Shaft Torsion and Diameter Calculator

How is a shaft sized for torsion?

In a circular shaft carrying torque the shear stress is zero at the centre and highest at the surface: τ = T × (d / 2) / J. For a solid shaft the polar second moment of area is J = π × d⁴ / 32, so the stress becomes τ = 16 × T / (π × d³). With a known allowable shear stress the required diameter follows directly: d = ∛(16 × T / (π × τ)).

For a hollow shaft with inside to outside diameter ratio k, replace d³ with d³ × (1 − k⁴). Material near the centre carries little stress, so a hollow shaft transmits more torque for the same weight.

Angle of twist

The relative rotation between the two ends is θ = T × L / (G × J) in radians. G is the shear modulus, about 80 GPa for steels. On long shafts and in precision drives the twist can be excessive even when the stress is acceptable.

Example

A solid shaft carries 500 N·m with an allowable shear stress of 40 MPa. Then d³ = 16 × 500 000 / (π × 40) = 63 662 mm³ and d = 39.9 mm. The next standard size, 40 mm, can be chosen. Over a length of 1 m the angle of twist is 1.43 degrees.

Limits

Real shafts usually carry torque together with bending, keyways and shoulders raise the stress locally, and the load varies as the shaft turns. This tool gives static torsion on a uniform section only. Bending calls for a combined calculation and fluctuating load calls for a fatigue calculation.