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

The gradient:

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)

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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

A better version of that joke:

A mathematician and a physicist were asked the following question: Suppose you walked by a burning house and saw a hydrant and a hose not connected to the hydrant. What would you do?

P: I would attach the hose to the hydrant, turn on the water, and put out the fire.

M: I would attach the hose to the hydrant, turn on the water, and put out the fire.

Then they were asked this question: Suppose you walked by a house and saw a hose connected to a hydrant. What would you do?

P: I would keep walking, as there is no problem to solve.

M: I would disconnect the hose from the hydrant and set the house on fire, reducing the problem to a previously solved form.