Phenomenological model of elastic materials
Yeoh model prediction versus experimental data for natural rubber. Model parameters and experimental data from PolymerFEM.com The Yeoh hyperelastic material model[ 1] phenomenological model for the deformation of nearly incompressible , nonlinear elastic materials such as rubber . The model is based on Ronald Rivlin's observation that the elastic properties of rubber may be described using a strain energy density function which is a power series in the strain invariants
I
1
,
I
2
,
I
3
{\displaystyle I_{1},I_{2},I_{3}}
Cauchy-Green deformation tensors .[ 2]
I
1
{\displaystyle I_{1}}
I
3
{\displaystyle I_{3}}
reduced polynomial model .
Yeoh model for incompressible rubbers [ edit ] Strain energy density function [ edit ] The original model proposed by Yeoh had a cubic form with only
I
1
{\displaystyle I_{1}}
W
=
∑
i
=
1
3
C
i
(
I
1
−
3
)
i
{\displaystyle W=\sum _{i=1}^{3}C_{i}~(I_{1}-3)^{i}}
where
C
i
{\displaystyle C_{i}}
2
C
1
{\displaystyle 2C_{1}}
shear modulus .
Today a slightly more generalized version of the Yeoh model is used.[ 3]
n
{\displaystyle n}
W
=
∑
i
=
1
n
C
i
(
I
1
−
3
)
i
.
{\displaystyle W=\sum _{i=1}^{n}C_{i}~(I_{1}-3)^{i}~.}
When
n
=
1
{\displaystyle n=1}
neo-Hookean model for incompressible materials.
For consistency with linear elasticity the Yeoh model has to satisfy the condition
2
∂
W
∂
I
1
(
3
)
=
μ
(
i
≠
j
)
{\displaystyle 2{\cfrac {\partial W}{\partial I_{1}}}(3)=\mu ~~(i\neq j)}
where
μ
{\displaystyle \mu }
shear modulus of the material.
Now, at
I
1
=
3
(
λ
i
=
λ
j
=
1
)
{\displaystyle I_{1}=3(\lambda _{i}=\lambda _{j}=1)}
∂
W
∂
I
1
=
C
1
{\displaystyle {\cfrac {\partial W}{\partial I_{1}}}=C_{1}}
Therefore, the consistency condition for the Yeoh model is
2
C
1
=
μ
{\displaystyle 2C_{1}=\mu \,}
The Cauchy stress for the incompressible Yeoh model is given by
σ
=
−
p
1
+
2
∂
W
∂
I
1
B
;
∂
W
∂
I
1
=
∑
i
=
1
n
i
C
i
(
I
1
−
3
)
i
−
1
.
{\displaystyle {\boldsymbol {\sigma }}=-p~{\boldsymbol {\mathit {1}}}+2~{\cfrac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}~;~~{\cfrac {\partial W}{\partial I_{1}}}=\sum _{i=1}^{n}i~C_{i}~(I_{1}-3)^{i-1}~.}
For uniaxial extension in the
n
1
{\displaystyle \mathbf {n} _{1}}
principal stretches are
λ
1
=
λ
,
λ
2
=
λ
3
{\displaystyle \lambda _{1}=\lambda ,~\lambda _{2}=\lambda _{3}}
λ
1
λ
2
λ
3
=
1
{\displaystyle \lambda _{1}~\lambda _{2}~\lambda _{3}=1}
λ
2
2
=
λ
3
2
=
1
/
λ
{\displaystyle \lambda _{2}^{2}=\lambda _{3}^{2}=1/\lambda }
I
1
=
λ
1
2
+
λ
2
2
+
λ
3
2
=
λ
2
+
2
λ
.
{\displaystyle I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=\lambda ^{2}+{\cfrac {2}{\lambda }}~.}
The left Cauchy-Green deformation tensor can then be expressed as
B
=
λ
2
n
1
⊗
n
1
+
1
λ
(
n
2
⊗
n
2
+
n
3
⊗
n
3
)
.
{\displaystyle {\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{\lambda }}~(\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\mathbf {n} _{3}\otimes \mathbf {n} _{3})~.}
If the directions of the principal stretches are oriented with the coordinate basis vectors, we have
σ
11
=
−
p
+
2
λ
2
∂
W
∂
I
1
;
σ
22
=
−
p
+
2
λ
∂
W
∂
I
1
=
σ
33
.
{\displaystyle \sigma _{11}=-p+2~\lambda ^{2}~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{22}=-p+{\cfrac {2}{\lambda }}~{\cfrac {\partial W}{\partial I_{1}}}=\sigma _{33}~.}
Since
σ
22
=
σ
33
=
0
{\displaystyle \sigma _{22}=\sigma _{33}=0}
p
=
2
λ
∂
W
∂
I
1
.
{\displaystyle p={\cfrac {2}{\lambda }}~{\cfrac {\partial W}{\partial I_{1}}}~.}
Therefore,
σ
11
=
2
(
λ
2
−
1
λ
)
∂
W
∂
I
1
.
{\displaystyle \sigma _{11}=2~\left(\lambda ^{2}-{\cfrac {1}{\lambda }}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}
The engineering strain is
λ
−
1
{\displaystyle \lambda -1\,}
engineering stress is
T
11
=
σ
11
/
λ
=
2
(
λ
−
1
λ
2
)
∂
W
∂
I
1
.
{\displaystyle T_{11}=\sigma _{11}/\lambda =2~\left(\lambda -{\cfrac {1}{\lambda ^{2}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}
Equibiaxial extension [ edit ] For equibiaxial extension in the
n
1
{\displaystyle \mathbf {n} _{1}}
n
2
{\displaystyle \mathbf {n} _{2}}
principal stretches are
λ
1
=
λ
2
=
λ
{\displaystyle \lambda _{1}=\lambda _{2}=\lambda \,}
λ
1
λ
2
λ
3
=
1
{\displaystyle \lambda _{1}~\lambda _{2}~\lambda _{3}=1}
λ
3
=
1
/
λ
2
{\displaystyle \lambda _{3}=1/\lambda ^{2}\,}
I
1
=
λ
1
2
+
λ
2
2
+
λ
3
2
=
2
λ
2
+
1
λ
4
.
{\displaystyle I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=2~\lambda ^{2}+{\cfrac {1}{\lambda ^{4}}}~.}
The left Cauchy-Green deformation tensor can then be expressed as
B
=
λ
2
n
1
⊗
n
1
+
λ
2
n
2
⊗
n
2
+
1
λ
4
n
3
⊗
n
3
.
{\displaystyle {\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda ^{2}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\cfrac {1}{\lambda ^{4}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}~.}
If the directions of the principal stretches are oriented with the coordinate basis vectors, we have
σ
11
=
−
p
+
2
λ
2
∂
W
∂
I
1
=
σ
22
;
σ
33
=
−
p
+
2
λ
4
∂
W
∂
I
1
.
{\displaystyle \sigma _{11}=-p+2~\lambda ^{2}~{\cfrac {\partial W}{\partial I_{1}}}=\sigma _{22}~;~~\sigma _{33}=-p+{\cfrac {2}{\lambda ^{4}}}~{\cfrac {\partial W}{\partial I_{1}}}~.}
Since
σ
33
=
0
{\displaystyle \sigma _{33}=0}
p
=
2
λ
4
∂
W
∂
I
1
.
{\displaystyle p={\cfrac {2}{\lambda ^{4}}}~{\cfrac {\partial W}{\partial I_{1}}}~.}
Therefore,
σ
11
=
2
(
λ
2
−
1
λ
4
)
∂
W
∂
I
1
=
σ
22
.
{\displaystyle \sigma _{11}=2~\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{4}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}=\sigma _{22}~.}
The engineering strain is
λ
−
1
{\displaystyle \lambda -1\,}
engineering stress is
T
11
=
σ
11
λ
=
2
(
λ
−
1
λ
5
)
∂
W
∂
I
1
=
T
22
.
{\displaystyle T_{11}={\cfrac {\sigma _{11}}{\lambda }}=2~\left(\lambda -{\cfrac {1}{\lambda ^{5}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}=T_{22}~.}
Planar extension tests are carried out on thin specimens which are constrained from deforming in one direction. For planar extension in the
n
1
{\displaystyle \mathbf {n} _{1}}
n
3
{\displaystyle \mathbf {n} _{3}}
principal stretches are
λ
1
=
λ
,
λ
3
=
1
{\displaystyle \lambda _{1}=\lambda ,~\lambda _{3}=1}
λ
1
λ
2
λ
3
=
1
{\displaystyle \lambda _{1}~\lambda _{2}~\lambda _{3}=1}
λ
2
=
1
/
λ
{\displaystyle \lambda _{2}=1/\lambda \,}
I
1
=
λ
1
2
+
λ
2
2
+
λ
3
2
=
λ
2
+
1
λ
2
+
1
.
{\displaystyle I_{1}=\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=\lambda ^{2}+{\cfrac {1}{\lambda ^{2}}}+1~.}
The left Cauchy-Green deformation tensor can then be expressed as
B
=
λ
2
n
1
⊗
n
1
+
1
λ
2
n
2
⊗
n
2
+
n
3
⊗
n
3
.
{\displaystyle {\boldsymbol {B}}=\lambda ^{2}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\cfrac {1}{\lambda ^{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\mathbf {n} _{3}\otimes \mathbf {n} _{3}~.}
If the directions of the principal stretches are oriented with the coordinate basis vectors, we have
σ
11
=
−
p
+
2
λ
2
∂
W
∂
I
1
;
σ
22
=
−
p
+
2
λ
2
∂
W
∂
I
1
;
σ
33
=
−
p
+
2
∂
W
∂
I
1
.
{\displaystyle \sigma _{11}=-p+2~\lambda ^{2}~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{22}=-p+{\cfrac {2}{\lambda ^{2}}}~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{33}=-p+2~{\cfrac {\partial W}{\partial I_{1}}}~.}
Since
σ
22
=
0
{\displaystyle \sigma _{22}=0}
p
=
2
λ
2
∂
W
∂
I
1
.
{\displaystyle p={\cfrac {2}{\lambda ^{2}}}~{\cfrac {\partial W}{\partial I_{1}}}~.}
Therefore,
σ
11
=
2
(
λ
2
−
1
λ
2
)
∂
W
∂
I
1
;
σ
22
=
0
;
σ
33
=
2
(
1
−
1
λ
2
)
∂
W
∂
I
1
.
{\displaystyle \sigma _{11}=2~\left(\lambda ^{2}-{\cfrac {1}{\lambda ^{2}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}~;~~\sigma _{22}=0~;~~\sigma _{33}=2~\left(1-{\cfrac {1}{\lambda ^{2}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}
The engineering strain is
λ
−
1
{\displaystyle \lambda -1\,}
engineering stress is
T
11
=
σ
11
λ
=
2
(
λ
−
1
λ
3
)
∂
W
∂
I
1
.
{\displaystyle T_{11}={\cfrac {\sigma _{11}}{\lambda }}=2~\left(\lambda -{\cfrac {1}{\lambda ^{3}}}\right)~{\cfrac {\partial W}{\partial I_{1}}}~.}
Yeoh model for compressible rubbers [ edit ] A version of the Yeoh model that includes
I
3
=
J
2
{\displaystyle I_{3}=J^{2}}
W
=
∑
i
=
1
n
C
i
0
(
I
¯
1
−
3
)
i
+
∑
k
=
1
n
C
k
1
(
J
−
1
)
2
k
{\displaystyle W=\sum _{i=1}^{n}C_{i0}~({\bar {I}}_{1}-3)^{i}+\sum _{k=1}^{n}C_{k1}~(J-1)^{2k}}
where
I
¯
1
=
J
−
2
/
3
I
1
{\displaystyle {\bar {I}}_{1}=J^{-2/3}~I_{1}}
C
i
0
,
C
k
1
{\displaystyle C_{i0},C_{k1}}
C
10
{\displaystyle C_{10}}
C
11
{\displaystyle C_{11}}
When
n
=
1
{\displaystyle n=1}
neo-Hookean model for incompressible materials.
The model is named after Oon Hock Yeoh. Yeoh completed his doctoral studies under Graham Lake at the University of London .[ 4] Freudenberg-NOK , MRPRA (England), Rubber Research Institute of Malaysia (Malaysia), University of Akron , GenCorp Research, and Lord Corporation .[ 5] Melvin Mooney Distinguished Technology Award from the ACS Rubber Division .[ 6]
^ Yeoh, O. H. (November 1993). "Some forms of the strain energy function for rubber". Rubber Chemistry and Technology 66 (5): 754–771. doi :10.5254/1.3538343 . ^ Rivlin, R. S., 1948, "Some applications of elasticity theory to rubber engineering", in Collected Papers of R. S. Rivlin vol. 1 and 2 , Springer, 1997.
^ Selvadurai, A. P. S., 2006, "Deflections of a rubber membrane", Journal of the Mechanics and Physics of Solids , vol. 54, no. 6, pp. 1093-1119.
^ "Remembering Dr. Graham Johnson Lake (1935–2023)". Rubber Chemistry and Technology . 96 (4): G2–G3. 2023. doi :10.5254/rct-23.498080 . ^ "Biographical Sketch" . ACS Rubber Division. Retrieved 20 February 2024 .^ "Rubber Division names 3 for awards" . Rubber and Plastics News . Crain. 27 October 2003. Retrieved 16 August 2022 . 
