Mechanics and Structural Analysis MCQ With Answers for JKSSB Junior Engineer (Civil)

Technical · 2026-10-27 · 6 min read

Mechanics and structural analysis carried 15 marks in the board's earlier Junior Engineer (Civil) syllabus. It lists forces and moments, bending moment and shear force in determinate beams, stress and strain, principal stresses, bending theory, torsion, buckling of columns, slope and deflection, and the analysis of trusses. Our guide to the 2026 notice shows where this section fits. This page gives a revision list, worked problems and solved MCQs.

Quick revision list

  • Elastic constants: E = 2G(1 + μ) and E = 3K(1 − 2μ), with Poisson's ratio about 0.3 for steel
  • Beams: support reactions, shear force and bending moment diagrams, and the point of contraflexure where the bending moment changes sign
  • Bending theory: M ÷ I = σ ÷ y = E ÷ R, with section modulus Z = I ÷ y
  • Shear in beams: the maximum shear stress is 1.5 times the average for a rectangle and 4/3 times the average for a circle
  • Deflection: cantilever with end load WL³ ÷ 3EI, simply supported beam with central load WL³ ÷ 48EI, and with a uniform load 5wL⁴ ÷ 384EI
  • Columns: Euler's load π²EI ÷ Le², and the slenderness ratio Le ÷ r
  • Torsion: T ÷ J = τ ÷ r = Gθ ÷ L, with J = πd⁴ ÷ 32 for a solid shaft
  • Trusses and indeterminacy: a plane pin-jointed truss is determinate when m = 2j − 3, and moment distribution uses stiffness 4EI ÷ L and a carry-over factor of one-half

Worked numerical problems

  • Beam moment: a simply supported beam of 6 m span carries 10 kN/m. Each reaction is 30 kN, and the maximum bending moment is 10 × 6² ÷ 8 = 45 kN·m
  • Cantilever deflection: a 3 m cantilever carries a 20 kN end load, with EI = 20,000 kN·m². The fixed-end moment is 60 kN·m, and the tip deflection is 20 × 3³ ÷ (3 × 20,000) = 0.009 m, or 9 mm
  • Bending stress: a beam 200 mm wide and 400 mm deep has Z = 200 × 400² ÷ 6 = 5.33 × 10⁶ mm³. Under 40 kN·m the stress is 40 × 10⁶ ÷ 5.33 × 10⁶ = 7.5 N/mm²
  • Euler load: a column hinged at both ends, 3 m long, with E = 200 GPa and I = 5 × 10⁶ mm⁴, buckles at π² × 200,000 × 5 × 10⁶ ÷ 3,000² = 1,097 kN
  • Thin cylinder: internal pressure 2 MPa, diameter 1,000 mm and thickness 10 mm give a hoop stress of 2 × 1,000 ÷ 20 = 100 MPa and a longitudinal stress of 50 MPa
  • Torsion: a 50 mm solid shaft under 1 kN·m has τ = 16T ÷ πd³ = 40.7 MPa

Effective length of columns

Euler's formula uses the effective length, which depends on the end conditions. A column hinged at both ends has an effective length equal to its actual length. If both ends are fixed it is half the length, if one end is fixed and the other hinged it is about 0.7 times the length, and if one end is fixed and the other free it is twice the length. The buckling load is inversely proportional to the square of the effective length, so a fixed-fixed column carries four times the load of a hinged-hinged one of the same size.

Common mistakes

  • Writing WL ÷ 8 for the central-load moment (it is WL ÷ 4) or wL² ÷ 4 for the UDL moment (it is wL² ÷ 8)
  • Using the wrong moment of inertia for a rectangle: it is bd³ ÷ 12 about the centroid, not bd³ ÷ 3
  • Mixing up the hoop stress pd ÷ 2t with the longitudinal stress pd ÷ 4t
  • Forgetting that Euler's formula holds only for slender columns

Practise

Work the solved questions below, then read the Junior Engineer (Civil) guide for the dates and the marks for every subject. The other subject pages in this series cover fluid mechanics, RCC, and construction management.

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

Practice questions

Q1. Young's modulus is the ratio of:

  1. Strain to stress
  2. Stress to strain in the elastic range
  3. Shear stress to shear strain
  4. Load to area
Show answer

Stress to strain in the elastic range — E = σ ÷ ε within the proportional limit.

Q2. The Poisson's ratio of steel is about:

  1. 0.3
  2. 3
  3. 0.03
  4. 1.3
Show answer

0.3 — Most metals have Poisson's ratio between 0.25 and 0.35.

Q3. The relation between the elastic constants E, G and μ is:

  1. E = G(1 − 2μ)
  2. E = 3G(1 + μ)
  3. E = G ÷ (1 + μ)
  4. E = 2G(1 + μ)
Show answer

E = 2G(1 + μ) — E = 2G(1 + μ) and E = 3K(1 − 2μ) link the three moduli.

Q4. The shear force diagram of a simply supported beam under a uniformly distributed load is:

  1. A parabola
  2. A horizontal line
  3. A straight inclined line
  4. A rectangle
Show answer

A straight inclined line — Shear force changes linearly with distance under a UDL.

Q5. The maximum bending moment of a simply supported beam of span L with a central point load W is:

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

WL ÷ 4 — The moment is greatest at the centre and equals WL ÷ 4.

Q6. The maximum bending moment of a simply supported beam of span L with a UDL w per unit length is:

  1. wL² ÷ 8
  2. wL² ÷ 4
  3. wL² ÷ 2
  4. wL² ÷ 12
Show answer

wL² ÷ 8 — The moment is greatest at midspan and equals wL² ÷ 8.

Q7. The maximum bending moment of a cantilever of length L carrying a UDL w per unit length is:

  1. wL² ÷ 8
  2. wL² ÷ 4
  3. wL²
  4. wL² ÷ 2
Show answer

wL² ÷ 2 — It occurs at the fixed end.

Q8. The maximum shear force in a cantilever with a point load W at the free end is:

  1. WL
  2. W ÷ 2
  3. W, constant along the length
  4. Zero
Show answer

W, constant along the length — Shear force equals the load at every section between the load and the support.

Q9. At the point of contraflexure of a beam:

  1. The shear force is zero
  2. The bending moment changes sign
  3. The deflection is maximum
  4. The slope is zero
Show answer

The bending moment changes sign — The bending moment passes through zero and the curvature reverses.

Q10. The shear force at a section is the:

  1. Rate of change of bending moment
  2. Rate of change of load
  3. Integral of the deflection
  4. Square of the moment
Show answer

Rate of change of bending moment — V = dM ÷ dx.

Q11. The moment of inertia of a rectangle of width b and depth d about its centroidal axis parallel to b is:

  1. bd³ ÷ 3
  2. bd³ ÷ 36
  3. db³ ÷ 12
  4. bd³ ÷ 12
Show answer

bd³ ÷ 12 — I = bd³ ÷ 12. The value bd³ ÷ 3 is about the base.

Q12. The section modulus of a rectangular section of width b and depth d is:

  1. bd² ÷ 12
  2. bd³ ÷ 12
  3. bd² ÷ 6
  4. bd² ÷ 3
Show answer

bd² ÷ 6 — Z = I ÷ (d ÷ 2) = bd² ÷ 6.

Q13. The flexure formula is:

  1. T ÷ J = τ ÷ r = Gθ ÷ L
  2. M ÷ I = σ ÷ y = E ÷ R
  3. M ÷ y = σ ÷ I
  4. σ = M × y × I
Show answer

M ÷ I = σ ÷ y = E ÷ R — The second form given is the torsion formula.

Q14. The maximum shear stress in a rectangular beam section is the average shear stress multiplied by:

  1. 1.5
  2. 1.33
  3. 2
  4. 1
Show answer

1.5 — The parabolic distribution peaks at the neutral axis at 1.5 times the average.

Q15. The deflection at the free end of a cantilever of length L with a point load W at the free end is:

  1. WL³ ÷ 8EI
  2. WL³ ÷ 48EI
  3. WL⁴ ÷ 8EI
  4. WL³ ÷ 3EI
Show answer

WL³ ÷ 3EI — δ = WL³ ÷ 3EI.

Q16. The maximum deflection of a simply supported beam with a central point load W is:

  1. 5WL³ ÷ 384EI
  2. WL³ ÷ 3EI
  3. WL³ ÷ 48EI
  4. WL³ ÷ 8EI
Show answer

WL³ ÷ 48EI — The 5 ÷ 384 form with wL⁴ is for a uniformly distributed load.

Q17. The maximum deflection of a simply supported beam with a UDL w per unit length is:

  1. wL⁴ ÷ 8EI
  2. 5wL⁴ ÷ 384EI
  3. wL⁴ ÷ 48EI
  4. 5wL³ ÷ 384EI
Show answer

5wL⁴ ÷ 384EI — δ = 5wL⁴ ÷ 384EI at midspan.

Q18. Euler's buckling load for a column hinged at both ends is:

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

π²EI ÷ L² — A fixed-fixed column carries 4π²EI ÷ L², and a fixed-free one π²EI ÷ 4L².

Q19. The effective length of a column fixed at one end and free at the other is:

  1. L
  2. 0.5L
  3. 0.7L
  4. 2L
Show answer

2L — The free end sways, so the effective length is twice the actual length.

Q20. The hoop stress in a thin cylinder with internal pressure p, diameter d and thickness t is:

  1. pd ÷ 4t
  2. pd ÷ t
  3. pd ÷ 2t
  4. 2pd ÷ t
Show answer

pd ÷ 2t — The longitudinal stress is half of that, pd ÷ 4t.

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