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 = -\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 $$I_m \rightarrow \infty$$, 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} ~. $$ 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} $$