Difficulty: Medium
Correct Answer: (τ_max^2 / 4G) × Volume of shaft
Explanation:
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:
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:
Shear distribution: τ(r) = (τ_max / R) * r with R = D/2.Energy density: u(r) = τ(r)^2 / (2G) = (τ_max^2 / (2G R^2)) * r^2.Integrate over annulus: U = ∫ u(r) dV = ∫_{r_i}^{r_o} (τ_max^2 / (2G R^2)) * r^2 * (2π r L) dr.Simplify integration ⇒ U = (τ_max^2 / 4G) * (π (r_o^2 − r_i^2) L) = (τ_max^2 / 4G) × Volume.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
Discussion & Comments