Torque capacity of a plate clutch

Notation: $r_o$ = outer radius, $r_i$ = inner radius of the friction face, $\mu$ = coefficient of friction, $F$ = total axial force, $p$ = intensity of pressure.

Consider a thin ring of radius $r$ and width $dr$. Area $= 2\pi r\,dr$; normal force $dF = p\cdot2\pi r\,dr$; friction force $= \mu\,dF$; friction torque $dT = \mu\,dF\cdot r = 2\pi\mu p\,r^2dr$.

   ro ____
     / ri \   ring of radius r, width dr
    |  O  |   dF = p 2 pi r dr
     \____/

(a) Uniform pressure ($p$ constant)

Axial force:

$$F = \int_{r_i}^{r_o} p\,2\pi r\,dr = \pi p\,(r_o^2 - r_i^2)$$

Torque:

$$T = 2\pi\mu p\int_{r_i}^{r_o} r^2dr = \frac{2\pi\mu p}{3}(r_o^3 - r_i^3)$$

Eliminating $p$:

$$T = \frac{2}{3}\,\mu F\,\frac{r_o^3 - r_i^3}{r_o^2 - r_i^2}$$

The mean radius is $R_m = \dfrac{2}{3}\dfrac{r_o^3-r_i^3}{r_o^2-r_i^2}$, so $T = \mu F R_m$.

(b) Uniform wear

Wear is proportional to pressure times rubbing velocity, i.e. $p\,r\omega$. For uniform wear $p\,r = C$ (constant), so pressure is greatest at the inner radius: $p_{max} = C/r_i$.

Axial force:

$$F = \int_{r_i}^{r_o} \frac{C}{r}\,2\pi r\,dr = 2\pi C\,(r_o - r_i)$$

Torque:

$$T = 2\pi\mu C\int_{r_i}^{r_o} r\,dr = \pi\mu C\,(r_o^2 - r_i^2)$$

Eliminating $C$:

$$T = \mu F\,\frac{r_o + r_i}{2} = \mu F R_m$$

Design assumption

Uniform wear is used for design of clutches already run in. A new clutch follows uniform pressure, but after initial wear the pressure redistributes; the uniform wear torque is smaller (e.g. for $r_i = 0.6r_o$ it is about 2% lower), hence safer. Uniform wear also gives the maximum pressure at the inner radius, $p_{max} = F/[2\pi r_i(r_o-r_i)]$, which must not exceed the allowable value of the lining. For $n$ friction surfaces multiply the torque by $n$.