Strain Energy MCQ Quiz - Objective Question with Answer for Strain Energy - Download Free PDF

Last updated on Jun 21, 2025

Latest Strain Energy MCQ Objective Questions

Strain Energy Question 1:

Which of the following factors primarily influences the toughness of a material?

  1. Surface finish
  2. Grain size
  3. Yield strength
  4. Elastic modulus

Answer (Detailed Solution Below)

Option 2 : Grain size

Strain Energy Question 1 Detailed Solution

Explanation:

Toughness of a Material

  • Toughness is a critical mechanical property of materials that defines their ability to absorb energy and plastically deform without fracturing. It is an essential property for materials used in structural and mechanical applications where resistance to impact and the ability to withstand sudden loads without breaking are necessary. Toughness is measured as the area under the stress-strain curve of a material, representing the total energy absorbed up to the point of fracture.

Grain Size

  • Grain size plays a crucial role in determining the toughness of a material. Materials are made up of small crystals or grains, and the boundaries between these grains are called grain boundaries. The size of these grains significantly affects the mechanical properties of the material, including toughness. According to the Hall-Petch relationship, reducing the grain size increases the yield strength and toughness of the material. Smaller grains provide more grain boundaries, which act as barriers to dislocation motion, enhancing the material’s ability to absorb energy before fracturing.

Smaller grain sizes improve toughness by:

  • Restricting Crack Propagation: The grain boundaries hinder the growth of cracks, making the material more resistant to fracture.
  • Enhancing Energy Absorption: More grain boundaries require more energy to overcome, leading to higher toughness.
  • Improving Uniformity: Finer grains result in more uniform stress distribution, reducing the likelihood of weak points where fractures can initiate.

Strain Energy Question 2:

The maximum energy stored at elastic limit of a material is called

  1. Proof resilience
  2. Modulus of resilience
  3. Resilience
  4. Bulk resilience

Answer (Detailed Solution Below)

Option 1 : Proof resilience

Strain Energy Question 2 Detailed Solution

Explanation:

Proof resilience

  • It is the energy absorbed per unit volume by a material up to the proof stress (yield point under certain conditions).

  • It represents the material's ability to absorb energy without permanent deformation beyond the proof stress.

  • Less commonly used term compared to modulus of resilience.

Additional InformationModulus of resilience

  • It is the maximum energy per unit volume that a material can absorb without permanent deformation, i.e., up to the elastic limit or yield point.

  • Calculated as the area under the stress-strain curve up to the elastic limit.

  • Important in design for materials subjected to sudden or impact loads.

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Resilience

  • General term for the ability of a material to absorb energy when deformed elastically and release it upon unloading.

  • It can be considered as the total energy absorbed before failure (elastic + plastic), but often loosely used.

Strain Energy Question 3:

If the strain energy absorbed in a cantilever beam in bending under its weight is X times greater than the strain energy absorbed in an identical simply supported beam in bending under its own weight, then the magnitude of X is

  1. 16
  2. 12
  3. 6
  4. 8

Answer (Detailed Solution Below)

Option 4 : 8

Strain Energy Question 3 Detailed Solution

Explanation

Strain Energy General Formula:
\(U = \int \frac{M^2}{2EI} \, dx \)

For Cantilever Beam (under uniformly distributed load w):
Bending Moment at a section x from fixed end:

\(M(x) = \frac{w}{2}(L - x)^2 \)

\(U_{\text{cantilever}} = \int_0^L \frac{1}{2EI} \left[ \frac{w}{2}(L - x)^2 \right]^2 dx = \frac{w^2 L^5}{40EI} \)

For Simply Supported Beam (under uniformly distributed load w):
Bending Moment at section x:

\(M(x) = \frac{wLx}{2} - \frac{wx^2}{2} \)

\(U_{\text{ss}} = \int_0^L \frac{1}{2EI} \left( \frac{wLx}{2} - \frac{wx^2}{2} \right)^2 dx = \frac{w^2 L^5}{320EI} \)

Ratio of Strain Energies (X):

\(X = \frac{U_{\text{cantilever}}}{U_{\text{ss}}} = \frac{\frac{w^2 L^5}{40EI}}{\frac{w^2 L^5}{320EI}} = \frac{320}{40} = 8 \)

 

Strain Energy Question 4:

The strain energy stored in a spring, when subjected to maximum load, without suffering permanent distortion, is known as 

  1. Proof stress
  2. Modulus of resilience
  3. Proof resilience
  4. Impact energy

Answer (Detailed Solution Below)

Option 3 : Proof resilience

Strain Energy Question 4 Detailed Solution

Explanation:

Strain Energy in a Spring:

  • Strain energy is the energy stored in a body or structure due to deformation caused by applied forces. For a spring, this energy is stored when the spring is subjected to a load, and it is released when the load is removed. The strain energy stored in the spring is a result of elastic deformation, and the spring returns to its original shape once the load is removed, provided the material does not exceed its elastic limit.
  • When a spring is subjected to a maximum load without suffering permanent distortion or plastic deformation, the energy stored is termed as the proof resilience. Proof resilience is the maximum strain energy a material can store within its elastic limit.

Formula for Strain Energy in a Spring:

The strain energy stored in a spring, also known as elastic potential energy, can be calculated using the following formula:

U = ½ × F × δ

Where:

  • U = Strain energy (Joules)
  • F = Force applied to the spring (Newtons)
  • δ = Deflection or displacement of the spring (meters)

Alternatively, in terms of the spring constant (K):

U = ½ × K × δ²

Where:

  • K = Spring constant (N/m)
  • δ = Deflection (m)

Proof Resilience:

  • Proof resilience is the maximum amount of strain energy stored in a material per unit volume without causing permanent deformation. It is a measure of the material's ability to absorb energy elastically. For a spring, it is the strain energy stored when the spring is subjected to its maximum elastic load.

The formula for proof resilience is given by:

Proof Resilience = σ² / (2 × E)

Where:

  • σ = Maximum stress within the elastic limit
  • E = Modulus of elasticity

Strain Energy Question 5:

If elastic strength increases 3 times, then Proof Resilience:

  1. increases 3 times
  2. decreases 9 times
  3. decreases 3 times
  4. increases 9 times

Answer (Detailed Solution Below)

Option 4 : increases 9 times

Strain Energy Question 5 Detailed Solution

Concept:

Proof Resilience is the strain energy stored per unit volume when a material is deformed up to its elastic limit. It depends on the square of the elastic strength (proof stress).

Given:

  • Initial elastic strength = \(\sigma_e\)
  • Elastic strength increases by 3 times → New elastic strength = \(3\sigma_e\)
  • Young's modulus (E) remains constant

Step 1: Recall Proof Resilience formula

\( U = \frac{\sigma_e^2}{2E} \)

Step 2: Calculate new Proof Resilience

With new elastic strength \(3\sigma_e\):

\( U_{\text{new}} = \frac{(3\sigma_e)^2}{2E} = \frac{9\sigma_e^2}{2E} = 9U \)

Conclusion:

When elastic strength increases 3 times, Proof Resilience increases by 9 times because it is proportional to the square of the elastic strength.

Top Strain Energy MCQ Objective Questions

Maximum energy that a given component can absorb without undergoing any permanent deformation upto elastic limit is known as:

  1. Proof Resilience
  2. Resilience
  3. Hardness
  4. Toughness

Answer (Detailed Solution Below)

Option 1 : Proof Resilience

Strain Energy Question 6 Detailed Solution

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Explanation:-

Resilience

  • 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 resilience is also defined as the capacity of a strained body for doing work on the removal of the straining force.
  • It is the property of materials to absorb energy and to resist shock and impact loads.
  • It is measured by the amount of energy absorbed per unit volume within elastic limit this property is essential for spring materials.
  • The resilience of material should be considered when it is subjected to shock loading.


Proof resilience

  • The maximum strain energy, stored in a body, is known as proof of resilience. The strain energy stored in the body will be maximum when the body is stressed upto the elastic limit. Hence the proof resilience is the quantity of strain energy stored in a body when strained up to the elastic limit.
  • It is defined as the maximum strain energy stored in a body.
  • So, it is the quantity of strain energy stored in a body when strained up to the elastic limit (ability to store or absorb energy without permanent deformation).


Modulus of resilience

  • It is defined as proof resilience per unit volume.
  • It is the area under the stress-strain curve up to the elastic limit.

RRB JE ME 16 15Q SOM Chapter 2 Hindi - Final images Q1c

 

Toughness:

  • Toughness is defined as the ability of the material to absorb energy before fracture takes place.
  • This property is essential for machine components that are required to withstand impact loads.
  • Tough materials have the ability to bend, twist or stretch before failure takes place.
  • Toughness is measured by a quantity called modulus of toughness. Modulus of toughness is the total area under the stress-strain curve in a tension test.
  • Toughness is measured by Izod and Charpy impact testing machines.
  • When a material is heated it becomes ductile or simply soft and thus less stress is required to deform the material and the stress-strain curve will shift down and the area under the curve decreases thus toughness decreases.
  • Toughness decreases as temperature increases.

Hardness:

  • Hardness is a measure of the resistance to localized plastic deformation induced by either mechanical indentation or abrasion. 
  • Hardness Testing measures a material’s strength by determining resistance to penetration.
  • There are various hardness test methods, including Rockwell, Brinell, Vickers, Knoop and Shore Durometer testing.

26 June 1

When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue.

The slow and continuous elongation of a material with time at constant stress and high temperature below the elastic limit is called creep.

The strain energy of a beam is ____.

  1. Independent of shear force in the beam
  2. Independent of bending moment in the beam
  3. Same as total potential energy
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Strain Energy Question 7 Detailed Solution

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The elastic strain energy stored in a member of length s (it may be curved or straight) due to axial force, bending moment, shear force and torsion are summarized below:

Axial Force, P

\({U_1} = \mathop \smallint \limits_0^s \frac{{{P^2}}}{{2AE}}ds\)

Bending, M

\({U_2} = \mathop \smallint \limits_0^s \frac{{{M^2}}}{{2EI}}ds\)

Shear Force, V

\({U_3} = \mathop \smallint \limits_0^s \frac{{{V^2}}}{{2AG}}ds\)

Torsion, T

\({U_4} = \mathop \smallint \limits_0^s \frac{{{T^2}}}{{2GJ}}\)

 

The strain energy of a beam

  • Depends on the shear force in the beam
  • Depends on the bending moment in the beam
  • It is different than the potential energy

For a linearly elastic structure, Which of the following principles states that where external forces only cause deformation, the complementary energy is equal to the deformation energy?

  1. Castigliano's Second Principle
  2. Maxwell Reciprocal Principle
  3. Muller breslau Principle
  4. Castigliano's First Principle

Answer (Detailed Solution Below)

Option 4 : Castigliano's First Principle

Strain Energy Question 8 Detailed Solution

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Explanation:

Castigliano’s first theorem-

  • For linearly elastic structures, where external forces only cause deformations, the complementary energy is equal to the strain energy.
  • For such structures, Castigliano’s first theorem may be stated as the first partial derivative of the strain energy of the structure with respect to any particular displacement gives the force causing the deflection at that point
  • This first theorem is applicable to linearly or nonlinearly elastic structures in which the temperature is constant and the supports are unyielding.

F1 Abhayraj Shraddha 15.12.2020 D1

\(\begin{array}{l} \frac{{\partial U}}{{\partial \Delta }} = P\\ \frac{{\partial U}}{{\partial \theta }} = M \end{array}\)

Castigliano’s second theorem-

  • The first partial derivative of the total internal energy in a structure with respect to the force applied at any point is equal to the deflection at the point of application of that force in the direction of its line of action.
  • The second theorem of Castigliano is applicable to linearly elastic (Hookean material) structures with constant temperature and unyielding supports.

 

F1 Abhayraj Shraddha 15.12.2020 D1

\(\begin{array}{l} \frac{{\partial U}}{{\partial P}} = \Delta \\ \frac{{\partial U}}{{\partial M}} = \theta \end{array}\)

Resilience of material is important, when it is subjected to

  1. Thermal stresses
  2. Shock loading
  3. Fatigue
  4. Wear and tear

Answer (Detailed Solution Below)

Option 2 : Shock loading

Strain Energy Question 9 Detailed Solution

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Explanation:-

Resilience

  • It is the property of materials to absorb energy and to resist shock and impact loads.
  • It is measured by the amount of energy absorbed per unit volume within elastic limit this property is essential for spring materials.
  • The resilience of material should be considered when it is subjected to shock loading.

Proof resilience

  • It is defined as the maximum strain energy stored in a body.
  • So, it is the quantity of strain energy stored in a body when strained up to the elastic limit (ability to store or absorb energy without permanent deformation).

Modulus of resilience

  • It is defined as proof resilience per unit volume.
  • It is the area under the stress-strain curve up to the elastic limit.

RRB JE ME 16 15Q SOM Chapter 2 Hindi - Final images Q1c

26 June 1

When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue.

The slow and continuous elongation of a material with time at constant stress and high temperature below the elastic limit is called creep.

Additional Information

Thermal Stress – Stress caused due to the change in temperature

Fatigue – Fatigue occurs when a structure is subjected to cyclic loading.

Wear and tear – This is a deterioration or damage happen to something naturally has over time through normal day to day use.

Modulus of toughness

  • It is the ability to absorb energy up to fracture.
  • From the stress-strain diagram, the area under the complete curve gives the measure of modules of toughness.

RRB JE ME 16 15Q SOM Chapter 2 Hindi - Final images Q1d

The strain energy stored in a body due to suddenly applied load compared to when it is applied gradually is

  1. Same
  2. Twice
  3. Half
  4. Four times

Answer (Detailed Solution Below)

Option 4 : Four times

Strain Energy Question 10 Detailed Solution

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Concept:

In gradual loading, the loading starts from zero and increases gradually till the body is fully loaded, while in sudden loading, the load is suddenly applied on the body.

04.11.2017.029

\({σ _{gradual}} = \frac{P}{A}\)

\({σ _{sudden}} = \frac{{2P}}{A}\)

\({\rm{Strain\;energy}}\left( {\rm{U}} \right) = \frac{{{{\rm{σ }}^2}{\rm{V}}}}{{2{\rm{E}}}}\)

Calculation:

Given:

σgradual = σ, σsudden = 2σ.

\({\rm{Strain\;energy}}\left( {\rm{U}} \right) = \frac{{{{\rm{σ }}^2}{\rm{V}}}}{{2{\rm{E}}}}\)

\({{\rm{U}}_{{\rm{gradual}}}} = \frac{{{{\rm{σ }}^2}{\rm{V}}}}{{2{\rm{E}}}}{\rm{\;}},\)

\({{\rm{U}}_{{\rm{suddenly}}}} = \frac{{{{\left( {2{\rm{σ }}} \right)}^2}{\rm{V}}}}{{2{\rm{E}}}} = \frac{{4{{\rm{σ }}^2}{\rm{V}}}}{{2{\rm{E}}}}\)

\(\frac{{{{\rm{U}}_{{\rm{sudden}}}}}}{{{{\rm{U}}_{{\rm{gradual}}}}}} = \frac{{4{{\rm{σ }}^2}{\rm{V}}}}{{2{\rm{E}}}} \times \frac{{2{\rm{E}}}}{{{{\rm{σ }}^2}{\rm{V}}}} = 4\)

The strain energy stored in a body due to suddenly applied load compared to when it is applied gradually is four times.

The capacity of a strained body for doing work on the removal of the straining force is called

  1. Strain energy 
  2. Resilience
  3. Proof resilience
  4. Impact energy

Answer (Detailed Solution Below)

Option 2 : Resilience

Strain Energy Question 11 Detailed Solution

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Concept:

  • The energy which is absorbed in the body due to the straining effect is known as strain energy (U). 

\({\bf{U}} = \frac{1}{2} \times {\bf{\sigma }} ~\times~{\epsilon} = \frac{{{{\bf{\sigma }}^2}}}{{2{\bf{E}}}} \times {\bf{vol}}\)

  • Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence the resilience is also defined as the capacity of a strained body for doing work on the removal of the straining force
  • The strain energy stored in the body will be maximum when the body is stressed up to the elastic limit
  • Resilience is also defined as the energy absorbed by a component within an elastic region. 
  • The proof resilience is the maximum quantity of strain energy stored in a body when strained up to the elastic limit
  • Modulus of resilience is defined as proof resilience of a material per unit volume
  • Impact energy is defined as the energy absorbed by the component just before its fracture. It is also called as toughness

The strain energy stored by an elastic member subjected to bending is given by

  1. ∫ M2 dx/4EI
  2. ∫ M2 dx/EI
  3. ∫ M2 dx/3EI
  4. ∫ M2 dx/2EI

Answer (Detailed Solution Below)

Option 4 : ∫ M2 dx/2EI

Strain Energy Question 12 Detailed Solution

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Concept:

The strain energy is the energy stored in a body due to its elastic deformation.

The most general formula for strain energy is:

U = \(\frac{1}{2}\) x F x δ 

Where, F is the applied force and δ is deformation

However, when stress is proportional to strain, ϵ, the strain energy formula is:

U = \(\frac{1}{2}\) x V x σ x ϵ 

Where, V is the volume of the material.

Based on the above formulas, the strain energy in the material due to different types of forces/moments is given as:

1. Strain energy due to bending is:

U = ∫ \(\frac{M^2}{2EI}\)dx

2. Strain energy due to torsion is:

U = ∫ \(\frac{T^2L}{2GJ}\)dx

U1 and U2 are the strain energies stored in a prismatic bar due to axial tensile forces P1 and P2 respectively. The strain energy U stored in the same bar due to combined action of P1 and P2 will be

  1. U < U1 + U2
  2. U = U1U2
  3. U = U1 + U2
  4. U > U1 + U2

Answer (Detailed Solution Below)

Option 4 : U > U1 + U2

Strain Energy Question 13 Detailed Solution

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Concept:

Strain energy stored in prismatic bar due to axial load is:

\(U = \frac{{{P^2}L}}{{2AE}}\)

Calculation:

\({U_1}\left( {{P_1}} \right) = \frac{{P_1^2L}}{{2AE}}\)

\({U_2}\left( {{P_2}} \right) = \frac{{P_2^2L}}{{2AE}}\)

\(U\left( {{P_1} + {P_2}} \right) = \frac{{{{\left( {{P_1} + {P_2}} \right)}^2}L}}{{2AE}} = \frac{{\left( {P_1^2 + P_2^2 + 2{P_1}{P_2}} \right)L}}{{2AE}}\)

\(U\left( {{P_1} + {P_2}} \right) = \frac{{P_1^2L}}{{2AE}} + \frac{{P_2^2L}}{{2AE}} + \frac{{2{P_1}{P_2}L}}{{2AE}} = {U_1} + {U_2} + \frac{{{P_1}{P_2}L}}{{AE}}\)

U > U1 + U2

The deflection is ‘δ’, strain energy ‘U’ and load ‘W’ on a truss. These are related by

 

  1. \({\rm{\delta }} = \frac{{\partial {\rm{U}}}}{{\partial {\rm{W}}}}\)
  2. \({\rm{\delta }} = \frac{{{\partial ^2}{\rm{U}}}}{{\partial {{\rm{W}}^2}}}\)
  3. \({\rm{\delta }} = \frac{{{\partial ^3}{\rm{U}}}}{{\partial {{\rm{W}}^3}}}\)
  4. \({\rm{\delta }} = {\left( {\frac{{\partial {\rm{U}}}}{{\partial {\rm{W}}}}} \right)^2}\)

Answer (Detailed Solution Below)

Option 1 : \({\rm{\delta }} = \frac{{\partial {\rm{U}}}}{{\partial {\rm{W}}}}\)

Strain Energy Question 14 Detailed Solution

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Concept:

The two theorems of Castigliano's for structural analysis are as follows:

Castigliano’s 1st theorem: The partial derivative of total strain energy of the system with respect to any particular deflection at a point is equal to the force applied at that point in the same direction as that of the deflection.

\(\frac{{\partial {\rm{u}}}}{{\partial {\rm{\delta }}}} = {\rm{w}}\)

F1 Neel Madhu 02.04.20 D11

Castigliano’s 2nd theorem: The partial derivative of strain energy of the system with respect to load at any point is equal to deflection at that point.

\(\frac{{\partial {\rm{u}}}}{{\partial {\rm{w}}}} = {\rm{\delta }}\)

Also, the partial derivative of strain energy of the system with respect to couple at any point is equal to slope at that point.

\(\frac{{\partial {\rm{u}}}}{{\partial {\rm{M}}}} = {\rm{\theta }}\)

Where,

u is the strain energy of the system.

Note:

These theorems are valid in both the beams and truss, but in truss strain energy is only due to axial loads.

What will be the strain energy stored in the metallic bar of cross sectional area of 2 cm2 and gauge length of 10 cm if it stretches 0.002 cm under the load of 12 kN?

  1. 10 N-cm
  2. 12 N-cm
  3. 14 N-cm
  4. 16 N-cm

Answer (Detailed Solution Below)

Option 2 : 12 N-cm

Strain Energy Question 15 Detailed Solution

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Concept:

The strain energy stored in the metallic bar is given as,

\(E = \frac12 P.δ\)

Where

P = load, δ = deflection or stretches 

Calculation:

Given:

δ = 0.002 cm, P = 12 kN = 12000 N

The energy stored in the metallic bar will be,

\(E = \frac12 P.δ\)

\(E = \frac12 \times 12000\times 0.002\)

E = 12 N-cm

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