Gent hyperelastic model

Stress-strain curves under uniaxial extension for Gent model compared with various hyperelastic material models.

For uniaxial extension in the ${\displaystyle \mathbf {n} _{1}}$-direction, the principal stretches are ${\displaystyle \lambda _{1}=\lambda ,~\lambda _{2}=\lambda _{3}}$. From incompressibility ${\displaystyle \lambda _{1}~\lambda _{2}~\lambda _{3}=1}$. Hence ${\displaystyle \lambda _{2}^{2}=\lambda _{3}^{2}=1/\lambda }$. Therefore,

${\displaystyle I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=\lambda ^{2}+{\cfrac {2}{\lambda }}~.}$

The left Cauchy-Green deformation tensor can then be expressed as

${\displaystyle {\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{\lambda }}~(\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\mathbf {n} _{3}\otimes \mathbf {n} _{3})~.}$

If the directions of the principal stretches are oriented with the coordinate basis vectors, we have

${\displaystyle \sigma _{11}=-p+{\cfrac {\lambda ^{2}\mu J_{m}}{J_{m}-I_{1}+3}}~;~~\sigma _{22}=-p+{\cfrac {\mu J_{m}}{\lambda (J_{m}-I_{1}+3)}}=\sigma _{33}~.}$

If ${\displaystyle \sigma _{22}=\sigma _{33}=0}$, we have

${\displaystyle p={\cfrac {\mu J_{m}}{\lambda (J_{m}-I_{1}+3)}}~.}$

Therefore,

${\displaystyle \sigma _{11}=\left(\lambda ^{2}-{\cfrac {1}{\lambda }}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.}$

The engineering strain is ${\displaystyle \lambda -1\,}$. The engineering stress is

${\displaystyle T_{11}=\sigma _{11}/\lambda =\left(\lambda -{\cfrac {1}{\lambda ^{2}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.}$

For equibiaxial extension in the ${\displaystyle \mathbf {n} _{1}}$ and ${\displaystyle \mathbf {n} _{2}}$ directions, the principal stretches are ${\displaystyle \lambda _{1}=\lambda _{2}=\lambda \,}$. From incompressibility ${\displaystyle \lambda _{1}~\lambda _{2}~\lambda _{3}=1}$. Hence ${\displaystyle \lambda _{3}=1/\lambda ^{2}\,}$. Therefore,

${\displaystyle I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=2~\lambda ^{2}+{\cfrac {1}{\lambda ^{4}}}~.}$

The left Cauchy-Green deformation tensor can then be expressed as

${\displaystyle {\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda ^{2}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{\lambda ^{4}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}~.}$

If the directions of the principal stretches are oriented with the coordinate basis vectors, we have

${\displaystyle \sigma _{11}=\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{4}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)=\sigma _{22}~.}$

The engineering strain is ${\displaystyle \lambda -1\,}$. The engineering stress is

${\displaystyle T_{11}={\cfrac {\sigma _{11}}{\lambda }}=\left(\lambda -{\cfrac {1}{\lambda ^{5}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)=T_{22}~.}$

Planar extension tests are carried out on thin specimens which are constrained from deforming in one direction. For planar extension in the ${\displaystyle \mathbf {n} _{1}}$ directions with the ${\displaystyle \mathbf {n} _{3}}$ direction constrained, the principal stretches are ${\displaystyle \lambda _{1}=\lambda ,~\lambda _{3}=1}$. From incompressibility ${\displaystyle \lambda _{1}~\lambda _{2}~\lambda _{3}=1}$. Hence ${\displaystyle \lambda _{2}=1/\lambda \,}$. Therefore,

${\displaystyle I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=\lambda ^{2}+{\cfrac {1}{\lambda ^{2}}}+1~.}$

The left Cauchy-Green deformation tensor can then be expressed as

${\displaystyle {\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{\lambda ^{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\mathbf {n} _{3}\otimes \mathbf {n} _{3}~.}$

If the directions of the principal stretches are oriented with the coordinate basis vectors, we have

${\displaystyle \sigma _{11}=\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{2}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~;~~\sigma _{22}=0~;~~\sigma _{33}=\left(1-{\cfrac {1}{\lambda ^{2}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.}$

The engineering strain is ${\displaystyle \lambda -1\,}$. The engineering stress is

${\displaystyle T_{11}={\cfrac {\sigma _{11}}{\lambda }}=\left(\lambda -{\cfrac {1}{\lambda ^{3}}}\right)\left({\cfrac {\mu J_{m}}{J_{m}-I_{1}+3}}\right)~.}$

The deformation gradient for a simple shear deformation has the form[4]

${\displaystyle {\boldsymbol {F}}={\boldsymbol {1}}+\gamma ~\mathbf {e} _{1}\otimes \mathbf {e} _{2}}$

where ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2}}$ are reference orthonormal basis vectors in the plane of deformation and the shear deformation is given by

${\displaystyle \gamma =\lambda -{\cfrac {1}{\lambda }}~;~~\lambda _{1}=\lambda ~;~~\lambda _{2}={\cfrac {1}{\lambda }}~;~~\lambda _{3}=1}$

In matrix form, the deformation gradient and the left Cauchy-Green deformation tensor may then be expressed as

${\displaystyle {\boldsymbol {F}}={\begin{bmatrix}1&\gamma &0\\0&1&0\\0&0&1\end{bmatrix}}~;~~{\boldsymbol {B}}={\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}={\begin{bmatrix}1+\gamma ^{2}&\gamma &0\\\gamma &1&0\\0&0&1\end{bmatrix}}}$

Therefore,

${\displaystyle I_{1}=\mathrm {tr} ({\boldsymbol {B}})=3+\gamma ^{2}}$

and the Cauchy stress is given by

${\displaystyle {\boldsymbol {\sigma }}=-p~{\boldsymbol {\mathit {1}}}+{\cfrac {\mu J_{m}}{J_{m}-\gamma ^{2}}}~{\boldsymbol {B}}}$

In matrix form,

${\displaystyle {\boldsymbol {\sigma }}={\begin{bmatrix}-p+{\cfrac {\mu J_{m}(1+\gamma ^{2})}{J_{m}-\gamma ^{2}}}&{\cfrac {\mu J_{m}\gamma }{J_{m}-\gamma ^{2}}}&0\\{\cfrac {\mu J_{m}\gamma }{J_{m}-\gamma ^{2}}}&-p+{\cfrac {\mu J_{m}}{J_{m}-\gamma ^{2}}}&0\\0&0&-p+{\cfrac {\mu J_{m}}{J_{m}-\gamma ^{2}}}\end{bmatrix}}}$