chapter 7. viscoelastic models - seoul national...

Chapter 7.

Viscoelastic Models

Polymer has linear viscoelasticity. It is desirable to understand the

representation of linear viscoelastic processes by certain model

systems in order to gain insight into relaxation behavior.

Two broad classes of models

(1) Mechanical Analogues : combination of element (spring and

dashpot)

(2) Molecular Models : predicted on the basis of molecular

parameters

7. Viscoelastic Models

A. Mechanical Elements Tensile elongation experiment

Pure Hookean Spring - Purely elastic

𝝈𝟎 = 𝑬𝜺𝟎

𝝈𝟎 : instantaneous stress

𝜺𝟎 : instantaneous strain

𝑬 : Young’s modulus

Spring Element

𝑬

𝝈

- Purely elastic


Fluids : simple linear viscous dashpot

Viscosity : Newton’s Law

𝝈 = 𝜼𝒅𝜺

𝒅𝒕

𝜺(𝒕) =𝝈𝟎𝜼𝒕

𝜼 : viscosity of fluid

Dashpot(piston) Element

𝜼


Viscoelastic Materials(such as polymers) poorly represented by

either the spring or the dash pot

Combination of the spring and dashpot

𝑬 : the instantaneous tensile modulus

𝜼 : the viscosity of the liquid in the dashpot

𝑬

𝜼

𝝈 = 𝑬𝜺

𝝈 = 𝜼𝒅𝜺

𝒅𝒕

𝒅𝜺

𝒅𝒕=𝟏

𝑬

𝒅𝝈

𝒅𝒕+𝝈

𝜼

𝜺 = 𝜺𝒆𝒍𝒂𝒔 + 𝜺𝒗𝒊𝒔𝒄𝒐𝒖𝒔

𝑑

𝑑𝒕(𝑯𝒐𝒐𝒌𝒆′𝒔 𝒍𝒂𝒘) Viscous element


Creep experiment

The model is subjected to an instantaneous constant stress σ0.

Thus equation becomes

since dσ/dt is zero. Integration of equation from time 0 to some time t

yields after division by σ0

𝒅𝜺

𝒅𝒕=𝝈𝟎𝜼

𝜺(𝒕)

𝝈𝟎=𝜺𝟎𝝈𝟎

+𝒕

𝜼

𝒅𝜺

𝒅𝒕=

𝝈𝟎

𝜼

𝒕

𝟎

𝜺

𝜺𝟎 (stress is constant)

𝜺 𝒕 = 𝜺𝟎 +𝝈𝟎

𝜼𝒕 (𝜺𝟎는 적분상수, 𝜺𝟎 =

𝝈

𝑬𝟏)

Response of spring 𝜺𝟎

𝝈𝟎 =

𝟏

𝑬 =D

𝝈𝟎 로 나누면,


𝒕

𝐷0

𝐷(𝑡) 1

𝜂

𝑫(𝒕) = 𝑫𝟎 + 𝒕/ 𝜼

Creep compliance

Strain instantaneously to 𝜺𝟎, 𝝈 𝒕 ?

Stress gradually decreases

Stress Relaxation 𝜺 = 𝜺𝟎 constant

After application of strain ⇒ 𝒅𝜺

𝒅𝒕= 𝟎

𝒅𝜺

𝒅𝒕=1

𝑬

𝒅𝝈

𝒅𝒕+𝝈

𝜼 ⇒

1

𝑬

𝒅𝝈

𝒅𝒕= −

𝝈

𝜼

𝒅𝝈

𝝈= −

𝑬

𝜼𝒅𝒕 𝝈(𝒕) = 𝝈𝟎𝒆

−𝑬𝜼𝑡

𝝉 =𝜼

𝑬 Define a relaxation time


𝒕 = 𝝉

𝑬(𝝉) = 𝑬𝟎 𝒆−1

𝑬(𝝉)

𝑬𝟎=1

𝒆≈ 0.36

relaxation time(𝝉) : Relaxation elastic constant가

t=0에서 36% 정도까지 되는 데 걸리는 시간

Decay exponentially

𝑬(𝒕) =𝝈(𝒕)

𝜺𝟎= 𝑬𝟎 𝒆

−𝒕/𝝉

𝑬(𝒕)

𝑬𝟎

𝟎. 𝟑𝟔𝑬𝟎

𝟏 𝒕/𝝉


Figure 7-3b. Maxwell body behavior using stress relaxation conditions; log-log plot

-1 0 1

- 5

- 4

- 3

- 2

- 1

0

Act like a spring

Viscous liquid

dashpot에 의한 effect

𝑬 𝒕 = 𝑬𝟎𝒆−𝒕/𝝉 ≅ 𝑬𝟎(𝟏 −

𝒕

𝝉)

𝒕/𝝉 → ∞ ⇒ 𝑬 ∞ = 𝟎

＊Linear polymer 는 Maxwell model로 많이 취급한다.


1. Introduction

1. Maxwell Element

Dynamic Experiment.

Response to sinusoidal stress on Maxwell model

𝝈(𝒕) = 𝝈𝟎𝒆𝒊𝝎𝒕

𝒅𝜺

𝒅𝒕=𝟏

𝐄+𝒅𝝈

𝒅𝒕+ 𝝈

𝜼

𝒅𝜺

𝒅𝒕=𝝈𝟎𝐄𝒊𝝎𝒆𝒊𝝎𝒕 +

𝝈𝟎𝜼𝒆𝒊𝝎𝒕

On integration (𝒕𝟏 → 𝒕𝟐)

𝜺 𝒕𝟐 − 𝜺 𝒕𝟏 =𝝈𝟎

𝑬𝒆𝒊𝝎𝒕𝟐 − 𝒆𝒊𝝎𝒕𝟏 +

𝝈𝟎

𝜼𝒊𝝎𝒆𝒊𝝎𝒕𝟐 − 𝒆𝒊𝝎𝒕𝟏


𝜺 𝒕𝟐 − 𝜺 𝒕𝟏 =𝟏

𝑬 [𝝈 𝒕𝟐

− 𝝈 𝒕𝟏 ] +

𝝈𝟎𝜼𝒊𝝎

[𝝈 𝒕𝟐 − 𝝈 𝒕𝟏

]

𝜺(𝒕𝟐) − 𝜺(𝒕𝟏)

𝝈(𝒕𝟐) − 𝝈(𝒕𝟏)=𝟏

𝑬−

𝒊

𝝉𝝎

where

D*:complex tensile creep compliance.

𝑫′: storage compliance

𝑫′′:𝟏

𝜼𝝎=

𝑫

𝝉𝝎, loss compliance

𝝉 = 𝜼/E,

≡ 𝜼𝐃

E = static modulus

D*

(Complex tensile compliance)

𝑫∗ = 𝑫− 𝒊𝑫

𝝎𝝉

𝑫∗ = 𝑫′ − 𝒊𝑫′′

storage loss compliance

𝜼 = 𝝉/D


Complex tensile modulus

𝑬∗ =𝟏

𝑫∗ =𝟏

𝑫 −𝒊𝑫𝝎𝝉

=𝟏

𝑬−𝟏 −𝒊𝑬−𝟏

𝝎𝝉

𝑬∗ =𝑬𝝉𝟐𝝎𝟐

𝟏 +𝝎𝟐𝝉𝟐+

𝒊𝝉𝝎𝑬

𝟏 + 𝝎𝟐𝝉𝟐= 𝑬′ + 𝒊𝑬′′

𝑬∗ =𝑬𝝉𝝎

𝝎𝝉 − 𝒊ｘ

𝝎𝝉 + 𝒊

𝝎𝝉 − 𝒊

storage modulus + loss modulus


𝐭𝐚𝐧𝜹 =𝑬′′

𝑬′=

𝟏

𝜸𝝎

𝑬′ =𝑬𝝉𝟐𝝎𝟐

𝟏 + 𝝉𝟐𝝎𝟐=

𝑬

𝟏 + 𝟏 𝝎𝟐 𝝉𝟐

Low frequency

𝝎𝝉 → 𝟎 𝑬′ → 𝟎 High frequency

𝝎𝝉 → ∞ 𝑬′ → 𝑬

(loss tangent)


-2 -1 0 1 2 𝐥𝐨𝐠𝝎𝝉

𝑬′ Storage modulus

𝑬′′ =𝑬𝝎𝝉

𝟏 + 𝝎𝟐𝝉𝟐=

𝑬

𝝎𝝉 +𝟏𝝎𝝉

Loss modulus

𝝎𝝉 → 𝟎 𝑬′′ → 𝟎 𝝎𝝉 → ∞ 𝑬′ → 𝟎

𝐥𝐨𝐠𝝎𝝉

𝑬′′ loss modulus


𝑬′ =𝑬𝝉𝟐𝝎𝟐

𝟏 + 𝝉𝟐𝝎𝟐 =𝑬

𝟏 + 𝟏 𝝎𝟐 𝝉𝟐

Storage modulus

2. Voigt Element

𝜂

𝜎(𝑡)

𝐸

Constraint : strain in the same for spring and dashpot

𝝈 𝟏 = 𝑬𝜺 𝒕

𝝈 𝟐 = 𝜼𝒅𝜺

𝒅𝒕

𝝈 = 𝝈𝟏 + 𝝈𝟐

𝝈 𝒕 = 𝜺 𝒕 𝑬 + 𝜼𝒅𝜺

𝒅𝒕

𝝈𝟎 = 𝜺 𝒕 𝑬 + 𝜼𝒅𝜺

𝒅𝒕

𝝈𝟎𝜼 =

𝒅𝜺

𝒅𝒕+ 𝜺 𝒕 𝑬

𝜼 =𝒅𝜺

𝒅𝒕+ 𝜺

𝝉

𝝈 𝒕 = 𝝈𝟎 Creep Expt

𝒅𝜺

𝒅𝒕+𝜺

𝝉=𝝈𝟎𝜼

𝒆𝒕 𝝉 𝒅𝜺

𝒅𝒕+𝜺

𝝉𝒆𝒕 𝝉 =

𝝈𝟎𝜼𝒆𝒕 𝝉

𝜏 =𝜂

𝐸


𝒅𝜺

𝒅𝒕+𝜺

𝝉=𝝈𝟎𝜼

𝒆𝒕 𝝉 𝒅𝜺

𝒅𝒕+𝜺

𝝉𝒆𝒕 𝝉 =


𝒅

𝒅𝒕𝜺𝒆𝒕 𝝉 =


𝜺 𝒕 =𝝈𝟎𝝉

𝜼𝟏 − 𝒆−𝒕 𝝉

𝒅 𝜺 𝒕 𝒆𝒕 𝝉 𝜺(𝒕)

𝜺(𝟎)

=𝝈𝟎

𝜼 𝒆𝒕 𝝉 𝒅𝒕 =𝝈𝟎𝜼𝝉

𝒕

𝟎

𝒆𝒕 𝝉 𝟎

𝒕

On integration by using

Integrating factor 𝒆𝒕 𝝉


𝝉𝜼 =

𝟏

𝑬= 𝑫

𝜺(𝒕)𝝈𝟎 = 𝑫 𝒕 = 𝑫 𝟏 − 𝒆−𝒕 𝝉

𝝉 =𝜼𝑬 ∶ 𝑹𝒆𝒕𝒂𝒓𝒅𝒂𝒕𝒊𝒐𝒏 𝒕𝒊𝒎𝒆

Voigt model 은 creep에만 사용

𝒔𝒕𝒓𝒆𝒔𝒔 𝒓𝒆𝒍𝒂𝒙𝒂𝒕𝒊𝒐𝒏 𝜺 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 = 𝜺𝟎 𝝈 𝒕 = 𝑬𝜺𝟎 (𝒔𝒑𝒓𝒊𝒏𝒈 만 생각)

𝒅𝜺

𝒅𝒕= 𝟎 𝝈 𝒕 = 𝜺𝟎𝑬 ⇒ 𝑬 𝒕 = 𝑬

𝒕

𝑫(𝒕)


Sinusoidal stress 𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝑬𝒙𝒑𝒕𝒔, 𝝈 𝒕 = 𝝈𝟎𝒆

𝒊𝝎𝒕

𝝈𝟎𝒆𝒊𝝎𝒕 = 𝑬𝜺 𝒕 + 𝜼

𝒅𝜺(𝒕)

𝒅𝒕

Find solution of this equation in the form of

𝜺 𝒕 = 𝜺𝟎𝐞𝐱𝐩 (𝒊𝝎𝒕)


Substitute this into equation and solve for 𝜀0

𝝈𝟎𝒆

𝒊𝝎𝒕 = 𝑬𝜺𝟎𝒆𝒊𝝎𝒕 + 𝜼𝒊𝝎𝜺𝟎𝒆

𝒊𝝎𝒕 = 𝑬 + 𝒊𝜼𝝎 𝜺𝟎𝒆𝒊𝝎𝒕

∴ 𝜺𝟎 =𝝈𝟎

𝑬 + 𝒊𝜼𝝎=

𝝈𝟎𝑬(𝟏 + 𝒊𝝎𝝉)

=𝑫𝝈𝟎

𝟏 + 𝒊𝝎𝝉 𝝉 =

𝜼𝑬 𝒓𝒆𝒕𝒂𝒓𝒅𝒂𝒕𝒊𝒐𝒏 𝒕𝒊𝒎𝒆

𝑫∗ =𝜺(𝒕)

𝝈(𝒕)=𝜺𝟎𝒆

𝒊𝝎𝒕

𝝈𝟎𝒆𝒊𝝎𝒕

=𝜺𝟎𝝈𝟎

=𝑫(𝟏 − 𝒊𝝎𝝉)

(𝟏 + 𝒊𝝎𝝉)(𝟏 − 𝒊𝝎𝝉)

=𝑫

𝟏 + (𝝎𝝉)𝟐− 𝒊

𝝎𝝉𝑫

𝟏 + 𝝎𝝉 𝟐 = 𝑫′ − 𝒊𝑫′

𝑫′ =𝑫

𝟏 + (𝝎𝝉)𝟐 ← 𝒔𝒕𝒐𝒓𝒂𝒈𝒆 𝒄𝒐𝒎𝒑𝒍𝒊𝒂𝒏𝒄𝒆: 𝒔𝒑𝒓𝒊𝒏𝒈 에 대한 𝒄𝒐𝒏𝒕𝒊𝒃𝒖𝒕𝒊𝒐𝒏

𝑫′′ =𝝎𝝉𝑫

𝟏 + 𝝎𝝉 𝟐 ← 𝑳𝒐𝒔𝒔 𝒄𝒐𝒎𝒑𝒍𝒊𝒂𝒏𝒄𝒆: 𝒅𝒂𝒔𝒉𝒑𝒐𝒕 에 대한 𝒄𝒐𝒏𝒕𝒊𝒃𝒖𝒕𝒊𝒐𝒏


-1 0 1 𝐥𝐨𝐠𝝎𝝉

𝑫′

𝑫𝒐𝒓

𝑫′′

𝑫

2

𝐷′𝐷

𝐷′′𝐷

Frequency 가 적을 때는 spring 만에 의한 힘이고, 어느정도 frequency가 되면

Dashpot에 저항이 있고, 더욱 커지면 dashpot에 의한 energy loss 가 커진다.

Deficiencies of two element model

1. Only one transition. Real polymer는 glass, rubber, rubber liquid

2. Decay in modulus is rapid for the models (real polymer는 느리다.)


𝑫′ =𝑫

𝟏 +𝝎𝟐𝝉𝟐 𝑬′ = 𝑬

𝑫′′ =𝑫𝝎𝝉

𝟏 + 𝝎𝟐𝝉𝟐 𝑬′′ = 𝛚𝜼

Experiment Maxwell Element Voigt Element

Creep

(constant stress) 𝑫 𝒕 = 𝑫 + 𝒕/𝜼 𝑫 𝒕 − 𝑫(𝟏 − 𝒆−𝒕 𝝉 )

Stress relaxation

(constant strain) 𝑬 𝒕 = 𝑬𝒆−𝒕 𝝉 𝑫 𝒕 = 𝑬

Sinusoidal

dynamic

experiments

𝑫′ = 𝑫

𝑫′′ = 𝟏/𝜼𝝎

𝑬′ =𝑬𝝎𝟐𝝉𝟐

𝟏 + 𝝎𝟐𝝉𝟐

𝑬′′ =𝑬𝝎𝝉

𝟏 + 𝝎𝟐𝝉𝟐

𝑫′ =𝑫

𝟏 + 𝝎𝟐𝝉𝟐

𝑫′′ =𝑫𝝎𝝉

𝟏 + 𝝎𝟐𝝉𝟐

𝑬′ = 𝑬

𝑬′′ = 𝝎𝜼


3. Maxwell-Wiechert Model

Parallel connection of Z Maxwell Models


Z elements

Stress relaxation

𝜺 𝒕 = 𝒄𝒐𝒏𝒔𝒕 = 𝜺𝟎

For each element 𝑑𝜺(𝒕)

𝑑𝒕= 𝟎

𝟏

𝑬𝟏

𝑑𝝈𝟏𝑑𝒕

+𝝈𝟏𝜼𝟏

= 𝟎

𝟏

𝑬𝟐

𝑑𝝈𝟐𝑑𝒕

+𝝈𝟐𝜼𝟐

= 𝟎

𝟏

𝑬𝒛

𝑑𝝈𝒛𝑑𝒕

+𝝈𝒛𝜼𝒛

= 𝟎

(a)

Total stress 𝝈 = 𝝈𝒊𝒛𝒊=𝟏 (b)


Integrating each of eqn (a)

Where 𝝈(𝟎)𝒊 is the stress on the 𝒊 th element at time equal to zero

∴ 𝑬 𝒕 =𝝈(𝒕)

𝜺𝟎=

𝝈(𝟎)𝒊𝜺𝟎

𝒆−𝒕𝝉𝒊 =

𝒛

𝒊=𝟏 𝑬𝒊

𝒛

𝒊=𝟏𝒆−𝒕𝝉𝒊

𝝈𝟏 𝒕 = 𝝈(𝟎)𝟏𝒆−𝒕𝝉𝟏

𝝈𝟐 𝒕 = 𝝈(𝟎)𝟐𝒆−𝒕𝝉𝟐

𝝈𝒛 𝒕 = 𝝈(𝟎)𝒛𝒆−𝒕𝝉𝒛

∴ 𝝈 = 𝝈(𝟎)𝒊𝒆−𝒕𝝉𝟏

𝒛

𝒊=𝟏



- Total modulus is the summation of

the response of the individual

elements.

- The flexibility of this model is

reproducing real viscoelastic

behavior can be easily demonstrated.

- Its behavior reproduces the two

transitions observed in real polymers.

Glassy

Transition

Rubbery

Viscous


Glassy transition

Crosslinked

- It is possible to replace one of the

Maxwell elements in the Maxwell –

Weichert model with a spring.

- The stress would decay to a finite

value in such a model rather than 0

and would approximate the behavior

of cross-linked polymers.

𝑬 𝒕 = 𝑬𝟏𝒆−𝒕𝝉𝟏 +

𝝈𝟎

𝜺𝟎


Dynamic Exp.

Sinusoidal stress 𝝈𝒋 𝒕 = 𝝈𝟎𝒋𝒆𝒊𝒘𝒕 for the j th element

For the j th element,

𝒅𝜺𝒋(𝒕)

𝒅𝒕=

𝟏

𝑬𝒋

𝒅𝝈𝒊𝒅𝒕

+𝝈𝒋

𝜼𝒋=

𝟏

𝑬𝒋𝝈𝟎𝒋𝒊𝒘𝒆

𝒊𝒘𝒕 + 𝝈𝟎𝒋

𝜼𝒋𝒆𝒊𝒘𝒕

= 𝝈𝟎𝒋𝒆𝒊𝒘𝒕

𝒊𝒘

𝑬𝒋+𝟏

𝜼𝒋=

𝝈𝒋(𝒕)

𝜼𝒋𝒊𝒘τ𝒋 + 𝟏

𝒘𝒉𝒆𝒓𝒆τ𝒋 = 𝜼𝒋

𝑬𝒋


Let the soln. be 𝜺 𝒕 = 𝜺𝟎𝒆𝒊𝒘𝒕, 𝒕𝒉𝒆𝒏

𝜺𝟎𝒊𝒘

𝜺(𝒕)𝒆𝒊𝒘𝒕 =

𝝈𝒋 𝒕

𝜼𝒋(𝒊𝒘τ𝒋 + 𝟏)

∴ 𝝈𝒋 𝒕 = 𝜺 𝒕 𝒊𝒘𝜼𝒋𝟏 + 𝒊𝒘τ𝒋

Total Stress,

𝝈(𝒕) = 𝝈𝒋𝒁

𝒋=𝟏𝒕 = 𝜺(𝒕)

𝒊𝒘𝜼𝒋𝟏 + 𝒊𝒘τ𝒋

𝒛

𝒋=𝟏

∴ Complex Modulus 𝑬∗ 𝒕 = 𝝈(𝒕)

𝜺(𝒕)

𝒊𝐰𝛈𝐣 𝟏 − 𝒊𝐰τ𝐣

𝟏 + 𝒊𝐰τ𝐣(𝟏 − 𝒊𝒘τ𝒋) (𝒘의 함수)

𝒛

𝒋=𝟏

= 𝑬𝒋𝒘

𝟐τ𝒋𝟐

𝟏 + 𝒘τ𝒋 𝟐+

𝒁

𝒋=𝟏𝒊

𝑬𝒋𝒘τ𝒋

𝟏 + 𝒘τ𝒋 𝟐

𝒛

𝒋=𝟏

= 𝑬′ + 𝒊𝑬′′

𝒘𝒉𝒆𝒓𝒆 τ𝒋 = 𝜼𝒋

𝑬𝒋



If Z = 2,

𝑬′ = 𝑬𝟏𝒘

𝟐τ𝟏𝟐

𝟏 + 𝒘𝟐τ𝟏𝟐 +

𝑬𝟐𝒘𝟐τ𝟐

𝟐

𝟏 + 𝒘𝟐τ𝟐𝟐

𝑬′′ =𝑬𝟏𝒘τ𝟏

𝟏 + 𝒘𝟐τ𝟏𝟐 +

𝑬𝟐𝒘τ𝟐𝟏 + 𝒘𝟐τ𝟐

𝟐

If E1 = E2, 𝜼1 =𝜼𝟐인 경우

한 개가 2E1, 2 𝜼1 으로

Transition 이 1개로 표시된다


σ (t)

4. Voigt – Kelvin Model

- Generalized the Voigt Model which results from

connecting Z Voigt Model in series.

- Compliance functions are easily calculated while

the modulus fits are rather complicated.

Creep experiment

σ (t) = const = σ0

- Compliance can be calculated for each element

σ 𝒕 = σ𝟎 = 𝑬𝒋ε𝒋 𝒕 + η𝒋𝒅ε𝒋 𝒕

𝒅𝒕 𝒋 = 𝟏, … , 𝒁

By integration,

ε𝒋 𝒕 =σ𝟎

𝑬𝒋𝟏 − 𝒆

− 𝒕

τ𝒋 (𝒋 = 𝟏, 𝟐,… , 𝒁)

or ε𝒋(𝒕)

σ𝟎= 𝑫𝒋 𝒕 = 𝑫𝒋(𝟏 − 𝒆

−𝒕

τ𝒋)

Total strain

𝜺 𝒕 = ε𝒋(𝒕)𝒁𝒋=𝟏 …………………… (1)

and

𝑫 𝒕 =ε(𝒕)

σ𝟎= 𝑫𝒋 𝒕 = 𝑫𝒋(𝟏 − 𝒆

−𝒕

τ𝒋) …(1a)



Dynamic Exp.

By applying sinusoidal strain to the model the strain expression

for the 𝒋 th element becomes

ε𝒋 𝒕 = ε𝟎𝒋 · 𝒆

𝒊𝒘𝒕 ⃪ (excitation 가하면)

∴σ 𝒕 = 𝑬𝒋ε𝒋 (𝒕) + η𝒋𝒅ε𝒋(𝒕)

𝒅𝒕

= 𝑬𝒋ε𝒋𝒆𝒊𝒘𝒕 + η𝒋𝒊𝒘ε𝟎𝒋 · 𝒆

𝒊𝒘𝒕 = ε𝟎𝒋ε𝒋 𝒕𝒆𝒊𝒘𝒕 𝑬𝒋 + 𝒊𝒘η𝒋

∴ ε𝒋 𝒕 = σ(𝒕)

𝑬𝒋+𝒊𝒘η𝒋……………………………. (2)

Substitution eqn (2) into eqn (1)

ε 𝒕 = σ 𝒕 𝟏

𝑬𝒋+𝒊𝒘η𝒋

𝒛𝒋=𝟏 = σ 𝒕

𝟏

𝑬𝒋+(𝟏+𝒊𝒘η𝒋

𝑬𝒋)

𝒛𝒋=𝟏

→ 𝑫𝒋

→ 𝝉𝒋


𝑫∗ =ε(𝒕)

σ(𝒕)=

𝑫𝒋

𝟏 + 𝒊𝒘𝝉𝒋

𝒛

𝒋=𝟏

(𝟏 − 𝒊𝒘𝝉𝒋)

(𝟏 − 𝒊𝒘𝝉𝒋)

= 𝑫𝒋

𝟏 + (𝒘𝝉𝒋)𝟐− 𝒊

𝑫𝒋𝒘𝝉𝒋

𝟏 + (𝒘𝝉𝒋)𝟐

= 𝑫′ − 𝒊𝑫′′

For Z = 2

𝑫′ =𝑫𝟏

(𝟏 + 𝒘𝝉𝟏)𝟐+

𝑫𝟐

𝟏 + (𝒘𝝉𝟐)𝟐

𝑫′′ =𝑫𝟏𝒘𝝉𝟏

𝟏 + (𝒘𝝉𝟏)𝟐+

𝑫𝟐𝒘𝝉𝟐𝟏 + (𝒘𝝉𝟐)

𝟐 log w


B. Distribution of Relaxation & Retardation times.

- For the Maxwell – Weichert model, it could be easily calculated.

Three spring constant (𝟏𝟎𝟏𝟎, 𝟏𝟎𝟖, 𝟏𝟎𝟔) are calculated with relaxation

time (𝟏𝟎𝟐, 𝟏𝟎𝟒, 𝟏𝟎𝟔)

유한갯수에선 연속적

으로 표시되지만

무한일 경우에는

continuous 하게 표시


If Z is large

E(t) = 𝑬𝒊𝒆−𝒕

τ𝒊

𝒁𝒊=𝟏 Various 𝑬𝒊’s are replaced by a continuous function E(τ).

Summation in the above equation can be replaced by integration.

E(t) = E(τ)∞

𝟎 𝒆−

𝒕

τd τ …… distribution of relaxation times

(완화계수 τ 인 modulus E → E (t), t 시간의 E)

- To apply for all time scales define

H(τ) = E(τ)∙ τ

∴ E(t) = 𝑯(τ)τ

∞

𝟎𝒆−

𝒕

τ𝒅τ

= 𝑯(τ)𝒆−𝒕

τ𝒍𝒐𝒈τ= +∞𝒍𝒐𝒈τ = −∞ 𝒅lnτ

Creep 완화 실험인 경우 → retardation time spectrum

𝑯(τ) 𝒅lnτ = Relaxation time spectrum ( = concentration of Maxwell elements

with relaxation times between lnτ and lnτ + 𝒅τ)


- Tobolsky has suggested that the stress

relaxation modulus of NBS polyisobutylene:

it is composed of a “Box and Wedge”

H(τ) for NBS polyisobutylene

- In this particular case, Tobolsky gives the

pertinent parameter the following values:

M = 8.9 × 𝟏𝟎𝟑 Pa 𝒔𝟏

𝟐,

𝑬𝟎 = 7.2 × 𝟏𝟎𝟓 Pa, τ𝟏 = 𝟏𝟎−𝟏𝟐.𝟓 , τ𝟐 = 𝟏𝟎−𝟓.𝟒 s,

τ𝟑 = 𝟗. 𝟔𝟓 × 𝟏𝟎−𝟐𝟔 𝑴𝑾𝟑.𝟑𝟎s,

τ𝒎 = 𝟏. 𝟎𝟔 × 𝟏𝟎−𝟐𝟎 𝑴𝑾𝟑.𝟑𝟎s


- Clearly the wedge is independent of molecular weight and

gives rise to the primary transition.

- The box portion of the spectrum generates the rubbery plateau and

rubbery flow regions of the master curve.

- These regions are strong functions of the molecular weight, and the

box portion of the spectrum mirrors this fact in the dependence of

τ𝟑 and τ𝒎 on molecular weight.


C. Molecular theories - Isolated polymer chain (고체) 은 dilute polymer solution 에 의해 (서로

interaction 이 없음) long freely orienting molecule이 된다.

= Hookean entropy springs (polymer chain 의 spring 화)

- Spring 의 Restoring force

𝒇 = 2kT𝒃𝟐 𝒓= 𝟑𝒌𝑻∆𝑿

𝒓 𝟐 , ∆x = distorted amount

(𝒃𝟐 =𝟑

𝟐𝒓 𝟐 ), 𝒓 𝟐 : mean square end to end distance


- Molecular theory subdivides the

polymer molecule into subunits or

just long enough, so that the

distribution of 𝒓 of each of these

units is Gaussian (real polymer

molecules in dilute solution).

- Submolecule 의 질량은 Bead 에 다 차

지되고, spring 에는 질량이 없다고 생각

한다.

- Consider the response of the

system to a un-directional

perturbation in x-direction.

(Spring – Bead Model)

submolecule


- Effective spring constant (random 을 일직선상에 있다고 생각한다) in x-

direction allows to portray the system as a 1-dimension chain.

Linearization (선형화)

- Subdivide the polymer molecule into Z submolecules z – springs, z +1

beads

- Restoring force on each bead

𝑿𝟏 만큼 변형, (𝑿𝟏- 𝑿𝟎) 에 의한 힘

force


𝒇𝟎𝑿 = −𝟑𝒌𝑻

𝒂𝟐𝑿𝟎 − 𝑿𝟏

k(𝑿𝟏 − 𝑿𝟎) k(𝑿𝟐 − 𝑿𝟏),

k = spring constant

K((𝑿𝟐-𝑿𝟏) - (𝑿𝟏 −𝑿𝟎)) = k(𝑿𝟐-2𝑿𝟏+𝑿𝟎)

𝒇𝟎𝑿 = −𝟑𝒌𝒕

𝒂𝟐𝑿𝟎 − 𝑿𝟏

𝒇𝟏𝑿 = −𝟑𝒌𝒕

𝒂𝟐−𝑿𝟎 + 𝟐𝑿𝟏 − 𝑿𝟐 ………………… (1)

𝒇𝒊𝑿 = −𝟑𝒌𝒕

𝒂𝟐−𝑿𝒊−𝟏 + 𝟐𝑿𝒊 − 𝑿𝒊+𝟏 1 ≤ i ≤ z-1

𝒇𝒛𝑿 = −𝟑𝒌𝒕

𝒂𝟐−𝑿𝒛−𝟏 + 𝑿𝒛


Where 𝒇𝒊𝑿 = force on the 𝒊 th bead in x-direction

𝑿𝒊 = amount by which the bead 𝒊 has been displaced from its

equilibrium position.

𝒂𝟐 = m. s. end-to-end distribution of the submolecule

탄성력 외에 점성력이 작용

- An additional force acts on the molecule due to viscous nature of the

medium.

- Beads move in the manner of sphere through a viscous solvent drag force

on each bead is

𝒇𝒊𝑿′ = ρ

𝒅𝑿𝒊

𝒅𝒕= ρ𝑿 𝒊

- where ρ is segmental (submolecular) friction factor


- Assume the forces due to the acceleration of beads are small, then

Elastic force = Viscous force

ρ 𝑿 𝟎 = − 𝟑𝒌𝑻

𝒂𝟐(𝑿𝟎 − 𝑿)

ρ 𝑿 𝒊 = − 𝟑𝒌𝑻

𝒂𝟐(−𝑿𝒊−𝟏 + 𝟐𝑿𝒊 − 𝑿𝒊+𝟏) (2)

ρ 𝑿 𝒛 = − 𝟑𝒌𝑻

𝒂𝟐−𝑿𝒛−𝟏 − 𝑿𝒛 1 ≤ 𝒊 ≤ z-1


- Linear first order differential equation → Matrix

[𝑿 ] = -B[A][X]………(2a)

where [𝑿 ] =

𝑿 𝟎𝑿 𝟏...𝑿 𝒛

[x] =

𝑿𝟎

𝑿𝟏...𝑿𝒛

and B = 𝟑𝒌𝑻

𝒂𝟐ρ

- These are diff. eqn resulting from coupling of the motions of beads.

(coupling 되지 않는 것으로 만든다.)

- To solve these equations, define a new set of coordinates, q𝒊,

made up of a linear combination of x𝒊’s

(z+1)× (z+1) square matrix

coupling 되어 있으므로 서로 영향을 미친다


- New coordinates will be define as

𝒒𝒊 = 𝑸𝒊𝒋𝑿𝒋𝒋 or in matrix notation (technique of normal coordinate)

[q] = [Q] [X] ……… (3)

Where [q] & [X] are column matrix, [Q] is a (z+1) × (z+1) square matrix

- [X] 를 [q] 로 바꾸어서 각각 bead 운동이 다른 운동에 영향을 미치지 않도록

바꾸어 보자.

- Define [Q] in such a way that it will bead to the solution of (2)

[𝑿 ] = -B[A][X] (2a)

- Express eqn (2) in terms of normal coordinates of time derivatives.


- To diagonalize the matrix [A], there exists another matrix [Q] such that

[𝑸−𝟏][A][Q] = [Λ] (diagonal matrix)

[𝑸−𝟏][Q] = [I]

premultiply [𝑸−𝟏] to eqn (2a), (2)식에서 [𝑿 ]=-B[A][X]

[𝒒 ] ← [𝑸−𝟏][𝑿 ] = -B [𝑸−𝟏] [A] [Q] [𝑸−𝟏] [X]……(4)

Λ Identity

If we set [𝑸−𝟏][𝑿 ] = [𝒒 ] , and [q]

[𝑸−𝟏][X] = [q]

[𝒒 ] = -B[Λ][q]……………………………(5)


- the 𝒑 th equation in this set reads 𝒒 𝒊 = -Bλ𝒊q𝒊

- Since the normal coordinate approach merely represents a

linear transformation of the real coordinates, the motion of the polymer

represented by all the q𝒊′s will be identical to the motion of the polymer

represented by all the X𝒊′s.

𝒒𝒊 = 𝑸𝒊𝒋𝑿𝒊

𝒋

- Our problem becomes the rather simple case of finding a diagonal

representation of the (z+1)×(z+1) matrix [A]

diagonal matrix [Λ]


𝒒𝒑 = −𝑩 𝝀𝒑 𝒒𝒑

𝝀𝒑 = 𝟒 𝒔𝒊𝒏𝟐𝒑𝝅

𝟐 𝒛 + 𝟏 𝒑 = 𝟏~𝒛

Which can be directly integrated to give

𝒒𝒑 𝒕 = 𝒒𝒑 𝟎 𝒆−𝑩𝒕𝝀𝒑 = 𝒒𝒑 𝟎 𝒆−𝒕/𝝉𝒑

𝝉𝒑 =𝟏

𝑩𝝀𝒑 , 𝒑 = 𝟏~𝒛

B = 𝟑𝒌𝑻

𝒂𝟐ρ


𝒒𝒑(𝟎) is the volume of the normal coordinate at time zero, that is, at the

application of a perturbation, and 𝒒(𝒕) is the volume of the coordinate

at time t.

The coordinate response is exponential and the system response is the sum

total of all the coordinate responses.

Although boundary conditions corresponding to these real experiments are

more complicated in terms of normal coordinates, it can be shown that the

relaxation times that arise in a stress-relaxation experiment are just one-half

as large as those calculated above

𝝉𝒑 =𝟏

𝑩𝝀𝒑 𝒑 = 𝟏~𝒛


Therefore, 𝝉𝒑 will be need to denote a relaxation time and is given by

𝝉𝒑 =𝟏

𝟐𝑩𝝀𝒑=

𝟏

𝟖𝑩𝒔𝒊𝒏𝟐(𝒑𝝅

𝟐 𝒛 + 𝟏)

𝝀𝒑 = 𝟒𝒔𝒊𝒏𝟐(𝒑𝝅

𝟐 𝒛 + 𝟏)

Stress relaxation behavior for bead and spring model

𝑬 𝒕 = 𝑬𝒑𝒆−𝒕/𝝉𝒑

𝒛

𝒑=𝟏

𝒇 =𝟑𝒌𝑻

𝒂𝟐∆𝒙

𝝈 =𝒇

𝒃𝟐=

𝟑𝒌𝑻

𝒃𝟐𝒂𝟐∆𝒙

𝒇: 𝒆𝒍𝒂𝒔𝒕𝒊𝒄 𝒇𝒐𝒓𝒄𝒆 𝒃𝟐: 𝒂𝒗𝒈. 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂


𝑬 𝟎 =𝝈

𝜺=𝟑𝒌𝑻

𝒃𝟐𝒂

𝒂𝒃𝟐 =𝟏

𝒄𝒛

𝑬 𝟎 = 𝟑𝒌𝑻𝒄𝒛 = 𝑬𝒑

𝒛

𝒑=𝟏

Assumed that all of the individual 𝑬𝒑’s are equal

𝑬𝒑 = 𝟑𝒄𝒌𝑻

We have 𝝉𝒑’s and 𝑬𝒑’s of Maxwell Weichert model. We may write

𝑬′ 𝒘 = 𝟑𝒄𝒌𝑻 𝝎𝟐𝝉𝒑

𝟐

𝟏 + 𝝎𝟐𝝉𝒑𝟐

𝒛

𝒑=𝟏

𝑬 𝟎 : 𝑰𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒕𝒆𝒔𝒊𝒍𝒆 𝒎𝒐𝒅𝒖𝒍𝒖𝒔

𝒛: 𝒔𝒖𝒃𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝒏𝒖𝒎𝒃𝒆𝒓 𝒄: 𝒑𝒐𝒍𝒚𝒎𝒆𝒓 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏


D. Applications of Flexible Chain Models to Solutions 𝒂𝟐, 𝝆 : may not be easily evaluated → eliminating segmented friction factor

𝒂𝟐: mean square end-to-end distance of sub molecule

𝝆 : segmental friction factor

Solution viscosity in excess of solvent viscosity can be thought as arising

from the dissolved polymer, Excess viscosity is sum of viscosity of each

element

𝜼 − 𝜼𝒔 = 𝑮𝒑𝝉𝒑

𝒛

𝒑=𝟏

= 𝒄𝒌𝑻 𝟏

𝟐𝑩𝝀𝒑

𝒛

𝒑=𝟏

=𝒄𝒂𝟐𝝆

𝟐𝟒

𝟏

𝒔𝒊𝒏𝟐 𝒑𝝅/𝟐(𝒛 + 𝟏)

𝒛

𝒑=𝟏

𝝉𝒑: 𝐫𝐞𝐥𝐚𝐱𝐚𝐭𝐢𝐨𝐨𝐧 𝐭𝐢𝐦𝐞

𝒔: 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝜼: 𝐬𝐡𝐞𝐚𝐫 𝐯𝐢𝐬𝐜𝐨𝐬𝐢𝐭𝐲 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝑮𝒑: 𝐬𝐡𝐞𝐚𝐫 𝐦𝐨𝐝𝐮𝐥𝐮𝐬

B = 𝟑𝒌𝑻

𝒂𝟐ρ


For small values of 𝒙, 𝒔𝒊𝒏𝒙 ≅ 𝒙

≅𝒄𝒂𝟐𝝆

𝟐𝟒

𝟒𝒛𝟐

𝝅𝟐∙𝝅𝟐

𝟔 ≈

𝒄𝟐𝒂𝟐𝝆𝒛𝟐

𝟑𝟔

𝝉𝒑 =𝒂𝟐𝝆𝒛𝟐

𝟔𝝅𝟐𝒌𝑻𝒑𝟐 𝒇𝒐𝒓 𝒑 < 𝒛

∴ 𝝉𝒑 =𝟔(𝜼 − 𝜼𝒔)

𝒄𝒌𝑻𝝅𝟐𝒑𝟐

Excess viscosity and solution concentration are known or are easily

measured


E. Zimm Modification Velocity gradient of solvent would be perturbed deep inside a coiled

polymer molecules. → shielding

Zimm : Hydrodynamic shielding

Relaxation times according to Zimm treatment

𝝉𝒑 =𝟏. 𝟕𝟏 𝜼 − 𝜼𝒔

𝒄𝒌𝑻𝑲𝒑


Hydrodynamic

shielding

: tightly coiled

polymer

conformation

Polystyrene


F. Extension to Bulk Polymer No solvent is present in the bulk state

𝝉𝒑 =𝟔𝜼 − 𝜼𝒔𝒄𝒌𝑻𝝅𝟐𝒑𝟐

대신𝟔𝜼

𝑵𝒌𝑻𝝅𝟐𝒑𝟐

What molecular motion does 𝜼 depend upon?

𝜼 = 𝑮 𝒕 𝒅𝒕

∞

𝟎

= 𝒕𝑮 𝒕 𝒅 𝒍𝒏 𝒕

∞

−∞

Critical length of entanglement (polystyrene) = 30,000

𝝆𝟎 : friction factor shorter than critical relaxation time 𝝉𝒄 𝝆 : friction factor longer than 𝝉𝒄

𝜼 : steady flow shear viscosity of Bulk polymer


𝝉𝒑 =𝝆𝟎𝒂𝟐𝒛𝟐

𝟔𝝅𝟐𝒌𝑻𝒑𝟐 (𝝉𝒑 < 𝝉𝒄), 𝝉𝒑 =

𝝆𝒂𝟐𝒛𝟐

𝟔𝝅𝟐𝒌𝑻𝒑𝟐 (𝝉𝒑 ≥ 𝝉𝒄)

𝒍𝒐𝒈 𝝆 𝝆

𝟎 = 𝟐. 𝟒 𝐥𝐨𝐠 𝑴 𝑴𝒄

𝑴𝒄: 𝒄𝒊𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒎𝒐𝒍.𝒘𝒕 𝒇𝒐𝒓 𝒆𝒏𝒕𝒂𝒏𝒈𝒍𝒆𝒎𝒆𝒏𝒕

𝜼 = 𝑮𝒑𝝉𝒑

𝒛

𝒑=𝟏

=𝑵𝒌𝑻𝝆𝒂𝟐𝒛𝟐

𝟔𝝅𝟐𝒌𝑻

𝟏

𝒑𝟐

𝒑𝒄

𝒑=𝟏

∴ 𝜼 ∝ 𝐌𝟑.𝟒

N (number of chain per cubic centimeter) is inversely proportional to mol.wt

Z (number of submolecules per polymer molecules) is directly proprtional to mol.wt


G. Reptation Motion of polymer molecules has reptation(creeping) in terms of a wormlike

motion of polymer molecules through the matrix formed by its neighbors.

with obstacles

in a tube


Figure 5.9 The chain is considered within a tube (a) Initial position (b) The

chain has moved to the right by reptation (c) The chain has moved to the left,

the extremity / choosing another path, 𝑰𝟐𝑱𝟐. A certain fraction of the chain,

𝑰𝟏𝑱𝟐, remains trapped within the tube at this stage(54).


According to De Gennes

𝒗𝒄 = 𝝁𝒕𝒖𝒃𝒆𝒇 ---- ①

Let n be the number of monomer unit per molecules

To obtain the same velocity 𝒗𝒄 with various length molecules (various of n)

Force must be directly proportional to n

𝒗𝒄

𝝁𝒕𝒖𝒃𝒆= 𝒇 ∝ 𝒏, 𝝁𝒕𝒖𝒃𝒆 =

𝝁𝟏

𝒏 ----- ②

𝒇: 𝒔𝒕𝒆𝒂𝒅𝒚 𝒇𝒐𝒓𝒄𝒆, 𝒗: 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝝁𝒕𝒖𝒃𝒆:𝒎𝒐𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝒊𝒏 𝒕𝒖𝒃𝒆

(Under a steady for chain moves with a velocity)

Knowing the molecular mobility allows one to calculate the diffusion constant

through the Nerst-Einstein equation.

𝒂 = 𝒌𝑻𝝁𝒕𝒖𝒃𝒆 =𝒌𝑻𝝁𝟏

𝒏 =𝒂𝟏

𝒏 ------ ③

𝒌:𝑩𝒐𝒍𝒛𝒎𝒂𝒏𝒏′𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, 𝝁𝟏: 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒐𝒇 𝒄𝒉𝒂𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝟏 𝒊𝒔 𝒂𝒍𝒔𝒐 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒐𝒇 𝒄𝒉𝒂𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉


From the diffusion study

𝑫 = 𝒙𝟐𝟐𝒕

The direct combination of equation ③ and ④ gives

𝝉𝒎𝒂𝒙 ∝𝑳𝒕𝒖𝒃𝒆𝟐

𝒂=𝑳𝒕𝒖𝒃𝒆𝟐

𝒂𝟏∝ 𝒏𝟑

Viscosity : 𝜼 = 𝝉𝑬

∴ 𝜼 ∝ 𝒏𝟑 It is known from experiment that viscosity actually varies as the 3.4 power of

molecule weight.

𝒙: 𝒂𝒗𝒈. 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒂 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝒎𝒐𝒗𝒆𝒔 𝒊𝒏 𝒂 𝒎𝒆𝒅𝒊𝒖𝒎 𝒐𝒇 𝒅𝒊𝒇𝒇𝒖𝒔𝒊𝒐𝒏 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑫

--- ④

(m𝒂𝒙. 𝒓𝒆𝒍𝒂𝒙𝒂𝒕𝒊𝒐𝒏 𝒕𝒊𝒎𝒆)

𝑳𝒕𝒖𝒃𝒆𝟐 : tube length ~ directly proortion to the polymer chain length n

Max relaxation time is predicted to depend on chain length to the third power


Diffusion coefficient, D, of a chain in the entangled polymer matrix depends

on the molecular weight M

𝑫 ∝ 𝑴−𝟐 Diffusion coefficient in bulk system range from 𝟏𝟎−𝟐 to 𝟏𝟎−𝟔 ㎠/㎱

Ex. PE : 𝟏 × 𝟏𝟎𝟒 g/mol

D = 𝟏 × 𝟏𝟎𝟓㎠/s at 176℃

PS : 𝟏 × 𝟏𝟎𝟓 g/mol

D= 𝟏 × 𝟏𝟎−𝟏𝟐 ㎝/s at 175℃

chapter 7. viscoelastic models - seoul national...

Documents