Advanced elasticity/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

$\boldsymbol{\sigma} = \rho~\boldsymbol{F}\bullet\cfrac{\partial \psi}{\partial \boldsymbol{E}}\bullet \boldsymbol{F}^T = 2~\rho~\boldsymbol{F}\bullet\cfrac{\partial \psi}{\partial \boldsymbol{C}}\bullet \boldsymbol{F}^T$

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

$\boldsymbol{S} = 2~\rho_0~\cfrac{\partial \psi}{\partial \boldsymbol{C}}$

where $\boldsymbol{S}$ 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 $\boldsymbol{C}$. This also mean that the free energy must depend only on the principal invariants of $\boldsymbol{C}$ which are

$\begin{align} I_{\boldsymbol{C}} = I_1 & = \text{tr}(\mathbf{C}) = C_{ii} = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 \\ II_{\boldsymbol{C}} = I_2 & = \tfrac{1}{2}\left[\text{tr}(\mathbf{C}^2) - (\text{tr}~\mathbf{C})^2 \right] = \tfrac{1}{2}\left[C_{ik}C_{ki} - C_{jj}^2\right] = \lambda_1^2\lambda_2^2 + \lambda_2^2\lambda_3^2 + \lambda_3^2\lambda_1^2 \\ III_{\boldsymbol{C}} = I_3 & = \det(\mathbf{C}) = \lambda_1^2\lambda_2^2\lambda_3^2 \end{align}$

In other words,

$\psi(\boldsymbol{C}) \equiv \psi(I_1, I_2, I_3)$

Therefore, from the chain rule,

$\cfrac{\partial \psi}{\partial \boldsymbol{C}} = \cfrac{\partial\psi}{\partial I_1}~\cfrac{\partial I_1}{\partial\boldsymbol{C}} + \cfrac{\partial\psi}{\partial I_2}~\cfrac{\partial I_2}{\partial\boldsymbol{C}} + \cfrac{\partial\psi}{\partial I_3}~\cfrac{\partial I_3}{\partial\boldsymbol{C}} = a_0~\boldsymbol{\mathit{1}} + a_1~\boldsymbol{C} + a_2~\boldsymbol{C}^{-1}$

From the Cayley-Hamilton theorem we can show that

$\boldsymbol{C}^{-1} \equiv f(\boldsymbol{C}^2, \boldsymbol{C}, \boldsymbol{\mathit{1}})$

Hence we can also write

$\cfrac{\partial \psi}{\partial \boldsymbol{C}} = b_0~\boldsymbol{\mathit{1}} + b_1~\boldsymbol{C} + b_2~\boldsymbol{C}^2$

The stress-strain relation can then be written as

$\boldsymbol{S} = 2~\rho_0~\left[b_0~\boldsymbol{\mathit{1}} + b_1~\boldsymbol{C} + b_2~\boldsymbol{C}^2\right]$

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

$\boldsymbol{\sigma} = 2~\rho~\left[a_2~\boldsymbol{\mathit{1}} + a_0~\boldsymbol{B} + a_1~\boldsymbol{B}^2\right]$

where $\boldsymbol{B}$ 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 $W(\boldsymbol{F})=\hat{W}(I_1,I_2,I_3) = \bar{W}(\bar{I}_1,\bar{I}_2,J) = \tilde{W}(\lambda_1,\lambda_2,\lambda_3)$, then

$\begin{align} \boldsymbol{\sigma} & = \cfrac{2}{\sqrt{I_3}}\left[\left(\cfrac{\partial\hat{W}}{\partial I_1} + I_1~\cfrac{\partial\hat{W}}{\partial I_2}\right)\boldsymbol{B} - \cfrac{\partial\hat{W}}{\partial I_2}~\boldsymbol{B} \cdot\boldsymbol{B} \right] + 2\sqrt{I_3}~\cfrac{\partial\hat{W}}{\partial I_3}~\boldsymbol{\mathit{1}} \\ & = \cfrac{2}{J}\left[\cfrac{1}{J^{2/3}}\left(\cfrac{\partial\bar{W}}{\partial \bar{I}_1} + \bar{I}_1~\cfrac{\partial\bar{W}}{\partial \bar{I}_2}\right)\boldsymbol{B} - \cfrac{1}{3}\left(\bar{I}_1~\cfrac{\partial\bar{W}}{\partial \bar{I}_1} + 2~\bar{I}_2~\cfrac{\partial\bar{W}}{\partial \bar{I}_2}\right)\boldsymbol{\mathit{1}} - \right.\\ & \qquad \qquad \qquad \left. \cfrac{1}{J^{4/3}}~\cfrac{\partial\bar{W}}{\partial \bar{I}_2}~\boldsymbol{B} \cdot\boldsymbol{B} \right] + \cfrac{\partial\bar{W}}{\partial J}~\boldsymbol{\mathit{1}} \\ & = \cfrac{\lambda_1}{\lambda_1\lambda_2\lambda_3}~\cfrac{\partial\tilde{W}}{\partial \lambda_1}~\mathbf{n}_1\otimes\mathbf{n}_1 + \cfrac{\lambda_2}{\lambda_1\lambda_2\lambda_3}~\cfrac{\partial\tilde{W}}{\partial \lambda_2}~\mathbf{n}_2\otimes\mathbf{n}_2 + \cfrac{\lambda_3}{\lambda_1\lambda_2\lambda_3}~\cfrac{\partial\tilde{W}}{\partial \lambda_3}~\mathbf{n}_3\otimes\mathbf{n}_3 \end{align}$

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

Proof 2:

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

Proof 3:

To express the Cauchy stress in terms of the stretches $\lambda_1, \lambda_2, \lambda_3$ recall that $\cfrac{\partial \lambda_i}{\partial\boldsymbol{C}} = \cfrac{1}{2\lambda_i}~\boldsymbol{R}^T\cdot(\mathbf{n}_i\otimes\mathbf{n}_i)\cdot\boldsymbol{R}~;~~ i = 1,2,3 ~.$ The chain rule gives $\begin{align} \cfrac{\partial W}{\partial\boldsymbol{C}} & = \cfrac{\partial W}{\partial \lambda_1}~\cfrac{\partial \lambda_1}{\partial\boldsymbol{C}} + \cfrac{\partial W}{\partial \lambda_2}~\cfrac{\partial \lambda_2}{\partial\boldsymbol{C}} + \cfrac{\partial W}{\partial \lambda_3}~\cfrac{\partial \lambda_3}{\partial\boldsymbol{C}} \\ & = \boldsymbol{R}^T\cdot\left[\cfrac{1}{2\lambda_1}~\cfrac{\partial W}{\partial \lambda_1}~\mathbf{n}_1\otimes\mathbf{n}_1 + \cfrac{1}{2\lambda_2}~\cfrac{\partial W}{\partial \lambda_2}~\mathbf{n}_2\otimes\mathbf{n}_2 + \cfrac{1}{2\lambda_3}~\cfrac{\partial W}{\partial \lambda_3}~\mathbf{n}_3\otimes\mathbf{n}_3\right]\cdot\boldsymbol{R} \end{align}$ The Cauchy stress is given by $\boldsymbol{\sigma} = \cfrac{2}{J}~\boldsymbol{F}\cdot \cfrac{\partial W}{\partial \boldsymbol{C}}\cdot\boldsymbol{F}^T = \cfrac{2}{J}~(\boldsymbol{V}\cdot\boldsymbol{R})\cdot \cfrac{\partial W}{\partial \boldsymbol{C}}\cdot(\boldsymbol{R}^T\cdot\boldsymbol{V})$ Plugging in the expression for the derivative of $W$ leads to $\boldsymbol{\sigma} = \cfrac{2}{J}~\boldsymbol{V}\cdot \left[\cfrac{1}{2\lambda_1}~ \cfrac{\partial W}{\partial \lambda_1}~\mathbf{n}_1\otimes\mathbf{n}_1 + \cfrac{1}{2\lambda_2}~ \cfrac{\partial W}{\partial \lambda_2}~\mathbf{n}_2\otimes\mathbf{n}_2 + \cfrac{1}{2\lambda_3}~ \cfrac{\partial W}{\partial \lambda_3}~\mathbf{n}_3\otimes\mathbf{n}_3\right] \cdot\boldsymbol{V}$ Using the spectral decomposition of $\boldsymbol{V}$ we have $\boldsymbol{V}\cdot(\mathbf{n}_i\otimes\mathbf{n}_i)\cdot\boldsymbol{V} = \lambda_i^2~\mathbf{n}_i\otimes\mathbf{n}_i ~;~~ i=1,2,3.$ Also note that $J = \det(\boldsymbol{F}) = \det(\boldsymbol{V})\det(\boldsymbol{R}) = \det(\boldsymbol{V}) = \lambda_1\lambda_2\lambda_3 ~.$ Therefore the expression for the Cauchy stress can be written as $\boldsymbol{\sigma} = \cfrac{1}{\lambda_1\lambda_2\lambda_3}~ \left[\lambda_1~\cfrac{\partial W}{\partial \lambda_1}~\mathbf{n}_1\otimes\mathbf{n}_1 + \lambda_2~\cfrac{\partial W}{\partial \lambda_2}~\mathbf{n}_2\otimes\mathbf{n}_2 + \lambda_3~\cfrac{\partial W}{\partial \lambda_3}~\mathbf{n}_3\otimes\mathbf{n}_3 \right]$

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

$\boldsymbol{S} = \lambda~\text{tr}(\boldsymbol{E})~\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{E} ,$

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.