Derivation of the torsion equation

Assumptions

1. The shaft is straight, of uniform circular cross-section, and the material is homogeneous and isotropic.
2. The material obeys Hooke's law; stresses stay within the elastic (proportional) limit.
3. Plane transverse sections remain plane and circular after twisting; radii remain straight.
4. The twist is uniform along the length and the torque acts in a plane perpendicular to the axis.
5. Shear strain varies linearly with the distance from the axis.

Derivation

  fixed end                free end
     |  A ------------- B     line AB before twisting
     |  A ------------- B'    after twisting
     |<------- L ------->|    arc BB' = r*theta

A line AB on the surface of the shaft of radius $r$ and length $L$ moves to AB' when the free end twists through angle $\theta$ (radians). The arc BB' $= r\theta$, and the shear strain (small angle) is

$$\gamma = \frac{BB'}{L} = \frac{r\theta}{L}$$

By Hooke's law in shear, $\tau = G\gamma$, so at the surface

$$\frac{\tau}{r} = \frac{G\theta}{L}$$

At a general radius $\rho$, $\tau_\rho = G\rho\theta/L$, hence shear stress varies linearly with radius: zero at the axis and maximum at the surface.

The torque on a ring of area $dA = 2\pi\rho\,d\rho$ is $dT = \tau_\rho\,\rho\,dA$.

$$T = \int \tau_\rho\,\rho\,dA = \frac{G\theta}{L}\int \rho^2\,dA = \frac{G\theta}{L}J$$

where $J = \int\rho^2\,dA$ is the polar second moment of area. Combining:

$$\boxed{\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}}$$

Section properties

• Polar modulus $Z_p = J/r$ (so $T = \tau_{max}Z_p$). For a solid shaft $Z_p = \pi d^3/16$.
• Torsional rigidity $= GJ$, the torque needed to produce a unit angle of twist per unit length ($T/\theta = GJ/L$).

Power transmitted: $P = T\omega = \dfrac{2\pi N T}{60}$ with $N$ in rpm.