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{align}

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{align}$$

The isoparametric map is
 * $$\begin{align}

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{align}$$

Plugging in the numbers, we get
 * $$\begin{align}

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{align}$$

Therefore, the derivatives with respect to $$\xi$$ are
 * $$\begin{align}

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{align}$$

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{align}

x_{,\xi} = \frac{\partial x}{\partial \xi} &= 0.20131 \\ y_{,\xi} = \frac{\partial y}{\partial \xi} &= -0.20131 ~. \end{align}$$

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.

The Maple code for this calculation is shown below.