Question 5 of 6: Single-Surface Disk Clutch (Uniform Wear)

Problem 5: Single-Surface Disk Clutch (Uniform Wear) (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. A single friction surface, molded lining, sized under the uniform-wear assumption.

Find. Outside and inside diameters, and the transmitted power.

Approach. Apply the uniform-wear model ($p\,r = \text{const}$, maximum at the inner radius), express torque and axial force in terms of $r_o$ with the fixed ratio, solve for $r_o$, then get the power from $T\omega$.

1. Uniform-wear law. The pressure varies as $p(r) = p_{\max}\,r_i/r$ (largest at $r_i$). Integrating pressure and friction over the annulus: $$F = 2\pi p_{\max} r_i (r_o - r_i), \qquad T = \pi\mu p_{\max} r_i\,(r_o^2 - r_i^2).$$
2. Insert the ratio. With $r_i = 0.577\,r_o$, the torque becomes $T = \pi\mu p_{\max}\,(0.577)(1-0.577^2)\,r_o^3 = \pi\mu p_{\max}(0.385)\,r_o^3$.
3. Solve for the radii. $r_o = \left[\dfrac{100}{\pi(0.25)(1.2\times10^6)(0.385)}\right]^{1/3} = 0.0651\ \text{m}$. Hence $$\boxed{\,d_o \approx 130\ \text{mm},\quad d_i \approx 75\ \text{mm}\,}$$
4. Axial actuating force. $F = 2\pi(1.2\times10^6)(0.0375)(0.0651-0.0375) = 7.79\ \text{kN}$ (cross-check: $\mu F r_m = 0.25(7790)(0.0513) = 100\ \text{N}\!\cdot\!\text{m}$ ✓).
5. Power transmitted. $P = T\omega = 100\times\dfrac{2\pi(750)}{60} = 7854\ \text{W} = \boxed{7.85\ \text{kW}}$.

The specified ratio $d_i/d_o = 0.577 \approx 1/\sqrt{3}$ is not arbitrary — it is the value that maximises the torque a uniform-wear clutch of given outer radius can carry, so this design is running at the optimal proportion for its lining.