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revision 1858 by gross, Fri Oct 3 03:57:52 2008 UTC revision 1859 by gross, Wed Oct 8 03:03:37 2008 UTC
# Line 53  solution.setTolerance(TOL) \\ Line 53  solution.setTolerance(TOL) \\
53  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\
54  velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\  velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\
55    
56  \subsection{Benchmark Problem}  % \subsection{Benchmark Problem}
57    %
58  Convection problem  % Convection problem
59    
60    
61  \section{Temperature Cartesian}  \section{Temperature Cartesian}
# Line 105  Update level set function; advection and Line 105  Update level set function; advection and
105  Rayleigh-Taylor instability problem  Rayleigh-Taylor instability problem
106    
107    
108  \section{Drucker Prager Model}  % \section{Drucker Prager Model}
109    
110  \section{Isotropic Kelvin Material \label{IKM}}  \section{Isotropic Kelvin Material \label{IKM}}
111    As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into
112    an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
   
113  \begin{equation}\label{IKM-EQU-2}  \begin{equation}\label{IKM-EQU-2}
114  D_{ij}=D_{ij}^{el}+D_{ij}^{vp}  D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
115  \end{equation}  \end{equation}
116  with the elastic stretching  with the elastic strain given as
117  \begin{equation}\label{IKM-EQU-3}  \begin{equation}\label{IKM-EQU-3}
118  D_{ij}^{el}=\frac{2 \mu} \sigma'_{ij}  D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \sigma\hackscore{ij}'
119  \end{equation}  \end{equation}
120    where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
121    If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
122  \begin{equation}\label{IKM-EQU-4}  \begin{equation}\label{IKM-EQU-4}
123  D_{ij}^{vp}=\sum_{q} D_{ij}^{q}  D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
124  \end{equation}  \end{equation}
125    where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
126  \begin{equation}\label{IKM-EQU-5}  \begin{equation}\label{IKM-EQU-5}
127  D_{ij}^{q}=\frac{1}{2 \eta^{q}} \sigma'_{ij} \mbox{ with } \eta^{q}=\eta^{q}_N \left(\frac{\tau}{\tau_t^q}\right){\frac{1}{n^{q}}-1}  D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
128    \end{equation}
129    where $\eta^{q}$ is the viscosity of material $q$. We assume the following
130    betwee the the strain in material $q$
131    \begin{equation}\label{IKM-EQU-5b}
132    \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
133    \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
134  \end{equation}  \end{equation}
135    for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
136    Notice that $n^{q}=1$ gives a constant viscosity.
137  After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:  After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
138  \begin{equation}\label{IKM-EQU-4}  \begin{equation}\label{IKM-EQU-6}
139  D_{ij}^{vp}=\frac{1}{2 \eta^{vp}} \sigma'_{ij}  D\hackscore{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij} \mbox{ with } \frac{1}{\eta^{vp}} \sum\hackscore{q} \frac{1}{\eta^{q}} \;.
140  \end{equation}  \end{equation}
141    With
142    \begin{equation}\label{IKM-EQU-8}
143    \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
144    \end{equation}
145    one gets
146    \begin{equation}\label{IKM-EQU-8b}
147    \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
148    \end{equation}
149    With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
150    \begin{equation}\label{IKM-EQU-8c}
151    \tau \le \tau\hackscore{Y} + \beta \; p
152    \end{equation}
153    which leads to the condition
154    \begin{equation}\label{IKM-EQU-8d}
155    \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
156    \end{equation}
157    Therefore we modify the definition of $\eta^{vp}$ to the form
158    \begin{equation}\label{IKM-EQU-6b}
159    \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
160    \end{equation}
161    Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3} and~\ref{IKM-EQU-6b} to get
162    \begin{equation}\label{IKM-EQU-10}
163    D\hackscore{ij}'=\frac{1}{2 \eta\hackscore{eff}} \sigma\hackscore{ij}' \mbox{ with }
164    \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu}+\frac{1}{\eta^{vp}}
165    \end{equation}
166    The deviatoric stress needs to fullfill the equilibrion equation
167  \begin{equation}\label{IKM-EQU-1}  \begin{equation}\label{IKM-EQU-1}
168  -\sigma'_{ij,j}+p_j=F_j  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
169  \end{equation}  \end{equation}
170    where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
171  \begin{equation}\label{IKM-EQU-2}  \begin{equation}\label{IKM-EQU-2}
172  -v_{i,i}=0  -v\hackscore{i,i}=0
173  \end{equation}  \end{equation}
174    After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
175  \begin{equation}\label{IKM-EQU-3}  \begin{equation}\label{IKM-EQU-1ib}
176  \sigma_{ij}=\sigma'_{ij,j}-\frac{1}{d} p \delta_{ij}  -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}
177  \end{equation}  \end{equation}
178    

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