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

\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 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 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).


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!Proof 1:
 * The second Piola-Kirchhoff stress tensor for a hyperelastic material is given by
 * 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}}~. $$
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!Proof 2:
 * To express the Cauchy stress in terms of the invariants $$\bar{I}_1, \bar{I}_2, J$$ recall that
 * To express the Cauchy stress in terms of the invariants $$\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} $$
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!Proof 3:
 * To express the Cauchy stress in terms of the stretches $$\lambda_1, \lambda_2, \lambda_3$$ recall that
 * To express the Cauchy stress in terms of the stretches $$\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] $$
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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.