Derive the torsional equation

Torsional equation: $\dfrac{T}{J} = \dfrac{\tau_s}{R} = \dfrac{G\varphi}{L}$, where $T$ is the torque, $J$ the polar moment of inertia, $\tau_s$ the shear stress at the surface (radius $R$), $G$ the modulus of rigidity, $\varphi$ the angle of twist and $L$ the length.

Assumptions

1. The shaft is straight, of uniform circular section (solid or hollow) and of homogeneous, isotropic material.
2. Plane sections perpendicular to the axis remain plane and circular after twisting; radii remain straight.
3. The twist is uniform along the length and the stress is within the elastic limit (Hooke's law).
4. Shear stress is proportional to shear strain; the shaft is loaded only by torque.

Derivation

  fixed end                 free end
  |---------------------------|  A -> A' (twist)
  |   O ======================|=> torque T
  |      L          B         B'
  |
  section: OB = R, BB' = R*phi, AB' tilts by gamma

Consider a shaft of length $L$ fixed at one end and twisted by torque $T$ at the other. A line AB on the surface moves to AB'. Let $\varphi$ be the angle of twist and $\gamma$ the shear strain at radius $R$. From the geometry:

$$BB' = R\varphi = L\gamma \Rightarrow \gamma = \frac{R\varphi}{L}$$

For a point at radius $r$, the shear strain is $\gamma_r = \dfrac{r\varphi}{L}$.

From Hooke's law in shear, $\tau = G\gamma$:

$$\tau_r = G\frac{r\varphi}{L} \quad\Rightarrow\quad \frac{\tau_r}{r} = \frac{G\varphi}{L} \quad(1)$$

so the shear stress varies linearly with radius, and at the surface $\dfrac{\tau_s}{R} = \dfrac{G\varphi}{L}$.

Consider an elemental ring of radius $r$ and thickness $dr$, area $dA = 2\pi r\,dr$. The force on it is $\tau_r\,dA$ and its moment about the axis is $\tau_r\,dA\cdot r$. The torque resisted by the entire section is

$$T = \int \tau_r\,r\,dA = \int \frac{\tau_s}{R}\,r^2\,dA = \frac{\tau_s}{R}\int r^2\,dA$$

Since $J = \int r^2\,dA$ (polar moment of inertia):

$$T = \frac{\tau_s}{R}J \quad\Rightarrow\quad \frac{T}{J} = \frac{\tau_s}{R} \quad(2)$$

Combining (1) and (2):

$$\boxed{\frac{T}{J} = \frac{\tau_s}{R} = \frac{G\varphi}{L}}$$

Polar moment of inertia: solid shaft $J = \dfrac{\pi D^4}{32}$; hollow shaft $J = \dfrac{\pi}{32}(D^4-d^4)$. Then $\tau_{max} = \dfrac{16T}{\pi D^3}$ for a solid shaft and $\varphi = \dfrac{TL}{GJ}$, where $GJ$ is the torsional rigidity.