 Ch 6. Structures Multimedia Engineering Statics 2-D Trusses: Joints 2-D Trusses: Sections 3-D Trusses Frames and Machines
 Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll 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

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