Thin-walled circular tube in torsion: For permissible shear stress f_s, a thin tube of mean diameter D and uniform wall thickness t transmits a torque T equal to which of the following?

Introduction / Context:Thin-walled closed sections (like tubes) resist torsion primarily through a uniform shear flow around the median line. For circular thin tubes, this gives a compact closed-form torque formula in terms of mean diameter D, wall thickness t, and allowable shear stress f_s.

Given Data / Assumptions:

• Thin-walled circular tube; thickness t ≪ D.
• Uniform shear stress around the median line at value f_s.
• Saint-Venant torsion, elastic range.

Concept / Approach:

For thin closed sections, T = 2 A_m q, where A_m is the area enclosed by the median line and q is the shear flow. For a wall of thickness t, q = τ t = f_s t. For a circle, A_m = π (D/2)² = π D² / 4. Substitute to obtain the torque capacity.

Step-by-Step Solution:

Verification / Alternative check:

Units: D² t f_s has dimensions of (length³ * stress) = force * length, i.e., torque. Numerical checks against thin-tube torsion formula confirm the 1/2 factor.

Why Other Options Are Wrong:

Option (d) misses the 1/2 factor (overestimates T). Options (a), (b), (e) are not the correct thin-tube torsion expressions for a circular section.

Common Pitfalls:

Using solid-shaft formula T/J = τ/R for thin tubes; mixing polar moment J for solid sections with thin-wall torsion theory.

Final Answer:

T = (π D² t f_s) / 2