Question 3 of 6: Single-surface disk clutch designed on uniform wear (20 marks)

Question 3: Single-surface disk clutch designed on 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 flat annular clutch facing sized on the uniform-wear model, with the lining properties and the diameter ratio prescribed:

Find. The outside and inside diameters $d_o$ and $d_i$ of the friction annulus, the axial clamp force $F$ that produces the specified peak pressure, and the power transmitted at 750 rev/min.

Approach. Adopt the uniform-wear pressure law, which pins the peak pressure at the bore, integrate the friction moment over the annulus to get torque as a function of $r_o$ alone once the ratio $k$ is fixed, solve for $r_o$, then recover the clamp force from the same pressure law and the power from $P=T\omega$.

1. Adopt the uniform-wear pressure distribution. Wear rate is proportional to the rate of frictional work per unit area, $\mu p V = \mu p \omega r$, so a lining that has bedded in to a uniform axial wear rate must satisfy $$p\,r=\text{constant}=p_{\max}r_i$$ The peak pressure therefore occurs at the inner edge of the annulus, the smallest radius in contact, and the design condition is $p\left(r_i\right)=p_{\max}=1.2$ MPa. (The alternative uniform-pressure model applies only to a brand-new, perfectly rigid facing; uniform wear is the conservative and standard design basis, and it is what the question specifies.)
2. Clamp force from the pressure law. Integrating the axial pressure over the annulus with $p=p_{\max}r_i/r$, $$F=\int_{r_i}^{r_o}\!p\left(2\pi r\right)dr=2\pi p_{\max}r_i\int_{r_i}^{r_o}\!dr=2\pi p_{\max}r_i\left(r_o-r_i\right)$$ Notice how the $1/r$ pressure law cancels the $r$ in the area element, which is what makes the uniform-wear algebra so much simpler than the uniform-pressure case.
3. Friction torque for one surface. Each ring of area $2\pi r\,dr$ contributes a friction force $\mu p\left(2\pi r\,dr\right)$ acting at radius $r$: $$T=\int_{r_i}^{r_o}\!\mu p\left(2\pi r\right)r\,dr=2\pi\mu p_{\max}r_i\int_{r_i}^{r_o}\!r\,dr=\pi\mu p_{\max}r_i\left(r_o^{2}-r_i^{2}\right)$$ Equivalently $T=\mu F r_m$ with $r_m=\left(r_o+r_i\right)/2$ — the whole facing behaves as though its friction force acted at the arithmetic mean radius, a useful check at the end.
4. Introduce the prescribed ratio and solve for the outer radius. Writing $r_i=k\,r_o$ collapses the torque expression to a single unknown: $$T=\pi\mu p_{\max}\,k\left(1-k^{2}\right)r_o^{3}$$ With $k=0.577$ the geometric factor is $k\left(1-k^{2}\right)=0.577\left(1-0.333\right)=0.3849$, so $$r_o=\left[\frac{T}{\pi\mu p_{\max}k\left(1-k^{2}\right)}\right]^{1/3}=\left[\frac{1.00\times10^{5}}{\pi\left(0.25\right)\left(1.2\right)\left(0.3849\right)}\right]^{1/3}=\left(2.756\times10^{5}\right)^{1/3}=65.09\ \text{mm}$$ Hence $r_i=0.577\times65.09=37.56$ mm and $$\boxed{\ d_o=130.2\ \text{mm},\qquad d_i=75.1\ \text{mm}\ }$$ In practice these would be rounded to standard stock, say $d_o=130$ mm and $d_i=75$ mm, which changes the torque capacity by well under 1 %.
5. Recognise what the ratio 0.577 is. Differentiating the geometric factor, $\dfrac{d}{dk}\left[k-k^{3}\right]=1-3k^{2}=0$ gives $k=1/\sqrt3=0.5774$. The prescribed ratio is therefore the torque-maximising one: for a given outside diameter and peak pressure, no other bore extracts more torque from the facing. That is why the value keeps reappearing in clutch problems, and it is worth stating explicitly rather than treating 0.577 as an arbitrary number.
6. Axial clamp force. Substituting the radii back into the force integral: $$F=2\pi\left(1.2\right)\left(37.56\right)\left(65.09-37.56\right)=\boxed{\ 7.80\ \text{kN}\ }$$ This is the load the pressure plate springs or the hydraulic piston must supply. Cross-check through the mean radius: $r_m=\left(65.09+37.56\right)/2=51.3$ mm, so $\mu F r_m=0.25\left(7796\right)\left(51.3\right)=1.00\times10^{5}\ \text{N}\cdot\text{mm}=100\ \text{N}\cdot\text{m}$ — the required torque exactly, confirming the radii and the force together. The mean pressure over the annulus is $F/\left[\pi\left(r_o^{2}-r_i^{2}\right)\right]=0.878$ MPa, i.e. 73 % of the 1.2 MPa peak, which is the signature of the uniform-wear distribution.
7. Power transmitted. The clutch transmits the stated torque at the stated speed: $$\omega=\frac{2\pi n}{60}=\frac{2\pi\left(750\right)}{60}=78.54\ \text{rad/s},\qquad P=T\omega=100\times78.54=7854\ \text{W}$$ $$\boxed{\ P=7.85\ \text{kW}\ \left(10.5\ \text{hp}\right)\ }$$ Note that this is the power carried once engaged and slipping has ceased; the heat generated during engagement is a separate, and often governing, thermal calculation.