Single-surface disk clutch — uniform-wear design and transmitted power (20 marks): 22-Mec-B1 Advanced Machine Design, December 2019, Q3

Question 3: 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. A single friction face working against a molded lining, with the diameter ratio fixed by the designer.

Given data — Problem 3
Quantity	Symbol	Value
Torque to be transmitted	$T$	120 N·m
Rotational speed	$N$	500 rev/min
Maximum lining pressure	$p_{\max}$	1.5 MPa
Coefficient of friction	$\mu$	0.30
Diameter ratio	$k=d_i/d_o$	0.70
Number of friction surfaces	—	1

Find. The outside and inside diameters of the friction annulus, the required clamping force, and the power transmitted.

Figure 3.1 — Single-surface disk clutch. Left: the friction annulus, 123.9 mm outside by 86.7 mm inside. Right: the uniform-wear pressure distribution, peaking at 1.5 MPa at the bore and falling as 1/r to 1.05 MPa at the outer rim.

Approach. Under the uniform-wear assumption the product $p\,r$ is constant across the annulus, which pins the pressure distribution to $p_{\max}$ at the bore; integrating the friction moment over the annulus then gives torque as an explicit cubic in the outer radius once the ratio $k$ is imposed.

Set up the uniform-wear pressure law. Wear rate is proportional to $p\,v$ and $v\propto r$, so uniform axial wear requires $p\,r=\text{constant}$. The constant is fixed at the inner radius, where the pressure is highest: $$p(r)\,r = p_{\max}\,r_i \quad\Longrightarrow\quad p(r)=p_{\max}\frac{r_i}{r}$$ This is why a uniform-wear clutch always runs its peak pressure at the bore, and why making $r_i$ too small is self-defeating.
Integrate for the clamping force. With $\mathrm{d}F = p(r)\,2\pi r\,\mathrm{d}r = 2\pi p_{\max} r_i\,\mathrm{d}r$, $$F=\int_{r_i}^{r_o} 2\pi p_{\max} r_i\,\mathrm{d}r = 2\pi p_{\max} r_i\left(r_o-r_i\right)$$
Integrate for the friction torque. Each ring contributes $\mathrm{d}T=\mu\,r\,\mathrm{d}F$, so for one friction surface $$T=\int_{r_i}^{r_o}2\pi\mu p_{\max}r_i\,r\,\mathrm{d}r=\pi\mu p_{\max}r_i\left(r_o^{2}-r_i^{2}\right)$$
Impose the diameter ratio and solve for $r_o$. Putting $r_i = k r_o$ with $k=0.7$ turns the torque equation into a cubic in $r_o$ alone: $$T=\pi\mu p_{\max}k\left(1-k^{2}\right)r_o^{3}$$ $$\pi\times 0.30\times 1.5\times 0.7\times\left(1-0.49\right)=0.5047\ \text{N/mm}^{2}$$ $$r_o=\left(\frac{120\times 10^{3}}{0.5047}\right)^{1/3}=\left(237\,760\right)^{1/3}=61.95\ \text{mm}$$ so that $r_i = 0.7\times 61.95 = 43.37$ mm.
Report the diameters. Doubling the radii, $$\boxed{d_o = 123.9\ \text{mm},\qquad d_i = 86.7\ \text{mm}}$$ In practice one would specify the next convenient stock lining, say $d_o = 125$ mm with $d_i = 87.5$ mm, which keeps the ratio at 0.70 and adds a small margin.
Clamping force required. Substituting into the force integral, $$F = 2\pi\times 1.5\times 43.37\times\left(61.95-43.37\right)$$ $$\boxed{F = 7.60\ \text{kN}}$$ A useful cross-check uses the mean radius $r_m=(r_o+r_i)/2 = 52.66$ mm: the uniform-wear torque is exactly $T=\mu F r_m = 0.30\times 7596\times 52.66 = 120.0\ \text{kN}\cdot\text{mm}$, which closes on the specified torque.
Power transmitted. At the stated speed, $$\omega=\frac{2\pi N}{60}=\frac{2\pi\times 500}{60}=52.36\ \text{rad/s}$$ $$P = T\omega = 120\times 52.36$$ $$\boxed{P = 6.28\ \text{kW}}$$
Sanity-check the lining duty. The mean pressure over the annulus is $$p_{\text{mean}}=\frac{F}{\pi\left(r_o^{2}-r_i^{2}\right)}=\frac{2p_{\max}r_i}{r_o+r_i}=1.24\ \text{MPa}$$ comfortably below the 1.5 MPa peak at the bore, and the rubbing speed at the mean radius is $v = \omega r_m = 2.76$ m/s — both well inside the normal range for a molded lining.

Question 3 — results
Quantity	Value
Outside diameter, $d_o$	123.9 mm (specify 125 mm)
Inside diameter, $d_i$	86.7 mm (specify 87.5 mm)
Mean radius, $r_m$	52.66 mm
Clamping force, $F$	7.60 kN
Angular velocity, $\omega$	52.36 rad/s
Power transmitted, $P$	6.28 kW
Mean lining pressure	1.24 MPa (peak 1.5 MPa at the bore)

Topic. Axial-contact (disk) clutch sizing on the uniform-wear assumption.

Key relations. $p\,r = p_{\max}r_i$; $F = 2\pi p_{\max}r_i(r_o-r_i)$; $T = \pi\mu p_{\max}r_i(r_o^{2}-r_i^{2}) = \mu F r_m$ per friction surface with $r_m=(r_o+r_i)/2$; with $r_i = kr_o$, $T = \pi\mu p_{\max}k(1-k^{2})r_o^{3}$; $P = T\omega$.

Why this works. A new clutch pressed together by a rigid plate starts with roughly uniform pressure, but wear removes material fastest where $pv$ is greatest, i.e. at the outside. Within a few hundred engagements the pressure redistributes until wear is uniform across the face, which requires $pr$ constant. Uniform wear is therefore the state a clutch spends its working life in, and it predicts a lower torque for the same $p_{\max}$ than the uniform-pressure model — making it the conservative and standard design basis. Because $T$ scales as $r_o^{3}$, the design collapses to a single cube root once the ratio $k$ is chosen.

Common pitfall. Mixing radii and diameters inside the cubic — the ratio $k$ is the same for radii and diameters, but $r_o$ must be a radius when the torque constant is evaluated. The second pitfall is multiplying by two friction surfaces: a single-surface clutch has one rubbing interface, not two, so the torque here is not doubled. Note also that $k = 0.7$ is imposed by the question; the ratio that maximises torque for a given $p_{\max}$ and $r_o$ is $k = 1/\sqrt3 = 0.577$, so this design gives up about 11 % of the torque available from that outside diameter in exchange for a cooler, more lightly loaded bore.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §16-5 (disk clutches, uniform wear); Juvinall & Marshek, Fundamentals of Machine Component Design, 6th ed., §18.

Question 3 of 6: Single-surface disk clutch — uniform-wear design and transmitted power (20 marks)

Question 3: Single-surface disk clutch — uniform-wear design and transmitted power (20 marks)