# Gent hyperelastic model

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The Gent hyperelastic material model is a phenomenological model of rubber elasticity that is based on the concept of limiting chain extensibility. In this model, the strain energy density function is designed such that it has a singularity when the first invariant of the left Cauchy-Green deformation tensor reaches a limiting value $I_{m}$.
The strain energy density function for the Gent model is
    $W=-{\cfrac {\mu J_{m}}{2}}\ln \left(1-{\cfrac {I_{1}-3}{J_{m}}}\right)$
where $\mu$ is the shear modulus and $J_{m}=I_{m}-3$.
In the limit where $J_{m}\rightarrow \infty$, the Gent model reduces to the Neo-Hookean solid model. This can be seen by expressing the Gent model in the form
    $W=-{\cfrac {\mu }{2x}}\ln \left[1-(I_{1}-3)x\right]~;~~x:={\cfrac {1}{J_{m}}}$
A Taylor series expansion of $\ln \left[1-(I_{1}-3)x\right]$ around $x=0$ and taking the limit as $x\rightarrow 0$ leads to
    $W={\cfrac {\mu }{2}}(I_{1}-3)$
which is the expression for the strain energy density of a Neo-Hookean solid.
Several compressible versions of the Gent model have been designed. One such model has the form (the below strain energy function yields a non zero hydrostatic stress at no deformation, refer for compressible Gent models).
    $W=-{\cfrac {\mu J_{m}}{2}}\ln \left(1-{\cfrac {I_{1}-3}{J_{m}}}\right)+{\cfrac {\kappa }{2}}\left({\cfrac {J^{2}-1}{2}}-\ln J\right)^{4}$
where $J=\det({\boldsymbol {F}})$, $\kappa$ is the bulk modulus, and ${\boldsymbol {F}}$ is the deformation gradient.

## Consistency condition

We may alternatively express the Gent model in the form
    $W=C_{0}\ln \left(1-{\cfrac {I_{1}-3}{J_{m}}}\right)$
For the model to be consistent with linear elasticity, the following condition has to be satisfied:
    $2{\cfrac {\partial W}{\partial I_{1}}}(3)=\mu$
where $\mu$ is the shear modulus of the material. Now, at $I_{1}=3(\lambda _{i}=\lambda _{j}=1)$,
    ${\cfrac {\partial W}{\partial I_{1}}}=-{\cfrac {C_{0}}{J_{m}}}$
Therefore, the consistency condition for the Gent model is
    $-{\cfrac {2C_{0}}{J_{m}}}=\mu \,\qquad \implies \qquad C_{0}=-{\cfrac {\mu J_{m}}{2}}$
The Gent model assumes that $J_{m}\gg 1$

## Stress-deformation relations

The Cauchy stress for the incompressible Gent model is given by
    ${\boldsymbol {\sigma }}=-p~{\boldsymbol {\mathit {I}}}+2~{\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}=-p~{\boldsymbol {\mathit {I}}}+{\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}~{\boldsymbol {B}}$

### Uniaxial extension

For uniaxial extension in the $\mathbf {n} _{1}$-direction, the principal stretches are $\lambda _{1}=\lambda ,~\lambda _{2}=\lambda _{3}$. From incompressibility $\lambda _{1}~\lambda _{2}~\lambda _{3}=1$. Hence $\lambda _{2}^{2}=\lambda _{3}^{2}=1/\lambda$. Therefore,
    $I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=\lambda ^{2}+{\cfrac {2}{\lambda }}~.$
The left Cauchy-Green deformation tensor can then be expressed as
    ${\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{\lambda }}~(\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\mathbf {n} _{3}\otimes \mathbf {n} _{3})~.$
If the directions of the principal stretches are oriented with the coordinate basis vectors, we have
    $\sigma _{11}=-p+{\cfrac {\lambda ^{2}\mu J_{m}}{J_{m}-I_{1}+3}}~;~~\sigma _{22}=-p+{\cfrac {\mu J_{m}}{\lambda (J_{m}-I_{1}+3)}}=\sigma _{33}~.$
If $\sigma _{22}=\sigma _{33}=0$, we have
    $p={\cfrac {\mu J_{m}}{\lambda (J_{m}-I_{1}+3)}}~.$
Therefore,
    $\sigma _{11}=\left(\lambda ^{2}-{\cfrac {1}{\lambda }}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.$
The engineering strain is $\lambda -1\,$. The engineering stress is
    $T_{11}=\sigma _{11}/\lambda =\left(\lambda -{\cfrac {1}{\lambda ^{2}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.$

### Equibiaxial extension

For equibiaxial extension in the $\mathbf {n} _{1}$ and $\mathbf {n} _{2}$ directions, the principal stretches are $\lambda _{1}=\lambda _{2}=\lambda \,$. From incompressibility $\lambda _{1}~\lambda _{2}~\lambda _{3}=1$. Hence $\lambda _{3}=1/\lambda ^{2}\,$.
Therefore,
    $I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=2~\lambda ^{2}+{\cfrac {1}{\lambda ^{4}}}~.$
The left Cauchy-Green deformation tensor can then be expressed as
    ${\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda ^{2}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{\lambda ^{4}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}~.$
If the directions of the principal stretches are oriented with the coordinate basis vectors, we have
    $\sigma _{11}=\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{4}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)=\sigma _{22}~.$
The engineering strain is $\lambda -1\,$. The engineering stress is
    $T_{11}={\cfrac {\sigma _{11}}{\lambda }}=\left(\lambda -{\cfrac {1}{\lambda ^{5}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)=T_{22}~.$

### Planar extension

Planar extension tests are carried out on thin specimens which are constrained from deforming in one direction. For planar extension in the $\mathbf {n} _{1}$ directions with the $\mathbf {n} _{3}$ direction constrained, the principal stretches are $\lambda _{1}=\lambda ,~\lambda _{3}=1$. From incompressibility $\lambda _{1}~\lambda _{2}~\lambda _{3}=1$. Hence $\lambda _{2}=1/\lambda \,$.
Therefore,
    $I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=\lambda ^{2}+{\cfrac {1}{\lambda ^{2}}}+1~.$
The left Cauchy-Green deformation tensor can then be expressed as
    ${\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{\lambda ^{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\mathbf {n} _{3}\otimes \mathbf {n} _{3}~.$
If the directions of the principal stretches are oriented with the coordinate basis vectors, we have
    $\sigma _{11}=\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{2}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~;~~\sigma _{22}=0~;~~\sigma _{33}=\left(1-{\cfrac {1}{\lambda ^{2}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.$
The engineering strain is $\lambda -1\,$. The engineering stress is
    $T_{11}={\cfrac {\sigma _{11}}{\lambda }}=\left(\lambda -{\cfrac {1}{\lambda ^{3}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.$

### Simple shear

The deformation gradient for a simple shear deformation has the form
    ${\boldsymbol {F}}={\boldsymbol {1}}+\gamma ~\mathbf {e} _{1}\otimes \mathbf {e} _{2}$
where $\mathbf {e} _{1},\mathbf {e} _{2}$ are reference orthonormal basis vectors in the plane of deformation and the shear deformation is given by
    $\gamma =\lambda -{\cfrac {1}{\lambda }}~;~~\lambda _{1}=\lambda ~;~~\lambda _{2}={\cfrac {1}{\lambda }}~;~~\lambda _{3}=1$
In matrix form, the deformation gradient and the left Cauchy-Green deformation tensor may then be expressed as
    ${\boldsymbol {F}}={\begin{bmatrix}1&\gamma &0\\0&1&0\\0&0&1\end{bmatrix}}~;~~{\boldsymbol {B}}={\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}={\begin{bmatrix}1+\gamma ^{2}&\gamma &0\\\gamma &1&0\\0&0&1\end{bmatrix}}$
Therefore,
    $I_{1}=\mathrm {tr} ({\boldsymbol {B}})=3+\gamma ^{2}$
and the Cauchy stress is given by
    ${\boldsymbol {\sigma }}=-p~{\boldsymbol {\mathit {1}}}+{\cfrac {\mu J_{m}}{J_{m}-\gamma ^{2}}}~{\boldsymbol {B}}$
In matrix form,
    ${\boldsymbol {\sigma }}={\begin{bmatrix}-p+{\cfrac {\mu J_{m}(1+\gamma ^{2})}{J_{m}-\gamma ^{2}}}&{\cfrac {\mu J_{m}\gamma }{J_{m}-\gamma ^{2}}}&0\\{\cfrac {\mu J_{m}\gamma }{J_{m}-\gamma ^{2}}}&-p+{\cfrac {\mu J_{m}}{J_{m}-\gamma ^{2}}}&0\\0&0&-p+{\cfrac {\mu J_{m}}{J_{m}-\gamma ^{2}}}\end{bmatrix}}$