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Torsion MCQs – Part 2

15 questions. Tap an option to check it instantly, then read the solution.

Question 16Easy
A shaft rotating at N r.p.m. under a torque T transmits a power of:
Full explanation

Answer: A. TπN/30 N·m/s

P = Tω = T × 2πN/60 = πNT/30 W (N·m/s).

Question 17Easy
A shaft is subjected to a bending moment M and a torque T simultaneously. The ratio of the maximum bending stress to the maximum shear stress developed in the shaft is:
Full explanation

Answer: C. 2M/T

σb/τ = (32M/πD³)/(16T/πD³) = 2M/T.

Question 18Medium
For a permissible shear stress fs, the torque transmitted by a thin tube of mean diameter D and wall thickness t is:
Full explanation

Answer: A. (πD²/2)·t·fs

Bredt: T = 2·Am·t·fs = 2(πD²/4)·t·fs = (πD²/2)·t·fs.

Question 19Medium
Pick the incorrect statement. The torsional resistance of a shaft is directly proportional to:
Full explanation

Answer: D. moment of inertia of the shaft section

T = GJθ/L — it is the polar moment of inertia J, not the (planar) moment of inertia I.

Question 20Easy
The angle of twist of a circular shaft under a torsional moment T is given by:
Full explanation

Answer: B. TL/(GJ)

From T/J = Gθ/L: θ = TL/(GJ).

Question 21Easy
The assumption that a plane section before twisting remains plane after twisting is valid for:
Full explanation

Answer: D. only hollow or solid circular cross-sections

Only circular sections (solid or hollow) twist without warping; non-circular sections warp out of plane.

Question 22Easy
Which of the following sections is best in torsion?
Full explanation

Answer: B. Hollow circular

Stress is low near the centre, so removing that material (hollow circle) gives the largest J and Zp per unit weight.

Question 23Medium
A rectangular bar is subjected to torsion. The maximum shear stress will occur:
Full explanation

Answer: C. at the middle of the longer side

Shear stress is zero at the corners and the centre; it is maximum at the mid-point of each longer side.

Question 24Medium
A torsion member is made from two concentric thin tubes welded at the ends to rigid discs so that both twist as a unit. The radius of the outer tube is 2r and of the inner tube r. If the shear stress in the outer tube is T, the shear stress in the inner tube is:
Full explanation

Answer: D. 0.5T

Same θ/L and G, so τ ∝ radius: τinner = T × r/2r = 0.5T.

Question 25Easy
A circular shaft is subjected to a torque T that is half the bending moment applied. The ratio of maximum bending stress to maximum shear stress is:
Full explanation

Answer: B. 4

σb/τ = 2M/T = 2M/(M/2) = 4.

Question 26Medium
A circular shaft ABC fixed at A has diameter D over AB (half its length) and diameter D/2 over BC (the other half), with torque T at the free end C. If the rotation of B relative to A is 0.1 rad, what is the rotation of C relative to B?
Full explanation

Answer: C. 1.6 rad

Same T and length; θ ∝ 1/d⁴ → θBC = 2⁴ × 0.1 = 1.6 rad.

Question 27Medium
A hollow shaft is subjected to torsion. Which description correctly shows the shear stress variation along its radius?
Full explanation

Answer: C. Zero in the hole; across the wall, linear from a non-zero value at the inner surface to the maximum at the outer surface

τ = Tρ/J for ri ≤ ρ ≤ ro: τ(ri) = τmax·ri/ro ≠ 0, rising linearly to τmax. No stress in the hole.

Question 28Medium
A solid circular shaft is subjected to a pure torsional moment. The ratio of maximum shear stress to maximum normal stress at any point is:
Full explanation

Answer: C. 1 : 1

Pure shear τ gives principal stresses ±τ at 45°, so τmax : σmax = 1 : 1.

Question 29Medium
Two shafts of the same length and material are joined in series. If the ratio of their diameters is 2, the ratio of their angles of twist will be:
Full explanation

Answer: D. 16

In series both carry the same T; θ ∝ 1/d⁴ → ratio 2⁴ = 16.

Question 30Medium
A circular shaft can transmit a torque of 5 kN·m. If the torque is reduced to 4 kN·m, the maximum bending moment that can be applied to the shaft is:
Full explanation

Answer: C. 3 kN·m

Equivalent torque √(M² + T²) must not exceed 5: M = √(25 − 16) = 3 kN·m.