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

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

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?

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/mm2?
a) 47.746 N/mm2
b) 34.15 N/mm2
c) 48.456 N/mm2
d) 71.02 N/mm2

Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 47.746 N/mm2.

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/mm2?
a) 1.19mm
b) 2.14mm
c) 3.45mm
d) 4.77mm

Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 47.746 N/mm2
So stretch = stress x length / E = 4.77mm.
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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/mm2
b) 120 N/mm2
c) 125 N/mm2
d) 150 N/mm2

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm2.

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/mm2?
a) 312.5 N-m
b) 314500 N-mm
c) 245.5 N-m
d) 634 N-m

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm2
Strain energy = σ2V/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/mm2?
a) 100 N-m
b) 132 N-m
c) 148 N-m
d) 143.2 N-m

Explanation: Stress = 60,000 / 400π = 47.746
Strain energy = σ2V/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 the 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

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/cm2?
a) 240kgf/cm
b) 242kgf/cm
c) 264kgf/cm
d) 270kgf/cm

Explanation: First stress = load /area, then the strain energy will be calculated as
Strain energy = σ2V/2E.

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