Question

The C-clamp shown in Fig. 2 uses a double threaded square screw with minor diameter 32 mm and pitch of 4 mm. The frictional coefficients are 0.2 for the threads and 0.15 for the collar. The collar has a friction diameter of 50 mm. i. Determine the maximum force P to be applied to the handle at a radius 5 cm from the screw centreline if the clamping force is F-50 N Evaluate whether self-locking occurs. ii. 5 cm

Answer 1

Given that d = 32 mm P = 4 mm Thread = 0.15 collar = 0.20 dm = d - P/2 = 32 - 4/2 = 30 alpha = 14.5 lambda = tan^-1(L/pi dm) = tan^-1 (32/3.14 times 30) lambda = 18.72 degree alpha_n = tan^-1 (tan alpha cos lambda) = tan^-1 (tan (14.5) cos (18.72)) alpha_n = tan^-1 [(0.258) (0.947)] = tan^-1 (0.244) alpha_n = 13.71 Now using eqin = 100 times 30/2 [0.15 times 3.14 times 30 + 4 times cos (13.17)/3.14 times 30 times cos (13.71) - 0.15 times 4] + 100 times 0.2 times 5/2 = 1500 [18.014/90.8682] + 50 = 347.36 T = 347.36 N - mm clamping Force F_C = T/50 = 347.36/50 = 6.9472 F_C = 6.9472