Strength of Materials MCQ Quiz - Objective Question with Answer for Strength of Materials - Download Free PDF

Last updated on Mar 27, 2025

Strength of Material or SOM or mechanics of solid is a very important subject in both mechanical and civil engineering. This is very high weightage subject across all engineering exams be it GATE or SSC JE or any State JE/AE level exam. Strength of material includes topics from Stres and Strain, Shear force and bending moment (SFD/BMD), deflection, strain energy etc. This article covers top objective questions of strength of material in MCQ format. This provides free strength of material objective questions with detailed solution PDF. It includes detailed explanation on how to solve Strength of material questions with PDF.

Latest Strength of Materials MCQ Objective Questions

Strength of Materials Question 1:

Which strain measuring technique is best suited for high-temperature environments?

  1. Electrical resistance strain gauge
  2. Photoelastic coating
  3. Optical fiber strain sensors
  4. Brittle lacquer

Answer (Detailed Solution Below)

Option 3 : Optical fiber strain sensors

Strength of Materials Question 1 Detailed Solution

Explanation:

Strain Measuring Techniques for High-Temperature Environments

When selecting a strain measuring technique for high-temperature environments, it is critical to consider the durability and accuracy of the sensors under extreme conditions. Here, we analyze the given options:

  1. Electrical resistance strain gauge (Option 1)

    • While widely used for various applications, electrical resistance strain gauges have limitations in high-temperature environments due to potential drift in resistance and insulation degradation.

  2. Photoelastic coating (Option 2)

    • Photoelastic coatings are useful for visualizing strain distribution but are not ideal for high temperatures due to thermal sensitivity and potential for delamination.

  3. Optical fiber strain sensors (Option 3)

    • Optical fiber strain sensors are highly suited for high-temperature environments due to their ability to withstand extreme temperatures and provide accurate measurements over a wide temperature range.

  4. Brittle lacquer (Option 4)

    • Brittle lacquer is used for detecting strain by cracking patterns but is limited in high-temperature applications as it can degrade or perform inconsistently under thermal stress.

Conclusion

Analyzing the options, it is clear that optical fiber strain sensors (Option 3) are the most suitable for high-temperature environments due to their robustness and accuracy in such conditions, thus making them the best choice.

Strength of Materials Question 2:

The main disadvantage of brittle coating method is:

  1. No visual output
  2. Reusability
  3. Fragility and surface preparation
  4. Cost

Answer (Detailed Solution Below)

Option 3 : Fragility and surface preparation

Strength of Materials Question 2 Detailed Solution

Explanation:

Brittle Coating Method

The brittle coating method is used to detect strain in materials by applying a brittle lacquer that cracks under stress. However, this method comes with certain disadvantages, primarily:

  • Fragility and surface preparation are major concerns.

  • It requires extensive surface preparation to ensure proper adhesion of the coating.

  • The brittle nature of the coating means it can easily crack or damage during handling and application.

  • Additionally, the method cannot be reused once the coating has cracked, requiring a new application for each test.

Analyzing the Given Options

  1. "No visual output." (Incorrect)

    • The brittle coating method provides visual output in the form of cracks, which indicate areas of stress.

  2. "Reusability." (Incorrect)

    • The method is not reusable; once the coating cracks, it needs to be reapplied for another test.

  3. "Fragility and surface preparation." (Correct)

    • The coating is fragile and requires significant surface preparation.

    • Proper adhesion of the coating is crucial, which involves extensive surface preparation.

    • The fragility and the need for surface preparation make this a significant disadvantage.

  4. "Cost." (Incorrect)

    • While cost can be a factor, it is not the primary disadvantage compared to fragility and surface preparation.

Strength of Materials Question 3:

The gauge factor of a strain gauge is defined as:

  1. Ratio of strain to voltage
  2. Ratio of change in resistance to strain
  3. Ratio of resistance to current
  4. Change in temperature per unit strain

Answer (Detailed Solution Below)

Option 2 : Ratio of change in resistance to strain

Strength of Materials Question 3 Detailed Solution

Explanation:

Gauge Factor of a Strain Gauge

The gauge factor of a strain gauge is a crucial parameter that indicates the sensitivity of the strain gauge. It is defined as the ratio of the relative change in electrical resistance to the mechanical strain. When a strain gauge is subjected to strain, its resistance changes, and the gauge factor quantifies this relationship. Mathematically, it is given by the formula:

  • Gauge Factor (GF) = (ΔR / R) / ε

Analyzing the Given Options

  1. Option 1: "Ratio of strain to voltage" (Incorrect)

    • The gauge factor is not defined in terms of voltage. While voltage may be involved in the measurement process, it is not part of the gauge factor definition.

  2. Option 2: "Ratio of change in resistance to strain" (Correct)

    • This option correctly describes the gauge factor as the ratio of the change in electrical resistance (ΔR) to the mechanical strain (ε).

  3. Option 3: "Ratio of resistance to current" (Incorrect)

    • This option refers to Ohm's law (V = IR), which is unrelated to the definition of the gauge factor.

  4. Option 4: "Change in temperature per unit strain" (Incorrect)

    • The gauge factor does not involve temperature change. It is specifically related to the change in electrical resistance due to mechanical strain.

Strength of Materials Question 4:

In two-dimensional photoelastic analysis, which parameter is directly measured?

  1. Principal stress direction
  2. Strain energy
  3. Maximum shear stress
  4. Stress concentration factor

Answer (Detailed Solution Below)

Option 3 : Maximum shear stress

Strength of Materials Question 4 Detailed Solution

Explanation:

Two-Dimensional Photoelastic Analysis

In two-dimensional photoelastic analysis, the main objective is to visualize the stress distribution within a material subjected to external forces. This is done by observing patterns of light (fringes) through a polarized light setup. The key measurable parameter in this technique is the stress difference at any point, which can be related to the maximum shear stress.

  • Stress-optic law: The stress difference (and hence maximum shear stress) is proportional to the fringe order observed.

  • Fringe patterns indicate the loci of points with the same maximum shear stress.

Analyzing the Given Options

  1. "Principal stress direction." (Not directly measured)

    • While the principal stress direction can be inferred from the orientation of fringes, it is not the parameter directly measured.

  2. "Strain energy." (Not measured)

    • Strain energy is not directly measurable through photoelastic analysis. It is a derived quantity based on stress and strain.

  3. "Maximum shear stress." (Correct answer)

    • The maximum shear stress is directly related to the fringe order observed in the photoelastic analysis.

    • The higher the fringe order, the higher the maximum shear stress at that point.

  4. "Stress concentration factor." (Not directly measured)

    • The stress concentration factor is a derived quantity that indicates how much stress is amplified around discontinuities. It is not directly measured in photoelastic analysis.

Strength of Materials Question 5:

Which one of the following methods gives full-field stress visualization?

  1. Strain gauge
  2. Brittle coating
  3. Photoelasticity
  4. Load cell

Answer (Detailed Solution Below)

Option 3 : Photoelasticity

Strength of Materials Question 5 Detailed Solution

Explanation:

Full-field Stress Visualization Methods

Full-field stress visualization is a technique used to observe and analyze the distribution of stress across a material or structure. This method is crucial in understanding how different areas of the material respond to applied loads and can help in identifying potential points of failure. The given options are various methods used in stress analysis, but only one provides full-field visualization.

  1. Strain Gauge

    Strain gauges are devices used to measure the strain (deformation) in a specific area of an object. They do not provide full-field visualization as they only measure strain at discrete points where they are attached.

  2. Brittle Coating

    Brittle coating is a method where a brittle lacquer is applied to the surface of a structure. When stress is applied, the coating cracks, indicating stress concentration areas. However, it does not provide a continuous full-field visualization, only indicating regions of high stress.

  3. Photoelasticity

    Photoelasticity is an experimental technique for measuring the full-field distribution of stress. It utilizes polarized light passed through a transparent material under stress, creating a pattern of fringes that represent the stress distribution. This method provides a complete visual representation of the stress field, making it the correct answer.

  4. Load Cell

    Load cells are transducers that convert force into an electrical signal. They are used to measure load or force applied to an object. Similar to strain gauges, load cells do not provide full-field visualization as they measure force at specific points.

Based on the explanations above, the method that gives full-field stress visualization is option 3: Photoelasticity.

Top Strength of Materials MCQ Objective Questions

A tensile test is performed on a round bar. After fracture, it has been found that the diameter remains approximately same at fracture. The material under test was

  1. Mild steel
  2. Cast iron
  3. Copper
  4. Aluminium

Answer (Detailed Solution Below)

Option 2 : Cast iron

Strength of Materials Question 6 Detailed Solution

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Explanation:

Ductile material fails along the principal shear plane as they are weak in shear and brittle material fails along with principal normal stress.

EKT Free Test1 images Q3

In Brittle materials under tension test undergoes brittle fracture i.e their failure plane is 90° to the axis of load and there is no elongation in the rod that’s why the diameter remains same before and after the load. Example: Cast Iron, concrete etc.

But in case of ductile materials, material first elongates and then fail, their failure plane is 45° to the axis of the load. After failure cup-cone failure is seen. Example Mild steel, high tensile steel etc.

The room-temperature stress (σ) -strain (ϵ) curves of four materials P, Q, R, and S are shown in the figure below. The material that behaves as a rigid perfectly plastic material is

F1 Ateeb Madhu 12.07.21  D1

  1. P
  2. Q
  3. R
  4. S

Answer (Detailed Solution Below)

Option 4 : S

Strength of Materials Question 7 Detailed Solution

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Explanation:

Perfectly Plastic Material:

For this type of material, there will be only initial stress required and then the material will flow under constant stress.

The chart shows the relation between stress-strain in different materials. 

Stress-Strain Curve

Type of Material or Body

Examples

quesImage8214 Rigidly Perfectly Plastic Material

No material is perfectly plastic

F1 A.M Madhu 24.04.20 D1

Ideally plastic material.

Visco-elastic (elasto-plastic) material.

F2 A.M Madhu 15.05.20 D1

Perfectly Rigid body

No material or body is perfectly rigid.

F2 A.M Madhu 15.05.20 D2

Nearly Rigid body

Diamond, glass, ball bearing made of hardened steel, etc

F2 A.M Madhu 15.05.20 D3

Incompressible material

Non-dilatant material, (water) ideal fluid, etc.

F1 A.M Madhu 24.04.20 D2

Non-linear elastic material

Natural rubbers, elastomers, and biological gels, etc

If a part is constrained to move and heated, it will develop

  1. Principal stress
  2. Tensile stress
  3. Compressive stress
  4. Shear stress

Answer (Detailed Solution Below)

Option 3 : Compressive stress

Strength of Materials Question 8 Detailed Solution

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Explanation:

  • Change in the temperature causes the body to expand or contract.
  • Thermal stress is created when a change in size or volume is constrained due to a change in temperature.
  • So an increase in temperature creates compressive stress and a decrease in temperature creates tensile stress.

If a piece of material neither expands nor contracts in volume when subjected to stresses, then the Poisson’s ratio must be

  1. Zero
  2. 0.25
  3. 0.33
  4. 0.5

Answer (Detailed Solution Below)

Option 4 : 0.5

Strength of Materials Question 9 Detailed Solution

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Explanation:

F1 S.S M.P 23.09.19 D6

ϵv = ϵx + ϵy + ϵz

\( {ϵ_x} = \frac{1}{E}\left[ {{σ _x} - ν \left( {{σ _y} + {σ _z}} \right)} \right] \)

\({ϵ_{\rm{y}}} = \frac{1}{{\rm{E}}}\left[ {{σ _y} - ν \left( {{σ _x} + {σ _z}} \right)} \right] \)

\({ϵ_{\rm{z}}} = \frac{1}{{\rm{E}}}\left[ {{σ _z} - ν \left( {{σ _x} + {σ _y}} \right)} \right]\)

Total strain or volumetric strain is given by 

\( {ϵ_v} = \frac{1}{E} [ {σ_x} + {σ_y} + {σ_z} ](1-2ν) \)

There will be no change in volume if volumetric strain is zero.

ϵv = 0 ⇒  ν = 0.5

A steel cube, with all faces free to deform, has Young’s modulus, E, Poisson’s ratio, ν, and coefficient of thermal expansion, α. The pressure (hydrostatic stress) developed within the cube, when it is subjected to a uniform increase in temperature, ΔT, is given by

  1. 0
  2. \(\frac{{{\rm{\alpha }}\left( {{\rm{\Delta T}}} \right){\rm{E}}}}{{1 - 2{\rm{v}}}}\)
  3. \(- \frac{{{\rm{\alpha }}\left( {{\rm{\Delta T}}} \right){\rm{E}}}}{{1 - 2{\rm{v}}}}\)
  4. \(\frac{{{\rm{\alpha }}\left( {{\rm{\Delta T}}} \right){\rm{E}}}}{{3\left( {1 - 2{\rm{v}}} \right)}}\)

Answer (Detailed Solution Below)

Option 1 : 0

Strength of Materials Question 10 Detailed Solution

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Explanation:

Since all the faces are free to expand the stresses due to temperature rise is equal to 0.

Mistake Point

If the cube is constrained on all six faces, the stress produced in all three directions will be the same.

∴ thermal strain in x-direction = -α(ΔT) = \(\frac{{{\sigma _x}}}{E} - \nu \frac{{{\sigma _y}}}{E} - \nu \frac{{{\sigma _z}}}{E}\)

σx = σy = σz = σ

\(\sigma = - \frac{{\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2\nu } \right)}}\)

The reactions at the rigid supports at A and B for the bar loaded as shown in the figure are respectively:

F1 Shubham B 14.4.21 Pallavi D4

  1. 20/3 kN, 10/3 kN
  2. 10/3 kN, 20/3 kN
  3. 5 kN, 5 kN
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 20/3 kN, 10/3 kN

Strength of Materials Question 11 Detailed Solution

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Concept:

F1 Ateeb Madhu 17.12.20 D1

Let RA and RB be the reaction at support A and B respectively.

Free body diagram of the system is:

F1 Ateeb Madhu 17.12.20 D2

\(R_A=\frac{Pb}{L}\;\&\;R_B=\frac{Pa}{L}\)

Calculation:

Given:

F1 Shubham B 14.4.21 Pallavi D4

As per figure P = 10 kN, a = 1 m and b = 2 m.

\(R_A=\frac{Pb}{L}\)

\(R_A=\frac{10\times2}{3}=\frac{20}{3}\;kN\)

\(R_B=\frac{10\times1}{3}=\frac{10}{3}\;kN\)

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

Strength of Materials Question 12 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 loading and unloading response of a metal is shown in the figure. The elastic and plastic strains corresponding to 200 MPa stress, respectively, are

F1 Sumit.C 24-02-21 Savita D14

  1. 0.02 and 0.01
  2. 0.02 and 0.02
  3. 0.01 and 0.01
  4. 0.01 and 0.02

Answer (Detailed Solution Below)

Option 1 : 0.02 and 0.01

Strength of Materials Question 13 Detailed Solution

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Explanation:

F1 Sumit.C 24-02-21 Savita D14

Elastic recovery/strain: The strain recovered after the removal of the load is known as elastic strain.

Plastic strain: The permanent changes in dimension after the removal of load is known as plastic strain.

The load is removed when the stress was 200 MPa and the corresponding strain was 0.03

After the removal of load, the body recovered and the final strain found was 0.01.

∴ Elastic strain = 0.03 - 0.01 ⇒ 0.02 and Plastic strain = 0.01 respectively.

A rigid body is very slowly dropped on another body and a deflection δst occurs in the second body. If the rigid body be placed suddenly, the value of the impact factor will be:

  1. 0
  2. 1
  3. 2

Answer (Detailed Solution Below)

Option 4 : 2

Strength of Materials Question 14 Detailed Solution

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Explanation:

Strain energy:

When a body is subjected to gradual, sudden, or impact load, the body deforms and work is done upon it. If the elastic limit is not exceeded, this work is stored in the body. This work-done or energy stored in the body is called strain energy.

Strain energy = Work done.

Case-I:

F1 Ashik Madhu 16.10.20 D5

When a rigid body is slowly dropped on another body, it is a case of gradual loading:

Work-done on the bar = Area of load-deformation diagram \(⇒\frac{1}{2}\;×\;P\;×\;δ l\)

Work stored in the bar = Area of resistance deformation diagram

\(⇒\frac{1}{2}\;×\;R\;×\;δ l\)

\(⇒\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\;\;\;[\because R=\sigma A]\)

We can write;

\(⇒\frac{1}{2}\;×\;P\;×\;δ l=\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\)

\(\sigma_{gradual}=\frac{P}{A}\)

Case-II:

F1 Ashik Madhu 16.10.20 D6

Work-done on the bar = Area of load-deformation diagram ⇒ P × δl

Work stored on the bar = Area of resistance deformation diagram

\(⇒\frac{1}{2}\;×\;R\;×\;δ l\)

\(⇒\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\;\;\;[\because R=\sigma A]\)

We can write;

\(P\times\delta l=\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\)

\(\sigma_{sudden}=\frac{2P}{A}\)

\(\therefore \frac{\sigma_{sudden}}{\sigma_{gradual}}=2\)

maximum stress intensity due to suddenly applied load is twice the stress intensity produced by the load of the same magnitude applied gradually.

Impact Loading:

When a load is dropped from a height before it commences to load the body, such loading is known as Impact loading.

The ratio of the stress or deflection produced due to impact loading to the stress or deflection produced due to static or gradual loading is known as the Impact factor.

\(IF=\frac{\sigma_{impact}}{\sigma_{gradual}}=\frac{\Delta_{impact}}{\Delta_{gradual}}\)

\(IF=\frac{\sigma_{sudden}}{\sigma_{gradual}}=\frac{\Delta_{sudden}}{\Delta_{gradual}}=2\)

deflection due to sudden loading is twice that of gradual loading.

If the cross-sectional area of the bar is 15 m2 then find the stress acting in the section BC?

F1 Tabrez 11.12.20 Pallavi D13.1

  1. 0.002 N/mm2
  2. 0.2 N/mm2
  3. 2 N/mm2
  4. 2 N/m2

Answer (Detailed Solution Below)

Option 1 : 0.002 N/mm2

Strength of Materials Question 15 Detailed Solution

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Concept:

Stress at any section of the bar is given by,

\(stress, \sigma =\frac{{Load ~(P)}}{{Cross-sectional ~area~(A)}}\)

Calculation:

Given:

F1 Tabrez 11.12.20 Pallavi D14

Load in section BC, P = 30 kN (compressive), 

cross-sectional area, A = 15 m2 = 15 × 106 mm2

\(stress~ in ~section ~BC, \sigma =\frac{{30~\times~10^3}}{{15~\times ~10^6}}=0.002~N/mm^2\)

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