Question

Assuming the pulley at E is of negligible radius, determine the moment M required for equilibrium of the 3 member pin connected frame. Remember from statics the tension in all parts of a rope or cable is the same. Also determine the normal stress in CD if its cross section area is 0.5 in and the shear stress in pin B ifit's diameter is 3/8" and is in double shear 2.5 3' D 300 lb 60° 60°

Answer 1

FBD Take moment about E sigma ME = 0 T times r - 300 times r = 0 T = 300 lb sigma Fy = 0 FE - T - 300 = 0 FE = 600 lb FBD of member BDE sigma MB = 0 FCD times 3 sin 60 - FE times 4.5 cos 60 degree = 0 Therefore FCD = 779.42 lb sigma fx = 0 Bx = FCD = 779.42 lb sigma Fy = 0 By = FE = 600 lb member AC sigma MA = 0 M - FCD times 2 sin 60 = 0 m - 779.42 times 2 sin 60 = 0 Therefore m = 1350 lb ft Normal stress in CD sigma_CD = FCD/A = 779.42/0.5 = 1558.84 Psi Reaction at B = R_B squareroot Bx^2 + ?By^2 = squareroot (779.42)^2 + (60 D) = 983.614 lb Shear stress at B, tau_B = RB/2 (pi/4 d^2) = 983.64/2 (pi/4 times (3/8)^2)