# Nonlinear finite elements/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
    ${\begin{aligned}I_{\boldsymbol {C}}=I_{1}&={\text{tr}}(\mathbf {C} )=C_{ii}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}\\II_{\boldsymbol {C}}=I_{2}&={\tfrac {1}{2}}\left[{\text{tr}}(\mathbf {C} ^{2})-({\text{tr}}~\mathbf {C} )^{2}\right]={\tfrac {1}{2}}\left[C_{ik}C_{ki}-C_{jj}^{2}\right]=\lambda _{1}^{2}\lambda _{2}^{2}+\lambda _{2}^{2}\lambda _{3}^{2}+\lambda _{3}^{2}\lambda _{1}^{2}\\III_{\boldsymbol {C}}=I_{3}&=\det(\mathbf {C} )=\lambda _{1}^{2}\lambda _{2}^{2}\lambda _{3}^{2}\end{aligned}}$
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
    ${\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{\sqrt {I_{3}}}}\left[\left({\cfrac {\partial {\hat {W}}}{\partial I_{1}}}+I_{1}~{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}\right){\boldsymbol {B}}-{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2{\sqrt {I_{3}}}~{\cfrac {\partial {\hat {W}}}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {2}{J}}\left[{\cfrac {1}{J^{2/3}}}\left({\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {B}}-{\cfrac {1}{3}}\left({\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {\mathit {1}}}-\right.\\&\qquad \qquad \qquad \left.{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+{\cfrac {\partial {\bar {W}}}{\partial J}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {\lambda _{1}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {\lambda _{2}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {\lambda _{3}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}$
(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 ${\begin{aligned}{\cfrac {\partial W}{\partial I_{1}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{1}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{1}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{1}}}\\&=I_{3}^{-1/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}=J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}\\{\cfrac {\partial W}{\partial I_{2}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{2}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{2}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{2}}}\\&=I_{3}^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}=J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\\{\cfrac {\partial W}{\partial I_{3}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{3}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{3}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{3}}}\\&=-{\cfrac {1}{3}}~I_{3}^{-4/3}~I_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}-{\cfrac {2}{3}}~I_{3}^{-5/3}~I_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}+{\cfrac {1}{2}}~I_{3}^{-1/2}~{\cfrac {\partial W}{\partial J}}\\&=-{\cfrac {1}{3}}~J^{-8/3}~J^{2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}-{\cfrac {2}{3}}~J^{-10/3}~J^{4/3}~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\\&=-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\end{aligned}}$ 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 ${\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[\left(J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+J^{-2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\\&\qquad 2~J~\left[-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\right]~{\boldsymbol {\mathit {1}}}\end{aligned}}$ or, ${\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[{\cfrac {1}{J^{2/3}}}~\left({\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {1}{3}}\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {\mathit {1}}}-\right.\\&\qquad \left.{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+{\cfrac {\partial W}{\partial J}}~{\boldsymbol {\mathit {1}}}\end{aligned}}$
    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 ${\begin{aligned}{\cfrac {\partial W}{\partial {\boldsymbol {C}}}}&={\cfrac {\partial W}{\partial \lambda _{1}}}~{\cfrac {\partial \lambda _{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{2}}}~{\cfrac {\partial \lambda _{2}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{3}}}~{\cfrac {\partial \lambda _{3}}{\partial {\boldsymbol {C}}}}\\&={\boldsymbol {R}}^{T}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {R}}\end{aligned}}$ 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.