

First, the coordinates of joints A, B, C and D need to be found.
From the figure it can be seen that ΔABC is an equilateral triangle with each side of 5 feet. The height of the equilateral triangle will be equal to 5 cos30^{o}, or 4.330 feet. Thus, the coordinates each joint will be,
A = [0, 0, 0] B = [5, 0, 0]
C = [2.5, 0, 4.33] D = [2.5, 5, 4.33]
Since, there are only three unknown forces at joint D, the force analysis of the truss will begin at joint D. Recall, with 3D trusses, three equations can be used at each joint. 


Let the force in member AD, BD and CD be labeled as F_{AD}, F_{BD} and F_{CD} respectively.
Expressing each force acting at joint D in the vector notation gives,
W = 250 k
F_{AD} = F_{AD} (2.5i + 5 j + 4.33k) / 7.071
F_{BD} = F_{BD} (2.5i + 5j + 4.33k) / 7.071
F_{CD} = F_{CD} 5j / 5
All forces have to be in equilibrium, giving
ΣF = 0
W + F_{AD} + F_{BD} + F_{CD} = 0 


Equating the i, j and k components gives,
ΣF_{x} = 0
0.3536 F_{BD}  0.3536 F_{AD} = 0
F_{BD} = F_{AD}
ΣF_{z }= 0
0.6124 F_{AD} + 0.6124 F_{BD}  250 = 0
By substituting F_{AD} = F_{BD}
0.6124 F_{AD} + 0.6124 F_{AD}  250 = 0
F_{AD} = 204.1 lb = F_{BD}
ΣF_{y} = 0
0.7071 F_{AD} + 0.7071 F_{BD } + F_{CD} = 0
F_{CD} = 288.6 lb
Members AD and BD are in tension and member CD is in compression. 