# 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 (36)

(80)\begin{aligned} \psi \left(\bar{\lambda}_1, \bar{\lambda}_2, \bar{\lambda}_3, J \right) &= \psi_{\text{vol}}(J) + \psi_{\text{iso}}\left( \bar{\lambda}_1, \bar{\lambda}_2, \bar{\lambda}_3 \right)\\ &=\bulk \, V(J) + \sum_{i=1}^3 \omega(\bar{\lambda}_i). \end{aligned}

where

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

with the consistency condition

(82)$2 \secondlame = \sum_{j=1}^N m_j \alpha_j \quad \text{with} \quad m_j \alpha_j > 0$

obtained from comparison with the linear theory.

Tip

For $$N=1, \alpha_1 = 2, m_1 = \secondlame$$ the Ogden model (80) simplifies to

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

which is the same as the decoupled Neo-Hookean energy function (59). Also for $$J=1, N=2, \alpha_1 = 2, \alpha_2 = -2, m_1 = \secondlame_1, m_2 = -\secondlame_2$$ and by using $$J=1=\lambda_1 \lambda_2 \lambda_3 = \bar{\lambda}_1 \bar{\lambda}_2 \bar{\lambda}_3$$ constrain, we can simplify (80) as

\begin{aligned} \psi \left( \bar{\lambda}_1, \bar{\lambda}_2, \bar{\lambda}_3, J=1 \right) &= \frac{m_1}{2} \left(\bar{\lambda}_1^2 + \bar{\lambda}_2^2 + \bar{\lambda}_3^2 - 3\right) - \frac{m_2}{2} \left(\bar{\lambda}_1^{-2} + \bar{\lambda}_2^{-2} + \bar{\lambda}_3^{-2} - 3\right) \\ &= \frac{\secondlame_1}{2} \left(\mathbb{\bar{I}}_1 - 3\right) + \frac{\secondlame_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 (62) to yield the second Piola-Kirchoff tensor,

(83)\begin{aligned} \bm{S} = \frac{\partial \psi}{\partial \bm{E}} &= \frac{\partial \psi_{\text{vol}}}{\partial J} \frac{\partial J}{\partial \bm{E}} + \frac{\partial \psi_{\text{iso}}}{\partial \bm{E}} =\bm{S}_{\text{vol}} + \bm{S}_{\text{iso}}\\ &= \left(\bulk_p J \, V' - p \, J \right) \bm{C}^{-1} + \sum_{i=1}^3 S_i^{\text{iso}} \hat{\bm{N}_i} \hat{\bm{N}_i}^T. \end{aligned}

where the pressure-like variable $$p$$ is defined via (61) and we have used (29) and (32). By employing $$\frac{\partial J}{\partial {\lambda}_i} = J {\lambda}_i^{-1}$$ and relation (36)

(84)\begin{aligned} S_i^{\text{iso}} = \frac{1}{\lambda_i}\frac{\partial \psi_{\text{iso}}}{\partial \lambda_i} =\frac{1}{\lambda_i}\frac{\partial \bar{\lambda}_k}{\partial \lambda_i} \frac{\partial \psi_{\text{iso}}}{\partial \bar{\lambda}_k} &=\frac{J^{-1/3}}{\lambda_i} \left(\delta_{ik} -\frac{1}{3}\lambda_i^{-1}\lambda_k \right) \frac{\partial \psi_{\text{iso}}}{\partial \bar{\lambda}_k} \\ &=\frac{J^{-1/3}}{\lambda_i} \left(\delta_{ik} -\frac{1}{3}\bar{\lambda}_i^{-1}\bar{\lambda}_k \right) \frac{\partial \psi_{\text{iso}}}{\partial \bar{\lambda}_k}. \end{aligned}

in which

(85)$\frac{\partial \psi_{\text{iso}}}{\partial \bar{\lambda}_k} = \frac{\partial \omega (\bar{\lambda}_k)}{ \partial \bar{\lambda}_k} = \sum_{j=1}^N m_j \bar{\lambda}_k^{\alpha_j - 1}$

Tip

To derive an equivalent numerically stable form of the isochoric part of (83) we rewrite the $$\psi_{\text{iso}}$$ by substituting $$\bar{\lambda}_i = J^{-1/3}\lambda_i$$ as

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

then

(87)\begin{aligned} S_1^{\text{iso}} = \frac{1}{\lambda_1}\frac{\partial \psi_{\text{iso}}}{\partial \lambda_1} &= \frac{1}{\lambda_1} \sum_{j=1}^N \frac{m_j}{\alpha_j} \left[ \left(\alpha_j \lambda_1^{\alpha_j -1} \right)J^{-\alpha_j/3} - \frac{\alpha_j}{3} J^{-\alpha_j/3} \lambda_1^{-1} \left( {\lambda}_1^{\alpha_j} + {\lambda}_2^{\alpha_j} + {\lambda}_3^{\alpha_j} \right)\right] \\ &= \frac{1}{\lambda_1} \sum_{j=1}^N m_j \left[ \lambda_1^{\alpha_j -1} - \frac{1}{3} \lambda_1^{-1} \left( {\lambda}_1^{\alpha_j} + {\lambda}_2^{\alpha_j} + {\lambda}_3^{\alpha_j} \right)\right] J^{-\alpha_j/3} \\ &= \frac{1}{\lambda_1^2} \sum_{j=1}^N \frac{m_j}{3} \left[ 2\lambda_1^{\alpha_j} - {\lambda}_2^{\alpha_j} - {\lambda}_3^{\alpha_j} \right] J^{-\alpha_j/3} \\ &= \frac{1}{\lambda_1^2} \sum_{j=1}^N \frac{m_j}{3} \left[ 2\left(\lambda_1^{\alpha_j} -1 \right) - \left({\lambda}_2^{\alpha_j} -1 \right) - \left({\lambda}_3^{\alpha_j} -1 \right)\right] J^{-\alpha_j/3} \\ &=\frac{1}{\lambda_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\lambda_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 $$(\lambda_i, \hat{\bm{N}_i})$$ are computed by solving the eigen problem of Green-Lagrange strain tensor $$\bm{E} \hat{\bm{N}_i} = \lambda_i^E \hat{\bm{N}_i}$$ with $$\lambda_i = \sqrt{1 + 2 \lambda_i^E}$$ and and $$\ell_i = \log \lambda_i = \frac 1 2 \operatorname{\tt log1p}(2\lambda_i^E)$$. Following the above approach we have

(88)\begin{aligned} S_2^{\text{iso}} = \frac{1}{\lambda_2}\frac{\partial \psi_{\text{iso}}}{\partial \lambda_2} =\frac{1}{1 + 2\lambda_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^{\text{iso}} = \frac{1}{\lambda_3}\frac{\partial \psi_{\text{iso}}}{\partial \lambda_3} =\frac{1}{1 + 2\lambda_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 $$S_i^{\text{iso}}$$ (87), (88) into (83) 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

(89)$\bm{\tau} = \bm{F}\bm{S}\bm{F}^T = \bm{\tau}_{\text{vol}} + \bm{\tau}_{\text{iso}} = \left(\bulk_p J \, V' - p \, J \right) \bm{I} + \sum_{i=1}^3 \tau_i^{\text{iso}} \hat{\bm{n}_i} \hat{\bm{n}_i}^T.$

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

Tip

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

(90)\begin{aligned} \tau_1^{\text{iso}} &= \lambda_1 \frac{\partial \psi_{\text{iso}}}{\partial \lambda_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^{\text{iso}} &= \lambda_2 \frac{\partial \psi_{\text{iso}}}{\partial \lambda_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^{\text{iso}} &= \lambda_3 \frac{\partial \psi_{\text{iso}}}{\partial \lambda_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 $$(\lambda_i, \hat{\bm{n}_i})$$ of Green-Euler strain tensor $$\bm{e} \hat{\bm{n}_i} = \lambda_i^e \hat{\bm{e}_i}$$ with $$\lambda_i = \sqrt{1 + 2 \lambda_i^e}$$ and $$\ell_i = \log \lambda_i = \frac 1 2 \operatorname{\tt log1p}(2\lambda_i^e)$$.

Note

It should be noted that the Ogden model can be implemented with single field displacement as described in Neo-Hookean isochoric model (66) which in that case we have

(91)\begin{aligned} \bm{S} = \frac{\partial \psi}{\partial \bm{E}} &= \frac{\partial \psi_{\text{vol}}}{\partial J} \frac{\partial J}{\partial \bm{E}} + \frac{\partial \psi_{\text{iso}}}{\partial \bm{E}} =\bm{S}_{\text{vol}} + \bm{S}_{\text{iso}}\\ &= \bulk J \, V' \, \bm{C}^{-1} + \sum_{i=1}^3 S_i^{\text{iso}} \hat{\bm{N}_i} \hat{\bm{N}_i}^T. \end{aligned}