### Equilibrium configurations of elastic torus knots (n,2)

**Starostin, E ***(University College London)*

Thursday 06 December 2012, 11:30-11:45

Seminar Room 1, Newton Institute

#### Abstract

We study equilibria of braided structures made of two elastic rods with their centrelines remaining at constant distance from each other. The model is geometrically exact for large deformations. Each of the rods is modelled as thin, uniform, homogeneous, isotropic, inextensible, unshearable, intrinsically twisted, and to have circular cross-section. The governing equations are obtained by applying Hamilton's principle to the action which is a sum of the elastic strain energies and the constraints related to the inextensibility of the rods. Hamilton's principle is equivalent to the second-order variational problem for the action expressed in reduced strain-like variables. The Euler-Lagrange equations are derived partly in Euler-Poincare form and are a set of ODEs suitable for numerical solution.

We model torus knots (n,2) as closed configurations of the 2-strand braid. We compute numerical solutions of this boundary value problem using path following. Closed 2-braids buckle under increasing twist. We present a bifurcation diagram in the twist-force plane for torus knots (n,2). Each knot has a V-shaped non-buckled branch with its vertex on the twist axis. There is a series of bifurcation points of buckling modes on both sides of each of the V-branches. The 1st mode bifurcation points for n and ną4 are connected by transition curves that go through (unphysical) self-crossing of the braid. Thus, all the knots turn out to be divided into two classes: one of them may be produced from the right-handed trefoil and the other from the left-handed. Higher-mode post-buckled configurations lead to cable knots.

It is instructive to see how close our elastic knots can be tightened to the ideal shape. For the trefoil knot the tightest shape we could get has a ropelength of 32.85560666, which is remarkably close to the best current estimate. Careful examination reveals that the solution is free from self-intersections though the contact set remains a distorted circle.

#### Video

**The video for this talk should appear here if JavaScript is enabled.**

If it doesn't, something may have gone wrong with our embedded player.

We'll get it fixed as soon as possible.

## Comments

Start the discussion!