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 Solve for Truss Reactions Force Diagram The force in the top member 6-7 needs to be determined when the 1,500 kg car is at joint 2. First, solve for reactions R1 and R5 by applying both the moment and force equilibrium equations for the whole truss.      ΣM1 = 0      -4 (14.72) + 16 R5 = 0      R5 = 3.68 kN      ΣFy = 0      R1 + R5 - 14.72 = 0      R1 = 11.04 kN Cut Truss at Required Member Cut 1 Diagram Since, the member 6-7 needs to be determined, that member must be included in the cut. However, there is is usually more than one location to make a cut. It is easiest to make a cut where only three or less members are cut. Thus, a vertical cut through members 6-7, 6-3 and 2-3 was done as shown at the left. Notice that there are only three unknowns, which can be solved for with the three equilibrium equations.      ΣFy = 0      11.04 - 14.72 - F63 cos45 = 0      F63 = -5.204 kN      ΣM1 = 0      -4 (14.72) - 4cos45 F63 - 4sin45 F63 - 4 F67 = 0       -58.88 - 2 [(4) (0.7071) ( -5.204)] = 4 F67      F67 = -7.360 kN  compression For this cut location, you need to use at least two of the three equilibrium equations. The number of equations may be reduced by other cut locations. Alternate Cut Cut 2 Diagram By carefully choosing where the cut is made, the number of calculations can be reduced. For example, if a cut is made through members 6-7, 3-7, 3-8, and 3-4 it is possible to solve for the member force F67 with one equilibrium equation even though four members are cut. Since all unknowns, except F67, go through joint 3, the moment about joint 3 has only one unknown.      ΣM3 = 0      -8 (11.04) - 4 F67 + 4 (14.72) = 0      F67 = -7.360 kN  compression The solution is identical to the solution for the previous cut.