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STATICS - CASE STUDY SOLUTION

     


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Top View at Point B


FBD of Joint B

 

 


Example of Space Truss

 

Since all members are symmetrical around the inter-stage 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,

     ΣFB = 0

The free-body 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 inter-stage structure so each joint will have to withstand a vertical load of 300 kN.

     FT = 300 k


Truss Diagram

Use the location of points B and E to define the unit directional vector of FBE.

     Ex = 0.75 sin22.5 = 0.2870 m
     Ey = -0.75 cos22.5 = -0.6929 m
     Ez = 1.0 m

     Bx = 0 m
     By = -1 m
     Bz = 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 FBE in i, j, k format is

 

The member force FBD is similar to FBE except the x component is reversed. The magnitudes will be the same:

     FBD = FBE (-0.2645i + 0.2830j + 0.9217k)

If forces are summed in the z direction, ΣFz = 0, only one unknown remains, FBE. Solving for FBE gives

     2 FBE 0.9217 + 300 = 0

     FBE = -162.7 kN   compression

     
   
 
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