Mixed Hyperelasticity#

Ratel solves the momentum balance equations using unstructured high-order finite/spectral element spatial discretizations.

Initial configuration#

In the total Lagrangian approach for the incompressible Neo-Hookean hyperelasticity problem, the discrete equations are formulated with respect to the initial configuration. In this formulation, we solve for displacement and pressure $$\bm{u} \left( \bm{X} \right), p$$ in the reference frame $$\bm{X}$$. The notation for elasticity at finite strain is inspired by [Hol00] to distinguish between the current and initial configurations. As explained in the Continuum Mechanics section, we denote by capital letters the reference frame and by small letters the current one.

Weak form#

We multiply the following strong form

(107)#\begin{aligned} \nabla_X \cdot \bm{P} + \rho_0 \bm{g} &= 0 \quad \text{in} \quad \Omega_{0} \\ - \frac{\partial V}{\partial J} - \frac{p}{\bulk - \bulk_p}&= 0 \quad \text{in} \quad \Omega \end{aligned}

by test functions $$(\bm{v}, q)$$ and integrate by parts to obtain the weak form for finite-strain incompressible hyperelasticity: find $$(\bm{u}, p) \in \mathcal{V} \times \mathcal{Q} \subset H^1 \left( \Omega_0 \right) \times L^2 \left( \Omega_0 \right)$$ such that

(108)#\begin{aligned} \int_{\Omega_0}{\nabla_X \bm{v} \tcolon \bm{P}} \, dV - \int_{\Omega_0}{\bm{v} \cdot \rho_0 \bm{g}} \, dV - \int_{\partial \Omega_0}{\bm{v} \cdot \left( \bm{P} \cdot \hat{\bm{N}} \right)} \, dS &= 0, \quad \forall \bm{v} \in \mathcal{V}, \\ \int_{\Omega_0} q \cdot \left( -V' - \frac{p}{\bulk - \bulk_p} \right) J \, dV &= 0, \quad \forall q\in \mathcal{Q}, \end{aligned}

where $$\bm{P} \cdot \hat{\bm{N}}|_{\partial\Omega}$$ is replaced by any prescribed force/traction boundary condition written in terms of the initial configuration. This equation contains material/constitutive nonlinearities in defining $$\bm{S}(\bm{E})$$, as well as geometric nonlinearities through $$\bm{P} = \bm{F}\, \bm{S}$$, $$\bm{E}(\bm{F})$$, and the body force $$\bm{g}$$, which must be pulled back from the current configuration to the initial configuration.

Newton linearization#

To derive a Newton linearization, we need the Jacobian form of the (108): find $$(\diff \bm{u}, \diff p) \in \mathcal{V} \times \mathcal{Q}$$ such that

(109)#\begin{aligned} \int_{\Omega_0} \nabla_X \bm{v} \tcolon \diff \bm{P} dV &= \text{rhs}, \quad \forall \bm{v} \in \mathcal{V}, \\ \int_{\Omega_0} q \cdot \left( \diff L J + L \diff J \right)dV &= 0, \quad \forall q \in \mathcal{Q} \end{aligned}

where $$L = -V' - \frac{p}{\bulk - \bulk_p}$$ and

(110)#\begin{aligned} \diff \bm{P} &= \diff \bm{F}\, \bm{S} + \bm{F} \diff \bm{S}, \\ \diff L J + L \diff J &= J \frac{\partial L}{\partial \bm{E}} \tcolon \diff \bm{E} + J \frac{\partial L}{\partial p} \diff p + L \frac{\partial J}{\partial \bm{E}} \tcolon \diff \bm{E} \end{aligned}

with $$\diff \bm{F} = \nabla_X\diff \bm{u}$$ and

$\diff \bm{E} = \frac{\partial \bm{E}}{\partial \bm{F}} \tcolon \diff \bm{F} = \frac{1}{2} \left(\diff \bm{F}^T \bm{F} + \bm{F}^T \diff \bm{F} \right).$

The linearization of the second equation of (110) is

(111)#\begin{aligned} \diff L J + L \diff J &= \left( -V^{''} \diff J - \frac{\diff p}{\bulk - \bulk_p} \right) J + L J (\bm{C}^{-1} \tcolon \diff \bm{E}), \\ &= \left( - J^2 \, V^{''} - J \, V' - J \frac{p}{\bulk - \bulk_p} \right) \bm{C}^{-1} \tcolon \diff \bm{E} - J \frac{\diff p}{\bulk - \bulk_p} \end{aligned}

The linearization of the second Piola-Kirchhoff stress tensor, $$\diff \bm{S}$$, depends upon the material model.

Deriving $$\diff\bm{S}$$ for isochoric Neo-Hookean material

For the Neo-Hookean model (46), we derive split

(112)#$\diff \bm{S} = \underbrace{\frac{\partial \bm{S}_{vol}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{vol}^u} + \underbrace{\frac{\partial \bm{S}_{iso}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{iso}} + \underbrace{\frac{\partial \bm{S}_{vol}}{\partial p} \diff p}_{\diff \bm{S}_{vol}^p},$

then,

(113)#\begin{aligned} \diff \bm{S}_{vol}^u &= \left( \bulk_p \diff J \, V' + \bulk_p J V^{''} \diff J - p \diff J \right) \bm{C}^{-1} + \left(\bulk_p J \, V' - p \, J \right) \diff \bm{C}^{-1}, \\ &= \left( \bulk_p J^2 V^{''} + \bulk_p J \, V' - p J \right) (\bm{C}^{-1} \tcolon \diff \bm{E}) \bm{C}^{-1} + \left(\bulk_p J \, V' - p \, J \right) \diff \bm{C}^{-1} \end{aligned}
(114)#$\diff \bm{S}_{vol}^p = -J \bm{C}^{-1} \diff p,$

and

(115)#\begin{aligned} \diff \bm{S}_{iso} &= -\frac{2}{3}\mu J^{-2/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \left(\bm{I}_3 - \frac{1}{3} \mathbb{{I}_1} \bm{C}^{-1} \right) \\ &- \frac{1}{3} \mu J^{-2/3} \left( 2 \trace \diff \bm{E} \, \bm{C}^{-1} + \mathbb{{I}_1} \diff \bm{C}^{-1} \right), \\ &= -\frac{4}{3}\mu J^{-2/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \bm{C}^{-1} \bm{E}_{dev} - \frac{1}{3} \mu J^{-2/3} \left( 2 \trace \diff \bm{E} \, \bm{C}^{-1} + \mathbb{{I}_1} \diff \bm{C}^{-1} \right) \end{aligned}

where

$\diff \bm{C}^{-1} = \frac{\partial \bm{C}^{-1}}{\partial \bm{E}} \tcolon \diff \bm{E} = -2 \bm{C}^{-1} \diff \bm{E} \, \bm{C}^{-1} .$

Note

If we use single field isochoric model (50) the linearization of the volumetric stress becomes

(116)#$\diff \bm{S}_{vol} = \left( \bulk J^2 V^{''} + \bulk J \, V'\right) (\bm{C}^{-1} \tcolon \diff \bm{E}) \bm{C}^{-1} + \bulk J \, V' \diff \bm{C}^{-1}.$

$$\diff \bm{S}_{vol}^p = 0$$ and we only need to solve the first equation of (108).

Deriving $$\diff\bm{S}$$ for isochoric Mooney-Rivlin material

For the Mooney-Rivlin model (58), we derive split

(117)#$\diff \bm{S} = \underbrace{\frac{\partial \bm{S}_{vol}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{vol}^u} + \underbrace{\frac{\partial \bm{S}_{iso}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{iso}} + \underbrace{\frac{\partial \bm{S}_{vol}}{\partial p} \diff p}_{\diff \bm{S}_{vol}^p},$

where $$\diff \bm{S}_{vol}^p$$ and $$\diff \bm{S}_{vol}^u$$ are the same as (114), (113) (or (116) if we use single field stress (62)) and the isochoric part is

(118)#\begin{aligned} \diff \bm{S}_{iso} &= -\frac{4}{3}(\bm{C}^{-1} \tcolon \diff \bm{E}) \left( \mu_1 J^{-2/3} + 4 \mu_2 J^{-4/3} \right) \bm{C}^{-1} \bm{E}_{dev}\\ &- \frac{1}{3} \left( \mu_1 J^{-2/3} + 2 \mu_2 J^{-4/3} \right) \left( 2 \mathbb{{I}_1}(\diff \bm{E})\, \bm{C}^{-1} + \mathbb{{I}_1} \diff \bm{C}^{-1} \right) \\ &- \frac{8}{3}(\bm{C}^{-1} \tcolon \diff \bm{E}) \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{E})\bm{I}_3 - \bm{E} \right) \\ &+2 \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{E})\bm{I}_3 - \diff \bm{E} \right) \\ & + (c_1 + c_2) \bm{C}^{-1} - \frac{4}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{E}) + 2 \mathbb{{I}_2}(\bm{E})\right) \diff \bm{C}^{-1} \end{aligned}

where

(119)#\begin{aligned} c_1 &= \frac{16}{9} \mu_2 J^{-4/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \left( \mathbb{{I}_1}(\bm{E}) + 2 \mathbb{{I}_2}(\bm{E}) \right). \\ c_2 &= -\frac{4}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{E}) + 2 \mathbb{{I}_1}(\bm{E}) \mathbb{{I}_1}(\diff \bm{E}) - 2 \bm{E} \tcolon \diff \bm{E} \right). \end{aligned}
Deriving $$\diff\bm{S}$$ for isochoric Ogden material

Similar to the Neo-Hookean model we have

(120)#$\diff \bm{S} = \underbrace{\frac{\partial \bm{S}_{vol}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{vol}^u} + \underbrace{\frac{\partial \bm{S}_{iso}}{\partial \bm{E}} \tcolon \diff \bm{E}}_{\diff \bm{S}_{iso}} + \underbrace{\frac{\partial \bm{S}_{vol}}{\partial p} \diff p}_{\diff \bm{S}_{vol}^p},$

where $$\diff \bm{S}_{vol}^p$$ and $$\diff \bm{S}_{vol}^u$$ are similar to volumetric derivative of Neo-Hookean and Mooney-Rivlin models and the isochoric part is

(121)#\begin{aligned} \diff \bm{S}_{iso} &= \sum_{i=1}^3 \diff S_i^{iso} \hat{\bm{N}_i} \hat{\bm{N}_i}^T + S_i^{iso} \left( \diff \hat{\bm{N}_i} \hat{\bm{N}_i}^T + \hat{\bm{N}_i} \diff \hat{\bm{N}_i}^T\right) \end{aligned}

For example computing $$\diff S_1^{iso}$$ from (71) gives

(122)#\begin{aligned} \diff S_1^{iso} &= -\frac{2\diff \beta_1}{\beta_1^3} \sum_{j=1}^N \frac{m_j}{3} \left[ 2\operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ &+ \frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j \alpha_j}{3} \left[ 2 \diff \ell_1 \exp(\alpha_j \ell_1) - \diff \ell_2 \exp(\alpha_j \ell_2) - \diff \ell_3 \exp(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ &- \frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j \alpha_j}{9} \left[ 2 \operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} (\bm{C}^{-1} \tcolon \diff \bm{E}) \end{aligned}

To compute $$\diff \beta_i$$ we differentiate $$\bm{C} = \sum_{i=1}^3 \beta_i^2 \hat{\bm{N}_i} \hat{\bm{N}_i}^T$$ as

(123)#$\diff \bm{C} = \sum_{i=1}^3 2 \beta_i \diff \beta_i \hat{\bm{N}_i} \hat{\bm{N}_i}^T + \beta_i^2 \left(\diff \hat{\bm{N}_i} \hat{\bm{N}_i}^T + \hat{\bm{N}_i} \diff \hat{\bm{N}_i}^T \right)$

and used the fact that the eigenvectors are orthonormal i.e., $$\langle \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle = \delta_{ij}$$ so that $$\langle \hat{\bm{N}_i}, \diff \hat{\bm{N}_i} \rangle = 0$$, we can multiply (123) from the left and right by $$\hat{\bm{N}_i}$$

(124)#$\langle \hat{\bm{N}_i}, \diff \bm{C} \hat{\bm{N}_i} \rangle = 2 \beta_i \diff \beta_i \Rightarrow \diff \beta_i = \frac{1}{\beta_i} \langle \hat{\bm{N}_i}, \diff \bm{E} \hat{\bm{N}_i} \rangle$

By differentiating $$\bm{E} \hat{\bm{N}_i} = \beta_i^E \hat{\bm{N}_i}$$ we will arrive at

$\diff \bm{E} \hat{\bm{N}_i} + \bm{E} \diff \hat{\bm{N}_i}= \diff \beta_i^E \hat{\bm{N}_i} + \beta_i^E \diff \hat{\bm{N}_i}$

taking the inner product of above equation with $$\hat{\bm{N}_j}, \, j\neq i$$ simplifies to

\begin{aligned} &\quad \langle \hat{\bm{N}_j}, \diff \bm{E} \hat{\bm{N}_i} \rangle + \langle \diff \hat{\bm{N}_i}, \bm{E} \hat{\bm{N}_j} \rangle = \beta_i^E \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \\ &\quad \langle \hat{\bm{N}_j}, \diff \bm{E} \hat{\bm{N}_i} \rangle + \beta_j^E \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle = \beta_i^E \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \\ &\quad \langle \diff \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle = \frac{1}{\beta_i^E - \beta_j^E} \langle \diff \bm{E} \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \\ &\quad \diff \hat{\bm{N}_i} = \sum_{j \neq i} \frac{1}{\beta_i^E - \beta_j^E} \langle \diff \bm{E} \hat{\bm{N}_i}, \hat{\bm{N}_j} \rangle \hat{\bm{N}_j} \end{aligned}

and finally the linearization of $$\ell_i = \log \beta_i$$ is

$\diff \ell_i = \frac{\diff \beta_i}{\beta_i}.$

which complete $$\diff\bm{S}$$ for Ogden model.

Current configuration#

Similar to what we have shown in (85), we can write the first equation of (108) in the current configuration (see (86) ) which yields to the Jacobian form

(125)#\begin{aligned} \int_{\Omega_0} \nabla_x \bm{v} \tcolon \left( \diff \bm{\tau} - \bm{\tau} \left( \nabla_x \diff \bm{u} \right)^T \right) dV &= \text{rhs}, \quad \forall \bm{v} \in \mathcal{V}, \\ \int_{\Omega_0} q \cdot \left( \diff L J + L \diff J \right)dV &= 0, \quad \forall q \in \mathcal{Q} \end{aligned}

where the second equation is similar to (109) and we can use (92)

$\diff\bm\tau - \bm\tau\left( \nabla_x \diff\bm u \right)^T = \left( \nabla_x \diff\bm u \right) \bm\tau + \bm F \diff\bm S \bm F^T.$

to compute the linearization of the first equation in current configuration for different material models.

Representation of $$\bm F \diff\bm S \bm F^T$$ for isochoric Neo-Hookean material

Based on the split (112), we can derive

(126)#$\bm{F} \diff \bm{S}_{vol}^u \bm{F}^{T} = \left( \bulk_p J^2 V^{''} + \bulk_p J \, V' - p J \right) \trace (\diff \bm \epsilon) \bm{I}_3 - 2 \left(\bulk_p J \, V' - p \, J \right) \diff \bm \epsilon.$
(127)#$\bm{F} \diff \bm{S}_{vol}^p \bm{F}^{T} = -J \diff p \bm{I}_3,$

where $$\diff \bm \epsilon$$ is defined in (90) and we have used (97) and (98). The isochoric part can be simplified as

(128)#\begin{aligned} \bm{F} \diff \bm{S}_{iso} \bm{F}^{T} &= -\frac{2}{3} \mu J^{-2/3} \left( 2 \trace (\diff \bm \epsilon) \bm{e}_{dev} + \trace(\diff \bm e) \bm{I}_3 - \mathbb{{I}_1} \diff \bm \epsilon \right), \end{aligned}

where $$\mathbb{{I}_1} = \trace(\bm{b})$$.

Note

If we use single field isochoric model the linearization of the volumetric stress becomes

(129)#$\bm{F} \diff \bm{S}_{vol} \bm{F}^{T} = \left( \bulk J^2 V^{''} + \bulk J \, V'\right) \trace (\diff \bm \epsilon) \bm{I}_3 -2 \bulk J \, V' \diff \bm \epsilon.$

$$\diff \bm{S}_{vol}^p = 0$$ and we only need to solve the first equation of (125).

Representation of $$\bm F \diff\bm S \bm F^T$$ for isochoric Mooney-Rivlin material

The $$\bm{F} \diff \bm{S}_{vol}^p \bm{F}^T$$ and $$\bm{F} \diff \bm{S}_{vol}^u \bm{F}^T$$ in (117) are the same as (127), (126) (or (129) if we use single field formulation ). The isochoric part can be derived as

(130)#\begin{aligned} \bm{F} \diff \bm{S}_{iso} \bm{F}^T &= -\frac{4}{3}(\trace (\diff \bm \epsilon)) \left( \mu_1 J^{-2/3} + 4 \mu_2 J^{-4/3} \right) \bm{e}_{dev}\\ &+ \frac{2}{3} \left( \mu_1 J^{-2/3} + 2 \mu_2 J^{-4/3} \right) \left( \mathbb{{I}_1} \diff \bm \epsilon -\mathbb{{I}_1}(\diff \bm{e})\, \bm{I}_3 \right) \\ &- \frac{8}{3}(\trace (\diff \bm \epsilon)) \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{e})\bm{b} - \bm{b} \bm{e} \right) \\ &+2 \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{e})\bm{b} - \bm{b} \diff \bm \epsilon \bm{b} \right) \\ & + (c_1 + c_2) \bm{I}_3 + \frac{8}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\bm{e}) + 2 \mathbb{{I}_2}(\bm{e})\right) \diff \bm \epsilon \end{aligned}

where

(131)#\begin{aligned} c_1 &= \frac{16}{9} \mu_2 J^{-4/3} (\trace (\diff \bm \epsilon)) \left( \mathbb{{I}_1}(\bm{e}) + 2 \mathbb{{I}_2}(\bm{e}) \right). \\ c_2 &= -\frac{4}{3} \mu_2 J^{-4/3} \left( \mathbb{{I}_1}(\diff \bm{e}) + 2 \mathbb{{I}_1}(\bm{e}) \mathbb{{I}_1}(\diff \bm{e}) - 2 \trace (\bm{b} \bm{e} \diff \bm \epsilon) \right). \end{aligned}

and we have used (100), (101), and

\begin{aligned} \bm{F} \bm{C}^{-1} \bm{E}_{dev} \bm{F}^T &= \bm{F}^{-T} \left( \bm{E} - \frac{1}{3} \mathbb{{I}_1}(\bm{E}) \bm{I}_3 \right) \bm{F}^T = \bm{e}_{dev}. \\ \bm{F} \bm{E} \bm{F}^T &= \frac{1}{2} \left( \bm{F} \bm{C} \bm{F}^T - \bm{F} \bm{F}^T \right) = \bm{b} \bm{e}. \\ \diff \bm{E} &= \frac{1}{2} \left( \bm{F}^T \diff \bm{F} \bm{F}^{-1} \bm{F} + \bm{F}^T \bm{F}^{-T} \diff \bm{F}^T \bm{F} \right) = \bm{F}^T \diff \bm \epsilon \bm{F}. \\ \bm{E} \tcolon \diff \bm{E} &= \trace(\bm{E} \diff \bm{E}) = \trace(\bm{E} \bm{F}^T \diff \epsilon \bm{F}) = \trace (\bm{F} \bm{E} \bm{F}^T \diff \epsilon) = \trace(\bm{b} \bm{e} \diff \bm \epsilon ). \end{aligned}
Representation of $$\bm F \diff\bm S \bm F^T$$ for isochoric Ogden material

Based on the split (120), the $$\bm{F} \diff \bm{S}_{vol}^p \bm{F}^T$$ and $$\bm{F} \diff \bm{S}_{vol}^u \bm{F}^T$$ are the same as (127), (126) (or (129) if we use single field formulation ) and the isochoric part is

(132)#\begin{aligned} \bm{F} \diff \bm{S}_{iso} \bm{F}^{T} &= \sum_{i=1}^3 \beta^2_i \diff S_i^{iso} \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \beta_i S_i^{iso} \left( \bm{F} \diff \hat{\bm{N}_i} \hat{\bm{n}_i}^T + \hat{\bm{n}_i} \diff \hat{\bm{N}_i}^T \bm{F}^T \right) \\ &= \sum_{i=1}^3 \left(\diff \tau_i^{iso} - 2 \frac{\diff \beta_i}{\beta_i} \tau_i^{iso} \right) \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \tau_i^{iso} \bm{A}_i \end{aligned}

where for example $$\diff \tau_1^{iso}$$ can be written

\begin{aligned} \diff \tau_1^{iso} &= \sum_{j=1}^N \frac{m_j \alpha_j}{3} \left[ 2 \diff \ell_1 \exp(\alpha_j \ell_1) - \diff \ell_2 \exp(\alpha_j \ell_2) - \diff \ell_3 \exp(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ &- \sum_{j=1}^N \frac{m_j \alpha_j}{9} \left[ 2 \operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \trace \left(\diff \bm \epsilon \right) \end{aligned}

and

$\bm{A}_i = 2 \frac{\diff \beta_i}{\beta_i} \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \diff \left( \hat{\bm{n}_i} \hat{\bm{n}_i}^T \right) + \left( \left( \nabla_x \diff \bm{u} \right) \hat{\bm{n}_i} \hat{\bm{n}_i}^T + \hat{\bm{n}_i} \hat{\bm{n}_i}^T \left( \nabla_x \diff \bm{u} \right)^T \right)$

where $$\tau_i^{iso}$$ is defined in (74) and we have used $$\bm{F} \hat{\bm{N}_i} = \beta_i \hat{\bm{n}_i}$$ and $$\tau_i^{iso} = \beta_i^2 S_i^{iso}$$. Note that the linearization of principal stretches, $$\beta_i$$, and eigenvectors, $$\hat{\bm{n}_i}$$, are similar to initial configuration but are written in terms of Green-Euler tensor i.e.,

$\diff \beta_i = \frac{1}{\beta_i} \langle \hat{\bm{n}_i}, \diff \bm{e} \hat{\bm{n}_i} \rangle.$
$\quad \diff \hat{\bm{n}_i} = \sum_{j \neq i} \frac{1}{\beta_i^e - \beta_j^e} \langle \diff \bm{e} \hat{\bm{n}_i}, \hat{\bm{n}_j} \rangle \hat{\bm{n}_j}.$

Perturbed Lagrange-multiplier method#

The weak form (108) is not obtained by variation of a Lagrangian and in general, its linearization will not be symmetric. However, by considering the strain energy of the form

(133)#$\psi \left(\bm{C} \right) = \psi_{vol}(J) + \psi_{iso}(\bar{\bm{C}}) = \frac{\bulk}{2} \left(U(J)\right)^2 + \psi_{iso}(\bar{\bm{C}})$

we can write a two fields energy functional as

(134)#\begin{aligned} \Pi (\bm u, p) &= \int_{\Omega_0} \left[ \psi_{iso}(\bar{\bm{C}}) - p \, U(J) - \frac{1}{2}\frac{p^2}{\bulk - \bulk_p} + \frac{\bulk_p}{2} U^2\right] \, dV - \Pi_{\text{ext}} (\bm u) \\ \Pi_{\text{ext}} (\bm u) &= \int_{\Omega_0}{\bm{u} \cdot \rho_0 \bm{g}} \, dV + \int_{\partial \Omega_0}{\bm{u} \cdot \bar{\bm{t}}} \, dS \end{aligned}

where $$p$$ is the Lagrange-multiplier and $$\bar{\bm{t}}$$ is traction applied on the boundary. Finding the stationary conditions with respect to $$\bm{u}$$ and $$p$$ by taking Gateaux derivative

$D_{\diff \bm u} \Pi (\bm u, p) = 0, \quad D_{\diff p} \Pi (\bm u, p) = 0.$

where $$\diff \bm u$$ and $$\diff p$$ are virtual displacement and pressure and can be seen as the test functions $$\bm v, q$$, we will arrive at

(135)#\begin{aligned} \int_{\Omega_0} \bm F \underbrace{\left( \bm{S}_{iso} + (\bulk_p U - p) \, J \, U' \, \bm{C}^{-1} \right)}_{\bm S} \tcolon \nabla_X \bm v \, dV &= L_{\text{ext}} (\bm v) \\ \int_{\Omega_0} \left(- U(J) - \frac{p}{\bulk - \bulk_p} \right) q \, dV &= 0 \end{aligned}

where, $$L_{\text{ext}} (\bm v) = \int_{\Omega_0}{\bm{v} \cdot \rho_0 \bm{g}} \, dV + \int_{\partial \Omega_0}{\bm{v} \cdot \bar{\bm{t}}} \, dS$$ and we have used

\begin{aligned} \frac{\partial \psi_{iso}}{\partial \bm u} \cdot {\diff \bm u} &= \frac{\partial \psi_{iso}}{\partial \bm E} \tcolon \frac{\partial \bm E}{ \partial \bm u}{\diff \bm u} = \bm{S}_{iso} \tcolon {\diff \bm E} = \bm{S}_{iso} \tcolon \text{sym} \left(\bm F^T \diff \bm F \right) \\ \frac{\partial U}{\partial \bm u} \cdot {\diff \bm u} &= \frac{\partial U}{\partial J} \frac{\partial J}{\partial \bm u} {\diff \bm u} = U' {\diff J} = J \, U' \, \bm{C}^{-1} \tcolon \tcolon {\diff \bm E} = J \, U' \, \bm{C}^{-1} \tcolon \text{sym} \left(\bm F^T \diff \bm F \right) \end{aligned}

where $$\diff \bm{F} = \nabla_X \bm v$$. The Jacobian for problem (135) can be written as

(136)#\begin{aligned} \int_{\Omega_0} \nabla_X \bm{v} \tcolon \left(\bm F \diff \bm{S} + \diff \bm{F} \bm{S} \right) dV &= \text{rhs}, \\ \int_{\Omega_0} q \left( -\diff U - \frac{\diff p}{\bulk - \bulk_p} \right) dV &= 0, \end{aligned}

where

(137)#\begin{aligned} \diff \bm{S} &= \diff \bm{S}_{iso} + \diff \bm{S}_{vol}^u + \diff \bm{S}_{vol}^p, \\ \diff \bm{S}_{vol}^u &= \left(\bulk_p U' \diff J \right) J U' \bm{C}^{-1} + \left(\bulk_p U - p \right) \left(\diff J \, U' \bm{C}^{-1} + J\, U^{''} \diff J \bm{C}^{-1} + J U' \diff \bm{C}^{-1} \right), \\ &= \left[\bulk_p \left( J U' \right)^2 + \left(\bulk_p U - p \right) \left( J U' + J^2 U^{''} \right) \right] \left( \bm{C}^{-1} \tcolon \diff \bm E \right) \bm{C}^{-1} \\ &+ \left(\bulk_p U - p \right) J U' \, \diff \bm{C}^{-1}\\ \diff \bm{S}_{vol}^p &= - dp J U' \, \bm{C}^{-1} \\ \diff U &= U' \diff J = J \, U' \, \bm{C}^{-1} \tcolon \diff \bm E, \end{aligned}

where $$\diff \bm{S}_{iso}$$ is derived for Neo-Hookean, Mooney-Rivlin and Ogden in (115), (118) and (121). To complete the derivation we only need $$U(J)$$ function with condition

$U(J) = 0 \quad \text{if and only if} \quad J = 1.$

For the volumetric strain energy function of the form given in (43), if we choose $$V(J) = \frac{1}{4} \left(J^2 - 1 - 2 \log J \right)$$, from $$\bulk V(J) = \frac{\bulk}{2} U^2(J)$$, we will have

$U(J) = \pm \sqrt{2 V} = \frac{\sign(J-1)}{\sqrt{2}} ( \underbrace{J^2 - 1 - 2 \log J }_{A})^{1/2}$

where the derivatives are

\begin{aligned} U' &= \frac{\partial U}{\partial J} = \sign(J-1) \frac{J^2 - 1}{J \sqrt{2}} A^{-1/2}, \\ U{''} &= \frac{\partial^2 U}{\partial J^2} = \sign(J-1) \frac{1}{J^2 \sqrt{2}} \left((J^2 + 1) A^{-1/2} - (J^2 - 1)^2 A^{-3/2} \right). \end{aligned}