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Shaft Torsion Calculator (Angle of Twist & Shear Stress)

Calculate a circular shaft's angle of twist and peak shear stress under an applied torque, θ = T·L ÷ (J·G), for a solid or hollow cross-section, plus its factor of safety against an allowable shear stress.

Max shear stress (at outer surface)
20.37 MPa
Angle of twist
0.589°
Factor of safety
3.93
% of allowable shear stress
25.5%
Polar moment of inertia (J)
613,592.3 mm⁴

Where this falls

25.5 falls in the "Comfortable margin (factor of safety > 3)" range.

How it works

A torque applied about a circular shaft's own axis twists it through an angle θ = T·L ÷ (J·G), where T is the applied torque, L is the shaft length, G is the material's shear modulus, and J is the cross-section's polar moment of inertia — J = π·(Do⁴ − Di⁴) ÷ 32 for a shaft of outer diameter Do (an inner diameter Di of 0 gives the solid-shaft case). The same torque produces a shear stress that peaks right at the outer surface, τ_max = T·(Do/2) ÷ J, which is what actually governs whether the shaft survives.

That peak stress is compared against an allowable shear stress you supply — your material's design limit, which varies by material, keyways, surface finish, and the design code you're following, so it's never invented here — to give a factor of safety (allowable ÷ actual) and a percentage of that limit. A factor of safety below about 2 leaves little margin; below 1 means the shaft is already over its allowable stress.

A hollow shaft of the same outer diameter has less material resisting twist (a smaller J) than a solid one, so it twists further and reaches a higher peak stress under the same torque — which is also why hollow shafts are only worth using when the weight savings matter more than that extra twist and stress.

FAQ

Why does removing material from the center of a shaft (making it hollow) only increase stress a little, even though it saves a lot of weight?

Material near a shaft's axis barely resists torsion in the first place — shear stress in torsion is zero at the center and rises linearly to a maximum at the outer surface, so the material removed to make a shaft hollow was doing the least work per unit mass. That's why a hollow shaft can shed a meaningful fraction of its weight while its peak stress and twist rise by comparatively little, the classic reason drive shafts and bicycle frames are often tubular rather than solid.

Where do I get my material's allowable shear stress?

From the applicable design code or your material's datasheet — it depends on the specific alloy, any keyways or stress concentrations in the actual part, the safety factor your application requires, and whether you're checking against yield or fatigue. This calculator takes it as a direct input rather than guessing a number, since no single figure is correct for every material and application.

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