PROBLEM 3.31
Determine the largest allowable diameter of a 3-m-long steel rod (G
through 30° without exceeding a shearing stress of 80 MPa.
77.2 GPa) if the rod is to be twisted
SOLUTION
L
3 m,
TL
,
GJ
c
T
30
523.6 10 3 rad,
180
GJ
Tc GJ c
,
L
J
JL
(80 106 )(3.0)
(77.2 109 )(523.6 10 3 )
80 106 Pa
G c
,
L
5.9374 10 3 m
c
L
G
5.9374 mm
d
2c
11.87
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291
PROBLEM 3.35
300 N · m
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.
D
200 N · m
C
48 mm
0.9 m
B
44 mm
A
1.2 m
40 mm
1m
SOLUTION
(a)
Angle of twist between B and C.
TBC
c
J BC
B/C
(b)
200 N m, LBC
1
d
2
2
c4
TL
GJ
1.2 m
27 109 Pa
0.022 m, G
367.97 10 9 m
(200)(1.2)
(27 109 )(367.97 109 )
24.157 10 3 rad
B /C
1.384
Angle of twist between B and D.
500 N m, LCD
J CD
(0.024)4 521.153 10 9 m 4
2
(500)(0.9)
31.980 10 3 rad
(27 109 )(521.153 109 )
C /D
B /D
2
0.9 m, c
1
d
2
TCD
0.024 m, G
27 109 Pa
c4
B /C
C /D
24.157 10
3
31.980 10
3
56.137 10 3 rad
B /D
3.22
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295
PROBLEM 3.38
60 mm
TB ! 1600 N · m
The aluminum rod AB (G 27 GPa) is bonded to the brass
rod BD (G 39 GPa). Knowing that portion CD of the
brass rod is hollow and has an inner diameter of 40 mm,
determine the angle of twist at A.
D
36 mm
C
TA ! 800 N · m
250 mm
B
375 mm
A
400 mm
SOLUTION
Rod AB:
G
27 109 Pa, L
T
800 N m c
J
c4
2
TL
GJ
A /B
Part BC:
G
39 109 Pa
T
800
B /C
TL
GJ
Part CD:
c1
c2
J
C/D
Angle of twist at A.
1600
2
0.400 m
1
d
2
0.018 m
(0.018) 4
164.896 10 9 m
(800)(0.400)
(27 109 )(164.896 10 9 )
L
0.375 m, c
1
d
2
2400 N m, J
2
1
d1
2
1
d2
2
2
TL
GJ
c24
0.030 m
c4
(2400)(0.375)
(39 109 )(1.27234 10 6 )
71.875 10 3 rad
2
(0.030)4
1.27234 10 6 m 4
18.137 10 3 rad
0.020 m
0.030 m, L
c14
0.250 m
(0.0304
0.0204 )
2
(2400)(0.250)
(39 109 )(1.02102 10 6 )
A
A /B
1.02102 10 6 m 4
15.068 10 3 rad
B/C
C /D
3
105.080 10 rad
A
6.02
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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298
PROBLEM 3.51
A
Aluminum
12 in.
1.5 in.
B
T ! 12.5 kip · in.
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.
Brass
2.0 in.
18 in.
C
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation
Cylinder AB:
c
J
B
Cylinder BC:
c
J
B
Matching expressions for
B:
1
d
2
c4
2
TAB L
GJ
1
d
2
c4
2
TBC L
GJ
0.75 in.
for each cylinder.
12 in.
L
0.49701 in 4
TAB (12)
(3.7 106 )(0.49701)
1.0 in.
2
(1.0) 4
L
18 in.
TBC (18)
(5.6 106 )(1.5708)
6.5255 10 6TAB
TAB
TBC
TAB
6.5255 10 6TAB
1.5708 in 4
TBC
Equilibrium of connection at B :
B
T
TBC
2.0463 10 6TBC
2.0463 10 6TBC
3.1889 TAB
0
T
12.5 103
(1)
12.5 103 lb in.
(2)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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311
PROBLEM 3.51 (Continued)
4.1889 TAB
Substituting (1) into (2),
TAB
(a)
TBC
9.5159 103 lb in.
Maximum stress in cylinder AB.
AB
(b)
2.9841 103 lb in.
12.5 103
TABc
J
(2.9841 103 )(0.75)
0.49701
4.50 103 psi
AB
4.50 ksi
BC
6.06 ksi
Maximum stress in cylinder BC.
BC
TBC c
J
(9.5159 103 ) (1.0)
1.5708
6.06 103 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
312