\chapter{Homogenization of Thermal Properties of porous Ceramic Materials Using Meshfree Method with Distance Field}
\section {Asymptotic Homogenization}
\paragraph In previous chapter different homogenization methods, specifically asymptotic expansion homogenization were explained and also solution structure for meshfree method were introduced. In this chapter we are going to combine both of these methods to find thermal conductivity tensor of a porous ceramic material. Applying asymptotic expansion homogenization and following the steps which was explained in chapter 2, temperature field can be expressed as follows:
\begin{equation}
\label{Asymptotic}
T(x,y)=T^0(x)+\varepsilon (T^1)(x,y)
\end{equation}
$T^0$ is the macroscopic or global (homogenized) temperature and $T^1$ is the microscopic or local temperature. The macroscopic quantities are functions of the macroscale (x) only as shown in equation (\ref{Asymptotic}). According to \cite{M.Shabana} the governing equation of heat conduction problem is described by:
\paragraph
\begin{equation}
\label{heatconduction}
\int_\Omega k_{ij}\frac{\partial T}{\partial x_j}\frac{\partial T}{\partial x_i} d\Omega = \int_\Omega f\delta T d\Omega + \int_\Gamma h\delta T d\Gamma \forall\delta T
\end{equation}
Substituting equation (\ref{Asymptotic}) into the governing equation (\ref{heatconduction}) and using averaging technique for a periodic function $\Psi$ in the same way with the standard homogenization procedure, the micro-macro coupled problem can be resolved into macroscopic and microscopic problems:
\begin{equation}
\label{si}
\lim\int_\Omega\psi(\frac {x}{\varepsilon})d\Omega = \int_\Omega \frac{1}{|Y|}\int_Y \psi(y)dYd\Omega
\end{equation}
\begin{equation}
\label{microscopic}
\int_\Omega k_{ip}\frac { \partial \phi^j}{\partial y_p}\frac{\partial \delta T^1}{\partial y_i} dY = \int_\Omega k_{ij}\frac{ \partial \delta T^1}{\partial y_i} dY \forall\delta T^1
\end{equation}
\begin{equation}
\label{macroscopic}
\int_\Omega k_{ij}^H (\frac{\partial T^0}{\partial x_j})(\frac{\partial \delta T^0}{\partial x_i}) d\Omega = \int_\Omega f^H\delta T^0 d\Omega + \int_\Omega h \delta T^0 d\Gamma
\end{equation}
\paragraph
By solving the microscopic equation (\ref{microscopic}) using the periodic boundary condition, we obtain the characteristic function associated with heat conduction in the microstructure due to the mismatch of the thermal conductivities of the constituents. In equation (\ref{macroscopic}), the homogenized thermal conductivity tensor $k_{ij}^H$ and the homogenized internal heat source $f^H$ are calculated by the integration over the whole domain and rigorously defined as follows:
\begin{equation}
\label{Homogenized}
k_{ij}^H =\frac {1}{|Y|}\int_Y (k_{ij}-k_{ip}\frac{\partial \phi^j}{\partial y_p})dy
\end{equation}
\begin{equation}
\label{internal heat}
f^H = \frac{1}{|Y|}\int_y fdy
\end {equation}
These procedure which is used by Yasser M.Shabana and Nadotake Noda and is solved by Finite Element method. Using the same approach, we solved the problem by applying the Meshfree method.
\paragraph
\section{Representative Volume Element and Periodic Boundary Condition}
\paragraph
As first step, we need to define the representative volume element and the boundary condition. The RVE is usually regarded as a volume V of heterogeneous material that is sufficiently large to be statistically representative of the composite,or effectively include a sampling of all microstructural heterogeneities that exist in the composite. This is the general principle which is adopted, and it leads to the fact that the RVE must include a large number of the composite microheterogeneities such as: grains, inclusions, voids, fibers, etc. It must however remain
small enough to be considered as a volume element of continuum mechanics \cite{determination}. Here, RVE is selected as it was shown in Fig.\ref{RVE}.
\paragraph
After selecting the RVE, periodic boundry conditions should be applied. Usually in mathematical models and computer simulations a set of boundary condition is used to simulate the large system by modeling the small part that is far from its edge and since it is supposed that, most of the heterogeneous mirostructures have almost the same pattern which is repeated through the whole material, it can be called periodic boundary conditions. Figure \ref{sample} shows how periodic boundary condition is defined in this problem.
\begin{figure}[htb]
%\vspace{3pt}
%\hrule
%\vspace{3pt}
\begin{center}
\includegraphics[width=80mm]{sample.eps}
\end{center}
%\vspace{-4mm}
\caption{Periodic Boundary Condition}
%\vspace{3pt}
%\hrule
%\vspace{3pt}
\label{sample}
\end{figure}
According to this figure:$$
\phi\mid_{x=-a}=\phi\mid_{x=a}
$$
$$
\phi\mid_{y=-b}=\phi\mid_{y=b}
$$
$\phi$ can be any function on boundaries.
\paragraph
\section{Solution Structure for Periodic Boundary Condition}
\paragraph
Expression(\ref{shape_function2}) can be introduced as a solution structure for this problem. The first term provides approximation of $\phi$ inside the RVE and satisfied the homogeneous Dirichlet boundary condition on RVE's boundary and other terms are responsible for treatment of periodic boundary condition:
\begin{equation}
\label{shape_function2}
\phi^j=\omega\sum\limits_{i=1}^n C_i^J\chi_i+ C_{n+1}^J\chi_{n+1}+C_{n+2}^J\chi_{n+2}+C_{n+3}^J\chi_{n+3}+C_{n+4}^J\chi_{n+4}
\end{equation}
\begin{figure}[htb]
\begin{center}
\begin{tabular}[5]{ccccc}
\includegraphics[width=24mm]{w0.eps} &
\includegraphics[width=24mm]{w1.eps} &
\includegraphics[width=24mm]{w2.eps} &
\includegraphics[width=24mm]{w3.eps} &
\includegraphics[width=24mm]{w4.eps} \\
(a) & (b) & (c) & (d) & (d) \\
\end{tabular}
\caption{Solution structure for periodic boundary condition: (a) function $\omega$ which satisfied the homogeneous Dirichlet boundary condition; (b) shows quadratic basis function in X direction; (c) quadratic basis function in Y direction;(d) basis function of third order of polynomial in X direction;(e) basis function of third order of polynomial in X direction.}
\label{random}
\end{center}
\end{figure}
\paragraph
We can transfer $\omega$ inside the summation operator, so we will have:
\begin{equation}
\label{shape_function3}
\phi^j=\sum\limits_{i=1}^n \omega C_i^J\chi_i+ C_{n+1}^J\chi_{n+1}+C_{n+2}^J\chi_{n+2}+C_{n+3}^J\chi_{n+3}+C_{n+4}^J\chi_{n+4}
\end{equation}
\paragraph
$\xi$ is defined in this way:
\begin{equation}
\label{xi}
\xi=
\begin{cases}
\omega\chi_i& \text {$i=1,...n$},\\
\chi& \text {$i=n+1,.....n+4$}
\end{cases}
\end{equation}
\paragraph
then $ \xi$ is substituted in equation (\ref{shape_function3}), so final shape function will be expressed as:
\begin{equation}
\label{shape_function4}
\phi^j=\sum\limits_{i=1}^{n+4} C_i^J\xi_i
\end{equation}
\section{Approach}
\paragraph
So, up to here, a shape function is produced which satisfied both homogenous Dirichlet and periodic boundary conditions, now we could substitute this shape function into the microscopic heat conduction problem equation. Therefore, equation (\ref{microscopic}), will turn to:
\begin{equation}
\label{microscopic2}
\int_\Omega k_{ip}\frac { \partial \phi^j}{\partial y_p}\frac{\partial \xi_k}{\partial y_i} d\Omega = \int_\Omega k_{ij}\frac{ \partial \xi_k}{\partial y_i} d\Omega
\end{equation}
\paragraph
This equation could be considered as a system of algebraic equation like $[A][C]=[B]$ in which:
\begin{equation}
\label{a}
a_{km}=\int_\Omega k_{ip}\frac { \partial \xi_m}{\partial y_p}\frac{\partial \xi_k}{\partial y_i} d\Omega
\end{equation}
\begin{equation}
\label{b}
b_k=\int_\Omega k_{ij}\frac{ \partial \xi_k}{\partial y_i} d\Omega
\end{equation}
\paragraph
Solving this algebraic equations system, will result in finding $C_i^J$ coefficients which can be substituted in shape function equation and find the $\phi^j$ from equation (\ref{shape_function4}). Finally using the calculated shape function in equation (\ref{Homogenized}), we will get the homogenized thermal conductivity. The homogenization procedure is summarized in Fig\ref{procedure}.
\paragraph
\paragraph
\paragraph
\begin{figure}[htb]
%\vspace{3pt}
%\hrule
%\vspace{3pt}
\begin{center}
\includegraphics[width=100mm]{procedure.eps}
\end{center}
% \vspace{-4mm}
\caption{ Summary of homogenization procedure}
%\vspace{3pt}
%\hrule
%\vspace{3pt}
\label{procedure}
\end{figure}
\paragraph
\paragraph
\section{Benchmarks}
\paragraph
Since, there are no related experimental results available in this field in terms of thermal conductivity tensor, for checking the consistency of the proposed method, we set up some numerical experiments to demonstrate this momentous. The goal, initial data and results are explained precisely in each experiment.
\paragraph
\subsection{Test 1 : Homogeneous material with no porosity}
\paragraph
The goal of this experiment is to demonstrate that the proposed approach can reconstruct the originally specified material properties in the absence of porosity. We consider a rectangular as with this initial data:
k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $a$ = $b$ = 1.
In this case, since we do not have any heterogeneity, so we should expect to get the same result as the homogenous thermal conductivity. performing homogenization for this problem, we get $ k^H $ = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$.
\paragraph
\subsection{Test 2: Material with rectangular pattern of the circular holes}
\paragraph
The goal of this experiment is to demonstrate that selection of different RVE's for one material, will not affect the homogenized thermal conductivity tensor. If we consider a domain with rectangular pattern of circular holes, Fig \ref {test_2}, there are many possibilities of selecting different RVE's and in between there are two possibilities for rectangular RVE's, one is a square with a hole in center and another one is a square with quadrant in corners. The initial data for both cases are:
k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, \\$R$ = 0.25, $a$ = $b$ = 1.\\
Fig \ref{test_2}, shows the RVE's and also shape functions of both cases $\phi^j$ , which are generated regarding to the selected geometry and distance function which were explained in previous chapter.\\ Although we have two different RVE's here, but since we are dealing with one material, we expected to get the same thermal conductivity for each case. The obtained result for first RVE which was a square with a hole in center is: $k^H$ = $\left[
\begin{array}{cc}
0.906622 & 0 \\\sum
0 & 0.906622 \\
\end{array}
\right]$,
and result for second RVE which was a square with quadrant in corners is: $k^H$ = $\left[
\begin{array}{cc}
0.908391 & 0 \\
0 & 0.908391 \\
\end{array}
\right]$.
% \paragraph
\begin{figure}[htb]
\begin{center}
\begin{tabular}[4]{cccc}
\includegraphics[width=30mm]{RVE_1-2.eps} &
\includegraphics[width=30mm]{RVE_1-3.eps} &
\includegraphics[width=30mm]{fi1-1.eps} &
\includegraphics[width=30mm]{fi2-1.eps} \\
(a) & (b) & (c) & (d)\\
&\\
\includegraphics[width=30mm]{RVE_1-5.eps} &
\includegraphics[width=30mm]{RVE_1-6.eps} &
\includegraphics[width=30mm]{fi1-2.eps} &
\includegraphics[width=30mm]{fi2-2.eps} \\
(e) & (f) & (g) & (h) \\
\end{tabular}
\caption{Two different RVE's and their shape functions for a material with rectangular pattern of the circular holes}
\label{test_2}
\end{center}
\end{figure}
\begin{figure}[htb]
\begin{center}
\includegraphics[width=100mm]{test2.eps}
\end{center}
\caption{Convergence study for test 2: case one is a square with a hole in center and case two is a square with quadrant in corners}
\label{con}
\end{figure}
\paragraph
As it can be seen, the results are in good agreement. The very close results, also, indicate the consistency of the proposed method. Furthermore, in order to make sure that the results are reliable, we did convergence study and check the convergence of results by increasing the degree of polynomials of basis function. Convergence graph can be seen in Fig \ref{con}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Test 3: Applying standard simulation software approach (SSSA)}
\paragraph
As a matter of fact, not only Asymptotic Expansion Homogenization (AEH) could be applied to find homogenized thermal conductivity of porous ceramic materials, but also other approaches like those which are used by standard simulation software packages could be used. The goal of this experiment is to solve the problem with other approach rather than AEH and compare the results. To do so, first we are going to explain how standard simulation software approach works, then using this approach, solve the problem both by ANSYS and Meshfree method.
The heat flux for nonhomogeneous material in two dimension could be written as:
q = -$\left[
\begin{array}{cc}
k_{11} & k_{12}\\
k_{21} & k_{22} \\
\end{array}\right]$
$\left [
\begin{array}{c}
T_x\\
T_y\\
\end{array}\right]$
So, we have:
\begin{equation}
\label {xconduction}
q_{x} = -(k_{11} T_x - k_{12} T_y)
\end{equation}\\
and
\begin{equation}
\label {yconduction}
q_{y} = -(k_{21} T_x - k_{22} T_y)
\end{equation}\\
which $T_x$ and $T_y$ are partial derivatives of temperature.
Therefore, for performing homogenization we have to divide the problem in two parts, first applying the temperature gradient in $X$ direction and found $\bar{q_x}$ and $\bar{q_y}$ with applying these boundary condition:
$$ T\mid_{x = 0} = 0 $$ and $$T\mid_{x =2 a} = 1 $$ and assume adiabatic condition at $Y$ direction. In this case, $T_y$ become zero and we can find $k_{11}$ and $k_{21}$ from these equations:
\begin{equation}
\label {xconductionave1}
q_x = - k_{11} T_x
\end{equation}
\begin{equation}
\label {xconductionave2}
q_y = - k_{21} T_x
\end{equation}
In next step, the problem is solved by applying the temperature gradient in $Y$ direction with these boundary conditions:
\paragraph
$$
T\mid_{y=0}=0 $$ and $$ T\mid_{y=2b}=1 $$ and assume adiabatic condition at $X$ direction. In the same way, $T_x$ become zero, so we can find $k_{12}$ and $k_{22}$ from these equations:
\begin{equation}
\label {yconductionave1}
q_x = - k_{12} T_y
\end{equation}
\begin{equation}
\label {yconductionave2}
q_y = - k_{22} T_y
\end{equation}\\
This is the general approach which is used by standard simulation software packages. Fig.\ref{scheme} also shows the schematic of the approach.
\begin{figure}[htb]
\begin{center}
\includegraphics[width=100mm]{schem.eps}
\end{center}
\caption{General schematic of standard simulation packages Approach }
\label{scheme}
\end{figure}
\paragraph
Now, we are going to solve our problem in ANSYS. In order to have comparable results, we choose the same RVE such as the previous test. A rectangle with a hole in center ,Fig \ref{test_2}.b, and these initial data:
\\
k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $R$ = 0.25, $a$ = $b$ = 1. Then, the selected geometry was meshed with the ANSYS standard options. As it was explained, we applied temperature gradient in two steps in X and Y direction.
\begin{figure}[htb]
\begin{center}
\includegraphics[width=100mm]{ansysi.eps}
\end{center}
\caption{ANSYS User Define Result Window}
\label{ansys}
\end{figure}
\paragraph
After solving the problem, the contour of temperature distribution was shown as in Fig \ref{ansyst}, but temperature distribution is not enough for us, because we need to find heat flux in X and Y direction according to equation \ref{xconduction} and \ref{yconduction}.
\begin{figure}[htb]
\begin{center}
\includegraphics[width=100mm]{ansyst.eps}
\end{center}
\caption{Temperature Distribution in Y direction}
\label{ansyst}
\end{figure}
As a result, we tried to find a way that enables us to manage the ANSYS results. "User Defined Result" button is used for this purpose. Clicking on this button, will open a new window such as Fig \ref{ansys}, which gave us this ability to control showing the results in more details. Using this option, we can get the heat flux distribution in X and Y direction.
\paragraph
For finding the homogenized thermal conductivity tensor, the heat flux should be integrated over the domain such as below:\\
$ \bar{q}_x = \int q_xdx /A$ and $ \ \bar{q}_y = \int q_ydy /A$.
But since ANSYS did not support integrating the results in this way, so we had to export the information of each element into the Excel file and averaged them discretely to get the averaged amount of heat flux according to the following formulation:
$ \bar{q}_x = (\sum \limits_{i=1}^n q_{xi})/n $.\
where "n" is the number of the elements. We solved the problem for both cases and found the averaged heat flux in X and Y directions and applied equations (\ref{xconductionave1}), (\ref{xconductionave2}), (\ref{yconductionave1}) and (\ref{yconductionave2} to find the thermal conductivity tensor. Here are the results of this test:
\\
$k^H$ = $\left[
\begin{array}{cc}
0.9523 & 0 \\
0 & 0.9541 \\
\end{array}\right]$
and at last, using the same approach, we solve the problem with Meshfree method and get the following result as homogenized thermal conductivity tensor:
$k^H$ = $\left[
\begin{array}{cc}
0.908626 & 0 \\
0 & 0.908393 \\
\end{array}
\right]
\\$
First, comparing this result with the previous ones shows the 4.8\% differences between result of meshfree method and ANSYS as one of the most powerful finite element packages. This is because of the ANSYS limitations in performing the integration over the physical domain for homogenization purposes and once more, pointed the weakness of finite element based packages for finding the accurate physical properties of material.
Second, we can see that this result are almost identical with the results in test 2 and show that, although two different approaches were used in these two tests, but we got the same result for one macrostructure.
\section{More Complicated Patterns}
\paragraph
Since porous ceramic materials have both isotropic and anisotropic microstructure, we try to find homogenized thermal conductivity for both cases by testing different RVE's. So, in the following, results for more complicated pattern is presented and it is tried to accommodate anisotropic RVE's as well. Periodic boundary condition is applied in all the cases. Each example is solved in meshfree (AEH), (SSSA) and ANSYS and results brought for comparison. In addition, convergence study for each case has been done and the graphs are given.
\subsection{ Example 1: square with three diagonal circle}
\begin{figure}[htb]
\begin{center}
\begin{tabular}[3]{ccc}
\includegraphics[width=30mm]{RVE_1-7.eps} &
\includegraphics[width=30mm]{e1f3.eps} &
\includegraphics[width=30mm]{e1f32.eps} \\
(a) & (b) & (c)
\end{tabular}
\caption{RVE and shape function for example 1}
\label{Example_1}
\end{center}
\end{figure}
\paragraph
The initial data for this case is : k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $R$ = 0.25, $a$ = $b$ = 1.
Meshfree AEH result for example 1 is:
$k^H$=$\left[
\begin{array}{cc}
0.743341 & 0.021001 \\
0.021001& 0.743341 \\
\end{array}\right]$. As result shows, $k_{11}$ and $k_{22}$ are equal so there is equal thermal conductivity in X and Y directions. This can be related to the symmetry of the geometry of this RVE and the equal amount of material distribution in each direction. The convergence graph is shown in Fig \ref{con1}. Meshfree (SSSA) result are for this case is: $k^H$=$\left[
\begin{array}{cc}
0.756382 & 0.018629 \\
0.018521 & 0.756014 \\
\end{array}\right]$.
Ansys result for this geometry is: $k^H$=$\left[
\begin{array}{cc}
0.857957 & 0.057042 \\
0.055536 & 0.865689 \\
\end{array}\right]$.
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[2]{cc}
\includegraphics[width=80mm]{GID3.eps} &
\includegraphics[width=80mm]{GID33.eps} \\
(a) & (b)\\
\end{tabular}
\caption{Convergence graph for example 1}
\label{con1}
\end{center}
\end{figure}
Comparison between the results show that unlike ANSYS result, the meshfree results for both AEH and SSSA are identical. This is because of ANSYS limitation in performing integration over domain.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{ Example 2: square with triangle inside and holes}
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[3]{ccc}
\includegraphics[width=30mm]{RVE_1-8.eps} &
\includegraphics[width=30mm]{e2f51.eps} &
\includegraphics[width=30mm]{e2f52.eps} \\
(a) & (b) & (c)\\
\end{tabular}
\caption{RVE and shape functions for example 2}
\label{Example_2}
\end{center}
\end{figure}\
\paragraph
The initial data for this case is : k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $R$ = 0.25, $a$ = $b$ = 1.
Meshfree AEH result for Example 2 is: $k^H$=$\left[
\begin{array}{cc}
0.428356 & 0 \\
0 & 0.478928 \\
\end{array}\right]$.
\ As it is obvious here, since we have smaller porosity coefficient than the previous example, the thermal conductivity become less too. Also, in this example, the thermal conductivity in X and Y direction are not exactly the same but close to each other and the off diagonal components are zero which means that $k_{12}$ is zero. This situation shows an orthotropic behavior. The convergence graph is shown in Fig \ref{con2}. Also the SSSA result is: $k^H$=$\left[
\begin{array}{cc}
0.430363 & 0 \\
0 & 0.483600\\
\end{array}\right]$. The ANSYS result for this RVE is: $k^H$=$\left[
\begin{array}{cc}
0.557629 & 0.001405 \\
0.005423 & 0.631368 \\
\end{array}\right]$.
Comparing these three results, show that the meshfree results for both AEH and SSSA are almost the same, while ANSYS result is much different and it could be because of the same reason which was explained before.
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[1]{c}
\includegraphics[width=80mm]{GID5.eps}\\
\end{tabular}
\caption{Convergence graph for example 2}
\label{con2}
\end{center}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example 3: square with rectangle inside}
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[3]{ccc}
\includegraphics[width=30mm]{m1.eps} &
\includegraphics[width=30mm]{e4f71.eps} &
\includegraphics[width=30mm]{e4f72.eps} \\
(a) & (b) & (c)\\
\end{tabular}
\caption{RVE and shape function for example 3}
\label{Example_3}
\end{center}
\end{figure}
\paragraph
The initial data for this case is : k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $a$ = 0.2, $b$ = 0.8.
Meshfree AEH result for example 3 is: $k^H$=$\left[
\begin{array}{cc}
0.401291 & 0\\
0 & 0.815216\\
\end{array}\right]$.
In this example, again we have larger amount of material in X direction than Y, so the $k_{11}$ is lager than $k_{22}$ and since we have different amounts in X and Y direction, therefore this material can be considered anisotropic and also since the off diagonal terms are zero, it is orthotropic as well. The convergence graph is shown in Fig \ref {con3}. Meshfree SSSA result is:$k^H$=$\left[
\begin{array}{cc}
0.402193 & 0\\
0 & 0.817064\\
\end{array}\right]$.
The Ansys result is: $k^H$=$\left[
\begin{array}{cc}
0.473586 & 0.003258\\
0.00031 & 0.961689\\
\end{array}\right]$. Close results can be seen in both meshfree approaches and rough result for ANSYS. Results also showed the anisotropy of homogenized thermal conductivity very well. It is one of the privilege of using meshfree method homogenization because other methods like "role of mixture" which frequently used in this kind of problem does not show the anisotropy.
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[2]{cc}
\includegraphics[width=80mm]{GID71.eps}&
\includegraphics[width=80mm]{GID72.eps}\\
(a) & (b) \\
\end{tabular}
\caption{Convergence graph for example 3}
\label{con3}
\end{center}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example 4: square with ellipse inside}
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[3]{ccc}
\includegraphics[width=30mm]{m3.eps} &
\includegraphics[width=30mm]{e6f91.eps} &
\includegraphics[width=30mm]{e6f92.eps} \\
\end{tabular}
\caption{Example 4}
\label{Example_4}
\end{center}
\end{figure}
\paragraph
The initial data for this case is : k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $a$ = 0.3, $b$ = 0.8.
Meshfree method for example 4: $k^H$=$\left[
\begin{array}{cc}
0.709890 & 0.109712\\
0.109712 & 0.561575\\
\end{array}\right]$. Again, the role of orientation is obvious in this RVE and it has different amounts of thermal conductivity in each direction, so it is anisotropic. This RVE is comparable to RVE in example 3, the difference in the slope of the pore result in changing the off-diagonal components from 0 to 0.109712. The convergence graph is shown in fig \ref {con4}. Meshfree SSSA result is: $k^H$=$\left[
\begin{array}{cc}
0.71228 & 0.111013\\
0.110487 & 0.563759\\
\end{array}\right]$. \\ The Ansys result is: $k^H$=$\left[
\begin{array}{cc}
0.837510 & 0.007002\\
0.055770 & 0.72824\\
\end{array}\right]$. \\
In this example, also, ANSYS result is different from two meshfree approaches.
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[2]{cc}
\includegraphics[width=80mm]{GID9.eps}&
\includegraphics[width=80mm]{GID99.eps}\\
\end{tabular}
\caption{Convergence graph for example 4}
\label{con4}
\end{center}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{ Example 5: N shape}
%\begin{figure}[htbp]
%\begin{center}
%\begin{tabular}[3]{ccc}
%\includegraphics[width=30mm]{RVE_1-9.eps} &
%\includegraphics[width=30mm]{e3f61.eps} &
%\includegraphics[width=30mm]{e3f62.eps} \\
%(a) & (b) & (c)\\
%\end{tabular}
%\caption{RVE and shape function for example 5}
%\label{Example_3}
%\end{center}
%\end{figure}\
%\paragraph
%Meshfree result for Example 5 is : $k^H$=$\left[
%\begin{array}{cc}
%0.159681 & 0.096483 \\
%0.096483 & 0.483326 \\
%\end{array}\right]$.
%An inclined beam can be seen in the geometry of this RVE that make the material to show anisotropic behavior. As results show, not only the $K_{11}$ and $K_{22}$ are different, also the off diagonal components are not zero and it describes the situations when properties depend on the direction.
%The convergence graph is shown in Fig \ref{con3}. The ANSYS result for this RVE is: $k^H$=$\left[
%\begin{array}{cc}
%0.285847 & 0.179534 \\
%0.145664 & 0.8706876 \\
%\end{array}\right]$.\\
%\begin{figure}[htbp]
%\begin{center}
%\begin{tabular}[2]{cc}
%\includegraphics[width=80mm]{GID6.eps}&
%\includegraphics[width=80mm]{GID66.eps}\\
%\end{tabular}
%\caption{Convergence graph for example 5}
%\label{con5}
%\end{center}
%\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Example 6: H shape}
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[3]{ccc}
\includegraphics[width=30mm]{m2.eps} &
\includegraphics[width=30mm]{e5f81.eps} &
\includegraphics[width=30mm]{e5f82.eps} \\
\end{tabular}
\caption{RVE and shape function for example 6}
\label{Example_6}
\end{center}
\end{figure}
\paragraph
The initial data for this case is : k = $\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]$, $a$ = 0.3, $b$ = 0.8.
Meshfree method for example 6 is: $k^H$=$\left[
\begin{array}{cc}
0.168535 & 0.020295\\
0.020302 & 0.211362\\
\end{array}\right]$.
An inclined beam can be seen in the geometry of this RVE that make the material to show anisotropic behavior. As results show, not only the $k_{11}$ and $k_{22}$ are different, also the off diagonal components are not zero and it describes the situations when properties depend on the direction. The convergence graph is shown in Fig \ref{con6}. Meshfree SSSA result for this example is:$k^H$=$\left[
\begin{array}{cc}
0.160483 & 0.021491\\
0.0202466 & 0.211704\\
\end{array}\right]$. The Ansys result for example 5 is :$k^H$=$\left[
\begin{array}{cc}
0.432033 & 0.0581818\\
0.0353719 & 0.7049917\\
\end{array}\right]$. In all of these example, the meshfree AEH and SSSA results were almost identical, while the ANSYS results were different. Although all the results were approximated, but since two of them were more close to each other, therefore, we can conclude that those two, are more accurate. It was also concluded that the rough results of ANSYS is because of the limitation of this simulation software package in performing integration over the domain.
\begin{figure}[htbp]
\begin{center}
\begin{tabular}[2]{cc}
\includegraphics[width=80mm]{GID8.eps}&
\includegraphics[width=80mm]{GID88.eps}\\
\end{tabular}
\caption{Convergence graph for example 5}
\label{con6}
\end{center}
\end{figure}
\\
\section{ Realistic Microstructure Obtained From the Micrograph }
\paragraph
In this part, we are going to accomplish our commitment in the research and apply our method on realistic geometry of microstructure. In this regard, we got a microstructure which is obtained from the micrograph. We used the TaC 1800 micrograph and applied our method to find it's homogenized thermal properties.
\paragraph
As it is shown in Fig \ref{heat}, we need to convert the gray picture to binary (black and white) picture. It can be done by the help of software like Coral Photo Paint, Irfan View and Paint. Since Meshfree method works with distance functions in constructing the domain, we need to have the exact information of each point and their distance to the boundary. So, here, it is needed to convert the binary image to some kind of image that have the facility to give us the distance information. We used the " Skeleton " software for this mean. The obtained data from the " Skeleton " software have been used in the code, and rest of the work is similar to previous examples. Finally we got homogenized thermal conductivity tensor as:
\paragraph
$K^H$=$\left[
\begin{array}{cc}
0.430363 & 0\\
0& 0.483600 \\
\end{array}\right]$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=100mm]{heat.eps} \\
\caption{Scan\&Solve thermal analysis performed from a SEM micrograph of surface of TaC sample.}
\label{heat}
\end{center}
\end{figure}