consent of McGraw–Hill Education.
PROBLEM 10.11
Under normal operating conditions,
the electric motor exerts a torque of
2.4 kN ⋅ m on shaft AB. Knowing
that each shaft is solid, determine the
maximum shearing stress in (a) shaft
AB, (b) shaft BC, (c) shaft CD.
SOLUTION
(a) Shaft AB:
3
1
2.4 10 N m, 0.027 m
2
AB
T cd=×⋅ ==
36
33
2 2(2.4 10 ) 77.625 10 Pa
(0.027)
AB
Tc T
Jc
τππ
×
= = = = ×
77.6 MPa
(b) Shaft BC:
1
2.4 kN m 1.2 kN m 1.2 kN m, 0.023 m
2
BC
T cd
= ⋅− ⋅= ⋅ = =
36
33
2 (2)(1.2 10 ) 62.788 10 Pa
(0.023)
BC
Tc T
Jc
τππ
×
= = = = ×
62.8 MPa
(c) Shaft CD:
31
0.4 10 N m 0.023 m
2
CD
T cd=×⋅==
36
33
2 (2)(0.4 10 ) 20.929 10 Pa
(0.023)
CD
Tc T
Jc
τππ
×
= = = = ×
20.9 MPa
consent of McGraw–Hill Education.
PROBLEM 10.12
In order to reduce the total mass of
the assembly of Prob. 10.11, 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 be
increased.
PROBLEM 10.11 Under normal
operating conditions, the electric
motor exerts a torque of 2.4 kN ⋅ m
on shaft AB. Knowing that each shaft
is solid, determine the maximum
shearing stress in (a) shaft AB,
(b) shaft BC, (c) shaft CD.
SOLUTION
3
2Tc T
Jc
τπ
= =
3
3 63
6
2 (2)(1.2 10 ) 9.8415 10 m
(77.625 10 )
T
c
πτ π
−
×
= = = ×
×
33
21.43 10 m 2 42.8 10 mc dc
−−
=×==×
42.8 mm
consent of McGraw–Hill Education.
PROBLEM 10.13
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
43
max max
,,
22
Tc J cT c
J
ππ
ττ
= = =
Rod AB:
max
3
1
15 ksi 0.75 in.
2
(0.75) (15) 9.94 kip in.
2
cd
T
τ
π
= = =
= = ⋅
Rod BC:
max
3
1
8 ksi 0.90 in.
2
(0.90) (8) 9.16 kip in.
2
cd
T
τ
π
= = =
= = ⋅
The allowable torque is the smaller value.
9.16 kip in.T= ⋅
consent of McGraw–Hill Education.
PROBLEM 10.14
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
10 kip in.T= ⋅
is applied at A, determine the
required diameter of (a) rod AB, (b) rod BC.
SOLUTION
3
max max
2
,,
2
Tc T
Jc
J
π
τπτ
= = =
(a) Rod AB:
max
10 kip in. 15 ksi
τ
=⋅=
T
33
(2)(10) 0.4244 in
(15)
0.7515 in.
c
c
π
= =
=
2 1.503 in.dc
= =
(b) Rod BC:
max
10 kip in. 8 ksiT
τ
=⋅=
32
(2)(10) 0.79577 in
(8)
0.9267 in.
c
c
π
= =
=
2 1.853 in.dc= =
consent of McGraw–Hill Education.
PROBLEM 10.15
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:
6
all
1
85 10 Pa 0.030 m
2
cd
τ
=×==
3
all
all all
36 3
2
(0.030) (85 10 ) 3.605 10 N m
2
J
Tc
c
τπ
τ
π
= =
= ×= × ⋅
Pipe CD:
6
all 2 2
1
54 10 Pa 0.045 m
2
cd
τ
=×==
( )
12
4 4 4 4 64
21
66 3
all
all 2
0.045 0.006 0.039 m
(0.045 0.039 ) 2.8073 10 m
22
(2.8073 10 )(54 10 ) 3.369 10 N m
0.045
cct
J cc
J
Tc
ππ
τ
−
−
= −= − =
= −= − = ×
××
== =×⋅
Allowable torque is the smaller value.
3
all
3.369 10 N mT=×⋅
3.37 kN m⋅
consent of McGraw–Hill Education.
PROBLEM 10.16
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
43
max max
2
2
Tc T
J cc
J
π
τπτ
= = =
(a) Rod AB:
3 63
6
3
(2)(1250) 15.915 10 m
(50 10 )
25.15 10 m 25.15 mm
c
c
π
−
−
= = ×
×
=×=
2 50.3 mm
AB
dc= =
(b) Rod BC:
3 63
6
3
(2)(1250) 31.831 10 m
(25 10 )
31.69 10 m 31.69 mm
c
c
π
−
−
= = ×
×
=×=
2 63.4 mm
BC
dc= =
consent of McGraw–Hill Education.
PROBLEM 10.17
The solid shaft shown is formed of a brass for which the
allowable shearing stress is 55 MPa. Neglecting the effect of
stress concentrations, determine the smallest diameters dAB and
dBC for which the allowable shearing stress is not exceeded.
SOLUTION
6
max
3
max 3max
55 MPa 55 10 Pa
22
τ
τπτ
π
= = ×
= = =
Tc T T
c
Jc
Shaft AB:
3
36
1200 400 800 N m
(2)(800) 21.00 10 m 21.0 m
(55 10 )
AB
T
c
π
−
= −= ⋅
= =×=
×
minimum 2 42.0 mm
AB
dc= =
Shaft BC:
3
36
400 N m
(2)(400) 16.667 10 m 16.67 mm
(55 10 )
BC
T
c
π
−
= ⋅
= =×=
×
minimum 2 33.3 mm
BC
dc= =
PROBLEM 10.18
Solve Prob. 10.17, assuming that the direction of TC is reversed.
PROBLEM 10.17 The solid shaft shown is formed of a brass for
which the allowable shearing stress is 55 MPa. Neglecting the effect
of stress concentrations, determine the smallest diameters dAB and
dBC for which the allowable shearing stress is not exceeded.
SOLUTION
PROBLEM 10.19
Under normal operating conditions, a motor
exerts a torque of magnitude
1200 lb in.= ⋅
F
T
at F. Knowing that
8 in.,=
D
r
3 in.,=
G
r
and
the allowable shearing stress is 10.5 ksi in
each shaft, determine the required diameter of
(a) shaft CDE, (b) shaft FGH.
SOLUTION
all
3
3
1200 lb in.
8(1200) 3200 lb in.
3
10.5 ksi =10,500 psi
22
or
τ
τpτ
p
= ⋅
= = = ⋅
=
= = =
F
D
EF
G
T
r
TT
r
Tc T T
c
Jc
(a) Shaft CDE:
33
(2)(3200) 0.194012 in
(10,500)
p
= =c
0.5789 in. 2
DE
c dc= =
1.158 in.
DE
d=
(b) Shaft FGH:
33
(2)(1200) 0.012757 in
(10,500)
p
= =c
0.4174 in. 2
FG
c dc= =
0.835 in.
FG
d=
consent of McGraw–Hill Education.
PROBLEM 10.20
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
all
12 ksi
τ
=
Shaft FG:
3
all
,all all
3
10.400 in.
2
2
(0.400) (12) 1.206 kip in.
2
τp
τ
p
= =
= =
= = ⋅
F
cd
J
Tc
c
Shaft DE:
3
,all
3
, all
10.450 in.
2
2
(0.450) (12) 1.7177 kip in.
2
4.5 (1.7177) 1.189 kip in.
6.5
pτ
p
= =
=
= = ⋅
= = = ⋅
E all
G
F EF
D
cd
Tc
r
T TT
r
Allowable value of
F
T
is the smaller.
1.189 kip in.
F
T= ⋅
consent of McGraw–Hill Education.
PROBLEM 10.12
In order to reduce the total mass of
the assembly of Prob. 10.11, 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 be
increased.
PROBLEM 10.11 Under normal
operating conditions, the electric
motor exerts a torque of 2.4 kN ⋅ m
on shaft AB. Knowing that each shaft
is solid, determine the maximum
shearing stress in (a) shaft AB,
(b) shaft BC, (c) shaft CD.
SOLUTION
3
2Tc T
Jc
τπ
= =
3
3 63
6
2 (2)(1.2 10 ) 9.8415 10 m
(77.625 10 )
T
c
πτ π
−
×
= = = ×
×
33
21.43 10 m 2 42.8 10 mc dc
−−
=×==×
42.8 mm
consent of McGraw–Hill Education.
PROBLEM 10.13
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
43
max max
,,
22
Tc J cT c
J
ππ
ττ
= = =
Rod AB:
max
3
1
15 ksi 0.75 in.
2
(0.75) (15) 9.94 kip in.
2
cd
T
τ
π
= = =
= = ⋅
Rod BC:
max
3
1
8 ksi 0.90 in.
2
(0.90) (8) 9.16 kip in.
2
cd
T
τ
π
= = =
= = ⋅
The allowable torque is the smaller value.
9.16 kip in.T= ⋅
consent of McGraw–Hill Education.
PROBLEM 10.14
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
10 kip in.T= ⋅
is applied at A, determine the
required diameter of (a) rod AB, (b) rod BC.
SOLUTION
3
max max
2
,,
2
Tc T
Jc
J
π
τπτ
= = =
(a) Rod AB:
max
10 kip in. 15 ksi
τ
=⋅=
T
33
(2)(10) 0.4244 in
(15)
0.7515 in.
c
c
π
= =
=
2 1.503 in.dc
= =
(b) Rod BC:
max
10 kip in. 8 ksiT
τ
=⋅=
32
(2)(10) 0.79577 in
(8)
0.9267 in.
c
c
π
= =
=
2 1.853 in.dc= =
consent of McGraw–Hill Education.
PROBLEM 10.15
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:
6
all
1
85 10 Pa 0.030 m
2
cd
τ
=×==
3
all
all all
36 3
2
(0.030) (85 10 ) 3.605 10 N m
2
J
Tc
c
τπ
τ
π
= =
= ×= × ⋅
Pipe CD:
6
all 2 2
1
54 10 Pa 0.045 m
2
cd
τ
=×==
( )
12
4 4 4 4 64
21
66 3
all
all 2
0.045 0.006 0.039 m
(0.045 0.039 ) 2.8073 10 m
22
(2.8073 10 )(54 10 ) 3.369 10 N m
0.045
cct
J cc
J
Tc
ππ
τ
−
−
= −= − =
= −= − = ×
××
== =×⋅
Allowable torque is the smaller value.
3
all
3.369 10 N mT=×⋅
3.37 kN m⋅
consent of McGraw–Hill Education.
PROBLEM 10.16
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
43
max max
2
2
Tc T
J cc
J
π
τπτ
= = =
(a) Rod AB:
3 63
6
3
(2)(1250) 15.915 10 m
(50 10 )
25.15 10 m 25.15 mm
c
c
π
−
−
= = ×
×
=×=
2 50.3 mm
AB
dc= =
(b) Rod BC:
3 63
6
3
(2)(1250) 31.831 10 m
(25 10 )
31.69 10 m 31.69 mm
c
c
π
−
−
= = ×
×
=×=
2 63.4 mm
BC
dc= =
consent of McGraw–Hill Education.
PROBLEM 10.17
The solid shaft shown is formed of a brass for which the
allowable shearing stress is 55 MPa. Neglecting the effect of
stress concentrations, determine the smallest diameters dAB and
dBC for which the allowable shearing stress is not exceeded.
SOLUTION
6
max
3
max 3max
55 MPa 55 10 Pa
22
τ
τπτ
π
= = ×
= = =
Tc T T
c
Jc
Shaft AB:
3
36
1200 400 800 N m
(2)(800) 21.00 10 m 21.0 m
(55 10 )
AB
T
c
π
−
= −= ⋅
= =×=
×
minimum 2 42.0 mm
AB
dc= =
Shaft BC:
3
36
400 N m
(2)(400) 16.667 10 m 16.67 mm
(55 10 )
BC
T
c
π
−
= ⋅
= =×=
×
minimum 2 33.3 mm
BC
dc= =
PROBLEM 10.18
Solve Prob. 10.17, assuming that the direction of TC is reversed.
PROBLEM 10.17 The solid shaft shown is formed of a brass for
which the allowable shearing stress is 55 MPa. Neglecting the effect
of stress concentrations, determine the smallest diameters dAB and
dBC for which the allowable shearing stress is not exceeded.
SOLUTION
PROBLEM 10.19
Under normal operating conditions, a motor
exerts a torque of magnitude
1200 lb in.= ⋅
F
T
at F. Knowing that
8 in.,=
D
r
3 in.,=
G
r
and
the allowable shearing stress is 10.5 ksi in
each shaft, determine the required diameter of
(a) shaft CDE, (b) shaft FGH.
SOLUTION
all
3
3
1200 lb in.
8(1200) 3200 lb in.
3
10.5 ksi =10,500 psi
22
or
τ
τpτ
p
= ⋅
= = = ⋅
=
= = =
F
D
EF
G
T
r
TT
r
Tc T T
c
Jc
(a) Shaft CDE:
33
(2)(3200) 0.194012 in
(10,500)
p
= =c
0.5789 in. 2
DE
c dc= =
1.158 in.
DE
d=
(b) Shaft FGH:
33
(2)(1200) 0.012757 in
(10,500)
p
= =c
0.4174 in. 2
FG
c dc= =
0.835 in.
FG
d=
consent of McGraw–Hill Education.
PROBLEM 10.20
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
all
12 ksi
τ
=
Shaft FG:
3
all
,all all
3
10.400 in.
2
2
(0.400) (12) 1.206 kip in.
2
τp
τ
p
= =
= =
= = ⋅
F
cd
J
Tc
c
Shaft DE:
3
,all
3
, all
10.450 in.
2
2
(0.450) (12) 1.7177 kip in.
2
4.5 (1.7177) 1.189 kip in.
6.5
pτ
p
= =
=
= = ⋅
= = = ⋅
E all
G
F EF
D
cd
Tc
r
T TT
r
Allowable value of
F
T
is the smaller.
1.189 kip in.
F
T= ⋅