Strength of Materials MCQ With Answers for JKSSB Junior Engineer (Mechanical)

Technical · 2026-10-21 · 6 min read

Strength of materials questions test stress and strain, elastic constants, beam bending, torsion and columns. Most are formula-based and quick to solve once the standard results are memorised. This page gives a revision list and then solved MCQs with explanations.

Quick revision list

  • Elastic constants: E = 2G(1 + ν), steel has E of about 200 GPa, and Poisson's ratio is lateral strain divided by longitudinal strain
  • Beams: bending equation M/I = σ/y = E/R, section modulus Z = I/y, and I = bd³/12 for a rectangle and πd⁴/64 for a circle
  • Torsion: T/J = τ/r = Gθ/L
  • Columns: slenderness ratio, and Euler's load π²EI/Le² with Le = L for hinged ends, L/2 for fixed ends and 2L for a cantilever
  • Thin cylinders: hoop stress pd/2t and longitudinal stress pd/4t

Worked numerical problems

  • Axial stress: a 25 mm diameter rod carries a pull of 50 kN. The area is π/4 × 25² = 490.9 mm², so the stress is 50,000 ÷ 490.9 = 101.9 MPa
  • Elongation: the same rod is 2 m long and E = 200 GPa. The elongation is σL ÷ E = 101.9 × 2,000 ÷ 200,000 = 1.02 mm
  • Simply supported beam: a 6 m beam with a uniform load of 10 kN/m has reactions of 30 kN each and a maximum bending moment of wL²/8 = 10 × 36 ÷ 8 = 45 kN·m
  • Cantilever: a 2 m cantilever with a 5 kN load at the free end has a maximum moment of 5 × 2 = 10 kN·m at the fixed end
  • Bending stress: a 100 mm × 200 mm rectangular beam has I = 100 × 200³ ÷ 12 = 6.67 × 10⁷ mm⁴ and a section modulus of 6.67 × 10⁵ mm³. A moment of 10 kN·m gives a stress of 10⁷ ÷ 6.67 × 10⁵ = 15 MPa
  • Torsion: a solid shaft of diameter 50 mm carries a torque of 1 kN·m. The shear stress is 16T ÷ (πd³) = 16 × 10⁶ ÷ (π × 125,000) = 40.7 MPa
  • Euler column: a pinned column 3 m long with E = 200 GPa and I = 2 × 10⁶ mm⁴ buckles at π²EI ÷ L² = about 439 kN
  • Thin cylinder: with p = 2 MPa, d = 500 mm and t = 10 mm, the hoop stress is pd ÷ 2t = 50 MPa and the longitudinal stress is 25 MPa

Shear force and bending moment diagrams

For a simply supported beam, find the reactions first by taking moments. The shear force changes by the value of a point load as you pass it and varies linearly under a uniform load. The bending moment is the area under the shear force diagram, and it is greatest where the shear force passes through zero. For a cantilever, start from the free end, where both are zero.

The stress–strain curve

For mild steel, the curve shows the proportional limit, the elastic limit, the yield point, strain hardening, the ultimate strength and the fracture point. Ductile materials show a long plastic region and necking. Brittle materials such as cast iron break with little plastic deformation. The area under the curve is a measure of toughness.

Failure theories in brief

The maximum principal stress theory suits brittle materials. The maximum shear stress theory (Tresca) and the distortion energy theory (von Mises) suit ductile materials. Fatigue failure occurs under repeated loading, even when the stress is below the yield strength.

Common mistakes

  • Using the wrong units, such as mixing mm and m, or N and kN
  • Taking the section modulus from the wrong axis for a rectangular section
  • Forgetting that the column's effective length depends on the end conditions
  • Using the diameter in formulas that need the radius

Questions candidates ask

Which formulas must I memorise? Stress, strain, Hooke's law, bending equation, torsion equation, Euler's load and the bending moment results for standard beams.

What units should I use? Keep N and mm together, or kN and m together, and do not mix them.

Practise

Solve the MCQs below, then try the full-length mocks and read the exam guide. More topics are on the JE Mechanical guide.

By Shivam · Written with AI assistance and fact-checked · Questions on this page are original · Editorial policy · Report an error

Practice questions

Q1. Strain is:

  1. Measured in metres
  2. A dimensionless quantity
  3. Measured in N/m²
  4. Measured in Newtons
Show answer

A dimensionless quantity — Strain is the ratio of change in length to original length.

Q2. Hooke's law holds good up to the:

  1. Upper yield point only
  2. Ultimate stress
  3. Breaking point
  4. Proportional (elastic) limit
Show answer

Proportional (elastic) limit — Stress is proportional to strain only within this limit.

Q3. Young's modulus is the ratio of:

  1. Lateral strain to longitudinal strain
  2. Longitudinal stress to longitudinal strain
  3. Pressure to volumetric strain
  4. Shear stress to shear strain
Show answer

Longitudinal stress to longitudinal strain — E = σ / ε.

Q4. Young's modulus of mild steel is about:

  1. 2,000 GPa
  2. 20 GPa
  3. 200 GPa
  4. 2 GPa
Show answer

200 GPa — Steel is about 2 × 10⁵ MPa.

Q5. Poisson's ratio is the ratio of:

  1. Lateral strain to longitudinal strain
  2. Shear stress to shear strain
  3. Longitudinal strain to lateral strain
  4. Volume change to original volume
Show answer

Lateral strain to longitudinal strain — It is taken as a positive number.

Q6. The relation between Young's modulus E, modulus of rigidity G and Poisson's ratio ν is:

  1. E = G(1 − 2ν)
  2. E = 3G(1 + ν)
  3. E = G / (1 + ν)
  4. E = 2G(1 + ν)
Show answer

E = 2G(1 + ν) — For steel, with ν ≈ 0.3, G is about 0.38 E.

Q7. The relation between Young's modulus E, bulk modulus K and Poisson's ratio ν is:

  1. E = K(1 + 2ν)
  2. E = 3K(1 − 2ν)
  3. E = K / (1 − 2ν)
  4. E = 2K(1 + ν)
Show answer

E = 3K(1 − 2ν) — As ν approaches 0.5, K becomes very large, which describes an almost incompressible material.

Q8. A bar of length 2 m, area 100 mm² carries a 10 kN pull. With E = 200 GPa, the elongation is:

  1. 10 mm
  2. 0.1 mm
  3. 1 mm
  4. 0.01 mm
Show answer

1 mm — δ = PL/AE = (10,000 × 2) / (100 × 10⁻⁶ × 200 × 10⁹) = 0.001 m.

Q9. A simply supported beam of span L carries a central point load W. The maximum bending moment is:

  1. WL/8
  2. WL
  3. WL/2
  4. WL/4
Show answer

WL/4 — The maximum moment occurs at the centre.

Q10. A simply supported beam of span L carries a uniformly distributed load w per unit length. The maximum bending moment is:

  1. wL/2
  2. wL²/4
  3. wL²/8
  4. wL²/2
Show answer

wL²/8 — It occurs at mid-span.

Q11. A cantilever of length L carries a uniformly distributed load w per unit length. The maximum bending moment is:

  1. wL²/8
  2. wL²/2
  3. wL
  4. wL/2
Show answer

wL²/2 — It occurs at the fixed end.

Q12. A cantilever of length L carries a point load W at its free end. The maximum bending moment is:

  1. WL/2
  2. WL/4
  3. WL
  4. WL²
Show answer

WL — It occurs at the fixed end.

Q13. The simple bending equation is:

  1. M = I × σ
  2. T / J = τ / r
  3. σ = P / A
  4. M / I = σ / y = E / R
Show answer

M / I = σ / y = E / R — It links bending moment, stress and radius of curvature.

Q14. The section modulus of a beam cross-section is:

  1. I divided by A
  2. I multiplied by y
  3. I divided by the distance of the extreme fibre from the neutral axis
  4. A divided by I
Show answer

I divided by the distance of the extreme fibre from the neutral axis — Z = I / y_max.

Q15. The moment of inertia of a rectangle of width b and depth d about its centroidal axis parallel to b is:

  1. bd² / 6
  2. bd³ / 12
  3. b³d / 12
  4. bd³ / 3
Show answer

bd³ / 12 — This is the standard formula for a rectangle.

Q16. The moment of inertia of a solid circular section of diameter d about a diameter is:

  1. πd³ / 32
  2. πd² / 4
  3. πd⁴ / 32
  4. πd⁴ / 64
Show answer

πd⁴ / 64 — The polar moment of inertia J is πd⁴/32.

Q17. The torsion equation is:

  1. σ = Eε
  2. T / J = τ / r = Gθ / L
  3. P / A = σ
  4. M / I = σ / y
Show answer

T / J = τ / r = Gθ / L — It applies to circular shafts.

Q18. The slenderness ratio of a column is:

  1. Effective length divided by the least radius of gyration
  2. Load divided by area
  3. Length divided by area
  4. Length divided by diameter squared
Show answer

Effective length divided by the least radius of gyration — Columns with a high slenderness ratio buckle easily.

Q19. Euler's crippling load for a column with both ends hinged is:

  1. π²EI / 4L²
  2. π²EI / L²
  3. 4π²EI / L²
  4. π²EI / 2L²
Show answer

π²EI / L² — The effective length equals the actual length.

Q20. The effective length of a column with one end fixed and the other free is:

  1. L / 2
  2. 2L
  3. 0.7 L
  4. L
Show answer

2L — Such a column is the weakest of the standard end conditions.

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