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A hyperelastic or Green elastic material^[1] is a type of constitutive model for ideally elastic material for which the stress–strain relationship derives from a strain energy density function. The hyperelastic material is a special case of a Cauchy elastic material.

For many materials, linear elastic models do not accurately describe the observed material behaviour. The most common example of this kind of material is rubber, whose stress-strain relationship can be defined as non-linearly elastic, isotropic and incompressible. Hyperelasticity provides a means of modeling the stress–strain behavior of such materials.^[2] The behavior of unfilled, vulcanized elastomers often conforms closely to the hyperelastic ideal. Filled elastomers and biological tissues^[3][4] are also often modeled via the hyperelastic idealization.

Ronald Rivlin and Melvin Mooney developed the first hyperelastic models, the Neo-Hookean and Mooney–Rivlin solids. Many other hyperelastic models have since been developed. Other widely used hyperelastic material models include the Ogden model and the Arruda–Boyce model.

Hyperelastic material models

Saint Venant–Kirchhoff model

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

$${\displaystyle {\begin{aligned}{\boldsymbol {S))&={\boldsymbol {C)):{\boldsymbol {E))\\{\boldsymbol {S))&=\lambda ~{\text{tr))({\boldsymbol {E))){\boldsymbol {\mathit {I))}+2\mu {\boldsymbol {E)){\text{.))\end{aligned))}$$

${\displaystyle \mathbin {:} }$

${\displaystyle {\boldsymbol {S))}$

${\displaystyle {\boldsymbol {C)):\mathbb {R} ^{3\times 3}\to \mathbb {R} ^{3\times 3))$

${\displaystyle {\boldsymbol {E))}$

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

${\displaystyle \lambda }$

${\displaystyle \mu }$

${\displaystyle {\boldsymbol {\mathit {I))))$

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

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

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

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

Classification of hyperelastic material models

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^[5]
   • Neo–Hookean model^[1]
   • Buche–Silberstein model^[6]
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 ${\displaystyle (\lambda _{1},\lambda _{2},\lambda _{3})}$:

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

Stress–strain relations

Compressible hyperelastic materials

First Piola–Kirchhoff stress

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

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

${\displaystyle {\boldsymbol {F))}$

${\displaystyle {\boldsymbol {E))}$

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

${\displaystyle {\boldsymbol {C))}$

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

Second Piola–Kirchhoff stress

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

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

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

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

Cauchy stress

Similarly, the Cauchy stress is given by

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

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

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

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

Incompressible hyperelastic materials

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:

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

${\displaystyle p}$

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

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

Expressions for the Cauchy stress

Compressible isotropic hyperelastic materials

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

$${\displaystyle W({\boldsymbol {F)))={\hat {W))(I_{1},I_{2},I_{3})={\bar {W))({\bar {I))_{1},{\bar {I))_{2},J)={\tilde {W))(\lambda _{1},\lambda _{2},\lambda _{3}),}$$

$${\displaystyle {\begin{aligned}{\boldsymbol {\sigma ))&={\frac {2}{\sqrt {I_{3))))\left[\left({\frac {\partial {\hat {W))}{\partial I_{1))}+I_{1}~{\frac {\partial {\hat {W))}{\partial I_{2))}\right){\boldsymbol {B))-{\frac {\partial {\hat {W))}{\partial I_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]+2{\sqrt {I_{3))}~{\frac {\partial {\hat {W))}{\partial I_{3))}~{\boldsymbol {\mathit {1))}\\[5pt]&={\frac {2}{J))\left[{\frac {1}{J^{2/3))}\left({\frac {\partial {\bar {W))}{\partial {\bar {I))_{1))}+{\bar {I))_{1}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{2))}\right){\boldsymbol {B))-{\frac {1}{J^{4/3))}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]+\left[{\frac {\partial {\bar {W))}{\partial J))-{\frac {2}{3J))\left({\bar {I))_{1}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{1))}+2~{\bar {I))_{2}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{2))}\right)\right]~{\boldsymbol {\mathit {1))}\\[5pt]&={\frac {2}{J))\left[\left({\frac {\partial {\bar {W))}{\partial {\bar {I))_{1))}+{\bar {I))_{1}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{2))}\right){\bar {\boldsymbol {B))}-{\frac {\partial {\bar {W))}{\partial {\bar {I))_{2))}~{\bar {\boldsymbol {B))}\cdot {\bar {\boldsymbol {B))}\right]+\left[{\frac {\partial {\bar {W))}{\partial J))-{\frac {2}{3J))\left({\bar {I))_{1}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{1))}+2~{\bar {I))_{2}~{\frac {\partial {\bar {W))}{\partial {\bar {I))_{2))}\right)\right]~{\boldsymbol {\mathit {1))}\\[5pt]&={\frac {\lambda _{1)){\lambda _{1}\lambda _{2}\lambda _{3))}~{\frac {\partial {\tilde {W))}{\partial \lambda _{1))}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {\lambda _{2)){\lambda _{1}\lambda _{2}\lambda _{3))}~{\frac {\partial {\tilde {W))}{\partial \lambda _{2))}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {\lambda _{3)){\lambda _{1}\lambda _{2}\lambda _{3))}~{\frac {\partial {\tilde {W))}{\partial \lambda _{3))}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned))}$$

Proof 1

The second Piola–Kirchhoff stress tensor for a hyperelastic material is given by

$${\displaystyle {\boldsymbol {S))=2~{\frac {\partial W}{\partial {\boldsymbol {C))))}$$

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

$${\displaystyle {\boldsymbol {\sigma ))={\frac {1}{J))~{\boldsymbol {F))\cdot {\boldsymbol {S))\cdot {\boldsymbol {F))^{T}={\frac {2}{J))~{\boldsymbol {F))\cdot {\frac {\partial W}{\partial {\boldsymbol {C))))\cdot {\boldsymbol {F))^{T))$$

where ${\displaystyle J=\det {\boldsymbol {F))}$. Let ${\displaystyle I_{1},I_{2},I_{3))$ be the three principal invariants of ${\displaystyle {\boldsymbol {C))}$. Then

$${\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C))))={\frac {\partial W}{\partial I_{1))}~{\frac {\partial I_{1)){\partial {\boldsymbol {C))))+{\frac {\partial W}{\partial I_{2))}~{\frac {\partial I_{2)){\partial {\boldsymbol {C))))+{\frac {\partial W}{\partial I_{3))}~{\frac {\partial I_{3)){\partial {\boldsymbol {C))))~.}$$

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

$${\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C))))={\frac {\partial W}{\partial I_{1))}~{\boldsymbol {\mathit {1))}+{\frac {\partial W}{\partial I_{2))}~(I_{1}~{\boldsymbol {\mathit {1))}-{\boldsymbol {F))^{T}\cdot {\boldsymbol {F)))+{\frac {\partial W}{\partial I_{3))}~I_{3}~{\boldsymbol {F))^{-1}\cdot {\boldsymbol {F))^{-T}~.}$$

Plugging into the expression for the Cauchy stress gives

$${\displaystyle {\boldsymbol {\sigma ))={\frac {2}{J))~\left[{\frac {\partial W}{\partial I_{1))}~{\boldsymbol {F))\cdot {\boldsymbol {F))^{T}+{\frac {\partial W}{\partial I_{2))}~(I_{1}~{\boldsymbol {F))\cdot {\boldsymbol {F))^{T}-{\boldsymbol {F))\cdot {\boldsymbol {F))^{T}\cdot {\boldsymbol {F))\cdot {\boldsymbol {F))^{T})+{\frac {\partial W}{\partial I_{3))}~I_{3}~{\boldsymbol {\mathit {1))}\right]}$$

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

$${\displaystyle {\boldsymbol {\sigma ))={\frac {2}{\sqrt {I_{3))))~\left[\left({\frac {\partial W}{\partial I_{1))}+I_{1}~{\frac {\partial W}{\partial I_{2))}\right)~{\boldsymbol {B))-{\frac {\partial W}{\partial I_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]+2~{\sqrt {I_{3))}~{\frac {\partial W}{\partial I_{3))}~{\boldsymbol {\mathit {1))}~.}$$

For an incompressible material ${\displaystyle I_{3}=1}$ and hence ${\displaystyle W=W(I_{1},I_{2})}$.Then

$${\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C))))={\frac {\partial W}{\partial I_{1))}~{\frac {\partial I_{1)){\partial {\boldsymbol {C))))+{\frac {\partial W}{\partial I_{2))}~{\frac {\partial I_{2)){\partial {\boldsymbol {C))))={\frac {\partial W}{\partial I_{1))}~{\boldsymbol {\mathit {1))}+{\frac {\partial W}{\partial I_{2))}~(I_{1}~{\boldsymbol {\mathit {1))}-{\boldsymbol {F))^{T}\cdot {\boldsymbol {F)))}$$

Therefore, the Cauchy stress is given by

$${\displaystyle {\boldsymbol {\sigma ))=2\left[\left({\frac {\partial W}{\partial I_{1))}+I_{1}~{\frac {\partial W}{\partial I_{2))}\right)~{\boldsymbol {B))-{\frac {\partial W}{\partial I_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]-p~{\boldsymbol {\mathit {1))}~.}$$

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

$${\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C))))={\frac {\partial W}{\partial I_{1))}~{\frac {\partial I_{1)){\partial {\boldsymbol {C))))={\frac {\partial W}{\partial I_{1))}~{\boldsymbol {\mathit {1))))$$

In that case the Cauchy stress can be expressed as

$${\displaystyle {\boldsymbol {\sigma ))=2{\frac {\partial W}{\partial I_{1))}~{\boldsymbol {B))-p~{\boldsymbol {\mathit {1))}~.}$$

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

$${\displaystyle {\begin{aligned}{\frac {\partial W}{\partial I_{1))}&={\frac {\partial W}{\partial {\bar {I))_{1))}~{\frac {\partial {\bar {I))_{1)){\partial I_{1))}+{\frac {\partial W}{\partial {\bar {I))_{2))}~{\frac {\partial {\bar {I))_{2)){\partial I_{1))}+{\frac {\partial W}{\partial J))~{\frac {\partial J}{\partial I_{1))}\\&=I_{3}^{-1/3}~{\frac {\partial W}{\partial {\bar {I))_{1))}=J^{-2/3}~{\frac {\partial W}{\partial {\bar {I))_{1))}\\{\frac {\partial W}{\partial I_{2))}&={\frac {\partial W}{\partial {\bar {I))_{1))}~{\frac {\partial {\bar {I))_{1)){\partial I_{2))}+{\frac {\partial W}{\partial {\bar {I))_{2))}~{\frac {\partial {\bar {I))_{2)){\partial I_{2))}+{\frac {\partial W}{\partial J))~{\frac {\partial J}{\partial I_{2))}\\&=I_{3}^{-2/3}~{\frac {\partial W}{\partial {\bar {I))_{2))}=J^{-4/3}~{\frac {\partial W}{\partial {\bar {I))_{2))}\\{\frac {\partial W}{\partial I_{3))}&={\frac {\partial W}{\partial {\bar {I))_{1))}~{\frac {\partial {\bar {I))_{1)){\partial I_{3))}+{\frac {\partial W}{\partial {\bar {I))_{2))}~{\frac {\partial {\bar {I))_{2)){\partial I_{3))}+{\frac {\partial W}{\partial J))~{\frac {\partial J}{\partial I_{3))}\\&=-{\frac {1}{3))~I_{3}^{-4/3}~I_{1}~{\frac {\partial W}{\partial {\bar {I))_{1))}-{\frac {2}{3))~I_{3}^{-5/3}~I_{2}~{\frac {\partial W}{\partial {\bar {I))_{2))}+{\frac {1}{2))~I_{3}^{-1/2}~{\frac {\partial W}{\partial J))\\&=-{\frac {1}{3))~J^{-8/3}~J^{2/3}~{\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{1))}-{\frac {2}{3))~J^{-10/3}~J^{4/3}~{\bar {I))_{2}~{\frac {\partial W}{\partial {\bar {I))_{2))}+{\frac {1}{2))~J^{-1}~{\frac {\partial W}{\partial J))\\&=-{\frac {1}{3))~J^{-2}~\left({\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{1))}+2~{\bar {I))_{2}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)+{\frac {1}{2))~J^{-1}~{\frac {\partial W}{\partial J))\end{aligned))}$$

Recall that the Cauchy stress is given by

$${\displaystyle {\boldsymbol {\sigma ))={\frac {2}{\sqrt {I_{3))))~\left[\left({\frac {\partial W}{\partial I_{1))}+I_{1}~{\frac {\partial W}{\partial I_{2))}\right)~{\boldsymbol {B))-{\frac {\partial W}{\partial I_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]+2~{\sqrt {I_{3))}~{\frac {\partial W}{\partial I_{3))}~{\boldsymbol {\mathit {1))}~.}$$

In terms of the invariants ${\displaystyle {\bar {I))_{1},{\bar {I))_{2},J}$ we have

$${\displaystyle {\boldsymbol {\sigma ))={\frac {2}{J))~\left[\left({\frac {\partial W}{\partial I_{1))}+J^{2/3}~{\bar {I))_{1}~{\frac {\partial W}{\partial I_{2))}\right)~{\boldsymbol {B))-{\frac {\partial W}{\partial I_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]+2~J~{\frac {\partial W}{\partial I_{3))}~{\boldsymbol {\mathit {1))}~.}$$

Plugging in the expressions for the derivatives of ${\displaystyle W}$ in terms of ${\displaystyle {\bar {I))_{1},{\bar {I))_{2},J}$, we have

$${\displaystyle {\begin{aligned}{\boldsymbol {\sigma ))&={\frac {2}{J))~\left[\left(J^{-2/3}~{\frac {\partial W}{\partial {\bar {I))_{1))}+J^{-2/3}~{\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)~{\boldsymbol {B))-J^{-4/3}~{\frac {\partial W}{\partial {\bar {I))_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]+\\&\qquad 2~J~\left[-{\frac {1}{3))~J^{-2}~\left({\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{1))}+2~{\bar {I))_{2}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)+{\frac {1}{2))~J^{-1}~{\frac {\partial W}{\partial J))\right]~{\boldsymbol {\mathit {1))}\end{aligned))}$$

or,

$${\displaystyle {\begin{aligned}{\boldsymbol {\sigma ))&={\frac {2}{J))~\left[{\frac {1}{J^{2/3))}~\left({\frac {\partial W}{\partial {\bar {I))_{1))}+{\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)~{\boldsymbol {B))-{\frac {1}{J^{4/3))}~{\frac {\partial W}{\partial {\bar {I))_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]\\&\qquad +\left[{\frac {\partial W}{\partial J))-{\frac {2}{3J))\left({\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{1))}+2~{\bar {I))_{2}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)\right]{\boldsymbol {\mathit {1))}\end{aligned))}$$

In terms of the deviatoric part of ${\displaystyle {\boldsymbol {B))}$, we can write

$${\displaystyle {\begin{aligned}{\boldsymbol {\sigma ))&={\frac {2}{J))~\left[\left({\frac {\partial W}{\partial {\bar {I))_{1))}+{\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)~{\bar {\boldsymbol {B))}-{\frac {\partial W}{\partial {\bar {I))_{2))}~{\bar {\boldsymbol {B))}\cdot {\bar {\boldsymbol {B))}\right]\\&\qquad +\left[{\frac {\partial W}{\partial J))-{\frac {2}{3J))\left({\bar {I))_{1}~{\frac {\partial W}{\partial {\bar {I))_{1))}+2~{\bar {I))_{2}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)\right]{\boldsymbol {\mathit {1))}\end{aligned))}$$

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

$${\displaystyle {\boldsymbol {\sigma ))=2\left[\left({\frac {\partial W}{\partial {\bar {I))_{1))}+I_{1}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)~{\bar {\boldsymbol {B))}-{\frac {\partial W}{\partial {\bar {I))_{2))}~{\bar {\boldsymbol {B))}\cdot {\bar {\boldsymbol {B))}\right]-p~{\boldsymbol {\mathit {1))}~.}$$

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

$${\displaystyle {\boldsymbol {\sigma ))=2{\frac {\partial W}{\partial {\bar {I))_{1))}~{\bar {\boldsymbol {B))}-p~{\boldsymbol {\mathit {1))}~.}$$

Proof 3

To express the Cauchy stress in terms of the stretches ${\displaystyle \lambda _{1},\lambda _{2},\lambda _{3))$ recall that

$${\displaystyle {\frac {\partial \lambda _{i)){\partial {\boldsymbol {C))))={\frac {1}{2\lambda _{i))}~{\boldsymbol {R))^{T}\cdot (\mathbf {n} _{i}\otimes \mathbf {n} _{i})\cdot {\boldsymbol {R))~;~~i=1,2,3~.}$$

The chain rule gives

$${\displaystyle {\begin{aligned}{\frac {\partial W}{\partial {\boldsymbol {C))))&={\frac {\partial W}{\partial \lambda _{1))}~{\frac {\partial \lambda _{1)){\partial {\boldsymbol {C))))+{\frac {\partial W}{\partial \lambda _{2))}~{\frac {\partial \lambda _{2)){\partial {\boldsymbol {C))))+{\frac {\partial W}{\partial \lambda _{3))}~{\frac {\partial \lambda _{3)){\partial {\boldsymbol {C))))\\&={\boldsymbol {R))^{T}\cdot \left[{\frac {1}{2\lambda _{1))}~{\frac {\partial W}{\partial \lambda _{1))}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {1}{2\lambda _{2))}~{\frac {\partial W}{\partial \lambda _{2))}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {1}{2\lambda _{3))}~{\frac {\partial W}{\partial \lambda _{3))}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {R))\end{aligned))}$$

The Cauchy stress is given by

$${\displaystyle {\boldsymbol {\sigma ))={\frac {2}{J))~{\boldsymbol {F))\cdot {\frac {\partial W}{\partial {\boldsymbol {C))))\cdot {\boldsymbol {F))^{T}={\frac {2}{J))~({\boldsymbol {V))\cdot {\boldsymbol {R)))\cdot {\frac {\partial W}{\partial {\boldsymbol {C))))\cdot ({\boldsymbol {R))^{T}\cdot {\boldsymbol {V)))}$$

Plugging in the expression for the derivative of ${\displaystyle W}$ leads to

$${\displaystyle {\boldsymbol {\sigma ))={\frac {2}{J))~{\boldsymbol {V))\cdot \left[{\frac {1}{2\lambda _{1))}~{\frac {\partial W}{\partial \lambda _{1))}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {1}{2\lambda _{2))}~{\frac {\partial W}{\partial \lambda _{2))}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {1}{2\lambda _{3))}~{\frac {\partial W}{\partial \lambda _{3))}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {V))}$$

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

$${\displaystyle {\boldsymbol {\sigma ))={\frac {1}{\lambda _{1}\lambda _{2}\lambda _{3))}~\left[\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2))}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]}$$

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

$${\displaystyle {\boldsymbol {\sigma ))=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2))}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}-p~{\boldsymbol {\mathit {1))}~}$$

Some care is required at this stage because, when an eigenvalue is repeated, it is in general only Gateaux differentiable, but not Fréchet differentiable.^[8][9] A rigorous tensor derivative can only be found by solving another eigenvalue problem.

If we express the stress in terms of differences between components,

$${\displaystyle \sigma _{11}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))}~;~~\sigma _{22}-\sigma _{33}=\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2))}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))))$$

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

$${\displaystyle \sigma _{11}-\sigma _{33}=\sigma _{22}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))))$$

Incompressible isotropic hyperelastic materials

For incompressible isotropic hyperelastic materials, the strain energy density function is ${\displaystyle W({\boldsymbol {F)))={\hat {W))(I_{1},I_{2})}$. The Cauchy stress is then given by

$${\displaystyle {\begin{aligned}{\boldsymbol {\sigma ))&=-p~{\boldsymbol {\mathit {1))}+2\left[\left({\frac {\partial {\hat {W))}{\partial I_{1))}+I_{1}~{\frac {\partial {\hat {W))}{\partial I_{2))}\right){\boldsymbol {B))-{\frac {\partial {\hat {W))}{\partial I_{2))}~{\boldsymbol {B))\cdot {\boldsymbol {B))\right]\\&=-p~{\boldsymbol {\mathit {1))}+2\left[\left({\frac {\partial W}{\partial {\bar {I))_{1))}+I_{1}~{\frac {\partial W}{\partial {\bar {I))_{2))}\right)~{\bar {\boldsymbol {B))}-{\frac {\partial W}{\partial {\bar {I))_{2))}~{\bar {\boldsymbol {B))}\cdot {\bar {\boldsymbol {B))}\right]\\&=-p~{\boldsymbol {\mathit {1))}+\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2))}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned))}$$

${\displaystyle p}$

$${\displaystyle \sigma _{11}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))}~;~~\sigma _{22}-\sigma _{33}=\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2))}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))))$$

${\displaystyle I_{1}=I_{2))$

$${\displaystyle {\boldsymbol {\sigma ))=2{\frac {\partial W}{\partial I_{1))}~{\boldsymbol {B))-p~{\boldsymbol {\mathit {1))}~.}$$

${\displaystyle \lambda _{1}=\lambda _{2))$

$${\displaystyle \sigma _{11}-\sigma _{33}=\sigma _{22}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1))}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3))))$$

Consistency with linear elasticity

Consistency with linear elasticity is often used to determine some of the parameters of hyperelastic material models. These consistency conditions can be found by comparing Hooke's law with linearized hyperelasticity at small strains.

Consistency conditions for isotropic hyperelastic models

For isotropic hyperelastic materials to be consistent with isotropic linear elasticity, the stress–strain relation should have the following form in the infinitesimal strain limit:

$${\displaystyle {\boldsymbol {\sigma ))=\lambda ~\mathrm {tr} ({\boldsymbol {\varepsilon )))~{\boldsymbol {\mathit {1))}+2\mu {\boldsymbol {\varepsilon ))}$$

${\displaystyle \lambda ,\mu }$

$${\displaystyle W={\tfrac {1}{2))\lambda ~[\mathrm {tr} ({\boldsymbol {\varepsilon )))]^{2}+\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon ))^{2}\right)))}$$

${\displaystyle \mathrm {tr} ({\boldsymbol {\varepsilon )))=0}$

$${\displaystyle W=\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon ))^{2}\right)))}$$

${\displaystyle W(\lambda _{1},\lambda _{2},\lambda _{3})}$

$${\displaystyle {\begin{aligned}&W(1,1,1)=0~;~~{\frac {\partial W}{\partial \lambda _{i))}(1,1,1)=0\\&{\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j))}(1,1,1)=\lambda +2\mu \delta _{ij}\end{aligned))}$$

If the material is incompressible, then the above conditions may be expressed in the following form.

$${\displaystyle {\begin{aligned}&W(1,1,1)=0\\&{\frac {\partial W}{\partial \lambda _{i))}(1,1,1)={\frac {\partial W}{\partial \lambda _{j))}(1,1,1)~;~~{\frac {\partial ^{2}W}{\partial \lambda _{i}^{2))}(1,1,1)={\frac {\partial ^{2}W}{\partial \lambda _{j}^{2))}(1,1,1)\\&{\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j))}(1,1,1)=\mathrm {independentof} ~i,j\neq i\\&{\frac {\partial ^{2}W}{\partial \lambda _{i}^{2))}(1,1,1)-{\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j))}(1,1,1)+{\frac {\partial W}{\partial \lambda _{i))}(1,1,1)=2\mu ~~(i\neq j)\end{aligned))}$$

Consistency conditions for incompressible I₁ based rubber materials

Many elastomers are modeled adequately by a strain energy density function that depends only on ${\displaystyle I_{1))$. For such materials we have ${\displaystyle W=W(I_{1})}$. The consistency conditions for incompressible materials for ${\displaystyle I_{1}=3,\lambda _{i}=\lambda _{j}=1}$ may then be expressed as