GATE 2023 AG – Question 55
A single-disc clutch transmits10 kW at 1400 rpm. Mean axial contact pressure.07 N/mm² and friction coefficient.25. Diameter/lining face-width ratio is 8. Under uniform wear, find face width in mm, to two decimals.
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Correct answer: 26.75 to 26.90
Explanation
Interpret diameter as mean friction diameter $D_m=8b$,so mean radius4b and radii4.5b,3.5b. Contact area is $\pi(r_o^2-r_i^2)=8\pi b^2$ and axial force $W=.07(8\pi b^2)$. Two working faces transmit $T=2\mu W(4b)=64\pi\mu p b^3$. Required torque is $10000(60)/(2\pi1400)=68.20926$ N·m=68209.26 N·mm. Solving gives **b=26.86 mm**. The mean-diameter and mean-pressure conventions are needed because the printed wording does not identify an outer diameter or inner-edge maximum pressure.