1. What is the strain energy stored in a body when the load is applied with impact?
a) σE/V
b) σE2/V
c) σV2/E
d) σV2/2E

2. What is the value of stress induced in the rod due to impact load?
a) P/A (1 + (1 + 2AEh/PL)1/2)
b) P/A (2 + 2AEh/PL)
c) P/A (1 + (1 + AEh/PL)1/2)
d) P/A ((1 + 2AEh/PL)1/2)

Explanation: The value of stress is calculated by equating the strain energy equation and the work done equation.

3. What will be the stress induced in the rod if the height through which load is dropped is zero?
a) P/A
b) 2P/A
c) P/E
d) 2P/E

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2)
Putting h=0, we get stress = 2P/A.

4. A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the instantaneous stress (E=210GPa)?
a) 149.4 N/mm2
b) 179.24 N/mm2
c) 187.7 N/mm2
d) 156.1 N/mm2

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2)
Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm2.

5. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous stress induced in the vertical bar if E = 200GPa?
a) 50.87 N/mm2
b) 60.23 N/mm2
c) 45.24 N/mm2
d) 63.14 N/mm2

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )
Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2.

6. A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the strain (E=210GPa)?
a) 0.00089
b) 0.0005
c) 0.00064
d) 0.00098

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2)
Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm2
As strain = stress / E, thus, strain = 187.7 / 210,000 = 0.00089.

7. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous elongation in the vertical bar if E = 200GPa?
a) 0.245mm
b) 0.324mm
c) 0.452mm
d) 0.623mm

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )
Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2
Elongation = stress x length / E = 60.23 x 1500 / 200,000 = 0.452mm.

8. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the strain energy stored in the vertical bar if E = 200GPa?
a) 2.045 N-m
b) 3.14 N-m
c) 9.4 N-mm
d) 2.14 N-m

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )
Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2.
Strain energy stored = stress2 x volume / 2E = 60.232 x 2525000 / (2×200,000) = 2.045 N-m.

9. The maximum instantaneous extension, produced by an unknown falling weight in a vertical bar of length 3m. what will be the instantaneous stress induced in the vertical bar and the value of unknown weight if E = 200GPa?
a) 100 N/mm2
b) 110 N/mm2
c) 120 N/mm2
d) 140 N/mm2

Explanation: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.

10. The maximum instantaneous extension, produced by an unknown falling weight through a height of 4cm in a vertical bar of length 3m and of cross section area 5cm2. what will be the instantaneous stress induced in the vertical bar and the value of unknown weight if E = 200GPa?
a) 1700 N
b) 1459.4 N
c) 1745.8 N
d) 1947.5 N

Explanation: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.
As, P( h + δL) = σ2/2E x V
So P = 1745.8 N.

11. An unknown weight falls through a height of 10mm on a collar rigidly attached to a lower end of a vertical bar 500cm long. If E =200GPa what will be the value of stress?
a) 50 N/mm2
b) 60 N/mm2
c) 70 N/mm2
d) 80 N/mm2

Explanation: Stress = E x strain = E x δL/L = 200,000 x 2 /5000 = 80 N/mm2.

Sanfoundry Global Education & Learning Series – Strength of Materials.

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