Small-Amplitude Oscillatory Shear
Apply a sinusoidal shear strain
γ(t) = γ0 sin(ωt)
(2-45)
The shear rate is therefore
γ̇(t) = γ0 ω cos(ωt) = γ̇0 cos(ωt)
(2-46)
The stress is also sinusoidal with the same frequency, but leads the strain
by phase angle δ.
σ(t) = σ0 sin(ωt + δ)
(2-47)
Figure 1: Oscillatory Shear. The stress leads the applied strain by phase
angle δ.
1
Small-Amplitude Oscillatory Shear
BOLTZMANN SUPERPOSITION
Using the Boltzmann Superposition Principle
Z t
G(t − t0 )γ̇(t0 )dt0
σ(t) =
(2-8)
−∞
with s ≡ t − t0 , ds = −dt0 , t0 = t ⇒ s = 0 and t0 = −∞ ⇒ s = ∞
Z ∞
σ(t) =
G(s)γ0 ω cos(ω[t − s])ds
0
Z ∞
Z ∞
σ(t) = γ0 ω
G(s) sin(ωs)ds sin(ωt)+γ0 ω
G(s) cos(ωs)ds cos(ωt)
0
0
Define Storage Modulus
∞
Z
0
G (ω) ≡ ω
G(s) sin(ωs)ds
(2-65)
G(s) cos(ωs)ds
(2-66)
0
and Loss Modulus
00
Z
G (ω) ≡ ω
∞
0
Thus the stress is
σ(t) = γ0 [G0 (ω) sin(ωt) + G00 (ω) cos(ωt)]
(2-49)
Using the formula for the sine of a sum
σ(t) = σ0 sin(ωt + δ) = σ0 [cos(δ) sin(ωt) + sin(δ) cos(ωt)]
Defining
σ0
γ0
0
G = Gd cos(δ)
(2-50)
G00 = Gd sin(δ)
(2-51)
Gd ≡
The ratio
G00
= tan(δ)
G0
2
Small-Amplitude Oscillatory Shear
HOOKEAN SOLID
Hooke’s Law
σ = Gγ
(2-52)
For oscillatory shear
γ(t) = γ0 sin(ωt)
(2-45)
σ(t) = γ0 G sin(ωt)
(2-53)
For the solid
Comparing with the general form
σ(t) = γ0 [G0 (ω) sin(ωt) + G00 (ω) cos(ωt)]
(2-49)
The solid has
G0 (ω) = G
and
G00 (ω) = 0
δ = tan−1 (G00 /G0 ) = 0
meaning that the stress is perfectly in phase with the strain.
NEWTONIAN LIQUID
Newton’s Law σ = η γ̇
(2-54)
γ̇(t) = γ0 ω cos(ωt)
(2-46)
σ(t) = ηγ0 ω cos(ωt)
(2-55)
For oscillatory shear
For the liquid
The liquid has
G0 (ω) = 0
and
G00 (ω) = ηω
δ = tan−1 (G00 /G0 ) = π/2
meaning that the stress is 90◦ out of phase with the strain (the stress is
in phase with the rate of strain).
3
Small-Amplitude Oscillatory Shear
BASIC PHYSICS - THE MAXWELL
MODEL
G(t) = G0N exp(−t/λ)
Storage Modulus
∞
Z
0
G (ω) ≡ ω
G(t) sin(ωt)dt
(2-65)
0
0
G (ω) =
ωG0N
∞
Z
exp(−t/λ) sin(ωt)dt =
0
G0N (ωλ)2
1 + (ωλ)2
and Loss Modulus
Z
00
∞
G (ω) ≡ ω
G(t) cos(ωt)dt
(2-66)
0
00
G (ω) =
ωG0N
Z
∞
exp(−t/λ) cos(ωt)dt =
0
G0N ωλ
1 + (ωλ)2
Figure 2: Storage and Loss Modulus for the Maxwell Model.
The real power of the oscillatory shear experiment is that we can probe the
viscoelastic response on different time scales by applying different frequencies
ω.
4
Small-Amplitude Oscillatory Shear
COMPLEX MODULUS, VISCOSITY AND
COMPLIANCE
COMPLEX MODULUS
G∗ (ω) ≡ G0 (ω) + iG00 (ω)
Magnitude
Gd ≡
p
σ0
= |G∗ | = (G0 )2 + (G00 )2
γ0
Phase
tan(δ) =
(2-58)
(2-59)
G00
G0
COMPLEX VISCOSITY
η ∗ (ω) ≡
G∗ (ω)
iω
η ∗ (ω) = η 0 (ω) − iη 00 (ω)
Magnitude
|η ∗ | =
p
|G∗ (ω)|
(η 0 )2 + (η 00 )2 =
ω
(2-64)
η 0 = G00 /ω
(2-61)
η 00 = G0 /ω
(2-62)
Phase
tan(δ) =
η0
η 00
COMPLEX COMPLIANCE
J ∗ (ω) ≡
(2-63)
1
= J 0 (ω) − iJ 00 (ω)
G∗ (ω)
5
Small-Amplitude Oscillatory Shear
RC-3 polybutadiene Mw = 940, 000, Mw /Mn < 1.1, Tg = −99◦ C
Figure 3: Storage and Loss Moduli of RC-3.
Time period for one cycle is 2π/ω.
For ω = 10−4 rad/sec, 2π/ω ∼
= 1 day.
6
Small-Amplitude Oscillatory Shear
VISCOSITY, PLATEAU MODULUS AND
STEADY STATE COMPLIANCE
G00 (ω)
η0 = lim
= lim η 0 (ω)
ω→0
ω→0
ω
Z ∞ 0
2
G (ω)
η0 =
d ln ω
π −∞ ω
Z
2 ∞ 00
0
GN =
G (ω)d ln ω
π −∞
0
Z ∞
G (ω)
AG ≡ lim
=
G(s)sds
ω→0
ω2
0
(2-73)
(2-74)
Proof:
lim sin(ωs) = ωs
ω→0
∞
Z
0
G (ω) ≡ ω
G(s) sin(ωs)ds
(2-65)
0
lim
ω→0
G0 (ω)
ω2
1
=
ω
Z
∞
G(s)ωsds
0
Recall that
JS0
1
= 2
η0
∞
Z
G(s)sds
AG = Js0 η02
(2-75)
G0
= lim
ω→0
(G00 )2
INTERRELATIONS
JS0
(2-33)
0
J0 =
G0
(G0 )2 + (G00 )2
J 00 =
G00
(G0 )2 + (G00 )2
G0 =
J0
(J 0 )2 + (J 00 )2
G00 =
J 00
(J 0 )2 + (J 00 )2
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Small-Amplitude Oscillatory Shear
INTEGRATION OF LOSS MODULUS TO
GET THE PLATEAU MODULUS
G0N
2
=
π
Z
∞
G00 (ω)d ln ω
−∞
RC-3 polybutadiene Mw = 940, 000, Mw /Mn < 1.1, Tg = −99◦ C
Figure 4: Loss Modulus of RC3 at 26◦ C.
8
Small-Amplitude Oscillatory Shear
DISSIPATED ENERGY PER CYCLE
Dissipated Energy (WORK) per unit volume
Z
W = σ γ̇dt
(2-56)
σ(t) = σ0 sin(ωt + δ)
(2-47)
γ̇(t) = γ0 ω cos(ωt)
(2-46)
Z
2π/ω
σ0 γ0 ω sin(ωt + δ) cos(ωt)dt
W =
0
sin(ωt + δ) = cos(δ) sin(ωt) + sin(δ) cos(ωt)
Z
W =
2π/ω
σ0 γ0 ω cos(δ) sin(ωt) cos(ωt) + sin(δ) cos2 (ωt) dt
0
2π/ω
cos(2ωt)
t sin(2ωt)
W = σ0 γ0 ω − cos(δ)
+ sin(δ)
+
4ω
2
4ω
0
cos(0) = cos(4π) = 1,
sin(0) = sin(4π) = 0
1−1
2π/ω 0 − 0
W = σ0 γ0 ω − cos(δ)
+ sin(δ)
+
4ω
2
4ω
W = πσ0 γ0 sin(δ)
Recall
G00 = Gd sin(δ) =
σ0
sin(δ)
γ0
W = πγ02 G00
The loss modulus is proportional to the dissipated energy.
9
(2-51)
(2-57)