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GATE Strength of Materials: practice questions

GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)

Original practice questions written to the published syllabus · not past papers
10worked here, free
20written to this exam in this subject
9syllabus units

About GATE

GATE — Agricultural Engineering (AG) is a postgraduate admission exam run by IITs and IISc on rotation (2026: IIT Guwahati), for candidates in India. Computer-based, 3 hours, 100 marks (15 General Aptitude + 85 technical: 25 one-mark and 30 two-mark questions). MCQ, MSQ and NAT; negative marking on MCQ only (−1/3 for one-mark, −2/3 for two-mark); no partial credit on MSQ. On-screen virtual calculator.

Source: gate2026.iitg.ac.in.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Strength of Materials in GATE

No separate strength-of-materials section, but machine design questions use torsion of shafts, combined bending and torsion, and shear of pins and keys (shear-pin sizing for a breaking torque).

How members carry load: stress, strain, bending, torsion, deflection and buckling, used in machine and structure design.

Our 20 GATE questions in this subject: 2 easy, 7 medium, 11 hard, across 9 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to GATE and are not shown on any other exam's page.

Question 1 · hard

A round steel implement bar tapers uniformly from 40 mm to 20 mm diameter over a length of 1 m and carries an axial load of 60 kN. With E = 200 GPa, its elongation is closest to:

  1. 0.424 mm
  2. 0.955 mm
  3. 0.239 mm
  4. 0.477 mm
Answer and working

D. 0.477 mm

For a linearly tapered circular bar, δ = 4PL/(πE d₁ d₂) = 4 × 60,000 × 1000/(π × 200,000 × 40 × 20) = 0.477 mm. Using the average diameter of 30 mm gives 0.424 mm, which underestimates because stress rises faster than linearly towards the small end. Using the small-end area (20 mm) throughout gives 0.955 mm, and the large-end area (40 mm) gives 0.239 mm.

Question 2 · medium

A solid round steel axle of a trailed implement, 50 mm in diameter, carries a bending moment of 1.5 kN·m. The maximum bending stress is closest to:

  1. 61.1 MPa
  2. 244.5 MPa
  3. 122.2 MPa
  4. 12.22 MPa
Answer and working

C. 122.2 MPa

For a solid circle Z = πd³/32 = 12,272 mm³; σ = 32M/(πd³) = 122.2 MPa. 16M/(πd³) is the torsion formula's form and halves the answer. 16M/(πd³) gives 61.1 MPa, 64M/(πd³) (the I constant) gives 244.5 MPa, and 12.22 MPa is a decimal slip.

Question 3 · hard

A 2 mm thick steel strip (E = 200 GPa) of a conveyor is bent around a drum of 1 m radius. The maximum bending stress in the strip is:

  1. 400 MPa
  2. 100 MPa
  3. 200 MPa
  4. 0.2 MPa
Answer and working

C. 200 MPa

σ/y = E/R with y = t/2 = 1 mm: σ = 200,000 × 1/1000 = 200 MPa. Using y = t (the full thickness) gives 400 MPa. 0.2 MPa takes 200 GPa × 0.001 = 0.2 GPa and reports it as MPa.

Question 4 · easy

A tractor delivers 45 kW through its 1000 rev/min PTO shaft. The torque in the shaft is closest to:

  1. 2,700 N·m
  2. 795.8 N·m
  3. 429.7 N·m
  4. 45.0 N·m
Answer and working

C. 429.7 N·m

T = 60P/(2πN) = 60 × 45,000/(2π × 1000) = 429.7 N·m. Dropping 2π gives 2700 N·m; using 540 rev/min gives 795.8 N·m. 45.0 N·m divides power by speed with no unit conversion (45,000/1000).

Question 5 · medium

A solid steel shaft of a thresher must transmit a torque of 500 N·m with an allowable shear stress of 50 MPa. The minimum diameter is closest to:

  1. 37.1 mm
  2. 46.7 mm
  3. 18.5 mm
  4. 14.7 mm
Answer and working

A. 37.1 mm

τ = 16T/(πd³) → d = (16 × 500,000/(π × 50))^(1/3) = 37.1 mm. Using 32 (the bending constant) gives 46.7 mm. 18.5 mm is the radius, not the diameter, and dropping the 16 ((T/(πτ))^(1/3)) gives 14.7 mm.

Question 6 · medium

A tractor's PTO is switched from 540 to 1000 rev/min to drive the same implement at the same power. The torque in the driveline becomes:

  1. 1.85 times the original torque
  2. The same as the original torque
  3. 0.29 times the original torque
  4. 0.54 times the original torque
Answer and working

D. 0.54 times the original torque

At constant power T ∝ 1/N, so T₁₀₀₀/T₅₄₀ = 540/1000 = 0.54. A lower torque allows a lighter driveline for the same power, which is why high-power implements use the 1000 rev/min standard. 0.29 = 0.54² is a squared-ratio slip.

Question 7 · hard

In Rankine's formula P = σ_cA/[1 + a(L_e/k)²], the constant a is theoretically equal to:

  1. π²E/σ_c
  2. σ_c/E
  3. E/(π²σ_c)
  4. σ_c/(π²E)
Answer and working

D. σ_c/(π²E)

Rankine combines 1/P = 1/P_c + 1/P_e with P_c = σ_cA and P_e = π²EA/(L_e/k)². Rearranging gives a = σ_c/(π²E). In practice a is taken from tests, e.g. about 1/7500 for mild steel in Indian texts.

Question 8 · hard

A thresher drum shaft carries a bending moment of 3 kN·m together with a torque of 4 kN·m. Using the maximum shear stress theory with an allowable shear stress of 60 MPa, the required solid shaft diameter is closest to:

  1. 75.2 mm
  2. 69.8 mm
  3. 84.1 mm
  4. 87.9 mm
Answer and working

A. 75.2 mm

T_e = √(3² + 4²) = 5 kN·m. d = (16T_e/(πτ))^(1/3) = (16 × 5 × 10⁶/(π × 60))^(1/3) = 75.2 mm. Using torque alone gives 69.8 mm (unsafe); adding M + T = 7 kN·m overestimates. Adding M + T gives 84.1 mm, and using the equivalent bending moment (4 kN·m) with the shear allowable in 32M/(πτ) gives 87.9 mm.

Question 9 · hard

A 500 N weight falls 20 mm onto a collar at the lower end of a vertical steel rod 1.5 m long, area 300 mm², E = 200 GPa. The maximum instantaneous stress is closest to:

  1. 3.33 MPa
  2. 96.0 MPa
  3. 1.67 MPa
  4. 191.9 MPa
Answer and working

B. 96.0 MPa

σ = (W/A)[1 + √(1 + 2AEh/(WL))] = 1.667[1 + √(1 + 2 × 300 × 200,000 × 20/(500 × 1500))] = 96.0 MPa. The approximation √(2WhE/(AL)) = 94.3 MPa is close. 2W/A is a sudden load with no drop height. W/A = 1.67 MPa is the static stress, and 191.9 MPa uses 2W/A in front of the bracket.

Question 10 · hard

A solid circular shaft is twisted so that its surface shear stress is τ. The strain energy stored per unit volume of the whole shaft is:

  1. τ²/(4G)
  2. τ²/(2G)
  3. τ²/(6G)
  4. τ²/(4E)
Answer and working

A. τ²/(4G)

Shear stress rises linearly from zero at the centre; integrating τ(r)²/(2G) over the circle gives τ_max²/(4G) per unit volume, half the value if the whole section were at τ. A hollow shaft stores more per unit volume because its material is more uniformly stressed. τ²/(6G) borrows the 1/3 averaging factor of a rectangular beam in bending, and τ²/(4E) uses E in place of G.

A study plan for Strength of Materials in GATE

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (55% of the 20 questions): Torsion of circular shafts (5, 25%); Bending and shear stresses in beams (3, 15%); Theories of failure (3, 15%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Combined stresses and Mohr's circle (2, 10%); Deflection of beams (2, 10%); Strain energy (2, 10%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Columns: Euler and Rankine formulas (1, 5%); Shear force and bending moment diagrams (1, 5%); Stress, strain and Hooke's law (1, 5%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Strain energy (2 hard of 2), Columns: Euler and Rankine formulas (1 hard of 1), Shear force and bending moment diagrams (1 hard of 1).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Bank the easy marks: Torsion of circular shafts (2 easy of 5).

    Do these early in revision so they are fast on the day.

  6. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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