 -818
8 0 - I ,098 - 1,046 - 1,046
a -872 +0. 91 5 +872 0
Using the values of S and s in Eq. (d), \ve find
-872
X = - +0:915 = +953 lb
Having the value of "Y, \ve may no\\' fill in cohunn 4 of the table, and then
fronl Eq. (a) the axial force for each bar of the given truss may be calculated
and recorded in colunln 5. It \vill be noted that \ve obtain zero for the force
in the fictitious bar a, \vhich scrves as a partial check on the calculations. In
this particular case, \ve obtain a further check by observing that the results in
column 5 satisfy the conditions of symmetry in Fig. 2.5S".
PROBLEMS
1 Determine the axial forces in the bars of the complex truss supported and loaded
as sho\vn in Fig. 2.59. Each inclined bar nlakes an angle of 30° \vith the horizontal.
2 Using Henneberg's method, make a complete analysis of the COJTlplcx truss sup-
ported and loaded as sho\vn in Fig. 2.60.
Hint Only the portion CDEF of this truss is of complex form.
F
G
FIG. 2.59
p

ARTICLE 2.8
97
p
p p
FIG. 2.60
r
50'
-I
A
(a)
p
FIG. 2.61 (b)
p
p
,
5'
,
-T
I
I
10'
-+
10'
FI G. 2.62
p

98
STATICALLY DETERMINATE PLANE TRUSSES
FIG. 2.63
3 Using rvlax\vell diagran1s in conjunction \\,ith Hennebcrg's method, make a com-
plete analysis of each of the cornplex trusses supportcd and loaded as 5ho\\'11 in
Fig. 2.61.
4 l\1ake a cornpletc anal ysis of the complex truss supported and loaded as sho\vn in
Fig. 2.62. 10 reduc:: this system to a sin1plc truss, remove the bar .t. and sub-
stitute a bar ,7 as indicated by the dashed line.
5 J\1ake a cornplctc analysis of the lattice truss supported and loaded as sho\\'n in
Fig. 2.63. All diagonals are il1cJined at 45° \vith the horizontal, and the sInaU
subdivisions are squa"res.
I lint To reduce this truss to a silnple truss, it is necessary to rClllOVC the
bars .\0 and :y and add the bars t1 and b.
2.9 METHOD OF VIRTUAL DISPLACEMENTS
In rnany cases, the principle of virtual displacements, as discussed in Art. 1.10,
can be used to advantage in the analysis of statically dctenninate trusses. To
find the axial force in a given bar of a truss by this n1ethod, \ve imagine that
bar to be removed and its action on the rcst of the truss representcd by t\VO
equal and opposite forces S. In this \vay, \ve obtain a nonrigid systeI11 \vith
onc degree of frccdoIn to \vhich the principle of virtual displacelnents can be
applied. 1
As a first examplc, consider the simple truss sho\vn in Fig. 2.64a, and let it
be requircd to find the axial force in the bar AIJ. Replacing this bar by
collinear forces S at A and 1), \ve obtain a nonrigid systen1 with one degree of
freedom as sho\vn in }'ig. 2.6411. .Lo\. virtual displaceITIcnt of this system can be
defined by an infinitesimal horizontal displacement 0 of the upper portion of
the truss. We see from the figure that under this displacement each of the
applied loads PI, P2, P3, . . . produces a \vork Pi , \vhile the force S at ])
1 The use of the principle of virtual displacements in the analysis of trusses \vas introduced
by Otto Mohr. See his papers, Beitragc zur Thcorie des Fachwerks, Z. Architekten-/llgr.
Ver. Hc1l11/0Ver, 1 H74, p. 5U9, and 1875, p. 17; also Bcitrag zur -rheoric des Fach\verks,
Zi'Vi/il1grnieur, 1885, p. 284.

ARTICLE 2.9
99
produces a \vork -S 0 cos a. The force S at A produces no \vork because
its point of application does not move. l--hus, the equation of virtual \vork
for the systenl becomes
(PI + ]>2 + Pa) 0 - S () cas ex = 0
(a)
from \vhich \ve find S = (/.1 1 + P2 + P3) sec a.
To obtain the force S i:t the vcrtical bar F[), \VC procced in a similar manner,
replace this bar by collinear forces S at 1) and /.', and obtain the nonrigid
systenl sho\vn in Fig. 2.64c. In this case, a virtual displacement of the system
is best defined by the infinitcsin}al angle of rotation 08 of the upper portion of
the truss as sho\\'n. lOhe corrcsponding displacements of points G, E, and If',
respectively, are 211 08, h 08, and a lJ8, and the equation of virtual \vork beconles
1\211 58 + [>2h 08 + Sa 08 = 0
(b)
from \vhich S = - (h/a) (2ft + P2).
From the preceding examples, \ve see that, in general, the determination of
the axial force in anyone bar of a tI'\.1ss by the mcthod of virtual \vork neces-
sitates the evaluation of thc virtual displacements of the points of application
of an activc forces. If there are forces at alrnost every joint of the truss, thc
amount of calculation rcquired to deternline analytically all necessary virtual
displacements Inay be considerable. In such cases, displacement diagrams as
discussed in Art. 1.10 \vill be found helpful.
V\rhile there is usually little advantage in employing the method of virtual
\vork in the analysis of simple or compolmd trusses, it is oftcn valuable in the
case of a complex truss. By \\'ay of illustration, let us consider the complex
truss sho\vn in Fig. 2.65a. If \VC succeed in finding the force in anyone bar
of this truss by the method of virtual \vork, the rcnlainder of the analysis can
be made \vithout difficulty by the method of joints. In this case, we select
the bar }/C, replace it by collincar forces S at l' and C', and obtain the nonrigid
P 3
r a -1
G liT
h
rt
,.
rl
h
-1 6 r
PI
G
I G
H PI I
68
I
F P2
,D P 3
B
Pi
P3
P 2
P2
A
(a)
(b)
(c)
FI G. 2.64

100 STATICALLY DETERMINATE PLANE TRUSSES
B B
A A
c
F D F D
E
PI P2 P2
(a) (b) (c)
FI G. 2.65
system sho\vn in Fig. 2.65b. A displacement diagran1 for this system (con-
structed as explained in .A.rt. 1.10) is sho\vn in Fig. 2.65£". Using the dis-
placements Ot , OC , od , or rather their projections c, (1, h (Fig. 2.65c) on thc
corresponding forces .5, PI, P2 (Fig. 2.65 b), \VC obtain the equation of virtual
\vork,
Sc - Pia - })2b = 0
from \vhich
S _ PIa + j)2b
"- - - - - --
c
The displaccnlcnt diagram nlust, of course, be constructed to scale and the
distances a, b, and c rncasurcd fron1 it.
Besides displaccn1ent diagrams of the kind sho\vn in Fig. 2.65£:, there is
another graphical method for detcrmining a set of compatible displacements
for the hinges of a truss from \vhich onc bar has been removed. To dcn10n-
strate this method, \ve consider first the simple exanlple in Fig. 2.66. Here
\VC have a rigid triangle ABC supported by t\VO hinged bars so that the systcm
c
FI G. 2.66
,,\ ,/
\', I /
\\ 1/
1f
o

ARTI CLE 2.9
101
has one degree of freedom. In such a case, a virtual displacen1ent of the
system can be dcfined by an infinitesin1al angular displacement 08 around the
instantaneous center O. l'he corresponding lincar displacements A/f;, BBh
and ('L\ of the hinges A, /3, and C arc proportional to the radii OA, OB, and OC
and perpendicular to them, respecti vely, as sho\vn. If \ve rotate each of these
displacen1ent vectors clock\vise by 90°, \\'c obtain the points A', B', and C '
on the corresponding radii as sho\vn. 'I'he figure A' B'C ' , so obtained,
is geometrically sinlilar to the figure ABC, since by' virtue of thc ratios
AA': RB': ec' = OA: OR: OC, corresponding sides of the t\VO figurcs are
parallel. This conclusion holds for the virtual displacement of any rigid
plane figure \\lith one degrce of freedom. Reversing the above procedure,
\ve conclude that a set of compatible virtual displacements for the joints of
any rigid planc system having one degree of freedom can be obtained by
assuming the rnagnitude of the displacelnent of one joint and then constructing
the proper similar figure like A'B'C' in Fig. 2.66.
Take, for example, the simlc truss already considered in fig. 2.64, and
suppose that the vertical bar BD is removed as sho\vn in Fig. 2.67a. In such a
case the truss can rotate around the hinge A, and \ve ean define a virtual
displacement of the systcm by any arbitrary linear displacement OSd of the
hinge [) normal to AD. Rotating this displacement vector clockwise by 90°,
,ve obtain the point I)' on the line AD. Corresponding displacements of the
other joints are no\\' obtained simply by constructing on AD' the figure
AI)'C' £'£' H'G' \vith its sides parallel, respectively, to those of the given truss.
Thus, GG' represents in magnitude the displacement of point G, HH ' that of
point H, etc. If the bar AI) of the truss is removed (Fig. 2.67/7), the virtual
displacement of the system is defined by an arbitrary horizontal displace-
ment OSd of point D; and, proceeding as before, \\'c obtain the similar figure
ABC' D' £' F'(;' H' as shown.
The foregoing procedure can be applied also in the case of a complex truss.
G
H
G
H
68d
FIG. 2.67
(a)
(b)

102 STATICALL Y DETERMINATE PLANE TRUSSES
A
c(
\
F
d 2
FI G. 2.68
PI
P2
For the truss sho\vn in Fig. 2.65 b, for example, \VC can define a virtual dis-
placement of the system by an arbitrary linear displacCll1cnt 8s b at right angles
to AB as sho\vn in Fig. 2.6H. Rotating this displacenlcnt vector by 90°, \VC
obtain the point B' on the line A B. BE being a rigid body, the corresponding
displacemcnt of E, as defined by EE', \vill be obtained by dra\ving the dashed
line B' £' parallel to BE. Then, having the displacement of F; corrcsponding
to the assumed displactmcnt of B, \ve obtain the corresponding displacclnent
of D, as defined by DD', by dra\ving £' D' parallel to £1), since E/) also is a
rigid body. Finally, the displaccITIcnt of C, as representcd by C('/, is found
by dra\\'ing B'(1/ and D'C ' parallel, respectively, to BG' and [JC'.
Having compatible displacements for all joints of the truss, the unkno\\.rn
force Scan no\\" be found by \\rriting the equation of virtual \\lork for the
system as before. In tl1is case, ho\vevcr, it can be noted fronl the figure that
since the displacen1ents have all been rotated by 90° fron1 their true directions,
the corresponding \\Torks of the various forces acting on the joints can be
obtained by taking their moments \vith respect to the points £', D ' , (". 'rhus,
for example, the \vork of the force Pion the true displacernent of the joint E
is identical \\lith the n10mcnt of PI, \vith respect to point £1, etc. LTpon using
this conclusion, \ve obtain the equation for determining S:
-};Pid i + Sd = ()
(c)
from \\Thich
S = 'l; Pid i
d
(d)
It should be noted that the valuc of S as givcn by Eq. (d) is not changed if all
the distances sllch as RB', EE', . . . are decrcased or increased in the salnc
proportion. Thus, any figurc AB'C'D' £' f" \vith sides parallel to those of the
figure ABCDEF can be uscd in \vriting Eq. (c) above.

ARTICLE 2.9
103
c
D
c
D
A
A
FI G. 2.69
(a)
(b)
The graphical constructions illustrated in Fig. 2.68 are also helpful in
detecting a critical fom1 of complex truss. Referring to Eq. (d), we see that
the truss of Fig. 2.68 can have an indetenninate form only if the distance d is
zero, i.e., if the linc fIll coincides \vith the line FC. We can gencralize this
observation as follo\\'s: If, upon removal of one bar of a complex truss, a
figure such as AB'C'D' E' F (r'ig. 2.68), defining compatible virtual displace-
ments of the joints of the remaining system, has all its sides (including the one
correponding to the removal bar) parallel to those of the given figure, then
the truss has a critical form and is statically indeterminate. Two such
examples are shown in Fig. 2.69.
PROBLEMS
Using the method of virtual displacements, find the axial force Sz in the bar x of
the complex truss supported and loaded as sho\vn in Fig. 2.70.
Ans. )( = o.
p
FIG. 2.70
2a
2a

10. STATICALLY DETERMINATE PLANE TRUSSES
FIG. 2.71
(a)
(b)
(c)
FI G. 2.72
2 LT sing the tncthod of virtual displacenlcnrs, prove that the cOlnplcx truss in Fig.
2.54a is COIllplctcly rigid, \\'hile the one in Fig. 2.54h is not.
- Test each of the complex trusses sho\vn in Fig. ,2.71 for rigidity, and idcnti fy those
\\'hich have a critical form.
4- Study the lattice truss sho\\'11 in Fig. 2.72, and prove that it has a critical fonn.
5 Prove that the statically dctcrnlinate complex truss sho\\'n in Fig. 2.54a \vill become
statically indeterminate if a hinge i introduced at the center.

Chapter 3
Influence lines
3.1 MOVING LOADS AND INFLUENCE LINES
Engineering structures in general and bridges in particular are frequently
submitted to the action of various systems of moving loads such as trucks,
trains, automobile traffic. rl'hese loads arc called live loads to distinguish them
from dead loads such as the \veight of a structurc itself. Dead loads are always
fixed in magnitude and position, \vhcrcas live loads, although fixed in magni-
tude, can have a variety of positions on the structure. One of the most com-
mon examples of live load \vith \vhich \ve have to deal in the design of railroad
bridges is represented by the \vheelloads frolll a pair of 213-ton loconl0tives,
follo\ved "by a unifonn train of 3 tons per lineal foot, as shown in Fig. 3.1.
This particular system of loads is knovln as a standard train or Cooper's E-60
loading. lhe \\,heels arc nunlbcrcd consecutively from left to right, and the
distances bet\\'cen them are as sho\vn. It must be understood that these
distances bct\vecn \vheels are fixed so that the systcm can move only as a unit.
105

108 INFLUENCE LINES
15
FIG. 3.1
The numbers directly 3bovc the \\,heels represent the load concentrations In
kips per rail or tons per track.
A tabulation of useful numerical data pertaining to the standard train above
and I<no\vn s a '!1I0'll/cut tahle for Cooper's E-60 loading is sho\vn on page 107.
Besides thc information given in Fig. 3.1, the first t\VO horizontal lines of this
table give the distance of each \\,'heel from either end of the locomotive system,
\.vhile the third and fourth lines give successive summations of loads up to and
including any chosen \vheel from either cnd of the system. 'rhc fifth horizontal
line gives the sum of moments, \vith respect to \vhecl CD, of all loads back to and
including any chosen ",heel. Thus, fo'r example, the sun1 of moments, \\,ith
rcspect to \vheeI CD, of aU loads back to and including \vheel (8), is 4,040 kip-ft.
l'he rernaining lines of te table give sums of momcnts, with respect to each
\vheel, of all loads bet,,'ecn that \vheel and any other chosen \\'hcel ahead of it.
Thus, for example, thc sum of moments of \vheels 0, . . . , @ \vith rcspect
to \\'heel @ is 12,500 kip-ft. To find this figure in the table, \ve follo\v
vertically do\vn\vard, under whecl @, to the bottom of the tablc and thcn
move horizontally to the left and read 12,500 under \vheel 0. Examples
illustrating the use of the monlcnt table \\rill be discussed later.
The analysis of structures under the action of live loads presents t\VO major
problems not cncountered in connection \vith dead loads. I.'irst, the moving
loads Inay have a dynamic effect on the structure, tending to produce vibrations,
shock, or other undesirable effects. Second, even the purely statical effect
of a systen1 of moving loads is continually changing o\ving to change in position
of the loads, and it beCOITICS necessary to consider the problen1 of ho\v to place
them on the structure in order to rcalize the most sevcrc stresses. It is only
this latter aspect of thc problem that \VC shall consider here. l
Sometimes \ve are able to determine by inspection hO\\f to place a given
systcn1 of loads on a structure so as to realize the most severe stresses. In
other cascs, it may be necessary to rcsort to mathen1atical criteria for maxima
and occasionally to trial-and-error procedures. In all cases, ho\vcver, the
problem can be greatly simplified by the use of influence 1i1les, which sho\\'
graphicaIIy just ho\v changing the position of a single load on a structure
influences various significant quantities such as reactions, bending moments,
1 For discussions of the dynarnic action of live loads on structures, sce C:hap. 12. See also
C. E. Inglis, "A 1 athcnlatical Treatise on 'V ibrations in Railway Bridges:' Cambridge
Univcrity Press, Nc\\' York, 1934.

 E

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< 0.. (l) Q.I
t 0 0. a.
 t.O (o? IJJ
I (fJ rn 
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'- . 0 II':>  B 3 u
L:&  U") rn C/)
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I t- U") s..
m 0 to <T.I  C'J t- ""-  0 r= 0 s..
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t- I!'J '1' :"j N ....
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 - ri N  e- XI r- W <":: N .-4 .-4
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.... rl ...... .... ...... !
Gt- e 0 0 0 0 0 0 0 0 0 I C' oc c :0 0
0  \ ..... 0 0 x {- "1" 00 rl 0 M (- C'J oc 0 t.O I t 
M r- oo "" CI'J.. 01 --..  1-1 t-.. "1'.. C'J --:.   If':- :J)..  ...,. 
M .... <T.I M 
J) t- to ;:;. -..- 0':0 00 t- '.:> "1" M .J ...... .-4
8 ..... ri ......
. ..,...-..- 0
0 0 0 0 0 0 0 -::> c o c 0 0 0 0 0 0 1
M 00 .-4  0-1 0 c- Ui 0 '" 0 0 , MI- ..... N C"':' 00 CC' 0  rl -
0 "1' .-   00.. I m c..:>.. m O" .. o r,t;  L1 to..   C) 
.... '<1' N I .. ..
N m t!> "" .-4 0 0) ! (   M N ......
CIJ N ...... ..... .... ..... .-4
C'
0 C- O 0 0 0 ,;:> 0 0 0 0 0 0 0 If) 0 II') N
& 0 0"> II"> !J:i \J') 0 0 t N 0 0 t.O r- l. "::' (.r) M to .,. N - .....
0 - N 0  '<1". "'1"..   ri.. L1 0.. t,,: m  "!. N  "!. r- M
......  "':' N C' "t) to M ...... 0 ;;t)  ljj .,. M N ....
N N C'.: .-4 ..... ...... rl ......
 1fup ud s " p" O"J B 1 u a W 0 1
107

108 INFLUENCE LINES
I
p
X-1
I B
r - I
I 0
(a)

I
a
112
IIc , I
-x 113
,
(6)
FIG. 3.2
R
I
 A  al t a2 
Pl P2 P3
b
I B
(c)
shearing forces, and deflections. Consider, for example, the cantilevcr beam
shown in Fig. 3.2a. In the design of this beam, we shall be especially interested
in the bending stresses at the built-in end, since this is the dangerous section.
Thus, a diagram sho\ving ho\\r the bending moment at A varies with change
in position of a load P on the beam will be helpful in deciding the most critical
position of that load. Such a diagram may be constructed as follo\\'s: First
let the load have any position on the beam as defined by the distance x from
the free end. Then the bending moment at the built-in end A is
Ma = -P(J - x)
and the factor - (I - x) by \vhich the load P is multiplicd is called the i11flue1/ce
coefficient for !vIa. Being a linear function of x, it can be represented graphicaJly
by the straight line ab as sho\\'n in Fig. 3. 2b! and this line is called the i11fiue1/cC
line for bending moment at A. Correspondingly, the diagram abo is called the
influence dingra111. The influence diagram for bending moment must not be
confused \vith a hel1dil1g-111U111cnt diagrmn for the beam. Whereas the latter
shows by each ordinate the bending moment at the corresponding section due
to ji1.ed loads, the infh1encc diagram sho\\'s by each ordinate that factor by \\,hich
a correspondingly placed load P must be multiplied to give the bending
1 We take the ordinates of the influence line do\vn if the influence coefficient is negative.

ARTICLE 3.1
109
mOlncnt at a fixed cross sectio1l. 1'hc ordinatcs of a bending-nl0n1cnt diagram
have the dimcnsion of force X length; the ordinates of an influence diagram for
bending moment, as sho\vn in l.'ig. 3.2b, have the dimension of length. Thus,
only \vhcn multiplicd by the load P do they give the proper dimension for
bending moment. Numerically, ho\\'evcr, the ordinates in f'ig. 3.2b Inay be
considcrcd as representing thc bcnding moment at A due to a correspondingly
placed 1I11;! load, and \ve shall often llse them in this sense.
c see at once from the influence diagram in Fig. 3.2b. that for maximum
bending moment at A thc load P should bc placcd in correspondcnce \\,ith the
largest ordinate, i .c., at the free end of the bcam, and that the corresponding
bending nloment at A is - PI. In this particular case such conclusions,
obtained \vith the aid of the influence diagram, are rather obvious; but for marc
complicatcd systenls, \VC shall find that this diagran1 can be of more sub-
stantial help. For cxarnplc, if \VC have a system of concentrated loads as
sho\\7n in Fig. 3.2e and if Yh Y2, and )'3 are the ordinates of the influence line
corresponding to a certain position of this system of loads on the beam, it
folJo\\'s fron1 simple statical considerations that the corresponding bending
moment at A is
Ala = - (PIYl + P2J'2 + P 3 Ya)
(a)
Since the ordinates )'1, )'2, and Ya \vill aU be increased by moving the system
of loads to the right, \ve conclude that the most critical condition of loading in
this case \vill be obtained by placing the load {)3 at the free cnd of the bearn.
Using the corresponding values of )'1, )'2, and Y3, \\'e obtain the desircd maxi-
ffiUZ11 bending rnoment at A from Eq. (i1).
If R is the resultant of the loads Ph P2, and P 3 and J'e is thc corresponding
ordinatc of the influcnce line, \ve have
RJ'c = PIYl + P2Y2 + PaYa
(b)
To prove this, \\'e nced only recall that the moment of the resultant \vith
respect to any point is cqua I to the algebraic sum of the corresponding momcnts
of its components. Equation (h) follo\vs directly from this statcment together
\vith the fact that thc ordinates )/1, Y2, Y3, and )'e arc proportional to thcir
distances from the built-in end of the beam. Such replacement of a system
of loads by their resultant, as in Eq. (b), aI\vays holds if the influence line is
rectilinear and is very useful in establishing criteria for the most critical
position on a structure of a complicated system of loads such as the standard
train in Fig. 3.1.
The influcnce diagram also can be llsed to advantage in dealing \vith uni-
formly distributed live load. Suppose, for example, that \VC \vish to calculate
the bending moment at the built-in end of a cantilever beam partly covered by

110 INFLUENCE LINES
 B
(a)
 A
I_
a
dx x --..
I
o
y
I
(b)
FI G. 3.3
---
b
a uniform load of intensity q as sho\vn in Fig. 3.3a. Considering such loading
as a scries of infinitesimal concentrated loads cach of magnitude q dx and at
the distance ..t. from the frcc end of the beam, \ve conclude, by the same
reasoning as used above foc the case of three loads, that the total bending
moment at A is
Ma = loa qy dx
(c)
where y is the ordinate of the influence line corcesponding to x. Since
loa qy dx = q loa y dx = q[areaJ
we conclude that the bending moment at A due to any partial distribution of
unifonn load on the beam is obtained simply by taking the area of the cor-
responding portion of the influence di2gram and multiplying it by the intensity
of load q. This fact is useful because it enables us to ascertain by inspection
of the influence diagram just what portion of the beam should be uniformly
loaded to obtain the most severe bending moment at a chosen section. In the
case illustrated in Fig. 3.3, of course, \ve obtain the greatest negative bending
moment at A by loading the entire span.
As mentioned ahove, we can construct an influence line for any quantity
that varies \vith change in position of load on a structure. Consider, for
example, the beam AB supportcd as sho\\'n in Fig. 3.4a. If a single load P
acts on this beam as shown, the tension induced in the tie rod CD will be
x
S = P - see a
a

ART' CLE 3.1
111
Here, again, the inRuence coefficicnt (xl a) see a, by \\,hich the load P must be
multiplied to give the corresponding tensile force S in the tie rod, is a lincar
function of .'t., and the corresponding influcnce line has the form sho\\'o in
Fig. 3.4b. We see that as .'\." varies [rom zero to I, the influence coefficients,
\vhich in this case arc pure nun1bers, vary from zero to (// a) see lX.
In the t\\'o preceding examples, the influence lines happcned to be rectilincar,
but they may not al\vays be so. Consider, for example, an influence line for
deflection 0 at the free cnd of a cantilever beam of uniform flexural rigidity
EI (Fig. 3.5). For a load P at the distance.\" from the built-in end, this
deflection is
[ .\"3.\"2 ]
() = P 31I + 2EI (I - x)
and the influence coefficient obviously is not a linear function of .r. Accord-
ingly, the influence line has the curvilinear form sho\vn in Fig. 3.5b. It \vill
be notcd that the ordinates of this line have the dimension of length + force,
e.g., inches per pound. Thus, when multiplied by the load P, they give the
proper dimension (length) for o. Since, for small deflections, the effects of
several loads on the beam \\,iU be independent of each othcr, we can conclude
at once that, for the loading sho\vn in Fig. 3.5 c, the deflection at the free end
p
c
x
(b) Y3

11 I
A' see a
/. a
 see C( lx 
a r 

PI P2 Pa
0
(6) (c)
FIG. 3.4 FI G. 3.5
I
(a)

112 INFLUENCE LINES
\vill he
[) = [)IYl + P2Y2 + P3Y3
(d)
\vhere )'1, )'2, and Y3 are the ordinatcs of the influence line corresponding to the
positions of the loads PI, P'l, and P3, respectively. Equation (b) Inust not be
used in this case because of the nonrectilinear charactcr of the influence line.
3.2 INFLUENCE LINES FOR BEAM REACTIONS
Influence lines can often be used to advantage in the detern1ination of sinlplc
beanl reactions. Consider, for cxao1ple, the simply supportcd beam sho\vn
in Fig. 3 .6a. I f a load P acts on this bean1 at the distance x fronl the right
support, the vertical reactions at A and Bare
R = p'
a I
( X )
Rb = P I - l
(a)
\\,hich \ve shall consider positive \vhen directed up\vard. "rhe factors .1.:/1
and 1 - x/I, by \vhich Pmust be multiplied to give the corresponding reactions,
are the influence coefficicnts and can bc representcd graphically by the influence
lines Ot? and ba as sho\\'11 in Fig. 3.6b and c.
p
rX1
(b)
(a)
( c ) x ----.!!
L

4+4 1
(d)

I'
FI G. 3.6

ARTICLE 3.2
113
(a)
p
t
C  a 
c
A
x-1
1--
(b)
x
b
FIG. 3.7
(c)
b
f
1
-L
.
c
YB
By using these influence lines and proceeding as explained in thc previous
article, \ve can obtain thc rtactions for any given system of loads on the beam.
For exarnplc, if there arc several loads PI, P2,. . . acting on the bealn and
)'1, )'2, . . . arc the corresponding ordinates of the influence line for Ra (Fig.
3 .6b), \ve may \\Trite
R =  ( 1),",/ . )
n , I. I
(b)
In the same \\'ay, denoting by y, y;, . the corresponding ordinates of the
influence line for lb (Fig. 3 .6c), \VC have
lb = 2; (PiJ'D
(hi)
If a portion of the span carries a unifonnly distributed load of intensity q,
each reaction \vill be obtained by nlultiplying by q the corresponding area of
the proper influcnce diagranl. Thus, for a uniform load that extends from B
to mid-span (Fig. 3 .6d), \ve obtain
1 I 1 3 I 3
Ra = q 4 2 = "8 ql Rb = q 4 i = R ql
\vhere tl12 apd fl12 are the shaded areas sho\\ln In Fig. 3.6/7 and c,
respectively.
In thc case of a simply supportcd bearn \,/ith overhang (Fig. 3.7 a), expres
sions (a) for the reactions due to a nloving load [> still hold. Ho\vever, since
the value of .'t' can no\\' vary fronl zero to I + a, \VC obtain the inRuence lines
sho\\;n in Fig. 3.7 band c. For values of of bct\vcen I and I + a, \\'e see
that the influence coefficient for Ra becomes greater than unity, while that

114 INFLUENCE LINES


for Rb becomes negative. .rhus, their SUlll re01alns constant and equal to
unity. "rhe negati ve ordinates for I
l) simply indicate that \\'hen the load is
on the overhang, this reaction is directed do\vn\\'ard.
Since the influence coefficicnts .r/ I and 1 - :r/ I in expressions (a) can be
regarded nUlllcricaIly as the reactions that \\'ould be produced at
4 and B by a
unit load in place of the load IJ
 \ve can devise a very sirnple nlethod of con-
struction of influence lines for reactions. As a first cxaInple, let us reconsider
the influence line. for the reaction R lJ in Fig. 3.7 c. If \VC ilnaginc a unit load
placed at B, the reaction Rb obviously has the nlagnitude of unity; \vhen the
unit load is at A, it is zero. "[hus, having the key points b and IT and keeping
in mind that the reaction is a linear function of x, \ve construct the straight lint

bae through these key points, and the diagranl is conlpleted.
1.he procedure above can be used also in nlorc cOlllplicated cases \vhere the
\vriting of analytical expressions for thc reactions olay becorne involved.
Consider, for example, the COIllpound beanl ACIJ supported as sho\vn in Fig.
3.8a. To construct the influence lines for the vertical reactions at A, B, and D
in this case, \ve proceed as fol1o\vs: Beginning \vith the reaction at A
 \VC
notc that this force nlust have magnitudes of unity and zero, respectively, as a
unit load successively takes positions at A and 13. Thus, through the key
points a and b (Fig. 3.8b) \ve can construct the straight line abc.. \.'.Then the
load is at [J, the reaction Ra is again zero, and \\re complete the influence
diagram \vith the straight line cd. The same procedure is used in the con-
struction of the influence lines for Rb (Fig. 3.8e) and 1(1 (Fig. 3.8d).


(b)


ABC D


II ndr
ala bt c
-r
 I
lJLlIU}I

 d
Ie
i ('
0, I
+
r I .


d


(a)


(c)


a


(d)


d
Lilliill l
c


FI G. 3.8




ARTICLE 3.2
115
A Ithough influ(ncc lines for reactions are hardly necessary for such simple
beams as those sho\\'n in Figs. 3.6 and 3.7, they may prove very helpful in
more complicatcd cases like that sho\vn in Fig. 3.8. In such a casc, \ve can
see clcarly from the influence diagrams just ho\v to place given loads on the
beam in ordcr to obtain maximuln values of the reactions. For example, under
uniform load, \\'c see that the Ill0St severc uplift on the reaction at A \'!ill be
obtained \vhen only that portion of the beam between Band J) is loaded.
Again, it is evident at a glancc that loads on the beam AC have no effect on the
reaction at D. For the maximum reaction at B, \ve must put as Illany loads as
possible on the entire beam, etc.
The principle of virtual displacements as discussed in Art. 1.10 can be used
to advantage in the construction of influence lines for reactions . Consider,
for example, the simply supported bcam ABC \\rith overhang loaded as sho\\'n
in Fig. 3. 9a. To find the reaction at A by the method of virtual displacements,
\ve rcn10ve the constraint at ./1 and rcplace it by a vertical force Ra as sho\vn
in Fig. 3.9/1. This leaves the rigid bar AC free to rotate around the fixed
hinge B. Defining such a virtual displacement of the systen1 by a linear
displacement M* of point A, \ve see that all other points on the axis of the
beam n1ust have displacements 'as sho\vn by the ordinates to the dashed line
A'C'. Thus, the equation of virtual \vork becomes
la AA' - P)' = 0
from \vhich
)'
Ra = P /tr (c)
\vhere ,)' is the displacement of the point of application of the load P. From
· Actually the displacements are int1nitesin1al circular arcs that we can consider as coincident
with their o\vn vertical projections; thus, the projection AA' for the arc AA", etc.
(a)
l-x=1 A

C
I
p
A ll
A'
--
A
C
C' c:'
--
B --- g
--
FI G. 3.9
(b)
Ra

116 INFLUENCE LINES
expression (c), \:ve conclude that by taking 4A' = unity, \ve can consider the
dashed line A'C" as the influence line for the reaction Ra. '['he negative
ordinates to the left of B indicate that this reaction \\fill be directed do\vn\vard
\vhen the load is on the overhang.
The procedure above can be llsed also in more complicated cases such as
that sho\vn in Fig. 3.1 Da, \vhere \ve have a statically dcternlinate COlTlpound
beam AG on four supports. To obtain an influence line for the vertical reac-
tion at (', \ve sirnply rmove the constraint at (' and make a positive (up\vard)
displacenlcnt (;C" of this point as sho\vn in Fig. 3.1 Db. By taking CC ' equal
to unity, the ordinates to the dashed lines defining the ne\v cont1guration of the
system \viB represent the corresponding influence coefficients for Re. In the
samc manner, removal of the constraint at G and a subsequent positive unit
displacernent of this point as sho\vn in Fig. 3. 1 Dc leads to the influence line
for Ro, as sho\vn by dashed lines. Thus, \\'e see that to obtain an influence
line for any statically determinate reaction, it is necessary only to relnove the
corresponding constraint and make a positive unit displacement of its point
of application. Then the nc\\' configuration of the systcm defines the required
influence line. It is left as an exercise for the reader to construct influence
lines for the reactions at Band }/ by this procedure.
'c have no\\'. discussed several rncthods of constructing influence lines for
reactions. 'Their application in the analysis of beams under various systems
of live loads can best be illustrated by an exanlple. For this purpose, \\re take
the sin1plc beam \vith overhang as sho\vn in Fig. 3.11 a and assurnc that \ve \vish
to evaluate the rnaximuffi reaction \vhich can be induced at B by the standard
train shc>\vn in Fig. 3. 1. V\l ith the aid of the influence I ine for lb (Fig. 3. 11 b) ,
\ve see that in this cas the nlost critical position of the train \vill be obtained
(a)
AB f CD In EF
'",AN  
at- b aJ.. - b ---
C' _ -0. 1 )'
'1"- ! --.
A B ...-----  "'-'-... E F
A'  C J) 0 
G

a
b
(b)
G

FIG. 3.10
(c)
R
A 
G'
C D E F _...-.........-...
nMn  - ------  G
E'

ART I CLE 3.2
117
(a)
C
a = 12'
1=30'
FI G. 3.11
(b)
by placing \vheel 0 at (.i' as sho\vn. rhat is, as the train rnoves slo\vly cicross
the bridge from right to left, the reaction Rb \vill have its largest value aftcr the
\vheel CD leaves the bridge and just as \vhcel 0 rcaches the cnd. Thls, the
question of placement of the loads having been disposed of, \ve arc ready to
consider the nUlnerical calculation of Rb for the chosen condition of loading.
I'his calculation can be made in two \\'ays : In one case, \\'c scale the influence
coefficients )'2, )'3, Y 4, . . . , Inultiply each by' the corrcsponding \\lheelload,
and use the formula
Ib = };(Pi')'i)
(d)
as explained on page 113. i\n casier and more accurate I11cthod, ho\vever, \vill
be to discard the influence diagram completely as soon as the critical position
of the loads has been selected and to calculate R" sirnply by equating to zcro
the algcbraic sum of nloments of all forces \vith respect to point A. Thus,
denoting by i\t[a the surn of moments of all loads on the beam with respect to
A, \\'e havc
Rb = 2:/\;/0
I
(e)
The numerical calculations indicated by Eq. (e) can be made in a very simple
manner by using 'the 1110'111(,11/ table for Cooper's E-60 loading as sho\\'n on
page 107. To use this table, \\'e first \\Trite expression (e) in the equivalent form
Rb = 2;,\;f)_ + ( !:i)
I
(e')
\vhcre 2;1\1@ denotes the sum of monlcnts of all loads on the beam \vi th respect
to \vhecl 0 and };P i denotes the summation of all loads on the beam, x being
the distance bet\veen A and wheel @. Taking the nurnerical values of j\I[0
and ];P i directly from the moment table and using I = 30 ft and ..t = 2 ft as

118 INFLUENCE LINES
sho\"n in Fig. 3.11 fl, \VC have
R - 4,520 + 19 8 X  = 163.5 ki p s
b - 30
l'his discussion assumes that the bridge consists of t\\'o parallel beams and
that there is a single track, i.e., that the load on each girder is the same as that
for one rail.
PROBLEMS
Construct an influence linc tor the axial force in the strut C!) supporting the hori-
zontal bealn A B as sh()\vn in Fig. 3.12.
2 Construct influence lines for the axial forces developed in the bars AF, UR, and
BD, \\'hich support the bean} ABC as sho\vn in Fig. 3.1 3.
3 Construct influence lines for the reactions at .4 and B of the sin1ply supported beam
sho\vn in Fig. 3.14. C sing a standard train (Cooper's E-60) , calculate the grcatest
possible value of R" if a == 10 ft, 1= 30 fr, and b == 12 ft.
Ans. i\llax Ib = 322 kips.
4 C:onstruct influcnce lines for the rcactions at /1, C, and I) of the compound beam
sho\vn in Fig. 3.15. I-Io\\' should a standard train be placed to rnake Ra a maxi-
n1um? "lhat is rhis Inaxirl1um value of Ro if t1, b, and I have the sanle numerical
values as in Prob. 3?
Ans. /lax Ra = 189.1 kips.
5 Construct influence lincs for the reactions at A, B, and C of the systcln sho\\rn in
Fig. 3.16.
p
A
B
b-1
p
B
x-1
A
D
I
E
F
FIG. 3.12
FIG. 3.13
CAB D
L a ! l !bJ
p
r-X-l BCD
 ".nJJr
 IJa+-b
FIG. 3.14
FIG. 3.15

ARTICLE 3.3
119
p
c
'i
D
FIG. 3.16
3.3 INFLUENCE LINES FOR SHEARING FORCE
In the case of a simply supported bcam (Fig. 3.17 a), an influence line for
shearing force at a given section 11111 can be obtained as follo\vs: \V'hen a load
P is to the right of the section 11111, \ve note that the shearing force at the section
is positive and numerically equal to the reaction R4. Likc\vise, when the load
is to the left of the section, the shearing force is negative and numerically equal
to the reaction Rb. Hence, \VC obtain the influence diagram for shearing force
at 11111 simply by taking thc shaded portions of the t\VO influence diagrams for
the reactions R4 and - Rb as shown in Fig. 3.1 7 b. Any ordinatc of this
diagram is numerically equal to the shearing force at the fixed section 11111 \vhen
a unit load has that position on the beam corresponding to the chosen ordinate.
As this load crosses the scction from right to Icft, thc shcaring force changes
abruptly from a maximum positive value b / I to a nlaximum negative value a/ /.
From the influence diagram (Fig. 3.17 b), it is evident that under the action
of a single load P the maximum shearing force at any chosen section will occur
\vith the load at that section (theoretically, an intinitestimal distance to one
side). . Like\vise, under uniform load, the numerical maximum of shearing
p
m r X-l
 I

(a) n
r a
b
T
1
I- f
(b) 1

FIG. 3.17

120 INFLUENCE LINES
force at a given section occurs \vhen the distribution is continuous bet\\l'ecn
that section and the more distant support. The magnitude of such shearing
force \vill be obtained as the product of the intensity of load q and that area
of the influence diagram corresponding to the loaded portion of the beam.
For example, in the case represented in Fig. 3.17, the uni form load should
extend from the sect:on 11ln to the support B, and the corresponding shearing
force at 11111 is
b b q b 2
Jl:nn = q 7 7. = 2/
The principle of virtual displacements can be used also in the construction
of influence lines for shearing forces in beams. To illustrate, we consider the
simply supported beam \vith overhang as sho\vn in Fig. 3.18a. Assuming that
we are interested in the shearing force at the section 1nn, \ve imagine the
resistance to transverse sliding at this section to be destroyed and replace it by
shearing forces V as shown. At the same time, the resistance to bending at
the section 11tn is assumed to remain intact. To realize such conditions, \ve
imagine that at the section 11111 one element of the beam of length d.t. is replaced
by t\VO links 111111' and 11n' as shov.'n in Fig. 3 .18b. This preserves the resistance
against relative angular displacement between the t\VO portions of the beam
and yet alIo\vs a small relative transverse displacement as sho\vn in Fig. 3.18e.
p
X1
c
v
A
v
a -+
b
(a)
c
I
:1 dx t;
(b)
n n'
(c)
11m.j! ; 11;1
FIG. 3.18 (d)

ARTI CLE 3.3
121
A

B C m
II !
D
n/)n.
E
D
F

a b
(a)
I
c
I
d
I
(6)
FIG. 3.19
(c)
A

B R- ::m

Thus \ve introduce one degree of freedom into the system and can define a
corrsponding virtual displacement by a small relative displacement 0 as shown
in Fig. 3.18d. 1 Noting that the two portions of the beam have to remain
parallel, \ve conclude that the displaced system must satisfy the condition
0111 :6n = a: b, and the configuration is completcly determined. Since only the
shearing forces V and thc applied load P produce \vork during the displacement,
the equation of virtual \vork becomes
Vo - Py = 0
from which
V = p

(a)
We see no\\' that the numbers represented by the ratios y /8 are influence
coefficients for shearing force at the section 1nn. Thus, the diagram in Fig.
3 .18d has the shape of an influcnce diagram, and to use it as such we need only
select a vertical scale that makcs 0 = 1.
The above procedure is pcrfcctly gencral and applicable to more complicated
cases. Consider, for cxample, the compound beam sho\vn in Fig. 3.19a, and
assume that \\'C rcquire an influence diagram for shearing force at the section
11111. To obtain this by the usual methods of statics \vill involve considerable
labor because of the necessity of first finding all reactions. Only the principle
of virtual displacements offers a method of approach by which \ve can ignore
the reactive forces. By this method, we simply induce a unit transverse sliding
1 It should be kept in Inind that we are discussing displacements of int1nitesimallnagnitude
and that \ve exaggerate the distortion only for the sake of clarity in the figure.

122 INFLUENCE LINES
displacement at the section 11111, keeping all other constraints (both external
and internal) intact. The corresponding configuration of the systcm as sho\vn
in Fig. 3.19 h dcfines an influence diagranl for shcaring force at the section 1111/.
"'lith the aid of such a diagram it is easy to sce ho\v to place given loads on the
beam in order to obtain maximum shearing force at this section. 'I-he dis-
tribution of unifonn load illustrated in Fig. 3.19c, for example, is selected to
give maximum negative shearing force at 11111.
In the case of a beam submitted to the action of a series of conc,ntrated
loads like the standard train in 'ig. 3.1, the condition of loading for olaxirnum
shearing force at a given section is not ahvays self-evident. Ho\vever, \vith
the aid of an influence diagran1, it is possible to develop gcneral criteria to
determine the most critical position of the loads in any given case. 1'hcse
criteria follo\v directly from a consideration of the fluctuations in shearing force
at a given section as a series of concentrated loads Inoves across the span.
To illustrate, lct us suppose that a standard train n10ves froln right to left
across the simply supported beam sho\vn in Fig. 3.20a. As it docs so, \ve see
from the influence line that the shearing force at the section 111n increases
stcadily until \vheel CD reaches this section, since the influence coefficients
under the loads \vill be increasing both in nunlber and in rnagnitude. As wheel
CD crosses the section, there \vill be an abrupt drop in the shearing force by an
amount equal to the load on this \vhcel; but, \vith further movement to the
left (Fig. 3 .20b), the positive shearing force at 11111 continues its gradual increase
since the positive ordinates of the influence line are increasing and the negative
ordinate under the '.vhccl CD is dccreasing. This condition prcvails until
\vheel G) crosses the section, \vhen there is again a corresponding abrupt drop
in the shearing force, follo\vcd by another gradual rise, etc. 1his indicates
that there \vill be as many n1axima of shearing forcc as there are \vheels and
that each maximum occurs just as some \vheel reaches the section. Thus, as a
fir.;;t criterion for 111QximU111 shearing force at tl chosen sectio1l, the systt1ll of loads
'Inust be placed so that one of the wheels is at this section.
To dcterlnine \vhich load of the series should be placed at the givcn section
in order to obtain a maximum shearing force, VlC begin \vith the first wheel
at the section and ccnsider thc change il1 Shetlrillg force as each ne\v \vheel is
moved up. The total increase in shearing force during an advance of the
loads is evidently
 V = dRa - dP
(3.1 )
\vhere dRa is the increase in the reaction R(J due to the advance and dP is the
increase of total load to the left of the section. If expression (3.1) is positive,
it indicates that the shearing force at the section has been incrcased by advanc-
ing the loads. Hence, the procedure is repeated until there is a change in sign
of expression (3:1), \vhich indicates that the greatest maximum of shearing

ARTICLE 3.3
123

r a 
C2X320@ ::J:0 @@
b
(a)
FI G. 3.20 (6)
r n
1
A
(a)
(6)
'-
-,
-
-'-
FIG. 3.21 (c)
w
A
force has been passed. Thus, we 11UlY regard a change in sign of expression
(3.1) as the secund criterion for 1na:d111U1Jl shearing force.
If no loads enter or leave the span during one of the advances discussed above,
the changes Ra and p may be calculated in a very simple manner. Suppose,
for exan1ple, that we are advancing the loads from the position sho\vn in
Fig. 3.21 a to that sho\vn in Fig. 3.21 c. From the influence line for Ra

124 INFLUENCE LINES
Pn
(a)
l
I'
(b)
FI G. 3.22
Q'
(c)
(Fig. 3.21 b), \VC see that if lV is the resultant of all loads on the span and a the
distance the loads arc advanced, the change io Ra \V ill be
If''1
R(j = J-J tan a = T
(b)
\vhere tan a = 1/1 :s the slope of the influence line. J.lso, \vhen no load
leaves the span, /1P is sirnply Pn, the load \vhich crosses th section. Thus,
expression (3.1) beconles
J/= Vd _ p
I n
(3.1a)
and a change in sign of this expression may be taken as the second criterion
for maximunl shearing force.
I..oads entering or leaving the span complicate the calculation of Ra. In
such a case (Fig. 3.22), let JJI be the resultant of all loads on the span before
the advance, Q' a load that enters the span a distance h, and P' a load that goes
beyond A a distance c. Then, for the change in Ra as the loads are advanced
fronl the position sho\\'n in Fig. 3.22a to that shovln in Fig. 3.22c, \\Fe have
aR a = J;1 - P' (I + l) + Q/
(c)
In this expression, the. first term lVa/ / represents the increase in Ra, if \ve
assunlC that there is sufficient overhang at the left to carry the load ])! \vhich

ARTI CLE 3.3
125
has gone beyond A, \\rhercas the second term simply corrects for the fact that
there is no such overhang. The third term represents the incrcase in Ra due
to the appcarance on the span of the load Q', \vhich \vas not included in W
rhe load I)' leaving the span also affects the expression for increase of total
load to the left of the section so that
p = P n - p'
(d)
Substituting expressions (c) and (d) into Eq. (3.1), we obtain
Wa P'c Q'b
 V = I - Pn - T + . /--
(3.1 b)
Since the ratios ell and bll are usually small compared with unity and since
they are of opposite signs, the last t\VO terms in expression (3 .1 b) can be
neglected, in which case it coincides exactly with expression (3.1a).
As an' application of the foregoing criteria, \\7C shall now determine the
maximum shearing force that a Cooper's E-60 loading can induce at the section
lnn of the simple beam shown in f'ig. 3.23. Beginning \vith wheel (Dat the
section and moving up wheel @, \ve have, by Eq. (3.1a),
213 X 8 .
 V = . - -70 -- - 1 5 = + 9.34 kl ps
m 213
 L  (2)(a)(4) @(7) (8): B
I j;,
n
(a) 20' 50'
228 2' 1
&
 L ei) (a)(4)(6) @(7)@(9)
(6) 20' 50'
228
.....
B
FI G. 3.23
(c)
20, -J
50'
.1

126 INFLUENCE LINES
This being positive, the shearing force has been increased, and \ve start
again \vith ,,,heel @ at the section and move up \\'heel 0. lhen, by Eq.
(3.In),
228 X 5 ,
d V =. 70 - 30 = - 1 3.7 ]\1PS
This being ncgati ve, the shearing force has been reduced by the second advance,
and we put \\'heel C!:I at the section for maximum shearing force there (Fig.
3 .23b). No\\', to calculate this maximunl shearing force, \\rc use the forrnula 1
V = Ra - 15. Calculating Ra \vith the aid of the moment table (page 107)
as explained in the preceding article, this becornes
v = AJ([o) + 228 X 2 _ 1 5 = 6,950 + 4  _ 15 = 90 8 k '
70 . 70 · IpS
It must be emphasized here that the procedurc above, i.e., the llse of Eq.
r.
(3.1) as a criterion for maximum shearing force, is applicable only to the case
of a simply supported beam \vithout overhang. In more complicated cases,
the influence line should ah\rays be dra\vn and used as a guidc in selecting the
most critical condition of loading.
PROBLEf\v1S
1 A simply supported girder has a clear span of 20 tt. \-Vhat maximum shearing
force can be induced by a standard train (Cooper's E-60), and at \vhat cross section
docs such shearing force occur?
Al1S. V mM = 75 kips.
2 A simply supported beam has a clear span of 90 ft. Const[ct the influence line
for shearing force at a cross section 20 ft fronl the left support. Ho\\r should
Cooper's £-60 loading be placed on the beanl to give maximum shearing force at
this section, and what is this maximum shearing f()rc?
Al1S. V Max = 126.9 kips.
3 i\ simply supported beam \vith an ovcralliengrh of 44 ft has a clear span of 32ft
with a 12-ft overhang on one end. Construct an influence line for shearing force
at a cross section that is 22 ft fronl either cnd of the hea(n. Ho\v should C..ooper's
E-60 loading be placed on the beam to give maximun1 shearing force at this section?
Evaluate this shearing force.
Ans. V mu = 57.1 kips.
4 J.\ssuming a uniform live load of 6 kips per lineal foot and a uniform dead load of
2 kips per lineal foot, \vhat maximum upli ft nlust be provided for at the support D
of the compound beam sho\vn in Fig. 3.IS? Assume I = 30 ft. (l = 10 ft, and
h = 20 ft.
1 We assume that wheel ( is just to the right of the section,

ARTICLE 3.4
127
5 C:onstruct an influcnce line for shearing force on the pin B of the systen1 sho\vn in
Fig. 3.15. \Vhat 111axitnuln shearing force can be induced in this pin by a Cooper's
E-60 loading? By a uniforrl1 live load of 6 kips per lineal foot? Use the same
nUI11crical data as in Proh. 4.
3.4 INFLUENCE LINES FOR BENDING MOMENT
The question of variation in bending moment at a chosen section of a beam or
girder \vith change in position of a system of moving loads is one of great
practical importance. Consider, for example, a simply supported beam AB
under the action of a moving load P as sho\\'n in Fig. 3. 24a. In such a case,
\ve conclude that so long as the load is to the right of a given section lUll, the
bending moment at this section \vill be laa. Likc\vise, when the load has any
position to the lcft of the section, the bending moment at the section will be
Rbb. In both cases, the bending moment is positive. Thus, as in the case of
shcaring force, \ve obtain the influence line for bending moment at the given
section 11/11 dire'ctly from the influence Jines for the reactions Ra and Rb. For
the p:>rtion of the beam to the right of the section, we use the influence line for
Ra and simply multiply each ordinate by the length a. Similarly, for the
portion of the bean1 to the lcft of the section, \ve use the influence line for Rb
and multiply each ordinate by the length b. Thus the complete influence
diagram for bending moment at the section 11/n is represented by the shaded
triangle (lcb sho\vn in Fig. 3. 24b. We see that the ordinates of an influence line
for bending momcnt ahvays have thc dimension of length. Thus, the product
bet\\een one of these ordinates and the load P has the proper dimension
(force X length) for bending moment.
From the constructions indicated in Fig. 3.24b, it can be seen that the maxi-

m
I
n
j-Xi
(i) j

(a)
a
b
/'
/'
/'
/'
/'
/'
/'
/'
/'
/'
/'
/'
-..... - C'" /'
-
1
b
1
b
FI G. 3.24
(6)
T
a

a

128 INFLUENCE LINES
mum ordinate of the diagram occurs under the scction 11111 and has the magnitude
ah/I. "rhus, for maximum bending moment due to a single concentrated load
1), the load should be placed at the section, and the corresponding bending
I110mcnt is Pah/l. For maximunl bending nlonlcnt due to uniformly distributed
load, the entire span should be covered. Since the area of the diagram is ah/2,
the corresponding magnitude of this maximum bending moment will be qab/2,
\vhcre q is the intensity of load. If \ve take a = b = //2, this reduces to the
\vell-known expression q12/8.
r-rhe influence line for bending moment in a beam can also be obtained by
using the principle of virtual displacements. Consider, for example, the
simply supported beam with overhangs as sho\\'n in Fig. 3.25a. To construct
an influence line for bending nloment at the section 11ln., we imagine that an
element of the beam at this section is replaced by an ideal hinge as sho'n in
Fig. 3.25 b. "rhis aIIo\\'s relative rotation bet\veen the t\VO portions of the
beam as sho\vn in Fig. 3.25c but at the same time keeps the resistance to shear
and direct tension intact. Thus, we obtain a system with one degree of
freedom as sho\vn in Fig. 3.25d. No\v to this movable system \VC apply at
any point a load P and at the hinge t\VO equal and opposite couples 1\1 as shown.
Obviously., the relation bet\veen P and At[ for equilibrium of this fictitious
system represents the desired relation bet\veen P and the bending moment at
the section 111n of the original beam in Fig. 3.2 Sa. This relation is found by
r-Xl
c A m (P) B D
I  t  I
n
(a) b 
a
11 I 1 2

i X 
(b)
p
(d)
c'
FI G. 3.25

ARTICLE 3.4
129
A U CiD
 q n
8
I
t
"/),$",,
F
G
r
(a)
FI G. 3.26
m-n
 G'
 + D F '
 ,_ G
B' E'
(b)
assurning a virtual displacen1cnt of the system as sho\vn In Fig. 3.25d and
equating to zero the corresponding virtual \vork. rhus,
A1 08 - Py = 0
\vhere 08 is the total angular displacement between the t\VO parts of the beam
and y is the vertical displacement of the point of application of the load P.
From this equation of virtual \vork, \ve find
1),\,
,vI = --.-:...
58
and \ve see that the lengths y /58 are the. influence coefficients for bending
moment at the section 11111. Hence, the diagram in Fig. 3.25d has the shape of
an influence diagram, and to use it as such \ve need only magnify the vertical
scale in the ratio I : o(). Thus, if 08 = T1.0' rad, \\'C use a magnification factor
of 10.
The conclusion above is general and can be used to great advantage in more
complicated cases. Consider, for example, the compound beam shown in
Fig. 3.260, and let it be required to construct an influence line for bending
moment at the section 11111. To accomplish this, ""C simply introduce an
imaginary hinge at this section (Fig. 3.26h) and make, bct\\'een the t\VO portions
of the beam BE, a relative angle of rotation in such a direction that positive
bending moment at 11111 \vill do positive ,\vork. The corresponding configura-
tion of the system, compatible \vith all remaining constraints, defines the
required influence line from \vhich any desired influence coefficient can be
scaled directly.
""ith the aid of the influence line in Fig. 3. 26b, we can see at once ho\v to
place given loads on the structure so as to obtain the maximum bending n10ment
,
at the section 11111. The maximum negative bending moment under the action
of a single load P, for example, occurs \vith thc load at the hinge B. Again,
under uniform live load, the \vorst condition \vill be realized \vhen the spans
AC and D} of the system are loaded, and this condition of loading also induces
negative bending moment at the section 11111.

130 INFLUENCE LINES
We shall consider no\\' the question of ho\v to place a series of concentrated
loads such as a standard locomotive on a simply supported girder so as to
induce the greatest possible bending moment at a given section. V\e begin
with the influence line AC'B for bending moment at a section 11117 as sho\vn in
Fig. 3.27 and a 110\\' the system of loads Ph . . . , Pn to move onto the span
from right to left. Then, for any position of the loads as sho\\'n in the figure,
the bending mornent at the section will be obtained simply by taking the stun
of moments produced by the individual loads. Thus,
M = PIY1 + P2Y2 + P3Ya + . . . + PnYn
(a)
\vhere YI, Y2, . . . are the influence coefficients as sho\vn in the figure. No\\"
imagine that the loads arc moved a short distance ox to the left. l'hen the
corresponding change oNl in bending nloment at the section can be expressed as
the algebraic sum of the individual changes in the ternlS PiYi of the exprcssion
(a) . Denoting by RI the resultant of all loads to the left of the section and by
R 2 the resultant of all loads to the right, \\'e can express the change oAJ as
follows,
oM = ox(R 2 tan a2 - Rl tan al)
(b)
\vhere al and az are the slopes. of the lines AC and BC, respectively, as 8ho\\'n.
Since tan al = b/l and tan az = a/I, expression (11) can be \\'ritten in the form
oAt! = ab ox ( R2 _ Rl ) (c)
J b a
In this expression, the terms R2/ band Rt/ a represent, respectively, the average
loads on the right and left portions of the span. Thus, \VC conclude froln
expression (c) that the bending moment at the section \vill continue to increase
with movement of the loads to the left so long as
R z R.
- > -.: (3.2)
b 11
i.e., so long as the average load on the right portion of the span is greater than
c
ab
I
B
FI G. 3.27
I:
a
b
:1
l

ARTICLE 3.4
131
L (i)
- -
10'
C?(3)(4)(5)
j:
(a)
20'
FIG. 3.28 (b)
that on the left. As soon as this condition is reversed, the bending moment
\vill begin to decrease.
If. during movement of the loads to the left, a nc\v load enters the span from
the right, R2/ b \vill be increased, \vhile Rl/ a \vill remain unaffected. Hence,
to reach a n1axirTIUnl bcnding mOITlent at 1111, further movement to the left is
indicated. Likc\vise, if a load leaves the span on the left, Rt/ a will be
decrased, and R2/ b \vill be unaffected, so that in this case also further move-
ment to the left is indicated. We conclude, then, that the entrance of a load
on the right or the exit of a load on the left cannot invalidate the conclusion of
the preceding paragraph and is not significant in determining the position of the
loads for greatest bending mornent at the section 11111. On the other hand,
each time a load crosses the section 11111 there \vill be a sudden increase in R 1/ a
and a corresponding decrease in R2/ b, \vhich may reverse condition (3.2).
Fronl this \ve conclude that the nl:1ximunl bending moment at 11t1l will occur
\virh some load at the section. The proper load \vill be that one \vhich, \vhen
moved across the section, reverses condition (3.2).
As a full scries of loads rnoves fronl right to left across the span, there may
be several reversals of condition (3.2). l"his simply means that the bending
moment at the section passes through several maxima, as \\'c have already
seen for shearing force (page 122). In such a case, each maximum (with
some load at the section) must be evaluatcd, and then the one that is numerically
greatest \vill be used as a basis for design. An example \\rill serve to illustrate
these ren1arks.
In Fig. 3.28, \VC have a short girder AB for \\,hich it is required to determine
the maximum bending moment at the third point under the action of a; Cooper's
E-60 loading (see Fig. 3.1, page 106). With \vheel CD just to the right of the
section (Fig. 3 .28a), expression (3.2) becomes
1 20 1 5
20>10

132 INFLUENCE LINES
\vhile, \vith the same \vheel just to the left of the section, \ve have
90 45
20 - 10
This condition of equality indicates that the bending mornent remains constant
as ",heel @ crosses the section and up to the time \vhen wheel CD lcavcs the
span. As \vhecl 0 comes up to the section (Fig. 3.28b), conditiqn (3.2)
becomes
109.5 > 30
-- -
20 10
which indicates that the bending moment is still increasing; but as this \vheel
crosses the section, condition (3.2) drops to
 < .
20 10
and the bending n10ment begins to decrease. Hence, for maximum bending
moment at the third point, \ve place the loads as sho\vn in Fig. 3.28b. l"hc
corresponding bending moment ill = 5 5 6.5 kip-ft can be found very casily
",ith the help of the Inoment table (page 107).
ABSOLUTE MAXIIUl BENDING MOMENT Previously, we have con-
sidered ho\v to position a given system of loads on a beam in order to realize
the maximum bending nlornent at some chosen section. We shall no\v consider
a method of determining that particular cross section of the beam for \vhich the
maximum bending moment is greater than for any other cross section. That
is, as a given system of loads moves across the span, \vhat is the maximum
bending moment that occurs in the beam and at \vhat cross section does it
occur? This particular cross section is caned the dangerous section, and the
maximum bending nloment occurring there is called the absolute 111axi11lU111
bellding 1110111C11t. It is, of course, for the case of a beam of uniform cross
section, the bending moment that should be used as a basis for design.
To establish a criterion of loading for such absolute maximum bending
moment, consider the beam in Fig. 3.29'1. "\'e conclude, by inspection, that
to obtain an absolute maximum bending moment \ve should place as many
heavy loads on the beam as possible. Then, for any assumed position of these
loads, \ve kno\\' that the tnaximum bending n10ment \vill occur under that load
\vherc the shearing force in the beanl changes sign. Let us assume in Fig. 3.29
that this load is Pi, and let R denote the resultant of all the loads on the span
and x its distance from the support B. Also, let 114 denote the sum of tnOnlents
of all loads to the left of ].J i ",ith respect to the point of application of Pi; and,

ARTICLE 3.4
133
(a)
r a
I
FI G. 3.29 (b)
finally, let a denote the distance betvlcen R and Pi as shown. With these
notations, the bending moment under the load Pi can be expressed as follo\vs:
Rx
M p = T (/ - x - a) - 1\11
(d)
We assume now that \VC can vary x slightly without any loads going on or off
the span. l'hen, \\,ithin these limits, M p can be regarded as a continuous
function of x \vhich \vill be a maximum when dMpj dx = O. This gives
R
7 (I - 2x - a) = 0
from which
/-x-a=x
(e)
This expression sho\vs that M p \"ill be an absolute maximum \vhen the load Pi
and the resultant of all the loads on the span are equidistant from the ends of
the beam. The foregoing conclusions can be summarized by the following
criterion: rrhe 11laxi111u11Z bending 1nU/l1ellt in a si111ply supported be((1Jl under a
series 0./ crn/celltrated loads occurs ul1dcr that load where the shear changes sign and
is al1 absolute 1J1axil1nl111 when the loads are so placed that the 11tid-point of the span
bisects the distance a between this load and the resu/taut R of all the loads on the beanl.
In using the criterion above for absolute maximum bending moment, it
must be kept in mind that as a given series of loads moves across the beam,
there may be several absolute maxima of bending moment as different wheels
occupy the middle region of the span. However, expericnce shows that the
section of absolute maximum bending moment, i.e., the dangcrous section, is
never far removed. from the middle cross section. Thus, we can often reduce

134 INFLUENCE LINES
the amount of labor in calculation by first determining the position of the loads
for maximum bending moment at the middle cross section. This done, the
loads can be shifted slightly to satisfy the criterion for absolute nlaximum
bending moment, anc then the moment can be evaluated. For all practical
purposes it can be taken for grantcd, without further investigation, that this
\vill be the greatest p3ssible absolute maximum bending moment.
PROBLEMS
L simply supported girder AB has a dear span of 60 ft. Under the action of a
standard Cooper's £-60 loading, \vhat is the maximum bending 1110mcnt that can
occur (a) at the mid-point of the span, (b) at a third point?
AllS. (a) 1,907.5 kip-ft. (b) 1,73R.7 kip-fro
2 A. sin1ply supported beam 30 ft long has a clear span of 24 ft \\.rith a 6-ft ovcrhang
at one end. 'lhat maxinlum bending rnonlcnt can be produced by a standard train
(a) at the nlid-point of the 24-ft span, (b) at the overhung support?
AllS. (a) +12.5 kip-ft. (b) -210.0 kip-ft.
3 Find the absolute I1laxinllll11 bending Olomcnt in the sirnple bcaln loaded as sho\vn
in Fig. 3.30. Ho\v docs this conlpare \vith the ll1aximl1tTI bending nlonlcnt at the
mid-point of the span?
4 Determine the absolute nlaxitnUnl bending moment in a sirnpl y supported bcatn
having a clcar span of 46 ft if the live load is a standard train (Cooper's E-60).
Ans. 1,243 kip-fro
5 Detcrn1inc the absolute maximum bending IllOlnent induced in the bCall1 C/) of the
system sho\\rn in Fig. 3.3 t.
18'
A
2 tons 2 tons
5 tODS
2 tons
5 tons
FI G. 3.30
FI G. 3.31
3.5 GIRDERS WITH FLOOR BEAMS
In those cases where long-span girders arc used in bridge construction, the Ii ve
loads arc rarely applied directly to the girder. Instead, as illustrated in Fig.
3.32a, the main girders AB carry several crossbeams at a, b, e, . . . \vhich arc
called floor bea1ns, and these in turn carry a series of stril1${erS ab, be, cd, . . .
by which the floor system of the bridge is supported. Those portions of the

ARTICLE 3.5
135
nlain girder bet\vecn the floor beams are called panels, and the points (1, b,
c, . . . arc called panel poilltJ. It follows from such construction that a load P
applied to any stringer vlill be transmitted to thc main girder only at the two
corresponding panel points. Such division of the load bct\veen t\VO pancl
points has no e-lfect upon the reactions at A and B, and the influence lines for
these quantities \vill be the same as for a girder without floor beams. In thc
construction of influence lines for shcaring force and bending moment, ho\\'cvcr,
it becomes necessary to modify slightly our prcvipus discussions.
v\r c begin \vith thc question of inRuence lines for shearing force. Since
loads can be transrnittcd to the girders only at pancl points, \ve conclude at
once that, for any condition of loading on the stringers, the shearing force in a
girder \vill be constant throughout anyone panel. rhus, \vithout ambiguity,
\ve may speak of the shearin${ force il1 a PtlllC/ instead of the shearing force at any
particular cross section of that panel. With this idea in mind, \ve shall no'}.'
consider the construction of an influence line for shearing force in the panel cd
of the girder sho\\'o in F'ig. 3.3 2a . For any position of the load P to the left
of panel point c, the shearing force in the panel cd is negative and numerically
equal to R b , \vhereas for any position of P to the right of d, it is positive and
equal to Ra. Thus, for the portions de and ac of the span, \\'e use directly the
corresponding portions d' c' and a' c' of the influence lines for R(l and R b , respec-
tively, as sho\vn in Fig. 3.3 2b.
]\T O \\' let the load P have any position within the panel cd as defined by
the distance x. Then, at c and d, the corresponding panel-point loads are,
p x-1
a b c d e
(a) A B
r:!-d d d
{-----_ lie for R
-- a
1 -----
(6) e'
a' }1
---
{
(c) 1 - - - - - InlJuellce lill e lor Ra _ e'
FIG. 3.32 a'

136 INFLUENCE LINES
respectively,
p = p
C d
and
jJ d = P d - x
d
(a)
\vhere d is the panel distance as sho\vn. LTsing the corresponding influence
coefficients )' c and )'a, we find that the shearing force in the panel is
x d - :r
J7 = 1'1) - + ')'dP .
.c d' d
(b)
This expression is a linear function of x and reduces to PYd \vhen x = 0 and to
pYc \vhen x = d. Hence, to obtain that portion of the influence line cor-
responding to positions of the load \vithin the panel ed, ,vc need only conne<.:t
the established points c' and d' by a straight line as shovln (Fig. 3.3 2b). By
the same procedure, \\'C obtain the influence diagram for shearing force in the
panel ab as sho\vn in Fig. 3.3 2e.
To construct an inAuence line for bending n10n1ent at some given cross
section 11111 of a girder \vith floor bean1s (Fig. 3.3 3a), \ve proceed in a sin1ilar
manner. Again ,,,e note that for any position of the load f) to the right of d,
we have at the section unl the bending moment laa, \vhile, for any position to
the left of point e, \ve have a bending moment Rbb. Hence, as for a girder
,,,,ithout floor beams, \,re obtain the portions a' c' and e'd' of the desired influence
p
a
b
c
Xl
d
e
(a)
"I
n
B
r- a
d
b
a
"-
.....
.....
......
.....
.......
..... /
........... ,-
....... ,-
..... /
...... ,-/
...... ,-
..... " "
...... ,-
..... i-. " "
c'j "-
,,; ........
.....
.......
.....
b
Y c
Yd
FIG. 3.33
(b)
a'
e'

ARTICLE 3.5
137
diagram (Fig. 3.33b) simply by multiplying by il and b, respectively, the cor-
respJnding ordinatcs of the influcnce lines for Ra and Rb. No\\', let the load
have any position \vithin the pancl as defined by the distance x. "[hen, as
before, the panel-point loads at c and d are given by expressions (£1); and using
these, together \vith thc established influence coefficients )'e and )'d, \ve find for
the bending moment at the section 1Nll
\ 1 } ) x + [ ) d - x
j / = ) ' ,.- . " d ----
mn "c d' d
(c)
Like expression (b), this is a linear function of x, reducing to PYd \vhen .\" = 0
and to PYe \vhen x = d. Accordingly, \ve conclude that the straight line c'd'
in Fig. 3.33 b completcs the influence diagram for bending moment at the
sectIon 1Jll/.
In general, \ve can state that in the case of a girder \-.rith floor beams, any
influence line can be obtained directly from the corresponding inRuence line for
a girder \\lithout floor beams. It is necessary only to connect by a straight
line the points of the latter influence line corresponding to the ends of the panel
under consideration. l'his foJlo\vs from the fact that if the load is applied at
any panel point, it is transmitted directly to the girder, \\'hercas for any position
of the load \vithin a panel, the shearing force and bending moment at a given
section vary linearly \vith change in position of the load.
In the case of a girder \vith floor beams the development of criteria for the
placcrnent of concentrated load systems to produce maximum effects does not
require much additional discussion. Since thc shearing forcc in an end panel
will al\vays be grcatcr than in any internlcdiatc panel and since the influence line
(Fig. 3.3 2c) in such a case has the sanlC form as an influence line for bending
moment, \ve may use the criteria already developed in Art. 3.4 for maximum
bending moment at a given section. Again, in the case of actual bending
monlcnt, the Inaxinlum \vill occur at one of the panel points since loads arc
transmitted to the girder only at such points. Thus, in this case, also, the
criteria already developed for maximum bending Inoment at a given section
are applicable, and the question of absolute maxinlum moment need not be
considered.
As an example illustrating the numerical calculation of maximum live-load
shearing force and maximum Jive-load bending nloment in the case of a girder
\\lith floor hearns, \VC assume that Fig. 3. 34a represents a single-track plate-
girder railroad bridge having a span of 88 ft and subdivided into four equal
panels as sho\vn. \\'e assunle also that the live load is a standard Cooper's
E-60 loading.
For shearing force in an end panel, \\'c havc the influence diagram sho\vn in
Fig. 3.3 2c and can use, as a criterion of maximun1, the reversal of condition (3.2)
as discussed on page 1 3 ] . \Vith \\Theel CD just to the right of panel point b

138 INFLUENCE LINES
bed
22'+ 22'+ 22'-+-- 22'
Rb
(a)
Ra
(b)
FI G. 3.34
Ro
Rb
(Fig. 3. 34a), expression (3.2) becornes
273 75
- -.- > -
66 22
\vhereas \vith this \\'heel just to the left of b, we have
243 105
-<---
66 22
1.hus, there is a reversal of condition (3.2) as wheel CD crosses the section, and
\\'c place this \\7hcel at b for maximum shearing force in the end panel ab. The
orrcsponding value of this shearing force is
b = Ra - Pa
where Ra is the reaction at A and Pa is the panel-point load at a. Using the
moment table for Cooper's E-60 loading (page 107), \ve find
R = 13,100 + . 348 X 5 = 168.5 ki p s
a 88
P 720 k o
a = 22 = 32.7 IpS
so that
Jlab = 168.5 - 32.7 = 135.8 kips
For maximum -bending morncnt, \vhich \vill occur at th panel point c, we
put wheel @ at this point (Fig. 3.34b) because, as this \vheel crosses mid-span,

ARTICLE 3.5
139
condition (3.2) changes from
1 80 I 23
44- > 44
to
150 IS3
44 < 4f
rfhe corresponding value of the reaction at .4 is (see moment table)
R. = !..!,70 0 + 291 8 ; 4 2 X 2 = 146.3 kips
and the bending n10rncnt at c then becomes
;\;Jc = 146.3 X 44 - 2,605 = 3,840 kip-ft
With the foregoing maxima of shearing force and bending moment, a trial
section can be chosen for the girder on the basis of \vhich more detailed
investigations of stresses can be made.
SomctirTIcs the n1aximum live-load shearing force in an intermediate panel of
a girder \vith floor beams \viJl be required. For such cases, a criterion for
placing the loads can be developed as follo\\'s: Referring to Fig. 3.35 and
assuming that \ve require the maxin1um shearing force in the panel be as a
standard train moves from right to left across thc span, \ve donote by W the
resultant of all loads on the girder and by x its distance froIn B as sho\\'n.
Likcvlise, let W" be the resultant of all loads \vithin the panel bt and J:' its
distance from the panel point c. Then, by statics, the shearing force in the
panel be is
W\' Hl / x'
l(. = j'- - d
(d)
If the loads are no\v advanced a distance x to the left, the shcaring force
becomes
J,'J/ V'
I/, = T (x + x) - ulT( x' + AX)
(l)
and \VC see that the change in shearing force is
a v = ( r -  ) ax
(f)
As long as this expression is positive, i.e., as long as
v V'
- > ---
I d
(3.3)
the shearing force in the panel be continues to increase as the loads advance

140 INFLUENCE LINES
FIG. 3.35

b
d
I
I
FIG. 3.36
11 " E
() 0
A B
16,-t12.l-12.J-16' 
from right to left. Thus, a maximum value of c \vill be reached only ,vhen
there is a reveral of condition (3.3), and from Fig. 3.35 \ve see that this \vill
occur just as some wheel enters the panel be. Consequcntly, for maxill1um
shearing force in the panel be, \\'e place at panel point e that \\rheel \vhich upon
entering the panel be reverses condition (3.3).
PROBLEMS
Ho\v should a standard train be placed on thc girder in Fig. 3.34 to produce a maxi-
mum shearing force in the panel be? Evaluate this Inaximunl shearing force.
Alls. 65.9 kips.
2 Evaluate the maximum bending moment at the panel point b of the girder in Fig.
3.34 if the loading is a standard Cooper's E-60 train.
Ans. 2,990 l<ip-ft.
3 Dctermine the greatest floor-beam rcaction (i.e., panel-point load) produced by
the passage of a C:oopcr's E-60 train across the girder shown in Fig. 3.34.
Alls. 105.3 kips.
4 Evaluate the maximunl live-load bending 1110mcnt at the center of the span in Fig.
3.35 if I = 50 ft and d = 10 ft. Assume a Cooper's E-60 loading. Conlpare
this \vith the nlaxinlUIT1 bending nloment at c or d.
5 Referring to Fig. 3.36, find the tnaxinlum live-load shearing force nd bending
mOlnent produced in the girder AB as a standard train (C:ooper's E-60) crosses the
span.
3.6 INFLUENCE LINES FOR THREE-HINGED ARCH RIBS
In discussing influence lines for three-hinged arches, \VC begin with the case of
a nonsymmetrical arch A CB (Fig. 3.37 a) and assume that a moving vertical

ARTICLE 3.6
141
load P acts directly upon the rib as sho\vn. This load induces reactions at the
supports A and B, each of \vhich \\'e resolve into t\VO components, one vertical
and the other directed along the line AB as sho\vn. .rhen, by equating to zcro
the algebraic sum of mon1cnts of all forces ,vith respect to the points A and B,
\\'e conclude that the vertical components Ra and lb of the reactions are the
same as for a 'simple beam of span I. Dcnoting by H' the components in the
direction of AB and projecting all forces onto a horizontal axis, \\'c conclude
that these components are cqual in magnitude and opposite in direction. The
horizontal projection of H' is called the horizontal thrust of the arch, and \ve
denotc it by H, that is, H = H' cos a. To determine the magnitude of H,
\ve consider the portion A C of the arch as a free body and equate to zcro the
algebraic sum of moments of all forces \\,ith respect to the hinge C. Thus, we
\vnte
R Q l1 - H'f cos a = 0
\vhere f is the 1)crtical rise of the arch as sho\vn in the figure. From this
equation \ve find
H' cas a = H = R 7 ( a )
. aj
(Jbserving that lai1 is the bending filoment that \vould be produccd at the
distance a fron1 thc left end of a simple beam of span I, \ve conclude that during
motion of the load [.J along the arch, the horizontal thrust H is al\vays pro-
portional to this sinlple-beam bending nlonlcnt. Hence, to obtain the influence
p
c
Rb
(a)
f""" _ 1
a ' -
., ..... '........1:_,---- b
b
a
FI G. 3.37
(b)

142 INFLUENCE LINES
line for H, \ve simply divide by f the ordinates of the corresponding influence
line for bending moment at C in a sinlplc beam, as sho\vn in Fig. 3.37 b.
If a system of vertical loads acts on the arch and it is required to find their
position to make Ii a maxinlum, \VC simply use the criterion already developed
for maximum bending moment at point C in a simple beam [see Eq. (3.2),
page 130]. To obtain the horizontal thrust H when vertical load is unifornlly
distributed along the span of the arch, \ve simply nlultiply the area ach of the
influence diagratn by the intensity q of this load. Since the ordinates of the
diagram arc pure numbers in this case, the area acb has the dimension of length,
and its product \vith intensity of load q has then the dimension of force.
I..et us consider no\\r the construction of an influencc diagram for bending
moment at sonlC chosen cross section D of the unsymmetrical arch in Fig.
3.38t1. If the position of this cross section is defined by its distance c from
the reaction Ra, the bending moment for the loading sho\\'n may be expressed
as follo\\'5:
1\tl d = Rac - HYd
(b)
The first term on the right side of this expression represents the corresponding
bending moment in a simple beam of span I, \vhilc the second term represents
the moment contributed by the horizontal thrust H. Hence, the influence
diagram for j\tf d is obtained by subtracting, from the ordinates of the influcnce
line for bending moment in a simple beam, the ordinates of the influence line
for H (Fig. 3.37 b) multiplied by )'(1. The result of such subtraction is sho\vn
by the shaded areas in Fig. 3.3 8b. This influence diagram can be representcd
in a some\vhat simpler fonn by plotting the ordinates from a horizontal base ab
as sho\vn in Fig. 3. 38c. To obtain this simplified diagranl, \\'c note that the
zero point e must lie on the vertical through E (Fig. 3.3 8a) \vhere the lines
BC and AD intersect. This follo\vs from the fact that \vhen the load P passes
through point E, the total reaction at A evidently passes through point D, and
l\l d vanishes. Obscr'/ing further that the points d and c are on verticals
through D and C and that at the left end of the diagram the ordinate of the
straight line ccd must be equal to c, \ve can readily make the simplified con-
structions sho\vn in Fig. 3. 38c.
The shearing force produced at the cross section D (fig. 3.39(1) by a vertical
load P on the arch is given by the equation
fI:, = Ra CDS <P - II' sin (<p -- ex)
D A. H sin (4) - ex)
- n IJ COS 'f' - -
cos ex
(c)
\\'here <p is the angle that the tangent to the center line at }) makes \"jth the
horizontal and ex is the angle bet\veen the horizontal and the line A B as shown.
l"hc first term on the right side of expression (c) represents the corresponding
simple-beam shearing force at D multiplied by cos 4>, while the second term

ARTICLE 3.6
p
(a)
c
b
-1 1 Ud
a
b
(b)
d
a  b
(c) c
FI G. 3.38
143
p
I
c---::J-a)
7' cos a
a
b
a
b
e
dl
c
(c)
FI G. 3.39
represents the contribution to shearing force made by the horizontal thrust Ii.
Thus, the ordinates of the influence diagram for shearing force at D are ob-
tained as differences bet\veen corresponding ordinates for shearing force in a sim-
ple beam multiplied by CDS 4> and for thrust H multiplied by sin (cp - a)/cos a.
These di ffcrcnces are sho\vn by ordinatcs of the shaded areas in Fig. 3.39 b.
Here, again, the influence diagram can be prcsented in simpler forn1 by plotting
the ordinates from a horizontal base line ab as sho\\'11 in Fig. 3. 39c. In this
case, 've obtain the simplified diagram by noting that the zero point e must lie
on the vertical through E obtained as the intersection between the line BC
and the line through A which makes the angle cp \vith the horizontal. This
follo\\'s from the fact that the total reaction at A is parallel to the tangent at
D \vhcn the locrd P passes through E. Having the zero point e and noting
further that the straight line ccd has the ordinate CDS 4> at the left end of the
span, \\'e can make the constructions in Fig. 3.3 9c \vithout difficulty.
Arches arc usually so designed that thcrc can be no direct transmission of
Jive load to the rib as assumed throughout the preceding discussion; instead,

U4 INFLUENCE LINES
the loads are applied to a system of floor beams supported by thc arch as
shown in Fig. 3.40a. Under such conditions, forces can be transmitted to the
arch proper only at the panel points I, J, K, . . . , and in the construction of
influence lines we must proceed in the samc manncr as for a girder with floor
beams (see Art. 3.5). That is, we construct first the influence lines as for an
arch without floor beams and then connect by straight lines those points \vhich
corespond to the extremities of the panel under consideration. For example,
to obtain the influence line for horizontal thrust H of the arch in Fig. 3.40a, we'
begin in Fig. 3.40b with the influence line acb, which assumes direct trans-
mission of the load P (see Fig. 3.3 7 b), and then join by a straight line the
points j and k corresponding to the panel points J and K as shown. Thus, the
p
a
f
a --
--1.--
6 1
lid
c -1_ _ _
---
(a)
b
a
(b)
b
., lid
a
b
(c)
jl
C08 C/I
b sm (-a)
f cos a
a
b
FIG. 3.40
(d)

ARTICLE 3.6
145
ordinatcs of the line q)kb rcpresent the required influence coefficients for
horizontal thrust l-l. "-rhc correctness of this procedure follo\vs directly frorTI
thc fact that II, as \ve already have seen from Eg. (a), is proportional to the
simplc-bean1 bending nlomcnt at C, and (' is contained \vithin the panel J [(.
If the floor beams are so arranged that the hinge C occon1es a panel point, the
influence line for H \vill be the same as for direct transmission of the load P.
The influence diagran1 for bending mOInent at the section fJ of the arch is
obtained in a sinliJar J1}anner as sho\vn in Fig. 3.40c. The portions acb and
adb of this diagram <ire first constructed on the assumption of direct trans-
mission of live load (compare \vith Fig. 3.38b), and then the pointsj, k and ;,
jl corresponding to the panel points I, f, [( are connected by straight lines as
p
D
F
c
G
E
(a)
I
D' F' A' N' C' B' G' E'
(b)
r c d-l
a' b'
c
+
d e
(c)
, 9
I
d
e
(d)
FIG. 3.41
'I

146 INFLUENCE LINES
sho\vn. l'he ordinates of the shaded portions of die diagram reprcsent the
required influence coefficients for bending moment at ]J.
The influence diagram for shearing force at the cross section D, obtained by
the same general procedure, is shown in Fig. 3.40d. It should be noted in
connection \\lith this latter diagram that, o\ving to the curvature of the arch rib,
the shearing force is not constant throughout the panel If as in the case of a
girdcr \vith floor beams.
Expressions (a), (b), and (c), developed for the case of a simple three-
hinged arch, hold also for the arch \vith overhangs as sho\\rn in Fig. 3.41 a.
Accordingly, from Eq. (a) \ve conclude that the influence line for horizontal
thrust H in this arch \vill be obtained by dividing by f the ordinates of an
influence line for bending moment at C in a corresponding compound beam
D' f"G' E', supported as sho\\'n in Fig. 3.41 b. To obtain this influence line,
\\'e use the method of virtual displacements and make at c (corresponding to C)
such a relative angular diplaccmcnt 11 08 as sho\vn in Fig. 3.41 c that aa' = a! f
and bb' = b / f. The ordinates of the shaded diagram so obtained represent
the required influence coefficients for horizontal thrust H of the arch. The
influence diagram for bending moment at N is obtained in a similar manner, as
sho\\'n in Fig. 3.41d. Here djl1l1g1t represents the influence line for bending
momcnt at f.l' of the corresponding compound beam (Fig. 3.41b), while dfcge
is simply the foregoing influence line for H \\Tith all ordinates multiplied by
Yn. In accordance \vith Eq. (b), the differences in ordinates obtained by
superposition of these t\VO diagrams represent the required influence coefficients
for bending moment at N in the arch of Fig. 3.41a. An influence diagran1 for
shearing force at N can be obtained in a similar manncr.
PROBLEMS
Construct an influence line for horizontal thrust H in a s)'lnmetrical three-hinged
arch for \vhich I = 60 fr and.r = 8 ft. (a) IT sing this fine, determine the magni-
tude of H produced by a unitorn1ly distributed load of intensity q = 1 kip/ft and
extending over the entire span. (b) Find the nlaximunl value of H produced by
a standard train (Cooper's £-60) assuming that the cro\vn C is a panel point.
An.':. (11) 56.25 kips. (b) 239 I<ips.
2 Assun1ing that the arch in Fig. 3.3 Sa has the forn1 of a parabola \\lith vertical axis
and vertex at r: and that (l = b = 1/2. that is. that the arch is symmetrical, find
the maxinlum positive bending 1l10ment at D duc to a uni [ortn live load of intensity q
per unit .1ength of span. Note that only the portion AE of the span should be
loaded. lhe tolknving nllnlcrical data are given: I = 60 ft, f = 15 ft, ( = 15 ft.
q = 1 kip/ft.
A."s. 67.5 kip-ft.
3 R.cferring to Fig. 3.39 and assuming, as in the preceding problenl. that the arch is
a symnlerricaJ parabola of span J = 60 ft and rise f = 15 ft, find the nunlerical
maximurn of shearing force at the quarter point D (c = 1/4) under the action of a

ARTICLE 3.7
147
c
FI G. 3.42
uniforn) live load of intensity q = I kip per fc)o[ of span. i\ssUn)c a dircct trans-
n1ission of load to the arch rib.
AllJ. 3.35 kips.
4 l..onstfilct influence diagra[l1s for horizontal thrust 1/, bending [T10mcnt i\..J d, and
shcaring force V d in the case of a symn1ctrical scn1icircular three-hinged arch of
span I as sho\vn in Fig. 3.42. ASSU[l1ing dircct tnlnstnission of a uniforn11)' dis-
tributed live load of intensity q = 1 kip per foot of span, find the maximul11 numer-
ical value of each of the forcgoing quantities. A.SSlln1e I = 60 fe, 4> = 30°.
Ans. 1/ = 15 kips, At d = -104 kip-ft. J/d = 8.06 kips.
5 Construct influence diagran1s for shearing force at Jt.w1 and for bending tnolncnt at Al
of the arch sho\vn in Fig. 3.41.
3.7 INFLUENCE LINES FOR SIMPLE TRUSSES
In the design of a truss \VC must find for each member the most unfavorable
position of live load and then evaluate thc corresponding axial force in that
mernber. For such investigations, \ve shall see that influence lines can be used
to advantage. We begin, for exatnple, \vith the sirnple truss sho\vn in Fig.
3.43a and aSSUIne that all loads are transmitted by n1eans of floor beams to the
joints A, C, I), . . . of the lo\\'cr chord. Then, to construct an influence line
for the axial force in an upper chord member such as EF, \ve assume a vcrtical
load P acting on the truss and make a section 11111 as shown. Considering that
portion of the truss to the left of this section as a free body and \vriting an
equation of momcnts with respect to joint D, \VC obtain
Ratl + Slh 1 = 0
from which
Sl = Roll
- hI
(a)
From this expression, \ve see that the comprcssive force Sl is proportional to
the bending moment Raa \vhich \\'ould occur at the cross section D of a corre-
spondingly loaded simple beam of span I. The same obscrvation holds also if
the load P is to the left of I). Hence, \\'e conclude that the influence line for
8 1 will be obtained by dividing by - h l the ordinates of an influence line for

148
FI G. 3.43
INFLUENCE LINES'
(a)
(b)
(c)
(d)
m
/'-...-... E 8,
h 2 Ra"
-
/ --
i --A
G -:::.. - - -
a
x
b
c
I
a
Ib
hi
l--
---d........'...... b
........
........, hI
""""j
d
h3
a
b
---
"-
"-
"
"
"-
a
b
c"
"
"
"
" l+c
"
" h 2
"-
"
"
"-
"-
"
"-
"
"
"
"-
"
"

ARTICLE 3.7
149
bending moment at the cross section 1) of a simple beam AR. This construc-
tion is sho\"n in Fig. 3.43 b, and adb is the desired influence line for axial force
in Ef'. Multiplying the load P by the ordinate y of this line, \vhich is a purc
number, \ve obtain the corresponding value of SI. In the saIne \vay, if a
uniforn11y distributed load extends ovcr the entire span, the corresponding axial
force 51 is obtained as the product of the intensity q of this load and the area abd
of the infll!cnce diagram. To find the maximum value of 51 that can bc
induced by a system of loads such as the standard train in Fig. 3.1, we proceed
in the same manner as to find the maximum bending moment at D of a simple
beam (see page 130).
An influencc line for the axial force in a lower chord member such as C'D
can be obtained in a similar manner. In this case, we use E as a moment center
and note that the force 8 3 is proportional to the simple-beam bending moment
at this point. Hence, the desired influence line will be obtained by dividing
by ha the ordinates of an influence line for bcnding monlcnt at E of a simple
beam of span /. This construction is shown in Fig. 3.43c, and \ve note that thc
influence coefficients are positive, indicating tension in the bar CD.
We shall now consider the construction of an influence line for axial force
in the diagonal ED of the truss in Fig. 3.43a. Assuming first that the load P
acts to the right of joint D and equating to zero the algebraic sum of moments
of all forces to the left of 11111 \vith respect to .point G, we obtain
S2 = Ra
h 2
(b)
Thus, so long as P is to the right of D, the ordinatcs of the influence line for S2
can be obtained simply by multiplying by c/ h 2 the ordinates of the influence
line for Ra. In this way we obtain the portion bd of the required influence line
as shown in Fig. 3.43d. When the load P is to the left of joint C, \ve consider
the equilibrium of the right portion of the truss; and an equation of moments
\vith respect to point G then gives
S2 = _ I5 b(/ + c)
h 2
Thus, for positions of the load to the lcft of C, the ordinates of the influence
line for 52 can be obtained by multiplying by - (/ + c) / h'J. the ordinates of the
influence linc for Rb. This construction is sho\vn in Fig. 3.43d by the straight
line ac, having at b the ordinate - (I + c) / h 2 . No\v, \vhen P is between C
and Dt \\'c follo\\' the same reasoning used in the case of a girder \vith floor
beams (see page 134) and conclude that to complete the diagram \Vc need only
connect the established points c and d by a straight line as sho\vn. We sec
that this influence diagram for a diagonal has a form simi lar to that for shearing
force in an intermediate panel of a girder \,ith floor beams (see Fig. 3.3 2b)
(c)

150 INFLUENCE LINES
and that the axial force S2 n1ay he either tension or compression depending on
the placement of the live load. For example, to realize the nlaximum tensile
force in £1) due to uniforn1ly distributed live load of intensity q, \\'c must load
only the portion gh of the span. '"fhcn the corresponding magnitude of 52 is
obtained as the product bet\vecn q and the area of the triangle gbd. Sinlilarly,
the maximum cOInpressivc force in t'D occurs \vhen the distributed load
extends only over the portion erg of the span. In finding the maximum value
of S2 due to a series of concentrated loads such as a standard train, ,ve proceed
in a rnanner similar to that used in finding maximum shearing force in an
intermediate panel of a girder \\'ith floor beams (see page 135).
The influence lines for vertical 111clnbcrs of a truss \vill usually he con-
structed in a manner simi]ar to that discussed above for diagonal members.
For example, if VlC rcquire an influence line for the axial force S in the vertical
bar J)f' of the truss in Fig. 3.44a, \ve make the section '/1111 as sho\\!n and then,
\vith G as a nl0nlcnt center, \\"rite equations of equilibrium for the left and right
portions of the truss as P is successively to the right of I and to the left of D.
This gives
.s =
Ra c
and
S = Rb '- -!-- (
c1
(d)
,7
froIn which \VC conclude that the required influcnce line for S \vill be obtained
froIll the influence li:1es for Ra and lb l110dificd as sho\vn in Fig. 3.4411.
The rnethod of virtual displacements can often be used to advanrage in the
construction of influence Jines for truss nlembers. Consider, for cxanlple, the
truss sho\vn in Fig. 3.45a, and assume that live load is transn1itted to the joints
(a)
H
Rb
a
C
I
./
,.,.
./
/'
,/
./
./
./
./
./
./
l+c
a
./
d ,.,../
'./
(b)
a
./
./ C
,..., -
",./ --------
.::::-------
b
FI G. 3.44

ARTICLE 3.7
1S1
x
(a)
a
b
p
I
1J
y
Ib
(b)
a8 _ _
t-.  .-- ...... ---- -- d
a
a'
(c)
(I
h 1 - _
I _---- d' ..............
)..- -
b'
FI G. 3.45
b
.... .... .... .... .... .... .... j h 1
--
I
of the lO\\lcr chord. lhen, to obtain the influence line for a member EF of
this truss, \\.'e jrnagine this bar to be removed and its actioQ on the rcmainder of
the truss represented by two equal but opposite forces 8 1 as shown. In chis
\\lay, we obtain a system \\lith one degree of freedom, as represented by the
possibility for relative rotation about joint [) bct\vccn the tVlO shaded rigid
portions. A virtual displacement of this system can be defined by the line adb
(Fig. 3.45b), the ordinates of \vhich rcprcsent the vertical deflections of points.
on the Jo\\'cr chord corresponding to a small relative angular displacement 08
bct\vecn the [",'0 rigid portions. If y denotes the deflection of the point of
application of the load P and d the shortening of the distance EF during the
assumed virtual displacement, the equation of virtual \vork becomes
SId + Py = 0
and we obtain
- P)'
51 = Ll
(e)
We see from this expression that 51 is proportional to y, and hence the deflection
line adb gives us the general shape of the required influence diagram. To
obtain the influence coefficients to scale, we need only divide the ordinates y
of this diagram by - £1. 10 do this, we note from the figure that for the
portion bd of the. line we can \\'rice y = xa 88/1 and also that A = o(J h 1 .

152 INFLUENCE LINES
Hence, for positions of the load Pto the right of D, the influence coefficients are
y _ xa
-  - - hIl
In Fig. 3.45c, then, the corresponding portion h'd' of the true influence line for
Sl must have the ordinate a/hI \vhen x = 1 as sho\vn. In the same manner, we
can sho\v that for the left portion of the truss the line a'd' must have at h'
the. ordinate h/ hI. The same procedure can be used in constructing the
influence line for any lo\ver chord member.
To obtain an influence line for the diagonal EIJ by the method of virtual
displacements, we first replace this bar by forces S2 as sho\vn in Fig. 3.46a.
Then a virtual displacemcnt of the system can be defined by a relative angular
displacement 08 bct\vccn thc two rigid shaded portions. This relative rotation
must take place about the instantaneous centcr G in such a manner that the
supported points A and B of the truss do not move vertically. To accomplish
this, we assume first that the support at B is removed and rotate the entire
truss as a rigid body about the hinge A. Small vertical displacements of the
]o\vcr chord corresponding to this rotation are indicated in Fig. 3.46h by the
straight line (lb 1 , and \ve see that, while b goes to b I , the instantaneous center
/""
h 2 ',E
L:-___A---
x
F
(a)
p
c
I
g
08 c()(J
9 I '-::::.. -=-'-1- -:;)_ _ _ _ _ _ _ d
9 -
(b)
a
b
--
c -_
--
-y (){J (I +c)
- 1
-
--
- -- b 1
(c)
c
,",__ h d'
....." -------
, ..... .....,.(+TIIIIllllll1lllill'I"
a' -
c,... - "
........
.......
.....
"
........
b'
f
(I + c)
..... h 2
- - --- ....-J
I
FI G. 3.46

ARTICLE 3.7
151
g goeS o g 1. . No\v, keeping thc left portion of the trus stationary, \VC rottc
the right portion about 1 by such an angle ()8 that pOInt R comes back to Its
original position, i.e., in Fig. 3. 46b, point b l conlCS back to b. lhc final
configuration of the bottom chord, then, is represented by the line acd b as
sho\vn. l)cnoting by y the vertical displacement of the load P corresponding
to the assumed virtual displacenlcnt of the system, the equation of virtual \vork
becom es
S2D. + r)y = 0
from \\.hich
- Py
52 =

(f)
\vhere A, as hefore, is the shortening of the distance £1) corresponding to 08.
Again, \ve see that 8 2 is proportional to y and conclude that the line aed b gives
us the general shape of the required influence line for S2.
To gi;t influcnce coefficients - yj  to scale, \ve note from Fig. 3.46b that
x
-)' = c {)8 7
and
 = o(J 11 2
Hence, for the right portion of the truss, the influence coefficients are
-.:-.
Ll - I 11 2
and the corresponding portion /7' d' of the true influcnce line has, at x = /, the
ordinate e/11 2 as sho\vn in Fig. 3.46c. In the same manner, \ve conclude that
the line a' c' fntlst have at b ' the ordinate - (I + e) / h z as shown. Comparing
Fig. 3.46c \vith Fig. 3.4 3d, \VC sce that the influence line obtained by the
method of virtual displaccnlcnts is identical \vith that previously obtained by
the method of sections.
In the analysis of trusses, \ve often use influence diagranls only to find the
most unfavorable positions of live load and then calculate the corresponding
axial forces in the I11cmbers \vithout further rcfertnce to these diagranls. For
such purpose, it is evident that \VC do not need the actual magnitudes of the
influence coefficients but only the general shapes of the diagranls, \vhich, as
we have just seen, can be obtained very easily by the method of virtual
displacements.
PROBLEMS
Referring to Fig. 3.43 and asslll11ing that live load is rransrl1itted to the lo\ver-chord
joints, find (t7) the n)\xitnlltn possihlc axial fiJrcc in cn \\,ith a uniformly dis-
trihlltcd live load of intensity q = I kip/It, (b) the InaxinllHl1 possible axial force
in R[) \vith a standard Coopcr's 1-:-60 loading. l\SStln1C that I = 120 ft, 113 = 18 fr,
c = 30 fr.
Ans. (a) 66.7 kips. (b) 80.2 kips.

154 INFLUENCE LINES
lO'101010lO10
FI G. 3.47
(0)
p
FI G. 3.48
(b)
2 Using the' rnethod of sections, construct influence diagrams for the axia] forces
51, S2, 53 in the truss sho\vn in Fig. 2.32. i\ssunlc that the live load is transmitted
to the upper-chord joints. \-"hat are the numerical maxima of these axial forces
for a unifonTI live load of intensity q = 1 kip/ft? .Assume a = 10 ft.
Ans. 51 = +33.75 kips, S2 = -26.25 kips, Sa = - 33.75 kips.
3 Using the method of virtual displaccrnenrs, construct influence lines for the axial
forces 51 and S2 in CD and 1)£, respectively, of the truss in Fig. 3.47. Assume
live load applied to the joints of the upper chord. For a live load of uniform inten-
sity q = 1 kip/ft, what arc the possible numerical n1axima of these axial f()rces?
Alls. 51 = 36.7 kips, 52 = -26.4 kips.
4 l\ssuming that live load is translnittcd to the joints of the upper chord, construct
an influence line for the middle vertical of the sinlp)c trusses sho\vn in Fig. 2.27,
page 72. Assume t1 = 10 ft, II = 12 ft. For a uniforn1 live load of intensity
q = 1 kip/ft, evaluatc the nlaxirnum value of this axial force.
Ans, S = - 5 kips.
5 C:onsrruct an influence line for axial force S in the n1iddle bar of cach of the simple
trusses shown in Fig. 3.48. ASSUJllC cPJ = 60°, cP2 = 45°.
3.8 INFLUENCE LINES FOR COMPOUND TRUSSES
A compound truss \vith subdivided panels like that sho\vn in Fig. 3.49a may
be considered as consisting of a basic simple truss (Fig. 3.49b) on \vhich are
superimposed several secondary trusses (Fig. 3 .49c). In the analysis of such a

ARTICLE 3.8
155
systcm under dead load, \ve determine first the axial forces in the bars of the
basic truss and then add algebraically the forces in the corresponding bars of
the secondary truss (see Art. 2.6). The same procedure can be used in the
construction of influence lines for the compound truss. We begin \vith the
influence lines for the bars of a secondary truss as sho\vn jn Fig. 3.49c. These
arc sho\\'n, respectively, in Fig. 3.49d for the vertical member, Fig. 3.4ge for
m
G H
1 p
I.. ,
h
B 2
(a) n
I
(b)
I ll
g1-
d
d+d
(d)
el
f - - - l
d/h
hI
 ttant/J
a.
(g)
(e)
f '-
-......
°2
b 2
ec
(f) I I
(h)
62---
-....
-
---
-
-....
FIG. 3.49

156 INFLUENCE LINES
the horizontal l11cmber, and Fig. 3.491 for either inclined member. With
these secondary diagran1s, the influence diagram for any bar of the compound
truss can easily be obtained. Consider, for example, the nlcmber CD. For
the corresponding member of the basic truss (Fig. 3.49b) the influence diagraln
is represCl1tcd by the trianglealctb1 sho\vn in Fig. 3.49g. No\v on the ordinates
of this diagram \ve must superimpose the ordinates of the diagram in Fig. 3.4ge.
Since - tan <p = idll1, this is accornplished sinlply by extending the line aIel
to point el and thcn connecting points (1 and d 1 by a straight line as sho\vn.
The figurc llle 1 d 1 b I obtained in this \\fay represents the required influence
diagratll for the member ("'D of the given con1potlnd truss.
To obtain the influence diagram for the lo\ver portion FD of a diagonal of the
compound truss, \VC construct first the influcnce diagram a2c 2 d 2 b 2 (Fig. 3.4911)
for the corresponding diagonal of the basic truss. Then, on the ordinates of
this diagram, \ve superimpose the ordinates of the secondary diagram in Fig. 3.491
as sho\vn. In this \vay, \ve obtain the shaded diagram a2c1.d2b2 (Fig. 3.49h),
the ordinates of \vhich rcpresent thc rcquired influence coefficients for the
membcr FD of the given compound truss. The upper portions of the diagonals
as \vell as the vcrticals and upper chord members of the compound truss \vill,
of coursc, have the same int1\lcnce lines as thc corresponding members of the
basic truss. rhus in Fig. 3.49/1, for example, the line a2c 2 d 2 b 2 may be used as
the influence line for the member GF of the compound truss.
In the case of a compound truss like that in Fig. 3.49tl, the influence diagrams
in Fig. 3.49g and h can also be constructed by direct application of the method
of sections as in the preceding article. If the load P is to the right of joint D,
\\..e consider the equilibrium of that portion of the truss to the left of the section
1111/ (Fig. 3.49a) and find that the lines bld l and b 2 d 2 must he the same as for
the basic truss (Fig. 3 .49b). Like\vise, if the load ]J is to the left of joint E,
\ve consider the equilibrium of that portion of the truss to the right of the
section 11111 and find the lines aiel and a2e2 as shown. Then, to complete the
construcions, it rcmains only to draw the straight lines e1d 1 and e 2 d 2 cor-
responding to positions of the load \vithin the panel EIJ.
Influence diagrams for various members of the cantilever truss shown in
Fig. 3.50a can be readily constructed by considering the compound beam in
Fig. 3.50b as the corresponding basic system. I..et us consider, for example,
the influcnce diagram for the upper-chord member CD of this truss. Making
a section 11111 as sho\vn and \vriting an equation of mon1cnts \vith respect to the
joint E, \ve conclude that for any position of the load P on the lower chord the
axial force S in the oar CD is obtained by dividing by - h the corresponding
bending moment at E' of the compound beam sho\vn in Fig. 3.50b. Hence,
the desi red influence diagram for S is obtained by dividing by h the ordinates of
an influence diagram for bending moment at E' as sho\\'n in Fig. 3.50c. Influ-
cnce diagrams for members of the lo\vcr chord of the cantilever truss can be
constructed in a similar manncr.

ARTI CLE 3.8
157
B
(a)
I' F' A' E' P J B' H' J'
Jb II ///// I . 12" rr/51.,
(b) a I a
c
(
i
j
(c)
a.
i
b 1 ,..,""
",,,,"'1'"
", ", (I + c)
c -
- e ","'" h I h
h ..,
l  a __ j
.  b
------- k
--al
(d)
(
FI G. 3.50
In the case of a diagonal member DE, the axial force Sl is obtained by using
the same section 1J/11 and \vriting an equation of moments \vith respect to point
G (Fig. 3 .50a). If the load P is to the right of joint ](, this equation gives
5. =
laC
- -----
hI
(a)
and \\'e see that the corresponding portion of the influence diagram for Sl can
be obtained by multiplying by - c/ hI the ordinates of an influence line for the
reaction RR of the compound bcam in Fig. 3.50b. This construction is repre-
sented by the line jhbk in Fig. 3.50d. If the load P is to the left of joint E,
\VC consider the equilibrium of the portion of the truss to the right of the

158 INFLUENCE LINES
section 11111 and obtain
1+ C
Sl = Rb-.-
hI
(h)
Thus, in Fig. 3.S0d, \ve obtain the portion ile of the required diagram by
multiplying by (I + c)! hI the ordinates of an influence line for the reaction Rb
h/\' p
K 3 \
8 1 \ G 3
r
h
lA
(a)
c d
a b
c
(b) e
a b
I
r---
(c)
a b
(d)
-
--I
.... I
I
I
I
I
e :  c.
f- t
; r -: : -.
a 1 ( h b
i 7 h3
-,..,..--"- 1
,..-
c _-
-- ..!!
h 2
a
b
(e)
FI G. 3.51

ARTICLE 3.8
159
of the compound bean1 in Fig. 3.50b. Finally, for positions of the load P
within the panel EI(, \ve dra\v the straight line ek, and the required influence
diagram ifekhj for Sl is completed.
In the case of a three-hinged arch under vertical load as sho\\!n in Fig. 3.5 I a,
\VC notc that the reactive forces at A and B differ from those for a simply
supported truss A B only by the presence of the horizontal thrust H. Hence,
\ve conclude that the axial force Si in any bar of the system can be obtained by
superimposing on the axial force Si, calculated as for a simply supported truss,
the additional axial force induced by the thrust action H. Thus, for the
system in Fig. 3.51a,
S. = Sl + II {:
It
S2 = S2 - H -
h 2
S H h
3 = Sa - -
113
Fronl these expressions, \ve see now that to obtain an inRuence diagram for the
axial force 51' in any bar of the three-hinged arch, \ve need only to construct
the corresponding Si influence line, as for a simply supported truss, and then
superimpose on it a certain modification of the H influence line sho\vn in Fig.
3.51 b. In Fig. 3.51c, for example, the line adb with negative ordinates is the
influence line for the axial force £1 in a simply supported truss, and acb is the
influence line for H \vith all ordinates multiplied by !d/ hi. The shaded areas
obtained by superposition of these t\VO diagrams represent the desired influence
diagram for the axial force Sl in the three-hinged arch. The influence
diagrams for S2 and S3 are obtained in a similar manner as sho\\'n in Fig. 3.51d
and e, respectively.
PROBLEMS
1 Construct influcnce diagrams for the axial forces Sh 8 2 , and 8 3 in each of the cotn..
pound trusses sho\vn in Fig. 3.52.
a a
Am
p
r
a
(a)
FI G. 3.52
p
(b)
f
a
L

160 INFLUENCE LINES
p
G
FI G. 3.53
2 Construct influence lines for 51, 82, 8 3 , S" of the cantilever bridge shown in Fig. 2.34,
page 77. The height of the truss is 15 ft, and the panel distances arc uniform at
15 ft each. What is the maximum value of 51 that can be produced by a Cooper's
E-60 loading?
Al1S. Sl = 449 kips.
Referring to Fig. 3.51, find the nlaxianun1 horizontal thrust H that can be produced
in the arch by a uniformly distributed live load of intensity q = 1 kip/ft. i\ssume
/ = 120 ft and f = 20 ft. What maximum thrust can occur in the arch under the.
action of a standard train?
Ans. II = 90 kips. H = 344.5 kips.
4 Construct an influcnce diagram for the vertical mernber AK of the three-hinged
arch sho\vn in Fig. 3.5 [a. What maximum compressive force can be induced in
this nlcmbcr by a uni form live load of intensity q = 3 kips/ ft? Assunle / = 120 ft,
f = 20 ft, h = 30 fr, and tan a = t.
Ans. 55.1<ips.
Construct influence diagrams for the axial forces 81, 8 2 , and Sa in the structure shown
in Fig. 3.53.
Hint
3
5
C..onstruct first the influence lines for Sit S2, and Sa as in a canti-
lever truss and then modify for the effect of the horizontal thrust
H as \vas done in Fig. 3.51c.

ChajJter 4
Statically determinate space
structures
4.1 CONCURRENT FORCES IN SPACE
If several forces in space have the same point of application, they can always
be reduced to a single resultant force by using the principle of the parallelogram
of forces as discussed in Art. 1.1. Consider, for example, the three concurrent
forces F 1 , F 2 , and P'3, represented by the vectors OA, OB , and OD in Fig, 4.1a,
Considering, first, only the vectors OA and DB and using the principle of the
paral1elogram of forces, \ve find their resultant OC as shown. Now, taking
this partial resultant together \\lith the remaining force F3 and again applying
the principle of the parallelogram of fores, we find their resultant as repre-
sented by the diagonal OE of the paralIelograIn ODEC. Obviously, this force
()£, \\,'hich \ve denote by R, is the resultant of the three given forces; and \vc
note that it is obtained as the diagonal of the parallelepiped formed on the three
given vectors. 1'he same resultant R can also be obtained as the closing side
Ci'E' of the space polygon of forces 0' A'C' £' sho\vn in Fig. 4.1 b. This foIlo\vs
161

162 STATICALL Y DETERMINATE SPACE STRUCTURES
from the fact that thc space figurc 0' A'C' E' is identical \vith thc space figure
OACE in Fig. 4.1a. Si:1cC the constructions sho\\'n in Fig. 4.1 can readily bc
extended to any number of concurrent forces in space, \ve conclude that the
resultant of such a system can ah\.'ays be obtained as the geometric sum of the
given forces and that its line of action \vill ahvays act through the point of
concurrence of the given forces.
It may be notcd here that if a system of concurrent forces in space (Fig.
4.1a) and the corresponding space polygon of forces (Fig. 4.1 b) are orthogo-
nally projected onto any plane, \ve obtain in the plane of projection a polygon
of forces the sides of \vhich arc. equal to the corresponding projections of the
given forces. From this fact, \\re conclude that any projection of the resultant
of the given forces is identical \vith the resultant of the corrcsponding projec-
tions of the forccs themselves. In the particular case \vherc the given system
of forces in space is in equilibrium, it folIo\vs that their orthogonal projections
on any plane \vill rcpresent a coplanar systerll of forces in equilibrium. As \VC
shall see later, this observation can often be used to advantage in dealing \vith
equilibrium of space systems.
From the forcgoing discussion, \VC can conclude at once that thc projection
on any axis of a resultant of several concurrent forces in space \vill be equal to
the algebraic sum of the projections of the given forces on thc sanle axis.
f'ronl this it follo\\'s that if the given forces are in equilibrium, the algebraic
sums of their projections on mutually orthogonal axes x, y, and z must vanish.
Thus \VC arrive at the ,vell-kno\vn equatio1ls of equili briu1Jl
2; Xi = 0
y; = 0
};Zi = 0
(4.1 )
where Xi, }i, and Zi ae the orthogonal projcctions of any force F i , and the
summations are understood to include all forces in the system.
On the basis of Eqs. (4. I) several general observations can be made that
will prove useful in later discussions. (1) l'hrec concurrent forces that do
not lie in one plane cannot be in equilibrium unless all three forccs arc zero.
To prove this statement, consider the three forces in Fig. 4.1a, and equate to
D
- - --
" ,/'" I E'
""
I
I
I
I
I
I
I
0' A'
B C C'
FIG. 4.1 (a) (b)

ARTICLE 4.1
113
A
z F;
n/ :
I I
I
N
1ft
FI G. 4.2
zero the algebraic sum of their projections on an axis normal to the plane A()B.
Then, since F: is the only force that has such a projection different from zero,
\\'e conclude that Fa = o. In the san1C \vay, by successively projecting the
forces onto axes that are normal, respectively, to the planes A()J) and BOD,
\ve conclude that }'2  0 and that Fl = O. (2) If t\\TO of four concurrent
forces that are not all in one plane are collinear, equilibrium can exist only if
the other t\VO forces are zero. l'he t\VO collinear forces, of coursc, must be
equal in magnitude and oppositc in direction. l'his statement may be proved
by projecting the systeln onto a plane normal to the line of action of the t\VO
collinear forces. 'fhen, in this plane of projection thcre are only two forces
that are not collinear, and t\VO such forces cannot be in equilibrium unless the}'
are both zero. (3) I f all but one of any number of concurrent forces in space
are coplanar, equilibrium can exist only if this odd force is zero. l'his is
proved by equating to zero the algebraic sum of the projection of all forces on
an axis normal to the coplanar forces. (4) If the kno\vn lines of action of all
but t\\'o of any number of concurrent forces in equilibrium are coplanar and the
magnitude of one of these t\VO is kno\vn, the magnitude of the other can ah\'ays
be found by projecting all forces onto an axis normal to the coplanar forces.
In dealing \vith space systems, the notion of 1nUI11ftlt of a force with respect 10
all axiJ is often useful. Io obtain the moment of a force Fi \vith respect to an
axis z (Fig. 4.2), \VC first project the force onto a plane MN that is normal to
the axis z and then take the moment of this projection f" \vith respect to the
point 0 \vhcrc the axis pierces the plane. This moment of the projection F'
with respect tQ the point 0 is equal to the doubled area of the triangle A'OB'
and is considered positive \vhen directed as sho\\Tn in the figure. frorn this
definition of moment of a force \\,irh respect to an axis, \ve see that the moment
vanishes if the force is parallel to the axis or intersects it. We see also that, jf
any systen1 of concurrent forces in space is in equilibrium, the algebraic sum
of their morncnts \vith respect to any axis must be zero, since, by the theorem
of moments,l the algebraic sum of moments of the given forces is equal to
1 S. Timoshcnkoand (). H. Young, UEngineering Mechanics," 4th cd., p. 180, McGraw-
Hi11 Book Company, Ne\\' York, 1956.

164 STATICALLY DETERMINATE SPACE STRUCTURES
the corresponding monlcnt of their resultant, and \"hen the forces arc in
equilibrium, the resultant vanishes.
Let us no \\1' turn our attention for a nlonlent to the purely geometri'cal
question of the complete constraint of a point in space. Referring to Fig.
4.3a, we see that the attachnlcnr of a point () to a rigid foundation by means of
t\VO bars OA and OR serves only to establish constraint of the point in the
plane AOB of the bars and that there remains the possibility of rotation of this
plane about the axis AB. 10 remove this possibility of rotation, a third bar
OC (Fig. 4.3b) that is not in the plane AOB of the other t\VO is necessary.
If all three bars by \vhich the point 0 is attached to the foundation lie in one
plane, complete constraint is not realized. In the case represented in Fig.
4.3c, for exatnple, the ends A, C, and B of the three bars all lie on one axis,
and there is the same unlimited freed em of rotation about this axis as in the
case represented in Fig. 4.3a. J.-\gain, in the case sho\vn in Fig. 4.3d, it is
evidcnt that considerable n10vement of point () in the direction nonnal to the
plane ABC of the bars can take place \vithout appreciable changes in the lengths
of the bars. Thus, in this case, also, \VC have unsatisfactory constraint of the
point 0 in space. Fronl this discussion, \\'c can conclude that complete and
satisfactory constraint of a point in space can always be realized by attaching
it to a foundation hy three bars the axes of \\,hich do not lie in one plane.
Let us assun1C no\v that an externdl force P is applied to the completely
constrained point 0 in Fig. 4.4. Under the action of this load, axial forces
o
o
II B
./
./
,/
./
./
./
A ./ (a) A
./
C
c
FI G. 4.3
o
B
B
A
A

ARTICLE 4.1
165
FIG. 4.4
p
A
B
\vill be induced in the three supporting bars, and accordingly each bar \vill
exert on the joint () a reaction S.. directed along the axis of the bar and repre-
senting in magnitude the corresponding axial force. Thus, at 0 \\'c have a
systcn1 of four ccncurrent forces that are in cquilihriuffi; and since aHlines of
action are given, \VC see that the only unkno\vn clements in the system are the
magnitudes 51, S2, and S3. flence, regardless of the magnitude or direction of
the applied force P, the corresponding rnagnitudes of the three reactions can
ahvays he found by using Eqs. (4.1). For this reason the systcrll is said to be
statically deternlilMte. If the point () is attached to the foundation hy more than
three bars, i .C., if there are redundant fLJpporf', the three necessary and sufficient
conditions of equilibrium represented by Eqs. (4.1) \vill be insut1icient to
determine the unkno\vn elements, and the systern is said to be statically il1dc-
tei!Jli1late. If there are fe\vcr than three supporting bars or three bars in one
plane, the system is nonrigid and \vill not rernain in equiIibriutn under the action
of an applied force P that does not coincide \vith the plane of the bars.
Confining our attention to statically dctenninate systems, \V'C shall no\v
consider various practicable methods of application of the conditions of
equilihriuln representcd by Eqs. (4.1). i\S a first exarnple, \VC take the simple
space structure sho\vn in Fig. 4.5. This sysrCITI consists of a strut 40 hinged
at 0 to a vertical \vall i\{;.\T and supported at A by guy \vires AB and AC' as
sho\vn. l}nder :he action of an external force }) at A, the analysis of this
systern can be Inade \vithout ditticulty by direct application of Fqs. (4.1).
\\re begin \vith a frce-body diagranl of point A, \vhich is acted upon by the
external force F), and the three rcactions 51, S2, and Sa, representing the axial
forces in the bars. \'\"C assunle each of these unkno\vn reactions to be directed
a\vay fronl the joint so that in our final results plus signs \vill indicate tension
and IT1inus signs corllpression. Equating to zero the algebraic StUll of the
projections of all forces at A on each of the orthogonal axes x, y, and z, directed
as sho\vn in the figure, Eqs. (4.1) becolnc, rcspecti vel y,
S ' 1 2 s 1 2 5 - 0
- l-. 2-1-g- 3-.
_fJ + T:S2 = 0
(a)
5 s -i s 0
"J-g- 3 - 1 2 

166 STATICALLY DETERMINATE SPACE STRUCTURES
z
II
A
x
p
FI G. 4.5
from which we find 8 1 = -7.20P, 8 2 = +4.33P, 8 3 = +3.47P. The nega-
tive sign for SI indicates that the strut is under compression, \vhereas the guy
\vires cach carry tension.
The same results can be found in another \\'ay by using the notion of moment
of force with respect to an axis. For example, if \ve equate to zero the
algebraic sum of moments of all forces \vith respect to the z axis, \\'c obtain
- P X 12 + -AS2 X 12 = 0
from which, as before, \\le find 8 2 = +4.3 3P.
Sometimes the analysis of a space structure can be handled to advantage by
working with one or more plane projections of the given system. Consider,
for example, the simple space system shown in Fig. 4.6a. We recall that, in
general, if a system of concurrent forces in space is in equilibrium, their
projections on any plane will also be in equilibrium. This idea can be used to
advantage in the present casc. Projecting onto the vertical plane AEB, we
E
8'
A
D
p 
B
(6)
c
8' (c)
e a
(a) p 
b ap
FI G. 4.6

ARTICLE 4.1
167
b'
I I I
I I I I f I
I '3" I I dill I
I / a"  :1
__ , I / I , I
- I / "
-- I bIt C I
J -...._ II "
___ + _n I
O -- I -
" 1 -- -- I
,-- --- ---- --  ---
I / ------- I
I I -- I
I / -- -..I 4"
, I
I I
,I
FIG. 4.7
2"
obtain the coplanar system sho\\'n in Fig. 4.6b, the analysis of \vhich is indicated
by the accompanying triangle of forces fbn. From this triangle, the compres-
sive force in the strut is seen to be 51 = .iP. In the same \vay, by projecting
the system onto the horizontal plane A(l) (Fig. 4.6c) and using the previously
deternlined value of SJ, the horizontal projection of which is equal but oppositc
to the force S' in Fig. 4.6h, \ve find the tensions in the t\\'o guy \vires to be
S2 = 53 = O.722P.
In dealIng \vith a system of concurrcnt forces in space, \ve may oftcn
encounter difficulty in detennining analyticaIly the projections or moments of
the forces \vith respect to various axes. In such cases, a graphical method may
be helpful. For example, the axial forces produced in the bars 01, 02, and
03 of the system in Fig. 4.7 can be found graphically by resolving the givcn
force P, defined by its orthogonal projections P' and P", into thrce components
acting along 01, 02, 03. ,""C begin \\'ith a determination of the linc of inter-
section of the plane defined by the bar Oland P \vith the plane defined by the
bars 02 and 03. One point on this line of intersection is obviously point O.
Another point on the sanle line is the point 11 obtained by using the intersection
of the lines 1 "4" and 2"3" in the horizontal projection. With the points 0
and 11, the required line of intersection is completely defined by its projections
O'n' and O"n". V\e next resolve the load Pinto t\VO components, one along
the axis of the bar 01 and one along the line 011, the projections of \\,hich \ve
have just found. The triangles of forces a' b' c' and ai' b" e" rcpresent the vertical
and horizontal projections, respectively, of the corresponding triangle of forces
{[be in space. The side ab of this triangle gives the component of P that acts

168 STATICALLY DETERMINATE SPACE STRUCTURES
along 01, and the side be , the component that acts along 011. Finally, since
the line 011 is in the plane 023, this latter cornponent of P can be resolved into
t\VO components bd and de that act, respectively, along the axes of the bars 03
and 02 as sho\\'n. With the orthogonal projections a'll, b'd' , d'e' and a"b" ,
b" d" , d" e" of the three components of P, the components themselves can be
obtained without difficulty. As already n1cntioned, the axial forces in the bars
can be found directly from thcse components. We see that in this case the bar
01 \\,jll be in tension, \\,hereas the bars ()2 and 03 are in compression.,
PROBLEMS
"Three bars of equal length , are joined together at A and supported at B, C, and D,
as sho\vn in Fig. 4.8. Find the axial force induced in each bar due to a vertical
load P at A, j f OB = OC = 0/) = /.
Ans. 8. = -2P/3, 8 2 = -2P/3, 8 3 = +P/3.
2 l.he legs of a tripod are of equal length , and arc supported at points A, B, and C,
which form an equilateral triangle \vith sides of length I in a horizontal plane. Find
the conlpressivc force S induced in each lcg by a vertical load P applied at the
apex of the tripod.
AlIs. S = -PIV6
3 A.ssuoling that the l:gs of the tripod in Prob. 2 arc supportcd on a frictionless
horizontal plane and that they are prevented from spreading by a string ABC, find
the tension induced in this string.
Au.'. S = +P/3V 6.
y
P'
1
A'
B
I I
I I
--1---- 1 -- -
I I
I I
e'l I
I
I
I
1 0 "
P"
x
1
FI G. 4.8
FI G. 4.9

ARTICLE 4.2
169
4 Detcrnlinc the axial fc)rces S h 8 2 , and S3 in the bars of the space SYStCIl1 arranged
and loaded as sho\vn in Fig. 4.9. Note that the distances O"B" and O"A" in the
horizontal projection arc equal. ASS\1Ine that  = 45°.
AnI. 8 1 = -P see a, 8 2 = -2P sec a, 53 = -[-1 V2 - tan 2 a.
, 11o\v \\,ill the axial forces SIt 8 2 , and 8 3 in Fig. 4.9 be changed if the external force P
is vertical instcad of horizontal as sho\vn?
4.2 SIMPLE SPACE TRUSSES: METHOD OF JOINTS
A system of bars in space, joined together at their ends in such a way as to
form a rigid space structure, is called a space truss. Many kinds of engineering
structures such as radar telescopes, transmission-line towers, and cranes arc of
such construction. In the fabrication of these structures, it is common practice
to make the connections at the joints by either rivcting or \velding. Of course,
the rigidity of this type of connection is bound to interfere to SOme extent \\lith
te free adjustment of the system to applied loads so that some secondary
bending of the bars will be induced. However, in many practical cases, the
presence of such secondary bending does not materially influcnce the primary
action of the structure, and the axial forces can usually bc calculated with good
accuracy by ignoring the effect of the rigidity of the joints. 1 'rhus \ve gen-
erally assume that the bars arc connected at their ends by ideal spherical hinges
even though such connections can never be realized in practice. In our
further discussion here, \ve shall al\vays assume such hinges.
By \vay of establishing a criterion of rigidity for space trusses \\Tith ideal
spherical hinges, we recall that the complcte constraint of a point in space
requires its attachment to a foundation by three bars the axes of \vhich arc not
in one plane. With this notion of complcte constraint of one point in space,
we can easily establish a method of assembling a system of bars in space so as
to form a rigid structure. Refcrring to Fig. 4.10, \VC begin \vith three bars
AE, BE, .and CE by \vhich the joint E is rigidly attached to a foundation
ABCD. \Vith the joint E completely constrained in space, it follows that by
the thrce bars EF, [)r', and Af the joint f'is completely constrained in spacc.
E and F no\v being fixed points like A, B, c., and ]J, \ve conclude that the joint
G can be rigidly connected to the rest of the structure by the three bars EG,
FG, and CG. Since this procedure may bc carried on indefinitely, we may statc
the fol1o\ving rule: A riJ{id space truss call always be fonned by attaching the
first joint to a .foundatio11 by 111eans of three bars that do 'llot lie in one plane aud
establishing each additional joint thereafter by three 1110re bars that do not lie il1 one
 In sorne cases, the presence of secondary bending in a space structure Inay greatly affect
Its behavior under load, and it bccort1cs necessary to consider this effect in detail. Several
such exalT1plcs arc discussed by A. F(ppl in his book "Vorlesungcn iiber technischc
h.1echanik," vol. II, p. 276, B. G. "rcuhner, Lcipzig,1912.

170 STATICALLY DETERMINATE SPACE STRUCTURES
F G
- - - - - - - -"..,-....1
E "..,"'" I
I
I
I
I
I
Ie
A
FIG. 4.10
plaue. Those three bars by \vhich any joint after the first is attached to the
existing structure may be joined either to the foundation or to previously
established joints. Any space system formed in accordance \vith this rule is
called a si1llple space truss. Several exanlples of such space trusses are sho\vn
in I,"ig. 4.11. In each case the joints have been set up in alphabetical order,
the first joint being attached to the foundation by three bars not in onc plane
and succeeding joints being attached to the existing structure in the same
n1anncr. It should be noted that in gcncral the rigidity of such structures is
not independent of their conncction to the foundation. 'That is, none of the
systcms sho\vn in Fig. 4.11 \vill reprcsent a rigid body if disconnected from
the foundation points A', B ' , (", . . . .
Suppose no\v that anyone of the space trusses in Fig. 4.11 is subjected to the
action of scveral applied loads. Under the influence of these external forces,
internal forces will be induced in the various bars, and the determination of
these internal forces constitutes the analysis of the truss. In dealing \vith such
problenls, we make the same idcalizing assulnptions that we did for plane
tnlsses; namely, (1) the bars themselves are weightless, (2) they are connected
at their ends by ideal hinges, and (3) external forces arc applied only at the
joints. On the basis of these assumptions, it follo\\'s that each bar \vill suffer
only simple tension or compression without bending, and thus the equal but
D
(a)
FIG. 4.11
H
F
A'
(b)
D
c
B'
(c)

ARTICLE 4.2
171
oppositc reactions that it exerts on the joints at its t\VO ends \vill be directed
along the axis of the bar. We have then at each joint a system of concurrent
forces in space, the lines of action of \\'hich are kno\\rn and to \vhich the three
conditions of equilibrium represented by Eqs. (4.1) can be applied directly.
The analysis of space trusses by \vhich \ve apply successively to each joint
the conditions of equilibrium representcd by Eqs. (4.1) is called the 111cthod of
joinu. To illustrate, let us consider the analysis of the space system arranged
and loaded as sho\\'n in Fig. 4.12. Beginning \vith the joint D and denoting by
51, 52, and 53, respectively, the axial forces in the bars 1, 2, and 3, \\'c see that
these three forces can be evaluatcd \vithout difficulty by one of the rncthods
discussed in the preceding article. Assuming then that 51, S2, and S3 have
been determined, \ve find at the joint C only three unkno\vns, nalTIely, S4, S 6,
and 56, and these three magnitudes can be found \vithout difficulty. When
this has been done, there \vill remain only three unkno\vn forces at B, and the
equilibrium of this joint can be considercd next. Finally, \VC may proceed to
the joint A, \vherc all forccs except 5 10 , 8 11 , and S12 \vill be kno\vn; and \vhen
these arc found, the analysis of the truss is completed.
In general, \ve see that to be successful \vith the method of joints \VC must,
at the beginning, find at lcast one joint of the truss to \vhich only three bars
not in one plane arc attached. 'rhen, when the forces in these three bars have
becn determined, there must be another joint where only three unkno\vns \\,ill
be encountered, etc., until the analysis is completcd. Recalling the method of
formation of a simple space truss \vherc each joint is attached to the existing
portion of the structure by three bars not in one plane, we see that the method
of joints must ahvays be applicable to such trusses. It is necessary only to
begin the analysis with the last joint added to the structure and then consider
the joints successively in the .reverse order from \vhich they \Vere sct up.
P2
3
1
FIG. 4.12
11
12


172 STATICALL Y DETERMINATE SPACE STRUCTURES
p
H F
FI G. 4.13
By such procedure \\le encounter only three unkno\vns at each nc\v joint, and
the analysis proceeds \vithout difficulty.
As a prclirninary step in the analysis of any space truss, \ve can often
simplify the problcIll by examining the systetn for ;l1acti't,( '11Jt?'lJlbers, i.e., bars,
\vhich, under a given condition of loading, do not carry cither tension or
con1:->rcssion. Consider, for exan1plc, the space truss constructed and loaded
as sho\vn in Fig. 4.13. Bcginning \\'ith the joint G, \VC see that there is no
external load here, and only three bars hot in one plane. Hence the axial
forces in these thrce bars must a1l be zero since, as we have already pointcd out,
three concurrent forces can be in equilibrium only if they are coplanar. '],he
same reasoning may he applied to the joint H. 'fhcn, since no forces are
exerted on the joint F by the bars H F and (; F, \Ve conclude that the three
remaining bars DF, BF, and ('F nlust also be inactive since they are three bars
not in one plane and since no exrernal force acts at F. Thus, the inactive
bars (represented in the figure by fine lines) can be removed, leaving, for
further analysis, only that portion of the structure representcd by heavy lines.
In Fig. 4.14 \\"e have another example of a simple space truss in \vhich several
of the bars arc inactive for the particular case of loading sho\vn. In this case
\\'C do not begin the analysis at once \vith a consideration of the joint }' to \vhich
only three bars are attached. Instead, let us consider the joint B at \vhich there
FIG. 4.14

ARTICLE 4.2
113
D
A
B
FIG. 4.15
c
is no external load and four bars, three of \vhich are in one plane. From this
fact it follo\\ls that the bar AB must be inactivc. Similar reasoning \\'ith
respect to the joints C, D, and E leads us to conclude that the bars BC, CD, and
DE also are inactive. L.et us imagine, then, that these inactive bars A B, BC,
CD, and DE arc removed from the structure. As soon as this is done, \ve see
that at each of the joints B, C, and D there remain only t\\'o bars which are not
collinear. From this \\'e conclude that the bars BB', BC', (lC', ('I)', DD', and
DE' are also inactive since t\VO forces can be in equilibrium only if they are
collinear in action. Finally, then, if all inactive bars (represented by fine
lines) arc removed from the system, \ve have for further analysis only thar
portion of the truss represented by heavy Jines.
Wc have alrcady notcd that thc rigidity of such simple space trusses as those
in Fig. 4.11 is not independent of their attachment to a foundation. Some-
times, as in the case of a dirigiblc, it is desirable to have a rigid space structure
independent of any foundation. "[0 assemblc a system of bars in space so as to
form a self-contained rigid frame\vork independe.nt of any foundation, \\'e can
proceed as follo\vs: Beginning \\'ith three bars in the fornl of a triangle ABC
(Fig. 4.15), \ve attach to these a fourth joint D by means of the three bars
AD, BD, and C/) that do not lie in one plane. In this \vay we obtain a rigid
tetrahedron that cannot bc distorted by any system of external forces applied
to its joints. 1 This tetrahedron represents the simplest form in which a system
of bars in space can be interconnected to make a self-contained rigid body.
Any such tetrahedron can readily be extended by attaching nc\v joints to the
existing systen1, each by means of three bars the axes of \vhich do not lie in one
plane. Several examples of more extended forms are sho\\'n in Fig. 4.16;
such systems also are called si11tple space trusses. In each case \Vc begin with a
triangle ABCl and establish succeeding joints in alphabetical order each by
means of three bars not in one plane.
vVhenever any self-contained simple space truss like those sho\vn in Fig.
4.16 is submitted to the action of a balanced system of external forces applied
1 \Ve neglect, of course, any slight change in shape that accompanies the small clastic
deformations of the bars under axia11oad.

17. STATICALLY DETERMINATE SPACE STRUCTURES
H
G
E H E
D E D
D
C
C F
A B A C
B
(a) (b) (c)
FI G. 4.16
to its joints, \\re can ahvays make a complcte analysis by thc method of joints.
This follo\vs at once from the manner of formation of such trusses; \\'c see
that in the process of analysis it is necessary only to consider the joints in the
reverse ordcr from \vhich they \vere set up in the formation of the system.
By \vay of illustration, let us consider the simple space truss sho\\'n in I;ig.
4.17a. This system has the form of a cube; at the corncrs A and H are applied
two equal and opposite forces P that act along the diagonal All, as sho\\'n.
That the cube as a \vhole is in equilibrium under the action of t\\'o such collinear
forces is self-evident, and we may proceed at once \vith the analysis. Bcginning
with the joint H and successively projecting all forces there onto axes coinciding
\\,ith HF, HE, and HG, we find that each of these bars carries an axial com-
pression equal to P / y3, as sho\\'n in f'ig. 4.17 b. Conditions at A being
identical \vith those at H, we conclude the same for the bars AB, AC, and AD.
Then, at each of the joints G and F., \\'e find four forces in equilibrium, three of
which are in. one plane. . Accordingly, \\'c conclude that CD and FD arc
p
D
B
(a)
FIG. 4.17
- PI-!3
(6)

ARTICLE 4.2
175
inactive bars, as sho\vn by dashed lines in Fig. 4.17 b. Also, we find that
GE == }..£ = + V2 P/y3 (tension), and GC = FB = -P/V3 (compres-
sion). This done, there remain at each of the joints C and B only three bars
\vith unknown axial forces, and we find CD = DB = + V2 P/V3, \vhile
BC = 0 and CE = BE = 0, as sho\vn by dashed lines in Fig. 4.] 7 h. Finally,
from a consideration of the conditions of equilibrium either at D- or at E, \ve
conclude that 1)£ = - P, and the analysis is completed.
Although the method of joints is always applicable in the analysis of a
simple space truss, \ve need not follow it too rigorously in all cases. That. is,
it is not always necessary to begin the analysis \vith the last joint and then
systematically proceed through the structure joint by joint until aJl axial forces
have been found. Instead, \ve can take a somewhat more general vie\v 'of the
problem as follo\vs: Since the joint-by-joint procedurc is al\vays applicable
and must lead to a definite and unique solution of the problem, \ve conclude
that a simple space truss is ah\'ays s/aticall)' deterI11;l1t1le; i.e., for a given condition
of loading there is onc and only one set of values for the axial forces that can
satisfy the conditions of equilibrium at: each and every joint. Hence, if by
some indircct procedure, such as guessing, \ve can succeed in finding a set of
values for the axial forces that satisfy the conditions of equilibrium at all
joints, we may rest asured that this constitutes the true solution of the prob-
lem. ConsiJer, for example, the simple space truss loaded as sho\vn in Fig.
4.18. In this system, the triangle ABC is equilateral \vich sides of length Q,
the triangle A' B'C' is equilatcral \vith sidcs of length la, and the distance
bct\veen the horizontal planes ABC and A' B'C' is a. If \\'e remove the bars
7, 8, and 9 fron1 this system, \ve see that the conditions at the joints A, B, and C
\vill bc identical. Accordingly, \ve conclude, after a consideration of the
joint C, that a complete solution, satisfying the conditions of equilibriun1 of
all joints, will be obtained by taking S7 = 58 = S9 = 0, S4 = S5 = S6 =
-2P/y3 (compression), and SI = S2 = Sa = - iP (compression).
p
p
p
c'
A'
FIG. 4.18
B'

176 STATICALL Y DETERMINATE SPACE STRUCTURES
p
FIG. 4.19
PROBLEMS
1 Examinc the simple space truss sho\vn in plan and elevation in Fig. 4.19, and
identi fy, by inspection, the bars that arc inactive under the action of a single vertical
load P applied at A as sho\vn.
2 In Fig. 4.11h, EGfH and ACBIJ are t\VO squares \vith parallel sides of lengths a
and 2a, respectively. The vertical distance bet\vcen their t\VO horizontal planes
is also 2(1. Make a complctc analysis of the upper story of this structure if there
is a vertical load P at each of the four top joints.
3 lake a complete analysis of the cube sho\vn in Fig. 4.16c under the action of t\yO
equal, opposite, and ,collinear forces P applied to the joints C and II.
4 Make a cOInplcte analysis of the space truss in Fig. 4.18 if thcre is a vertical load P
at Conly.
5 Makc a cornplete analysis of the space truss in Fig. 4.11b if a tcnsile force P is intro-
duced in a bar j-IG (not sho\vn) by means of a turnbuckle. For dimcnsions of the
structure, see Probe 2.
4.3 STATICALLY DETERMINATE CONSTRAINT OF A RIGID
BODY IN SPACE
In the general case, a system of forces acting on a rigid body may be neither
concurrent nor coplanar. Such a system of forces can always be reduced to a
Tcsultantforce and a rcsultant couple. Consider, for example, any single force F i

ARTICLE 4.3
177
applied at point A of a body (Fig. 4.20). This force can be transformed into
an equal and parallel force F; at the origin 0 together \vith a couple formed by
the given force F j at A and equal and opposite force F.;' at O. This is done
simply by introducing at 0 the two equal and oppositc forces F; and F:', \vhich,
being themselves in equilibrium, do not alter the action of the given force F i
at A. FolJowing the same procedure for each force of the system, we conclude
that, in general, any system of forces can be transformed into a systcm of forces
concurrent at 0 together \vith a system of couples the moments of \vhich are
equal to the corresponding moments of the given forces with respect to point O.
We can have equilibrium of two such systems only if the resultant force and
the resultant couple are both zero. For the resultant force to vanish, \ve must
have the algebraic sums of projections of thc givcn forces on the coordinate
axes x, y, and z equal to zcro; for the rcsultant couplc to vanish, we must have
the algebraic sums of moments of the given forces with respect to the three
coordinate axes equal to zero. Thus, the conditions of equilibrium for a
system of forccs in space maybe expressed as follows:
IX.. = 0
2;t: = 0
2;M y = 0
2;Zi = 0
2;M z = 0
(4.2)
2;M;z; = 0
These equations will determinc any six unknown elements pertaIning to a
completely general system of forces in equilibrium.
With the foregoing equations of equilibrium in mind, let us now considcr
the general problem of attaching a rigid body to a foundation in such a way
that it will be completely constrained in space. Consider, for example, the
rigid body having the fonn of a rcctangular parallelepiped as sho\\'n in ig.
4.21a. As \ve alrcady kno\v, to fix completely any point A of the body in
space, \ve need three bars the axes of which do not lie in one plane, as shown.
With such attachment to the foundation, it is evident that freedom of motion
of the body is limited to rotation about point A. Under such conditions any
other point in the body is frce to move on the surface of a spherc with center
at A and having a radius equal to the distance froIn A to the point in question.
II
FI G. 4.20 z
x

178 STAT'CALL Y DETERMINATE SPACE STRUCTURES
c
c
D
FI G. 4.21
(a)
(b)
Hence, further constraint of the body can be obtained by completely con-
straining a second point B in a plane \vhich is tangent to the sphere on \vhich
B \\'ould othcr\vise be free to nlove. This, \ve kno\v, can be accomplished by
two barJ at B arranged as sho\vn in the figure. Frcedon1 of Inotion of the body
is no\v lin1ited to rotation about the axis AB, and in this case any point in the
body can move only along the arc of a circlc the plane of \\'hich is normal to the
axis AB and the center of \vhich lies on this axis. lhus, at a third point C,
one bar, the axis of \vhich does not intersect A B, \vill complete the constraint of
the body in space. \Ve conclude then that at least .fix bars are necessary for
the complete constraint of a rigid body in space. It is not necessary, ho\vever,
for them to be arranged exactly as in Fig. 4.21a. For example, \ve can replace
one of the bars at A by a parallel bar at ]J and one of the bars at B by a
parallel bar at E (Fig. 4.21b) and still have complete constraint of the body.
It must not be concluded from the discussion above that six Dars \vill ah\rays
be adequate for the constraint of a rigid body in space. For example, if the
bars are all parallel or ]ic in parallel planes as sho\vn in Fig. 4.22a, it is evident
that there is sonle freedom for cnd\vise motion \vithout inducing appreciable
changes in the lengths of the supporting bars. Hence, six bars so arranged
cannot be said to furnish complete constraint. In Fig. 4.22b, \ve have another
case \vhcre six bars arc not satisfactorily arranged for the cornplctc constraint
of a rigid body. In this case four of the bars are attached to one point A,
and the other t\VO can be arranged in any nlanner \vhatsocver. Undcr such
conditions the axis of the bar at C' and the point A define a plane, and the axis
c
FI G. 4.22
(a)

ARTICLE 4.3
179
of the bar at B intersects this planc at some point D as sho\vn or is parallel to
it. Hence all six bars intersect the axis AD; and if extcrnally applied forces
givc a momcnt \vith respect to this axis, the six supporting bars \vill be unable
to develop a balancing moment, and the rigid body \\,ill rotate to some other
position. Such incomplete constraint of a rigid body in space is analogous to
that of a rigid body in a plane ,vhere the axes of three bars arc parallel or
intersect in one point (see page 58).
From the forcgoing discussion, v.rc conclude that a system of supports con-
sisting of six bars so arranged that their axes cannot all be intersected by a
straight line is always necessary and usuaJly sufficient l for the complete con-
straint of a rigid body in space. Any bars in excess of this number ,viII con-
stitute redundant constraints.
Let us assunlC no,v that either of the constrained bodies sho\\'n in Fig. 4.21
is subjected to the action of a system of external forces. Under the action of
these loads, axial forces \\fill be induced in the supporting bars, and each bar \vill
exert a reaction on the body at the point of attachment and coincident in
direction with thc axis of the bar. Considcring both active and reactive forces
together, \\'C have the gencral case of a system of forces in equilibrium. The
forces are not concurrent, coplanar, or parallel. .For such a system we have
six equations of equilibrium as represented by Eqs. (4.2), and Vie see that
these six equations will determine the magnitudes of the six reactive forces
exerted on the body by the supporting bars. Thus, the six bars that are both
necessary and sufficient for the conlpletc constraint of the rigid body also
represent a statically detcr111;nate system of supports. If there are more than
six supporting bars, i.e., if there are redundant constraints, the six equations of
statics wiH be insufficient to completely determine the reactive forces, and
the system of supports is said to be statically i71deler111i71t1te. In special cases
like those sho\\,'n in Fig. 4.22, where there are only six bars but so arranged
that their axes can all be intersected by a straight line, the determinant of the
six equations of equilibrium will vanish and the systcm of supports ,viII again
prove to be statically indeterminate.
In the analysis of a system of supports of a rigid body in space, \\'e first
replace the supporting bars by the rections that they exert on the body, thus
obtaining a free-body diagram. Since the six rcactive forces coincide \\,ith
the axes of the bars that exert them, their directions arc kno,vn, and, gcnerally
speaking, the six unknown magnitudes can al\\7ays be determined by Eqs. (4.2).
However, in the solution of practical problems, \ve shaH not always adhcrc
rigorously to the use of this system of equations. Since the body is in equi-
librium, it follows that the algebraic sum of the moments of all the forces with
respect to any axis (not necessarily x, y, or 21) must be zero and like\vise that
the algebraic sum of the projections of all forces on any axis must be zero.
1 In exceptional <;ascs, the determinant of Eqs. (4.2) may vanish even \vhen all six bars do
not intersect one axis; this \vill ahvays be an indication of incomplete constraint.

180 STATICALL Y DETERMINATE SPACE STRUCTURES
FI G. 4.23
E
r
3'
PI
p
'rhus, in \\'riting the equations of equilibriul11, it is usuaHy possible, by a proper
choice of axes, to obtain six equations in each of \vhich only one or t\VO of the
unkno\vn forces appear. In this \vay \ve avoid the difficulties connected \vith
the solution of six simultaneous equations \vith as Inany unkno\vns, and our
problem is greatly simplified.
To illustrate the application of the foregoing general discussion, Jet us con-
sider the rectangular parallelepiped ABCDEFGH supported and loaded as
sho\vn in Fig. 4.23. To make a free-body diagram, \ve replace the six sup-
porting bars by the rC2ctions \vhich they exert on the body, rerl1embering that
each force must act along the axis of the bar \vhich produces it. \\ assume
all bars to be in tension as sho\vn so that any negative signs obtained during
calculation \viJl simply indicate that the corresponding bars are in compression.
Writing first an equation of moments \vith respect to the vertical axis BF, \vhich
is parallel to or intersects all bars except 4, \ve obtain
-t5 4 X 8 - P X 4 = 0
froIn \vhich 54 = - *P. In the same \vay, by equating to zero the algebraic
sun} of moments of all forces \vith respect to the axis AE, \ve find S3 = +-l'>.
Since \ve already koo\v the value of S4, this latter result may also be obtained
directly by equating to zcro the algebraic sum of projections of aU forces on
the axis A ('.
No\\', taking moments of all forces, first \virh respect to the axis AB and
again \vith respect to the axis C'D', \ve conclude, successively, that
S6 = -Sa
and
Sl = -S2
Then, equating to zero the algebraic sum of the projections of all forces on the
axis AB, \ve obtain
-is! + 2 + P = 0
from which 51 = -52 = P.
Finally, 'Ne equate to zero the algebraic sum of l110mcnts of all forces \vith

ART. CLE ..3
181
respect to the axis A C and obtain
is! X 4 + i-S2 X 4 + -lS3 X 8 + S5 X 8 + P X 3 = 0
trom \\,hich, using the previously determined values of 81, S2, and S3' we find
S6 = -S6 = -i-P, and the analysis is completed.
Sometimes a problem' involving a general system of forces in space can be
simplified by introducing an equivalent loading. Consider, for example, the
rigid body supported by six bars as sho,vn in J:."'ig. 4.24a and subjected to a
horizontal force P in the end plane ABE. We begin by adding to the givcn
system at A the t,vo equal, opposite, and collinear forces P as sho\vn in the
figure. Two such forces being in equilibrium do not alter in any ,"'ay the
action of the system, but we may no,v consider that \ve have a force P at A
together \\rith a couple consisting of the force P at E and the opposite force P
at A. l'his couple, the moment of \vhich in the cnd plane of the body is Ph,
can be transferred to the opposite end of the body and represented by the
vertical forces Pbja at C and I) as sho\\'n in .Fig. 4.24b. The systeln of three
applied loads sho\\1n in Fig. 4.24h is equivalent in action to the single load P at
E in Fig. 4.24« and \vill induce in the supporting bars the same axial forces.
However, in f'ig. 4.24b \ve have at A, C, and D three sitnple systems of con-
current forces that can be analyzed by the most elementary procedures. By
inspection, \'le conclude that S3 = So = 0, while S.. = -S6 = Ph/a. Con-
sidering the remaining forces at A and projecting successively onto horizontal
and vertical axes, we find S 1 = P csc a, and S2 = - P cot a.
In Fig. 4.25 \ve have a plan and elevation 0'£ an equilatcral triangular slab
supported in a horizontal plane by six bars of equal length arranged as shown.
It is requircd to find the axial fi)rces induced in these bars by a couple of
moment Ai acting in the plane of the slab as shown. We begin with a free-
body diagram of the slab as indicated in the figure. Then, taking A B as an
axis of moments, we see that only the reactions Sa and S 4 have moments
different from zcro \vith respect to this axis. Hence, we conclude that these
r a -1
D
pxb
-
CI
6
FI G. 4.24

182
STATICALLY DETERMINATE SPACE STRUCTURES
FI G. 4.25
/o
E'-r--
/
forces, \\rhich are equally inclined to the vertical, must be equal in magnitude
and represent, respectively, tension in one bar and compression in the other as
indicated by the arro\vs in the figure. l By similar considerations of rnoments
about the axes AC and BC, \ve conclude that S6 and S6 as \vcll as Sl and Sz arc
equal in magnitade and directed as shown. No\v, by successively equating
to zero the algebraic sums of projections of all forces on the axes AD and BE,
\ve conclude that the six reactive forces must all be equal in magnitude.
Finally, then, by balancing moments \vith respect to a vertical axis through the
centroid of the triangle, \\'c obtain
S 2 vI " 1
6 s:n a X '3 T a = H
from \vhich S = V3 AJ/ (6a sin a), \vherc S is the magnitude of axial force in
anyone of te bars.
PROBLEMS
1 A strut AB is hinged to a vertical \vall at B and supported horizontally by t\VO guy
\vires arranged as sho\vn in Fig. 4.2(). Find the tension induced in each guy ,vire
by a vcrricalload P applied at A.
Ans. 51 = 3 .15P, 81. = 5.96P.
1 For expediency, we depart here from our usual rule of arbitrarily 3ssun1ing, to begin with,
that all bars are intension.

ARTICLE 4.4
183
c
p
A
FI G. 4.26
FIG. 4.27
2 An equilateral triangular slab is supported in a horizontal plane by six bars arranged
as sho\vn in Fig. 4.27. Each side of the triangle is of length a, and the vertical
bars also are of length (t. If the \veight of the slab is  \vhat must be the moment
of a couple AJ, acting in the plane of the slab as sho\vn, for the vertical bars to be
inactive?
All.':. Al = lVd/2 0.
3 1.Jslng the notion of equivalent loading, make an analysis of the system of supports
in Fig. 4.24(( if the load Pat E is vertical.
Ans. SJ = Sa = 8 5 = 0, .5 2 = S.. = - S6 = - P.
4 1\1al<e a complete analysis of each sysem of supports in Fig. 4.21 under the action
of a vertical load P at D. Assulnc in each case that the rectangular parallelepiped
has ditncnsions (1, (1, and 2n.
5 Analyze the systcm of supports in Fig. 4.23 under thc action of a vertical load P
at E.
A1/s. SI = S2 = -iP, 8 3 = S4 = 0, Sij = -S6 = -}P.
4.4 COMPOUND SPACE TRUSSES: METHOD OF SECTIONS
In the preceding article, \ve saw hO\\f a rigid body can be completely constrained
in space by means of six bars so arranged that their axes cannot all be inter-
sected by one straight line. On this basis, \ve conclude that any self-contained
simple space truss such as those sho\\ln in Fig. 4.16 can be rigidly attached to a
foundation in the same nlanner. Consider, for example, the system sho\vn in
Fig. 4.28, \vhere \ve have a simple space truss ABCDEf"GH attached to the
foundation by six bars arranged as shown. In this \vay, \VC obtain a rigid
and statically determinate structure capable of holding in equilibrium any
system of externally applied loads such as PI, P2, . . .. Such a structure \vill
be called a cun/poul1d space truss. In the analysis of a compound truss, we shall
generally find that the method of joints alone is inadequate. For example, in
this case (Fig. 4.28) \ve cannot begin the analysis by the method of joints

184 STATICALL Y DETERMINATE SPACE STRUCTURES
6
H
1
B
FIG. 4.28
because there is no joint to \vhich less than four bars are attached. Con-
sequently, \ve must first consider the entire portion ABCDEFGJ-f as a free
body and, using the six equations of equilibrium of the preceding article
[Eqs. (4.2)], find the axial forces in the six supporting bars. Such procedure
in the analysis of a space truss is usually called the 11Iethod of ectiolls. As
soon as \ve kno\v cornplctely the balanced systern of forces external to the
simple truss as a \vholc, the remainder of the analysis can be n1ade \vithout
difficulty by thc method of joints.
In Fig. 4.29, \ve have an example of a self-contained compound space truss.
To obtain this structure, \ve take t\VO silnple space trusses, as represented by
the shaded portions of the system, and interconnect them by six bars arranged
in accordance \\rith the requirements for complete constraint of a rigid body in
space. It must be self-evident, then, that the resulting system js rigid and
statically determinatc. HO\l.rever, under the action of a balanced system of
external forces, \ve again Inay be unable to n1ake a con1plcte analysis by the
method of joints. To begin the analysis in this case, \\,IC make a section cutting
the six nun1bcrcd bars and isolate one of the simple trusses as a frce body.
Then, \virh the help of the six available equations of equilibrium, \ve detern1ine
the axial forces in these six interconnecting bars, after \vhich the t\VO simple
trusses can be analyzed \vithout difficulty by the method of joints.
If desired, the sirnplc truss ABC'DEFGH of Fig. 4.28 can be considered as a
compound truss like that in Fig. 4.29. It is necessary only to regard it as t\VO
D
H
o
N
F
I
A
FI G. 4.29
c
hi
J

ARTICLE 4.4
185
simple tetrahedrons ABCD and EFGH interconnected by the six bars AE,
CE, [)£, CF, I)F, and CG. Accordingly, \\'e conclude that the axial forces in
these bars, also, can be found by the method of sections, leaving for analysis by
the method of joints only the t\VO tetrahedrons. In the same \vay, the sin1ple
space truss sho\vn in Fig. 4.11 a can be considered as a cOInpound space truss
consisting of the simple tetrahedron A BCl) attached to the foundation by six
bars arranged in a manner satisfying the conditions of complete constraint of a
rigid body in space. ()n the other hand, no such conclusion can be made
for the sin1plc space truss in Fig. 4.11 b. In the first place, the portion
ABCDEfr;H of this structure does not represent a rigid body; and, second, to
separate it froIn the foundation \VC should have to cut eight bars instead of six.
Under such circumstances, the method of sections cannot be used.
The identification of any space structure as a simple or compound truss \vill
ah\rays be an indication that it is rigid and statically determinate, i.e., that
under any given condition of loading there \vill be one and only one set of
values for the axial forces that can satisfy the conditions of equilibrium at each
and every joint. \Vhether or not \ve rigorously follo\v the method of joints
in finding this set of values is of no iruportance. Any procedure by \\Thich \VC
succeed in finding a solution that satisfies all conditions of equilibrium is a
valid procedure.
As an example, let us consider the analysis of the compound space truss
sho\\1n in Fig. 4.30. l"his structure consists of a rigid equilateral prism
ABCDEF, \\lith edges of lcngth I, that is attached to the foundation by six bars
arranged as sho\vn and then subjected to the action of a vertical load P at A.
The orthodox procedure in the analysis of this compound space truss \vould be
to find the axial forces in the six supporting bars by the method of sections and
then use the n1cthod of joints for the sin1ple truss A BC'DEF. \tVhile such a
procedure \vill involve no particular difficulty, it \viIl be somc\vhat more
advantageous in this particular case to find the forces in the supporting bars
by using the notion of equivalent loading, \vhich \vas discussed on page 181.
Introducing at (' t\\'o equal and opposite vertical forces P that, being in equi-
p p p P
2 2 2 2
P E P
F F
A A
6 6
p
(b)
(c)
FIG. 4.30

186 STATICALLY DETERMINATE SPACE STRUCTURES
p
86
-{2
3'
c
FI G. 4.31
p
librium, do not affect the system, we see that the given loading can be con-
sidered as a ver.ticalload Pat C together with a counterclockwise couple Pl12
in the end plane ABC of the structure. Transferring this couple to the end
plane DEI/ where \\'e represent it by two vertical forces P /2 at E and F, \ve
obtain the equivalent loading sho\\'n in Fig. 4. 30b. Regarding this notion of
equivalent loading, It must be emphasized here that the loading in Fig. 4.30b
is equivalent to that in Fig. 4. 3 Oil only in so far as the axial forces induced in the
supporting bars arc concerned. The forces induced in the other bars of the
system \\fill be completely different in the t\VO cases. From Fig. 4.30b, \ve
see by inspection that the supporting bars 2, 3, and 5 are inactive, while
the axial forces in the other three are as follo\vs: 51 = - P, S. = - P /2,
S6 = +P/2. Returning, no\\', to the true loading and replacing the six sup-
porting bars by the rcactions that thcy exert on the simplc truss ABCDEf', we
obtain, for further analysis, the systcm shown in Fig. 4.30e. This simple
truss, ho\vever, CCln also be considered as a cornpound truss consisting of the
rigid triangles ABC and DEF, \vhich arc interconnected by the six bars a, b,
c, d, e, and f, as sho\vn. As can be seen by inspection of the joints C and F,
the bars e and c of these six are inactive and can be removed from the system.
Making a section that cuts the four remaining bars and projecting onto the
plane of the triangle ABC all forces that act upon it, \\'e obtain a coplanar
system of forces in equilibrium as sho\\'n in Fig. 4.31. Successively equating
to zero the algebraic sums of moments of these forces with respect to the points
A, B, and C, we find
Sd = + vi P
Sf = - v p
and
Sb = - vi P
To find Sa, we rctum to Fig. 4.30e and use the method of joints. Projecting
all forccs at A onto an axis coinciding with AE, \ve obtain Sa = +P/V 3 .
The t\VO triangles 4BC and DEF may no\v be analyzed as any other plane
truss. In Fig. 4.3], for example, we see that S = S = - P/V3, \vhile
S = +P/V3.

ARTICLE 4.4
187
PROBLEMS
The con1pound space truss in Fig. 4.32 consists of a rigid square pyramid ARCED
attached to the foundation by seven bars arranged as sho\vn. Then, by mcans of
a turnbuckle f', a tensile force P is induced in the bar ec'. What axial forces will
be induced in the other bars of the systcm?
Al1S. AA' = ec' = +P, BB' = EE' = -P, I)C = DA = + 0 p,
DB = DE = - 0 P, AB = BC = CE = EA = +P/0,
AE' = AB' = EC' = 0, AC = -2P.
2 The rigid cube sho\\'n in Fig. 4.33 is formed by interconnecting the t\VO shaded
tetrahedrons by means of six bars the axes of \vhich cannot all be intersected by
one straight line. l\1ake a complete analysis of this system under the action of
t\vo equal, opposite, and collinear forces P acting along the diagonal AG as shown.
AlIs. 8 1 = 8 6 = +P/2 y3, S2 = S3 = Sol = S,. = -P/2 yl3.
D
E
B
FIG. 4.32
E' A' 
r-- I ---j/ I
p
E
A
A
c
B
2
H
G
D
F
p
FI G. 4.33
FI G. 4.34

188 STATICALLY DETERMINATE SPACE STRUCTURES
p
A
B
D
E
c
p
p
F
FI G. 4.35
FI G. 4.36
1 Prove that the sy;,tclu 0 f bars sho\vn in Fig. 4.34 constitutes a compound space
truss, and make a cotnplctc analysis of the systcrl1 under the action of t\vo cqual
and opposite forces P acting as sho\vn. rhe system has thc form of a rcgular
octahcdron.
4 J\llakc a cOJllplcte anal ysis 0 f the con1ponnd space truss loaded as sho\\'n in Fig.
4.35. /\sslIrl1c that ABC and A'B'C' arc equilateral trianglcs and that A'B'BA is
a square.
Aus. A'C' = B'C = +P/V2, C'A = ("B = -P/V2, A'A = B'B :::
+P/2, CA = CB = -P/2, AB = +P/2.
5 In Fig. 4.36, the triangles ABC and D£F arc equilatcral \'lith sidcs of lcngth a,
and the distance bet\veen their parallel planes is also t1. find the axial forccs in the
six bars that interconnect these t\VO triangles if t\VO equal and opposite collinear
forces P are applied to the system as sho\vn.
AIlS. 51 = 8 6 = -P/V2, 52 = -So = +P/V2, S" = -S3 =
+P/vZ.
4.5 GENERAL THEORY OF STATICALLY DETERMINATE
SPACE TRUSSES
In this article, \\'e shall consider the general probletn of ho\v to assemble a
systcm of bars in space so as to form a complctcly rigid space truss. We have
alrcady seen in Art. 4.2 that a si111ple space truss can be fornled by beginning
with a rigid foundation to \vhich the first joint is attached by means of three
bars not in one plane; thereafter, succeeding joints are attached to the founda-
tion or to previously established joints in the same manner. The truss in
jg. 4.37a has been formed in this \\Tay, beginning \vith joint A and setting up
the other joints in alphabetical order. lrom the rule of formation of a simple
space truss, it follo\vs that, bet\vecn the numbcr of members 111 and the

ARTICLE 4.5
189
number of joints j, there Inust exist the relationship
111 = 3 j
(4.3a)
since \ve use three bars for each joint.
While the foregoing rule of formation of a space truss is a very simple one,
it does not represent the only way in \vhich a system of bars can be assembled
to form a rigid space structure. For example, if we remove the bars CC'
and BD' of the simple truss in Fig. 4.37 a and introduce instead the bars DC
and EF as shown in Fig. 4.37 b, \ve obtain a C0111pOUl1d space truss like that
sho\vn in Fig. 4.2R. "rhat is, the portion ABCDEF(JH of the structure no\v
represents a self-contained simple space truss \vhich is attached to the founda-
tion by six bars the axes of which cannot all be intersected by one straight line.
Since, by the above rcarrangement of bars, \ve change neithcr the number of
bars 1fl nor the number of joints j, we conclude that the compound space truss
in Fig. 4.37b also satisfies Eq. (4.3a). Still another type of rigid space truss
can be derived from the simple truss in Fig. 4.37 a simply by changing the
directions of the diagonals in the two side panels as shown in Fig. 4.37c. In
this way, \ve obtain a system that, although it still satisfies the relationship
111 = 3j, can no longer be classed either as a simple truss or as a compound truss.
Many of our engineering structures are of this latter form, \vhich is called a
crnllplex .fpace trus.f.
From the preceding examples, we may draw the general conclusion that 3j
bars are always necessary and, \\,hen properly arranged, sufficient for the
rigid interconnection between themselves and the foundation ofj joints in space.
This observation holds also for any self-contained space truss that is attached
to a foundation by the six bars necessary and sufficient for the complete con-
straint of a rigid body in space. To demonstrate this, we recall fron1 Art. 4.2
the rue of formation of a self-contained simple space truss, by \vhich \ve begin
\\,ith three bars and three joints in the form of a triangle and attach succeeding
H
H
H
(a)
FI G. 4.37
(b)
(c)

190 STATICALL Y DETERMINATE SPACE STRUCTURES
joints, each by means of three bars not in one plane. Since, by this rule, \ve
use three bars for each joint except the first three, for \vhich there is only one
bar for each joint, \ve conclude that bct\vecn the total nurnbcr of bars 1)1 and
the total number of joints j of a self-contained simple space truss there rnust
exist the relationship
111 = i - 6
(4.3b)
Then the six additional bars required to conlplete the constraint of a rigid
body in space bring us to 'IJl = 3) - 6 + 6, or sirnply 111 = 3) as before.
In arranging a systcnl of bars to fOrIn a rigid space truss, it is often desi rabIe,
for purposes of utility, to avoid any obstruction of the inner space. R.egarding
space structures \\,ithout internal diagonals, it can be proved that if any self-
contained space trl:SS has the form of a closed polyhedron the plane faces of
\vhich are triangular or subdivided into triangles, then Eq. (4.3 b) \vill be
satisfied, and, generally speaking, the truss \vill represent a completely rigid
hody. To prove this staten1cnt, \ve begin \vith a gcneral thcorem of stere-
ometry that is due to Euler. This theorcrn states the relationship bct\vecn
the nunlocr of faces, edges, and apices' of any closed polyhedron and may be
established on the basis of the foIIo\ving reasoning: Bcginning \\,ith one face,
represented by any polygon, \ve see that \\'c have, to start \vith, an equal
number of edges and apices. No\\', \"hen \VC come to add a second face to the
first, there \vill be one of its edges and t\VO of its apices already cxisting.
Again, \\,hcn \VC C0;11e to add a thi rd face to the first and the second, t\VO of its
edges and three of its apices \vill already exist, etc. Thus, in general, the
addition of each face after the first cntails the addition of onc morc nc\v edge
than ne\\' apex; and as \ve proceed, the total number of edgcs gains on the total
nunlbcr of apices by exactly the nUfi1ber of faces that up to any rnornent have
been added to the fir,st, or starting, face. flence, \VC conclude that, at any
stage of construction, the number of edges must be equal to the nunlber of
apices plus the nunlber of added faces, i.e., all except the starting face. This
relationship holds Lntil \ve comc to thc last, or closing, face, the addition of
\vhich entails no nc',\' edges and no ne\v apices; thus, \ve acquire a second extra
face. Denoting by 1J1 the number of edges, by.} the nunlbcr of apices, and
by f the nunlbcr of faces, \ve have, then, for any closed polyhedron the
relationship
1J1 = .i + (I - 2)
(,1)
and this <:quation expresses the theorem of Euler mentioned above. The fore-
going arguments apply to the general case of any closed polyhedron. In
the particular case of a closed polyhedron the faces of \vhich are triangular or
dividcd into triangles, \ve can express a further relationship bct\veen the nUJ11-
ber of faces and the nun1ber of edges, i.e., bet\vcen.f and '/11. ]1:very edge
being common to t\VO faces and all faces being triangles, \ve see that there

ARTICLE 4.5
191
FI G. 4.38

(b)
(d)
must be exactly half as many edges as three times the number of faces. l'hat is,
'In = jf (b)
Eliminating f bet\\fCen Eqs. (a) and (b), \ve obtain 1JJ = 3j - 6 as expressed
by Eq. (4.3b). Thus, \ve conclude that any self-contained space truss the
bars of \\'hich represent the edges of a closed polyhedron having triangular
faces and being \\,ithout internal diagonals must satisfy Eq. (4. 3b). Several
examples of such structures are sho\vn in Fig. 4.38, and we see that they must
be classificd as complex trusses. As already stated, these forms have the
practical advantage that their inner space is free from obstruction; for this
reason, they are commonly employed in all kinds of structural work.
\\e have seen no\v that there arc three general classes of space trusses,
si111p!e, CU/1tPOl/.11 d, and cOl11plex, and that in each case 1n = 3j is a general
requirement of rigidity and complete constraint. We shall now discuss another
general significance of the relationship 1n = 3j. Consider, for example, any
completely constrained space truss comprised of 111 bars and j joints and
submitted to external loads appJicd only at the joints, as sho\vn in Fig. 4.39.
Under the action of such applied loads, axial forcs \vill, in general, be induced
in all the 111 bars of the system, and the determination of these internal forces
constitutes the analysis of the truss. If we are dealing \vith a simple or
compound truss, we 1<no\\' that the analysis can ah,vays be made by the method
of joints or by the method of sections; such procedures have already been
discussed in detail in Arts. 4.2 and 4.4. In the case of a complex truss, ho\v-
ever, these n1erhods of analysis may fail. In the present case, for example,
\\'e see that there arc fouf bars meeting at each joint, and so the method of
joints cannot be used. Like\vise, there is no possibility of employing the

191. STATICALLY DETERMINATE SPACE STRUCTURES
Pz
FIG. 4.39
method of sections because no section can be conceived \vhich cuts only six
bars that do not all intersect one straight line. Under such circumstances,
\ve must take a more general vie\\' of the problem as follo\vs: Replacing each
of the 111 bars by the t\VO equal but opposite reactions that it exerts on the joints
at its ends, we obtain j systems of concurrent forces in cquilibriuIn as sho\vn.
l'hen, for each joint \VC \vritc three equations of statics (k4\ i = 0, 2; }, = 0,
};Z.. = 0) and obtain 3j simultaneous equations involving 111 unkno\vns. '-'Te
see no\v that, if 111 = 3j, there are exactly as many unkno\vns as there are
equations and that in all but exceptional cases,l \vhere the determinant of these
equations vanishes, the system of equations of statics must yield a unique
solution to the problcrn. For this reason, any completely constrained space
truss that satisfies the condition 111 = 3j is said to be stlltically deter1l1illl1te.
rhat is, under any condition of loading, the axial forces in all bars can be
found from equations of statics alone, and there is no necessity to take account
of the elastic deforrnations of the bars. The solution of 3j simultaneous
equations involving as many unkno\vns is, of course, a problem in itself, but
\ve shall content ourselves for the moment \vith the kno\\rledgc that the axial
forces can be found by equations of statics alone if 111 = 3j.
If 111 > j, there \viII, of course, be more unkno\vn axial forces than there
arc independent equations of statics, and the system of 3j equations cannot
yield a unique solution. Accordingly, the truss is said to be statically il1de-
ter11Zi1Jate. Under such conditions, the clastic deformations of the bars ITIUst
be taken into account to determine the \vay in \vhich the internal forces adjust
themselves to mect the conditions of equilibrium of the joints. ()n the other
hand, if 111 < 3j, the structure is not rigid and \viII probably collapse under
the action of externally applied loads.
Returning to the case \vhere 1J1 = 3j, let us consider the exceptional trusses
for \\Thich the detcrminant of the 3j equations of statics vanishes and the
equations do not yield a definite solution for the axial forces. This circum-
stance \vill a!v.'ays be an indication that the truss is nonrigid and therefore
unsuitable for structural purposes. Such exceptions are called critical fOr/lis.
A critical fornl \vjIl be obtained, for example, in the formation of a simple
1 Such exceptional case,s \vill be discussed later.

ARTICLE 4.5
193
space truss j f \VC forget that those three bars by \vhich any joint is attached
to the rest of the systcm must not all lie in one plane. "l'hus the self-contained
simple truss in Fig, 4.40a has a critical form if the joint E lies in the plane
A BD, and \ve see that under such conditions therc is a limited frecdorn for
relative movement of this joint in the direction perpendicular to the plane
ABD, I..ike\vise, the compound space truss in Fig, 4.40b has a critical form
because the six bars by \\,hich the rigid tetrahedron AB('fJ is attached to the
foundation are so arranged that they all intersect the vertical axis BD or are
parallel to it. '"rhus, again, there is a limited freedom for rotation of the
tetrahedron around this axis,
Such critical forms as those sho\vn in Fig. 4.40 are easily detected by inspec-
tion and, in fact, can be avoided by careful observation of the rules of formation
of simple and compound space trusses, In the case of a complex truss, ho\\'-
ever, \\'e may be unable to discover a critical form by inspection, and it is for
such trusses that they are most likely to occur. A general method of detection
of critical form is based on a consideration of the 3j equations of statics for
the j joints of the system. If the determinant of these equations is different
from zero, \ve have a statically determinate system, and there is no critical
form. If the determinant is zero, the system is statically indeterminate, and
\ve do have a critical form.
'rhe actual evaluation of the above-mentioned detcnninant is, of course,
in1practicable. Ho\vever, since it depends only on the configuration of the
tfusS and not at all on ho\v the truss is loaded, it fo11o\\'s that if for any assumed
loading \ve can find, \vithout ambiguity, the axial forces in all bars, the deter-
minant evidently is not zero, and the truss is rigid. ()n the other hand, if
under an assutned loading \ve can discover some ambiguity regarding the
internal forces, the determinant evidently is zero, and the truss has a critical
form. In undertaking such an investigation, the siluplcst procedurc is to assume
a zero load at each joint. Thcn, one obvious solution satisfying the conditions
of cquilibrium at the joints is obtained by taking all bars \vith zcro axial forces.
If no other set of values different from zero can be found to satisfy the con-
ditions of equilibrium at the joints, the truss is rigid and statically determinate;
but if, under zero loading, a set of axial forces different from zero can be
found to satisfy the equilibrium of the joints, the solution is ambiguous, and
4
D
C A
A
FI G. 4.40
(a)
(6)

194 STATICALL Y DETERMINATE SPACE STRUCTURES
the truss has a critical fonn. Such analysis of a truss undcr zero load for the
purpose of detecting a critical form is kno\vn as the zero-load tcst.
As an example, \\'c consider the con1plex space truss sho\vn in plan and
elevation in Fig. 4.41. This system consists of a square frame ABCD sup-
ported in a horizontal plane by eight bars attached to the foundation as shown,
and \\re see that the condition 1fl = 3j is satisfied. No\v, \vith zcro load at
each joint, \ve assume an arbitrary tension in AR and, at the same tinle, an
equal conlpression in A F. Then, the resultant of these two forces at A \vill
lie in the plane of the square ABCD and \vill be perpendicular to its diagonal
AC. Hence, it can be balanced by a proper tension in /lB and an equal
compression in AI), and all conditions of equilibrium for the joint A arc
satisfied. Similar arguments can be made for the joints B, (1, and D. Thus,
\\'e conclude, finally, that undcr zcro loads a set of axial forces di ffercnt from
zero can exist in thc bars of the truss as sho\vn. Consequently, the solution
is ambiguous, and the truss has a critical fonn.
11hc nonrigidity of this system can also be seen physically by considering
the plan vie\v sho\\'n in Fig. 4.4Ic. from this figure, it is clear that the joints
A, B, C, ]) can move in or out along the diagonals of the square \vithout changes
in the length of any of the bars. l"hus, the truss can take the distorted fonn
as indicated, and ,ve have a nonrigid system unsatisfactory for practical use.
H
E
F
A B

(b)
H
E
G
FIG. 4.41
F

ARTICLE 4.6
195
ABC D
(a)
FI G. 4.42
PROBLEMS
i\pply the zero-load test to the complex truss sho\vn in Fig. 4.37f, and prove that
it is rigid and statically dctenninatc.
2 U sing the zero-load test, prove that the con1plcx space truss sho\\'n in plan and
clt.vation in Fig. 4.42 is rigid and statically dctern1inate not\vithstanding its general
rcscrnblancc to the nonrigid system sho\vn in Fig. 4.41.
3 Sho\v in gencral that a cornplcx space truss like those in Figs. 4.41 and 4.42 \vill
be rigid and statically deterrninatc if the regular polygon ABC!)1*.: . . . has an odd
number of sides, and nonrigid and statically indctcrrninatc if there is an even num-
ber of sides.
4.6 ANALYSIS OF COMPLEX SPACE TRUSSES
In the preceding article, \\!e have discussed the general problem of fonnation of
space trusses and a criterion for the detection of critical forms. We shall no\v
turn our attention to the probIen1 of analysis of various complex trusses that
have been proved to be rigid and statically determinate.
One of the most useful mcthods of analysis of space trusses, in general, is
the method of joints, \\'hich \ve have already discussed in Art. 4.2. This
method, it \vill be rernembered, is generally applicable only to simple trusses,
but there arc cases \vhcrc it can bc used successfully in the analysis of a conlplex
truss. l"he truss in Fig. 4.43, for cxample, is of this kind and can be completely
analyzed by the method of joints. We begin \vith joint A and find the axial
forces in the bars I, 2, and 3. rhen, replacing these bars by the reactions
that they exert on the rcmainder of the structurc at B, C, and D, \ve find that

196 STATICALL Y DETERMINATE SPACE STRUCTURES
A
FI G. 4.43
there are still four unl<no\vn axial forces at each of these joints. Ho\vevcr, at
I), three of the bars arc in one plane, and so the axial force in the fourth bar
B1) can be found by projecting all forces at fJ onto an axis normal to the plane
DC F. As soon as the force in BD is known, \\'e can proceed to the joint B,
\vhere there \vill no\\' be only three bars (BG, BE, and B(') with unknown axial
forces, and these can be found. lhen, kno\ving the action of BC on the joint
C, \\'e can consider the equilibrium of this joint and find the forces in ('£, CF,
and CD. After this, \VC return to 1) and find the remaining unkno\vn forces
there (Dr' and DC). Beginning \vith G, the same procedure may be employed
at the joints G, E, and F, and so on, throughout the truss until the analysis is
completed. It \vill be seen that success \vith this complex truss by the
method of joints rests on the fact that, \vhenever all but one of the unkno\\rn
forces of a concurrent systcm lie in one plane, this one force can alv..rays be
found by projecting the systern onto an axis nomlal to the plane defined by
the lines of action of thc other unkno\vn forces, no nlatter ho\\' many (see
page 163).
Sometirncs the analysis of a complex space truss can be greatly simplified
by taking advantage of the symmetry of the system. Consider, for example,
the structure sho\"n in Fig. 4.44, and suppose that there is a vertical load at
each joint and that all loads on anyone horizontal ring are equal. U ndcr such
conditions of symmetry, a complete analysis of the truss can be made in a very
simple manner. \\'e begin \vith the zero-load tcst for complete rigidity. At
each joint of the top ring there are four bars, three of \vhich lie in one plane.
Hence, undcr zero load, the bars of this ring must be inactive and can be
renloved from the system. This leaves only t\VO bars at each joint that are
not collinear, and \ve conclude accordingly that these bars also are inactive
and may be removed. l{epeating this reasoning for the joints of succeeding
rings, we conclude that no axial forces di ffcrent from zero can exist in the

ARTICLE 4.6
197
systenl under zero load. Hence, it is rigid and statically deterrninatc. "'[his
indicates that if under external loads different from zero we can find, by any
convenient 1l1CanS, a set of axial forces that satisfy the conditions of equilibrium
of the joints, \VC Inay be surc that \VC have the true solution.
I(ceping the foregoing gencral remarks in mind, let us consider now any
meridian AB(,-']) of the structure, as shown in Fig. 4.45a. Noting that the
external loads Ph P2, and Pa lie in the vertical plane of the meridian, \ve conclude
that this one rib of the structure can be in c<)uilibriun1 only if at each joint the
forces exerted by the bars entering from the two sides have' a resultant H which
lies in the plane of the meridian. Furthermore, \\'e see that this condition can
be realized by assuming that all diagonals are inactive and that all bars con1-
prisig anyone horizontal ring have equal axial fi)rces. Proceeding on this
basis, \ve isolate the rib ABCD and project all forces onto the plane of the
meridian as sho\vn in Fig. 4.45 b. l--hen, beginning with the joint A, \ve can
construct the polygon of forces for each of the joints A, B, and C, as sho\vn
in Fig. 4.45c. By this procedure, ,vc detenninc the axial forces St, 8 2 , and Sa
in the bars of thc rib as \vell as the resultants HI, H 2 , and Ha of the tVlO equal
horizontal ring-bar forces at each joint. 1'his done, \ve find the axial forces
in the ring bars thenlselves simply by resolution of HI, If 2 , and H3 into com-
ponents parallel to the corresponding bars as indicated by the equilateral
triangles in Fig. 4.45c. Since the conditions along each meridian arc identical,
the diagram of forces in Fig. 4.45,- constitutes a completc anlysis of the truss.
Frcqently, the analysis of a complex space truss can be sirnplified by
reducing it to the analysis of several plane trusses. Consider, for example,
the space truss sho\vn in Fig. 4.46a. This complex system is obtained by
(a)
FIG. 4.44 (b)

198 STATICALLY DETERMINATE SPACE STRUCTURES
PI
Ps A H 1
P2
P l
(c)
H 2
D
P 2
Pa
(a)
FIG. 4.45
removing from an otherwise rigid parallelepiped AA' BB' EE' FF' the bar EF'
and introducing instead the extra bar g in the system of supports. In the
analysis of this system under the action of a horizontal load P at D, we begin
at the joint F, where, by inspection, \\'e see that Sl = o. Then, proceeding
to the joint f'" and projecting all forces onto an axis coinciding with Ff"', \ve
conclude that 8 2 = o. We may no\\' consider the joint D, where \\'e find
that Sa is the only unkno\\'n force \vhich lies out of the plane ABFE. Hence,
by projecting all forces at IJ onto an axis norn1al to this plane, \ve find S 3 = - P.
Now, kno\ving the action of the bar 3 on the joint J)' and projecting all forces
at D' onto an axis coinciding \\fith ])D', \ve find 5.. = + P sec a. Continuing
in this \\fay and considering the equilibrium of the joints Band B' in succession,
we find Sr, = - P and S6 = P see {3. In this way \ve find the forces in all v/eb
members of the top panel.
To find the axial forces in the \veb members of the bottom panel, \ve must
first find the force in the bar c of the system of supports. 1"'his we accompish
by considering the entire parallelepiped as a free body and equating to zero
the algebraic sum of moments of all forces acting on it with respect to an axis
coinciding \\!ith AB. '"rhis gives
SfS cos 'Y X 2a + Pa = 0

ARTICLE 4.6
199
from \\,hich Se = - (P/2) see "y. Then, starting at E' and considering, in
succession, the joints £', E, C', C, and A' of the bottom panel, \ve find 5'7 = 0,
58 = +(P/2) see a, S9 = -P/2, 5 10 = +(P/2) sec a, and S11 = -P/2.
Finl1y, replacing the \\'eb members of the top and bottom panels by the
reactions that \ve have just found them to exert on the joints of the t\VO side
panels, \VC obtain for further analysis the t\VO plane trusses loaded as sho\vn
in Fig. 4.46b and c. l'he analysis of these plane systems \vill be a straight-
forward procedure and need not be discussed here. Although \ve have chosen
in the example a very simple case of external loading consisting of a single
force P, it should be understood that the sanle general procedure can be used in
case of a more elaborate system of loads.
In general, the resolution of a space truss into several statically dctern1inate
plane trusses represents a very practicable method of analysis. As another
example, let us consider the complex system loaded as sho\\'n in Fig. 4.47 tl.
This truss satisfies the condition 11l = 3j and, as can easily be demonstrated
by the zero-load test, is rigid and statically determinate. Hence, \ve conclude
that under the given loading there is one and only one set of values for the
axial forces \vhich can satisfy the conditions of equilibrium of the joints and
that any procedure by which we can obtain such axial forces \vill be a valid
onc. V\C begin by resolving the applied load at A into three cOlnponents that
coincide, respectively, with the lines AA 3 , AB, and AE as sho\\'n. Then,
making a separate analysis of the system for each of these components and
B
P tan a
....
D
F
A
p - C
"2 tan Ct
(b)
P tan Q D'
B'  F'
J
F
p  c' p ,.
"2 tan Q' 2 tan Ct
(c)
FI G. 4.46

200 STATICALLY DETERMINATE SPACE STRUCTURES
superimposing the results, \ve shall obtain the desired axial forces induced hy
the given load P.
Firsf, \VC consider only the component PI, \vhich coincides \vith A/la, anq
conclude that, under such load, all bars except AA 1, AlA 2, and A 2A 3 are inacti ve
and that these three bars each carry a compressive force numerically equal to
Pl. l'his done, \\'e consider next the con1ponent [>2 \vhich coincides \"ith AB.
Again, for this loading it can be seen that only those bars comprising the panel
ABA 3B3 \\,ill be active, and our problem reduces to the analysis of a plane truss
loaded as sho\\'n in }/ig. 4.47 b. \\, make this analysis and record the results
as sho\vn. In the san1C \\'ay, \ve conclude that, under the action of the force
Pa alone, only thc bars of the panel AEA 3 r;3 "..ill be active, and \ve again have
a simple problenl of 'analysis of a plane truss as sho\vn in Fig. 4.47c. l\:O\V, to
obtain the axial force in any bar under the original loading (Fig. 4.47a), \ve
simply add algebraically the results already obtained. Thus, f()r example, in
the bar AAt, \ve have a force S = -PI + Pa tan a; in the bar A 1 A 2 , \ve have
S = - PI + P2 tan a + 2P'J tan a; etc. If there are applied loads at the
other joints, they can be handled in the same \vay.
'[he nlethod of resolution of a space truss into several plane trusses can be
used to advantage in the analysis of the complex systetTl sho\vn in plan and
elevation in Fig. 4.48. 'fhis structure is seen to satisfy the condition 111 = 3j;
but, as a preliminary to its analysis under load, \ve nlust first rulc out the
possibility of a critical form. Using for this purpose the zero-load test, \ve
P
P2 P3
p'). 
C 0 S 0
a.. M
+

c. S
o..M

+
ts ts
C 2 S !

CW)
+
Ai B3 A3 E3
(b) (c)
FI G. 4.47

ARTICLE 4.6
(c)
(a)
FI G. 4.48
(b)
I \
A y B
I ,
, \
I \
, I \
A L. - _ _ - _ _ B'
G
FA
B C
201
c'
c'
begin by isolating the triangle AB(; as sho\\'n in Fig. 4.48c. Since all bars
that join this triangle at (; lie in one plane that intersects the plane of thc
triangle itself in a horizontal line A' 8' through (;, it follo\vs that any action
exerted on the triangle by such bars must lie along this linc of intersection of
the t\vo. planes. In the same \vay, all bars that join the triangle at A lie in
one plane that intersects the plane of the triangle along A A'; hence, the
resultant of any action of thcse bars on the triangle at A must lie along AA'.
Like\vise, any action on the triangle at B must lie along BB'. No\v, since
three coplanar forces can be in cquilibriunl only if they arc concurrent and since
the lines AB, AA', and BB' do not intersect in one point, \VC conclude that no
such forces can exist at A, B, and (;, 'fhis leaves the statically detenninate
triangle free fronl external forces, and hence no internal forces diffcrent from
zero can exist in the three bars A B, A G, and BG. What is true for the
triangle ABG holds for the other triangles like it, and \ve conclude accordingly
that, under zero load, no axial forces different froln zero can exist in the
system. 'rhus, the system docs not have a critical forJ11 and \vill be rigid and
statically determinate under any system of applicd loads.
I\Tow, let us consider an analysis of the system under the action of a vertical
load P applied at A as sho\vn in Fig. 4.49. As in the preceding exan1ple, \ve
first resolve this force into three components, PI coinciding \vith AA', P2
F'
Fl
E' J D'
F'
A'
G
B'

202 STATICALLY DETERMINATE SPACE STRUCTURES
c'
c'
p
(a)
(a)
F 1
E' J
D'
E' J
D'
F'
c'
F'
c'
(b)
A'
G
B'
(b)
A' G
B'
FI G. 4.49
FI G. 4.50
coinciding \vith A B, and P 3 coinciding \vith A }/. "rhen, making a separate
analysis of the system for cach of these components and superimposing the
results, \VC shall obtain the required axial forces induced by the given load P.
We begin \vith the conlponcnt PI acting along A A' as sho\\'n in Fig. 4.50.
Since this force does act along AA', our previous rcasoning regarding such
triangles as ABG (see Fig. 4.48c) is still valid, and \ve conclude accordingly
that the load PI is carried entirely by the bar AA'. "rhus, owing to PI, only
the bars sho\'n by heavy lines in .Fig. 4.50 are active, and the axial forces in
such bars can be found \vithout difficulty by the nlcthQd of joints.
No\v, considering only the component P 2, \vhich acts along AB (Fig. 4.5 1),
and proceeding as for the case of zero load, \ve again concludc that all bars
except those sho\vn by heavy lines are inactive. l"hen, to proceed further,
\ve isolate the triangle ABG again as sho\vn in Fig. 4.51c and consider its
conditions of equilibrium. In this case, \ve conclude that the resultant of P 2
and the reaction Ra acting along AA' must pass through the point B' \vhere
the kno\vn lines of action of Rb and lu intersect. Thus, the polygon of forces
in Fig. 4.51 d detcrmines the reactions Ra, R b , and Ro, and a complete analysis
of the triangle ABG can no\v be made \vithout difficulty by the method of joints.
Furthermore, since the adjacent triangles A/41J and BCH arc inactive, \VC
conclude that the reactions Ra and Rb represent, respectively, the axial forces
in AA' and BB'. Kno\ving these forces and applying the method of joints first
to the hinges A' and B' and then to the hinges G, L, and H, we find the forces

ARTICLE 4.6
203
in the remaining active bars as indicatcd hy heavy lines in Fig. 4.51. i\n
analysis of the SystCITI under the action of the component P3 that acts along /IF
can be made in the same manner. r-[hen, adding algebraically the axial forces
in any bar due to each of the three components PI, })2, and P: h we shall obtain
the requircd axial forces due to the given load P. If thcre are loads at the
other joints of the structure, the same general procedure may be used.
As a last example, let us consider the complex space truss loaded as sho\vn
in Fig. 4.52. lhis system, as \ve have already seen in Art. 4.5, is rigid and
statically determinate, and \ve may proceed directly \vith its analysis. As a
first step in this direction, \\'e consider the conditions of equilibrium of such
joints as B, C, D, and E, at each of \vhich there is no external load and only
four bars. Since the t\VO bars that support the ring ABClJE at B, for example,
define a plane that intersects the plane of the ring itself in' a horizontal line as
sho\vn, it f01l0\V5 that the resultant of the reactions exerted at B by the sup-
porting bars n1ust lie along this line. rhlls, projecting all forces at B onto a
horizontal axis normal to this line, \VC conclude that 51 = - S2. Similar
arguments can be l11ade \vith rcspect to the joints C, D, and E, and \ve conclude
accordingly that S1 = -S2 = S3 = -8 4 = So. No\v, kno\ving that 8 1 = So and
projecting all forces at A onto an axis that is normal to the plane A' AE', \ve
may determine S 1 and S 6 \vithout difficulty, after \vhich the remainder of the
system can easily be analyzed by the mcthod of joints. If therc arc extcrnal
loads at the other joints, \\'c n1ake a separate analysis for each load, as above,
and then superimpose the results.
F'
Ra Rb
c'
P2 I
I
I
I
A'L. - - - .j B'
R G
(1
(a)
Pi
(c)
E' J D'
F'
P2
c'
(d)
Ry
FIG. 4.51
(b)
A'
G
B'

204 STATICALL Y DETERMINATE SPACE STRUCTURES
(0)
A'
E'
D'
A'
C'
FI G. 4.52
(b)
B'
It will be noted that, in each of the last t\VO examples, \ve have managed
the analysis by taking advantage of the fact that \vhcnevcr a system of con-
current forces in space is such that the forces all lie in two intersecting planes,
the resultant of those forces in either plane must lie along the line of inter-
section of the two planes. This, of course, foHo\vs from the fact that, for
equilibrium, the resultant of the forces in one plane must be equal, opposite,
and collinear \vith the resultant of the forces in the other plane. In general,
this conclusion is very helpful in the analysis of complex space trusses.
PROBLEMS
Mal<e a complcte aalysis of the structurc in Fig. 4.44b under the action of a single
vertical load P applied to one joint of the top ring. For sin1plificarion, assume that
the storics arc of equal height, the ribs being straight and inclined by 60° with the
horizontal.
2 Make a cOlnplete analysis of thc systen1 in Fig. 4.46 under the action of a hori-
zontal force P at each of the joints B, I), and F, instead of only at n as sho\vf1.
For simplification, assume a = {3 = 'Y = 45°, that is, that all panels are subdivided
Into squares.
3 Make a complete analysis of the structure in Fig. 4.49 undcr thc action of a singlc
vertical load P at A as shown. Using the rcsults of this analysis, compute the
axial forcc in each bar of the systcnl \vhen there is a vertical load P at each joint
of the top ring.
4 Make a complete aalysis of the complex system in Fig. 4.52 if the load P at A
is vertical. Assume thc follo\ving numerical data: AB = BC = · . . = 10 ft,

ARTICLE 4.7
205
A' H' = R'C' = . . . = 15 fe, and Ii = 12.5 ft. Using the results of this analysis,
rompute the axial force in each bar of the systcm \\,hen there is a vertical load P at
each joint of the ring A BCDt'.
4.7 HENNEBERG'S METHOD
!\:ot infrequently, Henneberg's method of analysis can be used to advantage in
\vorking \vith complex space trusses. 'fhis method has already been discussed
in Art. 2.8 for the case of plane trusses, but here \ve shall revic\v briefly a
general outline of the procedure. Suppose, for example, that Fig. 4.5 3a
represents a complex space truss to which none of the methods of analysis
discussed in the preceding article seems to apply, but that by removing the bar
.t and substituting a bar a as sho\vn in Fig. 4.53b, \VC obtain a truss \vhich can
be analyzed by elementary methods. This may mean, for example, that the
substitution of the bar a for the bar .t reduces the system to a simple space
truss or to another con1plex truss \vhich can be more readily analyzed. This
fictitious truss \ve no\v analyze under cach of t\VO separate conditions of
loading: (1) the given system of loads PI, P2, . . . as sho\vn in Fig. 4.53b and
(2) t\VO equal and opposite unit forces acting along the axis of the removed bar
x as shc>\vn in Fig. 4.53c. Let S; be the axial force in any bar due to the P
loading (Fig. 4.53b) and s the axial force' in any bar due to the unit-force
loading (Fig. 4.53c). No\\', it is obvious that if \VC have forces of magnitude
j¥ instead of unit forces in Fig. 4.53c, the axial force in any bar \vill be simply
sX instead of s. Next \ve superimpose this latter condition of loading on that
in Fig. 4.53 b and conclude that the corresponding axial force in any bar of
the fictitious truss under the con1bined X and P loading \\rill be
S1 = S; + sJ\" (a)
In the particular case of the added bar a, then, \\Te have
Sa = S + s:X (b)
No\\', if \VC choose X of such magnitude as to make Sa = 0, the bar a becomes
P2
P2
(a)
FI G. 4.53
(b)
(c)

206 STATICALLY DETERMINATE SPACE STRUCTURES
inactive and can be rcnl0ved, leaving the fictitious truss identical \\lith the
given truss except that the bar x is replaced by forces '<.. Hence, \ve conclude
that the value of X \\lhich n1akcs Sa in Eq. (b) equal to zero represents the true
axial force in the bar .\' of the given truss (Fig. 4.5 3a). Proceeding in this
n1anner, \\rc \vnte
S + s:J'<. = 0
from \vhich
x=
S'
a
-,
Sa
(c)
Using this value of }( in Eq. (a), \ve may nov. r calculate the axial force Sa' in
any other bar of the given truss \vithout further di fficulty.
I f in the procedure above \ve should find s: = 0, the value of ,,'<. defined by
Eq. (c) becomes indeterminate or infinite depending on \vhcther S is zero or
different from zero, and the truss is statically indeternlinate. This idea can
sometin1es be used to advantage in test,ng a complex truss for critical form.
Let us no\\' consider the application of Hcnneberg's nlethod to several
particular cases. As a first example, \ve take the octahedron loaded as
shown in Fig. 4.54a. To define the configuration of this systenl, ,,'e assume
that ABCI) is a square \vith edges of length 8 ft, \vhilc each of the other
eight bars has a ler.gth of 9 ft. This makes the diagonals of the square
AC' = BD = 8 V2 ft, and the vertical diagonal EF = 14 ft. The equal and
opposite loads P are assumed to act along the diagonal A C of the square so that
the system as a \vhole is in equilibrium.
If \ve remove the bar EC (marked x) and substitute a bar AC (marked a),
\ve obtain a simple space truss as sho\vn in Fig. 4.54b. \\lithout difficulty, \ve
can nO\\l make a complete analysis of this fictitious truss under the action of
the applied loads P (Fig. 4.54b) and again under the action of t\\TO equal and
opposite unit forces acting along the axis EC of the rcnl0ved bar x (Fig. 4.54c).
E
E
E
A
c
F
F
F
(a)
FI G. 4.54
(b)
(c)

ARTICLE 4.7
207
TABLE 4.1
Bar S , /x s,
S.
J & I
(1) (2) (3) (4) (5)
1 0 + 1. 000 +O.397P +0.397P
2 0 - I. 000 -0.397P -0.3971'
.t. + J .000 +0.397P +0.397P
4 0 - 1. 000 -0.397P -0.3971'
5 0 +0.889 +O.353P +0,353P
6 0 +0.889 +0.353P +0.353P
7 0 +0.889 +0.353P +0.3531'
R 0 +0.889 +O.353P +0.353P
9 0 + J .000 +0.397P +0.397P
10 0 - I . 000 -0.3971' -0.397P
1 [ 0 + 1 . 000 +0.3971' +0.397P
12 0 - I . 000 -0.397P -0.397P
a +1' -2.515 -I.OOOP 0
r
The results of such analyses are sho\\'n in the second and third columns of
lablc 4.1. Using the values of S and s: from this table in Eq. (c) above, \ve
obtain
x = - p - - = +0 397P
-2.515 .
Having this value of X, \\'C may no\\' fill in column 4 of the table, and then the
values ot S i in column 5 are obtained from Eq. (a).
In using Henneberg's method, \VC naturally raise the question as to "Thich
bar of the system to remove and \\'here to place the added bar a. In general,
any bar can be taken as "", but the added bar a must then be so placed as to restore
the rigidity of the frame, since other\\'ise the fictitious truss \vould be staticaUy
indeterminate. l'here is usually more than one possibility for the added bar
a, and \\'e try to nlake a choice that \vill render the fictitious truss as easy to
analyze as possible. For example, by placing the bar a bet\veen the joints A
and (' in the example above, \ve obtained a very simple case \vhcrc all bars
except a \vcre inactive undcr the P loading. A bar a bet\veen the joints B and
D would have served our general purpose just as \vell, but then the analysis of
the system under the P loading \\'ould have been a little more complicated.
.l\.s a second example, let us consider the rectangular parallelepiped supported
and loaded as sho\\'n in Fig. 4.55. V\'e have already seen in Fig. 4.46 (see
page 199) that if the bar EF' of this truss is removed and a bar g introduced
bet\vcen the points E' and H', \ve obtain a. system which can be readily

208 STATICALLY DETERMINATE SPACE STRUCTURES
analyzed by resolution into plane trusses. (--1cnce, if \ve take the truss in
Fig. 4.46 as the fictitious truss, \ve already have the results of the analysis under
the P loading and can at once fill in column 2 of Table 4.2. 1his done, \VC next
consider the analysis of the fictitious truss (Fig. 4.46) under the action of t\VO
equal and opposite unit forces applied at E and [/' and acting along the line
joining these t\\'O points. This analysis can be carried out in about the saIne
manner as that alrcady discussed in connection with Fig. 4.46, and \ve give
simply the final results in column 3 of the table. Using no\\' the values of
S and s: from the last line of this table in Eq. (c), Vie find
x = + _P;- = +O.354P
2v 2
V\lith Y kno\vn, the values of Si as given in column 5 of the table can no\\' be
computed from Eq. (a), and the analysis is completed.
Sometimes, in order to reduce a given COIllplcx space truss to a form that can
be readily analyzed, it may be necessary to remove several bars Xl, X2, . . .
and substitute as many others th, (12, . . .. Then, in application of Henne-
berg's method, \VC proceed as before and make a conlplete analysis of the
fictitious truss under each of the follo\ving conditions of loading: (1) the given
P loading, (2) a pair cf unit forces replacing the bar Xl, (3) a pair of unit forces
rcplacing the bar ..2, etc. Then, denoting by S:: the axial force in any bar due
to the P loading and by .<, s/, . . . , the axial forces due to each pair of unit
F
1
F'
P
16
B
r E'
I
I
10' 21 I
I
Ig
I
A d I
I
H 'H'
10' l4 f
I
1 G
FI G. 4.55  10'-1

ARTICLE ..7
209
TABLE 4.2
Bar I s' , /x s,
s.
· i , ,
(I) I (2) (3) (4) (5)
J
1 0 0 0 0
2 0 -I -0.3541' -0.354P
3 -1' +1/yZ +0.250P -0.7501'
4 + VZ p -I -0.354P + 1 . 061'
5 -1' +I/y'i +0.250P -0.7501'
6 + yZl' -I -0.3541' + 1 . 061'
7 0 0 0 0
8 +P/V2 -J -0.354P +0.3531'
9 -1'/2 +1/Y2 +0.2501' -0.2501'
10 +1'/ vi -I -0.3541' +0.3531'
II -1'/2 +1/yZ +0.2501' -0.2501'
12 I - 31'/2 +vi +0. 5001' - 1 . OOP
13 +P/ y2 -I -0.3541' +0.3531'
14 -P/2 +1/yZ +0.2501' -O.250P
15 +P/VZ -1 -0.3541' +0.3531'
16 0 -1/yZ -0.2501' -0.2501'
17 -1'/2 +1/0 +0.250P -0.2501'
18 0 +1/yZ +0. 2501' +0.2501'
19 -1'/2 + I / V2 +O.250P -0.2501'
20 -1'/2 +1/yZ +0.2501' -0.2501'
21 +1'/2 -1/y'2 -0.2501' +0.250P
22 -P/yi +1 +0.3541' -0.3531'
23 +1'/2 - 1/ y'2 -0.2501' +0.2501'
24 -P/y2 +1 +0.3541' -0.3531'
25 0 0 0 0
26 -1'/2 0 0 -0.5001'
27 0 0 0 0
28 0 0 0 0
29 0 0 0 0
.t +1 +0.3541' +0.3541'
g +1'/2 -yZ -0.5001' 0
loads, we find by superposition that
Si = S; + SXl + S/X2 +
where "( h X 2, . . . are so chosen as to satisfy the equations
Sal = SI + SIX 1 + s::X 2 + . · . = 0
Sat = St + S2X 1 + S:X 2 +- . . · = 0
(d)
(e)
......................

210
-.
'.. -
....
'-'
....
-..
... ' 0
::: -
.... .--

C"';
--
... 
 ... "'-'"
1"'1
N
,-..,
...... 00
... '.. '-'
1"'1
t-:'
.... ... "-"""

... --
... -0
....,;,- '-'
::. .-..
... '.. ""
'-'I '-'
... .-..
... '.. 
 '-'
-..
...
..... ... ....-....
V)N
'-'
M

W
...J
ex)
<
to-
.-...
-
CQ '-'
Q...
t---.. r- 0 r...... 
C' 1"-- v-. t'- lr.
r -0 
00000
, I 1+
Q...Q...Q...
r-. 't' '<:)
0--.0
r..... o:t' 00
0-0
+1+
O""Q...
r-.... r-. t-.. o:t'
r--- - (..... ""
N - '"
00000
1++1
ooN
- f"J 
.,.. XI t'-.
N
,+ I
a
oo N
- t".j r'\
 co r-....
,.......
1+
-1"N
_ r'\
1:f'r---
o
+1
Q...Q...
.-rN
-_r'\
o:t'r---
1+1
- "'I r"" "1" 11"\
 
,....... r- r...... r-
r- - r......
f"J f"1
oocoo
+ + I 1

r- r--. r-....
r..... r-..
N
ooo
+ 1 I
:....:...
-TO
"" 11"\
f -.c
:...
r-..... 'r, 
t... ,....... '.r,
- 
coooo
1 +-t-++
Q...
t -0  0
0-0-0
,........-rC
00-
1+++
Q...Q...
.,..,.......
'" -
r t"J
c
co
+1
Q...
o t
'r, r-..
t"1
Q...c...
",.,..0
"""j lr. ""
 "J
00
+1
000
1++'

r-- 0
o C
r-- ,,....
00
++
Q....c.....c.....c.....
1'"- "" .,.. 0
t-- "I ' r, ""
r", ""'1
0000
I I ., I
c...
,......
""'1
o
+
.,..,......
.... ,...,..,
 t'--
o
+1
ooN
_ ".1 r'\
 00 r-....
N
I + I
.,.
-

N
r'\
r-...
+
-0 t-.. 00 0-. 0
 ro-...
.... ,.,..
...... t'-.
.-.
'-'
+1

-
-:r
00 N
t".j 
00 r......
,....J -
+1
Q...
,...... ,,.... 
r- t" 1 Lr.
0:''',
000
+- + +
000
ro-... 0
000
('., ,.....
I 1
:!:8
o
I

!8
o
N""
-
o:t'
.,..
00
N
00
f'j
I
c.....
o
""
N
o
+
-1"0
.,..8
000
8
1'"'1 ,....J
, I
o:t' -
.,..g
I
00 's;2
O
,...., 'I
I
8
o
1
8
o
"I
I
'-Or---oo()\O
- N

 Q...Q." Q... 0... o...Q... Q.,.
I..... va, 00 00 ...... 8 '"
"'. N 11"1 ...... ""0 ...... N
'" N '" ",", ...... -
0 0 0 0 C 0 0 0 00 0
I I + I I + + ++ +
0 8  8 00 8 0
0  '" 0
0 0 f"'1 C 00 0 0
t"i "f'"1" '" ,..,.,
+++ I I
8  88 00 0 8
- 1"" 0
0 '1" 00 00 0 0
r-", N
I +++ I I
8 00 0 a "" 08
'" 0 C  80
0 00 OON
 N - N"1" ""
I I +++
 coccg 0 0
- 880 0 8
 0
- - N 
I +++ I I I
--- co 00'""8 8 0
N oo 0
00 004"10 0 0
,.., - '"   ''''.
I +++
 00 000 0 -
-
- 88 8 0 0
 0 C
4"1 ,.,..
I ++ + I I I
00 8 8 .----- ----- 0 8 -----
o
4"1 o 0
00 0 0 NO 0 0
,... - N  '1" ,,,
I ++ + I
88 000  - 8
NO - C
000  0 0
1"" ,.,..
+++ I I I [
Q..0...Q... - - --.  0... Q..,
000 0  0 0
8 8 - 0 0
 0 c.
c - I"' ,.,.. 1"'.
++ I I
----
o...
''''' "" r---... 0 C
r...... '" 0 '" '.....
,..,., - ,....... ,....... 1""
o...o...o...
,....... ...... ''', r---... 0 C r-
o ". ". 0 '..... .,.., 0
t"- ...0 .- ,....... ,....... "'. ,.......
. . , .
oeoco
+ I + I 1
Q",o...c...
gsc",8
oc 0
. . , .. . .
0000000
I +++ I I I
o...o...
888
r---... 11"1 '"
" ""
"- "-
8 0
...... 8
0000
Q...
8
'"
-I-
"J "'
1+
. .
000
+ 1 I
Q..,0...Q..
000
I,.., '" 
N t" '"
. .
o ON
+ + +
Q.., Q.. -
v-. 0 I.....
NO'"
......
-Q.., 0...
...... ...... '"
...... '" '"

o
o
'.....
a
I
. . .
000
1++
. . .
000
1++
oce.
I 1 I
N . .,. lr,
,., N ,.... "'I N
...or-ooO\o 4"1
,... ,, ". 4"1 ,.,.. ,..,.. ,.,..
p; '
 
 1!3
Q.., Q..,
'" 11"10
'" NO
- ......
0 0 00
I + I
 .  
211

212 STATICALL Y DETERMINATE SPACE STRUCTURES
@
(a)
FI G. 4.56
(b)
(c)
As a specific example, let us consider the complex space truss consisting of
12 joints and 36 bars arranged as sho\vn in Fig. 4.56a. This is a rigid and
statically determinate system, but one that does not lend itself to ready analysis
by elementary methods. Ho\\rever, if \ve remove the four bars .'\:1, .:\'2, .\.3, and
X4 that form the top ring and introduce instead the horizontal constraints aI,
112, a3, and fl., as sho\vn in Fig. 4.5 6b, \ve obtain a simple space truss that can
readily be analyzed by the nlcthod of joints. \\ begin \vith the observation
that, under the action of any applied load Q at the joint A, only the bars
indicated in Fig. 4.56c by heavy lines \vill be active. Furthermore, o\ving to
the symmetry of the system, \\'C conclude that if \ve analyze this heavy-line
portion under the action of each of three unit-force components as shown in
Fig. 4.57a, b, c, respectively, \\'c can easily obtain fronl this the axial force
in any bar of the system due to any combination of loads applied to the joints
A, B, (', D of the top ring. For example, let us assume that the actual loading
on the given truss is limited to a single vertical load P at A. Then, by using
the results from Fig. 4.57a, the value of S; for each bar of the system \-,,'ill be as
sho\vn in column 2 of Table 4.3. Likc\vise, if \ve \\'ant the values of s due to
t\\'o equal and opposite unit forces replacing the bar Xl, \ve use the results frorn
Fig. 4.57 band c and obtain the values sho\vn in column 3 of this table. In
h h I r: II "' d ,,,, h . I 4 6
t e same \\'ay, t e va ucs lOr S, , Si , an si , as S o\\'n In co lImns to ,
are obtained. No,,', using the values frorn the table for the fictitious bars

ARTICLE 4.7
-1.414
lIb
(a)
FI G. 4.57
lib
(b)
213
- 2.828
lIb
(c)
'1I, t12, a3, and fl.. in Eqs. (e) on page 209, \ve may \vrite
SUI = -J) - 4"'\1 - \2 - Y4 = 0
SU2 = 0 - Yl - 4"'Y 2 - ""\3
SUl = 0 - .1\2 - 4.'.\3 - ""\4
Sat = - 2P - Xl - Xa - 4.X... = 0
from 'Nhich \\.e readily find
1)
,,\ 1 = - -
R
=0
=0
"\2
, P
'3 = + H
(e')
J\4 = - P.
2
Finally, using these values in Eq. (d), \VC obtain the desired axial forces Si as
recorded in colun1n 11 of the table. This cornplctes the analysis of the givcn
systcrr: under the action of a single vertical load P at A.
PROBLEMS
=0
L7 sing I Icnncberg's method, IHake a cOlnplete analysis of the complex space truss
loaccd as sh()\vn in Fig. 4.58. To obtain a truss easily anal yzed hy eletncntary
methods, remove the bar J.;C and substitute a fictitious bar A H'. A RCI) is a square,
and cach inclined bar rnakes an angle of 45° \vith the vertical.

214 STATICALLY DETERMINATE SPACE STRUCTURES
K
J
I
L
H
E
(a)
E
F
G
p
p
E
G
E'
G'
FI G. 4.58
(b)
FI G. 4.59
2 Repcat the analysis of the systctn sho\vn in Fig. 4.58 if there is a vertical load P
at l instead of a horizontal load Pat C as sh()\vn.
3 Using Hcnncbcrg' mcthod, make a complctc analysis of the cOlnplcx space truss
sho\vn in plan and elevation in Fig. 4.59, (a) under the action of a single vertical
load P at A as sho\vn, (/1) under the action of a vertical load P at each of the joints A,
B, C, and D.
4 Repeat the analysis of the systenl in Fig. 4.59, assulning a horizontal force 1> applied
at E and acting to the right.
5 Make a complete analysis of the con1plcx system sho\vn in Fig. 4.56l1 if there is
a horizontal forcc P applied at A and acting to the right in the dircction coinciding
\vith AR.

Chapter .5
General theorems relating
to elastic systems
5.1 STRAIN ENERGY IN TENSION, TORSION, AND BENDING
.10 a certain extent, all structural Inatcrials are elastic and deform slightly
under the action of applied loads. This means that the loaded configuration
of any structure, such as a truss or arch rib, is slightly different from its
unloaded configuration. Iio\\'cvcr, if the structure is statically determinate,
this extremely snlall change in configuration has no significant effect on its
geOtT1ctry, and it is {'ntirely justifiable to use the unloaded configuration as a
basis of calculation of internal forces. In the case of a statically indetertninate
system, ho\vever, such small defofrnations have a significant effect on the dis-
tribution of internal forces and must be taken into account. We therefore
turn our attention no\\/ to various relationships pertaining to clastic dcfortna-
tions of structural elcments under load.
\\lithin limits, most structural rnaterials can be considered as perfectly
elastic, and they can be assumed to obey Hooke's la \\'. In the case of a
215

216 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
p
T L
dP 1T
I
1 PI
0
P
FIG. 5.1 (a) (6)
prismatic bar in simple tension (}t'ig. 5 .la), this means that the load-dcflcction
curve (Ja is a straight linc as sho\vn in Fig. 5.1 b. ""hen such a bar elongates
under the action of a gradually increasing tensile load P, the load does \vork,
and this \vork is stored in the bar in the form of strain cnelJ!,)'. To calculate
the amount of this strain encrgy, \VC consider the state in \\,hich the load has
the magnitude PI, and the elongation the corresponding magnitude 01. '[his
statc is reprcsented in .Fig. 5.] b by the point b. If at this point an incren1ent
dP l is added, the elongation of the bar \vill increase by the amount dOl, and the
load Ph v.,hich was a]ready acting on the bar, \vill producc the \vork PI dOl.
This work is represented in the figure by the area of the shaded rectangle
bcde. The \vork of the slnall increment dP l of the load on the small dis-
placement dOl can be neglected as a small quantity of second order. '{'he total
work done by the load during its gradual increase from zero to the final value
P is obtained as the sum of all such elemental areas as that shown in Fig. 5.1 b
and is represented by the area of the triangle (Jag. 1'hus, denoting by U the
strain energy that is stored in the bar during its extension, \ve have
[) = Po
2
(a)
Using the kno\vn expression
0= 
AE
(b)
for the elongation of a prismatic bar of length /, cross-sectional area A, and
modulus of elasticity E, \\Fe can represent the strain energy (a) either as a
function of the load P or as a function of the elongation o. These t\VO fonns
for the strain energy are
P2/
U = 2AE
AE!
U=--
2/
(5.la)
(5.1 b)

ARTI CLE 5.1
217
The strain energy of torsion of a prismatic shaft (Fig. 5.2a) can bc obtained
fronl the torsion-test diagram sho\\'n in Fig. 5.2b. '''his diagram sho\\'s the
relation bct\vccn the t\visting moment, or torque, 7' and the corresponding
angle of t\vist cp. We see that \vithin the elastic range, the angle of t\vist is
proportional to the torque as represented by the straight line Oa. Again, the
shaded elemental area in the figure represents the \vork done by the torque
during an infinitesimal increase dcp of the angle of t\vist cp, and the area of the
triangle Dab represents the total \vork done by the torque as it increases
gradually fronl zero to T. '[his \\fork is equal to the strain energy stored in
the shaft during torsion, and we have
Tcp
u=-
2
(c)
Using, for the angle of t\vist, the kno\v n fonnula
11
cP =-
C
(d)
in \vhich I is the length of the shaft and C its torsiOl1alrigidity, I \\I.e obtain
() = T2 / (5.2a)
2C
or Ll = cp 2(,' (5.2b)
21
In the first of these fOffilulas, the strain energy is represented as a function of
the torque T and in the second as a function of the angle of twist cp.
In the case of plire be1lding of a prislnatic bar in a prillcipal plane (Fig. 5. 3a) ,
the angle (J of rotation of one end \vi th respect to the other is proportional to
1 See S. Tirnoshenko, "Strength of l\-Iatcrials," 3d cd., vol. I, p. 290, D. Van Nostrand
Company, Inc., Princeton N.J., 1955.
T
T T /
a
L
I dT
l r
T 0 4>
b
FI G. 5.2 (a) (b)

218 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
y
- e---\D
'I
a
/
o
8
L
db!
f
x
FIG. 5.3
(a)
the bending mon1ent 1\,1 as sho\vn in Fig. 5.3 b. Hence, \ve conclude that the
strain entrgy of bending, equal to the total \vork produced by the moment i\;l, is
Ll = 1\18
2
(t')
Using for 8 the knc\vn formula
...\;/1
() =-
EI
Cf)
in \vhich 1 is the length of the beanl and EI its flexural rigidity, \VC can represent
this strain encrgy in either of the follo\ving .t\VO forllls:
t i\1 2 1 (5.3a)
[j = 2EI
T 8 2 £1 (5.3 b)
or L/ =-
21
Again, \ve see that :he strain energy can be rcpresentcd ei ther as a function of
the acting forces, in this case the bending moment AJ, or as a function of the
quantity 8 defining the defornlation.
In the case of a prisnlatic beam subjected to the action of transverse loads in
a plane of symmetry, \ve have strain energy due to both bending and shear
deforrnation. Ho\vcvcr, for beams of usual proportions, it can be sho\vn that
the strain energy due to shear is small conlpared \vi th that due to bending, and
the former is usually neglected in structural analysis" Considering only
bending, \VC obtain the strain energy in an element of the beam of length dx
from Eqs. (5" 3) sin1pl y by substituting dx for I and dO / dx for all. rrhus, for
one clement,
i T r = )"f 2 dx
( v 2EI
dU = ! ( dO ) 2 d.t"
_ dx
(g)
(h)
1 For a con1plctc discussion of this question, see ibid., p. 318.

ARTICLE 5.2
219
rhcn, to obtain the strain energy in the entire beam, \ve have only to sun1
expressions (g) and (h) over the entire length I of the beam. Noting that, for
small deformation, 8  dy / d......, \VC obtain in this way
l} = (I }\/[2 dx
J 0 2EI
U = EI (l ( d 2 Y ) 2 dx
2 J 0 dx 2
( 5 . 4a)
(5.4b)
These fonnulas can be used also for a beam of variable cross section jf the
moment of inertia I can be exprcssed as a kno\vn function of x.
PROBLEMS
l\vo identical prisnlatic bearns, onc sinlpJy supported, the othcr \vith built-in ends,
are bent by equal concentrated loads transvcrsely applied at the middle. In \vhat
ratio are thc amounts of strain energy stored?
Ans. 4:1.
2 .A \vooden cantilever bCalTI having a rectangular cross section 8 in. deep and 5 in.
\vide and a length of 6 ft carries a uniform load of 200 Ib/ft. C:alculate the amount
of5train energy in thc bcaln if E = 1.5(10)6 psi.
Ans. U = 42 Ib-in.
3 COlnpare thc strain cncrgy of torsion of a prismatic bar of circular cross section
\vith that of pure bending of thc same bar if the t\visting monlent and the bending
tnOn1cnt arc equal in the t\VO cascs, chat is, if T = .1\1. .-\ssumc that E = 2.5 G.
Al1S. UTI U", = !.
5.2 PRINCIPLE OF SUPERPOSITION
We consider no\v the case \vhere a structural member is subjected to the
simultaneous action of several loads, as sho\vn, for example, in f"'ig. 5.4. In
such a case, if the bar is prismatic in form and the material obeys Hooke's la\v,
the total elongation \vill be
h = (/)1 + P2 + Pa)/ l + (/)2 + _ fa) (/ 2 - 1 1 ) + f1 S/3 - /2)
AE AE AE
_ Pli l + P 2 1 2 + Pa1a
- AE AE AE
(a)
\vhich is seen to be a linear function of the external forces. The first term
on the right side of Eq. (a), for example, is the elongation Pill/ AE produced
by the force PI acting alone. The second and third tenns, likewise, are the
elongations of the bar produced by the forces P2 and Pa, respectively. Thus,

2H GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
I  Pi t P2 ! P3 + 
11
J
1 2 (a)
P 1 d
13
12J ---  -:=-----
----
P3 (b)
FIG. 5.4 FIG. 5.5
\\"C see that the tota; elongation is obtained simply by summing up the elonga-
tions produced by the individual forces.
As a second example, Ict us consider bending of a simply supported beam
AB under the action of transverse loads ])1, P2, P3, as sho\vn in Fig,S .5a. In
calculating deflections of the bean1, \ve consider in Fig. 5,5 b the cOlill!{ate beanl
ab loadcd by the bending-moment area acdeb. It is kno\\'n 1 that thc bending
rnoments of this conjugate bcam give, to a certain scale, the deflections of the
actual beam. No\v, from simple statical considerations, it follo\vs that the
total moment area acdeb can be obtained by summing up the triangular areas
indicatcd in the figure by dashcd lines and representing, respectively, the
mOlnent areas for the individual loads. Thus, the total fictitious load acdeb
on the conjugate bcam is the sum of the fictitious triangular loads corresponding
to the individual forces Ph P2, a'nd P3 'acting on the beam AB. Repeating the
same reasoning in calculating the bending moment at any cross section of the
conjugate bcanl, \\'e conclude that the bending moment produced by the total
load acdeb is equal to the sunl of the bending moments produced by the COln-
poncnt triangular loads. Hence, the deflection at any cross section of the
actual beam ,,4 B is equal to the sum of the deflections produced at the same
cross section by the individual loads Pi, P2, and P3. Further, since these latter
deflections are proportional to the corrcsponding forces, it can be concluded
that the total deflection of any cross section of the beam A B \vill be a linear
function of the forces PI, P2, and P a and is obtained by summing up the cor-
responding deflections produced by the individual loads.
In all cases of compositc loading \vhere deRections arc linear functions of
the applied forces, the above conclusion holds, and \\te obtain the total deflection
of any point simply as the sUln of the deflections produced by the individual
forces. This statcment is called the principle of superposition.
1 S ' b ' d
ee , I ., p. 147.

ART. CLE 5.2
221
Under ordinary circumstances, a linear relationship bct\vccn deflections and
applied forces rests solcly on the assumption that the material of a structure
follo\vs Hooke's la\v. Ho\vever, thcre arc certain cases in \vhich this assun1p-
tion alone \vill not be sufficient, and the deflections will not be lincar functions
of the applied forces cven though the material does follo\\' Hooke's la\v. ()ne
such example \ve have in the case of a bar submitted to the simultaneous action
of axial and lateral forces. Considering, for example, the bending of the
beam AB loaded as sho\vn in Fig. 5.6a, \ve conclude that the deflection under
the load P is no longer proportional to the load. Specifically, it can be
represcnted by the formula 1
P[3 1
o  48£/1 - S/Scr
(b)
in \\!hich the first factor represents the deflection produced by the lateral load
if acting alone, and the second factor represents the effect on this deflection of
the axial comprcssive force S. 1'he magnitude of this effect depends on the
nlagnitude of the ratio S/Scr in \\Thich Scr = 7r 2 EI/ [2 is the critical load for
buckling of the beam in the plane of the figure. Since S is proportional to P,
expression (b) is no longer a linear function of the load, and the relation
bet\\'een 0 and P is no longcr linear. Representing this relation graphically,
\ve obtain the curve sho\\Tn in Fig. 5.6b. It is seen that the deflection is no
longer proportional to the load and that it begins to increase very rapidly as the
compressive force in the bcatn approaches the critical value. Thus, if \ve
joublc the load P, the deflection \vill al\vays be more than doubled, and the
principle of superposition does not hold.
As another example \\There the principle of superposition docs not hold, let
JS considcr the system of t\VO identical horizontal bars hinged together, as
)hown in Fig. 5.7a. Under the action of a verricalload P the bars \vill undergo
;orne extension, and the hinge C \vill move do\vn by an amount eC I = O.
ssuming that this deflection is small, \ve find for the corrcspondingly small
. See ibid., vol. II, p. 52.
p
B
r
b
A
!
2
p
I
2
o
FI G. 5.6
(a)
(6)

222 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
p
k ------- b
p
(a)
L
dP
,

p
(b)
(c)
FIG. 5.7
angle a and for the tensile force S in each bar the follo\ving values:
o
a7
1)
S-
2a
Considering, no\v, the unit elongation of each bar, we find
(c)
(d)
E=
Vl2+- - /
I
1 0 2
): 7 2
The same elongation, if \ve use Hooke's la\v, is
S
e=-
AE
and \ve obtain
1 0 2 S P PI
"2 /2 = AE = 2aAE - 20AE
from \vhich
O=/ 
and }) = o3AE
/3
(e)
(f)
(g)
Again \ve see that the deflection 0 is not proportional to the load although the
material follows Hooke's la\v. This case diffcrs from the preceding one in
that with increasing deflection the system becomes stiffer, and the relation

ART. CLE 5.3
223
bet\vccn 0 and P is reprcsented by the curve 8ho\\'n in }t'ig. 5.7 c. It is typical
of the behavior of a system of critical 101711 as discussed in Art. 2.7 (page 89).
Follo\ving the same reasoning as for the case of simple tension, it can be
sho\vn that the area Oabc() in Fig. 5.7 c represcnts the \\'ork done by the load
P during the deflection 0 and that it is equal to the strain energy stored in the
bars AC and CR. 'I'he expression for this energy is obtained by summing up
all such elemental areas as the shaded strip sho\vn in the figure. This gives
[] = (0 P do = (1!i (6 0 3 do = AEo"
Jo /3 Jo 4/ 3
1.) = IP_
4yAE
(5.5a)
or
(5.5b)
It is seen that these expressions for strain energy are no longer second-degree
functions of the displacCIllcnt or force such as \ve had in those examples for
\vhich the la\v of superposition holds.
Considering \vhat is characteristic of the examples reprcsented in Figs. 5.6
and 5.7, \ve see that in each case the action of the external forces is apprcciably
affected by smaH deformations that take place in the system. l"'he axial force
S in Fig. 5.6 produces only compression of the beam AB if acting alone.
Ho\vever, \vhen it acts in conjunction \vith the transversc load P, it produces
not only compression but also some additional bending. In the case repre-
sented in Fig. 5.7, the tensile forces S in the bars depend on the deformation
and are invcrsely proportional to the deflection 0 that takes placc. AI\vays
\vhen \ve have stich conditions that the action of external forces is affccted by
small defoflnations produced in the system, stresses and deformation \vill not
be JincC1r functions of thc applied forces, and the principle of superposition does
not hold. rhus, in conclusion, it can be said that the material must follo\v
Hooke's la\v if the principle of superposition is to bc applicable; but this
requirement alone is insufficient. We havc to consider also \vhcthcr or not
the action of the applied forces \vill be affected by small deformations of the
structurc. If such effect is substantial and must be taken into account, in
calculating intcrnal forces or stresses, thc principle of superposition does not
ho ld .
5.3 STRAIN ENERGY IN GENERALIZED FORM
In Art. 5.1, it \vas secn that expressions for strain cnergy in an elastic bar
\vcrc second-degree functions either of the applied forces or of the displace-
ments. It \-vill no\\' be sho\vn that the same conclusion holds equally \vell for
any elastic structure, provided that the principle of superposition can be
applied. Let us consider, for example, an elastic body supported as sho\\'n in

224 GENERAL THEOREMS RELATING TO ELASTIC SVS"rEMS
Pz
FI G. 5.8
(a)
T
1;.
_L
o ..
1>n
..j
(b)
Fig. 5.8a and subn1itted to the action of external forces ])1, P2, P3, . . . .
Since the amount of strain energy stored in the body depends not on the order
in which the forces are applied but only on their final magnitudes, \\'e can
simplify our further discussion by assuming tat all the forces are applied
simultaneously and then gradually increased in the same proportion. I'hcn,
if the principle of superposition holds, the displacements will be linear functions
of the forces, and during loading they increase in the same proportion as the
forces do. In calculating the work of any force P,h \\'e shall bc interested not
in the total displacement of its point of application but only in that component
in the direction of the force. IJct On denote this component of displacen1ent.
rrhen, during the assumed gradual loading , 0 , & increases in the sanle proportion
as P n, and the relation bet\vcen these two quantities can be represented by the
diagram sho\vn in Fig. 5.8h. Hence, \\'e conclude that the \\'ork produced by
the load Pn is P n o,,/2. 1he total work of all external forces, equal to the
strain energy stored in the deformed body, is obtained by summing up the works
of the individual forces, which gives 1
U = j-('p.o 1 + P 2 0 2 + P 3 0 3 + . . . )
(5.6)
The reactions la' R b , and Rc do not appear in this cxpression, because for the
conditions of support illustrated in Fig. 5.8, their \york is zero.
Since the disp]accments 01, 02, . . . arc homogeneous linear functions of
the forces PI, P 2 , , . . , it follo\vs that if these functions are substituted in
expression (5.6), \ve shall find the strain energy to be represented by a
homogeneous second-degree function of the external forces PI, ]J 2, . . . .
Likev.'ise, if \ve express the forces as linear functions of the displacemcnts and
then substitute these functions in exprcssion (5.6), we shall find that the strain
1 The statcment that the strain energy stored in an elastic body is equal to half of the
sum of thc products of external forces by the corresponding displacc[nents is sO[Ilctimes
called Claptyro71's thtOrml. It is mentioned in Latne's "Theory of Elasticity," seventh
lecture. See also Max\vcll, "Scientific Papers," vol. I, p. 598, and Todhunter and Pearson,
"History," vol. I, p. 578, Can1bridgc University Press, London,] 886.

ARTICLE 5.3
225
energy can also be represented as a homogeneous second-degree function of
the displacements 01, 02, . . .. Both forms of representation of the strain
energy will be useful in our further discussion.
It should be noted that the forcgoing conclusions regarding the degree and
homogeneity of the strain-energy function \"cre obtained on the assumption
that the principle of superposition holds. If this principle does not hold, the
strain energy \vi II no longcr be a second-degree function of the forces or of the
displacements, as can be seen, for example, from E<]s. (5.5) of the preceding
article.
Expression (5.6) for strain energy can be made more general by using the
notion of generalized forces. 1 Any group of statically interdependent forces
that can be completely defined by one symbol can be considered as a generalized
force. For example, in the case of axial extcnsion of a bar (Fig. 5. 9a), \\'e
have t\VO equal and opposite forccs acting along the axis of the bar, and one
symboll J defines entirely this pair of balanced forces. Again, in the case of
pure bending (Fig. 5.9/1), \VC have t\\'o equal and opposite couples that are in
equilibrium, and one synlbol At{ completely defines the system. In the case
of a transverse load acting on a beam (Fig. 5. 9c), \\rc have a group of three
forces that, being in equilibrium, are completcly defined by the magnitude P
of the applied load ° In all such cases wherc \ve can define a group of forces
by one symbol, \ve may treat that group as a generalized force.
In using the notion of gcncralized force, \ve havc to generalize also the notion
of displacement. In dealing \vith single forces (Fig. 5.8), \\'e have already
pointed out that not the total displacements of their points of application but
1 The notions of gnJeraliud force and generaliud displt1(nlttllt \vcre introduced by Lagrange
in his fan10us book "Mecanique analyrique," Paris, 1788. In application to the deforma-
tions of elastic systems, they \vere extensively used by Lord Rayleigh. Sec, for example,
his "Scientific Papers," vol. I, p. 255; "lOhcory of Sound," 2d ed., p. 91.
p 1
B B'
-- B
I 1
I I
I I
I I
I I
I I
I /
I I
I t
I I
I I
"
....1 II
'1:'
p
FI G. 5.9 (a)
G 8
MM
(b)
p
(c)

226 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
only those components in the directions of the forces should be considered,
since the \vork done by the forces depends only on such components. '-rhus,
if On is thc component of thc displacement of the point of application of the
force P,. in the direction of that force, \ve say that On is the displacement
corresponding to the force Pn. If a small increment on is given to this dis-
placement, the \\'ork done by the force P,. is
p" 8"
(a)
No\v, in using the notion of generalized force, the cor1'espol1di11K KellC1"a/ized
displacC'/l1e1Jt must be taken in such a \\'ay that the product of the generalized
force and the incrcment of the corresponding generalized displacement gives
the work. Take, for example, the case of simple tension (Fig. 5 .9a). It is
seen that the bar AB can be brought to any ne\v position AIRl by first moving
it parallel to itself to the position A lB' and then rotating it about the point AI.
During such displacement of the bar as a \vhole, the systenl of t\VO equal and
oppositc forces does not produce any \\lork. ""brk is produced only \vhen
the bar elongates by some amount o. rhis elongation, then, is the generalized
displacement corresponding to the generalized force consisting of t\\'o equal
and opposite forces P. If an increment ilo is given to the elongation 0, the
generalized force P produces \vork of the amount P ilo, \\'hich is similar to
expression (a) for the case of a single force. In the case of pure bending
(Fig. 5.9b), the work produced by the couples A1 is entirely defined by the
angle of rotation 8 of one end of the bar \vith respect to the other. Hence
this angle must be taken as the generalized displacement corresponding to the
generalized force represented by the t\VO equal and opposite couples. If 8
obtains an increment j,.8, the couples 1\;[ produce the \vork 1\J d8 and again \\'e
obtain an expression similar to expression (a). In the case of a transverse
load acting on a beam (Fig. 5. 9c), the group of forces ]J, Ra, and Rb docs not
produce any \vork \\Then the beam moves \\7ith its supports as a rigid body.
Work is produced only when the beam deflects and point C moves perpendicular
to AB. The deflection 0 of the point C \vith respect to the line AB joining the
ends of the beam axis is the gcneralized displacement in this case.
Sometimes the gencralized displaceInent corresponding to a chosen general-
ized force may not be entirely self-evident. Consider, for example, the simply
FIG. 5.10
Iqdx

. h _ .... _... .__ . .
g j.- x dx , I
x

ARTICLE 5.3 227
/flu 8 0 P L 8 b iff b
h ; (  B
A +
I I I 
FIG. 5.11 2 2
supported beam under uniform load as sho\vn in Fig. 5.10. If the intensity q
of the load is takcn as the generalized force, the corresponding generalized
displacement \vill be the area bet\vccn the chord AB and the deflection curve
as sho\\'n in the figure. To sho\\' this, \ve give to the deflection y of each
point on the axis of the beam a small increase y. Then the \\'ork done on
each such additional displacement by the corresponding clement of force q d:t.:
is q dx )', and the entire load \vill produce the \vork
q Jo' a.y dx
(b)
Now, observing that the integral Jo' a)' dx is an increment of the area between
the chord AB and the deflected axis of the beam, \ve conclude, by definition,
that such area is the generalized displacement corresponding to the generalized
force q. It is \vell to note in this case that \vhile thc gcneralized force has the
dimension of forcc + ICllgth, the corresponding generalized dispJaccment has
the dimension of length squared, so that their product has the proper dimension
for \'ork. Since thc \vork done by a generalized force on an increment of the
corresponding generalized displacement has ahvays the same form as expression
(a) derived for a single load, \ve conclude that the total \\fork produced by a
generalized force during a gradual loading has the same form as that for single
forces and is equal to half tht.. product of the final magnitude of the generalized
force and the final value of the corresponding generalized displacement. '[his
means that expression (5.6) derived for isolated forces can be used also if the
forces and the displacements arc generalized.
As an example of the application of exprcssion (5.6), let us calculate the
strain energy in a simply supported beam of prismatic form that is loaded by a
force P at the Iniddle and by the t\VO couples Ma and Mb appJied at the ends as
sho,,'n in Fig. 5 .11 . In such a case the system of forces acting on the beam
can be represented in tcrms of three generalized forces, namely, the load P
togechcr \\'ith its reactions, and each of the couples Nt a and NIb together \vith
their reactions. The corresponding generalized displacements \\"ill be the
deflection 0 under the load and the angles of rotation Ba and Bb of the ends of
the beam. The total strain energy stored in the bean1, from expression (5.6),
]S
u  -}(Po + M a 8a + MbBb)
(c)

228 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
Using no\\' the kno\vn expression for the deflection curvc, \ve have
1)/3 Ma/ 2 Atf b / 2
(j = 48£1 + "i6E'1 + 1 6£ /
P/2 ,\1a l A1bl
(}a = 16£ 1 + -3£1 + 6E i
Ob = Pl2 + ,\1[0 1 + i\1 b /
1 6£/ 6EI 3E/
(d)
Substituting these values of the displacements into expression (c), \ve obtain
u = 9i:j(J>2 +  PM. +  PM b +  M.2 + ll Mb 2
16 )
+ 7f Nlolvlb
(e)
It is seen that the strain energy is a homogeneous second-degree function of the
gencralized forces.
Solving Eqs. (d) for P, Ma, and J\tJ b and substituting them in expression (c),
we shall obtain the strain energy as a second-degree function of the generalized
di splacements.
PROBLEMS
i\l[ake a gcneral expression for the strain energy stored in the beam sho\vn in Fig.
5.11, assunling that P = 0, (n) expressed as a function of the end filaments 1\110 and
.lI b , (h) exprcssed as a function of the angles of rotation 8 0 and Ob.
I
Al1S. (11) U = 6El (AJ a 2 + Al o l\th + Alb 2 ).
2t'/
(b) U = '-,- (0 0 2 - Ou(}b + Ob 2 ).
2 .i\ thin steel strip of lcngth I and uniform flexural rigidity E/ is elastically bent into
a complete circlc by applied end mOlllents. C:Omputc the strain energy U that is
stored in the hoop.
Al1S. U = 21r2F;//1.
3 An elastic prismatic hcaln of length / has built-in ends and is subjected to the action
of a transverse load P at the middle and of such nlagnitudc that the deflection under
the load is o. i\ssuming that the deflected axis of the beam has the form of a full
\vavc of a cosine curve, calculate the strain energy stored in the beam.
A11s. lJ = EI7r 04 0 2 //3.
4 Using expression (5.6) and the notion of generalizcd force, cOlnpute the strain
cncrgy stored in the beam sho\vn in Fig. 5.10.
Ans. U = q2J6/240El.

ARTI CLE 5.4
229
5 An clastic body undergoes uniforn1 cornprcssion under hydrostatic pressure p.
I f the pressure p is taken as the generalized forl:c, \vhat \vill be the corresponding
generalized displacement?
5.4 CASTIGLIANO'S FIRST THEOREM
l"he principle of virtual displaceInents as discussed in Art. 1.10 for any ideaUy
connected system of rigid bodies can readily be extended to apply to an elastic
system. It is only necessary in this case to consider the virtual \vork of
intcrnal forces as \veJl as that of applied external forces. As a model of an
clastic body, \ve take a system of particles connected by springs as sho\vn in
Fig. 5,12, \vhere the springs rcprcsent the clastic constraints bet\veen the
various particles. Although not sho\vn in the figure, we assume that there
are suitable springs to rcsist relative rotation and sliding of the particles as
\\7ell as direct changes in lincar distances. Under the action of applied external
forces, the model \vill be elastically deformed, and the springs will exert
internal forces on the various particles of the systen1 as sho\vn in Fig. 5.12h.
Thus, \\'c obtain a very complex system of forces that, ho\vever, are in
equilibrium and therefore subject to application of the principle of virtual
displat'ements.
Iet Ph P2, . . . denote external forces applied to the body and let 01,
02, . . . be the corresponding displacements. We may no\v define a virtual
displacement of thc system by any set of sn1all changes f).o 1, d02, . . . in the
displacernents 01, 02, . . . , and the principle of virtual \vork gives
};P i f).Oj + f)."F = 0
(a)
\vhere  T denotes the \vork of the internal forces on the virtual displacements
of the particles. No\v, since \\'c assume an clastic body, the springs all obey
I-Iookc's la\v, and from this it £01l0\\7s that the \vork of internal forces on the
virtual displacements of the particles in Fig. 5.12h is equal to the change in
P l
P2\ P1.
P4
<
PI -*
*

Ps
///////////////////
FIG. 5.12
(a)
(b)

230 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
strain energy in the springs \vith reversed sign, that is,
D.T = -Ill}
(b)
Substituting -dU for D.T in Eq. (a), \VC obtain
!'P, D.Oi - dU = 0
(c)
To calculate the increment D. lJ, the strain energy must be rcprescntcd first
as a function of the displacements 01, 02, . . .. l'hen, in the usual \vay,
the increment of U due to the increments dO 1, D.02, . . . of the displacements is
al] af) aU
D.L" = - D.Ol + -- D.2 + --. D.03 +
aO l a0 2 a0 3
(d)
In applying the principle of virtual displacements, \ve can select in any
particular case a displaccmcnt that is mose suitable for the solution of a given
problem. For exanlple, \ve can imagine a virtual displacement that affects
only the displaccment 0,1, corresponding to the load Pn, and leaves the displace-
ments (h, 02, . . . unchanged. Undcr such conditions, D.OI, C102, . . . are all
zero except on, and expression (d) for the increment of (] reduces to
aU
D. U = aOn llo 11
Using this expression in Eq. (c), \ve obtain
aU
Pn 1l0,t - dOn D.o,. = 0
which, upon canceling '" reduces to
au
80'1 = Pn
.This equation states that if the strain energy of a deformed elastic body
is represented as a function of the displacements 01, 02, . . . , a partial
derivative of that function \vith respect to any chosen displacement gives the
corresponding force. This statement is usually referred to as Castigliano'J
first theore111. 1 Although derived above for the case of single forces Ph
P'l, . . . , it holds equally \vell for generalized forces and corresponding
generalized displacements.
As an example of the application of Eq. (5.7), let us consider a symnlctrical
system of three bars hinged together at fJ and loaded as sho\vn in Fig. 5.1 3.
IJenoting by (; the vertical displacement of the hinge D, \ve find the elongation
of each inclined bar to be 0 cos a, \vhile for the vertical bar it \vin be O. l\ssun1-
(t')
(5.7)
1 Alberto Castig;iano, "'rhcorcmc de l' equilibre des systcmcs clastiqucs ct ses applications,"
Paris, 1879. The English translation of this book \vas .nade by E. S. Andrc\vs, Scott,
Greenwood & Son, London, 1919.

ARTICLE 5.4
231
ing no\v that the cross-sectional areas and the [noduli of elasticity of all three
oars are equal, we find the strain energy of the system to be
L = AE (D 2 + 22 cos 3 a)
2/
(f)
No,v, using Eq. (5.7), \ve obtain
A};fJ
/ j - (1 + 2 CDS 3 a) = P
(g)
and from this \\'C find the a xial force in the vertical bar to be
s =  E[) =
I
p
+ 2 cos 3 a
(11)
Thus, by using Castigliano's first theorem, \\'e solve the statically indeterminate
problem.
A consideration of the argurnents used in the derivation of Castigliano' s
first theorem sho\vs that its validit)': is in no \vay dependent upon the principle
of superposition. 1he only requirement is that the system behave elastically.
For this reason, the theorem is of particular value in dealing \\'ith systems of
critical form like those discussed in Art. 5.2, for \vhich the principle of super-
position does not hold. As an example of this kind, \ve consider again the
case of t\VO bars arranged and loaded as sho\vn in fig. 5.7. Denoting by 0
the dcflection corresponding to the load P, \ve see that the elongation of each
bar \vi II be
dl = ylI2-+ o 2 _ I  0 2 (i)
2/
]'hcn, by using Eq. (5.1 b), we find that the strain energy of the system,
expressed as a function of the displacement 0, becomes
L} = 2 ( AE d/ 2 ) = 04 (j)
2/ 4/ 3
\vhich ,ve notc is not a second-degree function of o. Nevertheless, Eq. (5.7)
A
c
FIG. 5.13
p

232 GENERAL THEOREMS RELATltlG TO ELASTIC SYSTEMS
--j I
s
----:b --i-=.j
p
--
S x
FIG. 5.14 Y
applies, and \ve find, by differentiation,
AEo3
--=p
/3
from \\,hich
 -
o = / - . p
AE
(k)
This result is seen to agree \\..jth expression (1) obtained on page 222.
As a last example, let us considcr the case of a beam column sinlpIy sup-
ported at its ends and subjected to the simultaneous action of compressive
forces S and a transverse force P as sho\vn in Fig. 5. 14. Ie is desircd to find
the magnitude of the deflection 0 undcr thc load P. To sin1plify the problem,
we assume that the deflection curve can be representcd \vith sufficient accuracy
by a half sine wavc, that is,
. 11" .\..
'V = 5 sin T
1'hen the configuration of the system is entirely defined by 0, and the strain
energy of bending! ,vilJ be
v = £/ r l ( d . 2y ) 2 dx = E/7r 4 0 2
2 J 0 dx 2 4/ 3
(I)
Associated \vith this bending, \ve see that the t\VO ends of the beam come
together by the amount
i l ( dy ) 2 . 7r202
X = i - dx = -
o dx 4/
( 111 )
To find no\v the generalized force corresponding to the displacement 0, \ve
recall that by definition it is that quantity by which \ve must multiply the
increment Ao in order to obtain the corresponding virtual \\'orle l\1aking a
I It is unnecessary to consider the strain energy of direct cotnprcssion since it remains con-
stant and independent of o.

ARTICLE 5.4
233
small increase dd in 0, \ve see that this \\'ork is
P ao + S aX = (p + S  ) ao
Hence, the generalized force corresponding to 0 is P + S dX/ do. Now, using
expressions (I) and (111), Eq. (5.7) becomes
E {7r 4 = P + S 7r 2 0
2/ 3 2/
from \vhich
2P/3 I
0= --
1r 4 E/1 - SI2/7r 2 EI
(n)
Since the numerical factor 'Jr'. /2 = 48.7, \\'c see that this approximate result
con1pares very favorably \\'irh Eq. (b) on page 221.
PROBLEMS
Find the tensile force S in the vertical bar of the statically indeterminatc systcln in
Fig. 5.13 if its cross-sectional area is A \"hilc that of each inclined bar is A12.
Al1S. S = P/(l + cos 3 a).
2 ;\ prisn13tic beam \virh built-in ends has flexural rigidity FJ and carries a transverse
load P at the 111iddle. Assutning that the deflection line can he representcd \vith
sufficient accuracy by the cosine curvc
.v =  (I - cos 2;X )
calculate the Inagnitudc of the deflection 0 under the load.
Ans. 0 = PI3/21r411.
3 A rigid beam is supported hy thrce vertical clastic bars as shown in Fig. 5.15.
Each of the supporting bars has cross-sectional area A and modulus of elasticity E,
but the rniddlc bar is only half as long as the othcr t\vo. Calculate the Jatcral deflec-
tion 0 produced by the horizontal f()rcc P acting as sho\vn. ;\ssume in <''alculation
that the deflection 0 is a small quantity compared with the ovcrall dimensions of
the structurc.
/lus.  = 2PJ31 AE.
--ibr
--ibr
-tb
-..,
TT
If I
:1
FI G. 5.15

234 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS


5.5 CASTIGLIANO'S SECOND THEOREM


In the previous article, \ve used the expression for strain energy as a function
of the displacements and found that a partial derivative of this function \vith
respect to any gcncralized displacement gives thc corresponding generalized
force. No\\r lct tiS considcr the expression for strain energy as a function of
thc external forces and assume that the principle of superposition holds, in
\vhich case the strain energy is a homogeneous function of second degree.
Ullder sllch conditions, a pattia/ derivati'!}e of the strain energy with respect to one 01
the external iotees give; the displace111cnt corresponding to that loree. This state-
ment is usually called Castigliallo's second theOre111. 1
To prove the second theorem, let us consider the case of an clastic body
carrying the loads PI, P 2 , P 3 , . . . as shown in Fig. 5.8a. During the applica-
tion of these loads, deformations are produced, and a certain amount of strain
energy U is stored \vithin the body. I f, subsequently, one of the forces, say
Pn, receives an increment
Pn' some additional deformation of the body \\riIl
ensue, and the strain energy U \vill obtain an increment
T ae}

 () = iJP'l
P/I


so that the total strain energy becomes
(] + all
p
aP n "
Now, suppose that instead of introducing the increment
P,t after the
application of the loads PI, P2, P3, . . . , \VC reverse the procedurc and apply
first the infinitely small increment
Pn and aftcr\vard the loads PI, P2, P 3 , . . . .
Since the infinitesimal load
Pn produces an infinitesimal displacement, the
corresponding \vork is a small quantity of the second order and can be neglected.
Further, during the subsequent application of the loads Pl, P2, P 3 , . . . , \\'e
observe, by virtue of the principle of superposition, that the \vork of thcse
forces \\,il1 not be affected by the presence of the load
Pn and will be equal to
its previous value U. At the same timc the acting load
Pn rides through the
displacement On resulting from the application of the loads ])1, P2, P3, . . . and
produces the work
n On. Thus, in this case, the total \vork, equal to the
total strain energy stored in .the body, is


(a)


LT +
Pn On


(b)


Since the total amount of strain energy stored in the body does not depend
on the order in \vhich thc loads arc applied, we conclude that expressions (a)
I Alberto Castigliano, Trans. Acad. Sci. Turin, vol. 11, pp. 127-286 , 1876. See also his
treatise "Theorcrne de I'cquilibre des SYStenleS clasriques et ses applications."




ARTICLE 5.5
235
R
A
(a)
r
I
x-1

hI

FIG. 5.16 (b)
and (11) must be equal. 'rhus,
u +  g A.P" = U + A./)n 8"
from \\'hich
alj
ap" = 0'1
(5.8)
and the theorem is proved.
In Fig. 5.8a, PI, /J 2 , P3, . . . denote single forces; but the derivation holds
also if they are generalized forces and Ot, 02, 03, . . . are the corresponding
generalized displacements. Thus, \ve can state that the partial derivativc of
the strain energy \vith respect to any generalized force gives the corresponding
generalized displacement.
In the derivation of Castigliano's second theorem, \ve assumed that it is
possible to give an arbitrary increment to onc of the forces \vithout changing
the othcr forces. Thus, \ve consider these forces as independent. Such
forces as statically detern1inatc reactions do not satisfy this condition since
their rnagnitudcs arc not independent of PI, [>2, j) 3, . . . and can be found from
equations of statics. From this consideration it follo\\'s that the strain energy
LT in Eq. (5.8) must be represented as a function of statically illdepel1de'llt
external forces. Take, for example, the t\VO beanlS shown in Fig. 5.16. The
same forces Rand ,\;[ are acting in both cases, and from statics \VC conclude
that 1\1 = RI. "'rhus, the strain energy can be represented in either of the
follo\ving t\VO forms: t
or
R2/3
17 _
V - "6£1
1\;/ 21
U = - --
6El
(c)
(d)
If \\tC use expression (c), \\'C consider R as an independent external force and
AI as the reaction, as in the case of a cantilever beanl (f'ig. 5.16(1). In such a
1 Both forrnulas arc readily obtained fron1 the general expression (5.4'1) of the strain energy.

236 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
case the derivative of expression (c) \vith respect to 1< gives
a V R/3
dR" - 3EI
\vhich is the dcflection of the cnd A of the cantilever beam built in at B. If
expression (d) is used, \ve assun1e that M is the independent external force and
R is the reaction as in the case of a sinlply supportcd beam (Fig. 5 .16b).
Then, taking the derivative of expression (d) \vith respect to Ai, \ve obtain
au MJ
aAl - -3-El
which is the angle of rotation of the end R of the sitllply supported beam.
This example il1ustrates the significance of thc requircmcnt of representing
the strain cncrgy as a function of statical1y independent forces.
Castigliano's second theorem is very useful in calculating deflections of
beams. Take, for cX2mp]e, the cantilever beam bent by a force and a couple
applied at the cnd (Fig. 5.1 7 a) . 1'he bending moment at any cross section is
Nt = Ala - Px
(c)
and the strain energy stored in the bearn, as obtained from .formula (5 .4a), is
CJ = (l ht 2 dx (/)
) 0 2EI
The derivativc of this expression \vith respect to P gives the displacement cor-
responding to P, that is, the deflection 0 of the cnd A of the cantilever. l'hus,
(b)
1
r- 4
x n
(a)
x
A JI
P a
(c)
FIG. 5.17
(d)
-1(x-I/2}

ARTICLE 5.5
237
\ve obtain
al.J 1: l At[ aAl I 1: l
o = oP = 0 Elap dx = EI 0 (Ma - Px)(-x) dx
PI3 l\tf al 2
= 3EI - 2-E{ (g)
The derivative of expression (I) \vith respect to the generalized force Ma gives
the corresponding displacement, i.e., the angle of rotation 8a of the end A of
the cantilever, and \ve obtain
au 1: I M aM I 1: l
8a = (JAlfa = 0 £1 aM dx = EI 0 (1\;l a - Px) (1) dx
j\J[ a 1 PI2
- -Ei- - 2EI (h)
In this application of Castigliano's theorcm \VC did not calculate the final
expression for strain energy as a function of external forces but uscd it in its
general form (f), substituting the value of ,\;1 only aftcr diffcrentiation under
the integral signs in Eqs. (g) and (h). In this \vay a considerable simplification
of calculation is accomplished, especially if there arc several external forces.
It can be noted that thc partial derivatives aj\;f/ap and akl/aM a , \vhich
enter in the above calculations, have very simple meanings. The first of these
derivatives represents the fate of change of bcnding moment in the beam \vith
change of the load P. As sho\vn in Fig. 5. I 7 b, it can be visualized as the
bending-Inoment diagram for a unit load at the end of the beam. The second
derivativc, reprcsenting the rate of change of the bending nl0n1cnt ,\1 \vith
change of ,\;fa, is sho\vn in Fig. 5.17 c. Using for these derivativcs the nota-
tions lvl; and M, \VC can \vrite the expressions for the displacelnents in the
folIo\ving simplified forms:
o = J:. fOI MM dx (i)
8« = iI fol MM dx (j)
We see that in each case \ve have to integrate along the length of the beam
the product of the actual bending momcnt \vith the corresponding unit-load
bending momcnt. rhis conclusion can be extcnded and applied to beams
with any kind of lateral loading. If \ve have, for example (Fig. 5.18£1), a
sin1ply supported beam carrying a uniform load q and a concentrated load f>
at the middle, the derivative aJvllap is sho\vn in Fig. 5 .18b, and the deflection
at the middle is
_ 1 { l I. _ 2 1: l/2 ( p. q I QX 2 ) X .
o - EI J 0 J;llW p dx - EI 0 2" .t + "2 x - T 2 dx
Pl3 5 q/4
= 48£1 + 3"84 EI

238 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
r- 4
p
---1
(a)
FIG. 5.18 (b)
It is seen that by applying Castigliano's theorem, \ve obtain the displaceIllcnt
corresponding to any of the forces acting on the elastic system, but there are
cases whcrc \\'e need to find such displacements to \vhich thcrc arc no cor-
responding forces acting. In such cases \VC add to the actual forces fictitious
forces of infinitesimal magnitudes such that they do not change the actual
displacement. We then obtain the requircd displacements by differentiation
of the strain encrgy \vith respect to these added forces. "lake, for example,
the cantilevcr beam in Fig. 5.17 a, and assume that it is rcquired to calculatc
the deflection at the middle of the beam. Since therc is no corresponding force,
\ve assume an inflnitesirnal fictitious load Q applicd at the middlc. lhen the
rcquired deflection is
a [J 1 r l aM
01 = aQ = El } 0 ,\iI aQ dx
(k)
Since the added load Q is infinitesinlal, \VC use for M the prcvious expression
(e) . The derivative aAl/ aQ, representing the rate of change of the bending
moment M \vith change of the load Q, can be visualized by the unit-load
bending-moment diagranl11111p sho\vn in }"'ig. 5 .17d. Substituting these values
into exprcssion (k), \ve obtain
I { I ( ' ) 5P/3 1\1 0 / 2
01 = - EI } '/2 (lvI a - Px) x -"2 dx = 4- 8l1 - - 8 El
As a last example, \ve take the case of a slender cantilever beam of circular
cross section, the axis of \vhich is a circular quadrant of radius ]( in a horizontal
plane, as sho\vn in Fig. 5.19. Under the action of a vertical load P applied at
A, such a bar \vill be subjected to both bcnding and torsion, and \ve have for
any cross section, defincd by the angular coordinatc cp, a bending moment
jVl", = PR sin t/J
and a t\visting moment
(I)
T = PR(1 - CDS cjJ)
(111)

ARTICLE 5.5
239
FIG. 5.19
"
"
......
.......
.......
......
.......
__---p:c
A R
1"hus, neglecting the shear deformation associated \vith bending, the total
strain energy of the system becomes
u = (7f/2 Mt/J2  d + (,,/2 T2 d
J 0 2EI 4> J 0 2GJ <P
(11)
To find the deflection 0 corresponding to the load P, \ve have
o =  U =  ( 1(/2 ,Wtj> aj !. dcJ> +!i ( 1(/2 Tq,  T d4>
aP El Jo ap GJ Jo oP (0)
Substituting the values of j\;[ and TfjJ together \virh their derivatives from
Eqs. (I) and (111) and perfornling thc indicated integrations, \ve obtain
1rPR3 PR3
o = 4El + 4GJ (311" - 8)
(p)
In conclusion, \\'C note that the derivation of Castigliano' s second theorem
'.vas based on the principle of superposition. Hence, the expression for ;\train
energy U must be a homogeneous second-degree function of the acting forces.
If the principle of superposition does not hold and l.J is not a second-degree
function of the acting forces, Castigliano's second theorcm is not applicable.
To illustrate this point, let us consider the example sho\vn in Fig. 5.7. The
Hrain energy in this case is given by expression (5.5 b), \vhich is not quadratic
in P. Taking the derivative of that function with respect to P, \ve obtain
o U _ 1 3/ />/3
ap -}'\j AE
which is only one-third of the true deflection 0 as given by Eq. (f) on page
222. \\'e have already seen that \ve get the correct result in this case by
taking the strain energy U as a function of the displacement" 0 and using
Castigliano's first theorem.

240 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
PROBLEMS
1 Using Casrigliano's second theorem, find the angles of rotation at the ends of a
simply supported bcan1 AB of lcngth / and flexural rigidity Ef, due to a uniformly
distributed load of intensity q.
Hint Introduce equal infinitesimal moments A1 Q at each end of the beam
with respect to \vhich the strain energy V can be differentiated.
At/s. 'OA = 8B = qI3/24EI.
2 A continuous square fran1c ABCD built in at A is loadcd as sho\vn in Fig. 5.20.
Find the vertical deflection of point ]) if all three InCtl1bers have the samc EI.
Consider only strain energy of bending.
Ans. 0 = 5P/3/3EI.
3 A continuous frame ABC hingcd at A and supported by a roller at C is loaded as
sh()\vn in Fig. 5.21. Find the horizontal displaccn1cnt of point C if both 111cmbers
havc the same length I and the saine flexural rigidity Ef.
AlIS. 0 = 2PJ3/3EI.
4 Dctermine the horiwntal and vertical displacements 0" and 0" of the free end A
of the circular cantilever beam loaded in its o\vn plane as shown in Fig. 5.22.
Assume that the cross-sectional dimensions of the beam arc small con1pared \vith
[he radius r, so that the straight-bean1 formula (5.40) can be used in ca,lculating the
strain energy of bending. Neglect the strain energy due to direct and shearing
stresses.
Al1S. Oh = Pr 3 /2EI.. 0\1 = 7rPr 3 /4EI.
p
c r I B f. I -t e
T ,D P
I
I
I
I I
1 I I
I lA
I
I
B A
FI G. 5.20 FI G. 5.21
P
I
I
r I
, I
-------\t
FIG. 5.22
A
;'
</:/--
FIG. 5.23

ARTICLE 5.6
241
5 "I'hc curved cantilever sho\vn in Fig. 5.23 has a circular center line of radius Rand
subtcnds a central angle cx. On the free end B there act an axial force N, a shear
force Q, and a bending moment 1\-1, all in the plane of the figure. Develop a general
torn\ula for the angular rotation of the tangent to the elastic line at A.
AJRa QR2 i\TR2.
l111S. 0 = -- + - (1 - cos a) - -- (a - sIn a).
E/ £/ E/
5.6 THEOREM OF LEAST WORK
In the prc<:eding article \ve have considered applications of Castigliano's
second theorem to statically dcterminate systems and found that the displace-
rncnt of any point is obtained as the derivativc of the strain energy of the
system \vith respect to the corresponding force. .A.pplying the theorem in
the satne \vay to a statically indeterminate system, we conclude that the
derivative of the strain cnergy \vith respect to any redundant rcaction or
internal constraint must bc zero since it is the function of such a reactivc force
to prevent any displacelnent at its point of application. IIence, if X,  Z,
. . . dcnotc thc magnitudcs of redundant forces and if thc principle of super-
position applies, \\'e have
iJ [I
-- = 0
rJJ;
au = 0
aY
u = 0
az
(5.9)
\vhere U is a second-degree function in "(,  Z, . . .. We see that Eqs.
(5.9) may bc interpreted as conditions of a maximum or minimurn of the
strain-energy function U. By calculating the second derivatives and sho\ving
that they arc always positive, it can be concluded that \ve have the case of a
minimum. l'his means physically that the structurc \viII deform under load
in such a \vay as to keep its strain energy, equal to thc \vork of applied forces,
a minimum. l'his statement, caUed the thcOre111 01 least work, is very useful in
the analysis of statically indeterminate structures. We sin1ply derive the
expression 1r strain cnergy as a function of the redundant forces and then find
the nlagnitudes of these rcdundants from Eqs. (5.9). It is seen that thcre
\vill al\vays be exactly as many of these equations as there are redundants.
As an example, \ve take the case of a uniformly loaded beam bui It in at B
and supported by a roller at A as sho\vn in Fig. 5.24. For the determination
of the unkno\vn reactive forces , }b, j\t[b, in this case, \ve have only t\VO
equations of statics,
y = 0
M = 0
(a)
and thc problem is staticaUy indeterminate. Choosing the vertical reaction
 at A as the redundant reaction, \VC obtain the case of a simple cantilever
loaded as shown in Fig. 5.24b, \vhcrc the reactions at the built-in end have been

242 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
Jlb
r..
I
I
(0)

(b)
2
!!!.. - Y I
2 a
ql _ Ib
2 . X --j !II + ilC1J
 W IIIIII i I1I1III1 [111111111  111 11 ;;' Mb
FIG. 5.24 (c)
expressed in tern1S of 1:. No\v, applying the theorem of least \\'ork to this
system we Inay \Vf1tc
  = . (l ' \112 d x =  (l i\tl  (\I[ d x = 0
a fa a J a J 0 2EI E I J 0 a 1:
(b)
Substituting in this equation
qx 2
1\1 = Ynx - T
eJi\1
-x
aY a -
\ve find
{ Z ( qX2 ) Ya/3 ql4
Jo Kxx- 2 .,'dx=-T--g=O
(c)
\vhich gives
Y a = 'q/
i\s an alternative approach to this same problem, \ve may take the simply
supported beam sho\vn in Fig. 5.24c as the primary systenl and consider the
reactive mornent JIb as the redundant. l"'hcn, again applying the thcoreln of
least \\'ork, \ve see that the necessary equation for determining the magnitude
of Mb becomes
iJ U a ( l J\{2 dx 1 ( J d,,\tl
ai\t[b = aA,f b J 0 2£[ - = El J 0 1\1 akl b dx = 0
(d)

ARTICLE 5.6
243
l'he bending moment at any cross section, represented as a function of A-f h , is
( ql i\Jb ) q.\.2
j\I[= 1-T x- 2-
froln \vhich
a j\l[
a 1\4 b
x
I
Substituting these values into Eq. (d), \ve obtain
(l [( ql lV[o ) . Ql' 2 J .\' d . _
J 0 2 - -T x - T 1 :t - 0
\v hich gives
. . '1 12
1\4 0 = -
8
Sometiolcs the internal forces acting bct\\'cen t\va adjacent portions of a
structure \vill be chosen as the statically indeterminate quantities, rather than
the reactions at the supports. In such cases the thcorcrn of least \vork still
holds and can be uscd to advantage. Consider, for example, the synloletrical
rectangular ffalnc sho\vn in Fig. 5.25a. Taking a section through the hori-
zontal p]anc of symmetry 1111/ and considering the cquilibriulTI of the upper part
of the frame (Fig. 5,25 b), \ve can represent the action of the lo\vcr part
thereon by the vertical forces l P /2 and by the couples ..lIo, the magnitude of
\vhich cannot bc dctermined fronl statics. Frolll the SYOlmctry of the defornla-
tion sho\vn in Fig. 5.2Sa, \VC conclude that the cross sections on \vhich the
couples i\J n are acting in Fig. 5.25 b do not rotate. Hence the derivative, \vith
rcspect to At[ 0, of the strain energy stored in the upper part of the fralne Olust
vanish, and \VC can state, as before, that the nlagnitude of the starically indctcr-
Il1inatc quantity illo is such as to make the strain energy a tninimut11. In
1 FroIT1 symmctry it can he concluded that there \vill be no shearing forces in the plane 11/11.
p P
I I
2 _ -3:1
,f c
\ I h
\ II I 2
\
I I hlo ;r MO
nl I I n
I \ h
I , 2
-1
FI G. 5.25 P (a) (b)

244 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
calculating this strain energy, \\'c neglect the strain energy due to direct stress
and assume that the effect ofaxiI forces on bending can be neglected. l"'hus,
the btnding mOlnent for the portion A B of the frame is AJ 0, \vhile for the
portion BC it is i\1 o - r.\/2. The strain energy of the upper half of the
fran1e then is
LT = /\102/z + J_ {l/2 ( AIo _ P.t ) 2 d.\'
2Rl Ell J 0 2
i\.ppJying the principle of least \\fork, \VC no\\' obtain
aLl
aAI o
AJ 011 2 r 1/2 ( P:t: )
- - 1/-- + El J 0 1\1[ 0 - 2 dx = 0
\v hich gives
PI
ivf 0 = 8 [1 + (hi I) (It I 1)" )
It is seen that \vhcn the cross-sectional mon1ent of inertia 11 of the horizontal
hars of the fraIne is large in con1parison \vith the corresponding quantity / for
the vertical bars, the mon1ent /\1[1) becoo1es srnall, and the condition of bending
of the horizontal bars approaches that of siolply supported beams. ()n the
other hand, if II is small in comparison \vith 1, the moment l\l o approaches the
value PI/8 as for a bcan1 \\lith built-in ends.
As another example let us consider the circular ring sho\\rn in Fig. S. 26a and
detcrnline the moments 1\-1 0 and the increase in the vertical dian1etcr of the ring
under the action of forces P appLed as sho\vn. Proceeding as in the case of a
rectangular frame (Fig. 5.25) and considering the equilibrium of the upper half
of the ring, \VC conclude that
a [] 1 a  Trr /2 ,2
--- = - -- AJ ds = 0
ai\1 0 EI ai\1 o 0
(e)
Substituting in this equation
Pr
l\4 = i\J 0 - -2- (I - cos 4»
aAJ
a AI -
\ve obtain
( 7r /2 r Pr ]
) 0 /\,[ 0 - 2 (1 -, cas 4» d cp = 0
from \vhich
J>r ( 2 )
AIo = - 1 - - == 0.1 821>,
2 1r
<f)
To find the incrcase of the vertical diamctcr of the ring, \VC observc that
the forces P in Fig. 5.26(1 rcpresent the generalized force corresponding to the

ARTICLE 5.6
245
Alo
blo
FI G. 5.26
(a)
(b)
incrcase 0 of the diameter. I-Ience, this increase l is
_ aU _ 2 a r 1f /2 2
o - at> - EI aP J 0 M r dq,
4r r 1f /2 [ Pr ] r
= - £ 1) 0 1\;1 0 - 2 (1 - cos <1» "i (1 - cos q,) d q,
If \ve use the value of A10 from Eq. (I) above, this becomcs
0=
Pr 3  7:/2 ( 2 )
- - cos q, - - (I - cos cp) d <I>
EI 0 1r
Pr 3 ( 1T' 2 ) Pr3
= El 4 -;. = 0.149 EI
(g)
As another application of the method of least vlork, let us calculatc the
thrust H in the t\vo-hingcd arch under vertical load as sho\\'n in Fig. 5.27.
The vertical reactions  and  can be obtained in the usual \/ay from equations
of statics; but the horizontal thrust H is a statically indeterminate quantity,
and to find it \ve must use the theorem of least \vork, \vhich requires that
aU = 0
aH
(h)
In the case of a flat arch, the strain energy due to direct stress is an important
factor and cannot be neglected in comparison with the strain energy of bending.
Thus, we take
V = r lJ M2 ds + r B N2 d ( i )
Jo 2EI Jo ZAE
\vhere Al is the bending moment and N is the axial force at any cross section.
The length of the axis of the arch is dcnoted by s. l"he bending moment at
any cross section cn be considered as consisting of t\VO parts: onc representing
1 In this calculation \\..c assume that the strain energy due to direct and shearing stresses
can be neglected in con1parison \\'ith strain energy of bending. This is justifiablc only in
the case of a thin ring.

246
GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
YI
I
I

II f
\ J.. Y
A H --\---t - - - - H B x
 4 \ .j. i 
\ <p I
r1
\ I

FIG. 5.27
the moment M 0, calculated as for a simple beam (H = 0), and the other
representing the 1110ment due to the thrust H. Hence,
1\1  "'[0 - Hy
(j)
\vhere. y denotcs the ordinates of the axis of the arch as sho\\!n in the figure.
For a flat arch, the axial force N can be assumed, \vith good accuracy, as equal
to the thrust H. Then,
U _ {3 (Mo - Hy)2 ds + (8 H2 d!
- )0 2EI Jo 2AE
Substituting this expression in Eq. (h), \ve obtain
_ (8 (Mo - Hy)y d  + (8 H ds = 0
Jo El Jo AE
from which
( a  10)' £Is
[-l = ) 0 HI
r a y2 ds + (8 d s
JoEl ) 0 A E
This is a gcncral formula from \vhich H can be calculated if the loading and the
dimensions of the arch are given.
In the case of an arch of homogeneous material and uniform cross section,
E, I, and A arc constants and can be taken from under the integral. signs.
Then,
(k)
(s M oy ds
H = J o
foB y2 ds + ;2 foB ds
\vhere i denotes the radius of gyration of the cross section \\,ith respect to its
centroidal axis. The second tcrm in the denominator represents the influence
of the direct stress on the magnitude of H. If this influence is neglected, \ve
(I)

ARTICLE 5.7
247
obtain
(8 l\;loY ds
H = ) 0 . (1n)
{ B ')12 ds
J 0 .
An especially simple expression for H is obtained in the case of a parabolic
arch under the action of a load uniformly distributed along the horizontal span.
In such a case
y = x (I - x)
1\4 0 = qx (I - x)
2
Then, substituting into Eq. (111), we obtain
ql2
H=-
8f
\vherc j is the vcrtical rise as sho\\'n in the figure.
PROBLEMS
Find, by using the theorem of least \vork, the bending mOluents Ala and I\J b at the
ends of a unifonnly loaded beanl \vith built-in ends and of uniforrl1 flexural rigidity
EI.
Al1S. .A1a = kfb = -qI2/I2.
2 Solve the preceding problem if the intensity of load along the length of the bcan1
varies according to the linear la\v q = qQ.'\,!I.
Ans. '\;/a = -qol2/30, i\1 b = -q o I2/20.
3 Referring to Fig. 5.240, calculate the reactive moment i\tl b if, instead of the uni form
loading sho\vn, the beam carries an isolated load P at the distance f froIn A.
Al1S. '\;[b = PC(/2 - c 2 ) /2/ 2 .
4 I f, instead of the uni form loading sho\vn in Fig. 5.24a, the hcam A B is subjected to
an active moment Ala at the cnd At sho\v by the theorem of least \vork that the
corresponding rnon1cnt induced at B \vill be ,\1 6 = - ALI/2.
5 Considering only strain energy of bending, calculate the horizontal thrust H in
the case of a t\vo-hinged semicircular arch rib of span I that carries a vertical load P
at the Cf()\vn. .Assume a uniform flexural rigidity EI of the rib.
Ans. 1/ = P/1r.
5.7 THE RECIPROCAL THEOREM
Let us begin \vith a sirnple example, the bending of a cantilever beanl, and
consider t'o loading conditions, (I) a load 1 J at the free end (Fig. 5.28a) and
(2) a load Q at any point (' (Fig. 5 .28b). In the first case, the deflection at C

248 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
A C B
-
X
pI. I
(a)
(b) y
A Q r c =;f- -
u l
FIG. 5.28 C B x
is given by the equation
J)c 2
Oc = 6EI (3/ - c)
(a)
In the second case the deflection at ( is calculated as for a cantilever of length
c and is equal to Qc 3 /3El. The part AC of the beam remains straight and is
tangent to the deflection curve at C so that it has the slope Qc 2 /2El. Thus, the
total deflection at point /1 in this case is
Qc 3 Q.c 2 Qc 2
oa = 3Hl + 2El (I - c) = 6F:I (3/ - c)
(h)
Comparing expressions (a) and (b), \ve see that \vhen P = Q, thcse t\\'o
deflections are equal, that is, a load P, placed at point A, produces at point C
the same deflection as the load }>, placed at C\ produces at point A. * In a more
general case \vhen ]J and Q arc not equal, the deflections (a) and (b) are no
longcr equal, but the \vork donc by the force P on the corresponding displace-
ment (b) is equal to the \\lork done hy the force Q on the corresponding dis-
placenlcnt (a), that is,
POa = Qoc
Let us considcr no\\' a general case of an elastic body subjectcd to t\VO dif-
fercnt conditions of loading. In the first casc therc act forces PI and Pz (Fig.
5.29a), and in the second case the forces P3 and })4 (Fig. 5.29h). We shall
designate the displacements of the points 1, 2, 3, and 4 in the directions of the
forces for the first condition of loading by o, o, o, and o and for the second
condition of loading by O/, o', o', and o'. Assume, no\\', that all four forces
arc acting on the body simultaneously and that thc principle of superposition
holds. Then the displacements of points I, 2, . . . corresponding to the
forces 1\, P2, . . . arc obtained by superposition and are equal to o + lJ',
* l"his conclusion \\'as obtained by J" C. M ax\vc1J in considering deformation of trusses;
see Max\vell, HScienrific Papers,"' vol. I, p. 602.

ARTICLE 5.7
249
, + "
02 2"
The total strain energy stored in the body is
U = ![(<<5 + «5/)Pl + (o + «5')F)2 + (<<5 + 0/)P3
+ (o + o') P 4 ] (c)
This amount of strain energy does not depend on the order in \vhich the loads
are applied. Let us assume, for example, that the loads PI and P2 arc applied
first. Then, the corrcsponding anl0unt of strain cncrgy is
! (Pl0 + P20)
(d)
IJet us apply, no\\', the loads P3 and P4. Since the principle of superposition
holds, the displacements produced during the application of P3 and P4 \vill not
be affccted by the prescnce of the previously applied loads PI and P2 and \vill
be equal to o', o', o', and o' as bcfore. The \vork done by P3 and ]J.. during
their gradual application \vill bc
. (P30' + P40')
(e)
At the same time, the prcviously applied loads PI and P 2 \vill produce \vork, on
thc displacements o' and o', equal to
PIO' + P20'
(f)
Upon summing up expressions (d), (e), and (I), rhe total amount of strain
energy is obtained. Equating this encrgy to expression (c), \\'e obtain
P " + P " P r.' + P ,
lUI 2 U 2 = 3 U a 404
(5.10)
This equation states that the \vork donc by the forces of the first state of
loading (Fig. 5.29a) on the corresponding displacements of the second state
(Fig. 5.29b) is equal to thc work done by the forces of the second statc on the
corrcsponding displaccmcnts of thc first. This rcprescnts the so-called
reciprocal theOrC11l. 1
1 In its general form the reciprocal theorem has been proved by E. Betti, NufYUO Ci111t'ntn,
sere 2, vols. 7 and 8, 1872. See also Lord Rayleigh's paper in Pro(. UJ11do1l ,\tfath. Soc.,
vol. 4, pp. 357-368,1873.
P2
FI G. 5.29
(a)
(b)

250 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
'rhe foregoing derivation of the rcciprocal theorem holds also in the case of
generalized forces. .A,s an exanlplc, Jet us consider the t\va conditions of
loading of the cantilever beam sho\vn in Fig. 5.30. In the case of the load P
applied at thc end, ttc rotation of the end A is
PI2
0 0 = 2£1
Like\vise, the deflection at 4 under the action of the couple jf is
AII/
On = 2El
It is seen that these values of 8 0 and Oa satisfy the reciprocal theorem, i.e.,
AI8t) = POa
f\S another example, let us consider the case of a rigid slab having the fOrITI
of an equilateral triangle supported by thrce vertical and three inclined bars of
equal cross-sectional areas A and hinged at the ends (Fig. 5.3 I) . In one case,
the slab is loaded by J/ at the center (Fig. 5.31 a), and in another case it is
loaded by a couple jl acting in the plane of the slab (Fig. 5.3 1 b) . ()\ving to
the action of the load lV, the vertical bars \vill bc compressed by the 3Inount
lVh/3AE; and since the lengths of the diagonals rernain unchanged, the com-
pression of the vertical bars \vill be accompanied by latcral distortions in the
side planes of the structure as sho\vn in Fig. 5.  1 c. Consequently, there \\'i II
be rotation of the slab. In calculating the angle of rotation 0, \ve observe that
the middle point Jl bet\\'ccn the hinges 1111 and 111 Illoves along 11-111/1 by the
amount Vh2/3AEI, and thc angle of rotation of the slab is obtained by dividing
this displacement by the distance 1/2 y"3' of the point s frotll the center 0 of
the slab (Fig. 5.3 1 a). This gives
H l h 2 2 y3 2U ' 11 2
(J = 3AF;J -, = AE y3/ 2
()
A
J-
- I
t
(a)
FIG. 5.30
(b)
,0
\..J Itl
B'

ARTICLE 5.7
251
'\s a result of the loading sho\vn in Fig. 5.31 b, \ve shall obtain not only rotation
of the slab but also some vcrtical movement. To procced \vith this case, \ve
first replace the torque M by a statically equivalent systcm of three cqual
forces P coinciding \\'ith the three edges of the triangular slab. Thcn, since
1
3P _ = A1
2V3
\ve have
p = 21\1
IV?'
I'O\V each of these three forces P produces a compressive force Phi J in the
corresponding vertical bar as sho\vn in Fig. 5.31d. Hence, the vertical dis-
placement tJ of the slab will be
o = _ Ph 2 = 2kf /12 (g')
lAE f2AE y3
Considering the rotation (g) in the first case of loading and the vertical
displacement (g-') in the second case, \ve see that the reciprocal theorem is
satisfied, i.e.,
,\;[ () = lV tJ
l'he reciprocal theorem is especially useful in the construction of influence
lines for redundant reactions. l..et us consider, for example, a bcanl on thrce
I -,
n
I
h
1
(a)
Wh 2
3AEI
In r-- ' 8 I b-  I 1 . j
..,...------0-- -- - 7t  3AE r p
1 m ! 81,,'" ,nl
t " I
I ",
, I h
I J 1
I I
FIG. 5.31 (c) (d)

252 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
Yi X --1. IC 18_
(a) A 0
 u5Ju X
-II 1 2
11
lIb
FI G. 5.32
(b)
c
Y 0
_.  __ c____
B
A
supports as sho\vn in Fig. 5.32(1 and investigate the change in the redundant
reaction  at the intermediate support \vith change of the distance x defining
the position of a moving load P. In applying thc rcciprocal thcorem, \ve have
ahvays to compare t,vo conditions of loading of the given structurc. .As the
first state, \\'e take the acrualloading, sho\\'n in Fig. 5.3 2a; as. a second state,
\\'c select the fictitious loading shQ\vn in Fig. 5.3 2b, \vhcre the load P is
removed and in place of the reaction  a unit force acts. l-'his second statc is
statically determinate, and \ve can define the deflection curve A(..' by the
equation 1
_ ( l Ib ) 1 2 x r (/ + I ) 2 1 2 .2 1
Y - 6 (/1 + 1 2 ) EI 1 2 - 2 -.t
A sin1ilar expression can be established for the right portion BC'. rhe deflec-
tion under the unit load is obtained by substituting x = 11 in Eq. (11), \vhich
gl ves
(11)
o = . (I Ib)/ 1 2/ 2 2
c 3(1 1 + /2)£1
Applying no\v the reciprocal thcorem, \ve calculate first the \vork done by the
forces of Fig. 5.3 2a cn the corresponding displaccmcnts of Fig. 5.3 2b. Such
\vork cvidently is
(i)
}/ 0 - 1>1'
c c .
Considering no\\,' the \vork of the forces of Fig. 5.3 2b on the corrcsponding
displacen1ents of Fig. 5.3 2a, \ve find that this work is zcro since, for the unit
load in Fig. 5.3 2b, the .corresponding deflection in Fig. 5.3 2a is zero. Hence,
the rcciprocal theorcm gives
}oc - PJ' = 0
1 See "Strength of Materials," vol. I, p. 145.

ARTICLE 5.7
253
from which
_ Py _ 3P(/ 1 + 1 2 )£/
 - 1; - (1 Ib)/ 1 2 / 2 2 )'
(})
It is secn that \\Then the load P is n10ving along the bearn AB, the reaction at C
varies in the saine proportion as the ordinates y of the deflection curve A CB
calculated for a unit load at C. Thus, this deflection curve can be taken as the
influe11ce line for the reaction. I laving such an influence line, \VC can find
the reaction 1: for any system of vertical loads PI, P2, P 3 , . . . in the usual
\va y from the equation
y =  ( PI'}'t + P 2 Y2 + P 3 'V3 + . . . )
c oc' . .
\vhcrt )'1, Y2, . . . arc the ordinates of the influence line A C'B corrcsponding
to the forces PI, P2, . . . .
If \\re desirc an influence line for bending tll001ent i\tJ c at the intermediatc
support C of the beam in Fig. 5.3 2a, \VC first cut the beam at this section and
introduce the equal and opposite couples A1 c as sho\vn in Fig. 5.3 3a. . The
corrcsponding fictitious loading i sho\\rn in Fig. 5.33 b. I\TO\V, using the
reciprocal theorem, \ve find
"'1,(8 1 + 8'1.) - p"" = 0
from \\,hich
)'
,We = p. -----
8 1 + 8 2
Upon substituting the kno\vn expressions for )', 81, and 8 2 , \ve see that the
influence coefficient for the left span becomes
X(/ I 2 - X2)
-2/ 1 (/1 -+- 1 2 )
(k)
'[he same expression can be uscd for the right span if we interchange the
subscripts of 11 and 1 2 and measure x from B.
(a)
/c Ale
C  Q C B...ii. X
1 [12]
-x
C B
FIG. 5.33
(b)
A

254 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
If the dcrivation of an analytical expression for thc influence line is not
practicable as in the preceding examples, the line can al\vays be obtained
experimentaJJy by o1aking a tcst \\'ith a model of the actual structure. Sup-
pose, for example, that \ve desire an influence linc for the thrust H of the t\-\'o-
hinged arch sho\vn in Fig. 5. 34a. In the fictitious state of loading (J.'ig. 5. 34b),
\ve n1akc the support B movable and apply to the arch t\VO equal and opposite
unit forces as sho\vn. Let 0 be thc horizontal displacement of the support B
produced in this fictitious case and y the vertical deflection of point C. l'hcn
the application of the reciprocal thcorem gives
Ho - P" = 0
from \vhich
P)'
H=-
o
(I)
'This result is similar to that in Eq. (j) obtained for the previous problem.
Hence, \ve can state that the ratios )'10 for the case sho\vn in Fig. 5. 34b
represent the influence coefficients for the thrust H. It is evident that, by
making a model of the arch and loading it as sho\vn in Fig. 5. 34b, the influence
coefficients can bc obtained from the mcasured deflections J' and o.
The expcrimental detcrmination of influcnce lines may be of practical
intercst in those cases where \ve havc many redundant indeterminate
quantities. lake, for example, the rectangular frame built in at the supports
as sho\"n in Fig. 5.35a. '[his is a structure \vith three redundant constraints.
Removing the support at the right end and replacing it by rcactive forccs, we
obtain the three statically indeterminate quantities H,  and Il1b as sho\vn.
To obtain for each of these quantities an influence linc by experiment, \ve use
a modcl of the frame and give in each case to the cnd B of the modcl a dis-
placement correspondhlg to the force that \ve arc planning to determine. If \ve
ace intercsted, for example, in thc reaction H, \\'e give to the end B of thc modcl
a horizontal displacement 0 and at the same time by proper constraints \ve
p
A
(a)
1-.
I
1
A
FI G. 5.34 (b)

ARTICLE 5.7
255
r x
p
r- x
--r-----
y
A
H
A
H'
I
I
I
I
I
I
I
JIb I
I
I
bl--
v' I
FIG. 5.35
(a)
v
(b)
prcvent the cross section from both vertical and angular displacement. In
this manner \ve obtain thc dcformation shown in Fig. 5.35 b. Let. H', V', and
M be the forces \vhich \ve apply to the model at B to realize the abovc-
described displacement. No\v, applying the reciprocal theorem to the loading
conditions shown in Fig. 5.35a and band obscrving that only those displace-
mcnts in the fictitious state (Fig. 5.3 5 b) corrcsponding to the actual forces P
and H arc different froln zcro and that the displacements in Fig. 5.3 Sa cor-
responding to the forces H', J/', and j\1 are all zero, \ve obtain
Ho - Py = 0
from which
H= Py
o
Again, the influence coefficients for H are obtained by dividing the deflections
y by the displacemcnt 8, and all thcse quantities can be measured in making the
experiment \\,ith the model. Influence lines for V and A1b can be obtained in a
similar manner. 10 get the influence line for  \\'e have to give to the end B
of the model a vertical displacement and at the same time prevent any horizontal
or rotational displacen1cnt. In the case of Mb \ve must give to the cnd B a
rotation and at the samc time prevent any horizontal or vertical motion. The
measuremcnts of deflections in these two cases give the information required
for calculating the ordinatcs of the rcquircd influence lines. 1
PROBLEMS
Referring to the t\vo-span Oealll sho\vn in Fig. 5.3 2 and using expression (k), find
the maximum bending 1110tnent j\f c at the intermediate support if /1 = 40 ft, /2 = 20
ft, and P = 1,000 lb.
Al1S. Max i\J e = 5,125 ft-Ib.
1 Special apparatus, kno\vn as Begg's defo1711t1tor gaUKtS, are available for controlling the
deformations of Itlodcls. For a description of these instnlments, see J. Franklin Inst., March,
1927, p. 375.

256 GENERAL THEOREMS RELATING TO ELASTIC SYSTEMS
c
r-x --l
till
 I
(a)
FI G. 5.36

-i
T
h

I ---1 B
(b)
A
I ---1 B
(c)
2 Using the reciprocal theorem, derive an expression for the inRuence coefficient for
end moment Ala in the case of a beatn AB \vith built-in ends. 'I"here should a
load P be placed on the beam to produce the maximum value of this cnd nloment?
A11s. y = - (l2.t - 2/:t.:J. + :t. a ) //2, X = //3.
3 Referring to Fig. 5.36a and using the reciprocal theorenl, derive an expression for
the vertical reaction Rb for any position of the load P on the bearn. l"hc beam has
l1ni form flexural rigidity EI.
Px 2
Ans. Rb = - (3/- x).
2/3
4 Referring to Fig. 5.3 6b and using the reciprocal theorem, derive an expression for
the tension S in the vertical tie rod BC for any position of the load P on the beam.
The beam has uniform flexural rigidity EI and the tie rod has the same rnodulus of
elasticity E, length Jz, and cross-sectional area A.
AlIS. S = p'2 __''____ .
6 /3/3 + Jh/A
5 Referring to Fig. 5 J6c and using the reciprocal theorem, derive an expression for
the tension S in the lnclincd tie rod BC for any position of the load P on the beam.
The beam has uni form flexural rigidity HI and the tie rod has the same modulus of
elasticity E and cross-sectional area A.
A l1J . S _ Px 2 3/ - x .
- 6 (/ 3 /3) sin a + (2/// A) csc 2a

Chapter 6
Deflection of pin-jointed
trusses
6.1 APPLICATIONS OF CASTIGLIANO'S THEOREM
Castigliano's theorem, discussed in Art. 5.5, can be used to advantage in
studying the deflections of trusses, especially if the displacements of only a
fc\v joints are required. Consider, for cxample, the simple truss sho\vn in
Fig. 6.1a to \vhich the vertical loads P, Q, and R are applied, and assume that
it is required to detcrmine the vertical deflection of the middlc joint B of the
lo\ver chord . For this purpose, \Vc must dcri ve an expression for the strain
energy of the truss as a function of the given loads [J, Q, R and then calculatc
the partial derivative of this expression \vith respcct to the force Q acting at
joint B. lhc expression for strain energy in the case of a statically determinate
pin-jointed truss \vith prismatic membcrs is a vcry sin1ple one. Let S.. be thc
force in any bar i due to the actual loading (Fig. 6. I a) . Then, the strain
energy stored in this bar is
('.2 / .
)t 1
i/i-;E
(a)
257

258 DEFLECTION OF PIN-JOINTED TRUSSES
6
248
P Q R
(0)
1
FIG. 6.1
(6)
\vhere Ii is the length and A i the cross-sectional arca of the bar. Summing up
expressions (a) for all bars of the truss, we obtain 1
U = L 2£ (6.1)
In calculating the partial derivative \vith respect to the load Q, \ve obscrve
that in expression (6.1) only the quantities Si depend upon Q. Hence ordinary
differentiation gives the required deflection in the follo\ving form:
lJ =  () = \' _Stli aS i
aQ L., AiE aQ
(b)
Observing that the derivative as,/ aQ in this expression is the rate of change
of the force S1 in any mcn1ber i \vith respect to the load Q, \ve can determine
the magnitude of this derivative in a vcry simple manner by considering the
action on the truss of the unit load sho\\'n in Fig. 6.1 b. l)enoting by Si the
force produced in any bar i by this unit load, the magnitude of the derivative
8S i /8Q is evidently equal to the ratio Si: 1, and exprcssion (b) for the deflection
o can be \\'ritten in the following forn1:
 = \' Stli I
U '-' AtE Sj
(6.2)
in \vhich the factor Sd,/AiE in each term rcprcsents the clongation of the bar i
1 It is assumed that the tnatcrial of all bars is the same and that the modulus E is constant.

ARTICLE 6.1
259
under the actual loading and .< is an abbreviation for the ratio Si : I) obtained for
each bar from the fictitious loading sho\vn in Fig. 6.1 b. 1 Since, in this casc)
the truss is statically detern1inatc, the calculations of the quantities Si and s
can be made \vithout difficulty and the deflection 0 can then be calculated
from Eq. (6.2) provided that the dimensions of the bars and the modulus of
elasticity E arc given.
Assuming, for examplc) I J = Q = 16,000 lb, R = 8,000 lb, and taking the
lengths of the bars from column 2 of Table 6.1, \ve obtain for S.. and s the
values sho\\'n in columns 4 and 5, respectively, of the table. Using no\v the
cross-sectional areas, given in column 3, \ve obtain the figures in column 6.
l'he sum of this last column, multiplied by 10 3 and divided by the modulus E,
gives the rcquired deflection 0 as sho\vn by Eq. (6.2). Assuming that the
truss is of structural steel and taking E = 30 X 10 6 psi, \ve obtain
5,170 X 10 3 .
o = 30 X 10 6 = 0.1723 In.
"[he calculations in 1ablc 6.1 arc made \vith four significant figurcs; but
calculations \vith three significant figures, as given by an ordinary slide rule,
\\'ill have sufficicnt accuracy for practical purposes.
If the horizontal displacement of the joint B is required, \\'e add to the actual
loads P, Q, R an infinitesimal force T acting in the direction corresponding to
1 In our further discussion, \ve shall ahvays denote the ratio Si: 1 simply as s/, but it should
be kept in lnind that \vhen so used it luust be considered as a pure number, \vhich \vill be
called the inflllrT1C 1l1l11lber for the bar i.
TABLE 6.1
I I , Sdi II
Ii. in. Ai. in.:! Sit Ib I SaS,h "
I S. silt, .- S.
. Ad0 3 Ai I
I
(J) I (2) (3) (4) (5) (6) (7) (8)
I 250 0 -27,500 J -0. ()25 716.0 0.625 -2,387
2 150 3 J 6,500 0.375 309.4 -0.375 - 1,03 I
3 200 2 16,000 0 0 0 0
4 150 J J 6,500 0.375 309.4- -0.375 - I ,03 I
5 250 2 7, 500 0.625 585.9 0.625 1,953
6 300 4 -21,000 -0.750 1,181.0 0 0
7 250 2 12,500 0.625 976.6 -0.625 -3,255
H 150 J I 3, 500 0.375 253.1 0.375 844
9 200 2 8,000 ° 0 0 0
10 250 6 -22,500 - 0 . 625 585.9 - 0 . 625 1,953
II ISO 3 13,500 0.375 253.1 0.375 844

260 DEFLECTION OF PIN-JOINTED TRUSSES
6
6
(b)
FIG. 6.2
2 4 8
P Q R
(a)
(b)
FIG. 6.3
the required displacenlcnt (Fig. 6.2(7). "hcn, the derivative aLi / aT gives the
requircd horizontal displaccll1cnt as follo\vs:
o = L 1i ::;
(c)
Since T is an inhnitcsinlal force, the values of Si arc the same as those already
recorded in coluoln 4 of Table 6.1. ---rhe values of the derivatives as i / aT,
representing the rates of change of the forces Si \\'ith respect to the force T,
arc obtained from the loading condition sho\vn in Fig. 6.2b. For this latter
condition of loading, only the bars 2 and 4 arc activc. Hence, in expression
(c) only t\VO tcnl1S are di ffercnt from zero, and \VC obtain for thc requircd
horizontal displacelnent
5 = 5 2 / 2 + S  = ) 16,500 X 150 = O()  5 '
i4 2 £ A 4 E .. 3 X 30 X 1 06 . - In.
\vhich is evidently cqal to the sum of the elongations of the bars 2 and 4.
Sooletio1CS \VC have to find the angles of rotation of ccrtain membcrs of
a truss during dcflec:ion. l'hese angles can also be calculatcd by using
Castigliano's theorcn1. To illustrate the procedure of such calculation, Ict us
consider again the trus sho\\rn in Fig. 6.1a and find the angle of rotarion of the
bar 6 produced by the loads P, Q, and R. The generalized force corrcsponding
to this rotation is the couple AI applied as sho\vn in Fig. 6. 3a. 1'his follo\vs
fronl the fact that such a group of forces produces \\'ork only during rotation

ARTI CLE 6.1
261
of the bar 6 in the plane of the figure. If this bar moves parallel to itself Of
elongates, keeping its direction unchanged, the \\'ork of t\VO such equal and
opposite forces perpendicular to the axis of the bar vanishes. Since the
generalized force M corresponds to rotation of the bar 6, \ve obtain the angle
of rotation by calculating the partial derivative au/aN!. If \ve substitute fOf
U its exprcssion (6.1), the angle of rotation becomes
8 _ aU _ "Sill aS i
- a ,\1 -  A..E aA1
(d)
Assullling that the couple M is infinitesimal, the forces S.. \vill have the same values
as bcfore and are given in column 4 of Table 6.1. The derivatives as,'; aM,
representing the rates of change of the forces Si \vith respect to the couple Nl,
are obtained by applying to the truss the couple equal to unity as sho\\'n in
Fig. 6.3 b and I thcn calculating the forces s' produced in the members of the
truss by this couple. Thus,
as..
aM
s'
,
Since the right side of this equation has the dimension of force divided by
moment of force, the partial derivatives as;1 aA1 have the dimension of Icngth- 1 ,
and the angle 8 as obtained from exprcssion (d) is a pure number.
Instead of the loads I 116 as sho\\'n in Fig. 6.3 b, it is simpler to use unit
loads, as before, and then divide cOfrcsponding axial forces s' 16 by the length
16 to obtain the numerical values of the derivatives as../ai\;l. For the numerical
data already assumed, the values of s' 16 calculated for unit loads instead of
1 116 as sho\vn in Fig. 6.3 b arc given in column 7 of Table 6.1. By using this
column togethcr \vith columns 2 to 4, \VC calculate the numbers in column 8.
Summing up these numbers and dividing by E, \\'e obtain, from Eq. (d), the
required value of the angle of rotation 8. For stfuctural steel this value is
8 = J. \' SJ, , = _ _ 2, 11 0 ,_._ = _ 70 . 3 X I 0 - 6 d
E  Ai 5. 30 X 10 6 · ra
(e)
l'hc minus sign indicates that the bar 6 rotated in thc direction opposite to the
direction of the unit couple in Fig. 6.3 b.
Castigliano's theorem can be used also to calculate the change in distance
bet\veen t\VO joints of a truss during deflection. Assume, for example, that
it is requircd to find the change in distance bet\\'een the joints D and /: (Fig.
6.4a) produced by the action of loads P, Q, and R. l"'he gencralized force
corresponding to this change in distance is evidcntly a group of t\VO equal and
opposite forces T applied at I) and [/ and acting along the line J)f' as sho\vn.
Castigliano's theorcm then gives
 = a T = " S Ji aS i
aT  AiE al
(I)

262 DEFLECTION OF PIN-JOINTED TRUSSES
p
Q
(a)
R
FI G. 6.4
(b)
If \\,re assume again that l' is an infinitesinlal force, the forces Si arc those
produced by the actual loads P, Q, and l. The values of the derivatives
as i / aT arc obtained by calculating the forccs produced in the truss menlbcrs
by unit forces acting as sho\vn in Fig. 6.4b.
Reviewing the preceding exan1ples, \ve see that in each case \ve need only to
make t\VO simple statical analyses of the truss: one under the actual loading
and another under the action of a fictitious unit load or gcneralized force cor-
responding to the desired deflection or displaccment. Then, \vith a little
simple tabulation and the use of Eq. (6.2), the required deflection is casily
found.
PROBLEMS
U sing the data in Table 6.1, calculate the change in the distance l)F of the truss
sho\vn in Fig. 6.4, duc to the action of the applicd 10ads P, Q, R.
Ans. -O.002H in.
2 Each bar of the simple truss sho\'lJ1 in Fig. 6.5a has cross-sectional area Ai = 1 in. 2 ,
and modulus of elasticity l = 10(10 6 ) psi; a = 10 ",/2 in., and P = 10,000 lb.
Calculate the horizontal deflection Oc of joint C.
Al1S. Oc = 0.010 in.
3 l'he sinlple truss sho\vn in Fig. 6.5b has the form of a square ABCl) \vith sides
loin. long, the diagonal AC being horizontal. Calculate the decrease 0 in the
vertical distance un due to the applied load P = 10,000 lb. Each bar has cross-
sectional area Ai = 1 in. 2 and nlodulus of elasticity F: = 30(10 6 ) psi.
Ans. 0 = 0.0114 in.
4 Find the vertical deflection of joint IJ of the sitnple tnlss loaded as sho\vn in Fig. 6.6.
Eat:h bar has cross-sectional area Ai = 2 in. 2 , and E = 30(10 6 ) psi.
411S. Od = O.7R6 in.

ARTICLE 6.2
I.
a
../
FI G. 6.5
(a)
20 kips
c
263
B
a
c
(b)
50 kips 50 kips 50 kips
20'
10'
30'
FIG. 6.7
5 Find the horizontal displacement of the roller-supported joint B of the simple truss
loaded as shown in Fig. 6.7. Each bar has cross-sectional area Ai = 8 in. 2 , and
modulus of elasticity E = 30(10 6 ) psi.
A 11 S . Db = 4. 71 in.
6.2 MAXWELL-MOHR METHOD OF CALCULATING
DEFLECTIONS!
To explain this method, let us consider the truss sho\vn in Fig. 6.8a and calculate
the vertical deflection D of a joint C produced by the given loads P and Q.
lhis deflection can be obtained as the sum of thc small deflections due to
deformations of single bars of the truss. To find the dcflection at C due to the
elongation of onc single bar,2 say bar CD, let us consider the system sho\vn in
Fig. 6.8b. The bar CD is removed, and instead of the actual loads only a unit
load is acting at joint G'. In this \vay \ve obtain a movablc system consisting
1 1.he method ,vas developed by Max,vell in a paper published in Phil. A-fag., vol. 27, 1864.
This paper did not con1C to the attention of engineers, and 10 yeats later in a paper pub-
lished in Z. Architekttll-blg. Yer., 1874, p. 223, O. 1ohr developed the sante method by
using another ,yay of reasoning. Since l\1ohr's \vork ,vas done independently of l\1ax\vell's
and since the method ,vas accepted by engineers only after numerous applications sho\vn
by Mohr, it is customary to call it the Maxwell-Mohr 1nethod.
2 The other bars are considered as rigid and their lengths unchanged.

264 DEFLECTION OF PIN-JOINTED TRUSSES
of the t\\'o rigid porrions, shaded in the figure, \vhich can rotate. \vitn respect
to each other. J() ensure the cqui libri UIn of this system, the forces .< 01ust
be added. These forces are replacing the action of the removed bar ClJ on
the shaded portions of the truss and evidently arc equal to th force produced
in that bar by the unit load at joint ('. \\.c no\\-' have a sinlple geometrical
probleril, [0 find the deflection illS of the joint (' resulting from the Inovement
of the systcm by \\.hich the distance bct\vcen the joints C and D increases by a
snlall amount Al i . 10 solve this problen1, \\'C lIse the principle of virtual displace-
ments. Sjnce the unit force at [' and the forces .< together \vith the reactions
at the supports represent a SYStC111 of forces in equilibrium, their \vark on the
foregoing assuIlled small displacenlcnts must vanish; hence,
1 lS - .< AI.. = 0
(a)
rhe reactions at the supports do not entcr in this equation. They do not
produce \vork since the support ..:1 does not rnove and the support B rnaves
perpendicularly to the reaction at B. Franl Eq. (a), \\'C obtain
,
t / J,
L\u =  . -
t 1
(b)
Thus \VC see that the deflection of the joint (' is ohtained by n1tIltiplying the
change of distance A/ i bct\veen the joints C and D hy the ratio s/ I. Having
this relation, \\lC proceed no\v \vith the actual case of loading sho\vn in Fig.
6.8a. ()ur previous notations being used, the elongation of a bar nUlllbcr i,
\vhich is the bar Cl) in our case, is S')i/AiE. Fronl the relation (b), \VC no\v
conclude that the deflection of joint C due to the clongation of the bar i is
All = SJi .< (f)
AiE 1
Proceeding in the same \vay "rith all other hars and summIng up the cor-
4
E
8
(a)
E
FI G. 6.8
(b)
lIb

ARTICLE 6.2
265
responding snlall deflections,l \ve obtain the actual deflection of the joint C in
the foIIo\ving form:
 l SJi, l I '
U = - E s, = .s.
A ft , &
i J
(d)
It is seen that, for the determination of the deflection of any joint of a truss,
\ve need to find the elongations AI i of all n1cmbers of the truss under the actual
load (Fig. 6.8a) and the forces .< produced in the same members by a unit load
applied at the joint the deflection of \vhich \VC have to find (Fig. 6.8b). Both
these calculations can be easily perfornlcd in the case of a statically dcterminate
truss and can be presentcd in tabular fornl as \vas done in the prcvious article
(see Tble 6.1, page 259).
(omparjng expression (d) \virh expression (6.2), \\'C see that the lVlax\vell-
Mohr method gives the deflections in the form already obtained by applying
Castigliano's theorCll1, This second derivation is given here because it makes
it easier to grasp the purely geometrical character of the problem. It is
evident that expression (d) holds independently of \vhat causes the changes li
in the lengths of the bars. Sometimes \ve have to consider the elongations of
bars due to a rise in temperature from some specified temperature. If ai is
the coefficient of thermal expansion and t i is the tern perature increase, the
corresponding elongation of the bar is ailil... Son1ctinles the lcngth of a
har can be changed by using some nlcchanical device such as a rurnbuclde.
The effect of such a change on the deflection of a truss can also be calculated
froll1 Eq. (d), provided that the change in length ll.. is kn()\vn. Supcrposing
deflections produced in a truss by various causes, \VC can \vrite expression (d)
in a rnore general form,
 \' ( s i J i + I ) '
U =  AiE aiti i + i Si
(6.3)
in \vhich not only elastic elongations of the bars but also elongations due to
temperature changes and elongations produced by some special devices are
considered.
Sometimes it is required to find not only deflections due to changes in the
lengths of the bars of a truss but also deflections produced by some small
movell1cnts of the supports. These additional deflections can be rcadily found
from Eq. (6.3). \\e assume only that the hinges at the supports arc attached
to an immovable foundation by sOlnc fictitious bars, the changes in length of
\vhich are chosen in correspondence \virh the kno\vn displaccn1cnts of the
supports. Including these fictitious bars in the summation sho\vn by expres-
sion (6.3), \ve autoolatically take into account the effect on the deflections of
1 It is assumed {hat the principle of superposition holds for the slnall defofIn3tions produced
in actual trusses.

266 DEFLECTION OF PIN-JOINTED TRUSSES
Pi
P"
P 3
p..
l
(a)
FIG. 6.9 (b)
1
the kno\vn movements of the supports . For example, let us consider an
unsymmetrical arch \\..ith three hinges (Fig. 6.9a) and calculate the horizontal
displacelnent of the upper hinge C produced by the loads P 1, . . . , P 4, a
uniform increase in temperature i, and the slnall movements of the supports the
componcnts 1, 111 and 2, 1'/2 of \vhich arc sho\vn to an enlarged scale in the
figure. i\.ssuming that each of the hinges A and B is attached to an immovable
foundation by t\\'o fictitious bars, one vertical and onc horizontal, \VC can
consider that tI, 111 and 2, 112 represent the slnall shortcning of thcse bars.
No\\', considcring the samc arch again, \ve assume that a horizontal unit load
is applied at the upper hinge C (}t'ig. 6.9b) and calculate the reactions Vi, HI,
, H 2 and the forces s in the bars produced by this load. Applying, now,
expression (6.3), \\'c obtain the follo\\,ing value of the horizontal displacement
of the hinge C:
8 = L ( :i + atl;) s: + bHl + 711 VI - 2H2 - 712 V 2
The summation in this exprcssion includcs all actual bars of the arch, and the
last four tcrms correspond to the four fictitious bars. Since these bars arc
placed in such a nlanncr that thc displacements 1, 771, 2, 712 represent shorten-
ings of the bars, \ve take all thcse displacements \vith negative signs. We
take also \vith negative signs the reactions HI and  since, from the directions

ARTICLE 6.3
267
of these reactions in Fig. 6.9b, \ve conclude that they represent compressive
forces in the ficti tious bars.
PROBLEMS
Referring to the truss in Fig. 6.9, calculate the vertical conlponent of the deflection
of the hinge C due .to a unifonTl rise in temperature of 70°F. Assunlc that all bars
are of steel \vhose coefficient of thermal expansion is a = 0.0000065 (in./in.) /oF
and that a = 10 ft.
Al1S. 0" = 0.158 in., up.
2 Each bar of the three-hingcd arch tfUSS in Fig. 6.9 has cross-sectional area Ai =
2 in. 2 , modulus of elasticity 1£ = 30(10 6 ) psi, and a = 10 ft. I'he applied loads
PI = Pz = P3 = P4 = 10,000 lb. Calculate the vertical deflection of the hinge C
if, in addition to the applied loads, the support B undergoes sonlC scttletnent rela-
tive to the support A such that I = '11 = 0, t:z  '12 = 0.01 fr, and the tempera-
ture drops 40°F.
Ans. Oc = 0.223 in., do\vn.
6.3 GRAPHICAL DETERMINATION OF TRUSS DEFLECTIONS
From the discussion of the preceding article, \\ore see that the analytical cal-
culation of truss deflcctions rcquires consideration of a special loading of the
truss for cach joint \vhose deflection is desired. If the dcflections of many
joints are required, the foregoing calculations become tedious, and a graphical
determination of the displacements can be uscd to advantage.! We begin \vith
a vcry simple example consisting of a joint A attached to joints Band C by t\VO
bars 1 and 2, as sho\"n in Fig. 6. lOa. The displacements BB' and ec' of the
joints Band C and the changes in length of the bars 1 and 2 are given; it is
required to find the corresponding displacement of joint A. We first assume
that the bars arc disjointed at A and translate thern to the positions A' B' and
A" C' parallel to their initial positions and such that BB' and ec ' represent the
given displacen1ents of the joints Band C. In these nc\v positions, \ve keep
the joints B' and C' fixed and givc to the opposite ends of the bars the dis-
placements A' A and A" A' as sho\vn by hcavy lines. 1"hese latter displace-
ments are equal to the given changes in the lengths of the bars; that is, A' A
is the kno\\'n elongation of the bar I, and A" A' is the kno\vn compression of
the bar 2. No\\', to finish the construction, \ve have to bring the points A
and A' togcthcr by rotating the bar B ' A \vith respect to the center B' and the
bar (" A' \vith respect to the center C'. Since \\'C are dealing \\lith small
deformations and small anglcs of rotation, the arcs of the circles along \vhich
I Such graphical constructions for the deflections of trusses are called JJTilliot ditlgra1ns.
See Williot, Notation pratique sur la statique graphique, Pub/. sei. ind., 1877.

268 DEFLECTION OF PIN-JOINTED TRUSSES
"
"
"
B - -  - -" B'
Ai
A ____/\,
'........,,/A' ,
, "........ ,
,,( ..... .... '
'" ..... \
, ....'
\. ... ........ "
 " ......::b
_--- :AI
--
Ail, -
1 \
\
,
\
\
\
\. \
\ \
" '
\. \
\. \
, \
't
C'
(a)
FIG. 6.10
A
o \"..... 11 ",
\ ................. "
........ \
...... ,
.... \.
........ ,
A" tJ.1 2 ,-...
_ - - - At
---
(b)
the points A and A' travel during such rotations of the bars can be replaced
by pcrpcndiculars AA 1 and A' A 1 the intersection of \vhich gives the nc\\!
position A 1 of the joint A. Thus, the vector IfIf; represents the required dis-
placement of the joint .(4.
Sincc the elongations of the bars and the displacerncnts of the joints are vcry
small in conlparison \vith the lengths of the bars, it is neccssary to rcpresent
these quantities to a large scale and to make all constructions in a separate
diagram as shown in Fig. 6.10b. We take a pole 0 and layout to the chosen
scalc the given displa(,'Cments ()A' and OA" of the joints Band C. Then,
from the points A' and A" \ve draw the vectors /l and /2, sho\vn by heavy
lines and representing the kno,vn changes in length of the bars 1 and 2. I)ue
attention must be paid to the sign of these elongations. The bar 1 is increasing
in length; hence, ll is put in the direction fro In B to A . The bar 2 is decrcasing
in length; hence, 12 is put in the direction from A to (.... Finally, perpcndic-
ulars constructed at the cnds of the vectors ll and 12 intersect at point A I,
defining the required displacernent OA 1 of joint A. It \vill be noted that the
displacement diagrarn in f"ig. 6.1 Ob is identical \vith the portion AA ' AAIA;' A"
of Fig. 6.10a, except that it can be constructed to a much larger scale if
desired. Figure 6.10b reprcsents the Wi/fiot diagnl111 for the simple structure
BAC of Fig. 6.10a.
The foregoing procedure can be llsed in the construction of displacement
diagranls for all simple trusses formed in accordance \\,ith the rule given on
page 53. As a first example, we considcr the truss sho\vn in Fig. 6.1111 and
assume that its variou members are so strcssed that they have changes in

ARTICLE 6.3
269
lcngth t:a/ i as rccorded in Fig. 6.11 h, \vhcre the plus signs indicatc extension
and the minus signs compression. V\'e begin \vith the displacement of joint C
\vhich is connectcd to A and B by the bars 1 and 2. The hinges A and B being
immovable, points a and b coincide with the pole 0 as sho\\rn in Fig. 6.11&.
Then from a and b, \ve layout, to a suitable scale, the vectors /1 and t:a/ 2 ,
representing extcnsion of bar I and shortening of bar 2, respectively. Con-
structing perpendiculars at the ends of these vectors, \ve obtain the intcrsection
point c, which determines the displacement OC of the hinge C. No\v, having
points a and c in the diagram, we arc ready to detcrmine the displaccment of
joint I), \\Thich is connccted to A and C by the bars 3 and 5. Laying out from
a the vector 13 reprcscnting extension of bar 3, and from & the vcctor / b
representing shortening of bar 5, and then constructing perpendiculars to these
vectors, \VC obtain the intcrsection point d \vhich deternlines the displacement
Od of joint I). Continuing in this \vay for joint E and, finally, for joint F,
\ve obtain the complcted Williot diagram as sho\\'n in Fig. 6.11 c from \vhich
the displacement of each joint of the truss can be scaled.
As a second example, \ve take the case of a simply supported truss as sho\vn
in Fig. 6. I 2a and begin \vith a calculation of the length changes lh . . . , A/ 7
of all bars produced by the given load P. These calculations can be made
\vithout difficulty since the system is statically determinatc and all dimensions
arc kno\vn. Starting no\\' \vith joint A, \vhich is fixed, \ve assume, for the
beginning, that the bar AE rctains its horizontal position during deformation.
Hence, joint E has only a horizontal displacement equal to the elongation 12
of the bar 2. Knowing the displacements of the joints A and E, \ve can no\v
find the displacement of the joint B, \vhich is attached to these joints by the
bars I and 4. Proceeding as before, we start with an arbitrarily chosen pole
o (Fig. 6.] 2b) and mark points a' and e', \vhich correspond to the joints A and
E. Since joint A is fixed, a' coincides \vith the pole O. l"he vector Oe' , equal
A
3
D
Bar li' in.
1 + 0.05
2 - 0.05
3 +0.10
4 - 0.10
5 - 0.05
6 + 0.05
7 + 0.05
8 - 0.05
4 0, a, b 3
2 1\ 1 :
,C / I
X I
6/ "'6 I
, " I
"v'" I
'" I
,,/', I
I 'I
'" " I
'...... ',I
ye d 7
, ...
" "
" ...
, /
, "
" /
, '"
" '"
, '"
v
4
E
(a)
(b)
f
(c)
FIG. 6.11

270 DEFLECTION OF PIN-JOINTED TRUSSES
to the elongation l2 of the bar 2, is taken in thc direction from A to E,
indicating extcnsion of the bar. From points a' and c' \ve dra\\' in the proper
directions the vectors /l and A. I..., representing the shortening of bar 1 and the
elongation of bar 4, respectively. Nlaking the perpendiculars at the ends of
these vectors, \ve obtain the intersection point h', \vhich determines the dis-
placemcn: Ob' of the hinge B. Having no\v points e' and 11' on the diagram, \\'e
can determine the position of the point e' and the displacement Oc' of the joint
C, \vhich is attached to joints Band f; by the bars 3 and 5. For this purpose
\VC dra\v from points h' and e' the vectors 13 and /6, representing, respectively,
the shortcning of bar 3 and the elongation of bar 5. The intersection of the
perpendiculars dra\\'n at the ends of these vectors determines the position of
point e'. Proceeding further in the same \vay, \ve finally obtain the last
point d' of the diagram. Vectors dra\vn from the pole 0 to the points e', "',
c', and d' represent in both magnitude and direction the required displacements
of the corrcsponding hinges of the truss. By using then1 to some reduced
scale, we can construct the distorted shape of the truss as indicated by dashed
lines in Fig. 6.12a.
Since ',\'e started \\'irh an arbitrary assun1ption that the bar AE remains
horizontal, the deformed shape AB' C' I)' E' of the truss does not satisfy thc
condition at the support D. This support can move only horizontally, \vhereas
the displacement ]JD', obtained from the diagram in Fig. 6.12h, is not hori-
zontal. To satisfy the condition at the support, we must rotate the deformed
truss, indicated by dashed lines in }t"'ig. 6.12a, \vith respect to the hinge A by
such an amount that the point D' \viII reach the horizontal line AI). In this
\vay the conditions at both supports \vill be satisfied, and the true displace-
ments of the hinges \vill be obtained by geometrically adding to the displace-
ments found in Fig. 6.12h the displacements produced during such rotation.
These latter displacements can be found as follo\vs: Since the disorted shape
AB'C' I)' E' of the truss is actually very close to its initial shape, it can be
assumed \vith good accuracy that during rotation about the hinge A, the hinge
D' moves perpendicularly to AD, hinge C' moves perpendicularly to AC, etc.
Furthermore, the magnitudes of these displacements will be proportional to
the radii AD, AC', etc. "rhe rcquircd rotatory displaccment of the point D'
is evidently equal to the vertical component of the displacement DD' repre-
sented by the vector Od' in Fig. 6.12b. This rotatory displacement of D'
having been obtained, the corresponding displacements of the other hinges are
readily obtained from Fig. 6.12c, in \vhich Od" is taken equal to the vertical
component of Od' in Fig. 6.12b and the other points are obtained by making
the figure all' e" d" e" geometrically sin1ilar to the figure ABCDE of the truss
but rotated by 90° \vith respect to that figure. The vectors b"() , c"O , d"O ,
and e"O \vill then represent the rotatory displacen1ents of all hinges of the truss.
This follo\vs (rom the geometric similarity that makes the vectors perpendicular
to the corresponding radii and proportional to the Icngths of these radii.

ARTICLE 6.3
271
C'
8' B_ _ - - - - - - - - 7' ,
 C
/ " 3 I,
/, I'
/ '\ I \,
/ ,
/ ,
I 1 ,
I '
,
I D'
I --
I.
",
d'
....j
./'" I
I
I
I
I
I
1
I
I
I
I
I
I
I
I
I
,
I
I
I
I
./
./
",
",
....
,I"
",
./
./
6
",
",
",
/
tJ.,\ c'
1.........
I ........
I ........
I '.....
I '
I
I
I
I
....
,
2
(a)
p
(b)
b"
d"
t
I
I
I
I
I "
I ",

b' '........
b"
- - -  - - - - - - - --- - - - - - O,a"
(d)
(c)
FIG. 6.12
"rhus vector b"O in Fig. 6.12e is perpendicular to the radius AB in Fig. 6.12a,
and h"O: AB = d"O: AD; vector e"O is perpendicular to the radius AC, and
e" () : A C = d" () : AD; ctc. To simplify the geometric addition of the rotatory
displacements to thc displacements givcn by the diagram in Fig. 6.12h, \ve
superimpose Fig. 6.12e on Fig. 6.12b so that the polcs 0 coincide, as sho\vn
in Fig. 6.12d. Then, the total displacements of the hinges B, C, D, E will be
represented by the vectors b" hi , e" e ' , d" d ' , and e" e ' . "[0 see this, let us take,
for example, the hinge B. Its displacement, corresponding to the dashed-line
shape of the truss in Fig. 6.12a, is given in Fig. 6.12/1 by the vector Ob'.
The rotatory displacement of the same hinge is given by the vector h"O ; hence

272 DEFLECTION OF PIN-JOINTED TRUSSES
13 l;
/ I J
/1/ I
/ I I I
." I
::::: - - - 1- _ _ _ _ _ _ _ _ _1- __
I I
I I
I I
I I
I I
I I
I I
I
I
I
I
I
I
I
I
go
(a)  --/
I _/1
1/......- !
.,../1 I
e" e'
- - - - - -- - - - - - - ----R
II I
I I
I I
I I
I I
/ 
I I
II I
h" I I
h', , I I
1'1 I
I' II I
I '
I "II (b) 
I I' I
I ,
I I' I
I I I' I
I I I' I
I ' A I 6" I
I I' I
I I c' f  Ll12 I
I d I 5', I 11 7 I
o 1 r I
- -=---=- -=. -:-.. -=. -=. -= -= = =-  '- ----y-::- \- - -r d'
g'll I', \ I
I II'\. \ I
J,r II Al ,\ I
j , I  8 \', I
l;t A9 \ I
I b' I
I 2 I
------------------
"0 AIO 0, a'
d"
a"
(c)
FIG. 6.13
the geon1etric sum of these t\VO displaceIl1ents, giving the true displacenlcnt of
the hinge B, is the vector b" b' . Sinlilar reasoning holds for the othcr hinges..
In the case of a truss \vith many bars, it may bc anticipatcd that if \ve start
the construction of the displaccment diagram from the support, as \ve did in
Fig. 6.12, the constructions \viII extend farther and farthcr 3\Vay froo1 the pole
0, and the diagranl nlay bccon1c un\vieldy. In such cases, a more cOIn pact
diagram can be obtaintd and the unavoidable inaccuracies of dra\ving can bc
reduced by starring the construction fronl the middle of the truss. (onsidcr-
ing, for example, the truss in Fig. 6.1 3a, \ve begin \vith the bar AR. Assuo1ing
that the hinge A is immovab]e and that the hinge B moves only vertically, \ve
find fronl the displacement diagram (Fig. 6.] 3b) that point a' coincides \vith
the polc 0 and that point h' is vcrtically above 0 at a distance Db' equal to the
elongation 11 of the bar 1. Having the points a' and b', \ve obtain the points
corresponding to the othcr hinges in exactly the sarnc Inanner as in Fig. 6.12.
1 The foregoing rHethod of correcting the vVilliot diagraln to satisfy the conditions at the
supports is due to Otto Mohr. See Z;'V;iillgeniellT, vol. 33, p. 639, 1887. The corrected
diagram should be called a vViliiot-Mohr diagranl.

ARTICLE 6.3
273
Considering the portion of the truss to the right of thc bar AB, \ve obtain in
the diagran1 in Fig. 6.1 3 b the points e', d', and e'. For the left portion of the
truss \ve obtain the points I', g', and h'. Vectors dra \vn from the pole 0 to
these points give the displacements of the hinges of the truss during deforma-
tion jf the n1iddlc bar AB is kept fixed. The corresponding distorted shape of
the truss is indicatcd in J.ig. 6.1 3a by dashed lines. Since our assumption
regarding the bar AB \vas entirely arbitrary, the displacements obtained do
not satisfy the conditions at the supports, \vhich require that the hinge H be
imn10vable and that the hinge E move only horizontally. To satisfy these t.vo
conditions, we have to perform t\VO additional movements; first we have to
move the distorted truss parallel to itsclf by the amount H' H, so that the
condition at the left support H \vill be satisfied. Then, subsequently, we
rotate the truss \vith respect to hinge If by such an amount that the point E'
rcaches the horizontal line lIE in }"'ig. 6. I 3a. When the truss moves paraUcl
to itself, all hinges suffer the same displacement given in Fig. 6.13b by the
vector h'O , and this must be geometrically added to the previously obtair:ed
displaccments. This is readily accomplished by taking point h', instead of
0, as the pole in Fig. 6. I 3h. Thus, for example, to obtain the result of super-
imposing the parallel motion represented by the vector l10 on the previously
obtained displacement De' of the joint E, \ve have only to take the geometric
sum of these t\\'O vectors, \\,hich is given by the vector h' e' . Similar con-
clusions will be obtained for th"e other hinges. Hence, by taking point h',
insrcad of 0, as a pole, \ve accompJish the required geolnctric addition of the
translatory displacement 110. After this operation, the displacement of the
hinge E is represented by the vector h' e' . Now, to satisfy the condition at
the support E, \ve must rotate the truss with respect to the hinge H by such an
amount that the rotatory displacement e" h' of the hinge E, \vhen added
geometrically to hie' , makes the resultant displacemcnt e" c' of the hinge E
horizontal (Fig. 6.13 b). The corresponding rotatory displacements of the
remaining hinges of the truss are then obtained by constructing the diagram
e" hili" h" e" geometrically similar to the shape of the truss. The vectors a"tt' ,
h, c"e' , . . . will then give the true displacements of the hinges of the truss.
Having the total displacemcnts of the hinges, \ve can readily obtain, by
proje<;cion, the vertical components of the displacements of the lo\ver chord
joints as sho\\'n by the polygon hogoaodoco in Fig. 6.13c. Such a polygon is
called the deflectioll pO/YK01l for the truss.
From the prcceding examples it will be appreciated that in the case of simple
trusses there is no difficulty in constructing the displacement diagram cor-
responding to known changes in the lengths of all bars  In proceding \vith
this construction, it is necessary only to repcat \vith each ne\v hinge the con-
struction illustrated in I.jg. 6.10h. In the case of statically determinatc trusses
that arc compound or complex in form, some additional considerations arc
required, \vhich \ve shall no\\' illustrate by a final example. For this purpose,

274 DEFLECTION OF PIN-JOINTED TRUSSES
\ve. take the compound roof truss sho\vn in Fig. 6. I4a \vhcre tension rnembers
are marked \vith plus signs and cornpression mcrnbers \vith minus signs. To
construct the displacement diagran1 for this truss, \VC first disregard the \\"cb
mcrnbcrs on the right-hand side and consider F(;H as one triangle. This
nleans that tcmporarily \ve treat F(;, GH, and FH as single bars, the length
changes of \vhich \vill be obtained simply by summing algebraically the kno\vn
length changes in their component parts. In this manner, a simple truss is
obtained for \vhich the displacement diagraol can be readily constructed.
Such a diagrarn, constructed on the assumption that the hinge B renlains
immovable \vhile the bar liC retains its original direction, is sho\vn in Iig.
6. J 4b. Point b coincides \\,ith the pole 0, and point c is obtained by construct-
ing Oc parallel to BC' and equal to the kno\vn shortening of that bar. Proceeding
in this \vay, \ve finally obtain the points tl and h defining the displacements va
and Oh of the points of support A and H. '10 satisfy no\\' the actual support
conditions, \ve displacc the truss parallel to itsclf by the amount (10 so that the
hinge A coincides \vith the fixed support and then rotate the truss about this
support until the hinge H reaches the horizontal line through A in Fig. 6.14a.
l\S already explained, the gconletric addition of the translatory displacement
is obtained siolply by using t1 instead of 0 as a pole. l.he rotatory displacc-
mcnt of the hinge If is vertical and of such magnitude h' a that \vhcn geometri-
caJly added to the displacement all , it makes the resultant displacement h' h of
the hinge H horizontal. The corresponding rotatory displacerncnts of the
othcr hinges are then obtained, as before, by constructing the figure ,,' b'f'g' h'
('
\
\
\
I
I
,
,
\
\
\
,) f
".,...
 '" /'
,
,
- ' .-
h' -:. -, .. :: :,.... h .. " " ( c )
A::.: // '
,g ,
" I. ,,'
,,' ,
k'"
(a)
FIG. 6.14
hi h
- - - - - ,'"
"
,
,
..
"
Jl

ARTICLE 6.3
275
in Fig. 6.14/1 geomctrically similar to the truss but rotated by 90° as sho\vn.
The vectors measured on this diagrarll fron1 the priolcd points to the cor-
responding unprinlcd points give then the true displacenlcnts of the hinges A,
B, (', f), F., I:, G, H,
"J() finish the problt\nl, \VC have to consider the displacenlents of the hinges
I, J, [(, \vhich \vere disrcgardcd in thc previous constructions. For this
purpose, \VC make, in the usual \\ray, the displacement diagranl for the right-
hand half of thc truss, assurl1ing that the hinge G rernains stationary and that
the bar GI retains its direction. This diagranl is sho\vn in Fig. 6. J 4c, from
\vhich \ve find the displacClncnts Of and 011 of the hinges f' and H. Naturally,
these do not agree \vith the true displacements of these t\VO hinges as already
found in Fig. 6.14b, since \ve began the constructions on the basis of arbi-
trary assumptions regarding the hinges C and 1. Io corrcct the diagrao1, \VC
first translate the right-hand half of the truss by the amount i' 0 such that
j'O + OJ =]7, taken froo1 Fig. 6.14/1. This translation is accoll1plishcd
silnply by taking /' in Fig. 6.14c as a ne\v pole. Next, \ve Jnust rotate thc
right-hand half of the truss about F until the hinge H attains its true horizontal
displacenlcnt h' /1 , taken again from Fig. 6.14b. Thus, the required rotatory
dispJacenlcnt of H about f" is representcd in Fig. 6.14c by the vector I' h' , and
the corresponding rotatory displacements of the other hinges arc found by
constructing on .f' h' the figure /' R' h' i')' k', geometrically similar to the right-
hand half of the truss }/GHlfK. i\ chcck on the accuracy of the \vork \vill be
obtained at this point by noting that the vector I' h' in Fig. 6. 14c should be
perpendicular to the chord linc f11 in Fig. 6.14a.
Finally, the true displacements of all hinges of the truss \v ill be obtained by
scaling the vectors from the primed points to the corresponding unprimed
points. That is, the true displacement of joint D is representcd by the vector
d'd in Fig. 6.14b, that of joint J by the vector i'j in Fig. 6.14c, etc.
PROBLEMS
The bars of the sirnple truss sho\vn in Fig. 6.15 are of such cross-sectional areas
that they t'longare (or shortcn) 0.01 in. per 1,000 lh of axial force. Construct a
\\Tilliot diagrao1 and find therefrotn the horizontal and vcrtical COOlponcnrs of the
displacenlcnt of joint E.
An!. Oh = 0.017 in., Or = 0.080 in.
2 Construct a ""illiot-}\1ohr diagranl for the simple truss supported and loaded as
sho\vn in Fig. 6.16, assunling that each bar has a unit axial strain of n.oo 1. For
nlaxinnull accuracy, begin the constructions on the assumption that joint [) rCfilains
stationary and that I)C relnains vertical. \Vhat is the final vertical dcflection of
the hinge D?
AT!!. OtJ = 0.345 in.
3 Construct a "lilliot diagralll for the sill1ple tfusS sho\vn in Fig. 2.18, page 65,
assuo1ing that each bar has an axial strain of 0.00 I .

276 DEFLECTION OF PIN-JOINTED TRUSSES
A
c
E
B
a a
8' 8'
5001b 5001b P
FIG. 6.15 FIG. 6.16
4 Construct a Williot diagram for the sinlplc truss sho\vn in Fig. 2.19, page 65,
assuming that each bar has an axial strain of 0.001.
S Construct a Williot-J\ilohr diagratn for the compound truss sho\vn in Fig. 2.45,
page 85, if there is a vertical load P at each lower-chord joint and the cross-sectional
areas of the bars arc such that each bar has an axial strain of 0.001.
Hint First, construct a separate displacement diagram for each half of
the truss, assuming, for the left half, that joint C remains stationary
\vhile CD retains its direction and, for the right half, that joint E
rClllains stationary \vhile EF retains its direction.
6.4 METHOD OF FICTITIOUS LOADS
The graphical mcthod of finding displacements described in the preceding article
tcnds to become inaccurate in the case of a truss containing many bars. l"hc
unavoidable errors of construction increase rapidly \vith the number of graphical
operations, especially if it is ncccssary to determine the intersections of lincs
that are too nearly parallel. In practical applications \ve oftcn need to }<no\v
only thc vertical deflections of the chord joints of a truss. l'his Jimited
problem can be solved by using a mcthod of fictitious loads 1 similar to that
applied in calculating deflections of beams. Take, for example, the truss in
Fig. 6.17 a, and assume that the elongations of all bars produced by the given
loads PI, P2, . . . arc )(no\\'n. Then, to calculate the vcrtical deflections of
the joints resulting from an elongation t:"l i of one bar, say bar CD, \VC use the
method eXplained in Art. 6.2. Accordingly, the deflection 0". of any joint 111
(Fig. 6.17 b) is obtained from thc equation
s
0". = /i i (a)
1 The application of fictitious loads in calculating deflections of trusses \vas introduced by
O. 1\1ohr, Beitrag zur lheoric des Fachwerks, Z. Archiukten-ln?r.- VeT. Han1lQVtT, 1875,
p. 17. See also his book uAbhandlungen aus clem Gcbiete dcr tcchnischcn lechanik:'
p. 377, 1906.

ARTICLE 6.4
277
in \vhich s is the axial force in the bar Cl) due to a unit load at 111. Since all
bars, except (..'D, are considered rigid, the t\\'O portions of the truss, shaded in
Fig. 6.1 7 b, move as rigid bodies, rotating ,vith respect to each other about the
hinge 1n, and the vertical deflections of all joints are evidcntly given by the
corresponding ordinates of the diagram a111b in Fig. 6.17c. Observing that
the axial force s produced in the bar C[) by the unit load at 111 is equal to the
bending moment at 1n divided by the distance h, \\'e conclude, from Eq. (a),
that the deflection diagram in Fig. 6. I 7 c can bc considcred as the bending-
moment diagram for a fictitious beam A B acted upon by a fictitious load
dl i
h
(h)
as sho\vn in Fig. 6.17 d. In a similar manncr the deflections resulting from a
change in length /i of any other chord member of the given truss can be found.
Using, no 'v , the method of superposition, \ve conclude that the deflection of
the truss resulting from changes in length of all chord members can be cal-
culated for cach joint as the bending moment at the corresponding cross section
(a)
c
I
1
(b)
c 8 8
I I
II
b
(c) a  m
m
 A;I! /
FIG. 6.17 (d)

271 DEFLECTION OF PIN..JOINTED TRUSSES
of a sirnplc bean1 subjected to fictitious loads as defined by cxpression (b). *
Rcgarding the sign of the fictitious loads, \ve obscrve that compression of any
uppcr chord n1crnber or clongation of any lovicr chord membcr results in a
do\vo\vard deflection of the truss. 1"'his indicates that shortening of the upper
chord or clongation of the lo\ver chord implies a do\vn\vard or positive fictitious
load.
Let us consider, no\\', the effect of changcs in length of \veb Incmhers on
the deflection of a truss. If only onc bar, say the bar 1)£ of the truss' in Fig.
6.18a, changes in lcngth by the anl0unt /1' thc deflection of a joint 111 (Fig.
6.18b) is again obtained from Eq. (n). Applying this equation to all joints in
succession, \VC conclude that their deflections are obtaincd by multiplying by
d/.. the corrcsponding ordinatcs of the influence diagralll for thc bar IJE as
shown in Fig. 6.18c. To construct this influence diagram, \ve note that, for
any position of the unit Joad to the kft of joint E, the force .< (Fig. 6.18b) is
obtained from the equation
I xa ()
S i = -, r C
in \vhich the distances a and r are as sho\\'n in the figure. I f the unit load is to
the right of joint D, the force s:. is given by the equation
s:' = /  a (1 - 1) (0)
Thus, the influence diagram for s has the form shown in Fig. 6.18c. Upon
multiplying the ordinates of this diagram by A./.., \ve obtain the deflections of
the truss produced by the change in length A./.. of the bar DE (Fig. 6.18£1). In
all cases \vhere the influence diagram has the shape sho\\'n in Fig. 6.18c, the
shaded area in ig. 6.18d can be considered as the bending-moment diagram
for a sinlply supported beam loaded as sho\\'n in Fig. 6.18e. l-"his loading
consists of t\VO fictitious loads P and Q thc magnitudes of \vhich are obtained
from the conditions that the reactions at the ends of the beam are
R =  d/i
a rl
This gives
p = A/ i /2
r /3
and
lb = _ (a + I) 61 i
rl
and
Q = l i /1
r /3
(e)
\vhere 11, /2, and /3 are the distances indicated in Fig. 6.18e. As a check, we
calculate the bending n10ment for a point of the beam to the left of the load P
* These loads, as \'1e see, arc pure numbers, and the corresponding bending 1110ments have
the dimension of length, as should be the case.

ARTI CLE 6.4
279
and at a distance oX from the support A (Fig. 6.18e). This moment
ax
A.[ = - - f.
r I '
is seen to agree \\.rith expression (c) multipJied by AI,.. In a similar manner,
the bending moment for the portion of the beam to the right of the load Q is
represented by expression (d) multiplied by tJ./... l'his indicates that the
bending-moment diagram for the beam in "ig. 6.18e is identical with the
deflection diagram In .Fig. b.ISd. vVe conclude, then, that by introducing- for
each \veb Incmber the t\VO fictitious loads, defined by expressions (e), and
using the method of superposition, \\lC obtain the deflections of the truss due to
changes in length of all \veb membcrs by calculating the bcnding moments in the
corresponding simple beam carrying the above-mentioned fictitious loads.
Exprcssions (e) for the fictitious loads P and Q can be some\vhat simplified
as folIo\vs: Drawing verticals through the ends of the \veb member DE
(6)
I I-x
P3
x
(a)
(c)
a
r
(d)
a . AI;
-
r
r x
I
(e)
FIG. 6.18

280 DEFLECTION OF PIN-JOINTED TRUSSES
(Fig. 6.18b), \vcseefronlgconlctricsinlilaritythath 1 /r = la/1zandh2/r = 13//1.
Substituting these values in (f), \\'C obtain
l.
p = ._-
hi
and
Q = li
11 2
(f)
rhc directions of these fictitious loads in each particular case are readily
obtained fronl the signs of the ordinates of the influence diagram. \1\Te can
usc, also, the rule stating that the fictitious load at either end of a \vcb mernher
acts do\vn\vard if, at this cnd, tht, chord mernher and the \veb nlembcr cut hy a
section stich as secrion JJ in Fig. 6.18b have axial forces of the saIne sign.
The load acts up\vard if the axial forces are of opposite sign.
To illustratc the method, let us consider the truss sho\\'n in Fig. 6.19a.
lhc lengths and cross-sectional areas for the bars of this truss are given in
r 5 't- 1' 
I 1
)
(a)
24,0001b
(b)
12.85
(c)
13.08 12.85
20.19
25.26
nz
p
8
FIG. 6.19 (d)

ARTICLE 6.4
281
TABLE 6.2
,
Fictitious loads at joints multiplied by
I, A, AlE E/l,OOO
I S - - -.
In. in. 2 1,000 
, I I I
111 I 11 P q s
:
1 84.8 3.66 -11,310 -262 6. t 8 I
2 120 3.66 8,000 262 4.37
3 84.8 3.66 11,310 262 -6.18 6.18
4 120 4.98 -16,000 - 385 6.42
5 84.8 2.48 -11,310 -385 -9.08 9.08
6 120 3.66 I 24,000. 785 13.08
7 84.8 2.48 11,310 385 -9.08 9.08
8 120 4.98 - 32,000 -771 12.85
9 84.8 3.66 22,620 524 J2.36 - 12.36
10 120 3.66 16,000 524 8.73
11 84.8 3 . ()6 -22,620 -524 12.36
Tablc 6.2. l-"hcre are given also in this table the axial forces in the bars and
the elongations produced by a 24,OOO-lb load applied at the joint q. 10 find
the dcflections due to the corresponding changes in length of the chord mem-
bers, \ve calculate fictitious loads fronl expression (b). Since the chords are
parallel, all values of h are equal to 60 in. in this case. The values of these
fictitious loads, multiplied by t'/l,OOO, arc sho\vn in Fig. 6.19b and are given
also in the table. Assuming that these loads are acting on a simple bearn AB
(Fig. 6.19c), the bending moments at their points of application, divided by
E/l,OOO, give the deflections of the corresponding joints of the truss. (:00-
sidering, for example, the joint p, the corrcsponding bending moment for the
simple beam is
,\11 = 20.19 X 180 - 4.37 X 120 - 6.42 X 60 = 2,725
Hence the deflection of the joint p due to elastic deformation of the chord
mcnlbers is
o = 2 0M = 1,000 X 2,725 = 0 0908 .
E 30 X 106 . In.
To calculatc the deflections due to the elastic deformations of the \veb
members, \\'c apply to the cnds of cach \\1cb mcn1ber the fictitious loads found
from Eqs. (f), in ,,,hich, for this case, the distances hI and 11 2 are equal to
60/VZ = 42.4 in. 11he magnitudes of all these loads are given \vith the
propcr signs in Table 6.2. After sununation, \\'c obtain the fictitious loading
of the truss sho\vn in .Fig. 6.19d. The bending momcnt produced by thcse

282 DEFLECTION OF PIN-JOINTED TRUSSES
loads at the cross section of a fictitious beam, corresponding to joint p, is
Al = 13.33 X 180 - 21.44 X 60 = 1,11 3
and the corresponding deflection of the joint p of the truss is
 = 1,113 X 1,000 = 00371 .
u 1 30 X 10 6 · In.
The total deflection of the joint p is obtained by adding 01 to the previously
found deflection 0, which gives
01' = 0 + 01 = 0.128 in.
In thc forcgoing discussion of the effect of \veb Illcmbers on dcflection, \ve
considered a case in \vhich these membcrs \vere not vertical. 10 extend the
method to the case of vertical 'web members, \VC take the truss sho\vn in Fig.
6.20a and consider the vertical member CD. To apply our previous reasoning,
\ve assunlC, at the beginning, that the member CD is slightly inclined to the
vertical as indicated in the figure by the dashed line. l'hen, our prcvious
reasoning is applicable, and the fictitious loads at thc ends of the member are
given by Eqs. (e). Assuming, no\\', that the angle of inclination indefinitely
diminishes and approaches zero, the distances r, 1 2 , and 11 in Eqs. (e) become
equal to the distance c in Fig. 6.20a, \vhile 13 approaches zero. Thus, both
fictitious loads increase indefinitely; but their difference, equal to dli/ r,
approaches the value /i/ c, and their Inoment \\lith respect to joint C approaches
the magnitude (d/../1 3 )/ 3 = dl... The corresponding bending-moment diagram
for the fictitious beam is sho\vn in Fig. 6.20b. The discontinuity in this
diagram at the point of application of the fictitious couplc dl i indicates that the
deflection of the joint (' of the lo\ver chord is larger than the deflection of the
joint D of the upper chord by the amount equal to the elongation I i of the
vertical member CD.
Considering, no\\', the vertical membcr EF (Fig. 6.20a), \ve readily see
that change in length of this bar has no general influence on the deflection of
the truss. The deflection of joint }/ is obtained by complctely disregarding the
presence of the bar EF, and to obtain the deflection of joint E \ve have only to
add the elongation of the bar EF to the deflection of joint f.
To obtain the deflections produced by a change in Icngth of the n1iddle
vertical M!\T, \ve consider the influence diagram for this Illember as sho\vn
in Fig. 6.20c. 1 Multiplying the ordinates of this diagram by /.., \ve obtain,
in accordance \vith Eq. (a), the deflcction line of the lo\ver chord of the truss
as sho\vn in l'ig. 6.20d. l'his line can be considered as the bending-moment
1 It is assumed that the unit load moves along the lo\\'er chord of the truss. Hence, we
shall obtain the deflections of that chord.

ARTICLE: 6.4
(a)
(b)
(c)
(d)
FIG. 6.20 (e)
283
c
l,
'J'IIIIIIllj.JlJ:'flJ..Cl!1II:lljlllll'" I I ·
- - _ _ _ _ _ _ 2h . 2 81D a
---

.11/. I sin.. 
Jf
diagram for the fictitious bealTI AB loaded as shown in Ic'ig. 6.20e. From this
discussion, \VC see that the deflection of a truss due to changes in length of
vertical \vcb members can also be obtained as the bending moment produced
by certain fictitious loads, \vhich can be readily obtaincd in cach particular
case.
As a specific example dealing \vith vertical \\'eb members, let us consider
the truss shown in Fig. 6.21a and assume that the load is symmctrically dis-
tributed \vith respect to the vertical axis of symmctry of the truss. "rension
and compression members are indicated in the figure by plus and minus signs,
respectively, on the lcft side of the trnss, and the magnitudes of the stresses arc
so assumed that the unit elongation or contraction of each active member is
equal to 1/2,000. Considcring first the deflections due to chord membcrs and
using expression (b) for fictitious loads, \ve find that the loading on the
fictitious beam is that sho\\'n in Fig. 6.21 b. For the diagonals \ve use expres-

284 DEFLECTION OF PIN-JOINTED TRUSSES
(a)
1
6' . ._ 60' I
1 1 _..!._ -L --L-  1 1 1
1,000 1.00 0 1,000 1.000 1.000 1,000 1,000 1 ,000 1,000
(b)
-A-.
l
A
(c)
0.036 0.036 0.036 0.036 0.036 0.036 0.036 0.036 0.036 0.036
" ,,r
(d)
-A...
'*0 l

FIG. 6.21
(e)
Slons (f) and find
1
p= Q=
1 ,000
These fictitious loads act do\vn at the Jo\ver cnds of the diagonals and up at the
upper ends. After summation, \VC obtain the loading sho\vn in Fig. 6.21c.
Coming no\\' to the verticals and using the reasoning illustrated in Fig. 6.20b,
\\'e find that the corresponding loading on the fictitious beam consists of the
fictitious couples sho\vn in Fig. 6.20d. At the point of application of each
couple \VC havc an abrupt change in the bcnding moment by thc amount 0.036
in. equal to the shortening of the verticals. By calculating the bending
moments just to the left of the points of application of the fictitious couples, \\"e
obtain the deflections of the joints of the lo\ver chord of the truss. The bending
moments just to the right of the same points give the deflections for the joints
of the upper chord. Considering the deflections of the lo\\'er chord, \ve see
that instead of fictitious couples, \ve can take the fictitio\1s load sho\vn in
Fig. 6.21 e. This load produces reactions at the ends equal to 1/2,000, and thc
bending moments at the points corresponding to the joints of the lo\vcr chord

ARTICLE 6.5
285
are the safficas those produced by the fictitious couples of Fig. 6.2Id. Finally,
to gct the total deflection of any joint of the lo\ver chord of the truss in Fig.
6.2'la, \ve have to conlbinc the fictitious loadings sho\vn in Fig. 6.21 h, c, and e.
The deflection at the middle, obtained in this \yay, is
4.5 4 1 1
o = OO 360 - 1,000 180 + 1,000 360 + 2 OOO 360
= 0.9 + 0.36 + 0.18 = 1.44 in.
It consists of thrce parts, a deflection of 0.9 in. due to chord nlen1bers, a dcflec-
tion of 0.36 in. due to diagonals, and a deflection of 0.18 in. due to verticals.
PROBLEMS
Referring to the truss in Fig. 6.19, calculate the deflection of joint p if the 24-kip
load acts at this joint instead of at q. Use all numerical data for the truss as given
in "Iable 6.2.
Al1S. op= 0.175 in.
2 Refcrring to the truss in Fig. 6.21, calculate the deflection of each lo\\'cr-chord
joint due to a vertical load P = 10 kips acting at C. -\ssumc that each bar has a
cross-sectional area of 1 in. 2 and E = 30(10 6 ) psi.
Ails. Oc = 1.25 in.
6.5 ALTERNATIVE METHOD OF FICTITIOUS LOADS
"[he problem of finding the deflections of the joints of one chord of a truss can
be solved by considering only the ,mcmbers of that chord. This solution is
especially simple if the chord is a horizontal straight line. Take, for example,
the truss shown in Fig. 6.22a, and let it be required to find the vertical deflec-
tions of the upper-chord joints. It is evident that such displacernents \vill be
completely defined if \ve kno\v the changes in the lengths of the bars of the
upper chord together \vith their angles of rotation during deflection of the truS5.
Since the chord is a horizontal straight line, changes in the lengths of its bars
result only in horizontal displacements of the upper-chord joints. l-'hus,
vertical displacen1ents of these joints depend only on rotations of the uppcr-
chord rncmbcrs. I n calculating these rotations, \ve assumc that the changes
in the lengths of all bars of the truss arc 1<no\\'n. Then, the corresponding
changes in the angles of each triangle can be calculated \vithout difficulty, as
\vill be sho\vn later. Assuming for the rnoment that such changes in the
anglcs CPJ, CPz1 and CPa at an upper joint (Fig. 6.22(1) have been found, ,,'c obtain
by thcir sunl a small angle A8rn represcnting the angle bct\\'ecn the t\\'o chord
mClnbers at the joint 11/ after deflection. 11he corresponding deflection of the

286 DEFLECTION OF PIN-JOINTED TRUSSES
(a)
II
1 2
I
L
M r
x
..
- - - - _-n 8 m
--
---
--
t
M m l 2
--L
(b)
f:18 m
FI G. 6.22
(c)
1'10 m 1 2
I
1'18 m I)
I
upper chord, assuming that d()m is pOSitiVe, is shown in Fig. 6.22b. rhe
deflection for any point of the chord to the left of joint 111 is equal to 8m 1 2 xll,
\vhilc, for any point to the right of 111, it is 8m /1(1 - .t.) I I. \\e see no\v that
these expressions for deflections are identical \\Tith those for bending momcnts
produced in a simply supportcd beam by the fictitious load Om acting as shown
in Fig. 6.22c. l"hus \\'C conclude that by calculating the values of 6, OJ for each
joint j of the upper chord and using the method of superposition, the rcquired
deflections \vill be obtaincd as the bending moments for a simply supported
beam subjected to fictitious loads Oj. *
We have nOVl to develop a method of calculating the changcs in the angles of
a triangle if the changes in the lengths of its three sides are kno\\'n. Since \VC
arc dcaling \vith very srnall elongations, the V\Tilliot diagranl discussed in Art.
6.3 \vil1 be useful for this purpose. Consider, for example, a triangle ABC
(Fig. 6.23a) \vith sides of lengths 11, /f}., and 13 and altitude h, and let it be
required to find the changes in the angles due to kno\vn elongations J)./h d12,
and 13 of the sides of the triangle. To accomplish this, \\'c assume that the
joint A is fixed and that the bar AB retains its horizontal direction. Then, to
find the displacement ec ' of the vertex G', due to the kno\vn elongations I h
/2' and /3 of the sides of the triangle, \ve construct the usual Williot diagram
as sho\vn in Fig. 6.23b. If PI and P2 denote the lcngths of the perpendiculars
U' and C 2 C' in Fig. 6.23 b, the corresponding changes in the angles al and a2
... This method of calculating deflections of trusses \vas introduced by H. M iiller-Breslau,
z. AT(hitektn-Ingr.- VeT. Ilan'lOver, 1888, p. 605. See also his book "IJie graphischc Statik
der Baukonsrrucrionen," vol. II, pp. 1 and 96, 1904.

ARTICLE 6.5
287
at 13 and A in Fig. 6.23'1 arc t
A P2 P2 sIn al
.u.al = - = -- - - -
/2 h
and
PI PI SIn CX2
daz = It = Jz
Remembering, no\v, that the sum of the angles of a triangle n111st ahvays be
180 0 , \ve conclude that the required change in the angle aa at C is
P2 sin al + PI sin CX2
a3 = - (al + (2) = - ----- ---
Jz
(a)
. rhe numerator on the right side of this expression represents the sum b of the
horir,ontal projections of the perpendiculars PI and P'l in Fig. 6.23 b. j\'1easuring
this sum and dividing it by the height /1, \ve obtain the decrease in the angle Q'3.
Exprcssion (a) can be representcd in a sill1plc analytical form very useful for
numerical calculations. "", note that the length b is equal to the sum of the
horizontal projections of the lengths  and ('1('3 in Fig. 6.23b. The vertical
projections of the same lengths arc D 2 [) a and TJ;[53' I-Icnce,
(b)
b = D2D3 cot at + })ll)3 cot ct2
()bserving that the triangle ()CaR in Fig. 6.23b is similar to the traingle A('B
in Fig. 6.23a, \ve conclude that O/.J a = h /3/13' Then,
D D OD r D . h /3 J ( jj,/2 f1/a )
2 3 = 2 - {J 3 = d/2 sin Q'1 - -- - = 1 .- - __0
13 1 2 13
- - . h jj,/3 ( D./l D.13 )
DlDa = OD l - OD 3 = /l sIn a2 - - - = h - - -
13 11 13
1 The $[nall changes 6/ 1 and .1/'! in the lengths /, and /2 arc neglected in these txpressions.
D.
D'j
k-b--j
. c" I
Q 1-=-',- - - I
C2 1 P2 " of 1 I
--"a 2 -_
" I
C\

C  a.,
j -
D 2
A 0 B
1:1 .1l;l
(Cl ) (b)
FI G. 6.23

288 DEFLECTION OF PIN-JOINTED TRUSSES
Substituting these values in exprcssion (b)., \\'C obtain
b = h ( /2 - /'J ) cot at + h ( d/ 1 - /3 ) cot a2
1 2 13 II 13
and Eq. (a) bccon1cs
a3 = -  = ( /3 - /2 ) (ot a} + ( /3 - /l ) cot az (c)
h 13 /2 13 II
Upon introducing, fOf the stfesses in the bars, the notations
(f 1 = E  1 (f 2 = E  2 (f 3 = F  J
,.\'e can finally rcprcsent Eq. (c) in the follo\ving sinlple fornl:
E a3 = (0"3 - 0"2) cot a} + (O"a - 0"1) cot az (d)
Obviously, this expression for the change in a3 111ay be used also for the changes
in al and a2 by a proper change of subscripts. I-Icncc, for further reference,
\ve \vnte
I:. al = (0"1 - O"'l) cot (X:t + (0"1 - 0"3) cot lX2
E a2 = (0"2 - 0"1) cot a: + (0"2 - 0"3) cot a} (6.4)
E da a = «(13 - 0"2) cot a} + (0":1 - 0"1) cot a2
These three equations furnish a sinlple "ray of calculating the changes in the
angle of all triangles of a truss if the stresses in the bars are kno\vn. I n using
5' II J.....
Jm
--10'-+---10'--+ 5'
4 I p 8 Is
2
6
24,0001b
10
(a)
12,230
} .

45.260
E
(b)
23.240
E
34.250
E
6.110
E
32,000
E
20.910
E
FIG. 6.24 (c)
25,410
-y-
38.610
---g--

ARTICLE 6.5
289
TABLE 6.3
! I
r Fictitious loads multiplied by E
.
Bar It, Ai, S" CT i,
In. in. 2 b pSI I I I I
n q 111 P S
I
1 84.8 3.66 -11,310 - 3,090 -5,276
2 120.0 3.66 :3 ,000 +2,186 +4,372
3 84.8 3.66 11,310 + 3 ,090 +6,302
4 120.0 4.98 - 16,000 - 3,212 -4,954
5 84.8 2.48 -11,310 - 4,560 - 1 I, 1 20 -1,348
6 120.0 3.66 24,000 +6,560 + 13, I 20
7 84.8 2.48 11;310 + 4-,560 - 2,000 + 10,985
R 120.0 4.98 - 32,000 -6,425 - 23,590
9 84.8 3.66 22,(>20 +l>, J 80 + 12,605
10 120.0 3.66 16,000 +4,372 +8,744
I I 84.8 3.66 - 22,620 - 6, 1 80 -10,552
--
Total -12,230; -45,262\ +3,024 +32,0271 +19,729
thern, \ve have only to remember that 0"1, 0"2, and 0"3 denote the stresses in the
bars opposite to the angles Q}, Q2, and a3, respectively.
To illustrate the details in applying this altcrnative method of fictitious loads,
let us consider again the example of the prcceding article, repeated in Fig.
6.24a, and calculate the vertical deflections of the lo\\'er-chord joints 11 and q.
rhe lengths Ii of the bars, their cross-sectional arcas A i, and the axial forces Si
due to the loading sho\vn are recorded in the first three columns of Table 6.3.
After this, the values of the unit stresses 0" i = Si/A i are recordcd in the fourth
column of thc table. No\v using the first of Eqs. (6.4) and observing that
cot a2 = 0 and cot a} = 1, \ve obtain
E al = (- 3,090 - 2,186) 1 = - 5276
\vhich is rccorded under 11 in the table. In a similar \vay, \\'e find the values of
E a4, E da7, E dao, E das, and E dau as recorded under nand q in the
table. Since all these angle changes arc negative, the angles decrease, and \ve
obtain deflections do\\'n\vard. Summing up the angle changes at 11 and at q,
\ve obtain the corresponding fictitious loads for these points on the conjugate
beam as sho\vn in ig. 6.24b. Then, for the deflections of joints 11 and q,
\ve simply calculatc the bending moments at the corresponding points on the
conjugate beam. "raking E = 30(10 6 ) psi, \VC obtain in this \vay (see }tig.
6.24h)
t 23,240 .
Un = 30(-10 6 ) 120 = 0.093 In.
 34,250 7 .
q = 3 0(10 6 ) 120 = 0.13 tn.

290 DEFLECTION OF PIN-JOINTED TRUSSES
If the chord \\,hosc deflection is required is of polygonal form as, for example,
the lo\ver chord of the truss in Fig. 6.22a, \ve have to consider the deflections
of the bar chain sho\vn in Fig. 6.250. 'I'he dcflections due to changes in thc
angles between the bars of this chain can be calculatcd in exactly the same
manner as in the case of a straight chord. To show this, let us assume that
the angle bet\\.reen the t\VO chain mcrnbers at a joint 111 diminishes by the
amount A8m while the other angles remain unchanged. To find the corre-
sponding deflections, \ve assume first that the part A111 of the chain rcmans
immovable, \\rhile the part 111B rotates \vith respect to the hinge 1n and the end
B describes a small arc r !l8rn, the vertical projection of \\,hich is equal to
/2 A8m. After this rotation, the vertical deflections of the joints to the left
of 111 are zero, \vhile the joints to the right of 111 have up\vard deflections,
x - II
y = -/ 2 68m -r:;--
(e)
To satisfy the condition of zero deflection at the right support B, \ve now rotate
the bar chain as a rigid body \vith respect to joint A by such an amount that
the joint B makcs a vertical displacement equal to /2 8m' The corresponding
vertical displacements of other joints are then obtained from the equation
x
y = 1 2 A8m I
(f)
Superposing these deflections on the dcflections previously found, \\'C conclude
that expression (I) gives the total vertical dcflections for the part A11I of the
bar chain, while for the part 11lB the final deflections are
x-II ..t / - x
y = -/268m 1 2 + 1 2 8m r = A8m III - (g)
Expressions (f) and (g) coincide \vith those previously found for straight line
chords (see page 286). Hence, the vertical deflections of the joints of a bar
chain, due to changes in the anglcs bct\vccn the bars, can be found, as before,
by calculating the bending moments produced in a simply supported heam by
fictitious loads 8m'
In addition to the deflections produced by the changes dUrn in the angles,
there will also be deflections due to changes in the lengths of the bars, Refer-
ring to I"'ig. 6.250, lee us considcr the bar i and assume that it makes an angle <1>,'
\\rith the horizontal and has an elongation l i. lhen, if \\'e consider the left
portion of the bar chain as immovabJe, the right portion, due to the cJongation
I i, obtains a vertical displacement equal to
y = - /. sin At..
. ''I'I
(11)
To bring the joint B back to the Jevel A B, \\'c no\v rotate the chain as a rigid
body \vith respect to the joint A by an angle / i sin CP./ I. Then, the final

ART) CLE 6.5
291
 ll
A
I

1'1
(a)
x
r-
l, sin A.,
I '1",
1--
---
---
---
---
--- ""- --.. - .......
(b)
-.
l, sin"'.
, '1",
---L
---
..L.
M( !
T tan t/>i _
t 4? ta n t/>i
..Ji..
(c)
ul
FIG. 6.25
deflection for the left portion of the chain, due to the elongation f)./ it is
I . x
'Y =  . sin . -
. · . I
(i)
For the right portion, cOlnbining the displacement (h) \\lith the displacemcnt
(i), \ve find
I . .t' / ' / ' I - x
Y =  ' S in . -- - A . sin . = - . S in "", --
. .. a I '& '1'. '& '/"1 I
l'hesc deflections are sho\\'n in Fig. 6.25 h. The shaded area in this figure is
identical \vith the bendinb-moment diagram for the fictitious loads sho\vn in
Fig. 6.25c. Hence, the additional deflcctions, due to elongations of the chain
bars, can be obtained by calculating the bending momcnts produced in a simply
supported beam by the fictitious vertical loads
AI.. tan 4>t' .1h tan tP.. () ' )
and
Ii li
applied at the ends of each bar. In these expressions /i is taken positive
\vhen it reprcsents an elongation, and CPa' is positive if the bar; is rotated in a
countcrclock\vise direction \vith respect to the axis AB as \\'e pass along the
bar chain from A to B.
We may ndw return to the example in Fig. 6.24 and proceed \vith the calcula-
tion of the vertical deflections of the upper..chord joints. We have to consider

292 DEFLECTION OF PIN-JOINTED TRUSSES
here the inclincd bars 1 and 11 \vhich, together \\,ith the t\VO horizontal bars 4
and 8, form the bar chain attached to the supports A and B. Using the
numcrical data fron1 Table 6.3 and Eqs. (6.4), \VC calculate the changes in thc
anglcs along the upper chord exactly as \vas done bcforc for the lo\\'er chord.
Thcse changes E Aa2, E a5, E Aa3, E Llah E a9, E Aa7, and E AalO, com-
puted from Eqs. (6.4), are rccorded undcr 111, p, s in the table. Thc corre-
sponding fictitious loads for the joints 111, p, and s are then obtained by sunl-
mation as sho\vn.
To take care of the changes in the lcngths of the chain bars, \\'c use expres-
sions (j) for bars 1 and 1 I and obtain, for the rnagnitudes of additional fictitious
loads at 111 and s,
E / 1 _ _
---- - (Tl - 3,090
'.
and
E Ll/ ll
0._- ---- = 0'11 = 6, 180
/n
(k)
Adding these fictitious loads to those already obtained in Table 6.3, \ve get the
final loading of the conjugate beam as sho\vn in Fig. 6.24c. "fhcn, the required
vertical deflections of the joints 111, p, and .f, obtained as fictitious bending
moments in the conjugate beam, arc
25,410 .
Om = 30(10 6 ) 60 = 0.051 In.
25,410 6,110 .
op = 30(10 6 ) 180 - 3 0(10 6 Y 120 = 0.128 In.
 38,610 6 7 .
u. = 3 0(10 6 ) 0 = 0.0 7 In.
PROBLEMS
1 Using the method of fictitious loads, find the vertical deflection of joint C of the
sinlplc truss supported and loaded as sho\vn in Fig. 6.26. Each bar has cross-
sectional area Ai = I in. 2 and tnodulus of elasticity E = 30( 1 06) psi.
Al1S. oc = 0.067 in.
FIG. 6.26
A B
.lOO..-i-100" lOO..-llOO......j
10 kips

ARTICLE 6.5
293
30 kips
100" 1100" 1100"1100" 1100"
30 kips
FIG. 6.27
D
FIG. 6.28
p
p
p
p
p
100 kips
E
I
30'
I
----------------
40' 40' 40' 40'
40'
A
1-10'+ 10' 10'--j
P
FI G. 6.29
FI G. 6.30
2 Using the nlcthod of fictitious loads, find the vertical deflections of joints C and D
of the simple truss supported and loaded as sho\vn in Fig. 6.27. Each bar has
cross-sectional area Ai = 1 in. 2 and modulus of elasticity E = 30(10 6 ) psi.
Ans. On = Oc = 0.20 in., up.
3 The truss sho\vn in Fig. 6.28 is symnlctrically loaded \vith respect to the nliddlc
vertical CD so that each active IncJl1bcr has a unit axial strain of 0.00 I. Using
the mcthod of fictitious loads, calculate the vertical deflection of joint C.
Al1S. oc = 1.800 in.
4 Each har of the simple truss sho\vn in Fig. 6.29 is 10 ft long, 3 in. 2 in' cross sect:on,
and has a modulus of elasticity E = 30( 1 06) psi. Using the mcthod of fictitious
loads, find the vertical deflections of the lo\ver-chord joints (' and [) if the applied
load P = 30 kips.
A,lS. Oc = 0.0785 in" On = 0,1215 in.
5 Each bar of the truss shown in Fig. 6.30 has a cross-sectional area Ai = 5 in. 2 and
nlodulus of elasticity f; = 30(1{)6) psi. Using the method of fictitious loads, find
the vertical deflections of the lo\ver-chord joints C, D, and E.
Alls. c = 1.53 jn., On = 3.28 in., OJ.: = 2.60 in.'

Chapter 7
Statically indeterminate
pin-jointed trusses
7.1 GENERAL CONSIDERATIONS
In Chap. 2 \ve have already scen that any pin-jointed truss in one plane is
gcnerally rigid and stati{,(ll/y deteT1llillate if its j joints are interconnectcd bct\veen
thcmsclves and the foundation by 111 = 2j bars. If the bars exceed this
number, the available 2j simultaneous equations of statics are insufficient to
dctennine the internal forces, and the truss is said to bc staticall)' indeteT111il1ate.
Those bars in excess of the 2j which arc both necessary and sufficient for the
rigidity of a truss are usually called redundal1t 111nnhers. A sinlple example
of a truss \vith one rcdundant member is sho\vn in Fig. 7.1a. Here \ve have a
single joint C attached to a foundation by three bars arranged in one plane as
sho\vn. For simplicity, we assume the t\VO inclincd bars to be identical so
that the system is symmetrical \vith respect to thc vertical axis CD. If a
vertical load P is applied at C as sho\vn, some axial force \\rill be induced in
cach of the three bars; but since, for the joint C, \ve have only two equations of
equilibrium
294

ARTICLE 7.1
295
-,Y = 0
and
2;}1' = 0
(a)
there is evidently one redundant element, and the system is statically inde-
terminate. That is, so long as \ve consider the bars to be absolutely rigid, \ve
can assume any value X for the tcnsion in the vertical bar (Fig. 7. I b); and
then, using Eqs. (a), \\'c shall find that each inclined bar carries a tensile force
s = -.(l) - A) scc a
(b)
To find the true value of X, \ve must no\\' consider thc clastic deformations of
the bars. Denoting by 0 the elongation of the middle bar, \ve sce from Fig.
7. I b that the corresponding elongation of each inclined bar must bel
0' = 0 cos a
(c)
No\v, by using expression (b) for the axial force in an inclined bar and denoting
by A and A () the cross-sectional areas of the bars as sho\\'n in the figure,
condition (c) may be \vritten in the follo\ving form,
(P - )( )/ Xl
2 cos 3 aAE - AoE
froIT, \\,hich
1)
Y =
1 + 2(A/Ao) cos 3 a
Using this value of X in Eq. (b), we can no\v find thc corresponding value of
S for each inclined bar, and the analysis is complcted.
Revic\ving the foregoing procedure, \VC see that to solve the givcn statically
indeterminate problem, we first removed the rcdundant bar CD and replaced it
by the force X \vhich it excrted on the remaining statically determinate system.
Then this system (Fig. 7.1 b) was analyzed by the ordinary equations of statics
I This relation assurTICS that the elastic elongations of the bars are snlall conlpared ''lith thc
overall dimensions of the structurc.
A
B
A
B
FIG. 7.1
p
(a)

296 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
and all axial forces expressed in tcnns of the unkno\vn quantity X. Finally,
to find the true value of X, \VC established an additional equation (c) by taking
account of the clastic deformations of the bars.
The same procedure can be applied in more con1plicatcd cases. \\'c begin
with a proper selection of the members that \vill be considered as redund 4 nts.
Rcmoving these bars, \VC obtain a statically detcrminatc system, called:" the
prhllary systC'J11. This primary system can then be analyzed by the Jncthods
alrcady discussed in prcceding chapters, and \VC can rcadily find thc forces in
all bars and also the displacements of all joints. V\'e shall be especially
interested in the changes in distance bet\veen those joints corresponding to the
removed redundant members; for evidently these changes must be equal to the
changes in length of the corrcsponding redundant bars in the actual truss. In
this \\Tay \\'e obtain as many equations, similar to Eq. (f), as therc arc redundant
bars; and, fronl these equations, thc forces in thc redundant bars arc obtained.
The forces in the rcmaining bars are then found from statics.
From the discussion \ve see that the redundant bars must be so sclccted that
after their rernoval \ve obtain a rigid statically detcrminate truss. "rake, for
example, the statically indeterJninate system sho\vn in f'ig. 7.2. The truss
propcr has 14 bars, and in addition to this \\'c have to consider a vertical sup-
porting bar rcplacing the movable support at the right cnd of the truss and t\VO
bars replacing the immovable support at the left end. All together, then, \ve
have 17 unkno\vn forces, \vhile the doubled number of joints is 16. 1'hus, \ve
have a truss \vith onc rcdundant member. As a rcdundant mcmber \ve can
select, for cxan1plc, one of the diagonals in the middle panel of the truss. But
\ve cannot take the diagonal in onc of the other panels as a rcdundant bar, for
after removing such a diagonal \\'e obtain a nonrigid system.
In Fig. 7.3 \ve have another example of a truss \vith one rcdundant tl1embcr.
As the rcdundant membcr in this case, \ve can take anyone of the supporting
bars; but such a bar as DE cannot be selected, for after its removal \\'C obtain a
system having a critical form. "rhat is, \VC have the t\VO rigid shaded portions
supported at A and B and joined together by a hinge C like a threc-hinged arch;
but since the hinge C lies on the straight linc AB, it can have considerable
vertical mOVClncnt \vithout apprcciable changes in lcngth of any of the bars.
p
D
E
G
FIG. 7.2
FIG. 7.3

ARTICLE 7.2
297
D
A
p P
FI G. 7.4 B
There are, ho\vever, other possibilitics besides the rerTIoval of onc of the sup-
porting bars. For cxanlple, if \\'c remove thc bar CF, \ve obtain, as our
primary system, an ordinary three-hinged arch in \vhich thc hinge E does not
lie on the straight line AR. Hence, thcrc is no critical form in such case, and
the bar (1[/ can be chosen as thc rcdundant oar, if desired.
Figurc 7.4 represents anothcr systen1 \vith one redundant bar. As the
rcdundant in this case, \VC can takc the bar /IB or thc bar CD; but \VC cannot
sclect the bar CE, for after its renl0val we obtain a rigid frame ACF attached
to the foundation hy three bars AB, CD, and EF that intersect in one point.
Such a systcm, as \ve already kno\v (sce page 58), is not rigid.
7.2 TRUSSES WITH ONE REDUNDANT ELEMENT
As a first exanlple of a system \\'ith one rcdundant element, let us considcr the
truss on three supports as sho\vn in Fig. 7.5a. As the redundant element, \\'c
select the intermediate support C. Rernoving this support, \VC obtain a
statically dctcrminatc simple truss (Fig. 7.5 b) that can be rcadily analyzed.
Thcn the redundant reaction )( at the n1iddle support \viIl be dctermined from
the condition that the deflection of joint ("' due to the combined action of the
loads I't, P2, P3 and the rcaction )( rnust vanish. Using expression (6.2) for
this deflection, \\'e obtain
2; I i s  = 0
(a)
in \vhich l:1/ i = Sil if A iF; denotcs the change in length of any bar i due to the
loads Ph P2, P3 and the redundant reaction -,Y, and s is obtained by using the
unit loading 1 sho\vn in };ig. 7.5(; and represents the ratio of the force in any
bar j to the unit load at C that produces it. In calculating the values l i in
expression (a), "rc note that the force S.. in a bar i is given by the equation
Si = S; + s-,y (b)
1 rrhc unit load at C is taken in the up\vard direction, so \VC may consider the redundant
reaction X as positive in this direction.

298 STATICALLY INDETERMINATE PIN-JOINTED TRUSSES
in \vhich S is the force in any bar i produced by the kno\vn externallods Pt,
p 2, P 3, acting on the simple truss sho\\'n in Fig. 7.5 b, and s:.X is thc l force
produced in the saIne bar by the reaction Y. Then the requircd elongation of
any bar i is
Ai. = S  + s.tY l = ( S + lAT\ p .
t A .E l' · J 1
,
Ii
Pi = /l iE
(c)
where
(d)
Substituting expression (c) into Eq. (a), 'NC obtain
};(S + SX)PiS = 0
from \vhich
(e)
' S ' ,
6.J is iPi
Y = - 2; ( S;')2 Pi
VV ith the va]ue of the redundant reaction .t¥, the forces in the bars of the given
staticaJly indcterminatc truss (Fig. 7.Sa) are obtained from Eq. (b), and the
analysis is completed. T t is seen that the statically indeternlinate problem
(Fig. 7. Sa) is rcduced to the t\\'O statically dctcrnlinatc problerns sho\\'n in
Fig. 7.5 band c. The values S:, s, and Pi for all members of the truss can be
(7 .1 )
Pi
P2
P3
a)
x
PI
P2
P3
(b)
FIG. 7.5 (c)
1

ARTICLE 7.2
299
put in tabular fornl as \vas donc in 'lable 6.1, page 259. "]'hcn, \\Iith the aid
of such a table, the values of the SUlllS entering in Eq. (7.1) can be readily
calculated.
i\s a second example, let us consider the statically indeternlinate truss sho\\'n
in Fig. 7.6ll. The reactions in this case can be readily calculated from equations
of statics, hut the forces in the hars cannot be determined by statics alone
because the tfUSS contains one redundant member. L.et us take the vertical
bar BC"' as the redundant bar. '[hen, after relnoval of this bar, \ve obtain a
statically detcrminate simple truss (Fig. 7.6b) on \vhich, in addition to the
kno\vn loads 1\, 1)2., P"" there \viH act the t\va equal and opposite forces X
replacing the removed bar as sho\vn. '[he magnitude of )( is fotind froo1 the
condition that the change in the distance bet\vecn the joints Band C of the
statically dcterminate truss (Fig. 7.6b) must be equal to the change in length
of the vertical bar BC of the actual truss (Fig. 7.6(1). "[his change in the
distance BC'is k f1/ i s, in \vhich the SUffiIl1ation includes all bars of the statically
determinate truss (Fig. 7.6b) and the values of 'i are given by the equation
li = (S; + SA}Pi
(f)
in \vhich S denotes the force produced in any bar i by the given loads PI, P2, P3
and s is found, as before, by using the unit loading sho\vn in Fig. 7.6c.
PI
P2
P3
(a)
PI
P2
P 3
(b)
: r
FI G. 7.6 (c)

300 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
j kip 1
1
1
1
1
1
FIG. 7.7
1
j kip
B 24'
_ _ _ _ _ _ _ _ _ . _1_ , _ _ _ _ _ _ _' _
x X
-144' ----
The equation for calculating X then becomes
}; (S: + s;\') PiS = - Xpo
10
Po =-
AoE
(g)
(h)
\vhere
is the quantity defining the extensibility of the redundant bar BC. 1he minus
sign in the right-hand side of Eq. (g) f01l0\\'S from the directions of the unit
loads in Fig. 7.6c. With these directions, shortening of the distance BC
becomes positive, \vhich corresponds to compression of the vertical bar BC.
Solving Eq. (g) for X, \ve obtain
X "1;Ss Pi
_ t t
- - (s) 2 Pi + p'
(7.2)
In this example y.'e see again that calculation of the redundant force X is
reduced to two statically determinate problems, (1) the calculation of the
forces S and (2) the calculation of the forces .<. ""'hen all these forces are
found and the quantities Pi and po are dctermincd from the given dimensions of
the tcuss, thc sums entcring into ]q. (7.2) can be readily evaluated.
In each of the foregoing examples, \\7C used expression (6.2) as a basis of
evaluation of displacements. Ho\vever, \ve can come to the same results by
using the theorcn1 of lcast \vork as discussed in Art. 5.6. On this basis, for
example, the valuc X for the redundant reaction at C in Fig. 7.5 must be such as
to make the total strain encrgy of the statically indetcrn1inate system a
minimum. If \ve use expression (b) for the forces in the bars, the total strain
energy of the truss becomes
L (S + sX)2/i
U = ------..---.
2A i R
Equating to zero the derivative of this expression \\,ith respect to }( and using
the notation in expression (d), \VC obtain
2;(S: + SL\JPiS = 0

ARTICLE 7.2
301
\vhich coincides v.,ith our previous exprcssion (e) and leads to Eq. (7.]). In
the same \vay the total strain energy of the statically indetcrminate truss in
Fig. 7.6 becomes
u =  (S -t _ s l X)2/ i + X /o
 2A ,"£ 2A olJ
\vhcre the summation includes all bars of the primary system (Fig. 7 .6b).
Equating to zcro the derivative of this expression v.,ith respect to X and using
the notations in exprcssions (d) and (11), we obtain the previous exprcssion
(K), leading to Eq. (7.2).
As a specific example \\lith numerical data and all calculations, let us considcr
the t\vo-hingcd arch loaded as sho\vn in I'ig. 7.7 and calculate the magnitude X
of the horizontal thrust. 'I'he overall dimcnsions of the structure arc sho\vn
in the figure, and the lengths of the individual bars arc given in colulnn 2 of
'Etbic 7.1. I-iere also are given, for each bar, the cross-sectional arca A i, the
TABLE 7.1
It, Ai, Pi X lOG, S;, , (S) 2 S" X 6 ,
I in. 2 in./lb Jb Si is iPi 10 (S..)2PiX 10 6
In.
(I) I (2) (3) (4) (5) I (6) I (7) (8) (<;)
,
I 324 11.5 0.940 - 4,000 +0.59 0.35 - 2,220 0.329
2 227 11.8 0.641 - 3,000 +0.50 0.25 -962 O. 160
3 292 9.0 1 .081 + 3,900 -0.65 0.42 -2,740 o 454-
4- 259 11.5 0.750 0 -1. 16 I. 35 0 1 012
5 263 9.0 0.974 - 3,800 +0.47 0.22 - 1 ,740 0.214
6 220 11 ,8 0.621 - 6,200 +1.00 1.00 - 3,850 0.621
7 277 9.0 1 .02f) + 3,950 -0.64 0.41 - 2,600 0.421
8 234 11.. 0.678 + 3, I 50 -1.60 2.56 - 3,420 1.735
9 216 9.0 0.800 -2,750 +0.30 0.09 -660 0.072
10 216 11.8 0.610 - 8,600 +1.43 2.04 - 7,500 t .244
11 270 9.0 1.000 + 3,200 -0.55 0.30 -1,760 0.300
12 223 11. 5 0.646 +6,200 -2.05 4.20 - 8,21 0 2.712
13 192 9.0 0.711 - 1,150 +0.03 0.00 -24 0
14 216 11.8 0.610 - 9,700 +1.60 2.56 -9,470 1.560
15 277 9.0 1.026 + 1 ,400 -0.23 0.05 -330 0.051
16 216 11 .5 0.626 + 8,600 -2.43 5,90 -13,080 3.694
-117,132 I 29.158
17 180 9.0 0.667 -250 -0.14 0.02 +23 0.013
1;
- II 7, 109 i + 29 . I 7 1

302 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
quantity Pi, assuming E = 30(10 6 ) psi, and the axial forces S: and s. *
U sing the data from the table and forming the sums appearing in Eq. (7.1),
we find
x = - - 11 7 , 1 09 = 4 015 Ib
29.171 '
PROBLEMS
1 Find the horiwntal thrust X for the statically indeterminate truss sho\vn in Fig. 7.8a.
All dimensions are given in the figure, and the cross-sectional areas of the bars arc
as follo\vs: Ai = A.. = 5 in. 2 , A 2 = Ao = 3 in. 2 , and A3 = 2 in. 2
Al1S. X = (540.9/616.5)P.
2 l)etermine the axial force in the redundant horizontal bar of the truss sho\vn in
Fig. 7.8b if the cross-sectional area of this bar is A 6. Assume that the other bars
have the same dimensions as in the preceding problem.
Ans. X = .}40.9P/(300/A 6 + 616.5).
3 Determine the axial forces in all bars of tht. statically indeterminate systen1 sho\vn
in Fig. 7.90, assun1ing that the cross-sectional areas of all bars are equal.
/1 in! Take the axial force X in the diagonal bar () as the redundant
element.
Al1S. X = 0.854P.
* These axial forces can be found \virhout difficulty by constructing Max\vcll diagrams for
the statically dctenninate primary system obtained by removing the horizontal constraint
at B.
p
p
150"
150"
150"
150"
-----------1--
3 50"
--- 4 ------ t--
5 50"
_i__
6
(a)
(b)
FIG. 7.8
2
2
1
3
p
4
FIG. 7.9
(a)
(b)

ARTICLE 7.3
303
20 kips
r- 5'T- 5'T-5'
A
5'1- 5 ' -1
B
FIG. 7.10
c
D
4 Determine the rnagnitudc of the reactive force ,x: in Fig. 7 .5a if PI = P'l = Pa = 10
kips. The horizontal and vertical rncrllbcrs of the truss all have cross-sectional
areas Ai = I in.\ \vhilc the diagonal nlcmbers have Ai = 0 in. 2 .AlI bars have
the same n10dulus of elasticity, and the panels arc square.
Al1S. X = - 19.23 kips, up.
S All bars of the statically indeterminate truss sho\\'n in Fig. 7.10 have the saIne
cross-sectional area Ai = 1 in. 2 and the same Inodulus of elasticity E. Find the
axia] force S:r induced in the redundant bar C/J due to the 20-kip load acting as
sho\vn.
Alls. S = + 14,500 lb.
7.3 TRUSSES WITH SEVERAL REDUNDANT MEMBERS
rhe mcthod used in the preceding article for the analysis of statically inde-
tcrminate trusses \vith one redundant member can be rcadily extcnded to
trusses \vith t\VO or more redundant members. As a first cxample, let us
consider the simple truss \\'ith t\ro redundant supports as sho\vn in "'ig. 7.11n.
Removing these supports and replacing thcm by the rcactive forces X and Yas
sho\\rn in the figure, \VC obtain a statically dcterminate systelll on \vhich, in
addition to the given loads Ph P2, P3, the t\VO redundant reactions X and Yarc
acting. The rnagnitudes of these rcactions \vill no\\' be found from the con-
ditions that the deflections of the joints (..' and D n1ust vanish: For these
deflections, \VC use Eq. (6.2), \\Thich gives in this casc the folIo\ving equations:
 l i s = 0
2; J i s/ = 0
in \vhich the quantities s and s' arc obtaincd by using the unit loadings sho\vn
in Fig. 7.11 band c, respectively. The quantities li arc the actual elongations
of the bars of the system (Fig. 7.11a) produced by the givcn loads Pi, P2, P3
and the unkno\vn reactions }( and 1": The force in any bar i of the primary
system (Fig. 7.11 b) due to the given loads 1\, /)2, P3 is denoted, as beforc, by
S;. 'rhe forces produced in the same bars by the redundant rcactions X and Y
are obtained by using the unit loads sho\vn in Fig. 7.] 1 hand c and are equal to
(a)

304 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
(a)
:eNVyvV\/\
(b)
T/V\/\/\
(c)
FIG. 7.11
sx and .<' J respcctively. Hence, the total force in any bar i is
Si =S: + S;A + s' Y (b)
and the corresponding elongation is
/i = (S + sX + S'Y)Pi
Substituting into Eqs. (a), \VC obtain the follo\ving t\VO equations for cal-
culating ..tY and Y:
2; (S + sX + s' Y)pi.< = 0
''' (S ' + ' "(.. + " Y) " 0
 L i SiA. Si PiS.. = .
X}; (s) 2 pi + Y2;ss' Pi = - 2;S:S:Pi
XSS'Pi + Y};(S;')2Pi = -S:S'Pi
(c)
or
(d)
Having the values of S;, s, s/, and Pi tabulated, \VC can rcadily cvaluatc the
sums that appcar in Eqs. (d) and solve the equations for X and 1': l'hen upon
substituting these values in Eq. (h), \ve can calculatc the forces in all bars of
the truss. It is secn from this discussion that the analysis of a system \\'ith t\VO
redundant supports reduces to the analysis of three statically determinate
problems, i.e., to the evaluation of the forccs S;, j', and s'.
Equations (c) for calculating the rcdundant rcactions can be obtained also by
using the theorem of Icast work. The total strain energy of the system sho\vn

ARTICLE 7.3
305
in Fig. 7.11 a is
U = i1; (S + s;x + s;' Y) 2 Pi
.Equating to zero the dcrivatives of this expression \vith respect to ,LY and 1
\VC obtain Eqs. (c).
As a second example let us consider the statically indetcnninate truss sho\\fn
in Fig. 7 .12a. This truss has 83 bars. Adding to this number 6 clements of
constraint for the system of supports, \VC have all together 89 unkno\vn forces;
t\vicc the numbcr of joints is 86. l-Ience, thcrc arc three redundant members.
As these redundants, \ve sclect the t\VO horizontal bars ab and cd and the
horizontal constraint' at the support D. Renl0ving these three redundant
elements and replacing thcir actions by forces }{, J and Z, \ve obtain a statically
(a)
(b)
(c)
(d)
(e)
FIG. 7.12
'1
I
'1
1

306 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
determinate system consisting of the three simple trusses sho\vn in :Fig. 7.12b.
The forces prodlced in the membcrs of this statically detcrminate system by the
given Joads PI, P2, Pa \ve denote by S;.. The forces produced in the members
of the same systenl by the redundant forces X, 1 and Z are found, as before,
by using the unit loadings sho\vn in Fig. 7.12c, d, and e. -rhe corresponding
axial forces in any bar i \\'c denote, respectively, by s, s', and S". Then, the
total force in a bar i of the given system (Fig. 7.12a) is
S S ' + ' ,r + "1..:" + II' Z
; =. S.,,\. S'.l. r.
. , \ I ",
(e)4
The values of the thrce rcdundant forces X,  and Z are obtained, no\v, from
the conditions that the changes in distance bct\veen the joints a and b and the
joints c and d (Fig. 7 .12b) must be equal to the changes in length of the bars ab
and cd (Fig. 7 .12a), \v hile the horizontal displacement of the support at D
must vanish. [)cnoting by px and Pu the extcnsibility factors for the redundant
bars ah and cd and using Eq. (6.2) for displacements, \ve can rcprescnt the
foregoing conditions in the fornl of equations as follo\\'s:
 (5; + sX + s' Y' + s" Z) PiS = - )(Pr
(S + s}{ + <' y" + <" Z)PiS' = - Jpu
'"' (S ' + ' v + " 1 1' + ", 'l\ . II' - 0
 i J i \ S i S i LJ I P, S i -
(f)
The minus signs in the first t\VO equations result from the fact that, for the
dircctions of the unit loads in Fig. 7.12,' and d, positive displacements cor-
respond to shortening of the distances ab and cd and hence to comprsivc
forces in the redundant bars ab and cd. If the values Pi, s, S/, s" for all bars
of the primary system (Fig. 7.12b) are tabulated, the sums appcaring in Eqs.
(I) can be evaluatcd, and \ve obtain three lincar equations, \vith numcrical
coefficients, that can be solved for the unkno\vn redundant forces X, r: and Z.
From the foregoing discussion it may be seen that as the numbcr of redundant
forces increases, the number of cquations required for. their dctermination
incrcases also, and the problem of analysis beCOInes nlore and morc involved.
Somctimcs \VC find cases \\There each cquation contains only a fe\v of the
redundant forces; such equations can be solved \vithout n1uch difficulty evcn
though the total numbcr ofunkno\vns may be large. i\S an example of this
kind, lct us consider the truss \vith five rcdundant bars as sho\vn in Fig. 7.] 30.
Sclecting onc diagonal from each panel as a redundant bar and representing the
actions of these bars on the remainder of the truss by redundant forces X, 1':
z, . . . , \ve obtain as the primary system a statically determinate simplc
truss loaded as sho\vn in Fig. 7.13/1. 1'his systcm can be readily analyzed, and
the forcc in any bar i \\fill be
S S ' + ' X + " y "' Z
i = i Si Si + Si +

ARTICLE 7.3
307
"'[he quantity S; in this expression denotes the force produced in any bar i of
the prin1ary sYSteo1 (Fig. 7.13b) by the givcn loads PI, P2, Pa, \\'hile the
.. ,,, II' b . d b r b . h . 1 d .
quantltlcs Si, Si ,Si , . . . are 0 talnc , as Clore, Y uSIng sue unit oa Ings
as thosc sho\vn in Fig. 7.1 3c and d. Considering these loadings, '.ve see that
in each c3:se the applied unit loads produce axial forces only in those bars \vhich
comprise the corresponding panel. "'rhus, in Fig. 7 .13c we conclude that the
axial forces s are different from zero only for the bars of the first panel as
sho\vn by heavy lines. Likc\\'ise, in Fig. 7.1 3d, the forces S' vanish for all
members except those indicated by heavy lines, etc. With the foregoing
observations in mind, lct us no\v consider the equations for determining the
redundant forces. "1'hc first of these cquations \vil) be \vritten on the basis of
the fact that the change in the distance bct\vcen joints A and B of the simple
truss (Fig. 7.] 3b) is equal to the change in length of the rcdundant diagonal A B
in the actual truss (Fig. 7.] 3a). This change in the distance AB obviously
depends only on the deformations of the bars \virhin the first panel of the truss,
Pi P2 Pg
f
h
(a) 
 I - ,
PI P 2 P3
(b)
(c)
FIG. 7.13
(d)

308 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
and the forces in these bars depend only on the givcn loads PI, P2, Pa and the
redundant forces X and 1 The remaining redundant forces do not affect the
bars of the first panel and need not be considcred. Hencc', the first equation
becomes
i=5
)' (5: + sX + s'Y)pjs:, = -XPz
'-'
i;::l
(g)
This equation contains only t\VO unkno\vns, X and } and the summation
includes only the five bars of the first panel. For the remaining bars, as
already mentioned, s vanishes.
The second equation is obtained by equating the change in thc distance
bct\veen joints (' and D (Fig. 7.1 3 b) to the change in length of the diagonal
CD in the complete truss (Fig. 7.1 3a). This change in the distance CD
depends only on the deformations of the bars \vithin the second panel of the
truss (Fig. 7.13/;), and these deformations in turn depend only on the lateral
loads PI, 12, F'3 and on the redundant force Y in the second panel, together \\,ith
the redundant forces X and Z in the adjacent panels. lIenee, the second
equation becomes
 (S ' + 'v + " y + ''' Z) '' } 7
.&J i Si"'\. Si Si -J PiSi = - Pu
(h)
\vhere the summation includes all bars of the second panel. This equation
contains three unkno\vns, X,  and Z.
Similar equations can be \\'rittcn for the third and fourth panels. The fifth
equation \vill again have only t\\'o unkno\\!ns, as Eq. (g). lhus, \\.'c obtain
a system of five equations, t\\'o of \vhich have t\VO unkno\vns each, \vhile the
rcmaining three have three unknowns each. Such a systcm of equations can be
rcadily solved. 'rhcy arc similar to three-1J1fnl/el1t equations as uscd in the
analysis of continuous hearns, and the same mcthods of solution can be applied.
A method of successive approximations is somcrin1es uscfuJ in the analysis of
a statically indeterminate systcm \vith n1any redundant elements. For
example, to obtain a first approximation XI, }, £1, . . . for the unkno\vn
forces X, t: Z, . . . in the system above, we assume that the .shear in each
panel is equally divided between the t\VO diagonals of that pane1. Then,
conling to Eqs. (g), (h), . . . and noting that the most important tcrms are
those having for coefficients the summations of squares like 2;(s) 2,  (s') 2,
. . . , \ve replace the unkno\vns X, 1: Z, . . . in the remaining tcrms by their
first approximations Xl, 11, Zl, . . . and obtain in this \vay the equations
}:S;JPi + [«) 2 pi + p] X + };ss' Pi Yi = 0
!:S;s' Pi + ss' PiX l + r (s') 2 pi + Pu] Y + 2;s' s" PiZl = 0 (i)
. ..... ............. ..............
Each of these cquations contains only one unkno\vn and can be easily solvcd.

ARTICLE 7.3
309
In this manner, \ve obtain a second approxilnation X 2, Y2, 2 2 , . . . for the
unkno\vn forces. Then, upon putting these nc\v values, instcad of Xl, lt,
Z 1, . . . , in Eqs. (i) and solving again for ,,\r, 1 Z, . . . , \\'e can calculate
a further approximation for the unkno\vns. Usually this process convcrges
fairly rapidly so that the third approximation \vill be satisfactory for all
practical purposes.
PROBLEMS
1 Evaluate the redundant reactions at the intenncdiate supports C and D of the truss
shon'n in Fig. 7.14 if thc load P = 20,000 lb. i\11 hars of the truss are identical.
Ans. Xc = 17,850 Ib, .Iy d = 3,600 lb.
2 Solve the preceding problcn\ if the rollcr supports at C and f) are replaced by
vertical bars such as those of \vhich the truss itself is con\priscd.
A1Js. Xc = 16,950 lb, .If D = 4,20') lb.
3 Find the Inagnitude of the vertical reaction X at the support G of the statically
indeterminate truss loaded as sho\\'n in Fig. 7.15. -[he din1cnsions of the truss arc
sho\vn in the figure, and each bar has cross-sectional area At = 2 in. 2 and modulus
of clasticity E = 30(10 6 ) psi.
Al1S. X = 35.8 kips.
4 Solve the precding problem if the support G settles 0.3 in. during loading of the
truss.
Ans. X = 21.0 kips.
5 Itakc a complete analysis of the statically indeterminate truss sho\vn in Fig. 7.1 3
if.P 1 = 1'2 = P3 = 20 kips, I = 45 ft, and 11 = 12 ft. Assume that each chord
member has a cross-sectional area of 3 ;n. 2 , each diagonal 1.5 in. 2 , and each vertical
1 in. 2 .J\11 bars have the same Inodulus of clasticity E.
Al1S. X = -17.7 kips, Y = -16.1 kips, Z = -6.5 kips,
- + 7.1 kips, . . . = + 16.1 kips.
p
FIG. 7.14
A
e
D
B
FIG. 7.15
10' --+-10'-+-10'-+-10'
A IB Ie ID
20 kips
E

310 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
7.4 ASSEMBLY AND THERMAL STRESSES IN STATICALLY
INDETERMINATE TRUSSES
From the preliminary discussion of statically indeterminate trusses given in
Art. 7.1, it is evident that the lengths of redundant bars must satisfy certain
geonlctrical requirenlcnts. The shape of a truss and the mutual distances of
all its joints arc entirely defined by the lengths of the nlembers of the pri111ary
SYS[e-Ill. Hcnce, a redundant member can be freely placed bet\veen t\VO joints
only if its lcngth is exactly equal to the distance bet\veen these joints. Other-
wise the bar \vill not fit its propcr place and can be forced into the primary
system only by applying some initial extension or cornpression. Under such
conditions, it is eVJdent that some axial forces, usually called asselJ1bly lorces,
\\,ill be induced in the bars during the formation of the truss. These internal
forces exist even in the absence of external loads and must be superimposed on
the axial forces produced by external loading in order to get the total force in
each bar.
Take, as a simple example, the statically indeterminate systenl sho\vn in
Fig. 7.1 £1, and assume that in consequence of srnall errors in the lengths of the
inclined bars the distance bet\vcen the hinges C and I), measured after assembly
of the inclined bars, is larger by an amount  than the lcngth of the vertical bar.
Then, to put this latter bar in place, \ve must some\vhat extend it. Thus, after
assembly it \vilI pull up on the joint C and induce compression in the inc]incd
bars \vhile remaining under some tension itself. If Y is the magnitude of the
tensile force in the vertical bar, after assembly, the compressive forcs in the
inclined bars \viIl be equal to X/2 cos a, and the corresponding up\vard dis-
placement of the joint C \viII be XI/2AE cos 3 a. The magnitude of the force
Y is no\v found from the condition that the vertical displacemcnt of the joint C
together \\,ith the elongation of thc vertical bar must be equal to the initial
discrepancy  in length. Hence,
Xl Xl
---- + -- = 
2AE cos 3 ex AoE
where A 0 is the cross-sectional arca of the vertical bar and A the cross-
sectional arca of an inclined bar. From this equation, \VC find
x = . AE __
l(i scc 3 ex + A/Ao)
Similar rcasoning can be applied in more complicated cases of statically
indeterminate trusses. Let us consider, for exalnple, the truss in Fig. 7.6a and
find the assernbJy forces that \\fill result from inaccuracies in the Icngths of the
bars. I .let i be the error in length of any bar i, considered positive if the bar
is too long, and .10 the error in length of the rcdundant bar BC. lhe effect of

ARTICLE 7.4
311
the errors i on the distance bet\vecn the joints Band C of thc primary systcm
(Fig. 7.6b) is found by using Eq. (6.3), from \vhich it follo\\ls that the above-
mentioned distance is shortcr than that theoretically dcsigned for by an anl0unt
,
 's.
I I
(a)
To take carc of this error and also of the crror o in the length of the redundant
bar BC, \ve have to produce some compression in this latter bar before placing
it bet\vcen the joints Band C. After assembly, then, the redundant bar \vill
be pushing on the joints Band (', and assembly forces \vill be induced through-
out the truss. If X denotes the assembly force in the redundant member, 1
the corresponding elongations of the mcmbers of the primary system arc sXP",
and the diminishing of the distance BC', due to these clongations, is 2; (s::) 2 XPi.
This change in distance, together \vith the elongation Xpo of the redundant
bar, must bc numcrically equal to the initial discrcpancy bet\vecn thc distancc
BC in the primary system and the length of the rcdundant bar Be'. Hence, \\'c
obtain
2; (s) 2XPi + Xpo = - ( iS + llo)
(b)
'["he minus sign on the right foI1o\\'s from the fact that \ve like to consider
positive X as tension, \vhile the expression in parcnthcses indicates ho\v much
the distance BC (Fig. 7.6b) is shorter than the length of the rcdundant bar.
Solving the equation for Y, ,ve obtain
x - }; iS:' + o
- -  (SD2Pi + P o
(7 . 3a)
If \\'C add this force to that produced by lateral loading of the truss [see ]1:q.
(7.2)], the total force in the redundant bar becomes
x = _  (S:Pi + i t:'. + o
(sD 2 Pi + Po
(7.3 b)
Assembly forces can be calculated in the sanlC \vay if the redundant force is
an extcrnal reaction. Take, for example, thc truss sho\\'n in Fig. 7.5a, and
assume that the middle support is too high by an amount o. I)cnotc also, as
before, by d i the error in length of any bar i. ()\ving to these errors, thc joint
C of the primary system (Fig. 7. 5 b) is displaced up\\'ard by an amount equal
to 2; iS, \"hcre the values of < are obtained from the unit loading sho\vn in
Fig. 7.5t. l'he requircd reaction \"ill be found fron1 thc condition that the
upv.rard deflection of the primary system produccd by X is equal to o - }"; iS,
Hencc,
.Iy (s) 2 pi = o - 2"; iS:'
1 Tcnsion considered positive.
(c)

312 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
and \ve obtain
x = .-: }; iS + Ao
};(sD 2 Pi
Adding this to the previously found reaction produced by lateral loading of the
truss [see Eq. (7.1) J, \Vc obtain for the tota I reaction at the intcrmcdate support
(7 . 4a)
x = _ :Z(S; Pi + .6;)s; - .6 0 p.4b)
2;(S)2Pi
If the middle support scttles by an anlount Ao, \ve simply take this quantity
\vith negative sign in these equations.
The equations derived for calculating assembly forces can be used also in
calculating forces produced in a statically indeterminate truss by a te1Jlperaturc
change. If the tenlperaturc of a bar j is increased by an amount ii above a
certain specified uniforrn tempcrature of the truss, the length of the bar increases
by the amount cxitJi, \vhcre ai is the coefficient of lincar therolal expansion for
the bar. Treating these thero1al elongations of the bars in the same \\'ay as
the crrors in length Ai in our preceding discussion, \VC can rcadily calculate the
thermal strcsses in statically indeterminatc trusses. laking, for example, the
truss in Fig. 7.6a and using Eq. (7. 3a), \VC find that the force produced in
the redundant bar BC by thc tempcrature changc is
X = - 2;aitJis + cx_ ot% (7.5)
};(S:)2Pi + Po
When \ve deal \vith statically indeterminatc systcms having several redun-
dant elcnlents, calculation of the combined effects of applied loads, crrors in
length of the bars, temperaturc changes, settlement of supports, etc., become
rather involved, and it is desirable to employ a systematic procedure in their
analysis. Such an approach \vill no\v be described, \vithout reference to any
particular system.
Given a statically indetcrminate system, \VC first sclcct the redundant
clements and denote the redundant forces by Xa, X b , Xc, . . . , JY:, and the
corresponding generalizcd displacements by a, A b , Llc, . . . , Az. "rhen, \\lith
thcse redundant eleffil'nts all removed, \VC nlake a simplc statical analysis of
the rcmaining prirnary system under the action of the external loads and also
undcr the action of a unit force in place of each rcdundant force. V\lith these
data at hand, all the displaccnlcnts resulting from various causes such as applied
loads, errors in length, thermal effects, etc., can bc evaluatcd separately and
\vithout difficulty by the I11cthods alrcady discussed in Chap. 6 dealing
\vith the deflection problenl. Let A: o A::., A', . . . dcnotc thcse various
displacernents corrcsponding to the redundant force JY r " and let ouv denote the
displacement corresponding to XII due to a unit force in place of Xt. Finally,
lct Au denote thc nct kno\vn displacemcnt corresponding to the redundant

ARTICLE 1.4
313
p= 10 kips
10 kips
D
5
E
1
F
Xa
X b
(a)
(b)
1
1
1
(d)
(c)
FIG. 7.16
clement Xu' rrhcn, by superposition, \\'C must have the algebraic sum of all
the displaccments corresponding to each rcdundant force )(u cqual to the kno\vn
final displacement .1 11 ' Thus, \ve Inay \vrite the follo\ving systen1 of equations,
analogous to Eqs. (b) and (c):
A a ' + A ' a ' + . + x ,  + V'  +
L.1 L.1 . aUaa .J\ hUab
+ X,Oa: = .1a
.1 + .1' + .
+ "¥aOba + );bObb +
+ }(:Ob: = .1b
(7.6)
.1; + .1;' +
. + XaO za + XbO zb +
+ Xzo u = z
This systen1 of linear algebraic equations represents the so-called "supcrposition
equations" for a statically indctcrminate systcm, and their solution givcs the
magnitudes of thc z redundant elements involved.
To illustrate thc application of Eqs. (7.6), let us considcr the truss \vith t\\,.o
redundant elemcnts as sho\vn in Fig. 7.16a. Choosing the bar Cf' and thc
support at B as the redundant clements, \ve denote the force in the bar ('}'
by Xa (tcnsion positive) and the reaction at the support by )(b (positive
up\\'ard). Positive displacement corresponding to "Y a \vill then be a decrcase
in the distance ("F, and positive displacement corresponding to .LY b \\fill be an
up\\'ard deflection of joint B. Three statical analyses of the primary system
undcr the action of the P loading an.d unit forces in place of each redundant

314 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
TABLE 7.2
Bar h, S/, st' si" S,'s"/ S&,s,"1a s, '1. s,"1a (si') 2" (si") 2/, si's"'/,
In. kips
(1) (2) (3) (4) I (5) (6) (7) (8) (9) (10) (II) (12)
! ,
I
1 100.0 + 10.00 + 1. 000 0.000 + I ,000 0+100 a +100 0 0
2 100.0 0 + 1 .000 0.000 0 0+100 0 +100 0 0
3 141.4 -14.14 -1.414 0.000 +2,828 a -200 a +283 0 0
4 141.4 0 -1.414 0.000 0 a - 200 a +283 0 0
5 100.0 + to. 00 - 1 .000 0.000 - I ,000 o -100 0 +100 0 0
6 100.0 - 20 . 00 -1.000 +1.000 + 2,000 - 2,000 -100 +100 +100 +100 -100
7 141.4- a + I .414 0.000 0 o +200 0 +283 Q 0
8 141.4 +14.14 + 1.414 -1.414 +2,828 -2,828 +200 -200 +283 +283 -283
}; +7,656 -4,8281 01-1001 + 1 ,532; + 383 ; - 383
element are no\\' madc as indicatcd in Fig. 7.16b, c, and d. The results of
these analyses are recorded in columns 3 to 5 of Table 7.2. The lengths of
the bars are given in column 2 of the table. Each bar has cross-sectional arca
A = 3.33 in. 2 and modulus of elasticity E = 30(10 6 ) psi; so AE = lOR lb.
The coefficient of thermal expansion for each bar is a = 6.5 (10 6 ) (in./in.) 1°F.
It is requircd to find the magnitudes of Xc. and X b due to the combincd
effects of the given P loading, a uniform rise in tempcrarure of dt = 50o",
an error in length of the rcdundant bar CF of !:ala = -0.10 in. (that is, the bar
is too short), and a settlcmcnt of the support B by the amount !:ab = -0.05 in.
We begin with a calculation of the displacemcnts corresponding to Xa and
X b due to various causes. Using Eq. (6.3) from Art. 6.2 and the data from
Table 7.2, \VC have
li = I S;; i = +76,560
!:a" = 2;a t I , .!/ = 0
a ,
8aa = 1: (:i = +15.32
, " I
8 00 = 1: SiEi = - 3.83
S ' " I
A' = \"' L!. = _ 4 8 28 0
Ub  AE '
!:a' = 2;a t hS' = - 32,500
" ' I
8 ba = I S;/ = - 3.83
 _ \' (s/) 21. = + 3 8 3
CJbb - L AE ·
All numerical values arc in microinches.
From the statement of the problem, we conclude that the net decrease in
length of the redundant bar CF is
XJ
a = - AE a - a !:at to - !:ala

ARTICLE 7.4
315
or, \vith the given numerical data,
.1a = - Xa - 32,500 + 100,000
and for the final displacement of the support B,
b = - 50,000
Thus, Eqs. (7.6) become
76,560 + 15.32X a - 3.83X b = -X a - 32,500 + 100,000
-48,280 - 32,500 - 3.83X a + 3.83X b = -50,000
\\,hich reduce to
16.32X a - 3.83X b = -9,060
-3.83X a + 3.83X b = +30,780
From these simultaneous equations, \ve find
Ya = + 1,740 lb
Yb = +9,7801b
PROBLEMS
Referring to the truss in Fig. 7.16 and using Eqs. (7.6) together \vith the data in
Table 7.2, find the magni tudes of the redundant forces Xa and X b due to the load
P = 10 kips if there arc no errors in length, no change in temperature, and no
scttlcrnent of the support at B.
Al1S. Xa = -2,260 lb, X b = + 10,340 lb.
2 Repeat the solution to the preccding problen1 if the ten)pcrature drops by the
amount .1t = - 50°F, the redundant bar is too short by the arnount l1la = -0.05
in., and the settlement of the support R is zero.
Al1S. Xa = + 1,740 lb, X b = + 5,860 lb.
3 Referring to Fig. 7.17, find the axial force X in the redundant bar x due to the
action of the vcrticalload P = 10 kips, a uniforn) rise in temperature of t = 60°F,
p
r
2'
L
p
2'-+ 1,--1
I
L
I-- 1 -L 1 --l- I ------j
.x
FIG. 7.17
FIG. 7.18

316 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
and an error in It:ngth of the redundant bar .\' by the amount D./ z = -0.05 in. (too
short). Each bar has cross-sectional area Ai = 1 in. 2 , modulus of elasticity E =
30(10 6 ) psi, and coefficient of thermal expansion a = 6.5 (10- 6 ) (in./in.) 1°F.
Al1S. X = 8,575 Ib, tension.
4 Referring [0 the truss sho\vn in Fig. 7.1 R, the follo\ving data arc given: P = 10 kips,
/ = 100 in. Al1 bars arc aluminum, for \\rhich E = 10(10 6 ) psi and a = 10(10- 6 )
(in./in.) 1°F. Eac:-t horizontal and vertical bar has a cross-sectional area of I in. 2 ,
and each diagonal bar -V2 in. 2 Calculate the magnitudes X, Y, Z of axial force in
the three redundant bars x, )', z due to the applied load P.
Aus. X = -2,100 Ib, Y = -1,800 Ib, Z = + 3,910 lb.
5 Repeat the solution to the previous problenl if, in addition to the applied load P,
there is a uni form rise in tenlpcrature of the truss by the amount t = + 50°F and
an error in length of the redundant bar .t of the amount D./ % = +0.10 in., that is,
the bar is too long.
Ans. X = -11,840 lb, }" = -9,440 lb, Z = - 3,720 lb.
7.5 INFLUENCE LINES FOR STATICALLY INDETERMINATE
TRUSSES
In preceding articles, \ve have discusscd the analysis of statically indcterminatc
trusses under thc action of stationary loads. If, as in thc casc of bridges, \ve
are dealing \vith moving loads and rnust consider various positions of these
loads, influence lines can be used to advantage. In general, the construc-
tion of influcncc lines for statically indeterminate trusses can bc grcatly
simplified by using the rcciprocal theorcm, discusscd in Art. 5.7, togcthcr \vith
deflection curves for trusses. The gcneral statement of the rcciprocal theorem
is represented by Eq. (5.10) (sec pagc 249), and its application in the construc-
tion of influcnce lines \vill no\\' be illustratcd by several cxamples.
As a first example, let us consider thc truss in Fig. 7 .19a having onc redun-
dant support, assuming that the moving loads arc transmitted to the joints of
the lo\ver chord. Taking the reaction )( at the intermediate support as the
redundant forcc, we construct the influence line for this forcc by applying the
reciprocal thcorem to the t\VO loading conditions sho\vn in t'ig. 7.19a and b.
In the first case a unit load acts at the joint 111 of the given truss and produces a
reaction X at the intcrmediate support C. In the second case the intern1ediatc
support is removed, and a unit load' is applicd at the joint C, l""'his second
case is statically determinate, and \\le can rcadily calculatc the axial forces and
elongations of all bars. Having such elongations, \ve calculate the deflections
of thc lovlcr-chord joints by using the mcthod of fictitious loads discussed in
Art. 6.5. Let Om and oc be thc deflections of the joints 111 and c" obtained in
this manncr. Thcn, observing that the forces of Fig. 7.19a produce on the
displacements calculated for Fig. 7 .19b the \\fork equal to 1 X Om - XO c , and
that the \\fork of the forces of Fig. 7 .19b on the corresponding displacemcnts

ARTICLE 7.5
317
of Fig. 7 .19a vanishes, \ve find that Eq. (5.10) becomes
1 X m - XOc: = 0
and \\re obtain
X = 1 m
c
(a)
Hence, for any position of the moving unit load in Fig. 7 .19a, the intermediate
reaction }{ is proportional to the deflection of the corrcsponding joint of the
truss in Fig. 7.19/1. Upon dividing these deflections by the deflection Or., \\'C
obtain the ordinates of the influence line acb for the rcaction }[ as sho\vn in
Fig. 7.1ge. By use of this influence line, the redundant reaction X for any
system of moving loads PI, P2, /)3, . . . can no\\' bc obtained from the equation
X = 2;Ym P m
Having the influence line for the redundant force X, \ve can readily obtain
the influence diagraols for other quantities by using the mcthods developed in
Chap. 3 for statically determinatc trusses. Take, for example, the influence
diagram for the reaction Ra. For any position of the unit load, this reaction
is obtained fronl the statics equation
Ra = I I I X m - X ';
(b)
The first tenn on the right side of this equation represents the rcaction for a
simple beam supportcd at A and B, \vhereas the second term takes care of the
intermediate reaction "Y. Thus, thc ordinatcs of the rcquired influence diagram
\vill be obtained by subtracting the ordinates of the influence line acb, diminished
in the ratio 1 2 11, from thc ordinatcs of thc influence line constructed as for a
simply supported beam AB. '"I 'his \vill be accomplished by drawing the
straight line bed as sho\\rn in Fig. 7 .19c and then reducing in the ratio /2// the
ordinates of the shaded portions of the diagram.
The bending moment for a cross section through thc joint 111 is evidently
obtained by subtracting the moment )(1 2 x m l I, due to the intcrnlcdiate reaction,
from the moment calculatcd for a simplc beam supportcd at A and B. Thus, to
obtain the influence diagram for bcnding nloment at 111, \\'e begin ,vith the
influence line acb for .tY t on \vhich \ve superimpose the triangle adb as shown
in Fig. 7.19d. This triangle represents the influence linc for siolplc-beam
bcnding monlcnt at 'IJl except that the straight line bde has the ordinate /11 2 at
tl instcad of the ordinate X m as it should have. I-Ience, to obtain the influcnce
coefficicnts for simplc-beam bending momcnt at 111, thc ordinates of thc triangle
adb nlust be l11ultiplied by x"J2/1. This, ho\vcvcr, is the same factor by \vhich
the ordinatcs of the X influence line must bc multiplied to obtain the bcnding
mon1cnt due to X. Hence, the ordinatcs of the shaded area in Fig. 7.19d,

318 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
FIG. 7.19
(a)
(b)
(c)
(d)
(e)
1
a
f
1
' 2
:Ii
b
a b
r
' 2 '. . c
I _---d
..L! --
f
b
a
17
I
I It ___---d
O--- e

ARTICLE 7.5
319
\vhen multiplied by .\'m/2/ I, represcnt the required influence coefficients for
bending nl0ment at 111.
The same diagram can be used also as the influcnce diagram for ax ial forcc
in the bar nq of the upper chord of the truss oppositc the hinge 111. To obtain
the force in this bar, we have only to divide the bending momcnt at 111 by the
distance hm. and change the sign. Hence the influence diagram for the above-
mentioned bar is obtained by multiplying the ordinates of the shaded area in
Fig. 7 .19d by the numerical factor -x m I2Ih rn l. In a similar manner, influence
diagran1s for the bars of the lower chord can be constructed.
Let us consider no\v a web member np (Fig. 7 .19a). If the unit load is to
the right of joint 111, \VC consider the equilibrium of the left portion of the truss.
Taking the moments of all forccs acting on this portion \vith respect to point
0, \ve find that the force in the \veb mernber l1P is
h
S = - Ra I
flu
(c)
Hence, for positions of the load to the right of joint 111, the influence diagram
for S is obtained by using the same shadcd areas already used for Ra in Fig.
7.19c, except that in this case the ordinatcs must be multiplied by -J 2 h/II1".
Whcn thc unit load is to the left of pancl p111 and at the distance x from the
support A, \ve obtain for S, instead of Eq. (c), the follo\ving equation:
s = - Ra  + 1 h + x
h n h n
(d)
i\ll nccessary constructions for the required influence diagram are indicated
in Fig. 7 .1ge.' For the portion of the span to the right of joint 111 \ve use, as
already mentioned, the same arcas as in Fig. 7.19c, \vhile for the portion to the
left of paneljnn the sccond term on the right side of Eq. (d) must be takn into
account. ,.\ shall accomplish this by dra \ving the line oal as previously
explaincd in Chap. 3 (see page 149). The ordinates included bet\veen the
straight Jines oaf and ocb, after multiplication by the factor 1 2 h/lh n , give us thc
second tcrm on the right side of Eq. (d). Hencc, the ordinates of the shaded
areas in Fig. 7 .1ge, after multiplication by this factor 1 2 h/lh n , give thc required
influence coefficients for axial force in the bar np. The proper signs of these
ordinatcs are rcadily determined from Eqs. (c) and (d) and are indicated in
Fig. 7.19t'.
For a second example, \ve considcr the two-hinged arch sho\vn in Fig. 7. 20a
and take the horizontal thrust H as the redundant force. Then, to construct
the influence line for H, \ve apply the reciprocal theorem to the t\\'o loading
conditions sho\\'n in Fig. 7.20a and b. In the first case, a unit load, I:1oving
along the upper chord, acts at the joint k. In the second case, the red'Jndant
constraint is removed, and t\VO equal but opposite unit forces are applied as

320 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
1
1
Hi Uk
---________J_l______
(a)
(6)
1
(c)
Ok
a b
r
X m
(d) 11m
1
.,..... -
x. x:
I I
a b
r I
xi
Xi III
11, 
(e) L - -
- --- 9
-- -
FIG. 7.20
sho\vn. This second case is statically determinate, and \ve can calculate the
elongations of all bars. With these elongations, the increase Oh in the distance
AB can be found by using Eq. (6.2), and the deflections of the upper-chord
joints can be found by applying the mcthod of fictitious loads discussed in
Art. 6.5. 1"hc deflection curve obtained in this manner is sho\vn in Fig. 7.20c.

ARTICLE 7.5
321
Applying, no\v, the reciprocal theorem, \VC obtain
-]-llJ/t + I X Ok = 0
H Ok
=1-
0"
\\, see that the ordinatcs of the required influcncc line for Hare obtaincd by
dividing by Oh the deflections of the upper-chord joints. This line is given by
the polygon acb in Fig. 7.20d. "he influence line for thc rcdundant forcc ]-l
having been obtained, the influence diagranl for any other quantity can be
readily constructed by using the methods developed in Chap. 3. In Fig. 7.20d,
for example, the influence diagram for bending monlcnt \vith rcspect to the
joint 111 is sho\vn. With such influence diagranls for bcnding II10mcnt, the
corresponding influence diagrams for axial forces in the chord members can
also be obtained v...ithout difficulty.
Let us consider no\\' the influencc line for the axial force S in a web member
11Jn (Fig. 7 .2ob) . Making a section pq through that membcr as sho\vn and
considering the equilibrium of the left portion of the truss, \ve obtain
Hence,
y. y. ( h. )
S = So - H  =  So  - H
hi hi Yi
(e)
where So is the force in the membcr 1Jlll \vhen H = 0, and the distances Yi and hi
are as indicated in Fig. 7.20b. Upon dra\ving the lines ae and hd as sho\vn in
Fig. 7.20e, \VC obtain the influence line a.fgb for the quantity Sch;1 )'i. Sub-
tracting from this the ordinates of the influence linc for H, \\le obtain the
shaded areas sho\vn in Fig. 7.20t'. The ordinates of these arcas multiplied by
)'i/ hi give us the rcquired influence coefficients for S. Influence diagrams for
other \veb members can be constructed in a similar manner.
As an cxample of a system \\'ith t\VO redundant clcments, lct us consider the
truss on four supports as shown in Fig. 7 21a. It is assumed that the moving
loads are transmittcd to the joints of the lo\ver chord. For the redundant
forces, we select the reactions )( and Y at the intermediate supports. In
addition to the actual loading condition sho\\'n in Fig. 7.2Ia, \ve consider the
t\VO cases sho\vn in Fig. 7.21 band c in \vhich the intermediate supports arc
removed and a unit load is applied at joint (,' and at joint [J, respectively. l'he
latter t\VO cases are statically determinate, and \ve can calculatc the dcflections
of the lo\\'cr chord for each of these cases by using the mcthod of fictitious
loads. Lct Oce, Oed, and OCfU denotc the deflections at C, D, and 111, respectively,
\vhen the unit load is acting at C, \vhile Ode, Odd, and Odm denote the deflections
produced at the same joints by the unit load at [). Assuming that the supports
C and D do not displace vcrtically under the action of the unit load in Fig. 7.2 J a
and applying the reciprocal theorem to Fig. 7.2In and b, \ve obtain
-XO cc - YO cd + 1 X Oem = 0
(I)

322 STATICALL Y INDETERMINATE: PIN-JOINTED TRUSSES
(a)
(b)
1
(c)
1
(d)
/I llli!III"
b
m
(e)
/I 
FIG. 7.21
Similarly, considering Fig. 7 .21a and c, \ve obtain
- ,:ro de - J70dd + 1 X OeJrn = 0
(g)
Solving Eqs. (f) and (g) for \ and }/Yand observing that Oed = Odc, from the
reciprocal theorClll, \ve find the follo\ving exprcssions for the redundant
reactions:
x = ---. d _ _.  _ Oed  d
.J    2 ucm    2 U m
UccOdd - Oed UccOdd - Ued '
(11)
1 7 = cc Ocd
{) {) () 2 Odm - 0 0 0 '} Oem
cc del - cd cc dd - ClJ"
(i)
It is seen that \vhen th unit load changes position along the truss (Fig. 7.2In),
only the quantities Oem and Odm in exprcssions (11) and (i) change in magnitude;

ARTICLE 7.5 323
the other quantities remain constant. Introducing the notations
. dd. = C 1
ccdd - cd2
. Oc c _ C 2
OccOdd - - Oc(12
Oed
ccOdd - Ocd 2
- ('3
(j)
\ve obtain
)( = Clcm - C3dTU
y = C2dm - C 3 0 cm
(k)
#-[he constants C1, C 2 , and ['3 are rcadily calculated from Eqs. (j) provided
that the deflcction curves for the loadings in Fig. 7.21 band c have been con-
structed. 'rhc expressions on the right sides of Eqs. (k) arc plottcd in Fig.
7.21 d and e, and the required influence coefficients for the redundant reactions
X and Yare givcn by the ordinates of the shaded arcas. "lith these construc-
tions, the influence diagram for axial force in any bar of the truss (Fig. 7.2IlT)
can be casily obtained . For this purpose, \VC observe that the force Si in any
member can be represented in the follo\ving form:
S S f I x " Y
; = l . - s. - s.
. I I I
(I)
",,-here S is the force in that member \"hen the redundant supports arc removed,
and s and s:' are the il1fluence 11l1111bers 1 for the same membcr calculated for the
unit loads shown in Fig. 7.21 band c, rcspecti vely . 1"hus, \ve see that the
required influence diagram for axial force in any bar can be obtaincd by com-
bining the ordinates of the influence diagram, constructcd as for a simply
supported truss, \vith those of the sX diagram and the s:' Y diagram. These
lattcr diagrams arc obtained simply by multiplying the ordinatcs of the shaded
areas in Fig. 7.21d and e by the influence numbers s and s:'.
PROBLEMS
1 Construct an influence line for the redundant reaction Xc in the truss sho\vn in
Fig. 7.22 if live loads are applied to the lo\\'er-chord joints. Each bar has the
saIne cross-sectional area Ai = 4 in. 2 and the same modulus of elasticity l'. Using
this influence line, find the maximum value of Xc due to a standard Cooper's E-60
train loading.
AllS. Max Xc = 98.3 kips.
2 V sing the data in the preceding problem, construct an inAucncc diagram for axial
force S in' the bar JF of the truss sho\vn in Fig. 7.22. v\lith the aid of this dia-
g ram, find the maXiITIUm value of S due to a standard Cooper's E-60 loading.
Alls_ Srnnx = - 23.6 kips.
1 'rhese influence numbers si', st", as dcfined on page 259, should not be confused \vith influ-
ence coefficients as defined,on page 108.

324 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
FIG. 7.22 Xc
r 6',6'1- 6't- 6' T 6'T 6'l
r
8'
tl
FIG. 7.23
D
FIG. 7.24
(b)
raT-aT a-t;-a-r a-1
r
h
&
(a)
.
3 Repeat the solution to Probe 1 if, instead of a rigid support at C as sho\vn in Fig.
7.22, there is a vertical steel bar C/) (not sho\vn) of cross-sectional area A 0 = 4 in. 2
and length 10 = ]0 ft. All other data re1l1ain the same as in Probe 1.
A'11s. Max Xc = 85.5 kips.
4 Construct an inRuence line for the axial force X in the redundant bar C}) of the
truss sho\vn in Fig. 7.23 if live load is applied to the lo\vcr-chord joints of the truss
proper. Each bar has the saIne cross-sectional area /1 i = 4 in. 2 and the same
l110dulus of elasticity E; the bar CD is 6 fe long. Using this influence line, find the
maximum value of )( due to a standard (=oopce s E-60 loading.
Al1S. X mnx = 40.8 kips, compression.
5 Construct the influence line for axial force Y in the bar 1J/ll of the truss sho\\'J1 in
Fig. 7.24 if a = 60 in., 11 = 80 in., and each bar has the sanle AF: value. Assume
that the moving loads are applied to the lo\vcr-chord joints.
Al1S. See Fig. 7.24b.

ARTICLE 7.6
325
7.6 STATICALLY INDETERMINATE SPACE STRUCTURES
I f a structure in space has 12 joints and n10re than 311 bars, including supporting
bars or their equivalent, the numbcr of equations that can bc obtained from
statics \vill bc insufficient to detcrn1ine the unkno\vn axial forccs in all mcrnbcrs
and the systcm is statically indeterminate. In analyzing such systcrns, the
elastic dcformations of thc bars must be considered. lhis can be accornplished
by using the methods already developed for plane trusses. Io illustrate the
application of thesc methods to space systems, Ict liS consider the simple
systel11 consisting of a single joint A attached to the foundation by four bars as
sho\\'n in Fig. 7.25. Such a systcm has one rcdundant bar. To define the
configuration, \ve assume that all bars are identical and have the length I = 5 ft,
that the bars I and 3 are in a horizontal plane /lBIJ, and that the bars 2 and 4
are in a vertical plane ACE bisecting the angle BAD. The force [:J applicd
at A acts in a vcrtical plane parallcl to the plane BCD and makcs an anglc
of 45 0 \vith the vertical bar AE. We select the bar AE as the redundant bar
and denote by ....Y the axial force in it. Then the forces in the other bars \\,ill be
found from the equation
Si = S; + <X
(a)
in \vhich S; is the force produced by the load P in any bar i of the primary
system, i.e., of the system that remains after removal of the redundant bar AE,
and s is the influence number for any bar i. These influence numbers are
obtained by applying at joint A of the primary systcm a vertical unit load and
calculating the corresponding axial forces in the bars. To obtain the redundant
force X by the theorem of Icast \vork, \ve makc the exprcssion for strain energy
_  Si 2 /i X 2 /0 _ I (  2 2 )
U -  2A i E + 2AoE - i L Si Pi + X Po
(b)
FI G. 7.25

326 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
\vhere thc summations are understood to include all bars of the primary system.
Sctting the derivative of this exprcssion \\,ith respect to X equal to zero, \ve
obtain
: = L SjS;Pi + Xpo = 0
(c)
Observing that in this case Pi is the same for all membcrs and substituting
exprcssion (a) for Si, \ve obtain
Sl
X = - (S;)2' -+ I (d)
The values of S; and s togcther \vith all calculations necessary to determinc X
are sho\vn in Table 7.3. The load P is taken equal to 1 ton, and Eq. (d) then
gives
x = - __.5_ = -0 570 ton
5. I 65 .
Finally, using Fq. (a), \ve obtain the values of Si as recorded in the last column
of the table.
Equation (c) for calculating LY can also be obtained by applying the Max\veJl-
Mohr method as discussed in Art. 6.2. In this case the \vork done on the
actual displacements by the forces corresponding to the unit load must vanish.
If it is desircd to considcr a change in tcmperature and possible errors i in
the lengths of the bars, \VC have only to substitute in Eq. (c) the expression
SiPi + cxtdi + Ll i instead of SiPi and Xpo + o:to/o + o instead of Xpo. In this
\\fay \ve obtain the equation
(SiPi + o:tdi + Ai)S + Xpo + o:t% + Ao = 0
from \vhich
(SPi + alih + i)S: + atolo + Ao
 (s) 2 Pi + Po
This equation can be used to calculatc the redundant force in any pin-connected
space structure \vith one redundant melnber. It is necessary only to extcnd
x=
(7.7)
TABLE 7.3
, , Sf , (s;) 2 S" tons
Sit tons s. 'S" tons
, , ,
1 -5 y2/6 IS 0.983 0.694 -0.705
--6
Z 5 0/6 . 1.967 2.777 0.230
3 0 --i 0 0.694- 0.475
4 -0.570
2; 2.95 4.165

ARTICLE 7.6
p
FIG. 7.26
327
a
a
a
29
A
r
-I
a
the summations to includc all rnembers of the primary system. Omitting
terms containing temperature fa" and length errors i, \VC see that Eq. (7.7)
coincides \vith Eq. (d).
As a marc complicated example, let us consider the space structurc sho\vn
in ig. 7.26 to \vhich a horizontal force P = 1 ton is applied. This structure
has 12 joints and 39 bars. Hence, thcre are three redundant bars . For these
redundants, we choose the threc horizontal diagonals and denote their axial
forces by X, 1': and Z, as sho\\'n. After rcmoval of these bars, \\'C obtain the
statically determinate primary system sho\vn in Fig. 7.27. rrhe forces S
produced in the bars of this systcm by the load P = 1 ton are given in the
second column of Table 7.4, page 328. We consider no\\' the unit loads acting
on the primary system as sho\vn in Fig. 7.27a to c and calculatc the cor-
d . . fl b '" ",. . I 5 f h bl
respon lng In uence nurn crs Si' Si , Si , gIven In co umns 3 to 0 t e ta e.
The force in any bar i of the given structure (Fig. 7.26) \vill now be represented
by the exprcssion
Si = S + sX + s'Y + S;"Z

328
... "..-....
"-
"- . .. 
"'" -
  ---
... ---
...
...  ,..,..,
-
...  ........
''';':
''';':
... .;.,-
..
....;,;
Vi
........
'vf
C'4
':;:--- --
co
...  .........
---
N
...--- --
''';': t"-
........
"'-/
C'4
--- "..-....
''';': '-0
'-" '-"
"-
... .;.,..
... .;.,..
.. '..
"'"
v

L.&J
..J
CO
«

v:r
.,-..
f'J
-
.........
000 0 0 00 0 ::>0 00000 000 0 0
- -
00 000 00 000 co 0 0 0 0 a 000
0 0 ,O 0 00000 o a ....r. 0 ....r. 1 o C"*, - :",:)1 -
--- - - ----
0 000 0 00 00 0 00 00 0 0 00 0 0
- -
S oSSo
00 => 00 000 00 0 o 0 0
.........
- -
------ - - - -
 f"J I ""J I f"1 osSo
>0  0 0 S'2° 00 0 0'>0 0
......... .........
- - - ,..,.., r
-
0 0 00 0 ooe 00 0000 0 0 00 0 0
- ---- - --
00 00 c 000 00 o 0 ....:I 0 ....:1 0 ....c :.:-. .... ....r.1 -
....r.. 0 "':;:1 0 ...., - .....I - .-4 r.' - ....:-:- - .-4'' ....r.. ....r.-. .... r.H,,*, - c:: r.1 -
-........
0 0000 0 0 .::> 0 0 0000 0 0 0 0 0 0
- - - -. - - - -
S IN . r" I
o 0 > IN >
0 00 0 0 00 0 - c 0 0>- -
-
......... ......... .........,
- - ......... -
I I - I
. -
S :N IN S S IN J "'"
'> I ,.... > I""'J f'J > SIN>
o 0> - :::-1>- .........,.........>......... ').- -
......... ........... "......... I
- - ......... - -.........-
I I - I ......... , I - I -  ,
-
----- -
- 00 0 0 ,r"J - 0 00 00 - 00 O-SI 0
I > I I
-
- N r"\ T ''''' \Q  XI 0- 0 - N r"\  '"'" \Q t"- oo 0- 0
- - - ...... - - - - - - N
,,-...
-
'-"
--
a
-
.........
---
0-
-
"..-.....
'''''
.........
-..

--
--
ff\
'-"
-..
N
---
,....
-
'-"

o 0 , 0 ....:':'
o 0 0 0 -f':1
C',*' -"',:,, - -r.'
c -. 0 M:-:1 -
M:''':'1 ..... .::-...... M: ':' --
"'f: - r.' - '0"*1 -
'"
o -f:, 0 "-';-:1-
I-
I
..")'
-I:'  '':1....1':' 1:1 -
..-:' ' ..",'
.:-1........ r1.-. . :1  r..........
I ...., I 
I t"J IN ' I r-
'> 1 ("I ''''1 >0 '>
0000 0 C O 0000
,   , 'T
,
\ "1 IN ' r-.1 "1
OCO 0 '> 00 O'>S >0 cooo '>
, , (",1
, ="
 I ,, >
O>s '>0 0000
OOO  00
.......... ' . "1
tr. I N
o 0 0 0 -f':1
...:.,.  ""(:1 -- .....:-.
:, - Cr., - "':1
0000
: "'I 1(")
I r-1 _ > _ >
> "I'
, - -
I I
s-
..........
r
IN Ir1
>->
..........,..........

o ....::, 0 ...:-:1 -
:', ...'':1-:':'''*' -
':-....!.: "":1 . 
, 'j
I "I
'>
,
I"',!
......... N
OO>-
..........
-f':' - f:' - ....:.:, - c"
,,*.  "'::. - ,'*, -\ ;;::
-:I "")1 - .:
C,'-:-' - ':':11:1
t--..
r:
I ,... 1 I t"J
»IN:r- 1
....................»
- -....................
I I - 
1 "'1
><S'S -
..........-,..........
I - '''''
, "')
>-
..........1
,,..)
"-

............
IN
>-
"'-I
S  I""
1 ("1 ,
-, - -
............ I ? "I
 .......... r.
I r I
IN 1 r-J
>- I "I "-
> >-
"'- 1 .......... I
'r. "'- ''''
I,..,
0 0 0 0 0 0 - "" S . 0 0 0 0 0
--
- -- - ---
r"'1 " '1" .,.., -0 t--.. 00 0- 0 "1 r#\  "" -c
"J N "" '"'I r"'J "J ("J """ N    f"',  #\ #\ 
319

330 STATICALL Y INDETERMINATE PIN-JOINTED TRUSSES
(a)
FIG. 7.27
(6)
(c)
and Eqs. (7.6) reduce to the following three equations for calculating the
redundant forces:
s (S; + sX + S'}T + s" Z) Pi + XP:r: = 0
S/ (S; + SX + s/ y + s" Z)Pi + YPu = a (e)
S" (S; + s:X + s' Y + S" Z) Pi + Zp: = 0
In these equations, pz, Pu, and pz denote the extensibility factors for the three
redundant diagonals. l"he magnitudes of the redundant forces will, of course,
depend on the cross-sectional dimensions of the bars. Assuming, for example,
that the extcnsibilities of all bars are equal, the quantities p cancel in Eqs. (t'),
and the numcrical values of all cocfficients in these equations arc obtained by
making summations of columns 6 to 14 of the table. In this \vay, Eqs. (e)
finally become
R8X + 47Y + 15Z = -22 y2
47X+ 32Y+ 112 = -12y2 (f)
15.lY + II Y + 82 = -4 V2
fronl which \\re find
x = -0.2327 yi ton
Y = -0.0215 y2 ton
Z = -0.0341 y2 ton
PROBLEMS
The sitnple space truss sho\\'n in Fig. 7.28 consists of t\vo panels ABCI) and ABEF
attached to a vertical \vall at points C, 1), };, F, the panel ABCl) being in a hori-
zontal plane. ,All bars have the san1e cross-sectional area A and the same modulus

ARTICLE 7.6
331
of clasticity E. Calculate the axial force X produced in the redWldant bar .\' by
the vertical load P = 1 kip applied at joint A as sho\vn.
Aus. X = + 56 lb.
2 ""hat thermal force X, \vill he induced in the redundant bar .t of the space truss in
Fig. 7.28 if there is a uniforrn rise in temperature of 50°F? lhe coefficient of
thermal expansion for each bar is a = 6.5 (10- 6 ) (in./io.) 1°F, Ai = 1 in. 2 , and
E = 30(10 6 ) psi.
Aus. X, = -1,295 lb.
3 rhe statically indeterrninate ball-jointed space truss sho\vn in Fig. 7.29 has the
forn1 of a square pyran1id \vith din1ensions a = loft and h = 16 ft, R, F, G, H
being rnid-points of the edges AI, BI, CI, ])/, respectively. A horizontal. force
P = 10 kips, acting parallel to AR, is applied at joint I as sho\vn. Calculate the
magnitudes of the axial forces X and Y in the redundant bars x and y, assunling
that the redundant bar y has a length error Au = -0.01 ft, that is, the bar is too
short. All bars have the same cross-sectional area Ai = 4 in. 2 and the san1C
modulus of elasticity E = 30(10 6 ) psi. rhe foundation points A, B, C, 1) arc
immovable.
Aus. X = +14,370 lb, Y' = +18,210 lb.
I
_" C - -- -----  -; D
.... "
- -
- .....
- .....
....- --
... .....
A ---------- B a
a
p
P-l kip
FIG. 7.28
FIG. 7.29

Chapter 8
Arches and frames
8.1 INTRODUCTION
Any plane-curved bar or rib, properly supported at its ends and so loaded that it
acts primarily in direct compression, rnay be called an arch. It is assumed that
the plane of curvature of the rib is also a plane of synu11ctry for each cross
section and that external forces applied to the arch act only in this plane.
Undcr such conditions, deformation \vill also take place in the plane of synl-
mctry, and the problern of analysis becomes a t\vo-dirncnsional onc. If the
cross sections of the rib arc unsynunetrical \vith respect to the plane of
curvature or if laads arc applied normal to this plane, there \vill be [\visting of
the loaded rih, and under such conditions it is not properly considered to he
an arch. Condltions leading to such torsion \vill be cxcludt'd from considera-
tion in a II further discllssion.
In (:hap. 1, (1 fe\\" cxanlplcs of statically d<.tcrminatc th1'l'c-l1ingtd archl''
\vere discussed. J Jerc \ve shall considcr only statically indeterminate arches
332

ARTICLE 8.1
333
c
Ra
Rb
f
Ha I
___1___
FI G. 8.1
(a)
I
(b)
such as two-hinged arche.r and hi1JgeleJs a1'ches as sho\vn in Fig. 8.1a and b,
respectively. In the case of such statically indetcnninatc arches, it is essential
that the ends A and B, \\,hether hinged or built in, remain immovable \vhen
loads are applied to the rib; other\vise, true arch action cannot bc dcveloped.!
The arch in Fig. 8.1 a \vhich has one end lo\ver than the other is called an
lI11SYJ111J1ctrical arch, \vhercas the onc in Fig. 8. I b is called a s.1'l1oJletrical arch.
The highest point C on the arch axis in each case is called the crOWll, and the
line AB joining the points of support is called the springing li1lc. The horizontal
distance bet\\'ccn supports, denoted by I, is called the span of the arch, and the
maximum ycrtical distance from springing line to arch axis, denotcd by f, is
called the rise of the arch.
"[he t\\'o-hinged arch (Fig. B.la) rcprcsents a structure \vith one redundant
reactive force. Since the points of support are irnrnovable, applied loads
produce, at both ends, not only vertical but also horizontal reaccions. Thus
\ve have four unkno\vn reactive forces, for the calculation of \vhich there are
only three equations of statics. l() obtain a fourth cquation, deformation of
the arch must be considered. We can take, for instance, the horizontal
reaction Hb as the rcdundant clement; then, its magnitude will be
determincd from the condition that the horizontal displacemcnt of thc
hinge B vanishes.
In the case of the hingelcss arch (t-"ig. 8. I b), \ve have six reactive elements,
four componcnts of force, and t\VO end momcnts. \\. can take as redundant
clements in this case the forces Ra, Ha and the moment j\l[a. Then, the
magnitudes of these quantities \vill be detcrnlined from the conditions
that the corresponding displacemcnts and rotation of the end A of the arch
must vanish.
As a general approach to thc problem of analysis of statically indeterminate
arches, \ve shall use the second thcorenl of Castigliano or thc theorem of least
\vork, as already developed in Arts. 5.5 and 5.6, respectively. This requires
that \VC first establish an expression for the total strain cncrgy storcd in the
arch rib under load. 10 do this, \ve consider an elelnent of the arch rib as
shown in Fig. 8.2 and denote by 1\1, I and N, respectively, the bending
1 Unexpected yielding of the abutruents due to poor foundation conditions has led to failure
of many arch structures.

33. ARCHES AND FRAMES
v
FIG. 8.2
moment, the shearing force, and the normal force at any cross section, positive
when directed as shown. Since the depth h of the cross section is usually small
compared Wltu ".. . -':q of curvature r of the arch axis, \VC may use for strain
energy of bending the expression
( . M2 dJ
Vb = J a 2EI
(a)
similar to chat used for straight beanls, \vhere s is the length of the center line
of the arch and I is the momenr. of incrtia of the cross section. Also, the
strain energy due to shearing forces V in Fig. 8.2 is usually small compared
with that due to bending and can be neglected. Finally, for the strain energy
of direct compression, \ve have
U c = r' ll2 ds
Jo 2AE
(b)
where s is rhe lcngth of the center line of the arch and A is the cross-sectional
area. Thus, neglecting the etfect of curvature of the arch axis and the cffect
of shear deformation, \ve find the total strain energy in the arch rib to bel
u = (3 A1 2 ds + (. N2 d s
)0 2EI Ja 2AE
(8.1)
In most practical cases, the cross-sectional area A and the moment of inertia I
vary along the length of the arch, and this variation must be taken into account
before the indicated integrations can be performed. Particular types of varia-
tions will be considered later in examples.
1 For a more complete expression for strain energy containing the effect of shear deforrnation
and curvature of the arch axis, see S. Tilnoshcnko, "Strength of 1\1atcrials," 3d cd., vol. I,
p. 382, D. Van Nostrand COlnpany, Inc., Princeton. N.J., 1955. Results obtained on the
basis of this more exact expression sho\v that for usual proportions of arches, Eq. (8.1) is
sufficiently accurate for practical applications.

ARTICLE 8.2
335
8.2 SYMMETRICAL TWO-HINGED ARCHES
Let us considcr a synunctrical t\'lo-hinged arch subjected to any vertical loading
as sho\vn in Fig. 8. J. Choosing the thrust H as the redundant reaction and
applying the theorenl of Icast \vork, \VC obtain
: = 10' Z  ds + loB  :Z ds = 0
(a)
Lct Ai/' and N' dcnote, respectively, the bending nl0nlcnt and axial con1pression
in the arch \vhen the hinge B is fret to nl0ve horizontally and H is not acting.
Then, the con1plctc expfesions for bending 1110Incnt and axial compression aft
i\J = iJ' - ll(f - y)
1\' = J.l\TI + H cos 4>
(b)
\vhcre f is the risc of the arch and cf> is the angle that the tangent to the center
line o1akcs \'lith thc x axis. \\lhen expressions (b) and their derivatives \vith
respecr (r) H arc substituted into Fq. ('1), it becolllcs
 8 }\;[' - H(f - y)
- , .. (f - v) ds
o Ef .
 :I l\T' + H cos 4> d 0
+ cas <p s =
o AE
(c)
Solving for H, \VC obtain
loB (A1'IEl) (f - y) ds - loB (N'IAE) COS cp dJ
H = - -- -_.- (8.2)
loB [(f - y)21EII tis + 10' [(cos 2 cp)/AE] cis
This expression holds for a symmctrical t\vo-hingcd arch of any shape and
under any condition of loading. For ordinary proportions of an arch, the
c
x
FJ G. 8.3
y

336 ARCHES AND FRAMES
second teflll in the numcrator of Eq. (8.2) is usually s111all compared \\lith the
first and can often be neglectcd.
As a specific example, let us assunle that the arch axis is a parabola defined
by the equations
4f.\02
Y= -
" 1 2
1
cos <P = - -- -
\ vi 1 + (8/x /12 )2
(d)
and that the cross-sectional area and nl0ment of inertia vary according to ,the
cxprcsslons
A = Ao
cos 4>
I = _! 0 _ -
cos cJ>
(e)
\vhere A 0 and 10 denote cross-sectional area and moment of inertia at the
cro\vn C. Finally, assuming a uniformly distributcd vertical load of intcnsity
q on the arch, \\'c have
M' = q2 (I _ 4;2 )
J.\T' = qx sin 
Cf)
Using exprcssions (d) to (f), the various intcgrals appearing in Eq. (8.2) can
rcadily be evaluated, and neglecting the second integral in the nunlerator, ""C
obtain, as a final rcsult,
\vherc
ql2
H= fl +13
_ 1 5 I 10 f _I 4/
(3-32 f3Ao tan T
( 8 . 2a)
(g)
-rhe first factor in Eq. (S.2a) reprcsents the thrust for a uniformly loaded three-
hinged arch. We see that this must be reduced in the ratio 1/(1 + (3) to
obtain the thrust for a t\\'o-hinged arch.
i\nother exanlplc of symmetrical deforrnation of a t\\'o-hinged arch (Fig.
A.3) is that produced by a unifornl temperature change. Let t be the increase
in tcmperature of the arch above the temperaturc at \vhich it fits freely bet\veen
the fixed supports at its ,ends, and a the coefficient of therrnal expansion of the
material. l'hen, in the case of free expansion, thc span length \\.'ould increase
by the anl0unt atl. In the actual case \\,here the ends of the arch are restrained,
a thrust H ",ill be developed of such n1agnitudc as to counteract the free
cxpansion atl. Applying the Castigliano theorem in this case, \ve obtain
ill)
- = atl
aH
(11)
The left-hand side of this equation can be obtained from the previous Eq. (c)

ARTI CLE 8.2
337
Xd
P
C x
D Rb
f
H_____1_____H
(a)
(b)
FI G. 8.4
simply by substituting 1\;[' = I'l' = O. Then, solving for H, \ve obtain
H aU
= 10 ' [ U  y)2 / EIj d-:+ 10 ' [( co 2--;)/ AE ld (8.3)
For the parabolic arch defined by Eqs. (d) and (e), this becomes
H _ 1 5 Ef., . atl
- -8f 2 / - -1+ 
(8.3a)
For an arch of rectangular cross section, the corresponding maxinlum and
minimum thernlal stresses at the 'cro\\'n \vill be
H 6Hf 15 F.h 2 at ( 6f )
CTr = - 4"1 0 + AJI = - R 1 21 2 "1 +- 1 + Ii
(8.3/1)
I is Setl1 that these them1al stresses increase rapidly \vith the ratio hlf and
may be very inlportant for thick flat arches.
10 obtain an illjluence liut: for the thrust H of a synlnletrical t\va-hinged arch,
\Vc use the reciprocal theorenl (see page 24<)}. Comparing the t\VO conditions
of loading sho\vn in Fig. 8.4a and b, \\'C conclude that
P'L' + HA = 0
,,-here '7.,' is considered positive do\\'n\vard. This gives
H = - p  (i)
A

338 ARCHES AND FRAMES
c
Q
FI G. 8.5
T "'''",,,,D I
f L xd ----1",
L y  " \ B
! ---- 1 \ =rUb
x
and \ve see that for the construction of an influence line, \ve need to calculate
the displacements  and IV produced by the unit loads acting as sho\\'n in Fig.
R.4h. The displacernent Ll can he obtained from Eq. (8.3) simply by sub-
stituting  for atl and taking H = t. Thus,
I _ 1, 8 Cf - y)  1, , cas  p.
 - f."1 d. + 4 L' dJ
0.; 0 . I:'
( ')
.1
In calculating deflcctions ', \ve consider the right-hand half of the arch a a
curved cantilever heanl built in at C and acted upon by the horizontal unit
force at B as sho\vn in Fig. 8.5. J)cnoting by 't'lJ and Vd the vertical deflections
of points Band D, \VC obtain the rcquired displacernents 't} from the equation
It' = '1,;' d - "t 'b
(k)
The deflection 't-'d \vi II he found by using the Castigliano theorem. l\pplying
an infiniresimal vertical force Q at D, \VC obtain, by differentiating the strain-
energy expression \vith respect to this force Q,
'l)r/ == f B .j}! aA! ds + f }/ !'l- aN dJ
c EI an c AR aD
......... , --
(I)
No\v, for the portion CD of the bar, the expressions for bending moment and
compressive force \vnl be
AJ = - [1 (f - y) + Q, ('\'d - x)]
l\Y = t cas cp - Q sin 4>
If \ve put Q = 0, these sanlC expressions hold for the portion DB of the bar.
Substitution of the appropriatc expressions for i\.l and 1\! and their derivatives
\vith respect to Q into Eg. (l) givcs, \vith Q = 0,
't-'d = ( D .(1 _ y) (Xd - x) ds _ r D o sin 4> ds
Jc HI Jc AF:
(111)

ARTICLE 8.2
339
Simply by substituting 1/2 for Xd in this expression, \ve obtain Vb, and Eq. (k)
finally becomes
V=
_ f R V - y) ( /I  x). ds + f D (/ - y) (d - x _ ds
c £1 c EI
+ JD B cos ;n  ds
(11)
Using this equation together \vith Eq. (j), \ve can, in each particular case, cal-
culate the ordinates 'L'/  of the influence line for thrust H.
As an example, let us consider the parabolic arch defined by lqs. (d) and
(e). Then, from the prcvious calculations [see Eq. (S.3a)], \ve have
8 1f2
 = 15 Elo (1 + ,8)
Considering the second integral appearing in exprcssion (11), \ve have
f D (f - _ Y).,(Xd -32 ds = L (Xd ( I
c EI Elo ) 0
4X 2 ) ( , d
- /2. Xd - X, x
_ f X d 2 ( 1 Xd2 )
- Elo '2 - 3/ 2
The first integral in expression (n) can bc obtained also from this simply by
taking 1d = /12. "fhe third integral is usually small and can be neglected.
Thus, approximately,
5 11 2 fXd2 ( 1 Xd 2 )
V = - 48 Elo +.El "2 - Jj2
and the ordinates of the influence linc are defined by the expression
vIS ( 5 I Xd 2 Xd 4 )
-  = 8( 1 + 8) 48 J - 211 + 3113
(8.4)
In Fig. 8.6b, this line is sho\\'n by thc curve Qlt:}b 1 . With this line, the
influence lines for bending moment and for axial force at any cross section [)
can be constructed \vithout difficulty. Denoting by !vI' the bending moment
for a simply supported beam A B of span I and defining the position of D by
the distances a, b, and c, \ve find that the bending nl0ment at the cross section
D of the arch, for any position of the moving load P, is
( All )
A1 = ,\iI' - He = e --c - II
It is secn that the moment A1 is obtained by multiplying by c the ordinates of

340 ARCHES AND FRAMES
p
(a)
f
H _______L___-
H
A

a
B
b
Cl
(b)
25 ,
128 (1+{3)
4t ---- - _____l___ - - - - - b 1
I.. a b 
(c)
]
T
tan
1
1
tan
1
FI G. 8.6
(d)
the shaded area in Fig. 8.6c, which represents the differences of the ordinates
of the At!' / c influence line and the H influence line.
The axial for«e for any position of the moving load is
;.V = H cos 4> - V; sin 4> = cos cp (H - V tan cp)
\vhcrc V is the shearing force at D for a simply supported beam. Hence,
lV is obtained by multiplying by cas cp the ordinates of the shaded area in
Fig. 8.6d, which represents the di tfcrences of the ordinates of the H influence
Jinc and the V; tan cp influence line.
PROBLEMS
'[hc sYJnnlctrical t\va-hinged circular arch sho\\'n in Fig. 8.7 a has a center line of
radius rand subtends a central angle 2a. l'he cross-sectional area A and moment

ARTICLE 8.3
341
of inertia I = Ak 2 are both constant along the length of the arch. Using the
second Castiglia no theoretn, dcvelop a general expression for the decrease  in
the chord distance AB due to thrust 11 acting as sho\vn.
A'Ils' d = (2Hr 3 jEl)F(a)[1 + (k2jr2)(;(a)1 ",herein
F(a) = al2 - ! sin 2a + a cos 2 a,
G(a) = (0./2 + i sin 2a)/F(a).
2 Referring again to the circular arch sho\vn in Fig. 8.7 b, dcternline the magni-
tude of the thrust Ii produced by a uniformly distributed vertical10ad of intensity q
on the arch jf the hinged ends A and R are restrained. Vsc F.q. (8.2), neglecting
the sc(ond ternl j n the nunlcrator.
qr 1/I(a) 1 .
411S. II = - - - - -- , \vhercln
2 F(a) 1 + (k/r)2G(a)
.p(a) = j sin 3 a + -a cos a cos 2a - t cos a sin 2a.
3 H.efcrring again to the ci rcular arch sho\vn in Fig. 8.7 b, determine the thrust II
produced by a uni form rise in temperature t if the coefficient of thermal expansion
for the arch is at and the hinged ends A and J1 are restrained.
A H = ! Ja ltl _ _ J _
ns.
2r 3 F(a) 1 + (k/r)2G(a)
C
(a) A  -- L -  B
I' H I H 'I
I " I / :
I , I
q
/
,
,
/
/
/
r ' I , r
... I I
... a I a /
'I
" I 'I'I

(b)
FIG. 8.7
8.3 SYMMETRICAL HINGELESS ARCHES
In general, a hingcless arch represents a structure having three redundant
elemcnt. We can consider as redundants the bending moment, shearing force,
and axial force at some chosen cross section of the arch. In the case of a
symmetrical arch, it is advantageous to takc this cross section at the crown C.
l'hen, jf the loading on the arch is also symmetrically distributed, the shearing
force at the cro\vn vanishes, and \\'e have to consder only t\VO redundant
quantities: the bending moment Ale and the axial force N c at the cro\vn as
sho\\'n in Fig. 8.8a. "rhese redundant elements \vill be obtained by using the

342 ARCHES AND FRAMES
c
x
(b)
(a)
FIG. 8.8
theorem of least \vork; this gives the equations
aC:
a 1'\1 c
_ 2 r . s A.I _ aAJ d,' = 0
In EI aA1 c .
aL,T
di\Tc
r s AI ai\J r B l\T 01\i
-- 2 J 0 £ 1 ai\r(. dJ' + 2 J 0 :4£ a l'lc d.r = 0
(a)
\vhere s denotes the length of the arc ('B in Fig. 8.8, that is, half the length of
the full arch axis.
l)enoting by A.J' and 1\!" the bending InOOlcnt and axial force at any cross
section due to the txternal loading only, the total bending Il10mcnt and axial
force at any cross section 1) become
/\1 = /\1[(' + l\f,;y + AI'
1\1 = l\T c cos cJ> + i\? I
(b)
Substituting these expressions and their derivatives \vith respect to AIr and l\;c
into Eqs. (a), \VC obtain
Ale fo' ;; + N e fog yy = - )' ;' ds
\ 1 r 8 'V ds ;\. (81,2 dJ + ;\.! r.'I O S2 4> d -
i i C J 0 "-£1 + i V (' JoEl - .n C J 0 A r; J
(c)
_ _ r 8 JI') ds _ r s / O  dJ
Jo EI Jo IdE
\\lhen the dinlcnsions of the arch and the type of cxternalloading arc kno\vn,
the magnitudes of the rcd'Jndant clen1cnts i\4(. and 1.\:(' can be calculatcd from
these equations.
Equations (c) can be sornc\vhat sinlplificd if \\le 010VC the origin of coordi-
natcs froIl1 the cro\vn C to a point 0 as sha\vn in Fig. 8.8/7 and then selcct the
distance d so that the ne\\<' ordinates .'VI = Y - d satisfy the condition
r 8 Yl dJ r ' y - d i
J 0 - RT = J 0 -EI- £ S = 0
(d)

ARTICLE 8.3
343
from \vhich
foB (y dsjEl)
d=
foB (dsjEl)
(8.5)
The point 0, defined in this way, is caJled the elast;c cel1ter of the arch. If \ve
use it as the origin, substitute .1'1 + d for y into Eqs. (c), and observe that
fo. (vt! El) ds vanishes, we obtain
. (M' dsjEl)
,\1 0 = ,\I e + Hod = - O  s
(dsl RI)
o
(8.6)
_ fo' M'y';EI) ds + foB [(N' cos q,)jAE] ds
Ho = N c =
foB (yNEl) ds + foB [(cos 2 q,)jAE] ds
rhese same results can be obtaincd also if \\'e irnagine a rigid brackct CO
attached to the cross scction C of the arch and introduce, instead of N! c and lV c ,
the statically equivalent system of forces Mo and Ho acting at the end 0 of
this bracket. Then, choosing /\1 0 and Ho as the redundant quantities and
procceding as before \vith the theorem of least \vork, Eqs. (8.6) \vill be
obtained directly. In this \vay, \ve obtain t\VO independent equations (8.6)
for the determination of the magnitudes of the redundant clcments 1\1 0 and Ho.
\\lith tbe aid of the reciprocal theorem, \\'e can see that the use of the clastic
center 0 nas a simple physical significance. Namely, the force Ho applied
at () produces no rotation of the cross section at ('; conversely. the monlent ,'10
produces no horizontal displacement of point O.
Let us consider no\v the case of a symmetrical hingcless arch subjected to a
uniform rise in temperature. Taking Ho and Mo as the redundant quantities
(Fig. 8.Sb), \ve observe that the magnitude of Ho must be such as to counteract
the horizontal displacement atl/2 of point C duc to the rise in temperature.
Also, from symnletry, it follo\\'s that there must be no rotation of the cross
section at thc cro\vn C. Thus, using the second Castigliano theorem, we have
aL!_ _ ( 8 ill _(J\! ds = 0
aA1 0 - JoEl 01\1 0
a Ll = ( 3 Alf A/ dr + ( 8 ]\v' alv ds = atl
a J-! 0 J 0 R 1 all o ' J () A E a H 0 2
(e)
\vhcrc
,"'1 = 1\;1 0 + Hoyt
.\r = Ho cos cJ>
(f)
Substituting expressions (f) and their dcrivatives Into Eqs. (e), \ve obtain

344 ARCHES AND FRAMES
/\I[ 0 = 0 and
Ho = _____ atl/2
/os (Y1 2 ds/ El) + /0' l (cos 2 cJ> ds) / AEJ
(8.7)
From this equation, the thrust due to a uniform ternperaturc change in any
symmetrical hingelcss arch can be calculated \vithout difficulty if the dio1cnsions
of the arch and the change in temperature are given.
In the development of Eqs. (8.6), \\'e assumed that the arch \vas symmetri-
cally loaded, in \vhich case the shearing force at the cco\vn cross section
vanishes. l...et us consider no\\' the case of antin1ctric loading such as that
sho\vn, for example, in Fig. 8.9a. From symmetry of the arch, \ve conclude
that at the cro\"n ( the bending moment and the axial compression produced
by the force P to the right of Care nun1crically equal and opposite in sign to
those produced by the force P to the left of C'. "[his mcans that under the
simultaneous action of the t\VO forces P, the bending mon1cnt and axial com-
pression at the cro\vn both vanish, and \ve have to consider at the cross section
(' only the shearing force V o as sho\vn in Fig. 8.9 i, . For the calculation of this
redundant quantity, \VC use the theorem of least \vork \vhich gives
a () ( 8 i\J a 1\4 ( 8 l\T a 1\:
avo = 2 )0 EI a.'v o ds + 2 Jo AE a  cis = 0
(g)
Since the strain ener6'Y is the same for both halves of the arch, \VC have to make
the integration only from C to B and double the result as sho\vn. Observing
from Fig. 8.9 b that
1\tJ = V 0..\'1 + /\I[ 1
N = - V o sin cp + 1\:1
(11)
and substituting into Eq. (g), \ve obtain
( B J10X 1 + A4' . d _ ( 8  V o sin cp + l\r I. d -
Jo EI Xl s}o AE sin cp s - 0
from \vhich
V o =
/0' ( M'xtlEI) ds + /0' [(iV' sin cJ»/AEJ ds
- --_.. ---- --- --
/0' (xNEI) ds + /0' (sin 2 cJ»/AE] ds
(8.8)
c p C P x
//v Xl
// 0
I
I

 Yt
FI G. 8.9 (a) (b)

ARTICLE 8.3
345

q
, . .
, I I
I I .
, I .
I . I
I . , q
" [  III r 1111 J 11111 [ II ] ! I "2
. : :
1 I ·
I I I
. . :
. I , q
: ,  I i III  II : II "2
I J Illl] I J II
qlq2
. I I
, I I
I I I
I I I
I It.
rrrr:rrnmmmmT1 1 q 1 -t- q2
llill.J.J.WJ.lL 
I I
I I
I I
I I
I I
I : q2- q,
I  -
2
I
I
I
UJ"
I
I
P p:

I 1
I
:  :
I <
P
2
(a)
FI G. 8.10
(b)
(c)
From this cquation, the shearing force J/o can be calculated in each particular
case of antin1ctric loading provided the dinlcnsions of the arch are given.
Having solutions for SYI11I11ctric and antinletric loadings, t:,c general case of
loading can he treated by splitting the load into t\VO parts, symmetric and
anrinlctric, as illustrated in Fig. R.1 0 for several particular cases. Some
applications of this method of treating the general case of loading \vill be sho\vn
later.
Iet us consider 110\V the construction of influcnce lincs for symmetrical
hingeless arches, for \vhich purpose \ve use the reciprocal theorem. 10 con-
struct the influence line for the thrust Ho, \ve have to compare the actual
condition of loading sho\vn in Fig. 8.11 a \\,'ith the fictitious loading sho\vn in
Fig. 8. 11 b. In thc first case, \ve have the vertical load }) at the point J) and
the unkno\vn redundant forces j\'[ 0, H 0, and /o at the cro\vn} In the second
case, the load P and the redundant quantities All () and J. 7 o arc removed, and
instead of Hn, t\\"o oppositc unit forces are applied.' ()\ving to the specific
choice of point () rsee Eq. (8.5)], the unit forces \vill not produce any rotation
of cross sections C. rhesc cross sections \vill move apart only by the amount
2 u o \vithout any rclative rotation or rclative sliding. Since the \vork of the
redundant I110nlcnt i\J 0 and of the redundant shearing force J70 on the corre-
sponding displacetnents produced by the unit fortes in Fig. 8.11 b vanishes, the
reciprocal theoretJ1 gives
Hollio + Ptd = 0
and \ve obtain
Vd
110 = -p--
2uo
(8.9)
l'he displacements 110 and 'Vd due to the unit loads acting as sno\vn in Fig. 8.11 b
1 To sh()\v these forces fHore clearly, a small distance bct\veen the extension bars OC is
shown in Fig. 8.11'1.

346 ARCHES AND FRAMES
can be calculated by using Castigliano's second theorem, giving
/. . B )' 1 2 . /. H COS 2 cJ>
//0 = EI dJ + -- 1 - E ds
C C / 
_ r R )'1 (Xl - .\'d) d + J B cos 4> sin  d
'lJd = j D El S D AE S
(R.9a)
In any particular case, \vhen the shape and dirncnsions of the arch arc given,
the displacements (8.9a) can be evaluated and substituted into Eq. (8.9) to
obtain the influence coefficients for the redundant quantity flo.
In a similar nlanner, influence Jines for the redundant rnomcnt j\l[ 0 and the
redundant shearing force I, can be obtained. In the case of 1\/[0, \ve compare
(a)
Yo
p
Ho 0 0 Ho
 j t;.-- 
Xl
(c)
X
(b)
Vo
1 1
V o
t
(d)
V d
FIG. 8.11

ARTICLE 8.3
347
the actual loading condition (Fig. 8. I Ia) \vith thc fictitious case sho\\'n in
Fig. B. I I e. In this latter case, the unit moments do not produce any relative
displacements of the points 0 but oniy rotate the cross sections C, one \virh
rcspect to the other, by the angle ZOo as sho\vn. ()\ving to this fact the \vork
of actual forces Hu and /o in Fig. 8.lltl on the corresponding displacements of
the fictitious case in Fig. 8.IIe vanishes, \vhcrcas the \\Lork of 1llo is A.J o (20 o ).
At the same tin1c, in thc actual case of a continuous arch, there is no relative
displacemcnt corresponding to the unit mOInents of the fictitious case, and the
\vork of these n10lnents vanishes. l'he reciprocal thcoreJn then gives
1l1028o + PVd = ()
and \\'e obtain
'L'd
i\1 o = - p.-
28 0
(8.10)
It is seen that the required ordinates of the influence line for Alo \vill be
obtail1cd as soon as the displacements 'L'd and 8 0 produced by the fictitious
loading in Fig. H.lle have been calculated. For this purpose, \VC again llse
the second Castigliano theorcm and obtain
8 0 = J B ds
c £1
(B. IDa)
J B Xl - Xd d
'L'd = - - s
f) EI
To obtain the influcnce line for V O , \\'C have to comparc the actual loading
condition (Fig. R.lla) \vith the fictitious case sho\vn in Fig. 8.lId. ()bserving
that in the later case only relative sliding 2vo of the cross sections at the cro\vn
is produced and that no relative rotation or separation occurs, \ve obtain from
the reciprocal theorem the equation
- J702vo + PVd = 0
from \\rhich
110 = fJ _'? d_
2vo
(8.11 )
For calculating the displaccrnents 'tt'd and 'tt'o produced by the unit forces In
Fig. 8.11 d, \VC again use the Castigliano theorcm and obtain
(B.\'1 2 (Bsin 2 cP
Vo = - ) c EI ds - J c .-AE- ds
''' d = - f B X l (x 1 - Xd) i . _ J lJ sin 2 cp d
v D El ( .\ D AE S
(S.lla)

348 ARCHES AND FRAMES
It is secn that, in all three cases, the ordinates of the required influence lines
arc proportional to the vertical deflections 't'd produced in the arch by the unit
loads applied as sho\vn in Fig. 8.11 b to d.
As a specific example, let us no\\o' consider influence lines for a hingcless
parabolic arch the central axis of \vhich is defined by the equation
4(x 2
V = -'-
. /2
(i)
and the cross-sectional variation by the expressions
A = . Ao
cas tP
I = -.
cos cP
( i)
Beginning \..rith Eq. (8.5), \ve find that for such an arch the elasti c ccnter is
located by the distance
(1/2 d
J( v x
d = . _ = f
f. 1/2 i l .
IX
o
(H. Sa)
\;\/ith this clastic center as an origin of coordinates ..\'}, .vI, \\'c have Xl = x and
)'1 = Y - d.
For the Ho influence line, \\'C use Eq. (8.9), in \vhich the values of uo and 'L'd
arc givcn by Eqs. (8.9a). For a first approximation to 11 0 , \\'C nlay neglect
the sccond terms containing cJ> in Eqs. (8.9a) and, \vith this simplification,
obtain
Uo =
r 1/2 )'1 2 . d. ! = _!2
}o Elo 45Elo
_ (1/2 Yl (Xl - J"'d) dXl =
)0 Elo
11 2 ( _ 4  / X _ d2 ) 2
- 48Elo 1 -,
(8.9b)
'L'(I =
Substituting these va)ucs into Fq. (8.9), \VC hnd
I/o = ] 5fl. ( 1 _ 4 X d 2 ) 2
64f /2
oJ
For the A-f 0 influence line, \VC use Eqs. (8.10) and (8.1 Oa), the later becoming
{ 1/2 d.t'l 1
0 0 = jo El = iEI
(8.9c)
r l/2 (Xl - Xd) dXl _ I ( I ) 2
'Vd = EI - - 2Elo _ 2 - Xd
- "Xd .--
When these values arc substituted into Fq. (8. to), \ve obtain
Mu = £ G - X<lY
(8.10b)
(8.1 Dc)

ARTICLE 8.3
u.
Having this expression for k10, \ve find the bending moment Me at the cro",'n C
from the equation
!VIc = Mo - Hod
(k)
For the V o influence line, Eqs. (S.IIa), neglecting again the second terms
containing 4>, become
_ ( 1/2 Xl 2 dXl = _ /3
Vo = Jo Elo 24El o
(8.11 b)
Vd = _ ( 1/2 Xl (Xl - .td) dXI = _ J__ (  + Xd 3 _ /2 Xd )
} X.I ETo Elo 24 6 8
and t:q. (8.11) then gives
V o =  ( 1 + - - 3 X d ) (8.11&)
2 /3 /
The influence Jines for H 0, M 0, V o , plotted from Eqs. (8. ge), (8.10&), and
(8.l1e), are shown in Fig. 8.120, b, and c, respectively. Also shown in Fig.
8.12d is the influence diagram for bending moment M r at the crown C as
defined by Eq. (k).
I.
I
2
I
2
t
(a)
(b)
(c)
(d)
FIG. 8.12
1 Ho
Mo
Yo
Me

350 ARCHES AND FRAMES
Comparison of arches of various shapes indicates that considcrable deviation
from the shape of the rch defined by Eqs. (i) and (j) produces only a small
effect on the ordinatcs of influence lines. This means that the influence lines
derived for parabolic arches can be used in an approximate analysis of arches
of other shapes. The use of these lines, hO\\Tevcr, in the calculation of strcsses
produced by dead load or tcmperature change may rcsult in errors of con-
siderable magnitude. l'he reason for these crrors is clearly scen from the
influence line in Fig. 8.l2d. If we assume, for cxample, that a dead load q/
is uniformly distributcd along the span, the bcnding moment Me at the crown
is obtained by multiplying by q the algebraic sum of the three shaded areas of
the influence diagram. It is evident that the accuracy \vith \\,hich this sum is
obtained will be much lo\ver than the accuracy \vith \\,hich the ordinates of the
individual H 0 and A10 influence lines are kno\vn.
PROBLEMS
lJ sing Eq. (8.7). develop a general formula for thrust H, in the hingcless
parabolic arch defined by Eqs. (i) and (j) due to a uniforln rise t in temperature.
The coefficient of thennal expansion for the material is ex, and V 10/ Ao = ko is the
radius of gyration of the cross section at the cro\vn.
45 Eloat
Anf. H, = 4/2 [1 + H(I/J) (k 2 /j2) r;n- 1 (4]il)f
2 Using the inAuence lines sho\\'n in Fig. 8. J 2 for the hingelcss parabolic arch defined
by Eqs. (i) and (j), calculate the bending moment ltJ c produced at the cro\vn C by
the dead \veight of an earth fill extending over the entire span and level \vith the
cro\vn of the arch. ,:ssume that the earth fill \veighs 'w Ib/ft 3 and that the arch
rib is 1 ft \vide.
Ans. "Ie = -w/:t/560.
3 Find the distance d defining the position of the clastic center 0 of the hingeless
circular arch sho\vn in Fig. 8.13. A.sSl1n1e that the cross-sectional area A and
moment of inertia I of the rib arc constant along its length.
Al1S. d = (ria) (a - sin a).
4 Using the first ofF.qs. (8.6) and referring to the hingeless circular arch in Fig. 8.13,
I
2
FIG. 8.13
y
I
.... I /
" I /
r.......  a a //r
... /
.... I /
..../

ARTICLE 8.4
351
calculatc the redundant monlcnt 1\1 0 due to a uniform load of intensity q extending
ovcr thc full span as sho\vn.
ql2
AJ/.c .\.1 0 = --- (a - sin a cos a).
I6a sIn 2 a
5 Uing Eq. (8.7) and referring to the arch in Fig. 8.13, calculate the tnagnitude of
the thrust Ho produced by a uni fonn rise in tctnpcrature of (0. The coefficient of
thermal expansion is at, and the radius of gyration ofthccrosssccrion is VI/A" = k.
AI/.f. Ho = F(a) (J/extt/r2), \vhcrc }'(a.) =
2a. sin ex
- --
a. 2 + ex sin ex cos a. - 2 sin 2 a + (k'l/r2) (a 2 + a sin a CQS ex)
8.4 NUMERICAL CALCULATION OF REDUNDANT ELEMENTS
The calculation of redundant quantities in arches requires, as \\'e have scen,
the evaluation of various intcgrals such as those appearing in Eqs. (8.5), (8.6),
(8.7), etc. In the case of parabolc and circular arches like those discussed
in the t\VO preceding articles, such integrals can be evaluatcd in an exact
manner. Ho\vever, \\le somctimes have morc complicated curves, or curves
for \\,hich no analytic expression exists, and only a set of ordinates is given
numerically to dcfine the shape of the arch axis. In all such cases, recourse
must be had to numerical mcthods of evaluating the intcgrals. r"'or this
purposc, the arch axis \vill be divided into a finite numbcr of scgn1cnts and the
intcgration replaced by a summation. In using this numerical method of
approximate integration, there arises the qucstion rcgarding the numbcr of
subdivisions that must be used in order to obtain the dcsired accuracy in the
result. This qucstion can best be ans\\'crcd and the accuracy of calculations
demonstrated \\'ith particular examples. To sho\v this, \ve select an example
the rigorous solution of \vhich can be found; then, the crror of the approximate
calculations can be readily sccn.
IJct us take a parabolic arch (f'ig. 8.8) for \\,hich
.f I
1 R
A Ao
= ----
CDS tJ>
I=-
cos 3 q,
(a)
and calculate the thrust 110 produced by a uniform rise in temperature. 'The
position of the clastic center is dctcnnined froIT1 Eq. (R.S), \vhich gives
f 112
, V cos 2 cp d.t
_ 0" .
d - - ---- -
r 1/2
}o cos 2 cp dx
/[1 -=- (//4/) t n-_ (4// 1 )]
4 tan- 1 (4.(/1)
(h)
The calculation of ,Ho no\v rcquircs the evaluation of the t\VO integrals in thc

352 ARCHES AND FRAMES
TABLE 8.1
DATA FOR THE PARABOLIC ARCH
GIVEN BY EQS. (0)

I cost .p
Y1 2
/2
yl2 COS 2 cJ>
-------
/2
sin 2 cb
X Yl
-
I /
o
-0.03920 0.001,37
0.001537
o
1 -0.03725 () . 00 1 3 88 2:')6 0.001383 1
16 "25--1 -21"
1 -0.03138 0.000984 64 0.000969 1
If lfE lf1>
f.r -0.02162 : 0.000467 206 0.000452 9
2"65 -26l)
1 -0.00795 0.000063 J G 0.000059 1
:r 1"1" T'f
5 +0.00963 0.000093 5G 0.000085 25
T1T .-!:I"r -2""A"T
a +0.03111 0.000968 64 0.000849 9
"8- 7'3- '7 "9'
7 +0.05650 o .003 192 26 0.002679 49
Dr 11" 0 lr O5-
1 +0.08580 0.007362 4 I 0.005890 1
 "3" i "3"
dcnominator of Eq. (8.7). .l'he first of these integrals is
(3 111 2 ds I ( 1/2
)0 . El . = Elo Jo .1'12 cos 2 cP dx
- 3iil o g + :j [ f - tan"; i 1(4//1) ]} (c)
and the second is
{ B COS 2 4> ds _ I (l/2, 2 . _ /2 _I 4/
)0 A""E - AoE Jo cos cP d.t - 8AoE f tan T
(d)
With these results, Eq. (8.7) gives
Eloat
Ho = 111 /2
( e)
\vhere
1JJ =
16
1 / [ / 1 ] 4k2 -1 4/
3" + 4/ 4f - ta n- 1 (4f/ /) + .(1 tan T
(I)
k = v IolA 0 being the radius of gyration of the cross section.
We shall no\\r repeat the analysis by nurnerical integration. For this pur-
pose, \\'e divide the length of the half span into eight equal parts and calculate
for each subdivision or segment the quantities given in Tablc 8.1 by using for a
parabola the kno\vn equations
4fx 2
'VI = -- - d
. /2
cos 2 4>
I
- I + 64f 2 x 2 J /4

ARTICLE 8.4
353
No\\' applying Simpson's rule to the integral (c), \ve obtain
1 r 1/2 /3 I .
RIo}o Y1 2 cos 2 4> dl' = 16E I X "3 [0.001537 + 4(O.0013H3
+ 0.000452 + 0.000085 + 0.002679) + 2(0.000969
+ 0.000059 + 0.000849) + 0.005890] = 0.004929 8fI
Proceeding in the same manner \\'ith the integral (d) and assuming a rectangular
cross section, \ve obtain
1 { l/2 / 1110 2
A;E)o C05 2 cp dx = Z;I oE 0.9273 = -2 4EI  0.9273
Substituting these calculated values of the integrals into Eq. (R. 7), \\'C obtain
HoI 3 ( 1 h02 . ) atl
8Elo 0.004929 + 3 -/2 0.9273 ="2 (g)
"[he second term in the parentheses represents the influence of the longitudinal
force N on the value of H u. I..et us consider the case of a very thick arch and
assume 110/1 = -b. l'hen, Eq. (,,) gives
H 0/ 3 atl
8E lo (0.004929 + 0.003816) - 2
and \\'e obtain
H) = 457.4 at{;lo
Taking 110// = }-,-, \ve find
H = 747 1 atEl o
} . . /2
(12)
(i)
To four significant figures, these results coincide exactly with the values given
by Eqs. (e) and (I). 1'his indicates that the subdivision of the half span into
eight parts and the use of Simpson's rule give a very high degree of accuracy.
If \VC divide the half span into only four parts, Simpson's rule gives Ho for the
foregoing example \vith an error of about 0.2 percent. This shows that it is
not necessary to divide the arch into many parts to obtain good accuracy by
numerical integration.
In applying Simpson's rule, \ve have to subdivide the center line of an arch
into equal parts. Sometimes it is sirnpler to have unequal subdivisions, and in
such case Simpson's rule cannot be used. Then \ve replace the integrations
by simple summations. 'e calculate the values of the quantities under the
integral signs for the center of each segment and sum up these values. To
sho\\' this kind of calculation, let us determine the thrust produced by a uniform

380 ARCHES AND FRAMES
A
61.3
0.984
FI G. 8.24
of the last three columns here in Eqs. (8.25), \ve find
lz, = f )'0 2 d; = 19,861 lyO = f "0 2 d; = 482,828
{3 = 10°20'
(0)
No\\", having the directions of the principal axes of the arch, \ve can calculate
the ne\v coordinates Xo, Yo of the rnid-point of each segment. Such values of
)'0 arc sho\vn in the first colunln of the third portion of Table 8.9. Follo\ving
this, the sunlmation of statical nl0mcnts of the quantities )'0 s/ I about sucessivc
division points 9', 8', 7', . . . arc computed and recorded in column 5 of the
third part of 1:1ble R. 9. U sing these values as nurnerators in the first of
Eqs. (8.26) and noting that the second ternl of the denolninator is neg]igible
compared \vith the first, \\le calculate the influence coefficients o for the
redundant quantity X 0 as recorded in colunln 6.
'''-''ithout sho\ving the calculations, values of the influence coefficients 170
and So for the rcdundant quantities y and A40, respectively, as computed from
Eqs. (8.26), arc sho\vn in the last t\VO colunlns of Table 8.9. The correspond-
ing influence lines, plotted from the data in the table, are sho\vn in Fig. 8.24.
By \vay of a check on the numerical calculations, it \vill be noted that the
values of o and 110 for the lcft end of the arch (point }') give as their rcsultant
a unit vertical force and that the mODlent of this force \vith rcspect to point A
balances the moment So as it should he.
PROBLEMS
Using the influcnce coefficients from rablc 8.9, calculate the values of the redundant
quantities Xo, Yo, A10 for the unsymnH:trical arch in Fig. 8.24 due to a unifonnly
distributed load of intensity q = I kip/ ft over the entire arch.
Ails. Xo = il.9 kips, Yo = 63.9 kips, Alo = 2,545 kip-ft.

ARTICLE 8.7
381
2 Repeat the calculation of Xo, 1'0, 1\;/0 obtained by the use of influence coefficients
in the preccding problem by direct application of Eqs. (8.24). Note that the
integrals appearing in the denominators of these exprcssions are alrcady available
in Table 8.9. 'The second teffll in the denominator of the first equation can be
neglected.
3 Calculate the valucs of "Yo, Yo that \vill be produced in the arch in Fig. 8.24 due to
a uni form risc in tcrnpcraturc of 4O°F. l'hc coefficient of thern1al expansion is
a, = 6(10- 6 ) (in./in.) /oF. A.ssun1e a modulus of elasticity  = 432(10 6 ) Ib/ft 2 .
Ans. Xo = 736 lb, Yo = -2.47 lb.
8.7 FRAMES WITHOUT HINGES
Equations (8.23) and (8.24), previously derived for arches, can be used also
for calculating redundant forces in frames such as those shown in Figs. 8.25
and 8.26. In such a case, the cquations can be some\vhat simplified by omitting
the terms representing the effect of axial force. Such a simplification is
justifiable if we observe that in the case of arches under the action of dead load
the ccntcr linc is usually very closc to the line of pressure and the stresses
produced by the axial force are of the same ordcr of magnitude as the maximum
bending stresses. Hence, the tenns representing the action of the axial force
may be of practical importance. In the case of frames, the prcssure line is
usually far from coinciding \vith the center line, and the effect of axial force is
negligible in comparison with that of bending moment. Aftcr this simplifica-
tion, Eqs. (R.23) become
 R {s. " d S } J ( 8 2 d s _
o J 0 X lY 1 EI + 10 JoY 1 EI -
( B A ' 1 ' dJ
Jo iV, )'1 "EI
J. 3 ds J. ' ds
Ro X1 2 ---. + Ho Xl)'l - =
o EI 0 RI
{If  1 ' ds
J 0 ,. Xl Ei
(8.27)
M ( 3 ds = _ ( 8 M' ds
o JoEl JoEl
In each particular case of a frame analysis, \ve start \vith the determination
of the coordinates of the elastic center by using Eqs. (8.22). Then, taking
the elastic center as the origin of ;t.I)' 1 axes, \ve calculate the integrals appearing
in Eqs. (8.27). laking fictitious masses dsj El and M' ds/ EI distributed along
the center line of the frame, we see that each of the above-mentioned integrals
has a simple physical meaning. Considering, for example, the first of these
equations, \\'e see that the first integral is the product of inertia of the distributed
fictitious masses ds/ El. The second integral is the moment of incrtia about
the Xl axis of the same masses. The integral on the right-hand side of the
equation represents the statical moment about the Xl axis of the distributed

382 ARCHES AND FRAMES
fictitious rnasses ,\1' ds/ EI, etc. The calculation of these intcgrals becomes
especially simple if the frame consists of prisnlatic members.
Let us no\\' take, as an example, the case of a frame of uniform cross section
as sho\vn in Fig. 8.25 tl and aSSUlne that / = 2h and c = h. 1he redundant
reactions Ra, ]ia, Ala at the support A \ve replace by the statically equivalent
system of forces Ro, Ho, AJ 0 applied at the elastic center 0 as sho\\'n in Fig.
8.25 h. l'hcn, for calculating the coordinates a and b of the elastic ccnter,
\ve use Eqs. (8.22). Since EI is constant in this case, it can be canceled,
and \ve conclude that the clastic center coincides \vith the centroid' of the
center line of the frame. Thus,
Jo' ds = h + I = 311
 8 /2
X ds = - = 211 2
o 2
 B h 2 5
- Y ds == /h +  == - 11 2
022
and \ve obtain, from Eqs. (8.22), a = 2h/3 and b = - 5h/6. The integrals
on the left-hand side of Eqs. (8.27), aftcr canceling the constant factor 1/ EI,
are
( 3 /h 2 /z3 hh 2 1
I XI - Jo Y1 2 ds = "36 + 12 + "9- = 4- Jz3
I _ ( 8 X t 2 ds =  + Ih2 + II 4h2 =  h3
1/1 10 12 9 9 3
 8 h Jz 2h 2 1 3
I X I "' 1 ds = -I - - - h - = - -- h
Xilli = 0 J 6 3 9 3
(a)
In calculating the intcgrals on the right-hand sidc of Eqs. (8.27), \\'c observe
that k!' denotes the bending fi1on1ents in the frame \vhen the redundant
quantities 1\-1 0 , Ro, I-Io are removed. Thus AJ' is represented by the shaded
triangle in Fig. 8.25 b, \vherc \ve consider bending moment to be positive \\'hen
it produces flexure concave to\vard the outside of the frame. Accordingly,
the restltant and the statical filoments of the fictitious distributed masses
j\;[' dJ are
Jo' M ' ds = -i Ph2
 8, PJz3
At[ '}'l ds = -
O' 12
Substituting the quantities (a) and (b) into Eqs. (8.27), \ve obtain
Jo' M' xl ds = _PI13
(b)
-4Ro + 3Ho = -I)
8Ro - 2Ho = 3P
6Nl o = Ph

ART. CLE 8.7 383
P
c
B
I .t
(a)
f
p
h L Ro
8
Xl
(b)
FIG. 8.25
A
1/1
from \"rhich
Ph
A10 = -
6
P
Ho =-
4
7P
Ro =-
16
Using these values of the redundant quantities, the complete bending-moment
diagram for the framc is easily constructed as sho\vn in Fig. 8.25a. A partial
check on this diagram may be obtained by noting that the algebraic sum of the
moment areas must vanish since there can bc no rclative rotation between the
fixed ends A and B of the frame.
Equations (8.24), cach containing only one unkno\vn, could also be used in
the analysis of frames. However, these equations, although useful in cal-
culating influence coefficients for unsymmetrical arches, have no advantage in
the analysis of frames under dead load. It is much simpler to solve the system
of Eqs. (8.27) for three unknowns than it is to determine the directions
of principal axes in the case of an unsymmetrical frame. 'fherefore, in our
further problems only Eqs. (8.27) will be used.
As a second example, let us consider the symmetrically loaded symmetrical
frame of uniform cross section sho\vn in 'ig. 8.26a. In this case, \\7C conclude
at once that the clastic center 0 ,",j II be on the axis of symmetry, and from
the second of Eqs. (8.22) \\.'c find that its distance from the x axis is 2//3.
1'hen, \vith the elastic center 0 as origin, the XlYl axes are evidently principal
axes, and omitting the constant factor 1/ EI, \VC obtain
IZll1l = 0
!c 8 /3
I = Y 1 2 ds = -
%1 0 3
!c If 7/ 3
I = X1 2 ds = - .
Ut 0 12
(c)

384 ARCHES AND FRAMES
p
p
T
I
A

PI
24

PI
24
(b)
(a)
FIG. 8.26
The bending moments M' produced by the external load P arc sho\vn by the
shaded areas in Fig. 8.26b, and \ve obtain
fo' M' ds = - .P/2
10"M'xl ds = --A PI3
fo' M'Yl ds = lr Pl3
(d)
Substituting the values (c) and (d) into Eqs. (8.27), \VC find
P
Ho = -
8
P
Ro = -
2
Afo = SPl
24
(e)
Having these values, \ve can easily construct the complete bending-moment
diagram for the frame as sho\\'n in Fig. 8.26a.
In discussing the symmetrical frame in Fig. 8.26, we used Eqs. (8.27)
derived for an unsymmetrical frame. To take advantage of the condition of
symmetry, it is preferable to use, for symmetric loading, Eqs. (8.6), and
for antimetric loading, Eq. (8.8) as derived for symmetrical arches in Art. 8.3.
Omitting the influence of the longitudinal force, these equations become
M (,q!.. = _ (3 M' ds
o 10 EI 10 EI
 If ds  ' I ds
Ho )'1 2 - = - M Yl ----
o El 0 EI
V o = 0
(8.28)

ARTICLE 8.7
385
for the case of sYlnmetrical loading and
Ho = /\;1 0 = 0
J l t If ds t a , ds
o X 1 2  = - kl "\'1 -
o EI 0 EI
(8.29)
for the case of antimetrical loading. In both cases, the cross section is made
at the cro\,'n of the arch (see Figs. 8.8 and 8.9) and the integrations are
c:(tended over one-half of the arch.
As an example, let us consider an unsymrnetrical loading of the previous
symmetrical framc as sho\\'" in t"'ig. 8.27a. Resolving this loading into sym-
metrical and antirnctrical parts as sho\vn in Fig. 8.27 band c, \ve obtain
r'  1 ' i _PC? + Pel
-Jo' (S-4 2
_ r' A1'Yl ds = _ Pc'll + Pc/ 2
10 12 12
r :f Pc 2 ( I ,. ) PC/2
- Jo 1\;['Xl ds = 4 2 -"3 + T
\\!hen these values are substituted into Eqs. (8.28) and (8.29), \ve find
l:J c 2 J)c
AtJ 0 = --- - + - -
6/ 3
Pc 2 Pc
Ho = - - + -
2/ 2 2/
61)c 2 ( I f ) 6!)c
V o = 7/2 2 - 3/ + 7T
69Pl
1,000
P
2
P
2
Ho
A
A
42Pl
1.000 (a)
FI G. 8.27
3lPI
1,000
(b)
(c)

386 ARCHES AND FRAMES
q
A ql
2
1/
+
:1
12'
x

I := 24 '
(a)
(b)
FI G. 8.28
For (" = /13, these results reduce to
71 J /
1\1 0 = h
54
P
Ho =--
9
v; _ 61P
o - 189
'rhe corrcsponding bending-momcnt diagram is shown in Fig. 8.27a.
In the preceding cxamples, \ve have considcred only frames of uniform cross
section. Franlcs consisting of prislnatic Inen1bcrs \virh different values of I
can be treated in a similar manner \\'ithout much difficulty. "[0 iIlustratc, \VC
considcr the examplc sho\\'n in Fig. 8.28. A symmetrical frame, the ends A
and B of \vhich are built in, carries a uniformly distributed load of intensity
q as shown. '] 'he cross-sectional moment of inertia of thc legs is II and that of
the raftcrs is 1 0 , \vhere II = 31 o.  can reduce the problem to that of a
fran1e of uniform cross section sinlply by multiplying Eqs. (8.27) by Elo. In
such a \yay, the diffcrcnce in the values of II and 10 \\Till be compensated for by
distributing along the legs the fictitious Inasses dJ InlI 1 and ,W' ds 1 0 111, instead
of ds and AI[' ds as \\'as done before.
To illustrate the actual calculations, \VC assun1e the numerical data for the
frame as sho\vn in :Fig. 8.28a. "rhen, the summation of fictitious masses
dJ 10/11 distribut ed along the center line becomes
2 V5 2 + 12 2 + * X 2 X 12 = 34 ft
'[he statical moment of the same nlasscs \vith respect to the x axis is
2 X 1 3 X 14.5 + -l X 2 X 12 X 6 = 425 ft 2
"rhus, from Eqs. (8.22), the coordinates of the elastic ccnter become
d = 12ft
b = _\2/- = 12.5 ft

ARTICLE 8.7
387
No,\r cutting the framc at A, \ve have the system of redundant elements
acting at the elastic ccntcr 0 as sho\vn in Fig. 8.28b. 1 The momcnts of
inertia of the fictitious masses \vith respect to the principal axes ..\'J, )'1 are
1%. = 2 C--- ? p + 4 X 3 122 ) - 34(0.5) 2 = 592.2 ft 3
I,,, = 2 (4 X 12 2 + __1 2) = 2,400 ft 3
Using for kl' the ordinatcs of the bending-tnom,ent diagram sho\vn in Fig.
8.28b, \ve no\\' consider thc fictitious masses J;f' ds /0//1 distributed along the
center line of the inclined mcmbers KD and DI. The centers of gravity of
these masses \vill be located at points F and E as sho\vn in Fig. 8.28b. Then
the integrals on the right-hand side of Eqs. (8.27), multiplied by RIo, become
f TJ 2 q J2
- K }\J'YI ds = 3" 8" 26 X 2t = 3,276q
f lJ
- K kI'x! ds = 0
f L 2 ql2
- M' ds = - - - 26 = - 1,248q
K 3 8
fr'
ft:.
Substituting the obtained numerical values togcthcr \vith the expression
1\;/ 0 - q/2/4, instead of Alo, into Eqs. (8.27), we obtain
H = 3,276q = 5 532 ft
o 592.2 . q
ql2 1,248q
Mo - 4 = - -34-- = -36.71q ft 2
,"'ith thcse values of the redundant quantities, the bending moments at the
cross sections A, K, G, D (Fig. 8.28a) will be
Alfa = - 36.71 + 5.532 X 12.5 = + 32.44q ft 2
,\1 k = - 36.71 + 5.532 X 0.5 = - 3 3 .94q ft 2
}\tifl = - 36.71 - 5.532 X 2 + 54 = +6.23q ft 2
i\;[d = - 36.71 - 5.532 X 4.5 + 72 = + 10.40q ft 2
The corfesponding bending-moment diagrarrl fOf the left-hand half of the
framc is sho\vn in Fig. B.28a.
In the case of frames having nonprismatic members, the problcm can be
1 To takc advantage of symmetry, ,\o'c leave the known vertical force ql/2 at A and com-
pensatc by putting at () the (noment Mo - q12/4 instead of A/o.

388 ARCHES AND FRAMES
reduced to that of a frame l:aving uniform cross-sectional moment of inertia
10 by using the modified fictitious loads ds 10/1 and M' ds 10/1, instead of ds
and M' ds as previously used.
PROBLEMS
I A rectangular frame ADFR built in at the supports A and B is uni fonnly loaded
along the horizontal member DF as sho\vn in Fig. 8.29. 'rhc moment of inertia
of the cross section of each leg is /1 and that of the member I)F is 10. Calculate
the bending moment at D and construct the bending-moment diagram for the frame
if h// = t and 10 = 21 1 .
Al1S. Md = -q/2/21.
2 A frame of the same dimensions as that shown in Fig. 8.28a is subjected on the left
side to the action of a horizontal load uni formly distributed along the vertical axis
as sho\vn in Fig. 8. 30a. Calculate the redundant quantities Ii 0, Atl o , V 0 at the
elastic center 0, and construct the bending-mon1ent diagranl for the frame.
Hint Divide the loading into symrnetrical and antirnetrical parts as
sho\vn in Fig. 8.30b and c and use Eqs. (8.28) and (8.29), derived
for symmetrical frames, to make the analysis.
Ans. Ho = 3.743q ft, ;\10 = 9.417q ft 2 , V o = -1.533q ft.
3 An unsymmetrical frame having the dimensions sho\vn in Fig. 8.31 is subjected
to the action of horizontal loading uni forrnly distributed along the vertitClI axis as
shown at the left side. The moment of inertia of the leg on the left is /0, and for
the other members as sho\vr. in the figure. Calculate the redundant quantities Ho,
F
/0 I
11 1 1 h
A 81
FI G. 8.29 I- I 
q
!!.
2
q
2
q
2'
D
q
2
(a)
(b)
(c)
FI G. 8.30

ARTICLE 8.7
-f
15'
S89
q C
-------- - 0 1
 A ----J---L______ B
I ... I ,. I
',13' I 13 ';';'
5 ' I ........ " ,. 1 5 '
 12'.... .. '.. I _.- "" 12' I
--________________J
I
I
I
I 24' I
, 1 18.37' I
24.7ki!sj__ ____ __l__-J__-
10.-4---12' 12'10' q,t>
FIG. 8.31
FI G. 8.32
27.9
I
,
18'
I
J 2' _1_ -.1 2 '_ __
1 ..-
5 ' -
...... .... I / / 3 '
13'...........// 1
o
6.40 kips
111
FI G. 8.33
Xl
7.31 kips
B
8.40 kips
1\1/0' R(. acting at the elastic center 0, and construct the bending-moment diagram
for the frame.
Hint To sin1plify the calculation of the integrals on the right-hand side
of Eqs. (8.27), cut the frame at B rather than at A.
Ans. J-I o = 4.488q ft, Mo = 31.14q ft 2 , Ro = 2.659q ft.
4 A circular arch of uni form cross section has built-in ends at A and B and is sub-
jected to the action of horizontal loading uniformly distributed \vith respect to the
vertical axis as sho\vn in Fig. 8..32. l"he dimensions of the arch arc sho\vn in the
figure. C:alculatc the redundant quantities HOt Mo, Ro at the elastic center 0, and
consrn:ct a bending-moment diagram for the arch.
Al1S. H 0 = 1. 7 30q ft, A10 = 3.43 4q ft', R 0 = 0.61 5 q ft.
5 The frame sbo\vn in Fig. 8.33 has built-in ends at A and B and uniform cross sec-
tion. It is subjected to horizontal loading uniformly distributed with respect to

390 ARCHES AND FRAMES
the vertical axis as sho\vn. Determine the reactions at the supports A and Band
construct a bending-momcnt diagram for the frame.
An.r. For intensity of load q = I kip/ft, the reactions and the ordinates
of the bending-mon1cnt diagram are sho\vn in Fig. 8.33.
8.8 FRAMES WITH HINGES
I f a framc or arch has onc hinge, say at the support A in Fig. 8.34, the bending
moment at A vanishes, and \\'c havc only t\VO redundant quantities, the
componcnts Ra and Ha of the reaction at A. Assuming that the hingc A is
immovable and considering only energy of bending, the equations for calculating
Ra and H a become
au
aR a =
au
aH  =
(8   }'l ds = 0
Jo £1 aR a
( 8 M _?l/ ds = 0
Jo E! aHa
(a)
The bending moment at any cross section of the frame is
M = Ra.'. + HaY + A1'
(b)
\vhcre, as before, M' denotes bending moments produced by the loads ,vhen
Ra and Ha are not acting. Substituting expression (b) into Eqs. (a), \ve obtain
r If xy (8 y2 ( 8 Al' y
Ra Jo Ei ds + Ha Jo El ds = - Jo - EI ds
r s x 2 r 8 xy ( 8 Al' x
Ra Jo EI ds + Ha Jo EI ds = - Jo EI ds
(8.30a)
We see that thc integrals on the left-hand side of these equations represent
moments of incrtia and products of inertia, \vith respcct to the .t, y axes, of
fictitious masses ds/ EI distributed along the center line of the frame. l,--hc
integrals on the right-hand side are the statical momcnts of fictitious masses
M' ds/EI. For practical applications, \ve usually multiply Eqs. (8.30a) by a
FI G. 8.34

ARTICLE 8.8
391
ITIIlTI I L I: \I ITITTrnrn  I! II , II q
;-- 24' 24'--
J
2.4 r-
24'
I
L.tt
10 II tl 6 '
I) = 210
Ha
x
Ra /I
(a)
(b)
FI G. 8.35
constant factor E/ o and \vork \vith the equations
(sIo {sIo
R(l Jo 7 xy ds + lla Jo 7 )'2 ds =
( 8 1 0 2 (  !  ." '_
Ra Jo I x ds + Hn Jo I X) d5 -
f 3/0
- 1\;[' y ds
o I .'
r slo
Jo 7 /\J'x £Is
(8.30b)
-rhcsc equations become especially simple in the case of framcs of uniform
cross section and franles consisting of prismatic membcrs.
As an example, let us considcr the casc of a frame having a hinge at the
support A and built in at B, the load being uniformly distributcd along the
horizontal span as sho\vn in Fig. 8.35,1. Taking x, y axes as sho\vn in the
figure and assuming that 11 = 2/ 0 , \ve find thc folIo\ving numerical values for
the integrals on the left-hand side of Eqs. (8. 30b) :
{slo c.
Ix = Jo 7)'2 d... = 34,293.5 It 3
I - {s 1 0 .2 d - 4 '1 9 60 -} C. 3
11 - Jo l''t S - J, .J It
J XY = fo.)x.Yds= -32,199.6ft 3
In calculating the integrals on the right-hand side of Eqs. (H. 30b), \\'e have
1\1' = -qx 2 /2; so
f 81 0
-  AtJ I,)' d s = - 5 36 610 q
o I. ,
ft 4
{s 10
- Jo -j i\1'x ds = R32,750q
Substituring the numcrical values of the various integrals into Eqs. (8.30b),
\VC obtain
ft 4
!-fa = 6.85q
ft
Ra = 23.96q
ft

392 ARCHES AND FRAMES
The crresponding bending-momcnt diagram is sho\vn in Fig. 8.35 b for
q = 1 kip/ft.
In the case of a symmetrically loaded symmetrical frame as shown in }1'ig.
8.36, we have only one unknown, the thrust H. Taking the origin of coordi-
nates at the hinge D and proceeding as before, the equation for calculating fl
becomes
H 1 B 10 2 d - f B 10 .1' { ' d
- y s - - - iV. Y S
D 1" D 1
(c)
Then, for the dimensions sho\vn in the figure, we obtain
f B 10 Y 2 ds = 4932 ft 3
D I '
The diagram for Nf' is sho\\'n on the right-hand side of the framc in Fig. 8.36,
and we obtain
- J: o M'y ds = 1 X 3 X 18 X. 2iP = 567P ft 3
Substituting these values of the integrals into Eq. (c), we find
567P
H = 4932 = 0.115P
,
The corresponding bending-moment diagram is sho\\'n on the left-hand side of
the frame in Fig. 8.36, assuming P = 1 kip.
In the case of a frame \\,ith two hinges, we again have only one redundant
quantity and need for its calculation only one equation in addition to equations
of statics. As an example, lct us consider a symmetrical frame of uniform
cross section as sho\vn in Fig. 8.37. I t is requircd to dctcrmine the thrust H
for the case \vhere a console at E transmits a couple Alo to the frame as shown.
20'
x
g
6'
f-
6'
_ _- 1
II ; 21 0 _ _ 18'
J
B-
FIG. 8.36
0.45
3.00P

ARTICLE 8.8
393
Rcmoving the support at A, \ve take the thrust H as the redundant quantity.
Its magnitude \vill be calculated [roIn the equation
: = It B  rJ ds = 0
(d)
\vhich states that the distance bet\veen the hinges A and B remains unchanged
during defornlation of the franlc. '"fhc bending monlcnr at any cross section is
J\tl = !-1 y + 1\1'
and, \virh constant EI, \ve obtain from Eq. (d)
(e)
j B j B
H y2 ds = - 1\'['1' ds
A . Ii.'
(f)
Using the dimensions sho\vn in the figure, \\le no\v find, for the integral on
the Icft,
j /l
A )'2 ds = 27,529 ft 3
To calculate the integral on the right, \\'c note that All' = All (}t/' and obtain
- JAB M'y d' = iMo26(24 + ¥) + Mo26(¥)
= 344.51\.1 0 ft 2
Finally, substi tuting the numerical values of the t\VO integrals into Eq. (1)
and solving for H, \ve obtain
H = _ 44.5i\,lo = 0 0 12 51 ,lE L t -l
27 529 . · 0 11
,
Using this valuc of H in Eq. (e), \ve can calculate the bending nl0fi1cnt for any
cross section of the frame.
5'
24'
A
H
B
FI G. 8.37
,U o
I Y
x
l


394 ARCHES AND FRAMES
p
q
FI G. 8.38
-f
5'
r'
12' 210
lA
PROBLEIIS
:\n unsynunctrical frame, hinged at /1 and built in at /1, has dimensions as sho\vn
in Fig. 8.3H. Find the redundant reactions Rn and Ilu at ,4 due to thl action of a
lIni fornll)' distributed load of intensity q = 1 kip/ft over the horizontal span.
Ails. R" = ) 1.301 kips, fIll = 2.684 kips.
2. Repeat the solution of the previous problcrn t()r the case of a concentrated vertical
load P = 10 kips applicd at f)'
A17J. R,J = 4. 7H I kips, I-/ u = 1.822 kips.
8.9 EFFECTS OF TEMPERATURE CHANGES AND SUPPORT
SETTLEMENT
Framcs \vith redundant constraints cannot frecly expclnd or contract \vith
change in temperaturc, and as a result of this, thcmal stresscs \\rill ensue.
To analyze these stresses in the case of a fratnc \vith built-in ends (Fig. R. 39),
\Vc assume first that the constraints at the support ,4 arc rClnoved so that the
frarne can expand freely  and then determine the thcrmal displacements ll" 7,,1 t,
and Of at the clastic center O. * The rcdundant quantities Ho, Ro, and .AIo,
produced by the temperature change, nlust evidently bc of such Inagnitudes as
to eJinlinate the thermal displacements, and \VC obtain, for their calculation,
* The displacctncnts tit and 'Lt arc takcn positive in the positive directions of the x!, )'1 \xes;
Ot is positive if clockwis.
FI G. 8.39
(a)
A

ARTICLE 8.9
395
the equations
_ a l..; = r 8  iJ 1\1 ds = _ 11
aH o 10 £1 aH o I
iJ[L = (II 1\ _ ds = + V,
aRo 10 EI aR o
a U { ,,1\1 a J\;f
al\tf = } 0 £1 aJl[ ds = - Ot
(a)
()bserving that thcrc are no loads acting on the frame, in this case, \VC have
1\1 = lv10 + HoYt + R O :t:l
and Eqs. (a) becomc
(b)
r.'J X('l { 8 )'1 2
Ro)o -EI- ds + Ho)o EI ds = - tit
Ro /0'  ds + Ho /0' 71? ds = + V,
Mo /0' ; = - 8 1
(8.31 )
\vhere s is the full length of the center line of the frame.
Wc shall no\\' apply these gencral equations to the particular case of the
synlnlctrical frame sho\vn in Fig. 8.28. Assuming a uniform rise in tempera-
ture of the frame by tOF, \ve obtain 1
11 t = - atl
Vt = 0
Ot = 0
l-Iencc Ro = A10 = 0, and \Vc have only to find Ho from the first of Eqs.
(8.31). After multiplying this equation by RIo and using thc numerical value
of Ix& from page 387, \ve obtain
H _ atlElo
o - ----
592.2
laking, for example, for concrete
a = 6(10- 6 ) (in./in.) jOF
E = 2.88(10 6 ) kips/ft 2
and assuming t = 60o"', 10 = 0.1728 ft\ \VC obtain
1-10 = 0.7261 kip
f-Iaving the thrust Ho, \VC can readily calculate the bending momcnts and
stresses produced by the assumcd change in temperature.
1 l'he fictitious bar A 0 does not change its length, and since for a uniform ternperature
change there is no rotation, the displaccrncnts of points A and 0 arc identical.

396 ARCHES AND FRAMES
In the case of an unsymmetrical frame such as that sho\\'n in Fig. 8.3 1, a
uniform temperature change produces both 11, and v, displacements; so for
calculating Ho and Ro, we Inust use the first t\VO of Eqs. (8.31). In this case
it should be observed that Xl and Yl are not principal axes, and the product of
inertia docs not vanish in Eqs. (8.31).
We have a problem similar to that of thcrmal stresses \vhen \ve considcr the
effect of shrinkage of concrete on the nlagnitudes of redundant reactions.
1.hc amount of shrinkage depends on many factors, but a satisfactory approxi-
Ination is obtained by considering its effect to be equivalent to that of a uniform
temperature drop of 60°F.
SOInetimes additional stresses in statically indeternlinate frames arc produced
by settlement of the supports. If the amounts of scttlcnlcnt are kno\vn, the
magnitudes of the corrcsponding rcdundant reactions can also be calculatcd
by using Eqs. (8.3 1) . let us take again, as an example, the frame sho\\'n in
Fig. 8.28 and assurne that the support A undergoes a vertical settlement .
O\\,ing to this, the prcviously calculated vertical reaction \vill be diminished by
an amount equal to the force Ro \\'hich, if applied at the elastic center 0, would
produce a do\vn\\'ard displacement  of this point. The magnitude of this
force is obtained from the second of Eqs. (8.31) by substituting  for v,.
Multip1ying this equation by Elo, we obtain
R - Elo
o - I
111
Taking, for example,
 = 0.75 in. = 0.0625 ft
10 = 0.1 728 ft"
and using the value of 1"1 from page 3 87, \VC obtain
E = 288,000 kips/ft 2
R = 0.0625 X 2R8, O 0 X 0.1728 = 1 296 k .
o 2 400 . 1 ps
,
acting do\vnward.
As another cxample of support movement, let us consider the case in which
the support A of the previous frame rotates in a counterclockwise direction by
a small angle 8 as sho\vn in Fig. 8.40. To determinc the redundant forces
H 0, Ro, Jt[ 0 corresponding to this rotation, \\'c observe that the end 0 of the
rigid bar .4.0 undergoes not only rotation 8 but also displacen1ents
/0
1l = - a8 v = - "2
in the XI and Yl directions. Substituting these values \vith the proper signs 1
1 In Eqs. (8.31), HOI Ro. Mo denote forces required to eliminate displacclncnts produced by
a tempcrature change. Here, the same notations are used for forces producing the known
displacerllents.

ARTICLE 8.9
397
FI G. 8.40 91
Xl
a = 12.5'
I -12' 1 ... 12'
2 2
into Eqs. (8.31) and multiplying by Elo, \ve obtain
0/
HJx 1 = - 8(1£/ 0 Rof = - Elo MoA = - OEl o
111 2
laking, for example,
8 = 0.005
E = 288,000 kips/ft 2
10 = O. i 728 ft 4
and using rhe numerical values of I XI and J UI from page 387, we obtain
0.005 X 12.5 X 49,770 = _ 5 252 I .
592.2 . (IpS
Ro = + 0.005 X 21 49 ,?7Q = 1.244 kips
,
Mo = - 0.005 /9,770 = - 7.318 kip-ft
Ho=
Having thcse quantities, \VC can rcadi Iy calculatc the bending moment at any
cross section of the frame. For example, at the support A \ve have
Nfa = -7.318 - 5.252 X 12.5 - 1.244 X 12 = -87.89 kip-ft
lhc corresponding bending-moment diagram is shown in Fig. 8.40.
PROBLEMS
Referring to the unsymmetrical frame shovln in Fig. 8.31 and using the numerical
data already given for this frame, calculate the magnitudes of the redundant quan-
tities Ho and lo duc to a uniform rise in temperature t = 40°F. i\ssume that
the coefficient of thermal expansion is ex = 6(10- 6 ) (in./in.)/oF and that l =
2.88(IOb) kips/ft 2 and 10 = 0.1728 ft 4 .
Ans. }/o = 186.4 Ib, Ro = - 31.8 lb.
Z Construct a bending-moment diagram for the frame sho\vn in Fig. 8.31 sho\ving
the effect of a I-in. vertical sctrlcn1cnt of the support A. Use the saine numerical
data given in the preceding problcn1.

398 ARCHES AND FRAMES
8.10 RINGS
lhe nlethod of analysis developed in preceding articles for arches and frames
can be used also in the case of closed rings. As sho\vn in Fig. 8.41 tl, a ring
on \vhich a system of kno\vn self-equilibrating forces Ph P2, Pa, . . . is acting
represents a structurc \vith three rcdundant forces. ''', can take, as redun-
dant, the axial force, the shearing force, and thc bending momcnt at any cross
section A. 'Ve then assume that the ring is cut at this section and that a rigid
bar AO is attached to one side of the section \vhile the othcr sidc is fixed as
sho\\rn in Fig. 8.41 b. Assurning that fictitious masses ds/ EI arc distributcd
along the center line of the ring, \VC detcrmine the position of the clastic ccnter
by using Eqs. (8.22). Then, proceeding as before, \VC replace the forces
acting on the cross section A by the statically equivalcnt system of forces Ho,
Ro, ,AIo applied at the clastic centcr O. The magnitudes of thcse rcdundant
forces \vill be detcrn1ined by using Eqs. (8.27) dcrived previously for hingeless
frames.
As an example, let us consider the rectangular synulletrical frame sho\\.'n in
Fig. 8.42a. We first make a cut at A and conclude that the elastic center 0
Pi
FIG. 8.41
(a)
I
1 1
1
/2
1 2 h
P 1
x
(b)
Xl
P
I
2 P
(a) (b)
FI G. 8.42

ARTICLE 8.10 399
is on the axis of symmetry of the frame. Then, \vith the notations
10
-- = a
II
10 = (3
I').
\\le have
( 81 0
J 0 } d.'" = I + 2{3h + al
 /llo
- l' ds = {3 h 2 + I h
o I .
where s denotes the full length of the center line of the ring . Finally, from the
second of Eqs. (8.22), the distance of the elastic center from the top of the
frame is found to be
(311 2 + 1/1
b = I ( ] + ex) + 2fj }1
Having the location of the elastic centcr, \ve no\v find that
/0" 7 Y1 2 d.f = I(h -' b)2 + 133 + l3h(h - 2b)2 + alb 2
The bending moments Al' produced by the extcrnal forces are sho\vn in Fig.
8.42b, and \ve obtain
( 3fu
Jo 7 kI' ds = Pc 2 + 2{3Pch + aPcl
( 8/0
Jo 7 A,1'y) ds = (3/)ch(h - 2b) - aPclb + } J c 2(h - b)
= Pcf{3h(h - 2b) - exlb + c(1I - b) J
Substituting the values of these integrals into Eqs. (8.27), multiplicd by Elo,
\\'e find
l\if = _ /J c + 2{3 h+_ 
o c 1(1 + a) + 2{3h
(311(11 - 2b) - alb + c(h - b)
Ho = -Pc J(h- .::.f,) 2 + &113/6 + j(3h(h - 2b )2 + alb 2
Having these values of 1\;[0 and 1-/ 0 , \ve can rcadily calculatc the bending
moment for any cross section of the frame and construct a bending-moment
diagran1.
i\s a second example, let us assume that the same rcctangular frame discussed
above is subjected to the action of lateral forces .-p as sho\\Tn in Fig. 8.43a.
The external reacrions at C and D then \vill be as sho\vn. Making a cut at
the section AB, \\'c find that the extcrnal loading produces bending rnoments
1\4' as sho\vn by the shaded areas in .Fig. 8.43b. For this loading, thcre \vill be

.00 ARCHES AND FRAMES
p P
2 2
1 1
h 1 2
Xl
1'2
-f
b
p C 10 D
I
(a) (b) (c)
FI G. 8.43
no axial force and no bending moment at A; so \ve have only one redundant
quantity Ro. To calculate its magnitude, \\'e use the second of Eqs. (H.27)
multiplied by Elo. lhe integrals entcring into this equation are
/0'  .1:1 2 ds = :; [61311 + /(1 + a) 1
( 8 1 0 , Phi
J 0 7 1\1 Xl dr = -f;f (/ + 3{3h)
using \vhich \ve obtain
11(1 + 3{3h)
Ro = -p /[6{3h + /( i-+)j
Having this valuc of Ro, \ve can readily construct a bending-moment diagram
of the shape sho\vn in Fig. 8.43c for any given dimensions of the frame.
i\.S a last example, \ve consider the case of an cUi ptical ring of uniform cross
section subnlitted to unifornl intcrnal pressure of intensity p, as sho\vn in
Fig. 8.44. Cutting the ring at the section AB, \ve conclude at once from
synunetry that the shearing force at this section vanishes and that the axial
forcc is equal to pal 'Thus, there is only onc redundant quantity i\;f 0 to be
considered. lreating pa as an external force applied at A (see page 387), \VC
a
c
,--
I
Xl
b
FI G. 8.44
Yl

ARTICLE 8.10
401
TABLE 8.10 VALUES OF a AND 13 FOR VARIOUS
PROPORTIONS OF AN ELLIPSE
a/b
0.9 I 0.8 I 0.7 I 0.6
0.5
0.4 I O.J
I I I
(\' 211" 5.97 5.67 5.38 5 . 11 4.85 4. ()O 4.39
{j 211" 6.65 i.18 I 7.93 9.09 10.94 14.24 21.31
have for the bending nlomcnt i\1' at any cross section n
i\J' =  p (x 1 2 + y 1 2 - a 2 )
l'hcn the integrals in the third of Eqs. (8.27) become
fo'ds = ah foB M' ds = -pa2ah + fp13a 2 b
in \vhich a and 13 are constants depending on the proportions of the ellipse and
are given in Table L 10 above. 'rhe value of i\J 0 is then obtained from the
equatIon
1 !c 8 IIp
itl o + pa 2 = - - j\t[' d.'; = - pa 2 - - - {3a 2 b
o:b 0 2 2 ab
PROBLEMS
(:onstruct a bending-rnornent diagram for the fran1e sh()\vn in Fig. 8.42 if a = t3 =
and / = II.
Alls. b = //2, A10 = -Pc(c 1- 3/)/4/, I/o = 3Pc(1 - C)/4/2.
2 A. circular ring of uniforrn cross section is suhjected to the action of forces P directed
along a vertical diameter Dl of the ring as sho\vn in Fig. 8.45. Calculate the
rnagnitudcs of the bending moments at A and at 1).
Alls. Ala = Pr/1f' - Pr/2, ill d = Pr/1f'.
P
D
A
Xl
FI G. 8.45

Chapter 9
Continuous beams
and frames
9.1 SLOPE-DEFLECTION EQUATIONS
Frame structures Ii ke that sho\vn in Fig. 9.1 a usually consist of straight
prismatic rnembcrs \vhich can be treated as bcams elastically constrained at the
ends. Such a member as ab, for example, can be isolated as sho\vn in Fig.
9.1 b and the elastic constraints at the ends replaced by reactive couples Ma
and j\tlb. In this \\'ay, \ve reduce the problem to that of a simply supported
bcam acted upon by lateral loads togcther \vith end moments. Under such
loading, the beam deflects and thc cnd tangcnts rotate by small angles (Ja and (Jb
as sho\vn. For our further discussion, it \\rill bc helpful to establish analytical
relationships bet\veen these angles, the end Inoments, and the lateral loading.
Such relationships are generally kno\vn as slope-defiection equtttions, and \ve shall
now consider their derivation.
Assuming that the beam is bent in a principal plane and neglecting shear
402

ARTICLE 9.1
403
Q
b _-----_c
I
I
I
I
\
\
\
\
--
- - -- \
\
I
,
/
I
(a)
d
w
e
FIG. 9.1
,
I
-.;..- d<p
/I
II
,1
, I
II
II Q
II
(b)
deformation, 1 the curvature of the elastic line at any cross section is given by the
\\rell-kno\\ln equation
1 1\1
r - EI
(a)
in \\thich ,\1 is the bending rnonlcnt and EI the flexural rigidity at the cross
section under consideration. In using this equation, \VC shall take the longi-
tudinal axis of the bcam as the x axis and consider the y axis positive do\vn\\'ard
as sho\vn. V\ shall also assume that deflections of the beam arc small so that
the length of an elcJTIcnt of the clastic line can be taken equal to the length
dx of its projection on the x axis. In such a case, the small angle del> bct\vecn
t\VO adjacent ross sections a fter bending is
dx i\.J d:....
del> = r = EI
(b)
\\lith this expression, the rotations of the end tangents can be calculated by the
rnethod of fictitious loads already used in the analysis of trusses (see page 276).
In Fig. 9.2a, for cxanlple, let the curve def be the bending-moment diagram for
a transversely loaded simple beam AB. If \ve assume that only one infinitcsi-
rnal elcrnent £Ix of this beam is elastic, \vhik the rest of it renlains absolutely
rigid, the clastic line t1cb \vill be as sho\vn in Fig. 9 .2b. lhis line consists
simply of t\VO straight portions ac and cb having bet\\'cen then1 the angle del>
1 For ffalllc Illcmbers of usual proportions, this effect is known to be very small. See
S. Tinloshcnko, "Strength of lVlatcrials," 3d cd., vol. I, p. 318, D. Van Nostrand COlnpany
Inc., Pri:lceton, N.J., 195 S .

404 CONTINUOUS BEAMS AND FRAMES
e
A
r:-= x -J  d_\

(a)
a Y- l- X{_?
1; d '
L__---c-_. ,!!6 (l-x)d<i>
(b) d ------L
FI G. 9.2
a d4>
(c) t- x
b

as defincd by Eq. (b). O\ving to such deflection, the left end of the beam, as
\\'e see from the figure, rotates by an infinitesin1al angle of the magnitude
1- x
dO a = -y- dcp
(c)
Considcring clock\vise rotations as pOSItIve, \ve take expression (c) with
positive sign. The end b of the beam rotates countcrclock\\,jse, and the cor-
responding angle of rotation is
.
dOb = - :. del>
I
(d)
Expressions (c) and (d) permit a very simple interpretation. Imagine a
simply supported beam on \vhich a fictitious load dt/J = M dx/ EI is acting as
shown in Fig. 9.2c. Thcn, exprcssions (c) and (tf) reprcsent the shearing
forces at the ends of this beam due to such a fictitious load.
I-Iaving the angles of rotation of the ends resulting from bending of one
element of the beam, we shall obtain the total angles of rotation by summing
up expressions (c) and (d) over the entire length of the beam. This gives
(Ja = 10 1 J J x dt/> = 10 1 J I x 1x
8" = _ [I  d = _ {l  kf dx
} 0 1 q. } 0 1 £1
(e)

ARTICLE 9.1
405
Observing that A1 dx is the area of an elemental strip of the bending-moment
diagram in Fig. 9.2a and that El is constant for beams of uniform cross section,
\vc conclude that to obtain the angles of rotation of the ends of such beams,
we need only consider the bending-moment area AdefB as a fictitious load
and then divide by EI the shearing forces produced at the ends of the beam by
this load.
Taking, for example, a uniformly loaded beam, we see that the bending-
moment area is a parabolic segment (Fig. 9. 3a) having the area
21 ql2 qJ3
A=--=--
3 8 12
If \\'C treat this area as a fictitious load, the corresponding reactions at the
supports are qJ3/24, and \ve obtain
qJ3
8a = 24EI
ql3
8b = - 24£/
(f)
In the case of a concentrated load P applied at a distance c from the right
support, the bending-moment area is represented by the triangle in Fig. 9.3 b.
Treating the area of this trianglc as a fictitious load, \VC obtain for the angles of
rotation at the supports
])c(12 - c 2 )
8a = 6/EI
8b =
Pc(l - c) (21 - c)
6lEI
(g)
I et us consider now th{ specific case of bending of a simply supported beam
AB by couples applied as sho\vn in Fig. 9.4a. In our further discussion these
couples will be called end 11101I1cnts, and \ve shall denote them by the symbols
Mob and }\;f ba , the two subscripts indicating the ends of the beam as sho\vn in
1-
I
.j
(a)
FIG. 9.3
(6)

406 CONTINUOUS BEAMS AND FRAMES
MOb
(a) (
M
f1: ab 8
 a
(b) 
l!'

M bo
8 6 B

FI G. 9.4
r-----------iL
(c)
Fig. 9.4a. Clock\visc end moments \viIl be considered as pOSItIve. This
means that \\,ith the conventional rule for sign of bending moment, the end
momcnt Mob is identical \vith the bcnding moment at A, \\rhile the end momcnt
I\1 ba is equal but opposite in sign to thc bending moment at B. Beginning \vith
the case \vhere only the moment A1ab is acting (Fig. 9.4b) and observing that
the corresponding bending-moment diagral11 is a triangle (Fig. 9 .4c), \ve obtain
MaJ
8a = 3EI
Mabl
8b = - -.'-
6El
(h)
We obtain similar formulas for the case \\'here the momcnt kf ba is acting alone.
Conlbining the actions of thc t\\'o end nloments (Fig. 9 .4a), \\le obtain
8 - M ab ! _ Alba
a - 3EI 6EI
8b = ,\;fba l _ Atlali
3EI 6EI
(9.Ia)
Solving these t,vo equations for Mab and 1\1 ba , \VC obtain
2E!
"'f ah = T ( 28 a + 8b)
2f."/
Alba = --j- (28b + 8a)
(9.1 b)
From these two equations the end momcnts can be calculated if the rotations
8a and Ob of the end tangents arc }{no\vn.
Let us consider nc\v a more general case in \vhich there are both end
moments and lateral load, acting as sho\vn in Fig. 9. 5a. The angles of rotation
produced by the end moments alone are given by expressions (9.1 a) . To
obtain the angles of rotation produced by the lateral load, \ve consider the
corrcsponding bcnding-monlent diagram as sho\vn in Fig. 9.5 b. Upon denot-
ing by c and d the distances from the cnds of the bcam to the vertical through
the centroid C of this bending-moment area and by A the magnitudc of the area,

ARTICLE 9.1
407
J lab ( A
..h...
! t

l!,\1 ba

(a)
FIG. 9.5
(b)
-d
-c
-1
A
the reactions at the cnds produced by such fictitious loading are
Ia = 4c
I
Rb = /ld
I
and the corrcsponding angles of rotation arc
.I4c
8(1 = HI!
Ad
(h = - Ell
(i)
(:om bining these ang les \vi th those prod uccd by the end I1l0ments [E<]s. (9.1 a) ],
\YC obtain thc follo\ving total angles:
o =  '\Jabl _ i\ l b f!! + Ac
a 3EI 6EI Ell
Ob = !\lf.1 _ ilJbl _ ..Ii.1
3 'I 6EI Ell
(9.2a)
Solving these equations for A1 a 'l and i\1 bn , \VC obtain
ill ah =
2EI 2A
T (20 a + Ob) - 1 2 - (2c - d)
2EI 2A
. T (20 b + 8a) + f2 (2£1 - c)
(9.2b)
i\.J ba =
From these equations the cnd monlents can be readily calculated if the rotations
of the ends and the bending-molnent diagram for the latcral load are kno\\'n.
Somctin1Cs, as a result of displa<.:eIncnts of the supports, a beanl, in addition
to bending, rotates as a rigid body, and \VC denote the angle of this rotation
by f)ulJ, considered positive if the rotation is clock\visc. rrhe total angles of

408 CONTINUOUS BEAMS AND FRAMES
rotation of the ends \\,ill then be
.Nlab l i\tfbal Ac
8a = 3EI - 6El + Ell + (jab
1\1 bal ;\1 abl Ad.
8b = 3El - 6El - Ell + Oab
(9.3a)
Solving these equations for the- cnd moments, \'le obtain
'" 2EI ( ) 2A ( d) 68abEI
; V1 ab = -T 28a + Ob - 72 2c - - -,- -
2EI 2A 69,wEl
,\tl ba = -T (20 b + 8a) + 72 (2d - c) - I
(9.3 h)
Equations (9.1) to (9.3) represent the required slope-deflection equations so
\videly used in the analysis of frame structures.
9.2 BEAMS WITH FIXED ENDS
Using the slope-deflecrion equations dcveloped in the preceding articlc, \VC can
readil y analyze various cases of statically indeterminate beams. V\Te shall
discuss here only a fe\v particular cases that arc of special intcrest in con-
nection \vith the analysis of frames. \;\ begin \vith a beam built in at one
end and sinlply supported at the other, as sho\vn in Fig. 9.61. Assunling that
the magnitude of the active end moment i\l ub is given, let it be rcquired to find
the corresponding magnitude of the reactive cnd moment A1f ba produccd at the
fixed end B and also the rclation between the active moment 1\1 ab and the angle
of rotation 8 a . Observing that the end B of the beam does not rotat( and
substituting 8b = 0 in the second of Eqs. (9.1 a), we find
,\tl ba = j 1\1 ab
(a)
This shows that if an active momcnt i\tf(Lb is applied at the end A, a reactive
moment A1 ba , equal to j-A1 nb , is produced at the fixed end B. This latter
moment is of particular interest in rigid-frame analysis and is called the carry-
over 1110111en t .
Substituting 8b = 0 in the first of Eqs. (9.1 b), \ve obtain
4£1
Mab = T 8a
(b)
Fronl this equation, the moment i\;lab IS rcadily calculatcd if the angle of
rotation Oa is known.

ARTICLE 9.2
.09
1U ab (

B f 1ba

(a)
(b) 
A ] B
r e j
(c)  f
FI G. 9.6
(d)
,
d
r
Pef
I

c
It is interesting to note that if both ends of the beam arc simply supported,
the relation bct\vecn the end n10ment A.f nb and the angle 8ab is
3EI
J\.1 ab = T 8a
(c)
rhis sho\vs that in calculating A4ab by, using the angle ()a the placing of a hinge
at B is equivalent to rcducing the flexural rigidity EI of the beam to the value
3EI/4,
Considering no\\' lateral loading, \\'e observe that in the case of a uniformly
loaded beam (Fig. 9.6h), the rcactive end moment Alba is found from the second
of Eqs. (9 .2a). Substituting into this equation
AJ ab = 0
()b = 0
ql3
A =--
12
I
d=-
2
\ve obtain
ql2
A1f ba = -
R
We note that this end moment has the same value as the bending moment at
the middle of a simply supported beam.
If a concentrated force P acts on the beam (Fig. 9 .6c), the bending-moment
diagram for the corrcsponding simply supported beam is represented by the

410 CONTINUOUS BEAMS AND FRAMES
triangle sho\vn in Fig. 9.6d, and \\'c obtain
A = Pel
2
d= '+f
3
The second of Eqs. (9 .2b), \vith 8b = 0 and ,\t[ob = 0, then gives
3Ad Pf(12 - f2)
'\![ba = J2 = 2/ 2
If the built-in end B of the beam suffers a vertical do\\rn,vard displacement
 without rotation, the cnd mornent kf ba produced by such displacement is
obtained by using Eqs. (9.3 b). Substituting in the first of these equations
j\lfab = 0, Ob = 0, A = 0, and 8ab = / " \\'e obtain
3
() =--
a 2 I
Substituting this value in the second of Eqs. (9. 3b), \\le find
3El
Atf ba = - T }
(d)
Applications of these results \vill be sho\\'n later in the analysis of frames.
In the case of a beam \\lith both ends built in (Fig. 9.7 a) and carrying any
lateral load, \ve use Eqs. (9.2b) to calculate the statically indcterminate end
JTIomcnts. Substituting (Ja = 0'1 = 0 into these equations, \ve obtain
ZA
jUab = -72 (2c - d)
2A
i\tf ba = -/2- (2d - c)
(9.4)
For any given lateral load, the bending-moment diagram for a corresponding
simply supported beam, Fig. 9.7/1, can be constructed, and its area A together
\\'ith the ccntroidal distances c and d can be calculated. Substituting these
quantities into Eqs. (9.4), \ve obtain the required end nlonlents. These
(a)
(b)
d
c
FIG. 9.7

ARTICLE 9.2
411
nl0rnents produced by lateral loads are called the fixed-cnd 11l0111el1ts and \\'iJl be
dcnoted in our further discussion by the symbols nab and mIba. Observing
that they arc takcn positive \vhcn they act clock\\'ise, \\'e conclude that mtab
is identical \vith the bcnding moment at the end A of the beam, \vhile mlbo is
numerically equal but opposite in sign to the bcnding moment at the end B.
In thc case of a load uniformly distributed along the entire length of the
beam, .4. = q13/12, c = d = 1/2, and Eqs. (9.4) become
ql2
m!ob= -12
ql2
TIba = 12
(9. 4a)
In the case of a concentrated load P applied at the distance e from the support
B, the bending-moment diagram is a triangle (Fig. 9 .6d), and \ve obtain
A = p"fe /2, c: = (I + e) 13, d = (I + f) /3. Substituting these values into
Eqs. (9.4), we find
Pe2j
nab=-T
Pef2
rrbo = T
(9.4b)
For any value of the distance e, the values of the fixed-end moments can be
readily calculated.
Let us consider now the bending of a fixed-cnd beam produced by nonequal
settlements of the supports. Assumc, for example, that the support B settles
by an anlount , \vhilc the support A remains stationary. Assume, also,
that no rotations of the ends of the beam occur during such displacement.
Then, in calculating the end moments produced by such settlement, \\'e use
Eqs. (9.3 b). Substituting 8 0 = ()b = 0 and A = 0 in these equations and
obscrving that Bob = I I, \\'c obtain
Alab = Alba = - l} l = - E I J Sab
1 2
(9.5)
Introducing the notation k = Ell/for the so-called "stiffness factor" of the
bean1 and using the notation mt for fixed-end moment, \VC can \\'ritc the slope-
deflection equations (9. 3 b) in the form
"lab = 2k(28 o + 8 b ) + mIab - 6k8 a b
i\tl ba = 2k(28 b + 8 Q ) + mlbu - 6k8ah
(9.6)
These equations \vill find a \vide appJication in thc analysis of framcs, as \vill
be shown latcr. For future referencc, expressions for fixed-end momcnts for
various cases of latcral loading of a beam A B arc recorded in Table 9.1 belo\v.
The first t\\'O cases have alrcady been calculated as represented by Eqs. (9.4a)
and (9.4:b) above. Vcrification of the remaining cases in the table is left as
an exercise for the reader.

412 CONTINUOUS BEAMS AND FRAMES
TABLE 9.1 FIXED-END MOMENTS FOR PRISMATIC BEAMS
 I -4
AIIII;IIII "'I'IIII
p
t--f -T e-j
r l ,
-1 c r- d --1

f-  . e-F
r c -t-- d --1
 A J; B
nab =
q/2
12
ql2
mLba = +-
12
mra =
Pe2j
--
/2
PeJ2
ffilba = + J2
q
mt b = - - [d ' (41 - 3d) - e 3 (4/ - 3e)]
a 12/2
q
mt b = + - [1 8 (41 - 3/) - c l (41 - 3c)]
Q 12/ 2
mtob =
QO/2
QO/2
ffil bo = +-
30
20
mtoe, = - Thq 0/ 2
mtba = + -rArq 0/ 2
/\'fd
;mall = + - (2c - d)
/2
Me
mIll!) = + - (2d - e)
/2
9.3 CONTINUOUS BEAMS
Equations (9 .2a), derived in Art. 9.1, can be used to advantage in the analysis
of continuous beams. Starting from the left end of the beam, \ve denote by
0, 1, 2, . . . the consecutive supports, by 11, 1 2 , . . . the lengths of the spans,
and by II, 1 2 , . . . the cross-sectional rnoments of inertia, assumed constant
along each span. This done, \ve consider any two adjacent spans In and In+l
as sho\\'n in Fig. 9.8d and construct the corresponding simple-beam bending-
momcnt diagrams for the lateral loading as sho\vn in Fig. 9.8b. The areas of
these diagrams are denoted by An and An+l and the positions of their centroids
C n and (,'n+1 by the distances Cn, d n and Cn+h d n + h respectively. Considering
bending of cach span\ \\'e must take into account not only the lateral loading
acting on the span but also the end moments representing the action of the
adjacent spans on the span under consideration. 1'he magnitudes of these
moments for the span In \ve denote by Al n - 1 ,p& and 1\{n,n-l, and for the span I n + J

ARTICLE 9.3 413
n-l LJ J n  I n+l
--D-  
(a) ,.. In + 'n+l --
(b)
d"
c,, d"+l
C,,+ 1 -..J
A"
A" + 1
(c)
M
:- 
M", ,,+1
!
M,,+ I, "
n+1\
-A.. J
FI G. 9.8
by 1\1 n ,"+1 and ;\1"+1.,t, as sho\vn in Fig. 9.8c. The equations for calculating
these mon1ents can be \vritten on the basis of the continuity of the elastic line.
From such continuity, it foJ)o\vs that the angle of rotation 8n.n-l of the right
end of the span In nlust be equal to the angle of rotation 8 n ,n+l of the left end
of the span /"+ 1. These angles of rotation \vill be found fron1 Eqs. (9.2,1),
\vhich give
() _ /[n,"-lJn _ A,[n-l.n 1 n _ And n
n."-1 - 3El n 6El n JnEln
8 _ jln.n+l111+1 _ Al n + 1 ,tJ"+1 + A n + 1 c n +l
n,"+1 - 3Eln+l 6Eln+1 In+1Eln+l
(a)
Let us introduce no\v the conventional bending tnornents /V[n-11 JV/ n , Aln+l at
the supports, instead of the cnd n1omcnts. From the assumed positive direc-
tions of the end nl0nlents (see page 406) and the conventional rule for sign of
bending moment it follo\vs that
'\;/11-1," - + AI,t-l
A.J n ,n+l - + Aif n
l\;["+1.n = - }\;["+1
i\;/n.n-l = - AI"
and exprcssions (a) become
8 n ,"-1 =
j\l n ln lV[n-l/n And n
- -3E lr - -6£l n - - lnE1n
8 j\;[n/n+l + A,fn+lln+l + _ An+lCn+l
",n+l = EI EI 1 EI
3 n+l 6 n+ln+l n+l

414 CONTINUOUS BEAMS AND FRAMES
Equating these t\VO angles, \ve obtain
A l l !'2 _ 2 A I ! ( In + In+l ) + .t1 1 In+l
n-l I H t. I I ,. n+l I
n n n+ 1 n+ 1
6A n d n
= -
I "..In
6A n + 1 C n +l
In+11n+1
(9.7)
which is thc \velI..kno\vn equatio11 of three 1110111e11ts. Such an equation can be
\\'rittcn for each intermediate support of a continuous beam; and if the ends of
the beam arc simply supported, the number of equations is equal to the nwnber
of statically indeterminate mornents at the supports. Hence, all unkno\vns can
be found from these equations.
If the left end of the beam is built in at the support 0, \\'e have an additional
condition stating that the left end of the beam does not rotatc. In our previous
notations, this means that 6 0 vanishes. Using the first of Eqs. (9 .2a), \ve find
that this condition becomes
}\t[O.1 / 1 _ ,\t{ 1,0 / 1 + A lCl = 0
3El l 6£1 1 !1El 1
Introducing bending moments again, instead of end moments, \\.'e obtain
2Mo + M 1 = _ 6A 1 cl
/}2
(9.8)
It is seen that a built-in cnd introduces an additional unkno\vn n10ment Alo,
but at the same time \ve have an additional equation (9.8). Hence, the
number of unkno\\'ns is agin equal to the nunlber of equations. An equation
similar to Eq. (9.8) can be \vritten fOf the right end of the beam if that cnd is
built in.
Sometimes, instead of the three-n10ment equation (9.7), an equation con-
taining as unkno\vns the angles of rotation of the cross sections at three con-
secutive SUppOfts can be used. Considering again t\VO consecutive spans
(Fig. 9.8c) and calcu]ating the cnd moments at the nth support by using :qs.
(9.6), \ve obtain
"1".n-1 = 2k l1 (28 n + 8"-J) + n,n-1 - 6k n 8n.n-1
Al".,.+J = 2k n + 1 (28 n + 8 n + 1 ) + lIn.n+l - 6k n + 1 e n ,n+l
rhen, observing that kln,n-l = - kln."+h \ve find
2k n 8 n-l + 48n (k n + k n + 1 ) + 2kn+18 n + 1
= - (n"1-1 + mIn,n+l) + 6(k I1 8 n ,n-l + k n + 1 8 n ,n+l) (9.9)
This is called the equation of three angles.

ARTICLE 9.3
415
As a first application of the three-moment equation (9.7), let us consider
the case of a beam carrying a uniformly distributed load on each span. If q,.
is the intensity of load on the span In' \ve have
C n = d n = In
2
An  qn l n 3
12
\\ obtain similar expressions for the span In+1. Equation (9.7) then becomes
M n - 1 In + 2M n ( !r:. + In+l ) + 1\1 n + 1 /"+1
In In 1,,+1 In+l
qJn 3 qn+1 /3 n+l (9.7a)
= - 4/ n - 4/ n + 1
Take, for example, a beam of uniform cross section on five supports and
with simply supported ends. In this case Mo = A14 = 0, and the moments
Mt, M 2 , M3 at the intermediate supports will be calculated from Eqs. (9.7a),
v.rhich become
2M 1 (/ 1 + 1 2 ) + 1\1[2/2 = _ ql 1 1 3 _ q2/2 8
4 4
1\1 1 1 2 + 2M 2 (12 + 1 3 ) + Mala = _ q 2 / 2 3 _ qa / a 3
4 4
1\;/2/3 + 2Ma(/a + I..) = _ qa / a 3 _ q4 1 4 3
4 4
If all spans are equal, II = /2 = 13 = I.. = /, and the equations give
/2
A1 1 = - 224 (15ql + llq2 - 3q3 + q..)
[2
M 2 = - 224 (-4ql + 12q2 + 12q3 - 4q.)
/2
1\1 3 = - 224 (ql - 3q2 + l1q3 + 15q4)
(b)
Taking, for example, q2 = q3 = q4 = 0 and only q] different from zero, \ve
obtain the case sho\vn in Fig. 9. 9a. In such case, the moments at the inter-
mediate supports, from Eqs. (b), are A1 1 = -'.l.,l:rQ 1 / 2 , M 2 = + "'5\q 1 / 2 ,
M3 = -ThQl12. The corresponding bending-moment diagram is sho\vn in
Fig. 9.9b.
If a load of constant intensity covers the entire length of the beam (Fig. 9. 9c) ,
ql = q2 = q3 = q4 = q, and Eqs. (b) give
M 1 = --hq/2 kl 2 = -T!Tq/2 A13 = -y,\qI2
The corresponding bending-moment diagram is sho\\'n in Fig. 9.9d.

416 CONTINUOUS BEAMS AND FRAMES
(a)
Q 111T111I.lli1l1JIIJl ..£1.... 4
z: 0 :1C 1 .A- 2 A.... 3
l---+-l-+-'-+-l
(b)
T
2
ql 1
8
J
(c)
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII [I !III: 11111111111.1
T
ql2
8
FIG. 9.9 (d) J
Ifa concentrated load Pn acts in a span In (Fig. 9.10a), the area An is repre-
sented by the triangle shown in Fig. 9.1 Oh and we obtain
An = PnCn/n
2
C n = t(ln + en)
d" = j(ln + In)
(c)
Such expressions for each conccntrated force acting on the beam must be used
in calculating the right-hand side of Eq. (9.7).
As an example involving concentrated forces, let us consider a beam of
uniform cross section on five supports and loaded as sho\vn in Fig. 9.11a.
"'rom symmetry \\'e conclude that M 1 = M 3, and \ve have to \vrite only t\\TO
equations . For the loads and distances indicated in the figure, we find
(a)
r n
nf- fn 
'n
n
-A..
en
1
(b)
dn
FIG. 9.10

ARTICLE 9.3
417
(a)
5 tons 4 ton9 4 tons 4 tons 4 tons
o  1 
t 9' -t6 '
18'21'
5 tons
!2LJa  4
6 'f. 6' +6 t 9' I"
21'-+- 18'
(b)
i......: + +
FIG. 9.11 (c)
1.1 n - 1 Un J 1J1" + )
  n+l)
 
in , In+l -1
FIG. 9.12 (a) (b)
Al = - ton-ft 2 , Cl = d. = 9 ft, ,At = 360 ton-fr 2 , C2 = d 2 = _'1.,/- ft. Sub-
stituting in Eqs. (9'.7), \VC ohtain
78M l + 21M 2 = _ ?/05
21AJ 1 + 84}\..[2 + 21i\J 1 = -6 X 360
6 X 360
2
\vhich give
i\.J 1 = - 17.0 ton-ft
i\I 2 = - I 7 .2 ton- ft
Using these values, the bcnding-nl0nlent diagralTI sho\vn in Fig. 9.11 b can be
constructed.
rhen the bending filoments at the internlcdiate supports have been cal-
culated, the reactions at the supports can be found from equations of statics.
Considering the support II (Fig. 9.12), \VC calculate separatc]y the reaction R7I
produced by bending of the span /'1 and of the span 1 1I - t - 1 and then take the sum
of these t\VO reactions. r-rhus, fron1 Fig. 9.12a, \VC conclude that the first part
of the reaction Nfl is
R' + i\1n-l - ./\1 n
n /n
(d)

418 CONTINUOUS BEAMS AND FRAMES
\vhere R represents the rcaction produced by latcral loads. Considering
Fig. 9.12b in the same nlanncr, \VC conclude that the second part of the
reaction at the support 11 is
R" + AJn+l - A'[n
n I
n+1
(c)
Taking the sum of expressions (d) and (e), \ve find for the total reaction at the
support 11
Rn = R + R' + 1\,1"-1 - A1n + 1\'/n+1 -: Ain
In In+1
(9.1 0)
\"here R + R is the 111agnitudc of the reaction at the support 11 that \vould
be obtained if the continuous beam \verc cut at all intermediate supports and
thus transformcd into a system of simply supported beams. It should bc
noted that if there are loads applied over the supports, they \vill be transmitted
directly to the supports, and reactions equal to these loads must then be added
to those calculated from Eq. (9.10).
Using Eq. (9.10) for the case sho\vn in Fig. 9.11, \ve obtain the foIlo",-ing
values of the reactions:
Ro = R4 = 1.56 tons
Rl = R3 = 7.43 tons
R 2 = 8.02 tons
"rhe corresponding shearing-force diagram is sho\vn in Fig. 9.11 I:.
In all prcvious examples, \\'C have assunled a continuous beanl of uniform
cross scction. In such a case, thc moments of inertia appearing in Eq. (9.7)
cancel, and the nlagnitudes of moments at the supports do not depend on the
cross-sectional dimcnsions of the beam. If thc cross-sectional ditncnsions of
the bealll in different spans arc different, \ve multiply Eq. (9.7) by an arbitrary
constant moment of inertia 1 0 * and introduce the notations
I 10 = /'
n In n
J 1 0 / '
n+l ' I - =n+1
n+l
A /0 A ,
n In = 11
An+lIo _ A' (I)
1 - n+l
"+1
"lith these notations, Eq. (9.7) becomes
,\tin-II:, + 2AIn (I: r + l: t + 1 ) + M n+l1:t+l
6A: t d n
=-/-
6A+lCn+l
1"+1
(9.11 )
-rhis equation docs not contain momcnts of inertia explicitly, and the problem
is reduced to that of a heam of uniforn1 cross section for \vhich the modified
lengths of spans I, I, . . . and the nlodified morncnt areas A, A, . . . are
calculated from Eqs. Cf).
* Usually \ve take for /0 the cross-sectional mOIncnt of inertia in one of the spans.

ARTICLE 9.3
419
I..et us consider no\v the effect of settlcnlcnt of supports on bending of a
continuous beam. Assuming that the vertical dispJacements of three con-
secutive supports 11 - ], 11, and 11 + 1 are n-l, L1n, and L1n+l and considcring
the t\VO consecutive spans sho\vn in }t'ig. 9.Rc as sirnply supportcd beanls, \ve
conclude that the angle of rotation of thc span In due to settlement of the
supports is (L1n - L1 n -l) /1 n. In the same n1anner for the span I n+ h \ve obtain
(L1nH - A n )/I,,+l. Using these angles in Eqs. (9.3a), \ve obtain, instead of
Eqs. (a) above, the follo\ving equations:
8 - A1n.n-l/n _ M n - 1 ,n/n _ And"" + An - An-l
n,n-l - 3El n 6El n InEln In
971.n+1 = l\111,n+1In+l _ lvl n +l,nln +l + _Ant+] + n+t - L1n
3Eln+l 6 Eltt+ 1 In+1Eln+l In+1
(g)
Again, introducing bending moments, instead of end nl0ments, and equating
the angles 8n,n1 and 8 n ,n+h \\'c obtain
\1 In + 1 1 ( In 1,1+1 ) M In+l 6A n d n
1 n-l _ 1 -- 2/ n / -- + - / + n+l / - = - -- 1 - / --
n 71 n+ 1 n+ 1 n n
_  An+1Cn+l + 6£ ( n - n-l _ n+l -  ) (11)
In+l/n+l In In+l
I f the beam has a constant cross section \vith moment of inertia I along the
entire length, \VC multiply Eq. (11) by I and obtain
A1n-l / n + 2M n (/n + 1 71 + 1 ) + 1\171+l/n+l = 6A l 71 d n
- -----
6A n + 1 c n +l
In+l
+ 6EI ( n - n-l _ n+l - An ) (9.12)
In In+l
If the moments of inertia in different spans are different \\'e multiply Eq. (11)
by an assumed constant moment of inertia 10. Then, using notations (f),
\ve reprcscnt the three-moment equation in the follo\ving form:
\1 I , + 2 .L { (I ' + I , ) + ,1, 1 I , 6Adn 6A: 1 + 1 C n + l
j I n-1 n lv, n n n+l lv. 1/+1 n+l = - I - I
n,,+1
+ 6E/o ( n - ,,-1 _ n+l - n ) (9.13)
In In+1
It is seen that in this case the three-moment equation has the same forrn as
before. Hence, the monlcnts at the supports can be readily calculated if the
magnitudes of scttlcnlent L1 ,a -h L1n, and n+ 1 are kno\vn.

420 CONTINUOUS BEAMS AND FRAMES
PROBLEMS
Find the bending moment All at the intcrn1cdiatc support of the t,va-span con-
tinuous beam of uniforn1 cross section sho\vn in Fig. 9 .13a if a uniformly distributed
load acts on the first span and the ratio J 2: /1 is equal to Ci.
q/2 1
A11s. ,\1/1 = - -
8 1 + a
2 Solve the previous problclTI if the ratio /2: /1 is equal to 'Y.
q/2 'Y
Ans. i\J 1 = - -- --,
8a+i'
3 Find :v[o and "11 for the continuous beam shown in Fig. 9.13b. Assume that
the cross section is uniform along the entire length and that the intensity of load is q.
ql2 2 + 4a ql2 2
A11s. 1\1'0 = - - ----, ,}f = - - -- --.
8 3 + 4a 1 8 3 + 4a
4 Solve the previous problem jf both ends of the beam arc built in (Fig. 9.13c) and
the cross section is uni form along the entire length.
ql2 2 + 3a q/2 2
Al1S. 1\;/0 = - 8 3(1+j 1\-1 1 = - "8 3(1 + a)'
ql2 1
/\1 2 = +83(1 +a)
5 A. beam of uoi fonn cross section rests on four supports and is loaded as sho\vn in
Fig. 9. 14a. Find the bending mon1cnts at the supports if the intensity of load is q.
ql2 4(a + (3)
Al1'. /\1 1 = -,
8 4(1 + a) (a + 13) - a 2
qJ2 2a
1\-1 2 = +-
8 4(1 + a) (a + 13) - a 2
(0)
1
l--+--al
2
.1
(b)
 1
al
2:j
FIG. 9.13
(c)
 1
al
f

ARTICLE 9.4
421
(a)
1 2
t--,-+- 1-+-(31
...A.. 3
-I
o 1 2 3
l--+-al-+--(jl
FIG. 9.14 (b)
6 Solve the previous problem if the middle span is loaded as sho\vn in Fig. 9.l4b.
ql2 2a 3 (a + 2(3)
Ans. All = - - ,
8 4 (1 + a) (a + (3) - a 2
£1/2 2a 3 (2 + a)
Al 2 =
8 4(1 + ex) (a + (j) - a 2
7 Solve the preceding problem if both ends of the beam are built in and {3 = 1.
£1/2 20: 3 £1/2 4a 3
AIlS. 1\11 = 1\-1 = - , 1\1 = Al = - - ·
o 3 8 3 (1 + 2et) 1 2 8 3 (1 + 20:)
9.4 BEAMS OF VARIABLE CROSS SECTION
Often, especially in reinforced-concrcte structurcs, beams arc deepcned near
their supports in ordcr to avoid the sharp reentrant corncr that other\vise
occurs at the junction of a beam \vith a column. In such a casc there arises
the question of ho\" the haunches affect the angles of rotation of the ends of the
beam under applied loads. To calculate these angles, we use Eqs. (e) of Art.
9.1. Multiplying and dividing these equations by the constant moment of
inertia /0 of the uniform portion of the beam, \ve obtain
8 0 =
[ l I - x M 10
J 0 I -EI; 7 dx
_ ( l  M 10 dx
) 0 J Elo I
(a)
()b =
In this manner the problem is reduced to that of a beam of uniform flexural
rigidity Elo. To takc care of the cross-sectional variation, \ve have only to
use AJ/o/I, instead of At/. This means that the fictitious load is given in this
case by the modified bending-nl0ITICnt diagram, obtained by multiplying the
bending moment at each cross section by the ratio 10/1.
Take, for example, the case sho\vn in Fig. 9.15a. A beam \vith reinforced
ends is bent by a load P applied at the middle. The modified bending-moment

422 CONTINUOUS BEAMS AND FRAMES
Al II PlJ o 
I I
(a) l l
I
c
(b)
b
FIG. 9.15
a
A D F B
10
(a) 11
r- M Xl 1
l
Mob D F --1
 10
(b)
(c)
f
FIG. 9.16
a
A
D
B
diagram is shown by the shaded area in Fig. 9.15 b. Along the reinforced
portions of the beam, of length XI, the ordinates of the bending-moment diagram
acb are reduced in a constant ratio 10: [1. The angles of rotation of the ends of
-the beam arc now obtained jf \ve divide by Elo the shearing forces produced at
the ends of the beam by the fictitious load represented by the shaded area in
Fig. 9. IS b. It is seen that o\\,ing to the reinforcement of the ends of the beam,
these angles of rotation will be diminished in the same ratio as the fictitious
load is diminished \vhen \\'f repJa'ce the bending-moment area neb by the
modified bending-moment area. Assume, for example, that X = t and
10//1 = -. Then, the fictitious load is reduced in the ratio 7 : 8, and the angles
of rotation wi II be diminished by one-cighth of the values obtained for a beam
having uni form flexural rigidity Elo.
In such cases as shown in Fig. 9.16, the ratio 10/1 varies along the length
of the haunches, and instead of the straight lines ad and eb sho\vn in Fig. 9.15 b,
\ve shall obtain certain curves. We always can trace these curves with

ARTICLE 9.4
423
sufficient accuracy by dividing the lcngth Al of a haunch into several parts and
calculating the ratio 10/1 for each segment. Naturally, the shape of the
modified bending-moment diagram \vill depend on the form of the haunches.
Assume, for example, that a rectangular beam \vith straight haunches (Fig.
9. I 6b) is bent by a couple lvl ab applied at the cnd A. lhen, the bending-
moment diagrarn is the triangle sho\\'n in Fig. 9. 16c. Since the depth of the
beam increases linearly \\lith the distance  and since 1 is proportional to the
cube of the depth, \\'e obtain
1 = 10 [ 1 + ( : - 1) l r
(b)
\\rhere 1 1 is the nl0mcnt of inertia at the support. laking, for exalT1ple,
11/10 = 10, A = 0.3, ,,,c find the values ofrhe ratio 10/1 as given in ro\v 3 of
Table 9.2 belo\v. '[he corresponding modified bending-monlcnt area is sho\vn
by the shaded portion of the triangle in Fig. 9.16c. Treating this area as a
fictitious load and using Eqs. (a), \ve can calculate the angles of rotation at
the ends of the beam.
To illustrate this kind of calculation by using Simpson's rule, \\'e begin \vith
the portion AD of the beam. The corresponding values of the function under
the integral sign in the first of Eqs.. (a) arc given in ro\\' 5 of lable 9.2. Thus
the contribution 0: of the portion AD of thc beam to the total angle of rotation
at A is
8' = ( O.al ! - x Ai 10 dx = Mab r O.3l .('-= )2 10 dx
a )0 I Elo I . RIo Jo /2 I
= M El a 0.3/ [0.4900 + 4(0.2807 + 0.1318) + 2 (0.1842)
-J o 4X3
+ 0.1000]
MaJ
= 0.0652 E1 0 -
TABLE 9.2
(I)  0 1 1 3
-- :f  T
Xl
(2) 1 - .t o . 7000 0.7750 0.R500 0.9250 1 . 0000
/
(3) 10 I . 0000 0.4673 0.2549 0.1540 O. 1000
7
(4) (I - X)2 0.4900 o . 6006 0.7225 0,. 8 5 56 1 . 0000
--- -------
/2
(5) (/0/1) (I - ,.') 2 0.4900 0.2807 O. 1842 0.1318 0.1000
--
/2

424 CONTINUOUS BEAMS AND FRAMES
Treating the portion FB of the beam in the same manner, \ve obtain
0" = A!ab [I  ! - _x) 10 dx = 0.0049 j\1ab l
o EIo ]0.71 1 2 I Elo
Along the middle portion DF, the cross section is constant, and \ve obtain
8'" = 1\-t ab [o.71 _(I _ X)2 dx = 0.1053 ¥abl
a Elo ]O.3l /2 Elo
The total angle of rotation at A then is
8 0 = 8 + 8' + 8" = 0.1754 i:'
Proceeding in a similar \vay with the second of Eqs. (a), we find for the angle
of rotation at the right end of the beam
8b = -0.1266 70'
In the case of beams having straight-line or parabolic haunches, the integra-
tions in Eqs. (a) can be made rigorously. However, the calculations arc
tedious and time-consuming. To simplify the analysis of beams of variable
cross section, special tables have been prepared giving values of the numerical
factors in the expressions for 8 0 and 8b for those shapes of haunches that are
most oftcn encountercd in practice. 1 For example, in the case of a symmetrical
beam \vith straight or parabolic haunches, the angles of rotation at the ends A
and B under the action of end moments k[ab and M ba can be readily calculated
from the equations
Oa = C 1 k[oo/ - C 2 kl ba !
Elo Elo
8 - C Mbal _ C Mab/
b - 1 Elo 2 Elo
(9. 14a)
in \vhich the constants C 1 and C 2 can be taken from Table 9.3.
1 Such tables \verc calculated by A. Strassner; see his book "Neue Methoden zur Statik der
Rahrncntrag\vcrke," 4th cd., vol. 1, Berlin, 1937. Follo\ving the work of Strassner,
W. Ruppel prepared an extensive set of tables for haunched bcaIns; see Trans. ASCE, vol. 90,
p. 152, 1927. A useful series of charts for beams of variable cross section are given in the
book by R. Guldan, t'Rahmenrrag\verke und I)urchlauftrager," Vienna, 1940, and also in
the book by James '1. Gere, HMoment Distribution," I). Van Nostrand Company, Inc.,
Princeton, N.J., 1963. In our further discussion, tables published by the Portland Cement
Association will be used; see "I-Iandbook of Frame Constants" and "Continuous Concrete
Bridges." For bean15 \vith abrupt changes in cross section, see Iowa Eng. Exp. Sia. Bull.
176, 1954.

ARTICLE 9.4
425
TABLE 9.3 CONSTANTS C 1 AND C 2 FOR SYMMETRICAL
HAUNCHED BEAMS
10 I
 - 0.60 0.30 0.20 0.15 0.12 0.10 0.08 0.06 0.05 0.04 0.03 0.02
II
(0) Bcan1S \vith straight haunches (Fig. 9.16h)
I
0.50 C 1 0.251 0.173 0.140 0.121 0.107 0.099 0.089 0.078 0.071 0.064 0.056 0.047
C 2 0.141 0. 106 0.091 0.082 0.076 0.071 0.065 0.059 0.055 0.050 0.045 o . 040
CI 1 0.264 1 °.198 1 ° .171 1 0. 154 1 0 . 144 1 0.136 1 0 .127 1 °--.117 1 °.112 1 0.106 1 °.099 1 ° .091
C 2 0.1470.1250. 1 1 50.1080.1040.1000.0960.0920.0890.0860.0830.072
C 1 1 0.2 71 1 0 . 212 1 0 . 18'7 1 0 . 172 1 0. 162 1 0. 155 1 0. 147 1 ° . 138 1 0 . 133 1 0. 128 1 0. 122 1 0. 114
C" 0.15 1 0.1380.1260.121 0.1170.1140.111 0.1070.105 0.1030.1000.096
C 1 1 0.278 1 °.226 1 °.204 1 °.191 1 °.182 1 °.175 1 °.168 I o.16° 1 0.156IOI'51 I o. 145 I o-. 139
('2 ,0. 155 O. 142 O. 136 0. 132 O. 1290. 127 0. 124 ° . 121 O. 120,0. 118 O. 1160. 113
C 1 1 0.286 1 °.241 1 °.222 1 °.21° 1 °.203 1 °.197 \ °.191 1 0.184 1 0.180 1 0.176 1 0' ]71 1 0.165
C'l ° . 1 5 8 O. 149 O. 144 0 . 142 ° . 1 39 0. I 3 8 ° . 1 36 ° . I 34 0 . 1 3 3 O. I 32 ° . I 30 ° . 128
C 1 1 0.29- 4 1 0 . 257 1 0. 241 1 0 .231 1 0 . 225 1 0. 220 1 0:-:£i5 I O' 09 1 0' 20 6 1 0. 202 1 0 . ] 98 1 0 .193
C 2 0 . 161 O. 1 5 5 ,  . 152 O. 1 50 O. 149 O. 148 ° . 147 ° . 145 0 . 144 0 . 143 (). 142 O. 14 I
0.40
0.35
0.30
0.25
0.20
0.15 C 1 0.303 0.274 0.261 0.2541°.249 0.245 0.241 0.237 0.234 0.231 0.228 0.224
C 2 0.163 0.160 0. 158 0 . 1 57[0. 156 0. 1 56 O. I 55 O. 154 O. 154 0.153 0.153 0.152
(b) Beams \vith parabolic haunches (Fig. 9.16a)
0.50 C 1 o . 273 o. 2 13 1 0 . 186 O. I 70 O. I 59]0. 15010. 141 O. 130 O. 123 O. 116 O. 108 0 .097
C'l O. 151.<>' 1 321.0. 122 O. 115 O. 11110. 107:0. 103 0.098 0.0940.090 (). 0860.080
- -
0.40 C 1
C'/,
0.35 C 1
C 2
0.30 C 1
C 2
0.25 C 1
C 2
0.20 C.
C 2
-
0.15 C J
C'l
!o.283 I o.233 I o.-ZIO I O' 196 ! 0. 187 1 °' 180 \ 0.17 2 1 0.162 1 0.157 1 0.151' 1 0.143 1 0.] 34
10 . I 56 O. 1 43 0 . I 36 0 . I 32 O. 129 O. 126 ° . I 23 O. 120 O. 11 7 0. I 1 5 ° . 1 I 2 O. 1 ° 7
1 0.2891 1 °.-243 1 °.223 1 °.21° 1 °.20210.195 1 0-.188 I to.179 I o.175 I O69 I o.162 I o.154
O. 158 0. 148 O. 143 O . 1 39 O. 137 10. 135 O. 133 O. 130 O. 128 0. 126 O. 123 0. 120
1 0.294 I o.254 I o.236 I o.225 1 -02i8 1 Io.212 1 0'205 I o.198 I o.193!0.188Io.182 1 0' 175
0.1 600. 153 0.149 0. 1460. 144 0. 143 O. 141,0. 139 O. 13710.1361°. 1340. 131
1 0. 300 1 °. 2()6 1 0' 250 1 °.241 1 0. 234 1 0. 229 1 0.224 ] 0. 217 1 0. 213 1 0 .209 ] 0. 204 ! 0. 197
° . ] 62 0 . 1 57 O . 1 54 0 . I 52 0 . 1 5 I O. 149 0 . 148 0 . 147 O. 146 0 . 144 0. 143 O. 14 I
1 0.306 1 0.27810.265 1 °.2571°.252 1 °.248 1 °.243 1 °.237 1 °. 234 1 °.231 1 °.226 1 0.221
O. 1 64 0 . 160.0. 1 58 O. I 57 .0 . 1 56 0. 1 5 5 ° . I 54 0. 1 5 3 O. 1 5 3 0. 1 52 ° . 15 I O. 1 50
o . 3 1 2 ° . 291 1 ° . 281 0. 274 0 . 270 0 . 267 -6  i 6 3- o 25 9 0 .. 25 70 . 254 -0 . 250 ° . 246
0.167 0. 163,0.162 0.161 0.161 0.1600.1590.1590. 159 0.158 0.158 0.157
I f there arc no ready rabIes for the shape of the beanl under consideration,
the constants f..\ and C'J. have to be calculated as explained above. These t\\'o
constants completely define the stiffness of a symmctrical beam and arc called
its angular coefficients.
If, in addition to end moments, there arc latcralload on the beam and also
some displaccnlcnts of the supports, rhe general expressions fot" the angles of

426 CONTINUOUS BEAMS AND FRAMES
rotation at the cnds of the bcam becomc
1\1 0b / A1bol Ae
8 0 = C 1 £1 0 - - C 2 . EIo + E iol + Sab
1\;[ ba l 1\;I ab l Ad
8b = C. EI-; - C 2 - -E1 0 - BIoi + 8ab
(9.14b)
in \\'hich Ac/I and -..4d/1 arc shearing forces at the ends produced by the
modified bending-moment area A, such as that sho\vn in Fig. 9.15 b, and 8ab
is the rotation of the beam due to displacenlents of the supports.
Solving Eqs. (9.14b) for the cnd moments .Nlab and 1\1 ba , \ve obtain
1\;[ab = lo ( _8_ + __ C 2 L _ C 1 + C 2 e ab )
I C 1 2 - C2 C 1 2 - C 2 2 C 1 2 - C 2 2
A eC l - dC 2
- }2 C 1 2 - C22
(e)
1 Elo ( C280 ('18 b C\ + C 2 )
'\1 ba = T [12 _ - Cl},2 + C12 _ C 2 2 - (\2 _ C  eob
A eC 2 - dC.
- /2 C 1 2 - C22
l'hen, introducing the notations
Elo C 1
- = k - = kbb
/ C. 2 - C 2 2
C 2 k
C 2 C 2= ab
1 - 2
(d)
and observing that the last terms in Eqs . (c) represent the end moments for the
case in ,vhich 0 0 = 8b = Sob = 0, that is, the fixed-end 111O'Jl1cnts, '\Fe can
re\vritc these equations in the follo\ving simplified form:
M ab = k [k bb 8 a + kabOb - (k bb + k ab ) 8 o b] + mtab
lWba = k[ k ab 8 a + kbb()b - (k bb + k ab ) Sab] + miba
(9.15)
l-'hcse are the slope-deflection equations for symmetrical beams of variable cross
section. In the case of prismatic beams, \VC have C 1 = j, C 2 = .:, k 'Jb = 4,
kab = 2, and Eqs. (9.1 5) coincide \vith Eqs. (9.6) obtained before.
"fa simplify the calculation of end monlents from Eqs. (9.1 5), tables or
charts for the constants kbb and kob have been prepared, cxanlples of \vhich are
sho\\'n in Figs. 9.17 and 9.18. "l'hc values of fixed-end n10ments for both
concentrated and unifornlly distributed lateral loads have also been tabulated
for various types of haunches. l
I n aU of the foregoing discussion, it has been assumed that thc beam is
symmctricaL 1n the case of a nonsymmctrjcaJ bean), three constants Ct, C 2 ,
and C a are rcquired to define its elastic properties. In such case, the angles
1 See "Handbook of FraIne Constants."

of rotation at A and B in the general case of loading arc
C 1\labl _ C \!bal + _Ac + e
1 Elo  Elo <'[ 01 ah
C ., Alb a'- _ C i\.labl _  d_ . + n
3 Elo 2 Elo Elol U"b
as
.
en (I)
!J
a> C)
,a 6
.sJ
:i.'d
: .J
o b
 'I)
::I
-;
>
FIG. 9.17
SO
28
26
oS 24
.,.
 22
eft i 20
J-8 18
.. 
c2 16
oC""
a 14
'a 'e; 12
:J  10
::I
 8
>
6
4
FI G. 9.18
ARTICLE 9.4
427
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
o
\ \ 
1\ \ \ \
\ \ \ "
\ \  \ - --
\ \ \ \
\ \ \ , 
\ \  '\. q
\ '\ o
\. "
\. '" o R " -- --- -- "- -
\ \ '"  ""- ""
, '"< - f--
\ " .....lJ.ti "'" ............"'-- " ......... ............
..... ,
'!'o..  -............ ....................... ........ ............ ............
r--..... ""'--
........ ........... .............. .............. .....-......., - --.. -- X = 0.50
to-.... """"'-  t---. ,
"- ........... --. -- - -- - r-- / 0.40
.20  --. -- 
r--- r-- - -- r-- -  v 0.30
-::... - - -- ....
- - - \ Y 0.20
- 
_ 0.10 :"":  .... 0.10
0
I _...;1 ::t--
. I . . .
0.04
0.06
0.08 0.10 0.12 0.14 0.20 0.40 0.60 0.80 1.00
Values of /0/1.
0.02
 \ \ \ \
\ \  \
\ , \ \ 1\
\ \ \ ' r\\ .
,..
\ \ \ \'Z;;- >-.
\ 1\ ' v
\ \   " '-..
,
\ " '- O.J.  " '" 
, .0 .......................... ............ ....... 
"- '- '-.. 
"'"  ............ :-........... ........  
. .............. r--....  - 0.150
 "'- "'-  .
" .....  ..... --- - --- r-- -- .
----.Q.20 r-- ---  ,0.40
--  r--
-- r-- --- -- r---  l'l/ 1
 r-- --- -- -  ./0.30
r--- - --
--- -   0.20
-.....J.).10 ..... " I
I .....  . 0.10
,.....;
-
I I I . , I I I I . I . I I' 
2
o 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.20 0.40 0.60 0.80 1.00
Values of /0//1
Oa =
(9.16 )
8b =

428 CONTINUOUS BEAMS AND FRAMES
Solving these equations for Mal) and j\;[ba and introducing the notations
C 1
C C C 2 = kbb
1 3 - 2
C 2
. = k b
C 1 C 3 - C 2 2 a
C 3
C - (' C 2 = kaa (e)
1 3 - 2
\\'c obtain
j\;/ab = k [k aa 8 a + kabOb - (k nb + kaa) G a b1 + mIab
A1l = k r k ab 8a + kbbO b - (k bb + k ab ) 8 ab ] + mtba
"fhese are the slope-deflection equations for nonsymmetrical beams of vadable
cross section. For symmetrical beams, \ve have ('3 = CJ, kaa = k bb , and
Eqs. (9.17) coincide \vith Eqs. (9.15) obtained previously.
10 illustrate the application of Eqs. (9.16) and (9.17), lct us consider the
unsymmetrical beam \\,ith parabolic haunches \\'hich is built in at B and simply
supportcd at A as sho\vn in Fig. 9.19(1. Assuming that an active end Inomcnt
Alab is applied at A, the corrcsponding reactive moment at B can be calculatd
from the second of Eqs. (9.16). Setting ()b = 0 in this equation and noting
that there are no latcral loads or settlement of the supports in this case, \ve
obtain
(9.17)
C 2
Moa = ("3 Mab
In a similar manner, by using Eqs. (9.1 7), \VC obtain
i\t/ (]a = k kob i\1 0b
00
(b)
2.5h
(a't
(c)
FIG. 9.19

ARTICLE 9.4
429
lhe factor
C b = C'2 = kab
a C 3 kaa
(f)
sho\ving thc fraction of the applied nl0I11cnt that is transmitted from /1 to B
is called the carry-o'ver jac!nr. In the saIne \vay, it can be sho\vn that the
carry-over factor fronl a pinned cnd B to a fixed cnd A is
C' ba = C 2 = kab (g)
C 1 kbb
l'hcse carry-over factors arc very useful in the analysis of haunchcd beams
and have been tabulated for various shapes of haunches.
'fhe angle of rotation of the end A of the beam loaded as sho\\'n in Fig. 9.1 9a
can be calculated fronl Eqs. (9.16). Noting in this case that there is no
lateral load and no support settlcn1cnt and that Ob = 0, \ve obtain
{J = ('lC a - C 2 2  ], 1
f1 C 3 Elo l ab
"\'c see that the angle of rotation ()a is invcrsely proportional to the factor
('3 ' - k
Jab = C C - aa
1 _.. 3 - C'}.2
(h)
\vhich is called the stijJ1Jess .factor for the cnd A of the beam. Similarly, \ve
find that the sri ffness factor for the end B is
('1 ( i )
Sba' = - C " C -- , C 2 = kbb
1 a - 2
If the beam carries latcralloads (Fig. 9.19b), the reactive 010nlcnt i\J ba at
the huilt-in cnd can be obtained by using Eqs. (9.17). If \ve substitute
(h = Alab = Oab = 0 in these equations, they reduce to
kkao()a = - Ttnb
...\1 ba = kkab()a + mtbu
Then, eliminating Oa and using notation (f), \ve have
k
ill!"a = - k all rrab + gnba = - Cabmt.ab + mlba
no
(9.1 8)
\\Ie see that for each particular case of loading, the reactive moment ,\I/ba can
be calculated very easily by using the tabulatcd values of the fixed-end nl0ments
and carry-over factor. 1
i\s an illustration of the numerical details, let us construct an influence line
for A.f 'la in the case of the beam in Fig. 9.19 b. For several values of the ratio
xl I defining the position of the load P on the bealn, values of the fixed-end
1 See ibid.

430 CONTINUOUS BEAMS AND FRAMES
TABLE 9.4
x/I
0.1 I 0.3 0.5 I 0.7 I 0.9
ml ah -0.0958 -0.2148 -0.1752 -0.0665 -0.0032 X PI
mt ba 0.0025 0.0443 0.1370 O. 1881 0.0954 X PI
M blJ 0.062 0.177 0.245 0.229 0.097 X PI
mOInents mtab and mI.bu taken from the tables are sho\vn in lincs 1 and 2 of
Table 9.4 above. Also from the tables the carry-over factor for the beam is
found to be Cub = 0.618. The values of Alba, calculated fron1 Eq. (9.18), are
sho\vn in linc 3 of Table 9.4. '"rhe influcnce linc for A1 ba , plotted frorn these
data, is sho\vn in Fig. 9.19c.
In conclusion, let us recapitulate the relations bct\vcen the various constants
that havc arisen throughout the preceding discussion of haunched beams.
First, the t111/{ult1l' coe.tficients C\, C 2 , C\, \vere introduced, and it \vas shown ho\v
thcse quantities can bc calculated for any particular shape of haunches. F 01-
lo\ving this came the factors k bb , k(Ib't kaa't \vhich appcar in the slopc-deflection
equations and \vhich are dcfined by notations (c).
In discussing a beanl built in at onc cnd and simply supported at thc other,
the notion of tarry-over factor, as defined by notations (f) and (g), \vas
encountered. .:Iso, in connection \vith the same problem, \VC rnet the idea of
stiffness factors as defincd by notations (h) and (i).
It is seen that all of the derived constants can be expressed in tcrms of the
angular coefficients C t , C 2 , C 3 . Ho\\'cvcr, in the tables of the Portland Cement
Association, the values of carry-over factors and stiffness factors arc given
instcad of the angular coefficicnts. In such a case, by using the tabulatcd
values, the angular coefficients can bc readily calculated from the follo\\'ing
equatIons:
I 1
C 1 = --- ._-
Sab 1 - CabC ba
1 1
C 3 =-.--.--
Sba 1 - CabC ba
(9.19)
C 2 = J..- C ba _  Cab
Sab 1 - CcabC ba Sba 1 - CabC ba
PROBLEMS
A syn)merrical bean) \vith straight haunches (Fig. 9 .O) has 10//1 = 0.20 and the
haunches have the lengths >../ = //4. I f the bearn is fixed at both cnds, calculate
the fixed-end .TIoments (a) due to a uni form lateral load of intensity q, (b) due to
a concentrated load P at nlid-span.
AlIS. (a) mlab = -mlba = -0.1 I 38q/2.
(b) ffiIab = -mIlia = -O.l708PI.

ARTICLE 9.5 431
 10 
../ /
(a)
'ab i I £1
2
---J
FIG. 9.20 (b) 
2 \Vhat end rnoments \vill be produced in the haunched he3111 described in the pre-
ceding problcrn jf the right-hand support B settles by an anlount  = 1 in. relative
to the support A? Assurnc that the span I = ] 6 ft.
Ans. Af alJ = /\4'm = -0.0668£/0/1.
3 .\ssurning that the synlnlctrical beam in Fig. 9.20b is built in at 13 and simply sup-
ported at A, calculate the reactive end moment lfbu if an active moment "'[(Ib is
applied at A.
(;'..
AI/s. Al bu = -= Alub.
C I
4 If the beanl in Fig. 9.20b carries a uniforn1 load of intensity q and A1ab = 0, calcu-
late the reactive mon1cnt Ah.., at B. l'he follo\ving nurnerical data are given:
A = 0.3, /0//1 = 0.1.
/ll1s. l"fba = O.238qI2.
5 Assulning that thc unsyn1metrical beanl in Fig. 9. I9b carries a uni fonnl y dis-
tributed load of intensity q, calculate the nlagnirude of the reactive end 1110lnent
.\lbll at the huilt-in end of the bCeun. \lalues of the angular coefficients f()r this
hean1 arc C. = 0.201, C 2 = 0.146, C 3 = 0.237.
/11/ S . i\ 'ha = O. I 59 if /2.
9.5 CONTINUOUS BEAMS OF VARIABLE CROSS SECTION
In discussing continuous bean1S of variable cross section, \\'c consider two
consecutive spans (Fig. 9.21) and denotc by 0 0 , 81, 02, . . . the angles of rotation
of the cross sections at the supports. Considering the span to the left and
appl ying the second of Eqs. (9.17), \ve obtain 1
l\1"".-1 = k n r k,t-t,rt8n-l + k:,nOn
- (k,n + k n - t . n ) 8n-1.n] + lIn.1&-1 (iT)
Similarly, the first of Eqs. (9.17) applied to rhe right-hand span gives
1\;'[,.,11+1 = k'l+l[k10rt + k1l.n-+-l0rt+J
- (kll.n+l + k:l)en.u+d + )lllt,PI+l (b)
1 Since the support 11 belongs to both spans n and II + I, \VC use the notations k:,n and
k:.l to distinguish the t\VO values of knl'

432 CONTINUOUS BEAMS AND FRAMES
(Jbserving no\\,' that at the support 11 \vc must have
ill n,1I-1 = - i\J n ,n+l
\ve obtain the three-angle equation
knkn-t,n(},t-l + (knk,n + kn+lkl) 8 PI + kn+lkn,n+l()n+l
= k n (k,T\ + kn-t,n)8n-J.71 + k'l+l (k n ,n+l + kl)en,n+l
- Jf(n,n-J - ffirn,n+l (9.20a)
\vhich can be \vritten for each intermediate support. If the cnds 0 and 1J1 of
the beam arc built in, \ve have 8 0 = 8m = 0, and Eqs. (9.20a) arc sufficient to
detcrmine all of the unkno\vns. [f the end 0 of the beam is sinlply supported,
the end monlcnt il[OI vanishes and the first of Fqs. (9.17) gives
o = kd'kooOo + k 01 (}. - (kilt + k uo )8ul] + mI OI
(9.20b)
Sinlilarly, if the cnd '/11 cf the bcaln is simply supported, the second of Eqs.
(9.17) gives
o = k m [k m - 1 "it8m-l + Ic narn 8rn - (k mm + km--l,m) 8 111 ,m-l]
+ Tl71l,m-l (9.20c)
rhus, in any casc, frOITl aJ of Eqs. (9.20), the angles of rotation 8 0 , fJ 1 , O 2 , . . .
can be calculated. f-1aving these angles, all cnd n10n1cnts can then be calculated
from the slope-dcflcction equations (9.17) and the analysis is complcted.
Instead of the three-angle equations, the three-nlOJnent equations can also
be derived. For that purpose, \ve use Eqs. (9.16), \vhich \\'e rc\vrite in the
fo 110 \\0' ing fonn:
C i\labl C ilbal
8n = 1 -£70- - 2 RIo + 'Ya + 8ab
C lv[bal C j\llabl
Ob = 3 - EI  - 2 -EI-;; - 'Yb + 8ab
Ac Ad
\vherein 'Ya = BIoi 'Yb = lifol
(c)
(d)
These quantities 'Ya and - 'Yb represent the angles of rotation of the ends of the
bcan1 produced by lateral loads alone. From Eqs. (9.17), \VC find that their
lU n - 1. n iU n. n .. I J/ n . n+ 1 lU n + 1 , n
C n - 1 n  C n n+l 
FI G. 9.21 f. 'n J I In+ 1 -I

ARTICLE 9.5
433
magnitudes in terms of fixed-end nl0rTICnts arc
I
'Yo = Elo (- C1mlab + C 2 ffi'rba)
I
'Yb = E10 (C3ba - C2nab)
(9.21)
Since the fixed-end monlcnts arc tabulated for many particular cases and the
constants Cl, C 2 , C a arc given by Eqs. (9.19), the angles)'a and 'Yb can be
readily calculated from the above equations.
Considcring no\\' t\\'o adjacent spans 11 and '/J + 1 and \\'riting the exprcssions
for the angle of rotation 8,, \ve obtain
C t l ,- n-l /n, _ (-'"n i\;/n-l.nln _ t& + e
3 Elo" .. Elon 'Yn n.lt-l
= C n+l Mn,n il /n+l _ C ,&+1 Al +l'tt/ n+l + n+l + 0
1 Elon+l 2 Elon+l 'Yn n,n+l
Introducing the notations
Elon.
J;=kn
Elon+1
--- = kn+l
In+l
and observing that AJ,&.n-l = - 111,.,n+l, \VC obtain finally
('2"' ( (.I'3 n Cl1&+1 ) ('2"+1
- - j\1 1 + - + -- ---- "1 1 + ---- ,\1 +1
k n- ,n k k n. n- k n. I.
n n n+1 n+1
- 'V PI + 'V ,.+1 - e + 0
- I n Inn, n-l ,&. n+l
(9.220)
. fhis represents the three-mon1cnt equation for the case of a continuous beam
of variablc cross section. It can bc \"ritten for each intcrmediatc support; so
in the case of a beam \\'ith sirnply supportcd cnds, \VC have as nlany equations
as thcre are unkno\\'n morncnts. If the cnds of the beam are built in, there
\vill be t\VO additional unkno\vn momcnts 1\1 01 and 1\4 m,ni-I, and \ve have t\\'O
additional conditions of constraint; narncly, the end cross sections 0 and tIll
do not rotate. Using Eqs. (9.16), \ve express these conditions as follo\vs:
C 1 }\tlol/l C I A11CJl + 1 +8 _ 0
1 -£1-;[ - 2 Elol 'Yo . 01 -
C m iW m . m - 1 /m _ C In Nl m - 1 ., .nlm - m + e - 0
3 E10m 2 Elom 'Ym m-l,m -
(9.22b)
Thus, all of Eqs. (9.22) again represent a sufficient number of equations for
calculating all statically indctcrminate momcnts.
Application of the thrce-angle equations (9.20) \\'ill no\\' be illustrated \vith
the analysis of a three-span bridge loaded as sho\vn in Fig. 9.22a. In thc case

434 CONTINUOUS BEAMS AND FRAMES
(b)
FIG. 9.22
1ft
o hi hi 
r- 36 ' r 1 I: .1" 2 36'-l

-594 -453
(a)
of large spans, the bridge usually consists of a deck of constant thickness
reinforced by ribs of variable depth so that \ve have to analyze beams of 1"
section. To use the tables, calculated for rectangular beams, \ve replace the
r-section beams by equivalent rectangular beams the depth of \vhich at the
supports is given oy the formula (see page 423)
3 -
 11
hi = ho -
10
(e)
in which J 0 and /1 are the ccntroidal moments of inertia of the 1" sections at
mid-span and at the supports, respectively. The shapes of the haunches of the
equivalent rectangular beam are takcn the same as those of the actual ribs.
In this case, \\'e have parabolic haunches \\,ith Xl = 1/2 and the depths ho = 2.5
ft, hI = 7.5 ft. Starting \vith the first span and using the Portland Cement
Association tablcs 1 \ve find the follo\ving constants:
$10
I
I
-0.0439qI1 2 I O.1622qll'
C 01 I
1. 260 I
C 10
S'01
mtOl
mtl0
0.266
6.45
30.58
Then the various constants entering into Eqs. (9.20) become
koo = SOl = 6.45 kot = k 10 = C 01 k oo = 1.26(6.45) = 8.12
ku I = S10 = 30.58 nOl = -56.9 kip-ft mt tO = 210.2 kip-ft
For the second span, \\'e find again from the tables 2
C u
C 21
$12
$21
12 21
0.784
0.784
22.83
22.83
-0.I099qI2 2 O. l099ql2 2
and the constants entcring into Eqs. (9.20) become
ku Il = Iu = 22.83 kI'l = ('12kllII = 0.784(22.83) = 17.89
k 22 11 = S21 = 22.83 mt 12 = - 570 kip-ft ml21 = 570 kip-fr
1 See "Handbook of Fran1C Constants," p. 13, table 26.
2 See ibid.., table 25.

ARTICLE 9.5
435
For the third span, the tables give
C'l2
Cu
$23
Sa2
mlu I mtu
-O.1622ql.1 I O.0439ql.1
0.266
1.260
30. 58
6.45
and the constants entering into Eqs. (9.20) become
k 22 III = .\"23 = 30.58 k 23 = k 32 = C 2 J22 III = 0.266(30.58)
= 8.13
k33 = .\'32 = 6.45 mt 23 = mt32 = 0
Equations (9.20) for the problem are no\v written as follo\vs:
k 1 (k oo 80 + k 01 8 1 ) = - filo!
k)k o1 8 o + (k 1 k Il I + k2ku II ) 8 1 + k 2 k 12 82 = - mt10 - mL 1 2
k 2 k L2 8 1 + (k 2 k 22 11 + k3k22III) 8 2 + k3k238 3 = - ffi!21 - ml 2 3
k a (k 23 82 + k 33 ( 3 ) = - mt32
(f)
Observing nOVl that
k 1 = Elo
/1
k 2 = Elo = k 1
1 2 2
Elo
k3 = - = k 1
13
\ve divide Eqs. (1) by k 1 and obtain, after substituting the numerical values
for the variou constants,
6.458 0 + 8.128 1 = 5:9
8.128 0 + (30.58 + X 22.83)81 + i X 17.8982 = - 21o t+ 570
. X 17.838 1 + (i X 22.83 + 30.5.8)8 2 + 8.138 3 = -2 0
8.1382 + 6.458 3 = 0
Solving these equations, \ve obtain
8 0 = _ 10.46
k 1
15.33
8 1 = k 1
82 =
22.24
k 1
() _ 27.95
3 - k 1
Finally, substituting these values of the angles of rotation and the values of the
constants into Eqs. (a) and (b) above, \\'e find
,l11) = - klu = - 8.12 ( 10.46) + 30.58 (1 5 . 3 3) + 210.2
= 594 kip-ft
1\;1 21 = - AJ 23 = .. [1 7 .89 ( 1 5 . 3 3) - 22.83 (22.24-)] + 570
= 453 kip-ft
l.he corresponding bending-nl0ment diagram is shown in Fig. 9.22b.

436 CONTINUOUS BEAMS AND FRAMES
Llet us no\v solve the same problem (Fig. 9.22) by using the three-moment
equations (9.22). First, using Eqs. (9.19), \ve calculate, for each span, the
angular coefficients C t , C 2 , C 3 and then, using Eqs. (9.21), the quantities 1'a
and i'b as foJlo\vs:
Span
c)
C 2
C a
'Ya
'Yb
---- -
I
I o . 23 30 0.0620 0.0492 q13 q13
0.0203 Efo 0.0107 Efo
II 0.1140 0.0893 0.1140 ql8 q/3
0.0223 Elo 0.0223 Efo
III 0.0492 0.0620 0.2330
With these numerical data, the three-moment equations (9.22a) become
(0.0492 + 2 X 0.114)/\;[10 + 2 X 0.08931\/[21
= 'O.Ol07q/12 + 2 X 0.0223q12 2
- O.0893M 12 + (0.114 + i X O.0492)kI 21 = O.0223q12 2
Introducing bending moments \vith conventional signs instcad of end moments,
\ve have
A1 10 = - A'11
1\1 21 = -A1 2
LvI 12 = 1\;1 1
Then \vith the given numerical values for 11 and 1 2 , we have
M 1 = - 590 kip-ft
1\1 2 = -450 kip-ft
Within the limits of accuracy of the calculations, these values coincide \\Tith
the values previously obtained by using the threc-angle equations (see Fig.
9.22b) .
As a second example, lct us consider the construction of influence lines for
the four-span bridge sho\vn in Fig. 9.23a. This beam has symmetrical
parabolic haunches \vith Xl = 1/2. The minimum depth in all spans is
110 = 30 in., and the depths at the intermediate supports are hI = 69 in.,
h 2 = 75 in., h3 = 69 in. The extremc cnds of the bridge are simply sup-
ported. Using the samc tables and formulas as in the preceding example, \\Fe
obtain the constants in the slope-deflection equations as sho\vn in Fig. 9.2 3a.
To construct inRuence lines for bending momcnts at thc supports, \VC start
\\rith loading the first span. I)ividing this span into tcn equal parts, \ve put
the concentrated load successively at each division point and calculate the
bending moments at the supports. The three-angle equations for any of these

ARTICLE 9.5
437
75"

79'
k 3 =1
16.5
 4.2. °"
67.5'
k4 :: 1.374
10.5
4.36
4.92
(a)
4-1
3 0:2.... rj9"
o
57.5' 79'
k. =1.374 k 2 -1
4.92 15.5
4.36 11.8
10.5
11.8
16.5 15.5
9.49 9.32
(6)
Influence line for M 1
4.23
10.67 10.67
(c)
Influence line for M 2
4.26
4.26
FI G. 9.23
loading conditions \vill be of the follo\vjn form:
k l (k oo 8o + k o1 8 l ) - - fiLo l
k 1 k o1 8o + (k 1 k ll I + k2kllll) 8 1 + k 2 k 12 82 = - mt10
k 2 k 12 8 1 + (k22I1 + k3k22111) 8 2 + kak238a = 0
k 3 k 23 8 2 + (kak33III + k4k33 1V ) 8 3 + k4k348. = 0
k4(k 34 8 3 + k 44 8,,) = 0
in which
k 1 = Elo
/1
k 1
k 2 = k3 = --
1.374
k. = k 1
and the other constants are given in Fig. 9.23a.

438 CONTINUOUS BEAMS AND FRAMES
In solving this system of equations, \ve begin \vith the first equation and
solve it for 8 0 . Then substituting this in the second equation, \ve obtain
( - k 1 Oo 102 + k l k l1 ' + k 2 k l1 11 )81 + k 2 k 12 82 = :: mtOI - mt 10
k 2 k 12 81 + (k 2 k z2 I1 + k3k221II) 82 + kak 23 8 3 = 0 (g)
k 3 k 23 82 + (k3k3III + k4 k 33 1 V) 8 3 + k..k u 8 4 = 0
k4 (k a4 8 3 + k 44 8 4 ) = 0
From the last of these equations, \\'c can calculate the ratio 8 3 /8 4 . Then,
substituting this in the third equation, we find the ratio 82/8 3 , and \virh this
value we find from the second equation the ratio 81/82. We see that all these
ratios arc independcnt of thc position of the load in the first span, and can be
readily calculatcd by using the constants given in Fig. 9.23a. Having these
values \ve use the first of Eqs. (g) to calculate the value of 8 1 which evidently
\vill be proportional to the magnitude of the expression on the right-hand side
of the equation. This right-hand side of the first of Eqs. (g) can be taken
from the ready tables for each position of the load in the first span. Proceeding
in the described manner and using the constants as given in Fig. 9.23, \ve find
8..
8 3 = -0.886
8 3
- -
8 2
-0.479
82
- -
8 1
-0.432
l'he first of Eqs. (g) then gives
0.0703
81 = k 1 (0.886mLOl - ffi'tlO)
(h)
Values of the fixed-end moments for various posirions of the load, obtained by
interpolation from the rabIes, are given in r-rablc 9.5 below, together \vith the
values of the quantiry in the parentheses in Eq. (h).
TABLE 9.5
-
II
m101
Ph
mtl0
Pit
O.886mtOl - mt 10
PI 1
Olk l
Pit
,WI
I'll
M 2
Pl 1
M3
Pit
x
O. 1 -0.077 0.019 -0.087 -0.0061 - 0 . 047 +0.021 -0.008
0.3 -0.121 0.130 -0.238 -0.0167 -0.128 +0.057 -0.023
0.5 -0.079 0.237 -0.307 -0.0216 -0.165 +0.074 -0.030
0.7 -0.026 0.231 -0.254 -0.0179 -0.136 +0.061 -0.024
0.9 -0.002 0.095 -0.094 -0.0068 -0.052 +0.023 - 0 . 009

ARTICLE 9.5
439
TABLE 9.6
Span I Cab I C&a I s.
Sb
k
C 1
C 2
C a
I 0.888 0.417 4.92 10.47 1.374 0.323 0.135 0.152
II 0.758 0.717 15.54 16.47 1.0 0.141 O. ]01 0.133
III O.71i 0.758 16.47 15.54 I .0 0.133 0.101 0.141
IV 0.41 i 0.888 10.47 4.92 1.374 0.152 0.135 0.323
Having the ratios of the angles of rotation and the value of 81, \\'C obtain
the values of the bending moments at the supports from Eqs. (a) and (b), \\7hich
in this case give
M 12 = k 2 (k l1 l1 (h + k 12 82) = k 2 81 (kl111 + k 12 :: ) = M 1
M 2a = k a (k 22 111 8 2 + k u 8a) = k a 82 (k22Il1 + ku :: ) = M 2
Ma4 = k4(kaa 1V 8 a + k a( 8 4 ) = k 4 8a (k aalV + kat :: ) = Ma
where Ml, 1\1[2, Ala are the bending moments at the supports. From these
equations, \ve obtain
M 1 = k 2 0 1 (15.5 - 11.8 X 0.432)
M3 = +O.181M 1
A1 2 = -O.450M 1
Using no\v the above calculated value of 01, the bending moments at the sup-
ports are readily obtained. l"heir values are given in the last three columns
of Table 9.5 above. This finishes all necessary calculations for positions of
the load in the first span.
We no\v consider loads in the second span and, for the purpose of illustration,
use the three-moment equations (9.22) instead of the three-angle equations
(9.20). We begin \\rith a calculation of the constants Cl, C 2 , C 3 by using
Eqs. (9.19) and of 1'0' 1'b by using Eqs. (9.21). The values, so obtained, for
all four spans are given in Table 9.6.
F or span I I, the quantities l' Q and 'Yb arc
1
'"fa = k 2 (-O.141mI12 + 0.101£m:21)
1
'Yb = k 2 (+O.133mt21 - O.lOlmt12)
(i)

440 CONTINUOUS BEAMS AND FRAMES
With these values of the constants, the three-moment equations (9.22) become
( 0.152 )
1 .3 74 + O. 141 At/ IO + 0.101 1\;/21 = - 0.141 mI 12 + O. 101 mt 21
-O.lOIM 12 + (0.133 + O.133)lW 2 1 + 0.lOlM 32
= O.133mI21 - O.IOlffrtI2
( 0.152 )
-0.lOIM3 + 0.141 + 1.37 4 M 32 = 0
Introducing bending moments
M 1 = -Al 10 = M 12
M 2 = -Jt1 21 = M 23
,\t[3 = - M 32
and solving the equations, we obtain, for any position of the load in the second
span, the values
,WI = -o.200n21 + 0.464n12
M 2 = -O.500mt21 + 0.240mt 12
M3 = -O.400M 2
Now, using the available tables for fixed-end moments, the bending moments
at the supports can be readily calculated. l'he values of these bending
moments, so obtained, are sho\\'n in Table 9.7.
It is seen that use of the three-moment equations resulted in considerable
simplification since, in this case \vith simply supported ends, we have only
three redundant quantities to calculate, \vhereas in using the three-angle
equations, \VC had to calculate five unknown angles of rotation at the supports.
In the case of a beam with built-in ends, the three-angle equations \\'ould hold
a like advantage over the three-moment equations.
We have no\\' considered the conccntrated load acting in each of the first
t\VO spans. O\ving to symmetry of the beam in this case, the results already
qbtained arc sufficicnt to construct the influence lines for all four spans. Such
influence lines for All and M 2 are sho".'n in Fig. 9.23b and c, respectively.
TABLE 9.7
x/I
mt12
mt21
,\1 1
A1 2
A13
0.1 -0.092/'/2 O.OO6P/2 -O.044P/2 -0.0251'1 2 +0.01OP1 2
0.3 -0.203P/ 2 0.065P1 2 -O.107P/2 -0.081P/:z +0.033PI 2
0.5 -0. 163Pl 2 O. I 82Pl:z -0. I 12Pl 2 -0. 130Pl 2 +O.052P/ 2
0.7 -O.055PI 2 O. 214PI 2 -0.069PI 2 -0. 120PI 2 +0.048PI 2
0.9 -0.004P/2 O.094P/ 2 -0.021PI 2 -0. 048P/ +0.019P/2

ARTICLE 9.6
..41
9.6 SIMPLE FRAMES WITH PRISMATIC MEMBERS
In the analysis of simple frames like those discussed in the preceding chapter,
the slope-dcflection equations derived in Art. 9.1 can also be used to advantage.
In discussing such frames here, \ve make the same simplification as before and
neglect the effect of axial forces on deformation of the frame.
As an example, lct us considcr the symmetrical frame ABC pinned at A and
(' and loaded as shown in Fig. 9.24a. Neglecting shortening of the bar BC
due to axial compression, there \vill be no displaccment of joint B, 2nd the
slope-deflection equations in the form (9.2) will apply. Using Eqs. (9.2a),
\ve obtain, for the member AB,
Mba l ql3
8b = - 3£1 - 24£1
and, for the member BC,
Mbc l
8 6 = 3 EI
Equating these angles and observing, from the equilibrium of joint B, that
''\;[ba = -lW bc , \ve obtain
2 kibei
"3 EI -
ql3
- 24£1
Since the end moment J.tl bc is identical with the bending moment M b , we find
qJ2
Mb = - -
16
The corrcsponding bending-moment diagram is shown in Fig. 9.24b.
It should be noted that the continuity condition at joint B is the same as was
used in the derivation of the three-moment equation, \vhich suggests the use of
this equation in the analysis of frames. Treating the frame ABC as a t\vO-
span continuous beam and using the three-Inoment equation [Eq. (9.7 a) ], we
FI G. 9.24
(b)

442 CONTINUOUS BEAMS AND FRAMES
obtain directly
_ ql3
4AJ b l = 4
which gives the same value for M b as that calculated above.
Having the magnitude of the bending rnon1ent j\'[b, \ve can find from statics
the axial compressive forces in the bars /lB and BC and the corresponding
small displacement of the rigid joint B, \vhich \vas neglected in the above
anaysjs. Upon introducing the additional angles of rotation of the t\VO bas,
due to this displacctnent, \\,'C can find a correction to the previously calculated
value of Alb by using Eqs. (9. 3a). This correction is usually sIllall and can bc
disregarded in most practical calculations.
In the case of a sytnmctrical frame loaded as sho\vn in Fig. 9.25 there \vill
be no latcral s\vay, and the three-moment equation can be applied in calculating
the bending moments at the rigid joints Band (t. LT sing Eq. (9.7) and observ-
ing from SYOlnletry that i\;f b = AJ c, \\'c obtain
( 1112 )
2i\1 b J; + / + l\tJbl =
'r'hen \\ith the notations
h
a = --
/
1 2
-="'(
1 1
\ve find
i\1 b = J\1 c =
3Pl 1
--
8 2a"'( + 3
3P/2
R
(a)
(b)
The corresponding bending-moment diagran1 is sho\vn in Fig. 9.25b.
If there is no symmctry, the top of the loaded frame \vill move laterally, and
this rnotion must be considcred in calculating the bending rnOInents at the rigid
joints. Take, as an exan1ple, the case sho\\'n in Fig. 9.26(1, \vhere a load P
is applied nonsymnlctrically. 'To solve the problem by using the three-
nl0ment equation, \vc: assurne first that there is a horizontal force H applied
as sho\vn and of such nlagnitude that the lateral s\vay of the frame is prevented.
P
I I
812 21c
1 2 1
II 11 h
A D1
FI G. 9.25 (a) (b)

ARTICLE 9.6 443
P 1.., t,
e1c H- - 1r
B i h ",,:-;..
H 1 2 ( "
I C b c
II /JI' iU'
b c
h
A 1 D
H H
hlb hI; 2 2
(a) (b) (c)
h h
FIG. 9.26
l--hen, proceeding as before, \ve obtain the bending moments 1\1 and 1\1; at the
rigid joints Band C fron1 the foHo\ving three-moment equations:
2M; ( h + i ) + AJ  =
1 1 1 2 /2
kf  + 2i\1 (  + 12 ) =
1 2 1 2 II
Pc (/ 2 - ( 2 )
/21
Pe(1 - e) (21 - e)
/21
(c)
"'lhen the bending moments k[ and 1\;/; have becn found, the axial forces
acting on the horizontal girder BC' can be calculated from equations of statics.
l"hese forces, assuming positive bending rnoments .;\J[ and 1\1;, arc indicated
in Fig. 9.26. The condition of equilibrium then gives
H _ 1\.J + l\J = 0
h h
and the force rcquired to restrain the fran1c against lateral s\\!ay is
,1 [ ' 11 [ '
II = _ 1 C - j .. b
h
(d)
To find the values }\J b and J\J(' of the bending nlomcnts \\"hen there is no
horizontal restraining force H, \ve considcr the auxiliary problelll sho'Nn in
Fig.9.26c. The required mOInents \vill cvidently he obtaincd by superimposing
the solutions for this case and that sho\vn in Fig. 9.26a. Denoting by ,\1'
and l\I[;' the bending moments at Band C for the case in Fig. 9. 26c, \VC obtain
NIb = M + AJ'
fc = i\tl; + A1;'
(e)
The bending of the franle in this case is of a vcry simple naturc since, [rom
symmetry of the franlc, the horizontal reactions of thc hinges A and D must

444 CONTINUOUS BEAMS AND FRAMES
each be equal to H/2. I-Iencc,
Al" = - !vI" = _ hH
b c 2
Substituting for H its value (d), \ve obtain
, W b " = - NI" = A;[ - ,\1
r 2
(f)
and }qs. (e) give
A1 b = i\1 = Alf + j\t[
r. 2
(g)
'[he sum of the monlents M and AJ; is obtained by adding together Eqs. (c),
\vhich gives
Ai' + M' 3Pe (1 - e)
b " = - 1[3 + 2 (11-/1) (1 2 // 1 ) ]
Hence, the requircd solution is
3Pe(1 - e)
Mb = Me = - 2/(3+'i( hll) (/ 2 /h)]
(11)
In the foregoing example, \ve obtained the rcquircd solution by supcrimposing
the solutions of t\VO separate problems as sho\\'n in Fig. 9.26a and -c. In the
first of these two problems, the lateral s\\"ay \vas restrained, and in the second,
the effect of latcral s\\'ay alone \vas considered.
Ho\\rever, by using thc slope-deflection equations, \\'C can \\'rite directly all
equations necessary to determine the bcnding tnomcnts at the rigid joints and
also the lateral s\vay of any given frame. Take, for example, the synuTIctrical
frame sho\vn in Fig. 9.27a and assume that, under the action of the load P,
the top of the frame undergoes a lateral displacement  to the right. l'hen
\vith the notations
Ell
k l = k3 = -
/1
k 2 = R/ 2
/2
A
8=-
/1
(i)
we \\'rite the foJlo\ving slope-deflection equations [see Eqs. (9.6)]:
i\l ab = 2k 1 0 b - 6k 1 fJ
l\1 ba = 4k 1 Bb - 6k 1 8
j\1bc = 2k 2 (20 b + Be) + mtbc
MCb = 2k 2 (20 c + Ob) + mtcb
k1cd = 4k 1 8c - 6k 1 8
Al dc = 2k 1 8c - 6k 1 8
(})

ARTICLE 9.6
445
We see that all end mOfilcnts arc expressed in terms of the three quantities
(}b, ()e, and 8. For their deternlination, \ve note first that
AIL bo = - At/be
,\If cb = - At[ cd
\vhich, after substitution from Eqs. (j), become
4(k 1 + k 2 )()b + 2k 2 8 c = 6k 1 8 - mIbc
2k z ()b + 4(k 1 + k 2 )()c = 6k 1 8 - filcb
(k)
ro obtain a third equation, \ve consider the equilibrium of the beam BC under
the action of horizontal forces transmitted to it by the legs of the frame. From
the action of the end moments on the vertical member AB (Fig. 9.27/7) \ve
conclude that there must act at A and B the horizonta 1 shearing forces
H = M ba + "lab
/1
(l)
SiInilar)y, for the member' CD, \ve obtain
H - J\;I cd + ,\l de
1 - --- --
/1
(111)
rhen front equilibriunl considerations of the beam BC, \\'e conclude that
l
II + H 1 = /1 (,\-t ho + MOb + "lcd + it-I ar ) = 0
Substi tuting for the cnd moments from Eqs. (j), \ve obtain
8b + Oe = 48 (11)
From Eqs. (k) and (11), the quantities (}b, 0(:, and f) can be determined. They
Br f p e--i
II -c.
,
,
I
1 1 I I
lA I I
1 2 D
Mab 9.5 12.7
(a) (b) (c)
FIG. 9.27

446 CONTI HUOUS BEAMS AND FRAMES
arc
8b = _ _mlbc + rrn ch + nr.b - Ttbc
2 (k 1 + 6k 2 ) 4 (2k l + k 2 )
8e =
mtbc + filch
2 (k 1 + 6k 2 )
nch - Ttb c
4(2k l + k 2 )
(0)
8=
mt bc + mcb
4(k 1 + 6k,.)
For the indicated loading, \ve have the fixed-end moments (see page 412)
Pc2j
mtbc = - -- I 2-
"
"
11 2 c
(''Yn' _
vl'-cb - -/2
and Eqs. (0) become
8b = _ ef(e - I) + Pel
2/ 2 2 (k l + 6k 2 ) 4/ 2 (2k 1 + k 2 )
o = l)ef(e - f) _ Pel
c 2/ 2 2 (k 1 + 6k 2 ) 4/ 2 (2k 1 + k 2 )
e = Pef(e - I)
4/ 2 2 (k 1 + 6k 2 )
(p)
Substituting these values into Eqs. (j), \ve find
Pel(e - f) Pef
J\tl ab = 2/22(1 + 6k 2 / k 1 ) + 2/ 2 (2 + k 2 /k 1 )
"IIbe = -A1 ba = Pcf(e - f) ]Jef
2/ 2 2 (1 + 6k 2 / k l ) 1 2 (2 + k 2 /k 1 )
J\il cb = -/\1 d = _ Pef(e - f) + _ Pef
c 2/ 2 2 (1 + 6k'}./k 1 ) 1 2 (2 + k 2 /k 1 )
,_ fJef(e - f) Pcf
ilfd - - --- - .-
c 2/ 2 2 (1 + 6k 2 /k 1 ) 2/ 2 (2 + k 2 /k 1 )
(q)
Using these formulas, the end n10ments can be calculated and the bending...
moment diagram constructed for any position of the load P. J t is seen that
the moments depend only on the ratio k 2 / k. but not on the values of these
quantities. lake, for example, II = 15 ft, /2 = 30 ft, k 1 = k 2 , f = 10 fe,
P = 10 kips. Then fron1 Eqs. (q), \ve obtain
At[ ab = 9.52 kip-ft
,\J cb = 20.6 kip-ft
i\1 bc = - 23.8 kip-ft
i\1 dc = - 12.7 kip-ft
and thc bending-moment diagram \\,ill bc as sho\vn in Fig. 9.27&.

ARTICLE 9.6
447
To calculate the latcral displacement  of thc top of the frame, ,,e use the
expression for the angle of rotation 0 as given hy the last of Eqs. (p). Thus,
f}ef(e - /)/ 1
 = lIe = 4k1/22(1 + 6k 2 / k 1 )
(r)
It is seen that \VC need not only the ratio k 2 / k 1 but also the value of k I to gct
the nunlerical value of d. Assume E = 288,000 kips/ft 2 , /1 = 0.80 ft 4 ; then
k 1 = 288,000 X 0.80
15
and froIn Eq. (r) \VC find d = 0.0093 in.
In numerical calculation of the deflection of a frame \vithout lateral restraint,
the follo\ving procedure is frequently used: 'Ve calculate first thc cnd
moments and shcar forces, assuming that there is no lateral displacement.
Then \\'C assume a certain latcral displacement, say  = 1 in., and calculate
the corresponding valus of the cnd moments and shear forces at thc joints.
Since the value of the lateral displacement \vas arbitrary, these shear forces
\vill not balance those calculatcd for the laterally restrained frame. But since
the shear forccs arc proportional to the assumed value of Ll, it is easy to find,
in each particular casc, a numerical factor by \vhich the assumed  must be
multiplied in order to attain the required equilibrium of the frame. To illus-
trate this Incthod of calculation, Ict us consider the unsymnlctrical franlc sho\vn
in Fig. 9. 28a. Considering first that the frame is laterally restrained and
assuming that k 1 = k 2 = -1k 3 = k, \\'C obtain, frolTI the slope-deflection equa-
tions (9.6), the foJ1o\ving expressions for the end moments:
i'lb = 2kO b + ffilnb
Aln = 4k8 b + mtba
l\Jc = 2k(20 b + 8e) + mIbc
i\1: b = 2k(20 c + 8 b ) + ncb
i\tf: d = 8k8 c + 4k8 d -= 6k8 c
o = 20 d + Be (}tl = --Oc
(s)
'rhcn, fronl the conditions
IlI ha = -Al be
Alco = -. \1 cd
\ve obtain
48b + Or. =
nba + Ubc
- -----
2k
(t)
8b + 5 (Je =
lnch
- ----
2k

448 CONTINUOUS BEAMS AND FRAMES
Taking 11 = /2 = 2/3 = 20 ft, P = 2Q = 20 kips, \\'c have
mtab = -19.2 kip-ft mtba = 28.8 kip-ft
mtbc = - 57.6 kip-ft mtcb = 38.4 kip-ft
and Eqs. (:) give
8 _ 4.8
8b = - c - T
8 _ 2.4
d- 
k
With these values of the angles of rotation at the joints, \VC obtain from Eqs. (s)
J\1 = - 9.6 kip-ft
kfa = - N[c = 48.0 kip-ft
(11)
j\;l:b = - l;d = 28.8 kip-ft
l"'hc horizontal forces transmitted from the vertical menlbers to the horizontal
beam BC' \vill then be
H = .1 0 X J3 + j\;fb + A;[a = 7 92 k .
20 I} . 1 pS
H j\tl;d k .
1 = t; = -2.88 IpS
"[hese horizontal forces, as sho\vn in Fig. 9. 28b, are not in equilibrium, and
\vithout lateral restraint, the beam BC \vould move to the right.
In the second step of calculation, \ve remove the latcral restraint as \vell as
the loads P and Q and assume that the bean1 BC is moved to the right by the
amount d = 0.1 ft. Then the angles of rotation of thc members \vill bc
 1
8 1 = /1 = 200
O 2 = 0
 1
e - -
3 - - - -----
13 100
Br glz P  l21c
.-- I
ill
13
Q 1
III D
1- A (a)
7.92 2.88
 ..--
B (b) C
O.OO229k 1 O.OO269k
.- III
FI G. 9.28 B (c) C

ARTICLE 9.6
449
For this case, the slope-deflection equations (9.6) become
A'[ = 2kO b - 6k81 = 2kO b - 0.03k
Al = 4k8b - 6k8 1 = 4k8 b - O.03k
M;' = 2k(28b + 8e)
Al = 2k(28 c + 8b)
N[; = 4k(29 c + 8d) - 12k8 3
o = 4k(28 d + 8e) - 12k8 3 8d = ---Bc + j8 3
Substituting these cnd mon1cnts into the conditions
(v)
j\;[  = - 1\;1 
Atl" = - Al"
cb cd
\ve obtain
48h + Be = 0.015
8b + 58e = 0.030
from which
8b = 0.00237
and Eqs. (v) give
M = -0.025 3k j\1 = - A,I = -O.0205k
AI[; = -M: = -O.0269k
8e = 0.00552
(w)
The corresponding axial forces acting on the bcam BC' arc sho\vn in Fig. <) .2Sc.
Since the magnitude of the displacement  = 0.10 ft \vas arbitrary, these forces
usually will not balance the previously calculated forces sho\vn in Fig. 9.28b.
To attain equilibrium, \\'C multiply the assumed displacement by a factor x.
Then the axial forces in Fig. 9 .28e \viII also be multiplicd by x, and the condition
of equilibrium becomes
(O.00229k + 0.00269k)x = 7.92 - 2.88 = 5.04 kips
from which \ve find
5.04
.'t = O.00498k
The true end moments, corresponding to the actual lateral displacement of the
fran1c, \vill thcn be obtained by multiplying the values (w) by x, \vhich gives
N[' = Xj\1 = - 25.6 kip-ft
J;1:' = - j\;l;' = xAt[ = - 20.75 kip-ft
' L l c ' a " = :t , { '" ;\ ' 1 " 27 2 J . Co
lV, - lVJ cb = Xlv. cd = - . {IP-It

450 CONTINUOUS BEAMS AND FRAMES
Superimposing these moments on those given by Eqs. (11), the accual end
moments arc obtained as follo\vs:
A1ab = -9.6 - 25.6 = - 35.2 kip-ft
A1 ba = - Mbr. = 27.25 kip-ft
A1cd = -M cb = -56.0 kip-ft
With these values of the end momets, the bending-moment diagram for the
frame can be constructed, if desircd.
PROBLEMS
1 Construct the bending-molT1ent diagran1 for the syrnrnerricaHy loaded symmetrical
frame sho\vn in Fig. 9.29. 1.hc follo\ving numerical data are given: P = 150 kips,
I = 12 ft, h = 30 ft, C = 3.5 ft, 10/12 = 2,/ 1 11 2 = i.
Al1S. A1ad = -,\'lab = -94.0 kip-ft, A1ba = -M bc = +21.4 kip-ft.
2 .A symmetrical frame like that sho'n in Fig. 9.25 has built-in ends at A and [J and
carries, instead of the concentrated load P, a uni formly distributed load of intensity
q along the beam BC. For the particular case \vhere 1 1 = 1 2 , find the bending
nlaments at the ends of the members.
Ans. A,f b = ,\;f c = -qJ2/6(2 + a), j\1a = lvld = -kfb/2,0: = hll.
3 Find the bending moment A'i b for the franlc shown in Fig. 9.30 if qo = O.
-1 Pe 2 (1 - e)
Ans. i\lb =, \vhere a = h//, l' = /2// 1 .
a*( + 1 /2
4 Find the bending moment i\J b for the frame loaded as sho\vn in Fig. 9.30 if P = o.
qo/2
Al1S. Alb = - , \vhere a = h/I, l' = / 2 // 1 .
20(1 + Ct'Y)
5 Find the bending nloments i\lI b and Me for the frame sho\vn in Fig. 9.26 if the girder
BC is unifonnly loaded along the right half of the span and the frame is free to
sway.
Ans. ,\l b = '\ll,  - (qf2)/8(3 + 2a,,), \vhere 0: = h//, *( = / 2 // 1 .
B
II
1 2
FI G. 9.29
I.
l
-I

ARTICLE 9.7
451
p
I
B 1 2
1 0
11 /1
1 A
FIG. 9.30
6 Find the bending nl0ments at A, 11, C, and lJ for the frarne sho\vn in Fig. 9.26 if
lateral s\vay is restrained and the columns are built in at A and D.
Al1s. "'ith a ... h / I and 'Y ;:; 1'1/ J 1,
2Pt'(1 - e) [al'(/ + e) + 2el
/2 (al' + 2) ( 3a l' + 2)
Ale = _ 2Pt(/ - e) [a"Y(2/ - e) + 2(1- r)]
/2 (al' + 2) (3al' + 2)
..lfa = - Jllb ,\J d = _ l\1.,
2 2
7 Solve the preceding problem if the load is uniforn11y distributed ctJong the right
hal f of the girder.
I-/inl Substitute qde, instead of P, in the ans\vers of the precedjng prob-
Jem, and integrate \\lith respect to e fro111 0 to 1/2.
Afb =
9.7 CONTINUOUS FRAMES WITH PRISMATIC MEMBERS
In the preceding article, several simple cases of frame \vere considered in
\vhich direct application of the slopc-deflection equations gives the required
solution. In practice, such systems as those sho\vn in Fig. 9.3 1 are often
encountered. rhese frames arc more complicated than those previously con-
sidered, and before ,,'riting the necessary equations, some preliminary dis-
cussion is desi rable. 1 For this purpose, \ve di vide the systems into t\VO groups.
In the first group, \VC include frames of the type sho\vn in Fig. 9.3 1 a \vhcre,
o\ving to the fact that the hinge a is fixed, displacements of all the joints arc
prevented, and only rotations of these joints need to be conidercd. For the
case sho\vn in .Fig. 9.31 b, both ends iT and f of the girder are free to move
horizontally, so as a result of bending of the members, latenll sway can take
1 l'hc slope-deflection equations and their application in (ratne analysis \vcre developed by
Axel Bcndixen; see his book "Die 1\1cthodc der Alpha-Glcichungen Lur Bcrechnung yon
Rahmenkonsrrukrionen," Berlin, 1914.

S2 CONTINUOUS BEAMS AND FRAMES
e
a b c d
d Ii b
3 1
II
6
(
(b) (c)
FIG. 9.31
place. The building frame sho\vn in Fig. 9.3 I c is also subject to latcral s\vay.
In the analysis of franlcs of the second class, the calculations are usually made
in t\VO steps. First, \ve assume that the lateral s\vay is prevented hy some
constraints and consider only rotations of the joints. After\vard, in the
second step, the effect of lateral s\vay can be studied separately.
In deriving equations for the analysis of frarnes restrained against side s\vay,
\VC take the structure sho\vn in Fig. 9.31c and consider a typical joint a attached
to the adjaccnt joints /;, c, £I, and e by the bars 1, 2, 3, and 4. Dcnoting by
k 1 , k 2 , ka, and k4 the sti ffness factors for the bars and by (J,., fh, . . . the angles
of rotation of the joints, \ve see that the end nlon1cnts for the bars that nlcer
at joint a \vill be [see Eqs. (9.6)]
AI! nb = 2k 1 (2Ba + fh) + mab
ilIac = 2k 2 (20 a + Be) + ;J1I(JC
,\t[ad = 2k3 (20 a + Od) + l1ad
A-fat = 2k4 (20 a + Of) + llae
(a)
where the last terms on the right-hand side of the equations represcnt the
fixed-end monlents produced by forces acting on the bars. In each particular
case of loading, these fix<:d-end Ihon1ents can be easily calculated or taken fronl
existing tables.
The end Inomets (17) represent actions on the bars; equal nlon1ents, but of
opposite signs, \vill rcpresent actions on the joint a, and from the condition of
equilibrium of this joint \ve conclude that
i\1 ab + j\1ac + i\l(ld + llioe = 0
(b)
Thcn substituting from Eqs. (a) for the momcnts, the follo\ving equation of
equilibrium for the joint a \viJI be obtained:
4(k 1 + k 2 + k3 + k 1 ) On + 2k 1 0 b + 2k 2 0 c + 2ka O d + 2k40
= - (llob + moc + nad + )rc(le) (9.2 3a)

ARTICLE 9.7
453
Such an equation can be \vrittcn for each joint of the franle, and \\'c shall obtain
as many equations as there are unkno\\'n anglcs of rotation. Thus, the prob-
lem is reduced to the solution of a system of linear algebraic equations.
Solving this system, \\,PC find the numerical values of the angles of rotation.
Substituting these into Eqs. (a), \ve then calculate the end nl0mcnts for all
members of the frame.
In the above deri vat ion of Eqs. (9.2 3a), \\'c considered a joint II at which four
mcmbers mcet and obtaincd an equation \\,ith five UnknO\\fns. Considering
joint b where only thrce members meet, \VC shall have only four unkno\vns, etc.
In the case of joint c, \\'e shall again get only four unkno\vns since the bar cf is
built in at f and the angle 0/ vanishes. If thcre \vere a hinge at f, the anglc 0/
\vould not vanish but the end momcnt j\ll/ c \vould, and \ve gct the equation
Hence,
o = 2k 6 (20/ + Oe) + ffil/ c
I mt lc
0 1 = - 2" Oc - .4 k 6
and Vle obtain
Aile/  2k 6 (28 c + 8/) + mI C1  3k68c --}m1 /c + fil cl
l'he equilibrium equation for the joint c then becolnes
4(k 6 + ik6 + k7 + k 2 ) + 2k 2 8 a + 2k 6 ()g + 2k 7 ()h
= - (mIca + mIca + mI ch + fil el - {-1t/c) (9.23b)
Comparing Eq. (9.23b) \vith the previously developed Eq. (9.23a), it can be
secn that if, in considering the equilibrium of a joint like c, \ve have a member
hinged at the far end, a modified stiffness factor k' = i-k should be used
instead of k (see page 409). Also, instead of a true fixed-end moment, a
modified fixed-end moment
mIJ = mt c / - ''mIIC
for a bar with one end fixed and the other simply supported appears on the
right-hand side of Eq. (9.23b).
The above equations are sufficient for the treatment of frames without side
s\\'ay. In the case of frames subject to side s\vay, \\'c have still to consider a
second step in the analysis. For this second step, \VC disregard the external
loads and consider only rotations of the menlbers due to lateral s\vay of the
frame. Then, instead of Eqs. (a), \ve shall have equations of the type (see
page 411)
A1,", = 2k(20 na + 8n) - 6k8m'
and in considering the equilibrium of joints, \\le shall obtain equations similar
to Eqs. (9.23). The only diffcrcnce \vill be on the right-hand side of thc

454 CONTINUOUS BEAMS AND FRAMES
equations where the fixed-cnd moments such as mt mn \\'iJl be replaced by
quantities - 6k i 8mn.
As an example of analysis of a continuous frame \vithout side s\vay, let us
consider the case sho\vn in Fig. 9.3 2a. Since the ends of the girder are built
in at a and e, there can be no latcral movement, and dcformation of the frame is
con1plctely defined by the three angles of rotation 8b, Be, 8". rrhc nccessary
three equations of the type (9.2 3a) \vill be
4(k 1 + k 2 + k 6 ) 8b + 2k 2 8 c = - (rrnba + mtbc. + ffi"lb/)
4(k 2 + k3 + k 6 )Oc + 2k'}.Ob + 2k38d = - (fiLch + mLcd + tmco) (c)
4(k 3 + k, + k 7 )8 d + 2k 3 8 c = - (mtdc + mtde + ffiLdh)
If the dinlensions of the structure and the acting loads are given, the values
of k i and film" can be calculatcd, and \\'e obtain froln Eqs. (c) thrce lincar
numerical cquations of the same kind as in the case of continuous beams.
Assume, for example, that
k 1 = k.
k 2 = k3 = k6 = k6 = k7 = ik 1
11 = 14 = h = I
1 2 = 1 2 = 21
(d)
Also, for a uniform load of intensity q covering the entire length of the girder
ae, \ve have
mtbc = m!cd = - m-t cb = - mtdc =
_ ql2
]2
_ ql2
3
(e)
;)Uab = Ttd = - )Uba = - mted =
"rhere are, of course, no fixed-end mornents for the laterally unloaded columns.
(a)
k6
g
h k7
e
H 3 d 4
7
(b) h
FI G. 9.32

ARTICLE 9.7
455
Equations (c) then become
ql2
8k 1 8b + k 1 8c = 4
8b + 69c + (Jd = 0
_ q/2
k t 8 c + 8k I 8d = 4
(I)
and \ve obtain
/2
f)b = - 8d = -"---
32k 1
8e = 0
rhc fact that Be = 0 could, of coursc, have been concluded in advance from
symmetry and the solution of Eqs. (f) some\\'hat simplified. Having the
angles of rotation, the end moments are obtained from equations such as Eqs.
(a), above, and \VC find
A.tf bc = - M dc =
_ q/2
48
_ 1 3 q /2
48
5ql2
" .f b = -/li d =-
a e 24
A1ab = -M cd =
MCb = -M cd = 3f2
ql2
Mb' = - M dh = -
16
With these values of the end moments, the bending-moment diagram for the
framc can be easily constructed.
If, instcad of built-in ends, therc are fixed hinges at a and e (Fig. 9.32a),
lateral motion is again restrained, but \ve use Eqs. (9.23b) for joints band d
and Eqs. (9.23a) for joint c. Then, instead of Eqs. (c), we get
(3k 1 + 4k 2 + 4k 6 )(h + 2k 2 8 c = - (mtba + mIbc + mrb' -jmtab)
4(k 2 + k3 + k 6 ) Oc + 2k 2 8 b + 2k 3 8d = - (mtcb + mIcd + mLco)
(4k 3 + 3k 4 + 4k 7 )8 d + 2k 3 0 c = - (mtdc + mt d6 + mtdh -mred)
(g)
These equations again can bc rcadily solved in each particular case.
let us consider no\v the case where the frame is free to move laterally.
,A.s an example, \VC take the frame in Fig. 9.32b and assume that it is uniformly
loaded only along the first t\VO spans. In this casc, the problem ,""ill be solved
in t\\'o steps. Assuming first that there is a lateral constraint at a and using

456 CONTINUOUS BEAMS AND FRAMES
Eqs. (g) and (d), \\,e obtain the system of equations
5q/2
7 k 1 8b + k 1 8 c = 24
ql2
k 1 8h + 6klf) + k 1 f)d = - T
k18c + 7k 1 8 d = 0
from \vhich
qJ2
8b = 0.0388 Ii;
qJ2
Oe = -0.0635 k;
(Jd = 0.00908 f2
No\v using equations similar to Eqs. (a), we can calculate aU end moments.
For the columns, we get
MI = 2M 1b = 2k 1 0 b
M dh = 2M hd = 2k 1 ()d
Mea = 21\1(1(: = 2kl B o
Then, the corresponding shearing forces at the tops of the columns (see Fig.
9.27 b for directions) will be
llbl = .3 k lz(Jb
H =  klOc
C(1 Jz
3k 1 0 d
Hdh - -.-
- h
and the total horizontal force transmitted to the girder, positive in the direction
from left to right, becomes
3k 1 3q/2
h (Ob + Be + OJ) = -0.0156 -h-
(h)
11he minus sign indicates that this forcc has the direction from right to left,
and to prcvent lateral movement, the force H nlust act at a as shown in Fig.
9.32b.
In the second step of calculation, \ve disregard the external forces and con-
sidcr the effect of lateral sway. Assuming that the girder ae moves to the
left by an amount , \ve find the follo\\ring angles of rotation for the members
1, 2, 3, . . . of the frame:
8 1 = 8 2 = 8 3 = 8 4 = 0
6.
8 6 =: 8 6 == 8 7 == e = - h
'[he slope-det1ection equations for the girder membcrs \vill have the form
M mn = 2k(28,,, + 8n}
iU nm = 2k(28 n + 8m)

ARTICLE 9.7_
457
and, for the colulnns, the form of the equations \vill be
l[m" = 2k(2fJ,,, + 8") - 6ke
M nm = 2k(28 n + 8m) - 6ke
(i)
Proceeding then as already explaincd and \\7riting equations of equilibriuo1 for
the joints b, c, d, \\'C obtain the equations
(3k 1 + 4k 2 + 4k 6 )(h + 2k 2 8 c = 6kr,8
4(k 2 + k3 + kG) (J(J + 2k 2 8" + 2ka8d = 6k 6 8
(4k 3 + 3k 4 + 4k 7 )8d + 2k3Bc = 6k 7 8
\Vith the prcviously taken geometrical proportIons (d), these equations
become
70 b + ()e = 3e
8b + 68c + 8d = 38
()c + 7fJd = 38
(j)
from \\,hich the three angles ()b, Dc, Del could be found. The end momcnts for
the columns \vill then be determined from equations of the form (i), and \ve find
IU bl = 2k18b - 3k 1 0
lv[co = 2k18r. - 3 k 1 8
,\tf dh = 2k18d - 3 k1e
I\lJb = k 1 8h - 3k 1 8
AJ oc = kt8e - 3k 1 8
,\1 hd = kiOd - 3k J 8
With these values of the end momcnts for the columns, v.'e calculatc thc hori-
zontal forces acting on the girdcr at. They arc as follows:
RbI = i\1 b L + l\1J  =  _kl (b - 28)
h h
Hro = 3kl( _  h _ 20) Hdh = 3kl -<_-_ e)
h
"{'hen the total horizontal force acting on the girder, positive in the direction
from left to right, is
H == Rbi + Reg + Hdh = }1? (Ob + Or. + Od - 69) (k)
To get the summation of the angles 8b + Oe) + 8 d , \\'C add together Eqs.
(j) and obtain
(h + Be + Od = le

458 CONTINUOUS BEAMS AND FRAMES
l'hen from Eq. (k). \VC find
H = - 3k -  9 e = 3k 39 tJ.
h 8 h 2 8
Taking this result together \\,ith the solution (h) obtained on the assumption of
lateral constraint, \ve obtain the latcral s\vay  fronl the condition of equilib-
rium of the girder (Ie \\'hich gives
-00156 q/2 +  1   = 0
. h h 2 R
From this equation, \ve find
 ql2
- - -8 - -
h - - 312k 1
vv'ith this value of 8, \ve obtain no\\" from Eqs. (j)
ql2
8b = 8 = 8d = - ---
c 832k 1
(I)
With these angles kno\vn, the end momcnts corresponding to the lateral
n10vement of the framc can be readily calculated. '[his finishes the second
step of calculation.
10 gct the complete valucs of the end moments, \\'e have to supcrimpose the
moments produced by lateral s\vay on those previously calculated for the frame
\vith lateral constraint. Assuming, for example, that q = I kip/ft and
I = 12 ft, and using the angles of rotation 8 b , Oc, 8d, calculated in the first step
of the analysis (see page 456),\\'c obtain
i\'f a = 34.8 kip-ft "vI c = -46.0 kip-ft
1\4:' d = - 17.0 kip-ft '\If;b = 35.3 kip-ft ( 111 )
J\;[c = - 6.57 klp-ft Me = 3.92 kip-ft
Using the angles (I), \ve obtain, as a result of side s\vay,
A/ I " -  1 " - A",' - 1 1 " _ A' I " _ A "" _ _ 3q/2
ba - , be - II'1cb - 't (Ii - IV. de - 'Ide - 832
= -0.519 kip-ft (11)
Adding the moments (111) and (11), \ve obtain total moments for the girder.
It is seen that in this case lateral s\vay has only a snlall effect on the end
moments.

ARTICLE 9.
459
PROBLEMS
Dra\\' the bcnding-IlloI11cnt: diagran1 for the unj forlnly loaded girder bac as shown
in Fig. 9.330. The ends band ( arc built in, and there is a rigid connection bet\vecn
beall1 and colulnn at 11. .A.SS11111C, tor the stiffness factors, k 2 = k3 = O.8k 1 and,
for the load, q = 1 kip/ fr.
/111";. See Fig. 9.33b.
2 (:alculate the cnd InOlncnts produced by the forces P and Q acting on the frame
sho\vn in Fig. 9.34. Numerical data are given as follo\\'s: Ii = I = 16 fr, f = 6 ft,
and k 1 = k 2 .
Ans. At ub = {-(-g-Qt' - Pd), i\1 oc = --f,i-g-Qe + f(ll-g-Qt' - Pd).
3 The t\vo-span building fral11c sho,,'n in Fig. 9.35 is subjected to the action of a
horizontal uniforn11y distributed load of intensity q. Construct a bending-moment
diagran1 if" = / = 12ft and q = I kip/ft. "[he stiffness factor k is the same
for all members.
AlIs. Alba = - NI be = - 2.25 kip-it, i\J cb = 3.94 kip-ft, i\1" = 6.94 kip-ft,
,\-led = -10.88 kip-ft, i\J ee '= -Aft/ = 8.25 kip-ft, l\J ab =
-26.44 kip-fr, i\ldc = -12.75 kip-ft, 1\1/ e = -11.44 kip-ft.

3
(a)
FIG. 9.33 (b)
-r- -
8.87 '
-L..
40.5
-,1
29.5
1
k 1 b c el
m
Q T
h
l- J t d ,1
k'l. l . l
c
FI G. 9.34 FI G. 9.35

.'0 CONTINUOUS BEAMS AND FRAMES
d k.=3 e
4 kip. I
k 3 -3
12'
c k& -- 4 
8 kips
k 2 =5
12'
b k6=7 i
8 kips
k l -6 18'
a hl
FIG. 9.36 I_ 24' .t
4 Construct a bending-moment diagraln for the horizontally loaded symmetrical
three-story building frame sho\vn in Fig. 9.36. l'he stiffness factors for the
various mcnlbers as sho\vn in the figure are given in kip-foot units. otc that the
lo\vcr ends of the colutnns arc hinged at 11 and h.
Ans. A1btl = - 160.0 kip-ft, A1 bc = -25.3 kip-ft, Af bg = 185.3 kip-ft,.
A1cb = -46.7 kip-ft, J\t'd = -7.86 kip-ft, A1f}/ = 54.6 kip-ft,
AJ de = -1\11 de = - 16.1 kip-ft.
9.8 MOMENT-DISTRIBUTION METHOD
We have seen in preceding articles that the analysis of frames \vith rigid ...joints
rcquircs the solution of a system of linear algebraic equations. Sometimes
these cquations have the same form as thc threc-moment equations used in the
analysis of continuous beams and can be readily solved. However, in manY'
cases, \VC obtain more compJicated systems of equations, and their solution
may be cumbcrson1e. In such cases, a nlethod of successive approxinlations 1
may be used to advantage. The particular mcthod to be described here is
lOne such rnethod was developed in connection \vith calculation of secondary stresses in
truses and is described in the book by O. l\ttohr, uAbhandlungen aus clem Gebiete cler
tcchnischen M cchanik," p. 429, 1906. fn this country the rnethod was first used by
S. Hardesty and is fully explained in thc book by J. A. L. Waddell, "Bridge Engineering,"
1916. Pfhe extension. of the Incthod to the analysis of highly statically indeterminate fran1c
structures is due to K. A. CaHsev, who used it in analyzing building fran1cs with and \vithout
lateral constraints. Scc 1tch. l..isf., Zagreb, no. 17-21, 1923. A Gcnnan translation of
this paper \vill be found in Publ. lntrrll. 4ssoc. Brill!!.r Structural E1/g., vol. 4, pp. 199-215, t 936.
FI'he final form of the n1cthod of successive approxiInations for fcalne structures was obtained
by H. Cross; see "frans. ASCE, vol. 96, 19J2. Solutions of many engineering problclns by
using methods of successive approxinlations arc discussed in the book by R. V. Sourh\\,cll,
&CReiaxation Methods in Engineering Science," Oxford University Press, Fair La\vn, N.J.,
1940.

ARTICLE 9.8
461
called 11l0111£l1f distri but;oll. It has a simple physical interpretation and has
found a \vide application in the analysis of statically indetero1inate fraole
structures.
oro describe the procedure, \VC consider the structure sho\vn in Fig. 9.37.
.\s prcviollsly pointed our (see page 452), the deformation of this frame,
consisting of four Inembers \vith built-in ends, is compJetely defined by the
angle of rotation (Ja of the joint a, and for the deternlination of this angle re
need only onc equation. 'fa \vritc this equation, \\Fe observe that the sum-
mation of moments acting on the joint a must vanish. Hence,
'\.[ab + i\tf ac + AJ ad + i\;[ae = 0
( tl)
Using, no\v, the slope-deflection equations, \VC find for the end moments at a
i'\-[ab :;:: 4k 1 8 a + =1Tlab
ill ad = 4k38a + D1t ad
Alae = 4k 2 0 a + grrac
1\1 ae = 4k 4 0 Q + mta
(h)
\\tith these expressions, Eq. (a) beconlCS
40 a (k 1 + k 2 + k3 + k 4 ) = - (1tab + nac + mlad + na6) = Ma
\\'here \ve introduce the symbol Ma for the sum of fixed-end monlents at a
\vith reversed sign. Fronl this equation, \l'e find
M a
9a = 4(k 1 + k 2 -tk + k 5
(9.24)
"lhen this is substituted back into Eqs. (b), they become
MOb = rrab + k 1 + k 2  ka + k4 1\1 0
k 2
Alae = mIac + k 1 + k z + ka + k. Ma
k3
Ilfad = ;lRad + k 1 + k 2 + k3 + k  Ma
Afa. = ;mo. + k 1 + h  ka + k4 M n
It is seen that each end 1110nlcnt at joint a consists of two parts: (1) the .fi'red-end
lIIOll/en!, \vhich in cach particular case of loading can be readily calculated or
taken fronl the tahles, and (2) a portion of the moment Ma, obtained as the
algebraic sunl of aJI fixed-end moments at joint a, but \vith opposite sign.
This moment, representing the total turning effort exerted on joint (1 \\rhile it
is locked, is called the unbalanced '1IlO111ent at the joint. V\'e see from Eqs. (c)
that o\ving to rotation of joint a, the unbalanced momcnt distributes itself among
the nlenlbers nlceting there in proportion to their stiffness factors kl, k z , . . . .
(t)

462 CONTINUOUS BEAMS AND FRAMES
a
FI G. 9.37
The ratios
. _ k 1
rb ------
a - k 1 + k 2 + k3 + k..
k3
rad = .- __m_ ------
k 1 + k 2 + k3 + k4
k 2
far.: = - - ------
, k 1 + k'l + k3 + k.
k4
rae = .-- -- -
k 1 + k 2 + k3 + k4
(d)
by \vhich the unbalanced momcnt Ma in Eqs. (c) is multiplied are callcd
distribution factors; they \\.rill be denoted by Tab, rae, . . .. We see that their
magnitudes are independcnt of the loading and can be readi Iy calculated \vhen
the dimensions of the mcrnbers arc kno\vn. It is seen also that only the ratios
of the stiffncss factors k cnter into this calculation. IIcncc, \\'C can usc,
instcad of the actual vah1cs of these factors, any set of nUll1bers proportional
to rhern.
With the introduced notions of unbalallccd 1110111Cllt and diJtribLltioll factors,
\VC can recapitulate the solution of the problen1 in Fig. 9.37 in a more con-
dcnsed form and give it a physical interpretation that \vill be helpful in further
discussion. V\C start the problem \\'ith the assunlption that joint (1 is locked
against rotation by some constraint and calculate, or take from the ready tables,
all fixed-end Inoments. l'he sum of these moments at joint a represents the
mon1ent that the Jock excrts on the joint to kecp it frorn turning. []ulockillg
the joint is equivalent to applying the unbalanced moment,\vhich is equal and
opposite to the moment of the constraint. '[his unbalanced moment \vill be
distributed 31nong the Il1embers n1eeting at joint a in proportion to thei r dis-
tribution factors (d). 1"he final end monlcnts at joint a arc then obtained by
superimposing the distributed momcnts on the prcviously calculated fixed-end
moments as sho\vn by Eqs. (c).
Equations (c) give us the end mornents at the joint tl. Regarding the end
mornents at the far ends of the members, \VC observe that \vhen a distributed
momcnt r abMa is applied to the end a of the member 1, a moment --r abMa is
carried over to the far cnd b of this member (see page 408), and thc final value

ARTI CLE 9.8
463
of thc end momcnt at b becomes
i\tf ba = mtba + j-rabMa
The far-end moments for the other members can be calculated similarly.
It \"as assumcd in the forcgoing discussion that the far ends of all membcrs
meeting at tl arc built in. If one of them, say member 1, is hinged at b, ,,'e
take this into account \vhile distributing thc unbalanced moment Ma and intro-
duce the 1110dified stijJllcSS factor 3kl/4, instead of k 1 , in calculating the dis-
tribution factors (d) (see page 409). It is also to bc noted that a 1110dified
Ilj.cti-eJl£f 1JIGrll1Cllt
'.YYT I _ m"Y 1 ,-,"""
lJll,ab - .Jll,ab - "!fv1lba
111USt be used if the far end b is hinged (see page 453).
rhe foregoing J110111cllt-diJl'ributioll procedurt used in the analysis of the frame
in Fig. 9.37 \vas especially simple since there \vas only onc joint a free to rotate.
Ho\\'ever, the method can be rcadily extended to the morc gencral case in
\\'hieh the rotations of scveral joints have to be considered. To illustrate, let
us reconsider the cxan1ple sho\vn in Fig. 9.3 2a and assumc first that the joints
b, c, d arc all three locked, so that \\re can calculate the t1xed-cnd moments and
the unbalanced InOInent M b , Me, and Md. No\v \ve unlock the joint c and
distributc the unbalanced momcnt Me among the members 2, 3, 6, in accordance
\vith Eqs. (c). 'Thus, to thc initial fixed-end moments at joint c, \ve no\\' add
the (orrections
JI;b = 1',-bM ('
AJd = rcdMo
1'\.1: 0 = rcgMc
(c)
J)uring the unlocking of joint l:, the adjaccnt joints b, d, and f{ arc considered
as fixed. Hence, mon1cnts equal to A1b/2, J;f;d/2, 1\1o/2 \vill bc carried over
to thc far cnds of the members 2, 3, and 6. rcspectively, \vherc they must be
added algebraically to the fixed-end moments at these ends. Considering next
the joint b, \ve have to distribute the moment Mb - }\1[;b/2, \vhcre the second
term represcnts thc monlent that \vas previously carried ovcr from joint c.
Thus, the distributed momcnts at bare
\la = fba(M b - i.j\.Jb)
Alc = rhc(M b - ill) (f)
.lJ1 = rb/(M b - 'I\II;b)
Similar equations for the distributed moments at the joint d can also be \vritten.
All these nlomcnts represent the first corrections to the initially calculated
fixed-end Illorncnts. To get a second set of corrections, \ve start again \vith
,oInt c.
I}uring unlocking of the joints band d, the moments AJc/2 and Al:/ c /2 were
carricd over to the joint c, and an unbalanced moment of the magnitude

464 CONTINUOUS BEAMS AND FRAMES
---(i\1c + j\tlc) \vas crcated at this joint. I)istributing this moment, \ve
obtain the second corrections to the end moments at joint c. Thcn \ve can
calculate corrections to the cnd moments at the joints band d, distributing the
unbalanced moments created at these JOInts during the unlocking of joint c.'
To calculate the third corrections, \VC again start \vith joint c and proceed as
before. The calculation of consecutive corrections must be carried far
cnough to make thcm ncgligibly small. 'rhe final cnd moments \\'ill thcn be
obtained \vith sufficient accuracy by adding to the first approximations all .of
the consecutive corrections.
Rccapitulating the described mcthod of successive approximations, \\'C see
that the process of calculation consists in repetition of very simple opcrations
of distribution of unbalanced momcnts and carrying over momcnts to the
adjacent joints. In each particular case, \ve assurne first that all joints, except
hinged ends, are locked and calculatc on this assumption the fixed-end momcnts
for all members of the structure. 'rhe sums of these lllomcnts for each joint,
taken \vith revcrsed signs, give us the unbalnced moments. 'Vc calculatc
also for each joint the distribution factors for the membcrs meeting at that
ioint. Aftcr this prcliminary \vork, \VC begin preferably with the joint at
\vhich the largcst unbalanced momcnt is found. V\'e unlock this joint and
distribute the moment bet\vecn the members meeting at the joint. In our
previous cxample it .ras joint c, and the distribution of the n10mcnt Me \\'as
accomplished by using Eqs. (e). No\v \ve carryover moments to the adjacent
joints and add to the unbalanced moments at these joints the momcnts cqual to
the carry-over mOInents, but \vith oppositc sign. l'he obtained sunlS \VC again
distribute, '.Ising equations sinlilar to Eqs. (e). vVhen the distribution of
moments is accon1plished at all joints, \\'c obtain the second approximations for
all end momcnts by adding the distributed moments to the prcviously found
fixed-end momcnts. To get the second corrections, \VC repeat the rTIoment-
distribution process, but this time \\'c distribute only the carry-over moments
remaining after the first cycle of calculation, etc.
Since \\'e ahvays rcpeat the same sinlple operations, it is not necessary to
\vrite c<luations \vhich \\'cre used in our previous discussion. Instead all
calculations can be put in propcr order on a sketch of the structure. 'fake,
for example, the continuous beam sho\vn in Fig. 9.38. i\ssuming that, at the
supports t1, b, and c, the cross sections of the bcam cannot rotatc and that at the
support d thc rotation is frec, \VC obtain the fixed-end moments in kip-feet,
as sho\vn in the first line of the table attached to the figure. In calculating the
distribution factors, v...e assume that kab = k bc = !kcd. lhen,
kab
rba = Tbe = --- = 0.5
kOb + k hc
kcb
Tcb = k + ak = 0.6
cb "{ cd
'Jk
red = k +Ck = 0.4
cb 'i cd

ARYl CLE 9.8
465
In this calculation \VC observe that the end d of the member cd is free to rotate
and introduce !k cd , instead of ked, in our fornlulas. rhe calculated values of
the distribution factors arc sho\\'n in the figure. We no\\' begin \vith joint c.
The unbalanced momcnt at this joint is
- (24.0 - 48.0) = +24 kip-ft
Distributing this momcnt between thc members cd and cb in proportion to their
distribution factors rc:d and rcb, \\'e obtain the distributed moments at c, equal
to +9.6 kip-ft and + 14.4 kip-ft. laking no\v joint b, \ve find that the fixed-
end moments at this joint balance each other, and we have to consider only the
moment - 7.2 kip-ft, equal and opposite to the carried-over moment 7.2 kip-ft
produced during unlocking of the joint c. 'fhis moment \viII be equally dis-
tributcd bet\\'cen the mcmbers be and ba, and we obtain the distributcd moments
at b each equal to - 3.6 kip-ft. Considcring no\v joint a and observing that
the end a is built in, \ve conclude that the correction momcnt - 3.6 kip-ft
acting at b produces the carry-ovcr moment -1.8 kip-ft at the fixed end. All
distributed moments are sho\\'n in the second and third Jincs of the table in
4'ig. 9.38, and the heavy horizontal lines belo\v these corrections indicate that
the first cycle of calculation is concluded. Thc second cycle \\'e again begin
\vith joint c. Considering the carry-over moment - 1.8 kip-ft created during
thc unlocking of joint b, \\'e have an unbalanced moment + 1.8 kip-ft at joint
c. Distributing this moment bct\veen the members cd and cb, \ve obtain the
corrections +0.72 kip-ft and + 1.08 kip-ft. Taking now joint b, \ve have to
distribute the momcnt -0.54, equal and oppositc to the carry-over moment
+0.54, \vhich givcs the corrections -0.27 and -0.27; and finally \ve obtain
the correction -0.14 at the built-in end. We sec that the corrections are
already small, and \VC can limit ourselves to t\VO cycles of calculations. The
third cycle of calculation is also indicated in the figure, and under the double
horizontal lines are given the final values of the end moments equal to the
algebraic sums of the initially calculated fixed-end moments and all the
corrections.
I-
12'
8'
16,OOOlb I d

-4-
+24.00 -48.00
+14.40 "'9.60
-1.80
+ 1 . 08 +.72
-.14
+.08 +.06
+37.62 -37.62
-24.00
-1.80 
-.14 
. -.01 .--
FIG. 9.38 -25.95
+24.00 - 24 . 00
-+7.20
-3.60 -3.60
+.54
-.27 -.27
+.04
-.02 -.02
+20.11 -20.11




466 CONTINUOUS BEAMS AND FRAMES
As a second example, let us considcr the continuous frame sho\vn in Fig.
9.32b and assunlC first that latcral movement is restrained. \c also aSsunle
that the proportions of the structurc arc those defined by Eqs. (d), page 454,
\vith J = 12 ft and q = 1 kip/ft. \I\l'ith these numerical data, \VC calculate
thc distribution factors as sho\vn in Fig. 9.39. "i staTt \\'ith a calculation of
the fixed-cnd Inomcnts, assunling that joints b, c, and d are locked and that
thcrc arc hinges at the ends (l and t. l'hc calculated values of these n10mcnts
in kip-feet arc given in the first horizontal line of the table above Fig. 9.39.
'fhe distribution of the unbalanccd moments \VC begin from joint c, at \vhich
the unbalanced nl0ment has the largest ntuTIcrical valuc. 1-1hc results of this
distribution arc given in the second horizontal }ine above joint c and in the
second vertical line to the right of colun1n cg. C:onsidering, no\\', joints b
and d, \ve distribute the Inoments there as sho\\'n jn the third horjzontal line
above joints band d and in the vertical lines to the right of columns lif and dlz.
I "his finishes the first cycle of calculation, the results of \vhich arc underlined
in Fig. 9.39. The second cycle \ve again begin \,'jth joint c, at \\'hich \\'c have
the unbalanced Inomcnt - (5.43 + 1 .15) . rhe results of the second distribu-
tion at this joint and also at the joints band d are again underlined in the figure.
'rhere is given in the figure also the third cycle of calculation. \\'hich gives only
very small corrections of end mOInents; thus, \VC conclude the calculations
\vith this cycle and put under the double lines the sums of the initially cal-
culated fixed-cnd n10nlcnts and of all corrections that give us the required end
n1()Jnents. The same nurncrical values \\Tere obtained previously by the dircct
solution of the slope-deflection equations (see page 458).
The mcthod of successive approxitnations can be used also in case the
continuous fratnc is free to nlove latcrally. I..et us takc again the example
sho\vn in Fig. 9.32b. The calculated end 010mcnrs in the columns produce on
the girder ae the horizontal shear action in the direction from right to left
+18.00 -48.00 +48.00 0 0 0
-8.00  -16.00 -16.00  -8.00_
+ 16.28 +10.B6  +5.43 +1.15 --4 +2.29 +3.42
-1.10 0+0- -2.19 -2.19 ---- -1. 10
+0.48 +0.31  +0. 16 +0.16 -+-- +0.31 +0.48
-0.06  -0.11 -0.11 ..- -0.06
-'16:99 -
0.03 0.02 +35.29 0.02 o . o_
.34.79 -45.97 -6,54 +3.93
Nit-
II ;;
.... ....
o:::
+ ...
...ICIIt !
II 0 0')  I o
0.....- 1 (")
.cc.i° l oo
- ....II
I I
11
d e
r de   r de -= 4
N 1
II  .....\N N
{';I M,O CD
.=3 c c.i oo 
- + +1+ +
h I
FIG. 9.39

ARTICLE 9.1
467
equal to
_ . Jib' + ,vl cg t__ i\J dh _ -11.19 + 18.30 - 2.62
2 h - 8
- 561 Ib
(g)
To prevent any lateral movement, a horizontal external force of the same
magnitude but acting from lcft to right must be applied. If there is no such
force and the frame is free to move laterally, it \vjIl move to the left by such
an amount as to produce shearing forces at the tops of the columns sufficient to
balance the above-calculated force (g). In discussing the end moments and
shearing forces produced by lateral movement alone, \ve assume first that the
joints b, c, and d do not rotate during this Inovcment. Then, the end moments
equal to - 614,8bh - 6k cg 8 cg , -;- 6edh will be produced at the ends of the
columns hi, t:J{, and dh [see Eqs. (i), page 457], In our case all these end
moments are equal since the sti ffness factors and the angles of rotation are
the same for all columns. Since \ve do not kno\v in advance the magnitude
of the displacement, \ve starr the calculation by assuming that it is of such
magnitude that
6k bJ 8bl = 6k co 8cu ::c 6hedh = + 1.000 kip-ft
"[he corresponding end moments are indicated in the first vertical lines to the
right of the columns in Fig. 9.40. It is seen that there are unbalanced moments
equal to 1.000 kip-ft at joint's b, c, and d. Distributing them in the usual \\fay,
\\'C finally obtain the values of te end moments \vritten under the double Jines.
Having these moments" \ve calculate the shearing forces in the columns and
find that the total horizontal force acting on the girder (lC in the direction from
0.167
0.357 0.238
-0.040
+0.017 +0.011
-0.002
+0.001 +0.001
+0.375 +0.375




-+-
0.333 0.333
0.119 0.119
-0.079 -0.079
+0.006 -to.OOG
-0.004 -0.004
+0.375 +0.375

..-.-



0.167
0.238 o. 357
-0.040
+0.011 +0.017
-0.002
+0.001 +0.001
+0.375 +0.375
a
b
r ba = 4 r be Q ,
r eb = i
c
r -1
cd- 3
d
r de -= t
e
3
r de =a,
.

f /
U')U')
Nt-
palCCI
00
+ I
OMQ)O
OMt-O
OMOOt-
 OOOQ
" I + / I I
C.
..CJ g 
o
.-40
g I +
'r.I
....
Q)
OCOP-tP-tO
OMP-tOln
ONoor-
Na. n QOOO
I + + + I
oC
... 0  it)
ON r...
o...c 00
h '7  9
000.........0
0C?0lt)
ONOOt-
 . It . . .
n ooo9
....
....0 0
o
o
.
<:>
I
FIG. 9.40

468 CONTINUOUS BEAMS AND FRAMES
right to left is
3 (0.875 + 0.750) kip-ft = 406 Jb
12
It is seen that to balance the horizontal force given in Eq. (g) we have to take
6k bl 8bl = 6k cll 8 cQ = 6k dh 8 dh = - 1 .000 X M-t kip-ft
= -1.38 kip-ft
instead of the previously aSSUtl1ed value of 1 kip-ft. Hence, the cnd moments
due to lateral sway will be obtained by multiplying the values given in 'ig. 9.40
by the factor - 1.38. In this manner, \\'e find
l\.1 ba = lUbe = ,Web = Mcd = M do = M dfJ = -0.375(1.38)
= -0.518 kip-£t
These same results \verc obtained before by direct solution of the equations
of equilibrium of the joints (see page 458).
PROBLEMS
Using the n1cthod of nlonlent distribution, construct a bcnding-nl0mcnt diagram
for the continuous beam sho\vn in Fig. 9.4Ia. Assuo1e P = 40 kips and stiffness
factors as sho\vn in the figure.
Al1S. ,lIb = -20.75 kip-ft, Ale = - 37.6 kip-ft, AId == + 18.3 kip-ft.
2 Using the 111cthod of moment distribution, construct a bending-rnomcnt diagram for
the continuous beam shown in Fig. 9.41 b. .ssume P = 16 kips, q == 2 kips/ ft,
and nloments of inertia It = 1 2 = itla.
,,111S. A1a = - 25.95 kip-fr, AIb = - 20.11 kip-ft, i\1 c = - 37.62 kip-ft.
3 Using the moment-distribution n1cthod, calculate all cnd morncnts for the frame
structure sho\vn in Fig. 9.42ll if 311 olco1bers have the same flexural rigidity EI.
Ans. Atf ab = -110.5 kip-ft, j\ftl(J = + 78.8 kip-ft, l\.1 bc = -42.4 kip-ft,
AJt.b = -21.2 kip-ft, Alr..d = -37.0 kip-ft, ,\l db = -Afdt =
-9.0 kip-ft.
4 -rhe fran1e structure shown in Fig. 9.42b is so loaded that the fixed-end moments
in the girder are mIbc = - 20.0 kip-tt, mIcb = +24.0 J<ip-ft, and the modified fixed-
cnd moment at b is fJTla = lIba - -tffil ab = + 50.0 kip-ft. The stiffness factors
 p
 k .1- 7 1/4-A- 7k'/
1---10'-+-10'+10--1
FIG. 9.41 (a)
p

ARTICLE 9.9
469
5'-+-5'
15'
d e_ r
5'
b-+
5'
L
c
10' 1
10'
k 2k 1
d e
(b)
80 kips
a
(ll)
FI G. 9.42
FI G. 9.43
.1
br=--__ ;1 I
r - , /fP
I J I
, , -
I 2
, d I
1 , -.L
,
La 1
10 kips
40"
(a)
b
r 123001n.4
25(¥'
1 a /1 =500(:.4
for the various mernhers are as sho\vn in the figure, \\'here k = 1,000 kip-ft.
Calculate the nlagnitudc of the side s\vay of the frame.
t111S.  = 0.0] 79 fr to the left.
5 The simple fran}c sho\\'l1 in Fig. 9.43a has uniform flc;xural rigidity Jl = 10(10 6 )
Ib-in. 2 for all three nlembers and I = 100 in. l'he colUlnns arc huilt in at t1 and d.
Calculate the rnagnitude of the side s\vay that \vill be produced by a horizontal
force P == 60.0 lb acting at joint ( as ShO\VrL
AnJ. 6. = 0.101 in.
6 The footing at a of the fralne sho\vn in Fig. 9.43b is elastically restrained so that
it rotates D.OOOI rad per 30 kip-in. of nlonlent Jl ab . Using the n\cthod of t110rncnt
distribution, find the nlagnitudes of the end rnotnents produced by a vertical force
P = 10 kips acting as sho\vn.
AJ1s. Alab = - 16.8 in.-kips, j\.{bQ = - 53.6 jo.-kips, AJ bc = + 53.6 in.-
kips, Al cb = - I 41.2 in. - kips.
9.9 ANAL VSIS OF BUILDING FRAMES
In the design of building frames we have to analyze complicated structures of
the kind sho\vn in Fig. 9.44. '[he solution of the problem is usually divided
into t\VO parts, (1) a calculation of end rnOlI1cnts produced by vertical loads,
assurning that there are no lateral movenlcnts, (2) a calculation of end nlomcnts

470 CONTINUOUS BEAMS AND FRAMES
m n 0 p
a b c d
e f g h
i j k I
FIG. 9.44
produced by lateral movemcnts, the latter being produced, not only by lateral,
but also by. vertical loads.
If there arc no latcral nl0vcmenrs, \ve apply the method of nlonlent dis-
tribution described in the preceding article. LTsing the kno\vn dinlensions of
the structural clen1cnts, \ve calculate the distribution factors for the mClnbers
meeting at each joint. '\Ie calculate also the fixed-end moments produced
by actual loads. "'.Thcn this prclinlinary \\'ork is finished, \ve start \vith
distribution of unbalanced monlents and proceed in exacdy the sarne manner
as already described in the preceding article.
If the load is appJied in onc span only, the end moments produced by it
dirninish rapidly as \ve consider joints farther and farther a\vay from the
loaded span. Using this fact, \ve can silnpJify the caJculations by considering,
instead of the entire structure, only a portion of the structure adjacent to thc
loaded member. .Lssurne, for example, that for the case sho\vn in Fig. 9.44,
the load covers only one span ab. Then, an approxilllatc solution is obtained
if \ve consider only the portion of the structure sho\vn in Fig. 9.45a and assun1e
that the upper joints Jl1, 11, 0, the lo\vcr joints e, f, .'{, and joint d arc fixed.
-\ssurning that the fixed-end moments and the distribution factors have the
numerical values indicated in the figure, the results of the first cycle of calcula-
tion \vill be as underlined by the heavy lines in the second and third lines in the
figure. l"'here arc sho\vn in the figure, also, the calculations of the second and
third:ycles, and under the double lines the final values of the end nl0ments arc
gIven,
If r:1orc accurate values of thc end mor11ents at the joints a and b arc required,
the etlect of rotations of the joints 'IJl, 77, e, and f on these Inornents can be
readily taken into account. 'rhis calculation is sho\vn in Fig. 9.45 b. \,,
start by \vriting in the end mornents i\J ab = - 13.14, .""fae = + 7 .89, and
i\J a ,,, = + 5 .26, obtained from the previous calculation. 1'he carry-over
moments at joints 111 and e arc then +2.63 and + 3.95, respcctively, as sho\vn
in the first vertical line in the figure. Hence, at joints ?Jl and e therc arc the
unbalanced moments - 2.63 and - 3.95. Unlocking these joints and assullling

ARTICLE 9.9
471
that 1t"a = i and rea = !a-, \VC distribute the momcnts and find the carry-over
moments -0.44 and -0.46 at joint a as sho\\'n in the second vertical line.
Distributing, no\v, the corrcsponding unbalanced moment + (0.46 + 0.44)
at a, \\'C finally obtain under the doublc lines the required corrected values of
the end momcnts at joint a. In a sin1iJar manner the corrected values of the
end moments at joint b can be calculated. When all calculatioris, made on the
assumption that there are no lateral movements of the structure, are finished
and the numerical values of all end moments for columns are obtained, \\re
calculatc the horizontal forces acting on cach floor of the building.
The calculation of the end moments produced by horizontal forces \vill now
be explained by a simple example of a symlnetrical three-story frame loaded
as sho\\'n in Fig. 9.46. \\Ie begin with a discussion of the three auxiliary
problems sho\vn in Fig. 9.47, in each of \vhich we aSSUlne a rclative horizontal
displacement in one story only. In the case sho\\'n in Fig. 9.47 a, \ve assume
first that the upper part of the framc is displaced horizontally \vithout any
bending. To produce such a displacement, \ve apply a propcr horizontal force
at b, and also couples arc applied at joints band f of such magnitude as to
m n 0
-24.50 k-it 24.50 k-ft 0 0 0
-3.20  -6.39 -lO.6f>  -;).33
+ 15.11  +7.56 +1.07  ..2. 13 + 1 . 70
-1.13  -2.25 -3.75  -1.88
+0.62  +0.31 .0.38  +0.75 .f 0 . 60
"'1::- -0.09  -0.18 -0.30  -0.15
1\ -'0.05 23.5f> -13.25 "0.06 ..0.05
f: -13.14 -4.42 2.35
"a m MIC;l
II II
&: ...
".c
6 roo = 2\ TbclK Tcl> =  r cd = 2\
a Tab Q n
b c
M I
<0 00 I
. .
I:'-lO
t .
jo)
It
...2
-13.14
:: +0.49
 :   " -12.65
t.'; 0   :
+ I · + ... Tab -- &
d tD"'t- a
!'t).,.(\)
 0 0 r--: ::
t I "1 
II
'"
...a
0/)1:: an  M (1)
II In M 0 00
oo
" + <4
..,, (D <:> N I r:r;
\I (\) an ..... CX;1
"'-  ...; 0 I I
.....c t I I I
.., I  
 u") ,::I C\I t-
" c:e:
!:IIoOQ
...4.J i + 
11':1 -
(1) 
MO
i I
Mf
II
...
e
e
f
g
(a)
(b)
FI G. 9.45

472 CONTINUOUS BEAMS AND FRAMES
4,0001b k dh = 3 h
d
k cd = 3 12'
8,000 Ib keg = 4
c g
k be = 5 12'
8,0001b k bf = 7
b (
kab = 6 16'
a 1 e
FIG. 9.46
0.051 t1 h 0.392 t;2 h 1.184 1;3
.. II "
0.051 0.392 1.184
h=12'
i 0.521 c 0.051 2.386 c 0.392 1.184
.. . 
.. g . g ..
h12' 0.470 1.994 0.392
1.336 b .. 0.470 2.229 b 1.994 0.417 b 0.392
( . (  f
k I(
 0.235
0.866 0.025
a e a e
(a) (b) (c)
FI G. 9.47
prevent any rotation of thesc joints. If l is the magnitude of the displace-
ment, the end Illoments at the tops of thc columns of thc first story \viII be
1
l\1 ba = J\J Ie = - 3k ab _ I.
"It
For our further calculation let us assume son1C nurncrical value for these
ITIOmcnts, say i\I',a = ilfe = - 9 kip-ft. No\\' \ve assume that any furthcr
latcral movenlcnts of t:1C joints of the frame arc preventcd by special con-
straints, and \\'c unlock the joints b ilnd .f, In calculating the end Illon1cnts
rcsulting froIll rotations of the joints, "'C oLserve froIn symrl1ctry that there
\viJ1 be inflection points at the Iniddles of the girders. \\.' can assume hinges
in these points and considcr only one-half of the fralnc as sh()\vn in Fig. 9.48a.
lhc nun1bers proportional to the stiffness factors of the girders, indicated in
Fig. 9.47, nUlst no\v be rnultiplicd by 2, and three-quarters of thesc values
111ust be used in calculating the distribution factors. The values of these

ARTICLE 9.9
.73
factors are indicated in fig. 9.48a. Starting calculations with joint b, at \vhich
\ve assumed an unbalanced moment of 9 kip-ft, \ve obtain as usual the values of
all end moments. They are \vritten under double lines in Fig. 9.48a. Having
these moments, we rcadily obtain, from equations of statics, the shcaring forces
in all columns and the horizontal forces acting on the girders. These latter
forces in kips are as indicated in Fig. 9.4-7a. Combining them, we obtih the
horizontal forces that must be applied at joints b, e, and d to keep the trame
in the assumed displaced position \vith all joints unlocked. These forces arc
also indicated in Fig. 9.47 a.
Considering the problem in Fig. 9.47 b, we assume that, by a proper applica-
tion of horizontal forces and couples at the joints b, c, g, and f, the upper story
of the frame is displaccd \vith rcspect to the lo\\'cr story so that only the
.columns be and fg are bent. - If d2 denotes the amount of horizontal displace-
ment, the end moments for these columns are
d2
M bc = MCb = M ,g = M g1 = -6k bc Ii
Taking for these moments an arbitrary. value, say - 10 kip-ft, and assuming
that there is no lateral movemcnt of the frame from its displaced position, we
unlock the joints and calculate as usual the resulting end moments and horizontal
shear forces. These moments are given in Fig. 9.48b, and the shear forces
and horizontal forces required to produce the assumed lateral displacement
2 are sho\\'n in Fig. 9.47 b. The case sho\\'n in Fig. 9.47e can be treated in a
d 'd" -I
...11) 8t
 000 tOO +0 . 072 h
";0 <fr:fci I · 2 Q.I
. ... +0.076
t l t !

    I   I 
91'
tlUt
-.  .
"'N
NO OON
C: 0 000
... I + I
  .... ...
0(0)
... 00 Or-
....0 00 . .
00
.. . ... I  +
It) II) N II) :t ..')
g
....OOOO
.. , . . . .
t  t!
...1
'-A
c
-0.482 (J
Gj -0.021
I -0.503

tJtt
g!88
0-(0)0000...)
.... . . I ... t .. I
I

0... O NN I
\.') 0 o o eft
 00 00
ci 00 Q
. I . .
oon ,., S
OCi !
00 o eft
caN 0 'f
. . 
00  eftr- (O)...."P I
00 lID Q) 8C: gg
01/') .. ') (0)
ON ... 0 00 O
... .. ... . ... I I .
t
0 N  N
  ... gs

  0 0'"
... I I "+
...
'6' -H
6
+4. 72 ,
=- :;2_. iO 
+o.oos
 4.836
I

(0)
(b)
Q
FI G. 9.48
d 'ellt - I
-0.553 h
NO -:O-25-'
I
.. -0.001
-0.690
nf:
.
...1 'qr - A
c
+3.750 (J
 +0.104
I +0.007
 +3.921


... '6f - H
6
15.250 ,
-0.821
-0.036
 -0.002
I --1.391
...1,
a
o M I M .... 0 on N  CIO d
0....0 oor-
--: ....
\OONOOOOI") .
, r'i tit i ' :
'eI" - I
+3.214 h
'0.061
. 0. 003
+3.278
0\0 N.... 0 en....... I
g r- 0 N 001")
ON 0 00=0
10'" "'0 ? 001")
. ... t I I . I
M on ... V)....
"f'M g: c:g !OJ
... ...
NO 9 00.... .
.. . I .. 
!tt!t
..1
'ell =: A
C
'2.572 lJ
: --=-0.4 02-
-0.018
.."2 +2.162
-c..
N:IO II)N gf! I
S  o r,Q
-;Or;' 'i 000
. . .
N co N'
..  g2
N
00 0 °1°
I .. ... .

'6f - fA
b -0.563 ,
+0.088
10.00.4
 -0 ."71
.
...J
(c)
a

474 CONTINUOUS BEAMS AND FRAMES
simi lar manncr. Assuming for the end moments in columns cd and gh the
value - 6 kip-ft, \\'c obtain thc values of the end mon1ents and horizontal shear
forces as sho\vn in Figs. 9.48c and 9.47c, respectively.
After solving the three auxiliary problems in Fig. 9.47, \\'C can, by a proper
combination of these solutions, obtain the requircd solution for the horizontal
forces sho\vn in Fig. 9.46. It is evident that if the displacements 1, 2, and
3, \vhich \vcrc assumed in the auxiliary problems, are changed in a certain
proportion, the corresponding horizontal forces \vill change in the same
proportion. Assumc, no\v, that the displacements are lX, 2Y, 3Z, and sclect
the values of x, J', and z in such a \vay that the superposition of the horizontal
forces corresponding to these displacemcnts \vill give us the actual horizontal
forces. To accomplish this, \VC have only to multiply the horizontal forces
given in Fig. 9.47a, b, and c by x, )', and z, respectively, and after superposing
them put the obtained resultant forces equal to the actual horizontal loads in
Fig. 9.46. In this manner \ve obtain the equations
0.051x- O.392y+ 1.184z = 4
- 0.521.1: + 2. 3 86y - 1.57 6z = 8
1.336x - 2.229y + 0.417z = 8
(a)
\\'hich give
x = 26.61
)'= 13.63
z = 6.75
To calculate the actual cnd moments for the members of the frame in Fig.
9.46, \ve have to multiply the mon1cnts given in Fig. 9.48/1, b, and c, rcspectively,
by x, y, and z and then supcrimpose the obtained results. In this manner \\'e
obtain
jv1ba = - 6.928 X 26.61 + 1.882 X 13.63 - 0.202 X 6.75
= - 160.1 kip-ft (b)
To check this result, \ve observe that in our case the total horizontal load
4 + 8 + 8 = 20 kip-.f is equally divided bct\vecn the t\VO supports a and e.

Hence, thc truc value of i\;{ba is - lOX 16 = - 160 kip-ft. For the remaining
end moments in the columns, \VC obtain, by using equations similar to Eq. (b),
the follo\ving valucs: kf bc = -25.3 kip-ft, i\1 cb = -46.7 kip-ft, A1cd = -7.8
kip-ft, /\;[dc = -16.1 kip-ft.
We have considered herc a frame \vith three stories (Fig. 9.46), but the
same method can be used for any number of storics. As the number of stories
incrcases, the number of cquations of cquilibriunl sinlilar to Eqs. (a) naturaJly
increases, and the amount of arithmetical \vork requircd Inay become vcry
grcat. However, to obtain a satisfactory approxin1ate solution, \\'C do not
nccd to consider an stories of the framc simultaneously. The effcct on the
cnd moments of the lateral distortion assumed in one story dies out rapidly

ARTICLE 9.10
475
as \ve go farther and farther from the distorted story. Hence, \\'e can get a
satisfactory approximation by considering cach riIl1t: only three cor.secutive
stories. In analyzing lateral deflections of a building franlc like that sho\vn
in Fig. 9.44, \VC start fronl thc top of the building and consider the three upper
stories. i\ssuming that thcrc are hinges at the mid-points of the colurnns in
the third story frool the top, \ve obtain a problem sirnilar to that in Fig. 9.46,
and \\'C can calculate all end 111001(.'nts after solving a systenl of three equiIibriuIll
equations similar to Eqs. (a). The results obtained in this \vay \vill not be
exact, but they \vill be accurate enough for the top Roor. V\'e can then rcmove
the uppcr story, replace its action on the remaining portion of the frarr:e by the
forces and thc mOOlcnts that have alrcady been calculatcd, and consider again
only thrce consccutive stories by assunling hinges at the nlid-points of the
columns of the fourth story from the top, etc.
9.10 FRAMES WITH NONPRISMATIC MEMBERS
'[he moment-distribution method used in the preceding articles for fracncs
\vith prismatic members can be extended to the analysis of fraes \vith
members of variable cross section. In using the method, \ve have to perform
thrce operations in \vhich thc variation in cross section must be taken into
account. rrhey are (1) calculation of fixed-cnd moments, (2) calculation of
distribution factors, and (3) calculation of carry-over factors.
To illustrate the procedure, let us consider the example sh()\vn in Fig. 9.49.
'"his three-span high\vay bridge \vith span ratios 3 : 4: 3 consists of a continuous
girdcr \vith straight-line haunches rigidly attached to vertical col'Jnlns of
constant cross section. It is required to dra\v the bending-moment diagranl
for a uniforrn load q = 1 kip/ft distributcd along rht, first t\VO spans as sho\vn.
\Ve start, by using the tables, 1 \\rith a deternlination of the clastic characteristics
for each n1Ctnuer. The Incmbcr ab has at the end b a straight haunch of lcngth
0.4/ 1 and a depth of 4.5 ft = I.Sh o . For these proportions, \ve find frorn the
tables
Carry-over
factors
Sri tfness
factors
Fixed-cnd 1110nlents
C n /)
C ha
Sa
S tHJ
llab
)Tl.lla
0.996
0.403
5.180
12.82
-0.0541qJ2
O. : 5 5 4q /2
()bscrving that the membcr a!J is hinged at t1, \ve must usc, instead of the
stiffncss factor 't)[h a modified stiffness factor .\'a given by the equation
Jf1 = sbu(l - ('ab(J'ba) = 12.82(1 - 0.996 X 0.403) = 7.67
I See HI-Iandbook of Franlc Constants."

478 CONTINUOUS BEAMS AND FRAMES
ho = 3'
ho = 3'
ho = 3'
36'
h
52.5'
70.0'
52.5'
618 653
(b)
FI G. 9.49
and, from Eq. (9.18), the modified fixed-end nl0ment
m7.Q = ba - CabmLab = (0.1 554 + 0.0541 X 0.996) q/t 'l
= 576.9 kip-ft
For the symmetrical middle-span member be, \VC find from the tables
I
-0.1074q/2 2 I +0. 1074qJ 2 2
Cbe
e cb
Sb. I S.b
14.27 r 14.27
mIbc
mlcb
0.753
0.753
The member cd has the same characteristics as the member ab, except that
the fixed-end n10ments vanish since there is no lateral load on this span. Thus,
for this member, we have
C cd
Cdo
Sed
Sde
mtcd
I mtdc
I
o
0.403
0.996
12.82
5.180
o
Observing that the cnd d is hinged, \ve use for this member the modified
stiffness factor Sd = sc = 7.67, as already obtained for the member nb.
Regarding the vertical columns, we assume that along the portions be and cg

ARTICLE 9.10 477
their flexural rigidities arc infinite. Then, from the tables, \ve have
Cbl
CIf,
S,,/
SIb
0.480
0.715
6.82
4.57
Since the lo\\'cr ends of the columns are hinged, \ve use, for each of these
members, a modified stiffness factor
S:'h = s/ = sb/(1 - Cb/C/b) = 6.82(1 - 0.480 X 0.715) = 4.48
For the rigidity constants k, we take for the first and third spans
E10
k 1 = k3 = - =
11
and for the middle span,
_ Elo _ 3
k 2 - - - --
1 2 4
For the vertical colunlns, \ve have
k  = k r = E10 = 5_2:5 = 1 46
.. 0 h 36 .
To calculate the distribution factors, we observe that in the case of beams
of variable cross section the relation between the end nl0ment Nl ah and the
corresponding angle of rotation 8a is given by the equation (see page 429)
Elo
At/ ab = 8a 7-; Sab = 8 a k n s ab
Hence, the distribution factors \\,ill be proportional to the quantItIes ks.
U sing now for joint b the numerical values
k 1 = 1 .ba = 7.67 k1s ba = 7.67
k 2 = 0.75 Sbe = 14.27 k 2 s bc = 10.70
k.s = 1.46 t;b/ = 4.48 k 4 s bj = 6.54
\\'e find T,k i s r1ln = 24.91, and the distribution factors become
7.67 3
Tba = 24 .91 = O. 08
6.54 2 3
rbl = 24.91 = O. 6
rbo =
10.70 = 0429
24.91 ·
From symmetry, \ve conclude that the same values hold for joint c.

..78 CONTINUOUS BEAMS AND FRAMES
a
b
c
d
..576.9 -526.3 1-526.3
-170.0 ..c. -225.8 -162.1
+36.8 +51 .2 . +38.6
-12.5  -16.6 -11.9
1'3.84 +5.4 . '.4. 02
-1.30 -
< -1.72 -1.24
I---- -
O. 40 +0.56 . +0.42
-0.14  -0.18 -0.13
1-----
+0.04 0.06
(a) +618.0 -653.0 +325.0 -175.4
FIG. 9.50
(b)
d
a
-149.8
Having the distribution factors, \Vc can no\\' begin the calculation of end
moments by the method of moment distribution. 'rhcsc calculations are put
in tabular form in Fig. 9. 50. V\ start \\lith the calculation of the fixed-end
moments, the values of \vhich are sho\vn in the first horizontal line under the
girdcr «hed. 'The distribution of moments is startcd \\lith joint c, \\There the
unbalanced moment is largest . Using the previously calculated distribution
factors, the results of the distribution of the unbalanced moment Me = - 526.3
\vi II bc
M: b = -526.3 X 0.429 = -225.8 kip-ft
/W;d = - 526.3 X 0.308 = -162.1 kip-ft
M;/ = - 526.3 X 0.263 = -138.4 kip-ft
These distributed n10ments are sho\vn in the second line under joint c in
Fig. 9.50a and at the top of the colulnn ch in Fig. 9.5Gb. The moment
C cb ,\1:b = -0.753 X 225.8 = - 170.0 \vill be carried over to the joint b as
sho\\Tn, and there the unbalanced moment, equal to
- 576.9 + 526.3 + 170.0 = 119.4 kip-ft
\\.,ill exist. The results of distribution of this mornent are shown in the third
line under joint b in Fig. 9.50a and at the top of the column hi in Fig. 9.50b.
rhe heavy lines under the distributed monlents at the joints band c indicate
that the first cycle of mon1ent distribution has been completed.
To calculatc further corrections, \ve again start \vith joint c, to which the
moment of 38.6 kip-ft has becn carried over as indicated. Distributing the
corresponding unbalanced moment of - 38.6 kip-ft at c and the unbalanced

ARTICLE 9.10
479
momcnt of + 12.5 kip-ft created at b, the second cycle of calculation is com-
pletcd as indicated by the second set of hcavy lincs under thc joints band c.
Further calculations sho\v that the corrections diminish rapidly, and after four
cycles, they bccorne negligible. Summing up the initial fixed-end moments
and all consecutive corrections, \ve obtain for the final end moments thc values
sho\vn in Fig. c). 50. l"hc corrcsponding bending-nlon1cnt diagram is sho\vn
in Fig. 9.49b.
..\s another example of application of the moment-distribution mcthod in the
analysis of structures having n1embers of variablc cross section, let us consider
the previously treatcd syn1metrical thrce-span bridge sho\vn in Fig. 9.22.
U sing the given elastic constants and observing that therc are hinges at the
supports 0 and 3, \ve find, as in the preceding example,
J;o = SlO( I - C 10 C'01) = 20.34
mIo = mIlo - C01n01 = 28 1.9 kip-ft
Then, the constants necessary for calculating the distribution factors are
1Clnber k s ks
01 1.0 20.34 20.34
12 0.5 22.83 II .42
and these factors becomc
20.34
1'10 = . - -- = 0 640
31.76 .
11.42
1"12 = 3 1.76 = 0.360
All calculations of consecutive corrections arc shown in Fig. 9.51. It is seen
that after four cycles, \VC get results coinciding \vith those prcviously obtained
on page 435.
o

@B 1 


3

.-281. 9 -569.7 +569.7
-160.8 ..... -205.1 -364.6
...287.1 .. 161. 5  ",,126.6
-35.7 1E -45.6 -81. 0
.22.9 +12.8  +10.1
-2.8  -3.6 -6.4
+1.8 +1.0 .. +0.8
-0.2 .c -0.3 -0.5
+0.1 +0.1
+593 , 8 -452.5
FI G. 9.51

Chapter' 10
Matrix methods in structural
analysis
10.1 FORCE AND DEFORMATION METHODS
The various mcthods of analysis of statically indetern1inatc systems that have
been used in prcceding chapters fall into t\\'o distinct classi f.cations. In the
analysis of arches and frarncs in Chap. H, for cxarnple, the procedure \vas as
follo\vs: First, all redundant constraints \vere removed and replaced by the
corresponding redundant forces (or nlomcnts). 'fhc n1lgnitl1dcs of these
forces \vere then found by using the theorern of least \\'ork based on a con-
sideration of the strain energy in the structure. 1\ similar procedure \vas used
in Chap. i in the analysis of statically indetenninate trusses. l"his general
approach is called the lI/fthod of jorccJ.
In the analysis of continuous heams and frames in Chap. 9, a sonlc\vhat dif-
ferent procedurt \vas used. In this case, \VC calculated first the angles of
rotation of the joints (deformations) and considered the redundant forces only
later. The three-angle equation used in the analysis of continuous beams
480

ARTICLE 10.1
481
2L
t.- X
u..-I
R I V
Ug
FIG. 10.1
represents again the same kind of approach. Such procedurc is called the
111cthod o.f defoT111at;O'lls.
10 illustrate, on the sarne example, the distinction bet\\'cen the t\VO methods,
let us consider the statically indctcrminatc plane truss sho\vn in Fig. 10.1. 1
Here, a load I), dcfined by its components Px and P lI , is supported by five
prismatic members hinged togcther at A and to a rigid foundation at their
upper ends. Since the number of bars is greater than the number of equations
of equilibrium for the joint A, the problem is evidently statically indeterminate.
In general, if the hinge A is attached to the foundation by 11 bars, all in onc
plane, the number of rcdundant bars \vill be 11 - 2. Then, to detcrmine the
corresponding redundant forces .IY l , X 2 , X 3 , . . . , X,a-2 by a method of forces,
\ve \vritc the cxprcssion for the strain encrgy of the system as a function of
these forces and, by using the theorem of least work, obtain the necessary
equations:
au
aX l = 0
at)
--=0
aX 2
au
-----=0
d.L'<n-2
(a)
Each of these equations \vill contain all of the redundant forces, so \\,ith the
increase in he number of bars, the solution of Eqs. (a) becomes more and more
cumbersome.
To solve the same problem, Navicr suggestcd the usc of a method of dis-
placements. The deformation of the system in Fig. 10.1 is completely deter-
mined if've kno\v the horizontal and vertical components u and v, respectively,
of the displacemcnt of the hinge A produced by the load P. Assuming that
these displacements ace small, the elongation of any bar i \vill then be
/i = v sin ai - 11 COS exi
and the corresponding axial force in the bar becomes
S EA i ( . )
i = T v sin (Xi - tl COS (Xi
EA i ( . ) .
= -,- v sin ai - 1t COS ai sin ai
/1
(b)
11his is sometimes called Navicr's problem. See his book "Resume des leons . . . ,"
2d ed., p. 345, Paris, 1833.

482 MATRIX METHODS IN STRUCTURAL ANALYSIS
Writing no\v the t\VO equations of equilibrium for the hinge A, \ve obtain
i=n i=n
\' \' })%11
v  Ai sin 2 (Xi cos ai - U  Ai cos 2 ai sin a l = E
i=l ;=1
i=n i=n
\' A . 3 \' A . 2 P yh
v '-' i sIn CXi - U  i SJn ai COS ai = - E
;=1 i=l
(c)
From these t\VO equations, the unkno\\,'ns u and 'V can be readily calculated in
each particular case. Aftcr this, substitution of 11 and v into expression ( h)
gives us the force Si in any bar of the system. It is seen that for this problem,
direct consideration of the dcformation of the system results in a substantial
simplification of the solution, especially if there arc a large number of bars since,
indepcndently of that npmber, \\'c havc to solve only t\VO equations [Eqs. (c)"J.
In a sin1ilar \vay, dircct consideration of the defonnations simplifies the
analysis of a continuous bcarn on many supporrs. If \VC renlOVC alJ interme-
diate supports and consider the corresponding reactions }[h }( 2, Jy. 3, . , . as
the redundant quantities, the theorem of least \vark yields a system of equa-
tions (a), each of \vhich contains all of the unkno\\'ns. lhus, the solution of
the problcm bccomes very cumbersoo1e if the number of spans is large. A
great improvement in thc solution of this problenl is attained by considering
each span of the continuous beam as a simple beam on t\\'o supports and
calculating the angles of rotation of the ends of such beams. l"hen, from the
condition that at each intermediate support these angles for t\\'o adjacent spans
must be equal, the kno\vn three-angle equations (or three-moment equations)
are obtained (see page 414). Such equations are much simplcr than Eqs. (a)
because no one of them contains nlore than three unkno\\'ns.
Anot,cr example in \vhich the method of dcforolations resulted in a great
silnplification is represented by the systcln sho\vn in Fig. 9.37, \vhere four
mcmbers are rigidly joined together at a and built in at their far ends. Neg-
lecting the effect of axial forces in the bars, this systcnl has seven redundant
reactive clenlents, and for their determination, the theorenl of least \\'ork \\'ould
give seven equations. Again, the problem \vas greatly simplificd by con-
sidering the deformation of the structurc. This deformation is complctely
defined by the angle of rotation ()a of the joint &1 produced by the appJicd Joads.
When the magnitude of this angle is found (sec Eq. (9.24)], the cnd moments
for all the nlell1bers can be rcadily calculatcd from the slope-deflection equa-
tions. Thus, by considering deformations first, \ve need only onc equation
\vhich \vas \vritten on the basis of thc equilibrium of the end mOOlents at joint a.
It is not to be concluded from the foregoing discussion that, in the analysis of
a statically indeterminate system, a method of deformations is al\vays superior
to a method of forces. For example, in the case of a simple truss having one

ARTICLE 10.1
483
redundant rcaction and tcn joints (see Fig. 7.5), the nlethod of dcforo1ations
described above \vould beC0J11C very cumbersome, \vhereas the mcthod of forces
used in Chap. 7 is extremely simple.
In dealing \vith highly statically indetenninate systems, \VC usually find that
rcgardless of \vhethcr \VC use a method of forces or a n1cthod of deformations,
it becomes necessary to so)ve a large nurnber of simultaneous linear a1gcbraic
equations \\rith as many unkno\vns. \t\lithout regard to any particular probleol
of structural analysis, let us no\v consider a systen1 of such equations:
tlUXt + tl12X2 +
+ i11rXn = £.'1
a21Xt + (hZ.'r2 + . + a2n.\'n = f2
(10.1)
am lX1 + arn2X2 +
+ tlmnXn = C m
Thcoretically, such a system of linear algebraic equations can aI\vays be
solved, but the process of solution becomes cumbersorne as the ntunber of
equations increases, and to simplify the tcchnique of this solution, the notation
of 111i1trix algt'brn 1 \viJI no\\' be introduced. 'rhus, in nlarrix notation, Eqs.
(10.1) may be \\'ritten in the condensed form
(1 ", "
.\'1 CJ
..\"2 C2,
-
.\"11 C m
(IO.la)
au a12
t121 (122
.
l1 111 1 {11112
111n
a2n
or, still morc briefly,
faij] [.l'rl = [Ci]
(to.lb)
or sin1ply
/1.1." = t
(IO.le)
Each array of numbers (or symboJs) in the brackets of expression (IO.la) is
called a 1l1atrix. l'he nUITtbers (or symbols) themselves arc ca lIed ele1JltlltJ,
and \"hen there arc 111 rows and 11 COIl/UtIlS, the matrix is said to be of order
'/11 X n. V\lhcn there is only one column or one ro\\' of clements in the matrix,
it is called a cohl'lJlU 'l,ector or a row vector. It is understood that the matrix
[aij] in (10.1 a) operates on the column vector [.\"1'1 in such a \vay as to produce
the left-hand side of the system of equations (10.1) above. 1his brings us to
1 For a discussion of the fundamentals of matrix algebra, see A. C. Aitken, U Dctcrnlinants
and 1\1atrices," Intcrscicnce Publishers, Inc., New Y(>rk, 1958.

484 MATRIX METHODS IN STRUCTURAL ANAL YSIS
the necessity to learn some of thc rules of nlatrix algebra \\,hich \vill be dis-
cussed briefly in the next article beforc proper attention can be given to the
treatrnent of structural problcIllS by 111atrix I11ethods.
Before procceding \vith this, ho\vcver, the rcadt'r should undcrstand that the
use of Inatrix n1cthods in structural analysis holds no particular Inagic, nor docs
it represent any great advantage over the nlcthods discussed in prcceding
chapters so long as numerical calculations arc to be nlade by hand. Its rcai
advantage lies in the fact that it lends itself particularly ,veIl to the use of the
electronic digital cOlnputcr and thereby opens the door to the analysis of
structural problems that \vould other\vise be too involved and complex to cope
\vith by desk-calculator techniques. In the limited space available here, \\'e
shall be unable to disclose the full po\\'er of the olatrix approa,ch, but it is hoped
that the sinlple exanlplcs to be discussed ,vi II give the reader enough fami liarity
\vith the nlcthod to enable hin1 to study the literature on the subject 1 to better
advantage.
10.2 ELEMENTS OF MATRIX ALGEBRA
J.-\s pointed out in the preceding article, a 111atrix is any rectangular array of
numbers, sYlnbols, or elnnellts. It is not reducible to anything Sitl1pler and
nlust be rcgarded as an entity in itself. rhen, likc numbers or algebraic
symbols, tnatriccs, undcr certain conditions, may be added, subtractcd, rnulti-
plied together, or one divided by another. The rules for pcrfornling these
operations consti tute '/Jlt1tri.r al gebrtl.
To add t\va n1atriccs, \VC silnply add corresponding elenlcnts to ohtain the
Clell1ents of the matrix sum. "[his is possible only if the t\\'O matriccs to be
added arc of the sanle order 11/ X II. This rule of addition is stated symboli-
cally as follo,vs:
laijJ + [b ij ) = [aij + b,-j]
(a)
Specifically, if [t7ij] and [bill arc both of 2 X 2 ordcr,
[::: :::] + [::
/;12 J [ all + 1>11
b'l2 - .721 + b'21
.112 + b12 ]
t 1 22 + b 2 1.
Similarly, the rule of tnatrix subtraction is
[t7ijJ - f biJ = [aij - bijl
(b)
1 For a general discussion of Illatrix n1lthods in structural analysis and a bibliography on
the subject, see J. H. Argyris, On the Analysis of Complex Elctstic Structures, Appl. ,\Juh.
Rtv., vol. ] 1, no. 7, 1958.

ARTICLE 10.2
485
Wc see fronl this that t\\'O matrices can be said to be equal only if they arc equal
element by clement.
From the rule of nlatrix addition, there follo\vs at oncc a rule of scalar
multiplication: namely, to multiply a given matrix by a scalar number, \\'C
simply multiply each clcment of the matrix by this scalar number. In symbolic
fornl, this rule of scalar n1ultiplication becomes
X[aij] = [XaijJ
(c)
\vhere X is the scalar factor.
Io obtain thc product AB of t\VO matrices A and B, \ve have the follo\ving
rule: The element of the ith ro,",' and thc jth column of the product 11/atri.'l' is
obtained by multiplying the ith ro\\' of A into rhe}th column of B, clement by
elenlcnt, and summing the products so obtained. If A is of the order 111 X 11
and B is of the order 11 X q, the product matrix (' \vill be of the ordcr 111 X q.
Statcd in symbolic form, the element Cij of the product nlatrix C = AB \vill be
k-n
Cij = 2: aikbkj
k-l
(d)
As a speci fic example,
[ all t112
a'll a'/.2
] [ bu bU' ]
t113
a23 b'll b 22
b 31 /1 32
= [ auh ll + a 12 b 21 + 1113b31
tl 21 b u + tl 22 b 21 + a 23 b 31
a U b 12 + a 12 b 22 + a 13 b 32 ]
11 21 h 12 + a2b22 + a 23 b 32
It must be notcd that such multiplication is possible only if the number of
columns of A is equal to the number of rows of B. It is also necessary to
observe that nlatrix multiplication is not commutative, that is, AB r5- BA.
For instancc, in the above example \vherc the matrix product A H \vas of order
2 X 2, the matrix product BA \vould be of order 3 X 3. \\Then both AB and
BA exist, BA is called the reversed product of AB. 'l'his makes it necessary
to distinguish bct\veen prnnultiplicat;(Jll and pOSl111ultiplicatiul1. In the above
example, it is said that B is premultiplicd by A, or A is postmultiplicd by R.
Returning no\v to the matrix expression (1 O.la), let us apply the multiplica-
tion rule to the left-hand side of this expression, i.e., \VC prcmultiply the column
vector [Xj 1 by the matrix [a'-i]' \\Then \VC do this, \ve see that \ve obtain the
left-hand sidc of Eqs. (10.1). This explains the reason for the multiplication
rule as stated above.
For the present, \\'c defer any discussion of the qucstion of division of one
matrix by another, but therc is one more useful operation in matrix algcbra

486 MATRIX METHODS IN STRUCTURAL ANALYSIS
kno\vn as transposition. 1.'he tftlllSpOSe of a matrix A, denoted by A', is obtained
by re\vriting the matrix A so that its ro\vs become co]unlns, taken in the same
sequence, and vice versa. l'hus, symJoJically,
[aij]' = [aji]
Specifically, if
(e)
A = [ all a12 a 13 ]
thl a22 a23
[ an t121 ]
then A' = au (122
a13 il23
Consideration of the rule of multiplication together \\rith that of transposition
sho\vs that the transpose of a matrix product must be thc reversed product of the
individual transposes. That is, .
(AB)' = B'A'
(1)
l'his is kno\\'n as the rcversalrule.
Before proceeding further, \ve shall mention a fe\v particular types of
matrices that \vill bc commonly met \vith in applications. l'hc matrix 1
l=[  ]
(g)
is callcd a ul1it 111atr;x of 3 X 3 order. In general, it is a squarc rnatrix of
ordcr n X 1/ having all elcmcnts zero except those on the principal diagonal,
\vhich runs from top lcft to bottom right. In matrix algebra, the unit matrix I
corresponds in every \vay \vith the idea of unity in ordinary algebra.
If a unit matrix I is multiplied by a scalar number , \ve obtain
AI = [  J
(11)
\vhich is kno\vn as a scalar 111atrix.
A matrix of the form
A _ [ 1 2 : ]
. «3
(i)
is calJed a diago1lal 1natrix. l-he unit matrix and the scalar matrix arc, of
courSe, special cases of a diagonal matrix.
1 The dots indicate zero elcrncnts.

ARTICLE 10.2
487
l"herc arc various other special matrices, but the oncs introduced so far \vill
sufficc for our present purposes. In summary, \VC have the rectangular 1natr;x
of order 1JJ X 11, thc square 111t1trix of order 11 X 1/, the row vector and the
C0I1/11111 vector, the unit 1J1atr;:t:, the scalar 111atrix, and the diag011a/1J1atrix.
l{eturning, again, to the system of equations (10.1) and \\rriting them in the
matrix form Ax = c, \ve state by definition that the solution can be expressed
in the follo\ving form:
C
.t = A = A-Ie = Rc
(j)
I 'his brings us to the idca of dividing one matrix by anothcr, or, {norc properly,
of finding the reciprocal R of a given matrix A. 'fhis process is called ;llVerS;011.
To accomplish it, \ve Jook for a matrix R such that RA = I, \"here I is the unit
matrix. It is to be noted at this point that since a system of simultaneous
equations ,,,ill have a unique solution only if the number of equations is equal
to the number of unkno\vns, A \\rill al\vays be a square matrix. Other\vise,
the idca of inversion has no meaning.
Various procedurcs have been devised for the inversion of a given square
matrix. One such nlethod \vill no\v be bricfly described. 1 f"'irst, it is neces-
sary to introduce the notion of the adjugatc, or adjoint, of a given matrix A \vhich
is \vritten ad} A. 'l'his adjugate is defined as the transpose of another matrix
(' formed out of the cofactors of the elements t1ij of the given matrix A. This
formation of the adjugate of a givcn matrix is iJlustrated by the foJIo\ving
example.
Let the given matrix be
[ '11 h1 Ct ]
A = (12 b 2 C2
d3 b a C3
(k)
l'hen the matrix C, formed out of the cofactors of A, \vill bc
[ Ib2C31
c = -lb1c31
Ib 1 c 21
-lthCal
lalc31
-l a l"21
la 2 b 3 1 ]
-la 1 b 3 1
la 1 b 2 1
\vhcre the detcr1Jli1/tll1t
/b 2 c:d
b 2 c2 1
- = b1. C 3 - c 2 b a
b 3 C3 I
1 See Aitken, trp. cit., p. 5 I.

488 MATRIX METHODS IN STRUCTURAL ANALYSIS
is called the cofactor for the elenlcnt at,
I I hI Cl I I
-lb 1 c 3 = - b = - (b 1 C 3 - c 1 b 3 )
, 3 C3
the eof(u;tor for the element 112, etc. In general, to find the cofactor for any
elernent aij of the given n1arrix, \ve cross out the ro\v i and the colunln j, for
exanlple, \vith i = 2, j = I,
hi Cl ]
b a C3
and write the determinant of the remaining clements taking the sign plus or
l11inus according to rhe follo\ving rule: If the ro\v nlunbcr i and the colurnn
number j are both odd or e'L'ell, the sign of the cofactor is plus; if the ro\V
number is odd and rhe column nUlnber is e'l)tlJ, or 'l.Jice 'V'trsa, the sign of the
cofactor is nlinus.
Having obtained the cofactor 111atrix (' above in accordance \\.ith this rule,
the adjugate of A, defined as the transpose of (', becomcs
[ I bzcal
adj /1 = - jt12C3r
1L1 2 b a !
-lb 1 c 3/
I'll e 31
- 111 1 b 3 1
Ih 1 e 21 ]
-l a lC2!
lalb'll
(k')
v\lhen the adjugate of a given square I1latrix A has bcen fOrIl1cd, it can be
demonstratcd 1 that
A (adj A) = (ad; A)4 = JA 1/ (I)
wherc IA I is the detir"li71al1t of /1 and / is the unit matrix. Dividing expression
(I) through by IA I  0,
1 (ad; A) _ (ad! A ) == / = IA
IAI - IAI
Thus,
ad j /1 .
R = -'71!- = 1411 ad, A
(111)
is the required il1'verJC of ,t;4 .
FoJlo\ving the rules for inverting any given squarc rnatrix, it can be easily
demonstrated that the inverse of any diagonallnatrix \vill be obtained simply
1 See ibid., p. 53.

ARTICLE 10.2 489
hy inverting each individual element along the principal diagonal. Thus, if
5
1

.,
J..
then rAJ-l =
1
-g-
[/1] -
3
1

Having no\\' the method of inversion of a square Inatrix, \VC may illustrate
the Il1atrix solution of a system of simultaneous equations Ii kc Eqs. (10.1).
C:onsidcr, as a simple example, the equations
3x + 2y - z = 4
x - y + 2z = 5
- 2x + y - z = - 3
In [l1atrix notation, these equations \vill be \vritten in the form
[-: -: =iJ[u=[-n
Follo\ving the rules outlined ahove, \VC find the adjugate of the square coeffi-
cient nlatrix A to he
[ -I
ad; A = - 3
-1
- -; ]
-7 -5
'10 find the dctcrminant of /'1, \\'c expand it by co factors of the clen1cnts of the
first ro\v 1 and obtain
jAr = 3(-1) - 2(3)
1(-1)--3-6+1--8
Finally, using Eq. (j), \VC obtain
[ .t' ] [ - 1
:: = -i -3
 -]
I
-5
-; ] [ ; ] = -i [ -: ] = [  ]
- 5 - 3 - 24 3
-7
That is, .r = 1, Y = 2, z = 3 represents the required solution. This siIllple
exan1ple embodies many of the operations of tnatrix algebra discussed above,
and the rcadcr should be sure that all steps arc clear to him beforc proceeding
fu fthcr.
Refore proceeding to some applications of matrix nlt.thods in the anal ysis of
structures, \\'c shall Incntion a fc\\' examples of \\rjting algebraic cxpresions
1 See ibid., p. 3 .

490 MATRIX METHODS IN STRUCTURAL ANALYSIS
in matrix notation. L1ke, for exao1ple,
c = a1b l + a 2 b z + . . . + c1 n b",
(11)
To \vrite this . . form, let us define t\\'o coIuInn vectors /1 and B as
In matnx
at hI
t12 b z
a3 "3
11 - and B=
an
b n
Then, transposing the column vector A into a ro\v vector A' and perforrning
the multiplication A' B, \ve obtain
b l
b z
[at tl2
an]
= 'lib l + "2/1 2 + . . . + tl7ih1Z
b n
\vhich is the right-hand side of Eq. (n). Thus, in nlatrix notation, Eq. (11)
can be \v ri tten
c = A'B (n / )
..gain, let us take
C = (111.\'IYl + tl22 X 2Y2 + t133X3Ya
(0)
in connection \vith \vhich \ve define the folIo\ving Illatrices:
[ Xl ]
\' = '2
.'\:3
[ VI ]
Y = ) . '2
)'3
[ tI U . . ]
.4
- : 1122 113
l'hen, transposing the column vector .\' into a ro\\' vector x' and perforoling the
nlultiplication x'A)', \VC obtain
[Xl
[ all . . ] [ Yl ]
Xz X3] . t7Z2 . ..v 2 -
. 4133 y'3
[(111.\'1)'1 + (722.\'2)'2 + a33.\'3)'3J
From this, \ve see that Eq. (0) may be cxprcssd in matrix forn1 as
c = x'Ay
(0 ' )

ARTICLE 10.3
.91
PROBLEMS
Find the matrix sum A + B if
[ 3 2 I J
A= 014
203
and
B = [ 
-2
o
-3
1
n
2 Find the nlatrix product AB if A and B arc as given in the preceding problem.
l 6 -5
Ails. AB = -7 1
-2 3
In
3 Find the matrix product BA of the t\VO Inatrices given in Probe 1.
[ 6 4
Al1J. B4. = 13 -1
-4 -3
]
4 \,\rrite the transpose of each of the n1atrices given in Prob. 1.
5 C;ivcn the square n1atrices
I
A = 1
. 1
2
3
5
3 ]
5
12
R = l i
-9
9
-3
-n
sho\v that H is the adjugatc 0 f A and find the matrix product A B.
411S. AB = 3/.
10.3 APPLICATION OF MATRIX METHODS TO PLANE
TRUSSES
To illustrate the use of matrix nlethods in the aIlalysis of pin-jointed trusses,
let us begin \\.ith the deflection problem as previously discussed in Chap. 6.
In Fig. 1 a.2a, a si:nplc truss having 111 nlcmbers is subjected to the action of a
set of extcrnalloads P, and it is requircd to find the vertical deflection of joint j
due to this loading. To accomplish this, \ve denote by Si the axial force in any
1
FIG. 10.2
(a)
(b)

492 MATRIX METHODS IN STRUCTURAL ANAL VSIS
bar j duc to the P loading (Fig. 10.2a) and by Jij that duc to a vertical unit force
at joint j (Fig. 10.217). \\'e have then, for calculating the vertical deflection
of joint j, the follo\ving exprcssion [sec Eq. (6.2), pagc 258] :
ic;:m
Il j = L SijPiSi
i-I
(10.2)
\vhere Pi = li/Ai E is the flexibility factor for the bar i.
If \ve \\rant such a deflection for each of 11 chosen joinrs of the truss, \\'C
must calculatc values of Sij for a vcrtical unit force at each of these 11 joints.
L.et us supposc that this has becn done and that \ve arrange these il{fluc1JCC
11lJ111be-rs in the form of a matrix of order 111 X 1/ as follo\vs:
Sl1 S12 SIn
[Sij] -
S21
J22
S2n
(a)
Sml Sm2 Smn
This will thcn be called the }!,c01l1elT)' 111atrix for the truss or simply the S 111otrix.
Next, lct us arrange the flexibility factors Pi for the 111 mcmbers in the fom1
of a diagonal matrix,
pu
P2'1.
r Pi] -
P3:J
(b)
P1IInI
\vhich is called the flexibility 1J1fltrix for the truss or simply the p 1llatrix.
Finally, assuming that the axial forces Si produced by the givcn P loading
have been calculated, let these bc arranged in the form of a column matrix,
Sl
S2
[Si] -
(c)
Sin
\vhich is called the load 111al.,-;1' or sinlply the S nlatrix.
No\\' recalling the rule of nlatrix tnultiplication frotn the prcceding article,
\ve see that Eqs. (10.2) abovc can be expressed by the matrix formula
 = SipS
(10.3)

ARTICLE 10.3
493
That is, if \\'c transpose the s matrix (a), postmultiply it by the p matrix (b),
and then postmultiply this product by the S nlatrix (c), \ve obtain the full set
of Eqs. (10.2) for the deflections of the 11 chosen joints of the truss.
To iIlustrate, let us assume that 111 = 3 and 11 = 2. Then Eq. (10.3)
becomes
[ PH .
[  2 1 ] = [ 511 521 J:U ] . P22
u 512 522 S 32 . .
PJ [:]
_ [ SlIPl1
.f 1 2P 11
S21P22
52 2P 2 2
S31P33 ] [ : ]
S32P33 S3
_ [ 511PllSl + S21P22S2 + S31P33S3 ]
J'12PllSl + S22P22S2 + S32P33QC)3
1'hc matrix product (10.3) is caIled the A lfultri:t:. We see that it represents
a very compact \\fay to express the deflection formulas (10.2) for all of the
chosen 11 joints. In addition to this, as \ve shall see later, the actual matrix
multiplication with numerical values can be made in a systematic manner so
that the danger of calculation errors is reduced to a mininlum. Also, several
different loading conditions can be handled all at oncc, simply by filaking a
rectangular S matrix \vith a colunln for each different loading condition that it
is desired to study.
As an example, let us consider the simple truss sho\vn in F:g. 10.3 \\,hich has
111 = 9 members. We assume that it is required to find the vertical deflections
of the two upper-chord joints a and b, under the action of t\VO separate loading
conditions as sho\\'n in Fig. 10.3a and b. The nun1bering of the nlembers is
sho\vn in Fig. 10. 3a, \vhcreas the dimensions of the truss 2re shown in Fig.
a
6 b
(a)
1 kip
a
(c)
FIG. 10.3
a
b
40 11 -+-40"+40 11
(6)
1 kip
b
(d)

494 MATRIX METHODS IN STRUCTURAL ANAL YSIS


IO.3b. Each bar has a cross-sectional area Ai = I in. 2 and a modulus of
elasticity E = 30(10 3 ) ksi. ,\\Jc begin \vith a calculation of the axial forces Si
in the nine members under the t\va loading conditions sh()\vn in Fig. 10. 3a and
b. This set of axial forces \vill constitute the S Inatrix for the problclll.
Sin1ilarly, \ve calculate the axial forces in all bars for each of the unit loads
acting at the joints Cl and b as sho\vn in Fig. 10. 3c and d. l'his set of axial
forces \vill constitute the s matrix for the problcn1. Finally, \ve calculate the
flexibility coefficient Pi = It/ AE for each mcn1bcr and \vritc do\vn the diagonal
flexibility matrix, l'hesc three basic 1l1atriccs for the problt'111 arc as follo\vs:


8 4
-10 -5
-3 3
4 R
Sii = .k 5 -5
-8 -4
0 0
4 8
-5 -10
4 4 10
5 -5 0
3 3 0
:0 4 R 10
Pi = -- 5
 - -5 0
E (> i -
4 -4 0
3 9 0
4 8 10
5 -10 0


No\\-' using Eq. (10.3), the required deflections jj,j are found by matrix multi-
plication as foIlo\v
:
Do = 10 [8 -10 -3 4 S -R 0 4 -5]
9£ 4 -5 3 8 -5 -4 0 8 -10
4 4 10
5 -5 0
-' 3 0
4 8 10
X .5 -5 0
4 -4 0
3 9 0
4 8 10
5 -10 0




ARTICLE 10.3
495
(a)
(b)
FIG. 10.4
l\1aking the indicated multiplications in t\VO steps, \VC obtain
 _ IO [32 -50 -9 16 25 -32 0 16 -H]
. 9£ 16 -25 9 32 -25 -16 0 32 - ;iJ
4 10
-5 0
3 0
8 10 ]
-5 0 IO [ 860 640
X - 9£ 1,417 800
-4 0
9 0
8 10
-10 0
l'hus, \vith }; = 30(loa) ksi, the deflections due to the load P = 9 kips acting
as sho\\'n in Fig. 10. 3tl arc.
, .
a = 0.0318 In.
, .
b = 0.0525 In.
\vhiJe those due to the load Q = 10 kips in Fig. 10.3b arc
" .
a = 0.0237 In.
" .
.1b = 0.0296 In.
Let us proceed no\\' to a consideration of a statically indeterminate truss
having Ul rnemblrs, II of \\'hich arc redundant 1 (Fig. 10.4). For the dctern1ina-
rion of the axial forces ..\1, A\2, . . . , \n in the 1/ redundant bars, \ve can use
Eqs. (7.6) on page 313. i\sSU[l1ing that there are no thcrnlal effects or
assembly errors, these equations \\.ill take the follo\ving fonn:
- al = t5 11 ./\ 1 + t5 12 "Yz +
- 2 === O:n,\r 1 + 022'Y2 +
+ t5 1n ;Yn
+ 02'lJY'
(10.4)
- 11 = t5n.lY 1 + t5,.2.1Y 2 + + t51n"\T
1 Note that 11 no\\' refers to the number of redundant oars and not to the f1l1lnbcr of joints.

496 MATRIX METHODS IN STRUCTURAL ANALYSIS
\vhere jj.i is the generalized displacement, corresponding to }(j, caused by the
P loading and Oji is the displacenlcnt, corresponding to Xj, caused by unit forces
acting in place of X j.
In matrix form, this set of equations beconles
6 1 5 11 012 01n Xl
6 2 021 22 02n X 2
-
. . . .
6n Ont 0,&2 Onn X n
or, more simply,
- 6 = oX
The solution of Eqs. (IO.4b) is
X - jj.- - 0-1
" - - - -
"
\vhere 6 = [s'pS] and  = [s' ps ]
( I o. 4a)
(10.4b)
(d)
(e)
from Eq. (10.3). Substituting expressions (t) into Eq. (d), \veobrain, for the
axial forces in the redundant bars, the matrix formula
X = - [S'pSJ-l[S'p£]
(10.5)
When the forces in the redundant bars arc kno\vn from Eq, (10.5), the
forces in the remaining bars can be found from the expression
Si = £i + .\'i)X 1 + .\\2X 2 + . . . + Sin"\'n
(f)
or, in Inatrix form,
S = s + sX
(g)
St1hstituting for X its expression fronl (10.5), \VC obtain
s = s - sf.\" pSJ-1 rJ' psl
or, introducing a unit matrix I so as not to have  appear t\vice,
s = rI - J(.\"pJ)-l(J'p)]S
(10.6)
1 4 his represcnts a final matrix formula for the axial forces in all bars of a
statically indeterminate plane truss under given loading conditions.
As an example of the application of Eq. (10.6), let us consider the statically

ARTICLE 10.3
497
P ::.10 kips
10 kips
-1
8.66'
1
10 kips
(a)
(b)
FIG. 10.5
indetenninate truss sho\vn in Fig. 10.5. l'his plane truss has seven bars, of
\vhich three are seen to be redundant. As the redundants, \\'c choose the bars
5, 6, 7, so that the prinlary system \vill be as sho\\'n in,Fig. 10.5 b.
A.s a first step, \ve rnake a statical analysis of this prinlary systcrn under the
actlon of he given P loads and )ikc\vise under the action of a pair of unit forces
in place of each redundant bar. 'fhen \vith cross-sectional areas A 1 = A" =
/1 6 = 5.00 in. 2 , A 2 = Aa = 8.66 in. 2 , andA 6 = A7 = 2.89 in. 2 , the geometry
matrix, the flexibility matrix, and the load matrix become
+ y3 0
2 2
y3 0 1
--- - --
2 2
Y3 1 0
- -
s = 2 2
1 y3-
+ -- 0 --
2 2
+1 0 0
0 +1 0
0 0 +1
2 - 5y3
2 + 5y3
2 -5
p= 2 s= -5
2 0
3 0
3 0
\vhere, for convenience, \ve have taken E = 1 ksi.

498 MATRIX METHODS IN STRUCTURAL ANALYSIS
No\\! transposing J to s' and postn1ultiplying it by p, and thcn by s, \VC obtain
1
2
V-f
_ r--__
2
V3
----
2
1
o
- V}
V3
---
2
   ]
003
[ 1 - VI - VI
[S'pS] = - t 3 _ -
o
o
1
2
x
2
o
V3
2
o
o
1
=[  J
1
2
1
o
o
o
o
1
o
Since this result is a diagonal rnatrix, it is easily inverted (see page 489), and
\VC have
[S'pS]-l = [  ! : ]
. . 1
K
Premultiplying this by the s nlatrix, \\'C obtain next
+! V)- 0
---
2 2
V3 0 I
-.- - -- 1
2 2 -
V3 6
s(s' pJ)-l = - _.-- 0 1
2 2 5
+! 0 V"3 1
-
2 2 -
5
1 0 0
0 1 0
0 0 I
10 - 1 2 VI 0
- 10 VI 0 -12
- 10 VI -12 0
I 10 0 -I2V3
- --
120 20 0 0
0 24 0
0 0 24

ARTICLE 10.3 499
Postnlultiplicarion of this by s' p gives
J(.r' pS)-lS' p
46 30 6 10 20 108 0
-- -- - -----
V3 VI V3
30 42 )0 6 60 0 -36
- ------ - --
VT V3 V3
6 30 42 30 60 - 36 0
- ---
V "3- V3 V3
1 10 6 30 46 20 108
- - --- 0 --
120 V3 V3 V'j
20 60 60 20 40 0 0
- --
V3 V3
72 0 -24 0 0 72 0
--
V3
0 -24 0 72 0 0 72
--
V3 Finally, subtracting this from the unit matrix I and complcting the operations
indicated in F<). (10.6), \ve obtain
74 30 6 -10 -20 108 0 15
-... --
vI V3 V3 V3
30 78 -- 30 6 60 0 36 15
--
V3 V3 V3 V3
6 -30 78 30 60 36 0
-- - -5
V)" V'3- V3
1205 = -10 6 30 74 -20 0 108 -5
--
vI vI ,,/3-
-20 60 60 -20 80 0 0 0
VI V3
72 0 24 0 0 48
- 0 0
vI
0 24 0 72 0 0 48 0
V3
-423.5
692.8
- 706.4
- -400.0
400.0
- 480.0
0

500 MATRIX METHODS IN STRUCTURAL ANALYSIS
Thus, in l<ip units, the final values of the axial forces in the bars become
I
-4.00 I
o
Bar
2 345
6
7
s
-3.53 +5.78 -5.89 -3.33 +3.33
As a partial check on the calculations, these results can easily be shown to
satisfy the conditions of equilibrium at each joint.
PROBLEMS
U sing a matrix fornullation of the problem, find the axial forces that \viIl be induced
in the bars of the truss in Fig. lO.Sa by a unifortn rise in temperature of 50°F. The
coefficient of thernlal expansion for the bars is at = 6.667 (10- 6 ) (in.fin.) JOF, and
rhe modulus of elasticity is E = 30(10 6 ) psi. l'hc external loads are rernoved.
[-liut Replace the [natrix [SipS] in Eq. (10.5) by the rnatrix lXt6t[s'/J.
Al1S. 8 1 = S. = -6.67 kips, S2 = Sa = -23,1 kips, S5 = +16.7 kips,
56 = S7 = + 17.3 kips.
2 Using nlatrix fonnulation, find the axial forces induced in the bars of the plane truss
sho,vn in Fig. 10,6. .-\11 bars have the sanlC modulus of elasticity R, and their
r- 90" -T-160"-t-9011
FIG. 10.6
L
90 kips
cross-sectional areas are as follo\vs: 44. = A4 = 7.5 in. 2 , A 2 = A3 = 10.0 in. 2 ,
A7 = 6.4 in. 2 , A6 = ,,4 5 = 8.0 in. 2 l'hc dimensions of the truss and the applied
loads are sho\vn in the figure.
Ans.
Ba r I I 2 I 3 I 4 I 5 I 6 I 7
S 1+51.61 -4.55\ +35.5) +21.6/ +51.4\ +51.41-15.4
10.4 MATRIX ANALYSIS OF CONTINUOUS BEAMS
1'he discussion of statically indeterminate trusses in the preceding article
represents an illustration of Il1atrix forn1ulation of a force 111ethod of analysis.
As an illustration of rnatrix fornlulation of a defor1Jlat;on 111ethod, \ve shall
discuss the problem of continuous beams. Although a matrix treatment of

ARTICLE 10.4
501
FIG. 10.7
Mb Y
 .- -
I 8 0
· l
 ba
8b
I
this problem can be made quite general, \ve shall limit the discussion here to the
case of a continuous beam on unyielding supports. By so doing,' attention can
be more clearly focused on the essential features of the matrix method.!
We begin \vith the case of a simply supported beam ah of uniform cross
section subjected to end mOlnents (positive \vhen clock\vise) and lateral loads
as sho\vn in Fig. 10.7. For this case, the slope-deflection equations have the
form [see Eqs. (9.6), page 411J
MOb = k( 48 a + 20 b ) + mIab
Alba = k(20 a + 40 b ) + mIba
(10.7)
\vhcre mIab and mIba arc the fixed-end moments (see page 411) and k = EI//
is the stiffness factor for the bean1. These equations may be written in matrix
fom1 as follo\vs:
[ "Jab ] = [ 4k 2k ] [ 00 ] + [ mt ab ]
Alba 2k 4k Db mLba
(lO.7a)
In this expression, the 2 X 2 matrix
[( = [ 4k 2k J
2k 4k
(a)
is called the stijJ11eSJ 111atrix for the beam.
This notion of the stiffness matrix for a beam of uniform cross section can
readily be extended to the case of a bm of variable cross section. Referring
to the slope-deflection equations (9.17), \ve conclude that for a nonprismatic
beam
K = [ kkaa kkab J
kkab kkbb
(b)
where the values of kaa, k ab , kbb are defined by expressions (e) on page 428, and
k = Elo/ /, 10 being the moment of inertia of the smallest cross section. To
avoid confusion \\,ith subscripts in later discussion, \\'e introduce at this point
1 See S. U. Benscoter, Matrix Analysis of Continuous Beams, Trans. ASCE, vol. 112,
p. 1109, 1947.

502 MATRIX METHODS IN STRUCTURAL ANALYSIS
fiC\\' notations for the stiffness values in expression (b) and \vrite
[( = r. J
\v here Ka = kkac ]( b =-= kkbb R = kkab
(c)
(d)
v\lith thesc notations, the matrix equation (10.7£1) beconlcs
[ AJab J = [ 1(a
i\tJ ba . R
 J [ Oa J + [ :1Tl.ab J
Kb Ob . .mha
(10.7/7)
If AJ ab = 0, that is, if \VC have a beam \vith only one end nl0J11ent A/f ba ,
Eqs. (lO.7b) give
o = KIJO a + ROb + nab
A.J ba = R t1 a + ](b8 b + mba
1.'hen, upon elinlination of fJ a oct\veen these t\VO equations, \\'c obtain
KII](b - 12 ( l )
i\I ba = - Ka {h + m7:bfJ - /(0 mnh
rrhus, in such a case,
\vherc
j\;J'Ja = K8b + lrCl
I n _ KaKb - R2 d
\b - /(a - an
l
rra = gn.ba - } T tab
\a
(IO.7c)
(l')
rcpresent a 'IJlodttied s;iffne.r:s factor and a 'JJ1od ijied .fi:red-t'1ld 1J1mJlent for the end b
of the beanl \vhen there is no end moment at the far end t1, In \vorking \vith
continuous beanls, these notions of modified stitfncss factor and n10dified fixed-
end monl<..nt \viII greatly simplify the trcatnlcnt of a span having one end sirnply
supportcd.
Having the basic slope-deflection equations in Inatrix form (10.7 b), \\'c may
no\\" proceed to a consideration of continuous bcan1s. l'hc discussion is
intended to be perft'ctly general for a bean1 having 11 spans on 1/ + 1 unyielding
supports and \\..ith its t\\'O cnds either built in or SiIllply supportcd. Ho\\'cvcr,
to avoid cUIllbcrsoIl1c suhscript notations, \ve take a specific case of four spans,
the left end being simply supported and the right end built in as sho\vn in
Fig. 10.8. '['he lateral loads are assuIncd to be kno\vn; so all fixed-cnd
mon1cnts can be calculated or taken from the ready tables. .A.ssun1ing that
this has been done, \VC are ready to \vritc the slope-deflection equations (IO.7b)
for each of the four spans.

ARTICLE 10.4
SOl
1 01 1 12 123 '34
[  :;J [KI2 R12] [K R] [  ]
R 21 K 2t R 32 K 32
FIG. 10.8
For the left end of each span (except the first \vhere MOl = 0), the first 0 f
these equations gives, with 84 = 0,
iV/12 = [(128 1 + R 12 (J2 + mil2
M 23 = K 23 82 + R 23 8 3 + ml23
Ma., = [( 301 8 3 + mL34
(f)
or, in matrix form,
[ :: ] = [ I2 ::
M 34 0 0
o ] [ (J 1 ] [ mt 12 ]
23 82 + mt 23
I\. 34 8 3 mL34
(f')
\vhich is easily verified by performing the indicated matrix multiplication and
addition. In the same \\lay, for the right end of each span [except the last,
v.,here (J 4 = 0, and the first, for \vhich \VC use the modified equation (10.7 c) ],
Eqs. (10.7 b) give
1\1 10 = KO(Jl + m!o
1\1 21 = 12181 + K 21 ()Z + mL21
A1 32 = R 32 82 + K32 8 a + ;J1l 32
(g)
or, in matrix form,
[ AlIO ] _ [ Ko 0
M 21 - R 21 ](21
Jk(32 0 132
] [ ] [ ' ]
o 0 1 nl0
o 02 + mt21
J( 32 () 3 mt 3 2
(g')
In each of expressions (f') and (g'), it should be noted that the first subscript
on any of the Inatrix c]etnents indicates the support number and the second
subscript indicates \vhich span is involved. l"hus, K 23 is the stiffness factor
for the left cnd of the third span, mt21 is the fixed-cnd moment at the right end

504 MATRIX METHODS IN STRUCTURAL ANAL VSIS
of the second span, etc. Since there is only one R value for each span, the
order of subscripts in this case is not significant.
The coefficient matrices in Eqs. (.f') and (g') can be formed in the foIlo\\'-
ing manner: Referring to Fig. 10.8, \VC \vrite do\vn under each span the
stiffness matrix (c) for that span. Because \ve have already eIin1inatcd one of
the slope-deflection equations for each of the end spans, \\re have only one
element in each of the corresponding matrices, as sho\vn in the figure. In
\vriting these single-clement matrices for the end spans, \ve also ta)<c account
of the conditions of constraint at the supports 0 and 4. Thus, for the left-end
span, \\'e use the modified stiffness factor l(o as defined in expressions (e),
\vhereas for the right-end span, \ve use the standard sti tfness factor ](34.
()n having these stiff:lcSS n1atrices for the individual spans, the coefficient
matrix in Eq. (.f') is fonned simply by \vriting the upper ro\vs of the 2 X 2
single-span matrices as successive ro\vs of the final matrix, each ro\v staggered
one place to the right \vith respect to the preceding ro\\', and filling in the
blank spaces \vith zeros. 'I'he coefficient matrix in Eq. (g') is formed in
exactly the same nlanner, using the lo\ver ro\vs of the 2 X 2 single-span
matrices in Fig. 10.8.
\\ introduce no\\! the foJIo\ving column-vector notations:
[ M12] [ M 10 ] 8 = [::]
}\.1 ab = M 23 ,\.1 ba = ,\;I 21
i\ II 34 i\lf 3 2 (}a
r mLo ] (11)
[ 12 ]
Jrrah = n23 lrrba = l mL 2 1
ffi134 ffil32
and the stiffncss-tnatrix notations:
[ /\12 R 12 23 ] [ K' 0 J3J
10
Kab =  ](23 K ba = RI [(21 (i)
0 K34 132
With these notations, expressions «(') and (g'), representing Eqs. (f) and
(g), nlay be \vritten in the compact fortn
[1'\.[ abJ - r ](abl [OJ + (lIaJ)]
[ i\tl ba 1 - r leba ] r 0 J + f D1tba )
(10.8)
'fhcsc are the slope-deflection equations for any continuous beam, the supports
of \vhich are restrained against scttlclllcnt. It is to be pointed out that they
are exactly the san1C equations that \VC \\"ould \vrite in follo\ving the procedure
developed in Chap. 9, the only advantage of the matrix formulation being that
of compactness of notation.

ARTICLE 10.4
505
In general, \ve may choose any set of angles of rotation at the intermediate
supports, as defined by an arbitrary column vector 0, and then find, fronl Fqs.
(I O.), the corresponding end mOlnents. By so doing, the t\VO end nlOll1ents
found at each intcnnediare support \\rill not generally be in equilihriun1, and
some externally applied couple \vould be required at each support to hold the
beam in the assun1cd configuration. To find the true set of end moments, \ve
must no\\' iIl1pOSe the further condition that, at each intennediate support, the
end mOInents for each pair of adjacent spans must be equal in Inagnitudc but of
opposite sign. This condition \vill be expresscd by the Inatrix equation
[ill ab ] + [j\11 bu ]  0
(10.9)
Subtituting expressions (10.8) into Eq. (10.9), \ve obtain
[ Kl [01 = - [M I
(10.10)
\vhere
[ KIZ + ](;0
[ [(] - r K ab ] + r](ba! - R Z1
o
R 1Z
1(23 + [\21
R 32
o ]
R Z3
1( 34 + ]( 3 2
(j)
and
[ ;)11.12 + lltO ] [ M 1 ]
rM] = ffi7: 23 + 1121 = M 2
_ ;m 3 4 + mt:: i 2 M3
(k)
M i being the surnlnation of fixed-cnd moments at each intermediate support,
sin1ilar to the idea of unbalanced monlcnt previousI y defined in connection \vi th
n10ment distribution (see page 461).
Equation (10. 10) represents, In t11atrix form, the three-angle equations for a
continuous beam as represented hy Eqs. (9.9), page 414, or Eqs. (9.20), page
432, if \Vc ignore the 0 terms representing rigid body rotations of the individual
spans. The reader nlay verify this by carrying out the indicated n1atrix
multi plication of Eq. (10. 10) .
The K matrix defined by expression (.1) is called the stiffness 1natriJ.: for the
entire continuous heam. For the four-span beam considered here, it is a
three-by-threc square n1atrix, but in general, for a bcanl of II spans on 11 + 1
supports, it \vill be a square matrix of order (11 - 1) X (11 - 1). It ahvays
has tht same fonn, consisting of a principal diagonal \vith a superdiagollal
directly above and a sllhdiagonal directly bcknv, all other elcnlents being zero.
Such a n1atri x is called a continuant '7natrix. 1 l'his sti ffncss matrix for a con-
tinuous heam rnay al\vays be fOffilCd by \vriting the sun1 of the cnd sti tfness
factors at each support as successive eIenlents of the principal diagonal and the
R values for each span as successive elen1cnts of the super- and subdiagonals.
1 See Aitken, op. cit., p. I 26.

506 MATRI X METHODS IN STRUCTURAL ANALYSIS
The solution of Eqs. (10.10) \vill be
e = - [K]-l [M]
(10.11)
\\'here [f( 1- 1 is the inverse of [!{ J. l'hcn, upon substituting (10.1 I) into
Eqs. (10.8), the latter become
[AJabJ - [ab] - [](ab] [1<]-1 fM]
[i\l ha ] - [-rba 1 - [!(ba] [Kl-l[M]
(10.12)
These (natrix fOnTIulas Inay be regarded as representing the final solution of
the problenl; i.t:., they define all end monlents at the intemlcdiate supports of
the beam. Since filIal)] = - [i\.JbaJ, \VC need to use only one set of these
equations in the analysis of any given continuous bean1.
In the case of a continuous beam \vith built-in ends, \ve still have to calculate
the end nlornents A1 01 Cl:1d i\J/I,t'_l from the previously unused slope-deflection
equations for the end spans. These end nlonlcnts are given by the equations
\ 1 I 10
j.l OJ = ;)TI:Ol - K (mlO - i\1 1O )
10
(10.13)
AJ n,n-J = $n,ft-l - K 1i -l,n ()n:T1-1,n - AI[ 71-1, n)
11-1, n
.As a first example of the application of the nlatrix fonnulas (10.12), let us
consider the three-span continuous bea[n sho\vn in Fig. 10. 9a. . fhi s beam
16 kips
1,2001b/ft
(a) 3
k- 2
-!
15'
[ ] [: :] [ j
+ 53.16
(b)
FIG. 10.9
- 13.68

ARTICLE 10.4
507
has a unifonTI cross section, for \vhich \ve take 1 HI = 10 kip-ft 2 , and its t\VO
ends are simply supported. The dilTIensions of the beam and the lateral loads
together \virh the values of k = EI/ I for each span are as sho\\'n in the figure.
\\ begin the analysis by \\'riting under each span the stiffness matrix for
that span. Since the beam is of uniform cross section, expression (t1) applies,
but for the single-element matrices for the end spans, \ve replace k by 3k/4 to
account for the simply supported ends. With these matrices defined, \ve next
find froIn expressions (i) that
Kab = [  ]
and
K ba = [;  ]
(/)
l'hen, from expression (j), the sti ffness matrix for the entire beam becomes
}-' = [ 7 2 ]
\ 2 6
(111)
"10 invert this nlatrix (see page 488), \ve first \vrite the adjugatc of [(:
adi K = [_ -]
and calculate the determinant of ](:
(n)
IKI = 7
2
2
= 42 - 4 = 38
6
(0)
'"hen, the reciprocal of K becomes
[Kl- 1 = ad; [( =  [ 6 _ 7 2 ] (p)
I KI 3 R - 2
Using 1:1ble 9. 1, page 412, the fixed-end nloments are fi)und to be
mLo = + 5 n12 = - 10 m21 = + 10 113 = -15
all values being in kip-foot units. 'rhen, the sevcral column matrices appearing
in Eqs. (10.12) are
llab = [ -10 ]
-15
and Eqs. (10.12 becon1e
[ ,\I[ 12 J [ - 1 0 J [ 4 2 J 1 [ 6
i\1 23 = -15 - 0 2 38 - 2
[ +5 J
mIba = + 1 0
M = [=n
-J [ =n
[ '\;/10 ] [ 5 J [ 3 0 ] 1 [ 6 - 2 ] [ - 5 ]
1\1 21 = 10 - 2 4 38 -2 7 -5
(q)
I Since the results \vill not depend on the actual value of EI, \ve choose EI = 10 so that the
stiffness values \vill be simple ,vhole numbers, thus simplifying the nlatrix [nultiplications.

loa MATRIX METHODS IN STRUCTURAL ANALYSIS
FEIW - 56.9 210.2 - 570 +570 0 o FEIJI
Mli'Ebl 0 281.8 - 570 +570 0 o blFEbl
0 1 2 3
k k k
2
kaa ' 6.45 22.83 30.58 kaa
kab 8.12 17.89 8.12 k ba
kbb 30.58 22.83 6.45 kbb
[ 2036 ] [11.42 8.95 ] [ 2036 ]
8.95 11.42
FIG. 10.10
To evaluate these expressions, \ve first calculate the matrix product:
Then
[ J[] [ ] [J
I 6 - 2 - 5 I - 20 5 - 4
38 -2 7 -5 = 38 -25 = 38 -5
2 ] 5 [ -4 J 5 [ -26 J [ -3.42 J
2 38 -5 = 38 -]0 = -1.32
O J 5 r -4 J 5 [ - 12 J r - 1.58 ]
4 38 - 5 :;:r 38 - 28 = - 3.68
r
u
and
so that expressions (q) beCOIne
r Ai[ 12 J _
L At[ 23
[=:J [=:J - [-::]
- [+J - [=::J = L::J
[ M 1 0 ]
/\1 21
rrhcse arc the final results for the end moments, and the fact that \ve obtain
identical results, except for signs, from the t\VO sets of equations serves as a
partial check on the calculations. ] 'he corresponding bending-n10ment dia-
grarn for the beam is sho\vn in Fig. 10.9b.
l\S a second example, let us consider a matrix analysis of the three-span
beam of variable cross section sho\vn in Fig. 9.22, page 434. As a preliminary
step, "rc record, as sho\vn in Fig, 10.10, the fixed-end rnomcnts and the stiffness
factors kaa, kab, kbb for each span. 1 \\ have also recorded, under each span,
the value of £/ 0 /1 for that span, using the notation k = Elo/36.
Since the ends of the beam arc sirnply supported in this case, \VC n1llst next
calculate modified t1xcd-end mOlncnts and modified stiffness factors for the t\va
I These data are taken fron: page 4-34, where the problem was previously discussed.

ARTICLE 10.4
509
cnd pans. Using Eqs. (e) for this purpose, \ve have
, k Ol 2 8.12 ( )
Jt 10 = n.l 0 - k 0 0 :mo 1 = 210. - 6.45 - 5 6.9 = + 28 1 .8
The full set of modified fixed-cnd moments is recorded in Fig. 10.10 directly
above the beam. Similarly, for nl0dificd stiffness factors in the first and last
spans, \ve have
k' = k' = kooku - k 01 2 = (6.45) (30.58) - (8.12) 2 = 20 36
10 23 koo 6.45 ·
l'hen using expression (b) and suppressing the scalar factor k = Elo/36, the
sti ffr.css nlatrices for the individual spans are formed as sho\vn in Fig. 10.10.
\'/e are no\\' ready to begin the analysis. Using the data from Fig. 10.10
in Eqs. (i), \ve first find
Kab = [I.42
8.95 ]
20.36
[( = [ 20.36
ba 8 .95
I .42 ]
Then, using Eg. (j) and the rules for inversion (see page 488), \ve have
]( = [ 3 1. 7 8
8.95
8.95 ]
31.78
K-l - .  [ 31.7R
- 929.9 -8.95
-R.95 J
31.78
Finally, the first set of nlatrix formulas (10.12) becoITIes
[ A112 J = [ -570 ] _ r 11.42 8.95 ]
/\-1 23 0 0 20.36
1 [ 31.7R -8. 95 1 [ -288.2 ] [ -594 ]
X 9 29.9 -8.95 31.78 +570.0 = -453
These values of the end moments are seen to agree \vith those obtained on page
435. In the examples discussed above, the number of unkno\vns is snlall, and
there is no particular advantage in using the matrix nlethod. I-Io\vevcr, in the
case of a continuous beam having many spans, the nlcthod may be helpful
particularly if the electronic computer is to he used for the numerical solution
of the equations.
PROBLEMS
Repeat the analysis of the continuous beaIn of uniform cross section shown in
Fig. 10.9 if the ends of the beanl arc built in instead of simply support cd. .Li\.l1 other
data rcrnain unchanged,
2 Repeat the analysis of the three-span hcan1 sho\\'n in Fig. 9.22 if the middlc span
has a length of 36ft instead of 72ft. ,All other data rernain the saIne.
A11.f. AI 12 = - 115.9 kip-ft, /\-1 23 = - 113 .8 kip-ft.

510 MATRIX METHODS IN STRUCTURAL ANAL VSIS
10.5 MATRIX TREATMENT OF ARCHES AND FRAMES
'"fo sho\v some further aspects of Inat:"ix methods in structural analysis, let us
consider the problcI11 cf an unsymmetrical hingcless arch rib ,4 B as sho\vn in
Fig. IO.ILl. Taking the reactions at A as the redundant quantities and using
the theorern of least \vork, \\'C obtain, for calculating H(H Ra, AJ a, the system
of equations (a) and (/;) already given on page 372. Substituting expressions
(b) into E<]s. (a) and neglecting the terms containing the axial force 1.\t, \ve
obtain
Il / - r 8 ), . ds + R ( s :x y ds + H r s )'2 ds I (8 ;\..[I . r d! = 0
. (I JoEl a J 0 RI a ) 0 £1 I JoEl
;\.. 1 - r If . ds + R { ., ,2 _£!s + 1 - 1 r . " :ry ds + r s !\t{ I.'l' ds = 0
.. (I J 0 h'l a J 0 £1 a Jot'! J 0 t'j
r s d s r s X d s {s V d s [ 8 AI 'ds
Ala Jo ill + la Jo . f1. + f/(I Jo 'f;/ + Jo - EI = 0
(10.14)
l'his systenl of equations is not suitable for the practicaJ calculation of the
unkno\\'ns H q , 1(Jt i\-1 a , and, as already pointed out in A.rt. 8.6, \VC obtain a
great sin1plification by taking, instead of 1-/(1, R(l' ,\la, the statically equivalent
system of redundant reactions }{ 0, l, i\J 0, acting at the clastic center 0 of the
arch and directed along its principal axes as sho\vn in Fig. 10.11 c. By so doing,
\ve obtain, instead of the systcnl of equations (10.14), a systcrn of three
independent <:.quations, each of \vhich contains only one unkno\vn, Such a
transfortnation of Eqs. (10.14) can be made by proper manipulation of the
associated I11atrices, and this procedure \vill no\\' he explained.
A
y
A A.. 0+/.
x
----------
x y
/
Y Y ds
B
FIG. 10.11 (d)

ARTICLE 10.5
511
We begin \vith the foJIo\ving notations:
Q = ( 3 Lt!
% Jo E/
I = ( 8 r 2 d
% Jo E/
 = (3 xds
10 EI
I = (S: ds
y 10 EI
A = 10' ;;
I = (tJ :t:y d s
ZlI Jo £1
(a)
If, as sho\vn in Fig. 10. 11d, \ve give the arch axis a fictitious variable thickness
1/ EI, Vle see that the integrals (a) represent the statical moments, thc area, and
the moments and product of inertia of this shaded area with respect to the
chosen coordinate axes x and )' through the end A of the arch axis. Using
notations (0), Eqs. (10.14) become
/Jla + {"uRa + Q"M. = - 10' M'y ;:
/"u Ha + IuRa + QyMa = - 10' MIX -fA
Q"H. + Qu R . + AM. = - 10' M ' }
( 1 O. 1 4a)
. . .
Of, In matrIX notation,
[ /z 1 zu Q% ] [ Ha ]
1%1} 1u Qu Ra-
Qz QlI A Ma
1 M'y ds
EI
1 [/ s
1 M' ds
EI
(lO.14b)
No,,,, Jet us consider, in these equations, the matrix
[ l:z; IZ1J A ,, ]
F = /:z;y /1} 
Qz Qv
(b)
which is called the gC0111etry matrix for the arch. We see that it is entirely
independent of the loading on the arch and depends only on the shape and
proportions of the rib, i.e., only on the geometry of the arch. It is called a
sY/1lfl1etric 111atrix because its clements arc symmetrically arranged \vith respect
to the principal diagonal.
If, instead of axes x and )' through point A, \\'c take a parallel set of axes Xl
and Yl through the elastic center 0 of the arch, i.e., through the centroid of the
fictitious shaded area in Fig. 10. lId, the statical moments Q:r;l and Q111 \viII

512 MATRIX METHODS IN STRUCTURAL ANAL VSIS
vanish, and the corresponding geon1etry matrix I'l \vill have the simpler fann
[ lx,
F 1 = /7{'
I Z'tli 0 ]
IUt 0
o A
(c)
This nc\v matrix f\ can be obtained directly from the matrix F by introducing
a trallfOrI11ltion J11atrix l' and its transpose ']1/ as folIo\vs:
1 0 Qz 1 0 0
- A
T= 0 1 (b 1'" = 0 0
- A
0 0 1 Qz 
- 4 - A 1
Then f't = T FI"
(d)
(10.15)
To demonstrate this, \ve notc first from the parallel-axis theorem for
IT10nlents and products of inertia that
roe = Iz - Ab 2
I
1 111 = /11 - Aa 2
rZfl1 = I zu - l-lab
(e)
\vhere the coordinates of the elastic center 0 are
a = Qv
A
h = Qx
A
(f)
Substituting these values of t1 and b into Eqs, (e), \ve obtain
QZ2
lr = I. - 
I - A
Q'l
I VI = IJ1 - A
{.,u, = lzu - gQy (e')
No\v, pcrfomling the n1atrix fnultiplications indicated by Eq. (10.15), \ve
obtain
1 0 - Qz tz I zu Qz 0 0
A
F 1 = 0  Ixu 11) Qv 0 1 0
- A
0 0 1 Qz QlI A Qz Qv A
- A -If
I _ Qx 2 I - Q 0
% A %11 A
- I - Qz I- QJ! 0
%11 A 1) ,,4
0 0 A
the elements of \vhich are seen to agree \vith expressions (c').

ARTICLE 10.6
513
Referring no\\' to Fig. 10.11 c, let us rotate the axes Xl, Yl into the positions
Xo, Yo coinciding \vith principal directions for the fictitious shaded area in Fig.
10.11 d. 'l'hen, the product of inertia IroI/O \vill also vanish and the correspond-
ing geometry n1atrix becorTIcs
[ l:to 0 0 ]
I/o = 0 I Yo 0
o 0 A
(g)
"his I11atrix again can he obtained fronl the preceding rnatrix 1/ 1 by the use of a
ttaus! OrlJlill iOJJ 'J)J(tlri.r P and j ts transpose P' as fO]]0\\I5:
[ cos a
p = si {J
-sin {3 0 ]
cos {3 0
o 1
[ cos {3 sin {3 0 0 1 ]
p' = -n {3 CO {3
(h)
The cletnents of this matrix arc sinlply the direction cosines for the ne\v set of
axes. l'hen, as before,
I/o = P f'lP'
(10.16)
l"his transformation holds for any value of the angle tj, but to make the product
of inertia JrotJ'J vanish, \ve must take (see page 376)
tan 2{3 = 2I:l:1I !1_
III - I r ,
(i)
\.'"hen both tranSfC)rnlations indicated by Eqs. (10.15) and (10.16) have been
olade, the equations for calculating the redundant reactions LY O , 1"'0, i\J 0 in Fig.
J 0, J 1 C oecoIne
fro 0 0 "'\0 J M' j_d!
0 lvo 0 Yo J M'xu ds ( 1 O.14c)
- EI
0 ° lJ. i\/f 0 J 1' ds
EI
'10 illustrate the application of Fq (] O.14c), let us consider the rectangular
fraIne supported and loaded as sho\vn in Fig. IO.12a. The dimensions of this
fraITIC and the cross-sectional moments of inertia of its n1cn1bers are sho\vn in
the figure, and for sin1plicity \Vc assume Elo = 1 kip-fez. rhus, the fictitious
area corresponding to the center line of the frame \vill have the \vidth 1/ El
as sho\vn in Fig. 10. I 2b. laking axes x and)' through point A as shO\\Tn, the

514 MATRIX METHODS IN STRUCTURAL ANAL VSIS
- 12'
cl If)
T-
r 10
6' 10
LA
p
1
I
2/0 12'
81
(a)
(b)
c
c
A
Yo
FIG. 10.12
(c)
(d)
-i-12P
geometrical quantities defined by Eqs. (a) become
Q;r. = - 6 X 3 - 1 2 X 6 + 0 = - 90
Q;, = 0 + 12 X 6 + 6 X 12 = 144
A = 6 + 12 + 6 = 24
Ir = 6 X 6; + 12 X 6 2 + 6 X \2 = 576
I" = 0 + 12 X 1: 2 + 6 X 12 2 = 1440
rry = 0 - 12 X 6 X 6 + 0 = -432
(j)
Then, from Eqs. Cf), the coordinates of the elastic center 0 arc
144
a = -24 == 6.00
90
I , = - --- = -  7 
24 -. -
(k)
L\ssembling the quantities (j) in accordance \vith exprcssion (b), the gcon1etry
matrix for the franlc becornes
[ 576
[/ = -432
-90
-432
1 ,440
144
-90 ]
144
24
(/)

ARTICLE 10.5
515
Similarly, using the coordinates a and b froln Eqs. (k), the transformation
matrix 'r.., as defined by the first of expressions (d), becomes
T = [
o 3.75 ]
1 -6
o 1
(111 )
l'hen, for axes .\'1, )'1 through the clastic center 0 in Fig. 10.12e, the geon1ctry
Inatrix }/1 is found fron1 Eq. (10.15) to be
}/J
= [ ° 3.75] [ 576 -432 - 90] [] ° 7]
I -0 -432 I ,440 144 0 I
0 ] -90 144 24 3.75 -6
[ 238.5 108 2]
= 108 576
0 0
No\v, froIl1 Eq. (i)
2 X 108
tan 213 = 576 _ 238.5 = 0.640
and \ve find
6 = J 6°18'
cas (3 = 0.960
sin 13 = 0.281
\.\lith these values, the second transfoflna6on rnatrix P, as defined by the first
of expressions (h), hccon1cs
[ 0.960
P = .281
-0.281
0.960
o
7]
(11)
. fhen, frol11 Eq. (10.16) the gconlctry Inatrix for the principal axes Xo, Yo as
sho\vn in Fig. 10. I 2c is found to be
f'o
[ 0. Q 60
= O.2H I
o
-0.281
o . <) 60
o
0 ][ 23R.5
o 108
1 0
1 0 H 0 ] [ 0.960 0.28 1 0 ]
576 0 -0.281 0.960 0
o 24 0 0 I
= [ 35.3 60.0  ]
o 0 24
Having this geomctry matrix for the principal axes of the fraIlle, \VC are no\v
ready to consider the load matrix appearing on the right-hand side of Eq.

516 MATRIX METHODS IN STRUCTURAL ANALYSIS
(1 O.14c). To \vrite this Inatrix, \\!e consider the bcnding-n10ment diagram ,\;['
as sho\vn in Fig. lO.12d. 'fhc area of this diagram, divided by E/, is
v :::% ! B }J', ds = ! B i\;1' d = _ 36P
A EI D 2Elo
and its centroid is at J'V having the coordinates
Xu = Xl COS {3 + )'1 sin {3 = 6 X 0.960 + 5.75 X 0.281 - 7.38
)'0 = Yl cas {3 - .\'1 sin (3 = 5.75 X 0.960 - 6 X 0.281 - 3.R4
Hence, the load matrix beconlcs
f Al')' ds
.-El
f / !' Xo ds
E!
f 1\1' ds
F:/
[ 3 t 84 ]
- 3 6P  . 38
(0)
No\v, inverting the above matrix F 0, the solution of Eq. (1 O.14c) becomes
[ \ . T 0 ] 36P [ 2.82
Yo = - .
1000
i\J o '
I .64 : ] [  :: ] = / go [  :   ]
41. 7 1 ' 41 .70
that is,
..Y u = O.390P kips
Jf 0 = 0,43 7 P kips
/\1 0 = 1.50P ft-kips
10.6 MATRIX ANALYSIS OF CONTINUOUS FRAMES
Continuous frame structures such as building franles are Ii kely to be highly
statically indeterminate so that in their analysis \ve have to deal \vith a large
nunlber of unkno\vns. 'The only practicable \vay of solving such problems is
to have recourse to the electronic digital conlputer, and for this purpose a
matrix fonnulation of the problcIn is the BloSt advantageous. To illustrate a
matrix method for such problcIllS, \VC shall consider here a t\vo-story building
fran1e as sho\vn in Fig. 10,13. On the one hand, this frarne \vill not involve
so many unkno\vns as to Blake the discussion un\vieldy, yet, on the other hand,
it \vill be extensive enough to permit us to illustrate all the steps that \\'ould be
required in the analysis of a n1uch larger structure.
For simplicity, \ve assume that each Incmber has the same length I and the
S3Il1e flexural rigidity £1 so that the stiffness factors arc all equal, that is, k = EI/ I
is the same for all melnbcrs. As is usual practice, \VC also neglect the deforma-

ARTICLE 10.6 517
ar- '---t a Q
1:, 1 2
I
b
P b 3 4
I
FIG. 10.13 1
tions caused by axial forces and by shearing forces in the members and consider
only bending deformation. Undcr these assumptions, the deformation of the
frallle under load \vill be completcly defined by a set of six displacements:
namely, the horizontal displacements Oa and Ob of the two floors and the angles
of rotation 81, 82, 8 3 , 8 4 of the four rigid joints. When these six displacements
have been found, all cnd moments can be calculated fronl the slope-deflection
equations (9.6), page 411, and the problem is solved. We therefore introduce
the column vector}
[OJ] = {Oa Ob ()l (J2 0 3 ()4}
(a)
and select this set of displacements as the unkno\\'ns of the problem.
As a preliminary step to the calculation of the displacements (a), we first
consider the t\VO simple problems illustrated in Fig. 10.14. In Fig. lO.14a, \VC
give to the end A of a prismatic bcatn AB \vith built-in ends a displacement 0,
\vithout alIo\ving any rotation of the tangent at A or any movement at all of
the end B. Then, the reactions at A and B can easily be calculated by using
the slope-deflection equations (9.6) on page 411, and \ve find
ROb = I 2k 0
/2
6k
kfab = T 0
I2k
R ba = J2 0
6k
Alba = T 0
(b)
In Fig. IO.14b, the end A of the same beam is given an angle of rotation 8
\vithout allowing any lateral deflection of A or any movcn1ent at all of the end
B. Then, again using the sIopc-deflection equations (9.6) on page 411, \ve find
6k
Rab = - 8
/
j\1ab = 4kO
6k
R ba = T 8
Nf ba = 2kO
(b ' )
1 To save space, a colurnn vector is often \vritten in one horizontal line enclosed by braces
instead of brackets to indicate that it is not a ro\v vector.

518 MATRIX METHODS IN STRUCTURAL ANALYSIS
E/ =k
I
R ba
JJI !:ID
1
EI -k R ba
8 I 
I JU m:: I
12k 12k
1 2 1 2
(a) Ok 6k
1 I
FIG. 10,14
6k 6k
I I
(b) 4k 2k
The coefficients appearing in front of (, and 8 in Fqs. (b) and (b ' ) are seen to
represent the reactions, cr forces of constraint, at the ends of the beam \vhen the
displacements () and {) arc each equal to unity. "l'hese quantities are called the
.'itfflless influence coefficients fur the beam corresponding to each type of displace-
n1cnt. For convenience of easy reference, these stiffness coefficients are
recorded in matrix fonn under each beam in Fig. 10.14.
No\\r, let us return to the frarllc in Fig. 10.1 3, remove all applied loads, and
lock all joints against both translatiun and rotation. lhis done, \VC ren10ve
just one constraint corresponding to anyone of the six degrees of frcedom of
the systen1, say constraint j, and make there a unit displacement OJ = 1. *
This \vill result in some deformation of the structure consistent \vith the
renlaining artificial constraints, and \ve proceed to calculate the reaction cor-
responding to each of the six degrees of freedom. rhat is, \ve calculate the
system of external forces and couples necessary to hold the structure in the
assumed configuration defined by OJ = 1. In the generalized sense, \ve denote
such an external reaction at i by Saj regardless of \vhether it is a force or a
couple. Thus, we define the stiff1less influence coefficieNt Sa; as the external
reaction at i due to an irnposed unit displacement at j \vhen all other displace-
ments are held equal to zero. In our example there \vill be 36 of these stiffness
influence coefficients, and \ve no\\' set about their calculation, (llaking use of
the single-n1embcr stiffness coefficients sho\vn in Fig. 10.14.
Let us begin in Fig. lO.15a \vith a unit displacement oa = 1, that is, a unit
lateral displacement of the top floor, all other displacements being held equal
to zero. '[hen, the external forces required to hold the structure in this con-
figuration \viII act as sho\vn in the figure, 1 and their n1agnitudes \vill be as listed
beside the structure. In these calculations, \ve consider linear displacen1ents
tit l'his rnay be either a linear displacement or an angular displacement; i.e., \ve use the
symbol 8j in the sense of a gencralized displacement (see page 226).
1 Since the nlelnbers of the frarne are treated as inextensible, all of the horizontal force of
constraint is assumed to act (In the left-hand side of the frame.

ARTICLE 10.6
519
and forces positive to the right, negative to the lcft, and angular displacements
and couples positive when clock\visc, negative \\,hen countcrclock\vise. Con-
sider, for exanlple, the calculation of SbCJ' From Fig. IO.14a, we see that the
reaction at the bottom of each upper-story column in Fig. lO.15a is 12k/J2
acting to the left and that there are t\VO such columns; hence, Sba = - 24k/ /2
as sho\vn. Consider, again, the calculation of S4(J. From Fig. IO.14a, \\'e see
. that the reactive .moment at the bottom of the column 42 in Fig. IO.15a is
counterclock\vise and of magnitude 6k/ / and that there is only one column;
8 = 24k 24k
8ab 8,,---
aa 1 2 o ,2
24k S = 48k
8"a= -12 "" 1 2
6k 6k
8 10 =-T S"" SIb - T
6k 6k
S ::1-- S2b - T
2G 1
6k S3b = 0
S30 = - T
s = -  S4b = 0
40 I
(a) (b)
6k 6k
8 1 =-- 8 a2 8 ---
o I 02 I
6k 6k
8"l = --i- S b2 = ,-
Sbl 8 11 = 8k S2 S12 - 2k
8 21 - 2k 8 22 = 8k
8 31 - 2k 8 32 = 0
S.n IS 0 8"2 = 2k
(c) (d)
6k 6k
8a3 S =-- So.. 8 ---
03 1 04 I
8 M = 0 8 tH = 0
8 13 = 2k SI.. = 0
8 23 = 0 8 24 = 2k
8 33 - 12k S34 - 2k
(e) 8.. 3 = 2k (f) 8... = 12k
FIG. 10.15

520 MATRIX METHODS IN STRUCTURAL ANALYSIS
hence, S4a = - 6k/ /. ll1e reader should check the other values of Sij for
himself.
Next, in Fig. 10.15 b, \ve Inake a unit displacement b = 1, that is, a unit
horizontal disp]acerncnt of the n1iddle floor, holding all other displacements
equal to zero, Then, as before, using the stiffness coefficients fronl Fig.
lO.14a, \ve find the external reactions 8 ij as sho\\'o in the figure. I)eforn1ation
patterns corresponding to (h = 1, ()2 = 1, 0 3 = 1, (}.. = 1 and the correspond-
ing induced external forces are sho\vn in Fig. 10.15 c, £1, e, f, This corl1pletes
the calculation of the influence coefficients for the frame.
\\. no\\" assemble these stiffness coefficients in the form of a square rnatrix,
called the stijf17eJJ 1I1atrix for the structure. It is \vritten in the order a, b, I, 2,
3, 4, both as to r()\vs and as to colurnns, and becornes
24k 24k 6k ok 6k 6k
- ..- -', -7
/2 /2 / I
24k 48k 6k 6k 0 0
--
/2 /2 I J
6k 6k 8k 2k 2k 0
-,- I
Sij = (c)
ok 6k
I I 2k Hk 0 2k
6k a 2k 0 I2k 2k
I
6k 0 0 2k 2k I2k
/
It \vill be noted that this is a !;ynrlnetric 1I/41tri:r and that such sYlnmctry is to he
expected from the reciprocal theorem.
Having the stiffness influence coefficients as sho\vn in the nlatrix (c) ahove,
\ve nlay no\\' use the principle of superposition to calculate the external forces
required to hold the franlc in any configuration defined by an arbitrary set of
values of the displacen1cnts j. For example, the requi red external force at a
\\Tauld be
Fa :-= SaaDa + Sa1.JOb + SalEh + Sa2()2 + S a3 03 + Sa4 8 ..
The requi red external n10ment at 1 \vould be
i\-1 1 = Slaa + Slbb + S1101 + S1 2 82 -1- S 13 03 + SuO"
etc. Ho\vever, \VC are :ooking for a set of values of the displacements that
\vill result from the particular systen1 of external forces sho\\'n in Fig. 10.13,
i.e., the actual loading on the structure. Thus the true set of displacen1ents

ARTICLE 10.6
521
is defined by the SYStC01 of algebraic equations
SaaOa + SabOb + Sel181 + Su282 + Sa3 8 3 + Sa84 = Po
SbaOa + SbbOb + SbI(JI + Sb z 8 z + Sb3(J3 + Sb4 8 4 = P b
SlaOa + S1I)Ob + S118 1 + S12(J2 + 5 13 0 3 + SuO. = 0
S'2aOa + S2bOh + S 21 8 t + 8 22 8']. + S 23 03 + 8 21 .0 4 = Qa
53aa + 53bb + 5 31 8 1 + 5 32 8 2 + 5aa83 + 5 34 8 4 = i;2
54aa + 54bb + 5 41 8 1 + 5 42 8 2 + 5 43 8 3 + 5 44 8 4 = - i
(10.17)
\vhere qJ2/12 and - q/'J-/ t 2 represent, \vith reversed signs, the fixed-cnd
moments for the girder 34, that is, the unbalanced 7110111fntJ at the joints 3 and 4,
respectively. Introducing the matrix notation
{ ql2
("F i ] = Pa P b 0 Qa IT
_ q12 }
12
(d)
\\'hich is called the load 1J1atriJ:, Eqs. (10. 17), in matrix notation, become
r Sii] r OJ] = rf.]
(IO.17a)
"[he three matrices appearing in his equation are defined by expressions (c),
(tT), and (d), respectively.
Having Eq. (IO.17a), the displacen1ents are found froo1 the equation
[0;1 = [Siil- I [FiJ
(10.17b)
We note that this solution requires the inversion of the stiffness matrix [Sij]
[see expression (c)], and it is here that \ve need the help of the computer.
To illustrate the solution of Eqs. (10. I 7), let us no\\' introduce the follo\ving
numerical data:
I = 10 ft
EI .
k = -___. = 5 000 lo p- ft
I '
Pa = 5 kips
P b = 5 ki ps
Q = 2 kips
a = 2 ft
q l2
- = 8 kip-ft
12
(e)
\\/ith these data, the sti ffness 111atrix (f) beCOITICS
12 -12 - 30 -30 -30 -30
-12 24 30 30 0 0
rSijl 100 -30 30 400 100 100 0 (f)
-
-30 30 100 400 0 100
-30 a 100 0 600 100
-30 a 0 100 100 600

522 MATRIX METHODS IN STRUCTURAL ANALYSIS
and the load matrix (d) becomes
r F i ] = {5
5
o
4
8 -8}
(g)
Then, from Eq. (10.17 b), \ve have 1
45a 34.848 15.152 0.909
45b 15.152 11.515 0.091
01 I 0.909 0.091 0.324-
-
8 2 10,000 0.909 0.09 J -0.033
8 3 1. 364 0.636 -0.018
8 1 . 3 64 0,636 0,054
253.6
133.7
1 4.296
=--
10,000 6.868
11.93
8.216
0.909
0.091
-0.033
0.324
0.054
-0.018
1 . 3 64
0.636
-0.018
0.054
0.234
0.020
1 . 3 64
0.636
0.054
-0.018
0.020
0.234
5
5
o
4
8
-8
The deflections 0 0 and Ob are in fcet, \vhereas the rotations 01, (}2, 8 3 , fJ 4 are
in radians. lhus, 6 0 = 0.02536 ft, Ob = 0.01337 ft, lh = 0.0004296, fJ? =
0.0006868, 8 3 = 0.001193, ()" = 0.0008216.
As already pointed out, once \\'e have these displacements, all end moments
in the structure can be readily calculated from the slope-deflection equations
(9.6), and the analysis is complete. It will also be noted that the analysis can
be made for as many sets of extcmalloads as dcsi red simply by introducing a
rectangular load matrix (d) \vith a column for each set of loads. The stiffness
matrix (c) is independent of ho\v the structure is loaded. In this example, \ve
see that the matrix method is superior to any of the methods discussed in Chap.
9, provided that \ve can have the use of an electronic computer to invert the
sti ffness matrix (c).
1 The inversion of the stiffness matrix (f) has been made on the electronic computer.

Chapter 11
Suspension bridges'
11.1 PARABOLIC FUNICULAR CURVE
-\SSl1me that a llniforn1 and perfectly flexible cable fixed at points /1 and R
(Fig. 11.1) is submitted to the action of distributed vertical load. Then the
ordinate y of any point (' of the cable is obtained from the equation of Inoments
\vith respect to (' of forces to the left of this point, \vhieh gives
mz + H  x - Hy = 0
(a)
In this equation 9JtI: denotes the bending moment at the cross section 1111] of a
sirnply supported bcan1 of span I and carrying the load acting on the cable.
H is the horizontal con1ponent of the tensile force in the cable, and h is the
di tferencc in elevation of the ends of the cable. In the particular case \vhen
1 Originally published by S. 1in1o.shenko, Frtlllklill but., vol. 235, no. 3, Jviarch, nd no. 4,
April. 1943.
523

524 SUSPENSION BRIDGES
II
x
B
I
--
2
FIG. 11.1 y
the load of intensity w is unifonnly distributed along the horizontal projection
of the cable \ve have
Wl'
Wt = -t (I - x)
and Eq. (a) gives
wx h
y = 2H (I - x) + 7 x
(11.1)
\vhich sho\vs that the funicular cur\re in this case is a parabola \\,irh vertical axis.
If the ends of the cable arc on the same level, \ve obtain
wr
y = 2H (I - x)
(11. 2)
.l.t\pplying this equation to the Inid-point of the cable, \vhcrc the ordinate of the
funicular curve represents the sag f, \ve obtain
w/ 2
f = - Sfl
or
H = w/ 2
HI
(11.3)
These equations hold also in the more general case sho\vn in Fig. 11.1 iff is
measured fron1 the n1iddle of the line AB joining the ends of the cable. In
our further discussion the length s of the funicular curve \vill be required. It is
obtained froIn the equation
s = fOl (1 + y'2)1 dx
Developing the expression under the integral sign into a series and substituting
expression (11.2) for )', \ve obtain
=/ ( 1 + _ .f2 _ }2f4 + 256 f _" . )
s 3 /2 5 /,1 7 /6
In the case of flat parabolic curves, say f / / < T\r, \ve can take only the first

ART. CLE 11.2
525
t\VO tcrtns of the series and llse the approximate formula
( 8 /2 )
S = / 1 + 3" /2
([ 1.4)
To establish the rclation bct\\'cen the change in length of the curve and the
change in its sag, \ve di fferentiatc Eq. (11.4), \vhich gives
and
16f
s = .-- -. Af
3 /
3 I
ttf = - - - s
16/
(11.5)
(11.6)
To find the change D..f due to a rise in temperature of to, \\'e substitute
s = EtJ into Eq. (11.6) and obtain
3 /2 ( 8 /2 )
df = "16 f Et 1 + 3" J2
(I I .7)
l'hc clastic elongation of the cable is obtained from the equation
- f 8  ds d - f l  ( 12 ) d ,
D..S - A E d . S - _ A E 1 + y .t
o Col c .t 0 c J C
in \vhich Ac is the cross-sectional area of the cable and Ec its modulus of
elasticity. Substituting Eq. (11.2)' for y and integrating, \ve obtain
HI ( 16f2 )
ds = AcEc 1 + 3 -/2
(11 .8)
The corresponding change in sag, fronl Eq. (11.6), is 1
_ 3 H/2 ( 16/2 )
Af - 16 AcEc/ 1 + 3 [2
(11.9)
11.2 DEFLECTIONS OF UNSTIFFENED SUSPENSION BRIDGES
In the case of a suspension bridge of large span, the dead load unifornlly dis-
tributed on a horizontal plane is usually nlany times larger than that uniformly
distributed along the cables, and \\'e can assume \'lith sufficient accuracy that
the curve of the cable under the action of dead load is a parabola. Let us
consider no\\' deflections in the cable produced by live load. ,A.s a first
example \\'c consider the symnletrical case in \vhich the load of intensity p is
unifonnly distributed along the distance 2a of the span (Fig. 11.2). The solid
1 In the derivation of Eqs. (I 1.7) and (11.9) it is assumed that the change df in sag is small,
and its effect on the horizontal tension H is neglected.

526 SUSPENSION BRIDGES
H
H
FIG.11.2
==3
line indicates the shape of the cable under the action of dead load w only.
Let f wand Hie denote the corresponding values of the sag of the cable and of
the horizontal component of the tensile force in the cable. l'he length of the
cable then is
( 8 f w2 ) ( 1 W 2 / 2 )
S = / 1 + 3 72 = / 1 + 24 Hw2
(a)
After application of live load p, the shape of the cable \\rill be as sho\vn by the
dashed line ACI). It consists of t\VO parabolic curves A(' and CD having a
common tangent at C. The curve CD carries the load of intensity w + p,
and the curve AC' carries the load of intensity w. The distance 1 1 /2 of the
vertex 0 of this latter curve from the vertical through A \\,ill be found from
the condition that the total load along the portion CO of the curve ACO is the
same as the total load along the curve CD. IIence,
/1 I P
-=-+a--
2 2 w
(b)
We denote by J and H the sag of the cable and the tensile force in the cablc at
D after application of the live load. One relation bet\vecn these t\VO quantities
is obtained by making the equation of moments ",ith respect to point D, \vhich
gIves
_ 1 [ W/ 2 pa ]
f - ---- - + - (/ - a)
H 8 2
(c)
The second equation is obtained from the condition that the length of the cablc
remains unchanged! during application of the live load. Using for the length
of the curves AO, CO, and C/) the approximate equation (a), the condition of
1 The srnall influence on the deflection due to clastic elongation of the cable \vill be discussed
later.

ARTICLE 11.2
527
incxtensibility of the cable is
! ( 1 +  fw2 ) = ! ( 1 +  '2/12 )
2 3 /2 2 24Hz
a(w + p) [ a2('t!) + P)2 ] [ a2(w+ p) 2 J
- - - 1 + - ---' + a 1 +
'U' 6Hz 6H2
(d)
Introducing notations
 = 11
'tV
(e)
2a
- = z
/
(1)
\ve ohtain fronl Eq. (d)
H = HID VI+ 3112 + 3112Z2 - (211 2 + 11)23
(11.10)
Substituting this value of H into Eq. (c), \\'C obtain
f = f 1_+ 11(22 - __
. W VI +- 3nz + 317 2 22 - (2n 2 + 11)23
(11.11)
To find \vhat portion of the span fnust be loaded to produce the InaxilTIum
deflection at the middle of the bridge, \ve put equal to zero the derivative of
expression (II .11) \vith respect to z, \vhich gives the equation
(211 2 + 1l)Z4 - 2"11(11- l)z3 - 3(17 - l)z2 - 4z + 1 = 0
Solving this equation for several nurnerical values of 11, \VC obtain for z the
values given in Table 11.1. Substituting these values of z into Eq. (11.11), \ve
find the values of the sag.f of the cable after application of the live load. 'I'he
calculated ratios of the change in sag to the initial sag are given in the second
line of lble 11.1. Tn the case of long-span bridges, the ratio 11,/ "ll) is usually
TABLE 11.1
11
---- ----- -.---- - -
0 0.10 0.25 0.50 1 . OJ
z 0.333 0.322 0.306 0.289 0.253
f -Jut 0 0.0069 0.0151 0.0281 0.0456
fit;
Ii
H,v 1.047 1 . 112 1 .21 3 1 . 379

528 SUSPENSION BRIDGES
small,l say smaller than I: 4, and it may be seen from Table I 1.1 that the
deflection at the middle is of the order of one-hundredth of the sag !w or of
onc-thousandth of the span I of the bridge. Such deflections can be considered
as sufficiently small to make the use of any stiffening truss unnecessary.
To calculate the deflection at the middle due to clastic deformation of the
cable produced by live load, \ve use the approximatc formula (11.9) in which
H - Hw, instead of H, must be substituted. Then, the deflection due to the
elongation of the cable is
l:1! =  H t -'- ( .l!_ _ 1 ) ( 1 +  /2 )
] 6 AcEc! [i w 3 1 2
The values of the ratio HI Hw for various values of 11 calculated from Eq.
(11.10) are given in the last line of Table 11. I. Taking for a numerical
example n = -1-, Hw/AcEt; = 0.002,* fll = 0.1, \\'C find H/Hw = 1.112, and
Eq. (g) gives 1).1 = 0.00044/. l'his deflection must be added to the deflec-
tion 0.00151/, calculated from Eq. (11.11),to obtain the total deflection pro-
duced by live load.
l..et us consider now the deflection of the cable produced by a concentrated
force applied at the middle of the span. 'fhis force can be considered as a
load distributed along a very short distance, and Eq. (11.11) can be used also
in this casc. From notations (e) and (1) it follo\vs that
(g)
2pa P
nz==-=t/;
'WI Q
(h)
where 1/1 denotes the ratio of the live load J' to the dead load Q of the bridge.
Substituting t/! for nz and zero for z into Eq. (11.11), \VC obtain
1 + 2'"
1 = lID vI" + 31/1 + 31/12
(1.1.12)
In the case of long-span bridges, the concentrated load P is small in comparison
\vith the dead load Q of the bridge, and t/! is a small quantity. Developing then
the radical in the denominator of expression (11.12) into a series and taking
only the first three tenns of that series, \ve obtain
VI + 3f + 31/12  1 + it/! + it/!2
and Eq. (11.12) give<;
f = fw(l + i'" - -11/1 2 )
Hence, the deflection produced in the cable by a concentrated force applied at
1 In the case of the George Washington Bridge ovcr the Hudson River, this ratio is about
1 : 6.
· This value depends evidently on the allowable stresses produced by dead load.

ARTICLE 11.2
529
the middle is
t/!
df = f - fw = Iw 2 (1 - It/!)
(11.13)
Assuming, for example, t/! = 0.01* andfw// = 0.1, \ve obtain
f = 0.000489/
\vhich is a very snlall deflection. To find the deflection due to elastic elonga-
tion of the cable produced by a concentrated force, \VC calculate first the change
H - Hw in the horizontal tensile force. Equation (11.10) in this case gives
H = Hw vI + 3tf-+ 31/12  Hw(1 + t/! + -t/l2)
and \\'e obtain
H - Hw = lHID1/;(l + i1/;)
Substituting this for H in Eq. (11.9), \ve find
ilf =  Hw  t/; ( 1 + .1/t ) ( 1 +  fw2 ) /
32 AcEc f 4 3 /2
:For a small1/!, \vhich \\'C have in the case of large spans, this deflcction is a very
small one.
We discussed up to now the syrnmetrical case of loading (Fig. 11.2). The
case of a nonsymmetrical distribution of unifonn live load can be treated in a
similar manner. Let us consider now a general case of vertical live load
acting on a cable with both ends on the same level. The initial ordinates of
the funicular curve arc obtained from an equation of moments, \vhich gives
Wl w
y=-
Hw
(i)
In this equation, IDlw is the bending moment due to dead load calculated as for a
simply supported beanl, and Hw is the horizontal component of the tensile force
produccd in the cable by dead load. If live load is now applied, the bending
moment calculated as for a simple beam becomes IDlw + ID1 p , and the horizontal
component of cable tension becornes Hw + H p . Dcnoting by 71 the vertical
deflections of the cable, we obtain, ftom an equation of moments,
+ _ lJJl tD + WI"
Y 71 - [lID + H p
(j)
· In the case of the George Washington Bridgc, this value of 1/1 corresponds to a concentrated
load of 570 tons.

530
SUSPENSION BRIDGES
FIG. 11.3

Subtracting Eq. (i) from this equation, \ve obtain
Wl p - HpY
1/= .
Hw + H p
(11.14)
It is seen that the vertical deflections 17 can be readily calculated, provided
that \VC know the horizontal component H p of cable tension produced by live
load. This latter quantity can be found from geometrical considerations.
llct us consider no\\' an infinitely small element ab of the cable (Fig. 1 1. 3) .
Under the action of live load this element elongates somc\vhat and takes a ne\v
position a1b!. \\'t: denote by t and 17 the horizontal and the vertical components
of the small displacement of point a. '[he initial length of the clement is
obtained from the equation
ds 2 = dx 2 + dy2
(k)
The length of the same clement after application of live load is found from the
equatIon
(ds + ds) 2 = (dx + dt) 2 + (dy + d1/) 2
(I)
in \vhich ds is the elongation of the clement produced by live load. Neglect-
ing the snlall change in slope of the cable produced by live load, 1 \\'C put
ds = ds H p ds
AcEc dx
(111 )
Since H p dsj dx is that part of the tensile force in the cable \vhich is produced by
live load and which is usually much smaller than the part produced by dead
load, the unit elongation dsj ds is usually nluch smaller than one-thousandth.
In such a case (ds) 2 in Eq. (I) can be neglected. For the saIne reason,
and from the observation that the curve of the cable is a flat curvc, \ve neglect
also (d) 2. Then \ve obtain, from Eqs. (k) and (J),
ds ds = dx dt + dy d1J + j(d17)2
1 The error introduced by this omission ,viII be discussed latcr; see page 535.

ARTICLE 11.2
531
\\'hich gives
ds d" 1 dfJ
d = - !::ads - - dfJ - - - dTJ
dx dx 2 dx
Substituting expression (111) for !::ads in this equation and integrating, we obtain
H !c .2: ( dS ) 3 J; :e !c x
 =  -- dx - - Y'fJ' dx - -} TJ'2 dx
AcEc 0 dx 0 0
(11.15)
where the primes indicate derivativs \\,ith respect to x. With the values of
y' and TJ' which are encountered in long-span bridges, the value of  usually
does not exceed onc-thousandth of x. * At the ends of the cable,  vanishes,
and Ne obtain from Eq. (11.1 5)
H p (I ( dS ) 3 dx (l Y'TJ' dx + j_ ' 0 1 TJ'2 dx
AcEo 10 dx = 10 J
(11)
The integral on the left side of this equation for the assumed parabolic shape
of the cable can be readily evaluated, and \ve obtain
(l ( dS ) 3 dx = (L ( 1 + '2 ) 1 dx = I { ! (  + j2 ) ( 1 + 16 j2 ) i
J 1 dx 10 y 4 2 /2 /2
+ /1 j 1n [ Y + (1 + 1 Z.Y]} (0)
On the right side of Eq. (11), \ve make integration by parts. Observing that
TJ vanishes at the ends of the cable, and using Eq. (11.2), we obtain
/0 ' y' '11' dx = I y' '11 : - /0 ' y"'l1 dx = :L /ol '11 dx
. /0 ' '11'2 dx = ..I '11''11 I  - .. /0 ' '11"'11 d x = - . /Ol '11"'11 dx
(p)
Substituting expressions (0) and (p) into Eq. (11) and denoting the integral (0)
by L, \ve obtain
H p 1 W !c L d !c l " d
---- J = - fJ x -} fJ TJ x
AcEc Hw 0 0
(11.16)
rhis cquation, together \vith Eq. (11.14), gives the system of equations
sufficient for calculation of vertical deflections of the cable.
· The maxirTlum value of /x occurs near the suppons whcre 77' and y' usuaJly have thcir
largest nuruerical values.

532 SUSPENSION BRIDGES
Let us apply these equations to the above-discusscd case shown in Fig. 11.2.
In this case
I
IDl p = pax x < "2 - a
IDl p = pa:t: - !P ( x - 4 + a Y  + a > x >  - a
'11 = -- -- [ po.,,' - - x(l - . r)H p ] :.: < ! - a
Hw + H p 2Hw 2
17 = Hw  H p [pax - !P (x -  + aY - 21O x(/ - X)H p ]
I I
2 + a > x > 2 - tl
" wH p
'11 = --
Hw(Hw + H p )
" -pIJ VJ + wH p
'11 =- -
Hw(Hw + IIp)
I
x<-i- a
I I
2+ a >x>Z-a
Substituting these expressions into Eq. (11.16), assuming that the cable is
inextensiblc, and introducing our previous notations (e) and (f), \\'c obtain,
for calculation of H 1" the follo\ving quadratic equation:
(:Y + 2 ( Z: ) - 31lz - 3n 2 z 2 + 7lZ 3 + 2n 2 z 3 = 0
which gives for H p + Hw the same value as obtained from the previous
Eq. (11. 1 0) .
Sometimes Eq. (11.16) is simplified by on1itting the second term on the
right side and taking
H p IJ = .- (l'l1 dx
AcEc H 1D J 0
(11.17)
Considering an incxtcnsible cable and substituting for '11 its expression (11.14),
we obtain, from Eq. (11.17),
H p = JOIIIDl  =  ( I 9111' dx
!o y dx 21/ Jo
(11.18)
In the case shown in Fig. 1 I .2, Eq. (11.18) gives
H - llHw ( ' 3 3 )
p-- z-z
2
( 11.l8a)

ARTICLE 11.3
533
Applying this approximate formula to the numerical examples given in "[able
11.1, \ve find that the results obtained are in good agreement \\lith those given
in the last line of the table.
11.3 FUNDAMENTAL EQUATIONS FOR STIFFENED
SUSPENSION BRIDGES
It is seen from the preceding discussion that the deflection of the cable
produced by live .load is small only in the case of heavy long-span bridges.
()ther\vise, the deflections rnay be considerable. In order to reduce thes(
deflections, stiffening trusses are usually introduced. A simplest structure of
this kind, sho\vn in Fig. 11'.4, consists of a single-span cable stiffened by a
sinlply supported truss of constant cross section. It is assumed that by a
proper assembly, the dad load of the structurc, uniformly distributed along
the span, is enti rely transnli tted to the cable, \vhich takes the parabolic fonll
sho\1,'n in the figure by a solid line. A live load produces deflection of the
cable and of the truss as indicated in the figure by dashed lines. We assume
that both these deflections are equal. 1 The spacing of hangers is assumed
snlall as coolpared \vith the length of span so that the action of the hangers on
the cable and on the truss can be considered as continuously distributed along
thc span.
I..ct us consider first the case \\,here the structure is carrying only dead load.
. rhe truss does not suffer bending in this case, and the equation of nloments
for the forces to the left of a cross section 'IJlIl (Fig. 11,4) thcn gives
ID1w - Hw')' = 0
(a)
When live load is applied and deflections '1'J arc produc,ed, there \vill be bending
1 'fhat is, the elongation of the hangers and their srnall inclination to the vertical during
deforation are neglected in this discussion,
m
I
"2
x
x

""
" f
"-
"-
......
......
t...........
x 4---  ---
--- - l- ---------------
I 17
nl
FIG. 11.4 y

534 SUSPENSION BRIDGES
mon1cnt Ai acting in a cross section 1111/ of the truss, and the equation of
moments for the forces to the left of this cross section is
9Rw + lm p - (Hw + H p )()' + 11) - 1\1 = 0
Subtracting Eq. (a) from Eq. (b), we obtain
M = ID1 p - (flU' + H p )l1 - H p )'
(b)
(11.19)
From this equation, the bending moment at any cross section of the truss can be
calculated provided the horizontal component of the tensile force in the cable
and deflection 11 arc kno\\'n.
In the case of very rigid stiffening trusses, the deflections 11 can be neglected,
and \ve obtain
kf = 9Jp - Hpj'
(11.20)
lhis bending rnon1cnt is indepcndent of deflections and can be evaluated by the
methods used in the analysis of rigid statically indeterminate structures.
Investigations sho,"' that the stiffening trusses in large-span bridges arc usually
very flexible; and in calculation of bending monlents, recourse must be had
to the more complete equCltion (11.19), \vhich requires the calculation of
deflections 17 * of the truss . Using for the truss the di fferential equation of the
deflection curlc of a beam
d 2 11
EJ-- = -li1
dx 2
\ve obtain, by using expression (11. 19), the follo\ving equation for calculating 11 :
d 2 'rJ
EI dx 2 - (liw + IIp)'rJ = H p )' - 9Jl p
(11.21)
Thc quantity 9)1p in this equation can be readily calculated for any distribution
of Jive load over the span. The quantities)' and Hw are given by Eqs. (11.2)
and (11.3), and only the quantity H p is unknown. It depends on the deflec-
tions 11, and for its deternlination Eq. (11.16) of the preceding article is used.
Equation (11.21), togethcr \vith Eq. (11 .16), complctely defined the deflections
of the stiffening truss. In the solution of these equations, the trial-and-error
method is used. \\c assume a certain value for H p , for instance the value
obtained for the unstiffened cable, and \vith this value solve Eq. (11.21).
The obtained expression for 'rJ \ve substitute in the integrals on the right side
ofEq. (11.16). Since H p \vas taken arbitrarily, the result of this substitution
usually \viJl not equal the left side of Eq. (11.16), and it will be necessary to
· J. Melan was the first \\,ho jnjcated the importance of considering deflections in analysis
of suspension bridges; see his book "Theorie dcr eiscrncn Bogenbriicken und dcr Hangc-
briickcn," Leipzig, 1906; English translation by I). B. Steinman, Chicago, 1913. Melan's
theory has been \vidcly used in analysis of large-span suspension bridges in this country.

ARTICLE 11.3
535
repeat the calculation with a new assumed value of H p . r-rhese trial calcula-
tions are continued far enough to obtain H p \vith a sufficient accur2CY. l--he
procedure of this calculation \vith all details \vill be discussed in the next t\VO
articles.
NO\\T we shall discuss ho\v accurate Eqs. (11.16) and (1] .21) are and \vhat
is the magnitude of errors introduced in these equations by neglecting various
small quantities during their derivation. We begin \\,ith the discussion of
elongation of the cable. In the derivation of F.q. (111) of the preceding article,
we neglected the change in the defletion of the cable produced by live load.
Taking into account this additional deflection, \\re obtain
ild = ds H p [ ] + ( ,,+ ' ) 2 ] 1 ,..., d s H p ( !!!. + " , + 1 '2 )
S AcEc '71 ,..., AcEc d.t ) 'Y1 T 'Y1
Using this more accurate expression for dds, we obtain, instead of Eq. (11) of
the preceding article, the following equation:
AEc 10 1 ()3 dx = 10 1 y'rl' dx + t 10 1 1/'2 dx
_ H p (I ( d s ) 2 ( , , + 1 , 2 ) dx
AcEc Jo dx Y 71 'Y1
The last ternl on the right side of this equation represents the required correc-
tion. Since Hp/ AcEc is usually smaller than one-thousandth, \VC conclude
that the relative error in the magnitud of the right side of Eq. (n) due to the
use of the approximate expression (111) is of the order of one-thousandth, which
can be disregarded in a practical analysis.
I..et us consider nOVl the effect on the bending moment in the truss of
horizontal displacements  in the cable, \vhich \\'ere entirely disregarded in our
previous derivation. To take these displacements into account, v..'e observe
that the vertical distances bet\veen the solid-line and dashed-line curves,
marked by 1] in Fig. 11 .4, arc rn ore accurately equal to 'Y1 -  dy / d.",-. as sho\\'n
in Fig_ 11.3. We note also that each element of the load transmitted to the
cable, and approximately equal to -dx (Hw + H p ) d 2 y/dx 2 , has a horizontal
displacement t, which produces the change in moment of this clement equal to
d 2 y
-  dx (Hw + H p ) dx 2
With these t\VO considerations \VC obtain, instead of expression (11.19), the
follo\ving more accurate value of the bending moment:
dy
M = IDl p - (Hw + H p )l1 - HpY + (Hw + Hp)t -d.
( z d 2 y
- (Hw + H p ) Jo dx 2  dx
(11.22)

536 SUSPENSION BRIDGES
l'he required correction in the bending moment is represented by the last
two terms in this expression. 10 get a clearer idea regarding the magnitude
of this correction, let us calculate the intensity of the load acting on the truss.
This intensity is obtained as the second derivative with respect to x of the
bending moment (11.22), taken \vith oppositc sign, which gives
d 2 M d 2 y d271
- dx 2 = p + H p dx 2 + (Hw + H p ) dx 2
- (HtD + H p ) 1x (! :%)
(11.23)
The last term in this expression represents the correction due to horizontal
displacements of the cable.
The same expression for the intensity of the load on the truss is obtained
also in another manner by subtracting the intensity of the upward pull of the
hangers on the truss from the combined intensity w + p of the downward
loading. The intensity of the vertical loading on the cable at a distance x from
the left support is
d 2 ( dY )
q = - (Hw + H p ) dx 2 y + 71 -  dx
(c)
The upward pull transmitted to a length dx of the truss at x is the downward
pull on a horizontal length dx (1 + dt/dx) at x + t on the dcflected cable.
Hence, the required intensity of the load on the truss is
w + P - ( q + t q ) ( 1 + d t ) = w + P - q - _d. (qt) (d)
dx dx dx
Substituting for q its expression (c) and neglecting small terms of higher order,
we obtain the previous expression (11.23).
Substituting for t the first two terms in expression (11.15) into the last
term of expression (11.23), we find that the correction in the intensity of the
load on the truss, due to horizontal displacements , is
- (H tD + H p )  ('y')
= - (Hw + H p ) [ -  ( - Hp S'3 - Y'r/ )
Hw AcEc
+ y' ( 3Hp S'2 S " + _ w 11' - y1r," )]
AcEc Hw
where primes denote the first derivative \vith respect to '.
For flat curves we can take
S'  1 + i y ' 2
s" ",. - }ltD Y

ARTICLE 11.3
537
\vhich gives
- (Hw + H p ) - ('y')
dx
_ ( Hw + H p )'}!!. [ _ _H p -.. _ (S'3 + 3S'2 y '2) _ 2y'rl' 1
Hw AcEc J
+ (Hw + H p )r/'y'2 (e)
Fron1 our previolls discussion, \ve conclude that the first tenn on the right side
of this equation is of the order of one-thousandth of wand can be neglected
in practical calculation. The second term, (Hw + Hp)rl"y'2, can also be con-
sidered as small in con1parison \vith the term (Hw + l-lp)l1" representing the
effect on the intensity of the load of vertical deflections of the cable. 1-lence,
the total effect of horizontal displacements of the cable can be considered as
small and usually can be neglected in practical calculations.
In a similar manner the effect of extension of hangers on the magnitude of
bending moment can be investigated. l'hc calculation sho\vs that this effect
is also very sn1all and can be neglected. l
Let us consider nOVl the effect of shearing force on deflection of a stiffening
truss . For this purpose \ve take the di fferential tquation of the deflection
curve in the :)lIo\ving fonn:
d 2 .,., d 2 AI
£1-- = -j\;1 + 111£/- -
d.'t. 2 dx 2
(f)
in \\rhich the second term on the right side represents the effect of shearing
force on deflection, The magnitude of the factor 111 in this terrl1 depends on
the kind of structure used fCJr the stiffening truss. In the case of an I bearn,
\ve take
1
111 = Aw(;
(g)
\vhere Aw is the cross-sectional area of the \veb of the bearTI and G is the modulus
of elasticity cf the nlatcrial in shear.
In the case of a truss, as sho\vn in Fig. 11. 5, \VC ta ke 2
1
111 = --
AdE sin cp C08 2 <p
(h)
\vhere A d is the SUI11 of cross-sectional areas of the t\VO diagonals in a panel.
1 Such calculations \vere (nade by F. E. Turneaurc; sce Johnson, Bryan, and Turncaurc,
"!\'lodcrn Fralned Structures," 9th ed., vol. 2, p. 299, John \Viley & Sons, Inc., Nc\\' York.
1917.
2 See S. P. 'fimoshenko and J. rvL Gere: "Theory of Elastic Stability," 2d ed., p. 137,
McGra,,,'-I'Iil1 Book Company. Nc\v York, 1961.

538 SUSPENSION BRIDGES
--r
b
I
-
FIG. 11.5
a
Substituting for i\.f in Eq. (f) its expression (II. 19), \ve obtain
d217
RIfl + 111 (Hu. + Hp)l dx 2 - (Hw + H p )17
= - fireI' + HI'Y - mE! (p - z: w) (11.24)
This equation is of the san1e fonn as Eq. (11.21), and \ve see that the effect
of shearing force can be readily taken into account provided the factor 11/ is
kno\vn.
In the derivation of Eq. (11.16) the clastic elongation of the cable
alone \vas considered. The equation can be readily generalized and extended
to those cases in \vhich an elongation of the cable depends also on a change in
its ten1peraturc. Instead of Eq. (111) of the preceding article, \ve use in such a
case the equation
A d ds Hs ds i
L.1 S = -. - + £ S Et
AcEc dx
(i)
\vhere E is the coefficient of thermal expansion, and Hs is the horizontal con1-
ponenr of the tensile force produced in the cable by the combined action of live
load and temperature change. LTsing Eq. (i) and introducing the notation
f l ( dS ) 2
- dx = L 1
o dx
\ve obtain
Hs L 1 w f I d f l /1 d
4 E + Et 1 = H 17 J: - ! 17 11 x
 C J C U' 0 0
(11.25)
This equation, instead of Eq. (11 .16), must be used if \ve are considering a
simultaneous action of live load and temperature change.
11.4 ANALYSIS OF STIFFENING TRUSSES
Let us begin \vith the case in \vhich a single concentrated load P is acting on the
truss. Making the second derivative of Eq. (11. 21), \ve find that deflections

ARTICLE 11.4
539
FI G. 11.6
A I
HW+Hp J"
I
c
B
../ Hw + Hp
x
of the truss in this case are the sanle as those occurring in a simply supported
bean1 subjectcd to the conlbined action of an axial tensile force Elw + Hp, of a
unifornlly distributed up\vard lateral load of intensity H p '1.JJ/ Hw, and a con-
centrated load P as sho\vn in Fig. 11.6. In such a case, \vith the notation
Hw + H p = k 2
RI
(11.26)
the deflections produced by the load P in the part of the bcanl to the left of this
load (x < I - c) arc)
P sinh kc . Pcx
1]1 = - lilt' + H p k si nh kI sinh kx + (Hw + Hp)1
(11.27)
For the pOftion of the beam to the right of the load (x > I - c), the deflections
are
171 =
P sinh k(1 - c) .
- Hw- + H p k sinh kl sInh k(l - x)
+ ]J (I - c) (I - x)
(Hw + H p )/
(11.28)
The deflections produced by the up\vard puB are 2
1]2 =
_ .l! wl 2 r coSI _ (kill - kx) _  + xU - X) ]
Hw Httt + H p k 2 /2 cosh (kI12) k 2 / 2 2/2
(11.29)
The total deflections 1] of the truss are obtained by superimposing deflections
7'J 1 on deflections 1]2.
To determine the magnitude of tension H}H entering into Eqs, (11.27) to
(I 1.29), \ve use Eq. (11.1 7), \vhich is obtained from Eq. (11.16) by omitting
the second term on the right side. 3 Substituting expressions (11.27) to (11.29)
into Eq. (11.17) and performing the integration, \ve obtain, for calculating
1 Sec S. Timoshcnko, "Strcngth of laterials,t' 3d ed., vol. II, p. 42, D. \r an Nostrand
Company, Inc., Princeton, N.J " 1956.
2 Sec ;hid., p. 43.
;s This tcnn usually has only a slnall cffect on the 111agnitude of l-I p' This effect will be
discussed in the next article.

540 SUSPENSION BRIDGES
IIp, the equation
H [ flU) + H p J I ( 8 f ) 2 ( 12 24 I kl )]
p -- AcEc -' + 12 7 I - k 2 /2 + k 3 /3 tan1 I
= p { g i (1 -1) - f2f2 s:nhk/
X [sinh kf - sinh kc - sinh k(f - c)] } (11.30)
In the case of long-span bridges the quantity kl is usually a number of con-
sidcrahlc magnitude,l and all tenns in Eq, (11. 30) containing k are small
and, in a first approxin13tion, can he neglected. '[he term (}{w + H p )/ A(1c
can also be neglected as vcry srnall, and \ve obtain
HP=Pjl(I-7)
For" = 1/2, this gives
3 I
IIp = "2 p Hf
\\"e obtain the same result for sn1all t/I frorn Eq. (11.10), \vhich indicates that
by on1itting all terms cO:1taining k, \ve obtain, from Eq. (11. 30), for II p the
same value as in the case of an l1nstiffened suspension bridge.
Equation (11.30) can be used for calculating the influence line for H p . In
such a case \VC assume that P is a sn1allioad moving along the truss. 'I'hen H p
can be neglected in comparison \virh Ilu,t, kl  , V Hw/ EI, and \ve obtain
.!_ :. ( 1 _  ) _ sin l kl- sinh kc - si nh k (! - c)
_ .> 8f 2 I / k 2 1 2 sinh k/
H -} -.- --- --- .- . -- - - ..
p / Hw L 1 ( 8 f ) 2 ( 12 24 kl )
4c-E 7 + 12 7 I - k 2 1 2 + k 3 / 3 tanh I
The magnitude of H p depends not only on position of the load P but also on
the quantities kl, H w/ A cEe, and f/ I. In Fig. 11. 7 a is sho\vn the influence
line for H p calculated on the assumption that kl = 10, Hw/ AcEc = 0.002, and
f /' = O. I . For comparison there are sho\vn in the same figure by dashed line
the values of fI p for a nonstiffcned cable. It is seen that for the assumed value
of kl the stiffening truss has only a small effect on the Inagnitudc of H p .
Having H p and using Eqs, (11,27) to (1 I .29), \VC can calculate the deflections
of the truss. In Fig. I [ .7 b the deflection curve is constructed for the case
\vhere c = 0.75/. Since H p is neglected in conlparison \\'ith Hw, the deflections
1 In the case of the Arnbassador Bridge (Detroit), k/ = 9.52. In the case of the George
\"Tashington Bridge, after placing the planned sti ffening truss, k/ = 35.

ARTICLE 11.4
541
(a)
1.827 P
AI ,
r--p
AI
'B
(b)
O.0398Pl = M mM
I /
/
/
I
/
I
I
I
I
/
\ I
\ I
\ I
----......
,., .......
./ .....
/ ........
/ "
/ "
"
IB
(c)
FIG. 11.7
become proportional to P, and the principle of superposition holds. The
reciprocal theorcn1 also holds, and the deflection curve in Fig. 11. 7 b is the
influence line for the deflection at the quarter point ('. Using this line, \ve can
read:ly construct the influence line for bending moment at C. Neglecting lip
in comparison \vith H". in expression (11.19), \ve obtain
,\1 = imp - ]1 p Y - Hwll
l"he first two tenns on the right side of this equation give the bending moment
if the influence of deflections of the truss is disregarded. 1'he corresponding
influence line is given by the dashed line in Fig. 11.7 c. The last term on the
right side gives the effect of deflection of the truss on the bending moment.
Taking this into account, \ve obtain the full line in Fig. 11.7 c. It is seen that.
in this case the deflections have a very large effect on the bending moment and
cannot be disregarded.
In using the derived influence line for calculation of bending moment, it
must be noted that the increase of tension in the cable produced by live load
was neglected. l-Ience, using this influence line \vill give a satisfactory
result only if the live load is very small in comparison \vith the dead load. If
it is not small, the influence line \vill not give an accurate value of the moment
and can be used \vith sufficient accuracy only for a detern1ination of the limits

542 SUSPENSION BRIDGES
FIG. 11.8
A
Hw+Hp l
b
I
p B
Df--- a 1 Hw+Hp
\vithin \vhich the live load Inust be distributed to produce the maximun1 value
of the monlcnt. 1 Calculations of deflections and 010ments must then be
accomplished hy using Eqs. (11.27) to (11.30) in \vhich H p has been retained.
Assume, for example, that live load is distributed as sho\vn in Fig. 11. 8.
l'hen, the equation for calculating H p is obtained fronl Eq. (11.30) by substitut-
ing p dc, instead of P, and integrating the right side of the equation from c = a
to c = b, \vhich gives
II [ I-J ur + H p L I ( Rf ) 2 ( 12 24 h kl )]
'p -- 4 cEc 1 + 12 I 1 - k 2 / 2 + k 3 f3 tan -2-
= g l p { 31 ( [,2 - a ) - 2 (!J3 - 3) _ __!n [ b - ,1
'J. 1213 k 2 /2 I
_ cosh k b - cos h ka + cosh k(L - a) - osh (I  b) ] }
kl sinh k/
(11.31)
"fhe value of H p can no\V be calculated from this equation by successive
approximations. \,Vt start by onlitting all ternlS containing k, and \ve neglect
also the tenn \vith (Hw + IIp)/Acl('. In this \vay, \ve obtain the first approxi-
mation for II p, \vhich \v j II be close to the true value if kl is of considerable
magni tude, 2 say kl = 10. lc) get better accuracy, \vc use the approximate
value of H 11 to calculate k from Eq. (11 .26) and then substitute this value of k
into Eq. (11.31), \vhich gives then the second approximation for H p, \\'hich is
usually accurate enough for practical application. I f necessary, further
approximations can be calculated in the satne manner. Y\lhcn H p has been
calculated, the deflection curve \vill be found by using Eqs. (11.27) to (11.29)
together \vith the rI1ethod of superposition. To find the deflections for the"
portion AC of the truss (Fig. 11.8) \VC substitute p de for Pinto Eq. (11.27) and
integrate from c = a to c = b. This gives (for x < / - b)
pI f b ( sinh kc . h I, ex ) d
?71 = H 1 -: + I1  a - k / sinh kl sin 1?J: + J2 c
=_1__  [ coSh k a - cos kb . h k . +  b2 - (12 ) J
h T tc + H p k2/2 sinh k/ sin x 213
1 The use of influence lines in analysis of suspension bridges was first proposed by T.
Godard, AT/n, p()nts ChtltJSS((S, scr. 7, vol. 8, p. 105, 1894.
2 An example of such calculat:on is given in a paper by A, A. Jakkula, Pllbl. Inttro. Assoc.
Bridge Stn/ltllrn/ Eng., voL 4, p. 333, 1936.

ARTICLE 11.5
5.3
To obtain the complete deflection, \ve superimpose on this deflection the deflec-
tion 172 produced by the up\vard pull [Eq. (11.29) L giving, for .'t' < I - h,
_ _ _. ._P-' [ ' cOSh a -=---_c.os k . (b 2 - a 2 )x J
17 - 1]1 + 772 - Hu, + H p k2/2 sinh k/ sInh kx + 2/ 3 -
H p w/ 2 r cosh (kl/2 - kx) 1 .\"(1 - X) ]
- H-(l/w + r Ip) k 2 7z" cosh (kl/2) - k 2 J2 + 2/ 2
In a similar rnanncr the deflections in the portions ('[J and l)B of the truss can
be obtained. .-\ sirnpler method of calculating deflections is sho\vn in the next
article.
If a cornbined action of live load and tCIllpcrature change is considered, the
equation fi)r calculation of the additional horizontal tension H, is obtained by
substituting into Eq. (J 1.31) H$ for H p and HsI;,/ AlEc + ElL 1 for HpL/ A<:Ec
(see page 538).
u.s APPLICATION OF TRIGONOMETRIC SERIES
IN CALCULATING DEFLECTIONS
The deflection curve of a stiffening truss can be obtained also by using trigo-
nometric series. 1 In applying this method, not only is the calculation of deflec-
tior:s sirnplificd, out also the calculation of IIp can be made \vith better accuracy
since, \vithout much .complication, \ve can USt Eq. (11. 16) instead of the
sinlplified Eq. (II. J 7) used in the preceding article. Let us begin \vith thc
case of one concentrated force .acting on the stiffening truss (Fig. 11.6). 'The
differential equation (11.21) in this case bcconlcs, for x < I - c,
£1 d21] ( L J H ) H 4( (I ) Pc.\'
dx 2 - .1. w + p 77 = P j2- x. - x - T
(a)
and, for x ? I - c,
d 2 'f1 _ 4/. . _ , - c) (I - x)
£1 dx 2 - (Hw + H p )77 - H p /2 x(l - J:) J
(b)
-rhe right sides of these equations can be represented for the entire length I of
thc truss by the trigonornetric series
l . 7rX l . 27T"X b . 37rX +
11 sIn T + '2 sIn -,- + 3 sIn T ...
(c)
the coefficients h 1 , "2, 17 3 , . . . of \vhich (nust be calculated in the usual \vay
1 See S. Timoshenko, Tri7TJs. 4SCR, vol. 94, p. 37', 1930.

54. SUSPENSION BRIDGES
from the formula
2 [ 4f ( l . 111",.X
h m = 7 HI' J2 Jo x(l - x) sin I dx
Pc h l-c . 1117rX d P(/ - c) t ' (/ ) . 1111rX d ]
- - x SIn --- x - -- - x sin  x
I 0 I I l-c I
which gives
16H p f 2Pl . m7r(1 - c)
b m = ---- (1 - cos 1n7r) - - sin --
m 3 7f'3 m 2 ",.2 I
(d)
Since the series (c) is applicable for the entire length of the truss, the two
equations (a) and (b) can no\v be replaced by one equation,
d 2 
El dx; - (Hw + H,,)-q = L b m sin m;x
m!iS 1
(e)
"rhe solution of this equation, satisfying the end conditions, will no\v be taken
in the form of the series
QO
 . m1rX
11 =  am sin 1-
m=l
(I)
Substituting this series into Eq. (e) and equating, for each value of 111, the
coefficients of sin (m1rx/ f), we obtain
b m J2
a =- --
m EI11Z27f'2 + (Hw + H,,)/2
and the series (f) becomes
 h m / 2 sin (11t1rx/ J)
11 = - '-' EI(m21f'2 + k 2 / 2 )
m=l
\vhere k 2 , as before, denotes the value of the ratio (Hw + HI') / E/. Substitut-
ing expression (d) for b"., we finally represent the deflection of the truss by the
following series:
32H p fl2  sin (m7rx/f)
11 = - 7r3EI '-' 11t3(m2",.2 + k 2 J2)"
m E3 1.3.5. . . .
2P/3
+ £/1('2
eo
2:
m=1.2.3... .
1 . m1r(1 - c) sin m1rX
m 2 (m2r 2 + k2J2) sin J I
(11.32)

ARTICLE 11.5
ses
'[his is a rapidly converging series, and the calculation of deflections requires
less time than in the case when \VC are using Eqs. (11.27) to (11.29). Applying
this series to the numerical example of the preceding article (Fig. 11.7 b), we
have
c = 0.751
H p = 1.3411>
k 2 ....., Hw
-- EI
kl = 10
f = 0.11
and the series (11.32) gives the deflection curve
PI ( . 1l'X . 31rx
11 = -0.1384 I-lw 0.9102 sin T + 0.0196 sIn --I
. 51rx ) PI ( . 7rX
+ 0.0023 sin T + . .. + 0.2026 Hw 0.6434 sin T
. 27rx . 31f'x . 51f"x
+ 0.1793 sin T + 0.0416 sin T - 0.0082 sin T
. 67rx . 71rx )
- 0.0064 sin -j - 0.0024 sIn T - . . .
(g)
The number of terms which are sho\vn in the series (g) is sufficient to get
values of the deflections \vith an error smaller than 1 percent.
The obtained deflection curve is also the influence line for the deflection at
the quarter point C in Fig. 11.7 b. Using it as explained in the preceding
article, we can determine the portion of the truss which must be covered with
live load to produce maximum bending moment at C.
The equation of the deflection curve produced by live load uniformly dis-
tributed along a portion of the truss (Fig. 11.8) can be readily obtained from
Eq. (11.32); it is only necessary to substitute therein p de for P and integrate
from c = a to e = b. In this manner we obtain
'7 = 32H"f/2  sin _ (m1rx/l)
- 1r 3 EI  1113(m'l1f"2 + k 2 /2)
m=1.3,5,...
2p/4
+ 3 EI
GO
2:
m = 1,2,3.. . .
m7r(1 - b) m1f"(1 - a)
cos - cos-
I /. 1n1f"X
--- sin -
1n 3 (m 2 'fr2 + k 2 /2) 1
Observing that
8fH w
--=W
/2

546 SUSPENSION BRIDGES
and introducing notations
HI' = {3
Hw
1- h
I
= al
I-a
I
= lX2
(11.33)
we represent the deflection curve by the series
CIO
\' ,1111(' X
17 =  am sin I
m!!le 1
(11.34)
in \vhich
2/ 4 - ,Bw (1 - COS 1117r) + 1!( c os 1111('lXt - cos 1!!1(' a21
am = 1('3£1 - 1}13(1n'l + k 2 / 2 )
(11.35)
The calculation of these coefficients requires the values of {3 and of k which
depend, as \ve see from notations (11.26) and (11.33), on the horizontal tension
HI' produced by live load p. To get the equation for calculating H p , \\'e
substitute the series (1: .34) into Eq. (11.16), \\,hich gives
Hpl = w 21 ( a I + a3 + a o + . . . )
AcEc Hw 1(' 3 5
1r 2
+ 4/ (a1 2 + 2 2 a2 2 + 3 2 a:J2 + . ..) (11.36)
This equation can be solved by successive approximations. As a first approxi-
mation \\'e can take for fl p the value corresponding to an unstitfened and
inextensible cable, as \vas done in the preceding article. With this approximate
value of H p, \VC calculate {3 and k and find the coefficients at, a2, . . . from
Eq. (11.35). Substi ruting these values into the right side of Eq. (11. 36), we
shall find that the equation is not satisfied. We then repeat the calculation
\\'ith a son1e\vhat s01alIer value for HI'. From t\VO such calculations a suffi-
ciently accurate value of H; can be obtained by interpolation. Calculations of
this kind \vere made for the Ambassador Bridge in !)ctroit,l and comparison
of the obtained values of H p \vith those found from the simplified r. (11.17)
sho\vcd that the effect of the second term on the right side of Eq. (11.36) is
small. It \vas of the order of 1 percent for a live load p = 2,000 lb/ft covering
half of the span. This term takes care of nonuniformity in the distribution of
vertical puJI transmitted to the cable, and it must become more important in
the cases of more concentrated actions of live load. When the tension H p ,
from Eq. (11.36), and the deflections 11, from the series (11.34), have been
found, \ve calculate the maximum bending tnorncnt at C by using Eq. (11.19).
1 See Jakkula, op. (it.

ARTICLE 11.6
547
In the case of combined action of Jive load and tenlpcraturc change, the
Intthod of series can also he used. In sllch a case it \\riJJ onlv be neccssarv tc
J J
put on the left side of Eq. (11.36) H"L///lcEc + fILl, instead of 11pr,,/!.:4c/c'
and use the notations {3 = lI.// flU! and k 2 = (flIt' + 11$) /1:1 in fonnula (11.35).
11.6 THREE-SPAN SUSPENSION BRIDGES WITH SIMPLY
SUPPORTED STIFFENING TRUSSES
l'he Inethods of analysis of a single-span. suspension bridge discussed in the
t\VO preceding articles can be readily extended to the case of several spans if
each span is stiffened by a separate sin1ply supported truss. Let us consider.
for example, a syn1n1crrical three-span bridge as sho\vn in Fig. 11.9. ""t:
aSUtl1e that the cable can slide \vithout friction at the tops of the to\vers or
that the to\vers arc very flexible. l'hen the horizontal conlponent /-1 p of the:
tension produced in the cahle by live load is the san1e for all three spans, and in
its calculation \ve can again use Fq. (11.16). '1'h<: quantity IJ (see page 53])
in such a case is obtained by extending the integration over all three spans.
\vhich gives 1
r 1 ( d S ) 3 r 1. ( d S ) a
I, = }o(1-; £Ix + 2)0 dJ: d:r
(a)
.\pplying the trigonollletric series in our analysis and assuI11ing that live load is
applied on the middle span, \\'e can nse for the deflection of this span the
previously obtained series (11.34), the coefficients of \vhich arc calculated
from forITnIla (11.35), .),he side spans \vill be bent by the up\vard puJI of the
hangers, and frotn synllTIctry it tollo\\'s that thei r deflection curves \\'i II be
idcntic,11. <:onsidering the left-hand span and taking the origin of coordinates
at the left support, \ve can represent the deflections 171 of this span by the
foHo\ving series:
. 1I"'X . 27r.\' . 37r.\" ( ")"
1]1 = Cl sIn _ I + C'!, sIn . I - -1- (3 sIn -- I +... 11. J 7 )
1 1 1
rhc coefficients in this series are obtained frOITl forrnula (11. -' 5') by taking
1 \Vc assume that corresponding truss and <..'abJc spans are equal.
f
f
1
FIG. 11.9
A.
,
I
A AA
ll I
B
I
AA A
-.J I .
l -. ----, r--ll

548 SUSPENSION BRIDGES
p = 0 and substituting the quantities /1 /), 'lVI, and kIt instead of I, I, "'lV, and k,
for the side span, l'hen,
C," =
_ _!!t' I} Wt( 1 - cas 17r)
rr3EI11113(IJZz7r'l. + k t 2 /12)
(11.38)
Substituting the serie;, (11. 34) and (I 1. 37) and the quantity IJ, fron) Eq. (a) 'I
into Eq. (11 .16), \VC ohtain the follo\ving equation for calculating H p :
H pI", . _ 2 (I L (1m + 21  (m . )
--,' - -- 'lV - lWl' -
AcJc rrHw 111  111
1,a.5.... 1.3,5....
2 ( I) J 2 )
+ rr L am" + ....Cm 2
- - - - 1J1
4 I 11
1,2.:3, . . .
(11.39)
"-rhis equation can he solved by the trial-and-error method used in the previous
article for the solution of Eq. (11.36). "lhen H p has been found, the deflec-
tions of the trusses are calculated by using series (11,34) and (11.37), and the
moments r.:'c found from Eq. (11. 19).
If, instead of trigononlctric series, the expressions (11.27) to (1 1,29) arc used
for the deflection curves of the stiffening trusses, \ve apply the simplified
Eq. (11.17) in calculating H p. ()bserving that in this case the quantity I-J is
given by Eq. (a) and extending the integration on the right-hand side of Eq.
([ 1.17) over all three spans, \VC obtain for calculating H p the follo\ving equation:
H { ( }/t!' . +_ljp) + _ 1 ( 81 ) 2 [ ] _ _ I. + 24 tanh (kl/2) 1
p AcEc l 12 / k 2 / 2 k 3 / 3
+ _ ( ! ) 2 /1 [ 1 _ _ 12 +  tah _ ( kt/t/2) ]}
12 I] I k 1 2 / 1 2 k t 3 / 1 3
= P S{ {7 (1 - j) - k 2 /2 sn h II
X [sinh kl - sinh kc - sinh k(l - C)]} (11.40)
Comparing this equation \vith Eq. (1] .30) for a single-span bridge, ,tiC see
that the difference is represented by the last terrn on the left-hand side of
Eq. (1 1.40). Because of the presence of this tern1, the magnitude of IIp
beconlcs smaIler than in the case of a single-span bridge having the same
dinlensions as the middle span in Fig. 11.9.
Substituting p dc for J:> intoEq. (11.40) and integrating \vith respect to c,
an equation for fI p is obtained for the case in \vhich live load is distributed
along a portion of the n1iddle span of the bridge,

ARTICLE 11.7
549
11.7 THREE-SPAN SUSPENSION BRIDGE WITH CONTINUOUS
STIFFENING TRUSSl
Let us consider a syn1metrical three-span bridge \vith a unif()rm continuous
sti ffening truss as sho\vn in Fig. 11. J Oa. For the deflection curve of each
span, Eq. (11.2 I) \vill be used, in \vhich 9Jl 1 ) \vill no\v denote the bending
rnorTIcnt in a sirnply supported beam due to live load and the n10lnents Nfl and
./\I[ 2 at the supports A and B. \\'t= begin \vith the case in \vhich, in addition
to dead load wand 'tV., one concentrated force f) is applied at a distance c froln
the support B. The corresponding free-body diagrams for the three spans are
sho\vn in Fig. 1 I. lOb, c, and d.
rhe rnagnitudes of the bending il10ments AJ. and /\-1 2 \vill no\\" be found
from the conditions that at the supports A and B the deflection curves of the
t\VO adjacent spans have a common tangent. l)eAections produced by the con-
centrated load P and the up\vard pull of the hangers arc given by Eqs. (11.27)
to (11.29). To obtain deflections produced by a couple applied at the end of a
oear11, \\'c use Eq. (1 1.27) and aSSUITIC that the distance c indefinitely decreases
\vhilc the moment Pc remains constant and equal to i\ll. 'r'hen, Eq, (11.27)
gIves
AI ( X sinh kX )
." = Hw + H ; 7 - sinh k/
(11.41)
1 See the papcr by S. 'rioloshcnko and S. \.\Tay, Publ. bJtrll. Assoc, Bridge Structural Ellg..
vol. 2, p, 452, 1934.
(a)
( b ) 111
. }(1111111111'1111;1110;
Hw+Hp Hw+Hp
(Ill H p
Hw
"=
Itl'].
(d)
Hw Tllp
Hw + Hp
p
J.H 2 __-.J... wH p _
H",
Hw+Hp I
(c)
M 1
Hw+Hp r
FIG.11.10

550 SUSPENSION BRIDGES
and, by di tfercntiation, \\'C obtain
( d"l'f ) i\;[ ( kl ) lvII 6 ( kf )
dx x=sO = l(f[ + fl; Y 1 - s1 nh kl = 6£/ k 2 /2 1 - si nh kl
( 11.42)
( "I'f ) = _. j\.J, ( 1 _ kl )
d.'l' pal I(Hw + H p ) tanh kl
1\/[1 3 ( kl )
= - 3£1 k 2 f2 tanh kl - 1
(11.43)
'rhe slopes produced by the concentrated force P are obtained by di tferentiation
of Eqs, (11.27) and (11.28), \vhich gives
()_o = il :i- Fi (1 - :: {7) (11.44)
( d'YJ ) P [ I - c sinh k(l - c) J (11.45)
d. %==1 = - Hu: + I-i; .-y- - sinh kl
Finall y, the slopes produced by the up\vard pull arc obtained by di tfcrcntiation
of Eg. (11.29), \vhich gives
( d"l'f ) = _ ( TJ ) = _ H p _ __ (  _ tanh kl )
dx %=0 dx z=l Hw 24£1 k 3 /3 2 2
(11.46)
'fo silnplify our \vriting, \ve introduce the follo\ving notations:
kr i ( 1 - Si n kl ) = <P
k/2 ( tank kl - 1) = '"
24 ( kl kl )
k3/3 2 - tanh"2 = 8
(11.47)
','lith these notations, the slopes at the ends of the Iniddlc span produced by all
forces sho\vn in Fig. 11.1 Dc are
( d l1 ) '\If 21 1\1 1 1 H p '7.L,/3
dx XIt:jO = 6£7 <P + 3£1 t/; - HUt 24£/ 0
p ( c sinh kC )
+ Hw + H; 1 - sI nh kJ
( d'T1 ) _ 1\;[11 _ At[ 2/ \fI + H p . '11)/ 3 8
dx rr=l = 6El <P 3EI Hw 24£1
_ ____ __ [ 1- c _ sinh k(/- C) ]
HUI + H p I sinh k/

ARTICLE 11.7
551
Similarly, the slopes at the ends of the t\VO side spans can be obtained. It is
only necessary to substitute /1 and 'WI for / and wand to denote by CPI, "'h and 8 1
the corresponding values of the functions (11.47). The conditions of con-
tinuity at the supports A and B give us, then, the following t\VO equations:
2M 1 (/1/; + 1 1 1/;1) + M 2 /<{J = fJ; (h 3W 181 + 1 3 w8)
+ 6P ( sinh kc _ £ ) ( 11.48 )
k 2 sinh kl I
H
i\lllcp + 2j12(/t/1 + 1 1 "'1) = 4Hw (11 3W 18 1 + 13 w 9)
+ 6P [ sinh k(/ - c) _ _ !_ - C J (11.49 )
k 2 sinh k/ /
Frarn these equations, the mon1cnts i\.J 1 and i\J 2 at the supports can be calculated
if the additional horizontal tension H p produced by the live load P is kno\vn.
For calculation of H p , \VC use the 'simplified Eq. (11.17). 1"hen, extending
the integration over all three spans and using the notation
(l ( dS ) 3 (LI ( dS ) 3
l = Jo dx dx + 2 Jo dx d.t
\ve obtain
H p l. W ( l 'tV 1 ( 11 'Wl ( II
AcE  = Hw 10 77 d.t. + Hw )O 171 dx + fl: 10 772 dx
(11.50)
In this equation 71, 771, and 772 arc the deflections of the three spans produced by
the loads indicated in Fig. 11.10c, b, and d, respectively. Using Eqs. (11.27)
to (11.29) and (I J .41), \ve obtain for these deflections the follo\ving expressions:
For x < I - c,
P sinh kc . Pcx
11 = - Hw +- Hp k sinh kl Sinh kx + I (Hw + H p )
H p w/ 2 [ 1 ( kl )
- H Hw + H p k 2 / 2 - co sh -(kI12) cosh 2 - kx
_ . _. + I _ .; A J + All [  - X _ sinh .k(/ - ;r) J
k /.. 2/" Hw + H p / sinh kl
+ _ ( :t: _ sinh kX ) (11.51)
Ht/) + H p I sinh kl

512 SUSPENSION BRIDGES
For x > 1 - c,
= _ P sinh k(l - c ) . h k(1 _ ) + P( I - c) (I - x )_
71 Hw + HI' k sinh kl sIn x /(H w + HI')
H p wl 2 [ 1 ( kl )
- H.. H.. + H p k 2 /2 cosh (kI12) cosh 2' - kx
_ ..1_ +  (l- X) ] + M 1 [ 1- x _ sinh k(/- I ]
k 2 J2 2/ 2 HtD + H p 1 sinh kl
+ H.. 2 - Hp (; - ::: J) (I 1.52)
M 1 ( X sinh kx )
711 = Hw +H p 11 - sinh k/ 1
H p w l /1 2 [ 1 ( kll k )
- HfI) Hw + HI' k 2 / 1 2 cos h- ( k/ 1 /2)- cosh 2 - x
_  + x(/ J -=- ] (11.53)
k 2 / 1 2 2/]2
_ M 2 [ II - x sinh k(/ l - X) ]
'112 - -
HtJ) - H p 11 sinh kit
H p wl/t 2 [ 1 h ( kit k )
- Hw Hw + H p k 2 /12 cosh ( k/ 1 /2)- cos - i- - x
_  + %(/ 1 - X) ] (11.54)
k 2 / 1 2 21 1 2
Substituting these expressions into Eq. (11.50), we obtain
where
H [ r (Hw  Hp)L + ( W ) 2 ( . tanh kl _ .J. + E- )
P A cEc H tD k 3 2 k 2 12
( W 1 ) 2 ( 2 k/l 11 1 13 )]
+ 2 Hw k 3 tanh -2- - k 2 + 12
w { PC(1 - c) P. ·
= HI> 2 - k 2 sinh kl [smh kl - smh kc
- sinh k(/- c)] + (M 1 + M2)X}
k 2 ( WI )
X = - 8/ 3 + - 8 1 / 1 3
24 'tV
(11.55)
(11.56)
Equations (11.48), (11.49), and (11.55) define the three statica1Jy indeter-
minate quantities M 1 , M 2 , and H p .
In calculating these quantities, we determine first the value of M 1 + M 2 by
adding together Eqs. (11.48) and (11.49). Substituting this value into Eq.

ARTICLE 11.7
553
(] 1.55), \ve obtain
H { (fIw + I-I < 2T + < ( 81 ) 2 [ 1 _ 12 + t( k//2) J
p .4 cEc l 1 2 I k 2 / 2 k 3 1 3
+ ! ( f ) 2 ! <  [ 1 _ 2 < _ + 24 tanl_ (kl /2 ). J
6 Ir I .< k 2 / 1 2 k 3 / 1 3
-  ({y (0 + I 133 01) 2(/1/; + 11/; 1)+ l;}
= p Sf { .  ( 1 _  ) _ i!l k ' - sJ  kc - sin} 1 k(l_ - c).
/ 2 I / k 2 /2 sinh k/
X [1 + 2W+1 ) + icp J} (11.57)
From this equation, IIp can be calculated for any position of the live load P.
Substituting II pinto Eqs. (11.48) and (11.49), \ve can find the nlOlI1cnts at the
supports.
rro apply Eq. (1 1.57) for the construction of influenct lines, \VC neglect H p
in conlparison \vith Hw and assun1e that k 2  I Iw/ RI. \,\lith this valuc of k 2 ,
the value of H p can be readily calculated from Eq. (11.57) for any poition of
the load, and the influence line for lIT} can be constructed. 'Ikc, as an example,
the case in \vhich 11 = .1, '7.Dl = "U\ kl == 10,1,/1 = 0.1, Hu,/.,,/'AcEcl == 0.0043,
Substituting these nUI11crical values into Eq. (11.57), \ve obtain
.. H { I c ( C )
0.07)3 p = 0.8P 27 ] - /
_ 3.12 [ 1 _ sinh kc _ .sinh k( l - c) . J }
100 sinh kl sinh kl
From this equation, the influence line of H p for the n1iddk span is oltained as
sho\vn in Fig. 11.11. For comparison, there is sho\vn by a dashed line in the
ame figure the influcnce line calculated from Eq. (11.40) for the sti ffcning
truss \vith hingcs at the supports (Fig. 11. 9) . It is seen that by using a con-
tinuous truss, \ve increase the rigidity of the structure some\vhat and dirninish
the values of IIp, I-laving the influence line for lip, \VC can construct the
influence lines for AlII and .Al z by using Eqs. (11.48) and (11.49). The
1
1.020
FIG. 11.11

554 SUSPENSION BRIDGES
influence lines for deflections \vill thcn he obtained frorn Eqs. (11.51) to
(11.54), and those for bending moments, frorn Eq. (11.19). Having a]] these
lines, \ve can determine the most unfavorable position of Jive load.
l\ssume now that the live load is distributed along a portion of the middle
span as sho\vn in Fig. 11.8. ---rhcn to obtain an equation for calculating H p ,
\ve llse Eq. (11.57) and apply the method of superposition, Substituting p de
fOf P in Eq. (11.57) and integrating the right-hand side \vith rtspect to c from
c = a to c = b, \ve ohtain
Ii { {liu+ Jlp) L + . .  ( 8f ) 2 [ 1 _ _._. + 24 t__( kl/2) ]
p AcEcl 12 I k 2 J2 k 3 1 3
+ }_ ( 8f ) 2 11 r 1 _ 1 . + 24 t a/1 /2) ]
6 1 1 I k 2 / 1 'J. k 3 / 1 3
-  (?fY (0 + :1 'l3 01) 2 (/f + I; fl) + }
= 8pf { }/ (b2 a2) - 2(b3 _{/32 _  [ !J - a
1 2/ 3 k 2 /2 I
_ o I  cosh ka t cosh k(l - a )- cosh k (! - b) ]
kl sinh kl
X r 1 + 2(/f +  If+t;]} (11.58)
This equation, instead of Eq. (11.57), must be used for calculating H p jf live
load is distributed along a portion of the middle span. In a similar mannef,
equations can be dc.:ri vcd if the load is on one of the side spans.
If in addition to live-load action \ve have to take into account a temperature
change, the equation for calculating the change H g in the horizontal tension in
the cable is obtained by substituting (see page 53 H) HsI4/ AcEc + ElL 1
instead of HpL/ Ac;Ec into Eq. (11.50). This gives
H ( Hw --t H s)! + EtIJ1 +  ( .1 ) 2 r 1 - .
8 A E I I I 2 J k 2 / 2
r c
+ t an  1/ 2 ] + ! ( 1 ) 2 ! . 1_ [ 1 _ _1 + 24 tanh (kl1/2) ]
k 3 1 3 6 /1 / k 2 /12 k3J13
- .i ( Y Y (0 + :1 I;; 01) 2(/f + I;fl) + I }
= _ HU!ftLi  + Bpi { 3/ ( b 2 - a 2 ) - 2(b 3 - a 3 ) _J [ --- 
I 1 2/3 k 2 J2 J
_  os l - c<? .I._ a + cosh k (1 - a) - cosh !-=--/1 ]
kJ sinh kJ
X r 1 + 2(/f + I) + I ]} (11.59)

ARTICLE 11.7
555
From this equa6on, Hit can be calculated by trial and error, as has been done in
previous cases. 1ake, as a nurnerical exalnple, the case in \vhich I = 800 ft,
1 1 = 400 ft, 1/1 = 0.105, /1//1 = 0.0525, Hw = 3.667 X 10 6 lb. EI =
56.84 X 10 9 Ib-ft 2 , pI = 0.28361H w , f = 65 X 10- 7 , t = 60°F, Ac = 87.8
in. 2 , Ec = 29 X 10 6 psi, L/I = 2.5937, LII/I = 2.4975. For b = I and
for various values of the ratio a/ /, Eq. (11. 59) gives for Hs the values \\:ritten
in the second colunln of Table 11.2. The values of Hs for sInall arnounts of
live load are negative because the temperature risc of 60°, \vhich \\re have
assumed, expands the cable, allo\ving pt.ft of the dead load to be taken by the
sti £fening truss.
To obtain values of the tnOn1cnts /\1 1 and i\J 2 at the supports, \ve use Eqs.
(11.48) and (11.49). Substituting p de for P and integrating from c = a to
(  b, \ve obtain
2M 1 (/!/I + 1 1 !/11) + M 2 /1f' = 41f;,. (/ 1 3 W 1 9 1 + 1 3 'lL'9)
+ 6pl ( cOSh kb - cosh k'l _ /12 - Cl 2 )
k 2 kl sinh kl 2/ 2
lH 1 /<{J + 2M 2 (/t/1 + It!/lI) = 41f; (lt3W181 + /3w8) + l
X r cosh k. ! - _l_ =- _c _ o h k( '-=J.2 _ l_( - a) - ( b 2 - a 2 L ]
L k/ sinh kl 2/ 2
(11.60)
From these equations, "1 1 and ./\.1 2 can be calculated provided H p is found from
Eq. (11.58). Similar equations are obtained also \vhen a combined action of
live load and tcrnperature change is considered. 'fhc values of the moments
TABLE 11.2
,7 Ila l\-f 1 J i\1 2 /
- - - -
/ H"" EI HI
0 0.2430 - 0 . 0160 -0.0160
O. 1 0.2386 -0.0188 0.0030
0.2 0.2230 -0.0290 0.0320
0.3 0.1930 -0.0500 0.0500
0.4 0,1535 -0.0742 0.0520
0.5 O. 1082 -0.0960 0.0410
0.6 0.0625 - O. 1100 0.0180
0.7 0.0222 -0. 1085 -0.0070
o. -0. 0085 -0.0908 -0.0280
0.9 -0.0258 -0.0628 -0.0410
I -0.0305 -0.0445 -0.0445

556 SUSPENSION BRIDGES
0.28
0.20
0.16
0.12 -
0.08
1 0,04
::::. :x:9 0 -
-0.04
- 0.08 '
- 0.12 -
-0.16 -
-0.20
-0.24
0
0.2
0.4 0.6
I-a
---
I
0.14
0.12
0,10
0.08
0.06
0.0.1 1
0.02 I
..
i[
- 0.02
j -0.04
- 0.06
-0.08
-0.10
-0.12
1.0
0,24 - -. . . _.0 - --

t
0.8
FIG.11.12
calculated fronl these equations for our nUIllerical example arc given in the
third and fourth colunlns of Table 11,2. The variation of Hf;,/II w and of
/111/ EI and i\lt 2// EI \vith the length I - a of the loaded portion of the middle
span is sho\vn in Fig. 11.12. v\!e see that the nloment at the left to\vcr is
greatest \vhen the live load extends over 35 percent of the main span.
10 sho\v the effect of continuity of the sri ffening truss at the support, there
is given in Fig. 11. 12 also the curve for Hs calculated on the asslunption that
there arc hinges at the supports A and R. In Inaking this later calculation, \ve
have only to onlit in Eq. (11.58) the terms containing the function X, It is
seen that in the case of hinged spans, the H8 curve is somc\vhat ahove the
corresponding curve for a continuous stiffening truss.
11.8 STIFFENING TRUSS OF VARIABLE CROSS SECTION
The tncthod of trigonometric series previously applied (see Art. 11.5) to a
stiffening truss of uniform cross section can be extended also to cases in \\rhich
the stiffening truss has a variable cross section. 1 \\, start \vith the di ffcrcntial
equation for the deflection curve of the stiffening truss l Eq. (e), page 544],
d 2 fJ \' . Ul7rX
EI d.\,2 - (Hw + IIp)fJ = '- b m SIn I
(a)
1 1'his cxtnsion of the application of trigonOIl1etric series \vas 5ho\\'n by E. Stcuerman; see
S. Tirnoshcnko, J'r.,ns. ASCJ..;, vol. 94, p. 377, 1930.

ARTICLE 11.8
557
in \vhi<.:h
2 ( 1
b m = T Jo (Hpy - mL p ) dx
For the case of one concentrated load fJ (Fig. 11.6) \ve have (see page 544)
16H p f 2P/ . 11lrr(/ - c)
b m = ---- (1 - COS'lllTr) - -- sIn (b)
'lJZ 31r 3 l1Z 2 7r 2 1
For the case of distributed load, as in Fig. 11.8, \ve obtain b rn by substituting
p de for Pinto Eq. (b) and integrating frolll c = a to c = b, \vhich gives
2/2 { 8 f H p
b m = _ / 2 (1 - COS '!Jl1r)
11/ rr'
[ 1 11rr(/ - b ) o111rr(1 - a) ] } (c)
- p COS - I - - CDS 1
L-\pplying no\\' Eq. (a) to the case of a truss of variablc cross section, \ve asun1C
that the flexural rigidity of the truss is defined by the equation
EI = Rlo({J(x) (d)
in \vhich <p(x) is a kno\vn function of x, and take the solution of Eq. (a) in form
of the series
=0
 . 1111rX
'fJ =  am SIn  -r -
m ....1
(c)
Sunstituting expressions (d) and (e) into Eq. (a), \ve ontain
:.0
71" 2 C' l . ( . ) \' . 2 . 11171" .'t'
- 1 2 1"J o<,O.t  111 am sIn I
m=1
00
\' . 1117r X
- (Hw + IIp) L a,,. sIn I =
1n= 1
00
\' I . 1llrr:r
 J m SIn I
m=l
(f)
To calculate the coefficients t11, t12, '13, . . . of the series (e), \ve Inultipy Eq.
(f) by sin (i7r:r/I), \\.here i = I, 2, 3, . . . , and integrate from x = () to
X = J. In this manner, \ve obtain
00 .
27r 2 EI \' t) ( l ( ) . 1111r.\' . t7rX d
- f3 0  In''a m J 0 <,0 x sIn I sIn --r- x
m...l
- tl,(Hw + H p ) = b. (g)
J.i\fter calculation of the integral on the left side, \ve obtain a linear equation
containing the unkno\vn coefficients at, (12, aa, . . ,. In practical calculations
\ve take only the first fe\v terms of the series (e). Let n denote the number of
these ternlS. lhen, taking i = 1, 2, . . . , 1/, \ve obtain JJ linear equations

158 SUSPENSION BRIDGES
(g) from which 11 coefficients '11, (12, . . . , an of the series (e) can be calculated
for any assumed value of HI'. To determine H p , we use the previous Eq.
(11.36):
!!pL = 161 ( al++'" +n )
AcEc 1rl 3 n
1J"2
+ 4/ (a1 2 + 2 2 a2 2 + · . · + 1l 2 a n 2 ) (h)
l""his equation, together \vith 11 equations (g), is sufficient for the solution of the
problem. In each particular case, ,ve start \vith a calculation of the coefficients
b m for the assumed live-load distribution by using Eq. (b) or (c). Then Vle
calculate the integral on the left side of Eq. () for the assumed q;(x) and obtain
11 linear equations for the coefficients aI, (12, . . . , a,l. Taking no\\', as a trial,
a certain value for HI" \ve calculate these coefficients and substitute them into
the right-hand side of Eq. (h). Since the value of IIp \\'as taken arbitrarily,
the equation ,viII not be satisfied, and a second trial value of H p must be taken.
After two such trials the correct value of H p can be obtained \vi th sufficient
accuracy by interpolation. Such calculations show that considerable varia-
tions of flexural rigidity £1 along the span of the truss have only a small effect
on the magnitude of HI' and that a very good approximation for HI' is obtained
by replacing a variable flexural rigidity by its constant average value.
To illustrate the calculations required in the case of a truss of variable cross
section, let us assume that
EI = Elo sin ex + a ) (i)
To have a truss symmetrical \vith respect to the middle, \ve take
S7r + a = 1J" - a
Then
2a
s=l--
11"
(j)
and Eq. (i) gives
(£1)%=O,%-=l = Elo sin a
(£1)x-l/2 = Elo
(k)
Taking, for example, ex = 1r/6, \ve obtain s = j, and the flexural rigidity at the
ends of the truss is only half of that at the middle. The integral in Eq. (g) for
Ip(x) = sin Cx + a )
(I . ( S7f'X ) . m1rX . i7rx d
IS J 0 SIn T + a SIn T sIn T x
_ 2[(- l)m+i+l - 1] slmi cos ex (l)
- 1r [ (m - i) 2 - S2] [(1n + i) 2 - S2]

ARTICLE: 11.8
559
and Eq. (g) bccon1cs
_ 47rs cos . C!. Elo m  n If. ;111_ :(. (- I) m+i =nl1-_ _ ..
/2 -J  m [('111 _ i) 2 - S2] [ (111 + i) 2 - S2]
m=l
- ai(I-I w + HI')
- b .
- 1
Introducing the notations
41rs COS exElo
------ ----. -- = p,
f2(Hw + H p )
il1Z 3r ( - l)m+i+l - 1]
u _u --- -0 _____n-'--___ = N .
[(111 - i) 2 - S2] [('111 + i) 2 _ S2] ml
hi
Hw + H; = Ci
\VC represent Eq. (g) for calculation of the coefficients tlt, (12,
foIlo\ving form:
( 111 )
(11)
(0)
, 11" in the
m=n
lli + IJ. L aml\Tmi + Ci = 0
m=l
t = I, 2, . . . , 11
(p)
Assuming, for example, that ('i  'If' /6, s = i, and taking only four terms of the
series (e), \ve obtain for calculation of the four coefficients the follo\ving four
equations:
( 81 27 X 81 )
at +}J. 64 al - 16 X 140 03 + &1 = 0
aq + Jl ( l 02 - }. a. ) + &2 = 0
.. 3 5 40
( 243 81 2 )
a3 + Jl - i6 X 140 al + 640 03 + C3 = 0
( 81 31 X 81 )
a. + p. - 160 Q2 + 143 a4 + &4 = 0
(q)
It is seen that this system of equations is split into t\VO groups, one containing
the coefficients at, aa of odd order, and the other containing the coefficients
a2, Q.. of even order. The solution of thcse equations for any numerical values
of J.l., Ch C2, C3, &4 can be readily accomplished. For a truss of given dimensions
aoo for an assumed value of H p , \ve calculate the valuc of p. from Eq. (111).
The quantities CI, . . . , C4, for a given load distribution, are calculated from
Eq. (0). Then the coefficients at, .:. . , a. \vill be found from Eqs. (q).
Substituting these coefficients into Eq. (h), \\'c check how accurately the
assumed value of IIp satisfies this equation. With t\\'o trial calculations \ve
find the correct value of H p and the corresponding values of the coefficients.
Equation (r) then gives the deflection curve. The bending moments \viI! be
found from Eq. (11.19). Calculations made for a single-span bridge of the

580 SUSPENSION BRIDGES
Mm..
(a)
(b)
I I
1 2 :: 34.7 ft.(
,-r 1 3 =27.8ft 4
11 = 20.8 ft.
I 262.5 ft
T
x
FIG. 11.13
98.4 ft 164.1 ft
164.1 ft 98.4 ft
same dimensions as the middle span of the Manhattan Bridge and for live load
distributed from x = 0 to x = //4 sho\v that the horizontal tension H p cal-
cu]ated for the constant rigidity Elo diminishes only by about 1 percent when
the rigidity of the truss, given by Eq. (i), is reduced at the ends to one-half of
its value at the middlc.
Calculation of M max for a truss of uniform cross section shows that this
quantity varies along the span as shown in Fig. 11.13a. It is seen that a rational
design requires a reinforcement of the cross section of the truss near the quarter
points and not at the middle of the span, as \vas assumed in the preceding
example. A proper variation in flexural rigidity along the span may be
represented by the equation
EI = Elo ( 1 - a cos 41r X )
1 - a I
(r)
Substi tuting
qJ(x) = 1 ( 1 _ a cos 41r X )
1 - a /
into Eq. (g) and proceeding as in the previous example, we obtain the deflection
and the moments for a truss the flexural rigidity of which varies according to
Eq. (r).
Practically, the rigidity of a truss is not a continuous function of oX but
changes abruptly as s.own in Fig. 11.13h. Calculations made for this easel
show that whereas the moment of inertia of the cross section di ffers by + 20
1 Such calculations, made by using a certain generalization of the series method, 2re given
in the book by Hans H. Bleich, "Die Bcrechnung verankerter Hangebriicken," p. 63,
Springer-Verlag OHG, Berlin, 1935. This book contains several numerical examples com-
pletely worked out. lhc use of finite-difference equations in handling trusses of variable
cross section is shown in the paper by F. Stiissi, Publ. Intern. Assoc. Bridge Structural Eng.,
vol. 4, p. 531, 1936.

ARTICLE 11.8
561
percent of its average value, the values of All max for the middle and for the
quarter points differ from the corresponding values, calculated for the truss
of constant average cross section, only by 2.5 and 3. 5 percent, respectively.
We see that a considerable variation in the flexural rigidity along the span of
the truss produces only a secondary effect on the bending moments. Hence
the usual calculations based on the assumption of a truss of a constant average
cross section give satisfactory results.

Chapter 12
Structural dynamics
12.1 FREE VIBRATIONS: ONE DEGREE OF FREEDOM
In previous chapters, the behavior of structures or structural clements under
static loads only has been considered. Ho\vcver, there are many cases where
the response of a structure to moving or pulsating loads or to suddenly applied
loads must be investigated. Examples of such dynarnic problems arc illus-
trated by the behavior of buildings subjected to earthquakes, bridges subjected
to moving loads, and structures subjected to \vind gusts or bomb blasts.
Several important problems of this kind \vill be considered in this chapter.
Since any structure consists of distributed mass interconnected in some man-
ner by elastic constraints, it \vill perform vibrations if disturbed from its con-
figuration of equilibrium. In dealing \'lJth this problem of vibration of a given
structure, it is sometinles possible to represent the structure by a single rigid-
body rnass supported by an clastic n13ssless spring. Such an idealized rep-
resentation of the structure is called a'lJ1odel. Consider, for example, the case
562

ARTICLE 12.1 563
0 
 
(a) (b)
JV
,-
FIG. 12.1
(c)
(d)
of a simply supported prismatic beam that carries a heavy generator at mid-
span as sho\vn in Fig. 12.1a. Treating the elastic beam as a spring \vhich
supports the 1nass of the generator, \ve obtain the model of the system as sho\vn
in 'ig. 12.1 b. Assuming that motion of the mass is restricted to the vertical
direction, \ve have a system \vith une degree of freed0111; i.e., a single coordinate
x defines the configuration of the system. As a second example, we consider
the portal frame structure sho\vn in Fig. 12.1c, consisting of a very stiff heavy
beam supportcd by flexible vertical columns. Assuming that the beam is
completely rigid and that the columns are massless, we see that for horizontal
motion of the beam, the structure behaves like the spring-supported mass
shown in Fig. 12.1d, and again \ve have a model \\,ith one degree of freedom.
Let us now consider, in detail, the dynamic behavior of a spring-suspended
mass \vith one degrce of freedon1 as sho\vn in Fig. 12.2. Under the action
of the gravity force  the spring \vilI be extended by the amount
IV
.t = T
(a)
\vhere k, called the spring COllstant, denotes the load required to produce a unit
extension of the spring. If the \\'cight W is measured in pounds and extension
of the spring in inches, the spring constant k \vill have the dimension of pounds
per inch. "fhe quantity lIt determines the equilibrium position of the load as
sho\\'n in the figure,
By means of an impulse or suddcn application and removal of a vertical
force, vibrations of the system can be: produced. Such vibrations \vhich arc
maintained by the elastic force in the spring alone arc called free or 11atural
vibrations. In studying these vibrations, we shall use Ne\\'tons second la\\' of
motion stating that the product of the n1ass of a particle and its acceleration is
equal to the force acting in the direction of the acceleration. In this case, the
mass of the vibrating body is JV / g, \vhere g is the acceleration due to gravity.

564 STRUCTURAL DYNAMICS
1-
e
-1
,de
--r:
JV bet
x
r- -...,
I I
I
I I
L..__.J
FIG. 12.2 x
The acceleration of the body is given by the second derivativc of the dispJace-
n1ent .1:, fron1 the equilihriun1 position, \\-,ith respect to time and \vill be denoted
by.t;. "rhe forces acting on the vibrating body are the gravity force J acting
do\\rn\vard, and the force in the spring, \vhich, for the position of the \veighr
indicated in Fig. 12.2, acts up\vard and is equal to U" + k.t. Thus, the dif-
ferential equation of motion, in the case under consideration, becomes
JV
- £ = W - (V + kx) = - kx
rr
,.,
(b)
Introducing the notation
p2 = kg = . g
v d
(c)
Eq . (b) can be represented in the follo\ving fonn:
x + p2X = 0
r-[his equation \vill be satisfied if \ve take either
(12.1)
x = C 1 cos pt
or
x = C 2 sin pt
\vhere ('1 and C 2 are arbitrary constants.
obtain the general solution of Eq. (12,1).
x = C 1 cos pt + C 2 sin pt
By adding these two solutions, \VC
l'hus,
(12.2)
It is seen that the vertical motion of the \\,'cight W has a vibratory character
since cos 1Jl and sin pt repeat themselves after intervals of tinle 'T such that
p(t + T) - pt = 211" (d)
l"his interval of time is called the period of vibration. Its magnitude, from

ARTICLE 12.1
565
Eqs. (d) and (c), is
T =  = 2  = 2 Jf
It is seen that the period of oscillation of the suspended \veight W is the same
as that of a siri1ple pendulum the length of \vhich is equal to 5 8t and is independ-
ent of the magnitude of oscillations. To detenninc the period T, \ve have only
to calculate, or to obtain experimentally, the static deflection 8t. The number
of oscillations per unit time, say per second, is called the fr(qucl1cy of vibration
and \vill be denoted by f. Thus, from Eq. (12. 3a), \ve obtain
(12.30)
f=  =..!.. rg
T 21r '\j
(12.3h)
The vibratory motion represented by Eq. (12.2) is called si111ple han110nic
1noti(Jn. To dctennine the constants C 1 and C 2 in this equation, we must
consider the initial conditions of motion. Assume, for instance, that at the
initial moment (t = 0) the \veight J{l has initial displacC'111Cllt Xo from the equilib-
rium position and initial velocity Xo. '[hen, substituting t = 0 and x = Xo in
Eq. (12.2), \ve obtain C 1 = Xo. Likc\\'isc, substituting t = 0 and x = Xo in
the first time derivative of Eq. (12.2), \VC find C 2 = :i:o/p. \\'ith these values
of the constants C 1 and (J'2, Eq. (12.2) becomes
%0 .
x = Xo CDS pt + - sIn pt
p
(12.4a)
It is seen that this exprtssion for vibrations consists of t\\'o parts; one, depend-
ing on initial displacenlent, is proportional to cos pI, and the other, depending on
initial velocity, is proportional to sin pt.
Sometimes it is advantageous to represent thc solution (12.4a) in another
form by means of rotating vectors. Referring to Fig. 12.3, let OB be a vector
of length Xo \vhich rotates in the counterclock\vise dircction about the origin 0
\vith constant angular velocity p. Counting time from thc instant t = 0 \vhen
o
Xo cos Pt!
:0 sin Pt{
-- -- ----,;?C
;'
-
-------......
Xo
B
A
;'
FIG. 12.3
x

566 STRUCTURAL DYNAMICS
x
FIG. 12.4
Xo
o d--
t
this vector coincides \\lith the x axis, \ve see that at any time t it \vill Blake \vith
the x axis an angle pt, and its projection on the .r axis represents the first tenn in
Eq. (12.4a). No\v, at right angles to the vector DB , let us construct a second
vector BC of length io/p. '[hen its projection on the x axis represents the
second term in Eg. (12.4a), and \ve see that the projection on the .'t o axis of the
resultant vector DC' , of length A, represents the conlpletc solution given by
Eg. (12.4i1). Denoting the angle bet\vcen OB and OC by ex, \ve may no\v
\vrite expression (12.4,1) in the form
.to = A cos (pt - ex)
(12.4b)
\\there, by reference to Fig. 12.3,
A = {x02 + (y
(e)
1 Xo
a = tan- -
p.'t'o
(I)
The quantity A, representing the nlaximum displacement of the vibrating
\veight, is called the a111plitude of (vibration; it is readily calculated from Eq. (e)
\vhcn Xo and Xo are given. The quantity ex, defined hy Eq. (f), is called the
phase illlf!.le. v,, see frolll Fig. 12.3 that alp represents the time that elapses
after the initial nloment t = 0 until the vibrating \veight reaches the first
cxtrlnle position. 'J"his ti111f lag is also sho\vn in Fig. 12.4, \vhich represents
Eq. (I 2,4b) in graphic form,
Natural frequencies of vibrating systerns can often be calculated by using
the lav.r of conservation of energy, provided damping is negligible. Consider,
for example, the system sho\vn in Fig. 12.2. Neglecting the mass of the spring
and considering only the mass of the suspended body, the kinetic energy of the
systcrl1 during vibration is
1 = lV .t 2
g 2
(g)

ART. CLE 12.1
567
'[he potential energy of the systern in this case consists of t\VO parts: (1) the
strain energy of deformation in the spring and (2) the potential energy of the
\veight W by virtue of its position. Considering the strain energy first, the
tension in the spring for any displaccnlenr x fronl the equilibrium position is
k(o" + x), and the corresponding strain energy is k(oll' + x)2/2. For the
position of equilibrium (x = 0), this energy becomes kOte/'/2. Hence, the
energy stored in the spring during the displacement x is
k(o. + X)2 ko 2 kx 2 kx 2
- t2-- - - i t. = kO,t X + -2- = Wx + -- 2 -
The potential energy due to position of the \\"eight W diminishes during the
displaccnlcnt x by the amount J,V x. Hence, the total changt in the potential
energy of the system during the displacement X is
kx 2 kX2
V = Wx + -2- - Wx = 2-
(h)
It will be noted that this is simply the strain energy in the initially unstressed
spring extended by thc amount x.
Having expressions (g) and (11) and neglecting damping, \ve see that the
equation of conservation of energy for the vibrating system becomes
W x 2 kx 2
- - + -- = const
g 2 2
(i)
The magnitude of the constant on the right-hand side of this equation is
detennincd by the initial conditions of motion. Assuming that, for t = 0,
x = .1'0 and i: = 0, the initial total energy of the system is k.\'02/2, and Eq 0 (i)
becomes
W .i 2 kx 2 kX02
- - + ---. = --
g 2 2 2
( j)
When, during vibration, thc displacement .t becomes equal to Xo, the velocity x
becomes equal to zero, and the energy of the system consists of potential
energy only. When x beconles equal to zero, i .c., \\Then the vibrating body
passes through its Iniddle position, the velocity x has its maximum value, and
\ve obtain from Eq. (j)
.  = kxo_ 
g 2 2
(12.5)
rhus'l the maximum kinetic energy of the system in its middle position is (,.'qual
to the nlaxin1um potcntia I energy in an extreme position.
In all cases \\>,hcre it can be assun1cd that the vibration of a system is a simple
harmonic motion, Eq. (12.5) can be used to calculate the frequency. We

568 STRUCTURAL DYNAMICS
simply aSSUl11e that the motion is rcpresented by the equations
.'t' = .\'0 cos pt
x = - .'l'op sin pI
Then \ve see that for such simple harmonic rnotion
.t max = p:'to
(k)
Substi turing this relationship into Eq. (1 2.5), \ve find
kg
p2 = Hl
\vhich coincides \vith Eq. (c) obtained previously. Several examples of this
n1ethod of frcqutncy calculation \vill be sho\vn in the next article.
PROBLEMS
1 A \veight lV = 500 Ib is dropped through a height !J = I in. onto the lniddlc of a
sin1ply supported steel bealn having a span / = 8 ft and a cross-sectional I110I11cnt
of inertia I = 12 in." Neglecting the 111ass of the bean1, calculate the rnax ilnun1
deflcction of the bearn and the period T of the ensuing free vibrations.
/lJJs. Oml1x = 0.252 in., T = 0.051 t sec.
2 An elevator cage of \veight IV = 10 kips is supported by a flcxible t\visted steel
cable the upper end 0 f ,,,hich is un\vinding frolll a rotating drunl. As the cage is
being l()\vcred \virh unifornl vdoeity Vo = 3 fe/see, the drUtl1 suddenly stops \vhen
the length of ulHvound cable is J = 50 ft. ''''hat \\fill be the rnax inlum tensile stress
induced in the cable if its cross-sectional area is A = 2 in. 2 and its IT10duius of
elasticity is l' = 20(10 6 ) psi?
L411L O'nllLX = 29,300 psi.
3 C:alculatc the natural period of free lateral vibration of the portal fralne sho\\'n in
Fig. 12.lc if IV = 2 kips and the length of each vertical colulnn is I = 10 ft, Each
column is a channel section having a cross-sectional area A = 4 in. 2 , least radius
of gyration r = 0.60 in., and l = 30(10 6 ) psi. The columns are built in both
top and Dattoln. The cross beam is cornplctcly rigid.
Ails. T = 0.583 sec.
12.2 RAYLEIGH'S METHOD
In discussing free vibrations in the preceding article, \ve sirllplified the proble(T1
by neglecting the Inass of the spring in comparison \vith the mass of the sus-
pended body. In order to dctrminc the effect of such simplification on the
frequency of vibration and to calculate a closer approximation, a nlcthod
developed by Lord Rayleigh 1 \vill no\\" be discussed. In applying this 111cthod
son1e assurnption regarding the configuration of the vibrating system has to be
1 See his book "Theory of Sound," 2d ed., vol. I, pp. Ill, 287, 1894.

ARTICLE 12.2
589
made. rhe frequency of vibration \vill then be found from Eq. (12.5) of the
preceding article.
As a simple example of the application of l{ayleigh's method, we take again
the case represented in Fig. 12.2. LA\.ssuming that the mass of the spring is
small in comparison \'lith the mass of the vibrating body  the 1node of
vibratio1J \viII not be substantially affected by the mass of the supporting spring,
and, \\.'ith sufficient accuracy, it can be assumed that the displacement of any
clement of the spring at a distance e from the fixed end is the same as in the
case of a massless spring, i.e., proportional to the distance c from the fixed end
equal to
e
7 x
(a)
\vhere I is the natural length of the spring.
If the displacements arc not affected by the Inass of the spring, the expression
for the potential energy of the systeol \vill be the same as shown by the right
side of Eq. (12.5), and only the kinetic energy of the system has to be recon-
sidered. Letw denote the \veight of the spring per unit length. Then the
nlass of an clement of length de will be wdejg, and the corresponding kinetic
energy of the element, if \ve use expression (a), becomes
'tV ( xe ) 2
- - de
2g I
(b)
l'he total kinetic energy of the spring then becomes
w f l ( ;tC ) 2 _ .i: 2 wi
- - - de - - -
2g 0 I 2g 3
(c)
This kinetic energy must be added to that of the load  so that the energy
equation becomes
x : ( w + WI ) + kx 2 = kX02
2K 3 2 2
(d)
Comparing this v.,ith Eq. (j) of the preceding article, it can be concluded
that, in order to estin1atc the effect of the nlass of the spring on the frequency of
free vibrations of the system in Fig. 12.2, it is only necessary to add one-third
of the \"eight of the spring to the v.Tcight Jt" of the suspended load.
1'his conclusion, obtained on the assumption that the \veight of the spring
is very small in comparison \vith that of the load, can be used \vith sufficient
accuracy even in cases \vhcre the \veight of the spring is of the same order as
W Comparison \vith the rigorous solution shows that in the case wi = O.5
the error of the approxilnatc solution is about} percent, and for wi =  the
error is about f percent.
As a second example, let us consider the case of vibration of a beam of

570 STRUCTURAL DYNAMICS
de
01 r-c-j
 
· 4
FIG. 12.5

I
1
2
uniform cross section loaded at the middle (Fig. 12.5). If the ,,'eight 'u.'J of
the beam is small in comparison \virh the load  it can be assumed \vith
sufficient accuracy that the deflection curve of the beam during vibration has
the same shape as the static deflection curvc. Then, denoting by :t' the dis-
placement of the load W during vibration, the displacement of any clement wdc
of the beam, at a distance c from the support, \"iB be
3c/ 2 - 4c 3
X J3 (e)
and the kinetic energy of the entire beam \vill be
{l/2 w ( .. 3C/2 - . 4C3 ) 2 _ 17 i'2
2 Jo 2g X /3 de - "IT wi 2g
1'his kinetic energy of the vibrating beam must be added to the kinetic energy
Wx 2 /2g of the load concentrated at the middle in order to estimate the influence
of the \veight of the beam on the period of vibration; i.e., the period of vibration
\"ill be the same as for a massless beam carrying at the middle the load
(f)
W + iJw/
}1:xpression (f), developed on the assun1ption that the \veight of the beam is
smaJl in comparison \\J.ith that of the load f can be used in all practical cases,
even in the extreme case of an unloaded beam, \vhere W = O. In such a case,
if the assumption is tnadc that jiw/ is concentrated at the middle of the beam,
the accuracy of the approximate method is sufficiently good for all practical
applications. The deflection of the beatn under the action of the load jt wl
applied at the middle is
17 /3
lIt = 35 wI 48£/
Substituting this into Eq. (12.30), the period of natural vibration is
I ., J wl 4
T = 27r \j-g = 0.632 \J EIg
lhe exact solution for this case is
2 rwJ4 }WJ4
T = ;\j EIg = 0.637 \j EIg

ARTICLE 12.2
571
FIG. 12.6
r a -1. b---1
 I
-i>7 II II 
  f-- d d ; -1 1-- 7/
It is seen that for this cxtrClTIe case the error of the approximate solution is less
than J percent.
'rhe same nlcthod can be applied also in a Il10re general case \vherc the load
H.T is not placed at the n1iddle of the bearn (Fig. 12.6). rrhe statical deflection
under the load in this cas' is
lVa 2 h 2
Ot = 3/£1
(g)
and the spring constant is
3/£/
k = ----
" I ?
a" J.
(11)
If \\le neglect the 111ass of the bearn, the period of vibration, fronl Eq. (12. 3a),
\vill be
2 !f:t /W " a 2 h 2
T = 1i '\j Si = 21r '\j -3 IEj
(i)
'10 take the rnass of the bean! into account, \ve use the expressions
Vb I)
1'1 = - _L [a(/ + b) - N 1
. 6/£/
lJl tl'l1
X2 = 6/E1 [b(/ + a) - '112]
(j)
for the static deflections of the t\VO portions of the beam. l'hcn, denoting by
,t uH1X the maximum vclocity of the load  the velocities for the t\VO portions of
the beam \vill be
x
( .. ) .. 1
.\1 Innx = X mBX 5 ..
t
( ) J,' ?
X2 max = X mnx _::.
u'c
and the corresponding kinetic energy is
\vhere
U' 1RX [ (a ( ''. )  d + (b ( ,'2 ) 2 d'l1 ] =  1ax (at1 + (3b) (k)
..g Jo Ost Jo Ol't 2g
1 /2 23 ,7 2 8 al
a = "3 /J 2 + 105 b 2 - 1-, b 2
13 = _1 _ (I + a)2 + _ ! _ !J JI +a 
1 2 ,, 2 28 if'.! J 0 a 2

572 STRUCTURAL DYNAMICS
--fo take care of the tnass of the beam in calculating the period of vibration, \VC
have only to add the kinetic energy [Eq. (k)] of the beaII1 to the kinetic energy
V (x)  ax/ 2g of the \vcight lV. Then, instead of Eq. (i), \ve obtain
T = 27r / JlfT +-  a + iVb)t12b2
'\j 3 g/El
(I)
In the case of a cantilever beam \vith a \veight lJ/T at the free cnd (Fig. 12.7),
the same method can be used in calculating the period of vibration. Assuming
that during vibration the sIlape of the deflection curve of the beam is the saIne
as the one produced by a load statically applied at the cnd and denoting by x
the vertical displacclnent of the load J \VC see that the kinetic energy of
the cantilever of uniform cross section \vill be
f I 'lL'.i' Z ( 3e 2 1 - C 3 ) 2 33 .t . 2
- -- .- - --- de = - - 'wi -
o 2 K 2F 140 2 g
('111)
Thus, the period of vibration \vill be the same as for a massless cantilever
carrying, at the end, a \veight
lV + llo7.vl
This result \vas obtained on the assurnption that the \\'eight '"I.,vl of the bearn is
snlall in conlparison \vith v, but it also is accurate enough for cases \vherc 'lvl
is not small. i\pplying tf.e result (HZ) to the extrcnlC case \vhere lV = 0, \VC
obtain
3 3 wl/ 3
Ollt = 140 TEl
and the corresponding period of vibration is
T = 27T . = 3:7  i:Ji
l'hc exact solution for the saIne case is
27r /'1])/4
T = 3.5 1 5 \j III g
It is seen that the error of the approxiInate solution is about 1-. percent.
FIG. 12.7
de
rcJ:____  +
f-- r -t

ARTICLE 12.2
571
--- - ---
Xl -X-;i--X3---
FIG. 12.8
'Vi
W 2
Similar approxinlate solutions can be made also in the case of a portal frame
like that in Fig. 12.1 c, provided the cornpressive forces in the columns are
small in comparison \vith the Euler loads.
The described approximate method of frequency calculation can be used
also in the case of a beam supporting several loads WI, U'h U'3. If Xl, X2, Xa
are statical deflections under the loads, the potential energy of defonnation
stored in the beam is
v = -!(WIXl + W 2 X2 + J¥aX3)
(1/)
In calculating the frequency of the lo\vcst mode of vi bration, the static deflec-
tion curve, shown in Fig. 12.8, can be taken as an approximation for the
extreme configuration of the system during vibration. l'hen, assuming sinlple
harmonic motion, we see that the displaccrncnts of the loads WI, W 2, W 3
during vibration \vill be
Xl cos pt
X2 cos pt
.\'3 cos pt
(0)
The kinetic energy of the systen1 ecomes a rnaximum at the moment \vhen the
beam, during vibration, passes through its middle position. It \\rill be noted,
front expressions (0), that the numerical values of the load velocities cor-
responding to this position are PXh pXz, PX3, and the kinetic energy of the
systenl becomes
T = f; (W 1 X 1 2 + W 2 X 2 2 + W 3 X 3 2 )
Equating expressions (u) and (p) in accordance \vith Eq. (12.5), we obtain
(p)
., W 1 X l + W 2 X 2 + W a X 3
p- = g U'lX 2 + W2X22 + W 3 X 3 2
and the period of vibration is
T = 2?r = 21r I. W 1 .X1 2 + 'Jf2 X 2 2 + .3X32-
p  g(W1Xt + W 2 X 2 + W 3 X 3)
(12.6)
I t is seen that for calculating T, the statical deflections Xl, X2, .\"3 of the beam alone
are necessary. These quantities can casily be calculated by the usual methods.
If the beam has a variable cross section, a graphical method of getting the
deflections can be used to advantage. The effect of the \veight of the
beam itself also can be taken into account if desired. It is only necessary, for

574 STRUCTURAL DYNAMICS
this purpose, to divide the beam into several parts, the weights of \vhich,
applied at their respective centers of gravity, must be considered as concen-
trated loads.
PROBLEMS
1 Referring to the portal frame in Fig. 12.1c, \vhat portion of the total weight wi of
each vertical column should be added to the weight W of the rigid beam in order
to correct for the mass effect of the columns on the period of free lateral vibration
of the franle? Neglect the effect of axial compression in the columns on their
flexural rigidities, and assume built-in ends.
Ans. 13w/135.
2 A thin circular disk having mass moment of inertia Jo.,vith respect to a diameter is
rigidly attached to one end of a shaft of length / and weight 'U)J, as shown in Fig. 12.9.
lne shaft is supported in brings A and B, \vhich are loose enough to allow the
shaft to bend as sho\vn. For small amplitudes of rotational oscillations of the disk
about its horizontal diameter, the period of oscillation, neglecting the maSS of the
shaft, is T = 211'" Vlo lk , \vhere k = 3EIII, EI being the flexural rigidity of the
shaft. To correct this period for the mass effect of the shaft, \ve should use,
instead of the l11on1ent of inertia 1 0 , the increased moment of inertia 10 + aw/ 3 /g.
Calculate, by Rayleigh's I11Cthod, the proper value of the factor a.
Ans. a = rio.
3 Neglecting the weight of the canti lever I bearn shown in Fig. 12.10, and using
Rayleigh's method, Eq. (12.6), calculate the approximate period T of the funda-
mental mode of free vibration of the system. The following numerical data arc
given: v = 1,000 lb, I = 15 in.4, E = 30(10 6 ) psi.
Ans. T = 0.27 sec.
4 Using Rayleigh's method, calculate the approximate period of the fundamental mode
of free vibration of the beam carrying two weights W as shown in Fig. 12.11.
Neglect the distributed mass of the beam and assume W = 1,000 lb, E = 30(10 6 )
psi, 1 = 15 in..
Ans. T = 0.248 sec.
FIG. 12.9
A
I


4'
 
 
 5'-+--5'-+-5'
i
I"
4'
I
FIG. 12.10
FIG. 12.11

ARTICLE 12.3
575
12.3 FORCED VIBRATIONS: STEADY STATE
In preceding articles, it \vas assumed that vibrations \\fere produced by impart-
ing to the suspended mass some initial displacen1ent or initial velocity. Such
vibrations are called free vibrations. Because of various kinds of damping such
as friction, air resistancc, and imperfect elasticity of the spring, free vibrations
arc gradually dan1pcd out and soon disappear. \\'e shall no\v consider another
kind of vibrations produced by the action of an externally applicd disturbing
force; these are called .f()rced "l)ibrathl1u. In practice, periodic disturbing forces
produced by some unbalance of rotating Il1achines arc especially iIl1portant.
In Fig, 12,12a, such a disturbing force is represented by the centrifugal force of
an unbalanced rotor, I)cnoting by Pu the magnitude of this force and Illeasur-
ing the angle of rotation wi as sho\\'n in the figure, \ve obtain vertical and
horizontal components of the disturbing force equal to Po sin wt and Po cos wt,
respectively. If the fralTIe of the n1achine is rigidly attached to the foundation
(Fig. 12. I 2a), it \vill have no Illation and the total centrifugal force \vill be
transmitted to the foundation. ']0 diminish this transmitted force, a spring
mounting as sho\vn in Fig. 12.12b is somctin1es used. Assuming that thcre
arc rollers to prevent lateral tnOVCll1cnt of the fraIne lJ" \ve obtain a SYStcIl1
similar to that previously considered in Fig. 12.2, 'fo determine the pulsating
vertical force translnitted through the springs to the foundation, vertical vibra-
tions of the fran1c under the action of the disturbing force j>o sin wt nlust be
conidered. l'he required differential equation of motion \vil] be obtained if,
to the previously considered forces r see Eq. (b), page 564], the vertical dis-
turbing force \vill he added. Thus,
HI
- £ = J..JI - (If' + k:t:) + Po sin wI
g
(,1)
and \vith the notations
j)oR
qo =-
V
k
p2 = W'
(b)
,v
FIG. 12.12
(a)
(b)

576 STRUCTURAL DYNAMICS
the equation of motion (a) becomes
.;i& + p2J.' = qo sin wt
A particular solution of this equation is obtained by assuming
(12.7)
x = A sin wI
\vhere A is a constant. Substituting this into Eq. (12.7), \\'e find
A = --
p2 - w2
and the required particular solution becomes
qo .
x = - SIn wt
p2 - w2
(12.8a)
Adding to this the expression (12.2), representing free vibrations, \\'C obtain
the complcte solution of Eq. (12.7):
x = C 1 cos jJt + C z sin pt + -' 2 .3 0 - 2 sin wt (12.9)
p - w
It is seen that this solution consists of two parts: l"hc first t\\'o teons represent
free vibrations, already discussed in Art. 12.1, and the last term, depending on
the disturbing force, represents forced vibrations. It is seen that the latter
vibrations have the same period Tl = 21rlw that the disturbing force has.
Considering only these forced vibrations and using notations (b) for qo and
p2, Eq. (12.8a) takes the fann
o 1 ·
x = - k 1 2 I 2 sIn wt
- w,p
(12.8b)
In this expression, (Polk) sin wt represents the displacement that the disturbing
force \\fould produce if acting statically, and the factor I I (1 - w 2 I p2) accounts
for the dynamic nature of this force. The absolute value of this factor is called
the 111agnijicatioll factor. It depends only on the ratio wi p, \vhich is obtained
by dividing the frequency of the disturbing force, called the i1JlprCssedfrequcllcy,
by the natural freque11cy of the system. In Fig. 12.13, values of this factor,
denoted by {3, are plotted against values of the ratio wlp. It is seen that for
smaIl values of wlp, that is, when the impressed frequency w is small compared
\vith the natural frequency p, the magnification factor 13 is approximately
unity, and dynamic displaccn1cnts arc about the same as they \vould he in
the case of a purely statical action of the force Po sin wt. As the ratio w/p
approaches unity, the magnification factor and the amplitude of forced vibra-
tions rapidly increase and become infinite for w = p, that is, for the case when

ARTICLE 12.3
577
{3
4
1
3
2
FIG. 12.13
o
o
O.IS
1.0
1.5
2.0
2.5
3.0 w
p
the impressed frequency coincides \vith the natural frequency of the system.
This condition is called resonallce. 'J 'he infinite value of the magnification
factor at resonance indicates that the amplitude of vibration increases indefi-
nitely if there is no damping. In practice, \\'e al\vays have some damping, and
this kecps the amplitude finite at resonance, although it may bccon1c danger-
ously large.
\\Then the frequency of the disturbing force increases beyond the condition
of resonance, the Inagnification factor again becomes finite, and its absolute
valuc diminishes as w/p increases so that it approaches zero as sho\\'n in Fig.
12.13. This means that a high-frequency pulsating force produces forced
vibrations of very small amplitude, and in n1any cases thc body W may be
considered as remaining in1movable in space. The practical significance of
this situation \vill be discussed later,
Forced vibrations of a system like that in Fig. 12.12b can be produced not
onl}' by a periodic force but also by a periodic moven1cnt of the foundation.
Consider, for exan1plc, the spring-mass system sho\vn in Fig. 12.2, page 564,
and assume that the upper end of the spring supporting the \vcight U' perforn1s
a simple harmonic luotion
t = (l sin wt
(c)
in the vertical direction. Then, measuring displacement .'t' of the suspended
\vcight v from its equilibrium position, \\rhen  = 0, the elongation of the
spring at any instant t \vill be .'t' -  + Oah and the corresponding tension in
the spring is k(x - ) + W 1'hus, the equation of motion of the suspended
\veight becomes
W £ = W - k(x - ) - W = -k(x - )
g
(d)
Substituting for  its expression (c) and using the notations
akg
qo =-
W
kK
p2 = 
W
(e)

578 STRUCTURAL DYNAMICS
Eq. (d) becomes
oX + p 2 x = qo sin wt
\vhich is identical \\,ith Eq. (12.7). "[hus, it may be concluded that giving
the upper end of the spring a gr(Jlll1d 'Jllotio'fl a sin wt is equivalent to applying
directly to the suspended \\'cight JV a disturbing force ak sin wt. AU previous
conclusions regarding the solution of Eq. (12.7) hold also in this case, and \ve
again obtain steady-state forced vibrations defined by the equation
1 .
x = a _ 1 2/ 2 sIn wI
- w P
This coincides \vith Eq. (12.8b) for ak = Po.
As a first application of the foregoing theory, lct us consider a device, called
an oscillator, \vhich is used for experimentally determining the natural frequency
of vibration of various engineering structures 1 (Fig. 12.14). 1"his device
consists of t\\'o identical disks rotating in opposite directions \\,ith constant
angular velocities w as shov.rn in the figure. The bearings of the disks are
housed in a rigid fralne \\rhich is attached to the structure, the vibrations of
\vhich are to be studied. By attaching to the disks unbalanced "reights as
sho\\7n, \VC see that the centrifugal forces P produced during rotation of the
. disks give a resultant pulsating force 2P sin wt acting along the axis of symmetry
11111. lhis produces forced vibrations of the structure \vhich can be recorded
by proper instruments. By slo\vly changing the speed of the disks, \ve can
establish the number of revolutions per second at \vhich the amplitude of forced
vi brations of the structure becomes a maximum. Assuming that this occurs at
resonance, the natural frequency of free vibration of the structure is then equal
to the observed number of revolutions per second of the disks. Such oscil-
lators have been very useful in dctennining the natural frequencies of large
structures such as bridges, office buildings, and missile launching stands.
As a second example, let us consider an instrulncnt for nlcasuring the ampli-
tude of vertical vibrations. It consists of a \veight JV suspended inside a box
on rather flexible springs, as sho\vn in Fig. 12.15. If the points of suspension
A arc in1movable and free vertical vibration of the \vcight W is produced, the
equation of motion (12.1) can be applied. Assume no\\r that the box con-
(12.8c)
1 Such an osciUator is described in a paper by "Vt Spath, Z. Jt'tr. Deut. lng., vol. 73, p. 963,
1929.
FIG. 12.14
 m!
,.k !
n!

ARTICLE 12.3
FIG. 12.15
579
A
A
JV
-Lx
L-----.J T
taining the suspended ,"'eight is attached to a foundation performing small
vertical vibrations. In such a case, the upper ends A of the springs Inove with
the foundation, and forced vibrations of the weight W \vill be produced. Let
us assume that vertical vibration of the box is represented by :Eq. (c). l'hen
\ve have forced vibrations of W as represented by Eq. (I2.Sc), and it is seen
that \\'hen w is small compared \vith p, the displacement x is approxinlately
equal to t, and the suspended \veight W performs practically the same motion
as the box.
Consider no\v the case \vhen w is very large compared \vith p, that is, the
frequency of vibration of the box is high in comparison with the natural
frequency of the suspended \vcighr w: rhen, the amplitude of forced vibra-
tion, Eq. (12,Sc), becomes very small, and the \\7cight W can be regarded as
immovable in space. l'he box sin1ply moves up and down around it. Taking,
for instance, w = lOp, we find that the amplitude of forced vibrations is only
a/99. Thus, in such a case, vibrations of the box are scarcely transmitted to
the \veight W at all. This fact is utilized in various instruments for measuring
and recording the amplitude of vibrations. If a dial is attached to the top of
the box as sho\vn in Fig I 12.15 \vith its plunger in contact \vith the upper face
of the \veight  then, during vibration, the hand of the dial moving back and
forth \vill record the double amplitude of relative motion of W \\'ith respect
to the box. From Eq. (I2.Sc), this amplitude is equal to the maxin1um value
of the expression
x -  = a sin wt C _ '/P' - 1)
from \\,hich
W 2 /p2
(x - )max = a 1 _ w'/p' (12.10)
When p is small in comparison \vith w, the numerical value of this expression
is very close to the amplitude a of the vibrating body to \vhich the instrument

580 STRUCTURAL DYNAMICS
- I
A
a
FIG. 12.16
is attached. Such an instrument is very accurate for measuring amplitudes of
high-frequency vibrations and has proved very useful in power plants for
studying vibrations of turbogcncrators. 1'he springs of the instrument arc.
usually so chosen that the natural frequency is about 200 cycles/min. Then if
a turbogencraror runs at 1,800 rpm, \ve get the amplitude of forced vibration
\\' i th good acct1 racy.
To get a conlpletc record of the vibrations in Fig. 12.15, a cylindrical drum
rotating \\lith constant speed about its vertical axis can be placed inside the box.
Then a pencil attached to the \veight IV and pressing against the drum \vi II
produce a completc record of the relative motion, Eq. (12.10), bct\veen the
box and the \vcight J.t: Introducing horizontal springs in addition to the
vertical springs, \ve see that the same instrument can be used to measure
horizontal vibrations.
For recording vibrations of lo\v frequencies such as \ve encounter in large
heavy structures, the instrument sho\vn in Fig. 12.15 is not practical since, to
attain the required very lo\v natural frequency for the instrument, extremely
flexible springs must be used, and the static deflection bccotnes excessive.
"fhis difficulty is overcome by using an instrument like that sho\vn in Fig.
12.16. In such an instrument, we have
8.t =  GY
so that the natural frequency is
f = 2:-t = 2 ] 
By adjusting the ratio aj I, this frequency can be made as small as desired.
(f)
(g)
PROBLEMS
A large generator having total \veight W = 20 kips is mounted at the middle of t\VO
simply supported parallel I beanls, each of '\'hich is a standard 23-1b 8-in. rolled-
steel section \vith a clear span of 12 ft. The generator runs at 600 rpm and is out
of balance to the extent of 5 ]b at a radius of 10 in. '-""hat will be the alnplirude of
steady-state forced vibrations? Neglect the mass effect of the I beams.
A11s. Xma = 0.0036 in.

ARTICLE 12.4
581
6,-t3'
I
I
I
.x
+
FIG. 12.17
FIG. 12.18
ld o cos wi

) 
0)

5' --.t--- 5' --i
2 'The dial gauge of the vi brograph sho\\rn in Fig. 12.15 registers a relative amplitude
of 1: 0.08 in. during vibration of the foundation to \vhich it is attached. _Assuming
that vertical rnotion of the foundation is sio1plc harrnonic and has a frequency of
1,200 cycles/min, '.vhat is its true an1plitudc a? '[he static deflection of the sus-
pended \veight rv i:\ 03t = 1.00 in.
,/lI1S. tl = O.07H in.
3 ;\ steel beam havi:1g flexural rigidity EI = 30 X 4( 106) Ib-in. 1 is supported as
sho\vn in Fig.. J 2.17 and carries a \\'tight J.V = 600 Ib at' its fi'cc end. (onlpurc
the amplitude of steady-state forced vibrations of the \veight v if the support A
performs srnall vertical vibrations defined by the lxpression XA = tl sin wt, \vhere
t1 = 0.12 in. and w = 30 see-I. lhc support B does not Inove, and the Inass of
the beam can be nelectcd.
.4.1/J. ."l'max = 0.1 31 in.
4- .t'he simply supported beam in Fig. 12,18 consists of t\\'o 6- X 2-in. 13-1b steel
channel sections set back to back and supporting at mid-span a \veight Hi" = 12 kips.
(olnpurc the an}pJitude of tcady-tatc f()rced vibrations of Hr if a pulsating externaJ
moment Alo cos wt acts on the left end of the bcaJl1 as sho\\rn. l\ssutnc that
w = O.9p, \vhere p = V'g/-::;, and .."'[0 = 1,000 in.-Ih.
/ll1S. Xmll = 0.00457 in.
12.4 GENERAL CASE OF A DISTURBING FORCE
In the preceding article, only the stcady-state forced vibrations of a spring-
suspended mass produced by a periodic disturbing force Po sin wt \\"cre con-
sidered. \l cry often, cspecially in structural dynan1ics \ve may have to deal
\virh a disturbing force Q(t) that is not periodic. In such a case there \vill be
no steady-state forced vibrations, and \ve need to kno\\r the con1pletc response
of the svstem \vhich 'NiB consist of a 111ixturc of both frec and forced vibrations
..
caused by the disrurbjng force. l\ fi1cthod of obtaining a so]ution of the
differential equation in such cases \viJI no\\' be considered. 1
1 This Incrhod \\'as introduced by C. G. Stokes in a paper on diffraction. See HCollected
Papers," \'01. 2, p. 243. It was also \videly used by Lord Rayleigh in his book "Theory of
Sound," p. 74.

512 STRUCTURAL DYNAMICS
To explain the method, let us consider the system sho\vn in Fig. I2.19a
consisting of a spring-supportcd \veight U/ constrained to move vertically and
acted upon by a disturbing force Q(t). l'he magnitude of this disturbing force
per unit of mass, denoted by q (t), is rcprescnted as a function of time by the
ordinates of the curve 1nll in Fig. 12.19 b. Assuming that at the initial moment
t = 0, the weight W is at rest in its equilibrium position (..\'0 = 0, Xo = 0), it is
required to find the displacement of the body at any time t = (1. To calculate
this displacen1cnt, \ve imagine that the continuous action of the force q is
reprcsented by a series of impulses q dt one of \vhich is represented in Fig.
12.19b by the shaded strip. As a result of this one impulse occurring at the
instant t, the body obtains an incren1cnt of velocity
dx = q dt
(a)
\vhich states that the change in tnomentum per uni t mass is equal to the cor-
responding impulse. No\v considering this increment of velocity as an initial
velocity imparted to the body at the instant i, it follows from the solution for
free vibrations represented by Eq. (12.4a), page 565, that the displacement of
the body at some later time 11 \\,'ill be
d dx. ( ) q dt . ( )
.t = - SIn p t1 - t = --- sin Ptl - pt
p p
(b)
if no further impulses are considered. To obtain the displacement x of the
body at time t1 due to the continuous action of the disturbing force from t = 0
to t = tJ, it is only necessary to sum up expressions (b) over this time interval.
This gives
x = ! {h q sin (pt! - pt) dt (12.11)
P Jo
This expression includes both free and forced vibrations produced by the dis-
turbing force, and it is especially useful in studying the motion of a system due
to some kind of transient disturbance such as a wind gust or bomb blast. It
can be used even in cases \\1here the variation of q \vith time cannot be expressed
analytically but is only represented in graphical form. It is only necessary in
q(t)
m
n
o
t---f dt
t
FIG. 12.19
(a)
(b)

ARTICLE 12.4
583
such cases to evaluate the integral (12.11) by some numerical method of
quadrature such as Silnpson's rule.
If the \\'cight llr has some initial displacement Xo and some initial velocity
.to at time t = 0, the solution for :r at time t -= tl becomes
+ .t o . 1  tl . ( ) d
x = Xo cos Ptl - SIn pt} + -- q SIn ptt - pt t
P P 0
(12.1Ia)
as is readily seen by referring again to the solution (12.4a), page 565.
i\S an exarnple of the application of Eq. (12.11), let. us again consider the
case of a periodic disturbing force Q.o sin wt and assun1e .\'0 = .t o = O. Sub-
stituting q = '10 sin wi, \vhere qo = Qog/JV, into Eq. (12.11) and observing
that
sin wt sin (pt! - pt) = i[ cos (wt + pt - ptl)
- cos (wt - pt + Pit)]
\ve obtain, after integration,
qo ( . w. )
.t = ---;,- SIn wtl - - sin Ptl
p" - w p
(12.12)
The first tcrn1 in this result represents forced vibrations discussed in the
preceding article [see Fq. (12.8a)], and the second term represents free vibra-
tions. I f the assumed initial conditions (xo = 0, .t o = 0, \vhen t = 0) are
suhstituted into the general solution (12.9) on page 576, \ve find
C 2 = _  . _ qo __
P p2 _ w 2
and the solution (12.9) becorncs identical \vith the solution (12.12).
Let us no\\' consider the application of Eq. (12.1 I) in the case of an clastic
structure subjected to a blast pulse as sho\vn in Fig. 12.20. l'hc pulse is
assumed to begin \vith an initial value Qo and dirninish unifonnly \virh tirne
until it vanishes at time t = J" (Fig. 12.20b). lhus, as a function of tiole
C 1 = 0
and
Q(t) = Qo (1 - ) (c)
-1 x r- Q(t)
Q(t) r
JV --,
I
__J
, ,
I I
J I Qo
I I J
I I
1 I I
J I
0 T I t
- -.,
FI G. 12.20 (a) (b)

58t S'TRUCTURAL DYNAMICS
or, if \\'c consider the force q per unit of mass of the vibrating structure, \ve
have
Qog ( t ) ( t )
q(t) = W 1 - l' = qo 1 - T
(d)
The natura! period of the structure in free lateral vibration may be calculated
from Eq. (12. 3a) on page 565. Assuming that the colufilns arc built in top
and bottom and that the horizontal beam of \\Fcight W is completely rigid, the
lateral deflection of the top of the frame under the action of a horizontal force
P \vould be 0 = PI3/24EI, and we see that the spring constant is k = 24EI/ /3.
Thus, from Eq. (12.30), the period is
27r {W I W/3
T = P = 27r  kg = 27r  24EIg
From Eq. (12.11), \VC no\\' have for the displacement of top beam of the
frame at any time t = t I
(e)
qo {t1 ( t ) .
x = p J 0 1 - 'y sIn (ptl - pt) dt
(I)
This expression, of course, holds only for 0 < tl < T. 'or larger values
of tl, the blast pulse has vanished and \ve have to consider only free vibrations
of the structure corrcsponding to the displacement and velocity imparted to it
by the pulse.
Performing the integration indicated in Eq. (f), \ve obtain
qo ( tl + T . )
X = - 1 --- - cos Ptl - sIn Ptl
p2 l' 27rT 
(g)
Noting from Eqs. (d) and (e) that QO/p2 = Qo/k, Eq. (g) may be written
Qo ( t1 T . )
X = Ii 1 - -t - COS Ptl + Z;y sin Ptl
(12.13a)
and the corresponding velocity is, by differentiation,
x = o [p sin Ptl +  (cos Ptl - 1) ]
Equations (12.13) represent the cornpletc solution to the problem.
To get the time tl at \vhich the displacement x becomes a maximum, we set
the velocity x equal to zero and obtain
(12.13b)
1rtl 27r T
tan - = ---
T T
(h)
If the durarion l' of the pulse is large compared \vith the natural period T of the

ARTICLE 12.4
585
Q(t)
r
Qo
I
FIG. 12.21
o
r'C
T
.1
t
structure, tan (7rtl/r) is a large number, and the angle 7rtl/r approaches 1r/2.
'rhus, the time il at \vhich the deflection of the fralne becomes a maXiI11Ull1 \viII
he il  r/2. Substituting this value of t1 into Fq, (12,13a), \VC find
, Qo ( r )
.t max = k 2 - 2T
(i)
Frorn this ,ve see that "..hen the ratio r/T is small, the maXimUlTI displaccrnent
of the frame is approxinlately t\vice the static displaccnlcnt Qo/ k.
If the duration T of the pulse is small conlpared \vith the period r of the
systcrn, the velocity renlains positive during the tinlC t = 0 to t = 7", and
frool Eg. (12.13h), \VC obtain
X " 1 ' = .1 Qo _! ( 7r T ) 2 = QoE 7
. 2 k 1" T V 2
(j)
'rhe corresponding value of the displacement from Fq. (12.13a) is
, _ Qog 7 2
.tr - .  v 3
(k)
Having these values of displacement and velocity at till1C t = T and using then1
as initial displacement and initial velocity in Eg. (12.4a), \\'c obtain the dis-
plaCCITIent-tiIl1C equation for the ensuing free vibrations of the systern.
If the structure in Fig. 12.20£1 is subjected to a blast pulse of parabolic fornl
as sho\vn in Fig. 12.2], \ve have for the force per unit mass
QoS{ ( t ) 2 ( t ) 2
q(t) =. - 1 - - = qo 1 - --
J,V T T
(I)
instead of expression (J). In such case, Fq. (12.11) becon1cs, for 0 < II < '/:
qo r h ( t ) 2 ,
X = p J 0 1 - l' sIn (ptl - pt) dt

586 STRUCTURAL DYNAMICS
\vhich, after integration, gives
qo [ ( 2 ) 2. 2t1 t 12 ]
X = - ( 1 - cos P il ) 1 - -- + - SIn P il - -- + -
p2 p2'p. pT T T2
Noting that Qolp2 = Qo/ k \\-.hile p = 21r/T, this can be rcexpressed in the form
x = o [ (1 - cos 2:t 1 ) (1
_ J...  ) +  sin 21r t l
21r 2 T2 'TrT T
_ 2  + t 12 ]
T T2
(12.14)
We see again that if T is small compared \vith T, the maximum dynamic dis-
placenlent approaches the value 2QoI k when 11  T 12.
While the triangular and parabolic pulses sho\vn in Figs. 12. 20b and 12.21
may be considered as rough approximations to a real blast pulse, the actual
variation of force \virh time is more closely represented by the diagram sho\vn
in Fig. 12.22. Such a blast pulse is characterized by a very rapid risc in
magnitude to a maximum value Qrnax at time t = t a follo\vcd by a more gradual
exponential decay. This variation of force v.,ith time can be very realistically
expressed by the equation 1
Q (t) = Cte- at
(111)
\vhere C and a arc constants that can be chosen so as to give the desired
strength Qrnnx of the blast at the desired time t a . Setting the time derivative
of expression (111) equal to zero, \VC obtain
C(l - at)e- at = 0
from \vhich we see that Qrn"x occurs at time t a = 1/0. Substituting this value
of t into Eq. (111), we then find Qmnx = Clae. Thus, \vhen Qmax and ta. are
specified, the constants C and a can be determined.
1 See L. S. Jacobsen and R. S. Ayre, "Engineering Vibrations," p. 154, McGraw-Hili Book
Company, Ne\v York, 1958.
q(t)
Qmu
o t
FI G. 12.22 t a

ARTICLE 12.4
587
Introducing the notation
c'
c-
-w
the expression for the blast force per unit mass of the structure becomes
q (t) = ete- at
(n)
r>fhen, assuming that the structure is at rest in its equilibrium position at time
t = 0, Eq. (12.11a) becomes
x = !. {tl te-at sin (ptt - pt) dt
P 10
Noting that
sin (ptl - pt) = sin pt} cos pt - cos ptt sin pt
expression (0) can be integrated by parts, and \\re obtain finally for the dynamic
displacement of the structure
(0)
etle-at1 2ac e(a 2 - p2) .
X = ;72 + p2 + (a 2 + p2 ji (r ol . - cos Ptl) + p(a 2 + p2)2 Sm pt)
(12.15)
A good approximation to the maximum value of this displacement can be
obtained by evaluating x for il = T /2.
Equation (12.11) is also useful in finding the response of a structure to a
series of impulses. Assume, for example, that the \vcight W in Fig. 12.l9a
receives at the instants t = t', t = t". . . . increments of velocity Xl,
X2, . 0 .. Then at time t == tl, the displacement \\rill be
x = ! [AXI sin (ptl - pt') + AX2 sin (ptl - pt") + · · oJ (p)
p
If t', t", . . . are multiples of'T = 21f' / p, the amplitude of vibration will gro\v
\vith time 0
PROBLEMS
Using Eq. (12.11), develop a displacement-time equation for the response of the
system in Fig. 12.I9a to a disturbing force Q(t) = J:Vbt/K, where b is a constant.
Assume that Xo = 0 and i: o = 0 \vhen t = O.
Al1S. X = bt/p2 - (b/p3) sin pt.
2 A.n elevator cage of \\'cight JV is supported by a flexible steel cable ,vhich is unwind-
ing from a circular drum of radius r so that the cage is being lo\vered \vith constant
speed Vo. At a certain instant t = 0 a brake is applied to the rotating drull1 so
that it begins to slow do\vn \vith constant angular deceleration a = air. If the

588 STRUCTURAL DYNAMICS
lcngth of un\"ound cable at this instant, t = 0, is I and AR/ / = k, \vhat is the dis-
placement x of the cage thcreftcr from its equilibrium position at t = O?
Alls. x = Iot - t t1t 2 + (a/p2) (1 - cos pt), \\There 0 < t < vola.
3 -\ftcr the drunl conles to rest in the preceding problem, the cage \vill have sOlnc
ensuing free vibratio:1s as a mass suspended on the vertical clastic cable no\v fixed
at its upper end. '''hat ,,,ill be the anlplitude .l4 of t hese residual free vibrations?
Ans. A = CV2 a/p2) vi 1 - cos (p'Uo/a).
4 For the frame structure sho\vn in Fig. 12.20£1, the follo\ving numerical data are
given: v = 10 kips, k = 24EI/ /3 = 1 no ki ps/in, 'The structure is subjected to
a horizontally acting blast pulse like that sho\vn in Fig. 12.22 for \vhich Qml1x = 100
kips and t a = I/o = 0.10 sec. C:alculate, froln Eq. (12.15), the approximate
value of .t mMP
A1ls. .\''''AX  2.54 in.
S Using Eq. (12. I 1), calculate the complete response of the spring-supported \vcight
W in Fig. I2.19a to a disturbing force Q(t) = Qo cos wt, if Qog/JV = qo and
kg/W = p2. ASSU01C .\. = 0, X = 0, \vhen t = O.
C/o
Al1S. .\' =  -- (cos wt - cos pt).
p2 _ w2
12.5 NUMERICAL INTEGRATION
In the preceding article, the spring force \vas assumed to be proportional to
deflection, and on the basis of this assumption, the general solution (12.11) \vas
obtained. Sometimes elastic forces follo\v a more complicated nonlinear la\v,
in \vhich case a rigorous solution of the differential equation of motion becomes
difficult, and recourse must be had to some approximate method. One such
procedure, applicable in these cases, consists in using numerical step-by-step
integration. To explain this procedure, let us consider the case of free vi bra-
tions \\'ithout damping \vhere the differential equation of motion \vill have the
form
f = f(x)
(0)
If the initial conditions, i.e., the values of displacement x and velocity .t at
t = 0, are given, the change in x and i: \virh time can be calculated by using
step-by-step integration. By substituting the value of (x) 1c::0 into Eq. (a), \ve
can calculate the value of (x) t-o. 'rhen, kno\ving the initial acceleration, the
velocity and displacement at the instant t = 11, chosen very close to the time
t = 0, can be calculated on the assumption that the acceleration remains con-
stant during this short tin1e interval. If \VC denote by t,.t the small interval of
time between t = 0 and t = tl, the approximate values of (i) I=-t. and (x) t-It
win be obtained from the following equations:
Xl = Xo + fo Ilt
Xl = Xo +. j(Xo + Xl) Ilt
(12.16a)

ARTICLE 12.5
589
Next, substituting the value of Xl into Eq. (a), \VC shall obtain the value of .fl.
Then, by using this latter value, sti 11 better approximations for Xl and Xl can be
calculated frorn the equations
Xl = Xo + j-(xo + Xl) ilt
Xl = Xo+i(XO+Xl) ilt (12.16b)
A second approximation for Xl \vill no\\' be obtained by substituting this second
approximation for Xl into Eq. (a).
Taking a second step and using the obtained values of Xl, Xl, X], the magni-
tudes of X2, X2, £2 for the time t = t2 = 2 ilt can be calculated in exactly the
same manner as explained above, and so on. By taking the time inrervalilt
sufficiently small and nlaking the calculations for every value of t t\vice, in
order to get the second approximation, this method of numerical integration can
usually be used \\'ith sufficient accuracy for practical applications, especially if
\\'c are only interested in the motion for a comparatively short period of time.
To illustrate the procedure and to give some idea of the accuracy that can be
obtained \vith it, \\'e consider the case of simple harmonic vibration, for \\'hich
the equation of motion is
x = - p 2 x
(b)
Assuming that \vhen t = 0, x = Xo, X = 0, the exact solution of this equation is
x = Xo cos pt
x = -pxo sin pt
(c)
The results of the numerical integration, carried out as explained above, are
given in Table 12.1 belo\\'. In this case, the time intervals \\'ere taken to be
!::,.t = 1/ 4.p. Since the period of vi bration in this case is T = 27r / p, it is seen
that the chosen interval ilt is approximately equal to one-sixth of a quarter
period T/4. lhe first line of the table expresses the assumed initial conditions.
Starting no\\' \vith the first step, \ve obtain the first approximation for Xl and Xl
at the time t = ilt = 1/4p by using Eqs. (J 2.16a). l'he results so obtained
are sho\vn in the second line of the table. To get better approximations for
.tl and Xl, the calculations arc now repeated, using Eqs. (12.16b). These
results appear in the third Jine of the table. Proceeding in this manner, the
complete table \vas calculated. In the last two columns, the corresponding
values of cas pt and sin pt are sho\\7n. In this case they are proportional to the
exact solution (c), so that the accuracy of the numerical integration can be seen
directly from the table. We see that the velocities obtained from the numerical
integration have a very high accuracy. The error in the displaccments, hovl-
ever, is seen to be gro\"ing with time, and in the last line of the table it amounts
to about] percent of the initial displacement Xo. l'hese results \vere obtained
by taking only six intervals ilt in a quarter period. By increasing the number
of intervals, the accuracy can be improved, but at the same time, the amount
of labor in making the calculations becomes greater.

NUrJlERICAL INTEGRATION
I
I
I
I
590 STRUCTURAL DYNAMICS
TABLE 12.1
t
!
,
I
o I
-0. 2500pxo I
- 0 . 2461 p.'l'o
-0. 4884pxo I
-0. 4769pxo !
-0. 6966pxo
-0.6783p x o
-0. 8619pxo
-0.8378p x o

-0.9740p x o
-0.9457p x o
- I .0262pxo
-0.9954pxo
-1.0153pxo
-0. 9838pxo
x
.t.
I
I
---
At I O. 9687.ro
at o. 9692xo
--- -
2 t I O. 8774.to
2 At O. 8788.ro
- -- - -.-.- --- -----.
.' t I 0.732].%0
.' t 0.7344.%0
------
4 t I O. 5419 x o
4 At O. 5449xo
5 At I () . 3 I 84.To
5 At I O. 3220xo
6 At I 0 . 07 5 5,'o
6 6t 0 . 0794z o
----
7t -0.1719.o
7 At -0. 1680.ro
x
cos pt
sin pt
I
t . 0000 I 0 . 0000
u 096891 002474
- p 2 xo
-0. 9687p2xo
-0. 969Zp2xo
-0. 8774p2xo
-0. 8788p2xo
-0.732Ip2 xo
I -0.7 344p 2x o
-0. 5419p2.\'o
-0. 5449p 2 xo
-0. 3184p2xo
- 0 . 3220p2xo
-0.0755 p2.\'o
-0.0794p2 XQ
--- --- ---------
-0. 1719p2xo
-0. 1680p2 xo -0.1792
()
Xo
0.8776 I
oonl71
I
0.5403 I
0031531
0007071
0.4794
0.6816
0.8415
0.9490
0.9975
0.9840
By using the data froIl1 rable 12.1, the period of free vibration also can be
determined. It is seen froln the values of.1' that for t = 6 t, the displacement-
time curve has a positive ordinate +0.0794.t.o, \vhile for t = 7 6t, the ordinate
is negative and equal to - O.1680xo. The point of zero displacement deter-
mines the time required to make a quarter cycle of vibration. Using linear
interpolation, this time \:vill be found from the equation
r 0.0794 6.32 1.58
4 = 6 t + t 0.0794 + 0.1680 = 6.32 6t = 4p = "p--
1.he exact value of the quarter period is r/4 = I.S7Ip. Hence, the above-
calculated approximate value may be considered satisfactory.
The descri bed method of numerical integration can be used also in those
cases \vhere the clastic spring force is not proportional to x but is represented
by a more general function of x; see Eq. (a). As an example, let us consider
free vibrations of the system sho\vn in Fig. 12.23a, consisting of a particle of
nlass 111 attached to the mid-point of a tightly stretched \vire of lcngth 2/.
When the particle has a small displacement x from its middle position, the
tension in each branch of the \vire \vill be
s = So + .4/£ (yl2+X 2 - /)
(d)
where So is the initial tension \vhen x = O. Expanding the radical in Eq. (d)
by the binomial theorem and keeping only the first t\VO terms of this expansion

ARTICLE 12.5 591
T f(x)
\ j//
to t ..
X
So s
I /
1 I
I
, 0
X
FI G. 12.23 (a) (b)
on the assumption that x « I, \ve obtain
s  So + E x 2
2/ 2
(d')
If So is large, the second term in Eq. (d') can be neglected, and \ve have S  So
provided the ratio xl I is extreIl1ely small. l'his means that for very snlall
amplitudes of free vibration the equation of motion becomes
111.t = - 2;0 x
or \vith the notations
2S o
ko = .--
I
ko
po? = -
111
(e)
. + Po21' = 0
\vhich represents simple harmonic vibrations \vith the period
27r
To =-
po
If the second ternl In Eq. (d') is not neglected, the equation of motion
becomes
(1)
(g)
25 0 AE
111£ = - - x - - - x 3
/ /3
or using notations (e), together \vith the additional notations
AE
k -
1---
/3
{3 = k 1
ko
(h)
\VC obtain the equation of Inotion in the fonn
. = _P02(.t + (3x 3 )
(i)

592 STRUCTURAL DYNAMICS
We see that the restoring force f(.::) per unit mass of the particle in this case is
a cubic parabola as sho\vn in Fig. 12.23h. For very small values of x, the
curvc OA coincides very closely with its initial tangent, and \ve have practically
a linear restoring force. This leads to simple harmonic vibrations having the
period TO given by E<}. (g). For somc\vhat larger values of x, the curve OA
in Fig. 12.23 h deviates noticeably from its initial tangent, and \ve must use
the equation of motion (i) instead of (f). This f:q. (i) no longer represents
simple harmonic motion, and \\'e shall find that the period T depends on the
amplitude, becoming shorter and shorter for larger and larger amplitudes.
10 carry out a step-by-step nunlerical integration of Eq. (i), we assume the
follo\ving numerical data: 11tg = lib, ko = 2 Ibjin., k l = 4 Ib/in. 3 , and, as
initial conditions at t = 0, Xo = I in., Xo = O. Then, {j = k l / ko = 2 in.- 2 ,
and r. (i) takes the form
x = _p02(X + 2x 3 )
(i')
In this case it is more convenient to \vork \vith the quantities x, x/Po, f/p02,
and then replace the time steps tJ.t in Eqs. (12.16) by tJ. (pot) . Then, for
numerical calculation, Eqs. (12.16a), for the first approximation, arc \vritten
in the form
Xl .to fo ,
- = - + - tJ.(Pott
po po P02 i
1 ( XO Xl ) (
Xl = Xo + - - + - 6. Pot)
2 po po
(12.17a)
and Eqs. (12.16b), for the second approximations, in the form
Xl _ Xo + 1 ( .fO + £1 ) A ( )
- - - - - - L.l pot
po po 2 P02 P02
Xl = Xo + ! ( X O + Xl ) 6.(Pot)
2 po po
For the length of the steps tJ.(Pot), \\'e divide the basic period To, as given by
Eq. (g), roughly into 40 parts, i.e., about ten steps in the quarter period To/4.
'-[his gives tJ. (Pot) = pOTo/40 = 2.". / 40 = 0.157; thus, in round numbers,
6.(pot) = 0.15 will be used. In 'choosing this time step, it should be realized
that for amplitude Xo = 1 in., the period T will be considerably less than To,
and so we are probably dividing the true quarter period T/4 into only seven or
eight parts. l'his means that we can expect about the same degree of accuracy
from the numerical integration in this case that we had in the preceding example.
On the basis of the chosen interval tJ. (pot) = 0.1 5 and Eq. (i'), the numerical
calculations made from Eqs. (12.1 7) are shown in Table 12.2. After seven
steps, each repeated once, we find that the displacement x becomes negative,
indicating that we have completed a little more than a quarter cycle of vibration.
Using a linear interpolation between the last two values x = 0.153 and
x = -0.059, we find that x = 0 when .pot = 1.01. Hence, the completc
(12.17b)

ARTICLE 12.5
TABLE 12.2
Pot
o
0.15
0.15
0.30
0.30
----
0.45
0,45
0.60
0.60
0.75
0,75
0.90
0.90
1.05
1.05
I
I
l. 000 I
0.966 I
0.968
0.871 I
0.875
0.727 I "
0.734
--------
0.551 I
0.558
0.355
0.361 I
O. ]49 I t
O. 153
-0.062 I
-0.059 I
x
x
po
o
-0.450
-0.433
-0.866
-().803
-I. 177
- I .081
-1.361
-1 .262
- 1. 444
-1 . 363
-1.465
- I .408
- I .454
-) .424
x 3
1.000
0.902
0.907
I 0 . 66 t
0.670
_ I . - 0: 384
0.395
I ' 0.167
O. 174
I . 0 . 04 5
0.047
I 0.003
0.004
I 0 .000
I
.t
po2
i I
j -3.000
I - 2. 770 I
-2.782
I -2.193 I
-2.215
I - 1 . 49 5 I
-1. 524
I -0.885 I
-0. 906 -.
I -0.445 I
-0.455 i
I , -0. 15 5 I
-0.161
-0.062 I
.O + Xl
-- -- A (Pot)
2p02
-0.450
-0.433
-0.433
-0.370
-0.374
-0.278
-0.280
-0.181
-0. 182
-0.101
-0.102
-0.045
- 0 .046
-0.016
593
.i'O + Xl
- - A (Pot)
2po
-0.034
-0.032
-0.097
- 0 . 093
- O. 148
-0.141
-0.183
-0.176
-0.203
-0.197
... - - ---------......---...
-0,212
-0.208
-0.215
-0.212
period corresponds to PDT = 4(1.01) = 4.04, and with po = y ko/111 = y 2g
= y 772 = 27.8 see-I, \VC find
4.04
T = 27.8 = 0.1452 see
The exact value of T in this case can be calculated from an elliptic integral and
is T% = 0.1447 sec. \Vc see that the error in the period from the results of
the numerical integration is about 0.3 percent.
For very small amplitudes of vibration of the nonlinear system sho\vn in
Fig. 12.23a, the period calculated from Eq. (g) is
21r 6.28
TO = - = --- = 0.266 sec
po 27.8
Comparing this vlirh the above calculated period for an amplitude of I In.,
\ve see that the nonlinear effect on the period is quite significant.
I f necessary, the accuracy of numerical integration can be further improved
by using, instead of sin1plc equations' like (12.16) or (12.17), more elaborate
expressions l for calculating consecutive values of x and x. Recourse to such
methods should be had if the motion during a long interval of time is to be
studied.
1 See H. von Sanden, "Practical Mathematical Analysis," E. P. Dutton & Co., Inc., New
York, 1926.

594 STRUCTURAL DYNAMICS
12.6 GRAPHICAL INTEGRATION
In Inany practical cases of structural dynan1ics, the disturbing force, as a
function of time, may be given in the form of a curvc, as sho\vn in Fig. 12.24,
\vhich cannot be expressed analytically. In such cases, an approximate solu-
tion for the response of the systcn1 to the disturbing force can be obtained by a
method of graphical integration. 1 To accomplish this, \ve replace the given
Q(t), represented by the sn100th curve !lbcde in Fig. 12.24, by a step function
having constant ordinates Ql, Q2, Q3, . . . over equal intervals of time t as
sho\vn. 1'hen beginning at t = 0 and kno\ving the initial displacement '\'0 and
the initial velocity Xo, \ve calculate the motion of the mass produced by the
constant force Ql acting for the first time interval dt and find the displacement
Xl and the velocity i'1 at the end of this interval, i.e., at time t = dt. Then
\vith Xl and .\"1 as ne\\! initial displacement and nc\\' initial velocity, \ve again
calculate the motion of the systern under the action of the constant force Q2,
suddenly applied at tilllC t = t, and find the displacement X2 and the velocity
.t2 at the end of the second interval, i.e" at tin1c t = 2 t, ctc. In this \vay,
the complete Illation of the system under the influence of the step function
can be detern1incd.
\,. see that in order to carry out the procedure outlined above, it is first
necessary to study the response of the systen1 to the action of a suddenly
applied constant force Qi. To do this, \ve consider the systcm in Fig. 12.25a
and assume that Q(t) = Qt' = canst. Then the equation of motion becomes
711.t; = - kx + Qi
or \vith the notations
2 _ k
P - 111
Oi = i
(a)
1 'rhe method to be described is due to J. Larncon; see R(v. Uni'l.-'erseJle ,Wines, vol. 11,1935.
Q(t)
b
o
f
Qa
.
Qz r=
Ql
t
t
FI G. 12.24
d

ARTICLE 12.6
595
\ve obtain
.t; + px = P20i
(b)
l-ohe solution of this differential equation is
x = C L COS pt + C 2 sin pt + Oi (c)
\vhich can easily be verified by substituting back into Eq. (b). Differentiating
Eq. (c) once \vith respect to time, we obtain
x = - pC't sin pt + pC 2 cos pt (d)
No\\', assuming initial conditions of motion x = Xo, X = Xo, vlhen t = 0, \ve
find, for the constants of integration,
C 1 = Xo - 0,
XO
C 2 = .
p
Using these values, \ve find, with some rearrangement of terms, that Eqs. (c)
and (d) become
( ) Xo .
1" - 0, = Xo - 0, cos pt + Ii sIn pt
(e)
oX
- (xo - 0,) sin pt + Xo cos pt
p
p
Squaring and adding Eqs. (t'), \VC obtain
( X ) 2 ( XO ) 2
(x - 5,) 2 + P = (xo - 0,) 2 + p
(f)
x
A
%"
, ;I'
'pt....... ",/
\ ;I'
Q;' U .,'" Xo
k__ C
o
FI G. 12.25
(a)
(b)

596 STRUCTURAL DYNAMICS
l'his expression will be recognized as the equation of a circle of radius
 --- -
Xo 2
R = (xo - 8;) 2 + (p )
\vhich has its center at the point (Oi, 0) in the coordinate plane x versus x/po
'"'his plane is usually called the phase plauc and the circle is called the phase-pla11t
trajectory. Thus, we may conclude that the motion produced by a suddenly
applied constant force Qi is a simple harmonic vibration centered about a nc\v
equilibrium position x = Oi = Qi/ k instead of.t = o. For assumed values of
Qi/ k, Xo, and io, the phase-plane circle is shown in Fig. 12.25 b. To visualize
the motion from its representation in the phase plane, \ve need only to imagine
that a point i:l, starting from Ao at time t = 0, moves in the counterclockwise
direction around the circle \vith constant angular velocity p. Then at any
instant t, defined in Fig. 12.25 b by the angle pt, the displacement;\" and the
velocity i: are given by the coordinates of the moving point A.
Having this graphical phase-plane representation of the motion produced by
a suddenly applied constant force, \\FC may no\\" return to the step function
sho\vn in Fig. 12.24 and see ho,"' to trace out a phase-plane trajectory repre-
senting the response of a spring-supported mass to a series of constant-force
impulses. Referring to Fig. 12.26a, \ve let the point Ao, with coordinates _to
and xo/ p, represent the initial state of motion in the phase plane at time
t = o. r]hen \vith C 1 , having the coordinates (Ql/ k, 0), as a center, we con-
struct the circular arc A oA 1 \vith a central angle pdt, as sho\vn. Point A 1,
obtained in this \\'ay, represents the displacement Xl and the velocity Xl of the
disturbed mass at the end of the first time interval, i.e., at time t = At. No\\'
with C 2 , having the coordinates (Q2/k, 0), as a nc\v center, \ve construct the
circular arc A lA2 \vith central angle p dt as before and obtain the point A 2 ,
defining the displacement X2 and velocity X2 at the end of the second interval,
x
x
Q2
K
t
l:1t 2t 3t 4t 5l:1t 6.1t
(a)
c..
(b)
FI G. 12.26

ARTICLE 12.6 597
, in.
W 1.0
x
0.8
0.6
0.4
0.2
e 0 t, see
(a) (b)
FIG. 12.27
i .c., at time t = 2 t. Proceeding in this \vay, the construction in Fig. 12.26a
has been carried out to time t = 6 t on the basis of tjC step function in
Fig. ]2.24.
Having completcd the construction in Fig. 12.26a, the points A 0, A 1, A 2, . . .
can be readily projected onto a displacetnent-titTIe plane and the curve x = J(t)
constructed as sho\vn in Fig. 12.26b.
The described graphical procedure \vill give good accuracy if reasonable
care is taken in choosing the step curve to approxiInate tbe given disturbing
force Q(t). In general, this should be done in such a \\fay that the area under
each step is the same as that under the corresponding pcrtion of the given
snl00th curve representing Q(t). In choosing the time interval t, \ve must be
guided not only by variations in the given Q(t) curve but also by the natural
period T = 27r/p of the spring-suspended mass. In general, t must be a small
fraction of this period. If the disturbing force exhibits rapid variations over
part of its duration and is fairly constant over another part, it may be expedient
to vary the time intervals t accordingly. rrhere is no fundamental rcason for
the tin1e intervals to be equal.
As an example of the application of the above-described method of graphical
integration, let us consider the case of a \vater tank, sho\vn in Fig. 12.27, the
base of \vhich is subjected to a horizontal ground motion  = q (t) as repre-
sented graphically by the smooth curve in Fig. 12.27 b. 'rhe \veight of the
tank is  and the flexural rigidity of the columns is such that the natural period
of free lateral vibration of the tank is T = 0.48 sec. Considering this period
together \vith the fluctuations in the given ground motion \ve conclude that
t = 0.02 see \vill be a satisfactory tiIne interval for the problem, and we make
the step function as sho\vn in Fig. 12.27 b accordingly. Since p = 21('/T, the
angular steps in the phase plane \vi II be
 27r 2 0.02 1(' 150
P t = --;: t = 7r 0.48 = T2 -

598 STRUCTURAL DYNAMICS
Denoting by x the lateral displacement of the tank and by t the displacement
of the ground, the equation of motion becomes
w
- f == -k(x - )
g
or with the notation kg/W = p2,
of + p 2 x = p2
(g)
Comparing this equation with Eq. (b), we see that, in this case, the ground
displacement  corresponds exactly to the quantity 8 i = Qi/ k in the case of a
disturbing force. Hence, we may proceed with the phase-plane constructions
as already explained above. These constructions made on the basis of the
approximate step curve in Fig. 12.27 bare sho\vn in Fig. 12.280, and the cor-
responding displacement-time curve is shown in Fig. 12.28b. It \vas assumed
in making the constructions that the initial conditions of motion of the tank
were ..t.o = 0 and .to = 0 at t = O. \Ve note that the tank attains, within the
duration of the disturbance, a maximum absolute displacement x = 0.85 in. at
t = 0.17 see and is left \vith an ensuing free vibration having the amplitude
A = 0.925 in., which is the length of the radius vector OA 13 of the last point
A 13 of the phase-plane trajectory (see Fig. 12.28a). However, comparison of
the displacement-time curve x = f(t) \vith the ground-motion curve  = q(t),
also sho\vn in Fig. 12.28b, sho\\Ts that the maximulll relative displacement
between the tank and the ground is (x - ) max  I .13 in. and that chis occurs
at t = 0.16 sec. Naturally, this maximum relative displacement represents
the critical point of the response, since it wj)] determine the maximum bending
stresses induced in the columns.
x
X
1.0
\
,
"
\
\
\
.
0.08 0.12 , 0.9'" 0.24
, '
\ "
t
r, Ground motion
, ,
, \ ,--
, , ' \
I '.l \
I \
I '
, "
.
0.. ,
,
I
,
0.2 ,
,
,
,
0.6
0.8
o
(a)
(6)
FI G. 12.28

ARTICLE 12.6
599
i\S a second exanlplc, let us consider the simple portal frame sho\vn in
Fig. 12.29a \vhich is subjected to a horizontally acting rectangular blast pulse
of strength Qo = J ,492 Ib and duration 11 = 0.2 see as sho\vn in Fig. 12.29b.
Each of the t\\'O side columns is a stee] channel section built in at both ends and
having cross-sectional fi10rncnt of inertia I = 1. 54 in. 4 Hence, the lateral
deflection 5 under the action of a horizontal force P acting as ShO\Jln in Fig.
12.29c is
PI3
8 = 24£1
(h)
and the spring constant for the. structure bccon1cs
24£/ .
k = Ij- = 373 lbjln.
Then p = y'ki? Jl = 'V373 X -.\6( = 7.75 see-I, and'T = 27r/p = 0.813
sec.
Noting that the bending Jl10ments at t1 b, c, d in Fig. 12.29c have the magni-
nJde i\J = PI/4, Eg. (IT) can be expressed also in the form
{) = }y[ /2
6EI
(h')
rhcn, if the yield stress for the steel is such that the plastic moment for the
channel section is i\11 p = 40,000 in,-Ib, \ve find fronl Eq. (h') that
 = 4f p{  = .
u,) 61 3 In.
i.e., \vhen the displacen1cnt x of thc disturbed structure attains the value op,
the colun1ns begin to behave plastically and offer no further increase in
resistance to lateral disp]aceIl1ent.
Let us assume no\v that at t = 0 the frame stands vertically in its equilibrium
position, that is, XI) = 0 and .t o = O. rhcn, so long as the COIUll1nS rcolain
Q(t) Q(t) --j&r
JV = 2,40011>
--f--- d
y
1=12'
I x
1- 0 0.1 0.2 t, see
(a) (b) (c)
FIG. 12.29

600 STRUCTURAL DYNAMICS
x
F
B
o
x
p
FI G. 12.30
clastic, i.e., up to the displacement x = 01' = 3 in., the equation of motion is
of + p 2 X = p2 Qo
k
(i)
\vhere Qo/k = 1,492 + 373 = 4 in. Hence, with Co having the coordinates
(4, 0) in the phase plane (f'ig. 12.30) as a center, Vl"C construct the arc OA of
such length that point A has the ordinate x = 01' = 3 in. as sho\vn. lhc
corresponding subtended angle is pt = 75° = S1J'i12, and \vith p = 7.75 see-I,
this corresponds to t = O. I 67 sec. Hence, at this instant the structure begins
to behave plastically, and the disturbing force Qo has another 0.033 see to act,
.which corresponds to an angle p t = 15° in the phase plane. For this remain-
ing time that the disturbing force has to act, that is, for 0.167 < t < 0.200,
the equation of motion \vill be
W
-.f=-ko + Q o
g P
\vhich can be \vritten in the equivalent form
f + p 2 x = p2 ( 1° - 61' + x) = P 26 i
(j)
\\,here 0; determines the position on the x axis in Fig. 12.30 of the center C,
for -each arc of the phase-plane trajectory. In our example, we have Qo/ k = 4
in. and 01' = 3 in., so that i = 1 + x. Noting that we begin with x = 3 in.
and that a step p lJ.t = 15° \vi]] carry us to x  4- in., \VC take, for the average

ARTICLE 12.7
601
value of Oi during the intetval, (Oi) f\V = (4 + 5) /2 = 4.5 in. and obtain the
center C t as sho\vn. In this example, \ve have reached, at point B, the limit
of validity of Eq. (j), since the disturbing force vanishes at pt = 7r /2 = 90°,
and only one step is required.
After the disappearance of the disturbing force and until the structure comes
to a stop, the equation of motion \vill be
w
- .1' = -ko p
g
\vhich, for the purpose of phase-plane representation, can be \vritten in the
equivalent form
1 + p2.t = p2 (x - 6 p ) ;: P'J.5i
(k)
Succeeding centers C 2 , C 3 , C.., . . . are no\v determined from the expression
Oi = X - 01' = X - 3
Since these values of 0; again depend upon the displacement .'r, the average value
of Oi for each step should be used in locating the center C.. for that step. For
exarnpJe, to make the arc HC, \ve decide in advance to go from x = 4 in. at
point B to .t' = 5 in. at point C. Then for this step (Oi)av = (1 + 2) /2 = 1.5
in., and \ve choose the center C 2 accordingly. Proceeding in this \vay, the
constructions in .Fig. 12.30 have been continued until the trajectory BCDEF'
intersects the x axis at F, indicating that the frame has come to a stop, i' = O.
l'he ensuing free vibrations of the structure are represented in the phase plane
by the circle centered again at Co and having the radius Xp - 00 = 7 - .:f. = 3 in.
This indicates that, as a result of the blast pulse, thc structurc suffered some
permanent damage and vibrates about a nc\v distorted equiJibrium configuration
that is displaced 4 in. to the right of the original vertical equilibrium configura-
tion. This exanlplc illustrates ho\v the phase-plane construction may be used
to fo11o\\' out the response of a structure to a given disturbing force in both
elastic and plastic phases of its behavior. 1
12.7 STATICAL AND DYNAMIC STRESSES IN RAILS
L.ct us assume that a rail can be considered as a long bar continuously supported
by an clastic foundation. Then, cqsidering the simplest case of a single load
p (t"'ig. 12.31) and denoting by k rhe rnodulus of rhe foundation, i.e., the load
per unit length of rail required to produce a deflection of the foundat:on equal
1 Such extended applications of the graphical phase-plane 111cthod have been developed by
L. S. Jacobsen. See his paper On a General lethod of Solving Second Order Ordinary
Differential Equations by Phase-plane DispJacenlcnts, J. Appl. Af(c}z., Decembcr, 1952.

602 STRUCTURAL DYNMICS
p
x
FIG. 12.31
y
-0.4
-0.2
..-

"'- 0


 0.2
-
Jot
"'-
-e.
 0.4 -
0

=' 0.6
-
tJ$
>
0.8
1.0
0
FIG. 12.32
1
234
Values of {3x, radians
5
to unity, the differential cqu3tjon of the deflection curve is
J.i '"
t'[ ;; = -ky
(a)
U sing the notation
4/-
f3 = 'i 41
(b)
\ve \vritc the solution of Eq. (tl) in the form 1
p. . p
V = -- e-f3x ( cos (3x + sin I3x ) = - cp ( x )
. 2k . 2k.
(c)
'The corresponding deflection CUf\TC is shou'n by the curve <t>(x) in Fig. 12.32.
The rnaxin1uo1 deflection takes place under the load, and substituting x = 0 in
Eq. (c), \ve obtain
{3P
Y:z:c=O = 2k
(cf)
1 For the derivation of this solution see S. Tirnoshenko, "Strength of l\-JatcriaJs," 3d ed.,
vol. II, p. 2, D. Van Nostrand Coo1pany, Inc., Princeton, N.J., 1956.

ARTICLE 12.7
603
This expression can be used to caJculate the moduJus k of the foundation,
provided the dcflection of the rail under kno\\'n load has been measured. The
maximum deflection depends on the constants {3 and k. It increases \vith
decrease of the foundation modulus k and of the flexural rigidity EI of the rail.
For a 130-lb rail (I = 72.8 in. 4 ) and good track conditions, \ve may take
k  1,500 psi and {3  0.020 in. -1 For calculating bending stresses, we use,
for bending moment, the expression
y p P
lW = - EI dx 2 = 4{3 e-z(cos (3. - sin (3:t:) = 4{3 t/t(J:) (e)
The function t/J(x) is also represented graphically in Fig. 12.32. It is seen
that the maximum bending moment occurs under the load and has the value
p
lmax = -
4{3
(f)
For maximum stress in the rail, \ve then have
4
i\1 mox P P 4J4EI
(Tmax = Z = 4{3Z = 42 '\) T
(g)
\vhere Z dcnotes the section Inooulus of the rail.
In order to compare the stresses in rails, \vhich have geometrically similar
cross sections, Eq . (g) can be put in the follo\ving form:
}1 A -\Ii 4/ 4E
Umax = A 42- '\)T
(h)
in \\rhich A denotes the cross-sectional area of the rail. Since the second factor
in this expression remains constant for geometrically similar cross sections and
since the third factor does not depend on the dimensions of the rail, the maxi-
mum stress is inversely proportional to the area of the cross section, i.e.,
inversely proportional to the \\'cight of the rail per unit length.
An approximate value of the maximum pressure R on a tic is obtained by
multiplying the n1aximum deflection, Eq. (d), by the tie spacing I and by the
foundation rnodulus k. "[hus,
4 - -
J{ = IJ kl = P{31 = PI 4/_k_
2k 2 2 '\) 4EI
(i)
It may be seen from this that the pressure on the tie depends principally on the
tic spacing I. I t should also be noted that the modulus k of the foundation
appears in both Eqs. (g) and (i) as a fourth root. lIenee, an error in the
determination of k \viH introduce only a nluch smaJJcr error in the magnitudes
of (Tmax and R.

604 STRUCTURAL DYNAMICS
r- 66"--r-- 66 "-T- 66 "-1
"""""""""/////"//""""/"/"/0//""0,//,,8,,,,,,,,0,,,,///,,,,/,,,,//////,,//,,////
FIG. 12.33
"then several loads ar acting on the rail, as in the case of a set of locomoti vc
\vhecls, the bending Inornents and stresses in the rail can he obtained by super-
position. In Fig. 12.33, for example, the result of such superposition for the
hending moments produced by four equal and equally spaced \vheel loads is
sho\vn.
The above analysis \vas nlade on the basis of several simplifying assumptions
\vhich are only an approxinlation. 1\evcrthelcss, the results obtained have
proved very useful espcciaJIy as a guide to experimental investigations of
stresses in rails. 1
l'he dynamic deflection of rails and the dynan1ic stresses under the action
of the moving \vhcels of a locon10tive may become much larger than those
calculated on the basis of static fonnulas. Thtre are various causes \vhich Inay
produce such an increase in deflcction and stress, the principal ones being:
I Different kinds of irregularities in the shape of the \vhcel or
rail, such as flat spots on the rinl of the \vheel, lo\v spots on the rail,
and discontinuities at the rail joints
2 Variation in the forces acting on the rail caused by variable
spring forces on the \vhceI, the vertical component of the centrifugal
force of a countcr\veight, and the vertical component of the forces
in the connecting rods
In discussing dynarnic stresses produced in a rail by the effect of a lo\v spot
(Fig. 12.34), let 17 be the variable depth of the lo\v spot, a given function of x
nleaslJred along the J;ail. Let V/g denote the unsprung nlass per \vheel,
2k/ {3 the vertical load required to produce a rail deflection equal to unit}' r see
Eq. (d) J, and y the additional deflection of the rail under the \vheel due to the
dynamic effect of the lo\v spot. Then the vertical displacenlcnr of the ,vheel
1 See the paper by S, l'in10shcnko and B. F. Langer, Trans. ASlE, vol. 54, p. 277, 1932.

ARTICLE 12.7
605
due to both the dynamic deflection of the rail and the lo\v .spot is y + 1], and
the differential equation of Illation of the \vheel in the vcrtical direction is
HI d'2 (y + '17) 2ky
- dt 2 +  = 0
g fJ
(12.18a)
Dcnoting by tV the unifornl loconl0tive speed, \ve obtain
d'17 d'YJ dx efT]
- - - - - . -, v
£1t - dx dt - dx
d 2 1] d 2 'YJ
- - = - v 2
dt 2 dx 2
and Eq. (12.18a) can be put in the follo\ving form:
HI d 2 y 2ky lV? d 2 T]
- + ---- = - - 't) -
g dt 2 {3 g dx 2
(12.18b)
If the shape of the lo\\' spot and the speed v arc kno\\'n, the right-hand n1cmbcr
of Eq. (12.18b) can easily be expressed as a function of tinle, and \ve arrivc
at the equation of forced vibration as discussed in Lrt. 12.4, the solution of
,vhich can be easily ohtaincd in each particular case.
. i\s an example, Ict us consider the case of a lo\v spot as sho\vn in Fig. 12".34,
the shape of \vhich is given by the equation
x ( 21rX )
1] = 2 1 - cos ,-
(j)
in \vhich / is the lcngth of the la\\' spot and X the depth at the middle of its
length. Then the right-hand nlcmbcr of Eq. (12 .18b) becomes
H! . X 47r 2 21rX
- - - 'L,Z -- -- - cos . .-
g 2 /2 I
j\'lcasuring time from the instant \"hen the point of contact of the \vhecl
coincides \vith the beginning of the lo\v spot, as in Fig. 12.34, \ve have x = vt,
and Eq. (12.18b) becomes
lf l d 2 y + 2ky = _ .J ,2  2 cos ?!!!5'!.
 dt 2 {3 g 2 /2 ,
FIG, 12.34
-I
, x

606 STRUCTURAL DYNAMICS
or, using the notation
2kg ?
- = p"
/3W
\ve obtain
d 2 y 2X7r 2 'l;,2 27r'Vt
dt;- + p2y = - ---/2-- cas /-
To solvc this differential equation, \\'c use the method described in Art. 12.4.
Then proceeding as indicated by Eq. (12. I I), \VC obtain
2A7T' 2 V 2 Jc 11 27r'7)t.
'Y = - cas -- . -- SIn p( tt - t ) dt
. p/2 0 I
(k)
Performing the indicated integration and denoting by T = 21f / P the period of
free vibration of the \vheel on the rail and by ; 1 = 1/ tt' the tin1e it takes the
\\,'heel to pass over the 10\1,' spot, \ve obtain
x ( 21ftl 27T't 1 )
Y = 2-(i _ T12/T2) cos -;; - cas --; .
(12.19)
From this expression the additional dynamic deflection at any instant tt < Tl
can be calculated. It is seen that the deflection produced by the lo\v spot is
proportional to its depth ;\ and depends also on the ratio Tt/T. For various
values of this ratio, the dynamic deflections arc represented hy curves in Fig.
12.35. The abscissas give the position of the \vhcel along the lo\\! spot, and
the ordinates give the additional deflection expressed in terlns of A. ...s soon
as the \\,'hccl enters the 10\\' spot, the pressure on the rail, and conscquently the
deflection of the rail, begir.s to diminish (y is negative) \vhile the \vheel begins
to accelerate in a do\vn\vard direction. Then follo\vs a retardation of this
movement \vith corresponding increases in pressure and in deflection. From
the figure \\'e see that for Tl < T the maxirnunl prcssure occurs \vhen the \\Thecl
is approaching the far end of the 10\\7 spot. 'fhe ratios of this maxinlum
dynarnic deflection to the depth X of the lo\\' spot, calculated fron1 Eq. (12.19),
are as follo\vs:
Tl i .. ., 2 3 1
-" 2
I  "5 1f K -2-
T
Ymax!X I 0.33 0.65 1. 21 1. 41 1.47 1.45 1. 33
It is seen that the maximum value of this ratio is about equal to 1.5 \vhich
occurs \vhen the locornotive speed is such that TI/T  f, that is, \vhen the time
required for the \vhcel to cross the 10\\/ spot is equal to t\vo-thirds of the period

ARTICLE 12.7
607
-0.8\
-0.6\
- 0.4
-0.2\
O
0.2
O.4
0.6>"
0.8>"
LOX
1.2X
1.4h
1.6X
0 il
FIG_ 12.35
i 1
l
 I
l
II
l
;1
I
of natural vibration of the \vhccl on the rail. I t can be concluded that the
additional dynanlic pressure due to the lo\\' spot is about equal to the force
\vhich produces a static deflection of the rail equal to I.SX. Similar results can
be obtain(x! for 10\\' spots having other shapes. Calculations of this kind sho\v
that the ratio Ym:lx//X docs not depend suhstantially on the shape of thc lo\\! spot
provided it is represented by a continuous SIT100th curve. The calculation
nlade above can bc applied also to the case of a flat spot on the rim of the \\'heel,
bccuse such a flat spot produces exactly the same vertical movements of the.
\vheel as a lo\\! spot of the sarne shape on the rail.
In considering the additional \vhcel pressure of a locomotive due to the
centrifugal force Q of a counter\\'l'ight, \VC can use the theory of forced vibra-
tions as disclIssed in Art. 12.3 and multiply this force by a rnagnification factor.
Then the additional pressure on the rail, produced by the counter\\'cight, is
Q/ (] - ,2/722), \vhere 7 denotes the period of natural vibration of the \\Theel
on the rail, and 12 the tinlC of one revolution of the \vheel. In this calculation,
the :nass of the rail is neglected, \vhich is justified, because the time 12 is large
in comparison \virh the period of natural vibration '3 of the rail. l'he lov.rest
type of rail vibration is an oscillation of the rail in the vertical direction as a
rigid body. "fo get the period 13 of this vibration, \\e use the kno\vn equation
73 = 211"  .O;'

608 STRUCTURAL DYNAMICS
in \\7hich 08t is the statical deflection of the rail due to its o\vn \\'cight. laking,
for cxan1plc, a ] 30-lb rail and assuming k = 1,500 psi, \\'e obtain T 3 = 0.0157
sec, \\rhich is small in conlparison \vith the time TZ of one revolution of the \vheel.
12.8 LATERAL VIBRATIONS OF PRISMATIC BEAMS
rhe differential equation for the static deflection curve of an elastic pristnatic
beam is 1
d 4 y
EI d:l. 4 = q
\vhere q denotes the intensity of the distributed load and EI is the flexural
rigidity of the bean1- To adapt this equation to the case of free lateral vibra-
tions of a beam, \ve have only to use D'AleInbert's principlc and replace the
intensity q of lateral load by the inertia force per unit length of the beam.
The equation then becomes
a 4 y "tV a 2 .:v
EI - - = - - -.
ax" K at 2
\vhere 'tv/g is the nlass per unit length of the beam. lhc solution of Eq.
(12.20) is especially simple in the case of a beam \vith simply supported ends
as sho\vn in Fig. 12.36. In such a case, the vibrating beam has a sinusoidal
shape, and \ve can satisfy Eq. (12.20) and also the conditions at the ends /1 and
B by taking a solution in the form
(12.20)
. t7f'X
'V = cPi sIn - -
. I
(a)
\vherc i denotes the nuo1bcr of half \\raVeS in \vhich the vibrating beam is sub-
divided, and cPi is a function of time. Substituting expression (t1) into Eq.
(12.20), \ve obtain the equation
.. I)
cPi + P('<Pi = 0
(12.21)
in \\7hich \ve use the notation
I) i 4 7f'4 EI
pi M = /4- - w-
(12.22)
1"hc differential equation (12.21) represents simple harmonic motion as dis-
cussed in Art. 12. I, and it has the frequency
f,- = Pi = 1Ti 2 / EI15
1 27f' 2/ 2  W
(12.22a)
1 See "Strength of Materials," vol. I, p. 140.

ARTICLE 12.8
609
A
:t
l
B
l'
.
x
(a)
(b)
 ----- - ------ 
(c)
FI G. 12.36 (d)
Taking i = 1, 2, 3, . . . , \\re obtain vibrations of the beam in the various
{nodes sho\vn in Fig. 12.3 6b, c, d, . . .. It is seen that the frequencies of
vi bration in the various modes are proportional to the square of thc number i of
half \vaves and inversely proportional to the square of the length / of the beam.
For geometrically sin1ilar beams, the frequency of lateral vibration decreases
in direct proportion to the increase in the linear dimensions.
To discuss forced vibrations of beams, \VC denote by q(.t, t) the intensity of
SOIne distributed disturbing force and add this to the inertia force on the right-
hand side of Eq. (12.20) so that it becomes
a 4 y w ( 2 )'
EI ,- = - -' + q( x t )
ax" g dt 2 '
(12.23)
Taking the solution of this equation again in the form (a) and substituting back,
\ve obtain
. .
.. . t7rX . 17rX g ( )
CPi sin T + p,2CPi sin T = w q x, t
We now multiply both sides of this equation by sin (i1rx/ I) dx and integrate
over the length I of the beam. In this way, \\'c obtain
.. 2g (l ( ) . ;7rX d
CPa + Pi 2 cpj = wi 10 q x,t sIn T x
(12.24)
For any given distribution of the disturbing force q(x, t), the integral on the
right-hand side of Eq. (12.24) can be evaluated. Then using the general
solutions (12.17) developed in Art. 12.5, we obtain the solution for <Pi.
As an example, let us consider the case of a beam acted upon by a stationary

610 STRUCTURAL DYNt\MICS
A r-- c -i P sin wt
---D-
I
r- -----[
FIG. 12.37 u,
B
--A:.- -
.1
pulsating force ].J sin wt as sho\vn in Fig. 12.37. In this case the function
q(x, t) vanishes for all values of x except the point x = c, and \\'c get
( I ( ) . i1rx d P . . i1rc
)0 q x, t SIn T x = SIn wt sIn T
When this is substituted on the right-hand side of Eq. (1 2.24), it becomes
.. 2gP.. i7r'C
cPi + Pi 2 cJ>.. = wi SIn wt SIn T
(b)
and then using Eqs. (12.17), \ve obtain the general solution
2gP . i7rC t h. .
cPi = - _ I sIn - / sIn wt SIn (Pitl - Pit) dt
Pi W 0
.fter intcgration, this becomes
(c)
cPo = 2gP sin  7rC ( }_ wt _ w  in J !i t __ )
, wi I Pi 2 - W 2 Pi Pi 2 - w 2
(d)
Substituting this expression for CPi into Eq. (a), \\'c get the ith modc of vibration
of the bean1. I:'1king i = I, 2, 3, . . . and sun1n1ing the corresponding modes
of vibration, \ve get the con1plctc solution for the response of the bcam to the
disturbing force P sin wt. This is
)' =
ao
2g  \" sin (i7rc/ C)i_ (ilr!) sin wt
wi i..t P .2 - w 2
i--l t
_ 2gP  _  in J!7rc /J) si  Y7rx !l) sin p;t
wi '-' P . P .2 - W"
1"=1 r t
(12.25)
rhe first series, proportional to sin wt, represents forccd vibrations having
the sae frequency w/21r as the disturbing force itself. rrhc second series,
proportional to sin Pit, represents free vibrations having the frequcncies as
given by Eq. (12.22a). O\ving to various kinds of damping \vhich \vcre omit-
ted in the derivation, these free vibrations \vill gradually disappear and only
the forced vibrations \viJI remain.
vVe see that if w is very close to' anyone of the natural frequencies Pi, then
that term in the series becomes vcry large and \ve have a condition of resonancc.

ARTICLE 12.8
611
FIG. 12.38
A I vt--:{
nf;7
l-
I
gf
I
B
 --

To avoid the possibility of resonance completely, it is necessary that the
frequency of the lo\vest mode of vibration (i = 1) be several times larger than
the frequency of the disturbing force.
Ta ke, as an example, the case in \vhich the disturbing force acts at the middle
of the bearn (c = 1/2). Then, introducing the notation W 2 /P12 = a 2 , \\Fe obtain
from the first series in Eq. (12.25)
y = . 2P/  [ si n (rrx /!2 _ sin (37rx/ '2. + in (57rx//) _ . . . J sin wt
. 'ff4 EI 1 - a 2 3 4 - a 2 54 - a 2
For small values of a 2 , that is, for w «PI, the first term alone of this series
represents the steady-statc response \vith good accuracy, and we conclude
that the ratio of dynamic to static deflection is 1/ (1 - a 2 ). If, for example,
w = Pl/4, the dynamic deflection \vill be about 6 percent greater than the
static deflection.
i\s a second example, let us consider the case sho\"n in Fig. 12.38, \vhere a
constant force P 1110VCS along the beam \vith constant speed 'U. Measuring
tin1C from the instant \\Then the force P is at A, its position ,,,ill be defined by
the distance vt as sho\vn in the figure. Then, in such a case, the integral on
the right-hand side of Eq. (12.24) is
{ 1 ( ) . i1rx d P . i1rvt
J 0 q .t" t sIn T x = sIn - I
and the equation for determining rj)i becon1es
.. 2gP . i7rvt
cPi + Pi 2 rj)i = wi sIn T-
(e)
Comparing this \vith Eq. (b), \ve see that we have i1rv/1 in place of w, \vhercas
the factor sin (i-rrc/ l) vanishes. Introducing these changes into the general
solution (12.25), \ve obtain the follo\ving solution for the case of a constant
force P traveling at constant speed v along the beam:
v = 2KP  n J i1rxj 1 t s _ in (  'I)tjl)
 wl. Pi 2 - i2V2//2
r = 1
2gP  i7rv sin (i7rx/ l) sin Pit
- ivT  p .i - p .2 --t21r2 V2/ /2- (12.26)
i=l t ,

612 STRUCTURAL DYNAMICS
As before, the first series in this solution represents forced vibrations of the
beam, \vhercas the second series represents free vibrations.
As a particular case of interest, let us assume that the time required for the
force P to cross the span is just half the period r of the first mode of free vibra-
tion, that is, 21/v = T = 27r/Pl. In such a case, PI2 = 7r 2 v 2 /1 2 , and \ve see that
the first term in each series becomes infinite, but the difference is finite and
can be expressed in the form 1
Pgt 7rvt. 7rX
J'  -1V;rv cas - r sIn T
(f)
This expression has its maximum value \vhen vt = I, that is, just as the moving
force P reaches the far end of the bean1, when it has the value
Pg/ . 7rX
)'mnx = V2 SJn T
or, observing that
7rJ EIg 7r 2 V 2
P1 2 = J4 w - = /2-
we obtain
Pi 3 . 11".1'
Ymax = ;3E/ sIn T
()
This maximuIll dynamic deflection of the beam due to the traveling force is
seen to be about 50 ?crcent greater than the static deflection produced by the
sanle force when applied at the middle of the span.
In the case of bridges, the time required for a moving load to cross the span
is usually large in comparison with the period of the lowest mode of vibration.
In such a case the ratio 7f 2 v 2 /1 2 pt 2 is sroan and \vill be denoted bya 2 . Then,
taking again only the first term in each of the series in expression (12.26) and
assuming that in the most unfavorable case the forced and free vibrations may
come inco phase so that their amplitudes add together, \ve obtain, for the
maximum deflection,
2;{P ( 1 1f'V 1 )
)'max = wt p  7r 2 v 2 /1 2 + Pil /11 2 - 7r 2 v 2 /1 2
Introducing into this expression the factor ex as defined above, it becomes
2ft 8 1 '+ a 2P/8 1
Ylnnx = 1f'4. Elt - a 2 = ;:;mol - a
(12.27)
We see from this that for a slo\vly traveling force, the maximum dynamic
1 See S. Timoshenko, "Vibration Problems in Engineering," 3d ed., pp. 47, 4R, D. Van
Nostrand Company, Inc., Princeton, N.J., 1955.

ARTICLE 12.9
61:1
deflection of the beam is approximately in the ratio 1/ (1 - a) to the nlaxinlum
static deflection produced by the same force.
As a final example, let us consider the case of a pulsating force moving along
the beam \vith constant speed v. Such a condition, for exarnple, may occur in
the case of an imperfectly balanced locomotive \vhecl moving across a bridge.
In such a case the vertical component of the centrifugal force due to unbalance
is P cos wt, \vhere w is the angular velocity of the 'v heel. Proceeding \\,ith this
case in the same manner as in the preceding examples, 'ATe obtain for the
dynamic deflection of the beam
= P1 3_  sin ;7rX { in _( i7rv/1 + w)t _ sin (i7rv// - w)t
y 7r"EI,L I i 4 - (13 + io:)2 i" - ({1 - ia)2
t.. 1
- T [=i 2 a 2 n1_ {j) 2 + -i2a2 ; + f3Y 2]} ... (12.28)
\vherc a is the ratio of the period T of the lo\vest mode of vibration of the beam
to t\\lice the time Tl = /Iv that it takes the moving force to cross the span, and
{3 = T/r2 is the ratio of the fundamental period of the beam to the period
T? = 21r / w of the pulsating force.
When the period T2 of the pulsating force is equal to the period T of the beam
in its lo\vest mode of vibration, (3 = 1, and \ve obtain a condition of resonance.
The amplitude of the vibrations during rTIotion of the pulsaring force is gradu-
ally built up and attains its maximum at the time t = //'1,), \vhen the most
important term of the series (12.28) may be reduced to the form
1 2P/8 , 7rX .
 7r 4 Ei sin T sIn wi
and the maximum deflection becon1cs
1 2P/3 2T] 2P/3
.'YmRx =  1('4EI = r ;' 4£1
(12.29)
l'his sho\\'s that the maximum dynamic deflection of the beam is approximatcly
in the ratio 2Tt/1' to the static deflection produced by the force P. Since 1'1
is usually large in comparison \vith T, \VC see that the value of Ymnx may becon1e
many times larger than the maximum static deflecrion.
12.9 VIBRATION OF BRIDGES
It is known that a rolling load produces in a bridge a greater deflection and
greater stresses than the same load acting statically. Such an impact effect
of live loads on bridges is of great practical importance, and many engineers

614 STRUCTURAL DYNAMICS
r - /

FIG. 12.39 y x P
-IB

have \vorked on the solution of this problem. l There are various causes
producing impact effect on bridges, of \vhich the follo\ving will be discussed:
(I) live-load effect of a smooth-running load, (2) impact effect of the countec-
\\Tcights of locomotive driving \vheels, and (3) impact effect due to irregularities
of track and flat spots on \vheels.
In discussing the live-load effect of a smooth-running mass, two extreme
cases will be considered: (1) The mass of the moving load is large in com-
parison with the mass of the beam, girder, or rail bearer, and (2) the mass of
the moving load is small in comparison with the mass of the bridge. In the
first case, the mass of the beam can be neglected. Then, the deflection of the
beam under the load for any position of this load \vill be proportional to the
pressure R \vhich the rolling load P* produces on the beam (Fig. 12,39), and
can be calculated from the kno\vn equation
Rx 2 (1 - x) 2
y = 3 IE!
(a)
In order to obtain the pressure R on the beam, the inertia force
- (P / g) (d 2 y / dt 2 ) should be added to the rolling load P. Assuming that the
load P moves along the beam \vith constant speed v, \ve obtain
dy dy d 2 y d 2 y
dt = v dx -di 2 = v 2 "dx?
and the pressure on the beanl will be
( V2 d 2y )
R=P 1---
g dx 2
Substituting this into Eq. (a), Vie obtain
_ ( V2 d 2y ) X2(J - X)2
Y - P I - g d x 2 3/EI (12.30)
This equation determines the path of the point of contact of the rolling load
with the beam. An approximate solution 2 of Eq. (12.30) will be obtained by
(b)
ll0he early history of the subject is extensively discussed by Barre de Saint- \.Tenant in his
translation of A. Clebsch's book, HTheorie der Elastizitat fester Korper," Paris, 1883. See
"notc finale du paragraph 61," p. 597,
* P denotes the unspn1ng load, uch as weight of locomotive drivers and their axles, The
force trans(uitted to the axle through the spring is considered as constant.
2 Equation (12.30) and its approxilnatc solution werc establishcd by R. Willis; see Appendix
to the "Report of the Conunissioncrs . . . to Inquirc into the Application of Iron to Rail-
,yay Structures," London, 1849. "his report was reprintcd in P. Barlow, "Treatise on
the Strength of TilTlber Cast Iron and Malcabrc Iron," London, 1851.

ARTICLE 12.9
615
assuming that the path is the same as at zero speed and substituting
Px 2 (1 - X)2
3 IE!
for y on the right side of the equation. l",hen, it can be sho,"'n that y becomes
a maximum ,"Then the load is at the middle of the span, and the maximum pres-
sure IS
( V2 PI )
Rmnx = P 1 + g 3E I
The maximum deflection Od at the middle of the span increases in the same
ratio as the pressure, so that
( V2 PI )
Od = 0", 1 + g 3EI
(12.31)
This approximate solution, as compared \\,irh the results of an exact solution 1
of E<I. (12.30), is accurate enough for practical applications. The additional
term in the parentheses is usually small, and observing that the load P is the
unsprung weight of the \vheel and that the force transmitted to the axle through
the spring remains practically constant, it can be concluded that the live-load
effect of a smooth-nlnning load is not of great practical importance in the case
of short spans.
In the second case, \\Then the mass of the load is small in comparison v.,ith the
mass of the bridge, the moving load can be replaced, \vith sufficient accuracy,
by a moving force,2 and then the results given in the preceding article can be
used. Assume, for example, that for three single-track raihvay bridges \\rith
spans of 60, 120, and 360 ft, the lo\vest frequencies of vibrations are as sho\\'n
in the following tablc:
/
60 ft 120 ft 360ft
.-..._
/, cps 9 5 2
T 1 1 1
-
2/lv '9. TO T2"
11'he exact solution of Eq. (12.30) was obtained by G. G. Stokes; see his ul\1athenatical
and Physical Papers," (:amhridge, vol. 2, p. 1 79. 1'he solution can be obtained also by
numerical integration. Examples of such integration are given in a paper by E. Marquard,
/ngr.-Arch" vol. 23, p. 19, 1955.
2 The case of a rnoving force together with a uniformly distributed moving load has been
discussed by H. Steuding, Ingr.-Arch., vol. 5, p. 275, 1934.

816 STRUCTURAL DYNAMICS
Then, taking the velocity v = 120 ftl see, the ratio T / (21/ v) of the period of
the lowest type of vibration to t\vice the time II v for the load to pass over the
bridge \vill be as sho\\'n in the third line of the table. Considering the bridge
as a beam of uniform cross section and using Eq. (12.27), it can be concluded
that for a span of 60 ft, the increase in deflection due to the live-load effect
is about 12 percent and \viII be even less for the longer spans. From these
examples it can be concluded that the live-load effect of a smooth-running load
is not an important factor. i
Much more serious effects may be produced by pulsating forces due to
rotating counterweights of steam lccomotivcs. .I'he most unfavorable condi-
tion occurs in the case of resonance, \vhen the number of revolutions per second
of the driving wheels is equal to :he frequency of natural vibration of the
bridge. For shorter spans, the frequency of natural vibration is usuaJIy so
high that the resonance condition is impossible at any practical velocity. But
for larger spans, the resonance condition should be taken into consideration,
and the impact effect should be calculated from Eq. (12.29) of the preceding
article.
Let Pi be the maximun1 resultant pressure on the rail due to the counter-
\veights \\'hen the driving \vheels arc revolving once per second, and 11 the total
number of revolutions of the driving \\,heels during passage along the bridge.
Then, from Eq. (12.29) we obtain the following additional deflection due to
the impact effect:
2n 2P l 13
Omax = 2" -4 £ -- / (12.32)
T 11" 4
In the case of shorter spans, \ven the frequency of natural vibration is
considerably larger than the number of revolutions per second of the driving
\vheels, a satisfactory approximation will be obtained by using the first term
of the series (12.28).
It should be noted that all our calculations were based on the assumption of a
pulsating force moving along the bridge. In actual conditions we have rolling
1nasses which will cause a variation in the natural frequency of the bridge in
accordance with the varying position of the load. This variability of the
natural frequency is very beneficial because the pulsating moving load \viII no
longer be in resonance all the time during passage over the bridge, and its
cumulative effect wiIJ not be as pronounced as impJied by the above theory.
Irregularities such as lo\v spots on the rails, rail joints, flat spots on the
\vhcels may be responsible for considerable impact effect, ,vhich may become
especially pronounced in the case of short-span bridges. 1"hesc additional
dynamic effects justify the high impact factors usually applied in the design
of short-span bridges. By climinating rail joints, the dynamic effect on short-
t See S. Timoshenko, Vibration of Bridges, presented at American Society of Mechanical
Engineers, Dccenlber, 1927.

ARTICLE 12.10
817
span bridges and rail bearers can be substantially reduced and the strength
conditions improved.
The theoretical solutions of the preceding article, \vhich \vcre used in the
above discussion, indicate that the Jo\\rcst mode of "ibration usuaJJy has the
greatest importance. l"'his fact indicates that \ve can get a satisfactory approxi-
mation by considering only the first mode and treating a bridge as a system
\\,ith one degree of freedom. Simplifying the problem in this \vay, \ve shall
get for its solution an ordinary di ffcrential equation and can consider the mass
of the bridge and the rolling masses simultaneously. The equation becomes
a complicated one, but it can be treated by using step-by-step integration.
rhis Inethod is no\v \videly used in Russia in the extensive theoretical and
experimental investigations of bridge vibrations. 1
12.10 STRUCTURES SUBJECTED TO EARTHQUAKES
In discugsing the response of a structure to an earthquake disturbance, \\Fe
begin \vith the case of a simple rectangular frame as sho",'n in Fig. 12.40.
Assuming that the horizontal beam AB is absolutely rigid and neglecting axial
deformation of the vertical columns, \ve find that the spring constant k for
the structure is
24£/ J¥ 411- 2
k = --- = - ---
h 3 g 1'2
\vherc T = 27rjp = 271" yWjkg is the period of free vibration (see page 565).
()enoting by 11 the absolute horizontal displacement of the ground, and by v
the displacement of the top of the frame relative to the ground, the differential
equation of motion for the structure bccomes 2
(a)
W d 2
--- - ( u + v) = -kv
g dt 2
1 rhe description of the mcthod and its application in various cases is given in the book by
I. I. I(azey, "Dynanlical Analysis of Railway Bridges," Mosco\v, 1960.
2 The effect of damping is neglected in this discussion.
WIg B
1
h
FIG. 12.40 uf--

618 STRUCTURAL DYNAMICS
With the notation p2 = kgjW, this equation takes the form
d 2 v d 2 u
dt 2 - + p 2 v = - dt2
The complete solution of Eq. (b) is given by Eq. (12.11) of Art. ) 2.4, namely,
v = ! (La ( _ d21 ) sin (ptl - pt) dt
P ) 0 dt 2
(b)
or, \vith T = 2.". jp,
T h ta ( d2U ) . 211"
V = - - . - SJn - - (tl - t) df
211" 0 dt 2 T
(12.33)
The ground acceleration d 2 u/ dt 2 appearing under the integral sign in Eq.
(12.33) is obtained directly from accelograph records taken during actual
earthquakes. Using such an Q(Celo$!,Tdlll, the complete response v = /(t1) of
the structure to the earthquake can be calculated from Eq. (12.33). l"his
method of analysis of earthquake vibrations of a structure \vas first proposed
by Biot. l
For practical applications, the horizontal shear force S induced in the frame
during the disturbance is of particular interest. This shear force is S = kv,
\\1hich, using expression (12.33) for v and notation (a) for k, becomes
S = fV 21r (tJ ( _  2zi ) sin  (t1 _ t) dt = W a
g T J 0 dt 2 T g
a = '"'- ( h ( _ d 2 U ) sin 211" (tl - t) dt
T J 0 dt 2 r
(12.34)
where
(12.35)
is called the effective acceleration. During the earthquake, this effective acceler-
ation varies in a complicated manner, but to obtain the largest shear force, \ve
need only the maximum absolu:e value of a, which \ve denote by A. Then,
the maximum shear force can be expressed as
W
Smax = - A
g
(12.36)
A study of expression (1 2.35) sho\vs that this maximunl shear force in the
frame for a given earthquake depends on the natural period T of lateral vibration
of the frame. By changing the rigidity or mass of the structure, \VC change its
natural period T and therefore its response to the given earthquake, i.e., the
value of A. Representing A as a function of r, \\-'e obtain a so-called neee/cr-
1 See M. A. Biot, Proc. Natl. Acad. Sci., vol. 19, 1933; Z. Angtw. klath. A-ftch., vol. 14, 1934;
A Mechanical Analyzer for the Prediction of Earthquake Stresses, Bull. &iS11101. Soc. A,J1.,
vol. 31, p. 151, 1941; and Analytical and Experimental Methods in Engineering Seisnlology,
Trtrr.r. ASCE, vol. 108, p. 365, 1943.

ARTICLE 12.10
619
d 2 u
dt 2
FIG. 12.41
t
t1tioll spec/riffll. Having such a curve for a given earthquake, the filaximunl
shearing force induced in a given structure is readily obtained fron1 Eq. (12.36).
To actually obtain an acceleration spectrum for a given earthquake, defined
by its accelogram d 2 u/ dt 2 , requires laborious calculations. For each of a series
of chosen values of T, it is necessary to evaluate the effective acceleration a
from Eq. (12.35) over the duration of the earthquake and then select its
maximum absolute value. .rhis procedure, repeated over a suitable range of
values of'i, gives the values of A from \vhich the acceleration spectrum can be
plotted. Such calculations are usually made \vith the aid of a digital or analog
computer. Biot, for exanlple, used a simple torsional pendulum, the natural
period of \vhich could be suitably varied, to obtain acceleration spectra. 1
It should be noted that in the derivation ofEq. (12.33) the effect of damping
\vas neglected I Because of internal friction in the material and joints of a
structure, the amplitude of vibration and consequently the induced stresses vlil1
be somc\vhat reduced. Thus, the shear force S, calculated from Eq. (12.36),
must be considered only as an upper limit.
As an example of calculation of an acceleration spectrum, let us consider an
artificial earthquake having a sinusoidal accelogram \\,ith period 27r / W = 0.5
sec, amplitude O.lg, and duration of 2 sec, as sho\\,'n in r""ig. 12.41. In such a
casc,
d 2 u .
dt 2 = O.1g sIn wt
and expression (12.35) becomes
211" !c t1 ' . 27r ( )d
a = - O.lg - - Sin wt sin - tl - t t
T 0 'T
(c)
For the particular case \vhere the natural period of the structure coincides \\,ich
the period of the disturbance, that is, \vhen T = 27r / w, this becomes
211" !c t I . 27rt . 27r )
a = - 0.1 K --- sin . - sIn - - (t 1 - t dt
TOT T
\vhich, afrer integration, gives
( 1 . 27f'tl 1I"tl 21rt 1 )
tl = -O.Jg - sIn -- - -- cos --
2 T T T
(d)
1 For a description of this device see ibid., Bull. SiS1I1()/. Soc. All'. and Trans. ASCE.

820 STRUCTURAL DYNAMICS
O\\7ing to the factor 7rtl/T in the second term of this expression, it is evident that
the absolute maxinlUffi acceleration A occurs very nearly at il = 41r = 2.0 see,
the full duration of the earthquake, for \vhich value
lamaxl = A = O.lg 41r = 1.26g
Other points on the acceleration spectrum for chosen values of T can be cal-
culated from Eq. (c), and the conlplete spectrum is sho\\'n by the dashed curve
in Fig. 12.42.
Applying the artificial accelogram of Fig. 12.41 to the torsion pendulunl,
tuned to various periods T, Biot obtained values of the absolute maximum accel-
eration A as sho\\rn by the small circles in Fig. 12.42. It is seen that these
experimental points agree very closely \vith the theoretical curve. From this,
it may be concluded that the experimental method of obtaining the acceleration
spectrum is satisfactory for practical purposes.
In f"ig. 12.43, we have an example of a real earthquake \vhich occurred at
Ferndale, Calif., on Feb. 6, 1937. The accelogram sho\vn in Fig. 12.43a
represents the horizontal ground motion in the northeast-south\vest direction,
and Fig. 12.43b sho\vs the corresponding acceleration spectrum obtained from
the mechanical analog. This spectrum sho\vs a peak acceleration of 0.37 g
for a period T = 0.315 see, whereas the maximum recorded ground acceleration
\vas only O.039g. Thus, the maximum accelerations produced by the earth-
quake in buildings having the natural period T = 0.315 see are about 9.5 times
larger than the maximum ground acceleration itself.
The problem of stresses produced by earthquakes is most important in the
case of tall buildings such as sho\vn in Fig. 12.44a. Considering a simplified
case in which all floors are identical, \ve can replace such a structure by a
1.0
'9
f\
I 
.
i 
f .
I
: ,
, I
f ,
, ,
" c
\
\
, ,\
-.- - - - -
9 
I \
J 
;;d' --\ ot(}
)oo-_.. ........0.. --
..0....._
-
A
o
1.4
1.2
0.8
0.6
0.4
0.2
FI G. 12.42
o
o 0.2 0.4 0.6 0.8 1.0 1.2 1.4
T, see

ARTICLE 12.10
a
(a)

11""04
d

,.-4
d
621
N orthea8t-8outhwest
Seconds
0--- ---- -----5-------- --- 10 ----- ----- 15 - - --- -- - - 20--
A
"
0.4
(b)
0.2
0.3
- =
0.1
o
o
0.6
0.2
0.4
0.8
FI G. 12.43
X
I
- .-
I
I
I
, .'" .,. #'
I-
II
T
FIG. 12.44
(a)
---
1.0
1.2
1.6
2.0 T t see
1.4
1.8
x
x
c.. uS d
 -+-;,; x
S dx
inl--
.... I
x
.
j
_t.
(b)
c)
hypothetical prisnlatic bean1 of hon10gencous nlaterial that has elasticity only
in shear as sho\vn in Fig. 12.44b. Then, denoting by v the displacement at
any cross section 1n, the shearing strain at that cross section \vill be a,v/ ax,
and the corresponding shearing force \vill be
av
s = }J. --
ax
where, from Eq. (a),
11 = kh = 24£/
,.. h 2
(e)
(/)

622 STRUCTURAL DYNAMICS
IJct us consider no\\' the case of free vibrations of the bean1, assuming that
the foundation is inlmovable. Since the difference in shearing forces at the
adjacent cross sections 111 and 11 in Fig. 12.44b ll1USt equilibrate the inertia force
of the element 111'11 of the bean1, \ve obtain the equation of motion
(j2V w iJ 2 v
J.I.-------=O
(jx 2 g 8t 2
(g)
where w denotes the \veight per unit length of the hypothetical beam. Assum-
ing that vibrations of the beam are harmonic and proportional to sin pt, \\'C try
the solution of Eq. (g) in the fornl
v = X sin pt
(12)
in \\'hich )( is sOlne function of x defining the shape of the bearn during vibration.
Substituting expression (12) into Eq. (g), \ve get
d 2 X
c 2 - + P 2X = 0 ( 12.37 )
d:t: 2
in \vhich
c2 = f{
w
(i)
The general solution of Eq. (12.37) is
X = A cos px + B sin px
c e
(12.38)
To determine the constants A and B, \VC consider the conditions at the t\\'o ends
of the beam. At the ground (x = 0), the displacement v must vanish; hence,
A = O. At the free end of the beam (x = I), the shear strain (J7,;'/ ax must
vanish. l"'his requires that cos (pi/c) = 0; hence, pl/e = 7r/2, 37r/2, 51f'/2,
. . . ,and 'T = 21r /p = 41/ c, 41/3e, 4//5 c, . . .. "rhus, the assumed solution
(h) takes the form
B . ;7rX . i7ret
'V = i sIn 21 sIn n
(j)
\vherc i = 1, 3, 5, . . .. Taking i = 1, the first mode shape is seen to be a
quarter sine \vavc as sho\vn in Fig, 12.44b, and for i = 3 a three-quarter sine
wave as sho\vn in Fig. 12.44e. A general case of free vibration is obtained by
superposition of alilnodes, and \ve get
v=
l Xi sin Pit =
i &II 1,3,5
l
i = 1,3,5
B . ;1rX . i7rct
i sIn  21 sIn 21
(k)
Considering no\\' the case of forced vibrations due to earthquake-induced
ground motion li, the absolute displacement of a cross section 11l of the beam \vill

ARTICLE 12.10
623
be 1I + 'V, and instead of Fq. (g), \VC have the equation of motion
i)Z'l..-' W a 2
J.L - - - - -- ( u + v) = 0
i)x 2 g at 2
or, using notation (i),
a 2 v i) 2 v iJ 2 u
c2 - - -- - u_.
ox 2 at 2 - at?
(I)
\vhere a 2 u/ at 2 is the ground acceleration, given by an accelogram.
case, \ve take a solution of Fq. (I) in the form
I n this
( ) . 11rX
'Z.,'i = tPi t sIn -iT
( 111 )
\vhcre CPi is an undetennined function of time and i = 1, 3, 5, . . .. \\lhcn
expression (111) is substituted into Eq. </), \ve find that 4>i must satisfy the
equation
. 2 'I) 2 '
.. , 11rX . C Z"1r . l1rX
CPi sIn - iF + 4/ 2 CPi sIn 2/ =
iJ'l.u
- --
at 2
1\1ultiplying this equation through by sin (i7rx/2/) d.,,' and then integrating from
x = 0 to x = I, \\rc obtain
.. c 2 j 2 7r 2 4 (fl.u
CPi + W- cP, = - i7r at 2
(12.39)
This has the salnc forn1 as Eq. (b), and again \ve can use the solution (12.11)
of Art. 12.4. Substituting the quantity (2i2.".2/'4/ 2 for p2 and - (4/ in) a 2 u/ot 2
for q in that solution, \ve obtain
8/ J ta ( azu ) . ci1T'
CPi = ----;;;- 2 - _ 0 I) SIn _ 2/ (t1 - t) dt
cZ"7r 0 to.
()bserving that Ti = 41/ ie, and returning to Eq. ('111), \ve have
16/ 2 27r ( tl ( a2U ) . ci7r , i1rx
'l..-'i = c 2 i 3 1f"3 1'i J 0 - at2 sIn 2i (t1 - t) dt sIn 2/
(n)
representing the irh ITIode of response of the hypothetical beam to the ground
motion. The Iniddle factor of this expression is the effective acceleration for
the ith mode of vibration. IJenoting its maximun1 absolute value by Ai and
sunlffilng exprCSSlon (n) for all modes, \ve obtain for the complete absolute
n1aXIIDum response
16/ 2
V = ..---
C27r 3
I
i=1.3.5
Ai . i7rX
73 SIn. 2[
(0)

624 STRUCTURAL DYNAMICS
l"he corresponding shearing force at any cross section then becomes
av 8p.1 \' Ai i7rx
S = Jl ax = C21f2  12- cos 21
. i=l,::J,5
lhe maximum value of this shearing force occurs at the built-in end (x = 0)
and, with p. = wc 2 / g, hecomes
Smax = 8-:v 1  i
1f' g '-' t
i -1.3,0
(12.40)
Taking the values of Ai, corresponding to the periods Ti of the building in its
various modes of free vibration, from a given acceleration spectrum, the cor-
responding absolute maximum shearing force can be calculated from Eq.
(12.40). Comparing Eqs. (12.40) and (12.36), we see that for each mode of
vibration of the building there appears the factor 8/i 2 1f'2, indicating the portion
of shearing force associated \vith that mode.
In a similar manner, earthquake-induced vib'rations of the hypothetical beam
in Fig. 1 2.44b can be investigated also in the case where the 10\\Tcr end of the
beam is elastically attached to the ground. This case approximates the con-
dition of a building in \\,hich the columns of the first floor are intentionally made
more flexible than those above. Investigations have sho\\'n that such an
increase in flexibility of the first floor may be of value in the case of buildings
of comparatively smaJl height having their lo\vest period in the range of 0.2 to
0.3 sec. For taller buildings having fundamental periods of 1.0 sec or more,
such first-floor flexibility is not of much value.
Again, it must be repeated that damping due to internal friction in buildings
and in their foundations was not taken into account in the above theory.
'rhus, the calculated Smax is only to be regarded as an upper limit to the maxi-
mum shear force.
Regarding practical application of the theoretical analysis of earthquake-
induced vibrations of structures, the accepted point of vic\\' of engineers in
earthquake areas is that the design should be such that the structure \vill survive
the more frequent moderate ground motions \virhout damage, but in the more
rare event of a very strong earthquake, damage ,,'ouId be tolerated so long as it
is not a hazard to life. l 'fhe above theory based on consideration of elastic
deformations is applicable only in the case of moderate ground motion.
1 See G. W. Housner, Proe. ASCE, Eng. Meeh. Div. t 1955, pp. I09129, and Blume, New-
nlark, and Corning, "Dcsign of Multistory Reinforced Concrete Buildings for Earthquake
Motions," Portland Cement Association, 1961. For a bibliography on the subject, see
E. Rosenblueth, Appl. l\1rch. Rev., vol. 14, pp. 923-926, 1961.

Name index
Aitken, A. C., 483, 505
Andrc\vs, E. S., 230
Argyris, J. H., 4R4
Ayre, R. S., 5R6
Barlo\v, P., 614
Baticle, E., 367
Beggs, G., 255
Bendixen, A., 451
Benscoter, S. U., 501
Bctti, E., 249
Biot, M. A., 618
Blcich, Hans H., 560
BIUlnc, J. A., 624
Bryan, (. W., 537
{:aHscv, K. A., 460
Castigliano, A., 230, 234
Clebsch, A., 614
C:orning, L., 624
Cross, Hardy, 460
Culmann, C., 377
Emperger, F., 363
Eulcr, L., 190
Foppl, A., 169
Freyssicnt, E., 367
Gere, J. M., 424, 537
Godard, 1'., 542
Guldan, R., 424
Hardesty, S., 460
Hcnneberg, L., 93
Housner, G. W., 624
Inglis, C. E., 106
Jacobscn, L. S., 586, 601
Jakkula, A. A., 542
Johnson, J. B., 537
Kazey, I. 1.,617
Lagrange, J., 225
LarIle, G., 224
Lameon, J., 594
Langer, B. F., 604
Marquard, E., 6 J 5
Max\vcll, J. C., 68, 224, 248, 263
Mc1an, J., 534-
Mohr, 0., 98, 263,272,276,460
Miiller-Rreslau, H., 286
Navier, M., 481
Nc\vrnark, N., 624
Paris, A., 367
Pearson, M. A., 224
625

US NAME INDEX
Rayleigh, Lord, 225, 249, 568, 581
Ritter, W., 377
Rosenb]ueth, E., 624-
Ruppel, W., 424
Strassner, A., 424
Stiissi, F., 560
Todhunter, I., 224
Turneaure, F. E., 537
Saint- Venant, Barre de, 614
South\vcll, H. V., 460
Spath, W., 578
Steinman, D. B., 534-
Steuding, H., 6 15
Steuerman, E., 556
Stokes, G. G., 581,615
von Sanden, H., 593
Waddcll, j. A. L., 460
Way, S., 549
Williot, 267
Willis, R., 614

Subject index
Absolutc IllaxinlUrn mOInent, I 32
Accelcration spcctrurn, 618
Accclograrl), 618
Adjugate, 487
Arnplitude of vi bration, 566
Angular coefficients, 425
Arches, analysis of, 332
hingclcss, 34 I
influencc lines for, 140, 337, 345, 355
Inatrix analysis of, 510
numerical analysis of, 351
syrnrnetrical, 335, 34 I
thenna] stresses in, 336, 343
threc-hingcd, 27
t\'io-hinged, 335
unsymnlctrical, 371
Assernbly stresses, 310
Axial force, 2
Hearn constants, 424
HeaIns, conjugate, 220, 403
continuous, 412, 431, 549
deflection of, 236
\vith fixed ends, 408
haunchcd. 42 I I 431, 556
inAucnce lines for, 112, 119, 127, ]34
vibrations of, 608
Bending-nlOJllcnt diagrarns, 16, 39
Hetti's la \\', 249
Alast pulse, 583
Bo\v's notation, 68
Huilding franlcs, 469
Cables, 33, 523
Carry-over factor, 408, 429, 462
Casrigliano's theorenl, 229, 234, 257
Catenary, 30, 36
Clapeyron's theorcrn, 224
Complcx trusses, 87, 189, 195
Conlpound trusses, 77, 183
Conjugate beanl, 220, 403
Continuous hearns, 412, 431, 549
matrix analysis of, SOO
Continuous fr.lmes, 451, 516
Illatrix analysis of, 516
Cooper's loading, 105
Criteria, for Inaximurn mOInent, 133
for nlaxirnurll shear, 123
for truss rigidity. 91,103,193
Critical f()CIn, 89, 223
Deflection, of beams, 236
of (Cusses, 257, 276, 285
Deforrl1arion I11cthods, 480, 500
Displaccrncnt, generalized, 226
vi rtual, 45, 98
Distrihuted force, 29
f)isrribution factor, 463
Earthquakes, 617
Elastic center, 343, 373, 377
End mOll\ents, 405
Equi linri urn equations, 3, 10, 162, 177
Fictitious loads, 276, 285
Fixed-end I110Inent, 4 I 1, 426, 461, 502
Flexibility factor, 492
Flexibility matrix, 492
627

628
SUBJECT INDEX
Force, axial, 2
distributed, 29
generalized, 225
internal, 14
resultant, 1
Force u\cthods, 480
Force polygon, 2
Frames, building, 469
continuous, 451, 516
simple, 332,381,390,441
theCll1al stresses in, 394
Funicular curve, 30, 358, 523
Matrix inversion, 488
l\1atrix methods, 480
1\1ax\ycll diagratns, 65
Max\vcll-Mohr Jnethod, 263
1\1ethod, of dcforn13tions, 480, 500
of forces, 480
of joints, 62, 169
of least ,york, 241, 335
of sections, 70, 183
l\1odels for influence lines, 225
l\1omcnt distribution, 460, 475
Moment table, 107
Gcncra1ized displacclnent, 226
Generalized force, 225
Graphical integration, 594
Navier's problem, 481
Numerical integration, 351, 588
Haunched bcatns, 421, 4- 3 1, 556
Henneberg's method, 92, 205
I-lingdess arches, 341
Hooke's la\v, 215, 219
ParalleiograJO la\v,
Phase angle, 566
Phase plane, 596
Plane trusses, 52
Ideal systen1, 4-7
Influence coefficient, (08, 518
Influence lines, 105, 251
for arches, 140, 337, 345, 355
for bean1s, 112, 119, 127. 134
modc1s for, 225
for suspension hridges, 5 W, 553
for trusses, 147, 154, 316
Influence number, 259, 323
Integration, graphical, 594
J1mnerical, 351, 588
Intcrnal force, 14
Rails, stresses in, 601
Rayleigh's method, 568
Reactions, 5i
Reciprocal figures, 67
Reciprocal theorenl, 247
Resonance, 577
Resultant forcc, I
Rings, analysis ot: 398
Magnification factor, 576
Matrix algebra, 483, 484
fvlatrix analysis, of arches, 510
of continuous beams, 500
of continuous frames, 5 16
of trusses, 491
Secondary stresses, 56
Scttlelnent of supports, 312, 394
Shcar force diagratns, 16
Simple (raInes, 332, 381, 390,441
Simple truss, 52, 169
Sin1pson's rule, 353
Slope-deflection equations, 402, 426, 428
Space structures, 161, 325
Space truss, 169
Spring constant, 563
Standard train, 105
Starically determinate truss, 59, 88
Statically indcterminate truss, 59, 89, 294,
304, 3 2 5
Stiffening truss, 538, 549, 556
Stiffness factor, 429, 501
Stiffness rnatrix, 50 I
Lateral s\vay, 451
Least ,york, 241
theorern of, 241, 335
Live load, 105

SUBJECT INDEX
Straia energy, 215, 223, 334-
String polygon, 30, 358, 523
Structural dynamics, 562
Superposition, 219
Suspnsion bridges, 523
intluence lines for, 540, 553
Syrnmetrical arches, 335, 341
l'heorcrn, Castigliano's, 229, 234, 257
C!apeyron's, 224
of least ,york, 241, 335
reciprocal, 247
of three moments, 414
lhermal stresses, in arches, 336, 343
in franlcs, 394-
in trusses, 310
-rhrcc-angle t.uation, 414, 505
l'hrce-hinged arches, 27
Three-motncnt equation, 4- I 4
Truss rigidity, criteria for, 91, 103, 193
TnHscs, compound, 77, 183
complex, 87, 1 B9, 195
deflection of, 257, 27(), 285
general theory of, 85, 188
influence lines for, 147, ]54,316
matrix analysis of, 491
plane, 52
829
Trusses, sitnplc, 52, 169
space, 169
st(uically detcrnlinatc, 59, 88
statically indeternlinate, 59, 89, 294,
304, 325
stiffening, 538, 549, 556
thcrmal stresses in, 310
T\vo-hingcd arches, 33 S
Unhalanced nlonlcnt, 461
U nsynunctrical arches, 371
Vibrations, an...plitude of, 566
of bearns, 608
of bridges, 613
of buildings, 617
forced, 575
free, 562
Virtual displacements, 45, 9
Williot diagratns, 26i, 272
"'ork, definition of, 46
Zero-load test, 9 I, 194