---
title: "Torque capacity of a cone clutch"
subject: "Machine Design II"
chapter: 7
key: "semester_7/machine_design_ii:cone-clutch-torque-axial-force"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/cone-clutch-torque-axial-force
site: BCT Engineering
---

# Torque capacity of a cone clutch

Machine Design II · Chapter 7: Clutches and brakes

- Asked in: year not given, 6 marks
- Video: [Torque transmitting capacity of cone clutch](https://www.youtube.com/watch?v=WWEBt_DFCfA) — Mechanical Engineering Online Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/cone-clutch-torque-axial-force
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-7

## Question

Derive the expression for the torque transmitted by a cone clutch of semi-cone angle alpha, outer radius R and inner radius r, on the basis of uniform wear. Obtain the axial force needed for engagement. State the advantages of a cone clutch over a plate clutch and the condition on the semi-cone angle to avoid sticking.

## Short answer

Torque (uniform wear): T = μ F_a R_m / sin α, with R_m = (R+r)/2. Axial force for engagement: F_e = F_n(sin α + μ cos α). A cone clutch gives torque amplification due to wedge action; to avoid sticking, tan α > μ must hold.

## Answer

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.

```text
         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 $\mu = 0.35$, $\alpha > 19.3^\circ$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For uniform wear, the cone clutch torque can be written as
   A. T = mu F_n R_m, with R_m = (R + r)/2
   B. T = mu F_a R_m, with R_m = (R + r)/2
   C. T = mu F_n (R - r)/2
   D. T = mu F_n R_m sin(alpha)
   Answer: A. T = mu F_n R_m, and since F_a = F_n sin(alpha), T = mu F_a R_m / sin(alpha). (Key point: Cone clutch torque)

2. The axial force needed to engage the cone is
   A. F_e = F_n sin(alpha)
   B. F_e = F_n cos(alpha)
   C. F_e = F_n (sin(alpha) + mu cos(alpha))
   D. F_e = F_n (sin(alpha) - mu cos(alpha))
   Answer: C. During engagement friction opposes motion into the seat, adding mu cos(alpha) to the sin(alpha) term. (Key point: Engagement force)

3. Why does a cone clutch transmit more torque than a plate clutch for the same axial force?
   A. It has more friction surfaces
   B. Wedge action makes normal force F_n = F_a/sin(alpha)
   C. Its mean radius is always larger
   D. It uses higher pressure lining
   Answer: B. Wedge action multiplies the normal force by about 1/sin(alpha). (Key point: Wedge advantage)

4. To avoid sticking, the semi-cone angle must satisfy
   A. alpha < tan^-1(mu)
   B. alpha > tan^-1(mu)
   C. alpha = 45 degrees
   D. alpha < 5 degrees
   Answer: B. The semi-cone angle must be greater than the friction angle so the cone releases easily. (Key point: Sticking condition)

## Related in Machine Design II

- [Torque capacity of a plate clutch](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aplate-clutch-uniform-pressure-wear) — Machine Design II, chapter 7
- [Number of plates in a multiplate clutch](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Amultiplate-clutch-power-numerical) — Machine Design II, chapter 7

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