# Large Plastic Deformation Theory

1. Jun 17, 2014

### trabo

Hello everyone,

I'm studying the large plastic deformation theory based on the multiplicative decomposition of the deformation gradient, commonly noted as the matrix F, into an elastic and plastic part $F=F_ e F_ p$ (theory introduced by Simo and Miehe) The constitutive law is :
$-\dfrac{1}{2}L _ v(b _ e)=d\lambda \dfrac{\partial f}{\partial \tau} b _ e$ ​
where $b _ e=F_e ^tF_e$, $L _ v$ is the Lie derivative operator, $f$ the yield criterion ($f \le 0$), and $\tau$ the Kirchhoff stress.
Actually, I don't understand how this law can be found. I read some articles on the net and this is what I came to :
-First, we use the Clausius-Duhem inequality. This inequality is usually written with the Cauchy stress and the total deformation rate. In the large deformation therory based on the multiplicative decomposition of F, the Cauchy stress is replaced by the kirchhoff stress and the total rate deformation by $d=sym \Big(\dfrac{d}{dt} (F)F^{-1}\Big )$. I assume that these new quantities are the equivalent of the first ones when an eulerien description is used, but I'm not sure of that.
-After that, it is stated that $\tau=2 \dfrac{\partial \Psi}{\partial b_e}b_e$ (relation 1) where $\Psi$ is the free energy. It is asumed that it depends only on $b_e$ and an internal variable $\xi$.
-The dissipation function in the Clausius-Duhem inequality is expanded into
$\tau:\Big(-\dfrac{1}{2}L_v(b_e)b_e^{-1}\Big)-\dfrac{\partial \Psi}{\partial \xi} \dfrac{d\xi}{dt} \ge 0$ (relation 2)​
- Then the maximum plastic dissipation principle is used to state the law

What I don't understand is why we have relation 1 and how do we get to the above expression of the dissipation function (relation 2).
If you have any ideas

Regards.

Last edited: Jun 17, 2014