Question 3 of 6: Single-surface disk clutch

Question 3: Single-surface disk clutch (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. $T=100$ N·m, $N=750$ rpm, $p_{max}=1.2$ MPa, $\mu=0.25$, uniform-wear model, ratio $d_i/d_o=0.577$, single friction surface.

Find. Outside diameter $d_o$, inside diameter $d_i$, and the transmitted power.

Approach. Under the uniform-wear assumption the pressure peaks at the inner radius, $p\,r=p_{max}r_i$; integrate the friction torque over the annulus, substitute $r_i=0.577\,r_o$, solve for $r_o$, then get the clamp force and power.

1. Uniform-wear torque. With $pr=p_{max}r_i$ (constant), the capacity of one surface is $$T=2\pi\mu p_{max}r_i\!\int_{r_i}^{r_o}\!r\,dr=\pi\mu p_{max}r_i\left(r_o^{2}-r_i^{2}\right).$$
2. Insert the ratio. Let $k=r_i/r_o=0.577$: $$T=\pi\mu p_{max}k\,(1-k^{2})\,r_o^{3}.$$ Note $k=1/\sqrt3$ is exactly the ratio that maximises uniform-wear torque for a given $r_o$.
3. Solve for $r_o$. $$r_o=\left[\frac{T}{\pi\mu p_{max}k(1-k^{2})}\right]^{1/3}=\left[\frac{100}{\pi(0.25)(1.2\times10^{6})(0.577)(1-0.333)}\right]^{1/3}=0.0651\text{ m}.$$ Hence $$\boxed{d_o=130.2\text{ mm},\qquad d_i=0.577\,d_o=75.1\text{ mm}}.$$
4. Clamp (axial) force. $F=2\pi p_{max}r_i(r_o-r_i)=2\pi(1.2\times10^{6})(0.0375)(0.0651-0.0375)=\boxed{7.79\text{ kN}}$; cross-check $T=\mu F r_m$ with $r_m=(r_o+r_i)/2$ returns 100 N·m.
5. Power transmitted. $$P=T\omega=100\left(\frac{2\pi(750)}{60}\right)=\boxed{7.85\text{ kW}}.$$