18 mm
T
PROBLEM 3.2
For the cylindrical shaft shown, determine the maximum shearing stress caused by
a torque of magnitude T  800 N  m.
SOLUTION
Tc

; J  c4
J
2
2T
 max  3
c
2(800 N  m)

 (0.018 m)3
 max 
 87.328  106 Pa
 max  87.3 MPa 
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262
260
BeerMOM_ISM_C03.indd 260
12/23/2014 3:40:57 PM
PROBLEM 3.15
The allowable shearing stress is 100 MPa in the 36-mm-diameter steel rod AB and
60 MPa in the 40-mm-diameter rod BC. Neglecting the effect of stress concentrations,
determine the largest torque that can be applied at A.
SOLUTION
τ max =
Shaft AB:
J=
π
2
c4 ,
T=
π
2
τ max c3
τ max = 100 MPa = 100 × 106 Pa
c=
TAB =
Shaft BC:
Tc
,
J
1
1
d AB = (36) = 18 mm = 0.018 m
2
2
π
2
(100 × 106 )(0.018)3 = 916 N ⋅ m
τ max = 60 MPa = 60 × 106 Pa
c=
TBC =
1
1
d BC = (40) = 20 mm = 0.020 m
2
2
π
2
(60 × 106 )(0.020)3 = 1.754 N ⋅ m
The allowable torque is the smaller of TAB and TBC.
T = 754 N ⋅ m 
PROPRIETARY
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BeerMOM_ISM_C03.indd 273
273
12/23/2014 3:41:00 PM
PROBLEM 3.17
TB 1200 N · m
The solid shaft shown is formed of a brass for which the
allowable shearing stress is 55 MPa. Neglecting the effect of
stress concentrations, determine the smallest diameters dAB and
dBC for which the allowable shearing stress is not exceeded.
TC 400 N · m
A
dAB
750 mm
B
dBC
C
600 mm
SOLUTION
 max  55 MPa  55  106 Pa
 max 
Shaft AB:
Tc
2T

J
 c3
c 3
2T
 max
TAB  1200  400  800 N  m
c 3
(2)(800)
 21.00  103 m  21.0 m
 (55  106 )
minimum d AB  2c  42.0 mm 
Shaft BC:
TBC  400 N  m
c 3
(2)(400)
 16.667  103 m  16.67 mm
 (55  106 )
minimum d BC  2c  33.3 mm 
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277
275
BeerMOM_ISM_C03.indd 275
12/23/2014 3:41:01 PM
C
PROBLEM 3.21
40 mm
A torque of magnitude T  1000 N  m is applied
at D 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.
T 1000 N · m
A
B
D
100 mm
SOLUTION
TCD  1000 N  m
TAB 
(a)
Shaft AB:
rB
100
TCD 
(1000)  2500 N  m
rC
40
 all  60  106 Pa
 
Tc
2T
 3
J
c
c3 
2T


(2)(2500)
 26.526  106 m3
 (60  106 )
c  29.82  103  29.82 mm
(b)
Shaft CD:
d  2c = 59.6 mm 
 all  60  106 Pa
 
Tc
2T
 3
J
c
c3 
2T


(2)(1000)
 10.610  106 m3
 (60  106 )
c  21.97  103 m  21.97 mm
d  2c = 43.9 mm 
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281
279
BeerMOM_ISM_C03.indd 279
12/23/2014 3:41:02 PM
PROBLEM 3.34
40 mm
A
50 mm
(a) For the aluminum pipe shown (G  27 GPa), determine the torque T0
causing an angle of twist of 2°. (b) Determine the angle of twist if the
same torque T0 is applied to a solid cylindrical shaft of the same length
and cross-sectional area.
T0
2.5 m
B
SOLUTION
co  50 mm  0.050 m,
(a)
J 

2
(co4  ci4 ) 

2
ci  40 mm  0.040 m
(0.0504  0.0404 )
6
 5.7962  10 m 4
  2  34.907  103 rad
 
TL
GJ
L  2.5 m
G  27  109 Pa
GJ 
(27  109 )(5.7962  106 )(34.907  103 )

T0 
2.5
L
 2.1851  103 N  m
Area of pipe:
(b)
T0  2.19 kN  m 
A   (co2  ci2 )   (0.0502  0.0402 )  2.8274 m 2
Radius of solid of same area:
A
cs 
Js 
s 


2
 0.030 m
cs4 

2
(0.030)4  1.27235  106 m 2
T0 L
(2.1851  103 )(2.5)

 0.15902 rad
GJ
(27  109 )(1.27235  106 )
 s  9.11 
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294
292
BeerMOM_ISM_C03.indd 292
12/23/2014 3:41:11 PM
PROBLEM 3.38
60 mm
TB 1600 N · m
The aluminum rod AB (G  27 GPa) is bonded to the brass
rod BD (G  39 GPa). Knowing that portion CD of the
brass rod is hollow and has an inner diameter of 40 mm,
determine the angle of twist at A.
D
36 mm
C
TA 800 N · m
B
250 mm
375 mm
A
400 mm
SOLUTION
G  27  109 Pa, L  0.400 m
Rod AB:
T  800 N  m c 
J 

c4 

1
d  0.018 m
2
(0.018) 4  164.896  109 m
2
2
TL
(800)(0.400)

 71.875  103 rad
 A /B 
GJ
(27  109 )(164.896  109 )
Part BC:
G  39  109 Pa
L  0.375 m, c 
T  800  1600  2400 N  m, J 
 B /C 
1
d  0.030 m
2

2
c4 

2
(0.030)4  1.27234  106 m 4
TL
(2400)(0.375)

 18.137  103 rad
GJ
(39  109 )(1.27234  106 )
1
d1  0.020 m
2
1
c2  d 2  0.030 m, L  0.250 m
2
Part CD:
c1 
J 
C / D 
Angle of twist at A.


c  c   (0.030  0.020 )  1.02102  10 m

2
2
4
2
4
1
4
4
6
4
(2400)(0.250)
TL

 15.068  103 rad
GJ
(39  109 )(1.02102  106 )
 A   A/B   B/C  C/D
 105.080  103 rad
 A  6.02 
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298
296
BeerMOM_ISM_C03.indd 296
12/23/2014 3:41:12 PM
PROBLEM 3.41
Two shafts, each of 22-mm diameter are
connected by the gears shown. Knowing that
G = 77 GPa and that the shaft at F is fixed,
determine the angle through which end A rotates
when a 130 N ⋅ m torque is applied at A.
SOLUTION
Calculation of torques.
Circumferential contact force between gears B and E.
F =
TAB
T
= EF
rB
rE
TEF =
rE
TAB
rB
TAB = 130 N ⋅ m
TEF =
150
(130) = 177.3 N ⋅ m
110
Twist in shaft FE.
L = 300 mm, c =
π
1
d = 11 mm, G = 776 Pa
2
π
(0.011)4 = 23 × 10 −9 m 4
2
2
TL
(177.3)(0.3)
=
= 0.03 rad
ϕ E /F =
GJ
(77 × 109 )(23 × 10−9 )
J =
c4 =
Rotation at E.
ϕ E = ϕ E/F = 0.03 rad
Tangential displacement at gear circle
δ = rEϕ E = rBϕ B
Rotation at B.
ϕB =
Twist in shaft BA.
L = 0.35 m
ϕ A/ B =
Rotation at A.
rE
15
ϕ E = (0.03) = 0.0409 rad
rB
11
J = 23 × 10−9 m 4
TL
(130)(0.35)
=
= 0.0257 rad
GJ
(77 × 109 )(23 × 10−9 )
ϕ A = ϕ B + ϕ A/B = 0.0666 rad
ϕ A = 3.82° 
PROPRIETARY
MATERIAL.
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BeerMOM_ISM_C03.indd 299
299
12/23/2014 3:41:13 PM
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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309
BeerMOM_ISM_C03.indd 309
12/23/2014 3:41:22 PM
PROBLEM 3.53
The composite shaft shown consists of a 5 mm thick brass jacket
(G = 39 GPa) bonded to a 30 mm diameter steel core (G = 77 GPa).
Knowing that the shaft is subjected to 565 N ⋅ m 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
c1 =
Steel core:
J1 =
1
d = 15 mm
2
π
2
c14 =
π
2
(0.015)4 = 79.52 × 10−9 m4
G1J1 = (77 × 109 )(79.52 × 10−9 ) = 6123.04 N ⋅ m 2
ϕ
Torque carried by steel core
T1 = G1J1
Brass jacket:
c2 = c1 + t = 15 mm + 5 mm = 20 mm
π
L
(
) π
c24 − c14 = (0.024 − 0.0154 ) = 0.1718 × 10−6 m 4
2
2
9
G2 J 2 = (39 × 10 )(0.1718 × 10−6 ) = 6700.4 N ⋅ m 2
ϕ
Torque carried by brass jacket
T2 = G2 J 2
L
J2 =
T = T1 + T2 = (G1 J1 + G2 J 2 )
Total torque
ϕ
L
T
565
=
=
= 0.044 rad/m
L G1 J1 + G2 J 2 6123.04 + 6700.4
ϕ
(a)
Maximum shearing stress in brass jacket
τ max = G2γ max = G2C2
ϕ
L
6
= 34.37 × 10 N/m2
(b)
= (39 × 109 )(0.02)(0.044)
34.37 MPa 
Maximum shearing stress in steel core
τ max = G1γ max = G1C1
ϕ
L
= (77 × 109 )(0.015)(0.044)
= 50.82 × 106 N/m2
(c)
Angle of twist (L = 1.8 m)
ϕ = L
ϕ
L
50.82 MPa 
= (1.8)(0.044) = 79.2 × 10 −3 rad
= 4.53°

PROPRIETARY
MATERIAL.
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311
BeerMOM_ISM_C03.indd 311
12/23/2014 3:41:22 PM
PROBLEM 3.68
30 mm 75 mm
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  77.2 GPa, determine the power
being transmitted.
SOLUTION
Twist angle:
  2  34.907  103 rad
1
d1  0.015 m
2
1
c2  d 2  0.0375 m
2
c1 
J 


c  c   (0.0375  0.015 )

2
2
4
2
4
1
4
J  3.0268  106 m 4 ,
 
TL
GJ
T 
4
L  4m
GJ 
(77.2  109 )(3.0268  106 )(34.907  103 )

L
4
T  2.0392  103 N  m
f  120 rpm 
120
Hz  2 Hz
60
P  (2 f )T  2 (2)(2.0392  103 )  25.6  103 W  25.6 kW

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331
328
BeerMOM_ISM_C03.indd 328
12/23/2014 3:41:27 PM
T'
PROBLEM 3.87
60 mm
The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing
that the radius of the fillet is r  8 mm and the allowable shearing stress
is 45 MPa, determine the maximum power that can be transmitted.
30 mm
T
SOLUTION
 
KTc
2 KT

J
 c3
T 
 c3
2K
1
d  30 mm c  d  15 mm  15  103 m
2
D  60 mm, r  8 mm
D 60

 2,
30
d
r
8

 0.26667
30
d
From Fig. 3.32,
K  1.18
Allowable torque.
T 
Maximum power.
P  2 f T  (2 )(50)(202.17)  63.5  103 W
 (15  103 )3 (45  106 )
(2)(1.18)
 202.17 N  m
P  63.5 kW 
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350
348
BeerMOM_ISM_C03.indd 348
12/23/2014 3:41:35 PM