Question 4 of 6: Single-Surface Disk Clutch Design (Uniform Wear)

Question 4: Single-Surface Disk Clutch Design (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. $T=120$ N·m, $N=750$ rpm, $p_{max}=1.2$ MPa, $\mu=0.3$, uniform wear, $d_i/d_o=r_i/r_o=0.625$, single friction surface.

Find. Outside and inside diameters $d_o,d_i$; the required axial (clamp) force; and the transmitted power.

Approach. Under uniform wear $pr$ is constant, so $p$ peaks at $r_i$ with $p_{max}$; write the axial force and single-surface torque as integrals over the annulus, substitute $r_i=0.625\,r_o$, solve the resulting cubic for $r_o$, then $P=T\omega$.

1. Uniform-wear pressure model. Wear $\propto p\,r$ is uniform, so $p(r)=p_{max}\dfrac{r_i}{r}$ — the highest pressure is at the inner edge. Integrating the axial force and the friction torque over one face, $$F=2\pi p_{max}r_i(r_o-r_i),\qquad T=\pi\mu p_{max}r_i\left(r_o^2-r_i^2\right).$$
2. Insert the diameter ratio. With $r_i=0.625\,r_o$, $$T=\pi\mu p_{max}(0.625\,r_o)\big(r_o^2-0.625^2 r_o^2\big)=\pi\mu p_{max}(0.3809)\,r_o^3.$$
3. Solve for the outer radius. Substituting $T=120$, $\mu=0.3$, $p_{max}=1.2\times10^6$ Pa, $$r_o^3=\frac{120}{\pi(0.3)(1.2\times10^6)(0.3809)}=2.79\times10^{-4}\ \text{m}^3\ \Rightarrow\ r_o=0.0653\ \text{m}.$$ $$\boxed{d_o=130.6\ \text{mm}},\qquad d_i=0.625\,d_o=\boxed{81.6\ \text{mm}}.$$
4. Required clamp force. With $r_i=0.0408$ m, $r_o=0.0653$ m, $$F=2\pi(1.2\times10^6)(0.0408)(0.0653-0.0408)=\boxed{7.54\ \text{kN}}.$$
5. Power transmitted. At $N=750$ rpm, $\omega=2\pi(750)/60=78.5$ rad/s, $$P=T\omega=120(78.5)=\boxed{9.42\ \text{kW}}\ (\approx 12.6\ \text{hp}).$$