PROBLEM 3.5
60 nrur
iJO
rnnr
torque 7 = 3 kN .m is applied to the solid bronze cylinder shown.
Determine (a) the maximum shearing stress. (&) the shearing stress at
point D which lies on a lS-mm-radius circle drawn on the end of the
cylinder, (c) the percent ofthe torque carried by the portion ofthe cylinder
within the 15 rnm radius.
A
'l'
,,,,,
l.\
kN . rri s.
SOLUTION
(a) , =!4 = 30 mrn = 30x10-3rn
2
J = 1"r = 4eox
22
t0-3)4 = 1.27235x
t0{
ma
Z=3kN=3x103N
,-"' = !L-
(: x to3x:o x lq-3)
J
(b)
Pn =
15 mrn
-0D -Pr,-L
-L
(c)
r,-, =
"
r,
=
!282JD
f{tsx
= 70.736x r06 pa
1.27235x10-
=
15
x
T* = 70.7 MPa
{
rn =
{
10-3 m
(15x10-3X70.736x10{)
(30 x 10-3)
J,.Tr. t
'r=-ff=
35.4 MPa
r
rPhtn
tor;rq3s.368 x 106) = 187.5 N . m
lLxfio% = J8741ro
T
3 x l0r'
oo/oS
= 6.25%
6.2s%<
PROPR{ETARY II'IATERIAL. (o 2012 The lVlcGra'rv-Hill C'ompanies, Inc. All rights reserved. No part ol this Manual may be displayed.
itt any fomr or by any means, rvithout the prior rritten pennission of the publisher, or gsed beyond the limited
reproduced, or distributed
distribution to teachers and educators permitted by McGrarv-t{ill for their individual course preparation. A student using this manual is using it
rvithout pennission.
PROBLEM 3.13
Under nonxal operating conditions, the
electric rnotor exerts a 12-kip in.
torque at E. I(nowing that each shaft is
solid, cletermine the maxirnuln shearing
in (ru) shaft BC, (b) shaft CD, (c) shaft
DE.
!.2,5 in.
SOLUTION
(a)
From free body shown:
Shaft BC:
TN: =3
kiP'in
Tc
Tc
,]
lf4
T
2
(l)
t
c'
-L
2
2
'-;(t
(1r)
Shaft CD:
3
-cD
2in.
Sliall DE:
{
l-xl.75in.l
\2
)
* 4 =7 kip . in
From Eq.
(l):
2T 27kip.in
_
I =---i=---lr c' lr (l in.)'
r
= 4.46 ksi
{
Tt;ti =12 kiP' in
From Eq. ( l):
T
trr'
_.-----1- _2
_2
12
kip .in
r
lTlt
\j
| -x2.25in.1
[2
)
PROPRIETARY \{,4TERIAL. O 2012'I'he NtcCrarv-Hill Companies, lnc,
wilhout permission.
ksi
From fiee body shor.vn:
*;)p;so;
reproducetl, or distributed
= 2.85
From fiee body shor.vn:
\
{,15i4.
r
\t
Tr:, =3
(c)
kio'in
All rights
reserved. No part
olthis Manual
= 5.37 ksi
{
rnay be displayed^
in ani,L,rm or hy an) nreans.,witlrout the prior rvritten perrnissiorr oltlre publisher, or used be),ond the limited
PROBLEM 3.18
The solid rod BC has a diameter of 30 mm and is made of an
aluminum for which the allowable shearing stress is 25 Mpa.
Rod
is hollow and has an outer diameter of 25 mrn; it is
made of a brass fbr which the allowable shearing stress is 50
lB
MPa. Determine (a) the largest inner diameter of rod AB for
which the factor of safety is the same for each rod, (b) the
largest torque that can be applied atl.
SOLUTION
SolidrodBC:
r=!e J=Lca
J2
ruv
=
25
x
106 Pa
, = !,1=
)
T^, =
Hollow rod lB
i
lTt
fc3q1,
tu, = 50 x
Tutt
o.ol5 m
=
lf
x t06y = 132.536 N . m
;{o.ots)3(zs
106 Pa
= 132'536 N'm
ll
- -d"
2'
c? =
^ _9
Ll
-
= -(0.025) = 0.0125 m
2'
l
r^,=h =+ki -,i)t
_ -4
.,4 c)
Ll
-
2Tuncz
frtul
= 0.01254
(a)
(b)
-
(2X132'536X0.0125)
7r(50
x l0o)
ct = 7.59 x 10-3m = 7.59
Allowable
torque.
mm
= 3.3203x l0-ema
d1
= 2c1=
Tu,
15.1g
mm
= 132.5 N . m
{
<(
PROPRIETARY MATERIAL,
reproduced. t-rr distributed
(c) 2012 Thc ilIcGrarv-Hill (Jompanies, lnc. All
rights reserved. No part of this Manual may be displayed,
in any form or by any means, without the prior written pernrission o[ the publisher, or used beyonrJ the limited
distribution to teachers and educators permilted by McCrarv-Hill lor their individual course preparation. A student using this manual is
using it
rvithout permission.
PROBLEM 3.38
The aluminum rod
rod
lB
(G = 27 Gpa) is bonded to the brass
BD (G = 39GPa). Knowing that porlion CD of
brass rod is hollow and has an inner diameter
determine the angle of twist atl.
the
of 40 mm,
SOLUTION
Rod
lB:
= 27 x
G
T =
10e
Pa,
Z = 0.400 m
800N.m , = ld = 0.01g
rn
2
J = 1ro = L$118)a = | 64.Bg6x
22
t,ttD
rAtD
to-e m
(800x0.400)
\uvv.,\u'Tvv,
TL
-1.975 x l0*3 rad
GJ ei xrc\(164.g96 x to-e)
PartBC: G = 39x lOepa L=0.375 m..2 = ld = 0.030 m
----,
"
Z =800+1600 =2400N.m,
, =1,
=Lp.OtO1o =1.27234x10-6ma
(2400\(0.37s)
TL
' /\ '
(0-,,.=-=
r t''t
GJ (J9 x I oq\(t.zlzlq x lo-6) -18.137x10-,rad
c, =
'2
PartCtD:
!4,
= 0.020
I
t't = -d2 = 0.030
, =;t": rL
j=-=
Q,.,r,=
\
' lt '
Angle of tivist
atl.
GJ
m
rn, I
= 0.250
rl) = fto.ozoo-
(39
m
0.0204) = 1.02r02x
(2400x0.2s0)
x loqlt
.oz lo2
x lo-6;
r0{ma
15.068x10 3rad
Qt=Q,qrn+Qrtc*Qon
=
105.080
x
10-3 rad
Qt = 6.02" a
PROPRIET'4RY MATERIAL' @ 2012Tht NlcGraw-Hill Companies, lnc. All rights reserved. No part of
this Manual may be displayed,
reproduced. or distributed in any form or hy any nteans, witlrout the prior written permission of the publisher,
or used beyond the limited
distribution to teachers and educators pennitted by McGrarv-Hill tbr their individual course preparation. A student
usirrg this rnanual is using it
rvithout pennission.