Engineering Questions with Answers - Multiple Choice Questions

# MCQs on Strain Energy and Resilience

1 - Question

MCQs on Definition of Strain Energy
What is the strain energy stored in a body due to gradually applied load?
a) σE/V
b) σE2/V
c) σV2/E
d) σV2/2E

2 - Question

Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is __________
a) s2V/2E
b) sV/E
c) sV2/E
d) sV/2E
Explanation: Strain energy = s2V/2E.

3 - Question

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. advertisement

4 - Question

PL3/3EI is the deflection under the load P of a cantilever beam. What will be the strain energy?
a) P2L3/3EI
b) P2L3/6EI
c) P2L3/4EI
d) P2L3/24EI
Explanation: We may do it taking average Strain energy = Average force x displacement = (P/2) x PL3/3EI = P2L3/6EI.

5 - Question

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) 125 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 - Question

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 - Question

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 - Question

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 Young’s modulus of the material
Explanation: Strain energy per unit volume for solid = q2 / 4G.

9 - Question

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 - Question

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?
a) 1.1mm
b) 1.24mm
c) 2mm
d) 1.19mm
Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 470746 N/mm2 So stretch = stress x length / E = 1.19mm.

11 - Question

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) 14 N-m
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 = 15.9 N-m.

12 - Question

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.

13 - Question

MCQs on Resilience
The ability of a material to absorb energy when elastically deformed and to return it when unloaded is called __________
a) Elasticity
b) Resilience
c) Plasticity
d) Strain resistance
Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape.

14 - Question

The strain energy stored in a specimen when stained within the elastic limit is known as __________
a) Resilience
b) Plasticity
c) Malleability
d) Stain energy
Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

15 - Question

The maximum strain energy stored at elastic limit is __________
a) Resilience
b) Proof resilience
c) Elasticity
d) Malleability
Explanation: Proof resilience is the maximum stored energy at the elastic limit. Resilience is the ability of material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets. advertisement

16 - Question

The mathematical expression for resilience ‘U’ is __________
a) U = σ2/E x volume
b) U = σ2/3E x volume
c) U = σ2/2E x volume
d) U = σ/2E x volume
Explanation: The resilience is the strain energy stored in a specimen so it will be U = σ2/2E x volume.

17 - Question

What is the modulus of resilience?
a) The ratio of resilience to volume
b) The ratio of proof resilience to the modulus of elasticity
c) The ratio of proof resilience to the strain energy
d) The ratio of proof resilience to volume
Explanation: The modulus of resilience is the proof resilience per unit volume. It is denoted by σ.

18 - Question

The property by which an amount of energy is absorbed by material without plastic deformation is called __________
a) Toughness
b) Impact strength
c) Ductility
d) Resilience
Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it when unloaded.

19 - Question

Resilience of a material plays important role in which of the following?
a) Thermal stress
c) Fatigue
Explanation: The total strain energy stored in a body is commonly known as resilience. Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence the resilience is also define as the capacity of a strained body for doing work on the removal of the straining force.

20 - Question

A steel has its yield strength of 200N/mm2 and modulus of elasticity of 1x105MPa. Assuming the material to obey hookes law up to yielding, what will be its proof resilience?
a) 0.8 N/mm2
b) 0.4 N/mm2
c) 0.2 N/mm2
d) 0.6 N/mm2
Explanation: Proof resilience = σ2/2E = (200)2 / (2 x 105) = 0.2 N/mm2.

21 - Question

A 1m long bar of uniform section extends 1mm under limiting axial stress of 200N/mm2. What is the modulus of resilience for the bar?
a) 0.1 units
b) 1 units
c) 10units
d) 100units
Explanation: Modulus of resilience, u = f2/2E, where E = fL/δL Therefore, u = 200×1 / 2×1000 = 0.1units.

22 - Question

A square steel bar of 10mm side and 5m length is subjected to a load whereupon it absorbs a strain energy of 100J. what is its modulus of resilience?
a) 1/5 N-mm/mm3
b) 25 N-mm/mm3
c) 1/25 N-mm/mm3
d) 5 N-mm/mm3
Explanation: Modulus of resilience is the strain energy stored in the material per unit volume. u = U/v = ( 100 x 1000 ) / ( 10 x 10x 5x 1000) = 1/5 N-mm/mm3.

23 - Question

24 - Question

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

25 - Question

A tensile load of 60kN is suddenly applied to a circular bar of 4cm diameter. What will be the maximum instantaneous stress induced?
a) 95.493 N/mm2
b) 45.25 N/mm2
c) 85.64 N/mm2
d) 102.45 N/mm2
Explanation: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2.

26 - Question

A tensile load of 60kN is suddenly applied to a circular bar of 4cm and 5m length. What will be the strain energy absorbed by the rod if E=2×105 N/mm2?
a) 140.5 N-m
b) 100 N-m
c) 197.45 N-m
d) 143.2 N-m
Explanation: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2 Strain energy = σ2V/2E = 95.492 x 2×106π / (2x2x105) = 143238 N-mm = 143.23 N-m.

27 - Question

A tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm. What will be the instantaneous stress in bar?
a) 100 N/mm2
b) 120 N/mm2
c) 150 N/mm2
d) 250 N/mm2

28 - Question

2 tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy absorbed in the bar if E=1×105 N/mm2?
a) 312.5 N-m
b) 314500 N-mm
c) 1250 N-m
d) 634 N-m
Explanation: Stress = 2xload / area = 2×100,000/ (20×40) = 250 N/mm2 Strain energy = σ2V/2E = 250x250x20x40x5000/ (2×100,000) = 1250000 N-mm = 1250 N-m.

29 - Question

A steel rod is 2m long and 50mm in diameter. A axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous stress induced in the rod?
a) 101.89 N/mm2
b) 94.25 N/mm2
c) 130.45 N/mm2
d) 178.63 N/mm2
Explanation: Area = π/4 d2 = 625π Load = 100kN = 100×1000 N Stress = 2 x load / area = 2x100x1000 / (625π) = 101.86 N/mm2.

30 - Question

A steel rod is 2m long and 50mm in diameter. An axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous elongation produced in the rod if E=22GN/m2?
a) 0.0097 mm
b) 1.0754 mm
c) 1.6354 mm
d) 1.0186 mm
Explanation: Area = π/4 d2 = 625π Load = 100kN = 100×1000 N E=22GN/m2 = 200 x 109 / 106 = 200,000 N/mm2 Stress = 2 x load / area = 2x100x1000 / (625 π ) Elongation = stress x length / E = 101.86×2000 / 200000 = 1.0186 mm.

31 - Question

What will be the amount of axial pull be applied on a a 4cm diameter bar to get an instantaneous stress value of 143 N/mm2?
a) 50kN
b) 60kN
c) 70kN
d) 80kN
Explanation: Instantaneous stress = 2 x load / area Load = instantaneous stress x area / 2 = 143 x 400×3.14 / 2 = 60kN.

32 - Question

What will be the instantaneous stress produced in a bar 10cm2 in area ans 4m long by the sudden application of tensile load of unknown magnitude, if the extension of the bar due to suddenly applied load is 1.35mm if E = 2×105 N/mm2?
a) 67.5 N/mm2
b) 47 N/mm2
c) 55.4 N/mm2
d) 78.5 N/mm2
Explanation: The value of stress = load / area where area is 10cm2 and load can be calculated by stress strain equation.

33 - Question

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

34 - Question

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.

35 - Question

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

36 - Question

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.

37 - Question

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.

38 - Question

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.

39 - Question

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.

40 - Question

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.

41 - Question

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
Explanation: For gradually applied load, stress = load / area Load = stress x area = 160 x 700 = 112000 N = 112kN.

42 - Question

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.

43 - Question

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

44 - Question

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)
Explanation: The value of stress is calculated by equating the strain energy equation and the work done equation.

45 - Question

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
Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2) Putting h=0, we get stress = 2P/A. advertisement

46 - Question

A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the instantaneous stress (E=210GPa)?
a) 149.4 N/mm2
b) 179.24 N/mm2
c) 187.7 N/mm2
d) 156.1 N/mm2
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/mm2.

47 - Question

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 stress induced in the vertical bar if E = 200GPa?
a) 50.87 N/mm2
b) 60.23 N/mm2
c) 45.24 N/mm2
d) 63.14 N/mm2
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/mm2.

48 - Question

A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the strain (E=210GPa)?
a) 0.00089
b) 0.0005
c) 0.00064
d) 0.00098
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/mm2 As strain = stress / E, thus, strain = 187.7 / 210,000 = 0.00089.

49 - Question

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
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/mm2 Elongation = stress x length / E = 60.23 x 1500 / 200,000 = 0.452mm.

50 - Question

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 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
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/mm2. Strain energy stored = stress2 x volume / 2E = 60.232 x 2525000 / (2×200,000) = 2.045 N-m.

51 - Question

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/mm2
b) 110 N/mm2
c) 120 N/mm2
d) 140 N/mm2
Explanation: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.

52 - Question

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 5cm2. 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
Explanation: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2. As, P( h + δL) = σ2/2E x V So P = 1745.8 N.

53 - Question

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/mm2
b) 60 N/mm2
c) 70 N/mm2
d) 80 N/mm2