Calculator · Structural Mechanics
Angle of Twist Calculator
How far one end of a circular shaft rotates relative to the other under a steady torque. The torsional stiffness of the shaft is GJ/L, so the twist grows with torque and length and falls with the shear modulus and the polar second moment.
Inputs
Steel is about 79 GPa, aluminium about 26 GPa.
Results
| Polar second moment J | 613590 mm⁴ |
|---|---|
| Torsional stiffness GJ/L | 32439 N·m/rad |
| Angle of twist φ = TL/(GJ) | 0.036993 rad |
| Angle of twist φ | 2.1195° |
| Twist per metre | 1.413° per metre |
| Maximum shear τmax = T c / J | 48.892 MPa |
| Applied torque T | 1200 N·m |
Assumptions
- · Circular prismatic shaft; plane sections remain plane and radii remain straight.
- · Linear-elastic material and pure torsion (no warping restraint).
Open in the Lab
The Lab carries this shaft over as a single step and accumulates the twist across further steps and torques.
Technical reference
φ = T L / (G J)
kt = G J / L
τmax = T c / J
G = E / [2(1 + ν)]
- T
- internal torque carried by the shaft[N·m]
- L
- length over which the twist accumulates[m]
- G
- shear modulus of the material[Pa]
- J
- polar second moment of area of a circular section[m⁴]
- φ
- relative rotation of the two ends[rad]
- kt
- torsional stiffness of the shaft[N·m/rad]
This form assumes T, G and J are constant over the length. When any of them changes — a stepped shaft, a material change, a torque applied part way along — the shaft must be split and the contributions summed: φ = Σ TᵢLᵢ/(GᵢJᵢ). The Torsion Lab does exactly that.
Circular sections only. A non-circular section warps under torsion, so its torsion constant is not the polar second moment of area and this relation does not apply to it.
G is an independent elastic property here. For an isotropic linear-elastic material it is related to E and ν by G = E/[2(1+ν)], and the structural material layer derives it that way when you supply E and ν instead.
A large computed twist is a sign the shaft is far outside the small-rotation range the linear model assumes, not a prediction of what the real shaft would do.