User:Eml5526.s11.team6.joglekar/hwk6

=Problem 6.4=

Given
The vector field is given as follows -

$$\displaystyle (Eq. 6.4.2-1) $$ on the domain shown in figure--

Verify the divergence theorem.The curved boundary of the domain is a parabola.



To Find
To verify the Divergence theorem for the given vector field and the given domain as per Figure 6.4.2-1 We need to prove that : $$\int\limits_{\Omega }{\overline{\nabla }.}\overline{q}d\Omega =\oint\limits_{\Gamma }{\overline{q}.\overline{n}}d\Gamma $$ $$\displaystyle (Eq. 6.4.2-2) $$

Solution
We have, $${{q}_{x}}=3x^2y+y^3 $$  and secondly  $${{q}_{y}}=3x+y^3$$ as the given vector field over the domain given by Figure 6.4.2-1 Now we can write - $$\overline{\nabla }.\overline{q}=\frac{\partial {{q}_{x}}}{\partial x}+\frac{\partial {{q}_{y}}}{\partial y}=6xy+3y^2$$ Integrating over the given domain, we get LHS for $$ Eq.6.4.2-2 $$ as -

$$\int\limits_{\Omega }{\overline{\nabla }.}\overline{q}d\Omega = $$ $$\int\limits_{-2}^{2} \left[ {\int\limits_{0}^{4-x^2}{\left (6xy+3y^2 \right) }}dy \right] dx= \int\limits_{-2}^{2}{\left[ 3x(4-x^2)^2 + (4-x^2)^3 \right]}dx =117.029 $$

Verifying with| Wolfram Alpha

Thus : $$\displaystyle (Eq. 6.4.2-3) $$

To find RHS:

I)The computed boundary integral on the straight line AB is

$$\displaystyle \int\limits_{AB }{\overline{q }.}\overline{n}^{(1)}d\Gamma= \int\limits_{AB } \left ( \right )(-j)d\Gamma = \int\limits_{-2}^{2}-3xdx=0 $$

where $$ \overline n^{(1)}=-\overline j, d\Gamma=dx, y=0   on   AB $$

II)The integral is also calculated on the curved portion which is parabola with the same boundaries AB.

$$\displaystyle \int\limits_{AB }{\overline{q }.}\overline{n}^{(2)}d\Gamma= \int\limits_{AB } \left ( \right ) \left[ \begin{array}{c} {n_x \ } \\ {n_y \ } \end{array} \right] ds $$

where $$\displaystyle d \Gamma =ds  $$



Since normal vector is on the curved surface, we consider a small portion ds of the curve and then find outward normal to that portion of the curve. From the figure, it can be seen that outward normal is given as

$$ n=\left[ \begin{array}{c} {cos \theta \ } \\ {sin \theta \ } \end{array} \right]=\left[ \begin{array}{c} \frac{dy}{ds} \\ \frac{-dx}{ds} \end{array} \right] $$

Substituting the value of normal vector in the equation,we get,

$$ \int\limits_{AB } \left ( \right )ds $$

$$I= \int\limits_{AB } \left ((3x^2y+y^3)dy + (3x+y^3)(-dx) \right )$$

From the equation of parabola, $$ \displaystyle dy=-2xdx \quad and \quad y=4-x^2 $$

$$I= \int\limits_{AB } \left ((3x^2y+y^3)(-2xdx) + (3x+y^3)(-dx) \right )$$

$$ \displaystyle I=117.029 $$

Verifying with| Wolfram Alpha

$$\displaystyle (Eq. 6.4.2-3) $$

Thus divergence theorem is verified.

=Problem 6.9 Reproduction of problems solved in the FB=

Given
Consider the heat conduction problem given below modeled with 16 quadrilateral finite elements as shown in figure 6.9.1_1. The coordinates are given in meters. The conductivity is isotropic and $$ \displaystyle 5W^{0}C^{-1}.$$ The temperature T=0 is prescribed along edges AB and AD. The heat fluxes $$ \overline q =0 \quad and \quad \overline q=20 W m^{-1}$$ are prescribed on edges BC and CD,respectively.A constant heat source $$ \displaystyle s= 6Wm^{-1}$$ is applied over the plate.



To Find
Solve this problem using finite element code and get the temperature distribution and heat flux at the element Gauss points.

Solution
The codes used to solve this problem can be downloaded from the following site.

Codes

The program is written in MATLAB using several files for distinct operations of the analysis. In the main program file, these subfiles are recalled to plot the final result.





The problem can also be analysed using other softwares such as Abaqus. Following figure shows the temperature distribution in the element calculated using Abaqus.