Number of plates in a multiplate clutch

Data

$r_o = 100$ mm, $r_i = 60$ mm, $\mu = 0.12$, $p_{max} = 0.3$ N/mm², $P = 30$ kW, $N = 1200$ rpm. Uniform wear.

Torque to be transmitted

$$T = \frac{P}{\omega} = \frac{30\,000}{2\pi\times1200/60} = \frac{30000}{125.66} = 238.7\ \text{N m}$$

Axial force from the pressure limit

For uniform wear, maximum pressure is at the inner radius and $p_{max}\,r_i = C$:

$$F = 2\pi C(r_o - r_i) = 2\pi p_{max} r_i (r_o - r_i)$$ $$F = 2\pi\times0.3\times60\times(100-60) = 4524\ \text{N}$$

Torque per friction surface

Mean radius $R_m = \dfrac{r_o+r_i}{2} = 80$ mm.

$$T_1 = \mu F R_m = 0.12\times4524\times0.080 = 43.4\ \text{N m}$$

Number of friction surfaces

$$n = \frac{T}{T_1} = \frac{238.7}{43.4} = 5.5 \;\Rightarrow\; n = 6$$

Number of discs

Number of friction surfaces $= n = n_1 + n_2 - 1$, where $n_1$ and $n_2$ are the numbers of discs on the driving and driven shafts.

$$n_1 + n_2 = 7 \;\Rightarrow\; n_1 = 4,\ n_2 = 3$$

(or 3 and 4).

Actual axial force and pressure with $n = 6$

$$F = \frac{T}{n\mu R_m} = \frac{238.7}{6\times0.12\times0.080} = 4145\ \text{N}$$ $$p_{max} = \frac{F}{2\pi r_i(r_o - r_i)} = \frac{4145}{2\pi\times60\times40} = 0.275\ \text{MPa} < 0.3\ \text{MPa}$$

Answer: 6 friction surfaces (4 discs on one shaft, 3 on the other); axial force 4145 N; $p_{max} = 0.275$ MPa.