For a hollow shaft with external and internal diameters of 10 cm and 5 cm, the torsional section modulus is approximately:
Try it first: pick an option, then read the explanation.
Correct answer: A. 184 cm³
Asked in SSC JE 01/03/2017 (Morning)
Detailed Explanation
The torsional (polar) section modulus is J divided by the outer radius.
\[ Z_p = \frac{J}{D/2} = \frac{\pi(D^4 - d^4)}{16D} = \frac{\pi(10^4 - 5^4)}{16\times10} = \frac{9375\pi}{160} \approx 184\ \text{cm}^3 \]
Why “184 cm³” Is Correct
The torsional section modulus is about 184 cm³.
| 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
- B. 275 cm³: 275 cm³ does not follow from these data.
- C. 368 cm³: 368 cm³ is double the answer — using J/(D/4) or forgetting the 1/2.
- D. 536 cm³: 536 cm³ is not obtained from any correct formula here.
Key Point
Zp(hollow) = π(D⁴ − d⁴)/(16D); divide J by the OUTER radius.
Key Concepts for Students
- Solid 10 cm shaft: Zp = 196 cm³ — the hole removes only 6% of Zp but 25% of the area.
- Max τ = T/Zp.
- J = 920 cm⁴ here.
Practice more: this is question 12 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.