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

revision 2299 by gross, Wed Feb 11 06:48:28 2009 UTC revision 2300 by gross, Wed Mar 11 08:17:57 2009 UTC
# Line 431  After inserting equation~\ref{IKM-EQU-5} Line 431  After inserting equation~\ref{IKM-EQU-5}
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  With  and finally with~\ref{IKM-EQU-2}
435  \label{IKM-EQU-8}  \label{IKM-EQU-2b}
436  \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}  D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'

one gets
\label{IKM-EQU-8b}
\tau = \eta^{vp} \dot{\gamma}^{vp} \;.
437
438  With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve  The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}
439  \label{IKM-EQU-8c}  \label{IKM-EQU-8c}
440  \tau \le \tau\hackscore{Y} + \beta \; p  \tau \le \tau\hackscore{Y} + \beta \; p
441
442  which leads to the condition  with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.
\label{IKM-EQU-8d}
\eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .

Therefore we modify the definition of $\eta^{vp}$ to the form
\label{IKM-EQU-6b}
\frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})

443  The deviatoric stress needs to fullfill the equilibrion equation  The deviatoric stress needs to fullfill the equilibrion equation
444  \label{IKM-EQU-1}  \label{IKM-EQU-1}
445  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
# Line 472  where the index $i$ may depend on the lo Line 461  where the index $i$ may depend on the lo
461  \subsection{Solution Method \label{IKM-SOLVE}}  \subsection{Solution Method \label{IKM-SOLVE}}
462  By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form  By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
463  \label{IKM-EQU-3b}  \label{IKM-EQU-3b}
464  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)  \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)
465
466  where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.  and
467  Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get  \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}^{-'}
469
470    where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
471    \label{IKM-EQU-2c}
472    \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' -
473    \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
474
475    we have
476
477    \tau = \eta\hackscore{eff} \cdot \dot{\gamma}
478
479    where
480
481    \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
482    , \eta\hackscore{max}) \mbox{ with }
483    \eta\hackscore{max} = \left\{
484    \begin{array}{rcl}
485    \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
486    &\mbox{ if } \\
487    \infty & & \mbox{otherwise}
488    \end{array}
489    \right.
490
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
492  \label{IKM-EQU-10}  \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)  \mbox{ with }  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)
\frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
495
Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
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  \label{IKM-EQU-1ib}  \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}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
501
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.
504  In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ which
505  is a function of $\sigma\hackscore{ij}$ via $\tau$.  To get $\tau$ and $\eta\hackscore{eff}$ we need to solve the  If we set
506  non-linear equation  \label{IKM-EQU-44}
507    \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
508  \tau = \eta\hackscore{eff} \cdot \dot{\gamma}\hackscore{total} \mbox{ with }
509  \dot{\gamma}\hackscore{total} = \sqrt{ 2 \left( D\hackscore{ij}' +  we need to solve the nonlinear problem
\frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
\label{IKM iteration 5}

The Newton scheme takes the form
510
511  \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff}  \cdot \dot{\gamma}\hackscore{total}}{1 - \eta\hackscore{eff}'  \cdot  \dot{\gamma}\hackscore{total}}, \tau\hackscore{Y} + \beta \; p)  \eta\hackscore{eff} -  min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})
512  \label{IKM iteration 6}  , \eta\hackscore{max}) =0
513
514    We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
515    \label{IKM-EQU-45}
516    \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
517    \eta\hackscore{eff}^{(n)} -
518    \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }
519    {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
520    =\min(\eta\hackscore{max},
521    \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot  \eta'( \tau^{(n)} )  }
522    {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
523
524  where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ (?)). We have  where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
525    and $\tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.
526
527    Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
528    the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
529    In fact we have
530
531  \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'  \eta' = - \frac{\Theta'}{\Theta^2}
532  \mbox{ with }  \mbox{ with }
533  \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'  \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
534  \label{IKM iteration 7}  \label{IKM iteration 7}
535
536  \label{IKM-EQU-5XX}  As
537    \label{IKM-EQU-47}
538  \left(\frac{1}{\eta^{q}} \right)'  \left(\frac{1}{\eta^{q}} \right)'
539  = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}  = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}
540  = \frac{1-\frac{1}{n^{q}}}{\tau}\frac{1}{\eta^{q}}  = \frac{1-\frac{1}{n^{q}}}{\eta^{q}}\frac{1}{\tau}
541
542  Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}  we have
543  positive.  \label{IKM-EQU-48}
544    \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}
545