A multi-disc clutch employs 3 steel and 2 bronze discs having outer diameter of 300 mm and inner diameter of 175 mm. If the coefficient of friction is 0.25 and axial force on each pair of surfaces is 5 kN, then the torque transmitted (assuming uniform wear) is

Concept:

For uniform wean Theory

Torque \(= \mu W\frac{{\left( {{r_0} + {r_i}} \right)}}{2} \times n\)

Where n → no. of plates surfaces

Calculation:

Given: steel plates = 3, Bronze plates = 2 

Here we have to subtract 1 because we need the number of contacting surfaces

∴ n = 3 + 2 – 1 = 4

Now,

Torque \(= \mu W\frac{{\left( {{r_0} + {r_i}} \right)}}{2} \times n\)

\(T = 4 \times 0.25 \times 5000 \times \left( {\frac{{150\; + \;87.5}}{{2000}}} \right)\)

∴ T = 593.75 N-m