What shape does an elastic rod take when both ends are dragged to the same point?

Question

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.

Do you mean what happens when you pinch a rubber band? Perhaps a photo/drawing would help — AccidentalTaylorExpansion, Jun 01 '22 at 21:37
Neat question. The answer will depend on the angle of contact. In the extreme case where the flat ends are in full contact you obviously get a circle ( O-ring). I'm not quite sure how you can get the angle to be zero without applying a pinching force and thus having a line-segment contact region. — Carl Witthoft, Jun 02 '22 at 14:24
@CarlWitthoft The angle is whatever the ends want to be at. Imagine we're sticking pins in the ends and dragging them like that so they can still rotate freely. The thickness of the rod is negligible, so we don't have to worry about its effect on the final shape. That's why I brought up the paper example. — zucculent, Jun 03 '22 at 17:32
It’s going to depend on what you mean by “elastic.” A search term is buckling instability. I suspect, but haven’t shown, that quadratic deformation energy $U=\frac12 k\Delta x^2$ is minimized if the curvature of the rod is constant, which would make a circle. — rob, Jun 03 '22 at 18:03
It sounds like you are interested in the pure bending equation and its derivation. http://emweb.unl.edu/negahban/em325/11-Bending/Bending.htm — Stevan V. Saban, Jun 03 '22 at 19:29

score 3 · Answer 1 · answered Jun 17 '22 at 15:56

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.

score 0 · Answer 2 · answered Jun 03 '22 at 17:25

0

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.

answered Jun 03 '22 at 17:25

nicoguaro

1,671

This is a comment rather than a quantitative answer – Carl Witthoft Jun 03 '22 at 17:27
I don't know the final shape. The diagram only has a crude representation of what I expect the final shape to be like. – zucculent Jun 03 '22 at 17:32
@CarlWitthoft This partial answer is correctly posted as an answer. – rob Jun 03 '22 at 17:37
@zucculent, do you know the boundary conditions then? – nicoguaro Jun 03 '22 at 18:36

Narasimham · Answer 3 · 2022-07-28T22:35:42.530

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.

What shape does an elastic rod take when both ends are dragged to the same point?

3 Answers3