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

1. What is the strain energy stored in a body when the load is applied gradually?

a) σE/V

b) σE^{2}/V

c) σV^{2}/E

d) σV^{2}/2E

View Answer

Explanation: Strain energy in gradual loading = σ

^{2}V/2E.

2. What is strain energy?

a) The work done by the applied load In stretching the body

b) The strain per unit volume

c) The force applied in stretching the body

d) The stress per unit are

View Answer

Explanation: The strain energy stored in a body is equal to the work done by the applied load in stretching the body.

3. What is the relation between maximum stress induced due to gradual load to maximum stress the sudden load?

a) Maximum stress in gradual load is equal to the maximum stress in sudden load

b) Maximum stress in gradual load is half to the maximum stress in sudden load

c) Maximum stress in gradual load is twice to the maximum stress in sudden load

d) Maximum stress in gradual load is four times to the maximum stress in sudden load

View Answer

2Explanation: Maximum stress in gradual loading = P/A

Maximum stress in sudden loading = 2P/A.

4. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5cm long. What will be the stress in the rod if E=1×105 N/mm^{2} ?

a) 47.746 N/mm^{2}

b) 34.15 N/mm^{2}

c) 48.456 N/mm^{2}

d) 71.02 N/mm^{2}

View Answer

Explanation: Stress = Load/ area = 60,000 / (π/4 D

^{2}) = 47.746 N/mm

^{2}.

5. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the stress in the rod if E=1×105 N/mm^{2} ?

a) 1.19mm

b) 2.14mm

c) 3.45mm

d) 4.77mm

View Answer

Explanation: Stress = Load/ area = 60,000 / (π/4 D

^{2}) = 47.746 N/mm

^{2}

So stretch = stress x length / E = 4.77mm.

6. A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm. What will be the stress in bar?

a) 100 N/mm^{2}

b) 120 N/mm^{2}

c) 125 N/mm^{2}

d) 150 N/mm^{2}

View Answer

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm

^{2}.

7. A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy in the bar if E=1×105 N/mm^{2} ?

a) 312.5 N-m

b) 314500 N-mm

c) 245.5 N-m

d) 634 N-m

View Answer

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm

^{2}

Strain energy = σ

^{2}V/2E = 125x125x20x40x5000/ (2×100,000) = 312500 N-mm = 312.5N-m.

8. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the strain energy absorbed by the rod if E=1×105 N/mm^{2} ?

a) 100 N-m

b) 132 N-m

c) 148 N-m

d) 143.2 N-m

View Answer

Explanation: Stress = 60,000 / 400π = 47.746

Strain energy = σ

^{2}V/2E = 47.746×47.746×12,566,370 / (2×100000) = 143,236.54 N-mm = 143.2N-m.

9. A uniform bar has a cross sectional area of 700mm and a length of 1.5m. if the stress at elastic limit is 160 N/mm, what will be the value of gradually applied load which will produce the same extension as that produced by the suddenly applied load above?

a) 100kN

b) 110kN

c) 112kN

d) 120kN

View Answer

Explanation: For gradually applied load , stress = load / area

Load = stress x area = 160 x 700 = 112000 N = 112kN.

10. A tension bar 6m long is made up of two parts, 4m of its length has cross sectional area of 12.5cm while the remaining 2m has 25cm. An axial load 5tonnes is gradually applied. What will be the total strain energy produced if E = 2 x 106 kgf/cm^{2} ?

a) 240kgf/cm

b) 242kgf/cm

c) 264kgf/cm

d) 270kgf/cm

View Answer

Explanation: First stress = load / area , then the strain energy will be calculated as

Strain energy =σ

^{2}V/2E.

**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__.