Page Menu
Home
c4science
Search
Configure Global Search
Log In
Files
F78427069
readme.tex
No One
Temporary
Actions
Download File
Edit File
Delete File
View Transforms
Subscribe
Mute Notifications
Award Token
Subscribers
None
File Metadata
Details
File Info
Storage
Attached
Created
Tue, Aug 20, 19:44
Size
9 KB
Mime Type
text/x-tex
Expires
Thu, Aug 22, 19:44 (1 d, 23 h)
Engine
blob
Format
Raw Data
Handle
20039935
Attached To
rLIBMULTISCALE LibMultiScale
readme.tex
View Options
\documentclass[a4paper,11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[french]{babel}
\usepackage[T1]{fontenc} % accents codés dans la fonte
%\usepackage{layout}
\usepackage{a4wide}
\usepackage{amsmath}
\usepackage[usenames]{color}
\usepackage{url}
\usepackage{hyperref}
%\usepackage{comment}
%\usepackage{textcomp}
\usepackage{parskip}
\usepackage{stmaryrd}
\usepackage{tikz}
\usetikzlibrary{calc,patterns,snakes}
\setlength{\hoffset}{0pt}
\setlength{\oddsidemargin}{-1cm} % Marge gauche sur pages impaires
\setlength{\evensidemargin}{-1cm} % Marge gauche sur pages paires
\setlength{\marginparwidth}{0cm} % Largeur de note dans la marge
\setlength{\textwidth}{18cm} % Largeur de la zone de texte (17cm)
\setlength{\voffset}{0pt} % Bon pour DOS
\setlength{\marginparsep}{0pt} % Séparation de la marge
\setlength{\topmargin}{0cm} % Pas de marge en haut
\setlength{\headheight}{0cm} % Haut de page
\setlength{\headsep}{0cm} % Entre le haut de page et le texte
\setlength{\footskip}{1cm} % Bas de page + séparation
\setlength{\textheight}{25.5cm} % Hauteur de la zone de texte (25cm)
\begin{document}
\section{Getting linearized LJ}
We want to fit the parameters from an atomic lennard jones potential:
\
\begin{equation}
\phi(r) = 4 \epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right]
\end{equation}
But we want to model it in a thin sheet of 111 aluminum plane. Thus the pattern is
an hexagonal lattice with the inter-atomic distance $r_0 = \frac{\sqrt{2}}{2} a$ where $a$ is the
lattice parameter of the FCC structure of aluminum (i.e. $a = 4.0320$ \AA).
The atomic mass of the aluminum is $m_{al} = 26.98153860 g/mol$.
Also we want to take
into account the first neighbors only, so that the energy per atom at ground state is:
\begin{equation}
E = 6 \cdot \phi(r_0) = 24 \epsilon \left[ \left( \frac{\sigma}{r_0} \right)^{12} - \left( \frac{\sigma}{r_0} \right)^{6} \right]
\end{equation}
From this we can compute the length parameter $r_0$ so that we are at the minimum of energy:
\
\begin{align}
\frac{\partial E}{\partial r_0} &= 24 \frac{ \epsilon} {r_0} \left[ -12 \left( \frac{\sigma}{r_0} \right)^{12} + 6 \left( \frac{\sigma}{r_0} \right)^{6} \right] = 0 \\
& \Leftrightarrow 12 \left( \frac{\sigma}{r_0} \right)^{12} = 6 \left( \frac{\sigma}{r_0} \right)^{6} \\
& \Leftrightarrow 2 = \left( \frac{r_0}{\sigma} \right)^{6} \Leftrightarrow 2^{1/6} \sigma = r_0
\end{align}
So that this last relation between $r_0$ and $\sigma$ ensures the ground state equilibrium.\\
If we linearize this LJ potential we have:
\begin{equation}
f(r) = \phi''(r_0)(r-r_0) = \frac{72 \epsilon}{r_0^2} (r - r_0) = k (r - r_0)
\end{equation}
\section{The elastic tensor}
It can be demonstrated \cite{tadmor} that the elastic fourth order tensor can be derived to be:
\
\begin{align}
C_{ijkl}(r_0) &= \frac{1}{\Omega} \sum_{p=1}^6 \left[ \frac{\phi''(r_0)}{r_0^2} - \frac{\phi'(r_0)}{r_0^3} \right] X_i^p X_j^p X_k^p X_l^p \\
&= \frac{1}{\Omega} \sum_{p=1}^6 \left[
\frac{4 \epsilon} {r_0^2} \frac{\left[ 156 \left( \frac{\sigma}{r_0} \right)^{12}
- 42 \left( \frac{\sigma}{r_0} \right)^{6} \right]}{r_0^2} -
\frac{4 \epsilon} {r_0} \frac{\left[ -12 \left( \frac{\sigma}{r_0} \right)^{12}
+ 6 \left( \frac{\sigma}{r_0} \right)^{6} \right]}{r_0^3} \right]
X_i^p X_j^p X_k^p X_l^p \\
&= \frac{1}{\Omega} \sum_{p=1}^6 \frac{4 \epsilon} {r_0^4} \left[
\left[ 156 \left( \frac{\sigma}{r_0} \right)^{12}
- 42 \left( \frac{\sigma}{r_0} \right)^{6} \right] -
\left[ -12 \left( \frac{\sigma}{r_0} \right)^{12}
+ 6 \left( \frac{\sigma}{r_0} \right)^{6} \right] \right]
X_i^p X_j^p X_k^p X_l^p \\
&= \frac{1}{\Omega} \frac{4 \epsilon} {r_0^4} \left[
168 \left( \frac{\sigma}{r_0} \right)^{12}
- 48 \left( \frac{\sigma}{r_0} \right)^{6} \right]
\sum_{p=1}^6 X_i^p X_j^p X_k^p X_l^p \\
&= \frac{1}{\Omega} \xi(r_0) \sum_{p=1}^6 X_i^p X_j^p X_k^p X_l^p
\end{align}
With $X^p$ the position of atom $p$ within the 6 neighbors and $\Omega$ is the lattice volume.
\
\begin{align}
\xi(r_0) &= \frac{4 \epsilon} {r_0^4} \left( \frac{\sigma}{r_0} \right)^{6} \left[ 168 \left( \frac{\sigma}{r_0} \right)^{6} - 48 \right] \\
&= \frac{4 \epsilon} {r_0^4} \left( 2^{-1/6} \right)^{6} \left[ 168 \left( 2^{-1/6} \right)^{6} - 48 \right] \\
&= \frac{2 \epsilon} {r_0^4} \left[ 84 - 48 \right] = \frac{72 \epsilon} {r_0^4}
\end{align}
where we used the relation between $r_0$ and $\sigma$.
\begin{align}
C_{1111} &= \frac{1}{\Omega} \frac{72 \epsilon} {r_0^4} \sum_{p=1}^6 (X_1^p)^4 \\
C_{2222} &= \frac{1}{\Omega} \frac{72 \epsilon} {r_0^4} \sum_{p=1}^6 (X_2^p)^4 \\
C_{1212} &= \frac{1}{\Omega} C_{1122} = \frac{1}{\Omega} \frac{72 \epsilon} {r_0^4} \sum_{p=1}^6 (X_1^p)^2 (X_2^p)^2 \\
\end{align}
Let us then get interested in the sums $\sum_{p=1}^6 (X_1^p)^4$
$\sum_{p=1}^6 (X_2^p)^4$ and $\sum_{p=1}^6 (X_1^p)^2 (X_2^p)^2$.
For this let us assume that the orientation is that the X axis lattice is of length $r_0$
and the Y axis is of length $\sqrt{3}/2 r_0$ forming the lattice presented in figure
\ref{toto}.
Then the positions of the 6 atoms are:
$(r_0,0,0)$, $(-r_0,0,0)$,
$\frac{1}{2}(r_0,\sqrt{3} r_0,0)$ , $\frac{1}{2}(r_0,-\sqrt{3} r_0,0)$,
$\frac{1}{2}(-r_0,\sqrt{3} r_0,0)$ and $\frac{1}{2}(-r_0,-\sqrt{3} r_0,0)$.
the first sum is:
\begin{equation}
\sum_{p=1}^6 (X_1^p)^4 = 2 r_0^4 + 4 \frac{1}{2^4} r_0^4 = r_0^4 (2 + \frac{1}{4}) = r_0^4 \left(\frac{3}{2}\right)^2
\end{equation}
the second sum is:
\begin{equation}
\sum_{p=1}^6 (X_2^p)^4 = 2 \cdot 0 + 4 \frac{3^{4/2}}{2^4} r_0^4 = r_0^4 \left(\frac{3}{2} \right)^2
\end{equation}
And the third is:
\begin{equation}
\sum_{p=1}^6 (X_1^p)^2 (X_2^p)^2 = 2 \cdot r_0^2 \cdot 0^2 + 4 \cdot \frac{1}{2^4} r_0^4 3 = r_0^4 \frac{3}{4}
\end{equation}
Finally we have that:
\begin{align}
C_{1111} &= C_{2222} = \frac{1}{\Omega} \frac{72 \epsilon} {r_0^4} r_0^4 \left(\frac{3}{2}\right)^2 = \frac{1}{\Omega} 162 \epsilon \\
C_{1212} &= C_{1122} = \frac{1}{\Omega} \frac{72 \epsilon} {r_0^4} r_0^4 \frac{3}{4} = \frac{1}{\Omega} 54 \epsilon \\
\end{align}
\section{fitting the orthotropic law}
let us recall the orthotropic Hooke's law:
\
\begin{align}
\left(
\begin{array}{c}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{33}\\
\sigma_{23}\\
\sigma_{31}\\
\sigma_{12}\\
\end{array}
\right) = \Gamma
\left(
\begin{array}{cccccc}
E_1 (1-\nu_{23}\nu_{32}) & E_1 (\nu_{21} + \nu_{31}\nu_{23}) & E_1 (\nu_{31} + \nu_{21}\nu_{32}) & 0 & 0 & 0 \\
E_1 (\nu_{21} + \nu_{31}\nu_{23}) & E_2 (1 - \nu_{13}\nu_{31}) & E_2 (\nu_{32} + \nu_{12}\nu_{31}) & 0 & 0 & 0 \\
E_3 (\nu_{31} + \nu_{21}\nu_{32}) & E_3 (\nu_{32} + \nu_{12}\nu_{31}) & E_3 (1 - \nu_{12}\nu_{21}) & 0 & 0 & 0 \\
0 & 0 & 0 & \mu_{23}/\Gamma & 0 & 0 \\
0 & 0 & 0 & 0 & \mu_{13}/\Gamma & 0 \\
0 & 0 & 0 & 0 & 0 & \mu_{12}/\Gamma \\
\end{array}
\right)
\left(
\begin{array}{c}
\epsilon_{11}\\
\epsilon_{22}\\
\epsilon_{33}\\
\epsilon_{23}\\
\epsilon_{31}\\
\epsilon_{12}\\
\end{array}
\right)
\end{align}
with $\Gamma = \frac{1}{1 - \nu_{12}\nu_{21} - \nu_{23}\nu_{32} - \nu_{31}\nu_{13} - 2 \nu_{21}\nu_{32}\nu_{13}}$
In 2D with plane strain it reduces to:
\begin{align}
\left(
\begin{array}{c}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}\\
\end{array}
\right) = \Gamma
\left(
\begin{array}{cccccc}
E_1 & E_1 \nu_{21} & 0 \\
E_2 \nu_{12} & E_2 & 0 \\
0 & 0 & \mu/\Gamma \\
\end{array}
\right)
\left(
\begin{array}{c}
\epsilon_{11}\\
\epsilon_{22}\\
\epsilon_{12}\\
\end{array}
\right)
\end{align}
with $\Gamma = \frac{1}{1 - \nu_{12}\nu_{21}}$
So that we have the relations:
\
\begin{align}
C_{1111} &= \Gamma E_1 = \frac{1}{\Omega} 162 \epsilon \\
C_{2222} &= \Gamma E_2 = \frac{1}{\Omega} 162 \epsilon
\end{align}
which implies that $E_1 = E_2 = E$. We have consequently:
\begin{align}
C_{1122} &= \Gamma E \nu_{21} = \frac{E \nu_{21}}{1 - \nu_{12}\nu_{21}} = \frac{1}{\Omega} 54 \epsilon \\
&= C_{2211} = \Gamma E \nu_{12} = \frac{E \nu_{12}}{1 - \nu_{12}\nu_{21}} = \frac{1}{\Omega} 54 \epsilon \\
\end{align}
Then we have $\nu_{12} = \nu_{21} = \nu$. Let us summarize:
\begin{align}
C_{1111} &= C_{2222} = \frac{E}{1 - \nu^2} = \frac{1}{\Omega} 162 \epsilon \\
C_{1122} &= C_{2211} = \frac{E \nu}{1 - \nu^2} = \frac{1}{\Omega} 54 \epsilon
\end{align}
So that:
\begin{align}
\nu &= \frac{C_{1122}}{C_{1111}} = \frac{54}{162} = \frac{1}{3} \\
E &= \frac{1}{\Omega} \frac{8}{3} \cdot 54 \epsilon = \frac{1}{\Omega} 144 \epsilon \\
\mu &= \frac{1}{\Omega} 54 \epsilon = \frac{3}{8} E
\end{align}
Let us have $\Omega = \frac{r_0^3\sqrt{3}}{2} [m^3]$ with the z dimension taken as $r_0$.
So that let say we want a young's modulus of 68 GPa we have:
\begin{align}
E = 68 \cdot 10^9 N.m^{-2} \\
\epsilon = r_0^3\sqrt{3} \cdot 2.361111111 \cdot 10^8 Joules \\
\mu = 2.55 \cdot 10^{10} N.m^{-2} \\
\nu = 1/3
\end{align}
By choosing the lattice parameter $a = 4.0320$ we obtain $r_0 = \frac{\sqrt{2}}{2} a$ with
$a = 4.0320 \cdot 10^{-10}$ \AA, we obtain in real units
\begin{align}
a &= 4.0320 \cdot 10^{-10} \\
r_0 & = 2.8510545417441596184 \cdot 10^{-10} \\
E &= 68 \cdot 10^9 N.m^{-2} \\
\epsilon &= 9.4774964211648290153 \cdot 10^{-21} Joules \\
\mu &= 2.55 \cdot 10^{10} N.m^{-2} \\
\nu &= 1/3
\end{align}
By converting the distances to \AA~and the Joules to kcal/mol we have finally
\begin{align}
a &= 4.0320 [A] \\
r_0 & = 2.8510545417441596184 [A] \\
\sigma &= 2.5400008365880643002 \\
E &= 9.7874197348334701168 [kcal/mol/A^3] \\
\epsilon &= 1.3641211104312093062 [kcal/mol] \\
\mu &= 3.6702824005625512938 [kcal/mol/A^3] \\
\nu &= 1/3 \\
k &= \frac{72 \epsilon}{r_0^2} = 12.082988860820660663 [kcal/mol/A^2] \\
\rho &= \frac{m_{al}}{\frac{\sqrt{3}}{2} r_0^2 }
\end{align}
\end{document}
Event Timeline
Log In to Comment