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revision 3294 by gross, Fri Jul 23 07:10:03 2010 UTC revision 3295 by jfenwick, Fri Oct 22 01:56:02 2010 UTC
# Line 26  The following sections give a breif over Line 26  The following sections give a breif over
26    
27    
28  \section{Isotropic Kelvin Material \label{IKM}}  \section{Isotropic Kelvin Material \label{IKM}}
29  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  As proposed by Kelvin~\cite{Muhlhaus2005} material strain $D_{ij}=\frac{1}{2}(v_{i,j}+v_{j,i})$ can be decomposed into
30  an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:  an elastic part $D_{ij}^{el}$ and visco-plastic part $D_{ij}^{vp}$:
31  \begin{equation}\label{IKM-EQU-2}  \begin{equation}\label{IKM-EQU-2}
32  D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}  D_{ij}=D_{ij}^{el}+D_{ij}^{vp}
33  \end{equation}  \end{equation}
34  with the elastic strain given as  with the elastic strain given as
35  \begin{equation}\label{IKM-EQU-3}  \begin{equation}\label{IKM-EQU-3}
36  D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'  D_{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}_{ij}'
37  \end{equation}  \end{equation}
38  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).  where $\sigma'_{ij}$ is the deviatoric stress (Notice that $\sigma'_{ii}=0$).
39  If the material is composed by materials $q$ the visco-plastic strain can be decomposed as  If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
40  \begin{equation}\label{IKM-EQU-4}  \begin{equation}\label{IKM-EQU-4}
41  D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}  D_{ij}^{vp'}=\sum_{q} D_{ij}^{q'}
42  \end{equation}  \end{equation}
43  where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as  where $D_{ij}^{q}$ is the strain in material $q$ given as
44  \begin{equation}\label{IKM-EQU-5}  \begin{equation}\label{IKM-EQU-5}
45  D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}  D_{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'_{ij}
46  \end{equation}  \end{equation}
47  where $\eta^{q}$ is the viscosity of material $q$. We assume the following  where $\eta^{q}$ is the viscosity of material $q$. We assume the following
48  betwee the the strain in material $q$  betwee the the strain in material $q$
49  \begin{equation}\label{IKM-EQU-5b}  \begin{equation}\label{IKM-EQU-5b}
50  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{1-n^{q}}  \eta^{q}=\eta^{q}_{N} \left(\frac{\tau}{\tau_{t}^q}\right)^{1-n^{q}}
51  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'_{ij} \sigma'_{ij}}
52  \end{equation}  \end{equation}
53  for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\cite{Muhlhaus2005}.  for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau_{t}^q$, see~\cite{Muhlhaus2005}.
54  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
55  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:
56  \begin{equation}\label{IKM-EQU-6}  \begin{equation}\label{IKM-EQU-6}
57  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_{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'_{ij} \mbox{ with } \frac{1}{\eta^{vp}} = \sum_{q} \frac{1}{\eta^{q}} \;.
58  \end{equation}  \end{equation}
59  and finally with~\ref{IKM-EQU-2}  and finally with~\ref{IKM-EQU-2}
60  \begin{equation}\label{IKM-EQU-2bb}  \begin{equation}\label{IKM-EQU-2bb}
61  D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'  D_{ij}'=\frac{1}{2 \eta^{vp}} \sigma'_{ij}+\frac{1}{2 \mu} \dot{\sigma}_{ij}'
62  \end{equation}  \end{equation}
63  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}
64  \begin{equation}\label{IKM-EQU-8c}  \begin{equation}\label{IKM-EQU-8c}
65  \tau \le \tau\hackscore{Y} + \beta \; p  \tau \le \tau_{Y} + \beta \; p
66  \end{equation}  \end{equation}
67  with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.  with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau_{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.
68  The deviatoric stress needs to fullfill the equilibrion equation  The deviatoric stress needs to fullfill the equilibrion equation
69  \begin{equation}\label{IKM-EQU-1}  \begin{equation}\label{IKM-EQU-1}
70  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}  -\sigma'_{ij,j}+p_{,i}=F_{i}
71  \end{equation}  \end{equation}
72  where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:  where $F_{j}$ is a given external fource. We assume an incompressible media:
73  \begin{equation}\label{IKM-EQU-2bbb}  \begin{equation}\label{IKM-EQU-2bbb}
74  -v\hackscore{i,i}=0  -v_{i,i}=0
75  \end{equation}  \end{equation}
76  Natural boundary conditions are taken in the form  Natural boundary conditions are taken in the form
77  \begin{equation}\label{IKM-EQU-Boundary}  \begin{equation}\label{IKM-EQU-Boundary}
78  \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f  \sigma'_{ij}n_{j}-n_{i}p=f
79  \end{equation}  \end{equation}
80  which can be overwritten by a constraint  which can be overwritten by a constraint
81  \begin{equation}\label{IKM-EQU-Boundary0}  \begin{equation}\label{IKM-EQU-Boundary0}
82  v\hackscore{i}(x)=0  v_{i}(x)=0
83  \end{equation}  \end{equation}
84  where the index $i$ may depend on the location $x$ on the bondary.  where the index $i$ may depend on the location $x$ on the bondary.
85    
86  \subsection{Solution Method \label{IKM-SOLVE}}  \subsection{Solution Method \label{IKM-SOLVE}}
87  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
88  \begin{equation}\label{IKM-EQU-3b}  \begin{equation}\label{IKM-EQU-3b}
89  \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)  \dot{\sigma}_{ij}=\frac{1}{dt } \left( \sigma_{ij} - \sigma_{ij}^{-} \right)
90  \end{equation}  \end{equation}
91  and  and
92  \begin{equation}\label{IKM-EQU-2b}  \begin{equation}\label{IKM-EQU-2b}
93  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_{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma_{ij}'-\frac{1}{2 \mu dt } \sigma_{ij}^{-'}
94  \end{equation}  \end{equation}
95  where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With  where $\sigma_{ij}^{-}$ is the stress at the precious time step. With
96  \begin{equation}\label{IKM-EQU-2c}  \begin{equation}\label{IKM-EQU-2c}
97  \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' +  \dot{\gamma} = \sqrt{ 2 \left( D_{ij}' +
98  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}  \frac{1}{  2 \mu \; dt} \sigma_{ij}^{-'}\right)^2}
99  \end{equation}  \end{equation}
100  we have  we have
101  \begin{equation}  \begin{equation}
102  \tau = \eta\hackscore{eff} \cdot \dot{\gamma}  \tau = \eta_{eff} \cdot \dot{\gamma}
103  \end{equation}  \end{equation}
104  where  where
105  \begin{equation}  \begin{equation}
106  \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}  \eta_{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
107  , \eta\hackscore{max}) \mbox{ with }  , \eta_{max}) \mbox{ with }
108  \eta\hackscore{max} = \left\{  \eta_{max} = \left\{
109  \begin{array}{rcl}  \begin{array}{rcl}
110  \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\  \frac{\tau_{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
111  &\mbox{ if } \\  &\mbox{ if } \\
112  \infty & & \mbox{otherwise}  \infty & & \mbox{otherwise}
113  \end{array}  \end{array}
114  \right.  \right.
115  \end{equation}  \end{equation}
116  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 $\eta_{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form
117  \begin{equation}\label{IKM-EQU-10}  \begin{equation}\label{IKM-EQU-10}
118  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +  \sigma_{ij}' =  2 \eta_{eff}  \left( D_{ij}' +
119  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)    \frac{1}{  2 \mu \; dt} \sigma_{ij}^{'-}\right)  
120  \end{equation}  \end{equation}
121  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
122  \begin{equation}\label{IKM-EQU-1ib}  \begin{equation}\label{IKM-EQU-1ib}
123  -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})  -\left(\eta_{eff} (v_{i,j}+ v_{i,j})
124  \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+  \right)_{,j}+p_{,i}=F_{i}+
125   \left(\frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}   \left(\frac{\eta_{eff}}{\mu dt } \sigma_{ij}^{'-} \right)_{,j}
126  \end{equation}  \end{equation}
127  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
128  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.
# Line 133  If we set Line 133  If we set
133  \end{equation}  \end{equation}
134  we need to solve the nonlinear problem  we need to solve the nonlinear problem
135  \begin{equation}  \begin{equation}
136  \eta\hackscore{eff} -  min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})  \eta_{eff} -  min(\eta( \dot{\gamma} \cdot \eta_{eff})
137  , \eta\hackscore{max}) =0  , \eta_{max}) =0
138  \end{equation}  \end{equation}
139  We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem  We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
140  \begin{equation}\label{IKM-EQU-45}  \begin{equation}\label{IKM-EQU-45}
141  \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},  \eta_{eff}^{(n+1)} = \min(\eta_{max},
142  \eta\hackscore{eff}^{(n)} -  \eta_{eff}^{(n)} -
143  \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }  \frac{\eta_{eff}^{(n)} - \eta( \tau^{(n)}) }
144  {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )  {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
145  =\min(\eta\hackscore{max},  =\min(\eta_{max},
146  \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot  \eta'( \tau^{(n)} )  }  \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot  \eta'( \tau^{(n)} )  }
147  {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )  {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
148  \end{equation}  \end{equation}
149  where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$  where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
150  and $\tau^{(n)} =  \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.  and $\tau^{(n)} =  \dot{\gamma} \cdot \eta_{eff}^{(n)}$.
151    
152  Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate  Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
153  the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.  the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
# Line 155  In fact we have Line 155  In fact we have
155  \begin{equation}  \begin{equation}
156  \eta' = - \frac{\Theta'}{\Theta^2}  \eta' = - \frac{\Theta'}{\Theta^2}
157  \mbox{ with }  \mbox{ with }
158  \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'  \Theta' = \sum_{q} \left(\frac{1}{\eta^{q}} \right)'
159  \label{IKM iteration 7}  \label{IKM iteration 7}
160  \end{equation}  \end{equation}
161  As  As
162  \begin{equation}\label{IKM-EQU-47}  \begin{equation}\label{IKM-EQU-47}
163  \left(\frac{1}{\eta^{q}} \right)'  \left(\frac{1}{\eta^{q}} \right)'
164  = \frac{n^{q}-1}{\eta^{q}\hackscore{N}} \cdot \frac{\tau^{n^{q}-2}}{(\tau\hackscore{t}^q)^{n^{q}-1}}  = \frac{n^{q}-1}{\eta^{q}_{N}} \cdot \frac{\tau^{n^{q}-2}}{(\tau_{t}^q)^{n^{q}-1}}
165  = \frac{n^{q}-1}{\eta^{q}}\cdot\frac{1}{\tau}  = \frac{n^{q}-1}{\eta^{q}}\cdot\frac{1}{\tau}
166  \end{equation}  \end{equation}
167  we have  we have
168  \begin{equation}\label{IKM-EQU-48}  \begin{equation}\label{IKM-EQU-48}
169  \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{n^{q}-1}{\eta^{q}}  \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum_{q}\frac{n^{q}-1}{\eta^{q}}
170  \end{equation}  \end{equation}
171  which leads to  which leads to
172  \begin{equation}\label{IKM-EQU-49}  \begin{equation}\label{IKM-EQU-49}
173  \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},  \eta_{eff}^{(n+1)} = \min(\eta_{max},
174  \eta\hackscore{eff}^{(n)}  \eta_{eff}^{(n)}
175  \frac{\Theta^{(n)}  + \omega^{(n)}  }  \frac{\Theta^{(n)}  + \omega^{(n)}  }
176  {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })  {\eta_{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })
177  \end{equation}  \end{equation}
178    
179    
# Line 266  sets the relative tolerance for the effe Line 266  sets the relative tolerance for the effe
266    
267  \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setDruckerPragerLaw}  \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setDruckerPragerLaw}
268  {\optional{tau_Y=None, \optional{friction=None}}}  {\optional{tau_Y=None, \optional{friction=None}}}
269  Sets the parameters $\tau\hackscore{Y}$ and $\beta$ for the Drucker-Prager model in condition~\ref{IKM-EQU-8c}. If \var{tau_Y} is set to None (default) Drucker-Prager  Sets the parameters $\tau_{Y}$ and $\beta$ for the Drucker-Prager model in condition~\ref{IKM-EQU-8c}. If \var{tau_Y} is set to None (default) Drucker-Prager
270  condition is not applied.  condition is not applied.
271  \end{methoddesc}  \end{methoddesc}
272    
# Line 278  Sets the elastic shear modulus $\mu$. If Line 278  Sets the elastic shear modulus $\mu$. If
278  \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setPowerLaws}{eta_N, tau_t, power}  \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setPowerLaws}{eta_N, tau_t, power}
279  Sets the parameters of the power-law for all materials as defined in  Sets the parameters of the power-law for all materials as defined in
280  equation~\ref{IKM-EQU-5b}.  equation~\ref{IKM-EQU-5b}.
281  \var{eta_N} is the list of viscosities $\eta^{q}\hackscore{N}$,  \var{eta_N} is the list of viscosities $\eta^{q}_{N}$,
282  \var{tau_t} is the list of reference stresses  $\tau\hackscore{t}^q$,  \var{tau_t} is the list of reference stresses  $\tau_{t}^q$,
283  and \var{power} is the list of  power law coefficients $n^{q}$.  and \var{power} is the list of  power law coefficients $n^{q}$.
284  \end{methoddesc}  \end{methoddesc}
285    

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