Strain energy in torsion: for a hollow circular shaft (outer diameter D, inner diameter d) under elastic torsion, the total strain energy expressed using the maximum shear stress τ_max at the outer surface equals which of the following?

Introduction / Context:
Strain energy in torsion is vital for shock and fatigue considerations. For circular shafts, the shear stress varies linearly with radius, so the average energy density differs from the simple τ^2 / (2G) expression that assumes uniform shear. A compact and widely used result expresses total strain energy in terms of τ_max and the volume.

Given Data / Assumptions:

• Hollow circular shaft, outer diameter D, inner diameter d.
• Linear elastic torsion, Saint-Venant assumptions.
• Maximum shear stress τ_max occurs at r = D/2.

Concept / Approach:
Total strain energy U in torsion is U = (T * θ) / 2. Also, U = ∫(τ^2 / 2G) dV, but τ varies with r. Carrying out the radial integration for a circular shaft shows that the average energy density equals τ_max^2 / (4G), so the total is U = (τ_max^2 / 4G) * Volume.

Step-by-Step Solution:

Verification / Alternative check:
From U = T^2 L / (2 G J) and τ_max = T R / J; eliminate T and J to recover U = (τ_max^2 / 4G) × Volume.

Why Other Options Are Wrong:
(τ^2 / 2G) × Volume assumes uniform τ; not valid for circular torsion.Other coefficients (1/6, 1/8, 1) do not match the correct radial integration.

Common Pitfalls:
Using solid-shaft intuition incorrectly; forgetting τ varies with r; confusing τ_max with average shear.

Final Answer:

(τ_max^2 / 4G) × Volume of shaft