What shape does an elastic rod take when both ends are dragged to the same point? Suppose we have an ideal elastic rod of some kind, where the energy at a point along the rod is proportional to the square of the curvature, and we drag the ends of this rod so that they touch, and the rest of the rod is now making some kind of teardrop shape off to one side.
Like this:

Is there an equation for the exact shape the rod will take?
I think another way to approximate this is by dragging two opposite edges to touch and seeing the shape the cross-section of the paper takes.
I apologize if this question isn't very clear. I'm not familiar with the official terminology in this field.
 A: Assuming the rod is inextensible and elastic with an elastic energy proportional to the square of the curvature, the curve you are looking for is a particular case of Euler's elastica. Therein, check equation (12) with the parameters given for the "lemnoid" and Euler's figure (8). See also here, the case of the pseudo-lemniscate.
A: In the example that you present you already know the shape of the deformed configuration. If you compute a parameterization from the original configuration to the deformed one you would get "an equation". Now, I would guess that, in general, this displacement field would not satisfy equilibrium and it would need a body force to obtain such a shape.
If your equation is referring to the case where you have the flat case and you want to apply forces/displacements on the extremes to obtain a shape that is close to the one that you are showing, then you would need so solve for a non-linear partial differential equation subject to some boundary conditions. I general, you can't solve these differential equations analytically and you need a numerical methods such as the finite element method.
A: By Euler-Bennoulli Law,
$$ \kappa= \frac{M}{EI}=\frac{d\phi}{ds}=\frac{F y}{EI}=\frac{y}{c^2}$$
where $\kappa$ is curvature, $\phi$ slope, $M$ bending moment proportional to $y$ coordinate, $EI$ flexural rigidity, $F$ are two equal opposite forces applied at towards origin along x-axis and $c= \sqrt{EI/F},$ a length constant.
The deformations are elastic. the elastic rod rebounds back to original straight shape on removal of $Fs$.
$$ \frac{dy}{ds}=\sin \phi ;\;\frac{dx}{ds}=\cos\phi ;$$
The Elasticas are three types: progressive, regressive, stationary closed loop shapes which are relevant to this question.
The ode is intrinsic.
$$ \frac{d^2\phi}{ds^2}= \frac{\sin \phi}{c^2}$$
The deformed elastic shapes can be expressed in terms of elliptic integrals in closed form solutions, not given here. The numerical solution of ode for closed loops finds an approximate relation between loop length $smax$ and $c$ given here by trial and error,$(10,2.514)$ respectively for closed loops.
The semi-loops  are all geometrically similar and are capable of being defined using this specific $c/smax =0.2514 $ ratio obtained on Mathematica.. as a Boundary Value Problem.

EDIT1:
Taking the following ode
$$\phi'(s)=\frac{2 r(s)}{a^2}$$
with BCs $a=\sqrt 2,  c=1, \phi_{initial}=0.86053 $
and ode used $r$ in place of $y$
$$ \cos \phi=\frac{c^2-r^2(s)}{a^2} $$
still using trial/error $\phi_{initial}\; $  value we get one closed loop using code given below.

