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:
Try it first: pick an option, then read the explanation.
Correct answer: C. 2M/T
Asked in SSC JE 23/01/2018 (Morning)
Detailed Explanation
Compare bending stress and torsional shear stress at the surface of a solid shaft.
\[ \frac{\sigma_b}{\tau} = \frac{32M/\pi D^3}{16T/\pi D^3} = \frac{2M}{T} \]
Why “2M/T” Is Correct
The ratio is 2M/T.
| 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. M/T: Misses the factor 2.
- B. T/M: Inverted and misses the factor 2.
- D. 2T/M: 2T/M is the inverse relation.
Key Point
σb/τ = 2M/T because Zp = 2Z for a circular section.
Key Concepts for Students
- Equivalent torque √(M² + T²).
- Section modulus Z = πD³/32.
- Polar modulus Zp = πD³/16.
Practice more: this is question 17 in Torsion MCQs – Part 2. Try the whole set and check every answer instantly.