More worked problems.
Cantilever with uniform loading. Now there is a weight W/n at each point on the beam, not just a weight W at the end, so the potential energy changes to
We’ll evaluate the sum with an HP50g this time (simply because it has smaller fonts, to display the whole result at once)
Then we just substitute into y
substitute for m (= nx/L), take the limit as n -> infinity, and we get
(Note: I didn’t plan that ahead, and forgot that on the HP50g, they want you to use the variable x for taking limits. So before I take the limit, I swap X with Z, and N with X)
(I think the extra L I get in my denominator as compared to Wikipedia is that they define q=W/L; otherwise, it doesn’t seem their units work out)
Beam supported on both sides, but not rigidly (it is possible to bend there), with a weight directly in the middle.
We might not want to be too clever by half, but we barely need to do anything to solve this problem. We almost already solved it.
There’s a joke, not really a good one (in that it’s not actually funny) about a mathematician boiling water. How do you boil water in a pot that’s on the floor? Fill it up with water, put it on the stove, etc. How do you boil water in a pot that’s on a table? Put it on the floor, since that reduces it to a solved problem.
Anyway, for this beam problem, we know that the beam shape must such that one half of it is equal to the half from the cantilever problem, only flipped inside out (so the flat part is in the middle of the beam rather than against the boundaries) and upside down. We know this because the elastic energy in the beam is independent of orientation, and this sort of flip guarantees that the beam and its first derivative are continuous in the middle, and also the cantilever problem has slope but no second derivative at its end point, matching this problem’s other boundary condition.
The only thing is that the effective weight is reduced by half in the earlier formula (because, with two “cantilevered beams”, you get twice as much elastic energy per gravitational energy as before). (And of course L is replaced with L/2)
So the answer is