Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 10

Problem 1: Part 10
Assume that the velocities at nodes 5 and 6 are $\mathbf{v}_5$ and $\mathbf{v}_6$, respectively. Compute the velocity gradient at the blue point.

To compute the velocity gradient, we have to find the velocities at the slave nodes using the relation

\begin{bmatrix} v^x_{i-} \\ v^y_{i-} \\ v^x_{i+} \\ v^y_{i+} \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} $$

For node 1 of the element (global node 5), the $$\mathbf{T}$$ matrix is

\mathbf{T}_1 = \begin{bmatrix} 1 & 0 & 0.0866 \\ 0 & 1 & -0.05 \\ 1 & 0 & -0.0866 \\ 0 & 1 &0.05 \end{bmatrix} ~. $$

Therefore, the velocities of the slave nodes are

\begin{bmatrix} v^x_{1-} \\ v^y_{1-} \\ v^x_{1+} \\ v^y_{1+} \end{bmatrix} = \begin{bmatrix} 1.0866 \\ 1.95 \\ 0.9134 \\ 2.05 \end{bmatrix} ~. $$

For node 2 of the element (global node 6), the $$\mathbf{T}$$ matrix is

\mathbf{T}_2 = \begin{bmatrix} 1 & 0 & 0.05 \\ 0 & 1 & -0.0866 \\ 1 & 0 & -0.05 \\ 0 & 1 & 0.0866 \end{bmatrix} ~. $$

Therefore, the velocities of the slave nodes are

\begin{bmatrix} v^x_{2-} \\ v^y_{2-} \\ v^x_{2+} \\ v^y_{2+} \end{bmatrix} = \begin{bmatrix} 2.05 \\ 0.9134 \\ 1.95 \\ 1.0866 \end{bmatrix} ~. $$

The interpolated velocity within the element is given by

\begin{align} \mathbf{v}(\boldsymbol{\xi}, t) & = \cfrac{D}{Dt}[\mathbf{x}(\boldsymbol{\xi},t)] \\ & = \sum_{i- = 1}^2 \frac{\partial }{\partial t}[\mathbf{x}_{i-}(t)]~N_{i^-}(\xi,\eta) + \sum_{i+ = 1}^2 \frac{\partial }{\partial t}[\mathbf{x}_{i+}(t)]~N_{i^+}(\xi,\eta)\\ & = \sum_{i- = 1}^2 \mathbf{v}_{i-}(t)~N_{i-}(\xi,\eta) + \sum_{i+ = 1}^2 \mathbf{v}_{i+}(t)~N_{i+}(\xi,\eta) ~. \end{align} $$

Plugging in the values of the nodal velocities and the shape functions, we get
 * $$\begin{align}

v_x & = 0.27165 (1 - \xi) (1 - \eta) + 0.51250 (1 + \xi) (1 - \eta)\\ & + 0.22835 (1 - \xi) (1 + \eta) + 0.48750 (1 + \xi) (1 + \eta)\\ v_y & = 0.48750 (1 - \xi) (1 - \eta) + 0.22835 (1 + \xi) (1 - \eta)\\ & + 0.51250 (1 - \xi) (1 + \eta) + 0.27165 (1 + \xi) (1 + \eta)~. \end{align}$$

The velocity gradient is given by

\boldsymbol{\nabla} \mathbf{v} = v_{i,j} = \frac{\partial v_i}{\partial x_j} ~. $$

The velocities are given in terms of the parent element coordinates ($$\xi,\eta$$). We have to convert them to the ($$x,y$$) system in order to compute the velocity gradient. To do that we recall that
 * $$\begin{align}

\frac{\partial v_x}{\partial \xi} & = \frac{\partial v_x}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial v_x}{\partial y}\frac{\partial y}{\partial \xi} \\ \frac{\partial v_x}{\partial \eta} & = \frac{\partial v_x}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial v_x}{\partial y}\frac{\partial y}{\partial \eta} \\ \frac{\partial v_y}{\partial \xi} & = \frac{\partial v_y}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial v_y}{\partial y}\frac{\partial y}{\partial \xi} \\ \frac{\partial v_y}{\partial \eta} & = \frac{\partial v_y}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial v_y}{\partial y}\frac{\partial y}{\partial \eta} ~. \end{align}$$

In matrix form

\begin{bmatrix} \frac{\partial v_x}{\partial \xi} \\ \frac{\partial v_x}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} = \mathbf{J} \begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} $$ and

\begin{bmatrix} \frac{\partial v_y}{\partial \xi} \\ \frac{\partial v_y}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix} = \mathbf{J} \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix}~. $$

Therefore,

\begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} = \mathbf{J}^{-1} \begin{bmatrix} \frac{\partial v_x}{\partial \xi} \\ \frac{\partial v_x}{\partial \eta} \end{bmatrix} \qquad \text{and} \qquad \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix} = \mathbf{J}^{-1} \begin{bmatrix} \frac{\partial v_y}{\partial \xi} \\ \frac{\partial v_y}{\partial \eta} \end{bmatrix}~. $$

The isoparametric map is
 * $$\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}$$

The derivatives with respect to $$\xi$$ and $$\eta$$ are
 * $$\begin{align}

\frac{\partial x}{\partial \xi}&= 0.20131 + 0.018301 \eta & \frac{\partial y}{\partial \xi}&= -0.20131 - 0.018301 \eta \\ \frac{\partial x}{\partial \eta} &= 0.0683 + 0.018301 \xi & \frac{\partial y}{\partial \eta} &= 0.0683 - 0.018301 \xi \end{align}$$

Therefore,

\mathbf{J} = \begin{bmatrix} 0.20131 + 0.018301 \eta & -0.20131 - 0.018301 \eta \\ 0.0683 + 0.018301 \xi & 0.0683 - 0.018301 \xi \end{bmatrix}~. $$

Since we are only interested in the point ($$\xi,\eta$$) = (0,0), we can compute $$\mathbf{J}$$ at this point

\mathbf{J} = \begin{bmatrix} 0.20131 & -0.20131 \\ 0.0683&0.0683 \end{bmatrix}~. $$

The inverse is

\mathbf{J}^{-1} = \begin{bmatrix} 2.4837 & 7.3205 \\ -2.4837 & 7.3205 \end{bmatrix}~. $$

The derivatives of $$\mathbf{v}$$ with respect to $$\xi$$ and $$\eta$$ are
 * $$\begin{align}

\frac{\partial v_x}{\partial \xi}&= 0.5 + 0.018301 \eta & \frac{\partial v_x}{\partial \eta} &= -0.0683 + 0.018301 \xi \\ \frac{\partial v_y}{\partial \xi}&= -0.5 + 0.018301 \eta & \frac{\partial v_y}{\partial \eta} &= 0.0683 + 0.018301 \xi \end{align}$$

At ($$\xi,\eta$$) = (0,0), we get
 * $$\begin{align}

\frac{\partial v_x}{\partial \xi}&= 0.5 & \frac{\partial v_x}{\partial \eta} &= -0.0683 \\ \frac{\partial v_y}{\partial \xi}&= -0.5 & \frac{\partial v_y}{\partial \eta} &= 0.0683 \end{align}$$

Therefore,

\begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} = \begin{bmatrix} 0.74184 \\ -1.7418 \end{bmatrix}; \qquad \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix} = \begin{bmatrix} -0.74184 \\ 1.7418 \end{bmatrix}~. $$

Therefore, the velocity gradient at the blue point is

\boldsymbol{\nabla} \mathbf{v} = \begin{bmatrix} 0.74184 & -1.7418 \\ -0.74184 &1.7418 \end{bmatrix} $$

The Maple code for doing the above calculations is shown below.