ch05

5-1
CHAPTER 5
PROBLEM (5.1) A hollow steel shaft of outer radius c is fixed at one end and subjected to a torque
T at the other end , as shown in Fig. P5.1. If the shearing stress is not to exceed
all
, what is the required
inner radius b ?
Given: c = 30 mm , T = 2 kN
m,
all
= 80 MPa
SOLUTION
44
()
2
all Tc
cb
or
3
644
2(2 10 )(0.03)
80(10 ) [(0.03) ]b
or
6 6 4
80(10 )[0.81(10 ) ] 38.197b
Solving,
0.024 24b m mm
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PROBLEM(5.2) A solid shaft of diameter d = 2 in. is to be replaced by a hollow circular tube of the
same material, resisting the same maximum shear stress and the same torque T (Fig. P5.2). Determine the
outer diameter D of the tube if its wall thickness is t = D/25.
SOLUTION
max 4 4 4
( 2) ( 2)
32 ( ) 32
i
T d T D
d D D
or
3 3 4
23
[1 ( ) ]
25
dD
33
(2) (0.2836)D
3.046 .D in
c
b
.
d=2in.
DDi25
23
D
25
D
t

5-2
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PROBLEM(5.3) A solid shaft of diameter d and a hollow shaft of outer diameter D and thickness t
= D/4 are to transmit the same torsional loading at the same maximum shear stress (Fig. P5.2). Compare
the weights of these two shafts of equal length.
SOLUTION
max 34 4 3
16 ( 2) 16 1
( ) (1 )
32 16
i
T T D T
dD D D
or
33
15
( ), 1.0217
16
d D D d
Ratio of weight:
2 2 2
22
( 2) 0.75(1.0217 ) 0.783 D D d
dd
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PROBLEM (5.4) The circular shaft is subjected to the torques shown in Fig. P5.4. What is the
largest shearing stress in the member and where does it occur?
SOLUTION
Apply the method of sections between the change of load points:
50
EF
T N m
30
DE
T N m
120
CD BC
T T N m  
80
AB
T N m
We have
4 4 6 4
2[(0.025) (0.015) ] 0.534 10
h
Jm
Thus,
max :Tc J
33
2 2(80) 3.26
(0.025)
AB TMPa
c
36
2(120) 120(0.025)
4.89 , 5.62
(0.025) 0.534 10
BC CD
MPa MPa
2
D
i
D
D
d

5-3
66
30(0.025) 50(0.025)
1.40 , 2.34
0.534 10 0.534 10
DE EF
MPa MPa
So,
max 5.62 CD MPa
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PROBLEM (5.5) Four pulleys, attached to a solid stepped shaft, transmit the torques shown in Fig.
P5.5. Calculate the maximum shear stress for each segment of the shaft.
SOLUTION
Apply the method of sections between the change of load points:
1
CD
T kN m
2.5
BC
T kN m
2.5
AB
T kN m
Therefore,
3
max 2:Tc
3
3
2(2.5 10 ) 58.9
(0.03)
AB MPa
3
3
2(2.5 10 ) 101.9
(0.025)
BC MPa
3
3
2(1 10 ) 79.6
(0.02)
CD MPa
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PROBLEM (5.6) Redo Prob. 5.5, with a hole of 20-mm diameter drilled axially through the shaft
to form a tube.
SOLUTION
We have, applying the method of sections:
1
CD
T kN m
2.5
BC
T kN m
2.5
AB
T kN m
Hence,
max 44
2
()
Tc
cb
gives,
3
44
2(2.5 10 )(0.03) 59.7
[(0.03) (0.01) ]
AB MPa
3
44
2(2.5 10 )(0.025) 104.5
[(0.025) (0.01) ]
BC MPa
3
44
2(1 10 )(0.02) 84.9
[(0.02) (0.01) ]
CD MPa

5-4
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PROBLEM(5.7) A stepped shaft ABC is fixed at the left end and carries the torques
B
T
and
C
T
at sections B and C as illustrated in Fig. P5.7. Calculate the maximum shearing stress in the shaft.
Given: d1 = 2 in., d2 = 1 ¾ in.,
B
T
= 30 kip
in.,
C
T
= 12 kip
in.
SOLUTION
12 .
BC
T kip in
18 .
AB
T kip in
Thus,
3
max 2Tc
gives
3
3
2(18 10 ) 11.46
(1)
AB ksi
3
3
2(12 10 ) 11.4
(7 8)
BC ksi
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PROBLEM (5.8) A stepped shaft ABC with fixed end at A is subjected to the torques
B
T
and
C
T
at sections B and C (Fig. P5.7). Determine the maximum shearing stress in the shaft.
Given: d1 = 60 mm, d2 = 50 mm,
C
T
= 3 kN
m,
B
T
= 2 kN
m
SOLUTION
31
BC AB
T kN m T kN m   
Then,
3
max 2Tc
yields
3
3
2(3 10 ) 122.2
(0.025)
BC MPa
3
3
2(1 10 ) 23.6
(0.03)
AB MPa
________________________________________________________________________
PROBLEM (5.9) Determine the values of the torques
B
T
and
C
T
applied at sections B and C so
that each segment of the shaft shown in Fig. P5.7 is stressed to a permissible shear strength of
all
.
Given: d1 = 2 3/8 in., d2 = 2 in.,
all
= 14 ksi
SOLUTION
3
max
max 3
2,2c TT
c
Thus,
33
14(10 ) (1) 21.99 .
2
C BC
T T kip in
33
14(10 ) (19 16) 36.83 .
2
BA B C
T T T kip in

5-5
or
21.99 36.83 58.82 .
B
T kip in   
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PROBLEM (5.10) The torques
B
T
and
C
T
are applied at sections B and C of the stepped shaft
shown in Fig. P5.7. Determine:
(a) The maximum permissible value of the torque
all
T
, if the allowable tensile stress in part BC is
all
.
(b) The allowable compressive stress in part BC.
Given: d1 = 100 mm, d2 = 50 mm,
B
T
= 12 kN
m,
all
= 140 MPa
SOLUTION
The polar moment of inertias are
4 6 4
(100) 9.817(10 )
32
AB
J mm
4 6 4
(50) 0.614(10 )
32
BC
J mm
In tension and compression (Figs. 5.8 and 5.10):
max max Tc
J
(a) We have
66
(0.025)
; 140(10 ) 0.614(10 )
BC BC C
all BC
T C T
J
or
3.438
C BC
T T kN m  
(b)
12 3.438 8.562
AB B C
T T T kN m     
3
6
8.562(10 )(0.05)
9.817(10 )
AB AB
all AB
Tc
J
43.61 MPa
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PROBLEM (5.11) A solid axle of diameter d is made of cast iron having an ultimate strength of
in tension
u
, ultimate strength in compression
u
’ , and ultimate strength in shear
u
. Calculate the
largest torque that may be applied to the axle.
Given: d = 50 mm,
u
= 170 MPa,
u
’ = 650 MPa,
u
= 240 MPa
SOLUTION
Formula (5.14a):
at
45o
at
135o
max 170 MPa
min 650 MPa