Ch 8. Internal Loads Multimedia Engineering Statics Int. Forcesand Moments V and M Diagrams I V and M Diagrams II V, M, Load Relationships
 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 Free Body Diagram Form a free-body diagram and solve for the external reactions giving,      ΣMA = 0      L By - F L/2 = 0      By = F/2 = 300/2 = 150 lb      ΣFy = 0      Ay + By - F = 0      Ay = 300 - 150 = 150 lb Solution of Internal Forces Left Section If the beam is cut just to the left of the applied load, then the proper free-body diagram for the section to the left of the cut is as shown. Notice, it does not include the 300 lb force since it is cut just before the force. Sum the forces and moments to solve for the unknown internal loads:      ΣFy = 0      Ay - V = 0      V = Ay = 150 lb      ΣMA = 0      M - V L/2 = 0      M = LV/2 = 8 (150) / 2 = 600 lb-ft Right Section The problem can also be solved by using the right section after making the cut. The free-body diagram for the right side section is as shown at the left. Notice, the 300 lb force is included on this section since the cut is to the left of the force. Summing the forces and moments give,      ΣFy = 0      V + By - F = 0     V = 300 - 150 = 150 lb      ΣMB = 0      -V L/2 + F L/2 - M = 0      M = -150 (8/2) + 300 (8/2) = 600 lb-ft As expected, both methods obtain the same results, a shear force of 150 lb and a bending moment of 600 lb-ft.

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