This discussion focuses on the methodology for selecting internal forces in free body diagrams (FBDs) in structural analysis. Participants emphasize the importance of identifying external forces and moments before constructing the FBD, ensuring minimal inclusion of external forces to simplify calculations. The conversation highlights that while certain distributed loads may not be directly included in the FBD, their effects are accounted for through other external forces acting on the body. Specific examples, such as the triangular and rectangular distributed loads, illustrate how to approach these decisions effectively.
PREREQUISITESStudents and professionals in civil engineering, mechanical engineering, and physics who are involved in structural analysis and need to understand the selection of internal forces in free body diagrams.
Before you start addressing the internal forces, you should first have determined the unknown external forces and moments (if possible). Then, you draw a free body diagram that includes your cross section of interest. You try to choose a free body that includes as few of the external forces and moments as possible; this reduces the amount of work you need to do, but doesn't affect the answer.princejan7 said:Homework Statement
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When you have to calculate the internal forces at a point, how do you decide which forces are included in the free body diagram?
Yes. But its effect is captured by the other external forces that are actually acting on the free body you have chosen.For D, why isn't the rectangular bit of the distributed load included? Doesn't it affect D?
Chestermiller said:Yes. But its effect is captured by the other external forces that are actually acting on the free body you have chosen.
Only the insignificant part between B and D.princejan7 said:But the question says that D is located to the left of point B. Wouldn't that mean the rectangular distributed load is also actually acting on the body at D?
Chestermiller said:Only the insignificant part between B and D.
That's OK. Your free body diagram does not include the part to the left. And, as I said earlier, the effect of the rest of the triangular distributed load is accounted for by the reactions at B and A. To prove this to yourself, use the free body to the left of E instead of the one to the right of E, and see if you get a different result for the internal forces and moments at E.princejan7 said:oh ok, but then what about the diagram for point E?
The triangular distributed load is included in the diagram even though the important part is quite a bit to the left?