6.1 A solid circular steel shaft having an outside diameter of d = 0.75 in.

is subjected to a pure torque of

T = 650 lb-in. Determine the maximum shear stress in the shaft.

Solution
The polar moment of inertia for the shaft is

J (0.75 in.) 4 0.031063 in.4
32
The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (650 lb-in.)(0.75 in. / 2)
max 7,846.93 psi 7,850 psi Ans.
J 0.031063 in.4

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6.2 A hollow aluminum shaft with an outside diameter of 80 mm and a wall thickness of 5 mm has an
allowable shear stress of 75 MPa. Determine the maximum torque T that may be applied to the shaft.

Solution
The polar moment of inertia for the shaft is

J D 4 d 4 (80 mm)4 (70 mm) 4 1,664,062 mm 4
32 32
Rearrange the elastic torsion formula to determine the maximum torque T:
J (75 N/mm 2 )(1,664,062 mm 4 )
T allow 3,120,117 N-mm 3,120 N-m Ans.
c 80 mm / 2

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6.3 A hollow steel shaft with an outside diameter of 100 mm and a wall thickness of 10 mm is subjected
to a pure torque of T = 5,500 N-m.
(a) Determine the maximum shear stress in the hollow shaft.
(b) Determine the minimum diameter of a solid steel shaft for which the maximum shear stress is the
same as in part (a) for the same torque T.

Solution
The polar moment of inertia for the shaft is

J D 4 d 4 (100 mm)4 (80 mm) 4 5,796, 238 mm 4
32 32
(a) The maximum shear stress in the hollow steel shaft is found from the elastic torsion formula:
Tc (5,500 N-m)(100 mm / 2)(1,000 mm/m)
max 47.445 MPa 47.4 MPa Ans.
J 5,796, 238 mm 4

(b) The polar moment of inertia for a solid shaft can be expressed as

J d4
32
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
d4 T

32 (d / 2)
and simplify to
d 3 T

16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

To support a torque of T = 5,500 N-m without exceeding the maximum shear stress determined in part
(a), a solid shaft must have a diameter of
16T 16(5,500 N-m)(1,000 mm/m)
d3 3 83.891 mm 83.9 mm Ans.
(47.445 N/mm2 )

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6.4 A compound shaft consists of two pipe
segments. Segment (1) has an outside diameter of
200 mm and a wall thickness of 10 mm. Segment
(2) has an outside diameter of 150 mm and a wall
thickness of 10 mm. The shaft is subjected to
torques TB = 42 kN-m and TC = 18 kN-m, which act
in the directions shown in Fig. P6.4. Determine the
maximum shear stress magnitude in each shaft
segment.
Fig. P6.4

Solution

Equilibrium:
M x T1 42 kN-m 18 kN-m 0 T1 24 kN-m
M x T2 18 kN-m 0 T2 18 kN-m

Section properties:

J1 D14 d14 (200 mm) 4 (180 mm) 4 54,019,686 mm 4
32 32

J2 D24 d 24 (150 mm)4 (130 mm)4 21,551,281 mm 4
32 32

Shear stress magnitudes:

(24 kN-m)(200 mm / 2)(1,000 N/kN)(1,000 mm/m)
1 44.4 MPa Ans.
54,019,686 mm 4
(18 kN-m)(150 mm / 2)(1,000 N/kN)(1,000 mm/m)
2 62.3 MPa Ans.
21,661,281 mm 4

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6.5 A compound shaft consists of two pipe
segments. Segment (1) has an outside diameter of
10.75 in. and a wall thickness of 0.365 in. Segment
(2) has an outside diameter of 6.625 in. and a wall
thickness of 0.280 in. The shaft is subjected to
torques TB = 60 kip-ft and TC = 24 kip-ft, which act
in the directions shown in Fig. P6.5. Determine the
maximum shear stress magnitude in each shaft
segment.
Fig. P6.5

Solution

Equilibrium:
M x T1 60 kip-ft 24 kip-ft 0 T1 36 kip-ft
M x T2 24 kip-ft 0 T2 24 kip-ft

Section properties:

J1 D14 d14 (10.75 in.) 4 (10.02 in.) 4 321.4685 in.4
32 32

J2 D24 d 24 (6.625 in.)4 (6.0650 in.) 4 56.2844 in.4
32 32

Shear stress magnitudes:

(36 kip-ft)(10.75 in. / 2)(12 in./ft)
1 7.22 ksi Ans.
321.4685 in.4
(24 kip-ft)(6.625 in. / 2)(12 in./ft)
2 16.95 ksi Ans.
56.2844 in.4

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6.6 A compound shaft (Fig. P6.6) consists of brass segment (1) and
aluminum segment (2). Segment (1) is a solid brass shaft with an outside
diameter of 0.625 in. and an allowable shear stress of 6,000 psi. Segment
(2) is a solid aluminum shaft with an outside diameter of 0.50 in. and an
allowable shear stress of 9,000 psi. Determine the magnitude of the largest
torque TC that may be applied at C.

Fig. P6.6

Solution
Section properties:

J1 (0.625 in.) 4 0.014980 in.4
32

J2 (0.50 in.) 4 0.006136 in.4
32

Allowable internal torques:

J (6,000 psi)(0.014980 in.4 )
T1 1 1 287.621 lb-in.
c1 0.625 in./2
2 J2 (9,000 psi)(0.006136 in.4 )
T2 220.893 lb-in. controls
c2 0.50 in./2

Equilibrium:
The internal torque magnitude in each segment equals the external torque; therefore, T1 = T2 = TC. The
controlling internal torque is T2 = 220.893 lb-in.; therefore, the maximum external torque TC that may be
applied to the compound shaft is
TC 220.893 lb-in. 221 lb-in. Ans.

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6.7 A compound shaft (Fig. P6.7) consists of brass segment (1) and
aluminum segment (2). Segment (1) is a solid brass shaft with an
allowable shear stress of 60 MPa. Segment (2) is a solid aluminum shaft
with an allowable shear stress of 90 MPa. If a torque of TC = 23,000 N-m
is applied at C, determine the minimum required diameter of (a) the brass
shaft and (b) the aluminum shaft.

Fig. P6.7

Solution
The polar moment of inertia for a solid shaft can be expressed as

J d4
32
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
d4 T

32 (d / 2)
and simplify to
d 3 T

16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

Equilibrium:
For this shaft, the internal torque magnitude in each segment equals the external torque; therefore, T1 =
T2 = TC = 23,000 N-m.

Minimum shaft diameters:

(a) Brass shaft (1)
16T1 16(23,000 N-m)(1,000 mm/m)
d1 3 3 125.0 mm Ans.
allow,1 (60 N/mm2 )

(b) Aluminum shaft (2)

16T2 16(23,000 N-m)(1,000 mm/m)
d2 3 3 109.2 mm Ans.
allow,2 (90 N/mm 2 )

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6.8 A solid 0.75-in.-diameter shaft is subjected to
the torques shown in Fig. P6.8. The bearings shown
allow the shaft to turn freely.
(a) Plot a torque diagram showing the internal
torque in segments (1), (2), and (3) of the shaft. Use
the sign convention presented in Section 6-6.
(b) Determine the maximum shear stress magnitude
in the shaft.
Fig. P6.8

Solution
Equilibrium:

M x T1 10 lb-ft 0 T1 10 lb-ft
M x T2 10 lb-ft 50 lb-ft 0 T2 40 lb-ft
M x T3 30 lb-ft 0 T3 30 lb-ft

Section properties:

J1 J 2 J 3 (0.75 in.) 4 0.031063 in.4
32

Shear stress magnitudes:

T c (10 lb-ft)(0.75 in. / 2)(12 in./ft)
1 1 1 1, 448.7 psi
J1 0.031063 in.4
T2c2 (40 lb-ft)(0.75 in. / 2)(12 in./ft)
2 5,794.7 psi
J2 0.031063 in.4
T3c3 (30 lb-ft)(0.75 in. / 2)(12 in./ft)
3 4,346.0 psi
J3 0.031063 in.4

The maximum shear stress in the shaft occurs in segment (2):

max 5,790 psi Ans.

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6.9 A solid constant-diameter shaft is subjected to
the torques shown in Fig. P6.9. The bearings shown
allow the shaft to turn freely.
(a) Plot a torque diagram showing the internal
torque in segments (1), (2), and (3) of the shaft. Use
the sign convention presented in Section 6-6.
(b) If the allowable shear stress in the shaft is 80
MPa, determine the minimum acceptable diameter
for the shaft.
Fig. P6.9

Solution

M x 160 N-m T3 0 T3 160 N-m

M x 160 N-m 380 N-m T2 0 T2 220 N-m
M x 160 N-m 380 N-m 330 N-m T1 0 T1 110 N-m

The maximum torque magnitude in the shaft occurs in segment (2): Tmax = 220 N-m.

(b) The elastic torsion formula gives the relationship between shear stress and torque in a shaft.
Tc

J
In this instance, the torque and the allowable shear stress are known for the shaft. Rearrange the elastic
torsion formula, putting the known terms on the right-hand side of the equation:
J T

c
Express the left-hand side of this equation in terms of the shaft diameter D:

d4
32 d 3 T
d / 2 16

and solve for the minimum acceptable diameter:

16 T 16(220 N-m)(1,000 mm/m)
d3 14,005.635 mm3
(80 N/mm 2 )
d 24.1 mm Ans.

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6.10 A solid circular steel shaft having an outside diameter of 1.25 in. is subjected to a pure torque of T
= 2,200 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine:
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in a 6-ft length of shaft.

Solution
The polar moment of inertia for the shaft is

J (1.25 in.) 4 0.239684 in.4
32

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (2,200 lb-in.)(1.25 in. / 2)
max 5, 736.7 psi 5.74 ksi Ans.
J 0.239684 in.4

(b) The magnitude of the angle of twist in a 6-ft length of shaft is

TL (2, 200 lb-in.)(6 ft)(12 in./ft)
0.055072 rad 0.0551 rad 3.16 Ans.
JG (0.239684 in.4 )(12,000,000 psi)

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6.11 A solid circular steel shaft having an outside diameter of 35 mm is subjected to a pure torque of T =
640 N-m. The shear modulus of the steel is G = 80 GPa. Determine:
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in a 1.5-m length of shaft.

Solution
The polar moment of inertia for the shaft is

J (35 mm)4 147,323.515 mm4
32

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (640 N-m)(35 mm / 2)(1,000 mm/m)
max 76.023 MPa 76.0 MPa Ans.
J 147,323.515 mm 4

(b) The magnitude of the angle of twist in a 1.5-m length of shaft is

TL (640 N-m)(1.5 m)(1,000 mm/m) 2
0.081453 rad 0.0815 rad 4.67 Ans.
JG (147,323.515 mm 4 )(80,000 N/mm 2 )

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6.12 A hollow steel shaft with an outside diameter of 85 mm and a wall thickness of 10 mm is subjected
to a pure torque of T = 7,000 N-m. The shear modulus of the steel is G = 80 GPa. Determine:
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in a 2.5-m length of shaft.

Solution
The polar moment of inertia for the shaft is

J D 4 d 4 (85 mm)4 (65 mm) 4 3,372,303 mm 4
32 32

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (7,000 N-m)(85 mm / 2)(1,000 mm/m)
max 88.219 MPa 88.2 MPa Ans.
J 3,372,303 mm 4

(b) The magnitude of the angle of twist in a 2.5-m length of shaft is

TL (7,000 N-m)(2.5 m)(1,000 mm/m) 2
0.064867 rad 0.0649 rad 3.72 Ans.
JG (3,372,303 mm 4 )(80,000 N/mm 2 )

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6.13 A solid stainless steel [G = 12,500 ksi] shaft that is 72-in. long will be subjected to a pure torque of
T = 900 lb-in. Determine the minimum diameter required if the shear stress must not exceed 8,000 psi
and the angle of twist must not exceed 5. Report both the maximum shear stress and the angle of twist
at this minimum diameter.

Solution
Consider shear stress:
The polar moment of inertia for a solid shaft can be expressed as

J d4
32
The elastic torsion formula can be rearranged to gather terms with d:
d4 d 3 T

32 ( d / 2) 16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:
16(900 lb-in.)
d3 0.831 in.
(8,000 psi)

Consider angle of twist:

Rearrange the angle of twist equation:
TL 4 TL
J d
JG 32 G
and solve for the minimum diameter that will satisfy the angle of twist limitation:
32TL 32(900 lb-in.)(72 in.)
d4 4 0.882 in.
G (5)( rad/180)(12,500,000 psi)

Therefore, the minimum diameter that could be used for the shaft is
dmin 0.882 in. Ans.

The angle of twist for this shaft is = 5 = 0.087266 rad. To compute the shear stress in a 0.882-in.-
diameter shaft, first compute the polar moment of inertia:

J (0.882 in.)4 0.059404 in.4
32
The shear stress in the shaft is thus:
(900 lb-in.)(0.882 in. / 2)
max 6,681.12 psi 6,680 psi Ans.
0.059404 in.4

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6.14 A solid stainless steel [G = 86 GPa] shaft that is 2.0 m long will be subjected to a pure torque of T =
75 N-m. Determine the minimum diameter required if the shear stress must not exceed 50 MPa and the
angle of twist must not exceed 4. Report both the maximum shear stress and the angle of twist at
this minimum diameter.

Solution
Consider shear stress:
The polar moment of inertia for a solid shaft can be expressed as

J d4
32
The elastic torsion formula can be rearranged to gather terms with d:
d4 d 3 T

32 ( d / 2) 16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:
16(75 N-m)(1,000 mm/m)
d3 19.70 mm
(50 N/mm2 )

Consider angle of twist:

Rearrange the angle of twist equation:
TL 4 TL
J d
JG 32 G
and solve for the minimum diameter that will satisfy the angle of twist limitation:
32TL 32(75 N-m)(2,000 mm)(1,000 mm/m)
d4 4 22.46 mm
G (0.069813 rad)(86,000 N/mm 2 )

Therefore, the minimum diameter that could be used for the shaft is
dmin 22.5 mm Ans.

The angle of twist for this shaft is = 0.069813 rad. To compute the shear stress in a 22.46-mm-
diameter shaft, first compute the polar moment of inertia:

J (22.46 mm) 4 24,983.625 mm 4
32
The shear stress in the shaft is thus:
(75 N-m)(22.46 mm / 2)(1,000 mm/m)
max 33.7 MPa Ans.
24,983.625 mm 4

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6.15 A hollow steel [G = 12,000 ksi] shaft with an outside diameter of 3.50 in. will be subjected to a
pure torque of T = 3,750 lb-ft. Determine the maximum inside diameter d that can be used if the shear
stress must not exceed 8,000 psi and the angle of twist must not exceed 3 in an 8-ft length of shaft.
Report both the maximum shear stress and the angle of twist for this maximum inside diameter.

Solution
Consider shear stress: The polar moment of inertia for a hollow shaft can be expressed as

J D 4 d 4
32
The elastic torsion formula can be rearranged to gather terms with D:
D d T
4 4

32 D / 2
Rearrange this equation to isolate the inside diameter d term:
32 T D 32 T D
D4 d 4 d 4 D4
2 2
From this equation, the unknown inside diameter of the hollow shaft can be expressed as
32 T D
d 4 D4
2
For the hollow steel shaft, the maximum diameter that will satisfy the allowable shear stress is:
32(3,750 lb-ft)(3.50 in.)(12 in./ft)
d 4 (3.50 in.)4 2.6564 in.
2 (8,000 psi)

Consider angle of twist: Rearrange the angle of twist equation:

TL TL
J D 4 d 4
JG 32 G
Rearrange this equation to isolate the inside diameter d term:
32TL
d 4 D4
G
and solve for the maximum inside diameter that will satisfy the angle of twist limitation:
32 TL 32(3,750 lb-ft)(8 ft)(12 in./ft) 2
d 4 D4 4 (3.50 in.) 4 2.991 in.
G (3)( /180)(12,000,000 psi)
Therefore, the maximum inside diameter that could be used for the shaft is
dmax 2.66 in. Ans.

The shear stress for this shaft is = 8,000 psi. To compute the angle of twist in an 8-ft length of the
hollow shaft, first compute the polar moment of inertia:

J D 4 d 4 (3.50 in.)4 (2.66 in.) 4 9.817318 in.4
32 32
The angle of twist in the shaft is thus:
TL (3,750 lb-ft)(8 ft)(12 in./ft) 2
0.0367 rad 2.10 Ans.
JG (9.817318 in.4 )(12,000,000 psi)

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6.16 A compound steel [G = 80 GPa] shaft
(Fig. P6.16) consists of a solid 55-mm-
diameter segment (1) and a solid 40-mm-
diameter segment (2). The allowable shear
stress of the steel is 70 MPa, and the
maximum rotation angle at the free end of
the compound shaft must be limited to C
3. Determine the magnitude of the largest
torque TC that may be applied at C.

Fig. P6.16

Solution
Section properties:

J1 (55 mm)4 898,360.5 mm4
32

J2 (40 mm) 4 251,327.4 mm 4
32

Consider shear stress: The larger shear stress will occur in the smaller diameter segment; that is,
segment (2).
J (70 N/mm 2 )(251,327.4 mm 4 )
Tmax 879,646 N-mm
c 40 mm/2

Consider angle of twist: The rotation angle at C is the sum of the angles of twist in segments (1) and
(2):
TL TL
C 1 2 1 1 2 2
J1G1 J 2G2
Since T1 = T2 = TC and G1 = G2, this equation can be simplified to
Tmax L1 L2
3
G J1 J 2
and solved for the maximum torque:
(3)( / 180)(80,000 N/mm 2 )
Tmax 739,395 N-mm
800 mm 1,200 mm
898,360.5 mm4 251,327.4 mm4

Therefore, the magnitude of the largest torque TC that may be applied at C is

TC ,max 739,395 N-mm 739 N-m Ans.

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6.17 A compound shaft (Fig. P6.17) consists of brass segment (1) and
aluminum segment (2). Segment (1) is a solid brass [G = 5,600 ksi]
shaft with an outside diameter of 1.75 in. and an allowable shear stress
of 9,000 psi. Segment (2) is a solid aluminum [G = 4,000 ksi] shaft
with an outside diameter of 1.25 in. and an allowable shear stress of
12,000 psi. The maximum rotation angle at the upper end of the
compound shaft must be limited to C 4. Determine the magnitude
of the largest torque TC that may be applied at C.

Fig. P6.17

Solution
From equilibrium, the internal torques in segments (1) and (2) are equal to the external torque TC. The
elastic torsion formula gives the relationship between shear stress and torque in a shaft.
Tc

J
Consider shear stress: In this compound shaft, the diameters and allowable shear stresses in segments
(1) and (2) are known. The elastic torsion formula can be rearranged to solve for the unknown torque.
An expression can be written for each shaft segment:
J J
T1 1 1 T2 2 2
c1 c2
For shaft segment (1), the polar moment of inertia is:

J1 (1.75 in.)4 0.920772 in.4
32
Use this value along with the 9,000 psi allowable shear stress to determine the allowable torque T1:
J (9,000 psi)(0.920772 in.4 )
T1 1 1 9,470.8 lb-in. (a)
c1 (1.75 in./2)

For shaft segment (2), the polar moment of inertia is:

J2 (1.25 in.) 4 0.239684 in.4
32
Use this value along with the 12,000 psi allowable shear stress to determine the allowable torque T2:
2 J 2 (12,000 psi)(0.239684 in.4 )
T2 4,601.9 lb-in. (b)
c2 (1.25 in./2)

Consider angle of twist: The angles of twists in segments (1) and (2) can be expressed as:
TL TL
1 1 1 2 2 2
J1G1 J 2G2

The rotation angle at C is the sum of these two angles of twist:

TL TL
C 1 2 1 1 2 2
J1G1 J 2G2

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and since T1 = T2 = TC:
L L
C TC 1 2
J1G1 J 2G2

Solving for TC gives:

C
TC
L1 L
2
J1G1 J 2G2
rad
(4)
180
3,308.4 lb-in. (c)
12 in. 18 in.

(0.920772 in. )(5,600,000 psi) (0.239684 in.4 )(4,000,000 psi)
4

Compare the torque magnitudes in Eqs. (a), (b), and (c). The smallest torque controls; therefore, the
maximum torque that can be applied to the compound shaft at C is TC = 3,308.4 lb-in. = 276 lb-ft. Ans.

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6.18 A compound steel [G = 80 GPa] shaft
consists of solid 30-mm-diameter segments
(1) and (3) and tube segment (2), which has
an outside diameter of 60 mm and an inside
diameter of 50 mm (Fig. P6.18). Determine:
(a) the maximum shear stress in tube
segment (2).
(b) the angle of twist in tube segment (2).
(c) the rotation angle of gear D relative to
gear A.
Fig. P6.18

Solution
The torques in all shaft segments are equal; therefore, T1 = T2 = T3 = 240 N-m.

Polar moments of inertia in the shaft segments will be needed for this calculation. For segments (1) and
(3), which are solid 30-mm-diameter shafts, the polar moment of inertia is:

J1 (30 mm) 4 79,521.56 mm 4 J 3
32
For segment (2), which has a tube cross section, the polar moment of inertia is:

J2 (60 mm) 4 (50 mm) 4 658,752.71 mm 4
32

Shear stress in tube segment (2): Use the elastic torsion formula to calculate the shear stress caused by
an internal torque of T2 = 240 N-m. Note that the radius term used in the elastic torsion formula for the
tube is the outside radius; that is, c2 = 60 mm/2 = 30 mm.
Tc (240 N-m)(30 mm)(1,000 mm/m)
2 2 2 10.93 MPa Ans.
J2 658,752.71 mm 4

Angle of twist in tube segment (2): Apply the angle of twist equation to segment (2).
TL (240 N-m)(2,000 mm)(1,000 mm/m)
2 2 2 0.009108 rad 0.00911 rad Ans.
J 2G2 (658,752.71 mm 4 )(80,000 N/mm 2 )

Rotation angle of gear D relative to gear A: The angles of twist in segments (1) and (3) must be
calculated. Since both segments have the same shear modulus, polar moment of inertia, and length, they
will both have the same angle of twist:
TL (240 N-m)(300 mm)(1,000 mm/m)
1 1 1 0.011318 rad 3
J1G1 (79,521.56 mm 4 )(80,000 N/mm 2 )

Since gear A is the origin of the coordinate system for this problem, we will arbitrarily define the
rotation angle at gear A to be zero; that is, A = 0. The rotation angle of gear D relative to gear A is
found by adding the angles of twist for the three segments to A:
D A 1 2 3
0 0.011318 rad 0.009108 rad 0.011318 rad
0.031743 rad 0.0317 rad Ans.

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6.19 A compound steel [G = 80 GPa] shaft
consists of solid 40-mm-diameter segments
(1) and (3) and tube segment (2), which has
an outside diameter of 75 mm (Fig. P6.19).
If the rotation of gear D relative to gear A
must not exceed 0.01 rad, determine the
maximum inside diameter that may be used
for tube segment (2).

Fig. P6.19

Solution
The polar moment of inertia for shaft segments (1) and (3) is:

J1 J 3 (40 mm) 4 251,327 mm 4
32
The rotation of gear D relative to gear A is equal to the sum of the angles of twist in segments (1), (2),
and (3):
D 1 2 3
The angles of twist in segments (1) and (3) is
(240 N-m)(300 mm)(1,000 mm/m)
1 3 0.003581 rad
(251,327 mm 4 )(80,000 N/mm 2 )
Therefore, the angle of twist that can allowed for segment (2) is
D 1 2 3 0.01 rad 2 0.01 rad 2(0.003581 rad) 0.002838 rad

The minimum polar moment of inertia required for segment (2) is thus:
T2 L2 (240 N-m)(2 m)(1,000 mm/m)2
0.002838 rad J2 2
2,114,165 mm4
J 2G2 (80,000 N/mm )(0.002838 rad)
Since the outside diameter of segment (2) is D2 = 75 mm, the maximum inside diameter d2 is:

D24 d 24 (75 mm)4 d 24 2,114,165 mm4
32 32
d 2 56.4 mm Ans.

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6.20 The compound shaft shown in Fig. P6.20 consists of aluminum segment (1) and steel segment (2).
Aluminum segment (1) is a tube with an outside diameter of D1 = 4.00 in., a wall thickness of t1 = 0.25
in., and a shear modulus of G1 = 4,000 ksi. Steel segment (2) is a tube with an outside diameter of D2 =
2.50 in., a wall thickness of t2 = 0.125 in., and a shear modulus of G2 = 12,000 ksi. The compound shaft
is subjected to torques applied at B and C, as shown in Fig. P6.20.
(a) Prepare a diagram that shows the internal
torque and the maximum shear stress in
segments (1), and (2) of the shaft. Use the sign
convention presented in Section 6-6.
(b) Determine the rotation angle of B with
respect to the support at A.
(c) Determine the rotation angle of C with
respect to the support at A.

Fig. P6.20

Solution
Section properties: Polar moments of inertia in the shaft segments will be needed for this calculation.

J1 D14 d14 (4.00 in.) 4 (3.50 in.) 4 10.4004 in.4
32 32

J2 D24 d 24 (2.50 in.) 4 (2.25 in.) 4 1.3188 in.4
32 32

(a) Equilibrium

M x 950 lb-ft 2,100 lb-ft T1 0

T1 1,150 lb-ft

M x 950 lb-ft T2 0
T2 950 lb-ft

Shear stress:
T c (1,150 lb-ft)(4.00 in./2)(12 in./ft)
1 1 1 2,653.75 psi 2,650 psi Ans.
J1 10.4004 in.4
T2c2 ( 950 lb-ft)(2.50 in./2)(12 in./ft)
2 10,804.95 psi 10,800 psi Ans.
J2 1.3188 in.4

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(b) Rotation angle of B with respect to A:
1 B A B 0 B
Therefore,
TL (1,150 lb-ft)(9 ft)(12 in./ft)2
B 1 1 1 0.035826 rad 0.0358 rad Ans.
J1G1 (10.4004 in.4 )(4,000,000 psi)

(c) Rotation angle of C with respect to A:

2 C B C B 2
The angle of twist in shaft (2) is
T2 L2 (950 lb-ft)(6 ft)(12 in./ft)2
2 0.051864 rad
J 2G2 (1.3188 in.4 )(12,000,000 psi)
and thus, the rotation angle at C is
C B 2 0.035826 rad (0.051864 rad) 0.016038 rad 0.01604 rad Ans.

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6.21 A solid 1.00-in.-diameter steel [G =
12,000 ksi] shaft is subjected to the torques
shown in Fig. P6.21.
(a) Prepare a diagram that shows the internal
torque and the maximum shear stress in
segments (1), (2), and (3) of the shaft. Use
the sign convention presented in Section 6-6.
(b) Determine the rotation angle of pulley C
with respect to the support at A.
(c) Determine the rotation angle of pulley D
with respect to the support at A.
Fig. P6.21

Solution
Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments
will be needed for this calculation.

J1 d14 (1.00 in.)4 0.098175 in.4 J 2 J 3
32 32

(a) Equilibrium

M x 240 lb-ft 280 lb-ft 130 lb-ft T1 0

T1 90 lb-ft

M x 280 lb-ft 130 lb-ft T2 0

T2 150 lb-ft

M x 130 lb-ft T3 0
T3 130 lb-ft

Shear stress:
T c ( 90 lb-ft)(1.00 in./2)(12 in./ft)
1 1 1 5,500.4 psi 5,500 psi
J1 0.098175 in.4
T2c2 (150 lb-ft)(1.00 in./2)(12 in./ft)
2 9,167.3 psi 9,170 psi
J2 0.098175 in.4
T3c3 ( 130 lb-ft)(1.00 in./2)(12 in./ft)
3 7,945.0 psi 7,950 psi
J3 0.098175 in.4
The maximum shear stress in the entire shaft is max = 9,170 psi. Ans.

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(b) Rotation angle of C with respect to A:
The angles of twist in the three shaft segments are:
TL ( 90 lb-ft)(36 in.)(12 in./ft)
1 1 1 0.033002 rad
J1G1 (0.098175 in.4 )(12,000,000 psi)
TL (150 lb-ft)(36 in.)(12 in./ft)
2 2 2 0.055004 rad
J 2G2 (0.098175 in.4 )(12,000,000 psi)
T3 L3 ( 130 lb-ft)(36 in.)(12 in./ft)
3 0.047670 rad
J 3G3 (0.098175 in.4 )(12,000,000 psi)
The rotation angle of C with respect to A is found from the sum of the angles of twist in segments (1)
and (2):
C 1 2 0.033002 rad 0.055004 rad 0.022002 rad 0.0220 rad Ans.

(c) Rotation angle of D with respect to A:

The rotation angle of D with respect to A is found from the sum of the angles of twist in segments (1),
(2), and (3):
D 1 2 3
0.033002 rad 0.055004 rad ( 0.047670 rad)
0.025669 rad 0.0257 rad Ans.

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6.22 A compound shaft supports several
pulleys as shown in Fig. P6.22. Segments
(1) and (4) are solid 25-mm-diameter steel
[G = 80 GPa] shafts. Segments (2) and (3)
are solid 50-mm-diameter steel shafts. The
bearings shown allow the shaft to turn
freely.
(a) Determine the maximum shear stress in
the compound shaft.
(b) Determine the rotation angle of pulley D
with respect to pulley B.
(c) Determine the rotation angle of pulley E
with respect to pulley A.
Fig. P6.22

Solution
Section properties: The polar moments of inertia for the shaft segments will be needed for this
calculation.

J1 d14 (25 mm) 4 38,349.52 mm 4 J 4
32 32

J2 d 24 (50 mm)4 613,592.32 mm 4 J 3
32 32

(a) Equilibrium

M x T1 175 N-m 0
T1 175 N-m

M x T2 2,100 N-m 175 N-m 0

T2 1,925 N-m

M x 650 N-m 225 N-m T3 0

T3 425 N-m

M x 225 N-m T4 0
T4 225 N-m

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Shear stress:
T c ( 175 N-m)(25 mm/2)(1,000 mm/m)
1 1 1 57.041 MPa 57.0 MPa
J1 38,349.52 mm 4
Tc (1,925 N-m)(50 mm/2)(1,000 mm/m)
2 2 2 78.432 MPa 78.4 MPa
J2 613,592.32 mm 4
T3c3 (425 N-m)(50 mm/2)(1,000 mm/m)
3 17.316 MPa 17.32 MPa
J3 613,592.32 mm 4
T4c4 ( 225 N-m)(25 mm/2)(1,000 mm/m)
4 73.339 MPa 73.3 MPa
J4 38,349.52 mm 4

The maximum shear stress in the entire shaft is max = 78.4 MPa. Ans.

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(b) Angles of twist:
The angles of twist in the four shaft segments are:
TL ( 175 N-m)(750 mm)(1,000 mm/m)
1 1 1 0.042781 rad
J1G1 (38,349.52 mm 4 )(80,000 N/mm 2 )
T2 L2 (1,925 N-m)(500 mm)(1,000 mm/m)
2 0.019608 rad
J 2G2 (613,592.32 mm 4 )(80,000 N/mm 2 )
T3 L3 (425 N-m)(625 mm)(1,000 mm/m)
3 0.005411 rad
J 3G3 (613,592.32 mm 4 )(80,000 N/mm 2 )
T4 L4 ( 225 N-m)(550 mm)(1,000 mm/m)
4 0.040336 rad
J 4G4 (38,349.52 mm 4 )(80,000 N/mm 2 )

Rotation angle of pulley D with respect to pulley B: The rotation angle of D with respect to B is
found from the sum of the angles of twist in segments (2) and (3):
D / B 2 3 0.019608 rad 0.005411 rad 0.025019 rad 0.0250 rad Ans.

(c) Rotation angle of pulley E with respect to pulley A: The rotation angle of E with respect to A is
found from the sum of the angles of twist in all four segments:
E 1 2 3 4
0.042781 rad 0.019608 rad 0.005411 rad ( 0.040336 rad)
0.058098 rad 0.0581 rad Ans.

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6.23 A solid steel [G = 80 GPa] shaft of variable
diameter is subjected to the torques shown in Fig.
P6.23. The diameter of the shaft in segments (1) and
(3) is 50 mm, and the diameter of the shaft in
segment (2) is 80 mm. The bearings shown allow
the shaft to turn freely.
(a) Determine the maximum shear stress in the
compound shaft.
(b) Determine the rotation angle of pulley D with
respect to pulley A.
Fig. P6.23

Solution
Section properties: The polar moments of inertia for the shaft segments will be needed for this
calculation.

J1 (50 mm) 4 613,592.32 mm 4 J 3
32

J2 (80 mm) 4 4,021, 238.60 mm 4
32

(a) Equilibrium

M x 1, 200 N-m T1 0
T1 1, 200 N-m

M x 1, 200 N-m 4,500 N-m T2 0

T2 3,300 N-m

M x T3 500 N-m 0
T3 500 N-m

Shear stress:
T c (1, 200 N-m)(50 mm/2)(1,000 mm/m)
1 1 1 48.892 MPa 48.9 MPa
J1 613,592.32 mm 4
T2c2 ( 3,300 N-m)(80 mm/2)(1,000 mm/m)
2 32.826 MPa 32.8 MPa
J2 4,021,238.60 mm 4
T3c3 ( 500 N-m)(50 mm/2)(1,000 mm/m)
3 20.372 MPa 20.4 MPa
J3 613,592.32 mm 4

The maximum shear stress in the compound shaft is max 1 48.9 MPa Ans.

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Angles of twist:
The angles of twist in the three shaft segments are:
T1L1 (1,200 N-m)(0.7 m)(1,000 mm/m)2
1 0.017112 rad
J1G1 (613,592.32 mm4 )(80,000 N/mm2 )
T2 L2 (3,300 N-m)(1.8 m)(1,000 mm/m) 2
2 0.018464 rad
J 2G2 (4,021,238.60 mm4 )(80,000 N/mm2 )
T3 L3 ( 500 N-m)(0.7 m)(1,000 mm/m)2
3 0.007130 rad
J 3G3 (613,592.32 mm4 )(80,000 N/mm2 )

(b) Rotation angles:

A 0 rad
B A 1 0 rad 0.017112 rad 0.017112 rad
C B 2 0.017112 rad (0.018464 rad) 0.001352 rad
D C 3 0.001352 rad ( 0.007130 rad) 0.008482 rad Ans.

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6.24 A compound shaft drives three gears,
as shown in Fig. P6.24. Segments (1) and
(2) of the compound shaft are hollow
aluminum [G = 4,000 ksi] tubes, which
have an outside diameter of 3.00 in. and a
wall thickness of 0.25 in. Segments (3) and
(4) are solid 2.00-in.-diameter steel [G =
12,000 ksi] shafts. The bearings shown
allow the shaft to turn freely.
(a) Determine the maximum shear stress in
the compound shaft.
(b) Determine the rotation angle of flange
C with respect to flange A.
(c) Determine the rotation angle of gear E
with respect to flange A.
Fig. P6.24

Solution
Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments
will be needed for this calculation.

J1 D14 d14 (3.00 in.)4 (2.50 in.)4 4.117204 in.4 J 2
32 32

J3 d34 (2.00 in.) 4 1.570796 in.4 J 4
32 32

(a) Equilibrium

M x 14 kip-in. 42 kip-in.
35 kip-in. T1 0
T1 7 kip-in.

M x 14 kip-in. 42 kip-in. T2 0
T2 28 kip-in.

M x 14 kip-in. 42 kip-in. T3 0
T3 28 kip-in.

M x 14 kip-in. T4 0
T4 14 kip-in.

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Shear stress:
T c ( 7 kip-in.)(3.00 in./2)
1 1 1 2.55 ksi
J1 4.117204 in.4
Tc ( 28 kip-in.)(3.00 in./2)
2 2 2 10.20 ksi
J2 4.117204 in.4
T3c3 ( 28 kip-in.)(2.00 in./2)
3 17.83 ksi
J3 1.570796 in.4
T4c4 (14 kip-in.)(2.00 in./2)
4 8.91 ksi
J4 1.570796 in.4
The maximum shear stress in the compound shaft is max 3 17.83 ksi Ans.

(b) Angles of Twist: The angles of twist in the four shaft segments are:
TL (7 kip-in.)(60 in.)
1 1 1 0.025503 rad
J1G1 (4.117204 in.4 )(4,000 ksi)
T2 L2 ( 28 kip-in.)(6 in.)
2 0.010201 rad
J 2G2 (4.117204 in.4 )(4,000 ksi)
T3 L3 ( 28 kip-in.)(18 in.)
3 0.026738 rad
J 3G3 (1.570796 in.4 )(12,000 ksi)
T4 L4 (14 kip-in.)(12 in.)
4 0.008913 rad
J 4G4 (1.570796 in.4 )(12,000 ksi)

A 0 rad
B A 1 0 rad 0.025503 rad 0.0255 rad
C B 2 0.025503 rad ( 0.010201 rad) 0.01530 rad
D C 3 0.015302 rad ( 0.026738 rad) 0.01144 rad
E D 4 0.011436 rad 0.008913 rad 0.00252 rad

The rotation angle of flange C with respect to A is

C 0.01530 rad Ans.

(c) Rotation angle of E with respect to A:

E 0.00252 rad Ans.

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6.25 A compound shaft drives several
pulleys, as shown in Fig. P6.25. Segments (1)
and (2) of the compound shaft are hollow
aluminum [G = 4,000 ksi] tubes, which have
an outside diameter of 3.00 in. and a wall
thickness of 0.125 in. Segments (3) and (4)
are solid 1.50-in.-diameter steel [G = 12,000
ksi] shafts. The bearings shown allow the
shaft to turn freely.
(a) Determine the maximum shear stress in
the compound shaft.
(b) Determine the rotation angle of flange C
with respect to pulley A.
(c) Determine the rotation angle of pulley E
with respect to pulley A.
Fig. P6.25

Solution
(a) Equilibrium

M x 95 lb-ft T4 0 T4 95 lb-ft

M x 95 lb-ft 270 lb-ft T3 0 T3 175 lb-ft

Note: The internal torque in shaft segment (2) is the same as

in segment (3); therefore, T2 = 175 lb-ft.

M x T1 650 lb-ft 0 T1 650 lb-ft

Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2)
are hollow tubes with an outside diameter of 3.00 in. and an inside diameter of 3.00 in. 2(0.125 in.) =
2.75 in. The polar moment of inertia of segments (1) and (2) is:

J1 D14 d14 (3.00 in.) 4 (2.75 in.) 4 2.33740 in.4 J 2
32 32
Segments (3) and (4) are solid 1.50-in.-diameter shafts, which have a polar moment of inertia of:

J3 d34 (1.50 in.)4 0.49701 in.4 J 4
32 32

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Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula:
T c (650 lb-ft)(1.50 in.)(12 in./ft)
1 1 1 5,005.56 psi 5,010 psi
J1 2.33740 in.4
Tc ( 175 lb-ft)(1.50 in.)(12 in./ft)
2 2 2 1,347.6 psi 1,348 psi
J2 2.33740 in.4
T3c3 ( 175 lb-ft)(0.75 in.)(12 in./ft)
3 3,169.0 psi 3,170 psi
J3 0.49701 in.4
T4c4 (95 lb-ft)(0.75 in.)(12 in./ft)
4 1,720.3 psi 1,720 psi
J4 0.49701 in.4
The maximum shear stress in the compound shaft is max 1 5,010 psi Ans.

Angles of twist:
TL (650 lb-ft)(72 in.)(12 in./ft)
1 1 1 0.060067 rad
J1G1 (2.33740 in.4 )(4,000,000 lb/in.2 )
TL ( 175 lb-ft)(36 in.)(12 in./ft)
2 2 2 0.008086 rad
J 2G2 (2.33740 in.4 )(4,000,000 lb/in.2 )
T3 L3 ( 175 lb-ft)(54 in.)(12 in./ft)
3 0.019014 rad
J 3G3 (0.49701 in.4 )(12,000,000 lb/in.2 )
T4 L4 (95 lb-ft)(54 in.)(12 in./ft)
4 0.010322 rad
J 4G4 (0.49701 in.4 )(12,000,000 lb/in.2 )

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each
segment:
1 B A 2 C B 3 D C 4 E D
The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle
at pulley A to be zero (A = 0). The rotation angle at B can be calculated from the angle of twist in
segment (1):
1 B A
B A 1 0 0.060067 rad 0.0601 rad
Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation
angle of pulley B:
2 C B
C B 2 0.060067 rad (0.008086 rad) 0.051981 rad 0.0520 rad
The rotation angle at D is:
3 D C
D C 3 0.051981 rad (0.019014 rad) 0.032967 rad 0.0330 rad
and the rotation angle at E is:
4 E D
E D 4 0.032967 rad 0.010322 rad 0.043289 rad 0.0433 rad

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(b) Rotation angle of flange C with respect to pulley A: Using the rotation angles determined for the
system, the rotation angle of flange C with respect to pulley A is simply:
C 0.051981 rad 0.0520 rad Ans.

(c) Rotation angle of pulley E with respect to pulley A: Using the rotation angles determined for the
system, the rotation angle of flange C with respect to pulley A is simply:
E 0.043289 rad 0.0433 rad Ans.

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