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

Problem 1: Part 7

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

File:CBBeamMotion.png Figure 4. Notation for motion of the continuum-based beam.

Since the fibers remain straight and do not change length, we have

$\text{(5)}\begin{array}{cc}{𝐱}_{i-}(t)& ={𝐱}_i(t)-\frac{1}{2}{h}_i^0{𝐩}_i(t)\\ {𝐱}_{i+}(t)& ={𝐱}_i(t)+\frac{1}{2}{h}_i^0{𝐩}_i(t)\end{array}$

where ${𝐱}_i$ is the location of master node $i$, ${𝐩}_i$ 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

$\text{(6)}\begin{array}{cc}{𝐯}_{i-}(t)& ={𝐯}_i(t)-\frac{1}{2}\frac{\partial }{\partial t}[h_i^0{𝐩}_i(t)]={𝐯}_i(t)-\frac{1}{2}{h}_i^0{\mathit{\omega }}_i(t)\times {𝐩}_i(t)\\ {𝐯}_{i+}(t)& ={𝐯}_i(t)+\frac{1}{2}\frac{\partial }{\partial t}[h_i^0{𝐩}_i(t)]={𝐯}_i(t)+\frac{1}{2}{h}_i^0{\mathit{\omega }}_i(t)\times {𝐩}_i(t)\end{array}$

where ${\mathit{\omega }}_i$ is the angular velocity of the director.

From equations (5) we have

$\text{(7)}\begin{array}{cc}{𝐩}_i(t)& =-\frac{2}{h_i^0}[{𝐱}_{i-}(t)-{𝐱}_i(t)]=-\frac{2}{h_i^0}[(x_{i-}-x_i){𝐞}_x+(y_{i-}-y_i){𝐞}_y]\\ {𝐩}_i(t)& =\frac{2}{h_i^0}[{𝐱}_{i+}(t)-{𝐱}_i(t)]=\frac{2}{h_i^0}[(x_{i+}-x_i){𝐞}_x+(y_{i+}-y_i){𝐞}_y]\end{array}$

where $({𝐞}_x,{𝐞}_y,{𝐞}_z)$ is the global basis.

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

${\mathit{\omega }}_i={\omega }_i{𝐞}_z.$

Therefore,

$\text{(8)}\begin{array}{cc}{\mathit{\omega }}_i\times {𝐩}_i& =-\frac{2}{h_i^0}\left|\begin{array}{ccc}{𝐞}_x& {𝐞}_y& {𝐞}_z\\ 0& 0& {\omega }_i\\ x_{i-}-x_i& y_{i-}-y_i& 0\end{array}\right|=-\frac{2}{h_i^0}[-{\omega }_i(y_{i-}-y_i){𝐞}_x+{\omega }_i(x_{i-}-x_i){𝐞}_y]\\ {\mathit{\omega }}_i\times {𝐩}_i& =\frac{2}{h_i^0}\left|\begin{array}{ccc}{𝐞}_x& {𝐞}_y& {𝐞}_z\\ 0& 0& {\omega }_i\\ x_{i+}-x_i& y_{i+}-y_i& 0\end{array}\right|=\frac{2}{h_i^0}[-{\omega }_i(y_{i+}-y_i){𝐞}_x+{\omega }_i(x_{i+}-x_i){𝐞}_y].\end{array}$

Substituting equation (8) into equations (6), we get

$\text{(9)}\begin{array}{cc}{𝐯}_{i-}& ={𝐯}_i-{\omega }_i(y_{i-}-y_i){𝐞}_x+{\omega }_i(x_{i-}-x_i){𝐞}_y\\ {𝐯}_{i+}& ={𝐯}_i-{\omega }_i(y_{i+}-y_i){𝐞}_x+{\omega }_i(x_{i+}-x_i){𝐞}_y.\end{array}$

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

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

Then equations (9) can be written as

$\text{(10)}\begin{array}{cc}{v}_{i-}^x{𝐞}_x+v_{i-}^y{𝐞}_y& =[v_i^x-{\omega }_i(y_{i-}-y_i)]{𝐞}_x+[v_i^y+{\omega }_i(x_{i-}-x_i)]{𝐞}_y\\ v_{i+}^x{𝐞}_x+v_{i+}^y{𝐞}_y& =[v_i^x-{\omega }_i(y_{i+}-y_i)]{𝐞}_x+[v_i^y+{\omega }_i(x_{i+}-x_i)]{𝐞}_y.\end{array}$

Therefore, the components of the velocity vectors are

$\text{(11)}\begin{array}{cc}{v}_{i-}^x& =v_i^x-{\omega }_i(y_{i-}-y_i)\\ v_{i-}^y& =v_i^y+{\omega }_i(x_{i-}-x_i)\\ v_{i+}^x& =v_i^x-{\omega }_i(y_{i+}-y_i)\\ v_{i+}^y& =v_i^y+{\omega }_i(x_{i+}-x_i).\end{array}$

Then, the matrix form of equations (11) is

$\left[\begin{array}{c}{v}_{i-}^x\\ v_{i-}^y\\ v_{i+}^x\\ v_{i+}^y\end{array}\right]=\left[\begin{array}{ccc}1& 0& -(y_{i-}-y_i)\\ 0& 1& (x_{i-}-x_i)\\ 1& 0& -(y_{i+}-y_i)\\ 0& 1& (x_{i+}-x_i)\end{array}\right]\left[\begin{array}{c}{v}_i^x\\ v_i^y\\ {\omega }_i\end{array}\right]$

or

${\left[\begin{array}{c}{𝐯}_{i-}\\ {𝐯}_{i+}\end{array}\right]}^{\text{slave}}={𝐓}_i{\dot{𝐝}}_i^{\text{master}}.$