---
title: "Torque capacity of a plate clutch"
subject: "Machine Design II"
chapter: 7
key: "semester_7/machine_design_ii:plate-clutch-uniform-pressure-wear"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/plate-clutch-uniform-pressure-wear
site: BCT Engineering
---

# Torque capacity of a plate clutch

Machine Design II · Chapter 7: Clutches and brakes

- Asked in: year not given, 4+4 marks
- Video: [Torque transmitting capacity of single plate clutches and multi-plate clutches](https://www.youtube.com/watch?v=TwKnDXsuCoI) — 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/plate-clutch-uniform-pressure-wear
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-7

## Question

Derive expressions for the torque transmitted by a single-plate friction clutch (one friction pair) on the basis of (a) uniform pressure and (b) uniform wear. State which assumption is used for design and why.

## Short answer

Uniform pressure: T = (2/3) μF (r_o³ – r_i³)/(r_o² – r_i²). Uniform wear: T = μF (r_o + r_i)/2. Design uses uniform wear because it yields a lower, safer torque after run‑in, with maximum pressure at the inner radius.

## Answer

Notation: $r_o$ = outer radius, $r_i$ = inner radius of the friction face, $\mu$ = coefficient of friction, $F$ = total axial force, $p$ = intensity of pressure.

Consider a thin ring of radius $r$ and width $dr$. Area $= 2\pi r\,dr$; normal force $dF = p\cdot2\pi r\,dr$; friction force $= \mu\,dF$; friction torque $dT = \mu\,dF\cdot r = 2\pi\mu p\,r^2dr$.

```text
   ro ____
     / ri \   ring of radius r, width dr
    |  O  |   dF = p 2 pi r dr
     \____/
```

#### (a) Uniform pressure ($p$ constant)
Axial force:

$$
F = \int_{r_i}^{r_o} p\,2\pi r\,dr = \pi p\,(r_o^2 - r_i^2)
$$

Torque:

$$
T = 2\pi\mu p\int_{r_i}^{r_o} r^2dr = \frac{2\pi\mu p}{3}(r_o^3 - r_i^3)
$$

Eliminating $p$:

$$
T = \frac{2}{3}\,\mu F\,\frac{r_o^3 - r_i^3}{r_o^2 - r_i^2}
$$

The mean radius is $R_m = \dfrac{2}{3}\dfrac{r_o^3-r_i^3}{r_o^2-r_i^2}$, so $T = \mu F R_m$.

#### (b) Uniform wear
Wear is proportional to pressure times rubbing velocity, i.e. $p\,r\omega$. For uniform wear $p\,r = C$ (constant), so pressure is greatest at the inner radius: $p_{max} = C/r_i$.

Axial force:

$$
F = \int_{r_i}^{r_o} \frac{C}{r}\,2\pi r\,dr = 2\pi C\,(r_o - r_i)
$$

Torque:

$$
T = 2\pi\mu C\int_{r_i}^{r_o} r\,dr = \pi\mu C\,(r_o^2 - r_i^2)
$$

Eliminating $C$:

$$
T = \mu F\,\frac{r_o + r_i}{2} = \mu F R_m
$$

#### Design assumption
**Uniform wear** is used for design of clutches already run in. A new clutch follows uniform pressure, but after initial wear the pressure redistributes; the uniform wear torque is smaller (e.g. for $r_i = 0.6r_o$ it is about 2% lower), hence safer. Uniform wear also gives the maximum pressure at the inner radius, $p_{max} = F/[2\pi r_i(r_o-r_i)]$, which must not exceed the allowable value of the lining. For $n$ friction surfaces multiply the torque by $n$.

## Check yourself

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

1. For uniform pressure, what is the clutch torque?
   A. T = mu F (r_o + r_i)/2
   B. T = (2/3) mu F (r_o^3 - r_i^3)/(r_o^2 - r_i^2)
   C. T = pi mu p (r_o^2 - r_i^2)
   D. T = (2/3) mu F (r_o^2 - r_i^2)/(r_o^3 - r_i^3)
   Answer: B. Integrating dT = 2 pi mu p r^2 dr and eliminating p gives this expression. (Key point: Uniform pressure torque)

2. What is the condition for uniform wear and where is pressure greatest?
   A. p is constant; greatest at outer radius
   B. p r is constant; greatest at the inner radius
   C. p r is constant; greatest at the outer radius
   D. p/r is constant; greatest at the inner radius
   Answer: B. Wear is proportional to p r omega, so p r = C and p is greatest at r_i. (Key point: Uniform wear assumption)

3. The uniform-wear torque is
   A. T = mu F (r_o + r_i)/2
   B. T = mu F (r_o - r_i)/2
   C. T = mu F r_o
   D. T = (2/3) mu F r_o
   Answer: A. Eliminating C gives T = mu F R_m with R_m = (r_o + r_i)/2. (Key point: Uniform wear torque)

4. Why is the uniform wear assumption used in design?
   A. It gives larger torque
   B. A run-in clutch redistributes pressure, and the torque it gives is smaller, hence safer
   C. A new clutch follows uniform wear
   D. It removes need for a pressure limit
   Answer: B. After initial wear the pressure redistributes; the lower uniform-wear torque is the safer design value. (Key point: Design choice)

## Related in Machine Design II

- [Torque capacity of a cone clutch](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Acone-clutch-torque-axial-force) — 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

## Related in other subjects

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- [Tribology in manufacturing: friction, wear, lubrication](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Atribology-friction-wear-lubrication-manufacturing) — Manufacturing and Production Processes, chapter 2
- [Dynamic and wear load check, spur pair](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Abuckingham-dynamic-wear-check) — Machine Design I, chapter 9
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