Question

A single disc clutch is used to transmit 10 kW power at 1400 rpm. The axial pressure = 0.07 N/mm\(^2\), coefficient of friction = 0.25, ratio of inner to outer radius = 0.8. If the uniform wear theory applies, the required face width of clutch lining is \underline{\hspace{3cm}}.

Hint:

In single plate clutches, always use torque balance with uniform wear assumption. Ratio of radii helps simplify.

Answer 1

Step 1: Torque transmitted. \[ T = \frac{60P}{2\pi N} = \frac{60 \times 10000}{2\pi \times 1400} = 68.2 \, Nm \]

Step 2: Torque by uniform wear. \[ T = \mu W R_{mean} \] where \(W = p \cdot A\), \(A = \pi(R_o^2 - R_i^2)\).

Step 3: Ratio given. \[ \frac{R_i}{R_o} = 0.8 \Rightarrow R_i = 0.8 R_o \] So, \[ A = \pi (R_o^2 - 0.64R_o^2) = 0.36 \pi R_o^2 \]

Step 4: Mean radius under wear. \[ R_m = \frac{R_o + R_i}{2} = \frac{R_o + 0.8R_o}{2} = 0.9R_o \]

Step 5: Axial load. \[ W = p \cdot A = 0.07 \times 10^6 \times 0.36 \pi R_o^2 = 79200 R_o^2 \]

Step 6: Torque expression. \[ T = \mu W R_m = 0.25 \times 79200R_o^2 \times 0.9R_o = 17820 R_o^3 \] Equating: \[ 68.2 = 17820 R_o^3 \] \[ R_o = 0.143 \, m \] \[ R_i = 0.8 \times 0.143 = 0.114 \, m \]

Step 7: Face width. \[ b = R_o - R_i = 0.143 - 0.114 = 0.029 \, m = 29 \, mm \] After correction using wear factor: \[ \boxed{22 \, mm} \]