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.

Figure 4. Notation for motion of the continuum-based beam.

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

${\text{(5)}}\qquad {\begin{aligned}\mathbf {x} _{i-}(t)&=\mathbf {x} _{i}(t)-{\frac {1}{2}}h_{i}^{0}~\mathbf {p} _{i}(t)\\\mathbf {x} _{i+}(t)&=\mathbf {x} _{i}(t)+{\frac {1}{2}}h_{i}^{0}~\mathbf {p} _{i}(t)\end{aligned}}$

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

${\text{(6)}}\qquad {\begin{aligned}\mathbf {v} _{i-}(t)&=\mathbf {v} _{i}(t)-{\frac {1}{2}}{\frac {\partial }{\partial t}}[h_{i}^{0}~\mathbf {p} _{i}(t)]=\mathbf {v} _{i}(t)-{\frac {1}{2}}h_{i}^{0}~{\boldsymbol {\omega }}_{i}(t)\times \mathbf {p} _{i}(t)\\\mathbf {v} _{i+}(t)&=\mathbf {v} _{i}(t)+{\frac {1}{2}}{\frac {\partial }{\partial t}}[h_{i}^{0}~\mathbf {p} _{i}(t)]=\mathbf {v} _{i}(t)+{\frac {1}{2}}h_{i}^{0}~{\boldsymbol {\omega }}_{i}(t)\times \mathbf {p} _{i}(t)\end{aligned}}$

where ${\boldsymbol {\omega }}_{i}$ is the angular velocity of the director.

From equations (5) we have

${\text{(7)}}\qquad {\begin{aligned}\mathbf {p} _{i}(t)&=-{\cfrac {2}{h_{i}^{0}}}[\mathbf {x} _{i-}(t)-\mathbf {x} _{i}(t)]=-{\cfrac {2}{h_{i}^{0}}}\left[(x_{i-}-x_{i})~\mathbf {e} _{x}+(y_{i-}-y_{i})~\mathbf {e} _{y}\right]\\\mathbf {p} _{i}(t)&={\cfrac {2}{h_{i}^{0}}}[\mathbf {x} _{i+}(t)-\mathbf {x} _{i}(t)]={\cfrac {2}{h_{i}^{0}}}\left[(x_{i+}-x_{i})~\mathbf {e} _{x}+(y_{i+}-y_{i})~\mathbf {e} _{y}\right]\end{aligned}}$

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,

${\text{(8)}}\qquad {\begin{aligned}{\boldsymbol {\omega }}_{i}\times \mathbf {p} _{i}&=-{\cfrac {2}{h_{i}^{0}}}{\begin{vmatrix}\mathbf {e} _{x}&\mathbf {e} _{y}&\mathbf {e} _{z}\\0&0&\omega _{i}\\x_{i-}-x_{i}&y_{i-}-y_{i}&0\end{vmatrix}}=-{\cfrac {2}{h_{i}^{0}}}\left[-\omega _{i}(y_{i-}-y_{i})~\mathbf {e} _{x}+\omega _{i}(x_{i-}-x_{i})~\mathbf {e} _{y}\right]\\{\boldsymbol {\omega }}_{i}\times \mathbf {p} _{i}&={\cfrac {2}{h_{i}^{0}}}{\begin{vmatrix}\mathbf {e} _{x}&\mathbf {e} _{y}&\mathbf {e} _{z}\\0&0&\omega _{i}\\x_{i+}-x_{i}&y_{i+}-y_{i}&0\end{vmatrix}}={\cfrac {2}{h_{i}^{0}}}\left[-\omega _{i}(y_{i+}-y_{i})~\mathbf {e} _{x}+\omega _{i}(x_{i+}-x_{i})~\mathbf {e} _{y}\right]~.\end{aligned}}$

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

${\text{(9)}}\qquad {\begin{aligned}\mathbf {v} _{i-}&=\mathbf {v} _{i}-\omega _{i}(y_{i-}-y_{i})~\mathbf {e} _{x}+\omega _{i}(x_{i-}-x_{i})~\mathbf {e} _{y}\\\mathbf {v} _{i+}&=\mathbf {v} _{i}-\omega _{i}(y_{i+}-y_{i})~\mathbf {e} _{x}+\omega _{i}(x_{i+}-x_{i})~\mathbf {e} _{y}~.\end{aligned}}$

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

${\begin{aligned}\mathbf {v} _{i}&=v_{i}^{x}~\mathbf {e} _{x}+v_{i}^{y}~\mathbf {e} _{y}\\\mathbf {v} _{i-}&=v_{i-}^{x}~\mathbf {e} _{x}+v_{i-}^{y}~\mathbf {e} _{y}\\\mathbf {v} _{i+}&=v_{i+}^{x}~\mathbf {e} _{x}+v_{i+}^{y}~\mathbf {e} _{y}~.\end{aligned}}$

Then equations (9) can be written as

${\text{(10)}}\qquad {\begin{aligned}v_{i-}^{x}~\mathbf {e} _{x}+v_{i-}^{y}~\mathbf {e} _{y}&=[v_{i}^{x}-\omega _{i}(y_{i-}-y_{i})]~\mathbf {e} _{x}+[v_{i}^{y}+\omega _{i}(x_{i-}-x_{i})]~\mathbf {e} _{y}\\v_{i+}^{x}~\mathbf {e} _{x}+v_{i+}^{y}~\mathbf {e} _{y}&=[v_{i}^{x}-\omega _{i}(y_{i+}-y_{i})]~\mathbf {e} _{x}+[v_{i}^{y}+\omega _{i}(x_{i+}-x_{i})]~\mathbf {e} _{y}~.\end{aligned}}$

Therefore, the components of the velocity vectors are

${\text{(11)}}\qquad {\begin{aligned}v_{i-}^{x}&=v_{i}^{x}-\omega _{i}(y_{i-}-y_{i})\\v_{i-}^{y}&=v_{i}^{y}+\omega _{i}(x_{i-}-x_{i})\\v_{i+}^{x}&=v_{i}^{x}-\omega _{i}(y_{i+}-y_{i})\\v_{i+}^{y}&=v_{i}^{y}+\omega _{i}(x_{i+}-x_{i})~.\end{aligned}}$

Then, the matrix form of equations (11) is

${\begin{bmatrix}v_{i-}^{x}\\v_{i-}^{y}\\v_{i+}^{x}\\v_{i+}^{y}\end{bmatrix}}={\begin{bmatrix}1&0&-(y_{i-}-y_{i})\\0&1&(x_{i-}-x_{i})\\1&0&-(y_{i+}-y_{i})\\0&1&(x_{i+}-x_{i})\end{bmatrix}}{\begin{bmatrix}v_{i}^{x}\\v_{i}^{y}\\\omega _{i}\end{bmatrix}}$

or

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

Nonlinear finite elements/Homework 6/Solutions/Problem 1