A hydraulically operated multidisk plate clutch has an effec

A hydraulically operated multidisk plate clutch has an effective disk outer diameter of 6.5 in and an inner diameter of 4 in. The coefficient of friction is 0.24, and the limiting pressure is 120 psi. There are six planes of sliding present. Using the uniform wear model, estimate the limiting axial force F and the torque T. Let the inner diameter of the friction pairs d be a variable. Complete the following table: What does the table show?

Solution

a) Torque T = f * pi * p_max * r_i * (r_o^2 - r_i^2)

                   = 0.24 * pi * 120 * 2 * (3.25^2 - 2^2)   =   1187.522 pounds-in

      Axial Force F = T / ( f * (r_i + r_o) /2 )    = 1187.522 / ( 0.5 * 0.24 * (2 + 3.25) ) = 1884.9556 pounds

b) for di = 2 in => ri =2 in

     Torque T = f * pi * p_max * r_i * (r_o^2 - r_i^2)

                   = 0.24 * pi * 120 * 1 * (3.25^2 - 1^2)   =   865.1946 pounds-in

     for di = 3 in => ri = 1.5 in

     Torque T = f * pi * p_max * r_i * (r_o^2 - r_i^2)

                   = 0.24 * pi * 120 * 1.5 * (3.25^2 - 1.5^2)   =   1128.146 pounds-in

     for di = 4 in => ri = 2 in

     Torque T = f * pi * p_max * r_i * (r_o^2 - r_i^2)

                   = 0.24 * pi * 120 * 2 * (3.25^2 - 2^2)   =   1187.522 pounds-in

     for di =5 in => ri = 2.5 in

     Torque T = f * pi * p_max * r_i * (r_o^2 - r_i^2)

                   = 0.24 * pi * 120 * 2.5 * (3.25^2 - 2.5^2)   =   975.46 pounds-in

     for di =6 in => ri = 3 in

     Torque T = f * pi * p_max * r_i * (r_o^2 - r_i^2)

                   = 0.24 * pi * 120 * 3 * (3.25^2 - 3^2)   =   424.115 pounds-in