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Beam deflection boundary condition calculation

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    See attached pic.

    3. 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.
     

    Attached Files:

  2. jcsd
  3. Feb 10, 2014 #2

    SteamKing

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    You ever heard of the 'slope' of the beam being discussed in your class? The slope = dv/dx.
     
  4. Feb 10, 2014 #3
    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?
     
  5. Feb 10, 2014 #4

    SteamKing

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    Start with deflection. What's the first derivative of a curve represent?
     
  6. Feb 10, 2014 #5
    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: Feb 10, 2014
  7. Feb 10, 2014 #6

    SteamKing

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    What's the slope of a curve in calculus class?
     
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