Search
 
 

STATICS - CASE STUDY SOLUTION

     


Loading

 


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

     
   
 
Practice Homework and Test problems now available in the 'Eng Statics' mobile app
Includes over 500 free problems with complete detailed solutions.
Available at the Google Play Store and Apple App Store.