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Reactions caused from self weight of an inclined beam. Same as non-inclined beam?

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data
    I have a beam on an incline at angle θ. It is uniformly loaded with "W", which represents the self-weight of the beam. Reactions occur at each end. The beam is simply supported. The scenario is drawn out below:


    3. The attempt at a solution

    This problem seems so simple, but I am getting confused. Typically, when there is a uniformly distributed load on a horizontal beam, the reactions at R1 and R2 would be equal. In this scenario however, I would expect R1 to be carrying a greater load. The reason for this is because if the inclination of this beam were to increase to the point in which is was essentially a column, the entire weight would be acting on one end.

    So the basic question is, does R1y = R2y in this scenario?

    Sorry if this question seems trivial, but I'm having a bit of a brain cramp right now.

    Thanks in advance.
  2. jcsd
  3. Aug 4, 2010 #2
    Also just to add,

    Assume R2 to be a roller. You can also assume the pin at R1 to have zero moment. Therefore through summation of moments at point R1, the reaction at R2y would be WL/2, which through summation of forces would make R1y = WL/2 as well. However, this just doesn't seem right to me. I figured R1 would take a greater portion of the load. Does this make sense?

  4. Aug 4, 2010 #3
    The supports a sdrawn by Probert are indeterminate. However, If R2 is a roller resisting vertical force only, then R1y = R2y =WL/2. R1x = 0. What distinguishes this from a horizontal beam is that in the lower part of the beam has a normal compression force, and in the upper part of the beam is an axial tension force. You have to believe the mechanics. However, if the roller at 2 was R2y=0 and resisted only forces in the x direction, then there would be an axial compression throughout, with R1x=R2x. The unanswerable question is that real supports are not perfect. What difference does that make? Can you at least determine the upper and lower bounds of the reactions?
    Probert asks a really good question, but this is a standard case of 'common sense' being wrong.
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