Lecture 06 - Cosserat Rod

#math

We define what we call the rigid body Poisson bracket:

{f, g} (ω) = ω \cdot (\frac{\partial f}{\partial ω} \times \frac{\partial g}{\partial ω})

We want to prove the Jacobi Identity:

Proof

Let $h$ satisfy $\frac{\partial h}{\partial ω} = ω$ , then:

\begin{aligned} {f, g} (ω) & = - \frac{\partial h}{\partial ω} (\frac{\partial f}{\partial ω} \times \frac{\partial g}{\partial ω}) \\ = | \begin{array}{c} \frac{\partial f}{\partial ω_{1}} & \frac{\partial f}{\partial ω_{2}} & \frac{\partial f}{\partial ω_{3}} \\ \frac{\partial g}{\partial ω_{1}} & \frac{\partial g}{\partial ω_{2}} & \frac{\partial g}{\partial ω_{3}} \\ \frac{\partial h}{\partial ω_{1}} & \frac{\partial h}{\partial ω_{2}} & \frac{\partial h}{\partial ω_{3}} \end{array} | \\ = \frac{\partial (f, g, h)}{\partial (ω_{1}, ω_{2}), ω_{3}} \\ = {f, g, h} & (Nambu Structure) \end{aligned}

Which indeed does satisfy the Jacobi bracket.

Eulers Equation of Incompressible Fluid

$\begin{aligned} \partial_{t} u + u \cdot \nabla u & = - \nabla p \\ \nabla \cdot u & = 0 \end{aligned}$

Cosserat Rod

Rod $r (s)$ is embedded in spatial frame as the vector space spanned by the neighborhood of orthonormal triads $(e_{1}, e_{2}, e_{3})$ .
We have the body frame $(d_{1} (s), d_{2} (s), d_{3} (s))$ :

\begin{aligned} d_{i} \cdot d_{j} & = δ_{i j} \\ d_{i} \times d_{j} & = ϵ_{i j k} d_{k} \end{aligned}

We have $d_{i} = R e_{i}$ (that is the body frame is a rotation of the orthonormal triad), thus:

\begin{aligned} d_{i}^{'} & = R^{'} e_{i} \\ = R^{'} R^{- 1} d_{i} \\ = \hat{u} d_{i} \\ = u \times d_{i} \end{aligned}

Where $u_{i} = \frac{1}{2} ϵ_{i j k} d_{j}^{'} \cdot d_{k}$ .

Second set of strain associated with shearing and extensions

\begin{aligned} v & = R^{T} r^{'} \\ r^{'} & = d_{3} \end{aligned}

If the rod is inextensible then $| r^{'} | = 1$
If the rod is unshearable we have:

\begin{aligned} v_{1} & = v \cdot d_{1} = 0 \\ v_{2} & = v \cdot d_{2} = 0 \end{aligned}

The strain energy density function associated with the rod becomes:

J (v, u) = \int_{0}^{L} W (v - \hat{v}, u - \hat{u}) d s

where $\hat{v}$ are from the reference configuration.

Doing the variational calculus $\partial J = \int_{0}^{L} [\frac{\partial w}{\partial v} \cdot \partial v + \frac{\partial W}{\partial u} \cdot \partial u] d s$ , we obtain:

\begin{aligned} n^{'} & = 0 \\ m^{'} + r^{'} \times n & = 0 \\ {(\begin{array}{c} m \\ n \end{array})}^{'} & = - J (m, n) \nabla H \end{aligned}

Thus we obtain finally our Cosserat Rod equation:

\begin{aligned} n^{'} + u \times n & = 0 \\ m^{'} + u \times m + v \times n & = 0 \end{aligned}