CHAPTER 3
4
0
4
0
PROBLEM 3.1
(a) Determine the maximum shearing stress caused by a 4.6-kN ⋅ m torque T in
the 76-mm-diameter shaft shown. (b) Solve part a, assuming that the solid shaft
has been replaced by a hollow shaft of the same outer diameter and of 24-mm
inner diameter.
SOLUTION
(a)
Solid shaft:
c =
J =
τ=
d
= 38 mm = 0.038 m
2
π
2
c4 =
π
2
(0.038)4 = 3.2753 × 10 −6 m4
Tc (4.6 × 103 )(0.038)
=
= 53.4 × 106 Pa
J
3.2753 × 10−6
τ = 53.4 MPa 
(b)
Hollow shaft:
do
= 0.038 m
2
1
c1 = di = 12 mm = 0.012 m
2
c2 =
J =
τ=
π
2
(c − c ) = π2 (0.038 − 0.012 ) = 3.2428 × 10 m
4
2
4
1
4
4
−6
4
Tc (4.6 × 103 )(0.038)
=
= 53.9 × 106 Pa
J
3.2428 × 10−6
τ = 53.9 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.2
(a) Determine the torque T that causes a maximum shearing stress of 45
MPa in the hollow cylindrical steel shaft shown. (b) Determine the
maximum shearing stress caused by the same torque T in a solid cylindrical
shaft of the same cross-sectional area.
SOLUTION
(a)
Given shaft:
J =
π
2
(c − c )
4
2
4
1
π
(454 − 304 ) = 5.1689 × 106 mm4 = 5.1689 × 10−6 m4
2
Tc
Jτ
τ =
T =
J
c
−6
(5.1689 × 10 )(45 × 106 )
= 5.1689 × 103 N ⋅ m
T =
−
45 × 10 3
J =
T = 5.17 kN ⋅ m 
(b)
Solid shaft of same area:
(
)
A = π c 22 − c12 = π (45 2 − 30 2 ) = 3.5343 × 103 mm 2
π c2 = A or c =
J =
τ =
π
2
c4 , τ =
A
π
= 33.541 mm
2T
Tc
= 3
J
πc
(2)(5.1689 × 103 )
= 87.2 × 106 Pa
3
(0.033541)
π
τ = 87.2 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.3
Knowing that d = 1.2 in., determine the torque T that causes a
maximum shearing stress of 7.5 ksi in the hollow shaft shown.
SOLUTION
c2 =
1
1
d =
(1.6) = 0.8 in.
2 2  2 
c1 =
1
 1
d1 =   (1.2) = 0.6 in.
2
 2
J =
π
2
c = 0.8 in.
( c − c ) = π2 (0.8 − 0.6 ) = 0.4398 in
4
2
4
1
4
Tc
J
(0.4398)(7.5)
J τmax
T =
=
c
0.8
4
4
τ max =
T = 4.12 kip ⋅ in 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.4
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
c2 =
1
 1
d2 =   (1.6) = 0.8 in.
2
 2
c1 =
1
 1
d1 =   (0.9) = 0.45 in.
2
 2
π
(
)
c = 0.8 in.
π
c24 − c14 = (0.8 4 − 0.45 4) = 0.5790 in 4
2
2
Tc (9)(0.8)
=
τ max =
J
0.5790
J =
τmax = 12.44 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.5
A torque T = 3 kN ⋅ m is applied to the solid bronze cylinder shown.
Determine (a) the maximum shearing stress, (b) the shearing stress at
point D which lies on a 15-mm-radius circle drawn on the end of the
cylinder, (c) the percent of the torque carried by the portion of the cylinder
within the 15 mm radius.
SOLUTION
(a)
c=
J =
1
d = 30 mm = 30 × 10− 3 m
2
π
2
c4 =
π
2
−
−
(30 × 10 3 )4 = 1.27235 × 10 6 m 4
T = 3 kN = 3 × 103 N
τm =
Tc
(3 × 103 )(30 × 10 −3 )
=
= 70.736 × 106 Pa
J
1.27235 × 10−6
τ m = 70.7 MPa 
(b)
ρD = 15 mm = 15 × 10 −3 m
τD =
(c)
τD =
TD =
ρD
c
τ =
TD ρ D
JD
π
(15 × 10−3 )(70.736 × 10−6 )
(30 × 10 −3)
TD =
J Dτ D
ρD
=
π
2
τ D = 35.4 MPa 
ρ D3 τ D
(15 × 10−3 )3 (35.368 × 106 ) = 187.5 N ⋅ m
2
TD
187.5
× 100% =
(100%) = 6.25%
T
3 × 103
6.25%
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.6
(a) Determine the torque that can be applied to a solid shaft of 20-mm diameter without exceeding an
allowable shearing stress of 80 MPa. (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 equal to half of its own outer
diameter.
SOLUTION
(a)
Solid shaft:
c=
J =
T=
(b)
Hollow shaft:
1
1
d = (0.020) = 0.010 m
2
2
π
2
c4 −
π
2
(0.10) 4 = 15.7080 × 10 −9 m 4
J τ max
(15.7080 × 10−9 )(80 × 106 )
=
= 125.664
0.010
c
T = 125.7 N ⋅ m 
Same area as solid shaft.
2

1   3
A = π c22 − c12 = π c22 −  c2   = π c22 = π c2

 2   4
2
2
c2 =
c=
(0.010) = 0.0115470 m
3
3
(
c1 =
J =
T =
)
1
c = 0.0057735 m
2 2
π
π
c − c )=
(0.0115470 − 0.0057735 ) = 26.180× 10 m
(
2
2
4
2
τ max J
c2
4
1
=
4
4
(80 × 10 6)(26.180 × 10− 9)
= 181.38
0.0115470
−9
4
T = 181.4 N ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.7
The solid spindle AB has a diameter ds= 1.5 in. and is made of a steel with
an allowable shearing stress of 12 ksi, while sleeve CD is made of a brass
with an allowable shearing stress of 7 ksi. Determine the largest torque T
that can be applied at A.
SOLUTION
Analysis of solid spindle AB:
c=
1
d s = 0.75 in.
2
τ = Tc
T=
J
T =
Analysis of sleeve CD:
c2 =
π
2
π
Jτ
= τ c3
c
2
(12 ×10 3)(0.75) 3 = 7.95 × 103 lb ⋅ in
1
1
d = (3) =1.5 in.
2 o 2
c 1 = c 2 − t = 1.5 − 0.25 = 1.25 in.
J =
T=
The smaller torque governs:
π
2
(c − c ) = π2 (1.5 − 1.25 ) = 4.1172 in
4
2
4
1
4
4
4
Jτ
(4.1172)(7 × 103 )
=
= 19.21× 103 lb ⋅ in
1.5
c2
T = 7.95 × 10 3 lb ⋅ in

T = 7.95 kip ⋅ in 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.8
The solid spindle AB is made of a steel with an allowable shearing stress of
12 ksi, and sleeve CD is made of a brass with an allowable shearing stress
of 7 ksi. Determine (a) the largest torque T that can be applied at A if the
allowable shearing stress is not to be exceeded in sleeve CD, (b) the
corresponding required value of the diameter d s of spindle AB.
SOLUTION
(a)
Analysis of sleeve CD:
1
1
do = (3) = 1.5 in.
2
2
c1 = c2 − t =1.5 − 0.25 =1.25 in.
c2 =
J =
T=
π
π
c − c )=
(1.5 − 1.25 ) = 4.1172 in
(
2
2
4
2
4
1
4
4
4
J τ (4.1172)(7 × 10 3)
=
= 19.21 × 103 lb ⋅ in
1.5
c2
T = 19.21 kip ⋅ in 
(b)
Analysis of solid spindle AB:
τ=
Tc
J
π
J
T
19.21 × 103
= c3= =
= 1.601 in 3
c
2
12 × 103
τ
c =3
(2)(1.601)
π
= 1.006 in. d s = 2c
d = 2.01in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.9
The torques shown are exerted on pulleys A and B. Knowing that both
shafts are solid, determine the maximum shearing stress (a) in shaft AB,
(b) in shaft BC.
SOLUTION
(a)
Shaft AB:
T AB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m
τ max =
Tc
2T
(2)(300)
=
=
J
π c 3 π (0.015) 3
= 56.588 × 10 6Pa
(b)
Shaft BC:
τ max = 56.6 MPa 
T BC = 300 + 400 = 700 N ⋅ m
d = 0.046 m, c = 0.023 m
τ max =
Tc
2T
(2)(700)
=
=
3
J
πc
π (0.023)3
= 36.626 × 106 Pa
τ max = 36.6 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.10
In order to reduce the total mass of the assembly of Prob. 3.9, a new design
is being considered in which the diameter of shaft BC will be smaller.
Determine the smallest diameter of shaft BC for which the maximum value
of the shearing stress in the assembly will not increase.
SOLUTION
Shaft AB:
TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m
τ max =
Tc
2T
(2)(300)
=
=
J
π c 3 π (0.015) 3
= 56.588 × 106 Pa = 56.6 MPa
Shaft BC:
TBC = 300 + 400 = 700 N ⋅ m
d = 0.046 m, c = 0.023 m
τ max =
Tc
2T
(2)(700)
=
=
3
J
πc
π (0.023) 3
= 36.626 × 106 Pa = 36.6 MPa
The largest stress (56.588 × 106 Pa) occurs in portion AB.
Reduce the diameter of BC to provide the same stress.
Tc
2T
=
J
π c3
2T
(2)(700)
=
= 7.875 × 10− 6 m3
c3 =
πτ max π (56.588 × 10 6 )
TBC = 700N ⋅ m
τ max =
c = 19.895 × 10− 3 m
d = 2c = 39.79 × 10− 3 m
d = 39.8 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.11
Knowing that each portion of the shafts AB, BC,
and CD consist of a solid circular rod, determine
(a) the shaft in which the maximum shearing
stress occurs, (b) the magnitude of that stress.
SOLUTION
Shaft AB:
T = 48 N ⋅ m
1
d = 7.5 mm = 0.0075 m
2
Tc
2T
= 3
τ max =
J
πc
(2) (48)
= 72.433MPa
τ max =
π (0.0075)3
c=
Shaft BC:
T = − 48 + 144 = 96 N ⋅ m
c=
Shaft CD:
1
d = 9 mm
2
τ max =
Tc 2 T
(2) (96)
=
=
= 83.835 MPa
J π c3 π (0.009) 3
T = − 48 + 144 + 60 = 156 N⋅ m
c=
1
d = 10.5 mm
2
τ max =
Answers:
Tc 2 T
(2 × 156)
=
=
= 85.79 MPa
3
J πc
π (0.0105) 3
(a) shaft CD
(b) 85.8 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.12
Knowing that an 8-mm-diameter hole has been
drilled through each of the shafts AB, BC, and
CD, determine (a) the shaft in which the
maximum shearing stress occurs, (b) the
magnitude of that stress.
SOLUTION
1
d1 = 4 mm
2
Hole:
c1 =
Shaft AB:
T = 48 N ⋅ m
c2 =
J =
τ max =
Shaft BC:
π
π
c − c ) = (0.0075 − 0.004 ) = 4.5679× 10 m
(
2
2
4
2
4
1
4
τ max =
π
c2 =
τ max =
4
1
d 2 = 9 mm
2
π
c − c ) = (0.009 − 0.004 ) = 9.904× 10 m
(
2
2
4
2
4
1
4
−9
4
4
Tc 2
(96)(0.009)
=
= 87.239 MPa
J
9.904 × 10− 9
T = − 48 + 144 + 60 = 156 N ⋅ m
J =
−9
4
Tc2
(48)(0.0075)
=
= 78.810 MPa
J
4.5679 × 10− 9
T = − 48 + 144 = 96 N ⋅ m
J =
Shaft CD:
1
d2 = 7.5 mm
2
π
2
c2 =
1
d2 = 10.5 mm
2
(c − c ) = π2 (0.0105 − 0.004 ) = 18.691× 10 m
4
2
4
1
4
−9
4
4
Tc2 (156)(0.0105)
=
= 87.636 MPa
J
18.691 × 10− 9
Answers:
(a) shaft CD
(b) 87.6 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.13
Under normal operating conditions, the
electric motor exerts a 12-kip ⋅ in.
torque at E. Knowing that each shaft is
solid, determine the maximum shearing
in (a) shaft BC, (b) shaft CD, (c) shaft
DE.
SOLUTION
(a)
Shaft BC:
From free body shown:
TBC = 3 kip ⋅ in
τ=
τ=
(b)
Shaft CD:
Tc
Tc
2 T
=
=
π 4 π c3
J
c
2
3 kip ⋅ in
2
π 1

 × 1.75 in. 
2

3
(1)
τ = 2.85 ksi 
From free body shown:
TCD = 3 + 4= 7 kip ⋅ in
From Eq. (1):
τ=
(c)
Shaft DE:
2 T
π c3
=
2 7 kip ⋅ in
π (1 in .) 3
τ = 4.46 ksi 
From free body shown:
TDE = 12 kip ⋅ in
From Eq. (1):
τ =
2 T 2 12 kip ⋅ in
=
3
π c 3 π 1

 × 2.25 in. 
2

τ = 5.37 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.14
Solve Prob.3.13, assuming that a 1in.-diameter hole has been drilled
into each shaft.
PROBLEM 3.13 Under normal
operating conditions, the electric
motor exerts a 12-kip ⋅ in. torque at
E. Knowing that each shaft is solid,
determine the maximum shearing in
(a) shaft BC, (b) shaft CD, (c) shaft
DE.
SOLUTION
(a)
Shaft BC:
From free body shown:
c2 =
J =
τ =
(b)
Shaft CD:
J =
τ=
(c)
Shaft DE:
1
(1.75) = 0.875 in .
2
π
4
2
4
1
4
J =
τ =
4
4
Tc (3 kip ⋅ in)(0.875 in .)
=
J
0.82260 in 4
τ = 3.19 ksi 
TCD = 3 + 4 = 7 kip ⋅ in
1
(2.0) = 1.0 in .
2
π
π
c − c ) = (1.0 − 0.5 ) = 1.47262 in
(
2
2
4
2
4
1
4
4
4
Tc (7 kip ⋅ in)(1.0 in.)
=
J
1.47262 in 4
From free body shown:
c2 =
1
c1 = (1) = 0.5 in .
2
π
c − c ) = (0.875 − 0.5 ) = 0.82260 in
(
2
2
From free body shown:
c2 =
TBC = 3 kip ⋅ in
τ = 4.75 ksi 
TDE = 12 kip ⋅ in
2.25
= 1.125 in.
2
π
2
( c − c ) = π2 (1.125 − 0.5 ) = 2.4179 in
4
2
4
1
4
4
Tc (12 kip ⋅ in)(1.125 in.)
=
J
2.4179 in 4
4
τ = 5.58 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.15
The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and
8 ksi in the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress
concentrations, determine the largest torque that can be applied at A.
SOLUTION
τ max = Tc , J =
π
τ max = 15 ksi
c=
J
Rod AB:
T=
Rod BC:
π
2
π
2
π
2
c 3τ max
1
d = 0.75 in.
2
(0.75)3 (15) = 9.94 kip⋅ in
τ max = 8 ksi
T =
2
c 4, T =
c=
1
d = 0.90 in.
2
(0.90) 3(8) = 9.16 kip ⋅ in
T = 9.16 kip ⋅ in 
The allowable torque is the smaller value.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.16
The allowable shearing stress is 15 ksi in the steel rod AB and 8 ksi in the
brass rod BC. Knowing that a torque of magnitude T = 10 kip ⋅ in. is applied
at A, determine the required diameter of (a) rod AB, (b) rod BC.
SOLUTION
τ max = Tc , J =
J
(a)
Rod AB:
π
2
, c3 =
πτ max
τ max = 15 ksi
T = 10 kip ⋅ in
c3 =
2T
(2)(10)
= 0.4244 in3
π (15)
d = 2 c = 1.503 in. 
c = 0.7515 in.
(b)
Rod BC:
τ max = 8 ksi
T = 10 kip ⋅ in
c3 =
(2)(10)
= 0.79577 in 2
π (8)
d = 2 c = 1.853 in. 
c = 0.9267 in.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.17
The allowable stress is 50 MPa in the brass rod AB and
25 MPa in the aluminum rod BC. Knowing that a torque of
magnitude T = 1250 N ⋅ m is applied at A, determine the
required diameter of (a) rod AB, (b) rod BC.
SOLUTION
τmax =
(a)
Rod AB:
c3 =
Tc
J
J =
π
2
(2)(1250)
π (50 × 10 6)
c4
c3 =
2T
πτ max
= 15.915 × 10−6 m3
−
c = 25.15 × 10 3 m = 25.15 mm
dAB = 2c = 50.3 mm 
(b)
Rod BC:
c3 =
(2)(1250)
π (25 × 10 6 )
= 31.831× 10−6 m3
−
c = 31.69 × 10 3 m = 31.69 mm
d BC = 2c = 63.4 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
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 AB is hollow and has an outer diameter of 25 mm; it is
made of 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, (b) the
largest torque that can be applied at A.
SOLUTION
Solid rod BC:
τ =
Tc
J
J=
π
2
c4
τ all = 25 × 10 6 Pa
c=
Tall =
Hollow rod AB:
1
d = 0.015 m
2
π
2
c 3τ all =
π
2
(0.015) 3(25 × 10 6) = 132.536 N ⋅ m
τ all = 50 × 106 Pa
Tall = 132.536 N ⋅ m
1
1
d 2 = (0.025) = 0.0125 m
2
2
c1 = ?
c2 =
Tall =
J τ all π 4
τ
=
c 2 − c14 all
c2
c2
2
c14 = c24 −
(
2Tallc 2
πτall
= 0.01254 −
(2)(132.536)(0.0125)
−
= 3.3203 × 10 9 m4
6
×
π (50 10 )
3
c1 = 7.59 × 10 − m = 7.59 mm
(a)
(b)
)
d1 = 2c1 = 15.18 mm 
Tall = 132.5 N ⋅ m 
Allowable torque.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.19
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 that can be applied at A.
SOLUTION
Rod AB:
τ all = 75 × 10 6 Pa Tall =
Jτ all
c
1
d = 0.030 m
2
π
c =
Tall =
π
2
c 3τ all =
π
2
J=
2
c4
(0.030) 3(75 × 10 6)
= 3.181 × 103 N ⋅ m
Pipe CD:
c2 =
J =
Tall =
1
d2 = 0.045 m
2
π
c1 = c2 − t = 0.045 − 0.006 = 0.039 m
π
c − c ) = (0.045 − 0.039 ) = 2.8073× 10 m
(
2
2
4
2
4
1
4
4
−6
4
(2.8073 ×10 − 6)(75 ×10 6)
= 4.679 × 10 3 N ⋅ m
0.045
Tall = 3.18 × 10 3 N ⋅ m
Allowable torque is the smaller value.
3.18 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.20
The solid rod AB has a diameter dAB = 60 mm and is made of a
steel for which the allowable shearing stress is 85 Mpa. The
pipe CD, which has an outer diameter of 90 mm and a wall
thickness of 6 mm, is made of an aluminum for which the
allowable shearing stress is 54 MPa. Determine the largest
torque T that can be applied at A.
SOLUTION
Rod AB:
τ all = 85 × 10 6 Pa
Tall =
=
Pipe CD:
c=
1
d = 0.030 m
2
J τall
π
= c3 τall
c
2
π
2
(0.030)3 (85 × 106 ) = 3.605 × 103 N ⋅ m
τ all = 54 × 106 Pa
c2 =
1
d 2 = 0.045 m
2
c 1 = c 2 − t = 0.045 − 0.006 = 0.039 m
J =
π
π
c − c ) = (0.045 − 0.039 ) = 2.8073× 10 m
(
2
2
4
2
4
1
4
4
−6
4
−
Tall =
J τ all (2.8073 × 10 6 )(54 × 106 )
=
= 3.369 × 103 N ⋅ m
0.045
c2
Tall = 3.369 × 10 3 N ⋅ m
Allowable torque is the smaller value.
3.37 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.21
A torque of magnitude T = 1000 N ⋅ m is applied at
D as shown. Knowing that the diameter of shaft AB
is 56 mm and that the diameter of shaft CD is
42 mm, determine the maximum shearing stress in
(a) shaft AB, (b) shaft CD.
SOLUTION
TCD = 1000 N ⋅ m
TAB =
(a)
(b)
Shaft AB:
Shaft CD:
rB
100
TCD =
(1000) = 2500 N ⋅ m
rC
40
c=
1
d = 0.028 m
2
τ=
Tc
2T
(2) (2500)
6
= 3 =
= 72.50 × 10
J
πc
π (0.028) 3
c=
1
d = 0.020 m
2
τ=
Tc
2T
(2) (1000)
6
= 3 =
= 68.7 × 10
J
πc
π (0.020) 3
72.5 MPa 
68.7 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.22
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.
SOLUTION
TCD = 1000 N ⋅ m
r
100
(1000) = 2500 N ⋅ m
TAB = B TCD =
rC
40
(a)
Shaft AB:
τ all = 60 × 10 6 Pa
τ = Tc = 2T3
J
πc
c3 =
2T
πτ
=
(2) (2500)
π (60 × 106 )
c = 29.82 × 10 − 3 = 29.82 mm
(b)
Shaft CD:
= 26.526 × 10 −6 m3
d = 2 c = 59.6 mm 
τ all = 60 × 10 6 Pa
2
τ = Tc = T3
J
πc
c3 =
2T
πτ
=
(2) (1000)
= 10.610 × 10−6 m3
π (60 × 10 6 )
c = 21.97 × 10 −3 m = 21.97 mm
d = 2 c = 43.9 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.23
Under normal operating conditions, a motor exerts a
torque of magnitude TF = 1200 lb ⋅ in. at F.
Knowing that rD = 8 in., rG = 3 in., and the
allowable shearing stress is 10.5 ksi ⋅ in each shaft,
determine the required diameter of (a) shaft CDE,
(b) shaft FGH.
SOLUTION
TF = 1200 lb ⋅ in
r
8
TE = D TF = (1200) = 3200 lb ⋅ in
rG
3
τ all = 10.5 ksi = 10500 psi
τ =
(a)
or
c3 =
2T
πτ
Shaft CDE:
c3 =
(b)
Tc
2T
= 3
πc
J
(2) (3200)
π (10500)
= 0.194012 in 3
c = 0.5789 in.
d DE = 2c
(2) (1200)
= 0.012757 in 3
d DE = 1.158 in. 
Shaft FGH:
c3 =
π (10500)
c = 0.4174 in.
d FG = 2c
d FG = 0.835 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.24
Under normal operating conditions, a motor exerts a
torque of magnitude TF at F. The shafts are made of a
steel for which the allowable shearing stress is 12 ksi
and have diameters dCDE = 0.900 in. and
dFGH = 0.800 in. Knowing that rD = 6.5 in. and
rG = 4.5 in., determine the largest allowable value of
TF.
SOLUTION
τ all = 12 ksi
Shaft FG:
1
d = 0.400 in.
2
π
Jτ
TF , all = all = c 3τ all
c
2
c =
=
Shaft DE:
c=
TE , all =
π
2
(0.400)3 (12) = 1.206 kip ⋅ in
1
d = 0.450 in.
2
π
2
c 3τ all
π
(0.450)3 (12) = 1.7177 kip ⋅ in
2
r
4.5
TF = G T E TF, all =
(1.7177) = 1.189 kip ⋅ in
rD
6.5
=
TF = 1.189 kip ⋅ in 
Allowable value of TF is the smaller.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.25
The two solid shafts are connected by gears as shown and are made
of a steel for which the allowable shearing stress is 8500 psi.
Knowing that a torque of magnitude TC = 5 kip ⋅ in. is applied at C
and that the assembly is in equilibrium, determine the required
diameter of (a) shaft BC, (b) shaft EF.
SOLUTION
τ max = 8500 psi = 8.5 ksi
(a)
Shaft BC:
TC = 5 kip ⋅ in
τ max =
Tc
2T
=
J
π c3
c= 3
c= 3
2T
πτ max
(2)(5)
= 0.7208 in.
π (8.5)
d BC = 2c = 1.442 in. 
(b)
Shaft EF:
r
2.5
(5) = 3.125 kip ⋅ in
TF = D TC =
rA
4
c = 3
2T
πτ max
=3
(2)(3.125)
= 0.6163 in.
π (8.5)
dEF = 2 c = 1.233 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.26
The two solid shafts are connected by gears as shown and are
made of a steel for which the allowable shearing stress is
7000 psi. Knowing the diameters of the two shafts are,
respectively, dBC = 1.6 in. and d EF = 1.25 in., determine the
largest torque TC that can be applied at C.
SOLUTION
τ max = 7000 psi = 7.0 ksi
Shaft BC:
dBC = 1.6 in.
1
d = 0.8 in.
2
Jτ max π
= τ maxc 3
TC =
c
2
c =
=
Shaft EF:
π
2
(7.0)(0.8)3 = 5.63 kip ⋅ in
dEF = 1.25 in.
1
d = 0.625 in.
2
Jτ max π
= τ maxc3
TF =
c
2
c =
=
By statics,
TC =
π
2
(7.0)(0.625) 3 = 2.684 kip ⋅ in
4
rA
TF =
(2.684) = 4.30 kip ⋅ in
rD
2.5
Allowable value of TC is the smaller.
TC = 4.30 kip ⋅ in 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.27
A torque of magnitude T = 100 N ⋅ m is applied to shaft AB of the
gear train shown. Knowing that the diameters of the three solid
shafts are, respectively, d AB = 21 mm, d CD = 30 mm, and
d EF = 40 mm, determine the maximum shearing stress in
(a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
T AB = T A = T B = T
Gears B and C:
rB = 25 mm, rC = 60 mm
Force on gear circles.
FBC =
TB
T
= C
rB
rC
r
60
TC = C TB =
T = 2.4 T
rB
25
Shaft CD:
TCD = TC = TD = 2.4T
Gears D and E:
rD = 30 mm, rE = 75 mm
Force on gear circles.
Shaft EF:
FDE =
TD
T
= E
rD
rE
TE =
rE
75
(2.4T ) = 6T
TD =
rD
30
TEF = TE = TF = 6T
Maximum Shearing Stresses. τ max =
Tc
2T
=
J
π c3
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.27 (Continued)
(a)
Shaft AB:
T = 100 N ⋅ m
c=
τ max =
(b)
Shaft CD:
τ max =
Shaft EF:
(2)(100)
π (10.5 × 10 −3 )3
= 55.0 × 10 6 Pa
τmax = 55.0 MPa 
T = (2.4)(100) = 240 N ⋅ m
c =
(c)
1
d = 10.5 mm = 10.5 × 10 −3 m
2
1
d = 15 mm = 15 × 10− 3 m
2
(2)(240)
π(15 × 10 −3) 3
= 45.3 × 106 Pa
τmax = 45.3 MPa 
T = (6)(100) = 600 N ⋅ m
c =
τ max =
1
d = 20 mm = 20 × 10− 3 m
2
(2)(600)
π(20 × 10 − 3) 3
= 47.7 × 10 6 Pa
τmax = 47.7 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.28
A torque of magnitude T = 120 N ⋅ m is applied to shaft AB of
the gear train shown. Knowing that the allowable shearing stress
is 75 MPa in each of the three solid shafts, determine the
required diameter of (a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
T AB = T A = T B = T
Gears B and C:
rB = 25 mm, rC = 60 mm
Force on gear circles.
F BC =
TB TC
=
rB
rC
TC =
rC
60
TB =
T = 2.4T
rB
25
Shaft CD:
TCD = TC = TD = 2.4T
Gears D and E:
rD = 30 mm, rE = 75 mm
Force on gear circles.
F DE =
TD TE
=
rD
rE
TE =
rE
75
(2.4T ) = 6T
TD =
rD
30
Shaft EF:
T EF = T E = T F = 6T
Required Diameters.
τ max =
Tc
2T
=
J
π c3
c =3
2T
πτ
d = 2 c = 23
2T
πτ max
τ max = 75 × 10 6 Pa
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.28 (Continued)
(a)
Shaft AB:
TAB = T = 120 N ⋅ m
d AB = 2 3
(b)
Shaft CD:
Shaft EF:
π (75 × 10 6)
= 20.1× 10−3 m
d AB = 20.1 mm 
TCD = (2.4)(120) = 288 N ⋅ m
d CD = 2 3
(c)
2(120)
(2)(288)
π (75 × 10 6 )
= 26.9 × 10− 3 m
dCD = 26.9 mm 
TEF = (6)(120) = 720 N ⋅ m
d EF = 2 3
(2)(720)
3
π (75 × 10 )
= 36.6 × 10− 3 m
dEF = 36.6 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.29
(a) For a given allowable shearing stress, determine the ratio T/w of the
maximum allowable torque T and the weight per unit length w for the
hollow shaft shown. (b) Denoting by (T /w) 0 the value of this ratio for a
solid shaft of the same radius c2 , express the ratio T/w for the hollow shaft
in terms of (T /w) 0 and c1 / c2 .
SOLUTION
w = weight per unit length,
ρg = specific weight,
W = total weight,
L = length
w=
Tall =
(a)
T
= c12 + c22 τall
W

c1 = 0 for solid shaft:
(b)
c2
(T /w)h
= 1 + 12
(T /w)0
c2
(
ρ gLA
W
=
= ρ gA = ρ g π c 22 − c12
L
L
(
(
)(
)
)
2
2
2
2
J τ all π c 42 − c14
π c2 + c1 c2 − c1
=
τall =
τall
2
2
c2
c2
c2
(
)
c12 + c22 τall
T
=
(hollow shaft) 
w
2ρ gc 2
)
c 2τ all
T 
(solid shaft)
  = 2
w
pg
 0
c12 
T  T  
=
+
1

 
    
c 22 
 w   w 0 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.30
While the exact distribution of the shearing stresses in a
hollow-cylindrical shaft is as shown in Fig. a, an approximate
value can be obtained for τ max by assuming that the stresses
are uniformly distributed over the area A of the cross section,
as shown in Fig. b, and then further assuming that all of the
elementary shearing forces act at a distance from O equal to
the mean radius ½(c1 + c 2) of the cross section. This
approximate value τ 0 = T /Arm , where T is the applied torque.
Determine the ratio τ max/τ 0 of the true value of the maximum
shearing stress and its approximate value τ 0 for values
of c1/c 2, respectively equal to 1.00, 0.95, 0.75, 0.50, and 0.
SOLUTION
For a hollow shaft:
By definition,
τmax =
Tc2
2Tc2
=
J
π c42 − c41
τ0 =
T
2T
=
Arm
A(c2 + c1)
(
)
=
2Tc2
π
(
c22 − c21
)(
c22 + c12
)
=
2Tc2
(
A c22 + c21
c (c + c )
1 + ( c1/ c2)
τ max
= 2 2 2 21 =
τ0
c 2 + c1
1 + (c 1/c 2) 2
Dividing,
)

c1 /c2
1.0
0.95
0.75
0.5
0.0
τ max /τ0
1.0
1.025
1.120
1.200
1.0

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.31
(a) For the solid steel shaft shown (G = 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
(a)
π
π
1
d = 0.015 m, J = c 4 = (0.015)4
2
2
2
J = 79.522 × 10−9 m4 , L = 1.8 m, G = 77 × 109 Pa
c =
T = 250 N ⋅ m
ϕ=
ϕ=
(b)
ϕ =
TL
GJ
(250)(1.8)
= 73.49 × 10 −3 rad
9
−9
(77 × 10 )(79.522 × 10 )
(73.49 × 10 −3)180
ϕ = 4.21 ° 
π
π 4
1
d1 = 0.010 m, J =
c2 − c14
2
2
π
TL
−
J = (0.015 4 − 0.010 4) = 63.814 × 10 9 m 4 ϕ
2
GJ
(
c2 = 0.015 m, c1
ϕ =
)
(250)(1.8)
180
−
−
= 91.58 × 10 3 rad =
(91.58 × 10 3)
(77 × 10 9)(63.814 × 10− 9)
π
ϕ = 5.25° 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.32
For the aluminum shaft shown (G = 27 GPa), determine (a) the torque T
that causes an angle of twist of 4°, (b) the angle of twist caused by the same
torque T in a solid cylindrical shaft of the same length and cross-sectional
area.
SOLUTION
TL
GJ
ϕ =
(a)
T =
GJϕ
L
ϕ = 4° = 69.813 × 10−3 rad, L = 1.25 m
G = 27 GPa = 27 × 109 Pa
J =
π
2
(c − c ) = π2 (0.018 − 0.012 ) = 132.324 × 10 m
4
2
4
1
4
−9
4
(27 × 109 )(132.324 × 109 )(69.813 × 10− 3 )
1.25
= 199.539 N⋅ m
4
T =
(b)
Matching areas:
(
A = π c2 = π c22 − c12
T = 199.5 N ⋅ m 
)
c = c22 − c12 = 0.018 2 − 0.012 2 = 0.013416 m
J =
ϕ=
π
2
c4 =
π
2
(0.013416) 4 = 50.894 × 10− 9 m 4
TL
(195.539)(1.25)
=
= 181.514 × 10 −3 rad
GJ (27 × 109 )(50.894 × 109 )
ϕ = 10.40 ° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.33
Determine the largest allowable diameter of a 10-ft-long steel rod (G = 11.2 × 10 6psi) if the rod is to be twisted
through 30° without exceeding a shearing stress of 12 ksi.
SOLUTION
ϕ = 30° =
L = 10 ft = 120 in.
30π
= 0.52360 rad
180
τ = 12 ksi = 12 × 10 3psi
τL
TL
GJ ϕ
Tc GJ ϕ c Gϕ c
=
=
ϕ =
, T =
, τ =
, c=
GJ
L
J
JL
L
Gϕ
c =
(12 × 10 3)(120)
= 0.24555 in.
(11.2 × 106 )(0.52360)
d = 2 c = 0.491 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.34
While an oil well is being drilled at a depth of 6000 ft, it is observed that the top of the 8-in.-diameter steel
drill pipe rotates though two complete revolutions before the drilling bit starts to rotate. Using
G = 11.2 × 106 psi, determine the maximum shearing stress in the pipe caused by torsion.
SOLUTION
For outside diameter of 8 in., c = 4 in.
For two revolutions, ϕ = 2(2π ) = 4π radians.
G = 11.2 × 10 6 psi
L = 6000 ft = 72000 in.
From text book,
TL
GJ
Tc
τm=
J
ϕ =
Divide (2) by (1).
(1)
(2)
τ m Gc
=
ϕ
L
τm =
Gcϕ (11 × 10 6 )(4)(4π )
=
= 7679 psi
L
72000
τ m = 7.68 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.35
The electric motor exerts a 500 N ⋅ m torque on
the aluminum shaft ABCD when it is rotating at a
constant speed. Knowing that G = 27 GPa and that
the torques exerted on pulleys B and C are as shown,
determine the angle of twist between (a) B and C,
(b) B and D.
SOLUTION
(a)
Angle of twist between B and C.
TBC = 200 N ⋅ m,
c =
J BC =
ϕ B/ C =
(b)
LBC = 1.2 m
1
d = 0.022 m, G = 27 × 109 Pa
2
π
2
−
c 4 = 367.97 × 10 9 m
TL
(200)(1.2)
=
= 24.157 × 10 −3 rad
9
9
GJ
(27 × 10 )(367.97 × 10 )
ϕ B/ C = 1.384° 
Angle of twist between B and D.
TCD = 500 N ⋅ m,
π 4
LCD = 0.9 m, c =
1
d = 0.024 m, G = 27 × 109 Pa
2
π
(0.024)4 = 521.153 × 10−9 m4
2
(500)(0.9)
= 31.980 × 10−3 rad
ϕC/ D =
(27 × 109 )(521.153 × 109 )
J CD =
2
c =
ϕB /D = ϕB /C + ϕC /D = 24.157 × 10 −3 + 31.980 × 10−3 = 56.137 × 10−3 rad
ϕB /D = 3.22° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 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
twist between (a) C and B, (b) D and B.
SOLUTION
(a)
Shaft BC:
c =
J BC =
1
d = 0.015 m
2
π
4
c 4 = 79.522 × 10 −9 m 4
LBC = 0.8 m, G = 27 × 109 Pa
ϕ BC =
(b)
Shaft CD:
TL
(400)(0.8)
=
= 0.149904 rad
GJ
(27 × 109 )(79.522 × 10 −9 )
ϕBC = 8.54° 
1
π
d = 0.018 m
J CD = c 4 = 164.896 × 10− 9 m 4
2
4
LCD = 1.0 m TCD = 400 − 900 = − 500 N ⋅ m
c =
ϕ CD =
TL
( −500)(1.0)
=
= −0.11230 rad
GJ
(27 × 109 )(164.896 × 10−9 )
ϕ BD = ϕ BC + ϕCD = 0.14904 − 0.11230 = 0.03674 rad
ϕ BD = 2.11° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.23
Under normal operating conditions, a motor exerts
torque of magnitude TF = 1200 lb ⋅ in. at F
Knowing that rD = 8 in., rG = 3 in., and th
allowable shearing stress is 10.5 ksi ⋅ in each shaf
determine the required diameter of (a) shaft CDE
(b) shaft FGH.
SOLUTION
TF = 1200 lb ⋅ in
r
8
TE = D TF = (1200) = 3200 lb ⋅ in
rG
3
τ all = 10.5 ksi = 10500 psi
(a)
τ =
Tc
2T
= 3
πc
J
c3 =
(2) (3200)
= 0.194012 in 3
π (10500)
c3 =
2T
πτ
Shaft CDE:
c = 0.5789 in.
(b)
or
d DE = 2c
d DE = 1.158 in. 
Shaft FGH:
c3 =
(2) (1200)
= 0.012757 in 3
π (10500)
c = 0.4174 in.
d FG = 2c
d FG = 0.835 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.24
Under normal operating conditions, a motor exerts a
torque of magnitude TF at F. The shafts are made of a
steel for which the allowable shearing stress is 12 ksi
and have diameters dCDE = 0.900 in. and
dFGH = 0.800 in. Knowing that rD = 6.5 in. and
rG = 4.5 in., determine the largest allowable value of
TF.
4
SOLUTION
τ all = 12 ksi
0
Shaft FG:
1
d = 0.400 in.
2
π
Jτ
TF , all = all = c 3τ all
c
2
c =
=
Shaft DE:
c=
TE , all =
=
π
2
(0.400)3 (12) = 1.206 kip ⋅ in
1
d = 0.450 in.
2
π
2
π
c 3τ all
(0.450)3 (12) = 1.7177 kip ⋅ in
2
rG
4.5
TF =
T E TF, all =
(1.7177) = 1.189 kip ⋅ in
rD
6.5
TF = 1.189 kip ⋅ in 
Allowable value of TF is the smaller.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.25
The two solid shafts are connected by gears as shown and are mad
of a steel for which the allowable shearing stress is 8500 ps
Knowing that a torque of magnitude TC = 5 kip ⋅ in. is applied at
and that the assembly is in equilibrium, determine the require
diameter of (a) shaft BC, (b) shaft EF.
SOLUTION
τ max = 8500 psi = 8.5 ksi
(a)
Shaft BC:
TC = 5 kip ⋅ in
τ max =
Tc
2T
=
J
π c3
c= 3
c= 3
2T
πτ max
(2)(5)
= 0.7208 in.
π (8.5)
d BC = 2c = 1.442 in. 
(b)
Shaft EF:
r
2.5
(5) = 3.125 kip ⋅ in
TF = D TC =
rA
4
4c = 3 02T = 3 (2)(3.125) = 0.6163 in.
πτ max
π (8.5)
dEF = 2 c = 1.233 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.26
The two solid shafts are connected by gears as shown and are
made of a steel for which the allowable shearing stress is
7000 psi. Knowing the diameters of the two shafts are,
respectively, dBC = 1.6 in. and d EF = 1.25 in., determine the
largest torque TC that can be applied at C.
SOLUTION
τ max = 7000 psi = 7.0 ksi
Shaft BC:
dBC = 1.6 in.
1
d = 0.8 in.
2
Jτ max π
= τ maxc 3
TC =
c
2
c =
=
Shaft EF:
π
2
(7.0)(0.8)3 = 5.63 kip ⋅ in
dEF = 1.25 in.
1
d = 0.625 in.
2
Jτ max π
= τ maxc3
TF =
c
2
c =
=
By statics,
TC =
π
2
(7.0)(0.625) 3 = 2.684 kip ⋅ in
4
rA
TF =
(2.684) = 4.30 kip ⋅ in
rD
2.5
Allowable value of TC is the smaller.
TC = 4.30 kip ⋅ in 
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 3.27
4
A torque of magnitude T = 100 N ⋅ m is applied to shaft AB of th
gear train shown. Knowing that the diameters of the three soli
shafts are, respectively, d AB = 21 mm, d CD = 30 mm, an
0
d EF = 40 mm, determine the maximum shearing stress i
(a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
T AB = T A = T B = T
Gears B and C:
rB = 25 mm, rC = 60 mm
Force on gear circles.
FBC =
TB
T
= C
rB
rC
r
60
TC = C TB =
T = 2.4 T
rB
25
Shaft CD:
TCD = TC = TD = 2.4T
Gears D and E:
rD = 30 mm, rE = 75 mm
Force on gear circles.
Shaft EF:
FDE =
TD
T
= E
rD
rE
TE =
rE
75
(2.4T ) = 6T
TD =
rD
30
TEF = TE = TF = 6T
Maximum Shearing Stresses. τ max =
Tc
2T
=
J
π c3
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.27 (Continued)
(a)
(b)
Shaft AB:
Shaft CD:
T = 100 N ⋅ m
c=
1
d = 10.5 mm = 10.5 × 10 −3 m
2
τ max =
(2)(100)
= 55.0 × 10 6 Pa
−3 3
π (10.5 × 10 )
T = (2.4)(100) = 240 N ⋅ m
c =
τ max =
(c)
Shaft EF:
τmax = 55.0 MPa 
1
d = 15 mm = 15 × 10− 3 m
2
(2)(240)
π(15 × 10 −3) 3
= 45.3 × 106 Pa
τmax = 45.3 MPa 
T = (6)(100) = 600 N ⋅ m
c =
τ max =
1
d = 20 mm = 20 × 10− 3 m
2
(2)(600)
π(20 × 10 − 3) 3
= 47.7 × 10 6 Pa
4
0
τmax = 47.7 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.28
A torque of magnitude T = 120 N ⋅ m is applied to shaft AB
the gear train shown. Knowing that the allowable shearing stre
is 75 MPa in each of the three solid shafts, determine th
required diameter of (a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
T AB = T A = T B = T
Gears B and C:
rB = 25 mm, rC = 60 mm
Force on gear circles.
F BC =
TB TC
=
rB
rC
TC =
rC
60
TB =
T = 2.4T
rB
25
Shaft CD:
TCD = TC = TD = 2.4T
Gears D and E:
rD = 30 mm, rE = 75 mm
Force on gear circles.
F DE =
TD TE
=
rD
rE
TE =
rE
75
(2.4T ) = 6T
TD =
rD
30
Shaft EF:
T EF = T E = T F = 6T
Required Diameters.
Tc
2T
=
τ max =
J
π c3
4
0
c =3
2T
πτ
d = 2 c = 23
2T
πτ max
τ max = 75 × 10 6 Pa
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.28 (Continued)
(a)
Shaft AB:
TAB = T = 120 N ⋅ m
d AB = 2 3
(b)
Shaft CD:
Shaft EF:
π (75 × 10 6)
= 20.1× 10−3 m
d AB = 20.1 mm 
TCD = (2.4)(120) = 288 N ⋅ m
d CD = 2 3
(c)
2(120)
(2)(288)
6
π (75 × 10 )
= 26.9 × 10− 3 m
dCD = 26.9 mm 
TEF = (6)(120) = 720 N ⋅ m
d EF = 2 3
(2)(720)
3
π (75 × 10 )
= 36.6 × 10− 3 m
dEF = 36.6 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.29
(a) For a given allowable shearing stress, determine the ratio T/w of th
maximum allowable torque T and the weight per unit length w for th
hollow shaft shown. (b) Denoting by (T /w) 0 the value of this ratio for
solid shaft of the same radius c2 , express the ratio T/w for the hollow sha
in terms of (T /w) 0 and c1 / c2 .
SOLUTION
w = weight per unit length,
ρg = specific weight,
W = total weight,
L = length
4
0
w=
ρ gLA
W
=
= ρ gA = ρ g π c 22 − c12
L
L
(
)
L
L
(
)(
)
2
2
2
2
J τ all π c 42 − c14
π c2 + c1 c2 − c1
=
τall =
τall
Tall =
2
2
c2
c2
c2
(a)
T
= c12 + c22 τall
W

c1 = 0 for solid shaft:
(b)
c 21
(T /w)h
1
= + 2
(T /w)0
c2
(
(
)
c12 + c22 τall
T
=
(hollow shaft) 
w
2ρ gc 2
)
c 2τ all
T 
(solid shaft)
  = 2
pg
 w 0
c12 
T  T  
  =   1 + 2  
c2 
 w   w 0 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.30
While the exact distribution of the shearing stresses in a
hollow-cylindrical shaft is as shown in Fig. a, an approximate
value can be obtained for τ max by assuming that the stresses
are uniformly distributed over the area A of the cross section,
as shown in Fig. b, and then further assuming that all of the
elementary shearing forces act at a distance from O equal to
the mean radius ½(c1 + c 2) of the cross section. This
approximate value τ 0 = T /Arm , where T is the applied torque.
Determine the ratio τ max/τ 0 of the true value of the maximum
shearing stress and its approximate value τ 0 for values
of c1/c 2, respectively equal to 1.00, 0.95, 0.75, 0.50, and 0.
SOLUTION
For a hollow shaft:
By definition,
τmax =
Tc2
2Tc2
=
J
π c42 − c41
τ0 =
T
2T
=
Arm
A(c2 + c1)
(
)
=
2Tc2
π
(
c22 − c21
)(
c22 + c12
)
=
2Tc2
(
A c22 + c21
c (c + c )
1 + ( c1/ c2)
τ max
= 2 2 2 21 =
τ0
c 2 + c1
1 + (c 1/c 2) 2
Dividing,

c1 /c2
1.0
0.95
0.75
0.5
0.0
τ max /τ0
1.0
1.025
1.120
1.200
1.0
4
)
0

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.31
(a) For the solid steel shaft shown (G = 77 GPa), determine the angl
of twist at A. (b) Solve part a, assuming that the steel shaft is hollo
with a 30-mm-outer diameter and a 20-mm-inner diameter.
SOLUTION
(a)
π
π
1
d = 0.015 m, J = c 4 = (0.015)4
2
2
2
−9 4
J = 79.522 × 10 m , L = 1.8 m, G = 77 × 109 Pa
c =
T = 250 N ⋅ m
ϕ=
ϕ=
(b)
ϕ =
TL
GJ
(250)(1.8)
= 73.49 × 10 −3 rad
(77 × 10 9 )(79.522 × 10− 9)
(73.49 × 10 −3)180
ϕ = 4.21 ° 
π
π 4
1
d1 = 0.010 m, J =
c2 − c14
2
2
π
TL
−
J = (0.015 4 − 0.010 4) = 63.814 × 10 9 m 4 ϕ
2
GJ
c2 = 0.015 m, c1
ϕ =
(
)
(250)(1.8)
180
−
−
= 91.58 × 10 3 rad =
(91.58 × 10 3)
9
−9
(77 × 10 )(63.814 × 10 )
π
ϕ = 5.25° 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by4McGraw-Hill
0 for their individual course preparation. A student using this manual is using i
without permission.
PROBLEM 3.32
For the aluminum shaft shown (G = 27 GPa), determine (a) the torque T
that causes an angle of twist of 4°, (b) the angle of twist caused by the same
torque T in a solid cylindrical shaft of the same length and cross-sectional
area.
4
0
PROBLEM 3.54
Solve Prob. 3.53, assuming that cylinder AB is made of steel, for
which G = 11.2 × 106 psi.
PROBLEM 3.53 The solid cylinders AB and BC are bonded together
at B and are attached to fixed supports at A and C. Knowing that the
modulus of rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi for
brass, determine the maximum shearing stress (a) in cylinder AB,
(b) in cylinder BC.
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation ϕ B for each cylinder.
Cylinder AB:
Cylinder BC.
π
1
d = 0.75 in. L = 12 in. J = c4 = 0.49701 in4
2
2
T L
T AB (12)
−
= 2.1557 × 10 6TAB
ϕB = AB =
GJ
(11.2 × 106 )(0.49701)
c=
π
π
1
4
4
d = 1.0 in. L = 18 in. J = c4 =
(1.0) = 1.5708 in
2
2
2
TBC L
TBC (18)
=
= 2.0463 × 10−6TBC
ϕB =
6
×
GJ
(5.6 10 )(1.5708)
c=
Matching expressions for ϕ B
2.1557 × 10 −6TAB = 2.0463 × 10 −6TBC
Equilibrium of connection at B : TAB + TBC − T = 0
TAB = 6.0872 × 10 3 lb ⋅ in
(2)
TBC = 6.4128 × 103 lb ⋅ in
Maximum stress in cylinder AB.
τ AB =
(b)
T AB + TBC = 12.5 × 103
(1)
2.0535 TAB = 12.5 × 103
Substituting (1) into (2),
(a)
TBC = 1.0535 T AB
TABc
(6.0872 × 103 ) (0.75)
=
= 9.19× 103 psi
J
0.49701
τ AB = 9.19 ksi 
Maximum stress in cylinder BC.
τBC =
TBC c (6.4128 × 10 3)(1.0)
=
= 4.08 ×10 3 psi
J
1.5708
τ BC = 4.08 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.55
Two solid steel shafts are fitted with flanges that 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 = 11.2 × 106 psi, determine the maximum
shearing stress in each shaft when a torque of T of magnitude
420 kip ⋅ ft is applied to the flange indicated.
PROBLEM 3.55 The torque T is applied to flange B.
SOLUTION
Shaft AB:
T = TAB , LAB = 2 ft = 24 in., c =
J AB =
π
c4 =
π
2
2
TAB LAB
ϕB =
GJ AB
TAB =
1
d = 0.625 in.
2
(0.625) 4 = 0.23968 in 4
GJ ABϕB (11.2 × 106 ) (0.23968)
−
ϕB
24
LAB
= 111.853 × 103ϕB
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD , LCD = 3ft = 36 in., c =
π
1
d = 0.75 in.
2
π
c 4 = (0.75) 4 = 0.49701 in4
2
2
T L
ϕC = CD CD
GJ CD
J CD =
TCD =
GJCD ϕC
(11.2 × 106 )(0.49701)
=
ϕ C = 154.625 × 103ϕ C
LCD
36
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.55 (Continued)
Clearance rotation for flange B:
ϕ B′ = 1.5° = 26.18 × 10 −3 rad
Torque to remove clearance:
′ = (111.853 × 10 3)(26.18 × 10− 3) = 2928.3 lb ⋅ in
TΑΒ
Torque T ′′ to cause additional rotation ϕ ′′ :
T′′ = 5040 − 2928.3 = 2111.7 lb⋅ in
′′ + T ′′CD
T ′′ = TΑΒ
2111.7 = (111.853 × 103 )ϕ ′′ + (154.625 × 10 3)ϕ ′′ ∴
ϕ ′′ = 7.923× 10−3 rad
′′ = (111.853 × 103 ) (7.923 × 10−3 ) = 886.21 lb ⋅ in
TΑΒ
−
′′ = (154.625 × 103 )(7.923 × 10 3 ) = 1225.09 lb ⋅ in
TCD
Maximum shearing stress in AB.
τ AB =
TAB c (2928.3 + 886.21)(0.625)
=
= 9950 psi
J AB
0.23968
τ AB = 9.95 ksi 
TCD c (1225.09)(0.75)
=
= 1849 psi
JCD
0.49701
τ CD = 1.849 ksi 
Maximum shearing stress in CD.
τ CD =
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.56
Two solid steel shafts are fitted with flanges that 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 = 11.2 × 106 psi, determine the maximum
shearing stress in each shaft when a torque of T of magnitude
420 kip ⋅ ft is applied to the flange indicated.
PROBLEM 3.56 The torque T is applied to flange C.
SOLUTION
Shaft AB:
T = T AB, L AB = 2ft = 24 in., c =
J AB =
π
c4 =
π
2
2
TAB LAB
ϕB =
GJ AB
TAB =
1
d = 0.625 in.
2
(0.625)4 = 0.23968 in4
GJ ABϕ B (11.2 × 106 ) (0.23968)
−
ϕB
24
L AB
= 111.853 × 103ϕ B
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD ,
π
LCD = 3 ft = 36 in., c =
1
d = 0.75 in.
2
π
c 4 = (0.75) 4 = 0.49701 in4
2
2
T L
ϕ C = CD CD
GJCD
J CD =
TCD =
Clearance rotation for flange C:
GJ CD ϕC
(11.2 × 10 6 )(0.49701)
=
ϕC = 154.625 × 103 ϕC
LCD
36
ϕC′ = 1.5° = 26.18 × 10− 3 rad
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.56 (Continued)
′ = (154.625 × 10 3)(26.18 × 10− 3) = 4048.1 lb ⋅ in
TCD
Torque to remove clearance:
Torque T′′ to cause additional rotation ϕ ′′ :
T′′ = 5040 − 4048.1 = 991.9 lb⋅ in
′′ + T ′′CD
T ′′ = TΑΒ
991.9 = (111.853 × 103 )ϕ ′′ + (154.625 × 103 )ϕ ′′ ∴ ϕ ′′ = 3.7223 × 10− 3 rad
′′ = (111.853 × 10−3 ) (3.7223 × 10−3 ) = 416.35 lb ⋅ in
TAB
−
−
′′ = (154.625 × 10 3 ) (3.7223 × 10 3 ) = 575.56 lb ⋅ in
TCD
Maximum shearing stress in AB.
τ AB =
TAB c (416.35) (0.625)
=
= 1086 psi
J AB
0.23968
τ AB = 1.086 ksi 
(4048.1 + 575.56)(0.75)
TCD c
=
= 6980 psi
J CD
0.49701
τ CD = 6.98 ksi 
Maximum shearing stress in CD.
τ CD =
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.57
Ends A and D of two solid steel shafts AB and CD are fixed, while
ends B and C are connected to gears as shown. Knowing that a
4-kN ⋅ m torque T is applied to gear B, determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
Gears B and C:
r
rB
φ B = C φC =
40
φC
100
φ B = 0.4φ C
(1)
 M C = 0 : TCD = rC F
(2)
 M B = 0 : T − T AB = rBF
(3)
Solve (2) for F and substitute into (3):
r
T − T AB = B TCD
rC
T = T AB +
100
T
40 CD
(4)
T = TAB + 2.5 TCD
L = 0.3 m, c = 0.030 m
Shaft AB:
φB = φB A =
TAB L
TAB (0.3)
T
=
= 235.79 × 103 AB
π
JG
G
4
(0.030) 6
2
(5)
L = 0.5 m, c = 0.0225 m
Shaft CD:
φC = φC D =
TCD L
JG
TCD (0.5)
π
2
= 1242 × 103
4
(0.0225) G
TCD
G
(6)
Substitute from (5) and (6) into (1):
φB = 0.4 φC :
235.79 × 103
TCD = 0.47462 = TAB
TAB
T
= 0.4 × 1242 × 103 CD
G
G
(7)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.57 (Continued)
Substitute for TCD from (7) into (4):
T = T AB + 2.5 (0.47462T AB)
T = 2.1865 T AB (8)
4000 N ⋅ m = 2.1865TAB
TAB = 1829.4 N ⋅ m
For T = 4 kN ⋅ m, Eq. (8) yields
TCD = 0.47462(1829.4) = 868.3 N ⋅ m
Substitute into (7):
(a)
Stress in AB:
τ AB =
(b)
TABc 2 TAB
2 1829.4
=
=
= 43.1 × 106
J
π c3
π (0.030) 3
τ AB = 43.1 MPa 
Stress in CD:
τCD =
TCDc
2 TCD
2 868.3
=
=
= 48.5 × 106
J
π c3
π (0.0225) 3
τ CD = 48.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.58
Ends A and D of the two solid steel shafts AB and CD are fixed,
while ends B and C are connected to gears as shown. Knowing that
the allowable shearing stress is 50 MPa in each shaft, determine the
largest torque T that may be applied to gear B.
SOLUTION
Gears B and C:
φB =
40
rC
φC =
φC
rB
100
φB = 0.4 φC
(1)
ΣM C = 0: TCD = rC F
(2)
ΣM B = 0: T − TAB = rB F
(3)
Solve (2) for F and substitute into (3):
r
T − TAB = B TCD
rC
T = TAB +
100
TCD
40
(4)
T = TAB + 2.5 TCD
L = 0.3 m, c = 0.030 m
Shaft AB:
φB = φB / A =
TAB L
TAB (0.3)
T
=
= 235.77 × 103 AB
π
JG
G
4
(0.030) G
2
(5)
L = 0.5 m, c = 0.0225 m
Shaft CD:
φC = φC /D =
TCD L
TCD (0.5)
T
=
= 1242 × 103 CD
π
JG
G
4
(0.0225) G
2
(6)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.58 (Continued)
Substitute from (5) and (6) into (1) :
φB = 0.4 φC :
235.79 × 103
TAB
T
= 0.4 × 1242 × 103 CD
G
G
TCD = 0.47462 TAB
(7)
Substitute for TCD from (7) into (4):
T = TAB + 2.5 (0.47462 TAB )
T = 2.1865 TAB
(8)
T = 4.6068 TCD
(9)
Solving (7) for TAB and substituting into (8),
 T

T = 2.1865  CD 
0.47462


Stress criterion for shaft AB:
τ AB = τ all = 50 MPa:
τ AB =
=
TABc
J
π
2
TAB =
π 3
J
τ AB = c τ AB
c
2
(0.030 m)3 (50 × 106 Pa) = 2120.6 N ⋅ m
T = 2.1865(2120.6 N ⋅ m) = 4.64 kN ⋅ m 
From (8):
Stress criterion for shaft CD:
τ CD = τ all = 50 MPa:
π
π
T c
TCD = c 3τCD = (0.0225 m)3 (50 × 106 Pa)
J
2
2
= 894.62 N ⋅ m
τ CD = CD
From (7):
T = 4.6068(894.62 N ⋅ m) = 4.12 kN ⋅ m
The smaller value for T governs.
T = 4.12 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.59
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 applied as shown, determine the
maximum shearing stress in the jacket.
SOLUTION
c =
Solid shaft:
JS =
1
d = 0.020 m
2
π
2
c4 =
π
2
(0.020) 4 = 251.33 × 10 −9 m 4
1
d = 0.040 m
2
c1 = c2 − t = 0.040 − 0.004 = 0.036 m
c2 =
Jacket:
JJ =
π
π
c − c ) = (0.040 − 0.036 )
(
2
2
4
2
4
1
4
4
= 1.3829 × 10−6 m 4
Torque carried by shaft.
TS = GJS ϕ/L
Torque carried by jacket.
TJ = GJ J ϕ /L
Total torque.
T = TS + T J = (J S + J J ) G ϕ /L
TJ =
∴
Gϕ
T
=
L
JS + JJ
JJ
(1.3829 × 10 −6 ) (500)
T =
= 423.1 N ⋅ m
JS + JJ
1.3829 × 10−6 + 251.33 × 10−6
Maximum shearing stress in jacket.
τ =
TJ c2
(423.1) (0.040)
=
= 12.24 × 106 Pa
JJ
1.3829 × 10−6
12.24 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.60
A solid shaft and a hollow shaft are made of the same material and
are of the same weight and length. Denoting by n the ratio c 1/ c 2,
show that the ratio Ts /Th of the torque Ts in the solid shaft to the
torque
Th
in
the
hollow
shaft
is
(1 − n 2) /(1 + n 2) if the maximum shearing stress is the same in
(a)
each shaft, (b) (1 − n)/ (1 + n 2) if the angle of twist is the same for
each shaft.
SOLUTION
For equal weight and length, the areas are equal.
(
)
π c02 = π c22 − c12 = π c22(1 − n 2)
Js =
(a)
π
2
c04 =
π
2
c0 = c2 (1 − n 2)1/2
∴
c24(1 − n 2) 2
Jh =
π
π
c − c ) = c (1 − n )
(
2
2
4
2
4
1
4
2
4
For equal stresses.
τ =
Ts c0
Tc
= h 2
Js
Jh
π 4 − 2 2
Ts
Jc
1 − n2
(1 − n 2 )1/2
2 c 2 (1 n ) c 2
= s2 =
=
=
4
4
2
1/2
2
2
1/2
π c (1 − n )c (1 − n )
Th
J hc0
(1 + n ) (1 − n )
1 + n2
2
2 2
(b)

For equal angles of twist.
ϕ =
T sL
TL
= h
GJ s
GJ h
4
2 2
π
c (1 − n )
Ts
J
(1 − n 2) 2 1 − n 2
= s = 2 24
=
=
π c (1 − n4 )
Th
Jh
1 − n4
1 + n2
2 2

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.61
A torque T is applied as shown to a solid tapered shaft AB. Show by
integration that the angle of twist at A is
φ =
7TL
12π Gc 4
SOLUTION
Introduce coordinate y as shown.
r =
cy
L
Twist in length dy:
dϕ =
Tdy
Tdy
2TL4dy
=
=
π
GJ
π Gc 4 y 4
G r4
2
4
2 L 2TL dy
=
4 4
ϕ = L
π Gc y
2TL

2L dy
π Gc4 L y4
2L
=
2TL 4  1 
2TL 4  1
1 
−
=
−
+ 3


4
3
4 
3
π Gc  3 y L
π Gc  24L
3L 
=
2TL 4  7 
7TL

=
π Gc 4  24L3  12π Gc 4

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 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 × 106 psi, determine the diameter of the wire for which the torsional
spring constant will be 4.27 lb ⋅ ft/rad.
SOLUTION
Torsion spring constant K = 4.27 lb ⋅ ft/rad = 51.24 lb ⋅ in/rad
K =
c4 =
T
ϕ
=
π Gc4
T
GJ
=
=
2L
TL /GJ
L
2 LK (2)(72) (51.24)
−
=
= 209.7 × 10 6 in4
6
G
π
π (11.2 × 10 )
d = 2c = 0.241 in. 
c = 0.1203 in.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.63
An annular plate of thickness t and modulus G is used to
connect shaft AB of radius r1 to tube CD of radius r2. Knowing
that a torque T is applied to end A of shaft AB and that end D
of tube CD is 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
ϕ BC =
T  1
1
 2 − 2 
4π Gt  r1
r2 
SOLUTION
Use a free body consisting of shaft AB and an inner portion of the plate BC, the outer radius of this portion
being ρ.
The force per unit length of circumference is τ t.
ΣM = 0
τ t(2 πρ) ρ − T = 0
T
τ =
(a)
Maximum shearing stress occurs at ρ = r1
τ
γ =
Shearing strain:
G
2π tρ 2
τ max =
=
T
2π tr12

T
2π GT ρ 2
The relative circumferential displacement in radial length dρ is
dδ = γ d ρ = ρ dϕ
dϕ = γ
dϕ =
(b)
ϕB /C =
=
r2
T dρ
 2π Gtρ
r1
=
3
T
2πGt

r2 d ρ
r1
ρ
T
dρ
2 π Gt ρ 2 ρ
=
3
ρ
dρ
=
T dρ
2 π GTρ 3
T  1  r2
−

2πGt  2 ρ2  r1
T  1
T  1
1 
1 
− 2 + 2  =
 2 − 2
2πGt  2r2
r2 
2r1  4πGt  r1

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.64
Determine the maximum shearing stress in a solid shaft of 12-mm diameter as it transmits 2.5 kW at a
frequency of (a) 25 Hz, (b) 50 Hz.
SOLUTION
(a)
(b)
f = 25 Hz
f = 50 Hz
c =
1
d = 6 mm = 0.006 m
2
T =
P
2500
=
= 15.9155 N ⋅ m
2π f
2π (25)
τ =
Tc
2T
2(15.9155)
=
=
= 46.9 × 106 Pa
3
J
πc
π (0.006) 3
T=
2500
= 7.9577 N ⋅ m
2π (50)
τ =
2(7.9577)
π (0.006) 3
= 23.5 × 10 6 Pa
P = 2.5 kW = 2500 W
τ = 46.9 MPa 
τ = 23.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.65
Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of
(a) 750 rpm, (b) 1500 rpm.
SOLUTION
c =
(a)
f =
T =
(b)
1
d = 0.75 in.
2
P = 75 hp = (75)(6600) = 495 × 103 lb ⋅ in/s
750
= 12.5 Hz
60
P
2π f
=
495 × 103
= 6.3025 × 103 lb ⋅ in
2 π(12.5)
τ=
Tc
2T
(2) (6.3025× 103 )
=
=
= 9.51× 103 psi
3
J
πc
π (0.75)3
f =
1500
= 25 Hz
60
T =
495 × 103
= 3.1513 × 103 lb ⋅ in
2π (25)
τ =
(2) (3.1513× 103 )
= 4.76 × 103 psi
π (0.75) 3
τ = 9.51 ksi 
τ = 4.76 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.66
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
τ all = 35 × 10 6 Pa
P = 0.375 × 10 3 W
f = 29 Hz
3
T=
P
0.375 × 10
=
= 2.0580 N ⋅ m
2π f
2π (29)
τ =
Tc
2T
=
J
πc 3
∴
c3 =
2T
πτ
=
(2) (2.0580)
= 37.43 × 10− 9 m3
6
π (35 × 10 )
c = 3.345 × 10 −3 m = 3.345 mm
d = 6.69 mm 
d = 2c
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.67
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
τ all = 7500 psi P = 100 hp = 660 × 103 lb ⋅ in/s
f =
1200
= 20 Hz
60
T =
Tc
2T
= 3
J
πc
∴ c3 =
c = 0.7639in.
d = 2c
τ =
2T
πτ
P
2π f
=
=
660 × 103
= 5.2521 × 103 lb ⋅ in
2π (20)
(2) (5.2521 × 10 3)
= 0.4458 in3
π(7500)
d = 1.528 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.68
Determine the required thickness of the 50-mm tubular shaft of Example 3.07 if is to transmit the same power
while rotating at a frequency of 30 Hz.
SOLUTION
P = 100 kW = 100 × 10 3 W
From Example 3.07,
τ all = 60 MPa = 60 × 106 Pa
c2 =
1
d = 25 mm = 0.025 m
2
f = 30 Hz
T =
J =
P
= 530.52 N ⋅ m
2π f
π
2
Tc2
2Tc2
=
J
π c42 − c14
(c − c )
τ =
2Tc2
= 0.0254 −
4
2
c14 = c24 −
4
1
πτ
(
)
(2)(530.52) (0.025)
= 249.90 × 10−9 m4
6
π (60 × 10 )
3
c1 = 22.358 × 10 − m = 22.358 mm
t = c2 − c1 = 25 mm − 22.358 mm = 2642 mm
t = 2.64 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.69
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 12-ft length. Using G = 11.2 × 10 6 psi, determine the power
being transmitted.
SOLUTION
ϕ = 2° = 34.907 × 10−3 rad
c2 =
1
d o = 1.5in.
2
π
(
)
c1 =
L = 12 ft = 144 in.
1
di = 0.6in.
2
π
c42 − c14 = (1.54 − 0.64 ) = 7.7486 in4
2
2
120
= 2 Hz
f =
60
J=
T =
GJϕ
(11.2 × 106 ) (7.7486) (34.907 × 10−3 )
=
= 21.037 × 103 lb ⋅ in
L
144
P = 2π fT = 2π (2)(21.037 × 103 ) = 264.36× 103 lb⋅in/s
P = 40.1 hp 
Since 1 hp = 6600 lb ⋅in/s,
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.70
The hollow steel shaft shown (G = 77.2 GPa, τ all = 50 MPa)
rotates at 240 rpm. Determine (a) the maximum power that can
be transmitted, (b) the corresponding angle of twist of the shaft.
SOLUTION
1
d2 = 30 mm
2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
π
c − c ) = [(30) − (12.5) ]
(
2
2
4
2
4
1
4
4
= 1.234 × 106 mm4 = 1.234 × 10−6 m4
τ m = 50 × 106 Pa
τm =
(a)
Tc
J
T =
τ mJ
c
−
=
(50 × 10 6)(1.234 × 10 6)
= 2056.7 N ⋅ m
30 × 10− 3
Angular speed.
f = 240 rpm = 4 rev/sec = 4 Hz
Power being transmitted.
P = 2π f T = 2π (4)(2056.7) = 51.7 × 10 3 W
P = 51.7 kW 
(b)
Angle of twist.
ϕ =
TL
(2056.7)(5)
=
= 0.1078 rad
GJ
(77.2 × 10 9)(1.234 × 10− 6)
ϕ = 6.17° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.71
As the hollow steel shaft shown rotates at 180 rpm, a
stroboscopic measurement indicates that the angle of twist of
the shaft is 3°. Knowing that G = 77.2 GPa, determine (a) the
power being transmitted, (b) the maximum shearing stress in the
shaft.
SOLUTION
1
d = 30 mm
2 2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
2
( c − c ) = π2 [(30) − (12.5) )]
4
2
4
1
4
4
= 1.234 × 10 6 mm 4 = 1.234 × 10− 6 m 4
ϕ = 3 ° = 0.05236 rad
ϕ =
TL
GJ
T =
GJ ϕ
(77.2 × 109)(1.234 × 10 6 )(0.0536)
=
= 997.61 N ⋅ m
5
L
−
(a)
Angular speed:
f = 180 rpm = 3 rev/sec = 3 Hz
Power being transmitted.
P = 2π f T = 2π (3)(997.61) = 18.80 × 103 W
P = 18.80 kW 
(b)
Maximum shearing stress.
τm =
Tc 2 (997.61)(30 × 10 −3)
=
J
1.234 × 10− 6
= 24.3 × 10 6 Pa
τ m = 24.3 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.72
The design of a machine element calls for a 40-mm-outer-diameter
shaft to transmit 45 kW. (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
(a)
f =
720
= 12 Hz
60
P = 45 kW = 45 × 103 W
T =
(b)
P
2π f
=
45 × 103
= 596.83 N ⋅ m
2π (12)
c =
1
d = 20 mm = 0.020 m
2
τ=
Tc
2T
(2)(596.83)
=
=
= 47.494 × 10 6 Pa
3
3
J
πc
π (0.020)
f =
1080
= 18 Hz
60
T =
45 × 103
= 397.89 N ⋅ m
2π (18)
τ =
Tc2
2Tc2
=
J
π c 42 − c14
c14 = c24 −
(
τmax = 47.5 MPa 
)
2Tc 2
πτ
c14 = 0.0204 −
(2)(397.89)(0.020)
= 53.333 × 10− 9
π (47.494 × 106 )
3
c1 = 15.20 × 10 − m = 15.20 mm
d 2 = 2 c1 = 30.4 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.73
A steel pipe of 3.5-in. outer diameter is to be used to transmit a torque
of 3000 lb ⋅ ft without exceeding an allowable shearing stress of 8 ksi.
A series of 3.5-in.-outer-diameter pipes is available for use. Knowing
that the wall thickness of the available pipes varies from 0.25 in. to
0.50 in. in 0.0625-in. increments, choose the lightest pipe that can be
used.
SOLUTION
T = 3000 lb ⋅ ft = 36 × 10 3 lb ⋅ in
1
d o = 1.75 in.
2
Tc
2Tc2
τ= 2=
J
π c24 − c14
c2 =
(
c14 = c24 −
)
2 Tc2
(2)(36 × 10 3 )(1.75)
= 1.75 4 −
= 4.3655 in 4
πτ
π(8 × 103 )
c1 = 1.4455 in.
Required minimum thickness: t = c 2 − c1
t = 1.75 − 1.4455 = 0.3045 in.
Available thicknesses: 0.25 in., 0.3125 in., 0.375 in., etc.
t = 0.3125 in. 
Use
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.74
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A, operating at a speed
of 1260 rpm, to a machine tool at D. Knowing that the
maximum allowable shearing stress is 8 ksi, determine
the required diameter (a) of shaft AB, (b) of shaft CD.
SOLUTION
(a)
Shaft AB:
P = 16 hp = (16)(6600) = 105.6 × 10 3 lb ⋅ in/sec
1260
= 21 Hz
60
τ = 8 ksi = 8 × 10 3 psi
f =
TAB =
τ =
P
2π f
=
105.6 × 103
= 800.32 lb ⋅ in
2π (21)
Tc
2T
= 3
J
πc
c =3
c =3
2T
πτ
(2)(800.32)
= 0.399 in.
π (8 × 103 )
d AB = 0.799 in. 
d AB = 2 c = 0.799 in.
(b)
Shaft CD:
r
5
TCD = C TAB = (800.32) = 1.33387 × 103 lb ⋅ in
rB
3
c = 3
2T
πτ
= 3
(2)(1.33387 × 103 )
= 0.473 in.
π (8 × 103 )
dCD = 2c = 0.947 in.
dCD = 0.947 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.75
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A operating at a speed
of 1260 rpm to a machine tool at D. Knowing that each
shaft has a diameter of 1 in., determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
(a)
Shaft AB:
P = 16 hp = (16)(6600) = 105.6 × 10 3 lb ⋅ in/sec
1260
= 21 Hz
60
P
105.6 × 103
=
= 800.32 lb ⋅ in
TAB =
2π f
2π (21)
f =
1
d = 0.5 in.
2
Tc
2T
=
τ=
J
π c3
(2)(800.32)
=
= 4.08 × 103 psi
π (0.5)3
c=
(b)
Shaft CD:
τ AB = 4.08 ksi 
TCD =
rC
5
T = (800.32) = 1.33387 × 10 3lb ⋅ in
rB AB
3
τ =
2T
(2)(1.33387 × 10 3)
=
= 6.79 × 10 3 psi
3
3
c
(0.5)
π
π
τ CD = 6.79 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.76
Three shafts and four gears are used to form a gear train
that will transmit 7.5 kW from the motor at A to a
machine tool at F. (Bearings 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
P = 7.5 kW = 7.5 × 103 W
Shaft AB:
f AB = 30 Hz
τ=
c3AB =
T AB =
τ all = 60 MPa = 60 × 106 Pa
P
7.5 × 103
=
= 39.789 N ⋅ m
2π f AB
2π(30)
Tc AB
2T
= 3 ∴
J AB π c AB
3
=
cAB
2T
πτ
(2)(39.789)
= 422.17 × 10 −9 m3
6
π(60 × 10 )
c AB = 7.50 × 10 − 3 m = 7.50 mm
Shaft CD:
d AB = 2 c AB = 15.00 mm 
P
7.5 × 103
=
= 99.472 N ⋅ m
2πf CD
2π(12)
f CD =
rB
60
f AB =
(30) = 12 Hz
rC
150
τ CD =
Tc CD
2T
2T
2(99.472)
3
= 3
∴ cCD
= CD =
= 1.05543 × 10− 6 m 3
πτ
J CD πc CD
π(60 × 10 6 )
TCD =
cCD = 10.18 × 10− 3 m = 10.18 mm
Shaft EF:
60
r
f EF = D f CD =
(12) = 4.8 Hz
rE
150
τ=
dCD = 2 cCD = 20.4 mm 
TEF =
7.5 × 10 3
P
=
= 248.68 N ⋅ m
2πf EF 2 π (4.8)
TcEF
2T
(2)(248.68)
3
=
∴ cEF
=
= 2.6886 × 10−6 m3
3
J EF
π cEF
π(60 × 10 6)
cEF =13.82 × 10 − 3 m =13.82 mm
d EF = 2c EF = 27.6 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.77
Three shafts and four gears are used to form a gear train that will
transmit power from the motor at A to a machine tool at F. (Bearings
for the shafts are omitted in the sketch.) The diameter of each shaft is
as follows: d AB = 16 mm, d CD = 20 mm, d EF = 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
τ all = 75 MPa = 75 × 106 Pa
Shaft AB:
c AB =
1
d AB = 0.008 m
2
π
TcAB
2T
=
3
J AB
π c AB
π
(0.008)3 (75 × 10 6) = 60.319 N ⋅ m
2
fAB = 24 Hz Pall = 2π fAB Tall = 2π (24) (60.319) = 9.10 × 103 W
Tall =
Shaft CD:
1
d CD = 0.010 m
2
π 3
π
Tc
2T
3
6
∴ Tall = cCD
τ = CD =
τ all = (0.010) (75× 10 ) = 117.81 N ⋅ m
3
π c CD
J CD
2
2
c CD =
fCD =
Shaft EF:
2
3
τ all =
cAB
τ =
rB
60
(24) = 9.6 Hz
f AB =
rC
150
c EF =
Pall = 2π fCD Tall = 2π (9.6) (117.81)= 7.11× 10 3 W
1
d EF = 0.014 m
2
π 3
π
(0.014)3 (75 × 106 ) = 323.27 N ⋅ m
2
2
r
60
(9.6) = 3.84 Hz
f EF = D f CD =
rE
150
Tall =
c EFτ all =
Pall = 2π fEF Tall = 2π (3.84) (323.27) = 7.80 ×10 3 W
Pall = 7.11× 10 3W = 7.11 kW 
Maximum allowable power is the smallest value.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.78
A 1.5-m-long solid steel of 48 mm diameter is to transmit 36 kW between a motor and a machine tool.
Determine the lowest speed at which the shaft can rotate, knowing that G = 77.2 GPa, that the maximum
shearing stress must not exceed 60 MPa, and the angle of twist must not exceed 2.5°.
SOLUTION
P = 36 × 103 W,
c=
1
d = 0.024 m, L = 1.5 m, G = 77.2× 109 Pa
2
Torque based on maximum stress:
τ =
Tc
J
T =
Jτ π 3
π
= c τ = (0.024)3 (60 × 10 6 ) = 1.30288 × 103 N ⋅ m
c
2
2
Torque based on twist angle:
ϕ=
TL
GJ
T =
τ = 60 MPa = 60 × 10 6 Pa
ϕ = 2.5° = 43.633× 10− 3 rad
GJ ϕ π c 4Gϕ
π (0.024)4 (77 × 10 9 ) (43.633 × 10 −3 )
=
=
L
2L
(2) (1.5)
= 1.17033 × 103 N ⋅ m
Smaller torque governs, so T =1.17033× 103 N ⋅ m
P = 2π fT
f =
P
36 × 103
=
2π T
2π (1.17033 × 103 )
f = 4.90 Hz 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.79
A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 30 Hz. Determine the maximum power
that the shaft can transmit, knowing that G = 77.2 GPa, that the allowable shearing stress is 50 MPa, and that
the angle of twist must not exceed 7.5°.
SOLUTION
c=
1
d = 15 mm = 0.015 m L = 2.5 m
2
τ = 50 × 10 6 Pa τ =
Stress requirement.
T =
τJ
c
=
π
2
τ c3 =
π
2
Tc
J
(50 × 10 6)(0.015) 3 = 265.07 N ⋅ m
ϕ = 7.5° = 130.90 × 10− 3 rad G = 77.2 × 10 9 Pa
Twist angle requirement.
ϕ =
T =
TL
2TL
=
GJ
π Gc 4
π
2
Gc 4ϕ =
π
2
(77.2 × 10 9)(0.015) 4(130.90 × 10− 3) = 803.60 N ⋅ m
Smaller value of T is the maximum allowable torque.
T = 265.07 N ⋅ m
Power transmitted at f = 30 Hz.
P = 2π f T = 2π (30)(265.07) = 49.96 × 10 3 W
P = 50.0 kW 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.80
A steel shaft must transmit 210 hp at a speed of 360 rpm. Knowing that G = 11.2 × 10 6 psi, design a solid
shaft so that the maximum shearing stress will not exceed 12 ksi, and the angle of twist in a 8.2-ft length must
not exceed 3°.
SOLUTION
Power:
P = (210 hp)(6600 in ⋅ lb/s/hp) = 1.336 × 106 in ⋅ lb/s
Angular speed:
f = (360 rpm)
Torque:
T =
Stress requirement:
τ =12 ksi, τ =
P
1.386 × 106
=
= 36.765 × 103 lb ⋅ in
2π f
(2 π)(6)
c = 3
Angle of twist requirement:
1 min.
= 6 Hz
60 sec
2T
πτ
= 3
2T
Tc
= 3
J
πc
(2)(36.765 × 103 )
= 1.2494 in.
π (12 × 103 )
ϕ = 3 ° = 52.36 × 10 − 3 rad
L = 8.2 ft = 98.4 in.
ϕ =
TL
2TL
=
GJ
π Gc 4
c = 4
2TL
(2)(36.765 × 10 3)(98.4)
= 4
= 1.4077 in.
π Gϕ
π (11.2 × 106 )(52.36 × 10− 3 )
The larger value is the required radius. c = 1.408 in.
d = 2.82 in. 
d = 2c
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.81
The shaft-disk-belt 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
τ = 9500 psi
τ =
Tc
2T
= 3
J
πc
P = 3 hp = (3)(6600) = 19800 lb ⋅ in/s
T =
π
2
c 3τ
Allowable torques.
3
in. diameter shaft:
c=
3
−
4
in. diameter shaft:
c = 83 in., Tall =
Statics:
Allowable torques.
3
  (9500) = 786.9 lb ⋅ in
2  8
TC = rC ( F1 − F2 )
rB
1.125
TC =
TC = 0.25TC
rC
4.5
TB ,all = 455.4 lb ⋅ in
TC ,all = 786.9 lb ⋅ in
Assume
TC = 786.9 lb ⋅ in
Then
TB = (0.25)(786.9) = 196.73 lb ⋅ in < 455.4 lb ⋅ in (okay)
P = 2π fT
(b)
π 3
TB = rB ( F1 − F2 )
TB =
(a)
π5 
5
in., Tall =   (9500) = 455.4 lb ⋅ in
16
2 16 
5
−
8
Allowable torques.
19800
P
=
2π TB
2π (196.73)
f AB =
TB ,all = 786.9 lb ⋅ in
TC ,all = 455.4 lb ⋅ in
Assume
TC = 455.4 lb ⋅ in
Then
TB = (0.25)(455.4) = 113.85 lb ⋅ in < 786.9 lb ⋅ in
P = 2π fT
f AB = 16.02 Hz 
f AB =
19800
P
=
2π TB
2π (113.85)
fAB = 27.2 Hz 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.82
A 1.6-m-long tubular steel shaft of 42-mm outer diameter d1 is to be
made of a steel for which τ all = 75 MPa and G = 77.2 GPa. Knowing
that the angle of twist must not exceed 4° when the shaft is subjected to a
torque of 900 N ⋅ m, determine the largest inner diameter d2 that can be
specified in the design.
SOLUTION
c1 =
1
d = 0.021 m L = 1.6 m
2 1
Based on stress limit:τ = 75 MPa = 75 ×10 6 Pa
τ =
Tc1
∴
J
J =
Tc1
τ
=
(900)(0.021)
= 252 × 10−9 m 4
75 × 106
Based on angle of twist limit:ϕ = 4 ° = 69.813 × 10− 3 rad
ϕ =
TL
GJ
∴ J =
TL
(900)(1.6)
=
= 267.88 × 10− 9 m 4
Gϕ
(77 × 109 )(69.813 × 10− 3 )
Larger value for J governs.
J =
π
2
J = 267.88 × 10− 9 m 4
(c − c )
4
1
c42 = c14 −
4
2
2J
π
= 0.0214 −
(2)(267.88 × 10−9 )
π
= 23.943 × 10−9 m4
c2 = 12.44 × 10 −3 m = 12.44 mm
d2 = 2c2 = 24.9 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.83
A 1.6-m-long tubular steel shaft ( G = 77.2 GPa ) of 42-mm outer
diameter d1 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 can rotate.
SOLUTION
c1 =
J =
1
1
d1 = 0.021 m, c2 = d 2 = 0.015 m
2
2
π
π
c − c ) = (0.021 − 0.015 ) = 225.97 × 10 m
(
2
2
4
1
4
2
4
−9
4
4
Based on stress limit: τ = 65 MPa = 65 × 10 6 Pa
τ =
Tc 1
Jτ
(225.97 × 10−9 )(65 × 106 )
=
= 699.43 N ⋅ m
or T =
J
G
0.021
Based on angle of twist limit: ϕ = 3° = 52.36×10 − 3 rad
ϕ=
TL
GJϕ
(77 × 109 )(225.97 × 10 −9 )(52.36 × 10 −3 )
=
or T =
GJ
L
1.6
= 569.40 N ⋅ m
T = 569.40 N ⋅ m
Smaller torque governs.
P = 120 kW = 120 × 103 W
P = 2π fT
so
f =
P
120 × 103
=
2π T
2π (569.40)
f = 33.5 Hz 
or 2010 rpm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.84
Knowing that the stepped shaft shown transmits a torque of magnitude
T = 2.50 kip ⋅ in., determine the maximum shearing stress in the shaft
3
when the radius of the fillet is (a) r = 81 in., (b) r = 16
in.
SOLUTION
D
2
=
= 1.33
1.5
d
D = 2 in. d = 1.5 in.
1
d = 0.75 in.
T = 2.5 kip ⋅ in
2
Tc
2T
(2)(2.5)
=
=
= 3.773 ksi
J
πc 3 π (0.75) 3
c =
(a)
r=
1
in. r = 0.125 in.
8
0.125
r
=
= 0.0833
d
1.5
K = 1.42
From Fig. 3.32,
τ max = K Tc = (1.42)(3.773)
τ max = 5.36 ksi 
J
(b)
r =
3
in.
16
r = 0.1875 in.
0.1875
r
=
= 0.125
d
1.5
From Fig. 3.32,
τ max = K
K = 1.33
Tc
= (1.33)(3.773)
J
τ max = 5.02 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.85
Knowing that the allowable shearing stress is 8 ksi for the stepped shaft
shown, determine the magnitude T of the largest torque that can be
transmitted by the shaft when the radius of the fillet is (a) r = 163 in.,
(b) r = 14 in.
SOLUTION
D = 2 in.
c =
1
d = 0.75 in.
2
τmax = K
(a)
r =
D
= 1.33
d
d = 1.5 in.
Tc
J
3
in.
16
τ max = 8 ksi
or
T=
πτ c 3
J τ max
= max
Kc
2K
r = 0.1875 in.
r
0.1875
=
= 0.125
d
1.5
From Fig. 3.32,
T=
(b)
r =
K = 1.33
π (8)(0.75)3
T = 3.99 kip ⋅ in 
(2)(1.33)
1
in.
4
r = 0.25 in.
0.25
r
=
= 0.1667
d
1.5
From Fig. 3.32,
T=
K = 1.27
π (8)(0.75)3
T = 4.17 kip ⋅ in 
(2)(1.27)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.86
The stepped shaft shown must transmit 40 kW at a speed of 720 rpm.
Determine the minimum radius r of the fillet if an allowable stress of
36 MPa is not to be exceeded.
SOLUTION
Angular speed:
 1 Hz 
f = (720 rpm) 
 = 12 Hz
 60 rpm 
Power:
P = 40 × 103 W
Torque:
T=
P
2π f
=
40 × 103
= 530.52 N ⋅ m
2π (12)
In the smaller shaft, d = 45 mm, c = 22.5 mm = 0.0225 m
τ =
Tc
2T
(2)(530.52)
=
=
= 29.65 × 10 6 Pa
J
π c3 π (0.0225)3
Using τ max = 36 MPa = 36 × 10 6 Pa results in
K =
36 × 106
τ max
=
= 1.214
τ
29.65 × 106
From Fig 3.32 with
90 mm
D
r
=
= 2,
= 0.24
d
d
45 mm
r = 10.8 mm 
r = 0.24 d = (0.24)(45 mm)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.87
The stepped shaft shown must transmit 45 kW. Knowing that the allowable
shearing stress in the shaft is 40 MPa and that the radius of the fillet is
r = 6 mm, determine the smallest permissible speed of the shaft.
SOLUTION
r
6
=
= 0.2
30
d
D
60
=
= 2
d
30
From Fig. 3.32,
K = 1.26
For smaller side,
c =
1
d = 15 mm = 0.015 m
2
τ=
KTc 2 KT
=
J
πc3
T =
π c3τ
2K
=
π (0.015)3 (40 × 106 )
(2)(1.26)
= 168.30 N ⋅ m
P = 45 kW = 45 × 103 P = 2π fT
f =
P
45 × 103
=
= 42.6 Hz
3
2π T
2π (168.30 × 10 )
f = 42.6 Hz 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.88
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.
SOLUTION
τ =
KTc 2 KT
=
J
πc3
T =
π c3τ
2K
1
d = 15 mm = 15 × 10− 3 m
2
D = 60 mm, r = 8 mm
d = 30 mm c =
D
60
=
= 2,
d
30
r
8
=
= 0.26667
d
30
From Fig. 3.32,
K = 1.18
Allowable torque.
T =
Maximum power.
P = 2π f T = (2π )(50)(202.17) = 63.5 × 10 3 W
π (15 × 10 −3) 3(45 × 10 6)
(2)(1.18)
= 202.17 N ⋅ m
P = 63.5 kW 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.89
In the stepped shaft shown, which has a full quarter-circular fillet, D = 1.25 in.
and
d = 1 in.
Knowing that the speed of the shaft is
2400 rpm and that the allowable shearing stress is 7500 psi, determine the
maximum power that can be transmitted by the shaft.
SOLUTION
D 1.25
=
= 1.25
d
1.0
1
r = ( D − d) = 0.15 in.
2
r
0.15
=
= 0.15
d
1.0
From Fig. 3.32,
K = 1.31
For smaller side,
c =
1
d = 0.5 in.
2
τ=
KTc
J
T =
π c 3τ
Jτ
=
Kc
2K
π (0.5)3 (7500)
= 1.1241 × 103 lb ⋅ in
(2)(1.31)
f = 2400 rpm = 40 Hz
T =
P = 2π f T = 2π (40)(1.1241 × 10 3)
= 282.5 × 103 lb ⋅ in/s
P = 42.8 hp 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.53 (Continued)
4.1889 TAB = 12.5 × 103
Substituting (1) into (2),
TAB = 2.9841 × 103 lb ⋅ in
(a)
Maximum stress in cylinder AB.
τ AB =
(b)
TBC = 9.5159 × 103 lb ⋅ in
TABc
(2.9841 × 103 ) (0.75)
=
= 4.50 × 103 psi
J
0.49701
4
Maximum stress in cylinder BC.
τ BC =
TBCc
=
τ AB = 4.50 ksi 
0
(9.5159 × 103 ) (1.0)
= 6.06 × 103 psi
τ BC = 6.06 ksi 
J
1.5708
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.54
Solve Prob. 3.53, assuming that cylinder AB is made of steel, fo
which G = 11.2 × 106 psi.
PROBLEM 3.53 The solid cylinders AB and BC are bonded togethe
at B and are attached to fixed supports at A and C. Knowing that th
modulus of rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi fo
brass, determine the maximum shearing stress (a) in cylinder AB
(b) in cylinder BC.
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation ϕ B for each cylinder.
Cylinder AB:
Cylinder BC.
π
1
d = 0.75 in. L = 12 in. J = c4 = 0.49701 in4
2
2
T L
T AB (12)
−
= 2.1557 × 10 6TAB
ϕB = AB =
6
GJ
(11.2 × 10 )(0.49701)
c=
π
π
1
4
4
d = 1.0 in. L = 18 in. J = c4 =
(1.0) = 1.5708 in
2
2
2
TBC L
TBC4(18) 0
=
= 2.0463 × 10−6TBC
ϕB =
GJ
(5.6 × 106 )(1.5708)
c=
−6
−6
Matching expressions for ϕ B
2.1557 × 10 TAB = 2.0463 × 10 TBC
Equilibrium of connection at B : TAB + TBC − T = 0
TAB = 6.0872 × 10 3 lb ⋅ in
(2
TBC = 6.4128 × 103 lb ⋅ in
Maximum stress in cylinder AB.
τ AB =
(b)
T AB + TBC = 12.5 × 103
(1
2.0535 TAB = 12.5 × 103
Substituting (1) into (2),
(a)
TBC = 1.0535 T AB
TABc
(6.0872 × 103 ) (0.75)
=
= 9.19× 103 psi
J
0.49701
τ AB = 9.19 ksi 
Maximum stress in cylinder BC.
τBC =
TBC c (6.4128 × 10 3)(1.0)
=
= 4.08 ×10 3 psi
J
1.5708
τ BC = 4.08 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.55
Two solid steel shafts are fitted with flanges that 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 = 11.2 × 106 psi, determine the maximum
shearing stress in each shaft when a torque of T of magnitude
420 kip ⋅ ft is applied to the flange indicated.
PROBLEM 3.55 The torque T is applied to flange B.
SOLUTION
Shaft AB:
T = TAB , LAB = 2 ft = 24 in., c =
J AB =
π
c4 =
π
2
2
T L
ϕ B = AB AB
GJ AB
TAB =
1
d = 0.625 in.
2
(0.625) 4 = 0.23968 in 4
GJ ABϕB (11.2 × 106 ) (0.23968)
−
ϕB
24
LAB
= 111.853 × 103ϕB
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD , LCD = 3ft = 36 in., c =
J CD =
π
c4 =
π
2
2
T L
ϕC = CD CD
GJ CD
TCD =
1
d = 0.75 in.
2
(0.75) 4 = 0.49701 in4
GJCD ϕC
(11.2 × 106 )(0.49701)
=
ϕ C = 154.625 × 103ϕ C
LCD
36
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 3.55 (Continued)
Clearance rotation for flange B:
Torque to remove clearance:
−3
4 ϕ B′ =01.5° = 26.18 × 10 rad
′ = (111.853 × 10 3)(26.18 × 10− 3) = 2928.3 lb ⋅ in
TΑΒ
Torque T ′′ to cause additional rotation ϕ ′′ :
T′′ = 5040 − 2928.3 = 2111.7 lb⋅ in
′′ + T ′′CD
T ′′ = TΑΒ
2111.7 = (111.853 × 103 )ϕ ′′ + (154.625 × 10 3)ϕ ′′ ∴
ϕ ′′ = 7.923× 10−3 rad
′′ = (111.853 × 103 ) (7.923 × 10−3 ) = 886.21 lb ⋅ in
TΑΒ
−
′′ = (154.625 × 103 )(7.923 × 10 3 ) = 1225.09 lb ⋅ in
TCD
Maximum shearing stress in AB.
τ AB =
TAB c (2928.3 + 886.21)(0.625)
=
= 9950 psi
J AB
0.23968
τ AB = 9.95 ksi 
TCD c (1225.09)(0.75)
=
= 1849 psi
JCD
0.49701
τ CD = 1.849 ksi 
Maximum shearing stress in CD.
τ CD =
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.56
Two solid steel shafts are fitted with flanges that 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 = 11.2 × 106 psi, determine the maximum
shearing stress in each shaft when a torque of T of magnitude
420 kip ⋅ ft is applied to the flange indicated.
PROBLEM 3.56 The torque T is applied to flange C.
SOLUTION
Shaft AB:
T = T AB, L AB = 2ft = 24 in., c =
J AB =
π
2
c4 =
π
2
4
0
1
d = 0.625 in.
2
(0.625)4 = 0.23968 in4
ϕB =
TAB LAB
GJ AB
TAB =
GJ ABϕ B (11.2 × 106 ) (0.23968)
−
ϕB
24
L AB
= 111.853 × 103ϕ B
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD ,
J CD =
π
LCD = 3 ft = 36 in., c =
c4 =
π
2
2
TCD LCD
ϕC =
GJCD
TCD =
Clearance rotation for flange C:
1
d = 0.75 in.
2
(0.75) 4 = 0.49701 in4
GJ CD ϕC
(11.2 × 10 6 )(0.49701)
=
ϕC = 154.625 × 103 ϕC
LCD
36
ϕC′ = 1.5° = 26.18 × 10− 3 rad
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.56 (Continued)
′ = (154.625 × 10 3)(26.18 × 10− 3) = 4048.1 lb ⋅ in
TCD
Torque to remove clearance:
Torque T′′ to cause additional rotation ϕ ′′ :
T′′ = 5040 − 4048.1 = 991.9 lb⋅ in
′′ + T ′′CD
T ′′ = TΑΒ
991.9 = (111.853 × 103 )ϕ ′′ + (154.625 × 103 )ϕ ′′ ∴ ϕ ′′ = 3.7223 × 10− 3 rad
′′ = (111.853 × 10−3 ) (3.7223 × 10−3 ) = 416.35 lb ⋅ in
TAB
−
−
′′ = (154.625 × 10 3 ) (3.7223 × 10 3 ) = 575.56 lb ⋅ in
TCD
Maximum shearing stress in AB.
τ AB =
TAB c (416.35) (0.625)
=
= 1086 psi
J AB
0.23968
τ AB = 1.086 ksi 
(4048.1 + 575.56)(0.75)
TCD c
=
= 6980 psi
J CD
0.49701
τ CD = 6.98 ksi 
Maximum shearing stress in CD.
τ CD =
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.57
Ends A and D of two solid steel shafts AB and CD are fixed, while
ends B and C are connected to gears as shown. Knowing that a
4-kN ⋅ m torque T is applied to gear B, determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
Gears B and C:
r
rB
φ B = C φC =
40
φC
100
φ B = 0.4φ C
(1)
 M C = 0 : TCD = rC F
(2)
 M B = 0 : T − T AB = rBF
(3)
Solve (2) for F and substitute into (3):
r
T − T AB = B TCD
rC
T = T AB +
100
T
40 CD
(4)
T = TAB + 2.5 TCD
L = 0.3 m, c = 0.030 m
Shaft AB:
φB = φB A =
TAB L
TAB (0.3)
T
=
= 235.79 × 103 AB
π
JG
G
(0.030) 4 6
2
(5)
L = 0.5 m, c = 0.0225 m
Shaft CD:
φC = φC D =
TCD L
JG
TCD (0.5)
π
2
= 1242 × 103
(0.0225) 4 G
TCD
G
(6)
Substitute from (5) and (6) into (1):
φB = 0.4 φC :
235.79 × 103
TCD = 0.47462 = TAB
TAB
T
= 0.4 × 1242 × 103 CD
G
G
(7)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.57 (Continued)
Substitute for TCD from (7) into (4):
T = T AB + 2.5 (0.47462T AB)
T = 2.1865 T AB (8
4000 N ⋅ m = 2.1865TAB
TAB = 1829.4 N ⋅ m
For T = 4 kN ⋅ m, Eq. (8) yields
TCD = 0.47462(1829.4) = 868.3 N ⋅ m
Substitute into (7):
(a)
Stress in AB:
TABc 2 TAB
2 1829.4
=
=
= 43.1 × 106
3
3
J
π c
π (0.030)
τ AB =
(b)
4
Stress in CD:
τCD =
TCDc
τ AB = 43.1 MPa 
0
=
2 TCD
3
=
2
868.3
3
= 48.5 × 106
τ CD = 48.5 MPa 
π c
J
π (0.0225)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.58
Ends A and D of the two solid steel shafts AB and CD are fixed,
while ends B and C are connected to gears as shown. Knowing that
the allowable shearing stress is 50 MPa in each shaft, determine the
largest torque T that may be applied to gear B.
SOLUTION
Gears B and C:
φB =
40
rC
φC =
φC
rB
100
φB = 0.4 φC
(1)
ΣM C = 0: TCD = rC F
(2)
ΣM B = 0: T − TAB = rB F
(3)
Solve (2) for F and substitute into (3):
r
T − TAB = B TCD
rC
T = TAB +
100
TCD
40
T4 = TAB 0+ 2.5 TCD
Shaft AB:
L = 0.3 m, c = 0.030 m
(4)
φB = φB / A =
TAB L
TAB (0.3)
T
=
= 235.77 × 103 AB
π
JG
G
(0.030) 4 G
2
(5)
L = 0.5 m, c = 0.0225 m
Shaft CD:
φC = φC /D =
TCD L
TCD (0.5)
T
=
= 1242 × 103 CD
π
JG
G
(0.0225) 4 G
2
(6)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.58 (Continued)
Substitute from (5) and (6) into (1) :
φB = 0.4 φC :
235.79 × 103
TAB
T
= 0.4 × 1242 × 103 CD
G
G
TCD = 0.47462 TAB
(7
Substitute for TCD from (7) into (4):
T = TAB + 2.5 (0.47462 TAB )
T = 2.1865 TAB
(8
T = 4.6068 TCD
(9
Solving (7) for TAB and substituting into (8),
 T

T = 2.1865  CD 
 0.47462 
Stress criterion for shaft AB:
τ AB = τ all = 50 MPa:
τ AB =
=
TABc
J
π
2
TAB =
π 3
J
τ AB = c τ AB
c
2
(0.030 m)3 (50 × 106 Pa) = 2120.6 N ⋅ m
T = 2.1865(2120.6 N ⋅ m) = 4.64 kN ⋅ m 
From (8):
Stress criterion for shaft CD:
τ CD = τ all = 50 MPa:
π
π
T c
TCD = c 3τCD = (0.0225 m)3 (50 × 106 Pa)
J
2
2
= 894.62 N ⋅ m
τ CD = CD
From (7):
T = 4.6068(894.62 N ⋅ m) = 4.12 kN ⋅ m
The smaller value for T governs.
T = 4.12 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
4
0
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
PROBLEM 3.59
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 applied as shown, determine the
maximum shearing stress in the jacket.
4
0
SOLUTION
c =
Solid shaft:
JS =
1
d = 0.020 m
2
π
2
c4 =
π
2
(0.020) 4 = 251.33 × 10 −9 m 4
1
d = 0.040 m
2
c1 = c2 − t = 0.040 − 0.004 = 0.036 m
c2 =
Jacket:
JJ =
π
2
( c − c ) = 2π (0.040 − 0.036 )
4
2
4
1
4
4
= 1.3829 × 10−6 m 4
Torque carried by shaft.
TS = GJS ϕ/L
Torque carried by jacket.
TJ = GJ J ϕ /L
Total torque.
T = TS + T J = (J S + J J ) G ϕ /L
TJ =
∴
Gϕ
T
=
L
JS + JJ
JJ
(1.3829 × 10 −6 ) (500)
T =
= 423.1 N ⋅ m
JS + JJ
1.3829 × 10−6 + 251.33 × 10−6
Maximum shearing stress in jacket.
τ =
TJ c2
(423.1) (0.040)
=
= 12.24 × 106 Pa
−6
JJ
1.3829 × 10
12.24 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.60
A solid shaft and a hollow shaft are made of the same material a
are of the same weight and length. Denoting by n the ratio c 1/ c
show that the ratio Ts /Th of the torque Ts in the solid shaft to t
torque
Th
in
the
hollow
shaft
(a)
(1 − n 2) /(1 + n 2) if the maximum shearing stress is the same
each shaft, (b) (1 − n)/ (1 + n 2) if the angle of twist is the same f
each shaft.
SOLUTION
For equal weight and length, the areas are equal.
(
)
π c02 = π c22 − c12 = π c22(1 − n 2)
Js =
(a)
π
2
c04 =
π
2
c24(1 − n 2) 2
∴
c0 = c2 (1 − n 2)1/2
Jh =
π
π
c − c ) = c (1 − n )
(
2
2
4
2
4
1
4
2
4
For equal stresses.
τ =
Ts c0
Tc
= h 2
Js
Jh
4
0
π 4 − 2 2
c 2 (1 n ) c 2
Ts
J sc2
1 − n2
(1 − n 2 )1/2
2
=
=
=
=
π c 4 (1 − n 4 )c (1 − n 2 )1/2
Th
J hc0
(1 + n 2 ) (1 − n2 )1/2
1 + n2
2
2 2

(b)
For equal angles of twist.
ϕ =
T sL
TL
= h
GJ s
GJ h
4
2 2
π
c (1 − n )
Ts
J
(1 − n 2) 2 1 − n 2
= s = 2 24
=
=
π c (1 − n4 )
Th
Jh
1 − n4
1 + n2
2 2

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.61
A torque T is applied as shown to a solid tapered shaft AB. Show by
integration that the angle of twist at A is
φ =
7TL
12π Gc 4
SOLUTION
Introduce coordinate y as shown.
r =
cy
L
Twist in length dy:
dϕ =
Tdy
Tdy
2TL4dy
=
=
π
GJ
π Gc 4 y 4
G r4
2
4
2 L 2TL dy
=
4 4
ϕ = L
π Gc y
2TL 2L dy

π Gc4 L y4
2L
=
2TL 4  1 
2TL 4  1
1 
−
=
−
+ 3


4
3
4 
3
π Gc  3 y L
π Gc  24L
3L 
=
2TL 4  7 
7TL
=
4 
3
π Gc  24L  12π Gc 4

4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.62
The mass moment of inertia of a gear is to be determined experimentally b
using a torsional pendulum consisting of a 6-ft steel wire. Knowing tha
G = 11.2 × 106 psi, determine the diameter of the wire for which the torsiona
spring constant will be 4.27 lb ⋅ ft/rad.
SOLUTION
Torsion spring constant K = 4.27 lb ⋅ ft/rad = 51.24 lb ⋅ in/rad
K =
c4 =
T
ϕ
=
π Gc4
T
GJ
=
=
2L
TL /GJ
L
2 LK (2)(72) (51.24)
−
=
= 209.7 × 10 6 in4
6
πG
π (11.2 × 10 )
c = 0.1203 in.
d = 2c = 0.241 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.63
An annular plate of thickness t and modulus G is used to
connect shaft AB of radius r1 to tube CD of radius r2. Knowing
that a torque T is applied to end A of shaft AB and that end D
of tube CD is 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
ϕ BC =
T  1
1
 2 − 2 
4π Gt  r1
r2 
SOLUTION
4
0
Use a free body consisting of shaft AB and an inner portion of the plate BC, the outer radius of this portion
being ρ.
The force per unit length of circumference is τ t.
ΣM = 0
τ t(2 πρ) ρ − T = 0
T
τ =
(a)
Maximum shearing stress occurs at ρ = r1
γ =
Shearing strain:
τ
G
2π tρ 2
τ max =
=
T
2π tr12

T
2π GT ρ 2
The relative circumferential displacement in radial length dρ is
dδ = γ d ρ = ρ dϕ
dϕ = γ
dϕ =
(b)
T dρ
T
=
ϕB /C =
r1 2π Gtρ 3
2πGt

=
r2

dρ
ρ
T
dρ
T dρ
=
2
2 π Gt ρ ρ
2 π GTρ 3
r2 d ρ
r1
T  1  r2
=
−
ρ3 2πGt  2 ρ2  r
1
T  1
T  1
1 
1 
− 2 + 2  =
 2 − 2
2πGt  2r2
r2 
2r1  4πGt  r1

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.64
Determine the maximum shearing stress in a solid shaft of 12-mm diameter as it transmits 2.5 kW at
frequency of (a) 25 Hz, (b) 50 Hz.
SOLUTION
c =
(a)
(b)
f = 25 Hz
f = 50 Hz
T =
1
d = 6 mm = 0.006 m
2
P
2π f
=
P = 2.5 kW = 2500 W
2500
= 15.9155 N ⋅ m
2π (25)
τ =
Tc
2T
2(15.9155)
=
=
= 46.9 × 106 Pa
3
J
πc
π (0.006) 3
T=
2500
= 7.9577 N ⋅ m
2π (50)
τ =
2(7.9577)
π (0.006) 3
4
= 23.5 × 10 6 Pa
0
τ = 46.9 MPa 
τ = 23.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.65
Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of
(a) 750 rpm, (b) 1500 rpm.
SOLUTION
c =
(a)
f =
T =
(b)
1
d = 0.75 in.
2
P = 75 hp = (75)(6600) = 495 × 103 lb ⋅ in/s
750
= 12.5 Hz
60
P
2π f
=
495 × 103
= 6.3025 × 103 lb ⋅ in
2 π(12.5)
τ=
Tc
2T
(2) (6.3025× 103 )
=
=
= 9.51× 103 psi
J
π c3
π (0.75)3
f =
1500
= 25 Hz
60
T =
495 × 103
= 3.1513 × 103 lb ⋅ in
2π (25)
τ =
(2) (3.1513× 103 )
= 4.76 × 103 psi
3
π (0.75)
τ = 9.51 ksi 
τ = 4.76 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.66
Design a solid steel shaft to transmit 0.375 kW at a frequency of 29 Hz, if the shearing stress in the shaft is n
to exceed 35 MPa.
SOLUTION
τ all = 35 × 10 6 Pa
T=
P
2π f
=
P = 0.375 × 10 3 W
f = 29 Hz
4
0
0.375 × 103
= 2.0580 N ⋅ m
2π (29)
τ =
Tc
2T
=
J
πc 3
∴
c3 =
2T
πτ
=
(2) (2.0580)
= 37.43 × 10− 9 m3
π (35 × 106 )
c = 3.345 × 10 −3 m = 3.345 mm
d = 6.69 mm 
d = 2c
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.67
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
τ all = 7500 psi P = 100 hp = 660 × 103 lb ⋅ in/s
f =
1200
= 20 Hz
60
Tc
2T
= 3
τ =
J
πc
c = 0.7639in.
T =
P
2π f
=
660 × 103
= 5.2521 × 103 lb ⋅ in
2π (20)
(2) (5.2521 × 10 3)
∴ c =
=
= 0.4458 in3
πτ
π(7500)
3
2T
d = 1.528 in. 
d = 2c
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.68
Determine the required thickness of the 50-mm tubular shaft of Example 3.07 if is to transmit the same pow
while rotating at a frequency of 30 Hz.
SOLUTION
P = 100 kW = 100 × 10 3 W
From Example 3.07,
τ all = 60 MPa = 60 × 106 Pa
c2 =
1
d = 25 mm = 0.025 m
2
f = 30 Hz
T =
J =
P
= 530.52 N ⋅ m
2π f
π
(c − c )
2
4
2
c14 = c24 −
4
1
2Tc2
πτ
τ =
Tc2
2Tc2
=
J
π c42 − c14
= 0.0254 −
(
)
(2)(530.52) (0.025)
= 249.90 × 10−9 m4
6
π (60 × 10 )
3
c1 = 22.358 × 10 − m = 22.358 mm
t = c2 − c1 = 25 mm − 22.358 mm = 2642 mm
4
0
t = 2.64 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
PROBLEM 3.69
While a steel shaft of the cross section shown rotates at 120 rpm, a
4
0
stroboscopic
measurement
indicates that the angle of twist is 2°
in a 12-ft length. Using G = 11.2 × 10 6 psi, determine the power
being transmitted.
SOLUTION
ϕ = 2° = 34.907 × 10−3 rad
c2 =
J=
1
d o = 1.5in.
2
π
c1 =
L = 12 ft = 144 in.
1
di = 0.6in.
2
π
c − c ) = (1.5 − 0.6 ) = 7.7486 in
(
2
2
4
2
4
1
4
4
4
f =
120
= 2 Hz
60
T =
GJϕ
(11.2 × 106 ) (7.7486) (34.907 × 10−3 )
=
= 21.037 × 103 lb ⋅ in
L
144
P = 2π fT = 2π (2)(21.037 × 103 ) = 264.36× 103 lb⋅in/s
P = 40.1 hp 
Since 1 hp = 6600 lb ⋅in/s,
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.70
The hollow steel shaft shown (G = 77.2 GPa, τ all = 50 MPa
rotates at 240 rpm. Determine (a) the maximum power that ca
be transmitted, (b) the corresponding angle of twist of the shaft.
SOLUTION
1
d2 = 30 mm
2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
π
c − c ) = [(30) − (12.5) ]
(
2
2
4
2
4
1
4
4
0
4
= 1.234 × 106 mm4 = 1.234 × 10−6 m4
τ m = 50 × 106 Pa
τm =
(a)
Tc
J
T =
τ mJ
c
−
=
(50 × 10 6)(1.234 × 10 6)
= 2056.7 N ⋅ m
30 × 10− 3
Angular speed.
f = 240 rpm = 4 rev/sec = 4 Hz
Power being transmitted.
P = 2π f T = 2π (4)(2056.7) = 51.7 × 10 3 W
P = 51.7 kW 
(b)
Angle of twist.
ϕ =
TL
(2056.7)(5)
=
= 0.1078 rad
GJ
(77.2 × 10 9)(1.234 × 10− 6)
ϕ = 6.17° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.71
As the hollow steel shaft shown rotates at 180 rpm, a
stroboscopic measurement indicates that the angle of twist of
the shaft is 3°. Knowing that G = 77.2 GPa, determine (a) the
power being transmitted, (b) the maximum shearing stress in the
shaft.
SOLUTION
1
d = 30 mm
2 2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
π
c − c ) = [(30) − (12.5) )]
(
2
2
4
2
4
1
4
4
= 1.234 × 10 6 mm 4 = 1.234 × 10− 6 m 4
ϕ = 3 ° = 0.05236 rad
ϕ =
TL
GJ
T =
GJ ϕ
(77.2 × 109)(1.234 × 10 6 )(0.0536)
=
= 997.61 N ⋅ m
5
L
−
(a)
Angular speed:
f = 180 rpm = 3 rev/sec = 3 Hz
Power being transmitted.
P = 2π f T = 2π (3)(997.61) = 18.80 × 103 W
P = 18.80 kW 
(b)
Maximum shearing stress.
τm =
Tc 2 (997.61)(30 × 10 −3)
=
J
1.234 × 10− 6
= 24.3 × 10 6 Pa
4
τ m = 24.3 MPa 
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.72
The design of a machine element calls for a 40-mm-outer-diamet
shaft to transmit 45 kW. (a) If the speed of rotation is 720 rpm
determine the maximum shearing stress in shaft a. (b) If the speed o
rotation can be increased 50% to 1080 rpm, determine the largest inn
diameter of shaft b for which the maximum shearing stress will be th
same in each shaft.
SOLUTION
(a)
f =
720
= 12 Hz
60
P = 45 kW = 45 × 103 W
T =
(b)
P
2π f
=
45 × 103
= 596.83 N ⋅ m
2π (12)
c =
1
d = 20 mm = 0.020 m
2
τ=
Tc
2T
(2)(596.83)
=
=
= 47.494 × 10 6 Pa
3
J
πc
π (0.020)3
f =
1080
= 18 Hz
60
T =
45 × 103
= 397.89 N ⋅ m
2π (18)
τ =
Tc2
2Tc2
=
J
π c 42 − c14
c14 = c24 −
(
τmax = 47.5 MPa 
)
2Tc 2
πτ
c14 = 0.0204 −
(2)(397.89)(0.020)
= 53.333 × 10− 9
π (47.494 × 106 )
3
c1 = 15.20 × 10 − m = 15.20 mm
d 2 = 2 c1 = 30.4 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.73
A steel pipe of 3.5-in. outer diameter is to be used to transmit a torque
of 3000 lb ⋅ ft without exceeding an allowable shearing stress of 8 ksi.
A series of 3.5-in.-outer-diameter pipes is available for use. Knowing
that the wall thickness of the available pipes varies from 0.25 in. to
0.50 in. in 0.0625-in. increments, choose the lightest pipe that can be
used.
SOLUTION
T = 3000 lb ⋅ ft = 36 × 10 3 lb ⋅ in
1
c 2 = d o = 1.75 in.
2
4
0
τ=
Tc2
2Tc2
=
J
π c24 − c14
(
2 Tc2
c14 = c24 −
πτ
)
= 1.75 4 −
(2)(36 × 10 3 )(1.75)
= 4.3655 in 4
3
π(8 × 10 )
c1 = 1.4455 in.
Required minimum thickness: t = c 2 − c1
t = 1.75 − 1.4455 = 0.3045 in.
Available thicknesses: 0.25 in., 0.3125 in., 0.375 in., etc.
t = 0.3125 in. 
Use
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.74
The two solid shafts and gears shown are used t
transmit 16 hp from the motor at A, operating at a spee
of 1260 rpm, to a machine tool at D. Knowing that th
maximum allowable shearing stress is 8 ksi, determin
the required diameter (a) of shaft AB, (b) of shaft CD.
SOLUTION
(a)
Shaft AB:
P = 16 hp = (16)(6600) = 105.6 × 10 3 lb ⋅ in/sec
1260
= 21 Hz
60
τ = 8 ksi = 8 × 10 3 psi
f =
TAB =
τ =
P
2π f
=
105.6 × 103
= 800.32 lb ⋅ in
2π (21)
Tc
2T
= 3
J
πc
c =3
c =3
4
0
(2)(800.32)
π (8 × 103 )
2T
πτ
= 0.399 in.
d AB = 2 c = 0.799 in.
(b)
Shaft CD:
d AB = 0.799 in. 
r
5
TCD = C TAB = (800.32) = 1.33387 × 103 lb ⋅ in
rB
3
c = 3
2T
πτ
= 3
(2)(1.33387 × 103 )
= 0.473 in.
π (8 × 103 )
dCD = 2c = 0.947 in.
dCD = 0.947 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.75
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A operating at a speed
of 1260 rpm to a machine tool at D. Knowing that each
shaft has a diameter of 1 in., determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
(a)
Shaft AB:
P = 16 hp = (16)(6600) = 105.6 × 10 3 lb ⋅ in/sec
1260
= 21 Hz
60
P
105.6 × 103
=
= 800.32 lb ⋅ in
TAB =
2π f
2π (21)
f =
1
d = 0.5 in.
2
Tc
2T
=
τ=
J
π c3
(2)(800.32)
=
= 4.08 × 103 psi
π (0.5)3
c=
(b)
Shaft CD:
TCD =
rC
5
TAB = (800.32) = 1.33387 × 10 3lb ⋅ in
rB
3
τ =
2T
(2)(1.33387 × 10 3)
=
= 6.79 × 10 3 psi
π c3
π (0.5) 3
4
0
τ AB = 4.08 ksi 
τ CD = 6.79 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 3.76
4
0
Three shafts and four gears are used to form a gear trai
that will transmit 7.5 kW from the motor at A to
machine tool at F. (Bearings for the shafts are omitted
the sketch.) Knowing that the frequency of the motor
30 Hz and that the allowable stress for each shaft
60 MPa, determine the required diameter of each shaft.
SOLUTION
P = 7.5 kW = 7.5 × 103 W
Shaft AB:
T AB =
f AB = 30 Hz
τ=
c3AB =
τ all = 60 MPa = 60 × 106 Pa
P
7.5 × 103
=
= 39.789 N ⋅ m
2π f AB
2π(30)
Tc AB
2T
= 3 ∴
J AB π c AB
3
=
cAB
2T
πτ
(2)(39.789)
= 422.17 × 10 −9 m3
6
π(60 × 10 )
c AB = 7.50 × 10 − 3 m = 7.50 mm
Shaft CD:
d AB = 2 c AB = 15.00 mm 
P
7.5 × 103
=
= 99.472 N ⋅ m
2πf CD
2π(12)
f CD =
rB
60
f AB =
(30) = 12 Hz
rC
150
τ CD =
Tc CD
2T
2T
2(99.472)
3
= 3
∴ cCD
= CD =
= 1.05543 × 10− 6 m 3
πτ
J CD πc CD
π(60 × 10 6 )
TCD =
cCD = 10.18 × 10− 3 m = 10.18 mm
Shaft EF:
60
r
f EF = D f CD =
(12) = 4.8 Hz
rE
150
τ=
dCD = 2 cCD = 20.4 mm 
TEF =
7.5 × 10 3
P
=
= 248.68 N ⋅ m
2πf EF 2 π (4.8)
TcEF
2T
(2)(248.68)
3
=
∴ cEF
=
= 2.6886 × 10−6 m3
3
6
J EF
π cEF
π(60 × 10 )
cEF =13.82 × 10 − 3 m =13.82 mm
d EF = 2c EF = 27.6 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.77
Three shafts and four gears are used to form a gear train that will
transmit power from the motor at A to a machine tool at F. (Bearings
for the shafts are omitted in the sketch.) The diameter of each shaft is
as follows: d AB = 16 mm, d CD = 20 mm, d EF = 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
τ all = 75 MPa = 75 × 106 Pa
Shaft AB:
c AB =
Tall =
f
1
d AB = 0.008 m
2
π
2
3
τ all =
cAB
π
τ =
TcAB
2T
=
3
J AB
π c AB
4
0
(0.008)3 (75 × 10 6) = 60.319 N ⋅ m
2
= 24 Hz P = 2π f T = 2π (24) (60.319) = 9.10 × 103 W
AB
Shaft CD:
1
d CD = 0.010 m
2
π 3
π
Tc
2T
3
6
∴ Tall = cCD
τ = CD =
τ all = (0.010) (75× 10 ) = 117.81 N ⋅ m
3
π c CD
J CD
2
2
c CD =
fCD =
Shaft EF:
AB all
all
rB
60
(24) = 9.6 Hz
f AB =
rC
150
c EF =
Pall = 2π fCD Tall = 2π (9.6) (117.81)= 7.11× 10 3 W
1
d EF = 0.014 m
2
π 3
π
(0.014)3 (75 × 106 ) = 323.27 N ⋅ m
2
2
r
60
(9.6) = 3.84 Hz
f EF = D f CD =
rE
150
Tall =
c EFτ all =
Pall = 2π fEF Tall = 2π (3.84) (323.27) = 7.80 ×10 3 W
Pall = 7.11× 10 3W = 7.11 kW 
Maximum allowable power is the smallest value.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.78
A 1.5-m-long solid steel of 48 mm diameter is to transmit 36 kW between a motor and a machine too
Determine the lowest speed at which the shaft can rotate, knowing that G = 77.2 GPa, that the maximu
shearing stress must not exceed 60 MPa, and the angle of twist must not exceed 2.5°.
SOLUTION
P = 36 × 103 W,
c=
1
d = 0.024 m, L = 1.5 m, G = 77.2× 109 Pa
2
τ = 60 MPa = 60 × 10 6 Pa
Torque based on maximum stress:
τ =
Tc
J
T =
Jτ π 3
π
= c τ = (0.024)3 (60 × 10 6 ) = 1.30288 × 103 N ⋅ m
c
2
2
ϕ = 2.5° = 43.633× 10− 3 rad
Torque based on twist angle:
TL
ϕ=
GJ
GJ ϕ π c 4Gϕ
π (0.024)4 (77 × 10 9 ) (43.633 × 10 −3 )
=
=
T =
L
2L
(2) (1.5)
= 1.17033 × 103 N ⋅ m
Smaller torque governs, so T =1.17033× 103 N ⋅ m
P = 2π fT
f =
P
36 × 103
=
2π T
2π (1.17033 × 103 )
f = 4.90 Hz 
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
PROBLEM 3.79
0
A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 30 Hz. Determine the maximum power
that the shaft can transmit, knowing that G = 77.2 GPa, that the allowable shearing stress is 50 MPa, and that
the angle of twist must not exceed 7.5°.
SOLUTION
c=
1
d = 15 mm = 0.015 m L = 2.5 m
2
τ = 50 × 10 6 Pa τ =
Stress requirement.
T =
τJ
c
=
π
2
τ c3 =
π
2
Tc
J
(50 × 10 6)(0.015) 3 = 265.07 N ⋅ m
ϕ = 7.5° = 130.90 × 10− 3 rad G = 77.2 × 10 9 Pa
Twist angle requirement.
ϕ =
T =
TL
2TL
=
GJ
π Gc 4
π
2
Gc 4ϕ =
π
2
(77.2 × 10 9)(0.015) 4(130.90 × 10− 3) = 803.60 N ⋅ m
Smaller value of T is the maximum allowable torque.
T = 265.07 N ⋅ m
Power transmitted at f = 30 Hz.
P = 2π f T = 2π (30)(265.07) = 49.96 × 10 3 W
P = 50.0 kW 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.80
A steel shaft must transmit 210 hp at a speed of 360 rpm. Knowing that G = 11.2 × 10 6 psi, design a soli
shaft so that the maximum shearing stress will not exceed 12 ksi, and the angle of twist in a 8.2-ft length mu
not exceed 3°.
SOLUTION
Power:
P = (210 hp)(6600 in ⋅ lb/s/hp) = 1.336 × 106 in ⋅ lb/s
Angular speed:
f = (360 rpm)
Torque:
T =
Stress requirement:
4
0
2T
Tc
= 3
τ =12 ksi, τ =
J
πc
1 min.
= 6 Hz
60 sec
P
1.386 × 106
=
= 36.765 × 103 lb ⋅ in
2π f
(2 π)(6)
c = 3
Angle of twist requirement:
2T
πτ
= 3
(2)(36.765 × 103 )
= 1.2494 in.
π (12 × 103 )
ϕ = 3 ° = 52.36 × 10 − 3 rad
L = 8.2 ft = 98.4 in.
ϕ =
TL
2TL
=
GJ
π Gc 4
c = 4
2TL
(2)(36.765 × 10 3)(98.4)
4
=
= 1.4077 in.
π Gϕ
π (11.2 × 106 )(52.36 × 10− 3 )
The larger value is the required radius. c = 1.408 in.
d = 2c
d = 2.82 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.81
The shaft-disk-belt 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
τ = 9500 psi
τ =
Tc
2T
= 3
J
πc
P = 3 hp = (3)(6600) = 19800 lb ⋅ in/s
T =
π
2
c 3τ
Allowable torques.
3
5
−
8
in. diameter shaft:
π5 
5
c=
in., Tall =   (9500) = 455.4 lb ⋅ in
16
2 16 
3
−
4
in. diameter shaft:
c = 83 in., Tall =
Statics:
TB = rB ( F1 − F2 )
TB =
(a)
Allowable torques.
π 3
3
  (9500) = 786.9 lb ⋅ in
2  8
TC = rC ( F1 − F2 )
rB
1.125
TC =
TC = 0.25TC
rC
4.5
TB ,all = 455.4 lb ⋅ in
TC ,all = 786.9 lb ⋅ in
Assume
TC = 786.9 lb ⋅ in
Then
TB = (0.25)(786.9) = 196.73 lb ⋅ in < 455.4 lb ⋅ in (okay)
4
0
P
19800
P = 2π fT
(b)
Allowable torques.
f AB =
TB ,all = 786.9 lb ⋅ in
2π TB
=
2π (196.73)
TC ,all = 455.4 lb ⋅ in
Assume
TC = 455.4 lb ⋅ in
Then
TB = (0.25)(455.4) = 113.85 lb ⋅ in < 786.9 lb ⋅ in
P = 2π fT
f AB = 16.02 Hz 
f AB =
19800
P
=
2π TB
2π (113.85)
fAB = 27.2 Hz 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.82
A 1.6-m-long tubular steel shaft of 42-mm outer diameter d1 is to b
made of a steel for which τ all = 75 MPa and G = 77.2 GPa. Knowin
that the angle of twist must not exceed 4° when the shaft is subjected to
torque of 900 N ⋅ m, determine the largest inner diameter d2 that can b
specified in the design.
SOLUTION
c1 =
1
d = 0.021 m L = 1.6 m
2 1
Based on stress limit:τ = 75 MPa = 75 ×10 6 Pa
τ =
Tc1
∴
J
J =
Tc1
τ
=
(900)(0.021)
= 252 × 10−9 m 4
75 × 106
Based on angle of twist limit:ϕ = 4 ° = 69.813 × 10− 3 rad
ϕ =
TL
GJ
∴ J =
TL
(900)(1.6)
=
= 267.88 × 10− 9 m 4
−3
9
Gϕ
(77 × 10 )(69.813 × 10 )
Larger value for J governs.
J =
π
J = 267.88 × 10− 9 m 4
(c − c )
2
4
1
c42 = c14 −
4
2
2J
π
= 0.0214 −
(2)(267.88 × 10−9 )
c2 = 12.44 × 10 −3 m = 12.44 mm
π
= 23.943 × 10−9 m4
d2 = 2c2 = 24.9 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.83
A 1.6-m-long tubular steel shaft ( G = 77.2 GPa ) of 42-mm outer
diameter d1 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 can rotate.
4
SOLUTION
1
1
0
c1 =
J =
2
π
d1 = 0.021 m, c2 =
d 2 = 0.015 m
2
π
c − c ) = (0.021 − 0.015 ) = 225.97 × 10 m
(
2
2
4
1
4
2
4
−9
4
4
Based on stress limit: τ = 65 MPa = 65 × 10 6 Pa
τ =
Tc 1
Jτ
(225.97 × 10−9 )(65 × 106 )
=
= 699.43 N ⋅ m
or T =
J
G
0.021
Based on angle of twist limit: ϕ = 3° = 52.36×10 − 3 rad
ϕ=
TL
GJϕ
(77 × 109 )(225.97 × 10 −9 )(52.36 × 10 −3 )
=
or T =
GJ
L
1.6
= 569.40 N ⋅ m
T = 569.40 N ⋅ m
Smaller torque governs.
P = 120 kW = 120 × 103 W
P = 2π fT
so
f =
P
120 × 103
=
2π T
2π (569.40)
f = 33.5 Hz 
or 2010 rpm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.84
Knowing that the stepped shaft shown transmits a torque of magnitud
T = 2.50 kip ⋅ in., determine the maximum shearing stress in the sha
3
when the radius of the fillet is (a) r = 81 in., (b) r = 16
in.
SOLUTION
D
2
=
= 1.33
1.5
d
D = 2 in. d = 1.5 in.
1
d = 0.75 in.
T = 2.5 kip ⋅ in
2
Tc
2T
(2)(2.5)
=
=
= 3.773 ksi
3
J
πc
π (0.75) 3
c =
(a)
r=
1
in. r = 0.125 in.
8
0.125
r
=
= 0.0833
d
1.5
From Fig. 3.32,
K = 1.42
τ max = K Tc = (1.42)(3.773)
J
(b)
r =
3
in.
16
r = 0.1875 in.
τ max = 5.36 ksi 
4
0
0.1875
r
=
= 0.125
d
1.5
From Fig. 3.32,
τ max = K
K = 1.33
Tc
= (1.33)(3.773)
J
τ max = 5.02 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.85
Knowing that the allowable shearing stress is 8 ksi for the stepped shaft
shown, determine the magnitude T of the largest torque that can be
transmitted by the shaft when the radius of the fillet is (a) r = 163 in.,
(b) r = 14 in.
SOLUTION
D = 2 in.
c =
1
d = 0.75 in.
2
τmax = K
(a)
r =
D
= 1.33
d
d = 1.5 in.
Tc
J
3
in.
16
τ max = 8 ksi
or
T=
πτ c 3
J τ max
= max
Kc
2K
r = 0.1875 in.
r
0.1875
=
= 0.125
d
1.5
From Fig. 3.32,
T=
(b)
r =
K = 1.33
π (8)(0.75)3
T = 3.99 kip ⋅ in 
(2)(1.33)
1
in.
4
r = 0.25 in.
0.25
r
=
= 0.1667
d
1.5
From Fig. 3.32,
T=
K = 1.27
π (8)(0.75)3
T = 4.17 kip ⋅ in 
(2)(1.27)
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 3.86
0 shaft shown must transmit 40 kW at a speed of 720 rpm
The4 stepped
Determine the minimum radius r of the fillet if an allowable stress
36 MPa is not to be exceeded.
SOLUTION
Angular speed:
 1 Hz 
f = (720 rpm) 
 = 12 Hz
 60 rpm 
Power:
P = 40 × 103 W
Torque:
T=
P
2π f
=
40 × 103
= 530.52 N ⋅ m
2π (12)
In the smaller shaft, d = 45 mm, c = 22.5 mm = 0.0225 m
τ =
Tc
2T
(2)(530.52)
=
=
= 29.65 × 10 6 Pa
3
J
πc
π (0.0225)3
Using τ max = 36 MPa = 36 × 10 6 Pa results in
K =
36 × 106
τ max
=
= 1.214
τ
29.65 × 106
From Fig 3.32 with
90 mm
D
r
=
= 2,
= 0.24
d
d
45 mm
r = 10.8 mm 
r = 0.24 d = (0.24)(45 mm)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.87
The stepped shaft shown must transmit 45 kW. Knowing that the allowable
shearing stress in the shaft is 40 MPa and that the radius of the fillet is
r = 6 mm, determine the smallest permissible speed of the shaft.
SOLUTION
r
6
=
= 0.2
30
d
D
60
=
= 2
d
30
From Fig. 3.32,
K = 1.26
For smaller side,
c =
τ=
1
d = 15 mm
4 = 0.015
0 m
2
KTc
=
2 KT
τ
T =
J
π c3τ
2K
=
πc3
π (0.015)3 (40 × 106 )
(2)(1.26)
= 168.30 N ⋅ m
P = 45 kW = 45 × 103 P = 2π fT
f =
P
45 × 103
=
= 42.6 Hz
3
2π T
2π (168.30 × 10 )
f = 42.6 Hz 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.88
The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing th
the radius of the fillet is r = 8 mm and the allowable shearing stress is 4
MPa, determine the maximum power that can be transmitted.
SOLUTION
τ =
KTc 2 KT
=
J
πc3
T =
π c3τ
2K
1
d = 15 mm = 15 × 10− 3 m
2
D = 60 mm, r = 8 mm
d = 30 mm c =
D
60
=
= 2,
d
30
r
8
=
= 0.26667
d
30
From Fig. 3.32,
K = 1.18
Allowable torque.
T =
Maximum power.
P = 2π f T = (2π )(50)(202.17) = 63.5 × 10 3 W
π (15 × 10 −3) 3(45 × 10 6)
(2)(1.18)
4
0
= 202.17 N ⋅ m
P = 63.5 kW 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.89
In the stepped shaft shown, which has a full quarter-circular fillet, D = 1.25 in.
and
d = 1 in.
Knowing that the speed of the shaft is
2400 rpm and that the allowable shearing stress is 7500 psi, determine the
maximum power that can be transmitted by the shaft.
SOLUTION
D 1.25
=
= 1.25
d
1.0
1
r = ( D − d) = 0.15 in.
2
r
0.15
=
= 0.15
d
1.0
From Fig. 3.32,
K = 1.31
For smaller side,
c =
1
d = 0.5 in.
2
τ=
KTc
J
T =
π c 3τ
Jτ
=
Kc
2K
π (0.5)3 (7500)
= 1.1241 × 103 lb ⋅ in
(2)(1.31)
f = 2400 rpm = 40 Hz
T =
P = 2π f T = 2π (40)(1.1241 × 10 3)
= 282.5 × 103 lb ⋅ in/s
P = 42.8 hp 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.90
A torque of magnitude T = 200 lb ⋅ in. is applied to the stepped shaft shown
which has a full quarter-circular fillet. Knowing that D = 1 in., determine th
maximum shearing stress in the shaft when (a) d = 0.8 in., (b) d = 0.9 in.
4
0
SOLUTION
(a)
1.0
D
=
= 1.25
d
0.8
1
r = ( D − d ) = 0.1 in.
2
r
0.1
=
= 0.125
d
0.8
From Fig. 3.32,
K = 1.31
For smaller side,
c =
1
d = 0.4 in.
2
KTc
2KT
=
J
πc3
(2)(1.31)(200)
=
= 2.61× 103 psi
3
π (0.4)
τ=
(b)
τ = 2.61 ksi 
1.0
D
=
= 1.111
d
0.9
1
r = ( D − d ) = 0.05
2
r
0.05
=
= 0.05
d
1.0
From Fig. 3.32,
K = 1.44
For smaller side,
c =
1
d = 0.45 in.
2
τ=
2KT
(2)(1.44)(200)
=
= 2.01× 103 psi
3
3
πc
π (0.45)
τ = 2.01 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
4
0
PROBLEM 3.91
In the stepped shaft shown, which has a full quarter-circular fillet, the
allowable shearing stress is 80 MPa. Knowing that D = 30 mm, determine
the largest allowable torque that can be applied to the shaft if
(a) d = 26 mm, (b) d = 24 mm.
SOLUTION
τ = 80 × 106 Pa
(a)
D
30
=
= 1.154
d
26
r=
From Fig. 3.32,
K = 1.36
Smaller side,
c =
1
d = 13 mm = 0.013 m
2
τ =
KTc 2 KT
=
J
π c3
T =
(b)
r
2
=
= 0.0768
d
26
1
( D − d) = 2 mm
2
D
30
=
= 1.25
d
24
π c 3τ
2K
=
1
r = ( D − d ) = 3 mm
2
π (0.013) 3(80 × 10 6)
4
(2)(1.36)
0
r
3
=
= 0.125
d
24
= 203 N ⋅ m
T = 203 N ⋅ m 
From Fig. 3.32,
K = 1.31
1
d = 12 mm = 0.012 m
2
π c 3τ π (0.012) 3(80 × 10 6)
=
= 165.8 N ⋅ m
T =
2K
(2)(1.31)
c =
T = 165.8 N ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0
PROBLEM 3.92
A 30-mm diameter solid rod is made of an elastoplastic material with τ Y = 3.5 MPa. Knowing that the elasti
core of the rod is 25 mm in diameter, determine the magnitude of the applied torque T.
SOLUTION
τ Y = 3.5 × 106 Pa, c =
1
(30 mm) = 15 mm = 0.015 m
2
1
(25 mm) = 12.5 mm = 0.0125 m
2
J
π
π
TY = τ Y = c3τ Y = (0.015)3 (3.5 × 106 ) = 18.555 N ⋅ m
c
2
2
3
 (0.0125)3 
ρ  4
4 
T = TY 1 − Y3  = (18.555) 1 −

3 
c  3
(0.015)3 

ρY =
T = 21.2 N ⋅ m 
= 21.2 N ⋅ m
4
0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limite
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using i
without permission.
PROBLEM 3.93
The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with τY = 21 ksi. Determine the magnitude T of the
applied torques when the plastic zone is (a) 0.8 in. deep, (b) 1.2 in. deep.
4
0
SOLUTION
τY = 21 ksi
TY =
τ YJ
c
= τY
= (21 ksi)
π
π
2
c3
(1.5 in.)3
2
TY = 111.3 kip ⋅ in
(a)
For t = 0.8 in. ρ Y = 1.5 − 0.8 = 0.7 in.
Eq. (3.32)
T=

4 
1 ρY3 
4
1 (0.7 in.)3 
=
⋅
−
TY  1 −
(111.3
kip
in)
1



3 
4 c3 
3
4 (1.5 in.)3 

T = 144.7 kip ⋅ in 
(b)
For t = 1.2 in.
T=
ρY = 1.5 − 1.2 = 0.3 in.

4 
1 ρY3 
4
1 (0.3 in.)3 
=
⋅
−
TY  1 −
(111.3
kip
in)
1



3 
4 c3 
3
4 (1.5 in.)3 

T = 148.1 kip ⋅ in 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
4
0