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Statics: Shear and bending moment, distributed load

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data
    ks4QbRZ.png


    2. Relevant equations
    ƩFy=0
    ƩM=0

    3. The attempt at a solution

    Finding the reaction at A:
    NBbavS8.png

    ƩFy=0: A + (L*w0)/6 - (L*w0)/2 = 0
    A = (L*w0)/3

    ƩMA=0: MA - [(L*w0)/2]*(L/3) + [(L*w0)/6]*((2L)/3) = 0
    MA = (w0*L2)/18

    For the shear and bending moment:
    uYe21ky.png

    ƩFy=0: (L*w0)/3 - (x*w0)/2 + (x*w0)/6 - V = 0
    V = (L*w0)/3 - (x*w0)/3

    ƩM=0: (w0*L2)/18 + [(x*w0)/2]*[(2*x)/3] - [(x*w0)/6]*(x/3) + M = 0
    M = -(w0*L2)/18 - (5/18)*w0*x2
     
    Last edited: Jul 28, 2013
  2. jcsd
  3. Jul 28, 2013 #2

    SteamKing

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    What happens to your equations for V(x) and M(x) when x = 0? Would these values of V and M match your statics equilibrium calculations?
     
  4. Jul 28, 2013 #3
    It looks to me like they do:
    V = (L*w0)/3 which balances out the force at A (V is pointing down)

    M = -(w0*L2)/18 = (w0*L2)/18 clockwise, which balances the equivalent counterclockwise moment at A.
     
  5. Jul 28, 2013 #4

    haruspex

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    x, I assume, is a distance from the wall.
    I don't understand how you obtained these equations. Can you spell that out more clearly?
     
  6. Jul 28, 2013 #5
    Yes, that is correct.
    I summed up the vertical forces in my diagram.

    (L*w0)/3 is the value I obtained for A in the beginning of the problem.
    (x*w0)/2 is from the triangle area formula (Bh)/2, where the base is distance x, and w0, the magnitude of the force, is the height
    (x*w0)/6 : also using area of a triangle, where the base is again x, and the height is (1/3)*w0 as indicated in the problem.


    V = (L*w0)/3 - (x*w0)/3 is what I get when I solve the ƩFy equation for V
     
  7. Jul 28, 2013 #6

    haruspex

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    I still don't get the reasoning behind it.
    Since the system is static, the total moment about any point on the beam is zero. The bending moment at X is the contribution from one side of the point X only. I would expect to see something like ##\int_{y=x}^L (y-x)F(y).dy##. Maybe you have a neat way of avoiding the integration, but if so I'm not seeing it.
     
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