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 .
Stress–strain curves for various hyperelastic material models.
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. In addition to being used to model physical materials, hyperelastic materials are also used as fictitious media, e.g. in the third medium contact method .
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 .
Classification of hyperelastic material models
Hyperelastic material models can be classified as:
phenomenological descriptions of observed behavior
mechanistic models deriving from arguments about the underlying structure of the material
hybrids of phenomenological and mechanistic models
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
(
λ
1
,
λ
2
,
λ
3
)
{\displaystyle (\lambda _{1},\lambda _{2},\lambda _{3})}
:
W
=
f
(
λ
1
)
+
f
(
λ
2
)
+
f
(
λ
3
)
.
{\displaystyle W=f(\lambda _{1})+f(\lambda _{2})+f(\lambda _{3})\,.}
Compressible hyperelastic materials
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
W
(
F
)
=
W
^
(
I
1
,
I
2
,
I
3
)
=
W
¯
(
I
¯
1
,
I
¯
2
,
J
)
=
W
~
(
λ
1
,
λ
2
,
λ
3
)
,
{\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
σ
=
2
I
3
[
(
∂
W
^
∂
I
1
+
I
1
∂
W
^
∂
I
2
)
B
−
∂
W
^
∂
I
2
B
⋅
B
]
+
2
I
3
∂
W
^
∂
I
3
1
=
2
J
[
1
J
2
/
3
(
∂
W
¯
∂
I
¯
1
+
I
¯
1
∂
W
¯
∂
I
¯
2
)
B
−
1
J
4
/
3
∂
W
¯
∂
I
¯
2
B
⋅
B
]
+
[
∂
W
¯
∂
J
−
2
3
J
(
I
¯
1
∂
W
¯
∂
I
¯
1
+
2
I
¯
2
∂
W
¯
∂
I
¯
2
)
]
1
=
2
J
[
(
∂
W
¯
∂
I
¯
1
+
I
¯
1
∂
W
¯
∂
I
¯
2
)
B
¯
−
∂
W
¯
∂
I
¯
2
B
¯
⋅
B
¯
]
+
[
∂
W
¯
∂
J
−
2
3
J
(
I
¯
1
∂
W
¯
∂
I
¯
1
+
2
I
¯
2
∂
W
¯
∂
I
¯
2
)
]
1
=
λ
1
λ
1
λ
2
λ
3
∂
W
~
∂
λ
1
n
1
⊗
n
1
+
λ
2
λ
1
λ
2
λ
3
∂
W
~
∂
λ
2
n
2
⊗
n
2
+
λ
3
λ
1
λ
2
λ
3
∂
W
~
∂
λ
3
n
3
⊗
n
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}}}
(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
S
=
2
∂
W
∂
C
{\displaystyle {\boldsymbol {S}}=2~{\frac {\partial W}{\partial {\boldsymbol {C}}}}}
where
C
=
F
T
⋅
F
{\displaystyle {\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}}
is the right Cauchy–Green deformation tensor and
F
{\displaystyle {\boldsymbol {F}}}
is the deformation gradient . The Cauchy stress is given by
σ
=
1
J
F
⋅
S
⋅
F
T
=
2
J
F
⋅
∂
W
∂
C
⋅
F
T
{\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
J
=
det
F
{\displaystyle J=\det {\boldsymbol {F}}}
. Let
I
1
,
I
2
,
I
3
{\displaystyle I_{1},I_{2},I_{3}}
be the three principal invariants of
C
{\displaystyle {\boldsymbol {C}}}
. Then
∂
W
∂
C
=
∂
W
∂
I
1
∂
I
1
∂
C
+
∂
W
∂
I
2
∂
I
2
∂
C
+
∂
W
∂
I
3
∂
I
3
∂
C
.
{\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
C
{\displaystyle {\boldsymbol {C}}}
are
∂
I
1
∂
C
=
1
;
∂
I
2
∂
C
=
I
1
1
−
C
;
∂
I
3
∂
C
=
det
(
C
)
C
−
1
{\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
∂
W
∂
C
=
∂
W
∂
I
1
1
+
∂
W
∂
I
2
(
I
1
1
−
F
T
⋅
F
)
+
∂
W
∂
I
3
I
3
F
−
1
⋅
F
−
T
.
{\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
σ
=
2
J
[
∂
W
∂
I
1
F
⋅
F
T
+
∂
W
∂
I
2
(
I
1
F
⋅
F
T
−
F
⋅
F
T
⋅
F
⋅
F
T
)
+
∂
W
∂
I
3
I
3
1
]
{\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
B
=
F
⋅
F
T
{\displaystyle {\boldsymbol {B}}={\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}}
and noting that
I
3
=
J
2
{\displaystyle I_{3}=J^{2}}
, we can write
σ
=
2
I
3
[
(
∂
W
∂
I
1
+
I
1
∂
W
∂
I
2
)
B
−
∂
W
∂
I
2
B
⋅
B
]
+
2
I
3
∂
W
∂
I
3
1
.
{\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
I
3
=
1
{\displaystyle I_{3}=1}
and hence
W
=
W
(
I
1
,
I
2
)
{\displaystyle W=W(I_{1},I_{2})}
.Then
∂
W
∂
C
=
∂
W
∂
I
1
∂
I
1
∂
C
+
∂
W
∂
I
2
∂
I
2
∂
C
=
∂
W
∂
I
1
1
+
∂
W
∂
I
2
(
I
1
1
−
F
T
⋅
F
)
{\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
σ
=
2
[
(
∂
W
∂
I
1
+
I
1
∂
W
∂
I
2
)
B
−
∂
W
∂
I
2
B
⋅
B
]
−
p
1
.
{\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
p
{\displaystyle p}
is an undetermined pressure which acts as a Lagrange multiplier to enforce the incompressibility constraint.
If, in addition,
I
1
=
I
2
{\displaystyle I_{1}=I_{2}}
, we have
W
=
W
(
I
1
)
{\displaystyle W=W(I_{1})}
and hence
∂
W
∂
C
=
∂
W
∂
I
1
∂
I
1
∂
C
=
∂
W
∂
I
1
1
{\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
σ
=
2
∂
W
∂
I
1
B
−
p
1
.
{\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
F
¯
:=
J
−
1
/
3
F
{\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
B
¯
:=
F
¯
⋅
F
¯
T
=
J
−
2
/
3
B
{\displaystyle {\bar {\boldsymbol {B}}}:={\bar {\boldsymbol {F}}}\cdot {\bar {\boldsymbol {F}}}^{T}=J^{-2/3}{\boldsymbol {B}}}
.
The invariants of
B
¯
{\displaystyle {\bar {\boldsymbol {B}}}}
are
I
¯
1
=
tr
(
B
¯
)
=
J
−
2
/
3
tr
(
B
)
=
J
−
2
/
3
I
1
I
¯
2
=
1
2
(
tr
(
B
¯
)
2
−
tr
(
B
¯
2
)
)
=
1
2
(
(
J
−
2
/
3
tr
(
B
)
)
2
−
tr
(
J
−
4
/
3
B
2
)
)
=
J
−
4
/
3
I
2
I
¯
3
=
det
(
B
¯
)
=
J
−
6
/
3
det
(
B
)
=
J
−
2
I
3
=
J
−
2
J
2
=
1
{\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
J
{\displaystyle J}
into the fray to describe the volumetric behaviour.
To express the Cauchy stress in terms of the invariants
I
¯
1
,
I
¯
2
,
J
{\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}
recall that
I
¯
1
=
J
−
2
/
3
I
1
=
I
3
−
1
/
3
I
1
;
I
¯
2
=
J
−
4
/
3
I
2
=
I
3
−
2
/
3
I
2
;
J
=
I
3
1
/
2
.
{\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
∂
W
∂
I
1
=
∂
W
∂
I
¯
1
∂
I
¯
1
∂
I
1
+
∂
W
∂
I
¯
2
∂
I
¯
2
∂
I
1
+
∂
W
∂
J
∂
J
∂
I
1
=
I
3
−
1
/
3
∂
W
∂
I
¯
1
=
J
−
2
/
3
∂
W
∂
I
¯
1
∂
W
∂
I
2
=
∂
W
∂
I
¯
1
∂
I
¯
1
∂
I
2
+
∂
W
∂
I
¯
2
∂
I
¯
2
∂
I
2
+
∂
W
∂
J
∂
J
∂
I
2
=
I
3
−
2
/
3
∂
W
∂
I
¯
2
=
J
−
4
/
3
∂
W
∂
I
¯
2
∂
W
∂
I
3
=
∂
W
∂
I
¯
1
∂
I
¯
1
∂
I
3
+
∂
W
∂
I
¯
2
∂
I
¯
2
∂
I
3
+
∂
W
∂
J
∂
J
∂
I
3
=
−
1
3
I
3
−
4
/
3
I
1
∂
W
∂
I
¯
1
−
2
3
I
3
−
5
/
3
I
2
∂
W
∂
I
¯
2
+
1
2
I
3
−
1
/
2
∂
W
∂
J
=
−
1
3
J
−
8
/
3
J
2
/
3
I
¯
1
∂
W
∂
I
¯
1
−
2
3
J
−
10
/
3
J
4
/
3
I
¯
2
∂
W
∂
I
¯
2
+
1
2
J
−
1
∂
W
∂
J
=
−
1
3
J
−
2
(
I
¯
1
∂
W
∂
I
¯
1
+
2
I
¯
2
∂
W
∂
I
¯
2
)
+
1
2
J
−
1
∂
W
∂
J
{\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
σ
=
2
I
3
[
(
∂
W
∂
I
1
+
I
1
∂
W
∂
I
2
)
B
−
∂
W
∂
I
2
B
⋅
B
]
+
2
I
3
∂
W
∂
I
3
1
.
{\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
I
¯
1
,
I
¯
2
,
J
{\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}
we have
σ
=
2
J
[
(
∂
W
∂
I
1
+
J
2
/
3
I
¯
1
∂
W
∂
I
2
)
B
−
∂
W
∂
I
2
B
⋅
B
]
+
2
J
∂
W
∂
I
3
1
.
{\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
W
{\displaystyle W}
in terms of
I
¯
1
,
I
¯
2
,
J
{\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}
, we have
σ
=
2
J
[
(
J
−
2
/
3
∂
W
∂
I
¯
1
+
J
−
2
/
3
I
¯
1
∂
W
∂
I
¯
2
)
B
−
J
−
4
/
3
∂
W
∂
I
¯
2
B
⋅
B
]
+
2
J
[
−
1
3
J
−
2
(
I
¯
1
∂
W
∂
I
¯
1
+
2
I
¯
2
∂
W
∂
I
¯
2
)
+
1
2
J
−
1
∂
W
∂
J
]
1
{\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,
σ
=
2
J
[
1
J
2
/
3
(
∂
W
∂
I
¯
1
+
I
¯
1
∂
W
∂
I
¯
2
)
B
−
1
J
4
/
3
∂
W
∂
I
¯
2
B
⋅
B
]
+
[
∂
W
∂
J
−
2
3
J
(
I
¯
1
∂
W
∂
I
¯
1
+
2
I
¯
2
∂
W
∂
I
¯
2
)
]
1
{\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
B
{\displaystyle {\boldsymbol {B}}}
, we can write
σ
=
2
J
[
(
∂
W
∂
I
¯
1
+
I
¯
1
∂
W
∂
I
¯
2
)
B
¯
−
∂
W
∂
I
¯
2
B
¯
⋅
B
¯
]
+
[
∂
W
∂
J
−
2
3
J
(
I
¯
1
∂
W
∂
I
¯
1
+
2
I
¯
2
∂
W
∂
I
¯
2
)
]
1
{\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
J
=
1
{\displaystyle J=1}
and hence
W
=
W
(
I
¯
1
,
I
¯
2
)
{\displaystyle W=W({\bar {I}}_{1},{\bar {I}}_{2})}
.Then
the Cauchy stress is given by
σ
=
2
[
(
∂
W
∂
I
¯
1
+
I
1
∂
W
∂
I
¯
2
)
B
¯
−
∂
W
∂
I
¯
2
B
¯
⋅
B
¯
]
−
p
1
.
{\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
p
{\displaystyle p}
is an undetermined pressure-like Lagrange multiplier term. In addition, if
I
¯
1
=
I
¯
2
{\displaystyle {\bar {I}}_{1}={\bar {I}}_{2}}
, we have
W
=
W
(
I
¯
1
)
{\displaystyle W=W({\bar {I}}_{1})}
and hence
the Cauchy stress can be expressed as
σ
=
2
∂
W
∂
I
¯
1
B
¯
−
p
1
.
{\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
λ
1
,
λ
2
,
λ
3
{\displaystyle \lambda _{1},\lambda _{2},\lambda _{3}}
recall that
∂
λ
i
∂
C
=
1
2
λ
i
R
T
⋅
(
n
i
⊗
n
i
)
⋅
R
;
i
=
1
,
2
,
3
.
{\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
∂
W
∂
C
=
∂
W
∂
λ
1
∂
λ
1
∂
C
+
∂
W
∂
λ
2
∂
λ
2
∂
C
+
∂
W
∂
λ
3
∂
λ
3
∂
C
=
R
T
⋅
[
1
2
λ
1
∂
W
∂
λ
1
n
1
⊗
n
1
+
1
2
λ
2
∂
W
∂
λ
2
n
2
⊗
n
2
+
1
2
λ
3
∂
W
∂
λ
3
n
3
⊗
n
3
]
⋅
R
{\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
σ
=
2
J
F
⋅
∂
W
∂
C
⋅
F
T
=
2
J
(
V
⋅
R
)
⋅
∂
W
∂
C
⋅
(
R
T
⋅
V
)
{\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
W
{\displaystyle W}
leads to
σ
=
2
J
V
⋅
[
1
2
λ
1
∂
W
∂
λ
1
n
1
⊗
n
1
+
1
2
λ
2
∂
W
∂
λ
2
n
2
⊗
n
2
+
1
2
λ
3
∂
W
∂
λ
3
n
3
⊗
n
3
]
⋅
V
{\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
V
{\displaystyle {\boldsymbol {V}}}
we have
V
⋅
(
n
i
⊗
n
i
)
⋅
V
=
λ
i
2
n
i
⊗
n
i
;
i
=
1
,
2
,
3.
{\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
J
=
det
(
F
)
=
det
(
V
)
det
(
R
)
=
det
(
V
)
=
λ
1
λ
2
λ
3
.
{\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
σ
=
1
λ
1
λ
2
λ
3
[
λ
1
∂
W
∂
λ
1
n
1
⊗
n
1
+
λ
2
∂
W
∂
λ
2
n
2
⊗
n
2
+
λ
3
∂
W
∂
λ
3
n
3
⊗
n
3
]
{\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
λ
1
λ
2
λ
3
=
1
{\displaystyle \lambda _{1}\lambda _{2}\lambda _{3}=1}
and hence
W
=
W
(
λ
1
,
λ
2
)
{\displaystyle W=W(\lambda _{1},\lambda _{2})}
. Following Ogden[1] p. 485, we may write
σ
=
λ
1
∂
W
∂
λ
1
n
1
⊗
n
1
+
λ
2
∂
W
∂
λ
2
n
2
⊗
n
2
+
λ
3
∂
W
∂
λ
3
n
3
⊗
n
3
−
p
1
{\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,
σ
11
−
σ
33
=
λ
1
∂
W
∂
λ
1
−
λ
3
∂
W
∂
λ
3
;
σ
22
−
σ
33
=
λ
2
∂
W
∂
λ
2
−
λ
3
∂
W
∂
λ
3
{\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
λ
1
=
λ
2
{\displaystyle \lambda _{1}=\lambda _{2}}
then a possible solution to the problem
requires
σ
11
=
σ
22
{\displaystyle \sigma _{11}=\sigma _{22}}
and we can write the stress differences as
σ
11
−
σ
33
=
σ
22
−
σ
33
=
λ
1
∂
W
∂
λ
1
−
λ
3
∂
W
∂
λ
3
{\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 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:
σ
=
λ
t
r
(
ε
)
1
+
2
μ
ε
{\displaystyle {\boldsymbol {\sigma }}=\lambda ~\mathrm {tr} ({\boldsymbol {\varepsilon }})~{\boldsymbol {\mathit {1}}}+2\mu {\boldsymbol {\varepsilon }}}
where
λ
,
μ
{\displaystyle \lambda ,\mu }
are the Lamé constants . The strain energy density function that corresponds to the above relation is[1]
W
=
1
2
λ
[
t
r
(
ε
)
]
2
+
μ
t
r
(
ε
2
)
{\displaystyle W={\tfrac {1}{2}}\lambda ~[\mathrm {tr} ({\boldsymbol {\varepsilon }})]^{2}+\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon }}^{2}\right)}}}
For an incompressible material
t
r
(
ε
)
=
0
{\displaystyle \mathrm {tr} ({\boldsymbol {\varepsilon }})=0}
and we have
W
=
μ
t
r
(
ε
2
)
{\displaystyle W=\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon }}^{2}\right)}}}
For any strain energy density function
W
(
λ
1
,
λ
2
,
λ
3
)
{\displaystyle W(\lambda _{1},\lambda _{2},\lambda _{3})}
to reduce to the above forms for small strains the following conditions have to be met[1]
W
(
1
,
1
,
1
)
=
0
;
∂
W
∂
λ
i
(
1
,
1
,
1
)
=
0
∂
2
W
∂
λ
i
∂
λ
j
(
1
,
1
,
1
)
=
λ
+
2
μ
δ
i
j
{\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.
W
(
1
,
1
,
1
)
=
0
∂
W
∂
λ
i
(
1
,
1
,
1
)
=
∂
W
∂
λ
j
(
1
,
1
,
1
)
;
∂
2
W
∂
λ
i
2
(
1
,
1
,
1
)
=
∂
2
W
∂
λ
j
2
(
1
,
1
,
1
)
∂
2
W
∂
λ
i
∂
λ
j
(
1
,
1
,
1
)
=
i
n
d
e
p
e
n
d
e
n
t
o
f
i
,
j
≠
i
∂
2
W
∂
λ
i
2
(
1
,
1
,
1
)
−
∂
2
W
∂
λ
i
∂
λ
j
(
1
,
1
,
1
)
+
∂
W
∂
λ
i
(
1
,
1
,
1
)
=
2
μ
(
i
≠
j
)
{\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}}}
These conditions can be used to find relations between the parameters of a given hyperelastic model and shear and bulk moduli.