Hyperelastic material

The simplest hyperelastic material model is the Saint Venant–Kirchhoff model which is just an extension of the geometrically linear elastic material model to the geometrically nonlinear regime. This model has the general form and the isotropic form respectively

${\displaystyle {\boldsymbol {S}}={\boldsymbol {C}}:{\boldsymbol {E}}}$

${\displaystyle {\boldsymbol {S}}=\lambda ~{\text{tr}}({\boldsymbol {E}}){\boldsymbol {\mathit {I}}}+2\mu {\boldsymbol {E}}{\text{.}}}$

where ${\displaystyle {\boldsymbol {S}}}$ is the second Piola–Kirchhoff stress, ${\displaystyle {\boldsymbol {C}}:{\rm {I\!R}}^{3\times 3}\rightarrow {\rm {I\!R}}^{3\times 3}}$ is a fourth order stiffness tensor and ${\displaystyle {\boldsymbol {E}}}$ is the Lagrangian Green strain given by

${\displaystyle \mathbf {E} ={\frac {1}{2}}\left[(\nabla _{\mathbf {X} }\mathbf {u} )^{T}+\nabla _{\mathbf {X} }\mathbf {u} +(\nabla _{\mathbf {X} }\mathbf {u} )^{T}\cdot \nabla _{\mathbf {X} }\mathbf {u} \right]\,\!}$

${\displaystyle \lambda }$ and ${\displaystyle \mu }$ are the Lamé constants, and ${\displaystyle {\boldsymbol {\mathit {I}}}}$ is the second order unit tensor.

The strain-energy density function for the Saint Venant–Kirchhoff model is

${\displaystyle W({\boldsymbol {E}})={\frac {\lambda }{2}}[{\text{tr}}({\boldsymbol {E}})]^{2}+\mu {\text{tr}}({\boldsymbol {E}}^{2})}$

and the second Piola–Kirchhoff stress can be derived from the relation

${\displaystyle {\boldsymbol {S}}={\cfrac {\partial W}{\partial {\boldsymbol {E}}}}~.}$

Hyperelastic material models can be classified as:

1) phenomenological descriptions of observed behavior

• Fung
• Mooney–Rivlin
• Ogden
• Polynomial
• Saint Venant–Kirchhoff
• Yeoh
• Marlow

2) mechanistic models deriving from arguments about underlying structure of the material

• Arruda–Boyce model
• Neo-Hookean

3) hybrids of phenomenological and mechanistic models

• Gent
• Van der Waals

Generally, a hyperelastic model should satisfy the Drucker stability criterion. Some hyperelastic models satisfy the Valanis-Landel hypothesis which states that the strain energy function can be separated into the sum of separate functions of the principal stretches ${\displaystyle (\lambda _{1},\lambda _{2},\lambda _{3})}$:

${\displaystyle W=f(\lambda _{1})+f(\lambda _{2})+f(\lambda _{3})\,.}$

If ${\displaystyle W({\boldsymbol {F}})}$ is the strain energy density function, the 1st Piola–Kirchhoff stress tensor can be calculated for a hyperelastic material as

${\displaystyle {\boldsymbol {P}}={\frac {\partial W}{\partial {\boldsymbol {F}}}}\qquad {\text{or}}\qquad P_{iK}={\frac {\partial W}{\partial F_{iK}}}.}$

where ${\displaystyle {\boldsymbol {F}}}$ is the deformation gradient. In terms of the Lagrangian Green strain (${\displaystyle {\boldsymbol {E}}}$)

${\displaystyle {\boldsymbol {P}}={\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}\qquad {\text{or}}\qquad P_{iK}=F_{iL}~{\frac {\partial W}{\partial E_{LK}}}~.}$

In terms of the right Cauchy–Green deformation tensor (${\displaystyle {\boldsymbol {C}}}$)

${\displaystyle {\boldsymbol {P}}=2~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\qquad {\text{or}}\qquad P_{iK}=2~F_{iL}~{\frac {\partial W}{\partial C_{LK}}}~.}$

If ${\displaystyle {\boldsymbol {S}}}$ is the second Piola–Kirchhoff stress tensor then

${\displaystyle {\boldsymbol {S}}={\boldsymbol {F}}^{-1}\cdot {\frac {\partial W}{\partial {\boldsymbol {F}}}}\qquad {\text{or}}\qquad S_{IJ}=F_{Ik}^{-1}{\frac {\partial W}{\partial F_{kJ}}}~.}$

In terms of the Lagrangian Green strain

${\displaystyle {\boldsymbol {S}}={\frac {\partial W}{\partial {\boldsymbol {E}}}}\qquad {\text{or}}\qquad S_{IJ}={\frac {\partial W}{\partial E_{IJ}}}~.}$

In terms of the right Cauchy–Green deformation tensor

${\displaystyle {\boldsymbol {S}}=2~{\frac {\partial W}{\partial {\boldsymbol {C}}}}\qquad {\text{or}}\qquad S_{IJ}=2~{\frac {\partial W}{\partial C_{IJ}}}~.}$

The above relation is also known as the Doyle-Ericksen formula in the material configuration.

Similarly, the Cauchy stress is given by

${\displaystyle {\boldsymbol {\sigma }}={\cfrac {1}{J}}~{\cfrac {\partial W}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~;~~J:=\det {\boldsymbol {F}}\qquad {\text{or}}\qquad \sigma _{ij}={\cfrac {1}{J}}~{\cfrac {\partial W}{\partial F_{iK}}}~F_{jK}~.}$

In terms of the Lagrangian Green strain

${\displaystyle {\boldsymbol {\sigma }}={\cfrac {1}{J}}~{\boldsymbol {F}}\cdot {\cfrac {\partial W}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\qquad {\text{or}}\qquad \sigma _{ij}={\cfrac {1}{J}}~F_{iK}~{\cfrac {\partial W}{\partial E_{KL}}}~F_{jL}~.}$

In terms of the right Cauchy–Green deformation tensor

${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~{\boldsymbol {F}}\cdot {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}\qquad {\text{or}}\qquad \sigma _{ij}={\cfrac {2}{J}}~F_{iK}~{\cfrac {\partial W}{\partial C_{KL}}}~F_{jL}~.}$

The above expressions are valid even for anisotropic media (in which case, the potential function is understood to depend implicitly on reference directional quantities such as initial fiber orientations). In the special case of isotropy, the Cauchy stress can be expressed in terms of the left Cauchy-Green deformation tensor as follows:[5]

${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~{\boldsymbol {B}}\cdot {\cfrac {\partial W}{\partial {\boldsymbol {B}}}}\qquad {\text{or}}\qquad \sigma _{ij}={\cfrac {2}{J}}~B_{ik}~{\cfrac {\partial W}{\partial B_{kj}}}~.}$

For an incompressible material ${\displaystyle J:=\det {\boldsymbol {F}}=1}$. The incompressibility constraint is therefore ${\displaystyle J-1=0}$. To ensure incompressibility of a hyperelastic material, the strain-energy function can be written in form:

${\displaystyle W=W({\boldsymbol {F}})-p~(J-1)}$

where the hydrostatic pressure ${\displaystyle p}$ functions as a Lagrangian multiplier to enforce the incompressibility constraint. The 1st Piola–Kirchhoff stress now becomes

${\displaystyle {\boldsymbol {P}}=-p~J{\boldsymbol {F}}^{-T}+{\frac {\partial W}{\partial {\boldsymbol {F}}}}=-p~{\boldsymbol {F}}^{-T}+{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}=-p~{\boldsymbol {F}}^{-T}+2~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}~.}$

This stress tensor can subsequently be converted into any of the other conventional stress tensors, such as the Cauchy stress tensor which is given by

${\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {P}}\cdot {\boldsymbol {F}}^{T}=-p~{\boldsymbol {\mathit {1}}}+{\frac {\partial W}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}=-p~{\boldsymbol {\mathit {1}}}+{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}=-p~{\boldsymbol {\mathit {1}}}+2~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}~.}$

For isotropic hyperelastic materials, the Cauchy stress can be expressed in terms of the invariants of the left Cauchy–Green deformation tensor (or right Cauchy–Green deformation tensor). If the strain energy density function is ${\displaystyle W({\boldsymbol {F}})={\hat {W}}(I_{1},I_{2},I_{3})={\bar {W}}({\bar {I}}_{1},{\bar {I}}_{2},J)={\tilde {W}}(\lambda _{1},\lambda _{2},\lambda _{3})}$, then

${\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{\sqrt {I_{3}}}}\left[\left({\cfrac {\partial {\hat {W}}}{\partial I_{1}}}+I_{1}~{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}\right){\boldsymbol {B}}-{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2{\sqrt {I_{3}}}~{\cfrac {\partial {\hat {W}}}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {2}{J}}\left[{\cfrac {1}{J^{2/3}}}\left({\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {B}}-{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]\\&\qquad \qquad +\left[{\cfrac {\partial {\bar {W}}}{\partial J}}-{\cfrac {2}{3J}}\left({\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right)\right]~{\boldsymbol {\mathit {1}}}\\&={\cfrac {2}{J}}\left[\left({\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\bar {\boldsymbol {B}}}-{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]+\left[{\cfrac {\partial {\bar {W}}}{\partial J}}-{\cfrac {2}{3J}}\left({\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right)\right]~{\boldsymbol {\mathit {1}}}\\&={\cfrac {\lambda _{1}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {\lambda _{2}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {\lambda _{3}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}}$

(See the page on the left Cauchy–Green deformation tensor for the definitions of these symbols).

Proof 1:

The second Piola–Kirchhoff stress tensor for a hyperelastic material is given by ${\displaystyle {\boldsymbol {S}}=2~{\cfrac {\partial W}{\partial {\boldsymbol {C}}}}}$ where ${\displaystyle {\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}}$ is the right Cauchy–Green deformation tensor and ${\displaystyle {\boldsymbol {F}}}$ is the deformation gradient. The Cauchy stress is given by ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {1}{J}}~{\boldsymbol {F}}\cdot {\boldsymbol {S}}\cdot {\boldsymbol {F}}^{T}={\cfrac {2}{J}}~{\boldsymbol {F}}\cdot {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}}$ where ${\displaystyle J=\det {\boldsymbol {F}}}$. Let ${\displaystyle I_{1},I_{2},I_{3}}$ be the three principal invariants of ${\displaystyle {\boldsymbol {C}}}$. Then ${\displaystyle {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}={\cfrac {\partial W}{\partial I_{1}}}~{\cfrac {\partial I_{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial I_{2}}}~{\cfrac {\partial I_{2}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial I_{3}}}~{\cfrac {\partial I_{3}}{\partial {\boldsymbol {C}}}}~.}$ The derivatives of the invariants of the symmetric tensor ${\displaystyle {\boldsymbol {C}}}$ are ${\displaystyle {\frac {\partial I_{1}}{\partial {\boldsymbol {C}}}}={\boldsymbol {\mathit {1}}}~;~~{\frac {\partial I_{2}}{\partial {\boldsymbol {C}}}}=I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {C}}~;~~{\frac {\partial I_{3}}{\partial {\boldsymbol {C}}}}=\det({\boldsymbol {C}})~{\boldsymbol {C}}^{-1}}$ Therefore, we can write ${\displaystyle {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}={\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {\mathit {1}}}+{\cfrac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})+{\cfrac {\partial W}{\partial I_{3}}}~I_{3}~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {F}}^{-T}~.}$ Plugging into the expression for the Cauchy stress gives ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~\left[{\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}+{\cfrac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}-{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T})+{\cfrac {\partial W}{\partial I_{3}}}~I_{3}~{\boldsymbol {\mathit {1}}}\right]}$ Using the left Cauchy–Green deformation tensor ${\displaystyle {\boldsymbol {B}}={\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}}$ and noting that ${\displaystyle I_{3}=J^{2}}$, we can write ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{\sqrt {I_{3}}}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+I_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~{\sqrt {I_{3}}}~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}$ For an incompressible material ${\displaystyle I_{3}=1}$ and hence ${\displaystyle W=W(I_{1},I_{2})}$.Then ${\displaystyle {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}={\cfrac {\partial W}{\partial I_{1}}}~{\cfrac {\partial I_{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial I_{2}}}~{\cfrac {\partial I_{2}}{\partial {\boldsymbol {C}}}}={\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {\mathit {1}}}+{\cfrac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})}$ Therefore, the Cauchy stress is given by ${\displaystyle {\boldsymbol {\sigma }}=2\left[\left({\cfrac {\partial W}{\partial I_{1}}}+I_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]-p~{\boldsymbol {\mathit {1}}}~.}$ where ${\displaystyle p}$ is an undetermined pressure which acts as a Lagrange multiplier to enforce the incompressibility constraint. If, in addition, ${\displaystyle I_{1}=I_{2}}$, we have ${\displaystyle W=W(I_{1})}$ and hence ${\displaystyle {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}={\cfrac {\partial W}{\partial I_{1}}}~{\cfrac {\partial I_{1}}{\partial {\boldsymbol {C}}}}={\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {\mathit {1}}}}$ In that case the Cauchy stress can be expressed as ${\displaystyle {\boldsymbol {\sigma }}=2{\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}-p~{\boldsymbol {\mathit {1}}}~.}$

Proof 2:

The isochoric deformation gradient is defined as ${\displaystyle {\bar {\boldsymbol {F}}}:=J^{-1/3}{\boldsymbol {F}}}$, resulting in the isochoric deformation gradient having a determinant of 1, in other words it is volume stretch free. Using this one can subsequently define the isochoric left Cauchy–Green deformation tensor ${\displaystyle {\bar {\boldsymbol {B}}}:={\bar {\boldsymbol {F}}}\cdot {\bar {\boldsymbol {F}}}^{T}=J^{-2/3}{\boldsymbol {B}}}$. The invariants of ${\displaystyle {\bar {\boldsymbol {B}}}}$ are ${\displaystyle {\begin{aligned}{\bar {I}}_{1}&={\text{tr}}({\bar {\boldsymbol {B}}})=J^{-2/3}{\text{tr}}({\boldsymbol {B}})=J^{-2/3}I_{1}\\{\bar {I}}_{2}&={\frac {1}{2}}\left({\text{tr}}({\bar {\boldsymbol {B}}})^{2}-{\text{tr}}({\bar {\boldsymbol {B}}}^{2})\right)={\frac {1}{2}}\left(\left(J^{-2/3}{\text{tr}}({\boldsymbol {B}})\right)^{2}-{\text{tr}}(J^{-4/3}{\boldsymbol {B}}^{2})\right)=J^{-4/3}I_{2}\\{\bar {I}}_{3}&=\det({\bar {\boldsymbol {B}}})=J^{-6/3}\det({\boldsymbol {B}})=J^{-2}I_{3}=J^{-2}J^{2}=1\end{aligned}}}$ The set of invariants which are used to define the distortional behavior are the first two invariants of the isochoric left Cauchy–Green deformation tensor tensor, (which are identical to the ones for the right Cauchy Green stretch tensor), and add ${\displaystyle J}$ into the fray to describe the volumetric behaviour. To express the Cauchy stress in terms of the invariants ${\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}$ recall that ${\displaystyle {\bar {I}}_{1}=J^{-2/3}~I_{1}=I_{3}^{-1/3}~I_{1}~;~~{\bar {I}}_{2}=J^{-4/3}~I_{2}=I_{3}^{-2/3}~I_{2}~;~~J=I_{3}^{1/2}~.}$ The chain rule of differentiation gives us ${\displaystyle {\begin{aligned}{\cfrac {\partial W}{\partial I_{1}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{1}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{1}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{1}}}\\&=I_{3}^{-1/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}=J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}\\{\cfrac {\partial W}{\partial I_{2}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{2}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{2}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{2}}}\\&=I_{3}^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}=J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\\{\cfrac {\partial W}{\partial I_{3}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{3}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{3}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{3}}}\\&=-{\cfrac {1}{3}}~I_{3}^{-4/3}~I_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}-{\cfrac {2}{3}}~I_{3}^{-5/3}~I_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}+{\cfrac {1}{2}}~I_{3}^{-1/2}~{\cfrac {\partial W}{\partial J}}\\&=-{\cfrac {1}{3}}~J^{-8/3}~J^{2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}-{\cfrac {2}{3}}~J^{-10/3}~J^{4/3}~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\\&=-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\end{aligned}}}$ Recall that the Cauchy stress is given by ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{\sqrt {I_{3}}}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+I_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~{\sqrt {I_{3}}}~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}$ In terms of the invariants ${\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}$ we have ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+J^{2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~J~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}$ Plugging in the expressions for the derivatives of ${\displaystyle W}$ in terms of ${\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}$, we have ${\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[\left(J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+J^{-2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\\&\qquad 2~J~\left[-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\right]~{\boldsymbol {\mathit {1}}}\end{aligned}}}$ or, ${\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[{\cfrac {1}{J^{2/3}}}~\left({\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]\\&\qquad +\left[{\cfrac {\partial W}{\partial J}}-{\cfrac {2}{3J}}\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)\right]{\boldsymbol {\mathit {1}}}\end{aligned}}}$ In terms of the deviatoric part of ${\displaystyle {\boldsymbol {B}}}$, we can write ${\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[\left({\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]\\&\qquad +\left[{\cfrac {\partial W}{\partial J}}-{\cfrac {2}{3J}}\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)\right]{\boldsymbol {\mathit {1}}}\end{aligned}}}$ For an incompressible material ${\displaystyle J=1}$ and hence ${\displaystyle W=W({\bar {I}}_{1},{\bar {I}}_{2})}$.Then the Cauchy stress is given by ${\displaystyle {\boldsymbol {\sigma }}=2\left[\left({\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+I_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]-p~{\boldsymbol {\mathit {1}}}~.}$ where ${\displaystyle p}$ is an undetermined pressure-like Lagrange multiplier term. In addition, if ${\displaystyle {\bar {I}}_{1}={\bar {I}}_{2}}$, we have ${\displaystyle W=W({\bar {I}}_{1})}$ and hence the Cauchy stress can be expressed as ${\displaystyle {\boldsymbol {\sigma }}=2{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\bar {\boldsymbol {B}}}-p~{\boldsymbol {\mathit {1}}}~.}$

Proof 3:

To express the Cauchy stress in terms of the stretches ${\displaystyle \lambda _{1},\lambda _{2},\lambda _{3}}$ recall that ${\displaystyle {\cfrac {\partial \lambda _{i}}{\partial {\boldsymbol {C}}}}={\cfrac {1}{2\lambda _{i}}}~{\boldsymbol {R}}^{T}\cdot (\mathbf {n} _{i}\otimes \mathbf {n} _{i})\cdot {\boldsymbol {R}}~;~~i=1,2,3~.}$ The chain rule gives ${\displaystyle {\begin{aligned}{\cfrac {\partial W}{\partial {\boldsymbol {C}}}}&={\cfrac {\partial W}{\partial \lambda _{1}}}~{\cfrac {\partial \lambda _{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{2}}}~{\cfrac {\partial \lambda _{2}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{3}}}~{\cfrac {\partial \lambda _{3}}{\partial {\boldsymbol {C}}}}\\&={\boldsymbol {R}}^{T}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {R}}\end{aligned}}}$ The Cauchy stress is given by ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~{\boldsymbol {F}}\cdot {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}={\cfrac {2}{J}}~({\boldsymbol {V}}\cdot {\boldsymbol {R}})\cdot {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}\cdot ({\boldsymbol {R}}^{T}\cdot {\boldsymbol {V}})}$ Plugging in the expression for the derivative of ${\displaystyle W}$ leads to ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~{\boldsymbol {V}}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {V}}}$ Using the spectral decomposition of ${\displaystyle {\boldsymbol {V}}}$ we have ${\displaystyle {\boldsymbol {V}}\cdot (\mathbf {n} _{i}\otimes \mathbf {n} _{i})\cdot {\boldsymbol {V}}=\lambda _{i}^{2}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}~;~~i=1,2,3.}$ Also note that ${\displaystyle J=\det({\boldsymbol {F}})=\det({\boldsymbol {V}})\det({\boldsymbol {R}})=\det({\boldsymbol {V}})=\lambda _{1}\lambda _{2}\lambda _{3}~.}$ Therefore, the expression for the Cauchy stress can be written as ${\displaystyle {\boldsymbol {\sigma }}={\cfrac {1}{\lambda _{1}\lambda _{2}\lambda _{3}}}~\left[\lambda _{1}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]}$ For an incompressible material ${\displaystyle \lambda _{1}\lambda _{2}\lambda _{3}=1}$ and hence ${\displaystyle W=W(\lambda _{1},\lambda _{2})}$. Following Ogden[1] p. 485, we may write ${\displaystyle {\boldsymbol {\sigma }}=\lambda _{1}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}-p~{\boldsymbol {\mathit {1}}}~}$ Some care is to required at this stage because, when an eigenvalue is repeated, it is in general only Gateaux differentiable, but not Fréchet differentiable.[6][7] A rigorous tensor derivative can only be found by solving another eigenvalue problem. If we express the stress in terms of differences between components, ${\displaystyle \sigma _{11}-\sigma _{33}=\lambda _{1}~{\cfrac {\partial W}{\partial \lambda _{1}}}-\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}~;~~\sigma _{22}-\sigma _{33}=\lambda _{2}~{\cfrac {\partial W}{\partial \lambda _{2}}}-\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}}$ If in addition to incompressibility we have ${\displaystyle \lambda _{1}=\lambda _{2}}$ then a possible solution to the problem requires ${\displaystyle \sigma _{11}=\sigma _{22}}$ and we can write the stress differences as ${\displaystyle \sigma _{11}-\sigma _{33}=\sigma _{22}-\sigma _{33}=\lambda _{1}~{\cfrac {\partial W}{\partial \lambda _{1}}}-\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}}$