Supplemental Materials
Large Amplitude Oscillatory Shear (LAOS) measurements to obtain constitutive equation model parameters: Giesekus model
of banding and non-banding wormlike micelles
A. Kate Gurnon and Norman J. Wagner1
The following is the full derivation for the analytic solutions to the Giesekus constitutive model for the first, second and third
harmonics.
Identities used in this derivation are given here:
1
( cos(a + b) + cos(a − b) )
2
1
cos(a)sin(b) = ( sin(a + b) − sin(a − b) )
2
1
sin(a) cos(b) = ( sin(a + b) + sin(a − b) )
2
1
sin(a)sin(b) = ( cos(a − b) − cos(a + b) )
2
cos(a) cos(b) =
cos(−a) = cos(a)
sin(−a) = − sin(a)
and:
1
During oscillatory shear deformation, strain is defined as:
γ = γ 0 sin(ωt )
(A.1)
And strain rate:
γ = γ 0ω cos(ωt )
(A.2)
The stress response during dynamic oscillatory deformation is expressed as an expansion in both strain and frequency. We use the
definition for the stress during oscillatory shear as proposed by Hyun et al. (submitted) eq. A.3. The shear stress response is expanded
using a Taylor expansion in both strain and frequency.
j
∑ ∑γ
τ12 (t ) =
j ,odd n ,odd
j
0
⎡⎣G'jn (ω ) sin ( nωt ) + G''jn (ω ) cos ( nωt ) ⎤⎦
(A.3)
We extend this definition for the total stress tensor. Where A and B are the coefficients for the stress in an expansion of strain and
frequency associated with the in-phase and out-of-phase components.
j
τ(t ) = ∑∑ ⎡⎣ A(n j ) sin ( nωt ) + B(n j ) cos ( nωt )⎤⎦
j
n
(A.4)
In this derivation, we utilize the derivative of the stress and have defined it here for convenience:
2
j
dτ(t )
= ∑∑ ⎡⎣nω A(n j ) cos ( nωt ) − nωB(n j ) sin ( nωt ) ⎤⎦
dt
j
n
(A.5)
Where:
( j)
( j)
⎡τ 11,
τ 12,
0 ⎤
n
n
⎢ ( j)
⎥
( j)
0 ⎥
τ = ⎢τ 12,n τ 22,n
( j) ⎥
⎢ 0
0 τ 33,
n⎦
⎣
A
( j)
n
( j)
⎡ A11,
n
⎢ ( j)
= ⎢ A12,n
⎢ 0
⎣
( j)
A12,
n
( j)
22, n
A
0
0 ⎤
⎥
0 ⎥
( j) ⎥
A33,
n⎦
B
( j)
n
( j)
⎡ B11,
n
⎢ ( j)
= ⎢ B12,n
⎢ 0
⎣
( j)
B12,
n
( j)
22, n
B
0
0 ⎤
⎥
0 ⎥
( j) ⎥
B33,
n⎦
In addition, the storage and loss moduli associated with the first harmonic from the Maxwell model are defined in terms of the
nomenclature used above.
G =
'
11
(1)
A12,1
γ0
G =
''
11
(1)
B12,1
γ0
(A.6)
We separate the stress tensor components into shear and normal stress contributions. Given the argument made in A.7 and A.8, during
oscillatory shear the even oscillating frequencies (terms where n=even) of the expansion contribute to the normal components of the
stress while the odd oscillating frequencies (terms where n=odd) contribute to the shear components of the stress [Hyun et al. (in
press)].
σ [ −γ (t ), −γ (t )] = −σ [γ (t ), γ (t )]
(A.7)
N1 [ −γ (t ), −γ (t )] = N1 [γ (t ), γ (t ) ]
(A.8)
3
The amplitude of the stress (also known as the intensity) at a particular harmonic, n is proportional to the strain raised to the value of
the harmonic (n), eq. A.9. This requires that to leading order in strain the intensity of the stress is always proportional to the oscillating
harmonic (n) as discussed in Nam et al. (2008) and defined here:
for τ12
Im ∝ γ 0m , m = {1,3,5...
2,4,6... for Ni for i = 1,2
(A.9)
Therefore, to leading order in strain, we write the expansion for the shear and normal stresses up to O(γ04, ω5). The derivation shown
here utilizes this definition as it expands in both frequency (n) up to n=3 and strain amplitude (j).
(1)
(1)
(3)
(3)
(3)
(3)
τ 12 (t ) = A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,3
sin ( 3ωt ) + B12,3
cos ( 3ωt ) + O ( γ 05 , ω 5 )
(0)
(0)
(2)
(2)
(2)
(2)
τ 11 (t ) = A11,0
sin ( 0 ) + B11,0
cos ( 0 ) + A11,0
sin ( 0 ) + B11,0
cos ( 0 ) + A11,2
sin ( 2ωt ) + B11,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
(0)
(0)
(2)
(2)
(2)
(2)
τ 22 (t ) = A22,0
sin ( 0 ) + B22,0
cos ( 0 ) + A22,0
sin ( 0 ) + B22,0
cos ( 0 ) + A22,2
sin ( 2ωt ) + B22,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
(0)
(0)
(2)
(2)
(2)
(2)
τ 33 (t ) = A33,0
sin ( 0 ) + B33,0
cos ( 0 ) + A33,0
sin ( 0 ) + B33,0
cos ( 0 ) + A33,2
sin ( 2ωt ) + B33,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
(A.10)
Simplifying,
(1)
(1)
(3)
(3)
(3)
(3)
τ 12 (t ) = A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,3
sin ( 3ωt ) + B12,3
cos ( 3ωt ) + O ( γ 05 , ω 5 )
(0)
(2)
(2)
(2)
τ 11 (t ) = B11,0
+ B11,0
+ A11,2
sin ( 2ωt ) + B11,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
(2)
(0)
(2)
(2)
cos ( 2ωt ) + O ( γ 04 , ω 4 )
τ 22 (t ) = B22,0
+ B22,0
+ A22,2
sin ( 2ωt ) + B22,2
(A.11)
(0)
(2)
(2)
(2)
τ 33 (t ) = B33,0
+ B33,0
+ A33,2
sin ( 2ωt ) + B33,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
The Giesekus constitutive model is written in Cartesian coordinates for laminar shear flow:
dτ 11 τ11
α
12 = 0
+
+ τ 112 + τ122
− 2γτ
λ
λG
dt
(
)
(A.12)
4
dτ 22 τ 22
α
2
+
+ τ122 + τ 22
=0
λ
λG
dt
(A.13)
dτ 12 τ 12
α
22 − Gγ = 0
+
+ (τ 11 + τ 22 )τ 12
− γτ
dt
λ
λG
(A.14)
dτ 33 τ 33
α
+
+ τ 332
=0
dt
λ
λG
(A.15)
(
)
The coefficients An and Bn in equation A.4 are solved for first by substituting eq. A.1, A.2, A.4 and A.5 into eq. A.12-A.15. For each
equation, terms having equal oscillating frequencies are collected. Then, terms having similar phases (sine or cosine) are separated
into a set of two equations from which we solve simultaneously for the coefficients An and Bn using Mathematica®.
Upon inspection, A.15 has a trivial solution, τ33,n = 0. As defined earlier in eq. A.9 the normal components of the stress tensor (τ11, τ22,
τ33) have contributions only from the even oscillating frequencies while the shear stress component (τ12) has contributions only from
the odd oscillating frequencies. It is straightforward then to argue that eq. A.12 and A.13 will only have terms consisting of even
oscillating frequencies. And by the same argument, we see that eq. A.14 only has contributions from odd oscillating frequencies.
5
We begin by collecting terms having frequency equal to zero (n=0) in eq. A.12:
0ω A
(0)
11,0
(0)
A11,0
(0)
B11,0
⎛
α ⎜
cos(0ωt ) − 0ω B11,0 sin(0ωt ) +
sin(0ωt ) +
cos(0ω t ) +
λ
λ
λ G ⎜⎜
⎝
(0)
11,0
B
λ
(0)
B11,0
λ
+
α
2λ G
(( A
(B )
+
(0)
11,0
(
(0)
11,0
2
λG
α
) + (B ) + (B ) − ( A )
2
(0)
11,0
2
( 0)
11,0
2
(0)
11,0
2
(( A
(0)
11,0
) + (B )
2
(0)
11,0
2
2
) + (( B
(0)
11,0
) −(A )
2
(0)
11,0
2
2
) (cos(0ωt)) + A
(0)
11,0
(0)
11,0
B
⎞
⎟
sin(0ωt ) ⎟ = 0
⎟
⎠
)=0
=0
)
(0)
⎛
B11,0
α⎞
(0)
⎜1 +
⎟=0
B11,0
⎜
G ⎟
⎝
⎠
For which the trivial solution is B11,0 is 0.
For n=0 in eq. A.13 to leading order in γ0 the solutions is the same: B22,0=0.
Given this, the definitions for the stress written in eq. A.11 are simplified further.
(1)
(1)
(3)
(3)
(3)
(3)
τ 12,n (t ) = A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,3
sin ( 3ωt ) + B12,3
cos ( 3ωt ) + O ( γ 05 , ω 5 )
(2)
(2)
(2)
τ 11,n (t ) = B11,0
+ A11,2
sin ( 2ωt ) + B11,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
(2)
(2)
(2)
τ 22,n (t ) = B22,0
+ A22,2
sin ( 2ωt ) + B22,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
τ 33,n (t ) = 0
6
Terms having frequency ωt or for n=1 contribute solely to the linear shear stress response. In the linear regime it is well known that
the Giesekus model reduces to the Maxwell model. We expect, the Maxwell model to be retained from eq. A.14 given the definitions
in eq. A.6.
Collecting terms with oscillating frequency ωt:
(1)
(1)
ω A12,1
cos(ω t ) − ω B12,1
sin(ω t ) +
(1)
B12,1
λ
cos(ω t ) +
(1)
A12,1
λ
sin(ω t ) − Gγ 0ω cos(ω t ) = 0
(A.16)
We separate the cosine (eq. A.17) and sine (eq. A.18) terms.
ωA
(1)
12,1
cos(ω t ) +
−ω B
(1)
12,1
(1)
B12,1
sin(ω t ) +
λ
cos(ω t ) − Gγ 0ω cos(ω t ) = 0
(1)
A12,1
λ
sin(ω t ) = 0
(A.17)
(A.18)
We solve A.17 and A.18 simultaneously, the solution is given as:
(1)
A12,1
=
Gγ 0 De2
1 + De2
(A.19)
(1)
=
B12,1
Gγ 0 De
1 + De 2
(A.20)
7
Eq. A.6 gives the definitions for the storage and loss moduli of the Maxwell model. In addition, we calculate the magnitude of the
first harmonic shear stress or, │G*12,1│ or the magnitude of the stress response from the first harmonic as shown in Macosko, C.W.
Rheology Principles,Measurements and Applications (1994).
G =
'
11
(1)
A12,1
G11'' =
γ0
Gγ 0 De
2
G11' = 1 + De
2
γ0
GDe 2
1 + De 2
G11'' =
GDe
1 + De 2
1/2
'' 2
11
( ) + (G )
G11* = ⎡ G11'
⎢⎣
γ0
Gγ 0 De
2
G11'' = 1 + De
γ0
G11' =
(1)
B12,1
2
⎤
⎥⎦
1/ 2
*
11
G
⎡⎛ GDe 2 ⎞ 2 ⎛ GDe ⎞ 2 ⎤
⎥
= ⎢⎜
+⎜
2 ⎟
2 ⎟
⎢⎣⎝ 1 + De ⎠ ⎝ 1 + De ⎠ ⎥⎦
1/2
G11*
G11*
⎛ 2 2
⎞
G De (1 + De 2 ) ⎟
⎜
=
⎜
⎟
2 2
+
1
De
⎝
⎠
GDe
=
1/2
1 + De 2
(
(
)
)
(A.21)
8
Terms having frequency 2ωt or for n=2 are the first nonlinear terms from the Giesekus constitutive equation, they contribute to the
first normal stress difference. Where:
(2)
(2)
(2)
τ11,2 (t ) = B11,0
+ A11,2
sin ( 2ωt ) + B11,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
(2)
(2)
(2)
τ 22,2 (t ) = B22,0
+ A22,2
sin ( 2ωt ) + B22,2
cos ( 2ωt ) + O ( γ 04 , ω 4 )
The τ2 term in eq. A.12 and A.13 is expanded below. For a particular strain and frequency combination we calculate this term to
leading order in strain and separate the sine, cosine and constant terms for convenience here. Only keeping O(γ0 2, ω2) terms.
(τ )
(1) 2
12,1
(τ )
2
(τ )
2
(1)
12,1
(1)
12,1
((
)
(
)
)
(1)
(1)
= ⎡ A12,1
sin(ωt ) + B12,1
cos(ωt ) ⎤
⎣
⎦
2
(( A ) sin (ωt) + ( B ) cos (ωt) + 2 ( A B ) sin(ωt) cos(ωt))
⎛ A
⎞
) + ( B ) ) (( B ) − ( A ) )
⎜ ((
⎟
=
+
( cos(2ωt ) ) + ( A B ) sin(2ωt )
=
2
(1)
12,1
(1)
12,1
⎜
⎜
⎝
2
2
(1)
12,1
(1)
12,1
2
2
2
2
(1)
12,1
(1)
12,1
2
(1)
12,1
(1)
12,1
2
(1)
12,1
2
(1)
12,1
⎟
⎟
⎠
(A.22)
and:
(
)
2
(2)
(2)
(2)
τ11,2
sin ( 2ωt ) + B11,2
cos ( 2ωt ) + O γ 04 , ω 4 ⎤⎦
= ⎡⎣ B11,0
+ A11,2
(
2
τ11,2
= O γ 04 , ω 4
2
)
(
)
(2)
(2)
(2)
+ A22,2
τ 222,2 = ⎡⎣ B22,0
sin ( 2ω t ) + B22,2
cos ( 2ω t ) + O γ 04 , ω 4 ⎤⎦
(
τ 222,2 = O γ 04 , ω 4
2
)
Eq. A.12 has an additional term, below.We simplify and collect like terms having an oscillating frequency of n=2.
9
(
(1)
(1)
(1)
γ τ12,1
= ( γ 0ω cos(ωt ) ) ( A12,1
) sin(ωt ) + ( B12,1
) cos(ωt )
ωγ 0
(1)
(1)
⎡⎣ A12,1
) ( cos(2ωt ) + 1) ⎤⎦
( sin(2ωt ) + 0 ) + ( B12,1
2
(1)
(1)
(1)
ωγ 0 B12,1
ωγ 0 A12,1
ωγ 0 B12,1
t
sin(2
)
ω
=
+
+
(
)
( cos(2ωt ) )
2
2
2
(1)
γ τ12,1
=
(1)
γ τ12,1
)
Neglecting higher order terms in strain for eq. A.12 the frequency 2ωt or n=2 terms are as follows:
2ω A
(2)
11,2
cos(2ω t ) − 2ω B
(2)
11,2
((
) (
)
sin(2ω t ) +
) ((
(2)
B11,0
λ
) (
+
(2)
B11,2
λ
)
cos(2ω t ) +
)
(2)
A11,2
λ
sin(2ω t )
2
2
(1)
(1)
⎛ A(1) 2 + B (1) 2
⎞
−
B
A
12,1
12,1
12,1
12,1
⎟
α ⎜
(1)
(1)
sin(2ω t ) ⎟
+
+
B12,1
( cos(2ω t ) ) + A12,1
⎜
2
2
λG ⎜
⎟
⎝
⎠
(1)
(1)
(1)
⎛ ωγ 0 B12,1 ωγ 0 A12,1
⎞
ωγ 0 B12,1
sin(2ω t ) ) +
−2 ⎜
+
cos(2ω t ) ) ⎟ = 0
(
(
⎜
⎟
2
2
2
⎝
⎠
(
)
(A.23)
Again, we separate those terms having cosine (eq. A.24) and sine (eq. A.25) contributions into two separate equations. These two
equations are then solved simultaneously for coefficients A(2)11,2 and B(2)11,2. As discussed in Nam et al. (2008) the constant terms, eq.
A.26, are the constant value the first normal stress difference oscillates about during oscillatory shear.
⎛
(2)
B11,2
α ⎜
(2)
2ω A11,2
cos(2ω t ) +
cos(2ω t ) +
λ
λ G ⎜⎜
(( B
( 2)
A11,2
λ
sin(2ω t ) +
α
λG
(1) 2
12,1
2
⎝
( 2)
−2ω B11,2
sin(2ω t ) +
) −(A )
(1) 2
12,1
(( A
(1)
12,1
)
) (cos(2ωt )) ⎞⎟ − ωγ B
⎟
⎟
⎠
)
0
(1)
12,1
( cos(2ω t ) ) = 0
(1)
(1)
B12,1
sin(2ω t ) − ωγ 0 A12,1
( sin(2ω t ) ) = 0
(A.24)
(A.25)
10
⎛
(2)
B11,0
α ⎜
+
λ
λ G ⎜⎜
(( A
(1)
12,1
) + (B )
2
(1)
12,1
2
2
⎝
(2)
(1)
B11,0
= λωγ 0 B12,1
⎛
α⎜
− ⎜
G⎜
⎝
(( A
(1)
12,1
) ⎞⎟ − ωγ B
⎟
⎟
⎠
(1)
12,1
0
) + (B )
2
(1)
12,1
2
2
)
=0
(A.26)
⎞
⎟
⎟
⎟
⎠
The solution for A(2)11,2 and B(2)11,2 is here:
(2)
11,2
A
=
((
)
B
))
(1 + De ) (1 + 4De )
Gγ De (α (1-5De ) + 2 (1 + De )( −1 + 2De ) )
B =−
2 (1 + De ) (1 + 4De )
(A.27)
2
2
=−
2
2
0
(2)
11,2
(2)
11,0
(
Gγ 02 De3 3 1+De2 + α De2 − 2
2
2
2
2
2
2
2
(A.28)
Gγ 02 De2 ( −2 + α )
(
2 1 + De2
)
(A.29)
Phase angle for τ(2)11,2
11
−
(
tan δ
(2)
11,2
(2)
B11,2
)= A
(2)
11,2
( (
) (
)(
2 (1 + De ) (1 + 4De )
Gγ De ( 3 (1+De ) )
(1 + De ) (1 + 4De )
Gγ 02 De2 α 1-5De2 + 2 1 + De2 −1 + 2De2
2
=
2
0
(
tan δ
2
2
(
)=−
−2 + α + 2De2 − 5α De2 + 4De4
)
−2 + α + 2De2 − 5α De2 + 4De4
(2)
tan δ11,2
=−
(
2De 3-2α + 3De2 +α De2
((
2
3
2
(2)
11,2
2
) (
))
2
)
2De 3 1 + De2 +α De2 -2
))
(A.30)
For eq. A.13 we solve for A(2)22,2 and B(2)22,2 as before, collecting terms oscillating with a frequency of 2ωt.
(2)
(2)
2ω A22,2
cos(2ωt ) − 2ω B22,2
sin(2ωt ) +
⎛
α ⎜
+
λ G ⎜⎜
⎝
(( A
(1)
12,1
) + (B )
2
(1)
12,1
2
2
) + (( B
(1)
12,1
(2)
B22,0
λ
+
) −(A )
2
(1)
12,1
(2)
B22,2
λ
2
2
cos(2ωt ) +
(2)
A22,2
λ
) (cos(2ωt)) + ( A
sin(2ωt )
(1)
(1)
12,1 12,1
B
)
⎞
⎟
sin(2ωt ) ⎟ = 0
⎟
⎠
(A.31)
Eq. A.32, A.33 and A.34 give the equations consisting only of the cosine, sine and constant terms from A.31.
2ω A
(2)
22,2
(2)
B22,2
⎛
α ⎜
cos(2ω t ) +
cos(2ω t ) +
λ
λ G ⎜⎜
⎝
(( B
(1)
12,1
) −(A )
2
(1)
12,1
2
2
) ( cos(2ωt)) ⎞⎟ = 0
⎟
⎟
⎠
(A.32)
12
(2)
−2ω B22,2
sin(2ω t ) +
⎛
(2)
B22,0
α ⎜
=−
λ
λ G ⎜⎜
(2)
A22,2
λ
(( A
(1)
12,1
) + (B )
2
(2)
=−
B22,0
(1)
12,1
2
(A
2G (
(1)
12,1
) + (B )
2
(1)
12,1
2
α
λG
(( A
(1)
12,1
)
)
(1)
sin(2ω t ) = 0
B12,1
(A.33)
) ⎞⎟
⎟
⎟
⎠
2
⎝
α
sin(2ω t ) +
)
(A.34)
Eq. A.35 and A.36 give the solution for A(2)22,2 and B(2)22,2 when equations A.32 and A.33 are solved simultaneously.
(2)
22,2
A
(2)
22,2
B
(2)
22,0
B
=
=
α Gγ 02 De3 ( De2 − 2)
(1 + De ) (1 + 4De )
2
2
2
(A.35)
α Gγ 02 De2 ( 5De2 − 1)
(
2 1 + De2
=−
) (1 + 4De )
2
2
(A.36)
α Gγ 02 De2
(
2 1 + De2
)
(A.37)
Phase angle for τ(2)22,2
13
α Gγ 02 De2 ( 5De2 − 1)
(
tan δ
(2)
22,2
(2)
B22,2
)= A
(2)
22,2
=
(
) (1 + 4De )
α Gγ De ( De − 2 )
(1 + De ) (1 + 4De )
2
0
3
2
(
tan δ
(2)
22,2
2
2 1 + De2
2
2
2
2
(A.38)
1 − 5De2
) = 2De
( 2 − De )
2
14
The first normal stress difference is defined as:
N1 = τ11,2 − τ 22,2
We calculate the coefficient for the sine contribution:
((
)
(
⎡ Gγ 2 De3 3 1+De 2 + α De 2 − 2
0
(2)
(2)
A11,2
− A22,2
=⎢
2
⎢
1 + De 2 1 + 4De 2
⎣
(
(2)
(2)
A11,2
− A22,2
)(
3Gγ De (1+De )
=
(1 + De ) (1 + 4De )
2
0
2
(2)
11,2
−A
(2)
11,2
−A
A
A
(2)
22,2
(2)
22,2
(
)(
)
)
⎡ α Gγ 2 De3 De 2 − 2 ⎤
0
⎥
−⎢
⎥ ⎢ 1 + De 2 2 1 + 4De 2 ⎥
⎦ ⎣
⎦
(
2
2
2
3Gγ 02 De3
(2)
(2)
A11,2
− A22,2
=
(2)
(2)
A11,2
− A22,2
3
)
) ) ⎤⎥
(1 + De )(1 + 4De )
Gγ De ( 3De )
=
(1 + De )(1 + 4De )
Gγ De ( (1 + 4De ) − (1 + De ) )
=
(1 + De )(1 + 4De )
Gγ De ( (1 + 4De ) ) Gγ De ( − (1 + De ) )
=
+
(1 + De )(1 + 4De ) (1 + De )(1 + 4De )
(2)
(2)
A11,2
− A22,2
=
2
2
2
0
2
2
2
2
0
2
2
2
2
0
2
2
2
Gγ 02 De
2
0
2
−
2
2
2
Gγ 02 De
(1 + De ) (1 + 4De )
2
2
1
(2)
(2)
A11,2
− A22,2
= γ 02 G11'' (ω ) − γ 02 G1''1 ( 2ω )
2
(A.39)
15
We calculate the coefficient for the cosine contribution:
(2)
11,2
−B
(2)
11,2
−B
(2)
11,2
−B
(2)
11,2
−B
B
B
B
B
(2)
22,2
(2)
22,2
(2)
22,2
(2)
22,2
(2)
(2)
− B22,2
B11,2
(2)
(2)
− B22,2
B11,2
( (
) (
)(
⎡ Gγ 2 De 2 −α 5De 2 − 1 + 2 1 + De 2
0
= ⎢−
2
⎢
2 1 + De 2 1 + 4De 2
⎣
(
((
)( −1 + 2De ) ) ⎤⎥ ⎡⎢ α Gγ De ( 5De − 1) ⎤⎥
−
⎥ ⎢ 2 1 + De
) (1 + 4De ) ⎥⎦
)
⎦ ⎣ (
2
2
0
2
2
2
2
2
)( −1 + 2De ))
=
2 (1 + De ) (1 + 4De )
Gγ De ( −1 + 2De )
=−
(1 + De )(1 + 4De )
Gγ De ( (1 + 4De ) − 2 (1 + De ) )
=−
(1 + De )(1 + 4De )
Gγ De (1 + 4De ) Gγ De ( −2 ) (1 + De )
=−
−
(1 + De )(1 + 4De ) (1 + De )(1 + 4De )
Gγ De ( 2De )
Gγ De
=−
+
(1 + De ) (1 + 4De )
Gγ 02 De 2 2 1 + De 2
2
2
0
2
0
2
2
2
2
2
2
2
2
2
2
2
2
0
2
2
2
2
2
0
2
2
2
0
2
2
0
2
2
2
2
2
2
2
1
(2)
(2)
− B22,2
= −γ 02 G11' (ω ) + γ 02 G11' ( 2ω )
B11,2
2
(A.40)
16
We calculate the coefficient for the constant contribution:
(2)
11,0
B
−B
(2)
22,0
=
(2)
(2)
− B22,0
=
B11,0
−Gγ 02 De 2 ( −2 + α )
(
2 1 + De 2
)
+
α Gγ 02 De2
(
2 1 + De 2
)
Gγ 02 De2
(1 + De )
2
(2)
(2)
(1)
− B22,0
= γ 0 A12,1
B11,0
(2)
(2)
− B22,0
= γ 02 G11' (ω )
B11,0
(A.41)
This derivation of the Giesekus model gives the coefficients for the τ(2)11,2 and τ(2)22,2 the difference of which is the first normal stress
difference. Below this difference is expressed using the solutions from A.39-A.41:
N1 = τ11,2 − τ 22,2
(
) (
)
(
)
(2)
(2)
(2)
(2)
(2)
(2)
sin ( 2ωt ) + B11,2
cos ( 2ωt )
− B22,0
+ A11,2
− A22,2
− B22,2
N1 = B11,0
1
1
⎡
⎤
⎡
⎤
N1 = γ 02 G11' (ω ) + ⎢G11'' (ω ) − G11'' ( 2ω ) ⎥ γ 02 sin ( 2ωt ) + ⎢−G11' (ω ) + G11' ( 2ω ) ⎥ γ 02 cos ( 2ωt )
2
2
⎣
⎦
⎣
⎦
(A.42)
This solution is in agreement with Ferry (1980) and Nam et al. (2008) (eq. A.43):
⎡ Gγ 02 De2 2Gγ 02 De2 ⎤
Gγ 02 De2 ⎡ Gγ 02 De Gγ 02 De ⎤
sin
2
ω
+
−
+
+
t
) ⎢−
⎢
⎥ (
⎥ cos ( 2ωt )
2
1 + De2 ⎣1 + De2 1 + 4 De2 ⎦
1 + 4 De2 ⎦
⎣ 1 + De
1
1
⎡
⎤
⎡
⎤
N1 = γ 02 G ' (ω ) + ⎢G '' (ω ) − G '' ( 2ω ) ⎥ γ 02 sin ( 2ωt ) + ⎢−G ' (ω ) + G ' ( 2ω ) ⎥ γ 02 cos ( 2ωt )
2
2
⎣
⎦
⎣
⎦
N1 =
(A.43)
17
Terms having frequency 3ωt or for n=3 exist in eq. A.14 and are the first nonlinear shear stress terms from the Giesekus constitutive
equation.
(1)
(1)
(3)
(3)
(3)
(3)
τ12,n (t ) = A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,1
sin (ωt ) + B12,1
cos (ωt ) + A12,3
sin ( 3ωt ) + B12,3
cos ( 3ωt ) + O ( γ 05 , ω5 )
A.14 has (τ11,n + τ22,n)τ12,n (A.44) and γτ11,n (A.45) terms. They are simplified below. We remind the reader that the summation is over
all frequencies and strains however we are only interested in contributions to leading order in strain (up to O(γ04,ω4)) for which we
only need to solve for the appropriate terms having contributions from the τ(2)11,2 , τ(2)22,2 and τ(1)12,1 terms.
(
⎛
⎞
(2)
(2)
(2)
(2)
(2)
(2)
⎜ ∑ τ 11, n + ∑ τ 22,n ⎟ ∑ τ 12, n = B11,0 + A11,2 sin(2ω t ) + B11,2 cos(2ω t ) + B22,0 + A22,2 sin(2ω t ) + B22,2 cos(2ω t )
n
⎝ n
⎠ n
(
)
(
)
(
)
(
)
) ( ( A ) sin(ωt ) + ( B ) cos(ωt ) )
(1)
12,1
(1)
12,1
(( A ) sin(2ωt ) + ( B ) cos(2ωt ) + ( A ) sin(2ωt ) + ( B ) cos(2ωt ) ) (( A ) sin(ωt ) + ( B ) cos(ωt ) ) =
(2)
11,2
(2)
11,2
(2)
22,2
(2)
22,2
(1)
12,1
1 (2) (1)
(2)
(1)
A11,2 A12,1 + A22,2
A12,1
( cos(ω t ) − cos(3ω t ) )
2
1 (2) (1)
(2)
(1)
B12,1 + B22,2
B12,1
( sin(3ω t ) + sin(ω t ) ) + B11,2
( cos(3ω t ) + cos(ω t ) ) + O γ 05 , ω 5
2
(2)
(1)
(2)
(1)
(2)
(1)
(2)
(1)
B11,0
A12,1
sin(ω t ) + B11,0
B12,1
cos(ω t ) + B22,0
A12,1
sin(ω t ) + B22,0
B12,1
cos(ω t ) +
+
1 (2) (1)
1 (2) (1)
(2)
(1)
(2)
(1)
B11,2 A12,1 + B22,2
A12,1
B12,1 + A22,2
B12,1
( sin(3ω t ) − sin(ωt ) ) + A11,2
2
2
(
)
(
(
)
(
(2)
(2)
(2)
11,2 = ( γ 0ω cos(ωt ) ) B11,0
γτ
+ ( A11,2
) sin(2ωt ) + ( B11,2
) cos(2ωt ) + O (γ 04 , ω 5 )
)
(
)
(
(A.44)
)
)
⎡
(2)
(2)
A11,2
) ( cos(3ωt ) + cos(−ωt ) )
( sin(3ωt ) − sin(−ωt ) ) + ( B11,2
⎣⎢
2
(2)
11,2 = ωγ 0 ⎢ B11,0
γτ
cos(ωt ) +
11,2
γτ
(1)
12,1
⎤
+ O γ 04 , ω 5 ⎥
⎦⎥
(
(2)
(2)
⎡ (2)
) ( cos(3ωt ) + cos(ωt ) )
A11,2
( sin(3ωt ) + sin(ωt ) ) + ( B11,2
+ O γ 04 , ω 5
cos(ωt ) +
= ωγ 0 ⎢ B11,0
2
⎢⎣
(
)
)
⎤
⎥
⎥⎦
(A.45)
Finally, for n=3 eq. A.14 is used and we collect all terms having frequency 3ωt.
18
(1)
(1)
(3)
(3)
(3)
(3)
ω A12,1
cos(ωt ) − ω B12,1
sin(ωt ) + ω A12,1
cos(ωt ) − ω B12,1
sin(ω t ) + 3ω A12,3
cos(3ωt ) − 3ω B12,3
sin(3ω t ) +
⎛
(1)
B12,1
λ
cos(ω t ) +
(1)
A12,1
λ
sin(ω t ) +
(3)
B12,1
λ
cos(ω t ) +
(3)
A12,1
λ
sin(ω t ) +
(3)
B12,3
λ
cos(3ω t ) +
(3)
A12,3
λ
sin(3ωt )
1 (2) (1)
1 (2) (1)
(2)
(1)
(2)
(1)
A11,2 A12,1 + A22,2
A12,1
A12,1 + B22,2
A12,1
( cos(ωt ) − cos(3ωt ) ) + B11,2
( sin(3ωt ) − sin(ωt ) ) ⎞⎟
2
2
⎟
1 (2) (1)
(2)
(1)
(2)
(1)
(2)
(1)
⎟
ω
ω
ω
ω
+
+
+
+
+
+
A
B
A
B
sin(3
t
)
sin(
t
)
B
B
B
B
cos(3
t
)
cos(
t
)
(
)
(
)
⎜
⎟
11,2 12,1
22,2 12,1
11,2 12,1
22,2 12,1
2
⎝ 2
⎠
(2)
(2)
⎛ (2)
( A11,2 ) ( sin(3ωt ) + sin(ωt ) ) + ( B11,2 ) ( cos(3ω t ) + cos(ω t ) ) ⎞
−ωγ 0 ⎜ B11,0 cos(ωt ) +
⎟⎟ = 0
⎜
2
⎝
⎠
+
(
(1)
(2)
(1)
(2)
(1)
(2)
(1)
B (2) A12,1
B12,1
A12,1
B12,1
sin(ω t ) + B11,0
cos(ωt ) + B22,0
sin(ωt ) + B22,0
cos(ωt ) +
α ⎜ 11,0
⎜
λG ⎜ 1
(
)
(
)
(
)
)
(A.46)
Eq. A.46 is separated into equations explicitly consisting of sine and cosine functions. Eq. A.49 and A.50 consists of the first
correction terms for the linear response in frequency (n=1).
(3)
3ω A12,3
cos(3ωt ) +
(3)
B12,3
λ
cos(3ωt ) +
ωγ
α
(2)
(1)
(2)
(1)
(2)
(1)
(2)
(1)
(2)
A12,1
A12,1
B12,1
B12,1
− A11,2
− A22,2
+ B11,2
+ B22,2
) cos(3ωt ) ) = 0
( cos(3ωt ) ) − 0 ( ( B22,2
(
)
2λ G
2
(
)
(A.47)
−3ω B
(3)
12,3
sin(3ω t ) +
(3)
ω A12,1
cos(ω t ) +
(3)
B12,1
λ
(3)
sin(ω t ) +
−ω B12,1
(3)
A12,3
λ
cos(ω t ) +
(3)
A12,1
λ
ωγ
α
(2)
(1)
(2)
(1)
(2)
(1)
(2)
(1)
(2)
B11,2
A12,1
A12,1
B12,1
B12,1
+ B22,2
+ A11,2
+ A22,2
) sin(3ω t ) ) = 0
( sin(3ωt ) ) − 0 ( ( A22,2
(
)
2λ G
2
(
sin(3ω t ) +
)
(2)
(1)
(2)
(1)
(2)
(1)
(2)
(1)
(2)
⎞
⎞
⎛ (2) B22,2
⎞
A11,2
A12,1
A12,1
B12,1
B12,1
+ A22,2
+ B11,2
+ B22,2
α ⎛ ⎛ (2) (1)
(2)
(1)
B12,1
+
+
cos(ω t ) ) ⎟ − ωγ 0 ⎜ B22,0
⎜ ⎜ B11,0 B12,1 + B22,0
(
⎟
⎟⎟ ( cos(ω t ) ) = 0
⎜
⎟
⎜
⎜
⎟
λG ⎝ ⎝
2
2 ⎠
⎠
⎝
⎠
sin(ω t ) +
(2)
(1)
(1)
( 2)
(1)
(2)
(1)
⎞ ωγ 0 (2)
⎞
A11,2
B12,1
A12,1
A12,1
+ A22,2 B12,1
− B11,2
− B22,2
α ⎛ ⎛ (2) (1)
(2)
(1)
A12,1
sin(ω t ) ) ⎟ −
A22,2 ( sin(ω t ) ) = 0
+
⎜ ⎜ B11,0 A12,1 + B22,0
(
⎟
⎟
⎟
λ G ⎜⎝ ⎜⎝
2
2
⎠
⎠
(A.48)
(A.49)
(A.50)
Eq. A.47 and A.48 are solved simultaneously for coefficients A(3)12,3 and B(3)12,3 (eq. A.51 and A.52).
(3)
12,3
A
(3)
12,3
B
=
=
(
α Gγ 03 De4 −21 + 30De2 + 51De4 + 4α ( 4 − 17De2 + 3De4 )
(
) (1 + 9De )(1 + 4De )
( −3 + 48De + 33De − 18De + α ( 2 − 48De
4 (1 + De ) (1 + 9De )(1 + 4De )
4 1 + De2
α Gγ 03 De3
3
2
2
2
(A.51)
2
4
3
)
6
2
2
2
+ 46De4
))
(A.52)
19
Eq. A.49 and A.50 are solved simultaneously for coefficients A(3)12,3 and B(3)12,3 (eq. A.53 and A.54).
(3)
12,1
A
=
(
α Gγ 03 De4 −21 − 41De2 − 8De4 + 4α ( 4 + 7De2 )
(3)
B12,1
=−
(
4 1 + De2
(
)
) (1 + 4De )
3
2
α Gγ De 9 + 11De − 10De + 2α ( −3 − De + 8De
3
0
3
2
4
(
4 1 + De2
2
4
(A.53)
))
) (1 + 4De )
3
2
(A.54)
The final solution for the magnitude of the third harmonic stress contribution is given in eq. A.55:
τ
*
12,3
Gγ 03
=
2
12,3
A
+B
2
12,3
Gγ 03
⎡
α De ⎢
9 − 12α + 4α 2 + 9De2
=
4 ⎢ 1 + De2 3 1 + 9De2 1 + 4De2
⎣
3
(
)(
)(
)
⎤
⎥
⎥
⎦
1
2
(A.55)
We can take a limit for small De numbers where the solution is:
*
τ12,3
Gγ
3
0
=
2
2
+ B12,3
A12,3
Gγ
3
0
=
α
⎡⎣(2α − 3)2 ⎤⎦
4
1
2
(A.56)
The graph below is the solution for the third harmonic stress intensity in the limit of small De as a function of the nonlinear parameter,
α:
20
0.30
0.25
*
3
|τ 12,1|/(Gγ 0)
0.20
0.15
0.10
0.05
0.00
0.00
0.25
0.50
0.75
1.00
α
The ratio of the out-of-phase (B(3)12,3) and in-phase (A(3)12,3) components gives a prediction for the tangent of the phase angle for the
third harmonic:
(
α Gγ 03 De3 −3 + 48De2 + 33De4 − 18De6 + α ( 2 − 48De2 + 46De4 )
(
)
(3)
tan δ12,3
=
(3)
B12,3
(3)
A12,3
) (1 + 9De )(1 + 4De )
α Gγ De ( −21 + 30De + 51De + 4α ( 4 − 17De + 3De ) )
−
4 (1 + De ) (1 + 9De )(1 + 4De )
( −3 + 48De + 33De − 18De + 2α (1 − 24De + 23De ))
=
De ( −21 + 30De + 51De + 4α ( 4 − 17De + 3De ) )
=
(
3
4 1 + De2
3
0
4
2
(
)
(3)
B12,3
(3)
A12,3
2
4
2
2
4
3
2
(3)
tan δ12,3
=
2
2
2
4
2
2
6
4
)
2
4
4
(A.57)
21
We calculate the nonlinear parameter, Q0, using the ratio of the third harmonic (eq. A.55) to the first harmonic (eq. A.21):
⎡
α Gγ De ⎢
9 − 12α + 4α 2 + 9De2
⎢ 1 + De 2 3 1 + 9De2 1 + 4De2
4
⎞⎛ 1 ⎞
⎣
⎟⎜ 2 ⎟ =
⎟⎝ γ0 ⎠
Gγ 0 De
⎠
1/ 2
1 + De 2
3
0
(3)
⎛ τ 12,3
Q0 = ⎜ (1)
⎜ τ 12,1
⎝
3
(
)(
)(
(
(
(
Q0 =
α De2
(
4 1 + De2
)(
(
)(
)(
) ⎤⎥
) ⎥⎦
⎡ 9 − 12α + 4α 2 + 9De2
⎢
⎢⎣ 1 + 9De 2 1 + 4De2
) (
)(
)
1
2
)
2
2
2
⎡
α De ⎢ 9 − 12α + 4α + 9De 1 + De
Q0 =
4 ⎢ 1 + De2 3 1 + 9De2 1 + 4De2
⎣
2
)
⎤
⎥
⎥
⎦
) ⎤⎥
⎥⎦
1
1
2
2
(A.59)
22
0.35
Third harmonic stress
Q
Q0 and |τ(3) |/(GWe3)
12,3
0.30
0.06
0.05
Q0
0.04
0.03
0.25
0.20
0.15
0.10
0.05
0.02
0.00
0.01
-0.05
-2
0.00
10
-2
10
-1
10
0
10
1
10
2
10
-1
10
0
10
De
De
23