Lectures 11 and 12

Topics to be Covered
• Polymer Melt Rheology
• Introduction to Viscoelastic Behavior
• Time-Temperature Equivalence
Chapter 11 in CD (Polymer Science and Engineering)
Polymer Melt R h e o l o g y
δ
z
Newton’s Law
τxy
The most convenient way to describe deformation
under shear is in terms of the angle θ through which
the material is deformed;
δ
tan θ = = γ xy
z
θ
Then if you shear a fluid at a constant rate
v0
y
˙ v
δ
γ˙xy = = 0
z
z
dx 
dv
d
γ˙xy =
=  
dz dt  dz 
Just as stress is proportional to strain in
Hooke‛s law for a solid, the shear stress is
proportional to the rate of strain for a fluid
τxy ∝ γ˙xy
τxy = ηγ˙xy
Newtonian and Non - Newtonian
Fluids
.
.
τxy = ηa (γ) γ
Shear
Stress
Slope = ηa
Shear
Thinning
Newtonian Fluid
(η = slope)
Shear Thickening
Strain Rate
When the chips are
down, what do you use?
Variation of Melt Viscosity
with Strain Rate
Log ηa
(Pa)
Zero Shear Rate Viscosity
5
4
3
2
1
0
-3
-2
-1
0
1
2
. 3 -1 4
Log γ (sec )
SHEAR RATES ENCOUNTERED IN PROCESSING
Compression
Molding Calendering
100
101
102
Extrusion
Injection
Molding
103
Strain Rate (sec-1)
104
Spin
Drawing
105
How Do Chains Move
- Reptation
D ~ 1/M2
η0 ~ M3
Log ηm + constant
Variation of Melt Viscosity
with Molecular Weight
Poly(di-methylsiloxane)
Poly(iso-butylene)
Poly(ethylene)
Poly(butadiene)
Poly(tetra-methyl p-silphenyl siloxane)
Poly(methyl methacrylate)
Poly(ethylene glycol)
Poly(vinyl acetate)
Poly(styrene)
ηm = KL(DP)1.0
ηm = KH(DP)3.4
Redrawn from the data of G. C.
Berry and T. G. Fox, Adv. Polym.
Sci., 5, 261 (1968)
1
2
3
4
Log M + constant
5
Entanglements
Short chains don‛t entangle but long ones do - think of
the difference between a nice linguini and spaghettios,
the little round things you can get out of a tin (we have
some value judgements concerning the relative merits of
these two forms of pasta, but on the advice of our
lawyers we shall refrain from comment).
Entanglements
Viscosity - a measure of the frictional forces
acting on a molecule
ηm = KL(DP)1.0
Small molecules - the viscosity varies directly
with size
At a critical chain length chains start to become
tangled up with one another, however
Then
ηm = KH(DP)3.4
Entanglements and the Elastic
Properties of Polymer Melts
d0
d
2.0
160°C
d/d0
180°C
1.8
Depending upon the rate at
which chains disentangle relative
to the rate at which they stretch
out, there is an elastic component
to the behavior of polymer melts.
There are various consequences
as a result of this.
1.6
200°C
1.4
1 .2
1.0
10-1
10-2
100
101
102
Strain Rate at Wall
Jet Swelling
103
(sec-1)
Melt Fracture
Reproduced with permission from J. J. Benbow,
R. N. Browne and E. R. Howells, Coll. Intern.
Rheol., Paris, June-July 1960.
Viscoelasticity
If we stretch a crystalline solid,
The energy is stored in the
Chemical bonds
If we apply a shear stress to
A fluid,energy is dissipated
In flow
VISCOELASTIC
Ideally elastic
behaviour
Ideally viscous
behaviour
Viscoelasticity
Homer knew that the first
thing To do on getting your
chariot out in the morning
was to put the wheels back
in.
(Telemachus,in The Odyssey,
would tip his chariot against
a wall)
Robin hood knew never to
leave his bow strung
Experimental Observations
•Creep
•Stress Relaxation
•Dynamic Mechanical Analysis
Creep and Stress Relaxation
Creep - deformation
under a constant load
as a function of time
5lb
5lb
5lb
5lb
TIME
Stress Relaxation constant deformation
experiment
5lb
4lb
3lb
TIME
2lb
Redrawn from the data of W. N. Findley, Modern
Plastics, 19, 71 (August 1942)
Creep and
Recovery
50
2505 psi
2305 psi
2008 psi
40
Strain
Creep(%)
2695psi
30
(c) VISCOELASTIC RESPONSE
Creep
Recovery
1690 psi
20
Permanent
} deformation
Stress
applied
10
1320 psi
1018 psi
0
0
2000
4000
6000
Time (hours)
8000
Stress
removed
Time
S t r a i n v s. Time Plots - Creep
(a) PURELY ELASTIC RESPONSE
(b) PURELY VISCOUS RESPONSE
Strain
Strain
Permanent
Deformation
Stress
applied
Stress
removed
Time
Shear
Stress
applied
Strain
(c) VISCOELASTIC RESPONSE
Creep
Recovery
Permanent
} deformation
Stress
applied
Stress
removed
Time
Shear
Stress
removed
Time
Redrawn from the data of J.R. McLoughlin and A.V.
Tobolsky. J. Colloid Sci., 7, 555 (1952).
Stress
Relaxation
σ (t)
E(t ) = ε
0
10
600C
Log E(t), (dynes/cm2)
The data are not usually reported as a
stress/time plot, but as a
modulus/time plot. This time
dependent modulus, called the
relaxation modulus, is simply the time
dependent stress divided by the
(constant) strain
400C
9
920C
1000C
8
1100C
Stress relaxation
of PMMA
0
112 C
1200C
800C
1150C
1250C
7
1350C
0.001
0.01
0.1
1
10
Time (hours)
100
1000
Amorphous Polymers - Range
of V i s c o e l a s t i c B e h a v i o u r
Liquid
Semi- Like
Solid
Vs
Rubbery
Leather
Soft Like
Hard Glass
Glass
T
Tg
V i s c o e l a s t i c Properties 0f
Amorphous Polymers
Log E
(Pa)
10
Glassy
Region
9
8
7
Cross-linked
Elastomers
6
5
4
Rubbery
Plateau
Low
Molecular
Weight
3
Temperature
Melt
Measured over
Some arbitrary
Time period - say
10 secs
V i s c o e l a s t i c Properties 0f
Amorphous Polymers
Stretch sample an
arbitrary amount,
measure the stress
required to maintain
this strain.
Then E(t) = σ(t)/ε
Log E(t) (dynes/cm2)
Glassy
Region
Glass
Transition
10
9
Rubbery
Plateau
8
7
6
5
Low
Molecular
Weight
High
Molecular
Weight
4
3
- 10
-8
-6
-4
-2
0
+2
Log Time (Sec)
Log E(t) (dynes/cm2)
Time
Temperature
Equivalence
Glassy
Region
10
9
8
7
6
5
4
3
- 10
Log E
(Pa)10
9
8
7
-8
5
4
3
Glass
Transition
High
Molecular
Weight
-6
-4
-2
Cross-linked
Elastomers
6
Rubbery
Plateau
Low
Molecular
Weight
Glassy
Region
0
+2
Log Time (Sec)
Rubbery
Plateau
Low
Molecular
Weight
Temperature
Melt
Relaxation in Polymers
First consider a hypothetical isolated
chain in space,then imagine stretching
this chain instantaneously so that there
is a new end - to - end distance.The
distribution of bond angles (trans,
gauche, etc) changes to accommodate
the conformations that are allowed
by the new constraints on the ends.
Because it takes time for bond rotations
to occur, particularly when we also add
in the viscous forces due to neighbors,
we say the chain RELAXES to the new
state and the relaxation is described
by a characteristic time τ.
How quickly can I do
these things?
Amorphous Polymers - the Four
Regions of Viscoelastic B e h a v i o r
Log E
(Pa)
10
GLASSY STATE - conformational
changes severely inhibited.
9
Tg REGION - cooperative motions
of segments now occur,but the
motions are sluggish ( a maximum in
tan δ curves are observed in DMA
experiments)
8
7
6
5
4
Low
Molecular
Weight
Polymer
Melt
3
Temperature
RUBBERY PLATEAU -
τt
becomes
shorter,but still longer than the
time scale for disentanglement
TERMINAL FLOW - the time scale for disentanglement becomes
shorter and the melt becomes more fluid like in its behavior
S e m i c r y s t a l l i n e Polymers
•Motion in the amorphous domains
constrained by crystallites
•Motions above Tg are often more
complex,often involving coupled
processes in the crystalline
and amorphous domains
0.30
α
0.25
0.20
0.15
0.10
LDPE
LPE
β
γ
0.05
0.00
-200
-150
-100
-50
0
50
100
150
Redrawn from the data of H. A. Flocke,
Kolloid–Z. Z. Polym., 180, 188 (1962).
•Less easy to generalize - polymers
often have to be considered individually
- see DMA data opposite
Topics to be Covered
• Simple models of Viscoelastic Behavior
• Time-Temperature Superposition Principle
Chapter 11 in CD (Polymer Science and Engineering)
Mechanical and
Theoretical
Models of
Viscoelastic
Behavior
GOAL - relate G(t) and J(t) to
relaxation behavior.
We will only consider LINEAR
MODELS. (i.e. if we double
G(t) [or σ(t)], then γ(t) [or
ε(t)] also increases by a factor
of 2 (small loads and strains)).
TENSILE
EXPERIMENT
SHEAR
EXPERIMENT
Stress Relaxation
σ (t)
E(t ) =
ε0
τ (t)
G(t) =
γ0
ε (t)
D(t) =
σ0
J(t) =
Creep
γ (t)
τ0
Linear Time Independent Behavior
1
E=
D
1
G=
J
Time Dependent Behavior
1
E(t ) ≠
D(t)
1
G(t) ≠
J(t)
Simple Models of the Viscoelastic
Behavior of Amorphous Polymers
Keep in mind that simple
creep and recovery data for
viscoelastic materials looks
something like this
Strain
(c) VISCOELASTIC RESPONSE
Creep
Recovery
Permanent
} deformation
Stress
applied
Stress
removed
Time
Simple Models of the Viscoelastic
Behavior of Amorphous Polymers
Glassy
Region
Log E(t) (dynes/cm2)
While stress relaxation
data look something like
this;
Glass
Transition
10
9
Rubbery
Plateau
8
7
6
5
Low
Molecular
Weight
High
Molecular
Weight
4
3
- 10
-8
-6
-4
-2
0
+2
Log Time (Sec)
Simple Models of the Viscoelastic
Behavior of Amorphous Polymers
Extension
l0
Hooke's law
σ = Eε
∆l
σ
I represent linear
elastic behavior
σ = Eε
Simple Models of the Viscoelastic
Behavior of Amorphous Polymers
Viscous flow
v0
y
v
Newtonian fluid
.
τ xy= ηγ
σ =η
dε
dt
S t r a i n v s. T i m e f o r S i m p l e
Models
PURELY ELASTIC RESPONSE
σ = Eε
PURELY VISCOUS RESPONSE
σ =η
Strain
dε
dt
Strain
Time
Stress
applied
Permanent
Deformation
Stress
removed
Shear
Stress
applied
Shear
Stress
removed
Time
Maxwell Model
Maxwell was interested in creep and stress relaxation and
developed a differential equation to describe these properties
Maxwell started with Hooke‛s law
Then allowed σ to vary with time
Writing for a Newtonian fluid
Then assuming that the rate of strain
is simply a sum of these two contributions
σ = Eε
dσ
_ = Εdε
_
dt
dt
_
σ = η dε
dt
dε
1 dσ
_ = σ
_+ _
_
η Ε dt
dt
M A X W E L L M O D E L - Creep a n d R e c o v e r y
Strain
Creep and
recovery
dε
_ = σ
_
η
dt
0
t
Time
Strain
Recall that real viscoelastic
behaviour looks something
like this
A picture representation
of Maxwell‛s equation
0
t
Time
Maxwell Model -S t r e s s R e l a x a t i o n
dε
_ = σ
_ + 1_ dσ
_
η Ε dt
dt
In a stress relaxation experiment
Log (Er /E0 )
0
-1
dε
_ = 0
dt
Hence
dσ
_ = σ
_ dt
σ
η
σ = σ 0 exp[-t/τ t ]
Where
_
τt = η
Ε
Relaxation time
-2
-3
-4
-5
-6
-2
-1
0
1
Log (t/τ )
M a x w e l l M o d e l -S t r e s s R e l a x a t i o n
Log E(t)
(Pa)
10
Real data looks like this
9
8
7
6
The Maxwell
model gives
curves like this,
OR like this
5
4
3
Time
MAXWELL MODEL
 t
σ (t) σ 0
E(t ) =
exp − 
=
 τ t 
ε0
ε0
Voigt Model
Maxwell model essentially assumes a uniform distribution
Of stress.Now assume uniform distribution of strain VOIGT MODEL
Picture representation
Equation
__
σ(t) = Eε(t) + η dε(t)
dt
(Strain in both elements of the
model is the same and the total
stress is the sum of the two
contributions)
σ1
σ2
σ = σ 1 + σ2
Voigt Model - Creep and Stress Relaxation
Gives a retarded elastic response but does not allow
for “ideal” stress relaxation,in that the model
cannot be “instantaneously” deformed to a given
strain.
But in CREEP σ = constant, σ0
σ1
σ2
__
σ(t) = σ 0 = Eε(t) + η dε(t)
dt
dε(t)
σ0
__ + ε(t)
__
_
τ t‘
η
dt
=
S train
RETARDED ELASTIC RESPONSE
σ = σ 1 + σ2
σ
ε(t) = _ 0 [1- exp (-t/τ t‘ )]
Ε
τ t‘ - retardation time (η/E)
Strain
Time
t1
t2
Summary
What do the strain/time plots look like?
Problems with Simple Models
I can’t do creep
•The maxwell model cannot account for
a retarded elastic response
And I can’t do stress relaxation •The voigt model does not describe stress
relaxation
•Both models are characterized by single
relaxation times - a spectrum of
relaxation
times would provide a better description
NEXT - CONSIDER THE FIRST TWO
PROBLEMS
THEN -THE PROBLEM OF A SPECTRUM OF
RELAXATION TIMES
Four - Parameter
Model
EM
Elastic + viscous flow + retarded elastic
Eg CREEP
σ0
_0 + σ
__0 t + _
ε = σ
ΕM ηM
Ε M [1- exp (-t/τ t )]
EV
ηV
Retarded or
Anelastic Response
Strain
Elastic
Response
Stress
applied
Stress
removed
ηM
Permanent
} deformation
Time
Distributions of Relaxation and
Retardation Times
We have mentioned that although the
Maxwell and Voigt models are seriously
flawed, the equations have the right
form.
What we mean by that is shown
opposite, where equations describing the
Maxwell model for stress relaxation and
the Voigt model for creep are compared
to equations that account for a
continuous range of relaxation times.
Stress Relaxation
E(t ) = E0 exp (− t / τ t )
∞
E(t ) = ∫ E(τ t )exp(− t / τ t )dτ t
0
Creep
D(t) = D0 [1 − exp(− t / τ ′t )]
∞
D(t) = ∫ D(τ t′)[ 1 − exp(− t / τ t′)] dτ ′t
0
These equations can be obtained by assuming that relaxation occurs at a rate that
is linearly proportional to the distance from equilibrium and use of the Boltzmann
superposition principle. We will show how the same equations are obtained from
models.
The Maxwell - W i e c h e r t Model
dε
1 dσ
_ = σ
_1 + _
_1
η 1 Ε 1dt
dt
σ
1 dσ
_2
2 _
= _
+
η 2 Ε 2dt
σ
1 dσ
_3
3 _
= _
+
η 3 Ε 3dt
E1
E2
E3
η1
η2
η3
Consider stress relaxation
dε
_ =0
dt
σ 1 = σ 0 exp[-t/τ t1 ]
σ 2 = σ 0 exp[-t/τ t2]
σ 3 = σ 0 exp[-t/τ t3]
Distributions of Relaxation and
Retardation Times
Stress relaxation modulus
E(t) = σ(t)/ε 0
σ(t) = σ 1 + σ 2 + σ 3
σ 01
σ 02
σ 03
__
__
E(t) = __
exp
(-t/τ
)
+
exp
(-t/τ
)
+
01
02
ε0
ε0
ε 0 exp (-t/τ03)
Or, in general
E(t) =
Σ E n exp (-t/τtn )
where
σ 0n
__
En = ε
0
Similarly, for creep compliance combine voigt elements to obtain
D(t) =ΣDn [1- exp (-t/τ tn‘ )]
Distributions of Relaxation and
Retardation Times
Log E(t) (Pa)
Example - the Maxwell - Wiechert Model With
10
n = 2
Log E(t) (dynes/cm2)
Glassy
Region
Glass
Transition
10
9
6
4
Rubbery
Plateau
8
8
2
7
0
6
-2
5
E(t) =
Melt
4
0
1
-8
-6
-4
-2
0
+2
Log Time (Sec)
2
3
Log time (min)
4
Σ E n exp (-t/τtn )
n=2
3
- 10
-1
Time - Temperature
Superposition Principle
Log E(t) (dynes/cm2)
Recall that we have seen that
there is a time - temperature
equivalence in behavior
Glassy
Region
10
9
8
7
6
5
4
3
- 10
Log E
(Pa)
10
9
8
7
6
5
4
Glass
Transition
Rubbery
Plateau
Low
Molecular
Weight
-8
3
High
Molecular
Weight
-6
-4
-2
Glassy
Region
Cross-linked
Elastomers
Rubbery
Plateau
Low
Molecular
Weight
Melt
Temperature
This can be expressed
formally in terms of a
superposition principle
0
+2
Log Time (Sec)
Time Temperature Superposition
Principle - C r e e p
T 3 > T2 > T1
ε(t) T
σ0
G'
T3
T2
T1
Master Curve at T0 0C
log t
ε(t) T
σ0
log a T
0
T0
[log t - log a T]
T
Stress Relaxation Modulus
dynes/cm2
1011
Stress
Relaxation
Data
Log Shift Factor
Time Temperature
Superposition
Principle - S t r e s s
Relaxation
SHIFT FACTOR vs
TEMPERATURE
+8
+4
0
-80.8°C
1010
-4
-76.7°C
109
-80
-40
0
Temperature
-74.1°C
-70.6°C
108
40
-65.4°C
-58.8°C
-49.6°C
-40.1°C
-0.0°C
107
106
Master Curve at 250C
+25°C
+50°C
105
-80°
10-2
100 102
-40°
-60°
10-14 10-12 10-10 10-8
10-6
10-4
-20° -10°
0°
25°
10-2 10-0 10+2
Time (hours)
80
0C
Relaxation Processes above T g
- the WLF Equation
From empirical observation
-C1 (T - Ts )
_________
Log a T =
C2 + (T - Ts )
For Tg > T < ~(Tg + 1000C)
Originally thought that C1 and C2 were universal constants,
= 17.44 and 51.6, respectively,when Ts = Tg. Now known
that these vary from polymer to polymer.
Homework problem - show how the WLF equation can be
obtained from the relationship of viscosity to free
volume as expressed in the Doolittle equation
Semi - Crystalline Polymers
NON - LINEAR RESPONSE TO STRESS.SIMPLE
MODELS AND THE TIME - TEMPERATURE
SUPERPOSITION PRINCIPLE DO NOT APPLY
Temperature
Tm
Tg
Amorphous melt
Rigid crystalline domains
Rubbery amorphous domains
Rigid crystalline domains
Glassy amorphous domains