Deformation of Polymers

4 Lectures for Paper C3 Prof. J.C. Tan

Hilary Term 2016

1

Department of Engineering Science

University of Oxford

Contents of the lectures

1. Introduction2. Preliminaries (B5 lectures): Large

deformations & rotations, Stresses in the presence of finite deformations and rotations

3. Elasticity of polymers: Linear elasticity; Hyperelasticity; Phenomenological & physically-based models

4. Viscoelasticity: Linear viscoelasticity in 3D; Relaxation and retardation spectra; Thermal activation; Glass transition …

2

1. Introduction• The subject of these lectures is ‘materials

modelling’ applied to the deformation of polymers.

What does this mean?

• ... the business of describing quantitatively, via

mathematics, the 3D deformation that occurs in polymer solids under the action of applied loads.

3

Why is this subject important in Engineering?

• Designers of manufacturing processes and of products need to be able to predict quantitatively in advance the deformation that will occur under applied loading.

• Materials engineers, developingand selecting materials for particular applications, need to predict how the (3D) deformation of a polymers depends on their composition and structure.

4

O-rings, gears, impellers, …

Orthopaedic implants (UHMW-PE knee meniscus)

Computer-based engineering analyses

(RiRo) “Rubbish input gives rubbish output”

• So it is essential to feed FE analyses with good models of the material behaviour!

5

• Engineers nowadays use finite element (FE) numerical analysis to predict the complex patterns of stress and strainthat usually arise in 3D ...

• But as always, such computer-based analyses follow the general rule of computer programs:

FE model of an artificial knee

implant

6

How do we create mathematical models of material behaviour?

• Where does the information come from?

• It comes to us from two sources at two

different length-scales, which comprise the “macro-world” and the “micro-world”

This is the everyday world you are familiar with, visible with the naked human eye!

(a) A few known physical laws applying at this length scale (e.g. Newton’s laws, the laws of thermodynamics...)

(b) Our idealisations of the material (a single continuum, assume some known symmetry...)

(c) Experimental data of specific materials, specific test conditions & parameters, specific samples...

“Phenomenological” Models

7

(I) “Macro-world” > 1 mm à few m

Handouts P.4-5

(II) “Micro-world” < 1 mm(a) Using microscopy and other advanced

characterisation tools (eg. EMs, X-rays, neutrons) to gain insights into materials

composition and microstructure

(b) Known physics and chemistry provide useful theories to devise more powerful models that capture more aspects of material behaviour

(c) Guide us in making predictions of how the material is likely to behave in the macro-world

“Physically-based” Models

8

Synchrotron radiation(Diamond Light Source @ Harwell)

TEM

What polymers are?• Polymers are materials consisting of very long

molecules, made up of many repeating chemical units (the monomer units), covalently bonded together.

9

Amorphous (e.g. PS, PMMA, LDPE, atactic PVC)

Semicrystalline (e.g. HDPE, Nylon, isotactic PVC)

Handout P.5

Molecular topology

10

branched

crosslinked

lineare.g. linear PE (“high density polyethylene” or HDPE)

e.g. branched PE with 6 - 10 branches per 100 carbon atoms (“low density polyethylene” or LDPE)

e.g. cross-linked polyethylene

What are the key features of the chemical architecture of a polymer?

1. Chemical composition(s) of its monomers

2. The number of monomers in each molecule (the ‘degree of polymerisation’ = n)

3. The manner of covalent connection between the monomers (the ‘molecular topology’)

11

* C

H

H

C

H

H*n

Polyethylene (PE)(2 main-chain covalent bonds (nb0), 28 amu: molar mass M0 = 28 g/mol)

C O* C

O

O *n

CH3

CH3

Polycarbonate(6 main-chain covalent bonds, M0 = 254 g/mol)

Stretched out lengths of polymers

12

Let the number of main-chain bonds per monomer be nb0

Then the total number of such bonds per molecule is

nb = n nb0

Let the mean main-chain bond length, as projected onto the axis of the

molecule when stretched out straight, be bp

Then the stretched out length of one monomer, L0, and that of the whole molecule, L, are respectively

0 b0 p 0 b p; L n b L nL n b= = =

a polyethylene (PE) molecule

bp

Molecule

axis

(1.1)

Handout P.7

n

b

Structure of polymer crystals

13

Its unit cell

the chains pack parallelalong c-axis

looking down c-axis

(254 pm)= 0.254 nmc−

• The crystals are tiny (10–50 nm thickness)

e.g. A lamella-shaped single crystal of polyethylene (PE)

Th

ickn

ess

c-a

xis

The molecules are aligned normal to the lamella-shaped crystal

looking down c-axis

TEM

“Microstructures” of spherulites in PE

14

In a polymer crystallised from the melt, instead of lamellaeforming independently, stacks of them grow outwards together from each nucleus, forming a ‘spherulite’.

200 μm“Maltese crosses” visible under cross-polarisers

Handout P.7

Sizes of typical structural features of polymers

15

Typical size of polymer-based engineering

components/products in the “Macro-world”:

a few millimetres (mm) up to several metres (m)

“Micro-world”

Handout P.8

vs.

2. Preliminaries (B5 Revision)

16

2.1 Large deformations and rotations

• We must use a mathematical description that does not suffer from restriction to small strains

• Require a formulation that is valid for arbitrarily large deformations and rotations

The vector displacement:

u = x ─ X

Assume it is a continuum:the new position x is a smooth function of the old position X

Handout P.9

Deformation gradient F

17

• It follows that we can arbitrarily pick another point in the material, initially at position Q0, close to P0,

• ... track it to its new positionQ, by applying a transformation F to their initial vector separation dX

to obtain their new vector separation dx

componentd d ; term or in ds di ij jx F X= =x F X

Fij =

∂xi∂X j

= ∇0xF is called the Deformation GradientIt fully specifies the deformation and rotation of the sample of material as it transforms from B0 to B

Handout P.10

Rigid body rotation R

18

F can always be expressed as the product of an orthogonal matrix R and a symmetric matrix U or V:

= =F VR RU• R corresponds to a pure rigid body rotation

• U and V correspond to pure deformations

• The eigen values of U and V are equal (i.e. differ by only

a rotation), and they are called the Principal Stretches

1 2 3, ,λ λ λnew lengthoriginal length

λ =Definition:“stretch”

Left Cauchy Green tensor B

19

“Principle of Material Frame Indifference”:• Stresses that give rise to the deformation must be

independent of any rigid body rotation R

• Consequently, to describe the deformation we do not use F itself, but instead we could use U or V

• ... but actually it turns out to be more convenient to use instead the left Cauchy Green tensor B or rightCauchy Green tensor C defined

T T ; = =B FF C F FHere we shall use:

The eigen values of both B and C are2 2 21 2 3, ,λ λ λ

(3x3 Symm.)

Handout P.11

True stress in 3D

20

Stresses in the presence of finite deformations and rotations

Δa

ΔPn

0aa Δ →

Δ=ΔPt

Assuming a continuum, we definetraction vector t as the local force

per unit area, in the limit

Cauchy Stress Tensor σ

component te or in rms i ij jt nσ= =t σn

This is a measure of true stress; it relates the traction vector t to the direction n of the normal of the plane on which it is acting:

σij is the i-th component of the force per unit current area acting on a plane whose normal is currently in the j-th direction.

Handout P.14

t

n ( i )( j )

3. Elastic Deformation of Polymers

21

3.1 The general case

WQ

U S,

; QU Q W STΔ

Δ = Δ + Δ Δ ≥

Helmholtz Free Energy (per unit initial volume) A ≡ U ─ TS

Under some circumstances solid polymers are elastic in their mechanical response, particularly at • very low temperatures, and • at high temperatures (>Tg) if

they are crosslinked or sufficiently entangled

Polymer sample as a thermodynamic system with U and S at absolute T

First and Second Laws of Thermodynamics:

• the equality applies to a reversible process = “Elastic”• the inequality applies to an irreversible process

Rate of work done (power supplied) by applied loads:

Handout P.15

!W ≥ !U −T !S = !A+S !T

T

Elastic deformation of a polymer at constant temperature

22

We define a material as elastic if deformation under applied loads is a reversible process (thus “=” sign).

But at constant temperature

Rate of change in the free energy

Power supplied to the system

To complete a model, we need to express in terms of σ, and in terms of B. HOW to establish such relationships?

!W = !A+S !T

!T = 0

!W = !A !W

!A

Linking Cauchy stress σ to B

23

... By reading handouts P.16 to P.20, we can show that:

0 2 Aρ

ρ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

σ BB

( )2 21 2 3

1 tr ; tr tr ; det 2

I I I⎡ ⎤= = − =⎢ ⎥⎣ ⎦B B B B

(3.7)

(3.11)

Cauchy stress

tensor (symmetric, σij = σji)

• Isotropic invariants of B applicable to isotropic materials

density ratio

• Note that we have not yet made any restriction to small deformations• Eqn. (3.7) is valid for hyperelastic material (large elastic deformations)

Essentially the 3 eigen values of B

3.3 Hyperelasticity: Phenomenological modelling

24

0 1 2 3 3

old volume 1 1 1new volume det I

ρρ λ λ λ

= = = =B

• A material that is elastic even at large deformations is called “hyperelastic”, i.e. rubbery

Handout P.23

What happens to the volume during deformation?

(3.21)

Since mass is constant, the density ratio after deformation is

new lengthold length

λ =Principal stretch: 2 2 2

3 1 2 3detI λ λ λ= =B3rd isotropic invariant

Hyperelastic “rubbery” polymers

25

Experimentally, hyperelasticity is seen in polymers onlyunder certain conditions:

• at temperatures above the glass transition, T >Tg (rubbery)

• when the polymer is amorphous• when there is connectivity between the molecules

(chemical crosslinks or entanglements providing ‘physical crosslinks’).

- Under these conditions, the molecules move very freely past each other, but they are reluctant to move apart from each other (volume conserved!).

Handout P.23

G << K and ν ~0.5

26

• In hyperelastic polymers, the change of volume (K) meets much more resistance than change of shape (G), i.e.

• e.g. Typically G ~ 1 MPa, while K ~ 1 GPa• Hence ν ~ 0.5, i.e. material is:

INCOMPRESSIBLE(constant volume)

G≪K

ν = 12

1− 2G3K

⎛

⎝⎜

⎞

⎠⎟

1+ G3K

⎛

⎝⎜

⎞

⎠⎟

!12

(3.20)

In light of these observations, there are twoconsequences for modelling hyperelasticity...

Handout P.24

Modelling rubbery polymers

27

(a) I3 =1 (see eqn (3.21)), since incompressible A depends on only I1 and I2 and the density ratio is unity

(b) The hydrostatic component of stress (–p) is indeterminate from the constitutive model alone, and must be found from the boundary conditions (BCs).

Thus for a known deformation state, the Cauchy stress is obtained from either:

( ) ( )1 2 1 2 3, , ,2 or 2A I I A

p pλ λ λ∂ ∂

= − = −∂ ∂

σ B σ BB B

I I

(3.22)

1 2 3 1λ λ λ =

Handout P.24

Cauchy stress expressed in terms of the isotropic invariants

28

For the models expressed in terms of the invariants, it is

helpful to evaluate first their differentials with respect to B

21

1 2 22 A A AI p

I I I⎡ ⎤⎛ ⎞∂ ∂ ∂

= + − −⎢ ⎥⎜ ⎟∂ ∂ ∂⎝ ⎠⎣ ⎦σ B B I (3.25)

Handout P.25

Hydrostatic stress -p (pressure) to be found

from the boundary conditions (BCs)

Phenomenological hyperelastic models

29

The parameters C0, C1, C2, CG, Im, µn and αn are material parameters, to be found empirically by fitting the models to experimental data.

Handout P.25

Example problems

30

Handouts P.26-27

neo-Hookean model vs. experiments

31

• Relatively simple to use as it involves only one material parameter C0

• Predicts that the Cauchy stress is a linear function of B

• Limitations?

- It cannot fit real data for rubbery polymers to within accuracy acceptable for many purposes, beyond a nominal strain of about 20% (λ ~ 1.2)

( )0 1 3A C I= − Table P.25

λ ~ 1.2

Tension

Compression

02C p= −B Iσ (3.26)

Handout P.28

expt.

neo-Hookean vs. uniaxial tensile data

32

Data are plotted as nominal stress (s) versus nominal strain (λ ─ 1)

Nominal stress s is found from true stress σ by exploiting the incompressibility:

1111s

σλ

= (3.30)

There are two systematic errors: (1) over-estimate of stress at intermediate values of λ, and (2) under-estimate of stress at highest value of λ.Potential solutions...?1) Mooney-Rivlin; 2) Gent

nominal strain λ − 1

0 1 2 3 4 5 6 7

nom

inal

stre

ss (M

Pa)

0

1

2

3

4

5

6

7

Exp.

Strain softening

Strain stiffening

Handout P.29

Natural rubber

Example problem 3

33

Handout P.30

Hyperelasticity: Physically-based model

34

Chemical structures of some common polymers that are hyperelastic at room temperature (RT)

* C

H

H

C C

H

C

H

H*

n

CH3

* C

H

H

C

H

C

H

C

H

H*n

1,4 polyisoprene (‘natural rubber’)

1,4 polybutadiene

* O Sin

CH3

CH3

*

poly(dimethyl siloxane) (PDMS ‘silicone rubber’)

C* C

H

H

C

H

C

H

H*

n

Cl

1,4 polychloroprene (‘neoprene’)

• These are all flexible molecules with Tg < RT• They are liquids at RT until they are chemically crosslinked

(e.g. using sulphur) at discrete points, • ... creating a 3D molecular network that cannot flow, and

hence forms a soft solid.

Handout P.31

Entropic elasticity

35

• But even though crosslinked, there is vigorous Brownian motion above Tg, as they would in the liquid state

• Between crosslinks, the molecular chains move freely relative to each other

• The network changes shape with negligible cost in its internal energy U

• The free energy of deformation Acan be considered entirely entropic in origin (recall: A = U – TS)

A! −TS

U≪TSSome resistance to

deformation comes entirely from the resistance of the system to reduction of its entropy

Handout P.32

Crosslink

Rapid changes of shape

Physical model assumptions

36

r

e2

e3

e1

Kuhn segment • Model derived following the approach first suggested by Kuhn

• Imagine each long molecule has a set of rigid segments (nK, LK)

• ... connected together by universal joints as if it were a freely-jointed chain à “Kuhn chain”

• For such an “imaginary” chain molecule, the direction of each segment is, by definition, totally uncorrelated to the directions of the other segments.

LK = length of each Kuhn segmentnK = no. Kuhn segments per chainr = instantaneous end-to-end separation (a broad distribution); the directions of r will be random.

a Kuhn chain

Crosslink(or entanglement)

Handout P.33

nKLK

3D random walk

37

• In the absence of external forces (subscript 0) acting on the network, there will be equal numbers of bonds pointing in all directions

• The path r from one end of the chain to the other is a 3D “random walk”. This allows us to do some statistics on it, c.f. handout P.34...

2 2K K0

r n L= (3.34)NB. <…> denotes avg. over the whole population of chains

• It is of interest to compare the unperturbed (i.e. coiled up & compact configuration) root-mean-square end-to-end distance rrms

to the stretched-out length of the chain L:

2rms K K K K K0 rms

; ; hence Lr r L n L L n nr

≡ = = = (3.35)

Handout P.34

maxλ

..with externally applied stresses..

38

e2

e3

e1

i-th Kuhn segment

f

-f

• When stresses are applied to the polymer in the macro-world, in the micro-world this corresponds to microscopic forces being applied to each individual chain via the crosslinks.

• The forces perturb the chain from its random shape, and thereby reduce its entropy

Crosslink -or-Entanglement

• Derivation of the relationships between magnitude of applied force fand end-to-end distance r, and the final link to continuum Cauchy

stress σ ... See full details in your Handouts P.36-40!

Handout P.35

Externalforces

r

“Inverse Langevin” Model

39

1max

max

3NkT pλ λ

λ λ− ⎛ ⎞⎛ ⎞

= −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

σ B IL (3.49)

where

is the Langevin function

approximated using the Padé expression

is the rms stretch λ1

max K max3n Iλ = =

113 3

trI= = B(3.46)

(3.35)

(3.48)

Maximum stretch

N chains per unit volume

Handout P.41

Hydrostatic stress -p (due to intermolecular

potentials; vdW) to be found from BCs, as this

model assumes incompressibility (K>>G)

L −1 x( )! 3x1− 1

3x2

1− x2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Relating to real molecular dimensions

40

2 2 2 2K K b K K b p0

; rmsr r n L C n b L n L n b∞= = = = =

• To use the Inverse Langevin model to predict the behaviour of a particular rubbery polymer, we need to relate the parameters of the freely-jointed Kuhn chain to the real molecular dimensions

• ... by equating the (a) unperturbed rms end-to-end distance, and (b) stretched-out length, respectively:

(a) Kuhn Real

(3.53)

• If we call Mc the molar mass of the chains between crosslinks, then from the definition of Avagadro’s number NA, N and nb are bothdetermined by Mc thus

cA b b0

c 0 ; MN N n n

M Mρ

= =

(b) Kuhn Real

(3.54)

Characteristic ratio

Handout P.43

Total # bonds between crosslinks

monomer

Bond length

crosslink

# chains per unit volume

Stretched out lengths of polymers

41

Let the number of main-chain bonds per monomer be nb0

Then the total number of such bonds per molecule is

nb = n nb0

Let the mean main-chain bond length, as projected onto the axis of the

molecule when stretched out straight, be bp

Then the stretched out length of one monomer, L0, and of the whole molecule, L, are respectively

0 b0 p 0 b p; L n b L nL n b= = =

a polyethylene (PE) molecule

bp

Molecule

axis

(1.1)

Handout P.7 – RECAP

n

b

Example problem 4

42

Handout P.45

4. Linear Viscoelasticity of Polymers

43

Handout P.46

• Polymers frequently deviate from purely elastic behaviour.

• They are then said to be “inelastic”, in which deformation under applied loads is an irreversible process and hence the net work done is positive

• Net work done is dissipated as heat, or locked inside the material as a permanent change in (micro)structure.

• One form of inelasticity is linear viscoelasticity, occuring at small strains of typically < ~0.005 (0.5%)

• It behaves as a time-dependent, but linear system obeying the Boltzmann Superposition Principle (similar to an electric circuit containing only resistors, inductors and capacitors, see B5/B8).

(4.1) ΔW > !A"∫ dt = 0

• Let the Infinitesimal strain tensor ε becomes vanishingly small relative to unity

• Dilation (vol. strain)

• Deviatoric strain tensor

3D linear elasticity ( )

44

2K GΔ= +σ eI

3Δ

≡ −e ε I

11 22 33trΔ ε ε ε= = + +ε (3.16)

(3.18)

(3.19)

3D equation of linear elasticity for modelling small strains

Bulk modulus Shear modulus

Mean vol. strain

Handout P.20-22 – RECAP (B5, A3)

Cauchy stress

!A"∫ dt = 0

Mean stress Deviatoric stress

3D equation of linear viscoelasticity

45

Handout P.47

For an isotropic linear viscoelastic polymer, the 3D equation of linear elasticity - eqn (3.19) – is then replaced by the corresponding 3D equation of linear viscoelasticity

( ) ( )0 0

d d( ) d 2 dd d

t tt K t u u G t u u

u uΔ

= − + −∫ ∫eσ I (4.2)

(3.19)2K GΔ= +σ eI

time-dependentCauchy stress

3D stress response is fully determined by 2 material time-dependent functions:K(x) = Bulk stress relaxation modulus functionG(x) = Shear stress relaxation modulus function

time step

K(x) G(x)

Convolution integral:

Expressing σ(t) in terms of σm & σ’

46

Handout P.47

Using * to represent a convolution integral, we can express

equation (4.2) in terms of the mean stress σm and deviatoric stress σ’ :

( ) ( )0 0

d d( ) d 2 dd d

t tt K t u u G t u u

u uΔ

= − + −∫ ∫eσ I (4.2)

(4.3)Hydrostatic response

Deviatoricresponse

Experimental results for polymers always show much less time-dependence in the hydrostatic response than in the deviatoric response (e.g. a factor of ~1000 in G(x) relaxation vs factor ~2 in K(x) at Tg).

s =σmI+ ʹs where σm = K ∗ !Δ ; ʹs = 2G∗ !e

Spring-dashpot analogy

47

kRk

yF

c

1 2

1 11 2 2 R

dd 1; ; d Δ d

F FyF F F F k yt c k t

= + = + =

Handout P.49-50

Zener model: Linear viscoelastic behaviour similar to that of a polymer is exhibited by Hookean springs (k) and Newtonian dashpots (c) coupled suitably in series or parallel.... Revise your B5 & B8 notes ...

Zener model captures qualitatively all the key features typical of polymers, e.g. in tensile tests (creep, stress relaxation, dynamic tests/cyclic deformation ...)

The response of this model is defined by three equations:

(4.8)

Defining the two time constants of the system:

( )R

R

Δ;

Δy F yk kc

k kτ τ τ

+= = (4.9)

Relaxation time Retardation time

; F yτ τ>(input = y) (input = F)

!

Zener model

48

Rd dd dy FF yF k yt t

τ τ⎛ ⎞+ = +⎜ ⎟⎝ ⎠

Handout P.50

Combining equations (4.8), and employing time-constant definitions (4.9), gives a single differential equation governing the response of the Zener model:

(4.10)

Now we consider three example cases.

Relaxation time Retardation time

Case 1: Creep and Recovery

49

A step load F0 is applied at time 0 and removed at time tu

( ) ( )( )

( )( ) ( )

0 uR R R

u0 u

R R

1 1 1 1 exp 0Δ Δ

1 1 exp exp .Δ

F

F F

ty F t tk k k k k

t t ty F t tk k k

τ

τ τ

⎧ ⎫⎛ ⎞ ⎡ ⎤⎛ ⎞⎪ ⎪= + − − − < ≤⎜ ⎟⎨ ⎬⎢ ⎥⎜ ⎟⎜ ⎟+ + ⎝ ⎠⎪ ⎪⎣ ⎦⎝ ⎠⎩ ⎭

⎧ ⎫⎡ ⎤⎛ ⎞ ⎛ ⎞− ⎛ ⎞⎪ ⎪= − − − − >⎜ ⎟ ⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎢ ⎥⎝ ⎠⎝ ⎠⎪ ⎪⎝ ⎠ ⎣ ⎦⎩ ⎭

disp

lace

men

t y (m

m)

0.0

0.5

1.0

1.5

2.0

time t (s)0 1 2 3 4 5 6 7

forc

e F

(N)

0.0

0.5

1.0

Handout P.51-53

(4.11)

Input

Response

Let:

R 1 N/mm; Δ 1 N/mm; 1 Ns/mm 1 s; 2 s.y Fk k c τ τ= = = ∴ = =

creep

recovery

0 0 u1 N; 1 mm; 2 s.F y t= = =

The resulting displacement is:

F0

y0

tu

Case 2: Stress Relaxation

50

forc

e F

(N)

-1

0

1

2

3

time t (s)0 1 2 3 4 5 6 7

disp

lace

men

t

y

(mm

)0.00.51.01.5

( )

( ) ( )

R 0 u

u0 u

Δ exp 0

Δ exp exp .

y

y y

tF k k y t t

t ttF k y t t

τ

τ τ

⎡ ⎤⎛ ⎞= + − < ≤⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞−= − − − >⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Handout P.52

A step in displacement of magnitude y0, is applied at time 0 and reversed at time tu.

(4.12)

The required force is:

Input

Responserelaxation

y0

tu

0 u1 mm; 2 s.y t= =

recovery

R 1 N/mm; Δ 1 N/mm; 1 Ns/mm 1 s; 2 s.y Fk k c τ τ= = = ∴ = =

Case 3: Dynamic Mechanical Response

51

displacement y (mm)-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

Forc

e F

(N)

-2.0-1.5-1.0-0.50.00.51.01.52.0

2 2

R 2 2 2 2Δ Δ

*; *1 1

y y

y y

k kF k y F y

ω τ ωτ

ω τ ω τ

⎛ ⎞ ⎛ ⎞ʹ ʹ́⎜ ⎟ ⎜ ⎟= + =

⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Handout P.52

A continuous oscillating displacement is applied0* expy y i tω=

The required oscillating force is where* ' "F F iF= +

(4.13)

F

y

time t (s)0 1 2 3 4 5 6 7

F (N

)

or

y (m

m)

-2.0-1.5-1.0-0.50.00.51.01.52.0

Fy

Input

Response

1 rad/sω =

F'

F"δ

y*

F*

ΔW = F dy!∫ = area enclosed by the ellipse > 0

The tangent of the phase angle peaks at a frequency

Case 3b - Variation with frequency ω

52

Handout P.53-54

frequency ω (rad/s) (log scale)0.01 0.1 1 10 100

forc

e F

(N)

0.0

0.5

1.0

1.5

2.0

2.5

|F"|

|F'|

frequency ω (rad /s) (log scale)0.01 0.1 1 10 100

tan δ

0.0

0.1

0.2

0.3

0.41/τy1/τF

• The patterns of response seen here are similar to those of polymers

in the linear viscoelastic regime (but only qualitatively)

• The one systematic discrepancy between the response of the particular spring-dashpot Zener model and the observed response of polymers is that it is too localised in the time/frequency domains.

tan /F Fδ ʹ́ ʹ=1/ y Fω τ τ=

yτLoss

tangentplot

Generalised Maxwell model

53

kRk1 k2 kj

F y

c1 c2 cj

1 1 R1

dd 1; ( 1.. ); d d

N j jj N N

j j j

F FyF F F j N F k yt c k t+ +

== + = + = =∑

Handout P.55

A truer representation of the viscoelasticity of a polymer would be given by a spring-dashpot model with multiple arms, such as

N + 1 arms (j = 1..N)

Replacing equation (4.8), the equations of the model are then:

(4.14)

The j-th relaxation time is now /j j jc kτ =

The relaxation spectrum of the model: , ( 1.. )j jk j Nτ⎡ ⎤ =⎣ ⎦

N1 2 N+1

log10 τj (s)-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

Gj (

GPa

)

0.00

0.05

0.10

0.15

0.20

0.25

0.30polystyrene at 100oC

Shear relaxation spectrum

54

Handout P.55-57

The shear “relaxation spectrum” of such a generalised Maxwell model is given by the set of pairs of:

, ( 1.. )j jG j Nτ⎡ ⎤ =⎣ ⎦

Example data: shear stress relaxation modulus G(t) for an amorphous polystyrene at 100°C, near, but just above its Tg~95°C

Zenermodel

GU

GR~1MPa

The discrepancy is very clear ... we need a shear relaxation spectrum

13 elementsPS

Shear stress relaxation modulus, G(t)

55

R1

d d where ( 1.. )d d

N j jj j

j jG G j N

t tσ σ γ

σ σ γτ=

ʹ ʹʹ ʹ= + + = =∑

R10

( ) exp N

jj j

tG t G Gσγ τ=

⎛ ⎞ʹ= = + −∑ ⎜ ⎟⎜ ⎟

⎝ ⎠

Handout P.56

The equations governing the shear response would then be (by analogy with equations (4.14) and replacing y by a shear angle γ (= 2 x tensorial shear strain e) and F by the corresponding shear stress σ’ and k by shear modulus G):

(4.15)

(4.16)

Applying the model to the case of shear stress relaxation (γ(t) is a step function at t = 0, of height γ0), the shear stress relaxation modulus is then

by analogy

with eqn

(4.12)

Prony series

Dynamic shear modulus G*

56

Handout P.59

The shear relaxation spectrum given above for polystyrene at 100°C couldbe used to predict the dynamic shear

modulus G* governing the oscillatory

response of polystyrene in shear, via the analogue of equations (4.13)

( ) ( )2 2

R 2 2 2 21 1

***

where ; 1 1

N Nj j j j

j jj j

G G iG

G GG G G

σγ

ω τ ωτ

ω τ ω τ= =

ʹ ʹ́= = +

ʹ ʹ́= + =∑ ∑+ +

log10ω (rad/s)

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Com

plex

she

ar m

odul

us (G

Pa)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

G'

G"

(4.21)

Storage modulus

Loss modulus

by analogy with eqn (4.13)

Case 3: Dynamic mechanical response

57

displacement y (mm)-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

Forc

e F

(N)

-2.0-1.5-1.0-0.50.00.51.01.52.0

2 2

R 2 2 2 2Δ Δ

*; *1 1

y y

y y

k kF k y F y

ω τ ωτ

ω τ ω τ

⎛ ⎞ ⎛ ⎞ʹ ʹ́⎜ ⎟ ⎜ ⎟= + =

⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Handout P.52 (RECAP)

A continuous oscillating displacement is applied0* expy y i tω=

The required oscillating force is where* ' "F F iF= +

(4.13)

F

y

time t (s)0 1 2 3 4 5 6 7

F (N

)

or

y (m

m)

-2.0-1.5-1.0-0.50.00.51.01.52.0

Fy

Input

Response

1 rad/sω =

F'

F"δ

y*

F*

ΔW = F dy!∫ = area enclosed by the ellipse > 0

Example problem 5

58

Handout P.61

How are viscoelastic data obtained?

59

• ... spanning such a broad range of times; there is no single technique capable of achieving this.

• We may exploit the phenomenon of time-temperature equivalence (TTE)

4.2 The role of temperature• Some of the earliest data for an uncrosslinked

sample of the amorphous polymer polyisobutylene (PIB):

• Tg ~ -20°C

• Tensile stress relaxation modulus E(t) (in dynes/cm2 = 0.1 Pa) vs time for a wide range of T (log scales on both axes)

• The data exhibit time-temperature superposition: with suitable horizontal shifts, they all superimpose to form a single “master curve”

* C

H

H

Cn

CH3

*

CH3

Glassy << Tg

Rubbery ~Tg

Elasto-viscous fluid >> Tg

Handout P.62

Time-temperature superposition (TTS)

60

The master curve for PIB obtained by time-temperature superposition of these curves (in previous slide), onto the curve for 25°C (reference temperature T0).

Glassy << Tg

Rubbery ~Tg

Elasto-viscous fluid >> Tg

The factor by which the time scale for a particular curve has to be reduced/expanded to give a fit to the reference

temperature curve (here T0

=25°C) is these days called aT(but called K298 in the graph)

Handout P.63

aT

Linking E(t) to G(t)

61

• In and above the glass transition Tg region, K >> G

• As such the Poisson’s ratio is ν ~ 0.5 (i.e. incompressible)

• and hence

• It follows that under these conditions, the tensile relaxation modulus function E(t) is related simply to the shear relaxation modulus function G(t):

2 (1 ) 3E G Gν= + ≈

( ) 3 ( )E t G t≈

Handout P.64

Time-temperature equivalence (TTE)

62

( ) ( ) ( )0 0, , or log , log log ,TT

tG t T G T G t T G t a Ta⎛ ⎞

= = −⎜ ⎟⎝ ⎠

( ) ( )0 for all j T jT a T jτ τ=

• The observation of time-temperature superposition (TTS) in the E(t) data shows, to a good approximation that,

• ...not only E(t) but also the shear relaxation modulus G(t) at any temperature T is related to that at a reference temperature T0 by a simple expansion or contraction of the time-scale.

In terms of equations, this means: (4.23)

Experimental data further suggest that the same aT function also applies to all τj :

Expansion / contraction factor

(4.24)Relaxation timeat any temp T

Handout P.64

Reference temperature

Vogel-Tamman-Fulcher (VTF) equation

63

This graph shows lnaT plotted versus 1/T (absolute). Two points are notable:

(a) The data do not give a straight line relation (i.e. do not fit the Arrhenius relation, when T > Tg).

(b) The data are fitted by the alternative 2-parameter calculated line included in the graph.

( ) ( )0 ln T

C CaT T T T∞ ∞

= −− −

(4.25)1000/T

3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4

lna T

0

5

10

15

20

25

30

35

calculated experiment

PIB

Reference temperature

e.g. the best-fit values of material parameters in PIBare:

1446 ; 120.5 CC K T∞= = − °

Handout P.65

Glass transition temperature Tg

64

Recall: the glass transition temperature Tg of an amorphous material is the temperature

below which it cannot maintain structural equilibrium, on a given experimental time-scale

The value of Tg is defined in terms of the volume V or enthalpy H measured during heating or cooling, and its values are rate dependent

• As shown in plot, faster coolingleads to a higher measured Tg.

• Because glass is not in thermodynamic equilibrium, its structure gradually evolves with time (e.g. polymer at BàC).

• ‘Fictive temperature’ Tf =temperature at which the glass is believed to have the same structure in equilibrium

Cooling ratesq1 > q2

Equilibrium line

Handout P.67

Physical ageing in glassy polymers

65

• The viscoelastic properties also vary with time in the glassy state (<Tg)

Shown here are tensile creep

curves for PVC (Tg = 83°C) obtained after rapid cooling from 90°C to 20°C, and waiting (‘ageing’) for varying times te before testing.

• Note how the creep curves shift to the right with increasing ageing time• ...there appears to be time – ageing-time (te) superposition going on.

( ) ( )e e e0 for all j jt a t jτ τ=

Each relaxation time after ageing for a time te is related by the same factor ae to its corresponding value after ageing for a time te0

(4.27)

Handout P.68

Secondary relaxations of polymers

66

• The viscoelastic process visible in the glass transition (Tg) region is often labelled the ‘primary relaxation’ of the polymer.

• But viscoelasticity is also prominent in other regions of the temperature scale. These are called the ‘secondary relaxations’

They differ from the primary relaxation in that:(i) they are not usually associated with evidence of ‘physical ageing’

(i.e. not linked to microstructural evolution).(ii) They are usually found, to reasonable accuracy, to show time-

temperature superposition, with the relaxation times conforming to the Arrhenius relation as follows

(4.28)

0exp hence ln T

c c caT T T

τ ⎛ ⎞∝ = −⎜ ⎟⎝ ⎠

where 0HcR

=Δ Activation energy (also denoted as Q)

Handout P.69

Example problem 6

67

Handout P.70

Non-linear viscoelasticity

68

• With increase in stress, the viscoelasticity shown by a polymer becomes increasingly non-linear

• i.e. the relation between stress and strain is time-dependent, butthe material ceases to behave as a linear system.

• It no longer obeys the Boltzmann Superposition Principle

• Many of the questions of (i) how to model them in the macro-world, and (ii) how to explain them in terms of what is going on in the micro-world form part of the ongoing research activities... (DPhil)

Exp. data from tensile creep and shear creep tests on PMMA at 30°C

Handout P.71