TAHA Engineering Toolsby TAHA Knowledge Lab

Mechanical Engineering

Shaft Torsional Stress Calculator

Torsional shear stress and angle of twist of a round shaft, or the diameter needed.

Inputs

Calculate from
A round shaft with twisting torque T applied at each end, diameter D and length L.TTLDτmax at surface
Equal and opposite torques T twist a shaft of diameter D over length L. Shear stress grows from zero at the centre to its maximum at the surface; the ends rotate by the angle of twist θ. Diagram not to scale.

From the torque and shaft power calculator if you know power and speed.

Material property from the supplier or a materials handbook (for carbon steels it is roughly 80 GPa). Needed only for the angle of twist.

Result

Maximum shear stress (at the surface)
39.7887
Polar second moment of area J
251,327mm⁴
Polar section modulus J ÷ (D/2)
12,566.4mm³
Working
J = π (D⁴ − d⁴) ÷ 32 = π × (40⁴ − 0⁴) ÷ 32 = 251,330 mm⁴ τ = T × (D/2) ÷ J = 500 N·m × 20 mm ÷ 251,330 mm⁴ = 39.789 MPa

The surface shear stress is 39.7887 MPa. Compare it with your allowable shear stress.

Educational estimate — check the assumptions below.

About this calculator

A shaft that transmits torque is twisted, and the material resists with shear stress. The stress is zero at the centre and highest at the surface. This calculator uses the classical torsion formula for round shafts — solid or hollow — to find that maximum stress and, if you give the length and shear modulus, how much the shaft twists. It can also size a solid shaft for an allowable shear stress.

Formula

τmax = T × (D/2) ÷ J    J = π (D⁴ − d⁴) ÷ 32

Solid shaft: τmax = 16T ÷ (πD³)    Dmin = ∛[16T ÷ (πτallow)]    θ = T L ÷ (G J)

T = torque, D = outside diameter, d = bore (0 for solid), J = polar second moment of area, L = length, G = shear modulus, θ = angle of twist (radians).

Worked example

  • Solid 40 mm shaft carrying 500 N·m: τ = 16 × 500 ÷ (π × 0.040³) = 39.8 MPa.
  • Over 1 m with G = 80 GPa: J = 251 327 mm⁴; θ = 500 × 1 ÷ (80 × 10⁹ × 2.513 × 10⁻⁷) = 0.0249 rad = 1.42°.
  • Diameter for 500 N·m at an allowable 40 MPa: ∛(16 × 500 ÷ (π × 40 × 10⁶)) = 39.9 mm.

Assumptions and limitations

  • Pure torsion of a straight, round, uniform shaft in the elastic range. Bending, axial load, stress concentrations (keyways, shoulders, holes), fatigue and critical speed are not included — real shaft design must combine them (for example with a design code or the methods in machine-design textbooks).
  • The allowable stress and the shear modulus are your inputs; no material data or factor of safety is assumed.
  • Twist is calculated only between two points where torque is applied, with constant torque in between.

Safety note

Rotating shafts that fail can release large amounts of energy. Use guards and have critical shafts designed and checked by a competent engineer.

References

  • Torsion of circular shafts: τ = T r ÷ J, θ = T L ÷ (G J) (e.g. Hibbeler, Mechanics of Materials; Shigley’s Mechanical Engineering Design).