Molecular modelling of entangled polymer fluids under flow The ...

1.5. LINEAR RHEOLOGY 7
1.5.1 Linear oscillatory shear
Equation 1.5 can, in principle, be used to measure the relaxation modulus, G(t), directly.
However, the step strain is never completely instantaneous so measured data at
early times are unreliable and at long times the signal to noise ratio is weak, making the
terminal behaviour difficult to obtain. A more effective approach is to use a continuous
oscillating shear strain history. In complex notation this is expressed as
γ(t) = R(γ max exp(iωt)). (1.9)
Under this deformation history equation 1.5 gives
σ xy (t) =
∫ t
−∞
G(t − t ′ ) dγ
dt ′ (t′ )dt ′ . (1.10)
Substituting in the form of the oscillating shear history (equation 1.9) and changing
the variable of integration gives
(
σ xy (t) = R iωγ(t)
∫ ∞
0
)
exp(−iωs)G(s)ds . (1.11)
which can be rewritten as
σ xy (t) = R (γ(t)G ∗ (ω)) . (1.12)
where G ∗ (ω) is known as the complex modulus. Thus
G ∗ (ω) = iω
∫ ∞
0
e −iωt G(t)dt. (1.13)
When the complex modulus is written as G ∗ = G ′ + iG ′′ it can be seen that G ∗ consists
of a component which is in phase with the strain and one which is out of phase. The
in phase part, G ′ , is known as the storage or elastic modulus and the out of phase
part, G ′′ , is the loss or dissipative modulus. A perfectly elastic solid of modulus G 0
would have G ′ = G 0 and G ′′ = 0. In the case of a viscous liquid with viscosity η then
G ′ = 0 and G ′′ = ωη since σ xy is in phase with the shear rate.
For a viscoelastic
material both G ′ and G ′′ are functions of the applied frequency, ω. In general, the
loss modulus dominates at low frequencies, while the elastic modulus dominates at
high frequencies. The material crosses over from viscous behaviour elastic behaviour at
some intermediate frequency where G ′ = G ′′ . A simple form of the relaxation modulus,
the Maxwell model, where G(t) = G 0 exp(−t/τ) which is characterised by a relaxation
time, τ. The complex modulus for this model is given by,
G ′ (ω) = G 0
ω 2 τ 2
1+ω 2 τ 2 , G ′′ (ω) = G 0
ωτ
1+ω 2 τ 2 . (1.14)