How it’s calculated
Torque on a circular shaft produces shear stress that grows linearly from zero at the center to a maximum at the outer surface. The shaft also twists by an angle proportional to torque and length.
Example: a 50 mm solid steel shaft carries 1 kN·m. J = π × 50⁴ ÷ 32 = 613,592 mm⁴, so τ = 1 × 10⁶ N·mm × 25 mm ÷ 613,592 mm⁴ = 40.7 MPa. Over 1 m with G = 75 GPa, θ = 1000 × 1 ÷ (75 × 10⁹ × 6.136 × 10⁻⁷) = 0.0217 rad = 1.24°.
A hollow shaft is far more efficient: a 50/40 mm tube has 59% of the solid shaft’s J with only 36% of its material. Assumes elastic behavior, circular sections and no stress concentrations at keyways, shoulders or holes, which can raise local stress substantially.
Frequently asked questions
What shear stress is allowable?
It depends on material, code and loading. A common rule takes shear yield as about 0.5–0.58 of tensile yield, then applies a safety factor and stress-concentration factors for keyways and fatigue.
Does this work for square or rectangular shafts?
No. Non-circular sections warp, and τ = Tr/J does not apply. They need separate torsion constants.
How do I get torque from motor power?
T = P / ω, with ω = 2π × rpm ÷ 60. For example, 10 kW at 1,450 rpm is about 66 N·m.
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Sources
- Mechanics of Materials (Roylance), 2.3 Shear and Torsion — Engineering LibreTexts
- University Physics Vol. 1, 12.3 Stress, Strain, and Elastic Modulus (Table 12.1) — OpenStax
Formulas are taken from the free public references above. Results are provided “as is” for informational and educational purposes only. See our disclaimer.
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