Strength of Materials MCQ & Objective Questions
Strength of Materials is a crucial subject for students preparing for school and competitive exams in India. Understanding this topic not only enhances your conceptual clarity but also significantly boosts your exam performance. Practicing MCQs and objective questions helps you identify important areas, refine your problem-solving skills, and prepare effectively for your upcoming tests.
What You Will Practise Here
Key concepts of stress, strain, and their relationships.
Types of materials and their mechanical properties.
Fundamental formulas for bending, torsion, and shear.
Understanding of elastic and plastic deformation.
Diagrams illustrating stress-strain curves and failure modes.
Applications of the theory in real-world engineering problems.
Important definitions and terminologies related to material strength.
Exam Relevance
The topic of Strength of Materials is frequently included in the syllabus of CBSE, State Boards, NEET, and JEE exams. Students can expect questions that test their understanding of fundamental concepts, application of formulas, and interpretation of diagrams. Common question patterns include numerical problems, conceptual MCQs, and scenario-based questions that require critical thinking.
Common Mistakes Students Make
Confusing stress and strain definitions and their units.
Misapplying formulas for different loading conditions.
Overlooking the significance of material properties in problem-solving.
Failing to interpret stress-strain graphs accurately.
FAQs
Question: What are the main topics covered under Strength of Materials?Answer: The main topics include stress and strain, mechanical properties of materials, bending and torsion, and failure theories.
Question: How can I improve my performance in Strength of Materials MCQs?Answer: Regular practice of objective questions, understanding key concepts, and reviewing mistakes will enhance your performance.
Now is the time to take charge of your exam preparation! Dive into our collection of Strength of Materials MCQ questions and practice diligently to test your understanding and boost your confidence.
Q. A steel rod with a diameter of 10 mm is subjected to a tensile force of 20 kN. What is the stress in the rod?
A.
1273.24 MPa
B.
159.15 MPa
C.
2000 MPa
D.
500 MPa
Show solution
Solution
Stress = Force / Area. The area A = π(d/2)² = π(0.005)² = 7.85e-5 m². Stress = 20000 N / 7.85e-5 m² = 254647.91 Pa or 254.65 MPa.
Correct Answer:
B
— 159.15 MPa
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Q. For a simply supported beam with a point load at the center, what is the maximum bending moment?
A.
WL/4
B.
WL/2
C.
WL
D.
0
Show solution
Solution
The maximum bending moment for a simply supported beam with a point load at the center is given by M = WL/4, where W is the load and L is the length of the beam.
Correct Answer:
B
— WL/2
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Q. If a material has a Young's modulus of 200 GPa and is subjected to a tensile stress of 100 MPa, what is the strain?
A.
0.0005
B.
0.005
C.
0.05
D.
0.5
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Solution
Strain can be calculated using the formula: Strain = Stress / Young's Modulus. Thus, Strain = 100 MPa / 200 GPa = 0.0005.
Correct Answer:
A
— 0.0005
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Q. In a beam subjected to bending, what does the neutral axis represent?
A.
The axis where shear stress is maximum
B.
The axis where bending stress is zero
C.
The axis where axial stress is maximum
D.
The axis where deflection is maximum
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Solution
The neutral axis in a beam subjected to bending is the axis where the bending stress is zero.
Correct Answer:
B
— The axis where bending stress is zero
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Q. In a beam, what does the area under the shear force diagram represent?
A.
Bending moment
B.
Deflection
C.
Shear stress
D.
Axial load
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Solution
The area under the shear force diagram represents the bending moment at that section of the beam.
Correct Answer:
A
— Bending moment
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Q. In a bending moment diagram, what does a positive bending moment indicate?
A.
Beam is sagging
B.
Beam is hogging
C.
Beam is in tension
D.
Beam is in compression
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Solution
A positive bending moment indicates that the beam is sagging, meaning the top fibers are in compression and the bottom fibers are in tension.
Correct Answer:
A
— Beam is sagging
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Q. In a cantilever beam with a uniform distributed load, where does the maximum shear force occur?
A.
At the free end
B.
At the fixed support
C.
At the midpoint
D.
Uniformly distributed along the length
Show solution
Solution
The maximum shear force in a cantilever beam with a uniform distributed load occurs at the fixed support.
Correct Answer:
B
— At the fixed support
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Q. In a simply supported beam, what is the maximum bending moment at the center when a point load is applied at the center?
A.
PL/4
B.
PL/2
C.
PL
D.
0
Show solution
Solution
For a simply supported beam with a point load P at the center, the maximum bending moment is given by M = PL/4.
Correct Answer:
B
— PL/2
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Q. What is the critical load for a column with a length of 3 m, a moment of inertia of 0.0001 m^4, and a modulus of elasticity of 200 GPa?
A.
1.96 kN
B.
2.45 kN
C.
3.14 kN
D.
4.00 kN
Show solution
Solution
The critical load P = (π² * E * I) / (L²). Substituting the values: P = (π² * 200e9 * 0.0001) / (3²) = 1.96 kN.
Correct Answer:
A
— 1.96 kN
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Q. What is the critical load for buckling of a column fixed at both ends?
A.
π²EI / L²
B.
2π²EI / L²
C.
4π²EI / L²
D.
πEI / L²
Show solution
Solution
The critical load for buckling of a column fixed at both ends is given by P_cr = π²EI / L².
Correct Answer:
A
— π²EI / L²
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Q. What is the definition of yield strength?
A.
The maximum stress a material can withstand
B.
The stress at which a material begins to deform plastically
C.
The stress at which a material breaks
D.
The stress at which a material is fully elastic
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Solution
Yield strength is defined as the stress at which a material begins to deform plastically, meaning it will not return to its original shape.
Correct Answer:
B
— The stress at which a material begins to deform plastically
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Q. What is the effect of increasing the moment of inertia on the bending stress in a beam?
A.
Bending stress increases
B.
Bending stress decreases
C.
Bending stress remains the same
D.
Bending stress becomes zero
Show solution
Solution
Increasing the moment of inertia decreases the bending stress in a beam for a given bending moment.
Correct Answer:
B
— Bending stress decreases
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Q. What is the formula for calculating stress?
A.
Stress = Force / Area
B.
Stress = Area / Force
C.
Stress = Force * Area
D.
Stress = Area * Force
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Solution
Stress is defined as the force applied per unit area, hence the formula is Stress = Force / Area.
Correct Answer:
A
— Stress = Force / Area
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Q. What is the formula for calculating the bending stress in a beam?
A.
σ = My/I
B.
σ = F/A
C.
σ = Eε
D.
σ = VQ/It
Show solution
Solution
The bending stress in a beam is calculated using the formula σ = My/I, where M is the bending moment, y is the distance from the neutral axis, and I is the moment of inertia.
Correct Answer:
A
— σ = My/I
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Q. What is the formula for calculating the deflection of a cantilever beam with a point load at the free end?
A.
δ = PL³ / 3EI
B.
δ = PL² / 2EI
C.
δ = PL³ / 12EI
D.
δ = PL / EI
Show solution
Solution
The deflection (δ) of a cantilever beam with a point load (P) at the free end is given by δ = PL³ / 3EI.
Correct Answer:
A
— δ = PL³ / 3EI
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Q. What is the formula for calculating the moment of inertia for a rectangular beam?
A.
I = (b * h^3) / 12
B.
I = (h * b^3) / 12
C.
I = b * h^2
D.
I = (b * h) / 12
Show solution
Solution
The moment of inertia for a rectangular beam is given by the formula I = (b * h^3) / 12, where b is the base width and h is the height.
Correct Answer:
A
— I = (b * h^3) / 12
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Q. What is the formula for stress?
A.
Stress = Force / Area
B.
Stress = Area / Force
C.
Stress = Force * Area
D.
Stress = Force + Area
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Solution
Stress is defined as the force applied per unit area, hence the formula is Stress = Force / Area.
Correct Answer:
A
— Stress = Force / Area
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Q. What is the modulus of elasticity?
A.
The ratio of stress to strain
B.
The ratio of strain to stress
C.
The ratio of load to deflection
D.
The ratio of deflection to load
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Solution
The modulus of elasticity is defined as the ratio of stress to strain in the linear elastic region of a material.
Correct Answer:
A
— The ratio of stress to strain
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Q. What is the Poisson's ratio?
A.
The ratio of lateral strain to axial strain
B.
The ratio of axial strain to lateral strain
C.
The ratio of stress to strain
D.
The ratio of strain to stress
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Solution
Poisson's ratio is defined as the ratio of lateral strain to axial strain in a material subjected to uniaxial stress.
Correct Answer:
A
— The ratio of lateral strain to axial strain
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Q. What is the relationship between torque and shear stress in a circular shaft?
A.
τ = T / J
B.
τ = J / T
C.
τ = T * J
D.
τ = T + J
Show solution
Solution
The shear stress (τ) in a circular shaft is related to the torque (T) and the polar moment of inertia (J) by the formula τ = T / J.
Correct Answer:
A
— τ = T / J
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Q. What is the shear force at the midpoint of a simply supported beam with a uniform load?
A.
Zero
B.
Maximum
C.
Minimum
D.
Equal to the total load
Show solution
Solution
The shear force at the midpoint of a simply supported beam with a uniform load is zero.
Correct Answer:
A
— Zero
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Q. What is the shear stress formula in a beam?
A.
Shear Stress = V / A
B.
Shear Stress = A / V
C.
Shear Stress = V * A
D.
Shear Stress = V + A
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Solution
Shear stress is defined as the internal force (shear force V) divided by the area (A) over which it acts, hence Shear Stress = V / A.
Correct Answer:
A
— Shear Stress = V / A
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Q. What is the shear stress in a circular shaft subjected to a torque of 500 Nm and a diameter of 50 mm?
A.
15.92 MPa
B.
31.84 MPa
C.
63.68 MPa
D.
79.58 MPa
Show solution
Solution
Shear stress τ = (T * r) / J, where J = (π/32) * d^4. For d = 0.05 m, J = (π/32) * (0.05)⁴ = 4.91e-9 m^4. τ = (500 * 0.025) / 4.91e-9 = 31.84 MPa.
Correct Answer:
B
— 31.84 MPa
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Q. What is the unit of shear force in the SI system?
Show solution
Solution
The unit of shear force in the SI system is Newton (N).
Correct Answer:
B
— N
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Q. What is the unit of stress in the SI system?
A.
Pascal (Pa)
B.
Newton (N)
C.
Joule (J)
D.
Kilogram (kg)
Show solution
Solution
The unit of stress in the SI system is Pascal (Pa), which is equivalent to one Newton per square meter.
Correct Answer:
A
— Pascal (Pa)
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