Nonlinear finite elements/Objectivity of constitutive relations

Material frame indifference
An important consider in nonlinear finite element analysis is  material frame indifference or  objectivity of the material response. The idea is that the position of the observer frame should not affect the constitutive relations of a material. You can find more details and a history of the idea in Truesdell and Noll (1992) - sections 17, 18, 19, and 19A.

We have already talked about the objectivity of kinematic quantities and stress rates. Let us now discuss the same ideas with a particular constitutive model in mind.

Hyperelastic materials
A detailed description of thermoelastic materials can be found in Continuum mechanics/Thermoelasticity. In this discussion we will avoid the complications induced by including the temperature.

In the material configuration, a hyperelastic material satisfies two requirements:


 * 1) a stored energy function ($$W$$) exists for the material.
 * 2) the stored energy function depends locally only on the deformation gradient.

Given these requirements, if $$\boldsymbol{N} = \boldsymbol{P}^T$$ is the nominal stress ($$\boldsymbol{P}$$ is the first Piola-Kirchhoff stress tensor), then

\boldsymbol{N}^T(\mathbf{X}, t) = \boldsymbol{P}(\mathbf{X},t) = \rho_0~\frac{\partial W}{\partial \boldsymbol{F}}\left[\mathbf{X}, \boldsymbol{F}(\mathbf{X},t)\right] $$

Objectivity
The stored energy function $$W(\mathbf{X},\boldsymbol{F})$$ is said to be objective or frame indifferent if

W(\mathbf{X}, \boldsymbol{Q}\cdot\boldsymbol{F}) = W(\mathbf{X}, \boldsymbol{F}) $$ where $$\boldsymbol{Q}$$ is an orthogonal tensor with $$\boldsymbol{S}\cdot\boldsymbol{Q}^T = \boldsymbol{\mathit{1}}$$.

This objectivity condition can be achieved only if (in the material configuration)

W(\mathbf{X}, \boldsymbol{F}) = \hat{W}(\mathbf{X}, \boldsymbol{C}) ~; \boldsymbol{C} = \boldsymbol{F}^T\cdot\boldsymbol{F} $$ since $$\boldsymbol{C}_r = \boldsymbol{F}^T\cdot\boldsymbol{Q}^T\cdot\boldsymbol{Q}\cdot\boldsymbol{F} = \boldsymbol{C}$$.

We can show that

 Proof:

The stress strain relation for a hyperelastic material is

\boldsymbol{P} = \rho_0~\frac{\partial W}{\partial \boldsymbol{F}}\left[\mathbf{X}, \boldsymbol{F}(\mathbf{X},t)\right] $$ The chain rule then implies that

\boldsymbol{P}:\boldsymbol{T} = \rho_0~\frac{\partial \hat{W}}{\partial \boldsymbol{C}}:\left(\frac{\partial \boldsymbol{C}}{\partial \boldsymbol{F}}:\boldsymbol{T}\right) $$ for any second order tensor $$\boldsymbol{T}$$.

Now, using the product rule of differentiation,

\frac{\partial \boldsymbol{C}}{\partial \boldsymbol{F}}:\boldsymbol{T} = \frac{\partial }{\partial \boldsymbol{F}}\left(\boldsymbol{F}^T\cdot\boldsymbol{F}\right):\boldsymbol{T} = \left(\frac{\partial \boldsymbol{F}^T}{\partial \boldsymbol{F}}:\boldsymbol{T}\right)\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\left(\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{F}}:\boldsymbol{T}\right) $$ or,

\frac{\partial \boldsymbol{C}}{\partial \boldsymbol{F}}:\boldsymbol{T} = \left(\boldsymbol{\mathsf{I}}^T:\boldsymbol{T}\right)\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\left(\boldsymbol{\mathsf{I}}:\boldsymbol{T}\right) = \boldsymbol{T}^T\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\boldsymbol{T} $$ where $$\boldsymbol{\mathsf{I}}$$ is the fourth order identity tensor. Therefore,

\boldsymbol{P}:\boldsymbol{T} = \rho_0~\frac{\partial \hat{W}}{\partial \boldsymbol{C}}:\left(\boldsymbol{T}^T\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\boldsymbol{T}\right) $$ Using the identity

\boldsymbol{A}:(\boldsymbol{B}^T\cdot\boldsymbol{C}) = (\boldsymbol{C}\cdot\boldsymbol{A}^T):\boldsymbol{B} = (\boldsymbol{B}\cdot\boldsymbol{A}):\boldsymbol{C} $$ we have

\frac{\partial \hat{W}}{\partial \boldsymbol{C}}:(\boldsymbol{T}^T\cdot\boldsymbol{F}) = \left[\boldsymbol{F}\cdot\left(\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\right)^T\right]:\boldsymbol{T} \quad \text{and} \quad \frac{\partial \hat{W}}{\partial \boldsymbol{C}}:(\boldsymbol{F}^T\cdot\boldsymbol{T}) = \left[\boldsymbol{F}\cdot\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\right]:\boldsymbol{T} $$ Therefore, invoking the arbitrariness of $$\boldsymbol{T}$$, we have

\boldsymbol{P} = \rho_0~\boldsymbol{F}\cdot\left(\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\right)^T + \rho_0~\boldsymbol{F}\cdot\frac{\partial \hat{W}}{\partial \boldsymbol{C}} $$ Since $$\boldsymbol{C} = \boldsymbol{C}^T$$ we have

\left(\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\right)^T = \frac{\partial \hat{W}}{\partial \boldsymbol{C}} $$ which implies that

\boldsymbol{P} = 2~\rho_0~\boldsymbol{F}\cdot\frac{\partial \hat{W}}{\partial \boldsymbol{C}} \quad \text{or} \quad \boldsymbol{N} = 2~\rho_0~\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\cdot\boldsymbol{F}^T $$ Recall the relations between the 2nd Piola-Kirchhoff stress tensor and the first Piola-Kirchhoff stress tensor (and the nominal stress tensor)

\boldsymbol{S} = \boldsymbol{F}^{-1}\cdot\boldsymbol{P} $$ Therefore, we have

\boldsymbol{S} = 2~\rho_0~\frac{\partial \hat{W}}{\partial \boldsymbol{C}} $$ Also from the relation between the Cauchy stress and the 2nd Piola-Kirchhoff stress tensor

\boldsymbol{\sigma} = J^{-1}~\boldsymbol{F}\cdot\boldsymbol{S}\cdot\boldsymbol{F}^T = \cfrac{\rho}{\rho_0}~\boldsymbol{F}\cdot\boldsymbol{S}\cdot\boldsymbol{F}^T $$ we have

\boldsymbol{\sigma} = 2~\rho~\boldsymbol{F}\cdot\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\cdot\boldsymbol{F}^T \quad \text{or} \quad \boldsymbol{\tau} = 2~\rho_0\boldsymbol{F}\cdot\frac{\partial \hat{W}}{\partial \boldsymbol{C}}\cdot\boldsymbol{F}^T $$

We may also express these relations in terms of the Lagrangian Green strain

\boldsymbol{E} = \frac{1}{2}(\boldsymbol{F}^T\cdot\boldsymbol{F} - \boldsymbol{\mathit{1}}) = \frac{1}{2}(\boldsymbol{C} - \boldsymbol{\mathit{1}}) $$ Then we have

\frac{\partial W}{\partial \boldsymbol{C}} = \frac{\partial W}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial \boldsymbol{C}} = \frac{1}{2}~\frac{\partial W}{\partial \boldsymbol{E}} $$ Hence, we can write

The stored energy function $$W$$ is objective if and only if the Cauchy stress tensor is symmetric, i.e., if the balance of angular momentum holds.  Show this.