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Beam that is simply supported in the middle

  1. Oct 21, 2008 #1
    Hey Guys,

    I am working on a situation where I have a beam that is simply supported in the middle, and the two ends of the beam are .005" higher than the middle where the support sits. I am trying to figure out with a linearly increasing load what the force is to make the two ends ,which are sitting .005" higher, the same heigh as the middle. Basically I want to flatten the piece.

    EI = 731 lbf*in^2
    length = 3.522in
    width = .404in

    So this is what I have:

    I broke it down into two beams that are cantilevered

    So for each cantilever:

    load = (displacement * 81 * EI ) / (7*(Length/2)^4) Length/2 because I broken the simply supported beam into 2 cantilevered beams

    load = 4.399 lbf/in

    pressure = load/width of beam = 10.9 psi

    My problem is assuming the above is correct, why does it not work out the same when I break that linearly increasing load into the resultant force:

    resultant force for one of the cantilevered = .5 * L/2 * load = 3.872lbf

    So since we need 2 of those forces, one on each end 2/3rds of the way up the linearly increasing triangle, that would equal 7.744 lbf

    My problem is now if I say that force (7.744 lbf) / (L*width) = 5.442 psi

    5.442 psi doesn't equal the 10.9 psi???

    I know it is different but I did the same thing using a uniform pressure, where I broke the uniform pressure into two resultant forces and then divided those forces by the area and got the same pressure as the uniform pressure I had originally calculated.

    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Oct 22, 2008 #2
    Re: Statics

    Are both ends inclined? (Engineers love drawings so make one)
     
  4. Oct 22, 2008 #3
    Re: Statics

    Yes both ends are inclined. They are 0.005" higher than where the support it
     

    Attached Files:

  5. Oct 22, 2008 #4

    FredGarvin

    User Avatar
    Science Advisor

    Re: Statics

    You are confusing me with what you term a "linear increasing load." It looks like in one case you mean a simple distributed load and the other you are doing a distributed load that is a function of the distance down the length of the beam, i.e. a triangular shaped distributed load. I understand what you are doing and as a rough first guess I would have done what you did. What I don't understand is the second part you are comparing to.
     
  6. Oct 22, 2008 #5
    Re: Statics

    Yeah by linearly increasing load I mean a triangular distributed load that is increasing the further away from the support you get.

    So in the first part I have the load required for the triangular distributed load

    and then I found the resultant force of that triangular distributed load by using the formula
    .5*L*load. Load being in the form of (lbf/in)

    When I compare the pressure from the triangular distributed load and the pressure from the resultant forces, I get two different pressures.
     
  7. Nov 10, 2008 #6
    Re: Statics

    If you have triangular loads which increase towards the free ends on both sides of the middle support, then you have the equivalent of a cantilever with half the length.
    (Ref: http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_bc_cantilever.cfm)
    Let the Triangular load be 0 at the support, and p at the far end, then it is equivalent to adding a uniform load of p to the cantilever, and subtracting the inverse triangular load, which is p at the support and 0 at the far end.
    The deflections at the "free" end of the cantilever for the first and second cases are
    w1=pL4/8EI, and
    w2=pL4/30EI
    Subtracting
    w1-w2=pl4/EI*(1/8-1/30)=11pL4/120EI
    now
    L=1.761
    EI=731
    w1-w2=0.005
    we have
    11*p*L4/(120*EI)=0.005
    Solving for p,
    p=160039/38600=4.1461
    pressure
    =4.1461/.404
    =10.263 psi (at the far end where the load intensity is maximum)
    Very close to what you had at the beginning.
     
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