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PROBLEM 10.29
The torques shown are exerted on pulleys A and B. Knowing
that the shafts are solid and made of steel (G = 77.2 GPa),
determine the angle of twist between (a) A and B, (b) A and C.
SOLUTION
(a)
4 94
3
99
1
300 N m, 0.9 m, 0.015 m
2
(0.015) 79.522 10 m
2
(300)(0.9) 43.980 10 rad
(77.2 10 )(79.522 10 )
π
ϕ
−
−
−
=⋅= ==
= = ×
= = = ×
××
AB AB AB
AB
AB AB
AB
T L Cd
J
TL
GJ
2.52
AB
ϕ
= °
(b)
4 94
3
99
1
300 400 700 N m, 0.75 m, 0.023 m
2
(0.023) 439.57 10 m
2
(700)(0.75) 15.4708 10 rad
(77.2 10 )(439.57 10 )
0.89
2.52 0.89
π
ϕ
ϕ
ϕ ϕϕ
−
−
−
=+= ⋅ = ==
= = ×
= = = ×
××
= °
= + = °+ °
BC BC BC
BC
BC BC
BC BC
BC
AC AB BC
T L Cd
J
TL
GJ
3.41
AC
φ
= °
PROBLEM 10.30
The torques shown are exerted on pulleys B, C,
and D. Knowing that the entire shaft is made of
steel
( 27 GPa),G=
determine the angle of twist
between (a) C and B, (b) D and B.
SOLUTION
(a) Shaft BC:
4 94
9
10.015 m
2
79.522 10 m
4
0.8 m, 27 10 Pa
BC
BC
cd
Jc
LG
π
−
= =
= = ×
= = ×
99
(400)(0.8) 0.149904 rad
(27 10 )(79.522 10 )
BC
TL
GJ
ϕ
−
= = =
××
8.54
BC
ϕ
= °
(b) Shaft CD:
4 94
99
10.018 m 164.896 10 m
24
1.0 m 400 900 500 N m
( 500)(1.0) 0.11230 rad
(27 10 )(164.896 10 )
0.14904 0.11230 0.03674 rad
CD
CD CD
CD
BD BC CD
cd J c
LT
TL
GJ
π
ϕ
ϕϕϕ
−
−
= = = = ×
= =−=− ⋅
−
= = = −
××
=+= − =
2.11
BD
ϕ
= °
PROBLEM 10.31
The aluminum rod BC
( 26 GPa)G=
is bonded to the brass rod AB
( 39 GPa).=
G
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:
3
4 34 9 4
16 mm = 6 10 m
2
(6 10 ) 2.03575 10 m
22
100 N m
cd
Jc
T
ππ
−
−−
= = ×
==×= ×
= ⋅
Rod AB:
9
39 10 Pa, 0.200 m
AB AB
GL=×=
(a)
99
(100)(0.200) 0.25191 rad
(39 10 )(2.03575 10 )
AB
B AB AB
TL
GJ
ϕϕ
−
= = = =
××
14.43
B
ϕ
= °
Rod BC:
9
99
26 10 Pa, 0.300 m
(100)(0.300) 0.56679 rad
(26 10 )(2.03575 10 )
BC BC
BC
BC BC
GL
TL
GJ
ϕ
−
=×=
= = =
××
(b)
0.25191 0.56679 0.81870 rad
C B BC
ϕ ϕϕ
=+= + =
46.9
C
ϕ
= °
PROBLEM 10.32
The aluminum rod AB
( 27 GPa)=G
is bonded to the brass
rod BD
( 39 GPa).=
G
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
Rod AB:
9
44 9
3
/99
27 10 Pa, 0.400 m
1
800N m 0.018 m
2
(0.018) 164.896 10 m
22
(800)(0.400) 71.875 10 rad
(27 10 )(164.896 10 )
AB
GL
T cd
Jc
TL
GJ
ππ
ϕ
−
−
−
=×=
= ⋅==
= = = ×
= = = ×
××
Part BC:
9
4 4 64
3
/96
1
39 10 Pa 0.375 m, 0.030 m
2
800 1600 2400 N m, (0.030) 1.27234 10 m
22
(2400)(0.375) 18.137 10 rad
(39 10 )(1.27234 10 )
BC
G L cd
T Jc
TL
GJ
ππ
ϕ
−
−
−
=×= ==
=+= ⋅ == = ×
= = = ×
××
Part CD:
( )
11
22
4 4 4 4 64
21
3
/96
10.020 m
2
10.030 m, 0.250 m
2
(0.030 0.020 ) 1.02102 10 m
22
(2400)(0.250) 15.068 10 rad
(39 10 )(1.02102 10 )
CD
cd
cd L
J cc
TL
GJ
ππ
ϕ
−
−
−
= =
= = =
= −= − = ×
= = = ×
××
Angle of twist at A.
///
3
105.080 10 rad
A AB BC CD
ϕϕ ϕ ϕ
−
=++
= ×
6.02
A
ϕ
= °
PROBLEM 10.33
Two solid shafts are connected by gears as shown. Knowing
that
77.2 GPaG=
for each shaft, determine the angle
through which end A rotates when
1200N m.
A
T= ⋅
SOLUTION
Calculation of torques:
AB CD C
TT r
PROBLEM 10.34
Two solid steel shafts, each of 30-mm diameter, are
connected by the gears shown. Knowing that G = 77.2 GPa,
determine the angle through which end A rotates when a
torque of magnitude T = 200 N ⋅ m is applied at A.
SOLUTION
Calculation of torques.
AB DE
TT
consent of McGraw-Hill Education.
PROBLEM 10.35
Two shafts, each of
7
8
-in.
diameter, are
connected by the gears shown. Knowing that
6
11.2 10 psi= ×G
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:
AB EF E
EF AB
BE B
TT r
F TT
rr r
= = =
1.2 kip in. 1200 lb in.
6(1200) 1600 lb in.
4.5
AB
EF
T
T
= ⋅= ⋅
= = ⋅
Twist in shaft FE.
6
4
4 34
3
/63
17
12 in., in., 11.2 10 psi
2 16
757.548 10 in
2 2 16
(1600)(12) 29.789 10 rad
(11.2 10 )(57.548 10 )
EF
L cd G
Jc
TL
GJ
pp
ϕ
−
−
−
= = = = ×
= = = ×
= = = ×
××
Rotation at E.
3
/29.789 10 rad
EF
E
ϕϕ
−
= = ×
Tangential displacement at gear circle:
EE BB
rr
δϕ ϕ
= =
Rotation at B.
33
6(29.789 10 ) 39.718 10 rad
4.5
E
BE
B
r
r
ϕϕ
−−
= = ×= ×
Twist in shaft BA.
34
3
/63
8 6 14 in. 57.548 10 in
(1200)(14) 26.065 10 rad
(11.2 10 )(57.548 10 )
AB
LJ
TL
GJ
ϕ
−
−
−
=+= = ×
= = = ×
××
Rotation at A.
3
/
65.783 10 rad
A B AB
ϕϕϕ
−
=+= ×
3.77
A
ϕ
= °
PROBLEM 10.36
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
2
AB A
C AB A
CD AB
B
E CD A
EF CD
D
TT
r TT
TT
r nn
r TT
TT
rn
n
=
= = =
= = =
2
3
33
42
11
11
EF EF A
E EF
E EA
DE
D
CD CD A
CD
A AA
C D CD
CC
BC
B
AB AB A
AB
T l Tl
GJ n GJ
r Tl
rn
n GJ
T l Tl
GJ nGJ
Tl Tl Tl
nGJ GJ n
n GJ n
r Tl
r n GJ nn
T l Tl
GJ GJ
ϕϕ
ϕ
ϕϕ
ϕ
ϕϕϕ
ϕ
ϕϕ
ϕ
= = =
= = =
= =
=+= + = +
= = = +
= =
42
11
1
A
A B AB
Tl
GJ nn
ϕϕϕ
=+ = ++
consent of McGraw-Hill Education.
PROBLEM 10.37
The design specifications of a 1.2-m-long solid circular 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
34
750 N m, 4 69.813 10 rad, 1.2 m, 2
π
ϕ
−
= ⋅ = °= × = =T L Jc
6
9
90 MPa 90 10 Pa
77.2 GPa 77.2 10 PaG
τ
= = ×
= = ×
Based on angle of twist.
4
3
4493
2
2 (2)(750)(1.2) 18.0569 10 m
(77.2 10 )(69.813 10 )
ϕπ
πϕ π
−
−
= =
= = = ×
××
TL TL
GJ Gc
TL
cG
Based on shearing stress.
3
3
336
2
2 (2)(750) 17.441 10 m
(90 10 )
Tc T
Jc
T
c
τπ
πτ π
−
= =
= = = ×
×
Use larger value.
3
18.0569 10 m 18.0569 mmc
−
= ×=
2 36.1 mm
dc= =
PROBLEM 10.38
The aluminum rod AB
( 27GPa)G=
is bonded to the brass
rod BD
( 39GPa).G=
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
Rod AB:
9
44 9
3
/99
27 10 Pa, 0.400 m
1
800N m 0.018 m
2
(0.018) 164.896 10 m
22
(800)(0.400) 71.875 10 rad
(27 10 )(164.896 10 )
AB
GL
T cd
Jc
TL
GJ
ππ
ϕ
−
−
−
=×=
= ⋅==
= = = ×
= = = ×
××
Part BC:
9
4 4 64
3
/96
1
39 10 Pa 0.375 m, 0.030 m
2
800 1600 2400 N m, (0.030) 1.27234 10 m
22
(2400)(0.375) 18.137 10 rad
(39 10 )(1.27234 10 )
BC
G L cd
T Jc
TL
GJ
ππ
ϕ
−
−
−
=×= ==
=+= ⋅ == = ×
= = = ×
××
Part CD:
( )
11
22
4 4 4 4 64
21
3
/96
10.020 m
2
10.030 m, 0.250 m
2
(0.030 0.020 ) 1.02102 10 m
22
(2400)(0.250) 15.068 10 rad
(39 10 )(1.02102 10 )
CD
cd
cd L
J cc
TL
GJ
ππ
ϕ
−
−
−
= =
= = =
= −= − = ×
= = = ×
××
Angle of twist at A.
///
3
105.080 10 rad
A AB BC CD
ϕϕ ϕ ϕ
−
=++
= ×
6.02
A
ϕ
= °
PROBLEM 10.30
The torques shown are exerted on pulleys B, C,
and D. Knowing that the entire shaft is made of
steel
( 27 GPa),G=
determine the angle of twist
between (a) C and B, (b) D and B.
SOLUTION
(a) Shaft BC:
4 94
9
10.015 m
2
79.522 10 m
4
0.8 m, 27 10 Pa
BC
BC
cd
Jc
LG
π
−
= =
= = ×
= = ×
99
(400)(0.8) 0.149904 rad
(27 10 )(79.522 10 )
BC
TL
GJ
ϕ
−
= = =
××
8.54
BC
ϕ
= °
(b) Shaft CD:
4 94
99
10.018 m 164.896 10 m
24
1.0 m 400 900 500 N m
( 500)(1.0) 0.11230 rad
(27 10 )(164.896 10 )
0.14904 0.11230 0.03674 rad
CD
CD CD
CD
BD BC CD
cd J c
LT
TL
GJ
π
ϕ
ϕϕϕ
−
−
= = = = ×
= =−=− ⋅
−
= = = −
××
=+= − =
2.11
BD
ϕ
= °
PROBLEM 10.31
The aluminum rod BC
( 26 GPa)G=
is bonded to the brass rod AB
( 39 GPa).=
G
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:
3
4 34 9 4
16 mm = 6 10 m
2
(6 10 ) 2.03575 10 m
22
100 N m
cd
Jc
T
ππ
−
−−
= = ×
==×= ×
= ⋅
Rod AB:
9
39 10 Pa, 0.200 m
AB AB
GL=×=
(a)
99
(100)(0.200) 0.25191 rad
(39 10 )(2.03575 10 )
AB
B AB AB
TL
GJ
ϕϕ
−
= = = =
××
14.43
B
ϕ
= °
Rod BC:
9
99
26 10 Pa, 0.300 m
(100)(0.300) 0.56679 rad
(26 10 )(2.03575 10 )
BC BC
BC
BC BC
GL
TL
GJ
ϕ
−
=×=
= = =
××
(b)
0.25191 0.56679 0.81870 rad
C B BC
ϕ ϕϕ
=+= + =
46.9
C
ϕ
= °
PROBLEM 10.32
The aluminum rod AB
( 27 GPa)=G
is bonded to the brass
rod BD
( 39 GPa).=
G
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
Rod AB:
9
44 9
3
/99
27 10 Pa, 0.400 m
1
800N m 0.018 m
2
(0.018) 164.896 10 m
22
(800)(0.400) 71.875 10 rad
(27 10 )(164.896 10 )
AB
GL
T cd
Jc
TL
GJ
ππ
ϕ
−
−
−
=×=
= ⋅==
= = = ×
= = = ×
××
Part BC:
9
4 4 64
3
/96
1
39 10 Pa 0.375 m, 0.030 m
2
800 1600 2400 N m, (0.030) 1.27234 10 m
22
(2400)(0.375) 18.137 10 rad
(39 10 )(1.27234 10 )
BC
G L cd
T Jc
TL
GJ
ππ
ϕ
−
−
−
=×= ==
=+= ⋅ == = ×
= = = ×
××
Part CD:
( )
11
22
4 4 4 4 64
21
3
/96
10.020 m
2
10.030 m, 0.250 m
2
(0.030 0.020 ) 1.02102 10 m
22
(2400)(0.250) 15.068 10 rad
(39 10 )(1.02102 10 )
CD
cd
cd L
J cc
TL
GJ
ππ
ϕ
−
−
−
= =
= = =
= −= − = ×
= = = ×
××
Angle of twist at A.
///
3
105.080 10 rad
A AB BC CD
ϕϕ ϕ ϕ
−
=++
= ×
6.02
A
ϕ
= °
PROBLEM 10.33
Two solid shafts are connected by gears as shown. Knowing
that
77.2 GPaG=
for each shaft, determine the angle
through which end A rotates when
1200N m.
A
T= ⋅
SOLUTION
Calculation of torques:
AB CD C
TT r
PROBLEM 10.34
Two solid steel shafts, each of 30-mm diameter, are
connected by the gears shown. Knowing that G = 77.2 GPa,
determine the angle through which end A rotates when a
torque of magnitude T = 200 N ⋅ m is applied at A.
SOLUTION
Calculation of torques.
AB DE
TT
consent of McGraw-Hill Education.
PROBLEM 10.35
Two shafts, each of
7
8
-in.
diameter, are
connected by the gears shown. Knowing that
6
11.2 10 psi= ×G
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:
AB EF E
EF AB
BE B
TT r
F TT
rr r
= = =
1.2 kip in. 1200 lb in.
6(1200) 1600 lb in.
4.5
AB
EF
T
T
= ⋅= ⋅
= = ⋅
Twist in shaft FE.
6
4
4 34
3
/63
17
12 in., in., 11.2 10 psi
2 16
757.548 10 in
2 2 16
(1600)(12) 29.789 10 rad
(11.2 10 )(57.548 10 )
EF
L cd G
Jc
TL
GJ
pp
ϕ
−
−
−
= = = = ×
= = = ×
= = = ×
××
Rotation at E.
3
/29.789 10 rad
EF
E
ϕϕ
−
= = ×
Tangential displacement at gear circle:
EE BB
rr
δϕ ϕ
= =
Rotation at B.
33
6(29.789 10 ) 39.718 10 rad
4.5
E
BE
B
r
r
ϕϕ
−−
= = ×= ×
Twist in shaft BA.
34
3
/63
8 6 14 in. 57.548 10 in
(1200)(14) 26.065 10 rad
(11.2 10 )(57.548 10 )
AB
LJ
TL
GJ
ϕ
−
−
−
=+= = ×
= = = ×
××
Rotation at A.
3
/
65.783 10 rad
A B AB
ϕϕϕ
−
=+= ×
3.77
A
ϕ
= °
PROBLEM 10.36
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
2
AB A
C AB A
CD AB
B
E CD A
EF CD
D
TT
r TT
TT
r nn
r TT
TT
rn
n
=
= = =
= = =
2
3
33
42
11
11
EF EF A
E EF
E EA
DE
D
CD CD A
CD
A AA
C D CD
CC
BC
B
AB AB A
AB
T l Tl
GJ n GJ
r Tl
rn
n GJ
T l Tl
GJ nGJ
Tl Tl Tl
nGJ GJ n
n GJ n
r Tl
r n GJ nn
T l Tl
GJ GJ
ϕϕ
ϕ
ϕϕ
ϕ
ϕϕϕ
ϕ
ϕϕ
ϕ
= = =
= = =
= =
=+= + = +
= = = +
= =
42
11
1
A
A B AB
Tl
GJ nn
ϕϕϕ
=+ = ++
consent of McGraw-Hill Education.
PROBLEM 10.37
The design specifications of a 1.2-m-long solid circular 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
34
750 N m, 4 69.813 10 rad, 1.2 m, 2
π
ϕ
−
= ⋅ = °= × = =T L Jc
6
9
90 MPa 90 10 Pa
77.2 GPa 77.2 10 PaG
τ
= = ×
= = ×
Based on angle of twist.
4
3
4493
2
2 (2)(750)(1.2) 18.0569 10 m
(77.2 10 )(69.813 10 )
ϕπ
πϕ π
−
−
= =
= = = ×
××
TL TL
GJ Gc
TL
cG
Based on shearing stress.
3
3
336
2
2 (2)(750) 17.441 10 m
(90 10 )
Tc T
Jc
T
c
τπ
πτ π
−
= =
= = = ×
×
Use larger value.
3
18.0569 10 m 18.0569 mmc
−
= ×=
2 36.1 mm
dc= =
PROBLEM 10.38
The aluminum rod AB
( 27GPa)G=
is bonded to the brass
rod BD
( 39GPa).G=
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
Rod AB:
9
44 9
3
/99
27 10 Pa, 0.400 m
1
800N m 0.018 m
2
(0.018) 164.896 10 m
22
(800)(0.400) 71.875 10 rad
(27 10 )(164.896 10 )
AB
GL
T cd
Jc
TL
GJ
ππ
ϕ
−
−
−
=×=
= ⋅==
= = = ×
= = = ×
××
Part BC:
9
4 4 64
3
/96
1
39 10 Pa 0.375 m, 0.030 m
2
800 1600 2400 N m, (0.030) 1.27234 10 m
22
(2400)(0.375) 18.137 10 rad
(39 10 )(1.27234 10 )
BC
G L cd
T Jc
TL
GJ
ππ
ϕ
−
−
−
=×= ==
=+= ⋅ == = ×
= = = ×
××
Part CD:
( )
11
22
4 4 4 4 64
21
3
/96
10.020 m
2
10.030 m, 0.250 m
2
(0.030 0.020 ) 1.02102 10 m
22
(2400)(0.250) 15.068 10 rad
(39 10 )(1.02102 10 )
CD
cd
cd L
J cc
TL
GJ
ππ
ϕ
−
−
−
= =
= = =
= −= − = ×
= = = ×
××
Angle of twist at A.
///
3
105.080 10 rad
A AB BC CD
ϕϕ ϕ ϕ
−
=++
= ×
6.02
A
ϕ
= °
