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

Problem 1: Part 9

The $x$ and $y$ coordinates of the master and slave nodes are

$\left[\begin{array}{c}{x}_1\\ y_1\\ x_2\\ y_2\end{array}\right]=\left[\begin{array}{c}r\cos {\theta }_1\\ r\sin {\theta }_1\\ r\cos {\theta }_2\\ r\sin {\theta }_2\end{array}\right]=\left[\begin{array}{c}0.55000\\ 0.95263\\ 0.95263\\ 0.55000\end{array}\right]$

$\left[\begin{array}{c}{x}_{1-}\\ y_{1-}\\ x_{2-}\\ y_{2-}\end{array}\right]=\left[\begin{array}{c}{r}_1\cos {\theta }_1\\ r_1\sin {\theta }_1\\ r_1\cos {\theta }_2\\ r_1\sin {\theta }_2\end{array}\right]=\left[\begin{array}{c}0.50000\\ 0.86603\\ 0.86603\\ 0.50000\end{array}\right]$

$\left[\begin{array}{c}{x}_{1+}\\ y_{1+}\\ x_{2+}\\ y_{2+}\end{array}\right]=\left[\begin{array}{c}{r}_2\cos {\theta }_1\\ r_2\sin {\theta }_1\\ r_2\cos {\theta }_2\\ r_2\sin {\theta }_2\end{array}\right]=\left[\begin{array}{c}0.60000\\ 1.0392\\ 1.0392\\ 0.60000\end{array}\right]$

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

$\begin{array}{cccc}{N}_{1-}(\xi ,\eta )& =\frac{1}{4}(1-\xi )(1-\eta )& N_{2-}(\xi ,\eta )& =\frac{1}{4}(1+\xi )(1-\eta )\\ N_{1+}(\xi ,\eta )& =\frac{1}{4}(1-\xi )(1+\eta )& N_{2+}(\xi ,\eta )& =\frac{1}{4}(1+\xi )(1+\eta ).\end{array}$

The isoparametric map is

$\begin{array}{cc}x(\xi ,\eta )& =x_{1-}{N}_{1^{-}}(\xi ,\eta )+x_{2-}{N}_{2^{-}}(\xi ,\eta )+x_{1+}{N}_{1^{+}}(\xi ,\eta )+x_{2+}{N}_{2^{+}}(\xi ,\eta )\\ y(\xi ,\eta )& =y_{1-}{N}_{1^{-}}(\xi ,\eta )+y_{2-}{N}_{2^{-}}(\xi ,\eta )+y_{1+}{N}_{1^{+}}(\xi ,\eta )+y_{2+}{N}_{2^{+}}(\xi ,\eta ).\end{array}$

Plugging in the numbers, we get

$\begin{array}{cc}x& =0.12500(1-\xi )(1-\eta )+0.21651(1+\xi )(1-\eta )\\ & +0.15000(1-\xi )(1+\eta )+0.25981(1+\xi )(1+\eta )\\ y& =0.21651(1-\xi )(1-\eta )+0.12500(1+\xi )(1-\eta )\\ & +0.25981(1-\xi )(1+\eta )+0.15000(1+\xi )(1+\eta ).\end{array}$

Therefore, the derivatives with respect to $\xi$ are

$\begin{array}{cc}{x}_{,\xi }=\frac{\partial x}{\partial \xi }& =0.20131+0.018301\eta \\ y_{,\xi }=\frac{\partial y}{\partial \xi }& =-0.20131-0.018301\eta .\end{array}$

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

$\begin{array}{cc}{x}_{,\xi }=\frac{\partial x}{\partial \xi }& =0.20131\\ y_{,\xi }=\frac{\partial y}{\partial \xi }& =-0.20131.\end{array}$

Therefore,

${𝐱}_{,\xi }=0.20131{𝐞}_x-0.20131{𝐞}_y,\Vert {𝐱}_{,\xi }{\Vert }_{}=\sqrt{0.20131^2+0.20131^2}=0.28470.$

The local laminar basis vector ${\widehat{𝐞}}_x$ is given by

${\widehat{𝐞}}_x=\frac{{𝐱}_{,\xi }}{\Vert {𝐱}_{,\xi }{\Vert }_{}}=0.70711{𝐞}_x-0.70711{𝐞}_y.$

The laminar basis vector ${\widehat{𝐞}}_y$ is given by

${\widehat{𝐞}}_y={𝐞}_z\times \widehat{{𝐞}_x}=0.70711{𝐞}_x+0.70711{𝐞}_y.$

A plot of the vectors is shown in Figure 13.

File:NFE HW6Prob1 1.png 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);