Beam deflection boundary condition calculation

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SUMMARY

The discussion focuses on calculating beam deflection at specific points, x=L/4 and x=L/2, using boundary conditions. The participant correctly identifies that at x=0, the displacement v=0, leading to the conclusion that C2=0. The confusion arises regarding the slope of the beam, represented by the derivative dv/dx, which is stated to be zero at x=L/2 due to the load P acting upward. This indicates a misunderstanding of the physical implications of boundary conditions in beam deflection analysis.

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  • Basic calculus concepts, particularly derivatives
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Homework Statement



Find the deflection at x=L/4 and x=L/2 for the beam

Homework Equations



See attached pic.

The Attempt at a Solution



So I have the solution derived in class. Only 0<x<L/2 is derived because since the load on the beam is at L/2, the equation is valid for the entire beam since its symmetric (or something like that, if this isn't the correct explanation somebody tell me). My question concerns the boundary conditions. I know that at x=0, the displacement v=0. Hence you get C2=0. Now the second thing:

dv/dx=0 at x=L/2

I have no idea what dv/dx means, why its 0, and why its taken at x=L/2. Any help would be appreciated thanks.
 

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You ever heard of the 'slope' of the beam being discussed in your class? The slope = dv/dx.
 
ok, so what is the slope physically mean? Why is it 0 at x=L/2? It has something to do with the load P? I see that the derivative of moment is shear force, and the derivative of shear force is distributed force. But that has nothing to do with deflection right?
 
Start with deflection. What's the first derivative of a curve represent?
 
the slope, y/x. Why would dv/dx=0 at x=L/2? That makes no physical sense, because at x=L/2 there is a load P upward and it must deflect. Hence dv/dx can't be 0.
 
Last edited:
What's the slope of a curve in calculus class?
 

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