Continuum mechanics/Nonlinear elasticity

There are two types models of nonlinear elastic behavior that are in common use. These are :

• Hyperelasticity
• Hypoelasticity

Hyperelasticity

Hyperelastic materials are truly elastic in the sense that if a load is applied to such a material and then removed, the material returns to its original shape without any dissipation of energy in the process. In other word, a hyperelastic material stores energy during loading and releases exactly the same amount of energy during unloading. There is no path dependence.

If $\psi$ is the Helmholtz free energy, then the stress-strain behavior for such a material is given by

$\mathit{\sigma }=\rho 𝑭\bullet \frac{\partial \psi }{\partial 𝑬}\bullet {𝑭}^T=2\rho 𝑭\bullet \frac{\partial \psi }{\partial 𝑪}\bullet {𝑭}^T$

where $\mathit{\sigma }$ is the Cauchy stress, $\rho$ is the current mass density, $𝑭$ is the deformation gradient, $𝑬$ is the Lagrangian Green strain tensor, and $𝑪$ is the left Cauchy-Green deformation tensor.

We can use the relationship between the Cauchy stress and the 2nd Piola-Kirchhoff stress to obtain an alternative relation between stress and strain.

$𝑺=2{\rho }_0\frac{\partial \psi }{\partial 𝑪}$

where $𝑺$ is the 2nd Piola-Kirchhoff stress and ${\rho }_0$ is the mass density in the reference configuration.

Isotropic hyperelasticity

For isotropic materials, the free energy must be an isotropic function of $𝑪$. This also mean that the free energy must depend only on the principal invariants of $𝑪$ which are

$\begin{array}{cc}{I}_{𝑪}=I_1& =\text{tr}(𝐂)=C_{ii}={\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2\\ I{I}_{𝑪}=I_2& =\frac{1}{2}[\text{tr}({𝐂}^2)-(\text{tr}𝐂{)}^2]=\frac{1}{2}[C_{ik}{C}_{ki}-C_{jj}^2]={\lambda }_1^2{\lambda }_2^2+{\lambda }_2^2{\lambda }_3^2+{\lambda }_3^2{\lambda }_1^2\\ II{I}_{𝑪}=I_3& =\det (𝐂)={\lambda }_1^2{\lambda }_2^2{\lambda }_3^2\end{array}$

In other words,

$\psi (𝑪)\equiv \psi (I_1,I_2,I_3)$

Therefore, from the chain rule,

$\frac{\partial \psi }{\partial 𝑪}=\frac{\partial \psi }{\partial I_1}\frac{\partial I_1}{\partial 𝑪}+\frac{\partial \psi }{\partial I_2}\frac{\partial I_2}{\partial 𝑪}+\frac{\partial \psi }{\partial I_3}\frac{\partial I_3}{\partial 𝑪}=a_{0}1+a_1𝑪+a_2{𝑪}^{-1}$

From the Cayley-Hamilton theorem we can show that

${𝑪}^{-1}\equiv f({𝑪}^2,𝑪,1)$

Hence we can also write

$\frac{\partial \psi }{\partial 𝑪}=b_{0}1+b_1𝑪+b_2{𝑪}^2$

The stress-strain relation can then be written as

$𝑺=2{\rho }_0[b_{0}1+b_1𝑪+b_2{𝑪}^2]$

A similar relation can be obtained for the Cauchy stress which has the form

$\mathit{\sigma }=2\rho [a_{2}1+a_0𝑩+a_1{𝑩}^2]$

where $𝑩$ is the right Cauchy-Green deformation tensor.

Cauchy stress in terms of invariants

For w:isotropic hyperelastic materials, the Cauchy stress can be expressed in terms of the invariants of the left Cauchy-Green deformation tensor (or right Cauchy-Green deformation tensor). If the w:strain energy density function is $W(𝑭)=\hat{W}(I_1,I_2,I_3)=\overline{W}({\overline{I}}_1,{\overline{I}}_2,J)=\tilde{W}({\lambda }_1,{\lambda }_2,{\lambda }_3)$, then

$\begin{array}{cc}\mathit{\sigma }& =\frac{2}{\sqrt{I_3}}[(\frac{\partial \hat{W}}{\partial I_1}+I_1\frac{\partial \hat{W}}{\partial I_2})𝑩-\frac{\partial \hat{W}}{\partial I_2}𝑩\cdot 𝑩]+2\sqrt{I_3}\frac{\partial \hat{W}}{\partial I_3}1\\ & =\frac{2}{J}[\frac{1}{J^{2/3}}(\frac{\partial \overline{W}}{\partial {\overline{I}}_1}+{\overline{I}}_1\frac{\partial \overline{W}}{\partial {\overline{I}}_2})𝑩-\frac{1}{3}({\overline{I}}_1\frac{\partial \overline{W}}{\partial {\overline{I}}_1}+2{\overline{I}}_2\frac{\partial \overline{W}}{\partial {\overline{I}}_2})1-\\ & \frac{1}{J^{4/3}}\frac{\partial \overline{W}}{\partial {\overline{I}}_2}𝑩\cdot 𝑩]+\frac{\partial \overline{W}}{\partial J}1\\ & =\frac{{\lambda }_1}{{\lambda }_1{\lambda }_2{\lambda }_3}\frac{\partial \tilde{W}}{\partial {\lambda }_1}{𝐧}_1\otimes {𝐧}_1+\frac{{\lambda }_2}{{\lambda }_1{\lambda }_2{\lambda }_3}\frac{\partial \tilde{W}}{\partial {\lambda }_2}{𝐧}_2\otimes {𝐧}_2+\frac{{\lambda }_3}{{\lambda }_1{\lambda }_2{\lambda }_3}\frac{\partial \tilde{W}}{\partial {\lambda }_3}{𝐧}_3\otimes {𝐧}_3\end{array}$

(See the page on the left Cauchy-Green deformation tensor for the definitions of these symbols).

Proof 1:

The second Piola-Kirchhoff stress tensor for a hyperelastic material is given by $𝑺=2\frac{\partial W}{\partial 𝑪}$ where $𝑪={𝑭}^T\cdot 𝑭$ is the right Cauchy-Green deformation tensor and $𝑭$ is the deformation gradient. The Cauchy stress is given by $\mathit{\sigma }=\frac{1}{J}𝑭\cdot 𝑺\cdot {𝑭}^T=\frac{2}{J}𝑭\cdot \frac{\partial W}{\partial 𝑪}\cdot {𝑭}^T$ where $J=\det 𝑭$. Let $I_1,I_2,I_3$ be the three principal invariants of $𝑪$. Then $\frac{\partial W}{\partial 𝑪}=\frac{\partial W}{\partial I_1}\frac{\partial I_1}{\partial 𝑪}+\frac{\partial W}{\partial I_2}\frac{\partial I_2}{\partial 𝑪}+\frac{\partial W}{\partial I_3}\frac{\partial I_3}{\partial 𝑪}.$ The derivatives of the invariants of the symmetric tensor $𝑪$ are $\frac{\partial I_1}{\partial 𝑪}=1;\frac{\partial I_2}{\partial 𝑪}=I_{1}1-𝑪;\frac{\partial I_3}{\partial 𝑪}=\det (𝑪){𝑪}^{-1}$ Therefore we can write $\frac{\partial W}{\partial 𝑪}=\frac{\partial W}{\partial I_1}1+\frac{\partial W}{\partial I_2}(I_{1}1-{𝑭}^T\cdot 𝑭)+\frac{\partial W}{\partial I_3}{I}_3{𝑭}^{-1}\cdot {𝑭}^{-T}.$ Plugging into the expression for the Cauchy stress gives $\mathit{\sigma }=\frac{2}{J}[\frac{\partial W}{\partial I_1}𝑭\cdot {𝑭}^T+\frac{\partial W}{\partial I_2}(I_1𝑭\cdot {𝑭}^T-𝑭\cdot {𝑭}^T\cdot 𝑭\cdot {𝑭}^T)+\frac{\partial W}{\partial I_3}{I}_{3}1]$ Using the left Cauchy-Green deformation tensor $𝑩=𝑭\cdot {𝑭}^T$ and noting that $I_3=J^2$, we can write $\mathit{\sigma }=\frac{2}{\sqrt{I_3}}[(\frac{\partial W}{\partial I_1}+I_1\frac{\partial W}{\partial I_2})𝑩-\frac{\partial W}{\partial I_2}𝑩\cdot 𝑩]+2\sqrt{I_3}\frac{\partial W}{\partial I_3}1.$

Proof 2:

To express the Cauchy stress in terms of the invariants ${\overline{I}}_1,{\overline{I}}_2,J$ recall that ${\overline{I}}_1=J^{-2/3}{I}_1=I_3^{-1/3}{I}_1;{\overline{I}}_2=J^{-4/3}{I}_2=I_3^{-2/3}{I}_2;J=I_3^{1/2}.$ The chain rule of differentiation gives us $\begin{array}{cc}\frac{\partial W}{\partial I_1}& =\frac{\partial W}{\partial {\overline{I}}_1}\frac{\partial {\overline{I}}_1}{\partial I_1}+\frac{\partial W}{\partial {\overline{I}}_2}\frac{\partial {\overline{I}}_2}{\partial I_1}+\frac{\partial W}{\partial J}\frac{\partial J}{\partial I_1}\\ & =I_3^{-1/3}\frac{\partial W}{\partial {\overline{I}}_1}=J^{-2/3}\frac{\partial W}{\partial {\overline{I}}_1}\\ \frac{\partial W}{\partial I_2}& =\frac{\partial W}{\partial {\overline{I}}_1}\frac{\partial {\overline{I}}_1}{\partial I_2}+\frac{\partial W}{\partial {\overline{I}}_2}\frac{\partial {\overline{I}}_2}{\partial I_2}+\frac{\partial W}{\partial J}\frac{\partial J}{\partial I_2}\\ & =I_3^{-2/3}\frac{\partial W}{\partial {\overline{I}}_2}=J^{-4/3}\frac{\partial W}{\partial {\overline{I}}_2}\\ \frac{\partial W}{\partial I_3}& =\frac{\partial W}{\partial {\overline{I}}_1}\frac{\partial {\overline{I}}_1}{\partial I_3}+\frac{\partial W}{\partial {\overline{I}}_2}\frac{\partial {\overline{I}}_2}{\partial I_3}+\frac{\partial W}{\partial J}\frac{\partial J}{\partial I_3}\\ & =-\frac{1}{3}{I}_3^{-4/3}{I}_1\frac{\partial W}{\partial {\overline{I}}_1}-\frac{2}{3}{I}_3^{-5/3}{I}_2\frac{\partial W}{\partial {\overline{I}}_2}+\frac{1}{2}{I}_3^{-1/2}\frac{\partial W}{\partial J}\\ & =-\frac{1}{3}{J}^{-8/3}{J}^{2/3}{\overline{I}}_1\frac{\partial W}{\partial {\overline{I}}_1}-\frac{2}{3}{J}^{-10/3}{J}^{4/3}{\overline{I}}_2\frac{\partial W}{\partial {\overline{I}}_2}+\frac{1}{2}{J}^{-1}\frac{\partial W}{\partial J}\\ & =-\frac{1}{3}{J}^{-2}({\overline{I}}_1\frac{\partial W}{\partial {\overline{I}}_1}+2{\overline{I}}_2\frac{\partial W}{\partial {\overline{I}}_2})+\frac{1}{2}{J}^{-1}\frac{\partial W}{\partial J}\end{array}$ Recall that the Cauchy stress is given by $\mathit{\sigma }=\frac{2}{\sqrt{I_3}}[(\frac{\partial W}{\partial I_1}+I_1\frac{\partial W}{\partial I_2})𝑩-\frac{\partial W}{\partial I_2}𝑩\cdot 𝑩]+2\sqrt{I_3}\frac{\partial W}{\partial I_3}1.$ In terms of the invariants ${\overline{I}}_1,{\overline{I}}_2,J$ we have $\mathit{\sigma }=\frac{2}{J}[(\frac{\partial W}{\partial I_1}+J^{2/3}{\overline{I}}_1\frac{\partial W}{\partial I_2})𝑩-\frac{\partial W}{\partial I_2}𝑩\cdot 𝑩]+2J\frac{\partial W}{\partial I_3}1.$ Plugging in the expressions for the derivatives of $W$ in terms of ${\overline{I}}_1,{\overline{I}}_2,J$, we have $\begin{array}{cc}\mathit{\sigma }& =\frac{2}{J}[(J^{-2/3}\frac{\partial W}{\partial {\overline{I}}_1}+J^{-2/3}{\overline{I}}_1\frac{\partial W}{\partial {\overline{I}}_2})𝑩-J^{-4/3}\frac{\partial W}{\partial {\overline{I}}_2}𝑩\cdot 𝑩]+\\ & 2J[-\frac{1}{3}{J}^{-2}({\overline{I}}_1\frac{\partial W}{\partial {\overline{I}}_1}+2{\overline{I}}_2\frac{\partial W}{\partial {\overline{I}}_2})+\frac{1}{2}{J}^{-1}\frac{\partial W}{\partial J}]1\end{array}$ or, $\begin{array}{cc}\mathit{\sigma }& =\frac{2}{J}[\frac{1}{J^{2/3}}(\frac{\partial W}{\partial {\overline{I}}_1}+{\overline{I}}_1\frac{\partial W}{\partial {\overline{I}}_2})𝑩-\frac{1}{3}({\overline{I}}_1\frac{\partial W}{\partial {\overline{I}}_1}+2{\overline{I}}_2\frac{\partial W}{\partial {\overline{I}}_2})1-\\ & \frac{1}{J^{4/3}}\frac{\partial W}{\partial {\overline{I}}_2}𝑩\cdot 𝑩]+\frac{\partial W}{\partial J}1\end{array}$

Proof 3:

To express the Cauchy stress in terms of the stretches ${\lambda }_1,{\lambda }_2,{\lambda }_3$ recall that $\frac{\partial {\lambda }_i}{\partial 𝑪}=\frac{1}{2{\lambda }_i}{𝑹}^T\cdot ({𝐧}_i\otimes {𝐧}_i)\cdot 𝑹;i=1,2,3.$ The chain rule gives $\begin{array}{cc}\frac{\partial W}{\partial 𝑪}& =\frac{\partial W}{\partial {\lambda }_1}\frac{\partial {\lambda }_1}{\partial 𝑪}+\frac{\partial W}{\partial {\lambda }_2}\frac{\partial {\lambda }_2}{\partial 𝑪}+\frac{\partial W}{\partial {\lambda }_3}\frac{\partial {\lambda }_3}{\partial 𝑪}\\ & ={𝑹}^T\cdot [\frac{1}{2{\lambda }_1}\frac{\partial W}{\partial {\lambda }_1}{𝐧}_1\otimes {𝐧}_1+\frac{1}{2{\lambda }_2}\frac{\partial W}{\partial {\lambda }_2}{𝐧}_2\otimes {𝐧}_2+\frac{1}{2{\lambda }_3}\frac{\partial W}{\partial {\lambda }_3}{𝐧}_3\otimes {𝐧}_3]\cdot 𝑹\end{array}$ The Cauchy stress is given by $\mathit{\sigma }=\frac{2}{J}𝑭\cdot \frac{\partial W}{\partial 𝑪}\cdot {𝑭}^T=\frac{2}{J}(𝑽\cdot 𝑹)\cdot \frac{\partial W}{\partial 𝑪}\cdot ({𝑹}^T\cdot 𝑽)$ Plugging in the expression for the derivative of $W$ leads to $\mathit{\sigma }=\frac{2}{J}𝑽\cdot [\frac{1}{2{\lambda }_1}\frac{\partial W}{\partial {\lambda }_1}{𝐧}_1\otimes {𝐧}_1+\frac{1}{2{\lambda }_2}\frac{\partial W}{\partial {\lambda }_2}{𝐧}_2\otimes {𝐧}_2+\frac{1}{2{\lambda }_3}\frac{\partial W}{\partial {\lambda }_3}{𝐧}_3\otimes {𝐧}_3]\cdot 𝑽$ Using the spectral decomposition of $𝑽$ we have $𝑽\cdot ({𝐧}_i\otimes {𝐧}_i)\cdot 𝑽={\lambda }_i^2{𝐧}_i\otimes {𝐧}_i;i=1,2,3.$ Also note that $J=\det (𝑭)=\det (𝑽)\det (𝑹)=\det (𝑽)={\lambda }_1{\lambda }_2{\lambda }_3.$ Therefore the expression for the Cauchy stress can be written as $\mathit{\sigma }=\frac{1}{{\lambda }_1{\lambda }_2{\lambda }_3}[{\lambda }_1\frac{\partial W}{\partial {\lambda }_1}{𝐧}_1\otimes {𝐧}_1+{\lambda }_2\frac{\partial W}{\partial {\lambda }_2}{𝐧}_2\otimes {𝐧}_2+{\lambda }_3\frac{\partial W}{\partial {\lambda }_3}{𝐧}_3\otimes {𝐧}_3]$

Saint-Venant–Kirchhoff material

The simplest constitutive relationship that satisfies the requirements of hyperelasticity is the Saint-Venant–Kirchhoff material, which has a response function of the form

$𝑺=\lambda \text{tr}(𝑬)1+2\mu 𝑬,$

where $\lambda$ and $\mu$ are material constants that have to be determined by experiments. Such a linear relation is physically possible only for small strains.