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Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 10.21
A torque of magnitude
100 N mT= ⋅
is applied to shaft AB of the
gear train shown. Knowing that the diameters of the three solid
shafts are, respectively,
21 mm,
AB
d=
30 mm,
CD
d=
and
40mm,
EF
d=
determine the maximum shearing stress in
(a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
AB A B
T TTT===
Gears B and C:
25 mm, 60 mm
BC
rr= =
Force on gear circles.
60 2.4
25
BC
BC BC
C
CB
B
TT
Frr
r
T T TT
r
= =
= = =
Shaft CD:
2.4
CD C D
T TT T= = =
Gears D and E:
30 mm, 75 mm
DE
rr= =
Force on gear circles.
75 (2.4 ) 6
30
DE
DE DE
E
ED
D
TT
Frr
r
T T TT
r
= =
= = =
Shaft EF:
6
EF E F
T TT T= = =
Maximum Shearing Stresses.
max 3
2Tc T
Jc
τπ
= =
SOLUTION Continued
(a) Shaft AB:
100 N mT= ⋅
3
110.5 mm 10.5 10 m
2
cd
−
= = = ×
6
max 33
(2)(100) 55.0 10 Pa
(10.5 10 )
τπ
−
= = ×
×
max
55.0 MPa
τ
=
(b) Shaft CD:
(2.4)(100) 240 N mT= = ⋅
3
115 mm 15 10 m
2
cd
−
= = = ×
6
max 33
(2)(240) 45.3 10 Pa
(15 10 )
τπ
−
= = ×
×
max
45.3 MPa
τ
=
(c) Shaft EF:
(6)(100) 600 N mT= = ⋅
3
120 mm 20 10 m
2
cd
−
= = = ×
6
max 33
(2)(600) 47.7 10 Pa
(20 10 )
τπ
−
= = ×
×
max
47.7 MPa
τ
=
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 10.22
A torque of magnitude
120N mT= ⋅
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:
AB A B
T TTT===
Gears B and C:
25 mm, 60 mm
BC
rr
= =
Force on gear circles.
60 2.4
25
BC
BC BC
C
CB
B
TT
Frr
r
T T TT
r
= =
= = =
Shaft CD:
2.4
CD C D
T TT T= = =
Gears D and E:
30 mm, 75 mm
DE
rr= =
Force on gear circles.
75 (2.4 ) 6
30
DE
DE DE
E
ED
D
TT
Frr
r
T T TT
r
= =
= = =
Shaft EF:
6
EF E F
T TT T= = =
Required Diameters.
max 3
3
3
max
6
max
2
2
2
22
75 10 Pa
Tc T
Jc
T
c
T
dc
τπ
πτ
πτ
τ
= =
=
= =
= ×
consent of McGraw-Hill Education.
SOLUTION Continued
(a) Shaft AB:
120 N m
AB
TT= = ⋅
3
36
2(120)
2 20.1 10 m
(75 10 )
AB
d
π
−
= = ×
×
20.1 mm
AB
d=
(b) Shaft CD:
(2.4)(120) 288 N m
CD
T= = ⋅
3
36
(2)(288)
2 26.9 10 m
(75 10 )
CD
d
π
−
= = ×
×
26.9 mm
CD
d=
(c) Shaft EF:
(6)(120) 720 N m
EF
T= = ⋅
3
33
(2)(720)
2 36.6 10 m
(75 10 )
EF
d
π
−
= = ×
×
36.6 mm
EF
d=
consent of McGraw-Hill Education.
PROBLEM 10.23
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
5 kip in.
C
T= ⋅
is applied at C and that the assembly
is in equilibrium, determine the required diameter of (a) shaft BC,
(b) shaft EF.
SOLUTION
max
8500 psi 8.5 ksi
τ
= =
(a) Shaft BC:
3
max 3max
3
5 kip in.
22
(2)(5) 0.7208 in.
(8.5)
C
T
Tc T T
c
Jc
c
τpτ
p
p
= ⋅
= = =
= =
2 1.442 in.
BC
dc= =
(b) Shaft EF:
3
3
max
2.5 (5) 3.125 kip in.
4
2 (2)(3.125) 0.6163 in.
(8.5)
pτ p
= = = ⋅
= = =
D
FC
A
r
TT
r
T
c
2 1.233 in.
EF
dc
= =
consent of McGraw-Hill Education.
PROBLEM 10.24
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,
1.6 in.
BC
d=
and
1.25 in.,
EF
d=
determine the largest torque TC
that can be applied at C.
SOLUTION
max
7000 psi 7.0 ksi
τ
= =
Shaft BC:
3
max max
3
1.6 in.
10.8 in.
2
2
(7.0)(0.8) 5.63 kip in.
2
BC
C
d
cd
J
Tc
c
τp
τ
p
=
= =
= =
= = ⋅
Shaft EF:
3
max max
3
1.25 in.
10.625 in.
2
2
(7.0)(0.625) 2.684 kip in.
2
EF
F
d
cd
J
Tc
c
τp
τ
p
=
= =
= =
= = ⋅
By statics,
4(2.684) 4.30 kip in.
2.5
A
CF
D
r
TT
r
= = = ⋅
Allowable value of
C
T
is the smaller.
4.30 kip in.
C
T= ⋅
PROBLEM 10.25
For the aluminum shaft shown
( 27 GPa),G=
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
(a)
( )
3
9
4 4 4 4 94
21
993
4 69.813 10 rad, 1.25 m
27 GPa 27 10 Pa
(0.018 0.012 ) 132.324 10 m
22
(27 10 )(132.324 10 )(69.813 10 )
1.25
TL GJ
T
GJ L
L
G
J cc
T
ϕ
ϕ
ϕ
ππ
−
−
−
= =
= °= × =
= = ×
= −= − = ×
× ××
=
199.539 N m= ⋅
199.5 N mT= ⋅
(b) Matching areas:
( )
2 22
21
22 2 2
21
4 4 94
0.018 0.012 0.013416 m
(0.013416) 50.894 10 m
22
A c cc
c cc
Jc
ππ
ππ
−
= = −
= −= − =
= = = ×
3
99
(195.539)(1.25) 181.514 10 rad
(27 10 )(50.894 10 )
TL
GJ
ϕ
−
= = = ×
××
10.40
ϕ
= °
PROBLEM 10.26
(a) For the solid steel shaft shown, determine the angle of twist at
A. Use
6
11.5 10 psi.= ×G
(b) Solve part a, assuming that the
steel shaft is hollow with a 1.5-in. outer radius and a 0.75-in.
inner radius.
SOLUTION
(a)
44
24
1.5 in. (1.5 in.) 7.9522 in
2
(60 kip in.)(36 in.)
(11,200 kips/in )(7.9522 in )
0.024252 radians
p
φ
φ
= = =
⋅
= =
=
cJ
TL
GJ
1.390
φ
= °
(b)
44 4
24
[(1.5 in.) (0.75 in.) ] 7.4552 in
2
(60 kip in.)(36 in.)
(11,200 kips/in )(7.4552 in )
0.025869 radians
J
p
φ
φ
= −=
⋅
=
=
1.482
φ
= °
PROBLEM 10.27
Determine the largest allowable diameter of a 10-ft-long steel rod
6
( 11.2 10 psi)G= ×
if the rod is to be twisted
through 30° without exceeding a shearing stress of 12 ksi.
SOLUTION
3
30
10 ft 120 in. 30 0.52360 rad
180
12 ksi 12 10 psi
,, ,
L
TL GJ Tc GJ c G c L
Tc
GJ L J JL L G
p
ϕ
t
ϕ ϕϕ t
ϕt ϕ
= = = °= =
= = ×
= = = = = =
3
6
(12 10 )(120) 0.24555 in.
(11.2 10 )(0.52360)
c×
= =
×
2 0.491 in.dc= =
consent of McGraw-Hill Education.
PROBLEM 10.28
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
6
( 11.2 10 psi)G= ×
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
T
GJ L
TcGJcGc
J JL L
ϕ
ϕ
ϕϕ
τ
= =
= = =
1
2rev (2)(2 ) 12.566 rad, 4.0 in.
2
5000 ft 60,000 in.
ϕπ
= = = = =
= =
cd
L
63
(11.2 10 )(12.566)(4.0) 9.3826 10 psi
60,000
τ
×
= = ×
9.38 ksi
τ
=
SOLUTION Continued
(a) Shaft AB:
100 N mT= ⋅
3
110.5 mm 10.5 10 m
2
cd
−
= = = ×
6
max 33
(2)(100) 55.0 10 Pa
(10.5 10 )
τπ
−
= = ×
×
max
55.0 MPa
τ
=
(b) Shaft CD:
(2.4)(100) 240 N mT= = ⋅
3
115 mm 15 10 m
2
cd
−
= = = ×
6
max 33
(2)(240) 45.3 10 Pa
(15 10 )
τπ
−
= = ×
×
max
45.3 MPa
τ
=
(c) Shaft EF:
(6)(100) 600 N mT= = ⋅
3
120 mm 20 10 m
2
cd
−
= = = ×
6
max 33
(2)(600) 47.7 10 Pa
(20 10 )
τπ
−
= = ×
×
max
47.7 MPa
τ
=
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 10.22
A torque of magnitude
120N mT= ⋅
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:
AB A B
T TTT===
Gears B and C:
25 mm, 60 mm
BC
rr
= =
Force on gear circles.
60 2.4
25
BC
BC BC
C
CB
B
TT
Frr
r
T T TT
r
= =
= = =
Shaft CD:
2.4
CD C D
T TT T= = =
Gears D and E:
30 mm, 75 mm
DE
rr= =
Force on gear circles.
75 (2.4 ) 6
30
DE
DE DE
E
ED
D
TT
Frr
r
T T TT
r
= =
= = =
Shaft EF:
6
EF E F
T TT T= = =
Required Diameters.
max 3
3
3
max
6
max
2
2
2
22
75 10 Pa
Tc T
Jc
T
c
T
dc
τπ
πτ
πτ
τ
= =
=
= =
= ×
consent of McGraw-Hill Education.
SOLUTION Continued
(a) Shaft AB:
120 N m
AB
TT= = ⋅
3
36
2(120)
2 20.1 10 m
(75 10 )
AB
d
π
−
= = ×
×
20.1 mm
AB
d=
(b) Shaft CD:
(2.4)(120) 288 N m
CD
T= = ⋅
3
36
(2)(288)
2 26.9 10 m
(75 10 )
CD
d
π
−
= = ×
×
26.9 mm
CD
d=
(c) Shaft EF:
(6)(120) 720 N m
EF
T= = ⋅
3
33
(2)(720)
2 36.6 10 m
(75 10 )
EF
d
π
−
= = ×
×
36.6 mm
EF
d=
consent of McGraw-Hill Education.
PROBLEM 10.23
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
5 kip in.
C
T= ⋅
is applied at C and that the assembly
is in equilibrium, determine the required diameter of (a) shaft BC,
(b) shaft EF.
SOLUTION
max
8500 psi 8.5 ksi
τ
= =
(a) Shaft BC:
3
max 3max
3
5 kip in.
22
(2)(5) 0.7208 in.
(8.5)
C
T
Tc T T
c
Jc
c
τpτ
p
p
= ⋅
= = =
= =
2 1.442 in.
BC
dc= =
(b) Shaft EF:
3
3
max
2.5 (5) 3.125 kip in.
4
2 (2)(3.125) 0.6163 in.
(8.5)
pτ p
= = = ⋅
= = =
D
FC
A
r
TT
r
T
c
2 1.233 in.
EF
dc
= =
consent of McGraw-Hill Education.
PROBLEM 10.24
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,
1.6 in.
BC
d=
and
1.25 in.,
EF
d=
determine the largest torque TC
that can be applied at C.
SOLUTION
max
7000 psi 7.0 ksi
τ
= =
Shaft BC:
3
max max
3
1.6 in.
10.8 in.
2
2
(7.0)(0.8) 5.63 kip in.
2
BC
C
d
cd
J
Tc
c
τp
τ
p
=
= =
= =
= = ⋅
Shaft EF:
3
max max
3
1.25 in.
10.625 in.
2
2
(7.0)(0.625) 2.684 kip in.
2
EF
F
d
cd
J
Tc
c
τp
τ
p
=
= =
= =
= = ⋅
By statics,
4(2.684) 4.30 kip in.
2.5
A
CF
D
r
TT
r
= = = ⋅
Allowable value of
C
T
is the smaller.
4.30 kip in.
C
T= ⋅
PROBLEM 10.25
For the aluminum shaft shown
( 27 GPa),G=
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
(a)
( )
3
9
4 4 4 4 94
21
993
4 69.813 10 rad, 1.25 m
27 GPa 27 10 Pa
(0.018 0.012 ) 132.324 10 m
22
(27 10 )(132.324 10 )(69.813 10 )
1.25
TL GJ
T
GJ L
L
G
J cc
T
ϕ
ϕ
ϕ
ππ
−
−
−
= =
= °= × =
= = ×
= −= − = ×
× ××
=
199.539 N m= ⋅
199.5 N mT= ⋅
(b) Matching areas:
( )
2 22
21
22 2 2
21
4 4 94
0.018 0.012 0.013416 m
(0.013416) 50.894 10 m
22
A c cc
c cc
Jc
ππ
ππ
−
= = −
= −= − =
= = = ×
3
99
(195.539)(1.25) 181.514 10 rad
(27 10 )(50.894 10 )
TL
GJ
ϕ
−
= = = ×
××
10.40
ϕ
= °
PROBLEM 10.26
(a) For the solid steel shaft shown, determine the angle of twist at
A. Use
6
11.5 10 psi.= ×G
(b) Solve part a, assuming that the
steel shaft is hollow with a 1.5-in. outer radius and a 0.75-in.
inner radius.
SOLUTION
(a)
44
24
1.5 in. (1.5 in.) 7.9522 in
2
(60 kip in.)(36 in.)
(11,200 kips/in )(7.9522 in )
0.024252 radians
p
φ
φ
= = =
⋅
= =
=
cJ
TL
GJ
1.390
φ
= °
(b)
44 4
24
[(1.5 in.) (0.75 in.) ] 7.4552 in
2
(60 kip in.)(36 in.)
(11,200 kips/in )(7.4552 in )
0.025869 radians
J
p
φ
φ
= −=
⋅
=
=
1.482
φ
= °
PROBLEM 10.27
Determine the largest allowable diameter of a 10-ft-long steel rod
6
( 11.2 10 psi)G= ×
if the rod is to be twisted
through 30° without exceeding a shearing stress of 12 ksi.
SOLUTION
3
30
10 ft 120 in. 30 0.52360 rad
180
12 ksi 12 10 psi
,, ,
L
TL GJ Tc GJ c G c L
Tc
GJ L J JL L G
p
ϕ
t
ϕ ϕϕ t
ϕt ϕ
= = = °= =
= = ×
= = = = = =
3
6
(12 10 )(120) 0.24555 in.
(11.2 10 )(0.52360)
c×
= =
×
2 0.491 in.dc= =
consent of McGraw-Hill Education.
PROBLEM 10.28
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
6
( 11.2 10 psi)G= ×
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
T
GJ L
TcGJcGc
J JL L
ϕ
ϕ
ϕϕ
τ
= =
= = =
1
2rev (2)(2 ) 12.566 rad, 4.0 in.
2
5000 ft 60,000 in.
ϕπ
= = = = =
= =
cd
L
63
(11.2 10 )(12.566)(4.0) 9.3826 10 psi
60,000
τ
×
= = ×
9.38 ksi
τ
=
