A circular shaft of diameter d and length l is subjected to a torque T and a bending moment M. The ratio of maximum (torsional) shear stress to maximum bending stress is:
Try it first: pick an option, then read the explanation.
Correct answer: B. T/(2M)
Asked in SSC JE 25/09/2019 (Morning)
Detailed Explanation
Compare the two surface stresses of a solid shaft.
\[ \frac{\tau}{\sigma_b} = \frac{16T/\pi d^3}{32M/\pi d^3} = \frac{T}{2M} \]
Why “T/(2M)” Is Correct
The ratio is T/(2M).
| Combined bending M and torque T (solid shaft) | Formula |
|---|---|
| Bending stress | σb = 32M/(πD³) |
| Torsional shear stress | τ = 16T/(πD³) |
| Equivalent bending moment (max principal stress) | Me = ½[M + √(M² + T²)] |
| Equivalent torque (max shear stress) | Te = √(M² + T²) |
| Ratio σb/τ | 2M/T |
Why the Other Options Are Wrong
- A. T/M: Ignores the factor 2 between 32 and 16.
- C. T/(4M): Factor 4 is wrong — the constants differ only by 2.
- D. 2T/M: 2T/M inverts the factor.
Key Point
τ/σb = T/2M; σb/τ = 2M/T (Zp = 2Z for a circle).
Key Concepts for Students
- Polar modulus is twice the section modulus for a circle.
- Length l does not affect these stresses.
- Used in combined-loading problems.
Practice more: this is question 11 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.