ch05

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
 all 
c
b
.

2
Tc
(c 4  b 4 )
or
80(106 ) 
2(2 103 )(0.03)
 [(0.03)4  b4 ]
or
80(106 )[0.81(106 )  b4 ]  38.197
Solving,
b  0.024 m  24 mm
________________________________________________________________________
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
t
Di 
d=2in.
D
 max 
D
25
T ( d 2)
T ( D 2)

4
 d 32  ( D 4  Di4 ) 32
or
23 4
) ]
25
(2)3  D3 (0.2836)
D  3.046 in.
d 3  D 3 [1  (
5-1
23
25
D
________________________________________________________________________
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
Di 
d
 max 
D
2
D
16T
T ( D 2)
16T


3

1
d
( D 4  Di4 )  D 3 (1  )
32
16
or
15
d 3  D3 ( ),
16
D  1.0217d
Ratio of weight:
D 2  ( D 2) 2 0.75(1.0217d ) 2

 0.783
d2
d2
________________________________________________________________________
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:
TEF  50 N  m
TDE  30 N  m
TCD  TBC  120 N  m
TAB  80 N  m
We have
J h  2 [(0.025)4  (0.015)4 ]  0.534 106 m4
Thus,
 max  Tc J :
2T
2(80)

 3.26 MPa
3
 c  (0.025)3
2(120)
120(0.025)

 4.89 MPa,
 CD 
 5.62 MPa
3
 (0.025)
0.534 106
 AB 
 BC
5-2
 DE 
30(0.025)
 1.40 MPa,
0.534 106
 EF 
50(0.025)
 2.34 MPa
0.534 106
 CD   max  5.62 MPa
So,
________________________________________________________________________
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:
TCD  1 kN  m
TBC  2.5 kN  m
TAB  2.5 kN  m
Therefore,  max  2T  c 3 :
2(2.5 103 )
 58.9 MPa
 (0.03)3
2(2.5 103 )
 BC 
 101.9 MPa
 (0.025)3
2(1103 )
 CD 
 79.6 MPa
 (0.02)3
________________________________________________________________________
PROBLEM (5.6) Redo Prob. 5.5, with a hole of 20-mm diameter drilled axially through the shaft
 AB 
to form a tube.
SOLUTION
We have, applying the method of sections:
TCD  1 kN  m
TBC  2.5 kN  m
TAB  2.5 kN  m
Hence,
 max 
2Tc
 (c 4  b 4 )
gives,
2(2.5 103 )(0.03)
 59.7 MPa
 [(0.03)4  (0.01)4 ]
2(2.5 103 )(0.025)

 104.5 MPa
 [(0.025)4  (0.01)4 ]
2(1103 )(0.02)

 84.9 MPa
 [(0.02)4  (0.01)4 ]
 AB 
 BC
 CD
5-3
________________________________________________________________________
PROBLEM(5.7) A stepped shaft ABC is fixed at the left end and carries the torques TB and TC
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., TB = 30 kip  in., TC = 12 kip  in.
SOLUTION
TBC  12 kip  in.
TAB  18 kip  in.
Thus,  max  2T  c 3 gives
2(18 103 )
 AB 
 11.46 ksi
 (1)3
2(12 103 )
 BC 
 11.4 ksi
 (7 8)3
________________________________________________________________________
PROBLEM (5.8) A stepped shaft ABC with fixed end at A is subjected to the torques TB and TC
at sections B and C (Fig. P5.7). Determine the maximum shearing stress in the shaft.
Given: d1 = 60 mm,
d2 = 50 mm, TC = 3 kN  m, TB = 2 kN  m
SOLUTION
TBC  3 kN  m
TAB  1 kN  m
Then,  max  2T  c yields
3
2(3 103 )
 122.2 MPa
 (0.025)3
2(1103 )

 23.6 MPa
 (0.03)3
 BC 
 AB
________________________________________________________________________
PROBLEM (5.9) Determine the values of the torques TB and TC 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
 max 
2T
,
 c3
T
 max c3
2
Thus,
14(103 ) (1)3
 21.99 kip  in.
2
14(103 ) (19 16)3
TBA  TB  TC 
 36.83 kip  in.
2
TC  TBC 
5-4
or
TB  21.99  36.83  58.82 kip  in.
________________________________________________________________________
PROBLEM (5.10) The torques TB and TC 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 Tall , 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, TB = 12 kN  m,
 all
= 140 MPa
SOLUTION
The polar moment of inertias are

J AB  (100) 4  9.817(106 ) mm 4
32

J BC  (50) 4  0.614(106 ) mm 4
32
In tension and compression (Figs. 5.8 and 5.10):
Tc
 max   max 
J
(a) We have  all 
TBC CBC
T (0.025)
; 140(106 )  C
J BC
0.614(10 6 )
TC  TBC  3.438 kN  m
or
(b) TAB  TB  TC  12  3.438  8.562 kN  m
TAB c AB 8.562(103 )(0.05)

J AB
9.817(106 )
 43.61 MPa
 all 
________________________________________________________________________
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
 max    170 MPa
at   135o
 min    650 MPa
5-5
The critical criterion is thus
Tc
 170 MPa
J
gives
170(106 ) (0.025)3
 4.17 kN  m
Tmax 
2
________________________________________________________________________
PROBLEM (5.12) A hollow shaft is made by rolling a plate of thickness t into a cylindrical shape
and welding the edges along the helical seams oriented at an angle  to the axis of the member (Fig.
P5.12). If the permissible tensile and shear stresses in the weld are  all and  all , respectively, what is
the maximum torque that can be applied to the shaft?
 = 60o,  all = 100 MPa,
Given: D = 120 mm,
t = 5 mm,
 all
= 55 MPa
SOLUTION
  60  90  150o

J  [(0.06) 4  (0.055) 4 ]  5.984 10 6 m 4
2
From Eq. (5.14a) at   150o :
 x '   sin 300o  0.866
Tc
T (0.06)
100 106  0.866  0.866
J
5.984 106
T  11.52 kN  m
or
Similarly, Eq. (5.14b):
 x ' y '   cos300o  0.5
gives
T (0.06)
55 106  0.5
,
T  10.97 kN  m
5.986 106
Thus,
Tall  10.97 kN  m
________________________________________________________________________
PROBLEM (5.13) Redo Prob. 5.12, assuming that the helical seam is oriented at an angle  =
40 o to the axis of the member (Fig. P5.12).
SOLUTION
From solution of Prob. 5.12:
J  5.984 106 m4
5-6
From Equations (5.14) at   40  90  130o ,
 x '   sin 260o  0.985
Tc
T (0.06)
100  (106 )  0.985  0.985
J
5.984 106
or
T  10.13 kN  m
and
 x ' y '   cos 260o  0.174
55 106  0.174
T (0.06)
,
5.984 106
T  31.52 kN  m
Thus,
Tall  10.13 kN  m
________________________________________________________________________
PROBLEM (5.14) A solid steel bar having the shear modulus of elasticity G and diameter d is
subjected to a torque T. Determine the maximum shear strain in the member.
Given: d = 2 in., T = 20 kip  in., G = 12 x 106 psi
SOLUTION
2T 2(20 103 )

 12.73 ksi
 c3
 (1)3

12.73(103 )
 max  max 
 1060 
G
12(106 )
________________________________________________________________________
PROBLEM (5.15) A tube of outer diameter D and inner diameter d experiences a torque at its
 max 
ends T, as shown in Fig. P5.15. At this loading, from measurement of a strain gage positioned at an
angle  it is calculated that shear strain  =1155  . Determine the shear modulus of elasticity G of
the material.
Given: D = 100 mm ,
d = 80 mm,
T = 10 kN  m,  = 60o
SOLUTION
J

[(0.05) 4  (0.04) 4 ]  5.796 106 m 4
2
Equations (5.14b) with   30o gives
   max cos(60o )  0.5 max
Thus,

0.5Tc
G 
 J (1155 106 )

0.5(10 103 )(0.05)
 37.3 GPa
5.796 106 (1155 106 )
5-7
________________________________________________________________________
PROBLEM (5.16) Rework Prob. 5.15 for the case in which a strain gage positioned at an angle 
= 25° (Fig. P5.15).
SOLUTION
From solution of Prob. 5.15:
J  5.796 106 m4
Equations (5.14b) with   65o :
   max cos(130o )  0.643 max
Therefore,
G

0.643Tc
(0.643)(10 103 )(0.05)


 J (1155 106 ) (5.796 106 )(1155 106 )
 48 GPa
________________________________________________________________________
PROBLEM (5.17) An electric motor at A exerts a torque of TA on part AB of the shaft ABCD of
the assembly shown in Fig. P5.17. Torques TB , TC , and TD transmitted through a gears B and C,
and pulley D, respectively. Calculate the maximum shear stresses in the parts AB, BC, and CD of the
shaft.
Given: dAB = 2 in., dBC = 1½ in., dCD = 1½ in., TA = 25 kip  in., TB = 12 kip  in.
TC = 8 kip  in.,
Assumptions:
TD = 5 kip  in.
The entire shaft is solid. Friction in the bearings may be neglected.
SOLUTION
For a circular shaft :  max  Tc J  2T  c 3 .
1
d AB  1 in.
2
2(25  103 )

 15.92 ksi
 (1)3
Shaft AB: TAB  25 kip  in. c AB 
 AB 
Shaft BC:
2TAB
 c 3AB
1
d BC  7 8 in.
2
2(13 103 )

 12.35 ksi
 (7 8)3
TBC  13 kip  in. cBC 
 BC 
2TBC
3
 cBC
5-8
1
d CD  3 4 in.
2
2T
2(5 103 )
 CD  CD

 7.54 ksi
3
 cCD
 (3 4)3
________________________________________________________________________
PROBLEM (5.18) An electric motor A exerts a torque TA on the assembly shown in Fig. P5.17.
Shaft CD: TCD  5 kip  in. cCD 
The shaft is made of steel having an allowable shear stress  all . Determine the required diameters dAB ,
dBC , and dCD for each part of the shaft.
Given: TA = 3 kN  m, TB = 1.4 kN  m, TC = 1.0 kN  m,
TD = 0. 6 kN  m,  all = 80 MPa
SOLUTION
For a circular shaft   16T  d 3 or d  3 16T  . We have  all  80 MPa
Shaft AB: TA  3 kN  m
d AB
16(3 103 )

 0.0576 m  57.6 mm
 (80 106 )
3
Shaft BC: TBC  1.6 kN  m
d BC 
3
16(1.6 103 )
 0.0467 m  46.7 mm
 (80 106 )
Shaft CD: TCD  0.6 kN  m
dCD
16(0.6 103 )

 0.0337 m  33.7 mm
 (80 106 )
3
________________________________________________________________________
PROBLEM (5.19) Solve Prob. 5.18 for the case in which shaft ABCD (shown in Fig. P5.17) is
hollow with outer diameter d and inner diameter d/2 throughout its entire length.
SOLUTION
T (d 2)
16Td (16)

4
 4 d 4
[d  ( ) ]  [15d ]
32
2
6
3
T  14.726(10 )d
 all  80 106 
or
Shaft AB: TA  3 kN  m
d AB 
3
3(103 )
 0.0588 m  58.8 mm
14.726(106 )
5-9
(1)
Shaft BC: TBC  1.6 kN  m
d BC 
3
1.6(103 )
 0.0477 m  47.7 mm
14.726(106 )
Shaft CD: TCD  0.6 kN  m
dCD 
3
0.6(103 )
 0.0344 m  34.4 mm
14.726(106 )
________________________________________________________________________
PROBLEM (*5.20) The two intersecting shafts are connected by bevel (or conical) gears as shown
in Fig. P5.20. Calculate the maximum permissible torque TA that can be applied to shafts A.
Given:  = 71o , dA = 15 mm, dB = 20 mm
Assumption: The gears are made of heat-treated steel for which the allowable shear stress is  all = 120
MPa
*SOLUTION
Shaft A :  all  120 MPa
cA 
1
d A  7.5 mm
2
J A all  3

 c A all  (7.5  103 )3 (120  106 )
cA
2
2
 79.52 N  m
TA 
rA

Shaft B :  all  120 MPa
rB
cB 
1
d B  10 mm
2
J B all  3

 cB all  (10  103 )3 (120  106 )
cB
2
2
 188.5 N  m
r
 1 
TA  A TB  
 TB
rB
 tan  
TB 
Statics :
 1 

(188.5)  64.91 N  m
o 
 tan 71 
Permissible value of TA is the smaller of the three:
(TA ) all  64.91 N  m
5-10
________________________________________________________________________
PROBLEM (5.21) Consider the aluminum shaft AF having shear modulus of elasticity G
loaded by the torques applied at the sections B, D, E, and F as illustrated in Fig. P5.21.
Determine:
(a) The angle of twist at C
(b) The angle of twist at F.
Given: a = 100 mm, G = 28 GPa
SOLUTION
We have:
Jh 

[(0.025) 4  (0.015) 4 ]  534 10 9 m 4
2
Applying method of sections, as needed:
TEF  70 N  m
TDE  30 N  m
TBC  TCD  150 N  m
TAB  100 N  m
(a) C   (TL GJ ) gives:
C 
0.1
(28 10 )
9

2
(100  150)
(0.025)
4
 0.291103 rad  0.017o
(b) F  C  CF result in
0.1
(150  30  70)
(28 10 )(534 109 )
rad  0.089o
F  0.291103 
 1.562 103
9
________________________________________________________________________
PROBLEM (5.22) Pulleys A, B, C, and D are attached to a solid steel shaft with shear
modulus of elasticity G and transmit the torques as illustrated in Fig. P5.22. Determine:
(a) The relative angle of twist in degrees between pulleys A and D.
(b) The relative angle of twist in degrees between pulleys A and C.
Given: L 1 = 0.5 m, L 2 = 1.5 m,
G = 80 GPa,
L 3 =1 m
SOLUTION
Apply the method of sections:
TCD  1.5 kN  m
TAB  3 kN  m
TBC  3.5 kN  m
(a)  AD   (TL GJ ) gives
5-11
 AD 
103
(80 109 )

[
3(0.5)
3.5(1.5)
3(1)


]
4
4
(0.035)
(0.03)
(0.025) 4
2
1

(1.000  6.48  7.68)  0.105 rad  6o
40
(b)  AC 
1
(1.000  6.48)  0.044 rad  2.5o
40
________________________________________________________________________
PROBLEM (5.23) A brass rod AB is bonded to an aluminum rod BC as shown in Fig. 5.23. What is
the angle of twist at C? The shear moduli of elasticity for brass and aluminum are Gb and Ga , respectively.
Given: Tb = 2Tc = 40 kip-in., d 1 = 2d 2 = 4 in., L 1 = 2L 2 = 16 in., Gb = 6 x 10 6 psi,
Ga = 4 x 10 6 psi
SOLUTION
We have
TBC  20 kips  in.
Then
C   (TL GJ ) yields
20(103 )
16
1
2
]
(8  )
3 4 100
3
(2)
2
2
 23.34 103 rad  1.34o
C 
[
8
TAB  20 kips  in.
4

(4 106 ) (1)

________________________________________________________________________
PROBLEM (5.24) Determine the torques TA , TB, and TC of the aluminum shaft having shear
modulus of elasticity G = 28 GPa in equiibrium, as shown in Fig. P5.24. The maximum shear
stress in segment AB is 80 MPa, and the rotation is 0.02 rad clockwise as viewed at A with
respect to C.
SOLUTION
T 0 :
TB  TA  TC
We have
TBA  TA
So,
(1)
TBC  TA  TB  TC  TC
2TA
2T
; 80(106 ) 
3
c
 (0.05)3
TA  15.71 kN  m
or
Therefore,
 AD   (TL GJ ) :
 max   AB 
5-12
(15.71)2 TC (1.5)

]
4
4
(0.05)
(0.0375)
9 
(80 10 )
2
4
87.965(10 )  502.72(104 )  75.85(104 )TC
0.02 
103
[
or
TC  5.47 kN  m
Equation (1) results in then
TB  15.71  5.47  21.18 kN  m
________________________________________________________________________
PROBLEM (5.25) Redo Prob. 5.24, given that the relative angle of twist between A and C is
zero.
SOLUTION
From solution of Prob. 5.24:
TBA  TA  15.72 kN  m
TBC  TC
Thus,    (TL GJ ) gives
103
(15.71)2 TC (1.5)

]
4
4
(0.05)
(0.0375)
9 
(28 10 )
2
TC  6.63 kN  m
502.72  75.85TC ,
0
Then,
[
TB  TA  TC  15.71  6.63  22.34 kN  m
________________________________________________________________________
PROBLEM (5.26) The solid aluminum rod AB is bonded to the solid bronze rod BC to form
a stepped composite shaft that supports the torques T 1 and T 2 acting as shown in Fig. P5.26.
Determine the permissible value of the torque T all , if the angle of twist at the free end is not to
exceed 3.5 o .
Given: Aluminum rod AB: d 1 = 60 mm, L 1 = 0.4 m, G a = 28 GPa, T 1 = T,
( a )all = 205 MPa, ( a ) all = 110 MPa
Bronze rod BC : d 2 = 75 mm, L 2 = 0.6 m, G b = 41 GPa,
( b )all = 260 MPa, ( b ) all = 140 MPa
T 2 = 3T
SOLUTION
TBC  4T
We have TAB  T


J AB  (60) 4  1.272(106 ) mm 4
J BC  (75) 4  3.106(106 ) mm4
32
32
Therefore
( ) J
110(106 )(1.272)(106 )
T  a all AB 
 4.664 kN  m
c AB
0.03
5-13
T
( b )all J BC 140(106 )(3.106)(106 )

 11.596 kN  m
cBC
0.0375
Hence
TAB L1 TBC L2

Ga J AB Gb J BC
T (0.4)
4T (0.6)




(3.5)
9
6
9
6
28(10 )(1.272)(10 ) 41(10 )(3.106)(10 ) 180
 AC   AB  BC 
Solving,
T  2.047 kN  m  4.664 kN  m
Tall  2.047 kN  m
________________________________________________________________________
PROBLEM (5.27) Three pulleys are attached to a solid stepped shaft with shear modulus of
elasticity G and transmit the torques as illustrated in Fig. P5.27. Determine:
(a) The maximum shear stress  max in the shaft.
(b) The angle of twist
(c) The angle of twist
 BC
 BD
between B and C.
between B and D.
Given: d 1 = 1¼ in.,
d 2 = 1 in., L 1 = 20 in.,
L 2 = 25 in.,
TD = 3.5 kip  in.
TB = 4.5 kip  in., TC = 8 kip  in.,
G = 4.1 x 10 6 psi
SOLUTION
TBC  4.5 kip  in.
TCD  3.5 kip  in.
16TAB 16(4.5 103 )

 11.73 ksi
3
 d AB
 (1.25)3
 17.82 ksi
(a)  AB 
 max
(b) Shaft BC
J BC 
TBC

d4 

(1.25) 4  0.24 in.4
32
32
 4.5 kip  in.
LBC  20 in.
So
BC 
(c) Shaft CD
J CD 
TCD
TBC LBC 4.5(103 )(20)

 0.09146 rad  5.2o
GJ BC
4.1(106 )(0.24)

d 24 

(1.0) 4  0.098 in.4
32
32
 3.5 kip  in.
5-14
LCD  25 in.
CD 
TCD LCD
3.5(103 )(25)

 0.21777 rad  12.5o
6
GJ CD
4.1(10 )(0.098)
Hence
BD  BC  CD  0.09146  0.21777  0.12631  7.24o
________________________________________________________________________
PROBLEM (5.28) A solid steel shaft of diameter d and shear modulus of elasticity G is
subjected to end torques T that produce an angle of twist per unit length of  L . Determine:
(a) The maximum tensile stress.
(b) The magnitude of the applied torque.
Given: d = 1 in.,  L = 0.02 rad/ft, G = 12 x 10 6 psi
SOLUTION
 max  G max  Gc(d dx)
 12 106 (0.5)(0.02 12)  10 ksi
(a) Equation (5.14a):
 max   max  10 ksi
(b) T 
 max J
c
10 103 ( )(0.5)3

 1963 lb  in.
2
________________________________________________________________________
PROBLEM (*5.29) A disk is attached to a 1½ in.-diameter, 20-in.-long steel shaft with shear
modulus of elasticity G = 12 x 10 6 psi, as shown in Fig. P5.29. In order to achieve a desired
natural frequency of torsional vibration, the stiffness of the system is specified so that the disk
will rotate 2° under a torque of T = 10 kip  in. How deep (x) must a 1-in.-diameter hole be
drilled to meet this requirement?
*SOLUTION
   (TL GJ )
2
T (20  x) Tx


180 
GJ s
GJ h

90

20  x
x

]
4
4
4

(12 106 ) (0.75) (0.75)  (0.5)
2
10(103 )
[
65.7974  3.1641(20  x)  3.9385 x
or
x  3.348 in.
5-15
________________________________________________________________________
PROBLEM (5.30) A steel rod of diameter d and shear modulus of elasticity G is rotated at a
constant speed in a hole which imposes a frictional torque of 10 N  m/m of contact length (Fig.
P5.30). Determine:
(a) The maximum contact length L if the shearing stress in the shaft is not to exceed 120 MPa.
(b) The relative angle of twist between A and B.
Given: d = 5 mm,
G = 80 GPa
SOLUTION
(a) Tx  10 x
16T
16(10 x)
 max  3 ; 120(106 ) 
d
 (0.005)3
x  0.295 m  295 mm  L
(b)  AB  
L
0
(10 x)dx
10


GJ
80 109 ( )(0.0025) 4
2
x2
 2.0372
2
 88.64 10

L
0
xdx
L
 1.0186(0.295) 2
0
3
rad  5.08o
________________________________________________________________________
PROBLEM (*5.31) A tapered round aluminum bar with shear modulus of elasticity G and
length L is subjected to end torques T, as shown in Fig. P5.31. Determine the angle of twist  .
Given: d a = 25 mm, d b = 75 mm,
L = 2 m,
T = 80 N  m, G = 28 GPa
*SOLUTION
75 mm
x
O
r
80 N m
d=1 m
B
A
2m
From geometry:
d d 2

,
d 1 m
25
75
r 0.0125

,
r  0.0125 x
x
1
 r4
Jx 
 38.35 109 x 4
2
Thus,
5-16
80 N m
3
3
Tdx
80
dx
3 x


 74.5 10
GJ x 28(38.35)  x 4
3 1
1
 24.834 103 (  1)
27
3
 23.914 10 rad  1.37o
________________________________________________________________________
PROBLEM (5.32) If the top of a 200-mm-diameter steel well drill pipe having shear
modulus of elasticity G = 80 GPa rotates through 90° at a depth of 400 m, calculate the maximum
shearing stress in the twisted pipe.
SOLUTION
  90o   2 rad .

TL
GJ
 max 
Tc
J
(1,2)
Divide Eq.(2) by Eq. (1):
 max cG
cG

,
 max 


L
L
Substitute the given data
0.1(80 109 )
 max 
 31.4 MPa
400(2)
________________________________________________________________________
PROBLEM (5.33) What is the magnitude of the largest allowable torque that a solid steel shaft (G
= 11.5 x 106 psi) 6 ft long and 2 in. in diameter can transmit if the maximum shear stress and the total
angle of twist are limited to 14 ksi and 0.05 rad, respectively?
SOLUTION
J

2
(1) 4  1.571 in.4
From   TL GJ , we obtain
 GJ
0.05(11.5 106 )(1.571)
L
6 12
 12.55 kip  in.
T

We have
 max 
2T
 c3
from which
5-17
T
 max ( c3 )
2
14 103 ( )(1)3

 21.99 kip  in.
2
Therefore,
Tall  12.55 kip  in.
________________________________________________________________________
PROBLEM (5.34) The circular shaft of diameter d and length L shown in Fig. P5.34 is acted upon
by a distributed torque of intensity t(x) which varies linearly from zero at the free end to a maximum
value of to at the fixed end. Determine:
(a) An expression for the total angle of twist  A of the free end of the shaft in terms of t0,
d, L , and shear modulus of elasticity G.
(b) The value of  A in degrees for t0 = 500 N  m/m, G = 28 GPa, L = 2 m, and d = 50 mm.
SOLUTION
x
t ( x )  t0
L
x2
Tx   t ( x)dx 
t0
0
2L
Tx dx
t
 0
GJ
2 LGJ
2
16t0 L

3 d 4G
(a)   
(b)  
x

L
0
x 2 dx 
t0 L2
6GJ
16(500)(22 )
 19.402 103 rad  1.11o
4
9
3 (0.05) (28 10 )
________________________________________________________________________
PROBLEM (5.35) Redo Prob. 5.34, with the intensity per unit length of distributed torque held
constant: t(x) = t0 =200 N  m/m.
SOLUTION
Tx dx
(t x)dx t0 L2
 0

GJ
GJ
2GJ
2
16t L
 04
d G
(a)   
(b)  
16(200)(22 )
 23.28 103 rad  1.33o
4
9
 (0.05) (28 10 )
5-18
________________________________________________________________________
PROBLEM (5.36) As illustrated in Fig. P5.36, two steel shafts with shear modulus of elasticity G
are connected by gears and subjected to a torque T. Determine:
(a) The angle of rotation in degrees at D.
(b) The maximum shearing stress in each shaft.
Given: d1 = 40 mm, d2 = 30 mm, T = 400 N  m, G = 80 GPa,
rB = 150 mm, rC = 100 mm
L1 = 1 m,
L2 = 1.5 m,
SOLUTION
F
C
100 mm
150 mm
B
F
TCD  T  F (0.1)  400 N  m
F  4 kN
TAB  4 103 (0.15)
 600 N  m  1.5T
TL
600(1)

 0.0299 rad
GJ 80 109 (  )(0.02) 4
2
400(1.5)
CD 
 0.0943 rad
9 
4
80 10 ( )(0.015)
2
D  CD  1.5AB  0.392 rad  7.97 o
(a)  AB 
2TAB
2(600)

 47.8 MPa
3
c
 (0.02)3
2T
2(400)
 CD3 
 75.5 MPa
c
 (0.015)3
(b)  AB 
 CD
________________________________________________________________________
PROBLEM (5.37) Calculate the torque T required to produce an angle of twist of 2o at D of the
gear-and-shaft system shown in Fig. P5.36. The shaft is made of bronze having shear modulus of elasticity
G.
Given: d1 = 60 mm,
d2 = 50 mm, G = 40 GPa, L1 = 2 m, L2 = 2.5 m, rB = 180 mm
rC = 120 mm
SOLUTION
From solution of Prob. 5.36:
 D   CD  1.5 AB ,
TCD  T ,
5-19
TAB  1.5T
Substitute the given data into the above:
2
T
2.5
2

[
 1.5
]
4
180  (40 109 )  (0.025)
(0.03) 4
2
21.9324  T (0.0640  0.0370)
T  217.2 N  m
________________________________________________________________________
PROBLEM (*5.38) Determine the maximum torque T that may be applied to the composite shaft
in Fig. P5.38 if the allowable shear stresses are ( s ) all in the steel core and ( b ) all in the brass tube
and if the angle of twist at the free end is not to exceed 3°.
Given: c = 50 mm,
b = 30 mm,
L= 2m
Brass : ( b ) all = 60 MPa, G a = 40 GPa
( s ) all = 80 MPa,
Steel :
G s = 80 GPa
*SOLUTION
Jb 
Js 

2

2
(0.054  0.034 )  8.545 106 m 4
(0.03) 4  1.272 106 m 4
Substitute the given data into Eq. (5.23):
3
T (2)
2T


180  (40  8,545)  (80 1, 272) 443,560
T  11.62 kN  m
Introduce the given values into Eq. (5.22):
(0.05  40 109 )T
60 106 
,
T  13.31 kN  m
443,560
and
(0.03  80 109 )T
80 106 
,
T  14.79 kN  m
443,560
Thus,
Tall  11.62 kN  m
________________________________________________________________________
PROBLEM (5.39) Consider the composite shaft in Fig. P5.38 with shear modulus of
elasticities G s = 80 GPa and G b = 40 GPa. If the brass tube carries 1.2 times as much torque as
does the steel core, determine the ratio of the external to the internal diameter of the tube.
SOLUTION
We have
5-20
 TL   TL 

 

 GJ  s  GJ b
substituting the given data
TL
(1.2T ) L



(80  109 ) b 4 (40 109 ) (c 4  b 4 )
2
2
or
2.4b 4  c 4  b 4
or
c
 1.358
b
________________________________________________________________________
PROBLEM (5.40) A solid steel shaft is fixed at its ends A and B, as illustrated in Fig.
P5.40. Determine the value of the permissible torque T applied at section C if the allowable shear
stress is  all = 12 ksi.
SOLUTION
TA
4 in.
A
7.5 ft
3 in.
T
C
B
TB
4.5 ft
Apply Eqs.(5.20):
T
 0.655T ,
7.5
4
1
(0.75)
4.5
 16T  d 3 :
TA 
Using  all
0.655T 
 d a3 all
TB  T  TA  0.345T
 (4)3 (12 103 )

16
16
T  230 kip  in.
Similarly,
 d 3
 (3)3 (12 103 )
0.345T  a all 
16
16
T  184 kip  in.
Thus,
Tall  184 kip  in.
________________________________________________________________________
PROBLEM (5.41) The round shaft AB in Fig. P5.40 is fixed to rigid walls at both ends and
subjected at section C to a torque T. Segment AC is made of aluminum with shear modulus of
elasticity G a and segment CB is made of brass having shear modulus of elasticity G a .
Calculate the maximum shear stress in each part.
Given:
T = 100 kip  in., G a = 4 x 10 6 psi, G b = 6 x 10 6 psi
5-21
SOLUTION
Statics TA  TB  100 kip  in.
Geometry & deformations
TA a (100  TB )b

Ga J a
Gb J b
Introducing the given number values
TA (7.5)
(100  TA )(4.5)



(4 106 ) (4) 4 (6 106 ) (3) 4
32
32
25600  256TA  202.6TA
or
TA  55.8 kip  in.
TB  100  55.8  44.2 kip  in.
Hence,
16T
16(55.8 103 )
 4.44 ksi
( max )a  ( 3 ) a 
d
 (4)3
( max )b  (
16T
16(44.2 103 )
 8.34 ksi
)

d3 b
 (3)3
________________________________________________________________________
PROBLEM (5.42) Redo Prob. 5.41, given that shaft AB is made of bronze, for which shear
modulus of elasticity G = 6 x 10 6 psi.
SOLUTION
Using Eqs. (5.20):
T
100
TA 

 65.5 kip  in.
7.5
a Jb
4
1
(0.75)
1
4.5
b Ja
TB  100  65.5  34.5 kip  in.
Thus,
16(65.5 103 )
 5.21 ksi
 (4)3
16(34.5 103 )
( max )b 
 6.51 ksi
 (3)3
( max )a 
5-22
________________________________________________________________________
PROBLEM (*5.43) A solid round shaft with fixed ends is subjected to torques as shown in
Fig.
(a)
(b)
(c)
P5.43.
Determine the reactions in terms of the applied torques (T D, T C ) and dimensions (a, b,c, L ) .
A ssuming that T C > T D , sketch the variation of torque along the length of the bar.
Calculate the maximum shear stress in the shaft, if d = 75 mm, a = 300 mm, b = 200 mm,
c = 100 mm, and T C = 2T D = 6 kN  m.
*SOLUTION
(a)Equations (5.21)
bT
aT
TA 
TB 
L
L
are expressed for this problem:
T (b  c) TD c TC (b  c)  TD c
TA  C


L
L
L
TC a TD (a  b) TC a  TD (a  b)
TB 


L
L
L
TC
(b)
(1)
(2)
TD
TA
TB
A
C
a
T
D
b
B
c
TAC
TA
-TBD
-TB
x
-TCD
(c) Introducing the given values:
6(0.3)  3(0.1)
TA 
 2.5 kN  m
0.6
6(0.3)  3(0.5)
TB 
 0.5 kN  m
0.6
We have
TAC  2.5 kN  m
TCD  3.5 kN  m
Thus,
16(3.5 103 )
 max 
 42.27 MPa
 (0.075)3
TBD  0.5 kN  m
________________________________________________________________________
PROBLEM (5.44) Resolve Prob. 5.43, given that the torque TD acts in the opposite
direction to that shown in Fig. P5.43.
5-23
SOLUTION
(a) In Eqs. (1) and (2) of the solution of Prob. 5.43, replace TD by TD :
T (b  c)  TD c
TA  C
L
TC a  TD (a  b)
TB 
L
(b)
T
TAC
TA
D
A
C
B
-TB
-TCD
x
-TBD
(c) Introduce the given data:
6(0.3)  3(0.1)
TA 
 3.5 kN  m  TAC
0.6
6(0.3)  3(0.5)
TB 
 5.5 kN  m  TBD
0.6
TCD  2.5 kN  m
16(5.5 103 )
 66.43 MPa
 (0.075)3
________________________________________________________________________
PROBLEM (5.45) The motor seen in Fig. P5.45 produces T A = 1200 lb  in. of torque and is
required to deliver T D = 900 lb  in. of torque to a gear at D; the relative angle of twist of the shaft
Therefore,
 max 
AD is 0.8 o . The shaft is made of steel with a shear modulus of elasticity G. Determine the
frictional torques T B and T C at bearings B and C.
Given: d = 1¼ in., L 1 = L 2 = L 3 = 1 ft, G = 11.4 x 10 6 psi.
SOLUTION
 5
( ) 4  239.68 103 in.4
2 8
Statics
TB  TC  1200  900  300 lb  in.
Geometry & deformations
0.8
 AB  BC  CD 
180 
or
1200 L1 (1200  TB ) L2 900aL3


 13.963 103
GJ
GJ
GJ
3
3 11.4(239.68)10
3300  TB  13.963 10
12
J
5-24
(1)
from which
TB  121 lb  in.
Equation (1):
TC  179 lb  in.
________________________________________________________________________
PROBLEM (5.46) A solid circular shaft (Fig. P5.46) with fixed ends is subjected to a
uniformly distributed torque of intensity t(x) = t 0 N  m/m. Determine:
(a) The reactions at the walls.
(b) The maximum shear stress in the shaft.
Given: d = 30 mm, L = 4 m, t 0 = 60 kN  m/m
SOLUTION
t0
TA
TB
A
B
L
T
TA
x
-TB
Assume TA as redundant.
x
Tx  TA   t0 dx  TA  t0 x
0
(a) Deformation
Tx dx TA

GJ
GJ
TA L t0 L2


GJ 2GJ
A  

L
0
dx 
t0
GJ

L
0
xdx
1
TA  t0 L
2
1
TB  t0 L
Statics: TA  TB  t0 L,
2
Geometry:  A  0,
2T 2(60  4 2)

 22.64 MPa
 c3  (0.015)3
________________________________________________________________________
PROBLEM (5.47) Resolve Prob. 5.46, given that the intensity of the torque varies linearly,
t(x) = (x / L ) t 0 N  m/m, along the length of the shaft.
(b)  max 
SOLUTION
Assume TA as redundant.
5-25
x x
1
Tx  TA   ( t0 )dx  TA 
t0 x 2
0 L
2L
(a) Deformation
L
Tx dx TA L
t
T L t L2

 0  x 2 dx  A  0
GJ
GJ GJL 0
GJ 6GJ
1
TA  t0 L
Geometry:  A  0,
6
1
1
TB  t0 L
Statics: TA  TB  t0 L,
2
3
A  
2T 2(60  4 3)
 15.1 MPa

 c3  (0.015)3
________________________________________________________________________
PROBLEM (5.48) Two transmission shafts-one a hollow tube with an outer diameter of 100
(b)  max 
mm and an inner diameter of 60 mm, the other solid with a diameter of 100 mm are each to
transmit 200 kW. If both operate at f = 6 Hz, compute the highest shearing stresses in each.
SOLUTION
T
159 P 159(200)

 5.3 kN  m
f
6
Jh 

32
Thus,
[(0.1) 4  (0.06) 4 ]  8.54 106 m 4
h 
Tc 5.3 103 (0.05)

 31.03 MPa
Jh
8.54 106
16T 16(5.3 103 )

 27 MPa
d3
 (0.1)3
 h  s  1.15
 h  1.15 s
________________________________________________________________________
PROBLEM (5.49) Design a solid aluminum shaft to transmit 300 hp at a speed of n = 1500
rpm, if the maximum shearing stress in the shaft is limited to  max = 12 ksi. Assume a factor of
safety ns = 1.5.
s 
SOLUTION
63, 000 P 63, 000(300)

 12.6 kip  in.
N
1500
J
T
J d3
12.6  103

;


c  all
c
16 12  103 1.5
d  2 in.
or
T
5-26
________________________________________________________________________
PROBLEM (5.50) Redo Prob. 5.49, with the additional requirement that the angle of twist in
a 6-ft length of the aluminum shaft having shear modulus of elasticity G = 4 x 10 6 psi is not to
exceed 3 o .
SOLUTION
  3 (180  )  52.36 103 rad .
From solution of Prob. 5.49: T  12.6 kip  in.
 c 4 TL
12.6 103 (6 12)
Then J 


2
 G (52.36 103 )(4 106 )
or
c  1.289 in.
d  2.578 in.
This is larger than 2 in., found in solution of Prob. 5.49. Thus, use:
37
d 2
in.
64
________________________________________________________________________
PROBLEM (5.51) The compound shaft, shown in Fig. P5.51, is attached to rigid supports at
A and C. Segment AB is of aluminum and segment BC is of steel. Calculate the diameters da
and ds at which each material will simultaneously be stressed to its limit.
Given: Aluminum : Ga = 28 GPa, ( a ) all = 40 MPa
Steel : Gs = 80 GPa, ( s ) all = 100 MPa
SOLUTION
T 
TL  L
 ,


J c
GJ cG
Geometry & deformations: a  s or
(40 106 )2
(100 106 )3

da
ds
(28 109 )
(80 109 )
2
2
42d a  32d s ,
d s  1.3125d a
Statics
 d 3
T  Ta  Ts  (
 d 3
)a  (
)s
16
16
Substitute the given values:
(40 106 ) d a3 (100 106 ) (1.3125d a )3
10 103 

16
16
1  785.4da3  4439.4da3
d a  0.0576 m  57.6 mm,
d s  75.6 mm
5-27
Use
d a  58 mm
d s  76 mm
________________________________________________________________________
PROBLEM (5.52) The shaft shown in Fig. 5.22 of Section 5.8 is to transmit 400 kW at f = 15
Hz without exceeding a yield shear stress of 250 MPa or a twisting through more than 3 o in a length
of L. The shaft is made of steel with shear modulus of elasticity G. Using a factor of safety of ns ,
calculate the diameter of the shaft.
Given: G = 80 GPa, L = 2 m, ns = 1.4
Assumption:
Friction at the bearings is neglected.
SOLUTION
159(400)
 4.24 kN  m
15
Using Eq. (5.27), we find that the shaft size needed to satisfy the stress
specification is
 3 4240(1.4)
c  24.7 103 m  24.7 mm
,
c 
2
250(106 )
T
The size of the shaft required to fulfill the distortion specification is
obtained from Eq. (5.28):
all T
3o ( 180o )
(4240)2

;

L GJ
2
(80 109 ) c 4
or
c  33.7 103 m  33.7 mm
Thus the minimum allowable diameter of the shaft must be
d  67.4 mm
________________________________________________________________________
PROBLEM (*5.53) A high-speed turbine rotates at about f = 100 Hz and drives, by means of a
helical gearset, a 250-kW generator at 12.5 Hz, as schematically shown in Fig. P5.53. The main shaft
AB and drive shaft CD are made of steel having the allowable shear strength  all and shear modulus of
elasticity G. Determine the diameters dAB and dCD for both shafts.
Requirement: The angle of twist in a length L of the drive shaft is limited to
Given:
L = 0.6 m,
 all = 120 MPa,
G = 80 GPa,
all
all .
= 3o
Assumptions: The friction at bearings is omitted. Moderate shock occurs on the turbine-generator set
and will be disregarded.
*SOLUTION
Main Shaft AB. Using Eq.(5.25):
159 P 159(250)
TAB 

 397.5 N  m
f
100
Equation(5.27) is then
5-28

2
c 3AB 
TAB
 all

397.5
120(106 )
or
cAB  0.0128 m  12.8 mm ,
d AB  25.6 mm
Drive Shaft CD. Applying Eq.(5.25):
159 P 159(250)
TCD 

 3.18 kN  m
f
12.5
Equation (5.27) gives
T
 3
3,180
cBC  BC 
,
cCD  0.0256 m
2
 all 120(106 )
So dCD  2cCD  51.2 mm
From Eq.(5.28)
all TBC 3( 180o )
(3,180)2

;

4
L GJ
0.6
(80 109 )( cCB
)
or
cCB  0.0232 m and dCB  2cCB  46.4 mm
Use dCD  51.2 mm
________________________________________________________________________
PROBLEM (5.54) An outboard motor delivers 8 hp at 5000 rpm to the drive shaft of length L
(Fig. P5.54). The shaft is made of brass with modulus of elasticity G and an allowable shear stress  all .
Determine:
(a) The required diameter d of the shaft.
(b) The angle of twist  of the shaft.
Given:
L = 0.5 m, G = 39 GPa,
 all = 70 MPa
SOLUTION
Equation (5.25) gives
9550 P 9550(8 0.7457)
T

 20.49 N  m
n
5000
(a) Equation (5.7) with c  d 2 and J   d 4 32 :
16T
16T
;
d3
3
d
 all
Substitute the data
16(20.49)
d3
 0.0114 m  11.4 mm
 (70 106 )
 all 
(b) By Eq.(5.15):
5-29
TL 32TL

GJ  Gd 4
32(20.49)(0.5)

 0.1584 rad  9.08o
 (39 109 )(0.0114)4

________________________________________________________________________
PROBLEM (*5.55) The electric motor exerts a torque T on the shaft ABCD at a constant speed
and the pulleys at B and D transmit torques TB and TD , respectively , as shown in Fig. P5.55. The
angle of twist betwee A and E is limited to all = 2o. The shaft is made of steel having shear modulus
of elasticity G and allowable shear stress
 all .
Calculate the minimum required diameter d of the shaft.
L1 = 0.3 m, L2 = 0.6 m, L 3 = 0.2 m, T = 1.2 kN  m,
G = 79 GPa,  all = 80 MPa
Given:
TB = 500 N  m,
*SOLUTION
Torques are
T  TAB  1.2 kN  m
TBD  700 N  m
Strength requirement, Using Eq.(5.27):
J
 c 4 all
c  all 
T
2T
Solving,
2T
2(1.2 103 )
c3 

 all  (80 106 )
or
c  0.0212 m  21.2 mm
TDE  0
Deformation requirement. We have AE  2o  0.0349 rad
Hence,
ED  0
TBD L2 (700)(0.6) 420


GJ
GJ
GJ
TBA L1 1, 200(0.3) 360



GJ
GJ
GJ
DB 
BA
Then
EA  ED  DB  BA 
780 2(780)

GJ
 Gc 4
or
2(780)
2(780)
4
 GEA
 (79 109 )(0.0349)
 0.0206 m  20.6 mm
Use larger value of diameter:
d  2c  2(21.2)  42.4 mm
c
4
5-30
TD = 700 N  m
________________________________________________________________________
PROBLEM (5.56) The electric motor delivers, through a solid shaft AE having a diameter d , 4
hp to the gear B and 3 hp to the gear D at a speed of n (Fig.P5.55). Calculate:
(a) The maximum shear stress in segment AB of the shaft.
(b) The maximum shear stress in segment BD of the shaft.
Given: d = ¾ in., n = 1500 rpm
Assumption: Friction at bearings C and E is neglected.
SOLUTION
(a) Torques are
63000 PB 63000(4)

 168 lb  in.
n
1500
63000 PD 63000(3)
TD 

 126 lb  in.
n
1500
Condition of torque equilibrium gives
TBA  TB  TD  294 lb  in.
TB 
TDB  TD  126 lb  in.
So
( max ) AB 
16TAB 16(294)

 3.55 ksi
 d 3  (3 4)3
16TBD 16(126)

 1.521 ksi
 d 3  (3 4)3
________________________________________________________________________
PROBLEM (5.57) Reconsider the motor, gears, and shaft assembly illustrated in Fig. P5.55, where
(b) ( max ) BD 
a solid shaft AE transfers 5 kW to gear B and 3.5 kW to gear D at a speed of 2200 rpm. The shaft is
made of B steel bar with a diameter d, allowable shear strength  all and shear modulus of elasticity G.
Determine:
(a) The minimum required diameter dmin .
(b) The angle of twist between the two gears,
Given: G = 79 GPa,
Assumption:
 all
= 50 MPa,
 BD
.
L2 = 0.4 m
Friction at bearings C and E is disregarded.
SOLUTION
(a) Torques are
9550 PB 9550(5)

 21.71 N  m
n
2200
9550 PD 9550(3.5)
TD 

 15.19 N  m
n
2200
By torque equilibrium condition:
TBA  TB  TD  36.9 N  m
TDB  TD  15.19 N  m
TB 
Inasmuch as TBA  TDB , design is on the basis of TBA .
5-31
Using the torque on formula, we obtain
16TAB
16(36.9)
d min  3
3
 all
 (50 106 )
 0.01554 m  15.54 mm
TDB L2 32 L2TDB

GJ
G d 4
32(0.4)(15.19)
 0.01343 rad  0.77o

9
4
(79 10 )( )(0.01554)
________________________________________________________________________
PROBLEM (*5.58) Figure P5.58 schematically illustrates an all-wheel-drive concept using two
(b) DB 
differentials, one for the front axle and another for the rear axle. The reardrive shaft is made of steel
tube with outer diameter D and allowable shear strength  all , and transmits 250 hp at 1000 rpm to the
rear differential. Determine the maximum permissible inner diameter d of the rear driveshaft.
Given: D = 3 in.,  all = 8 ksi.
*SOLUTION
63000hp 63000(250)

 15.75 kip  in.
n
1000
T ( D 2)
16TD
 all 

4
4

(D4  d 4 )  (D  d )
32
Solving,
16TD
d  4 D4 
T
 allow
Introducing the given data
16(15.75)(3)
 2.67 in.
d  4 (3)4 
 (8)
________________________________________________________________________
PROBLEM (5.59) Determine the required diameter d1, for the segment AB of the shaft shown in
Fig. P5.59, if the permissible shear stress is  all = 50 MPa and the total angle of twist between A and C
is not to exceed  AC = 0.02 rad.
Given:
G = 40 GPa,
TB = 2.5 kN  m,
TC = 0.5 kN  m ,
L1 = 3 m, L2 = 1 m,
SOLUTION
2.5 kN  m
2 kN  m
30 mm
d1
C
B
A
3m
1m
5-32
0.5 kN  m
d2 = 30 mm.
Apply the method of sections:
TAB  2 kN  m
TBC  0.5 kN  m
Then,
J  d13 T
2 103



c
16  all 50 106
d1  0.05884 m  58.84 mm
Also,
   (TL GJ ) :
or
103
0.5(1) 2(3)
0.02 
[
 4 ]
4
(0.04)
d1
9 
(40 10 )
32
6
78539.82  195312.5  4
d1
Solving
d1  84.66 103 m  84.66 mm
We therefore use
d1  85 mm
________________________________________________________________________
PROBLEM (5.60) Redo Prob. 5.59 for the case in which the torque applied at B is TB = 3 kN  m.
SOLUTION
40 mm
3 kN m
2.5 kN  m
d1
C
B
A
3m
1m
Apply the method of sections:
TAB  2.5 kN  m
TBC  0.5 kN  m
We have
J  d13 T
2.5 103



c
16  all 50 106
d1  63.38 103 m  63.38 mm
Also,
0.5 kN  m
   (TL GJ ) :
or
5-33
103
0.5(1) 2.5(3)

]
4
4
(0.04)
d
9 
1
(40 10 )
32
7.5
78539.82  195,312.5  4
d1
0.02 
[
Solving
d1  89.52 103 m  89.52 mm
Use
d1  90 mm
________________________________________________________________________
PROBLEM (5.61) The motor shown in Fig. P5.61 delivers 95 hp at 600 rpm. Gears B and C
transmit 55 hp and 40 hp, respectively, to operating machine tools. The shaft is made of steel having
allowable shear stress  all = 10 ksi and shear modulus of elasticity G = 11.5 x 10 6 psi. Determine:
(a) The required diameter of segments AB and BC of the shaft.
(b) The corresponding angle of twist in degrees between the ends A and C of the shaft.
SOLUTION
63, 000hp 63, 000(95)

 9975 lb  in.
N
600
63, 000(40)

 4200 lb  in.
600
TAB 
TBC
16T
16T
; d3 
3
d
 all
16(9975)

,
d AB  1.719 in.
 (10 103 )
16(4200)

,
d BC  2.139 in.
 (10 103 )
(a)  all 
3
d AB
3
d BC
T L
T L
TL
1

[ AB 4 AB  BC 4 BC ]
GJ G ( 32) d AB
d BC
32(12)
9975(6) 4200(9)

[

]
6
 (11.5 10 ) (1.719) 4 (2.139) 4
 10.629 106 (6854.261  1805.718)
(b)  AC  
 92.047 103 rad  5.27o
________________________________________________________________________
PROBLEM (5.62) Determine the maximum power in kilowatts that may be transmitted across the
steel shaft in Fig. P5.62 at a speed n, if the shearing stress is not to exceed
corresponding total angle of twist
 AD
between ends A and D?
5-34
 all .
What is the
Given: a = 3 m, d = 50 mm,
Assumption:
n = 600 rpm,
G = 80 GPa,
 all = 56 MPa
Friction at bearings B and C is neglected.
SOLUTION
 all d 3
16T
,
T

d3
16
6
3
(56 10 ) (0.05)
T
 1373.8 N
16
From Eq. (5.25):
Tf 1373.8(600 60)
P

 86.4 kW
159
159
Also,
TL
1373.8(3  3  3)
 AD  

GJ (80 109 )  (0.025) 4
2
3
 251.88 10 rad  14.43o
 all 
________________________________________________________________________
PROBLEM (5.63) Calculate the diameter d1 of shaft AB in Fig. P5.63 if the maximum shearing
stress in each shaft is not to exceed  max with a safety factor of ns.
Given: d2 = 60 mm, rB = 150 mm, rC = 100 mm,  max = 90 MPa, ns = 1.5
SOLUTION
90
 60 MPa
1.5
 ( d 23 ) (60 106 ) (0.06)3
 all

 2.53 kN  m
16
16
 all 
TCD
F
0.1 m
C
TCD 2.53 103

 25.3 kN
0.1
0.1
Thus,
0.3
TAB  25.3( )  3.795 kN  m
2
FC 
From  all  16T  d13 :
d13 
16(3.795 103 )
,
 (60 106 )
d1  68.55 103 m
Use
d1  69 mm
5-35
________________________________________________________________________
PROBLEM (5.64) A stepped shaft of diameters D and d is subjected to torque T (see Fig.
P5.64). The shaft has a fillet of radius r. Calculate:
(a) The maximum shear stress in the shaft for r = 2 mm.
(b) The maximum shear stress in the shaft for r = 6 mm.
Given: D = 80 mm, d = 60 mm,
T = 1.2 kN  m
SOLUTION
(a)
D 80

 1.33
d 60
r
2

 0.033
d 60
Figure 5.24: K  1.82 .
2T 2(1.2 103 )
 nom  3 
 28.31 MPa
c
 (0.03)3
 max  1.82(28.31)  51.52 MPa
(b)
r
6
D

 0.1,
 1.33; K  1.4
d 60
d
 max  1.4(28.21)  39.63 MPa
(Fig. 5.24)
________________________________________________________________________
PROBLEM (5.65) A stepped shaft consisting of solid circular parts with diameters D and d
transmits 100 kW (Fig. P5.64). The two parts are joined with a fillet of radius r. If the shaft is made of
brass with allowable shear strength  all , what is the required speed n ?
Given:
D = 60 mm,
d = 50 mm,
r = 10 mm,
 all
= 40 MPa
SOLUTION
r 10
D 60

 0.2

 1.25; K  1.2 (from Fig. 5.24)
d 50
d 50
For smaller part of the shaft, c  d 2  25 mm :
16T
16T
 max   all  K ( 3 ) ; 40(106 )  1.2[
], T  818 N  m
d
 (0.05)3
Equation (5.25) is then
159 P
159(100)
T
; 818 
, f  19.4 Hz
f
f
Thus
n  60 f  60(19.4)  1164 rpm
________________________________________________________________________
PROBLEM (5.66) A stepped steel shaft is to transmit 200 hp from an engine to a generator at a
speed of 150 rpm. If the allowable shearing stress is 14 ksi, determine the required fillet radius r at the
juncture of the 4-in.-diameter part with the 5-in. diameter part of the shaft.
5-36
SOLUTION
T
 all
63, 000 P 63, 000(200)

 84 kip  in.
N
150
16T
(14 103 ) (4)3
K
;
K

 2.094
d3
16(84 103 )
and
D 5
  1.25
d 4
r
Refer to Fig. 5.24:  0.025 .
d
Thus,
r  0.025(4)  0.10 in.
________________________________________________________________________
PROBLEM (5.67) Given the maximum allowable shearing stress in a stepped shaft (Fig.
5.64) of 8 ksi, calculate the maximum power in horsepower (hp) that may be transmitted at 240
rpm.
Given: D = 4 in., d = 2 in.,
r = 3/16 in.
SOLUTION
r 3 16
D 4

 0.094
 2
d
2
d 2
From Fig. 5.24: K  1.47 .
We have
 d3
16T
 all  K 3 ,
T  all
d
16 K
3
3
(8 10 ) (2)
T
 8549 lb  in.
16(1.47)
TN
8549(240)
P

 32.57 hp
63, 000
63, 000
________________________________________________________________________
PROBLEM (*5.68) The circular shaft shown in Fig. P5.68 is subjected to torques at C and D.
Shear module of elasticity of the brass and steel are G b = 40 GPa and G s = 80 GPa, respectively.
Based on a stress concentration factor of K = 1.5 at the step C in the shaft, determine:
(a) The angle of twist D at D.
(b) The maximum shearing stress in the shaft.
*SOLUTION
Apply the method of sections:
TCD  200 N  m
TAB  TBC  300 N  m
5-37
(a) D   (TL GJ ) gives
200(0.4)
300(0.1)
300(0.5)


]
4
4
4
4

(109 ) 80(0.02) 40(0.025) 40[(0.025)  (0.015) ]
2
 (3.98  1.22  7.03)103  4.27 103 rad  0.245o
D 
1
[
2T
 c3
2(200)

 15.92 MPa
 (0.02)3
( CD )max  1.5(15.92)  23.88 MPa
(b) ( CD ) nom 
We also have
Tc
300(0.025)


J
[(0.025) 4  (0.015) 4 ]
2
 14.05 MPa
________________________________________________________________________
PROBLEM (5.69) Consider the stepped shaft with the shear modulus of elasticity G and loaded as
( AB ) max 
shown in Fig. P5.69. Assuming that the applied torque produces a 1 o rotation of the free end A and a stress
concentration factor at the step of K = 1.2, determine the maximum shearing stress at section B of the shaft.
Given:
d 2 = 2d 1 = 3 in., L = 2L 1 = 7.2 ft, T B = -3T A , G = 4 x 10 6 psi
SOLUTION
1.5 in.
TA
3TA
3 in.
A
B
3.6 ft
C
2TA
7.2 ft
Apply the method of sections:
TBC  2TA
TAB  TA
Using    (TL GJ ) :
TA (12)
1
3.6
2(7.2)

[

]
4
180  (4 106 )  (0.75)
(1.5) 4
2
3
9.1386 10  TA (11.3778  2.8444)
TA  1.071 kip  in.
Thus,
2TA 2(1.071)

 1.616 ksi
 c3  (0.75)3
( max ) B  1.2(1.616)  1.939 ksi
( nom ) B 
5-38
________________________________________________________________________
PROBLEM (5.70) A circular shaft consisting of diameters D and d with a groove of radius r
is subjected to a torque T, as shown in Fig. P5.70. Determine:
(a) The minimum yield strength in shear  y required for the shaft material.
(b) The power that can be transmitted by the shaft at a speed of n = 800 rpm.
Given: D = 40 mm, d = 35 mm, r = 2 mm, T = 60 N  m
SOLUTION
(a) We have
D 40
r
2

 1.14

 0.057
d 35
d 35
Hence, by Fig. 5.25, K  1.6 .
We have


J  (d ) 4  (35) 4  147.3(103 ) mm 4
32
32
So
Tc
60(0.02)
y  K
 1.6[
]  13.04 MPa
J
147.3(109 )
(b) From Eq.(5.25),
Tn
60(800)
P

 5.03 kW
9550
9500
________________________________________________________________________
PROBLEM (5.71) The grooved shaft of diameters D and d illusrated in Fig. P5.70 is made
of heat treated steel with allowable shear stress  all . Determine the maximum torque and
corresponding horsepower that can be transmitted by the shaft at a speed of f = 20 Hz.
 all = 40 ksi
Given: D = 1 ¼ in., d = 5/8 in.,
r = 1/8 in.,
SOLUTION
We have
1
D 1 4
r 18


 0.2
d 58
d 58
From Fig. 5.25 , K  1.31
Tc
16T
 max   all  K ;
40(103 )  1.31[
]
J
 (5 8)3
or
T  1.464 kip  in.
Using Eq.(5.26):
1050 P
1050 P
T
; 1, 464 
f
20
Solving
P  27.9 hp
5-39
________________________________________________________________________
PROBLEM (5.72) The circular shaft consisting of diameters D and d carries a torque T, as
shown in Fig. P5.70. The shaft is made of steel having allowable shear strength  all . What is the
radius r of the smallest size groove that can be used to transmit the torque?
Given:
D = 75 mm, d = 30 mm, T = 180 N  m,  all = 50 MPa
SOLUTION
We have
16T
;
d3
K  1.47.
 max   all  K
Solving
50(106 )  K
16(180)
 (0.03)3
Hence, for
D 75

 2.5, K  1.47
d 30
Fig. 5.25 gives
r
 0.12,
r  3.6 mm
d
Comment: Note, as a check that, ( D  d ) 2  22.5 mm  3.6 mm
________________________________________________________________________
PROBLEM (5.73) A circular elastoplastic shaft of shear yield strength  y , shear modulus of
elasticity G, diameter d, and length L is twisted until the maximum shearing strain is 9000  (Fig.
P5.73). Calculate:
(a) The magnitude of the corresponding angle of twist
(b) The value of the applied torque T.
Given:
d = 60 mm,
L = 0.5 m,
G = 70 GPa,

 y = 180 MPa
SOLUTION
 max L
9000 106 (0.5)
 0.15 rad
c
0.03
 y 180 106
y  
 2600   9000 
G 70 109
and shaft is yielded.
(a)  

From similar
triangles:
0
30

2600 9000
or
0  8.67 mm
9000 
2600 
60 mm
20
5-40
(b) Elastic core. Use Eq.(5.32a):
03
T1 
y 

(0.00867)3 (180 106 )  184 N  m
2
2
Outer part. Use Eq.(5.32b):
2 3
2
T2 
(c  03 ) y 
(0.033  0.008673 )(180 106 )
3
3
 9.93 kN  m
Thus,
T  T1  T2  10.114 kN  m
Alternatively, use of Eq.(5.33) together with Eq.(5.31) yield the same result.
________________________________________________________________________
PROBLEM (5.74) A circular shaft of diameter d and length L is subjected to a torque T, as shown
in Fig. P5.73. The shaft is made of 6061-T6 aluminum alloy (see Table D.4), which is assumed to be
elastoplastic. Determine:
(a) The radius of the elastic core  0 .
(b) The angle of twist  .
Given: d = 2 in.,
L = 3.6 ft,
T = 40 kip  in.
SOLUTION
G  3.8 106 psi,
 y  20 ksi (Table B.4)
(a) For partially plastic shaft, using Eq.(5.33):

3T
6T
( 0 )3  4 
 4 3
c
Ty
c  y
Substituting the given values

6(40 103 )
( 0 )3  4 
 0.180
1
 (1)3 (20 103 )
0  0.565 in.
(b)  y 

0
L

y
G
,

y L
G 0
20 10 (3.6 12)
 0.4024 rad  23.06o
6
3.8 10 (0.565)
3
________________________________________________________________________
PROBLEM (*5.75) A 2-m-long circular shaft of 40-mm diameter is twisted such that only a 10-mmdiameter elastic core remains, as illustrated in Fig. P5.75. Upon removal of the load, calculate:
(a) The residual shear stresses.
(b) The residual angle of twist.
Assumption: The shaft is made of an elastoplastic material for which  y = 140 MPa and G = 80 GPa.
5-41
*SOLUTION
(a) Stress distribution for given condition is shown in Fig. a.
140 MPa
40 mm
T
93.5 MPa
T=0
10 mm
=
+
46.2 MPa
T
186.2
MPa
res=26.77o
=40.11o
’=13.34o
(b) Elastic-rebound
stresses
(a) Elastic-plastic
stresses
(c) Residual stresses
4  c3
1 
 yp [1  ( 0 )3 ]
3 2
4 c
3
2 (0.02)
5
T
(140 106 )[1  0.25( )3 ]  2.34 kN  m
3
20
Unloading of T causes elastic stresses (Fig. b):
Tc 2T 2(2.34 103 )
 '  3 
 186.2 MPa
J c
 (0.02)3
Residual stresses are sketched in Fig. (c).
Use Eq.(5.33): T 
(b)  y 
0
L

y
G
,

y L
G 0
140 106 2
 0.7 rad  40.11o
9
80 10 0.005
Rebound rotation:
TL
2.34 103 (2)
'

 13.34o
JG  (0.02) 4 (80 109 )
2
Thus,
res  40.11 13.34  26.77o

________________________________________________________________________
PROBLEM (5.76) A hollow shaft has an outside radius c and an inner radius b and is made of an
elastoplastic material (Fig. P5.76).
(a) Show that the yield torque and ultimate torque of the shaft are, respectively,
5-42
Ty 
yJ
c

y
c
(Jo  Ji )
(P5.76a)
and
J
J
2
4
Tu   y (c 3  b3 )   y ( o  i )
3
3
c b
(P5.76b)
The polar moment of inertia of the shaft is J =(c 4 - b4 )/2 = J o - Ji.
(b) What is the ratio of Tu to Ty , for the case in which c = 2b?
SOLUTION
yp
(a)
yp
c
b
Onset of yield
 y J  y ( 2)(c 4  b 4 )
Ty 

c
c
4
4
 y (c  b )

Q.E.D.
2c
Fully plastic deformation. Use Eq. (5.32b):
c
2
Tu  2 y   2 d    y (c 3  b3 )
b
3
(b) For b  c 2 :
c4
15
)
 y c3
2c
16
32
3
2
c
7
Tu 
 y (c 3  )   y c 3
3
8
12
Ty 

 y (c 4 
Thus,
Tu Ty  56 45,
Tu  1.244Ty
5-43
________________________________________________________________________
PROBLEM (*5.77) A hollow circular steel tube with the shear modulus of elasticity G and yield
stress in shear  y is subjected to a slowly increasing torque T at its ends. The tube has a length L, an
outside diameter 2c, and an inside diameter 2b, as shown in Fig.P5.76. Calculate:
(a) The yield torque Ty and the yield angle of twist  y .
(b) The ultimate torque Tu and the ultimate angle of twist
and (b) Plot a graph of T versus
.
u . Using the results obtained in parts (a)
(c) The residual stresses  res and the residual rotation res , when the shaft is unloaded.
 y = 160 MPa
Given:
D = 80 mm,
d = 60 mm,
L = 2 m, G = 80 MPa,
*SOLUTION
J

(c 4  b 4 ) 

(0.044  0.034 )
2
2
6
 2.7489 10 m4
yJ
160 106 (2.7489 106 )
 11 kN  m
c
0.04
 L  L 160 106 (2)
y  y  y 
c
Gc 80 109 (0.04)
 0.1 rad  5.73o
(a) Ty 

(b) Using Eq. (5.32b):
2
2
Tu 
 y (c 3  b 3 ) 
(160 106 )(0.043  0.033 )
3
3
 12.4 kN  m
At onset of yield at r = b, the deformation is plastic:
 L 160 106 (2)
u  y 
bG 0.03(80 109 )
 0.1333 rad  7.64o
The T   diagram is as shown below.
T
Tu
Ty
y u

5-44
Tu 12.4 103 (0.04)

 180.4 MPa
J
2.7489 106
b
30
 min   max  (180.4)  135.3 MPa
c
40
TL
12.4 103 (2)
' u 
JG 2.7489 106 (80 109 )
 0.1128 rad  6.46o
(c)  max 
res  7.64o  6.46o  1.18o
160 MPa
24.7 MPa
Tu
T=0
b
=
+
c
20.4 MPa
Tu
135.5
MPa
u=7.46o
180.4
MPa
res=1.18o
’=6.46o
Plastic stresses
Rebound stresses
Residual stresses
________________________________________________________________________
PROBLEM (5.78) A hollow circular shaft of outer radius c , inner radius b, and length L is made
of an aluminum alloy 2014-T6 that is taken to be elastoplastic having a yield strength in shear  y and
shear modulus of elasticity G, and carries a torque T (Fig. P5.76). Determine:
(a) The yield torque Ty of the shaft.
(b) The ultimate torque Tu of the shaft
(c) The ultimate angle of twist u and residual angle of twist res when the shaft is unloaded.
Given:
b = 1¼ in.,
c = 2 in.,
L = 5 ft ,
G = 4.1 x 106 psi
SOLUTION
We have
J
(a)  y 
Ty c
J

2
(c 4  b 4 ) 
; 32 

2
(24  1.54 )  17.18 in.4
Ty (2)
17.18
or
Ty  274.9 kip  in.
5-45
 y = 32 ksi
(from Table B.4)
(b) Using Eq.(5.32b), with  0  b and T2  Tu :
2 3 3
2
Tu 
[c  b ] y 
[(2)3  (1.5)3 ](32)
3
3
 310 kip  in.
(c) At onset of yield at r = b, deformation is plastic:
 yL
32(103 )(5 12)
u 

bG (1.250)(4.1106 )
 0.3746 rad  21.67o
Elastic rebound rotation
TL 310(103 )(5 12)
'

GJ (4.1106 )(17.18)
 0.2641 rad  15.13o
Thus
res  21.67 15.13  6.54o
________________________________________________________________________
PROBLEM (*5.79) A solid shaft of radius c (Fig. P5.79a) is made of a strain-hardening material
having a    diagram that can be approximated as shown in Fig. P5.79b. Determine the torque T
required to develop an elastic core in the solid with radius of 0.
Given: c = 15 mm, 0 = 10 mm
*SOLUTION
Refer to P5.79b:
 1 60(106 )


0.005
or
1  12(109 )
Also
 2  60(106 ) 90(106 )  60(106 )

  0.005
0.01  0.005
or
 2  6(109 )  30(106 )
(1)
(2)

From Fig.(a):
 max 
c
  1.5(0.005)

 0.0075


   max  (0.0075)  0.5
c
15
5-46

c
Figure (a)
max
Introducing  into Eqs.(1) and (2):
1  6(109 )  ,
 2  3(109 )   30(106 )
Equation (5.30) is therefore
c
T  2   2 d 
0
 2 {
0.01
0
6(109 )  3 d   
0.015
0.01
[3(109 )   3(106 )  2 ]d }
6(109 )(0.01) 4 3(109 )
30(106 )

(0.0154  0.014 ) 
(0.0153  0.013 )}
4
4
3
 2 {15  30.469  23.75}
 434.9 N  m
 2 {
________________________________________________________________________
PROBLEM (*5.80) The fixed-ended shaft shown in Fig. P5.80 is made of an elastoplastic material
for which the shear modulus of elasticity is G and the yield stress in shear is  y . Assuming that the angle
of twist at step C is
Given:
a = 2 m,
C
= 0.2 rad, calculate the magnitude of the applied torque T.
b = 1.5 m,
d1 = 100 mm,
d2 = 60 mm,
G = 80 MPa,
 y = 240 MPa
*SOLUTION
100 mm
A
y
60 mm
Tc
a=2 m
b=1.5 m
C
B
240 106
 3000 
G
80 109
c 0.05(0.2)
( max ) AC 

 5000 
a
2
0.03(0.2)
( max )CB 
 4000 
1.5
y 

Both segments are yielded and partially plastic.
Segment AC
0 
c y
 max
c
6
0.05(3000 10 )
 30 mm
5000 106
 c3
 (0.05)3
Ty 
y 
(240 106 )  47.124 kN  m
2
2

max
5-47
o
y
Use Eq. (5.33):
4
1 3
TAC  (47,124)[1  ( )3 ]  59.44 kN  m
3
4 5
Segment BC
c
0.03(3000 106 )
0  y 
 22.5 mm
 max
4000 106
Ty 
 c3
y 
 (0.03)3
(240 106 )  10.179 kN  m
2
2
Use Eq. (5.33):
4
1 22.5 3
TBC  (10,179)[1  (
) ]  12.14 kN  m
3
4 30
Total applied torque is therefore
T  TAC  TBC  71.58 kN  m
________________________________________________________________________
PROBLEM (5.81) Resolve Prob. 5.80 for the case in which d1 = d2 = 80 mm and a = 2b =
2.2 m.
SOUTION
80 mm
T
C
A
1.1 m
2.2 m
B
240 106
y 
 3000 
80 109
c 0.04(0.2)
( max ) AC 

 3637 
a
2.2
c 0.04(0.2)
( max )CB 

 7273 
b
1.1
Both segments are yielded and partially plastic.
Segment AC
0 
y 
c y
 max
 c3
0.04(3000  106 )

 33 mm
3637 106
y 

(0.04)3 (240 106 )  24.13 kN  m
2
2
Use Eq. (5.33):
4
1 33
TAC  (24,130)[1  ( )3 ]  27.66 kN  m
3
4 40
Segment CB
5-48
o 
c y
 max
Use Eq. (12.2c):

0.04(3000 106 )
 16 mm
7237 106
4
1 16
(24,130)[1  ( )3 ]  31.66 kN  m
3
4 40
T  TAC  TCB  59.32 kN  m
TBC 
Thus
________________________________________________________________________
PROBLEM (5.82) Determine the ultimate load (Tu) for the shaft illustrated in Fig. P5.80.
Given:
d 1 = 2 d 2 = 2 in.,
a = 3 ft,
b = 3 ft,  y  22 ksi
SOLUTION
2 in.
1 in.
T
A
B
C
a
b
Apply Eq. (5.31):
2 c3
2 3
(Tu ) AC 
y 
(1) (22 103 )  46.08 kip  in.
3
3
2
(Tu )CB 
(0.5)3 (22 103 )  5.76 kip  in.
3
Thus,
Tu  (Tu ) AC  (Tu )CB  51.84 kip  in.
________________________________________________________________________
PROBLEM (5.83) A rectangular bar of width b, depth a, and length L is fixed at one end and
carries a torque T at the other end, as shown in Fig. P5.83. Determine:
(a) The maximum shearing stress in the bar.
(b) The angle of twist  A at end A.
Given:
a = 75 mm,
b = 30 mm,
L = 0.5 m,
T = 600 N  m,
G = 39 GPa
SOLUTION
(a)
a 75

 2.5
b 30
Table 5.1:
 max 
T
600

 34.72 MPa
2
 ab (0.256)(0.075)(0.03) 2
(b) Table 5.1:
TL
A 
 ab3G
5-49
600(0.5)
 14.44 103 .
3
9
(0.263)(0.075)(0.03) (39 10 )
 14.44 103 rad  0.828o
________________________________________________________________________
PROBLEM (5.84) A torque T is applied to the steel bar having the allowable shear strength  all

(Fig. P5.83). Determine:
(a) The required dimension b, for a bar of rectangular cross section, with a = 2b.
(b) The required dimension b, for a bar of square cross section, with a = b.
Given: T = 2.5 kip in.,  all = 8 ksi
SOLUTION
(a) With a=2b, from Table 5.1:
T
T
 max 
,
b3 
2
 (2b)b
2 max
Substituting the given data,
2.5 103
b3 
,
b  0.86 in.
2(0.246)(8 103 )
(b) With a=b, by Table 5.1:
T
T
 max 
,
b3 
2
 (b)b
 max
Thus
b
3
2.5 103
 1.145 in.
(0.208)(8 103 )
________________________________________________________________________
PROBLEM (5.85) A circular, a square, and a rectangular shaft are subjected to torques at their
ends (Fig. P5.85). The shafts have the same cross-sectional area A and the same allowable shear
strength  all . Determine the ratios Tc / Ts and Tc / Tr of the largest torques that may be applied to the
shafts.
SOLUTION
A  Ac  As  At  a 2
Circular Shaft:
 c 2  a 2 , c  0.5642a
 J 

Tc  all c   all c3   all (0.5642a)3
c
2
2
3
 0.282 max a
5-50
Square Shaft (Table 5.1):
 c 2  a 2 , c  0.5642a
T
Ts
 all  s 3 
 a (0.208)a3
Ts  0.208 max a3
Rectangular Shaft (Table 5.1):
T
Tr
 all  r 2 
a
 ab
(0.267)(3a)( ) 2
3
3
Tr  0.089 max a
Hence
Tc
Tc
 1.36
 2.12
Ts
Tr
Note that, the circular shaft can support 35% and 120% more torques than a
square and a rectangle with sides 6:1 ratio, respectively.
________________________________________________________________________
PROBLEM (5.86) An ASTM –A36 structural steel bar of an equilateral triangular cross-sectional
shape of sides a and length L , is fixed at one end and twisted on the other end, as illustrated in Fig.
P5.85. Determine the value of the largest torque T that can be carried by the bar.
Requirements: Angle of twist of the bar is limited to all = 5o. A safety factor of ns = 2.5 with
respect to failure by yielding is to be used.
Given a = 40 mm,
L = 3 ft,
G = 79 GPa,
 all
= 145 MPa
(by Table B.4)
SOLUTION
We have  all   y ns  145 2.5  58 MPa
From Table 5.1:
20T
20T
 all   A  3 ; 58(106 ) 
,
T  185.6 N  m
a
(0.04)3
and
46.2TL

46.2T (3 12)
all  4 ; 5( ) 
aG
180 (0.04) 4 (79 109 )
or
T  10.61 N  m
Tall  10.61 N  m
Hence
________________________________________________________________________
5-51
PROBLEM (5.87) A shaft having an elliptical cross section of semi axes a by b and length L is
made of aluminum alloy with the shear modulus of elasticity G, and carries a torque T as shown in Fig.
P5.87. Calculate:
(a) The largest permissible torque that can be transmitted by the shaft, if the allowable stress in shear is
 all .
(b) The angle of twist  of the shaft.
(c) Power (in kW) transmitted by the shaftat a speed of f.
Given: a = 30 mm, b = 15 mm, L = 2.5 m,
f = 20 Hz,
G = 26 x 106 psi,
 all = 60 MPa
SOLUTION
(a) Maximum shear stress (Table 5.1):
2T
2Tall
 max   all  all2 ; 60(106 ) 
 ab
 (0.03)(0.015)2
or
Tall  636.2 N  m
(b) Angle of twist (Table 5.1):
(a 2  b2 )Tall L (0.032  0.0152 )(636.2)(2.5)
all 

 a3b3G
 (0.033 )(0.0153 )(26 109 )
 0.2404 rad  13.77o
(c) Power:
Tall f (636.2)(20)

 80.03 kW
159
159
________________________________________________________________________
PROBLEM (5.88) The ASTM-A242 high-strength steel bar having a length L an equilateral cross
Pall 
section of sides a is subjected to torques T1 , T2 , and T3
(a) The maximum shear stress  max in the bar.
as illustrated in Fig. P5.88. Determine:
(b) The rotation of one end relative to the other end of the bar.
Given: G = 79 GPa (by Table B.4),
a = 25 mm,
L1 = 0.6 m,
T2 = 50 N  m, T3 = 25 N  m
SOLUTION
From the condition of torque equilibrium:
TAC  T1  75 N  m
TBC  25 N  m
(a) Maximum shear stress (by Table 5.1):
20T
20(75)
 max  3AB 
 96 MPa
a
(0.025)3
(b) Angle of twist (from Table 5.1):
5-52
L2 =1.4 m,
T1 = 75 N  m
46.2 46.2

 TL
a 4G a 4G
46.2

(75  0.6  25 1.4)
(0.025) 4 (79 109 )
 0.1198 rad  6.86o
 AB  
________________________________________________________________________
PROBLEM (5.89) The ASTM-A48 gray cast iron square bar of sides a and length L with fixed
ends is acted upon by a torque T at section C, as illustrated in Fig. P5.89. Determine:
(a) The support reactions.
(b) The angle of twist c at C.
(c) The maximum shear stress in the part AC of the rod.
Given: G = 28 GPa (by Table B.4), a = 50 mm, L = 1.5 m,
T = 100 N  m
SOLUTION
(a) Equation (5.13) and condition of torque equilibrium give
T
100
TB 

 40 N  m
1  ( LCB LAC ) 1  (1.5)
TA  T  TB  100  40  60 N  m
(b) Refer to Table 5.1. For a b  1 :
T L
40(0.6 1.5)
C  B 4BC 
 a G (0.141)(0.05) 4 (28 109 )
 0.00146 rad  0.084o
(c)  AC 
TA
60

 2.31 MPa
3
 a (0.208)(0.05)3
________________________________________________________________________
PROBLEM (5.90) A bronze bar that has length L, elliptical cross section of semiaxes a by b (see
Table 5.1), and built-in ends carries a torque T at a section C (Fig. P5.89). Calculate:
(a) The reactions at supports.
(b) The angle of twist C at C.
(c) The maximum shear stress in part CB of the bar.
Given: G = 41 GPa (from Table B.4), a = 20 mm,
b = 10 mm,
SOLUTION
(a) From Equation (5.13) and equilibrium, we have
T
150
TB 

 60 N  m
1  ( LB LC ) 1  (1.5)
TA  T  TB  150  60  90 N  m
5-53
L = 3 m,
T = 150 N  m
(b) Table 5.1 gives
(a 2  b 2 )TB LBC
C 
 a 3b3G
(0.022  0.012 )(60)(0.6  3)

 (0.02)3 (0.01)3 (41109 )
 0.0524 rad  3o
2TB
2(60)

2
 ab  (0.02)(0.01) 2
 19.1 MPa
________________________________________________________________________
PROBLEM (5.91) A bar having a length L and an equilateral triangular cross section of sides a
(c)  BC 
(see Table 5.1) is fixed at both ends, as shown in Fig. P5.89. The bar is made of structural steel having a
shear yield strength of  all . What is the largest torque Tall that may be applied to the bar at section C to
the bar , based on a safety factor of ns with respect to failure by yielding?
Given:  y = 21 ksi (by Table B.4), a = 1½ in., L = 5 ft,
ns = 1.4
SOLUTION
By Eq. (5.13) and equilibrium condition of torque, we have
T
T
TB 

 0.4T
1  ( LCB LAC ) 1  (1.5)
TA  T  TB  T  0.4  0.6T
Refer to Table 5.1:
20Tmax
21 20(0.6Tall )
 max 
;

3
a
1.4
(1.5)3
or
Tall  4.2195 kip  in.
________________________________________________________________________
PROBLEM (5.92) A uniform thin-walled shaft of length L and rectangular cross section having
mean width a by mean depth b is fixed at one end and subjected to a torque T at the free end, as shown
in Fig. P5.92. The shaft is made of sheet metal with a shear modulus of elasticity G. Determine the
smallest required wall thickness t if the shearing stress and the angle of twist per unit length are not to
exceed  all = 40 MPa and all = 0.03 rad/m, respectively.
Given: a = 2b = 100 mm, T = 1.4 kN  m, G = 28 GPa
Assumption: The effect of stress concentration at the corners is disregarded.
SOLUTION
t
50 mm
Am  0.1 0.05  5 103 m2
100 mm
5-54

T
1.4 103
; 40 106 
2 Amt
2(5 103 )t
or
t (5 103 )  17.5 106
Solving:
t  3.5 mm
Equation (5.42):

T

L 4 Am2 G

ds
t
1.4 103[2(0.1)  2(0.050]
0.03 
4(5 103 )2 (28 109 )t
or
t (25 106 )  12.5 108
Solving: t  5 mm
Thus,
tall  5 mm
________________________________________________________________________
PROBLEM (5.93) Rework Prob. 5.92, assuming that the cross section of the shaft is a square (a =
b = 80 mm) box of uniform thickness t.
SOLUTION
t
Am  (0.08)2
80 mm
80 mm

T
1.4 103
; 40 106 
2 Amt
2(0.08) 2 t
or
t (6.4 103 )  17.5 106
Solving: t  2.73 mm
Equation (5.42):

T
0.08T

(4  0.08)  2
2
L 4 AmGt
AmGt
5-55
0.03 
0.08(1.4 103 )
(0.08)4 (28 109 )t
or
t (4.096 105 ) 4  13.33 108
t  3.255 mm
Solving:
Therefore,
tall  3.26 mm
________________________________________________________________________
PROBLEM (*5.94) The brass tube having a modulus of elasticity in shear G, length L, and an a x
a square cross section of uniform thickness t is twisted as shown in Fig. P5.92. The allowable yield stress
in shear and the angle of twist of the tube are  all and all , respectively. Determine the largest torque
and the power that can be transmitted by the tube at a speed of n .
Given: a = 1 in.,
t = ¼ in.,
L = 9 ft, G = 5.6 x 106 psi,
all
= 15o
*SOLUTION
Area bounded by center line is Am  a 2  (1)(1) in.2
So,
T
T
 all 
;
6 103 
1
2 Amt
2(1)( )
4
or
T  3 kip  in.
Similarly,
all 
So
TL
4 Am2 G

ds TLa TL
 2 
t
AmGt Gt

T (9 12)
15(
)
180 (1) 2 (5.6 106 )( 1 )
4
or
T  3.39 kip  in.
Tall  3 kip  in.
Hence
The corresponding power is
Tn
3, 000(200)
 9.52 hp
P

63, 000
63, 000
5-56
n = 200 rpm,
 all
= 6 ksi,
________________________________________________________________________
PROBLEMS (5.95 and 5.96) The aluminum alloy shafts with shear modulus of elasticity G and
cross sections shown in Figs. P5.95 and P5.96 are each subjected to a torque of T. Neglecting the effect
of stress concentration, calculate, for each shape:
(a) The maximum shear stress.
(b) The angle of twist in a 2-m length.
Given: T = 2 kN  m, G = 28 GPa
SOLUTION (5.95)
Am  (100  3)(200 1)   (50  2)2
 22.917 103 m2
(a) Apply Eq. (5.41),   T 2 Amt :
2 103
 AC 
 10.91 MPa
2(22.917 103 )(4 103 )
4
 AB   CD   AC ( )  14.55 MPa
3
 BD   AC (2)  21.82 MPa
(b)  
TL
4 Am2 G

ds
t
or
2 103 (2)
2(199)  (48) 97
[

 ]
3 2
9
4(22.917 10 ) (28 10 )
3
4
2
6
 68.003 10 (218.869)
 14.88 103 rad  0.85o

SOLUTION (5.96)
B
3 mm
4 mm
A
60o
103.93 mm
2 mm
60o
C
120 mm
Am  12 (120 103.93)  6.236 103 m2
(a) Equation (5.41),   T 2 Amt :
5-57
2 103
 40.09 MPa
2(6.236 103 )(4 103 )
  AB (2)  80.18 MPa
4
  AB ( )  53.57 MPa
3
 AB 
 AC
 BC
(b)  
TL
4 Am2 G

ds
t
or
2 103 (2)
120 120 120
[


]
3 2
9
4(6.236 10 ) (28 10 ) 4
3
2
 119.39 103 rad  6.84o
________________________________________________________________________
PROBLEM (5.97) The cross section of a thin-walled structure made of a brass with shear

modulus of elasticity G = 39 GPa is approximated as shown in Fig.P5.95.If the angle of twist per unit
length is limited to 0.4° per meter of length, calculate the maximum allowable torque Tall and the
corresponding maximum shearing stress  all .
SOLUTION
From solution of Prob. 5.95:
Am  22.917 103 m 2

ds
 218.869
t
Apply Eq. (5.42):

T
ds

2

L 4 Am G t
Tall (218.869)
0.4

180  4(22.917 10 3 ) 2 (39 109 )
Solving,
Tall  2.619 kN  m
Use Eq. (5.41),   T 2 Amt :
 all   BD 
2.619 103
 28.6 MPa
2(22.917 103 )(2 103 )
________________________________________________________________________
PROBLEM (5.98 and 5.99) The aluminum shafts with shear modulus of elasticity G and cross
sections illustrated in Figs. P5.98 and P5.99 each carries a torque of T. Disregarding the effect of stress
concentration, determine, for each shape:
(a) The maximum shear stress.
(b) The angle of twist in a 3-m length.
Given: T = 1.5 kN  m, G = 28 GPa
5-58
SOLUTION (5.98)
A
6 mm
60o
194 mm
B
97 mm

194 168
2
2
3
 31.068 10 m2
Am 
194 mm
(97) 2 
4 mm
C
168 mm
(a) Apply Eq. (5.41),   T 2 Amt :
1.5 103

 6.04 MPa
2(31.068 103 )(4 103 )
4
  AB ( )  4.02 MPa
6
 AB
 AC
(b)  
TL
4 Am2 G

ds
t
or
1.5 103 (3)
194  97

[2

]
3 2
9
4(31.068 10 ) (28 10 )
4
2 6
122.395
6
 41.63 10 (122.395)  5.095 103 rad  0.29o
SOLUTION (5.99)
Am 

4
(97) 2  7.386 103 m 2
(a) Use Eq. (5.41),   T 2 Amt :
1.5 103
  AC 
2(7.386 103 )(4 103 )
 25.4 MPa
  AC (2)  50.8 MPa
 BC
 AB
(b)  
TL
4 Am2 G

ds
t
or
1.5 103 (3)
97  97
[2 
]
3 2
9
4(7.386 10 ) (28 10 ) 4 2 6
 0.0544 rad  3.12o

5-59
________________________________________________________________________
PROBLEM (5.100) The cross section of an airplane fuselage made of an aluminum alloy with
shear modulus of elasticity G = 28 GPa is approximated as shown in Fig.P5.98. If the angle of twist
per unit length is limited to 0.4° per meter of length of the fuselage, calculate the maximum allowable
torque Tall and the corresponding maximum shearing stress  all .
SOLUTION
From solution of Prob. 5.98:

Am  31.068 103 m 2
ds
 122.395
t
Use Eq. (5.42):

T
ds

2

L 4 Am G t
Tall (122.395)
0.4

180  4(31.068 10 3 ) 2 (28 10 9 )
from which
Tall  6.17 kN  m
Use Eq. (5.41),   T 2 Amt :
 all   AB   BC 
6.17 103
 24.82 MPa
2(31.068 103 )(4 103 )
________________________________________________________________________
PROBLEM (*5.101) Structural aluminum tubing of equilateral triangular cross sectio n of sides
a with a uniform thickness t carries a torque T as shown in Fig. P5.101. Determine the shear stress at
a point A away from the corners.
Given: a = 3.5 in.,
t = 5/16 in., T = 6 kip  in.
Assumption: Effect of stress concentration at corners is neglected.
*SOLUTION
Refer to Fig.(a):
1
t  e tan 30o
2
60o 60o
b
60o
3b
2
30o
t/2
e
Figure (a)
5-60
or
and
e
t
5 16

 0.271 in.
o
2 tan 30
2 tan 30o
b  a  2e  2.952 in.
Am 
1
3
3
(b)( b) 
(2.952) 2  3.791 in.2
2
2
4
Hence
A 
T
6 103

 2.53 ksi
2tAm 2( 5 )(3.791)
16
________________________________________________________________________
PROBLEM (5.102) A thin-walled tube of length L possesses the cross section made from sheet
metal having thickness t (Fig. P5.102), the modulus elastcity in shear G, and allowable shear stress
 all . Neglecting the effect of stress concentration, determine:
(a) The maximum torque Tmax that can be applied to the tube.
(b) The corresponding angle of twist max of the tube.
Given:
t = 1/16 in.,
L = 15 in.,
 all
= 10 ksi,
G = 11.5 x 10 6 psi
SOLUTION
(a) Area bounded by center line
3
3 3
Am  2( )  (2  )( )  2.4375 in.2
4
4 4
Tmax
T
 all 
;
10 
1
2tAm
2( )(2.4375)
16
Tmax  3.047 kip  in.
or
TL
ds
(b)  
2

4 AmG t
(3.047 103 )(15)(2  2  2  3 4  2  2 1 )
4

4(2.4375) 2 (11.5 106 )(1 16)
 0.0274 rad  1.57o
________________________________________________________________________
PROBLEM (5.103) A hollow tubing having the cross section shown in Fig. P5.103 is fabric ated
from a sheet metal of uniform thickness t. Calculate:
(a) The minimum thickness tmin required for the tube.
(b) The angle of twist in a length L of the tube.
Given: a = 50 mm, r = 25 mm, L = 2.5 m, T = 1.5 kN  m, G = 79 GPa,
SOLUTION
5-61
 all
= 60 MPa
Am  2ra   a2  2(25)(50)   (25)2  4, 463.5 mm2
(a)  all 
T
;
2 Amt
60(106 ) 
1,500
2(4, 463.5 10 6 )tall
or
tall  0.0028 m  2.8 mm
(b)  
TL
1,500(2.5)

2
4 AmGt 4(4, 463.5 10 6 ) 2 (79 109 )(0.0028)
 0.21273 rad  12.19o
________________________________________________________________________
PROBLEM (5.104) A thin-walled hollow tube of contstant thickness t having the cross section
shown in Fig. P5.103 is acted upon by a torque T. The material has the shear modulus of elasticity G.
Determine:
(a) The maximum shear stress  max in the tube.
(b) The angle of twist  in a length of L of the tube.
Given: a = 2 in.,
r = 1 in.,
L = 5 ft,
t = ¼ in.,
T = 10 kip  in.,
G = 11.5 x 10 6 psi
SOLUTION
Am  2ra   a2  2(1)(2)   (1)2  7.142 in.2
(a)  max 
(b)  

T
10(103 )

 2.8 ksi
2 Amt 2(7.142)(1 4)
TL
4 Am2 G

ds TL (2 r  2a )

t
4 Am2 Gt
10(103 )(5 12)[2 (1)  2(2)]
4(7.142) 2 (11.5 106 )(1 4)
 0.0105 rad  0.602o
________________________________________________________________________
PROBLEM (5.105) An unequal-leg, L6 x 4 x ¾ in.,steel angle of length L is twisted as shown in
Fig. P5.105. Based on a safety factor n s with respect to failure by yielding in shear, determine:
(a) The largest torque Tall that can be applied to the angle.
(b) The maximum angle of twist max .
Given:
L = 4 ft,
t = ¾ in.,
n s = 1.5,
G = 11.5 x 10 6 psi,
2
A = 6.94 in. (by Table B.12)
SOLUTION
5-62
 y = 21 ksi (from Table B.4),  y = 21 ksi
3
in.
4
A 6.94
a 
 9.253 in.
b 34
a 9.253

 12.337
    0.333
b
34
We have t  b 
(a) The maximum torque (see Table 5.1):
T
Tall
21
 max  all 2 ;

 ab
1.5 (0.333)(9.253)(3 4) 2
Tall  24.265 kip  in.
or
(b) The maximum angle of twist (see Table 5.1):
T L
24, 265(4 12)
all  all 3 
 ab G (0.333)(9.253)( 3 )3 (11.5 106 )
4
3
o
 77.89(10 ) rad  4.46
________________________________________________________________________
PROBLEMS (5.106 and 5.107) The ASTM-A36 structural steel shafts having a shear modulus
of elasticity G and cross sections shown in Figs. P5.106 and P5.107 each are acted upon by a torque T.
For each shape, determine:
(a) The maximum shear stress.
(b) The angle of twist in a 1.2-m length.
Given: T = 4 kN  m, G = 79 GPa (from Table B.4)
Assumption: The effect of stress concentration at the corners are omitted.
SOLUTION (5.106)
Am 

2
(47.5) 2 

2
(49) 2  7.313 10 3 m 2
(a) Use Eq. (5.41),   T 2 Amt :
Left half
4 103
L 
 54.7 MPa
2(7.313 103 )(5 103 )
Right half
5
 R   L ( )  136.8 MPa
2
(b)  
TL
4 Am2 G

ds
t
or
5-63
4 103 (1.2)
47.5
49
[
 ]
3 2
9
4(7.313 10 ) (79 10 )
5
2
3
o
 30.34 10 rad  1.738

SOLUTION (5.107)
2 mm
2 mm
A
B
25 mm
150 mm
99 mm
mm
C
Am 
152.01 mm

2

(97) 2 
(74) 2 
2
197  147
(150)
2
 49.783 103 m2
74 mm
D
3 mm
(a) Use Eq. (5.41),   T 2 Amt :
4 103
  BD 
 20.08 MPa
2(49.783 103 )(2 103 )
2
  CD   AC ( )  13.39 MPa
3
 AC
 AB
(b)  
TL
4 Am2 G

ds
t
or

4 103 (1.2)
152.01  (99)  (74)
[2


]
3 2
9
4(49.783  10 ) (79 10 )
3
2
2
372.94
 2.29 10
3
rad  0.131
o
________________________________________________________________________
PROBLEM (5.108) The cross section of an aircraft structure made of an aluminum alloy having
shear modulus of elasticity G = 28 GPa is approximated as shown in Fig. P5.107. If the maximum
permissible angle of twist per unit length is 0.4° per meter of length of the fuselage, determine the
maximum allowable torque Tall and the corresponding maximum shearing stress  all .
SOLUTION
From solution of Prob. 5.107:
ds
Am  49.783 103 m 2
 t  372.94
Use Eq. (5.42):
5-64

T
ds
2

L 4 Am G t
Tall (372.94)
0.4

180  4(49.783 10 3 ) 2 (28 10 9 )

or
Tall  5.13 kN  m
Apply Eq. (5.31),   Tall 2 Amt :
 max
5.13 103
  AC   BD 
2(49.783 103 )(2 103 )
 25.76 MPa
End of Chapter 5
5-65