Question 5 of 6: Single-Surface Disk Clutch (Uniform Wear)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — December 2014 — 07-Mec-B1, Advanced Machine Design. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; answer any THREE of Part II (Problems 3–6). All six problems are solved here as a complete study resource.
Given. A single friction surface, molded lining, uniform-wear model.
Given data
Torque
$T=100$ N·m
Speed
$n=750$ rpm
Max lining pressure
$p_{\max}=1.2$ MPa
Friction coefficient
$\mu=0.25$
Diameter ratio
$d_i/d_o=0.577$
Find. The outside and inside diameters and the transmitted power.
Single-surface disk clutch: contact over the annulus between $r_i$ and $r_o$.
Approach. In the uniform-wear model the peak pressure occurs at the inner radius, and the axial force and torque follow from integrating the constant $pr$ product over the annulus. The given ratio $0.577$ (the torque-maximising value $1/\sqrt3$) lets us solve directly for the radii, then the power from torque and speed.
Uniform-wear torque relation. With $p\,r=p_{\max}r_i$ constant, the torque on one surface is $$T=\pi\mu\,p_{\max}\,r_i\,(r_o^2-r_i^2)=\pi\mu\,p_{\max}(0.577)(1-0.577^2)\,r_o^3.$$
Solve for the radii. Substituting $T=100$, $p_{\max}=1.2\times10^6$, $\mu=0.25$: $$r_o=\left(\frac{T}{\pi\mu p_{\max}(0.577)(1-0.577^2)}\right)^{1/3}=65.1\ \text{mm},\quad r_i=0.577\,r_o=37.6\ \text{mm}.$$ Hence $$\boxed{\;d_o=130.2\ \text{mm},\qquad d_i=75.1\ \text{mm}.\;}$$
Axial clamping force (check). $F=2\pi\,p_{\max}\,r_i(r_o-r_i)=7.79\times10^{3}\ \text{N}$, and $T=\mu F r_m$ with $r_m=(r_o+r_i)/2$ recovers 100 N·m — consistent.
Power transmitted. $$P=T\,\omega=T\,\frac{2\pi n}{60}=100\times\frac{2\pi(750)}{60}=7.85\times10^{3}\ \text{W}=7.85\ \text{kW}.$$
