The core idea
Twist a shaft and every cross-section shears against its neighbour. The shear stress peaks at the outer surface: τ = T·r / J — torque T (N·mm), outer radius r (mm), and J, the polar second moment of area (mm⁴). For a solid round shaft J = π·d⁴ / 32, so the diameter enters to the fourth power: doubling the diameter multiplies torsional strength 16×.
The twist angle is φ = T·L / (G·J) over length L (mm), with G the shear modulus (steel ≈ 79,000 MPa). A 20 mm steel shaft carrying 50 N·m sees only ~32 MPa of shear but twists 2.3° over a metre — small per metre, yet that compliance is why long shafts whiplash and vibrate.
Torsion is the strongest case for hollow shafts: material at the centre barely stresses, so it contributes mass but little strength. An OD 30 / ID 22.4 mm tube uses exactly the same steel area as a solid 20 mm rod yet has 3.5× the J — the opposite trade-off from bending, where hollow sections pay a smaller but real premium.
Real-world example
Every car on the road demonstrates both facts. Drive half-shafts and propeller shafts are tubes — maximum J per kilogram, so they spin smoothly at 3,000 rpm without whipping. Meanwhile the stub axles that carry the wheels are chunky solid steel: they are short, weight matters less, and a solid section shrugs off shock loads and corrosion better at the exposed wheel end.
Common pitfall
Using the bending formula I = π·d⁴/64 when the load is torsion. It is off by exactly 2× (the correct polar value is π·d⁴/32) — an error that halves every calculated stiffness and margin. Related trap: for equal area, hollow beats solid in torsion, but for equal outer diameter solid wins on pure strength; a tube only pays off when weight or material cost is part of the budget.