a) Determine the normal stress in bar BC. The cross sectional area of the bar is A = 10 in².

Bar BC is a two-force member: Reaction C is along CB.  Triangle ABC is a 3-4-5 triangle:

C_x = 0.8 C

C_y = 0.6 C

Resultant of distributed load:  W = 300(8) = 2400 lb

Moment equilibrium about point A: 3C_x - 4W = 0, C_x = 4W/3 = 3200 lb

C = C_x/0.8 = 3200/8 = 4000 lb

Normal stress in BC = C/A = 4000/10 = 400 psi, compressive.

b) Determine the average shear stress in the pin at A. The diameter of the pin is d = 0.25 in, and the pin acts in double shear.

Reactions at A:

A_x = C_x = 3200 lb

A_y = W - C_y = 2400 - 0.6(4000) = 0

Shear force in pin = A_x/2 = 1600 lb

Average shear stress in pin = 1600/[p(0.125)²] = 32,595 psi (high stress)