If D and d are the external and internal diameters of a circular shaft, its polar moment of inertia is:
Try it first: pick an option, then read the explanation.
Correct answer: D. π(D⁴ − d⁴)/32
Asked in SSC JE 29/01/2018 (Evening)
Detailed Explanation
For a solid circle, J = πD⁴/32 (twice the diametral I = πD⁴/64). A hollow section is the outer circle minus the hole.
\[ J = \frac{\pi D^4}{32} - \frac{\pi d^4}{32} = \frac{\pi}{32}(D^4 - d^4) \]
Why “π(D⁴ − d⁴)/32” Is Correct
The polar moment of inertia is π(D⁴ − d⁴)/32.
| 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. π(D⁴ − d⁴)/2: Divides by 2 — far too large.
- B. π(D⁴ − d⁴)/4: Divides by 4 — too large.
- C. π(D⁴ − d⁴)/64: π(D⁴ − d⁴)/64 is the moment of inertia about a diameter, not the polar moment.
Key Point
J(hollow) = π(D⁴ − d⁴)/32; I(diameter) = π(D⁴ − d⁴)/64.
Key Concepts for Students
- J = Ix + Iy = 2I for a circle.
- Polar modulus = J/(D/2).
- Use radii: J = π(R⁴ − r⁴)/2.
Practice more: this is question 6 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.