Question

The crane hook shown below is exposed to a force of 5000 lbf...

Answer 1

Given that crane hook is exposed to a force of 5000 lbf where b, = 1 in b0 = 2.5 is data is bi = 1in b0 = 0.25in R0 = 6in Ri = 2in h = uin p = 5000lbf (Fos) = 2 Eccentically calculated as Rw = [bi + 60/2] + h/[biR0-B0R/h]loge[R0/Ri]-[bi-b0] Rw = [1 + 0.25/2] times 4/[1.6-0.25*2/4]loge[6/2]-[1-0.25] = > 2.5/[1.35*1.098612]0.75 Rw = 3.2869138 in \r\n Rw = Ri + h(bi + b0) /3(bi + b0) = > 2 + 1.6 = B.6in Know that Eccentricity e = R-Rw = 3.6-3.2869138 = 0.3130862 in Calculation of Bending stress: - Calculated Area = A = (Ra-Ri) (bi + b0) /2 = 2.5 Bending moment plb = PR mb = 18000 lbf in Bending stress ab = Mh/Aer ab = 14797.891 lbf/in direct tensile stress: - a = P/A = 2000lbf/in^2 \r\n Total stress due to load 5000 left alpha = al + ab = 2000 + 1797.404 = 16797.891 lbf/in^2 calculation of required strength of material st = a*Fos = 16797.891*2 = 33594.9752 lbf/in^2 Material selected must have atleast strength of 33594.9782 lbf/in^2