Yeoh model prediction versus experimental data for natural rubber. Model parameters and experimental data from PolymerFEM.com

The Yeoh w:hyperelastic material model[1] is a phenomenological model for the deformation of nearly w:incompressible, w:nonlinear w:elastic materials such as w:rubber. The model is based on Ronald Rivlin's observation that the elastic properties of rubber may be described using a w:strain energy density function which is a power series in the strain invariants $I_1, I_2, I_3$.[2] The Yeoh model for incompressible rubber is a function only of $I_1$. For compressible rubbers, an dependence on $I_3$ is added on. Since a polynomial form of the strain energy density function is used but all the three invariants of the left Cauchy-Green deformation tensor are not, the Yeoh model is also called the reduced polynomial model.

## Yeoh model for incompressible rubbers

The original model proposed by Yeoh had a cubic form with only $I_1$ dependence and is applicable to purely incompressible materials. The strain energy density for this model is written as

$W = \sum_{i=1}^3 C_i~(I_1-3)^i$

where $C_i$ are material constants. The quantity $2 C_1$ can be interpreted as the initial w:shear modulus.

Today a slightly more generalized version of the Yeoh model is used.[3] This model includes $n$ terms and is written as

$W = \sum_{i=1}^n C_i~(I_1-3)^i ~.$

When $n=1$ the Yeoh model reduces to the neo-Hookean model for incompressible materials.

The Cauchy stress for the incompressible Yeoh model is given by

$\boldsymbol{\sigma} = -p~\boldsymbol{\mathit{1}} + 2~\cfrac{\partial W}{\partial I_1}~\boldsymbol{B} ~;~~ \cfrac{\partial W}{\partial I_1} = \sum_{i=1}^n i~C_i~(I_1-3)^{i-1} ~.$

### 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

$\sigma_{11} = -p + 2~\lambda^2~\cfrac{\partial W}{\partial I_1} ~;~~ \sigma_{22} = -p + \cfrac{2}{\lambda}~\cfrac{\partial W}{\partial I_1} = \sigma_{33} ~.$

Since $\sigma_{22} = \sigma_{33} = 0$, we have

$p = \cfrac{2}{\lambda}~\cfrac{\partial W}{\partial I_1} ~.$

Therefore,

$\sigma_{11} = 2~\left(\lambda^2 - \cfrac{1}{\lambda}\right)~\cfrac{\partial W}{\partial I_1}~.$

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

$T_{11} = \sigma_{11}/\lambda = 2~\left(\lambda - \cfrac{1}{\lambda^2}\right)~\cfrac{\partial W}{\partial I_1}~.$

### 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

$\sigma_{11} = -p + 2~\lambda^2~\cfrac{\partial W}{\partial I_1} = \sigma_{22} ~;~~ \sigma_{33} = -p + \cfrac{2}{\lambda^4}~\cfrac{\partial W}{\partial I_1} ~.$

Since $\sigma_{33} = 0$, we have

$p = \cfrac{2}{\lambda^4}~\cfrac{\partial W}{\partial I_1} ~.$

Therefore,

$\sigma_{11} = 2~\left(\lambda^2 - \cfrac{1}{\lambda^4}\right)~\cfrac{\partial W}{\partial I_1} = \sigma_{22} ~.$

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

$T_{11} = \cfrac{\sigma_{11}}{\lambda} = 2~\left(\lambda - \cfrac{1}{\lambda^5}\right)~\cfrac{\partial W}{\partial I_1} = T_{22}~.$

### 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

$\sigma_{11} = -p + 2~\lambda^2~\cfrac{\partial W}{\partial I_1} ~;~~ \sigma_{11} = -p + \cfrac{2}{\lambda^2}~\cfrac{\partial W}{\partial I_1} ~;~~ \sigma_{33} = -p + 2~\cfrac{\partial W}{\partial I_1} ~.$

Since $\sigma_{22} = 0$, we have

$p = \cfrac{2}{\lambda^2}~\cfrac{\partial W}{\partial I_1} ~.$

Therefore,

$\sigma_{11} = 2~\left(\lambda^2 - \cfrac{1}{\lambda^2}\right)~\cfrac{\partial W}{\partial I_1} ~;~~ \sigma_{22} = 0 ~;~~ \sigma_{33} = 2~\left(1 - \cfrac{1}{\lambda^2}\right)~\cfrac{\partial W}{\partial I_1}~.$

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

$T_{11} = \cfrac{\sigma_{11}}{\lambda} = 2~\left(\lambda - \cfrac{1}{\lambda^3}\right)~\cfrac{\partial W}{\partial I_1}~.$

## Yeoh model for compressible rubbers

A version of the Yeoh model that includes $I_3 = J^2$ dependence is used for compressible rubbers. The strain energy density function for this model is written as

$W = \sum_{i=1}^n C_{i0}~(\bar{I}_1-3)^i + \sum_{k=1}^n C_{k1}~(J-1)^{2k}$

where $\bar{I}_1 = J^{-2/3}~I_1$, and $C_{i0}, C_{k1}$ are material constants. The quantity $C_{10}$ is interpreted as half the initial shear modulus, while $C_{11}$ is interpreted as half the initial bulk modulus.

When $n=1$ the compressible Yeoh model reduces to the neo-Hookean model for compressible materials.

## References

1. Yeoh, O. H., 1993, "Some forms of the strain energy function for rubber," Rubber Chemistry and technology, Volume 66, Issue 5, November 1993, Pages 754-771.
2. Rivlin, R. S., 1948, "Some applications of elasticity theory to rubber engineering", in Collected Papers of R. S. Rivlin vol. 1 and 2, Springer, 1997.
3. Selvadurai, A. P. S., 2006, "Deflections of a rubber membrane", Journal of the Mechanics and Physics of Solids, vol. 54, no. 6, pp. 1093-1119.