What curve does a rod form when bent to intersect 3 or more points?

Question

Suppose that we have a sufficiently thin, flexible cylindrical rod of length $L$ made from a homogeneous, isotropic material, and that initially [at rest?] the central axis of the rod is a straight line segment.

We select $n$ points, $\mathbf x_1,\ldots,\mathbf x_n$ so that the sum of the distances between consecutive points is no greater than $L$ - i.e.

$$\sum_{i=1}^{n-1}\Vert\mathbf x_i-\mathbf x_{i+1}\Vert\le L$$

Assuming that energy (due to tension, stress, and such) is minimized, if the rod is deformed so that its cenral axis intersects each of the points $\mathbf x_1,\ldots,\mathbf x_n$ (and neither endpoint is located at one of the points, probably), what is the curve formed by the central axis of the rod?

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This is not a homework question, it's for a gardening project. I just figure that, since structural engineering and differential calculus have simultaneously existed for at least two centuries, this exact problem - or a nearly identical problem - has probably been extensively studied already.

Answer 1

The shape is that of a cubic spline. Each segment between two points is a 3rd order polynomial with matching deflection and slope at the nodes.

Between nodes $i$ and $i+1$ with distance $h_i = {\bf x}_{i+1} - {\bf x}_i$ the shape is described by a parameter $t=0\ldots 1$

$$ y(t) = (1-3 t^2+2t^3) {\bf y}_i + t^2 (3-2 t) {\bf y}_{i+1} + h_i \left( t (1-t)^2 {\bf y'}_i + t^2 (t-1) {\bf y'}_{i+1}\right) $$

The nodal displacements are zero ${\bf y}_{i}=0$ and ${\bf y}_{i+1}=0$, and the nodal slopes are unknown ${\bf y'}_i$ and ${\bf y'}_{i+1}$.

To actually deform the rod, you need some kind of loading, and since the nodes are fixed, you need to apply torques at each node to deform it. The torque might be zero in the middle and some value at one or both ends, or it might be defined at every node.

For nodes $i=1\ldots n-1$ the nodal torque $\tau_i$ relates to the slopes with

$$ \frac{\tau_i}{E I} = \frac{2}{h_i^2} \left( 3 ( {\bf y}_{i+1} - {\bf y}_i) - h_i ( {\bf y'}_{i+1} - 2{\bf y'}_i ) \right) $$

for the last segment, the torque is

$$ \frac{\tau_n}{E I} = \frac{2}{h_{n-1}^2} \left( 3 ( {\bf y}_{n-1} - {\bf y}_n) + h_{n-1} ( {\bf y'}_{n-1} + 2{\bf y'}_n ) \right) $$

Where $E\,I$ is the flexural rigidity of the rod.

The above is a system of $n$ equations to be solved for the $n$ unknown slopes. Note that typical static equilibrium requires the net torque to be zero $\sum_i M_i = 0$