Constitutive model for ideally elastic material
Stress–strain curves for various hyperelastic material models.
A hyperelastic or Green elastic material.
For many materials,
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 .
.
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
S
=
C
:
E
S
=
λ
tr
(
E
)
I
+
2
μ
E
.
{\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}}}
where
:
{\displaystyle \mathbin {:} }
is tensor contraction,
S
{\displaystyle {\boldsymbol {S}}}
is the second Piola–Kirchhoff stress,
C
:
R
3
×
3
→
R
3
×
3
{\displaystyle {\boldsymbol {C}}:\mathbb {R} ^{3\times 3}\to \mathbb {R} ^{3\times 3}}
is a fourth order
stiffness tensor
and
E
{\displaystyle {\boldsymbol {E}}}
is the Lagrangian Green strain given by
E
=
1
2
[
(
∇
X
u
)
T
+
∇
X
u
+
(
∇
X
u
)
T
⋅
∇
X
u
]
{\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]\,\!}
λ
{\displaystyle \lambda }
and
μ
{\displaystyle \mu }
are the
Lamé constants
, and
I
{\displaystyle {\boldsymbol {\mathit {I}}}}
is the second order unit tensor.
The strain-energy density function for the Saint Venant–Kirchhoff model is
W
(
E
)
=
λ
2
[
tr
(
E
)
]
2
+
μ
tr
(
E
2
)
{\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
S
=
∂
W
∂
E
.
{\displaystyle {\boldsymbol {S}}={\frac {\partial W}{\partial {\boldsymbol {E}}}}~.}
Classification of hyperelastic material models
Hyperelastic material models can be classified as:
phenomenological descriptions of observed behavior
mechanistic models deriving from arguments about 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
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})\,.}
Stress–strain relations
Compressible hyperelastic materials
First Piola–Kirchhoff stress
If
W
(
F
)
{\displaystyle W({\boldsymbol {F}})}
is the strain energy density function, the
1st Piola–Kirchhoff stress tensor
can be calculated for a hyperelastic material as
P
=
∂
W
∂
F
or
P
i
K
=
∂
W
∂
F
i
K
.
{\displaystyle {\boldsymbol {P}}={\frac {\partial W}{\partial {\boldsymbol {F}}}}\qquad {\text{or}}\qquad P_{iK}={\frac {\partial W}{\partial F_{iK}}}.}
where
F
{\displaystyle {\boldsymbol {F}}}
is the
(
E
{\displaystyle {\boldsymbol {E}}}
)
P
=
F
⋅
∂
W
∂
E
or
P
i
K
=
F
i
L
∂
W
∂
E
L
K
.
{\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 (
C
{\displaystyle {\boldsymbol {C}}}
)
P
=
2
F
⋅
∂
W
∂
C
or
P
i
K
=
2
F
i
L
∂
W
∂
C
L
K
.
{\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
S
{\displaystyle {\boldsymbol {S}}}
is the
second Piola–Kirchhoff stress tensor
then
S
=
F
−
1
⋅
∂
W
∂
F
or
S
I
J
=
F
I
k
−
1
∂
W
∂
F
k
J
.
{\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
S
=
∂
W
∂
E
or
S
I
J
=
∂
W
∂
E
I
J
.
{\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
S
=
2
∂
W
∂
C
or
S
I
J
=
2
∂
W
∂
C
I
J
.
{\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.
Cauchy stress
Similarly, the
Cauchy stress
is given by
σ
=
1
J
∂
W
∂
F
⋅
F
T
;
J
:=
det
F
or
σ
i
j
=
1
J
∂
W
∂
F
i
K
F
j
K
.
{\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}~.}
In terms of the
Lagrangian Green strain
σ
=
1
J
F
⋅
∂
W
∂
E
⋅
F
T
or
σ
i
j
=
1
J
F
i
K
∂
W
∂
E
K
L
F
j
L
.
{\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}~.}
In terms of the
right Cauchy–Green deformation tensor
σ
=
2
J
F
⋅
∂
W
∂
C
⋅
F
T
or
σ
i
j
=
2
J
F
i
K
∂
W
∂
C
K
L
F
j
L
.
{\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}~.}
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:
[7]
σ
=
2
J
∂
W
∂
B
⋅
B
or
σ
i
j
=
2
J
B
i
k
∂
W
∂
B
k
j
.
{\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
J
:=
det
F
=
1
{\displaystyle J:=\det {\boldsymbol {F}}=1}
. The incompressibility constraint is therefore
J
−
1
=
0
{\displaystyle J-1=0}
. To ensure incompressibility of a hyperelastic material, the strain-energy function can be written in form:
W
=
W
(
F
)
−
p
(
J
−
1
)
{\displaystyle W=W({\boldsymbol {F}})-p~(J-1)}
where the hydrostatic pressure
p
{\displaystyle p}
functions as a
Lagrangian multiplier
to enforce the incompressibility constraint. The 1st Piola–Kirchhoff stress now becomes
P
=
−
p
J
F
−
T
+
∂
W
∂
F
=
−
p
F
−
T
+
F
⋅
∂
W
∂
E
=
−
p
F
−
T
+
2
F
⋅
∂
W
∂
C
.
{\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}}}}~.}
This stress tensor can subsequently be
which is given by
σ
=
P
⋅
F
T
=
−
p
1
+
∂
W
∂
F
⋅
F
T
=
−
p
1
+
F
⋅
∂
W
∂
E
⋅
F
T
=
−
p
1
+
2
F
⋅
∂
W
∂
C
⋅
F
T
.
{\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
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
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
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}}}}
Incompressible isotropic hyperelastic materials
For incompressible
is
W
(
F
)
=
W
^
(
I
1
,
I
2
)
{\displaystyle W({\boldsymbol {F}})={\hat {W}}(I_{1},I_{2})}
. The Cauchy stress is then given by
σ
=
−
p
1
+
2
[
(
∂
W
^
∂
I
1
+
I
1
∂
W
^
∂
I
2
)
B
−
∂
W
^
∂
I
2
B
⋅
B
]
=
−
p
1
+
2
[
(
∂
W
∂
I
¯
1
+
I
1
∂
W
∂
I
¯
2
)
B
¯
−
∂
W
∂
I
¯
2
B
¯
⋅
B
¯
]
=
−
p
1
+
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 {\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}}}
where
p
{\displaystyle p}
is an undetermined pressure. In terms of stress differences
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
I
1
=
I
2
{\displaystyle I_{1}=I_{2}}
, then
σ
=
2
∂
W
∂
I
1
B
−
p
1
.
{\displaystyle {\boldsymbol {\sigma }}=2{\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}-p~{\boldsymbol {\mathit {1}}}~.}
If
λ
1
=
λ
2
{\displaystyle \lambda _{1}=\lambda _{2}}
, then
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
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.
Consistency conditions for incompressible I 1 based rubber materials
Many elastomers are modeled adequately by a strain energy density function that depends only on
I
1
{\displaystyle I_{1}}
. For such materials we have
W
=
W
(
I
1
)
{\displaystyle W=W(I_{1})}
.
The consistency conditions for incompressible materials for
I
1
=
3
,
λ
i
=
λ
j
=
1
{\displaystyle I_{1}=3,\lambda _{i}=\lambda _{j}=1}
may then be expressed as
W
(
I
1
)
|
I
1
=
3
=
0
and
∂
W
∂
I
1
|
I
1
=
3
=
μ
2
.
{\displaystyle \left.W(I_{1})\right|_{I_{1}=3}=0\quad {\text{and}}\quad \left.{\frac {\partial W}{\partial I_{1}}}\right|_{I_{1}=3}={\frac {\mu }{2}}\,.}
The second consistency condition above can be derived by noting that
∂
W
∂
λ
i
=
∂
W
∂
I
1
∂
I
1
∂
λ
i
=
2
λ
i
∂
W
∂
I
1
and
∂
2
W
∂
λ
i
∂
λ
j
=
2
δ
i
j
∂
W
∂
I
1
+
4
λ
i
λ
j
∂
2
W
∂
I
1
2
.
{\displaystyle {\frac {\partial W}{\partial \lambda _{i}}}={\frac {\partial W}{\partial I_{1}}}{\frac {\partial I_{1}}{\partial \lambda _{i}}}=2\lambda _{i}{\frac {\partial W}{\partial I_{1}}}\quad {\text{and}}\quad {\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j}}}=2\delta _{ij}{\frac {\partial W}{\partial I_{1}}}+4\lambda _{i}\lambda _{j}{\frac {\partial ^{2}W}{\partial I_{1}^{2}}}\,.}
These relations can then be substituted into the consistency condition for isotropic incompressible hyperelastic materials.
References
^ , Dover.
.
.
.
.
.
^ Y. Basar, 2000, Nonlinear continuum mechanics of solids, Springer, p. 157.
^ Fox & Kapoor, Rates of change of eigenvalues and eigenvectors , AIAA Journal , 6 (12) 2426–2429 (1968)
^ Friswell MI. The derivatives of repeated eigenvalues and their associated eigenvectors. Journal of Vibration and Acoustics (ASME) 1996; 118:390–397.
See also