I believe this problem is becoming more and more complex than was intended, but this is what happens when a problem is underspecified.
Xyius said:
What do you mean by "equally stiff"?
By that, I mean each
elastic support will deform (compress) the same amount under the same load, that is, they have the same effective 'spring' constant, k.
What is the difference between scenario 1 and 3?
Scenario 1 is a rigid beam on elastic supports. Scenario 3 is a rigid beam on rigid supports.
Would the reactions still be mg/3 even if a is not equal to b?
For scenario 1, yes.
How would I go about showing this?
The supports deflect equally under the rigid beam's weight , and thus, the force in the supports must be equal. If the forces in the supports are not equal, then the the compressions of each support would not be equal (visualize the beam rotating about the left support, causing greater compression and thus greater load on the right support than the middle support), and equilibrium cannot be maintained, making such rotation and unequal compressions of the supports not possible, for supports of the same 'spring constant'.
I am assuming that my professor is referring to scenario 3 so I am going to go with it is not solvable. I just don't know how to show it other than summing the forces of each point where the supports are and showing that there are too many unknowns.
I guess that's what Spinnor said a few posts ago..the problem is not solveable unless you make some assumptions.
Just out of curiosity, how can it be not solvable? Shouldn't there exist a solution since the reactions DO occur and are each producing some sort of force? Or do you mean its not solvable without going into more advanced topics?
For scenario 3, I meant to say that there are an infinite number of solutions, and thus no specific solution, as long as the equations of equilibrium are satisfied. For example, the left and right supports might have a force reaction of mg/2 each, and the middle support reaction be 0. Or all support forces could be mg/3. Or 0 at the left and right supports, mg at the center support. They all satisfy equilibrium. For scenario 1 and 2, there is a specific solution as noted. But for case 3, you get into infinities...yes, the reactions do produce some sort of force in the real world, but it depends on what is happening in real life...completely rigid beams and rigid supports do not exist in our Universe.
You sure?