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:
Try it first: pick an option, then read the explanation.
Correct answer: B. 4
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
Use σb/τ = 2M/T with T = M/2.
\[ \frac{\sigma_b}{\tau} = \frac{32M/\pi d^3}{16T/\pi d^3} = \frac{2M}{T} = \frac{2M}{M/2} = 4 \]
Why “4” Is Correct
The ratio is 4.
| 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. 2: 2 is the ratio when T = M.
- C. 3: 3 does not follow.
- D. 1.5: 1.5 does not follow.
Key Point
σb/τ = 2M/T.
Key Concepts for Students
- Bending stress uses Z = πd³/32.
- Torsion uses Zp = πd³/16.
- Combine for Me and Te.
Practice more: this is question 25 in Torsion MCQs – Part 2. Try the whole set and check every answer instantly.