Linear elastic materials for large strains (Ciarlet model)

In [19] it is explained that substituting the infinitesimal strains in the classical Hooke law by the Lagrangian strain and the stress by the Piola-Kirchoff stress of the second kind does not lead to a physically sensible material law. In particular, such a model (also called St-Venant-Kirchoff material) does not exhibit large stresses when compressing the volume of the material to nearly zero. An alternative for isotropic materials is the following stored-energy function developed by Ciarlet [17] ($\mu$ and $\lambda$ are Lamé's constants):

$$\Sigma = \frac{\lambda}{4}(III_C - \ln III_C -1) + \frac{\mu }{4}(I_C - \ln III_C -3).$$ (295)

The stress-strain relation amounts to ( $\boldsymbol{S}$ is the Piola-Kirchoff stress of the second kind) :

$$\boldsymbol{S}= \frac{\lambda}{2}( \boldsymbol{C} -1) \boldsymbol{C^{-1}} + \mu (\boldsymbol{I}-\boldsymbol{C^{-1}}) ,$$ (296)

and the derivative of $\boldsymbol{S}$ with respect to the Green tensor $\boldsymbol{E}$ reads (component notation):

$$\frac{d S^{IJ} }{d E_{KL} } = \lambda ( \boldsymbol{C}) C^{{-1}^{KL}} C^{{-1}^{IJ}}+[2 \mu - \lambda ( \boldsymbol{C} -1)] C^{{-1}^{IK}} C^{{-1}^{LJ}}.$$ (297)

This model was implemented into CalculiX by Sven Kaßbohm. The definition consists of a *MATERIAL card defining the name of the material. This name HAS TO START WITH ”CIARLET_EL” but can be up to 80 characters long. Thus, the last 70 characters can be freely chosen by the user. Within the material definition a *USER MATERIAL card has to be used satisfying:

First line:

• *USER MATERIAL
• Enter the CONSTANTS parameter and its value, i.e. 2.

Following line:

• $E$ (Young's modulus).
• $\nu$ (Poisson's coefficient).
• Temperature.

Repeat this line if needed to define complete temperature dependence.

For this model, there are no internal state variables.

Example:

*MATERIAL,NAME=CIARLET_EL
*USER MATERIAL,CONSTANTS=2
210000.,.3,400.

defines an isotropic material with elastic constants $E$=210000. and $\nu$=0.3 for a temperature of 400 (units appropriately chosen by the user). Recall that

$$\mu= \frac{E}{2(1+\nu)}$$ (298)

and

$$\lambda=\frac{\nu E}{(1+\nu)(1-2 \nu)}.$$ (299)