SOM Chap  3 (Torsion) Question Bank (2)

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Transcript of SOM Chap  3 (Torsion) Question Bank (2)

AARUPADAI VEEDU INSTITUTE OF TECHNOLOGY, PAIYANOOR
&
V.M.K.V. ENGINEERING COLLEGE
(II YEAR / IV SEMESTER)
(CBCS 2013 REGULATION)
STRENGTH OF MATERIALS
QUESTION BANK
UNIT III
PARTA
1. Write torsional equation.
=
=
Where, T Torque N/mm J Polar moment of inertia mm4 Shear stress N/mm2
R Radius mm C Modulus of rigidity N/mm2 Angle of twist rad l length  mm
2. Write the polar modulus for solid shaft and circular shaft.
For solid shaft,
Polar modulus,
Where, J =
D4
For hollow shaft,
Polar modulus,

Where, J =
[ ]
3. Write an expression for the angle of twist for a hollow circular shaft with external
diameter D, internal diameter d, length l and rigidity modulus G.
Where,
J (polar moment of inertia) =
[ ]
D External diameter, mm
d Internal diameter, mm
T Torque transmitted, Nmm
C Modulus of rigidity, N/mm2
Angle of twist, rad
l Length of shaft, mm
4. Define torsional rigidity.
The torsional equation is,
=
Since C, l, J are constant for a given shaft, (angle of twist) is directly proportional to T
(torque). The term CJ is known as torsional rigidity and it is represented by K.
5. Calculate the maximum torque that a shaft of 125mm diameter can transmit, if the
maximum angle of the twist is 1 in a length of 1.5m. Take C=70103 N/mm2.
Given data:
Diameter, D = 125mm
Angle of twist, = 1
Length, l = 1.5 m = 1500 mm
Modulus of rigidity, C = 70103 N/mm2
To find: Maximum Torque, Tmax

Solution:
Torsional Equation
T =
=
T = Tmax = 19.01106 Nmm
6. Differentiate between closecoiled and opencoiled helical springs.
Close coiled helical springs Open coiled helical springs
Adjacent are very close to each other Large gap between adjacent coils
Only tensile load can carry Tensile and compressive loads can carry
Helix angle is negligible Helix angle is considerable
7. A helical spring is made of 4mm steel wire with a mean radius of 25mm and number of
turns of coil 15. What will be deflection of the spring under a load of 6N? Take C=80103
N/mm2.
Given:
d = 4mm
R = 25mm
n = 15
W = 6N
C = 80103 N/mm2
Solution:
Axial deformation,
=
=
4 9 mm

8. Give shear stress and deflection relation for closecoiled helical springs.
Deflection, =
Shear Stress, =
9. The stiffness of spring is 10N/mm. what is the axial deformation in the spring when a load
is 50N is acting?
Given:
K = 10N/mm
W = 50N
Solution:
K =
=
=
= 5mm.
10. An open coiled helical spring of mean radius of coil of 20cm and helix angle of 12 is
subjected to an axial load of 10N. What is the bending moment in the coil?
Given:
R = 20 cm = 200mm
= 12
W = 10N
Solution:
Bending Moment, M = WR sin
= 10200sin 12
M = 415.82 Nmm

11. The stiffness of a spring is 10N/mm and the axial deflection is 10mm. What is the axial
load on the spring?
Given:
K =10 N/mm
= 10mm
Solution:
K =
=> W = K = 1010 = 100 N.
12. Write the equation for the deflection of an open coiled helical spring subjected to an axial
load W.
Deflection, =
[
]
13. Write down the equation for shear strain energy of a close coiled spring.
Maximum Shear Stress, =
14. Write the expression for vertical deflection of a close coiled helical spring due to vertical
load W.
Deflection, =
15. What is meant by stiffness? What is the formula for the stiffness of a close coiled helical
spring subjected to an axial load?
The stiffness of the spring is defined as the load required producing unit deflection.
Stiffness, K =
N/mm.

16. What are the assumptions made in torsion equation?
The material of the shaft is homogeneous, perfectly elastic and obeys Hookes Law.
Twist is uniform along the length of the shaft.
The stress does not exceed the limit of proportionality
The shaft circular in section remains circular after loading
Strain and deformation are small.
17. Why hollow circular shafts are preferred when compared to solid circular shafts?
The torque transmitted by the hollow shaft is greater than the solid shaft.
For some material, length and given torque, the weight of the hollow shaft will be
less compared to solid shaft.
18. Write down the expression for power transmitted by a shaft.
Power,
Where, t Torque in kNm
N Speed in rpm
P Power in Kw
19. Write down the equation for the maximum shear stress of a solid circular section in
diameter D when subjected to torque T.
For Solid Shaft
Torque, T =
=
Where, Shear stress in N/mm2
T Torque in Nmm
D Diameter in mm

20. Write down the expression for torque transmitted by hollow shaft.
T =
[
]
Where, T torque in Nmm
Shear Stress in N/mm2
D Outer diameter in mm
d Inner diameter in mm
PARTB
1. A solid circular shaft transmits 75kW at 200rpm. Find the shaft diameter if the twist in the
shaft is not to exceed 1 in 2m length of the shaft and the shearing stress is limited to
50N/mm2. Take G=100GN/mm
2.
Given data:
P = 75 KW
N = 200 rpm
= 1
= 0.017 rad
L = 2 m = 2000 mm
= 50 N/mm2
C or G = 100 GN/m2
= 100109 N/m
2 = 10010
3 N/mm
2
To find:
Diameter of the shaft
Solution:
We know that,
Power, P =
75 =
T = 3.58 kNm = 3.58 103 Nm

Torque, T = 3.58 106 Nmm
Shear stress and angle of twist both are given.
First Case: Considering the shear stress ()
Torque, T =
3.58 106=
D = 71.4 mm
Second case: Considering angle of twist ()
Where, J =
=
D = 80.9 mm
From the above two cases, we find that suitable diameter for the shaft is 80.9mm (i.e.
greater of two values).
Result:
Shaft Diameter, D = 80.9mm

2. A hollow shaft of diameter ratio 3/8 is required to transmit 588kW at 100rpm. The
maximum torque exceeds the mean by 20%. The shear stress is limited to 63N/mm2 and the
twist should not be more than 0.0081rad. Calculate the external diameter required
satisfying both the conditions. Take G=84Gpa. Length 3m.
Given:
P = 588kW
N = 110rpm
=
d = 0.375 D
Tmax = 1.2 Tmean
= 63 N/mm2
= 0.0081 rad
G or C = 84 GPa = 84 109 N/m
2
= 84 N/mm2
l = 3m = 3000mm
To find:
External Diameter (D)
Solution:
We know that,
Power, P =
588 =
T = 51.04 kNm
= 51.04 Nm
= 51.04 Nmm
We know that, Tmax = 1.2 Tmean
= 1.2 51.04 Nmm
Tmax = 61.2 Nmm
Shear stress and angle of twist both are given.
First Case: considering the shear stress ()
Torque, Tmax =
[
]

61.2 = [
4 ( 4)
]
61.2106 =
[ ]
External Diameter, D = 171.5 mm
We Know that, d = 0.375D = 0.375 171.5
Internal Diameter, d = 64.3 mm
Second Case: Considering angle of twist ()
Where J =
[ ]
[ ]
=
D4d4 = 2.7109
D4 (0.375 D) 4 = 2.7 109
D4 [ ] = 2.7 109
External Diameter, D = 229.09 mm
We know that, d = 0.375D
Internal Diameter, d = 85.90 mm
From the above two cases, we find that external diameter of the shaft is 229.09 mm and
internal diameter is 85.90 (i.e. greater of the two values)
Result:
1. External Diameter, D = 229.09 mm
2. Internal Diameter, d= 85.90 mm

3. A shaft is required to transmit power of 300kW running at a speed of 120rpm. If the shear
strength of the shaft material is 70N/mm2. Design a hollow shaft with inner diameter equal
to 0.75 times the outer diameter.
Given Data:
P = 300 kW
N = 120 rpm
= 70 N/mm2
d = 0.75 D
To find:
Inner Diameter (d) and outer diameter (D)
Solution:
We know that,
Power, P =
300 =
T = 23.87 kNm = 23.87 103 Nm
Shear stress is given. For hollow Shaft
Torque, T =
[
]
23.87 106 =
[
]
23.87106 =
[ ]
External Diameter, D = 136.45 mm
We know that, d = 0.75 D
= 0.75 136.45
Internal Diameter, d = 102.33 mm
Result:
1. External Diameter, D = 136.45 mm
2. Internal Diameter, d = 102.33 mm

4. A solid shaft A of 50mm diameter rotates at 250rpm. Find the power that can be
transmitted for a limiting shear stress of 60N/mm2 in the steel. It is proposed to replace A
by hollow shaft B of the same external diameter but with the limiting shear stress of
75N/mm2. Determine the internal diameter of B to transmit the same power at the same
speed.
Given:
Diameter, D = 50mm
Speed, N = 250 rpm
Shear stress, = 60 N/mm2
To find:
Power, P
Hollow Shaft B:
External Diameter, D1 = 50 mm
Shear stress, = 75 N/mm2
To find:
Internal Diameter, (d)
Solution:
We know that,
Power, P =
Shear stress is given. So,
Torque,
=
= 1.47 106 Nmm
= 1.47 103 kNmm = 1.47 kNm
T = 1.47 kNm
Substitute T value in equation
Power, P =
P = 38.48 kW.
Solid shaft is replaced by hollow shaft. Torque transmitted by the solid shaft is equal to the
torque transmitted by the hollow shaft

Torque transmitted by hollow shaft [considering shear stress]
Torque, T =
[
]
T =
[
]
Equating,
1.47 kNm =
[
]
1.47 106
Nmm =
[
]
1.47 106 = 0.294 [504 d4]
d4 = 1.25 10
6
Internal Diameter, d = 33.4 mm
Result:
i) Power, P = 38.48 kW
ii) Internal diameter, d = 33.4 mm
5. A solid shaft is subjected to a torque of 100Nm. Find the necessary shaft diameter if the
allowable shear stress is 100N/mm2 and the allowable twist is 3 per 10 diameter length of
the shaft. Take C=1105 N/mm
2.
Given Data:
T = 100 Nm = 100 Nmm
= 100 N/mm2
= 3
= 0.052 rad
l = 10 D
C = 1105 N/mm
2
To find:
Diameter of the solid shaft, D
Solution:
Shear stress and angle of twist both are given.

First Case: Considering shear stress ()
100103 =
D = 17.20 mm
Second Case: Considering angle of twist ()
Where J (polar moment of inertia)
=
=
D = 12.51mm
From the above two cases, we find that suitable diameter of the shaft is 17.20mm (i.e. greater
of two values).
6. A close coil helical spring of round steel wire 10mm in diameter has a mean radius of
120mm. The spring has 10 complete turns and is subjected to an axial load of 200N.
Determine (i) deflection of the spring (ii) maximum shear stress in the wire and (iii)
stiffness of the spring. G=80 kN/mm2.
Given data:
d = 10 mm
R = 120 mm
n = 10
W = 200N
G = 80 kN/mm2 = 8010
3 N/mm
2
To find:
i) Deflection of the spring ()
ii) Maximum shear stress in the wire ()
iii) Stiffness of the spring, (K)

Solution:
i) Deflection of the spring, ()
=
or =
=
= 276.48 mm
ii) Maximum shear stress in the wire ()
=
or =
=
= 122.2 N/mm2
iii) Stiffness of the spring, (K)
K =
=
0.723 N/mm
Results:
i) Deflection of the spring () = 276.48 mm
ii) Maximum shear stress in the wire () = 122.2 N/mm2
iii) Stiffness of the spring, (K) = 0.723 N/mm

7. A closed coil helical spring of 10cm mean diameter is made up of 1cm diameter rod and
has 20 turns. The spring carries an axial load of 200N. Determine the shearing stress.
Taking the value of modulus of rigidity = 8.4 104 N/mm
2, determine the deflection when
carrying this load. Also calculate the stiffness of the spring and the frequency of free
vibration for a mass hanging from it.
Given Data:
D = 10cm 100mm
d = 1cm 10cm
n = 20
W = 200N
C = 8.4 104 N/mm
2
To find:
The deflection,
Stiffness, K
Solution:
=
=
= 38.10 mm
K =
=
= 5.25 N/mm
Result:
i. The deflection, = 38.10 mm
ii. Stiffness, K = 5.25 N/mm

8. A closely coiled helical spring is to carry a load of 500N. Its mean coil diameter is to be 10
times that of the wire diameter. Calculate this diameter if the maximum shear stress in the
material in the spring is to be 80N/mm2.
Given Data:
W = 500N
D = 10d
= 80 N/mm2
To find:
Mean coil diameter
Solution:
We know that,
=
80 =
d2 =
d2 = 159.15 mm2
d = 12.62 mm
D = 10d
= 126.2 mm
Result:
Mean coil diameter, D =126.2 mm

9. The stiffness of the close coiled helical spring is 1.5N/mm of compression under the
maximum load of 60N. Its mean coil diameter is to be 8 times that of the wire diameter.
The maximum shearing stress produced in the wire of the string is 125N/mm2. The solid
length of the spring (when the coil are touching) is given as 5cm. Find
i. Diameter of wire
ii. Mean diameter of coil and
iii. No.of coils required.
Take C = 4.5 104 N/mm
2.
Given Data:
K = 1.5 N/mm
W = 60N
= 125 N/mm2
l = 5cm 50mm
C = 4.5104 N/mm2
D =8d
To find:
Diameter of wire, d
Mean diameter of coil, D
No.of coils required, n
Solution:
K =
=
= 40 mm
=
d2 =
= 9.78 mm2

d = 3.12 mm
D = 8d (by given)
D = 25.02 mm
L =Dn
n =
n = 0.64
Result:
Diameter of wire, d = 3.12 mm
Mean diameter of coil, D = 25.02 mm
No.of coils required, n = 0.64
10. A composite shaft consists of copper rod of 25mm diameter enclosed in a steel tube of
external diameter 45mm and 5mm thick. The shaft is required to transmit a torque of
1100Nm and both the shafts have equal lengths, welded to a plate at each end, so that their
twists are equal. If the modulus of rigidity for steel as twice that of copper, find (i) shear
stress developed in copper (ii) shear stress developed in steel.
Given Data:
Diameter of copper road, dc = 25 mm
External diameter of steel tube, d0 = 45 mm
Internal Diameter of steel tube,
di = d0 2 Thickness = 45  25
di = 35mm

Total torque, T = 1100 Nmm
= 1100 103 Nmm
Shaft length are equal, ls = lc = l
Modulus of rigidity for steel = 2 Modulus of rigidity of copper
Cs = 2 Cc
To find:
1. Shear stress developed in copper, c
2. Shear stress developed in steel tube, s.
Solution:
We know that,
For copper road,
For steel tube,
From equation
c =
From equation,
s =
For composite shaft, angle of twist are same
c = s

Tc =
=
[
]
=
[ ]
Tc = 0.075 Ts
We know that,
Total torque, T = Tc + Ts
T = 0.075 Ts + Ts
T = 1.075 Ts
1100 103 = 1.075 Ts
Ts = 1023 103 Nmm
Torque transmitted by steel tube, Ts = 1023 103 Nmm
Torque transmitted by copper rod,
Tc = 0.075 Ts
= 0.0751023 103
Torque transmitted by copper rod, Tc = 76.72 103 Nmm
We know that,

For copper rod,
For steel tube,
From the equation,
c =
=
=
c = 25N/mm2
From the equation,
s =
=
[
]
=
[ ]
s = 90.17 N/mm2
Result:
1. shear stress developed in copper, c = 25 N/mm2
2. Shear stress developed in steel tube, s = 90.17 N/mm2