Polar moment of inertia of a hollow circular shaft: If D and d are the external and internal diameters of a circular shaft (D > d), write the polar moment of inertia J about its centroidal axis in terms of D and d.

Introduction / Context:
The polar moment of inertia J of a circular shaft governs its torsional stiffness (θ = T * L / (G * J)) and shear stress distribution (τ_max = T * R / J). Accurate J is essential for shaft strength and twist calculations.

Given Data / Assumptions:

• Hollow circular cross-section with outer diameter D and inner diameter d.
• Centroidal (axisymmetric) torsion; linear elastic material.
• No warping or noncircular effects (Saint-Venant torsion for circular sections).

Concept / Approach:

For a circular section, J equals the sum of second moments about two orthogonal centroidal axes: J = I_x + I_y. For a solid circle of diameter D, J_solid = (π/32) * D^4. A hollow circle is the difference between the outer solid and the inner void.

Step-by-Step Solution:

Verification / Alternative check:

Setting d = 0 recovers the solid-shaft formula J = (π/32) * D^4, confirming consistency. As d → D, J → 0, which also makes physical sense.

Why Other Options Are Wrong:

• (π/64) * (D^4 − d^4) underestimates stiffness by a factor of 2.
• (π/32) * (D^2 − d^2) has wrong exponent; J scales with diameter^4.
• (π/16) * (D^4 − d^4) and (π/8) * (D^3 − d^3) are incorrect constants or exponents.

Common Pitfalls:

• Using section modulus or area in place of polar moment for torsion problems.
• Mixing radius and diameter; remember D = 2R so J = (π/2) * R^4 = (π/32) * D^4 for solid shafts.

Final Answer:

J = (π/32) * (D^4 − d^4).