Shear stress distribution in a circular shaft under torsion\n\nWhen a circular shaft is subjected to pure torsion, how does the shear stress vary from the center to the outer surface?

Introduction / Context:
Torsion of circular shafts is fundamental in power transmission. The internal shear stress distribution dictates allowable torque and sizing for shafts and axles.

Given Data / Assumptions:

• Circular (solid) shaft under uniform torsion T.
• Linear elastic behavior; Saint-Venant torsion assumptions.

Concept / Approach:
Shear stress varies linearly with radius: τ(r) = T r / J, where J is the polar second moment of area. Thus, τ = 0 at r = 0 and τ = τmax at r = R (outer radius).

Step-by-Step Solution:
Polar second moment for solid circular section: J = π D^4 / 32.Shear stress: τ(r) = T r / J → proportional to r.At r = 0: τ = 0; at r = R: τ = T R / J = τmax.

Verification / Alternative check:
Strain compatibility leads to angle of twist increasing linearly with radius, which implies linear shear strain and hence linear shear stress with radius.

Why Other Options Are Wrong:
Options claiming maximum at the center contradict τ ∝ r. Uniform distribution would require nonphysical material behavior under torsion.

Common Pitfalls:
Confusing normal (bending) stresses with torsional shear; mixing solid and thin-walled tube formulas (still linear from inner to outer radius for solid).

Final Answer:
zero at the centre to maximum at the circumference