Does R1y = R2y with a Beam on an Inclined Plane?

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SUMMARY

The discussion centers on the reactions at supports R1 and R2 of a uniformly loaded beam on an inclined plane at angle θ. The participants conclude that under the given conditions, R1y equals R2y, both equating to WL/2, where W is the weight of the beam and L is its length. The confusion arises from the expectation that R1 would carry a greater load due to the inclination, but the mechanics dictate that the load distribution remains equal. The analysis emphasizes the importance of understanding the mechanics of inclined beams and the behavior of supports.

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Probert
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Homework Statement


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:

LoadScenario.jpg


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.
 
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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?

Thanks
 
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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