# Ogden Isochoric-split#

The strain energy of Neo-Hookean and Mooney-Rivlin are written in terms of invariants of $$\bm{C}$$. The decoupled Ogden strain energy function is in terms of modified principal stretch (28)

(64)#\begin{aligned} \psi \left(\bar{\beta}_1, \bar{\beta}_2, \bar{\beta}_3, J \right) &= \psi_{vol}(J) + \psi_{iso}\left( \bar{\beta}_1, \bar{\beta}_2, \bar{\beta}_3 \right)\\ &=\frac{\kappa}{4} \left( J^2 - 1 -2 \log J \right) + \sum_{i=1}^3 \omega(\bar{\beta}_i). \end{aligned}

where

(65)#$\omega(\bar{\beta}_i) = \sum_{j=1}^N \frac{m_j}{\alpha_j} \left(\bar{\beta}_i^{\alpha_j} -1 \right)$

with the consistency condition

(66)#$2 \mu = \sum_{j=1}^N m_j \alpha_j \quad \text{with} \quad m_j \alpha_j > 0$

obtained from comparison with the linear theory.

Tip

For $$N=1, \alpha_1 = 2, m_1 = \mu$$ the Ogden model (64) simplifies to

$\psi \left( \bar{\beta}_1, \bar{\beta}_2, \bar{\beta}_3, J \right) = \frac{\kappa}{4} \left( J^2 - 1 -2 \log J \right) + \frac{\mu}{2} \left(\bar{\beta}_1^2 + \bar{\beta}_2^2 + \bar{\beta}_3^2 - 3\right)$

which is the same as the decoupled Neo-Hookean energy function (43). Also for $$J=1, N=2, \alpha_1 = 2, \alpha_2 = -2, m_1 = \mu_1, m_2 = -\mu_2$$ and by using $$J=1=\beta_1 \beta_2 \beta_3 = \bar{\beta}_1 \bar{\beta}_2 \bar{\beta}_3$$ constrain, we can simplify (64) as

\begin{aligned} \psi \left( \bar{\beta}_1, \bar{\beta}_2, \bar{\beta}_3, J=1 \right) &= \frac{m_1}{2} \left(\bar{\beta}_1^2 + \bar{\beta}_2^2 + \bar{\beta}_3^2 - 3\right) - \frac{m_2}{2} \left(\bar{\beta}_1^{-2} + \bar{\beta}_2^{-2} + \bar{\beta}_3^{-2} - 3\right) \\ &= \frac{\mu_1}{2} \left(\mathbb{\bar{I}}_1 - 3\right) + \frac{\mu_2}{2} \left(\mathbb{\bar{I}}_2 - 3\right) \end{aligned}

which is the Mooney-Rivlin isochoric model.

We differentiate $$\psi$$ as in the isochoric Neo-Hookean case (44) to yield the second Piola-Kirchoff tensor,

(67)#\begin{aligned} \bm{S} = \frac{\partial \psi}{\partial \bm{E}} &= \frac{\partial \psi_{vol}}{\partial J} \frac{\partial J}{\partial \bm{E}} + \frac{\partial \psi_{iso}}{\partial \bm{E}} =\bm{S}_{vol} + \bm{S}_{iso}\\ &= -p J \bm{C}^{-1} + \sum_{i=1}^3 S_i^{iso} \hat{\bm{N}_i} \hat{\bm{N}_i}^T. \end{aligned}

where the pressure $$p$$ is defined via (45) and we have used (21) and (24). By employing $$\frac{\partial J}{\partial {\beta}_i} = J {\beta}_i^{-1}$$ and relation (28)

(68)#\begin{aligned} S_i^{iso} = \frac{1}{\beta_i}\frac{\partial \psi_{iso}}{\partial \beta_i} =\frac{1}{\beta_i}\frac{\partial \bar{\beta}_k}{\partial \beta_i} \frac{\partial \psi_{iso}}{\partial \bar{\beta}_k} &=\frac{J^{-1/3}}{\beta_i} \left(\delta_{ik} -\frac{1}{3}\beta_i^{-1}\beta_k \right) \frac{\partial \psi_{iso}}{\partial \bar{\beta}_k} \\ &=\frac{J^{-1/3}}{\beta_i} \left(\delta_{ik} -\frac{1}{3}\bar{\beta}_i^{-1}\bar{\beta}_k \right) \frac{\partial \psi_{iso}}{\partial \bar{\beta}_k}. \end{aligned}

in which

(69)#$\frac{\partial \psi_{iso}}{\partial \bar{\beta}_k} = \frac{\partial \omega (\bar{\beta}_k)}{ \partial \bar{\beta}_k} = \sum_{j=1}^N m_j \bar{\beta}_k^{\alpha_j - 1}$

Tip

To derive an equivalent numerically stable form of the isochoric part of (67) we rewrite the $$\psi_{iso}$$ by substituting $$\bar{\beta}_i = J^{-1/3}\beta_i$$ as

(70)#$\psi_{iso}({\beta}_i) = \sum_{j=1}^N \frac{m_j}{\alpha_j} \left[\left({\beta}_1^{\alpha_j} + {\beta}_2^{\alpha_j} + {\beta}_3^{\alpha_j} \right)J^{-\alpha_j/3} -3 \right]$

then

(71)#\begin{aligned} S_1^{iso} = \frac{1}{\beta_1}\frac{\partial \psi_{iso}}{\partial \beta_1} &= \frac{1}{\beta_1} \sum_{j=1}^N \frac{m_j}{\alpha_j} \left[ \left(\alpha_j \beta_1^{\alpha_j -1} \right)J^{-\alpha_j/3} - \frac{\alpha_j}{3} J^{-\alpha_j/3} \beta_1^{-1} \left( {\beta}_1^{\alpha_j} + {\beta}_2^{\alpha_j} + {\beta}_3^{\alpha_j} \right)\right] \\ &= \frac{1}{\beta_1} \sum_{j=1}^N m_j \left[ \beta_1^{\alpha_j -1} - \frac{1}{3} \beta_1^{-1} \left( {\beta}_1^{\alpha_j} + {\beta}_2^{\alpha_j} + {\beta}_3^{\alpha_j} \right)\right] J^{-\alpha_j/3} \\ &= \frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j}{3} \left[ 2\beta_1^{\alpha_j} - {\beta}_2^{\alpha_j} - {\beta}_3^{\alpha_j} \right] J^{-\alpha_j/3} \\ &= \frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j}{3} \left[ 2\left(\beta_1^{\alpha_j} -1 \right) - \left({\beta}_2^{\alpha_j} -1 \right) - \left({\beta}_3^{\alpha_j} -1 \right)\right] J^{-\alpha_j/3} \\ &=\frac{1}{\beta_1^2} \sum_{j=1}^N \frac{m_j}{3} \left[ 2(e^{\alpha_j \ell_1}-1) - (e^{\alpha_j \ell_2}-1) - (e^{\alpha_j \ell_3}-1) \right] J^{-\alpha_j/3} \\ &=\frac{1}{1 + 2\beta_1^E} \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} \end{aligned}

where the eigen pair $$(\beta_i, \hat{\bm{N}_i})$$ are computed by solving the eigen problem of Green-Lagrange strain tensor $$\bm{E} \hat{\bm{N}_i} = \beta_i^E \hat{\bm{N}_i}$$ with $$\beta_i = \sqrt{1 + 2 \beta_i^E}$$ and and $$\ell_i = \log \beta_i = \frac 1 2 \operatorname{\tt log1p}(2\beta_i^E)$$. Following the above approach we have

(72)#\begin{aligned} S_2^{iso} = \frac{1}{\beta_2}\frac{\partial \psi_{iso}}{\partial \beta_2} =\frac{1}{1 + 2\beta_2^E} \sum_{j=1}^N \frac{m_j}{3} \left[ -\operatorname{\tt expm1}(\alpha_j \ell_1) +2 \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ S_3^{iso} = \frac{1}{\beta_3}\frac{\partial \psi_{iso}}{\partial \beta_3} =\frac{1}{1 + 2\beta_3^E} \sum_{j=1}^N \frac{m_j}{3} \left[ -\operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) + 2 \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \end{aligned}

substituting the new definition of $$S_i^{iso}$$ (71), (72) into (67) gives the stable form of the second Piola-Kirchhoff stress for Ogden model.

The Kirchhoff stress tensor $$\bm{\tau}$$ for Ogden model is given by

(73)#$\bm{\tau} = \bm{F}\bm{S}\bm{F}^T = \bm{\tau}_{vol} + \bm{\tau}_{iso} = -p J \bm{I}_{3} + \sum_{i=1}^3 \tau_i^{iso} \hat{\bm{n}_i} \hat{\bm{n}_i}^T.$

where we have used $$\bm{F} \hat{\bm{N}_i} = \beta_i \hat{\bm{n}_i}$$ which $$\hat{\bm{n}_i}$$ are eigenvectors of $$\bm{b}$$ and $$\tau_i^{iso} = \beta_i^2 S_i^{iso}$$.

Tip

Similar to the initial configuration, we can compute the $$\bm{\tau}_{iso}$$ in stable form by using the stable coefficients

(74)#\begin{aligned} \tau_1^{iso} &= \beta_1 \frac{\partial \psi_{iso}}{\partial \beta_1} = \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} \\ \tau_2^{iso} &= \beta_2 \frac{\partial \psi_{iso}}{\partial \beta_2} = \sum_{j=1}^N \frac{m_j}{3} \left[ -\operatorname{\tt expm1}(\alpha_j \ell_1) +2 \operatorname{\tt expm1}(\alpha_j \ell_2) - \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \\ \tau_3^{iso} &= \beta_3 \frac{\partial \psi_{iso}}{\partial \beta_3} = \sum_{j=1}^N \frac{m_j}{3} \left[ -\operatorname{\tt expm1}(\alpha_j \ell_1) - \operatorname{\tt expm1}(\alpha_j \ell_2) + 2 \operatorname{\tt expm1}(\alpha_j \ell_3) \right] J^{-\alpha_j/3} \end{aligned}

and computing the eigen pair $$(\beta_i, \hat{\bm{n}_i})$$ of Green-Euler strain tensor $$\bm{e} \hat{\bm{n}_i} = \beta_i^e \hat{\bm{e}_i}$$ with $$\beta_i = \sqrt{1 + 2 \beta_i^e}$$.

Note

It should be noted that the Ogden model can be implemented with single field displacement as described in Neo-Hookean isochoric model (50) which in that case we have

(75)#\begin{aligned} \bm{S} = \frac{\partial \psi}{\partial \bm{E}} &= \frac{\partial \psi_{vol}}{\partial J} \frac{\partial J}{\partial \bm{E}} + \frac{\partial \psi_{iso}}{\partial \bm{E}} =\bm{S}_{vol} + \bm{S}_{iso}\\ &= \frac{\kappa}{2} (J^2 -1)\bm{C}^{-1} + \sum_{i=1}^3 S_i^{iso} \hat{\bm{N}_i} \hat{\bm{N}_i}^T. \end{aligned}