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Beam under a distributed load

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A simple beam is under a distributed load q=c*sine(n*pi*x/L)? if there are two pivots at the end points supporting it, what will be the reactionary force on each one of them?

    Here, L is the length of the beam and x=0 is the leftmost point. 0[tex]\leq[/tex]x[tex]\leq[/tex]L

    The figure looks like this.

    ___________________
    ^............................^

    2. Relevant equations



    3. The attempt at a solution

    I cannot figure out how to approach to this problem. can you please help me ?
     
  2. jcsd
  3. Nov 14, 2008 #2

    PhanthomJay

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    How's your calculus (better than mine, I hope)? Start by integrating the load distribution from 0 to L to solve for the total load, which acts at the centroid of the sinusoidal load distribution. Then what?
     
  4. Nov 15, 2008 #3
    yes, but the problem is what to do with 'n'? it can change also....wat if n is odd and n is even?
     
  5. Nov 15, 2008 #4

    PhanthomJay

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    Yes, good point, that n makes it more difficult. When n is an integer greater than 1, the distributed load curve crosses the x axis, so integrating the load curve from end to end will not help in determining the reactions. It looks like you have to perform separate integrations between n segments that are each (1/n)L in length, then place the load at the centroid of each section to get the end reactions. There's probably a formula to calculate this, but I don't know what it is.
     
  6. Nov 15, 2008 #5
    When n is an integer greater than 1, the distributed load curve crosses the x axis can u tell me in detail this point
     
  7. Nov 15, 2008 #6

    PhanthomJay

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    it crosses at q=0, that is, when sin(n)(pi)x/l = 0, which occurs at x=0, and l/2 when n=2, at x=0, l/3, and 2l/3 when x=3, and in general, at x=0, l/n, .......(n-1)l/n.
     
  8. Nov 15, 2008 #7
    can u tell me where will be the centroid means how to calculate centroid for this case?
     
  9. Nov 16, 2008 #8
    where is the centroid for this case?
     
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