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 Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll

 STATICS - CASE STUDY SOLUTION Support Free Body Diagram (FBD) of Truss When solving any truss, the reaction forces at each support should be determined before trying to calculate individual member loads. Starting with a FBD of the entire truss, the basic moment equilibrium equation gives,      ΣM1 = 0      2.5(4/3 + 4) + 5(4/3 + 4) + 7.5(4/3 + 4) - 10R = 0      R = 8.0 kips Due to symmetry, the reaction forces at both the left and right wall are equal,      RL = RR = R = 8 kips Joint 1 Joint 1 With the reactions known, it is now possible to analyze joint 1. Note that both F31 and F21 are drawn away from the joint, representing tension. The two unknowns forces, F31 and F21, can be found using the two force equilibrium equations,      ΣFx = 0      - F21 - F31 cos40 = 0      ΣFy = 0      F31 sin40 + R = 0 Substituting known values and solving gives,      F31 = -8/sin40 = -12.45 kips      F21 = -F31 cos40 = 9.534 kips Joint 2 Joint 2 The value of F21, 9.534 kips, was determined at joint 1 and can now be used at joint 2. The force in member F32 is easy to determine since all members and loads are perpendicular. Summing the forces in the vertical gives,      ΣFy = 0      F32 = 1.333 kips Since there are only two horizontal member forces, they must be equal,      ΣFx = 0      F21 = F42 = 9.534 kips Joint 3 Joint 3 The value of  F31, -12.45 kips, and F32, 1.333 kips, were determined from joint 1 and joint 2, respectively. It is now possible to determine forces, F53 and F43, at joint 3,      ΣFx = 0 = cos40 [-F53 - F43 + (-12.45)]      F53 = -F43 - 12.45      ΣFy = 0 = sin40 [F53 - F43 - (-12.45)] - 4 - 1.333 Eliminating F53 gives,      0 = -2F43 - 4/sin40 - 1.333/sin40       F43 = -4.148 kips and      F53 = 4.148 - 12.45 = -8.302 kips Joint 4 Joint 4 Now that F42, 9.534 kips, and F43, -4.148 kips, are known, joint 4 can be analyzed. Due to symmetry, the left member forces are the same as the right member forces. There is only one unknown force, F54. Using ΣFy = 0 gives,       F54 + 2(-4.148) sin40 - 1.333 = 0      F54 = 6.666 kips Summary Final Loads (kips) in Truss Members As shown by the force diagram, the largest tensile force is 9.534 kips and the largest compressive force is -12.45 kips. The best truss angle, θ, that minimizes the member loads can be determined by using the truss simulation. The best angle is the largest angle possible in the simulator, 65o. Thus a steep truss reduces the load. However, the cost is higher and buckling problems will also need to be considered.

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