Isotropic Kelvin Material

As proposed by Kelvin [15] material strain $ D\hackscore{ij}=\frac{1}{2}(v\hackscore{i,j}+v\hackscore{j,i})$ can be decomposed into an elastic part $ D\hackscore{ij}^{el}$ and visco-plastic part $ D\hackscore{ij}^{vp}$:

$\displaystyle D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}$ (129)

with the elastic strain given as

$\displaystyle D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'$ (130)

where $ \sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $ \sigma'\hackscore{ii}=0$). If the material is composed by materials $ q$ the visco-plastic strain can be decomposed as

$\displaystyle D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}$ (131)

where $ D\hackscore{ij}^{q}$ is the strain in material $ q$ given as

$\displaystyle D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}$ (132)

where $ \eta^{q}$ is the viscosity of material $ q$. We assume the following betwee the the strain in material $ q$

$\displaystyle \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{1-n^{q}}$    with $\displaystyle \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}$ (133)

for a given power law coefficients $ n^{q}\ge1$ and transition stresses $ \tau\hackscore{t}^q$, see [15]. Notice that $ n^{q}=1$ gives a constant viscosity. After inserting equation 6.43 into equation 6.42 one gets:

$\displaystyle D\hackscore{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}$    with $\displaystyle \frac{1}{\eta^{vp}} = \sum\hackscore{q} \frac{1}{\eta^{q}} \;.$ (134)

and finally with 6.40

$\displaystyle D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'$ (135)

The total stress $ \tau$ needs to fullfill the yield condition

$\displaystyle \tau \le \tau\hackscore{Y} + \beta \; p$ (136)

with the Drucker-Prager cohesion factor $ \tau\hackscore{Y}$, Drucker-Prager friction $ \beta$ and total pressure $ p$. The deviatoric stress needs to fullfill the equilibrion equation

$\displaystyle -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}$ (137)

where $ F\hackscore{j}$ is a given external fource. We assume an incompressible media:

$\displaystyle -v\hackscore{i,i}=0$ (138)

Natural boundary conditions are taken in the form

$\displaystyle \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f$ (139)

which can be overwritten by a constraint

$\displaystyle v\hackscore{i}(x)=0$ (140)

where the index $ i$ may depend on the location $ x$ on the bondary.



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