CHAPTER 3
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has been replaced by a hollow shaft of the same outer diameter and of 24-mm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 m 4
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
(c
2
4
2
)
− c14 =
π
2
(0.0384 − 0.0124 ) = 3.2428 × 10−6 m 4
Tc (4.6 × 103 )(0.038)
=
= 53.9 × 106 Pa
−6
J
3.2428 × 10
τ = 53.9 MPa 
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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 =
J =
π
2
π
(c
4
2
− c14
)
(454 − 304 ) = 5.1689 × 106 mm 4 = 5.1689 × 10−6 m 4
2
Tc
τ =
J
T =
T =
Jτ
c
(5.1689 × 10−6 )(45 × 106 )
= 5.1689 × 103 N ⋅ m
−3
45 × 10
T = 5.17 kN ⋅ m 
(b)
Solid shaft of same area:
(
)
A = π c22 − c12 = π (452 − 302 ) = 3.5343 × 103 mm 2
π c 2 = A or c =
J =
τ =
π
2
c4, τ =
A
π
= 33.541 mm
2T
Tc
=
J
π c3
(2)(5.1689 × 103 )
= 87.2 × 106 Pa
π (0.033541)3
τ = 87.2 MPa 
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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 2 =   (1.6) = 0.8 in.
2
2
c1 =
1
1
d1 =   (1.2) = 0.6 in.
2
2
J =
π
2
(c
4
2
)
− c14 =
π
2
c = 0.8 in.
(0.84 − 0.64 ) = 0.4398 in 4
Tc
J
Jτ max
(0.4398)(7.5)
=
T =
c
0.8
τ max =
T = 4.12 kip ⋅ in 
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PROBLEM 3.4
Knowing that the internal diameter of the hollow shaft shown is d = 0.9in.,
determine the maximum shearing stress caused by a torque of magnitude
T = 9 kip ⋅ in.
SOLUTION
c2 =
1
1
d 2 =   (1.6) = 0.8 in.
2
2
c1 =
1
1
d1 =   (0.9) = 0.45 in.
2
2
(
)
π
c24 − c14 = (0.84 − 0.454 ) = 0.5790 in 4
2
2
(9)(0.8)
Tc
=
=
0.5790
J
J =
τ max
π
c = 0.8 in.
τ max = 12.44 ksi 
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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
c4 =
2
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)
ρ D = 15 mm = 15 × 10−3 m
τD =
(c)
τD =
TD =
ρD
c
τ =
TD ρ D
JD
π
2
(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
TD
187.5
× 100% =
(100%) = 6.25%
T
3 × 103
6.25%
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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
c4 −
2
π
2
(0.10) 4 = 15.7080 × 10−9 m 4
Jτ 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 c 2
 2   4

2
2
(0.010) = 0.0115470 m
c2 =
c=
3
3
1
c1 = c2 = 0.0057735 m
2
(
J =
T =
π
2
)
(c
4
2
τ max J
c2
)
− c14 =
=
π
2
(0.01154704 − 0.00577354 ) = 26.180 × 10−9 m 4
(80 × 106 )(26.180 × 10−9 )
= 181.38
0.0115470
T = 181.4 N ⋅ m 
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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
J
T =
Analysis of sleeve CD:
c2 =
π
2
T=
π
Jτ
= τ c3
c
2
(12 × 103 )(0.75)3 = 7.95 × 103 lb ⋅ in
1
1
d o = (3) = 1.5 in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25 in.
J=
T=
The smaller torque governs:

π
(c
2
4
2
)
− c14 =
π
2
(1.54 − 1.254 ) = 4.1172 in 4
Jτ
(4.1172)(7 × 103 )
=
= 19.21 × 103 lb ⋅ in
c2
1.5
T = 7.95 × 103 lb ⋅ in
T = 7.95 kip ⋅ in 
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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
(c
2
4
2
)
− c14 =
π
2
(1.54 − 1.254 ) = 4.1172 in 4
(4.1172)(7 × 103 )
Jτ
=
= 19.21 × 103 lb ⋅ in
1.5
c2
T = 19.21 kip ⋅ in 
(b)
Analysis of solid spindle AB:
τ=
Tc
J
π
19.21 × 103
J
T
= c3 = =
= 1.601 in 3
τ
2
c
12 × 103
c=3
(2)(1.601)
π
= 1.006 in.
d s = 2c
d = 2.01in. 
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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:
TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m
τ max =
Tc
2T
(2)(300)
=
=
3
J
πc
π (0.015)3
= 56.588 × 106 Pa
(b)
Shaft BC:
τ max = 56.6 MPa 
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
τ max = 36.6 MPa 
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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 =
2T
(2)(300)
Tc
=
=
3
J
πc
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 =
2T
(2)(700)
Tc
=
=
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
(2)(700)
=
= 7.875 × 10−6 m3
π (56.588 × 106 )
TBC = 700N ⋅ m
c3 =
2T
πτ max
τ max =
c = 19.895 × 10−3 m
d = 2c = 39.79 × 10−3 m
d = 39.8 mm 
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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
J
πc
(2) (48)
=
= 72.433MPa
π (0.0075)3
c=
τ max
τ max
Shaft BC:
T = − 48 + 144 = 96 N ⋅ m
c=
Shaft CD:
τ max =
Tc
2T
(2) (96)
= 3 =
= 83.835 MPa
J
πc
π (0.009)3
T = − 48 + 144 + 60 = 156 N⋅ m
c=
Answers:
1
d = 9 mm
2
1
d = 10.5 mm
2
τ max =
Tc
2T
(2 × 156)
= 3 =
= 85.79 MPa
J
πc
π (0.0105)3
(a) shaft CD
(b) 85.8 MPa 
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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:
τ max
2
(c
4
2
)
− c14 =
π
2
(0.00754 − 0.0044 ) = 4.5679 × 10−9 m 4
Tc2
(48)(0.0075)
=
= 78.810 MPa
J
4.5679 × 10−9
π
(c
2
)
c2 =
1
d 2 = 9 mm
2
π
(0.0094 − 0.0044 ) = 9.904 × 10−9 m 4
2
Tc
(96)(0.009)
= 2 =
= 87.239 MPa
J
9.904 × 10−9
4
2
− c14 =
T = − 48 + 144 + 60 = 156 N ⋅ m
J =
τ max =
Answers:
π
T = − 48 + 144 = 96 N ⋅ m
J =
Shaft CD:
1
d 2 = 7.5 mm
2
π
(c
2
4
2
)
− c14 =
π
2
c2 =
1
d 2 = 10.5 mm
2
(0.01054 − 0.0044 ) = 18.691 × 10−9 m 4
Tc2 (156)(0.0105)
=
= 87.636 MPa
J
18.691 × 10−9
(a) shaft CD
(b) 87.6 MPa 
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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
=
=
3
π
J
c4 π 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
2 7 kip ⋅ in
=
3
π c 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
3
π c π 1

 × 2.25 in. 
2

τ = 5.37 ksi 
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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 =
τ =
(b)
Shaft CD:
(c
2
)
− c14 =
π
2
2
(c
4
2
)
− c14 =
π
2
(1.04 − 0.54 ) = 1.47262 in 4
From free body shown:
τ =
TCD = 3 + 4 = 7 kip ⋅ in
Tc (7 kip ⋅ in)(1.0 in.)
=
J
1.47262 in 4
c2 =
τ = 3.19 ksi 
1
(2.0) = 1.0 in.
2
π
τ =
J =
1
(1) = 0.5 in.
2
(0.8754 − 0.54 ) = 0.82260 in 4
From free body shown:
J =
Shaft DE:
4
2
c1 =
Tc (3 kip ⋅ in)(0.875 in.)
=
J
0.82260 in 4
c2 =
(c)
1
(1.75) = 0.875 in.
2
π
J =
TBC = 3 kip ⋅ in
τ = 4.75 ksi 
TDE = 12 kip ⋅ in
2.25
= 1.125 in.
2
π
(c
2
4
2
)
− c14 =
π
2
(1.1254 − 0.54 ) = 2.4179 in 4
Tc (12 kip ⋅ in)(1.125 in.)
=
J
2.4179 in 4
τ = 5.58 ksi 
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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 =
Rod AB:
τ max = 15 ksi
T =
Rod BC:
π
π
Tc
, J = c 4 , T = c3τ max
J
2
2
π
2
π
2
The allowable torque is the smaller value.
1
d = 0.75 in.
2
(0.75)3 (15) = 9.94 kip ⋅ in
τ max = 8 ksi
T =
c=
c=
1
d = 0.90 in.
2
(0.90)3 (8) = 9.16 kip ⋅ in
T = 9.16 kip ⋅ in 
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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 =
(a)
Rod AB:
Tc
π
2T
, J = , c3 =
J
2
πτ max
T = 10 kip ⋅ in
c3 =
τ max = 15 ksi
(2)(10)
= 0.4244 in 3
π (15)
c = 0.7515 in.
(b)
Rod BC:
T = 10 kip ⋅ in
c3 =
d = 2c = 1.503 in. 
τ max = 8 ksi
(2)(10)
= 0.79577 in 2
π (8)
c = 0.9267 in.
d = 2c = 1.853 in. 
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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
τ max =
(a)
Rod AB:
c3 =
Tc
J
J =
π
2
c4
c3 =
2T
πτ max
(2)(1250)
= 15.915 × 10−6 m3
π (50 × 106 )
c = 25.15 × 10−3 m = 25.15 mm
d AB = 2c = 50.3 mm 
(b)
Rod BC:
c3 =
(2)(1250)
= 31.831 × 10−6 m3
6
π (25 × 10 )
c = 31.69 × 10−3 m = 31.69 mm
d BC = 2c = 63.4 mm 
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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 × 106 Pa
c=
Tall =
Hollow rod AB:
1
d = 0.015 m
2
π
2
c3τ all =
π
2
(0.015)3 (25 × 106 ) = 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 =
Jτ all
π 4
τ
c2 − c14 all
=
c2
2
c2
c14 = c24 −
(
2Tallc2
πτ all
= 0.01254 −
(2)(132.536)(0.0125)
= 3.3203 × 10−9 m 4
6
π (50 × 10 )
c1 = 7.59 × 10−3 m = 7.59 mm
(a)
(b)
)
Allowable torque.
d1 = 2c1 = 15.18 mm 
Tall = 132.5 N ⋅ m 
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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
τ all = 75 × 106 Pa Tall =
Rod AB:
c=
Tall =
1
d = 0.030 m
2
π
2
c3τ all =
π
2
Jτ all
c
J=
π
2
c4
(0.030)3 (75 × 106 )
= 3.181 × 103 N ⋅ m
Pipe CD:
c2 =
J =
Tall =
1
d 2 = 0.045 m
2
π
(c
2
4
2
)
− c14 =
c1 = c2 − t = 0.045 − 0.006 = 0.039 m
π
2
(0.0454 − 0.0394 ) = 2.8073 × 10−6 m 4
(2.8073 × 10−6 )(75 × 106 )
= 4.679 × 103 N ⋅ m
0.045
Allowable torque is the smaller value.
Tall = 3.18 × 103 N ⋅ m
3.18 kN ⋅ m 
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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 × 106 Pa
Tall =
=
Pipe CD:
c=
1
d = 0.030 m
2
π
Jτ all
= c3 τ all
2
c
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
c1 = c2 − t = 0.045 − 0.006 = 0.039 m
J =
Tall =
π
(c
2
4
2
)
− c14 =
π
2
(0.0454 − 0.0394 ) = 2.8073 × 10−6 m 4
Jτ all (2.8073 × 10−6 )(54 × 106 )
=
= 3.369 × 103 N ⋅ m
0.045
c2
Allowable torque is the smaller value.
Tall = 3.369 × 103 N ⋅ m
3.37 kN ⋅ m 
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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)
= 3=
= 72.50 × 106
J
π c π (0.028)3
c=
1
d = 0.020 m
2
τ=
Tc
2T
(2) (1000)
= 3=
= 68.7 × 106
J
π c π (0.020)3
72.5 MPa 
68.7 MPa 
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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
TAB =
(a)
Shaft AB:
rB
100
TCD =
(1000) = 2500 N ⋅ m
rC
40
τ all = 60 × 106 Pa
τ =
Tc
2T
= 3
J
πc
c3 =
2T
πτ
=
(2) (2500)
= 26.526 × 10−6 m3
π (60 × 106 )
c = 29.82 × 10−3 = 29.82 mm
(b)
Shaft CD:
d = 2c = 59.6 mm 
τ all = 60 × 106 Pa
τ =
Tc
2T
= 3
J
πc
c3 =
2T
πτ
=
(2) (1000)
= 10.610 × 10−6 m3
π (60 × 106 )
c = 21.97 × 10−3 m = 21.97 mm
d = 2c = 43.9 mm 
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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
TE =
rD
8
TF = (1200) = 3200 lb ⋅ in
rG
3
τ all = 10.5 ksi = 10500 psi
τ =
(a)
2T
Tc
= 3
J
πc
c3 =
2T
πτ
Shaft CDE:
c3 =
(2) (3200)
= 0.194012 in 3
π (10500)
c = 0.5789 in.
(b)
or
d DE = 2c
d DE = 1.158 in. 
Shaft FGH:
c3 =
(2) (1200)
π (10500)
c = 0.4174 in.
= 0.012757 in 3
d FG = 2c
d FG = 0.835 in. 
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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
= all = c3τ all
c
2
c=
TF , all
=
Shaft DE:
c=
TE , all =
π
2
(0.400)3 (12) = 1.206 kip ⋅ in
1
d = 0.450 in.
2
π
2
c3τ all
π
(0.450)3 (12) = 1.7177 kip ⋅ in
2
r
4.5
TF = G TE TF , all =
(1.7177) = 1.189 kip ⋅ in
rD
6.5
=
Allowable value of TF is the smaller.
TF = 1.189 kip ⋅ in 
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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 =
c=
Tc
2T
=
J
π c3
3
c=
3
2T
πτ max
(2)(5)
= 0.7208 in.
π (8.5)
d BC = 2c = 1.442 in. 
(b)
Shaft EF:
TF =
c=
rD
2.5
TC =
(5) = 3.125 kip ⋅ in
rA
4
3
2T
πτ max
=
3
(2)(3.125)
= 0.6163 in.
π (8.5)
d EF = 2c = 1.233 in. 
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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, d BC = 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:
d BC = 1.6 in.
1
d = 0.8 in.
2
π
Jτ max
TC =
= τ max c3
2
c
c=
=
Shaft EF:
π
2
(7.0)(0.8)3 = 5.63 kip ⋅ in
d EF = 1.25 in.
1
d = 0.625 in.
2
π
Jτ max
TF =
= τ max c3
c
2
c=
=
By statics,
TC =
π
2
(7.0)(0.625)3 = 2.684 kip ⋅ in
rA
4
TF =
(2.684) = 4.30 kip ⋅ in
rD
2.5
Allowable value of TC is the smaller.
TC = 4.30 kip ⋅ in 
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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, dCD = 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:
Gears B and C:
Force on gear circles.
TAB = TA = TB = T
rB = 25 mm, rC = 60 mm
FBC =
TC =
Shaft CD:
Gears D and E:
Force on gear circles.
rC
60
TB =
T = 2.4 T
rB
25
TCD = TC = TD = 2.4 T
rD = 30 mm, rE = 75 mm
FDE =
TE =
Shaft EF:
TB
T
= C
rB
rC
TD
T
= E
rD
rE
rE
75
TD =
(2.4 T ) = 6 T
rD
30
TEF = TE = TF = 6T
Maximum Shearing Stresses. τ max =
Tc
2T
=
J
π c3
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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 × 106 Pa
π (10.5 × 10−3 )3
T = (2.4)(100) = 240 N ⋅ m
c=
τ max =
(c)
Shaft EF:
τ max = 55.0 MPa 
1
d = 15 mm = 15 × 10−3 m
2
(2)(240)
= 45.3 × 106 Pa
−3 3
π (15 × 10 )
τ max = 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 × 106 Pa
τ max = 47.7 MPa 
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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:
Gears B and C:
Force on gear circles.
TAB = TA = TB = T
rB = 25 mm, rC = 60 mm
FBC =
TC =
Shaft CD:
Gears D and E:
Force on gear circles.
rC
60
TB =
T = 2.4 T
rB
25
TCD = TC = TD = 2.4 T
rD = 30 mm, rE = 75 mm
FDE =
TE =
Shaft EF:
TB
T
= C
rB
rC
TD
T
= E
rD
rE
rE
75
TD =
(2.4 T ) = 6 T
rD
30
TEF = TE = TF = 6 T
Required Diameters.
τ max =
c=
2T
Tc
=
J
π c3
3
2T
πτ
d = 2c = 2 3
2T
πτ max
τ max = 75 × 106 Pa
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PROBLEM 3.28 (Continued)
(a)
Shaft AB:
TAB = T = 120 N ⋅ m
d AB = 2 3
(b)
Shaft CD:
Shaft EF:
d AB = 20.1 mm 
TCD = (2.4)(120) = 288 N ⋅ m
dCD = 2 3
(c)
2(120)
= 20.1 × 10−3 m
π (75 × 106 )
(2)(288)
π (75 × 106 )
= 26.9 × 10−3 m
dCD = 26.9 mm 
TEF = (6)(120) = 720 N ⋅ m
d EF = 2 3
(2)(720)
= 36.6 × 10−3 m
3
π (75 × 10 )
d EF = 36.6 mm 
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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,
ρ g = specific weight,
W = total weight,
L = length
w=
Tall =
(
)
(a)
T
= c12 + c22 τ all
W

c1 = 0 for solid shaft:
(b)
(T /w) h
c2
= 1 + 12
(T /w)0
c2
W
ρ gLA
=
= ρ gA = ρ gπ c22 − c12
L
L
(
(
)(
)
)
2
2
2
2
π c24 − c14
π c2 + c1 c2 − c1
Jτ all
τ all =
τ all
=
2 c2
2
c2
c2
(
)
c12 + c22 τ all
T
(hollow shaft) 
=
2ρ gc2
w
c2τ all
T 
(solid shaft)
  =
2 pg
 w 0
c12 
T  T  
=
1
+

 
    
c22 
 w   w 0 
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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 τmax 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 + c2 ) 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 /c2 , respectively equal to 1.00, 0.95, 0.75, 0.50, and 0.
SOLUTION
For a hollow shaft:
τ max =
τ0 =
By definition,
Tc2
2Tc2
2Tc2
2Tc2
=
=
=
4
4
2
2
2
2
J
π c2 − c1
π c2 − c1 c2 + c1
A c22 + c12
(
)
(
)(
)
(
T
2T
=
Arm
A(c2 + c1)
τ max
c (c + c )
1 + (c1 /c2 )
= 2 2 2 21 =
τ0
c2 + c1
1 + (c1 /c2 ) 2
Dividing,
)

c1 /c2
1.0
0.95
0.75
0.5
0.0
τ max /τ 0
1.0
1.025
1.120
1.200
1.0

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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)
c=
1
d = 0.015 m,
2
−9
J =
π
2
c4 =
π
2
(0.015) 4
J = 79.522 × 10 m , L = 1.8 m, G = 77 × 109 Pa
T = 250 N ⋅ m
ϕ =
ϕ =
(b)
4
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
1
π 4
c2 = 0.015 m, c1 d1 = 0.010 m, J =
c2 − c14
2
2
π
TL
J = (0.0154 − 0.0104 ) = 63.814 × 10−9 m 4 ϕ
2
GJ
(
ϕ =
)
(250)(1.8)
180
= 91.58 × 10−3 rad =
(91.58 × 10−3 )
9
−9
π
(77 × 10 )(63.814 × 10 )
ϕ = 5.25° 

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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 =
GJ ϕ
L
ϕ = 4° = 69.813 × 10−3 rad, L = 1.25 m
G = 27 GPa = 27 × 109 Pa
J =
π
(c
2
4
2
)
− c14 =
π
2
(0.0184 − 0.0124 ) = 132.324 × 10−9 m 4
(27 × 109 )(132.324 × 109 )(69.813 × 10−3 )
1.25
= 199.539 N⋅ m
T =
(b)
Matching areas:
(
A = π c 2 = π c22 − c12
T = 199.5 N ⋅ m 
)
c = c22 − c12 = 0.0182 − 0.0122 = 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° 
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PROBLEM 3.33
Determine the largest allowable diameter of a 10-ft-long steel rod (G = 11.2 × 106 psi) if the rod is to be twisted
through 30° without exceeding a shearing stress of 12 ksi.
SOLUTION
L = 10 ft = 120 in.
ϕ = 30° =
30π
= 0.52360 rad
180
τ = 12 ksi = 12 × 103 psi
TL
GJ ϕ
Tc GJ ϕ c Gϕ c
τL
, T =
, τ =
, c=
=
=
ϕ =
GJ
L
J
JL
L
Gϕ
c=
(12 × 103 )(120)
= 0.24555 in.
(11.2 × 106 )(0.52360)
d = 2c = 0.491 in. 
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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 × 106 psi
L = 6000 ft = 72000 in.
From text book,
TL
GJ
Tc
τm =
J
ϕ =
Divide (2) by (1).
(1)
(2)
τ m Gc
=
L
ϕ
τm =
Gcϕ (11 × 106 )(4)(4π )
=
= 7679 psi
72000
L
τ m = 7.68 ksi 
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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, LBC = 1.2 m
c=
J BC =
ϕ B/C =
(b)
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
GJ
(27 × 109 )(367.97 × 109 )
ϕ B /C = 1.384° 
Angle of twist between B and D.
TCD = 500 N ⋅ m, LCD = 0.9 m, c =
J CD =
ϕC /D
π
2
c4 =
π
1
d = 0.024 m, G = 27 × 109 Pa
2
(0.024) 4 = 521.153 × 10−9 m 4
2
(500)(0.9)
=
= 31.980 × 10−3 rad
(27 × 109 )(521.153 × 109 )
ϕ B /D = ϕ B /C + ϕC /D = 24.157 × 10−3 + 31.980 × 10−3 = 56.137 × 10−3 rad
ϕ B /D = 3.22° 
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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 )
ϕ BC = 8.54° 
1
π
d = 0.018 m
J CD = c 4 = 164.896 × 10−9 m 4
2
4
= 1.0 m TCD = 400 − 900 = −500 N ⋅ m
c=
LCD
ϕCD =
(−500)(1.0)
TL
=
= −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° 
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PROBLEM 3.37
The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB
(G = 39 GPa). Knowing that each rod is solid and has a diameter of
12 mm, determine the angle of twist (a) at B, (b) at C.
SOLUTION
Both portions:
c=
1
d = 6 mm = 6 × 10−3 m
2
π
π
c 4 = (6 × 10−3 )4 = 2.03575 × 10−9 m 4
2
2
T = 100 N ⋅ m
J =
Rod AB:
(a)
ϕ B = ϕ AB =
Rod BC:
GAB = 39 × 109 Pa, LAB = 0.200 m
TLAB
(100)(0.200)
=
= 0.25191 rad
G AB J
(39 × 109 )(2.03575 × 10−9 )
GBC = 26 × 109 Pa, LBC = 0.300 m
ϕ BC =
(b)
ϕ B = 14.43° 
TLBC
(100)(0.300)
=
= 0.56679 rad
GBC J
(26 × 109 )(2.03575 × 10−9 )
ϕC = ϕ B + ϕ BC = 0.25191 + 0.56679 = 0.81870 rad
ϕC = 46.9° 
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PROBLEM 3.38
The aluminum rod AB (G = 27 GPa) is bonded to the brass
rod BD (G = 39GPa). Knowing that portion CD of the
brass rod is hollow and has an inner diameter of 40 mm,
determine the angle of twist at A.
SOLUTION
G = 27 × 109 Pa, L = 0.400 m
Rod AB:
T = 800 N ⋅ m c =
Part BC:
π
(0.018) 4 = 164.896 × 10−9 m
2
2
TL
(800)(0.400)
=
=
= 71.875 × 10−3 rad
9
−9
GJ
(27 × 10 )(164.896 × 10 )
J =
ϕ A/B
π
1
d = 0.018 m
2
G = 39 × 109 Pa
c4 =
L = 0.375 m, c =
T = 800 + 1600 = 2400 N ⋅ m, J =
ϕ B /C
1
d = 0.030 m
2
π
c4 =
π
(0.030) 4 = 1.27234 × 10−6 m 4
2
2
(2400)(0.375)
TL
=
=
= 18.137 × 10−3 rad
GJ
(39 × 109 )(1.27234 × 10−6 )
1
d1 = 0.020 m
2
1
c2 = d 2 = 0.030 m, L = 0.250 m
2
c1 =
Part CD:
ϕC / D
Angle of twist at A.
π
(
)
π
c24 − c14 = (0.0304 − 0.0204 ) = 1.02102 × 10−6 m 4
2
2
(2400)(0.250)
TL
=
=
= 15.068 × 10−3 rad
9
−6
GJ
(39 × 10 )(1.02102 × 10 )
J =
ϕ A = ϕ A/B + ϕ B/C + ϕC/D
= 105.080 × 10−3 rad
ϕ A = 6.02° 
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PROBLEM 3.39
The solid spindle AB has a diameter d s =1.5 in. and is made of a steel with
G = 11.2 × 106 psi and τ all = 12 ksi, while sleeve CD is made of a brass
with G = 5.6 × 106 psi and τ all = 7 ksi. Determine the angle through end
A can be rotated.
SOLUTION
Stress analysis of solid spindle AB:
c=
1
ds = 0.75 in.
2
τ =
Tc
J
(12 × 103 )(0.75)3 = 7.95 × 103 lb ⋅ in
2
1
1
c2 = do = (3) = 1.5in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25in.
J =
π
T =
The smaller torque governs.
Deformation of spindle AB:
(c
2
)
− c14 =
π
2
(1.54 − 1.254 ) = 4.1172in 4
T = 7.95 × 103 lb ⋅ in
c = 0.75 in.
ϕ AB =
π
2
c 4 = 0.49701 in 4 , L = 12 in., G = 11.2 × 106 psi
TL
(7.95 × 103 ) (12)
=
= 0.017138 radians
GJ
(11.2 × 106 )(0.49701)
J = 4.1172 in 4 , L = 8 in., G = 5.6 × 106 psi
ϕCD =
Total angle of twist:
4
2
Jτ
(4.1172)(7 × 10−3 )
=
= 19.21 × 103 lb ⋅ in
c2
1.5
J =
Deformation of sleeve CD:
Jτ
π
= τ c3
c
2
π
T =
Stress analysis of sleeve CD:
T =
TL
(7.95 × 103 ) (8)
=
= 0.002758 radians
GJ
(5.6 × 106 ) (4.1172)
ϕ AD = ϕ AB + ϕCD = 0.019896 radians
ϕ AD = 1.140° 
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PROBLEM 3.40
The solid spindle AB has a diameter d s = 1.75in. and is made of
a steel with G = 11.2 × 106 psi and τ all = 12 ksi, while sleeve
CD is made of a brass with G = 5.6 × 106 psi and τ all = 7 ksi.
Determine (a) the largest torque T that can be applied at A if the
given allowable stresses are not to be exceeded and if the angle of
twist of sleeve CD is not to exceed 0.375°, (b) the corresponding
angle through which end A rotates.
SOLUTION
Spindle AB:
c=
J =
Sleeve CD:
π
2
c4 =
π
2
L = 12 in., τ all = 12 ksi, G = 11.2 × 106 psi
(0.875)4 = 0.92077in 4
c1 = 1.25 in., c2 = 1.5 in., L = 8 in., τ all = 7 ksi
J =
(a)
1
(1.75in.) = 0.875 in.
2
π
2
(c
4
2
)
− c14 = 4.1172 in 4 ,
G = 5.6 × 106 psi
Largest allowable torque T.
Ciriterion: Stress in spindle AB.
Critrion: Stress in sleeve CD.
Jτ
c
τ =
Tc
J
T=
(0.92077)(12)
= 12.63 kip ⋅ in
0.875
T=
Criterion: Angle of twist of sleeve CD
T =
Jτ 4.1172 in 4
=
(7 ksi)
c2
1.5 in.
T = 19.21 kip ⋅ in
φ = 0.375° = 6.545 × 10−3 rad
φ =
TL
JG
T =
JG
(4.1172)(5.6 × 106 )
φ =
(6.545 × 10−3 )
L
8
T = 18.86 kip ⋅ in
T = 12.63 kip ⋅ in 
The largest allowable torque is
(b)
Angle of rotation of end A.
φ A = φ A / D = φ A / B + φC / D = 
Ti Li
=T
J i Gi

Li
J i Gi


12
8
= (12.63 × 103 ) 
+
6
6 
 (0.92077)(11.2 × 10 ) (4.1172)(5.6 × 10 ) 
= 0.01908 radians
ϕ A = 1.093° 
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PROBLEM 3.41
Two shafts, each of 78 -in. diameter, are
connected by the gears shown. Knowing
that G = 11.2 × 106 psi and that the shaft
at F is fixed, determine the angle through
which end A rotates when a 1.2 kip ⋅ in.
torque is applied at A.
SOLUTION
Calculation of torques.
Circumferential contact force between gears B and E.
F =
TAB
T
= EF
rB
rE
TEF =
rE
TAB
rB
TAB = 1.2 kip ⋅ in = 1200 lb ⋅ in
TEF =
6
(1200) = 1600 lb ⋅ in
4.5
Twist in shaft FE.
L = 12 in., c =
J =
ϕ E /F =
π
2
4
c =
ϕB =
−3 4
  = 57.548 × 10 in
2  16 
δ = rEϕ E = rBϕ B
rE
6
(29.789 × 10−3 ) = 39.718 × 10−3 rad
ϕE =
4.5
rB
L = 8 + 6 = 14 in.
ϕ A/B =
Rotation at A.
4
ϕ E = ϕ E/F = 29.789 × 10−3 rad
Tangential displacement at gear circle.
Twist in shaft BA.
π7
TL
(1600)(12)
=
= 29.789 × 10−3 rad
−3
6
GJ
(11.2 × 10 )(57.548 × 10 )
Rotation at E.
Rotation at B.
1
7
d =
in., G = 11.2 × 106 psi
2
16
J = 57.548 × 10−3 in 4
(1200)(14)
TL
=
= 26.065 × 10−3 rad
GJ
(11.2 × 106 )(57.548 × 10−3 )
ϕ A = ϕ B + ϕ A/B = 65.783 × 10−3 rad
ϕ A = 3.77° 
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PROBLEM 3.42
Two solid shafts are connected by gears as shown. Knowing
that G = 77.2 GPa for each shaft, determine the angle
through which end A rotates when TA = 1200 N ⋅ m.
SOLUTION
Calculation of torques:
Circumferential contact force between gears B and C.
TAB = 1200 N ⋅ m
Twist in shaft CD:
c=
J =
ϕC/D
Rotation angle at C.
TCD =
1
d = 0.030 m,
2
π
c4 =
π
F =
TAB
T
= CD
rB
rC
Twist in shaft AB:
L = 1.2 m, G = 77.2 × 109 Pa
(0.030) 4 = 1.27234 × 10−6 m 4
2
2
(3600)(1.2)
TL
=
=
= 43.981 × 10−3 rad
GJ
(77.2 × 109 )(1.27234 × 10−9 )
ϕC = ϕC /D = 43.981 × 10−3 rad
ϕB =
c=
J =
ϕ A/B
Rotation angle at A.
rC
TAB
rB
240
(1200) = 3600 N ⋅ m
80
Circumferential displacement at contact points of gears B and C.
Rotation angle at B.
TCD =
δ = rCϕC = rBϕ B
rC
240
(43.981 × 10−3 ) = 131.942 × 10−3 rad
ϕC =
80
rB
1
d = 0.021m, L = 1.6 m, G = 77.2 × 109 Pa
2
π
c4 =
π
(0.021)4 = 305.49 × 109 m 4
2
2
(1200)(1.6)
TL
=
=
= 81.412 × 10−3 rad
GJ
(77.2 × 109 )(305.49 × 10−9 )
ϕ A = ϕ B + ϕ A/B = 213.354 × 10−3 rad
ϕ A = 12.22° 
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PROBLEM 3.43
A coder F, used to record in digital form the rotation
of shaft A, is connected to the shaft by means of the
gear train shown, which consists of four gears and
three solid steel shafts each of diameter d. Two of the
gears have a radius r and the other two a radius nr. If
the rotation of the coder F is prevented, determine in
terms of T, l, G, J, and n the angle through which end
A rotates.
SOLUTION
TAB = TA
TCD =
rC
T
T
TAB = AB = A
rB
n
n
TEF =
rE
T
T
TCD = CD = A2
rD
n
n
TEF lEF
T l
= 2A
GJ
n GJ
ϕ
r
Tl
= E ϕE = E = 3 A
rD
n
n GJ
ϕ E = ϕ EF =
ϕD
TCDlCD
T l
= A
GJ
nGJ
1
Tl
T l
T l 1
ϕC = ϕ D + ϕCD = 3 A + A = A  3 + 
GJ  n
n
n GJ nGJ
ϕCD =
1 
ϕ
rC
Tl  1
ϕC = C =
 4 + 2
rB
n
GJ  n
n 
T l
T l
= AB AB = A
GJ
GJ
ϕB =
ϕ AB
ϕ A = ϕ B + ϕ AB =
TAl  1
1

+ 2 + 1 

4
GJ  n
n

 
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PROBLEM 3.44
For the gear train described in Prob. 3.43, determine
the angle through which end A rotates when
T = 5lb ⋅ in.,
l = 2.4in.,
d = 1/16in.,
G = 11.2 × 106 psi, and n = 2.
PROBLEM 3.43 A coder F, used to record in digital
form the rotation of shaft A, is connected to the shaft
by means of the gear train shown, which consists of
four gears and three solid steel shafts each of diameter
d. Two of the gears have a radius r and the other two a
radius nr. If the rotation of the coder F is prevented,
determine in terms of T, l, G, J, and n the angle
through which end A rotates.
SOLUTION
See solution to Prob. 3.43 for development of equation for ϕ A.
ϕA =
Data:
Tl 
1
1 
1+ 2 + 4

GJ 
n
n 
T = 5 lb ⋅ in, l = 2.4 in., c =
n = 2, J =
ϕA =
π
2
c4 =
π 1 
1
1
d =
in., G = 11.2 × 106 psi
2
32
4
−6 4
  = 1.49803 × 10 in
2  32 
(5)(2.4)
1
1 

1 + +  = 938.73 × 10−3 rad 
−6 
4 16 
(11.2 × 10 )(1.49803 × 10 ) 
6
ϕ A = 53.8° 
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PROBLEM 3.45
The design of the gear-and-shaft system shown
requires that steel shafts of the same diameter be
used for both AB and CD. It is further required
that τ max ≤ 60 MPa, and that the angle φD through
which end D of shaft CD rotates not exceed 1.5°.
Knowing that G = 77 GPa, determine the required
diameter of the shafts.
SOLUTION
TCD = TD = 1000 N ⋅ m
For design based on stress, use larger torque.
TAB =
rB
100
TCD =
(1000) = 2500 N ⋅ m
rC
40
TAB = 2500 N ⋅ m
Tc
2T
=
J
π c3
2T
(2)(2500)
c3 =
=
= 26.526 × 10−6 m3
πτ
π (60 × 106 )
τ =
c = 29.82 × 10−3 m = 29.82 mm , d = 2c = 59.6 mm
Design based on rotation angle.
Shaft AB:
ϕ D = 1.5° = 26.18 × 10−3 rad
TAB = 2500 N ⋅ m,
L = 0.4 m
TL
(2500)(0.4) 1000
=
=
GJ
GJ
GJ
1000

ϕ B = ϕ AB = GJ

Gears 
ϕC = rB ϕ B =  100  1000  = 2500




rC
GJ
 40  GJ 
ϕ AB =
Shaft CD:
TCD = 1000 N ⋅ m,
L = 0.6 m
TL
(1000) (0.6) 600
=
=
GJ
GJ
GJ
2500 600 3100
3100
= ϕC + ϕCD =
+
=
= π 4
GJ
GJ
GJ
G2c
ϕCD =
ϕD
c4 =
(2) (3100)
(2)(3100)
=
= 979.06 × 10−9 m 4
π Gϕ D
π (77 × 109 )(26.18 × 10−3 )
c = 31.46 × 10−3 m = 31.46 mm,
Design must use larger value for d.
d = 2c = 62.9 mm
d = 62.9 mm 
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PROBLEM 3.46
The electric motor exerts a torque of
800 N ⋅ m on the steel shaft ABCD when it
is rotating at constant speed. Design
specifications require that the diameter of
the shaft be uniform from A to D and that
the angle of twist between A to D not
exceed 1.5°. Knowing that τ max ≤ 60 MPa
and G = 77 GPa, determine the minimum
diameter shaft that can be used.
SOLUTION
Torques:
TAB = 300 + 500 = 800 N ⋅ m
TBC = 500 N ⋅ m,
TCD = 0
τ = 60 × 106 Pa
Design based on stress.
τ =
Tc
2T
=
J
π c3
c3 =
2T
πτ
=
(2)(800)
= 8.488 × 10−6 m3
6
π (60 × 10 )
−3
c = 20.40 × 10 m = 20.40 mm,
Design based on deformation.
d = 2c = 40.8 mm
ϕ D/ A = 1.5° = 26.18 × 10−3 rad
ϕD / C = 0
(500)(0.6) 300
TBC LBC
=
=
GJ
GJ
GJ
(800) (0.4) 320
T L
= AB AB =
=
GJ
GJ
GJ
620
620
(2)(620)
= ϕ D / C + ϕC / B + ϕ B / A =
= π 4 =
GJ
π Gc 4
G2c
ϕ C/ B =
ϕ B/ A
ϕ D/ A
c4 =
(2)(620)
(2)(620)
=
= 195.80 × 10−9 m 4
π Gϕ D/ A π (77 × 109 )(26.18 × 10−3 )
c = 21.04 × 10−3 m = 21.04 mm,
Design must use larger value of d.
d = 2c = 42.1 mm
d = 42.1 mm 
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PROBLEM 3.47
The design specifications of a 2-m-long solid circular transmission shaft require that the angle of twist of the
shaft not exceed 3° when a torque of 9 kN ⋅ m is applied. Determine the required diameter of the shaft,
knowing that the shaft is made of (a) a steel with an allowable shearing stress of 90 MPa and a modulus of
rigidity of 77 GPa, (b) a bronze with an allowable shearing of 35 MPa and a modulus of rigidity of 42 GPa.
SOLUTION
ϕ = 3° = 52.360 × 10−3 rad, T = 9 × 103 N ⋅ m
L = 2.0 m
2TL
2TL
TL
=
∴ c4 =
based on twist angle.
4
π Gϕ
GJ
πc G
2T
2T
Tc
τ =
= 2 ∴ c3 =
based on shearing stress.
πτ
J
πc
ϕ =
(a)
Steel shaft:
Based on twist angle,
τ = 90 × 106 Pa, G = 77 × 109 Pa
c4 =
(2)(9 × 103 )(2.0)
= 2.842 × 10−6 m 4
π (77 × 109 )(52.360 × 10−3 )
c = 41.06 × 10−3 m = 41.06 mm
Based on shearing stress,
c3 =
d = 2c = 82.1 mm
(2)(9 × 103 )
= 63.662 × 10−6 m3
π (90 × 106 )
c = 39.93 × 10−3 m = 39.93 mm
d = 2c = 79.9 mm
d = 82.1 mm 
Required value of d is the larger.
(b)
Bronze shaft:
Based on twist angle,
τ = 35 × 106 Pa, G = 42 × 109 Pa
c4 =
(2)(9 × 103 )(2.0)
= 5.2103 × 10−6 m 4
9
−3
π (42 × 10 )(52.360 × 10 )
c = 47.78 × 10−3 m = 47.78 mm
Based on shearing stress,
c3 =
(2)(9 × 103 )
= 163.702 × 10−6 m3
π (35 × 106 )
c = 54.70 × 10−3 m = 54.70 mm
Required value of d is the larger.
d = 2c = 95.6 mm
d = 2c = 109.4 mm
d = 109.4 mm 
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PROBLEM 3.48
A hole is punched at A in a plastic sheet by applying a 600-N
force P to end D of lever CD, which is rigidly attached to the
solid cylindrical shaft BC. Design specifications require that
the displacement of D should not exceed 15 mm from the
time the punch first touches the plastic sheet to the time it
actually penetrates it. Determine the required diameter of
shaft BC if the shaft is made of a steel with G = 77 GPa and
τ all = 80 MPa.
SOLUTION
T = rP = (0.300 m)(600 N) = 180 N ⋅ m
Torque
Shaft diameter based on displacement limit.
ϕ =
δ
r
=
15 mm
= 0.005 rad
300 mm
ϕ =
TL
2TL
=
GJ
π Gc 4
c4 =
2TL
(2)(180)(0.500)
=
= 14.882 × 10−9 m 4
π Gϕ π (77 × 109 )(0.05)
c = 11.045 × 10−3 m = 11.045 m
d = 2c = 22.1 mm
Shaft diameter based on stress.
τ = 80 × 106 Pa
c3 =
2T
πτ
=
τ =
Tc
2T
=
J
π c3
(2)(180)
= 1.43239 × 10−6 m3
π (80 × 106 )
c = 11.273 × 10−3 m = 11.273 mm
Use the larger value to meet both limits.
d = 2c = 22.5 mm
d = 22.5 mm 
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PROBLEM 3.49
The design specifications for the gear-and-shaft system
shown require that the same diameter be used for both
shafts, and that the angle through which pulley A will
rotate when subjected to a 2-kip ⋅ in. torque TA while
pulley D is held fixed will not exceed 7.5°. Determine
the required diameter of the shafts if both shafts are made
of a steel with G = 11.2 × 106 psi and τ all = 12 ksi.
SOLUTION
Statics:
Gear B.
ΣM B = 0:
rB F − TA = 0 F = TB /rB
ΣM C = 0:
Gear C.
rC F − TD = 0
TD = rC F =
n=
rC
5
= = 2.5
rB
2
Torques in shafts.
TAB = TA = TB
Deformations:
ϕC /D =
ϕ A/B
Kinematics:
ϕD = 0
rC
TA = nTB
rB
TCD L
GJ
T L
= AB
GJ
TCD = TC = nTB = nTA
nTAL
GJ
T L
= A
GJ
=
ϕC = ϕ D + ϕC/D = 0 +
rBϕ B = −rCϕ B
ϕB = −
ϕ A = ϕC + ϕ B /C =
nTAL
GJ
rC
ϕC = −nϕC
rB
ϕB
n 2TA L
GJ
n 2TA L TA L (n2 + 1)TA L
+
=
GJ
GJ
GJ
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PROBLEM 3.49 (Continued)
Diameter based on stress.
Tm = TCD = nTA
Largest torque:
τm =
c=
Tmc
2nTA
=
J
π c3
3
2nTA
πτ m
=
3
τ m = τ all = 12 × 103 psi, TA = 2 × 103 lb ⋅ in
(2)(2.5)(2 × 103 )
= 0.6425 in., d = 2c = 1.285 in.
π (12 × 103 )
Diameter based on rotation limit.
ϕ = 7.5° = 0.1309 rad
ϕ =
c=
(n 2 + 1)TA L (2)(7.25)TAL
=
GJ
π c 4G
4
(2)(7.25)TA L
=
π Gϕ
Choose the larger diameter.
4
L = 8 + 16 = 24 in.
(2)(7.25)(2 × 103 )(24)
= 0.62348 in., d = 2c = 1.247 in.
π (11.2 × 106 )(0.1309)
d = 1.285 in. 
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PROBLEM 3.50
Solve Prob. 3.49, assuming that both shafts are made of a
brass with G = 5.6 × 106 psi and τ all = 8 ksi.
PROBLEM 3.49 The design specifications for the gearand-shaft system shown require that the same diameter
be used for both shafts, and that the angle through which
pulley A will rotate when subjected to a 2-kip ⋅ in.
torque TA while pulley D is held fixed will not exceed
7.5°. Determine the required diameter of the shafts if
both shafts are made of a steel with G = 11.2 × 106 psi
and τ all = 12 ksi.
SOLUTION
Statics:
Gear B.
ΣM B = 0:
rB F − TA = 0 F = TB /rB
ΣM C = 0:
Gear C.
rC F − TD = 0
TD = rC F =
n=
rC
5
= = 2.5
rB
2
Torques in shafts.
TAB = TA = TB
Deformations:
ϕC /D =
ϕ A/B
Kinematics:
ϕD = 0
rC
TA = nTB
rB
TCD L
GJ
T L
= AB
GJ
TCD = TC = nTB = nTA
nTAL
GJ
T L
= A
GJ
=
ϕC = ϕ D + ϕC/D = 0 +
rBϕ B = −rCϕ B
ϕB = −
ϕ A = ϕ C + ϕ B /C =
nTAL
GJ
rC
ϕC = −nϕC
rB
ϕB
n 2TA L
GJ
n 2TA L TA L (n2 + 1)TA L
+
=
GJ
GJ
GJ
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PROBLEM 3.50 (Continued)
Diameter based on stress.
Tm = TCD = nTA
Largest torque:
τm =
c=
Tmc
2nTA
=
J
π c3
3
2nTA
πτ m
=
3
τ m = τ all = 8 × 103 psi, TA = 2 × 103 lb ⋅ in
(2)(2.5)(2 × 103 )
= 0.7355 in., d = 2c = 1.471 in.
π (8 × 103 )
Diameter based on rotation limit.
ϕ = 7.5° = 0.1309 rad
ϕ =
c=
(n 2 + 1)TA L (2)(7.25)TAL
=
GJ
π c 4G
4
(2)(7.25)TA L
=
π Gϕ
Choose the larger diameter.
4
L = 8 + 16 = 24 in.
(2)(7.25)(2 × 103 )(24)
= 0.7415 in., d = 2c = 1.483 in.
π (5.6 × 106 )(0.1309)
d = 1.483 in. 
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PROBLEM 3.51
A torque of magnitude T = 4 kN⋅ m is applied at end A of the
composite shaft shown. Knowing that the modulus of rigidity
is 77 GPa for the steel and 27 GPa for the aluminum,
determine (a) the maximum shearing stress in the steel core,
(b) the maximum shearing stress in the aluminum jacket, (c)
the angle of twist at A.
SOLUTION
c1 =
Steel core:
1
d1 = 0.027 m
2
J1 =
π
2
c14 =
π
2
(0.027) 4 = 834.79 × 10−9
−9
9
G1J1 = (77 × 10 )(834.79 × 10 ) = 64.28 × 103 N ⋅ m 2
Torque carried by steel core.
T1 = G1J1ϕ /L
Aluminum jacket:
c1 =
J2 =
1
d1 = 0.027 m,
2
π
(c
2
4
2
)
− c14 =
π
2
c2 =
1
d 2 = 0.036 m
2
(0.0364 − 0.027 4 ) = 1.80355 × 10−6 m 4
G2 J 2 = (27 × 10 )(1.80355 × 10−6 ) = 48.70 × 103 N ⋅ m 2
9
Torque carried by aluminum jacket. T2 = G2 J 2ϕ /L
T = T1 + T2 = (G1J1 + G2 J 2 ) ϕ /L
Total torque:
ϕ
L
(a)
=
T
4 × 103
=
= 35.406 × 10−3 rad/m
G1 J1 + G2 J 2 64.28 × 103 + 48.70 × 103
Maximum shearing stress in steel core.
τ = G1γ = G1c1
(b)
ϕ
= 73.6 × 106 Pa
73.6 MPa 
= 34.4 × 106 Pa
34.4 MPa 
Maximum shearing stress in aluminum jacket.
τ = G2γ = G2c2
(c)
= (77 × 109 )(0.027)(35.406 × 10−3 )
L
Angle of twist.
ϕ
L
= (27 × 109 )(0.036)(35.406 × 10−3 )
ϕ =L
ϕ
L
= (2.5)(35.406 × 10−3 ) = 88.5 × 10−3 rad
ϕ = 5.07° 
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PROBLEM 3.52
The composite shaft shown is to be twisted by applying a
torque T at end A. Knowing that the modulus of rigidity is 77
GPa for the steel and 27 GPa for the aluminum, determine the
largest angle through which end A can be rotated if the
following allowable stresses are not to be exceeded:
τ steel = 60 MPa and τ aluminum = 45 MPa .
SOLUTION
τ max = Gγ max = Gcmax
ϕall
L
Steel core:
τ all
Gcmax
L
for each material.
τ all = 60 × 106 Pa , cmax =
ϕall
L
Aluminum Jacket:
=
ϕ
=
60 × 106
= 28.860 × 10−3 rad/m
(77 × 109 )(0.027)
τ all = 45 × 106 Pa , cmax =
ϕall
L
=
1
d = 0.036 m, G = 27 × 109 Pa
2
45 × 106
= 46.296 × 10−3 rad/m
(27 × 109 )(0.036)
Smaller value governs:
ϕall
Allowable angle of twist:
ϕall = L
L
1
d = 0.027 m, G = 77 × 109 Pa
2
= 28.860 × 10−3 rad/m
ϕall
L
= (2.5) (28.860 × 10−3 ) = 72.15 × 10−3 rad
ϕall = 4.13° 
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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:
c=
1
d = 0.75 in.
2
L = 12 in.
π
c 4 = 0.49701 in 4
2
T L
TAB (12)
= 6.5255 × 10−6TAB
ϕ B = AB =
6
GJ
(3.7 × 10 )(0.49701)
J =
Cylinder BC:
c=
J =
1
d = 1.0 in.
2
π
c4 =
π
L = 18 in.
(1.0) 4 = 1.5708 in 4
2
2
TBC L
TBC (18)
ϕB =
=
= 2.0463 × 10−6TBC
GJ
(5.6 × 106 )(1.5708)
Matching expressions for ϕΒ
6.5255 × 10−6TAB = 2.0463 × 10−6TBC
TBC = 3.1889 TAB
Equilibrium of connection at B :
TAB + TBC − T = 0
(1)
T = 12.5 × 103 lb ⋅ in
TAB + TBC = 12.5 × 103
(2)
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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
τ AB = 4.50 ksi 
Maximum stress in cylinder BC.
τ BC =
TBC c (9.5159 × 103 ) (1.0)
=
= 6.06 × 103 psi
J
1.5708
τ BC = 6.06 ksi 
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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.
1
π
d = 0.75 in.
L = 12 in. J = c 4 = 0.49701 in 4
2
2
TAB L
TAB (12)
=
= 2.1557 × 10−6 TAB
ϕB =
GJ
(11.2 × 106 )(0.49701)
c=
1
π
π
d = 1.0 in. L = 18 in. J = c 4 = (1.0)4 = 1.5708 in 4
2
2
2
T L
TBC (18)
= 2.0463 × 10−6TBC
ϕ B = BC =
6
GJ
(5.6 × 10 )(1.5708)
c=
Matching expressions for ϕB
2.1557 × 10−6TAB = 2.0463 × 10−6TBC
Equilibrium of connection at B : TAB + TBC − T = 0
TAB = 6.0872 × 103 lb ⋅ in
(2)
TBC = 6.4128 × 103 lb ⋅ in
Maximum stress in cylinder AB.
τ AB =
(b)
TAB + TBC = 12.5 × 103
(1)
2.0535 TAB = 12.5 × 103
Substituting (1) into (2),
(a)
TBC = 1.0535 TAB
TABc (6.0872 × 103 ) (0.75)
=
= 9.19 × 103 psi
J
0.49701
τ AB = 9.19 ksi 
Maximum stress in cylinder BC.
τ BC =
TBC c (6.4128 × 103 )(1.0)
=
= 4.08 × 103 psi
J
1.5708
τ BC = 4.08 ksi 
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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 =
π
1
d = 0.625 in.
2
π
c 4 = (0.625) 4 = 0.23968 in 4
2
2
T L
ϕ B = AB AB
GJ AB
J AB =
TAB =
GJ ABϕ B (11.2 × 106 ) (0.23968)
ϕB
−
LAB
24
= 111.853 × 103ϕ B
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 in 4
(11.2 × 106 )(0.49701)
GJ CD ϕC
ϕC = 154.625 × 103ϕC
=
36
LCD
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PROBLEM 3.55 (Continued)
Clearance rotation for flange B:
ϕ ′B = 1.5° = 26.18 × 10−3 rad
Torque to remove clearance:
′ = (111.853 × 103 )(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 × 103 )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 
TCDc
(1225.09)(0.75)
=
= 1849 psi
J CD
0.49701
τ CD = 1.849 ksi 
Maximum shearing stress in CD.
τ CD =
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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 = TAB , LAB = 2ft = 24 in., c =
π
1
d = 0.625 in.
2
π
c 4 = (0.625)4 = 0.23968 in 4
2
2
T L
ϕ B = AB AB
GJ AB
J AB =
TAB =
GJ ABϕ B (11.2 × 106 ) (0.23968)
ϕB
−
LAB
24
= 111.853 × 103ϕ B
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD ,
J CD =
π
c4 =
LCD = 3 ft = 36 in., c =
π
2
2
T L
ϕC = CD CD
GJ CD
TCD =
Clearance rotation for flange C:
1
d = 0.75 in.
2
(0.75) 4 = 0.49701 in 4
GJ CD ϕC
(11.2 × 106 )(0.49701)
ϕC = 154.625 × 103 ϕC
=
LCD
36
ϕC′ = 1.5° = 26.18 × 10−3 rad
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PROBLEM 3.56 (Continued)
Torque to remove clearance:
′ = (154.625 × 103 )(26.18 × 10−3 ) = 4048.1 lb ⋅ in
TCD
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 
Maximum shearing stress in CD.
τ CD =
TCDc
(4048.1 + 575.56)(0.75)
=
= 6980 psi
J CD
0.49701
τ CD = 6.98 ksi 
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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:
φB =
rC
40
φC =
φC
rB
100
φB = 0.4 φC
(1)
 M C = 0 : TCD = rC F
(2)
 M B = 0 : T − TAB = rB F
(3)
Solve (2) for F and substitute into (3):
T − TAB =
rB
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
=
= 235.79 × 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
π
2
TCD (0.5)
(0.0225) 4 G
= 1242 × 103
TCD
G
(6)
Substitute from (5) and (6) into (1):
φB = 0.4φC :
235.79 × 103
TCD = 0.47462 = TAB
TAB
T
= 0.4 × 1242 × 103 CD
G
G
(7)
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PROBLEM 3.57 (Continued)
Substitute for TCD from (7) into (4):
T = TAB + 2.5 (0.47462 TAB )
T = 2.1865 TAB (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)
TAB c 2 TAB
2 1829.4
=
=
= 43.1 × 106
3
J
π c
π (0.030)3
τ AB = 43.1 MPa 
Stress in CD:
τ CD =
TCDc
2 TCD
2 868.3
=
=
= 48.5 × 106
3
3
J
π c
π (0.0225)
τ CD = 48.5 MPa 
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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 =
rC
40
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):
T − TAB =
rB
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
(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)
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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),
 TCD 
T = 2.1865 

 0.47462 
Stress criterion for shaft AB:
τ AB = τ all = 50 MPa:
τ AB =
=
TABc
J
π
TAB = τ AB = c3τ AB
2
J
c
π
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:
τ CD =
From (7):
TCD c
π
π
TCD = c3τ CD = (0.0225 m)3 (50 × 106 Pa)
2
2
J
= 894.62 N ⋅ m
T = 4.6068(894.62 N ⋅ m) = 4.12 kN ⋅ m
The smaller value for T governs.
T = 4.12 kN ⋅ m 
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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
(c
2
4
2
)
− c14 =
π
2
(0.0404 − 0.0364 )
−6
= 1.3829 × 10 m 4
Torque carried by shaft.
TS = GJ S ϕ /L
Torque carried by jacket.
TJ = GJ J ϕ /L
Total torque.
T = TS + TJ = ( J S + J J ) G ϕ /L
TJ =
∴
Gϕ
T
=
L
JS + JJ
(1.3829 × 10−6 ) (500)
JJ
= 423.1 N ⋅ m
T =
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 
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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 c1 / c2 ,
show that the ratio Ts /Th of the torque Ts in the solid shaft to the
torque
Th
in
the
hollow
shaft
is
(a)
(1 − n 2 ) /(1 + n 2 ) if the maximum shearing stress is the same in
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 =
2
c24 (1 − n 2 ) 2
Jh =
π
(c
2
4
2
)
− c14 =
π
2
c24 (1 − n 4 )
For equal stresses.
τ =
Ts c0
Tc
= h 2
Js
Jh
Ts
Jc
= s 2 =
Th
J hc0
(b)
π
c0 = c2 (1 − n 2 )1/2
∴
c 4 (1 − n 2 ) 2 c2
2 2
π c 4 (1 − n 4 )c (1 − n 2 )1/2
2
2 2
π
=
1 − n2
(1 − n 2 )1/2
=
(1 + n 2 ) (1 − n 2 )1/2
1 + n2

For equal angles of twist.
ϕ =
Ts L
TL
= h
GJ s
GJ h
Ts
J
= s =
Th
Jh
c 4 (1 − n 2 )2
2 2
π c 4 (1 − n 4 )
2 2
π
=
(1 − n 2 ) 2 1 − n2
=
1 − n4
1 + n2

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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
2L
ϕ = L
2TL4 dy
2TL 2L dy
=

4 4
π Gc y
π Gc 4 L y 4
2L
2TL4  1 
2TL4
=
−
=


π Gc 4  3 y 3 L
π Gc 4
=
1 
 1
+ 3
−
3
3L 
 24L
2TL4  7 
7TL
=
4 
3
π Gc  24L  12π Gc 4

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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 Gc 4
T
GJ
=
=
2L
TL /GJ
L
2 LK
(2) (72) (51.24)
=
= 209.7 × 10−6 in 4
6
πG
π (11.2 × 10 )
c = 0.1203 in.
d = 2c = 0.241 in. 
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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 (2πρ ) ρ − T = 0
T
2π t ρ 2
τ =
(a)
Maximum shearing stress occurs at ρ = r1
γ =
Shearing strain:
τ
G
τ max =
=
T
2π tr12

T
2π GT ρ 2
The relative circumferential displacement in radial length d ρ is
d δ = γ d ρ = ρ dϕ
dϕ = γ
dϕ =
(b)
ϕ B/C =
=

r2
r1
T dρ
T
=
3
2π Gt
2π Gt ρ

r2
r1
dρ
ρ3
dρ
ρ
T
dρ
T dρ
=
2
2π Gt ρ ρ
2π GT ρ 3
=
T  1  r2
−

2π Gt  2 ρ 2  r1
1 
1 
T  1
T  1
− 2 + 2  =
 2 − 2
2π Gt  2r2
2r1  4π Gt  r1
r2 

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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
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)
= 23.5 × 106 Pa
π (0.006)3
τ = 46.9 MPa 
τ = 23.5 MPa 
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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
2π (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 
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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 × 106 Pa P = 0.375 × 103 W
T=
τ =
P
2π f
=
f = 29 Hz
0.375 × 103
= 2.0580 N ⋅ m
2π (29)
Tc
2T
=
J
π c3
∴
c3 =
2T
πτ
=
(2) (2.0580)
π (35 × 106 )
= 37.43 × 10−9 m3
c = 3.345 × 10−3 m = 3.345 mm
d = 2c
d = 6.69 mm 
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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 =
τ =
1200
= 20 Hz
60
Tc
2T
=
J
π c3
c = 0.7639in.
T =
∴ c3 =
2T
πτ
d = 2c
P
2π f
=
=
660 × 103
= 5.2521 × 103 lb ⋅ in
2π (20)
(2) (5.2521 × 103 )
= 0.4458 in 3
π (7500)
d = 1.528 in. 
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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 × 103 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 =
P
= 530.52 N ⋅ m
2π f
π
(c
2
− c14
)
c14 = c24 −
2Tc2
= 0.0254 −
J =
4
2
πτ
τ =
Tc2
2Tc2
=
J
π c24 − c14
(
)
(2)(530.52) (0.025)
= 249.90 × 10−9 m 4
π (60 × 106 )
c1 = 22.358 × 10−3 m = 22.358 mm
t = c2 − c1 = 25 mm − 22.358 mm = 2642 mm
t = 2.64 mm 
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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 × 106 psi, determine the power
being transmitted.
SOLUTION
ϕ = 2° = 34.907 × 10−3 rad
c2 =
J =
1
d o = 1.5in.
2
π
(c
2
4
2
)
− c14 =
c1 =
π
2
L = 12ft =144 in.
1
di = 0.6in.
2
(1.54 − 0.64 ) = 7.7486 in 4
120
f =
= 2 Hz
60
T =
GJ ϕ
(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
Since 1 hp = 6600 lb ⋅ in/s,
P = 40.1 hp 
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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
d 2 = 30 mm
2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
(c
2
4
2
)
− c14 =
π
2
[(30) 4 − (12.5) 4 ]
= 1.234 × 106 mm 4 = 1.234 × 10−6 m 4
τ m = 50 × 106 Pa
τm =
(a)
(50 × 106 )(1.234 × 10−6 )
τ J
Tc
T = m =
= 2056.7 N ⋅ m
J
c
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 × 103 W
P = 51.7 kW 
(b)
Angle of twist.
ϕ =
TL
(2056.7)(5)
=
= 0.1078 rad
GJ
(77.2 × 109 )(1.234 × 10−6 )
ϕ = 6.17° 
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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 2 = 30 mm
2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
(c
2
4
2
)
− c14 =
π
2
[(30)4 − (12.5)4 )]
= 1.234 × 10 mm 4 = 1.234 × 10−6 m 4
6
ϕ = 3° = 0.05236 rad
(a)
ϕ =
TL
GJ
T =
GJ ϕ
(77.2 × 109 )(1.234 × 10−6 )(0.0536)
=
= 997.61 N ⋅ m
L
5
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 =
Tc2
(997.61)(30 × 10−3 )
=
J
1.234 × 10−6
= 24.3 × 106 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,
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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 × 106 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 c24 − c14
c14 = c24 −
(
τ max = 47.5 MPa 
)
2Tc2
πτ
c14 = 0.0204 −
(2)(397.89)(0.020)
= 53.333 × 10−9
π (47.494 × 106 )
c1 = 15.20 × 10−3 m = 15.20 mm
d 2 = 2c1 = 30.4 mm 
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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 × 103 lb ⋅ in
1
do = 1.75 in.
2
Tc
2Tc2
τ= 2=
J
π c24 − c14
c2 =
(
c14 = c24 −
)
2Tc2
(2)(36 × 103 )(1.75)
= 1.754 −
= 4.3655 in 4
πτ
π(8 × 103 )
c1 = 1.4455 in.
Required minimum thickness: t = c2 − c1
t = 1.75 − 1.4455 = 0.3045 in.
Available thicknesses: 0.25 in., 0.3125 in., 0.375 in., etc.
Use
t = 0.3125 in. 
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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 × 103 lb ⋅ in/sec
f =
1260
= 21 Hz
60
τ = 8 ksi = 8 × 103 psi
TAB =
P
2π f
=
105.6 × 103
= 800.32 lb ⋅ in
2π (21)
τ =
2T
Tc
=
J
π c3
c=
3
c=
3
2T
πτ
(2)(800.32)
= 0.399 in.
π (8 × 103 )
d AB = 2c = 0.799 in.
(b)
Shaft CD:
TCD =
c=
d AB = 0.799 in. 
rC
5
TAB = (800.32) = 1.33387 × 103 lb ⋅ in
rB
3
3
2T
πτ
=
3
(2)(1.33387 × 103 )
= 0.473 in.
π (8 × 103 )
dCD = 2c = 0.947 in.
dCD = 0.947 in. 
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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 × 103 lb ⋅ in/sec
f =
TAB =
1260
= 21 Hz
60
P
2π f
=
105.6 × 103
= 800.32 lb ⋅ in
2π (21)
1
d = 0.5 in.
2
2T
Tc
τ =
=
J
π c3
(2)(800.32)
=
= 4.08 × 103 psi
3
π (0.5)
c=
(b)
Shaft CD:
TCD =
τ =
τ AB = 4.08 ksi 
rC
5
TAB = (800.32) = 1.33387 × 103 lb ⋅ in
rB
3
2T
(2)(1.33387 × 103 )
=
= 6.79 × 103 psi
π c3
π (0.5)3
τ CD = 6.79 ksi 
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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 =
TAB =
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
πc AB
c3AB =
2T
πτ
(2)(39.789)
= 422.17 × 10−9 m3
π(60 × 106 )
c AB = 7.50 × 10−3 m = 7.50 mm
Shaft CD:
d AB = 2c AB = 15.00 mm 
P
7.5 × 103
=
= 99.472 N ⋅ m
2πfCD
2π(12)
fCD =
rB
60
f AB =
(30) = 12 Hz
rC
150
τ CD =
TcCD
2T
2T
2(99.472)
3
= 3
∴ cCD
= CD =
= 1.05543 × 10−6 m3
πτ
J CD
πcCD
π(60 × 106 )
TCD =
cCD = 10.18 × 10−3 m = 10.18 mm
Shaft EF:
f EF =
τ=
dCD = 2cCD = 20.4 mm 
rD
60
fCD =
(12) = 4.8 Hz
rE
150
TcEF
2T
= 3
J EF
πcEF
3
∴ cEF
=
TEF =
P
7.5 × 103
=
= 248.68 N ⋅ m
2πf EF 2 π (4.8)
(2)(248.68)
= 2.6886 × 10−6 m3
6
π(60 × 10 )
cEF = 13.82 × 10−3 m = 13.82 mm
d EF = 2cEF = 27.6 mm 
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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, dCD = 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 =
1
d AB = 0.008 m
2
π
2
c3AB τ all =
π
2
τ =
Tc AB
2T
=
J AB
π c3AB
(0.008)3 (75 × 106 ) = 60.319 N ⋅ m
f AB = 24 Hz Pall = 2π f ABTall = 2π (24) (60.319) = 9.10 × 103 W
Shaft CD:
1
dCD = 0.010 m
2
π 3
π
2T
Tc
τ = CD = 3 ∴ Tall = cCD
τ all = (0.010)3 (75 × 106 ) = 117.81 N ⋅ m
2
2
J CD
π cCD
cCD =
fCD =
Shaft EF:
60
rB
(24) = 9.6 Hz
f AB =
150
rC
cEF =
Pall = 2π fCDTall = 2π (9.6) (117.81) = 7.11 × 103 W
1
d EF = 0.014 m
2
π
3
cEF
τ all =
π
(0.014)3 (75 × 106 ) = 323.27 N ⋅ m
2
2
r
60
f EF = D fCD =
(9.6) = 3.84 Hz
rE
150
Tall =
Pall = 2π f EFTall = 2π (3.84) (323.27) = 7.80 ×103 W
Maximum allowable power is the smallest value.
Pall = 7.11× 103 W = 7.11 kW 
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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
π
= c3 τ = (0.024)3 (60 × 106 ) = 1.30288 × 103 N ⋅ m
c
2
2
Torque based on twist angle:
ϕ=
TL
GJ
T =
τ = 60 MPa = 60 × 106 Pa
ϕ = 2.5° = 43.633 × 10−3 rad
GJ ϕ π c 4Gϕ
π (0.024)4 (77 × 109 ) (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 
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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 × 106 Pa τ =
Stress requirement.
T =
τJ
c
=
π
2
τ c3 =
π
2
Tc
J
(50 × 106 )(0.015)3 = 265.07 N ⋅ m
ϕ = 7.5° = 130.90 × 10−3 rad G = 77.2 × 109 Pa
Twist angle requirement.
ϕ =
T =
TL
2TL
=
GJ
π Gc 4
π
2
Gc 4ϕ =
π
2
(77.2 × 109 )(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 × 103 W
P = 50.0 kW 
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PROBLEM 3.80
A steel shaft must transmit 210 hp at a speed of 360 rpm. Knowing that G = 11.2 × 106 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 =
c=
Angle of twist requirement:
P
2π f
3
2T
πτ
=
=
1 min.
= 6 Hz
60 sec
1.386 × 106
= 36.765 × 103 lb ⋅ in
(2π )(6)
3
Tc
2T
= 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
=
π Gϕ
4
(2)(36.765 × 103 )(98.4)
= 1.4077 in.
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. 
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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
=
J
π c3
P = 3 hp = (3)(6600) = 19800 lb ⋅ in/s
T =
π
2
c3τ
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)
5
π5
in., Tall =   (9500) = 455.4 lb ⋅ in
16
2  16 
5
−
8
Allowable torques.
P
19800
=
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 =
P
19800
=
2π TB
2π (113.85)
f AB = 27.2 Hz 
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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
d1 = 0.021 m L = 1.6 m
2
Based on stress limit: τ = 75 MPa = 75 × 106 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
(c
4
1
c24 = c14 −
− c24
2J
π
J = 267.88 × 10−9 m 4
)
= 0.0214 −
(2)(267.88 × 10−9 )
c2 = 12.44 × 10−3 m = 12.44 mm
π
= 23.943 × 10−9 m 4
d 2 = 2c2 = 24.9 mm 
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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
2
(c
4
1
)
− c24 =
π
2
(0.0214 − 0.0154 ) = 225.97 × 10−9 m 4
Based on stress limit: τ = 65 MPa = 65 × 106 Pa
τ =
Tc1
Jτ
(225.97 × 10−9 )(65 × 106 )
or T =
=
= 699.43 N ⋅ m
J
G
0.021
Based on angle of twist limit: ϕ = 3° = 52.36×10−3 rad
ϕ =
TL
GJ ϕ
(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 
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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 = 18 in., (b) r = 16
in.
SOLUTION
D
2
=
= 1.33
d
1.5
D = 2 in. d = 1.5 in.
1
d = 0.75 in.
T = 2.5 kip ⋅ in
2
2T
(2)(2.5)
Tc
=
=
= 3.773 ksi
3
J
πc
π (0.75)3
c=
(a)
r =
1
in. r = 0.125 in.
8
r
0.125
=
= 0.0833
d
1.5
K = 1.42
From Fig. 3.32,
τ max = K
(b)
r =
Tc
= (1.42)(3.773)
J
3
in.
16
τ max = 5.36 ksi 
r = 0.1875 in.
r
0.1875
=
= 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 
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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
3
transmitted by the shaft when the radius of the fillet is (a) r = 16
in.,
(b) r = 14 in.
SOLUTION
D = 2 in.
c=
1
d = 0.75 in.
2
τ max = K
(a)
r =
D
= 1.33
d
d = 1.5 in.
Tc
J
3
in.
16
τ max = 8 ksi
or
T =
Jτ max
πτ c3
= 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.
r
0.25
=
= 0.1667
d
1.5
From Fig. 3.32,
T =
π (8)(0.75)3
(2)(1.27)
K = 1.27
T = 4.17 kip ⋅ in 
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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 × 106 Pa
3
3
J
πc
π (0.0225)
Using τ max = 36 MPa = 36 × 106 Pa results in
K =
τ max
36 × 106
=
= 1.214
τ
29.65 × 106
From Fig 3.32 with
D 90 mm
r
=
= 2,
= 0.24
d
45 mm
d
r = 0.24 d = (0.24)(45 mm)
r = 10.8 mm 
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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
d
30
D
60
=
=2
d
30
From Fig. 3.32,
For smaller side,
K = 1.26
c=
1
d = 15 mm = 0.015 m
2
τ =
KTc 2 KT
=
J
π c3
T =
π c3τ
2K
=
π (0.015)3 (40 × 106 )
(2)(1.26)
P = 45 kW = 45 × 103
f =
= 168.30 N ⋅ m
P = 2π fT
P
45 × 103
=
= 42.6 Hz
2π T
2π (168.30 × 103 )
f = 42.6 Hz 
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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
π c3
T =
π c3τ
2K
1
d = 15 mm = 15 × 10−3 m
2
D = 60 mm, r = 8 mm
d = 30 mm c =
60
D
=
= 2,
30
d
8
r
=
= 0.26667
30
d
From Fig. 3.32,
K = 1.18
Allowable torque.
T =
Maximum power.
P = 2π f T = (2π )(50)(202.17) = 63.5 × 103 W
π (15 × 10−3 )3 (45 × 106 )
(2)(1.18)
= 202.17 N ⋅ m
P = 63.5 kW 
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PROBLEM 3.89
In the stepped shaft shown, which has a full quarter-circular fillet, D = 1.25 in.
d = 1 in.
Knowing that the speed of the shaft is
and
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
1.0
d
From Fig. 3.32,
For smaller side,
K = 1.31
c=
1
d = 0.5 in.
2
τ =
π c3τ
KTc
Jτ
=
T =
2K
J
Kc
T =
π (0.5)3 (7500)
(2)(1.31)
= 1.1241 × 103 lb ⋅ in
f = 2400 rpm = 40 Hz
P = 2π f T = 2π (40)(1.1241 × 103 )
= 282.5 × 103 lb ⋅ in/s
P = 42.8 hp 
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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 the
maximum shearing stress in the shaft when (a) d = 0.8 in., (b) d = 0.9 in.
SOLUTION
(a)
D 1.0
=
= 1.25
0.8
d
1
r = ( D − d ) = 0.1 in.
2
0.1
r
=
= 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
π c3
(2)(1.31)(200)
=
= 2.61 × 103 psi
π (0.4)3
τ =
(b)
τ = 2.61 ksi 
D 1.0
=
= 1.111
0.9
d
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 =
2 KT
(2)(1.44)(200)
=
= 2.01 × 103 psi
π c3
π (0.45)3
τ = 2.01 ksi 
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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 =
1
( D − d ) = 2 mm
2
From Fig. 3.32,
K = 1.36
Smaller side,
c=
1
d = 13 mm = 0.013 m
2
τ =
2KT
KTc
=
J
π c3
T =
(b)
r
2
=
= 0.0768
d
26
D
30
=
= 1.25
d
24
From Fig. 3.32,
r =
π c3τ
2K
=
1
( D − d ) = 3 mm
2
K = 1.31
c=
T =
π (0.013)3 (80 × 106 )
(2)(1.36)
= 203 N ⋅ m
T = 203 N ⋅ m 
r
3
=
= 0.125
d
24
1
d = 12 mm = 0.012 m
2
π c3τ
2K
=
π (0.012)3 (80 × 106 )
(2)(1.31)
= 165.8 N ⋅ m
T = 165.8 N ⋅ m 
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PROBLEM 3.92
A 30-mm diameter solid rod is made of an elastoplastic material with τ Y = 3.5 MPa. Knowing that the elastic
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
ρY =
T =

ρ3  4
4 
(0.0125)3 
TY 1 − Y3  = (18.555) 1 −

3 
c  3
(0.015)3 

= 21.2 N ⋅ m
T = 21.2 N ⋅ m 
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PROBLEM 3.93
The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 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.
SOLUTION
τ Y = 21 ksi
TY =
τY J
c
= τY
π
2
c3
π
(1.5 in.)3
2
TY = 111.3 kip ⋅ in
= (21 ksi)
(a)
For t = 0.8 in. ρY = 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 
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PROBLEM 3.94
The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa. Determine the magnitude T of the
applied torques when the plastic zone is (a) 16 mm deep, (b) 24 mm
deep.
SOLUTION
c = 32 mm = 0.032 m
τ Y = 145 × 106 Pa
TY =
π
π
Jτ Y
= c3τ Y = (0.032)3 (145 × 106 )
c
2
2
= 7.4634 × 103 N ⋅ m
(a)
t P = 16 mm = 0.016 m
ρY = c − tP = 0.032 − 0.016 = 0.016 m
4  1 ρY3  4
1 0.0163 
3 
=
×
−
T = TY 1 −
(7.4634
10
)
1

 4 0.0323 
3  4 c3  3


= 9.6402 × 103 N ⋅ m
(b)
T = 9.64 kN ⋅ m 
t P = 24 mm = 0.024 m
ρY = c − tP = 0.032 − 0.024 = 0.008 m
T =
4 
1 ρY 3  4
1 0.0083 
3 
=
×
−
TY 1 −
(7.4634
10
)
1




3 
4 c3  3
4 0.0323 

= 9.9123 × 103 N ⋅ m
T = 9.91 kN ⋅ m 
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PROBLEM 3.95
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with G = 11.2 × 106 psi and τ Y = 21 ksi. Determine the
maximum shearing stress and the radius of the elastic core caused by the
T = 100 kip ⋅ in.,
application
of
torque
of
magnitude
(a)
(b) T = 140 kip ⋅ in.
SOLUTION
c = 1.5 in., J =
(a)
π
2
c 4 = 7.9522 in 4 , τ Y = 21 ksi
T = 100 kip ⋅ in
τm =
Tc (100 kip ⋅ in)(1.5 in.)
=
J
7.9522 in 4
τ m = 18.86 ksi 
Since τ m < τ Y , shaft remains elastic.
c = 1.500 in. 
Radius of elastic core:
(b)
T = 140 kip ⋅ in
τm =
(140)(1.5)
= 26.4 ksi.
7.9522
Impossible: τ m = τ Y = 21.0 ksi  
Plastic zone has developed. Torque at onset of yield is TY =
Eq. (3.32):
T =
J
7.9522
(21 ksi) = 111.33 kip ⋅ in
τY =
c
1.5
4 
1 ρY3 
TY 1 −

3 
4 c3 
3
T
140
 ρY 
 c  = 4 − 3 T = 4 − 3 111.33 = 0.22743


Y
ρY = 0.6104c = 0.6104(1.5 in.)
ρY
c
= 0.6104
ρY = 0.916 in. 
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PROBLEM 3.96
It is observed that a straightened paper clip can be twisted through several revolutions by the application of a
torque of approximately 60 mN ⋅ m. Knowing that the diameter of the wire in the paper clip is 0.9 mm,
determine the approximate value of the yield stress of the steel.
SOLUTION
c=
1
d = 0.45 mm = 0.45 × 10−3 m
2
TP = 60 mN ⋅ m = 60 × 10−3 N ⋅ m
TP =
4
4 Jτ Y
4 π
2π 3
= ⋅ c3 TY =
TY =
c τY
3
3 c
3 2
3
τY =
3TP
(3)(60 × 10−3 )
=
= 314 × 106 Pa
2π c3
2π (0.45 × 10−3 )3
τ Y = 314 MPa 
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PROBLEM 3.97
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with τ Y = 145MPa. Determine the radius of the elastic core
caused by the application of a torque equal to 1.1 TY, where TY is the
magnitude of the torque at the onset of yield.
SOLUTION
c=
ρY
c
=
1
d = 15 mm
2
3
4−3
T
=
TY
T =
3
4   ρY  
TY 1 − 
 
3   c  
4 − (3)(1.1) = 0.88790
ρY = 0.88790 c = (0.88790)(15 mm)
ρY = 13.32 mm 
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PROBLEM 3.98
For the solid circular shaft of Prob. 3.95, determine the angle of twist caused
by the application of a torque of magnitude (a) T = 80 kip ⋅ in.,
(b) T = 130 kip ⋅ in.
SOLUTION
1
1
d = (3) = 1.5 in. τ Y = 21 × 103 psi
2
2
c=
L = 4 ft = 48 in. J =
Torque at onset of yielding:
Tc
J
τ =
τY J
TY =
(a)
T =
c
=
2
c4 =
π
2
(1.5)4 = 7.9522 in 4
τJ
c
(21 × 103 )(7.9522)
= 111.330 × 103 lb ⋅ in
1.5
T = 80 kip ⋅ in = 80 × 103 lb ⋅ in
Since T < TY , the shaft is fully elastic. ϕ =
TL
GJ
(80 × 103 )(48)
= 43.115 × 10−3 rad
(11.2 × 106 )(7.9522)
ϕ =
(b)
π
3
T = 130 kip ⋅ in = 130 × 10 lb ⋅ in
ϕY =
T > TY
3
4   ϕY  
T = TY 1 − 
 
3  ϕ  


TY L
(111.330 × 103 )(48)
=
= 60.000 × 10−3 rad
6
GJ
(11.2 × 10 )(7.9522)
T
ϕY
= 34−3
=
TY
ϕ
ϕ =
ϕ = 2.47° 
ϕY
0.79205
=
3
4−
(3)(130 × 103 )
= 0.79205
111.330 × 103
60.000 × 10−3
= 75.75 × 10−3 rad
0.79205
ϕ = 4.34° 
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PROBLEM 3.99
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with G = 77.2 GPa and τ Y = 145 MPa. Determine the angle
of twist caused by the application of a torque of magnitude (a)
T = 600 N ⋅ m, (b) T = 1000 N ⋅ m.
SOLUTION
1
d = 15 mm = 15 × 10−3 m
2
c=
Torque at onset of yielding:
τ =
TY =
(a)
π c3τ Y
2
=
π (15 × 10−3 )3 (145 × 106 )
2
= 768.71 N ⋅ m
T = 600 N ⋅ m. Since T ≤ TY , the shaft is elastic.
ϕ=
(b)
Tc
2T
=
J
π c3
T = 1000 N ⋅ m.
TL
2TL
(2)(600)(1.2)
=
=
= 0.11728 rad
GJ
π c 4G π (15 × 10−3 ) 4 (77.2 × 109 )
ϕ = 6.72° 
T > TY A plastic zone has developed.
3
T 
4   ϕY  
ϕY
= 3 4 − 3 
TY 1 − 
 
3  ϕ  
ϕ
 TY 


T L
2T L
(2)(768.71)(2.1)
ϕY = Y = Y4 =
= 0.15026 rad
GJ
π c G π (15 × 10−3 ) 4 (77.2 × 109 )
T =
ϕY
(3)(1000)
= 34−
= 0.46003
ϕ
768.71
ϕY
0.15026
=
= 0.32663 rad
ϕ =
0.46003
0.46003
ϕ = 18.71° 
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PROBLEM 3.100
A 3-ft-long solid shaft has a diameter of 2.5 in. and is made of a mild steel that is assumed to be elastoplastic
with τ Y = 21 ksi and G = 11.2 × 106 psi. Determine the torque required to twist the shaft through an angle of
(a) 2.5°, (b) 5°.
SOLUTION
L = 3 ft = 36 in., c =
J =
(a)
π
2
π
c4 =
2
1
d = 1.25 in., τ Y = 21 × 103 psi
2
(1.25) 4 = 3.835 in4
(3.835)(21 × 103 )
Jτ Y
=
= 64.427 × 103 lb ⋅ in
1.25
c
τY =
TY c
J
ϕY =
(64.427 × 103 )(36)
TY L
=
= 53.999 × 10−3 rad = 3.0939°
GJ
(11.2 × 106 )(3.835)
TY =
ϕ = 2.5° = 43.633 × 10−3 rad
ϕ < ϕY
ϕ =
T =
The shaft remains elastic.
TL
GJ
GJ ϕ
(11.2 × 106 )(3.835)(43.633 × 10−3 )
=
L
36
= 52.059 × 103 lb ⋅ in
(b)
ϕ = 5° = 87.266 × 10−3 rad
ϕ > ϕY
T =
T = 52.1 kip ⋅ in 
A plastic zone occurs.
3
4 
1ϕ  
TY 1 −  Y  
3 
4 ϕ  


3

4
1  53.999 × 10−3  
3 
= (64.427 × 10 ) 1 − 


3
4  87.266 × 10−3  


= 80.814 × 103 lb ⋅ in
T = 80.8 kip ⋅ in 
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PROBLEM 3.101
For the solid shaft of Prob. 3.99, determine (a) the magnitude of the torque T
required to twist the shaft through an angle of 15°, (b) the radius of the
corresponding elastic core.
PROBLEM 3.99 The solid shaft shown is made of a mild steel that is
assumed to be elastoplastic with G = 77.2 GPa and τ Y = 145 MPa.
Determine the angle of twist caused by the application of a torque of
magnitude (a) T = 600 N ⋅ m, (b) T = 1000 N ⋅ m.
SOLUTION
1
d = 15 mm = 15 × 10−3 m
2
ϕ = 15° = 0.2618 rad
c=
ϕY =
(a)
Lγ Y
Lτ
(1.2)(145 × 106 )
= Y =
= 0.15026 rad
c
cG
(15 × 10−3 )(77.2 × 109 )
Since ϕ > ϕY , there is a plastic zone.
τY =
TY =
TY c
2T
=
J
π c3
π c3τ Y
2
=
π (15 × 10−3 )3 (145 × 106 )
2
= 768.71 N ⋅ m
3
3

4 
1ϕ   4
1  0.15026  
T = TY 1 −  Y   = (768.71) 1 − 
 
3 
4 ϕ   3
4  0.2618  



= 976.5 N ⋅ m
(b)
T = 977 N ⋅ m 
Lγ Y = ρY ϕ = cϕY
ρY =
cϕY
ϕ
=
(15 × 10−3 )(0.15026)
= 8.61 × 10−3 m
0.2618
ρY = 8.61 mm 
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PROBLEM 3.102
The shaft AB is made of a material that is elastoplastic with
τ Y = 12 ksi and G = 4.5 × 106 psi. For the loading shown, determine
(a) the radius of the elastic core of the shaft, (b) the angle of twist at
end B.
SOLUTION
(a)
Radius of elastic core.
τ Y = 12 × 103 psi
c = 0.5in.
TY =
Jτ Y
π
π
= c3τ Y = (0.5)3 (12 × 103 )
2
2
c
= 2356.2 lb ⋅ in
T = 2560 lb ⋅ in > TY (plastic region with elastic core)
T =
ρY 3
c
c
(b)
=4−
3
ρY
3
4 
1 ρY 

TY 1 −
3 
4 c3 


3T
(3)(2560)
= 4−
= 0.74051
2356.2
TY
ρY = (0.9047)(0.5)
= 0.9047
ρY = 0.452 in. 
Angle of twist.
L = 6.4 ft = 76.8 in.
ϕY =
G = 4.5 × 106 psi
TY L
2T L
(2)(2356.2)(76.8)
= Y4 =
= 0.4096 radians
JG
π c G π (0.5) 4 (4.5 × 106 )
ϕY
ρ
= Y
ϕ
c
ϕ =
ϕY c 0.4096
=
= 0.4527 radians
ρY
0.9047
ϕ = 25.9° 
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PROBLEM 3.103
A 1.25-in.-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with
τ Y = 18 ksi and G = 11.2 × 106 psi. For an 8-ft length of the shaft, determine the maximum shearing stress
and the angle of twist caused by a 7.5 kip ⋅ in. torque.
SOLUTION
c=
1
d = 0.625 in., G = 11.2 × 106 psi, τ Y = 18 ksi = 18000 psi
2
L = 8 ft = 96 in. T = 7.5 kip ⋅ in = 7.5 × 103 lb ⋅ in
TY =
Jτ Y
π
π
= c3τ Y = (0.625)3 (18000) = 6.9029 × 103 lb ⋅ in
c
2
2
T > TY : plastic region with elastic core
γY =
cϕY
L
∴ τ max = τ Y = 18 ksi
∴ ϕY =
τ max = 18 ksi 
Lγ Y
Lτ
(96)(18000)
= Y =
= 246.86 × 10−3 rad
c
cG
(0.625)(11.2 × 106 )
4 
1 ϕ Y3 
TY 1 −

3 
4 ϕ 3 
1
1
ϕ
=
=
= 1.10533
3
ϕY
3T
(3)(7.5
10
)
×
34 −
34 −
TY
6.9029 × 103
T =
ϕ = 1.10533ϕY = (1.10533)(246.86 × 10−3 ) = 272.86 × 10−3 rad
ϕ = 15.63° 
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PROBLEM 3.104
A 18-mm-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with
τ Y = 145 MPa and G = 77 GPa. For a 1.2-m length of the shaft, determine the maximum shearing stress and
the angle of twist caused by a 200 N ⋅ m torque.
SOLUTION
τ Y = 145 × 106 Pa, c =
TY =
1
d = 0.009 m, L = 1.2 m, T = 200 N ⋅ m
2
Jτ Y
π
π
= c3τ Y = (0.009)3 (145 × 106 ) = 166.04 N ⋅ m
c
2
2
τ max = τ Y = 145MPa 
T > TY (plastic region with elastic core)
ϕY =
T =
TY L
2T L
(2)(166.04)(1.2)
= Y4 =
= 251.08 × 10−3 radians
GJ
π c G π (0.009)4 (77 × 109 )
4 
1 ϕ3 
TY 1 −

3 
4 ϕY3 
3
 ϕY 
3T
(3)(200)
=4−
= 0.38641
  = 4−
166.04
ϕ
T
 
Y
ϕ =
ϕ
0.72837
=
ϕY
= 0.72837
ϕ
251.08 × 10−3
= 344.7 × 10−3 radians
0.72837

ϕ = 19.75° 
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PROBLEM 3.105
A solid circular rod is made of a material that is assumed to be elastoplastic. Denoting by TY and φY,
respectively, the torque and the angle of twist at the onset of yield, determine the angle of twist if the torque is
increased to (a) T = 1.1 TY , (b) T = 1.25 TY , (c) T = 1.3 TY .
SOLUTION
T =

4
1 ϕY 3 
TY 1 −

3
4 ϕ 3 

ϕY
3T
= 34−
TY
ϕ
(a)
T
= 1.10
TY
ϕ
=
ϕY
(b)
T
= 1.25
TY
ϕ
=
ϕY
(c)
T
= 1.3
TY
ϕ
=
ϕY
3
3
3
or
ϕ
=
ϕY
1
3
4−
3T
TY
1
= 1.126
4 − (3)(1.10)
ϕ = 1.126 ϕY 
1
= 1.587
4 − (3)(1.25)
ϕ = 1.587 ϕY 
1
= 2.15
4 − (3)(1.3)
ϕ = 2.15 ϕY 
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PROBLEM 3.106
The hollow shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. Determine
the magnitude T of the torque and the corresponding angle of twist
(a) at the onset of yield, (b) when the plastic zone is 10 mm deep.
SOLUTION
(a)
At the onset of yield, the stress distribution is the elastic distribution with τ max = τ Y .
c2 =
J =
1
1
d 2 = 0.030 m, c1 = d1 = 0.0125 m
2
2
π
(c
2
4
2
τ max = τ Y =
)
− c14 =
π
2
(0.0304 − 0.01254 ) = 1.2340 × 10−6 m 4
TY c2
Jτ
(1.2340 × 10−6 )(145 × 106 )
∴ TY = Y =
= 5.9648 × 103 N ⋅ m
J
c2
0.030
TY = 5.96 kN ⋅ m 
TY L
(5.9643 × 103 )(5)
=
= 313.04 × 10−3 rad
GJ
(77.2 × 109 )(2.234010−6 )
ϕY =
(b)
ρϕ
L
ϕ =

ρY = c2 − t = 0.030 − 0.010 = 0.020 m
t = 0.010 m
γ =
=
ϕY = 17.94° 
ρY ϕ
L
= γY =
τY
G
τY L
(145 × 106 )(5)
=
= 469.56 × 10−3 rad
G ρY
(77.2 × 109 )(0.020)
Torque T1 carried by elastic portion:
τ = τ Y at ρ = ρY .
τY =
T1ρY
J1
ϕ = 26.9° 
c1 ≤ ρ ≤ ρY
where
J1 =
π
2
(ρ
4
Y
− c14
)
π
(0.0204 − 0.01254 ) = 212.978 × 10−9 m 4
2
Jτ
(212.978 × 10−9 )(145 × 106 )
T1 = 1 Y =
= 1.5441 × 103 N ⋅ m
ρY
0.020
J1 =
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PROBLEM 3.106 (Continued)
Torque T2 carried by plastic portion:
c2
2
T2 = 2π ρ τ Y ρ d ρ = 2π τ Y
Y
=
ρ3
3
c2
=
ρY
2π
τ Y c23 − ρY3
3
(
)
2π
(145 × 106 )(0.0303 − 0.0203 ) = 5.7701 × 103 N ⋅ m
3
Total torque:
T = T1 + T2 = 1.5541 × 103 + 5.7701 × 103 = 7.3142 × 103 N ⋅ m

T = 7.31 kN ⋅ m 
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PROBLEM 3.107
For the shaft of Prob. 3.106, determine (a) angle of twist at which
the section first becomes fully plastic, (b) the corresponding
magnitude T of the applied torque. Sketch the T − φ curve for the
shaft.
PROBLEM 3.106 The hollow shaft shown is made of a steel that
is assumed to be elastoplastic with τ Y = 145 MPa and
G = 77.2 GPa. Determine the magnitude T of the torque and the
corresponding angle of twist (a) at the onset of yield, (b) when the
plastic zone is 10 mm deep.
SOLUTION
1
d1 = 0.0125 m
2
c1 =
(a)
1
d 2 = 0.030 m
2
For onset of fully plastic yielding, ρY = c1
τ = τY ∴ γ =
TP = 2π
=
τY
G
=
ρY ϕ
L
=
c1ϕ
L
Lτ Y
(25)(145 × 106 )
=
= 751.295 × 10−3 rad
c1G
(0.0125)(77.2 × 109)
ϕf =
(b)
c2 =

c2
c1
2
τ Y ρ d ρ = 2π τ Y
ρ3
3
c2
=
c1
2π
τ Y c23 − c13
3
(
)
2π
(145 × 106 )(0.0303 − 0.01253 ) = 7.606 × 103 N ⋅ m
3
From Prob. 3.101,
ϕY = 17.94°
Also from Prob. 3.101,
ϕ f = 43.0° 
TP = 7.61 kN ⋅ m 
TY = 5.96 kN ⋅ m
ϕ = 26.9°
T = 7.31 kN ⋅ m
Plot T vs ϕ using the following data.
ϕ , deg
T kN ⋅ m
0
17.94
26.9
43.0
>43.0
0
5.96
7.31
7.61
7.61
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PROBLEM 3.108
A steel rod is machined to the shape shown to form a tapered solid shaft to which
torques of magnitude T = 75 kip ⋅ in. are applied. Assuming the steel to be
elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi, determine (a) the radius of the
elastic core in portion AB of the shaft, (b) the length of portion CD that remains fully
elastic.
SOLUTION
(a)
c=
In portion AB.
TY =
T =
ρY
c
=
1
d = 1.25 in.
2
JABτ Y
π
π
= c3τ Y = (1.25)3 (21 × 103 ) = 64.427 × 103 lb ⋅ in
2
2
c
ρ3 
4 
TY 1 − Y3 
3 
c 
3
4−
3T
=
TY
3
4−
(3)(75 × 103 )
= 0.79775
64.427 × 103
ρY = 0.79775c = (0.79775)(1.25) = 0.99718 in.
(b)
For yielding at point C.
ρY = 0.997 in. 
τ = τ Y , c = cx , T = 75 × 103 lb ⋅ in
T =
J Cτ Y
π
= cx3τ Y
cx
2
cx =
3
2T
πτ Y
=
3
(2)(75 × 103 )
= 1.31494 in.
π (21 × 103 )
Using proportions from the sketch,
1.50 − 1.31494
x
=
1.50 − 1.25
5
x = 3.70 in. 
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PROBLEM 3.109
If the torques applied to the tapered shaft of Prob. 3.108 are slowly increased,
determine (a) the magnitude T of the largest torques that can be applied to the shaft,
(b) the length of the portion CD that remains fully elastic.
PROBLEM 3.108 A steel rod is machined to the shape shown to form a tapered solid
shaft to which torques of magnitude T = 75 kip ⋅ in. are applied. Assuming the steel
to be elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi, determine
(a) the radius of the elastic core in portion AB of the shaft, (b) the length of portion
CD that remains fully elastic.
SOLUTION
(a)
The largest torque that may be applied is that which makes portion AB fully plastic.
c=
In portion AB,
TY =
1
d = 1.25 in.
2
Jτ Y
π
π
= c3τ Y = (1.25)3 (21 × 103 ) = 64.427 × 103 lb ⋅ in
2
2
c
For fully plastic shaft, ρY = 0
T =
T =
(b)
4 
1 ρY3  4
TY 1 −
= T
3 
4 c3  3
4
(64.427 × 103 ) = 85.903 × 103 lb ⋅ in
3
For yielding at point C, τ = τ Y ,
c = cx ,
τY =
Tcx
2T
=
Jx
π cx3
cx =
3
2T
πτ Y
=
3
T = 85.9 kip ⋅ in 
T = 85.903 × 103 lb ⋅ in
(2)(85.903 × 103 )
= 1.37580 in.
π (21 × 103 )
Using proportions from the sketch,
1.50 − 1.37580
x
=
1.50 − 1.25
5
x = 2.48 in. 
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PROBLEM 3.110
A hollow shaft of outer and inner diameters respectively equal to 0.6 in.
and 0.2 in. is fabricated from an aluminum alloy for which the stressstrain diagram is given in the diagram shown. Determine the torque
required to twist a 9-in. length of the shaft through 10°.
SOLUTION
ϕ = 10° = 174.53 × 10−3 rad
c1 =
1
1
d1 = 0.100 in, c2 = d 2 = 0.300 in.
2
2
γ max =
c2ϕ
(0.300)(174.53 × 10−3 )
=
= 0.00582
L
9
γ min =
c1ϕ
(0.100)(174.53 × 10−3 )
=
= 0.00194
L
9
z=
Let
γ
=
γ max
T = 2π

c2
c1
ρ
z1 =
c2
ρ 2τ d ρ = 2π c23
I=
where the integral I is given by
c1 1
=
c2 3

1
1/3

1
z1
z 2τ dz = 2π c23 I
z 2τ dz
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
I =
Δz
Σ w z2 τ
3
where w is a weighting factor. Using Δ z = 16 , we get the values given in the table below.
z
1/3
1/2
2/3
5/6
1
γ
0.00194
0.00291
0.00383
0.00485
0.00582
I =
τ, ksi
8.0
10.0
11.5
13.0
14.0
z 2τ , ksi
0.89
2.50
5.11
9.03
14.0
w
1
4
2
4
1
wz 2τ , ksi
0.89
10.00
10.22
36.11
14.00
71.22
←⎯
⎯ Σwz 2τ
(1/6)(71.22)
= 3.96 ksi
3
T = 2π c23I = 2π (0.300)3 (3.96) = 0.671 kip ⋅ in
T = 671 lb ⋅ in 
Note: Answer may differ slightly due to differences of opinion in reading the stress-strain curve.
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PROBLEM 3.111
Using the stress-strain diagram shown, determine (a) the torque that
causes a maximum shearing stress of 15 ksi in a 0.8-in.-diameter solid
rod, (b) the corresponding angle of twist in a 20-in. length of the rod.
SOLUTION
(a)
τ max = 15 ksi
c=
1
d = 0.400 in.
2
γ max = 0.008
From the stress-strain diagram,
z =
Let
γ
γ max
T = 2π
I=
where the integral I is given by
1

c
0
=
ρ
c
ρ 2τ d ρ = 2π c3
1
 z τ dz = 2π c I
2
3
0
 z τ dz
2
0
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
I =
Δz
Σ w z2 τ
3
where w is a weighting factor. Using Δ z = 0.25, we get the values given in the table below.
wz 2τ , ksi
γ
τ, ksi
z 2τ , ksi
w
0
0.000
0
0.000
1
0.00
0.25
0.002
8
0.500
4
2.00
0.5
0.004
12
3.000
2
6.00
0.75
0.006
14
7.875
4
31.50
1.0
0.008
15
15.000
1
15.00
z
54.50
I =
(0.25)(54.50)
= 4.54 ksi
3
T = 2π c3I = 2π (0.400)3 (4.54)
(b)
γ max =
ϕ =
←⎯
⎯ Σwz 2τ
T = 1.826 kip ⋅ in 
cϕ
L
Lγ m
(20)(0.008)
=
= 400 × 10−3 rad
c
0.400
ϕ = 22.9° 
Note: Answers may differ slightly due to differences of opinion in reading the stress-strain curve.
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PROBLEM 3.112
A 50-mm-diameter cylinder is made of a brass for which the stressstrain diagram is as shown. Knowing that the angle of twist is 5° in a
725-mm length, determine by approximate means the magnitude T of
torque applied to the shaft
SOLUTION
ϕ = 5° = 87.266 × 10−3 rad
γ max
1
d = 0.025m, L = 0.725 m
2
cϕ
(0.025)(87.266 × 10−3 )
=
=
= 0.00301
L
0.725
z =
Let
c=
γ
γ max
T = 2π

c
0
=

ρ
c
ρ 2τ d ρ = 2π c3
1
 z τ dz = 2π c I
2
3
0
where the integral I is given by I =
1
 z τ dz
2
0
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
I =
Δz
 wz 2τ
3
where w is a weighting factor. Using Δz = 0.25, we get the values given in the table below.
γ
z
τ , MPa
z 2τ , MPa
w
wz 2τ , MPa
0
0
0
0
1
0
0.25
0.00075
30
1.875
4
7.5
0.5
0.0015
55
13.75
2
27.5
0.75
0.00226
75
42.19
4
168.75
1.0
0.00301
80
80
1
80
←  wz 2τ
= 283.75 × 106 Pa
283.75
I =
(0.25)(283.75 × 106 )
= 23.65 × 106 Pa
3
T = 2π c3I = 2π (0.025)3 (23.65 × 106 ) = 2.32 × 103 N ⋅ m
T = 2.32 kN ⋅ m 
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PROBLEM 3.113
Three points on the nonlinear stress-strain diagram used in Prob. 3.112 are (0,0), (0.0015,55 MPa), and
(0.003,80 MPa). By fitting the polynomial τ = A + Bγ + Cγ 2 through these points, the following
approximate relation has been obtained.
T = 46.7 × 109 γ − 6.67 × 1012 γ 2
Solve Prob. 3.113 using this relation, Eq. (3.2), and Eq. (3.26).
PROBLEM 3.112 A 50-mm diameter cylinder is made of a brass for which the stress-strain diagram is as
shown. Knowing that the angle of twist is 5° in a 725-mm length, determine by approximate means the
magnitude T of torque applied to the shaft.
SOLUTION
ϕ = 5° = 87.266 × 10−3 rad, c =
γ max =
Let
z=
1
d = 0.025 m, L = 0.725 m
2
cϕ
(0.025)(87.266 × 10−3 )
=
= 3.009 × 10−3
0.725
L
γ
ρ
=
γ max c
T = 2π

c
0
ρ 2τ d ρ = 2π c3
1
 z τ dz
2
0
The given stress-strain curve is
2
τ = A + Bγ + Cγ 2 = A + Bγ max z + Cγ max
z2
T = 2π c3
 z ( A + Bγ
1
2
0
max z
)
2
+ Cγ max
z 2 dz
1
 1
2
= 2π c3 A z 2 dz + Bγ max z 3 dz + Cγ max
0
 0
1
1 2 
1
= 2π c 2  A + Bγ max + Cγ max

4
5
3


Data:
A = 0,
1
A = 0,
3

1

 z dz
4
0
B = 46.7 × 109 , C = −6.67 × 1012
1
1
Bγ max = (46.7 × 109 )(3.009 × 10−3 ) = 35.13 × 103
4
4
1 2
1
Cγ max = − (6.67 × 1012 )(3.009 × 10−3 )2 = −12.08 × 103
5
5
T = 2π (0.025)3(0 + (35.13 × 103 − 12.08 × 103 ) = 2.26 × 103 N ⋅ m
T = 2.26 kN ⋅ m 
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PROBLEM 3.114
The solid circular drill rod AB is made of a steel that is assumed to be elastoplastic
with τ Y = 22 ksi and G = 11.2 × 106 psi. Knowing that a torque T = 75 kip ⋅ in. is
applied to the rod and then removed, determine the maximum residual shearing stress
in the rod.
SOLUTION
c = 1.2 in.
π
L = 35 ft = 420 in.
π
c 4 = (1.2) 4 = 3.2572 in 4
2
2
(3.2572)(22)
Jτ
= 59.715 kip ⋅ in
TY = Y =
1.2
c
J =
Loading:
T = 75 kip ⋅ in
T =
ρY3
c
3
ρY
c
4 
1 ρY3 
TY 1 −

3 
4 c3 
=4−
3T
(3)(75)
= 4−
= 0.23213
59.715
TY
= 0.61458,
ρY = 0.61458c = 0.73749 in.
Unloading:
τ′ =
Tρ
J
At ρ = c
τ′ =
(75)(1.2)
= 27.63 ksi
3.2572
At ρ = ρY
τ′ =
(75)(0.73749)
= 16.98 ksi
3.2572
where
T = 75 kip ⋅ in
Residual:
τ res = τ load − τ ′
At ρ = c
τ res = 22 − 27.63 = −5.63 ksi
At ρ = ρY
τ res = 22 − 16.98 = 5.02 ksi
maximum τ res = 5.63 ksi 
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PROBLEM 3.115
In Prob. 3.114, determine the permanent angle of twist of the rod.
PROBLEM 3.114 The solid circular drill rod AB is made of steel that is assumed to
be elastoplastic with τ Y = 22 ksi and G = 11.2 × 106 psi. Knowing that a torque
T = 75kip ⋅ in. is applied to the rod and then removed, determine the maximum
residual shearing stress in the rod.
SOLUTION
From the solution to Prob. 3.114,
c = 1.2 in.
J = 3.2572 in 4
ρY
c
= 0.61458
ρY = 0.73749 in.
After loading,
γ =
ρϕ
L
∴ ϕ =
ϕload =
During unloading,
Lγ
ρ
=
Lγ Y
ρY
=
Lτ Y
ρY G
L = 35 ft = 420 in.
(420)(22 × 103 )
= 1.11865 rad = 64.09°
(0.73749)(11.2 × 106 )
ϕ′ =
TL
GJ
ϕ′ =
(75 × 103 )(420)
= 0.86347 rad = 49.47°
(11.2 × 106 )(3.2572)
(elastic)
T = 5 × 103 N ⋅ m
Permanent angle of twist.
ϕperm = ϕload − ϕ ′ = 1.11865 − 0.86347 = 0.25518
ϕ = 14.62° 
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PROBLEM 3.116
The solid shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The torque is
increased in magnitude until the shaft has been twisted through 6°; the
torque is then removed. Determine (a) the magnitude and location of the
maximum residual shearing stress, (b) the permanent angle of twist.
SOLUTION
ϕ = 6° = 104.72 × 10−3 rad
c = 0.016 m
γ max =
γY =
ρY
c
=
(0.016)(104.72 × 10−3 )
cϕ
=
= 0.0027925
0.6
L
τY
G
=
γY
γ max
145 × 106
= 0.0018782
77.2 × 109
0.0018
=
= 0.67260
0.0027925
π
π
c 4 = (0.016)4 = 102.944 × 10−9 m 4
2
2
Jτ
π
π
TY = Y = c3τ Y = (0.016)3 (145 × 106 ) = 932.93 N ⋅ m
2
2
c
J =
At end of loading.
Unloading:
At ρ = c
At ρ = ρY
Residual:
(a)
Tload =
4 
1 ρ Y3  4
1


TY 1 −
= (932.93) 1 − (0.67433)3  = 1.14855 × 103 N ⋅ m
3 

3 
4 c  3
4


T ′ = 1.14855 × 103 N ⋅ m
elastic
τ′ =
T ′c
(1.14855 × 103 )(0.016)
=
= 178.52 × 106 Pa
J
102.944 × 10−9
τ′ =
T ′c ρY
= (178.52 × 106 )(0.67433) = 120.38 × 106 Pa
J c
ϕ′ =
(1.14855 × 103 )(0.6)
T ′L
=
= 86.71 × 10−3 rad = 4.97°
GJ
(77.2 × 109 )(102.944 × 10−9 )
τ res = τ load − τ ′
ϕ perm = ϕload − ϕ ′
At ρ = c
τ res = 145 × 106 − 178.52 × 106 = −33.52 × 106 Pa
τ res = −33.5 MPa
At ρ = ρY
τ res = 145 × 106 − 120.38 × 106 = 24.62 × 106 Pa
τ res = 24.6 MPa
Maximum residual stress:
(b)
ϕperm = 104.72 × 10−3 − 86.71 × 10−3 = 17.78 × 10−3 rad
33.5 MPa at ρ = 16 mm 
ϕperm = 1.032° 
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PROBLEM 3.117
After the solid shaft of Prob. 3.116 has been loaded and unloaded as
described in that problem, a torque T1 of sense opposite to the original
torque T is applied to the shaft. Assuming no change in the value of ϕY ,
determine the angle of twist ϕ1 for which yield is initiated in this second
loading and compare it with the angle ϕY for which the shaft started to
yield in the original loading.
PROBLEM 3.116 The solid shaft shown is made of a steel that is
assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa.
The torque is increased in magnitude until the shaft has been twisted
through 6°; the torque is then removed. Determine (a) the magnitude
and location of the maximum residual shearing stress, (b) the permanent
angle of twist.
SOLUTION
From the solution to Prob. 3.116,
c = 0.016 m, L = 0.6 m
τ Y = 145 × 106 Pa,
J = 102.944 × 10−9 m 4
The residual stress at ρ = c is
τ res = 33.5 MPa
For loading in the opposite sense, the change in stress to produce reversed yielding is
τ1 = τ Y − τ res = 145 × 106 − 33.5 × 106 = 111.5 × 106 Pa
(102.944 × 10−9 )(111.5 × 106 )
T1c
Jτ
∴ T1 = 1 =
0.016
J
c
= 717 N ⋅ m
τ1 =
Angle of twist at yielding under reversed torque.
ϕ1 =
T1L
(717 × 103 )(0.6)
=
= 54.16 × 10−3 rad
GJ
(77.2 × 109 )(102.944 × 10−9 )
ϕ1 = 3.10° 
Angle of twist for yielding in original loading.
γ =
ϕY =
τY
G
=
cϕY
L
Lτ Y
(0.6)(145 × 106 )
=
= 70.434 × 10−3 rad
cG
(0.016)(77.2 × 109 )
ϕY = 4.04° 
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PROBLEM 3.118
The hollow shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The
magnitude T of the torques is slowly increased until the plastic
zone first reaches the inner surface of the shaft; the torques are
then removed. Determine the magnitude and location of the
maximum residual shearing stress in the rod.
SOLUTION
1
d1 = 12.5 mm
2
1
c2 = d 2 = 30 mm
2
c1 =
When the plastic zone reaches the inner surface, the stress is equal to τ Y . The corresponding torque is
calculated by integration.
dT = ρτ dA = ρτ Y (2πρ d ρ ) = 2π τ Y ρ 2 d ρ
T = 2π τ Y

C2
ρ2 dρ =
C1
=
Unloading.
2π
τ Y c23 − c13
3
(
)
2π
(145 × 106 )[(30 × 10−3 )3 − (12.5 × 10−3 )3 ] = 7.6064 × 103 N ⋅ m
3
T ′ = 7.6064 × 103 N ⋅ m
J =
π
(c
2
4
2
)
− c14 =
π
2
[(30)4 − (12.5)4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4
τ1′ =
(7.6064 × 103 )(12.5 × 10−3 )
T ′c1
=
= 77.050 × 106 Pa = 77.05 MPa
J
1.234 × 10−6
τ 2′ =
(7.6064 × 103 )(30 × 10−3 )
T ′c2
=
= 192.63 × 106 Pa = 192.63 MPa
J
1.234 × 10−6
Residual stress.
Inner surface:
τ res = τ Y − τ1′ = 145 − 77.05 = 67.95 MPa
Outer surface:
τ res = τ Y − τ 2′ = 145 − 192.63 = −47.63 MPa
Maximum residual stress:
68.0 MPa at inner surface. 
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PROBLEM 3.119
In Prob. 3.118, determine the permanent angle of twist of the rod.
PROBLEM 3.118 The hollow shaft shown is made of a steel that is
and
assumed to be elastoplastic with τ Y = 145 MPa
G = 77.2 GPa. The magnitude T of the torques is slowly increased
until the plastic zone first reaches the inner surface of the shaft; the
torques are then removed. Determine the magnitude and location of
the maximum residual shearing stress in the rod.
SOLUTION
1
d1 = 12.5 mm
2
1
c2 = d 2 = 30 mm
2
c1 =
When the plastic zone reaches the inner surface, the stress is equal to τ Y . The corresponding torque is
calculated by integration.
dT = ρτ dA = ρτ Y (2πρ d ρ ) = 2πτ Y ρ 2 d ρ
T = 2πτ Y
=

c2
c1
ρ2 dρ =
2π
τ Y c23 − c13
3
(
)
2π
(145 × 106 )[(30 × 10−3 )3 − (12.5 × 10−3 )3 ] = 7.6064 × 103 N ⋅ m
3
Rotation angle at maximum torque.
c1ϕmax
τ
= γY = Y
L
G
ϕmax =
Unloading.
τY L
Gc1
=
(145 × 106 )(5)
= 0.75130 rad
(77.2 × 109 )(12.5 × 10−3 )
T ′ = 7.6064 × 103 N ⋅ m
J =
ϕ′ =
π
(c
2
4
2
)
− c14 =
π
2
[(30)4 − (12.5)4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4
(7.6064 × 103 )(5)
T ′L
=
= 0.39922 rad
GJ
(77.2 × 109 )(1.234 × 10−6 )
Permanent angle of twist.
ϕperm = ϕmax − ϕ ′ = 0.75130 − 0.39922 = 0.35208 rad
ϕperm = 20.2° 
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PROBLEM 3.120
A torque T applied to a solid rod made of an elastoplastic material is increased until
the rod is fully plastic and them removed. (a) Show that the distribution of residual
shearing stresses is as represented in the figure. (b) Determine the magnitude of the
torque due to the stresses acting on the portion of the rod located within a circle of
radius c0.
SOLUTION
(a)
ρY = 0, Tload =
After loading:
Tc
2T
2(Tload ) 4
=
=
= τY
3
3
J
πc
π c3
4 ρ
τ ′ = τY
3 c
τ′ =
Unloading:
4
3
4ρ 

= τ Y 1 −
3c 
c

τ res = τ Y − τ Y
To find c0 set,
τ res = 0 and ρ = c0
T0 = 2π

c0
0
ρ 2τ d ρ = 2π
 ρ3 4 ρ4 
= 2πτ Y 
−

 3 3 4c 

(3/4) c
0
(3/4) c
0
at ρ = c
ρ
Residual:
0 =1−
(b)
4
4π 3
2π 3
TY =
c τY =
c τY
3
32
3
4c0
3
∴ c0 = c
3c
4
c0 = 0.150c 
4ρ
 dρ
3 c
 1  3 3  4  1  3 4 
= 2πτ Y c3    −     
 3  4   3  4  4  


ρ 2τ Y 1 −
27  9π
9
τ Y c3 = 0.2209 τ Y c3
= 2πτ Y c3  −
=
 64 256  128
T0 = 0.221τ Y c3 
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PROBLEM 3.121
Determine the largest torque T that can be applied to each of the two
brass bars shown and the corresponding angle of twist at B, knowing that
τ all = 12 ksi and G = 5.6 × 106 psi.
SOLUTION
L = 25 in., G = 5.6 × 106 psi, τ all = 12 × 103 psi
τ max =
ϕ =
(a)
(b)
T
c1ab 2
or
T = c1ab 2τ max
TL
c2ab3G
or
ϕ =
a = 4 in., b = 1 in.,
a
= 4.0
b
c1Lτ max
c2bG
From Table 3.1: c1 = 0.282,
From (1):
T = (0.282)(4)(1)2 (12 × 103 ) = 13.54 × 103
From (2):
ϕ =
a = 2.4in.,
(1)
(2)
c2 = 0.281
T = 13.54 kip ⋅ in 
(0.282)(25)(12 × 103 )
= 0.05376 radians
(0.281)(1)(5.6 × 106 )
b = 1.6in.,
a
= 1.5
b
From Table 3.1: c1 = 0.231,
ϕ = 3.08° 
c2 = 0.1958
From (1):
T = (0.231)(2.4)(1.6)2 (12 × 103 ) = 17.03 × 103
T = 17.03 kip ⋅ in 
From (2):
ϕ =
(0.231)(25)(12 × 103 )
= 0.0395 radians
(0.1958)(1.6)(5.6 × 106 )
ϕ = 2.26° 
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PROBLEM 3.122
Each of the two brass bars shown is subjected to a torque of magnitude
T = 12.5 kip · in. Knowing that G = 5.6 × 106 psi, determine for each
bar the maximum shearing stress and the angle of twist at B.
SOLUTION
(a)
L = 25 in.,
G = 5.6 × 106 psi,
a = 4 in,
b = 1 in,
From Table 3.1:
τ max =
ϕ =
T = 12.5 × 103 lb ⋅ in
a
= 4.0
b
c1 = 0.282,
c2 = 0.281
T
12.5 × 103
=
= 11.08 × 103
(0.282)(4)(1) 2
c1ab 2
τ max = 11.08 ksi 
TL
(12.5 × 103 )(25)
=
= 0.04965 radians
(0.282)(4)(1)3 (5.6 × 106 )
c2ab3G
ϕ = 2.84° 
(b)
a = 2.4 in.,
From Table 3.1:
τ max =
ϕ =
b = 1.6 in.,
a
= 1.5
b
c1 = 0.231,
c2 = 0.1958
T
12.5 × 103
=
= 8.81 × 103
2
2
(0.231)(2.4)(1.6)
c1ab
τ max = 8.81 ksi 
TL
(12.5 × 106 )(25)
=
= 0.02899 radians
3
(0.1958)(2.4)(1.6)3 (5.6 × 106 )
c2ab G
ϕ = 1.661° 
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PROBLEM 3.123
Each of the two aluminium bars shown is subjected to a torque of
magnitude T = 1800 N ⋅ m. Knowing that G = 26 GPa, determine
for each bar the maximum shearing stress and the angle of twist at
B.
SOLUTION
T = 1800 N ⋅ m
(a)
τ max =
ϕ =
c1 = 0.208,
c2 = 0.1406
T
1800
=
= 40.1 × 106 Pa
2
2
c1ab
(0.208)(0.060)(0.060)
TL
(1800)(0.300)
=
= 0.011398 radians
3
c2ab G
(0.1406)(0.060)(0.060)3 (26 × 109 )
a = 95 mm = 0.095 m,
From Table 3.1:
τ max =
ϕ =
G = 26 × 109 Pa
a
= 1.0
b
a = b = 60 mm = 0.060 m
From Table 3.1:
(b)
L = 0.300 m
b = 38 mm = 0.038 m,
c1 = 0.258,
τ max = 40.1 MPa 
ϕ = 0.653° 
a
= 2.5
b
c2 = 0.249
T
1800
=
= 50.9 × 106 Pa
2
2
c1ab
(0.258)(0.095)(0.038)
TL
(1800)(0.300)
=
= 0.01600 radians
3
c2ab G
(0.249)(0.095)(0.038)3 (26 × 109 )
τ max = 50.9 MPa 
ϕ = 0.917° 
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PROBLEM 3.124
Determine the largest torque T that can be applied to each of the
two aluminium bars shown and the corresponding angle of twist at
B, knowing τ all = 50 MPa and G = 26 GPa.
SOLUTION
L = 0.300 m, G = 26 × 109 Pa, τ all = 50 × 106 Pa
τ max =
ϕ =
(a)
T
c1ab 2
or
TL
c2ab 4G
or
a = b = 60 mm = 0.060 m,
From Table 3.1:
(b)
c1 = 0.208,
T = c1ab 2τ max
ϕ =
(1)
c1Lτ max
c2bG
(2)
a
= 1.0
b
c2 = 0.1406
From (1):
T = (0.208)(0.060)(0.060)2 (50 × 106 ) = 2246 N ⋅ m
From (2):
ϕ =
(0.208)(0.300)(50 × 106 )
= 0.01422 radians
(0.1406)(0.060)(26 × 109 )
a = 95 mm = 0.095 m, b = 38 mm = 0.038 m,
From Table 3.1:
T = 2.25 kN ⋅ m 
ϕ = 0.815° 
a
= 2.5
b
c1 = 0.258, c2 = 0.249
From (1):
T = (0.258)(0.095)(0.038) 2 (50 × 106 ) = 1770 N ⋅ m
From (2):
ϕ =
(0.258)(0.300)(50 × 106 )
= 0.01573 radians
(0.249)(0.038)(26 × 109 )
T = 1.770 kN ⋅ m 
ϕ = 0.901° 
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PROBLEM 3.125
Determine the largest allowable square cross section of a steel shaft of length 20 ft if the maximum shearing
stress is not to exceed 10 ksi when the shaft is twisted through one complete revolution. Use
G = 11.2 × 106 psi.
SOLUTION
L = 20 ft = 240 in.
τ max = 10 ksi = 10 × 103 psi
ϕ = 1 rev = 2π radians
τ max =
T
c1ab 2
(1)
TL
c2ab3G
(2)
ϕ =
Divide (2) by (1) to eliminate T.
ϕ
τ max
=
c1ab 2 L
cL
= 1
3
c2bG
c2ab G
c1Lτ max
c2Gϕ
Solve for b.
b=
For a square section,
a
= 1.0
b
From Table 3.1,
c1 = 0.208,
c2 = 0.1406
b=
(0.208)(240)(10 × 103 )
(0.1406)(11.2 × 106 )(2π )
b = 0.0505 in. 
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PROBLEM 3.126
Determine the largest allowable length of a stainless steel shaft of
3 3
×
8 4
-in. cross section if the shearing stress
is not to exceed 15 ksi when the shaft is twisted through 15°. Use G = 11.2 × 106 psi.
SOLUTION
3
in. = 0.75 in.
4
3
b = in. = 0.375 in.
8
a=
τ max = 15 ksi = 15 × 103 psi
ϕ = 15° =
τ max =
ϕ =
Divide (2) by (1) to eliminate T.
Solve for L.
ϕ
τ max
=
L=
15π
rad = 0.26180 rad
180
T
c1ab 2
(1)
TL
c2ab3G
(2)
c1ab 2 L
cL
= 1
3
c2bG
c2ab G
c2bGϕ
c1τ max
a
0.75
=
=2
0.375
b
Table 3.1 gives
c1 = 0.246, c2 = 0.229
L=
(0.229)(0.375)(11.2 × 106 )(0.26180)
= 68.2 in.
(0.246)(15 × 103 )
L = 68.2 in. 
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PROBLEM 3.127
The torque T causes a rotation of 2° at end B of the stainless steel bar
shown. Knowing that b = 20 mm and G = 75GPa, determine the
maximum shearing stress in the bar.
SOLUTION
a = 30 mm = 0.030 m
b = 20 mm = 0.020 m
ϕ = 2° = 34.907 × 10−3 rad
ϕ =
TL
c2ab3Gϕ
T
∴
=
L
c2ab3G
τ max =
T
c2ab3Gϕ
c bGϕ
=
= 2
c1L
c1ab 2
c1ab 2 L
30
a
=
= 1.5.
20
b
From Table 3.1,
c1 = 0.231
c2 = 0.1958
τ max =
(0.1958)(20 × 10−3 )(75 × 109 )(34.907 × 10−3 )
= 59.2 × 106 Pa
(0.231)(750 × 10−3 )
τ max = 59.2 MPa 
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PROBLEM 3.128
The torque T causes a rotation of 0.6° at end B of the aluminum bar
shown. Knowing that b = 15 mm and G = 26 GPa, determine the
maximum shearing stress in the bar.
SOLUTION
a = 30 mm = 0.030 m
b = 15 mm = 0.015 m
ϕ = 0.6° = 10.472 × 10−3 rad
ϕ=
c2 ab3Gϕ
TL
∴
=
T
c1 L
c2 ab3G
τ max =
c2 ab3Gϕ c2 bGϕ
T
=
=
c1 L
c1ab 2
c1ab 2 L
a 30
=
= 2.0
b 15
From Table 3.1,
c1 = 0.246
c2 = 0.229
τ max =
(0.229)(15 × 10−3 )(26 × 109 )(10.472 × 10−3 )
(0.246)(750 × 10−3 )
= 5.07 × 106 Pa
τ max = 5.07 MPa 
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PROBLEM 3.129
Two shafts are made of the same material. The cross section of shaft A is a
square of side b and that of shaft B is a circle of diameter b. Knowing that the
shafts are subjected to the same torque, determine the ratio τ A /τ B of maximum
shearing stresses occurring in the shafts.
SOLUTION
a
= 1, c1 = 0.208 (Table 3.1)
b
T
T
=
τA =
2
0.208b3
c1ab
A.
Square:
B.
Circle:
c=
Ratio:
τA
1
π
=
⋅
= 0.3005π
τ B 0.208 16
1
Tc
2T
16T
b τB =
=
=
2
J
π c3 π b3
τA
= 0.944 
τB
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PROBLEM 3.130
Shafts A and B are made of the same material and have the same
cross-sectional area, but A has a circular cross section and B has a
square cross section. Determine the ratio of the maximum shearing
stresses occurring in A and B, respectively, when the two shafts are
subjected to the same torque (TA = TB ). Assume both deformations
to be elastic.
SOLUTION
Let c be the radius of circular section A and b be the side square section B.
For equal areas,
π c2 = b2
TAC
2T
= A3
J
πc
Circle:
τA =
Square:
a
=1
b
τB =
b=c π
c1 = 0.208
from Table 3.1
TB
T
= B3
c1ab 2
c1b
Ratio:
τ A 2TA c1b3 2c1b3 TA
T
=
=
= 2c1 π A
3
3
TB
τ B π c TB
π c TB
For TA = TB ,
τA
= (2)(0.208) π
τB
τA
= 0.737 
τB
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PROBLEM 3.131
Shafts A and B are made of the same material and have the same crosssectional area, but A has a circular cross section and B has a square
cross section. Determine the ratio of the maximum torques TA and TB
that can be safely applied to A and B, respectively.
SOLUTION
Let c = radius of circular section A and b = side of square section B.
For equal areas π c 2 = b 2 ,
c=
Circle:
τA =
b
π
TAc
2TA
=
J
π c3
∴
TA =
π
2
c3τ A
Square:
From Table 3.1,
c1 = 0.208
τB =
TA
T
= B3
c1ab 2
c1b
π c3τ
B
Ratio:
TA
= 2 3
TB
c1b τ B
For the same stresses,
τB = τ A ∴
∴
TB = c1b3τ B
π ⋅ b3 τ
1 τA
2 π 3/2 B
=
=
3
c1b τ B
2c1 π τ B
TA
1
=
TB
(2) (0.208) π
TA
= 1.356 
TB
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PROBLEM 3.132
Shafts A and B are made of the same material and have the same length
and cross-sectional area, but A has a circular cross section and B has a
square cross section. Determine the ratio of the maximum values of the
angles ϕ A and ϕ B through which shafts A and B, respectively, can be
twisted.
SOLUTION
Let c = radius of circular section A and b = side of square section B.
For equal areas,
π c2 = b2 ∴ b = π c
Circle:
γ max =
Square:
From Table 3.1,
Ratio:
For equal stresses, τ A = τ B
τA
G
=
cϕ A
L
c1 = 0.208,
∴
ϕA =
Lτ A
cG
c2 = 0.1406
τB =
TB
TB
=
2
c1ab
0.208 b3
ϕB =
0.208 b3τ B L 1.4794 Lτ B
TB L
=
=
bG
0.1406 b 4G
c2ab3G
∴
TB = 0.208 b3τ B
ϕ A Lτ A
bG
bτ
τ
=
⋅
= 0.676 A = 0.676 π A
cG 1.4794Lτ B
cτ B
ϕB
τB
ϕB
= 0.676 π
ϕA
ϕB
= 1.198 
ϕA
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PROBLEM 3.133
Each of the three aluminum bars shown is to be twisted through
an angle of 2°. Knowing that b = 30 mm, τ all = 50 MPa, and
G = 27 GPa, determine the shortest allowable length of each
bar.
SOLUTION
ϕ = 2° = 34.907 × 10−3 rad, τ = 50 × 106 Pa
τ =
For square and rectangle,
T
c1ab2
Divide to eliminate T; then solve for L.
(a)
Square:
L=
(b)
a
= 1.0
b
From Table 3.1,
c=
cGϕ
τ
Rectangle:
L=
TL
c2ab3G
ϕ
c ab2 L
= 1 3
τ
c2ab G
L=
=
c2bGϕ
c1τ
c1 = 0.208, c2 = 0.1406
L = 382 mm 
1
Tc
TL
b = 0.015 m τ =
ϕ =
2
J
GJ
Divide to eliminate T; then solve for L.
(c)
ϕ =
(0.1406)(0.030)(27 × 109 )(34.907 × 10−3 )
= 382 × 10−3 m
6
(0.208)(50 × 10 )
Circle:
L=
G = 27 × 109 Pa, b = 30 mm = 0.030 m
ϕ
JL
L
=
=
cGJ
cG
τ
(0.015)(27 × 109 )(34.907 × 10−3 )
= 283 × 10−3 m
50 × 106
a = 1.2b
a
= 1.2
b
From Table 3.1,
(0.1661)(0.030)(27 × 109 )(34.907 × 10−3 )
= 429 × 10−3 m
(0.219)(50 × 106 )
L = 283 mm 
c1 = 0.219, c2 = 0.1661
L = 429 mm 
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PROBLEM 3.134
Each of the three steel bars is subjected to a torque as shown.
Knowing that the allowable shearing stress is 8 ksi and that
b = 1.4 in., determine the maximum torque T that can be applied
to each bar.
SOLUTION
τ max = 8 ksi, b = 1.4 in.
(a)
a = b = 1.4 in.
Square:
a
= 1.0
b
c1 = 0.208
From Table 3.1,
τ max =
T
c1ab 2
T = c1ab 2τ max
T = (0.208)(1.4)(1.4)(1.4) 2 (8)
(b)
Circle:
1
b = 0.7 in.
2
2T
π
Tc
=
=
T = c3τ max
3
J
2
πc
c=
τ max
T =
(c)
T = 4.57 kip ⋅ in 
π
2
(0.7)3 (8)
Rectangle:
a = (1.2)(1.4) = 1.68 in.
From Table 3.1,
c1 = 0.219
T = 4.31 kip ⋅ in 
a
= 1.2
b
T = c1ab 2τ max = (0.219)(1.68)(1.4)2 (8)
T = 5.77 kip ⋅ in 
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PROBLEM 3.135
A 36-kip ⋅ in. torque is applied to a 10-ft-long steel angle with an L8 × 8 × 1 cross section.
From Appendix C, we find that the thickness of the section is 1 in. and that its area is,
15.00 in2. Knowing that G = 11.2 × 106 psi, determine (a) the maximum shearing stress
along line a-a, (b) the angle of twist.
SOLUTION
a =
(a)
A 15 in 2
=
= 15 in., b = 1 in.,
t
1 in.
Since
a
1
b
> 5, c1 = c2 = 1 − 0.630 
b
a
3
or
c1 = c2 =
1
0.630 
= 0.3193
1−
3 
15 
T = 36 × 103 lb ⋅ in; L = 120 in.; G = 11.2 × 106 psi
T
Maximum shearing stress:
τ max =
c1ab 2
τ max =
(b)
a
= 15
b
Angle of twist:
36 × 103
= 7.52 × 103 psi
2
(0.3193)(15)(1)
ϕ =
TL
c2ab 2G
ϕ =
(36 × 103 )(120)
= 0.08052 radians
(0.3193)(15)(1)3 (11.2 × 106 )
τ max = 7.52 ksi 
ϕ = 4.61° 
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PROBLEM 3.136
A 3-m-long steel angle has an L203 × 152 × 12.7 cross section. From
Appendix C, we find that the thickness of the section is 12.7 mm and
that its area is 4350 mm2. Knowing that τ all = 50 MPa and that
G = 77.2 GPa, and ignoring the effect of stress concentration,
determine (a) the largest torque T that can be applied, (b) the
corresponding angle of twist.
SOLUTION
A = 4350 mm 2 b = 12.7 mm a = ?
a =
Equivalent rectangle.
A 4350
=
= 342.52 mm
b
12.7
a
= 26.97
b
b
1
c1 = c2 = 1 − 0.630  = 0.32555
a
3
(a)
τ max =
T
c1ab 2
τ max = 50 × 106 Pa
T = c1ab2τ max = (0.32555) (26.97 × 10−3 ) (12.7 × 10−3 ) 2 (50 × 106 )
= 70.807 N ⋅ m
(b)
ϕ =
T = 70.8 N ⋅ m 
TL
(70.807) (3)
=
3
c2ab G
(0.32555)(26.97 × 10−3 )(12.7 × 10−3 ) (77.2 × 109 )
= 0.15299 rad
ϕ = 8.77° 
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PROBLEM 3.137
An 8-ft-long steel member with a W8 × 31 cross section is subjected to a 5-kip ⋅ in. torque.
The properties of the rolled-steel section are given in Appendix C. Knowing that
G = 11.2 × 106 psi, determine (a) the maximum shearing stress along line a-a, (b) the
maximum shearing stress along line b-b, (c) the angle of twist. (Hint: consider the web and
flanges separately and obtain a relation between the torques exerted on the web and a
flange, respectively, by expressing that the resulting angles of twist are equal.)
SOLUTION
a 7.995
=
= 18.38
b 0.435
Tf L
1
c1 = c2 = 1 − 0.630 b = 0.3219
ϕf =
a
3
c2ab3G
Gϕ f
Gϕ
T f = c2ab3
= Kf
where K f = c2ab3
L
L
K f = (0.3219) (7.995) (0.435)3 = 0.2138 in 3
a = 7.995 in., b = 0.435 in.,
Flange:
)
(
a = 8.0 − (2)(0.435) = 7.13 in., b = 0.285 in.,
Web:
a
7.13
=
= 25.02
b
0.285
b
Tw L

c1 = c2 = 1 1 − 0.630  = 0.3249
ϕw =
3
a
c2ab3G
Gϕ w
Gϕ
Tw = c2ab3
K w = c2ab3
= Kw
where
L
L
K w = (0.3249) (7.13) (0.285)3 = 0.0563 in 4
ϕ f = ϕw = ϕ
For matching twist angles:
T = 2T f + Tw = (2 K f + K w )
Total torque.
Gϕ
T
=
,
2K p + K w
L
Tf =
Tf =
KfT
2K f + K w
,
K wT
2K f + K w
(0.2138)(5000)
(0.0563) (5000)
= 2221 lb ⋅ in; Tw =
= 557 lb ⋅ in
(2) (0.2138) + 0.0563
(2) (0.2138) + 0.0563
Tf
2221
= 4570 psi
(0.3219)(7.995)(0.435) 2
τ f = 4.57 ksi 
Tw
557
=
= 2960 psi
2
c1ab
(0.3249)(7.13)(0.285)2
τ w = 2.96 ksi 
(a)
τf =
(b)
τw =
(c)
Gϕ
T
TL
=
∴ ϕ =
L
2K f + K w
G(2 K f + K w )
ϕ =
Tw =
Gϕ
L
c1ab
2
=
where L = 8 ft = 96 in.
(5000)(96)
= 88.6 × 10−3 rad
(11.2 × 10 )[(2)(0.2138) + 0.563]
6
ϕ = 5.08° 
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PROBLEM 3.138
A 4-m-long steel member has a W310 × 60 cross section. Knowing that
G = 77.2 GPa and that the allowable shearing stress is 40 MPa, determine
(a) the largest torque T that can be applied, (b) the corresponding angle of twist.
Refer to Appendix C for the dimensions of the cross section and neglect the
effect of stress concentrations. (See hint of Prob. 3.137.)
SOLUTION
W310 × 60, L = 4 m, G = 77.2 G Pa, τ all = 40 MPa
For one flange:
From App. C,
Eq. (3.45):
c1 = c2 =
Eq. (3.44):
φf =
Tf L
3
c2ab G
=
a = 203 mm, b = 13.1 mm, a/b = 15.50
1
0.630 
= 0.320
1−

3
15.50 
T f (4)
0.320(0.203)(0.0131)3 (77.2 × 109 )
(1)
φ f = 355.04 × 10−6T f
For web:
From App. C,
Eq. (3.45):
c1 = c2 =
Eq. (3.44):
φw =
a = 303 − 2(13.1) = 276.8 mm, b = 7.5 mm, a/b = 36.9
1
0.630 
= 0.328
1−

3
36.9 
Tw (4)
0.328(0.2768)(0.0075)3 (77.2 × 109 )
φw = 1.354.2 × 10−6Tw
(2)
Since angle of twist is the same for flanges and web:
φ f = φ w:
355.04 × 10−6T f = 1354.2 × 10−6Tw
T f = 3.814Tw
(3)
But the sum of the torques exerted on the two flanges and on the web is equal to the torque T applied to the
member:
2T f + Tw = T
(4)
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PROBLEM 3.138 (Continued)
Substituting for T f from (3) into (4):
2(3.814Tw ) + Tw = T
From (3):
T f = 3.814(0.11589T )
Tw = 0.11589T
T f = 0.44205T
(5)
(6)
For one flange:
From Eq. (3.43):
T f = c1ab 2τ max = 0.320(0.203)(0.0131) 2 (40 × 106 )
= 445.91 N ⋅ m
Eq. (6):
445.91 = 0.44205T
For web:
Tw = c1ab 2τ max = 0.328(0.2768)(0.0075)2 (40 × 106 )
T = 1009 N ⋅ m 
= 204.28 N ⋅ m
Eq. (5):
204.28 = 0.11589T
(a)
Largest allowable torque:
(b)
Angle of twist: Use T f , which is critical.
Eq. (1):
Use the smaller value.
φ = φ f = (355.04 × 10−6 )(445.91) = 0.15831 rad
T = 1763 N ⋅ m 
T = 1009 N ⋅ m 
φ = 9.07° 
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PROBLEM 3.139
A torque T = 750 kN ⋅ m is applied to a hollow shaft shown that has a uniform 8-mm
wall thickness. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
Detail of corner.
1
t = e tan 30°
2
t
e=
2 tan 30°
=
8
= 6.928 mm
2 tan 30°
b = 90 − 2e = 76.144 mm
Area bounded by centerline.
a =
1
3
3 2
3
(76.144)2
b
b=
b =
2 2
4
4
= 2510.6 mm 2 = 2510.6 × 10−6 m 2
t = 0.008 m
τ =
T
750
=
= 18.67 × 106 Pa
−6
2ta (2)(0.008) (2510 × 10 )
τ = 18.67 MPa 
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PROBLEM 3.140
A torque T = 5 kN ⋅ m is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
T = 5 × 103 N ⋅ m
Area bounded by centerline.
a = bh = (69) (115) = 7.935 × 103 mm 2
= 7.935 × 10−3 m 2
At point a:
t = 6 mm = 0.006 m
τ =
T
5 × 103
=
2ta (2) (0.006) (7.935 × 10−3 )
= 52.5 × 106 Pa
At point b:
τ = 52.5 MPa 
t = 10 mm = 0.010 m
τ =
T
5 × 103
=
= 31.5 × 106 Pa
−3
2ta (2)(0.010)(7.935 × 10 )
τ = 31.5 MPa 
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PROBLEM 3.141
A 90-N ⋅ m torque is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
Area bounded by centerline.
a = 52 × 52 − 39 × 39 +
π
4
(39)2 = 2378 mm 2 = 2.378 × 10−3 m 2
T = 90 N ⋅ m

τa =
T
90 N ⋅ m
=
−3
2ta
2(4 × 10 m)(2.378 × 10−3 m 2 )
τ a = 4.73 MPa 
τb =
T
90 N ⋅ m
=
−3
2ta
2(2 × 10 m)(2.378 × 10−3m 2 )
τ b = 9.46 MPa 
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PROBLEM 3.142
A 5.6 kN ⋅m torque is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the
shearing stress at points a and b.
SOLUTION
Area bounded by centerline.
a = (96 mm)(95mm) +
π
2
(47.5mm)2 = 12.664 × 103 mm2
= 12.664 × 10−3 m 2
t = 5 mm = 0.005m 
At point a,
τ =
5.6 × 103
T
=
= 44.2 × 106 Pa
−3
2at
(2)(12.664 × 10 )(0.005)
τ = 44.2 MPa 
t = 8 mm = 0.008m
At point b,
τ =
T
5.6 × 103
=
= 27.6 × 106 Pa
2at
(2)(12.664 × 10−3 )(0.008)
τ = 27.6 MPa 
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PROBLEM 3.143
A hollow member having the cross section shown is formed from sheet
metal of 2-mm thickness. Knowing that the shearing stress must not
exceed 3 MPa, determine the largest torque that can be applied to the
member.
SOLUTION
Area bounded by centerline.
a = (48)(18) + (30)(18)
= 1404 mm 2 = 1404 × 10−6 m 2
t = 0.002 m
τ =
T
2ta
or
T = 2taτ = (2)(0.002)(1404 × 10−6 )(3 × 106 )
T = 16.85 N ⋅ m 
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PROBLEM 3.144
A hollow brass shaft has the cross section shown. Knowing that the
shearing stress must not exceed 12 ksi and neglecting the effect of stress
concentrations, determine the largest torque that can be applied to the
shaft.
SOLUTION
Calculate the area bounded by the center line of the wall cross section. The area is a rectangle with two semicircular cutouts.
b = 5 − 0.2 = 4.8 in.
h = 6 − 0.5 = 5.5 in.
r = 1.5 + 0.1 = 1.6 in.
π 
a = bh − 2  r 2  = (4.8)(5.5) − π (1.6) 2 = 18.3575 in 2
2 
T
τ max =
τ max = 12 × 103 psi
tmin = 0.2 in.
2atmin
T = 2atmin τ max = (2)(18.3575) (0.2)(12 × 103 ) = 88.116 × 103 lb ⋅ in
T = 88.1kip ⋅ in = 7.34 kip ⋅ ft 
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PROBLEM 3.145
A hollow member having the cross section shown is to be formed from sheet metal of
0.06 in. thickness. Knowing that a 1250 lb ⋅ in.-torque will be applied to the member,
determine the smallest dimension d that can be used if the shearing stress is not to
exceed 750 psi.
SOLUTION
Area bounded by centerline.
a = (5.94)(2.94 − d ) + 1.94 d = 17.4636 − 4.00 d
t = 0.06 in., τ = 750 psi, T = 1250 lb ⋅ in
T
τ=
2ta
a=
17.4636 − 4.00d =
d =
T
2tτ
1250
= 13.8889
(2) (0.06) (750)
3.5747
= 0.894 in.
4.00
d = 0.894 in. 
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PROBLEM 3.146
A hollow member having the cross section shown is to be formed from sheet metal of
0.06 in. thickness. Knowing that a 1250 lb ⋅ in.-torque will be applied to the member,
determine the smallest dimension d that can be used if the shearing stress is not to
exceed 750 psi.
SOLUTION
Area bounded by centerline.
a = (5.94)(2.94) − 2.06 d = 17.4636 − 2.06 d
t = 0.06 in., τ = 750 psi, T = 1250 lb ⋅ in
τ =
T
2ta
a=
17.4636 − 2.06d =
d =
T
2tτ
1250
= 13.8889
(2) (0.06) (750)
3.5747
= 1.735 in.
2.06
d = 1.735 in. 
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PROBLEM 3.147
A hollow cylindrical shaft was designed to have a uniform wall thickness of
0.1 in. Defective fabrication, however, resulted in the shaft having the cross
section shown. Knowing that a 15 kip ⋅ in.-torque is applied to the shaft,
determine the shearing stresses at points a and b.
SOLUTION
Radius of outer circle = 1.2 in.
Radius of inner circle = 1.1 in.
Mean radius = 1.15 in.
Area bounded by centerline.
a = π rm2 = π (1.15) 2 = 4.155 in 2
t = 0.08 in.
At point a,
τ =
T
15
=
2ta (2)(0.08)(4.155)
τ = 22.6 ksi 
t = 0.12 in.
At point b,
τ =
T
15
=
2ta (2)(0.12)(4.155)
τ = 15.04 ksi 
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PROBLEM 3.148
A cooling tube having the cross section shown is formed from a sheet of stainless
steel of 3-mm thickness. The radii c1 = 150 mm and c2 = 100 mm are measured to
the center line of the sheet metal. Knowing that a torque of magnitude T = 3 kN ⋅ m
is applied to the tube, determine (a) the maximum shearing stress in the tube, (b) the
magnitude of the torque carried by the outer circular shell. Neglect the dimension of
the small opening where the outer and inner shells are connected.
SOLUTION
Area bounded by centerline.
(
)
a = π c12 − c22 = π (1502 − 1002 ) = 39.27 × 103 mm 2
= 39.27 × 10−3 m 2
t = 0.003 m
T
3 × 103
=
= 12.73 × 106 Pa
2ta
(2)(0.003)(39.27 × 10−3 )
(a)
τ =
(b)
T1 = (2π c1t τ c1) = 2π c12 t τ
= 2π (0.150) 2 (0.003) (12.73 × 106 ) = 5.40 × 103 N ⋅ m
τ = 12.76 MPa 
T1 = 5.40 kN ⋅ m 
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PROBLEM 3.149
A hollow cylindrical shaft of length L, mean radius cm , and uniform
thickness t is subjected to a torque of magnitude T. Consider, on the
one hand, the values of the average shearing stress τ ave and the angle
of twist ϕ obtained from the elastic torsion formulas developed in
Sections 3.4 and 3.5 and, on the other hand, the corresponding values
obtained from the formulas developed in Sec. 3.13 for thin-walled
shafts. (a) Show that the relative error introduced by using the thinwalled-shaft formulas rather than the elastic torsion formulas is the
same for τ ave and ϕ and that the relative error is positive and
proportional to the ratio t/cm . (b) Compare the percent error
corresponding to values of the ratio t /cm of 0.1, 0.2, and 0.4.
SOLUTION
Let c2 = outer radius = cm +
1
2
t and c1 = inner radius = cm − 12 t
π
(c
2
4
2
)
− c14 =
π
(
)
c22 + c12 (c2 + c1)(c2 − c1)
2
1
1 
π
=  cm2 + cmt + t 2 + cm2 − cmt + t 2  (2cm ) t
2
4
4 
J =
1 

= 2π  cm2 + t 2  cmt
4 

Tc
T
τm = m =
 2 1 2
J
2π  cm + t  t
4 

TL
TL
=
ϕ1 =
 2 1 2
JG
2π  cm + t  cmt G
4 

Area bounded by centerline.
a = π cm2
τ ave =
ϕ2 =
(a)

Ratios:
T
T
=
2ta
2π cm2 t
TL
ds TL(2π cm /t )
TL
=
=

2 
2 2
4a G t
4(π cm ) G
2π cm3 t G
)
(
2π cm2 + 1 t 2 t
1 t2
T
τ ave
4
1
=
×
=
+
T
τm
4 cm2
2π cm2 t
(
)
2π cm2 + 1 t 2 cmtG
1 t2
TL
ϕ2
4
1

=
×
=
+
TL
ϕ1 2π cm3 tG
4 cm


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PROBLEM 3.149 (Continued)
2
(b)
τ ave
ϕ
1 t
−1 = 2 −1 =
τm
ϕ1
4 cm2

t
cm
0.1
0.2
0.4
1 t2
4 cm2
0.0025
0.01
0.04
%
0.25%
1%
4%

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PROBLEM 3.150
Equal torques are applied to thin-walled tubes of the same length L,
same thickness t, and same radius c. One of the tubes has been slit
lengthwise as shown. Determine (a) the ratio τ b / τ a of the maximum
shearing stresses in the tubes, (b) the ratio ϕb /ϕ a of the angles of twist
of the shafts.
SOLUTION
Without slit:
Area bounded by centerline. a = π c 2
τa =
T
T
=
2ta
2π c 2t
J ≈ 2π c3t
With slit:
ϕa =
a = 2π c, b = t ,
c1 = c2 =
TL
TL
=
GJ
2π c3t G
a
2π c
=
>> 1
b
t
1
3
τb =
3T
T
=
2
2π ct 2
c1ab
ϕb =
3TL
T
=
3
2π ct 3G
c2ab G
(a)
Stress ratio:
τb
3T
2π c 2t
3c
=
⋅
=
2
τa
T
t
2π ct
(b)
Twist ratio:
ϕb
3TL
2π c3t G 3c 2
=
⋅
= 2
ϕa
TL
2π ct 3G
t
τ b 3c

=
t
τa
ϕb 3c 2
= 2 
ϕa
t
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PROBLEM 3.151
The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft.
Knowing that the top of the 8-in.-diameter steel drill pipe (G = 11.2 × 106 psi)
rotates through two complete revolutions before the drill bit at B starts to operate,
determine the maximum shearing stress caused in the pipe by torsion.
SOLUTION
TL
GJ ϕ
T =
GJ
L
Tc GJ ϕ c Gϕ c
τ =
=
=
J
JL
L
ϕ =
ϕ = 2 rev = (2)(2π ) = 12.566 rad,
c=
1
d = 4.0 in.
2
L = 5000 ft = 60000 in.
τ =
(11.2 × 106 )(12.566)(4.0)
= 9.3826 × 103 psi
60000
τ = 9.38 ksi 
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PROBLEM 3.152
The shafts of the pulley assembly shown are to be
redesigned. Knowing that the allowable shearing
stress in each shaft is 8.5 ksi, determine the smallest
allowable diameter of (a) shaft AB, (b) shaft BC.
SOLUTION
(a)
Shaft AB:
TAB = 3.6 × 103 lb ⋅ in
τ max = 8.5 ksi = 8.5 × 103 psi
J =
c=
(b)
Shaft BC:
π
2
3
c4
2TAB
πτ max
τ max =
=
3
2T
Tc
=
J
π c3
(2)(3.6 × 103 )
= 0.646 in.
π (8.5 × 103 )
d AB = 2c = 1.292 in. 
TBC = 6.8 × 103 lb ⋅ in
τ max = 8.5 × 103 psi
c=
3
2TBC
πτ max
=
3
(2)(6.8 × 103 )
= 0.7985 in.
π (8.5 × 103 )
d BC = 2c = 1.597 in. 
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PROBLEM 3.153
A steel pipe of 12-in. outer diameter is fabricated from
1
4 -in. -thick plate by welding along a helix which forms an angle
of 45° with a plane perpendicular to the axis of the pipe.
Knowing that the maximum allowable tensile stress in the weld
is 12 ksi, determine the largest torque that can be applied to the
pipe.
SOLUTION
From Eq. (3.14) of the textbook,
σ 45 = τ max
hence,
τ max = 12 ksi = 12 × 103 psi
1
1
d0 = (12) = 6.00 in.
2
2
c1 = c2 − t = 6.00 − 0.25 = 5.75 in.
c2 =
J =
τ max =
T =
π
(c
2
Tc
J
4
2
)
− c14 =
T =
π
2
[(6.00)4 − (5.75)4 ] = 318.67 in.
τ max J
c
3
(12 × 10 )(318.67)
= 637 × 103 lb ⋅ in
6.00
T = 637 kip ⋅ in 
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PROBLEM 3.154
For the gear train shown, the diameters of the three solid shafts
are:
d AB = 20 mm dCD = 25mm d EF = 40 mm
Knowing that for each shaft the allowable shearing stress is
60 MPa, determine the largest torque T that can be applied.
SOLUTION
Statics:
TAB = T
TCD
T
= AB
rC
rB
TCD =
rC
75
TAB =
T = 2.5T
rB
30
TEF
T
= CD
rF
rD
TEF =
rF
90
TCD =
(2.5T ) = 7.5T
rD
30
Determine the magnitude of T so that the stress is 60 MPa = 60 × 106 Pa.
Shaft AB:
Tc
J
c=
1
d AB = 10 mm = 0.010 m
2
TAB = T =
Shaft CD:
c=
Tshaft =
π
2
c=
(60 × 106 )(0.010)3
T = 94.2 N ⋅ m
1
dCD = 12.5 mm = 0.0125 m
2
TCD = 2.5T =
Shaft EF:
π
Jτ
= τ c3
2
c
τ =
π
2
(60 × 106 )(0.0125)3
T = 73.6 N ⋅ m
1
d EF = 20 mm = 0.020 m
2
TEF = 7.5T =
π
2
(60 × 106 )(0.020)3
T = 100.5 N ⋅ m
The smallest value of T is the largest torque that can be applied.
T = 73.6 N ⋅ m 
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PROBLEM 3.155
Two solid steel shafts (G = 77.2GPa) are connected to a coupling disk B
and to fixed supports at A and C. For the loading shown, determine (a) the
reaction at each support, (b) the maximum shearing stress in shaft AB, (c) the
maximum shearing stress in shaft BC.
SOLUTION
Shaft AB:
1
d = 25 mm = 0.025 m
2
π
π
T L
= c 4 = (0.025) 4 = 613.59 × 10−9 m 4 ϕ B = AB AB
2
2
GJ AB
T = TAB , LAB = 0.200 m, c =
J AB
TAB =
Shaft BC:
(77.2 × 109 )(613.59 × 10−9 )
GJ AB
ϕB =
ϕ B = 236.847 × 103 ϕ B
0.200
LAB
1
d = 19 mm = 0.019 m
2
π
π
T L
ϕ B = BC BC
= c 4 = (0.019) 4 = 204.71 × 10−9 m 4
2
2
GJ BC
T = TBC , LBC = 0.250 m, c =
J BC
TBC =
GJ BC
(77.2 × 109 )(204.71 × 10−9 )
= 63.214 × 103ϕ B
ϕB =
0.250
LBC
Equilibrium of coupling disk.
T = TAB + TBC
1.4 × 103 = 236.847 × 103ϕ B + 63.214 × 103ϕ B
ϕ B = 4.6657 × 10−3 rad.
TAB = (236.847 × 103 )(4.6657 × 10−3 ) = 1.10506 × 103 N ⋅ m
TBC = (63.214 × 103 )(4.6657 × 10−3 ) = 294.94 N ⋅ m
(a)
TA = TAB = 1105 N ⋅ m 
Reactions at supports.
TC = TBC = 295 N ⋅ m 
(b)
Maximum shearing stress in AB.
τ AB =
(c)
TABc (1.10506 × 103 )(0.025)
=
= 45.0 × 106 Pa
J AB
613.59 × 10−9
τ AB = 45.0 MPa 
Maximum shearing stress in BC.
τ BC =
TBC c
(294.94)(0.019)
=
= 27.4 × 106 Pa
J BC
204.71 × 10−9
τ BC = 27.4 MPa 
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PROBLEM 3.156
In the bevel-gear system shown, α = 18.43°. Knowing that the
allowable shearing stress is 8 ksi in each shaft and that the system is in
equilibrium, determine the largest torque TA that can be applied at A.
SOLUTION
Using stress limit for shaft A:
τ = 8 ksi,
TA =
c=
1
d = 0.25 in.
2
π
π
Jτ
= τ c3 = (8)(0.25)3 = 0.1963 kip ⋅ in
2
2
c
Using stress limit for shaft B:
τ = 8 ksi, c =
From statics,
1
d = 0.3125in.
2
TB =
π
π
Jτ
= τ c3 = (8)(0.3125)3 = 0.3835 kip ⋅ in
2
2
c
TA =
rA
TB = (tan α )TB
rB
TA = (tan18.43°)(0.3835) = 0.1278 kip ⋅ in
The allowable value of TA is the smaller.
TA = 0.1278 kip ⋅ in
TA = 127.8 lb ⋅ in 
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PROBLEM 3.157
Three solid shafts, each of
3
4
-in. diameter, are
connected by the gears shown. Knowing that
G = 11.2 × 106 psi, determine (a) the angle
through which end A of shaft AB rotates, (b) the
angle through which end E of shaft EF rotates.
SOLUTION
Geometry:
rB = 1.5 in., rC = 6 in., rF = 2in.
LAB = 48 in., LCD = 36 in., LEF = 48 in.
Statics:
TA = 100 lb ⋅ in
TE = 200 lb ⋅ in
Gear B.
ΣM B = 0:
−rB F1 + TA = 0
−1.5F1 + 100 = 0
F1 = 67.667 lb
Gear F.
ΣM F = 0:
−rF F2 + TE = 0
−2F2 + 200 = 0
F2 = 100 lb
Gear C.
ΣM C = 0:
−rC F1 − rC F2 + TC = 0
−(6)(66.667) − (6)(100) + TC = 0
TC = 1000 lb ⋅ in
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PROBLEM 3.157 (Continued)
Deformations:
c=
For all shafts,
J =
Kinematics:
1
d = 0.375 in.
2
π
2
c 4 = 0.031063 in 4
ϕ A/B =
TAB LAB
(100)(48)
=
= 0.013797 rad
GJ
(11.2 × 106 )(0.031063)
ϕ E /F =
TEF LEF
(200)(48)
=
= 0.027594 rad
GJ
(11.2 × 106 )(0.031063)
ϕ C /D =
TCD LCD
(1000)(36)
=
= 0.103476 rad
GJ
(11.2 × 106 )(0.031063)
ϕC = ϕC/D = 0.103476 rad
rBϕ B = rCϕC
(a)
rC
6
(0.103476) = 0.41390 rad
ϕC
rB
1.5
ϕ A = ϕ B + ϕ A/B = 0.41390 + 0.01397 = 0.42788 rad
rFϕ F = rCϕC
(b)
ϕB =
ϕF =
ϕ A = 24.5°

rC
6
ϕC = (0.103476) = 0.31043 rad
rF
2
ϕ E = ϕ F + ϕ E/F = 0.31043 + 0.027594 = 0.33802 rad
ϕ E = 19.37°

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PROBLEM 3.158
The design specifications of a 1.2-m-long solid transmission shaft require that the angle of twist of the shaft
not exceed 4° when a torque of 750 N ⋅ m is applied. Determine the required diameter of the shaft, knowing
that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of
77.2 GPa.
SOLUTION
T = 750 N ⋅ m, ϕ = 4° = 69.813 × 10−3 rad,
L = 1.2 m, J =
π
2
c4
τ = 90 MPa = 90 × 106 Pa G = 77.2 GPa = 77.2 × 109 Pa
Based on angle of twist. ϕ =
c=
Based on shearing stress. τ =
c=
Use larger value.
TL
2TL
=
GJ
π Gc 4
4
2TL
=
π Gϕ
4
(2)(750)(1.2)
= 18.06 × 10−3 m
π (77.2 × 109 )(69.813 × 10−3 )
Tc
2T
=
J
π c3
3
2T
πτ
=
3
(2)(750)
= 17.44 × 10−3 m
π (90 × 106 )
c = 18.06 × 10−3 m = 18.06 mm
d = 2c = 36.1 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
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PROBLEM 3.159
The stepped shaft shown rotates at 450 rpm. Knowing that r = 0.5in.,
determine the maximum power that can be transmitted without
exceeding an allowable shearing stress of 7500 psi.
SOLUTION
d = 5 in.
D = 6 in.
r = 0.5 in.
D
6
= = 1.20
d
5
r
0.5
=
= 0.10
d
5
K = 1.33
From Fig. 3.32,
1
d = 2.5 in.
2
KTc
2KT
τ =
=
J
π c3
c=
For smaller side,
T =
π c3τ
2K
=
π (2.5)3 (7500)
(2)(1.33)
= 138.404 × 103 lb ⋅ in
f = 450 rpm = 7.5 Hz
Power.
P = 2π f T = 2π (7.5)(138.404 × 103 ) = 6.52 × 106 in ⋅ lb/s
Recalling that 1 hp = 6600 in ⋅ lb/s,
P = 988 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
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without permission.
PROBLEM 3.160
A 750-N ⋅ m torque is applied to a hollow shaft having the cross section shown
and a uniform 6-mm wall thickness. Neglecting the effect of stress concentrations,
determine the shearing stress at points a and b.
SOLUTION
Area bounded by centerline.
a =2
π
2
(33) 2 + (60)(66) = 7381 mm 2
= 7381 × 10−6 m 2
t = 0.006 m at both a and b,
Then at points a and b,
τ =
T
750
=
= 8.47 × 106 Pa
2ta
(2) (0.006) (7381 × 10−6 )
τ = 8.47 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.
PROBLEM 3.161
The composite shaft shown is twisted by applying a torque T at end A.
Knowing that the maximum shearing stress in the steel shell is
150 MPa, determine the corresponding maximum shearing stress in the
aluminum core. Use G = 77.2 GPa for steel and G = 27 GPa for
aluminum.
SOLUTION
Let G1, J1, and τ1 refer to the aluminum core and G2 , J 2 , and τ 2 refer to the steel shell.
At the outer surface on the steel shell,
γ2 =
c2ϕ
ϕ γ2
τ
∴
=
= 2
L
L
c2
c2G2
At the outer surface of the aluminum core,
γ1 =
Matching
ϕ
L
c1ϕ
ϕ γ1
τ
∴
=
= 1
L
L
c1
c1G
for both components,
τ2
c2G2
Solving for τ 2 ,
τ2 =
=
τ1
c1G1
0.030 27 × 109
c2 G2
⋅
⋅
⋅ 150 × 106 = 39.3 × 106 Pa
τ1 =
9
0.040 77.2 × 10
c1 G1
τ 2 = 39.3 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
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PROBLEM 3.162
Two solid brass rods AB and CD are brazed to a brass sleeve EF.
Determine the ratio d 2 /d1 for which the same maximum shearing stress
occurs in the rods and in the sleeve.
SOLUTION
Let
c1 =
1
d1
2
Shaft AB:
τ1 =
Tc1
2T
=
J1
π c13
Sleeve EF.
τ2 =
Tc2
2Tc2
=
J2
π c24 − c14
and
(
2T
2Tc2
=
3
π c1
π c24 − c14
For equal stresses,
(
c2 =
1
d2
2
)
)
c24 − c14 = c13c2
Let x =
c2
c1
x 4 − 1 = x or
x = 41 + x
Solve by successive approximations starting with x0 = 1.0.
x1 =
4
2 = 1.189, x2 =
x4 =
4
2.220 = 1.221, x5 =
x = 1.221
4
2.189 = 1.216, x3 =
c2
= 1.221
c1
4
4
2.216 = 1.220
2.221 = 1.221 (converged).
d2
= 1.221 
d1
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
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