CHAPTER I

PROBLEM 1.1
1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing thatrf, = 30 mm and d2 = 50 mm, find the average normal stress in
the mid section of (a) rod AB, (b) rod BC
SOLUTION
rod AB Fo^ce : Pr 60 x/o3 ys/ Te*s/o*
Area-' A= ^ cl,* = ■?" (*° *'°~J / "= 70*.
S£ */L>"6 *l'
Nor iMj.i' stress •
706. 8£ *
6*„ = S4*.*? MPa
'AB
foci BC
Force- P = 60X/0* - fc)(\Z£x\0*) " -WOX/Oa N
Am*-' Ar }J,1 = f (50xi0-3)* -
I. c?£3J'x'/0"3 *►,*
blo^naf srress : 6"^ -
- 1^0 * 10*
= - <?6.77*/04 Pq
1. ^C3i" •/cJ'3

PROBLEM 1,2
1.2 Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that the average normal stress must not exceed 150 MPa in either
rod, determine the smallest allowable values of the diameters^, and d^.
SOLUTION
rod A8
£+re<u: &. = ISOX/O*?*.
'AS
if «,
p • A = £
6»
-3
d. = «.SS x/0"3 *
Cf. - A A- 6 M»)
rod 8C
Force P - £0*jOJ - <2)(l2S*/03) c
5+rtfSS : S^ = - ISOxlO6 ?a
c- - £ - Jt£.
j * - Jfc£ W^f- ftt> *i°* }
da = 40. J6 xfo"3 vo
-S
cf4 = HO.Z m«

PROBLEM 1.3
30 in.
25 in
1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that dt = 1.25 in and d2 = 0.75 in., find the normal stress at the
midpoint of (a) rod AB, (b) rod BC.
SOLUTION
CCQ Aid AB
P = 13 + 10 = Zl ki'pt
A * f rf,1 - f 0-W)* * I.M74 t»*
(b^> rod BC
p - /o lops
A - -f J/ - f-(0.7^)* = 0.44/2 ;**
*e " A * 0.4413
- 29.6 ksi
PROBLEM 1.4
10 kips
1.4 Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that the normal stress must not exceed 25 ksi in either rod, determine
the smallest allowable values of the diameters d. and d-,.
SOLUTION
p r )2 -*■ lo * 2% kips
6~Aa * as k*'>
A^fd/-
n*B TTd,
rod BC: p ^ io wcpa S^ - -?s ki,- A& = -f d^
4= on Win.
d/r ^
^Cac
gga-**'"'"'

1.5 A strain gage located at C on the surface of bone AB indicates that the average
normal stress in the bone is 3 80 MPa when the bone is subjected to two 1200-N forces
as shown. Assuming the cross section of the bone at C to be annular and knowing that
its outer diameter is 25 mm, determine the inner diameter of the bone's cross section
alC.
SOLUTION
o - A -. * 0-
GeoKn crv^ ' A - i+ ( a*,"" - 4? j

PROBLEM 1.6
1.6 Two steel plates are to be held together by means of %-in.-diameter high-
strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the
average normal stress must not exceed 30 ksi in the bolts and IS ksi in the spacers,
determine the outer diameter of the spacers which yields the most economicaJ and safe
design.
SOLUTION
A"V e&.oVi boi^f Joc**Wom "He, oppe* -pJa4e is polled ©Jew^ Ly +A* 4&t*iJc
fot^ce Pb of -fta bofh A+ 4^e sct^e +i'*e -Re speu.es pushes
r^A/f oSo^e upward iw»'4n a eo^nr^ssf^e fWce ^. J* o^Je^ +a
For He soacer
E"^y ^T'Kig P^ (3th or R
ff.4' = f si (d," - <V)
j.*- j;* |4- - c ■♦§->*'
cfj * 0*402;*

PROBLEM 1.7
1.7 Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that
each pin has a 10-mm diameter, determine the maximum value of the average normal
stress in link BD if {a) 6 = 0, (ft) $ = 90
■20 k-N
SOLUTION
Use ba/* AB£
as "Free oooly
HO k»
(a) 9*0
(OASo s.'n 300?oW) - CO.3^0 cos 36') F** - O
Fbd - 17.3-2 */o3 N
(fe> 6 - ?o6 (o.4«ro cc5 3ef}(.2o*io*) - (o.3oo cos 3o° ^ FQ ■» o
a* -
- 3o *io* W
-c
00 6- =
A
R
2<*OxfO
Be _ ~30KJQ^
72.2 Mffet
360 x /O
-*
r -23.3*/o - S3.3MF^

PROBLEM 1.8
20 kN
1.8 Each of the four vertical links has an 8 x 36- mm uniform rectangular cross
section and each of the four pins has a 16- mm diameter. Determine the maximum
value of the average normal stress in the links connecting (a) points B and D, (ft) points
Cand£.
SOLUTION
Vsz W<* ABC *« a JVe« fe*Jy.
J Rao
D
R
et
XMC =
2MB=o
For r " IS-*"*'0* ^ ^»*>*< £ff '* '" fi*-»f«**''^
A/e4 area of o-e J>,V> k "for + e**.o* » (o^^fo.OW-ftOiC ^
Area -P©/» o«e -P,Vifc i*> Co~/>reiS/b" = (o.0OS )(o.03O

PROBLEM 1.9
0.5 in.
1.8 in.
1.9 Two horizontal 5-kip forces are applied to pin B of the assembly shown.
Knowing that a pin of 0,8-in. diameter is used at each connection, determine the
maximum value of the average normal siress (a) in link ABt [b) in link BC
SOLUTION
Law of Smes
Force +riVnftjfe
&&.
lo
L i'*» k Aft i*« a. 4*e^ s »*o* »* e^% b e^
M»«;*J» SloKw ett p/-» /.r^ - C/-S- O.gKO-^"^ = Q.S
W S+iress ;« A8
* 1.
in
^-tS- ^F = ^ *•■
Link BC is a t-o^pressi'oi r^e^te^
Cross sec+UMo^ dre* is A" 0.S)(0-O - 0.9/h1
(b) S^« ;« 8f 6ic = l5t = ~ l**s* r -7.96 tor

PROBLEM 1.10
1,10 The frame shown consists of four wooden members. ABC, DEF. BE. and CF.
Knowing that each member has a 2 x A -in rectangular cross section and that each pin
has a Vi - in. diameter, determine the maximum value of the average normal stress (a)
in member BE, (b) in member CF.
SOLUTION
:|M1 II.
USi** fiw+tVe. -fVia^c cl& -fir«* bodiy
Use me^be^ "D£F as -Fre* body
.*£
Dx
4*
^o 3-
/-^
o T f
Z MF = o -G.oy$ F^e ) - (3o + ^) t)x = o
(30/$ F«^ -(isity -o
Fee zr - 2Z6'0 A
S+ress in canpi^es^icm me^it** QE
hve<x h ~ ^ w * ti'<i - 8 i»>
AWn - (COC^o-o^ r 7.0,V*
to 5
r - ^qr- - 7-s-o _ iA-7 i

PROBLEM 1.11
/.// For the Pratt bridge truss and loading shown, determine the average normal
stress in member BE, knowing that the cross-sectional area of that member is 5.87 in2.
SOLUTION
12ft Use ewfiVe. -fwss &± -?v^* taJy
Use por+i'on df +WSS +o "H\e -P«£t of a sfic+i'ov*
coT+it^ r»e»*ib«*£ BDj B£^ a»i^ CE.
+t ZFy = o
12
|«0 - SO - t£ Fge = O -- F6s - 50 Jc.i
p*
6"«* = -5* = ^- - s.52 ^;
'er
^.37
PROBLEM 1.12
fl D
HI) ]citM .SO kips .SO kips
1.12 Knowing that the average normal stress in member CE of the Pratt bridge truss
shown must not exceed 21 ksi for the given lo'ading, determine the cross-sectional area
of that member which will yield the most economical and safe design. Assume that
both ends of the member will be adequately reinforced.
SOLUTION
(Jce en+iVe "bross <*s +We body
DlMH = o
'ipS $0 kip*
^ * r
Ace
f\ - For _ 90
Fctf - 9o k,"ps
*C€
6W
51
f.21? ;**

PROBLEM 1.13
1.13 A couple M of magnitude 1500 N-misapplied to the crank of an engine. For
the position shown, determine (a) the force P required to hold the engine system in
equilibrium, (b) the average normal stress in the connecting rod BC, which has a 450-
mnr uniform cross section.
SOLUTIQN
Use piifoA j roe| dtud CvasiU
wa.J!# pe&cJt<ov\ H **■*<} lot*****
r^o-c^io** Ay <a.*d Ay.
60 mm
fc>J 8C
.0 ZMA = O
tody- No+S- "H1A+ rod i«
ot Torc« F^c. 'S known, Dv*w +k« 4<>*ce
2 ~ -[Too
%00
GO'
- Zo%m$\ ^M
_p _
H
So
M
Co,)
Fac _ 3°M*
H
Co
.3
P = 17. 86 *IO
p - n, sc /cKI -*
Fgc " IS.6*3 x/o3, KJ
V5*0 X}0 hn
(b)
6^ - -4/.f Mft

PROBLEM 1.14
1.14 Two hydraulic cylinders are used to control the position of the robotic arm ABC.
Knowing that the control rods attached at A and D each have a 20-mm diameter and
happen to be parallel in the position shown, determine the average normal stress in (a)
member AE, (b) member DC
— 300 mm
400 mm
SOON
SOLUTION
Use vy\emke-r ABC
150 mm 200 mm
0OOM
?> 2TMa » O (0. l-S-o) I FAE - CO. £oo)(soo) = O F^- *f*/c? N
. Rtc- . L+XtQ*
- !2..73*/0*fk
Ae~ A 3l*/6*/o'
d Use cow, kt'neof ^ei^oe^s ABC ojaA 8FD <« fVee koeJy,
F^s-ISbo N
Av^ecc V s»J D3 fc A= ^J1"3 fC^ox/o-*)* = 3ty.l4>*lQ* *?
S4-r*ss m iroe/ *DG * Si
(b)
4.77 *jo Pa
6*D<3r =r - ^.77 M?«.

PROBLEM 1.15
8 mm
1,15 The wooden members A and B are to be joined by plywood splice plates which
will be fully glued on the surfaces in contact As part of the design of the join! and
knowing that the clearance between the ends of the members is to be 8 mm, determine
the smallest allowable length L if the average shearing stress in the glue is not to exceed
800 kPa.
SOLUTION
100 mm
Thane or*, fz>or sepa^*^e wtocs c& Q*0*~ i£sc&A
sk
e/**\ft«
5+
ress i
* «Jje
t ■* goo //oa P^
*-*
A = -E- , ^x/o3 m /5-v/0-»
H
O.I
iT9 */©"** -
L * 2j? +• j«y * w^/ro^ 8 r 30& *M
PROBLEM 1.16
SOLUTION
1.16 Determine the diameter of the largest circular hole which can be punched inlo
a sheet of polysiyrene 6-mm thick, knowing that the force exerted by the punch is 45
kN and that a 55-MPa average shearing stress is required to cause the material to fail.
**
f<^u */u
"i
As*
-irdt - ■£■
d = 4-3.4 mm "^

PROBLEM 1.17
r—I
1.17 Two wooden planks, each Ve - in. thick and 6 in. wide, are joined by the glued
mortise joint shown. Knowing lhat the joint will fail when the average shearing stress
in the glue reaches 120 psi, determine the smallest allowable length d of the cuts if the
joint is lo withstand an axial load of magnitude P = 1200 lb.
SOLUTION
Glue
f t^T1"
8 In.
Se^e^ sor-faxes car/y "Hie -h>-f*P
J(o<mJ P* \*oo A.
An* A- (7)&) J - f J
C A
A. \
J5 i. I3QO
3d" \ZO
d = /.easi
do
PROBLEM 1.18
0.25 in.
1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into
which a 0.6-in.-diameter hole has been drilled. Knowing that the shearing stress must
not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the
largest load P which may be applied to the rod.
SOLUTION'
p .
s ■
A, - it tit s TT(o.ftXo.4>
*= 13.57 k.'ps
po^ d-Pu^;«u^ Aa. = TTeJt =• tt ( l.cV0.25^ - /. 2 5*6 1mx

PROBLEM 1.19
1.19 The axial force in the column supporting the timber beam shown is P = 75 kN
Determine the smallest allowable length I. of the bearing plate if the bearing stress in
ihe timber is not to exceed 3 0 MPa
SOLUTION
s
olv\'r\« toiA L *
l3
L
£-
lS*lo
z
-3
L = /7S.6 m*>
PROBLEM 1.20
1.20 An axial load P is supported by a short W250 x.67 column of cross-sectional
area A = 8580 mm2 and is distributed to a concrete foundation by a square plate as
shown. Knowing that the average normal stress in the column must not exceed 150
MPa and that the bearing stress on the concrete foundation must not exceed 12,5 MPa,
deiermine the side a of the plate which will provide the most economical and safe
design.
SOLUTION
* j.as? * /oe n
Ab - a* = x.
p .
o^r

PROBLEM 1.21 '-^i Three wooden planks are fastened together by a series of bolts to form a
column. The diameter of each bolt is Vi in. and the inner diameter of each washer is Va
in., which is slightly larger than the diameter of the holes in the planks. Determine the
smallest allowable outer diameter d of the washers, knowing that the average normal
stress in the bolts is 5 ksi and that the bearing stress between the washers and the
planks must not exceed 1.2 ksi.
SOLUTION
Gb = -£- .'- Te*s-7e -fWe in bo?f P = 6h A = ■(&)(oJ\<l£3S} ' C.W l?fki'p$
lVa*SnffV*' m&fde oli'd^e4ev^ * di = "<f '*, outside cA^efa^ = G0O
£
C«t^'n«| 3.^*1
A* « £(<*.*- di4>)
O^
« /M ia/ -
61 '
d* ■ d* 4 Tie; = Cf) + Vt^To—r L¥3*3 "
cL = 1.117 |„
PROBLEM 1,22
1.22 Link /lit, ofwidthi = 2 in. and thickness / = V* in., is used to support (he end
of a horizontal beam. Knowing thai the average normal stress in the link is - 20 ksi and
that the average shearing stress in each o? the two pins is 12 ksi, determine (a) the
diameter d of the pins, (b) the average bearing stress in the link.
SOLUTION
Rod A8 is in cor press''^n-
p = -O'A ^(-M^(+)- IO k»>*
p-: rp - £
7 TT V~T?r 1^0^
^ 6; =
oLt " (1.030^0.P«5
3 8. S Arsr

PROBLEM 1.43
20 kN
1.8 Each of the four vertical links has an 8 x 36- mm uniform rectangular cross
section and each ot'the four pins has a 16- mm diameter.
1.23 For the assembly and loading of Prob. 1 8, determine (a) the average shearing
stress in the pin at B, (b) the average bearing stress at B in link BD, (c) the average
bearing stress at B in member ABC, knowing that this member has a 10 x 50-mm
uniform rectangular cross section,
SOLUTION
Ose ba^ ABC o-s a. -rVee body
-0.02S-*
O.O^o
A
6
I F.
6D
Fe
ce
(#■1 Ske<a^ pin a+ B
4 " - t
A = J* s (o*o/6)(o.oo&)? l28*/o"4*>1
; - iB» * l0-^**'*.?1^ = IS€.Vx/d* U7.0 MPa.
s;
Cc^ BeAiM*^ in A8C a+ B
-6
A = it r (o.0f&)(o.oio) r KO*to~ h1
e;-
I6ox/oc
503 Mfii

PROBLEM 1.24
20 kN
1.8 Each of the four vertical links has an 8 x 36- mm uniform rectangular cross
section and each of the four pins has a 16- mm diameter.
1.24 For the assembly and loading of Prob. 1.8, determine (a) the average shearing
stress in the pin at C, {b) the average bearing stress at C in link CE, (c) the average
bearing stress at C in member ABC, knowing that this member has a 10 x 50-mm
uniform rectangular cross section.
SOLUTION
Use bar ABC ccs cc SV«c tocL
11
O.OZS" -*
■ao«to
6
v R
BO
1
r.
ce
I M8 - O - (0.0<+o)Fce - (0.0ZS)(2Q*iOl) - O
Vce- -)P.rx/0
(a) Shear **i pin <=t+ C
-ft i
FcT .
-c
31. I y/O
3/. I MPs-
(b) Sean* w 3 in -liftk Cf a+C
* A \a8*IO~*
te^ B^^g ;« ABC «i C
Fee _
48.* MPa
<3w =
A
78. / MPa.

PROBLEM 1.25
0.5 in.
L3in.„
1.9 Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing
that a pin of 0 8-in. diameter is used at each connection, determine the maximum value
of the average normal stress (a) in link ABt (b) in link BC.
1.25 For the assembly and loading of Prob, 1.9, determine (a) the average shearing
stress in the pin at A, (A) the average bearing stress at A in member AB.
SOLUTION
Use ioiviT 8 o.% -?Ve«. \>Qc\sJt
J
7
o U-.'ps
Law/ of S tries
Force t^aw^fe
Fp
»B
•%c
O
5^ *KT s;^ &>■ " ««'« V
R>« = 1.33oS"kii
AG
ps
a+ A 2
(b) 6>ect**in* stress cc\ A in i^e^tc^ AB
r JE*fi
2A,
7.2S to.'
/*»
^ =
8.3o
13.3o fesi

PROBLEM 1.26
0.5 in.
1.8 in.
1.8 in.
1.9 Two horizontal 5-kip forces are applied to pin #of the assembly shown Knowing
lhat a pin of 0.8-in. diameter is used at each connection, determine the maximum value
of the average normal stress (a) in link AB, (b) in link BC.
1.26 For the assembly and loading of Prob. 1.9, determine (a) the average shearing
stress in the pin at C, (b) the average bearing stress at C in member BC. (c) the average
bearing stress at B in member BC
SOLUTION
\0 Km
10
Sin *i5°
5i"
Si« V
Fa-- 8.965*8 k;
P*
Let") Ske<*rVn<j stress in pi'm <^f C X- -^j^
£*
2.1 CSS
Wio.SoZf)
- t~°lZ
lb)
o.*f
- 22.4
6L =
=■ i/.a/
8^2 fcs."
cb
6i»
_ Re
" A
W.H te;
*
Mil ks.

PROBLEM 1.27
16 mm
1.27 Knowing that5 = 40° and P = 9 kN, determine (a) the smallest allowable
diameter of the pin at B if the average shearing stress in the pin is not to exceed 120
MPa, (b) the corresponding average bearing stress in member AB at #, (c) the
corresponding average bearing stress in each of the support brackets ai B.
SOLUTION
A
con isose^es TWtxi^Je. \*Ji4-ti
angles s^oton hev^.
Use jomf A <x£ a. +r£tf ioy^/
P
Lai*J of SmfiS a.pfi)le4
•\o -Poire* +ri'fitn*J,e
- _E
AS_ —
Fa,
3'*h20* S'ir\\l0O~ $i*>50'
AS
.Sin 20*
sin ;?o°
A©
J,*,., F^- M.73 *IOs H
-c a
1*1
(a) AJ^aJe
v - F^as. , Fab __ %R
TTZ TT(l^oxjO& )
(W Be&r'"n« stress in A8 <4 A.
Af id-- {0.0l&)(\LHS*9o%) • l$3.26*/o~* ^
^w At ISS^c^cr4 '
IMS mm
IS^.^MPa.
61 ^ I5l = iS^Xl^SJ^i . 10.0*10* W.OMPcl
b A \37. Hxlb-*
)o~^x

750 mm
1.28 Determine the largest load P which may be applied at A when 6 = 60 , knowing
that the average shearing stress in the 10-mm-diameter pin at B must not exceed 120
MPa and that the average bearing stress in member AB and in the bracket at B must not
exceed 90 MPa..
Geot^efrv ' "\Wov»i.P* ABC i*
O^ i sosc-Pes "fulfily^Jc fc/»+V)
Use joint A as -fr-ee. oo<iy
P
Force
Ivow of si me* c*.pplieJ ~h
P _ F*a_ _ 3v.
siv%3o
5 m 3o*
AS
■a _ _Rie. *■* 3** - cr
" sm So° **
XT snea^i'nn S+I^SS i* pin <d" B iS Cr»; fi'e«-/
F*a " 3 At = (O(78.SrtW0"' )(|aov/Qt) = ia.26>->*/03 N
IP kea^in* STr-eys i* w>e»*v Lft^ A8 «^" bi^*-fr^e+ at A is c^i'+i'c-ta/
Ab- to! » (0.016 )(o.O/o) - I^O^/O"4 m*"
Fab * Ab€L " •(2i*c>*tO'c.)(eio*-lo%) = 2l.4.*fo* M
MJ/oHjdoJt Fas is He 5*«i/«rf, ;.e. /V-Vox/o* W
= 8.3/ v/oJ A/ 2.3>iW

PROBLEM 1.29
125 mm
75 mm
1.29 The 6- kN load P is supported by two wooden members of 75 * 125- mm
uniform rectangular cross section which are joined by the simple glued scarf splice
shown. Determine the normal and shearing stresses in the glued splice.
SOLUTION
p * 6 x/o3 M
8 = ?o°- 70° ^ 2o
G"= S£f kPa.
PROBLEM 1,30
125 mm
"5 mm
70*
1.30 Two wooden members of 75 * 125- mm uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that the maximum allowable
tensile stress in the glued splice is 500 kPa, determine (a) the largest load P which can
be safely supported, (b) the corresponding shearing stress in the splice.
SOLUTION
, Je§L , <™**I*~*K^«'*K *******
_n
vm
S* = Stoo * io3 ^
" " cos1© COS1 ^O
(M
t1 = IS^.O kfix.

■Ht-r.:'"."■!!■■
PROBLEM l.3l
P cos*e
1.31 Two wooden members of 3 * 6- in. uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that the maximum allowahle shearing
stress in the glued splice is 90 psi, determine (a) the largest load P which can be safely
applied, (b) the corresponding tensile stress in the splice
SOLUTION
9 * <=ro^ - to° = So*
A0 = (3KO = l* '«*
r sin ?e " s.'* ICO0 ^^°
ga^Q cos* So*
= 75".S
P = 3«o A
6" = 7&.S" £>st'
PROBLEM 1.32
1,32 Two wooden members of 3 * 6- in. uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that P = 2400 lb, determine the normal
and shearing stresses in the glued splice.
P- TLHoo A
SOLUTION
e = 9o° - no* - so°
6"- «Sff. I pa/
r1 - ,£ a,-- ,2© - i^^b-irL/.oo0 ^ zs ?

PROBLEM 1.33
1.33 Aceniricload Pis applied to the granite block shown. Knowing that the resulting
maximum value of the shearing stress in the block is 2 5 ksi, determine (a) the magnitude
of P, (b) the orientation of the surface on which the maximum shearing stress occurs,
(c) the normal stress exerted on that surface, (d) the maximum value of the normal stress
in the block.
SOLUTION
Aar (&KO - 3S \S r^-- 2.rk»;
(b) sin 20* i ze * *to° e = *r* -*
(O
fcn
Sr*£
'V
2 A* (sOOO
-2.rks;
PROBLEM 1.34
6 in.
6 in.
1.34 A 240- kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the orientation
of the plane on which each of ihese maximum values occurs.
SOLUTION
- 2.
a-t e-- o*

PROBLEM 1.35
ftBS?3!ffl9W*^*^5'WSH?$?
1.35 A steel pipe of 300- mm outer diameter is fabricated from 6- mm- thick plate by
welding along a helix which forms an angle of 25 ° with a plane perpendicular to (he axis
of the pipe. Knowing that a 250- kN axial force P is applied to the pipe, determine the
normal and shearing stresses in directions respectively normal and tangential to the weld.
SOLUTION
= 5~SH x{o~s m
C»S
%B r
-^roK/o3
7=3
=-37./ *\o
6 6"=-37J MPa
t = =?- ** 2e> = -^ox/03 gl- ^y
-'\1.2B x/o1
-£ = /7.2S MflsL
PROBLEM 1.36
1.36 A steel pipe of 300- mm outer diameter is fabricated from 6- mm- thick plate by
welding along a helix which forms an angle of 253 with a plane perpendicular to the axis
of the pipe. Knowing that ihe maximum allowable normal and shearing stresses in
directions respectively normal and tangential to the weld are o = 50 MPa and x = 30
MPa, determine the magnitude P of the largest axial force that can be applied to the
pipe.
SOLUTION
d0 - 0*300 to fo = ^©r Cc\5"Om
ti = f0 " fc = O. ISO - 0.0O6 = Q% /4Y to
= S-SHx/o
-s
^
e - ^'
BaseJ
B^s«J
'■*
cos2 a
e°s* ZS-
r « £■ *.'* 3e
2Ao
Ps
Sm*ifl
e^

PROBLEM 1.37
1.37 Link BC is 6 mm thick, has a width w = 25 mm, and is made of a steel with a
480- MPa ultimate strength in tension What was the safety factor used if the structure
shown was designed to support a 16-kN load P ?
SOLUTION
Use bar AC D a5 a -£ve*
body cl^J not« +-*i A/f
(480) Fec - (Soo)P = ©
*fffo
**So
bHi^Je. J>o*J -Tor (>e^ SC F, - Si A
Fj * ^%Ok101)(0.O06)(0.O2S)~ -J1.*)CPH
PROBLEM 1.38
-600 mm
1.38 Link SC is 6 mm thick and is made of a steel with a 450-MPa ultimate strength
in tension. What should be its width w if the structure shown is being designed to
support a 20-kN load P with a factor of safety of 3 n
SOLUTION
Use \><xr ACD flA'a $Vet loo^y
anel hole *k**l i^e^te^ 8
?,. \/'
pi
FoV* A Tfr.e-4-ov tft
BC
HSo
V80
SI; 45 o H/O6
-t
GC.G7 * /O ^
F~oe* a recTci^^oroi^ -secriam A = w£ or
iv -
- A _ \CC.C7*IQ
-6
-1

PROBLEM 1.39
15 in.
1.39 Member ABC, which is supported by a pin and bracket at C and a cable BD, was
designed to support the 4-kip load P as shown. Knowing thai the ultimate load for
cable BD is 25 kips, determine the factor of safety with respect to cable failure.
SOLUTION
Ose me*wk*r ABC «■* *.
Fvee. \tody a*el *crfe
c*s 4o*)C30'«^ *(.?•*'* ^X'5>'0 " (f» coi 3o'X'5''^
•-12 in.
Folc+ov of -sarfcfy -fe>^ c«<AJe BO F.S. -
PROBLEM 1.40
«+*-12in.
1.40 Knowing that the ultimate load for cable BD is 25 kips and that a factor of safety
of 3.2 with respect to cable failure is required, determine the magnitude of the largest
force P which can be safely applied as shown to member ABC.
SOLUTION
0s« meml^er ABC <*-s o.
Free, body aw hoTe
*W>o-force ivtC"^ ker.
(PcosHO°)(3o,v') +l?s;«Hcr)(lSi*) -(FZbc<n3o?X)S-i^
- CF6b at* So*)0a "^ r c

1.4 m
PROBLEM 1.41 1-41 Members AB and AC of the truss shown consist of bars of square cross section
made of the same alloy. It is known that a 20- mm- square bar of ihe same alloy was
tested to failure and that an ultimate load of 120 kN was recorded If bar AB has a 15-
mm- square cross section, determine (a) the factor of safety for bar AB, (b) the
dimensions of the cross section of bar AC if it is to have the same factor of safety as bar
AB
SOLUTION
Lemo-l-k ert meeker A3
ha = 7o.7£ * + 0.* u = O-SS *"
Use entire jv^ss as a. £r*t kocly A* T ^y
\.H A* - (0.7f)(28) = 0 A^ IS kN
Ay* *2 UN
■IS kN
Ay - 28 =0
^21^ =o.
Use joi*t A olS Tree tody
t Ay -ZF„ = o
A£
2^£ F -A - n
0.8S h*e - ** - O
0.7*T
2«kK>
■tZFy - O
Ay-F^^Ftt= o
F-- «-iS^zls »*w
Ac
For He -fes+ k><*f
F©<r -fie irwibTe^'ft^
A - (o.ozo)x - voo*fo"c w1
P„ * 130 */03 N
S,» =
- "ft
'g0"/ql =• 3oowcnt
4ooWo
(a> F*, bar A8 F.S." #- r &A , (30O*foM(O-Q*)*
Be Fk I7K/03
(Wl For bar AC
3.t7
6*u 300 XfO*
(X = /£.2?y/o"' ^
l€.S?9
Min

PROBLEM 1.42
- 0.75 m -
0.4 m
1.4 m
9?MC = 6
+ tZFy = o
1.42 Members AB and AC of the truss shown consist of bars of square cross section
made of the same alloy. It is known that a 20-mm-square bar of the same alloy was
tested to failure and that an ultimate load of 120 kN was recorded. If a factor of safety
of 3.2 is to be achieved for both bars, determine the required dimensions of the cross
section of (a) bar ,45, (b) bar AC,
SOLUTION
LendH o? weaker AB
Use en+\V« +ri»i^ as a -free fco^/
Ay - 28 * O
^v*^g kN
list jomi A *,$ -pree loocly
*ZF„ = o
»Mtf
*tT^»c> Ay-Fte-AiFAB =
R = 28 - lMX'7l . *a *W
-6
For He fes+ feat- A = (o.ctfo^* =■ 4oo *70~* wx -py - I^O */Oa W
Fo^ +*J»e to*Te*VA<l
A
6i= 4-*- =
= 3oo*to' Pec
F.S.
B« F*a F«
a. = )3. V7 */o*3 r*
(eO Pot* m*e>wlef A8
13.4-7 mim
(b) Fo*r me^te^ AC RS. =
F*.
FA
b = \1.G\ */£>
-i
M
14.6/Him

PROBLEM 1.43
■10 kN
120 mm
■20 kN
1.43 The two wooden members shown, which support a 20-kN load, are joined by
plywood splices fully glued on the surfaces in contact. The ultimate shearing stress in
the glue is 2 8 MPa and the clearance between the members is 8 mm. Determine the
factor of safely, knowing that the length of each splice is L = 200 mm.
SOLUTION
Tkev* a.<re 4- s%.p*y*«^t. a*to>s of aloe,* EqjcX
a-P^e a^ecx wtuyf -fv-ou*<iSKn.'+ 10 k^ of &\tc*+
Lew^fk off s^rce L= 7JI + C wUm. J) - $Z+^h dfqi'oe ***J C '
cjCeaAaoce. 2 = -k(L- c} - i(o.zoo -o.oos) r o.o<?6*t.
Are*, of <jiU f\- Jtw = (0.09Cyo./2o^) =r ll.SZyla***-
7 P /ox/o*
PROBLEM 1.44
JDkX
20 kN
1.43 The two wooden members shown, which support a 20-kN load, are joined by
plywood splices fully glued on the surfaces in contact, The ultimate shearing stress in
the glue is 2.8 MPa and the clearance between the members is 8 mm.
1.44 For the joint and loading of Prob. 1.43, determine the required length L of each
splice if a factor of safety of 3.5 is to be achieved.
SOLUTION
<*K>e a^ftft. mus*f +ta»vsiwvi"t \0 ItM tf shea.*
P« = fuA * -T«iw
i*
_ R
3S*x/0'
-3
rtfw "(a.s»gofcXo./aoj
r /OV. IT xJO *
^/G.3 x/o " m
2J6 MM

PROBLEM 1.45
24 kips
1,45 Three 7 - in.-diameter steel bolts are to be used to attach the steel plaie shown
to a wooden beam. Knowing that the plate will support a 24-kip load and that the
ultimate shearing stress for the steel used is 52 ksi, determine the factor of safety for this
design.
SOLUTION
For ec^k loJ+ A = ?d' = $ (I)' = 0MH\% i*1
Pv - A?0 = (a.Mi*)(s*)* 2Wi *<■>*
Per loJt ? -- ^ ' 8 *>'
F_S. r £-.- -2|*S2L r 2.27 ~m
PROBLEM 1.46
1.46 Three steel bolts are to be used to attach the steel plate shown to a wooden beam.
Knowing thai the plate will support a 24-kip load, that the ultimate shearing stress for
the steel used is 52 ksi, and that a factor of safety of 3.37 is desired, determine the
required diameter of the bolts,
SOLUTION
For e*cU V>oH P = -^ = 8 U<f*
Reject P0 r(RS-)P ^ (3.37)^8).- ZCIC k*fs
r - &
A - fLr-§^ * O.S^C ;»x
= 0.2»25* in.

PROBLEM 1.47
1.47 A load P is supported as shown by a steel pin which has been inserted in a short
wooden member hanging from the ceiling The ultimate strength of the wood used is
60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is ] 50
MPa in shear. Knowing that the diameter of the pin is d = 16 mm and that the
magnitude of the load is P = 20 kN, determine (a) the factor of safety for the pin, (b)
the required values ofb and c if the factor of safety for the wooden member is to be the
same as that found in part a for the pin.
-5
SOLUTION
P = 20 ktt - 2©*io* N
(*) Pi*: A^d1 = f (O.0I6) *2Ol.06*lo~»:
Double *W ?= Ja ^mft-
= GO. SI'? *JO* N
K5>- - f> ' *o*ioa " 3-0/ ^
40 mm
6" = ^
°u A
b -~ ol -
- Pu
K/(t»-«s/)
wkere n/r 40 mm r O.OVO vn
Vi&v
h - *l- I
WlW
P„ r fO.3/9 *lc? N -Po
PooUe s^eo^g ea.ok cured, h A = WC
2A " awe
C r 2wt " (aXo.0Votf>.S,Wo*)
-s

40 mm
1.47 A load P is supported as shown by a steel pin which has been inserted in a short
PROBLEM 1.48 wooden member hanging from the ceiling. The ultimate strength of the wood used is
60 MPa in tension and 7,5 MPa in shear, while the ultimate strength of the steel is 150
,•-.._ MPa in shear.
1.48 For the support of Prob. 1.47, knowing that b = 40 mm, c = 55 mm and d = 12
mm, determine the allowable load P if an overall factor of safety of 3.2 is desired.
SOLUTION
Based on oloobh s\e+* f« pfw
= mf)(.O.Ottf(\SO*l&) - 33.93 x/O* N
Based oh "hemSi'oM i'cn WOoef
?„ - A<50 = w(b-J)G0
= {o.oho)(o.oho- o.o\z\go»id1)
* G7.2 xlO3 W
Based on eleoUfe s/ie«^ i* "+*« woo^
= 33. o x/o3 hi
Use smJJesi Py - 33.0 x/oJ |U
Al/ow-bA P* ^ = 3f;;'/0-3 ' 10.31 */o* N
10.SI kW <-•

PROBLEM 1.49
Top view
1.49 In (he structure shown, an 8- mm- diameter pin is used at A, and 12- nun-
diameter pins are used at B and D> Knowing that the ultimate shearing stress is 100
MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two
links joining B and Dt determine (he allowable load P if an overall factor of safety of
3.0 is desired.
8 mm<^J~~
T '
■=±L
U-200 min-«4--l80 mm -<A j2 mm
j3l
J.
~5F
B
20 mm TP
8 mm'
Front view
12 mm
I fl
*6 mm
\D
Side view
SOLUTION
S+a-kcs '. Use ABC as f*et body.
2"MS = O O.Zo Fa - OJ* P = C
ZMA= O 0.2o FiD-0.38pr: o
!P- 7ST Rk»
-& *■
A F.S. " 3.o
P-- fFA ^ 3.73 *I0* N
Based ot* cfo.uk(e ske<^ ''n pi*S an 8 **d D
F - ^SA - (aXfo6*/Qc)0l3.lo»Jo"4) ^T.Sztyvo^N
8D F.S. 3-0
P - -ft Pad * 3.97 */o* M
Based o* compress iotn in
F«r one jfuk A = (0.0*0X0.008) = l46*/o"c**-
F * g6»* teM«b«/o«XiCo»fcr') r ^7y/0sM
rao F.S. - 3.o
P = Tf fbt> =" 14. of *Jo* N
AMo-JJe vJue of P is a^iied" - ?*
3.12*10* N

PROBLEM 1.50
Top view
U-200 inin-O*-180 mm-w| ,2
1.49 In the structure shown, an 8- mm- diameter pin is used at A, and 12- mm-
diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100
MPa at all connections and that the ultimate .normal stress is 250 MPa in each of the two
links joinings and D, determine the allowable load P if an overall factor of safety of
3.0 is desired
1.50 In an alternative design for the structure of Prob. 1,49, a pin of 10-mm-diameter
is to be used at A. Assuming that all other specifications remain unchanged, determine
the allowable load P if an overall factor of safety of 3,0 is desired.
8 inin
<c
3=S_
.t3l
T3T
fl
T
SOLUTION
5+a+Vcs :
Ose ABC as free body.
6
8 min-*-
Front view
12 mm
fa B
-8 mm
iD
Side view
0.2o—V|i-0-l8
2"MS = O O.2o Fa ' O.'S P = O
Bcsed ow alouLle sKes^ i* pi* A
1 ~ JT.23$y/o3 A/
p = ^ FA - *••* *'oJ N
Based ow JooUe stie<*^ in piVs <d 8 <W D
F r -*£* -_ (aXlOO*/QC)(ll3JO»lO"4) rrT.S^fy/o^N
P = TT Fbo - 3.97 */o3 M
Based on Com press to vi (k> i.-ks 8D
Fo,r out *i„k A = (0.0*O XO.OO 8) = )£0*/0
P = TT Fsd - 14. of * JO* AJ
AfPo^Ue value of P is *»wJAsr+ A ?- 3-*f7 »-/0*A»
2.77 kM

PROBLEM 1.51
8F,
fte
1.51 Each of the steel links AB and CD is connected to a support and to member BCE
by ~2 - in.- diameter steel pins acting in single shear, Knowing that the ultimate shearing
stress is 24 ksi for the steel used in the pins and that the ultimate normal stress is 60 ksi
for the steel used in the links, determine the allowable load P if an overall factor of
safety of 3.2 is desired. (Note that the links are not reinforced around the pin holes.)
SOLUTION
Use feCE as -Pi^ee tod
IM8r O
P = f Fco
13 P = o P = ;§-5»
2MC - O
A = $*!* = %(±)1 = o.i9&ss-;Mv
R, - SiA - (6o)fo.wO= 7.Sb kips
F - ^8<7 A

F«.*
ct>
ei
J
PROBLEM 1.52 1.51 Each of the steel links AB and CD is connected to a support and to member BCE
by -j - in- diameter steel pins acting in single shear. Knowing that the ultimate shearing
stress is 24 ksi for the steel used in the pins and (hat the ultimate normal stress is 60 ksi
for the steel used in the links, determine the allowable load P if an overall factor of
safety of 3,2 is desired, (Note that the links are not reinforced around the pin holes,
1.52 An alternative design is being considered to support member BCE of Prob, 1,51
in which link CD will be replaced by two links, each of { x i-in. cross section, causing
the pins at C and D to be in double shear, Assuming that all other specifications remain
unchanged, determine the allowable load P if an overall factor of safety of 3.2 is
desired.
SOLUTION
Use roe^Le^ 3CS gl$ -pr«e lody
q ' s ZM8--o SFC0-2op = o p--£Fc0
C £
So^eJ on 'fetsi'oit /* -PiVtk AB
A r (t-J)t = O-OCi^ o./as-i'H1"
&*,J fe^ 'Jti*fct /B-7is.^rn*//€B,+)uTe» H£ =i4*TUM kips
tovvespon&W} i>WiW+C iofiJ £>„ sfysjdKj^; f^-J-^ = 3.^/6 Kips
8«-sei o« pi*s cuf C q^J D i« dooUe. shear
A = f^ =^(-t)^ ©.WS^m1,
F„» 2f„A = &XM)(o.l16SS) = ?.W/8 Kips
0>ase<£ o* tension m Sinks 13C
A=(b-«04 = 0--fcKi)* o.o6^r,'^ C***M)
U<P+»**H AwJ-Fb* JtUks Bt i's stoxJfed-j i.e. Fu = 7^<9 fc.p*
Ac+uJ wl^J* Jtoul h s^-dtferf^ i.e. P0 = 3.0o k/'ps
AiPowJJe W fer sfruc/ure T * ^ - JjJ2" r 0.132 ftp
P - ^38 A. "*

PROBLEM 1.53
1.53 Each of the two vertical links CF connecting the two horizontal members^/)
and EG has a 10 * 40- mm uniform rectangular cross section and is made of a steel with
an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20-
mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins connecting them to
the horizontal members.
SOLUTION
Use member- EFFG a.% -taeboJy.
IM. F
OAo
-rt*-
6.K-
24 kN
9lMe = e>
0.4O FCF -t<XGS)(im*IOl) * O
7f\'W
CF
8a,sed on fens?®*) in J^ ks CF
c , j. r i c, . F„ _ W.gV* */o3 _ , „, .
F-oUrofWety f^S. -- — = .-^__- = 2.42 —

PROBLEM 1,54
250 mm
1.53 Each of the two vertical links CF connecting the two horizontal members AD
and EG has a 10 * 40- mm uniform rectangular cross section and is made of a steel with
an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20-
mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins connecting them to
the horizontal members.
1.54 Solve Prob. 1.53, assuming that the pins at C and F have been replaced by pins
with a 30- mm diameter.
SOLUTION
Use member E"FG as -fWe hoJy.
Ffl*
t B
a:
c*
I
&Ho
-*+*-
i>ZME = o
MkM
■U kN
FCF = 3*1*10* N
A = Ct-Jlt - (0,0*o - O.osoXo.oio^ 100^/0* w (om* JUO
Fu ' 2 S^A = ttX4oo*/o')Ooo*l0"c) = gO.O*/o* n
BclScJ ov\ doo\oie. sheas \v\ pins
A - ^J*" * ^(o.OSo)1, " 106.t& */'->"* mE
Fu = ZftA = 0iK\£O*\0*){y06.U*l0m*)* 21^.06 *lO* N
AcJ-ua-J F0 i's s*&JJ*r v*Xe, U. Fv ~ go.o */C3 W
F^tf^^ *f sorely
RS. = ^
= goo*Jo* - ^ 0s5-
CF
3^ v/o1

PROBLEM 1,55
,X-
/. SS A steel plate ^ - in. thick is embedded in a horizontal concrete slab and is used
to anchor a high-strength vertical cable as shown. The diameter of the hole in the plate
is j - in., the ultimate strength of ihe steel used is 36 ksi , and the ultimate bonding
stress between plate and concrete is 300 psi- Knowing thai a factor of safety of 3.60
is desired when P = 2.5 kips, determine (a) the required width a of the plate, (b) the
minimum depth b to which a plate of that width should be embedded in the concrete
slab. (Neglect the normal stresses between the concrete and the lower end of the plate )
SOLUTION
A a -fo-m
F.s = -Bi - 6i(a-i)t
P P
■a. - A * iEliZ 1 ♦ <3.<pX2.Q
A
tt = 0.3oo Wsi'
F-S. = ^
So Ivi'n* Ti>v* b
b - 8-0-5" in

^
1.55 A steel plate ^- in. thick is embedded in a horizontal concrete slab and is used
to anchor a high-strength vertical cable as shown. The diameter of the hole in the plate
is j - in., the ultimate strength of the steel used is 36 ksi, and the ultimate bonding
stress between plate and concrete is 300 psi.
1.56 Determine the factor of safety for the cable anchor of Prob. 1.55 when P = 3
kips, knowing that a = 2 in. and b = 7,5 in.
SOLUTION
3 kips
i \
vv
m
Boused on +ens/ow *'"i p-^aTC
Pu = eruA
Based oh sne<xr between pPafe anc/ co*oc**efe siUt
A= peKw«W * ©UpH = 2(a.+ £)b = ^a + jjp )(7.5^
A - 34. G9 in2" 'K* " O.3oo ksi
Pu = ^A = C0.3oeO(3fc6?) = 10.4! kips
jr e - JBi- - J.O»HI _ o 117

PROBLEM 1.57
1.8 m
M.57 A 40-kg platform is attached to the end B of a 50-kg wooden beam AB, which
is supported as shown by a pin at A and by a slender steel rod BC with a 12-kN ultimate
load , (a) Using the Load and Resistance Factor Design method with a resistance factor
0=0.90 and load factors yD= 1,25 and yL= 16, determine the largest load which can
be safely placed on the platform, {b) What is the corresponding conventional factor of
safety for rod BC ?
SOLUTION
'a,
—^t
tv; i v,
9rMA = o (a.«0|P - 2A -tr, - i.z 1aTz ■■ P--{Tf, + £K
For J\Ve. J>o&Jiflfl "WJ = Yr\q Tl£ = O
w
ki'cJi
».-4f
rDpD + rt?L = ?p0
1 yl " ;.c
P * P» + Pl t
F.s. ^
9.81
3<S2. A3
P
13*10*
N
O8Sx/03
.7(3

PROBLEM 1.58
*1.58 The Load and Resistance Factor Design method is to be used to select the two
cables which will raise and lower a platfonn supporting two window washers, The
platform weighs 1601b and each of the window washers is assumed to weigh 195 lb
with his equipment. Since these workers are free to move on the platform, 75% of their
total weight and ofthe weight of their equipment will be used as the design live load of
eachcable.. (a) Assuming a resistance factor <p= 0.85 and load factors yD = 1.2 and
Yl = 1-5, determine the required minimum ultimate load of one cable, (b) What is the
conventional factor of safety for the selected cables?
SOLUTION
r0 - —
Convftrt"!-^^ r^cfo^ of Safe iy
P ' % + Pt r i* 8o + 0.74* x Z* IV r 373.S- JU

PROBLEM 1.59
A B
2.25 m
2.25 m
memb!rD^ knn^oSfhan.di0ading Sh°Wn' determine the average normal stress in
member Dfit knowmg that the cross-sectional area of that member is 2500 mm2.
SOLUTION
Us'.v^ *>e.+lio<J tff joiVh fo -F;nd Me»i be*-Payees
J01V+ B * A3 <W BD a^e ze^o fwc« v*«mW&.
m
ISO
130 Vn
Faq _ ISO
3.7.T " 3
Ad
(compression )
rAc
2?5"
to
t»r
6y SiwiiA^ Trictnoi'fiS
2,35- ' 2.75-
f>F = 135 iW Cco^p)
3
= /3r*/<r w
A^e*,' A^ - 7SOO m^,1* - ?SOQ»lQ<' *i
\ZS*IC

PROBLEM 1.60
1.60 Link AC has a uniform rectangular cross section | in. thick and 1 in. wide.
Determine the normal stress in the central portion of that link.
1201b
t)IMa = o
2^2.
Ayve«. of Pmk ^C'
Stress J« A'rtt AC :
SOLUTION
Use Hie. pi)a4e. -foat+k**-
wi-fk •jvo poxJ-eys as c_
-JVee tody. No-k +k«L.+
He CcXXq +e>iai'o*
causes at Uoo<Plo-i*
0* "Hie bocJy.
2
\J?oA
lO'.K
A - / ;n * i ^ = o. j 25- /* *■
6>t- ^ = - J§j^ r l0Mf.; = l.osi kS;
PROBLEM 1.61
15 mm
Steel
1.61 When the force P reached 8 kN, the wooden specimen shown failed in shear
along the surface indicated by the dashed line. Determine the average shearing stress
along that surface at the time of failure.
Area. \oe.in<t sheared
A= qo*iMf» ISmi* ~ 136"o *i*,X = 135*0 w/O'6
tV)
90 mm
Wood
SI.
fr gvio3 K)
* - £ *i£^ * ****>■ P-^93Hfe
PROBLEM 1.62
SOLUTION
1.62 Two wooden planks, each 12 mm thick and 225 mm wide, are joined by the dry
mortise joint shown. Knowing that the wood used shears off along its grain when the
average shearing stress reaches 8 MPa, determine the magnitude P of the axial load
which will cause the joint to fail.
-16 mm
h
16 mm
P- XA * (8></oS()^y ID"*) = 1.2%*l& N - %# WW
isiOmm
25 mm 225 mm
T

PROBLEM 1.63
4(H) II
1.63 The hydraulic cylinder CF, which partially controls the position of rod DEt has
been locked in the position shown. Member BD is § in. thick and is connected to the
vertical rod by a j - in.- diameter bolt. Determine {a) the average shearing stress in
the bolt, (b) the bearing stress at C in member BD.
SOLUTION
Use ine^be^ BCD as a. -Fv*ee ho^jj o.\*<£
1 vio+e +li oA AS is a. -f>vo -foirce wewbeA
~ 2S-X. in.
t>rM£ = 0
<*>£> a* 75°
4oo a;„ 74'°
-ZF,--o
~(lc*slo9)(ttooai« 7^°)-(7*^2o0)(HoO ^7S°) =0
3.3CC78 /^8 - 27^..S5"= o -V R% =■ 8%&J¥1 '&.
~ jrf TAB + Cx 4 ^o cos 7S» - o
C* - 4!iKfc*Q - ^oo cc* 75'
78.3*-Pi.
AB
* Cy ~ HOQ s.'n 75° = T>
a^a^ai + 400 3.^ 7^- - u?f.^i4.
A = f d* - ^(S)1, - o. now ;^
^ A
(tl Sea^'r^ sfVess cJ- C ''" ^e^bes- BCD, -
9= 11*7.2 ^i.
**£
0.2^*3 7i"
- «5;// x/O p*.' - 6". H kst'

PROBLEM 1.64
1001) lb
£•)
500 lb
P
A
Zkt
1.64 A 2 - in.- diameter steel rod AB is fitted to a round hole near end C of the
wooden member CD. For the loading shown, determine (a) the maximum average
normal stress in the wood, (b) the distance b for which (he average shearing stress is
90 psi on the surfaces indicated by the dashed lines, (c) the average bearing stress on
the wood.
SOLUTION
(<Z) Ma4tii"0*t nor*\&J) stress i'o "Hie wood
w
I.82T
2f£
^ -E.
looo
7.4/
iti
Jfc " (*)(*}
r 2CC7
i°
*/
PROBLEM 1.65
15 mm on __
, UUmm
1.65 Two plates, each 3-mm thick, are used to splice a plastic strip as shown.
Knowing that the ultimate shearing stress of the bonding between the surfaces is 900
kPa, determine the factor of safety wilh respect to shear whenP = 1500 N.
SOLUTION
B©*d ott^cc" (See -ft*ore,)
- ISO©*1*'*" - \&OC>*IQe'**
Tu " 7k% ■- U)^ox|ocX9ooWo3) - 270© M
wv\i U i to e.~r£v-3
■ p
PROBLEM 1.66
5.5 in.
3.5 m.
1.66 Two wooden members of 3.5 * 5.5-in. uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that the maximum allowable
shearing stress in the glued splice is 75 psi, determine the largest axial load P which can
be safely applied.
SOLUTION
A0= L3.s)(s.s)- I<?.W i«*
r «-■£*■
©g*j© = j&* 5i'n #©
5 m |HO'

1.67 A steel \ooyABCD of length 1.2 m and of 10-mm diameter is placed as shown
around a 24-mm-diameter aluminum rod AC. Cables BE and DF, each of 12-mm
diameter, are used to apply the load Q. Knowing that the ultimate strength of the
aluminum used for the rod is 260 MPa and that the ultimate strength of the steel used
for the loop and the cables is_480 MPa, determine the largest load Q which can be
applied if an overall factor of safety of3js desired.
SOLUTION
10 mm
* Q
■Aft
m
An
& = iF;,
Z • I Fa* - I=ac = o
|.fQ-F»c=o /
Bosesf a- s-ke^-U of1 cJ»/e 8F
Q&seJ o* ^Trc^i^Tti of roe/ AC

PROBLEM 1.68
A
6 in.
1.68 Link AC has a uniform { x 7 - in, uniform rectangular cross section and is
made of a steel with a 60-ksi ullimate normal stress. It is connected to a support at A
and to member BCD at C by f-in.-diameter pins, while member BCD is connected to
a support at B by a ^ -in.-diameter pin, All of the pins are in single shear and are made
of a steel with a 25-ksi ultimate shearing stress. Knowing that an overall factor of
safety of 3.25 is desired, determine the largest load P which can be safely applied at D.
Note that link AC is not reinforced around the pin holes.
SOLUTION
t)TMs - O CGUF^-lo V = o
4lFr-o 8, + | F^ - P * o
g^ , p. ffeos-a?) r - Q.CU67 P
B - J V> V * \.t\UdPJ pr0.76£?8>B
frseJ o« .srfre«3+t, ^ p,Vi *,+ C ■ AP;„ - "? J* = ?(!)*" = 0. llotf//
©0 - li/Ap.v, = (SSOfa. $?**>) ^ /.-?l7i- fc.'p.
T* - (o„7oS28)(l.Ws) - 1.3535" k>
Ad J* J Pj ,'s H« StoicJjesi- ?o - 0. 9dO k.'p,
A)h»Me v<Jo< ft* p: P =
FS
Q.9DQ
2.2S
r 0.2.77 ki'p • 211 A

PROBLEM 1.69
I0L-N
50 mm
30 mm
1.69 The two portions of member AB are glued together along a plane forming an
angle ^with the horizontal. Knowing that the ultimate stress for the glued joint is 17
MPa in tension and 9 MPa in shear, determine the range of values of 6 for which the
factor of safety of the member is at least 3.0.
SOLUTION
A„ - (0.03o)(o,OSO*) r \.$0*lO'% mU
p= /o*/o*N P0 *(F-S)P =■ 3o*itf ti
^r St co6L°
CCS
8 - Z7.m
Q £ X2.1al
<?0= & sinB^O - J^ s,v*e
Pj 30*10-*
20 s GH. I6B 0 - 32.03" 9 ^ 33.08"
M
€.wtj»
zi.iT * a £ sz.o*
.—»
PROBLEM 1.70
10 kN
7. 70 The two portions of member AB are glued together along a plane forming an
angle 0with the horizontal. Knowing that the ultimate stress for the glued joint is 17
MPa in tension and 9 MPa in shear, determine (a) the value of 0for which the factor
of safety of the member is maximum, (b) the corresponding value of the factor of
safety. (Hint: Equate the expressions obtained for the factors of safety with respect to
normal stress and shear.)
30 mm
su
-1°
SOLUTION
Ao = (o.ozo)(p.oso) - \.$o*lo *
A-+ He ojEk-Kt^L>i-* &ngJe (FS-V "(^"s-)f
'per*© wi6J
3" fcoj *e
HI

PROBLEM 1.C1
Element n
Element 1
1.C1 A solid steel rod consisting of n cylindrical elements welded
together is subjected to the loading shown. The diameter of element i is denoted
by d{ and the load applied to its lower end by P(> with the magnitude Pt of this
load being assumed positive if P; is directed downward as shown and negative
otherwise, {a) Write a computer program which can be used with either SI or
U.S. customary units to determine the average stress in each element of the
rod. (/>) Use this program to solve Probs. 1.1 and 1.3.
SOLUTION
It i$ t'hesun-] oj the forces applied
1o ilnat element tfnd all [ovJer o>r
k-i
w'
AVZKQGE STRESS IN £LE/nE>/T I :
Area = A; = -LTTdf:
P.ROGRm OUTPUTS
Ave stress = ±L
Problem 1.1
Element Stress (MPa)
Problem 1.3
Element Stress (ksi)
1 84.883
2 -96.766
1 22.635
2 17.927

PROBLEM 1.C2
20 kN
1.C2 A 20-kN force is applied as shown to the horizontal member ABC.
Member ABC has a 10 X 50-mmuniform rectangular cross section and is
supported by four vertical links, each of 8 X 36-mm uniform rectangular cross
section. Each of the four pins at A, B, C, and D has the same diameter d and
is in double shear, {a) Write a computer program to calculate for values of d
from 10 to 30 mm, using 1-mm increments, (1) the maximum value of the
average normal stress in the links connecting pins B and Dt (2) the average
normal stress in the links connecting pins C and E, (3) the average shearing stress
in pin B, (4) the average shearing stress in pin C, (5) the average bearing stress
at B in member ABC, (6) the average bearing stress at C in member ABC. (A)
Check your program by comparing the values obtained for d = 16 mm with
the answers given for Probs. 1.8, 1.23, and 1.24. (c) Use this program to find
the permissible values of the diameter d of the pins, knowing that the
allowable values of the normal, shearing, and bearing stresses for the steel used are,
respectively, 150 MPa, 90 MPa, and 230 MPa. id) Solve part c, assuming that
the thickness of member ABC has been reduced from 10 to 8 mm.
SOLUTION
?-20kN
h
A
fOKCES in UNKS F.B.X>t*G«*H OF A£C
2Fuki)Y^o' 2Fte(£C)~P(f\C)^0
~C^Q ^D-■?{*$/*&& (TENSION)
^ Z/\= 0:2F (£C)-?(fi&) =0
Q
*■■£.
0,25 W
O.H- m
*2F,
&£>
Link bd
FC£=. ?u\B)/z(Btr) CcoMr)
u5L
(3) Pill B
te
z
'B
- ^/cmyfj
(5)8£WNb 57RE5S £T g
Thickness cf jitemhw AC-t^c
(V LiNk CE
(U) PlN C
UNDE% PlN j&
C* = Z t~S£>
to" tfsc ^7
(CONTINUED)

PROBLEM 1.C2 CONTINUED
PROGRAM OUTPUTS
INPUT DATA FOR PARTS (a), (b), (c)i P = 20 kN, AB = 0.25 m, BC = 0.40 m,
AC = 0,65 m, TL = 8 mm, WL = 36 mm, TAC = 10 mm, WAC = 50 mm
Sigma BD Sigma CE Tau B Tau C SigBear B SigBear C
10. GO
11 ,00
12.00
13.00
14.00
15.00
16,00
78,13
81 , 25
84 ,64
88,32
92.33
96,73
101.56
■21.70
21.70
■21.70
■21,70
■21.70
-21 .70
■21.70
rrw
18 .00
19.00
20.00
21 .00
22.00
23.00
24.00
25.00
26.00
27.00
28.00
29.00
30.00
lOfc.yi
112.85
119.49
126,95
135.42
145 .09
"2TT7TT
■21,70
■21.70
■21.70
-21.70
■21.70
-21.70
-21.70
-21.70
-21 .70
-21.70
-21.70
-21.70
-21.70
66.82
/l.b!J
63 .86
57.31
51 .73
46.92
42.75
39.11
35.92
33 ,10
30.61
2 8.3-8
26. 39
24.60
22.99
79.5 8
65.77
55.26
47.09
40.60
35.37
31.08
St
216.67
27. b4
24.56
22 .04
19.89
18.04
16,44
15,04
13.82
12.73
11 .77
10.92
10 .15
9.46
8.84
203.12
1!J1.1U
180.56
171.05
162.50
154.76
147.73
141.30
135.42
130.00
125.00
120.37
116.07
112.07
108.33
125.00
113.64
104. 17
96.15
89.29
83.33
78.13 4H-
—!'S . SJJ
,(b)
69.44
65.79
62.50
59.52
56 .82
54 .35
52.08
50.00
48.08
46.30
44.64
43.10
41.67
(c) ANSWER: 16 mm 5 d z 22 mm
CHECK: For d = 22 mm, Tau AC = 65 MPa < 90 MPa O.K.
INPUT DATA FOR PART (d): P = 20 kN, AB = 0.25 m, BC = 0.40 m,
AC = 0,65 m, TL = 8 mm, WL = 36 mm, TAC = 8 mm, WAC = 50 mm
d
10.00
11.00
12.00
13 .00
14.00
15.00
16.00
17.00
18.00
19.00
20.00
21.00
22.00
23.00
24.00
25.00
26.00
27,00
28,00
29.00
30.00
Sigma BD Sigma CE Tau B Tau C SigBear B SigBear C
-21.70
-21.70
-21,70
-21,70
-21.70
-21.70
-21.70
-21.70
-21
-21
-21
70
70
70
-21.70
-21.70
-21
-21
-21
-21
-21
-21
-21
-21
70
70
70
70
70
70
70
70
79.58
65.77
55.26
47.09
40.60
35.37
31.08
27-54
24,56
22,04
19.89
18.04
16.44
15,04
13,82
12,73
11.77
10.92
10.15
9.46
8.84
"22 5
213
203
193
184
176
169
162
156
150
145
140
135
69
82
12
45
66
63
27
50
25
46
09
09
42
156.25
142.05
130.21
120 . 19
111.61
104.17
97.66
91 .91
86.81
82.24
78 .13
74,40
71,02
67 .93
65.10
62.50
60.10
57.87
55.80
53 .88
52.08
(0
CHECK: For d = 22 mm, Tau AC = 8
(d) ANSWER : 18 mm <; d z 22 mm
L25MPa<90MPa O.K.
(d)

PROBLEM 1.C3
0.5 in.
1.8 in.
1.C3 Two horizontal 5-kip forces are applied to pin B of the assembly
shown. Each of the three pins at A, B, and C has the same diameter d and is
in double shear, (a) Write a computer program to calculate for values of d from
0.50 to 1.50 in., using 0.05-in. increments, (1) the maximum value of the
average normal stress in member AB, (2) the average normal stress in member
BC, (3) the average shearing stress in pin At (4) the average shearing stress in
pin C, (5) the average bearing stress at A in member AB, (6) the average
bearing stress at C in member BC, (7) the average bearing stress at B in member
BC. (b) Check your program by comparing the values obtained for d = 0.8 in.
with the answers given for Probs. 1.9, 1.25, and 1.26. (c) Use this program to
find the permissible values of the diameter d of the pins, knowing that the
allowable values of the normal, shearing, and bearing stresses for the steel used
are, respectively, 22 ksi, 13 ksi, and 36 ksi. (d) Solve part c, assuming that a
new design is being investigated, in which the thickness and width of the two
members are changed, respectively, from 0.5 to 0.3 in. and from 1.8 in. to
2.4 in.
SOLUTION
Forces in members Ab *a/x> bc
frz.e 8_0x>y% pmvj5
2'P
(l)fttyt/We..STR*\<> f-A/ AR
fas Width r W
%& ^B
Q)PIN A
£*=(FAB/2)/(ndy*)
(5)BFfi'R?NG STRESS Pit A
51% Bear ft - Fm/dt
O) fiEflRiNC STR-feSS AT ft
IN MEMKL& Ur
FROiH FORCE T%if\N6LE
Sin 15° ~ 5 in 60°' Sfa7S°
F^2F(5miisy$in 75a)
(z) A\/e. stress /n Br.
FBC
Fi3c
(h-) pin c
tc={F5c/2.)/(TcllAJ
Si's Bear C= F£C/etb
Be
(CONTINUED)

PROBLEM 1.C3 CONTINUED
PROGRAM OUTPUTS
INPUT DATA FOR PARTS (a), (b), (c): P = 5 kips, w= 1,8 in., t = 0.5 in.
D
in.
0,500
0,550
0.600
0,650
0 .700
0,750
O.BQO
0.850
0.900
0.950
1.000
1 . 050
1.100
1 .150
1.200
1.250
1.300
1 ,350
1.400
1 .450
1 .500
SIGAB
kei
SIGBC
kei
TAUA TAUC SIGBRGA SIGBRGC SIGBRGB
kei kei ksi kei kei
11
11
12
12
13
13
14
262
713
201
731
310
944
.641
-9
-9
-9
-9
-9
-9
-9
962
962
962
962
962
962
962
35.863
32.603
29.886
27,587
25.616
23.. 909
22.414
17.932
16,301
14.943
13.793
12.808
11.954
11.207
15.412
16.268
17.225
18.301
19.521
20.916
962
962
962
962
962
962
962
962
962
,962
.962
.962
.962
.962
.450
754
,164
,660
,227
.852
.524
,236
.983
,758
.557
.378
.217
.071
900
047
324
708
177
717
,316
,964
653
,377
.132
.912
.715
,537
T7
16
15
14
13
13
12
12
11
11
10
10
10
9
T2T
268
412
,641
944
,310
.731
.201
,713
.262
.845
.458
,097
.761
21.096
19.924
18 .875
17.932
17.078
16.301
15.593
14.943
14 .345
13.793
13.283
12.808
12,367
11.954
10.548
9.962
9.438
8.966
8.539
8 . 151
7.796
7.471
7,173
6.897
641
404
183
977
(b)
(cj ANSWER: 0.70 in. idil.lO'm. -^fl (p)
INPUT DATA FOR PART (d): P = 5 kips, w = 2.4 in., t = 0.3 in.
D
in.
SIGAB
ksi
SIGBC
kei
TAUA TAUC SIGBRGA SIGBRGC SIGBRGB
kei kei kei kei ksi
29.886
27,169
24 .905
22.989
21.347
19.924
18.679
17 . 580
16,603
15.729
14.943
14.231
13.584
12.994
12.452
11.954
11.495
11 ,069
10.674
10.305
9.962
285
282
450
754
164
660
227
852
524
236
983
758
557
378
217
071
918
900
047
324
708
177
717
316
964
653
377
132
912
715
537
32.536
30,502
28.708
27.113
25.686
24.402
23 .240
22.183
21.219
20.335
19.521
18.771
18.075
17.430
16.829
16.268
35.160
33.206
31.459
29.886
28.463
27.169
25.988
24.905
23.909
22 .989
22.138
21,347
20.611
19.924
(d) ANSWER: 0.85 in. <> d 5 1.25 in.
(d)

D
A 1
~r ^
1
15 in.
L
HPHMKS^'
~"h
B
■* 18 in.-
wxy
4*— l2in.—
PROBLEM 1.C4 1.C4 A 4-kip force P forming an angle a with the vertical is applied as
shown to member ABC, which is supported by a pin and bracket at C and by
a cable BD forming an angle fi with the horizontal, (a) Knowing that the
ultimate load of the cable is 25 kips, write a computer program to construct a table
of the values of the factor of safety of the cable for values of a and 0 from 0
to 45°, using increments in a and (3 corresponding to 0.1 increments in
tan a and tan £. (b) Check that for any given value of a the maximum value
of the factor of safety is obtained for 0 = 38.66° and explain why. (c)
Determine the smallest possible value of the factor of safety for fi = 38.66°, as well
as the corresponding value of a, and explain the result obtained,
SOLUTION
(a) DRbvtf E b, DtQGRfiM OF ABC:
C~ F- p f5S\r\<X + 30 Cb5X
IS Co5/5 + IZ 5/0/5
F.St= Fuit/F
If),
0
ALPHA
0.000
5.711
11.310
16.699
J21.801
126,565
30.964
34. 992
38.660
41 . 987
45.000
3
2
2
2
2
2
2
2
2
2
2
0
125
991
897
837
805
795
803
826
859
899
946
5.71
3
3
3
3
3
3
3
3
3
3
3
358
214
113
049
014
004
013
036
072
116
166
11.31
3.555
3.402
3.295
3 ,227
3.190
3.179
3 .189
3.214
3.252
3 .298
3.351
VALUES OF
16.70
3.712
3 .552
3 .441
3.370
3.331
3 .320
3.330
3.356
3 .395
3.444
3.499
BETA
21.80
3.830
3.666
3 .551
3.477
3.438
3.426
3.436
3.463
3 .503
3.554
3.611
FS
26.56
3.913
3.74 5
3.628
3.553
3.512
3.500
3 .510
3.538
3.579
3 .631
3.689
30.96
3.966
3.796
3.677
3 .600
3 .560
3.54 7
3 .558
3.586
3.628
3 .680
3 .739
34 .99
3.994
3 .823
3.703
3.626
3 .585
3.572
3.583
3 .611
3.653
3.706
3.765
*—--^y~
38.66
4 .002
3 .830
3. 710
B .633
J.592
J.579
J.590
) .619
5.661
i .713
1.773
41.99
3.995
3.824
3 .704
3.627
3.586
3.573
3.584
3.612
3 .655
3.707
3.767
45.00
3.977
3.807
3 .687
3.611
3.570.
3 . 558)
3.568
3 .596
3 .638
3.690
3.750
TO)
(6) Whenfl^ 3$, 66J fanfi ~ 0.8 and cable BD is .perpendicular
tt f/ie lever any) Be.
(C) F,S.= 3S79 for <y- 26-6° ; P it perpendicular jfe. /^e
lever at-rn AC
NOTE:
Thz vaht Fc. =3,57? is ike Sent/fat 6f theM-lves of F$.
c^rre.*,pandm^ to f& - 3 8,66 ° afld the largest $f ih^s-e
Core* spending to 01- 26.6*. The point <¥-ZG.£,~ [h = 38.bC
is a "laddie point'' or *'mmimay" of the function FtZf&jfi)*

PROBLEM 1.C5
1.C5 A load P is supported as shown by two wooden members of
uniform rectangular cross section which are joined by a simple glued scarf splice.
(a) Denoting by <7V and TUt respectively, the ultimate strength of the joint in
tension and in shear, write a computer program which, for given values of
a, b, P, o-y.and tv, expressed in either SI or U.S. customary units, and for
values of a from 5 to 85° at 5° intervals, can be used to calculate (1) the normal
stress in the joint, (2) the shearing stress in the joint, (3) the factor of safety
relative to failure in tension, (4) the factor of safety relative to failure in shear,
(5) the overall factor of safety for the glued joint, (b) Apply this program,
using the dimensions and loading of the members of Probs. 1.29 and 1.32,
knowing that <rv = 1.26 MPa and Ty = 1.50 MPa for the glue used in Prob. 1.29,
and that try = 150 psiand t„ = 214 psifor the glue used in Prob. 1.32. (c)
Verify in each of these two cases that the shearing stress is maximum for
a = 45°.
SOLUTION
Draw ihe f\B.c\ iag ram of /Oiver mem be r;
4/2IP" - 0: F- Pjt/ic< -o F - P5mO(
Area, ~ ccb/s/'n <¥
Norma/ stress ;
0-=Ji = (?/<xb) s^x
Arecc
£3) Fi\4or tension (nor»xil
Fsn - Cr0/(?
F< S. for sheAr;
F55 - Vu/V
OXGALL PS,:
F^ - The smaller Or fsH and F5S>
(CONTINUED)

PROBLEM 1.C5 CONTINUED fR O&Rf^ O'JTPOlS
For ?rr... \.Z6j\, P = 6 k N
(X.-K5 mm , b r 75 nmi, Oi - 70°, (7 r U6 MP^ ■<? ^ /. yO MPa,
ALPHA SI
5
10
1 5
z5
30
35
40
45
50
55
60
65
10-
75
80
85
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
0000
G(MPa)
0.0049
0.0193
0.0429
0.0749
0.1143
0.1600
0.2106
0.2644
0.3200
0.3756
0.4294
0.4800
0.5257
0.5651
0.5971
0 .6207
0.6351
TAU(MPa)
0.0556
0.1094
0.1600
0.2057
0.2451
0.2771
0.3007
0.3151
0.3200
0.3151
0.3007
0 .2771
0.2451
0.2057
0.1600
0.1094
0.0556
259
65
29
16
11
7
5
4
3
3
2
2
2
2
2
2
1
FSN
1782
2905
3899
8301
0229
8750
9842
7649
9375
3549
9340
6250
3968
2296
1101
0300
9838
26
13
9
7
6
5
4
4
4
4
4
5
6
7
9
13
26
FSS
.9942
.7053
3750
.2925
1191
4127
9883
7598
6875
7598
9883
4127
1191
2925
3750
7053
9942
26
13
9
7
6
5
4
4
3
3
2
2
2
2
2
2
1
FS
.9942
.7053
3750
2925
1191
4127
9883
7598
9375
3549
93i40
6250
3968
2296
1101
0300
9838
**(c)
<*>)
For Prob, l.:7: P^ ZhOO \b
ALPHA SIG(psi) TAU(psi) FSN FSS FS
5.0000 1.0128 11.5765 148.1018 18.4857 18.4857
10.0000 4.0205 22.8013 37.3089 9.38 54 9.38 54
.0000 8.9316 33.3333 16.7942 6.4200 6.4200
..j.0000 15.5970 42.8525 9.6172 4.9939 4.9939
2 5.0000 23.8142 51.06 96 6.2988 4.1904 4.1904
30.0000 33.3333 57-7350 4.5000 3.7066 3.7066
35.0000 43.8653 62.6462 3.4196 3.4160 3.4160
4Q...QQQ.0 ..55.-QS01 6 5.6538,. }-122& 3.2 595 2.7228 «$ (8)
45 .0000 66 .6667 66.6667^ 2.2b00 T72TDTJ A .^!bUU -^fifr*)
50.0000 78.2432 65.6538 1.9171 3.2595 1.9171 ^ J
55.0000 89.4680 62.6462 1.6766 3.4160 1.6766
60.0000 100.0000 57.7350 1.5000 3.7066 1.5000
65.0000 109.5192 51.0696 1.3696 4.1904 1.3696
70.0000 117.7363 42. 8525 1.274 0 4.9939 1.2740
75.0000 124.4017 33.3333 1.2058 6.4200 1.2058
80.0000 129.3128 22.8013 1.1600 9.3854 1.1600
85.0000 132.3205 11.5765 1.13 36 18.4857 1.1336

PROBLEM 1.C6
8 mm
Top view
U-200 mm-O--180 mm-^j ..
1.C6 Member ABC is supported by a pin and bracket at A and by two links
which are pin-connected to the member at B and to a fixed support at D. (a) Write
a computer program to calculate the allowable load P& for any given values of (1)
the diameter dl of the pin at A, (2) the common diameter d2 of the pins at B and D,
(3) the ultimate normal stress vu in each of the two links, (4) the ultimate shearing
stress vu in each of the three pins, (5) the desired overall factor of safety F.S. Your
program should also indicate which of the following three stresses is critical: the
normal stress in the links, the shearing stress in the pin at A, or the shearing stress
mm in the pins at B and D. (b and c) Check your program by using the data of Probs.
1.49 and 1.50, respectively, and comparing the answers obtained for P^ with those
given in the text, (d) Use your program to determine the allowable load PM, as well
as which of the stresses is critical, when d{ = d2 = 15 mm, au = 110 MPa for
aluminum links, ry = 100 MPa for steel pins, and F.S. = 3,2.
SOLUTION
Front view
8 mm
12 mm ^-
Side view
r" alb-**
/do
zoo
/BO
B
5
f*
0) Fortfw d, ofjivj^ F^z^/fsW*1?/*)* P-,=
ZT ^ "-^' *% fs rbe smaller 0f P an>\?
V) OX de6ir^ ^f«// fa ' ?s U the smalkr 0f p ^ p *
\i <3 < Ptf , stress is critical in /inks
\l ^V <7^ ai)<* ?\< \i ti^ss /s Critical )n pin ft
lf Pi < ^ and P2 < ?t , stmt /5 critical in pins
PtZQ&QW OUTPUTS
(6) ?r?>b. Lhq. Df-)m: 'c/^Sm^d^^mitij^j =2 SO MPa^= lOOMPafi^&Q
P^-3,72 A.N. Stress in pih A /s critic*/ ^
(fc) Pngfc./.fP.aflTfi: ^ -/Dm 4rjzin^0*o- 25"flMPay^ too MP*,FSr 3,fl
Fill-5'.J7kN.; Stress In pins 3 and D /s critical ^
(d) mif^ dt=dzr ISmm^G^UOMPcL, T^-fGCiMPa, F*$»* 3.1_
Pal(- ^79 kN> Stress hi Uhks h critical ^
pinj& cund £)

CHAPTER 2

A PROBLEM 2.1
SOLUTION
w s-fr
2.1 A steel rod is 2.2 m long and must not stretch more than 1.2 mm when a 8.5 kN
load is applied to it. Knowing that E = 200 GPa, determine (a) the smallest diameter
rod which should be used, (b) the corresponding normal stress caused by the load.
M 6"
* " e% " (aoo**o*)(i.a*io-*) "
77.<?2*/o"e Ml
10^.) MPft.
PROBl EM 2 2 2.2 A 4.8-ft-long steel wire of { -in. diameter steel wire is subjected to a 750-lb
tensile load. Knowing that E = 29 x 106 psi, determine (a) the elongation of the wire,
(b) the corresponding normal stress.
SOLUTION
Co-) L- t-S ft = 57.6 i». A=^dt,= ^{f)l-^f.087 * to'3, in*
§a £t <7*oK*7Q 30.3V/O-5;, =0.0303 1, -
PROBLEM 2.3
SOLUTION
2.3 Two gage marks are placed exactly 10 inches apart on a £ -in.-diameter
aluminum rod with E= 10.1 x [06 psi and an ultimate strength of 16 ksi. Knowing
that the distance between the gage marks is 10.009 in. after a load is applied,
determine (a) the stress in the rod, (b) the factor of safety.
(a/) S - 10.001 - 10.000 = 0.009 ;».
1
L
-f ••• *■?
M
Si
PS » — ?
tt
^.09
10
».7€0
CT. O^x/O pa.
- 1.0* kSi*
PROBLEM 2.4
SOLUTION
2.4 A control rod made of yellow brass must not stretch more than 3 mm when the
tension in the wire is 4 kN. Knowing that E = 105 GPa and that the maximum
allowable normal stress is 180 MPa, determine (a) the smallest diameter that can be
selected for the rod, (b) the corresponding maximum length of the rod.
A-£
4*10*
-c
%o * 10*
= X2.2Z2*lo~- *
A--$^* d*J$f - J WtM.jM ^5=^ , s,3Z*i
•5.3<? »w*i
00 s.|t
P " Tv|o*
, ISO vn

PROBLEM 2.5
SOLUTION
2.5 A 9-m length of 6-mm-diameter steel wire is to be used in a hanger. It is noted
that the wire stretches 18 mm when a tensile force P is applied. Knowing that E=200
GPa, determine (a) the magnitude of the force P, (b) the corresponding normal stress
in the wire.
(a) A - $ dz = £(o.oog)z = ZZ.zTtxio'* ^L
(k)
* AS
AT P
V " L = *
- //.3/*/os W = IL3I 1<U
PROBLEM 2.6
SOLUTION
(a)
2.6 A 4.5-ft. aluminum pipe should not stretch more than 0,05 in, when it is subjected
to a tensile load. Knowing that E = 10.1 * 106 psi and that the allowable tensile
strength is 14 ksi,, determine (a) (he maximum allowable length of the pipe, (b) the
required area of the pipe if.)he tensile load is 127,5 kips.
s = EL ■ / «
EAS _
*-£
A- *-
II -
in
•7.1/
(Ml
PROBLEM 2.7
SOLUTION
(a)
2.7 A nylon thread is subjected to a 8.5-N tension force. Knowing that E ~ 3.3 GPa
and that the length of the thread increases by 1.1 %, determine (a) the diameter of
the thread, (b) the stress in the thread.
L loo
■fc -=■ <?o. 901
3 Ae
<b>
2.# A cast-iron tube is used to support a compressive load. Knowing that E = 10 x
106 psi and that the maximum allowable change in length is 0,025 percent, determine
(a) the maximum normal stress in the tube, (b) the minimum wall thickness for a load
of 1600 lb if the outside diameter of the tube is 2,0 in.
f-feT- <>■*">*
PROBLEM 2.8
SOLUTION
(CO
^.S"ksi-
)6QQ
im
^ a:- **- -_ a.^_ fitted * 3j,s, ;^ ■> A = ..78*7 ;.
t = i(i-«/^s i(5.c?.- l-7*«*7 ) « O. /o77 ;«.

rA 2.9 A block of 10-in. length and 1.8x1.6 in. cross section is to support a centric
>W PROBLEM2.9 compressive load P. The material to be used is a bronze for which £=14 x l06psi.
Determine the largest load which can be applied, knowing that the normal stress must
SOLUTION not exceed I8ksiand that the decrease in length ofthe block should beat most 0.12
percent of its original length.
CoMi'Je^M^ <yQJ)owcJoSe efye&s 6" - IS fat - lgy/os p&;
P * 6*A « 03*/O3)US8)* *57.3 */o* <*6
PROBLEM2 10 2,l° A "'^ t.ensile load wN1 beapplied toa 50-m length of steel wire with E= 200
GPa. Determine the smallest diameter wire which can be used, knowing that the
normal stress must not exceed 150 MPa and that the increase in the length of the wire
SOLUTION should be at most 25 mm.
Consf'cJen'na o^^ovjcX}^ S+*"fiSS <S" = ISO ^lOu P&.
A * f J* Jl'-fiK r -J ^»10^ , ,0.70^*,

^PROBLEM 2.11
2,5 m
2.11 The 4-mm-diameter cable BC h made of a steel with £=200 GPa. Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, find the maximum load P that can be
applied as shown.
3.5 m
SOLUTION
Lee - V G* + f*7 = 7-3UI m
Ose lo«^ AV3« as a.-Pree body
9IM, = o 3.6-P - (^(-i_.
,0 m—A
(W-«
\
r
A = ?J* * -UriOtOoi)*- - \z.Ste*io
-c ^
Const deri'*** &}Jre*j o^iJe e^o^ettj'ro^ § = £ * /° •*»
AE" " L*. "7-31111
PROBLEM 2.12
P = |30 kips
2.12 Rod BD is made of steel (E = 29 x 106 psi) and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member BD
is 0.02P. If the stress must not exceed 18 ksi and the maximum change in length of
BD must not exceed 0.001 times the length of ABC, determine the smallest diameter
rod that can be used for member BD.
SOLUTION
Feo ■=■ O.02 P * (0.0O(»3o) * l£&?s * Z.Cxicr'JL
S *
F5o
■ A - 5ft - 2al
IIM* i^
54 in.
<• _ Soi-M,. a - FmJ-w> _ C?.CWosX£^) _ a 0«/:j
Las*&*~ ove* gov/e^Ai A - 0.14444* iM
= 0.^x9 ;w.

PROBLEM 2.13
30 mm
2,13 A single axial load of magnitude P= 58 kN is applied at end C of the brass rod
ABC. Knowing that E = 105 GPa, determine the diameter d of portion BC for which
the deflection of point C will be 3 mm.
SOLUTION
c " ^ A; E " £" 1 A* A*c
- 3.13SHX103, wf%
6c
•ftc
0.8
3.7337*70
-,$ -
2>,lZhH«lo
1 ~ ZW.ltxtO'6, •»*
A* ■ S £ A *c%R£* = lmtmi*«*~) r /6^vro-»
te 1 TT
7T
I*
= 16.S"2. *»*
2.W Both portions of the rod ^4SC are made of an aluminum for which E= 73 GPa.
Knowing that the diameter of portion BC is d = 20 mm, determine the largest force P
that can be applied if aal, = 160 MPa and the corresponding deflection at point C is not
to exceed 4 mm.
SOLUTION
AAS= $(0.030)*~ 706.3fex|O"6 »^
Aat - iKo.oao)1'* 3l*/.u */c?'6ml
6"- £ •'. P« A6T
Po^-fTo^i A 8
M . 0.8
?* EStC^t ^)" - (73w/e/%*">-»)
-I
706.86V/0
-«.
3W*/t *7e>
F"0
P* 5o.$*t<? U - £fc.3 k/vj

PROBLEM 2.15
li-in. diameter
1-in. diameter
Is-in. diameter
2.15 The specimen shown is made froma l-in.-diameter cylindrical steel rod with two
1,5-in.-outer-diameter sleeves bonded to the rod as shown. Knowing that£=29 * 106
psi, determine (a) the load P so that the total deformation is 0.002 in., (b) the
corresponding deformation of the central portion BC.
SOLUTION
(a)
Lk
tew
AR
fct
CD
L, »*.
3
3
2
-l
CT, in
US
l.o
U5"
A,V
1.767/
0.7SS^
1.7671
L/A , i*-1
U»3I8
3.*m
1- 131 S
G.o%% -
5^»n
UO S^ =
A*£
S^^^)"- U25fx/o"5 i«
PROBLEM 2.16
LP
i
2.16 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa.
Knowing that the magnitude of P is 4 kN, determine [a) the value of Q so that the
deflection at A is zero, (b) the corresponding deflection of B.
SOLUTION
0.4 m
0.5 m
20-wm diameter
Force in welter $B is T tension
£"Aw (7o*(O*0( 3 !**_ /6 xJo" < )
= 12.15G *(o
-6
*n
60-mm diameter
(b) S^O - Sao r SB " ^a^CvJo"4^ - 0-0722 «* -rf

$ PROBLEM 2.17
0.4 m
0.5 in
2.17 The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that P
= 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point S.
SOLUTION
20-mm diameter
60-mm diameter
% * ^A6 + S
<Sc
U>> Se - S
m
A6C= iJflc2' - 1J(o.06o)* * Z.&ZMXIO
-3 2
PAB = P ^ 6 *lo3 N
fca Aa„E, ' Isi^icho-^Oo^io*)
A«.E (Z.SZn'loYloylo')
-c
1
I
«c
- •fro,*! * |o* v^ - - 0.Q1W r^v^

PROBLEM 2.18
Brass: E = 105 GPa
2m
Steel: £ = 200 GPa 1 |B
50 kN
3 m
2.5 m
W
100 k-N
2.18 The 36-mm-diameter steel rod ABC and a brass rod CD of the same diameter are
joined at point C to form the 7.5-m rod ABCD. For the loading shown, and neglecting
the weight of the rod, determine the deflection of (a) point C, (b) point D.
SOLUTION
A r f dz - 5(0.O3£)l= l.0|787x/o~3 ^
PorTl'ovt
AB
BC
CD
P.
ISO krJ
100 ICK/
loo kN
U
;?*>
3 »*i
?.5^
E*
^oo GP*
26© GPa
1 05 G Pa
PiLx /A £*,
(.47^ x|o'3 m
I.H7M* lo"1 m
2.33^/d"5 m
(a,1* §c = £AB *$«,
(W ^ * §c 4 §eo
- 1.474**0* + 1.474 vfc>"s
- 2.<? 4S x fcT 3 + 2.33*9 Wo"3
=- S.l%l*\o's m * ,5". X? wm
PROBLEM 2.19
64
15.0 in.
2.19 The brass tube AS (E = 15 * 106 psi) has a cross-sectional area of 0.22 in2 and is
fitted with a plug at A. The tube is attached at B to a rigid plate which is itself attsched
at C to the bottom of an aluminum cylinder (E = 10.4 x 106 psi) with a cross-sectional
area of 0.40 in2. The cylinder is then hung from a support at D, In order to close the
cylinder, the plug must move down through 54 in. Determine the force P that must
be applied to the cylinder.
L#f 15+^ - /5.047 in fa 0.22 (n1,
Lo- '$".«, A^^o-io;^ £«■ IM*lofcpsi
^ -
PL
CP _
? ds^
-c
J£*
kpAco (/a4"lo4)(o.4o)
- 3.605? ^O* P
To+J JeMeo+i1
SA - S
*6
^ r (4-^<?7 */d* t 3.eojg*fo'c)P .-. ? - 5".74 x/o* JL
a
= S".74 ki'pj

PROBLEM 2,20
B
1.2 m
2.20 A 1.2-m section of aluminum pipe of cross-sectional area 1100 mm2 rests on a
fixed support at A. The 15-mm-diameter steel rod BC hangs from a rigid bar that rests
on the top of the pipe at£. Knowing that the modulus of elasticity is 200 GPa for steel
and 72 GPa for aluminum, determine the deflection of point C when a 60 kN force is
applied at C.
SOLUTION
0.9 m
Pipe A3'* LA0«l.2*; Ew* "laxic^ fAj Aw*lioo^» noo^d'*^
PROBLEM 2.21
2.21 The steel frame (£ = 200 GPa) shown has a diagonal brace BD with an area of
1920 mm2. Determine the largest allowable load P if the change in length of member
p BD is not to exceed 1.6 mm.
SOLUTION
L&c TV^l+ fi2 = 7.8/0 m , £0t* ^oo xjo*1 ft
- ^ftc-^-Bc
>fcc
H" A
h
Be
•8C
- 12.G7*lor N
Use joi'*f B <=<-£ a.-fWe bo^v* -"•XF*-
O
7.8lo

PROBLEM 2.22
223 kN
2.22 For the steel truss (B = 200 GPa) and loading shown, determine the deformations
of members AB and AD, knowing that their cross-sectional areas are 2400 mmz and
1800 mm2, respectively.
SOLUTION
Shdri'cs • Rea,ctio«$ we llH kk) upn/«/d *A A 4 C.
Use joi'«+ A as a. W boay: 4f ZSy - O 114 - ^7; F^ =0
*t.7"7
FAR = ^^./o kw
^ZR = o
R
. f
to
H.m "a
* o
114 WW
Me^U^ A^:
"Afi
>*>
Fj^Lto _ Ga?.t*icO(f-*>
-s
E A«, (2e>e>xio,)0*oox|or*')
= %ro3 yl&~ hx - ^.c>3^
V PROBLEM 2.23
2.23 Members AB and 5C are made of steel (£=29 * 106 psi) with cross-sectional
areas of 0.80 in2 and 0.64 in2, respectively. For the loading shown, determine the
elongation of (a) member AB, (b) member BC.
SOLUTION
Oat.)
U9^ JT+ 5* ^7.X|o|t. - 93.72 I*
Use joj^+" A as a Free loaJy
rzFy = o
R. - 2* = ©
Z3/o ' AB
Fas " 43-74 ^ = 43.7* WC? A.
(Is)' Use Jom'T- 8 45 A -h*e« bod^
8 .Fie ^2^=° ^c - 7^ Rw s °
R.
Ro
' 2* fc.>s
^ IFb.
p^. i£}tol, 33.^ k;p s 33.CO xto3 A
7.Slo
*' E ^ " (Wm^Xo.6*)

PROBLEM 2.24
2.24 Members AB and CD are 1 £ -in.-diameter steel rods, and members BC and AD
are \ -in.-diameter steel rods. When the turnbuckle is tightened, the diagonal member
AC is put in tension. Knowing that E- 29 x 106psiand h =4 ft, determine the largest
allowable tension in AC so that the defomiations in members AB and CD do not
exceed 0.04 in.
SOLUTION
%z ' §*o * 0-04 ;» faz lift - 4$ ;„, - LCD
Ac^d*^ f(i.U5)*« O.IWoZn1-
q - Fc»Ltu
Use JdiVT C && «- t^*< koa?vj *~'
4zFy =o
f - -^r
° •'- ^ - $ F-
F*e x $(lW&*iOS)= 30.0 *lc? A.
3o. o k;ps
R
Ac

PROBLEM 2.25
2.24 Members AB and CD are 1 £ -in.-diameter steel rods, and members BC and AD
are j -in.-diameter steel rods. When the tumbuckle is tightened, the diagonal member
ACis put in tension. Knowing that E= 29 * 10* psi and h = 4 ft, determine the largest
allowable tension in AC so that the deformations in members AB and CD do not
exceed 0.04 in.
2.25 For the structure in Prob. of 2,24, determine (a) the distance h so that the
deformations in members AB, BC. CD and AD are all equal to 0.04 in., (b) the
corresponding tension in member AC.
SOLUTION
0*« ^diVi 3 <xs d_ -Iree to^L
u.
F*& - TT Fet
b
For. eejo^i* defor-MAri'ons
'«c
EA,
7'
Aft
b* " Aac ^j*1- * "3JT
b - 3ft - 3£ m.
h r | t . i(3) - 3.S6 -H- = 46.S ,V
c|ab
7/a " 7
(^ Se+fm-i S,
- O.C* i'n.
Sec
- Fecb . p - g Ac& Sec _ ft?vro*)y^lT(Q,Q4)
E A^ "" ** ' t ~ 36

PROBLEM 2.26
2.26 Members ABC and Z>£Farejoined with steel links (£ = 200 GPa). Eachofthe
links is made of a pair of 25*35-mm plates. Determine the change in length of (a)
member BE, (b) member CF.
(S_
180 mm
280 nun
L
@
a
9
SOLUTION
lfi
k* k-240mm^ ,SkN
c
e •
K
ra
I—»-
F«
he
ISkW
Use member A8C as ft. -f/^e bo«ly
(0.XCo)(l8v|O3) - (0. iao)Fcp= O
CF
!>z:mc = o y.wo )(\t»\a%) + Co./so)f6€ ^o
A - « Kd.wXo.a3£) */.tTk/O-*
1 1.
-6
w §c. . S£s-- \&£%;£l*r <7.»»^ - ao,7«-,

PROBLEM 2.27
0.36 m
deJ
P = 5 fcN
2.27 Each of the links^B and CD is made of aluminum (E = 75 GPa) and has a cross-
sectional area of 125 mm2. Knowing that they support the rigid member BC,
determine the deflection of point E.
&
SOLUTION
R,
rA&
0.20 m
-—0.44 m -|
i
'CD
Use *t£t^ te/1 BC a.2 o,
fir* ft. Woel
r
5*10* k
FA8 ' 3.437&X/0* tf
S> ZMB = o (O.tff) Fco - (o.2o)Cswo*)«o
Foir JUfcs AB a*d CD A- I^^Tmho3" » /24"x/o"& r» *
§AS, £^ (WoMCo^^ = cw r s
EA (TSxic^X^T^o-* )
Ac
-t.
OftW^.no^ efi'ft^ut*.^
- 113.5" */0_t raW
§f= sc+ 4©
-&
- \01.S*(Q~ m = O.lofS'
/»ihn

jf PROBLEM 2.28
D
2.28 Link BD is made of brass (E= 15 * 10* psi) and has a cross-sectional area of
0.40 in2. Link CE'is made of aluminum (£ = 10.4 x 10* psi) and has a cross-sectional
area of 0.50 in2. Determine the maximum force P that can be applied vertically at
points if the deflection of A is not to exceed 0.014 in.
9.0 in.
5.0 in.
m
M
SOLUTION
6.0 in.
E
f
8
Ev'
lee
Use. wa^cey ABC
».0in. ■-]
9 ZMc = o S P -7Frf = o , F« *o.&CT6 P
-c
O.G4/OXIO" P T
"De+ofT'whdiA T^ia^^t
S* - Si
-Pec
-C
m - -3. *? 85S >* £° P
P -
O.OI4
3.flW£a*'/0
-4
3.5*1 */?3 ii = 3.SI k:
p»

PROBLEM 2.29
SOLUTION
///<///,
M for S ~ gT
2.29 A homogeneous cable of length L and uniform cross section is suspended from
one end. (a) Denoting by /?the density (mass per unit volume) of the cable and by £
its modulus of elasticity, determine the elongation of the cable due to its own weight.
(b) Assuming now the cable to be horizontal, determine the force that should be
applied to each end of the cable to obtain the same elongation as in part a.
T** - w/ei^Lt eH portion be/o»*> 4-U poi^j
Eh EA E J
, X „„/_ = a-^-
7/ff
2.30 Determine the deflection of the apex .4 of a homogeneous circular cone of height
/i, density /?, and modulus of elasticity E, due to its own weight.
SOLUTION
eie^«*\t
L.£7 L - PA^Iios
o* He
s = t£.ax
3F 2
I., fee-
hv

PttORI FM 2 Si 2'^' T*ie volume °*" a tens'!e specimen is essentially constant while plastic
deformation occurs. If the initial diameter of the specimen is du show that when the
diameter is d, the true strain is el = 2 \n(d{/d).
SOLUTION
J-f fkt \Johw\9 \e> c**4+a*-* -f c^L - if 6l Le
L . £? _ /J,
k - A. = (ii\x
to dl u /
2.32 Denoting by fthe "engineering strain" in a tensile specimen, show that the true
PROBLEM 2.32 strajn jg ^ = |n(i + ^
SOLUTION

PROBLEM 2.33
5 mm 5 mm
2.33 An axial force of 60 kN is applied to the assembly shown by means of rigid end
plates. Determine (a) the normal stress in the brass shell, (b) the corresponding
deformation of the assembly.
20inm
5 mm-
mm
20 mm V \ /
SOLUTION
Steel core
E = 200 GPq
Brass shell
E = 105 GPa
250 mm
S
AtE-»
. ft/.
r-b - L
ASE. s L
P • Pi + P6 * i^Ab4 EA^f
*•'
^A,+ EkAs
As = (O.MoXo.OJuO - */OOWo"fc m*
/^ » (0.030 Xo.oa©1) - (aoaoYaoacO-- £00*10'*^
— = £ *-
60V/0"
= 452.83 WO
-c
(|a5yio<*X^o><io-ft)4(5oox|o<,y4oox/o'4 J
00 6i = £b£ r (lo$,wo,X<'«.&S*/o'c,)8- M7.5"v/04f«,
(M S " Lfi « (^0Wo"*Xw.83WcTfi) =• 1/3.2 x/o'4^
* £>. I 13%. w*h

PROBLEM 2.34
5 mm 5 mm
20inin V
5 mm
20 nun
k^ri^-5mm
Steel core
E = 200 GPa
Brass shell
E = 105 GPa
2.34 The length of the assembly decreases by 0.15 mm when an axial force is applied
by means of rigid end plates. Determine (a) the magnitude of the applied force, (b)
the corresponding stress in the steel core.
SOLUTION
250 mm
s - ^
L

* 2,35 The 4.5-ft concrete post is reinforced with six steel bars, each with a 1 g- -in.
diameter. Knowing that£s = 29 x l0"psiand£'c = 4.2x 106 psi, determine the normal
stresses in (he steel and in the concrete when a 350-kip axial centric force P is applied
to the Dost.
SOLUTION
Ps r f *rWai* oavWeeJ ty He SIX stecf foJs
s . PSL
§ " "EX
p . EA,.S
P = FAS
P
. 2.
^6* )*-£*** - ^^S" ^
£ ' -;
- 3So y to*
= -2S7.G.7V/0
-c
6"* = £*£ - (^-/O6)^87.6?x/O"c)- -*.3*v/0*ps;. -8.34 fci

PROBLEM 2.36
Brass core
(E = 105 GPa)
2.36 An axial centric force of magnitude/1 = 450 kN is applied to the composite block
shown by means of a rigid end plate. Knowing that h = 10 mm, determine the normal
stress in (a) the brass core, (b) the aluminum plates.
Aluminum plates
(E = 70 GPa)
Rigid
end plate
300 mm
SOLUTION
Let f\, - par+iow ox ay\o.J( i-cn/xe ceur*\'ej( 1©^
s« Si.
pfc =
£bA4S
pr Pb + P^ (£i,Ah+ 5.AO?-
Ab = (60)(<ro) - ^<foO m^ r ^OOx|C-' ***■
A«.3 0?X6kXl°)r noo*»?~ ' t2oo»to-e'*y
a -
4£o*<°* ^ /.33<?3*/o'3

PROBLEM 2.37
Brass core
(E = 105 GPa)
2.37 For the composite block shown in Prob. 2.36, determine (a) the value of h if the
portion of the load carried by the aluminum plates is half the portion of the load
carried by the brass core, (b) the total load if the stress in the brass is 80 MPa.
Aluminum plates
(E = 70 GPa)
Rigid
end plate
300 mm
SOLUTION
Let R, - po*"h©** tn oxxcj to1*.*. caw-Aftd
(e*^ G"\ve.A
Pl
5AS
L
r*- L
T.
8O0
*• 2 lo * io'
&)(&)
- J 5"
mi*
1\- i?k r ?6v|03 M
Ps Pb + Pa. " ^BXIO3 ^
28S WKJ

PROBLEM 2.38
Uft
2.35 The 4.5-ft concrete post is reinforced with six steel bars, each with a 11 -in.
diameter. Knowing that £, = 29 x 10* psi and £<. = 4.2 * 10* psi.delermine the normal
stresses in the steel and in the concrete when a 350-kip axial centric force P is applied
to the post.
2.38 For the post of Prob. 2.35, determine the maximum centric force which may be
applied if the allowable normal stress is 20 ksi in the steel and 2.4 ksi in the concrete.
SOLUTION
Cone****.' £,. -
£k 4.^*10*
-£lt.HZ*IO
-t
S** a.Jt*fe<r v'a.i^c ao^c*s
£ - *■ - s£"7/.H3x/o
.4
;e.J Ly six $+eeJl roJ&
F = Pc+ P* * (EcAt + aAa^fc
Ac. - f ^~ As = f{i«)x- ^^4 r Xf«.5 /n"

PROBLEM 2.39
IP
2,39 Three steel rods (£= 200 GPa) support a 36-kN load P. Each of the rods AB and
CD has a 200-mm! cross-sectional area and rod EF has a 625- mm2 cross-sectional
area. Determine the (a) the change in length of rod EF. (b) che stress in each rod.
I
^
500 mm
400 mm
SOLUTION
Use. t-vieh-.be^ B£D *-s «. $r«e body
■*» .p ta.
t4
StWC€
H« *-AB
EA.a >
*~Afl " ^ei
cmiJ A^ ** A
co^
p« * n* + pct - p * o
p- 2pA8 + p£f
'*F FA,
***
CO
Since, ptfi'nr^ Aj Cj a^J1 tr atAtf -fi'xW Sg "" §a*j ?fl = Sco i Sp = <SeF
Since ki e^ W«»r B£"D **5 rTgi'tff
EAac " E/\ff
P*« ^-*& - Pgplg* . p . A** . 4^t p - g^o ^jo -p
PA.. FA« " r« ' /^p Lab 6F " '-9r -^ ~fi
- 0. ZSC Pep
6,25 * &o r*r
P= 2PA, + P£P - ttVo.JwOPo* pfF- i.^/a?^
Pag * Pc* - (0..2.S6 V23.S/OW03) =■ CO*?*!*/ N
/at £ - S - (ga^^O^oo »«>"'? _ 7C owV«
hi
0\f
= 0.076 2 ***
»*» * 6U
Re_ r £.<*&& __ So^rf/o'ft,* 30-SMft
A** *°° >^°"
*«-£ -tSItsS - -".I.,c'Ax «.-MIW

PROBI EM 2 40 2*40 A brass bolt (£b = 15 x 10* psi) with a £ -in. diameter is fitted inside a steel tube
(£, = 29* 106 psi) with a f -in. outer diameter and j- in wall thickness. After the nut
has been fit snugly, it is tighened one quarter of a full tum. Knowing that the bolt is
single-threaded with a 0. lrin, pitch, determine the normal stress (a) in the bolt, (b) in
the tube.
SOLUTION
The wo^emenf of He not cJ<m$ -Hie JboJV" a"f+«** a. <fu«ur + dv j-wn ^ &?o<y*P
Le-4 Pt»/t ~ avirt./ fensi/e -Fo^c* i'a -H« bo/f"
*U ' f «*' • f (*)' = O.HOW »'
Aw. • f CJ."- J;') ' f Of)"-(f)') ' O.MW I.'
O.O*^ r 7.2*3/ xio'^ + {.HOGO'IO*? - 8.C4-8/*/04 ?

I
PROBLEM 2.41
8 In. 10 in. 101
2.41 Two cylindrical rods, CD made of steel (£ - 29 * 10s psi) and AC made of
aluminum (E = 10.4 x 10* psi), are joined at 5 and restrained by rigid supports at A
and D. Determine (a) the reactions at A and D, (b) the deflection of point C.
SOLUTION
AS:
l4-in. diameter l|-in. diameter
P " *A > I-** = 8 in
/«
14
3*8 " I?
R>
- 0.773*6 */©'* ft
CO: p = ft - ;>w|o* - |4x jo3 = ft-33*/o3
L^ /O ,'* A = ^^C = 30.625^ - 2.0739 m1.
-6
Since poivf I> ccc**o-{ j-via^e feJr*Jri¥€. "h* A S*o ~ O
£«t) U<*07$**/d* ft - ^2.73S>/03^ O ft * //.«*fO* JL.
ft = 3Z*IQ% - ft * '20.0* *[o* A *-
-■*
- Limztto'* ft - n.m%*\6
' £.AC0 (a-mo'Xa.own
^^

PROBLEM 2,42
SOLUTION
8
2.42 A steel tube (E - 200 GPa) with a 32-mm outer diameter and a 4-mro thickness
is placed in a vise that is adjusted so that its jaws just touch the ends of the tube
without exerting any pressure on them. The two forces shown are then applied to the
tube. After these forces are applied, the vise is adjusted to decrease the distance
between its jaws by 0.2 mm. Determine (a) the forces exerted by the vise on the tube
at A and D, (b) the change in length of the portion BC of the tube.
SO mm 60 mm 80 mm
For H« +Jbe <$i^A»-Zt
=: 351.86 mrn" = 3$!.$(,* to"" r*1
AS :
BC:
CO
? - PA . L:0.080 m
c , ZL - g»(o.oao)
36g*/0~5 /?A
? - RA + 42* lo\ L * 0.08O m
S*c fA ~C2°a*i0'>)(33'/.86*/O-<)
P r R* + /2 "/O^j L - O.O80
To+*i>; %AD - SAe 4 5^+5^ - 3^/ovv/d'r?A + Gi.3«s*/o
(a)-0.ft*to"*- 3.Hlotxio~*#i + ci.3$**jo'c » RA * -.7$. 6 wo1 N
* -76.C i<N
T?fc r R. + l2x/0"
- -0.031* **>

PROBLEM 2.43
SOLUTION
2.42 A steel tube (E = 200 GPa) with a 32-mm outer diameter and a 4-mm thickness
is placed in a vise that is adjusted so thai its jaws just touch the ends of the tube
without exerting any pressure on them. The two forces shown are then applied to the
tube. After these forces are applied, the vise is adjusted to decrease (he distance
between its jaws by 0.2 mm. Determine (a) the forces exerted by the vise on the tube
at A and D, (b) the change in length of the portion BC of the tube.
2.43 Solve Prob. 2,42, assuming that after the forces have been applied, the vise is
adjusted to decrease the distance between its jaws by 0.1 mm.
80 mm 80 mm 80 mm
P. '
* 8
■*_—«
c v
—-*
fir.
HZ kU 2>o kh>
For He +uhe dlL - A* -It
*n
AS : V - PA . i r O.OSOm
>Aft *
E A LZoo«io4 Kssi.gtf */o-c
= U3£8 *IO~n RA
6C : V ~~ r?A + </2* io^ L -- 0.08O
^6t ■
*->
= /./36SV/Q*'' ft + W.l*tL*lot'
Co: P - R* + /2*/o\ L - O.OSo
Due -J*> +ke *wo\y€*^«^t of +ke jaw« S^ r - 0. H «*» * - O. l*/0* »*»
(<x^ -0.|x|c?"s= 3.4lo4*/o~Tf?A + C/.S8ax/cT4 ft * - HT&SpIO* N
= - 47- 3 W W
- - 35„3"xlO w
* - 3^.3 kM
P0 ■= ft* 4- id*jo*
(b) S*, = (l.\Z£SxfO~'*)(>-,H7.*>Z1xl<?) + H1.1HC *lb~c - - 6,4£*/o"6 *
^r _<J.0?$C£»,^

PROBLEM 2.44
2.44 Three wires are used to suspend the plate shown. Aluminum wires are used at
A and B with a diameter of £ in, and a steel wire is used st C with a diameter of
7=T in. Knowing that the allowable stress for aluminum (£ = 10.4 x 10* psi) is 14
ksl and that the allowable stress for steel (£ - 29 * 10* psi) is IS ksi, determine the
maximum load P that may be applied.
SOLUTION
By svMwterrv PA = P8 > awd SA - £B
STTairt /rt QucM wire.
Del
"/x/O3
A#B
^ =
^ .
-i
* ,'39. G* it.

PROBLEM 2.45
8 in.
IK.
10 in.
i_
g -%fr
L 1 1 JD
2.¥5 The rigid bar AD is supported by two steel wires of ^ -in. diameter (E = 29 *
I06 psi) and a pin and bracket at D, Knowing that the wires were initially taught,
determine [a) the additional tension in each wire when a 220-lb load P is applied at
D, (A) the corresponding deflection of point D.
SOLUTION
L&+ 9 We +i« ^-f/e-A erf b*^ ASCO
Sc = 2v e
* ' AiH
p - £"A s»r - (y?woO Tfe-f 0* e">
rftp* i * ~ :
= io6-77>^/oJ e
t> 2M* -- o /2 p6£ f w p^ - 3c P = o
-3
Ox) P6e - Cl06.77^/o* yM^^/o"1 ) ' 2o¥.S A
Fc - (^IWMo'Xl.Wvlo'1)- «7.6 A
W So * 36 © * (acyuiiarx/o-3') * g^/xio*** i*
O- <>6<?/ m.

PROBLEM 2,46
2.46 The steel rods5£andCZ>eachhaveadiameterof { in.(£»29* I0*psi). The
ends are (hreaded with a pitch of 0.1 in. Knowing that after being snugly fit, the nut
at 5 is lightened one full turn, determine [a) the tension in rod CD, (b) the deflection
of point C of the rigid metober ABC
SOLUTION
L&i & be ilt t&\*~\-ic*\ of fc>*.^
But %a t %
W
- p** ^gg
U# =' 7:5"-ft- - 90 .*. rSfo = O./ !i
ytf
1?.
I 60
Sc
— -— *■* r^i
FA
CCt
CO
z.
to
V
Lto - eft * 72 ;rt^ Ac* = 0.3065 ;«*•
- 1.23^72 xlo* 0
DTHA - o gP^ - /o p« - o
- 0-0373 in

2.47 The rigid rod ABCD is suspended from three wires of the same material, The
cross-sectional area of the wire at B is eqnal to half of the cross-sectional area of
the wires A and C. Determine the tension in each wire caused by the load P.
SOLUTION
p- I-p-lp
^ ^L^^rf8
FROM T#£ D£FO(^Mff-^OA/ 0/A4G/iAA>
jrPB'Ip9>' %s?p Per0'*00?
P^ £»L(JL) =» ^p P* 0,S7£P
CHFCU %,
p^P+P^hCC&P . OK

PROBLEM 2.48
2.48 The rigid bar ABCD is suspended from four identical wires. Determine the
tension in each wire caused by the load P.
SOLUTION
Let 0 k He sJaf* of b«r ABCD af+e^
Sfc: SA 4 LB
§c: §4 + 2LQ
p0. ££ §0 - 4* s, 4 m± g
(n
£>XMA = O LPa 4- £LPC + 3^-Px, - 2.LP = o
■v.
*A jf lo EA
G 5* 4- m l& ^
) ^^^Jfa^e*^^-*^
= -Lp
-*.?$
E*
La
s*
&
r J- £2
to £a
Pft = ££_x££ ^ £4-j-.E£ * 4p
6 Ji **» £"A >e '° EA *
rc ^ lo £A -* *• EA lo
P t Q*.-L ft? 4 3J£.J-££ „ lp

2.49 A steel railroad track (E = 200 GPa, a = 11.7 x lOVC) was laid out at a
PROBLEM 2.49 temperature of 6°C. Determine the normal stress in the rails when the temperature
reaches 48 °C, assuming that the rails (a) are welded to form a continuous track, (/>)
SOLUTION are 10 m long with 3-mm gaps between them.
vi"0x jo"11.

PROBLEM 2.50
lin.
■+—»
h-2.5in-
Brosscore
E = 15 X l^psi
a = 11.6 X 10-fl/°F
Aluminum shell
E = 10.6 X lOVi
a = 12.9 X l<r6/°F
2.50 The aluminum shell is fully bonded to the brass core, and the assembly is
unstressed at a temperature of78°F. Considering only axial deformations, determine
the stress when the temperature reaches 180 °F (a) in the brass core, (b) in the
aluminum shell.
SOLUTION
AT - l$o -7£ = \ok F
lei "Pfc Le '-fJitf "fe^Si'-P* wee Je^eJafie.J
\vi +k* k>t»&$ cove
in The ctJ\jv*\i <\ o»»v\
r -&• 4 0/«.(AT\
O/c - 0!b = 1.3* lO"* /'F
f
Toraal1* ■"-*"»"''X1"1
0s xiofc)(o.*>«s»* ^ (lo.^x/o'
?w * 1-12 305" */o3 H

PROBLEM 2.51
5 mm 5 mm
20 mm V^ \
5 mm
*&&£-
Steel core
E = 200 GPa
Brass shell
E = 105 GPa
20 mm
5 mm
250 mm
2.51 The brass shell (ab = 20.9 x lO^C ) is fully bonded to the steel core (a, = 11.7
x 10" /°C). Determine the largest allowable increase in temperature if the stress in the
steel core is not to exceed 55 MPa.
SOLUTION
Ley" ps = a.xt'ciJf ■Wee cie^e/oped \n +l»e. s+eei cc*a^
Fo* eqm'I> d rio** *>H zqw -U+aJ -Fo/ee .+)ie
Stilus Ej - JL +04CAT1
+ Otb(AT^
3
E±>Ab
5s - eb
^ .o^ --£. + *b(AT>
-5/-IS
+
e7.)Ps = ^^S)(AT^
-i .L-i
£5Afc EtAt "" (2ooxiolX4>oo*IO"*) (\a£x\o°>XSooxtO~<')
= Sl.SS'xio'1 tf

PROBLEM 2.52
2.52 The concrete post (Ee = 25 GPa and ac = 9.9 * 10VC) is reinforced with six
steel bars, each of 22-mm diameter (EB = 200 GPa and aa = 11.7 x 10"6/°C).
Determine the normal stresses induced in the steel and in the concrete by a
temperature rise of 35 °C.
SOLUTION
Ma+oU. r §. = ^ A 4 o^ATO = - Bu + ofe(Ar)
Pc = 2K CI x/O* N
6c
- 0.39/> */Oc P<*_
0.391 MPa
-*
^ - - £ -i%£&r* r - ^7-"°' r«- - -™ "p-

12 in.
15 in.
PROBLEM 2.53
lj-in. diameter
steel
2j-in. diameter
bmS3
i
2.53 A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel (£9 = 29 x 10* psi, ^ = 6,5 x lO'V°F) and portion
flCismadeofbrass^^lSx 106psi,ab= 10,4 x lOVF). Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions AB and BC
by a temperature rise of 65°F, (b) the corresponding deflection of point B.
SOLUTION
F^ee "Hie****** exfl^^Si'on
Sko^fttviivM ©tae. +<» induced to^lJv^ssiVe +o*Ac^ *P
PL
QSl.
E.A
6c
Iff 9
-*
- SSZ. G^x/O P
-5
(^ G;
AB
P
Aag
_. -3SJft*J£? .- -zi.i*tos
1.227 a.
>Sl
= -*i./ ksi-
(W S>R * +
Pi
Aft
=- +
EsAte
(as:
- L»o&(An
(w*
:>»v^*y^) + 02X^*1°'^) - + &Mno* w t
i.e. 3.£*rf*jcT*M t

1.1 in
1.3 m
<JX^
PROBLEM 2.54
2.54 A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of brass (£b=l 05 GPa,ab = 20.9 x lOVC) and portion BC
is made of aluminum (£,. = 72 GPa, a, = 23.9 x 10"V°C). Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions AB and BC
by a temperature rise of 42°C, (b) the corresponding deflection of point B.
60-min diameter
; ^—-40-mm diameter
C
.3. *•
Z8*7H*JOSM*
SOLUTION
Fire*. Tke^w*,-? e/pa*%i'«M
=r 2.2 705 V /O
-3
fn
*r
r,
I. \ P
+
up
F
Of
|8.o7f Wo"1 P = 5.2705""IO"*
s-«'
4
*■-£--
i2£dL2-*Ja? , -/©o.o>'«>'RL*-ioo.oMFk
I. 2S"6& x^o"1
•Ml
(\oSx\o* X2.M7**vio-*i
— *■ ,5"oo x /o
-6
*i = -0-5*00 ^^
i.e. O.foo <*v*v v

PROBLEM 2.55
0.75 In.
Aluminum shell
2.55 The assembly shown consists of an aluminum shell (£a = 10.6 x 10 psi, #a =
12.9 x 106/°F) fully bonded to a steel core (£s= 29 x 106 psi, aa = 6.5 x loVF) and
is unstressed. Determine (a) the largest allowable change in temperature if the stress
in the aluminum shell is not to exceed 6 ksi, (b) the corresponding change in length
of the assembly.
SOLUTION
£ =
A,
f-f 0Cs(4r^ = J *oWAT)
/0.6 X/O'
- 0.q333&x/O
-S
At r inr.^t df
\,Z\C?> *YO
-3
Oir

PROBLEM 2.56
12 In.
It-in. diameter
15 in.
tffeeP
2^ -in. diameter
bfASS
2.53 A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel (£„ = 29 x 106 psi, as = 6.5 x 10"*/°F) and portion
BC is made of brass (Eb = 15 x 106 psi, o^ = 10.4 x 10"6/°F), Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions AB and BC
by a temperature rise of 65°F, (b) the corresponding deflection of point B.
2,56 For the rod of Prob. 2,53, determine the maximum allowable temperature change
if the stress in the steel portion AB is not to exceed 18 ksi and if the stress in the brass
portion CB is not to exceed 7 ksi.
SOLUTION
P = S%AW = O (2xio%)0. «?0 - - zi.oto x/os A
BCt 6ie- -7/ft,* = -7*/oV».-4 Ajt* polite" » $(*.;&)* = 3.?76/ m1
P = Sec A*. * C-7^|03X^^76/) * - Z7-Z33 *lOs A
UefoPwtA&d* Joe "h> P
-3
II
- - 13.004-x/O
Fi^ee + n e^*t a-/ €y pa^ 5 Tom
= (aw *io-t)(Ar^
AT - SS.Z* F
-3
=- o

PROBLEM 2.57
0.5 mm
-jh— 0.35 m
0.45 m
Bronze Aluminum
A = 1500 mm2 A = 1800 mm2
E = 105 GPa E = 73 GPa
a - 21.6 X 10-VC a = 23.2 X 10"VC
2.57 Determine (a) the compressive force in the bars shown after a temperature rise
of 96°C, (A) the corresponding change in length of the bronze bar.
SOLUTION
Cd)cJ«Ae Sfee iWw-J
-s
rv>
=r \.1Z% > ID
SrioiTT^^ |fl« awe "T» t*doceJi co^ jflr^SSiVe to^o* 'P
Bjt\ 1* Ve^*\s of" P
s>' &*♦ ?* --(^^)?
•(
AfcFt, AA£c vAb^ ' A«£"c
0.35" ! O.V5"
^>^
(A! E«|WM"\V\a
* zn kv
• 72S\76*(o"C - 4g3.Z<r/ */0~*
= 0-2*US* m^

PROBLEM 2.58
0.5 mm
-H h— 0.35 ra
2.58 Knowing that a 0.5-mm gap exists when the temperature is 20 °C, determine (a)
the temperature at which the normal stress in the aluminum bar will be equal to -90
MPa, (b) the corresponding exact length of the aluminum bar.
SOLUTION
Bronze Aluminum
A = 1500 nun8 A«1500mina
E ■ 105 GPa E = 73 GPa
a = 21.6 X 10-VC a - 23.2 X KrVC
S^ = -«?ovj0t Pa. A^- l8ao><to'c ***
P
**-
PL,
i
£bAfc EAAA
*-
- (fg^xj^Vo.SS'V, _ (U^IO* Xo.HS^
= (0.36T)fel.6>'lo'')AT + (0.^)^3.3 */0-4 ) AT
=■ IS.oox/o"' (A'O
?*>«•"
f in
!&.0Oy(O" (&T) ~ L4/<*79*/o~
Ar= 7S.6 *c
(ci)
T"hrf - T^u + AT
= jo 4 7s.* = ?s.e °c
(U Sa -
UoUAD - £^
- CP-WX 23.2*10 JC78.6; (73y/o^X.86o^to-)
- SZo.5S*IO~C- SSH.79»tQ~£ -
-4
^6^.78 x/o * ^
U*~* - La.4- $<*_ = o.ifh* *6r.?s*/o
-4
V*
- O.^o^6 m
45*0. 0 2£6 r»m

PROBLEM 2.59
0.02 in.
-12in.-
10in.-
•mmmmmm
Aluminum Stainless steel
A = 2.8 in2 A = 1.2 in2
E = 10.4 X 106psi E = 28.0 X 106psi
a = 13.3 X 10-6/"F a = 9.6 X 10"VC
p A. EL
2.59 At room temperature (70°F) a 0.02-in. gap exists between the ends of the rods
shown. At a later time when the temperature has reached 320°F, determine (a) the
normal stress in the aluminum rod, (b) the change in length of the aluminum rod.
SOLUTION
AT =■ 3ZO - 7o *• 2SO af
=■ 0? K/3.3 ^/o'^OlSbV 0°X^-^ * lo"c X*So)
- 63.T*JO~3 in = O.OC37 ^.
%? - 0.06^9 - 0.02. - 0.0^3? .V
-(,
1^
/O
(a)
P= £l.8£7 "JO* ^L
(=r - - -£. _ 6i.atS7»/o-
= - ;?2.o°i its;
(^
§a =
UoC(Ar) - £U.
- ^K.s.s^X^ - %^.t)^)
=- 3**.?o *lo~s - ZS.W */0
-3
O.OI44I ,'^

PROBLEM 2.60
Brass
r
2.60 A brass link (£b = 15 x 106 psi, flj, = 10.4 * 106/°F) and a steel rod (Es = 29 x
106ksi, #. = 6.5 x 10"6/°F) have the dimensions shown at a temperature of 65°F. The
steel rod is cooled until it fits freely into the link. The temperature of the whole
assembly is then raised to 100°F. Determine (a) the final normal stress in the steel
rod, (b) the final length of the steel rod.
0.005 in
2 in.
■*Lti
-J1-5™- SOLUTION
,^^_10in.^j 1.25-in diameter ^ M ti+j cK^S.'o**
team
Steel
Section A-A
AT= IOO - 65" = 35* °F
B««s #«fe (£T)t =■ oL^ATYl)^ (lo.qxio'-'X^y/oW 3. 6**/or" i*
O.oos + 2.375*10"* - 3.6**v/o"s - 3.63S"*/©'"* i*^
S
i>ow*e"- Haw -He b**aa 4V* k
ft I. L /q N - £k - P 0t°
= /I/. 1/ X/0'* P
r
* - 7.£Tw/os ^si ~ -7.i3"jAi
= ID. 0O46"? I*.

PROBLEM 2.61
2.61 Two steel bars (Et - 200 GPa and <*, = 11.7 * 10"6/°C) are used to reinforce a
brass bar (Eb - 105 GPa, a^ = 20.9 x 10"6/°C) which is subjected to a load P = 25 kN.
When the steel bars were fabricated, the distance between the centers of the holes
which were to fit on the pins was made 0.5 mm smaller than (he 2 m needed. The steel
bars were (lien placed in an oven to increase their length so that they would just fit on
the pins, Following fabrication, the temperature in the steel bars dropped back to
5 inin room temperature. Determine (a) the increase in temperature that was required to fit
the steel bars on the pins, (b) the stress in the hrass bar after the load is applied to it.
40 mm
ST- Lets AT o.sxto'* iUoo)(n.'r«/o-cX&T)> at =
O-SxlO-* = (O(ll.7x|o-C)UT)
AT - 2/.3S8 *C 2/.VCL •
(b) Once a.sse^.isJeJj a + e*»i*-P* -fo^te P* «fe^ci«^)9 i* He s+e«^ *.**/
e?o*a<Jfo -r-Kit sieeJ? a,*J conft^<s_of" -Hie b¥**ss.
E".fc«3<^i«« of a+eei.' As ■ CO(sX<to"i* MOO wm*" « ¥oO"/0"* ^
FL
^ - id
P ^a.oo^
"-T- S 26* */0'
-•»
-1
-*
Bwtis: 6L*= -^ - - 3-*"*'°* = -/7.<gW04P*--l4.6S M?cl
Tto fAese s+resscs **us4 b« tjAdeA fU s-fr*ss«« d*>c *» +U S5" /fW /od

P^oLPeh^ 2.GI Cof\h'r\0&d
Fov "Hie cctxJeJ JoclJ} +*)te addi'Titt*uJ Je^orme^T\o^ t's He s^t^c tw*
Ai*fj*.c€****£* Aifio^ >et 1\ «W Pb U He aJJ*!;***,) -forces
A*\j-e.£oped i* *H« a4«e>C a*#f W/ass^ ire*pec+tv/e-£y»
/Of
Si x I8.S6 **0* - W.C»*K>fc =■ ^CZxIcf ft. * 3.G8 tfP« -

PROBLEM 2.62
40 mm
2.61 Two steel bars (Es = 200GPa and a; = 11.7 x 106/°C) are used to reinforce a
brass bar (Eb = 105 GPa, flj, = 20.9 x 10,6/°C) which is subjected to a load P = 25 kN.
When the steel bars were fabricated, the distance between the centers of the holes
which were to fit on the pins was made 0.5 mm smaller than the 2 m needed. The steel
bars were then placed in an oven to increase their length so that they would just fit on
the pins, Following fabrication, the temperature in the steel bars dropped back to
room temperature. Determine (a) the increase in temperature that was required to fit
the steel bars on the pins, (A) the stress in the brass bar after the load is applied to it.
2.62 Determine the maximum load P that may be applied to the brass bar of Prob.
15 mm 2.61 if the allowable stress in the steel bars is 30 MPa and the allowable stress in the
brass bar is 25 MPa.
SOLUTION
See So)oh~0* f-o PROBLEM 3.6' f» ob+c.i* Me -fttukr/cA^c** s+V^Ses
AAP***A-Pe sf^esses: <5^= 30 MP* J <$ZjUf * ZS HP*.
6~s = Zo-ZJl.oZ- 7.97 Klp«
G%* «5"+ l"f-^8 * 37.6S HP*
Correspond in* a\y&.» J/*Xte. s+fift-iiA*
£3 ~ ^ ~ 200x10*> * 37.&SWO
El * S » ^-f&*'°U 377.* *k>-6
ft5" * lo *
£ ^S^g-TWO
Py * £LAb£ - Cior- /o-,Xtoo*/o"*>)C3^a5-*/o"*) * z.s\\*io* N
P * 9* + Pk • 3pI84^/Os + ZfU*lo3 - £.70*lD*hJ
5 * S.7o kYl •+

PROBLEM 2 63 2*^ ln a standarc^ tensi'e test a steel r°d or" J -'n- diameter is subjected to a tension
force of 17 kips. Knowing that v= 0,3 and E = 29 x 106 psi, determine (a) the
x|-in. diameter elongation of the rod in an 8-in. gage length, (b) the change in diameter of the rod
17 kips f " 17 kips
-8 in,-
SOLUTION
§»c ~ k** ■" (8.oVl74.1f WO~4) - 7.80v/o*\m - 0.0O78O ,*.
120 mm
PRORI FM 2 64 2,<** A standard tension test is used to determine the properties of an experimental
plastic. The test specimen is a 15-mm-diameter rod and it is subjected to a 3.5 kN
tensile force. Knowing that an elongation of 11 mm and a decrease in diameter of
0,62 mm are observed in a 120-mm gage length, determine the modulus of elasticity,
the modulus of rigidity, and Poisson's ratio of the material.
SOLUTION
,15-mm diameter £ = JI J* s S (,Sf r 176^^^*- ^.7I^|0"* ^
P = Z.S x/o1 hi
**r {* = 753" « '"■Wv/o-1
Sy = - p.ta •"*-»
5y q/.6£7*(o

PROBLEM 2.65
l64(JkN
2.65 A 2-m length of an aluminum pipe of 240-mm outer diameter and lO-mm wall
thickness is used as a short column and carries a centric axial load of 640 kN.
Knowing that E = 73 GPa and v = 0.33, determine (a) the change in length of the pipe,
(A) the change in its outer diameter, (c) the change in its wall thickness.
SOLUTION
<Mo-K^(*-o^ „._, ^
r .c --.Pi-
=r -2-43 **n
£ L
-3
- - /. 7.1 £3 * /O
-»
PROBLEM 2.66
„^>-
60 mm
*
f
^
■;
J-i
%
ar"
2.66 The change in diameter of a large steel bolt is carefully measured as the nut is
tightened. Knowing that E = 200 GPa and v= 0.29, determine the internal force in the
bolt, if the diameter is observed to decrease by 13 urn.
.-3
SOLUTION
V ' -
*** "^ ^t^^" 7"7./3,/o
-c
F- S"„A * (m.43*lo6 Xa-3a7*iO"s ) * 422*fOs W

PROBLEM 2.67
^=?
£* *
2.67 An aluminum plate (E = 74 GPa, v = 0.33) plate is subjected to a centric axial
load which causes a normal stress o. Knowing that before loading, a line of slope 2:1
is scribed on the plate, determine the slope of the line when o » 125 MPa.
SOLUTION
•* " £"
j. * . 3(l- Q.ooaSS'7'0 --«-.
PROBLEM 2.68
SOLUTION
2.68 A 600 lb tensile load is applied to a test coupon made from -fa in. flat steel plate
(£= 29 * 106 psi, v= 0,30). Determine the resulting change (a) in the 2.00-in. gage
length, (A) in the width of portion AB of the test coupon, (c) in the thickness of portion
AB, (d) in the cross-sectional area of portion AB.
KOOili
"V
■ 2.0 in.-
/Fa
b\
**- a o-eaias* ,1"c ^ ? -m.
6,. f * JM^ . CM. 07,10-
(o> S„ * L^e, - C?.0XC£3.O7wi0•',) " \.$ZH v/O"* i^.
?y i£t-- 2/ey = -Co.30^(66?.07y/d')- - !.<?*.<;;?* lo"
(J) Pi * Wt * Wo(H-*y)to0 + £2)
■ «R) Hi
~ - /?. 4/ */D
-6
1*1

PROBLEM 2.69
a;. = 6 ksi
lin.
2.69 A 1-in, square is scribed on (he side of a large steel pressure vessel. After
pressurization the biaxial stress condition at the square is as shown. Knowing that£
= 29 x 106 psi and v~ 0,30, deiermine the change in lengih of (a) side AB, (b) side BC,
(c) diagonal AC.
SOLUTION
<rs = L2 ksi
I
zimo*
[lX»|o*-to.SoKfi-*rf^
M %A& * CAB ^£w * O.oo Xss/:7a*io"*} = sr/,7 ^/o"6, i*.
= -J ( I + 352 x^Y1 + ( I + M.t "(6* T
(Kc\ - V? Ac -(fic)B - 307*10* i*.
La,tJe siJez w'nji cij bj a*J c as zbo&n
1 2 >i
C r Ct + 0
- 307 X IO
. 4
i^„

PROBLEM 2.70
cr„ — 6 ksi
ff,. = 12 ksi
Js -
2.69 A 1-in, square is scribed on the side of a large steel pressure vessel. After
pressurization the biaxial stress condition at the square is as shown. Knowing that.fi
=29 x 106 psi and v= 0.30, determine the change in length of (a) side AB, (b) side BC,
(c) diagonal AC.
2.70 For the square of Prob. 2.69, determine the percent change in the slope of
diagonal DB due to the pressurization of the vessel.
Tn-e Cnaweifc, i* s/ort« is,
CaXc
ccJlco J/os
s't a"2- ~ fa ~ a
o v*
-i
- <35/.72*70'
" 82.7£ x|o
-6

PROBLEM 2.71
2.71 A fabric used in air-inflated structures is subjected to a biaxial loading that results
in normal stresses ax = 120 MPa and a = 160 MPa. Knowing that the properties of
the fabric can be approximated as E =■ (ft GPa and e= 0.34, determine the change in
length of (a) side ABt (b) side BC, (c) diagonal AC.
SOLUTION
6X = \Zoxlo* Pa, %= o^ 6i - l4o*io*?b
-fi
= ^-(-*6; -v^ *6;Vj
" 87*lo?
a)
/,37o/ v/o~3
g La,k»eJ _s\«ta <3Tp vnglif fn'anjfe ABC as a, t, «-^J c
2c Jc = 2ad«, + 2fc J&
<fc. = f- da + £ db
But 0.-= /O0 KB, la- lOOv*^ C ~ V IOO<-+7fi/!') = llf W"
^°-^ S/ir = 0.07S4 w.m at * Sfc = 0-1370 y*^

PROBLEM 2.72
ft
50 mm
A
P\
P
1
240 mm 600
_
2.72 The brass rod AD is fitted with a jacket that is used to apply an hydrostatic
pressure of48MPa to lhe250-mmportion5Cofthe rod. Knowing that E = 105 GPa,
and v= 0,33, determine (a) the change in the total length AD, (b) the change in
diameter of portion SCof the rod,
SOLUTION
s; - 6*2 - - p - - w*\o' ?c , Sy - o
-c
I OS" * 10*
r - 3oQ.X<f v/o"
■or
rr^I-fo.33)(-Vg*/cO + tf -fo.SlK-^-'/c*)]
= - Sol.71 y io
-4
<Sy - LSy " (JlWoX-3ol.71*10*) t -0.072H w*
$*= Ss - <J6. ^ (SoX- 3o£.z<} rto~c) - " 0.O/S3I ">*
**1*H
PROBLEM 2.73
2.73 The homogeneous plate ASCD is subjected to a biaxial loading as shown. It is
known that az = a0 and that the change in length of the plate in the x direction must be
zero, that is, 4=0. Denoting by E the modulus of elasticity and by i>Poisson's ratio,
determine (a) the required magnitude of a,, (b) the ratio cj, / ez.
ff? = 6".
% = ° >
s* - o
e.-Ms<-*6;-z^)* ±te -»<50

PROBLEM 2.74
2.74 For a member under axial loading, express the normal strain £? in a direction
forming an angle of 45° with the axis of the load in terms of the axial strain e% by (a)
comparing the hypothenuses of the triangles shown in Fig, 2,54, which represent
respectively an element before and after deformation, (b) using the values of the
corresponding stresses <7'and <7X shown in Fig. 1,40, and the generalized Hooke's law.
Fi'j M<L(«I
to
SOLUTION
%/
i-^V
CM
Fi> Z-SI
i-«*
L^i/S
l + 6r
to
Ffo 1.40, (&)
l-i>£*
^(h£')
f^A Before dcfo^»^»,4i'ov\
Nej-Pefrf" S^tfairet «.s swujj 4£* - 2£»c-^^^
W
p'- ^^2
SI' *y'
j - -v JP
2E ^*
- 1>

PROBLEM 2.75
2,75 In many situations it is known that the normal stress in a given direction is zero,
for example az = 0 in the case of the thin plate shown. For this case, which is known
as plane stress, show that if the strains ^ and ey have been determined experimentally,
we can express axt <7y and et as follows:
£x + ve v
l-v2
ev + ve
'•s"I77('x + ',)
trs SOLUTION
e; = o
Mo'+ipJwiM 0 s) \»y V a^itfj adding +©fl?)
&=^-v6^ + ^ ft)
l-v
. e
PROBLEM 2.76
2,76 In many situations physical constraints prevent strain from occurring in a given
direction, for example ez = 0 in the case shown, where longitudinal movement of the
long prism is prevented at every point. Plane sections perpendicular to the
longitudinal axis remain plane and the same distance apart. Show that for the this
situation, which is known as plane strain, we can express az, ex and ey as follows:
<rM = v((rx + <ry)
', = 7K1" ">* - "0+ "K] £y = 7K1- "X- '(1 + ")'J
SOLUTION

PROBLEM 2.77
2.77 Two blocks of rubber, each of width w = 60 mm, are bonded to rigid supports
and to the movable plate AB. Knowing that a force of magnitude P= 19 kN causes
a deflection S= 3 mm, determine the modulus of rigidity of the rubber used,
SOLUTION
s*tr*ss t*
W*
X A
? = l?*/03 M
-3
/0.8*/O w,
85714
g * * - ^m^ - ">-**«>• * * »■* »?*■
PROBLEM 2.78
2.78 Two blocks of rubber, for which G = 7.5 MPa, are bonded to rigid supports and
to the movable plate AB. Knowing that the width of each block is w = 80 mm,
determine the effective spring constant, k = P/6, of the system.
SOLUTION
Ccmgv'ds/' "f^e u^pe^ GjfocM of polo bet*. The
r - *?
A
P r 2At
n
k
.+.
\*
WoKr^ +Ut r * ST ^
7^r "?* 4^c>^ w-^vicV § = hT1
h 0.03JT
CPx./O3 kW/w

PROBLEM 2.79
St)
CA^i*
-3
s+
re*s
80+ r = -jj- --•
2.79 The plastic block shown is bonded to a fixed base and to a horizontal rigid plate
to which a force Pis applied. Knowing that for the plastic used G = 55 ksi, determine
the deflection of the plate when P = 9 kips.
Com si"el e**
;* ? - 9**0* Jl
Trf =^TJ^ * O.OOS500G
PROBLEM 2.80
P
2.80 A vibration isolation unit consists of two blocks of hard rubber bonded to plate
AB and to rigid supports as shown. For the type and grade of rubber used raU = 220
psi and G = 1800 psi. Knowing that a centric vertical force of magnitude P^ 3.2 kips
must cause a 0.1 in. vertical deflection of the plate AB, determine the smallest
allowable dimensions a and b of the block.
SOLUTION
Conner "r-U YvAW VJ00U ©•» He *\^ k+. J4
Carries a 5^^iV«i ■fo/-c« e^oc^l "fw xP~
A . _£. - 3.gx|Q3 . „
Use k - ^.«ia •'* *«J H- 2XO f«i
in
■**
7 " a
^00
=r 0. 1223.3.
r 0.8/B i
Irt

PROBLEM 2.81
2.82 An elastomeric bearing (G = 0,9 MPa) is used to support a bridge girder as
shown to provide flexibility during earthquakes. The beam must not displace more
than 10 mm when a 22 kN lateral load is applied as shown. Determine (a) the smallest
allowable dimension b, (b) the smallest required thickness a if the maximum allowable
shearing stress is 420 kPa,
SOLUTION
r 5^-3S/ */G **>*
A - ftoOw^Xt)
b =
A . 52.38 <k/q-
2oo
2oo
2M
i^wv
Bo+ r^ J
T 466. £7 */0^
PROBLEM 2.82
2.#2 For the elastomeric bearing in Prob, 2,81 with b = 220 mm and a = 30 mm,
determine the shearing modulus G and the shear stress rfor a maximum lateral load
P = 19 kN and a maximum displacement 3= 12 mm,
SOLUTION
hreu. A - {.Zoo ^)(2ZO *>^ ~ 44*/o* h^
- 44 x ;o_s h**
r^--S^- «/.ti,io-P*
4M>|or
oke^rvnci 6T^ai
«\
,^A
*■ l.08o Mfe. ^
T
o.q

fi *2.83 Determine the dilatation c and the change in volume of the 200-mm length of
PROBLEM Z.83 the ^ shown jf ^ ^ ^ ;s made of stee, with £=200 GPa and v - 0.30, (&) the
, rod is made of aluminum with £ = 70 GPa and v= 0.35.
j" s2-mm diameter
-16 kN * :-*- ^ . 46 k.N SOLUTION
P r 4£*/Os /^ 6i = ^ * 121.Ol *)Ofc P«, 6> = Q = O
lb) A(c»fci.n^* e - — -J ^,^A ~ &l7*lG
JO vTlO*

PROBLEM 2.84
Fvo* PCOSLFM 2.£3
jWc-knes* * it \*
E - 29*10*
v - 0.3o
*2.84 Determine the change in volume of the 2-in. gage length segment AB in Prob.
2.68 (a) by computing the dilatation of the material, (6) by subtracting the original
volume of portion AB from its final volume.
rt-
filtt) lb
600ll>
SOLUTION
Voider V0 = AU = Gp.^/arKa.ool =r o.(96^* ;**
Cb")
^-f -oSr3T * '«"«>>■
ev = <s-« = o
.-€
T/rt ol!w€^at"o^s w*«y* tM#ler &. € OO >f& -|-«aoi'-& foA.of our* -
J*«3+t» L= U+S,. ~ £4 /.32YWO"* = 2.O0I3Z4 ',„.
-4
lw
^o)o*t *\T - Lu/t * OLOUflG^S?
i*
A2T - V- ^ - aoczncffsi -o.ot^r = ics**iom' }*

*2.85 A 6-1 n. diameter solid steel sphere is lowered into the ocean to a point where
PROBLEM 2.85 the pressure is 7.1 ksi (about 3 miles below the surface). Knowing that E = 29 * 106
psi and v= 0.30, determine (a) the decrease in diameter of the sphere, (A) the decrease
SOTiiTIOV in volume ofthe sphere, (c) the percent increase in the density of the sphere.
■ s
p^ *- SoJtiA sphere *2£ * J <$? * £ (6.*>)* ^ l)3.o«»7
Likewise £L * £t = -97.?3x/o"'
** -e =■ 2?3. 79 v/©~*
^-A^QQ^ = (Z13.79 *lcT*) (too 7*) = O.0VW%
fa

PROBLEM 2.86
-85 mm-
(ry = -58 MPa
E= 105 GPa
u = 0.33
135 mm
*2.86 (a) For the axial loading shown, determine the change in height and the change
in volume of the brass cylinder shown, (b) Solve part a assuming that the loading is
hydrostatic with a% = ay= az = - 70 MPa.
SOLUTION
h0- 1ST mm * O.I35" *«
to) 6^ = O 3 <5j = - S3*fc>* Pa. ^ 6^ * O
AV» ' h„£y =■ 03£" r**0(-5*5X38 x/o-') * - 0.0746 m», -^
e =
^Csws, +o ="4^ - (^'ff''^
-6
- -187. g/>a^
(b) €i = <S; = <S*Z * -/Ox/O6?*,
6; +. s; + 61 =» - */o x/ofc p«.
%s
e =
1£e = (7C&.06W2 >«*>*)(-£3<Wo~6>) r - .SX/ ^w,3

PROBLEM 2.87
|P
*2.87 A vibration isolation support consists of a rod A of radius R{ = j -in. and a tube
B of inner radius R2 = 1 in. bonded to a;. 3-in.-long hollow rubber cylinder with a
modulus of rigidity G = 1.8 ksi. Determine the largest allowable force P which may
be applied to rod A if its deflection is not to exceed 0.1 in.
SOLUTION
3.0in. Vttlolae^ cyJlindt* R| ^ ^ ^ Rl
? =
anr 'h
SJ^a^'M Je£>^ **■'•* o^ yrcjitjto^^^
Jf
To+eJ deform * *■'"*«
«\
2-irGh ^ R,
-P =
277fillS

PROBLEM 2.88
IP
3.0 in.
*2.88 A vibration isolation support consists of a rod A of radius R] and a tube B of
inner radius R2 bonded to a 3-in.-long hollow rubber cylinder with a modulus of
rigidity G= 1.6 ksi. Determine the required value of tbe ratio R^/R^ if if a 2-kip force
P is to cause a 0.12-in. deflection of rod A.
SOLUTION
Id Vs be a- v»A<o.Jf co£>r.^i'neL"f e , Over 4-K< V>ofJo*J
P Shearing S+acsS T* *<J"i»^
L A *nr'h
r = i ,_£—
il - r
2ttSK ^
rp4
2-nGh
2 A
r
J^rP = r£rUi Pi-^1?,)
2TT£h ^ r7
Vt P Z x- to3

PROBLEM 2.89
*2.89 A composite cube with 40-mm sides and the properties shown is made with
glass polymer fibers aligned in the x direction. The cube is constrained against
deformations in the>> and z directions and is subjected to a tensile load of 65 kN in the
x direction. Determine (a) the change in the length of the cube in the x direction, (b)
the stresses ax, ay, and av
Ex = 50 GPa vr = 0.254
Ey= 15.2 GPa vx'y = 0.254
Ez = 15.2 GPa v%l} = 0.428
SOLUTION
S\ress--)t> - s+ra*V» e.^ocuTto*\s av^
x ' /<.. E.
E.
(W
(ft
^=^f 0)
E.
Ey E>
G)
,__^_V^ ^
E.
*y E>
E, E
-¥ &
Tlie C4»*v*fr"A«'*»T GotoJli fife** cu^e Ey - O <^*J £z - o .
(7)
£* ^ " izt ^ s IP^ " %" °-W86"» s °'077m ^
j Sy * <T3 » 0.13^93 6^
Q.?3I42 6>
Co^-h in f€ d

Problem Z g<? co^-KwoeJ
= 75*6.18 */0
-6
PROBLEM 2.90
*2.90 The composite cube of Prob, 2.89 is constrained against deformation in the z
direction and elongated in the x direction by 0,035 mm due to a tensile load in the x
direction,. Determine (a) the stresses ax, ay, and az, (A) the change in the dimension
in the y direction.
Er=50GPa v„= 0.254
V
E* =
15.2 GPa
15.2 GPa
SOLUTION
£x
£y
e2
v* = 0.254
vZ9 = 0,428
Ex Ey E2
Ex
Ex
£*a
6^
Ey Ez
E E
^y Ljz
O
O
(0
M
Cs-)
V V
xy yx_
E\ ~ E..
Ey~ Ez
E. Er
F^o* Mut^cm C3) 0 * - ^ SV t 4- 6T.
e , 2^^ s-. = (o.as"/KMT.a) = Q. 0772/6 6;
ft)
en
GO
Cow+t*ige^

P^otP^w, Z^O ^i'hu«J
Prow Ge^ QosTiovi CO *>l tU
6y = 0
*« £<* -£s. = i^-fes-.
" &^" O-^Szl ' &!'" (P^^Xo.077XI6)Js;
*-*
** *• fe'^^T - ^'^
(a) 6* - C^0*10*1 XS7SX jot*) iw.c^S- */o3 ^ = 44.6 MP*. —
0.^8039
5"y * O ^
6^ r (#.077'Zlt M+MfriO*) =- 3.446 */Ofc f>«, = S.ttSMf*. «*
- - 3*3-73 */0~4
Sy * A.^ = (40m*X- 3^.73 v|D"6) ' -O.Ol« *,», **

PROBLEM 2.91
SOLUTION
G =
Z(\+»)
or
*2.9l Show that for any given material, the ratio G/B of the modulus of rigidity over
the modulus of elasticity is always less than \ but more than j- [Hint: Refer to Eq.
(2.43) and to Sec. 2.13.]
| - 2Cn-^
PP V ' **S "•""« bo<Jrw9<S
i
©^
*£**
PROBLEM 2.92
SOLUTION
K =
(al
(^
3(1-2*0
ant
*2.92 The material constants E, G, k, and vaxe related by Eqs. (2.33) and (2.43).
Show that any one of these constsnts may be expressed in terms of any other two
constants. For example, show that (a) k = GE/(9G - 3E) and (b) v= (3k - 2G)/(6k +
2G).
+■ %> -
-, EL
2&
k =
20 + *M
gg
Jl - aQ + >^
& " 3(1- 3UO
31<- 2G- - 2S +£k
^ r 3k - 2G-
ZE&
18(9- GF
£k +■ 16-

PROBLEM 2.93
3 m.
2.93 Two holes have been drilled through a long steel bar that is subjected to a centric
axial load as shown. For P = 6.5 kips, determine the maximum value of the stress (a)
at A>(b) at B.
SOLUTION
J * 3 - £ = 2.5*0 ;*
Ahtff = dt * t^o^} -- 1.25" i^
*"&■ "■"»
-fe.'-Sfr = ** <~
(to AJ ^oic 8 r = iO-^* °-7-S\ J- 3-1.5"= 1.5" ih
£* -^^ - O.S Fr«<* F»3 3.CHa \< - 2./o
PROBLEM 2.94
3 in.
2.94 Knowing that aM - 16 ksi, determine the maximum allowable value of the
centric axial load P,
SOLUTION
Af ho9e A ^=HM »'"
dl - 3-4 -- 2-so m.
A.*f =■ dt - UsoX±") - i.ar;^
A 2.S& J J
^"*" A
•wt
ps
M hole 6 r--£0«O - 0-7i-;Mj ^=3-/.$* = /.S",V
A«f = Jt = 0-5-)(4) - 0.74" ;«\
K 3.(o
P= 5-.7I k;,

2.95 Knowing that, for the plate shown, the allowable stress is 125 MPa, determine
the maximum allowable value ofP when (a) r= 12 mm, (&)r= 18 mm.
SOLUTION
d Goo
- £oo
X. _ 12
- 0.?
(a) V^-l*^ j - ^
SL- = K x
>**.*
15 mm
(k^ T^ /gwWj J- -g^ * 0.30; Fr^ F/j 2.C* Jb Kc 1.75
p , A^ e f^o^)()^Wj b ^.3m/03n w rt3
2.96 Knowing that P = 38 kN, determine the maximum stress when (a) r = 10 mm,
(b) r = 16 mm, (<?) r = 18 mm.
SOLUTION
- 2.0D
_|) _ ISO W.W
= O. I 667
Fret Fig 2.C4 W |<T = 2.06
120 mm
^K
15 mm
)»M^
<S" - KE
-= 87-Oy/ofcP& = £7.0 MPa,
V
16
(O f= 12^ £ = ■fe - 0.3O
F~~ Ra «.Ct b K - 1.75"
6-- -('-^?o-^- 75.1 W^P«. >- 73.9 HP,

PROBLEM 2.97
2,97 Knowing that the hole has a diameter of| -in,, determine (a) the radius ry-of the
fillets for which the same maximum stress occurs at the hole A and at the fillets, (A)
the corresponding maximum allowable load P if the allowable stress is 15 ksi,
SOLUTION
Ft*- He circot** hok V^CaXi")- O.iSTS U
<j = *t-\ * SMS i* %•*%£&-* O.oSt?
AMef - dt - (3.6257f*)» »-SS*4 in*
Fro* Fij ^.CIol Kv»fe * 3-82.
°",*,f A«t
? r A^ 9 (-^£^ - 7. zs k.>.
D = ^ iWj oi * a.6* ;*
5-41 •»-«>
PROBLEM 2.98
2.98 For P = 8,5 kips, determine the minimum plate thickness / required if the
allowable stress is 18 ksi.
SOLUTION
A+ He \xoh- YV = i ih dA= B.2- l.o - I.a i«
o.HP
F^om F.> 2-6* a K ' 2. XI
K?
***- a^ - ift - r ak*
l*T
MH«W X)=2.Z>% <*8* l-C i* ^ = -££ s (.37,5
K = 1.70
6- r K£ - K£
Fro* F.-j 2.Gf k
t- 0.87 iw.

PROBLEM 2.99
50 mm
t = 15 mm
r = 6 mm
75 mm
2.99 (a) Knowing that the allowable stress is 140 MPa, determine the maximum
allowable magnitude of the centric load P. (b) Determine the percent change in the
maximum allowable magnitude of P if the raised portions are removed at the ends of
the specimen.
SOLUTION
D - I£- . USQ L= Azzl ^ o.,*
■> d SO »»>
-6 i_
A*.-* = t J? US'XSO) - 7SQ **? = 75*>*lO~ r»
(bl W;fli0*f )r&.tseJ? secure*** l^- Loo
PROBLEM 2,100
• 15 mm
rt = 10 mm
2.100 A centric axial force is applied to the steel bar shown. Knowing that aA is
135 MPa, determine the maximum allowable load P.
SOLUTION
18 mm
ff - J1E.
p z A^ 6^ . (i.og«fcr%sr»/oO _ 0s w s ^ w
'l»l*#6
fS. ^- Oik
SjpuJtfet1 \Safve, -Ft^ *P co^-|v^vfs
"P = S£ fcW

PROBLEM 2.101 2'101 The 30~mm s<Iuare bar AB nas a knglh £ - 2.2 m; it is made of a mild steel that
is assumed to be elastoplastic with E = 200 GPa and aY = 345 MPa. A force P is
applied to the bar unlil end A has moved down by an amount Sm. Determine the
maximum value of the force P and the permanent set of the bar after the force has
been removed, knowing that (a) 4 ■ 4,5 mm, (b) Sm = 8 mm,
SOLUTION
Sr - S--S'
ft>) Sn * 8** > Sy P** fclO.S*/©J KJ = 310.i"- kU -*
p M 2.102 The 30-mm square bar AB has a length I -2,5 m; it is made ofmild steel that
PROBLEM 2.I02 ^ aMum8(| ,0 be elasloplesllc with E - 200 GPa and a, = 345 MPa. A force P is
applied lo the bar and then removed Id give it a permanent set Sp. Determine the
mexlmum value of lhe force P and ihe maximum amount 6m by which the bar should
be stretched if the desired value of dp is (a) 3.5 mm, (b) 6.5 mm.
SOLUTION
A - (&>)(.&*)* yOO mm* « ?OOX/0"4 ***
vVKeo %m eytee^a Sy ^ H>us prodjc/vi* «- pe^^a-we.*!"
(bl Sf * G.S
T

PROBLEM 2.103
a :
- t -in. diameter
60 in.
11 in.
a
t
^
ft
2.103 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29
x 106 ksi and <^ = 36 ksi. After the rod has been attached to a rigid lever CD, it is
found that end C is J -in. too high. A vertical force Q is then applied at C until this
point has moved to position C". Determine the required magnitude of Q and the
deflection d, if the lever is to snap back to a horizontal position after Q is removed.
SOLUTION
Since +l»€ Vod AB is to be s+i^e-fe^eJ p^<r^a^^^Jifi
He peek -fWc-e iVi+A« poA is P — Pf m/-L»*«
'Di/ir/n* 0* Jo* J in* fke Sp^Vn^btfcJc «J~ B f*
S&P* e = ■§£■=&•• ^vffs*- o.iii7^—
PROBLEM 2.104
a :
-T-in. diameter
60 in.
11 in.
Q
ULi
*>
c ft
©
6
EL
2.103 Rod ^S is made of a mild steel that is assumed to be elastoplastic with E = 29
x 106 ksi and ^ = 36 ksi. After the rod has been attached to a rigid lever CD, it is
found that end C is f -in. too high. A vertical force Q is then applied at C until this
point has moved to position C". Determine the required magnitude of Q and the
deflection 6} if the lever is to snap back to a horizontal position after Q is removed.
2.104 Solve Prob. 2.103, assuming that the yield point of the mild steel used is 50 ksi.
SOLUTION
Since He **J Aft i& 4u lac s+v*tei*ej pe^vnaKi«*vh0y,
pY . A6^ = *(4Wso} * A.SM k.>-.
R«£e^W*« \& He •fv^ee body ©lid*''**'* e*r A^e^CD
- 23. -p - fa* Kg".S3*) - * r ft I, ... **
Fro** 4te cHero^iMtfiTi'o^ o»i<wj/*,m

1.2 m
0.8 m
PROBLEM 2.105
1
2.105 Rods AB and BC are made of a mild steel that is assumed to be elastoplastic
with E = 200 GPa and <% = 345 MPa. The rods are stretched until end has moved
down 9 mm. Neglecting stress concentrations, determine (a) the maximum value of
the force P, (b) the permanent set measured at points A and B after the force has been
removed.
SOLUTION
. 45-mm diameter
-35-mm diameter
% -
to* P** * A^6V » Imincr^iwto*)
<V1 sr^UJc S'- gflfc.£(£♦£>>
aoo kio**
«?c2./ X/O-*
,o-* )
-s
A+po»V+A Sp = S* ~ S ~ 9 **w\- 2-6 3 i*im ~ £.37 hM
At p^i*f B • Wo yl«MiH« in BC J nt^t« Sp * ©
1.2 in
0.8 in
PROBLEM 2.106
■ 45-inm diameter
2.105 Rods AB and BCare made of a mild steel that is assumed to be elastoplastic
with E = 200 GPa and <% = 345 MPa, The rods are stretched until end has moved
down 9 mm, Neglecting stress concentrations, determine (a) the maximum value of
the force P, (b) the permanent set measured at points A and B after the force has been
removed,
2.106 Solve Prob. 2.105, assuming that the yield point of the mild steel used is 250
MPa,
SOLUTION
■ 35-nun diameter
A** f (Corf's 7G2.MoC^ A^-ICojMtf* WTWirfcTV
^ ^ eA*b EA.c E^S« A«c;
.-3

PROBLEM 2.107
D
* *flfl C$>
T
800 mm
800 mm
T
5+v\*,'« im c^fcA* AD <t"^ CF 2 £AD - £
2.107 Each of the three 6-mm-diameter steel cables is made of an elastoplastic
material fin- which Oy = 345 MPa and E = 200 GPa. A force P is applied to the rigid
bar ABC until the bar has moved downward a distance 6=2 mm. Knowing that the
cables were initially taut, determine (a) the maximum value of P, (b) the maximum
stress that occurs in cable AD, (c) the final displacement of the bar after the load is
removed. {Hint: In part c, cable BE is not taut.)
SOLUTION
For eack c«AJ« A = ^(o.OQtf = ^«-X74x-/o"c y^
€f* 3HS"*;ofc
-1
, A
- _1L — 3rv"*^
7£-*l&
fe,= ^°*'°
-a
Pa© * Pcf - A 6i, =■ (^.m^O^X^ylo') = 7*0L&*tos N
fc^r ^utJikr.'u*^ of U^ ABC ?*£>+■ Pb£ + fU ~ P = O
0*) ? * Rid + P&e-I- P^ -" (7.06&5~^ <?.7W5"+ 7.06 35" )></03 M
Since
(b) 6"Afi * tSo*lGf- Pet
*3.<? I/K)
Affft^ uhtomJtM P - O
Fo.r eeft/iJi k>s\ut*\ Pad = TJ*= " ^
of CaJaAfi At) <»-^ei CR S I*ce -f-k&e Caries We*^ nCi/e*
S = Sa© - Sep * O

PROBLEM 2.108
m* l
800 mm
800 mm
1
2.107 Each of the three 6-mm-diameter steel cables is madeofanelastoplastic
material for which <% = 345 MPa and E = 200 GPa. A force P is applied to the rigid
bar ABC until the bar has moved downward a distance 3= 2 mm. Knowing that the
cables were initially taut, determine (a) the maximum value of P, (A) the maximum
stress that occurs in cable AD, (c) the final displacement of the bar after the load is
removed. (Hint: In part c, cable BE is not taut.)
2.100 Solve Prob. 2.107, assuming that the cables are replaced by rods of the same
cross-sectional area and material. Further assume that the rods are braced so that
they can carry compressive forces.
SOLUTION
R*r CftcJi rod A - f-Co.006)2" - 2Z.X1H x/O-**,1
Sfram in roAs AD ol*A CF: £& =■ E^
7. **«
Lad " 16©Ow*i
I.T.S x/O
-3
Soo
ZSo x/o
5+ram m roe* BE : cTB(? =* — =
Si'hce f«> < *r j &** - ^^ap r C*Oo*/o,'Kl.25*/o'"J) =" ZSOxlO* Pa.
Forces: PftD = PcF =■ A6*Ao= (as.^xlo^K^o*^) - 7.o6g5*/o* N
Pee- AG*ai -' (22.27lt*lcri)(3y£xtot' ) -- 9.75*5*/O1 A/
For eiWA'bWum of txstr ABC PAD + Pw + Pt> - P - O
(a) P » P*o + Pft6 ^Rf = (T.OGSS"* <r.7SHS + 7.068S)*lo*V - 23.«? *U
(b^ 6*^ r ZSO MO4 P* = ZSo HP*
For e^*ii.ViUM P'- P,D' 4 P^ + Pc; =■ 11.137*10* S'
8o4 p - p* = o p' =■ P -- 23.$?k/o3 A/
S
* _
23. 8? * /o:
£<?o / /O
-3
Per*t a,«oci*T «»s p/flrce«*>e-t r ot o&^
ho
-3
-3
0.3/O XfO ht
* O.ZIO won

PROBLEM 2.109
190 mm
2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-
sectional area of 1750 mm2, Portion AC is made of a mild steel with E = 200 GPa and
0{ = 250 MPa, and portion gc is made of a high-strength steel with E = 200 GPa and
0{ = 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C,
190 mm
SOLUTION
*CS
T <J- C fo ctmse yieJJt'\A+ of A^
•Ae^Ac
Zooxxo*
c* Lea OJ^o
N
bor^ion AC yi*J?J%m
Fca = *>c * *P " 437,£xJ0*-t7SV/0jW-s -S27.S «l<? V
* 0.272 *n*n
2ib MP*,
ft*
>Sc " A * nsoxio
- s4*«. .. tWu. ... ps; r - £ ±* = . r
nr
FA
Us
At
P' = T75x/03 = P^ - P« * 2P^ PA1^ 457.5 "/0s */
ht
£p * Sw - §' » O.Ml77Wo"S- 0-2*Mtf*/o'3 * O.0l7t&*lo~**
~ 0-OZ7 M^

PROBLEM2.110
190 mm
190m
2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-
sectional area of 1750 mm2. Portion A C is made of a mild steel with E = 200 GPa and
(^ = 250 MPa, and portion BC is made of a high-strength steel with E = 200 GPa and
(^ = 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
2.110 For the composite rod of Prob. 2,109, if P is gradually increased from zero
until the deflection of point C reaches a maximum value of Sm = 0.3 mm and then
decreased back to zero, determine, (a) the maximum value of P, (b) the maximum
stress in each portion of the rod, (c) the permanent deflection of C after the load is
removed.
SOLUTION
Dispose e.m?i t «T C is %», ~ O.SOvn^. T^e. co^«^esrf»o*vJ,i^«i s+i^.Vt «^
2t>o* /o'
Iol> Forces- F^. - A6V " (>7^W0"4 )(2S"£>*/tffc") " 437..T*/efa |V
(W Sfreuses: AC 6^ * O^ac * 2«iD MP* -*
(C^ DefjPteTion o.*w( raroM -to*- Q*JouJi\ n*
- ' CO - rAc / " ~
*t
?* * Sih-S' * 0-3* m., - O.U87H *r> » 0.03/ m^ -

PROBLEM 2.1U
14 in.
2.111 Two tempered-steel bars, each "fif-in. thick, are bonded to a 2 _m- mild-steel
bar. This composite bar is subjected as shown to a centric axial load of magnitude P.
Both steels are elastoplastic with E= 29 * 106 psi and with yield strengths equal to 100
ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually
increased from zero until the deformation of the bar reaches a maximum value Sm =
0.0H in. and then decreased back to zero. Determine (a) the maximum value of P, (b)
the maximum stress in the tempered-steel bars, (c) the permanent set after the load is
removed.
SOLUTION
Fo/1 +k* m''J(J £*fee>r
A, - HX^ * i«00 *"*'
Vi
F©*- He U^pe^e* sfeei Az - 2 (£ XO * 0.7£T ,V*
Tot«J aveo. - A = A, 4-A^ - 1.75" in
%^ < Sw ^ Sn Tk<* wJU s+eeJ yfelJs * Te^pereJ sieeJ »s ef+stic,
(a) Farces P, - A, &n = O.doX&WO1)- ^0 x/O3 A.
p . £"Aa S» . (a°iyfpa)Co-75X6-^) ^ M ,4 y/0* ^
2 L 14
p: p1+pa -. nz.wvio* JL
\12J k'ps
Od) Stresses
s; =
. a _
(pi Pe.rWA»e*f s*+ SP - %^~ S' ~ O.OH - O.OZoW
- o. ooqoc r*.

PROBLEM 2.U2
14 in.
2.111 Two tempered-steel bars, each -j^-in, thick, are bonded lo a 7 -in. mild-steel
bar. This composiie bar is subjected as shown (o a centric axial load of magniiude P.
Boih sieels are elastoplastic with E= 29 x 106 psi and with yield strengths equal 10 100
ksi and 50 ksi, respeciively, for (he lempered and mild sieel.
2.112 ForthecomposiiebarofProb. 2.111, if Pis gradually increased from zero to
98 kips and (hen decreased back (o zero, determine (a) the maximum deformation of
the bar, (b) the maximum stress in ihe tempered-s(eel bars, (c) the permanent set after
tlie load is removed.
M.-JJ sietJ? A, - (±)CO = loo ;**-
Te^peveci s+eej> Aa ~ 2(-&)(2) r 0.15 m*-
To\cJ $orce ±0 ylM He ~w JJ s4eei>
6/1 r -T •"■ fy r A6V, * ll.7$-)(£o*io*)* S7.?o*\<?&.
V> -Py
■Me^
tV^e r»^i
J>J s+^i*
y
ie
5 .
L^f P, = -Pore
e cctr^te
re*/ fcv y^y'JJ s"ftfe'
«0
Pt = -Fo/ve cc^iT^/e^ ty -fewpes*ol sieeJ?
P, e A,<r, - [.\.oo)(so*io>) - j^/01 A
fc.1
S,
S^ - §' - 0.02010 - 0.o2?o3 =• O.OO387 iV

PROBLEM 2.U3
C
2.113 Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is
assumed to be elastoplastic with E = 200 GPa and <% = 250 MPa. Knowing that the
force F increases from 0 to 520 kN and then decreases to zero, determine (a) the
permanent deflection of point C, (A) the residual stress in the bar.
SOLUTION
a = 120 mm
440 mm
-C
A - Moo ^w «■ izo&y/o v*
F«~« h> yieJJ pc4-i*i* AC : PAc = A 6> = ( IPOO x/Cf^faro *JO**)
F
C6
= 3oo *iQ3N
"Foi^ eai/ii.'l)^** P + I eft ~ Pac ~ °
ci
- - 2£o x /o* m
**.= -
FA (Cfcaoyio*1 )0^oo x-io"*)
A
a™"«* -=-183.83.2 *\QC ?*.
c TA £A
CF.-Rc'U
eft
HA
r** UA + EA J " "?A~
r4c UUca
O.4*o
Pea = Pac!- F = 3>7B.lKi"(0*-S*0«LO
,3 -
4I.3/SWO M
Me -
^^ -
(CO
-3
^ - 6 SI*. HIV
'eo^r-ts

PROBLEM 2.114
2.113 Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is
assumed (o be elastoplastic wilh E= 200 GPa and <\ = 250 MPa. Knowing that llie
force F increases from 0 to 520 kN and then decreases to zero, delermine (a) the
permaneni deflection of point C, (b) the residual stress in the bar.
2.114 Solve Prob. 2.113, assuming that a - 180 mm.
SOLUTION
fc = \ZoO m/ * izoo x \o~t" *C"
R».rce 4o yieJfJ o*Ai<*. AC - PAc. - AS^ - (l3oo*IO~*"}CaSfc> ^|OcV)
= 3OO ^lO-1 W
'Ac
r£ft
Fee * P,c - F - Zoo*lQ3>-SZoylcr'
c HA (2oo*io'»Xl2oc,*7crc)
L)n A**^ tiw
£> Eh
Q'- FLca ^OK/^^Co.^o-^.t^l 3.,
P«i" P*i-F r 3o7.^73K/o'i-r-2o y /0S - - 2/2. 727 *70* M
, r f^ e 307.2 73^/0* ._ ^0tfly/o* ft
eg
EA
S*P * Sc - S*' - 0.23*333 */o'3- 0.230 H$r*IQ* - O.oo 78S */o~3 **>
= 0.OO728 *"i -^

PROBLEM 2.115
P'
14 in.
2.1U Two tempered-steel bars, each ^6-in, thick, are bonded to a 7 -in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of magnitude P.
Both steels are elastoplastic with E=29 x 106 psi and with yield strengths equal to 100
ksi and 50 ksi, respectively, for the tempered and mild steel.
*2.115 For the composite bar of Prob. 2.111, determine the residual stresses in the
tempered-steel bars ifP is gradually increased from zero to 98 kips and then decreased
back to zero.
Temp***J s+e*^ At * fiXitXa) = 0.7S;»
T*W • A ~~ A, <-Aa - |.7<T m1-
■fotrce 4* y.'ftAJ +4
SOLUTION
/
ft, = A6^ - (l.7SX5"0x|Cf) = 27.£O*l0* J\a.
Ua+ P, " 4-orce c*rW««f Wy iw.'/i sree*
?a. * Wee o*rritJ ty +e**p*/€eJ sfe*^
P + ?* " P fa * P"1^ r q**'°*- ■To WO* = ^S*/os A
6- ^ S, - ^WQ'
1 A, " 0.75-
6f*/o psi'

PROBLEM 2.116
14 in.
jgto.
2.111 Two tempered-steel bars, each -^ -in. thick, are bonded to a 7 -in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of magnitude P.
Both steels are elastoplasiic with E= 29 x l06psi and with yield strengths equal to 100
ksi and 50 ksi, respeciively, for the tempered and mild slcel.
*2.II6 For the composite bar in Prob. 2.111, determine the residual stresses in the
tempered-steel bars if P is gradually increased from zero until the deformation of the
bar reaches a maximum value S„ = 0.0 H in. and is then decreased back to zero.
SOLUTION
Fo^ He *M sfeei* A, r (^X^ - LOO \»x
Fo^ fAe tempered s-feeJ* A2 = £(j|)(> ^ " 0.75" .'«'
23 y [O1
To+^i* a^€A :
A = A. + A, = 1.75-/1
Sri < §m < Sn The. kM sfeey yields - Tempered s^eeJl is e&sf<c.
Fences P, " A, 6^ -(/.oo^o>/o3) - 52>*/o3 j%
S+
resses
6T. -
- £ -
A,
Y\
So x /o j>«> 1'
3 A i.7r
~ (8.78 ksi1

PRORI fm 7 2.117 A uniform steel rod of cross-sectional area A is attached to rigid supports and
is unstressed at a temperature of 8°C. The steel is assumed to be elastoplastic with <%
= 250MPaandG = 200GPa. Knowing that a= 11,7 x lOVC, determine the stress
in the bar (a) when the temperature is raised to 165°C, (b) after the temperature has
returned to8°C.
SOLUTION
.acv^e "fo cause yie.-r-ciincj
(AT),-- f^--^ ^r/D^ rr = loC.*3g"C
Boi AT =- 165" - 2 - /5"7 ' <2
C^o-P.n^ (AT)' = iSl°C
s' - s; + s; = - £| + l*ut)' , 0
6-' - -£ - - E«(AT)'
A
= - (SOOWO'X IL7W£?*fc)( IS-?") « - 367.38 */Oc p*
- H7.4 MP& —

PROBLEM 2.118
2.118 A narrow bar of aluminum is bonded to the side of a thick steel plate as
shown. Initially, at 7", = 20°C, all stresses are zero. Knowing that the temperature
will be slowly raised to T2 and then reduced to Tu determine (a) the highest
temperature T2 that does not result in residual stresses, (b) the temperature T2 that
will result in a residual stress in the aluminum equal to 100 MPa. Assume a, ~ 23.6
x \W°C for the aluminum and a, = 11.7 * lO-VC for the steel, Further assume that
the aluminum is elastoplastic, with E = 70 Gpa and ^ = 100 MPa. (Hint: Neglect
the small stresses in the plate.)
SOLUTION
Dei eirumVi* iet~>ve**&j-u\r* cliAH^e -fo cause yieJUiV**
(_AT 1 = ^ _ '°° */g*
U0.OTC
A"/^"eV yi'eJUin*
= %^ +. LotfAT) =r Lc6(AT)
CooJP.Vi
3
T(\e imivfcr*/ STress «*3
2 6^ (*KlOO*|Ofc>>
At-
ECot^-cU ) ' (70 *IO* )CZ3;tf- II-1 X'©c)
& 290- »°d
U>)
7\ = 17 + Al * 20 + JW.T
*£0. I, w C
If 71 > 2£0-4 *C ^ file aJtsmtnom L<f wiJJ rnost A>«^j

PROBLEM 2.U9
A = 0.70 in2 A= 1.0 in2
2.119 The steel rod ABC is attached to rigid supports and is unstressed at a
temperature of 38°F. The steel is assumed elastoplastic, with a? = 36 ksi and £ = 29
x 106 psi. The temperature of both portions of the rod is then raised to 250 °F,
Knowing that a= 6.5 x 10"6/°F, determine (a) the stress in portion AC, (b) the
deflection of point C.
SOLUTION
D ete^ •*"«? AT -h> cause y ieJJtl*\4 i'w AC .
- 7^ " Jr5ft + L„ ot(AT) = o
E Aac E A*a
A+ yiei^
"**
(AT) =
Las E.cC-
Ac*
)
Ac^V
(AT), = A*tGV
UoE cL
U +U\ _Lo.7oYtt*lo%} (7 j^x
= 15^.735 °F
Acf*J AT - zso - 38 = m°F > (at), „•. yrcJJ:^ ^c^.
P = SyA** (36x/o3Xo.7o}* ^.^/O3!
'Ac.
Sc =
PL
EAc
? - L^otfAr)
O.OIXI75 - O.o'^a?:*.
= ~Q.QQ1\\G ■"*
Sc" 0.007 iz i\

PROBLEM 2.120
A-0.TOin* A = 1.0m2
2.229 The steel rod ABC is attached to rigid supports and is unstressed at a
temperature of38°F. The steel is assumed elastoplastic, with a^ = 36 ksi and E = 29
x 106 psi. The temperature of both portions of the rod is then raised to 250 °F,
Knowing that a= 6.5 x 10"6/°F, determine (a) the stress in portion AC, (b) the
deflection of point C.
*2.220 Solve Prob. 2.119, assuming that the temperature of the rod is raised to 250
°F and then returned to 38 °F.
P
SOLUTION
(.C owaTfA'/i' )
Dc-fe^^'^* AT "T"'3 Cfl^se yi'^cjina im AC .
- £±£ - Lks. + LA6o/(AT) = o
E Aa^ E Acg
2- + il
.7o i.o
UbEoL \ A*_ Ac*|" (XlX^^o'XC^^o"* ) ^°-
tAT)y - LQ.o&Vi x/o~3 Vw. 2.*io*) = /«.7&T ° F
Aotu^ AT - ZSO-38 = 212. °F > (4r)y A y**cAJr«j oau/s.
EAcs
- O.Ol2i7£ - 0.oi?;W - - 0.007MC in
GdO Vi n
*
3
P' -
AT = 212 "F
AT'
p' = ^T
*K
&.062?>r|O
= 3
^.o^ZI xio
■=T
212
^3 - 3V. 7£7*/o3 A.
(cO Resi^oJ S+v«ss in AC
- 13.95* Usi
- - O.OIC38I +■ O.OIW2 = O-00BII .'*
Sep = Sc + St - -O. OOIIIC + a002l|| - - 0.0O47I .'*

PROBLEM 2.121
i
AX>
1
T
2,121 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5*6-
mm rectangular cross section and made of a mild steel that is assumed to be
elastoplastic with E = 200 GPa and ^ = 250 MPa. The magnitude of the force Q
applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m,
determine (a) the value of the normal stress in each link, (b) the maximum deflection
of points.
SOLUTION
5i-«.4-.cs : FMC * o O.6«to(<a-P«)- 2.W PAO = o
De-FoT^-fVo- : %A = 2.W &j §6 - <l& - ac*o B
G*Ao r Sa. =• 310.6 xlo1 6
^ =■ Be ^ us «/o* e
-ro»v>
ofeuT^C:
Q = Pee + £& p« = p«+ q-ns p'
AS
QY r ( 317.Ofe */oM(*o¥.S?*/0"4) - #55.-2 *loa bJ
Siftte ds ^GO //o3 > Q, j J».*|f AD yleJUa. 6"„D =' 2S& MP*
P«e - 27,97 */0*W
Yv,

PROBLEM 2.122
2.121 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5x6-
mm rectangular cross section and made of a mild steel that is assumed to be
elastoplastic with E = 200 GPa and tfv = 250 MPa. The magnitude of the force Q
applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m,
determine (a) the value of the normal stress in each link, (b) the maximum deflection
of point B.
2.122 Solve Prob. 2.121, knowing that a = 1.76 m and that the magnitude of the
force Q applied at B is gradually increased from zero to 135 kN.
SOLUTION
E.Pa*h'c A n * JL s t*s
A = (.S7.S)(t ) = 2*S**S* - TZSvicT* m*
i-Al> I * 7
- 79.2*10* d 6-ge = S& r 352 */o1 0
A
Fro^ S+*,fic$ Q.5 P*e+ ^ Pap = Pa? + »-SacCPAi>
QY = (|78.£Z*io4 ^/o.^*^-4) = Utf.»6v/o3 N
§a - 1.74 S * /. 322 *lo's *
1.3?* •*»*%

PROBLEM 2.123
2.121 The rigid bar ABC is supported by two links, AD and BE, of uniform 37,5*6-
tnm rectangular cross section and made of a mild steel (hat >is assumed to be
elastoplastic with E = 200 GPa and <% = 250 MPa. The magnitude of the force Q
applied at B is gradually increased from zero to 260 kN. Knowing (hal a = 0.640 m,
deiermine (a) the value of the nonnal stress in each link, (b) the maximum deflection
of points. &"iU
*2.123 Solve Prob. 2.121, assuming that the magnitude of the force Q applied at B is
gradually increased from zero to 260 kN and then decreased back to zero. Knowing
that a = 0.640 m, determine (a) the residual stress in each link, (b) the final deflection
of point B the residual stress in each link. Assume that the links are braced so that
they can carry compressive forces without buckling.
SOLUTION
See SoPoh'o* -k> PRoftLHM S./ai -Poir He no^J s4reiS€S »* z*,c\ Jl\*lr
ion ot 0ot>*
Tke «AvS + i'c OLV\cJys(s Qive* i« f-U so^+i'o^ -k* PROBLEM 2./*| cLfpJ;**
Q. = 3/7.06*/o* 9'
. Q.
^ 317.06 WO6 317,06^/0* o*^.*^
Sao * 3lo.L*\0* 6 - {3lOX*(oei)(Slo.oZ*iQe')* ZS^.lo y lo* ?<x
6^
-t
JAD.rei ~ VAD ^AO
GbEjI^j - £>ee &ee
^ -M.7o Mfc^
U4.Sv/oe- I Of. ^v/o'
l?.3«t MP*-
(lo) SB p * §a - 5j
( „
-c

PROBLEM 2.124
2Hkips
I
12 in.
18 in.
1.5 m.
-2.25 in.
§ =
PL
EA
€T= -?■
P
A
28 kips
1
2.124 The aluminum rod ABCiE = 10.1 * l06psi), which consists of two cylindrical
portions AB and BC, is to be replaced with a cylindrical steel rod DE (E = 29 * 106
psi) of the same overall length. Determine the minimum required diameter d of the
steel rod if its vertical deformation is not to exceed the deformation of the aluminum
rod under the same load and if the allowable stress in the steel rod is not to exceed 24
ksi.
SOLUTION
PL*.
i. -
h^B
' E
j-AS
A*
3/376 ih
Sfe«J> ro*P
A =
S - 0-03/376
PL _ (71
E S "" (^7V/o6)(0.03/376)
- 0.72317 ;**"
A = -£ -
- 1.1467 (W
X4x 101

PROBLEM 2.125
36 mm 28 mm
2.125 A 250-mm-long aluminum tube (E = 70 GPa) of 36-mm outer diameier and 28-
inm inner diameier may be closed at both ends by means of single-threaded screw-on
covers of 1,5-mm piich. With one cover screwed on light, a solid brass rod (E = 105
GPa) of 25-nim diameter is placed inside the tube and (he second cover is screwed on.
E^] Since the rod is slighily longer than the tube, it is observed that the cover must be
forced against the rod by rotaiing it one-quarter of a turn hefore it can be tightly
closed. Determine (a) the average normal stress in the tube and in the rod, (b) the
deformations of the lube and of the rod.
a
SOLUTION
E^A^/
5HWRf*(o" P
§* = ■£ fur* * 1.5*
M»t <-
37^* JO**1 ^
0.375~ mi -
8.88 IS" xicr* f3 + q.$SO£*lcT*P = "375* x/O"*
P =
O.S7i~y(o"
.-.a
tMS/r + H.as&rXio""1)
(^ $Wc
^W-
. , .E_ r J?7>3^a
,. P _ __ 27.3o3*|Q3
Av*J
4<?O.8?x/0-
81 17,30$ *IO M
r €7.? *lO*P*. =■ 6 7.1 MPa
mm
tv»M

PROBLEM 2.126
28 mm
2.125 A 250-mm-long aluminum tube (E = 70 GPa) of 36-mm outer diameter and 28-
mm inner diameter may be closed at both ends by means of single-threaded screw-on
covers of 1.5-mm pitch. With one cover screwed on tight, a solid brass rod (£ = 105
GPa) of 25-mm diameter is placed inside the tube and the second cover is screwed on.
Since the rod is slightly longer than the tube, it is observed that the cover must be
forced against the rod by rotating it one-quarter of a turn before it can be tightly
closed. Determine (a) the average normal stress in the tube and in the rod, (b) the
deformations of the tube and of the rod.
2.126 In Prob. 2.125, determine the average normal stress in the tube and the rod,
assuming that the temperature was 15 ° C when the nuts were snugly fitted and that the
final temperature is 55° C. (For aluminum, a = 23.6 * 106/°C; for brass, a = 20.9 x
SOLUTION 107°C)
AT ' 5S ~ IS *■ *> -c
.'7
-6
S^ - -^~T + Lo(AT) * - - ?C°'^ ^ + (o,2Co)(*ow6<)CHo)
$* * ^■Ai^ «■/-5"**i * Q.21S »m - 375" ^/o"6 *n
13,73**10^ P - 3¥3xfo"C p = 25, 34^ */o3 W
^+Ae
^-£r-^^—-&^p-- -*■***•

PROBLEM 2.127
y
14 in.
(U
ic)
2.127 The block shown is made of a magnesium alloy for which E = 6,5 x 10 psi and
v = 0,35, Knowing that ax = -20 ksi, determine (a) the magnitude of ay for which the
change in the height of the block will be zero, (b) the corresponding change in the area
of the face ABCD, (c) the corresponding change in the volume of the block,
SOLUTION
Sy -°
Sf -- f { S-j - vC^ - o
(ft,) Sy - v<5^ = (0.3S"Y-20x/fis)
=■ -7x/o3(5»; = - 7 k*i
p- . j-/e ve ^ - -3£>*i°* -fas**-7 */«=*;
- - 3.7 */o~3
A.+ A* * Ly(i+£y)i,(t + £^-- LXL2 ( i + f„ +-£,. + tO
= -H.^S */£>
-3
l»t
- fr.OOWt \r\%
Sr
St<
L
IS C^*v*
,-k.vf
A<2/* i.;(AA W G^rtf-Mawo-1} r -t.&s^io
-3 . J
= -0.0O€S5" »V

PROBLEM 2.128
I- 45 in.—
O.oHHIC
B'
.B
2.128 The uniform rods AB and BC are made of steel and are loaded as shown.
Knowing that E = 29 x 106 psi, determine the magnitude and direction of the
deflection of points when #=22°.
SOLUTION
O.oo£>73 _
"^ 5 = "b.04H<?6 "" 0-^16 Cj> " 2.5"/ 3^*
S - -/ ft; (WJG * + 0.00673 * - 0.0^55" in.
PROBLEM 2.129
45 in. •
B
25 kips
m
Area = 0.8 in2
-Area = 1,2 m2
25 in.
I
f!7.>Wi? S
2.129 Knowing that E = 29 * 106 psi, determine (a) the value of #for which the
deflection of point Sis down and to the left along a line forming an angle of 36° with
the horizontal, (b) the corresponding magnitude of the deflection of B.
SOLUTION
SBc_ = S cos 36"
p - E A gc Sac _
JAB
p« *
e-A*c5
Ac o»*t
•3 ^
*»»- ;;;;%£ l - ■•*" e=63-°
F= 25*|0* ="V(4I7-0*»*IO3 %Y ±{$\%„7.o*lO-% % ^ = «N8.3a*/C?S
S =
= 0.0273 m.

PROBLEM 2.130
Ei»,
t>Vkl
DeM«e+
2.130 The uniform wire ABC, of unstretched length 21, is attached to the supports
shown and a vertical load P is applied at the midpoint B. Denoting by A the cross-
sectional area of the wire and by E the modulus of elasticity, show that, for d« I, the
deflection at the midpoint B is
T
6=l\
SOLUTION
Use aAp^o^ti^A>r»x>uv
Sr*TlC:
2Fy--o
K6m0
PI
SI
iv. © - P = o
2S
S*B~ ~A^
Pi
?A£S
Fro^ +Ke ^ 10 lit TW*v\*-/e
* -Ei3
AC
ABS
Pi3
A&
S« 1
7
A£

PROBLEM 2.131
3/UN
ft-**
Zoo
ee
6
75^
AD
2.131 The steel bars BE and AD each have a 6x l8-mm cross section. Knowing that
E = 200 GPa, determine the deflections of points A, B, and C of the rigid bar ABC,
SOLUTION
Use lTfoiel bar A8C 6^ a. £ re*- fcooL
±5^Me=o (7^PAt> -(3oo)(3-0= o
Pad = I*-* feM
&A * ^b ~ rrA ~ CZOOxfo'X/Ogy/cT6)
/*ft " 7Sx/a'3
e =
-s
- 216. So y /o"c+ {Boo X /o"3 )(7.111 x y | o~z 1

PROBLEM 2.132
3.2 kN
e-e
3.4VW
2.131 The steel bars BE and AD each have a 6x 18-mm cross section. Knowing that
E = 200 GPa, determine the deflections of points A, By and C of the rigid bar ABC.
2.132 In Prob, 2.131, the 3.2-kN force caused point C to deflect to the right. Using a
=11,7 x 10 /°C, determine the (a) the overall change in temperature that causes point
C to return to its original position, (b) the corresponding total deflection of points A
and B.
SOLUTION
Use rfg»c) ABC «ls o. ■£*«« Wody
-PA* + 3.3. 4- PAo - o
Pec = IC UU
Soo
Sa - S.
Pa. L,
. 1 A» *-A»
'AD
-f
7$
EA
+ LtocK*n
Ao
= 237. 04*/cfc 4 4.£8*/0"d(AT)
- (237.04 * 10""% <f,'CB.*iO*GAr'> -(l.a5')C^6.30 ^/d& +• 4-6& WO^O]
-C
-C
= -57.7 *£
Sa * 0.03**!*^ —

PROBLEM 2.133
SOLUTION
2.133 A hole is to be drilled in the plate at A. The diameters of the bits available to
drill the hole range from Tfco 27 mm inG-mm increments, (a) Determine the diameter
d of the largest bit that can be used if the allowable load at the hole is not to exceed
that at the fillets, (b) If the allowable stress in the plate is 145 MPa, what is the
corresponding allowable load P?
^^1, d 12mm
112,5 mm
T ~ <"f »i*n A= IS Mi
2. le>
r
A " T>-Zr
A+ +U M« : hu = (V- Zr)tj
wliw« *£) - H2.S ^^ Y" = r^dt'da of ZsirtJe
t"-)2
****.
Wo& Jo«~
T mni
IS M»V|
2' H*
a 7 t**v\
r
4.5" hw»
7.-S"^s
(0.3*^
I3.S^k
d=T>-2*
103,5"^^
^.S'*.^
^\.^ mm
25*, $ *««*
r/d
OX>t35
0.077
0. 115
OJSS
K
J?. 87
2.1S
7X1
Z.S1
Mn«t
i^wd** *?
1170^^
KmxioV1
1026*10*^
R*
£2.7 Wo3 M
£I.7W<>3 W
i'lU/rf N
57.1* lo6 N
AW We -P

PROBLEM 2.134
SOLUTION
2.134 (a) For P ~ 58 kN and d - 12 mm, determine the maximum stress in the plate
shown, (b) Solve part a, assuming that the hole at A is not drilled.
d 12 mm
112.5 mm
Use Fig- ?. £4 Cl fa*" v&Joes o^ K
ITTF^ - °'^7,
K* a.*o
re . *.
A^f = 0^ Kills'- U"l - (2^6 ^- - IS?o£x|0~ to
l3se F;3. 2%tf b
£ . ± -
0.12
7^ ,>
£ X.
K - Z*\o
(&."> W;H f,eJ?e ^J PJA+s
<W ^ 134.7 MP*
<SL* ' 135. S MP^

PROBLEM 2.135
2 in.
4 in.
2 in.
3 in.
2 in.
3 in.
So */o"
SOLUTION
£^y - 0.0O2S * Sr
2.135 The steel tensile specimen ABCD(E = 29 x I06psi and ay - 50 ksi) is loaded
in tension until the maximum tensile strain is e = 0.0025. (a) Neglecting the effect
of the fillets on the change in length of the specimen, detennine the resulting overall
length AD of the specimen after the load is removed, (b) Following the removal of
the load in part (b), a compressive load is applied until the maximum compressive
strain is e = 0.0020. Detennine the resulting overall length AD after the load is
removed.
- O, OO 172'4
YieJU
S"sc - SY ~ So *ic? p^>;
£&c ' £***, ~ fr * 0.0O2S" - O.001724 - 0.0oO776
>»c
Lftc £e = (4 Vo. 000776 s) " (X 0o3lO in.
(O
S
1
-o.oo-ko
E
6V
0
/
D
/
//
/'
-5,
/
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Be
£ = - O.OO^o
CT-5 aiVctt. DoiTima f*€,*MOi/aJ' oT "T"««
,'s Gr/E * O.OOI7ZH*
Tnd per »vns. •■» e^T s-J-^Ai'rt iki HC fi
£gc =■ - O.OOZo + O.QOMZH- O. 0OOX1C
- 0+)(-O.oooZ7£,) - - O. O0|lO*f ,"w. -
Notc +Lf p»r-tio«a A8 <a.**l CD
an
y pc^w^ertt e/e? ©/"t-vtvfi on.

2.C1 A rod consisting of n elements, each of which is homogeneous and
of uniform cross section, is subjected to the loading shown. The length of el-
j ement (' is denoted by L{, its cross-sectional area by Ah its modulus of
elasticity by Ej. and the load applied to its right end by P„ the magnitude Pt of this
Pt load being assumed to be positive if P, is directed to the right and negative
otherwise, (a) Write a computer program that can be used to determine the
average normal stress in each element, the deformation of each element, and the
total deformation of the rod. (b) Use this program to solve Probs. 2.17 and
2.18.
SOLUTION
FOR £40/-] FLEMEVT, E-^tE£
COMPUTE DEFORMATION
UPPAT& AXIhL LOAb P = P + P{
GOtfPVTE FOR EACH ELEMENT
T0TS\ L DEFORMATION:
UPPf\T£ THROUGH n ELEMENTS
PWGRfiM OUTPUT
Problem 2.17
Element Stress (MPa) Deformation (mm)
1 19.0986 .1091
2 -12.7324 -.0909
Total Deformation = .0182 mm
Problem 2.18
Element Stress (MPa) Deformation (mm]
1 98.2438 2.3391
2 98.2438 1.4737
3 147.3657 1.4737
Total Deformation = 5.2865 mm

2.C2 Rod AB is horizontal with both ends fixed; il consists of n
elements, each of which is homogeneous and of uniform cross section, and is
subjected to the loading shown. The length of element /' is denoted by L{, its cross-
sectional area by A,, its modulus of elaslicity by £,. and the load applied to its
right end by P^. the magnitude P{ of this load being assumed to be positive if
P; is directed to the righl and negative otherwise. (Note that P, = 0.) (a) Write
a computer program that can be used to determine the reactions at A and B,
the average normal stress in each element, and the deformation of each
element, (b) Use this program to solve Prob. 2.41.
SOLUTION
Wf* CONSIDER THB REbCTlOti AT B REDUNMS/T
AND RELEASE THE ROD AT b
COMPUTE fB WITH ff6 = 0
FOR EfiCH 'ELEMEN7} ENT£%
Li j &i ) *i
UPVfiTt A*\hL L0A»
P = PtPi
COMPUTE FOR BACH ELEMENT
UPOATB TOTAL DEFORM^TlOKf
COMPUTE fa DUB TO UNIT LOAO AT B
(j/ViTu-( = l/fic
Uf*D/\76. TOTfiL UN\T DBFQKMATioN
UNIT fB - UNIT fB +> UNIT ft'
$UP£RPQ<>\T\0N
FOR TOTAL DISPLACEMENT AT B = 2Ef>0
SOLVING?
CONTINUED

PROBLEM 2.C2 CONTINUED
FOR FAOI ELEMENT
(T =' 0". t Kg unit 07
PROGRAM OU7PU7
Problem 2.41
RA = -11.909 kips
RB = -20.091 kips
Element Stress (ksi) Deformation (in.
1
2
3
12.002
-6.128
-9.687
-.00923
-.00589
-.00334

2.C3 Rod AB consists of/? elements, each of which is homogeneous and
of uniform cross section. End A is fixed, while initially there is a gap S0
between end B and ihe fixed vertical surface on the right. The length of element
i is denoted by Lt, its cross-sectional area by Ah its modulus of elasticity by
Eh and its coefficient of thermal expansion by a,-. After the temperature of the
rod has been increased by AT", (he gap at B is closed and the vertical surfaces
exert equal and opposite forces on the rod. (a) Wriie a computer program that
can be used to determine the magnilude of ihe reactions at A and B, the
normal stress in each element, and the deformation of each element, (b) Use this
program to solve Probs. 2.53. 2.54. 2.57. and 2.59.
SOLUTION
WE COMPUTE THE VHPLAceMEMTS ftT B
A-SSUMIN6 THERE tS NO SUPPORT At 3.*
Eh/TER Li } fii , E{ , Qt f
£NtBR T&MPFtfATL/tfH CHANCE T
COMPUTB For BACH ELBMBHT
Si '- *i Li T
UPDATE TOT/\L PBFOPMATiOti
COMPUTB <fs DU£_ TO UNIT LOAO f\T B
Uhln f{ = Li/AiEt
DPQfSTB TOTAt UN\T DSFORMArioy
UNIT CB * ^W1T fB + ^NlT if
CO!V)PUTE RgACTIOAJS
FROM 5UP£PP0$iTt0N
CONTINUED

PR06PF[M OUTPUT
Problem 2.53
R = 25.837 kips
Element Stress (ksi] Deform. (10*-3 in.)
-21.054
-6.498
-3.642
3.642
Problem 2.54
R = 125.628 kN
Element Stress (MPa;
Deform.(microm]
1 -44.432
2 -99.972
Problem 2.57
R = 217.465 kN
500.104
-500.104
Element Stress (MPa] Deform.(microm)
-144.977
-120.814
242.504
257.496
Problem 2.59
R = 61.857 kips
Element Stress (ksi) Deform. (10*-3 in.]
-22.092
-51.547
14.410
5.590

PROBLEM 2.C4
A,. E,, (oy).
A2. £2. (°"v)2 Plate
0*n
2.C4 Bar .45 has a length L and is made of two different materials of
given cross-sectional area, modulus of elasticity, and yield strength. The bar is
subjected as shown to a load P which is gradually increased from zero until
the deformation of the bar has reached a maximum value Sm and then decreased
back to zero, (a) Write a compuier program that, for each of 25 values of 8m
equally spaced over a range extending from 0 to a value equal to 120% of the
deformation causing both materials to yield, can be used to determine the
maximum value Pm of the load, the maximum normal slress in each material, the
permanent deformation Sp of the bar, and the residual stress in each material.
(b) Use this program to solve Probs. 2.109, 2.111, and 2.112.
SOLUTION
fJOTB ; THE FOLIO\N\N6 ASSUMES (<Tr)f < (^
Q\<>PLACEMBNT INCREMENT
Ql$PL&CEM£rt7S AT YlBLOING
^-to.L/e, fB*t<ry)2L/e2
FOR EAthl DISPLACEMENT
if f„ < fA :
<r, -- &n Vu
IF fA * fm < Si ',
<r, = K>,
IF fm> £~B '
(Tt -fay), CT2 =(Ty)z
?m= A,^ f f\2<Tz
PERM/\tJ£NT DEFORMAVOflS , RFSipUHL STRESS
5LOPE OF Fl£S7 (ELASTIC) SEGMENT
SLOPS = (A,E, + fi-,_Ez )/L
Sp- f„- (P^l-OPE )
(<r,)hts -r,-(^Pm/(LSL0?e))
K)nr-^'(E^l(L$Lm))
CONTINUED

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2.C5 The stress concentration factor for a flat bar with a centric hole
under axial loading can be expressed as:
K. 3.00 -,, 3(|). 3.66(fJ-,. 53(0
where r is the radius of the hole and D is the width of the bar. (a) Write a
computer program that can be used to determine the allowable load P for given
values of /; D, the thickness / of the har. and the allowable stress cr.ji of the
material, lb) Use lliis program 10 solve Prob. 2.94.
SOLUTION
BUTE R
COMPUTE K
RD = 2.0 r/D
K - 3.00 -3.13 RDi- 3.66 Rp1-|.r3 KJ)3
COMPUTE AVERAGE 5TRESS
AUOWADlE LOAD
PROGRAM OUTPUT
Problem 2.94
Hole at A:
K = 2.573 P = 7.773 Kips
Hole at B:
K = 2.159 P = 5.559 Kips

2.C6 A solid truncated cone is subjected to an axial force P as
.shown.Wrile a compurer program that can be used to obtain an
approximation of the elongation of (he cone by replacing ii by n circular cylinders of
equal thickness and ot radius equal to the mean radius of the portion of cone
they replace. Knowing (hat (he exact value of the elongation of the cone is
(PL)/(2 ttc-E) and using for P. L, c. and £ values of your choice, determine
(he percentage error involved when the program is used with {a)
n -6, (ft) n = I2.(c)/i = 60.
SOLUTION
C
FOP, L * / TO n :
Lt - (I fay)(L/n)
rL = 2c-c di/L)
AREA;
A = rr r*
DISPLACEMENT :
£XAC7 DISPLACEMENT:
T - PL/(?.0Trc2E)
0 £XAcT /
PERCENTAGE ERROR;
PERCENT^ lO0(S-SBth^)/Si
EXACT
PROGRAM OUTPUT
n
€
12
60
Approximate
0.15852
0.15B99
0.15915
Exact
0.15915
0.15915
0.15915
Percent
-.40083
-.10100
-.00405

CHAPTER 3

PROBLEM 3.1
18 mm
3.1 Determine the torque T which causes a maximum shearing stress of 70 MPa m
the steel cylindrical shaft shown.
SOLUTION
T' - Tc
J. \t
T » £cV = ?(o.»«)*(7o*lo4) r G4/ ^.w
PROBLEM 3.2
18 mm
3.2 Determine the maximum shearing stress caused by a torque of magnitude T-
800 N-m.
SOLUTION
X
_-^ ^%cH
fc7.3 MPa
PROBLEM 3.3
3.3 Knowing that the internal diameter of the hollow shaft shown is d = 0.9 in.,
determine the maximum shearing stress caused by a torque of magnitude T= 9 kip-in.
SOLUTION
C- 0.8 in.
r.
J" 0..S7?O '** K5
PROBLEM 3.4
1.6 in.
3.4 Knowing that d = 1.2 in., determine the torque T which causes a maximum
shearing stress of 7.5 ksi in the hollow shaft shown.
SOLUTION
J = *(cJ-c/)*f(0.»'-flLO = 0.4398 «*

PROBLEM 3.5
24011 N ■ u
3.S {a) For the hollow shaft and loading shown, determine the maximum shearing
stress, (b) Determine the diameter of a solid shaft for which the maximum shearing
stress is the same as in part a,
SOLUTION
Cz -k^l'{i)(0-0^0) ~ CX030 ^ C - 0.030 m
- 1.0210 »*m"4 ***
crt t;
'***
Te. . (^400^0.03) _ nrs c. v._ 4 p
(W
r-7^
sT
7T ^ ¥
70JT MPcl
TTC.*
-3
C, - 27.8S*K> r»
d3 = 2C3* $5*.o»*io'\*
5S.2
htm
PROBLEM 3 6 ^*** ^ Determine the torque which may be applied to a solid shaft of 90-mm outer
diameter without exceeding an allowable shearing stress of 75 MPa. (6) Solve part a,
assuming that the solid shaft is replaced by a hollow shaft of the same mass and of 90-
SOLUTION mm inner diame(cr.
(ol) For- +lic so A J sU-H c = i J - CfKo.o?©N( = cot* *
^= I& ■.'. T= %I; (7ff»«|0ft,J(W3.|'(fxlcrt)= 110.7^ */0s A/-w>
»J C
I0.7W M- ki -*i
For e&Oa-P mA55«s +^e cv*oss $ecTioiA*L<t ci^eAS w»o4+ be «4</**f
J = X (c^ . C|">) = |9.3^37 *J0« «*
T ^ &J" = (7S)f|o4l(irSa37v/Q-c) . ;?;i.77*/o3 N.m
c,
o.ocscsit
25.8 kN-^

3.7 (a) For the 3-in.-diameter solid cylinder and loading shown, determine the
maximum shearing stress, (b) Determine the inner diameter of the hollow cylinder, of
4-in. outer diameter, for which the maximum stress is the same as in part a.
SOLUTION
Co,") sJ.-j sUH c. - iai ^ i(xo) - j.s* ;».
C»* JT 7TC3 7T(|.Os '*^5 Wt' ^
C, = \.mt1S i»
PROBLEM 3.8
SOLUTION
3.8 (a) Determine the torque which may be applied to a solid shaft of 0.75-in.
diameter without exceeding an allowable shearing stress of 10 ksi. (b) Solve part a,
assuming that the solid shaft has been replaced by a hollow shaft of the same cross-
sectional area and with an inner diameter eqnal to half its outer diameter..
(a) SoiiA skejt: c * i ^(i)Lo.!S)^ 0.3ZTi'*.
J-r Jc4 " |(o.S7S"V = 0.03/ 0G3 J** r^* >0 Ws."
T. ^J^JM, 0.8*8 k.p,lH or 8*8 JU*
(b) HoPb* sUSf
C,'ic - ^(o.37y) = <3. 433013 i*

PROBLEM 3.9
3.9 The torques shown are exerted on pulleys A and B. Knowing that each shaft is
solid, determine the maximum shearing stress (a) m shaft AB. (b) in shaft BC.
SOLUTION
J
ttc* -n (a.ois*)'*
- Se>.£$8*lQ6 Pol -- ^£.5 A^PfiL
SUW- 8C Tic= 300 +■ 4oo - 7oo AA*»
- 3C.C26*/o' Pa,
36.6 MPac
PROBLEM 3.10
3,10 The torques shown are exerted on pulleys A and B which are attached to solid
circular shafts AB and BC. In order to reduce the total mass of the assembly,
determine the smallest diameter of shaft BC for which the largest shearing stress in the
assembly is not increased.
SOLUTION
5UcM- AB:
'«
= 3oo Al-M d= 0.o3o*i c = 0.otf**o
dl - O.0H6*>, c* 0-33h
- 56.S8«x/ot Pet - 5S.6 MPs.
v . Tc . HT aXloo)
- 36.626 x/O4 "Pa = 3fc,6 MP«*
TAc -Pay****) STVC33 (j*6*S88*/0* PO acctlfi* m por-Ko^ A-g
-3
C - l9.S?^x/o ^
-3
<t = 2C •' 39.79 */0 *n
3?.« »*

PROBLEM 3.11
.500 Il> in
Mi) Hi
'dar = 0.75 in.
d.B = 0.6 in.
3.11 Knowing that each portion of the shaft AD consists of a solid circular rod,
determine (a) the portion of the shaft in which the maximum shearing stress occurs,
(b) the magnitude of thai stress.
SOLUTION
Sfc*Ft A0 : T = Moo K- in
C= id = 0.3O i*
X * Is* , -21
S^aFt 8C : T =-400 + IZOQ = SOO Jb./M
irn = 0.9 in.
c = *d - o.w .. 2^ = Jfc - |E ,fg|^ - w,r:
Skftft Cti- T« -4oo + i2oo +>5"ot? = I30O AM'»
■k«,*y
A^
>veKS
M sk*?+ Be
9o8<?ps
PROBLEM 3.12
500 Iti ■ ill
.100 iU ■ it:.
'rfDr = 0.75in.
3.12 Knowing that a 0.30-in.-diameter hole has been drilled through each portion of
shaft AD, determine (a) the portion of the shaft in which the maximum shearing stress
occurs, (b) the magnitude of that stress.
SOLUTION
Hote- c, = -jd, - Q.IS m
Shaft A8- T- Moo A-iY,
J--f(q1-cI1)=5(D.3o,'-0./r,|)
rfcD = 0.9 m
d.n = 0.6 in.
J= S&'-c/')* £(a37<r*-o./sM) r o.o3oa«;-"
/»
Ar\SK/*AS: (ct) iKft-pT A8
(b) 10.06 ks,'

PROBLEM 3.13
3.13 Under normal operating conditions, the electric motor exerts a torque of 2.4
kN-m at A. Knowing that each shaft is solid, determine the maximum shearing stress
(a) in shaft AB, (b) in shaft BC, (c) in shaft CD.
At. _ /Tr; - O.S kN ■ in
46 mm / '
Tj_, = 0.4k\.m
40 mm
SOLUTION
SWPt AB : Tag, - ^y/b5 N-*, c = i«J = 0.027 v^
r- « ? - ^5 - Sg^ - ™*«> P. 77.. HP. -
SU-H BC* 7^. = 2.w left* - \.x kV"* - /.3 fc^-*^ c^-^d * o*oz$*
**' ?= ^ * TOl^fi'3 = ft.7»*/o*fk «.«Mfc-
•tri
<^~±<A - 0.023 v*
c*> J- 7TCS it (0.033 y* -w-o /» ra.
^o.f MBt
PROBLEM 3.14
A/¥ Under normal operating conditions, the electric motor exerts a torque of 2.4
kN-m at A. In order to reduce the mass of the assembly, determine the smallest
diameter of shaft BC for which the largest shearing stress in the assembly is not
increased.
SOLUTION
UV*€ CX<^A TCW (vi^Vl *^0**^ She^^Vl
S+^-esses 'iw p<w«Kc>»is AB5 BC^ aid £T> <s$ H« sl»a.++. The f^+es*
\/edue ts ^L^ = 77.425" */C»&
<£ = Tc
ZT
J TTC
3-2=1 - &^*.a*/os)
c =
unL ' tt( 77.^*5'*/o*)
-%
C =r %\.H*>%\Q m J* ZC' 42.S*/0"*lr*i
4*.3
l*tMt

1
PROBLEM 3.15'
3.15 The allowable stress is 15 ksi in the 1.5-in.-diameter rod AB and 8 ksi in the
1.8-in.-diameter rod BC. Neglecting the eflfect of stress concentrations, determine the
largest torque that may be applied at A.
Steel
Brass
SOLUTION
T= *cVh
PROBLEM 3.16
Brass
3.16 The allowable stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod
BC. Knowing that a torque T= 10 kip-in. is applied at A, determine the required
diameter of (a) rod AB, (b) for BC.
SOLUTION
r- *
s - IT.
7rr>
c =
*s*
£• it-Ay "* IJ KS I
C ~ O.ISISm d = ZC r LSD 3 i*.
strf+ ecu 7- /o fcrp-in tu* - 8 i^i
ainsr? i«
.s. ft fro) _
C = O. 9^6 7 in

PROBLEM 3.17
90 mm
3.17 The solid rod AB has a diameter dAB = 60 mm. The pipe CD has an outer
diameter of 90 mm and a wall thickness of 6 mm. Knowing that both the rod and the
pipe are made of a steel for which the allowable shearing stress is 75 MPa, determine
the largest torque T which may be applied at A.
SOLUTION
T>M = IS *10* ?<x
To, -^
Rod Ae: c = ■£ el * o. 030 *
o--£c
Pipe C01 C2 = j^^O.OHSm C8a C4- t « OjOHS~-0.&>4> ■= o.oS')
J = |(ct*- C,*) - |(o.^£-"- CaS/4) = 2.807S x/o- w"
m
*n
AitPc»H/«tble +e*-|ve ,'s+U SKieiA?^ \;*Aw? (3.1**10? A/-**") 3.J8 kN*m
PROBLEM 3.18
90 mm
A/* The solid rod /IS has a diameter ds = 60 mm and is made of a steel for which
the allowable shearing stress is 85 MPa. The pipe CD has an outer diameter of 90 mm
and a wall thickness of 6 mm; it is made of an aluminum for which the allowable
shearing stress is 54 MPa. Determine the largest torque T which may be applied at A.
SOLUTION
Rod AS : rltfr S5"V/06 Pe, j c-£«J= 0,O3O *n
J= IC^-c/Jr JCo.o^-aoS?*) = 2.8073 *k>* **
h))o»*4ak9e io^o^e. U s^affe^ va.i'o* 77/* r 3.S^^o W*i*
3.37 (fW**» «^

\
PROBLEM 3.19 /
Aluminum
3.19 The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum
rodfiC. Knowing that a torque T= 1250 N-m is applied at ^.determine the required
diameter of (a) rod AB, (b) for BC.
SOLUTION
<?o J At3 : cs = g? <1*?L =■ ,5_ 7*5- * /o- m
-3
C- 25". IS"*|C? m = 2S".lS"*v,^
TT (ZiTx/O*)
C = 3/.C9X/o" fn - 3/.Cq m
t^j
eL* 2d * GS.H
!*»*>
PROBLEM 3.20
Aluminum
3..20 The w//d rod SC has a diameter of 30 mm and is made an aluminum for which
the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter
of 25 mm; it is made a brass for which the allowable shearing stress is 50 MPa.
Determine (a) the largest inner diameter of rod AB for which the factor of safety is the
same for each rod, (A) the largest torque that may be applied at A.
SOLUTION
KjrH&vio* fix c = i~c|= 0.0iS*
Ho^^ ^J AB : r^ = 50 */Ofc ftu
T^ = 132.53* w-*
c, =
c*.- c/
-R
£, = 7.5? * |o** ^ = 7.5"?
M**>
hiiw+.U* Wf*t 7L* - /32.5 A/-**

PROBLEM 3.21
SOLUTION
Ten - l«X> hi ■*
3.21 A torque of magnitude T= 1000 N-m is applied at Das shown. Knowing that
the diameter of shaft AB is 56 mm and the diameter of shaft CD is 42 mm, determine
the maximum shearing stress in (a) shaft AB, (b) shaft CD.
CO
Aft
40 mm
*- TT - TTC3 - TtoOio)' " &2.7MO
US Mf>*.
G8.7MPa
PROBLEM 3.22
SOLUTION
3.22 A torque of magnitude T= 1000 N*m is applied at Z) as shown, Knowing that
the allowable shearing stress is 60 MPa in each shaft, determine the required diameter
of (a) shaft AB, (b) shaft CD.
40 mm
T»6 r ^ Tco - ^Oooo) :2S0Ohhn
'SU-t+ A8 •■ f** = SO x/os Pa
T "HC* TT 'T It (go XIO*)
-3
C - 2% 22* to - #9.8* w*i
5U-P+ CD: r*i? = ^y/o; Pa
d * ?c - v51.G
kvi*»i
? r
. Tc . 2T
J TTC*
C - 2/.T7 v/o"3™ r ZLH1**
7T^ " TT(C0vi04)
O.Gio */o~4 to*
.-tj* 2C * H3.<* *.*,

PROBLEM 3.23
SOLUTION
3.23 and 3.24 Under normal operating conditions a motor exerts a torque of magnitude
7> = 1200 Ib'in. at F. Knowing that the allowable shearing stress is 10.5 ksi in each shaft,
for the given data, determine the required diameter of (a) shaft CDEy (b) shaft FGH.
3.23 rD = 8 in., rG - 3 in.
c si 7TC3 j
'I*
rar2X
c -irr
(flO SU-T4 CVE
* = g)^00? = O.i^oui^
" tt(|oSo<p)
(fcO Shaft FGH
3 _ <a)0*oo)
C =■ 0.„4l74 ^
^D« = %C t
\.\S* w.
- 0.O1HS1 in3
PROBLEM 3.24
SOLUTION
TF - /20O A-in
3.23 and 3,24 Under normal operating conditions a motor exerts a torque of magnitude
7>= 1200 Ib'in. atF. Knowing that the allowable shearing stress is 10.5 ksi in each shaft,
for the given data, determine the required diameter of (a) shaft CDE, (A) shaft FGH.
3.24 rD = 3 in., rG = 8 in.
r _ Tc ...2T
J ^C3
SU-P+ CDS
c -
IT
<■-$£& >"»'^'
C - 0-30/05" /n
dDff -- 2c = 0.CO2. in
C * 0*417*/ .'*.
<V~ 2c r o.wrm

PROBLEM 3.25
SOLUTION
Sk*f+ F&: c - id r 0.400 m
a ? (aWoftuO * 1.7/77 k.p-ih
3.25 Under normal operating conditions a motor exerts a torque of magnitude Th at
F. The shafts are made of a steel for which the allowable shearing stress is 12 ksi and
have diameters of dCDE = 0.900 in. and dFGH = 0.800 in. Knowing that rD = 6.5 in. and
rc = 4.5 in., determine the largest torque TF which may be exerted at F.
TF^= Mt'-7I77>> " '•'*? k.«.i«
80 m
50 m
PROBLEM 3.26
3.26 The two solid shafts are connected by gears as shown and are made of a steel
for which the allowable shearmg stress is 60 MPa. Knowing that a 600 N-m-torque
Tc is applied at C, deiermine the required diameter of (a) shaft BC, [b) shaft EF.
SOLUTION
SU-H 21' Tc = Goo M-*^ %^- tox/o'fo.
c3 - 2JL rz fcX60°
TT-T TT(6oy|0c)
= C.3CG** 10'6 **
-5
c r 13.53 wo" m - IS^*^ clg =2c-Z7. \ w*n
Shaft EF: Tf - ^Tc = fM«oo} = 375" KJ-*
C J 7TC*
So

PROBLEM 3.27
90
50
3.27 The two solid shafts are connected by gears as shown and are made of a steel
for which the allowable shearing stress is 50 MPa. Knowing that the diameters of the
two shafts are, respectively, dBC = 40 mm and dEF = 32 mm, determine the largest
torque Tc which may be applied at C.
SOLUTION
c ,t(; ShafV Ac: r*.^-5"0><K/P«_, c-j[d = 0.0Z&
fo
T*3 & £ \r-^ ^^{OjQZafCSo^O^
SWf+ DF* r^r^^Pa, c=£J* 0.W6
bPPovcJoPe vJse of t is He SMA/Se^ /.e. TF - 6"/5" AAi*t
m
F/'oho SfedYcs
T = Ik- - so
0.5 ii
PROBLEM 3.28 3.28 In the bevel-gear system shown a = 18.43°. Knowing that the allowable
shearing stress is 8 ksi in each shaft, determine the largest torque T^ which may be
applied at^4.
SOLUTION
5h«f* A: r» 8 ks> c = *d » CUT*
TA " O..U78 k^/n - 1*7.8-/fe-i*

PROBLEM 3.29
3.29 (a) For a given allowable stress, determine the ratio The of the maximum
allowable torque Tand the weight per unit length w for the hollow shaft shown, (b)
Denoting by {Vw\ the value of this ratio computed for a solid shaft of the same radius
c2> express (he ratio 77w for the hollow shaft in terms of (771*% and cxlc2.
SOLUTION
IT.
IW fet'te,Mf.< = czr^
0+S.)
PROBLEM 3.30
330 While the exact distribution of the shearing stresses in a hollow cylinder shaft
is a shown in Fig. (1), an approximate value may be obtained for *„„, by assuming the
stresses to be uniformly distributed over the area A of the cross section, as shown in
Fig. (2), and then further assuming that all the elementary shearing forces act a distance
from O equal to the mean radto rm = l/a(c, + c7) of the cross sectioa This
approximate value is t0 = T/Arm, where T is the applied torque. Determine the ratio
T™Ai of the true value of the maximum shearing stress and hs approximate value t0
for values of c,/c2 respectively equal to 1.00, 0.95. 0.75, 0.50, and 0.
SOLUTION
Fo'»Li*JJm skat*: r, - TCz - 2Tc*
- ZTc* —
. zr
K
c* +c,
\ + (Ci/^y-
gTc,
e,/c2
««r/f.
l,o
LO
0.95
/.o25
O.lS
/J2o
0-5"
1.2oo
O-O
JvO)

PROBLEM 3.31
2.ii) N in
3.31 (a) For the solid steel shaft shown (C = 77 GPa), determine the angle of twist
at A. (b) Solve part a, assuming that the steel shaft is hollow with a 30-mm outer
diameter and a 20-mm inner diameter.
SOLUTION
T ■ %SO AJ.k, <y * Ik
0*
-a
(77^/Oi?)(7^25>'/o-^
<t> = C73.^<?y/Q-3 ^8P
TT
r T3^<**/0 y<u*
4.2/
(fcO
J
-1 *
<?*&
9~-
(77*io''X«.8N*|o
PROBLEM 3.32
3.32 For the aluminum shaft shown (G= 3.9 x i06psi), determine (a) the torque T
which causes an angle of twist of 5°, {b) the angle of twist caused by (he same torque
T in a solid cylindrical shaft of the same length and cross-sectional area.
SOLUTION
T =
- 'J
jq
-3
0.5 in.
A * 7T(C/- c,*) SoU sU+ A = 7TC*
c*= c*-c*= OJSX-O.S* • 0.3)25* i"nl
-3
V^fttff
I3.0O

PROBLEM 3.33
B
■»»■■!', 'I1.
3.33 The ship at A has just started to drill for oil on the ocean floor at a depth of
5000 ft. Knowing that the top of the 8-in.-diameter steel drill pipe [G = 11.2 *
|06psi) rotates through two complete revolutions before the drill bit at B starts to
operate, determine the maximum shearing stress caused m the pipe by torsion.
SOLUTION
5000ft
9- i
GS
' " L
JL ~ L
v- 0/.g»;ot)(l3.5*6;('J.o)
L = >5"0O£>-ff = GOooO ,'*
Coooo
?,3* Ks.*
PROBLEM 3.34
3.34 Determine the largest allowable diameter of a 3-m-long steel rod (G = 77 GPa)
if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa.
SOLUTION
L=3h <p - -2ojr _ S22.G */o"Sf«L X^ 2o*/o4 Pcl
ro - Lt X - GOT? <T - Ik - GJ"^c _ Q(pC „ Jt
~Oi*\o*){s2i.i*iom*
-3
PROBLEM 3.35
0.9 m
3.35 The torques shown are exerted on pulleys A and B. Knowing that the shafts are
solid and made of aluminum (G = 77 GPa), determine the angle of twist between (a)
A and B, (b) A and C
SOLUTION
0.75 m
CO,) T^e - ZOO W-rrjj LA9 =0.^m} ^^JrO.^ry
^^ G-nT " (77*/OfX7%S2a*fO-T
<V> Ttc - 20O + HQQ * 7oO W-v*j Lac " 0.7i*m i C* = \A = 0.0*3
Wi
0
fit -
- TicU . (7ooV(o.7f) = |5i5./; -* ^
GJ".c * (77 k/©')^..*^*/*^)
-j
<?Ac = %s + <«««. = SI-W° r«l ' 3.^2

PROBLEM 3.36
30 mm
3.36 The torques shown are exerted on pulleys B, C and D. Knowing that the entire
shaft is made of steel (G = 27 GPa), determine the angle of (wist between (a) C and
B, (b) DandB.
30 mm
36mm
SOLUTION
36mm ^-Shott BC '- C = £d - 0.01$ to
L8£ r O.Sm; G=27*/0* fk
T^o = 40O-9OO = -5*00 K)-W
L^= i.o hi
. TL _
PROBLEM 3.37
36 mm
30 mm
3.37 The sond brass rod AB(G = 3*> &Pc lis bonded to the solid aluminum rod
BC(G= Zl 6 P<t). Determine the angle of twist (a) at /?,(*) at A.
SOLUTION
(.so N ■ in
Sfe«ft 8C: C-^d = O.OJXr^ L = 0.32OH,
& = 27*10* Pa, T - ISO W***
(a) $?B * <Pgt r 12.^37 x JO"3 nut =5 0.7*1 * -*
(W % ~- <PBc + <pA6 r 27.447 *lo-***J = KS73* <■*

PROBLEM 3.38
300 lb-in.
3JS The brass rod AB (G =5.6 * 10* psi) is bonded 10 the aluminum rod BC
(G = 3-9 * 10* psi) ■ Knowing that each rod is solid, determine the angle of twist
(a) at B, (/>) at C
SOLUTION
X lO t*l7
T- 30O ik- in
rt) - <p - TLa« _ (300 n^
SUf+ 8C: G» 3,?*|o4p<.' lte* &£t *7£ ;*.
dW , Z±* r , (Z°o)(?*>
TftC St, J" (3^ *io6)(G. ias«fwo"3 >
- Q^O'hra.JL - SI. 7 *
r 0.4/7 4-0.^^3 = L320i^J> r 7^6*

PROBLEM 3,39
rc = 60 mm
3.39 Two solid steel shafts (G = 77 GPa) are connected by the gears shown.
Knowing (hat the radius of gear B is r B = 20 mm, determine the angle through which
end A rotates when TA = 75 N-ra
SOLUTION
-r - ii
r ^ I**. - Tex,
- ^ -r-
Yi, IA«
rv*
LC0 -- O.HoO *
-a
CiVcu^ifW'e-MV^r J i'sp/*ce *■*«■■*■. I <xT CQ*T6.i f* poi'lti or ae^/S
S = rc <pft -- r* 4?8
^Aer fc* r J fo.°'°V - /5;7ogx/o^ w* LAe r 0-Soo *■>
Po+J-fon *i A <jPA r <& * C^, * i38.7 //0***«J r 7.9*/° -*

PROBLEM 3.40
rc= 60 mm
3.40 Solve Prob, 3.39, assuming that a change in design of the assembly resulted in
the radius of gear B being increased to 30 mm.
& - 77 GPo-j fft r 30m* ^ TA * 75* W* r»
SOLUTION
F r Ik
X
CO
T - -^-T
TA6 * iA - 75" Ai-W!
T<
CD
(7S) = ISO W-m
Jcd= ?cj r £ (0.012 ^= 3?.S72^/o<' ►»* ■ L^ = O.Votf ^
i>e aT C
Twist ■'* .sU-H AB
& =77*10* TV (Jj^Xk r-J^i^l^ = St.OOtxlO* ^

PROBLEM 3.41
SOLUTION
3.41 Two shafts, each of j - in. diameter, are connected by the gears shown
Knowing that G = 11.2 * 10 psi and that the shaft at Fis fixed, determine the angle
through which end A rotates when a 750 Ib-in. torque is appbed at A.
CiVc^'wf ev*p-* n'*/ C<>* T*-c4"
F =
. T», _ T.
SE
rB
-I
6-^ 7M.3x|o4X3I.065h/0's)
JM.W*io'* r-u4
^ <pf r ^-O^as-wo"* ") = 3o.6<o*tor* r+J
La, = C + S^ ll.'*, JfiA - 3/.0S3*{dS .V
T^L6A _ OsoKtn
G^ ~ Ol.2*,*)t*i.O&i*r5 * M.7/3K/o-srJ
3o.£Co*to'2* 27.713 */o~3 * ^tS73j*to'3 r*^
= 3.12

PROBLEM 3.42
SOLUTION
3.41 Two shafts, each of j - in. diameter, are connected by the gears shown
Knowing that G = 11.2 * 10 * psi and that the shaft at F is fixed, determine the angle
trough which end A rotates when a 750 Ib-in. torque is applied at A.
3.42 Solve Prob. 3.41, assuming that after a design change the radius of gear B is 4
in. and the radius of gear E is 3 in.
U r.
Tfr r 7T fAa ' f O^o)
- -5"6P.5 A-/H
L«rJiV j^sI6"*B {(o.^ST* 3/. 663 x/o"' i** G '-11*2*10' ?Si
«
FC
r -Tk^ei r _.^1L€^*?.
-3
Gf Jp« (U.2 xlO* XSI.063 */o'3 )
= /2. 935" x/o"-5 r*^
\ *M*e+tuJ! JUp/eLceme^t cut cieW ci'rcJe § - fB§B - fB<Pe
*ft r £ % - -f (/?.7«5"/«s) * eJ.7ol«/0"3 rW
<fi
=• 23. 7/3 ^o" r*J
f?«T *.K <»vt out A
<ft = #6 + <fc,

PROBLEM 3.43
SOLUTION
T»6 -- Tk
'a
T - iT -
10
J.*3 A coder F, used to record in digital form the rotation of shaft A, is connected
to the shaft by means of the gear train shown, which consists of four gears and thtte
solid steel shafts each of diameter d. Two of the gears have a radius r and the other
two a radius nr. I f the rotation of the coder F is prevented, determine in terms of 71,
/, G, J, and n the angle through which end A rotates,
10
- IgMgF _
n*GJ
MGvJ
\ W n 1
PROBLEM 3 44 ^.43 A coder F, used to record in digital form the rotation of shall A, is connected
to the shaft by means of the gear train shown, which consists of four gears and three
solid steel shafts each of diameter d. Two of the gears have a radius r and the other
SOLUTION two a radius wr if the rotation of the coder Fis prevented, determine m terms of T,
, . /, G, J, and n the angle through which end A rotates.
See Z'o/jitOY] To PROBLEM 3.44 For the gear train described in Prob. 3.43, determine the angle through which
3.H3 fot devi«JPofl we*+ $£ end^ rotates when r= 0.75 Nm, / = 60 mm, d = 4 mm, G = 11 GPa, and n = 2.
<& * ££0 + ^ + »W
DaW T* 0.?f */*, i = 0.0<O », c--|J= O.00?m) G = 77*/o,P«.
»-tn^£:a:W'^*^°a'-"'"0-'^
- /.74?

3.45 The design specifications of a 2-nvlong solid circular transmission shaft require
PROBLEM 3.45 tnat tne angle of twist of the shaft not exceed 3° when a torque of 9 kN-m is applied.
Determine the required diameter of the shaft, knowing that the shaft is made of (a) a
steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77 GPa,
SOLUTION (t) a bronze with an allowable shearing stress of 35 MPa and a modulus of rigidity of
42 GPa.
rvi
(a) SW&Uf+: f= 9o*/06 P*., G-- 77*/0* P*
C= H/.06 */O"3 rv> - H/* 06^ ch 2o - gJ?. / roA7
Re^ir*^ vA-rue of ej is +Jie <P*A*e^; cf * 82.1 urn* **
M grorv7t sta-ft - r - 35V io6 pa , G =■ 42 wo9 Pet
7 "n(42*|o,X^.3£o*/o-s)
C - 47.7SWo"3 kn ■= ^7.78«m d - 2c - ?5".€, Hn*i
Require J vajPuc <*f e( i*s He ia^e^ d r 10*1.4 *» -«

- 3.46 The design specifications of a 4-ft-long solid circular transmission shaft require
U 3'4* that the angle of twist of the shaft not exceed 4° when a torque of 6 kip-in. is applied.
Determine the required diameter of the shaft, knowing that the shaft is made of a steel
SOLUTION wft*1 ^ allowable shearing stress of 12 ksi and a modulus of rigidity of 11.2 * 10 6 psi
Tr £ U^iV, = 6000 JL-?* G * l\.% xioc psi
~ " 7TC* C 7T£ ~ TT(l*oo«0 " w"5,»s '*
C - O. C83 ivi d ■* Zc * I.3C6 in.

PROBLEM 3,47
SOLUTION
T^»T0
3.47 The design of the gear-and-shaft system shown requires that steel shafts of the
same diameter be used for both AB and CD, It is further required that r^ i 60 MPa
and that the angle <pD (hrough which end D of shaft CD rotates not exceed 1.5°.
Knowing that G = 77 GPa, determine the required diameter of the shafts.
lOOO f/-Vh
40 mm
For desert L«,*ee) ©»a S-rV^fcSSj l«c
^ 0" ' TTC>
'*■<!£■ 38JS* • *•««-«
Sews
GJ
j_QOO
" <3rJ
- 122. V000 = ^<3Q
SUf+ co
71ft * IOOO rV*rV^ , L* O-G ii
9o - <P* + ^D
"T*
GO"
GvJ
(V = falC3100^ g C*fo'°<0
W " G--JC*
-•J
7T G- <p0 tt(77 x/o* X*^'8 Wo*3 0
C- 3I.H& Wo"* M = 3I.H M(M ^

PROBLEM 3,48
3.48 The electric motor exerts a torque of 800 N-m on the steel shaft ABCD when
it is roiating at constant speed. Design specifications require that the diameter of the
shaft be uniform from A to D and that the angk of twist between A and D not exceed
1,5°. Knowing that rmMt s 60 MPa and G = 77 GPa, determine the minimum diameter
shaft that may be used.
300 n .i>
/
SOLUTION
T6t* SOO W-*i
Desijw \mscJ w s+ress *£" ~ Go*/oAP&.
vj " 7TC3
C ^ 20Jto*ld rvi = ZO.*+&m*«
cl = 2c = ¥©. 8
hotvj
<P
D/C
=■ O
9^6
= T^j~f^ - C5~oo )(Q>g) - £QO
GO"
&J-
S-J"
dL* - TA»L»a _ (goo K O-<0 _ 3^0
^* " QrT ' GO" " GT
CM r (2l<M©>
^Xc*o)
-^
TG-^ " TC77^/c^)(^.i2y^o-3 )
^ I ^5". go */o *n
C - Zl.OHvlO" **, = 3.1.OH mm
Design mysf us« .Pcty^^ vaJijg. ©7 of
o* = ^C " W.I **i

PROBLEM 3.49
3.49 The sotid cylindrical rod SC is attached to the rigid lever AB and to the fixed
support at C. The vertical force P apphed at A causes a small displacement A at point
A. Show that the corresponding maximum shearing stress in the rod is
2 La
where 4 is the diameter of the rod and G its modulus of rigidity.
SOLUTION
dhftic CO +o tesftiOfi A'8 «.&
-froin wJti'eJ) ^P = arc Sin -gj-
"Ht« hTo^iwtJiM ^keaWiM STr**S iw PoeJ 8C fS
For sm*J5P ^
<3V€.&"1
A
*irfc*r« A- ** -f-
^** " HIT

PROBLEM 3.50
3,50 and 3.51 The solid cylindrical rod BCof length L = 24 in. is attached to the
rigid lever AB of length a= 15 in. and to the support at C. When a 100-lb force P is
applied at A. design specifications require that the displacement of A not exceed 1 in.
when a 100-lb force P is applied at A. For the material indicated determine the
required diameter of the rod.
3,50 Steel: *■„,= 15 ksi, G= 11.2 * l0*ps>-
SOLUTION
At Hie clSPokJcXU fwrtf a*jie S'w f = J - ^ = 0.O6667
T ~ Pa cos 9 = (/OO )(/£") cos 3,8226° = 14*6.7 7A- ,'«
8
-fu/i'sf
- 3^.603 *Jo" i**
C " TrTJi7aj</o4)(o.oe67^)
Base^ o« stress H » ^ " ^T •'
Use -P«r*ie^ \f*Jvt fo^ o/esr^M c-0.3*?7*/V» 6~.Zc =• 0.837 /n.
Tr6-g>
c = o.3w ;«.
PROBLEM 3.51
3.50 and 3.51 The solid cylindrical rod BC of length i = 24 in. is attached to the
rigid lever AB of length a = 15 in. and to the support at C. When a 100-Ib force P is
applied at At design specifications require that the displacement of A not exceed I in.
when a 100-lb force P is applied at A. For the material indicated determine the
required diameter of the rod.
3.51 Aluminum: Ta= 10 ksi, G = 3.9 * 10*psi.
SOLUTION
Af He *MowJ>h +wcs-f a«q/e siw%<f* al-^s O.oslqi
T * Pa <**<¥ - (/0o)O*Ws 3.22iW' = 1496. 7-#/n
ir(3.9xio^Ko-OCO/6)
2r
7TC/6000 ]
Use iai^e^ i/*.A>c ■£"* s/eafj* C - 0.-S"4*** i'k
C " 0-Vf7 ih
cU2c- low fn. ^

PROBLEM 3.52
SOLUTION
3.52 A4-:kN-m torque T is applied at end ^4 of the composite shaft shown. Knowing
that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum,
determine (a) the maximum shearing stress in the steel core, (&) the maximum shearing
stress in the aluminum jacket, (c) the angle of twist at A.
72 mm
54 mm
S+eei tore- C, -iA{ - 0.02,7 W
Aluminum jacket
Torcfoz. carried by sle*-^ co^t TJ - GyJJ ^/L
AP<W*Jwi j«*J<«,+ : c, = ^r<f# " 0,«7i«^ C2 = £J2 * 0.036 m
Ar
T
H-*/0
.^
= 35". W£ */o5 r*j| /*
?T - G(T - G%Ct& * (71*IO*i)(o.oZ)Y$SAo€> */o*3)
= 73.6*/oc Pa. 73-4 tfPcc
^ S2r - GtC^ = (21xiO*)(o.<%C)(3£Ao6>to-3)
- 31A */o* Rd 3<*,4 MPa
(C) A*g*e o? -K/rtf
- 3-.07-

PROBLEM 3.53
3.53 The composite shaft shown is to be twisted by applying a torque T at end A.
Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the
aluminum, determine the largest angle through which end A may be rotated if the
following allowable stresses are not to be exceeded rlleel = 60 MPa and Tliumkmm 45
MPa.
SOLUTION
rM = gk
*"**
G1 Cm««
L
54 mm
Steel core
Aluminum jacket
.-&
= 72. iS**/o v^J ' f./3

PROBLEM 3.54
3.54 The composite shaft shown consists of a 0.2-in.-thick brass jacket (G = 5.6 *
10 psi) bonded to a 1.2-in.-diameter steel core (G = 11.2 * 10*psi). Knowing that
the shaft is subjected to 5-kip-in. torques, determine (a) the maximum shearing stress
in the brass jacket, (b) the maximum shearing stress in the steel core, (c) the angle of
twist of end B relative to end A.
SOLUTION
0.2 in.
•Steel core
B/^ss J*.$&fc+'
J r \cs* 'i(o-&)H - 0.703575 ;/
Tofai ftY^e 7 = 7,^7^ (g, j; + &.JJ-2-
T JxlO*
L &, J, + Gr^3
-5
?. 2%OOKIO*+ ?.*rfi3Oirl0'
- /.OS1** */o* ^o/Ah
r 4.73 *JO* psi 4.72. ifsi*
(b^ Mftvi*"U* S»eet*Vn«i stress iin slet^ Co^e
- 7. os */o* yosi- 7-02 ksi
Cc) A«gfc o?-fwis+ (L= &fy ^72 u">
<p = L ■£ * (70(1.0542 "JO"*) = 75.9*/0~S v*ct
- 4.35"

PROBLEM 3.55
3.55 The composite shaft shown is to be twisted by applying the torques shown.
Knowing that the modulus of rigidity is 11.2 x I0*psi for the steel and 5.6 * 106psi
for the brass, determine the largest angle of twist of end B relative to end A if the
following allowable stresses are not to be exceeded r^ = 15 ksi and i^ = 8 ksi.
Brass jacket
0.2 in.
" Steel core
SOLUTION
T
&2 --
L
ft* -
•for G*.cJ\ ht^e^i'*-/
15*000
-s
r ^.^3^1 */0~ nJi/l*
L Cll-2x/04X0.6)
gooo
SmeJJef- \Jc^$oe aot/e/^s
- 7.37°

PROBLEM 3.56
3.56 Two solid steel shafts are fitted with flanges which are then connected by fitted
bohs so that there is no relative rotation between the flanges. Knowing that G = 77
GPa, determine the maximum shearing stress in each shaft when a 500 N-m torque is
applied to flange B.
SOLUTION
Shaft A 8
Ji* = %C\ !(0-Oi£y = 71.5* *lo"* r»*
-T _ G^JU rt _ (77^10^(7^.53*10*)^
= Jo.2a<r*/cf <fi*
Sh*ft CD
T =!"<*, L^a^, c=id-aoi«mJ j^*fc*r f(o.oit>y
J^ r IGV. 316* JO-1 M**
'<* uT ^c 57^ ^ r w^**'© <fi.
TA8 = (lO-aoS"Wo3X20.555 */o'S) - 20127 N-m
Tco - (14. I08w/o3)(^0.r65 >io'S) - 21©.. 13 N-wx
3/.? MPa.

PROBLEM 3.57
SOLUTION
3.57 and 3.58 Two solid steel shafts are fitted with flanges which are then connected
by bolts as shown. The bolts are slightly undersized and permit a 1.5 "rotation of one
flange with respect to the other before the flanges begin to rotate as a single unit.
Knowing that G -11 GPa, determine the maximum shearing stress in each shaft when
a 500 N-m torque T is applied to the flange indicated.
3.57 The torque T is applied to flange B.
- Tag Ljin
T - Tco> Leo = 0.^*^ c = iol =0.01**^ J^,-|cl**^(o.oi3)y
T^v« i<? remove d'to/vuM. = T^ *(^0.2o£^c|o"sX'&6.l&*lo",^,) = 2G7.H Kl-r^
Tor<ty* T° to e«^e <jJ.'u* J foK+id* <$> t T" = SOO - %&7.[ 1 - 232. 83 K/-^
T" - T " j. T""
,232.83 = O-2o£*l03 4- HJoS^IO*)^" ■*■ <p"r 7.^76^v/o'Sr*J
Trt" - Uo.2oS*to%)(*t.&&?xlOmtyi r 9-7,73 N-m
Teo" * (/M.lo8*/0*)(^S7«ir>/c> ) * 135.10 M-
hi
Maximo** sliea/'i'i'j S"f rc5^ i*
^o ^
2k£
- I4.7S v/o* P*.
V^^Mfit

PROBLEM 3.58
SOLUTION
$M*h AS
3.57 aid 3.58 Two solid steel shafts are fitted with flanges which are then connected
by botts as shown. The botts are slightly undersized and permit a 1.5 "rotation of one
flange with respect to the oiher before the flanges begin to rotate as a single unit.
Knowing that G = 77 GPa, determine the maximum shearing stress in each shaft when
a 500 N*m torque T is applied to the flange indicated.
3.58 The torque T is applied to flange C.
i**i
** L
A^~ „, . (VxtfKllSZKtO*)
.AS
0.6
<s>*
S\>«f+ CD
CO
, Lo^O.fm, c=£J =0.012**^ Jco = fc1*^ Ko.ois)
Jcd - W. 8<=,6 Wo"1 W
Tec' ^Jcb 0 r !Jl&±ll$W£!!& & = K./o8>/o3<a
*jJ;h*o«Jt rddtio* <pH- T"- -Too-36?.35* - /3<2>.£.T *A*»
TH *T* +T/
Co
-i
/30.£S" - (I0.2O5*'/O3 + |MJo8*to8 )<$" Cp" • 5.Z737 *IO ' raj
TV = (Mof* IO%)(S.%137* io"*) - ^4.84 A/.i^
^ r ££ r (ff.WXo.Ojr) , ,0.3q)(,0'pft 10.34 MP*.-
Jc
^.m* *(&

PROBLEM 3.59
r = 40 mm
200 mm
3.59 At a time when rotation is prevented at the lower end of each shaft, a 50-N-m
torque is applied to end A of shaft AB. Knowing that G - 11 GPa for both shafts,
determine (a) the maximum shearing stress in shaft CD, (b) the angle of rotation at A.
SOLUTION
Let Ta - To^fcie *ffJ;tJ *$ A - SO KJ*i*
Tcb - -fo*4*e i* sl*"H CD
£jcJ-t'c s
TA-TA8- FrA »o
X
CD
" A(TA-TA,W#(T,-TA^
Gear A
Geo/- C
GOab
<?*'%&'$&
&4
AnjPes of fw;s+ (J)A =
GJ^a 3 3 &JCD
TCD r | (So-26,on) •= \5.w far*
4^:% .■■&X'*<"> , 47. MO* A - W.I Wc
(b^ An^J'e o^ roi&AiaM a4 A
ft

PROBLEM 3.60
3.59 At a time when rotation is prevented at the lower end of each shaft, a 50-N*m
torque is applied to end A of shaft AB. Knowing that G = 77 GPa for both shafts,
determine (a) the maximum shearing stress in shaft CD, (b) the angle of rotation at A.
3.60 Solve Prob. 3.59, assuming that the 50-N'm torque is applied to end C of shaft
CD.
SOLUTION
eJ Jt C
Tcd - h-'-'i^e '" s.W+ CO
7*8 = -frji^jue i* ,sUf-f A8
i;-Teo-rcF^o
5o tJ-»\
Gear A
G-e«/- C
_ rA
. 1
= Stir = 4
G-JJ
8
= 3 le
Arises of1 +w»s+ <#. •= 5tL <pA
Sai - LA T, -T,
&0U " * ^ ex*
TCb ■ OA71C Tc = (o**7K)(&) * *3.<?8 W-m
Tab" K5^ -23.9S) r 31.03W-h
(cO M*yjci^0wt 5 kaftan* 3TY**3 in sh^H CD
*• • ¥ - f^#S ■ 7°-7*°'p* ■ 7" MP-
*
, 3^ r ZJkL ? caY3t.osXo>ao?^ s 20^7^,0%^
W*s ttGc«* " Tr(77*fc>,)(>oo7S),'
LI £9

PROBLEM 3.61
3.61 The steel jacket CD has been attached to the 40-mm-diameter steel shaft AE by
means of rigid flanges welded to the jacket and to the rod. The outer diameter of the
jacket is 80 mm and its wall thickness is 4 mm. If 500-N*m torques are apphed as
shown, determine the maximum shearing stress in the jacket.
SOLUTION
5oiU sUf+-* c *i«l - Q.OZO ^
JoeW : C2 - j[d - 0.040 m
Ct- Ca- t - O.oy-O-O.Ooi = 0.036 *n
J> " ?(<V-c/*)- f (o.oto- 0.036") - 1.38X9 */<56 m*
3.62 The mass moment of inertia of a gear is to be determined experimentally by
using a torsional pendulum consisting of a 6-ft steel wire. Knowing that G = 11.2 x
10 6 psi, determine the diameter of the wire for which the torsional spring constant will
be4.27lb-ft/rad.
SOLUTION
To**io« spr.'M cW**t K" 4.27 A-ft/r*»l - .Sl^ifcW^J
-4
C = 0.120% ih-
<J »2c = 0.2Y/ ,'».

PROBLEM 3.63
Co
3.63 A solid shaft and a hollow shaft are made of the same material and are of the
same weight and length. Denotmg by n the ratio c,lc2, show that the ratio TjTh of the
torque Ts m the solid shaft to the torque Th in the hollow shaft is (a)
VI - w /(1+rt ) if the maximum shearing stress is the same in each shaft, (A)(l -
«*)/(! + w2) if the angle of twist is the same for each shaft.
ci="c2 SOLUTION
_ T»c«.
I - n
j; = K^-<;>£^0-^
^r Sfe
77^
/+ n
GrT*
PHMWIWHII Jim. WWWHIjff
7 = °\
y-L
7^2L
3.64 A torque T is applied as shown to a solid tapered shaft AB. Show by
integration that the angle of twist at A is
771
SOLUTION
r* fix
w,
y
\2trGc'
,1
A& 3hOViV.
^ W £$rN ir&c* y*
T \ ttGc4 T " tTEc' \ y*
- 2Z
^i"3y']A
-ttg-
^2 *•*■* 3/-3S
- 3T11 f7 1 _ TTL.
77GC (W^J " WTTG-C*

PROBLEM 3.65
SOLUTION
Use afi
3.65 An annular plate of thickness / and modulus of rigidity G is used to connect
shaft AB of radius r, to tube CD of inner radius rv Knowing that a torque T is applied
to end A ofshaft^-Sand that endDof tubeCDis fixed, (a) determine the magnitude
and location of the maximum shearing stress in the annular plate, (b) show that the
angle through which end B of the shaft rotates with respect to end C of the tube is
t>cJt
&"h'w
The -forte pev* uwft *
J'e*\«i*Hi ot c.\recounte^^c/z *
is J*t. Vv
T M » o \,
tfcCa^^ - t = o
r^-
x
sk
eai^i«
x
G- " XB-St/o'
^rt^y
CO
d4> = r-ste
« /o ^
d<J> =
X
de
** £
1
T te

PROBLEM 3.66
SOLUTION
3.66 An annular aluminum plate (G = 27 GPa), of thickness / = 6 mm, is used to
connect the aluminum shaft AB, of length I, = 90 mm and radius r, = 30 mm, to the
aluminum tube CD, of length L2= 150 mm, inner radius r2 = 75 mm and 4 mm
thickness. Knowing that a 2500-N-m torque T is applied to end A of shaft AB and that
end D of tube CD is fixed, determine (a) the maximum shearing stress in the shaft-
plate-tube system, (b) the angle through which end A rotates. (Hint; Use the formula
derived in Prob. 3.65 to solve part b.)
vlt _ %r
7TC
W,
TT(0.O30)3
S*.<* MP*
RPo,+e EC (See PKofiLfM 3.C5 ^ J^Vcotaw)
T ZS°Q___
r--
2.H tf^1-" xtKg^sX0-^0)
- = 73.7 x/O6, ft
7V7 Mffeu
1*1
r *
- Tci.
SUff AB
II 4-9.3 rfl^-C ^* ^
2T = 73-7 MP**
d> = TM* _ ZTU* to)(*S&cQ(0.0'to)
?9eJte 6C (See PPofcL^M 3.65" *^ Je^WftaO
toCxiti&Y.
0.006) [0.03^*- 0.67**1
-1
=■ U 146 */o~* roJ
S^*?* CO
.3
- LZlo xio r*J
T«4*J r*+-AV>* *-^*e <p •- <Pto + <ftc + <ko * *•*' K/°"S "^
- O.SVO

PROBLEM 3 67 3*67 Using m allowable stress of 55 MPa> design a solid steel shaft to transmit 10
kW at a frequency ofl 5 Hz.
SOLUTION
r^ = £S~«io* pa p= io*/o* Wj ^« 15" Hz.
PROBLEM 3.68 3.68 Using an allowable stress of 5 ksi, design a solid steel shaft to transmit - hp at
a speed of 1725 rpm.
SOLUTION
C - O.I325" /* d = 2c ~ 0.^6i" i* -*
PRO „M 3 fi9 3.69 Design a solid steel shaft to transmit 100 hp at a speed of 1200 rpm, if the
maximum shearing stress is not to exceed 7500 psi.
SOLUTION
tcJJL ~ 1SOO ps. P= lOO lip ~ G&OXlO* &-!*>/&
£ r &°?. r Zo Hz T » -£• - _6CO»to» - ^j&tti xJO* i^i'i.
Cr 0.76 S** in cl-2c - l.«8 In. **
PROBLEM 3.70
3.70 Design a solid steel shaft to transmit 0.375 kW at a frequency of 29 Hz, if the
shearing stress in the shaft is not to exceed 35 MPa
SOLUTION
2ii, = 35" Wc?6 Pa. P = 0.375%^ -V f = 2W 4z
^ ^ " -15 -"• C - W = TT(3f*K>^ = 37.43^ m*
-3

problem 3.71
3.71 A hollow shaft is to transmit 250 kW at a frequency of 30 Hz. Knowing that
the shearing stress must not exceed 50 MPa, design a shaft for which the ratio of the
inner diameter to the outer diameter is 0.75.
2Tf " Z"n(3o^
SOLUTION
tat- 5o *i& Pa- P ^ Z50v la? W f = 3o Hz
c3-
teZ^jz
Ct = 2*?. I^x/o'no - 2^.12 »i»v>
PROBLEM 3.72
SOLUTION
Cz--£a/a- Sin
1 £o
3.72 One of two hollow drive shafts of an ocean liner is 125 ft long, and its outer and
inner diameters are 16 in. and 8 in., respectively. The shaft is made of a steel for which
tm = 8500 psi and G = 11.2 x 106 psi. Knowing that the maximum speed of rotation
of the shaft is 165 rpm, determine (a) the maximum power that can be transmitted by
the one shaft to its propeller, (b) the corresponding angle of twist of the shaft.
-2
(cl) P ■= ZttFT - 2v(2.1S)(6.Ho*Z*/cf) T ll.0.7¥»c/o* /A-m/s
to qp
Uo.-^v:jbc a».»a/s
=- ie.78 *<o3 k
66oo AWs Ap
^ r
OL2*/Oc)(£o3J_&3
r*.<
g. IS

problem 3.73
<?
TL
3.73 While a steel shaft of the cross section shown rotates at 120 rpm, a
stroboscopic measurement indicates that the angle of twist is 2° in a 4-m length. Using
G = ll GPa, determine the power being transmitted.
SOLUTION
75mm Twfcf <xn^\e ty -= 2° - 3H.1a? y\o~* roJi
J= 3.on&g * lO~c
W\
L= t*
1 ' L 4
PROBLEM 3.74
3.74 Determine the required thickness of the 50-mm tubular shaft of Example 3.07,
if it is to transmit the same power while rotating at a frequency of: Ho Hz-
SOLUTION
He. 3.07 P t /oo kl^/ * jOOx/^W
T =
2rrf
- - 397.ff<? W-M
r-Ia
TC:
T- Ifr* - cMN)
c, -
23.// X/o"S * t VC4- C, = K?<* >|Cf*hn - W /nm

f = ^ = 12 ^ C =-Jral = 0.75*.'*
T = -£-. 3 *K,**?% r ^^^ it. '«
J TTC* IT (0.75)3
PROBLEM 3.75
(a)
(«
3.75 The design of a machine element calls for a 1.5-in.-outer-diameter shaft to
transmit 60 hp. (a) If the speed of rotation is 720 rpm, determine the maximum
shearing stress in shaft a. (b) If the speed of rotation can be increased 50% to 1080
rpm, determine the largest inner diameter of shaft b for which the maximum shearing
stress will be the same in each shaft.
SOLUTION
7.^3 kil
&)
f = mo r lgMi
t-*& ■ iSasf "■«« «^*»
r - l£> -
arcz.
TTCC^-C^)
C8 - 0. 5&11 in
0 7sv- UX3.,ro^*/o*Xo.tt-'> .
lC(7.9*5£*/o*)
PROBLEM 3.76
60 mm
3.76 A steel pipe of 60-mm outer diameter is to be used to transmit a torque of 350
N-m without exceeding an allowable shearing stress of 12 MPa. A series of 60-mm-
outer-diameter pipes is available for use. Knowing that the wall thickness of the
available pipes varies from 4 mm to 10 mm in 2-mm increments, choose the lightest
pipe that can he used.
SOLUTION
tjt - 60xfoQ Vol C^a^z - 0.08Q **
^ s ^ - -SSi = O.o3o<- &ysroHo;oso^ = ,
C, - 22.43 v/o"3 £ = C2-C, :3o^-«,13iw. = 7.S7 ***,
(?e*|uire*/ Hi^r»*SS t *> 7. 5^ /v)rt?. A**it*.lit size t = X^

3.77 A steel drive shaft is 6 ft long and its outer and inner diameters are respectively
PROBLEM 3.77 equal t0 2<25 in. and 1.75 in. (a) Knowing that the shaft transmits 240 hp while
rotating at 1800 rpm, determine the maximum shearing stress, (b) Using G = 11.2 *
10 psi, determine the corresponding angle of twist of the shaft.
SOLUTION
C( = iM, » O.&S^j ca=i^B l.'as";nj L' 6*1- * 72 m
PRORI FM ^ 7« ^*"* Knowing that the allowable shearing stress of the steel to be used is 7500 psi,
determine [a) the smallest permissible diameter of a shaft which must transmit 15 hp
while rotating at 2000 rpm, (b) the corresponding angle of twist in a 4-ft length of the
SOLUTION shaft(G= 11.2 x |06psi).
t+ - 1SOO pv ^ f = 2^9. - 33.323 W*
P- 15-i.p * OSKUOO) ' °rt*\o* A't^/s
1 " *Tif " *TT(35.3S3,)
taj c j- ire3 ^ t( 75-00 ^
fbJ J"= ?C* - f(o.3*asV - ^. 576 Wo" ,*
<« ^ Ik _ (472^ V*m _ ,3>My/0
- 5". 3a
•red

PROBLEM 3.79
150 mm
150 mm
60 mm
3.79 Three shafts and four gears are used to form a gear train which will transmit 7.5
kW from the motor at A to a machine tool at F. (Bearing for the shafts are omitted in
the sketch.) Knowing that the frequency of the motor is 30 Hz and that the allowable
stress for each shaft is 60 MPa, determine the required diameter of each shaft.
SOLUTION
? = l.S kW = 7.£*lo* IV
SUH AB :
27rfv„r
^"** 2TTfAa
2ti<s<0
r * ^ ££ -•• & %k
j.
AQ
™S
C3 a)(3?.^) -.^.nwo1
SUH CD:
£o
T"co ^ TAft /5-0 <■ av /
Zhtz
X
CO
'ft
%?
_ Tc^, 2/T
<£* -*$Z
Cj =
-3
C^* |0-i8xfo~ »v» ^ lo.i£w*n
JCD = 2C«r £O.H*
**•?
<
Slwcfr ET=:
Cem " 13.83.* lo"* * 1^.83 *~v
«*»*
'6F
;e.£*S6 */o~' **
^Qr* 27"6 M~

PROBLEM 3.80
150 mm
150 mm
60 mm
f»= 2MH*
**
3.80 Three shafts and four gears are used to form a gear train which will transmit
power from the motor at A to a machine tool at F. (Bearing for the shafts are omitted
in the sketch.) The diameter of each shaft is as follows: dAB = 16 mm, dCD = 20 mm,
dEF = 28 mm. Knowing that the frequency of the motor is 24 Hz and that the allowable
shearing stress for each shaft is 75 MPa, determine the maximum power that can be
transmitted.
SOLUTION
tin = IS MPa. •= 7S*\oc fa
SU-ft AB: CA& = i<^ar 0.008 w
Y - Tew _ ZT
m
Staff CD- Crt4cLB O.O'O
f*
"W^lo^-- J (o.oio?(7i-*/o*)- in.«/N-H,
'fto
f.
. r,
]f-(^) ^ <U «e
StaW £ F: C6P - 5- d^ - 0.014 »
T^ = \c£?*t - |(o.OI'f^(75V/oO - 323.27 N-m
rep Te Tcs
_6o,
/50
(9.6) =• 3.*<f Ha
?^ ^ 7-/M<*V r 7.11 fcv

SU+ic*:
Fv F,
3.81 The shafl-disk-helt arrangement shown is used to transmit 3 hp from point A
to point D. (a) Using an allowable shearing stress of 9500 psi, determine the required
speed of shaft AB. (b) Solve part a, assuming that the diameters of shafts AB and CD
are respectively 0.75 in. and 0.625 in.
SOLUTION
Y - Tc _ _.2T t - i/V
i-, Tm » f(i )S(V«> ) - 786."? A- i.
TB = Ta(F;-Fj Tc= ^(F,-F^
Ta = Is-X
r« '£
_ l.as~
X « O.X3-Tc
Assume
2.TT(/?4.73}
to hfhuAh to^ocs TS^ifr 786/»A^ j T^ ^ nSS^lL-M,
Assume Tt ~ 4ST.9-A.iM Tten Ta *^o.Ps-)rt5T.9) * I13.8T-fl-i-»
-=: 786-^ .A- iM
P-2ttPT
p P _ )q*>o
=r 27.7 Hz

PROBLEM 3 82 3*®2 A steel s^^ must transm*t *^0 kW at a speed of 360 rpm. Knowing that G
= 11 GPa, design a solid shaft so that the maximum stress will not exceed 50 MPa and
the angle of twist m a 2.5-m length will not exceed 3°.
SOLUTION
p -- isoxio* w -p= ^r ■= 6 w
Go
z
Design *fov s+v^rss ^,'i t - *5o MP*. = sTOx/O* Vc,
C = 37.00 xtO~z vn
Design -R>^ ahj/e of -r-^isf Jp,**^ Cp r 3& - .S2,M **oT iTaJ
rn- It - 2IL * r1-2Er ftMS»'*W>xfo»)ft.O . |C7,>Twn4^
C - 55. fo > /6"1*
-*
Dse 9cL*^er VeJos. C- 37.00^/0 to - 37.0wm j o) - 2c = 7^,0 m^ -^
PROBLEM 3.83 3*83 A steel snflft of 1.5-m length and 48-mm diameter is to he used to transmit 36
kW between a motor and a machine tool. Knowing that G =11 GPa, determine the
lowest speed of rotation of the shaft at which the maximum stress will not exceed 60
SOLUTION MPa and tne ^^ of twist ^ not exceed 2i$°_
P = 3£Wo3 ■WJ c = i<J * 0.0Z4^J L' l.-fwj G- = 77*'o9 P
Toi^vt d*,?*** o^ t*i<^«!i»*.Uyri s TV-ess ^ ~ <oO H'fta. ~ COx/o'Pa
Je <p - 2.5"* * 4.3.£33 ^/o** i^
rt)- XL " X= ^^ r ^&Q? Tr(O.OZ+)tf(l7*IO*Yts£33*toi)
^ ' <W " * L ' ZL ' (J)G-S)
~ I.IC730 */£?* ^»^

PROBLEM 3.84
SOLUTION
3.84 A 1.5-in,-diameter steel shaft of length 4 ft will be used to transmit 60 hp
between a motor and a pump. Knowingthat G= 11.2 * 10 psi, determine the lowest
speed of rotation at which the shearing stress will not exceed 8500 psi and the angle
of twist will not exceed 2°.
C= 4 J - 0.1S i.
L = 4fi* * 48 «.
*J TIC*
+v,'s-| a.^Je J;^-i Cp ~ 2° - 3 V. <?d7 xlo"* r*«<
P= 2.TT-FT u/fcft^t P* 60 kp - (Co)(C(,C>c>) - 3?6 * /O3 M.\„/s
= 15157 Hz
^3W v^*i
PROBLEM 3.85
3.85 A1,6-m-long tubular shaft of 42-mm outer diameter dt having the cross section
shown is to he made of a steel for which rflU = 75 MPa and G = 77 GPa. Knowing that
the angle of twist of the shaft must not exceed 4° when the shaft is subjected to a
lorque of 900 N'm, determine the largest inner diameter d2 which can be specified in
the design.
di -- 42**
SOLUTION
6«.sc^ o* s+re&s Arv.-f 2^-75 M/V = 75"*/c>* P*
rt - TL . r - _Z4. .
.-9
* 7F T
-^
C* = /2.MM *io * = la.^f »^
da r 2c*- ' ^.<*
*Wlf>

PROBLEM 3.86
3.86 A 1,6-m-long tubular steel shaft (G = 77 GPa) of 42-mm outer diameter d{ and
30-mm inner diameter d2 is to transmit 120 kW between a turbine and a generator.
Knowing that the allowable shearing stress is 65 MPa and that the angle of twist must
not exceed 3°, determine the minimum frequency at which the shaft may rotate.
d1 = 4-2 >t»n
h^
SOLUTION
J Ci <Xo;u
? - \xo icy) =

PROBLEM 3.87
120 mm
£
«*
I2.Q
3.87 The stepped shaft shown rotates at 450 rpm, Knowing that r = 10 mm,
determine the maximum power that can be transmitted without exceed ing an allowable
shearing stress of 45 MPa.
SOLUTION
"~ '-2 > J^s °-'°, F~- R3- *-3* *-" L25
r= KTc . ZKT
ti/^
<T
TTC
■P = ^ro rp^ * 1.5" Hz
PROBLEM 3.88
3.88 The stepped shaft shown rotates at 450 rpm. Knowing that r = 4 mm,
determine the maximum power that can be transmitted without exceeding an allowable
shearing stress of 45 MPa.
SOLUTION
vt "D - l2o wfi
For S^4^ sU-H" C * i J = O.QSO * r = i^£ = ££1.
' 2K CzXl.SSj
-P = 4S"0 rpM r 7.5" Hz

PROBLEM 3.89
T' 2.5 in.
1.25 in.
3.89 Knowing that the stepped shaft shown must transmit 60 hp at a speed of 2100
rpm, determine the minimum radius r of the fillet if an allowable stress of €060 psi is
not to be exceeded.
SOLUTION
f * ^ = 3^ Hz
-ft r 2a£- r 2
Froir^ Fij 3.32
^ J" TTC*
4. O-lS
r
= O. IS d * (O.OSO^/.fcri'O * 0.^2,5" iv).
PROBLEM 3.90
2-5 in.
1.25 in.
3.90 The stepped shaft shown must transmit 60 hp. Knowing thet the allowable
shearing stress in the shaft Is eooo psi and lhat the radius of the fillet is r = 0, 25 in.,
datermine the smallest permissible speed of the shaft.
SOLUTION
/.as-
_D
IS.
Z.oo
?
. K.Tc sxt
Fro^ F,«j. ^.32 K = UZ&
For s^Jlter 5UH C = j[ J » 0.625*
J
TTC;
P
P = 27if7
TTC1? = T(Q-42g^(gOoo,l _ |# g^ y/03 £. ,„
- Go l«,

PROBLEM 3.91
3.91 A 25- N*m torque is applied to the stepped shaft shown which has a full
quarter-circular fillet. Knowing that D = 24 mm, determine the maximum shearing
stress in the shaft when (a) </= 20 mm, (/>) d = Zi.**m.
SOLUTION
r = ±(D - d)
(a.
Full quarter-circular fillet
extends to edge oflarger shaft
Rot« Fi<j 3. 32. K =■ 1.34
F™~ F^ 3. 32 K - '-42 .. F«r s*i«lfe* shift £*%<£* O. oiog*.
PROBLEM 3.92
J.92 In the stepped shaft shown, which has a full quarter-circular fillet, D = 1.5 in.
and d = 1.2 in. Knowing that the speed of the shaft is 1800 rpm and that the
allowable shearing stress is 8000 psi, determine the maximum power that may be
transmitted by the shaft.
SOLUTION
**■?#«<>. .as
r = hD- d)
F/n^ FiV 3-32
Full quarter-circular fillet
extends to edge of larger shaft
f<=" I.SI
For •s^o.JJer- sA^-ff C - \ et - O. & in
P-^tt-FT^ 2-n(3oXao7JO - 31.0.6x/o3JL-i«/s
330, G* fa3 Jft '» ^
^oo Jt-iw/s/hp
59.2 hp

PROBLEM 3.93
3.93 In the stepped shaft shown, which has a full auarter-circular fillet, the allowable
shearing stress is 12 ksi. Knowing that D = 1>2? i*, determine the largest allowable
torque that may be applied to the shaft if (a)d= 1.1 in,, (b)d= 1.0 in.
SOLUTION
r=\{D-d)
t?*a r IT ksi * \7ooo pS;
i*>
*'#-••■
o<\
r*\(T>-J)* i(\.2S-\A) » O.07S* i«
Full quarter-circular fillet
extends to edge of larger shaft
For- s*n«Jt/eA sfc«cH*
P 0.075*
F^o-* Fi
I'j 3-32. r<= /.40
"iW ■ O.SS'm
Fo^ 4*«J!/eA si,^ C » ^ */ r OSOl*

PROBLEM 3 94 '*'* ^ 54-mm-diameter solid shaft is made of mild steel which is assumed to be
elastoplastic with rY = 145 MPa. Determine the maximum shearing stress and the
radius of the elastic core caused by the application of a torque of magnitude (a) 4
SOLUTION kN-m,(&)5kNia
C = j[ of = 0.027 m Ty * H5 * I0£ Pa
Compote TY T* * ^Carr *5(o.0.27)3<W*lo6) ^ H.4«3v/o* N/-*
(jDi^ T = 4-.OX |03 U-* < Ty eJWic f> = C - 27»im -*
T-|Tr(|-^
ycV = 0.M8O C * (0. SlSolfO.oa?') = o.OzZW m - 23.4 m*. -*
PROBLEM 3.95 '*'^ ^ 1.5-in.-diameter solid shaft is made of mild steel which is assumed to be
elastoplastic with rr = 21 ksi. Determine the maximum shearing stress and the radius
of the elastic core caused by the application of a torque of magnitude (a) 12 kip-in., (b)
SOLUTION ,8 kip-in.
C = id - CIS m rr * *\ fc»i
Compo+e "TV Tf - ^= f C3fr * J(d>.-)OV2 0 = «.<!/£ fc.p. .V.
(a) T - 12 fe.p. ,^ -c yr ejPashc a =c • 0J1S m. *•
S+rew is £i*y = Z* - 2\ Us/ -^
T^Tr(|-^

PROBLEM 3.96
SOLUTION
fy - S.5vl06 Pa.
-T - J"^ - I
T-§TrO-
PROBLEM 3.97
SOLUTION
C -- -£ J - Q.OZ
'p - 3 'r 3
r 27TC3
PROBLEM 3.98
c = 1,25 xxi.^s^
^-^c
Tr =
(a*) ir - 0.6 i* ,
T'STr(l
(W\ tr= l-o .* ;
T=^rr0-
3.96 A 30-mm-diameter solid rod is made of an elastoplastic material with vY = 3.5
MPa. Knowing that the elastic core of the rod is 25 mm in diameter, determine the
magnitude of the torque applied to the rod.
£- %J * 0.015 ** Pr z -k^r- 0.O|£5* w
c*lr r J (o.o/s-Y($.s*io6) * is.5r.5- M-^
tf.33333
V^H ff l^-^OJj- ^(o.33S33?] ? zul ^^ ^
3.97 It is observed that a straightened steel paper clip can be twisted through several
revo lutions by the application of a torque of approximately 0.8 lb-in. Knowing that the
diameter of the wire used to form the paper clip is 0.04 in., determine the approximate
value of the yield stress of the steel.
**. Tf - o.g A-m
<3)^ , 47-7 to; -.
2tt (o. e*)3
^^ AM The solid circular shaft shown is made of a steel which is assumed to be
-~\^^"' elastoplastic with ty-2\ ksi. Determine the magnitude T of the applied torque when
\ f the plastic zone is (a) 0,6 in. deep, (ft) 1 in. deep.
j i
SOLUTION
= UTS .'* tV - 21 ks/
^ - ft1*, * * 0-^m<) - 64. 427 ki>.,'n
yOr = C-ir = 1.2?-0.6° = 0.6T.H , -£ = ^f * Q-S^o
- i-^V ff64.lt27)[l-4(o.«o)3] - 82.9 kip-i« -«
fy*C-tr • MS-IO - O.aS'm ^r ^fp = O.Qoo
■$ £') - ^ (ci.«7 )[ / - ^ fc.*»}3 ] -- gff.7 k>- m -I

PROBLEM 3.99
c = 1.25 in.
3.98 The solid circular shaft shown is made of a steel which is assumed to be
elastoplastic with ry = 21 ksi. Determine the magnitude T of the applied torque when
the plastic zone.is (a) 0.6 in. deep, (b) I in. deep.
3.99 For the shaft and loading of Prob. 3.98, determine the angle of twist in a 4-ft
length of shaft.
SOLUTION
c* 1.25" in, %* a/ & -- zi»to% p*.^ 6*11.2*10*^
<yB -£_ , 7^°/'^ r /SS.^/o'S^ c 7.93
' L
(bl £r- l.o-i
pr~- C-tr - /.ar-/-o = 0.2f ^ * -2:
2£
- 0.200
PROBLEM 3.100
3.100 A torque T is applied to the 20-mm-diameter steel rod AB. Assuming the steel
to be elastoplastic with G = 77 GPa and ry = 145 MPa, determine (a) the torque T
when the angle of twist at A is 25 , (b) the corresponding diameter of the elastic core
of the shaft.
L" I.S *
G=77*/o ■'
SOLUTION
C = ^d - O.OIO m ^
^r (77*/oi^o.©Jo)
(cO Tr|Ty(l-i^)^|C^7.77)[/-i(^f737)3J - 283 */■»
(t) & , ^ = O.G4737 /^* 0.GH737 c * (0.61757)fa.ow)
-1
7

PROBLEM 3.101
3.101 A 18-mm-diameter solid circular shaft is made of a material which is assumed
to be elastoplastic with G = 77 GPa and ry = 145 MPa, For a 1,2-m length, determine
the maximum shearing stress and the angle of twist caused by a 200 N-m torque.
SOLUTION
TV- i^x/o'Pi ; c= -Jto* ?o~Ceftn, L - 1.2* j T= Zoo M-^
T^ £Ty(l-*£)
($)** 4- JX^.M^L,a3^
^ : 0.7ZZ37
CO'- —
11%%1 ' 0.12837
-3
3417^0 r*J = /?.73
PROBLEM 3.102
SOLUTION
3.102 A solid circular rod is made of a material which is assumed to be elastoplastic.
Denoting by rY and <pY respectively, the torque and angle of twist at the onset of yield,
determine the angle of twist if the torque is increased to {a) T = 1.17}., (b) T =
L25Tr, (c)T=\.3TY.
£,
& fT-$
Lc£\ -f * l-lo
v
£ !
V-ft)0-3)
- 2,/r

PROBLEM3 103 3'103 A 0.75-in.-diameter solid circular sr^ is rnadeofaniaterial which is assumed
to be elastoplastic with G = 11.2 x I06 psi and r, = 21 ksi. For a 5-ft length of the
shaft, determine the maximum shearing stress and the angle of twist caused by a 2-
SOLUTION kipin. torque.
C = j[d "- 0.325" m j &- 11.2 Wo* p*;;, rr = 2/ Us," * 2looo^9)-
L = -b'ff. - Go iv T - 2 *>*!* - 5 Wo* ^4./^
TrriOr, f C*t/ - f(Q.Z?st (%\ooo) ^ /.73WWO* -A./W
r = 7, (/ - +£S v--
-^ - i _ = i _ aito

PROBLEM 3,104
30 mm "0 mm
3.104 A hollow steel shaft is 0.9 m long and has the cross section shown. The steel
is assumed to be elastoplastic with vY = 180 MPa and G = 77 GPa. Determine the
applied torque and the corresponding angle of twist (a) at the onset of yield, (b) when
the plastic zone is 10 mm deep.
SOLUTION
(0^
J -
■^d2 - O.OSS"^ ^ c, "- t^i r 0.0/5"^.
■= 11-71 kl0**»
HI
<Pr Z
TrL Ol.TrM^X^)
-3
£3 (77«tO,K2.^T7?>«JO"c)
r 60.11 y|o r*J ~ 3.W
M t - O.OlO m Pr = C4 - £ = 0.03.5"- 0.£/0 =■ 0. 0£5*^
r * £$ = &M *yY =&■
L i_ 'Or
x = ry & f>^ ^
%-- ^A »we j-,.-^;-<:,«)
.-1
J( = ^(0.025-r-O. 0/5"") * £34.07*10" V*
- ^(igo^io'X0-03^3' O.02S*) r 10.273 * to* M-m
•3
ToT A/ ■fura.y*
T^T-^*
3. MS*la 4 10.2*73 */0* r
w.rc x/o3 *;•*,
)f-U kW-in
—

PROBLEM 3.105
30 mm 70 mm
3.105 A hollow steel shaft is 0.9 m long and has the cross section shown. The steel
is assumed to be elastoplastic with ty = 180 MPa and G = 77 GPa. Determine (a) the
angle of twist at which the section first becomes fully plastic, (b) the corresponding
magnitude of the applied torque,
SOLUTION
__ I
(ck) Fo/- omsef of -folly p^ksi/c yieJ*fi'n*j p^ - C,
?>- V • Vs- 2i - ^$ - <=LiP
3 r i4.2<* kN-ho
PROBLEM 3.106
^J B- £ C jP
J./0tf A shaft of mild steel is machined to the shape shown and then twisted by
torques ofmagnitude 45 kip-in. Assuming the steel to be elastoplastic with zy=21 ksi,
determine (a) the thickness of the plastic zone in portion CD of the shaft, (b) the
length of the portion BE which remains fully elastic.
SOLUTION
(0.) In po*4i'ow CD C = j«| = 1.00 I w
-£^0.71^3, pr*(Qjt2BKi-*o)* 0.713 .'^ ty - C-^ ^0.227 ;„ -*
(lb) For N/ieiJi'n* J" pom+ F 7" - % _, C -" Ce J T * VO k-p in

PROBLEM 3.107
A \ B
3.107 The magnitude of the torque T applied to the tapered shaft of Prob. 3.106 is
slowly increased. Determine (a) the largest torque which may be applied to the shaft,
(A) the length of portion BE which remains fully elastic.
SOLUTION
(cU The ia^jes-f -f-o/^ue Whi'C'M v*my be <xpp$izj +» Ht
sW-f-V vnctke-s po^i©* CD ti/Wy p.fos'frc.
T~f Tr(l-^f) "i?? * £(33.^7) = *Su?g^k.p.Mrf W.Okr'>» -*
ttD For yUfJinj a+ p©.'rff £", * a 2^ ; C » Cr j T* Y3.TO2 fc/a-iV
Lfir " 7rrr - 17^3 - J-3336* '*
Ds«*«ei p^po^+toos t^tt*-, 4-^-e sWW/>
t
J
i.*t-
i-xs-
UlooG,
- |.0O
x -
-
2.31
^
V
,
/M

PROBLEM 3.108
3.108 Considering the partially plastic shaft of Fig. 3,38c, derive Eq, (3,32) by
recalling that the integral in Eq, (3,26) represents the second moment about the raxis
of the area under the x-p curve.
The s-frcaS is HJ" std*j* ©^ -H,e J-rH*.
uJ»er< JlA - ?. sip a*J X= 2*J
r - i, -1,
22V r3 -ir

PROBLEM 3.109
3.109 Using the stress-strain diagram shown, determine (a) the torque which causes
a maximum shearing stress of 15 ksi in a 0.8-diameter solid rod, (A) the corresponding
angle of twist in a 20-m. length of the rod.'
0 0.002 0,004 0.006 0.008 0.010 y
SOLUTION
F***^ "He stress- st**-'* «iV«ir<5u^
I--f Ivz^
*^ Az - 0.^5" ^«. ««^
2
0
0.25
O.S
0.75"
1.0
r
0,000
0.0O2
0.OO+
aoe>£
tXflo*
-r; ks;
O
8
ia
m
15
z% fa
O.OflO
0.500
3.000
7.S75
15". ooo
W
>
f
3
4-
1
vV^% tfs.'
0,00
2.oo
6,0O
3l.5to
\5"-00
*—/( rVl --
■j".o° ^
Zm/zV
Ic ^r)(si^V = ^ fef
Woffe. : Afvs^ez-s w*y olr-PfeA sA^Af-fly Joe. -Jo eirpfe^^ots ©T-opiAi*<\
!w recViA^ +A«, s4r-e^3- s-frfet* co/v^.

PROBLEM3.110
0 0.002 0.004 0.006 0.008 0.010 y
3*110 A hollow shaft of outer and inner diameters respectively equal to 0.6 in. and
0.2 in, is fabricated from an aluminum alloy for which the stress-strain diagram is given
in the sketch. Determine the torque required to twist a 9-in. length of the shaft
through 10°.
SOLUTION
c, = id, - o.ioo *'*, cz — £<4 =■ b.300 m
X*+ - Sf& r(Q.3ooX/7V.^Wcr^ = QooSSZ
T= Z*<£ ftJp - W?I ***** = 2ir<i3X
wV\e#«. H« mfe^ir*^ J is qiV<*v ky X = j 2 't dz
Let
iM*.y C2
I 3 |Z«^
2
i/3
i/z
2/3
5>fe
1
T
O.oor?**
O.0o2?|
0.003*3
C*.oo*/"35
0. ooSzl
r, Irs.'
J,o
l<?.o
u.r
l3.o
it.O
zV, k*;
o.si
ZSo
Jill
^©3
If.O
*,n}
w
1
Y
2
*t
1
4z. = j- *
W2Z%U%i
o.#?
10. DO
10.22
34.11
14-co
VZ QCf The V
„—. z. ,
Ifcdi
m

PROBLEM 3.111
d = 50 mm
3.111 A 50-mm-diameter cylinder is made of a brass for which the stress-strain
diagram is as shown. Knowing that the angle of twist is 5° in a 725-mm length,
determine by approximate means the magnitude T of the torque applied to the shaft.
SOLUTION
£> - S° - 8?.2te*i°~* roJ
C - -M = Q.OZS ^ L ^ 0.72S r«
'"* * L " CUT
£\j&&oJkJ. i>v»n <x method of no^e^c.^/ tniejt^riO^\, It Sr^s<»n 5
rt>Je js ost4 . -hke )h+e^r*-ti*>* fip^fM^l*. fs
wh&*2- w is ct wer^A"/"**^ -P^c^u^. UsTw-i Az = O.Z5'J we «=jeT -Hi* \j^J^^.s
0.001 0.002 0.003 r
z
0
0.25"
o.s*
0.75"
l.o
r
D
O.O0O7S"
O.OOlS-
o.ootzc
O.oo3ol
r, MPo.
O
3o
S3*
75*
SO
Z1^, MP^
0
U%1S
\%.1S
42. n
So.
w
1
4
2
4
1
WZ'V, HPa.
O
l.S
XI. 5"
I6S.7T
So.
/83. Aj *
-
ZTwzV

IVsWj 1V>UUI J, J J WITH), OllU \\J,\J\jj,ovlvia aj. u; utuiig iiiv puijiuuw, - *A . „
cy through these points the following approximate relation has been obtained.
r=46.7x I09r-6.67x 1012 y2
SOLUTION Solve Prob. 3.111 using the relation, Eq. (3.2) and Eq. (3.26).
Ld- z - X - ^"
*W C
ir -= A + 8r + cr1 = A + BYU z. * ciCz1-
T^ 2Ti(o.oar^^ o 4 3s-.i3*|o3- i?,o8v/o3) - 2.Z& *loz W-^

PROBLEM 3.113
25 mm
10 m
1x3
on
3.113 The solid circular drill rod AB is made of a steel which is assumed to be
elastoplastic with ty = 160MPaand G = 77Gpa. Knowing that a torque T= 5 kN-m
is applied to the rod and then removed, determine the maximum residual shearing
stress in the rod.
SOLUTION
C - O.CKL5" w
% r i& . Jc*^ - J(o.o3a)sOfiO"io*) r 3.9*7* ios M-hi
£=H-
ST , 4.i|l^!^ = 0.iao„
T,
3_<U7*fc*
\
-3
WKet'*
J
*o3.72 x/o* l^o.
At ^ ^
r
"ft** l&o*/oe - ll£"-08*/O6 -
- 43.7 MrV
44.<* MPo.

PROBLEM 3.114
3.114 The solid circular shaft AB is made of a steel which is assumed to be
elastoplastic with G = 11,2 * 10 psi and ry - 21 ksi. The torque T is increased until
the radius of the elastic core is 0,25 in. Determine the maximum residual shearing
stress in the shaft after the torque T has been removed.
SOLUTION
C = O.IT.V
Pr " 0-2S ,'„.
0.75 in.
Af t*el ef Ioax! nm '•
- \8.383 k:p .Vj.
Tn
€ Stresses CL^t
w^ Tf - Z\ Ksi ^t f> - O.ZZ i«
'\ - tr = 3.1 kV «t /a r O.-JS" /*.
A* fi--o.tr;* f- 'M*™£xt) - 9. ^ k».-
'^>,* = 't-r * <i ksi «T /a
7W?^* tla^gc Jjrlr\A or\J}o*J)'\*<\ ~T ~ " l8*'SS3 J<»y»\V
RttiJotJ si messes a.^e 4oJ*W ty a^/^^ "£^ ■= '7^0lw/ — "T
A+ f = 0.357* ^ - 21 - <7.7S" - 11.75" to,'
At /o * A 75- ;* 7^ " 2.1 - 27. 7V = - 6. 7T fcs;
0.7T
/°
-6.7V

PROBLEM 3.115
30min
3.115 The hollow shaft AB is made of a steel which is assumed to be elastoplastic
with vY= 145 MPa and G = 77 GPa. The magnitude T of the torque is sbwly
increased until the plastic zone first reaches the inner surface; the torque is then
removed. Determine (a) the maximum residual shearing stress, (b) the permanent
angle of twist of the shaft.
SOLUTION
Voa-dlf\
m : wV*e* pY re.t**Ae.$ i^ne^ ■s<jw\Fft^e.J +^ev% 7^ = "£y
• ct
%»-< ~ (o.Oi*X77* to**)
= l3o.77 vlo" r*J
7.4^3
-^
"St. 23*/o" r»J - -S".3«*J
«?esrJ^^: r^ r r(
r«4 - iJ«J
<M <ftw - #'
(a) A+ yO=C^ t^* I4SHO* - 174. N*/0* = - 2<7. \H >)04 P^
At p - C, li^ = l*/r*JO* - lo*.4<?*/o
40.S- MP^
(k) Cppe^= I3o-77xlo's- 94.73 WO"3 = 36.SH*io-* r*4
^.o^0

PROBLEM 3.116
0.6 m
16 mm
c
J
At
3.116 The solid shaft shown is made of a steel which is assumed to be elastoplastic
with ty = 145 MPa and G = 77 GPa. The torque T is increased in magnitude until the
shaft has been twisted through 6° and then removed. Determine (a) the magnitude and
bcation of the maximum residual shearing stress, (b) the permanent angle of twist of
the shaft.
SOLUTION
C= 0-0 16*1 op - Ga - |o4.7:?*fo'V*-l
0.0O1823I
= ^C*1^ £ Co.oie^* 102,944 '/d"* */
no
A+ f = fr X'-- f£ £ - (-l7».«*/o* (0.G7H33)- l3o.38x/o4 R
~y G--T C^W'XW.-Wx/a*)
At />= C 2^ ' l«Vlo'- l7*.52*lo' - -33-52*10*" fit
= -3315" MPa.
<P~-s ~ IMJ^/o"1- S6.9W /JO*3 = 17,32 *tO*r«J t I.Oir

3.113 The solid circular drill rod AB is made of a steel which is assumed to he
PROBLEM 3.117 elastoplastic with Ty = 160 MPa and G = 77 GPa. Knowing that a torque T= 5 kN-m
is applied to the rod and then removed, determine the maximum residual shearing
stress in the rod.
SOLUTION 3.117 In Prob. 3.113, determine the permanent angle of twist of the rod.
Fro* He SofAto* -fo PROBLEM 3.113
Per»*\a^6V.r twist a.n<>ife
Cp r Cp^ -,<p r I.H7/S-/-OS-S3 =■ D.4130*^ - 53.7° -^
.__ .-. |— -_.. - Ll--—-~—---■-, ——-.. xr —-p-._iii_L'--^— --—— r ' ' '—' .__J__ _ __^ „ 1
3.114 The solid circular shaft AB is made of a steel which is assumed to be
PR OUT EM 3 118
elastoplastic with G - 11.2 * 10 psi and ry = 21 ksi. The torque T is increased until
the radius of the elastic core is 0.25 in. Determine the maximum residual shearing
stress in the shaft after the torque T has been removed.
SOI 1ITION 3.118 In Prob. 3.114, determine the permanent angle of twist of the shaft.
fvo*n Ue SoJItA-t** +- VRoSLeM 3.t\H } £ =• 0-")S" ,\^ J"= O.*»97o/ ,V^
d) , - &*ffi^ - c \tono*rU -- 10.31°
T GT (ll.**lo3Xo.T77oO

PROBLEM 3.U9
TiTr
(al
"To Tx'nJ C-
3.119 A torque T applied to a solid rod made of an elastoplastic material is increased
until the rod becomes fully plastic and then is removed, (a) Show that the distribution
of residual stresses is as represented m the figure, (b) Determine the magnitude of the
torque due the stresses acting on the portion of the rod located within a circle of radius
SOLUTION
Lofl-d inq
Dv\io&A Inq \
gesr«fiift.l
- X£ „ «? T". T 2( W)
TTC
TC;
S^r -* fr C
*-' ^-jK£ =^0-g)
<0 T - 2-rr £y r 4. = n (|C ^Vy 0 - II H
= 2.TT

PROBLEM 3.120
SOLUTION
3.116 The solid shaft shown is made of a steel which is assumed to be elastoplastic
with tt = 145 MPa and G = 77 GPa. The torque T is increased m magnitude until the
shaft has been twisted through 6° and then removed. Determine (a) the magnitnde and
location of the maximum residual shearing stress, (b) the permanent angle of twist of
the shaft.
3.120 After the solid shaft of Prob. 3.116 has been baded and unloaded as described
in that problem, a torque T, of sense opposite to the original torque T is applied to
the shaft. Assuming no change in the value of fa, determine the angle of twist fa for
which yield is initiated in this second loading and compare it with the angle fa for
which the shaft started to yield in the original loading.
Fro** fk< &t>tot\o* -h> PROBLEM 3. 116 C= 0*0 16^ L^ 0.& *
% - N5* *loc ?aJ J = IdZ.WH'vId'' n*
Tke rest Jo <J s+ress ok f> = C is 7^ ^ &$^S MPa.
opposite, setose. "He 6»/iA*^e /*■» stress +»
J-
1 C 0.01C
= 717 W--w»
. Xt . (7-17. xlo'&)(o>fe)
-3
- 3- '/ •■

3.114 The solid circular shaft AB is made of a steel which is assumed to be
elastoplastic with G= 11.2 x 106psiand tY = 2\ ksi. The torque T is increased until
the radius of the elastic core is 0.25 in. Determine the maximum residual shearing
stress m the shaft after the torque T has been removed.
3.121 After the solid shaft of Prob. 3.114 has been loaded and unloaded as described
in that problem, a torque T, of sense opposite to the original torque T is applied to the
shaft. Assuming no change in the value of ty, determine the magnitude 71, of the
torque T, required to initiate yield in this second loading and compare it with the
SOLUTION magnitude Tr of the torque T which caused the shaft to yield in the original loading.
Fro* He SoJ>o±it»A -k PROBLEM 3. 1/4 c - 0.7S" i" t L- 2^ In.
% - Q.\ Usi > <J r 0.4^70/ In*
Fo«^ J'ofl.di'n* \\a "hXe opposite fiensfj -He alia*i*c iV» s+rejs +«.
'" ~ '" T| ' cT ' oris *.W.U,f,K -«

PROBLEM 3.122
1.8 in.
(a) ■
3.122 Knowing that the magnitude of the torque T is 1800 lb-in., determine for each
of the aluminum bars shown the maximum shearing stress and the angle of twist at end
B. Use G = 3.9 x 106 psi.
SOLUTION
T - JSOO A-i* , L- 36 i*
(gO 0_ ^ (.3 ;rt j t . o.6 .'* £ r L| - 3
d 1.8
<-> . TL _ (/goo)(3Q
=r K^S^/O r-w/ - 7.3/
Gap
r-T4
(/gOO U 36)
-3
C^ab^G- * <0.l4©6Xkfc)0-o)*(3.^*|o')
$.2*lO (*J r 6.7?

PROBLEM 3.123
3.123 Using tm= lOksi, determine for each ofthe aluminum bars shown the largest
torque T which may be applied and the corresponding angle of twist. Use G = 3,9
x 106 psi,
SOLUTION
t^ - 10 k^t G-%.ciWo'f%; * 3.^^/0* k*;
rvo*v T«JJ« 3. I C, :0.^7j ^=0.563
^reifc ■"• T=c,at'«W
?r
TL . 0-7SO){36^
ctaL*G
= )5&.PXIQ
-%
(^-03) (36")
TL
rp ^—Li - K"~~" '*•-"- 1 ^ _
~ C^a^G- (o-/«h* Xlo^.o^m Wo5 ) "
3G.6 v/o"3 rA«/
= 7.8*

35 mm
50 mm
400 mm
r>
r
vw»x
C, OlW
3.124 Knowing that 7= 800 N-m, determine for each of the cold-rolled yellow brass
bars shown the maximum shearing stress and the angle of twist at end B. Use G - 39
GPa.
SOLUTION
**i*wi ~ O.05O v*^
F^TUfe 3. / J C, - £.2og3 CA = 0. Hot
(0^08")(0.05-0)(O.OS*0")
r O.SSS"
Q>) <X* ?0 to*, * CK07O * j t r 3S*> * O.OSS"*^ ■£ s 22 a Z,0
T goo
^ -
^ ^ 37.9x/o'Pst " 37. <* MP.
9 r c^s
Co.X2^XO-07c»X«>^SSy^).|o€»)
^V WO~ Ta.fi/
- o.QSi

PROBLEM 3.125
3.125 Using r„i = 50 MPa, determine for each of the cold-rolled yellow brass bars
shown the largest torque T which may be applied and the corresponding angle of twist.
Use G = 39 GPa.
SOLUTION
"£**»/ -
X
c^w1
T" c.at*'^ - (o.2o&Xo.os,oXo.o5*o,)t(5o>-/o6)^ l3oo w.h
= |.3oo kM-^
(bl <X = 7c> ^ ^ O.OTor* ^ L^ &S**,M* 0.035**1 ^ ■£ = || = 2. O
Fro* Ta,Ue 3. I c, " 0.24£ ^ Ct = O. 2%*\
tl Goes*) (0^00}
cj) =
-i
c^alo^Gr " Co,2i<?Xo-o9o)(o.oS5-,)3{31' y /o*?

PROBLEM 3.126
3.126 A 2-kip-in. torque T is applied to each of the steel bars showa Knowing that
r,i, = 6 ksi, determine the required dimension b for each bar.
SOLUTION
(c)
• 3
7TC* T b3
(tl si**.**: CL-k f- =/.o. F^*. rr^Ue S.J ct - O.SoS
*7*l AX "' 2
T
* ca*
b - 1.170 tn
T_T • \a- T_ e
" cab* " z^b*
2<:,r^ «X0.^6)(4)
r OX^ ;n3
br 0.S7S ,'•*
PROBLEM 3.127
(c)
A127 A 300-N-m torque T is applied to each of the aluminum bars shown. Knowing
that ra|, = 60 MPa, determine the required dimension b for each bar.
SOLUTION
T - Zoo M■ * %*, = Goxio* Pa,.
r - Is. - ax - ^ r
b= 2%H*ioS * '- 2*-4 thin -
^
S^t>fl,»r<r
: a= b , t- - ^o- F^w "uJJe 3J c, = O. zos
? -
; b
c.ab* c(b3
T
t3^
3co
C,fKV ~(O.2o3)(&0x|O
-t r 2ir.0Mv/o'6 hnS
-3
b* 2%.c\y\o ^ = 22. <1 **
T_ __ T__ % ,3._T _ 3oo
£
:. \? =
*-* "c.ab^ ~ 2Ci t3 - b" "^pg7v ' tzVo.af^(6o>/o*l
=: /O.IGV/O ^3
b *ZL7*l$ ^ -31.7 *.*.

PROBLEM 3.128
20 mm
3.128 The torque T causes a rotation of 2 ° at end B of the stainless steel bar shown.
Knowing that G = 77 GPa, determine the maximum shearing stress in the bar.
SOLUTION
ty. - 3>0 mm * O. OZti hn ]q = %0 mv* ■= O. OZO «n
^r cat* * cTqFmT ~ c,i_
f1 = J^ = 1.5 F^o*. 7it/c 3./ e( r o.«/ Ct*0. y»^8
r - (DJ'?>fS)CO>^0^77VfO<lKgV.9g7x'fQ"3) = q0 g 0« p^
(0.231)07501
- 6^.2 HPsc
PROBLEM 3.129
t
ft
1
-«—b—"
6, cirol e
c=H
3.129 Two shafts are made of the same material. The cross section of shafts is a
square of side b and that of shaft B is a circle of diameter ft. Knowing that the shafts
are subjected to the same torque, determine the ratio of the maximum shearing stresses
occurring m the shafts.
A, s1 o*k £^i } c, = 0.ZO2 (Vjkh 3.1)
-tf
. T
* e.aw* ■ aioFp
Ta
■J JL .
J TTC3 " TT t3
o.3^5r ^ <x<m
3.130 Determine the largest allowable square cross section of a steel shaft of length
4 mif the maximum shearing stress is not to exceed 100 MPa when the shaft is twisted
through one complete revolution. Use G = 77 GPa.
PROBLEM 3.130
SOLUTION
Cp ' lit rW3
C,^» ->
-j
- t.m^io ^ = uz%% *.*>»

PROBLEM 3.131
A ;
3.131 Shafts A and B are made of the same material and have the same cross
sectional area, but A has a circular cross section and B has a square cross section.
Determine the ratio of the maximum torques TA and TB which may be safely
applied to A and B, respectively.
SOLUTION
V«"i" C = ^aoIi'us of ci^cora^ sec^io* A aj^Jl \d = stJc
T„ -
7i* cfc'-fc
~to
X b3 y
ra*rA
2c, WIT 2ia
TL--J
- /.3iT&
PROBLEM 3.132
3.132 Shafts .4 and £ are made of the same material and have the same length and
cross sectional area, but A has a circular cross section and B has a square cross section.
Determine the ratio of the maximum values if the angles (f>A and <pB through which
shafts A and B, respectively, may be twisted.
SOLUTION
L©+ c ~ poJltos or cvVeJ/ft^ secno* A av\J[ b - siWc
cvue; T_- %-. ^ .; ft* -^
tk^
t;
71
C,ab'
% - c£ab*G "
o./foc. VG-
T8 = O.^og i?2i
M7?4 Lfo
bG
r 0.
tr*
„G7C |^- *',OX?6Jli 4%
4*
S r 0.676-/? * M^8

PROBLEM 3.133
3.133 A 1.25-m-long steel angle has an L 127.x 76 x 6.4 cross section. From
Appendix C we find that the thickness of the section is 6.4 mm and that its area is 1252
mm2 . Knowing that rall = 60 MPa, G = 77 GPa, and ignoring the effect of stress
concentrations, determine (a) the largest torque T which may be applied, {b) the
corresponding angle of twist.
SOLUTION
A = \?S1 *»** b - &-M **, =
00 = ^ * W1 = ,,r-6 "^ *
0. Oo&4 ►*
O. 1^5*6 *i
C,«C4=^(l -O.65o£) - O.SZZS
T
.-. T - ct<xbz?U.
If -
■2.
a>) $> =
Ctab^G- ~ C2<xio3G- " C*bG- " tG-
PROBLEM 3.134
L4X4Xf
3./.W A 3000 lb-in. torque is applied to a 6-ft-long steel angle with a L 4 x 4 x -
cross section. From Appendix C we find that the thickness of the section is \ in. and
8
that its area is 2.86 in 2 . Knowing that G = 11.2 x 106 psi, determine (a) the maximum
shearing stress along line a-a> (b) the angle of twist.
A = 2.86 inzJ b = f in = 0.37^ ,Vi j (XTk'-'0^k ' 7'£l7 ,V
SOLUTION
T * 7$£ ~- *°-*1 C< * C* -~ kO - °^° k) = °-323(
0-37S
T
3ooo
KM*: l» 6 ft. « 7* ;,
" 3.5"/

PROBLEM 3.135
3,135 An 8-ft-long steel member with a W 8 x 31 cross section is subjected to a 5
kip-in. torque. From Appendix C we find that the thickness of the section is | in. and
that its area is 2.86 in2. Knowing that G = 11.2 x 10* psi, determine (a) the
maximum shearing stress along line a-a, (b) the maximum shearing stress along line
b-b, (c) the angle of twist. (Hint: Consider the web and flanges separately and obtain
a relation between the torques exerted on the web and a flange, respectively, by
expressing that the resulting angles of twist are equal.)
W8X31
SOLUTION
a 7. ffS_
. .T,L
=■ 13.3*
7.(3
Ci = C* = «0 - 0.630 £) * O.WH (J),
7; e Ciab*£& a ^fi*
K, = (o.W^X/.is^.mO4 = O.0«3 i«¥
TF -
(CL)
F C2X0.2l3.a>a^^S * CQJ>-2/3S)HXoJ*3
TP 222/
F " c,abl (6.32rtX7.nsX0w<#55-)
r*-=
T*
«y
^S^7
L ^ + 1^
, cp - TL
GGNt+KtO
2<**o par = Z96 kSf
<P =
(5b«°y%l
(//.a*io^[ro&).#3»H0-«33
= 3S.6 xio* rU - S.Ot

PROBLEM 3.136
J
W250X58
3.136 A 3-m-long steel member has an W 250 * 58 cross section. Knowing that G
= 77 GPa and that the allowable shearing stress is 35 MPa, determine (a) the largest
torque T which may be applied, (b) the corresponding angle of twist. Refer to
Appendix C for the dimensions of the cross section and neglect the effect of stress
concentrations, (See hint ofprob, 3,135.)
SOLUTION
FJ>
'«,»»y •' a-- 203 "^ b~ te.rmr^ "^ r I^O*
|^_ 103—>]
r
\^777rrrzziZZ\
k
til Jfr-S
T
KF - (o.zwtXo.zosXo.oiSsf = isy.szxio'01 w>v
Web - a - 237 -ft)(l3.S^= 22S ^ ^ b ~- 3 mno
Km/ -t*
16 7-5 3
Tw ' QafcV = C0.33^)(0^25)(o.w>8t(3ls-y/o4) = /6*.Z5 W-*
37. 5V
ZLZH
r<
L
T
r -- 726* A/-M
9 =
It r. (ya^Xs-oo") , /6i,Qy)oV^: 5-.7f
( ?lfr + & )G (3f6.6 >W« )(77Kl6qJ

PROBLEM 3.137
Gmm
60 mm
-C -
T
5ta &)(o.oo&)(Tiai»io-')
3.137 and 3.138 A 750-N-m torque T is applied to a hollow shaft having the cross
section shown. Neglecting the effect of stress concentration, determine the shearing
stress at points a and b. Th/e-kn^ = Gmm.
SOLUTION
Arect hounded kg ce*f"€«/* fine
= 733/ *;o" *>*
^ *.<*7*/<?*P* - 8.*7Mfru
750
PROBLEM 3.138
3.137 and 3.138 A 750-N'm torque T is applied to a hollow shaft having the cross
section shown. Neglecting the effect of stress concentration, determine the shearing
stress at points a and b. Tti re (chess B $ mm.
SOLUTION
De-fa,.*J' of corner
e -
t *= e +a*i 3c1
t
f *
r
2ta
■fc ■» 0.0(3 8 iv,
— 252 j- - |8.67^I06 ?CL
rr 6.128 m*

PROBLEM 3.139
a 0.2 in.
3.5 in.
3.139 and 3.140 A 50-kip-in. torque T is applied to a hollow shaft having the cross
section shown. Neglecting the effect of stress concentration, determine the shearing
stress at points a and b.
SOLUTION
■0.2 in.
r 12.40 './
.8) I
5.S*
0.3 in.
i_
Af po*'*f o- £ = 0.2 /•
3.&
r*
«5&
2tCL * C3Xo^X'8.^o^
g.t? ks;
M p»i*n + Id £ - 0.3 ■**,
r".-
^o
at a czXo.syis.to^
- ^.-53 fcs,
PROBLEM 3.140
T
3.139 and 3.140 A 50-kip-in. torque T is applied to a hollow shaft having the cross
section shown. Neglecting the effect of stress concentration, determine the shearing
stress at points a and b.
E^r
SOLUTION
5 in.
Tfe
3
M 8
Sin.
L_
— 3in.-
Atea. \>Q^4&i ioy centerJ?r»e
= /2-7I9 in1
T £b
tr-
7*86 fcsi'
Of*
T
SO
2i& (ttyp.37sr"Kt*',>i<0
= ,5\34 ks;
4^
\^z% -H

PROBLEM 3.141
50 mm "
50 mm
JJ,
I—I
10 mm
X =
7±CL
3.141 and 3.142 A hollow member having the cross section shown is formed from
sheet metal of 2-mm thickness. Knowing that the shearing stress must not exceed 3
MPa, determine the largest torque which may be applied to the member.
^"T SOLUTION
J 110 mm
H*
4*
4»
4o
t = 0-009, hn
- 2.<fT M-m
-*0
PROBLEM 3.142
— 50 mm *
50 mm
r
~7
20 mm
-I
20 mm
t = O.OOl ho
■£- = ^n_
at a
3.141 and 3.142 A hollow member having the cross section shown is formed from
sheet metal of 2-mm thickness. Knowing that the shearing stress must not exceed 3
MPa, determine the largest torque which may be applied to the member.
SOLUTION
Area. beondeJ hy
n
3o
3d
IB
IS

PROBLEM 3.143
7
2 in.
2 in.
2 in.
1
3in,-
3.143 and 3.144 A hollow member having the cross section shown is to be formed
from sheet metal of 0,06 in. thickness. Knowing that a 1250 Ib-in, torque will be
applied to the member, determine the smallest dimension d which may be used if the
shearing stress is not to exceed 750 psi.
SOLUTION
Ob - LS.WXMH) - 7.0G<A * I7 4«6 - £06 J
T
t =
MH
sy\
T
».*?*
I2.TO
(ZXo.C* ^756)
- /3,sss<?
1.9*
d r i£ZSZ . , 73i-
i^
2. <**
PROBLEM 3,144
2"eH
2 in.
2 in.
3.143 and ■?./¥■# A hollow member having the cross section shown is to be formed
from sheet metal of 0,06 in. thickness, Knowmg that a 1250 Ib-in. torque will be
applied to the member, determine the smallest dimension d which may be used if the
shearing stress is not to exceed 750 psi.
SOLUTION
t -- 0.06 m 5 -£* 7.TO fSlj T"8 '?5"o -?t«M
-3 in. H
"tf *
T
1.^1
2
at a
^ 2££
'2S-0
S.1t
7.4636-v.a>d -(^o^Oto^o
3. 888^
d -
- O.'M*/
2.H- 4

PROBLEM 3.145
2.4 in.
Af p* ill CL
3.145 A hollow cylindrical shaft was designed to have a uniform wall thickness of
0.1 in. Defective fabrication, however, resulted in the shaft having the cross section
shown. Knowing that a 15-kip*in, torque T is applied to the shaft, determine the
shearing stress at points a and b.
SOLUTION
^WK'tis of ao\e.r~ ci>cPe - /*2 in
QaJiivs of" inner C\*-cJe " I. ' i*»
Meatt radi'i*t z LIS *,»,
-?*1
2t<X '\£tO*ObX%\*s\
r-
_ t .
\s
zmx'&Yoj-zX1*'1^)
- 2ZQ ksi
- iS.oi k*;
PROBLEM 3.146
(<D t ''~
3.146 A cooling tube having the cross section shown is formed from a sheet of
stainless steel of 3 mm thickness. The radii c, = 150 mm and c2 - 100 mm are
measured to the centerline of the sheet metal. Knowing that a torque of magnitude
T = 3 k N-m is applied to the tube, determine (a) the maximum shearing stress in the
tube, (b) the magnitude of the torque carried by the outer circular shell. Neglect the
dimension^f the small opening where the outer and inner shells are connected.
SOLUTION
Af-eA bounded \>y ce^^e^-ff^e
= Z1.21 *ICT* m*"
t -- o. ooz n
12.73 x/O6 P^ s ia.73 MPa.
a>*> t; - (2irc.tr.O * 2irc*iz'
~ 2T (o. /5o)*(<5.oo3)(t4.73 */o6 ") - 5,4t> */oS M-
^
= ^.40 WW**v

PROBLEM 3.147
3.147 A cooling tube having the cross section as shown is formed from a sheet of
stainless steel of thickness t. The radii c, and c2 are measured to the centerline of the
sheet metal. Knowing that a torque T applied to the tube, determine in terms of 7", c„
c:, and / the maximum shearing stress in the tube.
SOLUTION
Avtu. kaOrideJ k>v C£^"ie^i i'rt€
1
d- ttCc,1- C,M
su
e^i ntfi st^is
•'n3
+1
t*
T
T
2tCL nrtCcS-cJ)
PROBLEM 3.148
T'
{o)
%
(b)
T
3.148 Equal torques are applied to thin-walled tubes of (he same length L, same
thickness /, and same radius c. One of the tubes has been slit lengthwise as shown.
Determine (a) the ratio r/ra of the maximum shearing stresses in the tubes, (A) the
ratio <fa/<pa of the angles of twist of the shafts.
SOLUTION
fKrea. l>ou*Je*\ tw ce^e/\Jv*i<. * Q-- ire
X
2t<X 2irc*t
<Rt =
. TL _ TL
GO"
piTetG-
c, = c2 = i
T . 3T
u =
c,abl
- 3TL
fe*G- nrct*G-
b ' t
ZvdtG
3c
±
2^
:>>

PROBLEM 3.149
3.149 A hollow cylindrical shaft of length L, mean radius cm ,and uniform thickness
t is subjected to torques of magnitude T. Consider, on the one hand, the values of the
average shearing stress r(W and the angle of twist ^obtained from the elastic torsion
formulas developed in Sees. 3.4 and 3.5 and, on the other hand, the corresponding
valuesobtairKxl from the formulas developed mSec. 3.13 forthin-wallcd hollow s^^
(a) Show that the relative error introduced by using the thin-wall-shail formulas rather
than the elastic torsion formulas is the same for rave and <J> and that the relative error
is positive and proportional to the square of the ratio tlcm. (A) Compare the percent
error corresponding to values of the ratio //coequal 0.1, 0,2 and 0.4,
SOLUTION
Lfii C2 = ot^br radios ~ C*+ "££ OuaJ C, = '«m<^ rVj.us
<p. =
T
TL
TL _
r _ r
d = 77-CM
L»*t. =>
2fcCL *TT C^t
rt, - TL Lis . T^3ttC»70
TL
XTTC^tS-
= c„-it
RJ-i
e>S
<», rarc?t&' tl " c"~
*. , _ x£
* ?"' =
tcj
CtM
^
^c*"
O.I
0,OotS
0.2S7.
0.?
O-OI
1%
O.V

PROBLEM 3.150
Tg = 12.5 kip ■ in.
3.150 For the soiid brass shaft shown, determine the maximum shearing stress in (a)
portion AB, (b) portion BC.
SOLUTION
Tr-3.5iriP-iD. A6 ' T= l?.S"-3.r= 9 kp-wi 3 C = i*l--0.87r.V
r - l£ . II -£M_ - 5? ££ w- ~«
6C - T= 3.5 /e.>-iV. , C^of- Q.Clfl*
<- * vj TTCS TU062S? ^ ,3 *5'
PROBLEM 3.151
3.151 Knowing that G = 5.6 * 106 psi for the soiid brass shaft shown, determine the
angle of twist at point C.
SOLUTION
TB = 15.5 kip ■ in.
^..-3..5Wip.m. A8- T^ 12.5" -3.5 = 9 fc> I*, . c=-^j»0.815-.'n.
¥*'
T«L
*i >-*a
mcs°
-3
- 5?.^ */0 r^
<ft
8c
- T^L
«£.
C3.srKa**>
-i
G-a (5.6 x-jo^Xo. 2S«U S J
- SZ.tfoWo /^
M&+ ^cjj^c .f -Kk3+ Cpat - (Rg - IWWO'V<l#/ - 0.587

3.152 The stepped shaft shown rotates at 900 rpm. Knowing that r^, = 42 MPa,
determine the maximum power which can be transmitted if the radius rof the fillei is
80 mm SOLUTION
D - ICO
WM
cfl = #0 ^.
C - -f-J : lO mw ■ O.OtO M
7 ' T^* ' 2-°
<T ire3
• T
p
7= ft - °-Z5
= ZirFT r (jhrX10(3.^0* 10s ^ = 332ylozW = 33* kW

PROBLEM 3.153
3-153 The long, hollow, tapered shaft AB has a uniform thickness/. Denoting by G
the modulus of rigidity, shown that the angle of twist at end A is
TL c. + c.
k-
*A T '-B
AxGt c\c\
SOLUTION
Fv*om qeo*\eTry
3
-f-euA <X ~ ■
J* f^cL CA- C*
d-c,
- TL* f j L-) TL* _£(Q>-Ol Cc-c^'
/ 7 TL (c»»-q»0
fTTfo-OtG 6 4 <Vj ' ^7l(C6-C,)feCAtCQ2'
TL (c^G-)

PROBLEM 3.154
3.154 Two solid steel shafts, each of 30-mm diameter, are connected by the gears
shown. Knowing that G = 77 GPa, detennine the angle through which end A rotates
when a 200-N-m torque T is applied at A.
SOLUTION
CuScoMariotA of •haloes
Ci^-cow\Tese<*i\'oiJ/ contact fc^e be-fwe*^
T"off " Y~ T^fi = gf" (2oo) - 300 M-m
Twi'ST in S^aff D£
0.2 m
<P.
-3
—rrr r ^- ^7 *i° v*J-
- ^
°* Gr J"otf (77^10* )(7<7.^X>'/0"'* )"
rs oT a/' «ovi
Tvist i"f> shift AS
Lw * O. I +■ 0.2 * 0.4 4 0.3, = 0.9 m 0 TAa - 77.^2 */o
- Tic Lab _
<fi
>a
i2?5LL(^ll_ = „.3,6^r.J
6J"« (.77x/o'»)(7?.5Wx|o-'f)
=■ 3-79° -*

PROBLEM 3.155
SOLUTION
3.155 The angle of rotation of end A of the gear-and-shaA system shown must not
exceed 4°. Knowing that the shafts are made of a steel for which raU = 65 MPa and
G = 77 GPa, determine the largest torque T which can be safely applied at end A,
Ca.&cuJi/cd'ta* of -fz>^i>e rWnO
Co* TA*T Tvf-ta. P
Tml. - Tl
0.2 m
F-
Off.
Since iW^e*^ -fo^^e occ^s i'a s^rT.
S+r*ss i* sk*ff tils
TTC
TTC
T-"C2^ c'? » f(o.oisf(t£*io*) = tn.i N-m
<a
J- * f C1 = Ko.OlsV r -M.02 fid"1 *f , U£ - 0.S"* , 1W- '■? T
T^v'sf m s/i-iPf A8 : LA6 - O.I 4 0.7.4 0.4 4 OJ% - 0.9 m
=r 33o.7/*7o"fi T
E^V^ Sdo.ltrlcr4, T* 6?.*i3WO"3 : T= *//./ AM*

PROBLEM 3.156
3.156 A sheet metal strip of width 6 in. and 0.12 in. thickness is to be formed into a
tube of rectangular cross section. Knowing that tm = 4 ksi, determine the largest
torque that may be applied to the tube when(a) w = 1.5 in., (b)w= 1.2 in., (c) w =
1 in.
SOLUTION
oUp-Hi el ■"■ | - W
Are*. koiMeJee* ki ce^+c^/n-c & - Wol * W (•£ - vJ )
PROBLEM 3.157
3.157 Two solid brass rods /IS and CD are brazed to a brass sleeve £7^. Determine
the ratio d2/dt for which the same maximum shearing stress occurs in the rods and in
the sleeve.
SOLUTION
T' Let C, = -^ J, Gu*d Cz = -idt
6' J, 'He?
t-t*r e<iu*.^f sitr«5ses - - —; » —

PROBLEM 3.158
3.158 One of the two hollow steel drive shafts of an ocean liner is 75 m long and has
the cross section shown, knowing that G = 11 GPa and that the shaft transmits 44
MW to its propeller when rotating at 144 rpm, determine (a) the maximum shearing
stress in the shaft, (b) the angle of twist of the shaft.
320 mm
590 mm
SOLUTION
L -- IS *
. m ^
? = w mv - 4-4* ios yr
- ^. _ 32j
C ■= — =
^ Zlo ^^ - o. S^o
m^
fflL> r- l£* - i^^Zil^lX^^^l r g3.«**lo4 Pa.- «3.^ MP*.

PROBLEM 3.159
0.5 In.
Kl[> III.
T - 3 k^.v
>
3.159 The shafts is made of a material which is elastoplastic with rr= 12.5 ksiand
G = 4 x 10 psi. For the loading shown, determine (a) the radius of the elastic core
of the shaft, (b) the angle of twist at end B.
SOLUTION
C - O.S in
T= |t,(i- + ^ .-. |S4, H-ar, if.a%^).-o.siso7
£r .
0. £«*.?)£
L --= 6-Ft- * 72 I*.
^(0^3lg)(0.S): 0.3V7 ;«.
G-- */*io4^s: - *+* io3 k$'
<p,
Hk , 3TrL , «)(a-^4)(7a) = QmHHn ^
JG
ttchG tt (as)1^*!©*)
<P C T pY/C 0.6 13 i*
3.160 If the 3 kip-in. torque applied to the shaft of Prob. 3.159 is removed,
PROBLEM 3.160 determine (a) the magnitude and location of the maximum residual shearing stress in
the shaft, (b) the permanent angle of twist of the shaft.
SOLUTION
fVo^ +-Vi« soSvh'o* of "PROBLEM 3JS9 , <ft+ He eic* of -Po*Ji*hi T~ 3 kt'p • i«
£t = 0.^3/3 j qfj^ = O.&tlOH ir*-l
S+rtsse.5 ?s° *t/>=o , r= 13.5" ks,' *,if>:fY ^ ?* \ZSks: 4.+ J0TC.
ATT " ^o.^3i«)(-i^^g') - /o.^ Ms; <U p- fr .
A+ p = C Ti« * /2*£*-l£.:?# * - 2-78 ksi *m
CPoe^ "= <P*«I + ^ r 0.C41O4 -Q.SS0OH - 0.099 »^i

PROBLEM 3.161
T
Aluminum
3.161 The composite shaft shown is twisted by applying a torque T at end A,
Knowing (hat the maximum shearing stress in the steel shell is 150 MPa, determine the
corresponding maximum shearing stress in the aluminum core. Use G = 77 GPa for
steel and G = 27 GPa for aluminum.
SOLUTION
LeT Qx j J , ^ c~^ 'i-i r«W +■» "B\e txJv**.\'r,o»* Co*>« .
r- L
5. - •«
f.
c,G
So/yi/n1 "Tor iL
r* " c, e, *"

PROBLEM 3.C1
-Element n
Elemeni 1
3.C1 Shaft AB consists of n homogeneous cylindrical elements, which
can be solid or hollow. Its end A is fixed, while its end B is free, and it is
subjected to the loading shown. The length of element i is denoted by Lf, its outer
diameter by ODt, ita inner diameter by IDit ita modulus of rigidity by Gh and
the torque applied to ita right end by T„ the magnitude T, of this torque being
assumed to be positive if T, is observed as counterclockwise from end B and
negative otherwise, (Note that /Df - 0 if the element is solid.) (a) Write a
computer program that can be used to determine the maximum shearing stress in
each element, the angle of twist of each element, and the angle of twist of the
entire shaft, (b) Use this program to solve Probs. 3.9, 3.35, 3.37, 3.150, and
3.151.
SOLUTION
FOR £feH cyiljHD£.icftL plfmFtor, £/-/rirh
^c 3
tvi ,
si > Ti
Problem
Element
3.9 <x.-h<i 313/*
Maximum Stress
(MPa)
Angle of Twist
(degrees)
1.0000 56.5884 2.5265
2.0000 36.6264 0.8887
Angle of twist for entire shaft - 3.4152
Problem
Element
3.37
Maximum Stress
(MPa)
Angle of Twist
(degrees)
1.0000 33.9531 0.8314
2.0000 19.6488 0.7413
Angle of twiet for entire shaft • 1.5726
Problem
Element
3 .150 27>J 3./S~/
Maximum Stress
(ksl)
Angle of Twist
(degreesI
1.0000 9.1266 3.5857
2.0000 -8.5526 -3.0002
Angle of twist for entire shaft - 0.5855

3.C2 The assembly shown consists of n cylindrical shafts, which can be
solid or hollow, connected by gears and supported by brackets (not shown).
End A, of the first shaft is free and is subjected to a torque T0, while end Bn
of the last shaft is fixed. The length of shaft A,£, is denoted by £,,, its outer
diameter by ODlt its inner diameter by tDh and its modulus of rigidity by Gt.
(Note that ID, = 0 if the element is solid.) The radius of gear At is denoted by
a„ and the radius of gear Bt by b{. (a) Write a computer program that can be
used to determine the maximum shearing stress in each shaft, the angle of twisi
of each shaft, and the angle through which end A{ rotates, (b) Use this program
to solve Probs. 3.21, 3.39, 3.41. 3.42, and 3.154.
SOLUTION
LL CD. ,VL £;
Cc**pi>7*x J,: ~ (77*7)1.0I)i ~ '2>i )
7*% « TL l^l)x)Jji
COtopurj /?6-r&T/0" ^T TM£ "^ £W GF £*cti ^^fiF^
£?7##r WT# ftH£Ur * PW^ ANO Up D*TZ~
Problem 3,21
Shaft No, Max.Stress (MPa) Twist Angle (degrees)
1.0000 68.7420 1.4615
2.0000 72.5013 0.7707
Angle through which Al rotates - 3.388 °
Problem 3.39
Shaft No. Max.Stress (MPa) Twist Angle (degrees!
1.0000 47.7465 1.7764
2.0000 82.8932 2.0560
Angle through which Al rotates « 7,945 °
Problem 3.41
Shaft No, Max.Stress (ksi) Twist Angle (degrees)
1.0000 9.0541 1.3587
2.0000 12.0722 1.3175
Angle through which Al rotates - 3.115 °
Problem 3.42_
Shaft No. Max.Stress (ksi) Twist Angle (degrees)
1.0000 9.0541 1.3587
2.0000 6.7906 0.7411
Angle through which Al rotates = 1.914 °
Problem 3.154
~ Shaft No. Max.Stress (MPa) Twist Angle (degrees)
1.0000 37.7256 1.6843
2.0000 56.5884 1.4036
Angle through which Al rotates » 3.790 •

PROBLEM 3.C3
Element n
Element 1
3.C3 Shaft AB consists of n homogeneous cylindrical elements, which
can be solid or hollow. Both of its ends are fixed, and it is subjected to the
loading shown. The length of element i is denoted by Lh its outer diameter by
ODh its inner diameter by IDh its modulus of rigidity by Gh and the torque
applied to its right end by T;, the magnitude T{ of this torque being assumed to
be positive if T( is observed as counterclockwise from end B and negative
otherwise. Note that ID, ^ 0 if the element is solid and also that 71, = 0. Write a
computer program that can be used to determine the reactions at A and B, the
maximum shearing stress in each element, and the angle of twist of each
element, Use this program (a) to solve Prob. 3.56, (b) to determine the maximum
shearing stress in the shaft of Example 3.05.
SOLUTION w^ Con&pffl Ttff £JrflcT/aA? fi~r 8 fiS
Compute <&„ \**/7/i 7~#-a:
F&e ffcH ELfHEtT ftf7&L
L
oo.
to
'I
* , n
0**1* 71 - T3-$
T* 7+ t£
T^Ul - T(oj>j3)lJL
Compute (S>^ pur rz> unit 7~&#&oir &~r g
on/7 7nuL * otiJzJi
unit Ml - LilGl-k
UMT&&(C) =VM7~£fe/£)i- UnitF^i
/=&# 7Z>7*L QN&if- #t s t* at- smb.
77-&W
/£yg £-A*# fiw&vr : /t-fflx Swiss: 7hwt 7ftvL «■ 7fiDc +7^(u»>7™^)
PjZo6fi*^ Output
Problem 3.56 TA =
TH =
Element tau max (MPa)
l -39.588
2 31.670
-0.290 kN*m
-0.210 kN*m
Angle of Twist (degrees)
-1.178
1.178
Problem 3.05
TA m -51.733 lb*ft
TB - -38.267 lb*ft

PROBLEM 3.C4
lOfyC/^6 • 7^, > X,
3.C4 The homogeneous, solid cylindrical shaft AB has a length L, a
diameter d, a modulus of rigidity G, and a yield strength ry. It is subjected to a
torque T that is gradually increased from zero until the angle of twist of the
shaft has reached a maximum value (f>m and then decreased back to zero, (o)
Write a computer program that, for each of 16 values of tf>m equally spaced
over a range extending from 0 to a value 3 times as large as the angle of twist
at the onset of yield, can be used to determine the maximum value Tm of the
torque, the radius of the elastic core, the maximum shearing stress, the
permanent twist, and the residual shearing stress both at the surface of the shaft
and at the interface of the elastic core and the plastic region. (&) Use this
program to obtain approximate answers to Probs. 3.95, 3.113, 3.159, and 3.160.
'M
fr
SOLUTION
/ S~ "N
HCi)
<■ *
\ -
fiT OfrSfT OF- Y/f-LO
4v
£6>.(0
£0.(2)
7fl
<t»
UNj-fBOW ftlfiSTU)
*D-
V % J
T-T-
ft
Y
/C
y= TWO AT &^ <zz
\NHfrf$>tyYl 7^ U&r f&6) ft oSi fG(x)
7fi> r> ~ 'a K
7? 2
Y
CONTINUED

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PROBLEM 3.C5
1—2,-4
3.C5 The exact expression is given In Prob, 3.64 for the angle of twist
of die solid tapered shaft AB when a torqne T is applied as shown, Derive an
approximate expression for the angle of twist by replacing the tapered shaft by
n cylindrical shafts of equal length and of radius r, « (n + i! - |)(c/n), where
i- 1.2 n. Using for Tt £,, G, and c values of your choice, determine the
percentage error in the approximate expression when (a) n = 4, (b) n - 8,
(c) n = 20, (d) n = 100.
F&ort /^£p/S. %.£* f~X*cr £***tfitiFtSf&»'>
SOLUTION
^ 77^
o<e.
<T~ (/XTty/^tcy
= &*f8Sta^p*
COEFFICIENT of TL/GcA4
Exact coefficient from Prob. 3.64 is 0.18568
Number of elemental dinks = n
n
approximate
exact
4
8
20
100
0.17959
0.18410
0.18542
0.18567
0.18568
0.18568
0.18568
0.18568
percent error
-3.28185
-0.85311
-0.13810
-0.00554

PROBLEM 3.C6
3.C6 A torque T is applied as shown to the long, hollow, tapered shaft
AB of uniform thickness t. The exact expression for the angle of twist of the
shaft can be obtained from the expression given in Prob. 3.153. Derive an ap~
proximate expression for the angle of twist by replacing the tapered shaft hy
n cylindrical rings of equal length and of radius r; = (n + i ~ \)(c/n), where
i " 1.2 n. Using for 7", Lt G, c and t values of your choice, determine
the percentage error in the approximate expression when (a) n <« 4, (b) n = 8,
(c) n *= 20, (rf) n - 100.
SOLUTION
/S SArtftLL fifHO W*-" CAN V*&' t 1* TWJF -frS/*)eA/£?f
2*c -■
U
A<fi =
- T/K.)
6J,
COEFFICIENT of TL/OtCn3
Exact coefficient from Prob. 3.153 is 0.05968
Number of elemental disks = n
n
4
8
20
100
approximate
0.058559
0.059394
0.059637
0.059681
exact
0.059683
0.059683
0.059683
0.059683
percent error
-1.8B3078
-0.483688
-0.078022
-0.003127

CHAPTER 4

PROBLEM 4.1
2in.2ln.2in.
1 A
R
——i •—•—
M*
2 in. f
~1.5 in. (
2 in.
23 kip • In.
f <*L*
m #r
i- 1- if
4.1 and 4.2 Knowing that the couple shown acts in a vertical plane, determine die
stress at (a) point A, (b) point A
SOLUTION
2
■E
®
BC
T
For recA-&*Qje I ~ "i^kn
(W) ye = o.fs;»
>*
r - ** r - fag^O-7^ , - 0-«*>
ksi
?8.8iTY
PROBLEM 4.2
M = !»K) N ■ in
30 mm
40inm
4.1 md 4.2 Knowing that die couple shown acts in a vertical plane, determine the
stress at (a) point A, (b) point B.
SOLUTION
\Ti - -i oli * 1-5" wt*. ^ = j e)» B ^° ^"t
FT - - ^ - CgboKo.Olo)
* * I 85.903^lo*
- - 1/6.V */Oe Pa " - Mfc1* MP*.
= - 27.Z*to* P«_ s -87.3 MP*.

\
PROBLEM 4.3
43 The wide-flange beam shown is made of a high-strength, low-alloy steel fix which
Of = 345 MPa and Ou = 450 MPa. Using a factor of safety of 3.0, determine the largest
couple that can be applied to the beam when it is bent about the z axis. Neglect the
effect of fillets.
SOLUTION
P
-7*1-
250 mm —*J igj^
I,-- itLh3+ A J
&-1
- /3/. 706. */c?' m*V
I - I, + 1» + Jj «■ 5^/. "76 »/o'^ r ,?<?/.7* »/o"* v^'
s-^
wJl
ere.
c - 4~ = /8o ~»t * O. i£o ^
K* - g* r *f^>! , |^OHOCl>A
FS.
M
^w
= S** . (/&>*/oc )(*?/■ 7g»jo'ft) . ^43y/^w^
0- /So
£<r3> I^W-m

PROBLEM 4.4
IS mm
250 mm —*\ to j.
18 mm
ly* !,.+ I, + Ij
4J The wide-flange beam shown is made of a high-strength, low-alloy steel fix which
Oj - 345 MPa and % - 450 MPa. Using a factor of safety of 3.0, determine the largest
couple that can be applied to the beam when it is bent about the z axis. Neelect the
effect of fillets,
4.4 Solve Prob. 4.3, assuming that is bent about the y axis.
SOLUTION
c
/
0
Jb
A
M6.9os y/oc m^,4
- 23.438*/o' m^v
r 3.1 * 10° ^^g
I3 = I, r 23.438 ^^"
4C.9i>3 */0~C M¥
C -
ZS"o
1*1 - 125 "** - O. (55
e -
I
ISo * /ofc P*
M
Sj#X 0soy/oOi96.9o3*;o"t)
o.ns
= 56,3 X/0^ N-hn ^ St.SM'to
PROBLEM 4.5
0.1 in.
4.5 Using an allowable stress of 16 ksi, determine the largest that can be applied to
each pipe.
SOLUTION
C - O.Q l*
_ Mc
(b)
0.5 in.
(W I = $ (o.-7V" 0.5") - |3r</f */o",*i«,r
=r $.11 fc'p-

PROBLEM 4.6
y
80 mm
mm *\
c-* S2 = 40 m,
6. JJa
4.6 A nylon spacing bar has the cross section shown. Knowing that the allowable
stress fix the grade of nylon used Is 24 MPa, determine die largest couple M, that can
he applied to die bar.
SOLUTION
0.04O rw
»*1 (VI
M
6*1 „ (^*/0'Xs.agW»lO"* )
0_O^o
= 2.3S kM- ch

PROBLEM 4.7
4.7 and 4.8 Two W 4 * 13 rolled sections are welded together as shown. Knowing
that fix the steel alloy used <^a36ksi and au = $&kai and using a fiictor of safety of
3.0, detentiine the largest couple that can be applied when the assembly Is bent about
the z axis.
KZT^El
SOLUTION
c
3T*
Z.o>2>
ex.
?.Dft
1 A
du
P^f elites o-f W 4*13 ruiifJ section
See App***!** 13
A*-e^ = 3.83 ,'„* DepH:: <U6 •«
Ix = M.3 ■IftT
For owe roJjfeaf secTi'cn-j MO*ieit o~F iKter-+i'*L. *uto</f ff^fOtS GL-&* <**
!„ -- IK + A cil r //-3 + (5.S3X5.08)* -- 27.*7m¥
H«* — qjj
r. Mc
J?.^ fc.p-irt.
PROBLEM 4.8
fc —
4.7 aad 4.8 Two W 4 x 13 rolled sections are welded together as shown. Knowing
that for the steel alloy used a, - 36 ksi and % - 58 ksi and using a fiictor of safety of
3.0, determine the largest couple that can be applied when the assembly Is hem about
the z axis.
SOLUTION
X
W
rt
^080
Woper+ies tff W M*l3 ^J!ie^ secfi'o«
See Apf>e*Jir< 6
Atreec= 3.83 m* H/;J+li = 4.06O In
Iy - 3.86 in*
For otrtC Pel/eel secA\0m i^©«v,ewiT oT fuevfjfaL a.boO'i axTS c-t is
It " Ij + Aal* - 3.E6 * (3.S3)C3-03o^ « 11.6*3 ,V
d s- K/i'dfJi =r 4-.060 m
^ r Is.
* T^r - tt.333 **.'
_ _ Mc
M^ - £*I r fl?,3Bfl)C8*.tt«) w |g?> , k. j¥i
H.Oto

PROBLEM 4.9
4.9 through 4.11 Two verticol forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion BC of the beam.
SOLUTION
. 4 kN I 4 kN
Al fil C\
3* 2 a "•*
nn»M
300 mm 300 mm
<?8/.7 4- 125*0
- - 2-33^ mih
I, * T,-+A,J,t'^ <W.*fiA*/6, + ftSl.7)02.1*V]1, - XO7.3S*l0* wmv
I* * wi *>*»** £(&>)(«? * £5. *o4*|03 ^"
d^ ■ I /L- -51s/- l**$-(-Z-'83lO) * 10. 144. w,^
y+Bp - 25" + 2.33V * 27.33** »m - 0.04733*/ *»
? <v
m^
y
= £7.8 MP*.

PROBLEM 4.10
4.9 through 4. II Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compresalve stresses in portion AC of the beam.
3 in, 3 in. 3 in.
SOLUTION
6 in.
2 in.
1 > kips 1.5 kips
0
©
©
I
A
)8
18
35
S.
1
Ay.
90
is
/o*
40 In.
—60 In.-^f. «^
40 in.
*
'
.-CD
X
s -® -
Vo ^
f OS
36
= 3 ;.
I = I, + Iz = \56 *• 78 - Zo<J l»"
y+r - Sin s/k.+ B - 3 m.
P
I2'C m
73 in
]«• a-*!
I" M
M- Pa- = US)(Ho)-- 000 k.'p.iV
p 2
XoS-
6u---^- -^f^ • «-M *•••

PROBLEM 4.11
U- 8 in. -*|
4.9 through 4.11 Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and uompressive stresses in portion BC of the beam.
SOLUTION
r
*
®
<!
Q>
d>
(D
2"
A
S
£
4
IS
y.
7-5"
4
o.i"
A*.
£0
2.*+
fl
8&
20 in.
20 tn.
zaxx
1h
P.
Necj-f-wJ axis P;*s ^.77 8 ;M
oJoeW «. 4-Ke loouS^..
4.77ft
-J—
1^
It' iktA' + ^^-A^O^ (^(0.778^
I = I( +• It 4 I3 r ^<?.<?4 +■ al.43 4 73.i"Y * 155"./C in*
P
M
a.
j
o
M - Pa - ■ o
V\ = Pol * «5Ka<0 - 5oo k:P-m.
** 1 I5S". IC
6U * - ^fet • - (^0o)^778) x ^ Vfl ks;

PROBLEM 4.12
100 mm
4.12 Two equal and opposite couples of magnitude of Afe 15 kN'tn are applied to the
channel-shaped beam AB. Observing that the couples cause the beam to bead in a
horizontal plane, determine the stress (a) at point C, (b) at point D, (c) at point E.
24 mm
150 mm
SOLUTION
o
£
m
T
A, i***1
2.400
3060
24t>0
78&o
Xd,*"*w
^0
A^», <*»w
45.<Nl03
:mv?*'oa
©
3C.5W
X -
aas^vjc**
=■ S6.374 *.**
©
\y
<S>
Jt » (00-36.37V' 63.626 nm
- 0.63G££ *n
Jfi - 30-36.37* or - 6.37Y »*■*>
<J. * Jo- 36,37*
3.62£ *»*
d* - 36. 374 -uS" ■» a).37V **n
M - /Jk/O3 M-m
(cO,
. Point C: g ,.!ifcu(^)(-0'P^) g
C.5187^/0*
00
Point D: <*--*&■- (iS^'X^aO _
e.5"i8?xio
83.7 */oc fit
- S3.7 MP*
-/'K.'M/o* Pa
l4.C7*lo* f*.
: H.67 MP*-

PROBLEM 4.13
\ i 15 mm
■15 mm T
45 mm
4.13 Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 8 kN • in, determine the total force acting on the top
4.
mm
\*—75 mm —*\
«i
Si
a* i3
Jt
SOLUTION
The rWesS clTs+^'tJ^iow &>/ftv* He e*+i>*
c-^ross seaJnoi* is giVen fcy "Hie bewdrng S"\v«4s
1
Wne^e y is a. coor>aii\a?re w\\-)n \-\~s ovs'gfn om
+ke heoj-rai olxiS anc/ X is He ^omeiff df ivuhrfi^
at He £r\4-i>e ci^oss se&hi'&naJP ae^-ecu The -fe^ce
o« +*»e slicuJe^ iV c«J?ci>^eyW 4Vtfw +L,'$ stress
Ji's+nbufioii. Over aw area. e.Pew***'!' aM He-fe^e '3
Tf>« -hsvaJ T&*c% o»o +4i€ shouted <a^eet i's fkeui
F=1_JF - -^JA = -^y«/A e -£jV
vAft*e y* i\s 4-Ke cew+^fW«i coo^iV-ufe <sf -He
I3 - I, t /.O33fix/ofe M^
eattS
A"
3D mi*,
iz&xicr* k>1
A* - ftsX'Sl - 11 J5 mrv,1
1 " 2. iff *I0"*
- - !?3„8x|03 N - - 123.8 kti

PROBLEM 4.14
-15 mm
45 mm
15 mm
75 mm
CD
@
&.
J
4.14 Knowing that a beam of the cross section shown is bent aboufra vertical axis and
that the bending moment Is 4 tN * m, determine the total force acting on the shaded
portion of the lower flange.
15 mm SOLUTION
cv^oss, secA~{av\ is ofve* ty "Hie hev\ACr\^ stv^ss
Whe^e y is a. coo^dinatre. wi-H* i+.s ovs'^i'n o«
-fke heufra^ akTs a«d X is He iviowe^if of Inertia.
of +4ie enitVe cwss sfic^'fiwa-f ai^e^.. Tlte Tt>^e
ovi +ke shaded (V c«Jculled •fira** +k»$ s+ress
Jrs+f-tbah'on. Ovev aw are««. ePe**»*T d4 He'-fe^e is
dF^ er,dA =, -J^JA
Tfc«? -kAoJ fotre o« -h^ic sk^ded expect Is Hen
i+t^t<J«^ coarfwli'n^ire o"^ "M*e,
yAfc^e, y <'j> 4-Ke ecu-
skaderi per-fioi* and A* <S i4s <av*eA-.
3 ~ £Os)<U)s
la - I, - 0.6"3"7 3 x|0~<
I = I, + Iz 4 I3 ~ I.OC1Z *IQ*
Q>£Zl2ft*lo* vw**
0.0I2S* x/o* ^^v
WJW
f. 0672 "/o"t ^^
tz
cz
7-Jf *
* _
.1-
r*37,T
J
A
r
F-
Wo* to"* ►n>"
^ £7.9 */03 N
37^ kit

PROBLEM 4.15
4.13 Knowing that a beam of the cross section shown is beffl about a horizontal axis
and that the bending moment is 3.5 kip-in., determine die total force acting on the
shaded portion of the beam.
SOLUTION
0.4 In.
0.8 In. 0.8 In, 0.8 In.
**"- ©
©
<b
oTin. The styms elte+^fe^i©" o^ftv He en-hVc
jo.4in. cross seeA-ton is giVc* ty "Hie bettelt'ng sivess
wkere y is a- coe^diindl wilU i+s ovs'gin o«
■fke ne^frai axi's a*W X »'* He mameift* <*f in*rhX"
Jtrl-ribofto*, 0v<«r aw area. exe»w««'l" tfM H«'-fi**c« 'S
W^*Mre y* i\s +lie cew+ewJaJ ctfdi'wt'hATe <sf -H»e
sketJeJ aof^-iout ctnj A* i-s i-fs en***..
->a*
- A

PROBLEM 4.16
y
4.15 Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 3.5 kip-in., determine the total fibrce acting on the
shaded portion of the beam.
« '»"
™c
$0.4
t »4*—*
0.8 in.
t
}(M
4.16 Solve Prob. 4.15, assuming that the beam is bent about a vertical
the bending moment is 6 kip-in.
SOLUTION
and that
0.8 in. 0.8 In. 0.8 in.
IS.
©
The s+f«5 eirt+^fc^iow otffcr He e*+iV«
c-foss secA-io* is o/ven ty "Hie henAin* STVess
fke heuf (*fti axis a«fi/ X t's He wowei^f rff i«arh'«.
of He en+iVe c»**ss stcfconeJl a^^o^m The fo+ce
€K,
*^-
J 0.1ZS2 r

PROBLEM 4.17
— ■ — r. 1.5 ii
4.17 Knowing that a beam of the cross section shown Is bent about a horiynmal axis
and that the bending moment is 6 kip*in., detemiinc the total fibrce acting on the shaded
portion of the beam.
0.25
Z>
1.0 in.
0.25 In. SOLUTION
0.25 in.
<a
25 In,
oxtfi
«Xftf g
rac:
.1. I
The stress elfs+^ki/K©* o\t&r -He e«\+i>*
cross $eeA-ioH* i's p;Ven ty "Hie bendi'n* s+v^ss
-Po^muJci
flie neuffAi axis ar\A X t's the m^menf af ruetHY*.
6T H«? cn+iVc citess se&h'anajQ £t^-ee^m The to+ce
Jt's+ritoh'on. Ov'w avv av^A. eye****"!" tfM -H)e4vifice /s
TJ>« Wfti f*>*\:e o* -Hie stiA^eJ area. ;$ flieui
©
©
•s 0.7(S7S i«y
jy*A* = ^Aft+ ybAb
= (&,»1SX0.S)(0.ZS) + to.S;)(o.XS)(\.o) = 0.23H38;h3
F = MiA* = (£A°-!'I3^ = i.w fc.-
x
o-iisTr
ip5

PROBLEM 4.18
J .JL
rn i
- ■ (- 1 « In
4.17 Knowing that a beam ofthe cross section shown is bent about a horizontal axis
and that the bending moment is 6 kip*in., determine the total force acting on the shaded
portion ofthe beam.
4.18 Solve Prob. 4.17, assuming that the beam is bent about a vertical ■**■ and that
the bending moment is 6 kip-in.
1.5 In.
I
SOLUTION
„„ J 1.0 in. V
"f
0.35 in.
cross secJ-io** i's g/ve* Ly Hi* henA%n* s-Vv^ess
®vj
®-l
Wheire y is ex. cocr-di'iMxA'e. wi-f-d v+5 ortVi'n o«
+J»e newfVai «*i*s <xnA X i's He *io»ie*ff" <jf i*e«Hy«.
eff He en+iV* emss secfii'onaJl a^ec^. The fo^ce
ov\ +J»e s^cwle^ f.f c^?ct>J'flL'lefl/ 4W>iw -Hi's cs+reas
eJfS+rikofion, Ovev av\ are*. eP&*i*{ JA Htf -fiance '3
haJeJ por4-!ot4 ct^J A* <S f-fs a*"*4«»
*>
5
=
©
-■
©
I -1, -1.
=r OJ&7S w
I
X "* 0,43 75"
h x 0,4375- *"5/ ^fs-

PROBLEM 4.19
48 mm
4.19 and 4.20 Knowing that for the extruded beam shown the allowable stress is 120
MPa in tension and 150 MPa In compression, detcnnine the largest couple M that can
be applied.
36 mm
SOLUTION
A tm**
33 U
\>Tj >n*»i
30
J£Ll
Mft*H
- 3%*BQ
IZZSOI
NeJ+vftJ? axis 4'i*a 55". 04 *»*\ <xWt Wo4foi<w».
I. = \5.^ + A J* = &(H*Xllf + («X^X7.0*)* = 3.7<73*/oe ^*
aP'
f«Mft|
On £
J* Mr^ r-ri^7 - 8*ZS*lQ N-
0.01096
m
WU: Cowf^,'.n M'^'TlJiy^'^^ 7-67*'0' "•*>
M* 7.a*iJ W-* - 7."*7 \<w-^

PROBLEM 4.20
-80 mm-
4.19 and 4.20 Knowing that for the extruded beam shown the allowable stress is 120
MPa in tension and 150 MPa in compression, determine the largest couple M that can
be applied.
SOLUTION
M
(
£
A t *»«
\0%O
3ZHO
J*>> """i
21
3&
AyB>> *»w
ShZlo
tlXoo
T 3Z*o
yl(+ ? -3^> **i»n - - O.O30 *-»
I * I, + I, -" 7$tM*lo'»mH =• 758 J< *£>"* m1
SI * J^
IMH^FI
Wa-tW* co^ress.owx M" r~m 3.7*«8*l& N*^
0. 030
CW*e He s*JAw *s Mj/ M«w* 3.7** *loE M-w, = 3.*?? UW*v*

PROBLEM 4.21
0,5 in. 0.5 in. 0.5 in.
h
4.21 Knowing that for the extruded beam shown the allowable stress is 12 ksi in
tension and 16 ksi in compression, determine the largest couple M that can be applied.
0,5 in.
1.5 in.
_L.
T
1.5 in,
1.5 in,
SOLUTION
fo
to
®
roi^
®
A
2.25
2.25
4-i'o
y*
\.25
o.us
Ay.
;?.*ia5
0.5M5
3.375"
V - ""^T r 0.75 in
The nejtai! a*f5 i;es 0.75 iw. «,k6ve Uth>^.
I = I, + Ie r 1.^375 i«H
isl-j-g
Top • COh^flresSi'dKi
M-|¥
ii- \.%S " *0-" Kip • ih
M, GOO.513 7jj _ 2SS x
- IK
Ou&Se H« Soulier AS Mai/ Ma// - #0. 4 Up. in.

30 mm
PROBLEM 4.22
— 80 mm —
40 mm
4,22 The beam shown is made of a nylon for which the allowable stress 24 MPa in
tension and 30 MPa in compression. Determine the larjgesi couple M that can be applied
to the beam.
SOLUTION
d = 60 mm
G>
)
©
M
z
1200
3600
yBJ»*
45
15"
i08OOO
IZgOoo
Y£
3too
" 33" ►*!»"»
Ii = 7fetA3 + A2J4* = ii Woyso^+dXoo^Cao)1- •= S?o*to *,*,'
990 * /O *»
•SI- 14*
Top. +&*sfon su*c M - ^ *j -. - nbO N.-^
t? it * ,- M C3QXI06Xg?'?6x^"eT) «4a U y»
' 0.02>5
Choose s*i*JJPef v/c/oe
M = g^ M- **

PROBLEM 4.23
30 mm
-80 mm-
40 mm
4.22 The beam shown is made of a nylon for which the allowable stress 24 MPa in
tension and 30 MPa in compression. Detennine the largest couple M thai can be applied
to the beam.
d = Sa
mm
4.23 Solve Prob. 4.22, assuming that d = 80 mm.
SOLUTION
G)
I
>
®
7
A _, *■*"*
ZHoo
7. ooo
4Ho0
a/°jm'*
65
2J"
Avo, *^
15*^000
5oooo
^O&OOO
- 204000
y*^ * SO-^fi.83^ 33.1ft **m = 0.0391$**
yw6t - - 46.82 ***, » -0.04CS2: m
I, s Tkt.K3 + A^ = ^f8^3o)34(^0O)0«-iS)1 = 0.173 MHO4*.**
I* I, + I, - *-3¥2 */Oc taMf ^ £3¥2*/o"C*,*
t. '*.
I*«»l**
T0
P-
ffinSi'
St 0^ Si«€
fJ< M ■ WW^' - '^^
|V\
8otf"
o«*» - compress (On
„ , t3o«to*\(.7M2*l&) g ,.sto|w/o»A/.
O.O^f6g^
m
£ho&3e &>i4jPjPer Vas^l
M» Li"OI V/C75 M-^ - J. 5b I kW-Kh

PROBLEM 4.24
1.6 in.
"(
4.24 Knowing that for the beam shown the allowable stress is 12 ksi in tension and 16
ksi in compression, determine the largest couple M that can be applied.
SOLUTION
A, = (U)(as)-- i.28 ux
A = >.** - o.iu? « o.m3 ihl
n/, - ©*H ,h Ji - srr 3^ 0.7127 .„
Y - Z ^3 (i.SS )Co.H^ -(0.39^71(0. g/^a) „ . .
2A " 0.SS73 " m*
^+^ - 0.2-0.433* - 0-3l4<?m, JW - -0. 4*83li'^
Top* Te"*3<<^L si^e rl - — *i|&3 l«*o" k«*» iw
O.^MI
p- ih
Glioos^. -He. s^w^ire^ i/ft/ot
M * 1.37a k.>-m.

PROBLEM 4.25
<*>"> f = %*
4.25 Knowing that ^ - 24 ksi for the steei strip AB, determine (a) the largest couple
M that can be applied, (*) Ihe corresponding radius of curvature. Use£-29 « 10s psi.
SOLUTION
i- AW3* A(*V*y- mo^- *.♦
e* ±(£) * cows'i«
e-.y
0*1
g*r _ te*/x/QlyH)?.ov{o"<)
C 0- 073 7JT
«• 105.5 jfl>.;»
p , EC „ U*«<Q*Yo.OT37S) . ,,3,-s ;,
(bl
PROBLEM 4.26
4.26 Straight rods of 0.30-in. diameter and 200-ft length are sometimes used to clear
underground conduits of obstructions or to thread wires through anew conduit The rods
are made of high-strength steel and, for storage and transportation, are wrapped on
spools of 5-ft diameter. Assuming that the yicid strength is not exceeded, determine (a)
the maximum stress in a rod, when the rod, which was initially straight, is wrapped on
a spool, (b) the corresponding bending moment in the rod. Use £= 29 * 106 psl.
SOLUTION
v- -kef = ji(o.&>) » o.is ■:*
D = *5~H - Go <'* yo * *£D * Bo u.
/°
So
r

PROBLEM 4.27
Jl.
0.06 In.
n
T
0.005 In.
I_<W. jj
4.27 It is observed that a thin steel strip of 0.06-in. width can be bent htto a circle of
3
4 - in. diameter without any resulting permanent deformation. Knowing that E = 29
* I06 psi, determine (a) the maximum stress hi the bent strip, (ft) the magnitude of the
couples required to bend the strip.
SOLUTION
r o.sis
PROBLEM 4.28
nktp-in.
Jl
EX
3^103
4LW A3kip-m.a>iiplei3appliedtothesteel bar shown, (a) Assuming that the couple
is applied about the z axis as shown .determine the maximum stress and the radius of
curvature of the bar. (A) Snive part a, assuming that the coupie is applied about the y
axis. Use E - 29 x 10s psi.
SOLUTION
& » 4h = i(i.s)- ■0.75' ;*
^.^3/3
P
- ^O? tf/G>
-6
_ I
\1rt
,5= if ,iW_jg^Lr i4.«l«io*P.; - ».t; fa;
J-
EJ
3 xyoJ
■= f. f3S"x/o"a ,'«"'

4.29 A couple of magnitude Af is applied to a square bar of side a. For each of the
PROBLEM 4.29 orientations shown, delermine the maximum stress and the curvature of the bar.
(b)
SOLUTION
(CO
'■*
AXIS
T
BM_i
^7
m
(K
COCV5
-L = Ji
f EX
I« Hr b^3 = itoto? =
ix
'fVtf
EX
- M£
X
n_.
i^
Ml
- ^
a*
ix
a
12 M
fa1*
po»" one -jWaw<!».'€ T*ne wo^ne*"
joe^"T"i^ (aJso^T ifs trt_5e i-S
I2 = J. = n
M Q./V1
ovm, " <v
6M
^—
J3-
Ea'

PROBLEM 4.30
y
4JO A 24 kN-m couple is applied to ihe V200 * 46.1 rolled-steel beam shown, (a)
Assuming that the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the beam. (A) Solve part a, assuming that the couple
is appiied about the y axis. Use E - 200 GPa.
SOLUTION
for W ZOO* H£>.\ Volt eel s4«ci se^ f'j-
'5-3 * lo''W
5 « 15/ WO"C no*
Sv - 'SI * la* WJ
(a) M- = ZH kv-** - M*to* W-^»
6>
M
^WO1
S H4S*KT*
= s53.fi ^/o' Pa. - sf3.6 MP«_
= 2.637 x/©'* r*"'
^o =■ 379 m
S " l£f */o'4
'58.? *«>' Pa. * /£«.<? MP**.
ZH*IO*
M _
7.8V *jo"3 **,"
P
XI.S «n

PROBLEM 4.31
4.31 (a) Using an allowable stress of 120 MPa, determine the largest couple M that
can be applied to a beam of the cross section shown, (b) Solve part a, assuming that the
cross section of the beam is an 80-mm square.
10 mm
SOLUTION
(oJ\ I« I, + 4TZj.uA e^e. 1, fa Tne wowe>-T of
lnev°Kfl>. erf a* SO-rww S^oq.v*€ Ctw ±z b "He
10 mm
-80 mm
5 mm
—II-
5 mm
Wim
I = X,+ 4X* = 3.8Z</0C^^ =■ 3-82*/o" v«\ c = 5d^ = O.o^o^
0. 0S"O
^ w = 7.17 Wlsi-i*i
M
.. SX xCUQ^ofc)(3.V^3^o-^ g |0-M)f|flpw.h , |0.?H M^ ^
0.01°

PROBLEM 4.32
Go '
h,
o«r
4«J2 A portion of a square bar is removed by milling, so that its cross section is as
shown. The bar is then bent about its horizontal diagonal by a couple M. Considering
the case where h = 0.9A,), express the maximum stress in the bar in the form am — Ao&
, where o0 is the maximum stress that would have occurred if the original square bar had
been benl by Ihe same couple M, and determine the value of A.
SOLUTION
_ I
kW^KV- 1U8 - 4h> - W
c -- h
ft = il^ - Mh
3M
6^ -
o *■'*<* i' r\o_X sq oo /~R,
3M
(4K,-3)uH.
3M
h03
k
Cffco-OXo-iKX^C)
= 0.76'O
k = o.Tfo

PROBLEM 4.33
4.32 A portion of a square bar is removed by milling, so that its cross section is as
shown. The bar is then beni about its horizontal diagonal by a couple M. Considering
the case where h = 0.9A,, express the maximum stress in the bar in the form a„ = k<%
, where o0 is the maximum stress thai would have occurred if the original square bar had
been bent by the same couple M, and determine the value of k,
4.33 In Prob. 4.32, determine (a) the value of A for which the maximum stress am is
as small as possible, (b) the corresponding value of*.
SOLUTION
1= 41, * ZIZ
3
c = h
i- ^X- h-
IS metXi rviutvi aA
* &\^tf-V] - o
f V0>-<3j>l=o
ir-awNM
fo*- He
Qf\ A 1
,W
5<»tM.ir-e.
M5 13
b = k
6~„ =
_ M^. . 3_M
OWI
k- O.W

PROBLEM 4.34
434 A couple M will be applied lo a beam of rectangular cross section which is to be
sawed from a iog of circular cross section. Determine ihe ratio d/b, for which (a) the
maximum stress a„ wiil be as small as possible, (b) the radius of curvature of the beam
will be maximum.
SOLUTION
I = tU^ c=|J
— - 6 bd
bol3
f 5 nofi-^c-t mJm
iO is )*i <x^ / no J 1*1 w n «.^ _L i s **t <>-9G i^J^n
(D2 - d*> de

PROBLEM 4.35
SOLUTION
4.35 For Ihe bar and loading of Example 4.01, determine (a) the radius of curvature
A (*) the radius of curvature p1 of a transverse cross section, (c) the angle between the
sides ofthe bar which were originally vertical. Use£= 29 x i06psi and u=029
FruM Exu^ph 4.0/ V\ - 30 klp-iw _, I ~ \.o^Z iw*
^ t= & \%Z$,o^ - ^^'O'.--- f* ,007 -.,.
<M
-6\ , _J
^ - 2>^r =r (0.29M^3*/o~6) ,v' = mW' f>'=3Hlo m.
-G
moolius yo* 3H70
PROBLEM 4.36
SOLUTION
4.36 For the aluminum bar and loading of Sample Prob. 4.1, determine (a) the radius
of curvature p* of a transverse cross section, (b) Ihe angle between the sides of the bar
which were originally vertical. Use£= 10.6 x I06 psi and v= 0.33.
X - J± - \03.&x)O3
f ~ El ' (10.6*10* )(CL. 97)
I - W.W ',■/
\A ~ 103. 8 U-p - i*
= 1SS y/o
- 4
lh
(M -J-, ^ oj£ r (a33)(7SJTv/o-4 ) - 249W6
/o' * 40/0 i* ^ 334 ff
PROBLEM 4.37
y
4.37 A W 200 x 31.3 rolled-sleei beam is subjected lo a couple M of moment 45
kN-m. Knowingthat E = ZOo<5fe / t>= 0.29, determine (a) the radiusof curvature
p, (b) the radius of curvature p- of a transverse cross section.
SOLUTION
W -fe = ^ =
4£~*/o*
- 7/7*7o"5 ^"'
(^ £,
r -2>»
_L -
^ = (O.^T7-'l7v/o-3) ^ ^07v/o5^' ^>'- 48/ ^

PROBLEM 4.38
4.38 it was assumed in Sec. 4.3 that the normal stresses ay in a member in pure
bending are negligible. For an initially straight elastic member of rectangular cross
section, (a) derive an approximate expression for ay as a function ofy, (b) show that
(^)mm a -(^2p)(ot)mm and, thus* lnBt 05, can be neglected in all practical situations,
(ffmf.- Consider the free-body diagram of the portion of beam located below the
surface of ordinate y and assume the distribution of the stress az is still linear.)
SOLUTION
De*.»+e r^e wi'd-f-U tff +Jic be*x>i by
b ojaA *H*»
Cos -^ « /
sj -■-£ s.| ^ <*, Jy *- A £ 6i J, - -^£ <r„ *
Bo*
s^-tsA^
Tn€ M <C^(*w o>"« ^e^A O^. 6^
occ i^^s <a-T v = 'O
y
*p

PROBLEM 4.39
cut *£
21
£L
m.
>\.S
10
439 and 4.40 Two brass strips are securely bonded to an aluminum bar of 30 *
30-mm square cross section. Using the data given below, determine the iargest
permissi ble bending moment when the composite member is bent about a horizontal
axis.
Aluminum Brass
Modulus of elasticity: 70 GPa 105 GPa
Allowable stress: 100 MPa 160 MPa
SOLUTION
Use a/uw*ifiijwi as i*k« ^e+e^t^c^ w\e-Ae.\r'.r>.y
11 - hj£* * \0S/7o r 1.5- ;» WJ5
3
*~-ir
AJlo^no*' r> = Lo s j - isTw-o - O.o/S" ^ } <S* = jog */o' fix
M r
J.G27 x/o w-^
Br^ss • D- \-S'j y- ^U^^ 0.OZ.I w* ^ S* ^ \Qox/Oc9a
(l&o*lO<-)(2W-GS.>t0"'')
M =
1.240 */o3 W-*,
M- I. ^^o x/o5 y.ho ^ i.a*to UW-iw

PROBLEM 4.40
6 mm
30 mm
6 mm
4.39 and 4.40 TwoJwbb' strips are securely bonded to apt alurmmiw bar of 30 *
3 0-mm square cross section. Using the data given betow, determine the iargest
permissi ble bending moment when the composite member is bent about a horizontal
axis.
Modulus of elasticity:
Allowable stress:
Aluminum
70GPa
100 MPa
Brass
105 GPa
160 MPa
SOLUTION
V\
a.vY£
©
J&.
ML
Use aJcov*.i •> o**> a.s 4-ke ircre^^^e w^e/^
IH ~ I. O ih aJywM-^ j»^v
m - E^JE^r \os~/io =■ /.5" ;* Was
Fo^ 4"ri-tf ")"<"an5-Fo^wcc/ section
I, -- £ t>2^3 = L£(a»)b*,? - io/.2jr*/o* h^ I, = I, - 5-8.8< wo3 *„'
I = I, + 1^ + Ij - £18.97 *lo~ **? -z 2\2.H~!*\o'' */
M ^ £1
*y
' J
r £/ *,*, r O.0£/ m 6"' lOOx/O'Pft
Brtf,ss : n = f.5" ^ J = 16"^ - O.O/^Mj 6"- /60 v/o€ Pa.
(,1.5 Ko-o'f 1
Choose T^e. s^^JlJer- vaA->e
]A- LOH$*lo> M*^ ■= /.o^3 kW-*n

PROBLEM 4.41
Brass
Aluminum
1
4.41 and4.42 For the composite bar indicated, determine the permissible bending
moment when the bar is bent about a vertical axis.
4.41 BarofProb.4.39
r
-30 mm-
A™1" SOLUTION
30 mm
6 mm
T
M
©
IE
t.O
i.r
Use alio*** no** as re+e^emce m^c^'aJ
h- Et/^r )OS / 70 = 1.5* ."* Ws;
r - H±u k3
, _l^(3^X3o)a - 67.S*io*»
3 y
I3" I, r XOm2?*IO*
ynni
-9
M
M^ Ogoxfo^hoiwO r 7;?0 ^
(|.o)(o.o/s-)
hn
Choose -He S*i4JPJe^ vaJjc
n = IZo N-r*

PROBLEM 4.42
4.41 and4*42 For the composite bar indicated, determine the pennissible bending
moment when the bar is bent about a vertical axis.
4.42 BarofProb. 4.40
SOLUTION
(D
©
_&
n = £*,/£*. = /0SY70 - /.s" ;„ t^55
Y)
l.o
T - "' U U1
- Jjf (*>(*>? *
/7, i i 1 1-5"
IS.-S"* lo3, wi
' \/-, \3
IX
\z
Ix ~- I, = 13.5 >fo-
Vn m
Is I, + It + Is - iZZ.ZSxtO2, k
IV)
* _
r*t **
138.2.5"*/O *n
teW|2^| Y\*
ny
■*i
Brass- V) = l-S", j = 15^ = O.OfS^J <S * i&OxiO* P«-
M
£liooS<? He s^AJJer ^JoJjt M - g53~ W-•*>

PROBLEM 4.43
4.43 and 4.44 Wooden beams and steal plates are securely belted together lo form
the composite members shown. Using the data given below, determine the largest
permissible bending moment when the oomposite beam is bent about a borizontal
axis.
Modulus of elasticity:
Allowable stress;
10 in,
Wood
2x10*03*
2000 psi
Steel
30 * 10s psi
22ksi
3 In. t 3 in.
£in.'
«0<vS
©<!»
®
SOLUTION
Use wood us "H& re4e/-e^c« wiA,+e/\'ft.r
If) -= \*Q in Wood
Is ■= 2, - 2^0 ••«"
M - G I
"y

PROBLEM 4.44
SOLUTION
4.43 and 4.44 Wooden beams and steel plates are securely bolted together to form
the composite members shown. Using the data given below* determine the largest
permissible bending moment when the composite beam is bent about a horizontal
axis.
Wood Steel
Modulus of elasticity: 2 * 106 psi 30 x 10s p9i
Allowable stress: 2000 psi 22 ksi
f*| = \^\
Use vJoogA as +Ke v*e-feire»nee witk&ytscJf
D - l-O Xa wood
n » E*/E'v * 3o /x =■ \S in &+ecJ?
J^ * J, -" 1034.4 in*
Wood:
Choose He s^JJ^sjUoe M - €>&T*/©3 A'* » CW 'hp-iV

PROBLEM 4.45
Aluminum
I*—24 mm —+\
6 mm
6 mm
4A5 and 4.46 A copper strip (Ec = 105 OPa) and an aluminum strip (E„r 75 GPa)
are bonded together to form the composite bar shown. Knowing that the bar is bent
about a horizontal axis by a couple of moment 35 N-m, determine the maximum
stress in (a) the aluminum strip, (A) the copper strip.
SOLUTION
0
HM.
CO<fS
Use flJo
* copper
tp
®
I
■ A.»w«
\4M
Y\A> »«m
shs:^
y*j ***»
i
3
Y>Ay0-> **?
I7oo.«
W*M
7. r J|~g , s.so »
(b) Copper ms(/.ifNi ^y * -SSnm - - O.oo^w

PROBLEM 4.46
Copper
avis
£L
3T
t.o
I.M
4.45 and 4.46 A copper strip (£^-105 OPa) and an aluminum strip (£L-75GPa)
are bonded together to form the composite bar shown. Knowing that thebar is bent
about a horiasontal axis by a couple of moment 35 N*in, determine die maximum
stress in («) the aluminum strip, (b) die copper strip.
SOLUTION
Use aJi>«*iViji» A5 rk« fe'ft/'evtea ^Ur^'J
n - I.o fw «.i't/UM,rtv>fcv\
©
A j W>W
2.J6
72
2IC
100.8
.Ycj wm
7.5
1.5
frAVo, ww*
IC*0
I5-U8
I77i. a
* WW.A tin*4
Ta 314.8
Copper * 0r l»1 j ^ " " ^-^^ww * - O.O&Si^ol *«

PROBLEM 4.47
h—6 in.—H
10 m.
^5x£in.
4.47 ud 4.48 A 6 * 10-ia timber beam has been strengthened by bolting to it the
steel straps shown. The modulus of elasticity is 1.5 * 10* psi for the wood and 30
* 10* psi for the steel. Knowing that the beam is benl about a horizontal axis by a
couple of moment 200 kip-in., detennine the maximum stress in (a) the wood, (b)
the steel.
SOLUTION
Ose wood as Hie ftre-t^^r.<z fr*j^e/'*,)
n - I.O i* wooeP
TVa*>s*fo^^ed seofr
Ol
®
r
A
6o
2.S
nA
Go
50
110
y*
S.S
OAS
mAy*
33o
3W.S
% -" -^j^1 =• 3. lit iv,
I,-- ■g-tJi.1 * n,A,J> ^(sXlof+(l.o)^(?.38{)1 r
W
a-xiS
/i'c:
I - I, * I.'
I*
zsz.t ;«*
(<v> Wood : .fi = J.o J = I0.S- 3.IN - 7.386 i'«

PROBLEM 4.48
■ 61n.~
2XTin.
2X4 in.
4.47aad4.48Atix iO-in. timber beam has been strengthened by bolting to it the
steel straps shown. The modulus of elasticity is i.5 * iO*psi for the wood and 30
* 10* psi for the sleei. Knowing that the beam is benl about a horizontal axis by a
couple of momenl 200 kip-in., determine the maximum stress in (a) the wood, (A)
the steel.
SOLUTION
Ose y/ooA as -He reference rmiTei**
n * £&/£* - 3&//.S * Zo ;» «+e*/
TV*, * Stov^ e*H secTt'e*w
®
©
&
r
A
Go
0-7i
o/w
r\A
£0
IS-
IS
9o
y-
5
1
1
nA£
Zoo
IS
IS
330
Tie ntyTr*J a*rs A'es 3.667 i'n. aiwe He ko-/foi*\
I, r j^V-V r)A^- ^(.XlO?+te0Kl.a38^ - G06.7 in"
V ^ = ^W*M.-tf ■ f (4)fr?+0*ta.«7f . Ml.7 i
(cO Woo J .* H ' /.O^ yr 10-3.U1 * 6.333 ;*
5-, .Atbl t.(^X^X-^^) = 17.67 ks;

PROBLEM 4,49
4^9 aad 4JSQ For the composite bar indicated, detennine the radius of curvature
caused by Ihe couple of moment 35N-m.
4.49 Bar of Prob. 4.45
Aluminum
JFjF~
24 mm —*■!
6 mm
6 mm
SOLUTION
Set SoJAto* io PROBLEM HAS -rV H*. c«Je^A+i*» o\ 1
j H 2S -i
PROBLEM 4.50
Copper
4.49 aad 4.50 For the composite bar indicated, determine the radius of curvature
caused by the couple of moment 35N-m-
4JS0 Bar of Prob. 4.46
9 mm
3 mm
SOLUTION
See S6Mi\>" 4° fl?6eLBM H. H£
35"
-/
m
45i and 4.32 Forthe composite beam Indicated, detennine the radius of curvature
PROBLEM 4.51 cnaed by the couple of moment 200 kip-in.
r4.31 Beam of Prob. 4.47
6 in.—H
I SOLUTION
See sJjti'oh -h> ?f?o6tFH 4-^7 -fV ca^A^ioh oi J.
10 Id.
^r
^316 .** r 783 ff,
* 5i «mf 4.52 For the compos He beam indicated, determ ine the radius of curvature
caused by the couple of moment 200 kip'in.
4.52 Beam of Prob. 4.48
SOLUTION
PROBLEM 4.52
U—6 In.—*j
10 In.
2xfin.
|ta. p = 62*5 .** * 5*tf fi.

PROBLEM 4.53
16-mm diameter
'80
x T
453 Aconcrctcslabisrciniorc^i^io^m^an^ersteeirodapiacedon 180-mra
centers as shown. The modulus of elasticity is 20 GPa for concrete and 200 OPA for
steel. Using an allowable stress of 9 MPa for the concrete and of 120 MPA for the
steel, determine the largest allowable positive bending moment in a portion of slab
1 m wide.
SOLUTION
n - £6 " *o <&*. "
180
Locate He rieo-hf-d) &.*{$
130 X j - (|00- x)(?.0/£>6x/o3) = O
X - 31.317 hun ; 100-X - 6Z.eo3 ^
= M.OI3 *'°£ *^ ~ l|.^3V/o"'^^
7"
■* loo
,0*-*l
0-o) (©.OS? 31*7)
w>
Choose j*^€ *hmi//«* vaXoz
TleaWe is-Hi* oi^ftLfe. PoSi'fiVe mo^e*! for a /SO ***i u/fcJ* .se^frfoM.
For ^ I m - /OOO «* w/udf^ i^Jfiply ty -^ - ^.5T6
M - (^.^Ofottf^WC?) '- \1\1Z«IC? Ur» - 11.73 kkj-wv ~4

PROBLEM 4.54
16-mm diameter
100 mm
4JS3 A concrete slab is reinforced by 16-mm-diamctcr steel rods placed on 180-mm
centers as shown. The modulus of elasticity is 20 GPa for concrete and 200 GPA for
steel. Using an allowable stress of 9 MPa for the concrete and of 120 MPA for the
steei, detennine the largest allowable posilive bending moment in a portion of slab
1 m wide.
4.54 Solve Prob. 4.53, assuming that the spacing of the 16-mm-diameter rods is
increased to 225 mm on centers.
SOLUTION
= /o
<s>«'s
r)As
±{oo
1
Co^etAe* a. sec-h'a* 225 *** wide wiW» o»k sfeey <oA.
l/S.$"*x ¥ P.0 106 x - Zoi.ot*to3 - o
S.U.g W x X - TTvTTTFi *
I- £(MS)XS + ^.oiO&x/o3 (loo- xV
- 3
n.;
6-1
.7o5 *10c ^^^ -
M =
161 " I i I - ■ ' - ry
S-feeJf: ft-\0 , y = ^.72? * * 0.0*57*7 « , 6** l2o */06 P*,
1 (loXO.O«r?Sn^
Choose He s»*JJer Y*Xe M* 2.137© * lo* W-m
The- <*hove is -Hie eJ$o*t*X)e posi'tWc mo*it*\ to*- ^ Z7£ ** icftJe 5£*t~<«m.
Fo^ a U = lOOO^m sce+i^j KjJlKpJj Jy ^^^ - H.4MM4

PROBLEM 4.55
450 mm
4.55 The reinforced concrete beam shown is subjected to a positive bending
moment of 175 kN-m. Knowing that the modulus of elasticity is 25 GPa for Ihe
concrete and 200 GPa for the steel, determine (a) the stress m the steei, (A) the
maximum stress in the concrete.
22- mm
diameter
50 mm
SOLUTION
n =
25SPfi.
2.0
|—- ISO ->|
CQt\S
X
4oo
HrA,
J
SoIv/im* -for X
x -
As •= ¥• Jd* • fflflXk)1 * \.SiaS*lo* mm1
Locate the n^oifaJl <**{s
I25V + /2./C?*/0SX - 4.3657 k/oc = o
- I2J6#*/C? 4-7(1?-/6f*/o*)i 4 (4KiasX**3457x tf>*
° ■ I
(a) S\eeP: j = - 2^S^S m^ * -0.245^^
(bl Cotter**!* : ^ - I W.ST m^ = O. Iff'tSS" n*

PROBLEM 4.56
Soo
22-mxn
diameter
4.55 The reinforced concrete beam shown is subjected to a positive bending
momenl of 175 kN-m- Knowing that the modulus of elasticity is 25 GPa for the
concrete and 200 GPa for the sleel, determine (a) the stress in the steel, (b) the
maximum stress in the concrete.
4^6 Solve Prob. 4.55 assuming that the 450-mm depth of the beam is increased
to 500 mm.
SOLUTION
T
H&o
*AS
So/cinj -fbr X
X =
X = l&Z+Wtnmj »K5"0-X'- 283.BJ *»i
1.3623 *L°'3
(b) C
OVSCJf*'
kz:
y
- ItG.II m» = 0.166 11 m
6--- (l-9l(l,,^^ltt'ti = -a/.3*/oef>«.= -3».3MP<f
/. 36 ?3 */©•

PROBLEM 4.57
30 in.
1-m, _^^
diameter^ S^^..
5 in.
4.57 Knowing that the bending moment in the reinforced concrete beam shown is
+ 150 kip-ft and that the modulus of elasticity U 3.75 * i0*psi for the concrete and
30 x 10* psi for the steel, determine (a) the stress in the steel, (b) the maximum
stress in the concrete.
SOLUTION
24 in.
in
- £* , 3o*IQ'
- 8.0
k
12 in
3o
,~A
2.5 in.
1
r~
©
AXfs
©
a
.£
<£>
£t 3.75*jO«>
^As - 25*. 133 ■'**
Loc&A-e +ke n^l-^i .sow's
6*-1 + 175". 133 X -3^.C<? = o
_ -175.133 + 7(175.133^ *-{4)COC^.61)
X -
'6.^- X '
\3
- 0.22S >Vi.
Iar flAsJa* - te51 l33X/6-X7^r - G£*S7. I ,V
s _ r(fr0ttllooX-lC.yO (fl.
S083-5"
(b) Concrete n* 1.0^ j s S*„:ttS ;«
#0S3.5*

PROBLEM 4.58
-£ -in. diameter
9
2 m.
MH
ayts
nA»
14
Solve -fc>^ X x =
4L58 A concrete beam is reioforced by three sieel rods placed as shown. The
modulus of elasticity is 3 x iO* psi for the concrete and 30 x iO6 psi for the steel.
Using an allowable stress of 1350 psi for the concrete and 20 ksi for the steel,
determine the largest allowable positive bending moment in the beam.
SOLUTION
nA3 ■= |S.o4o .v*"
4**" + IS.O^O ^ - %SZ.SQ = O
14 - X ^ 7-?<W in
G.O&S \«.
16"/ - |^| ■■
M- ^
"^
^M^e.-fe: |r\ =: I.Oa Wl* £.oo5 ^ 1S"I - *3S*<9 p»<
M
= ^?2l!Z|f^_= 3M./d»A.
(Lo )(6.0O5")
|V1
= 3^9 /**-;*>
£+e*i : n- |o; Ijl - 7. ^5* ; 6" = zo x/o2 p5r

PROBLEM 4.59
■J -in. diameter
aXi'S
4.58 A concrete beam is reinforced by three steel rods placed as shown. The
modulus ofelaslicity is 3 x I0*psi for the concrete and 30 * 10* psi for the steel.
Using an allowable stress of 1350 psi for the concrete and 20 ksi for the steel,
determine the largest allowable positive bending moment in the beam.
4.59 Solve Prob. 4.58, assuming that Ihe widlh of the concrete beam is increased
to 10 in.
SOLUTION
^ + \$.oHo X - *«.5*G = o
So^e: -W ^
X =
= - I3-Q4Q + J Qg^oVS ^toXasg-j-g ^ _
/4 - x - g.471 .v
I =^0o)x3 4 n^N-X^* ^(lo^siwO1*- l\S.Oio)(g.H7l)7'
= 1857.? ;**
■5"--5*9 i'«
Si -
IU±^
t*» =
6-J
rny
Sfeei*: n - /o |y I ~ 2.47/ .^ 16*1 - ^ox/o3
Choose +U SWiJti&r \te,fo<. XA-* 4ZS>& *lO* A-."^
- 3G.C kJp.-pf:
Fst'

PROBLEM 4.60
H
nAs
a-*
SOLUTION
6s =
<5"s
1 -
X
4.60 The design of a reinforced concrete beam is said to be balanced if the
maximum stresses in the steel and concrete are equal, respectively, to the allowable
stresses a, and o> Show that to achieve a balanced design the distance x from the
top of the beam to the neutral axis must be
d
where Ev and £, are the moduli of elasticity of concrete and sieei, respectively, and
d is the distance from the lop of the beam to the reinforcing steel.
h MQsl->0
X
x
^
- n
4*1 +
- h
£s<s;
X
! +
Ec6k

PROBLEM 4.61
^T*: *
4.60 The design of a reinforced concrele beam is said to be balanced if the
maximum stresses in Ihe steel and concrete are equal, respectively, to the allowable
stresses a, and ac.
4.61 For the concrete beam shown, the modulus of elasticity is 3.5 x i0* psi for the
concrete and 29 * 10* psi for the steel. Knowing that b = 8 in. and d = 22 in., and
using an allowable stress of 1800 psi for the concrete and 20 ksi for the sted,
determine (a) the required area^, of Ihe steel reinforcement if the design of the
beam is lo be balanced, (b) the largest allowable bending moment. (See Prob. 4.60
for definition of a balanced beam.)
SOLUTION
fa
V>A:
15 I
6L =
Mv
— , —^— _ n x "
6c.
A s
X
* 61 ' l 8.^857
I goo
- 2.34/0
Locate Y\e>A-r&4 <xxCs
(cu)
- ^.3S3^
i*
I - ifc>XS+- *A.(d-xV" = i (S)(^3?8? + (3.Z&57)C3.^&3rXi^.Co2.>ii'
6"
_ Y) My
X
|M
1 ^
Co^cv^+e-* n~ l.o ,y ~ <7.3?2> i-i 6" - ISO© /»3»'
S+ee^ * n * S-^gi"7 |y| - |?.Co2 ^ 6" *-?0 *<oa p;
M -= i-277 */o3 /<•>.;a - Jo&.^r k»>-4*

PROBLEM 4.62
Aluminum 1
Brass 1
Steel 1
Brass 1
Aluminum
y&-
0.5 in.
0.5 in.
-■ 0.5 in.
0.5 in.
0,5 in.
[—1.5 in.—*|
VI
AXIS
<D
<D
■a
42.
U>
4.62 and 4.63 Five metal strips, each of 0.5 * 1.5-in. cross section, are bonded
together to form the composite beam shown. The modulus of elasticity is 30 * 10*
psi for the steel, 15 * {0s psi for the brass, and 10 * 10* psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by couples of moment 12
kip-in., determine (a) the maximum stress in each of the three metals, (A) the radius
of curvature of the composite beam.
SOLUTION
« - £S - 30* IP* _ 3 . . i Q
lO *|0<
1.0
'l-ff
l.O
y, = %. , _!£<!<£ = |.S-
in = i*^ in 0.^1^*^.'^j»*(.,
3
fo) A-?u~>^: <5~ =
ntX =, O-O X 1^0-^") = 6^6 fcll-
X- 'S75~
6"
^ JELii* = 0-^03 Ylo.7r) a 6. n ks{
2,\%TS-
W 4; -
L _
M
iCt/lo*
/°
- I g£3 in « IS\. 1 ft
- ^4X.,s7 *JO * iV*

PROBLEM 4.63
r—1.5 in.-
axis
<D
<£>
H
3.o
•l.o
-feff
4.62 and 4.63 Five metal strips, each of 0.5 * 1.5-m. cross section, are bonded
together lo form the composite beam shown. The modulus of elasticity is 30 * 10*
psi for the steel, 15 * 10* psi for the brass, and 10 x 10* psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by couples of moment 12
kip-in., determine (a) the maximum stress m each of the three metals, (£>) the radius
of curvature of the composite beam.
SOLUTION
Use cd/owno~> as "Hie VCTev^e^t-e w«rtW«l
- %g0.s)(a.sf+@.o)(o.i:)0-°Y= WGI in*
I3, Bl^X -- Ijf 0s)(o.S? - 0.0W» ;»♦
I,. "» X2 =• C.ZC3
J ft
J
i * I( * Z^<^ m
X = Zlx = ^.0*3*;«
Brass *
238.-8*? X|0"C i*"'

PROBLEM 4.64
Brass
Aluminum
0.8 in.
4.64 The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semi circular cross sections. The modulus of elasticity is i5 x 10*
psi for the brass and 10 * 10* psi for the aluminum. Knowing that the composite
beam is bent about a horizontal axis by couples of moment 8 kip-in., determine the
maximum stress (a) in the brass, (6) in the aluminum.
SOLUTION
Potr eo-e^n S6i«ici>c/« r = O.S m^
A - Jr1" Loorthn* y0 = jg - Ote& S 0.33153 in
Iu.. * T^" T 0. I608^J«"
- O.OWiSJ in*
Use -J>.W
r«i55
T
A>1
\.Qo£3\
\.Oo£Z
tnA.iV
i.sisn
ypj i«
0.33c?5"3
-0. 33953
viAyo, in"2.
- 0.3^33
0./70&7
9 ^ OJ7067 . 67„
& 2,5*132.7
0.067^/ im &.laove -He
J, » 0.337-T3- 0.0^7?/ = O.ZU£2'i« J J2 = 0.33953+ 0.067?I - OMolW m
I( r K?,X + n.Ad,*- (/.S)^.^^^^)^C/^^0-^y3/)Co.27/62)Z = 0-l7S6^ ,V
I - I, + Jt * o.3<7o^v ;**
(a> firttsst W-Z.5^ J = 0.8-0.067^1 - 0.73^0^ •'*.
° jC " 0.3^o^h * *<mS> KSl
UO AJta*;*^'
yn= t.o y = -O.S - O.OC71I - -O.Z&III ih

PROBLEM 4.65
4.65 A steel pipe and on aluminum pipe are securely bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the steel and 70
GPa for the aluminum. Knowing that ihe composite beam is bent by couples of
moment 500 N-m, determine the maximum stress (a) in the aluminum, (A) in the
steel.
SOLUTION
Use ecJwyinOv* as Hie yerev/ie^cs r<na,je\f\*.J?
,1
sw-. I, = r?, ^OrJ'-O - (3.o)l(\iH-10") - iso.g^xj63 «„'
AJ>
l>M 1 Itl/M.
1= X/ +-T^ r 181-73 x/c^ v»«>S * |S/..73y/o"^ hn*
6-^iifck- C^O^QQKo.Q^) , ^3x/0Cfa - S2.Z M
1 1*1.75* ft?*-1
Pa
PROBLEM 4.66
3mm
4.65 A steel pipe and an aluminum pipe are securely bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the steel and 70
GPa for the aluminum. Knowing that the composite beam is bent by couples of
moment 500 N-m, determine the maximum stress (a) in the aluminum, (b) in the
steel.
4.66 Solve Prob. 4.65, assuming that the 6-mm-thick inner pipe is made of
aluminum and that the 3-mm-thick outer pipe is made of steei.
6 mm
10 mm SOLUTION
n . & s 212. , s.o

V-
PROBLEM 4.67
100 mm
50
_i
o>ii
©
®
1
J
4.67 The rectangular beam shown is made of a plastic for which the value of the
modulus of elasticity in tension is one half of its value in compression. For a bending
moment M= 600 N-m, determine the maximum (a) tensile stress, (6) compressive
stress.
SOLUTION
Y) ~ ^ o* flie j&flSi'o** Side or '/\e.j\\to) o~et\:
Locate heu/ftnfi.7 a.* fc.
i>*t*\
^ 6. 15" MPcl
^ * ' S ' l*fAzl *" = °- otAiHzl *
(b^ tohv?»*33i/« stress •

PROBLEM 4.68
SOLUTION
*4.68 A rectangular beam is made of a material for which the modulus of elasticity
is E, in tension and Ee in compression. Show that the curvature of the beam in pure
bending is
1 M
where
ErI
E_= AE&
'-~(V^+VO
Us* Et ^s "Hie ^efe^e^cc y^oJ^Jos.
The* Et » n£*
K>X
-(^-y^ - o
*---*
•fa y = (h-x)
fe-x -
P
■M
M
^t -I+w^, EV X
i*r-li ew-* _Z_ - la" ^
\r*,**»
Er I - £i I
ia
bh
•ft.
n
4Etec
w
(-/eIT^ + i)^ " (Vet + -fey-

4.69 Knowing that M = 250 N*m, determine the maximum stress in the beam
shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.
SOLUTION
w S-&0-10
ft*©* F^_ ^.3* K* |.87
r
PROBLEM 4.70
>8 mm
A
4.70 Knowing that the allowable stress for the beam shown is 90 MPa, determine
the allowable bending moment M when ihe radius r of the fillets is (a) 8 mm, (b)
12 mm.
SOLUTION
* Q.OZO m
- Z oo
<■» * - K - °-*
- K^
0,15-
— J^j*"v> -
4o
Ivtiv)
* 0.3i
Fro* Rj 4.31 k> I.S4T

PROBLEM 4.71
4.71 Semicircular grooves of radius r must be miiied as shown in the sides of a
steel member. Using an allowable stress of 8 ksi, determine the largest bending
moment
that can be applied to the member when the radius r nf the semicircular grooves is
(a) J in., (b) J in.
SOLUTION
&l) d - T>- 2t -- H.S -&Xi) r 3.75 in.
a
37^ * '^°,
~J- -^r- - O.I©
c = ■£ = ;.87^ /„
3.75
PROBLEM 4.72
4.72 Semicircular grooves of radius r must be miiied as shown in the sides of a
sleel member. Knowing that M = 4 kip-in., determine the maximum stress in the
3 3
member when (a)r='$ in., (6)f = ^" in.
SOLUTION
(a) d» = D-2r- V-5~(2V|)- 3.75 .V
d ~ 3.75 ' UA°J di 3.75 " ai°
FVo*. Rg. ^.37 K = 2.01
t = -kd - 1.875 ,V
Fr^ Fi*9 H.ZZ K = 1.61

PROBLEM 4.73
SOLUTION
D= ISO mwjC|: \0Otom)
* 73 The allowable stress used in the design of a sleel bar is 80 MPa. Determine
the largest couple M that can be applied 10 Ihe bar (a) if the bar is designed with
grooves having semicircular portions of radius r = 15 mm, as shown in Fig. a, (b)
if the bar is redesigned by removing ihe material above the grooves as shown in Fig.
b.
M
M
£ -
\SO
loo
\S
* I. So
- 0.15*
For- Coift+iqifir»T\OVl (&) j F13
1.32 ^iVtt K,- lip.
4.31 J.^es Kb ^ 1.^7.
r = Tat^3 * Mux***)*
<«>
00
l.^x/o6, m' -
-£.
\SxjO~ M
KO
6*- *£*
M- =
_ 6LI _. £8o*/o*)(l-S'xlO"*) _
mm
Kc
(l-WKo.os' )
r I. ^5 kN-*n
M
Kc " (i.57)Cao5o)
r l„53*lc? W-m * /.£3 kU-M

PROBLEM 4.74
SOLUTION
For boln cotnfVao^ixTi'oiftS
B- ISO Mm, d - \QO m«^
d loo
4.74 A couple of moment A/= 2 kN-m is to be applied to the end of a steel bar.
Determine the maximum stress in the bar (a) if the bar is designed with grooves
havingsemicircularportionsofradiusr=10mjm,asshowninFig. a, (6) if the bar
is redesigned by removing the material above the grooves as shown in Fig. b.
18 mm
(a)
R3 4-32 gw« Ke,= IZI
4.3i jives KbT
00
1*1 °r' " I*
C = ^ J - SO »*i*vi
W1W
I..79
-6 v
- o.oS
m
|-S"*/0
.5" XlO

ArffS
PROBLEM 4.75
4.75 A bar of rectangnlar cross section, made of a steel assumed to be eiastoplasdc
with ty = 320 MPa. is subjected lo a couple M parallel to the z axis. Determine the
moment M of the couple for which (a) yield first occurs, (b) the plastic zones at the top
and bottom of the bar are 5 mm thick.
SOLUTION
M,
r 6VT b (3;?o*/o'X3.37WO^ = ^
Y ~ C
o.oa7.s
jy = C - t = 7.5 -£"*n^ - ^.5m*v, = <X0G£5*m
- KlHH^ [l-ilM/J r *°Z *-
W)
PROBLEM 4.76
'fcf
■1
>
Of
4.75 A bar of rectangular cross section, made of a steel assumed to be elastoplastic
with Oy - 320 MPa. is subjected to a coupie M parallel to the z axis. Detennine the
moment M of the couple for which (a) yield first occurs, (£>) the plastic zones at the top
and bottom of the bar are 5 mm thick.
4.76 Solve Prob. 4.75, assuming that the coupie M is parallel to they axis.
SOLUTION
loft I»jkWi**£G5Xwt- ^'^'oW-^*/*^
C = ji\r\ = G *w = 0.006 en
1 ,r C O.006
(t") t = 5 *»*, Zr=C-£-£-s5"= / wm

PROBLEM 4.77
4.77 The prismatic bar shown, made of a sleel assumed to be elastoplastic with
Oy ~ 42 ksi, is subjected to a couple M parallel to the x axis. Determine the moment
A/of the couple for which (a) yield first occurs, (6) the elastic core of the bar is 0.1 in.
thick.
SOLUTION
C - J_)f\ - 0.\S m
MJ T *^ - oTTF - °-*«ff *t--
r 283.5" A-/M
0.1
0
r~
'
0..
o
1
uo ?v =
M. =
= i(™.s{\-£{%£)*]= n<n.sH.i
PROBLEM 4.78
4.77 The prismatic bar shown, made of a steel assumed to be elastoplastic with
aT = 42 ksi, is subjected lo a coupie M parallel lo the x axis. Determine the moment
A/of the couple lor which (a) yield first occurs, (b) the eiastic core of the bar is 0.1 in.
thick.
4.78 SoiveProb. 4.77, assuming that the couple M is paraHei lo the z axis.
SOLUTION
(a) I -= -jL bin3: £(©.3)(fc45)* r ;?.272lx|cTs »V
C = ih * 0.22S m
M>
gyI __ WK^.^gly/Q-*) -
0.225"
<*yis
, 0.Z2S"
O.OS j
(t) Xr - ^te = l(o-'1 r 0.06 ."n.
1—0.3 -J

PROBLEM 4.79
U-0.15«
1
= £, f
&
2k
c
4.79 A solid square rod of side 0.75 in. is made of a steel that is assumed lo be
eiastoplastic with £= 29* 10*psiand oj-=40ksi. Determine the maximum stress and
the radius of curvature caused by a 4 kip-in. couple applied and maintained about an
axis parallel to a side of the cross section.
SOLUTION
M
1/
" G - 037? " ?-gi*S kp-m.
M-|MrC'-i*
^ = /i^X--|^Il^ =«^i
/°r
, ££ - Ca<?K;QtXo*375^ _ ^7 K
i/O */03
83
|M
Yr
PROBLEM 4.80
4.80 The prismatic rod shown is made of a sleei that is assumed to be eiastoplastic
with E = 200 GPa and ar - 280 MPa. Knowing that couples M and M' of moment
525 Nm are applied and maintained about axes parallel to lhe.y axis, determine (a) the
thickness of the elastic core, (b) the radius of curvature of the bar.
SOLUTION
-^^y 3 - ^f^ - 0.3263? , yr = 0.32C32 c r :?.?3fiS ^

PROBLEM 4.81
4.80 The prismalic rod shown is made of a sleei that is assumed to be elasloplasiic
with E = 200 GPa and ar = 280 MPa. Knowing that couples M and M' of moment
525 N-m are applied and maintained about axes parallel to lhe.y axis, determine (a) the
thickness of the elastic core, (b) Ihe radius of curvature of the bar.
4.81 Solve Prob. 4.80, assuming that the couples M and M' are appiied and
maintained about axes parallel lo Ihe x axis.
SOLUTION
^
_ Or I _ j'ZSO^(06)(20.n^to'>)
Q-OLZ
= 4-83.8*1 M-m
M v|M,(|-i^I) or %-fe- 2$
J&__ I ■=> C?XS2S)
= 0.7/097
jyY = 0.11 oil c - 10.932 *»n
tear* = 2jV - 21. «? H*
(b) £>
/°
_ E^ _ (aoo^fo^X/^^^y/o"3) _
/°* ft
^o //o6
7.81
hn
PROBLEM 4.82
4.S2 A solid square rod of side 0.5 in. is made of a steel that is assumed to be
elaatopiaslic with E = 29 * 10s psi and Or - 42 ksi. Knowing that a couple M is
applied and maintained about an axis parallel lo a side of the cross section, determine
the moment M of the couple for which the radius of curvature is (a) 5 ft, (ft) 2 ft.
SOLUTION
* -L
■0.Si«,
C ■= "£k ~ 0.25 /o,
c
My
— - £75 Jl-K
£• =
£
£
75.6? i*.
(Ol yO r 6" ft ' 6t? in.
(U /D= 2ff. - 2M .V

ayfs
PROBLEM 4.83
c
■40 mm
60 mm
4.83 and 4.84 A bar of the cross section shown is made of a sleel thai is assumed to
be elastoplastic with E - 200 GPa and ar = 240 MPa. For bending about the z axis,
determine the bending moment at which (a) yield firsi occurs, (6) the plastic zones at
the top and bottom of the bar are 20 mm thick.
SOLUTION
(a) I = £ ^ r ii (MoYwrt3 = 7J?0 *to* w = 7*o y/o1 ^
c = -^lo * 30 *», * 0.030 m
M„ * SlI - f^otip^7go^/Q'f) . £76 y/o* N-^>
-5". 16 UN-»\ -^
0.03 o
Gr
Aa' .-
ao
Mtw)
.
<r *•
^C
I—io**-!
- 48 *lOz M
A - f flO M-^ 6.6C7*,*, - O.0O6667 *n
(WM- 2 (ff.y, 4 r?2yj = ^[(l^x/o3)(o.O^>>) + (Mg/|03XaO06^7)]
= £32 *\0* tf-*i T 2. £3 kW-r»

&J(\S>
10 mm -*■
PROBLEM 4.84
c
20 mm
t
20 mm
t
20 mm
♦
■*- 10 mm
20 mm
4.83 and 4.84 A bar of the cross section shown is made of a steel that is assumed to
be elasloplaslic with E = 200 GPa and ay = 240 MPa. For bending about the z axis,
determine the bending moment at which (a) yield first occurs, (b) the piastic zones at
the lop and bottom of the bar are 20 mm thick.
SOLUTION
(0\ I«&t " ~kWr ^ftoXfcO? r 12Q*tO% tor*"
I t 'IXOiiO3'- 12.33*10* - 10&M1 v\o* mm*
- 70 6.G7x/O"% wi **
C - ^ h = 30 ^ = O. 030
K
6-,
- 6VI _ C^py|Ot)(7o6.67>'/o'"r)
C " 0.030
= 5. £533 x 10* N.»i * S.« kK)* wi
^
'/.
/> ■- '' ■ ■'
s
-A,.-
XV
■
<
*C
^
5>^
^"^ - vx
*jz - f (lO«-0 * CCC7*»* - Q.Q066G7 r*
r 8.00 *|o* N-w - g.oo kO-iw

PROBLEM 4.85
1.5 in. -—
y
^ *
t
3 in.
*
*
3 in.
t
\
3 in,
+
*- 1.5 in.
3 in.
4.85 and 4.86 A bar of the cross section shown is made of a sled that is assumed to
be elasloplastic with E = 29 * 10* psi and Oy = 42 ksi. For bending about the z axis,
determine the bending moment at which (a) yield first occurs, (ft) the plastic zones felt
the top and bottom of the bar are 3 in. thick.
SOLUTION
id) I, = £UK5+ A.J.* • &&)(*?+ (3X3)&f =r S1.1S tV
I*= iHA* B iaC6"^ - |3.5 }**
I4 '- I, = *Z7jT i*"
M^ = —— - ^—j^= - I.7S?, K.|?-ivi -^
^
axis
R,
*,
R, - 6;A, - twX3X3^ = 37g kiP
y, - I.-5"+ 1.5" = 3.0 ,'vn
UO M^ ^(t?,y, v?,y^ r X[C37SX3.*>H 081X1.0^ r ^ 6 kr> • in

1.5 in. —
PROBLEM 4.86
y
^_
-1
3 in.
*
3 in.
*.
+
3 in.
♦
— 1.5 in.
3 in.
4.85 and 4.86 A bar of the cross section shown is made of a steel that is assumed to
beelastoplasticwilh£ = 29 x 106 psi and aT = 42 ksi. For bending about the z axis,
determine the bending moment at which (a) yield first occurs, 0) the plastic zones at
the top and bottom of the bar are 3 in. thick.
SOLUTION
I* ~ I, r I7S.S >'">
I -- I, + 1;
C = H.5" ,'n.
- SS7.1S .*»
(40(3^7.75)
f.5*
3 33^ k,p,,-,
6V
tt.y»s
JL
\?G
^ - 1.5 + 1-5 = 3 i"
(b) M - 2(£,J, +^yA)~ J?[(7S6K3WN.rtfl-ol] = ^V fc>-i*

PROBLEM 4.87
40 mm'
60 mm
4.87 through 4.90 For the bar indicated, detennine (a) the folly plastic moment Mp,
(b) the shape factor of the cross section.
4.87 BarofProb. 4.83
SOLUTION
F*s»* ?RoQLEK\ 4.83 £ ' ZOO &?<*. a»J &f = 2.HQ MP*,.
A, -" (40K30 ") = llOO m*,*"
Q - o;a1
* (;wo */o£ X1200 */cf6>)
d r 30
i*ii»i r 0.O3O **■»
rf
_ ^ „„..„■
R
(a) Mp = Rel - C^Sgx/o3 )(o.o3o) -- S.QHylO^hJ.v* - *.64 kW-^
C ~ SO >-n*v> - 0.03o wi
^ " C " O.o3o
.03o
k -
HY
5:7^

PROBLEM 4.88
10 mm
20 mm
20 mm
20 mm
4J»7 through 4-90 For the bar indicated, determine (a) the f>Hy plastic moment Mp,
(b) the shape factor of the cross section.
4.80 BarofProb. 4.84
SOLUTION
Fro** PROBLEM ^.8^ B' %OQ SP* a*el 6^ = ZHo Mpft..
20 mm
10 mm
**i*
- Q.OZO ^
9% « c;ax
«-—— -
*
—>«
—»*
—>
—>
~2l
3 V
X *■ I*.* - I«aU - 7^0Wol - 13.33 */o3 r 706.(S7 */o3 h
» 7o6.67 * /o"1 ^
C r £h • 30 »« t 0.03© ^
k»
Mr * £(53J
l.<J¥3

PROBLEM 4.89
y
1.5 in. —
4.87 through 4.90 For the bar indicated, determine (a) the fully plastic moment Mpt
(b) the shape factor of the cross section.
4.89 BarofProb. 4.85
3 in.
SOLUTION
*C
3 in.
3 in.
4
3 in.
1.5 in.
F^*~ ?Ro8L.£M 4.85" E = ^x/O* psl a^\ <S; t ^ lft.\
(?, - 6JA, - C^JteXs^ r 572 k:,
^, - 1.5 +• LsT - 3.o ■-
1 y
1?.
—*
*t
Hf ^ 2 (K,^, + 1?At)0 - *f(37*X3.0)+ (372X0.75")] r 2S3^ ^>*i.
I, - I, - ^7.75 ;«*
J - J, + ia +1* - igs.-r ;*T
k =
1835
175?. 3
I. 611

PROBLEM 4.90
y
1.5 in,
4J87 throngh 4.90 For the bar indicated, determine (a) the foliy plastic moment Mp,
(b) the shape factor of the cross section.
4.90 Bar of Prob. 4.86
3 m.
3 in. SOLUTION
3 in.
1.5 in.
Ws
<xx\$
Ml
R^SrA.^X^Xs")* JSC kip
^, * US* 1-5 * 3.o j*
^2= iO-S)- 0.7i" .*.
Mp = Z(*tj, +?&)' A[(7SO(3.oMl89Xo.7sY] - Y8lVT J6ri*
(^ J, - iUK* 4 A,J* ^ ^fcXi)5* (tfXsXfi^ • \1S.S iV
I » I, * Ia * I3 - 3.57.75- iV
Mp - 4819. J
p- i*
k-
M,
3331-
W*

PROBLEM 4.91
lOinm
u
f
50 mm
)
30 mm
10 mm
■*- 10 mm
30 mm
NA
mm
U J
4.91 and 4.92 Determine the plastic moment Mp of a steel beam of the cross section
shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
To-f^P cure* A - (So)(<1o) - {3a)(3o) - 3^00
tA - 1800 ^
I8QQ _
36
Wlt»n
fc
P 1
Kh >
P.
A, - t50)(36) = iSOOw^ ^ - ig^
Ax = (So)(m)* 7fto^ yz = 7*M
A3 = Wo)(3o)- 600^ y3 • x<\ m^
A,^, = 3ZH x/o* *,*,'
3 7
7*?. 2 x /o~* i*i
A.y. + *•& * ^ + Ai(Jf * 7?.;mo3
H.ol WM'V*

PROBLEM 4.92
4.91 and 4.92 Determine the plastic moment Mp of a steel beam of the cross section
shown, assuming the steel to be eiastoplastic with a yield strength of 240 MPa.
10 mm SOLUTION
25 mm
■25 mm-*
10 mm 10 mm
•A
ayt'
^
V/.
z
K^%
\
X
ToU). are* A - {ZS)l\& 4. &.)(lo){3f) • ISO ^
R, = SrA,= (2t0*IO*Xo.O3S')(0.O\oV GOv/01 f^
y, = 30-M.l.S " G.Z5 w - O.OO&iS m
Pj. B 6rA» = {2Vo*r|Ofc,>(o.<«o)(aQil«) = 54 Wo1 N
j% ~ Tlio.oitZS'} ' 0.OO5&ZS ^
& = i/ - O*0llS7S *i
Mp - Ry, + Payt * *?»y,
-- (tof(C>3X0.006Z£S)4(^y|Os)(>.C7c)562^+ (||4* lo3)(o.O/l«7^)

PROBLEM 4.93
0.6 in.
4.93 and 4.94 Determine the plastic moment M of a steel beam of the cross section
shown, assuming the sleel io be dastoplasiic with a yield strength of 36 ksi.
0.6 ln.-
0.8 in. SOLUTION
i.2in.
_i
0.4 In.
—r
h
A
ctvea A =( I. *?Xo* f^ •+ (o.£)0**O » 1.44 m2
iA = 0.72 >^
■iA _ o.i* _ . - .
X - -:— - - "-< m.
V> 0-6
A2 - iA * a?a .V ^ % = i (o.h ) * a z ^
MP - SyCAy, * VyJ
- 20,7 fop. m
PROBLEM 4.94
|*-4in.-J
jin-
4.93 and 4.94 Determine the plastic moment M. of a sleel beam of the cross section
shown, assuming the steel to be elastoplastic with a yield strength of 36 ksi.
SOLUTION
To+J area.' A - «X£ ) * (i)(3) + (i)({') = *.S \^
iA r MS i«*
A, » ^.ooi^ 5,^.7^
A* * o.xs \f*} y»*o.25^
2 in.
3 in.
^2
KA_J11&£
ffi-4^XNsJ
'.SO m*
A*y,=
Ah* l.oo i^0 5* =^'7,5j
Mp * 6VCAJ, + A2y* * A3yj + A^)
r (3^(l.,Sb + O.0«S" + ).£WS"*2.7S^ - 2I|.S VCr\*
- £75* i*3

PROBLEM 4.95
4.95 A thick-walled pipe ofthe cross section shown is made of a steel that is assumed
to be elastoplastic with a yield strength ar. Derive an expression for the plastic
moment Mp ofthe pipe in terms of clt c2, and or.
SOLUTION
PROBLEM 4.96
SOLUTION
4.96 . Determine the plastic moment Mp of a thick-walled pipe ofthe cross section
shown, knowing that cx — 60mm, c2 - 40 mm, and ar = 240 MPa.
See fj»e SoA^i'di h> PRo8LE"M *J.*S"-Por cAe*iV<wkovi c/9 He
fevtfo ^ M.,

PROBLEM 4.97
4.97 and 4.98 For the beam indicated a couple of moment eqnal to the fully plastic
moment Mp is applied and then removed. Using a yield strength of 240 MPa,
determine the residual stress aty = 30 mm.
4.97 Beam of Prob. 4.83
■40 mm'
SOLUTION
60mm Mp = £•&** kM-w. ($«« SoLOJtotj -f* P(?o6L£M H. 67 *)
I - 7XO * /O"* ^
C ~ O. OZO wi
5' -. J^y , j^£ .f^.c-so
**i*fi
,U>AC
f*
)IW&
2Ho MP*
+
^
^
:^
-?• 3^ MP*
7r
\zo MP*.
UUloAOlMG.
RgSlQOAf. STpgSSES
61.' 6*'- <5i = 360Yto'-Mo>io* * Uo^io'p. =r I20MP*.

PROBLEM 4.98
10 mm
*C
U
20 mm
20 mm
20 mm
20 mm
^—, ,
«
<• ■
I * •!
. ^
>
.. fc.
10 mm
740 HP*.
LOADING-
4.97 and 4.98 For the beam mdicated a couple of moment eqnal to the fiilly plastic
moment Mp is applied and then removed. Using a yield strength of 240 MPa,
determine the residual stress at y = 30 mm.
4.90 Beam of Prob. 4.84
SOLUTION
Mp - £.l£ kKJ.^ CSae SolOtiow -fe» proSLBM H. 88 )
I r 70C.C7 x \o tv, c ff 0.030 »,
uf
= £ .
706.67*!©'1
-^ 3^.4 MPa
106.H MPa.
UUt^AD IW£
<4
ee-soaAL stresses
ffre* s 6' -6^5 3HC.Hslo<'- 2HO>lOL ? IO£.<i*l06f* = |0«.<* Mfl*.

PROBLEM 4.99
1.5 in.
3 in.
h-f
3 in.
3 in.
-i
3 in.
1.5 in.
4.99 and 4.100 For the beam indicated a couple of moment eqnal to the fully plastic
moment Mp is applied and then removed. Using a yield strength of HZ ksi, determine
theresidnal stress aty-4.5 in.
4.99 Beam of Prob. 4.85
SOLUTION
Mf = £835 U-'f\* (See SoLonorj ■£» problem 4. 89")
J - 19$.S in*
ft' - H-«»y , .M.c
° " 2 ' X
& " ISS.5
- 57.7 ks;
LC>AD
■
|M&
42 w;
+
^
=^
^ 67.7 ks/
35.7 ks,*
UNLOADING REStDOAL STRESSES
6"w3 " "S* - 6V r 67."? - 42 =: ^5.7 JCS,"

1.5 in. -*
PROBLEM 4.100
y
+
3 in.
*
3 in.
*
3 in,
*
— 1.5 in
3 m.
4.99 and 4.100 For the beam indicated a couple of moment eqnal to ihe fully plastic
moment Mp is applied and then removed. Using a yield strength of 42 ksi, determine
the residnal stress at y = 4.5 in.
4.100 Beam of Prob. 4.86
SOLUTION
.Mp - liS\efr£ k.'p.i* (See S&LVT\otJ £o PRo8LeM 4.?o)
C = 4.5" i>
I ' 3S7.1S iV,
I ■ I
£
o^
5 C
LOAD
-*
KJG
<te ks;
5* 6o.62 ksi
UUtoADlMS
8.6? ks.
RCSlD0ALSrP£S5£S
ffr* * <* - s;
Go.C? - </* • |8.G* l«i

(GO
W
fc*
PROBLEM 4.101
■40mm'
4.101 and 4.102 A bending couple is applied to the bar indicated, causing plastic
zones 20-mm thick to develop at the top and bottom of the bar. After the couple has
been removed, determine (a) the residual stress aty=30 mm, (6) the points where the
residual stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the bar.
4.101 BarofProb.4.83
SOLUTION
60 mm
See solution -ro p«o8U?m ^..gs -&/• bevwJrnj t©ufJ« &*<\
E - 2oo GPa.
J -~ IZO-^IO"'1 m
3^6.TxJO& ffl. =: 3M6.7 WP*
_, M5*.G >* IO* P«.= MS".^ MP*-
6^ = 6:'- &r-= 346.7 - 24o = 106.7 MP*.
Af y c
2to Mrt^
<?«> = 6""-6"T * 115". 6 - 2Ho
SHoMR*
LOADING
-*-3Mt,7 «P«
-I2*/.H M-Pcl
lot.-7 MP^
6m - O
'r*\
UMU>ADlM&
F
\ m.7 MP*
RSStODAL STRESSES
16; _ (72oxio'M^4owoQ r
^o*
<J
P
6T^s " -IMA x/O* fft

PROBLEM 4.102
10 mm
*C
w
20 mm
20 mm
4.101 and 4.102 A bending couple is applied to the bar indicated, causing plastic
zones 20-tnm thick to develop at the top and bottom of the bar. After the couple has
been removed, determine (a) the residual stress at.y = 30 mm, (b) the points where the
residual stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the bar.
4.102 BarofProb. 4.84
20 mm
SOLUTION
20 mm
10 mm
Sec SOLUTION io PRoSLEM 4.8** -fix* ketokw* cc>u|>A and
M - $. OO k M' i*»
E = Zoo Gffe.
I - 706.67 kId* *H
yr - JOm.*, = O- OV£? >*%
<5y - ZHo H?*.
C = O. O3o m
(e0 <S - -j-
, M# ,
At
2MO tf Pa ■*■
Loading
Cm = 6*'-6; - 337.* -^o = <?%6 MP*
S,^ *S'"-<5V ' H5.X-2iO ~ - 1*6.-8 MP*.
33?.* HP* r-» rUMft*.
- i?4.5HrV
WCS MR*.
- "M.C HPa.
residual stresses
UML0ADIN&
CM
&« - o
M
?-er.
r O
-s
2/.2*/0 *\ ■= 21.^ *«*
G^ At J " .yr
(7o€..c7x/a"'T)Cz'te"f/ot)

PROBLEM 4.103
1.5 in.
3iiL
4.103 and 4.104 A bending couple is applied to the bar indicated* causing plastic
zones 3-in. thick to develop at the top and bottom of the bar. After the couple has been
removed, determine (a) the residual stress at y - 4.5 in., (b) the points where the
residual stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the bar.
4.103 Bar of Prob. 4.85
CO
(b)
3 in.
3 in.
-f
3 in.
SOLUTION
See Solution i» pffoet-Ef^ 4.85T -f> W;ra cwtf
'* o.-.d
1.5 in.
M - 2646 ^ • m ^yr r US ),
^ If!
1 lSg-5"
A4
2
~ c
I - 188.-5" ;«y C = */.£" ,V
63. H tfe.*
£3.17 - VZ * 2M7 its,-
a,l.0<>- 42* - - £o.S4 Ws.'
eM = er'-Sr =-
*—i
0.
m
>
►
>
-*? GS.n to.-
LOADING
vA " 2646
2o.?v k*<"
-tun "a-'
RestoDAL stresses
-,?C>77 ;*,
- 173J -W

PROBLEM 4.104
3 in.
4.103 and 4.104 A bending couple is applied to the bar indicated, causing plastic
zones 3-m. thick to develop at the top and bottom of the bar. After the couple has been
removed, determine (a) the residnai stress at y = 4.5 in., (b) the points where the
residual stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the bar.
4.104 BarofProb. 4.86
1.5 in.
W
3 in.
3 in.
4
3 m.
SOLUTION
1.5 in.
See. SoLOlfofJ to PftoflteM 4. ff£ W be^U^ ooopJt and
(dt) 6" -
' - Mc _
J » 357.75* ;,
a5"7.7S"
c = f.S" ,V
5<?-43 ks.'
/7.4S *s,-
19.3 1-42. - -^. n ^;
-J?51.43^; (-^ 17.43 Vftt
I
A+ X*c
357-75"
6^ - <T- <SV
►
►
»
+
LOADING
-JM.l1"? jfti
UMLOADIMG-
-\7.43 *»•'
rssiddal stresses
^° ' ^ 47 *S"
CO A+
M
P
3./S
t^
o.*S.
y<
r -3. |Si'm , Oj 3. /Si*.
£**« - - 2Z.W ks«
res

PROBLEM 4.105
SOLUTION
J, . Mr
ft ' EI ->
rr) =
- i± .
HY
Ml
♦4.185 A rectangular bar that is straight and unstressed is bent into an arc of circle of
radius p by two couples of moment AS. After the couples are removed, it is observed
that the radius of curvature of the bar is fa. Denoting by pj the radius of curvature of
the bar at the onset of yield, show that the radii of curvature satisfy the following
relation
Pr
\
e
i-
3/7
2py
1
l-3
( \2
_P_
\Py)
s | Mr ( \'s$) l«+ w <W<
#■
3 -Zvn
±1
J. .
fi
/° EI /=> El
-rt'-M- #'-
PROBLEM 4.106
4.106 A solid bar of rectangular cross section is made of amaterial that is assumed
to be elastoplastic. Denoting by Mr and Pr, respectively, the bending moment and
radius of curvature at the onset of yield, determine (a) the radius of curvature when a
couple of moment M-1.25Mr is applied to the bar, (b) the radius of curvature after the
couple is removed. Check the resulta obtained by using the relatioin denved in Prob.
4.105.
SOLUTION
I
m -
ft) £-
EI J
M'- |M,(l-±|;\) U+
w - -^ = /. ^r
^ - V3- Zm ' 0.7O7U
p = 0.7O7II ^V /
/° EX • p
J- - JEL
l.*sr
/° ■ /»■• 0.70111 f>f fr
fit ' 6.0? ^

PROBLEM 4.107
4.106 A solid bar of rectangular cross section is made of a material that is assumed
10 be elastoplastic. Denoting by UT and Pr, respectively, the bending moment and
radius of curvature at the onset of yield, determine (a) the radius of curvature when a
couple of moment A/= 1.25A/r is applied to the bar, (b) the radius of curvature after the
couple is removed. Check the results obtained by using the reiatioin derived in Prob.
4.105.
4.107 Solve Prob. 4,106, assuming that the moment ofthe couple applied to the bar
is \MMj.
SOLUTION
(Art
Pf
. H
EI j
M ^|My0-il>) ^
m
I.¥0
m
m % -
M
« £ **0-i&)
M
**■
/°
0. W7Z
/°<
Q.g3C07
1.40
/* * /. /**/°r

PROBLEM 4.108
4.108 The prismatic bar shown is made of a steel that is assumed to be elastoplastlc
and for which E = 200 GPa. Knowing that the radius of curvature of the bar is 2.4 m
when a coupie of moment M = 420 N*m is applied as shown, detennine (a) ihe yield
strength or of the steel, (b) the thickness of the elastic core of the bar.
SOLUTION
Gri
G>^
- a c U 3 rElc* J
~ e; he* (i - £ |^ )
Sr [ I - 7So*to~*' 6yz] - 373. MY */oc

PROBLEM 4.109
40 mm. B
tr (MPa)
/
y
1
A
(
30
mm
0.005
0.010 €
4.189 The prismatic baiAB\s made of an aluminum aiioy for which the lensile stress-
strain diagram is as shown. Assuming that the o-e diagram is the same in compression
as in tension, determine (a) the radius of curvature of the bar when the maximum stress
is 250 MPa, (b) the corresponding value of the bending moment. (Hint: For part b, plot
a versus y and use an approximate method of i ntegratioiL)
SOLUTION
(<0 6"„ - ZSO MP*. * ZSo-*lo* Pa.
C - j^h - 30^^ - O.C30 ^
t> ^ qo **. •= 0.040 i-,
Be^ai/i^ temple
rc rc r'
M =-^ ye-k-ly = 2fc\jkUy ■ ^fcc* j i) W\ do - 2bc* J
D 16*1 Jo
Eva-Pu^f« T us''w# a. v*eHtoel ©"F ni>m«w*,P ivita^vwrVe*. Xr Simpson 3
J -- ^ Z WUI6-I
Ha v»i-
u
0
o.^s
o.s
0.7S"
\.oo
lei
0
O.OOIC
0.00ZZ
O.OC4S
O, OO&f
\GTl,(MP^
o
/ lo
1 80
Z2S
ISO
uksl^p^
0
Z7.J
^o
I6 8.7JT
2S*>
W
1
*
51
4
1
wulcl/MP^
0
110
lgo
C7S
ZSo
iZiS' *■
^ WOlorl
Jr ,co.xyx^io g /o/.zs-MP^* /o/^rwo'P*
Mr (XVo-Oloyaogo^Clol.^-to*) T 7.^*?>/os W-k ^ 7.2<f kM-*i

PROBLEM 4.110
40 mm
a (MPa)
/
A
'
" (
30
mm
4.110 For the bar of Prob. 4.109, determine (a) the maximum stress when the radius
of curvature of the bar is 3 m, (b) the corresponding value of the bending moment. (See
hint given in Prob. 4.109.)
SOLUTION
(d\ f> * 3 *> j C = 0-030 *,»* = O. OSO *->
*t
- -£. Q.Q5Q
o.o\o
Fi^o'*i co^v«. S**, ~ £"75" MPol.
0 0.005 0.010 €
fc « - 6M^ ~ - €^ U w ke**i U *
^
vAe** -M»« inte***' J" 13 gtVe*. loy J U 1*5*1 dlo
J = 4^ Z wots-l
Wrie^e w is a w«,i**AtY>m 4**^*/^ D&in+ AU r 0.2.5* ^e ge*t
u
o
0.25*
O.S"
O.lS
l.oo
1*1
o
o.ooar
O.04£b
0.0O75*
O.oioo
lcrl,(^lV
0
'GO
2^t
*6$
XlS*
u|Sl(Mfi$
0
MO
127
|9*W
27S"
Vu
r
H
2
4
1
woleUMflt)
o
IGO
2S"4*
7<?g
21S
IHS7 *
Jr (o.*sXn»7) , ,„_, Mp^ s ,.23.9WO'fii
■Z"W0|6"I
M = C?)^*>Ho.o««V(i«3^xfo*)* s.72*to3 «•* ■= 8.WkU-
>v\

PROBLEM 4.111
<r(lcsi)
50
40
30
20
10
0
.4
1,2 in.
4.111 The prismatic bar AB is made of a bronze alioy for which the tensile stress-
strain diagram is as shown. Assuming that the ^diagram is the same ia compression
as in tension, determine (a) the maximum stress in the bar when the radius of curvature
of the bar is 100 in., (b) the corresponding value of the bending moment (See hint
given in Prob. 4.109.)
SOLUTION
(a) f> - IOO l* f h = O.S in y C = 0.6 in.
£ r -£- r Q'G- ■= 0.006
FV>o«-\ -r-l^e curve €""*, ~ ^ 3 W&t ^*
E, » - €w^ r ~ €* U w kc** L) *
-it
0.004 0.008 f
Bcn^in* Couple
M *-J ye-kWy * 2tJ v Iff I J- ■ *ic*J u)sldo = 2bc* J
viW* -flic iw+ejr^ J" fa g,\,«v\ by $ D ^ ^0
u
o
OAS
O.S
0-75"
I.DO
lei
0
O.OOIS"
O.OoZ
O.0D4S
o.oot
te^ks;
o
25
sc
40
MS
U M,, Ics.-
0
G.*2i
IS
So
M3 •
\AJ
1
4
*
4
1
WO \6"|jUsi
O
ZS
3£
IJO
MS
/X.T *"
J'
M
(°y > I8.C7 ks,-
= (2XO-8)(O.Gs)l63.67) = |O.Z£" k.p-i".
Zwo|6-l

PROBLEM 4.112
<r(ksi)
60
40
30
20
10
0
4.JJ2 For the bar of Prob. 4.i i i, determine (a) the radius of curvature of the bar
when the maximum stress is 45 ksi, (b) the corresponding value of the bending
moment. (See hint given m Prob. 4.109.)
SOLUTION
6^ * W Ift.-
C = O.Q i»
/
/
/
^
/» ~ 7vS" in,
O.OOS
0.0/3333 iV'
0.004 0.008 €
s, * - €«^ - - £* u w kc^ft u) * •g
-i
MJ-j^U' ^tf ykJJy " ^fcc1" J u)<s1do * 2 be2 J
-I
Wnevt m; is a w€-i*^ATV** 4vw&i»/'r ZJi/n* &0 ~ O.Z£ we. ge"f
u
0
0.2S*
0.5
0.75-
r.o
!£l
O.
O.OO 51
O.ODl
o.oofe
O.OOS
Id.lcs."
0
32
38
43
45
ulcl, Wi-
O
S-
n
3?.?S"
46*
W
\
H
3
4
5
wu Id^ W*.-
0
3*
38
/a*?
45
X4*f *
\At
u\S\
j__ (o.*y»> B ^.33 ks.-
M^ UXo.8)(0.€^to«>.S3i * \U7 kip.
iv»

PROBLEM 4.113
4.113 A prismatic bar of rectangular cross section is made of an alloy for which the
stress-strain diagram can be represented by the relation, e=ktf for a> 0, and e=~\k<f\
for a < 0. If a couple M is applied to the bar, show that the maximum stress is
l+2«Mc
°ms
3n
SOLUTION
For £ = KS'\ e„= K61
eca
AP
be'
c ■ '« c
TfaC
be1 'si
6,
2n+ I tfc
' 3n T

PROBLEM 4.114
4.114 A prismatic bar of rectangular cross section is made of an alloy for which the
stress-strain diagram can be represented by the relation e = kd1. If a couple M is
applied to the bar, show that the maximum stress is
1 Mc
<7„ =
m
9 I
SOLUTION
c
Far £:Ks\ e.* - K6*„
Tk
>w
a
W
be1
£eca/U>
'a c
be* '31
r 2n+ l ^6
W;+»t n = 3

PROBLEM 4.115
fit) kN
60 kN
150 mm- r>-J 60 kN
mm
4.115 Determine the stress at points A and P
60-kN loads are applied at points I and 2 onlj
SOLUTION
"P = ISO kW " 130*10* N
A+A«^8 * = --£ =^^£-! sr-8.33x/^p^
V - /20 fc|0 * l20*/0*
I- Ta^3 = u O^y^o)5 * /^.rg^o****'' = /o3.cs*/o'e
Kiel = 13 (2 >*>»t ~ O. 120 **
120 mm v 90 mm
PROBLEM 4.116
fl() kN
60 kN
130 mm
150 mm
120 mm x 90 mm
4.116 Determine the stress at points A and B, (a) for the loading shown, (b) if the
60-kN loads are applied at points 2 and 3 are removed.
SOLUTION
= - 8.33 MP*, -a
P: iGo kw * CO^/o* A/

PROBLEM 4.117
30 mm-
4.117 Knowing that the magnitude of the horizontal force P is 8-kN, determine the
stress at (a) point A, (b) point B,
24 mm
_-* •»-
45 mm
SOLUTION
e - hs - a = 33 ^^ - o. 033 v^
wn
15 mm-
= -|02.«>r(0* P«. =-i02.g MP«t
PROBLEM 4.118
4.118 The vertical portion of the press shown consists of a rectangular tube having
a wali thickness t ~ % in. Knowing that the press has been lightened until P = 6 kips,
determine the stress (a) at point A, (b) at point B.
4 in.
SOLUTION
A- (3X4,1 - (Oft) =■ Grt*
H* Pt * fe^ia)* 72 k.>.iV

PROBLEM 4.119
4.118 The vertical portion of the press shown consists of a rectangniar tube having
a wall Ihickness f = 2 in. Knowing that the press has been lightened until P = 6 kips,
determine the stress (a) at point A, (b) at point B.
4.119 Solve Prob. 4.118, assuming that the wall thickness of the vertical portion of
~~r the press is t = f in.
3 in.
J.
B
4 m.
C = 2 iV.(i
SOLUTION
€-- 10+2 * 12 .^ M - Pe - (G^i?^ * 72 lc.jp- Tv,
4 ll£ -
X
4-.687S- T 1.56^ ~ ,fc*AH RS'
,, \ fir _ £ ^ M£ ~—S <7g)(*> = - |3 78 *s,-
PROBLEM 4.120
^80 mm
60mmv ^W*
4720 As many as three axial loads each of magnitude P = So kN can be applied to
the end of &Sft3b3rf rolled-steei shape. Determine the stress at point A, (a) for the
loading shown, (b) if loads are applied at points i and 2 only.
W £00* 31.3
SOLUTION
For- W 2O0*3l.3 foJJtJl «+«e^ sIia^c
A-- ^000 w^ - M.oo<Wo~s m1
c = i J - 2 teio1) r ioS M»n - 0. las' *v,
I r 31.4 * I0C -imv = 3LH x/0"t*n'1
(a) Ce^-frrc i'ofc^
39 - So + 50 +50 = 15b kK) = /5fc*/o* |sJ
^ r "¥ ' - <fe>,|03 -=-S7.S*lol P* - -3*5 MP* —
2pt ^50 + 50 = 100 UrJ = toOM03M
r^ - ££ Mc ;oowo3 (4.oxio*'Xo-ios') _
4.0 * CO'1
31.* v)0

PROBLEM 4.121
/SO nun
60 mm
4.121 As many as three axial loads, each of magnitude P = &> W , can be applied
to the end of a Wfee& roiled-sleel shape. Determine the stress at points, (a) for the
loading shown, (/>) if loads are applied at points 2 and 3 only.
w zoo* 31.a
SOLUTION
Fc.r a W 7O0*%\.% ro/i«d sfte7 sh*f«
(i?) Etce-*.4rt'c .A>*#fcn* e = SO i»m-» - 0.030 t*
\0O x
€T --RE + MS- -
g ^•,°r:,^'") g-^;^;
« nac
PROBLEM 4.122
4/22 An oflset h must be introduced into a solid circular rod of diameter d.
Knowing that the maximum stress after the ofBet is introduced must not exceed four
times the stress in the rod when it was straight, determine the largest ofBet that can be
used.
SOLUTION
For ceviT^'C JrooLAm*. 6^ = "T-
G;
VftV\
€. * *S

PROBLEM 4.123
SOLUTION
4.123 An of^ frmttst he mtroduced into a niet^
2-mm wail thickness. Knowing that the maximum stress after the ofBet Is introduced
must noi exceed four times the stress in the rod when il was straight, determine the
largest offset that can be used.
c^Ar <?
t*\*\
, 3. "
1 = 5(^-^/^^(1^-1^)- 3.2C73
CA tt)t>o.S33
X
PROBLEM 4.124
12 kip;,
4.124 A short column is made by nailing two 1 * 4-in. planks to a 2 * 4-in. tlmbar.
Determine the largest compressive stress created m the column by a 12-kip load
applied as shown at the center of the top section of the timber if (a) the column is as
described, (£>) plank 1 is removed, (c) both pJanks are removed
SOLUTION
UWi
e>
12 k^<
\
i
\7 k.y,
(a*
W
CO
n * 4 in CV>SS S e*TV©w. A > ftYf)* /6 ;,1
(t) Ec.c.e*«4rOc JvJfiA^l Hi* * 3 '"i cros5 Se^fiVw. A s (f)(3)'~ 12 in
c = (i)(s) * '•£ •■* e * i.5"- ro * o.s .v
CO C«vHr,*c A*J,Vi^: 4,V* 2 >\ c^ss st^*^ A=(f)(0* 8 I't1

ISOiSi
PROBLEM 4.125
P
4.125 The two forces shown are applied to a rigid plate supported by a steel pipe of
140-mm outer diameter and !20-mm inner diameter. Knowing that the allowable
compressive stress is 100 MPa, determine the range of allowable values of P.
SOLUTION
A^ 3-(«C-d4*)- $(moNwo*)» M.o8«Me?**N H.om*\d% »
C - % tfl« - 10 «,», = O. 070 ^
A+ A R - -JE. - Ha - U5>'l6*)+1:> fj3.fr fo3- Q.o?pyo.o7o)
- - I45.6J * I06 + 48/.OS P = -\oo*\0< /. P~ <?«.*« |o* N
= 7a.l«w|0* - 970.75" P » -/00»I0* -*- P = 177.3*/o5 V
<?*L8 kli <" P * 1.77.3 i-rKJ -•
4.126 The two forces shown are applied to a rigid plate supported by a steel pipe of
l40HnmouterdiameteraiKll20-mminnerdianieler. Determine the range of allowable
values of P for which all stresses in the pipe are compressive and less than 100 MPa..
\S0VS
PROBLEM 4.126
IP
SOLUTION
C = ^i9 " 70*~* - O.O70 ^
F - isox/o' + P^ M = (o.oioKtso*/<?*)- o.o^o p r i3.S"»Wo"*- o.o«rP
HT * °* " A J " " 4.081 *lo-» ~ ?_C7<?*/©-e
" - |4?.6lx/Oc + M8/.03P = -lOOxlO*" .*. P? 94.ff>/0* N
6A- -l**5\6/*/o6 + 12L03P « o p-303WoJrV
o^ stress A^rlj ad" A TV. 8 k)J £ P £ 3&3 kV
8 * " A + I r " .*t,oB^Kio-3 S.C79 x JO"*
- 7?.IS3"*/0< - 77©.75 P = -too y|0' pr fflm3*[C? V
6T»* 72J5Cx/0t»<?70.?rP 'O P* 71-3 "/o^N
Bwei OM 5T*"eSS A*.'+s 4,+ 8 7V.3 W* P* 177.3 feU
S^seJ o* U\l$l~'M 9*3 few* P^ 117.3 ItV -*
A4 8

PROBLEM 4.127
4.127 A milling operation was used to remove a portion ofa solid bar of square cross
ection. Knowing that a = 1.2 in., d = 0.8 in., and o„ «= 8 fcsi, determine the largest
magnitude P of the forces that can be safely applied at the cenleri of the endsof the bar.
SOLUTION
K " audi + cot* ' (
SI _ 8
b * I 53 adJ
., 63^-a - 0-8) - , r0u ; -
P ^
K 2.6of
r 3.07 ICi'ps
PROBLEM 4.128
4.128 A milling operation was used to remove a portion ofa solid bar of square cross
section. Forces of magnitude P = 4 kips are applied at the centers of the ends of the
bar. Knowing that a = 1.2 in. and od = 8 ksi, determine the smallest allowable depth
d of the milled portion of the rod.
SOLUTION
*'#-»
«-- £* Jdt . £ + Pec « ^ + PikiM s ^ + 1%J)
or C Ax* -ff J - 3P = O

PROBLEM 4.129
PI
2 in.
WW//'//,.
%
's
4.129 Three steel plates, each of 1 * 6-in. cross section, are weided together to form
a short H-shaped column. Later, for architectural reasons, a 1 -in. strip is removed from
each side of one of the flanges. Knowing that the ioad remains centric with respect to
the original cross section, and that the allowable stress is 15 ksi, determine the largest
force P, (a) which could be applied to the original column, (6) which can be applied to
the modified column.
SOLUTION
A = (3)0X0 = '* ■'«*■
6*a-£ ■■ V-tfA^ktsKn) * wok;?* —t
y - SA.v«
CD
©
®
2
A/.«*
G
£
4
16
y0j;»i.
3.5
O
-3,-5
Ay*, ^
3L\.o
0
- IH. 0
l.o
7.Q
16
- 0AZ7S i.
Tke c&tafiroti A'-es 0.^375 <"> tVoik +ke iWeip<n>T of f/ic tv^i>.
M " Pe- wU^ e * 6.WW in
6.0765^/

PROBLEM 4.130
4.130 A steel rod is welded to a steel plate to form the machine element shown.
Knowing that the allowable stress is 135 MPa, datermine (a) the largest fibrce P that
can be applied to the element, (b) the corresponding location of the neutral axis.
Given: Centroid of the cross section is at C and /= 4195 mm4.
6-mm diameter
J M8mi"~l- J SOLUTION
3 mm ^
13.12 mm
Section a-a
e
•13. IZ—»r»|0.SB
J
y
I - 4115 *•** - 4115 x|o"W r**
J
v - -L , ex _
P * J§L - I35*IQ4
-*
(qOi3fZYo.ol3i2)
HH5 x-io-'1*
«• ^3. Iff8 */© m"Z
6"= o
er
^
3 £ - H^ = £ fe^ s
A
Ae
A X
-o
=■ 3.%cl*\o ** * 3.87 w*» -*
(82-27 v/o-')(O.OlSia)
Tta yie^4/*i ©.yfi J?iea 3.g<? k« +3 +ke rfjU of +lte ce^i^r*/
0*1 J7.0) *m */° "Me ^fglit" <»f -/-/te -?iw« ©J acrh'o^ of" fie Soeds

PROBLEM 4.131
4.131 Knowing that the allowable stress is 150 MPa in section a-a of the hanger
shown, determine (a) the iargesi vertical force P that can be applied at point A, (b) the
corresponding location of the neutral axis of section a-a.
SOLUTION
CD
1200
Woo
y>.
So
70
AyOJ *►*
3£,xjo*
8*/ * IC?
MO* to1
y , 1A&.
Y.
^400
20 mm
Section a-a
— SO *V( rr\
The ce^+^orJ A«* SO *^ -h> +Ve r^Lf ft-T -rU
Je-R- eei^e erf H« se^-Wow.
**t«i
KP
*=£--§*
A J Z*too*icr* l.36o xiO'<>
= 3.7*£5"//o3 nT7-
P =
_ 6" _ l^ox/o*
K 3.72£SV/oS
(b) /.ot*-Ti'<»* of net-nra* a/ fs *
qo.3 MO1 W * 40.3 M
<S= o
Z ' A
V =■ -i- =• — !T _ - G.30 */o w, =- 6.3o *,►*,
^ AC- (^Woo »|o"*)(O.090)
or «5"6.3o *m* f*-o^v +Le Jef+ e-^je

PROBLEM 4.132
4.131 Knowing that the allowable stress is 150 MPa in section a-a of the hanger
shown, determine (a) the largest vertical force P that can be applied at poioi A, (b) the
corresponding location of the neutral axis of section a~a.
4.132 SolveProb. 4.131, assuming thai the vertical foreeP is applied at point B,
40 mm
-60 mm-
40 mm
20 mm
Section a-a
SOLUTION
Yo ZA
t: SO »>wi
TJie cc^ + rot'tfl £it% SO mm +o +Ke ^V.* ©^
e ** J"0 - l?0 =■ -70 ^ r ~0.070 m
<2>
2
Aj *n»*i
/Zoo
24oo
Jfp j htm
3o
70
A.Voj^*
36x/o3
g<* V |c>S
l20)f/03
Ia" TkC^K^ + O^^X^o)1- r £20X10**0
I = I, + It - \.ZiO x/0* ta.
-%^^1 « -*.,«„,*
-I,
* - l.«t«flg»toi r 16.5 */o3 hf
*• *-?* = °
o - J
|-3toy|°'< - -8.lo*lo-V « -8.lo«*
Mei/+r»/f «wia i;es S"0 - «. lo » fl.1 «... £*>», irf+ «*J§e..

PROBLEM 4.133
4.133 Knowing that the clamp shown has been tighlened until P = 75 lb, delermine
in section a-a (a) the stress at poinl A, (b) Ihe stress at point , (c) the location of the
neutral axis.
SOLUTION
R^T
A, iS
O.o&H
0.1o24
0.lt64
^'
lv^
o.3a
AyPJ i*:
0.ou<ft*8
0,03277
0.07S85
v ^ z*y»
r.
■2A
Section
-4 U
16 in.
B end t net c<wp/fe M - Pe.
I, - ^(0.**)(0./6)5+ (0.064)0.7*- 0.^739 )* t ^.013 xjcf3 i*v
I - J, + Ix • 1.W*I0"
fcO Sfrcss *f per-,+ A - J - 0.»-O.M7aT - 0.3*61 ,'v,
Oft A . ' T*
- 76 _ (7S-X-£3gg/)(o.3*Ci)
A X " A 2
- '£.(? */os pit
v* A i A x o.iuy 9.<?6</*/o-s
= -7.S7 *lo* f%t - -7.S7 toia
*•*-#•-£-¥ = °
^r"ftr=
^SHx/o"*
(o.;664X-^.3acO
e^ - J.
j: " A
r -O.OW7 i*n
T^e *eu4v«-P tfuy-,'5 Jtas 0.4*78? - 0.0*57 - 0.44S i«. *<>as/c

PROBLEM 4.134
4.134 A vertical force P of magnitude 20 kips is applied at a point C located on the
line of symmetry of the cross section of a short column. Knowing that y = 5 m.,
determine (a) the stress at point A, (b) the stress at point B, (c) the location of the
neutral axis.
SOLUTION
fort
-3 in.-
©
2 in.
M
76
To"
~ 3. S i'ki
-3 in-
A
lin.
00
2 in.
4 in.
2 in.
Ucce^-fr.^.^y of A a J e - -6""- 3.S - I.;? ;.
I, r A&^ + OsKi.a)1 - zx.ts V
I b I, + Ia r S7.S&1 ;ny
(A) Sfr«is «^ A CA = 3.8 in
b' A 1 *0 + 57.SS7 °^7C *5'
to S+v^s3 «,-f B
CR - £ - 3,S - 2. 2 i
&
^ --2- - -P^^ r -22 _ ffo W.3X3.3) c .,.,,2 /,St-
Jto
<57.86-7
= 0
57.867
6" = -£ .Zeo,
A J
6-= 0
go. _ j_
cx^ -i
= 2.111
Ae C^xi.n
hieo + r^i axis. Acs ?.MI/ ■'«. £«jrW ce^WJ o** 3.g -£.</**
.389
m
A«
j l*»*<A
/. 38"?;^ .f/w*.. po »vf A

PROBLEM 4.135
IP
4.135 A vertical force P is applied at a poinl C located on the line of symmelry of Ihe
cross section of a short column. Determine the range of values ofy for which lensiie
stresses do not occur in the coiumn.
SOLUTION
Locale cfin+rof d
-3in.-
-3 in.-
<D
Z
A>1
S
Z<>
$> ^
2
Ay, «^
75
2 in.
4 in.
2 in. W 2 in.
lln,
- 1 - " »
Ecce*trtc'i4y tff ^?o*.#/ fi = ^y - 3.8 in..
e ^ JL , ^7-**7 =r 0.76/ ;*
If sfress ^4 13 e^udi.
y - <V7£J + 3.8 = y.£6\ in.
2 A
e =
j_ s?.<&?
AC,
ftoiffl.!")
L2>lS |w
y « -1.315 + 3.8 = 2.^25" ;n.
A*s^e^: S.HW iV> < j < t.Sfc/;«.

PROBLEM 4.136
4.136 The four bars shown have the same cross-sectional area. For the given loadings,
show that (a) the maximum compressive stresses are in the ratio 4:5:7:9, (b) the
maximum tensile stresses ore in the ratio 23:5:3. (Note: the cross section of the
triangular bar is an equilateral triangle.)
SOLUTION
Sir-ess es
M A
A+ 8
a
°A A I
■A
A A,
e-fc
I*-
A,= a\ I,= lira"', c, - cer |a;
fA, = ttc' = a
e *?a.
- x
3*
e- c

PROBLEM 4.137
P'
40 mm
4.137 The C-shaped sleel bar is used as a dynamometer to determine the magnitude
P of the forces shown. Knowing that the cross section of the bar is a square of side 40
mm and that strain on the inner edge was measured and found to be 450 )i, determine
the magnitude P of the forces. Use £ = 200GPa.
SOLUTION
I = ifto)ftof* 2l3.33*/o3 mm'* = 2iS.33*/0
C " Xo ^ - 0.020 M
KP
^,2+Mc r £ + Pec .
D A I A I
V - -L + IS. -
K " A + ~ '
Icooxjo"
+
t0.l6&)(o.Q2o)
21*3-33* lo~*
= 10.Qo y
lorn
-\
P -
-E. .
10*10'' r 9.06 X/O* A/ = %00 UU
\G.OV x lo*

PROBLEM 4.138
f
-10 in.-
U
%
A * 10.0 in2
I, =273 in4
4.130 A short iength of a rolled-steel column supports a rigid plate on which two
loads P and Q are applied as shown. The strains at' two points A and B on the center
iines of the outer faces of the flanges have been measured and found to be
^=-400* 10-4ui./ui. eB= -300x lO^inVin.
Knowing that £ = 29 x 10* psi, determine the magnitude of each load.
SOLUTION
S+vesses *r A *.»j 8 -Fro^ stv^r'n <?4.ges
BewJ'nj coupJe M - £P - 6 &
C - <S In,
ft - -.E.^ - P+_Q (€P" 6<OCg)
- £.7 x lO5 =■ - a*?0?89 P + 0.06121 Q
SoUi^e* (O and GO siw,^yf tan eoOs J\^
P - W-Z x/O3 A = 4^2 k.'ps
Q. - 57.3 V /03 A r 57.3^5
(I)
Of)

PROBLEM 4.139
-10 in.-
t
ItaBET
u
A = 10.0 in2
It = 273 in4
4.138 A short length of a rolled-steel column supports a rigid plate on which two
loads P and Q are applied as shown. The strains at two points A and B on the center
lines of the outer faces of the flanges have been measured and found to be
eA = - 400 x 10* in./in. eB = -300 x 10* in./in.
Knowing that E = 29 * 10* psi, determine the magnitude of each load.
4.139 Solve Prob. 4.138, assuming that the measured strains are
eA = -350 x 10* in./in. efl = -50 x 10* in./in.
SOLUTION
Stresses aA A a*«A B -fVo^ s+^a<V *<taes
5*A "E£A - (29vio6)(-35'o^/o^)--/o./S"«/03 p'.
R
Be*«K*^ couple M- £9-&Q
Sol
VI V\-
Hoj^lO3 - 0.00*789 P -O.J?o?*<7Q.
WB " A a ' io.o ^73
0)
(?)
$ i rv» j>f-f A»n e&\is -fy

PROBLEM 4.140
25 mm
30 nun
90m
4.140 An eccentric axial force P is applied as shown to a steel bar of 25 x 90-mm
cross section. The strains at A and B have been measured and found to be
eA=+3SQii 6i, = -70u
Knowing that £=200 GPa, determine (a) ihe distance d, (b) the magnitude of the force
P.
I5inm
SOLUTION
= 1.51875 k|CT* v»*
S"A - £eA- U^io^y^*^4) r 7o*/o* P^
'Ci.
A A x
A 2
(A)
<?^5 x lO3 M
M - - TV -•. of - - £
o,o^s-
"^5T r 6-0So- - So"-
tk")
T = lis ku.>*>

PROBLEM 4.141
25 mm
30 mm
90 m
15 mm
4.140 An eccentric axial force P is applied as shown to a steel bar of 25 * 90-mm
cross section. The strains at A and B have been measured and found to be
eA = +350 n et = -70 n
Knowing that E = 200 GPa, determine (a) the distance d, (b) the magnitude of the force
P.
4.141 Solve Prob. 4.140, assuming that the measured strains are
eA = +600 u eB - +420 u
SOLUTION
Vt ? |5+ MS + So = 90 vn-*,
Stresses 4vo«^ sf/vtt'ii 343 es oh A <&.«*! "8
°* a r
*,, £- **
(I)
CO
(*\
M ^-PJ
■ A = -— - _ "l?/^
P 2*3 *to*
S * to *-i * *5***i*i
(^
P * 2*3 kN

PROBLEM 4.142
P
4.142 The shape shown was formed by bending a thin sleel plate. Assuming that the
thickness / is small compared to the length a of a si de of the shape, determine the stress
(a) al A, (b) al B, (c) at C,
SOLUTION
C
t
d
-I
U_L
2VU
R
A I 2<tt £ top " Sat
ip.
6 A J " *«t ita» ' at
6;
a.
Area- A^C^fcXf) = *ftt -> C * $S

PROBLEM 4.143
h = 40 mm
4.143 The eccentric axial force P acts at point D, which must be located 25 mm
below the top surface of the steel bar shown. For P = 60 kN, determine (a) the depth
d of the bar for which the tensile stress at point A is maximum, (A) the corresponding
stress at points.
— 25inm
20 mm
SOLUTION
A = bd
c • *«f
*>-$ +
I '"khdls
e = i-d-a
Pec
I
A fe 1 <* d* J " »» 1 «* «i'i
(fit) 'Dep+h «l $»* «v,^y,"»,w^ S"A . D,"ff e*e~t! «d« wlfU r«pe<^t +o ol.
- O
d = 3a = 7f
GOx/O
<#>*/£'
PROBLEM 4.144
= 25 mm
4.143 The eccentric axial force P acts at point D, which must be located 25 mm
below the lop surface of the steei bar shown. For P = 60 kN, determine (a) the depth
d of the bar for which the lensi le stress at point A is maximum, (b) the corresponding
stress at points.
4.144 For the bar and loading of Prob. 4.143, determine (a) the depth d of the bar
for which the compressive stress at point B is maximum, (A) the corresponding stress
at point B.
I = fcbds
20 mm
SOLUTION
A - bd
r- 9 Pec
b ( <* d* j b c a d«-J
^Sk _ _E_S" JL _ !2a] - o J = Go. = ISO ~~
Se =
(k^
6 to * 10
/5bx/o-* (/SOx/Cr5)1

PROBLEM 4.145
M = 400 lb ■ in,
4.145 through 4.147 The couple M is applied lo a beam of the cross section shown
in a plane forming an angle p with the vertical. Determine the stress at (a) points, (b)
point B, (c) point O.
SOLUTION
-3 * t
tfi
-$
0.4 m.
00
Cc^
10 2^ J^~ «57.6*<eT» C.^Ojr/o-»
=- -?.&6)r*0*p»,- * -«?.&6/C5,'

PROBLEM 4.146
M =
A
16 mm
16 mm
3(X) N ■ m
'■.-CimBm1
(Pp
= 60
-.•'..i
4.145 throngh 4.147 The couple M is applied to a beam of the cross section shown
in a piane forming an angle pwith the vertical. Determine the stress at (a) point/*, (b)
point B, (c) point C>.
(-«-40 mm-*|"*-40 inm-»-|
D
SOLUTION
J^r £ (saX»)3 = /. 36.533 >«>'
3 9
Zl&.tS'fO"* mH
I. 3C533KA? * *o
> = > " ' tf
" 16 **!•
ZA = -2B- -Z
Mj.
3oocm3<59: 257.SJ A)*™
0 s VO *i**>
- - 3,37 v/O^P* =■ -3.37 HPa.
/^ fir -- ^fr x My2P Ps^CHsx/o-3) (35*7.8! X-'/oxfo-3)
' 3.37*/0fcP«_ = 3.37 Mpec

PROBLEM 4.147
0 = 30°
4.145 thnwgh 4.147 The couple M is applied to a beam of the cross section shown
m apiane forming an angle /? with the vertical. Determine the stress at (a) point A, (b)
point 5, (c) poinl 0.
0.5 in.
X
SOLUTION
M = 250 kip ■ in
Sin.
L 0.5 in,
——*-
- 90--aS3 in"
Welo: Iz = 7*(<>.sV<?)3 - IS.225 .V,"
W*/f: I* - (?)(?<?. 353") * IS. 2*5" * 1^8.8^ i«¥
ly - (2Xai-33S) + O.oZoZS'- 42.&S7 m*
K
+ frZy .
VIS.*** tt2.C«7
PIS. 81 + <«.<«■> ~ ■'7-,6*»'

PROBLEM 4.148
0 = 20°
K* = 10 kip -in. |p
4-148 through 4.150 The couple M is applied to a beam of the cross section shown
m a plane forming an angle /? with the vertical. Determine the stress at (a) point A, (b)
point B, (c) point Q.
SOLUTION
Locale ee*vTirt>i*l
2 in. 4 in.
®
©
0
Z
A, ,\x
\Q>
2<f
Z. i/i
-1
2
Az,;**
16
O
TUe cc^)+irotJ( J. as. «**" p*'"^T C
3. _
C/d
64 ,*«
- - 4 i*
I2 I,
= . C^syVO v (ZAioi)W r 0-^, jrsi
88
6t
tcl $ =
Iz
^
8S 6H

PROBLEM 4.149
H = 9 kN - in.
60 mm 40 mm
1 t=
U—100 mm —*\
120 mm *
4.140 through 4.150 The couple M is applied to abeam of the cross section shown
m a plane forming an angle /? with the vertical. Oetermine the stress at (a) point A, (b)
point B, (c) point D.
SOLUTION
M.
(<7*(t?3 ) co /£"
g.6<?3-3 * 10* N*^
1.62667* lo'
i"5.3 XlO* ?+.
£513 MPa.
4
I. C 2.667 k/o"« ^ 3667 if to*"*
« -5T. 3 */o6 Pa. = -£513 MP*,

PROBLEM 4.150
y
M =900 lb. in.
P =30°
r = 0.8 in,
4.140 through 4.150 The couple M is applied to a beam of the cross section shown
in a plane forming an angle /?with the vertical. Determine the stress at (a) point A, (b)
point Bt (c) point D.
SOLUTION
■/ $
i*
J6 =• 0.8 - O.S3*S"3 r 0.46047 ,V

PROBLEM 4.151
W310 X 38.7 15°
M= l6kN -in
4.151 through 4.153 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal plane, (6) the maximum tensile stress in the beam.
SOLUTION
for VJ ZlO * S8.~7 ro)hJi s+W e<wy«.
I, = ss: i x i ofc *„*v = as*, i x/o"6 tw+
^r 7-r?*'°b*^":r 7-27*/o~&^*
165 mm
M2 r (j£x/0*) cos \S* - l5.4-.£r*/G& M*r*
(o.) 4** 0
- L
^ e t y./K^^
7,£7x|o*6
/5'
3.l36r
3> * "B?.3*
(V) Mctxt^t;*^ Tev\St'y« STrc55 octoi/s «.f po»"*"f f
F " la Iy ^S".U/o-c + 7.17*to'*
- ~7S.lt IDC ?a. - 75*. I MP*.

PROBLEM 4.152
S150 X 18.8
M » 1,5 IcN ■
4.151 uurongh 4.153 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal plane, (b) the maximum tensile stress in the beam.
SOLUTION
For S ISO *|*.£ roJJeJ e+e«i sU^-e
T
152 mm
1, ■= <?.tt*io*
VtntMl
4 _
*7.1/ */o~" w,
w°
M2 = ./*^""/o3^ *•'* ?o
M-
( l.S* IO%) c»s 2o'
2e - -ZA ~ - 2a =■ Zx, X(£)(S50 * H2.X ^
* - 0,i"/3o3*/£>S N<^
7 X, 0.7«2xlOT*
0.732 *|C>"
<* = 88.*' - 76° -
18.2/
(.b J Ma.xi" •""""* 4e«\s»".P« iTres-S occ^irs <a-T poT-t 1)
I,
<?. I / w I O-6
- 2o.<t*ioc fa.

PROBLEM 4.153
A 2p°
C8 X 11.5
2.26 in.
0.571 in,
M
V
25 Si'r. 7.C
Mz - 2S c»s 2o
4.151 through 4.153 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal plane, (b) the maximum tensile stress in the beam.
SOLUTION
For CS^iUS aJJW sheJ sl^«
Iz = 1.5.t .'«* ^ I,-- 32.6 ;M*
^e = Zp - 0-4 .V j
Ye = fa ' %-%& - O.S7\ t i.ea^ m.
^0^ zA -r -MfM
(a) +«n g? ' ^L +flrt © r iiil ■/»„ ?0° r O.OM737
W*
32. C

PROBLEM 4.154
M = 400 N ■ ni
18.57 mm
I9. = 281 X 103 mm4
7.,= 176,9 X lO^mm*
4.154 throngh 4.156 The couple M acts in a vertical plane and is applied to a beam
oriented as showa Determine (a) ihe angle that the neutral axis forms with the
horizontal plane, (6) the maximum tensile stress in the beam.
SOLUTION
I7«- I7£.<? X/05 *W = J7&-9 xio''1 ^"
.Ye' *-18-57 «n^ ^ Zc = *S" *~i
M2' = 400 cos SO° - 3*/<i, 41 N-^
C( = 3o°- l?.97° = 10.03°
*.yi**viui^ ■fe«s/rf< ^4-^es-S occurs a-t pomt" E
= 36.36*10' + f7.7<?v/ofc = 54.* ^lo' ft
* S4.X MPc

PROBLEM 4.155
4.154 through 4.156 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal plane, (b) the maximum tensile stress in the beam.
o Iy" 6-7<> •""
.4
\
4 in.
/ \ 2m/£j\ 4in.
7j,. = 6,74 in4
(..= 21.4 in4
4 in.
M^' = - 25 si« 45
M2' = ?£ cos 45° -=
SOLUTION
° - - I7.C7S k.p.i"«
to.} 4m f r i: +a„ 9 * -2if4 4a* (-%>'") = 2.1751
^
6.14
9 = - 73.5**
* ;n..r
(L>^ ^ay i»mom }e»tsi'A .stress occy/5 &-4 pofm
6;
. MiId, M,4 ( l7.C7gK-0.g5l , (-l7.67gK-5.HO
I*
V
O.X065 + S.23S
g.^y Irs/

PROBLEM 4.156
4.154 through 4.156 The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms with the
horizontal plane, (b) the maximum tensile stress in the beam.
M =
SOLUTION
23.33 mm
1^ = ^(aoYso)3 4 (^l3oX3oX*o.-*s.s3^
- I- II * IOfc ^m4 •= L// v/c?'6 ^
rj ' *{ isT ^3o^6°^ + iOXso/3] = S.ho*io*
****? - (T.VoWo"6 vw,'
ye ~ - 23.33 m*>,
Ze = SO m
«9
(«0 +«*# = il +*•, © =■ JJJjilSll -f*„ ^ = o. 09SZZ2
- 76-1^ */ofc + IS;? 83 x/o* r q$.0»lo* Pa
- ?-5.O MPft.

PROBLEM 4.157
"4.157 and 4.158 The couple M acts in a vertical plane and is applied to a beam of the
cross section shown. Determine the stress at point A.
SOLUTION
6 in.
Iy = 8,7 in4
7I=24.5in4
Ip = +8.3 in4
Y (8.7j 8.3) »**
£F* 7.*? ^
F? t 8.3 ,V
a> -*-^j ^*
-/• -
Hu = M si« 0^ - (GO) ■■■* *S.J?- =" 23.SV *.>-/«
K, - M co.s Q» * (Co) ccft^S.a* - 5S.\S k'p.U
UA 5 A&0*6- f ** ^ 6" ' -S.<M&*2S.2*.«- I. OS *,\*a*.2* r -4.03 In.
v*A - 2kCo*d^ - ^A «■> ©», r - ),09c»42*.20 + 3.?2 SM23.r - Q. £SZ in.
6- - - M'u* + Muvii _
23.06 S.W
- 10.16 k's,'

10 mm
—10 mm
40min
10 mm
Iy = 1.894 Xl^min4
l% = 0.614 X 106 mm4
1^= +0.800 X106 mm4
*4.157 and 4.158 The couple M acts in a vertical plane and is applied to a beam of the
cross section shown. Determine the stress at points.
SOLUTION
Osihg MoWxs C('fc)*; de+er^'W -Hie pAVict'jJa-f
E 0.2^ o )*/ob Mm4
i-0^45" * lob ^M'
- &
Tan /CcV " —— * ■ : e - l-<<5
M^ - M tos 9„ - (l.a.xl03)c*s ^67° - /.Ogl^x/o* A/-v*>
^A r J* ^s 9* " 2A *'" 9* T ^S c^ 25167°- 45" sf* 25*. 67° - */. 07 **
V^ ~ ^ cos &w, + ^A si'* ©m * ^S" cos 2S".67° + ^5" s.> 2S".67° * So. OS" *,»,
- II3.0XI06 Pa.
1I3.0 Mfct

PROBLEM 4.159
0
v
M = IS kip
*4.159 A 4 x 10-in. timber has been trimmed to form a beam of the cross section
shown. Knowing that the couple M acts in a vertical plane, determine the stress at point
A .
1 jn SOLUTION
lin.
1 in. 3 in. 1 in. 1 in. 3 hi. 1 in,
Iy = 291 in4
lin, U^'ftG, Hon^ S CiVje -JcTe/^V^ ■H €
or i vi <?/-"!".' o,.
lin. p^i'\AC-'po^ a/es a.v\d Cr'^c- pay ^ &»*■ * ^3 i
Iz = 39.3 in4
E (165. i^o) i**
fan 2d,
. £1 - n.s
EF " |35.8r
*■ 0.I7S73
0*r S.<?7"
27.85" I* v
R. r yiF^-^ F22- - V I25.&s* + ;?;e.5x -
Iv r |G£JS - 127.85 = 37.3o ,'„*
Iu = 165^5+127.85 r 2^3.0 ,V
0A * JA cos 9^ + 7,5,'^ - ZcasS.O?" +SS^S'.0?0 - Z.^ISH in
V* t Z* o>s 9^ - yA s.v ©* * 5"c*s 5.07° - £ s.'n £o? a - <f. go4/ ,n
Mv/ - (£ co5 5*.07° * \H.<W k.>-»*
Hj - 15 si* 5;o7* - I.S2C k.y>. m
^ - - M^U' + -M^ - - 0H-?^fa-M34^ + (1.330(4.3041 _ _0 ^3 Ws-
A Iv I., 37.3 0 2<?3. o

PROBLEM 4.160
P = 4kN
R= 125 mm
4.160 A rigid plate of 125-mm diameter is attached lo a solid 150 * 200-mm
rectangular posi, wiih the center of the plate directly above the cenler of the post. If a
4-kN force P is applied at E with 0= 30°, determine (a) Ihe stress at poinl A, (b) the
stress at point B, (c) the point where the neutral axis intersects line ABD.
SOLUTION
p ^ 4*/D5 A/ (compression}
M2 = - PR c»s3o*
V
A &
-3 *
A = (i?oo)f/S'o') ■=■ 3ov/o w^ * 30x/0" y*
(a") SA
ao /jo
,u . _]> _ MxZo 4. fo** , H*|03 .. C-^5oX7Jk/d*) (-433.X
- -233 */o3 P* = "^ *?*•
looxto
ra)
lo
-6
6V = o
ZA * 75" tw
- •?
XA*
£ + Hj&
MZ(A
-V33 7 3ox*o"3 ^SJfxio-*

PROBLEM 4.161
IP = 4 kN
R = 125 mm
8
4.160 A rigid piate of 125-mm diameter is attached to a solid 150 x 200-mm
rectangular post, with ihe center of the plate directly above the center of the post. If a
4-kN force P is applied at £ with 0= 30°, determine (a) the stress at poinl A, (b) the
stress at point B, (c) the point where the neutral axis intersects line ABD.
4.161 InProb. 4.160, determine (a) Ihe vaiue of 6 for which the stress at breaches
its iargest value, (b) the corresponding vaiues of ihe stress at A, B, C, and D.
P- 4*ioa M
PR- (^/oPJO^/o*}* Soo W-m
My =■ -fR COS 3 -r-SbQ cos 0
X»k ~ /0(9 *"«n
^j = — 7$ »W*M
Poir cr "h» »<
x* * iz
a. »via./i wun^
>^U Z^ 2D . >C^ X,
pS" ft - ^2b&oSc> RX&_6l* 0 7
" ' ^ u 'S
A"" A Ik !«. 3o*te* ^.ZT^to-* " loo y lo"*
-(-0./3333+ 0.53333+ O.SOo) *lo C ?a ~ D.loox|oc P^ - 7&0 k?o,***
6*e-(-0. 13333 + O.S1333 - 0.3oo)>fot P^ - 0 JooWo* Pa. - |60>P^-*
6"c ^ (-0-/33*3 *C 4-bVlo'Pa. r - 133.3 )f?a. -*
61" (-0.13M3 - 6.<<ra333 - 0.3a»)*/°'p4. ' -0.96?x/o'fc*. *-9C7k^a-*»

PROBLEM 4.162
4,162 Thetubeshownhasauniform wail thickness of 0.5 in. For the given loading,
determine (a) the stress at points A and B, (b) the point where the neutral axis
intersects WntABD.
SOLUTION
AJU
how* .
5 in.
6 kips
iz = AfeXs?-4(*W^ w.s&s.v
I, - Tkf»0?- iSrW^l* - 8.SS33 ,V
A - (z)(s)-&)(<*) ^ 7.0 .v,1
(
CC0 Gl=£-4^+-^^ r J£ ^ fitting-^ +<n.sYLS)=^Mto ^
7
«o.,S"«3
^5^33
e6. £ -i%* + 14* r f -«gK3gl +4^ -w^a;
i*
2o.5a%
9.S233,
(b) i-e-4 p©i*f l4 be -hi* po^f \*rlies-<z T"/»e neoW** aXi'4 iVi+e/jffafa <Af3.
Zu - J.5"
Si> = o
0, £ .^.+ «tSL
jfr-acf-yv
;?o.583 J Z5* f^uKi-f)
&.5- ( 7 8-5M3
"l.^-l in.
Answer^ Su^tf-il )ia, <djo*« poi^f A,

PROBLEM 4.163
4.162 The lube shown has a uniform wall thickness of 0.5 in. For the given loading,
determine (a) the stress at points A and B, (b) the point where the neutral axis
intersects line ABD.
4.163 Solve Prob. 4.162, assuming that the 6-kip force at point E is removed.
SOLUTION
5 in.
AJJ
In,
6 kips
A r OKO - COC-0 » 7 .V
(?*4i-»Ji"Ai^T -port* «i»i«»l Cow»/ts
H7; -W.f)(3^ + (7.S)(Z) * 7.S^'p.iM.
Mj- -d.5)(aU(i.s)(e) * ^k;p.k
Ca) & = -J -
A —
, -£_ N^ + Jii t
£
A
It.
7
.fL
7
X0.5-S3 £.S"S3a
= ?.?SJtti
=• I.IU kai"
(V>) U"t poi*4 K b* He point wk*/*. He otu+iTAy «*»cfs i*"l«^3ee-K BC.
" -«■ -Vj
* ?
* ^ V 2Z A ' ^6" I 20. ^«a 7 j U'
1.5" *■ 0.7 IS" * S.fclf m

PROBLEM 4.164
4.164 An axial load P of magnitude 50 kN is applied as shown to a short section of a
W 150 * 24 rolied-sleel member. Determine the largest distance a for which the
maximum compressive stress does not exceed 90 MPa.
SOLUTION
AJJ
,-J
y- o."ci z - axes*.
fV W ISO * 2H roJJevr * W secf-'d.
A- 3o^o ^m1 ^ 306o*Jcr4 ^*"
It- IS.H *JO* ^ - 1^.4xlO*th,y
Jy - f.83*/oG *»<** ~ i.83*Jo~~»~4
h*-i* -*°" ,
b
Zfi ~ ^ ~ £"! mi
*fe
P * 50 y |03 N
M^ = -Pa
6i s .£ - £kSL + 4
X,
i
(JA = -?o "JO* ?*
-J
*1. • *f'SHf • *j
. 1.88*10^ (-3.7S«to'K-ft>*to"') + Sox/p3 , 9o ^
1. S3 * p
£{♦
«51. *io
Zoco y to'
2X.1>%% + 16. 3^0 - 9©C * 10
-j
£o v(o:
3C gx/<> na - 3G. 8

PROBLEM 4.165
J
VA////A
t
*
1
S*u
** w
pffl
T
I
°* A I2
4.165 An axial load P of magnitude 30 kN is applied as shown to a short section of a
C 150 x 12.2 roiied-steel channel. Detennine the largest distance a for which the
maximum compressive stress is 60 MPa.
SOLUTION
id 2- <=wes «-s sAo^"
bf r ^8 ****
ly * O.A74*10* ww4 « 0.27* *IO"4 v**
Liwe tff &*f /0M tft -We*. P _
^r s - a ?f - x - i £* = io. 15" wi*m
P - 30*/0S A/
Mz » -Pa*
6; c -tovlo* ?a*
ZA s X = l?.7
mM
**
M.
r if I3a5t T ■£■ -ft
Ya i X, A
1
.-*
+ 5ox
*\
_ 5.3>ryfo"c r m oM _ i^.ggi + (Soiv/o* r - L866X/63 W-»w
7c/io-3 C J

PROBLEM 4.166
4.166 A horizontal load P is applied lo Ihe beam shown. Knowing that a = 20 mm and
that the tensile, stress in the beam is not to exceed 75 MPa, determine the iargest
permissible load P.
SOLUTION
©
®
Dimensions in mm &0
©
©
I
A, MM
7.000
\zoo
SZOo
$2 W***i
10
'10
_kS,_™^_
7.0 * ID3
-\2* 10*
■'.** 10s
Y -
2A:
- 0.406*7* lo"6 v^
Iy " w(aoX"">f + lfc(»0X*>f - 2.0X&7 *|0* v*v*4 r *.
0367 xlcf'^*
-^

PROBLEM 4.167
4.167 A horizontal load P of magnitude 100 kN is applied to the beam shown.
Determine the largest distance afor which the maximum lensiie stress in the beam does
not exceed 75 MPa.
taP*.
20
20
^
v
^
Z^
^ y
%5
^IRSIIBb
20
Dii
tie
asio
p
80
isinr
nm
20*
SOLUTION
©
W
©
0
®
7.
P\^t*\*fi
2000
120 0
3^oo
%v*
10
- 10
A£ »»
#0*IO3
- l2*io3
s*io3
y--
5A
OL
ZA
_ 3x|Q3
8enJ.'rt3 tallies M* - N/pP My- -Ap
R^ poi^t A 96 - 5o
dim y - — x, 5 mM
= ^.t>K7xlcTci 3, ^ + ,^37 . 7S"7*lo* - - I.HWx I03 N-*i
>-0 ^ J
- 75^/0 f
— *.
loo v to3
17.1/ */e>* h
/7.//
f*iKr

4.168 A beam having the cross section shown is subjected to a couple Mq which acts
in a vertical plane. Determine the largest permissible value of the moment A/„ of the
couple if the maximum stress in the beam is not lo exceed \2 ksi. Given: Jy = Jz=l 1.3
in', .4 = 4.75 in1, *„„-0.983 in. (Hint: By reason of symmetry, the principal axes form
anangleof45°wirathecoordinateaxes. Use the relations/miB= ^4^ . andA. +/
mm I11U) ™"1 "*m»
-/, + ;,
5 in. SOLUTION
M0 - \A0 Bin W "= 0.707/1 H.
Mv * Hp C0*HS* -= 0.lOl\ Ho
-v^* * ly + X* - XHVl * "-3 4 11.2 - 4.5*1 ' IS.OI m¥
U6 * ^ ccft^0 +■ Z6 *M ^S" = -3.57 co*MS°-h 0.<?3 s<« 4S* = - t.8££ In
6i -
- MvOf + My* =.0.7*7//
M,
0*707/1 Mp||-("l"|^) + "%^rl r o.^1^4.Ho
2?. I kv>-m
O.f/34
14
0.11W

u
PROBLEM 4.169
y Mo
0.5 m.
0.5 in. ■
1,43 in.-
/
/
■5in.-
5 in.
4.168 A beam having the cross section shown is subjected to a couple Mq which acts
in a vertical plane. Determine the largest permissible value of the moment A/Oof the
couple if the maximum stress in the beam is net lo exceed 12 ksi. Given: Jy = lz ~ 11,3
in4, .4 =4.75 in2, £^ = 0.983 in. (Hint: By reason ofsymmetry, the principal axes fonn
anangleof45° wilh the coordinate axes. Use the relations /min= A k^ and^+I^^
= /« + /,.
4.169 Solve Prob. 4.168, assuming that the couple Reacts in a horizontal piane.
SOLUTION
M0 - M0 cos 4S° - 0.7O71I MD
Mv = -M> sin ¥5° * -0.70711 Mo
1^ - AkJl =C4.7sXo.983)ft - "«f.5^ in*
■ WAr
= Iv * I, - I™ = M.3 + 11.3 - H.S1 = It.ol ;.
UD - yD *«* ^ + ^d *f* *+S° = 0.^3 to* «*r* 4 C- 3.57 si^ IS" )-- -r 1.866 k
^»
MvUt
M0Vp __
-J'vup 4- iT'^J v0 3 0.^70711
M.-
0.707.1 M.[.i^£l4-^] - 0.1IWM
_ 12
>y9 • 1*1

PROBLEM 4.170
^
10 mm-
■ 70 mm-
40 mm
10 mm
4.170 The Z section shown is subjecled lo a couple Mq acting in a vertical plane.
Determine the largest permissible value of the moment A/0 of Ihe couple if the maximum
stress is not to exceed 80 MPa. Given: /„ = 2.28 x 106 mm4, 7mi0 = 0.23 * 106 mm4,
principal axes 25.7° v and 64.3° '.
SOLUTION
40 mm
u
10 mm
lo- Jw*." 0.23*lob»^4- 0.23*10"'' y^*
e - gh.3°
0.23*(a-6
<p = S7.22°
Pomis A a*J 6 a** -P*r+Aesf -Fro** +A« neJ^«J axi's*
- - 5L 0^ ^^
N/g * ' 2B cos 6V.*S° - JB s.V6V.3° *. (-S5)cos fc«r.V -(-^fi^6^.S°
~ A- J?«5\ 3 7., m^
Ok =— +
$o*lo
H
= 93.$f v /O3 M
8o *loc
0.23 */0
iOf.C xfo*
733 N-W)

PROBLEM 4.171
4.170 The Z section shown is subjected lo a couple Mq acting in a vertical plane.
Determine ihe largest permissible value of the moment M0 of the couple if the maximum
stress is not lo exceed 80 MPa. Given: /M = 2.28 x 10s mm\ JmiB = 0.23 * io6 mm4,
principal axes 25.7° ^ and 64.3° /.
4.171 Solve Prob. 4.170, assuming thai the couple Mq acts in a horizontal piane
SOLUTION
lo " Iw r ?A3Mofc^4- 2.23 * \oc ^
Mv= M0 cos Ct.5*
9 - £*/.3°
_ I
+** f * 4^ *■*"e
0.23*10* , 6y.3'r o.^o^ei
Z28xLO_t
d> - I).84°
_ _ ^4. ©2. p?m
=■ 38.38 w»*
°0 " t + "t " O^3x|o-fe 2.28 */D"
_L\/ -Lu
80xloc - £Q.H$*103 Mo
M^ - 1,823 * \Oz N-v^ = I.3H3 UKJ-v* -

PROBLEM 4.172
0 v y\
4.172 An extruded aluminum member having the cross section shown is subjected to
a couple Mo acting in a vertical plane. Determine the largest permissible value of the
moment A/0 of the couple if the maximum stress is not to exceed 12 ksi. Given: 1^ =
0.957 in , Imio = 0.427 in4, principal axes 29.4° •■ and 60.6° ^.
SOLUTION
M„= M, si* W.4-°
9'= 2<?. H
M^» ' M0 t« 2?.Y
yA* -0.7S .'* , ZA - - 0.7T.V
UA t j/A <-* 27.4" + 7* s.>ny* r - /. 07.16 ,v
r MvUa WaV* fH.cos 27.r)<r 1-oaiO , (M„ s;~ ^.V°)(-o^m)
- 1.^381 M(
M„ -
si
\z
\.*\3Z\ ' 1.1*8 1
£>.!<? fcTsi"

PROBLEM 4.173
^ 4.173 A beam having the cross section shown is subjected to a couple M<, acting
in a vertical plane. Determine ihe largesi permissible value of the moment A/0 of the
couple if the maximum stress is nol to exceed 100 MPa. Given: Iy = Jz = 6736 and /
SOLUTION
..b = 60 mm
h-- I
4 r* t
U4 £a &. it
xj*r n ' ^T * o-»8o^io6 ^
fe = 60 mm
Pvinci'pa/. <ayes are symMeTry &./eS .
Ii.
y
.V-
12
X
JfrJ.
O.SIOtlO***? * Q-SHoxlQ **
1
© = HS' fa? * & +«. 6 - |=7i~£^ #' , 3
Poioi A: U*-o. va - - 2o Vz ^
* Iv Ij 0./8O xlD"'
"■V
61
- Joo*|oc_
«ll.»Mo» ' - MIJIMO' " 9.00 •«.'*»
-*
fio
20
°fi T^ J„ O.Stoxfo-' "* o. J So
•toft *
- IIIJI x(0 Ho
6"o _ |ao*ft>*
w unwo--5 ni;n *io-3
«JOO N:**

PROBLEM 4.174
4.174 A couple M„ acting in a verlical plane is applied to a W 12 x 16 rolled-steel
beam, whose web forms an angle 6 with the vertical. Denoting by o& the maximum
stress in Ihe beam when (9=0, determine ihe angle of inclination 6 of ihe beam for
which ihe maximum stress is 2 05,.
SOLUTION
For W \Z * 16 rJJe J. sfe^i setii'a*
P01V.+ & it r^^lAfiSt" ttoi^.-/-|ic r\ejV*i ayTs,
2t(> + ±t+*"B>
Far 6 r O
+«* 9 -" ^ * ^-IH'';^ ' 0.08*13 © -- H.7o
4.175 Show that, if a solid rectangular beam is bent by a couple applied in a piane
containing one diagonal of the rectangular cross section, the neutral axis will lie aiong
Ihe other diagonal.
SOLUTION
M2 - \A co&B
1, - -k U*
TlwS r>eu>!r«<f 0^*1* passes -Hiirtx^li w^e^ A

PROBLEM 4.176
4.176 A beam of unsymmetric cross seaion is subjected to a couple M„ acting in the
vertical xy plane. Show that the stress at point A , of coordinates y and z, is
±y±z *yz
where ly, I„ and /^denote ihe moments and product of inertia of the cross section with
respect to centroidal axes, and Mz the momenl of the couple.
SOLUTION
^-
too^ «'^^t<^S Y cl^*? Z . S;^c€ +"r,e, OlvTa-/
My- $z6;dA ' C.S-yz^A \+ Ca£'z*!dA
r —£&■ c
- - I.C, -j3, l?C
C = + -£*£!*
^^--r^
6"a -
M,

PROBLEM 4.177
4.177 A beam of unsymmetric cross section is subjected to a couple M,, acting in the
horizontal xz piane. Show that Ihe stress at point A is
*~ 'TT¥My
where Ip I„ and /„ denote ihe moments and product of inertia of the cross section with
respect to centroidal axes, and My the moment of ihe coupie.
SOLUTION
Coord ikaf«S \j olwoI "2. S i vnie. 4-itc uL^v«>i
Z7 wy.'H He
= -IeC, -X»»C, = d
-2« %tC. + .I,Ct
j;h
V-L-j -i-2: "*" -*-Y* ' ^-"2.
~L^ -La. - XyZ
jr

PROBLEM 4.178
tcO
MK* ?2t
4.178 (a) Show thai, if a vertical force P is applied at poinl A of the section shown, the
equation of the neutral axis BD is
( ^ (- \
X
\k2zJ
x +
\Kj
where kz and kx denote the radius of gyration of Ihe cross section with respect to the z
x axis and the x axis, respectively, (b) Further show that, if a verticai force Q isappiied
at any point iocated on iine BD, the stress at point A will be zero.
SOLUTION
A
H2 - 7 P^
Ak,*-
--£
A
iM^B+(#J2l
K,*"
= o rf £ A
es on
h Cl/Iw J a,Xi*S,
'♦(^* + ($
X,
Z = °> ^X^^Z * "'
Go)
= Pz£-
M2 = -Px*
..£ + K*> Kelt - £
= O
t'y C^VAMotA, TV\>w-v Ptftl^t («.")

4.179 (a) Show lhat the stress al comer A of the prismatic member shown in Fig. (a)
wiil be zero if the vertical force P is appiied at a point located on the line
X Z
+ r"~7=1
b/6 ' h/6
(b) Further show thai, if no tensile slress is lo occur in ihe member, the force P musi be
applied at a point iocated within the area bounded by ihe line found in part a and the
three similar lines corresponding to the condition of zero stress at B, C, and D,
respectively. This area, shown in Fig.(6), is known as the kern of the cross section.
SOLUTION
I2* 7^ lob3 IK= £ bhs A-bV.
•z - -i x - -^
■6- --£-■*- M*** - ^Za
w*
1 -
At
A+
A -
-SI
*2
£hb8
| - J&L.
bite
p o t~ * T
po ivv T
£
F
I,
■4> p*>;4i
h/6 J
4- +■
Z =■ o
* - a i
2 _
.^
.".
= b/e
7F - h/c
xe
w"»+Jm* fke portion **«Hc«W 7* j a. -tenstfe
a^e r eL*4if ic*j.

PROBLEM 4.180
4.180 For the curved bar and loading shown, determine the stress at point A when
(a) r, = 30 mm, (b) r, - 50 mm.
75 N- in „'^N 75 N.
20 mm
SOLUTION
h _ «o
30 mm
R -
j?«"4 ' iu «
STASIS wi«i
30
A r (2oXa<0 * £0° ^ - t>t>o*io
-C 1.
- -VS. e> MPo.
1? -
J* -ft $* so
- S1.W0Z1
£ - f - R ~ 0.55*773 ^m
VA = S?.*moZ*? - So - 9. fN0*7 »h*i
*r--±L& frS-KT-fJo^ _ W.<?>/04
A " ken ' {6oo«to^)(o.&V3*\o~*)(fc*L° J

PROBLEM 4.181 jjf1 S^""1 "" "d ioadine shown- *«»'* «* *«- - Points ^ and B
20 mm
30 mm
SOLUTION
/\ - (30X20) » 6oo_^^* = 600 x id"6 wi1"
R =
*o.
5*$ ~J0„ ^f
- H1.3ZCI *,,
r = ±.(r, + pfc "} « So »*<->
**i»*i
or*
= 4^. £26 I- 4© -
C^ooxio^Xo-^^^^/cJ'^C^x/o*3)
k^r-.T- = -^s.^^fk
- - *3.3 MR*.
>
£75-)(-/f>.673f «*>"*)
(£»oGv I0"6X0-6739^10"* )(Go * jo"S )
33.0 »IQ& ?«.
= 33.0 MP<*. -

PROBLEM 4.182
4.182 For the curved bar and loading shown, determine the stress at point A when
(a) h =2.5 in., (b) h = 3 in.
5 kip . in. «~ S kip ■ in.
SOLUTION
(b>
R =
Jt ^5
*S '-M
~r— - 3-G&G7
e ^ r -R - 0-1^33 ;«.
y - ■ 3.60*7-2.5 ^ l./o67 ,"n
V) . 3
R^
3 r*«
*§f
S.gOH** In.
yft .- 3.8^? - a.y - 1.3©*** ;« *V * ^-^ ■*
(SXl.3^ )
(3.CXO.IWDUS)

4.183 For the curved bar and loading shown, determine the stress at points A and B
PROBLEM 4.183 when h = 2.75 in.
SOLUTION
h - 2.75"m. r, = ZS m > f% - £.25 i«
^ ^J 2.5 in. ^ a (,.^(3.75} = 3.30 f-i* M r 5" fc>- m.
5 kip. In. J 5 lap. in. L
r = i(r, *rj: 3.8?^ fh. e = r-P - o. less- i*.
" Ae/; (3.3oXo.i6g5X^-s-)
^B = 3.704r- ^"-25 - - /-54S5" /«. T6 - £*.Z5" ■>
°* Ae<ra t3.3o)(o.iMj)(s-.a5) *

PROBLEM 4.184
4.184 The curved bar shown has a cross section of 40 x 60 mm and an inner radius
r, - 15 mm. For the loading shown determine the largest tensile and compressive
stresses.
60 mm
40 mm SOLUTION
120N.ii
e -
A+ r
s - - M^
Aey*
Y t 30.73G- 15" " 15".736 km
~ 1-S tVMVN
_— ——rr - -1?.«r?*/t>*P<i

PROBLEM 4.185
60 mm
4,185 For the curved bar and loading shown, determine the percent error introduced
in the computation of the maximum stress by assuming that the bar is straight.
Consider the case when (a) rx ~ 20 mm, (b) rl = 200 mm, (c) r, = 2 m.
40 mm
SOLUTION
Vt ^ MO
A - f.6oW: IVQOk* - XVOOXIO* r^
M
\Z0 M.m
120 N • m
- 6.22 xfo" **,*
- 7.^ * /Ofi Pa = 7.5" MPa.
h
R =
^
Y\ - GO
Ho
A. %
^r M^.-g^ , («p)C-l4.1of<WQ-?) = _|lmWo<p^
°- Aer (:Woawo^YS.s*MWor,toowo-s )
= -//. ^6 *P«
ev-^of- =
-/l.</2£ - (-7.5")
x I0O7- ~ -^.4 V0
1
01)
(U
(c^i
^^
Id
Zoo
Xooo
V9 |W*M
go
Mo
2oHo
Kj *»i»^
S6,Ho?t
%\^.V\IC
2e>t%elZlto
?,'**
40
2£o
2o2o
e^*.*
^f^
0.6074
O.Ofefeo
-n.w
-7.^8*
-7.546
%eir*o^
-34.V ?-
S.6^o
O.Q V.

PROBLEM 4.186
4.186 Steel links having the cross section shown are available with different central
angles /?. Knowing that the allowable stress is 15 ksi, determine the largest force P
that can be applied to a link for which /?= 90°.
0.25 in.
0.6 in.
SOLUTION
0,6 in.
^Ccjjc* 5€cTi6ta -force 4-» ©. ta'c.' ■■-
Ae.
M
p'
•< 7^
9*-*
45'
\!
At po/A-f A 4^-e fe^s-'-Ce STVesa »S
Trie Levi efi net coup/v Is. M ~ -'Pou
F»^+k& fec4*«o*«nJ>«i/L 5ec4Vo^j "Hie nftJ-fvA-P
s
A Aet*. A *e>* a v er. ' h
'* "A Ae^, A Aetr, A x ~ e«\
Wne^e
P =
_ A6*
K
er,
CW
i & = R - f\
Da+ft.: (r - .0.9 m0 /*,= 0.6 i'hj'. ft= L2 m Vj'OJG;!^. lo; = 0.25!:in.
A^ (O-as'Uo-O r 0.75 /«* R— fiTTi ' 0.'S&«5l|V
a - 0.«T.^O.«C«Z = O.03H32 Jfo^ yA * O.Scrca - 6.£ = J0-2t5<SP ^
O- = O.^.'C I - cos «fS°"^ r 0.X.6 36O */n

PROBLEM 4.187
-^
\
\
\
T
4.186 Steel links having the cross section shown are available with different central
angles p. Knowing that the allowable stress is 15 ksi, deiermine the largest force P
that can be applied lo a link for which fi~ 90°.
4.187 Solve Prob. 4.186, assuming that/?= 60°.
SOLUTION
0.6 in.
coofufe Sy sfe*U a£.& He ee»«+ir*oi^
or -Hie oross SvcAriofk AS.
(X- T C| - Cos|)
TWe tei*<^'ttek ccxjpfo 7&",M.'" - Pa*
0.6 in.
Aft '
^ r A" * 'Ae>,
vJn&^e.
?»
- AG;
A Aeif, A
©mJ. yA = R - r,
K
<& - q.?.r"0.««6Z = OU03W33 &„, yA ft O.S€^6X - 0*6 ■ 0.ac$M
<X = o.*! (I - cos 30°) - 0. I205S m.
2. £526
0.831 k.'ps
est Pt.

PROBLEM 4.188
r = 20
&
P = 3 kN
25 mm
25inm
■a
4.188 The curved portion of the bar shown has an inner radius of 20 mm. Knowing
ihat the line of action of the 3-kN force is located at a distance a = 60 mm from the
vertical plane containing the center of curvature of the bar, determine the largest
compressive stress in the bar.
SOLUTION
Red 'J ce fKe i * 4-e^tftJ* ft*r*ce.s
+© c*. -force - coopJi sys*f«.i^ cdc +•&<
M- PCa-f-r)
Pt»r -Hie r^W^O/^ 5€<=^>o« +ta neotv«.-P
—k
JiVi -^
AJ>
So
Tn-
e- ^-R
r/re:
6-. a - £. - Ky» T _ ? _ pCai+rVy*
A A Aer, A Aeif,
,-k£
Tt
wfH ^ » R - it.
R= vug- = 30.S2&ii.vm<ft , e~ 3^-£,-5©,«^&2 - 1.6713. ^*
a- - 6(9 ^o a-v ? = 92.6" ^«- j R - r, - \o.%z&8 m^

= 20
P = 3 kN
25 mm
SOLUTION
|—a
Reduce +Ke in-Wv^i £xrce.s
TV«*yA wv»* Yre©| <a.a^a&s Sec&at AS
cen+r o id oT He S e eft a* . Thtf
t>final IAA £p t/p ^« IS
Far -He i^6aWi^c>'IU/> se^ho* "He n«i/t
R-s\
Aiso e=F-R
CU+ PO l *»T A ft X*f "»S <R i \/^v\ few
^A
Tt
A Aer, " A Ae/.
A wf+k ^/A - R - r,
= -k£
as
DcuW h r &5* *n*i , V\ * 2© •*<« ^ V*a » *T «W, f> = 32.5* h^
JW STL J
v*
P - 3*/03 M- v^
G*h = - /5*0*/os f.
«*.
f> 3*|o=>
Ct = ^3*27 - 32.5 r GO.9 Mm

PROBLEM 4.190
-&.
Q
©
2 in,
4.190 Three plates are welded together lo form the curved beam shown. For the
given loading, detennine the distance e between the neulral axis and the centroid of Ihe
cross section.
0.5 in.
0,5 in. SOLUTION
L I--II
0.5 in.
R -
^fc*k - ^A
ZM"
ft
r
a
3.5-
p.vff"
(D
®
©
7
b
5.
0-5
^
h
015"
2
0.3
FA
A b'fljjjr
1.5
l.o
I.P
3-5
o,M6av«
o.zssw
O. 1740.23
0.^&2M4g
¥*
3.25
t.5
5.75
A/1
^.375
H-5
5.75
15. 135
3.5
0.*62¥6&
e. = 7 - R -
= -y,05«/^ lii^ F ~-~r H. 5*1*3 .V

PROBLEM 4.191
2 in.
4.191 Three piates are welded together to form the curved beam shown. For M = 8
kip-in., determine the stress at (a) point A, (6) point B, (c) the centroid of the cross
section.
SOLUTION
*S*<M
***.
- 44*
P;
&
G>
CD
V*
3>
&5
o^+
<D
<2>
i
©
?
b
*
0-5
2
h
OS
2
OS
. A
1.5
l.o
I.P
3S
MlJfc
amMY&z
to.ZrWW
G..1740JZ3
O.fftZtCft
7
3.25
*\.s
5.75
A?
^.275
M-5-
5*. 75
ls:»as
R
3.5
0.*£2W&
e. * v1 -1?
= ^Loswta i^ ft ^ils*- 4.3*1*3 in.
to) yA -- R-rt =
H-.05S12 - 3 =? 1.0-5817 in-
«;= -
Aer,
(-3)0.053*0
(3.5X0.26330(3)
3.06 fesi'
-El*
J^efi (3.5X^^33 0(6)
5- , . m&- r - (-«X-n*w) s -?.si to,-
= R - Y1 « - e
Aer
_ Me
M
-8
A r
(3.SXH-32I431
= 0.5*1 fcsi

PROBLEM 4.192
4.192 and 4.193 Knowing that M= 20 kN-rn, determine the stress at (a) point A, (b)
points.
I
US+~<'' A V*"*\[ A ' 150mm
T 135 mm
SOLUTION
36 mm
R r
r .-
:a
ib,h
ZA;fc
2A,
r b, k ^
TAt
1 kt&.'fe'
©
©
fl*+
CD
2
k^ro"!
I0S
3&
K**
^5"
|3S
A KtW
H86O
42feo
77Zo
L 0 iki
n.zzsz
1^-..1S^
H7..2TW
^|W».
\12S
1QIS
Ai^ *"«"»
S3S.35 vfo3
1X75".75 */Os
7114./ Mo*
R-
<?72o
47. 774 "7
3S0
/7„89* wit*.
- ZH.H. |y|ott
Wo
- /£ i 7- -5" mm
Co.") ^ ~ "R - r,
Aer,
6 ^e^l
2os.CoC -S3o - -m.S^^
(;?ox/o*)6-ia*.3i**lo
-3N

PROBLEM 4.193
4.192 and 4.193 Knowing lhat M= 20 kN-m, determine the stress at (a) point A, (b)
point 5.
CD
W~\
150 mm
135 mm
45 mm
SOLUTION
R r
36 mm
r
33 o
^A
2i;L*
V -
7Ai v\-
TA.
2 fc^s* ?M*^
FAi
©
©
5
kj;vyiM
3G
log
VjjVnw
1SS
45
A^*^
M»6o
4-86o
Wo
b^h ^^ ww,
23.|0£7
IS 8332
■3*^31**
f; M"l
R 17.5"
3^7.5
Air, *•«*
\.oS~?o£*ioi'
1.4? 445"*/D*
2.SS 15" /lo1
a»-sS£r**-^ ^-^^^ *^~
e - ? -R = 12.8S^
M - Xok/o! K/-vr>
la) ^A * R-/\ ■= 2^?.cis - is"o - ^.6/S" w^
- - 106-1 k|c?6 P<
£L
-fOfe. I MPa_
6 Aey\

PROBLEM 4.194
16 mm
12 mm
4.194 The curved bar shown has a circular cross section of 32-mm diameter.
Determine the iargest couple M that can be applied to the bar about a horizontal axis
if the maximum stress is not to exceed 60 MPa.
SOLUTION
6wy Otc <JlTS <?-+ A
= ' A eft I ^""* ^re-t
Also A = ttc* = ttCis^ = goi.zSM^
D-yf«.: J/A = £- T, - ^,4*<71-I2 » 13.48^1**^
M .. UQa, r^ = i°7-8 W-*

PROBLEM 4.195
, 50II)
4.195 The bar shown has a circular cross section of 0.6-in. diameler. Knowing thai
a = 1.2 in., deiermine the stress at (a) point A, (b) poinl B.
0.8 in.
SOLUTION
c = x«* = o. s ;«. in - o,s + 0.3 = o.g ;«.
e. •= r-R r o. oam ^.
A = TC*" = ttCo.s')2' * 0.2S^74/M2
V\ = -? (a + y > = -So ( i.A f o.g) =-loo &. i«
^e-R-r2 - 0.7703/- L| * -©.^fn ,"*.
^ 6., £.-^_,^ (-.00X0.^08 0 w G.i1*|oV
A Ae<\ 0.m7V (0,Z827,/)f>.o3»i?YP'5)
^ 6.7H Us,'
= -a.45- kv.

PROBLEM 4.196
,50 Hi
0.6 in.-
0.5 in.
Sob.
4.196 The bar shown has a circular cross section of 0.6-in. diameter. Knowing that
the allowable tensile stress is 8 ksi, determine the largest permissible distance a from
the line of action of the 50-lb forces to the plane containing the center of curvature of
the bar.
50 fh
s>&
SOLUTION
e = ^ J* 0.3 ;* 3 r ' o.sr+o.% » 0.8 u
ft - il ? -v -/1 ^ - £* J * itl^a + Vow^-o.^3
= 0.77081 ,'ki. e*r-G - 0.0^11 r»i.
A = TTC1 * TTCo.3)1* Q.Z2ZV4 in*
M * - P CoL+iO
yA - R-/\* O.7708i -O.JT = O.X70SI iV
^ X ^eir, a Aer, 7T L' * ef\ J
K - ^A s (3*10*^0.^*7^ . 45-^-ss
? So
a - x.3w - o.s ■* i.ssff ;**

PROBLEM 4.197
I 2,5 kN
4.197 The split ring shown has an inner radius r, = 20 mm and a circular cross section
of diameter d=32 mm. For the loading shown, determine the stress at (a) point A, (b)
point B.
SOLUTION
^ T 2 O •*.*,
, Vi- r, + d * ££
mm.
_ J.
r -
4 Vrl-c* ] = i[^ + /i^
n
I6Z j
■p - %.s*io* N
e = r - R - i.s?53* ^
M - Pr =(2.s"Vict,K36>';o~s"j - ^o M-^
(Ol\ Ptfirtf A 2 JA = R-f, * 3H.1=H.T - 3o - 14.13*5
b*~ A ~Aer,
Z.SviO"
ftcACW.iz^sxto"*)
lb) Poi>+ B • j^ R^^ 3*m*S - 5* - - 17. S7ST ^
Cl°X-l7-&S5*(0'*)
A Aeir2 " goM-W^er" " (8o*f.2?y|o_c)fuS75S-«lo"*X^»'lo-*)
n.HO*to' ?CL
\7.Ho MPa

PROBLEM 4.198
2.5 kN
4.198 The split ring shown has an inner radius r, = {6 mm and a circular cross
section of diameter d=32 mm. For the loading shown, determine the stress at (a) point
A, (b) poinl B.
SOLUTION
e * f -R - ;?. use *•*»
a
P * 2.5WO* fj
(a) PofflT A: jA r R-v*f r 2%*SiH - IS * 13.85** *i
=• -*3.3/io*?ft * -43.3 MPa.
CM fi>i^rt8 * j/B - R-^ - Z9.8S6* - 48 - -I*. 1*36 «■*. .
g , -P _.Mj/6 _. ^xio6, (80^)8.f*3&^lQ"3)
s~ A Afiftt. 8oH.aSVio-fc~ (Sc^.X5*yio-t)(2.l436vlo"sX48^IO"s)
r 1¥.«3*/©*P«. = /*.*& Mpsc -

PROBLEM 4.199
4*199 Knowing that the machine component shown has a trapezoidal cross section
with a = 3.5 in. and b = 2.5 in., determine the stress at (a) poinl A, (6) point B.
\ SO kip- in. SOLUTION
Locate ce*+^0(<sf
6 In. 4 In
0
r
V*
7.5
IS
f4in
6
8
Av\".»5
£3
Go
123
R*
±hc(fc, + lO
7 - l-jf * €.8333 in.
L(3.SMoWi*X«)] in f - CSV3.5-- *.* )
Gert ja ' R - r, ' 6.SS78- 4 * 3.3*78 ,y
6.3878 m
(bl yfi - R-*v r
(T.--!l^
Aer,

PROBLEM 4.200
4.200 Knowing that the machine component shown has a trapezoidal cross section
with a = 2.5 in. and b = 3.5 in., detennine the stress at (a) point A, (b) point B.
m kip . in.
ifl
SOLUTION
T
a
6m. 4 in.
^
A>1
is
IS
r, ■«.
A?,'***
Hi"
84
\2*
R-
±Wb^tO
r - — - f4 I fc& * iiri
t(z^<0 - (?.5X*ffl At-If - fe)(*-s- ^
(CO _yA - R- ft =■ 2.7ICS In,
" £.7|68 in
Ae*\ (i^Ko-HWKtn
(^ jfe, = R - r2
B Aev\
--3.5 *32 ;„.
(jg&X-3.g&3^
r- S_24 ks

4.201 For the curved beam and ioading shown, determine the stress at (a) point A, (b)
point B,
SOLUTION
250 N ■ m 250 N ■ in
36 mm
R -
40 mm _ I
Section a-a
JVOd,^U^
®
1%
1
A,****
*?oo
-
ss
A?, ^^J
27 *IG*
IG.5"*lo*
Y3.S*/o3
[(4oX«5) - froX*")] Ai "ff- - (3*M**o - ?cO
e - r - R - 1.47*5" *~
p\ --ZSo VJ-^
<tt ^
&
Aev\
e-1%
-s
GS.6! y|o6 P*
5 r--SiS_ r
(<?doylc?-6X»-47^x/o"s)(ar>«/0
18. I3T3L mm

PROBLEM 4.202
4.202 For the crane hook shown, determine the iargest tensile stress in section a-a.
SOLUTION
l-K-l
60 mm
Section a-a
©
z
A,w#n
\oso
150
ISOO
r,*.-
(So
So
. — 4
£3 */o3
Go*fOs
1 i>2» * 10*
^ _ iosvjo:
- £g.333 mm,
18 oo
R =
iVjMb.*^
_ (0.0(60? ( 35* J-gg)
M«.xiwu"fi fe«Si'-/e .S-fr-CJJ OCCURS «4 poi>t A
^ - J? - r, - 23. S7&
- 63.2 72 *w.
W)>*1 .
g. _ JP My, _. i5*|o3 -(].QlS*lo3)tM.%7&*(o3 )
= g^.7 xlOc V*.
£4.7 MP*.

PROBLEM 4.203
4.203 and 4.204 Knowing that M= 5 kip-in., determine the stress at (a) point A, (b)
point B.
2,5 hi.
SOLUTION
Use ^o^nui^ -R^ "htetpezo'V
(b/a- ^i)^T» -V)(b,-bO
(^)^\^_+0A
[(2.S)(^-(0)ft)]^f -(3^(2.r-o)
let) y„ - R - f, - 0.8^548 m,
A " Ae/\ " {$.is)(o.iS'*sz)(2)
= ;?. 2<4SH% m.
~ -3.65" ^s/
'6
tV7s)(o.l5HSsO(5>
3.7^ k-5i

PROBLEM 4.204
4.203 and 4.204 Knowing lhat A/= 5 kip-in., determine the stress at (a) point A, (b)
pointfi.
2.5 in.
SOLUTION
A = i (a.s)C*} -- 3-75 /**
r - ^ + 7L - H.ooooo in
b,: fij V * m. ^ kt - ^.5" m. V\ - S" i>.
Use \o+-+*oje^ -to/' "frflpftoial,
R =
4h2(b, + bO
a
^ (O.SK3)t(o4-*.g')
(a) ^ - R- r, -
AeV\
l.85<f£6 m
(3.7Stfo. I ¥53*0(0
- 3.51 ks,"
(b) ^B - R- rt r - i./fs"^ ;
*v
ff5 =
_ M.Vi
(5X-M^5"3^ ^
Aex, cs.75-Xo-'H53«*H^^
^v Jo ksi

PROBLEM 4.205 ^ ^^ ** * = " ^ detCrmine *" Stre8S at ^ P°int A> <*> P°int *•
P| SOLUTION
3.6 in. i _ o . \ « -
.-b - 3 .«3. h- 3.6 1*.^ ^( = 4;^.
.Redact 'seciittn -p6^ce4 -fv <*. -fo^ce- cou»p^€ Sisie.** aJt -j-fie c&r/o-'iJ
P - 3.S kip* M = Pf r CS^X^S.) * /8-a. Kp-/,
■M
For .a, frtctr\c\oJ!a.r Seci'tovi R -
*h
-2.
'IS, A /":
.0^(3-0 , 5.07007 i.
Li. o 7-c _ ,
3.Q "" V
6- ._£ - M.ye _ _ ail Qa.gY-a.5Jffl3) _

PROBLEM 4.206
4.206 Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
R =
In
f \
ynj
vr2,
f \
yrj
where A is the total area of the cross section.
SOLUTION
7A _ A
R =
2£±JA ~ ^J^^~
A
M^t " m^k^T3
m
.fe -fno-f for e^c^i ye c4&*ifl/« 0 7^^ ~ J "
it

PROBLEM 4.207
SOLUTION
i 1ft
r
—V
\
\ ccoe/3
4.207 throngh 4.209 Using Eq.. (4.66), derive the expression for R given in Fig. 4 79
for
*4.2©7 A circular cross section
Dae. po.* <=cv coo /eu w&.Te. A «a.s sUo-jw
V" = f - C cos/9>
J/* * ■* C dm p <k/2
■■"■ , i
^ = (" ■'c'Cl-e^g) JA . „ f* p"-ClcosV - (f "-cT)
-J V - o v - C
&
cos^e
- Zr &\ + 2C SiryB
2
- 2(rfc-cm>ra
W^
X
i
An
-i fp -cl -him iyS
r + c
A =
= 2r (ir-cO 4- 2o (o-o") - ifVr1 - efc -(J - o )
7TC
TTC
^i4 2irr -n^t-c1-
" z
r + 77*--
Vr" -c-
r +
/p—■

PROBLEM 4.208
SOLUTION
I
h-i-M
l
i*- w
A = ^Ct,* WH
R--
A
4.207 through 4.209 Using Eq.. (4.66), derive the expression for R given in Fig. 4.79
for
4.208 A trapezoidal section
w = Co + c,r
W = b &4 r = Y\ *-^ W - b2 ojf ^ = *V
t, - c0 + c, r,
ba' c0-t-c,v\
b( - K = c/*\-0 - -c,h
f*t>, - f,b, = Ci-Oc. = he,
|Tz
I I
4^(b,+ bO.
$IE (TA-^Mi1^ - h(b,-b2)

PROBLEM 4.209
SOLUTION
h « i th
R -
A
^throogh 4.289 UsingEq.. (4.66), derive the expression for J? given in Fig. 4.79
4J.09 A triangular cross section
for
w - cb + qr
b - Co* c»r\
o = ce + c, i\
b = C(C^-^) * - C.r,
Ce =■ - q r2 =
_ b£
5*^'^-^*
v.
-- CoJtv, ^ + c,(rt-r,)
4-U,
KHVO
i^
- I

PROBLEM 4.210
*4.210 For a curved bar of rectangnlar cross section subjected lo a bending couple
M, show thai the radial stress at the neutral surface is
M
SOLUTION
a^Te
R nj
°a_~ Aeir
- M _ Ml?
Ae Aer
«cfs } fht pes v/■**■* + -Fo»"ce is
and compute the value of or for the curved bar of Examples 4.10 and 4.11.
(/7/iir: consider the free-body diagram of the portion of the beam located above the
neutral surface.)
/*
_«
t
- GVkR ^ cos/9
MLR
Ae
0-&--K)
or e<^u'>f I pri't/i*
F, - SMsm4 ^ o
2$fcl?«,vf - *J^« (i~£-^J^ = o

M„ - V
4.211 A single vertical fibrceP is applied to a short steel post as shown. Gages located
at A, B, and C indicate the following strains:
eA = -500 fi eB = -iOOO fi ec = -200 p
Knowing that E = 29 * 10* psi, determine (a) the magnitude ofP, (b) the line of action
of P, (c) the corresponding strain at the hidden edge of the post, where x = -2.5 in. and
z = -1.5 in.
SOLUTION
A. - <fX*"> * is ;**
XA - - 2.S Vh s X
B =r ^.r ?«
Xt ~ X. O '*'
2a'=
1.5" fh
^ 2a = /.5" I'rtj 2c--/.5'in3
6a
6k
—-E.
- £.5* m.
- /-£" m
-14.5- Jwf
- S"-8 fcs,'
^A * -J ' I
^*3* + J^ ^ - 0.O6E&7 P - OJ3333' Mx - O.o* Nz
0)
6k --.-? - 14*2? + I^t = -0.06667P - O. \333SHy + O.Og Mz
J.y -1-7-
S*c "- -^ -i^ + M^fr- = -0.06667P f 0.13.333 tfx + O.Og Ht (?)
M^« -9o.fc£5 k-.y.i^ f = 152.25 fe.'ps
p "
X
2 ~
37
is't.ar
O.S7J fw.
6; =
--£ - t^p 4 i^E_ r - O.OUtf ? f 0-13333 My -O.Og M«
= S.70 ksi

PROBLEM 4.212
4.212 The couple M, which acts in a vertical plane (fi=* 0), is applied to an aluminum
beam of the cross section shown. Determine (a) the stress at point A, (b) the stress al
point Bt (c) the radius of curvature of the beam. Use E = 72 GPa.
SOLUTION
Ld)e.P a*es ^y cu*e] 2 as show*, om +k-c ske^tk.
= 0. /053S3 */o'4 m^4 - 0.l.o.Ts33x/D~* *V
10 mm
Hz - 3O0 N-rrt
M* = O
<C1 f'- &- ■
£Jr '"' P '' Mz
' i, 0.WM3W*' =-70/1 MP*.
EJ2 _ (72*/o?)fo./c^&33^P~6) _
3oo
25*. H vn

4.213 The couple M is applied lo a beam of the cross section shown in a plane forming
an angle ^=15° with the vertical. Determine (a) the stress at points, (b) the stress at
point B, (c) the angle that the neutral axis forms with the horizontal.
SOLUTION
La>Vi-e! ajfts sj awcsl Z aj$ sl-iow^ on "Hie .ske+c^li.
I, - ^((o)LSo)3 + z-j-20o)CioY
- O.l0$%%Zx\0<' y«J* - 0. 105333 x io4^
10 mm
= 0. 02,5233 */of m^y - 0.0253 33*/o~C **
My - 3oa si* 15° = 77. &S N-r*
(CL)
6;
z- - Mz-Y* + J^lI* - _ (lg?-7-3)fov/o-3) (77.6SX-Syfo's)
Iz
X
ftO
6* - - HzYb 4 M,Zfl - - (?**?■ 78 Y-SVIQ-*; (77.&rX-'5*jO'aJ
fi " lz Xy 6- |o5XS3x/o-c 0.0^333 */0-4
^
0. 025"8a3*/O"
cp - *f7.7

PROBLEM 4.214
4001b
r = 0.3
4.214 Determine the maximum stress in each of the two machine elements shown.
SOLUTION
Far £*ch c#$e N =■ Ofao }(&.$) * fooo JfLt'w
c =• o. 75 ;*.
Pro*, Fr3 4.3£ tf - L76*
F/b-w R3 4.3/ tf = I.SO
All dimensions in inches
*^fl|

PROBLEM 4.215
4.215 The four forces shown are applied to a rigid plate supported by a solid steel post
of radius a. Determine the maximum stress in the post when (a) all four forces are
applied, (b) the force at D is removed, (c) the forces at C and D are removed.
SOLUTION
For *. &&hd etrtv-ptr sccfiVn £ vn«6v$ et
(CO Ce^W 4We F = 4P , M„ ^ Mz ^ O
F
<fP
e-=-r — is-- = -'■«^
(t>) Fo^c*. a.4 X> \"3 ^ewso\/eoi.
F = 3?, Mx = - Pa., Mt * O
F = 2P M„ ■* - Pa , Mz = - Pa.
IT fit

4.216 In order lo increase corrosion resistance, a 0.08-in.-thick cladding of aluminum
PROBLEM 4.216 has been added lo a steel bar as shown. The modulus of elasticity is 29 x 10*psi for
steel and 10.4 x i06psi for aluminum. For a bending moment of 12 kJp-in., determine
(a) the maximum stress in the steel, (6) the maximum stress in the aluminum, (c) the
radius of curvature of the bar.
SOLUTION
Use STee^ ^-s -He irefe/e/:^ maAef'ojf
*W - I 10^- |* - ^ - 0.3S8i>
- ^(l.StYl.M? + 0.35^6--^[^)f(.S)3 -0-^Xl.3^f] - 0.4333.T ;„ *
(a) y5 = Hp - o.67 ?„
5- - tl^. r (1? Vo.c?) _
to
ca
O.V3&3S
0.7S i
2.35 ksf
6— 0."** o.3S«MK2^1 , 7.S€*s/
M
I " O.H3 22S
<WH x/o"c ' "
rn
28.3 -Ff.

PROBLEM 4.217
5°/!/'
W200 X 19.3 A
4.217 A couple M of moment 8 kN-m acting in a vertical plane is applied to a
W200* 19.3 rolled-steel beam as shown. Determine (a) the angle that the neutral axis
forms with the horizontal plane, (/>) the maximum stress in the beam.
SOLUTION
For W ZOO* |<?.3 **))** s±<><>? <*zT'or
\JS */o4 "*^ -z \,is > lort ^
V =
Wi*n
Mz ~ (,?/io3') cos 5"° = 7.fsuxia3 m-m
(a.) fee* (p " ■=£ f A^ Q =
-Li- - - M*i% ic^(rS^ - - ».2G*?
I. \S K fO"
ve^1
.hor. "*
(M M^jo'^J^ +fii«fti^e s+i^ss ocaJ'hS a& potv\t I)
77.6 ^/O6 Pa- = 7T.6 MPo.

u
PROBLEM 4.218
120 mm
3s39 10 mm
4.218 Three 120 * 10-nim steel plates have been welded together to form the beam
shown. Assuming that the steel is elastopiastic with E = 200 GPa and Oy = 300 MPa,
determine (a) the bending moment for which the plastic zones at the top and bottom of
the beam are 40 mm thick, (b) the corresponding radius of curvature of the beam.
SOLUTION
120 mm
10 mm
. v—•
0^
10 mm
30O MPft.
1 T~
t?.-*"
K +
R* *■
A,= (l^)(|o):|20O^1 J?, r GrA, ^(30OX|O£)(ia0O*/O"6> ^feOv/C? M
20 m
.-»
(b> A .- &
EjV (^6ox/0T)('30KfO"S)
3ook|Oc
/"-^'

PROBLEM 4.219
M,
M,
M\
^^<Ji
D
4.219 Two thin strips of the same material and same cross section are bent by couples
of the same magnitude and glued together. After the two surfaces of contact have been
securely bonded, the couples are removed. Denoting by at the maximum stress and by
/?, the radius of curvature of each strip while the couples were applied, determine (a) the
final stresses at points A, B, C, and D, (b) the final radius of curvature
SOLUTION
Le.f b = wi'JfVi and t = 4kicknes5 of ©n £ sj~<r*f>
LoaA'tn^ one sfrip M - M,
6. = -^
/°' EI, "
13. M,
El3
Af+e* Mi is *fpjP\eel +© ea-^1* of+k& S+ft'pS.
fke ^fresses 'o.^ +"lio^ <7/Ven in -Hie sker^M
Af+e^ g^'^Jj *^i*s toopJle is pe*«o*e*I.
M* = *M, , l'=^t(2t?-- |tt4
C r t. 7~U stresses r^e^c^J ©t^«
FT1- - lily. = ^HY - -L-'^Mjf
la} RVwJ sfres-ses : 6J = -ff, - ("aO r "i^
a. _ M1 2K
3 Mi J-J.
let.) F;*J> vtejrcs
i

PROBLEM 4.220
3.25 in.
4.220 Knowing that the hollow beam shown has a uniform wall thickness of 0.25 in.
determine (a) the largest coupte that can be applied without exceeding the allowable
stress of 20 ksi, (b) the corresponding radius of curvature of the beam.
3.25 in.
SOLUTION
E ■= lo.G>/o* pi
^ * = tic ., tA=^± = .t*°X^0^ = go. I *p.lV
(bl
r
■VKAy
p
.6ZT
61
PROBLEM 4.221
10 40 10 40 10
Dimensions in mm
4.221 A beam of the cross section shown is extruded from an aluminum alloy for
which E = 72 GPa. Knowing that the couple shown acts in a vertical plane, determine
(a) the maximum stress in the beam, (b) the corresponding radius of curvature.
SOLUTION
For oO^es* recJ-AH^Pe ' b - | ic 1**1 3 h ~ 40 wm
Fo/- one c*A'ojf he^Tawi^J'c * o " ^O t*tv, ^ Vi w ^?
4
r m ^kio3

PROBLEM 4.222
4.222 For the machine element and loading shown, determine the stress at point A,
knowing that (a) ft = 0.9 in., (b) h - 1.5 in.
200 lb ■ in.
SOLUTION
Ceo Vi - o.i \» t r, = 0.5;«. v; = i.f ;*»
A = (o.foMo.i) - 0.5"*/ ;«l
R =
b
0.1
J«r*
T\
s '-*
0.S7MM ;„.
0-5"
200 ll>. in.
M = -Zoo &'i*
e = r - R - o. 075 8 9 in
j Jk r R _ r. " 0.37WM
(-20oX0.^7411 ")
(W
* Aef, (o.rv)05.o758^X«>.5)
- 3.65" ^s?
R -
>"fc
I.0S2O2 m.
£ - r - R * 0. IC7?8 >V
M^ - 4©o A- ?*. y* ~ R - ^ - 0.55U02 ;w
/\ef
(0.10)10. IC7*f&X^S)
i.54o UsC

PROBLEM 4.C1
[*~b = 60mm-»-|
4.C1 Two aluminum strips and a steel strip are to be bonded together to
form a composite member of width b = 60 mm and depth h = 40 mm. The
modulus of elasticity is 200 GPa for the steel and 75 GPa for the aluminum.
Knowing that M - 1500 N * m, write a computer program to calculate the
maximum stress in the aluminum and in the sleei for values of a from 0 to 20 mm
using 2-mm increments. Using appropriate smaller increments, determine
(a) (he largest stress that can occur in the steel, (b) the corresponding value
of a,
SOLUTION
Y^AHt^Qfi^eO ^jt^t/^M (Fill. STF£l) "H ~
b -A 1
^HSI^ll'- W"^
I
Foz c*-=Q TO ZO ■>»<*■> USW6 2-7»m jN7£G.vAL$ CGfstpLWt »_, J CC,.„ , QZ
b=60ram h = 40 mm M * 1500 N.m
Moduli of elasticity: Steel = 200 GPa Aluminum = 75 GPa
pR06kftM ourror
a
mm
0.000
2.000
4.000
6.000
8.000
10.000
12.000
14.000
16.000
18.000
20.000
I
nT4/10A6
0.8533
0.7088
0.5931
0.5029
0.4352
0.3867
0.3541
0-3344
0.3243
0.3205
0.3200
sigma
aluminum
MPa
35.156
42.325
50.585
59.650
68.934
77.586
84,714
89.713
92.516
93.594
93.750
sigma
steel
MPa
93.750
101.580
107.914
111.347
110.294
103.448
90.361
71.770
49.342
24.958
0.000
:3Ei:iiita!
Find 'a' for max steel stress
and the corresponding aluminum stress
6.600
6.610
6.620
0.4804
0.4800
0.4797
62.447
62.494
62.540
111.572083
111.572159
111.572113
Max Steel Stress = 111.6 MPa occurs when a
Corresponding Aluminum stress - 62.5 MPa
6.61 mm

PROBLEM 4.C2
4.C2 A beam of the cross section shown, made of a steel that is assumed
(o be elastoplastic with a yield strength o> and a modulus of elasticity E, is
bent about the x axis, (a) Denoting by yY the half thickness of the elastic core,
wnte a computer program to calculate the bending moment M and the radius
of curvature p for values of yy from \d to g d using decrements equal to 3 iy.
Neglect the effect of fillets. (£>) Use this program to solve Prob. 4.218.
SOLUTION
——.—— ^
X
/E>*? YfBiOi*£> /w me WW 62-%
V
U— 0C-—H
y (^
C-j
^7 F=f
I
^
SV^feS" RTJONCT*OM OF \n&& fl/wo F^ fl/'/^
rr— *
B£fifDM£> f*70*'*&*'7'
M=2? *L*
ti
t?<*9iUS of=- COM V/fTUtiG
CONTINUED

PROBLEM 4.C2 - CONTINUED
i,
i. M
~Ar
iV
r/
(T
7
■
:
fa7
^
[
Ll
BSHOtot Moment
a.- i %
1
/*=-?£: Ra
-»T^
> •*
*£%P/A!S QF*r!jnVf>TVl</r ljy* e' 0 t _Lp
^ **
r.
Fe&ztfi* : HFY in fspefesMtt* pott, cl^ a^d R^ /^>/£ -^ - / -To "7,
cwru-7* M = z zr&^-u fcK. ^s I to<9 #*& fi = -jz—.j 7**^ &&*>?
IffPtiT NWf&fCftt- VALUES rtf-'O ftU* Pfc&6&Srte
- 3/
fp&?£flM aoyPur
For a beam of Prob 4.218
Depth d = 140.00 ram
Thickness of flange tf •> 10.00 mm
Width of flange bf = 120.00 mm
Thickness of web tw = 10.00 mm
I = 0.000011600 m to the 4th
Yield strength of Steel sigmaY = 300 MPa
Yield Moment MY = 49.71 kip.in.
yY(mm) M(kN.m)
For yielding still in the flange.
70.000 49.71
65.000 52.59
60.000 54.00
For yielding in the web
60.000 54.00
55.000 54.58
50.000 55.10
45.000 55.58
40.000 56.00
35.000 56.38
30.000 56.70
25.000 56.97
rho (m)
46.67
43.33
40.00
40.00
36.67
33.33
30.00
26.67
23.33
20.00
16.67

PROBLEM 4.C3
0.8 0.4 1.6 0,4 0.8
Dimensions in inches
4.C3 An 8 kip * in. couple M is applied to a beam of the cross section
shown in a plane forming an angle 0 with the vertical. Noting that the centroid
of the cross section is located at C and that the y and z axes are principal axes,
write a computer program to calculate the stress at A, B, C, and D for values of
0 from 0 to 180° using 10° increments. (Given: Iy = 6.23 in4 and Iz = 1.481 in4.)
SOLUTION
I hi PUT c c>o n & / Atft-r&s & -f f.. B c,D
*0
.'^O/vtJ&M&M-rs &/=- M.
A8, =
M £/fi ft M^ - M Cosfi
^TE? ^SS AJetfl-
'??:
2^ Xu
P£Tu/ZM
&£? Turn
pfeaSfifi-hsi & u TfoT
Moment of couple M = 8.00 kip-in.
Moments of inertia: Iy « 6.23 in"4
Iz - 1.481 inA4
beta
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
Coordinates of points A, B, D, and E
Point A
Point B
Point D
Point E
A
ksi
-7.565
-7.896
-7.987
-7.836
-7.446
-6.830
.007
,001
,843
,569
.216
0.174
558
895
144
267
230
004
565
z(l) = 2: y(l)
z(2) = -2: y(2)
z(3) = -1: y(3)
z(4) « 1: y(4)
- - Stress at Points
B D
ksi ksi
-7.565
-7.004
-6.230
-5.267
-4.144
-2.895
-1.558
-0.174
1.216
2.569
3.843
5.001
007
830
446
836
987
896
565
565
673
548
193
621
846
895
794
578
284
0.049
1.381
670
879
970
909
669
227
565
= 1.4
= 1.4
= -1.4
= -1.4
E
ksi
7.565
7.227
6.669
5.909
4.970
3.879
2.670
1.381
0.049
-1.284
-2.578
-3.794
-4.895
-5.846
-6.621
-7.193
-7.548
-7.673
-7.565

PROBLEM 4.C4
4.C4 Couples of moment M = 2 kN • m are applied as shown to a
curved bar having a rectanguiar cross section with h = 100 mm and
b = 25 mm. Write a compuler program and use it to calculate the stresses at
poinis A and B for values of the ratio rjh from 10 lo 1 using decrements of
1, and from 1 lo 0.1 using decrements of 0.1. Using appropriate smaller
increments, determine the ratio rjh for which Ihe maximum stress in the curved
bar is 50 percent larger than the maximum stress in a strainght bar of Ihe same
cross section.
SOLUTION Writ* h~ loo^^ b^Zf «t*h, M --££#• -*
fdLloUJ^^ t-hTA-Tfofy dp SfC ^,/Sj K£Y/A/ 7A<i? t'Z&LL&kJtrtg*
rz=h + r,
R=h/J*(rrrf)
®
f/rbtWM: rort
rf s /60O To /0° f\T - fCO pEc.Q*tnpNTS
fasojmunT*-! ration %/v^^r
PTfDgFAw a O 77*07-
M = Bending Moment = 2. kN.m h = 100.000 in.
Stress in straight beam = 48.00 MPa
A = 2500.00 mmA2
rl
mm
1000
900
800
700
600
500
400
300
200
100
100
90
80
70
60
50
40
30
20
10
rbar
mm
1050
950
850
750
650
550
450
350
250
150
R
mm
1049
949
849
749
649
548
448
348
247
144
e
mm
0.794
0.878
0.981
,112
.284
518
,858
,394
.370
,730
sigmal
MPa
-49.57
-49.74
-49.95
-50.22
-50.59
-51.08
-51.82
-53.03
-55.35
-61.80
sigma2
MPa
46.51
46.36
46.18
45.95
45.64
45.24
44.66
43.77
42.24
38.90
150
140
130
120
110
100
90
80
70
60
144
134
123
113
102
91
80
68
56
42
730
170
.685
.299
8.045
8.976
10.176
11.803
14.189
18.297
-61.80
-63.15
-64.80
-66.86
-69.53
-73.13
-78.27
-86.30
100.95
138.62
38.90
38.33
37.69
36.94
36.07
35.04
33.79
32.22
30.16
27.15
rl/h
10.000
9.000
8.000
000
.000
000
000
000
000
000
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
Find rl/h for (sigma max)/(sigma straight) =1.5
52.70 103 94 8.703 -72.036 35.34 0.527
52.80 103 94 8.693 -71.998 35.35 0.528
52.90 103 94 8.683 -71.959 35.36 0.529
Ratio of stresses is 1.5 for rl = 52.8 mm or rl/h = 0.529
ratic
033
.036
.041
046
.054
.064
.080
.105
.153
.288
.288
.316
,350
.393
,449
.523
.631
.798
.103
.888
-1.501
-1.500
-1.499
[ Note: The desired ratio rl/h is valid for any beam
having a rectangular cross section. 1

PROBLEM 4.C5
*-E
-*.—
—*■
|
*,|
,,(
—h-~\
1- *i—
,
<L
r
la,
a.
gfiSf
4.C5 The couple M is applied to a beam of the cross section shown,
(a) Write a computer program that, for ioads expressed in either SI or U.S.
customary units, can be used lo calculate the maximum tensile and compressive
stresses in the beam, (b) Use this program lo solve Probs. 4.1, 4.10, and 4.11.
SOLUTION
//y 'PUT: &£*#£? M4. MoMB/^T M
P&H 77= / 7b> ■>?,' /tm7PR h^ amp hy,
Aflffffl* A_, h
->) TiT)
(Pr/ht)
O-^- «■»-,+(**,-,)/* +" h»/%
loco-Ttef &p <z&wrrtc>/& ft&ovt- &f\:£
2'» ^/frRM
I* Ji-41
(P0r>T/
(P#i»?)
COMPUTATION o^ &7&E8S&S
H
n>P
H-$
i i_~
Z&7TOM
q
(P£I"T)
(ffem)
Sfe //sx-r f>/}/£ par? fi24#T0ors ***? P&Sufrt ht, ¥■,&,#.//
CONTINUED

PROBLEM 4.C5 - CONTINUED
Problem 4.1
Summary of Cross Section Dimensions
Width (in.) Height (in.)
2.00 2.00
6.00 1.50
2.00 2.00
Bending Moment = 2 5.000 kip.in.
centroid is 2.750 in. above lower edge
centroidal Moment of Inertia is 28.854 in*4
Stress at top of beam - -2.383 ksi
Stress at bottom of beam = 2.383 ksi
Problem 4.10
Summary of cross Section Dimensions
Width (in.) Height (in.)
9.00 2.00
3.00 6.00
Bending Moment ■ 600.000 kip.in.
Centroid is 3.000 in. above lower edge
Centroidal Moment of Inertia is 204.000 in"4
Stress at top of beam - -14.706 ksi
Stress at bottom of beam - 8.824 kai
Problem 4.11
Summary of Cross Section Dimensions
Width (in.) Height (in.)
4.00 1.00
1.00 6.00
8.00 1.00
Bending Moment - 500.000 kip.in.
centroid is 4.778 in. above lower edge
Centroidal Moment of Inertia is 155.111 in*4
Stress at top of beam ■ -10.387 ksi
Stress at bottom of beam - 15.401 ksi

PROBLEM 4.C6
4.C6 A solid rod of radius c = 1.2 in. is made of a steel that is assumed
to be elasioplastic wiih E - 29,000 ksi and ay = 42 ksi. The rod is subjected
to a couple of moment M that increases from zero lo the maximum elastic
moment MY and then lo the plaslic moment Mp. Denoting by yy the half thickness
of ihe elaslic core, write a computer program and use it to calculate the
bending moment M and the radius of curvature p for values of yy from 1.2 in. to 0
using 0.2-in. decrements. (Hint. Divide the cross section into 80 horizontal
elements of 0.03-in. height.)
T,
SOLUTION My, $j<?^(HZ&l)\tl.Zi^ = J~7 top- '»
Mp ~ \ |-^2= ftzJiu)! (/.zs„f- ?4.s^r:,hl
C?HSlV£fi Top faLF of &QQ
AfT C - WUM&fK op- &.£-/ffwrS //v TOP #/fJ.T
&V\=-H£fd7>4 os= Eacv fLfwe^r: Ah- y-
(Jt. = 5"7??i??r &N £L&/wef->T
^-■?? (Ah)
^ i/ =?" 3y ^° TO
/06
^
(*}■{ o,S)Ah
%
2QO
£0 7Z> 206
- ST/?££S /<*> £A1S7>c c*>££~
■ _$772£<s //y -pLAST/c Za^ST
>
f?FPFf>7
PtfM/T 3y; -Mj WO p.
n
To
&=°
?/?£>£fftM our for
Radius of rod » 1.2 in.
Yield point of steel => 42 ksi
Yield moment =57.0 kip-in Plastic moment
Number of elements in half of the rod = 40
96.8 kip.in
For yY
For yY
For yY
For yY
For yY
For yY
For yY
1.20 in.
1.00 in.
0.80 in.
0.60 in.
0.40 in,
0.20 in.
0.00 in
M
M
M
M
M
M
M
57.
67
76
85
91
95
infinite
kip-in,
kip-in,
kip•in,
kip • in.
kip-in,
kip•in
Radius of curvature «= 828.57 in.
Radius of curvature => 690.48 in,
Radius of curvature = 552.38 in.
Radius of curvature « 414.29 in.
Radius of curvature = 276.19 in.
Radius of curvature » 13 8.10 in.
Radius of curvature = zero

PROBLEM 4.C7
-2 in-
C»-
2,5 in,
4.C7 The machine eiement of Prob. 4.204 is to be redesigned by
removing part of the triangular cross section. It is believed that the removal of a
small triangular area of width a will lower the maximum stress in the element.
In order to verify this design concept, write a computer program to calculate
the maximum stress in the element for values of a from 0 to I in, using 0.1-
f in. increments. Using appropriate smaller incremenis, deiermine the distance a
for which the maximum stress is as small as possible and the corresponding
% value of the maximum stress.
c Ot
A. SOLUTION &&£ /=/<$ V.79 Pfcf ?<??
h~ 3 -cl
ft fit ** (6,t&2)(h/z)
ffllilQ
Zb.
fl
■* %
r
P~ rx~(/f-^)
h*U + A*)
PRD6RM ovrrOT
a
in.
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
R
in.
3.855
3.858
3.869
3.884
3.904
3.928
3.956
3.985
4.018
4.052
4.089
sigmaD
ksi
-8.5071
-7.7736
-7.2700
-6.9260
-6.7004
-6.5683
-6.5143
-6.5296
-6.6098
-6.7541
-6.9647
sigmaB
ksi
2.1014
2.1197
2.1689
2.2438
2.3423
2.4641
2.6102
2.7828
2.9852
3.2220
3.4992
bl
O.00
0.08
0.17
0.25
0.33
0.42
0.50
0.58
0.67
0.75
0.83
rbar
4.00
4.00
4.01
4.02
4.03
4.05
4.07
4.09
4.11
4.14
4.17
e
0.145
0.144
0.140
0.134
0.127
0.119
0.111
0.103
0.094
0.086
0.078
Determination of the maximum compressive stress that is as small as possible
a R sigmaD sigmaB bl rbar e
in.
in.
ksi
ksi
0.620 3.961
0.625 3.963
0.630 3.964
ANSWER: When a =
-6.51198 2.642 5 0.52 4.07
-6.51185 2.6507 0.52 4.07
-6.51188 2.6591 0.52 4.07
625 in. the compressive stress is 6.51 ksi
0.109
0.109
0.109

CHAPTER 5

PROBLEM 5.1
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
A
V
l
*a
b
2k
L
& c
- y
L
o a
At sedio* 8
SOLUTION
Reo-c+to^fi
ZMC = O
nv o
F/*o»*i A
0
LA- bP = o A»» Ei
LC-aP-o C»^
+■08 O < *> < o-
4 2FT, s 0 ^fe _ V =
9lM, = o M-£kxr
M =
Ptx
Ck J.
k-L-x~*|£a.
f St^=o -M + ^•(L-x'l = o
L

PROBLEM 5.2
,*L
6
WL/a
- wl/2
5.1 through 5.6 Draw the shear and bcnding-momeni diagrams for the beam and
loading shown.
SOLUTION
wL
3ZMAr o
Ove*4 njl»oie bea-n O * x * L
6L -wL-|r o
*— 6=^"
A
. w .. ..
nrttim
i
M
?i.
ace
sec+rwA a+ X.
iv
WX
if^m
io<tcl by e<jui \/u/te*&
Coo eeniwtftrt .Pea el •
■rzr,--
t)ZMj.= o
^ - wx - V - o
V = W(t -y)
-^X + wxf + M - o
M- f(Ux-xM
Maki'mwh becttfft'n* r*io«*ie^T oce^i^S *■( X e «"«
8
M
M*,K

PROBLEM 5.3
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
SOLUTION
^rTTTTTT
V
Ar
Lay
♦tlF^o
t)2Mj = o
At x = i
I L
3 M
V
V = -
w0y
V = -
M

5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
SOLUTION
Ose perK** -f» We 1K»Uf of H« jeofi'oH <IS fkf
-Free kx>Jy.
Rep-taee ^t'sff/'t<jIs4 had by t<\HiVa/enf
MFj = o
i> TMj = o
V -- vi(L-*} -*
M ■ r
z
TJ.
us
IM\,

PROBLEM 5.5
5.1 through 5.6 Draw the shear and bending-momenr diagrams for the beam and
loading shown.
SOLUTION
teflLcti'ovxs of lev* ^ecAxct'*«»
£ o r> C t # | wf f *f JPocKpl*
F^o^ A to 8 O < x ^ a
V -~ £w(L-2a) -i
.-*<*-*)
F*t>w» B +o C
w
fca-*)
flnni^
x—^V
a < io < L-a
i ^
E
Jb
b=r
*-<*.
Wa.ce se^i'ow co*f «cT * . iRep^ace di*s4*« k*nec( /ocd' Ly e^i>iV. £ovic. xtxwl,
Fr-^M C +© X>
VtL,
L-a < y < L
A*
X *

PROBLEM 5.6
foad^i5'6 DnW ^ 3hCar "* *«***— ^agrams for the beam and
V
. WCL
M
A 8
SOLUTION
wa
Kx*
y
o -= x, <■ a.
|w*
V
Ua
-wa
+ fZFy = o
-W
c x>
l-x
Wa I |v
wa - wx - V = o
V^ w(a->n
0^! Mj = 0 -wax + (wx^ + M « o
M - w (ax-£*\
Froh* 8 +6 C
wa
t.
wcl- wa - V = O
t)IM3 = 0 -wax + waCx-f) + M = o M- £wa*
Frowi C -ft? ©
w
SI
♦tZl}»o
M
V
0
f |W(L-X^
L- a * x * L
V - \w (L-jO + wa -- o
V ' w(L-v.- a)
M = w[a(L-xVi(L-v^]
=■ o
wa

PROBLEM 5.7
SOLUTION
PROBLEM 5.8
SOLUTION
PROBLEM S.*9
SOLUTION
PROBLEM 5.10
SOLUTION
PROBLEM 5.11
SOLUTION
5.7 through 5.12 Determine the equations of the shear and bending-
for the beam and loading shown. (Place the origin at point A,)
See PROBLEM 5*, 1
5.7 through 5.12 Determine the equations of the shear and bending-
for the beam and loading shown. (Place the origin ai point A.)
See PROBLEM f.Z
5.7 through 5.12 Determine the equations of the shear and bending-
for the beam and loading shown. (Place the origin at point A,)
Set PRoSlsm 5.3
5.7 through 5.12 Determine the equations of the shear and bending-
for the beam and loading shown. (Place the origin at point A.)
See PROBLEM f.H
5.7 through 5.12 Determine the equations of the shear and bending-
for the beam and loading shown, (Place the origin at point A.)
See PffoBLFM r.5"
moment curves
momeni curves
moment curves
moment curves
moment curves
5.7 through 5.12 Determine the equations of the shear and bending- moment curves
PROBLEM 5.12 for the beam and loading shown. (Place the origin at point A.)
solution See PftOBLfiM S.G

75
Dimensions in mm
YU)
2JS
C. D
-0,85
-U5
?
5.13 and 5.14 Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (b) of the
SOLUTION
5?Mt = o 3 = O.SfkN
^ A V--I-5 kN j M = o
V
K-10O ^
M (M-^
ZFv = o -/^-V * <3
V - - t-S *W
A
n.a5u/
Y-150
y^\^.Z5
E P>
At X>
1.5
t)lMj-Oj (l75^/-5) - (7,5)(S.C5-) + M * O Msj/.W W*w
Co-)
r i
e
*|oo -+75 J* too 4
!)
ZF,= o
-1.5 + 3.C5* - /. 2 - V = O
M * 106.2f fcU
At 8
M = o

PROBLEM 5.14
200 N 200 N SOON 200 N
5.13 and-5.t4n>raw lhe"shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum atwoluie value (a) of the shear (b) of the
bending moment,
300
ICO
225 ' 300 '225
Dimensions in mm
too
c b
~lt>0
-300
SOLUTION
A+ 8
At E+
♦ tZFjS0
V - 200 W j M - O
V->?0OrO
V = ^O W
5ra=o - m- (.0^3^X300") = o
M (»-*0
At D*
M
y |* 300 +|*
o
«5-
+ [2F} - o) V +Soo- Zoo --a
-**s
V = ~3oo N
t>2M^O
(^
At C+
M_
^oo
5oo
i 11
V c?WV 2ot) +^1T-
5too
At A
M
Cf
vr^.*>
3oo -^aar+300 4**^*1
+1XFj = o
V - £00 - ZOO +S0O - £00 = O V = 160 U

PROBLEM 5.15
5.15 and 5.16 Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (b) of the
bending moment.
■ T - I SOLUTION
Iffy c j
13 (£.>«
\ZV\fs
C
'A to C
0 < TL < 6 ft
I3*
GV
+t^FY^ o -6 - 3x - V = O
J V =- - € - 3x k,ps.
H = -fix - /.i"vl k>#
C4o 6
-<?<> k,p.ft
Cf+ * * < <\ -F+
(j*
V
r v7
+i£F, ^o
t)2Mw--o
W
? -x
V - 20 = O
V r 3D Ki'ps
-M - 0*-XV3o> * O
M" Sox - 27o k^.ft-
C«0

PROBLEM 5.16
15 laps
5.15 and 5.16 Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (b) of the
bending moment.
2 ldps/1't
SOLUTION
V (k^
A
C
-a
& a
-23
M(k;PflO
+1ZFv*o -V-^x = O
V^-2* K
Pt
A» C V? - £ /<,>* M= -IGk.y ft
At D" i^kip*
A
I.
*
1
+ t IVy-O -8 -V = O
t)lM0ro (SXS ^ - M = O M * - «tt i«> f t
A* 8" * kir> Il5 k;Ps
•*-fZf^ = c> - g - 15 -V-O
v - - » k;PS --,
02M,»» -doX8) - Wis) - M « o
M * - 110 fc. ft. —
(b)

PROBLEM 5.17
3 k.N 3 kN
5.17 and SIS Draw the shear and bending-momenl diagrams for the beam and
loading shown, and delermine the maximum absolute vaiue (a) of the shear, (b) of the
bending moment.
SOLUTION
8 tJ^Mg = O
Dimensions in mm
V(kiT>
ZS5
& b e
o.vs
-3.V5"
(•jokes') - nso +(33o)fe)- roo£>4 = 6
A " ,2.55" IW 1
9ZMa - o -&oo)(z) - HS& -(7oo)C3)* loooS -o
A+ A V - j^ss" w M = o
A VC v = ;.5rw
At C -^ t)ZMc = o
M(*V)
1056"
-(3oo)(2SS 1 ■»■ M = O
C 4° E
A+ 3f
13
-3oo -4x00-,
4 M = O
M r CIS W-w
A+ D*
Z£5
3iy^ t)XMD=o
V
ftfto)(3)
- HSo + M » o
E +6 8
A+ E
"c!
At 8
L300 ^ M' 1035" N-*.
3.HS
Maying |M|" 11X5 W-*i

4001b
PROBLEM S.18
16001b
4001b
I
:■!■$•
c(
J \E ffi* \
,, »
12 in.
t
12 In.
12 in. 12 in.
Sin.
B Sin,
5.17 and S.18 Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (b) of the
bending moment,
SOLUTION
C= ' too A
*1?^ - 0 -¥00 - 160O + (Jry - ^0O = O
ifoo
e F
-**0O
-200O
M (A-i^
(?vr MOO &.
A +> E
E 4*. F
Ff-8
W A o J B M= o
A+ XT 4w DZMD r o
Mj
A+ D+
Hoo
V
J$
M
J M^ Woo ^-m-
tSoo
A4 E
Moo
n-
21oo
igoo
At F
Moo
M
H *-H20o -fb-i*
H- -V8oo A./n.

10 kN/m
PROBLEM 5.19
36 kN
10 IcN/m
l_^ i^a t^t ^.(-^ ^
0.9 m 0.9 m 0,9 m 0.9 m w
v(iuO
5.19 and S.20 Assuming the upward reaction of the ground to be uniformly
distributed, draw the shear and bending-momenl diagrams for the beam AB and
delermine the maximum value (a) of the shear, (b) of the bending moment,
SOLUTION
Ov/v wlioi* \>€«^ +iZ^Fy r O
w^ iskw/*,
M
CkM-i«>
13.1-5"
= 18 kW
- I2./5 kM-hn
AtoC
fflp
1 ' V
0 < X < O. 9 m
m M
is-toti
M x* C
4*?
15 x -/ox- V * O
V - 5x
x /.^ \ x
"a
M - 2.Sx*1
-0S*)| +Go>0£ -» M * o
M - 7.0ZS Irti- i*
X-0-4S
I5UWM
V
r*1v
4*2f;-o
is*
V r K5X-9
+ M =" O
M O"
t> -K 8
M" 7.5 x* - Tx + t.os =o
V • IS kN
M - H8.75 kW*wi

PROBLEM 5.20
3ldps
C
I .1
3ldps
D'
1.5 ft
1.5 ft
Vffci*)
<d
,u
I.*.
k C
<-l.8
3
-1.2
M ((4 ft
-Ul?5
5.19 and S.20 Assuming the upward reaction of the ground to be uniformly
distributed, draw the shear and bending-momeni diagrams for the beam AB and
deiemune the maximum value (a) of die shear, (b) of the bending moment.
SOLUTION
w - o. S k.yZ-Pr
+f 2Fy = o
J<-tf *t2Fr r o 0.8x - V = O
W C"
C fr 0
i3
* \.a ki>i 4 M *■ o.<* k.p-ff
f.i-ff-
c f*
ffrtnttit
fv
m 4frpj»o
0.*y - 3 - V * O
t?2MK*0 -(0.8 *)(£) + 3(x-'.S-)*M *0
M lie c**4e^ tff+l»e kt*** x= 3.75" H
Af C+ V r - I.8 fc.p M > O.? fe> ff
(fit) Ma^drvifl"!

PROBLEM 5.21
5.21 For the beam and ioading shown, detennine the maximum nonnal stress on a
transverse section at C,
10 kN
3kN/m 100 mm
SOLUTION
L)sin«i CB eci a. $rtc b«Jl<
4 v^ o ct5 a. -tree Dortv
200 mm $} ^ (^ = Q
'M - 7.26* 10s N-*
- CC6.7*/0"6 no3
WorwAjr stress
6"»
M
7.2**lo'
6" * 10.8? MP*.
10.8? */04 ft
PROBLEM 5.22
7501b 7301b
5.22 For the beam and loading shown, detennine the maximum nonnal stress on a
transverse section at the center of the beam
ISO lb.'ft
SOLUTION
C-'A k,
Sy Ktv^eTVv
V
12 in.
A= C - l£5o h.
Use M+ Uif «T
t)'IME = o
l&So
M = .5700 it .-ft -- 6g.*/x/o3 ii.Jrt.
Nor^ci stress
c H r ^'J*!0'*
6 = -u- - ^^ = Vo f3(

PROBLEM 5.23
30 kN 50 kN 50 kN 30 kN
5.23 For the beam and loading shown, determine the maximum normal stress on
section a~a.
SOLUTION
By 3yt*n*e4^y A = 3
i3» |5"°
JI0M
*o VV
No<r»**i stVeSS
if, W250X67
Us in* ^tfrf IvcW tff be^vm as -fVr* body
t)tv\T - o
Far W2S0*C7 S= &0 9Mos ^'
= 80^ * lo"4 y*3
PROBLEM 5.24
30 laps 30 laps
(i kips/it
5.24 For the beam and loading shown, determine the maximum nonnal stress due lo
bending on a transverse seclion al C.
SOLUTION
Wl8x76 L)se e^+i^e bcA^ as -£/ee tody
8ZM| - o
- \SA + 02.5X3O") +00)(30) 4 (6X7.^)^3-75) = O
A - SG.ZS kips
Use po<r*fio" AC as £tr*e boely
c£JF*V BcrJin. mome^ a7 C M - IfO.CXT *> ft
F<^ WlSx76 S - If6
3
in
Nofm*J sire*'* <S -
M
^ = ./■« *

PROBLEM 5.25
S.25 and S.26 For the beam and loading shown, delermine the maximum normai
stress on a transverse section at C,
25 25 10 1(1 10
kN kN kN kN kN
|c Jpjg fp J
SOLUTION
Use eivhVe beavn as TVec. bod-
e
S200 X 27,4
M A r 47.5 kN
cj Use p«r4iov\ AC as -Fireft lootfly
- (0.375" X47.5) + M = O
M - 17.8125" kW-w
Wo^ai s+rftSs CT= ^r ^'^y|f , TSXHO* ?* -7** MR*
¥7.5 V
For £ 2*0 x 71A
PROBLEM 5.26
8kN
3 kN/m
1.5 m—+ 2.1 m *
1 3 IcU/*
5.DJM
'■25 and 5.2tf For the beam and loading shown, delermine the maximum normal
stress on a transverse section at C-
SOLUTION
Use po^'hon C& as free body.
£) 2 Mr = o
W3iOX60
8kW
2.1
For W 310* 60 S= £5"/*/0*fc»*/
=■ SSix to'* >oJ
IWmaJ! sfress (T - -^— = ^ 6" ~ Z7.S * JO ra.
= 27.5" MP«x

W360 X64
5.27 Draw the shear and bending-momenl diagrams for the beam and loading shown
and determine the maximum noimal stress due to bending.
SOLUTION
5>ZMt =. O
8 = \£ kbi
C - 44 UN
A +o C"
Beading w<>i*fi*it
Af i*) 0-Oft<O +
M = o
31 UN- m
'-33
C If"
i)IH0=o
h^
vnax lMl - 3Z kWM ' 32*/03 N-
Tor foiicol sleeJ jecJion W 360* 6*1 S * 1030 * lO*5 w*i4
= )o3© * /o*fc *^*
r 31. / MPa.

■: k-x
iiiimnim
PROBLEM 5.28
4 kN/m
5.28 and 5.29 Draw the shear and bending-moment diagrams for the beam and
loading shown and delermine the maximum normal stress due to bending,
SOLUTION
(o.c)Ort + (o.iK*0 + (o.v)B - o
S100X
■600 mm-
-400 mm-
V (100
115 t)2M( =0
U.o)(z) +(o.£)C<0 - (o.4)C * o
C - 10 KN
A +° C 0< X * O.C m
I ¥ fcli//i»
h
3
-/.«
Af C M * -LW krt-v»
*~ M^dt TV ^2M,»o -*» + WxVfW/o)6<-o.4>>
< x M + M - *©
M - -2x* + 8x - C
Fov rt>)Ut{ sfeei sacfi'ow S /PC* //.5
Maniwow noTiwoJr STreS-S
6" =
- iMl - l.?2 *lo
n.cxjo
^- - 35.7 */o' P«
-- 38.7 MPo.

PROBLEM 5.29
5.28 and 5,29 Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to bending.
lft'2ft
SOLUTION
S12X35 C = 52.5 /c.y*
9ZMtr0
B - 27.s k;f>
Shzur
A +» C~ V = - ZS k;fs
Ch> tr V' 37.5- *,«
E4+» B V- -^.3" k;pS
At." C
At D
M * - 2.5* frp-W
25 M <-. _
1L£ S^ V?^:0
S7.5
3 —.
V + M = o
1 22.5
t)5"M6 - o -M * wito.5^ * o M^ iij^-It
ftjft* iMl = HS kfM - 5*o k.'p.jr,
il^ * J. fir ,M| ^°
5 = 3tf,4 «i'
H.I4 k5i

PROBLEM 5.30
5.30 Knowing thai W - 3 kips, draw the shear and bending-moment diagrams for
beam AB and determine the maximum normal stress due to bending,
SOLUTION
By ^y»nmeT»sj
CI w
2 kips I p 12 laps
A +q C"
C+ +• D"
D+ +• £"
EM. 8
A = B
A - 8 * 0.5 fc/p
V - 6.5" fps
V * - L5 k.>i
V* ■ 1.5 k^s
V - -0.5 V.>*
BeiJi
W
n4 rrto^en
At C ,
At
DJty
&S
M - 1.4" fc> H
A4 E M * 1.5 U.*P -F+ ^ a^Mwe-fy
max IMI - 3 k.'p-ff - 3G k.p.in
F»<r tolkA steeS sedio* W 13* 16
S^ 17.' 3
m
IMI
s
36 _
17. I
2.1/ ks*

5.3 J and S.32 Draw the shear and bending-momenl diagrams for the beam and
loading shown and determine Ihe maximum normal stress due to bending.
SOLUTION
A - 40 fcrisi
A +«» C"
DM*B
Bend itf<a m on* ten
V - Ho VU
V - 10 N
V ^ -JO *W
At C 12 M 2)?MC r o
c*
flo
i.V -J
V
+ M - o
ft - GO kfJ-m
A+ D
M<5
V
4o
24
D
t>MD = o
= o
im*1MI = 7aW». n - 7**/oJ W-m
For r*t)tJl sheJl sec+ion W 310/ 3S.7 S * 5¥<? */o3 ww*
1311 MPa

PROBLEM 5.32
S.Sl and S.32 Draw (he shear and bending-moment diagrams for the beam and
loading shown and determine the maximum nonnai stress due lo bending
30 k-N ■ m
C D'
W200 X 22.5
SOLUTION
-6A + UtoVs} + 3o = O
A - 2oW
OlMA - O
A +° C 0 < X < 2 »n
5
ipkv V
V - 2o--?x
-HO* 4 Wx^ 4 M r °
M=- #>x- fciV
Af C
A+ I)'
V= 2 WJ
M- 22 WW-w
A
!8kM
a ^*
2e>WW
7
2M/
8
yVH
F«a roJJeJ stee-P sedlo* W 20O* 22.£
Wort**/ 5Tres5 O = -^— - -—; —ZT "
2o -18 - V - o
V - 2 ky
Dzm0 = o
^ - - 4 j,^. ^
i3</.t> x/oc Pa * lS«/.0 MR*. -**

20 kN
PROBLEM 5.33
40kN
5.33 Detennine (a) the distance a for which the maximum absolute value of the
bending moment m the beam is as small as possible, (A) the corresponding maximum
normal stress due to bending,
(Hint: Draw the beading-moment diagram and then equate the absolute values of the
largest positive and negative bending moments obtained)
SOLUTION
W360X64 RfiAtf+iftW a+ B i)STMc = O
20 a - (iL1)(iu>) 4 (f.o)B * o
B * ZH - S au
X
Bedding Momt^-j ClT *D
0$==? ^M— °
lo
- M *
1.6 8 = O
IV '-a B " 3*** ~ 8a
Ben^f i n«j iMowe^t at C
Me= -2o a.
f
%o a - 38.<* - 8a
For IV 3Co* C4 roPPtd sf«i Wio*
M„* 27. W kw.»v>
5 = Io3o* fos **s
Nor**' sfett 6- - ^ * -ffiffiffi' = *«**»'* r ** M^

PROBLEM 5.34
500 kN 500 IcN
M(lOJ.m^
fe6,«»S"
-66.^»S
5.34 For the beam and ioading shown, determine (a) the distance a for which the
maximum absolute value of the bending moment in the beam is as smali as possible,
(b) the corresponding maximum normal stress due to bending. (See hint of Prob
5.33.)
SOLUTION
18mm ReAC+iO* G.+ A t) <L \\ = O
- A<a- + (5oo)(ol- o.S) - Soo(\ - a.) = O
Aa - iooo a - 750
A* 1000- 2g>
Bend t a.+ C i>7Mt = o
^ -Co.s )0ooo - ^ 4 Me= o
\\-- Soo -&f
\boo~W
1 iso vV
M.
Cf
M
'(l-fc>
5o<?
-l^p -(5t»>)(l-(0 = O
M0 - - «5"0O (l-a/J
E^va+t
- M, -- Mc
CL-^SCCo3 m - 8£€.&3 ** -*
For rtch^oh* ctois seUi*« S-£kb3 - ^■(W)(l«>3r UXC**yl0***

PROBLEM 5.35
5.35 Determine (a) the magnitude of the counterweight W for which the maximum
absolute value of the bending moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See hint of Prob, 5.33.)
SOLUTION
w
+ t IFj = 6
A-^W-^ + 8 - o
I* kip
El W12X18
r
1 3ft
3ft
3ft
3ft
B ^ Z- i
— 3
2-*
* 2
%
3 4_3 -1'
-t- MD = o
Ma - G - 3W
ETfUA-fe
- MD -- Mc
Mc * -e.o fc.p -R- M„ - - Zo kip -ft-.
w*v iMl c 2.0 k.yft ^ 2V k.>-iVf
For W /2*/C vo//eW stteJf s&M** S = /7.1 i»s
IM
Norw*J S+**S* ^ r -^ - f|-j
I.<K>4 ksi'

PROBLEM 5.36
1.2
0,8 Idps
C D1
rips
L.Sldps
■E 6
a 1,5 ft
1.2 ft
f
0.9 ft
5.36 For the beam and ioading shown, detennine (a) the distance a for which the
maximum absolute value of the bending moment in the beam is as small as possible,
(b) the corresponding maximum normal stress due to bending. (See hint of Prob.5.33).
SOLUTION
83X57 OSa.-US)(j^ -«.7Xi.O + (3jt)S = O
C - i.« v 0.2*32*0.
Bcwd;
*ia wtor/ie*
■I ac
o.«
^
^c
B6n«lin« imoiw**t «J D I0*'
I
5
V
2*M0 = o
- Me + O.^ & - O
M£ - I. 36 - 0.2 a
Assume - Mc = ME 0.8 ol - t.36 -.0.2a a = /.2£ -ft
Mc^= - /.0O8 fcp.-ft MF * l.tfo? k.'pff Mp- 0.-?/* /dp-**
Wax Itf) = Loos k.pff « l2.o<?£ fty-i*
No*w*J? 5+fess
6T
■= J*H - I ?■ ore
/.£&
7.40 *«,*

PROBLEM 5.37
5.37 and S.38 Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to bending
300N
\p ll M
M
253
-H3
51. C
-213
87
-3*2
*M.7
-3X3
300 n 2° mm SOLUTION
Ujl
J 30 mm Ne^c j^-f Mf:0 J„» fi
PjMe = o vfc I ]h
o.c H-(a.2K*n><-<k4>Ksoo) - o
N " 213.3^ ^
<tTF, - 0 Vc-#0-3oo * *i3.33 - O
V, - 126 -67 N
Ske«^: £toF V = 176.67 M-*
F 4o G- V * 86.67 A/-*i
G -k H V = -*i3.33 N- »n
B**«/>' iaa home^T *i F
^cdT" Mr -CO.2Kl24.C7) " ©
Severn* KnomMt a.4 £
M^- ^M. = o
Q 1 - M* * (0.2^213.33) = O
]t M& " "2.C7 N-m
213.33
1 j30° |ia4'c7
Free /?otfij ABC0E
+!>2Me = o -0.6 A + (o.<0ft»*W<UK3<w0
- (0.2.1036. CI") = O
^ A * 257. 79 V
+ -JzM»» 0 -t0.2tt3oc*>-(0.*tX30o} -(0.3)026.*?)
+ O.CD * ©
t)
H
257,7* V
|3w>
6 BerJi** moiMtn'f «f 8
i57.7*
?•
V
2l3.*3
C,
M8 = SI.S6 W-m
Bcwlin* wowenf ai C
-(a4)^5?.7«) +(0.Ofloo)
+ Mt * o
Mc r if?.ii /j.ki
- Hp-(o.ayai3.33)- o
Me= - £5-33 N-*i
Normal sfreS5
g - //'-gg c - n.n */o6 Bi
3x10 r 17.1* MP*.

v (k;f^
S.37 and S.SS Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to bending
SOLUTION
Free kxxfy AS +)2M9 = o
— 8* -4vB ^^ = °
V8 -- - u.afcpa
W12 X 40
-38.S
20.Z
-11.2
-it>C + (3*>C5)s o
A C = 49.2. K;r
IE Ol\ = 0
f - /1.3 fc.p*
A4-„ C
O < x < |o ff
't2F„ - o
i«u
r-*-4v Dm/Jo1*
2 -f.8x k:ps,
V- na-CHKio) *. m. & « 20.8 *;f*.
V = - I J. Z kip
M. -^
m*x |M| - 56 -Wf-W * 672 k:p-iw
For*
51.9

PROBLEM 5.39
b
M
5.39 A solid steel bar has a square cross section of side b and is supported as shown,
Knowing that for steel p - 7860 kg/mJ, determine the dimension A of the bar for
which the maximum normal stress due to bending is (a) 10 MPa, (b) 50 MPa.
solution WeiigH elensi'f^ V * p$
U+ L » +.4«J iewj+t of Wv*
C=D= £
If M %>* Mc - O
A^qD cm)♦ m - o
M-Jfr
= O
m*y
lM| =
WL
18
.1*
Fov a. s? uft^e secken S " ^ Id
islo^meii sfress
SoJ?\/e fo+ h
b'LVWn
ft1
m =■ 33.3 mm

PROBLEM 5 40 S4<f A *°^stecl rodofdiameterd is supported as shown, Knowing that for steel y
= 490 lb/ft3, determine the smallest diameter d which can be used if the normal stress
due lo bending is not to exceed 4 ksi.
d
1*1 SOLUTION U+ W- WvJw**jM
^ w = vr - ALr - fJ2LY
■L = ioft— 4 Pcactfow A.*f A
A - -i w
FR -flfft) ♦ «fl*) * " • o
Data: /_' /0-W- ' OAKlo^ = WO ,V>

PROBLEM 5.41
Pt/t
tPcJk>Y -W-
A B
5.41 Using the methods of Sec. 5^3, solve Prob, 5.1.
5.i through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
SOLUTION
t> T Mt = o
t)Z\AA' 0
A+ A+
A fa 8"
vt - o
v-
0
IA - tP
kc- <xr
^4
* x <■ a
s; «j*
* o
* o
M -
- o
A *£*
^ L
o
V0 - V. = O
B
Ve
, £b
e+ +«, c o.< *« l
V0 - VB - o V * - Ea.
i 'c I 'B Li. L.

PROBLEM 5.42
WL/2
5.42 Using the methods of Sec. 5.3, soive Prob, 5.2.
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
SOLUTION
±> I Me = 0
*' -w
r*
SI- wL-£ =o
A- *£
8--
, at
„ wL
- wx
V * VA - */x - A - wv -
JU V
VIa* 1 ml*** V\ occurs a.4 X - -J wle/^.
V- £ -o

PROBLEM 5.43
r^
5.43 Using the methods of Sec, 5.3, solve Prob. 5.3.
5.1 through 5.6 Draw the shear and bending-momenl diagrams for the beam and
loading shown.
SOLUTION
w - w0^.
VA = o
MA = o
w - -
MX
y
2L
V - -
Wo/:
M-M^X V^--S^J,

PROBLEM 5.44
• %
5.44 Using the methods of Sec. 53, solve Prob. 5.4.
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
SOLUTION
V = w L - wy -*
^- - V * w L * wx
M = -
wL*
+ wLy - ^

PROBLEM 5.45
|U-2*>
-#a-24)
5.45 Using the methods of Sec. 5.3, solve Prob. 5,5.
5.1 through 5.6 Draw the shear and bending-moment diagrams for the beam and
loading shown.
SOLUTION
Read-lows A r X> * -J[*/(L-2a.)
A+ A VA * A * *w(l- 2«.^ MA = o
A+oS o * x * ct w « o
«(V-l=) B +o c
a. « x "* L -a.
w =■ W
eho =■ - w (x - a ^
V r j| w(L-2a)- w(x-jCL)r £w(/.-;?*)
. . = ^w (Ux - x*- U+ a1)
M C
C 4» D
A+ x -
X
J.
* L-a
Mp = 0
M^ -
x 4,w(i.'y -
Vc-- -i*(L-?A)
= --jfW (L-^a.)
• «?-*■>
x* - of)
Mt r 4(L-2a^

PROBLEM 5.46
iWfcl
»-*
5.46 Using the methods of Sec. 5.3, solve Prob. 5.6.
5.1 Ihrough 5.6 Draw the shear and bending-momenl diagrams for the beam and
loading shown,
SOLUTION
A \o B
A » "D • wo.
O < x *" O. w t w
V - VA = - £ w J*,
— wax- iwx1
C +o X)
V » - w[ x -{L-cSl
x /• X
M - Mc = $ V U - \ - w£x - 0--^V|*U

PROBLEM S.47
1.5 kN 1.2 kN LSlcN
Dimensions in mm
V (W*0
.&*
2. IS
; e*o
-/.$-
M Ck/">
-tso
(-106 A
/06.55"
5.47 Using the methods of Sec. 5 J, solve Prob, 5,13,
5.13 and 5.14 Draw the shear and bending-moment diagrams for the beam and
loading shown, and delermine the maximum absolute value (a) of the shear, (b) of the
bending moment.
SOLUTION
moo)(l.s)-3ooc +(aasVL3W0as-)(t8) = o
C- 3.cs kw
A to c
C -k D
D *> E
E f> 8
8^ 0.85kM
V- -1.5 kv
Aireatfi of ske<*^
^iV^i^wn
ot she<*^ ^1^3
C4ot> ^VJk "'(2. (5X7^^ - ICI.arW.jv*
K. = Ma+^VJ* - O - ISO r - 150 W-«i
Mo = Mt + 5*vJx - ~IS04\CL2S - H.^AZ-m
hWi'wu* | Vl = Z IS kKi
NWit^otn IMI ~ ISO N-hn

PROBLEM 5.48
200 N 200 N .WON 200 N
300 225 300 225
Vfi Dimensions in mm
Vftrt
loo
&o}
f-2?-5)
-too
c-*o
Soo
-3oo
M (Ur^
67.5
MA * 37.5 A/-**
MG -- M04 £vjx
5.48 Using the methods of Sec. 5.3, solve Prob. 5.14.
5.13 and 5.14 Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (b) of the
bending moment.
SOLUTION
- M* - (a3X3*0 - (o.&zs)&o6) f (M*r&oo)
- (I.aO( ;*«?)*■ d
MA = 57.5 A/-w
VA - 2oo - zoo + 5oo - Zoo = o
VA - loo N
A +<> C V = \oo M
C +» D V * loo - *oO - - roo M
0 +o E V " -too -2*0 r - 3&°KJ
E-ft, 8 V"-3oo+5fco - 2od U
/\v^«t-S Un^lef SnCfti^ d T <a,*» ita ■*»
• 37 5 + 3o =r GZ5 A/*i*i
» Q7.S - 2?.S * ffs A/<*i
r -45 + 45 = O

PROBLEM 5.49
3 laps/ft
MOV*)
-1©fc."p--ft
5.49 Using the methods of Sec. 5.i, solve Prob, 5.i5.
5.15 and 5.16 Draw the shear and bending-momenl diagrams for the beam and
loading shown, and determine the maximum absolute vaiue (a) of the shear, (b) of (he
bending moment.
SOLUTION
02Nc-"0 -tA4 <3Xi8) -OXso) = o
A - -£ k,'p* i.e. C k.ys ^
+5?M, = o GC-C*fcs)-es0Cs<O - o
C - St k.'f» t
VA = - G k.>s
A+oC 0 * X < C -ft. w = - Steps/ff
0 0 I
Are<ts t'ncle/4 sJ»ect^ Ji'a^/^^
A 4o C (vir (£)(-6-2<X^

2 Itipvtt
PROBLEM 5.50
15 kips
-8
-*3
c-^f>
M (kif -FO
5.50 Using the methods of Sec, 5J, solve Prob. 5.16.
5.15 and 5.16 Draw Lhe shear and bending-momenl diagrams for the beam and
loading shown, and determine the maximum absolute value (a) ofthe shear, (b) of the
bending moment.
SOLUTION
V, - O
C +6 D V = - S -. fey*
D fe B V = - s ^ is -- - 2Z kipi.
Av*e&-s uneJ«»r .ske^*" m*.^vtt*^i
A 4u C S V •** ' (i^X-s") - - 16 l6>-fl
n£ > MA+S\/^y = o - \6 - -I* k.>-ft.

PROBLEM 5.51
240 mm 240 mm 240 mm
80 mm
120 N 120 N
7.2 W*« 7.2*)-**
looN
V(W)
I30>1 Iqo w
<*0
-a©
10©
^
,frfc»>
-Uo
C-33.0
£ 57 and S. 52 Draw the shear and bcnding-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (A) of the
bending moment.
SOLUTION
i)2MA= o
- 7. 2 - 7. 31 = O
A = ioo n
-fo.a«0Cw«O -to^ywo^-7.^
- 7 * + Q.7* B « O
Shear
A+oC V - \oo V
D fe B V:^°"'2o^ -/</o M
Air€&5 t^rle^ ske^ ^liftj/^v*
A +o C S v -4* - (o.av)f/oO «■ M W***i
D fe B iVdy '(pMY-lHo)r - 33.6 W-w
Me*- 2H + 7-a • SL 2 N-«n
Mjf * 3/.3 - V.8 r 2C.*J N-im
Hb* =■ 26. «J +7.^ = 33.G W-m
Me - 33.4 - 33.6 = O
M^^om |MlT 33.6 N-i*

PROBLEM 5.52
75 mm
200 mm 200 mm 200 mm
S. SI and S. 52 Draw the shear and bcnding-moment diagrams for the beam and
loading shown, and detennine the maximum absolute value (a) of the shear, (b) of the
bending moment.
SOLUTION
0.O75 Fer - (O.Z X3<kO -(0.£)(3*O - O
:c
2Ho
-rr
too 3oo
*
8
?
3oo
0X6)
GOO
30O
! (60^ I (60^
+tzi5*o
Ay - 3©0 -3oo = O
lva Goo N
Coo?ie at D H0 r (o.^rJCs-Zv/o')
A -ro C V r Coo N/
C +& 6 V - Sao - 3©o r 3oo /V
A^eas u**Je^ s«e*i^ «i*Mf4m
Mc r 0 + ISO -- /HO A/ *i
Mo - '^ + GO = 180 W->n
H0+ * ISO -^o * -6o A/.k,
MG - - So + Go = O
W I - 6oo N -*

5.53 Using the methods of Sec, 5,a, solve Prob. 5.21.
5-21 For the beam and loading shown, determine the maximum normal stress on a
transverse section at C,
SOLUTION
-3A +(1.5X10-) - (\.t)(i.2)(3) = o
A - 2.SS kU
OlMA = O
-6.s)G©) + 3C -ft.. itoaKs}* o
C = \H.6Z kv
-7-42
M(lrtJ.*)
A +0 t>*
D+-K» C
C +
B
V = 7.59-to ? -7.HZ kU
V - -1^7 * 14.02* G.feo K»
V * €.60 - (2.0(3) = O
A^*£ts ciM^le^ ,sJ>e<*/* dfa.*fft*w\
A^D ^v^y * ((.smss) - 3.3? kU-w
D +* C J> VJx - (LS)(-7.^)- -II.13 KW-m
C +0 B iv-lx * (4X*.*Yfc.<©^ - 7.*e fcu-m
M^ - O + "3.87 r 3.27 ItM.*
Mt - 3.87- »Ufc« -7.JUk|0-M
Me = ZZ6 - 7.ZC - o
IMcl = 7.?6 lef-* * 7.?6x/os W-^
F«** ired
cut ac or«L^ C^dS^ £€
-f;
on
r CCG.C7x|Os m*? - 6GC.67></»~' *>l
H
OfMfl
;s+
ires*.
e =
_ IMcl __ 7.26*/0S
C£6.6?*/0'
= lo-«9 MPs.

PROBLEM 5.54
750 lb 750 fb
150 lb/ft
5.54 Using the methods of Sec, 5 3, solve Prob, 5,22,
5.22 For the beam and loading shown, determine the maximum nonnal stress on a
transverse section at the cenier of the beam
SOLUTION
tfeae+t'oMs : C - A fey sy^mcVy
A * C = /£Sb Jfc.
vc- = /t$"o - WOso) = ioso ik
Vc* r |a£fc- 7-5"o * Zoo tk
M(JUV>
S7ou
Mc * S700 &-W
Av^ft^ owWe/' ■sl^eo/' eh«tjr«i*i
Me - -Sfoo 4 300 ~ ^7oo A-ft
P«»r rec4a*]ci4* c<*>» secfion S * ffch* * (fK3)C«)*- 72 m*

PROBLEM 5.55
30 kN 50 kN 50 kN ;30 kN
VOckrt
(«rt
So
!M
So
M (kM-w)
5.55 Using the* methods of Sec, 5.3, solve Prob. 5.23.
5.23 For the beam and ioading shown, determine the maximum normal stress on
section a-a,
SOLUTION
k h> C V = go kW
D -K* F V r ^o-^o * o
rWea$ aM«lc^ sViea^ ©lia.g^eiwi
A tb C $V^v = (tf&Xo.sV £* Hi-*
t> +» *E iV^ = o
M* = o.
Mo « 6»t + ¥o « |o«f kU-m
MK « IO*t fcTK* - *n r /0«J*/O* Hm
-*. a
F^ W 2£0*C7 S* SO? x IO hm* = SO^h/o *

PROBLEM 5.56
30 kips 30 laps
5.56 Using the methods of Sec. 5.3, solve Prob. 5.24,
5.24 For the beam and loading shown, determine the maximmn nonnal stress due to
bending on a transverse section at C.
W16X78
SOLUTION
A ' St>.2S k.'f»s
SWir A +« C V* 56.35 Jty*
Are«i u^olc/1 slice*/1 c^^« A'+aC. j\/dv* (56.ar)(9.5^
Mc - 0 4 140.«f - |Ho.62S k-p.lt r 1687.5 k.p - im
For W ISX76 r»ife<f a+eei s««+f
Mo^mAJ? s4<ess G"= it * IC>8/'^ r 11.56 If*.'
/+£

PROBLEM 5.57
5.57 and 5.58 Detennine (a) the equations of the shear and bending-moment curves
for the given beam and ioading, (b) the maximum absolule value of the bending
moment in the beam.
SOLUTION
W ~ W.
o-^
£•-•"-<".*¥
V = - wQx
+ c, - 4*
<Jx
M = o ^ ^o
M - o Jt x - L
Ct-o
W./.* . w.L"-
°f-f +
4 U,Z- " ' 3
*U
2. + GL
M
i s h^Axtm^M wK«re
u,
o -- - vv, x„ +
XL 3
?xN - Lx* 4 iLl - o
x*r
* (l±|U
-W.(o.'UacsU* w0(o.H224SQ3 w.Uo.'U^O
M-' - 2 + CZ. + 3 —
'WWUt
r O.OG.4IS W9L%

PROBLEM 5.58
w - u\, sin ^
S.57 and 5.58 Delermine (a) the equations of the shear and bending-moment curves
for the given beam and loading, (b) the maximum absolute vaiue of the bending
moment in fhe beam.
SOLUTION
M - o J x = O
\A- O J( X - L
w - W. L ][X *
V - ■ Cos ^ ■+ t-i
Ski
M -
^
w„l* .. try
V » o «i
L
X- 2
w.Lfc
= IT*

X S9 Determine (a) the equations of the shear and bending-moment curves for the given
beam and loading, (b) the maximum absoiute value of the bending moment in the
beam.
SOLUTION
4*. -w . -w^)±= -**
»/*
.3/1
O « -§ W0L4 Ct
3 L*
V* |wPL- §
2- WaX
3/»
w.
M< C, * |weLx-|.|^^
3 -•— 3 ^ ^
2
M -O- *f x^ L
£*L*
C,' jw.L
<V--?«u*

PROBLEM 5.60
-kw0
S. 60 For the beam and loading shown, determine the equations of the shear and
bending'inoment curves and the maximum absolute value of the bending moment in
the beam, knowing thai (a) k - 1,(6) k = 0,5,
SOLUTION
w„x*
^: kw,X - Cu^^i + C
ZL
Vs O ^ x * C»
C,= o
^6
le^x - Of^^L
M ..***!_ (ul0*£ + C%
€L
M V • \
V = w>„ x - ^J^*"
M =
w„x
wcx
M
d-jC i»k;^
M
a. 3L
\^l^z
, J.
o.^ k= i
M -
wfcxL
Wq XS
V = O **
X" =
fL
Atx-ft M-M^-^= *£* 0.03701 -U*
Mx= L M » O

PROBLEM 5.61
5.61 and 5.62 Draw the shear and bending-momenl diagrams for the beam and loading
shown and determine the maximum normal stress due to bending
9kN
12kN/»i
| | 1
fffiLi'iiim .-■-■hC^ii ---J&ili
0.9 m
W200 X 19.3
-3nr
V (kW)
[\
C E
a
<3-C *
-tf.3
M(kp.M^
1.7S
SOLUTION
(o.s)(<? ; - (j.sXsX'O +38:0
d - **?. 7 WW
8 V - 20.7 -(3)0a') - - iff. 3 kbi
e * r.72S--Ft 3- € * 1.275 H
2t>.7 IS.%
-K
Al^€Ct,S U^«*£^ Snea/- cfi'<^iA>»**
A 4* C ^Vjy = (0.9 X<0 * 3.1 M-t*
MA = o
Ma - 0 - *. / - -*. / J«/.»*
fl, »-*.!♦ 17. 2SZ1S r LISAS' k«-m
Mft " ^.75375 - ^.7S37ST = O
hW^ strew 6T • ^ • ^g^ - ^'^ P. * ■**.* MP* -

PROBLEM 5.62
- 16lcN/m
5.61 and 5.62 Draw the shear and bending-moment diagrams for the beam and loading
shown and determine the maximum normal stress due to bending.
SOLUTION
OZMB = °
02M* ■ O
S150X 18.6
V
a
-•?.% kv
C 4* S $VJy^ ftX-7.sO - -7.X kW-m
Ma * o
Ho T 0 + 8.83. * 8.52 k^.w*
Me- 7.4 -** - o
For S ISO* IS. & rJie«( s+eei se«+i©n S* t2O*/0Sm»? * IfO W© »»

PROBLEM 5.63
3 kips/ll
5.63 and 5.64 Draw the shear and bending-momenl diagrams for the beam and loading
shown and determine the maximum normal stress due to bending.
SOLUTION
10 in.
- 13 A + CffXBXs') - »a = °
A - \s w;Fs
B - ^ k;Ps
SV>
Cav'.
V. = \s k;**
VLr is - (gtfe-) * - **?*
C +* 8
v - - i k.-,
P*
Locate poinT "D w«evc V - o
*-** = 3"f*
Areas c^Wfe^ skto.^ aiV^vm
A 4* I> JV-4x r (iX-O^O - 37.S- *.)»■**
C -H> 8 5v4x ■- CAM) - -36 fc.).ft
Mfi r O + 37.S - 57.5- k.>^4
Mc = 37-S - i3.5- r 34 fc'fft
M6 * 2f - 3£ = - \Z fc> -ft
M«V.*w*» IMW 37.T klp-H * <*SO feip-in
F»*r rte4«,MjaAir c*>ss see-/?©* S " £ fcK* - (iW3)0c)* =r i:o in*
Normal sfress 6" - -^ r -jg— = 9 feaf -*

8 kips
W8X 31
M (kif .-H)
For W 8*31 ^oJjfe^l s+eei se<?fi'o*
5.63 and 5.64 Draw the shear and bending- moment diagrams for the beam and loading
shown and determine the maximum normal stress due lo bending.
SOLUTION
+>M0 = o
A = g k.ps
-CsU«)(0 MUD- O^KO " o
ne. - 43
A^eccs uv^e* skea* eli'*jirw,m
Afo£ $V«lx - &>0O(8) = IC kr7ff
C +» *D i V Jiie =- CgX-O - - z</ fcp.f^
o +- b Sv-»* * «XO * ** *rW
e _ G-e
8 " »t
e * 4ff
IMI
S
. JQ1. -
* 37.5
6.9* kn"

PROBLEM 5.65
S. 65 and 5.66 Draw the shear and bending-moment diagrams for the beam and loading
shown and determine the maximum normal stress due to bending.
i4la
-C.&
M(ifa>*)
\t.\s
SOLUTION
C+ V -- -?.** + /o.a - 7.8 k.y.
B ^a r 7.8 - OsXi.a) - - 6.6 k,yt
7.-3 fi-t
oi - 6.5 ft l*-*l - 5.5 -ft.
t\*e.aA 0<ndfi** S^CA* ati^/«ii^
A +o C $vJx -- &H-M) * -rz kr*+
Difi i VJU - (i)fr.5y-<.6)- -/SJS k.p.ft
Qcnii n^ Moments MA ~ O
Mc - O - 7.2. * -7.2 if.'p.fl.
MB = -7.^ + *5.3£- = IS.15 Kf-ft
MB - IS. IS - I8./5 - O
VWiwJw IM\ r I«-I5 kf.fi -- 2/7.8 k.y.iV
'of
p.p,
r = $(c/-c,M- W-3.S')-- **.** «
v
5 3.^m
83.?o
- 20. %o \v?
W*r««J cUift 6T = ^-~ 4ttt * '0.47 ks."
Z°.to

PROBLEM 5.66
S. 65 and S. 66 Draw the shear and bcnding-momenl diagrams for the beam and loading
shown and determine the maximum nonnal stress due lo bending.
800 lbfin.
8 in.
Oq*JI)
-?.29
2%.71*
I- 1-25 4
SOLUTION
+t>rMfl=o -;?oA + (GKM)ft*0 - o
A* £.7**it* A.
t)lM^o 20 B - UmV38)(«oo) = o
B • I5.6S* 10s it.
Shea^: VA - 6.73 xlo* A.
B" Va~ * £*72*(Ox - t?oX*oo) =- - <?_P8 Wo1 A
8"" Vf* -M^/oNG^Sxlo*)* 6.*M©*A
Loe*T«- P#'wf T^ w^e*^ \} - O
6.73. <?.*S
A^e4.s UrtJar slitct* «Ji'<M/*tw\
A+oD £vJx*(4KmX"W)* »*M*/o* A in
8 J* C S V^*r (iKaX^"/^)-- ZS.Cxio* A- »V
Ha- WMWa^-S&zwxto1 - -351£*/o* A-m
Mcr -2S.C*lo* + *5.6 *to3 • O
-|t ce*7'*iv| c*+ croft's SteTi'ow
7 T J-5- ■" *.£333 in. -Pro^ fc)o-rfo*K
6*
11.58 ksi'

PROBLEM 5.67
250 kN
J 50 kN
JF-
1
D
~&mK&~
2m
*—2m
J*
2m-«
ISO
V{MT>
5.67 *nd 5.68 Draw the shear and bending-moment diagrams for the beam and ioading
shown and detennine the maximum normal stress due to bending.
SOLUTION w - O
W410X 114 ^A " £° kti t
OJma = o
-**<?• -fcX^-fcKiS'O = o
Skw : V* * So kV
A +* C V * 5*0 VW
O \» B Vr -aoo + 3so = /S"o kw
A 4o C $V«U- (so^aV loo ktJ-**
O fog 5 VJv * (iso^a") = 3oo kw-^
Mc = M*-*S\A=be - 0+/00- loo W-»*
Mo5 Mt+$V«b<~ /oo-Moo= -%oo U\)'v^
Mb - M0 + £v Jx * -3«H ioo r o
For W4lo*|iq ro|/«J s4eei sec-fc©* Sy - 220o »/o3 m*N 2200*/d' **,*
S*>
0*O}
(-H«0
-aoo
IS"o
(3do)

PROBLEM 5.68
5.67 tnd 5,68 Draw the shear and bending-moment diagrams for the beam and loading
shown and determine the maximum normal stress due to bending,
2kN
SOLUTION
4 kNVui
•*— 600 mm
400 mm
*>lMg r o LoMQtV (o. i )(h ) ~(ox)Ca) = o
c 2£ R* - V'W i
S100* 11,5 pXMA = O (0.4)(f?.V&ffW>- (ri)£0 * O-
*>
jr ft t
2 kw
J*?
a
-J*.6 fcM
At x- 0.4 * Vtt ^ -5.6 *hl
£ 4© C 0.4^* X* 1.0^
- 1.12 kU-|r*
Fo^ 5 100 * II. S roi/e^i s+eei seo+ion S„ = 4?-6x/o* ^J - 4?.6x/o"c vw*
6^
TT-- ffig ' *""*«* ' *"M^

PROBLEM 5.69
5.69 Beam AS, of length L and square cross section of side a, is supported by apivol
at C and ioaded as shown, (a) Check thai the beam ism equilibrium, (b) Show thai the
maximum normal stress due lo bending occurs at C and is equal to w0Z,V(1.5aJJ.
SOLUTION
■w
^
©
M
a^LI ntutm
IMI » £"-^
Fo^ tfvar* cross seo+io*
V * k w0 L
-x ^o
IFs* o
ZM^o
0 = 0 etfiji/, p**"uw*
- v = -
V= Oj M * o
efM _ ,, - _ */**
he
2L
- + C - ^"^
M ~- -
2L
y 2L *
Ju* + f> fU ^3W of C
oA C
1^0"
°l" " T
a/ a5
*7 n* 'Kit Q* * 0-5a.^»
i a.
*7 q

PROBLEM 5.70
5.70 Knowing that rod AB is La equilibrium under the loading shown, draw the shear
and bending-moment diagrams and detennine the maximum normal stress due to
bending,
wQ = 50 lb/ft
Jin.
fcl
f^
1.2 ft
-1.2 ft
V (JPW
M (JMt)
SOLUTION
A +o C
O* %< \.7. -W
W - SO (l - ft) ■ So- HI. 6£7 x
H - - W r 4/. 667* - SO
= O ■+ 20.813 x%-- So* ~ j
M - MA + ^Valy
At x- l.a -tf, V -3o JL
C "h>. 8 Use £4MH>i»firy fionJif/enS.
NocM,J s+rtu e- -J-- E.s^^teY ' G.lS*tOp*t
= 6.15" fcsV

PROBLEM 5.71
'5.71 Beam AS supports a uniformly distributed load of 2 kN/m and two concentrated
loads P and Q, It has been experimentally deiennined that the normal stress due lo
bending on the bottom edge of the beam is -56.9 MPa at A and -29.9MPaatC.
Draw the shear and bending-moment diagrams for the beam and determine the
magnitudes of the loads P and Q.
2 kN/iii I I
(ft V
A C D |]
U •-}"" *■ ■* H
IS mm
n
36 mm
0.1m 0.1m 0.125 m
HoqN
C,i
Mf
A
C D
a
B
UCM
c
a
X$°
-22L2S- W•*
SOLUTION
M C Mca S€Tt
m*-- (vsis^o-'X-^-^ -"6.*5 *•■*»
221.23 - (o.lVSoo*) - 0.2P- O.VSO = O
1?» - lOO - SOO - 25*0 * O
VA - HSo M
Mt"- - H6.2C W-*
V0 -- 2S"o

* 5.72 Beam AS supports a uniformly distributed load of 1000 lb/ft and two
concentrated loads P and Q. ll has been experimentally determined thai the normal
stress due to bending on the bottom edge of the lower flange of the W i 0 x 22
roiled-steel beam is +2.07 lesi at D and +0,776 ksi at E. (a) Draw the shear and
bending- moment diagrams for the beam, (b) Determine the maximum normal stress
due to bending which occurs in the beam.
SOLUTION
For W 10* Z\ rJhA s+eei sech'o* S= 23.2 m$
BeniTrtj mo^e^"t& a.4 *D and B M=-SS
M6* (2S.2Xo.17CV IS.a|m « LSO k.p.ft
DsiriQ po^Tion Dfc as a. 4Vee body
.. r....... j*i, vDr ^ it--,
PROBLEM 5.72
» k.v/H
1000jWft
V
2ft
D E
—5 ft-*
" ■ " 4ftV-
3ft
W10X22
B
Mo •»
AIAill
0
ve
U &<>><* porTiow ACD as <*- -free bo*Uj
UlU^
- ' TV.
tis = °
Ust«* pov-fioo EFS cts, a. -fV^c co«L
V ——-n-E- -»B
+i rF"~o
Sr1£<^- Al» C Vr - | k.rs
C* V^ - / + s • H k.ps
F' V= f-OoVO ' -6 k.>»
F+ V - -6 + <? * 3 k;f*
F +• 8 V/ - S t.p«.
2P ■» 0X^0 W M, - 2 Vfi * o
P - I k.p
-*P+ C - teKO - VD - o
Q. r 3 k.p*
F^V6- (3VO- q = o
F - 9 it;rs
V (k.'r»)
a
ft
foe * Ho
U = V ff |0-y:6ft
Coiyfih\JeeJ

Probfewi S.17 Con+i*we4
A+o C
C+. &
F 4* B
**w
SenJi<
9^ She.a.tr cHio.^/'.AVvv
Mc * O - 2 - -2k.p.ff
M» - - Z * S « G fc> ft
Mr ■" 6 - ** * -«fc>.«
^
IA.
Wo-^n-ifiJ ST
r*s*
e -
S
23.0.
G.Z) ksi'
PROBLEM S.73
* 5.7J Beam -45 supports a uniformly distributed load of 8 fcN/m and two concentrated
loads P and Q. It has been experimentally delennined that the normal stress due lo
bending on the bottom edge of the lower flange of the W 200 *52 rolled-sieel beam is
100 MPa at D and 70 MPa at E, (a) Draw the shear and bending- moment diagrams
for the beam, (6)Determine the maximum normal stress due to bending which occurs
in the beam,
W200X52
SOLUTION
For W Zoo*SZ toJJeJ siecf se.ct*»*
8eh*li"«j iM»»*eATs a4 *D am J iT M - -S6"
0.1
+ fZ^ =o
con+fn J fid

Pwb.Pe^ S."73 co^\nOeJ
*-0.<tf—»k>3
♦v.
+ 2TIJ*o
u
• I'nj po'-'Vlow ETF8 AS &.ffit* koAy
Mi
fa
c
uuiiuu.
*0.3 »|*0.
MS"
■» 2 Ma :0
+ fzF^--o
-O.tiS P - (0.37S-)f0.75-)(ft")
- 6.15 V* 4- M,» * O
P = 13/-2*2. ktJ
A - 1*3.7*6 icW
- 0.7ffVf - Me * O
Q t HO.*. kN
V, - Q. -Co.75"V«*) f e = o
8 = C6.S67 *K/
SVie^: VA = 123.7^6 ky
Vc- = i23.7s-6-(o.^rX«)- \w.\sfkn
Vc+ r |2o. 153* - 131.«Z "- 11.067 kw
VF" - -11.067 -(I. ffXB}* -*3.067 kU
VF* = -23.667 - ^-2. r -C3.W? W
Va r -6S,276-(p.*r)(8V -66.*6?kW
Ar«A,s uneJe^ Sfiea'1 Wra*^*"
F h B i(p.45X»6&aei)-tt.«t7 ) « - at. A* W->»
Mc^ O + SH.83 « £%*8 *W'*
v(uu>
— 13L3.3
M
N
^yimi^w
1M^ r SH.M k\)*»s * 5^.88 W<>* N-*
C>rhn<l
i s+
-atre**
rr M, &M*lg r l01.7*lo'?K
S\7. xio
■ io?.a MP<w -*

PROBLEM 5.74
* S. 74 Beam-45 supports two concentrated loads P and Q. Ii has been experimentally
determined thai the normal stress due to bending on the bottom edge of the beam is
+15 MPa at C and +22 MPa al E. (a) Draw the shear and beriding-momeni diagrams
for the beam, (b) Delermine the maximum normal stress due to bending which occurs
in the beam.
SOLUTION
60 mm
-VM»
Ufciiftj portion AC fci a, fire* bocly
A - Vt ^ O
Vc * 720' N
=3
v<
4r^ =©
Uftj'n^ po^fion CDS' *s «. -fWt body
M ft/.nO
if
Isi'not porTion £ F 8 fi.% A 4V«« l»«<4*y
P «
I6 „
Areas u*#lev ftlieo.^ di^m**
A 4* D («45X74p) -- 36© M♦*
Dt» F fc>.O(*tM»0*-7a W*m
1M|
aao.- oa vt - Ms = o
v,-ft + s*o> B » mo n
N
orvAft
i s+
re*S
S*
36 o
H.4y|o
— = £f*/06Pft. * ^Mftc

PROBLEM 5.75
1.S kN 3.0 kN
40 mm
0.8 m 0.8 m 0.8 m
vcavn
(IM)
<
<\t
/
(-AM)
5.75 and 5.76 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 12 MPa,
SOLUTION
A « ?. M *W
-(flt*X!-8) - (U Y3.0 + A« D * O
Ct>rt^+rt>c*f sliest «»n<;l btf^di iri«% ^DMe*/"
^OO */0~* w^
~ %0O *\0 mm
h r 173. a m*

PROBLEM 5.76
5.75 and 5.76 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of i 2 MPa.
10 IcN/m
SOLUTION
Altj .-o
|MU, ; 3/. 25" fcjj.hn ra/.^M/o' AA**
S^* "sST r who* T ^.^,/o *
Y*w\

PROBLEM 5.77
5.77 and 5.78 For the beam and loading shown, design me cross section of the beam,
knowing thai the grade of timber used has an allowable normal stress of 1.75 ksi.
HH
!!•
SOLUTION
1,2 kips/ft
M.O
1 4 A
"•* r ^.37 inJ
Jr—
PROBLEM 5.78
S.77 and 5.78 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 1.75 ksi.
200 lb/ft
SOLUTION
V, (if)
(- 7S»o)
M(iu-m
-- So ic.>M«
S^ » 1.75" In.*
3o
- i7-i43 ;*i*
S*itV»a» ^bCaty- fb^ ibis's
V3 C3V17-N3) . s
b = ■ jT *^ 5-5.T ■« ^
-25"oo
b - 2.V In.

PROBLEM 5.79
3lcN/m
J, 79 and 5. #0 For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an aliowabie normal stress of i2 Mpa.
SOLUTION
O M6 - o
150 mm
M(kvM*)
-*.ȣ
Shear! VA = 2.7 kW
V6-- 2.*7 - U* KS") - - *-£" lew
\fc ---4.5*i 2. I - 3.6 Vw
Vc r 3.C-0.2K3) = O
q| -- 0.9 Ho ,?.**- ef - 1.5" I"
Aree^S On^l&r sne«r ci>r^e
M* - O + /.3 IS" * /.2J5 kW-*»
<5^ - W MP* * ««/o* Pq.
I ..
iSO */Q
^ ISO*
*»m

PROBLEM 5.80
S.79tndS,S0 For the beam and loading shown, design the cross section of the beam,
knowing thai the grade of timber used has an allowable nonnal stress of 12 MPa,
2.5 kN
2.5 kN
SOLUTION
By symmeT^
y
BrC
+ TZ^»o
0.6 m
0.8 m
-*.5"
8 + C + 2.5>2.S--<3KO - o
B - C - 6.S kV
She***: A +o B V = 2.*$* few
Vt- = 9 -«XO " -? *M
C +o D V" - 9 + 6.5" = -ZSkV
Arftc«.& u^de*'* sKeeor olr^v^^
A+o6 SvJy =(0.6)CUO r 1.5" JdJ-hn
8 +> E SV-AytfjaOOO = G.Kkh).*
C +» D SvJy - - t.S W->*
Betid 11^« Ht ant eV>T5
MB r O
IS *
M*» o
M£ = l.r + &.7S - 8.US' Hi-i^
M, » i. S"-i:s - o
May.V^ |M| - 8. X5 kV-w, *• 8.4£*/os fJ-i*i
S1j» - 1? MP* ^ W*/o* P*
£87. £ * /o * mJ * 687.-5** /o* »v»^s
F«** a. re or aw jo Air secTtom
^ -L LL«
S- *k»i
**_ (€Vtg7.^K|oi) _
loo
VWSxIo* m^

PROBLEM 5.81
90 kN
5.81 and 5.82 Knowing that the allowable normal stress for the sieel used is 160
MPa,.select the most economical metric wide-flange beam lo support the loading
shown.
0.6 m | 0.6 m
0.6 m
SOLUTION
■L>TMB = ©
Otvo
V (kU">
-3.CA + (3)0»oW2^fro) + (Utoo) =0
sv»
e^r
A4oB V ? l«o VM
D * E V ^ O - ?© ^ - ?t> fc-«
Av**«is UnJc/^ sY»e«^ diA.^/*^
0 ■!*=» E SvJ* " O.^-^ * -it? ItUr*
Me * l«- 144 * O
6ut * iGo MP*. -- /6o*/o< p^
SVto.pe
W Ht0*7*
W 4lo v6o
W 3CoxC1
W 310*7**
s Go3**3*;
lM6t>
l06t>
1036
I06o
Lia litest vim £/*.*ge k>ea<^
W HIO* GO 0 CO fcj/m.

PROBLEM 5.82
50 k-N/'m
0.8 m
0,8 m
vain
M (|AJ-»T)
5.81 and 5.82 Knowing that the allowable nonnal stress for the steel used is 160
MPa,,select the most economical metric wide-flange beam lo support the ioading
shown.
SOLUTION
B - \Zo kv
OlMa= O
3.3. t> - (0„%XZ-Z)CSo^ = o
sv,
€«/" -
Va-0
L
e
go
4o
VB- = 0 -to.8)(so) « -Ho kw
Vtt+ - -Ho + l3o = 8<? AV
Vt = Sto - C?.Y)(So> -HO k\J
Vo = * 40 + O * - Vo fcw
/So e - 1<?3
e - 1.6 m 2-H-e - o.g M.
6 fo E ^VJv-(4Vl.6V8o) r 64 W.*
E +o C S VJy - (i)(o.2X-*0 ^ - 16 IfW-.*,
Ci, O SV-ly* (6.g)(-*>) r -3**1/-^
M r 43 - /6 "- 3ilfeV-*i
h0 - 3a. - 3^
VW»i-^M iMl = 4g kU-i* t 4»*/oa M-Ki
6",jf * /Co MP* » ttoxfo4, ?*
SLtt
l4o v lo
-C 3
3co x/o* »««
.Sh
o.flfe
W 3/o"32.7
W 25"0*2S.4
■Sflci-W^
4/5
3o*8
3*U

PROBLEM 5.83
24 kip*
I
2.1S lap&>lt
5.83 and 5.84 Knowing that the allowable nonnal stress for the steel used is 24
Icsi,.select the most economical wide-flange beam to support the loading shown.
SOLUTION
9 ft-
8
15 ft
Shape
W 3o»ffl?
wri *w
WMk 104
>M t\* \o\
W It m 106
S CmO
2*1
*i3
2S*
Z21
2oW
A «■ f tf Vf*
Ve- - *f8-ftXfc70 r a.& krf«
vt - -o/7£ '-GsXa.^O * -4awp»
Areas onA&r- shear <}\*iyf*L**
A fe Bf SvJx *(iy^X««4»^)»3?o-6 k.p--jH-
MB - O + 320.6 = 3*0-6 Jc.> ff
Mc * 350.fi - 3*0.6 * 0
M^xiwt^ |m)- 33o.6 k^-ff r #SVS A./i - im
\HXI*2H @ SH AM -*

S.83 and 5.04 Knowing that the allowable nonnal stress for the steel used is 24
PROBLEM 5.84 lcsi,.select the most economical wide-flange beam to support the ioading shown,
0.5 Id^l't
M
A^f | rviift*
M
1.5 kip^/ft S0LUTI0N
<N - 0.5 + O^-^S")* * O.S + O.OS55 X
^'-w.-- 0.5 -^ 3.05556 x
AX
Y\* 0 - a:*?*'1- 0.00135*1 x*
6^, = 2? fc
Oi-,v ~
•*t(T»
SVi^pft
>NZl » H*
VJI%» 5?
W|6 *S1
WH*53
W 12*7;?
W lov &S
5 C-*')
21.£
SS.*
c\2.Z
77.%
37. H
"M[7
^*._, G7.s*,a
W i**^o eio A/ft

PROBLEM 5.85
5.85 and 5.86 Knowing thai the allowable nonnal stress for the steel used is 160
MPa,.select the most economical metric S-shape beam lo support the loading shown,
40 kN/m
v(uvn
M OdJ-M,l
-72.75
54.W5
75 IN SOLUTION
A -- 7t.as WW
O - 7Z.7S kW
SWcor* A to 8 V - 7**.^ kW
B to c V = 74.ar -73- = - o.is kw
Art*.s iM^le/1 s^ee^ ©u«.*/'4.<n
Bto C 5vJx'(0-<?W-O.7S-y--. 0.67S kKj^
C 4* D (iVl.8 X-0.75--7?.7S) = - 66- »S* W- *
Mbt o ± U.MS = 66.ft?5"kK/«i^
H0 r 66. IS" - 66. IS « e>
6jf »6o HPo. * I6oxioc?o.
S 310x17.3 e *7.$ I*> A*
" " Cum "
S 380 * 6^
S 3)0 * H?,3
S ZSo * SA
971
5<?3

PROBLEM 5.86
5.85 and 5.86 Knowing thai the allowable normal stress for the sleel used is 160
MPa,.seleci the most economical metric S-shape beam to support the loading shown.
SOkN
100 kN/ro
mmmu
-Wo
-ZS6.
■SWai
SOLUTION
A^cc«.&
.k.
<*•
B+&C (iXuVsnotso^r %s<> k\J*»i
o C* «t na t*io tneivf-s *
M,
Mc = -ZS* + ^6 « O
S^y " 16© MP*. * /eox/O4 Pa
S^ -
JM1 _ 25fcxto*
6U<
' l4^#r ».tfWo-*HS= IGOOWOmm
5 5"Jo * ?8.S
S(V^)
I 'ISO
UtS
Lfjkfey} S- see+i«*

PROBLEM 5.87
3bps/ft
Shape.
5 12x31.8
5 k>*3S
t>
5.47 and 5.5* Knowing that the allowable normal stress for the steel used is 24
ksi,,select the most economical S-shape beam lo support the loading shown,
SOLUTION
A r ? k.-rs
C ~ 71 k.y*
Cf»D V = -*? + 27 - '8 Kps
A*e«*: A +» 5 <1>C?K^ ' \%.s k/p-Ff
E +. S (t>(3X-Tl = - 13.5* fcp-ff
Mf = O -f is-5- * I3.S- k.> f*
MB - 13.5" -IS.? - o
Me. - O + 5** * ^1 k.p-ft
M*yim*n )Ml - 54 k.>.-ft " CM? k.p-i*
Rjb - V\ k»i
S =
£21
3V
27 i«a
SQ^
34.4
2*4

PROBLEM 5.88
S.87 and S98 Knowing that the allowable normal stress for the steel used is 24
ksi,.select the most economical S-shape beam to support the loading shown.
■*—►-■«—»+• 6 ft
2ft 2ft
V OupO
Sit
i.c
_,8 -3*.*
v\ (k.> .fO
7<.«
SOLUTION
E - 38.4 k«>«
SViear-: AV» 8 V- - V& ;W.p*
B +- C V * - MS + |6S\6 .- £?. G W.'pt
C +• t> v = 57.6 - 4$ r 9. c *;p*
O *■ £ V- ?.c-48 «• -3«.^W»f«
Areas: A 4o % ttX-Wi = - 76 Wp.ft
•B +o C G?XS7.0 «■ M£.a kip.-M
C +o t) (6K9.0 - 3"7.c k.pft
t> +- £ OOf-a*.**") ^ 7c.g k,*p.f+
Mc * - w *ur.2 - n.a fay ft
^a - H.2.+ S7.Z - 7€>.$ fcpft
Me * 74.8 - 76. 8 =■ o
M
(XXllMt>»vA
IM| * 96 Ify-ft = l/££ fey. in
6^/ r a4* fea.
^s* r
1H
H* .V
SW«.pe
S IS* 43. <*
$ i*» So
5 (»V)
So.*
Sl5x 4J?. <*

PROBLEM 5.89
20 kN 20 kN 20 IN
VGrtrt
3©
I©
*
VUkW-wO
5.89 Two metric rolled-steel channels are to be welded along their edges and used to
support the loading shown. Knowing thai the allowable normal stress for the sleel used
is i 50 MPa, determine the most economical channels that can be used.
SOLUTION
A t z r so *w
SV«o*: A 4* 8 V - 3e> kv
B+oC V * 2>o - la r \o k\J
C 4* E> V - to - 2o * -|o kw
D^ E V *- lo-So = - 3okW
Ar«^: A 4& 8 (O.WX&°^ r ,?0.£S W-*
B -*o C (6. Os^G^O r C75 W-*
C -*»» "^ ( A^Ot'o) ~ -6.75 kU'**»
*\ * 2o.Sts + 6.75- t 27 IrW- *x
h^r 27- £.?f * 2o.tStoJ-n
Mft/ifrt^ ImW 27 WvA h r 37*/o3 rJ*w
rd<" a. sedio** copse's fin* ©+ Kjo &k«*ifle&
^'* 6*// iso k /o*
Vat took cViayin*^ 5*= (a1 Xfcow/01 ) r ?0 x Jo* w*
-lo
-3©
2o.25"
sfc.
«pe.
£
C I8OX/4.G
SOoW)
19. a
LijKtcst e>n«.v> ne^/ 5 eeTfovi
C ISO* IM.6

PROBLEM 5.90
73 kN
•10 kN/m
V (W)
I'i.ts
C6.15
A 8 C
X>
po-r eJoutie &l«^^«7
F**" eetoi. oWweJ
5.90 Two metric rolled-steel channels are to be welded back lo back and used to
support the loading shown. Knowing thai the allowable normal stress for the steel used
is 190 MPa , determine the most economical channels that can be used.
SOLUTION
+t)2TM0 = o -3.cA + Clots') +(MXt*a>6>°'> * °
A= 11.IS few
02*MA= O 3.QV -(MX*V(2.7Xl.«K*0 * o
£> * 12.1S kU
B +i> C V - 74. ZS* - 75" » -0.7S"kW
Vr = -0.7S- -0.8X*>> «■ -iZ.nSkw
B to C CO.9K-O.7r") - - 0.675* fclJ-w,
C f* tl (±)(\.*V-o.ir-72.7s) * - G6./5*
Ma r 6 4 66.825 « £6.84T W- ^
Mc* Gt.gas"- o.6?rr r SC. ir fcu. *
Mccxim jm lM I = GCnS M-vn • ££M£*lo* /A*
Sn* = (iXssv.7*/oa) * 176.9 yl6* nfS
s\>
6.pe
C 3.00 x 27.^
S(lOlm»?)
85
7*1
lo'^l+tjt o)iA«n«^ Secfl'OM
C 2lo *ZZ

PROBLEM 5.91
2000 Ih
300 lb/ft I
{ _ f \
M^f-ff)
5.91 Two L 4 * 3 rolled-steel angles are bolted together to support the loading shown,
Knowing thai the allowable normal stress for the steel used is 24 ksi, determine the
minimum angle thickness thai can be used,
SOLUTION
6 in.
4 in,
+ ZFj = o
By syKi>«-,fe4vv/ A - C
A + C - Zooo - (O(aoo) = o
A = C - i'tOO Jk.
V8+ * luoo - Zooo * -faooib * - / Itip
A^eaS-' A-WB (^(3)0.1+ l> M.35 (6>.^+
8 *> C (iX^X- I - !.<=» ) -- - t.^ *+ "R
M1 *•* ,z.ns;S
Rv- $tC*\ov\ Comsr«4ir\^ oT +wo <*.*« le* S*;^ r *^—"
A¥
Po^ e^ck *^JPe S^v T CiY'^. J7S-") ~ 1.0 875" in'
Aw^ie %9e^\o¥\
L 4>3xi
S G^
1.99
Woo
Sma^McS+ cjHo^cXit +kiVkn&JS

PROBLEM 5.92
500 lb 500 lb
vcvcyo
e
-0.5
:0Urft>
- .©
-6.o
4 in.
s -
5.92 A steel pipe of 4-in. diameter is to support the loading shown. Knowing thai the
stock of pipes available have thicknesses varying from j" in. to i in. in £ - in.
increments, and that the allowable normal stress for the steel used is 24 ksi, determine
the minimum wall thickness t thai can be used.
SOLUTION
cctr ■ A -K B V= -Soo A - - O.Slclp
B to C V- -SOQ-Soo - - JoooJlz -16 Uy,
E fo c C4H-i.o*l - -H.o k.y.ft
Mft = o - 2. o" r - ?.x> k.y ft
Mft* -5.0-M-O - -G.6 k«>.ft
MflL»rimtfM .\Ml • Co ki'p«ft -" 72 Wi>*m.
- i r X ^ - C>
s - =
~?=— ~ —— = -5 in
'3
Dfti'rt* j «V iwc^et^e^'K +o^ tiJeftr«ir\
i'f in.

PROBLEM 5.93
Total load = 2 MN
5.93 Assuming the upward reaction of the ground lo be unifonnly distributed and
knowing that the allowable normal stress for the sleel used is 170 MPa, seiect the mosl
economical metric wide-flange beam lo support the loading shown,,
M (MM-*')
0.3751
SOLUTION
Dou^wav-J JiSTvi tw4eJ PoaJ W - —= « ,% MW/V*
Net dislriL-kd ipoJ ^er BC /.£ MN/v*
SVie^: VA = o
A*e*& = A -K> ft (i)(o.75^o.4 ) - o.«r MM- *
B ¥o f (4Yo.5X0*6)a O. ISO MW-^
C +o *o (-Jr)(^.Ky-o.6) =■ - o.aar MU-*
8ew«iw*v roe>*ieAT4 *
Ma *o
Mr * O + O.Wr = 0.225* MN-*i
Mc r 0.37S- O.IS'o * O.WS MW-i*
M„ * 0.22S" - 0.22S" ■= O
tWimt>»» HM " 0-S7S- MW-^ * 27$ *IO* N-i*
6jj » Po MPa. - l7c>x/o6 Pa
2.2.0$>< lo"* V** » 33.0& */0* »v,i^
Skape
W ^^oxiS-S*
\W 6lo« |oi
W 530" IS"o
W 4COM13
S6oVmS)
3Sio
2S3o
372 o
X4oo
W»

PROBLEM 5.94
240 Idps 240 kips
61
•Mo
Ho ZM
5.94 Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 24 ksi, select the most
economical S-shape beam lo support the loading shown.
SOLUTION
D>sW;V»>4e4 tticko* <l- -^ * to fc.p/M
eo.<r - Va = °
V6* - /Co - 2Ho * -20 !<•>£
Vt- - "to +(mX*°,> - so kt'|»»
Vc+ = SO" ^Vo r - lt>0 tC.'pt
vrt^c (4Xo(*o - *° *>"*■
C -W o (4HhS(+i6o-> ? -Mo ^p.-ft
MB * © + 3*o » 3*o fcp.tt
\MC ^ 246> t 80 * 3-2" k«'p-fr
M^vim** IM 1 y 32o k.>-$t r 3S4o W»p*in
|M| ^ 3S4Q
6^< * 24
6*^* HH fa
'tntw
ISO
Si*
*i»e
S2H**0
S *o*%
s.Ovi
175"
165
L4^l4es4 S-sk+peJ sec-ho^y

PROBLEM 5.95
5.95 sod 5.96 Detemiine the largest permissible value of P for the beam and loading
shown, knowing that the allowable normal stress is +80 MPa m tension and - 140 MPa
in compression.
SOLUTION
12 mm
RSBH _12 mm
0.2 m
-•— 0.4 m —*+• H
0,2 in
96 mm
n
P
i
c- X
-P
0.2 P
C ±» V> V * O - f> ^ - P
A^<^- A +* fc o.np
B i> C o
C +0 Ti - o. *P
MA = o
M& ■= o ■*- o.z P = o.^ P
Mc r 0.2P +0 * O.^P
Mc =■ 0.2P-0.5P * O
6
Lar-*c%T Y»e***(i/«. tew*irw* n^o^e-vf - 0
*P
1
J
©
16
*->*!
ft*+
0
(!)
21
A(«ME)
us*
1728
J(^^1
34
AJ {»*«»*)
?t»T3C
27648
<K*wO
*0
10
AdV**")
34r.c»/o*
T '
U*t.4*lo*
1 n<*
7ip c" M******
Bet*. C-/6m.n
i-M*fe..fIS ^=(^0^/p*y;|o.C^l66,)- 1495 M-M;
M r u'?$ W-M ! ■■'■
A-W^Jt **/>« of P WP" WIS
? -- 7W W — 7-43 kw -«

PROBLEM 5.96
5.95 and 5.96 Determine the largest permissible value of P for the beam and loading
shown, knowing that the allowable aormal stress is +80 MPa in tension and -140 MPa
in compression
96
1_t
KSKH 12 mm
' I [48mm
-Ik
12 mm
D '
0,25 m
0.15 m
a
c
-0
SOLUTION
O^Mc = o -0/75 A + o.S? - o./fP ~ o
A * 0.4«C7 P
O^MA = O 0.75 C - O.jtfP - 0.°l ? * o
C - |.«S»3 P
B fo C V - O.HU67 P- P *■ * 0-53S33 P
C +* D V -- - 0,5W3 P + I. 5333* P * P
Av*«tS- A +• 8 Uj")(M*«?P*> r a "667 P
B fo C CarV^-^ssV^-o.s^"? p
C+.D (.ojs^P » o./sP
S*i<('
4*: MA ro
Ma-0 *- 6. II 667 P t O.IU$7 ?
Mt - O. HC47 P - O-UCClf * -0./fP
M0 ~ -0.15 P + 0.15 P " O
p«"t
0)
r
ACcwrf)
ii5?
574
was
y (n*wl*l
5f
a</
Ay 6-^
|3SZ"
76osa
d(*>* J
10
ao
AjW)
23o4oo
ZHSLoo
I.ClWi")
• 13814
U^llfe
I" 5AJ*+ ?I » ITQOlt
iu«^
Top
PoTTi
X » iO^it s „ W6„/o» Hp? . «. 876-/€>-' .,*
- Ig*
<3 8
-0.I5P * - IMSS-ifO* P* %1l*lO* N

PROBLEM 5.97
S.97 Determine the largest pennissible unifonnly distributed load w for the beam
shown, knowing that the allowable normal stress is +12 ksi in tension and - 19,5 ksi
in compression,
SOLUTION
Sin.
Sin.
- 8 w
w
ID w
Vc* - -IOjw + 13 vw - 8w
A re<c& - A +• & i±Wt**»1 ^-32 *
HB - O - 35 v =■ -35 vu
\^e.-»3^w+t5-0lVr i8w
fV+
<3>
z
A 6V ^
\.C%7S
3-375"
y (i» ^
2X2S
\.\2S
Af (••»)
ff.«2T7
l.«9«9
£.328/
J "»
D.7S /
■
AJViV)
0.9 M 9?
0-9*W2
L 898*
1 C-M
0. 079 1
0.71 I *
0.7^/0
2.US
ScnAine* wombat *Viwi*ti
C«*"»p. *-■+ & *■*** C
A)R*/«AJe 1«J
1/^ - & 3 IOC mS
M - - cl/y
- CisfK £.39ofr^ ■= - 38.6*7 k.'p-iV
-(IU- l.**S«|S'ir 17. £1* k.'p i"M
- (-i9.5Va.3«ro6 )- HC.c i<;f. tn
6 * C - 3a w * - Xi. 4&1
B 18 u#» /7. 212,
W^ 0.87V fctp /in
S*J>Jes+ M* 0.87V fc>/.> - \o.W k.>/W.

PROBLEM 5.98
l.tfSi*
5.97 Determine the largest permissible uniformly distributed load w for the beam
shown, knowing that the allowable normal stress is +12 ksi in tension and - 19,5 ksi
in compression,
5.98 Solve Prob. 5.97, assuming that the cross section of the beam is reversed, with the
flange of the beam resting on the supports at B and C.
SOLUTION
2 A in.
Rowels
WftTi
fe + C - 3^w = 0
8 « C - 12 w
■A I—fin. SVteatT
VA = O
w
Vfe- - o - 8 w - -5
V6* - - S w •#■ |8 w = /O w
Vc* =■ low- 2o w - -low
Vc-» - -\0 W 4 |$ W r 8 V*»
VD - 2 */ - 8 *> • o
Mr = O - 3$ */ --33 w
Part
(2)
r
A(.V)
LUIS
i.CSTS'
z.yis
$(.->
I.8TS
0315
Ay(m*1
3.UH 1
3.7*»6-?
di>)
0.7S"
o.75
AJ*C-MM
LSlSt
I(.V^
0.711**
Cl 67*11
o.7f(0
V ~ 3.375 " »•'**'«■
J/y * /.-V343 ;*■
Top** v r /- *B7S" in
-t>/T.S")(-Mis*i3) = *?7^ci k>-i* •*-
Mfo*Me I00.A W
8* c -3?w» - n.nx.
£ IS w * *?-W
5n«Jie*4 W « 0.53 f Jc.-pAn - C.^ k-'p /#
W = /-56'1 k»>/i'h

PROBLEM 5.99
5,99 Beams AB, BC, and CD have the cross section shown and are pin-connected al B
and C. Knowing thai the allowable normai stress is +110 MPa in tension and -150
MPa in compression, determine (a) the largest permissible value of w if beam BC is not
to beoverstressed, (b) the corresponding maximum distance a for which the cantilever
beams AB and CD are not overstressed.
12.5 mm
1—200 mm-—I
SOLUTION
150 mm
I
12.5 mm
Ave**, o -Vo E o^ sltecw* 0fi'«.^i/stt,Mk
Mc * o -t- a.1** *> * £.48 w
Ekd;
AC—^
ZSoo
|*7i
H3TS"
3 OWi
156. V
75
A.Y(wf>
340635
rf<««^
3tf.M
HC.H3
AJ*WI1)
4,o«a*)o*
7.075 */o*
WW
I" ZA-J* + ZI " lO.WlWlo* t«m*
iM
3.5*1 g»Jb*
Y= *"*"*> = |2LM3
*f37y ' ^
LoCaT*
ow
■fop
y (♦*«)
HI.OT"
-121.43
r/y6o»^?) <g- <Jso do****)
- 57.47
f' *° '"'2f Be»d,\i3 W0^enf i.W^ Mr -ffl/y
S.Hgw - %C»k|o*
W * I.MS*to' N/t«
She** *t A Vjj = (a*-/3.-c) w
A*e^ A-M3 *f «k«*r- aUj**** ia(VA+Ve)r ^a(a+7.^w
ial 4,3i.fi-a. ■-. 1.8*7 « o
'. *3S
J L

PROBLEM 5.100
5.180 Beams AB, BC, and CD have the cross section shown and are pin-connected al
B and C. Knowing thai the allowable normal stressis +1 iO MPa in tension and - 150
MPa in compression, determine (a) the largest permissible value of P if beam SC is not
to be overstressed, (b) the corresponding maximum distance a for which the cantilever
beams AB and CD are not overstressed
12.5
I— 200mm-*j I
2.4 m 2.4 m 2.4 m
-J
12.5 mm
150 mm
SOLUTION
Area. 3 k> f *F sne«i** ©lt'*u»r«.*\,
2.M P
C**T*©M cenel mei^e^T erf i^e^'na.
1Vt
AW)
256o
437.S
y £k«0
1S*.2S
75
AyW)
\HoUS
53l?50
dC^
AJ* W1)
Z 073 Wo*"
ic-*^
O.OV4***
3.5*S*/o*
S - «ltfe
v -
43*r
7 I^/.WS M«t
]> TEAd* + Si * lo.ahio'^
Lociuii'
£*L
V<w*l
Ml.o7
1*1.43
I/.y ClbW) «- <J*<, (lo-V)
- S7.47
ISO '«•**
A J io w *. tie io«L«( F
Tensi««»f E*F: -6io«ic/li'.87.</7«/o*)= <?.cu*ios */•*
fir 4.0* ifW -^
Art*. A +* B tff SnCA'" Jfa-gAo
Oisi*
V\C€
-y.Olxld1 <*,*- _/3.|gHciO*
tfL* ^.2^ m

PROBLEM 5.101
5.101 Each of the three rolled-steel beams shown (numbered i, 2, and 3) is to carry a
64-kip load uniformly distributed over the beam. Each of these beams has a 12-ft span
and is to be supported by the two 24-ft rolled-steel girders AC and BD. Knowing that
the allowable normal stress for the steel used is 24 lesi, select (a) the most economical
S-shape for the three beams, (b) the most economical W-shape for the two girders.
SOLUTION
Bea^s ■' f Z. clv.A 3
P"TT * I ~T"
< If V I I
k i
12 H
3a,k;,
,4ft V Ck.>0
31
M»*!*ju* M - (i)(cXuU 76 icp-W ■= n5*£ k.>-!rt
(a) Use
SI*
ISZ
S A5-* M-^-9
* ¥S .*'
5k
^pe
S \S* HJ?.9
sc.v>
^.4
- 12
Bea*is AC a^d BC
132. |3* .3ek.fi
, r i" i
49
V
48
(rt$
(6
0m>
-It
~1*
3Zo
Av«tS un^e»r »nc«^ «|i'«*t/*w\
(b) Use W27-84
J*U
5h«pe
W3o* <7^
vir?*?*
Wfl» lo4
W 11 *lol
W 1-8 x jo&
SCinO
26<*
213
Z5S
Ul
QoH

PROBLEM 5.102
5.101 Each of the three roUed-steel beams shown (numbered 1, 2, and 3) is to cany a
64-kip load unifonnly distributed over the beam. Each of these beams has a 12-ft span
and is to be supported by the two 24-ft rolled-steei girders AC and BD. Knowing that
the allowable normal stress for the steel used is 24 ksi, select (a) the most economical
S-shape for the three beams, (b) the most economical W-shape for the two girders.
5.102 Solve Prob. 5.101, assuming that the 64-kip distributed loads are replaced by 64-
kip concentrated loads applied at the midpoints of the three beams.
SOLUTION
3*k,>
S„;, -
La) Use
\Ml 33CW
6Jt ' *4
§XD*6& -*
l^lAVt rt^U
^ ;**
*M - (0(o0« \<*z k.>.-ff -- j?3o«Jfc'p.;„
SUp
Sl<2*7o
SC.V)
Ml
103
Eectma AC a*J SD
3^ |3* i 3i D&'^s
W
6JL 3<
J i ; 1
Ul
(kipO
4- 8 -*|#*/^</3
(lit)
m
i&
0W>
-16
-V8
Av^eas JnJe^ sJifitftr cJiVa^*^
(m\(«) * m top ff
ZH
=■ 160 In
SKrt,p e
W 3o* t<*
W ^7» 8*
W M w 10**
W 21 * Jo 1
W |? * /Ofe
s ov)
;?£<*
fcl3
2Sg
J?«7
20 4
(t) Use W 27*S«J

PROBLEM 5.103
X74
V£Jfl/)
02O^
c
riao Beaws
0
■ p
■ / '" ■>
AB
* a. -9
MlW-*0 |7^^
JWokw
12c |T
|u^Ti
«M llM'f 1UO
V
If*
lao ku
M(W*rt
/N
U5%
C E *>
S. 103 A 240-kN load is to be supported at the cenler of the 5-m span shown. Knowing
thai the allowable normal stress for the steel used is 165 MPa, determine (a) the
smallest allowable length / of beam CD if the W 310 * 74 beam AB is not lo be
overstressed, (b) the W shape which should be used for beam CD. Neglect the weight
of both beams.
SOLUTION
£>*• ro)U<\ s+eei ftfcvfco* W 3\0*1H c$ fee*™ AS
Q« * U5 MP*. * /C5-X/0* pa
Mail = $$* - (/060x/o'*)(l65*k/ ) * \7H.1*ltf W-*
(p^ Beav^ AB
120 Q. r |7M. 9 - 1.4575 *
Ge©i««v4v^: 2et, + J?r5 J?=^*-2o. -- 2. 085 m-«
(bJ Beav^ CD Cw.;*lp.i*+ £)
Area C fo £ crT site*/' eliot^r***
=■ (i.omryiaow 125.1 m.m
^ ' 758. 2 *|os ■**
A*swe*
•SVmp<
W 4*0 "• SZ
WH© >"KJ
W3*o *57. 8
W 3/0 K 6t>
waro* G7
W2oov S4
S (10*^
<m
77*1 *-
s^
851
%0*\
«53

PROBLEM 5.104
,S4kN/m
S. 104 A uniformly distributed load of 84 IcN/m is lo be supported over the 5-m span
shown. Knowing that the allowable normal stress for the sleel used is i65 MPa,
determine (a) the smallest allowable length I of beam CD if the W 310 * 74 beam AB
is not to be overstressed, (b) the W shape which should be used for beam CD Nedect
the weight of both beams. ' ^^
S4lcN/m.
W310 X 74
VlW)
SOLUTION
Ft*- Y^olhA s-iee^ 5e_4i_^ WSloxTf of team AB
_r„,, - IC," MP*. - ics-*/-** p«.
M« = S 6^ = Clocoy/o"6Xi(;5>'/o')^n*T.9x/o1^-^
A = B =■ ?IO JfcN
8
_W/fAi
-Li-Llij:
*a*^ AB • A^ea. A -f o C erf slief- Jr^g i^a «o
r £ (1050 - M*_>*>
r
121.3 kM 1*1.3*1/
-t
S^yf<
V Ml© "52
WH-lOn 38.8
W 3.0 * 39
W 3/0 *3S.7
W goo » 51
8«ftt^ CD (hJp«M+ H *)
Av»ee«. C "fi> -IT ot 5ne«^ 0uAAr*.**\
i Cm**^^^ 1^1.3") ^ S»7.5S JfU.*
~_»#Ua \»om*A *~\ E M* 87.58 *NAm r 87.S8**o*Mih
<?H_ (b) Use */ 310x38.7 -—
637
Sl%
5*8*

PROBLEM 5.105
1*4* ft
P, ■ 24 Jr.'rs
P. • £ k->s
W = 0.7Sk,? /ff
Q.7S k»>/»|
1 t 1 L' H i
21 Wfi
6*1*
i
Kit* topi
W.81S"
\H.ns
■5.105 A bridge of length L = 48 ft is lo be built on a secondary road whose access to
trucks is limited lo two-axle vehicles of medium weight. It will consisi of a concrete
slab and of simply supported steel beams with an ultimate strength av = 60 ksi. The
combined weigh! of the slab and beams can be approximated by a uniformly distributed
load w - 0.7S kip/ft on each beam. For the purpose of the design, it is assumed that a
truck with axles located at a distance a = 14 ft from each other will be driven across the
bridge and that the resulting concentrated loads P, and P, exerted on each beam could
he as large as 24 laps and 6 kips, respectively. Determine the most economical wide-
flange shape for the beams, using LRFD with the load factors y0- 125, yL= 1,75 and
the resistance factor 0=0.9. [Hint It can be shown that the maximum value of \Mt\
occurs under the larger load when thai ioad is located to the left of the center of the
beam at a distance equal to aP2/2(Pl +■ P3).]
SOLUTION
De<U SqaJI : RA - £a = (±}(to)(o.7S } - IS k:?t
bred A-ft>£ of slie«^ ol.-^r*^ (j[YB)(l&) r 7.M fcpfi
X + C*. =■ 23. G + H - 36.6 4+
L-x-a. - 48-3tf.fi = //. <* ft".
SUe^/>: 4 ^ C V - 14.12f k-ps
O T-* B V^ -IS".875 k.'ps
Ar<a A4*C oacXrt.iaO r 3H.aar k«>-f+
S^,. r ^ Ma v VI Mc _ n.*s"Mag«)» 0.75X3*3')
vV27* 8*t
Sk«pc
V 3oX<»q
W J?7* 81*
W2q* M
WX/>/0l
W IS Mot
5 a^
zs*
Jh3 <*-
25*
111
7o^

PROBLEM 5.106
L = W -Vf
a « if ft
P, ~ M k:r
Pj* 6 (c.'ps
iv - o,7sk:P/if
■5.105 Abridge of lengths = 48 ft is lo be built on a secondary road whose access lo
trucks is limited lo two-axle vehicles of medium weight. It will consist of a concrete
siab and of simply supported steel beams with an ultimate strength oy = 60 ksi. The
combined weight of the slab and beams can be approximated by a uniformly distributed
load w - 0.75 kip/ft on each beam. For the purpose of the design, it is assumed that a
truck with axles located at a distance a = 14 ft from each other will be driven across the
bridge and that the resulting concentrated loads Pi and P3 exerted on each beam could
be as large as 24 kips and 6 kips, respectively, Determine the most economical wide-
flange shape for the beams, using LRFD with the load factors yD = 1.25, yL = 1.75 and
the resistance factor <p ~ 0-9- [Hint. It can be shown that the maximum vaiue of \ML j
occurs under the larger load when that load is located to the left of the center of the
beam at a distance equal to aPj/2(P, + P3).]
"5.106 Assuming that the front and rear axle loads remain in the same ratio as for the
truck of Prob. 5.105, determine how much heavier a truck could safely cross the bridge
designed in that problem.
SOLUTION
See. sJUi'e" -h> PffoBLfM 5.10S -for cchoiAi^ ©i flc foJJowinaZ
V\D - XScl2 fcp-in ML- 383/ fc>-iV
F<^ roMeJ sW section W27x84 S * 213 i»8
TL
1.75
t?^+n
Ml
ThercA-se
Z%% l
23.2 %
).2SZ
I + O. £32.

PROBLEM 5.107
I
-2.*kU
■5.107 A roof structure consisting of plywood and roofing material is supported by
several timber beams of iength L = 16 m. The dead ioad carried by each beam
including the estimated weight of the beam, can be represented by a uniformly
distributed load wD = 350 N/m. The live loads consist of the snow load, represented
by a uniformly distributed ioad wL = 600 N/m, and a 6-kN concentrated load P apphed
at the midpoint C of each beam. Knowing that the ultimate strength for tha timber used
is <*,= 50MPa and that the width of the beams isb = 75 mm, determine meinmimum
allowable depth h of the beams, using LRFD with the load factors yD= 1.2, yL= 1,6
and the resistance factor <p = 0.9.
SOLUTION
V^Lr £00 W/m t Q.& kVJ/Wj P= C IrW
8e**cU«* iHoMe^ .+C: II. 2 IfU-Hi s 11.2wo1 W-wi
Utfc load' RA« ^r(K)(0-C)4 C] r 7.8 j<w
■ 5Vift^ *JC V^ 7.*-Cl)feO " 3 kV
A^-e-A. *A fp C oT Sheets «JuMiNUn
(i)(8K7.8f 3) = HZ.X left/.*,
B*«dJv»^ wowe*t of C 43.5 Jrft/-* « Ml.?*/©3 *J.m
5>CT^ " (0-4X50 »lo«)
a |.«3«l7*ltr3M* = 1.85*7 «/©fc wm*

0.35- kN/*i
u u 1 n u
16 m
2.* kw
2.3 few
V
*
^Z» W
^^^ B
V c^\^
-?.ifci>>
nuuuim
1111
e |i—:b
vU*kM
M
• \A kw-Ki
■5.107 A roof structure consisting of plywood and roofing materiai is supported by
several timbar beams of length L - 16 m. The dead, load carried by each beam,
including the estimated weight of the beam, can be represented by a uniformly
distributed load w0 « 350 N/m. The live loads consist of the snow load, represented
by a uniformly distributed load wt = 600 N/m, and a 6-kN concentrated load P applied
at the midpoint C of each beam. Knowing that the ultimate strength for the timber used
is Oy = 50 MPa and that the width of the beams is b - 75 mm, detennine the minimum
allowable depth h of the beams, using LRFD with the load factors yD= i.2, yL= i.6
and the resistance factor 0=0.9,
■5.108 Solve Prob. 5,107, assuming that the 6-kN concentrated loads are replaced by
3-kN concentrated loads P| and P2 applied at a distance of 4 m from each end of the
beams.
SOLUTION
L - 16 rn, a* ^m, Wp- Z£o W/i* * Q.t£ kv/*
WL =■ 6Co NA» * OX kV/w, P * 3 kw
Av^efc. A -rt. C of £ lie a* e)rA.«r«iM
(f)(s)(3.8) r in JcM-».
Live foaJ: Rfc' i[(lcHo.O+34 3l=> 7.8 kN
Skea/ &+ £* 5".H - 3 - 3.V <cV
= i.«k>s* lor**.3 » jLiog*/oc *ms
h =
4
__ /6S _ /fe)(l.Mo»Kfo*)
TS
= 336
mm

PROBLEM 5.109
Me
♦ V.
c
ft.
ip
5.109 through 5.111 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown. (/>) Use the equation
obtained for M to determine the bending moment at point E and check your answer by
drawing the free-body diagram of the portion of beam to the right off.
A-|?
SOLUTION
Af p•;**■+ £ x - 2<x
IMA, o 3clC - a?= o C -- i?
PROBLEM5.110
5.189 through 5.111 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown, (b) Use the equation
obtained for A/ to determiite the bending moment at point E and check your answer by
drawing the free-body diagram of the portion of beam to the right of E.
SOLUTION
A- Jw«,a
JM
A+ p*?*^
x= U
^e T - ±*.(3Ai* +h«-i + i ""«A(*0
= i*a*
fc)Z>li r o 4aC -(olMIclw.) = O
i> 2 Mfi r o - Me + fjatf£w/„ft ) - o
C--3S-M4A

PROBLEM 5.111
tv0
5.109 through 5.111 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and ioading shown, (b) Use the equation
obtained for Ki lo determine the bending moment at point E and check your answer by
drawing the free-body diagram of the portion of beam to the right off.
Check
A = -^*0a
SOLUTION
At po.vl f x - 3a
r -±w.am —
4ZHB«o - n£ - f Cm.*)* ©
M6 ^-i^^
PROBLEM 5.112
*/*
JjL±
mi
6
5.112 through 5.114 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown, (b) Use the equation
obteined for M to determine the bending moment at poinl C and check your answer by
drawing the free-body diagram of the entire beaux
SOLUTION

PROBLEM 5.113
5.112 through 5.114 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown, (b) Use the equation
obtained for M to determine the bending moment ai point C and check your answer by
drawing the free-body diagram of the entire beam.
SOLUTION
W. X
Ch.ccW- +t>IX = o
B
A4 p©i*t 0 X '- 2*
60.
PROBLEM 5.114
5.112 through 5.114 (a) Using singuiariiy functions, write me equations defining the
shear and bending moment for the beam and loading shown, (b) Use the equation
obtained for A/to determine the bending momeni at point C and check your answer by
drawing the free-body diagram of the entire beam.
SOLUTION
-*
Cheek:
A+ poi'nT- C X = 2tfL
w.
Uuu^
&
H0-

PROBLEM S.115
&
P*
C?
5, US and X 77* (a) Using singularity functions, write the equations defining the shear
and bending moment for beam ABC under the loading shown, (b) Use the equation
obtained for M to determine the bending moment just to the right of point 5,
SOLUTION
m * - p<*-*y - pct<x-^>° -*
M ^ - o - P*. * - Pa, -«
R*
Pa
c
PROBLEM S.116
r
5.7/5 and S. 776(a) Using singularity functions, write the equations defining the shear
and bending moment for beam ABC under the loading shown, (b) Use the equation
obtained for M to determine the bending moment just to the right of point 5
SOLUTION
IMC* e> «*.)?+ *P- 2(Pfl.V* Safe »o
V= (R*-P) - P<v-a>*
^ -ip - P<rf-«.y
ft- -lr-p*->-
M r -4f>* - 1><y-a>' + pa + Pa.<x-a/
M ' - i Pa. - O + Pa. + Pa- ■ ^Pa- -**
*c

PROBLEM 5.117
25 kN/in
5.117 through 5.120 (a) Using singularity fiinctions, write the equations defining the
shear and bending momeni for the beam and loading shown, (b) Delermine the
maximum value of the bending momen i in the beam.
SOLUTION
8
3 SJ
friMcJvy 1?» - f?e
0.6 m
0.6 m
;0 - 62.5" - £S(x*~cU) + o - 40 - o x* = I.S m
PROBLEM 5.118
1.8 kN
1.25 kNVm I 1.26kN/m
5.117 through 5.120 (a) Using singularity fiinctions, write the equations defining the
shear and bending momeni for the beam and loading shown, (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
y Sy *>i i^\ efi^y f-a
1.2 m J \
0.6 m 0.6 m
^ ^ P.
W = LIS - /OS<x-J.3>° + 1.75 0-2.*>° * -$
M--0.fc2sV + o.«s<y-i.2>1- o.£*s-<x-2.</>*
+ 2.H x - /- 8 < x - /. S >' HA ^ -*
duf X - /- %

PROBLEM 5.119
20 kips
20 kips | | I 20 kips
5.117 through 5.120 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown, (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
A - l+o ki>».
2 ft 2 ft 2ft
V = W> - 2o<x-2>° - 2o<x-i/>° - 20<x-6>* k,p*
M - lox - :?<><*-;?>' - 2o<x-*r>' - *o<y-6>' kp-F+
VaJ^es. of V
A -h> B V* ¥» (c.p
8 ** C V * Ho-2o - Ho k|p*
C +* D V - 4o - Zo - 2c = o
D h> £ V * Ho-2o -Zo-2o * -^o k«y
A+ C X * 4-ft M * (MoW -(^VO - o - O r l?o k,p.ff
PROBLEM 5.120
5.117 throngh 5.120 (a) Using singularity functions, write the equations defining the
shear and bending moment for the beam and loading shown, (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
■1,5 m-«- < »|«
0.6 m 0.9 in
V = 40 - 48 <x- |.S>° - Go<V-3.o>° v €0<x-3.6>° k*>
M* ^oy - </8 <*-/.■£>' - 60<X-3.o>' +■ 6o<*-3-0* kM-m
Pt XM M(kK/-*0
A O O
B LS (4^0-^ * Go KKJ-m
C 3.0 (4oU3.o") -(H*W.S* '' 4S kW-^
D 3.6 (Mott^O- (*l«yzO -(€faXo.6^ - 7.* kW***
M
IMA*
60 4rtl-v*

PROBLEM S.l 21
111) kips | 25 kips
5.121 and S.122 (a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown, (b) Determine the maximum
normal stress due to bending.
10 kips
S6 x 12.5
15 in. 20 in.
10 hi.
SOLUTION
+5TMD- o
V= -lo + 25<X-/S>° - ?5<^-35>" + ;?o<V-<£o>" kip*
M - - /O* 4 25<X-'S>'- ^S<x-as>' + *o<x- to>' k.-p. |n
Ft /(ft) M(i&>-^
8 15 —0°X/s ) ~ - '-5"0 k;r '*
C 35 -0©\^^+ ^sXao) - \So k;p. i*
•D 6© -OoYco")4 Czechs') -(asXas') - - loo fc>-i"i
£ 7o -OoVTtO + tfsKSS^ -(23-Xas"> i-OtoYlo) r o ae^fcs
M&xi^JM |Ml * /So kip* in.
7.37

PROBLEM 5.122
24 kN 24 kN
2-1 kN | | I 24 kN
0.75 in.
5. 121 and 5.122 (a) Using singularity functions, write the equations defining the shear
and bending tnomenl for the beam and loading shown, (b) Determine the maximum
normal stress due to bending,
SOLUTION
W250X28.4 _3^ +(?.3S')(«H') -(l-5Y2^)
ra = so k;?&
O^MA r o ~(0.1S)(2H) -(LsXa^- (*.Js)0) + 3 f?ff - (S.7rt0W) = C>
Vs 3© - 2«y<y-c>.7s>0 - ZV<y-/.s>° - z*(v~2.2s$ 4 £<i<x-3>* |<M ^
8 0.7S" (3o)(<>.7£) ~ j?2.S kU-v*
C t.S &o)(\.s\ ~(Z<i\(o.Ts} = 27 I<W-m
O ?.2r (SoXJ-as^-tt^yi.s'J -(2«0(o.7s) = 13.5 W-*
E 3.0 (a«0(3.«>)-(M)(2.2s)-CJ*)(L5) -(2*0(0.7$) - -l*kW-v*
F 3.75 (3o)f3.75)-0wX3.6)- (mX3.W^ - (»»*)( I.r^ +(<0(o.-*-) c o *^
Fo\r roJlteA steef sco+Co* W25o x ^. 4 S- SoS^/o* w*m*
r 3oS * /o'fc w*
N»™*/ *+f«M €•« -tt * ~~ - *7.7>/o' P*. -" 87.7 MPa. -*

to kN
PROBLEM 5.123
80 IcN/m
5.123 and 5.124 (a) Using singularity (unctions, find the magnitude and location of the
maximum bending moment for the beam and loading shown, (b) Determine the
maximum normal stress due to bending,
SOLUTION
W530X 150 + 5Md - O
RB t mo kN
V = -10 + l4o<x-J>- -8o<v-2>' WW
B fo C V =■ - I© + Ho - i3o kw
V cAo^es Si*q#\ 6> B ftJ *4 poi^i E ( V * >fc ) b«+we<^ C a*^ O.
M- -lOx + Ho<x-l>' - 4o<x-£>X W-*
At pi B X* I Me--.<:ioX0 * - lo kti-to
M pi £ x*s.m -*
Fo«r W 530 * /5"0 S - 37XO x ZOS m*? * 37Zo^lDe' r*x*
No<M*«uf .stress
!M
2ZSX *lO
= C6.6 MP*

PROBLEM 5.124
40 kN/ni
5.123 and 5.124 (a) Using singularity functions, find the magnitude and location of the
maximum bending moment for the beam and loading shown, (b) Determine the
maximum normal stress due to bending.
. SOLUTION
| | 27 kN ■ in ess
'} JL S31° x 52 "*5 Mc a o
IS - 3-6 R„ + O-OCmX'M - 27 = O
V - 2<\.S - 4o <x- ia>' kV
P©m+ £>
V= a
RA - >29.3- kV
Zl.S - Vo (xD- 1.0 = o
M
c^onty^O^
JM I = 22.%7Z kK>*n a* X* l.<i*>l£ *«

PROBLEM 5.125
22.5 laps
L OCCtAioft tST point "&
Skctpe
waiii*
WUh^o
WI6* Ho
V W*MS
Wl2*5o
W lo y 6 8
S 6'«*>
*t.4
*«.<*
CM. 7
€2.1
C/l.T
75-. T
5.125 and 5.126 A beam is being designed to be supported and loaded as shown, (a)
Using singularity functions, cletermine the magnitude and location of the maximum
bending moment m the beam (ft) Knowing that the allowable stress for the steel to be
used is 24 ksi, find the most economical wide-flange shape that should be selected..
SOLUTION
+5ll\=0 -15 (?A j (7.S)65)C3HCu)fcH.S> - O
w = 3 *.>./«■ - -#
^.5* X0 = G ft.
,22.5<x-3>1 k;f-Ft
r 1^1.5 k.p• -ft" - \H£% ki'p- iw.
/2LS fcp.ft «i x r 6*f. -*
r 60.75 in1
Answer Wi6iti/o -^

PROBLEM 5.126
■3ft
V " 27. & -2My - /2<X-S>* - I3<x-J?>°
Ve- =■ 27. G -(*.<*)(£} -? (3.a fc>»
Va+ = 5?. C -ff.-nte") - IX = 1.5? *<>>'
Vc~ - 27.6 - (.ZMXn} - 17. - -|3.a fc.ps
5.125 and 5.126 A beam is being designed to be supported and loaded as shown, (a)
Using singularity functions, determine the magnitude and location of the maximum
bending moment in the beam (b) Knowing that the allowable stress for the steel to be
used is 24 ksi, find the most economical wide-flange shape that should be selected..
SOLUTION
RA * 27.6 kips
u.-i
V* o
O - 27.6 - 2.V X. - I* - O Xc - G.S"ft
M - 27.6 x - (.ax1 - /*<*-£>' - ia<x-l3>' k;p-ft
= 122.7 lap-"ft - 1472-4 le.p-i*.
[Ml
im.7 ky.-Ft of * = €.5* -Ft
•^mJk
147^.4
Sj, 2V
= C/.35
Shape.
W*l* 4</
W IS * So
W U k 4o
W IMx "43
\W 12 H50
W |o*6S
S UO
£1.6
SS.<*
64. 7
62.7
£4.7
75". 7
A^s^e^ * W 16 * 4o

PROBLEM 5.127
460 N/m
S.127 and 5.128. A timber beam is being designed to be supported and loaded as
shown, (a) Using singularity functions, determine the magnitude and location of the
maximum bending moment in the beam, (b) Knowing that the available stock consists
of beams with a 12-MPa allowable stress and a rectangular cross section of 30-mm
width and depth h varying from 80 tol60 mm in 10-mm increments, determine the most
economical cross section that can be used
SOLUTION
ISO M/»^ - O.W IcWA*
vv -
- O.HS
0.**8
.5 X ~ I.S
<X-I.S>' = 0.32.X - 0.&2<x- usy k\j/* = -'~
Ax
V - O.MS - 0.1C x* + O. (6<V-/.5>t kN
Locale p»iVf D u»J*et/^ V = O . Assume /.5 m < XD < 4
0 - 0.6^5 - o. 16 xj" + at&i-Xe-ls)*~
- O.C*/S" -J0-4^->r0*"4j^i4-yft*'- 0.48 X© * 0.3G
M * 0,6*5" x - 0.05333 X3 + O.Q5333 <X- l.5>3 VW-w
+ D Mj ' W ' w X"
» 0.872H W-m
i*i
M
»*>!->
M/ (O.W)feoW* V feflS3S3)£?. omsf +(0„QSZl?>)(o.SfS)sf
M,
0.372//*/O _< 3
r 72.67^8*10 *
72.67S&J./C? ^-0*
^
Vu.^-P^f"0^ - Vo.Si.
min
A+ tfext '-Parget /0-m»m iMC-rei^^f
h - /so
WW)

PROBLEM 5.128
500 N/m
5,127 and 5.128. A timber beam is being designed to be supported and loaded as
shown, (a) Using singularity functions, determine the magnitude and location of the
maximum bending moment in the beam, (b) Knowing that the available stock consists
of beams with a 12-MPa allowable stress and a rectangular cross section of 30-nun
width and depth h varying from 80 to 160 nun in 10-nun increments, determine the most
economical cross section that can be used
30 mm
!"T SOLUTION
1 OZMt = o
0.^- O„^o833<x-/-C>' IcW/th1*1-^
V = 0.220 - o.Sx + O. |o4 IC7 <x- l.6>*" kW
Locate
>t D wt»<
V = o
re pdiAf jj wde**e
■o. io«n67 kJ" - o-.m^ax^ + i. i4U7 - o
X =r °*g3333 :WCO.g^333)*-(4X<?JQ4f67yi,na-n
° CO(o-l0*J/C7)
Mr asso x - o.asx1, + 0.3*7 a«<x- 1.4>B vw-w*
M*^ r 0.776 kW-hi 4 X^ 1.1C.S2 to -^
Smm - —— - ^ y )0\ = 64.6<£*/D to - G4.€6x/o
rW --f ^ - H3.7 *">»>
to»i"

PROBLEM 5.129
5.129 through 5.132 Using a computer and step functions, calculate the shear and
bending moment for the beam and loading shown. Use the specified increments AL,
starting at point A and ending at the right-hand support.
9kN
AL = 0.3 m
12 kN/m
SOLUTION
C3.?)0»t - 3P„ + (l.sK3.oX^ = G>
B
V =
M •
Rfl = 21.1 kN
-9 +• 29.7<x-o.*?>" » I2<X- £>.<*>' kW -->
-9x + ^.7<y-0.9>' - <S<x-o.*>Z |<fcL
w>
X
m
0.0
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7
3.0
3.3
3.6
3.9
V
kN
-9.0
-9.0
-9.0
20.7
17.1
13.5
9.9
6.3
2.7
-0.9
-4.5
-8.1
-11.7
-15.3
M
kN»m
0.00
-2.70
-5.40
-8.10
-2.43
2.16
5.67
8.10
9.45
9.72
8.91
7.02
4.05
0.00

PROBLEM 5.130
J20VN
AL = 0.25 m
30 kN/m
5.129 through 5.132 Using a computer and step functions, calculate the shear and
bending moment for the beam and loading shown. Use the specified increments AL,
starting at point A and ending at the right-hand support.
SOLUTION
» o
2 m lm
w= ^<y-3>' - l2<*-3>'
X
m
0.0
0.3
0.5
0.8
1.0
1.3
1.5
1.8
2.0
2.3
2.5
2.8
3.0
3.3
3.5
3.8
4.0
4.3
4.5
4.8
5.0
5.3
5.5
5.8
6.0
V
kN
89.0
89.0
89.0
89.0
89.0
89.0
89.0
89.0
-31.0
-31.0
-31.0
-31.0
-31.0
-31.4
-32.5
-34.4
-37.0
-40.4
-44.5
-49.4
-55.0
-61.4
-68.5
-76.4
-85.0
M
kN»tn
0.0
22.3
44.5
66.8
89.0
111.3
133.5
155.8
178.0
170.3
162.5
154.8
147,0
139.2
131.3
122.9
114.0
104.3
93.8
82.0
69.0
54,5
38.3
20.2
-0.0

PROBLEM 5.131
5.129 through 5.132 Using a computer and step functions, calculate the shear and
bending moment for the beam and loading shown. Use the specified increments AL,
starting at point A and ending at the right-hand support.
3.6 kips/ft
SOLUTION
AL = 0.5 ft
i.akips/rt -OZM = o
-URA4 feJOaJas")-* GoYiYzXt'*^ r °
1.8 .. . L2_
&
= 3.C - 0.3 X +■ 0.3<*-£>
w = 3.6 - ilx + i^<x-t>
V - 15.3 - 3.CV + O.tfx1 - 0./5<x-C>* kips -
M^ 15.3*- UXl +0.05"y3 - aos<*-6>* fcp«fi
X
ft
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
V
kips
15.30
13.54
11.85
10.24
8.70
7.24
5.85
4.54
3.30
2.14
1.05
0.04
-0.90
-1.80
-2.70
-3.60
-4.50
-5.40
-6.30
-7.20
-8.10
-9.00
-9.90
-10.80
-11.70
M
kip*ft
0.0
7.2
13.6
19.1
23.8
27.8
31.1
33.6
35.6
37.0
37.8
38.0
37.8
37.1
36.0
34.4
32.4
29.9
27.0
23.6
19.8
15.5
10.8
5.6
0.0

PROBLEM 5.132
AL = 0.5 ft
5.129 through 5.132 Using a computer and step functions, calculate the shear and
bending moment for the beam and loading shown. Use the specified increments 4L,
starting at point A and ending at the right-hand support.
SOLUTION
4.5 kips/ft
W -
- M.S
x - ^<x-3>1 - 4.S<X-3>°
V - -0.1SX% + 0.75<y-3>1" + 4.S<*-3>' - £<x~4>'
M- -0.2S"*3 +0.25<;x-3>2 +?.?s-<;x-3>a" - G<y-^>'
X
ft
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
V
kips
0.00
-0.19
-0.75
-1.69
-3.00
-4.69
-6.75
-6.75
-12.75
-12.75
-12.75
-12.75
-12.75
M
kip*ft
0.00
-0.03
-0.25
-0.84
-2.00
-3.91
-6.75
-10.13
-13.50
-19.88
-26.25
-32.63
-39.00
k.'|0 3

PROBLEM 5.133
5 kN/m
3 kN/m
V-
X
m
0.00
0.25
0
0
1
1
1
1
2
2
2
2
3
3
3
3.
4,
4.
4.
4,
5,
.50
.75
,00
25
50
75
00
.25
50
75
00
25
50
75
00
25
50
75
00
2.83
2.84
2.85
1.5 m f 1.5 m
3kN
5.133 and 5.134 For the beam and loading shown, and using a computer and step
functions, (a) tabulate the shear, bending moment, and maximum normal stress in
sections of the beam from x - 0 to x = L, using the increments AL indicated, (b)
using smaller increments if necessary, determine with a 2-percent accuracy the
maximum normal stress in the beam..
SOLUTION
t)ZMD = o
ra - to. a kw
W200 X 22.5
L = 5 m
AL = 0.25 m
r o
&9C
\om% - 3x - 2<*-a>' - 3<x-3,s>° kW
V
kN
10.20
9.45
8.70
7.95
7
6
5
4
4
2
1
0
20
45
70
95
20
95
70
45
-0.80
-2.05
-6.30
-7.55
-P-8Q
■10.05
•11.30
12.55
•13.80
0.05
0.00
-0.05
M
kN'tn
0.00
2.46
4.72
6.81
8.70
10.41
11.92
13.26
14.40
15.29
15.88
16.14
16.10
15.74
15.07
13.34
11.30
8.94
6.27
3.29
-0.00
sigma
MPa
0
12
24
35
44.8
53.6
61.5
68.3
74.2
78.8
81.8
83.2
83.0
81.2
77.7
68.8
58.2
46.1
32.3
17.0
-0.0
W loo * 2a. r
OV\
-6
- S3,3*/o'ftt
= 33.3 MP*
16*164 83.3
16.164 83.3
16.164 83.3

PROBLEM 5.134
5.133 and 5.134 For the beam and loading shown, and using a computer and step
functions, (a) tabulate the shear, bending moment, and maximum normal stress in
sections of the beam from x = 0 to x = L, using the increments AL indicated, (ft)
using smaller increments if necessary, determine with a 2-percent accuracy the
maximum normal stress in the beam..
300 mm
lm
L = 6 m
AL = 0.5 m
SOLUTION
-«s +(0(5)+ fc.sXaXa«0 * °
the
kvi
M - -5x -v 4S<*-2>' - lo<x-0* * /o<x-s>*" V.VJ
w
X
m
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
6.00
V
kN
-5
-5
-5
-5
40
30
20
10
0
-10
-20
-20
-20
M
kN*m
0.00
-2.50
-5.00
-7.50
-10.00
7.50
20.00
27.50
30.00
27,50
20.00
10.00
0.00
sigma
MPa
0.0
-3.3
-6.7
-10.0
-13.3
10.0
26.7
36.7
40.0
36.7
26.7
13.3
0.0
|M^ = So V.\)
<Jt X - 4.0 v^
- ISO V \0% mrr?
ISO *LO
- c
W\
&L+ r
G-*
M
mo*4t
Zo *IO~
S 7so x(o
1MI

PROBLEM 5.135
2 kips/ft
5.13S and 5.136 For the beam and loading shown, and using a computer and step
functions, (a) tabulate the shear, bending moment, and maximum normal stress in
sections of the beam from x = 0 to x = L, using the increments AL indicated, (b)
using smaller increments if necessary, determine with a 2-percent accuracy the
maximum normal stress in the beam..
12 in.
SOLUTION
L = 5ft
AL = 0.25 ft
300 lb
ps
w--2-0.3<*-l*S>°-/.*<x- 3*5)° Wp/Pt
V r 3.«*J - #x + 0.8<x-/.S>' v \.2<y-3.s>1 ~ 0.3 <x - s.S>° tc.'
M = 3.S4v - x* + 0.*f<x-/-S>* +o.c<y-3.s>2' - o.S<y-3.5>'k.p-ft -
ImW 3.so4* k;P--W
- HS.&W kyf- i*
- 42 in*
X
ft
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
V
kips
3.84
3.34
2.84
2.34
1.84
1.34
0.84
0.54
0.24
-0.06
-0.36
-0.66
-0.96
-1.26
-1.86
-1.86
-1.86
-1.86
-1.86
-1.86
-1.86
M
kip»ft
0.00
0.90
1.67
2.32
2.84
3.24
3.51
3.68
3.78
3.80
3.75
3.62
3.42
3.14
2.79
2.32
1.86
1.39
0.93
0.46
-0.00
sigma
ksi
0.000
0.224
0.417
0.579
0.710
0.809
0.877
0.921
0.945
0.951
0.937
0.906
0.855
0.786
0.697
0.581
0.465
0.349
0.232
0.116
-.000
= 0.1s 1 ksr
2.10 0.12
2.20 0.00
2.30 -0.12
3,80 0.949
3.80 0.951
3.80 0.949

PROBLEM 5.136
48 kips/ft
S.J35 and 5.136 For the beam and loading shown, and using a computer and step
functions;, (a) tabulate the shear, bending moment, and maximum normal stress in
sections of the beam from x ~ 0 to x = L, using the increments 4L indicated, (b)
using smaller increments if necessary, determine with a 2-percent accuracy the
maximum normal stress in the beam..
■T SOLUTION
W12X30 +t)2*^D = O
AL-1.25 ft - 12.5" f?6 4 (lAS^fe©)^) + (SXloXs.* ) = O
2.5 ft 2.5 ft
RB = 36.8 ict'ps.
W r 4.S - (.6 <X-S>° k-ys/ft
V- -1.8X + 36.8<x-^.S>° + J.6<X-5>' k.'ps
X
ft
0.00
1.25
2.50
3.75
5.00
6.25
7.50
8.75
10.00
11.25
12.50
13.75
15.00
8.90
5.00
9.10
V
kips
0.0
-6.0
24.8
18.8
12.8
8.8
4.8
0.8
-3.2
-7.2
-11.2
-15.2
-19.2
0.32
-0.00
-0.32
M
kip*ft
0.00
-3.75
-15.00
12.25
32.00
45.50
54.00
57.50
56.00
49.50
38.00
21.50
0.00
57.58
57.60
57.58
sigma
ksi
0.00
-1.17
-4.66
3.81
9.95
14.15
16.79
17.88
17.41
15.39
11.81
6.68
0.00
17.90
17.91
17.90
S - 38.6 ;**
6"-
M
691. j_
38. G
- I7.<?l Jcsi

PROBLEM S.137
w = u"(»i;
S.137 and 5.138 The cantilever beamv4B, consisting of a cast-iron plate of uniform
thickness b and length L, is to support the distributed load w(x) shown, (a) Knowing
that the beam is to be of constant strength, express h in terms of x, L, and h0. (b)
Determine the smallest allowable value of h0 if £ =750 mm, ft = 30 mm, w0 = 300
kN/m, and a* = 250 MPa.
SOLUTION
W -=■ ~
V - -
- waxJ
XL
L
Ml -
s-
\y\\
\A/A X-
At X- 2-
ibb2 =
h - h.
h
■v
wax'
6lub
t)4«-: L =• 76"D w**i ? 0.7S w»
W0 = 3oo VM /^ r 3oo x/o3 N/**> j
3/t
_7 (300^(^X0.75)'
"1/ (25"0V/O*Xo-O3o")
-s
= I50 *|0 h^ = J^O knrv,

PROBLEM 5.138
u> = u>0 sin |£
_
tiff
iiu^m
I
L
W
JSK'BK
Ibi
S^::iSK«Bifc
Bm
d
\
*0
5.137 and 5.138 The cantilever beam v4i?, consisting of a cast-iron plate of uniform
thickness b and length L, is to support the distributed load w(x) shown, (a) Knowing
that the beam is to be of constant strength, express h in terms of x, L, and /j0. (ft)
Determine the smallest allowable value of /j0 if L =750 mm, b - 30 mm, w0 = 300
kN/m, and c^ = 250 Mpa.
SOLUTION
TTX
£=v=-^(i-o«^)
M =.^.i-
IT
(*-¥»
, 7TX \
M/--^y-^S.„^)
/- J_. ILL1- 2W»L /t, 2L _. 7T* N
Ca^ h> k[(f-f^|?)/('-f)]
l/*
76.-7
»**\

PROBLEM 5.139
5.139 and 5.140 The beam AB, consisting of a cast-iron plate of uniform thickness b
and length L, is to support the load shown, (a) Knowing that the beam is to be of
constant strength, express h in terms of *, Lt and A0- {b) Determine the maximum
allowable load if L ■ 36 in., h„ =12 in., b = 1.25 in., and 0^ = 36 ksi.
SOLUTION
. P
^
V
M - - Px
IMWPx
s-±-w.*
For ». recT«-n*fc//«iA c**oas scctYom
PROBLEM 5.140
5,139 and 5.140 The beamAB, consisting of a cast-iron plate of uniform thickness b
and length i, is to support the load shown, (a) Knowing that the beam is to be of
constant strength, express h in terms of x, L, and /i„. {b) Determine the maximum
allowable load if L = 36 in., ha =12 in., b = 1.25 in., and a^ = 36 ksi.
SOLUTION
vJ
1 '-*A.\
*
:-LLL
L
^
fcfi^iAj ^b>> £&tf
_ wL
•/*
S^lvt'n* -For vu»
W
* i K*b S
sv
©K^O

PROBLEM 5.141
5.141 and 5.142 The beam AB, consisting of a cast-aluminum plate of uniform
thickness b and length L, is to support the load shown, (a) Knowing that the beam is
to be of constant strength, express h in terms of jc, L, and h0 for portion AC of the beam.
(6) Determine the maximum allowable load if L - 800 mm, h0 = 200 mm b =25 mm
and oin =72 MPa.
SOLUTION
2
+)2Mj = o
(o
<x<^
EcjO&Tir*
±feK" -
Px
3 6un " *6i
3Px
2S*b
*U?
o*x<-^
ctce
S»iU3 J- -P P= 2^hl s Ca)C7gygyo.wYa^oora ^/b* u
x>^ -^ r«fu/
X- ty L-X
<b*>

PROBLEM 5.142
M0
-X
5 '
6^L
S.141 and 5.242 The beam AB, consisting of a cast-aluminum plate of uniform
thickness b and length L, is to support the load shown, (a) Knowing that the beam is
to be of constant strength, express h in terms of x, L, and /j0 for portion AC of the beam.
(b) Determine the maximum allowable load if L = 800 mm, h0 = 200 mm, b =25 mm,
and a,,, =72 Mpa.
JfcUx + M - o
For % » %
Rpv* a. ^offt^jo/ttc ca«ss secfVv* S • ^tn
Af x- 5 h» k -j§ W- KfbJi
Me * S*bk ff WWoQCawKo-W-= zq^ N-
3 o
M

PROBLEM 5.143
5.143 and 5.144 A preliminary design based on the use of a simply supported prismatic
timber beam indicated that a beam with a rectangular cross section 50 mm wide and 200
mm deep would be required to safely support the load shown in part a of the figure. It
was then decided to replace that beam with a buill-up beam obtained by gluing together,
as shown in part b of the figure, four pieces of the same timber as the original beam and
of 50 x 50- mm cross section. Determine the length / of the two outer pieces of timber
that will yield the same factor of safety as the original design.
SOLUTION
R
-- R« -
Z
a
D
\
V
M- C
v
0 < *<±
St
o* y\ -
\-Z
X
H
Ml-
'*n<wj£>
M6
1 =
2
;.2
Xj> r 0.3 *n

5.143 and S. 144 A preliminary design based on the use of a simply supported prismatic
timber beam indicated that a beam with a rectangular cross section 50 mm wide and
200 mm deep would be required to safely support the load shown in part a of the figure.
It was then decided to replace that beam with a built-up beam obtained by gluing
together, as shown in part b of the figure, four pieces of the same timber as the original
beam and of 50 * 50- mm cross section. Determine the length I of the two outer pieces
of timber that will yield the same factor of safety as the original design.
SOLUTION
R* - K6 - ^ = 0.4 w
Shear*. A+o C ' V = O.M w
k-fe.B V- -0-*r W
Area*" A "h> C (0-8V<vOw = 0.%2 */
C +o E (i)(0M)(O.H)*J - 0,oS iV
ArC Mc = 0.4o w
A4» C M = °-40 **
■J «■ /.<?00 m

PROBLEM 5.145
5.145 and 5.146 A preliminary design based on the use of a cantilever prismatic beam
indicated that a beam with a rectsngular cross section 2 in. wide andlO in. deep would
be required to safely support the load shown in part a of the figure. It was then decided
to replace that beam with a built-up beam obtained by gluing together, as shown in part
b of the figure, five pieces of the same timber as the original beam and of 2 * 2- in. cross
section, Determine the respective lengths /, and l2 of the two inner and two outer pieces
of timber that will yield the same factor of safety as the original design.
SOLUTION
P
i
K-
Px + M * O
|M 1 * P*
M:-P)
'tvXMA
- X
IN »
IMU .
IMI
&
car "
s
s
ft
A-U c
C -fo D
i
* 25
- .2.
Af 8 J Ml** Mv
Sc- i- b(tf - it3
*c
Z5
Sk = &3S-O.QS *G.GOft -*
j?a = 6.^-^5"= ^oo-Ff

PROBLEM 5.146
- /-Xfi-V - -Si X
5.145 and 5.146 A preliminary design based on the use of a cantilever prismatic beam
indicated that a beam with a rectangular cross section 2 in. wide and 10 ra deep would
be required to safely support the load shown in part a of the figure. It was then decided
to replace that beam with a built-up beam obtained by gluing together, as shown in part
b of the figure, five pieces of the same timber as the original beam and of 2 x 2- in. cross
section. Determine the respective lengths /, and 12 of the two inner and two outer pieces
of timber that will yield the same factor of safety as the original design.
A^ B IMl6» IMU.
A+ C \V\\^ |MUw(xt/*.*T>)*
A+ D |MIDr iML^Ofc/cas-)*
At b sB^u1' ^<w -- fv
A4* c s^ibk2-- it a)1 - -tb*
x ^i/? r 37rR

PROBLEM 5.147
5.147 A cantilevered machine element of cast aluminum and in the shape of a solid of
revolution of variable diameter d is being designed to support a horizontal concentrated
load P as shown, (a) Knowing that the machine element is to be of constant strength,
express d in terms of^, L, and d0. (b) Determine the maximum allowable value of F if
L - 300 mm, d0 = 60 mm, and o^ =72 Mpa.
SOLUTION
J 4)2^ = O M- Py * O
M- Py
6l« 61*»
^ c. - Tc ■ IT
= 4
t/a
J-^ •(¥£)
1/3
<A„~ [l )
?
31 L.
(32)(0.3oO "
PROBLEM 5.148
5.148 A cantilevered machine element of cast aluminum and in the shape of a solid of
revolution of variable diameter d is being designed to support a horizontal distributed
load w as shown, (a) Knowing that the machine element is to be of constant strength,
express d in terms of y, L, and d„. {b) Determine the smallest allowable value of d0 if L
= 300 mm, w - 20 kN/m, and a^ =72 MPa.
SOLUTION
s- |M
<51
'-«
3> £c"
E
Tii\<
For o. soJftJ cif-cJJti* c^os»s secff'ww
"3
CI - 2

PROBLEM 5.149
M
c?
b-L-x
P
S.149 A cantilever beam AB consisting of a steel plate of uniform depth h and variable
width ft is to support a concentrated load P at point A. (a) Knowing that the beam is to
be of constant strength, express ft in terms of x, L, and ft0, (ft) Determine the smallest
allowable value of A if
L = 12 in., ft0 = 15 in., F = 3.2 kips, and a± - 24 ksi.
SOLUTION
OZMj = o
pq-x)
A+ x= o
SoWt'^A -for n
b- k"
GPL
h-Vg£ -7
M*-f>(i-x>
S = H^
u- <spq-*i
<W<
'^ -/
PROBLEM 5.150
"fifes,
l*-L-y -*l
A+ x= o
5.250 A cantilever beam AB consisting of a steel plate of uniform depth h and variable
width ft is to support a distributed load w along its center lineAB. (a) Knowing that the
beam is to be of constant strength, express ft in terms of x, L, and ba. (ft) Determine the
maximum allowable value of w if L~ 15 in., ft0= 18 in., h = 0.75 in., and 0^ = 24 ksi
SOLUTION
I7
I I 2
IMl = *£*!*
S -
iM
6U< 2e>
Vvv W = -g^b;^ = fr»K"X*»V'r 0.36O *.'pA*

PROBLEM 5.151
40 kN/m
5.151 Two cover plates, each 7.5 mm thick, are welded to a W460 * 74 beam as shown.
Knowing that / = 5 m and b = 200 mm, determine the maximum normal stress on a
transverse section (a) through the center of the beam, (ft) just to the left of D.
■
^fc^|7,5mm SOLUTION
Ho lew A«
.afinn
160 kw
e>
VI
-O^Mj =o
(Co
MD * IIS ku-v*
-160 x -f Qfox)^ + M * O
Ml - 160 x -^xl ku.*
A4
eaw\
- fl?4. 2 * {D* mm*
N
D>Tl*l&
* 5+
rcss
C r *f7*7.S - 23C
<5" =
- i± _ 320*/o*
2«r&* io-fc
Al d s = into «io*m*? » miopia* ***
= 133. €. MP*.

PROBLEM 5.152
5.152 Assuming that the length and width of the cover plates used with the beam of
Sample Prob. 5.12 are, respectively, l = 4m and b = 285 mm, and recalling that the
thickness of each plate is 16 mm, determine the maximum normal stress on a transverse
section (a) through the center of the beam, (ft) just to the left of D.
SOLUTION
RA ^ 2a - ISO W
■*-*-<,
1
v\
A+ Cewfev* <*F bc&^H x = 4v*)
All* X r JL(S-inr £(3-*0 = Xk
2S0 If w V
M0- Sao ItU.^
- %So x + M - o
A+ 3)
m * 0 *1 /c- M £"oo */03
- IW.4WO'P*. * I43.«# MP^.

PROBLEM 5.153
160 kips
5.153 Two cover plates, each "J -in. thick, are welded to a W27 * 84 beam as shown.
Knowing that / = 10 ft and b = 10.5 in., determine tha maximum normal stress on a
transverse section (a) through the center of the beam, (b) just to the left of D.
1- SOLUTION
n 111.
At C x * if* Mcr 7*^ k/P .-H => ^fcVo fcy?, ;„
A4 D X^ ^-^- - 4 ft
M^- (80l(^ " 3?^k.>-ft --38*rO k^irt.
At ce^-re^ of bee* km JTZ -Tfa€*^ + ^-I/»AH
C = 2L1L + o.5i?o " 13. 8S£ ;„
fV.
O^tvi*
i S+
ress
M p.i*T T>
5 - 2i3 i**
No^wa/ S"f«re6S
ff, ^.
■213
(s. os /csr

PROBLEM 5.154
30 kips/ft
5 .
* * *!"■ \-b-\
5.154 Two cover plates, each ^ -in. thick, are welded to a W30 x 99 beam as shown.
Knowing that / = 9 ft and b = 12 in., determine the maximum normal stress on a
transverse section (a)
through the center of the beam, (b) just to the left of D.
SOLUTION
W30x99 . 3° kir*/fi t>5 M.T = ©
14© V\p-tt
^Hok.'i
V
- ^o * + Soy | V Kl- O
A+ ee*4e*" oi k«A^ V - 8-Pt
A+ po^t^, X^ ^06-*»)* 3.5-W
yip ^ &S6.25 ICfPf - 781S *•>•»*.
NorMAi stress O - "J * ^—74^ * *
- 74^8 Jh
5 = ZC°I i'«*
*--f-3£ - ^3 ks;

PROBLEM 5.155
40 kN/m
5.155 Two cover plates, each 7.5 nim thick, are welded to a W46O x 74 beam as shown.
Knowing that o& = 150 MPa for both the beam and the plates, determine the required
value of (a) the length of the plates, (ft) the width of the plates.
JL JLlL I t T I KMTSinm SOLUTION
TTTTTTTI
Li
\ZOkH
M - \&o x - 2o x* W-m
be«.t*-i
00/Kifs D ev>«f £"_> se+ M = M*v
x^ lto^Vico>-fty^X^ig ,.753 m
A+ ce*+e^ tff be&^ M- 32o ltw**n - 32o»/c? W.^
2133* lo* ^^S
133 ^ /o" * *»*
Re^«tVe*i Mawe^T or iAfiV*. X= Sc r 5o3,H v/O*" wim
B«f J • I
fc*eu*
^V*
- 333*/o' + go?. 2_ */0a k»
b - ^11 tn
M

PROBLEM 5.156
160 kips
5.156 Two cover plates, each "J -in. thick, are welded to a W27 x 84 beam as shown.
Knowing that a^ = 24 ksi for both the beam and the plates, determine the required value
of (a) the length of the plates, (b) the width of the plates.
u-i—I fta- SOLUTION
~ =. o
SO x + M = o
wo wr-ft
Set M* - M.*
80 *,-
\J -80x +
f * —T,
V M - SO X Wp-f-r
A+ D
S = z\z \
. s
SO X0 = 436 X0=" S.Z2S ft
£ - IS - 2X„ - V. 35 ft
A+ ee**er of te**x M * (&o)fa\ =■ 74© k.'p.ff * S&to Jr.'p.j«.
= it* 5*© + 185112 t
b = //. Sf /V

PROBLEM 5.157
30 kips/ft
\ ; j_
5. 157 Two cover plates, each "g" -in. thick, are welded to a W30 * 99 beam as shown.
Knowing that o^ - 22 ksi for both the beam and the plates, determine the required value
of (a) the length of the plates, (b) the width of the plates.
h-fc—| SOLUTION
W30X99 ^gofc^/W +)5Mj"-0
3p fcfci/W
f V^* if.
U
?40 ICp \;
S r 26*? In3
- f*?3.l67 k.'rff
2</o ±t/U<*o)*-C'»(IsX'*'>3.K7')
y =
cntio
Z.HX ff J I3.S^ ff
J? " X.- X,
I3-58- 2.HZ - //./6 -ff.
Cenle^ <$ be**. M - 9CO k.>-ff ■- I IS 2.0 lt/p-i»
Reef ire«f wfwei
Soto, sno + aJcbx^oW < ^y - *ax*«o^
b - 14.31 in

fKUJttLJWM 3.158
be applied at end a oi me oeam snown. negieui uic wciguus u» uic uuuu auu
plates.
12 X 225 mm
*
2.1m
.it
1.2 m
2.1
D
m
W310 X 60
-o.i»«rTP
SOLUTION
O ? t-L = ° -1.5 1^^ ;./P * o
Rc =■ /.4GCG7 p
Sleets'- A +o C
C +* B
V = - 0.1Wt7 ?
V * P
Be*«»Vt«i mowie^T* -*
A+ C
P^ * 125. f kN
I * Iu... + * ***
C - 4^ + \X = IC7 *u
= J ~ I6M X|0^ hn^S
~ /£// * /o_t mo3
««/«>*
Pal " J3C.6 VW
P r /;?£ 4 W

PROBLEM 5.159
5.159 For the tapered beam shown, and knowing that F = 150 kN, determine (a) the
transverse section in which the maximum normal stress occurs, (b) the corresponding
value of the normal stress.
SOLUTION
Rji " Rft ~ "o
i
F
•f SlZMj « o
For ©. Tapered Peeran eit/if a.r bfia
S r £fctf * £b(a*kvf
To 4-|V\eJ J't>C«.Tlot^ 07 /V»«\,XI tvii/i*^ E>&iteliirt
Vl r a + UX
i+»*SS
se
- IE
" b
CL- lOC
VJta.:
(a+ kx^3
= o
■k*
_ 300 -1*0
k
3oo
0.6
0.400 *~
•=- 300 ww/th
M,

PROBLEM 5.160
laOmm
5.160 For the tapered beam shown, and knowing that w = 160 kN/m, determine (a) the
transverse section in which the maximum normal stress occurs, (b) the corresponding
value of the normal stress.
20 mm
—| h- SOLUTION
rnfiTTi h t>2^ -- o
&
i*L
3
For -He f^et^eW fc>&a,t*\ Jl - <X ■# fey
ar |20 tnm V = 3<g^ '■ ?■? * 3^ w„/m
F«^ *. reol^n^uiA^ c^oss Sficf/'oui $ = ^bk1 s £ b (a+ kxY"
g o b (a + lw)1
Jiff*
- 3w 1 at *- 2ax - klxO = o
2a+KL "(*Xw«** ©*>)(**) - 0.11 m

PROBLEM 5.161
5.161 For the tspered beam shown, determine (a) the transverse section m which the
maximum normal stress occurs, (6) the largest concentrated load P that can be applied,
knowing that o^ = 140 MPa.
SOLUTION
20 mm
II- p
§\h ^a ~ #* = y
i
F
3
H
£>2 MT = o
M » ^ Co
*■* ■*
^
Bendiwq stress 6* = -r* =■ . . , .
J ob (a+ k* y*
sl§-_ _3P_ o> f x 7 3P J>-*k*^- x-2<a+kx)k
<^x " y> «fcrlca+k»)4S r b ""
Ca+ k^V
b (a+ kx>*
x* =
_ ^
n
e^>
M.
Px>
D«,"h
k-
3QQ- Uo
O.fe.
3&o iM»i/i^iy b= ?o -^

PROBLEM 5.162
S. 162 For the tapered beam shown, determine (a) the transverse section in which the
maximum normal stress occurs, (ft) the largest distributed load w that can be applied,
knowing that o^ = 140 MPa.
L* K2 *
Foi^ +li.e -h»-pev*e<J beam In = CL+ kx
a= /20 *»<" rv - 0s6
, 300- tfo
Fov* recfa-H^o)^ cross SC^+iov. S- gbk r gb(a+ IX I
atx " b Xaa+ky^J" b ( (tx+^y i
gwC ta+kxXL-2*}- ak(Lv-xx)l
= b t (ct+ i<x)a i
b C (<x^ kO3
gwC aL - fgqfkL^ 7 _
(.ex+kx^ ^
aL
02o )(t.^
** * ^a.+ kL ~ ^Xu°O + (3oo X"l-«)
* 0.2^ K.
hito«J*U vM* of M*
hHo^JoXJIt. \ZclXj^ of W/
= 1*7. ?t>S2 y\Os W»i*\

5.163 For the tapered beam shown, determine (a) the transverse section m which the
maximum normal stress occurs, (6) the largest concentrated load P that can be applied,
knowing that o^ = 24 ksi.
SOLUTION
I
3
M
*V
Fw a- +Ajjcv^ed loea^ In = a + k*
For ol recfavigvyia*' cA>ss se&Kon S = & k>h ' G *>(A+ WX )
Bending stress ^ ~ s ~ feCa+kv^
To -find JloceMu* tfT rm*yi^t»»i benJin^ stress set g* = O
l*i " b
d^ . 3P J C >
- If cl- kx
o
[a+kyp
DaU'. a=f.V, K-A^* 0-l«33 in/,
30
x„*-*
30 in.
0.|3S*S
in * a + kxw, = 8 in
XM 3o r

PROBLEM 5.164
5.164 For the tapered beam shown, determine (a) the transverse section in which the
maximum normal stress occurs, (6) the largest distributed load w that can be applied,
knowing that o^ = 24 ksi
SOLUTION
~Jh KK - Ra - ^wL
uiuu
M
3
M
V
L = GO ;„.
Fo^ He +ftne»rc*l fc>eav* to = ct+ l<y
For a ^et+awg^/**- cross Secfu»t S x ^ b h r 6^(0.+ ^*V~
Bendm, stress 6" » y » ^.__-
To 4rtn»l ioc*+ion tfi ni«.»ci*i»io>»*i loendi'n^ stress sef ^nr - O
^ 3w ? ai- + kLx 4ftx - akx* * 3kl* -v akx*-!
V> £ (a+ kx")» 5
= O
al
tnUO
if in.
h* « af kxw - H +(.^(15^ * S.OOirt.
s/aJ/ue of M^ - Sk»6"^/ r (-^-SQ)C.2V) - 180.0 k>-i*
X*(L-X^ (f&XfS^

PROBLEM 5,165
5.16S Draw the shear and bending-inoment diagrams for the beam and loading shown,
and determine the maximum absolute value (a) of the shear, (b) of the bending
moment.
2.5 kips/ft.
U) kips
SOLUTION
- IS PA +OZKO(2.5) 4<Xtt/5) = O
PA = 13 k.'ps
OZM* - o
S^ea^: V^ - 18 te/ps
Ve - /* - te )<*:?. s) T 3 \ar%
Cfot) V = S k;?s
DfeB V - 3 - /$" * - 15 fr.'f*
Aveas onJcr- slnea^ di^^a-vn
D 4b B 5vJy * teV-ia^ = - 7* ki>-ft
Mc - O + £3 - 63 k.p.f*
Ma - 12 - 72 = O
ImL. - 1% (o-p-ft

PROBLEM 5.166
150 kN
W460 X 113
A C
-£37
5.166 For the beam and loading shown, determine the maximum normal stress due to
bending on a transverse section at D.
SOLUTION
PA = 27* k^
48i?6 -(o.»XWd) -(UK/S©)- f«)fo«XftOr £>
&& - 237 W
SWf - MoC V^ 27<*
C +o O V= 17 9 - ISO -- l*«f kM
D to £" V =- 12* - 'So =■ -£i few
V4 = - 21 ^
VB = - XI - a^fto) r - ^37 UtJ
Are4.s ywele^ skect^ Ain^^A.**
A+oC ^VJ* r (0.8X277W 22S.2 kM-r»
VJo»-y^bf siiress 6" =
Ma*l»v>u^ ibe^WirtA (n^i^e^f o«J^s *t pe»f*TD
W roijfed sleej? «*+«»</> W fto * H3
5 =: %l\0O XIO3 **i*i* = ZHOO f/O"6 »*»*

PROBLEM 5.167
5.167 Determine (a) the distance a for which the maximum absolute value of the
bending moment in the beam is as small as possible, (b) the corresponding maximum
normal stress due to bending. (Hint: Draw the bending-moment diagram and then
equate the absolute values of the largest positive and negative bending moments
obtained.)
0.5 m.
M_l
T
0.75 m.
SOLUTION
a
t?t =
\2©
I
L- vc-d v
o
M.
£o
A C
M.
NV + la© (i*o- a) -= o
a
IWooG
S*.Ml
1120 l*o
L_ 2o 4— %a
±
*L Vk.1A\ »| J
M (SUinl ^3
■<*
MBoo - 120 a = H^oo - a
120 a2 = \HHooo. a-3t.6HI ,v>.
"i - - 6*3.0» jKWm
Mo^Jw,,;^ hocm&J stress
*!****
- /3.7? fcst
3..7a*l0*p*i
-£*3

PROBLEM 5.168
5.168 Draw the shear and bending-moment diagrams for the beam and loading shown,
and determine the maximum absolute value (a) of the shear, (6) of the bending
moment.
30 kN/m
24 kN
rf-J-aa-J-n"5*
W (kW-wO
2<».4
SOLUTION
4^ " "5i" %d r \3<f.¥
J r 1.4 m 3.2- of => \.« ^
A^Cft-S u^<=le^ sbca** diet^t^*^.
A+oV Sv.lv « (£)(i.V}(^;0 - ^.«# W.*
IV \ * S4 WW
May^*v iMl - 2?.f i^W- M

PROBLEM 5.169
3 kips/ft
-(5.75
5.169 For the beam and loading shown, determine (a) the maximum value of the
bending moment, (b) the maximum normal stress due to bending.
S10 X 25.4
SOLUTION
-/o(?t + t*5-)05)(3) = O
lo & -te.srXis)ft)r °
(?D = 2^75 ki'p.
Sti
ear
VAr O
M* = -6 + #>.25 * w.%£ kips
V = l«f.*5 - (101(3") - - Z.5.75 ,JCi>s
VD* * - I5V75 + 2V.7S * 9 k,>s
VB-
3 - (3Xa^ = o
kee-k-
-^ lo"e 3oe ~ >V:?.5
e -
H.7S ft
/o-e. = i'.wtt
Arec^S Unde^ sltfia^ ejr^/'Av^
A f0 C $V*ly 'fthOOC-O * -6 k,p.f+
CJoE Sv^y *(i)ft.75yw.30- 33. S 4 tip-ft
D ^ B 5vJx^ (i)teX*) = 13.5 k.p.-fV
B e«d i
vh
M*=°
Mc = O - 6 r - g fc,p.ft
MD ~ 27.g*-4/. 3¥ r -13.5 *,p-ft
M- = -it. 5* 1- 13. S" - o c/j^k*
M
0L/l»<n<>¥^
»m|*
27.8V fcf--ft —
- 334. I fcp-i«
Fo^ roJUeJL &leet se-+»'o" S 10*25.^
°^V " S
S - ;?</. 7 ;„*
M «.* I ww** nor rvKtyf STre^ s
33».l
2V.7
13.« ksi

PROBLEM 5.170
0.9 m 1.2 m 1.2 m
1.2 m
5.170.(a) Using singularity functions, write the equations defining the shear and
bending moment for the beam and loading shown, (ft) Determine the maximum value
of the bending moment in the beam.
SOLUTION
OZMfc = O
Re = 35 kW
V - - \Z + 35<x-0.(7>o - 2tJ<x-^.»>° - ia<y-3.S>a kK/
M = - 12* + 3S"<X-<9.?>' - 2V<*-:?. »>' - '2 <*-3.3>' kN-™
Pt C (* = O.^l Htr -(tf^o.*) r - |o.fi *W.m
P* D (x-a.l^ M0* -(is)(2.0 + S5 0.^") = 16.*'W-»n
Pf e* (y^ 3.3 m) m6 = -(11X3.3) +(^(2.0 - ewXi.iO
= 15*. G feM-i-h
Mo^iw^ |M I - : 16.S Jrn/.n*

PROBLEM 5.171
1.5 ktps/ft
5.171 For tha beam and loading shown, design the cross section of the beam, knowing
that the grade of timber used has an allowable normal stress of 1.75 ksi.
5.0 in.
SOLUTION
'-Z7.S<>
S^ =
'*W|M
MA r -21.5£ M
She***: A 4* B V - S.ZS k;?s
Be*^ to©*e*+* Hft = -37.S6 k.p-ft
Ma * - 27.56 + 18.38 « -?./*
Ma*iWiw IM U ^7.56 kp. ft - 330-7 k.p-i*.
><J*
\.7S
^V?
15.06 1V1.

PROBLEM 5.172
5.172 For the beam and loading shown, design the cross section of the beam, knowing
that the grade of timber used has an allowable normal stress of 12 MPa.
18 kN/iu
SOLUTION
+71^ = o
= o
0.9 m
0.9 m
v^l >l ' J _ ' 1
2MJ
SKWA
A C
% IfV ^
*
MoxiViUim b&nejin^ woiMei't occurs cck H>e ce^fei"
i)2Mt * o
Mt = ^6.73 W-* - 26.73 w/oa WAi
^Jrt €uT = l3MO~r 2.Wi*IO yc -- 3227.S-*/<?*.*

PROBLEM 5.173
40 k-N
2.2 IcN/m
7.2 M
\%.ol
S ■ -
5.173 Knowing that the allowable normal stress for the steel used is 160 MPa,.select
the most economical metric wide-flange beam to support the loading shown.
SOLUTION
S2Mt=0
Shear: VA = ZZM W
V8- " 22.« - (4.5-^2.Z\ - \3.0Z UYJ
Vs* - 13.OZ. - 40 = - ?€.M KM
Vc - -ZC^S - (Z.7)(l.Z) * -S7.1Z kW
htea.% OnAetf- sliea^ e*\a.^^en.v^\
Blot JVJ* ■ fttaJ'X-ZC.M- 3?.n)=-*>;36r*K/»
Ma - O -» 30.265" * 80*86$ kW-vn
S0.g6$~~ 80.86S" * O e^e.ks
5ria.pe
W 410 *3*.8
W 36o * 3^
W 3/o > 3S.7
W ZS'o * 44. %
VV 200 * £>
S OoW)
637
578
6"^ —
SSS
s\z
SOS * IO~ »* ~ SO&XfO V^r^
Use IV 3/D* 35.7

PROBLEM 5.174
Idps |
Bl
J1 kips/ft
J II III
C D
- 6ft
-4 t-
20 Idps
2ft 2ft
2° n kiPS m
4_i_^*yT
S. 174 Knowing that the allowable normal stress for the steel used is 24 ksi,.selectthe
most economical wide-flange beam to support the loading shown.
SOLUTION
2ft 2ft R* " f?F " i"0 fc.'pt.
M**i'i*if^ te«diM* vwo^c^l occots od ce^e^ o^ t«a^
•OZtfjS O ~(7)(S3)*-(S)(ZQ) + (\.s)(&0i) •* Mj * O
53
*
57
S^- )J- r^ r M0.7ffiV
Shape -
W $** * C*
W 21 x^
W IS * 76
W 14 x 77
WH»w
W 14 x <*6
S(l^
/$*
127
IMC
134
133
\s\
l>s« W^\x6^

PROBLEM 5.175
t
5.175 A machine element of cast aluminum and in the shape of a solid of revolution
of variable diameter d is being designed to support a distributed load w as shown, (a)
Knowing that the machine element is to be of constant strength, express d in terms
of K, L, and d0. (b) Determine the smallest allowable value of d0 if L =250 mm, w = 30
kN/m, and a^ =11 MPa.
WL.
2
Oat
2Gat
-OTMT = o
+
-^x + wx£ + M -
po>r
Ol SO
Jlic\
4
c = 4 I-Ic3
C^u-MTf i<>
e.»fc.J''A^ Cross secrVe**
i£3
33.
mt^i

PROBLEM 5.C1
+ Xn
< X
—*i-i*i
ip°
A A
i
'o ' ' *-
^n
I
»-
P.
'
t!
5.C1 Several concentrated loads Pi (i = 1,2 n) can be applied to a
beam as shown. Write a computer program that can be used to calculate the
shear, bending moment, and normal stress at any point of the beam for a given
loading of the beam and a given value of its section modulus. Use this
program to solve Probs. 5.23, 5.27, and 5.29. (Hint: Maximum values will occur
at a support or under a load.)
SOLUTION
REACTIONS flTft AN J B
• R.L-znpL-**
■nzn^o: Rj^ ^.•iyL
V
L)Z^-*)
\"tt-K*
Wt t/*E STEP FUNCTIONS (See bottom of page 318 of Uxt)
WC PC Five: fpx >CL TUIH S7P/=\ = ' SLSC STPft = 6
ITX> a^t-LTHEti STPB=I ELSE 5JP&-6
IFX>ZC- T^tcW jTp(l).= < ELSE STP(l).=J)
Q- = M/5j y/tare 5 fs o^rainfol from Append f Y C .
PRO&RflM OUTPUTS
Prob. 5.
RA=80.0
X
m
2.00
Prob. 5
Rl = 44
X
m
0.00
1.60
4.00
5.60
23
kN RB=
V
kN
0.00
.27
.0 kN
V
kN
-20.00
24.00
-16.00
-16.00
80,0 kN
M
kN.m
104.00
*2 = 16.0
M
kN.m
0.00
-32.00
25.60
0.00
Sigma
MPa
128.55
kN
Sigma
MPa
0.00
-31.07
24.85
0.00
4
«4
Prob. 5
Rl = 52
X
ft
0.00
1.00
3. 00
9.00
11.00
29
5 kips
V
kips
-25.00
27.50
2.50
-22.50
0.00
R2 = 22
M
kip.ft
o.oo
-25.00
30.00
45.00
0.00
5 kips
Sigma
ksi
0.00
-7.85
9.42
14.14
0.00

PROBLEM 5.C2 g q2 A timber beam is to be designed to support a distributed load and
up to two concentrated loads as shown. One of the dimensions of its uniform
rectangular cross section has been specified and the other is to be determined
so that the maximum normal stress in the beam will not exceed a given
allowable value o-j]. Write a computer program that can be used to calculate at
given intervals AZ, the shear, the bending moment, and the smallest acceptable
value of the unknown dimension. Apply this program to solve the following
problems, using the intervals AL indicated: (a) Prob. 5.75 (AZ, = 0.1 m), (b)
Prob. 5.79 (AL = 0.2 m), (c) Prob. 5.80 (AL = 0.3 m),
SOLUTION
REACT \ OrtS ftT A A HP B
Rfi ' Pt ^ + ^<%T^) - R&
U/Z USE STEP FUNCTIONS (See hotto^ *f ^ S<f8 of text)
FQK i -0 TO^
W£ DEFlNZ
IF x >a twen STPA~I E.L5Z 5TPp-0
IF * -> CL + L^HEN 5T?-6~i £L$ExSTB = 0
IT £ > X( TH?y STP} ~i £L$e ST?I ~ 0
I? X >
IT 7L
IT 7c > x
THEti
STpZ=r EL£E 5T?l= 0
5T/>3 -f el$£ sSTPi ^0
V=" fy STPf\ + 7?g STP3- "P ST?1 - P? STT2
M* RA (z-<i)?>TPfr +Re (z-a-L)STP3 - P, (Z-z,)5TP?
~% (x-Xz)STP2-$«T(X-Xf)z3TP5 ^(7i^fST?H-
IT M/f/W/i/ DiMZAlSfCA) is hi
From S-jth2-, vVe /,„•<> h = ]/€S/t
IF UNKh/QyJN V\NE0/$\ON /S t'
From $-±thz v« h*,L/e
t^6S/H
(CONTINUED)

PROBLEM 5.C2 CONTINUED
Prob. 5.75
RA = 2.40 kN RB - 3.00 kN
X V M H
m kN kN.m mm
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.40
2.40
2.40
2.40
2.40
2.40
2.40
2.40
0.60
0.60
0.60
0.60
0.60
0.60
0.60
0.60
-3.00
-3.00
-3.00
-3.00
-3.00
-3.00
-3.00
-3.00
0.00
0.000
0.240
0.480
0.720
0.960
1.200
1.440
1.680
1.920
1.980
2.040
2.100
2.160
2.220
2.280
2.340
2,400
2.100
1.800
1.500
1.200
0.900
0.600
0.300
0.000
0.00
54.77
77.46
94.87
109.54
122.47
134.16
144.91
154.92
157.32
159.69
162.02
164.32
166.58
168.82
171.03
173.21*^1
162.02
150.00
136.93
122.47
106.07
86.60
61.24
0.05
PROe^RM 9\)TP\J7S
Prob. 5.79
RA =»
X
m
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
2.70 kN RB =
V
kN
2.70
2.10
1.50
0.90
0.30
-0.30
-0.90
-1.50
-2.10
-2.70
-3.30
-3.90
3.60
3.00
2.40
1.80
1.20
0.60
0.00
M
kN.m
0.000
0.480
0.840
1.080
1.200
1.200
1.080
0.840
0.480
0.000
-0.600
-1.320
-2.160
-1.500
-0.960
-0.540
-0.240
-0.060
-0.000
8.10 kN
T
mm
0.00
10.67
18.67
24.00
26.67
26.67
24.00
18.67
10.67
0.00
13.33
29.33
48.00 <^|
33.33
21.33
12.00
5.33
1.33
0.00
Prob. 5.80
RA =
X
m
0.00
0.30
0.60
0.90
1.20
1.50
1.80
2.10
2.40
2.70
3.00
3.30
3.60
3.90
4.20
6.50 kN
V
kN
2.50
2.50
9.00
7.20
5.40
3.60
1.80
-0.00
-1.80
-3.60
-5.40
-7.20
-2.50
-2.50
0.00
RB =
M
kN.m
0.000
0.750
1.500
3.930
5.820
7.170
7.980
8.250
7.980
7.170
5 . 82 0
3.930
1.500
0.750
0.000
6.50 kN
H
mm
0.00
61,24
86.60
140,18
170,59
189.34
199.75
203,10 ^
199.75
189.34
170.59
140.18
86.60
61.24
0.06

PROBLEM 5.C3
w
_ J L
5.C3 Two cover plates, each of thickness tt are to be welded to a wide-
flange beam of length L, which is to support a uniformly distributed load w.
Denoting by o^, the allowable normal stress in the beam and in the plates, by
d the depth of the beam, and by Ib and Sb, respectively, the moment of inertia
and the section modulus of the cross section of the unreinforced beam about a
horizontal centroidal axis, write a computer program that can be used to
calculate the required value of (a) the length a of the plates, (6) the width b of
the plates. Use this program to solve Probs. 5.155 and 5.157.
SOLUTION
R
(q)R<louked Length of Plates
z?-Lx -h K ■= o
Sofrioj> il\t Q\Jadraticf x- £ -^*" *+k
Compuft PC ^d a* L -2X
(6) Required Wiclttt of Plate*
Af tfiidpctnt C ^ beam''
Set m = 2S&*:
K
~c
^=A
Compute .■ c -1 + ~cL
From <T^^ic ^„^e .J=-^£
^u) But ^r^ip)ak~Tk+2[^bi^bt(d±tf)
3 So1i/inj> for b ;y £ - . ^ (T-TL)
PEOfrttftM OUTPUTS
PROS, 5*155 1 W M 0x7 %(& -ISmra,
IfbOfiftfwi
Prob. 5.155
a = 4.49 m
b = 211 mm
VK03.5.1F7: W30X39 Cy- 2 2Aai
•HT--30/ops/ft, L--")feft, trS/9//7,
Prob. 5.157
a = 11.16 ft
b = 14.31 in

PROBLEM 5.C4
25 laps I 6ft I 25 laps
Ar! i_
E
*:-*! it
^ <■
i-8
Jf?e
3
£
(2C)
A
A
^ d. f'
— jc —*
i
s
13
(3
^ = P
R
x-^i
L
h
=3-
•i B
K
£>
5.C4 Two 25-kip loads are maintained 6 ft apart as they are moved
slowly across the 18-ft beam AB. Write a computer program and use it to
calculate the bending moment under each load and at the midpoint C of the beam
for values of x from 0 to 24 ft at intervals A* = 1.5 ft.
SOLUTION
Nomjig^L: Length of- beam ~ L - f8ft
Loads: V r^. = lp = IS k>'p*
Wt note that &<Lfz
(I) FRom uc = o. to x. =.. d '
■= P(L-z.)/l
/i Ol
(Z)FROM x.j=ci to
+) TMg-o: P(i-x)+P(L-X+d)-RAl~0
??A=r(2L-Zx+d)/L
Under p^ H^^z-Pd
Undtr rz ; Afz r ^ (x-. d)
(2^0 FRon X - Cl TO X_z L/l \
-~RR(L/z) -T(i-zxicL)
(ZB) FKon KrLll TO X :Liz -id:
(ZC) FROM &=-L/zid TO Z.-L:
(3) FRon X.-L TO z=lfd:
+!> £Mg- o: P(Z-x+d)-J?AL -=-0
(CONTINUED)

PROBLEM 5.C4 CONTINUED
PR06RP\n OUTPUT
P = 25 kips, L - 18 ft, D = 6 ft
X
ft
0.0
1.5
3.0
4.5
6.0
7.5
9.0
10.5
12.0
13.5
15.0
16.5
18.0
19.5
21.0
22.5
24.0
MC
kip.ft
0.00
18.75
37.50
56.25
75.00
112.50
150.00
150.00
150.00
150.00
150.00
112.50
75.00
56.25
37.50
18.75
0.00
Ml
kip.ft
0.00
34.38
62.50
84.38
100.00
131.25
150.00
156.25
150.00
131.25
100.00
56.25
0.00
0.00
0.00
0.00
0.00
M2
kip.ft
0.00
0.00
0.00
0.00
0.00
56.25
100.00
131.25
150.00
156.25
150.00
131.25
100.00
84.38
62.50
34.38
0.00

PROBLEM 5.C5
A
u
R
Wo,
i
vp
JB
R
B
-^H
5.C5 Write a computer program that can be used to plot the shear and
bending-moment diagrams for the beam and loading shown. Apply this
program with a plotting interval AL = 0.2 ft to the beam and loading of (a) Prob.
5.83, (ft) Prob. 5.125.
SOLUTION
"REACTIONS f\7 A /^Ni> B
DSiNk F3 Dift^Rf^M OF Blftrt •
OEMfl-0'- R^L- Pb- Mta(<X/z)
Rb =(l/L)(Pb + £«ra*)
= 0
**--
P -f 4tfa - /j>
lA/£ Uht STcP FU fit! IONS '(See b<rttAin of page 3*8 of tt*t)
StT n r L/&L , rOF c±OT0 <n : X ^ (A L) I
WE VEt/NC ; ITx>a TWN 5TPP\-\ ELSE SJP^-0
IF X> b THEN STP3 = / EL5E. 'STPV-Or
M r R^x - i-^crx2- +£«x(Z-a)'STPfl - .P(z-b)SrP3
LQCftTC ahD print (jc, V) /WJ> (x> M)
$f;H NtXT PP\GE:5 R7R f^OGRAM 0i'TPuTS
(CONTINUED)

PROBLEM 5.C5 CONTINUED
TROGRPitt CI\)T?\)7 TQZ?5<83_
PROBLEM 5.83
RA = 48.00 kips RB = 42.00 kips
Shear Diagram
x(ft)
Moment Diagram
x(ft)
(CONTINUED)

PROBLEM 5.C5 CONTINUED
PRogzm O^rpor FOR F5,125
PROBLEM 5.125
RA = 40.50 kips RB » 27.00 kips
Shear Diagram
x(ft)
Moment Diagram

PROBLEM 5.C6
-i*.
(a+tft^-a)
-&>
!_/_
e
L
J ft
5.C6 Write a computer program that can be used to plot the shear and
bending-moment diagrams for the beam and loading shown. Apply this
program with a plotring interval AL = 0.025 m to the beam and loading of Prob.
5.124.
SOLUTION
at
A ' I "3
WE U5Z STE-PP^ctioa/s (See Bottom oi fa^e 34& af £exfc)
SET 'Ti-L/AL, POR 1 = 0 70*1:' X=(£L)l
W£ jfe-Tint. \ IF x>a then: stpr = i tist jt^Pi-o
IF Xib TH<zN STpB=l ELSE S7P3=0
V- Rfl - <iA(X~a)5l?fr + wT(x-t>)STP&
PR06RAW OUTPUT ON Ntxy P^&E
(CONTINUED)

PROBLEM S.C6 CONTINUED
fftOfrRfln OUTPUT
PROBLEM 5.124
RA « 29.50 kips RB « 66.50 kips
Shear Diagram
x(m)
Moment Diagram
x(m)

CHAPTER G

PROBLEM 6.1
6.1 Three full-size 50 x 100-mra boards are nailed together to form a beam that is
subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force
in each nail is 400 N, determine the largest longitudinal spacing j that can be used
between each pair of nails.
SOLUTION
WM
&
fob
A
%
1
si ~ JfO rut*
= XSo*ic>~G
m
= V3. = (/5o^)(^OWO-Q ^ \3.333k/os NA*»
X
= 2 Ftoif
Z%M$*lO
<J- " 13.333 «/o*

PROBLEM 6.2
6.1 Three full-size 50 * 100-mm boards are nailed together to form a beam that is
subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force
in each nail is 400 N, determine the largest longitudinal spacing s thai can be used
between each pair of nails.
6.2 For the built-up beam of Prob. 6.1, determine the allowable shear if the spacing
between each pair of nails is s = 45 mm.
SOLUTION
/S - SO ****>
VQ. o r
r 28fc|3S*/04 *„*
^

PROBLEM 6.3
3 JS in.
¥//////■/?/<
*
s = 2 F««
ti
Y///?//;/,
0
BP
*ai^
~T
3 3
6.3 A square box beam is made of two "J" * 3.5-in. planks and two -j" x 5-in.
planks nailed together as shown. Knowing that the spacing between nails is ^ =
1.25 in. and that the vertical shear in the beam is V- 250 lbf determine (a) the
shearing force in each nail, (b) the maximum shearing stress in the beam.
SOLUTION
r -- £ KK - itKw;
A -- &)(.%) * Z.1S i«x
- 7.?6<t + P.^^7 r |0,2« in1
z™ zt L2>i-s?<&)0.s) ?

PROBLEM 6.4
3 3
6.4 A square box beam is made of two j x 3.5-in. planks and two 7 x 5-in.
planks nailed together as shown. Knowmg that the spacing between nails is 5 - 2
in. and that the allowable shearing force in each nail is 75 lb, determine (a) the
largest allowable vertical shear in the beam, (b) the corresponding maximum
shearing stress in the beam.
SOLUTION
- &(s)(sf- £&*)(%.$)* ' 39.578 .V
W/AW/t/A
A
(a)
5.
*
VQ
1.75 ^
^^^
S>
to
o.vs
T
a * i^on - 3.75- ,■„*
Si t 2.S -\ " *. m~ .'n.
^ = AE-* r WVys) = 7S A/im
v«-^ - f3T;%?(T5> *37*A
r»if
- 7.^6^ + 2. ?T7 - 10.266 in1

PROBLEM6.5
.#
W250X44.8
6.5 The beam shown has been reinforced by attaching to it two 12*175-mm
plates, using bolts of 18-mm diameter spaced longitudinally every 125 mm.
Knowing that the average allowable shearing stress in the bolts is 85 MPa,
determine the largest permissible vertical shearing force.
SOLUTION
Port
Too pi&T<
W XSo * *N. 8
got pf«+e
A 0*n
Z\oo
51X0
**
i(io<^)
a. J 2CA , 12
* -- IS*
1= TAJ*" + Zl * 152.So xid6 ^**
I5*.3o*yo"c mV
7r 3 ixJk/o-1
«8 Vfi V- ^-.--^'"•30</0"CY^-|y/0l)= lM.6w/o»N
80.6 W

PROBLEM 6.6
W250X44-8
6.5 The beam shown has been reinforced by attaching to it two 12 * 175-mm
plates, using bolts of 18-mm diameter spaced longitudinally every 125 mm
Knowing that the average allowable shearing stress in the bolts is 85 MPa
determine the largest permissible vertical shearing force.
6.6 Solve Prob. 6.5, assuming that the reinforcing plates are only 9 mm thick.
SOLUTION
tvt
Top pJ*CT*
waso*m.*
Z
A(w*M
\S1S
S7Zo
\S1S
A (O
* 137.5"
0
137. S
AJUioW)
A?-777
o
X^.lfl
si.sss
I (lo\~?)
0.0*1
71.1
o.ou .
■71.1*1
y/o_t ^
X = TAJ* + 2l r 130.6SV/04 ^ " 130.SB
^ £ .|*S*|o-»
X ID wi
_ VQ

PROBLEM 6.7
C8X 13.75
S10 X 25.4
y//////&$
6.7 tnd A* A column is fabricated by connecting the rolled-steel members shown
by bolts of j -in. diameter spaced longitudinally every 5 in. Determine the average
shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y
axis.
SOLUTION
< = (41 + (*.)«
y
5.
2's
- & + 0.30S=r JT. 303 in
" 0.5*33 i«
= -f - x r S.%oZ- O.S33 -- 4,770 in
of ,w+;»
R^t
Slo*2i~.W
CS-13.75*
2
A (inO
4.04
JCv)
H.77o
O
4.T7o
A J1 (in*1)
0
183.8V
1 6V)
1.53
1.5-3
1*7.06
I e ZAd*4 XI * 183. &^ 4 137.06 - 3'^-^ iV
Q. = A ^ -- (4.MVH.77o} - l<*. 371 .V
bit
r Phif B 4.6H<fr
Awi
O.MWr*
10.5^ k%i

PROBLEM 6.8
CIO X 25
6.7 and 6.S A column is fabricated by connecting the rolled-steel members shown
by bolls of ■£ -in. diameter spaced longitudinally eyery 5 m. Determine the average
shearing stress in the bolts caused by a shearing force of 30 kips parallel to the v
axis
SOLUTION
* J* t* XV
Pert
Top pJ**f<
Cio**s-
C totzf
H
AC.V^
7.35
dM
0
0
AJ'OVO
UZ.S&
I Cm*)
0,06
qi.a
0.OL
\u.sz
- 5.\%75 i.
I - ^Ad* + II - Z%2.SC> + IS*. 5" 2. r 4£5.OS m
Q J 4CS\e>& »
Fun r i^s - (i^(i.7S67)Cs^ - Y.3?4 k;r&
f.
Uit
r £±~£ - ^-393
Afc.it
o.4m%
<?.94 U%,'

PROBLEM 6.9
6.9 through 6.12 For the beam and loading shown, consider section n-n and
determine (a) the largest shearing stress in that section, (b) the shearing stress at
point a.
15 kips 20 kips 1*5 kips
0.6 in.
v(k.>s)
IS
(^
^A,
A, ~^ *•.•*
SOLUTION
lin.
i r-""°'-i i
10 in.
T
0.6 m.
P«A
Top FJrt*
Wet
B«t Fi*3
r
A GV>
3.3o
G
a (..)
4.7
0
4.7
A J2 GJ)
132.£4
0
?cr.os
I G»*)
O. IS
0, IS
21.66
I' ^Ad2 -♦ TI r *«.7* !«'
- 31.83 iV
t = 0.37* ;*
P^t
©
®
T
A<V>
G
*G^
4.7
A J (V)
22.3
3.£3
31.23
7,<r0 V*J
(V>
a£
A,
Q.- *2A^ * »8.8S i»3
t - 0.37S Jh
PftrT
7
AGnM
Q.lS
jOJ
4.7
4.2
Ay crt
0.63
2S.S3
f -
_ Og)(2g.83)
It ' [2%G. 74X0.375)
" G.70 As/

PROBLEM 6.10
10 kip? lokips
16 in. 12 in. 16
VtefO
-io
6.9 through 6.12 For the beam and loading shown, consider section n~n and
determine (a) the largest shearing stress in that section. (6) the shearing stress at
point a.
SOLUTION
£ro»% He s^c^c diV^ron V - 10 k;ps 0,+ n-rt.
(a)
rA',
>
■2
AC
Q~- A.y--,A»j»r (3i(i)(i.7sU ftKW?)(i)
4.«5 <*'
t • ±*± *
v. a
IW.
2
4^
ft*
t -
r =
Ay r en(*J(us
i + i * ' »«■
va (loys.s")
11 (iw.sasxn
* 2.**0 Its,"

PROBLEM6.il
6.9 through 6.12 For the beam and loading shown, consider section n-n and
determine (a) the largest sheanng stress in that section, (b) the shearing stress at
point a.
10 kN 40 m«iT"
150 nun
SOLUTION
At
V - lo kW
to)
^m^
©
75
I - I, + ^1,
u
- 2*.C?r«/Ofc 4 V[ O.OoMMO* 4- J?.fcS«»|0* ]
3?.5^y/o* »*»mH
37.58 v/o Ho'
Q. r Aty, + 2 A,y,
r 0»o)(75)(i7.5) 4 (2K$bXiO(«)
= 3C*rf.or*l0* ™«S - 3d.oS*to m
t * loo
•****
*■ O. IOO m
*,
It
(V)
^£/,w
'It"
s-SJaigsv • <wp*.Wk*
a - a, 5, f %k^
= 3CZ. S */o3 **s - 3oj?.S x/o-4 w3
(39.S8*lcT* )(o. loo) *° ^

PROBLEM 6.12
K-450 mm
*&
.%
,-©
w
6.9 through 6.12 For the beam and loading shown, consider section n-n and
determine (a) the largest shearing stress in that section, (b) the shearing stress at
point a.
SOLUTION
t - 6 mm
h 192 mm —A
I,5
I
- 4 1, + ZIZ + I, - 37.77x/Ofc
►viwi
37.77 » /o"' ^
(a)
T
J
V^
,-©
■©
t:
It
= 206.82*/03 ww3 - 206.82 WO"c m*
i^£l^gi^, „,.|,I0«P*. IN., MP*
-t *
t '
6 h/o"* m
Vq (l25xlOa)(f7^.C|x/Q"t)
<?C.9*|0*P«. - <*£.<» M?a,

PROBLEM 6.13
6.13 Two steel plates of 15 * 220-aun rectangular cross section are welded to the
W250 x 67 beam as shown. Determine the largest allowable vertical shear if the
shearing stress in the beam is not to exceed 100 MPa.
r 220 mm —H
~m
W250X67
—_ 15 mm
257 mm
SOLUTION
C*.>» c<j AeJie wofweyvt ot i*e*-ri
15 mm
* A r ^2 , Jf r l36
Port
Top pi*TC
W 250 x 67
Boi pAi+e
J
A U^O
330 0
33 Oo
fll (*iO
*»3C
0
136
AJl0oW}
61.036
0
£ 1. DS6
122.07?
I (10* ww*}
0.063
0.062
\o1.n<J
T =■ rAJ1, + ^I - 226.2 */©*• mi* - zx&.xtid*' *,*
,0
Par-t
lit/MA
-®
^
(1) T„p pl-k
© T.p «*.«
A^ (io***?")
4*J8.8
3*6.*
56.6
8^1. 8
S9/.8 x /o"6 mj
r.
v =
VQ
It
Itr^ _ (m.3«/oft)(g,9Wo"»)(looi«/oc) _ y/o»
Ml.* */0"*
a
- 226 kM

PROBLEM 6.14
W250 X 67
6.13 Two sleel plates ofl 5 * 220-mm rectangular cross section are welded to the
W250 * 67 beam as shown Determine the largest allowable vertical shear if the
shearing stress in the beam is not to exceed 100 MPa.
6.34 Solve Prob. 6.13, assuming that the two steel plates are (a) replaced by steel
plates of 10 x 220-mm rectangular cross section, (6) removed.
SOLUTION
Caic^xo^e »v\t>r^etftt of me/-M4 $ot- pa»-*t (.Ck.)
j 10 mm
* d« *!?+ f * i«s.s««
Pc^+
Top ©JL+e
2:
A(**rt
23.00
2*X0o
dl (*nm)
*
133.5-
0
Ad*(K>fc«^
0
7?. 4a
I (10 %«*
o.oib
o.oca
10 4.0«*
I - Z"AJ* + II = i9zn*/oe »MH r isi.m + io* V
Par*
Q Top oiJe
©Haif vcb
1
A (^
ZZoo
3203
1004
^ (^^1
133.5
St.4o
AjGo^M
?*3-7
386.4
736.7
CLr TAy r 736.7*|03 vm**3 - 736.7 */©"'
'J
M
-1
1 - tw r S.<? m*, * $.?*/0* M
tm«*
' xt
vntvi
/Of v JO"6 h?V
= 443* /O** wv3

PROBLEM 6.15
1.S kN 3.fi kN
6.15 Knowing that the allowable shearing stress for the timber used is 825 kPa,
check whether the design obtained for the beam indicated is acceptable and, if not,
redesign the cross section of the beam. Consider the beam of fa) Prob 5 75 (h)
Prob. 5.76.
40 mm (.^SOLUTION
0.8 m 0.8 in 0.8 m
+© PROQLEM 5.75"
|V) , ^ IA WW V» » 173.2 »«
^**
^- r f 6wVto-* " «^>-/^ P-- ' «<> ^
SZ$ kP*
De&tan is dtce^TcvMe/.
tO kN/in
120 mm (b) SOLUTION
jtoii'o* +o Pf?o8L£M v£".7C
lVL* - 25 kti Vi-- 36/ mm
A - bh * (Uo)(3Cl1 ? if3.32*/04 mi*
Fpr a. rec+fc^jw»A^ cross secft'ew £.+.** - a~~K
- ^.^rx/o"1 WM*
* ■ f * J^£^ = ^..
V> - 37*i m»v\

PROBLEM 6.16
\^
II-
1.2 kip/ft
6.16 Knowing that the allowable shearing stress for the timber used is 130 psi,
check whether the design obtained for the beam indicated is acceptable and, if not,
redesign the cross section of the beam. Consider the beam of (a) Prob. 5.77,(6)
Prob. 5.78.
id) SOLUTION
V*,^ - i (SHl.^ * 3.6 leC|»«
a = G. G7 ,V
Fov* ex recTaw a^-fa*^ sect To* 2**** " J
'm*.«
3
I3o
'»!
Desiah is <a.cce^TA,k>ie .
200 lb/ft
(k*) SOLUTION
oJoT.'ow TO PROBLEM 5". 7 8
WL* - lOOOife b^ MS in.
A - (bKxt^ *.b* * \7.^o ,V
Pe?r tf>- r^ecf A«gji«/ croiS secTio«
r.
- iM
*****
^ A
130 ^s/

PROBLEM 6.17
90kN
90 kN | | | 90 LN
6.17 Determine the average shearing stress in the web of the beam indicaled and
check whether the design obtained earlier for that beam is acceptable, knowing that
the allowable shearing stress for the steel used is 100 MPa. Consider the beam of
(a)Prob. 5.8l,(6)Prob. 5.82.
(a) solution
Frew H« so/u+iow -to ?roblew S.&l
WU,* ISO M
TVit %tJtc+t<A s«t^i«* *'& w hio * so
For Hat secT>'©»" t#~ 7.7 ^m J* Hoi wim
0.6 m f 0.6 m
0.6 m
-S t
AweL r tMd ' 3.\Z*to% mmX * 3.l3*/o~ k,
C - ^r^ r "L8°/IP* > r Sr.txt*' P* * 57.1 MP* * loo MP«
"* Awet 3.13 * |o-1
5°fcNA» (fe) SOLUTION
F*»m+k« joJffion -re PftoflLEM 5-8 2
Tit ithJttA «&+i>w ts W 2to * 2S.V
Fo^ + t\at seer*'©* tw«" 6.4** d * #60 mm
0.8 in
0,8 m
1
''**■ * _
So y /os
Aw^L ltd *IO
-4 - ¥».U|o' Pa s 4ft. I MP* ■< loo MP^

PROBLEM 6.18
24 ki
ips
I 2.7 Icln.v'j't
I II II I II I'll! I
6.18 Determine the average shearing stress in the web of the beam indicated and
check whether the design obtained earlier for that beam is acceptable, knowing that
the allowable shearing stress lor the steel used is 14.5 ksi. Consider the beam ot'(o)
Prob. 5.83,(6)Prob. 5.84
(.<*-) SOLUTION
Fro* He so-L4i'o*v -U pf?0BLeM S.&Z
V
«w*y
- 4* k;r*
Aw*t r tw<^ - (0.4to)C?6,7l) r tf.2«*m*-
-t.
WL.„
4?
3.«JF ks, <* n.S ks.
1.-5 kips/lt (t) SOLUTION
He s«/uT.*on +o ?RoBL£M 5.8M
F°r +M scdi'oh tw - 0.355 .V d= 17.^.*.
A^t = twJi ' (O.Z$$)(l?.<n) - G.39 .**
T.
= ?.S2 ksi <■ IV.5" k».

PROBLEM 6.19
V/2
6.19 A simply supported timber beam AB of rectangular cross section carries a
single concentrated load P at its midpoint C (a) Show that the ratio r„ lam of the
maximum values of the shearing and normal stresses in the beam is equal to hl2L,
where h and L are, respectively, the depth and the length of the beam. (b) Determine
the depth h and width b of the beam, knowing that L = 2 m, P - 40 IcN, r„ = 960
Uganda.- 12MPa.
SOLUTION
tons
TVs
-p/2
PL/V
(tf S ■ T^*
(O 6"m- *?*
■ft./- y>e
_ 3PL
^ftc/faii a vAt*.* sccf/an
'3 '*" "' " 6U U */ofc
So/ft'na eyoajio* C3) -p©* b
b--
exfo^icf)
3P . _____^___
= 17.7X/6
=■ 97.7 *.
-I
M

PROBLEM 6.20
Coo
6.20 A timber beam AB of length L and rectangular cross section carries a uniformly
distributed load w and is supported as shown, (a) Show that the ratio r_ lam of the
maximum values of the shearing and normal stresses' in the beam is equal to 2hIL,
where h and L are, respectively, the depth and the length of the beam, (b) Determine
the depth h and width b of the beam, knowing that L - 5 m, w =
8 kN/m, t„ * 1.08 MPa, and om~ 12 MPa.
SOLUTION
2
v/L- ^
see*.*
r - J. ^r * 3wL
en
A * bh (X)
(a)
Fro** be«c*('h^ W6»»en7 di'*.* ita**^
CO
- 6/. 7
CI.7 */o~4»v,

PROBLEM 6.21
6.21 and 6.22 For the beam and loading shown, consider section n-n and determine
the shearing stress at (a) point a, (b) point b.
200 kN 20O JcM
150 mm
0.73 m
0.75 m
50 mm
SOLUTION
8a * RB r Zoo kN
M *■
75 mm 75 mm 75 mm
koc^4c cW^ofcj a*J c«Jcji«j)> wo^e^i of >W-r-|-r«u
(a)
©
^
^
Pa^i
0
A («„0
22SOO
_V (tnm)
|J?S
AjOd'^M
as/.??
/687.S
<J Ct«i~< i
So
So
Ad*(loW)
28.13S"
55.25
23.438
r, _ l<87.5*ios _ 7C
<V A^ • (75H£oV)oo^ r 37£*|03mmH r 37rx/oCm"
t " ~7Sm», t 7S*l6% Kt
It * (TU68 Kr*\(.is*\c>
(fa)
^
7! J
t r 75-,
- 75** /d* ^
- 18. «Z MP*

PROBLEM 6.22
6.21 aod 6.22 For the beam and loading shown, consider section n-n and determine
the shearing stress at (a) point a, (b) point b.
SOLUTION
7.25 in.
Tin.-*-
Win.
8in.-
uoc^Te c&rttrot'tfl &tv:| ca/cj «vte wowewi ot fn^z-tV*..
tcH
f—
^
0-75
it
p**t
z
A(.V)
4.S75
lo.«7S
15". 7£
y(-'0
6.375
3.CZS
Ay (irs3)
33.^a
7?. IV
J(i«>
Ad'fr")
2*J.££"
H.OI
355*
IC.vM
0.23
*7.<S
47. S£
r " TA " IX.74- ' 4*G3' ,rt
1= rAd1+rx= s^.«* 47. g^ - «.*j? i'i
QAr Ay r (|)(/,r)C4,cil-o.7r) - 4.3CG ,V
cw
*.tll
Qfcr Ay * (fX^O-Ol- I.$) * 7.0** ;»'
t = 0.7S" .V

PROBLEM 6.23 °'*3 ,nd °-24 ^or ^a£ beam ^ loading shown, determine the largest shearing
stress m section n-n.
Zoo JcM zoo kH
150 mm
0.75 m
0.75 m
|-«—»4-«—»4-«—«^
75 mm 75 no 75 mm
SOLUTION
50mm At Secke* n-D V- 2&0 kM
Locate Ce*n"f"t^oi'<a and £«.<rc*»>rft-Te mow««7 tff i^tniV.
O
Jfi.
p**t
©
©
X
AUm)
H25"o
\\7St>
225"© o
y (**»,}
125"
2S
A\j(\dn»t)
lfo6.2r
281.25
U87.S 1
«*(Wi
50
^0
Ad'Cios*-)
2S.I2S"
,52.25-
l6o6^")
2.34V
23.43S
9, |4i, i^T^'^ 1Snmt
La^esT sheas i'*« s+»ress occurs on secA-fo* ■fk^ougU cet*"|"r©(W o*f
*H
1
-I-
75
_4_
62. 5
Q. = A^ * tis)(n?}C&.s) * £&s.vt*io
t " IS m»i ~ TTwJO" ^
= 17.61 MPo. -

PROBLEM 6.24 *-*3 an" 6-24 For ^ "eam ^d loading shown, determine the largest shearing
stress in section n-n.
10 in
L l^to.
,5 m. B *
ji*-
k
i 8m
bm—r
■ tlStn.
al
-A
.-
j 1.5 in.
-i*.
SOLUTION
A+ sec-Ka^ n-/i V - 2-5 k.'ps
Locaft cen4roid omJ CA*cji*-te Moment oT mer-fVa
T
Y
^
Part
®
2"
A(»*l>>
t.875
\o. %7*
1S".*?S
J(iO
6.37r
3.635
A5 <^
17M
d CO
7AW
1.006
A J* Cm")
M.oi
3S"-S6
0.23
M-7.6S
H7.86
y, ™
12.W
'• 4.631 „
ZA ~ iS.iS ~ ----- in
I* lAd1*?! = 3£« + 47. St * ^3.4^;«4
Lai^esf sliea*tn4 stress occurs o« i*c+i'o« +K^oyjK cev\4r-&fJ c$
e«rt+i>« cross sec+Jow.
n
f 4-tSi

PROBLEM 6.25
I CL\
(b)
8 A
6.25 Two W200 * 46,1 rolled steel sections are to be welded at A and B in either of
the two ways shown to form a composite beam. Knowing that for each weld the
allowable horizontal shearing force is 500 kN per meter of weld, determine the
maximum allowable shear in the composite beam for each of the two arrangements
shown,
to
<fc>
SOLUTION
For /oJJed stee.0 s«t+i'»< W20O*N6.l
- 211. 7 */C>c m*
o - 6"oo kw/t* -&•■ owe wtij. k*- 2 vJtJMs q^-tooo kVJ/*,
r 3S6 icN
= /51-V* x|o"c no*
QL
S^H.% x (O-*

PROBLEM 6.26
6.26 through 6.28 A beam having the cross section shown is subjected to a vertical
shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (6) the constant k in the following expression for the maximum shearing
stress
r -kV-
r- A
where A is the cross-sectional area of the beam.
SOLUTION
r.
OJ*n
VTl**yC,
A- *cl
For seivw c\<rc** As= jC
Tl,*.v occlh^ a.\ ce*4er wl-ier-e £ - #C
3 <"
It
2 ,3
r va = v-jc
^C-2c
3
V
fe. -U *■*"■««
PROBLEM 6.27
6.26 through 6.28 A beam having the cross section shown is subjected to a vertical
shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (6) the constant k in the following expression for the maximum shearing
stress
r -kV-
'max ~ A
where A is the cross-sectional area of the beam.
SOLUTION
For a +Ki'rt wcJIeJi drcoj/ar secriow A = ^TT»Utm
r\
For a Sft^i c\f~cO Xaf arc
J
- _ art*
A5 - TT^t,
t = 2U
Q* A4vr T"IXtt = 2C"t
A
k * X.°D

PROBLEM 6.28
l-M
6.26 through 6.28 A beam having the cross section shown is subjected to a vertical
shear V. Determine (a) the horizontal line along which the shearing stress is
maximum, (b) the constant k in the following expression for the maximum shearing
stress
where A is the cross-sectional area of the beam.
SOLUTION
A -- -kbh
IriU
36
r(y) =
For a cvi a.1 yocoTt'o* y
_ va .
it
mUim oT X Set
37 * °
K.»^(hy.-y^- ^rK-tthn-^-l^
K « | =■ 1-yoO

6.29 The built-up wooden beam shown is subjected to a vertical shear ol" 5 kN.
Knowing that the longitudinal spacing of the nails is s = 45 mm and that each nail is
90 mm long, determine the shearing force in each nail.
SOLUTION

PROBLEM 6.30
300
400
Dimensions in mm
»\\\W\\fA
D
B
0
<L30 The built-up wooden beam shown is subjected to a vertical shear of 8 kN.
Knowing that the nails are spaced longitudinally every 60 mm at A and every 2 5 mm
at B, determine the shearing force m the nails (a) at A, <b) at B. (Given: I, = 1.504 *
lO'inm'.)
SOLUTION
Sa - ZS >** - O.OZS ^
- ISO x/o-t r^3
- %$<* N
►TIM
p.' *
s* -
ft^fl
= Hl2S * ZO_t m1

PROBLEM 6.31
6.31 The built-up beam shown is made up by gluing together five planks. Knowing
that the allowable average shearing stress in the glued joints is 60 psi, determine the
largest permissible vertical shear in the beam.
_ / m SOLUTION
5 in. 2m. 5 in.
mm
,-j&
it
I, r ■kiSfa}* + (S)W(t)1 = 93.33 ,V
I*. r i^G^C*)* " 8S\ 33. ;„"
I - ^1, + I, - ^8.^6 in*
Fo/- eoc-k qPoeJ joi<^t t ~ 2 m.
PROBLEM 6.32
2 in.
^ in.^-H j-*-4in.-»-| \+- | in.
6.32 The built-up beam shown is made up by gluing together two ■y * 10-in. plywood
strips and two 2 x 4*in. planks. Knowing that the allowable average shearing stress
in the glued joints is 50 psi, detennine the largest permissible vertical shear in the
beam.
SOLUTION
va
m
t -
it
v Gt 3* xw

20 mm
V/////////A
6.33 Two 20 x 100-mm and two 20 * 180-mm boards are glued together as shown to
form a 120 x 200-mm box beam. Knowing thai the beam is subjected to a venical
shear of 3,5 kN, determine the average shearing stress in the glued joint (a) at .-I, (b)
at 5
SOLUTION
~ ST.. 613 w/o"*-
m
^m
t*s 0?K^ = Vo
- O.04O^
")
to

PROBLEM 6.34
6.34 Knowing that a W360 * 122 rolled-steel beam is subjected to a 250-kN vertical
shear, determine the shearing stress (a) at points, (b) at the centroid C of the section.
SOLUTION
For W360*|22; d~3G3 m^ > bf - 2£7«~, £f = 2t.7<?*«~>> £«: IS.O**
J - 36*5" ^/o* m^ = 3£S*to
-C 4
BW^-iM^—ima
<cO
t-Kl Mi
w - «* if r ^3 - gl.7Q . i-7r> £<: *»,**
&*-r Afc^,r 3S8.SXA>**»** = 388.8 */l>"* ^T4
U- tf » 2I-70,
* 2l.7*/o'* ^
IL
M
EZZ
£&-£*
',*$>
£*srt
A,- tftf -- US7)Cauo^= 5S7"? >™1
K~~ tw(i~ tfV Oa.oKlSl.«^- 2.017 mm%
y* = i(£-tO* 71.1-»
Qc= lAy - (S^77Yl7o.cr)+(2C77K^-'?') * lin.7v/oSM^3
= MI7.7X/0"' *»J
tt " tW - I3.0 MM - l3«/0'5fH

PROBLEM 6.35
Dimesiotu m mm
i
fez
V>V,V,V/fAVAWAl
P-&
£
6.35 and 6,36 An extruded aluminum beam has the cross section shown. Knowing
that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point
a, (b) point b.
SOLUTION
- 3.098 */o* •wi* = 3.o<tt*/6c m4
M Qj Ap^t * *A.y,
=■ 47. i7i*lo**»«a = 47. 172 >/o_t ™3
= ^.Xxio'Pfl. = 95".2 MP*.
fl»} Qt* A, J * GatXO(y7)* 27.97*10* m^3 = 21.17*10**?
tb* UUO T Umm = 0.012 *

Dimesions in mm
YY////MMZ?
;-®
6.35 and 6.36 An extruded aluminum beam has the cross section shown Knowing
that the verucal shear in the beam is 150 kN, determine the shearing stress at (a) point
a, (b) point b.
SOLUTION
- I. =i<f6 xio~A v**
w Q^ A.S, + *Aty>
= 31. 652* lo"6 m*
rl =
VQ«. _ Q£Q>|0SX3/633MO-t) _
I t^ * (1.^46 xio-'Xaoji')
0 1.6 x |fcc ?«.
r |Ol.6 MPCL
(b) Otr A,y, =^£K<:y^7)- \2.HSZ»ldtn»?
PROBLEM 6.37
OS in.h" <* "h 5 in.—»|- rf-»j
6.37 The vertical shear is 1200 lb in a beam having the cross section shown. Knowing
that d = 4 in., determine the shearing stress (a) at point a, (b) at point b.
K
SOLUTION
I, • £ (»fX0.5V + ^X^VSJS)1 - 2*. 167 ■,«*
!,= iCSM<^s ^ 106.67 i»#
I- 41, + Zlz r 3^6 in"
(cO Q* - 2A,y, + Axy,
(V>) Qb* A,y, =■ WKo.sX3.75")-- 7.5 .V tb-- O.S ,V
r V^ g C 1200)0-5) g „ , .

PROBLEM 6.38
OS
r
6.38 The vertical shear is 1200 lb in a beam having the cross section shown.
Determine (a) the distance d for which r„ = rh. (b) the corresponding shearing stress
at points a and b.
SOLUTION
A,^ O.S J \r>\ 5, - 3.75 i* tb* 0.5" ;n
Q^ Aay» + XQk = tto)fe7*tt)(i.*7fd)
= 40 + S.7S" «f
k* 5
IW.
(<* t» ^ -
Vq W«fc> + 3.7Sd)
IU
I (5)
8 + 0.7Sd - 3.7S el
d « f * 2.6S7 ;«.
75"
VJ
to I. »
Ik ' 4 tWO* - 106.67 ,-«i
I = 41, + 31, r aSS.Vr.'n*

PROBLEM 6.39
6.39 Knowing that a given vertical shear V causes a maximum shearing stress of 75
MPa in the hat-shaped extrusion shown, deiermine the corresponding shearing stress
(a) at point a, (b) at point b.
W- 40 mm —H
*J_I
4 mm
60 mm
SOLUTION
4 mm
20 mm 28 mm 20 mm
'Ai-
K
isssss^
- v
■mi
Neof^a/ txxts Ji'es 30 *■.*-, a,Uj< boffo**
^c-~
it T*~ re: Tk ■ iu
va
Oft. r (I4V4U28') = '^68 m^*
tw - 4 nn*vi
IS MPa.
if££8 £ ?r
426o'*f
41.4 MPa.
41.4 MP*,

PROBLEM 6.40
40 mm 12 mm 40 mm
10 mm
10 mm
r„ = ro mp^
6.40 Knowing that a given vertical shear V causes a maximum shearing stress of SO
MPa in a thin-walled member having the cross section shown, determine the
corresponding shearing stress (a) at point a. (b) at point b, (c) at point c.
SOLUTION
Q = UZXso^izX + to+i*)- IS*lo4 mm*
Q^+ 2Qb ■» (I2)f loins'* 5"^ - ^.6*10 »£
2*
o,.
o^
u
u
■ 'ft--'
- 12 .
M.sr
46' C
12
1*
U -
(0
12 -
0. %CH1
0.211%
O.Wo
^ =
^
r^
/2.23 MPa.
K 5*T MPa.
HC.7 MPc,

1.25 in.
1.25 ia.
1.25 in. 1.25 in.
CL* 0
6,41 and 6.42 The extruded beam shown has a uniform wall thickness of "g - in.
Knowing that the vertical shear in the beam is 2 kips, determine the shearing stress at
each of the five points indicated.
SOLUTION
t - Q-I2S" m &.f oil seoh'ons
V - 2 kP5
U It
Qt - fojJZi'XMZSX2^*) - 0.077CG m
.2C ^,l
Qc-
f<l IF " (i.aii».««r fe-^**'
Qc * Qj + COJJS-Kl.UsX^"1) " O.G0839
e " It ' (1.2**1 Ko-ii-
D
= 7.?fi j«;

6.41 and 6.42 The extruded beam shown has a uniform wall thickness of j - in.
Knowing that the vertical shear in the beam is 2 kips, determine the shearing stress at
each of die five pointa indicated.
SOLUTION
I r £ (a.SbX^so)* -^(znzKz.zsf r 1.2382 ;**
Qc = o
Qj r (0.iaS)(l.U?X42S'^ = °-07<*'° ^ tj- OJ2S in
Qt ' Qt'tO-WS^O-MS} ' O.O^lQS m* tc- O.Z5 in.
Qfa r Q&+ W^'.0625)fo.l2s)Cl.l875■,) - 0. 4l30g .'n* tt = 0.2S iM.
0*- Qb+aVo.nsXi.^sX^) ? 0.6oS3<? ;„3 t^O.SSm
r
1.25 in.
. «"
I
1.25 In.
1
PROBLEM 6.42
1 ■ * 1
LU
1.25 to. liSm.
tt
vq.
It*
V(\
Jt*
va.
it.
vo>
Itl
_ ff )(o.608Sl)
O.^sax^si
(ZK0.4I5O8)
o.^^xo.^n
_ (* KO. 0<*7*s)
(I.WS*.Xo.20
„ (zKo.oii/o)
Ci.*W*Ko-i*s')
r O
3.<?3 lev*
2-C7 k«,-
0.63 k%»"

6.43 Three 20 * 450-mm steel plates are bolted to foqr Ll52 * 152 x 19.0 angles to
form a beam with the cross section shown. The bolts have a 22-mm diameter and are
spaced longitudinally every 125 mm. Knowing that the allowable average shearing
stress in the bolts is 90 MPa, determine the largest permissible vertical shear in the
beam. (Given-]x =1896 * 10'mm4.)
x 450 mm
SOLUTION
Web*-
Ang-fe :
J^ = JL hSoYao? + (KoXlo)(21<;)Z -. 4 «7.3 wo* mJ
Iw. ± (*>)('&>')
a _
I - II.& */o* *."
S».«* *IO* WrV*
A - ^2©
J r
44.?. w~ ^-^^-HM.?- lU I »«
- ii.ewos &"^ao)(isai) ~ 1*7.f-x\o*m*
I r 2Ip -v Iw -» 41^ = lg<?6 x IOc vh^" - [8?£*/0~* ^¥
Q^ =■ t5>*©XlS©.n
?76*/0>* »nM*
Ab.^ * fdjl = 3C»)* 7 380. I h»^ - 3*>;l*/o"**,*
£-47.3e v|o* N/*vi

PROBLEM 6.44
50 mm 100 mm 50 mm
■K
m
6.44 A beam consists of three planks connected by steel bolts with a longitudinal
spacing of 225 mm. Knowing that the shear in the beam is vertical and equal to 6 kN
and that the allowable average shearing stress in each bolt is 60 MPa, determine the
smallest permissible bolt diameter that can be used.
SOLUTION
'Wtr
Part
z
A&W)
ISoo
ISoo
.9 (*^
SO
So
-So
Ay*6oe~«)
\*.7S
37.5o
75". 0o
Ido*^)
IM.OG
56.25"
-A «*
Pi-tt r ^t^"
MfV*
VQs
C»MS6 x/o-4 m*

10 in.
PROBLEM 6.45
3
6.4S and 6.46 Three planks are connected as shown by bolts of g -in. diameter spaced
every 6 in. along the longitudinal axis of the beam. For a vertical shear of 2.5 kips,
determine the average shearing stress in the bolts.
solution
Lt>c*.Tt rt*\JTV^ai a*i£.
2A * C*X5XlO +(!OCl<0= So \*x
4- fcWTiorN W6oX/.5)a - >T«.7 in1'
twif
r £L r &.79V/ r -J
Aut a."©*/
'? ^csi
PROBLEM 6.46
2 In. 2 in.
k*4-« 10 in. >4**j
10 in.
1l
6.45 and 6.46 Three planks are connected as shown by bolts of ■§ -in. diameter spaced
every 6 in. along the longitudinal axis of the beam. For a vertical shear of 2.5 kips,
determine the average shearing stress in the bolts.
SOLUTION
IA - KXaX'0^ + 6<>W •■ SO ,'«*■
v - gAv . ias. - <? r ■
I- 2.[^r(a^(»o)* + c?XioVi.5)*l
Q * (Z)(to)(i.s) * 3o ;«J
r r ^ T Y|S . M^l s 0.7^/ *,s

PROBLEM 6.47
50 mm!
200 mm
100 mm 100 mm
6.47 Three plates, each 12-mm thick, are welded together to form the section shown.
For a vertical shear of 100 kN, determine the shear flow through the welded surfaces
and sketch the shear flow in the cross section.
SOLUTION
Locale neo+t&t axi'ft
75". 77 wim-h
Y 2A *
TBBZEK
+ a. [A (WO*/ + W0*X7£77)*]

o —6W SO mm
MM!
HO mm 10 mm
PROBLEM 6.48 6.48 A plate of 2-mm thickness is bent as shown and then used as a beam. For a
vertical shear of 5 kN, determine the shearing stress at the five points indicated and
sketch the shear flow in the cross section.
k- 22 mm -^
1 * •' SOLUTION
=- (33.76 v/o* m*.* = 133.7i «/0"1 en*
GU = O
Qe r Qa+C?K2C)(i3) e -goo -^ - -SaO^lo"* r*?
^ VQ, C-5vio5)Ceoo"IO-'')

I—2ta.—4« •
1.2 in. 12 in.
6.49 A plate of 7 -in. thickness is corrugated as shown and then used as a beam. For
a vertical shear of 1.2 kips, determine (a) the maximum shearing stress in the section,
{b) the shearing stress at point B. Also sketch the shear flow in the cross section.
SOLUTION
Ud = /(.U?+U.O*- - 2.0 ;„
AaD - (o.aOU.o') r o.s i*x
Locu\e. fle^ST**** o^ci^ omA Co»*f>*j\*. Aid*** e*"/* erf iV\tf</*T^.
Part
AS
BT>
S>£
Z
A(.»
o.s
O.S
o.s
O.S
2.0
5C-^
0
0.8
0
AS g*o
0
0.4
0.4
0
0.%
Jti^
0.4
0.4
O.M
0.4
AdY^
0.080
0.0 80
O.CTto
O.Oto
O.SZo
l(i^)
*O.I067
*o.io;7
0.3/33
-- o. io67 ;„"
I- 2"AoT + fr
- 0. 5*333 ;„*
t*\
vws
0.5 in.
Q»c r (o.^)Co.2y)(o.2)-- 0.0*5- ;«s
r,B ^^ (i-nfo.Ms) r z<525 ta;
* It (O.SZ3Z)(0.2S)
<& QB r Q
>8
* 0.2 ;»*
t,fi" It ' (0.5233^(0.2?)
I. SO ksi"
r, r 0
^tJ.OZ ksp"

PROBLEM 6.50
6,50 A plate of thickness / is bent as shown and then used as a beam. For a vertical
shear of 600 lb, delermine (a) the thickness f for which the maximum shearing stress
is 300 psi, (b) the corresponding shearing stress at point E. Also sketch the shear llow
in the cross section.
SOLUTION
4.8 in
LBB-- <-
£F
-/4.SZ +
s.z ;«.
\*— 3 m. *\-~; *\* -3 iii.-*-f
2 m.
Ca/Ct»ieuifc I
TAR ^ (3t)(J?.V>* - 17.2* £
■AB
(Cd M t>eini C
GLr &*» + Qt* - (3£)0?.<O + <2.cfcM.aV- 10.3* £
r It ' L" tl * C3oo)(8^MtV 0.23IG8 ,n
T* -
O + &)(0.2%\tt)(2.H)*
It * &o.LH)(o.2%\61)'
I. CCg .'«'
£09 p%i'
201 p*,
Zoo ps,

PROBLEM 6.51
6.51 and 6.52 An extruded beam has a uniform wall thickness t. Denoting by V the
vertical shear and by A the cross-sectional area of the beam, express the maximum
shearing stress as ^ = A( V/A) and detennine the constant k for each of the two
orientations shown.
SOLUTION
to
(fc)
h"fa
A,^A, - at
J,- A.V.1-- atfc1 t $o?t
I-- 2J, * 41, - f a3t
. VQ . V-Bat r3 _v
T*" I«tV ($a»tiff T" at
- *£_£ - £5 J^ - I, Ji
S iat ' S A A
h - T
I, ' I, +A,d-
- ***** £<*** « ?,«'*
Q,« atlf ♦i,)s "K*
y - VO . y.-fc't
3oat 20 Gat " *> A
* A
ai
k»#'fc»o

PROBLEM 6.52
6.51 and 6.52 An extruded beam has a uniform wall thickness t. Denoting by V the
vertical shear and by A the cross-sectional area of the beam, express the maximum
shearing stress as v^ = k(V/A) and determine the constant it for each of the two
orientations shown.
SOLUTION
(c)
(b)
(a)
TTTHJEi
x
" J-Q't
I = 21, + <Jlt* \at
:IY 5 iJZ. r 3.U.
16 at 4 4a£ * A
J; • h ' ±V? a
I - 41, -- £a*t
L~* ' list) ' (§ a't^WJ
- lA .VL _ 3Vi _V_
2. A " * A
.- 2j§ ^
?./a

PROBLEM 6.53
j la. |m.
^J L,— 2ln.—^) |^-
2 in.
Sin.
>///////<
Ob
6.53 The design of a beam calls forconnecting two vertical rectangular^ *4-in. plates
by welding them to two horizontal "J x 2-in. plates as shown. For a vertical shear V,
determine the dimension a for which the shear flow through the welded surfaces is
maximum.
SOLUTION
Q. - (ztk) a. = a.
9 I 4.0H1^7 + ^a*"
m
4* -
Set 5*
= o
3a. * I (H.OHKC7 + *a4-)1' J
2ax ■ H.o*\Ul
a * 1.422 in.
PROBLEM 6.54
6.54 (a) Determine the shearing stress it point P of a thin-walled pipe of the cross
section shown caused by a vertical shear V. (b) Show that the maximum shearing stress
occurs for 0- 90° and is equal to 2V/A, where ^ is the cross-sectional area of the pipe.
SOLUTION
A, ■ 2rSt
Q** A/1 « Zvt hr4
t
A

PROBLEM 6.55
Plastic
6.55 Consider the cantilever beam AB discussed in Sec. 6.8 and the portion ACKJof
the beam that is located to the left of the transverse section CC ' and above the
horizontal plane JK, where K is a point at a distance >> <yy above the neutral axis (Fig.
P6.55). (a) Recalling that a, = Oy between C and E and a, = (OY,yr)y between E and
K, show that the magnitude of the horizontal shearing force H exerted on the lower face
of the portion of beam ACKJ is
H = $bar
2c-yY~
y
i\
(b) Observing that the shearing stress at K is
yrJ
Neutral axis
HH
lim
jj AW _ 1 ffl_
and recalling thatyr is a function of* defined by Eq. (6.14), derive Eq. (6.15").
U-b **
C
^?*
H
s r €*i, fo' 0gy<^
cm*
The STV^ess tfh'sivV t^ti"** is 4iVe«\ ^
For 'e^i^i^ib^i'ci*^ of kotri«&*+a,* -Voices cudiM o* AC^J
« Jfb<SV(Xc -JTr -#)
No+e H«,f Vy ''s A ^u»c\!ok> of x
r ,±M - J.s.f-2* hJC-J&O = - J-G- C I--£}■#
D#.«-f.-.f;-3 £ * * - t Mr (-i £&)
37 " _yYM/ " >js;bcu
(<o
3 P
ft)

PROBLEM 6.56
6.36 For a beam made of two or more materials with different moduli of elasticity,
show that Eq. (€.6)
r =
VQ
it
SOLUTION
£«
£
remains valid provided that both Q and / are computed using the transformed section
of the beam (see Sec. 4.6) and provided farther that I is the actual width of the beam
at the point where r is computed.
LeT Etot fc>e a reference m&doJtf
of eJ*.sii'6i'4y
Ac+a«./
k n»t iJ
+r^*sroM*ee< section. TUt ttw^.ViA
&+/«.s* etoa+v^Wo+i©* io "Hie cross
w/Ufcr* T is +ke vwewai«7 tff ••ierTi,«. of "f*i« 4 ir&v\a-f^A>ee/ cross See+YoK e^J
Tk< ^oA*«o«"fa^ .sheets'/na -Fivce oi/e»r >?tw«4li &x 1*
VQ
Zt
^_. VQ

12 mm
PROBLEM 6.57
K-150 mm-*J
6.57and 6.58 A composite beam is made by attaching the timber and steel portions
shown with boltsof 12-mm diameter spaced longitudinally every 200 mm. The modulus
of elasticity is 10 GPa for the wood and 200 GPa for thesteel. Determine the average
shearing stress in the bolts caused by a vertical shearing force of 4 kN. (Hint. Use the
method indicated in Prob. 6.56.)
SOLUTION
12 mm
Let E^ * Es * Zoo G?o,
%- »
«wT
2o
ezzzz
&x./rA
7.5
*»•*
VA'^Ms ©t +r<nrt54o/NH«J section
= 2 [o.O*.l4*/oS 30.S<fc>*/O*] + 7.7&6V/0*
Au+ * fJu« - (tJKhO* * M3. I m*,1' M3.I */o"c vn*"

PROBLEM 6.58
6 mm
-Hh-MOmn-HH*-
61
'V\
\\
^
XV'
^
S\\
2
^
\V
l« )to -*|*-w* -*l«-15*> -d
6.57 aud &5J A composite beam is made by attaching the timber and steel portions
shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus
of elasticity is 10 GPa for the wood and 200 GPa for the steel. Determine the average
shearing stress in the bolts caused by a vertical shearing force of 4 kN. {Hint. Use the
method indicated in Prob. 6.56.)
SOLUTION
Let woo«\ b«. \kt ^efe*"t,irt<:e v*oAvr\a.fl
Q. = (i4o")(9*)0tt+l*O - i.owa^/o'm^
0" I JT75.7 w/o"fc
3. I w*,1, * //SJv/0'c ^
6.73 WO6 P*
- 6.73 MPa. —"

PROBLEM 6.59
Aluminum
Steel
1.5 in.
6.59 and 6.60 A steel bar and an aluminum bar are bonded together as shown to form
a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the
modulus of elasticity is 29 * 10*psi for the steel and 10.6 x 10* psi for the aluminum,
detennine (a) the average stress at the bonded surface, (6) the maximum stress in the
beam. {Hint. Use the method indicated in Prob. 6.56.)
SOLUTION
aic»vur>t/»M
16-6*10*p%i
V\A On*)
3.0
7.|t>39
SCO
2.o
o.S
n Ay (•"•»*■)
cKO
a. 6335
Mfl>*(mO
2.Z525
I. (S3 in
_1
Y ZnA * 7J033 L,3W
3.MW
nf(.VO
0.3416
i. 3<a©
(a) A+ He bonjtfj su^c* Q^ (LrXaXo.Wtf) =r *.5!W,\,*
X ' It ' (X*N*/.0 '
T*~ It (tlWK/.5)

PROBLEM 6.60
Steel
Aluminum
1.5 in.
2 m.
lin.
6.59 and 6.60 A steel bar and an alununum bar are bonded together as shown to form
a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the
modulus of elasticity is 29 * 10*psi for the steel and 10.6 * 10* psi for the aluminum,
determine (a) the average stress at the bonded surface, (b) the maximum stress in the
beam. (Hint. Use the method indicated in Prob. 6.56.)
SOLUTION
V)~
ftw-V
n a,Xo**\\ njtn
s**9
J
M (in )
8.7PT*
1.5
j/Cinl nAyGft*)
as
0.75
i.»l»
+
9.707 V
9 - gn*? -
T * 2A "
17.161*
*t GO
0.2319
nAcJ'CU')
17. ICfg
<?.7o7W
" I.7t8^in
-7«* J- InAd* * ZViI =- S.7W% m*
O.*l*llo
2.4135
2.8S55-
nl GvO
2.7553
O. \XSo
2.24.0$
{CO A\ 7*le toft^eJ sorfa« Q. -
** It * Gs:7r/3)0.O "
0-S>0.268O
0.888 *s/
l.9©*3 f«s
00 Af +U
M5J*\ _
Q.? U.12S*)O.S)0.*3t*XklrS)* 3.H33 ,o
(4X3. H331
(^.7lH3)(l-5'5
U*£l Jai

6.61 through 6.64 Determine the location of the shear center O of a thin-walled beam
of unifonn thickness having the cross section shown.
SOLUTION
lj»T *f*s it*1 It*-I*r * -Wta*4 itjaitt1 * 2ia3
lo.'**<**?« Ita*
i-n - tftaJ
Past AB
A*t, y-f 'ft = ity b
It * $ta*t * 3««i*t
Fi.-SrjA -f^jy.J&^Jt-iv
Parf BD
D C
T77^
u
Q- Q«, + txa -* s-ia + ta.:
-&
TT= f ^({a' + ax)
#a>t v*
F-s J™ ■CJ*>*axU'
3aa
32a1
e« fa -*

6.61 through 6.64 Determine the location of the shear center Oof a thin-walled beam
of uniform thickness having the cross section shown.
SOLUTION
lot* &tfes 1= TI - £t(c«L+cb+Mlix
t&s/A "I ^_ N/Q. _ Vhy
kx-J T" Jt ' ZX
F.r^r^r ^t*c- ^j;-"A
- VkL X.T - VV»to.x
XI 2 {» HI
Part BD Q. r tx £ - 4 th*
i yyyyyM •£**
u3 it ax
21
vht xMb - V^tb"
51" * '• 41
41
e__ 3Cb*-rf.)
G(<x+fa) +.h

PROBLEM 6.63
6.61 through 6.64 Detennine the location of the shear centerO of a thin-walled beam
of unifonn thickness having the cross section shown.
[*- a~*\B SOLUTION
D E
F C
a
I-*L»- ail*)** A«t*<s it^1
[ Par+ AB Artx J -^ Q. « | at X
cr.ka.1^
F»
Pa^4 DE A * tx y r f Q= i«-tx
X> E
X
~* A*V- Jt 3|t«*t ' Wo*t

PROBLEM 6.64
6.61 through 6.64 Determine the location of the shear center O of a thin-walled beam
of uniform thickness having the cross section shown.
t
IT
a
4
H
V E
U
4
F*
If.
K
SOLUTION
Iw ' A£^f = |£a5 J. ZI
T
Mi
fe*
Q= Aj - £i (2a-j)(2«+,y)
6 "£ & V
ft^-t 08
ywstft
-4
Ux -J a,
L
-- ta(¥ + x)
- 5^ - gv
e- fa.

PROBLEM 6.65
50 kN
Dimensions in mm
5
F-w
F.
6.65 and 6.66 An extruded beam has the cross section shown. Determine (a) the
location of the shear center O, (b) the distribution of the shearing stresses caused by
a 50-kN vertical shearing force applied at O.
SOLUTION
I©a - (7oXOto)* + ±.{7o-)(cf- Lome*/©*"*-*
Id6 r ^0^0°°? " 0. 83333- 10 fc mm*
I ' II ? 3.76«*|o4 „*«
Part A8 : A = Id (^- lo}
T
Fl*
\ ?*A D8
I
V ' IL 2 Jo 3.7&.S* IOG
- 0.*flft6O
= *f«.7 V e - 48.7 ^m -*
Con+iOOec\

PGo&LSM G.ZS (conhnocj)
V= ^bx/Os M
r= o
\Q.*\o'c ^
P**-* ASj Pw-t B Q.r oQ - l^io3^1 ■= l3*io~~ »*r»} £=io*ia'
(VI 8D, %t~i B Q. * »»k|o-4 vn3 t * gwo"s ^
P«*4 8D> Po.„1 T> GL* 33*/o~6 ^ t = 4* lo"'*,
Rw+ D^ poi*t t> Q = 33 */<?"* mo* t * io>|cr» ^
* It C^Vio^Xiomo-*) " ^3^M0 r* ninr^
P*iVf W Q-- il£.S*IO'c ^ t * lO*iQ'%
m

PROBLEM 6.66
100
30 kN
6.65 and 6.66 An extruded beam lias the cross section shown. Determine (a) the
location of the shear center O, (b) the distribution of the shearing stresses caused by
a 50-kN vertical shearing force applied at O.
SOLUTION
m»n
1^-- ^(fiXlo^^ * a^oowoW
Dimensions in mm
F%
«
F.
F,
F.
1= 21 * 4-.50TC*/©4 mw'1
VQ QB- 3(Sox-}oo) = 72 Wo* *^s
r:
6 xt
Ft-i^.r^^ta^^
_* 1.72** yio* . 0 3gsS7
COTf(1^ficl

PRo6LEM 6.GZ (ctmh'*oeA
£ r £ *;o
-a
V*\
I3.3IKIO* 1V. - la-Sl MPo.
P^i V£ P*.v»4 T> Q = M.Z»lo'c M4
O > /o~ w*
T-t-lMlO* P-c = 7-<h MPo.
*fc.**/0*P«. = W-8-MV.
7tf.o*/o*P*. - 78.o MPa.
*■ ^-S;°:,l^:ft • --i"** - ■»■-»"*

67
6.67 and 6.68 An extruded beam has the cross section shown. Determine (a) the
location of the shear center O, {b) the distribution of the shearing stresses caused bv
a 25-kip vertical shearing force applied at O.
SOLUTION
I - 4* -&X±? ♦ <0ttX<|+ A-GKO* - ^-w in '
?*A AB A = f x 3 j = 4 ^ Q^Jt^x
Poi^f A x= o TT- o
Po.vt fc y= 3 t» £".11 ifti'
4 r
Point C v r o -£ - is.6* le%:
r

PROBLEM 6.68
6 in.
25 kips
P.
F.
F.
Bm4 AS
Q
667 and 6.68 An extruded beam has the cross section shown. Detennine (a) the
location of the shear center O, (b) the distribution of the shearing stresses caused by
a 25-kip vertical shearing force applied at 0.
SOLUTION
A5I|X; j^ a= Ay - X
Ux*l z' It (k.wxM ' 2-*05S*
Poiva Ax = o ^ - o -*
- 3 ^Ot-y1) « 7-4ym
y - 4 £- .3.3/ is." -*
?ari BD
R><*4 C

PROBLEM 6.69
6.69 through 6.74Determine the location of the shear center O of a thin-walled beam
of uniform thickness having the cross section shown.
2 in.
3 in.
4 in
A^ = L»*i = <£)(?) -- lz$;*%
SOLUTION
•-06 " -/H + 3 - Jin, n^
lot* i A«k**(£*'-«W= 3.7S" ,„*
I*m- AO^y^WC'Vl)1 -" 8./6G7;**
111
11
3MD - DM. -Ve = -7. F.ff^ r -0.7273 V
£ = 0.727 .'n.

6.69 through 6.74 Determine the location of the shear center 0 of a thin-wa
of uniform thickness having the cross section shown.
SOLUTION
Iod - iAaDV,* * i(a-K)(\.sy - o.stzsW
I =■ (2)(D. 78/aO* (XYo.S^XS-) ^ I.687S" it"
P«rt AB: A = -4 j Jr*y $= Aj -*
It ' (8Xl«75)(0.2T) ~ 3.375
3.37S" »■ '«^ 0-37rK3)
- 0.OS3SZ V
DM,, "* V^ Ve ^ F, (3s.^6o°)
Ve = ^X0-O8SSS)V(3 8'V *©•}

PROBLEM 6.71
6.69 through 6.74 Determine the location of the shear center O of a thin-walled beam
of uniform thickness having the cross section shown.
C mm
U-7
SOLUTION
Ltm.= 10 ~~ AM?i?t>Xi)= 43
Aft
O M*"!
I- tf X*s. ?£*/©'"UfrYai.S"*©8) • nw.S wo3 «m*
Port A8 A = tS -- CS
y - is j)m So' : -is
4= Ay' $sz
^r
3a
It
.70
- 70.2 V
e = 20.2 ►*«*

PROBLEM 6.72
6.69 through 6.74 Determine the location ol'the shear center O of a thin-walled beam
of uniform thickness having the cross section shown.
30 mm
40 mm
30 mm
40 mm
SOLUTION
L*> * V 3o* +H°* - SO y*\*« Ato = G*o XO - 3.00 ~~?
i
X
Par4 A*
6.
A
T
Z ' It - 1(4) " i
icov x
+!>MB* Oh,
2* K (^C5^7-©W/o»
Ve. - COF" (Ho) - 9.6V V

PROBLEM 6.73
6 69 through 6.74 Determine the location of the shear center Oof a thin-walled beam
ot unitorm thickness having the cross section shown.
SOLUTION
For u/liaie cf«S$ sec-tr«M f\- ZlTCtt
J - A a" • W* I BiJ- Wt
A » si = afit
7
s r aye -fe.*wf U
o(.
f &;*o(. - a
s»Vct
of.
f-
*c
But
VQ _ Met rx a\

PROBLEM 6.74
SOLUTION
6.69 through 6.74 Determine the location of the shear center O of a thin-walled beam
of uniform thickness having the cross section shown.
F*r a. tliin-M/nX/cJ boll**/ ci^c^Af cf^osS sc&ffo* Ar3ffO.£
J* a1 A - ;Wt
J * i J ' TtaH
Use pft/o*^ cooreJinaTC © -f*r pa^Tlfi.i' ciHSS Secri'd"
A ~ $t ' aBt S*a*: J**jH
a-
Q.x AJ-- a9UsfH^ =■ at (2s.noic.soc)
r a1* st-»2ot * a* is* 6
c^ it T T s7
o -J- J. 'fl
Va"t
iVa.
B^ Mc r Vc ,
e*ce
e* fa. -
273 *

PROBLEM 6.75
Sin.
D E
V
lin. i
10 m.
a~
, *■ r
3 in.
P.
F*
*-«
6.75 and 6.76 A thin-walled beam of uniform thickness has the cross section shown
Determine the dimension a for which the shear center O of the cross section is located
at the point indicated.
SOLUTION
?o.r\ AB
6
A= tx
- Si
<2=A^ * St*
2223
M
** it " X* i
Port DE
YS/A I
UxJ
55 5 ^
A-- t* j* 4.*.
L ' Tt It ' I
QrAJ=- 4£x
<JVx
r 3Vto.v
OIM. r 5IM. 0-- (lfl)f«.5 ^)-f8^ ^N"

PROBLEM 6.76
2 in.
I
f,
F»
6.75 and 6.76 A thin-walled beam of uniform thickness has the cross section shown
Determine the dimension a for which the shear center 0 of the cross section is locaied
at the point indicated.
SOLUTION
9aA AB> Let £ - US+ Ol as .s&om/*.
a = as * iifc-jKc^^ * it (c«-y)
r - it a J
1-5
u
F- ■ J *-"B £ ^^ty, * i^>v>J>
P^+ 80 Q. = Qi»4 tx j6
/6o
I
1
*»*f-&(cW*sO
<J 1.5
E
-fe* + 12 cl = o
CL= «J.S"-15 i 3.0° in. -«

PROBLEM 6.77
f*-50mm
v, «...
1
k-b
Like wise, -&«- Pa^+ D£
6.77 A thm-walled beam of uniform thickness has the cross section shown.
Determine the location of the shear center O of the cross section, knowing that
h = 80 mm.
SOLUTION
0.- Aj - it(ili(-(y)(*li1+^
6
-it(K-y)
_ h,? V
p - h»3 y
e -
h,a + h,* ♦ Hi
(goV+Cfiof+tto)3

PROBLEM 6.78
U-50mm-»l« ■
Id
50 mm-
R
F.
r
i~
Likewise, JW Pari DC
8
5
6.78 A thin-walled beam of unifonn thickness has the cross section shown.
Determine the dimension h for which the shear center 0 of the cross section is
located at a distance e = 25 mm from the center of the flange AB.
SOLUTION
Pw+ Ae A*(4K-yU
-*t(+Kt-/)
6 It 2* J
/..*M".
= $(Kj-*y*)|
'-K
- Hwo-wh
vth.'
in
h.'V
tf ♦ K * W
- o
0 - eF, -(b-e)F2 -(2A»-e) F»
^ , ^£^ *^V - fig*** ^g**** ^«o'J
h. * 74.2 ^ -*

PROBLEM 6.79
<S.79 For the angle shape and loading of Sample Prob. 6.5, check that \q dz = 0
along the horizontal leg of the angle and \q dy = F along its vertical leg
SOLUTION

PROBLEM 6.80
SOLUTION
6.80 For the angle shape and loading of Sample Prob. 6.5, (a) determine the
points where the shearing stress is maximum and the corresponding values of the
stress, (b) verify that the points obtained are located on the neutral axis
corresponding to the given loading.
Refe^Artj +o So.»yf>h TVot. 6.5
J J 4 to.* 4ta* J ^
J s ?«-
Vta1
*o ia
AA
h.r-.,..Ui ie, ^,3P(a^(^^ s 3P^ ( ft. , ^ 4 ^
^fa
A+ + Jie co^ne^
M fy -- ^ £ a'
a.
r 1 ?
fan y - il Vkwi © » ij-
© - Cp * 15-H.036 r 30.964*
TA
2a£
- !«■
e*» I.
- (^ + "4 -f*M ^.036')a. =• 0.6667 en.
z
3"
- i«t

PROBLEM 6.81
mm
*6.81 A cantilever beam AB, consisting of half of a thin-walled pipe of 30-mm
mean radius and 6-mm wall thickness, is subjected to a 1200-N vertical load.
Knowing that the line of action of the load passes through the centroid C
of the cross section of the beam, determine (a) the equivalent force-couple system
at the shear center of the cross section, (6) the maximum stress in the beam.
(Hint: The location of the shear center of the cross section was determined in
Prob. 6.74.)
SOLUTION
Fr-oi* "He soJpTi'om "k PRofiLCM 6.7V
e-- fa
I=fa*i
a - oft *•»»&
r If
^..
■ VQ
>£»*.
e - £*v
*"' #0.
e-* = f a
For ^rsio* of a. ft,cAa.n«o)ivr to/- C, - C^ - £ ( I - 0.i%O j I
'*•"••* ~c.it1 ' (t3mO(,*'U,ra*i<5y&Nfo-iy
- 2LTf Mp«
By Supe/'iieSfTioM
Z^ - «* * UJI « 2S.5£ MP«_

PROBLEM 6.82
•6.81 A cantilever beam AB, consisting of half of a thin-walled pipe of 30-mm
mean radius and '6-mm wall thickness, is subjected to a 1200-N vertical load.
Knowing that the line of action of the load passes through the centroid C
of the cross section of the beam, determine (a) the equivalent force-couple system
at the shear center of the cross section, (b) the maximum stress in the beam.
(Hinl: The location of the shear center of the cross section was determined in
Prob. 6.74.)
*6.82 Solve Prob. 6.81, assuming that the thickness of the beam is reduced to 5
mm.
SOLUTION
FvN3* -Hie soioT-iow +o flPoBLeM 6.7V
a = a11 s/n B
r-
VQ, _ Val
It
.-, H
k - iloi MPa.
e- £<x
X* fQ
e-x r -a
For -far^io*. tff a. recTaw-u £».* \»ar C, - Cx s jj £ ' ~ O.CSo^j
^^ ; - 30. H w/o* P*
= Sail Mfia.
E>y 3u^er P©*i 4ion
tL»- <5.01 4 30,11 r 3^.3 MPfiu

PROBLEM 6.83
27 kips
i
/
*6.83 The cantilever beam shown consists of an angle shape of g- - in. thickness.
For the given loading, determine the location and magnitude of the largesl
shearing stress along line A 'B' in the horizontal leg of the angle shape. The x'
and y' axes are the principal centroidal axes of the cross section and the
corresponding moments of inertia are/,- = 115.7 in' and /„ = 12.61 in*
SOLUTION
Use cootcAifMiA £- V «5 <SL>6*"|.
A = "§" V, I ^ ~ jf * in . ts | in
x' - X coa ^ - ^y s<*xyS
"Of
Ou< to V,
? = VyA^'-. (v^^ciXi^^^Iu^^^^^+c^s.^i
TAJ t, + fa r (ia.-jKi.ooo ■-. o.zsoy) ksi '
Due -K. N/y*
Vsi
■if. IM
tiVsD
o
Iff. 00
\
a.
S.oo
N
0
£
- 3.oo
*
-H.oo
\
to
-2.oo
IZ
o
V - \2 k%i ai corned
»**
*|£ =• -(w)(i^-^- (I- o.a*y>

PROBLEM 6.84
27 kips
2.05 InA
3.5S
D'
in. A
133° J
^ ^
\ \
\ lin.
\\ 00
/ x'
B'
-y'
■6.83 The cantilever beam shown consists of an angle shape of g" - in. thickness.
For the given loading, determine the location and magnitude of the largest
shearing stress along line A 'B' in the horizontal leg of the angle shape. The x'
and y' axes are the principal centroidal axes of the cross section and the
corresponding moments of inertia are/,. = 115.7 in4 and ly = 12.61 in*.
*6.84 For the cantilever beam and loading of Prob. 6.83, determine the location
and magnitude of the largest shearing stress along line A'D' in the vertical leg of
the angle shape..
SOLUTION
In veA+i'cA/ >Te*j use coo/dl^^T*
"3 =■ 4-
(G + x V I ■ £ +
Sue f. V,
Oi^e fc> Vj'
t;+-J; r, + r,
r ^.oco3 (G-x)(0.83S3I+ 0.1JIMO
= 0.04S5*? (C-x)(^8*?3V + O./34/^x)
x CO
r (i«ii ^
0
il.oo
i
\S. oo
*.
(O.oo
a
lS.00
*4
»9.o©
5*
7.oo
6
O
2L-K" 16 k»» *t *.* £ ii\ -**

PROBLEM 6.87
12 in.
*d *7The cantilever beam shown consists of a Z shape of -j - in. thickness. For
the given loading, determine the distribution of the shearing stresses along line
A'B' in tha upper horizontal leg of the Z shape. The x' and y' axes are the
principal centroidal axes of the cross section and the corresponding moments of
inertia are/, - 166.3 in* and/,. = 13.61 in4.
SOLUTION
V = 3 k.-f» fi • ZZ.$°
X-m Oppes hoKzor»Ta>P -fee usfi
A - i( « + *) m.
% * i (■**+*) in
*.' * X cot j8 t j &f*0
Doe +o V**
6' %t^
(IS.CI )( + >
* O.0/C6£65(6t x)[ C.C1/32 - 0.1113***}
X Gh>
ru^
-c
0
- s
-0.105
-H
-©.Wo
- 3
-O.loH
- 2.
o-ooB
- 1
OJto
o
0.12a

PROBLEM 6.88
12 in.
*d 87 The cantilever beam shown consists of a Z shape of \ - in. thickness. For
the given loading, determine the distribution of the shearing stresses along line
AW in the upper horizontal leg of the Z shape. The x' and y' axes are the
principal centroidal axes of the cross section and the corresponding moments of
inertia are/,. = 166.3 inland/,. = 13,61 in*,
"6.88 For the cantilever beam and loading of Prob. 6.87, determine the
distribution of the shearing stresses along \weB'D'm the vertical web of the Z
SOLUTION
V ~- 3 left fi = K.S*
A^ For par* AB' A = &KO - /•* ;»x
x * -3 *'*i , j * 6 i*.
x' * 96 taa/S + v s.v./S
Jf*
r, (vs.v./e)fo.s)6-3 coa>6 +
= (V»'"*» -Q.7ISS v I.7.MI - O-^ttf/] w p.3^^_ <).<,/< |<j y*
J * x* t
(ICC.3)(^
*.+
v/ W
t lk«i">
?l *
o
(,■287
(.2*67
± *
1.111
- 0.023*«*v*
* 4
O.tos
±C
o.«a

PROBLEM 6.89
50 mm
6,89 Three boards, each SO mm thick, are nailed together to form a beam that is
subjected to a 1200-N vertical shear. Knowing that tha allowable shearing force in
each nail is 600 N, determine the largest permissible spacing 3 between the nails.
SOLUTION
C*tc»i«At wi©m«*+ tif t*e/"(V«.
p«rf-t
Tap
MuUJe
Bottowi
Z
AU*M
75*00
SQoo
IS 00
cJ (,****)
IS
0
IS
Ad* (lo^)
42. 1*
0
BH.3S
I QPm?)
4.17
1.56
7.29
I = TAJ* + XI «■ 9I.C7 */D* *.*** * 9.1.67* to'* *"
0 . Vft - 0*ooX5«-g»A>") _ 343 M/0. w/m
T J <ll.£7x/0-4
F-.-I = $s
600
* VM>t^, —
<J. " 7.365 x/o*
.-s
a 81.5* JO"*** * 81.5**1* -^

PROBLEM 6.90
16 mm X 200 mm
$.90 The American Standard rolled-steel beam shown has been reinforced by
attaching to it two 16 x 200-mm plates, using bolts of 18-mm diameter spaced
longitudinally every 120 mm. Knowing that the allowable average shearing stress in
the bolts is 90 MPa, determine the largest permissible shearing force.
S310 X 52
SOLUTION
Par*
Tr piMe
$t\o«SX
6ot.pl4e
r
a &«n
3Xoo
G65o
3*oo
d(*»*)
o
ajVios*.*)
S2-M3
83.4*
IC4.86
I (lo***,4)
0.07
<*S.3
0.07
<?S\4«*
- ICO.5 w»»*
T I Q 5-116 x/o"*
- l?3.S fcW

PROBLEM 6.91
6.91 For the beam and loading shown, consider section n~n and determine the
shearing stress at (a) point a, (b) point b.
lta.-*| [*- SOLUTION
11 In.
A+ sed-i'o* n-n V - 12 ki'ipt.
|*-4ln.-*|
O"
©
Po*t
(I)
©
2:
A (i**")
W.
5 CO
M
1
Ay (^
16
2
X**
J f.O
2
1
AcTC*-)
3
21
"lii J}
5"-33
2.£7
S
Y = XA.V - i!.2M
I * TAJLl + 2I r ^+« = 3* in-*
W
m
i
3.SV
1
u it (ao(n X'SI*
(W
i
J.
Qb*
i •
^
Ai5ir
1 in.
it "
(H0O(3
00(0
6 in
2.2S *«/

PROBLEM 6.92
6.92 For the beam and loading shown, consider section n-n and determine (a) the
largest shearing stress in that section, (b) the shearing stress at point a.
0.3 in.
Skips
0.3 in.
SOLUTION
M sec*low lf>-* V= 8 Jc'ps
(a| Tm JWrjesT ske*rin4 s4vxs3 occurs on <=l scctio^ -fkvoudii -fl»< ceirryofel
cj'V +^e etrhVe cross secf/en.
U-4ln.-*J
£
^SV^)
<IK£
1
3
I*- Y
TVf
T
A(?h*>
1.2
t.£2
z.gz
,Y<"*
1.35
Q= Ay 6*0
3^2
2. 187
.£407.
£ ~ (2)(o.3) - O.C .'i,
^ It (27.385a)(o.6} *'/B **'
(W
V/M//tM
Q* = A«.J* = (L2)h.»S^ - 3.W2 ;„*
T*~ It " tt7.MS*)<0.0 *"'

PROBLEM 6.93
6.93 The built-up timber beam shown is subjected to a 6-kN vertical shear.
Knowing that the longitudinal spacing of the nails is * - 60 mm and that each nail is
90 mm long, determine the shearing force in each nail.
50 mm
150 mm
50 mm
B&fiHB
lOOmm^ffi
BSgjBs
1
100 mm
SOLUTION
12
w
50 mm J 50 mm
SO mm
CD
1
s 3Z.2HX MtD* win*
* \H.06ZS * \De *w*
PROBLEM 6.94
0.8
1.5
, 1.5 P-6
Dimensions fn inches
mi
z.o A
-4
V/////A
6.94 The built-up beam shown was made by gluing together several wooden planks.
Knowing that the beam is subjected to a 1200-lb vertical shear, determine the
average shearing stress in the glued joint (a) at A, (b) at B.
SOLUTION
I = z [ £ (o.sX4.$rts + £(7)ft>.8)J + (TtfaaYzc** ]
- 60.113'.**
£fc - 0.8 ;*.
(k* Ab= W^o.8') * 3.2 .'«z jtr £0 ;*.

PROBLEM 6.95
2 in. 2 in.
-H " h~ 6 in.—H K-
6 in.
©
©
®
lfc»Pt
6.95 A beam consists of three planks connected as shown by £ - in.-diameter bolts
spaced every 12 in. along the longitudinal axis of the beam. Knowing that the beam
is subjected to a 2500-lb vertical shear, determine the maximum shearing stress in
the bolts.
SOLUTION
LoGa.'Te nei/tf*J «-xi& ctntfl c&wpoTe Motne^i" <& icte^i^,.
?«A
©
I
AW)
12
12
36
^
3
1
3
A^ ;»
36
36
2*
dlCmVAoJ^i^)
O.U7
0.667
J. 333
5U.333
5". 333
3*
L (i^
36
4
36
76
Y-
2/\
SrfS^.SSa*
- 5cL* -
i-ujh-
I = IAcT+ZI - log ;„v
0.= A,y, - RK«K3-^.M3V 8i,J
Afcjf
O.iloif

6.96 An extruded beam with the cross section shown and a 3-mm wall thickness is
subjected to a 10-kN vertical shear. Determine (a) the shearing stress at point A, (b)
the maximum shearing stress in the beam. Also sketch the shear flow in the cross
section.
SOLUTION
30 mm
16mm
16 mm
Ttyssf >/S//t'S\
H.I*
tv\**i
3. _
i, = rf-Cja,U» =
* |03 kflj
?A/t
A OO
ISO
2o4
. *4
^68 .
5^
3o
15
0
A^ rum
SHoo
3*360
o
SH60
J(»*n")
II. 7*
18.03
Aoll6^*v.^M)
25*. £8
l-W
27.46
5U?S
I (/O***
O.I&S
l£3^
15:56
^= 2Air M6f r ,g^g
TA
WA.ft.
I - ZAd^ + Zl r lom*\o mm = 7©,<fS *io
QA= (Go^O'.-M ) - 2, 146*10"* m«s
-9 V
V)
2.144 * /O-'m'1
)
I/. <?2

PROBLEM 6.97
6.97 and 6.98 A thin-walled beam of unifonn thickness has the cross section shown.
Determine the location of the shear center O of the cross section.
SOLUTION
AOA- loo £
-60 mm
40 mm
O A A^ A 4- - I WAWA
Wt AB: A = tx j " 6o *.*> j^x i
GL- A^ - GO tx "^^
it rt I
|20tnM

PROBLEM 6.98
6.97 and 6.98 A thin-walled beam of uniform thickness has the cross section shown.
Detennine the location of the shear center O of the cross section.
SOLUTION
LA* '' i^ + 31 = £* i-- AAtt = St
= S3.75* £ ,■„«
kt°^.
r;
It
■hd ~ »i
I = ZIM + It* r 177. ^7 t in.
>. f
-Z.S-
PZM, -~£I M»
, Vt^^.^^sV^] - Vfc Clio-Hi?)
2 L 177.917 t
= <X£20Gl V
e r f I * " o.e£o£ /J - (.^65 /«.
flt*J FDe pass -H-ifooak potVif' k*» TAuSj +hese
ftD
FdC

PROBLEM 6.99
60 mm
45 mm
60 mm
45 mm
*ft
S-
Px
a.
■i
F,
6.99 A thin-walled beam of uniform thickness has the cross section shown.
Determine the dimension b for which the shear center 0 of the cross section is
located at the point indicated.
SOLUTION
?aA AS : A - tx j - Co ww
Q, = $\$ - Go t x **m
^. va . 6o Vx
.H*h
Aisznz!B
-- 4s
Gl- Aj = ^£x
1 mal
7T =
- V
ft'.
45 Vx
0^m0 - -OZK, 0 =■ «Xvs)F2 - eY^R
Note +tai" "He p<*i> tff FJ 4i*^es i&r*^ «. coop-fe .
Li'kewue^ .He p«(> D-f F2 -fences-
Tk< lines of *.©4i©* of -fka 4*^«-*£ m &DOGK*

PROBLEM 6.100
6.100 A thin-walled beam has the cross section shown. Determine the location of
the shear center O of the cross section.
Sin.
6 in.
SOLUTION
T
1
-ftk
x--
(M-a^u
AU
o.= Aj
v*
-1,,/i
e =
ttft
(0.7S-X4r(9)
t,h,** *A3 " (0.7S"X3^3 4 (0.70(6 )*
?. 37 /'h.
H
R
r*
- H
J V

PROBLEM 6.C1
-H*h-
I*
6.C1 A timber beam is to be designed to support a distributed load and
up to two concentrated loads as shown. One of the dimensions of its uniform
rectangular cross section has been specified and the other is to be determined
so that the maximum normal stress and the maximum shearing stress in the
beam will not exceed given allowable values 0^1 and Tav Measuring x from
end A and using SI units, write a computer program to calculate for successive
cross sections, from x = 0 to x ~ L and using given increments Ax, the shear,
the bending moment, and the smallest value of the unknown dimension that
satisfies in that section (1) the allowable normal stress requirement, (2) the
allowable shearing stress requirement. Use this program to design the beams of
uniform cross section of the following problems, assuming a-M = 12MPa
and TaM = 825 kPa, and using the increments indicated: (a) Prob. 5.75
(A* = 0.1 m), (b) Prob. 5.76 (Ax = 0.2 m).
SOLUTION
5ee solute at V5.CZ f°r the ^termination 0{ R p.p % y^ axAWQ
We recall that ' "'
V(x)r *RA £TpA+ R^STPg- ffSTPf - ?z t>T?2
^tiJJTu?T^ 5TF1> 3 ^ 3*?5,art STPL+ as* st^P
tuntpo/ii defined in VJ.CZ
0") To-frm&rr the ^i.i.0V)ti"Pf.f, K^H^ <T?T^ ^u^-eh^t?
\Jr Uhk-naWn dimension is /i;
^f, = /Wj/C^/ , fro* S^^-t^ wehd/e h0-h-\JJs/i
If unKvuavn rffmertiron Ls fc:
-^-jM"!/^,, , fro^ 5 = zth\ we have tQ-t~ £ S/h*
\)lt v5t 5f. (fjDy pa»e373- £--I.IJ - 3 IV)
< **<$,,
(CONTINUED)

PROBLEM 6.C1 CONTINUED
FRD6KAM OUTPUTS
Prob.
RA =
X
m
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
5.75
2.40 kN
V
kN
2.40
2.40
2.40
2.40
2.40
2.40
2.40
2.40
0.60
0.60
0.60
0.60
0.60
0.60
0.60
0.60
-3.00
-3.00
-3.00
-3.00
-3.00
-3.00
-3.00
-3.00
0.00
RB =
M
kN.m
0.000
0.240
0.480
0.720
0.960
1.200
1.440
1.680
1.920
1. 980
2.040
2 .100
2.160
2.220
2.280
2.340
2.400
2.100
1.800
1.500
1.200
0.900
0.600
0.300
0.000
3.00 kN
HSIG
mm
0.00
54.77
77 -46
94.87
109.54
122.47
134.16
144.91
154.92
157.32
159-69
162.02
164.32
166.58
168.82
171.03
173.21
162.02
150.00
136.93
122.47
106.07
86.60
61.24
0.05
HTAU
mm
109.09
109.09
109.09
109.09
109.09
109.09
109.09
109.09
27.27
27.27
27.27
27.27
27.27
27.27
27.27
27.27 ^
136.36 -*^f
136.36
136.36
136.36
136.36
136.36
136.36
136.36
0.00
Prob.
RA =
X
m
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
4.20
4.40
4.60
4.80
5.00
5.76
25.00 kN
V
kN
25.00
23-00
21.00
19.00
17.00
15.00
13.00
11.00
9.00
7.00
5.00
3.00
1.00
-1.00
-3-00
-5.00
-7.00
-9.00
-11,00
-13.00
-15.00
-17.00
-19.00
-21.00
-23.00
0.00
RB =
M
kN.m
0.000
4.800
9.200
13.200
16.800
20.000
22.800
25.200
27.200
28.800
30.000
30.800
31.200
31.200
30.800
30.000
28.800
27.200
25.200
22.800
20.000
16.800
13.200
9.200
4.800
0.000
25.0 0 kN
HSIG
mm
0.00
141.42
195.79
234.52
264-58
288.68
308.22
324.04
336-65
346.41
353.55
358.24
360.56
360.56
358.24
353.55
346-41
336.65
324.04
308.22
288.68
264.58
234.52
195.79
141.42
0.00
HTAU
mm
378.79 -^
348.48
318.18
287.88
257.58
227.27
196.97
166.67
136.36
106.06
75.76
45.45
15.15
15.15
45.45
75.76
106.06
136.36
166.67
196.97
227.27
257.58
287.88
318.18
348.48
0.00
The smallest allbwoile. value of h /S the largest of ifoe. values
Shawn /n The \o&\ two columns.
mm.
For Prob- 5H5, h - hG = /73.2
for ?roh. 5\76 /7 = fu •=■ 379 mm

PROBLEM 6.C2
6.C2 A cantilever timber beam AB of length L and of the uniform
rectangular section shown supports a concentrated load P at its free end and a
uniformly distributed load w along its entire length. Write a computer program to
determine the length L and the width b of the beam for which both the
maximum normal stress and the maximum shearing stress in the beam reach their
largest allowable values. Assuming a-dU = 1.8 ksi and tm = 120 psi, use this
program to determine the dimensions L and b when (a) P = 1000 lb and w = 0,
(6) P - 0 and w = 12.5 lb/in., (c) P = 500 lb and w = 12.5 lb/in.
SOLUTION
$t>th the mtyi'>n\jw\ S^ear and -//ie n\&xin)vm b<bisi<(^ twewy.i-
Occur at f\* M/e h^Vt
VA=P++vL
M, r PL+1 <#£
73 5#T!5rY THE ALLWH&LF tf&lNfri 5Tf?CS^ RE&QiRetfenfTS
L b(8b) " ft /,* * ^ Q// J
To* L = o} VA= PW bt>0t wWxk HA'0a^ b^^O-
r&a&
PRDGrRfiH QutVOTS
For P = 1000 lb, w = 0.0 lb/in.
Increment = 0.0010 in.
L = 37.5 in., b = 1.250 in.
For P = 0 lb, w = 12.5 lb/in.
Increment = 0.0010 in.
L - 70.3 in., b = 1.172 in.
For P m 500 lb, w = 12.5 lb/in.
Increment = 0.0010 in.
L = 59.8 in., b = 1.396 in.

PROBLEM 6.C3
6.C3 A beam having the cross section shown is subjected to a vertical
shear V. Write a computer program that, for loads and dimensions expressed
in either SI or U.S. customary units, can be used to calculate the shearing stress
along the line between any two adjacent rectangular areas forming the cross
section. Use this program to solve (a) Prob. 6.10, (6) Prob. 6.11, (c) Prob. 6.21,
(d) Prob. 6.23.
SOLUTION
A £nfer V «"d ike number n of
rectories-.
2. For i? I to n,j ent€rtht dimensions h^ and n^
3. Deter r^^t lfat area. f\i -h'hf o-f each rcvfa»j/e.
Mi Peferfltint the efefa+fon of tye Cehfraid &f each recfa-njle
and -Hie e/eraf\'on V of iln. Cenffofdl o4 fte trtlft StcU'on
■£ Deferrnine Me cer\troUcJ moment of inert fa <?f ^ ^*>«.
Section: ^ _ -
6> for each Surface ScparcdtoQ two rectan^fe^ c QnJ £+l>
dzkwfoe d* erf 'thz are**. bdow-fU&h svr-face,
7* Sekct for t; fht Sadler tf b 4hd i>;...
Tht she^r/Hj stress or) ftte Surface between line tecM^J^es
b and ii-f t3
t£ - V ^
'<- 1t.
(CONTINUED)

PROBLEM 6.C3 CONTINUED
|^ if. \nt.„
a
M
■<£-
<D
w
-*-
£0.5 i".
t-Sw>
PRC)<*$*M OUTPUTS
Problem 6.10
V=10.00 kips
YBAR of Section = 2.000 in
1= 14.583 in*4
Between elements 1 and 2:
Tau = 2.400 ksi
Between elements 2 and 3
Tau = 3.171 ksi
Between elements 3 and 4
Tau = 2.400 ksi
(a)
j?0Ow*t-
Problem 6.11
V= 10.00 kN
YBAR of Section = 75.00 mm
1= 39.580*10^-6 nT4
Between elements 1 and 2
Tau = 418.3 9 kPa
Between elements 2 and 3
Tau = 919.78 kPa
Between elements 3 and 4
Tau = 765.03 kPa
Between elements 4 and 5
Tau = 418.39 kPa
©
ZZ5nw-
IF
t
ffdtnM
56 hum
Problem 6.21
V=200.00 kN
YBAR of Section = 75.00 mm
1= 79.687*10"-6 m"4
Between elements 1 and 2:
Tau = ■ 18.82 MPa
Between1 elements 2 and 3
Tau *= 18.82 MPa
Between elements 3 and 4
Tau = 12.55 MPa
«4 (a)
frorO PC2I Pr°blem 6.23
$=75 m
V=200.00 kN
YBAR of Section = 75.00 mm
1= 79.688*10"-6 rrT4
Between elements 1 and 2:
Tau = 18.82 MPa
Between elements 2 and 3:
Tau = 19-61 MPa <

PROBLEM 6.C4
6.C4 A plate of uniform thickness t is bent as shown into a shape with
a vertical plane of symmetry and is then used as a beam. Write a computer
program that, for loads and dimensions expressed in either SI or U.S. customary
units, can be used to determine the distribution of shearing stresses caused by
a vertical shear V. Use this program (a) to solve Prob. 6.49, (ft) to find the
shearing stress at point E for the shape and load of Prob. 6.50, assuming a
thickness f = i in.
SOLUTION
For each element c*\ihe rY^M-iW 6('de.j
tVe campule (for L-l to -n)--
Lentf* of c/eti€^= L^fo ^Ulff ($. -^.J*
hc«i A-ftfcWht= A; - tl~* wh&rc t ~-I in.
Dkiamz from X am fo centroicf 0i e/e^^t-f. --(%• + %> \
biSiame frtm X**h h c^troid of sectioni L Z +'^
Not, fhatfn:0 <W fUt 3^,* ^+1*0
Moment ef itlfrtig pf Jcctiov about Ce^troidnj ^fj-
Com Ration flf Q q^prt,V^ P t/he^ yfress h des)red
■&-£\(f£-%y #htrt sun. extt*as To fhe afeas- tocakrf
beWeen ont end of section and point P
She&rtfin stress at Pi
It
tiQTc: &
%** fccuft or\ neotral axis, L¥e^ for fj-p= ^T,
FRQgrRBM OUT P (77 5
Part (a):
I = 0.5333 in"4
Taumax = 2 . 02 k^i
TauB = 1.800'ksi ^
Part (b):
I =» 22.27 inA4
TauE = 194.0 psi

PROBLEM 6.C5
2"
H
A*c
6.C5 The cross section of an extruded beam is symmetric with respect
to the x axis and consists of several straight segments as shown. Write a
computer program that, for loads and dimensions expressed in either SI or U.S.
customary units, can be used to determine (a) the location of the shear center
O, (b) the distribution of shearing stresses caused by a vertical force applied
at O. Use this program to solve Probs. 6.65, 6.68, 6.69, and 6.70.
SOLUTION
'COM pUT/+T/&^ UjUL &4T DC^iT FC/t 7&P
COM^U7tr leWCTH &P jf#C# &&£/>?J-*'S~
he
or? /&& £GU&i pAfcrr
Awe* ~ J-i f. //oo
Since oNiy 7t>r malf u>a$ oseo
"T*
'aid 'sn-euj J ^
CONTINUED

PROBLEM 6.C5 - CONTINUED
*T-*
r [%
RP6&&AM OUTPUT
c+/
Q,-Q
T
L 'ffoJf&Z^y J £
lOCfl7/QM Q fi= ^//£?AK C&A*T&l^
Peaces, stgooy a/Z/j/As
Fotz W^oif seen cm MOM&N7 r Z (A/&M&*-)
Prob. 6.65
T(i) X{i)
mm mm
1 10.00 70.00
2 6.00 70.00
3 10.00 0.00
4 10.00 0.00
Moment of inertia: Ix =
Junction Q
of segments
mm" 3
1 and 2 12000,000
2 and 3 33000.000
3 and 4 45500.000
Y(i)
mm
10.00
50.00
50.00
0.00
375995?
Tau
Before
MPa
15.96
73.14
60.51
Mi)
mm
40.000
70.000
50.000
mmA4 Shear = 50000 N
Tau Force in
After segment
MPa kN
26.60 2482.37
43.88 20888.54
60.51 27372.75
Moment of shear forces about origin:
+ counterclockwise
M = 2436.386 Nm
Distance from origin to shear center:
e =
48.728 mm
CONTINUED

PROBLEM 6.C5 - PROGRAM PRINTOUTS CONTINUED
Prob. 6.68
T(i)
in.
X{i)
in.
1 0.25 3.00
2 0.50 0.00
3 0-50 0.00
Moment of inertia: Ix =
Y{i)
in.
4.00
4.00
0.00
45.3328
Ui)
in.
3.000
4.000
in*4 Shear
25.000 kips
Junction
of segments
1 and 2
2 and 3
Q
inA3
3.000
7.000
Tau
Before
ksi
6.62
7.72
Tau
After
ksi
3.31
7.72
Force in
segment
kips
2.48
12.47
Moment of shear forces about origin: M =
+ counterclockwise
Distance from origin to shear center: e =
19.853 kip-in,
0.7941 in.
Prob. 6.69
i T{i)
in.
X{i)
in.
1 0.25 4.00
2 0.25 4.00
3 0.25 0.00
Moment of inertia: Ix »
Junction
of segments
1 and 2
2 and 3
Q
in" 3
2.000
3.875
Y{i) Ui)
in. in.
5.00 2.000
3.00 5.000
0.00
23.8331 in"4 Shear -
Tau Tau Force in
Before After segment
ksi ksi kips
3.36 3.36 0.91
6.50 6.50 6.80
10.000 ki
Moment of shear forces about origin:
+ counterclockwise
Distance from origin to shear center:
M = -7.273 kip-in.
e = -0.7273 in.
Prob. 6.70
L T{i)
in.
X{i)
in.
1 0.25 2.60
2 0.25 2.60
3 0.25 0.00
Moment of inertia: Ix =
Junction
of segments
1 and 2
2 and 3
Q
inA3
0.281
0.844
Y(i)
in.
0.00
1.50
0.00
1.6881
Ui)
in.
1.500
3.002
in"4 Shear « 10.000 kips
Tau
Before
ksi
6.66
20.00
Tau
After
ksi
6.66
20.00
Force in
segment
kips
0.83
11.65
Moment of shear forces about origin:
+ counterclockwise
4.332 kip-in,
Distance from origin to shear center:
e =
0.4332 in.

PROBLEM 6.C6
-ih-
Force exerf-e.4 w el't^c^t £
6.C6 A thin-walled beam has the cross section shown. Write a computer
program that, for dimensions expressed in either SI or U.S. customary units,
can be used to determine the location of the shear center O of the cross
section. Use this program to solve Prob. 6.100.
SOLUTION
I>f5JriJ>i'tibH ci shearing «pfflK.<frs )n element" L
Let V - t/he&r iv\ cross 5etiibh
T - Cevtfroielal moiv)^/- 0f if\txii'ck 0^5ec^\6ri
We h<x\l< for shaded are*.
It: 11 d
tffr
"1"
a.,
f
or
F. r
i
■red'*)* MFC*-*)*,
I
Th
ThlJ*i5fe^ oH^e forces Ti /?iusf h^eplvgl^t ft V atshwcQtrifer.
1
PROGRAM OUTPUT :
■ri. ^
Prob. 6.100
For element 1:
t = 0.75 in., a = 4 in., b = 0
For element 2:
t = 0.75 in., a = 3 in., b= 8 in.
Answer: e = 2.37 in.

CHAPTER 7

PROBLEM 7.1
4ksi
7.1 tfcrnagfc 7.4 For the given state of stress, detennine the normal and shearing
stresses exerted on the oblique lace of the shaded triangular dement shown. Use a
method of analysis based on the equilibrium of that element, as was done in the
derivations of Sec. 7.Z
SOLUTION
Stresses ,^^f
A Si nJ?o°
Are<is
3Aco»?o
forces I yA v-n ?0.
CA - 2Ac*#%a?0#-SAcos*o"!si*#>#- 3As«'rt20cosPo#-'/As.V2o05.^^)* = 0
?A ^3Acos^V«v?0#-SAe*$:fcW&o+ 3Ai;«to$mt?0* - fA a* #>%* A>* =■ o
PROBLEM 7.2
lOksi
7.1 threvga 7.4 For the given state of stress, detennine the normal and shearing
stresses exerted on the oblique fece of the shaded triangular element shown. Use a
method of analysts based on the equilibrium of that element, as was done in the
derivations of Sec. 72.
6 ksi
SOLUTION
10 ksi
10 A Cos IS*
If AtJn/S'
SA + 4Ac*s/.£Vm/s#+ ioAcosircoi;r^cAft/n^°s;«/r>+^ :A *.'« «■•*•* *r° = o
if A ¥ 1 Ac** If** IS* - /oAm«*im IS* - 6A*.*lf*c*s/S# - </A s/«/rtiui (T = o

PROBLEM 7.3
SOMPa
7.1 throng* 7.4 For the given state of stress, determine the normal and shearing
stresses exerted on die oblique owe of the shaded triangular element shown. Use n
method of analysis based on the equilibrium of that element, as was done in die
derivations of Sec. 1.%.
40MPa
SOLUTION
3flMP«.
2o AcvzSS*
"JoMfV
HoA*j«,sr1
JEfrrflta
G = So eoa"^5* - 4o «.'n**S5"Q - - o.52l MPa. —■
+ *5"F* o
fA - So A M ST*;* 5?" - 4o A sJv,55-°c^>S^0
7.1 zfaroagh 7.4 For the given state of stress, determine the normal and shearing
PROBLEM 7,4 stresses exerted on die oblique race of die shaded triangular dement shown. Use a
method of analysis based on die equilibrium of that element, as was done In die
derivations of Sec 7.Z
SOMPa
GO A cos So*
Forces
9oMPa
MP*
Stresses. A^cas
+MFr o
6" A - 9o A sin 30°cos 3a" - ?o^c«s 3d*am 30° + 60 A cos 30°cos 3o° « o
6" r ISO Sim 30" coa 30" - fiOcafi^O0 S" BZ.1 MP*. -*
f/5"F=0
?A + 9oA *i"r»^*©^ 3o*- <?oA c«s3e>"coxSo* - GoAc«s 3©V*3o* = o
t a lO (c«s*3a- -sirta3<3*) + Go teaSo'si* 3o° * 71.0 Mft* • -*

PROBLEM 7.5
12 bi
"' 8
|I|H|
ksl
18ksi
7.5 throagh 7.8 For the given state of stress, detennine (a) the principal planes, (b)
the principal stresses.
SOLUTION
<5y - is ks.- s;- - n k$i ^ -- ***;
- 3 * 17 Irti
PROBLEM 7.6
2ksl
7.5 through 7.8 For the given state of stress, detennine (a) the principal planes, (6)
the principal stresses*
SOLUTION
5, - Z ksi
§ • io k«;
«,»-3W
<M +■*«.= J% r PX-^ * 0.75O
26P - 3G.87* dr * l*.43% 108.43'
- e ± s Wi.-
Sw - // ksr
6w» * I fc»i

PROBLEM 7.7
7.5 throngli 7.8 For the given state of stress, detennine (a) the principal planes, (b)
the principal stresses.
SOLUTION
<Sk* -£o MP*
6^ r - Ho Mftt
2€>f =: ~7¥,05*
£,* -37.03*, ^?.S7'
cv>) er^ = 5dS ^(S^A)'* r„'
^C"-*^)*4^*
6*^„ * - 13.60 MPft.
PROBLEM 7.8
■IBM Pa
16 MP*
7.5 through 7.8 For the given state of stress, determine (a) the principal planes, (b)
the principal stresses.
SOLUTION
(5^r -4* MP*
^r-CBrtPo.
(a) fi«- %& •= -ZZ* = (tYr**) •= - i g7s
W TA« X»r e;.^ I&+HS |-SO
6U*- ~ -24 AlPa.

PROBLEM 7.9
7.9 thrmgh 7.12 For the given state of stress, detennine (a) the orientation of the
planes of maximum to-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
Si r la to
§*-l2fa
Ty=-8k&;
(a) ^ 2fti . - S^. . .^JS = -,87S-
,29, = - 61.13*
0S =-SO.-6*_, ■5^.0'*
PROBLEM 7.10
10 bi
2ksj
3 hi
7.9 throngs 7.12 For the given state of stress, determine (a) the orientation of Ihc
planes of maximum In-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
6>* -*1fti
<Oj* Id Us;
?y*-3k%;
U>> r^, *J(S^Sl)% + ^
2 -1 o \7
)N(-3)'
5 k».a

PROBLEM 7.11
7.9 tbroagh 7.12 For the given state of stress, determine (a) the orientatioa of the
planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
<£, =r - 4o MPo.
2iy = 33 MP*
(CO W. ZS, r - §L^Sl , - -G°+4°
2t^
(«XM)
o. ^asi
26.- i-S.^5*
Gl = 7. 97*. 97. T7° -**
(tt ^^(^S^V
--V^^T^r
- 36.V MPo. -*
CO 6' - €u* -- ^S. , zi|lie s -.sb MP*
PROBLEM 7.12
48MPa
16MPa
60MPa
7.9 2hroagh 7.12 For the given state of stress, determine (a) the orientation of the
planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
6£r-W HFfai
?vt-Co HP<x
GO
+a, -^.-^.-J^^U .o.
a*
**
0?**O
S333
20*- ZZ.Ol
^ - II. W*, K>*».0*e
(^ 2i« -
6,-eryv*
r - V
V^PT7^1 -
(c) 6- =
<5^
- Sst-* €v 16-fg
2 " «
C8 NPfl.
-1£MPa

PROBLEM 7.13
7.13 throngh 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
6* = - 4o Mp* <5^ go MP* r>y = Zo MPa.
6~A + <5y
0 MP*
^gfr = -5K> MP«.
e - 6; tgy 4 ffy - 6; CoS£© + V. s,„;?e
**
«¥*
- ^3 * S.'rt 3© * 2ij L*>$ 26
J - o " 7— CoJ *® " *J sm ^
(a> © = -2S° 2.G - -So*
10 - &0 cosC-SO*) + 2osi«t-So")
+ SO sin (-St? ) + ?0 oo&t-ZO9)
|0 +SO os (--SO*) - 30 5/ito')
6k* r
<* ^
-37.5 MP*
-25.4 MpA
J7. 5 MP*
6V - lo -5t> cps Ca<?°1 + Zo siM0o>i =
6^. * So +^o cos(Zo°) - 2o s/wteo-) =
- 30.I MPq.
3S\«I MPo.
50, I MP*

PROBLEM 7.14
SOMPa
50 MPa
7.13 through 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
6** - O
6j - -So Mflx
Tij = -So MP*
6*+6* - - 4o MPa
2
&-6^
r *0 MPa.
6*. - ^~^ + ^^ c*s *6 + ^ sin j?£>
^>y ~ - 5k^S sin ** + r*, coS^
6? - H^ " §lCSl c* 2* - r* s- ^
5,. r _ M-0 + HO cos (-£&•) - SDs'^(rS0o) *■ z«/.o MP* -*
'fry T " ^s.^C-^o0)^ cos (-$&') - - LS MP*. -*
Gy - - 4o - 40 cos(-5^oS) + JOifn G\#>") -r -/o^.O MP -^
-4o + Mo coi (aoO - £"0 511 (2o°)
- \rS MPc
- Go.T AlP*.
- €AS MP*

PROBLEM 7.15
8 ksi
7.13 throngh 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
6"„ =■ 8 ks»
<5j r - \7 ks<
r^= -c fcs«
2 2
Tv/
s r 6\^£, _ _6^_ Cto3j?6) _ ^ s.|i ^
rvy - - /O $,v, (So* ) -6m IrSoT ) *
«&•
- 2 - | O cos C-^0* ) + G Si** (-£©* ") "
l.oz ks."
- 13.0? ks,'
0° 20 * 20°
- ^ _ io cos C*o' ) +■ £ *f* (2^") " - ^.34 ks.'

PROBLEM 7.16
16 ksi
7.13 through 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (ft) 10°
counterclockwise.
10 ksi
SOLUTION
6* = o
6^-16 k«r
S^iSi » 8 fcs,
<SV * ^St + Sl^Sl Ct>& 2e 4 %, s;* 20
i;y -
2 + S it* (.-So* ^ - io a,-* (-5*^ = ;?o.So k»; -^
*,• -
«&■
(b^ © - lo° xe = Zo°
<5%, = 8 - 8c4>s(?°0>> 4 lOs;^(ao*) - 3.90 fcs.'
<Sy * 2 +3 c«*6io0>> - 10 co4 fro*) - 13.10 ks.

PROBLEM 7.17
1.8 M Pa
Ms"
7.17 and 7.18 The grain of a wooden member forms an angle of 15° with the vertical
For the state of stress shown, detennine (a) the in-plane shearing stress parallel to the
grain, (ft) the normal stress perpendicular to the grain.
SOLUTION
6* = - 3 MP*
9 * -IS*
Sj - - l.g MP*.
-fcy = O
U& ' - SO'
6^-6*.
(a-) r*y -- - °*~^ siw^ ♦ r*, s,-ft 2©
- - 0.300 MP«, -*
(fc> S„. = §<-*$* + ^C^L Cos ^ + ^ S(V, ^
- - 2.^ MPa. -*
PROBLEM 7.18
400 psi
7.17 and 7.18 The grain of a wooden member forms an angle of 15° with the vertical.
For the state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
fy *0
r^=- 4ao p*.-
SOLUTION
S, = o
B = - IS' 2B » - 3o°
(a) Xy = - g^"^ si*, £e + q^ co* 2©
~ -0 + .Moo co&C-zo")
(b) 6,. * S±S + -§LL& c^ ^ + ^ s/m ^e

PROBLEM 7.19
7.19 The centric force P is applied to a short post as shown. Knowing that the stresses
n plane a-a are a= -15 ksi and r= 5 ksi, detennine (a) the angle yfrthat plane a-a forms
with the horizontal, (b) the maximum compressive stress in the post.
on
fill-
SOLUTION
y/A
r^* o
s.
>«^ CJhy.
A
A
Are
ft + o- *±*~*
*>t>
:CtS
lb) Ps C«6"A setfiXsecfi)
A
-<5"
Coi'/G C*s* W.H1
=• 16.67 ki," -**
PROBLEM 7.20
7.20 Two members of uniform cross section 50 x 80 mm are glued together along
plane a-a, which forms an angle of 25° withthehorizontal. Knowing that the allowable
stresses for the glued joint are a= 800 kPa and r= 600 kPa, determine the largest axial
load P that can be applied.
50 mm
SOLUTION
G* -- o £, - o
For pf
A
f ' -(Q -?^s.\fiwi6 + VyUosQ- siVd ) - ^ 5;.G£0fosa+0
P =
3K
A r
si'h6Tc**65"a 5/n ^ cos *5"
httovcMc v&Joe of P /a He s»U/eA P-" 3.<?0y|0*N/ - 3.<?0 i<M

PROBLEM 7.21
100 kN
(a) irs "-D
7,21 Two steel plates of uniform cross section 10 * 80 nun are welded together as
shown. Knowing that centric 100-kN forces are applied to the welded plates and that
P= 25 * determine (a) the in-plane shearing stress parallel to the weld, (b) the norma)
stress perpendicular to the weld.
SOLUTION
irecL 07 ^
\OOVti
100 kN
Forces
P, - lOO S/n IS" - &
= 292:1 l©~' tof
"Au;
R r 42.26 JcM
ito ZF« r o F„ - ;oo Co<
PROBLEM 7.22
100 kN
7.22 Two steel plates of uniform cross section 10 * 80 mm are welded together as
shown. Knowing that centric 100-kN forces are applied to the welded plates and that
the in-plane shearing stress parallel to the weld is 30 MPa, detennine (a) the angle #
(b) the corresponding normal stress perpendicular to the weld.
SOLUTION
100 kN
ft - _
cos /&
forces
<co 2*FS - o 1^ - loo s»«/8 - o
2,0 vO r -——£--
(b\
F~ = (OO s.'nfi kU = /OOVO s.V/3 IS/
125x10* Sinficasfi
■WO^ * ^'^ -*
2 F„ = O F„ - /l>0 cos/3 ^ 0 R, = /0O cos |M.1$° =■ 96.'58 ^W
Av
tw - ——
$qo*JO
i-4
%%S'„7fl*l6 m"*"
Cos. J 4.3**
6-, fk , Jj^ffi!^, i/;7.3-/o4Pa - 117.3 MrV
Aw
%25.n*tHo

PROBLEM 7.23
7.23 The steel pipe AB has a 102-mm outer diameter and a 6-mm wall thickness.
Knowing that arm CD is rigidiy attached to the pipe, determine the principal stresses and
the maximum shearing stress at point H.
51 mm
SOLUTION
- d» - 12! -
■= s\ ***>
/V- = r0 -1 - *S
Mm
WtV7
= 4. i8SSx/0"fc k,"
I ' iJ =■ J?.o«*7 w*
\-£>rce.-.couf>Jt sws.4ent at ce^4e^ o"f +i>k« i^
Fw = IO v/oa N
1% =■ (lo«ial )(zoo*tom%) - Zooo l\/-m
T - M,
7.000 rJ-h*
51 */©"* ^
H
1
7*- r 4.185-S-xlO-6 X1.37x/0 H=
Fo»r Sfi^ici'rc-Pe A= ?rl ^ =
r, - -±
3TT
mm
H
2iu» JM.37WO*f //.M*to** 35'.3'?x/0^ Bt
6^- 6^-R -- -3s\3<**/o6 Pa = -3£¥ AlP*.
?L*. - 1? ' SS.M MP^

7.24 The steel pipe AB has a 102-mm outer diameter and a 6-mm wall thickness.
Knowing that arm CD is rigidly attached to the pipe, determine the principal stresses and
the maximum shearing stress at point K.
SOLUTION
ti* r.- i - hs~~
F* - 10 kW = lo* lo3 hi
tti* -Oo*totyiso*fc>*s j'- 1500 W-kx
At po/-»' k'j p*«.ce XocaA %-<KMi%. "*»
C^ C, - £l*to~z
m
B—1.--3= 16,1* '^ - ^.gg'rg!? - *..».*>' ?* - 34.ft MP.
rfti'M K -Tie* ow c«t^9i^ssro^ srJe at neoTv^ «c*.t's • 6y r - 36-££ HP*
To-ki s+wsses &A p«i>+ k 6i = o S^ - -36.56 MPo. } r*y -JW.37 MP«
6*^ « ^C^ + ^V -IS.23 MP*
K^p, * 6L. + ft - - !*.«« + 3o.46 - + 13. 18 MP« -»
K*;* = 6L. - R ' - 18.28 - So.4* - -M&.7W MPo. -^
f^ * R - 30.*U MP* -*

PROBLEM 7.25
T*
or$i£?fi
7.25 A 400-lb vertical force is applied at D to a gear attached to the solid one-inch
diameter shaft AB. Determine the principal stresses and the maximum shearing stress
at point H located as shown on top of the shaft.
SOLUTION
EfiSi'tfbJertT force - coop Je sys"lcw aj ceMe* <fi aiitfFf
T *■ (4ooKO ■ &>o A-im.
sl-fC - coigns* m* 2*£J> cowoii ;**
3 X O.Of9oft7 r
:*_ 4.07** ks>
^ -* 5W.V16 Its-'
6V* • £ (6; + Sy ) * \?.aas ks;
6*b" 6w*-■ R = -0.66/ fca,'

PROBLEM 7.26
7.26 A mechanic uses a crowfoot wrench to loosen at boll at £. Knowing that the
mechanic applies a vertical 24-lb force at A, determine the principal stresses and the
maximum shearing stress at point H located as shown on top of the J - in. diameter
shaft.
SOLUTION
Sri^f+ C^oSS secffo*: <A~ O.^Strt, C=-fcel» 0.^75* i*.
J - ^CH - 0.03l0t3 i«4
I-W* O.OISS3Z ^
*"*-T g-?' ^i^ - s.w^r;-M77h;
TV*.*»a*e*s* SKfft^ • Ai polfij H a*fv^ia «»J£ "f0 fv^nsve^s* aU^ i'a zero.
R«aui4M+ cypresses-' 6i-- 3.477 k*,^ § = O, ^ - ^.8*7 fcsi
s« * iCR+e;)r '.7^s
R - TC^J^T^V * -fuTStfTTw?* - S.378 Irs,'
5*a* 6^ + 'R - 5". II 4 tei

PROBLEM 7.27
7.27 For the state of plane stress shown, determine the largest value of 05, for which
the maximum in-plane shearing stress is equal to or less than IS ksi.
5,
4 ksi
12 fcsi
SOLUTION
6X - \Z kr>,
Lei U - *=*
* ? . TV*= 4 k*.'
% * 5w - 2o
It*
PROBLEM 7.28
1 8 ksi
10 ksi
7.28 For the state of plane stress shown, detennme (a) the largest value of tv for
which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the
corresponding principal stresses.
SOLUTION
r V~^£
= /2 kft.-
6"-* - i (<s; + 6}^ - » it*.-
61' 61* + P - 1+12 = /3te,'
St * 61* - R - 1-12 = - /I: It*,'

PROBLEM 7,29
7.29 Determine the range of values of at for which the maximum in-plane shearing
stress is equal to or less than SO MPa.
75 MPu
—" 40 MPu
SOLUTION
6* * ? , Sf = 7S MPa a ?TW -" Vc? MP*
lei u -- S^Sl $„* Sy +Zu
- "2^ - So MP*
R - Vu> V
U * ± V R1, - "V" ' ^ iso^^Ho^ - ± So MPc,
G* ■ €Ty +2u - 75 * (5tX3o) * 135" MP<i ^ IS MP*
PROBLEM 7.30
2 MPa
12 Ml'*
7.J0 For the state of plane stress shown, determine (a) the value of rv for which the
in-plane shearing stress parallel to the weld is zero, (6) the corresponding principal
stresses.
SOLUTION
6* * \Z MP* ^ 6^ = Z MP^ a r>y = ?
SLt - £(&+<%) r 7 KIP*
.6+.* 6*^c-> R = 7 + 5".773S" * 13.77 MP*
^ ■ 6U - R - 7- 5".773r ' I.Mt MPa.
OlA

PROBLEM 7.31
731 Solve Probs. 7.7 and 7.11, using Mohr's circle.
7.5 throagh 7.8 For the given state of stress, determine (a) the principal planes, (b)
the principal stresses.
7.9 throngh 7.12 For the given state of stress, detennine (a) the orientation of the
planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
<5„ = -£o MP*
6^ r - Ho MP*
&£& r _,$£> mpa
X S (-61,-2^ * (- 60MPA , -3TMrV)
Y-^j r^J.-C-VoMfVj 3^MP«)
0 - 7H.0£9
Qs = - ifi = - 37.03 * —
6w„ = &* + £ * -5©4S6.^ * -I3.e nP*_
0O - Ge +- ^* » 7.77"
Ge = ©^ f 45" r <?7.97*
rw = R = 36.V MPa.
%, ?3S MP*.

PROBLEM 7.32
48MPa
16MPa
732 Solve Probs. 7.8 and 7.12, using Moor's circle.
7.5 tkroagh 7.8 For the given state of stress, determine (a) the principal planes, {b)
the principal stresses.
7.9 through 7.12 For the given state of stress, determine (a) the orientation of the
planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
60MPa
SOLUTION
§; = 16 MP*. 6j--<i$ MP*
Zy --&> m*.
16 MPa
irt (miO
r 1.375
(B
D^
& / C
Y
^"Tn
o
t
X
F
A|
/
&a = ~i<* * " 50'?c
|S - 1*0 - <* * H8.67*
R - Y Zf1, + Px* = y 3Z1 4 Go"" r 58 Mp«
<56 = S*, ^ 6L. - R » - (4 - 68 - - 8*J M Pa.
- /e MP*.
r^ - ft
5
(MPA
5'= G
«•#«

PROBLEM 7.33
12 bi
18ksi
7.33 Solve Prob. 7.9, using Mbhr's circle.
7.9 throagk 7.12 For the given state of stress, detennine (a) the orientation of the
planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
6^ , $*±§*
Points
Y: (Sj^ W-Uk*;, 8 Its;)
C' (SU.o ) - (3k*'j o)
ran
ot «■
FX
CF
ot r Zg.ol*
15
= 0.5*333
6", - - 12 Iffti
=r 3 ksi
2*-" S ksi
9A r JL* r |cf.0«T
R - /IP^iTx1 - -/l£z + Sx * \7 k*;
r~» * r - 17 **; «*
6' = G~ - 3 ks.* -4

PROBLEM 7.34
734 Solve Prob. 7.10, using Mohr's circle.
iOksi
7.9 through 7.12 For the given state of stress, determine (a) the orientation of the
planes of maximum in-plane shearing stress, (&) the tnaximuni in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
-3ks
ta* ot -
FX
eB^ £* * 18.43°
* 0.75
rw - R «■ -s k*.'
G1 * 6^ =■ € Its.'
=• i-^;

PROBLEM 7.35
7.3S Solve Prob. 7.13, using Mohr's circle.
7.13 throvgh 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10°
counterclockwise.
SOLUTION
s; = - f o mpoo
°«*e ^^
6i=Qo Mpa. rlr^oMPa
**
/o MPc,
Points
X- (-WO MP* ^ -20 MP*)
(to MF^ 2oMpO
( lo Mffe, 0 )
Y
C
FX
2o
R = V FCl -v FXk = -fSo^Tzo1 = 53.8S MP*.
6*' = S** - /? cosCp - -S7.5" MPa —»
tf*3 3 - R StV> 3> - -2£W MP* <-»
fy - S-* ^ Rcoscp = £7.5" MP* --»
(b) 9: ICO ZB = JftO
6^ - 6U - £ cos Cf « -SO. I MPa. -*
£y - Rsi^ * -&5\1 MP* -**
6y * 6^ +Rcosf r 50.1 MP* -*
-T) CmpO
CnP^

PROBLEM 7.36
80 MPa
736 Solve Prob. 7.14, using Mohr's circle.
7.13 thro«fh 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (6) 10°
counterclockwise.
SOLUTION
6* - o
$r
_ 50 MPa
5Mt= JSa^L =
-86 MP*
- *0 Mpa
•T^ --5b MPa.
Poi«+*
Y
C
( o, So MfO
(-80 MPa, -£"0 MPO
* 6¥-03 MPa
(cO S ■ 35* 1 2$ ■ ^ *
qpx v5"/.3**-*So* * L3«r° 5
5/ = §«. + Rw*f-" «.0 MP*. -
6^--* 6U* f?c«* # r -|*.^ tfPo.
ryy * f r? 5/n $> * - 60.7 MPa. -
t*) (MP**
* (MP*)
(hW

PROBLEM 7.37
8ksi
737 Solve Prob. 7.15, using Mohr's circle.
7.13 through 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated through (a) 25° clockwise, (ft) 10°
counterclockwise.
SOLUTION
6T,c - $ ^m"
6J - - 12 (a;
%3 - -6 ks\
Cw'-^^ = -Zkti
X: (Ski/, 6 Its.')
y: (-t2u%;3 -cks.')
C: C-2 tei-j o)
2eP = 30.96'
T = °-<
ft _
= n.cc ksi
Cp - SO* - 30.94* r 19.09*
6"„»r 6** 4 Rc*s<j> * 9-02 far
'Zty * "R si* 9 - 3.80 fcl.* -*
(J) r 3o.9C* + *&* - 50M*
Gf ' 6^ + 1? Coi Cp r 5.3* 1*4.'
% * -Rsmgp - - 9.06 lfs.a -
•ti.ikal)
flrO
0*0
(fell)

PROBLEM 7.38
p
1 1
19
X- (o^-ioltt
Y: (16 k&i^ Ic
C-' (gtoi, o
Oksi
> ksi)
)
738 Solve Prob. 7.16, using Mohr's circle.
7J3 throngh 7.16 For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated throudi (a) 25° clockwise, (b) 10°
counterclockwise. '
Sj r 16 |^;
= 12.SI Jcsi
(a) 9 = 25* "> 29* SO* ")
<p- St-St'-So" - LS99
€y - Sw -v *Roo4 4> - ZQ.t\ k%;
(p = 5"/.3^°+ 20" r- 7/.3¥v
*J(k*rt
0»«)

PROBLEM 7.39
1.8 M Pa
739 Solve Prob. 7.17, using Mohr's circle.
7.17 and 7.18 The grain of s wooden member forms an angle of 15° with the vertical.
For the state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
SOLUTION
6"- - -3 MPa
• /.--fn
<S^ r - 1.8 MP*
-15°
6~ = ^f^- -~ -2.4 MPk
K, = O
0 = - ir* Z9' -3op
Cx - 0.6 MP*. R= 0.6 MP«.
rj (mP«^
S (Mfb.^
(V> 6V - S**-CX'co* 30* - -2.H - O.Cco*3o
O „
- 2.<?2 MPa.
PROBLEM 7.40
400 psi
7.40 Solve Prob. 7.18, using Mohr's circle.
7.17 and 7.18 The grain of a wooden member forms an angle of 15° with the vertical.
For the state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
SOLUTION
Tmj = 4oO ps.'
'«rfe
^ (MPO
V: (^ijliO r (°> Hoop*,)
(MM

PROBLEM 7.41
7.41 Solve Prob. 7-19, using Mohr's circle.
7.19 The centric force P is applied to a short post as shown. Knowing that the stresses
on plane a-a are <7= -15 ksi and r= 5 ksi, determine (a) the angle /fthat plane a-a forms
with the horizontal, (b) the maximum compressive stress in the post
-*
Fro** ib* Mtfl*** & ciVcJe.
* = W + 5A* toi ^
- IC.G7 *s,'
/3 = TS. 4

PROBLEM 7.42
50 mm
t*
6tr>S(f
7.42 Solve Prob. 7.20, using Mohr's circle.
7.20 Two members of uniform cross section SO * 80 mm are glued together along
plane rt-a, which forms an angle of 25° with the horizontal. Knowing that the allowable
stresses for the glued joint are a- 800 kPa and r= 600 kPa, determine the largest axial
load P that can be applied.
SOLUTION
$*--
V
o
0
9/A
' " Sin SO'
t
P/A
e= ^? (i + co<*50o>)
p.*
2A<5
(V^ f>*
+ cos 50*
I 4- c*a So*
f 5 3.<?oHo* N
- g)ft»K>">froO»lcE) „ tf#rMo* M
5** 5"0
CUos'i-j +Ae s^Jjer \t*Jj* *P < 3.1o*(<?V* 3.<?o kU

PROBLEM 7.43 7*43 Sobiz Prob. 7.21, using Mohr's circle.
100 kN
SOLUTION
7.21 Two steel plates of uniform cross section 10 * 80 mm are welded together as
shown. Knowing that centric 100-kN forces are applied to the welded plates and that
fi= 25 °, determine (a) the in-plane shearing stress parallel to the weld, (b) the normal
stress perpendicular to the weld.
£
A
loo *tox
(Iok(o-*)(8d*|o-^
*0 = °
- /25*IO*P* * nfMPa.
Fro** Moiir^3 c\rJ«
(cO r^ ~ £2.5 5.^5o° = 47. *? MP*.
= loi. 7 MPcu

PROBLEM 7.44
100 kN
a (*(V>
100 kN
7.44 Solve Prob. 7.22, using Mohr's circle.
7.22 Two steel plates of uniform cross section 10 * 80 mm are welded together as
shown. Knowing that centric 100-kN forces are applied to the welded plates and that
the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) the angle fi,
(b) the corresponding normal stress perpendicular to the weld.
SOLUTION
Ft^o*vt MoU** *s ciAtJe
- \2S>\0t Pa. » \7StfP*.
Go s;M zfi r ^ ,
0.48 fi » 14.3
Y
t
1
<— 6Q.-5 —*
« \
\^ i
3S
».
*
6"(»M
- 117.3 MP*.

PROBLEM 7.45 7.45 Solve Prob. 7.23, using Mohr's circle.
7.23 The steel pipe AB has a 102-mm outer diameter and a 6-mm wall thickness.
Knowing that arm CD is rigidiy attached to the pipe, determine the principal stresses and
the maximum shearing stress at point H.
51 mm SOLUTION
M, r (ioxioM^^x/o*1)- 20OO M*»*i
2>
Tor-ftr©^: T = M- -* Zooo hi-**
M
f
C - ^o - £t*lo ^
Be^Jn
■a*
i « J
<5T °
To+*f presses *A foi*T H G* = o > (^ - o r^ - 2V.3*7 + ll.o* r 3£ 31 MP*
MP*
R = 35". y* /ipft
6T(^Pft &■•» 'SL.-** - 3£".3<* MP* —»
ff,*;.* * 6U - R - - 35:5-1 MP«l —

PROBLEM 7.46
7.46 Solve Prob. 7.24, using Mohr's circle.
7.24 The steel pipe AB has a 102-mm outer diameter and a 6-nun wall thickness.
Knowing that arm CD is rigidly attached to the pipe, determine the principal stresses and
the maximum shearing stress at point K.
£!mm SOLUTION
ia _ alp _ lo2 _ c"i
vs = r»-1 = V5-
I * iJ " ?.°*2 7 mo'4 v* +
Foi^ce- Coi>pJfe sys^e^ at center tfT+tt«e /'« +J»e
p-?«i*e cortiAioiw* poinii H A,w J k
F* = lo *ICT N
y
Tors/on *
i
K
t
No+c H«iT Xoc*.J x-<*x\'$ is -f^keit a-^oha n€^M"t'v« aPohtJ 2 - JlxP-tcMoA.
tfen*Ug. I$l- __ r _____ . 3£m£L Mpa-
TA&S strew €5 <^f p0rRJ K 6; * O ^ ^ .- -3i-ftriPfi , ^ = 24-37 MP*
R = 7(^#yr^? - 3°_4<S MP*
GTChp^
- 12. 18 MP*. -*
r -4g.7*r MPcc -

PROBLEM 7.47
2 in.
Sh*.Ft deoss secfi'0"
Tc
J"
7rft«5 i/e^se SJieo^ :
7.47 Solve Prob, 7.25, using Mohr's circle.
7.25 A 400-lb vertical force is applied at D to a gear attached to the solid one-inch
diameter shaft ^fl. Determine the principal stresses and the maximum shearing stress
at point //located as shown on top of the shaft.
SOLUTION
V * 4oo Ji>. M - (HooXO * Moo J»t-tH
T *■ (4oo}(sO - goo Jfa.m,
el- I .'w C *£d * O-S.Vi.
J = ^c* - O.098PS" .V.1* X = 4 J = 0.0*n©»7 m"
0.01103 7 r
Si^w «^ poiVr M is ?e^o.
€; = 34.^46 ta j 6j = Oj -K, - 407* i<*;
AJcr
SI * 6L. - ^ - - 0.661 Ifef -*

Totr^i'
o\a -
PROBLEM 7.48 7.48 Solve Prob. 7.26, using Mohr's circle.
7.26 A mechanic uses a crowfoot wrench to loosen at bolt at £. Knowing that the
mechanic applies a vertical 24-lb force at A, determine the principal stresses and the
maximum shearing stress at point H located as shown on top of the X - in. diameter
shaft
SOLUTION
BefOiy/aJe^ii -fore* - cwplt sysie*^ <x+ Cewle^ erf sh*cf*f
Sh^ty C**o5S Se«Tiow- (J - 0.75",Vi. C = id" O-375'im
J r ?c" = 0.O3I06S ,V» * I = ^J= 0.OIS53* •h"
6eHj,nn: 6"-^£-r O^Xo-W) . 5.477*/o>/ = 5.V77 *,;
6".* * i(^ + ^/)r L738 Vs..'
- -/l.73a*" + 2.897* = 3.37« fesi
Rcsl>J+*w»+ s-\v«ssei :

7.49 Solve Prob. 7.27, using Mohr's circle.
7.27 For the state of plane stress shown, determine the largest value of ay for which
w mavimum in-Dlane shearing stress is equal to or less than IS ksi.
the maximum in-plane shearing stress is equal
SOLUTION
12 ksi

PROBLEM 7.50
8 ksi
10 ksi
Gort 1
I*- $ fes/—4*
*7,S0 Solve Prob. 7.2& using Mohr's circle.
7.28 For the state of plane stress shown, determine (a) the largest value of rv for
which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the
corresponding principal stresses.
SOLUTION
ter of +»i« M«Kr*» cfreie A« at
He i.V* /.X* of +<i
Tfce
(cO The -Pa.ir*M
of T*y »s ©k+«.iVie»l fn>v*
iWtiji* COX,
C\>") T^c ort'iAcipaf st"^S5es Avse

PROBLEM 7.51
7.51 Solve Prob. 7.29, using Mohr*s circle.
SOLUTION
7.29 Determine the range of values of ax for which the maximum in-plane shearing
stress is equal to or less than SO MPa.
V&r ■Hie Molirs oV^Je ^ p©t*««t
Circus r* R - SO MPa
Let" C, oe "Wva Xgca\Io%* of
circJe. Av\d CL te -f'A-cf"
or -He ^i>lit hoo5» one,
<v? - 50 mp«
.c.o* +■ r»ra = c,v*
CD** fo2 ^5o*
QO » 3D
Coo^.n^cs 0-T p*f«t C, A^e (^ 7S- S<0 = CO, «S MP<0
Likewise^ coarrf .'«*+« <*f poi«f Ct *^ * (oo 7S"+3o) - (O, I05 f|P«.)
Coor^i>4fe5 tff po{*t X, C <f5-■#>_, - ^o") * (iS'MPftj -*/0 MfO
CoorJ.'nJes *F p.?«t yx ( 105+ 30 , - Ho ) * (\3S MP*, - 4o wt>< 1
Tk« po.Vt (6*Wj-r^^ ***** SU o*+U jfc*e XfXt
Th^s 15 MP« £ ff* ^ V2>5 MP*

PROBLEM 7.52
7.52 Solve Prob. 7.30, using Mohr's circle.
SOLUTION
7.30 For the state of plane stressshown, determine (a) the value of tv for which the
in-plane shearing stress parallel to the weld is zero, (b) the corresponding principal
stresses.
P©;*+ X of MoUrvs e.vcJe
So fUf <% - A MPa. . TU
Co^t^i «*4 ts OT C. cur*
-Hurou^ li ISO* lor*V^5 JTim-C
CX 4*> CB^ M/te** 0f« <0.
* ^.77 MP*.
6i - 6^+12 * 7 + ff. 77 - 12.77 MP*
Si ' 6U - R " 7 - r.77 - 1.23 MP*.

PROBLEM 7.53
>2MPu
7.S3 Solve Prob. 7.30, using Mohr's circle and assuming that the weld forms an angle
of 60° with the horizontal
7.30 For the state of plane stress shown, determine (a) the value of i^, for which the
In-plane shearing stress parallel to the weld is zero, (b) the corresponding principal
stresses.
SOLUTION
at 6"* = ^1*1 mPa.
Wt'H t= Q.
* ^ ft Pa
R» 5 sec 60*
- ID MP*.
7^ - - £" f «* 6o°
= - 8.66 MP^
6; - 6L + r
= 7-/0 * -3 MP*

PROBLEM 7.54
25 MPa
6: * 25 + /o - ,33" MPft
iy * o + \i.vl - 17.32 h?Po.
7.54 thrmigh 7.57 Determine the principal planes and the principal stresses for the
state of plane stress resulting from the superposition of the two states of stress showa
SOLUTION
^ Y
S"x *.2o + Uo ws 6o*
" 10 MP«.
Txj ~ *o si* 6o*
- \7S2 MP*
■^^hPO
*V 49.11° ©«,- 24.4" 0br ||«r.4ft -
6^= <W + ft " ~T2.1I MP*
6t- 61^ - R ' 27„0<=t MPa. -

PROBLEM 7.55
7.54 through 7.57 Oetennine the principal planes and the principal stresses for the
state of plane stress resulting from the superposition of the two states of stress shown.
SOLUTION
+
Resoi+ a*t stresses
4ksi
6^ - i(s;+<s^ = ££/<*/
2&f= £7.38° 0fcr 33.6 9°
OtaO

PROBLEM 7.56
7.54 tfaroagh 7.57 Determine the principal planes and the principal stresses for the
state of plane stress resulting from the superposition of the two states of stress shown.
+
Re&tM*f«.n"l" s4reuses
SOLUTION
~ A^o --k 6a 00s Z^
- ^6L sm 2&
*XZ*$ &0 Si'*?©
4-fl *
aP =
R-
^" 6-,-q, g;+ $:&«*©
1 + cos 2©
^0
jm&r* x^' =
' V ^ + A6'-0** ^+ (^ ^ s- *e ^
= ^ 6; 7 1 + ^*a " GT01 cos e I
6^- §U + R - 6; + 6L cos 6

PROBLEM 7.57
7.54 through 7.57 Determine the principal planes and the principal stresses for the
state of plane stress resulting from the superposition of the two states of stress shown.
— J
■us* "^ftHiriN
>
!* .49HHI
IB
—
+
6* ' - T0 S.Vl 60° r - 3§ <£
**
r„ si-* 6o* -- ^ 2:
7?+j * ?<> co3 6o'
- J.
KmsoJf'ta^'t presses
% ' ° +
x ■= &t
•&, * r, 4 ^^
1? */(&5f7tJ~
**
• /
- 75 "£
2r„ 2 3
30" <
SOLUTION
+a„ 2©, =
6^-^ " --ft
--V5
2er r - go*
d„ = so-
Molirxs circle. W /?«el s4^+e. «sf s^i^ss
A = Co"

PROBLEM 7.58
738 For the state of stress shown, determine the range of values of gfcr which the
normal stress^- is equal to or less than 100 MPa.
90 MPs
SOLUTION
r 5 &?*!
ry * - Co MpA
6*«. * ite + ^V ^5 MP*.
- V^N 6o4 r 75 MPa
2Gp r -53.13°
Sfc = - 26. 545 *
&Y, < loo MP<*. -fo^ s+«J-e«. of s+^*«
CorreSf)o^e(in<| T^ Afc HBK of*
Mokr's e/rcie. F^t>"* ike ctVcie
R cos Z<p * LOO- V5 * 55 MP*
cos £$ r ^ t 0.73332
2cp r 42.S33' $ - «l.Hl7 •
©h * ec+<P *" - 26.56$° + *I.4I7* * -5. NS*
20K - ?9W 4 36o'- <f<p = -\o.%W> +360*- 85.46^°- «<f.oS7°
9K ' 132.02*
PerM/ss/ic ^aoje of © i* 6m * © - &*
-5. |.S* ^ 9 * l3*.o*° —

PROBLEM 7.59 7.59 For the state of stress shown, determine the range of values of 0 for which the
normal stress a; is equal to or less than 50 MPa.
_ 60 MPa
SOLUTION
& = 90 MPa > 6} - O
r^, = - 6e> MP*
4. oft - 3?*? _ c»x-*»i . ^
6** £ So MPa -Po^ s.-f^+es tff s-f^«s
R t^s ^ = so - nr * 5" MP«
2$ *■ 86.177" <p = 43.0*9 °
Z&u * 2£« + 3*o* - <t<p » 32.52»f0 + 36o* - I7*.3«- a- 2*o.lt<**
Pe^w»'**i* We. r«rt^e or 9 »'& 9i| tT © ^ ©*
l€„5^° S 0 ^ 110. 08S*

PROBLEM 7.60
7.60 For the state of stress shown, determine the range of values of0fcr which the
magnitude of the shearing stress %y is equal to or less than 8 ksi.
SOLUTION
65, * -K ksi'^ <5j - o
T*j r 6k*.'
ftjtfl * * **>' *»^ sKvito of s+r«s
CO*f-e,Z>po^e\ii\*i i<* elites H8K <xfid
0K - 0b + Cp = - IS.43S-+ 24.S6S"* *
AjP
So
9H * O * G>*
- *tS* - * & * S. l£°
e0 ^ e ^ ©v
,35" -■ e ^ IS*. 13°
2W< a * r>%.&°

PROBLEM 7.61
1120 MPa
7.61 For the element shown, determine the range of values of z^ far which the
maximum tensile stress is equal to or less than 60 MPa.
SOLUTION
6; = - 2° hp* <5j - - no hp*.
'20 MPa
R * 6L* - S^t - 130 MPa.
R*«3* of ^ - 12© MP* * r^ * l*o MPa.
PROBLEM 7.62
120 MPa
7.62 For the element shown, detennine the range of values of v^ for which the
maximum in plane shearing stress is equal to or less than 150 MPa.
SOLUTION
20 MPa
6"* ~- ~ 2o Kip*. 6} * - no MPc
^Cft-S^ ) = So MPa.
Se* ?^o~-?t~~^ R y '^° MP*
f? 4n a € oT ^
^
- HI.M MP* « tL, * HI. 4 MP<^

PROBLEM 7.63
7.63 For the state of stress shown it is known that the nonnal and shearing stresses are
directed as shown and that ax - 14 ksi, cj, = 9 ksi, and a^ = 5 ksi. Determine (a) the
orientation of the principal planes, (b) the principal stress o^, (c) the maximum in-
plane shearing stress.
SOLUTION
g; - w !«,■_, 6*^ <? wij $^ = i(<S,+^ V "-S" few
6"„.-rt - 6L* - 1? -- fc = &* - K,;„
But i4 Cs .4it/e* "Hat ^H^ Ti po«/Tii/e tA*s ^*y = "* ^ ^*'
*r.
(a) \^%&ri£^
-- ^r^
2©f = C7.3g°
S^ 33.4?"
- 6.5" k%i

PROBLEM 7.64
7.64 The Mohr circle shown corresponds to the state of stress given in Fig. xxaaadb,
page yyy. Noting that a, = OC + (CX1) cos (20p - 20) and that rxy =» (CX') sin (2ft,
- 2^), derive the expressions for crr. and ^ given in Eqs. (7.S) and (7.G), respectively.
[Hint: Use sin (A + B) - sin ^ cos B + cos -4 sin B
and cos (/4 + B) = cos ^ cos B + sin ^ sin B.]
CX' - CX
SOLUTION
CX' cos H&f - *CX
Cx' si> 20P - Cx stm 2©r =• r^
e;, = oc + ex' cos (Afep»3ei
r OC + CX' ( COS'^CoS^ + Sm ^e/S8t>^©')
= OC + CX' cos 2bp coiZp + CX' s,iVi 2£>f si* X©
- ^S + J§^iS cos ^© + r^ SlVl ^
r*
y -
cx' s.v.Cxap-^e') - £x' (s^ 2e^ cos 2d - cos 2^ si«Ze>}
- CX' s.Vi 2£f cos £€> - CXcos 2&p s,V» £©
=r TM cos Z© -
Si-€
* Si* 2©

PROBLEM 7.65
7.65 (a) Prove that the expression t^ay. - Tx.y>, where o^. , oy , and rx>. are
componenta of stress along the rectangular axes x' and y' , is independent of the
orientation of these axes. Also, show that the given expression represents the square of
the tangent drawn from the origin of the coordinates to Mohr's circle, (b) Using the
invariance property established in part a, express the shearing stress v^ in terms of a^
ay) and the principal stresses o^ and <w
(a) F+vn Mol»r s tiftA
6!,' = <SL« + ^CoS^a,
S/
6"*-e - R COS 26>,
oc2 = ok* + eft*
OK2 - OC* - CK
6"*$".* - 7^

7.66 For the state of plane stress shown, determine the maximum shearing stress
when (a) a, - 20 MPa, (b) a, = 140 MPa. {Hint: Consider both in-plane and out-of-
plane shearing stresses.)
SOLUTION
La) 61 - IHO MPa , 6"v * Zo Mp«
j My
r^y - SO MPa
- 20 MPc,
* -^ £04 + ^o* - loo M?A
6"b= &«~ -«? - So -/oo =■ - 2o MP* (-■*>
(b^ 6i « 1*0 MP* 4 6Jt ,4o MPa
•Cy - So MPa
6*~ - £(6; + 6p " iMo MP*
6L= 6L. + R r 22o MP* 6***.)
6i, - 6U - fc * So MP*.
6^ - O (*.;«")
C^^i ^ i(SL- <^J - So MP«
£_ - i(6L*-C-V no MP* -*
tl LmM
Cmp*1
^ "" ■" V -
s
/ /*
/ /
1 I
0 (6
\ I
\ \
\ \.
\ ^
;
c
<
\\
\
Al 6
ICmpo
SY
"y'
'""

7.67 For the state of plane stress shown, determine the maximum shearing stress
when(a) <$ = 40MPa,(6) <$s120MPa. (Hint: Consider both in-plane and out-of-
plane shearing stresses.)
SOLUTION
(«0 ff„T IHO MP«. 6^ 40 MrV
= 90 MPa
r 94.3^ MP*
6^= 6L. + 1?' 1^.34 MP* (*.**,">
6;= K«-fc« - 4.3M MP*. (^^
6"c = o
r^c^^^o - 405L-O* R « W.s*mP«
ft* -- SO MP*
00 6i r Ho MP* , 6^ - 130 MP*
61* * 4 ($"*+$} r \Z0 MPa
- -J lo^ go1- - go.cx MP*
6*-$«. + 1*ff 2IC62MP* U*^
6"b* 61*-R * HISS MP*
SI** * 61 - 2/0.62MP*
61r» - 61 - o
%3 " *0 tfP«
ri (mp^
-•—-* G«-fi»*
^ = *R - **.« Mft
Crnt^
(MflO
T^ - 4(6"^- 61-. V I0S.3I MP-c

7.68 For the state of plane stress shown, determine the maximum shearing stress
when(a) dt = 6ksi and<7j,= 18ksi, (b)^^ I4ksiand<7j,= 2ksi. (Hint: Consider both
in-plane and out-of-piane shearing stresses.)
SOLUTION
M 6; - 6 ksi <5j = is k*; ?*i = t k*;
R'/C^r + V
=r IO Us.'
6L = 6L« + ft - 13, + lo * ZZ ksc c^^o\
51, * 6U -R = 12- lo « Hr
6"c B ° C**"")
(ksi-)
(b^ Si - 14 /a.' <5J = Z k*i
= -/ 6* + 8* * lo Its.-
6"= o
6U., -- -2 If*;
01, r
8 k&!
(w*.*^
£►*»»* ^
'0 kt>\

6^ - 16 **■' 6.
7.69 For the state of plane stress shown, detennine the maximuin shearing stress
when(a)flt = 0ando5,=»i2ksi)(6)at='21 ksi and a, = 9 ksi. (Hint: Consider both
in-plane and out-ot-pJane shearing stresses.)
SOLUTION
Co.) S* - o j 6J = L? ks.'j T^ -- 8 Ju;
= /c» It*,-
- - 4 &SI
A ff
ChiO
?— = ■JeCk-.-si-..)5, ,0 ksi<
61* r ^ (<*«■
6"..' <S^t + 1? = 25 Its.- (>*^
6"b * 61c - R * 5 Ki>
61 - o 0~>* 1
61**. = ZS k*; J Gl;* - O
tf) Oft.'
c**^

PROBLEM 7.70
7.70 and 7.71 For the state of stress shown, determine the maximum shearing stress
when (a) at = 0, (b) a, = +45 MPa, (c) az = -45 MPa.
SOLUTION
6*„ * \oo MP* . <S"y - 2oH?c . t„ = IS MPa.
r GO MP«
r-^S,
R -V^O^V
= as ^P*
6"-. * S.„ * R - IMS" MP*
fMfW
6^* I4)? MP* , G^v. = -25 MP*j ^ - i (6U,-6U) -- BS MP*,
(fcrt 6", * + 4S" MP* _, 6; * I4S MP* , 5t = - 25" MP*
(C) Gr^-ISMPcc^ 6^ 145 MPa , 6** = -^MP*
6"^ - IHS" MP4j 6L. - - <tfHP« 2L» - jKS-,- CO -- 95- MP*.

7.70 ud 7.71 For the state of stress shown, detenntne the maximum shearing stress
when (a) oK = 0, (b) at - +45 MPa, (c) a, = -45 MPa,
SOLUTION
S* * ISO MPa 4 G, - 70 Mfa^ r^ - 75" W*
- no MP*
<5Tb * 6U - 1? * 25 MPa
/
/ /
I 1
[e
l
\ \
\ \
\
y
•s.
■■* *"—
^T"
-
—
„~
\s/
—!>
\A
ff
Chpo
(<0
(b)
CO
Sz- 0 , 6L* '^ M^s ^ ^M?a
6^« l*SMPa, S^O, -2U-i(G^-6LM^ ^ 97.5" MPa
5,» ^MPftj SL'HSMF^ 6^ 25 MP*
C^- 1^5 MPa, ^;,^5NPS W4(6L*-£^V 35" MPa
6, r - H$ MP*, Cffc- l<tt MP*, 6"t- ^5 MP«
G^i^mp^ C.V, * - «s"P«., r^'i(^-61v> laoMPA.

PROBLEM 7.72
7.72 and 7.73 For the state of stress shown, detennine the maximum shearing stress
when (a) ax = +4 ksi, (b) o, = -4 ksi, (c) az = 0.
SOLUTION
Vsi
Sv~ 7 k*< s Sj r Z ks,^ "^ - -G kar
7 ksi
0c*<'>
(V») 6"2 =-*tlfti'j 6^-11 k»i", (3"br-^^.-
(c) St = 0^ 6L» /I fa." , Gt = -Xfcw

PROBLEM 7.73
7.72 and 7.73 For the state of stress shown, detennine the maximum shearing stress
when (a) az = +4 ksi, (Jb) o7 - -4 ksi, (c) ax = 0.
SOLUTION
s; - s ks.-} 6^=10 ks/ ,, r^ - - e k
3(
6_ - KtfifO * 7.5- Us.
& )l + ?,;
6^ - 6*^ + ft = 14 leaf
(a} Gz " +4 fcs.-, 6^ - 14 ks;
6"b = I Us,-
ic) S, = 0
(iftO
6^* IH k%c t 61M = o,

PROBLEM 7.74
7.74 For the state of stress shown, determine two values of ay for which the maximum
shearing stress is 75 MPa.
SOLUTION
Sx -= -7o mPx
ttj - 4o MP,.
0 ^ gj-CTic
Sj r £u + 6,
6k^ = K^x * 6V} = 6* * °
ii- y u*+ «,*
u- iV^^- r^'
C^e I ^ - R - 75" MP* , o = ±f75r^~H6r - i €3.44 MP*
(lo^ i(J-'+ G_$.*W MP* 6"j * Po 4 6"* * ,56.83 MPa. -■
^o. - 6L. + R = G8.4</ MP* ^ 6"^ s 61* - R * - 81.« MPa
6a = O €*«« " 68.44 MRi j 61.--. - - S/.5*6 MP* tU^ *7S" MPa.
(ft") U * -63.44 M?c &j * 2U + G* - IH.S8 Mfcc 0^4^
S*. = i(S« + §)* -I33.4VMP*. 6^- 6^*+R = -5**44 MP*
Case ^ Assume 6^ * o , ?^ = i (G^- <S^O - 75 Mp*
JQ*+'%
* _
*7
6*+ u - 6*k
2U
u
-3© HRa.
R - VuN ^
■j. —
S^^u+<5; - - tSo MP^
50 MPa
*—e
6^* 6Tfc + 2R - -/5o +/oo - -^p MP.
ox.

PROBLEM 7.75
7.75 For the state of stress shown, determine two values of ay for which the maximum
shearing stress is 7.5 ksi.
SOLUTION
Gwt - i(<Si + S) ) - S"kvl>
Cxse I
tL*. * t? ^7.5" /ft; j u = i- 4.S k*i
(Ik*) u = -4.5 ksi % = 2u + S"* ' I ks», -*
6U*- 13 ksi ^ €„in = -X k*. ^ r^- i6w-<SLn V 7.SW a*.
Cfcse ^ Assume 6w-» " O <5~-* - 2.^*** = IS k&i * 6k
y
^U ^ 6T"6k " (5- IO ■ ****'
u - - I.I ksi S^ - 2o + S* = 7.8 k»i"
6*. = 61»+R ' 15 k*.' ^ St, • Sl« - R r 2.8 *»•'
6k^- is k%.' , 6*^ - o 71,* -.7.5" fc».' ^

PROBLEM 7.76
7.76 For the state of stress shown, determine the value of v^ for which the maximum
shearing stress is 80 MPa.
SOLUTION
£a " 6"^
e;^^ iter+6^ = 95 mpa
Assume <5L* = O 6^ =■ 2r^* =■ l&O MP*
R - G_^ - 61*. - I 6o - IS* = G5 MP«v
61* + *R
^b * 6"*. - 2R * "6° - '^° s 3o MP«. > o O.K.

PROBLEM 7.77
7.77 For the state of stress shown, detennine the value of t_for which the maximum
shearing stress is (a) 9 ksi, {b) 12 ksi.
SOLUTION
6"* = is kfei
Cj - £ ks,-
(w = ^ (6** + 6^ ) - ~\o*s k*
U
6"y- 6",
* - V.s-fcs,"
r)t^o
(a) Foir ?^= 9 ks/
ii'es c*a poiVT C. i-i'^es
•I I VIA i
Molar's ci^/e 6"ft- ^~«*
Corsesr>a» ds v> P^iVf A&.
= IS- ID.S = 7.£ £s(*
Qc*n
R-- u ks."
- - /.5-fe.*

PROBLEM 7.78
7. TV For the state of stress shown, determine two values of ay for which the maximum
shearing stress is 80 MPa.
SOLUTION
S„ = fo mPc 6* -- o ^2=6^ M?a
s;^ - 40X + O • 45-M^
60 MPa
<X*
Assume 6"^* 61 = 120 MP«
(mP^
^«°
AS3U^€ 61;* - St -" - 30 MP«
Sv r 6"».Ht - 61„-* + 2 *K«-k
- -So + («*&) = 13© ^Pc
?::-*0

7.79 For the state of stress shown, determine the range of values of t^ for which the
PROBLEM 7.79 maximum shearing stress is equal to or less than 60 MPa.
SOLUTION
Fof Mold's civ^cie or s+vtsses in Zx-pJum
<W " £(€**+ O - SO MP*
u -
S* - 6i
A^iu^e Si^, - 6y - |t>o MPa
= lOO -WV461 r -J?0 MP*
- So +SO = 80 MPa < <STV
-n* -- ± V** - ut
-Ho. MPa. « tU < 4o ^P«l
= 3o
TT"i tH9*S\

PROBLEM 7.80
*7.80 For the state of stress of Prob. 6.66, determine (a) the value of a, for which the
maximum shearing stress is as small as possible, (b) the corresponding value of the
shearing stress.
SOLUTION
Let u-- 2==ae
(5"j - 6* - #u
2
61. - 6w+R " iS^-u + -/u*" + T»ju
Assume '2tL-»* I* He ihy^mc *JieAWn<i a+rcss
8 = iT^V.r- SO MP*
£"*.- £,* + R =^ w> + Bo r ;uo
6^ <5«~- R ' ^o -So * £0
61,* * £*- u * No HP-,
MP*.
6TU,-,
o
r..^ * i(<S^-6"-.0* 4S,
u
35" " Yi^ + iy
* o
Tmo m/»4itnt'i«\ (
(6; - u )v r 6*** -< 2u6„ +/d*- = ^ + f^w
^ .- 5L^= "^fe^ - **.* MP*
S^= S, - 2c* » |40 - <™.-*> - 45.7 MP*.
l>- 4"UH MfV

PROBLEM 7.81
, 21 ksi
7.81 The state of plane stress shown occurs in a machine component made of a steel
with Of - 45 ksi. Using the niaximum-distortion-energy criterion, determine whether
yield occurs when (a) v^ = 9 ksi, (b) v^ = 18 ksi, (c) v^ = 20 ksi. If yield does not
occur, determine the corresponding factor of safety.
>*¥
36 ksi
SOLUTION
51 - 36 ka.'
61 = ^ / ks i'
^= o
Sir-^y r 7.S" kv
ft - TC^T^V = V(7.5)l + W = H.7/S ks;
6k- 6.^+R " Ho.^/S- Usi ^ <5"b ^ 6k«-ft - 16.875" W;
F.S. "
4s
3t.,<?77
I.2&7
6k - <£- + ft r 48 ksi' 5 6; =■ 61^ - ft = <? ks,-
F.S. -
4S
4?. 113
I.OI8
V ^ + Sb*" ' ^"*6"b r 4G.73* ks.* > fS-lft,- (/rJ'J.'^ occurs ">

PROBLEM 7.82
, 21 ksi
'*y
36 ksi
(*") t^r <? /».'
7.81 The state of plane stress shown occurs in a machine component ™<wk of a steel
with ** = 45 ksL Using the maximum-distortion-energy criterion, determine whether
yield occurs when (a) % = 9 ksi, (ft)t^=I8ksi, (c) ^ =« 20 ksi. If yield does not
occur, determine the corresponding factor of safety.
7.82 SoiveProb. 7.81, usmg themaxununi-shearing-^tresscriterion.
SOLUTION
S, - 3G k&i SH r j?/ /ft,"
6; - o
R'-ZC^jS^r^- - 11.7/5" Us/
61 - 6*~ + "R r uo.air u±i j 6; ^ &~ - i? - 14.S75 uv
61** * 3*977 ^s; , 61.-* ^ o
?t^ - 61^- 61-* = no.Zts !<•>; < ^ ^, (tioyfeJJ;^ )
FS. -
**s
4o.2(s
- I. 119
<fc>)
(Cfl
^ =■ I? ks; J? = ilr^Y + V" * l^-S **/
& - 61* + R r 42 **• , 61 - 6".* - R - 7 *».-
61^ * *2 ks; 5-.-n = o
221-*. T §!•*- 61.-*, * 4S k^i > ^5" /e-s; ^Y-'oU'*.. wc^s ")
% * 5o k%; R =■ VC^S^T* V" ^ ^'-^ *s;
61 * 6*„„ + 1? - ^. 86 k%; <5t - Slt - f? ^ 7. w fesr
61*. •= 4^.86 ks; 6*~* - O
^tU - 61h. - SI* * 4i.s* K* > ■*£" ift,- Cy.viUi'rtj «c^s)

PROBLEM 7.83
100 MPa
7.83 The state of plane stress shown occurs in a machine component made of a steel
with a, = 325 MPa. Using the maxiinum-shearing-stress criterion, determine whether
yield occurs when (a) 3, = 200 MPa, (b) a0 = 240 MPa, (c) a0 * 280 MPa. If yield
does not occur, determine the corresponding factor of safety.
SOLUTION
oq
» - <51
R - -/(£*~£t )* * -Hj*- - IOO MPq.
(GO 60- 200 MPa, 6^ * - ZOO MPcl
6"«. - 6**~ + R - - IOO MPa 61 - G^e - R * - 3oo MP*
<>"-**. - O , <S"w* * - Sc?o MP*.
0°) <50 - ;?Vo MPa. . 61- - - 2*Jo MPa
61- Sw. + R -- - Wo mP«
GTt * <SU -R = -3*o Mpc
6^ - O j 6*-.^ - -3Ho MP«-
^rL* * G^-Sw* * 3Vo MP*. > 2>ZS MP*. (Y.-tiJ^j ©eewO
CO <S, - *S0 MP*-, 6"** ~ -Z&o MPa,
G^ S^ + R = - *«o MP*. , Sb » 6L~ - R - - 3SO MP^.
6^ * o , SL-* = - 3So MPa,
2f« -" S— - 5^ - 3*o MP*. > 3X5 MP*- (Y.'*M.«3 *ct^ )

PROBLEM 7.84
O"o
100 MPa
7.83 The state of plane stress shown occurs in a machine component made of a steel
with Or=325MPa. Using the maximum-shearing-stress criterion, determine whether
yield occurs when (a) c& = 200 MPa, (b) a0 = 240 MPa, (c) a0 = 280 MPa. If yield
does not occur, determine the corresponding factor of safety.
7.84 Solve Prob. 7.83, using the maximum-distortion-energy criterion.
0q
SOLUTION
61** - - <5fl
R
=A^r+^; = ioo mp,
Co} Go = 200 MP*. 6^ = - 200 MPa.
6L - S« + 'R - - \oo MRx J 6"t * &~ - ft * -3oo MP*.
V SJ" ■* §fc* ~6"*6i - 2CV.5C MP*, -c 325 MP* (Wo j iWJm3 ">
00 S*0 -" 240 MPa. GU - -2Yo MP*
G.* Ce+'R = -Wo MPa. , <5i* 6L.-ft * -3tOMP,
Y^ -r €^~ <5^<S"b * 2V. <?7 MPs. < 3^ MP^ (No y i^-f.'nj )
(c^ S0' 2%o MP*. G^-~-Z2oMp4
6"^ * 6M + R ■« - (80 MRa. , 6*b -- 61^ - 1? * - 38o MPa
V s'*.*+ Sb* - 6"*$"b * 3^.24 MP* > 3*5" MP^ (tfeAUj occurs)

PROBLEM 7.85
7.85 The 38-mm-diameter shaft AB is made of a grade of steel for which the yield
strength is a,=250 MPa. Usingtheinaximum-shearing-sti^ criterion^
magnitude of the torque T for which yield occurs when P = 240 kN.
SOLUTION
-5 2.
6* - A
^ 11. G. * to* ?«. r ZW.C HP<=
% = °
I. lS4lx/o-3
= ££.£68 MP*. = &5:.££?»/0*fr
f- I^O >V, -fl
to/,5/£»i^
T =
J£
22.
J' |cN ^ i(¥Y" 204.7/xio1 *„" - 3_oH.7\*id* ^
c = -i ol - n ^io-5 ^
I =• a.—„—_ „a „ -7/7 ^/.^

PROBLEM 7.86
7-85 The 38-mm-diameter shaft AB is made of a grade of steel for which the yield
strength is o,=250 MPa. Using the maximum-shearing-stress criterion, detennine the
magnitude of the torque T for which yield occurs when F = 240 kN.
7.86 Solve Prob. 7.85, using the lnaximum-distortion-energy criterion.
SOLUTION
6*^ f.
-- -2M.&*fOCp«_ ■= ^i/.fi HPa-
<£, = £5
6k,e - itS.+$) - *««
R -/(^Sr)%^ , -fiSS+T*,*
e* + 6"b" - (Sis* - Is/ + ^-/ier/-*^ + AC * -r.
7
6;1 * 3^1 - ery*
r*r i(5/-er/)
2ij =■ ^jV »£)"■- an.6*- * 76.-8G7MP^ = 7£.8S7*/DgP«l
VOW
7-o^Si
01*
r.
_ Tc
T~ - ''•J *->y
?,
^ol - l9*/o" »m
T r
19 */<2
rr
8£8 l^*n

1.5 in.
7.87 The 1.5-in-diameter shaft ^B is made ofa grade of steel for which the yield
strength is a, = 42 ksi.. Using the maximum-shearing-stress criterion, determine the
magnitude of the torque T for which yield occurs when P - 60 kips.
SOLUTION
P - GO Id'ps
A* ?^* - ZO.SY* i.7ci !,■„•■
* A = 1.7*71
6TjS o
^ ■ 1
- - 33.953 k*;
- /2. 361 k=>;
Fro*** -J-^^sm
r«
- Is
C= id - 0.75";>
r-
_ n^v _
rr,
J- £c
0.497OI ;„*
P

1.5 in.
7.87 The 1,5-in-diameter shaft AB is made of a grade of steel for which the yield
strength is a, = 42 ksi.. Using the maximum-shearing-stress criterion, determine the
magnitude of the torque T for which yield occurs when P = 6Q kips.
7.88 Solve Prob. 7.87, using the maximum-distortion-energy criterion.
SOLUTION
P r <oD kips
A - id2- -- -ids)* * i.767; ;/■
Co
^ e " A * "i.767i
6> -" O
r - 33. <?5".S fc*.'
61 r &- + ft
sb = s;^ - r
% * # ^ ~ ^ T ^y«l- 33.15-3 a
= |V. 273 k%;
31i! - 6"/- Si
rro^ "hn^ii
0\A
v - Is r- ^-r-
S j ' c
2L
C * iJ r 0.75-.V J"? f C* * Jtf».7s)' - O.H<not .„"
T r
o.-?c
^..46 k/p /*.

PROBLEM 7.89
150 MPa
100 MPa
7.S9 and 7.90 The state of plane stress shown is expected to occur in a cast-iron
machine base. Knowing that for the grade of cast iron used o^= 160 MPa and obca
320 MPa and using Mohr's criterion, determine whether rupture of the component will
occur.
SOLUTION
St » O
Sy = - ISO MR*.
-ty - too MPs.
6U = i(ff-+0 ^-75-MP^
ri
^t] 7£z + \ooi- r 125" MP<
61' 6U -P * -^oo MP*
Gb (hW
I > r-
-2«> -loo
-|(W ■ ■
-Joo •
(mPO
,A S
CmPa^
^ = 1
I to 3^o
Wo ^o/i-f Jtre.

PROBLEM 7.90
50 MPa
7.S9 and 7.90 The state of plane stress shown is expected to occur in a cast-iron
machine base. Knowing that for the grade of cast iron used o^= 160 MPa and 0^ =
320 MPa and using Mohr's criterion, determine whether rupture of the component will
occur.
90 MPa
SOLUTION
61 = <?o MPc
&* - - SO MP*. , T„ * GS MP«
S^ - iCG**^ - 20 MP«.
R*-/t
gx-gy
)L+1V
€u_ = 6L
ft
'/S.S MP*.
$*b r 5L.- P '-7£S MP*
Ifeo 33»

PROBLEM 7.91
7.91 and 7.92 The state of plane stress shown is expected to occur in an aluminum
casting. Knowingthat for the aluminum alloy used om= 10 ksi and 0^=30 ksiand
using Mohr's criterion, determine whether rupture of die component will occur.
7ksi
Sksi
SOLUTION
6y = °,
Z.J = 7 k%,
Cut <5[»c
|0 "SO '

PROBLEM 7.92
2ksi
7.91 and 7.92 The state of plane stress shown is expected to occur in an aluminum
casting. Knowing that for the aluminum alloy used o^r =-= 10 ksi and o^c = 30 ksi and
using Mohr's criterion, determine whether rupture of the component will occur.
SOLUTION
6U = 7(6^+6^^- -6.S"J«.-
sj&-_ C-/g.»w) _ ,.l7>.
fiLCh,")
1?«p-»
>ptt»r*
.."J*
OCCO'T.
s.%t°i A *■

PROBLEM 7.93
7.93 The state ofplane stress shown will occur at a critical point in a cast pipe nude
of an aluminum alloy for which 0,^ = 15 MPa and ofo = 150 MPa. Using Mohr's
criterion, determine the shearing stress % for which railure should he expected.
80 MPa
TO
SOLUTION
r^= O
£. - - ?-
R ^ -/{SS&Y + 't? = f 4o* + r.c
6^ 6U + K
MP*
r0 = ±7^*" - *
S i^ce
-Pie^ !m
of M*k
&. .
16"*^
4H
cjvaA
. §k
6jt §<*.
-40 + R _ -HO- 9 . ,
7S /CO ' 7-T ISO
sb
/t(
/
/
75"
\£o
1
*-
<Sk
2t.cn
H0 ? i VG3f 331-*©* - ± M. I MPs

PROBLEM 7.94
7.94 The state of plane stress shown will occur in an aluminum casting that is made
of an alloy for which am= 10 ksi and a^ = 25 ksi. UsingMohr's criterion, determine
the shearing stress % for which failure should ba expected.
8 ksi
T0
SOLUTION
Sx - t Hi
Gj*o ,
f.
-J
^ -- $,t + R * (4+R) ^."
Since I SM* 1 ** r?^ stress po/^T
^*A nf IdoO <f\ df ay\l f ^
6L 6*4 - i
£ = ±T/R*-"f
61* 6L.-R-(f-R) *«;
't»T
^L»C
4+R _ 4-E -
|e> 2S"
/J. + J. ^ p r I . i + Jt
OV. "i
I? • S".4 2<? U*i
i^/Zif^f - * 3.C7 k*,;

PROBLEM 7.95
125m
6 kips
7.95 The cast-aluminum rod shown is made of an alloy for which o^.=8 ksi andobc
= 16 ksi. Using Mohr's criterion, determine the magnitude of the torque T for which
rupture should be expected.
SOLUTION
f +- z^ - 7*^^76 + Tp,*' i&.' 3 -£, •±-ltz*-sm<r?& k*;
Since 161* I * R , S+iress p*t*T
a. ies in 4 "fn <^u^ci ita^T- fc^u«T »01*
of 4H, »u«uj
5k. ^ ^L - i
6"l>T Cc#c
<Sfb r ^.VW4- .1? W
s.
s
* A y«f.s"ias"*- ^^74 - s.8o' fc*;
F^ f
Ct^S IOM
C = id - 0.62S* jVi
T =
j
(o.62r) - 0.23^* ;«

PROBLEM 7.96
7.96 The cast-aluminum rod shown is made of an alloy for which o^r = 70 MPa and
aw = 175 MPa. Knowing that the magnitude T of the applied torques is slowly
increased and using Mohr's criterion, determine the shearing stress ta which should be
expected at rupture.
%--
O
61 * Sl^-f -R
SOLUTION
i*^ t.o
*^c
&
&T
1?
70
R
gST
175
=• so MP*.
IZ,j I - So hPa.
«J

PROBLEM 7.97
1#! A inachuie component is made of a grade of cast iron for which a™- = 8 ksi and
obc-20ksi. For each ofthe states of plane stress shown, ai^
determine the normal stress a0 at which rupture of the component should be expected.
^o
(«>
<h)
(c)
SOLUTION
ft) 6^ 6-., g, » -i-s:
6L
s. _ a. r ,
8
_x
§L = i
2o
6a- G.c7 ks:
s
70
6"ft - S.8^7 k9l'
(w^
PROBLEM 7.98
SOLUTION
7.98 Determine the normal stress in a basketball of 9.5-in. diameter and 0.125-in.
wall thickness that is inflated to a gage pressure of 9 psi.

PROBLEM 7.99
SOLUTION
7.99 A spherical gas container made of steel has an 18-ft diameter and a wall
thickness of g" in. Knowing that the internal pressure is 60 psi, determine the
maximum normal stress and the maximum shearing stress in the container.
<J - i-a -Ff * z\c ;•
Y" * iol - t - 1C>7. 675" in.
fir fr E£ (&)t\Q7.&*S)
°' = Dl " 2-t " (zXo.375 )
SGIO ^%t' = g".C| k*."
PROBLEM 7.100
SOLUTION
7.180 The maximum gage pressure is known to be 8 MPa in a spherical steel
pressure vessel having a 250-mm diameter and a 6-mm wall diickness. Knowing that'
the ultimate stress in the steel used is <fy =■ 400 MPa, determine the factor of safety
with respect to tensile failure.
p = 2 MP* t 8 x/o* pa.
r = i <i - t * i<2*o - &
11^ ^ - O.llt w>
' z 2f 00^6 >icr* }
2t
6", iui
F.S. -- & - "*°
3-. 0 4
S3 MP*
PROBLEM 7.101
SOLUTION
7.101 A spherical pressure vessel of 900-mm outside diameter is to be fabricated
from a steel having an ultimate stress <^ = 400 MPa. Knowing mat a factor of safety
of 4 is desired and that the gage pressure can reach 3.5 MPa, determine the smallest
wall thickness that should be used.
p " Z.S MP<x,
Ca -
*.%
**■ at
I. 0I7S t - 7.8?S"
- 7.875" - 0.0 175" t
7.1H *^

PROBLEM 7.102
SOLUTION
7.102 A spherical gas container having a diameter of 5 m and a wall thickness of 24
mm is made of a steel for which £ = 200 GPa and v =0.29. Knowing that the gage
pressure in the container is increased from zero to 1.8 MPa, determine (a) the
maximum normal stress in the container, (b) the increase in the diameter of the
container.
Ad
^ £, - G5"K32«r"-£ */0~&) - I.CISi'lO w = \~6MZ m~\
PROBLEM 7.103
7.103 A spherical pressure vessel is 3 m in diameter and has a wall thickness of 12
mm. Knowing that for the steel used o# = 80 MPa, E = 200 GPa and v= 0.29,
determine (a) the allowable gage pressure, (b) the corresponding increase in the
diameter of the vessel.
SOLUTION
T - iJ -i;-- •£ (SoooV IX ' IH8S *»n
6", - 6"z = Sue - ?MP«v
P
8
(*«.*«;"!? p * ^ - «x«x«°l 3 ,_^0 Mftc
I48S
^(s*. -^O * -^ s; * ?-°-a? (s*idO - **_4 x*.
PROBLEM 7.104
U— 8m —*]
7.104 Whenfilledtocapacity,theunpressiuizcdstoragetajak*>)owflcontamswater
to a height of I S.SV*above its base. Knowing that the lower portion of the tank has
a wall thickness of 16 mm, determine the maximum normal stress and the maximum
shearing stress in the tank. (Density of water = 1000 kg/m')
16 m
SOLUTION
?= PS** = U°0O}{1.2l)(lS.S) - 151.OG x/o* Pa.
= 37.*? MP**
?-*<«* J,l~) -" i6r,' 18. TO MP*.

PROBLEM 7.105
SOLUTION
^ ' It '
^ £
65
51o
7.705 Determine the largest internal pressure that can be applied to a cylindrical
tank of 5.5-ft diameter and g" - in. wail thickness if the ultimate normal stress of the
steel used is 65 ksi and a factor of safety of 5.0 is desired
^ 13 k&.
Y* ' 32.37S*
* 0.25/ Jcsi' = 257
PROBLEM 7.106
7.106 The storage tank shown contains liquified propane under a pressure of 2
at a temperature of 100° F. Knowuigtriatu^tankhasadiameterofl2.6in.andawaU
thickness ofO.ll in, determine the maximum normal stress and the maximum shearing
stress in the tank.
SOLUTION
r - i J- £ s iOs.6)- o. 11 r g. 11 .w
p = 210 pt.' 3
°« " t " 0.11
z
PROBLEM 7.107
SOLUTION
7.107 The bulk storage tank shown in Fig. 1.H1 has an outer diameter of 3.3 m and
a wall thickness of 18 mm. At a time when the internal pressure ofthe tank is 1.5 MPa,
determine the maximum normal stress and the maximum shearing stress in the tank.
= 3.3 *» , t - l8*/o"s >*. V ■ £d - t - \.£%X *n
p = \.S MP*. v,

PROBLEM 7.108
A
7.108 A 36-in.-diameter penstock has a 0.5-in wall thickness and connects a reservoir
at A with a generating station at 3. Knowing that the specific weight of water is 62.4
lb/ft3, detennine the maximum normal stress and the maximum shearing stress in the
penstock under static conditions.
SOLUTION
- 2.16.67 pa.'
r,
M***
c^-of-*/*-*')
t 0.5
^6", - 3.7? fe*/
r 7.5S kst*
PROBLEM 7.109
A
7.109 A 36-in.-diameter steel penstock connects a reservoir at A with a generating
station at B. Knowing that the specific weight of water is 62.4 lb/ft3 and that the
allowable normal stress in the steel is I2.S ksi, detennine the smallest wall thickness
that can be used for the penstock.
SOLUTION
r £16.67 f\i
6\ = 12.5" if*.' - 1?.S ^lo^s,1
t »
t ' ?'
■& ' SS. 672.
t XlC-61

PROBLEM 7.UO
,600 mm.
7.110 The cylindrical portion of the compressed air tank shown is fabricated of 6-mm-
thick plate welded alongarwlix forming an anglep = 30° with the horizontal. Knowing
that the allowable stress normal to the weld is 75 MPa, determine the largest gage
pressure that can be used in the tank.
SOLUTION
1^ ">**
61^ i(6><r,y-*f
P'J-^-
PROBLEM 7.111
500 mm
7.1 U The cylindrical portion of the compressed air tank shown is fabricated of 6-mnv
thick plate welded along a helix forming an angle p = 30° with the horizontal.
Determine the gage pressuretWwill cause a shearing stress parallel to the weld of 30
MPa.
SOLUTION
if * £ J - fc r ± (Soo) - & = ZW
££
^
rr - -1- £^
3
f£
R =
s:-6i
- J- £JC
a " 1 t
^ V3 P
P-£^=3.<<IMP,
f3 «M*

PROBLEM 7.112
SOLUTION
7.112 The pressure tank shown has a g -in. wall thickness and butt-welded seams
forming an angle p = 20° with a transverse plane. For a gage pressure of 85 psi,
determine (a) the nonnal stress perpendicular to the weld, (b) the shearing stress parallel
to the weld.
^i - 2 6"| r 3357. 5" p^'
6".** i (6;+ O = 5036.2fp*/
R = ^"^ -" /672.7S p/
C^ S'w * 61,* - Pcos *rOu =r 375°>*? —
PROBLEM 7,113
SOLUTION
o( * 5 f+ = to i.
7.113 The pressure tank shown has a "g -in. wall thickness and butt-welded seams
forming an angle p with a transverse plane. Determine the range of values of p that can
be used if the shearing stress parallel to the weld is not to exceed 1350 psi when the
gage pressure is 85 psi.
- ij-t = So- # - Z9.CZS i.
&* = $ 6"i r 3357. 5 par
^/S^ = -53.53*
2/8b * + 53.53* 184 * 2C. 8
?y3c - -53.53° + Wo" - 126.47* |St « CS. *'
2&A » .53.53" * l?o°= 233.5S° jgj ff ||&.g'
R = S^k - |67g.75
rw = r sin 3/3 = -&*
C3.2°^/3* /'«.**>

PROBLEM 7,114
7.114 The pressure tank shown has a "g" -in. wall thickness and butt-welded seams
forming an angle p= 25° with a transverse plane. Determine the largest allowable
gage pressure, knowing that the allowable normal stress perpendicular to the weld
is 18 ksi and the allowable shearing stress parallel to the weld is 10 ksi.
SOLUTION
r-- -kJ-t - 3o-| = z'f.czs i
Sit t
7v
- 0. l?/fl -^
* 0.387 fe»." * 38 7 jp*;
* " (9. IKS'/ r
p = 387 jpsi*

PROBLEM 7.115
SOLUTION
7.115 The pipe shown was fabricated by welding strips of plate along a helix
forming an angle p with a transverse plane. Determine the largest value of p that can
be used if the normal stress perpendicular to the weld is not to be larger than 85
percent of the maximum stress in the pipe.
*** ft
K ^ " ft
S*w - 6"^ - "R cos 2/3
cos ^6 - - H (°.&$- #V - 0.4
^ - //3. 6* /3 =■ 5Q.%"
PROBLEM 7.II6
SOLUTION
7.116 ITiepirJC shown has adiameter of 600 mm and was fabricated by weldingstrips
of 10-mm-thick plate along a helix forming an angle p = 25° with a transverse plane.
Knowing that the ultimate normal stress perpendicular to the weld is 450 MPa and that
a factor of safety of 6.0 is desired, determine the largest allowable gage pressure that can
be used.
6"^ 61* + R cos 5o" - 0.^107 &~
S*.
GtL _ ^A"Q _
-*" " F.S 6
0.°l\Ol *jP • 7£
= 75" HP«.
P r
. (75Xl<0
(O.^lolUWi?)
" 2-SV MPc_

PROBLEM 7.117
20 ft
7.117 Square plates, each of 0.5-in. thickness, can be bent and welded together in
either of the two ways shown to form the cylindrical portion of a compressed-air
tank. Knowing that the allowable normal stress perpendicular to the weld is 12 ksi,
determine the largest allowable gage pressure in each case.
SOLUTION
d ' u f+ = \*4H ,v r- ±d - fc -- 7i.s ;*.
*.-¥
^
(tO 6T, - \2 kit
H t

PROBLEM 7.118
7.118 A torque of magnitude T = 12 kN • m is applied to ihe end of a tank
containuigcompressed air under a pressure of 8 MPa. Knowing that the tank has
a 180-mm inside diameter and a 12-mm wall thicknesa, determine the maximum
normal stress and the maximum shearing stress in the tank.
SOLUTION
d - ISO ***o T ~ i d =■ ^O YYiVn £s I^MM
-2T =
l£ - (ia»|Q>Vl02»lo"1'') _
Ve.ssor*
S.277 MP*
tj^ -|Tr &X2£} , 6oHP, ^.-If" 3oMP,
o"f sfjesses
A. 3o MPa
\S.277MP«
F
^. * 6^+R * C8.^MPa.
6**60 Mr\j 6y- 3o MPa, ^ =19.27? MP*
6k
)»*.»1t
si O
^ M.6t MP*. -*
C> * O
'•M»»
'•***.
- i (S"_- 61,^ = 3^.33. MP*

PROBLEM 7.119
7.119 The tank shown has a 180-mm inside diameter and a 12-mm wall
thickness. Knowing that the tank contains compressed air under a pressure of 8
MPa, determine the magnitude T of the applied torque for which the maximum
normal stress in the tank is 75 MPa.
SOLUTION
i, ~ \*X Ki*n
S - ^? r ^^ = CO MPc 6, - J£ = So MP*
'm**t
= n.06 k^-**
r - ^

PROBLEM 7.120
7.120 A pressure vessel of 250-mm inside diameter and 6-mm wall thickness is
febricated from a I.2-m section of spirally welded pipe AB and is with two rigid
end plates. The gage pressure inside the vessel is 2 MPa and 45-kN centric axial
forces P and P' are applied to the end plates.. Detcnnine (a) the normal stress
perpendicular to the weld, (b) the shearing stress parallel to the weld.
SOLUTION
r^ £ J - \2S wm
To- r+ t = /as+£ - /s
er- & * 12^2, „LC7 MP*
7VU>* s4res5tfi .: U^r+jJ.W1 5V T 2<?.83- 7.3Z£ r //.s** MP«
e -^ - - t';;;;lo^ - - *«*>**• p* - -?.** mp*
Otu^-Te^c^W & s *//.67 MPa.
fc^ - ifc*6j)* 26.585- Mp«
=• ^6.£85- - IS". OS I cos7o*
- 2\.H MPa
-- 14. 17 MPa.

PROBLEM 7.121
si! A pressure vessel of 250*nm inside diameter and 6-mrn wall thickness is
febncated from a 1.2-m section of spirally welded pipe AB and is with two rigid
end plates. The gage pressure inside the vessel is 2 MPa and 45-kN centric axial
forces P_ and P« are applied to the end plates.. Determine (a) the normal stress
perpendicular to the weld, (/>) the shearing stress parallel to the weld.
^cuJSmw' ?' l2°* aS&umin^ ** ^ maSnitude P of the two forces is
SOLUTION
St* ££ . fi^«il« m.67 mp-
To+oi s4ressci" Ioh^.'^-IiV^ 6; = 3^.833 - ;?4.S7o =■ - 4.037 MP*
C>t*~fere^-J % " HI, 6 7 MW
R - |^~*-| - 2Z.&5Z MP*
to 6/ -- 6"M - # cos 7o*
- U.oo MP*
(b) r,y - r? s.'n7o- -- ^.g^ *.■**>•
- XI. S MPa
4.ol7

7.122 The cylindrical tank AB has an 8-in. inside diameter and a 0.32-in. wall
thickness. Knowing that the pressure inside the tank is 600 psi, determine the
maximum normal stress and the maximum shearing stress at point K.
SOLUTION
r- - 4* = 4 ;„
77
ot^i o^ -
£ " 0.32. " 7^^° ps( = 7..SP k*t
^t = 2^ r .3.75" Jrt/
^'W/Z/A
A- ^
J 3 77
(3)(l|.0«0
C^.32*- -Vs) = ||.
OS
77. 48i ;■
6 * It
Z. \5 U*'t
* 7.50 !»!
K
T
3.7£ k%;
liUii}
S In e * •"
6T„ ' G; - 3,75" ks.'
(^ = 6^ r .7. 56 U%i
Gbr G,*-R =■ 2.77 U*!
6"^ - 0

PROBLEM 7.123
7.122 The cyluidrical tanked has an 8-in. inside diameter and a 0.32-in. wall
thickness. Knowing that the pressure inside the tank is 600 psi, determine the
maximum normal stress and the maximum shearing stress at point K.
7.123 Solve Prob. 7.122, assuming that the 9-kip force applied at point D is
directed vertically downward.
SOLUTION
1
To*1**
C*^0- ^37. ;~
^i* £6T, , 3.75 ks;
Sentltnj : I - i J" - 72. 4S i** C = ro » V. tZ m
A+ poi-it k, va/Jt = o
Su^^ct^y of .stresses "
*~ u.?ok%.'
Uoh^.'fi^i'n*^ 6^^ 6", - 3.7S+ 8.OS* - II.SO Us,'
SUd/ "Zjy - 2.68 l<rv
&„.= i( I/. 8o + 7..So^ ?.<$£ ks:

PROBLEM 7.124
30 mm
STEEL
tt = 4 mm
E, - 210 GPa
at - 12 X lOVC
BRASS
ti ■= 4 m m
^ - 105 GPa
ab - 20 X lOVC
7.124 A brass ring of 160-mm outside diameter fita exactly inside a steel ring of
160-mm inside diameter when the temperature of both rings is 5° C Knowing that
the temperature of the rings is then raised to 55° C determine (a) the tensile stress
in the steel ring, (b) the corresponding pressure exerted by the brass ring on ihe
steel ring.
SOLUTION
Le+ p be He cemfAcf pressure bc-ru/ee^
SVeeiW**,". I«UmrJ pressure p^ % = ^ (^
Ct?rrfiS pt?(^(J I H4 ST^AIM CSjf g - £ £t
St^Ai^ due +© +eiM^«^«-ttfir* cUw^e £sr " °U AT
To+Ai s+^.'ii ** = It +■ a* Ar
3 Al.- aire. - 2nr(|i*^A^
E.i>fch'w«j AL5 +o AU
r - "jf d - #0 him
28S.TIx|611 p- 40o*io'* p= l.^o' Pet
&<*-*-& + *"
(V>
p - /. 4oo MP«c

PROBLEM 7.125
30 mm
STEEL
tt ™ 4 mm
E, = 210 GPa
a, = 12 X 10-VC
BRASS
t (, •■ 4 m m
Ef, - 105 GPa
a6 - 20 x i<rVc
7.124 A brass ring of 160-mm outside diameter fita exactly inside a steel ring of
160-mm inside diameter when the temperature of both rings is 5" C. Knowing that
the temperature of the rings is then raised to 55 " C determine (a) the tensile stress
in the steel ring, {b) the corresponding pressure exerted by the brass ring on the
steel ring.
7.125 Solve Prob. 7.124, assuming that the thickness of the brass ring is (b = 6
mm.
SOLUTION
L&T p be fke confftcr pressure be+weevi
T"ke ^Ws. Suksen'pf s re-Pers h> "Hie sfeeJf
Whfl . Sjtsc^i|3'f b refers fo He br^ss rtVic,
SVe€Jl ^'"«*|*. Jnjerr.fii pressure p. 65 = *r-
r i ■ i ■ c- s* B£
Lorres poy\(Aii^a st^aim c^p- -p* - 2r£
TotaJ sWn £s = f^r ■*- o^ Ar
fc»E$
Cltc**»«|e\ frt JfttAgTU of C\^o u^T«^fiMC«
CotM^e*p©i*tfli «^ sTv^t'ws f^p ~ — jpj. £^7- - w^, AT
(0
E^ixt-TivtA ALS -fo Ai.^
fu^Ar--
-r 0L A/
(fe + £0 P s<**-°^ *T 00
it." C*«»*i - 6 > /£>
-a
Kl
Fro*n ef ft) \
(80*10*)
4 x\v'-
6s = 3£.0 MP*,
p - /. goo MPa,
t.

PDftnI FM _ „, 7-126 Anwth 7.129 For the given state ofplane strain, use the methods of Sec. 7.10
rKUBLtra /.izo to detennine the state of strain assodated with axes x' and y' rotated through the
given angle 8. ^^
SOLUTION
g*^gy = -360 >u g'-gj = -36^
£„. = g»* CJ + ^^ cos 26^ am 7&
^{-36o - 3&o coS(-fio°) + ^- sm(-to')} A - - 67o/t -^
£ , :r .& + &. _ g«-gJ co& 2© - ^* Sm 26
r {-36o- C-36&)cos(-6oa) - ^f2s.**iC-Cer)jA» * - 5o A -*
Yiy = -(£« - &,) sm 2© + r^ c«s ^e
-[-(-720-o) s.V, (-6o") + 300 t^(-^og)^ = -WHja ~*
7.126 throvgh 7.129 For the givensiate ofplane strain, use the methods of Sec. 7.10
PROBLEM 7.127 to determine the state of strain assodated with axes x' and y' rotated through the
given angle &
SOLUTION
£„=o fe^ " +32o>u ~JCyr-l°0/4 6- 30'
r , = & + gj + £x~ £y coS 2$ + lit s.*« ^
= { I to - 16© Cc* 6o° - i^? si„ 6o°}>4/ - 4 36..7/<
- f f.6o 4 |60 cos to* + ^ sinCo'l^jU - + 283
Yyy - - (e„ - ^) *;« z$ 4 Y!y tot ^e
A

„ M 7.126 thnmgh 7.129 For the given state of plane strain, use the methods of Sec. 7.10
rKUBULM 7.128 to detenniiw the state of strain assodated with axes jc and y' rotated through the
given angle &
£« - - ZOO M-, £j = + 4££> /Mj Vi, * 4200^ ( B r-2s*
SOLUTION
r I - 175 - G25 cui (-5©M + -^ s.>.(-SbO}/f = - OSS /# -*
£y - ^-"^ - 1*^1* Co^e _ >k SM 2<9
- \ - MS + 6*S* cos C-Sto-,> - ^ s i« (-50-^ ^ = + 5o3 // -*
"Yiy - - (ew- Cy) s.-- ^s + Xy cos ;26
7.126 tfciMgh 7.129 For the given state of plane strain, use the methods of Sec. 7.10
PROBLEM 7.129 to determine the state of strain associated with axes x' and y' rotated through the
given angle &
txr-SOOjU^ g^+JWO^ y^r O, a r |5
SOLUTION
* {-(Z5 - 37$" cos $o* + o] >« = -¥So/t
=■ £ - IC15* + 375" cos 3o" - o ^ /( - + .*oo //
cos ^6>

PROBLEM 7.130
7.130thnmgh 7.133 Forthegivenstateofplaneatrain, uscMohr'adrcletodetermine
the state of strain associated with axes x' and y' rotated through the angle 6.
£* ^ -720m, €y -- O , y*f = +300^4, 9 a -3o'
SOLUTION
^(.^
360
3£o
£ 6*0
C : (- 360/^ O)
0 r 2©'* <* r 6o'-2a.«* t 37.38°
£*' = Ewe - £ e«s /8 - - 360^ - 5?©/* c«p* 37. S&'
- - £7o /*
■+ ^cos fi - -36o/< ■+ Si^/coa 37.38"
- - So /(
Ik *-R 5.^/3 -- -3?<?/< s.^ 37.38*
PROBLEM 7.131
SOLUTION
\Co
7.130 thro«fh 7.133 For the given state of plane strain, use Mohr's circle to determine
the state of strain associated with axes x' and y' rotated through the angle &
£« - O
}Lo
£j r +2ZOM T^^-lOO/s Q= 30l
PAt+C-l pofnH
X ■ ( O , £to /*)
C : ( H>o/*j o )
£ (*0
R --/(Ito^)1^^)1 r /67.63A
£V r S«* " # CuS/« a '6oA " K7.C3// f* 42-<*'
- - 3C.7 >/
283/<

PROBLEM 7.132
7.130thnmgh7.133 Forthe given state of plane strain, use Mohr's circle todetermine
the state of strain associated with axes x' and y' rotated through the angle 0.
SOLUTION
£< * - SOO/< e, = ISO m VVy r + Zoo M B = ~ Z&>
LIS
ir(M)
-; I7S
PJtfffcei poi^l*
C- (- 175 //j O^
100
y
8(y«>
fi- 20 - oi - So'-%o<r' r <40.°<r
£** - £** - r?c«/S - -175> - £32.15)* c*»s ^?/
- £,5*3 /i —*
£y - £*« + Rc«s/S - -175"
+ 303 // -*
*^C= -R6r„/Sr -«?.?£> sii'to.iI° iCys-aiV
PROBLEM 7.133
SOLUTION
7.130 through 7.133 For the given state of plane strain, use Mohr's circle to determine
the state of strain associated with axes x' and y' rotated through the angle 0.
£y r -SOO/4 } £j ~ 4 ZSO /A, Ty = Z> , © = /6~ °
*r(//i
*(/0
**%5A >
Y: (+ a ro// cp )
C-* (-WSfi'l o )
R- 37£/<
*■ - VSOM
By * £*,* 4- t? coS 2© - - l^^-t 375" cos 3o"
~ 7.00 /A
jY!y = Rs.'« ?£ = 375 5^3o-

PROBLEM 7.134
SOLUTION
*r</0
7.134 through 7.137 The following state of strain has been measured on the surface
of a thin plate. Knowing that the surface of the plate is unstressed, detennine (a) the
direction and magnitude of the principal strains, (b) the maximum in-plane shearing
strain, (c) the maximum shearing strain, (Use v = -j)
£y - + l&o/* £y - - 48o>/ Yy - - GOd/A
C: (- |6o/<, o)
■On
£*-*J
^—-
£(m)
a«J -31.53+ ^o .- £8.*!
9^ = -2I.S8
R - •J($XOM)%*(3oo/4f " 43S.G .^
^ - -r=rrte^O --T^r(e» ♦ O = - £0«>>«-**>/t>
i-*
i- u
1/3
X/3
(GO A
877/y

PROBLEM 7.135
SOLUTION
K(*>
+ft* 2©pr
7.134 throngh 7.137 The following state of strain has been measured on the surface
of a thin plate. Knowing that the surface of the plate is unstressed, detennine (a) the
direction and magnitude of the principal strains, (b) the maximum in-plane shearing
strain, (c) the maximum shearing strain. (Use v = "3")
Far MoVir's oVele oT s"fr.su'*\ pA>T poiVf*
X : (- *6o/< ^ - 240^^
r : ( - c,oM ^ Zi°/<}
*»
480
» -;?.*
20, = - G7.38° 0b - -33.67*
sb - e^ - R » -i6o>i - 26^>A
42c
M
^f,^,^ * 2f? r SfcO A
T^r(«.*%^
2/3
160 M
2/3 v

PROBLEM 7.136
7.134 throngh 7.137 The following state of strain has been measured on the surface
of a thin plate. Knowing that the surface of the plate is unstressed, determine (a) the
direction and magnitude of the principal strains, (b) the maximum in-plane shearing
strain, (c) the maximum shearing strain. (Use v = ^)
£y r - HO M
SOLUTION
Sj = 7&0^
r^ - ^^qom
= -\.z
.y ol
y: (- Hom, - H*>m ^

PROBLEM 7.137
SOLUTION
7.134 through 7.137 The following state of strain has been measured on the surface
of a thin plate. Knowing that the surface of the plate is unstressed, determine (a) the
direction and magnitude of the principal stnuns, (6) the maximum in-plane shearing
strain, (c) the maximum shearing strain. (Use v = -3 )
60
(b>
\f(M^
£ <M\
y; (- ^00^ ^-><&7.Sm )
©b *-3o. 13"
ft'-fao/*)3- + (87.5/0*
=r |00. 8
£„ =
e—+ 1? * -as©/** loo.8/4 - -m.a /<
ff £^ - R - -2s*>/< - |oo.8/, - - 3i"l /*
^ /-. - v V /- - \ _ 1/3
^ * ",-„
z + ZSO /*
(^
£*♦* " Z5o/< £^v. - - 3SI/y
Y-^ - £M - £^ r 25"© /< + 3Sf /*
601 /<

PROBLEM 7.140
SOLUTION
in*)
7.138 thro«f h 7.141 The for given state of plane strain, use Mohr's circle to
determine (a) the orientation and magnitude of the principal strains, {b) the maximum
in-plane strain, (c) the maximum shearing strain.
£* r 4 400/, Ej r + ioo/^ Yy • +375 M
Y-" (4 2co//3 + IS7. S> )
X
ZiM)
-h*» ZBf • ■—**
37£_
*too-Zoo
- I.S7S
2©p r 61. W 6^ 30^*^ eb= \10.16
PROBLEM 7.141
SOLUTION
*T(*0
a.. *
7.138 tfaroagh 7.141 The for given state of plane strain, use Mohr's circle to
determine (a) the orientation and magnitude of the principal strains, (b) the maximum
in-plane strain, (c) the maximum shearing strain.
S* - + GC>J4 £y =■ +-ZHO/4 Ty - -5o/4
-SO
=■ 0.277778
£<»
*©, * I51W V 7.76° ©^=97.7*
fc) f6 = °
Y

PROBLEM 7.138
SOLUTION
PROBLEM 7.139
SOLUTION
7.138 through 7.141 The for given state of plane strain, use Mohr's circle to
detennine (a) the orientation and magnitude of the principal strains, (b) the maximum
in-plane strain, (c) the maximum shearing strain.
£* - - 90/A Sj ~ - ISO ^ /^ r + ISO /4
X: (-io^-75/f > y: (- 130^4 7S><>
C *• (- llo/t^ o 1
«*\
E0*1
4-i 2 V ft"
4o
3.7S
©*.- 37.53*
6b * \27.S3°
fc'V&O/lV^S,,^ - 77.6/t
£* = £*,* * R - -IIO/<+ 77.C/* = - 32.4yt/ -*
£b = £w* - R " - WOM-Tt^M - " »87.6 /< —»
(Y^ T^c^^ -2R* 155"-*/< -*
ccy et= o
£-•** O . ffc.nr -187.CA*
Y»» * €~* - ^ ~- O + )87.<>4 * 1*7-6^
7.138 throvgh 7.141 The for given state of plane strain, use Mohr's circle to
determine (a) the orientation and magnitude of the principal strains, (6) the maximum
in-plane strain, (c) the maximum shearing strain.
e„ =■ 4 37Sjll gy * + 75M Vfl " + MS/4
X: ( 3 75//, - «.S>i }. V: (7^ 6?. S> )
C: ( W/r, o)
£(/0
1*S
f Ev-e, 37<r-7s-
r ;?:?.«
a* ioi.sr
R *Jtiso/tSL+(&s#Y' = »«. S/<
CO £c= O £^ = 3*7. S/, ff^ "- O

The strains determined by use of the rosette shown during the test of a rocker
arm are:
7.142
t-v = -rwu ft ej =* +450 ft €3 = -75 ft
Determine (a) the in-plane principal strains, (b) the in-plane maximum shearing
strain..
6|--f<00/f
me (a)
SOLUTION
Gz -- iSo-
O.lS £„ +■ 0.25 Zj + O.HVioi r^ r £Oo/* 0)
£x tos1 9, + £y sin* Qz + YL 5;«6acasda = €t
0.75" fy + 0.25 £^ - O.4^ol Y^y = 4Sb>< Czl
<?
O
-75> to
So/v/irt^ CO/^^^I (^ S''^u-?T-ait«oUS j!
9
f- * 7*5
0=0
M
£». = 6tt^ 4- R -- 734 >r
£b - £w<- R - - 8<f.3 /t
4 of, 3 ^

PROBLEM 7.143
7.143 Detemiine the strain ^ , knowing that the following strains have been
determined by use of the rosette shown:
e, = +720 * 10* inJ'm. ^ = -180 x 10* wJ'm.
€3=*+120 x 10*inVin.
SOLUTION
0, = -15°
©3 * IS'
£* ou'd, 4 fcTj s.'i ©t + Y^ Sf'o ©t cos ©, = £,
0.9330 £„ + 0.066^ fj - 0.25 YVy - 720*10*'
0.75 £». + O.Z5 £3 + 0.4330^ = -l8o*/o"'
0.OC&99-* ■+ 0.9330 ej + 0.^5 Y^ =■ no^io'i
ex - 38o*/£>"* i'«/-„ , f. = ^fGOv/o"6 ;«/,« . ■/«-" 1339 y'G"' «'*/'*
(H
ce>
fe)
J "J
"J
-4
£rf * SSOx/O t„//n.

PROBLEM 7.144
7.144 The rosette shown has been used to determine the following strains at a
point on the surface of a crane hook:
^,"+420^ ej"-45^ et = +l65fi
(a) What shoald be the reading of gage 3? (b) Determine the principal strains and
the maximum In-plane shearing strain.
SOLUTION
(a) Gage* t o,^ 4 a*« 7op ap^f f*^ - i (£^ Bt\
Ga^ei I a-*vi 3 are aiso lo" tfis-t E^ - ■£ (£, + £»>
= 375>

PROBLEM 7.145
7.143 Detennlne the largest Itv-plane normal strain, knowing that the following
strains have been obtained by use of the rosette shown:
«:, = -50 x 10* inJin. e2 - +360 * 10"* InTla,
^=+3l5x|0*i!L/m
SOLUTION
O, » Hf
Q * - *S* Qs r O
O.-S* £* + 0.5 £j - 0.5 Yij - ^o v )o'c
'J
+ o
o
= 3'S"* /o
-t
(0
F^Otv\ ( ^ )
fw = Zi$*/o' U/im.
Fy.(l) - £?.6^ Txv = -SOvio"- 360x/Cf* r ^/OWo" m/,V,
E, 0 ) + £f <* >
*J
<Fk + £*- = £, + £t
%
2Gf = -52.0*
X%
6r ? - ZQ. O*

PROBLEM 7.146
3
60"
i V? Sh5W tha£lhc sum of A* **« atra«* ineasuremente made -with a 60* rosette
is independent of the orientation of the rosette and equal to
where ^, is the abscissa of the center of the corresponding Mohr'a circle for strain.
SOLUTION
1
60°
e. -
•*<*e
"*Z** cos «© 4- & si* 26
f^ 4 la-rJa COA fe& + ,2o°; + |rs.aC^e*tZo*)
r £
e*t/e
(0
Cz)
C3>

7.147 Using a 45° rosette, the strains e,, %, and Q have been determined at a
given point Using Mohr's clrcla, show that the principal strains are
(Hint: The shaded triangles are congruent)
'-u* « X (*1 + *j)± "!?[(*, " *2)2 + (*j " *3)2r
SOLUTION
Sine? *V*-^e afiireiyKows j o-^a/ 3 ^f-e 7°" a-pa.rT
+ C - *a*, - ft es

PROBLEM 7.148
200 MPa
7.148 The given state of plane stress is known to exiat on the surface of a machine
component Knowing that £ = 200 OPa and G = 77 GPa, determine the direction
and magnitude of the three principal strains (a) by determining the corresponding
state of strain [use Eq. (2.43), page <*« , and Eq. (2.38) page <*| ] and then using
Moor's circle far strain, (b) by using Mohr's circle for stress to determine the
principal planes and principal stresses and than determining the corresponding
strains.
150 MPa
SOLUTION
(*.\ S* * O 3 6^ • - 7oo*Iq<'?a. 9 ?^ = - ISO x(06 Pa
E * ZOO * to* Pa. &^ 17*loi Pcl
r - B -
EJ
v - — - I - O. W*
- t«w?
-I. Wfc
to
(MM
£b r fe-ave - R r - l£Z\ M -
6Lh " i (<Si + C* ) =■ loo MPa
= ISo-28 MPo.
&-- 6L. + R s
8<?.3
6T(MPO ^b « 6U-R • -2S0.3
MP*
= **o>it

PROBLEM 7.149
7.149 The following state of strain has been determined on the surrace of a cast-
iron machine element:
et - -720 x JO"6 inJia. e2 = -400 x lfr* UiTin.
r=+660x lo^rad
Knowing that £= 10* lO'psiandG = 4x I06psi,detenninetheprmcipaj planes
and the principal stresses (a) by detemuning the corresponding state of plane stress
[use Eq. 2.36, page ^ Eq. 2.43; page «* I; and the first two equations of Prob. 2.75,
page ww] and then using Mohr's circle for stress, (b) by using Mohr's circle for
strain to determine the orientation and magnitude of the principal strains and then
determining the corresponding stresses.
SOLUTION
G -
E_ _ )Q*iO<
> JL i -12—
- i - o.zs
l-v-
I - o.sr*
O.GG7 *fOA p*;
CoJ
^ " |5j* ^£> * ve*^ " IO.C47x/o' [-720«/cTfi + (o.m)(-1ooi«io"^
|-u
- - 2ma>.l fit;
6*2 ■ ■■fv- k^vE.) - |0.£67*to4[-*»o*/o"* + (0.25V-720-/O"*")"]
= -C/86^7 j*4.'
rj^o
^-ecpso
(b)
4r (io->
20t* -tff. I* ft,* -3*..|* ft^ 5"7.7°
t? = ^ZhfT^ r s^v/o'*
fOo-)
-c
£b * £* -R = -m.7v*/o-<
6L =
- nJ
- (et+ Ve^: - \oVOi
>s<

PROBLEM 7.150
S-Vyigl; rt-3
7.150 A single strain gage forming an angle fi= 30° with the vertical is used to
determine the gage pressure m the cylindrical steel {ank shown. The cylindrical wall
of the tank is g" in. thick, has a 36-in. inside diameter, and is made of a steel with
£ = 29 * 106 psi and v= 0J0. Determine the pressure in the tank corresponding to
a gage reading of 220 * IO^ul/ul
SOLUTION
♦S+resSes i« +U -tank ^joJJ
°* t
°j at
r„ ^o
= O
S^JviVtA Tt>y"
R - * (ew- 6,")= 0.325-
£j = £*•* - R c«s 60°
r .525"^ - Q.%1S $£ <j* fr?
. 0.3C2S ££
fct
r o.3«s r
36"? p*r

PROBLEM 7.151
ts.
So/i/in^ to/"
7.150 A single strain gage forming an angle (f= 30° with the vertical is used to
determine the gage pressure in the cylindrical steel tank shown. The cylindrical wall
of the tank is "g in. thick, has a 36-in. inside diameter, and is made of a steel with
E = 29* lO'psiand y-OJO. Determine the pressure in the tank corresponding to
a gage reading of 220 x lO^lnJin.
7.151 Solve Prob. 7.150, assuming that the gage forms an angle p= 60° with the
vertical.
SOLUTION
6- - ££
V f-t
% -- o
(±-^g = o-^If
r.
%
£.-. + £ o»s Co"

PROBLEM 7.1S2
7.152 A single strain gage is cemented to a solid 96-mm-dianieter aluminum shaft
at an angle f}= 20° with a line parallel to the axis of the shaft. Knowing that G = 27
GPa, determine the torque T corresponding to a gage reading of 400 A-
SOLUTION
r -
Tc
s; - ^ - o
Sketc-n Motives ov-cle w .slirA.Vt.
£** = ice * fj} - o
R. - -4 Y ""
Tc
2SJ
s;M 7fi =
7fGC%
G
jr. = e, -- o
Sohm* for T T- ?fJ fl =
Si* 4o'
- £.84 IcM-m

PROBLEM 7.153
46 mm
7.152 A single strain gage is cemented to a solid 96-mm-diameter aluminum shaft
at an angle 0- 20° with a line parallel to the axis of the shaft. Knowing that G = 27
GPa, determine the torque T corresponding to a gage reading of 400 M.
7.153 Solve Prob. 7.152 , assuming that the gage forms an angle 0= 60° with a
line parallel to the axis of the shaft.
J
\V
x.
61 = G, r O
T^5c^
r= I
£x = e^ - o
Ske-fcLi Moliir ^ ci>c-fc ~fo^ sixain.
Gct^c Jtrectiovi a i's. fi = Co° clockwitv
r ft"'**

PROBLEM 7.154
7.154 A centric axial force Panda horizontal force <& are both applied at point C
ofthe rectangular bar shown. A 45° strata rosette on the suriacc of the bar at point
A indiestes the following strains:
e,=-75x lO^inVin. 6i = *-300x WwJin. e3=+250* WwJm.
Knowing that £ = 29 * Itfpsi and y=0.30, determine the magnitudes of P and Q^.
SOLUTION
£, -- - IS WO"
£y r £j r ^° */°"
Y« = l£i ' £, - £s - H-Z5 «10
W
** r T^^e- + ^v"TrfVl"7Stf"">(aSto^
- o
r 7. 25" v/Oa p*i
*■«&
"?=• AG; ^ (zK6^7.as-vioO
G =
;?0+-»M
wj "* *t
^ T
m
29*\Oc , , t
-— — = II. l5Y*/0 pa'
3 i -- Z i
A t V
v . -Asi - -»)fr)g-wwgo g 87<t f|0, A
?»
Q, = V = 37. <7 *|03^], - 37.7 kips

PROBLEM 7.155
7.154 A centric axial force P and a horizontal force <& are both applied at point 0
of the rectangular bar shown. A 45° strain rosette on the surface of the bar at point
A indicates the following strains:
€y = -75 x 10"6wJln. ^ =-t-300-x WwJm. ^ = +250* lO^uuin.
Knowing that £ = 29 x lO^psiand v » 0.30, determine the magnitudes of P and Q,.
7.1SS Solve Prob. 7.154, assuming that the rosette at point A indicates the
following strains:
e, = -60x lO-'inVin. e^-MIO* lO^inTin. e^+200* lO^inTin.
SOLUTION
£* - e.
-60/IO
-4
<5y - £3 * ZOO WO
- f.
= O
% r ^■^+^^rT^f[Ao°f(a3](-^
= iT.Roo v/oJ f>s/'
£_
*
P - A6; r (* )(Ofeaoo • /o1}
.*
- 61.6 w/o^ilo - £<7.£ k,p5
V
i = 2.«.
a
re
= —jg—1 * — 3 := G0.7V/0 Jk.
\/ = £0.7*tO*JL - 60.7 if.^s

PROBLEM 7.156
7.156 The state of stress shown occurs in a steel member made of a grade of steel
with a tensile yield strength of 270 MPa. Determine the factor of safety with respect
to yield strength, using (a) the maximum-shearing-stress criterion, (b) the maximum
distortion-strength criterion.
SOLUTION
36 MPa
S«.* 6i* + tt * i2& MPa.,
S"b= &„-R - -54 MP* , 6,-0
let>
1 «mmc
r UG MP*.
6"^ r -S4 MP*
2^ =■ 6*^- 6~* = \So MP*
RS. -
270 MP^ C^» yfeAfin^ ^
£7o
(3o
= 1-500
(0 V Sj" + 6"/ - 6L5t - IS«7-31 MPcl * 27DMP^ CWo^ieJJ--^
5r , 27c?_
F.S. -
Vssv1^ is*-."
L e 8 8
PROBLEM 7.157
7.157 A spherical pressure tank has 1.2-m outer ajameter and a uniform wall thickness
of 10 mm. Knowing that the gage pressure is 1.25 MPa in the tank, determine (a) the
maximum normal stress, (ft) the maximum shearing stress, (c) the normal strain on the
surface of the tank. (Use E = 200 GPa and v= 0.30.)
SOLUTION
i* ,oyto*v«) r---fcJ-i = i(\.z)-io*K>% -- o-ffioM, p= /.as MP*
Fc
OiT O.
&f>
Ae^i'c cnJf fcLnfe' {jHciev i'»^r
evu^Jf piress.ot^«
6- - 6- * ££ = ili^K£^li , 36.1 -M-ftL
6i * o
^00 * ID'
r 121 X 10
-4
= 121//

PROBLEM 7.158
7.1S8 The strains determined by the use of a rosette attached as shown to the surface
of a structural member are:
€t = 220* lO^iiu/in. 6i = 425* lO^iiu/in. ej~480x iO^inVin.
Determine (a) the orientation and magnitude of the principal strains in the plane of the
rosette, (b) the maximum in-plane shearing strain.
SOLUTION
y
■x.-z
e, = Y5°
6L= O
S3 = -VS'
e„ cos2e, + fj s.»la, + ^siVie,cos©, - £,
£„ cos 9, 4 6y Sit-»*dt + T^y si'-ifi^ cos©4 s £t
£, 4 O + O = HZS * tOm& i«/i*
it*
«. £j- ~ z *Y*'
H80 x IO"4 /m Am
CO
(1)
(3 1
irfo-o
£«* « a fer * O = 350 ^^ i« A»
fOorO
= - 1.7333
8b * GO'
r I50 K \0'U in/in
(b) Yli^fm|./^» > r ^^
- 300 V/O"6 («/,rl

PROBLEM 7.159
7.159 For a state of plane stress it is known that the normal and shearing stresses are
directed as shown and that ^ = 5 ksi, ay= \2ksi, vail o^=* l&km. Determine (a) the
orientation of the principal planes, (b) the maximum in-plane shearing stress.
SOLUTION
'xy
R - 6T_ - 6Lfc = IS - *.S = Wfcs/
± 8.S3 k*;
+«* ^ =
P 6*-^
0k = -34.2° a 6C. -- 55*. 3"

PROBLEM 7.160
750 mm
7.160 The compressed-air tank AB has an inside diameter of 450 mm and a uniform
wall thickness of 6 mm. Knowing that the gage pressure in the tank is 1.2 MPa,
determine the maximum normal stress and the maximum in-plane shearing stress at
points a and b on the top of the tank.
SOLUTION
r - ^ e> = 27.5 **** £ - 6 w*f
TV*n*»e*S* sliep'''* t c O «Jr p«i'nTS <sl *-*d b-
P»;*.-j cl
M^ (Sxto%)(7So*(o% )* 37SO W-^
gs Jjc g(37SO)bai.h>») . 3ggMo
Aon3;fod.nAjP 6** 22.5 + 3.8S * 2£.38
Ci*c**Je*wto»jP S, = 45
6!wt* ifci+O* 35\6<* MPa
^ * 6*~* + R " ^. I MPa.
£.*6-p>—> = R "= ^-VO MP*.
Po.Vt t
6-^* 7.75 MPa
T*KJ stresses (MP^^i
S* - 2Z.S + 7.75 5 3o.^r
S^iCS^S,^ 37.^5 Mp^
R - |&^)l + r^ - 7.M87 MP«
&.,*= 6U + R * 45. I MPa. -•
^-.U-pi-O* R - 7.^ MP-l —

PROBLEM 7.161
7.160 The compressed-air tank ABh&sm inside diameter of 450 mm and a uniform
wall thickness of 6 mm. Knowing that the gage pressure in the tank is 12 MPa,
determine the maximum normal stress and the maxinium in-plane shearing stress at
points a and b on the top of the tank.
7.161 Solve Prob. 7.160, assuming that the 5-kN force applied at D is directed
vertically upward.
r = iA r ns *.<
Of^lOW
Point
SOLUTION
61 = fjr = 22.5" MP*
c, - Z2S ***« Cz = ZZ5~+& =231 •*«*,
T- (£"*los)(S00*lo_i ) = ZSoo H-n
Tv-Ani^^SC Sne«i* : "£ - O m points a. *Ai<si Jo.
Point la
M^ ^x|o^)(75o>./u9 ) - 37S"0 N-r*
cs - M£- C375oV^awio"1) . p
s " T"' ni.<ts*\o-— ' 3*88 MP*
To+«Jl stresses (MP«l^
U^iM.W 6; = 22.S -3.88= |*.« MP*
Circu*le^&4\«/f Sy - 45 MPa
S^e*." 2^ - - 1.2*2 Mpa.
6U - ^ (£"*+£,)" 31.8* MP*
6^-€ry
+ £J = IS.^5" MP.
S = S_ + ft »
'**»*
45"./ MPa -*
TU«„-,^> - R * '3.^5 MP*
6" - ^ - 7.75 MP*.
ToW stresses (MP*)
6; * 22.5 - 7.75" = 14.75
6-
^5-
1^ - - 1.2**
iL.^pj^.)* R* '-S-'s MP* -*

PROBLEM 7.162
7.162 The steel pressure tank shown has a 30-in. inside diameter and a "J -in. wall
thickness. Knowing that the butt-welded seams form an angle of 50° with the
longitudinal axis of the tank and that the gage pressure in the tank is 200 psi, determine
(a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the
weld.
SOLUTION
IT = -id =" IS in.
B £T - (gooXis^ .
0.376"
Sooo
f*\
R »
- g.-sv ..
- Zooo
>» r
5^0
(0.1 6i* r 6^ *■ R c«s loo
r 1^70 p»('

PROBLEM 7.163
7.163 A sqvareABCD of 2.4-in. side is scribed on the surface of a thin plate while the
plate is unloaded. After the plate is loaded , the lengths of sides AB and AD are
observed to have increased, respectively, by 540 x 10^ in. and 900 x iflr* in., while the
angle DAB is observed to have decreased by 360 * io-6 rad. Knowing that v = t ,
determine (a) the orientation and magnitude of the principal strains, (b) the maximum
in-plane shearing strain, (c) the maximum shearing strain.
SOLUTION
y
£ - Ah. - Afe
y Ax AB
J Ay ao
-4
ixv r dec^e^se /« y^i
t^ wti--^« <« '"'3^ flu^e DAB = 3CO xIQ t*J
- £
225 - US
- -J?.*J
6p = - 33.7
R* VC^r^^*- 0fN/ r w/o4
e* - e*- + R r m^k io
-c
tu ^ C«,i( " K
105 * (D-
A- eflo-*)
(2/3
- (^S^/cp"'4ilo5"> IO"cN) - -30O */o~4
T^ - 4*»«ff*/o"4 Smi* * -3^0 x io'
CO X,^ ^ e-* - £«.■« = 7"*£ */o"
R - *YU - 3*7.5*/or* j
Fvf do+fcJ MoUir^b oiVtJe

PROBLEM 7.164
7.164 For the state of plane stress shown, determine (a) the principal planes, (b) the
principal stresses, (c) the maximum shearing stress.
18ksi
SOLUTION
S, - 4 k*'" j Sj = s k»'"
(a) fc-2^ !&£-= 5-.00
2R,
(k*i-1
o. nfc fc*;

PROBLEM 7.165
1.2 MPa
3.0 MPa
— 0.375 MPa
7.165 The grain a wooden member forms an angle of 15° with the vertical. For the
state of plane stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
SOLUTION
<5* * -S.O MPa J S'j-- - 1.2. MPa.
'6*. - Cg \«-
vl .(mr*
) + ?»/■ - 0.975" MP*.
r Ok- 0^
2<9p - 22.62*
GO ^Luywi - J? wi 5A5# - C3.775- MPc
6*
26p + So*
.- -2.1© - O.S12 «■ -^613 MPa.
PROBLEM 7.166
SOLUTION
7.166 A cylindrical steel pressure tank has a 26-In. inside diameter and a
uniform 4" -in. wall thickness. Knowing that the ultimate stress of the steel used is
65 ksi, determine the maximum allowable gage pressure if a factor of safety of S.O must
be maintained.
r * i J » iK«v is ,x
>*«
F.S.
± - a.ZS ;-
13 k%/
<v- ¥
p r S* r ,(»a)(o-as^ , o.js fas; - 550 P*

7.167 For the state of stress shown, determine the maximum shearing stress when (a)
oy = +72 MPa, (b) ay = -72 MPa,
SOLUTION
5*= - SO MP* Tp = 21 MP* , 6?z = O
Co,) £, * 4 72 tfP«.
rJ(MPO
&*« - iC^O" ^G MP*
r*^(S^Y7-r^ - <kmp«.
6m* = &<.- .6L. + K * SI MPa
(b) '6^= -"75 MPa.
'»*«*9e.
= 6^ = 6U + R- ** ^p*
G^* 6; r G«-R • -81 MPa
CmPo

PROBLEM 7.C1
(a)
7.C1 A state of plane stress is defined by the stress components ax, ay,
and T^ associated with the element shown in Fig. P7.Cla. (a) Write a
computer program that can be used to calculate the stress components a>, oy, and
Tyy associated with the element after it has rotsted through an angle 8 about
the z axis (Fig. P7.C1*). (b) Use this program to solve Probs. 7.13 through
7.16.
SOLUTION P/?&<$/efiM f?&Ll.c>lu\t~t6 &&u*TlonS
<ry-<n,
F&toXfffl' o?,=
2
cr;+<nj
-r a cos 2& + 7J" .S'Vi;?^
Z *J
E€>l7A)tf<i*T. ^~?kJL Sfrta+T^eoSte
'w
J^Sf^fK
rx3
^WO <3
^*7 i^^nr <z>/?rw&o F&n. T^\;3V5rt-wo T^y
<»
Problem 7.13a
Sigma x = -40 MPa
Sigma y = 60 MPa
Tau xy « 20 MPa
Rotation of element
(+ counterclockwise)
theta = -25 degrees
Sigma x' = -37.46 MPa
Sigma y1 = 57.46 MPa
Tau x'y1 = -25.45 MPa
Problem 7.13b
Sigma x » -40 MPa
Sigma y = 60 MPa
Tau xy = 20 MPa
Rotation of element
(+ counterclockwise)
theta = 10 degreeB
Sigma x' = -30.14 MPa
Sigma y* =■ 50.14 MPa
Tau x'y = 35.89 MPa
Problem 7.14-a
Problem 7.14b
80 MPa
50 MPa
Sigma x = 0 MPa
Sigma y * -80 MPa
Tau xy * -50 MPa
Rotation of element
(+ counterclockwise)
theta * -25 degreeB
Sigma x1 = 24.01 MPa
Sigma y = -104.01 MPa
Tau x'y' = -1.50 MPa
Sigma x
Sigma y
Tau xy
0 MPa
-80 MPa
-50 MPa
Rotation of element
(+ counterclockwise)
theta = 10 degreeB
Sigma x' = -19.51 MPa
Sigma y* * -60.49 MPa
Tau X'y' = -60.67 MPa
CONTINUED

PROBLEM 7.C1 - CONTINUED
F%p£#ttt« Chr/=vr
12 ksi
8 ksi
Problem 7.15a
Sigma x « 8 ksi
Sigma y = -12 kBi
Tau xy = -6 kBi
Rotation of element
(+ counterclockwise)
theta = -25 degrees
Sigma x' = 9.02 ksi
Sigma y = -13.02 ksi
Tau x'y' = 3.80 ksi
Problem 7.15b
Sigma x = 8 ksi
Sigma y = -12 ksi
Tau xy * -6 ksi
Rotation of element
(+ counterclockwise)
theta = 10 degrees
Sigma xl =
Sigma y1 =
Tau x'y =
5.34 ksi
-9.34 ksi
-9.06 ksi
Problem 7.16a
Sigma x = 0 ksi
Sigma y = 16 ksi
Tau xy = 10 ksi
Rotation of element
(+ counterclockwise)
theta = -25 degrees
Sigma x1 = -4.80 ksi
Sigma y* = 20.80 ksi
Tau x'y* = 0.30 ksi
Problem 7.16b
Sigma x = 0 ksi
Sigma y = 16 ksi
Tau xy ■ 10 ksi
Rotation of element
(+ counterclockwise)
theta = io degrees
Sigma x' = 3.90 ksi
Sigma yr ~ 12.10 kBi
Tau x'y = 12.13 ksi

PROBLEM 7.C2
7.C2 A state of plane stress is defined by the stress components av av,
and t^ associated with the element shown in Fig. Pl.Cla. (a) Write a
computer program that can be used to determine the principal axes, the principal
stresses, the maximum in-plane shearing stress, and the maximum shearing
stress, (b) Use this program to solve Probs. 7.7, 7.11, 7.66, and 7.67.
SOLUTION
fef?. IS)
^v^ =
^* ^ve+ *
J^
- - C- - ff
'^l/e
-t ^?T,
Xy-Tv
^IM
5; —Cl/1
<o
7? cm y
^^(jn-p/tm)" R / ^^ (ov,t~ot~rh»#) ■=■ #
IF ^l..'>0 a?nd y~>0
'TVOy-
Vlrt
/s}rtchj.(]n "p/ane.)" ^ ' ^f^/'au/t-af-^^ttey' 2 >^^y
-Tplly
<a :
A»m*(m-04**yf? ', Z^frtt'Of-f*
kr
~M
pl£tYi£)— -Z?
P$0£(to^ OUTPUT
40MPa
35 MPa
Problems 7.7 AND 7.11
60MPa
Sigma x
Sigma y
Tau xy
-60.00 MPa
-40.00 MPa
35.00 MPa
Angle between xy axes and principal axes
( + counterclockwise )
Theta p = -37.03 deg. and 52.97 deg.
Sigma max = -13.60 MPa
Sigma min = -86.40 MPa
Angle between xy axis and planes of maximum in-plane shearing stress
( + counterclockwise )
Theta s ■ 7.97 deg. and 97.97 deg.
Tau max (in plane) = 36.40 MPa
Tau max = 43.20 MPa
CONTINUED

PROBLEM 7.C2 - CONTINUED
Fig. P7.66 and P7.67
Problem 7.66a: Sigma x <■ 140.00 MPa
" Sigma y = 20.00 MPa
Tau xy = 80.00 MPa
Angle between xy axes and principal axes
( + counterclockwise }
Theta p = 26.57 deg. and 116.57 deg.
Sigma max = 180.00 MPa
Sigma min = -20.00 MPa
Angle between xy axis and planes of maximum in-plane
in-plane shearing stress ( + counterclockwise )
Theta s = 71.57 deg. and 161.57 deg.
Tau max (in-plane) = 100.00 MPa
Tau max (out-of-plane) = 100.00 MPa
Problem 7.66b :
Sigma x <* 140.00 MPa
Sigma y - 140.00 MPa
Tau xy - 80.00 MPa
Angle between xy axes and principal axes
( + counterclockwise }
Theta p = 45.00 deg. and 135.00 deg.
Sigma max = 220.00 MPa
Sigma min = 60.00 MPa
Angle between xy axis and planes of maximum in-plane
in-plane shearing stress ( + counterclockwise )
Theta s = 90.00 deg. and 180.00 deg.
Tau max (in-plane) = 80.00 MPa
Tau max (out-of-plane) = 110.00 MPa
Problem 7.67a:
Sigma x
Sigma y
Tau xy
140.00 MPa
40.00 MPa
80.00 MPa
Angle between xy axes and principal axes
( + counterclockwise )
Theta p = 29.00 deg. and 119.00 deg.
Sigma max = 184.34 MPa
Sigma min = -4.34 MPa
Angle between xy axis and planes of maximum in-plane
in-plane shearing stress ( + counterclockwise )
Theta b = 74.00 deg. and 164.00 deg.
Tau max (in-plane) = 94.34 MPa
Tau max (out-of-plane) = 94.34 MPa
Problem 7.67b:
Sigma x
Sigma y
Tau xy
140.00 MPa
120.00 MPa
80.00 MPa
Angle between xy axes and principal axes
( + counterclockwise )
Theta p - 41.44 deg. and 131.44 deg.
Sigma max = 210.62 MPa
Sigma min = 49.38 MPa
Angle between xy axis and planes of maximum in-plane
in-plane shearing stress ( + counterclockwise )
Theta s = 86.44 deg. and 176.44 deg.
Tau max (in-plane) = 80.62 MPa
Tau max (out-of-plane) » 105.31 MPa

PROBLEM 7.C3
7.C3 (a) Write a computer program that, for a given state of plarte stress
and a given yield strength of a ductile material, can be used to determine
whether the material will yield. The program should use both the maximum-
shearing-strength criterion and the maximum-distortion-energy criterion. It
should also print the values of the principal stresses and, if the material does
not yield, calculate the factor of safety, (b) Use this program to solve Probs
7.81 through 7.84.
SOLUTION
<fc -far
*4-
Tv-^*
'J+Z
IF <Q fififf Ta /we s*tA£ sat/v, 7^^- * ^
f^flCTcK OP €&f&TY - Y—
jP f?vf/teAl < ^ //*> YJ^WJr/d C>CC v&£f /*rv
/^ftcj-of? of ^Fs-Ty-
*7
fdftOitLAL.
P/fg tffi M £)oTPu T
21 ksi
36 ksi
ProblemB 7.81a and 7.82a
Sigma x m 36.00 kBi
Sigma y » 21.00 kBi
Tau xy «■ 9.00 kBi
40.22 kBi
16.78 kBi
Sigmax «
Sigmin =
UBing the maximum-Bhearing-BtreBB criterion:
Material will not yield
F.S. * 1.119
UBing the maximum-distortion-energy criterion:
Material will not yield
F.S. = 1.286
CONTINUED

PROBLEM 7.C4
7.C4 (a) Write a computer program based on Mohr's fracture criterion
for brittle materials that, for a given stste of plane stress and given values of
the ultimate strength of the material in tension and in compression, can be used
to determine whether rupture will occur. The program should also print the
values of the principal stresses, (b) Use this program to solve Probs. 7.91 and 7.92
and to check the answers given for Probs. 7.93 and 7.94.
SOLUTION
<Gve=
<J^<^
R^
T*-<TH\*
y + r.
v^
^>X*v*
IF- %_ &r>o 57 /rwvf £.*>*& ,<.'i".v fl*-'7
<Z<TjT *~*
%>%r
OR
U<-
% > V ,
-&■-
t^f
FGfLvn.
a.
C°NS'VP£ J=b<JR-r# & ofj&p-T' qi= &'€. 7,'7,7
IF T„>o ftT->o T<:o
IF Tb> Cf2lT££ '■■>*?
CRITFIZ ION = T~ - \
P/?a£!?f\M outPut
7bi
8 ksi
Fig. P7.91
Problem 7.91 Sigma x = -8.00 ksi
Sigma y = 0.00 ksi
Tau xy » 7.00 ksi
Ultimate strength in tension = 10 ksi
Ultimate strength in compression > 30 ksi
Sigma max => Sigma a 4.06 ksi
Sigma min = Sigma b -12.06 ksi
Rupture will not occur
CONTINUED

PROBLEM 7.C5
7.C5 A stale of plane strain is defined by the strain components €x, cr
and y^. associated with the x and y axes, (a) Write a computer program that
can be used to calculate the strain components et., ey, and y,y associated with
the frame of reference x'y' obtained by rotsting the x and y axes through an
angle 6. (b) Use this program to solve Probs. 7.126 through 7.129.
SOLUTION
Pffij£1P&M FOLL-£>Kj>lh<£ ^CDL/fir-r/o^S
&(?•¥+) Km ^4^ + ^^~ a* ** + jf C *'»**
£ ux$
PfcddPfiM CWPUT
fr>{?.H) £"' = ^0- - &^_ &x€> -Ijr cos**
*j
^^ £„, ^^.^ <^
Problem 7.126
Epsilon x
Epsilon y
Gamma xy
-720 micro meters
0 micro meters
300 micro radians
Rotation of element, in degrees ( + counterclockwise }
Theta = -30 degrees
Epsilon x'= -669.90 micro meters
Epsilon y' = -50.10 micro meters
Gamma x'y' = -473.54 micro radians
Problem 7.127
Epsilon x » o micro meters
Epsilon y m 320 micro meters
Gamma xy = -100 micro radians
Rotation of element, in degrees ( + counterclockwise )
Theta » 30 degrees
Epsilon x1
Epsilon y*;
Gamma x'y':
36.70 micro meters
283.30 micro meters
227.13 micro radians
Problem 7.128
Rotation of element,
Epsilon x'
Epsilon y*
Gamma x'y1
Epsilon x » -800 micro meters
Epsilon y = 450 micro meters
Gamma xy - 200 micro radians
in degrees ( + counterclockwise )
Theta = -25 degrees
-653.35 micro meters
303.35 micro meters
-829.00 micro radians
Problem 7.129
Rotation of element,
Epsilon x « -500 micro meters
Epsilon y = 250 micro meters
Gamma xy » 0 micro radians
in degrees ( + counterclockwise )
Theta « 15 degrees
Epsilon x' = -449.76 micro meters
Epsilon y'= 199.76 micro meters
Gamma x*y*= 375.00 micro radians

PRORI VM 7 C6 7,C6 A state of Sttain 'a de^ned ty the strain comPonents €» ey and >V
associated with the x aady axes, (a) Write a computer program that can be used
to determine the orientation and magnitude of the principal strains, the maximum
in-plane shearing strain, and the maximum shearing strain, (b) Use this program
to solve Probs. 7.134 through 7.137.
SOLUTION Ptog/FPr .v /Tcji to »j t**6 £G<tnrrfaiv<:
**(*** ^ -^ RM^T^^^
£g>. I ?sd e^f eo^ R 6^ ^ - c
gHEGiciMt Sfrte-.,a/s /vftwirtitito M'piANfr s~/)&PteM* j;"1/"^'"
if £" ~> -? ~> *-' i Y r ■ - /* - ^.
f?QD£\?&*>\ pfcmrov.
Problem 7.134 EpBilon x » 160 micro meterB
EpBilon y = -480 micro meterB
Gamma xy = -600 micro radians
nu = 0.333
Angle between xy axes and principal axes (+ = counterclockwise)
Theta p = -21,58 degrees
Epsilon a = 278.63 micro meters
Epsilon b = -598.63 micro meters
Epsilon c = 159.98 micro meterB
Gamma max (in plane} = 877.27 micro radians
Gamma max = 877.27 micro radians
CONTINUED

PROBLEM 7.C6 - CONTINUED
Problem 7.135
EpBilon x * -260 micro meters
Epsilon y = -60 micro meters
Gamma xy * 480 micro radians
nu =■ 0.333
Angle between xy axes and principal axes
Theta p = -33.69 degrees
EpBilon a = 100.00 micro meters
Epsilon b = -420.00 micro meters
Epsilon c = 159.98 micro meters
Gamma max (in plane) = 520.00 micro radians
Gamma max = 579.98 micro radians
(+ = counterclockwise)
Problem 7.136
Epsilon x ■■
Epsilon y <
Gamma xy
nu = 0.333
-40 micro meters
760 micro meters
960 micro radians
Angle between xy axes and principal axes
Theta p = -25.10 degrees
Epsilon a = 984.82 micro meters
Epsilon b = -264.82 micro meters
Epsilon c = -359.95 micro meters
Gamma max (in plane) = 1249.64 micro radians
Gamma max = 1344.77 micro radians
(+ = counterclockwise)
Problem 7.137
EpBilon x * -300 micro meters
Epsilon y = -200 micro meters
Gamma xy = 175 micro radians
nu = 0.333
Angle between xy axes and principal axes
Theta p = -30.13 degrees
EpBilon a = -149.22 micro meters
EpBilon b ■ -350.78 micro meters
EpBilon c = 250.00 micro meters
Gamma max (in plane) = 201.56 micro radians
Gamma max = 600.77 micro radians
(+ » counterclockwise)

PROBLEM 7.C7
7.C7 A state of plane strain is defined by the strain components €„€r.
and Y^ measured at a point, (a) Write a computer program that can be used to
determine the orientation and magnitude of the principal strains, the maximum
in-plane shearing strain, and the maximum shearing strain, (b) Use this
program to solve Probs. 7.138 through 7.141.
SOLUTION
s&fzs/J
fFOC'^AM p£>j_io\<.„/-/£ £Ql>QTie>MS
/?-
T*-<Tu\S
'-^M-r/
y«v
fQl?$2) <%*£*»'*
K
xy
V*.
&Aj£AQi/tf S.TPWHS
MAX/MVM tH-PlAM£ ^tyffUliHt £TT?-QIAI
\tf#J?T#£& ?T IS 7#£ MrfW^W gAtpflktM £7/2/9//-/
**r/ <CA ^rtfiOjt
g : £> (f%&» -Z7724/S/)
"* £*>£*ts-a1 Xow
PL$r*i>
r^-^
PRC-Afrft <■" Pp / a/ ^ f. 17
Problem 7.138
EpBilon x = -90
EpBilon y « -130
Gamma xy ■ 150
Angle between xy axes and principal axes (+ « counterclockwise)
Theta p = 37.53 and -52.47 degrees
Epsilon a = -32.38 micro meters at 37.53 degrees
Epsilon b = -187.62 micro meters at -52.47 degrees
Epsilon c =• 0.00 micro meters
Gamma max {in plane) * 155.24 micro radians
Gamma max = 187.62 micro radians
CONTINUED

PROBLEM 7.C7 - CONTINUED
Problem 7.139
Epsilon x = 375
Epsilon y = 75
Gamma xy « 125
Angle between xy axes and principal axes (+ = counterclockwise)
Theta p = 11.31 and -78.69 degrees
Epsilon a = 387.50 micro meters at 11.31 degrees
Epsilon b = 62.50 micro meters at -78.69 degrees
Epsilon c = 0.00 micro meters
Gamma max (in plane) = 325.00 micro radians
Gamma max = 387.50 micro radians
Problem 7.140
Epsilon x = 400
Epsilon y = 200
Gamma xy = 375
Angle between xy axes and principal axes (+ = counterclockwise)
Theta p = 30.96 and -59.04 degrees
Epsilon a = 512.50 micro meters at 30.96 degrees
Epsilon b = 87.50 micro meters at -59.04 degrees
Epsilon c = 0.00 micro meters
Gamma max (in plane) - 425.00 micro radians
Gamma max = 512.50 micro radians
Problem 7.141
Epsilon x s 60
Epsilon y « 240
Gamma xy a -50
Angle between xy axes and principal axes (+ = counterclockwise)
Theta p = 7.76 and -82.24 degrees
Epsilon a = 243.41 micro meters at 7.76 degrees
Epsilon b = 56.59 micro meters at 97.76 degrees
Epsilon c = 0.00 micro meters
Gamma max (in plane) = 186.82 micro radians
Gamma max = 243.41 micro radians

PROBLEM 7.C8
7.C8 A rosette consisting of three gages forming, respectively, angles
0], 62, and 03 with the x axis is attached to the free surface of a machine
component made of a material with a given Poisson's ratio v. (a) Write a computer
program that, for given readings e,, e2, and £3 of the gages, can be used to
calculate the strain components associated with the x and y axes and to determine
the orientstion and magnitude of the three principal strains, the maximum in-
plane shearing strain, and the maximum shearing strain, (b) Use this program
to solve Probs. 7.142 through 7.145.
ond €„
SOLUTION
f//7f£' W'V
Ms t/Joz>, 11 ^r-=
fNTtn
o*&
Ll±£l
Rx
'
z
M4
■fc+V
C hi/
&> - J, a, -'_2sa_
<TMFri&H& ^Tzwns
AffiXjM'UM /-v-Z^A'f &/&A/1/M& g"772/Q/yy
Y , =2%
PPft/MArM &UTPV7
Problem 7.142
**.<**
07?t&2XO/S&'.'"
Sage
theta
degreeb
30
-30
90
K
- £, -<L
4:^*
'OUT- OF - Pt-AMT
epBilon
micro meterB
600
450
-75
* ^£
EpBilon x «
EpBilon y =
Gamma xy =
EpBilon a «
EpBilon b
725.000 micro meterB
-75.000 micro meterB
173.205 micro radians
734.268 micro meterB
-84.268 micro meterB
Gamma max (in plane) » 818.535 micro radians
CONTINUED

PROBLEM 7.C8 - CONTINUED
Problem 7.143
EpsiIon x
Epsilon y
Gamma xy
Gage
theta
degrees
-15
30
75
epsilon
in./in,
720
-180
120
379.808 in./in.—
460.192 in./in.
-1339.230 micro radians
Epsilon a - 1090.820 in./in.
Epsilon b = -250.820 in./in.
Gamma max (in plane} = 1341.641 micro radians
Problem 7.144
£A>T*r?
&^ fir-'v
#s
<£•_, ^ mo e^J
Gage
1
2
theta
degrees
0
45
135
epsilon
micro meters
420
-45
165
Epsilon x =
Epsilon y =
Gamma xy =
Epsilon a =
Epsilon b =
420.000 micro meters
-300.000 micro meters —
-210.000 micro radians
435.000 micro meters
-315.000 micro meters
Gamma max (in plane} = 750.000 micro radians
Problem 7.145
Epsilon x
Epsilon y
Gamma xy
Epsilon a
EpsiIon b
as
It II
Gage
1
2
3
315.000
-5.000
-410.000
415.048
-105.048
theta
degrees
45
-45
0
eps:
in
in./in.
in./in.
micro radians
in./in. —
in./in.
ilon
./in.
-50
360
315
Gamma max (in plane} = 520.096 micro radians

CHAPTER 8

PROBLEM 8.1
^-v
8.1 An overhanging W250 x 58 rolled-steel beam supports two loads as shown.
Knowing that P = 400 kN, a = 0.25 m, and o^ - 250 MPa, determine (a) the
maximum value of the normal stress am in the beam, (b) the maximum value of the
principal stress o^ at the junction of aflange and the web, (c) whether the specified
shape is acceptable as far as these two stresses are concerned..
M*-* ' H-00 kW * 'tooWo* Kf
i^fcown I„* 87.3kto4**1 Sx ' ^3-/o3^1
C = ie/ = /2C him jfc r C - fcf r "*-? »"**>■
« s.-i
*"***

PROBLEM 8.2
(kvT>
ZOO
-2oo
W
(kM-
\
8.1 An overhanging W250 x 58 rolled-steel beam supports two loads as shown
Knowing that P = 400 kN, a = 0.25 m, and a^ = 250 MPa, determine (a) the
maximum value of the normal stress c„ in the beam, (b) the maximum value of the
principal stress omax at the junction of aflange and the web, (c) whether the specified
shape is acceptable as far as these two stresses are concerned..
8.2 Solve Prob. 8.1, assuming that P = 200 kN and a = 0.5 m.
IML^ - (%ooY\diXo.£') - \oo*ic? i^-K"
C - ;J - 146 w»w ^ r c - tf - \\7.$ ww
^ c /OOWO* = m3v,0«P(,
-loo fctO-*"
G<?3 * fo_
« /^.3 MP^l
Jf r i(c-»yj * U9.7S "~<
Qt - Afjf - SZQ.Soticf^J? ^ 326.2oyfO~c ^a
(lo) 6^ = % + *R »
H3.C3 MP^
78*0 MP*.
Cc^
Si
nee 6"*** *"
ZSO MP*, l/v/250*5* ,-s cLccefrtcJ>h. -*

PROBLEM 8.3
frr^
+ 33.0
-SZooo
8.3 An overhanging W36 * 300 rolled-steel beam supports a load P as shown.
Knowing that P = 320 kips, a = 100 in., and o^ ~ 29 ksi, determine (a) the
maximum value of the normal stress <j„ in the beam, (b) the maximum value of the
principal stress o^ at the junction of a flange and the web, (c) whether the specified
shape is acceptable as far as these two stresses are concerned.
For W 3& x 300 toU*<\ s-rW We**,
A ' dG.lHm hf= le.tfSTiVi. tf- 1.680 Sw.
twso.Ww Xr-20300 iV s* = moiv1.
c<a si *
|M|
*—It _
3%ooO
UIO
- ;?s.s fc*r
(W
W
Af * Lfit =■ 27.<?8 m*
J* - i (e-v y^- 17.53 ■*.
Qb r Af^f ff H*?o.H^ ,**
r _ IVL,GU _ (ago Wio.^ _ .
^ I, t, ' ftoso©xo.no ~

PROBLEM 8.4
(fc«1
-f4oo
-3ZOQO
8.3 An overhanging W36 x 300 rolled-steel beam supports a load P as shown.
Knowing that P = 320 kips, a = 100 in., and o^ = 29 ksi, detennine (a) the
maximum value of the normal stress ct„ in the beam, (b) the maximum value of the
principal stress o^ at the junction of a flange and the web, (c) whether the specified
shape is acceptable as far as these two stresses are concerned.
8.4 Solve Prob. 8.3, assuming that P = 400 kips and a - 80 in.
(VI.
40O Jfc'ps
lMLv- tmx>)(2o} * 3ZOOO k.p-m.
For W 3& *30O roiied s\eej sec-ho*
<^- 3G.7t.Vi bf =■ ICG-SSiV £p=- /.G*o ,-n
tw- 0.9^ m. 3„* 20300 .*«" Sy ^ I MO .V
C - i«> ■ 18.37 irt yfc- c- tp - I&.6*? .*n
(a) Om - ^ )"no " *8*8 *s*
^r £ ^" SIMCa8-*^ *cz ^
^fr a (C4 yOr l7-s3 m.
" J»** " Cgo8oo^CQ-1^g) "
s:
+ R -
2?.7 ksr
Since 2?. 7 ks.' > §U/ W36xSOO .s ntff aaceptaJxle

PROBLEM 8.5
•250 kN 250 kN 250 kN
0.9 m 0.9 m 0.9 m 0.9 m
SIS
12S
■\*5
M
8.5 and 8.6 {a) Knowing that a^ = 160 MPa and r^ = 100 MPa, select the most
economical metric wide-flange shape that should be used to support the loading
shown. (£) Determine the values to be expected for a„. r^.and the principal stress
c^ at the junction of a flange and the web of the selected beam.
ffcfc^Jo^: 9^ 37<TktJ f ^ Pe*» B7S UN T
WL « 375T kltf
ImL« * 45c? tM-^
ivl cX p..*-/ c ia^ h/
H$o*t<y
\£0*[06
r 3SI2.3** loi w,^1
- ^SI^"/o"'^
5^«(pe
W g^O/176
IV 76o x 147
W £<?o x 12S"
w t\o*isC
VnJ S30 X 150
W *Ko *IS8
W 36o x 216
o^ Clo >■*<* )
53<?t>
44 lo
35"/ o
*«2 0
37^0
33 4 0
32 oO
Co.) Use
d - 6 7S ■fi*^
tf - l£».3o w*^
.1MS-E, ^°^°3 , ias.a*/o'p* r m.zMP*
t -
5K
IVl
*••» -
A.
35/OkIo"'
375" k/o*
M poi*f C
(G78MO-*Yn.7*(o~*)
.3
r 47. 3 »/»* Pi - 47.3 MPa -*
2^
V |2Sv/o . _
V 1^.76 */Ofc P«,
■** ~ A^ " (G78*(©"iK,,-7x/0",,>
= IS".76 MP-*
C-^d=^ = 33? m*n yb -- C-tf - 33?-16.3o r 3JW.7 mm
6L - -&SL-- ^2)(l2».^= Itt.o MP*
t? r 7(%)V+ ^* = -J(&-°)1 ■*'(<£• 7*'>* = G3-D MPcl
<^ * § + ^ * C/.0+63.O = 1^.0 MP*

PROBLEM 8.6
275 kN
8.5 and 8.6 (a) Knowing that o^, = 160 MPa and r^ = 100 MPa, select the most
economical metric wide-flange shape that should be used to support the loading
shown, (b) Determine the values to be expected for a„. r„.and the principal stress
fffn^at the junction of a flange and the web of the selected beam.
RB * Sot. 17 kw ! £ * 50*. \1 I
1.5 m
1.5 m
- 275 UV
r MI2.5 kW-v^
Vflclfl
ZVlM
Sti*i)€
W76<pv 147
W^O*' 1?5
IV 530 > /5~0
W 4^x 158
W 360* Z\C
SkCio1-.*?!
fMlo
35" lo
37£o
334 o
32 Oo
0*0 Use
d ■= 678 »w»t
€u*
S
-HI3.5 kw-ho
_ IVI—^ ^ wu _ X7gvio3
Aw ' ""JET " (67&*tcr*)(iL7*io~' )
^•5»'°I , 1,7.5*/©* A
= U7.5- MPv -*
- 3t.7*tOcP^ = 3^.7 MP*, -^t
6L* =
R
SS.1ZS + 6^. 2/C ~ UK 7 A^Pa,

PROBLEM 8.7
20 kips 20 kips
2 kips/ft
£.1T,! -1 L •
((Op*
-20
-So*
Zo
c--{l--
lo.W
-;?oo
-lo.M^r »v.
8.7 and «.« (a) Knowing that o^ =24 ksi and r^ =14.5 ksi, select the most
economical wide-flange shape that should be used to support the loading shown. (6)
Determine the values to be expected for om, I"*.and the principal stress omaxat the
junction of a flange and the web of the selected beam.
l?i> - 50 k. ps t
RA * SO top, f
IVL^ ■= So k,-ps
5K(t|0c
W £4v&2
W ^1 y 6X
W IS x76
W 16 V77
W 12 x 96
W/ /O » \\X
SG^
154
1X7
I4fc
134
103
tai
SL =
iM|^ _ THoo
U7
3o
toO Use
4*Zo.1<r in.
tf i 0.6I5" in.
yfc - C- £f = IO.H95 - 0.615 - <?.£S in.
gfca ^ T (TsStX'8^ - '7.71 fc»/
s;
•*»**.
% + R r
£.??£ + 7.^*6 r \«. *JS !&■'

PROBLEM 8.8
1.5 kips/ft
m (kirm
I5.i*f k,p.-Fl
tf.7 and 8.8 (a) Knowing that o^ =24 ksi and r^ =14.5 ksi, select the most
economical wide-flange shape that should be used to support the loading shown. (6)
Determine the values to be expected for am, r„,and the principal stress amaxat the
junction of a flange and the web of the selected beam.
Wl^ - II. 25" kips
lMlM^ = 21 fcp.ff -- 324 fcp-in
IM
It-***
§>
SVi^pe
W 12 v 1G
W JO x IS
W 8 x \S
W 6 x Zo
S ti«»)
n. i
13. 8
IS.^
13. 4
2^
M
s
W I o » 15"
tf" O.XlO In.
Blf!r* *3^ ws;
US'
-27 ic.p-ft
C
- 4.*o I?*,'
in
4.725 in.
6L-- -£-6„ ^(w|X^5) - «.* J»;
SL^ =" % + R =■ ^^ + /?. I - Z3.Z Irsi*

PROBLEM 8.9
90 kN
90 kN I I I 90 kN
—\ is°
u
1
1
fq°
D
B C
-«to
M (WJ•*)
8.9 through 8.14 Each of the following problems refers to a rolled-steel shape
selected in a problem of Chap. 5 to support a given loading at a minimal cost while
satisfying the requirement om&otll For the selected design, determine (a) the actual
value of cK in the beam, (b) the maximum value of the principal stress onlM at the
junction of a flange and the web.
8.9 Loading orProb. 5.81 and selected W410 x 60 shape.
Fro^ tYofcJe^ S. 8 1 €]jj r l€0 \A?o.
1mL*= i« kw-*v j c **<\ x>
d= Mot ~>«*; tf = ns-w, if- t2.8o****
C = "k tfl " 2.0%. S hvn
6"« -
IM
"W* —
«*/0
IST^-8 MPc^
yb = c - tf ~ 190.7 mw
Glr AfJ s (2X78)097.0 - 441 */o* *,„*
^.3 MPa-
wirv»
- -/fl.t* 4 24.3* = 75.6 MPcc
^ + R - V7/.6 + 75*. 6 = 147.2. MPo-

PROBLEM 8.10
SO ItN
100 kK'm
rnmum
-280
M (kW-*^
-2S&
8.9 through 8.14 Each of the following problems refers to a rolled-steel shape
selected in a problem of Chap. 5 to support a given loading at a minimal cost while
satisfying the requirement a„ <. a^ For the selected design, determine (a) the actual
value of a„ in the beam, (b) the maximum value of the principal stress omm at the
junction of a flange and the web.
8.10 Loading of Prob. 5.86 and selected S 510 x 98.3 shape.
Frol* IVotJPe^ S. 8 C 6Ij/- l£o MP*.
IK,.* - 256 \cV-v* <& p«m* 'B
IVl - SCO kv ai S
Fo^ S SlO * 98.3 roJJed s\eM section
d ~ 5"o8 "tin e>f ? IS'S mi*! if =* 20. X **■»»•»
"ML* 2S6»lOs
S^
I***** ..
yb = c - tf * A33.S
Af ' bf tf - 3*13. *>"»*
^ = ^ (c + yi} - 243. 9 iw^
- 131.3 WiV
^ r £o.V«T WP».
1 ~ Itw " (W*to-*Xl2.8*fo-*)
_ 61
? * R *
fiO.-'tf * 7.5T.04 - 135". 5" MP*.

m flcp-m
8.9 through 8.14 Each of the following problems refers to a rolled-steel shape
selected in a problem of Chap. 5 to support a given loading at a minimal cost while
satisfying the requirement a„ s t%m For the selected design, determine (a) the actual
value of a„ in the beam, (b) the maximum value of the principal stress o^ at the
junction of a flange and the web.
8.11 Loading of Prob. 5.83 and selected W27 x 84 shape.
IML, = 320.G fcp.-f+ - 38^7 Jop-t*
At 13- |VI ■= *3'.2S fc'ps
For IV 2? x g4 roXW S+eei see^'ow
dl - 26.71 in 3 bp = *?.9tfoU 3 £p - O.fi^o .V!
*3.3SSin.
S8H1
2.06 k»,-
M S 2i3
^t - C - £f - 12.7/5" m.
6*b - ^G"^ - i7. ?o k*i f *s.eo /ft/
Af * bftf - Lcf.96o)(o.4io) - G.S7«W ,V
Q. * Af J = g3.£><? in3
S + R -
f. 60, + *.7* &•' r \7.32 Wsia

PROBLEM 8.12
l.ri kips/ft
0.5
A
kip/ft
f f
i
■
Wi|iff?"!?a^ffWiR
■*
'
ps^BiBRPl-
is ft -m.
8.9 through 8.14 Each of the following problems refers to a rolled-steel shape
selected in a problem of Chap. 5 to support a given loading at a minimal cost while
satisfying the requirement a„ s % For the selected design, determine (a) the actual
value of a„ in the beam, (b) the maximum value of the principal stress amax at the
junction of a flange and the web.
8.12 Loading of Prob. 5.84 and selected W18 * 50 shape.
Kro-x Pvotie^ S.9* 61m r 2f k%/
\ML^ r 135 lcp.-P+ = |£20 /<■>• i» aH B
Wl^ - 18 k:pa <cf B
Jl - H.9«? in , fe>+ - 7. 49S\'Mj tf =O.S7o ,v,.
±w = O.SSS\-„ _, lz - £oO iV ^ Sz - gg.9 in*j C = £eJ - 8.9**r .V
/7, 07 W*.-
For \A/l8x^o 6^pe
§ -- *.£*$ Usi
yt "- c - tf r
Si - A ^ ,
Af ? tptp r 4. 275. iV
? = y Q - Ci3K37.^Q
b J„t^ (Soo)(0.3S5)
<W = % ■* R = 2-53S + 3.&SS - '7.3<? /rsJ
a = Af j
2.3C fesf
= si.x) ;**

PROBLEM 8.13
18 kips
v (k;fO
12
M(l6>-PO
8.9 through 8.14 Each of the following problems refers to a rolled-steel shape
selected in a problem of Chap. 5 to support a given loading at a minimal cost while
satisfying the requirement am s a^ For the selected design, determine (a) the aciual
value of am in the beam, (b) the maximum value of the principal stress o-ni4, al the
junction of a flange and the web.
8.13 Loading ofProb. 5.87 and selected S12 x 31.8 shape.
F^o** fV*tJ€wx 5". €7 (Sejt = ZH fcsi
IMLM - S* *,y-ft T ^2 Jty.;« J- C
A+ c Iv I - is 'fc.y*
For- SI2 *3/.3
tw ~ 0.35*0;.
-^
IM
I2 ' ^-^S i"i j 52 = S&M in
C - £ el - 6.00 m.
S* -
£<**
- 17. €o ksr
r =
VQ _ 0«')(lff.6lB^
Xztw 0JlS)f0.3SD">
%*fcO"?3 Jtft,'
R' 7(f )*"+ ^ =■ -/fcCWS* + 3.675* = g.g89 kv
6:
t^^t
§ + K.
g. 0<?3 + g.Sg<? = /£.9S ks/

PROBLEM 8.14
4S kips 48 kips
48 kips
2ft' 2ft
V(ki>»>
£7.C
l.fe
a c
-4»
8.9 through 8.14 Each of the following problems refers to a rolled-steel shape
selected in a problem of Chap. 5 to support a given loading at a minimal cost while
satisfying the requirement a„ <. ff»u for the selected design, determine (a) the actual
value of am in the beam, {b) the maximum value of the principal stress anax at the
junction of a flange and the web.
8.14 Loading of Prob. 5.88 and selected SI 5 *42.9shape.
Ue~\ 6.38 61« - ZH k%i
At T> lV\r 3S.4 *.p».
t* - 0.4H )Mj Jfc r 447 ;M f ^ sz r £<?.£ ;M3
M (■«.>■*)
6^ -
in
isa
r, -
^/t - C - tf - 6.87 8 ;«n
y - k (c + fa} T 7. i«*r;«
% - S.86- k*,'
f-<te
VQ
R - V(f? + %x' * V:8.w* * 7.7/
6*.
T?
JS.'SG + :il.7^ * ZO.& k&!

PROBLEM 8.15
8.15 Determine the smallest allowable diameter of the solid shaft ABCD, i
that Ik! = 60 MPa and that the radius of disk B is r - 80 mm.
T « WO N ■ ro
SOLUTION
2M^ = o T - TV = o P. £ = £^3 - 7.W N
- 3. ?S x (oa M
M8 = (X7s*to* )0e> »/o"5)
To*** ** c
fA 6 cf D
SC2.S N-m
Cn'Ti'cA/ seer/^"! J\*es aT p&i+f P
J" 3E,** _ (Vh»+tO--«
T r a " ^
M- S65.S" ^-ko I = 600 M-k,
£. 726 *IOc »f
C - 2O.S**/0"3 ,*
el* 2c* <+l.2><lcr3to r HL2*

PROBLEM 8.16
8.16 Determine the smallest allowable diameter of the solid shaft ABCD, knowing
that ig, = 60 MPa and that the radius of disk5 is r = 120 mm
SOLUTION
5M„'o r-fyro r-±--&£** •**"*»
5*10*
e
J?.5W
J>
T « 600 N ■ m
37?
&<X>
^ I20WO"
- ZS*lOs N
MB - (2.:rwo3)(0.ii&*/ds)
-- 375 M-w
ToiNycC
e
o
« As
&Jf poi"n*f D
M » 37S" N-«n , T- GOO K)-m
r.
«i/
c~
1*1
.-s
C - I 9.$8 "to * **
©I- 2c ~ 3<?.2*/o*J*v, ^ 31.Z
CVlMI

PROBLEM 8.17
8.17 Using the notation of Sec. 8.3 and neglecting the effect of shearing stresses
caused by transverse loads, show that the maximum normal stress in a cylindrical
shaft can be expressed as
SOLUTION
Mo^iV*^**-! bei'uKiV*! Sire's
J
61
M) + M]
)* + (j^ + A/,a + 7*)*
c
J
i. T«,
hJSU
2 XJ " J- /Myl+ Mi
C'rMAft - ^

PROBLEM 8.18
8.18 Use the expression given in Prob. 8.17 to determine the maximum normal stress
in the solid shafts, knowing that its diameter is 36 mm.
175 mm
SOLUTION
%.Un I L353
HCt_7 W-*
■Wees . kV
M2C= (0.l7S-^2.t67x|04)
- ^66.7 fJ-M
Bt-Jr
0*1 W0t*t (?**
+ Mj
70O W*h*
Goo W'W*
A+ poi'-tf C ^MV". * V3^orT~iiS77* = 5S3.3 tf-wi
M po.vf D 7M/f I^V ^ -Jloo* + 233.31 = 737.9 N-rn
R>i A"t £> 15 cri' I i'c*^
C - i ol -■ )2 mm * l8ylo~z *n
J- ^CV r 16^.90 */0"
f*M
* * l64.9d>*/o"
kv>
6L>* t[ t^T^V 4 VM,-trV*T*]
\8* IO
-3
\CH.1e»<0
77 J 737.9 + V 737.9 * + 600v ] - /SV.^w/O6 P^
\ ***.** MP-

PROBLEM 8.19
8.19 The vertical force Pi and the horizontal force P2 are applied as shown to disks
welded to the solid shaft AD. Knowing that the diameter of the shaft is 1.75 in. and
that r,u = 8 ksi, determine the largest permissible magnitude of the force P,.
SOLUTION
LeT P2 loc in ki'/>«.
?*-Wi =0 6P, - 8Pa r o P, = ^P2
Torque o</<V potion ABC T - 5 P2
iP*
Sending im iii.r.,2««.7x-» p-tawe
e
+Om *
* lOii
Bernef*** »*» v/CirTt'c*/ P44ne. t
3i,
F
Zoi.
Mc^ = 'o-iPt
Me2 -- 3P,
- 4P
C'r-'AictJl foini is j«"f ■/« "+k< Af-f of
Tt 8P2
M^ ^P,
Hz = £ P.
dU /.7^ *n C * ±J -- 0.875* iV, J~- %(o.27s)'
* 0.1?077 ,■*"
;^s - 873 Jh>.
P^ - 0.873 k;^s

PROBLEM 8.20
6
er\d\v\a lr> hov■-z«>*>ta/ pJltkme. . ''
8.19 The vertical force P, and the horizontal force P2 are applied as shown to disks
welded to the solid shaft AD. Knowing that the diameter of the shaft is 1.75 in. and
that ziu = 8 ksi, determine the largest permissible magnitude of the force P2.
8.20 Solve Prob. 8.19, assuming that the solid shaft AD has been replaced by a
hollow shaft of the same material and of inner diameter 1.50 in. and outer diameter
1.75 in.
SOLUTION
Le"!- Pa_ be ".r\ kip*
2"Msl>.,+ * 0 CP, -8Pt =o P, - £Pz
To^ai/*! (J^*"1 po^Ti'oin ABC T" ~ 8 1j
♦ Pt if.
C P.
10
Z»
|»,w
10 rn
OCutflitO* iV> Verti"c*X pj/avte. .
MS2= 3P,
= 3-jp.
C^it'CAJ? pof*f r"s JOS'/" +0 "He /*ff efT por'-tf C
c.= ±«L • aj?ff in.
Ci " a Ji, r 0.75O •"<
J = KCo^C,^ - 0.4237C ,V
Q.87S-
0.47376

PROBLEM 8.22
8.22 Assuming that the magnitudes of the forces applied to disks A and C of Prob.
8.19 are, respectively, P} = 1080 lb and P2 = 810 lb, and using the expressions given
in Prob. 8.21, determine the values of rHand tK in a section (a) just to the left ofS,
(6)justtotheleftofC.
SOLUTION
F^ov^ fV<?b. ?. 19, sk#H Ji'ame+ei^ = 1.75 i'm.
C^J = O. S75" '"«
J - f C* * 0.92077
To r^ we o«/e«r oot^Tio* A SC
8e*d
405 ?b
yoso M.in
|o#>&> I 11-3.M2 it. Iltt A-
tet> JuAT 4d +k< W* of poi^-f £
* loaajtfe
JviO tk - in
T^ 6480 ^1-iV,
~ GSSo j>3i*
"^j^tSX^^^ + ewjo] = 67K>ps-
T- W -$£■ - 2L
Soft = tV- %\.i*-z\.%0 - HC-40
Mc°a/&* 43« c*»44.*<* * 30O* A-i\*
* 0.^077 f r
■Shea/*
mome
Hi

PROBLEM 8.21
8.21 It was stated in Sec. 8.3 that the shearing stresses produced in a
shaft by the transverse loads are usually much smaller than those produced
by the torques. In the preceding problems their effect was ignored and it was
assumed that the maximum shearing stress in a given section occurred at point
H (Fig. P8.21a) and was equal to the expression obtained in Eq. (8.5), namely,
th = jVM2 + T2
Show that the maximum shearing stress at point K (Fig. P8.21£), where the
effect of the shear Fis greatest, can be expressed as
** =
J
^(Mcosflf + iicV+Tf
where P is the angle between the vectors V and M. It is clear that the effect of
the shear V cannot be ignored when tk z xh . {Hint. Only the component of M
along V contributes to the shearing stress at K.)
SOLUTION
(b)
Doe -t V
t = J * 2c
Doc +o T
Fo*' ex 6\VcjPc c*A across r|s eljVtweTei*
Fov* o. ctVcy/av section X r "5«J"
9 It (iJXfc) ' * ~
Since +Use £Kea^!n« srfvesses lio^t +^a same ©/Ye»Ma^Yo**
Be*Jm*i s"rr«iS a/r poiif K. Si ~ -*y— = ~^—
U = C Si** C* - C siwg-/g>) • Cc«/?
6** J
L) sift 4 M&liy* 5 ciVcy/e,
cross-secfio*

PROBLEM 8.23
100 mm
8.23 The solid shaft ABC and the gears shown are used to transmit 10 kW from the
motor Ki to a machine tool connected to gear D. Knowing that the motor rotates at
240 rpm and that r„, = 60 MPa, determine the smallest permissible diameter of shaft
ABC.
SOLUTION
90 mm
IP*IP* _
T =
_P__
397. 2<? W*^
G&uy A
FrA - T - O
F=X r 397.89
^s * L«F = OoOk/O^Yh^i^ « HHZ.I ti*r«
-»
C = IS.47«? x/O'^ho
oh *C - 37.0*/0*Si^ = 37,Ow>

PROBLEM 8.24
8.24 Assuming that shaft ABC of Prob. 8.23 is hollow and has an outer diameter of
50 mm, determine the largest permissible inner diameter of the shaft .
100 mm
90 mm
SOLUTION
PVovw. P^ok. 8.23
rot*/ e<r Tr&.i-' ."■He J Pr |OkW
VloW speeJ = WO irpwv = 4 Hx
tjf r CO MPa.
IV!
Ge*r A
Oi
*?o *iom
HHZi W
6eut«iiii^ Kio^e^-i 4 B He ~ Lw F = (|oo>'io"*)(4Mm) - 442.1 ^.^
C. 2* C " -£m
C0^ -i^or X.S'xiO"3**
c,4 - a"
■r-1
3?<?-GX5"x/0" - 151.HZ v/O
-^
232.25 x/O
- f
-j
Cz ~ ^I.*?C7k/o *vr,
ctr - 2C- - 43.<*3*IO"
►V*
* 43.<?

PROBLEM 8.25
8.25 The solid shaft AB rotates at 600 rpm and transmits 80 kW from the motor A/
to a machine tool connected to gear F Knowing that r^ = 60 MPa, determine the
smallest permissible diameter of shaft AB.
SOLUTION
CO Sec /**.vi
10 Wz
t- \%o -^|*_ 28o J 3 -
ft&F
-2*e»
T - -EL. - 8Q"'Qa - .,-,, , m
Seaf C Ft = yf
Ge*/> 1>
Foircca I'm Ve</-"ri,ce,> p^K**.
Mc, = (/Zo^o^^F^ ^ I33€.7 W-»i
Forces i* hovi zot^ia-/ pjrctvte
M0j^ 0^a^/Oi)(^Fov)- /7S2.C V-m
Mcj* all** - 764. O *.»
?* - j(Vm/* rV + T* )^
C "
£c» .GZngKHILU r ^iJ_ B 37.738 x/O-'
C = 38.8S"*/o m
al**C " £7.1 */©*•* - 37.7 »»>
f»n

PROBLEM 8.26
8.25 The solid shaft AB rotates at 600 rpm and transmits 80 kW from Ihe motor A /
|o a machine tool connected to gear F. Knowing lhat r^ = 60 MPa, determine the
smallest permissible diameter of shaft AB.
8.26 Solve Prob. 8.25, assuming that shaft AB rotates at 720 rpm
SOLUTION
Gectr C FL " vt
R =
IQ6/.Q
13. ZC* * loa N
Gea«r J) V- =
T
Fo, ]*£!£ . lwhl|0*N
routes in V^TiCft/ pAa.**.
M
M« * 477. V |J-»r»
<a
^MPJ * 63fi.fi N-in
AtC: 7Mjl+ ^2l+T* a I6G4. 3 N-
v»n
rU = WM/* m21 ♦ t* ")
*>%*■»«
*■«*■» _
886. ^
-6
to v to
— - 3i."*»*K>"fc »
-s
C - 5t7. l£'*£Osho
ol - Xc s S4.3x/O~\o - ^.3 *w^

PROBLEM 8.27
8.27 The solid shafts ABC and DEF and the gears shown are used to transmit 20 hp
from the motor M to a machine tool connected to shaft DEF. Knowing that the motor
rotates at 240 rpm and that r^ = 7.5 ksi, determine the smallest permissible diameter
of (a) shaft ABC, (b) shaft DEF.
SOLUTION
Zo hf> - UoKggoo) - 132 »/os inJfc/s
6 in.
T - -E- - IS*.* >Q*- 0c- ■ ,.
F<*s f= s¥1 = *7S-4 A-
to SJ»*P+ A8C
c - ac -iut —75^ " L'671 '"
C * O.^OSl /„ ^t ZC ■ I. g || in.
It) SUN DFF T* fDrcl) * (3.5 1(875-.^^ 3064 ;«.A
Be-ipfi*^ Mo^enT at £ M£ = ^Vs75*.4) - 3-5"o2. ■"*./!».
£> '7'- 75oo
C - 0.7337 in ol - Xe * I. 467 in.

PROBLEM 8.28
8.27 The solid shafts ABC and DEF and the gears shown are used to transmit 20 lip
from the motorMto a machine tool connected to shaft DEF. Knowing that the motor
rotates at 240 rpm and that ^=7.5 ksi, detennine the smallest permissible diameter
of (a) shaft ABC, (b) shaft DEF.
8.2B Solve Prob. 8.27, assuming that the motor rotates at 360 rpm.
SOLUTION
3&o vf^, ■=
Go
= C Wz.
6 in.
- JEL - \$z*io
Co.) SUf+ ABC T=^rr= \$a*M r 3Soi ]^L
Ge«f C Fco - -J - -#?J - 6'83,fi A.
£
ge«ii«5 wweuiH ad- 8 M& = (S)(5S3.C) - 4C&9 .V it.
r** jVmst1-
i r yc. , ia^n , V1^^^? s 0.77go6 lV
C " X
7-S"oo
- 2c - 1.582 i«.
(as-^ssax) - 20*3 ,vA
BeUi'n^ v*o***+ a* £ Me - (4X583.0- 2334 .VA
J
. 3
£ * ^ 7Soo
c = o.£4lo ;*
<J - 2c =■ i. as* m.

PROBLEM 8.29
8.29 The solid shaft AB rotates at 450 rpm and transmits 20 kW from the motor M
to machine tools connected to gears F and G. Knowing that r,,, = 55 MPa and
assuming that 8 kW is taken off at gearFand 12 kW is taken off at gearG, determine
the smallest permissible diameter of shaft AB.
Forces t*
m
43*S nil
*sr*N-&
e g
=1
nam 268*
|oo7.«i */-**
Forces iw ve>r--fic+J fJa**
AC D E 8
SOLUTION
■P - M2 -
T * 6<s
7.5 Mr
lor<f^ta On fleeter C o-~Jr £.
Te ■= 4|T0^ £54. 65" M-
Forces q*\ eiea^s
Jvi
* HZIH hi
S49
AC : T^ o
CD : T " lt<?.76 M-kv,
EB ; T « O
CrificftJ' po i*4 -P.es jVs4 H*n^\Jt*f D
My = Ioo7. °f A/-^
(■/M/+ H^+T1)^ -~ 10*7.3 M-n,
- /?.7£7 *<©-J
55"x io*
*vj
C- 23.2£ *to~ ^

PROBLEM 8.30
8.29 The solid shaft AB rotates at 450 rpm and transmits 20 kW from the motor A/
to machine tools connected to gears F and G. Knowing that r^, = 55 MPa and
assuming that 8 kW is taken off at gearFand 12 kW is taken off at gear G. determine
the smallest permissible diameter of shafts.
8.30 Solve Prob. 8.29, assuming that the entire 20 kW is taken off at gear G.
SOLUTION
*= ^- = 7.5 Hz
TotNjoe Applied o-t D
Tff » T0 * 4*4.4 I M-^i
Forces o^ ^ea/-* D a^A £
c = Jl - fJJ4.H.) - 7o,4 *■ ,j
Forces iVt nor• 2©MT«.-f pfa.**.
Foi^es iw verTic*^1 p/JWwe
f
7073.5
£&5M
#48.3*
53o.5->
met w\ot~ie*t-Ts
(V.M^+T1 )WMe - •llSC.H** ■» MJN.tl1, - 1046.3 M-Wi
Hat = K/M^rU

PROBLEM 8.31
8.31 The cantilever beam AB has a rectangular cross section of 150 * 200 mm.
Knowing that the tension in cable BD is 10.4 kN and neglecting the weight of the beam,
determine the normal and shearing stresses at the three points indicated.
SOLUTION
Mfcw
-2—1
it W
PB = /.75"1+ 1.8*
0.3 m
±_0
1.£ W
tforitodui Co~po**4 o-f Tor (££ )0°^ ) ~ <*. 6 *N
Section pe^pe/-Ties
c - O. loo »vi
M r«+ a * *-£ .& , -^\IM^^} . A„ Mft
r»» • o
Mrr-*b *-£♦£. -^-^ffiH,.lwW
M- p«{*+ C
^
^
JO£»
r r_ m , Cl^ 10^(75-0 >^) ^ _ aCoo Mp^
U-IJTO -J

PROBLEM 8.32
8.32 A 6-kip force is applied to the machine element AB as shown. Determine the
normal and shearing stresses at (a) point a, (b) point ft, (c) point c.
1.5 in.
1.5 in.'
SOLUTION
hess s O. 8 in.
"P - 6^35° ~ H. «if|4«J top? V= G s'-^r ^ 3.44/5" kip*
M- (G c.vSS-a)(l4) - ^co5S5*Xs) - LS".7H4 k.p-m.
flOAf poi^i &-
'PAtpoi^ Id
j^A^ poi^-i c
O o
^--a
« agp , z.os M
~> - 3y - 3 . 3. HH'^ _ « i r i> ■
6:
*,*
A
0

PROBLEM 8.33
8J3 A 6-kip force is applied to the machine element AB as shown. Determine the
normal and shearing stresses at (a) point d, (ft) point e, (c) point/
SOLUTION
Af
P* € C3&S9 - 4.9141 fe.'p* V- C im 35' r 3.*HT fe.'ps
M * (6 s.>5S")(0 - (6c«3S*K«1 r -I)."788 k.p-iV
I - ^(0.gX3.oV* - L8o i«"
Z. \S Use
to M
>"> A4 poul e
6L
- o
11.87 k%i
A
2.4
CO A+
)omT T
%i - O

PROBLEM 8.34
8.34 through 8.36 Member AB has a uniform rectangular cross section of 10 x 24 mm.
For the loading shown, determine the normal and shearing stresses at (a) point H, (b)
point K.
SOLUTION
rr„--o
8» = o
r3y (\ZO s^ $o°
- 9(60 s/* So") = o
A+ ™e secn'ow co*rTatv\iw^ pot rvfs Vl «^ K
P = M.S" t-5 So" t 3.S97 feM V- H.S" si* So" rr z.ZSkU
M= (4.5></0&)C40x|0"S Sirs $0°) r 90 H/-M.
^ - -2 -^ - 1 3.2S»io' - ,u o^ mo
(b) A+ p*,-4 K 6; - -f - T ■= - ?4oK|0-t - \,.5Jv/o->
- I ID.c> MPa
7C, =0

PROBLEM 8.35
8.34 through 8.36 Member AB has a uniform rectangular cross section of 10 * 24 mm.
For the loading shown, determine the normal and shearing stresses at (a) point H, (b)
point AT.
(no cos3**) I?* - to s;* 3o°)C*) * o
A^Fj = o
% - « * o
By =*? fcUt
B, = 2.S1& fclO
M He
9 - 9co5 3o°+ 2.5*fS 3m3o° = <?.093 fc(J
V - *f s;« So* - 2.5^g cos 3o*> =■ Z.25 ktJ
A = toy Z^o - ZHo *»S* - 2Ho*loc ***
I - fc(\o)CwY * \\>$z*id *.
-^
I.52*K? **
M A+ p»i-^ H
(tt A-* poi-4 K
^ - 3 V
^ " a A "
^ g.25">.ft>5
2 24ov(o-6
5» " A Y" '
ZHO v {o'c
«, = o
-37.S MP*.
\4. 06 MP<*
__ (qo)0^vlQ~s")
= - 131.6 MPa.

PROBLEM 8.36
8.34 through 836 Member AB has a uniform rectangular cross section of 10 * 24 mm.
For the loading shown, determine the normal and shearing stresses at (a) point H, (b)
point AT.
SOLUTION
DZM6 ' O
By * 7.S75 *W f
i_«m6
7.875*
P= 7.S75" cos 30° + l.«M84 si^So* r 7.79V kW
V = 7-«7Ssi>3o° -I.1t86 cos 3o° * 2.25 kU
(a) A+ PoM H S. = - £ = - zYillffi r - 3;?-5 MP"
r r Jy - j gas-*/**1 . ,. - MP
ftO Ai p«.;«f K
S7>
Lv
- O

PROBLEM 8.37
8.37 Two forces are applied to the bar shown. At point a, determine (a) the principal
stresses and principal planes, (b) the maximum shearing stress.
60 kips
h = 8 in.
Sksi *« S ks/
SOLUTION
A4 He sec4
Use Mohr i Ci>^>/e
6; * 6"- R * - K>.S«? ks,"
a er

PROBLEM 8.38
60 kips
h = 8 in.
838 Two forces are applied to the bar shown. At point b, determine (a) the principal
stresses and principal planes, (b) the maximum shearing stress.
SOLUTION
V - IO fc.'ps ^ p= CO k.'ps (compressi'oia^
M - CsX\o^ = SO k.p-iV.
A * (aXs) = £ .«*
A+ poiV} b
£,= O
Use MoH*xs ttwA
R - 7 H-t67*+ 1-37S"1 ~ ll.Sr&'r ks.'
G£ * 6^-R * - 23.I51 Ift; -
r^ - -R - ll.fft ks,*

PROBLEM 8.39
839 The billboard shown weighs 8,000 lb and is supported by a structural tube that
has a 1 S-in. outer diameter and a 0.5-in. wall thickness. At a time when the resultant
of the wind pressure is 3 kips located at the center C of the billboard, determine the
normal and shearing stresses at point H.
SOLUTION
A+ SccU
V - 5 k.>
S ecf (' o* p wpe^r J«.
cl0 - IS iV C0= £ Je * 7. S" r« d * C- t « 7. O ,"*
A - vicS^Cj*} = 22.7 77 i«l
J * U- I/98.GS. in"
*, S + ^ . ^X^) + ^S£«al , a-675* 0.2*8 - O.WICS.- -
sT It im.fci (^W.ii )(i.o)
f loft fc^. iwi
3ft& |^-^^L*
288 fc*p->

PROBLEM 8.40
8.40 The steel pipe AB has a 100-mm outer diameter and an 8-mm wall thickness.
Knowing that the tension in the cable is 40 kN, determine the nonnal and shearing
stresses at point H.
20 mm
SOLUTION
225 mm
i
7p
:j—r>- Zola;
175
Ho c«.s2o"r S'UM kW
J
Sec-T Ton pA?pe/-Tfes
r ' /\ Z.3H * /o-»

PROBLEM 8.41
10 in..
3500lb-in.
7501b
Uo<m*J S+*«s ok W
8.41 The axle of a small truck is acted upon by the forces and couple shown. Knowing
that the diameter or the axle is 1.42 in., determine the nonnal and shearing stresses at
point H located on the top of the axle.
SOLUTION
Tn« be*^ iin<=r yv%0^e^T e^usi'n* Y)t>rv*.*d STVC&S
ost c>oi •* t H i'4
C - "k d ~ O-TI ■'"»■
&--■*£ , .fcooo 1(0.7 0 r . ^K3>M? ,
= -z\~2> k*r
0.3<Hf6
PROBLEM 8.42
8.42 A 1.5-kip force and a 9-kip-in. couple areapplieilat the top of the cast-iron post
shown. Determine the normal and shearing stresses at (a) point H, (ft) point K.
0 kip ■ in.
SOLUTION
Cfict-v^eTf - %-S in.
A+ +U se^fro^ co^ift-.Vimc. poiVii^ H <*-*d K .
P = o
V * us ki,
d - 2.5 t« c = "k** = i.ZS in
A = TIC2"- <4.90<* ,"^ 2*$C<I - 1.^175" m" J" = 2X= 3.&?>$ iV
F>^ *. se^icwA Q.--|cS r '-30 5LI I*1

PROBLEM 8.43
3500 lb ■ in.
8.41 The axle ofa small truck is acted upon by the forces and couple shown. Knowing
that the diameter or the axle is 1.42 in., determine the normal and shearing stresses at
point H located on the top of the axle.
8.43 For the truck axle and loading of Prob. 8.41, determinetheprincipal stresses and
the maximum shearing stress at point H.
SOLUTION
-4 »
fi.2S J**."
^l-Sk*i
6T -. - ^* - /0.66" kft.-
R ' 1/(^)m+(c.23)a- - \S.34 ks."
eb - 61- R * -23.© k$r
r^ = 1? •= 12.24 !«#■

PROBLEM 8.44
9 kip ■ in.
CO.") Point H
> 3.3* l«
W
0>> Point K
8.80 lev
2.?3 Ift,"
j;
j« t
T
JW2 A 1.5-kip fbrce and a 9-kip-in. couple are allied at the top of the cast-iron post
showa Determine the normal and shearing stresses at (a) point H, (ft) point K,
8.44 For the post and loading of Prob. 8.42, determine the principal stresses and the
maximum shearing stress at (a) point //, (ft) point K.
SOLUTION
Fr°v* H.e soJotiou of Prob. 8.HZ
X
Sc - O
R* 3.3H ksr
SI = &4P - 3.3*YArw
6b B 6"c ~ R = ~ 3.5 -< lev
3.SY fc*,'
6^ - 6"c - R - - ?.4^ feft,-

PROBLEM 8.45
IMHotf.* TGb/ 1ootf*r«
8.45 The steel pipe AB has a 72-mm outer diameter and a 5-mm wall thickness.
Knowing that arm CDE is rigidly attached to the pipe, determine the principal stresses,
principal planes, and maximum shearing stress at poinj H.
SOLUTION
WepJfAcG fks. -ro/^t^S oA C a^J
E t>j aft &ct\si\J&.je*A.\ -hpirce -
coopJ^ S*j.s4etv, oJr T>.
FD = 9-3 = CW *
+ (sx/oi)Ci?o>./o"'")
x 1440 W-m
A4 fie
M - lC*to%)(iSo*tcF* } r <=IQO \\).to
Secff
0*
p^oiff
e/4ics -: cS0 =■ 72 w*v, cer i at " 36 mm Q = C0- t =• 3.1
-3 .. x
J - ZI - I. 1277 x /o~' m'
For WJf-p.^ Q = 4Cc^-C^)* U.2H3*(0*^ - ll.^3Klo-t^3
H
-*• SS.o MPA

PROBLEM 8.46
8.46 The steel pipe AB has a 72-mm outer diameter and a 5-mm wall thickness.
Knowing that arm CDE is rigidly attached to the pipe, determine the principal stresses
and the maximum shearing stress at point K.
SOLUTION
etn efu'lVft.J'evi to^cc - coop Jr.
•A
7^ r (^/oM^UoWo'*)
M - U+\c?V\so* loz) - <7oo aAm
J - 21 - '. '377 k/©~c m"
tvit*n
»5H.6HRi
5*^.6 MPft
T* " J ' M*77» /©-* 43*e nKa
K 27.3+(27.3 *
Sc - ~ H=T * -27.3 MP«_
r,
**>**
r f?
51. H nPa.

PROBLEM 8.47
8.47 Three forces are applied to4-in.-diameter plate that is attached to the solid 1.8-in.
diameter shaft.45. At point //, determine (a) the principal stresses and principal planes,
(b) the maximum shearing stress.
SOLUTION
2.5 kips
2-5 k
&
r
^ h
A * -rTc"" - ^.545 ,V
J* - 71 ~ I. 0306 in11
Gt* |c* * 0.486 in3
^> - l£ + va - CS"Xq.t) . rg.rKo.itf) _ c C7/. i^-
Xok.'p. in
-o»A &. se^i' Ci^/**
.fr.7/5" fe*i
e[
1 X
V
1*-H.7jS-*
(b»)
V
/ £.C?fc
y
Ca.^ 51* 6Te4 R - 3.7«? fcsi"
©* = 3S.7*-* ©fc^ \23.7'
6(!«0
(b) r^ » f? - 6. 15 ksf

PROBLEM 8.48
300 mm
8.48 A 13-kN force is applied as shown to the 60-mm-diameter cast-iron post ABD.
At point H, determine (a) the principal stresses and principal planes, (b) the maximum
shearing stress.
SOLUTION
A+ p*i*+ D F* ^o
O -12- - 5
=■ -l.oo Z t 0^75"
- 1.8 fc
kM-to
0.7SkW-i^
I £W
H
1.8 I^N-m
1.0 |(lJ»|M
Se.c-fio« p/viffe^'f>*£S
A* *C
z „
2.8??zfv/osfr,^
I - %CH - 634. 17*/o3 mm"
. !>£//<
Q. * ic* - 12.00 */o
3 ^ 8
Fo^ *- Se^i Ci'cx« vj*. - a
WH H ^TT fe'" Wp 87.J3 MP*
£|f T It " L2723X/0-* \63eW/O^60*ff>-3) " **<* M^
tt.tt MP*.
20. P4 MP«
W ^ r a r -H-J.565- MP*
5 6> 6*c4pr if.a MP* -*
(W) f«» R ■ *»S.l MPol

PROBLEM 8.49
8.49 Three forces are applied to the cantilever beam shown. Determine the normal and
shearing stresses at point H.
3 in
2 in.
A+ poi»+ H
SOLUTION
^+ Hi SCc4iw* Co<-»4ai>> i'ha pot*
Sec4i
4.".
33 kip-in
A ^ (HMO » *t .■«■■
^_JE + 4£ s --^ + ^^ « 0.375 «»;
2 A 2 2H
A ^ IT ~ JW T "7^
0.375 k»«

PROBLEM 8.50
&50 Three fbn^ are applied to the cantUever beam shown. Determine the normal and
shearing stresses at point K.
4 in.
c<?vf>M
So k;f*wi
SOLUTION
He
SeCT i ov\ prop crTics
Ij= £(0(4? - 33 .V
A4- point K
~ , _£ - Utf. tlii - - *L_ £-33^0 , <-
3o)(-^ __
3a.
= \.7*»a. ksi
22Z44
3W
£1
f -
(n(«n - ^ m1.
• ^.S rM.
Zk it ' Wo~ '
K
o.loH^ksi
"> 1.7<U ksi

PROBLEM 8.51
8.51 For the beam and loading of Prob. 8.49, determine the principal stresses and the
maximum shearing stress at point H.
SOLUTION
'0.1*75*:
§*„ - 0.375 Kv
fH - O.I25" k*i **~
cr(ks.-)
■O.I875*
Q-IZS ks:
0.375 k*,"
- 0.375" .
O.I87Sks;
SI =* Sc + R r o.wi3 iAi -*
ZL*. ~- R =■ 0.225 k*i -*
PROBLEM 8.52
5.52 For the beam and loading of Prob. 8. JO, determine the principal stresses and the
maximum shearing stress at point K.
SOLUTION
m in.
K
6e-' "HP r o-8->£ *»'
-ft..* - R - o.Hoz k%{

PROBLEM 8.53
8.53 Three forces are applied to a steel post as shown. Determine the normal and
shearing stresses at point A.
30 mm
%ok\i
SOLUTION
M^ (Sox/cPKtoono"*) = ^OoO /0-w
1
Stresses <^ p*i>t U
^-rl6.fe7Mto-J&£«MP«L-* 27.72M/3a = -|V.3S*MPa -
-« i
H I*t (?.C<ix|o-*)(60*icJ»)
H.3S MfV
i.g&hPfc

PROBLEM 8.54
30 mm
20 W
40 mm
100 mm
40 mm
60 mm
\70 W
8.54 Three forces are applied to a steel post as shown. Determine the normal and
shearing stresses at point K.
SOLUTION
At "B** section Cw\-Air\i'v\+ poiriT** H a«w/ k.
P- IZO JeW (ce~Y>resSio*)
H2 - (2o*i<f)Qoo»td*) - 2000 N/*m
- Iiooo W« km
S-Vresses -a* p©f*t k
\2Q*lO% QloooY&oxto*)
+ D
7.2 Mo"* g.44«/0-4
* -/C67 M?*.- 36.39tfP«. -f O =• - ^3./ MPa.
3 IVj
2 A
3 zo*i<? x 4-l7 MPa
1.2*10-^
54.T MP«.
HJ^MPh,

PROBLEM 8.55
9JSS Two forces are applied to the small post BD as shown. Knowing that the vertical
portion of the post has a cross section of 1.5 * 2.4 in., determine the principal stresses,
principal planes, and maximum shearing stress at point H.
SOLUTION
500 }\^ -|W«
Moment W»l (?"f s5"^)0 A--rVre.«
f - 3.2,^2 -f (6-0£
M ^
^ 4 k
3.25 £* O
» - C?26o £ -&-m
V2 r - gGOO ,#. ^ M2 -- -226© JL-in J M„ - -(H)Ue>oo ) r - 2*/ooO -ft- /V».
A- (i.sXa.V^ - 3.6 ,*„' Iz = ]4C?.<0(i.s? = o.£7£- ,-„
1 6001
2- 3.C
*« * I £ = I -I2!2 - ^ p-
1
T
^' 6*e.+- R * l5C)t P* "*

8.55 Two fijrces are applied to the small post BD as shown. Knowing that the vertical
portion of the post has a cross section of 1.5 x 2.4 in., determine the principal stresses,
principal planes, and maximum shearing stress at point H.
8.56 Solve Prob 8.55, assuming that the magnitude of the 6000-lb force is reduced to
1500 lb.
SOLUTION
Low pone*
F, - Cg°°*s'-7° - Ho Jk
7 - S.X5 Z * (6-l">±
M *
•c 7 k
3.ZS S O
Ho - 4Bo o
= -*260 k A-in
A+ +Le sec+i ow ConTO-lVl/Vl^ poll
A - (l.5X2.^W 3.6»Vix Jz - ii(^Kl-5)3 * 0.675 i^1*
H
t. 625 psi
i
6**= 6"t4R r HO p8l-
6*b =■ 61 - R * - 2784 f»s/
?Ur R = W67 p»;

PROBLEM 8.57
50 mm
0.5 lcN
8L57 Three forcesareappliedtothemachinecomponent^^jDas shown. Knowing that
die cross section containing point His a 20 x 40-mm rectangle, determine the principal
stresses and the maximum shearing stress at point H.
SOLUTION
2.5 kN
0.S khJ
sec^T to* Com *^«Vv{*\a poi'tor H
F;=--3W, ^r-askUj F.T-a.rkW
M*~Dj ^J T CO. i$o )(*&*} * 37S" W-w
1SNi
A- UoU^o)^ soo w,^1
= 800 * /D"d m
I» " &&)(&}*•: 2G.CC7*IOs m«h
r 2£. Ml yto'* wv
b« ' A" " XT $00*10-* ' K.AC7*/D-' ' ^ * * ^
s
4*6-
ftr J(™JZI)>' +C4.68701 = 13.0571 MP*
<5** 6-t4 R * 25.* MP* —
+«, 2dp - -gr- r ^ ^<_ - 0.3846
Gfc* 10.5° , B^ IO0.S*

PROBLEM 8.58
0 5kN
8.57 Three forcesareappliedtothemachinecomponentvi^jDasshown. Knowing that
the cross section containing point H is a 20 * 40-mm rectangle, determine the principal
stresses and the maximum shearing stress at point H.
8.58 Solve Prob. 8.57, assuming that the magnitude of the 2.5-kN force is increased
tolOkN.
O.SkW
10 kV
.3 lAl
1500 rt-M
SOLUTION
£<jt/i<JaJe#'i -fr^ce-coup-re sy'S'Tew^ &jt
section c&v\i*.t'*\ in* ooiwj H.
F* - -2>kW J FJ--0.SUU> Fx= -lo 1<W
Hy-Oj My- (0.i5oV|0oo?> * /^0Z> W-*i
Ma. » - (0.l5o)(5OO) - - 75" rJ-»*»
^ - Soo «lo~c i*i
6i?
A lt ' sootier* %.w7kp-»
* £ L^I - 3 I poop ^ ,Uc mPcl
2V-375 MP«
6^t ^- t? » -lo. l« MP«. -*
^*l 36P ^ "5" r 21^75- r L S"385"

PROBLEM 8.59
75 mm
150 mm
t = 13 mm
8.59 Three steel plates, each 13 mm thick, are welded together to form a cantilever
beam. For the loading shown, determine the normal and shearing stresses at points a
and 6.
SOLUTION
Gej(jiva.Je^ 4-oir£< - cooyJe system a,i
R, = <* Uw, 5 = - 13 k»J , Fj = O
Mx r (0.4oo)(lS*/o5") - 5200 W-i^
M2 - O
A- C?Vl50Xl3) + 0^X75-*0 = 4537 ^J"
'"or pd
4
mm
7. 33)S* lo"6 »,
31
T
For pdinf o- Qj, - O
For poirtt b
Qy * 0
L
J
HS
A* ■= (CoXlZ'i - 780 ww2
-& = o
J I*- -
Wp..--tl *■ "^-t^
- (52oo^(37.5x/o'*) (3frc>X- 1S»IQ3)
3.^3 03 v^D
7.3215 x |0-,fc
■= 57. t> MPct
= 3.3? MPa + 6. IS" MP*.
^47 MP*.

PROBLEM 8.60
8.60 Three steel plates, each 13 mm thick, are welded together to form a cantilever
beam. For the loading shown, determine the normal and shearing stresses at points d
ande.
SOLUTION
E^uC^A^e^T force - cookie, s^sfcvwi eA
5ecK©n coniammA Do i pits a. a*J b
9 kN
75 mm
150 mm
if = 13 mm
A - (^^1^X13^ (l3"X7S-20 ~ **537 m*,*
= 45 37 Wo"* *»*"
IM- 2 [tjO^X^* + (SSo)G%}($?.S'Csi'l + Ji(\*\(is-2CY - 3.<?3t>3*Jo6 ^M*
Iyr X[n^l3)0so/] 4- -fe(7S-2CY\?>)2 - 7.3ilSv/o*K.^ - 7. 3£/5*v/o"* Ki *
31
45
1*1 >*»
T
A4- pout e S£^ -^ - ^
For foini oi A* - (6oXi3) - 7S*> ^
x - 4S* ^ ^ = 3/
Qy ~ A* / = 35; / y /o3 rw3 =• SSI I * \xr€w?-
For- poi'rtf € Qy = O ^ Qy = O
M p*in4 * ^ , ^ - 2^
yi.^303x|0~* 7.SXl5>to-c 42.3_MKt-
Iv ~ i " 3.«»»5^iO-* " 7.3a/r«10"6
- \Z.1H MP*. —•
^ - o

PROBLEM 8.61
8.59 Three steel plates, each 13 mm thick, are welded together to form a cantilever
beam. For the loading shown, determine the normal and shearing stresses at points a
and 6.
8.61 For the beam and loading of Prob. 8.59, determine the principal stresses and the
maximum shearing stress at points a and b.
SOLUTION
OkN
75 mm
61= gfc.5 MP*. ^ = O
13kN 6*b"= 57. O MP&. fb - H.11 MPo-
Po'iot Ol
150 mm
t = 13 mm
fl.
5fc.£
fc = 1^ MP*. * H3.3S- MR,
6«. ' S. + 15 = S6.S MP*. -^
Tl^ - *R ^ M3.3 MP*_ -—>
PoM+ W
ir^
^57
6k*
^7,0
ZS.S MP*
R 'A^' + C^V = 30.03 MPa
6L« * $1 + R " 58. 5" MP^i -*
6^„ * 6*c - R *■ - /.53 mPa -*

PROBLEM 8.62
75 mm
8.60 Three steel plates, each 13 mm thick, are welded together to form a cantilever
beam. For the loading shown, determine the normal and shearing stresses at points d
ande.
8.62 For the beam and loading of Prob. 8. Co, determine the principal stresses and the
maximum shearing stress at points d and e.
SOLUTION
(Si * M3.S MP* tj* ^-S3 MPa
§a ' ^y- * 42.2. MP*
6i-« ^ 6; + R ^ 42.4 MP* —
st*
* "P - U^ MPd.
peiH+ e
12.7*/
6c * '^ - 6. 37 MP*
6^ - 6t*-R " W.t» MP*
f^ * <? r 6.37 nP«„

PROBLEM 8.63
8.63 Two forces are applied to a W8x 28 roiled-steel beam as shown. Determine the
principal stresses, principal planes, and maximum shearing stress at point a.
90 tolM W8 X 28
SOLUTION
Fo/- W 8 x 22 roU*J aieei set-fV
A - #.*£ .v,1 y d= s.oq ;*, bf - g.sss «v
A+ "Hie sec^*i*oi* co^T^tViriA poTrtTS a -st^d o.
P * - 70 kips, V* *o fc-fs
M- ft^O-ft-O^O = 117. S fcr;p. m
y -~J;J- tr - */.©S- 0.4*f r 3.525" ,V
j r IJi.^tf -" 4.03- 0.3-US' * 3.-7*75- ;„
Af ■= lo^-tf - (£.S3S)(o.MS) - 3.0 38* ;***
X =
fi-6^2 ks;
vgu _ (zo'yCit.s'io)
It* C<H.O\(0.38S}
* g. 263 k«i'
S^ <$~e - R - - 13.23 ks; -
•ru** * » *.*' fc»'"

PROBLEM 8.64
90 V W8X 28
y
At poKAi b
8.64 Two forces are applied to a W8 x 28 rolled-steel beam as shown. Determine the
principal stresses, principal planes, and maximum shearing stress at point 6.
SOLUTION
A - S.ZS ,^ d r & 06 in , bf = 6. S3-T .'v,
At Tne SSctJovi CoioTa-iwiVia poiiaTS CL a*tet b.
P ^ - <?© k.'p^ \i - 10 ktf%
VMSsSssS/SSW
P^+
r
A6V)
3.0383
1.016 1
5 00
A5 (*')
II.SHo
i.sn
13.35-1
3
ims ks.-
C-n -^t- t? * - 16.47 lf%* *m

PROBLEM 8.65
W310 X 60
9.65 Two forces Pi and P2 are applied as shown in directions perpendicular to the
longitudinal axis of a W310 * 60 beam. Knowing that P, - 25 fcN and P2 = 24 fcN,
determine the principal stresses and the maximum shearing stress at point a.
SOLUTION
Af T«C ScCTiOi CO»fa-<itii4A points CI a^\J b
Vx = - ^ few
\/y - - 25 khf
Fuv W 316> * 6o ro)itA s+e<?i se^+i'o*
*w>*t
t~=- 7.5"
*>1»V)
I*- IS^X/D6 jwn)"^ l2<?*/o'c h*^ X r |g.3x/o' ^ ^ IS.3*|o" *»'
*,= - ^ + 7S * -*G.S
**IW1
m * lo
!S.3*/o
-c
t<
ge*
■^v*
?g=^yi
^3E1
FT* ^=
S2.8*nMPa_- 4l.7oS"MP«. ■ II. HM MP*.
SneoMMa s"fr«ss <aT point Ct
(-3S^o>Xlg*SWC»"*Xl<H-?.r>'<o~>)
(a? xio"6, Xl3.1 */o"s>>
-6.2^5MPa+ 2.\ol M?*, - - ^. /gg MPa,
.^lUMM
*«*«
, iLiiW T StS-?z Mpe
M.I88
Rt -/(U^tf)l+(V.t8Sy - G-970 MP«
6^' 6U + £ = IS.£4 MP*
6*» » &« - £ * - mo M Rd
TTk^ * 1? - G.^7 MP*.

PROBLEM 8.66
W310 X 60
8.66 Two forces P, and P2 are applied as shown in directions perpendicular to the
longitudinal axis ofaW310x 60 beam. Knowingthat P, = 25 fcN and P2 = 24 fcN,
determine the principal stresses and the maximum shearing stress at point b.
SOLUTION
Mx - (USX**) r *S M-v*
Fo*- W3|O*60 iro-Med stee^ se^f'o*
ol - 3(53 ^m^ Lf - ^03 m^tj tf - 13. J
t^r 7.5*
Ij - Ig.S^/o4^^41 - /2.3*/d* ►m*
*s *
±7" '^ y'°*6
I'M**).
= -f8.Z3 MP*.
J3r-«; A3
rsr^-ara
Sdw^'i^ stress ©."f Por*»t b.
+ —
r- - -Mi
*2 I,i
•X **w
«w
s«.2S
Save - "^^ " "^-W M?«
6^ '- 6L.4 R - I.*?? MtV
6*^ - 61„-K - -^0.3 MP*
' **+•*
* £ - £G. I MP^

PROBLEM 8.67
8.67 A force P is applied to a cantilever beam by means of a cable attached to a bolt
located at die center of the free end of die beam. Knowing that P acts m a direction
perpendicular to the longitudinal axis of die beam, detennine (a) the normal stress at
point a in terms ofP, b, ht I, and p, (b) the values of fi for which the normal stress at a
is zero.
SOLUTION
1
a
I.- iW Whk"
e -
M>(h/g') _ My(b/Q
tfMy
€M
IX
ht1
-*>
M.
p - P s;n/9 r - P cos/2| ? ~ J k
(10
6- = o
h b
0 - W(£0

PROBLEM 8.68
8.68 A vertical force P is applied at the center of the free end of cantilever beam AB,
(a) If the beam is installed with the web vertical (ft= 0) and with its longitudinal axis
AB horizontal, determine the magnitude of the force P for which the normal stress at
point a is +120 MPa. (b) Solve part o, assuming that the beam is installed with 0=3°.
SOLUTION
For W ZSO * 44. & roJUeA *leei>
ion Con
My "- P-? cos /S j M^ - Pi s-vi/a
6--
M,
K
=>jr
S, Sy
ri/y ~
«*
1.2s
!hdr^ + °]~'=M*i<t»*s>-i fcw
p - *2o*io4 [cos 3° s.-*y T'
= 3?-7 *W

PROBLEM 8.69
*169 A 500-lb force P is applied to a wire that is wrapped around the bar AB as
shown. Knowing that the cross section of the bar is a square of side d =■ 0,75 in.,
determine the principal stresses and the maximum shearing stress at point a.
SOLUTION
<T- o
0s,\3 s^€^pos,"+i'o^ : X' 3.90S £ ? 3/^ ^S^ r 347o
■ Si
Ct
w*x
- 3*f70 p*; ~«*
CWo
Lw-^ ~ 3H7& p%{

PROBLEM 8.70
20 kN
6 mm
*8.70 A vertical 20-kN force is applied to end A of the bar AB, which is welded to an
extruded aluminum tube. Knowing that the tube has a uniform wall thickness of 6 mm,
determine the shearing stress at points a, b, and c.
SOLUTION
SO mm
Q. r (W)(<Vi)r 4Ll3Gx/o* mm1* * 4. l36*/tf"V^■
T _ Z\oo
t-
ZtCL ' UVG*lcr*X4.IS6*wrm)
T^m.»3 verse sliea^ : V - 20*10* N
Point b Q/= (35XOft7^
- 7. OS v/o »«»n " 7.
- 42.31 hPa.
-te>zz
IT
£>$"* /o'*hi*
?o;^ a
3Ea
Ne"l ske&</^*^ sivesS :
it (Zoo&wcryCt + tcr*)
- W.70 MPa
?of-,t* r-^.SI-O - 4*. 3 MP*
P»i*t t> r- 43.31- ii.7o r 3,0.6 Mft*
fc,-*+ c X * 42.51- 21.3^ " 21.0 MP*

PROBLEM 8.71
20 kN
50 mm
*8.70 A vertical 20-kN force is applied to end A of the bar A3, which is welded to an
extruded aluminum tube. Knowing that the tube has a uniform wall thickness of 6 mm,
determine the shearing stress at points a, bt and c.
*8.71 For die tube and loading ofProb. 8.70, determine die principal stresses and the
maximum shearing stress at point b.
SOLUTION
I - ^fcoYlOo)3- 7^(3g)(88)S- 2,OOS7k/oc w,*/
= Z.oo&lxto-'- **-*
T 2)oo
r -
2ta " ftX6«te-*)(H.l36x|o-)
42-31 MPa
IL^
TVawsve^se sAea/* ■' V - 20*10* }\)
■>> ^
Ne+ sJiWirin^ stress f - 43.3/- l|.7o - 3^.6'MP*
l.7o
65-7
6C"= £6"* 3-?. 85 MP«
6U^ Gl + R =• 77.7 MP* -*
T^ - t? - Mt.9 MP* -*

PROBLEM 8.72
10 in.
"\JL^0.15m.
T 9 kips
*& 72 Knowing that the structural tube shown has a uniform wall thickness of 0.3 in.,
determine the principal stresses, principal planes, and maximum shearing stress at fa)
point H, (b) point K. v
SOLUTION
A+ +^« Sec-Tiovt Cooi&mi'n*! poinTS H *-n*f K
V r <? k-'p» M -(9)(i°) = ^o k;p. i*.
T = (9X3- ais^ r 251 &S" k.'p. in
a =(5".7)(3.7) r 21. o<* ,■„"■
r =
^S-.6^
- 2.0X7 Us.'
H
I
2ta ' GXasXsi.oq)
Q* ^ (3^(001- (2.?yi.l)fo.8Sv) R
- 2.09S-S" in1
Lh - v L* It ' (it.iiszyo.s^ ■*. ©" Ks.
r *J(\l£&y+$.ozi)x=: £.607 ks.'
^— -«. -4. —^
J
I
I
1 -"*• ""''—
J
1
i
—an, *t. ..1
H
12.576 ks.-
*■'tor? ie%;
(t) Poi*+ K:
t * 1
6^ O ^ 2.0*7 + 4.3« - £.425" ks."

PROBLEM 8.73
40 kN
2.2 kN/m
V(ku^
J£*l
-?c.n«
l3.o2
-3?.*??
M 0^**0
go. 86 5"
5K*pe
W 36© x 3?
W Sio * 38.7
W 350 > * H. tf
W 2oo v Si
3(io*mp?)
Sl%
SW ■*-
•T3S
5i X
8.73(a)Knowingthat<fa" 165 MPa and r-i= 100 MPa, select the most economical
wide-flange shape that should be used to support the loading shown, (b) Determine
the values to be expected for am, r„, and the principal stress o^, at the junction of
a flange and the web of the selected beam.
-7.* R* + G?.aX7.*'Xs.O + WOO*. 7) * o
Vg r £?.<?* -(-?.2X**.S) » 13.02 kW
Vft - 13.02. -10' - 26.98 kV
Vc ~ - *C.**S-(?.*)(£ 7) r -32.92. KW
Ma3 D + 4(22.92 + '3- o2)(»KS) r 30.865" ^-*"
M* - o
S„,(-n - -^r— = y /D. - 490 v to m
J*«
= M9o Wo* wv^a
T^j W 3|o*38.7
a ~ 3 Id) wm
t# = 9.7
►Mm
ft - H*
1VL.
80.865**10*
r /47.3 Mp*,
1L*
32-72 */0'
* " *41« " (3iovjcr*)Cs:»*icf*)
= 19.3/ MPa -*
c * 4 ■* T IS? mm yk = c - t? * iss- 9.7 - \*S^ »*i.
^r ^^--(i^Xl^.3)- 138.1 MPa
V
(^g.^Svfo5)
Rry(&)L4 r/ - v/(£7\a$)l+(lS.O^ " 70.66, MP*
61^ = % + 'R « CT.-05 + 70.66 =■ 13^-7 MP<o
'tviMfb

8.74 Knowing that the shear and bending moment in a given section of a W21 * lol
PROBLEM 8.74 rolled-steel beam are, respectively, 120 kips and 300 kip ■ ft, determine the values
in that section of (a) the maximum normal stress am, (b) the principal stress amM at
the junction of a flange and the web.
SOLUTION
M - 2>oo ic;P -ft t 36oo k.'p i* V = Uo /t.ps.
Fow W Z\ * 10 I sUf><- cl - a/. 3C ;» bf - i2.2^o ;^ £^ = O.800 ;*y
Af - bftt ~ <?.S3* In1 j - ^ (c+y4 1 * 10.28 iV
Gi. * Afj -- IOI.07 ;**
^ T ^|_ , (U0XIO..O7) = ^;
R
. y(S)*- + V =" 7 7.334* + lo.O^^ = I3-M^i k*.*
G^ t & 4- 1R t 7.33C+ lO.^ai • n.76 fc»."

PROBLEM 8.75
'80 mm
100 mm
8.75 The 6JcN force is vertical and the force P is parallel to the z axis. Knowing that
r,n - 60 MPa, determine the smallest permissible diameter of the solid shaft AD.
SOLUTION
? = 7.-5" xio3, N
Forces ih r)o^iz.o«ta-f pietfie
|7Soo
Fo^cea i * Ve/"Ti'c«4-r p-f q. * e.
|60oo
Beimel i'^\«t rvto»v,C^.T^
At B M * -J SS2* + 240
= S7?>.\ N-v^
G^ic** See-'fi'oM is JOVT T° "H* <&?T of ^&a/- C
m
t*«
C * |<*.77kW
-3
ho

PROBLEM 8.76
8.76 The two 500-lb forces are vertical and the force P is parallel to the z axis.
Knowing that rM = 8 ksi, detennine the smallest permissible diameter of the solid
shaft AE.
SOLUTION
? * £££.67 Jk
Aft- T=o
CD-. T = CrtCsao} = Zo&o ivkJk.
DE". T = O
Forces in non'ZtJ^T^J p/K*w«.
333. SS & &£.£? 33i3S
Forces i* v/e^Ticoi1 p>rai*e
.'
6
1'
I
tfC&7 i*Jk
T- 2ooo i«.Jl M2 = 3S©o ;«•&
My* ^667 in A
£4
V46672-* 35-qo*+ ^ooo1
8 xfo*
- 0.77083 in*
C ^ 0.78<* in.
ol r ^C « 1.578 i
JH

PROBLEM 8.77
8.77 The solid shaft AE rotates at 600 rpm and transmits 60 hp from the motor M to
machine tools connected to gears G and H. Knowing that vM = S ksi and that 40 hp
is taken off at gear G and 20 hp is taken off at gear H, determine the smallest
permissible diameter of shaft AE.
SOLUTION
Toraoe, ot
8
U>i^ej^e& on Geoff*. C o~idi "D
Tc = !£T0 t 420/. 7 ,■«■ A
S^aJH lo/^e*
ec
CD
Tto - o
Aft
Fibers i* vc+ie** pjAne.
DC
O
Uia.7 v
lOSto.M 4&7
£1X7. S ih-{l
n
"3
Ztso.q
F&r S = Ji^I, |0Sb-l| A
Fo =
^±f^ = «*-. 2 A
J L
35* f. X
Ghn?.£
IW
. (8C8 in"

PROBLEM 8.78
120 mm
8.78 The motor M rotates at 300 rpm and transmits 30 kW to the solid shaft AB
through a flexible connection. Half of this power is transferred to a machine tool
connected to gear E and the other half to a machine tool connected to gear F .
Knowing that r-,=60MPa, determine the smallest permissible diameter of shafts.
SOLUTION
Vl O V i Z O n-T &/ 0/«t «*
1
3179
2CS3
°tSH. 9 N.*i
-r - -£- 3o*lO* _
OP^? CCS Oil O©*!
SVia'F'i 4-o^«e*i.
MA
AC
CO
DB
Tco * 4T7.S M-^>
G-eour -fcwcea
C^i + icA>? pdi'«T is Just "f» +U ivrf
Vm,' + M/ •» T1' - (124.1 W.h,
i = Xcs= VM..VHV4T
tlv
Gox |o**
S^WO-6 »?
C - 22. 85" WO"s m*
J r 2c * 45\7 x /o"3 m ■= ^r.7 ^

PROBLEM 8.79
.150 N
100 N
Dimensions in mm
8.79 Several forces are applied to the pipe assembly shown. Knowing that each
section of pipe has inner and onter diameters respectively equal to 36 mm and 42 mm,
determine the normal and shearing stresses at point H located at the top of the outer
surface of the pipe.
SOLUTION
? -O ^ V, -- loo N , ^ = O
M„ =-Co.4«oXiSo>) - - C7.5 hJ.y*
M, = - (o.X?bK|oo) t _ ^.s* M*^
£7-5 W-»
cL- H*
y*\m
*3 *i«o
A- ttCc^-c/)* 367.S7 ^ * 3GU?^o"tMl
Nat f„ * 3.3G - 0.6^ * Z.&Z MPcl -**
^O. -2. MPft.

PROBLEM 8.80
8.80 A vertical force P of magnitude 250 N is applied to the crank at point A.
Knowing that the shaft BDE has a diameter of 18 mm, determine the principal stresses
and the maximum shearing stress at point Hlocated at the top of the shaft, 50 mm to the
right of support D
SOLUTION
R - o J Vj = - 2S0 M }
125 mm
My - - (200 i.^^XlO^teo^ - 43.3^/ A/.rv,
I"* Jc** =51 /5"3 x/o3 ►«»/
37.S|
H3.6fr
R - 7 (S^T *(.47.^T = H3-.6G MP.
*V 3o-3
6k * 1*0'
IL-- R * 43.7 MP-t

PROBLEM 8.81
8.81 Knowing that the structural tube shown has a uniform wall thickness of 0.25 in.,
determine the normal end shearing stresses at the three points indicated.
6001b
, 5 in.
20 in.
N^S*
<So k.y- »Vi T ^3° U>'p-i*
Bend.
htf- 3,v hi - W.- 2t - Z.S" ,Vi.
Not^Mft,J airesses,
- - 16.wi W
6 " I. ~ ~
Cb) W. 338S- £.33S<T l5^63 **'
i rasas' g.S3»5"
-7. \o ks.'
(GO Point Q_ i's <a>i tfi/fsi^e Co«^v\«^* t+ - O
3ktf*
**£
3E
Due +o V,
A+ p««t fe U
-J2.M5
Q2t*(/.£X0.#to-*7S)-l-07»| ;rt3
A£ poivt't C
tzzm
r- 2.O230' ;«3
1.2 tcpTT
flzza
^£
Due +o Vz
I.WS
jzszza:
Q*u r C?.?O(0.a5 ^(1-375")- O.WS JwJ
<*t
Net ske^Cri'V STre^S M poiVb _k «"J- £
11= a?/£- O.GM * O.0H7 W —
1C* 1.256 U^" —

50 mm
50 mm
375 mm
8.82 Forthe post and loading shown, determine the principal stresses, principal planes,
and maximum shearing stress at point H.
SOLUTION
F* *50 cosSo° ~ 4S.3QI kN
V*- 43.3ol kW . \4 - -^ fcM
R, * - \ZO kW
H ] H373ol kV
Ponces
My: -0<i5 X^ST?) - - 9.375* M-vn
Mj - o
H2 r -(HS.SOI )(o.Z!S)~ - 16.238 Utf.-m
IC.238 kNJ-*»
A - (|ooK/5ro) = /S" h (os wM*
=■ IS ^(O'"4 **Z
Stresses <=^ poi»\"i H
4.33 MPa
H ' » A * l5MO"s
21S MPa,
H.S'i MP*
61 r -i6*» - 14-75- MPa
R - /(^f + 4.-J31 * 15.3? HPc
6"fc * <5t4 R ^ ZOA MPa
6*b T €Tfc- R * - 0.62 MP«
P ' -Cm
t<^ : 15.37 MP*.
- - 0. ^3G

PROBLEM 8.83
>J\
50 mm
50 mm
375 mm
1120 kN
75 mm
'75 mm
8.83 For the post and loading shown, determine the principal stresses, principal planes,
and maximum shearing stress at point K.
SOLUTION
Coi-xponeeiTS or -ha^ce M poi*W C
F„ r SO <los. 36" r 43. Sol WW
f\ r -so *.-„ 3o° = - 25 fcu 5 r - i£o Urj
P - ISO i<W (cowipreisfow )
Vx - HS.3oi M, Vz r - *5>w
MM-- -C?5 ^0.375) «■ - 7. 375 HJ.*,
Ma * -(^.3^0(0.375) - - \G.ZSS IrtJ- *n
l6»H
^.375-kV).* )6.^sg kf).*
H3.301 W
I
Cp^p*
- I5X/0"1 v»z
I2 - -^(looXlSb)* - ZS.tlSxto' *****
~ 2g.l2S x/O*1, mH
Forces
Stvesses ai -00 m4 K
2.5 MPa.
Lk * * A - 3" isDiicr*
51.3 MPo.
sfc. 2.5 MPa
R - Y(£py + (ZSy- - *5,77 MP,
6Tb t fft- R *■ - ,57. * MP*
S*T 2.8* Sfc* #.8*

PROBLEM 8.84
2500 it.
Coojfb.
Forces
8.84 Forces are applied at points A and B of the solid cast-iron bracket shown.
Knowing that the bracket has a diameter of 0. % *«., determine the principal stresses and
the maximum shearing stress (a) at point H, (ft) at point AT,
SOLUTION
AY f^e section CiP^Ttti'mV)* i3oi*4* H <x*\A K
y^ - - feoo A v/„ = o
A? ire* * o.6"o?6^,viu
I* ^C* - 30. |06kIO** m*
15 OO iL- i'n
11.1 ak*<* ff«. * ^^ 3 IZ.HSS'ks.-
_.?«* * ■i(s~*-6"~.v.1* 19. ^z ks/
H
2*.«7 ksi*
^l it.sta its,-
.£ 2s00
** * - -^ia- ^ -*.^q-irfp..
K
S*. * » ^^ = - 2.487 ksi
6~« =• G~« +1? - M.fci k"
6^* r &** - R - -11.1^ k's/

PROBLEM 8.C1
8.C1 Let us assume that the shear V and the bending moment M have
been determined in a given section of a rolled-steel beam. Write a computer
program to calculate in that section, from the data available in Appendix C, (a)
the maximum normal stress am, (b) the principal stress amaK at the junction of
a flange and the web. Use this program to solve parts a and b of the following
problems: N
(1) Prob. 8.1 (Use V = 400 kN and M = 100 kN • m)
(2) Prob. 8.2 (Use V - 200 kN and M = 100 kN • m)
(3) Prob. 8.3 (Use V ~ 320 kips and M = 32 X 103 kip ■ in.)
(4) Prob. 8.74.
SOLUTION
We &r\ttrj-^<L pi/tn Values cf V and M dbfain ~fro^ AwntfiV Q fht
Vahrtcf d> bjttj tt^3 J} <W S for-fheS'^n IMF shape-
\A/e Compute
PRP&RftM OUTPUT!»
Prob. 8.1
Given Data:
V = 400 kN, M = 100 kN.m
d = 252 mm, b£ = 203 mm
tf = 13.5 mm, tw = 8.6 mm
I = 87.30 (10*6 mm*4)
S = 693 .0 (10*3 mm*3)
Answers:
(a) SIGA = 144.3 MPa
(b) SIGM = 250.1 MPa
Tram A1 oh ric'r<i/e J
^--/^^^
Prob. 8.2
Given Data:
V = 200 kN, M = 100 kN.m
d = 252 mm, bf = 203 mm
tf = 13.5 mm, tw =" 8.6 mm
I = 87.30 (10*6
(10*3
mm* 4)
mm" 3)
S = 693.0
Answers:
(a) SIGA = 144.3 MPa ^
(b) SIGM = 172.7 MPa -<S$
Prob. 8.3
Given Data:
V = 320 kips, M = 32000 kip.in.
d = 36.74 in., bf = 16.655 in.
tf = 1.680 in., tw = 0.945 in.
I = 20300 iiT4, S = 1110 in*?
Answers:
(a) SIGA =28.8 ksi -^
(b) SIGM =28.5 ksi ^
Prob. 8.74
Given Data:
V = 120 kips,
d = 21.36 in.
tf = 0.800 in
I = 2420 in*4, S =
Answers:
(a) SIGA - 15.86 ksi
(b) SIGM = 19.76 ksi
M = 3600 kip.in,
bf = 12.290 in.
, tw = 0.500 in.
227 in*3

PROBLEM 8.C2
IF
8.C2 A cantilever beam AB with a rectangular cross section of width b
and depth 2c supports a single concentrated load P at its end A. Write a
computer program to calculate, for any values of x/c and y/ct (a) the ratios <rmsJ<rm
and ^min/^. where CTmax and (Train are the principal stresses at point K(x, y) and
<rm the maximum normal stress in the same transverse section, (b) the angle 0
that the principal planes at K form with a transverse and a horizontal plane through
K. Use this program to check the values shown in Fig. 8.8 and to verify that amK
exceeds <rm if x :£ 0.544c, as indicated in the second footnote on page 499.
IvL-x—A
9s
SOLUTION
S]f](^ fhe di$+r:(bofion of-th? normal 3i*tt&
Wktrt 0 - t^C_ - PZC
I 1
m
We U5e E<f-(&fypa$>!f9fi : ^ -j-^i1'^) ^
' ^ z a xc.
br, o
tl)r„ I - £*>&<$- If. _£ - <
(?- a y
rh
e
U$ifik Mohr'j circle, ive calculate.
R^iCioW?
^e.p-H^T
^
WW
9.
to^2b0-^i ^Izt
m
Q,
m
r
,trl
%/i in*/*) x/
A/(?T£
(2)
C3J
f^J
- iy+K £bhl- -J-y-ff ^1
For (dt > 0, the angle, dp is ^) which ]$ opposite ih vh^
Mas arbitrarily assumed in FtQ.P8,C2,
(CONTINUED)

PROBLEM 8.C2 CONTINUED PRO&flflM ft 0 T ? 0 T $
For x/c
y/c
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
= 2 :
Sigmin/Sigm
0.000
-0.010
-0.040
-0.090
-0.160
-0.250
-0.360
-0.490
-0.640
-0.810
-1.000
Sigmax/Sigm
1.000
0.810
0.640
0.490
0.360
0.250
0.160
0.090
0.040
0.010
0.000
Theta *
0.00
6.34
14.04
23.20
33.69
45.00
-33.69
-23.20
-14.04
-6.34
-0.00
For xy
y/c
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
c = 8 :
Sigmin/Sigm
0.000
-0.001
-0.003
-0.007
-0.017
-0.062
-0.217
-0.407
-0.603
-0.801
-1.000
Sigmax/Sigm
1.000
0.801
0.603
0.407
0.217
0.063
0.017
0.007
0.003
0.001
0.000
Theta^
0.00
1.61
3.80
7.35
15.48
45.00
-15.48
-7.35
-3 .80
-1.61
-0.00
To check fh&t Om)( > (^ // pL± Qjifif-r^ -w~ fm fhe oco^carn
-fbt X/c -O.SHH and -for z/c - 0,T^5 and observe fhat
^my/6^ e*cre<f$ / for several /a/u« of Q/c //i the ffrsh
ease, &vi d&es not wceed 1 in fht Second case.
For x/c =
y/c
0.30
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.40
0.544 :
Sigmin/Sigm
-0.700
-0.690
-0.680
-0.670
-0.660
-0.650
-0.640
-0.630
-0.619
-0.608
-0.598
Sigmax/Sigm
0.9997
1.0001
1.0004
1.0005
1.0005
1.0003
1.0000
0.9996
0.9990
0.9983
0.9975
Theta *)
39.92
39.72
39.51
39.30
39.09
38.88
38.66
38.44
38.21
37.98
37.74
For x/c =
y/c
0.30
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.40
0.545 :
Sigmin/Sigm
-0.698
-0.689
-0.679
-0.669
-0.659
-0.649
-0.639
-0.628
-0.618
-0.607
-0.596
Sigmax/Sigm
0.9982
0.9986
0.9989
0.9990
0.9990
0.9988
0.9986
0.9982
0.9976
0.9970
0.9962
Theta^
39.91
39.71
39.50
39.29
39.08
38.87
38.65
38.42
38.20
37.96
37.73

PROBLEM 8.C3
8.C3 Disks £>„ D2,..., D„ are attached as shown in Fig. P8.C3 to the
solid shaft AB of length L, uniform diameter d, and allowable shearing stress
Tgj,. Forces P1( P2,..., P„ of known magnitude (except for one of them) are
applied to the disks, either at the top or bottom of a vertical diameter, or at the left
or right end of a horizontal diameter. Denoting by rt the radius of disk D( and
by c, its distance from the support at A, write a computer program to calculate
(a) the magnitude of the unknown force P,, (b) the smallest permissible value
of the diameter d of shaft AB. Use this program to solve Probs. 8.75 and 8.76.
SOLUTION
/. Defcrirj/ne -fJie lynKniMn -force 7^ ^y €JO£lJ1"$ iv gerd
{he s\<1ii t( Hicir forces 7^ about the Zfixis-
Z. Defzrrniv\z the component* (K)[ ^4 (F ) • <rf all fangs*
3, Dei-ermine -hh-t components Aud+d A of rtactfin at f\
by so\i\Mt*rt§ Moments a booh tr/?*, B^//* and S^^/j'ty!
tf. Determine (M»). ,(M#).,W iorpt T, jost to the kft
of disk D- ; L
<l
(y/i^rY < > indicates a e>ir>yularitij function.
fn TA<? /him>?u'W7 Aiaw-ieJ&r d recivitzJL to -the (tft of ty ^
obUuned by fiW computing(J/c^from Ef (8,1)'.
(CONTINUED)

PROBLEM 8.C3 CONTINUED
-* 3
,CL>
& Re.<ULl!in^ ihtt J-^jr^ c-jni, fh\Jb, ih*t(JL\ -~ '>
IhiS is %* required cffdhieter Just f0 the feft of 4,skD[
7, The re<juire4 diameter JV<A to ike rfyhf of dt'sk D; }*
6tfalh&l toy rtflacMQ T^ith Vifl in the ccbofe commutation.
,8.77) e s$m.llest perrtys&iljlt /tf/ue of the. d\arr\t.hr of the
^hdit h +hz largest of fb* Valoe$ chtalnej for S
fAOGHRH OUTPUT5
Prob. 8.75
Length of shaft = 300 mm
TAU =60 MPa
For Disk 1
Force= 6.000 kN
Radius of disk = 75 mm
Distance from A in mm = 80
For Disk 2
Force* 0.000 kN
Radius of disk = 60 mm
Distance from A in mm = 180
Unknown force= -7.500 kN
AY= 4.400 kN, AZ= -3.000 kN
BY= 1.600 kN, BZ= -4.500 kN
Just to the left of Disk 1
MY=-240.Q0 Nm
MZ=-352.00 Nm
T= 0.00 Nm
Diameter must be at least 33.07 mm
Just to the right of Disk 1
T= 4 50.00 Nm
Diameter must be at least 3 7.47 mm
Just to the left of Disk 2
MY=-540.00 Nm
MZ=-192.00 Nm
T- 450.00 Nm
Diameter must be at least 39.55 mm
Just to the right of Disk 2
T= 0.00 Nm
Diameter must be at least 36.51 mm
0
in.
in.
in.
in.
ih.
in.
Prob. 8.76
Length of shaft= 28 in.
TAU (ksi)= 8
For Disk 1
Force = 0.500 kips
Radius of disk = 4
Distance from A= 7
For Disk 2
Force = 0.000 kips
Radius of disk - 6
Distance from A= 14
For Disk 3
Force = 0.500 kips
Radius of disk = 4
Distance from A= 21.0
Unknown force= -0.667 kips
AY= 0.500 kips, AZ = 0.333 kips
BY= 0.500 kips, BZ = 0.333 kips
Just to the left of Disk 1
MY= 2.3333 kip.in.
-3.5000 kip.in.
0.0000 kip.in.
Diameter must be at least 1.3 89 in.
Just to the right of Disk 1
T= 2.00 kip.in.
Diameter must be at least 1.437 in.
Just to the left of Disk 2
MY= 4.6667 kip.in.
MZ= -3.5000 kip.in.
T= 2.0000 kip.in.
Diameter must be at least 1.578 in.
Just to the right of Disk 2
T- -2.00 kip.in.
Diameter must be at least 1.5 78 in.
Just to the left of Disk 3
MY= 2.3333 kip.in.
MZ= -3.5000 kip.in.
T= -2.0000 kip.in.
Diameter must be at least 1.43 7 in.
Just to the right of Disk 3 *
T- 0.00 kip.in.
Diameter must be at least 1.389 in.
MZ =
T=

PROBLEM 8.C4
8.C4 The solid shaft AB of length L, uniform diameter d, and allowable
shearing stress t^ rotates at a given speed expressed in rpm (Fig. F8.C4). Gears
G,, G2,.... G„ are attached to the shaft and each of these gears meshes with
another gear (not shown), either at the top or bottom of its vertical diameter,
or at the left or right end of its horizontal diameter. One of these other gears
is connected to a motor and the rest of them to various machine tools.
Denoting by r-, the radius of gear G;, by c, its distance from the support at A, and by
Pi the power transmitted to that gear (+ sign) or taken off that gear (— sign),
write a computer program to calculate the smallest permissible value of the
diameter d of shaft AB. Use this program to solve Probs. 8.25, 8.29, and 8.77.
SOLUTION
/, tZ^izr CO in rpm and defermNe fre^eneu •§ _ oO/^Q •
2. For each ge.CLC, determine Thz foraot T* - F' /2TT;£
Where Pf /5 j^e po*/tr inp\jt ("*■) or ouipo-t £-)&!; fheg-ear.
3. For each gear, ^e/ermine ihe force F; - T; /t- eyerfc/
On f'lie^ear and its Components (Fa)- ccv\d_(F\. .
k. Determine FUe covriponenij; Ft. c\mA flg of (reaction at f\
1*^0'' ^^TiF^CL-c^O, *,; -±±(Ft).(L-ct.)
5". Determine (/Wj). , (rtz)L, and torcjot^ Just % the /eit
of gear Qi ;
TerZTk<Ci-ck>'
Where < > indicates <x ^mc^oUri'hf funchon.
(CONTINUED)

PROBLEM 8.C4 CONTINUED
£, The minlvYivvrt gf\a^eitr ci required to fhe /tit et (J'L (S
I: ~ o
7. Kecal/H^ ihaX J-i r?c¥a»d,r/,ust l^a'u (f). =1 KCC
* V ° i i n KcJL
Thi it 1/ie requires! ft i G/vieier jost to the left oL ftar^-
8.~Tfe reavirej diametcrju^t to fht r>^t^ -f ft ear &i '£
obtained k,Q replacing 7. \tith~~T In the abore. corny Nation .
9, the smallest permisS'dk v'alieot ihe diaw-e+tr of }hg
Sf^jt is -flie l£_nest c{ the \/a\vzs obtained for d; •
pRO&RfiM OUTPUTS
Prob. 6.25
Omega = 6 00 rpm
Number of Gears: 2
Length of shaft = 400 mm
Tau * 60 MPa
For Gear 1
Power input = 80.00 kW
Radius of gear= 60 mm
Distance from A in mm = 120
For Gear 2
Power input = -80.00 kW
Radius of gear= 6 0 mm
Distance from A in mm = 280
AY= 11.141 kN, AZ = 6.366
BY= 4.775 kN, BZ = 14.854
Just to the left of Gear 1
MY= 763.94 Nm
MZ=. -1336.90 Nm
T= 0.00 Nm
Diameter must be at least 50.75 mm
Just to the right of Gear 1
T=1273.24 Nm
Diameter must be at least 55.35 mm
Just to the left of Gear 2
MY=1782.54 Nm
MZ=-572.96 Nm
T=1273.24 Nm
Diameter must be at least 57.71 mm
Just to the right of Gear 2
T= 0.00 Nm
Diameter must be at ,least 54.17 mm
(CONTINUED)

PROBLEM 8.C4 CONTINUED
Prob. 6.29
Omega = 4 50 rpm
Number of Gears: 3
Length of shaft ~ 750 mm
Tau = 55 MPa
For Gear 1
Power input m -8.00 kW
Radius of gear« 60 mm
Distance from A in mm - 150
For Gear 2
Power input = 20.00 kW
Radius of gear=100 mm
Distance from A in mm = 375
For Gear 3
Power input = -12.00 kW
Radius of gear= 60 mm
Distance from A in mm = 600
AY= -0.849 kN, AZ= 4.386
BY= -3.395 kN, BZ- 2.688
Just to the left of Gear 1
MY= 657.84 Nm
MZ= 127.32 Nm
T= 0.00 Nm
Diameter must be at leaat 39.59 mm
Just to the right of Gear 1
T=-169.77 Nm
Diameter must be at least 4 0.0,0 mm
Just to the left of Gear 2
MY=1007.98 Nm
MZ= 316.31 Nm
T=-169.77 Nm
Diameter must be at least 46.28 mm
Just to the right of Gear 2
T= 254.65 Nm
Diameter must be at least 46.52 mm
Just to the left of Gear 3
MY= 4 03.19 Nm
MZ = 509.30 Nm
T= 254.65 Nm
Diameter must be at least 40.13 mm
Just to the right of Gear 3
T= 0.00 Nm
Diameter must be at least 3 9.18 mm
Prob. 8.77
Omega = 6 00 rpm
Number of Gears: 3
Length of shaft = 24 in.
Tau = 8 ksi
For Gear 1
Power input = 60.0,0 hp
Radius of.gear= 3.00 in.
Distance from A in inches = 4.0
FY= 0 *
FZ » 2.100845
For Gear 2
Power input = -40.00 hp
Radius of gear- 4.00 in.
Distance from A in inches = 10.0
FY= 1.050423
FZ = 0
For Gear 3
Power input - -20.00 hp
Radius of gear- 4.00 in.
Distance from A in inches « 18.0
FY- 0
FZ = -.5252113
AY=-0.6127 kips, AZ--1.6194 kips
BY=-0.4377 kips, BZ= 0.0438 kips
Just to the left of Gear i
MY= -6.478 kip.in.
MZ= 2.451 kip.in.
T= 0.000 kip.in.
Diameter must be at least 1.640 in.
Just to the right of Gear 1
T= 6.3025 kip.in.
Diameter must be at least 1.613 in.
Just to the left of Gear 2
MY= -3.589 kip.in.
MZ» 6.127 kip.in.
T= 6.3 03 kip.in.
Diameter must be at least 1.822 in.
Just to the right of Gear 2
T» 2.1008 kip.in.
Diameter must be at least 1.677 in.
Just to the left of Gear 3
MY- 0.263 kip.in.
MZ= 2.626 kip.in.
T= 2.101 kip.in.
Diameter must be at least 1.290 in.
Just to the right of Gear 3
T=> 0.0000 kip. in.
Diameter must be at least 1.189 in.

PROBLEM 8.C5
A-
M -W
^
T
!-*
I
1
He
T
8.C5 Write a computer program that can be used to calculate the
normal and shearing stresses at points with given coordinates y and z located on
the surface of a machine part having a rectangular cross section. The internal
forces are known to be equivalent to the force-couple system shown. Write the
program so that the loads and dimensions can be expressed in either SI or U.S.
customary units. Use this program to solve (a) Prob. 8.50, (b) Prob. 8.53.
SOLUTION
PZ66MM•. ^)?U I. = ht>//z 1^ hh~J/2
Not& Li /?/v^ 2 MUST 53?rwy 0<*tf of poLi r" /a/;
o/*
*f**h*/v *»» e-** *>%
(I)
(?)
-T,
Z. A
^ <f=ti% 7MC' fit-W? IS Of tfotfigofrjAi Sinew■f.ir)fy
?-
\i-
I3fc
CONTINUED

PROBLEM 8.C5 - CONTINUED
4 in.
Vu ~ - 3 kyu
V*-**:.
<flj
Force-Couple at Centroid
A*f -{3-£r*X/&». -*'»,)*-33 *&?f<h.
Problem 8.50
P - -24.000 kips
MY » -30.000 kip-in. MZ « -33.000 kip-in.
VY = 3.000 kipB VZ = 2.000 kips
+ + + + + + + + + + + + + + + + + + + + + + + + + + +
At point of coordinates: y » 2.000 in. z = -2.000 in,
aigma - 1.792 ksi
tau = 0.104 ksi
40 mm
PO/A>T //
Pa/WT // U~ ?D 'tow ^^SO'wm
Problem 8.53 porce-Couple at Centroid
P = -120000.00 N
MY = 2000.00 N-m MZ - 11000.00 N-m
VY - 50000.00 N VZ = -20000.00 N
+ + + + + + + + + + + + + + + + + + + + + + + + + + +
At point of coordinates: y = 20.00mm z = 30.00mm
sigma = -14.352 MPa
tau = 9.259 MPa

PROBLEM 8.C6
SOLUTION
8.C8 Member AB has a rectangular cross section of 10 X 24 mm. For
the loading shown, write a computer program that can be used to determine
the normal and shearing stresses at points H and K for values of d from 0 to
120 mm, using 15-mm increments. Use this program to solve Prob. 8.35.
£C>MPUT<E f?eACT/OH ttT'-fl*
(BO s?mr ~Jj . "*/H \ -
y^ - flees 30° +(9J&*)s''*3o*tS-rp)
P%D£Rf\M OUTPUT
Problem
d
mm.
0.0
15.0
30.0
45.0
60.0
75.0
90.0
105.0
120.0
8.35
Stresses
SigmaH
-43.30
-41.95
-40.59
-39.24
-37.89
-36.54
-2.71
-1.35
0.00
TauH
0.00
3.52
7.03
10.55
14.06
17.58
-7.03
-3.52
0.00
in MPa
SigmaK
-43.30
-65.39
-87.47
-109.55
-131.64
-153.72
-96.46
-48.23
0.00
TauK
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00

PROBLEM 8.C7
*8.C7 The structural tube shown has a uniform wall thickness of 0.3 in.
A 9-kip force is applied to a bar (not shown) that is welded to the end of the
tube. Write a computer program that can be used to determine, for any given
value of c, the principal stresses, principal planes, and maximum shearing stress
at point H for values of d from -3 in. to 3 in., using one-Inch increments. Use
this program to solve Prob. 8.72a.
A
7
To. TZt/SZ-'s*'
\
\.
- m~
■~f
^
rr
—* —«- -h
f
J
\U
r
l
V*
■*—
*■ •-* —*
0
—>
-r ' + **— -
\
g&toiw. 77 =
■— ■ ft
Mu
a)
^7o7»>.~
%.-rv
I
t " i
%
R
40
V T7-
%.„r^'E I
Bv
■ 4 fan (~zr— J > ^ ~
-* ^ Vv* / J anew
Rectangular tube of uniform thickness t « 0.3 in,
OutBide dimensions
Horizontal width a » 6 in.
Vertical depth b * 4 in.
Vertical load P » 9 kips; line of action at x = -c
Find normal and shearing streses at
Point H (x - d, y - b/2)
Problem 8.72 Program Output for Value of c « 2.85 in,
d sigma tauV tauT tauTotal sigmaMax sigmaMin tauMax theta p
in. ksi ksi ksi ksi keii ksi ksi degrees
00
00
00
00
00
00
00
12.58
12.58
12.58
12.58
12.58
12.58
12.58
-3.49
•2.33
.16
00
.16
33
.49
-2.03
-2.03
-2.03
-2.03
-2,03
-2.03
-2.03
-5.52
-4.35
-3.19
-2.03
-0.86
0.30
1.46
14.65
13.94
13.34
12,89
12.63
12.58
12.74
-2.08
-1.36
-0.76
-0.32
-0.06
-0.01
-0.17
8.36
7.65
05
61
35
30
-18.49
-16.00
-12.78
-8.73
-3
1
89
36
6.46
6.46

CHAPTER S

PROBLEM 9.1
C
aL_x __j
VM
E
9.1 through 9.4 For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the slope at the free
end.
SOLUTION
%> IM, = o
- M„+ M - O
M * M,
(b) J €> x = o
to jjfex.o
MI £ * M0v * C.
Ely = £M0X' +■ C,x + Ct
C2^ iMQL
(xa - J?Z.x + //)
Jk-r (L-x)X
ZEI
M0L*
y■ B ±k.(L- o)1 *
^A 2EX ZEX
J
M„L
S-&a-*> . -i^-a-o , -g
0A, &± -sj-

9.1 through 9.4 For the loading shown, determine (a) the equation of the elastic curve
fi>r the cantilever beam AB, (b) the deflection at the free end, (c) the slope at the free
end.
SOLUTION
EI^ = -PLx + ifx" + C,
[x'O, ^t ,0"] o = - 04 o -> C,
C, = o
v - £1! i
4/1 = -3k-(u-L) = -£i:
2EX l ' 2EX

PROBLEM 9.3
9.1 through 9.4 For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB> (b) the deflection at the free end, (c) the slope at the free
end.
SOLUTION
92Mr:(? twx>| 4- M ^ o
w
Ely r -^wx* 4 £*,L*x -v C
Cx-l, y-o]
la) fAw+tc corve
(V>) y © x = o
*--^ . - .»<■
*L'
6*T
gfx
SFI
y* sex ^

PROBLEM 9.4
9.1 through 9.4 For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the slope at the free
end.
SOLUTION
Mn
c
A
E
|wL
Ik
3
6
w„\
c.
HI
Jv if
1—
3
2L
'f
, _
A
-—
J
X
»
tf
5zm, = o
h±
■*««*■
5rMr = o
i«tt» - J,w„Ly * *£.£. + M - O
M - -g-WoL" + £w_Lx -
w„x
_;L
o--o + o-o*+ C,
[x^o^^-o] o--o-*o-o + o +C4
- Wot-" fi _ X
C, -- o
CL = o
(^ ^ €> v = Z.
>
vV. ('il^*. -Lj J+ -L-^)
EI
(i
\H ISO
/«° EI
_ Wj^ ( 1 _ JL + -L \ - _ _L WaLs

PROBLEM 9.5
9.5 For the beam and loading shown, determine (a) the equation of the elastic curve for
portion AB of the beam, (b) the slope at A, (c) the slope at B.
SOLUTION
M^
*»L
O 2 Mft * o
"c -1U ♦ (|«J.MO = o
** ' f »L
vj/tr
D
M
V
fwL
For por+io^ AB onij (0< 06 * L ^
£f = fw/Lx-4
EI H = ^""-* " a1"*
Jx1-
Lx= L, y = o]
o -- o - o + 0 + Ct Cz = o
Or £WLS- ^wl.* + C'L C' = ~ fc WL'
(W
4«k-o JSl'gt"-0-^;
a =
wl3
18 EI
*dL
* it*
x » L
£a = o

PROBLEM 9.6
9.6 For the beam and loading shown, determine (fl) the equation of the elastic curve for
portion BC of the beam, {b) the deflection at midspan, (c) the slope at B.
SOLUTION
i-—x Usi*w ABC as a. -Free fe^dy
0 "2M0 = o (f X#) - R.L t C^LX# } * o
wfe
s
w
Tin
T+5-
■ wl
EI
»■
4 wLx - j-wx1 - -^ w/.x
Eli?' £^xa- i-^x3 - ik"*1* + C,
^»
/o
EI s r iS"1** ~ ji***'1 - &wL**x ^C,x + C
2o
[x= L , ^y = o]
C2 -- o
OrO-O-O + O + C^
^ ***■* ** lt{i^^-i(^-aiL1^-iiLW]
£1
~ EI t 8b 38* *» at© J
nao ex
M
13. ^1+ l
(O ^ @ x - 0
$) *£(o-o-o-£
L3) r - -S
wL:
o EX

PROBLEM 9.7
9.7 For the cantilever beam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the deflection at B, (c) the slope at B.
SOLUTION
^ U&iita ABC «cs a 4>ee Wy
cft?T
**
2
Wt
i W
t
'd&
n
S J
x —-
ii
i,»
(cO E-f*.s+rc cu^s
(L> sj J x = £
CO g J y- t
>TlF, - o
I?. =o
Usinj AX <x& fl. "rV«e t*-\y ( Perti'«* AB o«^>
ft TMj = o - JfltL* + (wx)| + M - o
Ely - JwLV- ^wx* +C,y + Ct
JB " 321 El
1| wf-4*
384" £r
S W,L3

PROBLEM 9.8
9.8 For the beam shown with load P, determine (a) the equation of the elastic curve for
portion AC of the beam, (b) the slope at A, (c) the deflection at C.
m
M
D
SOLUTION
P«*f4;©* art A #?A - ^P
Fof jpoirfion AC owL ^ vsim f/ee laod^ ^^
OMj r O -jfP* + M .= O
M = iPx
EI ^ - *Pxv + C,
C,= -itFL1
EIv -- ^1V + C,x t C.
tv =0^-^-0] o-o+o+C^ C,, = o
Cai EiW«'e corn _y =■ jlj(£x5- j^L'x)
?L*
er *"" iter
(tf y art * = £
48 fr
o*
PI3 1

PROBLEM 9.9
9.9 and 9.10 For the beam and loading shown, (a) express the magnitude and location
of the maximum deflection in terms of w0, L, E, and /. (b) Calculate the value of the
maximum deflection, assuming that beam AB is a Wl 8 x 50 rolled shape and that w0 =
4.5 kips/ft, L = 18 ft, and E = 29 x 106 psi.
SOLUTION
Elffe • ^w.Ly*. Xtt^ +C|
EPa*f*e ccrve
0--0-0 + 0+C, C* = °
^ " EX"Z 3C LX «o L 3t© L * S
3* Er I ml* at l 36© 5
O - _ 7 waLJ
To Tin J J*flC<i i"e* of im«4umO%ia JeWecT'O^ S&T -J* * O
X^* 0.5*193 L
IS** - 30L*v + 7L* = O
>C * (« - VftU* ^ O.W7L*
or
= - 0.006^ -|^
W.L'
EJt
4Soo
L- 18-W - */C 'A
. (o.OQ652XS7S^falfcV _

y\
PROBLEM 9 10 ^*' an<* ^'^ For tne ^eam axi<^ loading shown, (a) express the magnitude and location
of the maximum deflection in terms of Wq, L, E, and /. (b) Calculate the value of the
maximum deflection, assuming that beam AB is a W18 x 50 rolled shape and that w0 =
,0 4.5 kips/ft, L = 18 ft, and E « 29 x 106 psi.
uiaLL?^Li£teaaJ x SOLUTION
M , - ^(K^ - ivO 4 Cv* 4 Cm
EIj . ^(tU^-^LX^T^xM ^Ctv+C?
\;y=o3-j = o] o=o-Ovo+o+Ct Cj.^0
To "F/'n^/ ^occti tow oF m&-x»r»iJ"n JervfecTio*
By Ne*/f»n- Rapkso* tnernoa Z1 « — Jf/Jz
2 « _Ar$- QJU&S, 0.4807 _, O.4807 V„ = 0-4807 L
J1*' lei I '• ii)

PR0BLEM9.il
9.11 (a) Determine the location and magnitude of the maximum deflection of beam AB.
(b) Assuming that beam AB is a W360 x 64, L = 3.5 m and E = 200 GPa, calculate the
maximum allowable value of the applied moment M0 if the maximum deflection is not
to exceed 1 mm.
SOLUTION
Usirw ewTiVc. beA**"* «A «t T^ee. boi<
in3
fc^yro]
D""-^-0^
O^HB * o
M0 - K*L^o
*,*£*
*>£
£
ft
[ye L^ y *o 3
-J
M<>- -£!*x + M* O
5M
0=o-oho+ Ct £?, = o
tc*v\ seT -"•■ - o
£>C- tx^ + iL*" - O
- O.HZZCSL
Solving W M* M. * o.O£^5 Ll
L = 3.5
y* - I *,*> * /o
-5
CM
M,
- (loo*<ot>Ki7gyicr&Xicr*>> m
4S.*3*/os W-^i = ^.3fcU.»n

PROBLEM 9.12
fr-°j ^ = °J
\>= Lj^sol
ft.
L
)
M.
9.12 (a) Determine the location and magnitude of the maximum absolute deflection in
^Bbetween^ and the center of the beam, (b) Assuming that beam AB is a W460 * 113,
M0 *= 224 kN-m and E = 200 GPa, determine the maximum allowable length L of the
—v - — - - — --» -- — ■ —-
beam if the maximum deflection is not to exceed 1.2 mm.
SOLUTION
OsTn* A 8 as a. -Pree b«A-y
?Ma '- o - 2.M0 - t^L * o
K* ■ " L
- Mft -v 2£!?x 4 M = O
M»
M
2 (L-*x'>
Mo
c
A
2H./L
5
*
[wLjyol
*
Ely = £(±LxMx*W.x + Ct
[i<--o^ = o] 0=0-0 + 0 + C* C£=C>
EIL
«*c " EXl
M.L*
1/3.
|J«1 ■ O.01CO37S" ^kt
DdW F » 2.00*10* P^j I r £$1*10* w* - ££6*io~c **
i ,JCgoo*fQ1K^6>'i6'4)0.g»r/b"*)T"
*fi-
6.0<J w

PROBLEM 9.13
M0 ■= 38 kN ■ in
W100 X 19.3
9.13 and 9.14 For the beam and loading shown, determine the deflection at ooini C
Use E = 200 GPa. ^
SOLUTION
a = 0.8 m
«■ L = 3.2 m
§ < -x, < & ^0 TMj - o
f^-
0"
A
r*- x ■>
a- * x, * L
5rtrr
o
(0
K./L
rr Jif - H. „
*S" «sH - y* x * M„ + M - o
a- * x * L
C4, o
[x-ft^ytf J EisfrWft
0 = O -* o + C,
e^4,'c e^ve G» 0-x*<* ^ -" m"[&xV ("UM^^O*]
. (3S*/o?)[CaK©.*)V8 + (3.^^.8)/3 - (3-^fo.s)*]
&*
(2ooicjt>< X^-77 "/d* )( 3.2)
12,75 */o"S m
12, 75" mw

PROBLEM 9.14
P = 20 kN
W150 x 18.0
9.13 and 9.14 For the beam and loading shown, determine the deflection at point C,
UseE=200GPa.
SOLUTION
Lei k* L-a
L = 3 m
O < 06 * a
ex & - I- (fc^
a * x < /-
[xrO, J.-o] £? ft) O t O 4 O + Ct Ct = O
IVsj'.yJ £wo«~j^) -£(iW)*c,a ♦ c8
= f [i^ti* + oj + J<a + C, C, = C, = o
Dot«.: P r 2o*/os N £--200 */o* p* I- rn^io6^^ ?.i?y|o"dH
P(ba3 + k3a- Lxa.k)
i.e ^&T ^m I

PROBLEM 9.15
k
~J
p
B
•72-
C
D
L/2
9.1S Knowing that beam AE is an S200 x 27.4 rolled shape and that P = 17.5 kN, Z, =
2.5 in, a « 0.8 m and E = 200 GPa, determine (a) the equation of the elastic curve for
portion BD, (b) the deflection at the center C of the beam.
SOLUTION
<X< X. * L-Ol
0 < oo < ct
EI & - M * ft
EI ft. = M » P&
-lx
Eljfc - P** «■ C,
EI ^ -- i?ax* + Csx + G
Pa" -iP*L
C^iP^-iPaL
EI & - ifV+C, (ii
Ely = £Px**Cx + Cz «1
[11=^ j=jl i P&.s ♦ (iPa -iP«d^ a. = i?6* - i-Pa't + C,
c, - £ p**
^ - ^-(iPax1 -. CiX + O
D«lK/« I - ^.Ivio'm^ = 23.^x/o"6 ykv* _, E - Zooxio'1 Pa.

PROBLEM 9.16
9.16 Uniformly distributed loads are applied to beam AE as shown, (a) Selecting the
x axis through the centers A and E of the end sections of the beam, determine the
equation of the elastic curve for portion AB of the beam. (ft) Knowing that the beam is
a W200 x 35.9 rolled shape and that L = 3 m, w - 5 kN/m, and E=200 GPa, determine
the distance of the center of the beam from the x axis.
W
SOLUTION
O <^ * f
S> 7 Mx = °
-C**^) + M * °
TTT
J r*
fc**
'¥
ei£ = 4
EI^ r atj^x1-* C,* +Ct
EI jjy =- J-wx3- ^wfr-^)» + C,
[«-*,&*&] ±»$tf + c. * £>&>* + o + c3 c, --c,
c, = ct = o
Sywit^ef^y tour\*|ftiAW Co»)tfll'r/oM £* S *5 i 3Gt ° -*
> c fX
- wjLI ( J— i ' 7
EI t 3Sf "" lo3£8 \o% i
v - 35 (Smio'Xs)^ - -317WeTfc*> ,e 0.3<?7*i*i

PROBLEM 9.17
9.17 through 9,20 For the beam and loading shown, determine the reaction at the roller
support.
tt»o, y*ol
W
/ F*
V
£x* <=>_, jr oQ
SOLUTION
A*.
M = - a wyl + ^a*
o , -£wL» + i*C * C, C,'i»/t*-4ftLx
(*-*>*« - (*-£>«*■
iff. = *«<■
1?A - I wt t

PROBLEM 9.18
m
c
J B' H
* L-x —
9.17 through 9.20 For the beam and loading shown, determine the reaction at the roller
support.
SOLUTION
f?e«.efi'o*\s a**e Static«JIa\i \<t\ek*.\&t*\in«ye .
t)THjr o - M 4 RsCl-xV Mo = o
£T £? * -M + ^(L-x)
£1 £ - - Mcy 4 1?fta*-ixO-v C,
0=-o-fo»O4G4Cz Cz=o
o 4 o- o -v C, C, = o
O^o"]
1
fx = O ^ sj ? O J
if?.
- JL 14

PROBLEM 9.19
L-y
w«
B
fe
9.17 through 9.20 For the beam and loading shown, determine the reaction at the roller
support.
SOLUTION
React.'
-M + r?a(L-x,> + iw0(L-x')|(L-x^
+ ^Mi(fO - O
- )?aa-^- $*[*»■*- ^x+-2f/1
4 xZ.v- ;>£*% x* ]
- l?8a-KV^(xJ.3L'x +*LS)
EI&- r^a-x^-^fx1- 3l3x+^lO
EI j - ft (iLx--fx») - ^(£x"'- i Llx4 + L1** ) 4 C.* + C
[x=o5^-o] - C, -- o

PROBLEM 9.20
9.17 through 9.20 For the beam and loading shown, determine the reaction at the roller
support.
SOLUTION
I>-°^-*»1
W0X//-
w
w,
£{i-0
--r-rrnT
R.
5
75?
=■ - w - -
(L-O
iff* * ,& «u j?„ .£»*£. t

PROBLEM 9.21
9.21 For the beam shown, determine the reaction at the roller support when w0 - 6
kips/ft.
SOLUTION
RecvCTi'c.iA's ewe st&."K c<*/A: \^a[t.\-t<fwvy.<tz..
£*«»>yt<>3
E
M
D
*5& y1
R*
C, = O
A.
C.'iwt^-i^L*
«?» = li(&XiO -- 3 fc.y* t

PROBLEM 9.22
fcL^rOl
4
S<*
S*L
9.22 For the beam shown, determine the reaction at the roller support when w0 = 15
kN/m.
SOLUTION
A .
* o
iS-a1-jtJ?.l» = o
ff. ■« £ •* l
* Co
9-7S" kM *

9.23 through 9.26 Determine the reaction at the roller support and draw the bending
moment diagram for the beam and loading shown.
SOLUTION
L
FI^'M = ft*
Elfe ' 5P*xl 4 C,
EI 3 - ifc.*1 * C,x +CL
^ < «. < L
EI JJ[ « M - Hx ■ P(x-V>
EI y - i^x'-iP^-^1+ CJX + C.
c< = o
&'ij3 "3 3
(0
IS)
(«)
C,= o
^ .£Pt -«
F*.~ (4\ wt+l> x» L
* - *PL

PROBLEM 9.24
9.23 through 9.26 Determine the reaction at the roller support and draw the bending
moment diagram for the beam and loading shown.
SOLUTION
Rec*.t4
o * y < \
^ ** * L
(3)
tv» 5, *a ^
O - o + o +. C C2 = o
iA&f + C, = iftdtf- o * c4 c,. cs
F*>~ CO , w&U x * £ Mt - R»(y) ■= ^wL1 = O.o*731 w/Lz -*
Fro- CO, to.-H»- L Mar ^L--*w(^ - (75i-i^L ~Sk»LX
=i - 0.07031 i^/L -^
'4w« M
W< L
M/
X ~ ^+ -Z.*\ r .11 L
Fro*. M> w.*+h X- X,

9.23 through 9.26 Determine the reaction at the roller support and draw the bending
moment diagram for the beam and loading shown.
SOLUTION
4S^=o £* + R8 = o ^ " - ft
0 <x < %
V\ » r?AX -V Ha - - Mo + RbL - Pa*
EX^ - -KX4 ft^*-***) + C,
EX;y =• -iMPx*+ f?(8(ii.X*-ivt^fC.x +^4
O + O+o+Oj^o C* - o
^'" 2

PROBLEM 9.25
9.23 through 9.26 Determine the reaction at the roller support and draw the bending
moment diagram for the beam and loading shown.
SOLUTION
React i
Of x
4 —J-x-fe-.
CO
CD
<3l
EL-il * *&*' * C.
EI ^ - i^x* +C,y + C,
o - o -i o + Ct
Fro*. (1^ wi4^ yr^ Mdr ^az - Two*'1-1 r 0.007031 w.i.8, -
= - 0.0*761 «U* —
L - O.M8t L
Xw t 0,SL 4 0. M8< Z. * O.618C L
- O.Oo8l43 WoLZ

W ^JwL
i
%
5
■0
M
9.27and 0.2J? Determine the reaaion at the roller support and the deflection at point C.
SOLUTION
Yeact\ons a**e s^a+icc»$X-\ i"cl«.7e,vM»VtctTe.
O * * *^
El£ - i^x'-^ +C.
EIJ * is*** -£*"<* + c«* 4 c*.
■5 < P6-* L C See. -£r<ee. kcdy Jiaa^w^. J
EI 4 - *^>c*-^L(K-^y +C*
EX
J
- J-
6f?„x3 ^wL,(x--^y ±Csx 4 Cy
C2, o
O - o + o ■» Ca =0
&*t>3°xi ^fttfef" *vtrV .ic^pfrvt^k - i-a&)E- >L$r+M c
C* -ifawL
3ft*
3*1
76* WC
7*8
wl
fc.L4*,0l i^^-iwL^ ^ C3 = O
S3 CH L A K**-
*%L»-*wL(^)»t (£WL*-^)L , ^wt' - o
(WVU'-U-#*£>*
3 ** V4
3?f
R.-fi-»U
^/L1* C j_ 41 11 \S 1 /±\* iL 1 .«]•
- /JU I IL \ wL.'
fl
w
^t'CHM EX
£v -
ov
"k
~ ^Cii* ,7U ■*»* /**• /

PROBLEM 9.28
9.27 mA 9.28 Determine the reaction at the roller support and the deflection at point C.
Cx'O, ;j=o1
0»,*M r L 1
i
SOLUTION
Re«*+r
El j* - M»y + iR,xlt^x* + C,
EI X - iMAX^i-Rx^+Awx*1 4C,x *Ci
Cx= o_, ||£ = o] o + o + o+C,=o C, - o
[y-Oj^ o] d + o + o + o t C4 = o Cj-o
e
r - C-i—L-
)wL'
wLJ
Cx^Lj^.-ft"]
R.» -Pi - -£wL
Ely,-- k*&Y* ifMMN £-(iV = * to-L'H*)1* HiM*)* •iWrV

9.29 and 9.30 Determine the reaction at the roller support and the deflection at point D,
knowing that a is equal to L/3.
SOLUTION
Rfi6.CTl*OhS Owe 3TA-Tl1CA>«/sJ \v\ J&4 Cfi«iW £ .
[vr^y^ol
EI& * M "- Rx
M- £A*>
EI J = £RaX3+C,x + Ct
EI&* ftx-M.
04o + Cz^o Cj=o
£f?Aas + C.cl 4 C, = £(?.«' + O + C3a-+ C, Ca= C, = o
xr*. | >
£J
f^-M.L'+[M.(L-t^-i-4¥L*^l
= lf(^+f-£) -ifc-^t

PROBLEM 9.30
9.29 and 9.30 Determine the reaction at the roller support and the deflection at point D,
knowing that a is equal to LB.
SOLUTION
Or^-o -h,-Fa-PeL -
D
MA - ff,L - Pcl
o <y * cl
M- Ma+J?a*
EI jj£ = M- MA+RAv
<x<^<L H^M^-v f?flx - P(x-<0
EI ^ » *MAx* + £l?Axa - ±P(*-<0* + Cax + C
l^^&vl o + o + c, = o
^ * o + o + Cz=o
Cx = L,^--ol iMAL* + i£)AL1-iPa-a? +0 + 0 = 0
= Paa(±L-±a>
</» " *i*7 £X *
[y= Oj ^- o]
C3 = C, = o
£1 I \t <■ ;p 3 J + /cl t ' j7 ' i 2187 EI >

PROBLEM 9.31
P
F
*»
f
1
■p
^
4?<-
-£PL
-t*
9.31 and 9.32 Determine the reaction at A and draw the bending moment diagram for
the beam and loading shown.
SOLUTION
El£* - MA + i?x
% sywme4r3 M6 = M* ' gFL^ —

PROBLEM 9.32
9.31 and 9.32 Determine the reaction at A and draw the bending moment diagram for
the beam and loading shown.
SOLUTION
Ke&c-Tt&ns ore, s4<5fT<'c'A'Wy i"nde+«*'»»*t"w«k.'fe.
Because erf s^»vi*>«4tw 4^ ~ ° **^ V" ^
Use po^u>" AC «£ fc>e&** (o< **<• if")
K
w
= -.0 58*
01
on
o= 0+0+0 + 0 + C< Ct=o
-£&f* R, . o RA -- *«U. t

PROBLEM 9.33
9.33 For the beam and loading shown, determine (a) the equation of the elastic curve,
(b) die slope at end A, (c) the deflection at the midpoint of the span.
SOLUTION
3
0 = O 4 O * O + Ct
C,= iw.Ll
fl
= M -
O - O+O+O+O-tCt
(eCl E-P^sf/'c coirve r
*•**-* In ^f £l*-
wttLs
3o £J
(fi 5fl»pe J- e«<J A.
^ xf{ &X^S-iix*^ *(*Y - A Wis
■ " -L*l - Cl WaL**
PT C
57t.0~ It© * IMM ~ 60 J
•SIGO EI
fet Wot** I
-S^feo EI *

PROBLEM 9.34
TTX
. W = Wq COS nr~
9.34 For the beam and loading shown, determine (a) the equation of the elastic curve,
(b) the slope at the free end, (c) the deflection at the free end.
SOLUTION
V , -^S;„JX , c.
C, = o
IT"
-OS
Ex.
3<-
- O
^= 2^j_ge^E - Tr'£ ♦ *»or-^f • + tCj-tt.")} —
IX-tJ

PROBLEM 9.3S
9.35 thorough 9.38 For the beam and loading shown, determine (a) the equation of the
elastic curve, (ft) the slope at end A, (c) the deflection of point C.
SOLUTION
OSMe * o
.-PW
R.T
T
E*& " ?* -P<x-ci>
^Ls -*P(L-«0* + C,L=o
6, e-^B"(bLfc-fc»7 - -i^(L'-f)
«*- S-o' *C,a - %L'.g«L^a
*
6c -£j£(L»-a*-fcO = -^{^ + ^t^-^-V\
3EIL

PROBLEM 9.36
[kpI, y.-o}
9.35 thorough 9.38 For the beam and loading shown, determine (fl) the equation ofthe
elastic curve, (b) the slope at end A, (c) the deflection of point C.
SOLUTION
jg - V , R, - *<x-4j>'
ei & * i-wL x*- - iw<*-4>s + c,
[x- o, ^o] O = O * O 4 O * Ct
Ct= o
^
°* " 3S7 EX
CO Def&ctlon d G ( x = ^ .n ^ 1
38t 76* ' £X
^
wL"
7S8 EI

PROBLEM 9.37
9.35 thorough 9.38 For the beam and loading shown, determine (a) the equation of the
elastic curve, (b) the slope at end A, (c) the deflection of point C.
_ M«
_ Mc
Dt-»,y=°3
Cvri,y_-oD
4
I
SOLUTION
O* X** ex M - £Ax
a< y < L M - R*x - M<
[x=L, y--o]
EI j=J ;i^v!-M.<v^)' ♦ C,
EI J - ± ft** - iM.<X-*>X " C,y -v Ct
o -- o - o + o + c Cz = o
X
t?A l* - i m a-«y + c, l + o =■ o
0,) SP.pe. of A ( 4*■ *t x = o }
a
, , -±Mo - o+ 3Ltl-lM
M<
GEIt
(sf-r")
3£I
£U-^ t

PROBLEM 9.38
9.3S thorough 9.38 For the beam and loading shown, determine (a) the equation of the
elastic curve, (b) the slope at end A, (c) the deflection of point C.
SOLUTION
y,, **> <*-£>'
MA = o
jijW.L
EI
J
4&
r<*-*/ + ih«°L»3 * c*x + c
ft = o
Mrl-w-^H^^^
37 UfcL*
/ i 37 \ K>L*
~ \WSX ~ \\Sl° ' EI
a, -
37 ia^L*
stto cr
w.L«
1^80 ei

PROBLEM 9.39
9.39 and 9.40 For the beam and loading shown, detennine (a) the deflection at end A,
(b) the deflection point C, (c) the slope at end D.
\
SOLUTION
V- -P
V ^-p + 2P- P
il
1
Rs
Rt
Bui M - o J x -o
(n
(3^
b"ft^=o]
-^Pa.3 + b - o + C,a ■+ Ct io
a£, + Ca^ ±ft*sW
Ejfc>-E.ftJ ZC.Ol* '{?<£ C, - "(s Pa.*
*-£f
0 + O - 0 + O
_ J* 1 - 3
Hi" "T
3 Pa3
<* BX
■*■*¥>■

PROBLEM 9.40
9.39 and 9.40 For the beam and loading shown, determine (a) the deflection at end A,
(b) the deflection point C, (c) the slope at end D.
SOLUTION
[y=A^5«0 L*sS*Jj*©1 o^ ^ M - - Mo
EI j - -iMy + iHo^-^N C,x + C*
aC, 4 C2 ' a:MDa
(t^ Defect"/©n A-t

PROBLEM 9.41
9.41 For the beam and loading shown, determine (a) the equation of the elastic curve,
(b) the deflection at point B, (c) the deflection at point C.
SOLUTION
C
M»
w
O < 56- * %
£
k<0L<L
w
*•■
*[
:kjfc
£wL
U-^
V
C See +Vee loodvi ?(r«.ftf«.^.)
j^»
[X^^g^j -040-0+C, = O
IV' »j ^' °3 -O+O-O + d? + C, = O

PROBLEM 9.42
9.42 and 9.43 For the beam and loading shown, determine (a) the equation of the elastic
curve, (b) the deflection at the midpoint C.
SOLUTION
jj^ * -w(>T> r -w<x-«l>° + w<x-3*>°
EI >f * i^AX3 -^t^<x-a>4' +3L-w<y-3a>'' + C,x -* C.
(00 E^OAjioVi cr exilic coi^e
Cb^ De-F^c-r
C-^a'

PROBLEM 9.43
r_y=o,y=©"3
9.42 aud 9.43 For the beam and loading shown, determine (a) the equation of the elastic
curve, (6) the deflection at the midpoint C.
SOLUTION
8j s^ivM^el^j RA r f?e
WOO" w - w<x*-ft.>0 + n{x-3a.y
-r V = f?ft - wx + w^x-et)1 - w<x-Say
M r MA + t?Av - iwxx t^w^-ay--£*<*-**>* «,;-M M<.= o
EI ^ ~ i^^X1"" fiWXS + t W/<X- *>* -^;^<V-3a>^ + C,
Ely -* fwAX^^wx'' +^h/<x-A>4 -2Lw<x-3a>* + C(X + ct
£x~°j V ■ 6 ^ O-O+O-O *0-vC4 = 0 Q=0
Lx-4*., y.-ol -^w^C^f-yqW^ft^+^wC^V-jifW^ 4 Ct<>0 = °
(b) D«Tie*ti«i *i £
(y J y = 2a ^
y<
wo.
E
^{i(^3!i^^A0r*o -fwl = -£*f'
*«. s it FT1

PROBLEM 9.44
9.44 For the beam and loading shown, determine (a) the equation of the elastic curve,
(b) the deflection at point A, (c) the deflection at point C.
SOLUTION
2. ^ "a1
a 4
Vl/Z *ti-/i
lHtb
It?
£ij = -£"*y + i«L<»c-iy + i^-jy -^w<^/.y +c,x -+C*
f oil*

PROBLEM 9.45
9.45 For the beam and loading shown, determine (a) the slope at end A, (b) the
deflection at the midpoint C. Use E = 29 * 106 psi.
4kiPs/ft |-~t» SOLUTION
X D.V^.'WaJ Ws: © W,6<) = W>o - Ux
S6X 12.5 (?) Wj(>0 - k*
? - 2k.>».
wOO - wD - kx 4 k<x-p4^
i^^-W - - 4 4 X- <X-4>'
to. d
k.p.-ft'-
fro© -J --ol 0-0*0 + O-tO 40 ^CtrO <?z = °
(R
Eii- 2**-§*, + *>,-*<x-<>,-<x-^ +c,
<=**
EX6at,°40 + d+0+0 ~ ZCSt*
\c:vMx
a**-
4451
_s
- <S.03 x/o"* vwt
(W DcfJfed,'o« 4 C (y d x* 4-tt.)
.. - 5?. 73 r _ ,3.^*Jo"S f+
HHS-1
O. 1G10 iw. 1

PROBLEM 9.46
9.46 For the beam and loading shown, determine (a) the slope at end A, (b) the
deflection at point C. Use E = 200 GPa.
20 kN
SOLUTION
W150 X 13.5
EIj£ ' M ? ?-8x - 6<*-0.f>*+G<x-(.*>*- *<><*-'-*>' kW.w,
FI^ ' f.^1 - ^x-o.*/)1 + ^<x- U>,-/o<y-l.X>t + C, W-^"
£Iy ' 1.63333 x* - £<x-o.*>\ £<x- L^y- ^<y- l.2)a + Qx ■» C4 lfU- *»*
rxn.6^ = o] O.GS33S)0.tf)*-i0.aV4. 4 (a*)*- £6?/0\ C.0.O + o - o
OJa-' E = 20o*\On Pa.
1 r 6.87-10*
- G>. %l»lo~*' ton*
EI =(^OOx/0',)(4.87y/o-c) =■ 1.37*/ v /o< N-** » /37V M-^1"
W SAy* *f A ( f|* «* X = o ^
££ 2** r O-0+O - O -S.ttoSO IfW- kV"
(W) Dc-Meeti".* J C C jf ^t ** \.Z m >
£I^t r (L63333Vl.a^*- j(0.8)" + O - O -(3.</©8© )£l-2) 4-o
=• - l.47*o. kW- ^3
j~^ - -1.071 xt6*»> - /.07I h.»,i

9.47 For the timber beam and loading shown, determine (a) the slope at end A, (b) the
deflection at the midpoint C. Use E = 1.6 x lo6 psi.
3.5 in.
1^ *i
SOLUTION
IT U^its - Forces iv\ kips ^ewv+As In
5.5 in. r j a
w(x) = 0.35o<y-^.£>"
^ ^ V= I.0562S - l<x-L-7S>° - <X35<*-3.5^
El$& - " M = l.*0«Sx - 2<x- l.7S>' - 0.l7S<y-3.5>* fcP-ft
EJTj* * Q.MIXS Kx - l<x-/-75>fc- O, OS833 <^-3.^>3 +C, k.> -Ff "
ET J ~ O.3O|04*xa- i<X-/-7b->3- O.OI*/5"g3<x-3.5>v 4 C, X + Ct kp-fl-*
C, - - 7.54 77? k.p. ffL
fX- 0.«*r(o*K48.«t) * 77. QH\1 k.pin1' =■ 531?./* fop.ft*'
EX j& ' O-O-O -7.5477* fc;p.4H*-
e _ 7.S177T _ . 14.00 x/D_i r^/ - !f.Oox/o"S rW ^ -*
Q*\ Defied u>« J C C^ a* x^ 3.5 -fO
Elj^ - (O.SoiO«l2X3.sf - iCl.75? - O -(7.5477^X3.5^ + O
V. 'f'2!7 - - J?*.37yJo"* -Ft - 0-3MO .'h 4- —i

PROBLEM 9.48
6.2 kN
0.9 m 0.9 m
9.48 For the beam and loading shown, determine (a) the slope at end A, (b) the
deflection at the midpoint C. Use E = 200 GPa.
SOLUTION
li*;4s: T-o^ce^» 'iw IcW, .lei4("lis <* vtene^s.
£
j
1,
w6n - 3<x- i.8>°
jk - -W(^ * -3<x-|.8>°
3£ * V =■ 5-GCC7-5<x-l.8>'-ff-2<X-3.t>*
El^jf r M " S.CU7 x - f <*-!.«>* - £.*<*-3.<£>' kM- m
Eljfe - 2.8333 X2- ±<X- |.8>* - 3. |<X-3.4>1 + C, WK>-^V
EI J * 0.^444 x1 - £<x- l.&y - I.OSS3 <x-3.G>3 + Cfx + C* Kl^1
[y^.^y^] (o.l'ttV'Ks.*}3--4(3.0*- /.0333(1.8^ + C.Gr.'O + o * o
C, * - 2?. 535 kW*w*
EI - CSoox/o'Xi^mo-*) - zs.zxio*- N •**■*- - *s;« x/03 JrtJ-i*il
El 3* = 0-0-0- Z?.£"S5 kV»-»**
8.
ZS.£ X 10»
-£73 */0"c - (5.373vto^raA
12, S37
ZS. 3 v 10*
-3

PROBLEM 9.49
*ki?t
350llVft
1.75 ft 1.75 ft
9.49 aud 9.50 For the beam and loading indicated, write a computer program and use
it to calculate the slope and deflection of the beam at intervals Ai, starting at point A
and ending at the right-hand support.
9.49 Beam and loading of Prob. 9.47 with A£ = 3.0 in.
SOLUTION
15.5in. Se« soJo'Tiovi "h> Ptfot. 9.47 £*■ +i»e
fTI^ - 0.903125 X7- - 1<X-I-1S>2 - 0.05*33<x-3.S>* -7.5477? W.'p^P *"
El^y - 0.3olOS2 X* - i<X- /.75>3 - O. 0l*S"83<x-^.S>V - 7.S4779 X k>-jY*
y G«")
o
3
£
^
1%
Iff
18
21
24
XI
So
33
Z&
3<*
•42
4S
M*
51
51
57
GO
cs
Gfc
&\
1%
IS
7*
3/
84
x(-JO
o
o.zs
o.s
o.is
l.o
1.25"
/.b~
I.7S-
2.0
2.2T
2.5
2.7S
3.0
3.25
3..S
3.75"
4.6
4.^5-
4.5
1.75
5.0
r.^r
5.5
5.7S"
Co
G.25
£.5
£.75
7.0
0 GflT^nuO
- H.oo
- 13.89
- 13.S3
- i3,Ofe
- 12.32
- 11.3*
- I».23
- 2.87
- 7.41
- 5.«?3
- 4,S7
- 3. 19
- 1. 3*
- 0rH%
O.SM
2.H
^.4^o
4-W
5.7<*
6.S6!
7-^2
1.11
<t.72
/0.47
M. fl
M.C2
I2.0O
I3.21*
12.32-
Vdtf'W
0
-3.H<?
-6.*3
- lo, H&
- 13.44
- icm
- i?. 11
- 2l.5i
- 23.S4
- 25.SU
^j&,5£
-27,5b
- 2 8.13
- 28.42
- Zft.37
- Zl.&o
- 27.3o
- 26. So
- as, oo
-23.41
- 2I.S&
- I?,4t
- I7./3
- IS.CI
- 11.91
- 9,o&
- 6.11
-3.07
£
^U>
0
- o.<m
- (9.0*3
-0-1*3
-0.1CI
-o.H7
- 0.22*1
- o. ars
-D.^82
-£>.So3
-0.3/8
-0.3So
-0.338
-0.341
- 0.3<rO
-0.334
-0.31%
-0.3/£
-0.3oo
-0.281
- 0.2S°f
- 0.234
- o.2o4
- 0.175*
- 0.143
- O.IO<*
- 0.073
-O.OS7
0

PROBLEM 9.50
3 lcN/in
fum
-1.8 m -*
6.2 kN
*-1.8m
9.49 and 9.50 For the beam and loading indicated, write a computer program and use
it to calculate the slope and deflection of the beam at intervals Ait starting at point A
and ending at the right-hand support.
9.50 Beam and loading of Prob. 9.48 with A£ = 0.3 m.
SOLUTION
W310 X 60
0.9 m 0.9 m
See SoJj-t-iovi 4© Pft>t. .?.4S -fc>^+fc«
€1^ r ^.g333 x* -i<x- l-S>* -3. )<x-3.C>1- - 22.535- kU-Mx
£1 nj - 0.1*M4x* ~i<x- l-S>4 ~ 1.03333 <x-3.fi>1 -«.S35x kU-*.*
x G>0
0
0.3
O.fc
0.1
\.Z
1.5
1.8
2. 1
2.4
2.7
3.0
3.3
3.6
3.<?
4.2
4.S
4.S
5". 1
514
9 (/o:fcr*J)
- 873
- vm
- 834
- 784
- 7(5
- 62G
-5"! 8
- B^o
- 24 b*
- "87
gl
25*1
43**
CO&
?S3
«72
9&o
\ol&
|o 35
V (to**)
o
- 0.261
- o.sic
- O.TJ"*
- o.fftS
- (.187
- 1.35"?
- I.4t5
- I.ST1I
- 1.64 1
- U*U
- 1.5? 1
- 1.M81
- l.33o
- 1.1X6
- 0.88Z
- 0.606
- 0.3©?
0

PROBLEM 9.51
9.51 through 9.54 For the beam and loading shown, determine {a) the reaction at the
roller support, (b) the deflection at point C.
SOLUTION
M*
V.
K*
J**
EX* - =t MAxL+ £(?Aya - iP<K-|>3 4 C,X + Ct
CX=*\> ^-°1 O+O + O+C, 'O
C = o
- j:
yt=5l^M.^t+ **.(*)'+ o + o + o}
Re^lfP t —
7 PZ\ ^
yc= us ex *

PROBLEM 9.52
9.51 through 9.54 For the beam and loading shown, determine (a) the reaction at the
roller support, (b) the deflection at point C.
Cx«l jV=(*^
Q 7 -2- &*
SOLUTION
^ Fj = ° *?» + *?& s ° ^ * - G»
Reaction* <**■*. s-ftfcK'CAJ^^ i'm^e-|-t^^«Vi<a+e.
o + o+o+C.-o C^=o
0+0+ 0+0*^=0 C2 = ^
II
Hi
C
Ma= m. -i^-L * "iM=
'" ex

9.51 through 9.54 For the beam and loading shown, determine (a) the reaction at the
roller support, (b) the deflection at point C.
SOLUTION
C,-o
PROBLEM 9.53
Elj£{ - M - f?„x - P<x-|>'-P<x-^>'
0+O+O4D + Ct=O
C, r *[UM)P-RAV , i(|P-f?.U*
c, - Hfp-f^k* - -iPL1
p.-fpt
ex

PROBLEM 9.54
9.51 through 9.54 For the beam and loading shown, determine (a) the reaction at the
roller support, (b) the deflection at point C.
u^o
-L/2
SOLUTION
EI^ - f(?Ay3 -^w/U^Y-f^4 4 C,x * C,
(b) De-Wee+io* rf C (y«* V-jO
^^El{(^^0(^5-^-(^+0-^w^i+o^
- Jaitl (-*i- J- _Li-\
" el 6,,*w ^^ is** '
«/1«f EX

PROBLEM 9.55
9 kips/ft
9.55 and 9.56 For the beam and loading shown, determine (a) the reaction at A, (b) the
deflection at C. Use E - 29 x 106 psi.
, -6 ft *
Ik.O^H:^
C B i W12 X 22
C*»u,y *■**-*
SOLUTION
k, 1J^1= ,^;pJ
at-p
k;f-ft*
wCx^ = U * - ^<x-65 - ls<x-6>'
EI3^* iR^x1-- u.o«5x-" + ur<*-6>a+ o.665S<x--«y + C, kip-ft*
Ely r * *?***- o.ousx*5"-* o.^^y-^y + a.0ia*<x-6>r 4 C* + Ca
+ (««?! - 72 l&XlO + O so
(8C4 - m)R* " 8lC4„g Rk= 14. HS* fcps t
EI- 0M*iosXibO * 4-5*1 x/o* k.>t«*- - 3»*f/7 k>>-Ft*
0>) DeUc^ieH *+ C (y «,+ X^ G^
= - 361.5 Jfe.>--Pta
V. = 0. I3<*2 in I

PROBLEM 9.56
w = 4.5 kipx/ft
■dLLLLi
9.55 and 9.56 For the beam and loading shown, determine (a) the reaction at A (b) the
deflection at C. Use E = 29 x 106 psi.
SOLUTION
D El
2.5 ft
2.5 ft ' 2.5 ft
W14X 22
2.5 ft
** * -wfr)* -H.5<x-2.S>%4.S<x-7.s>° k>pV*
ft, 4- ^.5<X-2.>5>' + 4-S<x-7.5>' k.'r»
El-^t - M •= R„x - z.xs^x-z.s^* a.as<x-7.5>* top--ft
Ev = O,
= o
o + o+o + o + Cz=o
Cg « 0
C, * 3o4.G<* - Jo tfA k;r-Ft*
^oo _ ±^)KA - X^coc* QA = 7-3833 Wlps t -*
C, = 3cW.^-(^o)(7.3«33) * -64.^5" k.p.-IH*
(M DefJP«c+.'«m «l+ C (^4 x^5ft)
EX^c r i (7.3833 VS^- ^|£(?.5)i| + o -(ci.<fr)C?) + © -- - I7S.76 tap--ft1
" - I7S.74. . _n-ag£6*|0T* "H v- s -O.OS2C 'm. -*
*foo76
*

PROBLEM 9.57
9.57 For the beam shown and knowing that P = 40 kN, determine (a) the reaction at E
(b) the deflection at C. Use E = 200 GPa.
\A B
HI
n£
0.5 m 0.5 m 0.5 m
W200
IX*0
0.5 m
[>=2j y=©3
4°. Ho tit>
D
SOLUTION
X 46.1 J J
+ tXr^ ^O (?A- 40-40-^0+ Pe - O
f?A/- 120- ftf Wl
+t>^~© -MA - 2o - 1o - Go + fcf?a = o
Re*.&T>»«* a/e 5-farios/My i ^o)eTe^«**.ware..
^:V" 1?A- Mo<^-0-S>a-^o<x-l>"- *Jo<X-l.5>'
EI0-- M* M, + 1?,x -^o<x-o.^'-4o<x-/>'-Vo<x-U5V
Rb
EIj* = M,x ^^l?Ax1-^<x-o.5>*-^o<x-i>t-^o<x-/.5)1+C(
EI j r 4m,/ * ***/ - f (y-oj>s- f <x- i>* - ^<x- i.s>s + C,y + Ct
Lx= ©, g£ = «1 0+0 + 0+0^0 + C,-C> Ct~0
l*s O, ^ * o ] O + O+04040 + O+C'O Cz - ^
Z.ccccl (?e = 3o + X4o- ifco = no £e = H1.Z5 ktit
R* r i^o - 4/.25 <= 78.IS k\*
= - 6.4SS3 kK>* ***
y£ - 0.7/0 *~* 1

PROBLEM 9.58
9.58 For the beam and loading shown, determine (a) the reaction at C, (b) the deflection
atS. Use E = 200 GPa.
14 kN/m
* r
[x=-s>^=o3
R4RM
SOLUTION
Forces In kM ^ -/tMg+^s »n iv».
W410X60 + ,^s0 ^-70+R^r D
MA = 8RC- J7>T kM-i*.
£I$ = M = MA + |?^-7xt+7<»c-5>4
kv
C,-o
d« o
[X.^jaol ^ M„fctf + ±fcA(g)S - £(%)*+ &(*? + O 4-0
3^(SR,- I7S)+ ^(7o-fO- *^* o
170.667 Rt = S^OO--5^ +a^251 = H«.3£ R^r |/.£36 kW f
MA - (SKM.6'36')- (75 - -8^.7/S kW-^
Da>«.: £ = 2oo*/o* p* I: 2/6 mio6**'1 = 216 ho"**4
£1 = l«oo*io*X^.i6i«to"4) = 43.2 x/o* N-*> - HZZoo K)-v^
(b^ OefUi^ fid- B (y *+ x^^)
EXy&= t(-ta.Vs)tS\% + t Ctt-MMtt^ - £ teV r -/So.5? kU**3
yB * ^-18 mw |* -**

PROBLEM 9.59
9.59 For the beam and loading shown, determine (a) the reaction at A, (b) the slope at C.
M»
c
ft
f
,5*
X
SOLUTION
EI ^ - MAx + i*Axx- M«,<x-t>' +C.
C,= o
C,= o
M»= iM„-i*L
[x=L, y roll iM^t i*ALM - iM.(x? = O

PROBLEM 9.60
9.60 For the beam and loading shown, determine (a) the reaction at A, (b) the deflection
at/).
SOLUTION
|*-a-«4-* 2o »
wOO - w<x-a>" - w ^x-3*y
[V'So, y r°3
■2o
O + o+o+o+C, = <-> d= o
£Mft<x + 12.5 Ri a3* 9.3333 *jcl* 0)
12.5 M,aN ^o.253Sf?fca* - 10 wcl'
6rt
So/vtn4 (H aj tz) Si'^jJ-faieo^S-Py
MA ~ - 1.3333 wal
f?A - I.R.80WCL t
2.
[4(-l.33S5)4 *(l.2O(^-A0^] iff r -0.9O7 *f
y0 . 0.9©7
EI
wptH
EI

PROBLEM 9.61
4ldps/ft
9.61 through 9.64 For the beam and loading indicated, determine the magnitude and
location of the largest downward deflection.
9.61 Beam and loading of Prob. 9.45.
SOLUTION
B X See *°^;°" +° ^°*»- **•43 for +Ue
S6x 12.5 de^W^i"^"! oT +Ke e<f^&fi'o^s useJ ^
Soiu^ L^ iferdufiOM- X^- 4.0 3.73 3.735 X^ 3.735 H. -^
orf/A,- 5.33 S.M*
Ely** ^(S.73^S-|ft."73S,),,+ iife.7as,iJ-ft6.8^)fe.7Sff) = -GO.Og kJp-pt5

PROBLEM 9.62
9.61 through 9.64 For the beam and loading indicated, determine the magnitude and
location of the largest downward deflection.
9.62 Beam and loading of Prob. 9.46.
SOLUTION
g^m Sec sJLrr.'oM +■= F^U. <?.</£ -Po^ U«.
W150 X 13.5
0.4 m 0.4 m
er - ei -- 1374 km-*/"
EIj - U3a33x*- ^y-oV + i^-'-^ -■£<*-»■*>* -3.4osoy l<N-*a
SoJU toy i+e^+ion X^ - 0-8 0.858 0.8S7 0.2S70 Xw = 0.2S7Dm -
Jtt/«l> * £, S8 -7.123 7.HJ
~ - I. <*! *W M* Hn

9.61 through 9.64 For the beam and loading indicated, determine the magnitude and
location of the largest downward deflection.
9.63 Beam and loading of Prob. 9.47.
3.5 in.
M
SOLUTION
"1.75 ft 1.75 ft
T
5-5,m See SoJ/oh'o* ~h TVoL ^.47 -ft^ +t*
£1 * 53<?.J2 k:r&*
£1 j6 * O.qoZlZSx' ' t<*-»-?0>*- O.OSff33<x-3.5\>"- 7.^779 lop-Pf--
Ely r 0.3©lo*ja y* - ^0-l-7s>*-0.oWS83><sx-3.s? - l.SHfl^ X Kp.f-P
To £rJ +U ioc^-fion aP waxC*^* ^ljSe*t gj = D„ Assume I-7S* tt^ 3.6"
^ *
■ ■"''■■ .. . I III || __|.-... ~ -:
3.SHO -Pfc
(2*Ko.O<*&S7S ")
Ely r (O.So|0HO(3.3^o')3 - £(3.3Wo-/.75)S -(7. SI 77*7 Ka.'ifo)
V - - /*:3?gg - - IJ?.<W>/o-* fl-
0.341 .'«. 4

PROBLEM 9.64
6.2 kN
3 kN/ni
0.9 m 0.9 m
9.61 through 9.64 For the beam and loading indicated, determine the magnitude and
location of the largest downward deflection.
9.64 Beam and loading of Prob. 9.48.
SOLUTION
W310X60 eleir.'v/^'f-i'o^ <sf "Hie e^^aTrows \Jsee\ Li*
■"3
ex - z^.s *ioa kw-*
EJ fe z 2-8333 X1 - t<x- !.«>* - 3.KX-3.;)1" - 22.535
FJ 3 * O.lHMHx"*'- -4 <X-I8>3 - l.03333<x- 3.6>s -22.S35*
£1^- 3.g333 X* -i (xM- (.8)* - ^?.53-T so F,
Jf/Jbc = /5".S , 15*. /J
- (o.^MHyj.aSS^-K*-*5*-'-*)' -U2.SSr)0.8£-S) c -W.Sol VW.w'
y* - 1.^8^ i -^
^r

PROBLEM 9.65
20kN/in
9.65 The rigid bar BDE is welded at point B to the rolled steel beam AC. For the
loading shown, determine (a) the slope at point A, (b) the deflection at point B. Use
£>200GPa.
W410 X 85
SOLUTION
Rk * o
EI & * 30<x-ls>*- *d<v-i.s>' -(iK*©K*-'-Om
+ C
+ c,x + c4
O+O + O+O+ d-'O C4=o
(eO Slope *+ A ( *fe ** * * ° *>
EI9A =• C, * ^s few.**
9.
4Sooo
0.7W xio"5 M
^OJ/Yx/G^rv^ ^
EIjB - (C.Xl.S^ r (hOH.S )r C?.S W.vJ
I.OTl x/cT* »*•*
/.07I *W*H T

PROBLEM 9.66
9.66 Rigid bars are welded to the steel rod AD as shown. For the loading shown,
determine (a) the deflection at point B, (b) the slope at end A. Use E = 200 GPa.
36kN
SOLUTION
\t>S
& \&
A>
** \
4v°-
DwH"&- Use krJ £>r ■po'x<S_> \r\ for /enaf-As.
3 m -H 60 mm *a , i . t
- lo.S<x-0.3>° + /0.S<x-o.^°
V
<W-»*) EX^** JS^-lSO-o^- I8<x-0.fi>z
<*#.
- lo.g<X-0.3>' + /©.S<X-0.«y + £#
(kuV) Elv - Cx3 - G<x-o.3>3-G<x-o.6>3
+ 0.9 C, + 0 = o
C, ' ~t.CZ UV-v*x
M Defice+i'»« ~t B (j J x^ 0.3)
EIy8- (0(0.3^-0-0 -O-^o -(i.£2)(o_3'> - -ajjtff W-m*"
El£A r C, - - l.« kW-l«*
0A « - -^~- - -7.&>*/t>*3 r«^
Js
- I.^OO V*l^ I
-3
©a = 7.^/oVul

PROBLEM 9.67
9.67 and 9.68 For the beam and loading shown, determine (a) the deflection at point
C, (b) the slope at end A.
I, SOLUTION, p
®* A
3^
®_<ft
^
X = Q.
i
243" eJ"
LoA^ih^ J : Case ^ a.= "3Jb--g- ' r ^ ,> ■"• "
Jc CEIL CEJILA3'^3' ^ ET
A ■ ■ ML - -(PL/3U . __L_Q^
* eel eex ' » ei
Ca) DeUec-h'on ait C ■
wsi.p.*+A: e,--Jfg-m-i-|tl- -fcff
J' *H3 CJ

9.67 and 9.68 For the beam and loading shown, determine (a) the deflection at point
C, (b) the slope at end A.
SOLUTION
Loading -I : XWm *)c*jyr\ J/oaJ P ad B
Use C&-S* 5" tff ApptMwfiV T) wJi + li
Fo* x-^o-j a',\je^ extsfi'c co^s*
^,\/e^ e«fisTic t^« is y ~ '
yA - ^ r-V. - ~ - _- . ~ Ol JT-T
CELL
€.EI L
»l EJ
P'-P, ** !r3 •>*** L-Z. , x^a -f
°A * " CEIL " »< EI
t^ nil i* + r , _ JL £42 . _±-£L! L- fii t
(^ Ue-Mect(0w at L yc - - ^ ^j * ^s eX - i^ ££ I
w sv *+ A ^ »^^ ♦ #f£ - ir^ ^

PROBLEM 9.69
9.69 and 9.70 For the beam and loading shown, determine (a) the deflection at the
midpoint C, (b) the slope at end A.
SOLUTION
s wr . i M/Z*
384 £J >
A L
M £1
d^niiuiuiii
MU
c
-®
/*F
/#?7
Uo^in^ II- Case 7 of Appellor "O.
J- M^t
' '€ £1
Wa 3t=i
w
irt M,
_ wl_*
Wi
12 i
(cO DeJjfec+i'oyi at C. yc
38* fl l« £1
&A 36 EX
I2Z EX
(b) S>hpe *A A.
2*4 El ' 36 £j.
7* EI ^*

PROBLEM 9.70
9.69 and 9.70 For the beam and loading shown, determine (a) the deflection at the
midpoint C, (£>) the slope at end A.
SOLUTION
L= Ha _, <x* Ci. 3 b*- Sou, x* Jta
h 6EXL C£I(4-0 *' **
Loading JE U<ul *A C C*4« 4 fl^ AppwMx V u/i+t /_- 4<*
Jc H8£I 43 EI 3 EI
, .£11 * rc^f - . Efl!
&
£J
C*,se -5" of AppevUix "D
Lt q^ > a. = 3aa fc> * a. } * * %<k c& p©.v>+ C

PROBLEM 9.71
9.71 and 9.72 For the cantilever beam and loading shown, determine the slope and
deflection at the free end.
SOLUTION
Load in* JL
J* 3 El " l» EX
*ex
_PL3
J > r ~3£X
J* J» + Of* " «EI 3 £1 ai EX

PROBLEM 9.72
9.71 and 9.72 For the cantilever beam and loading shown, determine the slope and
deflection al the free end.
■L/2—4*—L/2—-J
SOLUTION
CoufJe VI «* B.
Co.se -3 csF App&nclfx I> applied Ho poHi'o* BC.
v'- (PiKVO* - jl£Ls
y* - ys - u > *% 3 rx h £i » ex
Case I of Aflie*di*x D.
2eX i >
By super o&s it* on
e; - el!
£41
3£X
A - A' j. A" - J-£t** -Ll^ - E£
<-£
* = y*4 y>
2.2Jl -±£LL - _iZ PL3 ^ »7 PL3 i

PROBLEM 9.73
9.73 aud 9.74 For the cantilever beam and loading shown, determine the slope and
deflection at point C.
■ A
-. Ul
w =
i
B
L
I
.
P
fitl
-Ul »
SOLUTION
GET ~ ~ * EX
Jc SEX 8 EX
°B * 2EX - * EX
^ « , ?(L/ay _ j_ PL?
Po^Tfo" *BC fev^c»rws STV^M&iiT.
*
: J6 + ^ yB " ;w IT ^ EJ *8 £X
tJc - Be, + ©& - €> £T S ET " 2V EX " ^FX ^^

PROBLEM 9.74
9.73 and 9.74 For the cantilever beam and loading shown, determine the slope and
deflection at point C.
wL2
SOLUTION
W ^.pfiJluA Hu porfp'o^i AS.
C«-SC # oT Appev\J*,x J) &,ppi;ed -b porfio* AB.
Gft ' " mi ' ~ t? £r
jb ger ' i2« ex
a
48 £1
J* J.» W ;cts oaei n ex . 3&t BX
LocJ(\r\et XL '■ Coi^-fe^eiockwi'se Coop Jit. -j*- ^ff>J;eJ cc\ C
V
Er
an Er
J* 3.EX " ** EI
ft _ a * ft • l ™±: . -l »ti." - J- *>±:
3, - yc + «*. - 48 E3 + Zh EX 1* fX
E3 **» ex " ** er

PROBLEM 9.75
9.75 For the W360 x 39 beam and loading shown, determine (a) the slope at end A, (b)
the deflection at point C. Use E = 200 GPa.
S kN/m
SOLUTION
K-i
* * 2HEI *4 EI " el
y£ * -fig U*-*LX% + L3xl - - T^[(l^V.W3.^(l^)%(3^(l.3)l
ET
a - PiCi^-V _ (asy?.Q^t^.Ot __ 33,86/
.-«■ *
(^ Si*pe *4 A 9ft> - "-~"S + 33.8&I ^ _^5-s^|0-» ^
-J
(b) DcM»*+i«* *+ C
9A - Z.S$»I<3 v**\ "^3
2o.<?3*/ + 34.176 . „ . -3
— = - 2.7C3 */C2 *
2OH0O
* 2-7o

PROBLEM 9.76
14*1 kN
9.76 Forthe W410 x 46.1 beam and loading shown, determine (a) the slope at end A,
(b) the deflection at point C. Use E = 200 GPa.
SI) kN ■ m SOLUTION
B
i_* LU.'\s : Fortes \*
+ ^ B
6£l &FX "* EX
M
6EIL
(x3- L1/^ r -
80
-U^-^'UsVi = ^
6^r cs:*)
Lo'bd; n« H Noi*ie*t &.+ A
y* * %t=r " -a, ft. " " E
.353
3 EI
^c * EX
L©«**K«<1 JJX
3
3 FX HI
P= IfO ^M
6 ^ H
|S6>/o"6 f»7V
(a) Siope a.+ A
0. - €7.40^33.333-^8.75- = _ „ -i ^
^A 31 loo
9A * 0.60I*K>
-*
<» DrfJU,*;.* «f C y* '^+/3S- 3^.583 , -3.fe7x/o-»^,

PROBLEM 9.77
9.77 For the cantilever beam shown, determine the slope and deflection at end A. Use
E = 29x l^psi.
2.0 in.
M
SOLUTION
4.0 in.
r-—2 ft 4*
V
A
C '
UfUTS. - Forces. i^> W*f>*, -Pe^g-Rs m "ft".
Caie J fff Appeal** TX
A' - PL2 (\)CS)% . J2=£
°* " ZZZ ' Z&X ' EX
v' -—-£il *-iUii)? ,-ILl
647
* v v
%
Case 2 ot Appeal* D Ap|>J\'e<=i -ft* porf-'ho^ "BC
°8 4ex ^ ex TT
^B " VBX ""BEX " £X
ex
y* - j6 awe -gj v}^ez > rr
El
£t
EX
J* Y* + y* ei ex er
- 0.34O ,*«. |

PROBLEM 9.78
<D
A
e
®tt
f -if « 4> . '
L—3*+ -J
9.78 For the cantilever beam shown, determine the slope and deflection at point B Use
£ = 29x 106psi.
2.0 in.
\^*\ SOLUTION
40,in- 0«i4s: Forces i*. tpsj iU,34l* in-Ff.
u,i+k P ■= I k;P 3 L-6~-FY0 x = 3f+.
Ac le £*£
Aiji/3-fiMj He Sl'j* &fl - °'-"
Ja ~ 5ei - sex " ei
0,4 GET CCJ £1
B-
9a * 98' + es"
ex ex
EX
O^U:

PROBLEM 9.79
9.79 For the cantilever beam shown, determine the slope and deflection at end C. Use
E=200GPa.
3kN
3kN
SOLUTION
U*i,'|s * Forces i" ^M3 -Pe^<j4ks in I*).
sioox n.5
®
0.7S
&
■+T+-O.S
C(S.se_. I of Append iy. D <z.pr>J(ieJ\ ^> po/Tion AB.
v' - -£kl- (3)(o.is^ _ _ o.Ha.i8?r
^fi 3EZ ' SEX
ex
EI
6
Portion BC
c>c - Os ^j—
Uoa<rkn<a 31: CoMceto+rA-TeJ i*oa.tfl at C. Case I o~F Af9/>e*dix D
A " - £jj , (3Xl.as^ _ Z. 34375
CT* * 2£T "" xer " ex
J* " 3BX 3 EX EX
Bw 5i>p«iT" po* i.Tib*
yA - e>A + ©„ gp*-
D«FJe*f'o« a\C n| - - **6»*" -- -S.SZ WO* k, * 5.S3 **> 4 —

9.80 For the cantilever beam shown, determine the slope and deflection at point B. Use
E=200GPa.
3k;VJ SOLUTION
Dm'ts ' Force % i* kM^ Jw^Hs »V **
S100X 11.5
<t>
-0.75 m--
B
I
sVidww.
LoaJirw* X • Case \ of Appeal i/ ^
6 "XEX ' ^EI " " EX
w'-^£t!- - (Ofo.lS? - _ 0.8^3.75
JB ' SB J 3 EI EI
Loacii'riQ H : Case 3 ot App**4iy. T)
ns
~B EI £J £X
m « - - MLl - (L5Y6.7sV . __ o.HSLXSlS
J8 " 2 61 £T * EI
e6 - e6' + Ofe
EI

PROBLEM 9.81
9.81 aud 9.82 For the uniform beam shown, determine (a) the reaction at A, (b) the
reaction at B.
SOLUTION
I
©
6
Co*\ Si'detr "Qq as region cJa^T OjaA V-eyi-rVce
U»k * 3BX
\ *
3a c
] ^
1
r
o
6
M,
Lo^ir<\ IX C<x5e I *.ppf'teA +^ po^+lo^ Ad
^n - - PCl/3^ _ _ ± EL!
Qcln " 3 EI ~ Bt Ex
(y^jrr tychL + %•(§)„ r - £ fx
(>)m ■ SEr " si ex
5vpe^p69i-froM *.*^ cows*fr«'^t. y8 = Cyg^ + (Ye);n * ^y»)m ~ °
u
£p
5 r
X
3
Sf«J-fes
VlZF^o
OZM^o
^A - P - P + f ^ B °
f?e/ffM —

PROBLEM 9.82
9.81 and 9.82 For the uniform beam shown, determine (a) the reaction al A (b) the
reaction at B.
©
fU
«
#
@>
1 ' * if 4/ + 4/ ■ t
W
■vL.it,, V
A
X6
£
C ^ '
nr
.2
iA i— %l -^
SOLUTION
Befti^ is mdeTe^AMrtftf*: 4*. 4-i'rs.^ decree. Cowhide/1
R»L3
&W
3FX
8£I
M,
Loa^ih^ H r C«.se 2 of Ap/>«n«ljy *D
LoiulinjlB £a«c 2 of Afl»t**ta,D (Pw+ft»*C8)
l^t C5i « ex
^C-'IB SEI " '•*« EX
pow-fi'ow AC rcmA.ihs a+Nu'^bf'
3 ft * fl *3*f 5X $EZ 38V cr " ° ka ' 75* W£- T
EX * « 3*f 5X
Statics

PROBLEM 9.83
9.83 and 9.84 For the uniform beam shown, determine the reaction at each of the three
supports.
(D
j? c ®
yffir U—
Y^
x..—zzsv
T
I
2P
SOLUTION
CY )t , SJ2ti3 , J- ^j,1
Lofi^l in* JE C«tse £~ ot App*
- - M9 er
Ti
*
Superposition cl*A co^sfirai'ni yc •= (yc\ + Cyc)n + (yc.)m - °
X ^is . JL £t? . XEL* - i&£ . JL£l! - 0 0 * U p i
Stotrc s
R*
j i
2?
fc
-0 ?Me = O
K* - 3Z. y t
£P' P + ^P-^P + %E = <=>
3%
ft'sft

9.83 and 9.84 For the uniform beam shown, determine the reaction at each of the three
supports.
SOLUTION
V»epAtcc He JoA^ivt^ by iod^d'ngs J **^ -#-
^fBh 3EIL " 3FrL " 243 Ex
y& * (y>>j + <yA r °
S+d-fcs
vr ifj - o
*-r - 3^+ ^ = ° t^* lit —

PROBLEM 9.85
9.85 and 9.86 For the beam shown, determine the reaction at B.
SOLUTION
1=
i
©
R„
©
>
W
i
Jl-Jl
[tit mi a
l^a*Ji'*« H : Cft.se 3 of frpfiendiY T>
LcKxcKng HE •" C&,se 2 ol Appe^i'V T)
(y§)*
2£r
(e*)m- -
'm
Pup e/^poS / Ti'ori uv\e\ CotSTlNtinT
ye - (ys^ + (yo* + (yt)B = o
3£X Ke ££x r,s " gel r ^
il I? .1m wL^. _
o
o
en
fo)

PROBLEM 9.86
©
J
R.
oM"
4 ® i?°
4a V
8
9.85 aud 0.JW For the beam shown, determine the reaction at B.
SOLUTION
Sea.** ia seco^l clej^te i^de-re/WnaTe . Choose
^a c*.v\A M» as redu*««inT ^€6^.4w>w&.
l>)r jir .> (9a)*" 5Ir
__ia.
/ft\ - M.(l/a) - fcUr
(fib). - (QJ* - i *f
-3£X *B + J>£T U' + 8 EI ~
% * ^e)r 4 C©,), ^ fe6)„ =
EI
O
'&** + iiM» + *# =°
<rt
&

PROBLEM 9.87
8001b
9.87 The two beams shown have the same cross section and are joined by a hinge at C.
For the loading shown, determine (a) the slope at point A, (£>) the deflection at point B.
Use £==29 x I06rvji
SOLUTION
| 1.25 in.
T
6 in.
#00 &.
ft
e
goo it.
gOO Xb,
k [ C
-J P" Usmo ^e Wo^v ABC
1.25 in. J J
f?t = 5-33.33 it.
U&im4 CAw+i'ta'v tea.**
Case S Jt Appe^.V D P^ too A. /. » /g .V.., ^ lain.., t- 6 ,'*.
y* C5TL COC&teo*io* X IS)
w i- Pb*^ - CSQg VOX/3 f . _ 13.017* 10** \n
J° 3EIL (3X5.900K/OO08)
Adii+ioM siopc *.*k Jested" io» dv>e +o <*o*ev*e*t <$ fwi C
* Lac ' S
(a) Si*?* *A A 9, - ©; * ©a" - - 2.1695,io-*. 2.8«fi»|o'
(b> T>cWe*W ** B ^»y»+y/ = -lS.*l7*|0*3-3*.7lN/o*
- -47.7y/o"S i«, = 47:7 x/cT* ivi, 4- -

PROBLEM 9.88
9.88 A central beam BD is joined at hinges to two cantilever beams AB and DE. All
beams have the cross section shown. For the loading shown, determine the largest
allowable value of w if the deflection at C is not to exceed 3 mm. Use E = 200 GPa.
12 mm
A j
Hinge
0.4 m
8 (
0.4 m
7 D
0.4 m
V E
Hinge
0.4 m
24 mm
Ca-ses I 4,*A X o-f Appendix T)
?r
SOLUTION
Lei a - o.4w
Cai+t-Peve^ hed^A AS a^ CD.
Jc -- 3ez " sex " *» EX
8
& (Oct hioue.
*
384 £X
v " - v« - v -- - -U- ^
Data-' £> ZOO*IO* ?ol t I* j$(MK\zf = 3.456 WcT*^* 3.«*5£*fc>n ms
El =0?OOk|Oc,)(3.456*/o-*) - «/.* N-m*

2.2 m
4-, 2 m -*|«- 2,2 m -*]
3°
1
Jut
1
PROBLEM 9.89 9.89 Beam AC rests on the cantilever beam DE, as shown. Knowing that a W410 x
38.8 rolled-steel shape is used for each beam, determine for the loading shown (a) the
deflection at point B, (b) the deflection at point D. Use E = 200 GPa.
30kN/m
SOLUTION
Dm/Is 1 Fo^ei in kN^ Jew^Hs iv\ w>.
E"I^ (2O0*IO*)6:i7*lO**) =• 45.4 Wo' N-mx
= ZSHDQ kU-h^*-
For sjPope «*\*) Je-FiecTio^ oA C ^ *->se C«.se I
DC&-
e- ' Jet ■ vkisho*-) ' e- nu"l° *•*■
yc-. .££., <"£?*•»?. -i. m*tio» *
»ow «,+ B «LS5w^inA "fli*i poi^T C d»es /»o+ Mo^e.
Use C^e Q of V^lfx D. (YBY =-■£*£ T JSX^Uiiil
= ~S.7CHZ x/O"*
A«biiTi*oM«M eltT/eeTi6*\ «-"fr S etae -J» move w\«i*T ot -p©iV\*f C
To+ai* JePie&f/e»w a.1 B Yb ' (Y«)i + ^Ye)i * ~ lo.3£*lcTSrn = /o.38*>^ 1 «^
-1
r "23Jw/0 ^
£3. I w»m i

PROBLEM 9.90
20 kips 20 kips
9.90 Beam AD rests on beam EF as shown. Knowing that a Wl2 * 26 rolled-steel
shape is used for each beam, determine for the loading shown the deflection at points
5andC. UseE = 29 x 10*psi.
SOLUTION
For e^u»I Ji*Uv»ah rff be*** ABCO 4?A - 30 k.ps.
tff Lea*. 46CD. Us
777777
^!
///••
*" *'**" " ^-£-° ■ "^ - "ifife - - ■HL^-^W
Ho"1 ft
= 0.17/ i*h.i

PROBLEM 9.91
ft 91 For the loading shown, and knowing that beams AB and DE have the same flexural
rigidity, determine the reaction (a) at B, (b) at E.
SOLUTION
A
A
Re
p
Q.
*— <X.
3r
A
t-,,JL
E
Um'ts ' Forces in ki'pa ^ iev*4U in f+.
Kw teaw% AC8^ usi'^a C^se 4 »£ App •«<-/■*" t>.
Foir L»cav*\ DCE" utfnj Cose f if Appeal* "D-
Mct+«-^t"ng afcfjfecfVons <*-*f G
4-8 £1 48 £^r
P- Rc = 6-3.7CS s ?.OS2 fc.ps
Usi'i^ free bedy AC8 8M»--o £a R6 - al?c * o
u*;^ £*** W} dcc Om, = o ^t Re -1 (p. 10 = °
Rff T i(P-R^ = 1.0/6 K.>*

PROBLEM 9.92
D mSiinmfiii
O.Z m I..6 kN/r
I
j&
-4-mm diameter
t t t * t I
40-mm
diameter
■0.18 m
«—0.18 m
MO^~
D
P
©
A
T...
9.92 Knowing that the rod ABC and the wire BD are both made of steel, determine (a)
the deflection at B, (b) the reaction at A. Use E = 200 GPa.
SOLUTION
Le+ Fqo ke +t\e 4etA*ioM 'm vCe-« BD- Tlae
O-trtel VJ (, 11- )
Lamina £; C^e 4 o-f Appeal* "D.
to), - ^ WL*
/{lit
V
iUnU:
vk J y y i« >lr 1/
\A/
..L.:L: ■ *K * 'f ^
j;
DeFjP#*fiW a* B
-Sao = ^a' (;/ft)a + (:/ft)n
- FeoL3 jt w/L*
£f\ " W EX 38f El
v £A 4a ex ' ^ " 3«*f. ex
A* ^■J*"^?'- /2.56G**** l?.5^»,/<>t|h
EA '
L -" 0.36
v*»
W a |.c x|0* M/**
|S.2^6^P*" Pio - l3.<?2^>/0
-t
it - 1/7.74 IN/
CoV
. DrfJc^ko« 4S Sa = §^- (\t7.wKl*r&*lo''><I.Zi* IO"V r: o.0O«7*«i—
EA
<» £a = *c r * IwL" F»] - -i[06ooXo.iO- H7.7*f] - ^Ml

PROBLEM 9.93
r
''A
§
9.93 Before the load P was applied, a gap 60 = 0.5 mm existed between the cantilever
beam AC and the support at B. Knowing that £ = 200 GPa, determine the magnitude
of P for which the deflection at C is 1 mm.
A V-
60 mm
_i
60 mm
T
SOLUTION
Cort'M'cJcv po^iow AS of beavw AQC .
t*i* kac«j*-es forces *P «_***( Re ^
8 pius fk« cotjpAv Pol. Tfie JeWe<-ff©M ^4
Osi'n^ C<a-Ses \ awal 3 of
3EI
2 ft
(^*WP-#Ra = «*■
0)
8
?
■*R- Ol
'I?.
©
The ^etJ'ecTtf^ <*4 C Jepe^-S on 4-lie JeTO **»/«* TV o-
of
Fo^ J'oA^trt^ Xj i>Si** CtSC I c^ AppealX "D
3.ET
*
Sfl = 2EI
ft
J
Portion BC v*e^Ai*^s S+^At'd li+
BaU' F*2oo*IG* P«.
EI= 2.IC x loa N.**
hi Mi
O.0£Cf7 P - 0.0M/67 R»ft = /08 (lV
P - *.G3 «io3 W - 5.63 kU I --*
Rrj^ <S.« v/o* M

PROBLEM 9.94
60 kip ■ ft
9.94 Before the 60-kip*ft couple was applied, a gap, 60 = 0.05 in., existed between the
W16 x 26 beam and the support at C. Knowing that E ~ 29 x 106 psi, determine the
reaction at each support after the couple is applied.
SOLUTION
W16X26
M
©
17fr tag"
GO
R*
fl ion
3.3Z7
Units • Forces iV k.ps /ttij-f ms in Tr«
Lo»Jr*« I : C«Lse 7 of Appendix X)
^ " 6EIL ^X
w; + U M - GO I^.-ft ^ L* I3f+S y- G.5*ft
Lo4.#(in4 It : Case M of Appeal* 'O
,vx s All 3 da? ft
* 755-.07x/O_t Re
Deficc+i
t)2MB: O
Ra = CIS* frps, t
.3.
=1 O
■=- o

PROBLEM 9.95
30 11,
9.95 A 5/8-inch-diameter rod ABC was been bent into the shape shown. Determine the
deflection of end C after the 30-lb force is applied. Use E= 29 * 10* psi. and G = 11.2
x 106 psi.
SOLUTION
Let 30 tk> = P.
Con5iJitr +o ** S i f>« °t red A6.
™ " GJ CW "" sr
Co* s i de*'' be* el i* n 4 t>T A B
3sr
Co* si clef ben J inn of SC
_ PL3
(Cas* /. AP«.0 )
to)
m
3er
Ye " (ye)x +(ys\ 4 fya^ar
7*MoJ 3 /
y0 * o./?7r/M.i
*•' -
2n-5tf*ics*
/ 317.
V 167.

PROBLEM 9.96
b = 0.4 m
9.96 Two 24-mm-diameter aluminum rods are welded together to form the T-shaped
hanger shown. Knowing that E = 70 GPa and G = 26 GPa, detemiine the deflectionat
(a) end A, (ft) endfl-
ISON
C
J
le-fc
P
/
£
SOLUTION
Consi'tfle^ t©^Vo*\ oi pod CD
Com Si'&iev beMtflrn^ o"f t^od CD
(l8o K) t f>^
r=?fl
(a«L Ada tO
If
-^
Consider beweli'n^ ot ^©el p&v-|-|*o*\ AC
(Ja)m
Pa3
3HX
a.
S>i £i>i> tf/1 poS ( Ti\»tA .
- pj„a*k _J£ - -Oil
- v ? sT 3H 2£Z J
-1 i
■-" hnY
GJ" = £46-87 N-kx EX- WHO.OZ N-nx
Ol - 0.£ m J k - 0.4m
- 31.2 •»■» 1
•x Id w\
17/27 „» I1

PROBLEM 9.97
9.97 and 9.98 For the uniform cantilever beam and loading shown, determine (a) the
slope at the free end, {b) the deflection at the free end.
SOLUTION
S * L-+L -- ^L
e*/A * a- -i£3
By ^aco^of woi»i«^T"**■€&. ■flto/****.
^a - **/• • t ex
-**

PROBLEM 9.98
9.97 and 9.98 For the uniform cantilever beam and loading shown, determine (a) the
slope at the free end, (b) the deflection at the free end.
1~
fe-P- fain.
A
SOLUTION
B
Ha r -i CyL*
uUl
.1 i^
~ *> EI
EX
a - a - x w>i3
B<
seco^« wowe^
1
V* r *
A _ (4., V_-L»a4?\- . I *j>k
A - Vs-L J^ „ ej. J- -^~ ^

9.99 and 9.100 For the uniform cantilever beam and loading shown, determine (a) the
slope at the free end, (b) the deflection at the free end.
SOLUTION
Xt » a+la. = fa
A. - A + A * - *B^- * E^-- - ^"-^
U/A * A*, + At*a
EI
VB r *•■/* * " 2 EX

9.99 and 9.100 For the uniform cantilever beam and loading shown, determine (a) the
slope at the free end, (b) the deflection at the free end.
SOLUTION
?J^ce lr-e-Fe^e^e H*artge*+ of A. BA - °
A,' -i(&*0 * -iff
x^ - a + ia, - ^a.
8«j Second moMewt-ai^e*. T"Jieore»M
t*/„ ' A,x, + Azx4 = (-f^K-ig^
6 £1
Js " *-u« ' 6 Ex

PROBLEM 9.101
9.101 and 102 For the uniform cantilever beam and loading shown, determine (a) the
slope at point B, (b) the deflection at C.
M/er
S. J
SOLUTION
A - (¥U\(L^ . X Mat
Ht Lei M.» I " x EI
2 EX
©a = ©a + ©8M = O
X
ex
£r

PROBLEM 9.102
9.101 and 182 For the uniform cantilever beam and loading shown, determine (a) the
slope at point B, (b) the deflection at C.
U—U2—4*~-L/2
M/ez
SOLUTION
Place ve+e.^ewce 'Wn^ewt <x?t &* ©a - ^
*» *tT5r a* I 3Z ex
A* * "aViS K» ) 3* er
3B/A * A, +At+A3 - -^ IsT
*ic - -a a. 3 *-

PROBLEM 9.103
2(> kN/m
9.103 For the cantilever beam and loading shown, determine (a) the slope at point A,
(b) the deflection at point A. Use E = 200 GPa.
SOLUTION
Drills' Fo^eea (*> kMj -Je^a-Us i^ wv
E = ZOO * 10* ft
W250 X 28.4
H.3S
I.S4C
!tef. •***>.
-c
I - 40.0x10^ *^ = 40.0M/0"" w>
EI
^OOO
*J.95""/o"3 *i"'
A, ' iC^%*(o"*)(a.aV S.W5*to
-i
3 - 4-
ifc*) - 0.7333
t*7
X2 ■ ^(s.tI * 0.G7S m
ec - ^+ a
fe/A
0A ♦ A, + A:
O
&L * £.%X*\o
-i
-A*.
- o -o + A.x, + A**t

PROBLEM 9.104
1.2 kips
Cil W8X 13
lt>sM/££
-\.SoH7
3. ass-* l^
io* m/ei
iref. +*
9.104 For the cantilever beam and loading shown, determine (a) the slope at point A,
(b) the deflection at points. UseE=29x l^psi.
SOLUTION
£ ■= Zcfy\OcfS't ~ 2^*10* k*<*
A, = i(:/.5047x/O*4)(loN) • -7.S23r*/cTa
X, r i(lo) = 3.3333 4H
IF ' UX7W) U."Ulfc*lo it
©e/A T Ai4 AJr * I'd*!0'
- 0.13l&
-t
e. - 9. - e
-i
t/A
^.G7x/o" v»**l

PROBLEM 9.105
9.195 For the cantilever beam and loading shown, determine (a) the slope at point C,
(b) the deflection at point C. Use E = 29 * 10* psi.
4 ldjyfc
I.Skif*
-3.71**-
O-'SfcSS
r«f. +»*
3.0 in.
"te/i
SOLUTION
L)rti"T^ : Force* iVi k.ps > J'eui-t-Hi* "^ tt.
e - 29*/o6 ps,v s ^^x/o1 Jt«;
- £oo.74k/p--f*1~
M
Mx- - - Q=£ifiJ = .s^a^^-1
Pi
£1 goo.7*f
A, T 4(- I.W'°"*)^ - - 0.4<?*337*/cf
/ - £• a = 0.-2,333? 4*
ace
Be * ©a + &
^
'C/A
- H.M *to* r*A
-c/A
(l.^3S3 K- 3.7^5 kio"?v*
-P+
-s
y<
- - e,7/ */o
yA + CO ft*) + tc/vi
^ + c - 5:?z wo'3 = -<5:22wo_i -fr
-3

PROBLEM 9.106
1.1 kips 1.1 kips 1.1 taps
9.106 Two C 6 x 8.2 channels are welded back to back and loaded as shown. Knowing
that E = 29 x 106 psi., determine (a) the slope at D, (b) the deflection at D.
SOLUTION
C6X8.2 E~ Zlvio'-pi.i - 29 WO* k*.-
I - GWia.M = ZQ>.Z iui"
- 5"H74 kt*p. ff-
M
er ' ex " ex
<H
_^^"
*i
6,
't>M
x, - 4(0* * ?*.
I^frp. +«.^l
e„ r a v ©.
TWA
- ^.ev>*/o"3 v^*j.
y» 7 ±o/n r 2S.o2*lcTl -f+ = 0.3OO m 4/
-s
^ zs.oiYit's a

PROBLEM 9.107
9.107 For the cantilever beam and loading shown, determine the deflection and slope
at end D caused by the couple M0.
M/ec
Mo
SEX
SOLUTION
Draw ■£= Ji'a.««v^ai*^ -
A ^ - HA
^ EI
Qc* s A,*A£ + A,
1L M0 a
" ~6 IT
w- "(S^)fe*o-(|it^o-(^)rt^. -f-^
EX
~ 6 £T
*
- t - * IS Ma1,
'DM
« EI

PROBLEM 9.108
9.108 For the cantilever beam and loading shown, determine the deflection at (a) point
B, (b) point C
f_J_ WO-' V I WCL*
SOLUTION
Co.") DeMee/f;*- ai B,
36 er
ex
Cb) DefPec+.**« *4 C.
te/A - A,(a+^cO + A^Ca+fa^-v A3(a-ia)
-(*#V*0 = -%*g
y* = t
/y»
V7 ntft
72 £T

PROBLEM 9.109
9.109 Two cover plates are welded to the rolled-steel beam as shown. Using E ~ 29 x
106 psi., determine (a) the slope at end C, (b) the deflection at end C.
4 Idps/ft
.1.2 kips
W16 X 26
/o* M/EI
o;go*/S
SOLUTION
Tor W 16*26 t-oHzJi s+ce^ sech'o^
Fo^- T^t ■fw* coses-
A +o B EI, r(2<?*/O*)C3o|+;?<70-V)= 17. »S"I HO*&?•'*»
A, = - iCfl-^oA^ie^X^) " - 4.os/k/o~3
ft'-S^-'-^M-^*-1
»rf- +
A2^ -i(o.£73?-O.30:Z3Xte>"1X3) --0.4$7/to"3
ii. r __ (MKOC6) = . 0 60^x,o3 -ft-'
GO-'- ©e -= &a + ^c/a ? ° + A,-*A1. + Ai - - S.i3*to'* *+A —
* O + O - (l.0Sr>./o'3,)(G) -(o.H3T*<o3XO - C/.2o*?Wo"*X7-S)

PROBLEM 9.110
20 kN-in
W250 X 22.3
IO* M/EI
-3.%oZ
Hi
1 1
p
*-O.G —
1.1
-*oa k
9.110 Two cover plates are welded to the rolled-steel beam as shown. Using £ - 200
GPa, determine (a) the slope at end A, (ft) the deflection at end A.
SOLUTION
U«i4s'• Forces iV» kW^ J'ett^-ths i* m.
E = ZOO*{0« Pa
ri,(78K|oe W-w1- ' ^7SokM-^z
d - *£* * £ ' n^S *"""
Ad* = lO.0£Z*tOG mm1*
- H9.o3</t>* **,* r 4<rfco3*lo~4v*H
- 1805 klO.*1-
-2.iloSo
Dr&w M /£2 Jt\a.^ct^\ by p4*4$
A i= B ^--s§£r * -3.4Gc**fO-s m-1
B -U C
£1 " S?&o
EX " 9*0^
4
- -2.03T8 *|0~ i*"1
Jl* - t3o)Cl^)" . -^2030^/0" w
Hx • (ZX9SO.S) "
PJa.ce y^^e^^fe*^ce +«-ia^«ia1*
A,^ C-S.HtoaxlcT1)^*)* -2.7S82"/o"S
(a^ SjPope «,+ A £A - ^-©a/c = 0-(A, + AA+AO *" .6. ld*/0***l
(M Defjfe^.^yv at A

PROBLEM 9.111
9.111 through 9.114 For the prismatic beam and loading shown, determine (a) the
slope at end A, (b) the deflection at the center C of the beam.
SOLUTION
■£^*.
pi
ReocfiowiS RA " r?jj - ^;F
(Ol) SPo/>e J" A- 9A - ©c - ©^
^A - ° I4EI
ft = -rf-?^
16 BT
(V) Defied ;^ «,4 C
*=-*.** - AC^ = -Cliffy

9.111 through 9.114 For the prismatic beam and loading shown, determine (a) the
PROBLEM 9.112 slope at end A, (b) the deflection at the center C of the beam.
ip
-*p
fcA/C
*
A
c
B
^Ittti
SOLUTION
A _ k . i(J-£k\L - J-£L*
©A " ©c- S*/c = (9- A,-A,
U<* fx 6 + 44 ex s

PROBLEM 9.113 9.111 through 9.114 For the prismatic beam and loading shown, determine (a)
slope at end A, (b) the deflection at the center C of the beam.
M/£I
Kx -*|
r
M=
Er
SOLUTION
S^*\***T^C ft./ OeCf*n dvii^ -fG«<;\(V>« .
(a.) SLpe cd A 0A
(b^> DeWe^'w *+ C

PROBLEM 9.114
9.111 throngh 9.114 For the prismatic beam and loading shown, determine (a) the
slope at end A, (b) the deflection at the center C of the beam.
SOLUTION
Til
L/2
\B <
fc
■1
*—
d
jm
L/2 *
Qe^tAio*\s R"A ~ f?e * WO.
DCwcliri^ moment
Dr
Cv^ AB
M - wco< - ^wcl
M r "£ wet
£1
HI
A.*iTAw<t
k
'^ P
«»<rTs
■b
EX
M* X .**&.
X W/CL
3 .
A»»
i*
6 ex
M* EX
£-A)*Jr*f(L-2^
(a) SA>pe *+ A. GA - ec - Q^ = 0 - ( A, + Aa + A3 >>
-L wa? . X *£&* _ X ^i&Vl -?« ^ - *!£l(X\ . Xrt \
- z 3
X,? 3 * -> " - *
3 Ex ^ EI lL ^ '
1 W.a
3 EX

PROBLEM 9.115
9.115 For the beam and loading shown, determine (a) the slope at end A, (b) the
deflection at the midpoint of the beam. Use E= 200 GPa.
wu = 54 kN/iti
■ 2.4 m-
-2.4 m-
W460 X 52
SOLUTION
E * 2oo *tO* P*
k- Yh - ZZ.S kU/m2'
For A K» C M * Ra x - i kx*
A4 C
Aa-- - ^(MAJLtWo^ftH)* -0.7335-6 */o"3
(a) Siope <J A GA - 8C - Bc/a * O - CA, + A^
- - 4.4olS"*ilo~s + 0.713S6 */£>"* " - 3.C7K/tfs y^«J. -*1
-3
r - 5.63 v^O »n
5*. £3
WW
I

PROBLEM 9.116
9.116 For the beam and loading shown, determine (a) the slope at end 4. (*>) the
deflection at the midpoint C of the beam. Use E = 29 * 106 psi.
1.6 ktp.s l.fi kips
SOLUTION
. a
din. 9ln.
—18 in.—*-]
ios M/£I ;/'
A, * i (o.mssv/o^Xig) - ^^v/o'
-*
(^ S$ope Ji A 9A - ec - 8ca * O -CA.+ A^ = - ff.^x-fo"3 **«,
= 0./5</2 .*n J,

PROBLEM 9.U7
9.117 and 9.118 For the beam and loading shown, determine (a) the slope at end A,
(b) the deflection at the midpoint C of the beam.
SOLUTION
21\
C
'^..
>ref. +**»
©t=o
A,- *feV- *f£
* EX
■ 1 El *

PROBLEM 9.118
9.117 aud 9.118 For the beam and loading shown, determine (a) the slope at end A
(b) the defleaion at the midpoint C of the beam.
V
... 29
?
SOLUTION
tVa*v V^ t^3 c<^' rr^; J» e^itoi^S -
P.P*ce KiC. 6ft = o
r - iL fib!
: 18 El ^
ref. ***

PROBLEM 9.119
9.119 For the beam and loading shown, determine (a) the slope at end A, (b) the
deflection at the midpoint C of the beam. Use E = 29 * 106 psi.
17,5 kips 17.5 kips
2/) kip-ft I I 20 kip-ft
SOLUTION
-4ft-
2ft "2ft
D
W16 X 31
■4ft-
£ ^«f, 1*W.
I = 375" in "
- 753*1 k.p-ft*
■P.-J lift.
J?A - Re - 17.5 ttj*
* Ma =■ -*o + (n-5)(4)* -JZd + 7o kp.-Ft
- 2o/f r \4e. - -2o ■* 7o k.y - -ft
A, r i(70.V4) ^ 1**9/EXi
i A3r -&*)£■) s -Wo/EI
P<P&*e re-Pe<r«nce •famjew't «■+ C. flc - °
(a) SPope o* A. 9A^ 9C - A/,
©A r O -(A,+ M2+A3) r - |6£>/£T
- _ Jfo - ~ - 2 i/«? x/o
755-21
-3
r&
W.
(b) D«Wec+.on a> C lyj - tA/c
^ JUL- - , i/^Mo^ ft * 0.1/33 ;«. I
7«S3i

PROBLEM 9.120
9.120 For the beam and loading shown and knowing that w = 8 kN/m, determine (a) the
slope at end A, (b) the deflection at the midpoint C of the beam. Use E = 200 GPa.
40 kN ■ in
40 k-N ■ m
ritiiuiHnm
VA c db/
-5 m-
SOLUTION
-5 m ■
W310 X 60
-4 _ •*
M/EI
A,'
mm mil
aT^
Hi
( rx
I = I2^vl06 ***** - 129 x/6" ^
= ZS80O KM-m*"
A+ x -- £
M - 20tf - HO - /oo
A," a(7.iS|«?y/0~*)(-£') - l9.380x/0'* X, s (§£SV 3.3333 m
Azr -0.55OMU5-) * -1.15ZO*\cT* X^ti)^* 2.S"»"
A**-i(&.87«oXO T -fi.^OoWo"4 Xj»(f)fr)* 3.75**i
toi Siope «A A. 6A - ©t - 8C/A * O - (A, + K * Aa)
= (l9.38©*to"*V3.3333}- f7.753oWcTSX2.5) - (G.H£6o>/0-' Xa/75^
-3
21. o * i© ^
21.6 **m I

PROBLEM 9.121
9.121 For the beam and loading shown, determine (a) the slope at end A (b\ the
deflection at point A. Use E = 29 * 106 psi.
35 kips
^ kips/ft
SOLUTION
E = M^la'psi = 2<?*/o3ks,-
W14X53 X - 5V| i'h
D/^it^J tending ffU&A^a,*^ oj po^-T*
^ g ffftffi - 0.^8736 xte- ft-
EI
A, - Ci^(0-^w36*io','*)(6,i- ^.g^awio-*
&
rtf. W,
Pietce. y^fe^e^ctt «.+ j-y^^el^ po;<oH C
c^ et = e. 4 e^ * o
eA -- - eu ■ - a,- At- a*
- - ^J^axio" + 0.«Wia7*lO"* -» O. I65"2U*I0~S = - l.733Tx/o*S raj
8.£«n ^ lo"* -ft
^A ^ W~ t
B/C
3".3a.8 */o~3-f+ » 0.0639 in. ?

PROBLEM 9.122
hj:
<«—4*—1.7 m-
lm
-1.7 m
9.122 Knowing that P = 8 kN, determine (a) the slope al end A, (b) the defleaion at
midpoint C. Use E = 200 GPa.
SOLUTION
T £ - £00 wo" Pa
W200 X 19.3
- 3S£0 kW-iw*
Ss
^T rP'c bee*
J(0'
Pa - #a T P+S * S + -S" - 13 k*/.
Over A6 H - - Px « - 8x
Om*t SC M « - 3 X + IS (X- I")
Dtraw
El
(flu) Siopfi eA A.
Li
'r^^a*** by pa.*^iFS
0^=0
0. --C1*5
EX ~ £*■ ' *"* ~
£1
3S*o
S.I2S»lO rtjk
= - tA|5, + AsO
-*
- I. OI6x /o" n
/. G/6. »m»*o

PROBLEM 9.123
9.123 For the beam and loading of Prob. 9.120, determine the value of w for which the
deflection is zero at the midpoint C of the beam. Use E = 200 GPa.
I _J__T
5 m-
■Wk-N-m
B
SOLUTION
■5 m
M/£I
3Sw
re
f\ +»"»
Defied
d C is 2ev*o
1 ^7310 X 60 J J
M - Sw* - 46 - 4-
w/X
I A+ v = S *>
a floVs) _ - *QQ
H* T ~ £2 ' EX
A U M'S-*>Y*\ - ^0.333 ^v
M3 " ~ a*- rj A ' " ex
yp - |(0 - 3.3333 v^
y4 » ^(^ - 3.75" «
PJci.ce me^enee To-viae* T oA C.
t*/c * y* - y«. r °
r O
A,x1 + Aax\ * Aji^
£r
ex ex
w -
ISO-3I
EI
3.S+ kk)/^

PROBLEM 9.124
9.124 For the beam and loading of Prob. 9.122, determine the magnitude of the forces
P for which the deflection is zero at end A. Use E = 200 GPa.
r. i
AU-j-mmmm '■»
10 fcN
D
•-1.7 m-
lm
M/ei
-1.7 m-»4*—*
lm
W200 X 19.3
EX
EI
rtF. Um.
-A/e
- t
S/c
= O
SOLUTION
RA - RB - P+-S ( P ;* loO
Ov/e«r AB M - - Px leu.**,
= ^(x-n - pco
A+ y * 2.7 *
n* s.s - po)
A, •-££■<«> --*££
RP
A,(i + |-/.7)+ A3(i + i->7K A3(|) - A.^-i.7)-A9(i-i.i) - o
A,(0 + As(0 + A,(4^ = o
7.22S- _ 1.7 P Q.33333 P _ -,
£X EI ~Er "
P = ^f-' 3'SS M

PROBLEM 9.125
M/EX
.- X,-*
^
_*,.-«.
§
fN
EI
-A.
-A.
'9.125 A uniform rod AE is supported at two points B and D. Determine the distance
a from the ends of the rod to the points of support if the downward deflections of points
A, C, and E are to be equal.
SOLUTION
Le+ w * wetjH fe* o~;+ -Pew^-U erf ^oat.
#■* r £© * i*-*-
0*^ BCD M r - iwxl+ ^L^-^
EI * * ex h fr
t'X
£X
8 Er
A,x, * At*i s o
Ler U = Ct/<. . Diwde Lj ^
HO3 - 30 + f = O
f- - 0-«3 ar 0.223 Z.

9.126 A uniform rod AE is supported at two points B and D. Determine the distance
a for which the slope at ends A and E is to be zero.
SOLUTION
keT w = we\jW per ilnil ifcit*4i\ erf ^Otfl.
Ove^ A8 M = -iwx"1
OMe«- 8C3) M * -i^^1 * ^wL (x-<0
Ma. j. j*iii£ - -L _^Ll
EI " 2 EI ' « EJ
"« * er* » ' It EI
A - -IflM-^ r - JL laLL*
M2 - 3 ^£x >X ^ EJ
•^ = 9t- 0C/A -- O - (A.+ A^ r o
_ J. wUf.-3aV . X ^A! - 0
L«iT O - £- a«*r a>vtJe Ay HfjTjfT
/ " *U * f
£* 0.31) a^ 0.51/ L

9.127 through 9.130 For the prismatic beam and loading shown, determine (a) the
deflection at point D, (£>) the slope at end A.
SOLUTION
Reactions: Ra ' "]f ^ i ^3 r ~l* *
O
ra*)
BJ
ift<lt« "^ •
*3
A, ~ "5 K& EI *& ' ™ £1
a A
-O/A
/62 £J
L
A k -
gi EJ
8 er
•B/A
" i^a. e-j 3 V i* ex '
8\ EX '
(&) Shpe a+ A
L

PROBLEM 9.128
9.127 through 9.130 For the prismatic beam and loading shown, determine (a) the
deflection at point D, (b) the slope at end A.
11 Pit
Mo- —
M/CX
SOLUTION
Om, » o
*^r.
R« = o
D^A
v^
J.
acirawv.
#e$W«*e.*. TBt^^T *-t A.
A r a (.3 eF A * ' 1 ft
t - / afL'VJ.m r -
, - + *£ f
EX
JLfL't
5i3 EI
(Id) SJr©f><? J A
0,
, . j^tt r Lift!"
«' FT

PROBLEM 9.129
9.127 through 9.130 For the prismatic beam and loading shown, determine (a) the
deflection at point D, (b) the slope at end A.
if>
-Sp
M/£r
SOLUTION
a - J./-J-£tViN U. ELt
A4 - zV. IfcEJ'^ *» i " 158 EX
PPace reFe^e^ce 4av\«e^ *"t A.
+ (-£*£)(*♦**)
- -£_£ti - -L- £41
' »» EX
£o/a " \3i f7X >\3 3T J - ill. BI
t *• + - -L_Pil X/A£L*N - .-£-£L*
A
- - z££JL
3 PA1
>a& EX

PROBLEM 9 130 9*127 through 9.130 For the prismatic beam and loading shown, determine (a) the
deflection at point D, (b) the slope at end A.
Wei
SOLUTION
M/a
A,- aU e5T iL I* £-r
A2~ ^i t Ei /L 2t ex
ria-ce refe/'e^ce ■f«L«*ei'»T *A A.
5,* *l
9,
36 fX ,a° Q"
36© £T
•34° EI
Ed/a
CA
3*1? EI
<M
a Z- *+£

PROBLEM 9.131
-i.5 kips/ft
III!
|
8.5 kips
^2.5 ft-4-
5.0 ft-
M/ffI &
9.131 For the beam and loading shown, determine (a) the slope at point A, (b) the
deflection at point C. Use E = 29 x 106 psi.
SOLUTION
W10X45 J r 2^8 i"mv
ICSM -OI^-O - 7.5" /?A 4 C^YksX^') 4 (s'-^Xs.o) = o
#fl - 2Z.5HZ k;?s f
CW AC M= Z7.SH2X. - 2.^5 X1 &p-ft
O
M t. ,
-^^.
d
EI
is.
633.?*
A,'i(«W)' gr
PJ<s.ce reference ^A*\«e*t a^ A.
U = A<(^)+A5(4f) ~= ^
(a) Siope J A
/og.62 _ ,_ , -s .
SL 375* _ /^r\ g/1.f* _ _ IZo.U _ _ W.i£ „ UUl . .s P,
:y-r
ci

PROBLEM 9.132
9.132 and 9.133 For the beam and loading shown, determine (fl) the slope at point At
(b) the deflection at point D. Use E = 200 GPa.
UOkN |20kN
SOLUTION
A, K
W250 X 44.8
E - 3O0 > 10* Pa
I = 71. \ X /O* *»H - "7 LI* ID
»»n
r 14330 klJ-tvi*
ex
*3
a,- ±(tno.o - ^
az
A2= (f§V<^ =
£sm ? A» ^ + °"S ^ + Aa(3 + 0.7b")
4 a.uo = m*-
ET
m.
(al Siope ^-i A
a toA. . IMS _ _ U3.7r _ _ *33.7r
°A " ~ £ " ~ 6ET " El " I 4*2©
0^ DeWe^fi'on d" D.
- f *» +
B/A
|g7.5-
EX
=: -J.70x/0" ir«J.
£l(*&n r - mir
ex
£"*
* - m«? r -/«.«>*wo-a^
)i". 03 **m
i

PROBLEM 9.133
9.132 and 9.133 For the beam and loading shown, determine (a) the slope at point A,
(b) the deflection at point D. Use E = 200 OPa.
!26kN/m
63 kN/tn
W460 X 113
SOLUTION
£ • 200*10* P^
I - SSL* /ofc to**'* =■ SS6*IQ~C ^*
EI - (jtOOyl09)(SS6*tcT<') = lll-2*/0fc N-*.*
- j i )%oo krJ- *->*■
O 2"Mb- O
-v.fft +050(2. a X3.s^ + (63)^,00-0 = o
RA * 2H7.SS ktf. \
R& 173.25- ifU V
H
IJt - _ Jl O^fc H?,?^ _ _ 2 7421 x lo"5
EX H| 2oo
EX T " 2 I II Zoo
*\
^.4,7.7G k fO~3 **"'
1.37id) x/c?"5 m"1
Al = i |^(fc* ) * _ Z.OiOltv /o-*
A3 - i ^(«) T 3,770^+x/o"s
Ah" a ec"&0 * ~ l.*>±"»* lo-*
Plate r*+e^e*ce 4»wje*"l" *-+ A
taM T A,(X^S-$S) + Aa(2,75) + A»^.HM*7) + A,(l.6S.) = 13.8*1*16
t./« - A,(o.7M&) + At(o.sO = ?m7tH7*tcT* *
(ct"> Slop? *f A
ft/J Defied r«H ^3) > - tD/A - £ ^
■= f/./ff M I

PROBLEM 9.134
9.134 For the timber beam and loading shown, determine (<j) the slope at point A, (b)
the deflection at point D. Use E = | .5 x 106 psi.
1.2 kips/ft
3.5 in.
M
lft 2ft
¥*ef. +**v
SOLUTION
JT I = jjf(3.J)(7.5)s = I23.047.*«*
17.5 in. __ , ,
j^ fc - 1.5" *.0* pc/ * I.S-rf /t>J kv*
£1 5 )SH.S7*IO Wp-lV' s |Mn ^p-P**"
RB ' 2.2 k.>.
W, - (iZ.OC7> = !*.&> k-V-H
A, "- i(7Vl?-C ) - C8.C k.p-ff''
A,. * imO-^-O = - i?- 8 M*""
t ** A.
A - - la*
- + *p t-
£1 tB/A * A/(7-|)+A1(7-lW A3(7-{) - HO.o Jop-PT
EItt
C^P^ *-'.«<*"i' ^
EIy0 - 52.(33- 4^10.o) - - 26.q38 .k»rft*
*
2C. 433
1281.7
- ZO.€3v/o~S -W " 0-24* in. i

9.135 aud 9.136 For the beam and loading shown, determine (a) the slope at point A
(b) the deflection at point D.
SOLUTION
pjp.
ft.ce ^
if.**"1-
<*> e, -- L
Mil' " 48 £J
*c/,_ , L PI*"
ff» --o.

PROBLEM 9.136
9.135 and 9.136 For the beam and loading shown, determine (a) the slope at point At
(b) the deflection at point D.
M/ei
fe /-ef. +ftrt
SOLUTION
A - x /s.wi'-x , r A *iL*
n\ 2 V8 g-x J l(, EI
A - X /XwLN , =. _ _L J4ii*
2 ' " S ^ a £ I ' L * £ X
A - X /iwL'U . _ _LJSii!
in/A * A, 5 + At ^
f£ £1 24 £X
Ra=|^
- 3z ej " « fx ia* ex r us er
CV>) Drfieefiow cxf D.
m EI " a ' **8 £ r
yo
12* er
w* et

PROBLEM 9.137
50kNI
B? , C
00 kN/m
■2.5 m-
•2.5 m-»+*
D
9.137 For the beam and loading shown, determine (a) the slope at point C, (b) the
deflection at point D. Use E - 200 GPa.
SOLUTION
E* * ZOO y to* ?<x
* - ics'no'6 ^
M/£I
1.5 m
SI*/EX
W310 X 74 X " 165 * I O *"
= 33O0O IcW-w*"
+05" He = o -*^R* 4-(^alCx^)-CCoYi.s:Xo.75>)- o
RA r ii.s ku
A,- tiff Ms)- ^^
EX /v~ ' £^
-I25"/£X
75"
EX
A
i *+ C
*A/fc r A, (§-*) + A.(a.s-*|.*.s")
i7/.8?r
wj
G - ***■ -I?Ll2i"- 3V.37S~
c " *- " SEX ' ' ET
= -m&-- '.o*Mo-~j
'3>/G
t*-».
0O -OefJt^+.-w 4 D
yD * ©t **t + t
■D/c
_ /^.37r\/LgN _ 37-96B?r
£~X
gf-*3
33o6o
'- - - i?.*?/*/©'3, w>
5.7 1 m^ J

ponm v-M o 110 9.138 For the beam and loading shown, determine (a) the slope at point B, (b) the
PROBLEM 9.138 deflection at point A. Use E = 29 x I0« psi.
b
*6ft-
2.5 kips/t>
TTTT7
-i—:—15 ft A
W16X40
16 kips SOLUTION
HI = (*1*to,)(SlS,> a 15.0*2 */0* fc,p- i«*
- I64 3l<* k.p.ff*-
^ = ic?.ss kN
M,= 0a.3S->Ciff") * iK.as" fcp--H
A," i (I25-ZS)0S )/£l a l38*.3TSV£r
Aa -- -±(s»Ka* 'iOsVfr- -i4o6.*s /«■
r - WU.SUS /EI
y* - m/c + —— t
£~J
- 18.43 x /o's ff = 0.2^1 in. i

PROBLEM 9.139
9.139 For the beam and loading shown, determine (a) the slope at point D, (b) the
deflection at point E. Use E = 29 x 106 psi.
20 kips I
211 kips
J 4 kips
n
SOLUTION
VJireuw beit^tf fl4 wto^c^
i A<^4\r-
^a^v*\ OlS
n
ft 4* 3 ft 4* 3 ft*C— 5 ft —A
W12 X 40
CcO Siope a$ V
3>uv*\ cn "Kt/o <sir«.*tr«.iM£ i One ■£*>/• He Pa iV
A,v>^t owe
As - 1(^X7*") = -\is/ex
E - 29xi*>cps; - Zfxfo* ks,-
tA/or A,(^s) 4 A,(s)* - zio/ex -R-.
tA/D . _ Z7Q_ . _30
£1
tg/i, T A»(f'0 7 -S$Z.Z%% /EI - - *L 343g*/cT* -pf
(Wi t>eM«<,|,'ow at £
Ye s *-»e ^& + *b/d
- - (£Xo.H$of<t>.lo~%) - %3HSB *(Q~* - - ll. IS * IO~* f i

PROBLEM 9.140
9.140 Knowing the beam AD is made of a solid steel bar, determine the (a) slope at
point B, (c) the deflection at point A. Use E = 200 GPa.
3 kN/m
SOLUTION
-I k
30 mm
30 mm
T
E = Tpo x /o" P*.
mm
■t ^.^Wcr'1 >»V
EI - (zoo * to *)(&•?.£ xio'*)- is soo N"»^v
RD * O.QC7S kH
M, = (O.C^S'Ko.S) * 0.33375" kO-m
A,- i(O.S53W)&.rV£j r 0.0834 375"/Fr
Ai - i (-o.w&syo.nsi/er * - o.007*1 w/£z
t©/s ' A,(f-0.5)+ As(fC«^)tO-aO = O. Q2HZ1S-/ZI
<o) SP.Pe .4 8 ©u t - £* , - o.oa^r r _ o.<m7g*
1 * L o.s" ex i~x
> r *A/G " ^*B ®R

PROBLEM 9.141
9.141 and 9.142 For the beam and loading shown, determine (a) the slope at end A, (b)
the slope at end B, (c) the deflection at the midpoint C.
SOLUTION
H/El
A t J. (£l*-\(±\ , -L
L M0L
ET
" lfc ex"
(cO Siope *i A
14 FX
6S = eA + ee/„ - eA + A,+a, + a3
16 EXT + 8 £T * « £X +8"
Ex:
ex
_L M0L.
"8

PROBLEM 9.142
9.141 and 9.142 For the beam and loading shown, determine (a) the slope at end A (b)
the slope at end B, (c) the deflection at the midpoint C.
M/er
SOLUTION
'm - aUgi/U J It El
A =: _L/±Br\/Jr\ - J- £L*
^a. *(.* EI A* ) " SH El
t.«- a, (w*v *.(!£)
id) S&ope tkA A
A - - *■" - -^ £4T
(M SApe Jt 8
^8 ^ ©a+ ft/, r 0A 4 A,+ At
- - Jl£L\ j-£^- +j.ET
u,k* Mik)- (ifcf^uo- ilk1
*
'C/A
" ^tB/A
$S EX *^6 fJ" '

PROBLEM 9.143
9.143 For the beam and loading shown, determine the magnitude and location of the
maximum deflection.
SOLUTION
- H,/£I
i- ^'A
Dr«w |ji. <£i&.^a.**.
+ w±r\ - - J- HaL*
b>* — ■ 4 £r
ft, * 0, + 0** - ^a* ** =0
X MoL _ J. M'X/ _ 0
** - 3 u
t,A* A*(i*«>
J- H0Xtf-
G EIL
MqXk
GETL

PROBLEM 9.144
W/£X
X*
9.144 through 9.147 For the beam and loading shown, determine the magnitude and
location of the largest downward defleaion.
9.144 Beam and loading of Prob. 9.129
SOLUTION
B/A
Ref
+o Ha SoJjKtm of Pirofc. 7. US
l3
i? -_Lp + .3 Pi3 ft . . i PL1
ra - h r j ^bm - las TT j w* " us £i
Q* a 9*
0
V/A
= _-a-£Lfc + a
/as ex M"
"if?IF * TK^ex J*<
Tf-f3 L - O.H33 L
"ft'L '
- -1- Pfr* - W5 PL*
2k
3
- 0.00477-^ I

PROBLEM 9.14S
M/£±
9.144 through 9.147 For the beam and loading shown, determine the magnitude and
tocation of the largest downward deflection.
9.145 Beam and loading of Prob. 9.130
SOLUTION
"6^: ^ 1 i t EI ,L ** Ex
Q:
^■siL
**--ki-
n/ei
PXa.ce. irefe/-e^cc T^^ge^T «T A
tft/* - A,x, + Az*t
36. ££
Mo EI
36& EI
A - A f*tV - - -1 -^u*
"l/ /A " Al
'«/*
LU
i%
0*-?u« +i
= o
36o eJT
'K/A
A3Xa + A**,, - 0.0O357S?
6"!
*r
Jf* r £*/* ~ l %
8/A
■o.ooas7^^- -(o.msiX^^I
r - 6.QOQ&2
0.0OC« ^i

PROBLEM 9.146
9.144 throngh 9.147 For the beam and loading shown, determine the magnitude and
location of the largest downward deflection.
9.146 Beam and loading of Prob. 9.132
«W SOLUTION
W250 X 44.8
Retev-T.W 4*> +k« S>oJJT\on "fo P/oL
^. as
"^
Mi ex
h-\
'*/A
EI ~ wzio ku.^
+ 7**.?. 5"
Ba _, ,.ja3.7y.
dcTi'eo+i'tfM- Assume +^*l ^ -Pi'es be-fwee*
C ou*A X>.
e« - e. + s
fcr/A
- - us.7r 4 _fLSL . Co u _ 0
£X W « -
A, (o + o.s) + Am (io 1
Y«
Xv
'WM
'ft/A
2..8/3S
I4W.O
-i
= - /S.Mxji) * ^ /. rn
**m

PROBLEM 9.147
126kN/m 63^^
W460 X 113
M/£J
9.144 through 9.147 For the beam and loading shown, determine the magnitude and
location of the largest downward deflection.
9.147 Beam and loading of Prob. 9.133
T SOLUTION
Fyvoh^ He SoJloit'o* +o P^ot. 9.133
Er ^ /II2.00 kW- ^\ RA ^ Z12-SS kN
rxV ^
o
6K ~ 9A -r 9u/A - -3. |<H5 x/O-1 + (I.O904 y/- O. f88&4? yfcs )*/oJ s o
SoJUn^ -Po^ X* X^ - Z. IS907 ^ XK--2.H~x -*
Aft =-6-1838^-7 *t6s X*3, ? - 1.S4837* (0's, x4 r ■£ *< * O.S3t77 *
tK/A = A5Xs- * A6 X<
-3
Z.SC^C *tO *n
•ft/A
« ZStfCxlo'* - %'*1°? (.15.82** to*) * - l./S'x/o"' *
<t. l^ *«* I

PROBLEM 9.148
IP
■A - -^ *£
9.148 For the beam and loading of Prob. 9.135, determine the magnitude and location
of the largest upward deflection in span AC.
SOLUTION
KL -- fc
a
, x JPL1
D**<w M/£X «lr4*v*«* . Lt^f K be Joe*.4i*©i*
oT v^a-ot i w* o *t ortT /et 7-1*01* -
a< - o, * eu - i± ^ ■» a
- o
wke.re
*S EX
1 £U* . i £il
2 eX 4*4 £T
"= ^ *-
t»/« * A(iff0) - (-il^M^SS-O- -°.«.3«f|S
y
•*«*.
O.oiS;^ J£

PROBLEM 9.149
|
-L 4—L/2-
D
\Ajei
M»/fJ
9.149 For the beam and loading of Prob. 9.136, determine the magnitude and location
of the largest downward deflection in span .45.
SOLUTION
Fro*. soUw oV Pi^ob. •?. 144
n - **
U -
i>*\
e* - ©a * ©
K/A
<VA'+*V
L iiii3 . 3. Wll u1 _ -L MIL3 U3
~ ti EI (6 ET « £X
0 =■ O. H2/.S"3S" X* = 0.4*15" L
* (io*- ^uM
W
(.*
**«
* w ' S
£T
0.OO<T*U ff*
WL"

PROBLEM 9.150
DJ^W.'om «,+ A
*9.150 The cantilever AB is a beam of constant strength. It has a rectangular cross
section of uniform width b and variable depth h. Express the deflection at end A in
terms ofP, L and the flexural rigidity E/0 at B. (Hint: Since the beam is of constant
strength, Mc/I has a constant value along AB.)
SOLUTION
>
Se^cii^ev ^ow€^f M ■= - Px
F
or *~ Com 3+ ^"t" sVv^e«A4|n o
Mc
CQV*
- - PL- /.k\»*
2*a/b r J * ex ^
FT_ _ 1 * trT ^/3
a,-J,
3 £■Jo
EX
* p*3 \
3/2

PROBLEM 9.151
9.151 through 1.154 For the beam and loading shown, determine the reaction at the
roller support.
M
M/£T
SOLUTION
rx
ir-ef. -Kiw

9.151 through 1.154 For the beam and loading shown, determine the reaction at the
roller support.
SOLUTION
Remove. SopPo+T B a.*)A j-t^etf r?g 4S. Y^e<JvJt\Ja.viT.
A.' *gW - i&
** * A.(fO*Aa("W±0
t?e ' if P T

PROBLEM 9.153
9.151 through 1.154 For the beam and loading shown, determine the reaction at the
roller support.
M/£I
SOLUTION
Renoov/e .SfpporT A a«J +r*»/T ^ <as redoinga*."T.
DvW M/£T Ji
"a ~ 3 ^ 8 £-j /. 2. ) * ^ f J
iref. +i
4r»
0
3 rr 38H j="t
FT
- O
*» - iar wt f

PROBLEM 9.154
9.151 through 1.154 For the beam and loading shown, determine the reaction at the
roller support.
*\/ET
vi/ez
SOLUTION
Rev^ewe SiJfporT 8 Audi +re*T "R3 a.s r«<Jt>*\^**i\
Ace xoA.tfK*ai by e^ui'ywcwi Sftowm «4 <*7t
Use 'p«vts «.£ s/i0u^n.
A - -L/'-J.^V z . JL-JfifeL*
IV0
L*
- * EI 3° EX - °
3 ei
Ra * JrWj. - 0. 275" «*L T
A -
a-F. +
re-f. t«*.

PROBLEM 9.155
9.155 and 9.156 For the beam and loading shown, determine the reaction at each
support.
Mn SOLUTION
P-Pft.ce reference +*.*y + e*\ oA A.
A
^
!
4^
tM - -i(^XaX%W i(i#X«yo- -f^-i#
4-
U- -KifcXa^ + 4(i.|r^C4) ■ -**£ **%£
= a
**■*»*
S+A^l C!
I M,
1 O - X^!* .
M« .
iH.r
* T ]

PROBLEM 9.156
9.155 and 9.156 For the beam and loading shown, determine the reaction at each
support.
SOLUTION
f(e,w\ovC SUftpcwrT S o.^ei comSi'cUv K^ «.* red *jv\ J«t^t".
Co« afdef
!*.
Px*.ce rft-PeiT*.*ce ■K^^r <si A.
r +
HIIlllllll
t?»
M./Fi
A3* 4(i^¥
)-*
A
M/fJ
3 ^ * ET I
±
6 EX
A ail3
& ET
y\/ez
■ + A,(4-"^+ A,(*0
' w er -9 ex
* IT er *"* er
^ f i + w er
- J- s*±f
21 EX
'Q/A "
N/ ~ t - L/g-
m ex
•F? = O
j?e ,iwM -*
4 wL- i w/L
3 ^
•£w/L J/

PROBLEM 9 157 **157 and 9,JSS Determine the reaction at the roller support and draw the bending
moment diagram for the beam and loading shown.
60 kN 411 kN
J J
■1.5 m-4*-1.5m-4*-1.5 m -*]
SOLUTION
Ka^o\/< sodocmA'oJt A /«*d ■(■reed* fi* as t*<lu*det*t.
M/fX
M (kM-frO
A
C ft \
e
-73.1
£1
ET
= {30.375' (?A * W - ISo}^ - O
RA - 37,037 WM t -*
=■ - 73.3 kW- w -*

PROBLEM 9.158
3 kips/It
2ft 2ft
M/rr
»9-S*»
9.157 and P./50 Determine the reaction at the roller support and draw the bending
moment diagram for the beam and loading shown,
SOLUTION
(senoov/e SOppo^T d\ "B clwI TireaT Kg AS fedm*daM«
?* + J\
MW**
^ p**-*-
A,- ±(f*)(/<0 = S$&
©A = ©
= o
£1 £r fi
MB -- o
MD = (?. 71)00 ■ n-^ kl°-^
Mc r («m)(4>- (lo >(*)■■* \«m3 kW.^
-36. I

PROBLEM 9.159
9.159 For the beam and loading shown, determine the spring constant k for which the
bending moment at B is MB = -wL2/lO.
SOLUTION
W
Urn
J-*
a
*ref. t**
W
ft
■ — — — ,
"ii*L
r6"
Ik.'*^ W botfjy A8C +tZFy =0
F= J: *L
A, ■ a U f I ^ 3" EX
H2- H 2- CJ J*- 6 ex
Place reAreice "ta^e-iT «T 8. ©B ~ O
i»o ex
f--«cj
e
H4
£T

PROBLEM 9.160
9.160 For the beam and loading shown, determine the spring constant k for which the
force in the spring is equal to one-third of the total load on the beam.
VJ
1 * v * it
* * * * * i
|WL
««
A*^
2 wL^
/ 11
ZEI
SOLUTION
Sp«*£ Wee Fr ;£ (2wO * |wL
R, + F - Z*L 4 J^ r O
+ tjFr°
Draw 7=y eiiA,<*iT <*.**■- ey p«^TS-
7* £1
7 Ls
eA r o

PROBLEM 9.161
9.161 For the cantilever beam and loading shown, determine (a) the deflection at point
B, (b) the slope at point B.
M/er
SOLUTION
Use. Motwe^T a.re&. t^errioA.
"f o.T A . ^ = O
L*
*
e„ = e. + ©c
e«
$& - ^a + LOA + tB/A = t6M
A,- <g&W« i^
(a) dAteo^o* *i S
- 7« £j sF ex - in e.i

9.162 For the beam and loading shown, determine (a) the slope at point A, (b) the
deflection at point D.
SOLUTION
?J>
cwre
reference -fd-wtf ev>f <a,T A.
+ t - l(±\ - ' M«^
5" MoL
<k) De-FW,on a+ D

PROBLEM 9.163
0.5 m 0.3 m 0.3 m 0.5 m
»-|-«—»4*—»4*
9J$3 The rigid bars BF and DH are welded to the rolled-steel beam AE as shown.
Knowing that c = 0.4m, determine for the loading shown (a) the deflection at point B,
(b) the deflection at the midpoint C of the beam. Use E = 200 GPa.
SOLUTION
Usi'w* joiwt (j c*& o. rree body
c W100 x 19.3
I oO
#>
n
^D
r%
2 FfiH^ - loo * o j-MJ : So ku
Ponces iw kW_ Lena-fks i* w
M - 5"Ox- 5o<x-o_S>' - 5o<x-i. I>"
kv). ^
F«r HI^/B ^ x ? 0.5 vm
£IyB =('¥^°-s^ -O-o + o - o -Gi.7-Si(o.O = 0.ICG7 W-**
Foir £IjtJ y - o,gk,
Elyc r(¥X0L^-(¥)(0.S)1-O-^)(0.a^^O -(1.7-5 )(o.s) 4 <>
= - O. 8417 kr^-M3
Fo%r W lOoxn.S **oJJfeJ slc<ri se^i'o^ J- f.77«/O4 mm"* - 4.77*/o-t^H
EI- (CIookIo-V^^?^!©-') r ^SH * io* M*^ = W* IaJ*v^

PROBLEM 9.164
—i-"T—IT !_
M'(1
L/2 -
<%
■ ft i—**""'
-. JJ2-
B
—*■
-<"
V M y
,rjrx
vr'v
9.164 For the beam and loading shown, determine the deflection at point A.
SOLUTION
Express ioadiV^ i* -feri^s sf sin40*f«-/iTy
[x = o , V - o ]
O 4 D-+ C, - O C, - O
?"o
dg. V - -^x'**".<y-fc>'
C.= o
[y = o^ M - ol
0+0 + Ct. = O
21
Ct - 5Z7T w° u
?6o
EXyA - o + o + o
Va r "it* EX

PROBLEM 9.165
9.165 For the beam and loading shown, determine (a) the reaction at C, (b) the
deflection at point B. Use E= 29 * 106 psi.
W12X40
M/*r
SOLUTION
Pxo^cc reference -/-«>i 3 e «t 4 A. 6*-^
jc~ fr + ©Ai- * k/. - o + o * [ A,(f-iO + Ai(i^-i--s')]
** W A,(f-8V A,(|-8VA,(1.8)
- ZW5
. g^ _ g/<W = _ IHS.tf-*
£1 ,S"£X
EX
O.O S7^ mt J,

PROBLEM 9.166
St) IIVlii.
9.166 For the loading shown, knowing that beams AC and BD have the same flexural
rigidity, determine the reaction at 5.
SOLUTION
C©"tSidev Hie two be^v^s .shown Ideiow.
/
2*
Ci .
a
Let t?c ke tLe
bastes AC &vj BCD.
0 ci
*
App^j'iM^ C^-ses I **<* ^ (sT Appeal x "O 4j c<t*Ti^e^e^ be*.*** AC
y<
P^q3 iv a"
SET ' $EI
App/«wift^ £<a,se 4 *f Ap/5e^i<?|(x O "ft* Si'wpiL soppor-feA .^«avm "BCD.
_ &L3
^ " *w £X
3 f I g EX ~ 4* EX
(\&aa + L1 )Rt - Giva"
37$. XI Jk
UsiVa Leftm BCD <^S ^ +re< &«(m
i)?Mp--^ -l?BL+*t£*o t?8*-i*** /M.tfAt -*

PROBLEM 9.167
10 mm
K M
9.167 Beam DE rests on the cantilever beam AC as shown. Knowing that a square rod
of side 10 mm is used for each beam, determine the deflection at end C if the 25-N-m
couple is applied (a) to end E of beam DE, (b) to end C of beam AC. Use E - 200 GPa.
i SOLUTION
120 mm
160 mm
J EHlOmm
ci \ "T £ * 2oo*/o" fk
EI - 166.C67 N- hi*
25 N ■ in
^-K 1
• L ^\ (c0 Coop/e AppJteA -fo be*** 1
.ISO P - Z&= O
O
^
Fo^ be«j* ABC cWh> +A'e ^ JU
EI
*3
r««*
by p
a^
f».
Irt -*(*- (?.l8o—J
<^3.
€1 I&C.GC7
^X IC4.667 ^
A,- 4c0m>*/o'*Xo.iO - &y/0
-i
-&
*,/«
y* * > + ^ e* + tcM
i
^
-0-Soom
M/ffX
25>J.>w ^ Coopfe AfpJIleJi +* ee«vw. AC
Draw S- Ji'*.^*-*
25
£1 " r&C.6«7
lS"ox/0"3 *"'
-t5to*|©
/L = (-/rOw/0*Sy0.3o) r-4!T*/0"'
>
£C/A = As(o.»0 - -G.lS*ltfz m

PROBLEM 9.168
9.168 For the beam and loading shown, determine the value of P for which the
deflection is zero at end A of the beam. Use E = 29 x 106 psi.
S6 x 12.5
2.5 ft 3.75 ft 3.75 ft 2.5 ft
M, /EX
M2/£2
*-ef -K* C
SOLUTION
9c = o
FX •* Wp- ft1
*3= i(-^^-S)--i^
EX
> - yo * 4* - i«/c = °
A, (2.S+ z.s) +- Az(;?.s + I.*?*") 4 A3 ( 3-?.0
A, («.s^ - A, ('.3?^ - o
At(2-5) 4- AiCS.5 > * Aa(l.«<C7^ = O
7Q.3/2S 33./J37S1 P 5*. 3 08335 P ,.
Er Er EX " U
f r X. 4S kips

PROBLEM 9.169
9.169 For the beam and loading shown, determine the deflection (a) at point D, (b) at
point E.
M/ei
SOLUTION
eiA*T Avjd -*^ olf
i«L<\^ft.w^S.
A«- ^6CJ J^J Iff £j
A - J./.EL Vk\ - X £tl
Aa. ■ a ^3*r )U J - 18 fi
-D/A
A,(i-4)
i PL3
~ 33H ex
- x si . -l. ZL2 - x 2£
ftt) DeM«*fi«n J- "0
- O.OI7M1 ^?
W D«Wec+i\»* <£ E
o.oiitt f£

PROBLEM 9.170
9.170 For the beam and loading shown, determine the magnitude and location of the
largest downward deflection.
M/Fr
SOLUTION
m/£x
Spope *,t A
6** -
-aM
B* - ©A + ©it/A = 6A + A, + A4 - o
** - f + U V3 « JL 51 *~
*■ a/'-ll^.na/un- J2_ a£ + -ilL Bi . J&L J2ii.
MM 9Lt & x W. ,733 £Li
= -ao2095^jr

PROBLEM 9.171
M/
A
EX
^//
y
a,
//
4
Hob
Ik;
A
_ M»a
EXL
9.171 For the beam and loading shown, determine (a) the value ofa for which the slope
at end A is zero, (b) the corresponding deflection at point C.
SOLUTION
Lef lo * L - a
^6 = .Ya + ^ ©A + £&M = O 4 O -* £a/4 = O
11 £IL
J. H^b1
* en.
UM-- A,(f+iO + A,(|b)
1 Mi. of „ j M,tt*fe t 4-gUa
s
EXL
L)
+ 3o* - 2 - O
SojLi'n^ 4W u : d - 0.733.0S"
L-O.
= O.TSZOJ"
0.7S£<?S (/.-O
"■ I. ^*oT
na
A, '"**£ * -a-08*3'* Er
EJT
O.OVLSZ gji1 i
=■ o

PROBLEM 9.172
9.172 A hydraulic jack may be used to raise point B of the cantilever beam ABC.
Knowing that after the 20-kN load is applied, point C is to have the same elevation as
point At determine (a) how much B should be raised, (b) the reaction at B after point B
has been raised and the 20-kN load has been applied. Use E = 200 OPa.
SOLUTION
W130 X 23.8
\.8P8
fc
-Go Kfo
1*
A*- iC-^C^ - -?° frw-i**
kW-*^
A, = 75 k^-m1
Aa* ^(-60^(1.8^ r -54 UW-w."*
EIts/A - I.a A, 4 1.7 Aj 4 o.cAt
= C.^S**/o~ i^i
^ 6.95"
Minn
(to £R - 4C a *tf

PROBLEM 9.C1
P,
M*
(:
fl
Aj-v-
tf*
r*
-/
rL-
a.
ct- ,t
*T>-°, r-S=0
9.C1 Several concentrated loads can be applied to the cantilever beam
AB. Write a computer program to calculate the slope and deflection of beam
AB from x = 0 to x == L, using given increments Ajc. Apply this program with
increments Ax = 50 mm to the beam and loading of Probs. 979 and 9,80.
SOLUTION
FCR BACH LOAP, BhiJEK
)
Pi > Cf
COMPUTE REACTION AT A
FOR I
k*
V*
COMPUTE
= 1 '
-V
= MA
SLOFt
TO NUM&ER
Pi
-P(c
AND
LQfitbS
DEFLECTION
USE MFTWOD OF IHTFG$/\Ti : i:
STARTING WiTM "X=0 AND UPDATING
THROUGH IMCR£M5/VT53 SUPERPOSE:
(I) DUB TO REACTJOW /)T A:
t?1 0U£ TO EACN /.OAD WITH C(- < Ti :
0-(i/^(P('A.o)h-c(')z
THr cowst^ wts of
/WTfGtfjmo^ 5<3^L 2&RQ
CONTINUED

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PROBLEM 9.C2
3.5 ktps/ft
I
«,
t
a
J
ElgO ^k--^<x-^
9.C2 The 22-ft beam AS consists of a W21 X 62 rolled-steel shape and
supports a 3.5 kips/ft distributed load as shown. Write a computer program and
use il to calculate for values of a from 0 to 22 ft, using 1-ft increments, (a) the
slope and deflection at D, (b) the location and magnitude of the maximum
deflection. Use E = 29 X I06 psi.
SOLUTION
EN1ER LOAD ur^ LFNGTh L, <k
COMPUTE REACTION AT A
Rfl - ur (L-a)/(2.o L)
COMPUTE SLOPE ANO DEFj.ECri(W AT 0
USING SINGULARITY FUNCTIONS ;
•2 W , . . . 3
FROM BOUNDARY COUOniOfJS:
c, = o
^.--^o-^-f v
COMPUTE LOCKTiOM AN/D M/tGM iT'JQg Of
MAXIMUM DEFLBCTlQN
MAX'MiWi u, AT 0 = 0;
z
IF "X
AS5L/V)£ v *. A: x
■rt^ax
/na.^
£ <K
If
>*>** < ^
i RA7l + c, =o
THEW
I
7
»i*x
_ /-2.0 C,
J/max " f •Rp^m** +^
0*«*
BE6/W VV/TH * = <*■
IMdREASE "* BV SMALL AMOUNT
UNTIL 9 /S APPROXIMATELY O
CONTINUED

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PROBLEM 9.C3
Ai I 1/ V * * * ♦ ^
'• A ^ A A
Z\*r
(X
AT X-0.
-4fcs
0
/, THE O/JiTAiJTS OF
Ih/Tlr&KATiOfJ ARE
9.C3 The cantilever beam AB carries the distributed loads shown. Write
a computer program lo calculate the slope and deflection of beam AB from
x = 0 to x = L using given increments Ax. Apply this program with
increments Ax = 100 mm, assuming that L = 2.4 m, w = 36kN/m, and (a)
a = 0.6 m, (b) a = 1.2 m. (c) a = 1.8 m. Use E = 200 GPa.
SOLUTION
CoNWte reactiow at A
COMPILE SlQPE. AMD P£FLFCT|QM
(J5£ EQUftTiOfJ Cf ELfi.St >C CURVE.
STATING lA/lT/V X^O AWb IVPDftT/^o THROWN]
INCREMENTS 5CPER POSE;
(l ) PlV£ TO RF^^TlOWS AT A
CO Due to Load ur
es-(l/El)(^>J)
1*3^ D(/E TO LOAD Z.U-
1F y, < a.
XF "X > A.
CONTINUED

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PROBLEM 9.C4
9.C4 The simply supported beam AB is of constant flexural rigidity El
and carries several concentrated loads as shown. Using the Method of
Integration, write a computer program to calculate the slope and deflection at points
along the beam from x = 0 to x = L using given increments Ax. Apply this
program to the beam and loading of (a) Prob. 9.14 with Ax = 0.25 m, (b) Prob.
9.15 with Ax = 0.05 m, (c) Prob. 9.132 with Ajc = 0.25 m.
<K
FOR LOftS P;,
A I*
Jt7 A
SOLUTION
FOR FACri LOAD > ENTE R Pf ,
COMMIX REACTION F\l A
FOR i = / TO NUMBER LOAD5:
M/l = MA + Pi Ai
LOAD - LOAD 4- P;
THEM:
R6= MA/L
Rh= LOAD - RB
COMPUTE SLQpE AME DEFLE^TiQ/J
5"7ARTik;6 WITH * - O AMb typDAT|AJG
THROUGH IMCRPMEWIS , SUPER poS£ ;
(|) DUE TO ftPAO)0W At A
F*om Boundary cojw/tiows
q - <:, * °
c.^a -
5(L-*t)'4V'
(2^ Due to loa o s - constant p^T
COM5T( ='f RALZ
FOft / 7~0 NVMBC-8 LOA 05
C0NS72 - -^Pj (L-<xi)3 + CousTz
rHCM , TOTAL COW7R\8uTtOW FOR O^V7AwT|
COUST =(\/FtVCO»JST| + COUsTl)
(3) DUR TO LOADS - r?i"MAlMWC> PAK7
TF 7 < ^
Note; Ra FOR tOAD f>t-
CONTINUED

PROBLEM 9.C4 CONTINUED
P*06«4M OUTPUT
Problem 9.14
Problem 9.15
X
m
.000
.250
.500
.750
1.000
1.250
1.500
1.750
2.000
2.250
2.500
2.750
3.000
theta
rad*10**3
-6.058
-5.831
-5.150
-4.014
-2.423
-.719
.757
2.007
3.029
3.824
4.392
4.733
4.847
y
mm
.000
-1.496
-2.878
-4.033
-4.847
-5.235
-5.225
-4.875
-4.241
-3.379
-2.348
-1.202
.000
Problem 9.132
X
m
.000
.250
.500
.750
1.000
1.250
1.500
1.750
2.000
2.250
2.500
2.750
3.000
3.250
3.500
3.750
4.000
4.250
4.500
4.750
5.000
5.250
5.500
5.750
6.000
theta
rad*l0**3
-8.703
-8.615
-8.351
-7.911
-7.296
-6.505
-5.538
-4.483
-3.428
-2.373
-1.319
-.264
.791
1.802
2.725
3.560
4.307
4.967
5.538
6.021
6.417
6.725
6.944
7.076
7.120
y
mm
.000
-2.168
-4.293
-6.329
-8.234
-9.962
-11.472
-12.724
-13.713
-14.438
-14.900
-15.098
-15.032
-14.706
-14.138
-13.350
-12.365
-11.204
-9.889
-8.442
-6.886
-5.241
-3.531
-1.776
.000
X
m
.000
.050
.100
.150
.200
.250
.300
.350
.400
.450
.500
.550
.600
.650
.700
.750
.800
.850
.900
.950
1.000
1.050
1.100
1.150
1.200
1.250
1.300
1.350
1.400
1.450
1.500
1.550
1.600
1.650
1.700
1.750
1.800
1.850
1.900
1.950
2.000
2.050
2.100
2.150
2.200
2.250
2.300
2.350
2.400
2.450
2.500
theta
rad*10**3
-2.490
-2.485
-2.471
-2.448
-2.416
-2.375
-2.325
-2.265
-2.197
-2.119
-2.032
-1.936
-1.831
-1.716
-1.593
-1.460
-1.318
-1.172
-1.025
-.879
-.732
-.586
-.439
-.293
-.14 6
.00,0
.146
.293
.439
.586
.732
.879
1.025
1.172
1.318
1.460
1.593
1.716
1.831
1.936
2.032
2.119
2.197
2.265
2.325
2.375
2.416
2.448
2.471
2.485
2.490
y
mm
.000
-.124
-.248
-.371
-.493
-.613
-.730
-.845
-.957
-1.065
-1.168
-1.268
-1.362
-1.451
-1.533
-1.610
-1.679
-1.741
-1.796
-1.844
-1.884
-1.917
-1.943
-1.961
-1.972
-1.976
-1.972
-1.961
-1.943
-1.917
-1.884
-1.844
-1.796
-1.741
-1.679
-1.610
-1.533
-1.451
-1.362
-1.268
-1.168
-1.065
-.957
-.845
-.730
-.613
-.493
-.371
-.248
-.124
.000

PROBLEM 9.C5
I
W
% n tttv\
k—t—I
9.C5 The supports of beam AB consist of a fixed support at end A and
a roller located at point D. Write a computer program to calculate the slope
and deflection at the free end of the beam for values of a from 0 to £ using
given increments Aa. Apply this program to calculate the slope and deflection
at point B for each of the following cases:
L la w E Shape
(a)
12 ft
3m
0.5 ft
0.2 m
1.6 kips/ft
!8kN/m
29 x 10* psi
200 GPa
WI6 X 57
W460 X 113
SOLUTION
USE hPPENbW 0 AMD S^ff.R POSITION'
DETERMINE REf\C7\QN (\TD
DiyE TO DlSTK/eUTFO LOAD
or
'¥
p V " -
24
EI K
k
U,
QB - ©c
,j?j , 'fol(L-o)&D
DUE TO RSb(/hJD/\hJl LO^O:
/j
^
*'/?
3£\r
REOtWO/WT REACTION;
CpMPt/TE 5L0Pg fl^& DEFLECTION AT 6
SUPERPOSE ;
OUT TO DISTRIBUTES LO/\b :
e6~-~
h
due to Rp :
-2.
e,
*.
Pa'
2 EI
Pa3
x //_ - o. J
CONTINUED

3 w
13
n
0
a
ID
3
I I I I I ] I I I
ooooooi-ir-ot*jJi<TiCDOWcn
HPHOLOtDAUnb-JAUOiWU
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m
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I I I 1 I 1 I I I I I I
o o o o o o
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N) VC 1*1 <J1 O VD
Mpowwibuimsj
^ou^ioisiuai
o [Li
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tr
(OPMOO^Diaaia3~J^l<^0\LnLnifcifcUIUI(0(0h'P
ouiouiocnouioinouiouiocnouiocnomocno
Hi
I I I I I I I I
I t I I 1 I I I I I
PPHHPI\3MMMOJ
UUIIO[OHOOOPMUIJiO\-J(DON3UO'-JON)J>-JO
•Jw-opyimosjtnoi-jfflpfcicAp^asiDowiBi&p
^CD^oiMMmsjffloiWCDWIBWVDOlUKJiO-JCDWliHl)
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fl>
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O©C3OO©O©OH»H»N>lJJ.fc.<Tl-0V0l->Wma>M(_Pa>N>
OWWWUIMOtOtnlOVD-JvJvoNJNjUijidpaJIBPOim
OOWiD-JvlOlOi-JMPiliJiOibOIWlDlDftU'UJPWJO

PROBLEM 9.C6
P,
RJ J
t^
L
M*
« kWWWM
H - X f - /
9.C6 For the beam and loading shown, use the Moment-Area Method to
write a computer program to calculate the slope and deflection at points along
the beam from x = 0 to x ■ L using given increments Ax. Apply this program
to calculate the slope and deflection at each concentrated load for the beam of
(a) Prob. 9.76 with Ax = 0.5 m, (b) Prob. 9.116 with Ax = 3 in., (c) Prob.
9.119 with Ax = 0.5 ft.
SOLUTION
FOR EfitCti LOAD EhJTER P(- AMfc *t-
DETERMINE R£f\CTlON AT f\
DUB TO M0MBM1S AT FA/OS'
Due TO LOAbS PL :
FOR I * t TO NUMBER OF LOfthS
LQAb = Z.0A0 f P(-
/A/,)l = LOAD -ftfi
J)ET6RMlN£ SLOPE AT A
TO 6BT TAMGEtJllAL DEVMTiOfij AT 3
DU£ TO fy :
FOk i ~ ) TO NUM&Efi OF COAOS
sum im :
OeiBkMtKJB SLOPE /WO f}EFLeCT\OhJS
For -y - 0 To l suppose:
OVE TO M^ AM V
CONTINUED

PROBLEM 9.C6 CONTINUED
PrtGGMM 0U7PUJ
ftO FOR /\LL LObbi IA//TH A/ 4X
yy «P- (7-a^/U-O EI)
Problem 9.76
Problem 9.119
1
1
2
2
3
3
4
4
5
X
m
.000
500
.000
500
000
.500
000
.500
000
.500
000
theta
rad*1000
-.600962
-1.602564
-2.043269
-1.923077
-1.241987
.000000
1.241987
1.923077
2.043269
1.602564
.600962
y at x
mm
.000000
.574252
1.509081
2.524039
3.338675
3.672543
3.338676
2.524039
1.509082
.574253
.000000
Problem 9.116
X
ft
.000
.250
.500
.750
1.000
1.250
1.500
1.750
2.000
2.250
2.500
2.750
3.000
3.250
3.500
3.750
4.000
4.250
4.500
theta
rad*1000
-8.937931
-8.813793
-8.441380
-7.820690
-6.951724
-5.834483
-4.468966
-2.979310
-1.489655
.000000
1.489655
2.979310
4.468966
5.834483
6.951724
7.820690
8.441380
8.813793
8.937931
y at x
in.
000000
026690
.052634
077090
.099310
118552
134069
.145241
151945
.154179
151945
.145241
134069
.118552
,099310
077090
052634
026690
000000
X
ft
.000
.500
1.000
1.500
2.000
2.500
3.000
3.500
4.000
4.500
5.000
5.500
6.000
6.500
7.000
7.500
8.000
8.500
9.000
9.500
10.000
10.500
11.000
11.500
12.000
theta
rad*1000
-2.118621
-2.222069
-2.267586
-2.255172
-2.184828
-2.056552
-1.870345
-1.626207
-1.324138
-.993103
-.662069
-.331034
.000000
.331034
.662069
.993103
1.324138
.1.626207
1.870345
2.056552
2.184828
2.255172
2.267586
2.222069
2.118621
y at x
in.
000000
013051
026549
040146
053495
.066248
078058
.088577
.097457
,104408
.109374
.112353
.113346
.112353
.109374
104408
097457
.088577
078058
.066248
053495
.040146
.026549
.013051
.000000

32 kN
PROBLEM 9.C7
52 kN"
9.C7 Two 52-kN loads are maintained 2.5 m apart as they are moved
slowly across beam AB. Write a computer program to calculate the deflection
at the midpoint C of the beam for values of x from 0 to 9 m, using 0.5-m
increments. Use E = 200 GPa.
I
W460 X 113
\!
C
/?>*H
k
L
SOLUTION
ENJBR LOAQ P, BEAM LENGTH L AND SPACE
Between Lo^fis D
1A//LL SOLVE W\tH W0M£N7~/\R£/\ /V/£TM0£)
* DETERMINE QEFLFC7/QM AT c
4* For y - 0 to /_
* IF 0 *-y £/)•
J fJ/)l/F OWF Z.O/U TO UFT OF C
R6 - Py/L
fc = f^ " ^/ft
EI
A j | C
r
a
A
(Mp
t< -I
/^
X
t«,
B
i".
if v < -y £ l/2
rVflVE TWO LOADS TO UFT OF C
Rg = PX/L + F(Y-t>-)/L
tfflVfc OrJf /.CAD 70 /.EFT OF C AMD 0*>£
TO RISH7 OF C OR A7 C
CONTINUED

PROBLEM 9.C7 CONTINUED
ftp= Px/L + P(Tt-D)/L
B
|P if
A CJ. | i P IF (L/z.+ D)<V<L
f 7_0 i J HAVE ROTH Z.OADS TO R\6HT Of C
PKQ6RAK] OUTPL/T
X
m
.0
.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
RB
kN
.000
2.889
5.778
8.667
11.556
14.444
20.222
26.000
31.778
37.556
43.333
49.111
54.889
60.667
66.444
72.222
78.000
83.778
89.556
ThetaB
rad
.00000
.00315
.00624
.00921
.01200
.01456
.01998
.02499
.02947
.03331
.03639
.03859
.03980
.03989
.03876
.03629
.03235
.02684
.01963
YC
mm
.00000
1.17881
2.32839
3.41951
4.42296
5.30950
7.22872
8.94335
10.39493
11.52503
12.28492
12.66487
12.66487
12.28492
11.52503
10.39493
8.94335
7.22872
5.30950

PROBLEM 9.C8
M,
j n u i
Mt
st\
An:
a
ft,
C;
\n
M
w
* ( " LLtJ
I L J
/, THF CflMSTAJvT* OF
\NT['£.8f\Tl0tJ /\tf£ ?f#0
9.C8 A uniformly distributed load w and several concentrated loads Pt
may be applied to the cantilever beam AB. Write a computer program to
determine the reaction at the roller suppon and apply this program to the beam
and loading of (a) Prob. 9.57a, (b) Prob. 9.58a.
SOLUTION
THE BEAM IS iNDETFRMWATE
U$£ EQUATION OF ELASTIC Cl/R'^
£Nl£ R w AWO FOK BA^tf LOAD PL A^P c(-
Compute displacement /\t B doe to was
react low Ar A:
OOF TO ur
>Vt^2
F0£ i" - / TO NyMBE^ LOAfjS P,"
fO« DISPLACEMENT flT 3, SVPSRPOSE:
PVE TO RF^c7/0W /\T A
£r
/»
T**L + i ^L
PUB TO D^T^I^VTIrD 'OAb<
£>UE TO Pi
FOQ. L ~ \ TO MVMBFK LQftbS
Compi/tp displacs.i'ibni at b Owe r<? ^wit a^
L
(t
L
4'
"faUn " 3 ^
COMPUTE REACTION A
B
poo\)v/\f>Y CONDITIONS CiviS Ct^Cz = 0
CONTINUED

PROBLEM 9.C8 CONTINUED
PR06kf\y\ OUTPUT
Problem 9.57 (a)
Reaction at Roller Support = 41.2500 kN
Problem 9.58 (a)
Reaction at Roller Support = 11.5356 kN

CHAPTER 10

PROBLEM 10.1
10.1 Knowing that the torsional spring at B is of constant K and that the bar AB is
rigid, determine the critical load Pa.
SOLUTION
Lei 0 We -He o-^gie cX**,**. «n lew*/- AS-
M - K6 f X = L srn9 *> L©
= o
Ck-PL*)© = o
Pcr. ■■ K/L
10.2 Knowing that the spring at ^ is of constan t k and that the bar AB is rigid,
determine the critical load P„.
SOLUTION
F Let £ b« +k# a^^< clianej 2 a£ \?air A8.
e F t kx - kL s.'-id
i)IMfcrO FUosS - Px = o
kLzsi«©c»s© - Pls;^0 = o
Dii** Sio6 *i 9 «.^| COS Q & \
kCB -PlQ -- o

PROBLEM 10.3
10 J Two rigid bars AC and BC are connected as shown to a spring of constant k.
Knowing that the spring can act in either tension or compression, determine the
critical load Pa for the system.
Fc **
SOLUTION
Solvit C • +.FL = o i F^c cos o - F^8cos 0 = o
PROBLEM 10.4
10.4 Two rigid bars 4C and BC are connected by a pin at C as shown. Knowing
that the torsional spring at B is of constant K, determine the critical load Pa for the
system.
SOLUTION
UT 9 be He
Hs - KB
Ok^o
.. k'e
ke - FA l - o fa * *■*
fc«w AC t)ZMt= O
?c, i L 0 - iL FA - a
P.r =
- _B_ , K
fc
l_

PROBLEM 10.5
10.5 The rigid bar AD is attached to two springs of constant k and is in equilibrium
in the position shown. Knowing that the equal and opposite loads P and P' remain
horizontal, determine the magnitude Pa of the critical load for the system.
SOLUTION
Lex v6 &.y\d yc be +Jie ei«WecTi'o*3 of pof-i+s
n
en
) P
Fn - -U
yc = -yg
y« 1=1 = - J<,yc
F, ^ Ft = o Fc = - FB
&
ra ot*vl F, Vt
'fc +-o^*^ «t
CoOflA V
A <* *p
t>?H-Oj k 6f a ».-*£> ^)a cos© - P£ */ne = O P= ^f- cos©
Lei e — O o r
u
PROBLEM 10.6
10.6 A frame consists of four L-shaped members connected by four torsional
springs, each of constant K. Knowing that equal louds P are applied at points A and
D as shown, determine the critical value P„ of the loads applied to the frame.
LeT Q te T«* roTM"t'on
OT £a.cU L- Shaped
Pir\tiJ(t cUcmc.e across
fir* na
IS 2©
S^)K(2^

PROBLEM 10.7
10.7 The rigid rod AB is attached to a hinge at A and to two springs, each of
constant k = 2.0 kip/in., that can act in either tension or compression. Knowing that
h = 2.0 ft, determine the critical load.
SOLUTION
LeT 0 t>e +l« svndJJ \rct\-&,ri
0*"> a.fi«
M
Ft= kx^ *• 3kU©
F0* kx0 -* k^$
Ud-*: k^ 2.o k.p/;* , k = 2-Pi - 5V* ."■
P - i U°)Cm) r uo k;ps.
PROBLEM 10.8
10.8 If m = 125 kg, A = 700, and the constant of each spring is k =■ 2.8 kN/m,
determine the range of values of the distance d for which the equilibrium of the rigid
rod AB is stable in the position shown. Each spring can act in either tension or
compression.
SOLUTION
Ui 0 be He s^^i^ **W.o* tfT AS
Xt de F* kx * kd©
&*r
rn
» 7 k
*
a~ V k 'if ^.skio*)
oOMO
=n

PROBLEM 10.9
10.9 Determine the critical load ofa round woodtn dowel that is 48-in. long and has
a diameter of (a) 0.375 in., (b) 0.5 in. Use E = 1.6 * 106 psi.
SOLUTION
(CO C » iJ~ O.IS7S In 1= ■9c'* *" <?7o.7*/o"': m*
P - V**1 TTa(uWQ4Keno.7»J0~0 _ .,
Per " i i * (4«1** ~" ""*
P - T^SI _ ^T1^l.4>./o4X3.Q68Hfe>-3) _ 9, „ fli
PROBLEM 10.10
10.10 Determine the critical load ofa steel tube tliat is 5.0 m long and has a 100-
mm outer diameter and a 16 mm wall thickness, t se E - 200 GPa.
18 mm
.* i-
cr
SOLUTION
Co - £ J. - so
C^^ Co- t - So-14, * 34
Wl"i
3.8S"<? X/0*6 Kv,"
1= ^(c/-c/)^ ASflx/o1
i j—'~t = 3°Sxic> W r 365^ VW

PROBLEM10.il
1.1m
20 mm
1.1m
Brass
E = 120 GPa
p = 8740 kg/m3
10.11 Determine (a) the critical load for the brass strut, (b) the dimension d for
which the aluminum strut will have the same critical load, (c) the weight of the
aluminum strut as a percent of the weight of the b«vs strut.
SOLUTION
(a)Bm35 sirA I- y^aX^? = 13.333*/^ »M
13. 333 */o"* m
Aluminum
E - 70 GPa
p - 2710 kg/m3
p TT*5Jb it a0ao*<o*)03, 333 »iq-^
- 13.06, v /05 N - 13.06 kl)
s+rO*
.■._ '^PcrL1 _ Qgyis.o<../o»yuv-_„,.,..,., h
m*n
-- - &)tf)s(ffeW)E-°-<~-^*

prohi mion 10*12 A com')I*ssion member of 20 in. effectiv; length consists of a solid 1.0-in.-
j-Kunubia iv.iz diameter aluminum rod. In order to reduce the x 'eight of the member by 25%, the
solid rod is replaced by a hollow rod of the cross section shown. Determine (a) the
percent reduction in the critical load, (A) the vah e of the critical load for the hollow
rod. Use E= 10.6 x 106ps>-
SOLUTION
o JLl£i-__ Trafta6*io<0(o.otno&7>) . .- #„ . s „
oo
P.- P,

PROBLEM 10.13
10.13 Two brass rods used as compression members, each of 3-m effective length,
have the cross sections shown, (a) Determine the wall thickness of the hollow square
rod for which the rods have the same cross-sectional'area, (b) Using E = 105 GPa,
determine the critical load of each rod.
SOLUTION
(a} Sauoe a-eo. ~i}-(J0 ~ «* / r ko ~ »«•
Per. lip. , ^(.cs^yio^x^) ^ sg Ww r S8g w ^
S?^«^e r I* ^(bj1- kM * 7SG.'8£-y/os mv? - 736.gr* /o"1 ^
^JJjg^J^
40 mm
60 mm
P
|
HWI| -..=-£
ffl
60 mm
PROBLEM 10.14
10.14 A column of effective length £ can be made by gluing together identical
planks in each of the arrangements shown. Determine the ratio of the critical load
using the arrangement a to the critical load using the arrangement b.
SOLUTION
T ^ i ^
R
- n
~Z£I
JLM
*w.r^T a> I., . I, * ifct^V £«£)' * lirffctf - 3*- J*
o - TTar?X ^ 19 It'EoI*
l.<U\

PROBLEM 10.15
102 mm 102 mm
152 mm
10.15 A compression member of 7-m effective length is made by welding together
twoL152 * 102 x I2.7angles as shown. Usingi? = 200 GPa, detenninethe
allowable centric load for the member if a fector >f safety of 2.2 is required.
SOLUTION
An^Je L ISZ X lO^ x 12.7
:y " So.3 »v,*n
Xj r 2.6V v/ofi *>.
Two angles: X, ' ft)(7.aox;06)= \H.00*10° m**y
Iv*;* - Iy r I.I^WO^m'' «" 1.117^0''^*
Per * "JT" * (7 0)* ~ -370.^,0* W = 37©. S*
ku
F.S.
37Q. S"
PROBLEM 10.16
10.16 A column of 26-ft effective length is made J rom half a W16 x 40 rolled-steel
shape. Knowing that the centroid of the cross secti m is located as shown, determine
the factor of safety if the allowable centric load is ID kips. Use E = 29 * 106 psi.
SOLUTION
PJJ
J(/ r 28.7 .Vi\
Pc**
t\*EL
?^~-
A - (Wn.^ - s.io ;*"
I* - 4(51*^ -(X'?o)C8.ocr- /.SI)* - 35.57 ,h
c*
ES.
r-e - £*. - q*-S -
7.WS"

PROBLEM 10.17
TVhJ
10.17 A column of 22-ft effective length is to be made by welding two 9 * 0.5 in.
plates to a W8 x 35 as shown. Determine the allowable centric load if a factor of
safety of 2.3 is required. Use£=29 x 106psi.
SOLUTION
© W 8 *3S
Ij - n.C iy."
IK - \T7 .V
Iy - A(*VaS?+<«X^* %*)"]=■ 8L75S in*
W.6 4 6?tf3L753) * *<?6-/* 1*1
L * W4H
2cf ,**.
_ ^'EX _ 7rLr^wo^Q87.7s)
u*
ftl<
P«* 221
F.S. " 2.3
771.0 XIO3 A - 77/ k.j
335" fc.'ps

PROBLEM 10.13
10.1* A column of 3-m effective length is to be made by welding together two CI30
* 13 rolled-steel channels. Using £ ~ 200 GF a, determine for each arrangement
shown the allowable centric load if a factor of safety of 2.4 is required.
SOLUTION
Fo^ ih*vme^ C 130 x IS J\ =- (-7, D
^V»MA
(a) <*>)
I* - 3.70 wo1"
W^m^i
Iy- 0.26*-/o^*
x = is.;? *^
^7
2.4
9v. s ^
Ij = *[o.26tw/oc +07ioXHB-/^y] * 4.911 v/o
6 V
mm
Iw* * Ij = H.IUxiQ**** r q.lHxld***
?er , I*** , ir'ftoo^^.t//,/^? , |oW N b /07? ^
t3LO)1
- J^ - iiiZZ . ^ M
KS
2. H

PROBLEM 10.19
10.19 Knowing that P= 5.2 kN, determine the factor of safety for the structure
shown. Use E « 200 GPa and consider only buckling in the plane of the structure.
SOLUTION
22-mm diameter
Jo 1*7 B ' Fit-ok-v 4-oZ-ce ~Tr-ic*.v\Gi*(e.
~a&
SA
Pec - 1-Si52 ku U^>)
McnU A8: IM= ^V * ? (f"V r ^1^^'^^ >£ 15*3x/o"'
F.S.
* Fa6^ . _7>06Sf
Aft
3.10-?^
2.27
K"^* " *" " 2.575 3.
fc
* 3./3

PROBLEM 10.20
3,5 m
10.20 Members AB and CD are 30-mm-diameter steel rods, and members BC and
AD are 22-mm-diameter steel rods. When the tumbuckle is tightened, the diagonal
member AC is put in tension. Knowing that a factor of safety with respect to
bucklingof2.75 is required, determine the largest allowable tension in AC. Use E =
200 GPa and consider only buckling in the plane of the structure.
SOLUTION
*Joi*~f C
iZ-FJ^o F^-^T. =
o
"ac
T*
+ ZF^ =o
AC
r<1D 4. |&o8 '**
77c= 1.18*3 FCD
- If J.
Me.Us BC aJ AD*- Ifc . ?(%^ *(f )*• U.HW^1^. II.WWO* ^
F^=-E^- 2.32«3*^W
TiW= 2.77x/^A/
SmJk* voJo* S~ TM>«, go^-s TAC^ -- 2.77>**W-*77*W

PROBLEM 10.21
10.21 Each of the five struts consists of an aluminum tube that has a 32-mm outer
diameter and a 4-mm wall thickness. Using E = 70 GPa and a factor of safety of 2.3,
determine the allowable load P0 for each support condition shown.
|*o I l*o I l»o
2,0 m
(1)
(2)
Jcrt_
(3)
(4)
(5)
SOLUTION
Co ~ i4r 3^ * |4 *>*
- 3$. iSS^x/or* ^*
TT2 £1 = 77 * C 70 * /O* X^i" f 8*5 x /O'* >
p - Sis _ 2&&

PROBLEM 10.22
10.22 Two columns are used to support a block weighing 3.25 kips in each of the
four ways shown, (a) Knowing that the column of Fig. (1) is made of steel with a
1.25-in.-diameter, determine the factor of safety with respect to buckling for the
loading shown, (ft) Determine the diameter of each of the other columns for which
the factor of safety is the same as the fector of safety obtained in part a. Use E = 29
x 106 psi.
8ft
Effl RSH
SOLUTION
- O. MIMZ iV
(1)
(2)
(3)
9 =
VZEI
(4) r^ ~ L *
S1Z2 A - 3.12Z k.p. $or- o^« coJo***
|W
n
9 r
-pr^W r
5^£ff I.MS- fc.p.
P
p c, - hr. 3.732'
^.2,*?
P.
R
Cr(cA .
f
«*0 - (U,*Y
L1
(31 Le^/^ - 2.0
Jio " LfiX
= i
t^Ju )
1.76 » ,"r*^
4>* \.%>s i%.&

PROBLEM 10.23
10.23 A 25-mm-square aluminum strut is maintained in the position shown by a pin
support at A and by sets of rollers at B and C that prevent rotation of the strut in the
plane of the figure. Knowing that LM = 1.0 m, £^=1.25 m, and£CD = 0.5 m,
determine the allowable load P using a factor of safely with respect to buckling of
2.8. Consider only buckling in the plane of the figure and use E = 75 Gpa.
SOLUTION
p,
U2EI
'
ever
p -
Le = 0.7 Uo = (0.7yun = G.7 m
VoAioyo A6:
Portion CD: U - X L60 "tt.o)(o.s} - Lo ^
Le<)M<^.
- 1.0 ^
PROBLEM 10.24
10.24 A 32-mm-square aluminum strut is maintained in the position shown by a pin
support at A and by sets of rollers at B and C that prevent rotation of the strut in the
plane of the figure. Knowing that lM = 1.4 m, determine (a) the largest values of LBC
and LCD that may be used if the allowable load P is to be as large as possible, (b) the
magnitude of the corresponding allowable load if the factor of safety is 2.8. Consider
only buckling in the plane of the figure and use E = 72 GPa.
SOLUTION
I- iU3 = £ (SO&O*-- 87.381 x/o3 *m* = S7.3Slx/tP"'1
m
Efviv&Jl e^T ie/i^-rta
BC Le = 0.5 Let
CO Le = 2. Lco
fc 40oAXto
Lbc" r== L» -O^X^V i.^6 ^
O-S
L^'-^U --(o.^rXi-MV 0.4^ m

PROBLEM 10.25
10.25 Column ABC has a uniform rectangular cross section and is braced in the xz
plane at its midpoint C. (a) Determine the ratio bid for which the factor of safety is
the same with respect to buckling in the xz and yz planes, (b) Using the ratio found
in part a, design die cross section of the column so that the factor of safety will be 2.7
when P = \.% kips, I = 24 in., and E = 10.6 * 106 psi.
SOLUTION
r
Zi
I r fVe^b3
r " F.S 2.8 le*" " Kt(fcs';i«'
0.21406 m4
IT' (lO.&x/O*)
ia
beT
W
(a)
dfc? _ i! _ _g.2Wo& - i
W
^*
0.8S62S
_k - i.

PROBLEM 10.26
10.26 The aluminum column ABC has a uniform rectangular cross section with b =
£ in. andrf= £ in- The column is braced in the xz plane at its midpoint C and
camesacemncloadPofmagnitudel.Iklps. Knowing that a fector of safety of 2.5
is required, determine the largest allowable length I. Use E = 10.6 * 106 psi.
SOLUTION
III ' Ifc/ K 2.75VIO*
*t
J> ^ bdl3 - i(i)g)3^ 27.9.3 */<r\nH U = *L
* < fPw * / 7.75" MO*
PROBLEM 10.27
70.27 The uniform brass bar AB has a rectangular cross section and is supported by
pins and brackets as shown. Each end of the bur can rotate freely about a horizontal
axis through the pin, but rotation about a vertical axis is prevented by the brackets.
(a) Determine the ratio bid for which the factoi of safety is the same about the
horizontal and vertical axes, (b) Determine tht factor of safety MP = 1.8 kips, I => 7
ft, rf=1.5in., and£=15x I0*psL
SOLUTION
c*\
ir'£r
P -
lli) b * i J - 0.75" .'~
l_ » 7 -ff -8^ .«
^L
4t*^1
b-M

PROBLEM 10.28
W250 X 32.7
1&28 Column AB carries a centric load P of magnitude It kN. Cables BC and BD
are taut and prevent motion of point B in the xz plane. Using Euler's formula and a
factor of safety of 2.3, and neglecting the tension in the cables, determine the
maximum allowable length L. Use£ = 200GPa.
SOLUTION
W ZSOx 32.7 Iy* HS.^W^'m-S 42.9 xfo* **/
IjT 4.73v IOc*»rS- 4.73 x Ii5c v*H
p = 72 ./o1 W P„. w,.n - (R S.XP) -- /ST. t*lo N
QockJlina l'r> XZ-/>A»ie: Lt = O. 7 L
P
<* " (A7L*)
I - J_ [EXT . jL /fioo»/oOfti-73»to-';
Id. 7*
/-*)
vn
PROBLEM 10.29
10.29 An axial load P is applied to the 1.25-in.-square aluminum bar ABC as
shown. When P = 3.8 kips, the horizontal deflection at end C is 0.16 in. Using£ =
10.1 x lO'psi, determine (a) the eccentricity e of the load, (b) the maximum stress in
the rod.
SOLUTION
1.25 in.
25 in.
Le - ZL - SO ,*n Le - ZL ~- SO ,„.
g. HZJUfO*
0. V684?
= J£
M*fC
r £ [se*(|.076-02)- l] = l.io^S e
O.J4S) >«,
I. \o\% \.lo%%
6- r ■£- ♦ ids - .3-s^io* + (L»y^7Xo-<n) r s.<n ^irfpir - -sin **.'

PROBLEM 10.30
310 k.N
W250 X 58
«0
1030 The line of action of the 310-kN axial load is parallel to the geometric axis of
the column AB and intersects the x axis at x ~ e. Using E = 200 GPa, determine (a)
the eccentricity e when the deflection of the midpoint C of the column is 9 mm, (b)
the corresponding maximum stress in the column.
SOLUTION
p -JLl££ ,
-6 2.
L - G.S
f&£*/o"* i*t"
w
(^p -878.3k/o N
fk- 8TO.SMO*
e -
«?v/o
-s
-a
=- 13.24* to" * - 13. 2H *m
■t/
isao * /o~* iS5"x /o"^
= 41.78 xlo* + 37. 26 x/o6
l^ofyio'pa - 7toMr\ -*
PROBLEM 10.31
wax
10.31 The axial load P is applied at a point located on the jt axis at a distance e
from the geometric axis of the rolled-steel column BC. When P = S2 kips, the
horizontal deflection of the top of the column is 0.20 in. Using E = 29 * 106 psi,
determine (a) the eccentricity e of the load, (b) the maximum stress in the column.
SOLUTION
(av >^e[«c(af> l] =■ o.8o8.4e
e -
Jt
m»j»_
d.8o3l6
o.-ao
r 0.2.47 ,'»
- P + Mc _ £ + ±L - «^"to* + 34.693^/Q3 s Miuio**.-

PROBLEM 10.32
1.2 m
32-mm
diameter
1032 An axial load P is applied to the 32-mm-diameter steel rod AB as shown. For
P = 37 kN and e - 1.2 mm, determine (a) the deflection at the midpoint C of the rod,
(b) the maximum stress in the rod. Use E = 200 GPa.
SOLUTION
p„ - ^ ^ Tr'^/^y-r/.^./o-') = 70^ wo, M
9 37v/o*
<m ^ - e[*.c(|^)- i] ^i/ssi7e =(/.Mi7)a^
= 1.45*8 ^.m -4
'lH*i#l
r £ + M^ r 37*JO3 ^ (/PS".7rYl<;*JP"*j . 7g ^ x ,o' p«
$Qi.2S*IO~&
Sl.H7*lo
= 7gB<7 MPo.
PROBLEM 10.33
10.33 The line of action of the axial load P of magnitude 270 kN is parallel to the
geometric axis of the column AB and intersects the x axis at e ~ 14 mm. Using E =
200 GPa, determine (a) the deflection of the midpoint C of the column, (b) the
maximum stress in the column.
SOLUTION
L - 7. Z *
A- ££CO v*mx := £44o » /cTa w^
1. r (7,gx/o* M^ * I7.8X/0"* w,v
S^ ^ l 75" x /o* ww* =■ I IS x /o_c- •*> 3
W200X52
p - TTa£X 7TT(3QQx/Q'>)07.g)c/o-c)
~ G77.77 x/Oa rJ
£. - 27o»/Q* - a 3983G
P*. €77.77 *<o> ' 0-^836

PROBLEM 10.34
W200 X 52
10.33 The line of action of the axial load P of magnitude 270 kN is parallel to the
geometric axis of the column AB and intersects the x axis at e = 14 mm. Using E ~
200 GPa, determine (a) the deflection of the midpoint C of the column, (b) the
maximum stress in the column.
10.34 Solve Prob. 10.33 if the load P is applied parallel to the geometric axis of the
column AB so that it intersects the x axis at e = 21 mm.
SOLUTION
W loo-* SZ
A - GGGO nw * 6G£Ox/o"* m"1
S^ - 175*/©* "»** - 175" X \0C K>03
•«* " U2" = (7-*-)*-
- 677.77 x/G* 1^
£?o*<o3
671.11* to1
r- O.S9S36
(*1
GO
5.
M*^ ?(e + j^ ^O^x/o^X^l + (7.36Y/0'3) - \ozsq N-m

PROBLEM 10.35
0.25 hi.-J |*
1,75
2,5 ft
GO
10.35 An axial load P is applied at a point D that is 0.25 in. from the geometric axis
of the square aluminum bar BC. petermine (a) the load P for
which the horizontal deflection of end C is 0.50 in., (£>) the corresponding maximum
stress in the column. Use E~ IOA xto* Usi.
SOLUTION
L - 2.S ft » 3o»« Le * 2L - Co .■*.
p - TT^I - TT»Oo. I* 10^(0/78*57) . ^, £„, fc.
** " U1- (G°^ r
O.KMCU5
?= O.GHH Pcr * 13.2? »V,'f* -*1
Ow* /\X S.oCZS 0/7SI5/

PROBLEM 10.36
120 mm
t = 6 mm
10.36 A brass pipe having the cross section shown has an axial load P applied 5
mm from its geometric axis. Using E = 120 GPa, determine (a) the load P for which
the horizontal deflection at the midpoint C is 5 mm, (b) the corresponding maximum
stress in the column.
2.3 m
SOLUTION
I~ tJ (Co*- c/) = Z,SooSxtoc')„„H = 3.S0o^y*/O6
tv»
Cti'CcaS
J*
Vcr IT O +■ b -1

PROBLEM 10.37
120 mm
t = 6 mm
2.8 m
(a) y^eIsec(W|V \"]
cos\jm
/"it P
ft- ' 1 IT
10.36 A brass pipe having the cross section shown has an axial load P applied 5
mm from its geometric axis. Using E = 120 GPa, determine (a) the load P for which
the horizontal deflection at the midpoint C is 5 mm, (b) the corresponding maximum
stress in the column.
10.37 Solve Prob. 10.36, assuming that the axial load P is applied 10 mm from the
geometric axis of the column.
SOLUTION
L-- 2-8 m Lt * Z€ r*
r** - u. _-
SH%.S k/6*3 tJ t S2S. 8 fcW
O.MClQ
P= 0.28670 F> =r (5/. C fcW
tor
*~ ■ £♦ $*« $ftg> +-fi|^^lB W-,^..o,.5Mfk

PROBLEM 10.38
10.38 An axial loadP is applied at a point located on the x axis at a distance e =* 12
mm from the geometric axis of the W310 x 60 rolled-steel column BC. Assuming
that L = 3.5 m and using E = 200 GPa, determine (a) the load P for which the
horizontal deflection at end C is 15 mm, (b) the corresponding maximum stress in the
column.
W310 X 60
SOLUTION
A =
Sv - ISO WoVm1 = 180*10* m
/o~ft)
U* 2L* 7-0
- 737.3 x/os W = 737.* UW
<r P , Mc P , M - 3C>S3$>(Q% | q
=■ IO&.8 MR!. ^

PROBLEM 10.39
W310X60
£ , [f A^OS
10.38 An axial load P is applied at a point located on the * axis at a distance e = 12
mm from the geometric axis of the W310 x 60 rolled-steel column BC. Assuming
that L = 3.5 m and using E = 200 GPa, determine (a) the load P for which the
horizontal deflection at end C is 15 mm, (b) the corresponding maximum stress in the
column.
10.39 Solve Prob. 10.38, assuming that I is 4.5 m.
SOLUTION
SMgOX IO*Vnrr?- \%O*\0'<- v*Z
>^*e|XI&)-Q
seel
,-*e
1
L 77
Ji
T7 ^"oS 15+151
<M
P =■ 0:4^57 P^ ' aU.79 ItM
-] - o.qW7
Lb; bnw* - a T" " A Sy 15-<=id»io'* i sower* _^
= fc*. 8 MPcl

PROBLEM 10.40
y
1,2 in.
W12 X 50
10.40 The line ofaction of an axial loadP is parallel to the geometnc axis of the
column AB and intersects the jc axis at jc = 1.2 in. Using E = 29 x 106 psi., determine
(a) the load P for which the horizontal deflection of the midpoint C of the column is
0.8 in., (b) the corresponding maximum stress in the column.
SOLUTION
W I2y5b A= 1*7 ,h*, Ij- SG.3 .«*, S, r L3.CJ ,■„«
L'^fl-" 288 m U ? 288-iV-
Cos
2
4iTCCo&
e
*]*
C<0
•£ r [y ft^cas-^1;^^] r O.SHStl p= 0.5*18^ Per * 57-7 k.ps
ibi o^ A + x A % If."? IS.6! '^ **'

10.43 The steel bar AB has a | x f -in. square cross section and is held by pins
that are a fixed distance apart and are located at a distance e =0,03 in. from the
geometric axis of the bar. Knowing that at temperature T0 the pins are in contact
with the bar and that the force in the bar is zero, determine the increase in
temperature for which the bar will just make contact with point C if rf=0.ot in. Use
E = 29 x 1O6 psi. and the coefficient of thermal expansion a = 6.5 x 10 6/°F.
SOLUTION
A = CiVi^l * O.H0C2S- ;«'
EI- (^'h'/o'X'.6''7**'"/!")*-) " ^77£', -**■»"»*
, Z^r . ^ZZlLl s 7370 it.
c*- "" it
C2F
e = 0.03 in.
l/f r C^'d+ir - Cos" O-^V - Cos-'fO-Ts)- 0.72*73
T^"* I%(a7^73)]* = 0.21)70 P= O.^U7o Pc„ = 15*60.2 A.
(I) Siwijw'e CL-opiroxiVvA-Tjo^ by i<s no^i w« e.cc€*iT ^"ccf1-! -
To+J eJU^.'.n = OiLCATV |j*- r
l56<?-2
- £8.<*/°F
EA *e *Ur o
TiaTtJf e/«»rva *Ti'ow 07 ce^7f«ifl*/ 0-96 »S = Oi-L (AT 1
S r 6 ( p +** aT Cc>5 Pv ~ P s'" P*^
EA
E"Ac*.
otL

10.41 The steel bar AB has a f x f -in. square cross section and is held by pins
that are a fixed distance apart and are located at a distance e =0.03 in. from the
geometric axis of the bar. Knowing that at temperature T0 the pins are in contact
with the bar and that the force in the bar is zero, determine the increase in
temperature for which the bar will just make contact with point C if d =0.01 in. Use
£ = 29* lO'psi. and the coefficient of thermal expansion or =6.5 x 10"6/°F.
10.42 For the bar of Prob. 10.41, determine the required distance d for which the
bar will just make contact with point C when the temperature increases by 120 °F.
SOLUTION
e =0.03 in.
EX - e«»*tocXl.6^^1^Vfo-,) = ^77?' -ffc-mf
Tt'EI
R.i-
Vt^l) - 7370 4k
■ft.
fo«^ am iv^pfuved fie^vwa,* ah^Jy s/s in cjft/dl in* ecce^T/Vo"tm see.

PROBLEM 10.43
127 mm
127 mm
A = 3400 mm2
I = 7.93 X 10"6 m4
r = 48.3 mm
10.43 A 3.5-m-long steel tube having the cross section and properties shown is used
asacolumn. For the grade of steel used o^ = 250 MPa and £ = 200 GPa. Knowing
that a factor of safety of 2.6 with respect to permanent deformation is required,
determine the allowable load P when the eccentricity e is (a) 15 mm, (b) 7.5 mm.
p (Hinl: Since the factor of safety must be applied to the load P, not to the stress, use
Fig. 10.24 to determine Pr).
SOLUTION
3.5 m
A = 3MOO wo~c wiZ
Le - S.S
-*
U _ . 3.S_
<t$.3x/0
t, r 72.H£
C = ^ = 63.5 wm
9/A = 1^76" MP* - KH.7SV/04 Po-
Us.~> W~ of s<Wj P^ ^^ = isWw - m *i/
Usmj Fi3. lo.*f wi+fc L/r* 72.^6 «»d ec/r1^ o.zoms
p/A = US'. 2 MP* = I7SV2 *<0< P«-
P - (175*. 2 xlo'Xsvoo */o"' ) - SKx/o5 N
r J r J* J: -P. - &f*fo* - 221 *lc? IV * ^<f kW

PROBLEM 10.44
127 mm
127 mm
A = 3400 mm2
I = 7,93 X 10"6 m4
r = 48,3 mm
10.43 A 3.5-m-long steel tube having the cross section and properties shown is used
as a column. For the grade of steel used <% = 250 MPa and E = 200 GPa. Knowing
that a factor of safety of 2.6 with respect to permanent deformation is required,
determine the allowable load P when the eccentricity e is (a) 15 mm, (b) 7.5 mm.
(Bint: Since the factor of safety must be applied to the load P, not to the stress, use
Fig. 10.24 to determine Pr).
10.44 Solve Prob. 10.43, assuming that the length of the steel tube is increased to 5 m.
SOLUTION
3.5 m
Le. - S v*
C = -^ =G3.5
48. 3* A?
F? - 103.5*
(co e = is
ft (-VS.3)* O.VOg^f
5 ^ r 11^.75 MPa- « 11 2. IS * \0^ Pa,
P =- ( |l?.7S**/04)(3^00»/Oc ) " 383*/o^N
(b) e « 7-ff
€C _ C7.5")(C3.Q _
*•*> y.2 (tfg.3)*"
o.zoms
Ui.>5 F.y 1^.2*1 jiVes -j~ r J33.2.MP*.- 133. 2. * /o6 -pa.

PROBLEM 10.45
10.45 An axial load P is applied to the WlO x 30 rolled-steel column BC that is free
at its top C and fixed at its base B. Knowing that the eccentricity of the load is e =
0.5 in. and that for the grade of steel used ar = 36 ksi and E = 29 x 106 psi.,
determine (a) the magnitude of P of the allowable load when a factor of safety of 2.4
with respect to permanent deformation is required, (b) the ratio of the load found in
part a to the magnitude of the allowable centric load for the column. (See hint of
Prob. 10.43.1
SOLUTION
W IO x 3o
W10 X 30
A=^ 8.84 .v.*
r I 3 /- 3*1
•^
37
i«^,
v^
it
I go
7^7"
Le r ^L = ISO C~.
e, - 0.5" in
O. 7731
-£ r 10.47 ksi
P,
Le* ' OS©)*"
I47>S"
38.6
4u5" "
3S.G !<•*/»
ki'ps

PROBLEM 10.46
W10 X 30
10.45 An axial load P is applied to the WtO * 30 rolled-steel column AC that is free
at its top C and fixed at its base B. Knowing that the eccentricity of the load is e =
0.5 in. and that for the grade of steel used «rv = 36ksi and£=29x 106psi.,
determine (a) the magnitude of P of the allowable load when a factor of safety of 2.4
with respect to permanent deformation is required, (b) the ratio of the load found in
part a to the magnitude of the allowable centric load for the column. (See hint of
Prob. tO.43.1
10.46 Solve Prob. 10.45, assuming that the length of the column is reduped to 5.0 ft.
SOLUTION
^ * i-37 i«
J*r
ec
iaz , S7.4
i.a?
D*.'»3 R3 lo.ai £ * i**.-ro k%; Pt(/1.*oX«.S*) - 131.7 k»>*
w
e. -
(L>1
Tft>M
0.3^7

PROBLEM 10.47
10.47 A 55-kip axial load P is applied to a W8 x 24 rolled-steel column BC that is
free at its top C and fixed at its base B. Knowing that the eccentricity of the load is e
= 0.25 in., determine the largest permissible length I if the allowable stress in the
column is 14 ksi. Use E = 29 x 106 psi.
SOLUTION
<w*fL«* ft «-(*£)]
.sec
cos
I *Vr. I" eel p* ' ) (Ms<i«i L""» 'J ' *"AS"
lit"
I. I69SS
p r _£ .
u1
O-SSf/* IT* EX
_i
ss
- S2.12.ylo ivf
Lc = 229-7 ,V, ^ 3U
L - M4.8 i« * <?. S7 ff.

PROBLEM 10.48
10.48 A 26-kip axial load P is applied to a W6 * 12 rolled-steel column BC that is
free at its top C and fixed at its base B. Knowing that the eccentricity of the load is e
= 0.25 in., determine the largest permissible length L if the allowable stress in the
column is 14 ksi. Use E = 29 x 106 psi.
SOLUTION
Ut+a.: V = U ^'pS^ e = 0.25 -V
WGxia; A - 3.SS .«* \af - 4.ooo ,v
^t Jj* = &<**> >p Iy* 2.9*1 n^ V^* a«W3 i">.
sec
0.8GZOH
= O. Soil 7
L * ^ Q.3QII7 -it* £r r o.zoinii*(woooYwi) _ ^ 9/3 y/Qs /n<

PROBLEM 10.49
10.49 Axial loads of magnitude P = 84 kN are applied parallel to the geometric axis
of a W200 x 22.5 rolled-steel column AB and intersect the x axis at a distance e from
its geometric axis. Knowing that allowable stress aM « 75 MPa and E = 200 Gpa,
determine the largest permissible length I when (a)e = 5 mm, (b) e = 12 mm.
SOLUTION
D«£l- P = SHxlo* W E= zoo mo* Po,
W Zoo*Z7.% A * 286© w,*,*- - £g6o » fc>"* k^
bf - loa •»»* C = ^f - 5"/ *»«*
Ij * 1.43. x/oc ww^ =
1.43 *lo"" **
l^ - 22-3 mi
61« - <ST_ r 7S MP^ - 75*joc Px
" )
-0
A 6^
P
sec
-1 - ft «t(i
A 6".
—p
(IV£) -
ec
«*
«"<**:
<a^ e - S»\»>
sec
c^(f-/£)- 0.330D4 S-/C * '-^44
p T [£0-2^)] = 0.&17S7
p~. _£.
. , 2l£IL
0.61.757 l.e»
■ r- ^617^*7 T)x irr _ a6*757TTa-faoO»/O^X/-l*^v^") _
* P " £H >/o*
20.GI *x
Le " 4.rH *. L - /-e * 4.SH *
C»s (|Jf- ^ - 0.79 X) C f -/f r o. GS44* 4 35"
P.
2 i-
O. |746t
_ TT'CJ
Lt - — r — — —,——— =■ O.
Lc «■ 2.4)
S28
v*

PROBLEM 10.50
10.50 Axial loads of magnitude P = 580 kN are applied parallel to the geometric
axis of a W250 * 80 rolled-steel column AB and intersect the x axis at a distance e
from its geometric axis. Knowing that allowable stress aM = 75 MPa and E * 200
GPa, determine the largest permissible length L when (a) e = 5 mm, (b) e =* 10 mm.
SOLUTION
DJ&,: P = .5*80* /o3 w £ » 200* /*>* Pa.
W £5bxgO A = lo^OO wi«nl - l0200*|cffc ^*"
bf. -~ 255" ««
c - a
- \2~l.5 r>1tvi
^ .- ££"„0 ****
(a) e:^m
J
UV?^ ) ' {5Xms} [ szox\o* 1
-E- r [f-(/.07«OV)1 = 0.H7IOI
Le
S€C
3. VP^
1.0780*/-
c^ O.H7lo|
2
Tt'ESy
L,1
L - U • 8.31 v*
CCS)1
W e - 10 w** sec ^ yR, y - (/ottw>
p" T[f"(0-3^^^1 = 0.O44O&3
77* £1
«* O.O^O&S
?
2.54 k,
5BO * I

PROBLEM 10.51
10.51 A 12-klp axial load is applied with an eccentricity e = 0.375 in. to the circular
steel rod BC that is free at its top C and fixed at its base B. Knowing that the stock
of rods available for use have diameters in increments of -Jf in. from 1.5 in. to 3.0
in., determine the lightest rod that may be used if aM = 15 ksi. Use E = 29 * 106 psi.
SOLUTION
er
Tl'FJ 7,aQ?OooW
. i - Hi1. J- .. ^
A 6*f Tfcf*
iC
P = IZ Jc.ps
k;
p*
Us« A~Z-\Z£ m.
el(.V>
;?_25
■2.0
suas
A(.>n
3.^76
3. N/6
3.SH6
?c0^
S<».o7
J?4.S<»
3/,o<?
ec
i.assa
i.s
1.41/8
<W&si)
T.^6
|6.*r<*
//.9o
6w* -
*-
ll.<?0 hr
< irJrsr

PROBLEM 10.52
y
10.51 A 12-klp axial load is applied with an eccentricity e = 0.375 in. to the circular
steel rod BC that is free at its top C and fixed at its base B. Knowing that the stock
of rods available for use have diameters in increments of £ in. from 1.5 in. to 3.0
in., determine the lightest rod that may be used if 0^= 15 ksi. Use £ = 29 * lO'psi.
10.52 Solve Prob. 10.51, assuming that the 12-kip axial load will be applied to the
rod with an eccentricity e * £ d.
SOLUTION
E *;?<?* to4
u
P * 13 Icipa
WfafC
, £.
[
ec
+ ^r sec
ill
7VT5
A L
4- *f. o sec
(tflM
dfi^
Z.7.S
3.0
2.5
2.C25
A6V)
3„T?6
7.068
4.«?o9
5".4-/x
Per Oof*)
3*7.07
123.48
7?. 38
6L»(fa<)
£1.75
9. 3<7
15. %%
13.27
Us* d - HZ* iV
61^-- 13.3? fa,* * iSfai"

PROBLEM 10.53
10.53 An axial load of magnitude P = 220 kN is applied at a point located on the x
axis at a distance e~6 mm from the geometric axis of the wide-flange column BC.
Knowing that E = 200 GPa, chose the lightest W200 shape that may be used if o„i =
120 MPa.
SOLUTION
e - fe^rw c a ^ - ^ -
SKft,i
W^OOxHU
A(loV)
zz<\o
bf(M*o
13$
»oa
I,GtfV)
9*0!
3.3o
I, HZ
fV(»n»1
f£ O^CMfV)
Use
VJ Zoo* ZQ.Q, -
6-.
117.H MPa.

PROBLEM 10.54
10.53 An axial load of magnitude P = 220 kN is applied at a point located on the x
axis at a distance e = 6 mm from the geometric axis of the wide-flange column BC.
Knowing that E = 200 GPa, chose the lightest W200 shape that may be used if aM =
120 MPa.
10.54 Solve Prob. 10.53, assuming that the magnitude of the axial load isf=-345kN-
SOLUTION
P -- 3HS h to1 N
L- l.g *
L*2L*3.6
v*i
Pcr . SlJ& , *WfUy = ,^w ^ M
_ P
>r*»46
sec
Shane
w zoo*m.i
W Soto x 35:9
W »0O)t 3/.S
AOoV*)
33to
bp(lMW J
16&
)3M
j^do'V) ^c^
S.3o
7-64
4 JO
41.2
31.2
40.2
32. o
Y1*
ec _ e laf
1372.
cas.4
0.^7**
0.3M6
^5*5
172.6
Use W^ook 3r.*7
(Si-* lO^.S MPa.

PROBLEM 10.55
10.55 Axial loads of magnitude P = 175 kN are applied to a point located on the x
axis at a distance e = 12 mm from the geometric axis of the W250 x 44.8 rolled-steel
column AB. Knowing that Qy = £So MPa and E = 200 GPa, determine the factor of
safety with respect to yield. (Hint: Since the factor of safety must be applied to the
load P, not to the stresses, use Fig. 10.24 to determine Pt.)
SOLUTION
For W2S"0*4V.S A^SV^o*^ a ^- SS.I ^
Pr * A1J/A - tS"7«^**o"*")(!«.57x/oO ? ^)7xloa M ^ £"17 fe»>
R5-
- Ix , ^IZ r
i7r
a.?*"
PROBLEM 10.56
10.55 Axial loads of magnitude P = 175 kN are applied to a point located on the x
axis at a distance e = 12 mm from the geometric axis of the W250 x 44.8 rolled-steel
column AB. Knowing that ar * 2A"o MPa and E = 200 GPa, determine the factor of
safety with respect to yield. (Hint: Since the factor of safety must be applied to the
load P, not to the stresses, use Fig. 10.24 to determine Pr.)
10.56 Solve Prob. 10.55, assuming that e = 16 mm and P = 155 kN.
SOLUTION
Fo»r W50M1.8 A^ S1ZO *i*L i Cf ~ 3SJ ^^
Le = 3ZOO k*~ lt/\r - \o8.2G>
1?/A - *l. 17 hpa * S/.I7 M/m*
1* * ACRr//^ (S7Zo»to'*)tgl. 11*10"-') ? *t£**lo* N - •*€* *tf
F.S. =
Pr ^ 5KV
IS\S
3.00

PROBI EM 10 57 10.57 Using allowable stress design, determine the allowable centric load for a
column of 6.5-m effective length that is made from the following rolled-steel shape:
(a) W250 x 49.1, (b) W250 x 80. Use oY = 250 MPa and E = 200 GPa.
SOLUTION
** rK«(L/r)- ' (L«Xl3*.uV- " ^^ PG-
P*« r A$« - (MS-Ox/o-^CsS.^^/O* ) = 348x/Os N r 36g *W
(VO W250*go A^ Io^ooh /©-c k,*- yv^ C5:ox)o~s »v,
T T gT^F* ' ,0° < C: ^ - 0.7W7
F.S. * f + I (0.79577) -^ (0.77577)3 - i.qoa.o9
P.* - A 61* * (r©2oo*/cr6)(S1.82»'/dr6) * ^iMo'N - 9/€ kM
10.58 A W8 x 31 rolled-steel shape is used to form a column of 2l-ft effective
PROBLEM 10,58 length. Using allowable stress design, determine the allowable centric load if the
yield strength of the grade of steel used is (a) o? = 36 ksi, (b) Oy = 50 ksi. Use E =
29 x 106 psi.
SOLUTION
Steei-' £ - ^OOO lf«i' W g *3l A » ?. 13 in* ft,,** - ^.03 in
(a> 6V-36 *»; Cc'^^ --plfe) s „G.I0
R« f €^iA = (9-59^(^.13) - «"?.£ kips

PROBLEM 10.59
* = 0.28 in.-
10.59 A steel pipe having the cross section shown is used as a column. Using
allowable stress design, determine the allowable centric load if the effective length of
the column is (a) 18 ft, (b) 26 fr. Use ar - 36 ksi and E = 29 * 106 psi.
SOLUTION
(a) U - IS fr -- 2K
C0 " ^ - 3.0 in. Q - C- £ ' 2.72 ,Vv
E * ^000 Irs.'
Ct./srryssss,ut.l
- 2.0X47 ;«
F.5. - f+ 4(o-8Kot)-i(o.gHfiol)3 = /.^o82
W Le = 2*tt =■ 3/2 in U/^r /S4.0T7 > Cc
IT2 £ ttV^ooo^
6l» t-
t G.ZS its.'
PROBLEM 10.60
10.60 A column is made from half of a W360 x 216 rolled-steel shape, with the
geometric properties as shown. Using allowable stress design, determine the
allowable centric load if the effective length of the column is (a) 4.0 m, (A) 6.5 m.
Use Or = 345 MPa and E = 200 GPa.
J4 = 13.8Xl03mm2
lx = 26.0 X 106 mm4
Is = 142.0 X106 mm4
s 43,406 ^^
- 4 3. H£>£ Wo
-i
fw
''l20O*K>'')
a.
Rli- <&«A --(M3.6/x/Os)(l3.8x/o-3) r /5"6Sx/03V * J«2 fcW
R* = £*A * (45-.^rWoc)(t3.8^o-s)^ C33»K>SN * &33 ItW

R 10.61 A 3.5-m effective length column is made of sawn lumber with a 114 x 140-
PKOBLEM 10.61 mm cross section. Knowing that for the grade of wood used the adjusted allowable
stress for compression parallel to the grain ac = 7.6 MPa and E = 10 GPa, determine
the maximum allowable centric load for the column.
SOLUTION
$a*n Jlo*ber : c- O. 8 3 6~ - 7-6 MPa. Kctf - O. 3 E = /oooo MPa
L/J - 3.S//lVwo"* r 30.10
S? = 0.^18 78
<S1
U = JL±-Slfia J-H'&73- = O 88673 V = SfiiSLr O.S2347S"
Cp - U - 7U"-V r 0.374OS
7fttf2 A sawn lumber column with a 7.5 * 5.5-in. cross section has a 18-ft effective
PROBLEM 10,62 length. Knowing that for the grade of wood used the adjusted allowable stress for
compression parallel to the grain is ac~ 1220 psi and that £=1.3 * 10*psi,
determine the maximum allowable centric load for the column.
SOLUTION
Sa.wn io^ker-* C-0.8> 6L - IZ2o Pa, £ r |.3W06Ps/ k^- - 0.3
A r(7-£" )(£$")=■ 4/-2S"/*2* <A = S.S in. L- IS W =• */£ ;*
^ - &*£" _ (0.3Y|.-a,WQ<) _ ,c_ or

PROBLEM 10.63
t = 0.375 in.
4.0 in.
H—4.0 in.—»j
r
10.63 A compression member has the cross seetion shown and an effective length of
5 ft. Knowing that the aluminum alloy used is 2014-T6, determine the allowable
centric load,
SOLUTION
A - L^.of -(3.2^ - J.H37S i«*
ir Aft<i.«>y,-<3.Kri] * is.036 ,-**
- HO, 33 < £5 ^ ZQit-TC J^,'n<iM «i/o^
G*^i " 30."/ - O.asOViO - 30.7- (o„2%Xho.3$) = X/. 42 *s.
Rttf = GLrfA " (?»• ^KC^ 375) - 116.5 k.-ps
PROBLEM 10.64
rs
&^ ^M __ 15 mm
100 mm
T
■ 9 mm
10.64 A compression member has the cross section shown and an effective length of
1.55 m. Knowing that the aluminum alloy used is 6061-T6, determine the allowable
centric load.
SOLUTION
1 15 mm
T
!„ * 21m"1 I«r ^.73225* WOe m^'
Iv r *f -kOsXioof]-4 &OX*?)3 - 2. ro«2s"*/ofi ^
-c.
A r 2 060)05') ■»(***> 1 - 3€30 -^ = 3GSOWO" ^

PROBLEM 10.65
10.65 A column of 6.4-m effective length is obtained by connecting four 89 * 89 *
9.5-mm steel angles with lacing bars as shown. Using allowable stress design,
deteimine the allowable centric load for the column. Use oY = 345 MPa and E = 200
GPa.
P
1M
7
7Y.2
A 1
G.H
■■■■■>• Jtm^^*
Is -
V* " 7^oS3x lo~z
SOLUTION
a
7 cr "Y 3V5^/o4
=■ 106.9-7
AL- I6oo.wiyv,1
x - ^5.8 wh
X* - 1. f * lo^ K.m
d = /oo - X ~ 7*rf. 3_ ****
1= <#(M'+IWV ^[0600X7*0*+ l-nv/O*]
IT
^ JjT ^ 77.053 k,^ ■* 7?-0S"3*/os^
Le/r
gO.<?58 -< Q.
- 0.7 56S3
F.S. - f + f (0-7SG&s)-^(a75-€33)' = 1.8*63
Pat - SuA '0^.«3i»lO6)fe4oo*|O"*) - S3/x/o3 V - 831 HJ

PROBLEM 10.68
10.68 A column of 23-ft effective length is obtained by welding two |j -in. steel
plates to a W10 x 33 rolled-steel shape as shown. Using allowable stress design,
determine the allowable centric load for the column. Use or = 50 ksi and E = 29 *
106psi.
SOLUTION
Vor W 10x33
A= ?.7i i*\ d- ^.73 ■ « £>f- 7.760 i«.
umin -
r -
Sreej?
im'n
A = 9.7 M (JOCf-X^s") - 17. 007^ in*
Iy= I70 ■*- (*)■&■& )^-73^ ™ ^7.^7 i«V
U - 23 ft = *76 .V. ^ - ££L. r g,.os < Cc
F.S. - f ♦ %(o.^Z0S)-i(0.%iZosf = 1.10&7
TU ? $^A - (17. HTYl7.oo7ff") - 21* k."p»
U/r
PROBLEM 10.69
l:
■- i
1
'.
: ■;
i
&:>.
'■:
*
lf>.
■
'
'■ >
1
. *
i
-
If
S
140 mm
6fcfi
10.69 A rectangular column with a 4.4-m effective length is made of glued
laminated wood. Knowing that for the grade of wood used the adjusted allowable
stress for compression parallel to the grain is ac = 8.3 MPa and that E = 10 GPa,
determine the maximum allowable centric load for the column.
SOLUTION
6;- 8.3. H?^ E - lOOoo MP*.
L = 4.H** L/a| ^ 31.?m
U __ _ -^^ - 0.WB8M
= ^^ -,o.S«lM
--/^
<S* - S; Cp - (£.3X°* ^8^0 " 3. &84S- MP*?

PROBLEM 10.66
10.66 A column of21-ft effective length is obtained by connecting two CIO * 20
steel channels with lacing bars as shown. Using allowable stress design, determine
the allowable centric load for the column. Use or = 36 ksi and E = 29 * 106 psi.
-7,0 in.'
T -
SOLUTION
C\o*Zo A--6".S& ."m1
I*= 78.9 in
Ij= £.81 .V
• 3.S->|
waki - A - C?)(S"-S8)r 11.76 in l
Ix = eOC78.«0 -- 157. 8 .■«*
■ Y 11.76. - *-T'5 |M
— * r
k>fc -
0.67IG.S
F.S. = f + |(o.a/65)--i(o.£7Ks)3 = K8807
P.* - 6^A • (H.8*y 11.76 ^ - 17^.3 k.ps
f fttf 7 A compression member of 2.3-m effective length is obtained by bolting
together two 127 * 76 * 12.7-mm steel angles as shown. Using allowable stress
design, determine the allowable centric load for the column. Use a, = 250 MPa and
£=200GPa.
SOLUTION
S+fceJ?:
^TT'E _ /Zn%QlOo*(c?T
= I2S.CG
L 127*76 x 12.-7 MW "U^Uc giVei Aj *420 »*■** Ix - 3-93 * /Og *^s
R>/' coition
I.:n = Ix = 2.12*10'
h-iwt
* .
I, > J>
-4
= Z.lZvto****
u
^.a
= 5?6>93*k>~ i~>
Cc
Q.8 74SS"

6 mm
1
PROBLEM 10.70
\8mrn 8mm/
J \ I.,I. M mm »'■■' ' I-
6 mm
54 mm
mm
T"
10.70 An aluminum structural tube is reinforced by riveting two plates to it as
shown for use as a column of 1.7-m effective length. Knowing that all material is
aluminum alloy 20I4-T6, determine the maximum allowable centric load.
SOLUTION
b0 - 6 + % 4 34 + 8 + <© = 62 m^
h0 - % + 5H + 8 - lo ww
= 2.Sot <W!»«l ' Z.SOH»lo m
I, - ±[\ Kl - fc>.».;] ^A[(«X7«,f - Ms*)*!
~ l„32Go? x"/o: m
Ij r i[ ^t.1 - h,fc»;J * ^^oVttf-ev^)1] - L*l337-/o4
m*vi
T I.,
^ - 0EL - f I.ai337»lc
— r 22.0/3 *>*t t J??.0l3Wo' *
L^LT
t^i
L
1-7
22. o 13 »<
—5 - 77.23 > 55" •&*
Jo*5 ' J
c~ 373 Wo3
™£? - «.37 MP«
p^ r 6^ A -~ {CZSl*l0*)(2.SC#*lcr*) - \S£.2*l<?hJ - /^.5 feM

PROBLEM 10.71
10.71 A 280-kN centric load is applied to the column shown, that is free at its top A
and fixed at its hase B. Using aluminum alloy 2014-T6, select the smallest square
cross section that can be used.
280 kN
0.30 m
SOLUTION
2.07gS
b
A
-/iTa
■cdtoy
-(2
2 -
3.^4-
) MP^-^-^pjOo') P*
212b*- 3.^b - ZSO*to~* = o
4V.?*/©"3 ^>
L -
2.0785
^.Q7 3S"_ -. tir or < T£"
Answ* b * MH.^x/o"3 ^ - ^W.1**~v

10.72 A 16-kip centric load must be supported by an aluminum column as shown.
Using the aluminum alloy 6061-T6, determine the minimum dimension b that can be
used.
18 in.
SOLUTION
U * L - IS in
I X.-tn _ O
" if*.
A- Zh^ !_-- AftbXb^ * £b'
r aaa-i|ffi£ ks;
R* = S* A = (P.0^-I^)(2k2 1 ~- 4G.M b* - I5I7/S la WP
40.4 fc1 - II. 11/ la * 16 b * 7?X^T) " 0.85 3.*
L. - &?-£^f - .Cff.Syy r 73 0<t > CC AfcSv*wpf,a* not ve<r$*'eJ.
A^so^e "7: > £6
6- - gi&QQ- _ jifoo b__ . |3 ,,-j^ ^Si.
te
.zvi
O. 884 /*.
V^ " O.S8M r
b * 0.88*1 m.

PROBLEM 10.73
120 kN
2,25 m
10.73 An aluminum tube of 90-mm outer diameter is to carry a centric load of t20
kN. Knowing that the stock oftubes available for use are made of alloy 2014-T6 and
with wall thickness in increments of 3 mm from 6 mm to 15 mm, determine the
lightest tube that can be used.
SOLUTION
P = \ZO * /o3 M
P *«***,
90-mm outer
diameter
R
or
CA$c*}<dv ?jf $«r zouX +Lck«es*
S*# = 3U- (-585-(L/r) MP^ tf i-/V* * .ST
^av ioa MPcc ,-f l/^ > £"5
t
Kih->
G
9
I*
IS
*i
WW
3*
3C
33
3o
A
**>
22^o
2<H/
3s3**
I
|06w^
l.lDf
I.1CM
z.?a«l
2.53*/
V
mm
2^.78
2S.82
2l.?o
7.7.0*7
L/r
75". 56
7«.68
2(5.65
%%Ao
G+
MP*,
6^.1*
il.oi
SlJZo
53.7V
R*
kW.
/o3. 1
1^.7
\C8.Z
121.*
Since. Vat ^«Ji+ U ^rc~fe*> fU^ U° UWJ Dse £ = ^ hm

PROBLEM 10.74
Wd = 63.76
10.74 A 18-kip centric load is applied to a rectangular sawn lumber column of 22-ft
eflfective length. Using sawn lumber for which the adjusted allowable stress for
compression parallel to the grain is ac= 1050 psi and knowing that £=10* lO'psi.
determine the smallest cross section that can be used for the column if b = 2d.
SOLUTION
Junker C= 0.8 Klg- O^
St = l°S° pS1' Ir-lOxlO'p**
A
Assume** Cfp = 0.5
<5ur s 61 Cp = (/oSo)(o-f ) = SIS p&.
^
Pa/.
2 6Uf
- P 3 OQQ _ ? 4.8.68
°c6 a/ar
in.
3 * IP1"
H.N
757.<?p3<
0.70Z8
Cp =
1 ^ 6:6 /<5"c
£- - /( l+ (&*/&^V _ <W<5L
Rfeso^fs.of sitMiPdr + »-iVJs A^e suwiFwei^rteel
= o_5"&ol
Assumed Ct
0.5
0.535
Slef (p»;")
525
•53 3
S6L1S
JGO
3.11
4.00
L/*l
£3.16
67, M8
6£g(psi^
737.1
£58.8
£88.1
Ccg/gc
0.6X75
0.&SS1
0.6S7S
CKectod Cp
0.6'337
-0.0431
-O.0O13
0.S346 | 5: o [
ms^f-''
d= 4.01
■*i.

PROR1FM 1077 A column of 5.6-m eflfective length must cany a centric load of 2750 IcN.
Knowing that aY = 250 MPa and £ = 200 GPa, use allowable stress design to select
the wide-flange shape of 360-mm nominal depth that should be used.
SOLUTION
A > 41- ' ^O7^ * ...33-oV . ,S330 .^
p ^ 7T^£J
Mil1-
Try W3Cox^|£ A~ 2.7600 ^*,1 = Zl&OQ* lo"4" ^ 0y.
' ■ V 6; 1 XSohio - <25.€6
V* IO|k|o-* <-<^
F. S. = ^+ ^(0.44123.) ^(0.441231* ~ 1.8*1**
3*l£© HJ > ZlSo kW Use \A/34o*2l£

_ 10.78 A column of 4.6-in effective length must carry a centric load of 525 kN.
PROBLEM 10.78 Knowing that <% = 345 MPa and E = 200 GPa, use allowable stress design to select
the wide-flange shape of 200-mm nominal depth that should be used.
SOLUTION
1.12 U1
-**,
i > \.°<ijl.x = (i.^y.gts.to»x».cy = lo.89)(|0-<^ , lo>Mkto. _<
Ti E 1^(200*10*3
F.S. - f + ^(o.gflSMV ^(0.8415-4? = l.lOll
R* = $*A = (Il6.8)«/e>^(^860)./d^ r £g4 ^ > S2S M
Use W 20O x <*6. |

10.79 A column of 22.5-ft effective length must cany a centric load of 288 kips.
PROBLEM 10.79 Using allowable stress design, select the wide-flange shape of 14-in. nominal depth
that should be used. Use ar « 50 ksi and E -• 29 * 106 psi.
SOLUTION
Try W IM* 82 A " SW. I ma , I„.-w = /48 ;„*, ^ -- * HZ ,*«
C^^/1^ - -07.00
Use WH*W
PROBLEM 10.80
SOLUTION
10.80 A column of 17-ft effective length must cany a centric load of 235 kips.
Using allowable stress design, select the wide-flange shape of IO-in. nominal depth
that should be used. Use ^ = 36 ksi and E = 29 x 106 psi.
T*w w lo*S*J Ar '£"-& '** Ijr l°3 tV ^* 2.56 m
k , ^£2- , 79.69 « Cc ±$£ * _Z5i^_ , o.GS|^
P 2.56. Cc 136. lo
RS. = § + -§(°-Gs^c0--£(a.63H<0J = 1.872-1

PROBLEM 10.81
10.81 A centric load P must be supported by the steel bar AB. Using allowable
stress design, detennine the smallest dimension rfof the cross section that can be
used when (a) P= 108 kN, (b) P= 166 kN. Use <7r = 250 MPa and E = 200 GPa.
SOLUTION
1.4 m
r -
=• o. 288675" at
U
((-O P - /OS" |C>3 N A*sW»-e -^ > C
J. - Trtr <* ^
THE"
.-3-
-3
W\
h-i***!
o| r 30.125* io~*m r - 8.6^6 * /o'
U - U4 ICO.11 > I2r.44 *S d * 3o#|
Ck) ?=/££* iO* W Assume y- 7 C*
-3
^ r 33. .543 Wo"'* im
^ -
m
1. 68215"*/0
~*
t-t_
r* - I4Y.5S > I25-.66 ^
A -- 33. S"
t*\l*1

PROBLEM 10.82
10.82 Two _3i x Z i -in. angles are bolted together as shown for use as a
column of 8-ft effective length to carry a centric load of 41 kips. Knowing that the
j _ lt angles available have thicknesses of £ in., | in., and £ in., use allowable stress
h— 3^ in.—* *— 3^ *« -H design to determine the lightest angles that can be used. Use aT = 36 ksi and E = 29
MWf'.-. ^kjawtyrptjffipg
2j in.
C^.yigf rj&g^ = HK.,0
x 106 psi.
SOLUTION
Lfi r T6 - 133 c? > r
P-tf r S*A - (S.SO(«t.M } -- 35. 3 k.'ps < Ml ic<>3 t« n*T use.
T^ L 3i>^i> t i«. A r (z"i(Z7s) * S.So ;«*■
r - 0.70^ ;*
Use L 3j?*fci»4 .*.

PRnm fm n a* 10*83 Two ^ * ^ "in* ^Sl^ are ^Ited together as shown fin- use as a
rKUKULM io.83 column of 6-ft effective length to carry a centric load of 54 kips. Knowing that the
angles available have thicknesses of £ »"-» § in., and £ in., use allowable stress
design to determine the lightest angles that can be used. Use ar = 36 ksi and E = 29
* 106 psi.
SOLUTION
Y » A "-/ 4.2
2|- - 0.<m*
KS. - ■§■ + f (0.SI63S)-:£(o.S1633)* - i.a«8
Use L3jai>i m -*

PROBLEM 10.84
10.84 A square structural tube having the cross section shown is used as a column
of 3.1-m effective length to carry a centric load of 129 kN. Knowing that the tubes
available for use are made with wall thicknesses of 3.2 mm, 4.8 mm, 6.4 mm, and
7.9 mm, use allowable stress design to determine the lightest tube that can be used.
Use Oi- = 250 MPa and E = 200 GPa.
76.2 mm SOLUTION
[—76.2 mm—|
bto ~ 76.2. t*\»n b -
P - \w icy
b--*t
A ■ b.a - t >
U - 3. | w, P - \7* JO/ b^" 76.2 m-
T»w tJ 4.8^
b^ r 76.2. - ?.6 - GG.C *n*»
-*
I* -j)i\(76.Z^-(66.6)"*]*- I- I70o^»/£3C^MH * J. l706$"*/O-*m»
U _ 3.1
'-^ = ^..ai*to"*
ho
= 106. II < Cc
T*^ -- 0.84¥43
r-s- r 3 + f (o.8<W3)- £(o.8**m)s = l.<?o8*
- IIS. 6 kW « 13<? Jr^ Do »o+ use.
I^j t'CH^w*
bi = 76.3 - I2.S = 63. M mm
A - (76.X)i-(63.^)z - l.78£g8*/o3 m^1". /.7SGga»/d5v**
1= ia[(7Li^-^3-H1N1 = I.4C3IG* lo« «•*** = |. HC3J6 x /O"4 m"
I
Le/r .
Ce. "
r*
U ,
i
2.8.6iS"x/o"5 **
0.862*2
^ 28.6lS>lo-
F.S. = -f + |(0.«62l^-4(o.86XWy = I.IO'H
Rie r6Li<A r (M.2S"j«toe)Cl.78688Mfc>-*^ IH7.Oj.Io5 N
= 147.0 kKJ > |2^ kit
Use £ - £.4 m*^ -■

PROBLEM 10.85
|-«— 127 mm -~*\
t = 8 mm
178 mm
'10.85 A rectangular tube having the cross section shown is used as a column of
4.5-m effective length. Knowing that oY = 250 MPa and E = 200 GPa, use load and
resistance fector design to determine the largest centric live load that can be applied
if the centric dead load is 140 kN. Use a dead load factor yD= 1.2, alive load fector
yL = 1.6 and the resistance fector <p = 0.85.
SOLUTION
h0 -127^ be-= '78 *** K:= K-Zt '
= 4C2H ww1, T 4C:H WO'* ^'
V1 Sto.77i"»lO-3
gg.G3
V - 1- /IT - gg-^s f
7.00 * 10■*
0.<?<?74 * /.$"
X
2 _
o.WtS
V
0.1HJ
- 767,3 IcW
YpP» + YLR. - cpP^

PROBLEM 10.86
10.86 A column with a 19.5-ft eflfective length supports a centric load, with ratio of
dead to live load equal to 1.35. The dead load fector is Yd = 1.2, the live load fector
Yl = 1.6, and the resistance fector <p = 0.85. Use load and resistance fector design to
determine the allowable centric dead and live loads if the column is made of the
following rolled-steel shape: (a) Wl 0 x 39, (b) W 14 x 68. Use E = 29 x 106 psi and
0r=5Oksi.
SOLUTION
(a.) W IOxS^ A= I).■& in1
yj = \.w ;»
Ws =■ ws.\2
U/r /g
(b") Ia/ 14*6* A^ 20.0 ;*' r^s.fc m
Nc2 - 1.5206
Pfe * 93.7 kips
PL - S4_6 icrps
a?<3oo
l.«6i
Pu * A (0.&S8r 6"r - (tooKo.^ysV (s<^ - sfT/6 kip*
1> ' D
Pp* 183.? krp»
(1.2 )(l.3£ Pl) + |.C Pc <CXgS-)(5-|0 PL- 134.2 kips

PROBLEM 10.87
"10.87 The structural tube having the cross section shown is used as a column of 15-
ft effective length to carry a centric dead load of 51 kips and a centric live load of 58
kips. Knowing that the tubes available for use are made with wall thicknesses in
increments of rj in. from £ in. to |> in., use load and resistance factor design
to determine the lightest tube that can be used. Use oY = 36 ksi and E = 29 * 106 psi.
The dead load fiictor YD = 1.2, the live load factor yL » 1.6 and the resistance fiictor <p
- 0.85.
[> 6 in. *|
SOLUTION
Le = 15 -Pf - ISO in
fc,„.w P„ r £?p-*** = 0^X5.). 0.0(58) s l9Lt k.
9
o.ss
he ~ ha-Z± - S.S /h.
5
r 2. S^& \n
r^ JX . JK
v _ u^ / Cc 7&.6I rr&
ks - I go
•= 7C.6I
21COO
= O. S59/6 ■* 1.5 X/=0.738|5
!>*.
<3.-7WS"
P, - A (o.45g) 6V-(5.75)(o.658V (36^- l52.o Jc.ps </*/. % k:r
"1
*■ 0.2^6 .'-
- ^£
Try t - H .'* - 0.31X5 .>., b,- r 5'.37S
A * 7. lo^i^ J- 38.«H •»*, \r* 2.3*^ ,'n v"77-4'
X, -
2^L/
-3ft.
^ooo " O.SC8U * /-5
Xe* =~ C753GI
Po - (l.\o<fH)(o.£5S) ' (SO - 186.7 Ic.'ps > \%).2 kfS

'10.88 A column of 5.5-m effective length must carry a centric dead load of 310 kN
PROBLEM 10.88 and a centric live load of 375 IcN. Knowing that oj. = 250 MPa and E = 200 GPa, use
load and resistance factor design to select the wide-flange shape of 310-mm nominal
depth that should be used. The dead load factor yD = 12, the live load factor yL =
1.6 and the resistance factor <P =* 0.85.
SOLUTION
6^ ISO* 10*
"TVw W 3lo*Go A^ UTto***' r 7S*9Ok/0
^
-4 a.
.-3
Iy* 19.3 */o6 w^fi; ^ =■ iiiMT'rt.iK/o ^
c TTT TIE TKmfWiy*) |[ 2oox|«^
Xe* = 1-58^2
Too A4 lit* Do no+ osc .
-t t
TV* W3lo * 74 A - ^H8o *J- = 9wgo */o h
Po = (i?^80K/o'6)(;o.eS-8y*Sr5"lu(2S-ox/o<r^ = l232*/OsN
= 1232 kM > M43 Wv
Use W3lOv74

PROBLFM in 89 10.89 A sawn lumbar column with a 125-mm-square cross section and a 3.6-m
effective length is made of a grade of wood that has an adjusted allowable stress for
compression parallel to the grain uc = 9.2 MPa and a modulus of elasticity E = 12
GPa. Using the allowable-stress method, determine the maximum load P that can be
safely supported with an eccentricity of 50 mm.
SOLUTION
cf - \7S iw ■? O. \2$ * A ~- J* =■ \S.Q3S*tO'M v^ k * Jztz. * Z8. S
S;e = ^ '^ffigr0 • ** MP. «./« « 0..7.77
C I + OWS > _ //-( + <W& f- «S*AS ' = O.M1347

PROBLEM 10.90
10.89 A sawn lumber column with a 125-mm-square cross section and a 3.6-m
effective length is made of a grade of wood that has an adjusted allowable stress for
compression parallel to the grain ac = 9.2 MPa and a modulus of elasticity E = 12
GPa. Using the allowable-stress method, determine the maximum load P that can be
safely supported with an eccentricity of 50 mm.
10.90 Solve Prob. 10.89 using the interaction method and an allowable stress in
bending of 12.8 MPa.
A - dx = is.czsxio** *S
- n. s
SOLUTION
<SC - ^. 2 MP<* E = \QOOO MP<^ S«vw h* W- t^.8^ K* = O.Soo
K*£ - (°-3ooY.*,w^ .. -.. .^ ^^ =aq7l77
>efi
0-/<O
-, = ^ffff0^ - *M Mf*
C v V (W&) --//w<S£g/6;.\a _ <w& ^ o.Hi3*n
■J
ts.
Pec
i
ec
T
A6^fjC. XGJte
Yo.Q5a^aQ€<r)
34. 7*|0* M * 3*+. 7 kN/ -*

PROBLEM 10.91
1.0 in,
10.91 An eccentric load is applied at a point 1 in. from the geometric axis of a 2.2-
in.-diameter rod made of a steel for which ar = 36 ksi and E = 29 * 106 psi. Using
the allowable-stress method, determine the allowable load P.
2.2-in, diameter
SOLUTION
I ^ It* * I. |H*W \»*
48 in,
A - ITC* - 3.8o|3 ml
(29000)
36
t- }2C,\o > le/r
-I
^-('■••^[tos *^^r -
//.<*/ I^i'ps
PROBLEM 10.92
2.2-m. diameter
10.91 An eccentric load is applied at a point 1 in. from the geometric axis of a 2.2-
in.-diameter rod made of a steel for which o^ = 36 ksi and £ = 29 * 10* psi. Using
the allowable-stress method, determine the allowable load P.
10.92 Solve Prob. 10.91, assuming that the load is applied at a point 1.6 in. from
the geometric axis and that the effective length is 33 in.
SOLUTION
C = i<"l* I.I ivi. A = TIC2 r 3.SOI3 mX
I =- ^C1 - K WH Vh4 V = -fifc s O.SSO m
Le * 33 (V. L«/r» 33/0.SS-O ' GO
36
24,-/0
RS. - £ + J-(o.47S"8 )-i(o.4/5^)3 - l-83/e
9

PROBLEM 10.93
15 mm
152 mm
■ 152 mm ■
10.93 A column of 5.5-m effective length is made of the aluminum alloy 2014-T6
for which the allowable stress in bending is 220 MPa. Using the interaction method,
determine the allowable load P, knowing that when the eccentricity is (a) e = 0, (b) e
= 40 mm.
SOLUTION
\>. - IS* *~ k,= \>»-Zt -- 122 *,~
A r t>/-t>/ - S220 ^«1 r %Z20*Loc' **!*■
I r i(bb*-b;,,>) * Z$>.0Z*\O'
-2
372 »IQS
(<?7.7& )*-
V T Ji - S6.26 *.*, - -5~4. 24 > Jo"* tw
P
Vec

PROBLEM 10.94
15 mm
1-
132 mm
k
152 mm -
10.93 A column of 5.5-m effective length is made of the aluminum alloy 2014-T6
for which the allowable stress in bending is 220 MPa. Using the interaction method,
determine the allowable load P, knowing that when the eccentricity is (a) e = 0, (b) e
= 40 mm.
10.94 Solve Prob. 10.93, assuming that the effective length ofacolumn is 3.0 m.
SOLUTION
-1 a.
A - b^-br* ~ *M° ***** " fi^oov/o4^.'
4 *
1 A
L - 3-°
- sz.n * ss
(a> e * o
P- A6U* T {B^o^to6)(\?i.s^io') - 1048 wo7 M . /ova vw
(bl e - Ho •*« r Ho*(a*m
C = Ci^lS^ } * 76 ***" -- 76 > /o~3 tv,
P T £73 x /O* N * 673 kK/

PROBLEM 10.95
|P = .10.8 kips
2-in. diameter
10.95 An eccentric load P = 10.8 kips is applied at a point 0.8 in. from the
geometric axis of a 2-in.-diameter rod made of the aluminum alloy 6061-T6. Using
the interaction method and an allowable stress in bending of 21 ksi, determine the
largest allowable effective length L that can be used
SOLUTION
C - ^ c* - \.0 i*
Ao.
?ec
JL_ r | - Pec
PROBLEM 10.96
P = 10.8 Idps
2-in, diameter
10.95 An eccentric load P = 10.8 kips is applied at a point 0.8 in. from the
geometric axis of a 2-in.-diameter rod made of the aluminum alloy 6061-T6. Using
the interaction method and an allowable stress in bending of 21 ksi, determine the
largest allowable effective length I that can be used.
10.96 Solve Prob. 10.95, assuming that the aluminum alloy used is 2014-16 and
that the allowable stress in bending is 26 ksi.
SOLUTION
c * i el * I. o m. A - ire1- = 3. hil .V
e - o.z tw
Pec _
A 6^t J €^f t
Assume iVr > 55"
_ /
6"**t r
(L/r )«■

PROBLEM 10.97
P-lOSkN
25 mm
10.97 A rectangular column is made of sawn lumber that has an adjusted allowable
stress for compression parallel to the grain ac = 8.3 MPa and a modulus of elasticity
E = 11.1 GPa. Using the allowable-stress method, determine the largest allowable
effective length I that can used.
SOLUTION
180 mm
<S - 180 ^^ " O.lgO w t>= ZHO »»n - 0.240 vw,
A = V»JL - H3.2*|0-Sv^ E« "loo MPa
a.
C = tt - O.l^ow
t\ in
- ^ to,
Cp'**5*?4- o-"«6- ,
e^ zs wxk - o.ozs v^
.2*/*-* Z07.36X/0-' ^.^t^xiU ft*
s
. KceF
La-
KME<*1
6ce
wW^ Kifi r G.30O
kWF
«
= o,
.» /(°-3<
o^X II l°° ")
ccea
4. 81
*v\

PROBLEM 10.98
— 1ULI MS
10.97 A rectangular column is made of sawn lumber that has an adjusted allowable
stress for compression parallel to the grain ac = 8.3 MPa and a modulus of elasticity
£=11.1 GPa. Using the allowable-stress method, determine the largest allowable
effective length L that can used.
10.98 Solve Prob. 10.97, assuming that P = 85 fcN.
SOLUTION
dl * ISO ►*», - O. ISO v*
A = kef » tz.zxicr* v^~
A.
£= HlOO MPa.
= 207. Z&xlO'*- ^
ia
e ^ 25^ ^ 0.02S"
KM
c=^s o. 120 *n
MR
'a.
CP * f* r ^C^ = a 38^2 r J U+ * r S*^
•y - AjUL "Vf '**)*- ~ W^*^ C - O. g £>* saw* Am*U^
6" - JidL
r t*-
6*cfi
wWe ^ - 0.3O0
V- A^± - (o^^^XflLoo^ = s_n
*n

PROBLEM 10.99
4 in.
[■<—4 hi,—H
14 ft
10.99 A column of 14-ft effective length consists of a section of steel tubing having
the cross section shown. Using the allowable-stress method, determine the maximum
allowable eccentricity e if (a) P =* 55 kips, (b) P=3S kips. Use ox - 36 ksi and E =
29 x 106psi.
SOLUTION
A * bs.*- b> S.H375V I- ik(t>;~ t/)' W.osc ;«
r -
l£ - I; 4873 i«
A J ^ I ^ A
00 Pat * 35" fc.p*
«'cfe^-.^

PROBLEM 10.100
4 in.
10.99 A column of 14-ft effective length consists of a section of steel tubmg having
the cross section shown. Using the allowable-stress method, determine the maximum
allowable eccentricity e if (a) P = 55 kips, (b) P=35 kips. Use or = 36 ksi and E =
29xl06psi.
10.100 Solve Prob. 11.99, assuming that the effective length of the column is
increased to 18 ft and that (a) P = 28 kips, (b) P=IS kips.
SOLUTION
S+eei : 6Y - 3£ ksi fc - WOOO fcs."
_L b0 - <f.o i* btf* W-2t - s.*r.\
c - J?.o i*».
ff - _Jll£_ , JllCai£^X_ . 7.073G Us."
P.* fi*ec ^ P^ec «-.. P<^
- S*
X ' G*" A
A 1
(CO T>* * *2 H>*
(2»(28^ *- 51WS7SJ
^^--^)
ftO P^' IS fc
e -
?*
I?. Q3C
z*Y.\v)
b
0726-
1*
S.M37S"
1 »
I.25S in,

PROBLEM 10.101
p= 170 kN
10.101 The compression member AB is made of a steel for which ar= 250 MPa and
E = 200 GPa. It is free at its top A and fixed at its base B. Using the allowable-
stress method, determine the largest allowable eccentricity ex, knowing that (a) e =
0, (b) ey = S mm.
SOLUTION
SUcJf : S, r 7.SO MPo.
E - Zooooa MPc
A - (7£*lo%K$0*(ds) - 375© k/o"- k,1
T - -L
m - ^
U AU _ 76.31 ^ ,_ _
Cc " 12S.C6 T 0.fcOG5
--A"*] * S,[f---|]
f{ ^ " A " 17 J - <* L P - A
SJ =
_ 18I.2S*IQ
2S* /o-a
X
* ^ 37.5 -JO
P ? !7C3x)os N
-<i
- 31.25 x 10
-c
*n
\.lS?B»lo
.-&
- 46,87b~x|o'c v«*
(<x) eY- o
3/. 25 */o
-i
|o<?.S2*|0* |
- M.74 x|D
-s
170*10*
l*i
(bl
e^ * Sx/o'1*! e*- 3i.z5^/o6[-j^
S7fOy(o-4
M.7C ^^
32*10*
- O
t
]
8»icr
!Oy}o^
-3
6. 43 x lo~* ki
3750X/O-4
6. V3 miw
tK.875"* lo"
0

PROBLEM 10.102
P = no kN
50 mi
10.102 The compression member AB is made of a steel for which or = 250 MPa and
E = 200 GPa. It is free at its top A and fixed at its base B. Using the interaction
method with an allowable bending stress equal to 120 MPa and knowing that the
eccentricities ex and ey are equal, determine the largest allowable common value.
SOLUTION
Sfeei: $r = Z5Q MPa. E = ZOQOOO M?a
At (7S>'|0'^(:i''3x|0'1")r 3750 itlO* v^
vy - -/5 .- 1 h. 434" 10"* v*
I„r ]a(5dK/o"sX75>«/o*)- 1151**10* *»*
\r„-
-IX
- 21. C5I x /0"fc ""»
Le r ZL =Wo.5y)r|.|^ U/C, - l-to/U-lS-Mo-'* 76.21 < Cc
Lt/^v. rJML r 0 6065 FS.t f + -Kp-tfCA-J-ifec^S-y -'-8662
Cc 125". 66 * ° B
+ -pg»y
*&?
= /
w
.-H
S7.5"x/o
r*
l7Q»fPS
\aoy|ot V l-7S7fc*/o"
— +
781.25 * lO"*1 /
e ■=
|7C?lx(0'
(375"o >fcft >(lo^.3? * /o*
15. ^^6 e ■ I " 0.^6*
e - 7. 75" *{&'* to - 7* "75" m»*i

PROBLEM 10.103
9.2 kips
10.103 A sawn lumber column of rectangular cross section has a 7.2-ft effective
length and supports a 9.2 kip load as shown. The sizes available for use have b equal
to 3.5 in., 5.5 in., 7.5 in. and 9.5 in. The grade of wood has an adjusted allowable
stress for compression parallel to the grain ac = II80 psi and E = 1.2 * 106 psi. Use
the allowable-stress method to determine the lightest section that can be used.
SOLUTION
Sct^^ Aow\)oe^' Sc - /ISO psi"
C = o.s
Le- 7.% -rM r *£.*! ,«
A X*
<;e r 0.300
TU*
§Uf
_l . ec
e = 1.6
Irt
Ci = i (7-5) ■ 3.75" ,V A= 7.5- t
Ix = 7a k(7.S-y =■ ST-rS-t b
' 4 P-OfVO
7.5X *r. t?6 fe
= 3.2895 b
P.tf * 3-28^5 b6>
Gut -- ^Cp r 11 so CP
Pa* r (3.38^ b(ll«pC^* 38S2 b Cp (A.)
CJ^-M« P* -rV */f JW y/a^cs «* *».. See. -Kkle V*JLj.
w
Cia)
a..s
* s.s
7.5
^-5
a
go
3.5
S,5
7-T
7.S
6^/s
O.S007
1.3***
a.s^
XAII
CP
O.MSMt
0.7538
0.%%ll
0.%%%l
Vj>
(A)
5*^00
l6 2oo
ZStoo
378oo
Use
lo ~ s51S* »'«.

PROBLEM 10.104
9.2 kip
e = 1.6 in.
I0J03 A sawn lumber column of rectangnlar cross section has a 7.2-ft effective
length and supports a 9.2 kip load as shown. The sizes available for use have b equal
to 3.5 ia, 5.5 in., 7-5 in. and 9.5 in. The grade of wood has an adjusted allowable
stress for compression parallel to the grain «7C=1180 psi and £=* 1.2 * lO'psi. Use
the allowable-stress method to determine the lightest section that can be used.
10.104 Solve Prob. 10.103, assumingthat e = 3.2 in.
SOLUTION
c = o.g
Lt- 7. X -ft - BC.H ;„.
£ = 3-2. i« c =■ i(7.s) = 3.7* ;M
TU-
6u*
1 4 ec
A
0.x
A - 7-5 L>
A *
= 2.io^l b
P*« - 2.1067 b (51^
" ^-r~ t (3.2X3^
cl ~ 7.S" iw. o<- b^ w4ire>ket/ev is swAii«/
c* CL/«0* L^ (SCO* P
1 + £e/SL
Cp^
ac
5L ^IJIiSK^TW/K
P«lf r (^-^67^1 t (1180 C^ * Z4*& b Cp
CaJrcJ^ATe l^tf -tw o^// 4i>o«r VaA*s d? b . See *fo-l».l« bcjtouj.
P- ^ooit.
Use b ~ 5LS" i*.
b
cv'i
3-S
5". 5
7.S
9.S
d
C.v>)
3.s
£5
7.5
7.S
6ce/&
o.Sool
U2363
3.W
£;W?
Cp
O.HSHI
0.7S8S
o.saa*
0.838^
R*
W
37 80
\63lo
16540
ZOI00

PROBLEM 10.105
P = 32kN
1.2 m
10.105 A 32-kN vertical load P is applied at the midpoint of one edge of the square
cross section of the aluminum compression member AB that is free at its top A and
fixed at its base B. Knowing that the alloy used is 6061-T6, use the allowable-stress
method to determine the smallest allowable dimension d.
SOLUTION
A - «T
i--kr
'"/f 'g-»
C±J
e» W
P A ££c _ _p_. r$*Xte) . MP _ ^
Assume L/> > <SC ^" - ^
Ll " T5TT " Jv
13 L1
MS Pi
8
8
(tnSwW*
H.3 > C£
PROBLEM 10.106
P = 32 "kN
1.2 m
10.105 A 32-kN vertical load P is applied at the midpoint of one edge of the square
cross section of the aluminum compression member AB that is free at its top A and
fixed at its hase B. Knowing that die alloy used is 6061-T6, use the allowable-stress
method to determine the smallest allowable dimension d.
10.106 SolveProb. 10.105, assuming that the vertical loadP is applied at a corner
of the square cross section of the compression member AB.
SOLUTION
A T r J* a -is iJ4
L^ * X L * 2. H w>
If «C.
Asso
w,e
l_/r > GG
>-« =
CU/r)1
i->
6 - 3$I*to* ft
r -
B

PROBLEM 10.107
18 mm
10.107 A compression member made of steel has a 720-mm effective length and
must support the 198-kN load P as shown. For the material used aT = 250 MPa and
E = 200 GPa. Using the interaction method with an allowable bending stress equal
to 150 MPa, determine the smallest dimension d of the cross section that can be
used.
SOLUTION
Iv * £ (Ho«/or* yd -■ S. 3333* Jo"' d
$\-ee$: S^ - ^5"o MP* £ = 26oooo MP*
/Ixl - ^
e*| - fS who * igx/o"' vm
1 "e*
c.'7
£TTz6?£>OOOe>)
p . /IS
Z5o
- iZS'.QQ
Assume
= es.ss <" Cc
e^„.w- ^:[i-i(t^!:Y]-T^r7[i-iCaf(iMiy] - mi.simr
A61jr,£*4r.'«. Ij Sjl^^l;*.
(Hov[cr*yXMVS/ */»*) (3.33Vi*/D-Ma,j(ISO*/0')
Ml. 499 v lo
-3
-s
T* =0
<J* - HI. 181*10' J - 3.SCHOX10'
©I - 23.1 wh*i

PROBLEM 10.108
13 mm
10.107 A compression member made of steel has a 720-mm effective length and
must support the 198-kN load P as shown. For the msterial used a? - 250 MPa and
E = 200 GPa. Using the interaction method with an allowable bending stress equal
to 150 MPa, determine the smallest dimension d of the cross section that can be
used.
10.108 SolveProb. 10.107, assuming that the efTective lengthis 1.62 mand that the
magnitude P of the eccentric load is 128 kN.
SOLUTION
A = HoxlD'2, £ U - I.G2. wi
I„ - 7k(lrO»./o,1)afll =■ 5.3333*/o6 J
Ij - "A C^lo'*)*** - 3-3333 */o_* J*
S+ee^ : Sr r ;?SO MP* £ - 200OOO MP*
Asa^wne d > ^Owm- Hoyfd*^
T%
ew
Pe.x
I
(Ho*(o'* JXSMU, *toO (3 ^S^cs'J^Osox/o*)
a
2.ZOH *lo
-5
=: I
r O
-s
87.6 x/cf* iv, > ^ok/o"3 W!
J = 87. £ m**

PROBLEM 10.109
diameter d
1.5 m
?L:
ABv^
10.109 The eccentric load P has a magnitude of 85 kN and is applied at a point
located at a distance e - 30 mm from the geometric axis of a rod made of the
aluminum alloy 6016-T6. Use the interaction method with a 140-MPa allowable
stress in bending to determine the smallest diameter d that can be used.
SOLUTION
Assi/we L/V > £6
_ J
4
/r - s
B-35-/^/0"* -Pa.
t - JLcv = JL-tn
T ~ 7 A u
-Z.
A6>y
Fed
it Sj(, u-*:^
■V
£££.
■4*-- *§*,**a*j
= I
-s
TO. O X/O
I7.S6> |d**
►m
I. ff
17. <T*/o
-3
L4 x» £
d r 70.0 wtv. —
8S\7 > C£
-I

PROBLEM 10.110
diameter d
10.109 The eccentric load P has a magnitude of 85 kN and is applied at a point
located at a distance e = 30 mm from the geometric axis of a rod made of the
aluminum alloy 6016-T6. Use the interaction method with a 140-MPa allowable
stress in bending to determine the smallest diameter d that can be used.
10.110 Solve Prob. 10.109, using the allowable-stress method and assuming that the
aluminum alloy used is 2014-T6.
SOLUTION
1.5 m
Assume L/v > SS
c- A -ire1 rfd1" I* $** = £<>-
Br372x/o" Bt
i
Pec _
- <s
AS?* TBr*
Br*
€9 Pla + 32. PL:
ird'B
TrcPB
U**.J
7T (372x|0«) 2">T (373 WO*)
-a
-«• ^S _
/0.473 */£> XT 4- 2.5XSC* to
H r *0. |x/o"*

PROBLEM 10.111
10.111 A steel compression member of 5.8-m effective length is to support a 296-
kN eccentric load P. Using the interaction method, select the wide-flange shape of
200-mm nominal depth that should be used. Use E = 200 GPa, ar = 250 MPa and
oa\ - 150 MPa in bending.
SOLUTION
S\eeJ: E « ZOOOOQ Kip*, $Ty = 2$o Y\?o,
o
r„*l gg^ra^/oV^ V ^ |?r |05mM ' /OS *fO~3^
^-^:I»-i(^V]-^f['-K^r]^ 7
o MPa,
P
, Pe,.y
A 6^ t««+''',;* -L* ^«) t*-vJ.'*i
1^ *£*,*.**.
2^6 Ho* _(a^yios)(ia!D»to^)(i&r»ter')
*. ->" ' #v ' ' w " ' - 7573 v/o *i
- 7573 mm
Try W 200*5=* Ar 7SCOx/c»"fc ^ V= l<>£'*lcf*w , Tx - £i* I * W" m*
fj r Sl.qv |o-8Mj U/ty - MI.7S <" Cc
t ,py
- 0.^54 + 0.423^ = 0.?I93 -< I (eJio*«JO
—£ 4 ■■■■ ^-^ = I.OM7 > | (»»t*J!WO
Use W20O*S?

PROBLEM 10.112
e, = 70 mm
10.112 A steel column of 7.2-m effective length is to support an 83-fcN eccentric
load P at a point D located on the jc axis as shown. Using the allowable-stress
method, select the wide-flange shape of 250-mm nominal depth that should be used.
Use£ = 200GPa, ^ = 250MPa.
SOLUTION
U * 7. ^ iv,
"br -~ , *■ I4C.34 > Cc
l.*»2 (.L/tj)x ' 0-^* KlH6. 34 ■)*■
P x Pe*c
7T + TT-
83 * to* (g3*fo*X7o>nd*)Ool»<o'>)
rt * IS". I * /o~c
B
^ \%.2Z*toQ + SS.Sfiv/O'" * 5*JV MP* > **.&/ MP*
e^ire^ 4^«4
T*w W 2SO x SI
2.1.
So.%*id
T? - H3.U
(ff^)(<S^o ««' ) - G7&& *^
£- + Pe.C - g3*|ofr (g3»c^o^7o^to''>)(lQl^S^y/Q",^
A J, * 1H10»\0-* 18.8 x/o-c
^ M. H * |o' t -31.37 x toc - 4?.54 MP«, * SoAlYiP*.

PROBLEM 10.113
2.1 in.
10.113 A steel column of 21-ft effective length must cany a load of 82 kips with an
eccentricity of 2.1 in. as shown. Using the interaction method, select the wide-flange
shape of 12-in. nominal depth that should be used. Use E = 29 * 106 psi, or = 36 ksi.
and uM = 22 ksi in bending.
SOLUTION
Pec
r I. *J 21 + D. -.17 2. - I, G o I (nof ajV«^4 )
TVs/ W 17* So
■ 4 J^
g2!
to)C.?.iKfa«)
a er>,^ ' i*6>.^ "0i-7)w-of.-> ■ o^K^o
0.6/8 + 0. Ill » 0.73? LJh^)
TVs W 13 x 4o
^ = i. qs iia
N " J.33
l3o.5"7 > Ct
^ifi^^i- -: 8.7^ kit
6»#>"**•**■ S0.*»O(l7/i)») ^ 0.<U')03o.S7)i-
P x Pec . _82 + (gSHsu)^-'1.3*0
Use W/U"^

PROBLEM 10.114
Tr
T^y
10J14 A 43-kip axial load P is applied to the rolled-steel column BC at a point on
the x axis at a distance e « 2.5 in. from the geometric axis of the column. Using the
allowable-stress method, select the wide-flange shape of 8-in. nominal depth that
should be used Use £ = 29* It^psi. and £V = 36ksi.
SOLUTION
SieeJt: E - Ztooo ksi SY - 3£ *».-
U/r,
-*■ -■ 0.754
(not *ijUl"0
r^ - ?.oa
F.S. = L89S
>** =
i3.6S Uv'
J
F.S. - iS9$ &at - \3.7l ks/
ua> .
0.7^6
11.7
49.
!?-5/ K < I3.7IK. («i/o^
Use VJS*HO

PROBLEM 10.115
e — 20 mm ~*\
10.115 A steel tube of 80-mm outer diameter is to cany a 93-kN load P with an
eccentricity of 20 mm. The tubes available for use are made with wall thicknesses in
increments of 3 mm from 6 mm to 15 mm. Using the allowable-stress method,
determine the lightest tube that can be used. Assume E = 200 GPa, ox = 250 MPa.
SOLUTION
f„* iA.* 4o^ , r^ r0- t
80-mm outer
diameter
- ) — - o .—, . ^
1
vvw
3
6
9
12
IS
^
*^«
37
3^
31
2S
2 5
A
a.
72£
IS^S
7.0O1
256 4
%oCh
I
/04ww"
0.539
O. 94 1
i. ns
1.52 8
1.7&4
r
nn»Vl
27.2*
2<£.;?S
25.31
24. HI
2 3.59
U - 2.2 »
5W: ET**ooaoo MP* Cc r^X , ^*£°«>) = (2i:c£
U 2.2 _ _ U/r
-fry *-9-~ y ^f^ * "•*** C.
F.S. - l + lMnl-j^n)1 = I.SS5-
C,
r 0. 6 «l 11
£. + Pec r Wkiq* +il3^^(s^£lX^io22 , , „ a MPa >
^0O7>-|O
.ZtSXlGT'
MP,
F.S. - I. 8q©
0.717?
6"** - S8.3 MP*.
A J 25fc«Mor«- 1.51* */0"* 3 r,m
Use £ r 12 >»im

PROBLEM 10.116
e = 20 mm -*|
10.115 A steel tube of 80-mm outer diameter is to carry a 93-kN load P with an
eccentricity of 20 mm. The tubes available for use are made with wall thicknesses in
increments of 3 mm from 6 mm to 15 mm. Using the allowable-stress method,
determine the lightest tube that can be used. Assume E = 200 GPa, or = 250 MPa.
10.116 Solve Prob. 10.115, using the interaction method with P= t65 kN, e = 15
mm, and an allowable stress in bending of 150 MPa.
2.2 m
80-mm outer
diameter
SOLUTION
v; = iJo - 4o *m
w - r»- £
A - TrCr.'-r/') i^flO^-iV) r--if
t
rnhi
3
£
<?
«?
>S
**
KAM
37
34
31
29
2S
A
724
I3<tf
?ot>n
2S61
3,0 63
X
I6***i"
O.S3<>
£>.*6t
1.385
1.52?
l.lo*
V
Wi
X1.7*
It.??
25.31
2HMI
23.5<>
P r |££»/c>S A/
Sfeei: f. 2O000O MP. ^-/^ '/^JS^5 ' »**«
L«
£2
^ * 0.£*j7
r r zssi-io-' ' 8C-9X< c<-
F.5. - f + f(o.6*l7 ")-"£(<X£*i7 )3 = 1.885
^c^, ^ §.[i-K^Tj - ^[l-t(o.e*7)M = 100.1 mp<
'ec
l£S"*lo*
(Kg-y 10^ )^S MO"^)(40,/tfs)
= 0.8/5 + o. Si*t « 1.3** > / (M.t-^cw&O
A - 0-32*)(2oo7) * 2 647 ^Ml*
po x .„ U_ 2.2 _ _ ... .-. ^ /i U/r
-OlT
t*J2
Cc
- 0.7172
F.5 * l-81o
Pec
^c4r.'t * W. 3 MPa
A«kc*A* I^U-fc-v = (2SM*|cr^(9».S«\o*)
' = O.C^T + 0.432 = 1.0S7 > I 6**t -JiL/ftO
1r-a ^t )S" iw*^
F.S. * 1.8*1
? . Pec
k - J2J:
as.S*y|o-' * q5-*6 * C= (L«/r)/Oc = 0.74*2
€>, c*^.-* = *r.e^ MPa
= 0.S63 + 0.38? = O.<iS0 *. ) (^o>^)
Use £ = '5 fcM»*i

6.4 mm
PROBLEM 10.117
y
89 mm
64 mm ,
10.117 A column of 3.5-m effective length is made by welding together two 89 * 64
x 6.4-mm angles as shown. Using E = 200 GPa, determine the allowable centric
load if a factor of safety of 2.8 is required.
SOLUTION
Ohe aw^A
X = tS. 8
H»h
- 0.333*/o' + (^SsX^.M)1
= 0.686 "/o3 wiWfc
T^« a.n-^es Iy - C*Xo.68C*JO*) - \.%7Zx\oe **«** - l.3?Z»/o'c r*
ut
F.l
(F.v> U*
(Z*tf3.sy

PROBLEM 10.118
6.8 IcN
10.118 Member ^fl consistsof asingle C130 x 10.4 steel channel of length 2.5 m.
Knowing that the pins at A and B pass through the centroid of the cross section of the
channel, detennine the factor of safety for the load shown with respect to buckling in
the plane of the figure when 6 = 30*. Use Euler's formula with E = 200 GPa.
SOLUTION
IF, - O
+\T^o
AC
^ G.SkW
cos K**
- "FAt cos IS"* -v r^0 cos 3o" = O
a
Cos /S"
tr - F*tt cos 3o*
K„ =• C?S\378 JrN>
*s
-R
cr
Up1
72.32H xlo* rV= 7?."32H kW
AB
'Aft
PROBLEM 10.119
10.119 Supports A and B of the pin-ended column shown are at a fixed distance L
from each other. Knowing that at a temperature TQ the force in the column is zero
and that buckling occurs when the temperature is Tt = TQ + A7\ express AT in terms
of b, L. and the coefficient of thermal expansion a.
SOLUTION
LeT P tie "fli* Compressive -Po^ce i«o "He CoJ'u^i
Pc. - El^L - P ^ £Aoi(An
AT - TT*EI - 774£-i,Vi2

15-mm diameter
0,5 m
10.120 Knowing that a factor of safety of 2.6 is required, determine the largest load
P that can be applied to the structure shown. Use E = 200 GPa and consider only
buckling in the plane of the structure.
SOLUTION
BC: LK -■ t/i* + O.S *" ^ i.ii-Bo ^
- 7. 85"4 * JO-* **"*
F.s.
2.6
AS.
Uo T-/o-5*+ O.S* " 0.7o7ll »
*«,*#' "c
Pc+ _ ^.8IQg
p*.
Ot»i^iT O
"as
Be
J?. 6
+ f Z"FrO
3.77 3 Wtf
O.S
R. -
l.o
F.
o
Fee * O.79 0S7 FAa
F,„ + -^4— FBt - P > O
-5- r ar
0.7©7/l *° I. K So
O.707/1 F^4(o»*m:iiKd. 7^f7B»0 ^ - P * o
P r 1.06066 FA»
•* (iOCOCC ^(3.713 1 * q.OO IfpJ
"P < 7.3*7* ^c^- - 0.3^*4 V-tk77o ")* 6.**0 fcN
P^ * ^.00 i^w

PROBLEM 10.121
15 in.
10.121 The steel rod BC is attached to the rigid bar AB and to the fixed support at
C. Knowing that G = 11.2 * 106 psi, determine the diameter of rod BC for which the
critical load Pa of the system is 80 lb.
SOLUTION
G -- ll.Xx/o' p*<
j> ^^
%— - a v a/
4 _ TTOL*
32
T - Pi s."«<p » O
Kcp - P-P s^cp r ©
P =
KJL
c/-
jL
K - W7» W1 r FUA - (goals') - uoo
O. 32^ .'ki.

PROBLEM 10.122
e - 8xJo"3 wi
W200X.H '3
10.122 An axial load P of magnitude 560 kN is applied at a point on the x axis at a
distance e-8imm from the geometric axis of the W 200 * 46.1 rolled-steel column
BC. Using E = 200 -GPa, determine (a) the horizontal deflection of end C, (b) the
maximum stress in the column.
SOLUTION
Lc ■= XL - UK*.*,-) - 4.6 ^
W Zoo x HZ. \ A r 5860 ^* - 5*60**0^ ^l
p - JLl£E Tr'(2oo»ioq)Q5.3*/o"4)
£. _ s&d*jq*
rc/>
LM27Z7 x|©
7 r O. 3^36
-s
6:
*«**.
A +
. iSe*!* , i2Kl«2L , w.,,*,0< P* . W.iMPcc
^ 5260*10fc )SI * lo 6

PROBLEM 10.123
10.123 A column with the cross section shown has a 13.5-ft eflfective length.
Knowing that ar = 36 ksi. and E = 29 * 10* psi., use the AISC allowable stress
design formulas to determine the largest centric load that can be applied to the
column.
SOLUTION
A - 2A, + A2 = ttKi'lteucioXiv- *s .V"
r—8ta'—' Lc - 13-5" f+ - 162 i« fy* III.2? < Cfi
ht^ ~ O.S^ZC F.S. * £ + 4(0.8S*O-± (o. 88*£ )S ~ l.flia
P*« ■- ^A * (ll.49K$.sy- ^7.7 Je.pa
r /S.OI3 J«*
126.fo

PROBLEM 10.125
10.125 Bar AB is free at its end A and fixed at its hase B. Determine the allowable
centric load P if the aluminum alloy is (a) 6061-T6, (b) 2014-T6.
SOLUTION
A - (3o)0o") - 300 khh""= 3co x 10* ^
U = ZL - (Z\gs)* po m** ^ ^S*-*8
(bO GOC\ -T4 L/r * 64
r 87.tf» MVo-
P*# = 5*A *(87.«?»/o4V3oowo")- xfi.^fc/o' iv
P^f *6^A * (l07.3x|o4X3oo*/crc) = Z2.Z*\C? N - 32.3 W -*
30 rr
10 mm

PROBLEM 10.126
10.126 A sawn lumber column of 5.0 * 7.5-in. cross section has an effective length
of 8.5 ft. The grade of wood used has an adjusted allowable stress for compression
parallel to the grain ac= 1180 psi and a modulus of elasticity E= 1.2 * lO'psi.
Using the allowable-stress method, determine the largest eccentric load P that can be
applied when (a) e = 0.5 in., (b) e = 1.0 in.
SOLUTION
C - o. 3 *<^ = O.3oo
b = 7.S .-* , d* 5*.0 i^ C= ^ - 3.75" /*«
.A - fc>d -- C7.sXs.oV 37.5- .«** Iy;> A(s:oV7.srls- i7r.7g iw*
6te/6<. - S£>S/H8o =■ 0.7S3I
*♦ fig* » C
P.* - -
61//
***
(a)
e - o. s iVi
P**-
6So
,to.rK3.-7Q
37. S * I75V?c7
(V^ e - 1.0 ii
6 So
- \szio A -
\$*Z\ kips
WI70 A * 14.17 Iftps

PROBLEM 10.127
10.127 Two 4 x 3 x | -in. steel angles are welded together to form the column AB.
An axial load P of magnitude 14 kips is applied at point D. Using the allowable-
stress method, determine the largest allowable length I. Assume E = 29 x io6 psi.
and Of - 36 ksi.
SOLUTION
Owe a«^f L ^3*1 A =2.48 i*1"
I„r3.9£ .V*, Sx-LYG .'** Y\,~ 1.26 m. y » L2* .v.
3 in. 3in.
L, -- i.q* i*^
Two fluytS
V^ r 0. 879 ,V, 3 X^ 0.7*2 m
A =• CT>C^3) - f.96 ."»*
P ? ^ kips
E- 21qqc> it*;
S* ' A"* "ST - ^.96 + —"^ " s-™ *•
3€.
I 26,1
A«»*« T>CC ^' 1.92q^7 (^) " T.
L - /.¥*£■ - / TT1 (39000V" _ |fi7 o , > /i
ir*f
9* O
L- 167. J Cv = (I67.SJO. (77) r /99.S* (Vi. * /6.Y6 ft

PROBLEM 10.128
P = 32 Id pa
10.128 A compression membar of rectangular cross section has an effective length
of 36 in. and is made of the aluminum alloy 2014-T6 for which the allowable stress
in bending is 24 ksi. Using the interaction method, determine the smallest dimension
d of the cross section that can be used when e = 0.4 in.
2.25 hi.
SOLUTION
A r 2~ZS J>
C- iJ
e^O.^ ;*. Lt = 36 i'vi
Assort u/r^ > ix
£*.
A3>.
Pe<
Assise 1TK - Cv. _, />. A*2.2s
U//u.., - Via" U/d
Let x - ^ 4.096 X3 + l.^mx* - I
S<JU'«j 4^ X^ X r 0. 5"28n8, «J * V s K £=H in. -<■ ;.?fm,

PROBLEM 10.C1 10.C1 A solid steel rod having an effective length of 500 mm is to be
used as a compression strut to carry a centric load P. For the grade of steel
used E - 200 GPa and av = 245 MPa. Knowing that a factor of safety of 2.8
is required and using Euler's formula, write a computer program and use it to
calculate the allowable centric load PM for values of the radius of the rod from
6 mm to 24 mm, using 2-mm increments.
SOLUTION
ENTER RADIUS RAb ) EFFECTIVE L £ A/6T/-I L£
AND FACTOR 6F SAFETY FS
COMPOTB Rf\Q\US Of &YRATIQM
2.
A = TT RAb
__ I —- li'MIl
1 TT RA&
4
PETERhlW £ /lLLOIA//\f3LP tEU~R\C LOAD
C£IT1C*L ST^esi :
°"cr ^
TT*E
LFT 0" C^t/AL 5MALL&R OF ^r^^D^
p = <TA
Alr FS
P*o&ftAM 0VTPUT
Radius
of rod
m
.006
.008
.010
.012
.014
.016
.018
.020
.022
Critical
stress
MPa
"71.1
126.3
197.4
284.2
386.9
505.3
639.6
789.6
955.4
Allowable
load
kN
2.87
9.07
22.15
39.58
53.88
7 0.37
89.06
109.96
133.05
.024 1137.0 158.34
Below the dashed line we have:
critical stress > yield strength

PROBLEM 10.C2
/.fm.
g^^LIM6__/M j/Z- pLiMJF
Lp ~ 0.7 L
Bocku^c ikj X^ pLftVg
i^OSL
10.C2 An aluminum bar is fixed at end A and supported at end B so that
it is free to rotate about a horizontal axis through the pin. Rotation about a
vertical axis al end B is prevented by the brackets. Knowing that E -
10.1 X I06 psi, use Euler's formula with a factor of safety of 2.5 to determine
the allowable centric load P for values ofb from 0.75 in. to 1.5 in., using 0.125-
in. increments.
SOLUTION
ENJER B^ LENGTH L AMD FACfe* *F SAFETY PS
FOR, b = 0.7F *"0 /. 5" WKH O./zr INC8BM&WS
compete KAPjas o/= &r^/mo';
A = /.5~ fa
r =ji
r., -/ ^
Compute Critical ZtAESSFS
(<U =
(O. =
772E
(&7t/rJ
itze
? {0'sl/^)x
Let <r eqo/\l smaller, stress
COMPUTE flLLOI/VABLP CfWTRIC /.PAD
P.
all
pfl06KAIi OUTPUT
b
in.
.750
.875
1.000
1.125
1.250
1.375
1.500
Critical
stress
x axis
ksi
7.358
7.358
7.358
7.358
7.358
7.358
7.358
Critical
stress
y axis
ksi
3.6
4.9
6.4
8.1
10.0
12.1
14.4
Allowable
load
kips
1.62
2.58
3.85
4.97
5.52
6.07
6.62

PROBLEM 10.C3
4 m
A
JO/MT D:
3 ' *T
10.C3 The pin-ended members AB and BC consist of sections of
aluminum pipe of 120-mm outer diameter and 10-mm wall thickness. Knowing
that a factor of safety of 3.5 is required, determine the mass m of the largest
block thai can be supponed by the cable arrangement shown for values of h
from 4 m to 8 m, using 0.25-m increments. Use E = 70 GPa and consider only
buckling in the plane of the structure.
SOLUTION
COMPUTE MOMEMT OF IK/F5TIA
J = £( 0.0 fc*- COS*)
FOR, h * t TO % C5/M6 0.25" ItiCRctftuTs
Compute allowsbli= loads r^ ;'::."^^'~/?j
1V3
fa.),
it2 EI
* ;" *<-\r " ?,
n2FJ:
2 F = O Y'elds
Z* - A YIELDS
JOIMTB;
DETERMINE /M-LO^ASLP \A/
Wft|, EQUALS $MflLL[~R VALCti
Compute miss r>->
'y z
i.rw
Kn -
PROGRAM
h
m
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
w*,,
9.81
OUT PUT
Weight
critical
stress
AB
kN
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
455.11
Weight
critical
stress
BC
kN
269.7
286.6
303.4
320.3
337.1
354.0
37 0.8
387.7
404.5
421.4
438.3
455.1
472.0
488.8
505.7
522.5
539.4
mass
kg
7854.88
8345.80
8836.74
9327.66
9818.59
10309.52
10800.45
11291.38
11782.31
12273.24
12764.17
13255.10
13255.10
13255.10
13255.10
13255.10
13255.10

10.C4 An axial load P is applied at a point located on the x axis at a
distance e — 0.5 in. from the geometric axis of the W8 X 40 rolled-steel
column AB. Using E = 29 X I06 psi, write a computer program and use it to
calculate for values of P from 25 to 75 kips, using 5-kip increments, (a) the
horizontal deflection ai the midpoint C, (b) the maximum stress in the column.
SOLUTION
^7fR LErWiTH L^ ECtf:VTR\C\T Y e
COMPUTE CftfTK^L LOAD
p _ 77?E Ig
FOR P = 25* TO 7i~ i'^ /^RrM^MTS OF 5"
COtyPUJF Mfl/jMUM iTRg53
A (/ 2r/5 i » Peir '
PR06RAN) (9UTPl> f
Load
kip
25.0
30.0
35.0
40.0
45.0
50.0
55.0
60.0
65.0
70.0
75.0
maximum
deflection
in.
.059
.072
.086
.100
.115
.130
.146
.163
.181
.199
.219
maximum
stress
kips
3.29
3.99
4.69
5.41
6.14
6.88
7.65
8.43
9.22
10.04
10.88

PROBLEM 10.C5 10.C5 A column of effective length L is made from a ro| led-steel shape
and carries a centric axial load P. The yield strength for the grade of steel used
is denoted by o-y, the modulus of elasticity by E. the cross-sectional area of the
selected shape by A, and its smallest radius of gyration by r. Using the AISC
design formulas for allowable stress design, write a computer program that can
be used with either SI or U.S. customary units to determine the allowable load
P. Use this program to solve (a) Prob, 10.57, (b) Prob. 10.58. (c) Prob. 10.60.
SOLUTION
FNTEft PROPERTIES A, r
PETERN^I^E ALLOWABLE STRESS
c
IF
c ""
L/
°;..
J
rt
Ztt2£
2 Cc
TT2E
1.92. (L/r„)1
|F L/rf t. Cc
^ - ~
Ty /, W'
FS \ 2- Q1
CKCULh'Z ALLOWABLE LOAD-'
? ' ^ <T A
<?.!'
CONTINUED

PR06R/JM OUTPUT
Problem 10.57 (a)
Effective Length = 6.50 m
A = 6250.0 mm**2
ry = 4 9.2 mm
Yield strength = 250.0 MPa
E = 200 GPa
Allowable centroid load: P » 368.139 kN
Problem 10.57 (b)
Effective Length = 6.50 m
A = 10200.0 mm**2
ry = 65.0 mm
Yield strength = 250.0 MPa
E = 2 00 GPa
Allowable centroid load: P = 916.148 kN
Problem 10.58 (a)
Effective Length = 21.00 .ft
A = 9.130 in**2
ry - 2.020 in.
Yield strength = 36.0 ksi
E = 29000 ksi
Allowable centroid load: P = 87.566 kips
Problem 10.58 (b)
Effective Length = 21.00 ft
A = 9.130 in**2
ry = 2.020 in.
Yield strength = 50.0 ksi
E = 29000 ksi
Allowable centroid load: P = 87.452 kips
Problem 10.60 (a]
Effective Length = 4.00 m
A = 13800.0 mm**2
ry = 4 3.4 mm
Yield strength = 345.0 MPa
E = 200 GPa
Allowable centroid load: P = 1567.879 kN
Problem 10.60 (b)
Effective Length = 6.50 m
A = 13800.0 mm**2
ry = 43.4 mm
Yield strength = 345.0 MPa
E = 200 GPa
Allowable centroid load: P = 632.667 kN

PROBLEM 10.C6
10.C6 A column of effective length L is made from a rolled-steel shape
and is loaded eccentrically as shown, The yield strength of the grade of steel
used is denoted by aY> the allowable stress in bending by craM. the modulus of
elasticity by E. the cross-sectional area of the selected shape by A. and its
smallest radius of gyration by r. Write a computer program that can be used with
either SI or U.S. customary units to determine the allowable load P, using either
the allowable-stress method or the interaction method. Use this program to
check the given answer for (a) Prob. I0.111. (b) Prob. 10.112. \c) Prob. 10.113.
SOLUTION
EV1BK LjB, <rY) (Obe««l-...j>ex>e*
ED~\tK PROPERTIES (\} \ , 5^ , r^
PS1BRM\NB ALLOWABLE ST RESS
Cc-
fZTTzB
en.
IF L/f. > Cc
T7>£
<r„ -
IF L/r^ < Cc
n-s+ic&vrm
<T,
fOR /\U-0Wrt 5L£- STRESS MT7H0D
CoEf -
P-
A
<ki
^ + £
£.
'7
7
cotr
FOR INTER AdTlOM MBTHOP
COFF
<M*
/_
1.0
Co/T.F
-t
''-'All 'tenduM
CONTINUED

PROBLEM 10.C6 CONTINUED
PROGRAM GUJPUT
Problem 10.111
Effective Length = 5.80 m
A = 7560.0 mm**2
ry - 51.900 mm
Sx = 582000.0 mm**3
Yield strength « 250.0 MPa
E = 200 GPa
Using Interaction Method
Allowable load: P = 322.022 kN
Problem 10.112
Effective Length = 7.20 m
A = 7420.0 mm**2
ry = 50.300 mm
Sy = 185000.0 mm**3
Yield strength = 250.0 MPa
E = 200 GPa
Using Allowable-Stress Method
Allowable load: P - 97.781 kN
Problem 10.113
Effective Length = 21.00 ft
A = 11.800 in**2
ry - 1.930 in.
Sx - 51.90 in**3
Yield strength « 36.0 ksi
E = 29 x 10A3 ksi
Using Interaction Method
Allowable load: P = 86.722 kips

CHAPTER 1 I

PROBLEM 111 **'* ^etennuie ^ modulus of resilience for each of the following grades of structural
steel:
(a)ASTM A709 Grade 50: 0j- = 5Oksi
(6)ASTM A913 Grade 65: oY=* 65 ksi
SOLUTION (c) AS™ A?0^ Grade 100: oY = 100 ksi
(crt fjy - 50 ks.' =■ SO v/o* psi
M £Y - QSIcsi '- 66*(Os p*;
^r r *£"£" =7Tv7r—^r-r ~ 72-5 »*v*b/»n
PROBLEM 11 2 H'2 Determine the modulus ofresilience for each of the following aluminum alloys:
(a) 1100-H14: E = 70 GPa, ^ - 55 MPa
(6)2014-T6: £ = 72 GPa <* = 220MPa
SOLUTION <c> 6061-Y6: £ = 69 GPa ^ = 140 MPa
ipOi E - 70*10* Pa. 3 6V = ££*/06 Pa
<C) F = £<? * /o" Pa; Sr - MO*IOC ?<K

PROBLEM 11.3
SOLUTION
11.3 Determine the modulus of resilience for each of the following metals:
(a) Stainless steel AISI 302 (annealed): E = 190 GPa, or = 260 MPa
(b) Stainless steel AISI 302 (cold-rolled): E = 190 GPa oj- = 520 MPa
(c) Malleable cast iron: E = 165 GPa ox = 230 MPa
U* - ft - i^SoS^ » ni.-*-^ NW^ = ,77., W/..
<t0 E = l^vIO9 pa., <ST =" ^CWo' Pa.
(t) E ~ \CS* 10** fi^ €V " ZSOtlo' Pa
PROBLEM 11.4
SOLUTION
/2.4 Determine the modulus of resilience for each of the following alloys:
(a)Titanium: £=16.5* 10* psi: Oj.= l20ksi
(b) Magnesium E = 6.5 * IO6 psi: «% = 29 ksi
(c) Cupronickel (annealed): E « 20 x 106 psi: a, = 16 ksi
<c0
E = ic.^xyo'pS;
6^ - 120 *io4 p»i*
(U E «" G.-S'* HffXij $r r M*lO% p«/
a
£4.7 ,vii/;
. 3
' 2e (:OUo*/o<T

PROBLEM 11.5
0.006
0.14 0.18
11.5 The stress-strain diagram shown has been drawn from data obtained during a
tensile test of an aluminum alloy. Using E = 72 GPa, (a) determine the modulus of
resilience of the alloy, (b) the modulus of toughness of the alloy.
SOLUTION
TV»e cuje/-A,«i* o<r&Kn*.-re c*f +kt sircss-s'f
A
/"A'ft
.'j ^00 MP*. - &oo*ioc N/m*
PROBLEM 11.6
72-tf The stress-strain diagram shown has been drawn from data obtained during the
tensile test of a specimen of structural steel. Using E = 29 x 10* psi, (a) determine the
modulus of resilience of the steel, (b) datermine the modulus of toughness of the steel.
SOLUTION
(a) &r r £sY
U* = U
^- - i£erz - ^ft^ioMfo.oo^)1
0.021
10.002
0.2 0.25
(bO MoJuit/s of tostikneSS - ■/d+aJ' <3**ea 0*c\e*r fie
A2r 3k2«)(o.ar-o.ox.i )*■ 3,ai k.'psAV =■ 3. 4J i^ky/;*3

PROBLEM 11.7
11.7 The load-deformation diagram shown has been drawn from data obtained during
the tensile test of a specimen of structural steel. Knowing that the cross-sectional area
of the specimen is 250 mm2 and that the deformation was measured using a 500-mm
gage length, determine (a) the modulus of resilience of the steel, (b) the modulus of
toughness of the steel.
SOLUTION
S {mm)
*A p- C2.S kw
s- 18.75 M-m
- IS.7S" J
uf
- It
V
fS.7S"
125 >f lo-*
iso * to
A, - (62.S*loc)(ei£" lo'*} * G * /os M-^ » 6*/0* J
■ ZJ - Ur + A, 4 A4 + A3 - 7. 85" * ios J
J \y 125 * ID
V
63 MJ/^3

PROBLEM 11.8
0.104
D,
ZJr /54Q
V " 7.0484
11.8 The load-deformation diagram shown has been drawn from data obtained during
the tensile test of a 0.75-in.-diameter rod of an aluminum alloy. Knowing that .the
deformation was measured using a 16-in. gage length, determine (a) the modulus of
resilience of the alloy, (b) modulus of toughness of the alloy.
SOLUTION
V - f J*L
" ? (p«?0*('0 r 1.648& ,V
(a) MoJoJos of resilience.
PY - 30 k/ps, Sr - O.)o*i ,v
%%.\ i/i'tk/irt
7.06&4

PROBLEM 11.9
11.9 Using E *29 x 1Q* psi( determine (a) the strain energy of the steel rod ABC when
P = 8 kips, (b) the corresponding strain energy density in portions AB and BC of the rod.
SOLUTION
P^ S k:<
B e ft* fO* kin-
Pi - \Jl\ V - AL ,
U= uV
<r
«£
^2"
Pa fti'off
A8
6C
T
in.
O.C25
L
it
2H
3G
A
D..M1/B
V
7.3C3
0
u
lift- lt«*p
8<MX* WT*
t7€.M*\o'
-»
PROBLEM 11.10
27.2© A 30-in. length of aluminum pipe of cross-sectional area 1.85 in2 is welded to
a fixed support A and to a rigid cap B. The steel rod EF, of 0.75-in. diameter, is welded
to cap B. Knowing that the modulus of elasticity is 29 * 106 psi for steel and 10.6 * 106
for aluminum, determine (a) the total strain energy of the system when P - 10 kips, (b)
the corresponding strain-energy density in the pipe CD and in the rod EF.
SOLUTION
•Of
EF: A ' 5 al* * 0.441* t/-
EF: a- P^ = n°»">*yfr»>--- .: /S7.3S iv, A
MEr 2£"A (2>Cw*io*)(o.iwiO
CD: ^-l^~-^Pss u-ff*fe^^^^

PROBLEMll.il
11.10 A 30-in. length of aluminum pipe of cross-sectional area 1.8S in1 is welded to
a flxed support A and to a rigid cap B. The steel rod EF, of 0.75-in. diameter, is welded
to cap B. Knowing that the modulus of elasticity is 29 * lO'psi for steel and 10.6 x io6
for aluminum, determine (a) the total strain energy of the system when P~ 10 kips, (b)
the corresponding strain-energy density in the pipe CD and in the rod EF.
11.11 Solve Prob. 11.10, when P = 8 kips.
#*P SOLUTION
rrv r) - P**. - (-gooQ^(3o'>
U - ZA* + U£f - 168.8 ,'«.-/t -*
7VM
CD-' 6** -

PROBLEM 11.12
11.12 Using E = 200 GPa, determine (a) the strain energy of the steel rod ABC when
P~25 kN, (b) the corresponding strain-energy density of portions AB and BC of the rod.
20-mm diameter
16-mm diameter
C
SOLUTION
A*,- fW*- 314.14 *«*N 3/S.tt*|crsv^
9 - 25*10* hJ
II = X
£1L
2FA
*W5
(«0 tJ r 5.^68 + 5.213 - KU8 N-v* = 13.IS J
- 7?. 58 */o6 Pec
G
QC
- -£- - 35»/o3
= I2*t. XS x lO6 fls
u* = fr- ■ £S:'£\ - ****'* =*»■• "J/-.*
PROBLEM 11.13
77.23 The steel rod 45C is made of a steel for which the yield strength is Of = 250
MPa and the modulus of elasticity is E = 200 GPa. Determine, for the loading shown,
the maximum strain energy that can be acquired by the rod without causing any
permanent deformation.
20-mm diameter SOLUTION
16-mm diameter
P = SjA«.v= C^dxio'Xaoi.oGK/o"*^
,2
ZH.\S + W.lS - if?. 3 J

PROBLEM 11.14
10-mm diameter
■ 6-mm diameter
11.14 The steel rods AB and BC are made of a steel for which the yield strength is
ar = 300 MPa and the modulus of elasticity is E - 200 GPa. Determine the maximum
strain energy that can be acquired by the assembly without causing any permanent
deformation when the length a of toAAB is (a) 2 m, (b) 4m.
SOLUTION
u - Z
£!l
(CL")
L-a. r C-*. - 41"
T1 _ (3.iJ8at>«ftf yc?-^ (g.w^K/o?V(</i
- 4,£"S03 + ;?£. 4H6G
30.0 W-^ = 30.0 J
u -
PROBLEM 11.15
0.6-in. diameter
0.4-in. diameter
Use ^maJJc/* v/a /u€
11.15 Rod AB is made of a steel for which the yield strength is ox =* 65 ksi and the
modulus of elasticity is E=29 x I06 psi; rod fiC is made of an aluminum alloy for which
<jr = 40 ksi and E = 10.6 * 10*psi. Determine the maximum strain energy that can be
acquired by the composite rod ABC without causing permanent deformation.
E" - I04OO ks,*
SOLUTION
A** r * (o.**')1 - o. U566 in*
AB: fi^ - {&S)(0.)ZS-6C ) = 8./t79 fc.'r*
P= g. /67? k.-ps
=. 2J6.3 */0~* + '/C7.5'>/0"i -
(gj^ypf^

PROBLEM 11.16
A -
ALu;
11.16 Rod AB is made of a steel for which the yield strength is ar = 300 MPa and the
modulus of elasticity is E = 200 OPa. Knowing that a strain energy of 10 J must be
acquired by the rod when the axial load P is applied, determine the diameter of the rod
for which the factor of safety with respect to permanent deformation is six.
SOLUTION
et^
31
Y t E (ftboo * IO* )
=■ 13.73 wnn
PROBLEM 11.17
■^■-in, diameter
j -in. diameter
11.17 The rod ABC is made of a steel for which the yield strength is ox « 65 ksi and
the modulus of elasticity is E = 29 x 10*psi. Knowing that a strain energy of 90 in-lb
must be acquired by the rod as the axial load P is applied, determine the factor of safety
of the rod with respect to permanent deformation when a - 18 m.
SOLUTION
PT « 6yAw, t (GSOOoKo.I*»CS5-)- 127C3 A-
TJ . y Fr'l
■rj . 02763^0*8-18) + 02763^(18-)
W6SO
~ W& in-A
Es. =
li,
II
j»$*
4^g
H.<?8 -*

PROBLEM 11.18
11.18 The rod ABC is made of a steel for which the yield strength is or = 65 ksi and
the modulus of elasticity is E = 29 * 106 psl Knowing that a strain energy of 90 in • lb
must be acquired by the rod as the axial load P is applied, determine the factor of safety
of the rod with respect to permanent deformation when a = 18 in.
4 -in. diameter
4
j -in. diameter
11.18 SolveProb. 11.17, assuming that a =30 in.
- 2k
SOLUTION
0*763)'(48-3*0 (|276S)*(yO _ ,-.3 - . «
fl*
G. o4
PROBLEM 11.19
11.19 Show by integration that the strain energy ofthetaperedrod.45 is
1 P2L
where ^-i- is the cross-sectional area at end B.
SOLUTION
ZL
- ?t=-nrx <• X 'I.
u "J. ;?t"A ' ze \ ire* x"
2L
ZETIC
*1 **-
-(-pW) -
s i v
F5/.
4£A~;

PROBLEM 11.20
1.9c
1.7c
5 5 5 5 5
11.19 Show by integration that the strain energy of the tapered rod AB is
where A^ is the cross-sectional area at end B.
11.20 Solve Prob. 11.19, using the stepped rod shown as an approximation of the
tapered rod. What is the percentage error in the answer obtained?
SOLUTION
u ■ 2
ZEAjl
■ZE
A;
_ FZL «r
lo-nE ^ *x
= P*L
p'l r ' ,._!_
P'Z.
Elk.
r {s.tBS^ s O.ZISSL
\o Ftnc
% error ■= cT~ZS x - OfcS7i /fa
£A*:«

PROBLEM 11.21
2.10 in.
2.55 in.
2,85 in.
I
1,5 in. = 6 in. —-
11.21 Using£=!!10.6x lO'psi, determine by approximate means the maximum strain
energy that can be acquired by die aluminum rod shown if the allowable normal stress
is aM = 22 ksi.
SOLUTION
6**1/ - 220QO fs<-
Use Si^sdrt^s roJ*e "/» oo^poh "H<? i«-\-e.^r+/ Vi^ O.IS" i«
Seg,-i-itfn
2
3
k/d'O^
0.XZ675
0. 1231 I
O- II I II
^(\/^)0^^
0.4W44
0.^070
0.3076
O.ll/I
5" ' Z.7.C.ZS

PROBLEM 11.22
11.22 In the truss shown, all members are made of the same material and have the
uniform cross-sectional area indicated. Determine the strain energy of the truss when
the load P is applied.
SOLUTION
Jo i r\ r C
ZB
zE2k
Me^Wv*
BC
CD
1
F
-faP
L
ft'
A
A
A
F\/A
ft P*i/A
3 P2J!/A
1M?XA/A
■t?^ r O
AF..-P = O
ttc
s 2P
-F^-f Feo=o F^-JIP
- 3.S
£A
PROBLEM 11.23
11.23 In the truss shown, all members are made of the same material and have the
uniform cross-sectional area indicated. Determine the strain energy of the truss when
the load P is applied.
SOLUTION
-Foe - iFse = o
Mc^te**
BC
X
F
ft?
L
A
A
A
F*L/A
iP'iM
3?*JL/A
F„ = -£ P
<=D
D
A
p'l
HA

PROBLEM 11.24
11.24 In the truss shown, all members are made'©fithe same material and have the
uniform cross-sectional area indicated. Determine the strain energy or'the truss when
the load P is applied.
VLFj = o
R»= -
rs. O
+ -I F - o - F^ - i R
eo
= o
F^--
6
Joivt D
K.
*tZFjSo
+ a. To
F«>
FBO, p
rr O
Mew> te/>
Be
CD
6d
r
F
4?
p
L
i
714
A
A
A
A
F\/A
i?*$ /A
f?2i/A
V.732Pa*/A
*F ^ A
_ =^^-732^
= 2.37
£2 ^
FA

PROBLEM 11.25
1^^In the trUSS shown» ^ members are made of the same material and have the
uniform cross-sectional area indicated. Determine the strain energy of the truss when
the load P is applied
SOLUTION
F.„ -<—\
(?.
ep
+tZFj--o -P- «-|Fco= o
+ -TR,
F™
5" Tcd *
Me^te^
CB
CD
Bt>
Z
F
-fP
4P
L.
A
*A
2A
A
F2L/A
g* P*M
- -A. Jtp = _3 p
= G-253
£2 ^
F*

PROBLEM 11.26
0.75 m
c 120 kN
11.26 In the truss shown, all members are made of aluminum and have the uniform
cross-sectional area indicated. Using E = 72 GPa, determine the strain energy of the
truss for the loading shown..
SOLUTION
0.75 m
Jo mT C
n^
+*-ZF„=0 -
1.8
).9S
F^ ~^^d + Wo = o
QC
K.
>-
C WO
V 2oo
F«^ = 3^5- kW
Jo.«-V "D
+ f 2*Fy r O
GO
E* *o
F^ = - tts* *W
F»0 ■ 7-r m
*p
R
\D
A (icr%;
-* i
F*L/A (^/^
7.03 x /o'1
ZH.7Z k /c?'1
u -
!1G.IS xlO
ia-
oocn* 10" *)
4 4. I* v /£>"■
l.o/sx/o3 N-^ * /0I5 J
CO

PROBLEM 11.27
0.75 m
0,75
6l
>
CD
' * %oo
11.27 In the truss shown, all members are made of aluminum and have the uniform
cross-sectional area indicated. Using E = 72 GPa, determine the strain energy of the
truss for the loading shown..
11.27 Solve Prob. II.26, assuming that the 120-kN load is removed.
SOLUTION
To i' * t C
M? ^o
- o
Fat-
\,1S ra<L l.l* f^o -^O - O
Solufna (0 a*«/ C2) si^uJ^tcneooSst
F^ = 2&o ku
1-
Joint O
60
R.
+ tZFS"»
_ 0.7i ,- _
Ffto = lo<5 ^
60
u-- z
Fml
2.FA
£2.
ZE * A
Me^keir
Be
BD
CD
Z
FCkM")
2£o
loo
-ato
l c*a
MS
I--T
l.^ff
A (Id* J
\80t>
I2oo
30ao
, FaL/A (n/Vk^
73.23 x |o11
I 2.6""© x /o'*
i|3.94 v lotl
I2<?.67 v /o'4
u *
C2.X7;i*lo«)
<?oo A/.i^
loo J"

PROBLEM 11.28
11JS Taking into account only the eflfect of normal stresses, determine the strain
energy of the prismatic beam AB for the loading shown.
SOLUTION
^Mc = O
a—*■ ■* L
cxP t L(?B- o i?e..-^E.s ^Ej,
P
M* - Px
€
£
P* Xs1-1
O^ev* por+i'ott DB
M
c
""yaf
of
get
To+«i
U * U„ * U
4, t
AD
rDB
err
(«o
PROBLEM 11.29
11.29 Taking into account only the effect of normal stresses, determine the strain
energy of the prismatic beam AB for the loading shown.
SOLUTION
+t JFj -- o KK+Qi-ZP = o f?„ = £>a r p
Ove.r po^f."o« AD : M - RAX = Px
n - P'a.'a-*0
_ p'cJ
p<*

PROBLEM 11.30
1130 Taking into account only the effect of normal stresses, determine the strain
energy of the prismatic beam AB for the loading shown.
M
C
K
v w
UiJlIti
B
SOLUTION
^0 2.£i
.2
W
w2L
o
2) S"
jyo^r
PROBLEM 11.31
1131 Taking into account only the effect of normal stresses, determine the strain
energy of the prismatic beam AB for the loading shown.
u- ££■*>- j&S>*-xne*
SOLUTION
1? * ^
2 EI
.-&$w-****<)+*£ri&-*F*£t
w*l* r j l * j-i -
w2Ls
ZHo£X

PROBLEM 11.32
1132 Assuming that the prismatic beam AB has a rectangular cross section, show
that for the given loading the maximum value of the strain energy-density in the beam
is
u
Mmax = i5y
where U is the strain energy of the beam and V is its volume.
SOLUTION
M
vr
* I i i -* t v
-^
O^MK = 0
u- s:&j* - MW^-^vr
*£X
- ££
LI
61
•J* -
w
?i=x
u
•■"«•<
I
4 w»LV
2 EXV
.ti*^
w L c
II
•***
5cl
^(^
Ui
h*A^,
=■ IS*
15"
IL
V
V
JL'j
- w4I
*( £■
fOEZ
m
1
i3i
h

PROBLEM 11.33
11.33 Assuming that the prismatic beam AB has a rectangular cross section, show
that for the given loading the maximum value of the strain energy-density in the beam
is
Mmax g y
where U is the strain energy of the beam and V is its volume.
45 U_
8 V
SOLUTION
r?, = "iwL
M^c _ \ajL a
M
*■»*•&
n -
2fofX
SLI _
Omwn "
SJ
15" c1
«*»
•3 V
r*b*|

PROBLEM 11.34
11.34 Using E = 200 GPa, determine the strain energy due to bending for the
steel beam and loading shown.
160 kN
2.4 m —*
2.4 m
■4.8 m-
*■:
A
W310 X 74
B
iP
SOLUTION
Viow AC M-* iP*
^er "* r -sex
i1^
set a '0 mei
2 I ^
P*la
Bj sy^e-nry wCS - WWL mE1
U =
06o*jo»y (1.0*
C^tX^O^yjO
=)0S7f^ =-"'' *•* ' 8^J
PROBLEM 11.35
11.35 Using E = 200 GPa, determine the strain energy due to bending for the
steel beam and loading shown.
W310 X 74
^1.6m4*1.6m4-1.6m;
■4.8 m-
'LA
P
Xa-|i
SOLUTION
2ET
X&J 3 1. " QBI
OVCV pO'Ti'oii D t *'
U
porTiow Ufr - M - P*.
2er " zei
OE ="
By 5^ihim#t/ji Use " ^*o
6 EX
,«- a
D^+o.: P^o^o1^ aH.Uj F-- ^oox/o1* Pa.

PROBLEM 11.36
11.36 Using E = 29 * 106 psi, determine the strain energy due to bending for the
steel beam and loading shown.
8 kips |
S8 X 18.4
*
,r-L—k
e
SOLUTION
£>£^s = o -RAL - *P = o ft- ^ I
L
3
rj v~* $&**&#&?+
GFX
Ov** Beir-fi0«<i t>B
M* - T>v
*<£
kv-J
ft
-*.- r^jv -^s>vj*r ^
*-*
GCX
A. *
TofJ: U-U^U^ - |jj£(a+L,>
'i>e
(g0oo)z(3fcf (-73 + 3Q

PROBLEM 11.37
11.37 Using.E551.8 * l(rpsi, determine the strain energy due to bending for the
timber beam and loading shown.
16 kip ■ ft
A V
l_ U CL
X
M
■e
6
£A,
3gin.
-1 h
SOLUTION
Ove^ poHToh At) M ^ ±£*
O
R-*
Me
a- L
uAD, c
M'
xrx
■J* -
M*
Zei
tA^«
6HXLX
K
Udq " X ZEZ <** TElF \> * ** CEIL*
T>+J TJ - ILD + TJ
MJta**^)
*d * ^oa ^^j l*
(IC)a(H5 + GM _
(GX3U6 X/o")*
U , ^SS^\J\\ = 38.2 *lo- Wp.fi » 3*. 2 fl-ik
= HSZ ■'«• JL

PROBLEM 11.38
24-mm diameter
1138 Rod AC is made of aluminum (G = 73 GPa) and is subjected to a torque
T applied at end C. Knowing that portion BC of the rod is hollow and has an inside
diameter of 16 mm, determine the strain energy of the rod for a maximum shearing
stress of 120 MPa.
tat- ^
SOLUTION
Jfc^ iC^-C;^- K^**)* 2C.\38«l6"^^
rt
*6
(2CI.3*)1Ylto©*to"'')
=" 5.7*1 J
S.<?5/* J
PROBLEM 11.39
0.9 m
0.75 m
11.39 In the assembly shown torques 1A and TB are exerted on disks A and B
respectively. Knowing that both shafts are solid and made of aluminum (G = 73
GPa), determine the total energy acquired by the assembly. .
SOLUTION
Us r D.9 **
. u:U (300? (o.s)
u
** ' ^G-JAB ' {Z)(l2>xiO«Xlct.Stylo-*}
rz 6. 977 J
T^L^ C7oo)*(o.75l
i. ^ = D.7S k,
'*r *-* =
Uec * *G- J&, (2,>(73"ldt,>(l^.3*.S7*Jo■,,)
- S".7*& J
ToVJl 15^11/^+ Uac = fc.<n7 + S.7ZC =■ 12.7» J"

PROBLEM 11.40
?r
5000 ft
T =
11.40 The ship at A has just started to drill for oil on the ocean floor at a depth
of 5000 ft. The steel drill pipe has an outside diameter of 8 in. and a uniform wall
thickness of 0.5 In. Knowing that the top of the drill pipe rotates through two
complete revolutions before the drill bit at B starts to operate and using G = 11.2
x 10* psi, determine the maximum strain energy acquired by the drill pipe.
SOLUTION
cp - ttKa^ - 4tt v*A
L •- Sooo it - GOm/o1 m
f* — °
*/;.
J" - ?(C^-C/V I6G.H06 ,V
CrJ£
L
13
= r*L rf6Tg\a L 9 GJ<P*
2GJ*
V. 1_ / 2GJ
2L
PROBLEM 11.41
2.5 m.
25 ldp • in.
11,41. The design specifications for the steel shaft AB require that the shaft
acquire a strain energy of 300 ui-lb as the 25-kip-iii. torque is applied. Using G ~
11.2 x 106 psi, determine (a) the largest inside diameter of the shaft that can be
used, (b) the corresponding maximum shearing stress in the shaft.
SOLUTION
u -
L «
U
,T
3oo ;,*
36 in
_ T'L
ZGJ
- T*-L
A
—
"r 2S* kp-l* *ZS'*tO A I
>"\
Us'x/ol)lUO
* 3.3MB2. iw
dif- d" - If J"-" 2.S'- 1^(3.343:0 r 4.<rS7in iV
r
IF
J

PROBLEM 11.42
11.42 Show by integration that the strain energy in the tapered rod AB is
where J^ is the polar moment of inertia of the rod at end B.
SOLUTION
v L
J- \rH - *£*\ JL:."-
lr^
u
■i
J*
2G-X "J, Z&(*gxM
srr_
J. <3fl.(3L
2 I 4
T2L
^GJk,
2L
2&
« i %
v
.Jill
•J— + -M - -2- T'L
2L.
G-Jw,

PROBLEM 11.43
11.43 The state of stress shown occurs in a machine component made of a grade
of steel for which aT = 65 ksi. Using the maximum-distortion-energy criterion,
determine the range of values of or for which the factor of safety associated with the
yield strength is equal to or larger than 2.2.
SOLUTION
■'aiJt.
A (0+ 8^ * 4ks."
(0,-H) 7
(s^-O1* (&b-Ol +(€;-«; V = 2(fe)*
(|8.«+ lO.^O* + (-10.56- %)* + (€y-l$.S£)Z r >?(ff)1
897.97 + ( I U.S/+ 2U? Sy + <5^1) +(%*"-37,/2C£ * 3W.H1 r /7H£".27
Z&/ - l£<5> - «HI.«Z * o
6-, " ±^ i^C9^X^o =
6"r = 19.39 ksr ^ - D.39 fcsi*
- 11.39 Ifsi" ^ 61 * 19.39 Jrsi"

PROBLEM 11.44
11.44 The state of stress shown occurs in a machine component made of a grade
of steel for which or = 65 ksi. Using the maximum-distortion-energy criterion,
determine the factor of safety associated with the yield strength when (a) or ~ + 16
ksi, (b) <?j-=-16ksi,
S*.* i(°* O* 4 !<•.■
-s;
V-- me.:
+ r^
la) 6^ - <5y - \C> b;
(18.56+ |0.5"6)l+ (-I6.S6-/0* + Ofc-lS.sO*' * * (pf" V
F.S. - J?.33
S^7-<*7 + 70S*. 43 + G.SS- - f£f^
847.^7 + 29.59 + //<?**. 39 -
EHSo
F.S. - *.Ofc

PROBLEM 11.45
5 MPu
11.45 The state of stress shown occurs in a machine component made of a brass
for which or = 160 MPa. Using the maximum-distortion-energy criterion,
determine whether yield occurs when (a) az = + 45 MPa, (b) at = -45 MPa.
<5^*T ±U0O*ZO) =; 40 MP*
€T,.-5** _ 100- W _
Tw - Tff rift
40 t>»p*.
€Tfcr <S1„ + R -" 145 MR*
6; -6^c- R • ~-?5 MP«
(143 +35^*4 (-25- *is}% + ( 45- IVS^)1" < Z(.\^oY r iT/^oo
Ztqoo + 4900 + (OOOO - 4S80O * 5UOO (NoyiciJ )
(n<r +25^+ (-25+is")* + (-Hr-iHs? i ^"aoo
28<?<5o + 4oo 4 56/oo r &5^oO > 6"/^oo (Y.'eJU
OCCuiTS
)

PROBLEM 11.46
11.46 The state of stress shown occurs in a machine component made of a brass
for which ox =* 160 MPa. Using the maximum-distortion-energy criterion,
determine the range of values of os for which yield does not occur.
SOLUTION
75 MPa
6L,= i(loo+2oV GO MPa
ra * is mp«
Woo 4 (62S" + ^oS;+ 61*-) +- (6*,1 - 29© 6; + x\ozs a SUoo
s, = ^*^>' + cn(*X^ r G0± «.65-
G*z - U2.C5* MP* 5 - 2*&S HP**
- z.(>$ mp* < 6; < mxs mp*l

PROBLEM 11.47
11.47 Determine the strain energyofthe prismatic beam J5, taking into account
the effect of both normal and shearing stresses.
p _ Ma I ^ _ Mo *
M„L3
G£XL*
C-iJ
U ~
M«,ZL . 3 /^C
5 GU L
W;+ln I = Tat J*
+ ^
-S GtdL
£ bji:
■S- &WL
eT^L io & L* 5

PROBLEM 11.48
r<)
11.48 For the state of stress shown in Fig. a, determine the stresses in an element
oriented as shown in Fig. b. Compare the strain energy density in the given state
first by using Fig. a and then by using Fig. b. Equating the two results obtained,
show that
E
G~ 2(1+ y)
(a)
SOLUTION
U^ina MoViv .1 ciVtyfc
(b)
(a'i S»= o ^ 6-j = O j
•*i
u , -L (e;z + 6-/ - 2*$^ } + £ r^ -
2^
GO 6> - ■£,, Gy- -ro> X*y = °
E^u&.4
r
&* s
2fr"
•^ 'l
aCn**)
PROBLEM 11.49
*11.49 A vibration isolation support is made by bonding a rod A, of radius Rlt and
a tube B, of inner radius R2 to a hollow rubber cylinder. Denoting by G the
modulus of rigidity of the rubber, determine the strain energy of the hollow rubber
cylinder for the loading shown.
SOLUTION
2>—> Q.
+ - 2 R, ■= o
- r (a™o + q = o
r'
u -
U =
2.G 8TF*V*La&

PROBLEM 11.50
3.5 ft
SYA
P„
SVA
i*
k
L
£
£
11.50 The cylindrical block £ has a speed v0 = 16 ft/s when it strikes squarely the yoke
7
5Z> that is attached to the "g" -in.-diameter rods AB and CD. Knowing that the rods are
made of a steel for which a? = 50 ksi and E = 29 * 106 psi, determine the weight of the
block £ for which the iactor of safety is five with respect to permanent deformation of
the rods.
SOLUTION
At "Hie on%ei ot yiV-f*lt'«ft 4ke -Fo/-ce i*» ea.e.k ToJ- i-s
f - 6; A
y 5 R>Uft. 6;*A*l ^ 6VZAL
'AB
u
CO
2£"AAft
5am e
2F
2E
u„ .- ua + u
. C/Al
u. =(i^v.*XF-s.") r(^v;)(F.o
Dd-«.: a - 33. n -ft/sec* *■ Jge ;»/5cc* Sx ~ sa*ic? ps,^
A^ feT- -?(*)*' 0.£o\3* i*1 E= ^-/o'ps,
L = 3.5" ft * t^ ;« rs. - *5*
N/, r 16 ft/set r 19* i'» /sec
TAT -
ftK386KgO«/o^1(0-CQ'3*^<*g ^ .
0'fte;1(5")C^*/O*')
9.* 2 -ft.

PROBLEM 11.51
6yA
6"YA
■3.5 ft
£
^c
11.51 The 18-lb cylindrical block £ has a horizontal velocity v0 when it strikes
7
squarely the yoke BD that is attached to the "g -in.-diameter rods AB and CD. Knowing
that the rods are made of a steel fi>r which ffe = 50 ksi and E = 29 * iO'psi, determine
the maximum allowable speed v0 if the rods are not to be permanently deformed.
SOLUTION
Af +K<. onset en y ieJUirtg He-foirce i« e<*-^h v©J is
f * <s;a
6/A*/. _ 6V*AL
UCb = sft^e
- ^r'AL.
2£
^H"
ToW Umr 1L„ * UD -
(SV'AL
AB
U„ = i"V.' = if v.'
So/Vimg to«" V6
v.'-
ZqUm 2a 6V* A L
Dd-ct-.
3
A =
L
=■ 32.17 -Pf/sec8- r 586 i»/sec*, 6^ = 5*WoJ P*/
- J. 5" -ft -- -¥5 m. W- \S A.
v„ =
■* 305. £ ih Aec

PROBLEM 11.52
16-mm diameter
D
■L= 1.2 m
11.52 The uniform rod AB is made of a brass for which <% = 125 MPa and E =* 105
GPa. Collar D moves along the rod and has a speed v0 = 3 m/s as it strikes a small plate
attached to end B of the rod. Using a factor of safety of four, determine the largest
allowable mass of the collar if the rod is not to be permanently deformed.
SOLUTION
n i- x ■ 7T - P~l-__ (asttar/.-O
- |7. W3 J"
PROBLEM 11.53
16-mm diameter
D
-L = 1.2.
-J
11.52 The uniform rod AB is made of a brass for which or = 125 MPa and £ = 105
GPa. Collar D moves along the rod and has a speed v„=3 m/s as it strikes a small plate
attached to end B of the rod. Using a factor of safety of four, determine the largest
allowable mass of the collar if the rod is not to be permanently deformed
1L53 Solve Prob. 11.52, assuming that the length of the brass rod is increased from
1.2 m to 2.4 m.
SOLUTION
A4 onsei of yieUi^ P* - 6rA 6Y - l*S*/o4 *P«
= 35", 966 J
m *
U.
3S.lo£
Zv* ' (OfsV
99S* kj.

PROBLEM 11.54
r
4m
Bronze
E = 105 GPa
12-mm diameter
. Aluminum
E = 70GPa
2,^m fjL i 9-mm diameter
I- IS-r-0.6 m
11.54 Collar D is released from rest in the position shown and is stopped by a small
plate attached at end C of the vertical rod ABC. Determine the mass of the collar for
which the maximum normal stress in portion BC is 125 MPa.
SOLUTION
P* - 6"„ ABt = 7^5^ N
LT
p*/.
BC
m t-gc _
Q9SZ )a(g-g )
2£"ec Afic t£X7o*iO'*X63.6J7>ocrt )
- .17.750 J
>a
u
Ae.
M3.<?c>7 x/0"4 w*"
U,
P^ Lab
2S«Art * (2)f/Drx/O*)0l3.1o7*to"4)
^Bd * UAB * 28.33W J
-3
p.,
7*l3*/0 v*
Work of «/ef3t.t = U*v
715^
'•v.
0, . ?8.334

PROBLEM 11.55
12-mm diameter
Aluminum
E = 70 GPa
9-mm diameter
0.6 m
11.54 Collar D is released from rest in the position shown and is stopped by a small
plate attached at end Cofthe vertical rod ABC. Determine the mass of the collar for
which the maximum normal stress in portion BC is 125 MPa.
11.55 Solve Prob. 11.54, assuming that both portions of rod ABC are made of
aluminum.
SOLUTION
Pc^r+iov, BC : 6L. " 125"*/©6 ?c
Asc ' i (*0* = 63.&I7^- €3.Gl7x/o"4 ^l
- JE
1152. N
C
id.'.
oir»*espowel i *tg STV'a.i n energy
'Bt
A MS
00(70* jo* )(63.6/7 x/crc)
P^1 Lab (79S*")Z(0
7. 7JO J
T«+«J U^ ■= Uac ■» Umb ~ 33. Gil J
Corses poMeiin« exovt a&^ ioM A^ "i. "m A** * Un,
r 15.86/ J
^*
S.^ v/o~* m
7^2,
h r O.C + A^ * 0. Cog 45"
Pm '
Uv^ 33.GII
inn -
* h " CM»Ko.6o«iif )
= i".C3 k
:3

PROBLEM 11.56
ACE
11.56 The 100-lb collar G is released from rest in the position shown and is stopped
7 5
by plate BDF that is attached to the g- -in.-diameter steel rod CD and to the g - in.-
diameter steel rods AB and EF. Knowing that for the grade of steel used aM - 24 ksi and
E =<■ 29 * 106 psi, determine the largest allowable distance h.
8ft
SOLUTION
6UL S'cdL &„L
■m
E
^* e " e "
L » 8 ft * ?t i«
For e-uJi *-* CI - -^- ^ - ZL
046 ^ V„ *
C29KfO<-)(o.3Q6g6)(7'?.4WSM/Q)^ .
To-W U^ * UAt + DU + Uffr a i \ss, zi .vA
f«>/A"^ oli"stance i"fi h + Aw, _,
Tl «*, _^ ^0
TV ' loo -fk
11. £"83 .V».
to = IL58S - 11. ****** 16
-3
//.SO .'n.

PROBLEM 11.57
8 ft
11.5e The 100-lb collar G is released from rest in the position shown and is stopped
7 5
by plate BDF that is attached to the "jj" -in.-diameter steel rod CD and to the ^ - in.-
diameter steel rods AB and EF. Knowing that for the grade of steel used aM = 24 ksi and
E = 29 * 106 psi, determine the largest allowable distance h.
7
11.57 Solve Prob. 11.56, assuming that the "jf-in.-diameter steel rod CZ> is replaced
7
by a g" -in.-diameter rod made of a grade of aluminum for which aM - 20 ksi and E =
10.6 x I06psi.
SOLUTION
..L* S-fY * 96 i'M
B D
-fie
:«>
grPt
Err
If Sab =" ^^'O"3 p*i
^ (*y<r;,£^ - v^**^.
Fo^ e*<^ rod U - j^- ^A xl
£o<J CD: A^ = 3(£)1 r O.COI3* i'«'; E^? I0.ev,o4 p*;
Rods A6 a^ EF: Afta - A£r - 4{(j)* = 0.S0C80 ,V
11*4. £Z in-Jk.
*+ A^ ^ = :?^^ - ™« -
Vi * 7.l*i~3. - 79.^VSx/o
-3. _
7.37 -iV

PROBLEM 11.58
W150X XS.S
11.58 The steel beam AB is struck squarely at its midpoiut C by a 45-kg block moving
horizontally with a speed v0 = 2 m/s. Using E = 200 GPa, detennine (a) the equivalent
static load, (b) the maximum normal stress in the beam, (c) the maximum deflection of
the midpoint C of the beam.
SOLUTION
Ffo<^\ Appendix C y -&>^ W/ ISO* 13.5"
Ix - G.87* IO* •*.*♦ = 6.87if/Ofc m"
/C77T
L
U ^ iPJvJ -
t ulyj
|Vh1 , JEJLi M , £k
- T
(al
CO
P-*
'6^
4S
OOMl.t x|Cr*}
->' " R, ' 2a<H8*k>»
- 171.7 MP<*_
8.58 X yt?" m ~ &.£& W*n

PROBLEM 11.59
11-58 The steel beam AB is struck squarely at its midpoint C by a 45-kg block moving
horizontally with a speed v0 = 2 m/s. Using E = 200 GPa, determine (a) the equivalent
static load, (b) the maximum normal stress in the beam, (c) the maximum deflection of
the midpoint C of the beam.
11.59 Solve Prob. 11.58, assuming that the Wl50xt3Jrolled-steel beam is rotated by
90° about its longitudinal axis so that its web is vertical.
W150X 13.* SOLUTION
F^o^ Appe^WlxC^ -W W \So * 13.5"
777?T
^
I8.¥x/osm«s = l8.4«|o"4i^a
K;»ef-c euie^y T * i ^^ * iC^Xa*f -"^J
^
pj/.2
1 J"* > 4ft £
"33
o
M.
J3u
4S
(40(18.4* lO"*)
- 3ia MP*.
-a
?►,
7-C65"Mo:
^3.5"> /o" w, r 23.5 ».
*»^

PROBLEM 11.60
11.60 The post AB consists of a steel pipe of 3,5-in outer diameter and 0.3-in. wall
thickness. A 15-lb block C moving horizontally with a velocity v0 hits the post squarely
alA. Using£=29 * 106 psi, determine the largest speed v0 for which the maximum
normal stress in the pipe does not exceed 24 ksi.
777777
SOLUTION
Cot iJ0- aC3.S")- l-7r m.^ Q, - C - t = '-"^ - 0. 3 - 1.45 ■'«.
I - ^(C*« C/) r 3.8«H3 ,V R,s *fooo p*,*
■6T = &&
'»*»
M^I&. r (3-^^X^ooo^ __ 534o7A. .
P..*
«M
4^ - ni«.e« A.
48
u,
3£I
on-
'» ~W IS
Vw r ICtf.O ,V»/scc - 8.50 "ftAec
PROBLEM 11.61
/5?7?7
11.60 The post .45 consists of a steel pipe of 3,5-in outer diameter and 0.3-in. wall
thickness. A 15-lb block C moving horizontally with a velocity v0 hits the post squarely
iXA. Using E- 29 x 106 psi, determine the largest speed v0 for which the maximum
normal stress in the pipe does not exceed 24 ksi.
11.61 Solve Prob 11.60, assuming that the post AB consists of a solid steel rod of 3.5-
in outer diameter.
SOLUTION
c-iJ*
.75" I"
I; ^C** - 7. 3£££ .V,
^ , ^S } (X, I& , U^K^? s {oloz7 ^
6y Appe*\<4\y "D _> Case I
J* ' sex ' (gMfixio*- X7.3CCZ)
0. 3CS/7 /i
* 3 ° " "* ) ° " w * is t
zz I9G70 i*%/sec

PROBLEM 11.62
11.62 The 2-fcg block D is dropped from the position shown onto the end of a 16-mm-
diameter rod. Knowing that £ - 200 GPa, determine (a) the maximum deflection of end
A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the
rod.
SOLUTION
*=£(#V7 3(<V =■ 5.^7e^^--3.H7o^"V
p^y
? * ^y
I ^ I 5 J»n
c -
d -
%
Appctntfl ix X>
$*tO'* m
C«nse \
■A«,
r 0.6
l*i
J'
ri« Lab
(3 Vjtoo»/6,X3. 2.17*10'*)
-A*
CO-G-)*
#.<936t x|0 ^
03*
Work ©f otv^oppp^ wn^kl W$ Chf y* ) - U)(4.g/^(0.04O + ^"l
- 0.7848 + I^.Cay*
f«j^tTi w^ work <a>tt(4 e^e^^y
0.78H2 + I9.M y* - IWx/o1^1
y*T i
[
= \S.62q x>lo
-a
>V)
/ 5*. & 3 en m
to)
P* " (8.934/ v(oS)^l4T. CZ^x/o'*) r 13?. C6 W

PROBLEM 11.63
11.63 The 2-kg block D is dropped from the position shown onto the end of a 16-mm-
diameter rod. Knowing that £ = 200 GPa, determine (a) the maximum deflection of end
A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the
rod.
SOLUTION
■v-J
c - \ = 8 *>»» * 8x10"^
Omw Ab m «-P«x
u«= $'
J)
S--3
(o.tV
(GX^oxfo*X3.2i7ox<a~'t)
Pl
U„ =■ U» 4 U
t*»
J.
Aa
'8c ~
Use * uAe - ss-.w^'p*
It «l*rtl«*.
Ik* £P*y* - *.«ho */o»^
Wo^rW of dvopfe^ we-^U ^3 (h + y^ » ~ W^-SO^-MO + y,, )
- 0.7&MS + |<?.C$ y*,
E^ftTiwA work an<J energy
0.784$ f H.M y^ = Z.ZZHo*lo y^
(al y^r jL? 2.7825"xlo'"1 + -J (g.7^x<o-5)24 («*K3$"J . 2<is * 10"^ 1
(c^ g- , Jikk r (CWwq-Q __ ,57.6 v IO* ft,
= 16*7-6 MPcv

PROBLEM 11.64
11.64 The SO-lb block D is dropped from a height of 20 in. onto the steel beam AB.
Knowing that E= 29 * 106 psi, determine (a) the maximum deflection at point Et (b) the
maximum normal stress in the beam.
wax 13
SOLUTION
Sy = 9.91 m*
Jf
I
-a-*
= lO^^Sx/O'4, PM
Work <rf -fcJJ;^ wcC^t "W(h+ye> So^O + y^ - lOOO+SO^
.-3
= o
(GO
&>)
yez - /.o^/^93v/o"a - zo.smxio
- O. 145/ i^
PH = (r?57oo)(©.i'«'/,> - I3S3S" it.

PROBLEM 11.65
11.65 A 160-lb diver jumps from a height of 20 in. onto end C of a diving board having
the uniform cross section shown. Assuming that the diver's legs remain rigid and using
£= 1.8 x 106psi, determine (a) the maximum deflection at point C, (b) the maximum
normal stress in the board, (c) the equivalent static load.
SOLUTION
a
2.5 ft
L-&
9.5 ft
T 2.65 in.
{20 in. ^
I - A 0^(3.65y = 2*. 8/3 in4
4
M,-%t»
<aJ* L «
16 in.
. *£* Z.EX o G6X
£Er
*r*y„ - uw y„- —- r 3ei
- (3)0.8*loO(^.8l3)
IX**^ J* QlM VC 11 *f + %o ") J'
7l.i*<r8 y,
U* = ±R*y» * 35.7<?<?y**
Wo^lc of wti-jW - W(h+y*')« 0«6)(2O+ Yn") T 3^00 4 /£Oy
£<ju*4in^ 3Z00 -r /60y^ - 35".711 y^
^ - H. WHy* - $<7.3g8 - o
(al ^=ij 4.q4<i"f + V f.^9<r* + 00(21. 388^) ^ = ll.*?5
*vi
(O
to
T V*. %13,
SZ\o fs,

11.66 The 3-lb block D is released from rest in the position shown and strikes a steel
bar AB having the uniform cross section shown. The bar is supported at each end by
springs of constant 20 kips/in. Using E = Z1 * 106 psi, determine the maximum
deflection at the midpoint of the bar.
H—n
1.5 in.
0.75 in.
fc
fc
To+cJ V- U„ * Ua - UAC <- Uta
SOLUTION
k - Xo V:$%/\* r Zo*toz Ji/i*t
F^ *ri-Mj A, U^iP^i^^Jjf!
For Sp^j 8, UflHP0>-'if ^f
Por+fow AC 0^ L>ea», AC6 M= iR^X
Hk I2£x
EI
1= !& bd3' = ^(l.^X0-75)S t S2.121*i<S% U"
> " ^~ " |0</-1,',8x|0"i P- P-r ^^/x/O* y,,
W0*-k o-F -Fa/i';«3 u/e,"jli4 "W(K + yhft>) *(3)(z+y,J^ £ +- 3y^
E^t-J/nj £ + 3y„ - 4.7871 x l^yj"
y^ - 62G.C9**©"' y* - 1-25338 */o~3 - O
-3
35". 7 v/o"J .;„.
0.0357 ;n.

PROBLEM 11.67
11.67 The 3-lb block D is released from rest in the position shown and strikes a steel
bar AB having the uniform cross section shown. The bar is supported at each end by
springs of constant 20 kips/in. Using E =* 20 x I06 psi, determme the maximum
deflection at the midpoint of the bar.
11.67 Solve Prob 11.66, assuming that the constant of each spring is 40 kips/ia
I—I
1,5 in
0 J5 in.
SOLUTION
M
i
PwiioM AC of Lea** ACS M = sP*X
Hn'L
TJcJ V = U, + Ug + UM 4 UeB ^ ^ + -Bg^£
P* = io. ^tfc/o* y^
U
((HXHoxto*
U - (4r.^7?x I0"c Vlo.*7S6*(o*)*^ ~ 5:^7 8 » lo* y^
Work of Wi
Efu J-mj G + 3y^ « 5. W8 * jd* ^
v^ " SS).7o*tP~c yH - L/o3^v(o"s = o
-3
33.5" x*0 ;*
O. 033.T |vx

PROBLEM 11.68
SOLUTION
11.68 A block of weight Wis placed in contact with a beam at some given point D and
released. Show that the resulting maximum deflection at point D is twice as large as the
deflection due to a static weight FT applied at D.
he-igki h a.t©ve -Hie be**". - Tt»« wov l*
cJc*\e bj +Ke wergild fs
E^ucjf.'nj Work «n<f energy "W* ( h + $*> ) ? £ k>*
J1- k

PROBLEM 11.69
11.69 A block of weight Wis dropped from a height h onto the horizontal beam AB and
hits it at point D. (a) Show that the maximum deflection y„ at point D can be expressed
v S/sl
/A
r
where >>„ represents the deflection at D caused by a static load W applied at that point
and where die quantity in parentheses is referred to as the impact factor, (b) Compute
the impact factor for the beam and impact factor of Prob. 11.62.
11.62 The 2-kg block D is dropped from the position shown onto the end of a 16-mm-
diameter rod. Knowing that E=200 GPa, determine (a) the maximum deflection of end
A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the
rod.
SOLUTION
40 mm 1
Work 0f -PJA^ we*jW w**k - TtTU+y.,)
wher*. U 14 4^c Spring Co^»i^^t Sof A .IoaW afpj'{*e\ *-f poiV7 *D.
"w(k + y.*") * iky*1,
- ^Th - o
H 0.6 m —»-J
J* W J*
-k
w h*i/*e
FW Pw»t. 11.6*
Vi - Me? !*•« T MOx lo" v*
-s
J^ C3X2Doy|O*)(^2l7x|0-''

PROBLEM 11.70
k
J
P«
11.70 A block of weight Wh dropped from a height h onto the horizontal beam AB and
hits it at point D. (a) Denoting by ym the exact value of the maximum deflection at D
and by y£ the value obtained by neglecting the effect of this deflection on the change in
potential energy of the block, show that the absolute value of the relative error is (ym -
yfyy* never exceeds yJ2h. (b) Check the result obtained in part a by solving part a of
Prob. 11.62 without taking y* into account when determining the change in potential
energy of the load, and comparing the answer obtained in this way with the exact answer
to that problem.
11.62 The 2-kg block D is dropped from the position shown onto the and of a 16-mm-
diameter rod. Knowing that E=200 GPa, determine (a) the maximum deflection of end
A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the
rod.
////>
T7%}
SOLUTION
40mmi
whe«r«, k Cs. -r-i><
S»*V*\a
Co* sf <t**f- -for °- ^°"-^ +* P°'^ *&•
fe.*- Work ^ Wh
v1.
C_) Ft"**-*
Ji_
APfn>xi^K soW.oh: V- *3 * (Xi%t\\ r ire* w
r*i *>;>/« «**•* ' —js. &*. ~ "'^
-*
217* m ^
-6 _1-
2k

£^E
1
L
11.71 Using the method of work-energy, determine the deflection at poi
by the load P
SOLUTION
<W Ad M - ffcy » ££■*-
- P»l*a*
point £> caused
To+cJ U = LU + L)
JQc
" g£lLl
lit*.
iPS0= v Sd.^, £gb! i
PROBLEM 11.72
t
l-*X
1 L
Pa
T»UJl U =■ U^
11.72 Using the method of work-energy, determine the deflection at point D caused
by the load P
SOLUTION
i? - - £s-
0\Z£*~ po*-+io* OA H- - Pv
U« * \ Til J* • ziTZ^ \,"J"'
Tali
A8 ■
u
p
sex
3 EX
I

PROBLEM 11.73
11.73 Using the method ofwork-energy, determine the slope at point D caused by the
couple Mo.
Hi
L
M«
h>y
^
AM,
^
SOLUTION
L
^- ims-«
M*
* = m? \*%tk*
I
M =
Mo^
TAJ U - U-0 + U
i M0 9, r U
. MLfcfc>3
9 - 2JJ - M,(fts+^)
DB
n
&EIU
PROBLEM 11.74
11.74 Using the method ofwork-energy, determine the slope at point D caused by the
couple Mo.
SOLUTION
Ove/^ po^ fro* AB
M , - £U*
L>,
M*0-
ao
T»+*>P
U - LL
iM„0o * U
^l=X
GET
SEX*

PROBLEM 11.75
11.75 Using the method of work and energy, determine the deflection at point C
caused by the load P.
y\
-PL
TctJ U
SOLUTION
\z
^(i*-(m - *pv
**> £r
0
Vfcv*
ec u.c= 1%-s^' 3&{VJ*
_ j_ pv
e - 3M - J-EL? i
^c ■ f=> " /6 ££ *
PROBLEM 11.76
11.76 Using the method of work and energy, determine the deflection at point C
caused by the load P.
SOLUTION
S\»t*^^«-T<^>'c oea.'**\ and -fo£ttfJi*h<=i *?a ~ f?e ~ si i
a
2a
pv* + .EL-r^^^^i - 3 p
V
T*t*J TJ * LL* + U,
a a
iac
iPS^U
2-y sy Mt^effw
.a PV
_ 213 _ .3 Pa*
7v - n - J-pv
X--

PROBLEM 11.77
\*
11.77 Using the method of work and energy, determine the slope at point B caused
by the couple Mp.
Mo SOLUTION
Y\ - RAv - - %
ft- i.
Owe portion CB Dca - V "Jex *
^ M„©8 ^ U
3?
ex
PROBLEM 11.78
M
11.78 Using the method of work-energy, determine the slope at point A caused by the
couple Mq,
SOLUTION
*s L
\/
T*+*JP U -" U,
tm09, . a
K
M * ReV * i_ -
<W AC U^ i'lfe^
**• EX
a
' 3A EI

PROBLEM 11.79
11.79 The 12-mm-tHameter steel rod ABC has been bent into the shape shown.
Knowingthat E - 200 GPa and G - 77.2 GPa, determine the deflection of end C caused
1... *u„ i et\ m ii—
by the 150-N force.
SOLUTION
J * tC - K^)"-- 5.03^8 */03 mm
-i
- 2. 03£"8 * |o'n v*
I * £ J ^ l.oni kjo-
Vvi
^
*■
P = ISO N
P
Po^+io* AB : bending M =-'Px
rLAB mz ■pi r*"*8
- P2 Us3 _ U5Q)2 (Zoo*to-*1S
GET "(e)^OOK|0")C/.0/7^x lO''*)
HM
£
€
?L
&c
ec ^-AQ
J5n
ftP
-*■ X
E
2&T ~ Z&3
(2X77.? x IO * )(?.o35"8 * /O"1
0.S726S" J
Po^Tio* B<2 : M - - Px
= Q.t473£ J
i, L Lpc -M C 2U fi?K0.8< 757)
-i
= M.£7*lo"*m
II.S7 ^ V

PROBLEM 11.80
11.80 Two steel shafts, each of 0.75-in. diameter, are connected by the gears shown.
Knowing that G = 11.2 x lCpsi and that shaft DF is fixed at F, determine the angle
through which end A rotates when a 750-lb-in. torque is applied at A. (Ignore the strain
energy due to the bending of the shafts)
SOLUTION
Work- e.iae^#y e^aTion
i
PorKox A& of ikaft ABC
2H
IA« Ly»3
Oaq r 5" VT J " 2 l.~5~ ) " 3t.O£S*|D iv
(7SO)A0\)
= 8.89^ ^-J'fc
^*6 " 2GJ»a ' (2)Ol.awd*KSI.061v7o"s)
Porfio* BC of sUff A8C: Dgt = °
G*ar E Te = reFu - (4^ Iff©} * /OOO A-in
Fs*= -5s * t1 r ^ = ^° -"■
?«■+■"»* D£ of skrft t>£F.
Umr = o
UfF *
ToUft : U = U,B + L>B< + Upe + Dew
(L4<?7 Am'm
<ft
^ r (Jl^ii) . ^.^-^ * 5./,

PROBLEM 11.81
A
200 mm, "
11
4 a. *U-
11.81 The 20-mm-diameter steel rod CD is welded to the 20-mm-diameter steel shaft
AB as shown. End C of rod CD is touching the rigid surface shown when a couple Tj
is applied to a disk attached to shaft AB. Knowing that the bearings are self aligning and
exert no couples on the shaft, determine the angle of rotation of the disk when TB = 400
Nm. Use E - 200 GPa and G = 77.2 GPa. (Consider the strain energy due to both
bending and twisting in shaft AB and to bending in arm CD.)
SOLUTION
t)5"H
AB
= O
T F = T
^ = 3^^o-» r ■ SS3-S " v F.TISS8.5M
oeiad i't-\<H oT v^ot* CD*
S^h/o\^=- 7.gSH*/d't w/
S.0<?3 J"
F0a
Fa-
u =
BewJin^ of sU-^-f ADB
t)ZM^O -Fi.Ua. + FD b= O
t)^M, - O + FAiA0- FDb *o
V Ifo fHoo^^QO^o'1) c I3.|<?4 J
X&S * (zK77.?MO«X»r.7*8i-to-*
T*4-J: U" - r.09S+ O.I37+ is. 194 * 18.4^^ J"
£T6<J>e
u
% =
1^ M-O©

PROBLEM 11-82
TcfoJ
P2ol
11.82 A disk of radius a has been welded to end B of the solid steel sha&AB.. A cable
is then wrapped around the disk and a vertical force P is applied to end C of the cable.
Knowing that the radius of the shaft is r and neglecting the deformations of the disk and
of the cable, show that the deflection of point C caused by the application of P is
<?c =
PL'
3EJ
1+15
EdL
2\
GL2j
SOLUTION
T-
ov'sion
T- Pa
ut 2GJ"
DefiellM
3/
M =
*-w«i
P'a'L
ZG-J"
Pv
?*L3
cex
JELL (i
GJL*
)
^ 35r v.1
—I
t <SL*

PROBLEM 11.83
11.83 The thin-walled hollow cylindrical member AB has a noncircular cross section
of nonuniform thickness. Using the expression given in Eq. (3.53) of Sec. 3.13, and the
expression for the strain-energy density given inEq. (11.19) of Sec. 11.4, show that the
angle of twist of member AB is
TL rds
where ds is an element of the centerline of wall cross section and A is the area enclosed
by that centerline.
SOLUTION
'PO*» •■)U6,Ti't>M (3.-53^ 7? - 2+~A
St^c*-1^ enev^sj ere.* % i't y
u =
X&
SGz^A
«• Ai
Wov-k of T»r^e -" * T <? = j'j\ 4 T"
tzl
2GA
-H
is
8&A!
9r
MGA1 J t

PROBLEM 11.84
11.84 Each member ofthe truss shown hasauniform cross-sectional area^.Usingthe
method of work and energy, determine the horizontal defleetion of the point of
application of the load P.
SOLUTION
Me^Le^s A1S <a.*o«J BD aim. &/o -fe^c v^ei*^i>€^i,
'ToiviT A
F^'
3oi*i t> +•-FF* = o
o ^fP-Ft0 = o
h*-*
d
U= ^& Bz^^F,L
, 22 E£.
It. E]T
AI>
ZFy = o
+ T ^ Fy = o
- Fie - 3 £/i p = o
F„ - - f P
Mtw Le^
A6
6D
AD
CD
AC
£
F
0
o
-*p
p
3r
L
F*L
o
O
P*J?
?"
£2
PROBLEM 11.85
- if. £LL
l/.*5 Each member of the truss shown has a uniform cross-sectional area 4. Using the
method of work and energy, determine the horizontal deflection of the point of
application of the load P.
SOLUTION
Mew* be«^s AB^ ACj.o.J, C^ «■** ?e/u -h>^c« vr>e-lfl«^*.
Jol'wt 8
Be
F..
4-ZFy =o
P - * F«c = *
+ T TFr- o
- Pi* - # Fie = °
F- ' 5 P
BC
Fa» : " t-P
lEf\
IFY
YlorV ,? P = iPA - U
a P 8 £A *"* £A
Mew* be*
AS
AC
CD
BC
RD
X
F
O
o
0
^p
-4P
L
*
\t
a
W
3*
F'L
0
o
o
fp^
■iS-p^
"jPV

PROBLEM 11.86
11.86 Each member of the truss shown is made of steel; the cross-sectional area of
member BC is 800 mm1 and for all other members the cross-sectional area is 400 mm2.
Using E = 200 GPa, determine the deflection of point D caused by the 60-kN load
shown.
D 60kN SOLUTION
U.5 -
o.5~
1-3
EL * o
CD
60-^-rs F--o
r^* 3^5 JcW
F.a = 3*.jrkw
FBt- \Z.S kN
+ -IF, -& - PAC+ £?(w.s-Vo f.t - so fcu.
Ac ^ |."S
TJ -- 7 -E^- - J- 7 £^
u c %Bk - ze L A
MftwnLft^
CD
BD
A8
BC
AC
rCkW)
37.5
3o
27.S
l?.S"
3o
L ^
U3
1.2
1.3
O.S*
1.2
AOo*'^
Ho©
Hoo.
4oo
300
4do
F2L/A (M%0
3.4333*10ll
3.7 *iOA
2.M32S xlO1*1
0.0^71 x|0^
2.7 * 10"
n. 36 33*lt>,t
u*
12.3633* JO
r 30.908 J
4-PA - U
A . gj± . ftXSO-^08)
L0$O*lO r>j
LO30 *»** -»

PROBLEM 11.87
16 kijis
«Joi'f»t C
Zo
Be
Pc
u*
41H7*
11.87 Each member of the truss shown is made of steel and has a uniform cross-
sectional area of 3 in2. Using E = 29* 106 psi, determine the vertical deflection of the
point of application of joint A caused by the 16-klp load.
SOLUTION
412 hy -" O
-/fc -.gF* = O
15c 4 FAa - o
U - Z
2EA
^FA
ZF*L
F*l_ (k.y.iO
Fit * - to fcp*
F,B - 12 k.ps
Fg,c " 16 Kipi
A* 3 ,■*"■
4/H72
Q.43S34 k,'p. in.
IC

PROBLEM 11.88
11.88 Members of the truss shown are made of steel and have the cross-sectional areas
shown. Using E = 200 GPa, determine the vertical deflection of joint C caused by the
application of the 210'kN load.
1.5 m
SOLUTION
1.5 m
210 IcN
at
* O
JoimT B
(?.
LL=2
_lp _ jip
RAC = I"75 kti Ffc* - I7S icW
R
AS
-vTZFy = o
ns-wv Fft6~(i)(l70-- o
2EA
Me**V»e*
F2L/A (NVr^
M.SW7* to'*
133, *<m * *>'*■
A. =
i3a.8yt3»fo'v
33M.75 J
-3
= 3.19^/O'rv,
3.1*?
w»n

PROBLEM 11-89
11.89 Each member of the truss shown is made of steel and has a uniform cross-
sectional area of 5 in2. Using E = 29* 106 psi, determine the vertical deflection of the
point of application of joint C caused by the 15-kip load.
SOLUTION
Me^be^-a 8D AmJ AE an** 2&ro
Ft»r G^4-t>e +/Vss O MA -O
RD t 7* k;p*
Fo/- e *i u
J*of*iT C
Mi fe/i L/k^ £>f IOI \r\ T CI
F_ = - f?D - - i*. k.-,
ED
P5
4i rP -- o
■4->Z F„ =o
F^ T
■gF^/5-°
is ki'i
-2.5T
F^ = - 3<* JfVp*
6.5 ^ " ^Bc = ^
6.5
Jo.^-r t>
39 It-B
S+
t^Ano e^fce*
3^
U,r 2^ %FA * ^a ^ L
"Fka + ^iac =■ o
Me^be^
A6
BC
CD
DE
QD
AE
AD
1
F (16^
3C
2>G
- 3<*
-7*
O
0
3<?
L (;«)
73
72
78
7*
3°
3o
78
FlL(»e.y-iw,>
<?3 3i^
«3lV
II 8633
373 248
0
0
U8CSS
717 14 8
D*k~ £r zi*ios feSl-
Ur
7371*8
J.
a.
*A. = U
_ 2U,
(2X*.T**8)
15
- o, 3C£ ;*,, i

PROBLEM 11.90
11.90 Using the information provided in Appendix D, compute the work of the loads
as they are applied to the beam (a) if the load P is applied first, (b) if the couple M„ is
applied first.
My.
K
-I
Co.) RV-s-f •?j +1
P A
e*>
M.
SOLUTION
F^oi* App**j>y X*j C*.se *
^p
PL*
3fi
a
'AP
Fro»* fafpe^w Dj Ca5^ 3
X
AM
PL3"
M**-
©i
- SILL . P/ioL21 , Ma'L
£«■ £/rx 2£.T
(fc>) F.vs4 M0; +*e-i P
U = A4 + Ac 4 A,
=■ AP
M„&
i
a.■ y*p *' 'o^ap -*■ 2
M.a
• *—'Am
= £11 , M.PL', M'i.
£^x ^fx ^ei

PROBLEM 11.91
11.91 Using the information provided in Appendix D, compute the work of the loads
as they are applied to the beam (a) if the load P is applied first, (b) if the couple M„ is
applied first
SOLUTION
Ap^cnJi'x X> Cases I a*J 3
PCL/z¥ _ PL*
(aO Firs+ ?^ He* H,
^
BP
3EX
2^ EX
*** a ex 2 ex
yBM
M^Y r H.L'
0,
Ffl
2FX
M0L
EX
g£X
U r A,+A2+As
p H
^bh-^Yb^
U - A*+ A5 + Ac
^8£I
8 El ^ei

11.92 Using the information provided in Appendix D, compute the work of the loads
as they are applied to the beam (a) if the load P is applied first, (b) if the couple M„ is
applied first.
SOLUTION
Fi^o^ AppeneJi'x T> > Case H-.
1 , ELL
c$a= -
Fl3-
IQ£T
MaL
3EX
(CO Fi^sf P, tkeui M
1*^CP -*!•-&»
U r A, + A, + A
<?££! 16 £ I 6 HI
I* O^p "*r* 8^ -*1
U = A,, * A5 + /4t
_ £11 H.P11 , m:l
' ?6fX /6Fr 6 EI

PROBLEM 11.93
11.93 For the beam and loading shown, (a) compute the work of the loads as they are
applied successively to the beam, using the information provided in Appendix D, (b)
compute the strain energy of the beam by the method of Sec. II.V and show that it is
equal to the work obtained in part a.
SOLUTION
See
%e = G
Lilfc wise
Co) U.bei i^e -Forces VD a.*J ?e.
3EIL SeiL *56 CX
'EO " 768 £1
^— ^)o —J* Sdf -»l I^Set -J*— Scf
Le+ PD Le edited -&>*+.
U - Al+At4A3
U^ ■i?oS»*'R,S0* 4^-RS* ^ ^ 41^ * 5tt B*&L* +
^1 5
2 ■ D <3dj>
00 (?e&.cTtOuvS "RA ~ Kg
a!4
D c?D£ -* 2 'E^fif " £-ta EX r 7t* EX TJT'? CI
LN
<|3
w ex
p
Ove^ por+^ui Ad o<x«= ^- M - Px
3*H EI
Vl-V"
U
*£rx
jai>
PV J...A -
PL
■rt er

PROBLEM 11.94
11.94 For the beam end loading shown, (a) compute the work of the loads as they are
applied successively to the beam, using the information provided in Appendix D, (b)
compute the strain energy of the beam by the method of Sec. \t.H and show that it is
equal to the work obtained in part a.
SOLUTION
M Ufcei He forces ?B *«<! Pc
Z4 E"I
<; - ± JUL:
0=iPS,tp<; ,iPc , iiks4£ PWP« L* + J. Bit!
W.'H Pa - Pc ' P
EI
*l 3
Hi
TJ - (A , £ + i) P!L3 - J.SH1
2>t EX
0
\te<-
BC
ZtTI
M« Pv
TJ,
SC
•$.'#* •iSt'^-^-ift)'
2£T
4* ex
«i» 7 p^La
T.W U r UfcB 4. Uac r (j^sfe)^ ^ £
£X

PROBLEM 11.95
11.95 For the beam and loading shown, (a) compute the work of the loads as they are
applied successively to the beam, using the information provided in Appendix D, (b)
compute the strain energy of the beam by the method of Sec. It.¥ and show that it is
equal to the work obtained in part a.
SOLUTION
(Cl) La.\oet He coJpJeS MA a**) Mg
Appendix D^ C«.se ~1
3er
C©«*
ce.
_ MeL
K® get
H - ± M ft 4. M ft i J- M A -J- Ma*L a. J- MuMaL . _L J^
tL
% EX
U

PROBLEM 11.96
11.96 For the prismatic beam shown, determine the deflection at point D.
SOLUTION
A«W Wee Q
_ 3V _
Qveir •po^Tto^ AD
0\/e*- po^T tow D8
Set Q.= O
0
"I
M 3M
Jx =
ET
4 n 3a
J*
M = -Px
2H -
sxa.
- o
- /x . x . -l 4 x \£il - .£.£11 ^
■ 13 2H H 4 t6 JEj ' *f8 EJ
PROBLEM 11.97
11.97 For the prismatic beam shown, determine the deflection at point D.
SOLUTION
fa I
Add We Q cA poin+ D.
ill
2 EI
r1* M*
^
do. KX ft '
Otfe* p&r4ioM At) 0 * X * *£ H = - i***
- X(x-*->- _X4x\J*L!. - JZ_ £i! - Q oqtir? wt!
~ *l* 64 € H2 ) EX 38H EX ' v-VW*' EJT

PROBLEM 11 98 11.98 For the prismatic beam shown, determine the slope at point D.
SOLUTION
Add coupie \i0 o.i pom-) X).
u
- s
u Ma
J>
LM 3H
D8
5Mo " Jo £X 3M«
M = -Px
Wx
0< x « X
^ < v* L M^ -Px - Mo
Sef M0=o. 6b- ~L(-PxXo^v 4 ^iC-PxX-O-^
and - _ i
PROBLEM 11.99 11,W For *ie prismatic beam shown, determine the slope at point D.
SOLUTION
Add coupJe. Mo o\ poiV>4 D.
u* i
^x
SM-U " ^ EI 5MD *> ~ EI ^ StM,1'
M - -i«/yl
£!± - O
CVe^ forfiow At> 0<X-= J
0\/e^ p«K-+iow DB ![<*< L
M^-±*^-M. |&* -■
>L3. 7 iW.* ^

PROBLEM 11.100
a
M,
I
■a —*+*-
11,100 and 11.101 For the prismatic beam shown, determine the slope at point A.
SOLUTION
AeW CoOpJe H^ «.*t pot'*f A •
Ow^ pBr+.-«.. DB (0<v<UNl M - f?8v - .^ * tp* } f*t -- ^
Set M,= O e,r ^^(^k» V.-.f )Jx-.- M0b' %^f «»"
PROBLEM 11.101
11.100 and 11.101 For the prismatic beam shown, determine the slope at point A-
SOLUTION
QecJc
- Ho
QGIL
-t (3La* - 2a3<- £L3 ) ^

PROBLEM 11.102
-uv —»
0^<>/ Oo<r\<o*\ AD •
Q\/e\r po^iYon DE '.
11.102 For the prismatic beam shown, determine the deflection at point D.
SOLUTION
w
0\/cr porfi'oh E8 :
s
M = -^wu;
EX
• 30.
S = O
3Q
3Q
J_.
768 EI

PROBLEM 11.103
11.103 For the prismatic beam shown, determine the deflection at point Z).
SOLUTION
A«U -rWe Ql Jt poM t>.
Pa. I*
48 EI
i£ = -±v
3GL
_ r Po.Ll
Ove^ por-h'tM EB
2¥ Si
M = - Pu
aw = o
M
*>Q
EB, = O
^» * 90. 3d ^ W«3 2<**I '« £1 ^

PROBLEM 11.104
11.104 For the prismatic beam shown, determine the slope at poiut D.
SOLUTION
B k«U coupJe \A0 a4 poCw+ D.
GWU^s: 1?A- ^ , R^-wL-j
U = UAD + UDE 4 U« ,
L
P"U
SMo
= O
Ox/*** f>o«~4»o* AD: M= ^°x = o'' wi-+/» Mo" O
^ W wa W + ti / Pi ~ ««a
Owe^ poirT'iow Eft*. M =" - *
2" - o
3M0
3M<
= O
3M,
3M,
3 ft
w£3 ^ .i-**1
3«HfJ 4° " J8** EI

PROBLEM 11.105
11.105 For the prismatic beam shown, determine the slope at point D.
SOLUTION
7 M*
-o SIT Jx >
M - R*x -- -^x ■+ -^x
Over po#-h'ow fcO : Uto =■ \o ^__
i
_ 31 -
Po-L
24 el
2W r -fv
M- Rs*-P(**0 = f^Alv-^v/ - P(*#^- ^v-Pa
'» EX
(W po^fjo* EB: Mr -Pu
3M«
3cVp _ 0
3M<

PROBLEM 11.106
|8kN Q <f lSk-N/ra
■2.5i
11.106 For the beam and loading shown, determine the deflection at point B. Use E
= 200 GPa.
SOLUTION
W250X22.3 M . . r- h it 0 _Ll
Over AB
M = - Sx
EI " ('2oc>x(0,*)(2S.<?x|O't,) r £.7$ * 10* tf>»sx « £73o fefxJ-m**
PROBLEM 11.107
11.107 For the beam and loading shown, determine the deflection at point B. Use£
= 29 * 103 ksi.
1.5 laps
1.5 laps
SOLUTION
W8X 13
LW+»: -forces m^'pii iewj-Us i*H
E r WytlQ* V%i I ^ 3?. & ■Vi*
EI - (^"'^X^-O *" 1.1*^8 ^ IOC k-p./n1 = 7<»7S kip-ft*"
CW AS: M-iSx, Is r ° i^Jsr^'0
CW 6C : Mr -i.ff(v + 5^-l.5v/ - Qv * -3v - 7.S-Qv^ |g = -v
^M^, * \S(3v*+TS ')J* - (3Xils)% + (7S)(k-)(sf- H8.7S

PROBLEM 11.108
11.100 For the beam and loading shown, determine the deflection at points. Use£
=*29x KPksi.
SOLUTION
W8X13 IW+a: We* i* fe.ps, ienj+A* .* -ft
EI =■ (Mx 10*)($%£) - I.H3"/o4 fc.'p-i»* = 7*7S kip-ft*"
u- £
10
Ml
d>
Over por+,w. ^B o<x<^ M- -/.<r* - Qx
aw - _x
5*M^^ r 5/t'-^^^y - I.5^X\JK *(«&)(sf r C3.5
CW porfioM BC -S^x-io M^ - I.S* - /.S(x-£) - Gx
K * -3* + 7.5 - <3x ^S r ~*
fMg^, <;W-7.6~xU* = (3Yiy/o3-sn-arY^y/^5M
§ftr -L^ G2.S v£93.7sl = ^fP~ T 82.11*10'* ft r 0.^7 .h. 4 —

PROBLEM 11.109
160 kN
11.109 For the beam and loading shown, determine the slope at end A.. Use£ = 200
GPa
W310 X 74
SOLUTION
wff
™
* V
Unils." forces iy\ kWj ifcw^+lvi m tvt.
E* ZOOhIQ* ?a_i I: | £5 x /0 * ww*4 - Ififf * /O"* Vw*
EI^C^oo^/O^Xl^K/O"6) = 33*10* W'*^ 33*>0 ktf.*?|
fofc&tiohS : R4- 20-
■2.*»
(?«= 804 ^
-*.*
U = U« + Use - )o 3S* + Jo ^x«*/ ?9, - ^ -- — + -j^
Ove^ AB:
Set *V°
= i5
■H,
M " M„ + f?Ax - M* + Sox -f^x
J.*
3H,
- ^{WoXi^^f- Occui^\>.^ r
153.6
£X
M - f?flv r go^ + ^-i/,
2M - -J_v
EX
(I6.6667)(2.«Q3 _ 7C.3
3 £X ' £T
5* = ifi1"-^
76. g^ r
zso.t
6.98x/0-3 ir*./. "J

PROBLEM 11.110 77.770 For the beam and loading shown, determine the deflection at point C. Use£
= 2y x \v KSl.
j.Q, 8MT>S SOLUTION
JL UfiV+s- Forces i*o kxp j Vc**-Rs in "Pf.
S8X18.4 E=2?w^k5; I= j^ <n"
£1 = (W»lo')(S7.6> I.C7e>4M06fe:p.iV = I/Coo k.>-"ft*
u = il, * u
't©
u,
§= =
_ 27J . 3U,
3U.
3U
££
't>6 <* - s>a " 5q T 5>a T -3Q
Ow AC 0**<3 M--(4*iQ.>x ^g * - 5: x Set Q =■ O .
31L
^J(4rt^A . xIVa ^
£1
SO *
3£
EI
0*e* D8 0 * u < 3 M ^ - 8u i£ *
S>U
Sc- £+§I+°
St
tttoo

PROBLEM 11.111
8 kips
ric D
B
11.111 For the beam and loading shown, determine the slope at end A. Use E = 29
x 103 ksi.
SOLUTION
S8X18.4 £ r ^h/O* ks.-, I =57.6 ,*„*
EI= CWWO* )&7.£) = L&7oH*loa lc*'p.t«*-s IU60 k^--Ft*
TT - 7 r , ti r A - — - ^M 4. "^p*
S)Mi
El-.
EI
Ge* ' EI
+ o
- **
M6ftto
^ ^O
9KI*
3ZJ
?M-
DB -
- o
-S
3.07 x/O" ***

PROBLEM 11.112
11.112 Each member of the truss shewn is made of steel and has the cross-sectional
area shown. Using E = 200 GPa, determine the vertical deflection of joint C.
1,5 m
SOLUTION
1.5 m
U He Vez-rrcftx A06.A P. Ti,e v**-WcgJ
E *■ A 5>P
c - 2S - i T-Elk
5>P
-3**-*^-- o
f*Z 15
O
*a
Ms**te»
A8
Ac
0C
?
F
±P
4P
-*P
2F/aP
z
6'
-r
LW
3
z.r
2.5"
JVflrfV)
A 2 00
I20C.
|goo
FC^r/3P)LM
G*5 P
H4&.16 P
96^. 5*1 P
30ZC.Z1 P
v i(—p)^ (3^:^°"Q° - *■«"
o"3 »*»
- 3. n
Wi»\

PROBLEM 11.113
11.113 Each member of the truss shown is made of steel and has the cross-sectional
area shown. Using E = 200 GPa, determine the horizontal deflection of joint C.
1.5 m
SOLUTION
1,5 m
CM 4-Ve w*fc.JP We P. AJU| *
<i - 2H - 2. "5" Elk - J_<7 FL 2E
Jof«f S
Pin*
+frFj» o
Fw +
fFSt =
I Re - 3 F^ + Q. - O
0
Be
F>& r ~ S" F*. r *P~fQ
Metwtcir
AB
AC
BC
i"F
ip-fa
£p + #Q
-fp+ffc
s>F/3a
J*
u*y
3
2.5"
2.S
A(lo'V)
12oo
I80O
-H(>2JS P
I 08S\ 07 P
- 723.38 P
- |o7. 06 P
?OOK lO'

PROBLEM 11.114
11.114 and 11.115 Each member of the truss shown is made of steel and has the
cross-sectional area shown. Using E=29 x 106 psi, determine the deflection indicated.
11.114 Vertical deflection of joint C
SOLUTION
C*JJ +Ke >/eA<coJ 9ou*\ P. The v/e^f.'C,J
Geo^ef^j 4C - A1^ 3.75* - 1.7S" -ft - m .V
4ff * i/g ,*«. ^ 5f+= CO ;* 3.75 ft - ¥S"iVi.
<Xe*!rtT C
+ -FR-- 0 - ifi?. (r .
W7
K.
Go
7? **<*c
r„
- O
ff I^O
U7 **. 75" ^ * °
SoA/i'ng &imoti**eousJy FAc - 3.«P; FBc = - 3.75" P
Joi>*f 6
3.75 P
+ -*R**0 -F,e - ^FAC r o
75"
*a
F»A ^ - 3.oo P.
It?*
Member
A8
AC
BC
T
F
-3.WJ'f
3.2SP
-3.75P
5F/3P
-3.*>o
3.ar
-3.75
Ufh)
48
1*7
75"
A C»2)
H
Z
G
F(PF/^P)L/A
102,00 p
c; 7. «n P
175".78 P
</<>/.£<? P
s» -
qoHH P
(7© 1.61 )(7.5-«<o3')
2<? * /o6
O. £33 In. 1

PROBLEM 11.115
Jo I'm + C
11.114 and 11.115 Each member of the truss shown is made of steel and has the
(^ross-sectional area shown. Using E ** 29 * 106 osi. determine the deflection indicated.
11.115 Horizontal deflection of joint C.
7.5 kips SOLUTION
P. A,W
„ „. do*****-) iW«l Q oJ J<*i*i C * Tkc n^i'io"iW
3.75 ft -* J
^ pq ^a ^- Xi?A £ ^ A 3Gl
Geo*<4rsj AC * ■Jq'it.lS*' = 9.75- ft =■ 117 ,Vi
F*.
^
a
Be
4- ^F,= 0 --Jjf-FM -JgFi, +Q„-o
Fee * -3.7s-P - I.S6ZS a
(Jo/mT \->
^6c
Aa
= O
R
^F,=o *F*. - F„
Mcmte^
AS
AC
6C
2
F
-3.00 p- i.as'-Q
- 3.7^ p-/_5<^ro
-3F/3Q
-/.7S"
2.H37S"
L 0W>
4*
117
75
Atirf)
F(3F/3a)L/A
^5\00 p
^63.43 p
73. *4 P
S% 1.C7 P
s, -
S8LC7 P _ (^g|.67))(7.S-NlQ») _
O. »504 in. -•

PROBLEM 11.116
1.2 m
1.2 m
11.116 and 11.117 Each member of the truss shown is made of steel and has a cross-
sectional area of 500 mm2. Using E = 200 GPa, determine the deflection indicated
11.116 Vertical deflection of joint 0.
SOLUTION
Final T^e -fe*^%4-h of ca^-U k^twoe^ AS Snowvv.
e
FtE = &,-o M
Feo - -3.6 l<w
^F^ +- f F*D - 3.6 = o
+-2F* = o
s
j.
c f F«0
FeD ~ - 1.7S - 0.8SSS GL V\i
+ tS F^ = o
f F«o +.FM, - o
R.
f>o r -f^eo - <-°S + °-^
Member
A8
AD
BD
BC
CD
Z
,F
0o**n
6.2S + 0.8335 a
I.OS t O.SQ
- I.-7S - 0.33S3 Q
G.o
-3.C
3F/3&
0.833*3
o. s
-D.833*
O
0
u
<*0
2.o
2.4
2.0
l-ff
2.5"
wi+k Q- o
F(9F/5aU
Clo* N/-w">
I0.HK7
1.24
2.<ilt7
0
0
(f.593
§« = efc 2"F^f/5q)L ^
H.S<J5 * to*
(Xoo k^o** X-5"°o * '£>"* )
.-£

PROBLEM 11.117
1.2 m
11.116 and 11.117 Each member of the truss shown is made of steel and has a cross-
sectional area of 500 mm2. Using E = 200 GPa, determine the deflection indicated.
11.117 Horizontal deflection of joint 8.
SOLUTION
F*i'ncl t-«« * «.* «*V * ot £Acfci wie*-.bey* a.s sUwi^,
1.2 m
EA
^F
Joi'wL B
Join t C +1 7£y= o 1Fco - Y-8 * c>
Fo - -3.6 fer/
1 4-8 kW
+ - ZF„ = o
S n*
Fao - 3.c - a = o
5 r»0
I Ee - | KD - 4. 8 - O
Joi'nt "D
Fto * - 1.75 + 0.62SQ kW
F<
6D
Ad
F»h * - * F<
5-1 ap -
51 Fep + fM - o
K05 - o. 375(3
VI cm I e/
AB
AO
BD
SC
CD
z
F
lo* V
G.Z$ + 0.«SD
/.o5 + 0.375Q
- UTS -t £>.65SG>
6.0
-3.&
3F/3Q
0.625
- 0.37S"
0.£*5
O
0
L
C^")
2.o
2.4
Z.o
1-5
1.S
rT(^F/3Q)L
00* rJ*w0
7.81X5"
-O.'Wo
-^.IS75
O
0
*f.68b
s8 - J-rFCpF/?o-)L -
t ft
4.6SO HO*
{7.oOx\o*}(Soo xfO'* )
= O.OV&-8 Mim'

PROBLEM 11.118
M1.118 For the uniform rod and loading shown and using Castigliano's theorem,
determine (a) the horizontal deflection of point B, (b) the vertical deflection of point B.
SOLUTION
2H-
- -JfcsiViCp
GL Add d^mmo Jve^A O. d poiVt B .
M = Pa •* Gib
r PRs.Vk? + Q.R0 - cos 9?)
(a.^ SA =
n —* f*
DQ * £1
_ pgi
' Er
} (s.'« Op - Si'M<p e~s^") J<p = -^=- (-cos 9 - ^ &«w*<p)
^S? (_ c«s 5 + c^ ° - i s'^ J + i s;,°tc))
ET
fig
£1
(o+- I - 5 + o)
J.££i
* EI

PROBLEM 11.119
11.119 Two rods AB and BCof the same flexural rigidity EI are welded together at B,
For the loading shown, determine (a) the deflection of point C, (b) the slope of member
BC at point C.
SOLUTION
i2c = p. *■ a * Mi
+*^ F* = O P + Q. + (?to = o Rte - P + Q -
Me* bo AB*
§£ * hi M&A = o
Me^ke^ BC*.
M - Mc + r?cx- =■ Mt + (P + a + ^* \x
3d " * -»
3M6 " '7
M3-
rf*
3<0 ■ Ef J J ' 3« *
ft
Se.T Q. - O avw4 Mft * o
9
M " &*>£«*■ - £S>10-*)a - £$> - fU
EX
(a) D JW;<~ «f C Sc = g* + |g
3&
a EI
6 EI J

PROBLEM 11.120
11.120 A uniform rod of flexural rigjdity EI is bent and loaded as shown. Determine
(a) the horizontal deflection of point Dt (b) the slope at point D.
SOLUTION
ftefcrfrows 4 As 1?^Oj PAm = P*^ M„* Ma^
Hester AS : M - MA + f?Ay - M© + Py
■2£X
3MW " ei
Mentor BC M * MA 4 (?^ = f^-f Pi
5e+ Mp= O
2M - 0 3L£1 - i
2U* . J-f'nltt.L = -Lf^ppNf.NJv
3Md " ET~*
Me^fcer CD M = M„ + Py
- ££!
EI
5c*f M„ = o
2£1 -
* ais. suet 3Uj
^P " 3P 3? 5V
°» ^M, PM0 sn* ^ ' a ' CT
EJ

PROBLEM 11.121
11.121 A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine
(a) the vertical deflection of point D, (b) the slope of BC at point C.
M
SOLUTION
AeM d l»i*>»vw Twee Q a.4 p&tVtf *D ft.*«J Wu«*i n*y
Rc*e+.'o«s ** A : 9Ah'- P*- ^ 9Ay = Ql
3
^1T
2*1 = x
36) X
2tL .
■3Q
ET
1^*£{>&*-£L'<»^-If
3MC " £T
Me^Le*- CD M * P^
2" .
SO
= o
2l±
= O
5U* - 0
3Mt
- O
ion of poj""iT D
0.) 3^p< rf BC a C
P£!
EX

PROBLEM 11.122
11.122 A uniform rod of flexural rigidity £/is bentand loaded as shown. Determine
(a) the vertical deflection of point A, (b) the horizontal deflection of point A.
SOLUTION
Ove^ AB M* iPv + f Qv/
W = *v
•a.
2KL - J*
2
Se+ a = o
^ - rr(M &* ■ ilu^Xi^)^
" 'a EX
^•-^iC^^-^V^^
\* El
- o
(M V»o^i2*>*T*>T olerJefrT\oi^ erf" poui4 A
2&
151 tJ
er

PROBLEM 11,123
11.123 A uniform rod of flexuraj rigidity EI is bent and loaded as shown. Determine
(a) the vertical deflection of point B, (b) the slope of BC at point B.
SOLUTION
0>/e^ AS M - k Pv
2*. O 2* . o
3Q * PM«
213
3Q
!&& = o
2U
3M*
2* --*
dp ' >
2fX
# • if Mtf A - i$>-iV<* - fe[*-(W*]
- ft £j
3Ufc _ _l (L
3M„ EX
CM&A -^f P(*-*Vfcc = O
A - ilia . W* . r>

<4A
PROBLEM 11.124
1
S
V^-1R,
11.124 Determine the reaction at the roller support and draw the bending
diagram for the beam and loading shown.
moment
SOLUTION
Rei^oi/C. Soppoft B aj*JL adlA .rAeA.cTVo*i £a AS A. JfouJ,
30.
k
'2E3. X ft!
aH €X 48 EX
Ove* CB ■- M - ^aV
3<?tt V
ijW''''*.&&?"***
3ET
3
3tU . _L
J6 " U4 »*/ EX **8 ^t

11.125 Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
-^ -M*»}&'"°
5.FI

PROBLEM 11.126
11.126 Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
£
fib
SOLUTION
Revwov* So#po/-T A awd cl^X f«a.e.Ti'ou\ R* AS A xoft«J.
Por+iem AD 0<X<« M= RAX §^ * X
S.-
30,
3fl,
R*
_ ■■> M(L'-a.') _ iltULta) ,
L*
MA = o

PROBLEM 11.127
11.127 Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
*V
w
f
i_t_k
lL_t
SA -
o
9P.
Po^tfoii AC; o<*< % M = «?*» i& * *
^3 24) ft .,.* 7r
3W EI
c-.2Us.3tU x£ai! _7_^-o
** ' 3P* ^T r 3 EX "38H EI
K**lk«t~ t
<W AC M"= ]jj w/.x
Mt r 5^wLl =• O. 0273* i*//.1
CW CB M • ^wtCv*^ -i^v*
us
12%
- - 0.07031 wL1-*
7v '« ""
v^r^L
Mfc- ih^L(ih^k)-i<~h^
■^^-wL1 = 0.028SW wL*

PROBLEM 11.128
11.128 Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
Sa
i
%
SOLUTION
Remove support A and glA& trece+ia* R* «s a. JoiJi.
s - -2S -
po^fion Ad o«3*«£ m-rax IS" x
I>
3Mi .
</y *
EX
{texttO-ly
O.I7M PL
£(?,» EXJD ' '3fiA
~ 3fx ^J "si ex
Po»+U* DB 3 < X * L M- (?fcx - Pf*-^
&e*Jti«<$ *»o*eJs. Mc " f%(|) = £f PL =" O. 1723 PL --
MB * R.A - Hf) r "5? pL " ~ o.l¥S/ PL —

PROBLEM 11.129
11.129 For the uniform beam and loading shown, determine the reaction at each
support.
w
. L±
.f if r f r ■'
fc
T
\/ «*-
SOLUTION
-+ A **J
<?c
pr; t si. Hfe ^r fejo^^^> - lifter * *r-£r
Ml - X-y
W.
*S)P;
L K* 3 + w(5 * 'J UEI *ȣ
HEI
£X
' 3<?* ?(?* ^ t? ' « ^8£X ^
RA . - f wL = iwL 4
fc
- ^(-i«i.^+ i«L 3 tf^Lf
+f "Z Fj =■ o P^(?84l?t^LrO
R.* #u;L

PROBLEM 11.130 11.130 Three members ofthe same material and same cross-sectional area are used to
support the load P. Determine the force in member BC.
SOLUTION
t
ee
BC
*-g-Z^f -o
EA 3t?a
Fee &'^<p - F^o *'*<iP "O F«e = F«»
Fao Cos 9 + Fg£ cos 9 * 1?ft - P
F _ r r p- Rt
Tar, - r«» 2.COS9
MemL
c^
BD
ec
80 * rB£
(P-PBV2cc*g>
(p-Gb^A***
J?
f?ai>/£"A
^ = -P^/^£"Acos59 + R^jL/ZEAcos*? + R6P/EA = o
P - o _ „_£
I?*-
I + 2c.i3(p
r^* = Isr T
0£
I +?Ca51<P

PROBLEM 11.131
11.131 Three members of the same material and same cross-sectional area are used to
support the load P. Determine the force in member BC.
SOLUTION
poi^ C . AeU rea^h'** Fc as a. ^oo4.
Joint B.
\4e.w\ \>ef
BC
8D
8£
I
F
Ft
-V5P --SPC
■Pf Fc
3F/i>Ft
)
-7i
)
FpF^rt
Ft
-3P + 3FC
- P * Ft
-3P + 4F;
f Fco *Ft- P « o
FBt> = Va P - VS Ft
i 1 F„ = o
^F^»+ F„ = o
FBe - - P + Fc
^ -" ^(-3P^F^ = O
F„ = -fP

PROBLEM 11.132
11.132 Three members of the same material and same cross-sectional area are used to
support the load P. Determine the force in member BC.
SOLUTION
V = X
Yr*L
2EA " 2FA
S - -^ - -L_T F^ L
+tZF„ - O F.-P+|Fan--0 F^ #P-f Fe
5 ' So
±*IFX » o F^ + IFro-o
Fbc ~ - 3 P + 3 rc
Mev*Jber
8D
BE
5:
F
Fc
fp-fFc
-^P*|FC
2F/;>£
1
-f
' L
3*
1
r(3F/3E ) L
- gf P* + # 6*
-f RJ* 6"-»Lf
S^ ^(- f Pi - 6 F.i)= O Ft= iP Fa.-F. 'I?

PROBLEM 11.133
11.133 Three members of the same material and same cross-sectional area are used to
support the load P. Determine the force in member BC.
SOLUTION
Ffcc ky Jw4 Fg ac-4 *im^ on w^-ev^ be,/- BC *t-1 B.
3e 3Fe 3FeZ EA ■ EA^-^-3% L
Joiw+ C
+* ZF^ ^O
fFe. + fec-P- O
y
2_
FAC
= o
Mcv^ We^
AC
Be
CD
z
F
F6
2F/3F*
i
H
2.
L
J?
i
F(3FySWL
FBj>
-iPJ +iF.i
-Ci+fc^wi + frF.*
Pi>
s.*-(±**>3*(4**-)
5 + ii
Fife * F« r O.GJT* P

PROBLEM 11.134
11.134 Knowing that the eight members of the indeterminate truss shown have the
same uniform cross-sectional area, determine the force in member AB.
SOLUTION
oJt e**l A.
S>F* ~ 3*5k XEA
EA
2E
5Fa
L = o
80 ' F«e
' rto
F - --^F
r6D if •>
r - ip
P '
rBff - p + fFAB
- .£p _ £ rr
3 r ^ hA
Kd ^^Fa
~i
ofi
+fXFj = o
F« + f F0B - o
j^D=-|R,B r -#F, j
Meeker
AS
AD
AE
BD
Be
DE
2:
F
FA
"**
|P"fF,
-fF.
^FA
-*P+ FA
3F/5R
1
1
_ £
*t
-£
t
3.
1
L
£
#i
*i
*i
*J
4
FOF/^R^l.
FAi
S Fii
- J551 PO 4- i2£T C" J
8 W
-#Pi * F,J?
S.-^(-gPi*fRi^
fi.* £P
- FA - £? - 0.5*83 P

PROBLEM 11.135
11J35 Knowing that the eight members of the indeterminate truss shown have the
same uniform cross-sectional area, determine the force in member AB.
SOLUTION
Co\ v*iei*\bev* AB at &*A A «-*<4 /^ew-fflxe ivte^besA
** ■ 5f. ' s^21 EA - EA 2- H?B
J01V1 + B +^^y = O
P
F* ■*■
Toi'nt E"
3F* " S>F*
L = O
FL> - - ** Fi
-P- F« -*Go= O
'BE
MTi^o
f8^|F«^
ftc
FDff
A P.
B*
+-2FK = 0 -|FAt-F0ff- o
FB* - - P + f FA
rDe - -|p * f.
Joi-»+ P
fc
**o f F*»
+ tIFj: O
F.
*o
^ F" =0
S ' &*>
Pap ~ tf Fa
Me^loe/*
A3
fcD
AE
Bt>
B8
De
z
F
Fa
*R
fP-^F*
-fF,
-P+*R
-|Pv FA
3F/5F*
1 .
**
-51
-*
.2
1
4*
*f
^
4^
^
FOF/aF.U
- a? Pi t #W
g"5<
-■i^^s*
- 3 Pi + F. i
-?P-* + fF.^
S, - ^(-lP^^f^)--o F,* fP
F«* ^ *|P= O.C67P

PROBLEM 11.136
to
25 in.
10 in,
11.136 The steel bar v£0C has a square cross section of side 0.75 in. and is subjected
o a 50-lb load P. Using E = 29 * 106, determine the deflection of point C.
SOLUTION
Assume MewlfiT' *BD is c "iwo-Tofxs yvie^eet^.
t r _ i~sp Lg0 _
5*>Jb.
- ISoo ib-/h
'Bo_ 3.FA r (2XWlO*X3L*Ji6*/o-s )
- 0.5488 ii-A
Me^fee/- ABC
I - 7V (o.?s)(o.7sf s- K3C7X/0'3 ,'*»
R*4i«w AB M* - ISoofe r -ISO*
IT , C'°-M1-^ - jSSlf'VA-
,(lSo)XQo'>
^Jf^wo6"X"Z6.3.67M/o-4)C5)
r M.^O^ mJb.
R*4;«* BC : M - -So * U^ = $ °-r^ J
\/ -
3o
V*J
* P Sc r L/

PROBLEM 11.137
11.137 The steel bars BE and AD have each a 5 * 15-mm cross section. Assuming
that lever ABC is rigid and using E = 200 GPa, determine the deflection of point C.
2kN
240.mm
60 mm
be <x~j Ar>
U 2.EA
SOLUTION
60
60 Tge -(360^) - O
F&e -- to k\J.
FAB - 8 kvi
-= \.0OOO + O.&^oo
^PSC = U S,
= I.€4oo NJ-^*
)
ioc . tt e - ^U eatfi.*'***) _ . ,,. _-3
£v |os
.4«-x IO w = I-6 4*,^

PROBLEM 11.138
2kN
11.138 The steel bars BE and AD have each a 5 * 15-mm cross section and the steel
lever ABC has a square cross section of side 25 mm. Using E = 200 GPa, determine
the deflection of point C.
SOLUTION
~ae
rP
&o
F =■■ 8 JfW
Rsr Wcl^s BE om<l AD
r- ^jLtfi -
86 2£A " (D(^oo^io<;'^75x/o-e )
n - Rgtt^p - (g-)zC3oox/o-*)
A = S« 15 - 7S »«£ ~ 75Vio'6^1
XEA UX^oxlo^^Wcf'^
e
e«^
lokV/
ABC: X- fH*5)(25)3 " 32. SS^x/O3 ^ = 32.5f2*/o"* m
t J- s
Po»4i6M AB: M = ■^fie-%
<ser
0.35-3^ J
-por+;olA BCt m - 4s2
-0 *ei ^v xexu? Jo
1.4154 J
'at
r I.Oooo + O.6V00 + o.35"3«? + /. 4/56 — 3.4o*T J*
?.X)D
= 3.41 k.
HI

PROBLEM 11.139
11.139 Two solid steel shafts are connected by the gears shown. Using G= 11.2 x
106 psi, determine the strain energy in each shaft when a 24 kip-in. torque is applied
SOLUTION
h _ TcoU _ ft^teol . Q .n|.
Seel/* B
ca rv. " rt 5- T- fs
■AB
= Zo
m
U4a c Iklkfi. g (3B.M)*C*^ r = 0^23, iVfc-
To+*J IS- U^ + Uco » O.S"*33 4 O.H«*l) » I.OI^m./t.'ps,
PROBLEM 11.140
11.140 Two solid steel shafts are connected by the gears shown. UsingG^ 11.2 x
106 psi, determine the angle through which end D rotates when T*> 24 kip-in.
JC HeaUftO
SOLUTION
F*o*> ProL. II. I31?

PROBLEM 11.141
SOLUTION
11.141 (a) Determine the modulus of resilience of a grade of structural steel for
which or = 300 MPa and E = 200 GPa. (b) Determine the required yield strength of
an aluminum alloy for which E - 72 GPa if the modulus of resilience of the alloy is to
be the same as that of the structural steel.
(d) E= 200*10" P^ 6TY- Zoovio'po.
&n _ ( ZOO*tOi')1'
** ft— " /«*»^SM ^ A _*.*-_**
Uv =
Z2S ltJ/»S
PROBLEM 11.142
F^ s-r>.> AB^DB^Bf
77.742 A single 6-mm-diameter steel pin B is used to connect the steel strip BE to
two aluminum strips, each of 20-mm width and 5-mm thickness. The modulus of
elasticity is 200 GPa for the steel and 70 GPa for the aluminum. Knowing that for the
pin at B the allowable shearing stress is r,,, = 85 MPa, determine, for the loading
shown, the maximum strain energy that can be acquired by the assembled strips.
SOLUTION
fiw ' if « - TJ
- 4.8o6fexfO* N
A~(a©tfs)s too*** •* too•*lo*vf
(XX70 kIO",XIO0v/0's')
J"
U*. -" "Up*
■* :%.£**«« Maoo>. /o^Xi^o wo-4)
ToW: U* Una + Upb + U«c - £46 v/o"3 J c CgV6 J

PROBLEM 11.143
450 N
4SP»|
Hester- BC
11.143 The 18-mm-diameter steel rod BC is attached to the lever AB and to the fixed
support C. The uniform steel lever AB is 9 mm wide and 24 mm deep. Using E = 200
GPa, G = 77 GPa, and the method of work and energy, determine tbe deflection of
point/*
SOLUTION
Me^W AB
M =■ 450 x
Ma ■ \q% ro-^
^ i-i - tow x iv's - /fc* n-*
^ {Hso^L^ _ mso)x(zGoxio~*)
-a\«
= 0.7S"?38 J"
u
ec
TA-JL
7&j " (^(tjmo'X/o. soccer*)
U * UA6 + U8C - 10.68 1 J
- ?.9^I3 J
U
c - 211- (z)(\o.c&i) _ u_ «-*!,■<-* ^ 47 c i
£, - ^p " ^ icrn ' W/.5MO t*\ * H7.5 *w b
H50

PROBLEM 11.144
1.75-in, diameter
11.144 The 75-lb collar D is released from rest in the position shown and is stopped
bye plate attached at end C of the vertical rod ABC. Knowing that £=29* HPpsifor
both portions of the rod, determine the distance h for which the maximum stress in the
rod is 36 ksi.
SOLUTION
1.5-in. diameter
AS
fW."*iBC: A^= f Jjc* $0.5)'^ 1.761 IS In1
6^ * 30000 psi L6£ t G-ft = 12 ;«,
, fltfc - fssonYa<7t) , - . ..
I)
CC
Pqitt<i«»v*
H3i;om.I4,
3<*0S.3 i*A
±P*. * TJ S. - ^ - 4^^ - o.i-nw ,v
TaT (h* O * U
J3 0(T

PROBLEM 11.145
11.145 The 75-lb collar D is released from rest when h - 20 in. and is stopped by a
plate attached at end Cof the vertical md ABC. Knowing that E = 29 x KFpsi for both
portions of the rod, determine (a) the maximum deflection of end C, (b) the equivalent
static load, (c) the maximum stress that occurs in the rod.
1.75-in. diameter SOLUTION
Po.-r,^ AB: AA6- ^dj = f (I--I3Y* 2.HMM ;-'"
LA8 = g-Ft ■ ?6 in
1,5-in. diameter
P* I
Ssll^ -^ c*aj-M*n P.1
P*,W BC: A.t »f-»« * 3 0-5^* L7C7IS-.V L^ G-T+ - 7*m
U«t -
?Sl
Be-
pS ^
r 70^.48 x/o'9 Pj
U " iP^S** /7c?.776x/o3 S^
W<K-fe of ?«i/,«^ we.gkt W(M S>J = 75(20 4 S.J * /5"00+75§w
§m2 - 4/7. l8Sxto~' S„ - 8.3437 w/tf3 = O
§* r i[ 4n.i«"*i0"% y(m7J8s-x^*c^+ (*n(s.^37wo"5)5
(b) P.. * (£si.S5Z*to%)(om&ii'SS$) =■ 32irMl P^-Si^doA -*
M
6U
M^ii
1.7C7/S

PROBLEM 11.146
11.146 ThesteeIrodSChasa24-mmdiameterandthesteelcableJ45i)C^hasal2-
mm diameter. Using E = 200 GPa, determine the deflection of point D caused by the
12-kN load.
SOLUTION
0+>\>
tj0, ■= a F**~ Las ^ FfeLjc
13 - ^U„, + u8C
Lei P t* +U ioorl -L+ D
u. Feo Lgp ^fe. . Fgc^-ac ^F
T C A, 3P T PA- ^C
+t if; : o
fip * f P
3F«„ _ *
9? *
T 2 Fy = o '
Fsc-J-F^ -fp
3 Fee. ^
3P ' "3
^" H"' EAeo UJ £A«.
£
M h» ^ a* J
P = |2*|0* N
Lo0* G00«/O"S m
Let * ^o*/o'* m
E = Zoo * /o1 ?6
Aw * TO^O*" = H3.o?7 iwi* " l>3.o?7 Wo"6 mJ
S. -
ZOtpvfO
Li if
600 V(CT'
—7-V - I'.lllv/o »-i
r |. I | | miw\ -L
Hi
*?60 * /O"
— f ^ -
113. W lo" "» W-3<?k|

PROBLEM 11.147
11.147 The simply supported beam AB is struck squarely at D by a block of mass m
movinghorizontallywithavelocityvo. Show that the resulting maximum normal stress
a„ in the beam due to bending is independent of the location of point D
SOLUTION
Let Pw t* He etfu't^a.Jle.vA S+d-fc -Poftd «-t poi"^ V>
. RX
-B.«x
R&^+i©* Ad
To+«J U -
DB J0 «!fc_L J0 2.EJ 2eTL*J& 6 Ell.*
CflL1 GFIL
M
G>£X
I m„*- TV _- M„'L
M,
/3B£^%-
S-H**^
€•«' M
or G*- o* k.

PROBLEM H.C1
Element n Element /
Element 1
11.C1 A rod consisting of n elements, each of which is homogeneous
and of uniform cross section, is subjected to a load P applied at its free end.
The length of element i is denoted by L, and its diameter by d,, (a) Denoting
by E the modulus of elasticity of the material used in the rod, write a computer
program that can be used to determine the strain energy acquired by the rod
and the deformation measured at the free end. (fc) Use this program to
determine the strain energy and deformation of the rods of Probs. 11.9 and 11.12.
SOLUTION ^ F fi„p£
COMfWl&i pfO£f**iL £7't'&£%: '-\ i_ -
a
PzLl
ZF
Tc>Tfi<- S~-rft4itf £&/&<,*
JbTAL P£ jZo&M f) T/on
- ?0
Ipa-U : 4-
Pf?66&fiM OUTPUT
Problem 11.9
Axial load « 8.000 kips Modulus of
Element Length delta L Stress
in. in. ksi
1 24.000 0.022 26.08
2 36.000 0.022 18.11
Total Strein Energy * 176.24 in-lb
Total Deformation * 0.0441 in.
Problem 11.12
Axial load - 25.000 kN Modulus of
Element Length delta L stress
m mm MPa
1 0.00 0.497 124.34
2 1.20 0.477 79.58
Total Strain Energy - 12.1853 J
Total Deformation - 0.9748 mm
elasticity = 29 x 10"6 psi
Strain Energy Strain Energy Density
in-lb lb*ln./in."3
86.32 11.72
89.92 5.65
elasticity - 200 GPa
Strain Energy Strain Energy Density
J kJ/m"3
6.22 38.65
5.97 15.83

PROBLEM 11.C2
11 .C2 Two 0,75 X 6-in. cover plates are welded to a W8 X 18 rolled-
steel beam as shown. The lSOO-lb block is to be dropped from a height h = 2 in.
onto the beam, (a) Write a computer program to calculate the maximum
normal stress on transverse sections just to the left of D and at the center of the
beam for values of a from 0 to 60 in., using 5-in. increments, (b) From the
values considered in part at select the distance a for which the maximum norma)
stress is as small as possible. Use E » 29 X lO6 psi.
Fra 1500 lb
CT Vh E
f X 6 In.
SOLUTION
For ftp fl//a £8: wB*i£
J,*tf>f<»'
Sr/S'.Z in*
<\
OL-
A
ft
I
it
J-JM
Wq^/i
k
■r&B
POSIT/ON
F&R PCS: V/&?c/g pZu$€?&Su?iJ PL#7£S
k- im ~*l
4^0 7 W
3
fce?*^-r«- %w,h
J~ - 6/,J-h2(6*c>.7s)(ws) = 23%?2 tU
2 fft07+O.75; 4*2
<
M - P ^k WAVe£~ Ol =- tMF£V£MC(T C&PF/&FN7
-J^_ art ' —■'■ -~
SEP ft£/T P*?6£ Rotl Z>F7&faMtM*Ttow OR fc/,
7*n
f = JttuivQisr-'T £7B7l£ J.&m>
4H
W&&K QGM? SY W.tS W(h + ^rh)
2 U
Po?J7Mv 2
ore'.
"3
* - 2W&L^~ZW/>oL. {f\
•»*
PfcOdVftM Soi ut/ok, oR @ /^^ 'j^
fart <m. = c> /-& fc?}/>, j?7&7^ £"/'*?.:
T>/jr /5 7>fir oaT*)stcf o- For 2 \Q^ /»s S"tr*#L*- /is Poss/sif
CONTINUED

PROBLEM 11.C2 - CONTINUED
E
(*- a -A j)
+1
/
-J
L
z
4
5
i
2
Oi
=V<I
DC
«*[(i-i)~*+£*]l<*
P2D<£/2f)M OOTfVT
Beam - w 0x18 with two 6 by 0.75-ln. cover plates
h - 2 in. W = 1500 lb L = 120 in
a
in.
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
50.00
55.00
60.00
ystat
in.
0.00777
0.00770
0.00787
0.00812
0.00859
0.00930
0.01056
0.01220
0.01430
0.01718
0.02068
0.02496
0.03008
ymax
in.
0.1842
0.1844
0.1055
0.1805
0.1942
0.2033
0.2163
0.2334
0.2546
0.2799
0.3090
0.3419
0.3783
Use smaller increments to
18.33
18.34
18.35
0.00840
0.00840
0.00841
0.1919
0.1920
0.1920
Pmax
lb
35572
35544
35348
34834
33096
32509
30736
28706
26563
24436
22415
20550
18862
ffl
ksi
0.00
5.05
11.63
17.19
22.30
26.73
30.33
33,05
34.95
36.17
36.87
37.18
37.23
a2
ksi
21.46
21.44
21.32
21.01
20.45
19.61
18.54
17.32
16.02
14.74
13.52
12.40
11.38
<Tl - <T2
ksi
-21.46
-15.59
-9.69
-3.82 v
1.05 '
7.13
11.79
15.73
18.93
21.43
23.35
24.78
25.85
seek the smallest maximum normal stress
34259
34257
34255
20.657
20.667
20.677
20.665
20.664
20.663
-0.01
0.00
0.01
Max stress small as possible for a = I8.34in.
Smallest max stress - 20.67 ksi

PROBLEM 11X3 24 mm
•r
Hd
eq
24 mm
11 C3 The 16-kg block D is dropped from a height h onto the free end
of the steel bar AB. For the steel used <rat1 = 120 MPa and E = 200 GPa. (a)
Write a computer program to calculate the maximum allowable height h for
values of the length L from 100 mm to 1.2 m, using 100-mm increments, (b)
From the values considered in part a, select the length corresponding to the
largest allowable height.
A
SOLUTION
<£a = /*OMP*f
Oft - /£*£% j
7- d>
V2
Pot? J, s /0Oo^ 72> /^o&'w S~7W /<SK=>-»f^J
F/Pom ftza*. fA£ft pay: ?OS
fit/ox /
,m
-1W
[,+F^\ *»***. h*[(^-i)x-
w
Q^z- *>> 3s*>y*----Z- - "~ - h
Iwoyt) '/Me*. 5 me^%
1?£TV**-'
P&06B+?* £>U TP^T
m ■
L
mm
100
200
300
400
500
600
700
800
900
1000
1100
1200
Use
435
440
445
16.0 kg
ystat
mm
0.00946
0.07569
0.25547
0.60556
1.18273
2.04375
3.24540
4.84445
6.09766
9.46181
12.59367
16.35000
Problem
d » 24 mm
ymax
rati
0.167
0.667
1.500
2.667
4.167
6.000
0.167
10.667
13.500
16.667
20.167
24.000
smaller increments to
0.77883
0.80599
0.83378
3.154
3.227
3.300
11. C3
ff m 120 MPa G -
Pmax
N
2764.8
1302.4
921.6
691.2
553.0
460.8
395.0
345.6
307.2
276.5
251.3
230.4
seek the
635.6
628.4
621.3
Mmax
N-m
276.48
276.40
276.40
276.48
276.40
276.48
276.48
276.48
276.40
276.48
276.40
276.48
200 GPa
h
rati
1.301
2.269
2.904
3.205
3.173
2.007
2.109
1.076
-0.289
-1.980
-4.020
-6.385
largest height h
276.48
276.40
276.48
3.2316
3.2320
3.2317

PROBLEM 11.C4
11 .C4 The block D of mass m = 8 kg is dropped from a height
h = 750 mm onto the rolled-steel beam AB. Knowing that E = 200 GPa,
write a computer program to calculate the maximum deflection of point E and
the maximum normal stress in die beam for values of a from'lOO to 900 mm,
using 100-mm increments.
W150 X 13.5
F^&fc UN/7 lo&O IQT4r
'J
SOLUTION
<Z. - a,//aoa
b = L - o-
'4
/ +
f 4*0-? <?nav f
Problem 11.C4
Beam:
I m
L -
a
inn
100
200
300
400
500
600
700
800
900
W 150 x 13
6.87 xl0*-6
1.8 m h -
ystat
nun
0.0003
0.0011
0.0021
0.0033
0.0045
0.0055
0.0063
0.0068
0.0069
me*y
M*»,
P/&&T: <*-> fa.
f2fr7oa/-i
5
TlT4 S = 91
750 mm m *
ymax
mm
0.6775
1.2757
1.7946
2.2339
2.5936
2.0734
3.0734
3.1934
3.2334
.6xl0"-6
8 kg
Pmax
N
173.93
92.43
65.75
52.85
45.55
41.13
38.46
37,02
36.56
/£
m'3
g - 9.81 m/s"2
'»*Y
MPa
179.33
179.40
179.46
179.51
179.55
179.59
179.61
179.63
179.63
PlCQUlflCO &Y 7H£ J^lfiSS *5 tT MLiS 7-»fcoD6/* ^**«»*■
&£& ffcag, //./?7, faye l3/} f&ft <9 ease k>*/£&£
J-ft£&£Y D&WFfZ&O t£ r0ASS.74A*7 AHQ tf~ IS -#U* CCiNSm^i

PROBLEM 11.C5
11. C5 The steel rods AB and BC
are made of a steel for which
<rY = 300 MPa and E = 200 GPa. (a) Write a computer program to
calculate, for values of a from 0 to 6 m, using 1-m increments, the maximum strain
Imfl^l y 10-mm diameter energy that can ha acquired by the assembly without causing any permanent
MajP^i^^J*
*^_ ^^"*^
kT"^
6 m
&£&£#*
deformation, (b) For each value of a considered, calculate the diameter of a
, 6-inm (humuter uniform rod of length 6 m and of the same mass as the original assembly, and
>/^ the maximum strain energy that could be acquired by this uniform rod with-
*^^c out causing permanent deformation.
^*- SOLUTION
£/TT*tl\ ST r 30&MP* ^ £
= %**£&>. J* in
flgfA«8S %(W£>'t"T> G&**3^ if^oa&^t )*
K? S ******
For. <x = 0 7t> 6 m? S7£P
p zf c-
toy,
Fcxi ON/pofeM Rod of sqm& ^ccuz-aj?
Vol* o-iWe*^)
j \\ y. vol.
<**)*<- '
******»»%<**'
l-tL-eJCfiae*^
r^%("R*^)
Pz L
^*w~ 2*^*0
Pni*T cl} 0t i/di-, d,
HtTuflM
Problem 11C5
sigmaY = 300 MPa, Pm = 8482 N, L » 6 m,
a u Vol d New p
m J m*3 mm N
0.00 38.17 169.65 6.00 0482.30
1.00 34 .10 219.91 6.83 10995.58
2.00 30.03 270.18 7.57 13508.85
3.00 25.96 320.44 8.25 16022.12
4.00 21.88 370.71 8.87 10535.40
5.00 17.81 420.97 9.45 21048.67
6.00
13.74 471.24 10.00 23561.95
f 0
E ■ 200 GPA
newU
J
38.17
49.48
60.79
72.10
03.41
94.72
106.03

PROBLEM 11.C6
Sr
12 ft
~F20in.
2.65
•a
11 -C6 A 160-lb diver jumps from a height of 20 in. onto end C of a
diving board having the uniform cross section shown. Write a computer program
to calculate for values of a from 10 to SO in., using 10-in. increments, (a) the
maximum deflection of point C, (b) the maximum bending moment in the
board, (c) the equivalent static load. Assume that the diver's legs remain rigid
and use E - 1.8 X 106psi.
In.
16 In.
SOLUTION
\tf*MO/6
k-*^
A
e
A 5F
D/veR
r_
T>
ppS/7/OA> I
JOr) ■*»
W&£&£ ol = /MFLU&t^F toeFFtc/£t* T
Vt&RF f^. ~ f&WVfiLEMT OTITIC /6#D
>Vb
a--4-£/v =±3*-
%
&
&
r=g
PO<J 7/6" j?
f>KM7 a-, ^j
P/2C£RftH> OUTPUT
r+,
M^ V~
"m
a
in.
10
20
30
40
50
ym
in.
14.622
13.262
11.950
10.683
9.462
Jfl»
P£o&£flM SoloTton oP a. F&rz ^^ J^ht&z \K
/tD/2 a.-Join. 7t> So'tO. CT&P /£>/*.
js-olv? a Fen ^ 9 fcmo-M
O-
Pm
lb
757.7
002.6
855.6
919.1
996.4
Max M
kip*in.
101.532
99.519
97.536
95.583
93.661
sigma
psi
5422
5314
5208
5104
5001