Physics:Gent (hyperelastic model)

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

$W=-\frac{\mu J_m}{2}\ln (1-\frac{I_1-3}{J_m})$

where $\mu$ is the shear modulus and $J_m=I_m-3$.

In the limit where $I_m\to \mathrm{\infty }$, the Gent model reduces to the Neo-Hookean solid model. This can be seen by expressing the Gent model in the form

$W=-\frac{\mu }{2x}\ln [1-(I_1-3)x];x:=\frac{1}{J_m}$

A Taylor series expansion of $\ln [1-(I_1-3)x]$ around $x=0$ and taking the limit as $x\to 0$ leads to

$W=\frac{\mu }{2}(I_1-3)$

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 https://link.springer.com/article/10.1007/s10659-005-4408-x for compressible Gent models).

$W=-\frac{\mu J_m}{2}\ln (1-\frac{I_1-3}{J_m})+\frac{\kappa }{2}{(\frac{J^2-1}{2}-\ln J)}^4$

where $J=\det (\mathit{F})$, $\kappa$ is the bulk modulus, and $\mathit{F}$ is the deformation gradient.

Consistency condition

We may alternatively express the Gent model in the form

$W=C_0\ln (1-\frac{I_1-3}{J_m})$

For the model to be consistent with linear elasticity, the following condition has to be satisfied:

$2\frac{\partial W}{\partial I_1}(3)=\mu$

where $\mu$ is the shear modulus of the material. Now, at $I_1=3({\lambda }_i={\lambda }_j=1)$,

$\frac{\partial W}{\partial I_1}=-\frac{C_0}{J_m}$

Therefore, the consistency condition for the Gent model is

$-\frac{2C_0}{J_m}=\mu ⟹C_0=-\frac{\mu J_m}{2}$

The Gent model assumes that $J_m\gg 1$

Stress-deformation relations

The Cauchy stress for the incompressible Gent model is given by

$\mathit{\sigma }=-p{I}+2\frac{\partial W}{\partial I_1}\mathit{B}=-p{I}+\frac{\mu J_m}{J_m-I_1+3}\mathit{B}$

Uniaxial extension

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

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,

$I_1={\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2={\lambda }^2+\frac{2}{\lambda }.$

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

$\mathit{B}={\lambda }^2{\mathbf{𝐧}}_1\otimes {\mathbf{𝐧}}_1+\frac{1}{\lambda }({\mathbf{𝐧}}_2\otimes {\mathbf{𝐧}}_2+{\mathbf{𝐧}}_3\otimes {\mathbf{𝐧}}_3).$

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

${\sigma }_{11}=-p+\frac{{\lambda }^2\mu J_m}{J_m-I_1+3};{\sigma }_{22}=-p+\frac{\mu J_m}{\lambda (J_m-I_1+3)}={\sigma }_{33}.$

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

$p=\frac{\mu J_m}{\lambda (J_m-I_1+3)}.$

Therefore,

${\sigma }_{11}=({\lambda }^2-\frac{1}{\lambda })\left(\frac{\mu J_m}{J_m-I_1+3}\right).$

The engineering strain is $\lambda -1$. The engineering stress is

$T_{11}={\sigma }_{11}/\lambda =(\lambda -\frac{1}{{\lambda }^2})\left(\frac{\mu J_m}{J_m-I_1+3}\right).$

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,

$I_1={\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2=2{\lambda }^2+\frac{1}{{\lambda }^4}.$

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

$\mathit{B}={\lambda }^2{\mathbf{𝐧}}_1\otimes {\mathbf{𝐧}}_1+{\lambda }^2{\mathbf{𝐧}}_2\otimes {\mathbf{𝐧}}_2+\frac{1}{{\lambda }^4}{\mathbf{𝐧}}_3\otimes {\mathbf{𝐧}}_3.$

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

${\sigma }_{11}=({\lambda }^2-\frac{1}{{\lambda }^4})\left(\frac{\mu J_m}{J_m-I_1+3}\right)={\sigma }_{22}.$

The engineering strain is $\lambda -1$. The engineering stress is

$T_{11}=\frac{{\sigma }_{11}}{\lambda }=(\lambda -\frac{1}{{\lambda }^5})\left(\frac{\mu J_m}{J_m-I_1+3}\right)=T_{22}.$

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,

$I_1={\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2={\lambda }^2+\frac{1}{{\lambda }^2}+1.$

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

$\mathit{B}={\lambda }^2{\mathbf{𝐧}}_1\otimes {\mathbf{𝐧}}_1+\frac{1}{{\lambda }^2}{\mathbf{𝐧}}_2\otimes {\mathbf{𝐧}}_2+{\mathbf{𝐧}}_3\otimes {\mathbf{𝐧}}_3.$

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

${\sigma }_{11}=({\lambda }^2-\frac{1}{{\lambda }^2})\left(\frac{\mu J_m}{J_m-I_1+3}\right);{\sigma }_{22}=0;{\sigma }_{33}=(1-\frac{1}{{\lambda }^2})\left(\frac{\mu J_m}{J_m-I_1+3}\right).$

The engineering strain is $\lambda -1$. The engineering stress is

$T_{11}=\frac{{\sigma }_{11}}{\lambda }=(\lambda -\frac{1}{{\lambda }^3})\left(\frac{\mu J_m}{J_m-I_1+3}\right).$

Simple shear

The deformation gradient for a simple shear deformation has the form

$\mathit{F}=\mathit{1}+\gamma {\mathbf{𝐞}}_1\otimes {\mathbf{𝐞}}_2$

where ${\mathbf{𝐞}}_1,{\mathbf{𝐞}}_2$ are reference orthonormal basis vectors in the plane of deformation and the shear deformation is given by

$\gamma =\lambda -\frac{1}{\lambda };{\lambda }_1=\lambda ;{\lambda }_2=\frac{1}{\lambda };{\lambda }_3=1$

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

$\mathit{F}=\left[\begin{array}{ccc}1& \gamma & 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right];\mathit{B}=\mathit{F}\cdot {\mathit{F}}^T=\left[\begin{array}{ccc}1+{\gamma }^2& \gamma & 0\\ \gamma & 1& 0\\ 0& 0& 1\end{array}\right]$

Therefore,

$I_1=\mathrm{t}\mathrm{r}(\mathit{B})=3+{\gamma }^2$

and the Cauchy stress is given by

$\mathit{\sigma }=-p{1}+\frac{\mu J_m}{J_m-{\gamma }^2}\mathit{B}$

In matrix form,

$\mathit{\sigma }=\left[\begin{array}{ccc}-p+\frac{\mu J_m(1+{\gamma }^2)}{J_m-{\gamma }^2}& \frac{\mu J_m\gamma }{J_m-{\gamma }^2}& 0\\ \frac{\mu J_m\gamma }{J_m-{\gamma }^2}& -p+\frac{\mu J_m}{J_m-{\gamma }^2}& 0\\ 0& 0& -p+\frac{\mu J_m}{J_m-{\gamma }^2}\end{array}\right]$

See also

• Hyperelastic material
• Strain energy density function
• Mooney-Rivlin solid
• Finite strain theory
• Stress measures