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Deflection of a Beam using Double Integration

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Need to determine the maximum deflection of the beam, by using double integration, and EI is constant.


    Tried to draw this the best I could. It is a simply supported beam, with a concentrated couples of equal value (C) but opposite direction, with a distance between of (a).

    2. Relevant equations

    I pretty sure that my boundary conditions are: v=0 @ x=0 and v=0 @ x=3a

    and my max. deflection will occur at x=(3/2)a

    3. The attempt at a solution

    My problem is that I don't know how to get started, all the examples show only one concentrated moment. I'm assuming I have no reactions, because when I sum the moments around either side, the moments cancel out.

    I have tried using that M=C(x-a) and that didn't come out correct. So basically I'm looking for help in trying to get the moment function.

    I have also tried M=C where I get the following

    slope as a function of x = Cx/EI + C1
    deflection as a function of x = Cx^2/2EI + C1x + C2

    boundary conditions:
    v=0 @ x=0 so C2 = 0
    v=0 @ x=3a so C1 = (-3(C)(a))/(2EI)

    so then I come up with an answer of (-9(C)(a^2))/(8EI) for the max deflection using x=(3/2)a
    Last edited: Oct 23, 2007
  2. jcsd
  3. Oct 24, 2007 #2


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    Well, it's been way too long since I used the double integration method, but it does bring back some haunting memories. If you're looking at y" =M(x)/EI, my suggestion would be to calcualte M(x) using shear and moment diagrams. Looks to me like you've got no shear at all anywher along the beam, and just a constant moment of c in between the two applied couples, and no moment on either side of those couples to the support points.
  4. Oct 24, 2007 #3
    PhantomJay, that is correct. I'm just not sure how to come up with M(x). I have tried several things for it but still can't get the answer they give for Max Deflection. Which I know to be (5(C)(a^2))/(8EI).
  5. Oct 24, 2007 #4


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    Well, M(x) =0 between 0 and a; M(x) = C between a and 2a; and M(x) = 0 between 2a and 3a. Now double integrate and use proper boundary conditions , which is probably what you're trying to do. I think the key is to deternine the boundary conditons. I've been relying on tables and charts, so my calculus is a bit rusty at this point.
  6. Oct 18, 2009 #5
    What a coincidence, I found this post: Beam deflection by double integration method. It is exactly the same problem as this one.
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