The assembly is used to support the distributed load of w 5

The assembly is used to support the distributed load of w = 500 lb/ft. Determine the factor of safety with respect to yielding for steel rod BC and the pins at B and C. In tension, sigma_Y = 36 ksi; in shear T_Y = 18 ksi. The beam given in this video problem is supported by rod BC. Which is a two-force member. The pins connecting BC to the beam and the support at the wall are in \"double shear\", so the shear stress in the each pin is: 3.50 F_BC 14.1 F_BC 28.3 F_BC 7.07 F_BC

Solution

The area for double shear = 2 * (pi/4) * d²_pin  = 2* (pi/4) * 0.3² = 0.14137 in²

so, shear stress in each pin = Force in each pin/area of double shear = F_BC / 0.14137 = 7.07 F_BC