A solid circular shaft of diameter 100 mm is subjected to a torque of 8 kN·m. Determine the maximum shear stress induced in the shaft (in N/mm²).
Try it first: pick an option, then read the explanation.
Correct answer: D. 128/π
Asked in SSC JE 11/12/2020 (Evening)
Detailed Explanation
Convert the torque to N·mm and use τmax = 16T/(πD³).
\[ \tau_{max} = \frac{16T}{\pi D^3} = \frac{16 \times 8\times10^6}{\pi \times 100^3} = \frac{128\times10^6}{\pi\times10^6} = \frac{128}{\pi}\ \text{N/mm}^2 \]
Why “128/π” Is Correct
The maximum shear stress is 128/π N/mm² (about 40.7 MPa).
Why the Other Options Are Wrong
- A. 256/π: 256/π comes from using 32T/(πD³) — the bending formula, not torsion.
- B. 64/π: 64/π is half the correct value (e.g. 8T/πD³).
- C. 144/π: 144/π does not follow from these data.
Key Point
τmax = 16T/(πD³); 1 kN·m = 10⁶ N·mm.
Key Concepts for Students
- Bending uses 32M/πD³; torsion uses 16T/πD³.
- Shear stress is maximum at the surface.
- Answer ≈ 40.7 MPa.
Practice more: this is question 5 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.