Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 7

Problem 1: Part 7
Derive the equation (9.3.13) of the book chapter starting from equation (9.3.3).

Figure 4 shows the notation used to represent the motion of the master and slave nodes and the directors at the nodes.

Since the fibers remain straight and do not change length, we have
 * $$ \text{(5)} \qquad

\begin{align} \mathbf{x}_{i-}(t) & = \mathbf{x}_i(t) - \frac{1}{2} h_i^0~\mathbf{p}_i(t) \\ \mathbf{x}_{i+}(t) & = \mathbf{x}_i(t) + \frac{1}{2} h_i^0~\mathbf{p}_i(t) \end{align} $$ where $$\mathbf{x}_i$$ is the location of master node $$i$$, $$\mathbf{p}_i$$ is the unit director vector at master node $$i$$, and $$h_i^0$$ is the initial thickness of the beam.

Taking the material time derivatives of equations (5), we get
 * $$ \text{(6)} \qquad

\begin{align} \mathbf{v}_{i-}(t) & = \mathbf{v}_i(t) - \frac{1}{2} \frac{\partial }{\partial t}[h_i^0~\mathbf{p}_i(t)] = \mathbf{v}_i(t) - \frac{1}{2} h_i^0~\boldsymbol{\omega}_i(t)\times\mathbf{p}_i(t) \\ \mathbf{v}_{i+}(t) & = \mathbf{v}_i(t) + \frac{1}{2} \frac{\partial }{\partial t}[h_i^0~\mathbf{p}_i(t)] = \mathbf{v}_i(t) + \frac{1}{2} h_i^0~\boldsymbol{\omega}_i(t)\times\mathbf{p}_i(t) \end{align} $$ where $$\boldsymbol{\omega}_i$$ is the angular velocity of the director.

From equations (5) we have
 * $$ \text{(7)} \qquad

\begin{align} \mathbf{p}_i(t) & = -\cfrac{2}{h_i^0}[\mathbf{x}_{i-}(t) - \mathbf{x}_i(t)] = -\cfrac{2}{h_i^0}\left[(x_{i-} - x_i)~\mathbf{e}_x + (y_{i-} - y_i)~\mathbf{e}_y\right] \\ \mathbf{p}_i(t) & = \cfrac{2}{h_i^0}[\mathbf{x}_{i+}(t) - \mathbf{x}_i(t)] = \cfrac{2}{h_i^0}\left[(x_{i+} - x_i)~\mathbf{e}_x + (y_{i+} - y_i)~\mathbf{e}_y\right] \end{align} $$ where $$(\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z)$$ is the global basis.

In terms of the global basis, the angular velocity is given by

\boldsymbol{\omega}_i= \omega_i~\mathbf{e}_z ~. $$

Therefore,
 * $$\text{(8)} \qquad

\begin{align} \boldsymbol{\omega}_i\times\mathbf{p}_i & = -\cfrac{2}{h_i^0} \begin{vmatrix} \mathbf{e}_x & \mathbf{e}_y & \mathbf{e}_z \\ 0& 0 & \omega_i \\ x_{i-}-x_i& y_{i-}-y_i& 0 \end{vmatrix} = -\cfrac{2}{h_i^0} \left[ - \omega_i (y_{i-}-y_i)~\mathbf{e}_x + \omega_i (x_{i-}-x_i)~\mathbf{e}_y \right] \\ \boldsymbol{\omega}_i\times\mathbf{p}_i & = \cfrac{2}{h_i^0} \begin{vmatrix} \mathbf{e}_x & \mathbf{e}_y & \mathbf{e}_z \\ 0& 0 & \omega_i \\ x_{i+}-x_i& y_{i+}-y_i& 0 \end{vmatrix} = \cfrac{2}{h_i^0} \left[ - \omega_i (y_{i+}-y_i)~\mathbf{e}_x + \omega_i (x_{i+}-x_i)~\mathbf{e}_y \right] ~. \end{align} $$

Substituting equation (8) into equations (6), we get
 * $$ \text{(9)} \qquad

\begin{align} \mathbf{v}_{i-} & = \mathbf{v}_i - \omega_i (y_{i-}-y_i)~\mathbf{e}_x + \omega_i (x_{i-}-x_i)~\mathbf{e}_y \\ \mathbf{v}_{i+} & = \mathbf{v}_i - \omega_i (y_{i+}-y_i)~\mathbf{e}_x + \omega_i (x_{i+}-x_i)~\mathbf{e}_y ~. \end{align} $$

Let the velocity vectors be expressed in terms of the global basis as

\begin{align} \mathbf{v}_i & = v_i^x~\mathbf{e}_x + v_i^y~\mathbf{e}_y \\ \mathbf{v}_{i-} & = v_{i-}^x~\mathbf{e}_x + v_{i-}^y~\mathbf{e}_y \\ \mathbf{v}_{i+} & = v_{i+}^x~\mathbf{e}_x + v_{i+}^y~\mathbf{e}_y ~. \end{align} $$

Then equations (9) can be written as
 * $$ \text{(10)} \qquad

\begin{align} v_{i-}^x~\mathbf{e}_x + v_{i-}^y~\mathbf{e}_y & = [v_i^x - \omega_i (y_{i-}-y_i)]~\mathbf{e}_x + [v_i^y + \omega_i (x_{i-}-x_i)]~\mathbf{e}_y \\ v_{i+}^x~\mathbf{e}_x + v_{i+}^y~\mathbf{e}_y & = [v_i^x - \omega_i (y_{i+}-y_i)]~\mathbf{e}_x + [v_i^y + \omega_i (x_{i+}-x_i)]~\mathbf{e}_y ~. \end{align} $$

Therefore, the components of the velocity vectors are
 * $$ \text{(11)} \qquad

\begin{align} v_{i-}^x & = v_i^x - \omega_i (y_{i-}-y_i) \\ v_{i-}^y & = v_i^y + \omega_i (x_{i-}-x_i) \\ v_{i+}^x & = v_i^x - \omega_i (y_{i+}-y_i) \\ v_{i+}^y & = v_i^y + \omega_i (x_{i+}-x_i) ~. \end{align} $$

Then, the matrix form of equations (11) is

\begin{bmatrix} v^x_{i-} \\ v^y_{i-} \\ v^x_{i+} \\ v^y_{i+} \end{bmatrix} = \begin{bmatrix} 1 & 0 & - (y_{i-}-y_i) \\ 0 & 1 &(x_{i-}-x_i)\\ 1 & 0 & - (y_{i+}-y_i) \\ 0 & 1 &(x_{i+}-x_i) \end{bmatrix} \begin{bmatrix} v_i^x \\ v_i^y \\ \omega_i \end{bmatrix} $$ or

\begin{bmatrix} \mathbf{v}_{i-} \\ \mathbf{v}_{i+} \end{bmatrix}^{\text{slave}} = \mathbf{T}_i~\dot{\mathbf{d}}_i^{\text{master}} ~. $$