Yeoh hyperelastic model

The original model proposed by Yeoh had a cubic form with only ${\displaystyle I_{1}}$ dependence and is applicable to purely incompressible materials. The strain energy density for this model is written as

${\displaystyle W=\sum _{i=1}^{3}C_{i}~(I_{1}-3)^{i}}$

where ${\displaystyle C_{i}}$ are material constants. The quantity ${\displaystyle 2C_{1}}$ can be interpreted as the initial shear modulus.

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

${\displaystyle W=\sum _{i=1}^{n}C_{i}~(I_{1}-3)^{i}~.}$

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

For consistency with linear elasticity the Yeoh model has to satisfy the condition

${\displaystyle 2{\cfrac {\partial W}{\partial I_{1}}}(3)=\mu ~~(i\neq j)}$

where ${\displaystyle \mu }$ is the shear modulus of the material. Now, at ${\displaystyle I_{1}=3(\lambda _{i}=\lambda _{j}=1)}$,

${\displaystyle {\cfrac {\partial W}{\partial I_{1}}}=C_{1}}$

Therefore, the consistency condition for the Yeoh model is

${\displaystyle 2C_{1}=\mu \,}$

The Cauchy stress for the incompressible Yeoh model is given by

${\displaystyle {\boldsymbol {\sigma }}=-p~{\boldsymbol {\mathit {1}}}+2~{\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}~;~~{\cfrac {\partial W}{\partial I_{1}}}=\sum _{i=1}^{n}i~C_{i}~(I_{1}-3)^{i-1}~.}$

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

${\displaystyle \sigma _{11}=-p+2~\lambda ^{2}~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{22}=-p+{\cfrac {2}{\lambda }}~{\cfrac {\partial W}{\partial I_{1}}}=\sigma _{33}~.}$

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

${\displaystyle p={\cfrac {2}{\lambda }}~{\cfrac {\partial W}{\partial I_{1}}}~.}$

Therefore,

${\displaystyle \sigma _{11}=2~\left(\lambda ^{2}-{\cfrac {1}{\lambda }}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}$

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

${\displaystyle T_{11}=\sigma _{11}/\lambda =2~\left(\lambda -{\cfrac {1}{\lambda ^{2}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}$

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

${\displaystyle \sigma _{11}=-p+2~\lambda ^{2}~{\cfrac {\partial W}{\partial I_{1}}}=\sigma _{22}~;~~\sigma _{33}=-p+{\cfrac {2}{\lambda ^{4}}}~{\cfrac {\partial W}{\partial I_{1}}}~.}$

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

${\displaystyle p={\cfrac {2}{\lambda ^{4}}}~{\cfrac {\partial W}{\partial I_{1}}}~.}$

Therefore,

${\displaystyle \sigma _{11}=2~\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{4}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}=\sigma _{22}~.}$

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

${\displaystyle T_{11}={\cfrac {\sigma _{11}}{\lambda }}=2~\left(\lambda -{\cfrac {1}{\lambda ^{5}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}=T_{22}~.}$

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

${\displaystyle \sigma _{11}=-p+2~\lambda ^{2}~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{22}=-p+{\cfrac {2}{\lambda ^{2}}}~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{33}=-p+2~{\cfrac {\partial W}{\partial I_{1}}}~.}$

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

${\displaystyle p={\cfrac {2}{\lambda ^{2}}}~{\cfrac {\partial W}{\partial I_{1}}}~.}$

Therefore,

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

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

${\displaystyle T_{11}={\cfrac {\sigma _{11}}{\lambda }}=2~\left(\lambda -{\cfrac {1}{\lambda ^{3}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}$