Question 5 of 6: Single-surface disk clutch — uniform-wear design and transmitted power (20 marks)

Question 5: Single-surface disk clutch — uniform-wear design and transmitted power (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An annular molded-lining friction disk clamped between a driving and a driven plate, with one friction surface in contact.

Find. The outside and inside diameters of the friction annulus, the clamping force implied, and the power transmitted at 850 rev/min.

Approach. Under the uniform-wear assumption the product $pr$ is constant across the annulus, so the pressure peaks at the inner radius. Write the clamping force and the torque as integrals of that pressure distribution, impose the given radius ratio, and solve the resulting cubic in $r_o$ directly.

1. Set up the uniform-wear pressure distribution. Axial wear rate is proportional to $pV$, and $V = \omega r$, so uniform axial wear requires $p r = \text{constant}$. The largest pressure therefore occurs where $r$ is smallest: $$p(r)\,r = p_{\max}\,r_i \;\Longrightarrow\; p(r) = \frac{p_{\max} r_i}{r}$$ This is the physically correct model for a lining that has bedded in; the alternative uniform-pressure model applies only to a brand-new, perfectly flat facing.
2. Clamping (axial) force. Integrate the pressure over the annulus with $dA = 2\pi r\,dr$: $$F = \int_{r_i}^{r_o} \frac{p_{\max}r_i}{r}\,2\pi r\,dr = 2\pi p_{\max} r_i\,(r_o - r_i)$$
3. Torque capacity of one friction surface. Each annular element contributes a friction force $\mu p\,dA$ at radius $r$: $$T = \int_{r_i}^{r_o} \mu\,\frac{p_{\max}r_i}{r}\,(2\pi r\,dr)\,r = \pi \mu p_{\max} r_i \left(r_o^2 - r_i^2\right)$$
4. Impose the given radius ratio and reduce to a cubic. With $k = r_i/r_o = d_i/d_o = 0.6$, substitute $r_i = k r_o$: $$T = \pi\mu p_{\max}\,k\,(1-k^2)\,r_o^3 = \pi(0.25)(1.5\times10^6)(0.6)(1-0.36)\,r_o^3 = 452\,389\,r_o^3$$
5. Solve for the outside radius and report the diameters. $$r_o = \left(\frac{120}{452\,389}\right)^{1/3} = 0.06424\text{ m}, \qquad r_i = 0.6 r_o = 0.03854\text{ m}$$ $$\boxed{\;d_o = 128.5\text{ mm}, \qquad d_i = 77.1\text{ mm}\;}$$ For manufacture these would be rounded to a 130 mm outside and 78 mm inside diameter, which raises the torque capacity slightly and leaves the pressure below the limit.
6. Required clamping force. Substitute back into the force integral: $$F = 2\pi (1.5\times10^6)(0.03854)(0.06424-0.03854)$$ $$\boxed{\;F = 9.34\text{ kN}\;}$$ This is the spring or hydraulic force the release mechanism must overcome, and it sizes the diaphragm spring and the throw-out bearing.
7. Cross-check through the mean friction radius. For uniform wear the effective friction radius is the arithmetic mean, $r_m = (r_o+r_i)/2 = 51.39$ mm, so $$T = \mu F r_m = 0.25 \times 9338 \times 0.05139 = 120.0\text{ N}\!\cdot\!\text{m}\ \checkmark$$ The mean lining pressure is $F/A = 9338 / \left[\tfrac{\pi}{4}(128.5^2-77.1^2)\right] = 1.13$ MPa, safely below the 1.5 MPa peak — a useful reminder that the peak occurs only at the bore.
8. Power transmitted. At the rated speed, $$\omega = \frac{2\pi N}{60} = \frac{2\pi(850)}{60} = 89.0\text{ rad/s}, \qquad P = T\omega = 120 \times 89.0$$ $$\boxed{\;P = 10.68\text{ kW}\;}$$

It is worth noting how close the specified ratio sits to the optimum. For a fixed outside radius the torque function $k(1-k^2)$ is maximised at $k = 1/\sqrt{3} = 0.577$; the specified 0.6 delivers 99.8 % of that maximum, so the clutch is effectively as compact as a single-surface design can be. Pushing $k$ much higher (a narrow annulus) or much lower (a wide one with a small, high-pressure bore) would need a larger outside diameter for the same torque.