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) σE^{2}/V

c) σV^{2}/E

d) σV^{2}/2E

View Answer

Explanation: Strain energy in impact loading = σ

^{2}V/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

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

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 1000mm^{2} in section. What will be the instantaneous stress (E=210GPa)?

a) 149.4 N/mm^{2}

b) 179.24 N/mm^{2}

c) 187.7 N/mm^{2}

d) 156.1 N/mm^{2}

View Answer

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/mm

^{2}.

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 105cm^{2} 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/mm^{2}

b) 60.23 N/mm^{2}

c) 45.24 N/mm^{2}

d) 63.14 N/mm^{2}

View Answer

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/mm

^{2}.

6. A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm^{2} in section. What will be the strain (E=210GPa)?

a) 0.00089

b) 0.0005

c) 0.00064

d) 0.00098

View Answer

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/mm

^{2}

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

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/mm

^{2}

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 105cm^{2} 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

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/mm

^{2}.

Strain energy stored = stress

^{2}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/mm^{2}

b) 110 N/mm^{2}

c) 120 N/mm^{2}

d) 140 N/mm^{2}

View Answer

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

^{2}.

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 5cm^{2}. 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

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

^{2}.

As, P( h + δL) = σ

^{2}/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/mm^{2}

b) 60 N/mm^{2}

c) 70 N/mm^{2}

d) 80 N/mm^{2}

View Answer

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

^{2}.

**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__.

**Related Posts:**

- Practice Metallurgical Engineering MCQs
- Check Mechanical Engineering Books
- Apply for Metallurgical Engineering Internship
- Check Metallurgical Engineering Books
- Check Strength of Materials Books