A circular shaft of 20 mm diameter transmits torque. Find the safe torque if the shear stress is not to exceed 30 MPa.
Try it first: pick an option, then read the explanation.
Correct answer: C. 15000π N·mm
Asked in SSC JE 06/11/2024 (9 am–11 am)
Detailed Explanation
For a solid shaft, τmax = 16T/(πd³), so T = τπd³/16.
\[ T = \frac{\pi d^3 \tau}{16} = \frac{\pi \times 20^3 \times 30}{16} = \frac{240000\pi}{16} = 15000\pi\ \text{N·mm} \approx 47.1\ \text{N·m} \]
Why “15000π N·mm” Is Correct
The safe torque is 15000π N·mm.
| Quantity | Solid shaft (diameter D) | Hollow shaft (D outer, d inner) |
|---|---|---|
| Polar moment of inertia J | πD⁴/32 | π(D⁴ − d⁴)/32 |
| Polar modulus Zp = J/R | πD³/16 | π(D⁴ − d⁴)/(16D) |
| Maximum shear stress τ = T/Zp | 16T/(πD³) | 16TD/(π(D⁴ − d⁴)) |
| Angle of twist θ | TL/(GJ) | TL/(GJ) |
| Strain energy per unit volume | τ²/(4G) | (τ²/4G)·(D² + d²)/D² |
Torsion equation: T/J = τ/R = Gθ/L. GJ is the torsional rigidity; GJ/L is the torsional stiffness.
Why the Other Options Are Wrong
- A. 12500π N·mm: 12500π would need τ = 25 MPa.
- B. 20000π N·mm: 20000π would need τ = 40 MPa.
- D. 17500π N·mm: 17500π would need τ = 35 MPa.
Key Point
Solid shaft: T = (π/16) τ d³.
Key Concepts for Students
- Keep units consistent: N and mm give N·mm.
- Polar modulus Zp = πd³/16.
- Doubling d raises T eightfold.
Practice more: this is question 3 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.