Strength of Materials Questions and Answers – Impact Loading

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Impact Loading”.

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
View Answer

Answer: d
Explanation: Strain energy in impact loading = σ2V/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)
View Answer

Answer: a
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
View Answer

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

Answer: c
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
View Answer

Answer: b
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
View Answer

Answer: a
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
View Answer

Answer: c
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.
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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
View Answer

Answer: a
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
View Answer

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

Answer: c
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
View Answer

Answer: d
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.

To practice all areas of Strength of Materials, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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