# Which shape does an elastic rod take as its ends are getting closer?

Which shape does an idealized elastic rod take whose ends are moved towards each other? (The rod is supposed to be straight in relaxed state and to be deformable perpendicular to its direction, i.e. bendable. It's supposed to be not deformable in its own direction, i.e. not stretchable.)

Mathematically, this means:

How do I find in a systematic way (without guessing) the function $$f:[0,a] \rightarrow \mathbb{R}^+$$ that fulfills these conditions ($$a > 0$$):

1. $$f(0) = f(a) = 0$$

2. $$\int_0^a \sqrt{1+ f'(x)^2}dx = L$$ with $$L > a$$, i.e. the length of the function graph seen as a curve is $$L$$

3. $$\int_0^a |f''(x)|dx$$, i.e. the overall "curvature" (= "energy") of the curve is minimal

Not obviously false guesses (but of course strongly depending on the right definition of "curvature" resp."energy"):

• $$f(x)$$ a circle arc

• $$f(x)$$ a parabola

• $$f(x)$$ a half period of the sine function

Note that in the case of the circle arc you cannot bring $$a$$ to $$0$$ closer than $$a = L/\pi$$ without leaving the realm of functions $$y=f(x)$$ - for parabolas and sine functions you can.

Assuming that @Rahul's answer is correct and that - with the correct curvature and energy - the rod is shaped like a circle arc, this is what an oscillating rod would look like (roughly) after you release its ends (but constrain them to the horizontal line):

Note that this doesn't look at all like a sinoidal wave.

Note further that you cannot simply apply Fourier transformation to the semicircle because the endpoints are not fixed:

• If you're talking about a rod in the plane, then you don't want it to be the graph of a single variable function $f:[0,1]\to \Bbb R$. You want it to be a parametrised curve $f:[0, 1]\to \Bbb R^2$. You still want $f(0) = f(1)$, though, although this time that actually means that the ends of the rod meet at a single point. Then you want $\int_0^1\sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}dt = L$, and if $f$ is a parametrisation by arc length, then I believe that $\int_0^1 |f''(t)|dt$ is indeed the right quantity to minimise. – Arthur Nov 8 '18 at 12:49
• If I remember my lessons from the theory of materials correctly, the deflection of the point on a canitilever beam is proportial to $x^3$ with $x$ being the distance from the fixed point. So the shape of the beam is actually a cubic curve. And that formula was obtained starting with a number of extreme approximations, like "deflection of the beam is much smaller than the length of the beam". I guess, trying to find the shape of the beam in a general case without approximations would be a task of epic proportions. So good luck! :) – Oldboy Nov 8 '18 at 13:31
• The usual idealization is that an elastic rod's bending energy is proportional to $\int \kappa^2\,\mathrm ds$, where $\kappa$ is the curvature and $\mathrm ds$ is the arc length. Under this assumption the optimal shape is the arc of a circle. – Rahul Nov 8 '18 at 13:55
• @Rahul Are you sure about that? Remember that the end points can meet at an angle without that constituting any elastic stress energy. I have no idea about the mechanics here, but my personal intuition about bendy things in nature says that if you don't exert torque on the end points yourself to make them meet up straight on, they will meet at an angle (I'm curious as to what that angle is; under idealised conditions it would seem to me that it doesn't actually depend on the length or the stiffness of the rod). – Arthur Nov 8 '18 at 14:46
• Please note: In my question the endpoints are not supposed to meet! – Hans-Peter Stricker Nov 8 '18 at 15:51

The energy as defined in the question, $$\int_0^1|f''(x)|\,\mathrm dx$$, is the total variation of the slope $$f'(x)$$. If $$f$$ is concave*, as the figure suggests is desired, then $$f''$$ is negative and the energy is simply $$f'(0)-f'(1)$$. The optimal curve must be (two edges of) a triangle, otherwise one can reduce $$f'(0)$$ without changing $$f'(1)$$ by replacing the curve with the triangle $$ABC$$ where $$A=(0,0)$$, $$C=(1,0)$$, $$BC$$ has slope $$f'(1)$$, and $$B$$ is chosen so $$|AB|+|BC|=L$$. Among all triangles with $$|AB|+|BC|=L$$, the energy $$f'(0)-f'(1)=\tan\angle A+\tan\angle C$$ is minimized by an isosceles triangle with sides $$L/2$$, $$L/2$$, and $$1$$.