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All questions of Stress and Strain for Mechanical Engineering Exam

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A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, is
  • a)
    60 MPa tensile
  • b)
    80 MPa tensile
  • c)
    80 MPa compressive
  • d)
    60 MPa compressive
Correct answer is option 'D'. Can you explain this answer?

Neha Joshi answered
L = 2 m, t1 = 20° and t2 = 60°
Temperature change (∆t) = t2 − t1 = 40℃
E = 200 x 103 MPa
α = 10 x 10−6/℃
yielding of the wall (λ) = 0.2 mm
BB1 = δth = α(∆t)L
B1 B2 = Expansion restricted by supports
λ = BB2 = yielding of support
= −100[(2000 x 10 x 10−6 x 40) − 0.2]
= −60 MPa = 60 MPa compressive

If the rigid rod fitted snugly between the supports as shown in the figure below, is heated, the stress induced in it due to 20℃ rise in temperature will be (α = 12.5 x 10-6/℃ E = 200GPa)
  • a)
    0.07945 MPa
  • b)
    −0.07945 MPa
  • c)
    −0.03972 MPa
  • d)
    0.03972 MPa
Correct answer is option 'B'. Can you explain this answer?

Gate Funda answered
Expansion of rod due to temperature rise δ = α∆tL
= 12.5 x 10−6 x 20 x 0.5
= 1.25 x 10−4m
Force induced by the spring due to compression
F = kδ = 50 x 103 x 1.25 x 10−4 = 6.25 N
This stress is compressive for the bar,
∴ Stress = −0.07945 MPa

A rigid beam of negligible weight is supported in a horizontal position by two rods of steel and aluminum, 2m and 1m long having values of cross-sectional areas 1 cm2 and 2 cm2 and E of 200 GPa and 100 GPa respectively. A load P is applied as shown in the figure.
If the rigid beam is to remain horizontal then
  • a)
    the forces on both sides should be equal
  • b)
    the force on aluminum rod should be twice the force on steel
  • c)
    the force on the steel rod should be twice the force on aluminum
  • d)
    the force p must be applied at the center of the beam
Correct answer is option 'B'. Can you explain this answer?

Sarita Yadav answered
If the beam is to remain horizontal ⇒ δAB = δCD Let force on rod AB be P′
⇒ Force on rod CD = P − P′
⇒ 2P′ = P − P′
⇒ 3P′ = P
⇒ P′ = P/3(force on steel rod) Force on aluminum rod
⇒ Force on aluminum rod
= 2 x force on steel rod.

A rigid plate ABCD, 1 m2 is supported over four legs of equal lengths of same crosssection and same material. A vertical load 1200 N acts at point E such that
AM = LE = 0.5 m
AL = ME = 0.4 m
The reactions RA , RB , RC and RD are
  • a)
    RA = 360N, RB = 360N, RC = 240N, RD = 240N
  • b)
    RA = 240N, RB = 240N, RC = 360N, RD = 360N
  • c)
    RA = 360N, RB = 240N, RC = 360N, RD = 360N
  • d)
    RA = 240N, RB = 360N, RC = 240N, RD = 360N
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered
For equilibrium RA + RB + RC + RD = 1200 N ⋯ ①
ABCD is a rigid plate, under load, its plane is changed, it is not deformed. Say under load legs deform by δA , δB , δC , δD respectively.
Diagonal AC and diagonal BD remain straight. Centre of the plate lies on both diagonals AC and BD.
δO = deflection at centre of plates
Moreover within the elastic limit, deflection is proportional to the load (or reaction) so
RA + RC = RB + RD
= 1200 / 2
= 600N ⋯ ③
Now taking moments of the forces about the edge AB
1200 x ME = I x AC + RD x 1
1200 x 0.4 = RC + RD
But RC + RD = 480
Taking moments about edge AD
1200 x 0.5 = RB x 1 + RC x 1
RB + RC = 600
RB − RC = 120
RB = 360N
RC = 240N
RA = 600 − RC
RA = 600 − 240 = 360N
RB + RD = 600
RD = 600 − RB
= 600 − 360 = 240N
Finally the reactions are RA = 360N,
RB = 360N, RC = 240N, RD = 240N

An aluminum bar of 8 m length and a steel bar of 5 mm longer in length are kept at 30℃.If the ambient temperature is raised gradually, at what temperature the aluminum bar will elongate 5 mm longer than the steel bar? (The linear expansion coefficients for steel and aluminum are 12 x 10−6/℃ and 23 x 10−6/℃ respectively)
  • a)
    50.7℃
  • b)
    69.0℃
  • c)
    143.7℃
  • d)
    33.7℃
Correct answer is option 'C'. Can you explain this answer?

Sanvi Kapoor answered
L = 8 m Initially,
αs = 12 x 10−6/℃
αa = 23 x 10−6/℃
t1 = 30℃
Total elongation of aluminum-total elongation of steel
= (AA3) − (A1A2)
= (AA1) + A1A2 + A2A3) − (A1A2)
= AA1 + A2A3 = 10mm
L αal∆t − Lαs∆t = 10
8 x 103 x ∆t[23 x 10−6 − 12 x 10−6] = 10
t2 − 30° = 113.636℃
t2 = 143.636℃

Youngs modulus of elasticity and Poissons ratio of a material are 1.25 x 102 MPa and 0.34 respectively. The modulus of rigidity of the material is __________
  • a)
    0.9469 MPa
  • b)
    0.8375 MPa
  • c)
    0.4664 MPa
  • d)
    0.4025 MPa
Correct answer is option 'C'. Can you explain this answer?

Rajat Basu answered
Modulus of Rigidity Calculation:
1. Given Values:
- Young's Modulus (E) = 1.25 x 10^2 MPa
- Poisson's Ratio (ν) = 0.34
2. Formula for Modulus of Rigidity (G):
G = E / (2 * (1 + ν))
3. Calculate Modulus of Rigidity:
G = 1.25 x 10^2 / (2 * (1 + 0.34))
G = 1.25 x 10^2 / (2 * 1.34)
G = 1.25 x 10^2 / 2.68
G ≈ 0.4664 MPa
Therefore, the Modulus of Rigidity of the material is approximately 0.4664 MPa, which corresponds to option (c).

The independent elastic constants for homogeneous and isotropic material are
  • a)
    E, G, K, V
  • b)
    E, G, K
  • c)
    E, G, V
  • d)
    E, G
Correct answer is option 'D'. Can you explain this answer?

Vibhor Goyal answered
Only two independent elastic constant are required to express the stress-strain relation for a linearly elastic, isotropic material. We can find third elastic constant and Poisson’s ratio for the material.

Consider a bar of length L, breadth B and thickness t subjeced to an axial pull or tension P. The resulting volumetric strain will be equal to
  • a)
    ∈ (1 -2v)
  • b)
    2∈ (1 - v )
  • c)
    ∈ (1 + 2v)
  • d)
    3 ∈ v
Correct answer is option 'A'. Can you explain this answer?

Saranya Saha answered
Explanation:
The volumetric strain (ε_v) is defined as the change in volume per unit volume of the material. When a bar is subjected to axial tension, it experiences elongation in the axial direction and contraction in the lateral directions.

Formula:
ε_v = ε_x + ε_y + ε_z
Where:
ε_x = Strain in x-direction
ε_y = Strain in y-direction
ε_z = Strain in z-direction

Given:
ε_x = P/(A*E)
ε_y = -v*ε_x
ε_z = -v*ε_x
Where:
P = Applied load
A = Cross-sectional area of the bar
E = Young's modulus of the material
v = Poisson's ratio

Calculation:
ε_v = ε_x + ε_y + ε_z
ε_v = ε_x - v*ε_x - v*ε_x
ε_v = ε_x(1 - 2v)
Therefore, the resulting volumetric strain is equal to ε_x(1 - 2v), which corresponds to option 'A' - ∈ (1 - 2v).

For an Oldham coupling used between two shafts, which among the following statements are correct?
I. Torsional load is transferred along shaft axis.
II. A velocity ratio of 1:2 between shafts is obtained without using gears.
III. Bending load is transferred transverse to shaft axis.
IV. Rotation is transferred along shaft axis.
  • a)
    I and III
  • b)
    I and IV
  • c)
    II and III
  • d)
    II and IV
Correct answer is option 'B'. Can you explain this answer?

Vibhor Goyal answered
Oldham coupling is used to connect two shafts which are not on the same axis means they are not aligned to the same axis
So,
(1) Torsional load is transferred along shaft axis as both shafts are rotating member.
So, that statement is correct.
(2) A velocity ratio 1:1 between shafts is obtained using gears.
So, this statement is wrong
(3) Bending load is not transferred transverse to shaft axis as there is no transverse load.
(4) Rotation is transferred along shaft axis
So, this statement is correct

For a loaded cantilever beam of uniform cross-section, the bending moment (in N.mm) along the length is M(x)=5x2+10x , where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x=10 mm is
  • a)
    100
  • b)
    105
  • c)
    110
  • d)
    115
Correct answer is option 'C'. Can you explain this answer?

Nitin Joshi answered
Understanding the Problem
For a cantilever beam with a given bending moment function M(x) = 5x² + 10x, we need to find the shear force at a specific point (x = 10 mm) along the beam.
Key Concepts
- Bending Moment (M): Represents the internal moment that causes the beam to bend.
- Shear Force (V): Represents the internal force that acts perpendicular to the beam's axis.
Finding Shear Force
To find the shear force (V) at a position x, we can use the relationship between bending moment and shear force:
V(x) = dM/dx
This means we need to differentiate the bending moment function M(x).
Calculate the Derivative
1. Given M(x) = 5x² + 10x
2. Differentiate M with respect to x:
- dM/dx = 10x + 10
Evaluate Shear Force at x = 10 mm
1. Substitute x = 10 mm into the derivative:
- V(10) = 10(10) + 10
- V(10) = 100 + 10
- V(10) = 110 N
Conclusion
Thus, the magnitude of the shear force at x = 10 mm is 110 N, which corresponds to option 'C'. This calculation confirms the internal force acting on the beam at that point, which is crucial for understanding the beam's structural integrity.

Two circular shafts made of same material, one solid (S) and one hollow (H), have the same length and polar moment of inertia. Both are subjected to same torque. Here, θs​ is the twist and τs​ is the maximum shear stress in the solid shaft, whereas θH​ is the twist and τH​ is the maximum shear stress in the hollow shaft. Which one of the following is TRUE?
  • a)
    θs​=θh​andτs​=τH​
  • b)
    θs​>​θhandτs​>τH​
  • c)
    θs​<θh​andτs​<τH​
  • d)
    θs​=θh​andτs​<τH
Correct answer is option 'D'. Can you explain this answer?

Milan Ghosh answered
Understanding the Problem
Two shafts, one solid (S) and one hollow (H), are subjected to the same torque and have the same polar moment of inertia. The goal is to compare their twists and maximum shear stresses.
Key Concepts
- Twist (θ): The angle of rotation per unit length due to applied torque.
- Maximum Shear Stress (τ): The highest shear stress experienced in the material due to the applied torque.
Formulas for Twisting
- The twist for a circular shaft is given by:
θ = (T * L) / (J * G)
where T is torque, L is length, J is polar moment of inertia, and G is the shear modulus.
Solid vs. Hollow Shaft
- Both shafts have the same polar moment of inertia (J), length (L), and material (same G).
- The solid shaft will have a higher maximum shear stress compared to the hollow shaft, given the same torque.
Comparison of Twists
- Since both shafts have the same J and are subjected to the same torque, they will experience the same twist:
θs = θH.
Comparison of Shear Stresses
- The maximum shear stress in a solid shaft (τs) is higher than that in a hollow shaft (τH) because the solid shaft has more material resisting the torque.
Conclusion
Thus, the correct answer is:
- Option D: θs = θH and τs < τh.="" />
This indicates that while both shafts undergo the same amount of twist, the solid shaft experiences less maximum shear stress than the hollow shaft due to differences in geometry and material distribution.

Consider the following statements:
X. Two-dimensional stresses applied to a thin plater in its own plane represent the plane stress condition.
Y. Normal and shear stresses may occur simultaneously on a plane.
Z. Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
Which of the above statements are correct?
  • a)
    2 only
  • b)
    1 and 2
  • c)
    2 and 3
  • d)
    1 and 3
Correct answer is option 'D'. Can you explain this answer?

Dishani Desai answered
Understanding Plane Stress Condition
In the context of mechanical engineering, particularly in materials and structural analysis, the following explanations address the correctness of the statements provided.
Statement X: Two-dimensional stresses applied to a thin plate in its own plane represent the plane stress condition.
- This statement is incorrect.
- Plane stress conditions are typically relevant for thin plates subjected to loads in their own plane, but this does not mean that all two-dimensional stresses necessarily imply a plane stress condition.
Statement Y: Normal and shear stresses may occur simultaneously on a plane.
- This statement is correct.
- In mechanics, it is common for both normal and shear stresses to act on a plane simultaneously. This is a fundamental aspect of stress analysis in materials.
Statement Z: Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
- This statement is correct.
- In a plane stress condition, the thickness direction (perpendicular to the plane) experiences negligible stress, which results in zero strain in that direction.
Conclusion
- Based on the analysis above:
- Statements Y and Z are correct, while statement X is incorrect.
- Therefore, the correct answer to the question is option D (1 and 3), as it confirms that Y and Z are valid statements regarding stresses in a plane stress scenario.

Consider the following statements: The thermal stress is induced in a component in general, when
1. a temperature gradient exists in the component
2. the component is free from any restraint
3. it is restrained to expand or contract freely
Which of the above statements are correct?
  • a)
    1 and 2
  • b)
    2 and 3
  • c)
    3 alone
  • d)
    2 alone
Correct answer is option 'C'. Can you explain this answer?

Sanvi Kapoor answered
1. If a temperature gradient exists in the component, then there will be thermal stresses generated if the component is restrained to expand or contract freely.
2. If the component is free from restraints, no stresses will be generated. However, strain will be there.
3. Thermal stresses will be induced when it is restrained to expand or contract freely.
Ex.: fixed beam with a rise in temperature.

Two steel rods of 20 mm diameter are joined end to end by means of a turn buckle as shown in figure, other end of each rod is rigidly fixed. Initially, there is no tension in each rod. Effective length of each rod is 5 m. Threads on each rod has a pitch of 2.4 mm. Calculate the increase in tension when the turn buckle is tightened by one third of revolution. E = 208 GPa. If the temperature of the rods is increased, then at what temperature, tension in each rod is reduced to half of its magnitude. αs = 11 x 10−6/℃.
  • a)
    9.62℃
  • b)
    7.27℃
  • c)
    11.63℃
  • d)
    3.18℃
Correct answer is option 'B'. Can you explain this answer?

Sarita Yadav answered
A turn buckle is a type of nut having internal threads on two ends but in opposite directions. Ends of tie rods are inserted from both the sides as shown. By rotating the turn buckle, both the tie rods are tightened.
Length of each rod = 5 m Pitch of thread = 2.4 mm
Axial movement of rods on both the sides = 2.4/ 3 = 0.8 mm
Strain in each rod = 0.8 / 5000 = 0.16 x 10−3
E = 208000 N/mm2
σ, stress in each rod
= 0.16 x 10−3 x 208 x 1000 = 33.28 N/mm2
Area of cross–section of each rod,
A =π / 4 x 202 = 100 π mm2
Tension in each rod, T = 100π x 33.28
= 10455 N
= 10.455 kN
Tension in rod is to be reduced to half by increasing the temperature of the bar

If the Young’s modulus E is equal to bulk modulus K, then the value of Poisson’s ratio is
  • a)
    1/4
  • b)
    1/2
  • c)
    1/3
  • d)
    3/4
Correct answer is option 'C'. Can you explain this answer?

Suyash Patel answered
Understanding Young’s Modulus and Bulk Modulus
Young’s modulus (E) and bulk modulus (K) are crucial material properties used in mechanical engineering to describe how materials deform under stress.
Definitions
- Young’s Modulus (E): Measures the tensile or compressive stiffness of a material, defined as the ratio of stress (force per unit area) to strain (proportional deformation).
- Bulk Modulus (K): Measures a material's response to uniform compression, defined as the ratio of volumetric stress to the change in volume strain.
Relationship Among Material Properties
To understand the relationship between E, K, and Poisson's ratio (ν), we can use the following formula:
- K = E / [3(1 - 2ν)]
When E = K, we can substitute K in the equation:
- E = E / [3(1 - 2ν)]
This simplifies to:
- 1 = 1 / [3(1 - 2ν)]
Solving for Poisson’s Ratio
Rearranging the equation gives:
- 3(1 - 2ν) = 1
- 1 - 2ν = 1/3
Now, solving for ν:
- 2ν = 1 - 1/3
- 2ν = 2/3
- ν = 1/3
Conclusion
Thus, when Young’s modulus equals bulk modulus, the value of Poisson’s ratio is 1/3.
Correct Answer
- The correct answer to the question is option C: 1/3.

Chapter doubts & questions for Stress and Strain - Strength of Materials (SOM) 2025 is part of Mechanical Engineering exam preparation. The chapters have been prepared according to the Mechanical Engineering exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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