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?
Explanation: Strain energy when load is applied gradually = σ2V/2E.
2. Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is
Explanation: Strain energy = s2V/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) τ2/E x (1+ v)
b) τ2/E x (1+ v)
c) τ2/2E x (1+ v)
d) τ2/E x (2+ v)
Explanation: σ1=τ , σ2= -τσ3=0
U = ( τ2+- τ2-2μτ(-τ))V = τ2/E x (1+ v)V.
4. PL3/3EI is the deflection under the load P of a cantilever beam. What will be the strain energy?
Explanation: We may do it taking average
Strain energy = Average force x displacement = (P/2) x PL3/3EI = P2L3/6EI.
5. A rectangular block of size 400mm x 50mm x 50mm is subjected to a shear stress of 500kg/cm2. If the modulus of rigidity of the material is 1×106 kg/cm2 , the strain energy will be
a) a125 kg-cm
b) 1000 kg-cm
c) 500 kg-cm
d) 100 kg-cm
Explanation: Strain energy stored = τ2V/2G = 5002/2×106 x 40x5x5 = 125 kg-cm.
6. A material of youngs modulus and Poissons ratio of unity is subjected to two principal stresses σ1 and σ2 at a point in two dimensional stress system. The strain energy per unit volume of the material is
a) (σ12 + σ22 – 2σ1σ2 ) / 2E
b) (σ12 + σ22 + 2σ1σ2 ) / 2E
c) (σ12 – σ22 – 2σ1σ2 ) / 2E
d) (σ12 – σ22 – 2σ1σ2 ) / 2E
Explanation: Strain energy = (σ1ε1+ σ1ε1 ) / 2E
= (σ12 + σ22 – 2σ1σ2 ) / 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
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
Explanation: Strain energy per unit volume for solid = q2 / 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) W2L / 4AE
c) W2L / 2AE
d) WL / 4AE
Explanation: Deformation in the bar = WL / AE
Strain energy = W/2 x WL / AE = W2L / 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 = 2×105 N/mm2?
Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 470746 N/mm2
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 ) ?
b) 15.9 N-mm
c) 15.9 N-m
d) 14 N-mm
Explanation: Stress = 50,000 / 625π = 25.46
Strain energy = σ2V/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 = 2×105 N/mm2) ?
a) d143.23 N-m
b) 140.51 N-m
c) 135.145 N-m
d) 197.214 N-m
Explanation: Maximum instantaneous stress = 2P / A = 95.493
Strain energy = σ2V/2E = 143288N-mm = 143.238 N-m.
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.