Diff of /trunk/doc/user/Models.tex

revision 2370 by gross, Wed Mar 11 08:17:57 2009 UTC revision 2371 by gross, Mon Apr 6 11:48:32 2009 UTC
# Line 183  return the tolerance for the involved PD Line 183  return the tolerance for the involved PD
183   The following script \file{lit\hackscore driven\hackscore cavity.py}   The following script \file{lit\hackscore driven\hackscore cavity.py}
184  \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory  \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
185  illustrates the usage of the \class{StokesProblemCartesian} class to solve  illustrates the usage of the \class{StokesProblemCartesian} class to solve
186  the lit driven cavity problem~\cite{LITDRIVENCAVITY}:  the lit driven cavity problem:
187  \begin{python}  \begin{python}
188  from esys.escript import *  from esys.escript import *
189  from esys.finley import Rectangle  from esys.finley import Rectangle
# Line 281  Q^*u  + Q^*Q p & = & Q^*g \\ Line 281  Q^*u  + Q^*Q p & = & Q^*g \\
281  \end{array}  \end{array}
282
283  where $D^*$ and $Q^*$ denote the adjoint operators.  where $D^*$ and $Q^*$ denote the adjoint operators.
284  In~\cite{XXX} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible  In~\cite{LEASTSQUARESFEM1994} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible
285  to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)    to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)
286
287  The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have  The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
# 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~\ref{KELVN} 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 $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ 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  \label{IKM-EQU-2}  \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 425  betwee the the strain in material $q$ Line 425  betwee the the strain in material $q$
425  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
426  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
427
428  for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.  for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\cite{Muhlhaus2005}.
429  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
430  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:
431  \label{IKM-EQU-6}  \label{IKM-EQU-6}
432  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}} \;.  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}} \;.
433
434  and finally with~\ref{IKM-EQU-2}  and finally with~\ref{IKM-EQU-2}
435  \label{IKM-EQU-2b}  \label{IKM-EQU-2bb}
436  D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'  D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
437
438  The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}  The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}
# Line 445  The deviatoric stress needs to fullfill Line 445  The deviatoric stress needs to fullfill
445  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
446
447  where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:  where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
448  \label{IKM-EQU-2}  \label{IKM-EQU-2bbb}
449  -v\hackscore{i,i}=0  -v\hackscore{i,i}=0
450
451  Natural boundary conditions are taken in the form  Natural boundary conditions are taken in the form
# Line 544  we have Line 544  we have
544  \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}  \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}
545
546  which leads to  which leads to
547  \label{IKM-EQU-45}  \label{IKM-EQU-49}
548  \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},  \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
549  \eta\hackscore{eff}^{(n)}  \eta\hackscore{eff}^{(n)}
550  \frac{\Theta^{(n)}  + \omega^{(n)}  }  \frac{\Theta^{(n)}  + \omega^{(n)}  }

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