Strength of materials – Polar moment of inertia of a hollow circular shaft For a hollow shaft with outer diameter D and inner diameter d, what is the polar moment of inertia J?

Introduction / Context:The polar moment of inertia J quantifies a circular section’s resistance to torsion. For hollow shafts, material is distributed away from the center to increase torsional stiffness for a given weight.

Given Data / Assumptions:

• Hollow circular shaft, outside diameter D, inside diameter d.
• Linear elastic torsion; uniform material.

Concept / Approach:For circular sections, J equals the sum of second moments about two orthogonal axes through the centroid. Using the standard result for rings, J is the difference between the solid outer and inner polar moments.

Step-by-Step Solution:Solid circle polar moment: J_solid = (π/32) * D^4.Inner void polar moment: J_void = (π/32) * d^4.Hollow section: J = J_solid − J_void = (π/32) * (D^4 − d^4).

Verification / Alternative check:Boundary cases: if d = 0, J reduces to (π/32) * D^4 (solid case). If d → D, J → 0 as expected.

Why Other Options Are Wrong:(π/64) * (D^4 − d^4) is off by a factor of 1/2. Expressions with D^2 or D^3 are dimensionally incorrect. Using a plus sign increases J erroneously.

Common Pitfalls:Mixing polar moment J with area moment I, and forgetting that the hollow result is the outer minus inner solid contribution.

Final Answer:J = (π/32) * (D^4 − d^4)