Following the multiplicative decomposition of the deformation gradient, the Helmholtz free energy can be decomposed as
@f[
- \Psi(\mathbf{b}) = \Psi(\mathbf{J})_{\text{vol}} + \Psi(\overline{\mathbf{b}})_{\text{iso}} \, .
+ \Psi(\mathbf{b}) = \Psi_{\text{vol}}(\mathbf{J}) + \Psi_{\text{iso}}(\overline{\mathbf{b}}) \, .
@f]
Similarly, the Kirchhoff stress can be decomposed into volumetric and isochoric parts as $\boldsymbol{\tau} = \boldsymbol{\tau}_{\text{vol}} + \boldsymbol{\tau}_{\text{iso}}$ where:
@f{align*}
:= 2 \overline{\mathbf{b}} \dfrac{\partial \Psi_{\textrm{iso}}(\overline{\mathbf{b}})}{\partial \overline{\mathbf{b}}} \, .
@f]
-<h3> neo-Hookean materials </h3>
+<h3> Nseo-Hookean materials </h3>
The Helmholtz free energy corresponding to a compressible neo-Hookean material is given by
@f[
\Psi \equiv
\underbrace{\kappa [ \mathcal{G}(J) ] }_{\Psi_{\textrm{vol}}(J)}
- + \underbrace{\bigl[c_1 [ \overline{I}_1 - 3] \bigr]}_{\Psi(\overline{\mathbf{b}})_{\text{iso}}} \, ,
+ + \underbrace{\bigl[c_1 [ \overline{I}_1 - 3] \bigr]}_{\Psi_{\text{iso}}(\overline{\mathbf{b}})} \, ,
@f]
where $\kappa := \lambda + 2/3 \mu$ is the bulk modulus
and $\overline{I}_1 := \textrm{tr}\overline{\mathbf{b}}$.
The function $\mathcal{G}(J)$ is required to be strictly convex and satisfy the condition $\mathcal{G}(1) = 0$.
-In this work $\mathcal{G}:=\bigl[ \frac{1}{2} [ \widetilde{J}^{2} - 1 ] - \textrm{ln}( \widetilde{J}) ] \bigr]$.
+In this work $\mathcal{G}:=\bigl[ \frac{1}{2} [{J}^{2} - 1 ] - \textrm{ln}( {J}) ] \bigr]$.
Incompressibility imposes the iscohoric consraint that $J=1$ for all motions $\mathbf{\varphi}$.
The Helmholtz free energy corresponding to an incompressible neo-Hookean material is given by
@f[
\Psi \equiv
- \underbrace{\bigl[ c_1 [ I_1 - 3] \bigr] }_{\Psi(\mathbf{b})_{\textrm{iso}}} \, ,
+ \underbrace{\bigl[ c_1 [ I_1 - 3] \bigr] }_{\Psi_{\textrm{iso}}(\mathbf{b})} \, ,
@f]
$ I_1 := \textrm{tr}\mathbf{b} $.
Thus, the incompressible response of obtained by removing the volumetric component from the compressible free energy.
\Pi(\mathbf{\Xi}) := \int_\Omega \bigl[
\Psi_{\textrm{vol}}(\widetilde{J})
+ p[J(\mathbf{u}) - \widetilde{J}]
- + \Psi_{\textrm{iso}}(\mathbf{b}(\mathbf{u}))
+ + \Psi_{\textrm{iso}}(\overline{\mathbf{b}}(\mathbf{u}))
\bigr] \textrm{d}v
+ \Pi_{\textrm{ext}} \, .
@f]
291-304.
</ol>
+ @image html "setup.png"
+