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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 PART-A 1. Write torsional equation. = = Where, T – Torque –N/mm J – Polar moment of inertia – mm 4 τ – Shear stress – N/mm 2 R – Radius – mm C – Modulus of rigidity – N/mm 2 θ – 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 = D 4 For hollow shaft, Polar modulus,

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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

PART-A

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, N-mm

C Modulus of rigidity, N/mm2

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 N-mm

6. Differentiate between close-coiled and open-coiled helical springs.

Close coiled helical springs Open coiled helical springs

Adjacent are very close to each other Large gap between adjacent coils

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 close-coiled 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 N-mm

• 11. The stiffness of a spring is 10N/mm and the axial deflection is 10mm. What is the axial

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

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

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

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 kN-m

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 N-mm

D Diameter in mm

• 20. Write down the expression for torque transmitted by hollow shaft.

T =

[

]

Where, T torque in N-mm

Shear Stress in N/mm2

D- Outer diameter in mm

d Inner diameter in mm

PART-B

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

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 kN-m = 3.58 103 N-m

• Torque, T = 3.58 106 N-mm

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

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 kN-m

= 51.04 N-m

= 51.04 N-mm

We know that, Tmax = 1.2 Tmean

= 1.2 51.04 N-mm

Tmax = 61.2 N-mm

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 =

[ ]

[ ]

=

D4-d4 = 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 kN-m = 23.87 103 N-m

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 N-mm

= 1.47 103 kN-mm = 1.47 kN-m

T = 1.47 kN-m

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 kN-m =

[

]

1.47 106

N-mm =

[

]

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 N-m = 100 N-mm

= 100 N/mm2

= 3

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 N-mm

= 1100 103 N-mm

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 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 N-mm

Torque transmitted by steel tube, Ts = 1023 103 N-mm

Torque transmitted by copper rod,

Tc = 0.075 Ts

= 0.0751023 103

Torque transmitted by copper rod, Tc = 76.72 103 N-mm

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