Shear stress and twist in a round shaft under torque, with a check against the material's allowable shear stress.
Checks the stress against 0.577 × yield ÷ SF.
Empty = 2.0 when a yield strength is entered.
Example Result
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The calculated result will appear here after you press Calculate.
When a shaft transmits torque, every cross-section is twisted slightly relative to its neighbour, and the material resists with shear stress. The stress is zero on the axis and greatest at the outer surface, which is where torsion cracks start. Enter the torque, the shaft diameter (or outside and inside diameters for a tube), its length and the shear modulus of the material, and the calculator returns the maximum shear stress, the angle of twist and the polar moment of area. Add the material's yield strength and a safety factor to see how close the shaft is to its allowable shear stress. For a shaft that also carries bending loads from gears or pulleys, use the Shaft Diameter Calculator, which combines the two.
Select a solid or hollow shaft and SI or Imperial units.
Enter the torque, the shaft length between the torque input and the reaction, and the shear modulus (steel ≈ 79 GPa or 11.5 Msi).
Enter the diameter(s).
Optionally enter the yield strength and a safety factor to check the stress, then press Calculate.
By O.E. — Partner, chairman of the board and CEO of a metal products manufacturer for 21 years; the company exports to 45 countries.
Drafted with AI assistance; calculations checked against primary sources and tested with the site's own code. About FlowUtil
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τ_max = T × c ÷ J and θ = T × L ÷ (G × J), with T the torque, c the outer radius, J the polar moment of area (πD⁴/32 for a solid shaft, π(D⁴ − d⁴)/32 for a tube), L the length between the torque input and the reaction, and G the shear modulus — about 79 GPa (11.5 × 10⁶ psi) for steel and 26 GPa for aluminium. For a solid shaft τ = 16 T ÷ (π D³): doubling the diameter cuts the stress to one-eighth. The formula holds for round, homogeneous shafts loaded in the elastic range; the direction of the torque does not change the magnitude of the stress.
| Symbol | Meaning |
|---|---|
| T | Torque, N·mm or lb·in |
| J | Polar moment of area, mm⁴ or in⁴ |
| G | Shear modulus, MPa or psi |
| θ | Angle of twist, radians (× 57.3 for degrees) |
1Solid 50 mm steel shaft, 500 N·m, 1,000 mm long, G = 79 GPa, yield 250 MPa, safety factor 2.
J = π × 50⁴ ÷ 32 = 613,592 mm⁴; τ = 500,000 × 25 ÷ 613,592; τ_allow = 0.577 × 250 ÷ 2
τ = 20.37 MPa · θ = 0.59° · τ_allow = 72.1 MPa · utilisation 0.28
2The same amount of steel as a 60 mm tube with a 33.2 mm bore, under the same 500 N·m.
J = π (60⁴ − 33.17⁴) ÷ 32
τ = 13.0 MPa · θ = 0.31°
Material near the centre carries little stress, so moving it outwards is almost free strength — why drive shafts are tubes.
31.25 in steel shaft, 5,000 lb·in over 24 in, G = 11.5 × 10⁶ psi.
J = 0.2397 in⁴; τ = 5,000 × 0.625 ÷ 0.2397
τ = 13,038 psi · θ = 2.49°
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