Maxwell's Reciprocal Theorem

Hi all!
I'm working on an elasticity problem which, as I understand it, can be
solved using Maxwell's Reciprocal Theorem. The problem is as follows:
Given an inextensible ring subjected to a pair of diametrically opposing
forces, show that the area bound by the ring remains unchanged. The
ring can deform only by bending. Assume small deformation. (Note:
This is all the information I have...)
F /---\ F
\---/
Ring
My understanding of this theorem is that so long as the system is linear
and elastic, it can be used to relate one loading scenario to another.
In this particular case, we know nothing about the applied forces, other
than the fact they are equal and opposite and serve to stretch (or
rather bend) the ring.
Therefore, I expect the need for another generalized force (like gravity
or pressure or something) in order to make the theorem work.
Unfortunately, I really have no idea what that generalized force would be.
Anybody have any ideas how to approach this problem? I appreciate any
input!
Thanks,
Kevin
Reply to
Kevin Dressel
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Why do you need another force? It's a planar problem and the forces are equal and opposite.
Make use of the symmetry plane in the middle of the ring.
Reply to
Jeff Finlayson
I see your puzzlement, when I consider two experiments: 1) an elastic band initially laid out in a circle, can collapse to two straight adjacent lines containing no area if stretched. So evidently stretching is not to be countenanced. 2) I see that a circle contains more area for its circumference than any other shape. So if the shape is deformed by opposed diametric forces, its non circular shape would occupy less space if its perimeter remained constant in length.
Taking one and two together, I conclude: if a ring has constant perimeter length if distorted, it will include less area.
This conclusion is opposed to what you are asked to demonstrate, so I imagine that you have omitted some dimension of the puzzle that is needed to make sense of it.
Brian W
Reply to
Brian Whatcott
Well, first, let me thank you all for the responses!
As for the problem, however, I'm not sure how you could use the applied force and not contradict the reciprocal theorem. I thought the whole idea behind it was to use some other force to solve for the effects of the applied force. For instance, the only other example of using this theorem that I have seen involved an elastic cone for which we needed to measure the volume change due to gravity. In that particular case, the problem instructed to use hydrostatic pressure as the "phantom" force.
Frankly, I'm still at a loss here...
Regards, Kevin
Reply to
Kevin Dressel
Dear Kevin Dressel:
...
Kevin, buddy. This is surely going to hurt when you finally get it...
If the ring is not accelerating, and there is one force applied, then there is an equal and opposite applied as well. Note that your original diagram had two "F"s... one on the left and one on the right.
NOW do you get it?
David A. Smith
Reply to
N:dlzc D:aol T:com (dlzc)
** Actually there are 2 symmetry planes for this ring problem. You can use 1/4 of the ring and apply F/2 to it. Like this:
____ Guided end / (Pfiction) Well, first, let me thank you all for the responses!
OK, I see what you mean now. You want to apply a fictional force to use with an energy problem. I could figure out the deflections, but not sure how to get to the enclosed volume (area * unit depth).
Jeff F.
Reply to
Jeff Finlayson
Too late. It already hurts...
You're absolutely right, if the ring isn't accelerating (and it's not), than there has to be a counteracting force on the ring. However, I regard the system of two forces as the known applied force (it was, afterall, stated in the problem that the two forces acted on the ring). In my mind, then, there has to be some other force to which we can relate to area (as I am trying to prove that dArea = 0). What isn't my brain telling me here???
Reply to
Kevin Dressel
The funicular figure is a long thin double band, each of length Pi a. If central points are fixed, equilibrium form is a rhombus. The inextensional pure membrane situation gives no useful answer for concentrated forces!! For distributed pressure, we have circle,catenary etc. If inextensionality is dropped, you can solve normally with first order accuracy, like i.e., 1/a + M/EI=y''/(1+y'^2)^1.5.. no need for Maxwell-Betti theorem. Cheers.
Reply to
Narasimham G.L.

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