Hyperelastic materials are truly elastic in the sense that if a load is applied to such a material and then removed, the material returns to its original shape without any dissipation of energy in the process. In other word, a hyperelastic material stores energy during loading and releases exactly the same amount of energy during unloading. There is no path dependence.
If
ψ
{\displaystyle \psi \,}
Helmholtz free energy , then the stress-strain behavior for such a material is given by
σ
=
ρ
F
∙
∂
ψ
∂
E
∙
F
T
=
2
ρ
F
∙
∂
ψ
∂
C
∙
F
T
{\displaystyle {\boldsymbol {\sigma }}=\rho ~{\boldsymbol {F}}\bullet {\cfrac {\partial \psi }{\partial {\boldsymbol {E}}}}\bullet {\boldsymbol {F}}^{T}=2~\rho ~{\boldsymbol {F}}\bullet {\cfrac {\partial \psi }{\partial {\boldsymbol {C}}}}\bullet {\boldsymbol {F}}^{T}}
where
σ
{\displaystyle {\boldsymbol {\sigma }}}
Cauchy stress ,
ρ
{\displaystyle \rho }
F
{\displaystyle {\boldsymbol {F}}}
E
{\displaystyle {\boldsymbol {E}}}
C
{\displaystyle {\boldsymbol {C}}}
We can use the relationship between the Cauchy stress and the 2nd Piola-Kirchhoff stress to obtain an alternative relation between stress and strain.
S
=
2
ρ
0
∂
ψ
∂
C
{\displaystyle {\boldsymbol {S}}=2~\rho _{0}~{\cfrac {\partial \psi }{\partial {\boldsymbol {C}}}}}
where
S
{\displaystyle {\boldsymbol {S}}}
2nd Piola-Kirchhoff stress and
ρ
0
{\displaystyle \rho _{0}}
Isotropic hyperelasticity
edit
For isotropic materials, the free energy must be an isotropic function of
C
{\displaystyle {\boldsymbol {C}}}
C
{\displaystyle {\boldsymbol {C}}}
I
C
=
I
1
=
tr
(
C
)
=
C
i
i
=
λ
1
2
+
λ
2
2
+
λ
3
2
I
I
C
=
I
2
=
1
2
[
tr
(
C
2
)
−
(
tr
C
)
2
]
=
1
2
[
C
i
k
C
k
i
−
C
j
j
2
]
=
λ
1
2
λ
2
2
+
λ
2
2
λ
3
2
+
λ
3
2
λ
1
2
I
I
I
C
=
I
3
=
det
(
C
)
=
λ
1
2
λ
2
2
λ
3
2
{\displaystyle {\begin{aligned}I_{\boldsymbol {C}}=I_{1}&={\text{tr}}(\mathbf {C} )=C_{ii}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}\\II_{\boldsymbol {C}}=I_{2}&={\tfrac {1}{2}}\left[{\text{tr}}(\mathbf {C} ^{2})-({\text{tr}}~\mathbf {C} )^{2}\right]={\tfrac {1}{2}}\left[C_{ik}C_{ki}-C_{jj}^{2}\right]=\lambda _{1}^{2}\lambda _{2}^{2}+\lambda _{2}^{2}\lambda _{3}^{2}+\lambda _{3}^{2}\lambda _{1}^{2}\\III_{\boldsymbol {C}}=I_{3}&=\det(\mathbf {C} )=\lambda _{1}^{2}\lambda _{2}^{2}\lambda _{3}^{2}\end{aligned}}}
In other words,
ψ
(
C
)
≡
ψ
(
I
1
,
I
2
,
I
3
)
{\displaystyle \psi ({\boldsymbol {C}})\equiv \psi (I_{1},I_{2},I_{3})}
Therefore, from the chain rule,
∂
ψ
∂
C
=
∂
ψ
∂
I
1
∂
I
1
∂
C
+
∂
ψ
∂
I
2
∂
I
2
∂
C
+
∂
ψ
∂
I
3
∂
I
3
∂
C
=
a
0
1
+
a
1
C
+
a
2
C
−
1
{\displaystyle {\cfrac {\partial \psi }{\partial {\boldsymbol {C}}}}={\cfrac {\partial \psi }{\partial I_{1}}}~{\cfrac {\partial I_{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial \psi }{\partial I_{2}}}~{\cfrac {\partial I_{2}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial \psi }{\partial I_{3}}}~{\cfrac {\partial I_{3}}{\partial {\boldsymbol {C}}}}=a_{0}~{\boldsymbol {\mathit {1}}}+a_{1}~{\boldsymbol {C}}+a_{2}~{\boldsymbol {C}}^{-1}}
From the Cayley-Hamilton theorem we can show that
C
−
1
≡
f
(
C
2
,
C
,
1
)
{\displaystyle {\boldsymbol {C}}^{-1}\equiv f({\boldsymbol {C}}^{2},{\boldsymbol {C}},{\boldsymbol {\mathit {1}}})}
Hence we can also write
∂
ψ
∂
C
=
b
0
1
+
b
1
C
+
b
2
C
2
{\displaystyle {\cfrac {\partial \psi }{\partial {\boldsymbol {C}}}}=b_{0}~{\boldsymbol {\mathit {1}}}+b_{1}~{\boldsymbol {C}}+b_{2}~{\boldsymbol {C}}^{2}}
The stress-strain relation can then be written as
S
=
2
ρ
0
[
b
0
1
+
b
1
C
+
b
2
C
2
]
{\displaystyle {\boldsymbol {S}}=2~\rho _{0}~\left[b_{0}~{\boldsymbol {\mathit {1}}}+b_{1}~{\boldsymbol {C}}+b_{2}~{\boldsymbol {C}}^{2}\right]}
A similar relation can be obtained for the Cauchy stress which has the form
σ
=
2
ρ
[
a
2
1
+
a
0
B
+
a
1
B
2
]
{\displaystyle {\boldsymbol {\sigma }}=2~\rho ~\left[a_{2}~{\boldsymbol {\mathit {1}}}+a_{0}~{\boldsymbol {B}}+a_{1}~{\boldsymbol {B}}^{2}\right]}
where
B
{\displaystyle {\boldsymbol {B}}}
Cauchy stress in terms of invariants
edit
For w: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 w: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})}
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
3
(
I
¯
1
∂
W
¯
∂
I
¯
1
+
2
I
¯
2
∂
W
¯
∂
I
¯
2
)
1
−
1
J
4
/
3
∂
W
¯
∂
I
¯
2
B
⋅
B
]
+
∂
W
¯
∂
J
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 }}&={\cfrac {2}{\sqrt {I_{3}}}}\left[\left({\cfrac {\partial {\hat {W}}}{\partial I_{1}}}+I_{1}~{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}\right){\boldsymbol {B}}-{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2{\sqrt {I_{3}}}~{\cfrac {\partial {\hat {W}}}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {2}{J}}\left[{\cfrac {1}{J^{2/3}}}\left({\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {B}}-{\cfrac {1}{3}}\left({\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {\mathit {1}}}-\right.\\&\qquad \qquad \qquad \left.{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+{\cfrac {\partial {\bar {W}}}{\partial J}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {\lambda _{1}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {\lambda _{2}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {\lambda _{3}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}}
(See the page on the left Cauchy-Green deformation tensor for the definitions of these symbols).
Proof 2:
To express the Cauchy stress in terms of the invariants
I
¯
1
,
I
¯
2
,
J
{\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}
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}{\cfrac {\partial W}{\partial I_{1}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{1}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{1}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{1}}}\\&=I_{3}^{-1/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}=J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}\\{\cfrac {\partial W}{\partial I_{2}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{2}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{2}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{2}}}\\&=I_{3}^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}=J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\\{\cfrac {\partial W}{\partial I_{3}}}&={\cfrac {\partial W}{\partial {\bar {I}}_{1}}}~{\cfrac {\partial {\bar {I}}_{1}}{\partial I_{3}}}+{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\cfrac {\partial {\bar {I}}_{2}}{\partial I_{3}}}+{\cfrac {\partial W}{\partial J}}~{\cfrac {\partial J}{\partial I_{3}}}\\&=-{\cfrac {1}{3}}~I_{3}^{-4/3}~I_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}-{\cfrac {2}{3}}~I_{3}^{-5/3}~I_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}+{\cfrac {1}{2}}~I_{3}^{-1/2}~{\cfrac {\partial W}{\partial J}}\\&=-{\cfrac {1}{3}}~J^{-8/3}~J^{2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}-{\cfrac {2}{3}}~J^{-10/3}~J^{4/3}~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\\&=-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\end{aligned}}}
Recall that the Cauchy stress is given by
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 }}={\cfrac {2}{\sqrt {I_{3}}}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+I_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~{\sqrt {I_{3}}}~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}
In terms of the invariants
I
¯
1
,
I
¯
2
,
J
{\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}
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 }}={\cfrac {2}{J}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+J^{2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~J~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}
Plugging in the expressions for the derivatives of
W
{\displaystyle W}
I
¯
1
,
I
¯
2
,
J
{\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J}
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 }}&={\cfrac {2}{J}}~\left[\left(J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+J^{-2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\\&\qquad 2~J~\left[-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\right]~{\boldsymbol {\mathit {1}}}\end{aligned}}}
or,
σ
=
2
J
[
1
J
2
/
3
(
∂
W
∂
I
¯
1
+
I
¯
1
∂
W
∂
I
¯
2
)
B
−
1
3
(
I
¯
1
∂
W
∂
I
¯
1
+
2
I
¯
2
∂
W
∂
I
¯
2
)
1
−
1
J
4
/
3
∂
W
∂
I
¯
2
B
⋅
B
]
+
∂
W
∂
J
1
{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[{\cfrac {1}{J^{2/3}}}~\left({\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {1}{3}}\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {\mathit {1}}}-\right.\\&\qquad \left.{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+{\cfrac {\partial W}{\partial J}}~{\boldsymbol {\mathit {1}}}\end{aligned}}}
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}}
∂
λ
i
∂
C
=
1
2
λ
i
R
T
⋅
(
n
i
⊗
n
i
)
⋅
R
;
i
=
1
,
2
,
3
.
{\displaystyle {\cfrac {\partial \lambda _{i}}{\partial {\boldsymbol {C}}}}={\cfrac {1}{2\lambda _{i}}}~{\boldsymbol {R}}^{T}\cdot (\mathbf {n} _{i}\otimes \mathbf {n} _{i})\cdot {\boldsymbol {R}}~;~~i=1,2,3~.}
The chain rule gives
∂
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}{\cfrac {\partial W}{\partial {\boldsymbol {C}}}}&={\cfrac {\partial W}{\partial \lambda _{1}}}~{\cfrac {\partial \lambda _{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{2}}}~{\cfrac {\partial \lambda _{2}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{3}}}~{\cfrac {\partial \lambda _{3}}{\partial {\boldsymbol {C}}}}\\&={\boldsymbol {R}}^{T}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {R}}\end{aligned}}}
The Cauchy stress is given by
σ
=
2
J
F
⋅
∂
W
∂
C
⋅
F
T
=
2
J
(
V
⋅
R
)
⋅
∂
W
∂
C
⋅
(
R
T
⋅
V
)
{\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~{\boldsymbol {F}}\cdot {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}={\cfrac {2}{J}}~({\boldsymbol {V}}\cdot {\boldsymbol {R}})\cdot {\cfrac {\partial W}{\partial {\boldsymbol {C}}}}\cdot ({\boldsymbol {R}}^{T}\cdot {\boldsymbol {V}})}
Plugging in the expression for the derivative of
W
{\displaystyle W}
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 }}={\cfrac {2}{J}}~{\boldsymbol {V}}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {V}}}
Using the spectral decomposition of
V
{\displaystyle {\boldsymbol {V}}}
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 }}={\cfrac {1}{\lambda _{1}\lambda _{2}\lambda _{3}}}~\left[\lambda _{1}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]}
Saint-Venant–Kirchhoff material
edit
The simplest constitutive relationship that satisfies the requirements of hyperelasticity is the Saint-Venant–Kirchhoff material, which has a response function of the form
S
=
λ
tr
(
E
)
1
+
2
μ
E
,
{\displaystyle {\boldsymbol {S}}=\lambda ~{\text{tr}}({\boldsymbol {E}})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {E}},}
where
λ
{\displaystyle \lambda }
μ
{\displaystyle \mu }
![{\displaystyle {\begin{aligned}I_{\boldsymbol {C}}=I_{1}&={\text{tr}}(\mathbf {C} )=C_{ii}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}\\II_{\boldsymbol {C}}=I_{2}&={\tfrac {1}{2}}\left[{\text{tr}}(\mathbf {C} ^{2})-({\text{tr}}~\mathbf {C} )^{2}\right]={\tfrac {1}{2}}\left[C_{ik}C_{ki}-C_{jj}^{2}\right]=\lambda _{1}^{2}\lambda _{2}^{2}+\lambda _{2}^{2}\lambda _{3}^{2}+\lambda _{3}^{2}\lambda _{1}^{2}\\III_{\boldsymbol {C}}=I_{3}&=\det(\mathbf {C} )=\lambda _{1}^{2}\lambda _{2}^{2}\lambda _{3}^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3508ccb1fa33db50635d8f31163982b0ec0d0ba)
![{\displaystyle {\boldsymbol {S}}=2~\rho _{0}~\left[b_{0}~{\boldsymbol {\mathit {1}}}+b_{1}~{\boldsymbol {C}}+b_{2}~{\boldsymbol {C}}^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd7d81ac2f0d6f4eebd154d617b9df9e17da67c7)
![{\displaystyle {\boldsymbol {\sigma }}=2~\rho ~\left[a_{2}~{\boldsymbol {\mathit {1}}}+a_{0}~{\boldsymbol {B}}+a_{1}~{\boldsymbol {B}}^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35b19b12af043c759204fbe0fc18a57d6a0163c8)
![{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{\sqrt {I_{3}}}}\left[\left({\cfrac {\partial {\hat {W}}}{\partial I_{1}}}+I_{1}~{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}\right){\boldsymbol {B}}-{\cfrac {\partial {\hat {W}}}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2{\sqrt {I_{3}}}~{\cfrac {\partial {\hat {W}}}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {2}{J}}\left[{\cfrac {1}{J^{2/3}}}\left({\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {B}}-{\cfrac {1}{3}}\left({\bar {I}}_{1}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {\mathit {1}}}-\right.\\&\qquad \qquad \qquad \left.{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+{\cfrac {\partial {\bar {W}}}{\partial J}}~{\boldsymbol {\mathit {1}}}\\&={\cfrac {\lambda _{1}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {\lambda _{2}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {\lambda _{3}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\cfrac {\partial {\tilde {W}}}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c51b0c026b5a14d015b49862bfa1d157f8b0b76)
![{\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~\left[{\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}+{\cfrac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}-{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T})+{\cfrac {\partial W}{\partial I_{3}}}~I_{3}~{\boldsymbol {\mathit {1}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e14aca126ea518e5e3e0132e4fd899deadbc3ad)
![{\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{\sqrt {I_{3}}}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+I_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~{\sqrt {I_{3}}}~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb1ee0c6cf0b91150820746d66e9a5e38e7cfe65)
![{\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~\left[\left({\cfrac {\partial W}{\partial I_{1}}}+J^{2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~J~{\cfrac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4f30443899568890a1c8f543ec6a9524c996172)
![{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[\left(J^{-2/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+J^{-2/3}~{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-J^{-4/3}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\\&\qquad 2~J~\left[-{\cfrac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\cfrac {1}{2}}~J^{-1}~{\cfrac {\partial W}{\partial J}}\right]~{\boldsymbol {\mathit {1}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/942d309f640a99c84f20f0f4d241b013e1a933d6)
![{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\cfrac {2}{J}}~\left[{\cfrac {1}{J^{2/3}}}~\left({\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-{\cfrac {1}{3}}\left({\bar {I}}_{1}~{\cfrac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {\mathit {1}}}-\right.\\&\qquad \left.{\cfrac {1}{J^{4/3}}}~{\cfrac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+{\cfrac {\partial W}{\partial J}}~{\boldsymbol {\mathit {1}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcce371d66dc4cd58d17e2814c59455367e1eb2f)
![{\displaystyle {\begin{aligned}{\cfrac {\partial W}{\partial {\boldsymbol {C}}}}&={\cfrac {\partial W}{\partial \lambda _{1}}}~{\cfrac {\partial \lambda _{1}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{2}}}~{\cfrac {\partial \lambda _{2}}{\partial {\boldsymbol {C}}}}+{\cfrac {\partial W}{\partial \lambda _{3}}}~{\cfrac {\partial \lambda _{3}}{\partial {\boldsymbol {C}}}}\\&={\boldsymbol {R}}^{T}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {R}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7525fe32c1f006a7800f672d51035a43e2802c90)
![{\displaystyle {\boldsymbol {\sigma }}={\cfrac {2}{J}}~{\boldsymbol {V}}\cdot \left[{\cfrac {1}{2\lambda _{1}}}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{2\lambda _{2}}}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{2\lambda _{3}}}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {V}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/977230f6bcdfd24e95c31388332eadf83658fca3)
![{\displaystyle {\boldsymbol {\sigma }}={\cfrac {1}{\lambda _{1}\lambda _{2}\lambda _{3}}}~\left[\lambda _{1}~{\cfrac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\cfrac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\cfrac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98f1a3e0369450e0ee8ec83c6917e96874302e62)
