Subtopic: Power transmission systems – gear trains, differential, final drives and power take-off
Question Statement:
A single disc clutch is used to transmit 10 kW power at 1400 rpm. The axial pressure exerted on the contact surface is 0.07 N mm⁻² and the coefficient of friction is 0.25. Considering the ratio of diameter to face width of the clutch lining as 8 and assuming uniform wear theory, the required face width of friction lining in mm is __________________ (rounded off to 2 decimal places).
Official Answer Key:
26.75 to 26.90
Step-by-Step Solution & Explanation
Torque to be transmitted, T = P/ω = 10000/(2π×1400/60) = 10000/146.6 = 68.24 N·m = 68,244 N·mm.
Single-plate (“single disc”) clutches engage on TWO friction faces, so torque per face = T/2 = 34,122 N·mm.
With mean diameter D_m/face-width b = 8 (a standard clutch-lining design ratio), take lining inner/outer radii r₁=3.5b, r₂=4.5b (so r₂−r₁=b, mean radius R_m=4b). Using the given axial pressure p over the annular lining area: axial force W = p×π(r₂²−r₁²) = p×π×8b². Torque per face = μ×W×R_m = μ×p×π×8b²×4b = 1.7584×b³.
Solving 34,122 = 1.7584×b³ gives b³ = 19,405, so b ≈ 26.87 mm. Within official range 26.75–26.90.
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