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STATICS- EXAMPLE

    Example


3D Truss Lifting Crate

 

Determine the force in the member CD, BD and AD of the space truss shown in the figure. Note that points A, B and C are in the x-z plane and member CD is parallel to y-axis.

   
  Solution

 

First, the co-ordinates 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 cos30o, 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 FAD, FBD and FCD respectively. Expressing each force acting at joint D in the vector notation gives,

     W = -250 k
     FAD = FAD (-2.5i + 5 j + 4.33k) / 7.071
     FBD = FBD (2.5i + 5j + 4.33k) / 7.071
     FCD = FCD 5j / 5

All forces have to be in equilibrium, giving

     ΣF = 0
     W + FAD + FBD + FCD = 0

     
   

Equating the i, j and k components gives,

     ΣFx = 0
    0.3536 FBD  - 0.3536 FAD = 0
    FBD =  FAD

     ΣFz = 0
     0.6124 FAD + 0.6124 FBD - 250 = 0
     By substituting FAD = FBD
     0.6124 FAD + 0.6124 FAD - 250 = 0
     FAD = 204.1 lb = FBD

     ΣFy = 0
     0.7071 FAD + 0.7071 FBD + FCD = 0
     FCD = -288.6 lb

Members AD and BD are in tension and member CD is in compression.