# 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 ${\displaystyle x}$  and ${\displaystyle y}$  coordinates of the master and slave nodes are

${\displaystyle {\begin{bmatrix}x_{1}\\y_{1}\\x_{2}\\y_{2}\end{bmatrix}}={\begin{bmatrix}r\cos \theta _{1}\\r\sin \theta _{1}\\r\cos \theta _{2}\\r\sin \theta _{2}\end{bmatrix}}={\begin{bmatrix}0.55000\\0.95263\\0.95263\\0.55000\end{bmatrix}}}$
${\displaystyle {\begin{bmatrix}x_{1-}\\y_{1-}\\x_{2-}\\y_{2-}\end{bmatrix}}={\begin{bmatrix}r_{1}\cos \theta _{1}\\r_{1}\sin \theta _{1}\\r_{1}\cos \theta _{2}\\r_{1}\sin \theta _{2}\end{bmatrix}}={\begin{bmatrix}0.50000\\0.86603\\0.86603\\0.50000\end{bmatrix}}}$
${\displaystyle {\begin{bmatrix}x_{1+}\\y_{1+}\\x_{2+}\\y_{2+}\end{bmatrix}}={\begin{bmatrix}r_{2}\cos \theta _{1}\\r_{2}\sin \theta _{1}\\r_{2}\cos \theta _{2}\\r_{2}\sin \theta _{2}\end{bmatrix}}={\begin{bmatrix}0.60000\\1.0392\\1.0392\\0.60000\end{bmatrix}}}$

Since there are two master nodes in an element, the parent element is a four-noded quad with shape functions

{\displaystyle {\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

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle \xi }$  are

{\displaystyle {\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 ${\displaystyle \xi }$  and ${\displaystyle \eta }$  at that point are zero. Therefore, the values of the derivatives at that point are

{\displaystyle {\begin{aligned}x_{,\xi }={\frac {\partial x}{\partial \xi }}&=0.20131\\y_{,\xi }={\frac {\partial y}{\partial \xi }}&=-0.20131~.\end{aligned}}}

Therefore,

${\displaystyle \mathbf {x} _{,\xi }=0.20131\mathbf {e} _{x}-0.20131\mathbf {e} _{y},~~\lVert \mathbf {x} _{,\xi }\rVert _{}={\sqrt {0.20131^{2}+0.20131^{2}}}=0.28470~.}$

The local laminar basis vector ${\displaystyle {\widehat {\mathbf {e} }}_{x}}$  is given by

${\displaystyle {\widehat {\mathbf {e} }}_{x}={\cfrac {\mathbf {x} _{,\xi }}{\lVert \mathbf {x} _{,\xi }\rVert _{}}}=0.70711\mathbf {e} _{x}-0.70711\mathbf {e} _{y}~.}$

The laminar basis vector ${\displaystyle {\widehat {\mathbf {e} }}_{y}}$  is given by

${\displaystyle {\widehat {\mathbf {e} }}_{y}=\mathbf {e} _{z}\times {\widehat {\mathbf {e} _{x}}}=0.70711\mathbf {e} _{x}+0.70711\mathbf {e} _{y}~.}$

A plot of the vectors is shown in Figure 13.

 Figure 13. Laminar base vectors.

The Maple code for this calculation is shown below.

> # > # Shape functions > # > N1m := 1/4*(1-xi)*(1-eta): > N2m := 1/4*(1+xi)*(1-eta): > N1p := 1/4*(1-xi)*(1+eta): > N2p := 1/4*(1+xi)*(1+eta): > N := linalg[matrix](1,4,[N1m,N2m,N1p,N2p]); > # > # Compute local laminar basis vectors at the blue point > # > # Isoparametric map > # > x := x1m*N1m + x2m*N2m + x1p*N1p + x2p*N2p; > y := y1m*N1m + y2m*N2m + y1p*N1p + y2p*N2p; > # > # Derivative of Isoparametric map > # > dx_dxi := diff(x, xi); > dy_dxi := diff(y, xi); > # > # Jacobian evluated at blue point > # > dx_dxi_cen := subs(eta=0, dx_dxi); > dy_dxi_cen := subs(eta=0, dy_dxi); > # > # Local laminar basis vectors at the blue point > # > norm_dx_dxi_cen := simplify(sqrt(dx_dxi_cen^2 + dy_dxi_cen^2)); > ehat_x := vector([simplify(dx_dxi_cen/norm_dx_dxi_cen), > simplify(dy_dxi_cen/norm_dx_dxi_cen), 0]); > ehat_z := vector([0, 0, 1]); > ehat_y := crossprod(ehat_z, ehat_x);