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revision 2414 by gross, Mon Apr 6 11:48:32 2009 UTC revision 2415 by gross, Wed May 13 02:48:39 2009 UTC
# Line 401  later Line 401  later
401  % \section{Drucker Prager Model}  % \section{Drucker Prager Model}
402    
403  \section{Isotropic Kelvin Material \label{IKM}}  \section{Isotropic Kelvin Material \label{IKM}}
404  As proposed by Kelvin~\cite{Muhlhaus2005} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into  As proposed by Kelvin~\cite{Muhlhaus2005} material strain $D\hackscore{ij}=\frac{1}{2}(v\hackscore{i,j}+v\hackscore{j,i})$ can be decomposed into
405  an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:  an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
406  \begin{equation}\label{IKM-EQU-2}  \begin{equation}\label{IKM-EQU-2}
407  D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}  D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
# Line 465  By using a first order finite difference Line 465  By using a first order finite difference
465  \end{equation}  \end{equation}
466  and  and
467  \begin{equation}\label{IKM-EQU-2b}  \begin{equation}\label{IKM-EQU-2b}
468  D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'+\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}  D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'-\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}
469  \end{equation}  \end{equation}
470  where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With  where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
471  \begin{equation}\label{IKM-EQU-2c}  \begin{equation}\label{IKM-EQU-2c}
472  \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' -  \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' +
473  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}
474  \end{equation}  \end{equation}
475  we have  we have
476  \begin{equation}  \begin{equation}
# Line 490  where Line 490  where
490  \end{equation}  \end{equation}
491  The upper bound $\eta\hackscore{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form  The upper bound $\eta\hackscore{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form
492  \begin{equation}\label{IKM-EQU-10}  \begin{equation}\label{IKM-EQU-10}
493  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' -  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +
494  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)    \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)  
495  \end{equation}  \end{equation}
496  After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get  After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
497  \begin{equation}\label{IKM-EQU-1ib}  \begin{equation}\label{IKM-EQU-1ib}
498  -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})  -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
499  \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}-  \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
500  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}   \left(\frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}
501  \end{equation}  \end{equation}
502  Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a  Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
503  Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.  Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.

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