Section 3 Viscoelasticity; (still in scalar form; no tensors yet!)

In this section we consider the issue that our material may be viscoelastic. That is the material can exhibit both viscous and elastic properties at the same time. Viscoelasticity is important, particularly for polymers both in the solid and liquid state. The approach is mostly directed at the liquid (melt) state, however the mathematical analysis is also relevant to viscoelastic solids. Viscoelasticity is relevant to polymer melts, concentrated biotech solutions such as xantham gum. It is also relevant to certain foodstuffs (custard, sauces), personal products (shaving cream, lotions, shampoo), and a range of other materials, such as explosives, rocket fuel & certain natural materials. An understanding of viscoelasticity is useful in relation to help understand certain processing issues, in addition it is useful in characterising rheologically complex materials. This is important in terms of quality control. The mathematics involved in the modelling is relatively simple but needs a bit of thought. The modelling is highly suitable for setting Tripos questions!

Processing issues where viscoelasticity is important.

1 Processibility. pellets post processing Rheology

extruder die Extrusion line must work 24hrs/day

CET 2B. Section 3, Viscoelasticity-2011 1

2 Die swell

 = Di/Do

Newtonian  = 1.14

Do Di Viscoelastic fluid  = 1 – 5 big factor when making tubes

3 Extrusion Instabilities production rate

sharkskin – optically observed Usually bad, sometimes good

CET 2B. Section 3, Viscoelasticity-2011 2

Rheometers. Cone and Plate Rheometer

Torque

Cone angle = 14 0 1.5 mm

Rotation or oscillation 

Parallel Plate Rheometer

Torque

1 mm

Rotation or oscillation 

12 mm Couette Rheometer

Torque

30 mm

Rotation or oscillation 

20 mm 22 mm CET 2B. Section 3, Viscoelasticity-2011 3

Viscoelasticity measurement Parallel plate or cone and plate rheometers. A device where ideally the strain or stress is uniform.

A) Apply strain , measure stress. Controlled strain rheometer

Or B) Apply stress , measure strain. Controlled stress rheometer yield stress – TA Ares, Controlled strain rheometer. £20,000-50,000/machine

Upper shaft measures torque Torque measurement Force rebalance transducer

Sample Between h ~ 0.3-1 mm plates 25 – 50 mm stepper motor

To transducer Lower shaft to (1) Steady stepper Sample  (angular velocity) motor Mode 1 (2) Oscillat e  (angular frequency) Mode 2 γ (strain) To motor Mode 3

(3) Step strain CET 2B. Section 3, Viscoelasticity-2011 4

Different test geometries.

Parallel plate geometry Cone and plate geometry

h 

difficult to achieve close fit between plates without damaging equipment r tan  h

r r r r

r φ r φ φ strain γ  strain γ  h r tanα tanα r ω r ω ω ratestrain γ  ratestrain γ  h r tanα tan α radiuson dependboth dependboth radiuson both independant radiuson

γ γ

h r tan α

Rφ bottom plate R φ bottom plate

γ = rφ γ = rφ = φ h r tan α tan α

Stain rate depends on r Strain rate independent of r This is the best option, but often parallel plates are used

CET 2B. Section 3, Viscoelasticity-2011 5

Rheometers measure Torques; so we need to express Torque in terms of fluid shear stress. r1 r1

Cone and plate  = r  rdr  = 2  r2 dr

0 0

Force 2 3  r  3 1

r1

Parallel plate more tricky because  is a function of r.

CET 2B. Section 3, Viscoelasticity-2011 6

Rheological “ Rheometric” measurements.

We now examine the types of deformation that can be applied using, for example, a Rheometrics controlled strain rheometer.

a) Stress growth, steady shear, stress relaxation

B Strain  Start from rate 0 rest,  = 0  time A C Stress Newtonian  Stress relaxation

time

t = 0 t = t Stress growth 1 Steady state  t < 0  = 0   t > 0  =  o  t > t1  = 0 measure stress as a function of time for above strain rate history.

(A) Stress growth (Viscoelastic response)

(B) Steady shear (Non Newtonian behaviour)?

(C) Stress relaxation (Viscoelastic response)

Time dependence can be explored

Steady shear is the most used deformation, obtain “flow curve”

CET 2B. Section 3, Viscoelasticity-2011 7

b) Step strain . At t= o, instantaneously strain material by t , subsequently measure stress relaxation as a function of t . γ0 = γ0 t Step in strain

Strain  rate 0  time

0 Stress  Stress relaxation

time t = 0

Relaxation modulus G(t)

G(t) = (t) γ0

0 Elastic rubber

 Viscoelastic response 1/e Often exponential

t Characteristic relaxation time -3 3 10 – 10 s

CET 2B. Section 3, Viscoelasticity-2011 8

c. Oscillatory “rheometric” deformation Time dependence γ . γ γ0 . γ 0

 0

strain angular frequency  ti Apply  = o sin t Strain =  oe   ti then  = o  cos t Strain rate  = i  oe

measure  = o sin(t + ) variable o = max strain amplitude, typically 0.1 (10%)  = angular frequency, typically 10-2 – 103 rad/s ~ 1 hour to complete   limit of oscillator the experiment ti  =  o  cos t or  = i  oe

it +   = osin(t + ) or  =  oe

= o sin t cos  + o cos t sin   component of stress component of stress  in phase with  in phase with  Elastic bit! Viscous bit!

’ ’’ Defn  = G o sin t + G o cos t  α strain   So G' = o cos , G" = o sin Pa  o  o Storage Modulus Loss Modulus

Elastic Viscous

CET 2B. Section 3, Viscoelasticity-2011 9

 = G* γ Complex modulus G* (another way of saying the same thing).

it+ t  ' " oe = G = G + iG = it  t  oe   Elastic Loss Modulus Modulus  So G' + iG" = o e i   o now (e I = cos + i sin ) '  cos   So G = o G" = o sin N/m2 o o As before

So ti* 0 0  tcos''Gtsin'G or  0 eG

Another definition

For a given strain, γ if we know  and  from the rheometer 0 0 dimensions of viscosity 1 G'2 + G"2 2    Complex viscosity  = Pas 

For given , and known o. G’ :- Storage modulus G’’ :- Loss modulus * η :- Complex viscosity

These properties capture the viscoelastic properties of a material, but the values will depend on the test frequency ( time scale applied).

CET 2B. Section 3, Viscoelasticity-2011 10

Measure o and  using TA instruments rheometer or other instrument, then we know the following, for a given  1. G’ storage modulus. 2. G” Loss modulus. 3. *Complex viscosity

Storage Modulus 104

G’ *  G’ dominates G’ & G’’ (Pa) η* (Pa s) Loss Modulus

Cross over G’’

10-1 G’’ dominates 10-3 103 Angular frequency of oscillation  rads/sec

Rheometer measures Torque and from this we need shear stress Typical viscoelastic data that we wish to model. (See appendix) 1. Oscillatory Viscoelastic response. linear viscoelastic response ,γ

Frequency sweep

Strain sweep * 4  G’ G’ 10

G’’

G’’ linear non linear region 10 10-2 102 10-3 103 Strain angular frequency

at fixed frequency 0 ~ 1 rads/sec at fixed strain γ0

CET 2B. Section 3, Viscoelasticity-2011 11

2. Steady Shear

104 Linear region

Apparent Carreau viscosity  a (Pa s) shear thinning

10

. shear rate γ

Silly Putty; real data.

Silly Putty frequency sweep, strain = 1%, 160204

1000000

100000

10000 * / Pa.s 

1000

G' G" 100  G Pa, / ' G " / Pa,

10

1 0.01 0.1 1 10 100 Frequency / rad.s-1

CET 2B. Section 3, Viscoelasticity-2011 12

Modelling of viscoelasticity We are going to build a model that, eventually is going to be able to predict both the linear viscoelastic and non linear shear thinning behaviour of a viscoelastic material such as a polymer melt. Stage 1 The linear viscoelastic part Coupling of linear viscous and elastic elements The Maxwell element series coupling of elastic and viscous component

elastic constant g η viscosity

linear spring linear dashpot

Maxwell often favoured for stress relaxation The Voigt element. parallel coupling

viscous

γ = γ1 = γ2  = 1 + 2

.  = η γ Voigt often favoured for creep (constant stress experiments) We will follow Maxwell, but you should “play with” Voigt model.

The Maxwell Model (Differential form)

(, γ)

1γ1 2γ2

Local (1, 1) (22)   = g   =  1 1 2  2 entropy elastic

Coupling  = 1 = 2 Stress continuity viscous friction of chain

CET 2B. Section 3, Viscoelasticity-2011 13

 = 1 + 2 Strain additivity    1 = 2 =  Then  =  1 +  2 Governing ordinary differential equation relaxation times (s)

d d 1  = + or  = η (Pa s) dt dt g  g (Pa)

d d  1st order ODE g = + Maxwell equation dt dt  where the relaxation time of the element  is given by /g, (s) Example. Response of Maxwell element Spring and dashpot in series a..Steady shear

. .  = gγ γ . t  = ηγ d = 0 dt  t d  = 0 Then  =  dt 0 Linear response, Newtonian. We will have to make the model Non Newtonian later Maxwell model predicts Newtonian behaviour in simple shear.Most complex VE Fluids are shear thinning and so we will have to fix this later. b. Stress relaxation after steady shear d = -  t - t dt    0    t .  =  0 e γ0 d = - dt . . γ γ = 0   t 0 t0 -[t-t ]/  = 0 e 0  . -[t-t ]/ t  = ηγ0e 0 t0 All elastic  ∞ exponential decay -3 3 All viscous  0 time constant , 10 - 10

CET 2B. Section 3, Viscoelasticity-2011 14

c. Oscillatory motion. (Important and frequently used) Use complex notation it Variables (γ , ) Apply  t =  oe 0

i (t   ) Measure  (t) =  oe  it Strain rate  t = i oe

* i  t i  t  (t) = G  0 e =  G' + iG''  0 e

d t = iG' + iG"   ei t dt   o d d  Remember, g = + , so, Maxwell equation dt dt  Substitute for γ,  i t i t 1 it gi e = iG' + iG"   e + G' + iG "  e ] o o  o Yields 2 2 g  g * g G' = 2 2 , G" = 2 2 ,  = 1/2 1 +   1 +   1 + 22

 0 G’ 0  0 G’’ 0  0 η* = g = η *  ∞ G’ g  ∞ G’’ 0  ∞ η 0 Elastic domination Newtonian* Plateau G’  Elastic G’ & G’’ (Pa) η* (Pa s) Shear thinning

G’’ complex viscosity viscous cross over domination Loss angular frequency 

CET 2B. Section 3, Viscoelasticity-2011 15

A bit of maths. Differential equations versus integral equations.

The Maxwell Model (Integral form wrt strain rate) Differential equation  d d  + = g  = 1st order ODE  dt dt g Multiply by integrating factor d (et/) = et/ + et/ d  d d et  + et  = g e t  dt  dt  dt dt assume  = 0 at t’ =- current time t t  t'   ' '  e =  g e  t  dt - Past time t ' -t - t   ' ' Stress at current time  t =  g e  t dt strain rate at past time t’ - Maxwell equation in terms of past strain rate – current stress depends on past strain rate.

applied strain rate

 = 0 t’ = - ∞ t’ = 0 t’ = t’ t’ = t start the clock past time (variable) current time

.  (t) = g e-(t-t’)/ γ(t’)dt’

Fading memory

CET 2B. Section 3, Viscoelasticity-2011 16

Test strain rate equation Steady shear

.  .  0

t’ = - ∞ t’= 0 t’ = t past time current time

t ' t - t    ' '  t =  g e  t dt  t '  t - t   '  t = g o  e dt  t '   -t - t    g o  e  = g   o   -     ss =  o as before

t current time t is a constant

 (t) = ge-(t-t’)/γ. dt’ 0 -∞ t . -t/ t’/ = gγ0e e dt’ -∞

. -t/ t t’/ = gγ0e [ e ] -∞ . = gγ0.  = η g .  = ηγ Newtonian as before 0

The integral Maxwell strain rate equation gives the same result in steady and other shear deformations as the differential Maxwell equation.

CET 2B. Section 3, Viscoelasticity-2011 17

Integral Maxwell in term s of past strain rate. General deformations. . . γ2 .   γ1 0 strain rate

- ∞ t’ = t1 t’ = t2

t’ = t’= 0 t’ = t past time current time III stress II IV I

t -(t-t’)/ . I  (t) = ge γ (t’)dt’  = 0 -∞

t t -(t-t’)/ . -t/ t’/ II  (t) = ge γ1 dt’ = γ1 ge e dt’ 0 0

If t = t1 then  = 1

0 t1 t -(t-t’)/ ..-(t-t’)/ III  (t) = + ge γ1 dt’ + ge γ2 dt’

-∞ 0 t1

If t = t2 then  = 2 t 0 t1 t2 IV -(t-t’)/ . -(t-t’)/ . -(t-t’)/  (t) = + ge γ1 dt’ + ge γ2 dt’+ ge dt’ t2 -∞ 0 t1

Exponential decay

CET 2B. Section 3, Viscoelasticity-2011 18

Lets integrate again in order to obtain the Maxwell integral strain equation.

The Maxwell Model (Integral form wrt strain) We know, Stress at time t t ' t - t   d '  t =  g e ' dt  dt

Now udv = [ uv ] -  vdu '  t - t let u = ge   , dv = d integrate above by parts,

t  '  t ' t - t  g t - t   ' '  t = g e   -  e  t,t dt        where strain is given by strain strain

Two strain terms Define strain Strain = 0 at current time t t'  ' ""  , =   dtttt t where t =current time and t’ = past time Note when t’ = t  = 0 We are measuring strain from current time So Integral Maxwell equation – past strain

t ' g t - t  ' '  t = -  e  t, t dt  

Stress at time t Past strain history t’ . . γ(t,t’) = γ(t’’)dt t t’’

t’= t t’= t’

CET 2B. Section 3, Viscoelasticity-2011 19

 Test Maxwell strain integral equation. steady shear  = const

.  .  0

t’ = - ∞ t’= 0 t’ = t past time current time

Determine past strain history t '  . . ""' By definition  , =   dtttt t  t'  . ' " ' For steady shear  , = o  =  o  - ttdttt  t  . = -  o (t – t’ )   0 strain t =- ∞

strain increases t’ = t’= 0 t’ = t negatively current time backwards γ = 0 t - t’ )

Introduce new variable s, (you don’t have to do this, but it generally makes calculation simpler ).

Let s = t – t1 ds = - dt1

s =  s = t s = t – t1 s = 0

t’ = - t’ = 0 t’ = t1 t’ = t

CET 2B. Section 3, Viscoelasticity-2011 20

o g s  Integral strain equation t = +  e  o, s ds   '  Also, strain  =-  o t - t  =-  o s o g  s  t =-  o s e ds   o g  s /  o s   t =   o  s e  +  e ds       s  o  + g  o e      = g  o =  o Newtonian, as before.

t g -(t-t’)/  (t) = - e γ (t, t’)dt’

λ -∞

t g -(t-t’)/ .  (t) = λ e - γ0 (t - t’)dt’ -∞

t g . -t/ t’/λ  (t) = + λ γ0 e (t - t’)e dt’ -∞

integrate by parts

. .  = η γ0 = λ g γ0

Newtonian i!

The Integral strain Maxwell equation will predict the same results as the integral Maxwell strain rate equation and the differential Maxwell equation.

CET 2B. Section 3, Viscoelasticity-2011 21

Another example. Stress growth and stress relaxation. This one is a bit more challenging! The key to solving these problems is to be clear in your mind what the strain rate history is and then determine the correct strain history for the appropriate time domain that is of interest to you (or the examiner!)

.  II .  . 0 I γ0 III strain rate γ = 0

- ∞ t’ = t’= 0 t’ = 1t t’ = t sector 1 sector 2

If current time in sector 1 strain

- γ. (t – t’) 0

. . - γ0 t - γ0 t if current time in sector 2 - γ. t strain 0 1

γ = 0 - γ. (t – t’) + γ. (t – t ) . 0 0 1 - γ0 (t1 – t’)

0 t g -(t-t’)/ . g -(t-t’)/ . II  (t) = - e (-γ t)dt’ - e (-γ (t-t’))dt’ λ 0 λ 0 -∞ 0

0 t t 1 g -(t-t’)/ . g -(t-t’)/ . III  (t) = - e (-γ t )dt’ - e (-γ (t -t’))dt’ + (0) λ 0 1 λ 0 1 -∞ 0 t1

CET 2B. Section 3, Viscoelasticity-2011 22

Tackle our two further problems 1. Multiple relaxation times. One Maxwell element doesn’t usually fit the data well. 2. Non linear response. Maxwell elements are linear in steady shear and we know interesting fluids can, for example, shear thin.

1.Introduce spectrum of relaxation times The parallel coupling of Maxwell elements

g1 η1 λ1

 = Σ1 g2 η2 λ2

gi ηi λi

i ~ 3 - 8

i i = gi

The integral constitutive equation then becomes,

' multi-mode t i g  t - t i  t = -  i e    t,t' dt' Maxwell integral   equation in terms - i of past strain For oscillatory data the equations become.

2 2 ' gi  '' g   G  = i , G  = i i  2 2  2 2 i 1 +   i 1 + i 

 gii   =  1 i 2 2 2 1 + i 

CET 2B. Section 3, Viscoelasticity-2011 23

We now need to find the “best fit” g I , his is potentially a tricky problem. Use software on Rheometrics to get best fit.

storage and loss exp data modulus

G’ G’’ g i

Eqn model fit

angular frequency  i

ill posed problem, there are multiple answers

With a spectrum of relaxation times we can get a good fit to linear viscoelastic data.

Model x

Polydisperse material G’ G’’ gi x

x

Experimental data x

o ω 10-2 1 0-1 1 00 101

λ1 λ2 λ3 λ4

Choose a range of λ1 Perform least square fit for best fit, gi to fit G’, G’’ This requires software algorithms.

Multi-mode modelling is essential for realistic engineering predictions

Obtained giλi parameters

CET 2B. Section 3, Viscoelasticity-2011 24

2.Introduce non linear steady shear response

Mainly due to M H Wagner Rheol Acta 15, 136, (1976) one only Add a non linear “damping parameter” to constitutive equation. This is best done using the strain integral equation.

' ' t  t - t   k  t,t gi i   ' ' strain dependant  t = -   e e  t,t dt equation - i i damping factor This is the new bit!

'  k  t,t Modulus e

Modulus (always pos) dampi ng parameter k = 0 Maxwell k ranges from 0  1  Non linear t - t '  i   k  t,t gie   ' '  t =-    e  t,t dt        i  strain dep of material    Time dep of material  The equation has been “factored” in terms of a separated time and strain dependance.

Multi - mode integral strain dependant Maxwell equation with Wagner damping factor

CET 2B. Section 3, Viscoelasticity-2011 25

The effect of the damping parameter on different deformations   ' Steady shear  t =  o, always .  .  0

t’ = - ∞ t’= 0 t’ = t past time current time

  0 strain

t’ = t’= 0 t’ = t current time

. = - γ0 (t – t’) '  '  t,t = - o t - t 

Let s = (t – t’), substitution makes calc easier  use s =(t-t’) (o, s) = - o s (Optional) ds = - dt’  o  gi s i k o s  t = -  e e  o sds  i o i  gi  s t = -   e i sds   o  i 1  where i = + k o i integrate by parts

CET 2B. Section 3, Viscoelasticity-2011 26

  i   gi i  o   =   o   2 1 + ki  o        i  o stea d y shea r   =   o   2 1 + ki  o    k = 0 → Newtonian Check the form of steady shear equation Relaxation modes single mode or multi-mode Non linear k k = 0 k ≠ 0 a) single  , k = 0 k ~ 0 → 1   =  o Newtonian as before

Newtonian



. γ b) single , k  0    = o   2 1 + k    o  Max

decreases – not physically realistic  Stress

Non linear – good news

Newtonian . γ CET 2B. Section 3, Viscoelasticity-2011Strain rate 27

c) multiple , k = 0 

linear 3

 Stress 2

.  = Σ ηi γ

η2 1

η1

. γ Strain rate response is still linear

d) multiple , k  0     =  i o   2 good 1 + ki    o 

close to Power law

Non linear  Stress

λ1 λ2 λ3

3 -1 0 . 10 s Strain rate γ

operating range

This non linear response may match “Power law” Bingham …. Or other steady shear constitutive equation.

CET 2B. Section 3, Viscoelasticity-2011 28

We now have a model that describes LVE response and the non linear steady shear response. This is what we want!

Multi-mode Wagner Good for predicting LVE response Good for predicting steady shear shear thinning response

But how do we get the value that non linear parameter k?

Use step strain experiments

Step Strain

strain t’ = 0 t’ = t t’ = - ∞

 0 s = ∞ s = o s = t

s < t  = 0 -k γ(t,t’) s > t  = -o e recall

γ0 i t gi s i k0 t = -  e  0e ds -   stress at time t i i k 0 t i  =  0  i eget

Relaxation modulus at finite strain i  t k j t i  j tG = = e  ieg  j but small strain modulus given by

CET 2B. Section 3, Viscoelasticity-2011 29

 t i  t i G0 t = =  gie 0 tG  j  so ek = 0tG So, measure relaxation modulus at small and large strain and use above equation to get k.

Ratio gives k

Go(t) small strain log G(t)

G1(t) large strain

G2(t)

log (t)

CET 2B. Section 3, Viscoelasticity-2011 30

And finally ( in this section), Integral constitutive equation also solves another mystery in polymer science The Cox Merz Rule

shear thinning shear thinning η x x x x x a x x η* Complex Apparent x viscosity viscosity x η* x x x

. angular frequency (ω) steady shear rate γ (ω) LVE data Non linear behaviour

    Cox Merz rule   = a      where  =  Maxwell model with damping function predicts, i  i Complex viscosity.   =  no k in 2 2 equation 1 + i   Apparent viscosity   i a s =      2 1 + k    i o non linear   term equations are not identical but provided k > 0 and you have a spectrum of relaxation times, they are of similar form and give a close match.

Cox Merz Rule → is an accidental coincidence !

CET 2B. Section 3, Viscoelasticity-2011 31

Summary of this important section.

CET 2B. Section 3, Viscoelasticity-2011 32

Appendix 1 Rheological data. Golden Syrup A high viscosity Newtonian Fluid. Note. No Shear thinning, very little elasticity, G’ low, and Cox Merz obeyed.

Apparent visocosity

Golden Syrup at 25oC 100

ηa [Pa.s]

10

1 0.1 1 10 100 Shear Rate [s-1]

Golden Syrup at 25oC (1% Strain) 10000

1000 G"

100

η* 10 G'

1 110100 Frequency [rad/s]

CET 2B. Section 3, Viscoelasticity-2011 33

Golden Syrup at 25oC (Cox-Merz) 100 η*

ηa

10

1 110100 Shear Rate [s-1] / Frequency [rad/s]

CET 2B. Section 3, Viscoelasticity-2011 34

Appendix 2 Rheological data. A molten Polyethylene ( typical data for a commercial PE) A high viscosity viscoelastic Fluid. Note. Shear thinning, viscoelasticity, G’ and G’’ similar magnitude. Cox Merz obeyed.

Apparent and Complex viscosity; showing shear thinning and Cox Merz rule behaviour.

100000

Apparent 10000 Viscosity

Complex

Viscosity (Pas) Viscosity

Apparent and Complex Complex and Apparent 1000

0.01 0.1 1 10

Rate (s-1) and Frequency (rad/s)

Strain sweep showing linear regime

100000

10000

G' G'' Eta* 1000

G', G'' (Pa), Eta* (Pa.s) Eta* G'' (Pa), G', 100 0.1 1 10 100 1000 Strain (%)

CET 2B. Section 3, Viscoelasticity-2011 35

Frequency sweep showing viscoelastic response.

Data shows classic shear thinning of complex viscosity. Also scross over for G’, G’’ curves.

100000 100000 G'

10000 G'' 10000 G' Modelled 1000 G'' Modelled 1000 G', G'' (Pa) Eta* (Pa.s) Eta* 100 Eta*

10 100 Eta* Modelled 0.01 0.1 1 10 100 Frequency (rad/s)

CET 2B. Section 3, Viscoelasticity-2011 36