This problem is solved using the torsion equation, which relates the applied torque to the shear stress developed across a shaft's cross-section:

T/J = τ/R = Gθ/L

The three ratios describe, respectively, the torsional stiffness of the section, the stress distribution across the radius, and the angle of twist per unit length. For finding the stress at the outer fibre we only need the first two terms rearranged into τ = T·R / J. The shear stress varies linearly from zero at the shaft axis (neutral axis of torsion) to a maximum at the outermost fibre, which is exactly why designers check τ at the outer radius R = Dₒ/2.

Given data:

• Torque, T = 40000 N·m
• Polar moment of inertia, J = 0.004 m⁴
• Diameter ratio, Dᵢ/Dₒ = 0.8, with Dᵢ = 400 mm

Step 1 — Find the outer diameter: Dₒ = Dᵢ / 0.8 = 400 / 0.8 = 500 mm.

Step 2 — Find the outer radius: R = Dₒ/2 = 250 mm = 0.25 m.

Step 3 — Apply the torsion formula:

τₘₐˣ = T·R / J = (40000 × 0.25) / 0.004 = 10000 / 0.004 = 2500000 Pa.

Converting to MPa (1 MPa = 10⁶ Pa): τₘₐˣ = 2.5 MPa.

Why the other values are wrong: A result of 2500 MPa arises if one forgets to convert Pa to MPa (dividing by 10⁶) — it overshoots by a factor of a thousand and would exceed the strength of ordinary steel. A value of 250 MPa comes from a decimal-place slip in the same conversion. A value of 0.25 MPa results from mistakenly using the inner radius or otherwise mishandling the radius term. The consistent SI substitution above gives the correct 2.5 MPa.