# Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 9
## Problem 1: Part 9

Construct a laminar coordinate system at the blue point. (Use equations 9.3.16 and 9.3.17 from the book chapter). Assume that the blue point is at the center of element 5.
The $x$ and $y$ coordinates of the master and slave nodes are
    ${\begin{bmatrix}x_{1}\\y_{1}\\x_{2}\\y_{2}\end{bmatrix}}={\begin{bmatrix}r\cos \theta _{1}\\r\sin \theta _{1}\\r\cos \theta _{2}\\r\sin \theta _{2}\end{bmatrix}}={\begin{bmatrix}0.55000\\0.95263\\0.95263\\0.55000\end{bmatrix}}$
    ${\begin{bmatrix}x_{1-}\\y_{1-}\\x_{2-}\\y_{2-}\end{bmatrix}}={\begin{bmatrix}r_{1}\cos \theta _{1}\\r_{1}\sin \theta _{1}\\r_{1}\cos \theta _{2}\\r_{1}\sin \theta _{2}\end{bmatrix}}={\begin{bmatrix}0.50000\\0.86603\\0.86603\\0.50000\end{bmatrix}}$
    ${\begin{bmatrix}x_{1+}\\y_{1+}\\x_{2+}\\y_{2+}\end{bmatrix}}={\begin{bmatrix}r_{2}\cos \theta _{1}\\r_{2}\sin \theta _{1}\\r_{2}\cos \theta _{2}\\r_{2}\sin \theta _{2}\end{bmatrix}}={\begin{bmatrix}0.60000\\1.0392\\1.0392\\0.60000\end{bmatrix}}$
Since there are two master nodes in an element, the parent element is a four-noded quad with shape functions
    ${\begin{aligned}N_{1-}(\xi ,\eta )&={\cfrac {1}{4}}(1-\xi )(1-\eta )&N_{2-}(\xi ,\eta )&={\cfrac {1}{4}}(1+\xi )(1-\eta )\\N_{1+}(\xi ,\eta )&={\cfrac {1}{4}}(1-\xi )(1+\eta )&N_{2+}(\xi ,\eta )&={\cfrac {1}{4}}(1+\xi )(1+\eta )~.\end{aligned}}$
The isoparametric map is
    ${\begin{aligned}x(\xi ,\eta )&=x_{1-}~N_{1^{-}}(\xi ,\eta )+x_{2-}~N_{2^{-}}(\xi ,\eta )+x_{1+}~N_{1^{+}}(\xi ,\eta )+x_{2+}~N_{2^{+}}(\xi ,\eta )\\y(\xi ,\eta )&=y_{1-}~N_{1^{-}}(\xi ,\eta )+y_{2-}~N_{2^{-}}(\xi ,\eta )+y_{1+}~N_{1^{+}}(\xi ,\eta )+y_{2+}~N_{2^{+}}(\xi ,\eta )~.\end{aligned}}$
Plugging in the numbers, we get
    ${\begin{aligned}x&=0.12500(1-\xi )(1-\eta )+0.21651(1+\xi )(1-\eta )\\&+0.15000(1-\xi )(1+\eta )+0.25981(1+\xi )(1+\eta )\\y&=0.21651(1-\xi )(1-\eta )+0.12500(1+\xi )(1-\eta )\\&+0.25981(1-\xi )(1+\eta )+0.15000(1+\xi )(1+\eta )~.\end{aligned}}$
Therefore, the derivatives with respect to $\xi$ are
    ${\begin{aligned}x_{,\xi }={\frac {\partial x}{\partial \xi }}&=0.20131+0.018301\eta \\y_{,\xi }={\frac {\partial y}{\partial \xi }}&=-0.20131-0.018301\eta ~.\end{aligned}}$
Since the blue point is at the center of the element, the values of $\xi$ and $\eta$ at that point are zero. Therefore, the values of the derivatives at that point are
    ${\begin{aligned}x_{,\xi }={\frac {\partial x}{\partial \xi }}&=0.20131\\y_{,\xi }={\frac {\partial y}{\partial \xi }}&=-0.20131~.\end{aligned}}$
Therefore,
    $\mathbf {x} _{,\xi }=0.20131\mathbf {e} _{x}-0.20131\mathbf {e} _{y},~~\lVert \mathbf {x} _{,\xi }\rVert _={\sqrt {0.20131^{2}+0.20131^{2}}}=0.28470~.$
The local laminar basis vector ${\widehat {\mathbf {e} }}_{x}$ is given by
    ${\widehat {\mathbf {e} }}_{x}={\cfrac {\mathbf {x} _{,\xi }}{\lVert \mathbf {x} _{,\xi }\rVert _}}=0.70711\mathbf {e} _{x}-0.70711\mathbf {e} _{y}~.$
The laminar basis vector ${\widehat {\mathbf {e} }}_{y}$ is given by
    ${\widehat {\mathbf {e} }}_{y}=\mathbf {e} _{z}\times {\widehat {\mathbf {e} _{x}}}=0.70711\mathbf {e} _{x}+0.70711\mathbf {e} _{y}~.$
A plot of the vectors is shown in Figure 13.
* Figure 13. Laminar base vectors.
The Maple code for this calculation is shown below.