added some information regarding spectral method

added the information about the new glide system for hex
This commit is contained in:
Martin Diehl 2011-09-02 13:49:02 +00:00
parent fe9754a41c
commit 320ec4d0b4
12 changed files with 8020 additions and 9 deletions

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@ -75,6 +75,7 @@ slip & basal & $\left\{ 0001\right\} \left\langle 1\bar{2}10\right\rangle $ & 3
& prism & $\left\{ 10\bar{1}0\right\} \left\langle 1\bar{2}10\right\rangle $ & 3\\
& pyr \hkl<a> & $\left\{ 10\bar{1}1\right\} \left\langle 1\bar{2}10\right\rangle $ & 6\\
& pyr \hkl<c+a> & $\left\{ 10\bar{1}1\right\} \left\langle 2\bar{1}\bar{1}3\right\rangle $ & 12\\
& pyr & $\left\{ 11\bar{2}2\right\} \left\langle 11\bar{2}\bar{3}\right\rangle $ & 6\\
\midrule
twin & T1 & $\left\{ 10\bar{1}2\right\} \left\langle \bar{1}011\right\rangle $ & 6\\
& C1 & $\left\{ 11\bar{2}2\right\} \left\langle 11\bar{2}\bar{3}\right\rangle $ & 6\\
@ -109,6 +110,10 @@ and \ref{fig: twinning systems}.
\label{fig: dislocation slip pyramidal ca}%
\includegraphics{slipSystem_pyrCA}}
\quad
\subfloat[Pyramidal slip]{%
\label{fig: dislocation slip pyramidal ca}%
\includegraphics{slipSystem_pyr}}
\quad
\caption{
Dislocation slip systems considered for hexagonal lattice structure.}
\label{fig: dislocation slip systems}

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@ -214,13 +214,14 @@ follows:
Due to the order in which FFTW stores the array, the components of the 3D frequencies can be calculated by dividing a linear list $0,1,2,...N/2-1,\pm N/2,-(N/2-1),...,-2,-1$ by the sampling length in each dimension.
\section{Questions}
The inverse transform of $\mathcal{F}\left(\operatorname{div}\,(\tnsr P) \right)$ (FFTW, 3 transforms of rank 3) should give the divergence field in real space.
\section{Remarks}
The inverse transform of $\mathcal{F}\left(\operatorname{div}\,(\tnsr P) \right)$ (FFTW, 3 transforms of rank 3) gives the divergence field in real space.
My problem is, that the calculated divergence has imaginary parts after the inverse transform. That seems strange to me, as the divergence can also be computed in real space, e.g. using FDM.
If a complex-to-complex transform is done, $\hat{\tnsr P} (\vec \xi)$ possesses conjugate complex symmetry.
$\operatorname{div}\,(\hat{\tnsr P})(\vec xi) \right)$ should also have conjugate complex symmetry as it would lead to a real-only divergence after the inverse transform. Surprisingly, even for the simple derivative as given in formula eq.(\ref{div}), the complex conjugate symmetry is also not preserved. The multiplication changes the sign of the whole term and not only of the imaginary part.
$\operatorname{div}\,(\hat{\tnsr P})(\vec xi) \right)$ also has conjugate complex symmetry as it would lead to a real-only divergence after the inverse transform. It can be seen for the simple derivative as given in formula eq.(\ref{div}) that the complex conjugate symmetry is preserved.
As the tensor $\tnsr P$ is a real-only tensor, it should be possible to use a real-to-complex transform for the tensor field and a complex-to-real transform for the back transform of the divergence field. Than one of the dimensions should run from $0$ to $N/2+1$ only. But that implies, that the divergence in Fourier space has conjugate compley symmetry.
As the tensor $\tnsr P$ is a real-only tensor, it is possible to use a real-to-complex transform for the tensor field and a complex-to-real transform for the back transform of the divergence field. Than one of the dimensions should run from $0$ to $N/2+1$ only. That implies, that the divergence in Fourier space has conjugate complex symmetry.
The important think is that the highest frequencies are not correct for the DFT. Therefore, they should not contribute to the divergence calculation. This becomes obvious, as they can be multiplied with $\pm N/2$ (again divided by sampling length) in the case of an even number of data points. This shows, that they are not properly defined.
\end{document}
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@ -182,6 +182,7 @@ bookmarksnumbered=true,% % generate bookmarks with numbers
\begin{document}
\maketitle
% ----------------------------------------------------------------------------------------------------------------------------
\section{General Information}
At MPIE, the flexible crystal plasticity framework ``Düsseldorf Advanced MAterial Simulation Kit'' (DAMASK) is developed.
It consists of different constitutive models, homogenization schemes, and tools for post- and preprocessing \cite{Roters_etal2010}.
It has interfaces to different solvers to the mechanical boundary value problem.
@ -191,10 +192,16 @@ However, their use is limited to periodic boundary conditions due to the approxi
The spectral method implemented at MPIE uses a finite-strain formulation proposed in \cite{Lahellec_etal2001} that is written in terms of deformation gradient \tnsr F and Piola--Kirchhoff stress \tnsr P and can therefore be used to solve the mechanical boundary value problem in the reference configuration.
Calculations have shown that for inhomogeneous material convergence cannot be achieved any longer at strains larger than ca.~15--20 \%.
We presently believe that this is due to the fact that the regular mesh in the reference configuration is locally heavily deformed to an extent where single points cross the path of neighboring points.
To reach higher strains, a remeshing scheme should be implemented as follows.\begin{enumerate}
\item Write out the current state
\item Approximate the deformed configuration by a regular (undeformed, new) mesh
\item Translate the old state values to the new mesh
To reach higher strains, a remeshing scheme should be implemented as follows.
\begin{enumerate}
\item Write out the current state
\item Reconstruct current geometry
\item Approximate the deformed configuration by a regular (undeformed, new) mesh with potentially $a \ne b \ne c$
\item Translate the old state values to the new mesh (``The winner takes it all?'')
\end{enumerate}
\section{Remarks, Notes}
\begin{enumerate}
\item Which criterion? Even for converged solutions we get ``chess board patterns?''
\end{enumerate}
\bibliographystyle{unsrtnat}
\bibliography{Masterthesis}

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\title{Calculation of Mixed Boundary Conditions}
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\author{M.~Diehl}
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% main text
\begin{document}
\maketitle
\section{Residuum Method}
Make step of $\partial \tnsr P $ from current situation towards $\tnsr P_{wish} $ (leaving free components as they are).
This results in $\inc{\tnsr F} = \frac{\partial \tnsr F}{\partial \tnsr P} \cdot \inc{\tnsr P} $ which has potentially non-zero components at prescribed $F_{ij}$. Isolate those as residuum $\tnsr R$.
Use $\frac{\partial \tnsr P}{\partial \tnsr F} \cdot \tnsr R$ to calculate change of \inc{\tnsr P} to kill \tnsr R.
Adjust initial \inc{\tnsr P} by change and calculate $\inc{\tnsr F}= \frac{\partial \tnsr F}{\partial \tnsr P} \cdot \inc{\tnsr P_{corr}}$ which should be now free of fixed components.
Check wether $\inc{\tnsr F}= \tnsr R \cdot \tnsr U$ has $\tnsr R \approx \tnsr I$
\section{Reduction Method}
Calculate $\frac{\partial \tnsr F}{\partial \tnsr P}$ and convert into 9x9 matrix. Convert $\tnsr{\Delta P}$ and $\tnsr{\Delta F}$ into an vector of dimension 9.
Remove all entries from the stiffness matrix where \tnsr F is prescribed (and where $\tnsr{\Delta P}$ is undefined).
Invert the reduced matrix and fill the removed entries with 0.
For a linear behavior, ${\frac{\partial \tnsr F}{\partial \tnsr P}}_{red}:\tnsr{\Delta P}$ gives the exact change of $\tnsr{\Delta F}$.
% ----------------------------------------------------------------------------------------------------------------------------
\end{document}
\endinput