What is the torsion equation?

The torsion equation is T/J = tau/r = G theta/L, relating the torque T, polar moment of inertia J, shear stress tau at radius r, modulus of rigidity G, angle of twist theta, and length L of a circular shaft subjected to pure torsion.

Why are hollow shafts more efficient than solid shafts?

Torsional shear stress is largest at the outer surface and zero at the centre, so the material near the axis carries little torque. Removing it (hollow shaft) saves weight while keeping most of the torsional strength, giving a higher strength-to-weight ratio.

How is power related to torque?

Power P = 2 pi N T / 60 when N is in rpm and T in newton-metres, or P = T omega where omega is the angular velocity in rad/s. This lets you find the torque a shaft must carry from the power and speed of a machine.

Torsion of Circular Shafts — Torsion Equation, Power Transmission, and Examples

Shafts in motors, turbines, axles and drills transmit power by torsion — twisting. The torsion equation links the applied torque to the shear stress it produces and the angle through which the shaft twists. This article derives it and works through power-transmission examples.

Assumptions

The shaft is circular (solid or hollow) and the material is homogeneous and elastic.
Plane cross-sections remain plane and circular after twisting (true for circular sections).
The twist is uniform along the length, and radii remain straight.

Derivation

When a shaft of length L and radius R twists through an angle θ, a line on the surface shears by angle φ. From geometry, the surface shear strain is φ = Rθ/L. By Hooke's law in shear, the surface shear stress is τ = Gφ = G·Rθ/L. At a general radius r the strain (and hence stress) varies linearly:

τ / r = G θ / L     ...(1)

The torque is the sum of the moments of the shear forces about the axis: T = ∫ τ · r dA = ∫ (Gθ/L) r · r dA = (Gθ/L) ∫ r² dA = (Gθ/L)·J, where J = ∫r²dA is the polar moment of inertia. Hence:

T / J = G θ / L     ...(2)

Combining (1) and (2) gives the complete torsion equation:

T / J = τ / r = G θ / L

Polar Moment of Inertia

Shaft	Polar Moment J
Solid circular (dia D)	πD⁴ / 32
Hollow circular (D outer, d inner)	π(D⁴ − d⁴) / 32

The maximum shear stress is at the outer surface: τ_max = T·R/J = 16T/(πD³) for a solid shaft.

Power Transmission

P = 2πNT / 60   (N in rpm, T in N·m, P in watts)
P = T · ω       (ω in rad/s)

Worked Example 1 — Sizing a Solid Shaft

A solid shaft transmits 20 kW at 200 rpm. The allowable shear stress is 40 MPa. Find the required diameter.

T = 60P/(2πN) = 60 × 20 000 / (2π × 200) = 954.9 N·m = 954 900 N·mm
τ_max = 16T/(πD³) → D³ = 16T/(πτ) = 16 × 954 900 / (π × 40) = 121 580 mm³
D = (121 580)^1/3 = 49.5 mm → adopt 50 mm

Worked Example 2 — Angle of Twist

For the 50 mm shaft above, 1.5 m long, with G = 80 GPa, find the angle of twist under the working torque of 954.9 N·m.

J = πD⁴/32 = π × 50⁴ / 32 = 6.136 × 10⁵ mm⁴
θ = TL/(GJ) = (954 900 × 1500) / (80 000 × 6.136×10⁵) = 0.0292 rad = 1.67°

Worked Example 3 — Hollow vs Solid (Weight Saving)

Compare a solid 50 mm shaft with a hollow shaft of the same outer diameter and 30 mm bore for torsional strength.

J_solid = π×50⁴/32 = 6.136×10⁵ mm⁴; J_hollow = π(50⁴ − 30⁴)/32 = 5.341×10⁵ mm⁴
The hollow shaft keeps 87% of the torsional stiffness while removing 36% of the cross-sectional area — a large weight saving for a small strength loss.

Common Mistakes

Using the area moment of inertia I instead of the polar moment J.
Forgetting to convert rpm to rad/s (or using the 2πN/60 factor incorrectly).
Applying the circular-shaft theory to non-circular sections, where sections warp and the formula does not hold.