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

In summary: If the inclination of the beam increases so that the reactions at R1 and R2 become unequal, the beam will fail. The support at R1 will be unable to withstand the increased load and will fail.
  • #1
Probert
2
0

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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  • #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?

Thanks
 
  • #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.
 

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

What is R1y and R2y in the context of a beam on an inclined plane?

R1y and R2y refer to the vertical reaction forces at the two ends of the beam when it is placed on an inclined plane. These forces are perpendicular to the surface of the plane and are essential in determining the stability and equilibrium of the beam.

How is R1y related to R2y on an inclined plane?

R1y and R2y are equal in magnitude when the beam is in a state of static equilibrium on an inclined plane. This means that the beam is not moving or rotating and all forces acting on it are balanced.

Does the angle of inclination affect the relationship between R1y and R2y?

Yes, the angle of inclination does affect the relationship between R1y and R2y. As the angle increases, the magnitude of R1y and R2y also increases, but they remain equal to each other as long as the beam is in static equilibrium.

What is the significance of R1y = R2y in a beam on an inclined plane?

When R1y = R2y, it means that the beam is perfectly balanced and there is no net force acting on it. This is an important condition for the stability and safety of the structure.

How is the weight of the beam related to R1y and R2y on an inclined plane?

The weight of the beam is directly related to R1y and R2y on an inclined plane. These reaction forces are equal in magnitude to the weight of the beam, as the beam is in a state of static equilibrium. This relationship is crucial in determining the forces acting on the beam and ensuring its stability.

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