This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Definition of Strain Energy”.

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

a) σE/V

b) σE^{2}/V

c) σV^{2}/E

d) σV^{2}/2E

View Answer

Explanation: Strain energy when load is applied gradually = σ

^{2}V/2E.

2. Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is

a) s^{2}V/2E

b) sV/E

c) sV^{2}/E

d) sV/2E

View Answer

Explanation: Strain energy = s

^{2}V/2E.

3. In a material of pure shear stress τthe strain energy stored per unit volume in the elastic, homogeneous isotropic material having elastic constants E and v will be:

a) τ^{2}/E x (1+ v)

b) τ^{2}/E x (1+ v)

c) τ^{2}/2E x (1+ v)

d) τ^{2}/E x (2+ v)

View Answer

Explanation: σ

_{1}=τ , σ

_{2}= -τσ

_{3}=0

U = ( τ

^{2}+- τ

^{2}-2μτ(-τ))V = τ

^{2}/E x (1+ v)V.

4. PL^{3}/3EI is the deflection under the load P of a cantilever beam. What will be the strain energy?

a) P^{2}L^{3}/3EI

b) P^{2}L^{3}/6EI

c) P^{2}L^{3}/4EI

d) P^{2}L^{3}/24EI

View Answer

Explanation: We may do it taking average

Strain energy = Average force x displacement = (P/2) x PL

^{3}/3EI = P

^{2}L

^{3}/6EI.

5. A rectangular block of size 400mm x 50mm x 50mm is subjected to a shear stress of 500kg/cm^{2}. If the modulus of rigidity of the material is 1×106 kg/cm^{2} , the strain energy will be

a) a125 kg-cm

b) 1000 kg-cm

c) 500 kg-cm

d) 100 kg-cm

View Answer

Explanation: Strain energy stored = τ

^{2}V/2G = 500

^{2}/2×10

^{6}x 40x5x5 = 125 kg-cm.

6. A material of youngs modulus and Poissons ratio of unity is subjected to two principal stresses σ_{1} and σ_{2} at a point in two dimensional stress system. The strain energy per unit volume of the material is

a) (σ_{1}^{2} + σ_{2}^{2} – 2σ_{1}σ_{2} ) / 2E

b) (σ_{1}^{2} + σ_{2}^{2} + 2σ_{1}σ_{2} ) / 2E

c) (σ_{1}^{2} – σ_{2}^{2} – 2σ_{1}σ_{2} ) / 2E

d) (σ_{1}^{2} – σ_{2}^{2} – 2σ_{1}σ_{2} ) / 2E

View Answer

Explanation: Strain energy = (σ

_{1}ε

_{1}+ σ

_{1}ε

_{1}) / 2E

= (σ

_{1}

^{2}+ σ

_{2}

^{2}– 2σ

_{1}σ

_{2}) / 2E.

7. If forces P, P and P of a system are such that the force polygon does not close, then the system will

a) Be in equilibrium

b) Reduce to a resultant force

c) Reduce to a couple

d) Not be in equilibrium

View Answer

Explanation: The forces are not concurrent so the resultant force and couple both may be present. Thus the best choice is that forces are not in equilibrium.

8. The strain energy in a member is proportional to

a) Product of stress and the strain

b) Total strain multiplied by the volume of the member

c) The maximum strain multiplied by the length of the member

d) Product of strain and Youngs modulus of the material

View Answer

Explanation: Strain energy per unit volume for solid = q

^{2}/ 4G.

9. A bar of cross-section A and length L is subjected to an axial load W. the strain energy stored in the bar would be

a) WL / AE

b) W^{2}L / 4AE

c) W^{2}L / 2AE

d) WL / 4AE

View Answer

Explanation: Deformation in the bar = WL / AE

Strain energy = W/2 x WL / AE = W

^{2}L / 2AE.

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

a) 1.1mm

b) 1.24mm

c) 2mm

d) 1.19mm

View Answer

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

^{2}) = 470746 N/mm

^{2}

So stretch = stress x length / E = 1.19mm.

11. A tensile load of 50kN is gradually applied to a circular bar of 5cm diameter and 5m long. What is the strain energy absorbed by the rod ( E = 200GPa ) ?

a) S14N-m

b) 15.9 N-mm

c) 15.9 N-m

d) 14 N-mm

View Answer

Explanation: Stress = 50,000 / 625π = 25.46

Strain energy = σ

^{2}V/2E = 25.46×25.46×9817477 / (2×200000) = 15909.5 N-mm = 15.9N-m.

12. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the strain energy in the rod if the load is applied suddenly (E = 2×10^{5} N/mm^{2}) ?

a) d143.23 N-m

b) 140.51 N-m

c) 135.145 N-m

d) 197.214 N-m

View Answer

Explanation: Maximum instantaneous stress = 2P / A = 95.493

Strain energy = σ

^{2}V/2E = 143288N-mm = 143.238 N-m.

**Sanfoundry Global Education & Learning Series – Strength of Materials. **

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