Since all members are symmetrical around the interstage structure, all members will have the same load. Using the method of joints, the forces at joint B on the bottom ring is analyzed by summing the forces,
ΣF_{B}
= 0
The freebody diagram includes the external thrust load and all member
forces acting at joint B. Since Joint B is in static equilibrium, the
summation of all force vectors must equal zero. Since this is a 3D problem,
all forces will be represented in the vector i, j, k format.
The total thrust load of 2,400 kN is evenly distributed around the interstage
structure so each joint will have to withstand a vertical load of 300 kN.
F_{T} = 300 k
Truss Diagram
Use the location of points B and E to define the unit directional vector
of F_{BE}.
E_{x} = 0.75 sin22.5 = 0.2870 m
E_{y} = 0.75 cos22.5 = 0.6929 m
E_{z} = 1.0 m
B_{x} = 0 m
B_{y} = 1 m
B_{z} = 0 m
The length of member BE is
BE = ((0.2870  0)^{2} + (0.6929
 (1))^{2}
+ (1  0)^{2} )^{0.5}
= 1.085 m
The vector F_{BE} in i, j, k format
is
The member force F_{BD} is similar to F_{BE} except the x component is reversed. The magnitudes will be the same:
F_{BD} = F_{BE}
(0.2645i + 0.2830j + 0.9217k)
If forces are summed in the z direction, ΣF_{z}
= 0, only one unknown remains, F_{BE}. Solving
for F_{BE} gives
2 F_{BE} 0.9217 + 300 = 0
F_{BE} = 162.7 kN compression
