Torsion of Circular Shafts

Bending assumed plane sections remain plane; torsion assumes radii remain straight. Both assumptions produce a linear strain distribution, both are converted to stress by Hooke's law, and both are integrated over the section to give a formula. Seeing the parallel once saves learning the second topic from scratch.

1. 1Each cross-section rotates slightly more than the last, so twist accumulates along the length.
2. 2Assume radii remain straight: γ = rθ/L, shear strain proportional to radius.
3. 3τ = Gγ, so shear stress is linear in radius — zero at the centre, maximum at the surface.
4. 4Integrate the moment of those stresses over the section: T/J = τ/r = Gθ/L.

J = πd⁴/32 for a solid shaft — a fourth-power dependence, so a 20% diameter increase roughly doubles the torque capacity.

Since shear stress grows with radius, the core carries almost nothing. Boring out the inner half of the diameter removes about a quarter of the material and only about 6% of J.

It is the I-beam argument applied to a different loading: material near the axis does not earn its weight, so it is removed. Drive shafts, bicycle frames and aircraft structure all follow the same logic.

Power is P = Tω = 2πNT/60. At fixed power, doubling the speed halves the torque and allows a much smaller shaft — which is why turbines spin fast and why the low-speed side of a gearbox is visibly heavier than the high-speed side.

In practice the angle of twist often governs before shear stress does. Excessive wind-up spoils timing in an engine, positioning in a machine tool, and control response in a drive system.

Circular symmetry is what keeps the section plane under twist: every radius is equivalent, so distortion has nowhere to go. Non-circular sections warp out of plane, and the derivation collapses.

• A square or I-section warps; its corners lift out of plane.
• An open section is dramatically worse in torsion than a closed one of the same area — a slit along a tube destroys most of its torsional stiffness.
• Which is why a vehicle chassis is a closed box and why an open channel is never used to resist torque.
• Non-circular torsion needs its own theory, usually a membrane analogy or a numerical solution.

Pure shear on the cross-section is the same stress state as equal tension and compression on planes at 45° — which Mohr's circle makes obvious. Each material fails on the plane where it is weakest, and they disagree about which that is.

The fracture surface identifies which mechanism operated, which is why it is a standard exam question and a standard forensic observation.