Large Plastic Deformation Theory

[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.

 

[AndyW]

Thank you for your question regarding the constitutive law for large plastic deformation theory based on the multiplicative decomposition of the deformation gradient. Allow me to clarify some points and provide some insights on how this law is derived.

Firstly, you are correct in your understanding that the Cauchy stress and total deformation rate are replaced by the Kirchhoff stress and the symmetric part of the total deformation gradient, respectively, in the large deformation theory based on the multiplicative decomposition of F. This is because the Kirchhoff stress and the symmetric part of the total deformation gradient are the equivalent quantities in a Lagrangian description, while the Cauchy stress and total deformation rate are used in an Eulerian description.

Secondly, the relation \tau = 2 \dfrac{\partial \Psi}{\partial b_e}b_e is derived from the definition of the Kirchhoff stress as \tau = J^{-1}\sigma F^{-T}, where J is the determinant of the deformation gradient and \sigma is the Cauchy stress. By substituting JF^{-T} for \sigma, we can use the chain rule to obtain \tau = J^{-1}F^{-1}\dfrac{\partial \Psi}{\partial b_e}F^{-T}F^{-1}F^{-T}. Simplifying this expression using the definition of b_e=F^TF and the fact that F^TF is symmetric, we get \tau = 2 \dfrac{\partial \Psi}{\partial b_e}b_e.

Finally, the expression for the dissipation function (relation 2) is obtained by substituting the definition of \tau from relation 1 into the Clausius-Duhem inequality and simplifying the terms. This is a common approach in deriving constitutive laws from thermodynamic principles.

I hope this helps in clarifying your doubts. If you have any further questions, please do not hesitate to ask.