Clutch Torque Capacity Uniform Wear⚠ unverified
Mechanical / Clutches Brakes · Compute the torque capacity of a clutch assuming uniform wear
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Axial clamping force |
| mu | μ | — | 1.0 | Coefficient of friction between the facings (dimensionless) |
| Ro | Ro | m | 1.0 | Outer friction radius |
| Ri | Ri | m | 1.0 | Inner friction radius |
| n | n | — | 1.0 | Number of friction surfaces (pairs of contact). Default is 1 |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | T | N*m | Torque capacity, in newton-metres (N*m) |
The science & history
Understanding the Parameters
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Radii $R_o$, $R_i$ — the friction annulus runs between them. Under uniform wear the mean radius is their simple average, so the torque is as if all the friction force acted at $(R_o + R_i)/2$. A wider, larger-radius annulus gives more torque, but also more sliding speed and heat at the outer edge.
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Why uniform wear averages the radii — a running clutch wears fastest where the product of pressure and sliding speed ($p\,v = p\,\omega r$) is highest, i.e. at large radius. It wears until the pressure redistributes so that $p\,r$ is constant (highest pressure at the inner radius). That $p\,r = $ const distribution is exactly what makes the effective radius the arithmetic mean.
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Clamping force $F$ and surfaces $n$ — as in the general Disk Clutch Torque: capacity scales directly with clamp force and the number of friction interfaces.
Derivation (Approaching a Proof)
Under uniform wear, the wear rate $\propto p\,r$ is constant across the face, so
$$p(r)\, r = C \;\Longrightarrow\; p(r) = \frac{C}{r},$$
with the maximum pressure at the inner radius. The clamping force is the integral of pressure over the annulus:
$$F = \int_{R_i}^{R_o} p\,(2\pi r)\,\mathrm{d}r = 2\pi C\int_{R_i}^{R_o}\mathrm{d}r = 2\pi C (R_o - R_i).$$
The torque per friction surface is the integral of the friction force times its radius:
$$T_1 = \int_{R_i}^{R_o} \mu p\,(2\pi r)\, r\,\mathrm{d}r = 2\pi\mu C\int_{R_i}^{R_o} r\,\mathrm{d}r = \pi\mu C (R_o^2 - R_i^2).$$
Divide $T_1$ by $F$ to eliminate the constant $C$:
$$\frac{T_1}{F} = \frac{\pi\mu C(R_o^2 - R_i^2)}{2\pi C(R_o - R_i)} = \frac{\mu(R_o^2 - R_i^2)}{2(R_o - R_i)} = \frac{\mu(R_o + R_i)}{2},$$
using $R_o^2 - R_i^2 = (R_o - R_i)(R_o + R_i)$. Hence $T_1 = \mu F (R_o + R_i)/2$, and for $n$ friction surfaces
$$T = n F \mu\,\frac{R_o + R_i}{2}.$$
The clean arithmetic-mean radius is the hallmark of the uniform-wear model — the difference-of-squares factoring is what collapses the integral so neatly.
Dimensional check. $[T] = \text{N}\cdot\text{m}$. ✓
History and Development
The uniform-wear model is the standard basis for clutch and brake rating in Shigley and machine-design practice, preferred over uniform pressure because it matches a worn-in clutch and is conservative. It dates to the early analysis of automotive friction clutches and remains the design default.
Related Concepts: Disk Clutch Torque, Multiple Disk Clutch, Cone Clutch Torque, Clutch Facing Pressure, Clutch Wear Volume
Notes: Standard (conservative) design model. Uniform-pressure alternative: $T = \tfrac23 n\mu F(R_o^3 - R_i^3)/(R_o^2 - R_i^2)$. Identical to Multiple Disk Clutch ($R_{mean} = (R_o+R_i)/2$). Check facing pressure and heat separately.