## Q. 13.7

Pressure Capacity of a Cone Clutch

A cone clutch having an outside diameter $D$, inner diameter $d$, and width $w$ (Figure 13.18) transmits a torque $T$.

Find: The contact pressure and the actuating force on the basis of:

a. The uniform wear.

b. The uniform pressure.

Given: $D$=300 mm, $d$=280 mm, $w$=50 mm, $T$=150 N · m

Assumption: Coefficient of friction will be taken to be $f$=0.24.

## Verified Solution

a. The half-cone angle of the clutch equals (Figure 13.18)

$\sin \alpha=\frac{D-d}{2 w}=\frac{300-280}{2(50)}=0.2$

or

$\alpha=11.54^{\circ}$

From Equation (13.38), the maximum pressure is found as

$T=\frac{\pi f p_{\max } d}{\sin \alpha} \int_{d / 2}^{D / 2} r d r=\frac{\pi f p_{\max } d}{8 \sin \alpha}\left(D^2-d^2\right)$       (13.38)

$p_{\max }=\frac{8 T \sin \alpha}{\pi f d\left(D^2-d^2\right)}$      (a)

Inserting the given data, we have

$p_{\max }=\frac{8(150)(0.2)}{\pi(0.24)(0.28)\left(0.3^2-0.28^2\right)}=98 kPa$

The actuating force, applying Equation (13.37), is then

$F_a=\frac{1}{2} \pi p_{\max } d(D-d)$           (13.37)

\begin{aligned} F_a & =\frac{1}{2} \pi p_{\max } d(D-d) \\ & =\frac{1}{2} \pi\left(98 \times 10^3\right)(0.28)(0.3-0.28)=862 Pa \end{aligned}

b. Making use of Equation (13.41), the maximum pressure is

$T=\frac{\pi f p_{\max }}{12 \sin \alpha}\left(D^3-d^3\right)=\frac{F_a f}{3 \sin \alpha} \frac{D^3-d^3}{D^2-d^2}$     (13.41)

\begin{aligned} p_{\max } & =\frac{12 T \sin \alpha}{\pi f\left(D^3-d^3\right)} \\ & =\frac{12(150)(0.2)}{\pi(0.24)\left(0.3^3-0.28^3\right)}=94.6 kPa \end{aligned}         (b)

The actuating force, from Equation (13.40), is

$F_a=\frac{1}{4} \pi p_{\max }\left(D^2-d^2\right)$     (13.40)

$F_a=\frac{1}{4} \pi\left(94.6 \times 10^3\right)\left(0.3^2-0.28^2\right)=862 Pa$

Comment: The results show that uniform wear condition gives a pressure capacity of about 3.5% larger than that of uniform rate.