Beam on an inclined plane

[SeniorGara]

TL;DR

I analyzed a beam resting on two supports attached to an inclined plane. My equations suggest equal reaction forces, but my intuition says otherwise. Is my reasoning correct? What are the conditions when the beam starts sliding or rotating?

Hello!

I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below.

Here is how I wrote the condition of equilibrium forces:
$$
\begin{cases}
F_{g\parallel}=F_{t1}+F_{t2}, \\
F_{g\perp}=F_{r1}+F_{r2}
\end{cases}.
$$
On the other hand, the equilibrium of moments relative to the center of mass can be written as:
$$
\frac{1}{2}LF_{r1}=\frac{1}{2}LF_{r2},
$$
which of course leads to the equation:
$$
F_{r1}=F_{r2}.
$$
As I mentioned above, I'm not sure that this result is correct. So my question is: is everything fine and my intuition wrong, or did I make some mistake somewhere?If everything is correct, I have a second question: as alpha increases, the system reaches a point where the beam either starts to move along the plane or begins to rotate. What should the conditions look like for the first and the second case?

 

[Baluncore]

If everything is correct, I have a second question: as alpha increases, the system reaches a point where the beam either starts to move along the plane or begins to rotate.

You show two fulcrums that carry the load. The load will not slide over a fulcrum, that needs to be a roller.
If FR1 and FR2 are due to weight, then they should be vertical, not diagonal.

 

[wrobel]

https://en.wikipedia.org/wiki/Statically_indeterminate

 

[Steve4Physics]

View attachment 365169
Here is how I wrote the condition of equilibrium forces:
$$
\begin{cases}
F_{g\parallel}=F_{t1}+F_{t2}, \\
F_{g\perp}=F_{r1}+F_{r2}
\end{cases}.
$$

Note that $\vec {F_{g\parallel}}$ and $\vec {F_{g\perp}}$ are the wrong way round on your diagram. If you correct them, the above equations make sense.

On the other hand, the equilibrium of moments relative to the center of mass can be written as:
$$
\frac{1}{2}LF_{r1}=\frac{1}{2}LF_{r2},
$$
which of course leads to the equation:
$$
F_{r1}=F_{r2}.
$$

That part is ok.

The problem is that you have considered only the special case where the beam is not in a state of tension or compression. In this special case $F_{t1} = F_{t2}$. But there are other possibilities - this is what makes the system statically indeterminate (see @wrobel's post #3).

Minor edit.

 

[A.T.]

I have a second question: as alpha increases,

I assume alpha is the inclination angle.

I have a second question: as alpha increases, the system reaches a point where the beam either starts to move along the plane or begins to rotate. What should the conditions look like for the first and the second case?

To slide along both supports the beam must be longer than shown. It will happen when alpha > atan(static friction coefficient).

But what exactly do you mean by "begin to rotate"? Are you talking about the end of the beam reaching the upper support and falling off? Or are you talking about alpha > 90° and the beam sticking to one of the supports?