A single plate clutch, consisting of two pairs of contacting surfaces, is required to transmit 40kW power at 1560 rpm. The coefficient of friction between the contacting surface is 0.3 and the intensity of pressure is limited to 0.4 $\mathrm{N} / \mathrm{mm}^{2}$.The outer diameter of the friction disk is limited to 300 mm. If the service factor is 1.25, Determine,

i) The inner diameter of friction disk.

ii) The axial force required to engage the clutch.

Given Data:

Single Plate Clutch

Power $(P) =40 \mathrm{kW}$

Speed $(\mathrm{n})=1560 \mathrm{rpm}$

Coefficient of friction $(\mu)=0.3$

Pressure $\mathrm{Pmax}=0.4 \mathrm{N} / \mathrm{mm}^{2}$

Diameter $\left(D_{o}\right)=300=\left(r_{o}=150\right)$

Service factor $K=1.25$

Solution:

1) Number of friction plates in clutch (i)

i=2 for Single Plate clutch

2) Power Transmitted (P)

$ \begin{array}{l}{P=\frac{2 \pi N T}{60000} \therefore 40 \times 10^{3}=\frac{2 \pi \times 1560 \times T}{60000}} \\ {\therefore T=244853.75 Nmm=244854 N m m}\end{array} $

3) Design for Torque Transmitted. $\left[M_{t}\right]$

$\left[M_{t}\right]=K . T$

$\left[M_{t}\right]=1.25 \times 244854=306067.5 \mathrm{Nmm}$

4) Axial Trust

$ W=2 \pi P_{\max } r_{i}\left(r_{o}-r_{i}\right) $

$W=2 \pi \times 0.4 \times r_{i}\left(150-r_{i}\right)$

$W=2.5132 \times r_{i}\left(150-r_{i}\right)$-----------eq(1)

5) Torque Transmitted

$T=\frac{1}{2} i . \mu . W .\left(r_{o}+r_{i}\right)$

$\#$ Replace T with $\left[M_{t}\right]$ if Service factor is given.

$306067.5=\frac{1}{2} \times 2 \times 0.3 \times W\left(150+r_{i}\right)$----------eq(2)

From $Eq(1) \& \ (2)$ we get.

$\therefore r_{i}=140 \mathrm{mm} \quad \therefore D_{i}=280 \mathrm{mm}$

6) Axial Force required to engage the clutch

$W_{e}=2 \pi P_{\max } r_{i}\left(r_{o}-r_{i}\right)$

$W_{e}=2 \pi \times 0.4 \times 140(150-140)$

$\therefore W_{e}=3518.58 N$