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Beam bending, overhang with 2 concentrated loads

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    A beam supported at two locations is subjected to two equal loads at the end points
    Compute the central deflection W.

    Schematic:

    [itex]\downarrow[/itex].........................................[itex]\downarrow[/itex]
    ____________________________
    <--->Δ<-------><------->Δ<--->
    ...a..........L............L...........a

    The delta's are the supports the arrows the forces F and a & L distances

    2. Relevant equations

    M(x) = E*I*W"(X)

    3. The attempt at a solution

    At first I try to formulate the momentum along the beam. x = 0 at the center of the beam.

    I thought it would be zero at the points where the Forces are, and maximum at the center.
    So I came up with:

    M(x) = F(1+[itex]\frac{x}{a+L}[/itex])*(a+L) with x left from the center and

    M(x) = F(1-[itex]\frac{x}{a+L}[/itex])*(a+L) with x right from the center.

    w" = EI/M(x)
    Integrating this twice should yield the result:
    w'(x) = [itex]\frac{F}{2EI}[/itex]x[itex]^{2}[/itex] + [itex]\frac{F}{EI}[/itex](a+L)x + C

    But w'(0) = 0 so C = 0

    integrating once more gives:
    w(x) = [itex]\frac{F}{6EI}[/itex]x[itex]^{3}[/itex] + [itex]\frac{F}{EI}[/itex](a+L)x[itex]^{2}[/itex] + C

    solving for C with w(L)=0 gives: C = - [itex]\frac{FL^{2}}{EI}[/itex]([itex]\frac{a}{2}[/itex]+[itex]\frac{2L}{3}[/itex])

    which therefore should be the max deflection, because at x=0 deflection is max and w(0) = C

    The problem is I have no way to check my result, and I'm not quite sure of the result. Could someone check me and explain me where I went wrong (if so)
     
  2. jcsd
  3. Jan 21, 2013 #2

    haruspex

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    Could you explain how you arrive at that, please? I'm not saying it's wrong...
     
  4. Jan 22, 2013 #3
    With the argument that it should be zero at the force and maximum at the midpoint, I determined them to be linear in between, so that with negative x, M(x) goes from x= -a-l -> M(-(a+L))=0 to x=0 giving M(0)=F(a+L).

    Also I see that i did not integrate right, forgot some constants, but the problem lies mainly in the determination of the momentum along the beam, the rest is straightforward I guess.
     
  5. Jan 22, 2013 #4

    haruspex

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    I approached it by actually taking moments. For a point x left of centre, x < L, there's an anticlockwise moment F(x+L) from the support and a clockwise one F(x+L+a) from the load. Doesn't that make the bending moment Fa for all x between the supports? Beyond the supports it's F(L+a-x).
     
  6. Jan 22, 2013 #5
    I'm not sure how you determined these moments.. What do you mean by: "I approached it by actually taking moments". Did you divide the problem into halfs and used M = F*x,
    and then used that if there is a load F, the support gives a reaction force F?

    If this is what you did, it is still not obvious for me how you arrived at the moments from this.
     
  7. Jan 22, 2013 #6
    Ah never mind! Got it, thank you for your help!
     
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