Torque capacity of a cone clutch

Notation: $\alpha$ = semi-cone angle, $R$, $r$ = outer and inner radii of the cone face, $\mu$ = friction coefficient, $p_n$ = normal pressure, $F_n$ = total normal force on the cone surface, $F_a$ = axial force.

         axial force Fa
              |
        ______v_____
       /  \  alpha  \
      R    \ slant   r
       \____\_______/

Torque (uniform wear)

Take an elementary ring of radius $x$ and slant width $dx/\sin\alpha$. Area $= 2\pi x\,dx/\sin\alpha$. Normal force $dF_n = p_n\,2\pi x\,dx/\sin\alpha$. Friction torque:

$$dT = \mu\,dF_n\,x = \frac{2\pi\mu p_n}{\sin\alpha}x^2dx$$

For uniform wear $p_n x = C$:

$$F_n = \frac{2\pi C}{\sin\alpha}\int_r^R dx = \frac{2\pi C (R-r)}{\sin\alpha}$$ $$T = \frac{2\pi\mu C}{\sin\alpha}\int_r^R x\,dx = \frac{\pi\mu C (R^2-r^2)}{\sin\alpha}$$

Eliminating $C$:

$$T = \mu F_n R_m, \qquad R_m = \frac{R+r}{2}$$

The axial force transmitted by the cone face in steady running is $F_a = F_n\sin\alpha$, so $T = \dfrac{\mu F_a R_m}{\sin\alpha}$.

(For uniform pressure, $R_m = \dfrac{2}{3}\dfrac{R^3-r^3}{R^2-r^2}$ and $F_n = \pi p_n(R^2-r^2)/\sin\alpha$.)

Axial force for engagement

While the clutch is engaging, friction acts along the cone surface, opposing the motion of the cone into the cup. Resolving along the axis:

$$F_e = F_n(\sin\alpha + \mu\cos\alpha)$$

Hence the spring force needed to engage the clutch is $F_e = F_n(\sin\alpha+\mu\cos\alpha)$; once it is engaged and running, $F_a = F_n\sin\alpha$ is enough to hold it.

Advantages over plate clutch

• The wedge action gives a normal force $F_n = F_a/\sin\alpha$ much larger than the axial force, so for the same $F_a$ a cone clutch transmits more torque (about $1/\sin\alpha$ times).
• Smaller diameter for the same torque.
• No need for many plates.

Disadvantages: more expensive to make, needs close alignment, tends to stick.

Condition to avoid sticking

The cone must release easily. On disengagement the axial component of the normal force, $F_n\sin\alpha$, must overcome the friction along the slant, $\mu F_n\cos\alpha$, so

$$\tan\alpha > \mu, \qquad \alpha > \tan^{-1}\mu$$

(the semi-cone angle must be greater than the friction angle). The usual value $\alpha = 12.5^\circ$ ($\tan 12.5^\circ = 0.22$) suits linings with $\mu$ up to about 0.2, such as cork or leather. Linings with higher friction need a larger angle, up to about 20 degrees (for $\mu = 0.35$, $\alpha > 19.3^\circ$).