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:
Try it first: pick an option, then read the explanation.
Correct answer: C. 1 : 1
Asked in SSC JE 01/03/2017 (Morning)
Detailed Explanation
An element on the shaft surface is in pure shear: shear stress τ on the cross-section and the longitudinal plane, no normal stress.
\[ \sigma_{1,2} = 0 \pm \sqrt{0 + \tau^2} = \pm\tau \qquad \tau_{max} = \tau \]
Why “1 : 1” Is Correct
The maximum normal (principal) stress equals τ, on planes at 45° to the axis. Ratio 1 : 1.
Why the Other Options Are Wrong
- A. 1 : 2: 1 : 2 applies to uniaxial tension (τmax = σ/2).
- B. 2 : 3: 2 : 3 has no basis here.
- D. 2 : 1: 2 : 1 has no basis here.
Key Point
Pure torsion = pure shear: σ1 = τ, σ2 = −τ at 45°.
Key Concepts for Students
- Brittle shafts (chalk, cast iron) break on a 45° helix in tension.
- Ductile shafts shear straight across.
- Mohr's circle centred at the origin.
Practice more: this is question 28 in Torsion MCQs – Part 2. Try the whole set and check every answer instantly.