# Applications of Viscoelasticity

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18 Ιουλ 2012 (πριν από 5 χρόνια και 11 μήνες)

871 εμφανίσεις

1
A few facts:Polymers creep upon deformation
Polymers are elastic but also flow
Deformation are fully recoverable
Why plastic bags tear
when you are halfway home
τ
o
τ
t
τ
G
o
t
ε
Viscoelasticity
•Materials which exhibit elastic and viscous behavior
ε
τ
ε
What causes viscoelastic behavior?
Long polymer chains at the
molecular scale, make polymeric
matrix viscoelastic at the
microscale
Reference: Dynamics of Polymeric Liquids (1977). Bird, Armstrongand Hassager. John
Wiley and Sons. pp: 63.
Energy Storage + Dissipation
Structure/network
of an emulsion,
suspension, etc
Applications of Viscoelasticity
•Geology and mining
•Concrete technology
•Soil mechanics •Plastics industry•Tribology (study of lubrication)
•Food industry
2
Creep
ForceConstant
Stress Relaxation
Viscoelasticity
Constant deformation
This animation shows creep of a rod (60 cm long, 3 mm
diameter) of polymethylmethacrylate(PMMA) in
cantilever bending at room temperature, ~20 deg. C.
Images of the free rod end were captured after ~1, 10,
100, 1,000, 10
4, 105
seconds, 10
6
days), 10
7
seconds (almost four months).
Creep
Creep
ForceConstant
Kruip (Creep)
τ
o
τ
t
τ
G
o
t
Viscoelasticsolid
Viscoelasticfluid
γ
t
3
Spanning relaxatie(Stress relaxation)
Spanning relaxatie(Stress relaxation)
Constant deformation
Relaxatie(relaxation)
τ
γof ε
t
t
τ
ViscoelasticsolidViscoelasticfluid
How to model Linear Viscoelasticity??
Hooke’s law:
Newton’s law:
()()()
ttt
E
ε
τ
=
()
()
()
t
t
t
t
εη
ε
ητ

=

=
4
()()()
ttt
E
ε
τ
=
Linear elasticity:
Proportionality
Proportionality:The imposed strain yields a linear
proportional change in the response.
Ifresponse to strainjump
ε0
is
()
(
)
0
t
Et
τ
ε
=
Thenresponse to strainjumpΘ∗ε0
is
()
(
)
0
*
t
Et
τ
ε

()
()
()
t
t
t
t
εη
ε
ητ

=

=
IdealizedLinearViscousResponse
Linearviscosity:
HookeanSpringDashpot
σ
τ
τ
o

o
o
γ
γ
γ
2γo
2τ/η
τ/η
o
o
o
Proportionality also with dashpot?
σ
0
5
How to model viscoelasticity?
D2
η
τ
=
Viscous flow
Elastic deformation
τ=ηγ
τ=G*γ
.
BG
=
τ
Viscoelasticity
•Materials which exhibit elastic
and viscous behavior
ε
τ
ε
Elongation: ε
D(t)
: Creep Compliance
(Tensile creep compliance) E(t): Relaxation modulus
(Tensile relaxation modulus)
()
()
()
0
ε
τ
tE
t
=
()
()
0
τ
ε
tD
t
=
Shear: γ
J(t): Creep Compliance
(flexure creep compliance)G(t): relaxation modulus
(Flexure relaxation modulus)
()
(
)
0
τ
γ
tJ
t
=
()
(
)
()
0
γ
τ
tG
t
=
6
Maxwell (shear)
Spring and dashpot support the same τ
τ
τ
τ
==
dpsp
dpsp
γ
γ
γ
+=
dpsp
dpsp
dt
d
dt
d
dt
d
γγ
γ
γ
γ
γ

+=+==
η
τ
γ
=
dp

G
sp
τ
γ

=
,
τληγ
η
τ
ηγτ
τ
η
τ
γ

−=−=→+=
GG
1.
2.
3.
()
d
Dotrepresents
dt

G or E
η
τ
Maxwell (elongation)
Spring and dashpot support the same τ
τ
τ
τ
=
=
dpsp
dpsp
ε
ε
ε
+
=

d

d

d
εε
ε
ε
ε
ε

+=+==
η
τ
ε
=
dp

E
sp
τ
ε

=
,
τληε
η
τ
ηετ
τ
η
τ
ε

−=−=→+=
EE
1.
2.
3.
G or E
η
τ
Maxwell model: Creep
τ
o
τ
t
τ
G
o
t
} Perm. Deformation from dashpot
()
t
G
t
oo
η
τ
τ
γ
+=
G
η
τ
τ
λ
η
γ
τ

=
η
τ
τ
γ
+=
G

γ
After integration:
o
()
()
ητ
γ
t
G
t
tJ
o
c
+==
1
Definition of Creep compliance
()
(
)
ητ
ε
t
E
t
tD
o
c
+==
1
()
(
)
0
τ
γ
tJ
t
=
()
(
)
0
τ
ε
tD
t
=
7
Maxwell: Stress relaxation
ε
o
γ
o
λ
t
τ

t
E
η
τ
ηετ

−=
o
E
η
τ
()
()
()()
λ
η
εττ
τ
η
τ
t
E
t
t
t
t
Eee
E
t

==
⇔=+

00
0
Stress Relaxation Test
Viscous fluid
Time, t
Stress
Elastic
Viscoelastic
0
Stress
Stress
ε
o
γ
τ
o
λ
t
t
)exp()(
0
λ
ττ
t
t−=
Kelvin-Voigt
o
t
γ
t
τ
τ
Kelvin
G
η
, τ
γ
γ
η
τ
G+=

Creep Only!!!
ε
ε
η
τ
E+=

ε
ε
η
τ
E
+
=

Kelvin-Voigt (Creep)
η
εε
η
τ
E
t
+

=
Eersthomogeneoplossingzoeken:
λ
ε
η
εε
t
Ae
E
t

=⇔=+

0
8
λ
ε
t
Ae

= :oplossing homogene
Voorparticulieroplossingmoetgelden
:
()
λ
ε
t
t
ef

=
η
εε
η
τ
E
t
+

=
De particulieroplossinginvullenin Kelvin Voigt model
:
Levert:
()
()
()
λλλ
λλη
τ
t
t
t
t
t
t
e
f
efe
f
−−−
−+=
'
1E
η
λ
=
()
()
()
λλλ
λλη
τ
t
t
t
t
t
t
e
f
efe
f
−−−
−+=
'
()
λ
η
τ
λ
t
t
ef

−=
Na integratiegeeftdit
:
Met:
kruip bij en E ,
000
ττετ
η
λ
===
E
»
¼
º
«
¬
ª
¸
¹

¨
§
??
O
HH
t
texp1)(
0
ε
ε
η
τ
E+=

−−=
λ
εε
t
texp1)(
0
00
0
0
)(exp1)(
ττ
λτ
ε
ε
tJ
t
t=

−−=
Kelvin-Voigt (Creep)
ε
ε
η
ε
τ
EE
=
+
=

Kelvin-Voigt: Relaxation experiment
pure elastic response
τ
ε
o
ε
t
t
9
()
(
)
λ
/exptGtG

Δ
=
(
)
(
)
λ
/exp1)(tJtJ

Δ
=
00
)()exp()(
ε
λ
ττ
tG
t
t=−=
()
(
)
i
i
i
tGGtG
λ
/exp−Δ+=

Generalised Maxwell model
3
12
123
..............
n
tt
tt
ne
eeee
λλ
λλ
τ
τττττ
−−
−−
=++++
Generalised Maxwell model
(
)
(
)
i
i
i
tGGtG
λ
/exp−Δ+=

10
λi(s)
8.04x10-3
5.93x10-2
1.46x10-1
7.61x10-1
Gi(Pa)
3.00x105
4.83x105
2.98x104
1.04x102
()
(
)
i
i
i
tGtG
λ
/exp−Δ=

Generalised Maxwell model:
λi(s)
8.33x10-9
2.23x10-6
1.03x10-3
9.84x10-1
1.62x102
3.89x103
Gi(Pa)
6.13x10
8
3.06x10
7
5.40x10
5
4.12x10
4
5.33x10
4
1.46x10
5
3.38
(
)
(
)
i
i
i
tGtG
λ
/exp−Δ=

Generalized Voigt-Kelvin Model
Similar arguments can be applied for a generalized
Voigt-Kelvin model.

=
λ−
=
λ−
λ−
−=
τ
γ
=
−τ=γ
−τ=γ
n
1i
/t
i
o
n
1i
/t
io
/t
ioi
)e1(J
)t(
)t(J
)e1(J)t(
)e1(J)t(
i
i
i
4 Parameter Model
Viscous flowElastic deformation
τ=ηγτ=G*γ
.
Burgers model
11
4 Parameter Model
G1
G2
η1
η2
τ
4 Parameter Model
γ
η
γτ
ηη
τ
ηηη
τ


2
21
1
21
21
2
2
1
1
2
1
GG
G
GGGGG
+=+

+

+

+
G1
G2
η1
η2
τ
Creep response equation:
()

−++=

−t
G
ooo
e
G
t
G
t
2
2
1
211
η
τ
η
ττ
γ
γ
η
γτ
ηη
τ
ηηη
τ


2
21
1
21
21
2
2
1
1
2
1
GG
G
GGGGG
+=+

+

+

+
4 Parameter Model
12
Components
•Dashpot 1
–Molecular slip
–Permanent deformation
•Spring 1
–Elastic straining of bond angles and lengths
–instantaneous response
•Dashpot 2
–Polymer chain uncoiling and coiling
•Spring 2
–Restoring force brought by thermal agitation
G1
G2
η1
η2
G1
G2
η1
η2
3 parameter Network Models
Neither the Maxwell fluid nor the Kelvin-Voigt solid gives a
viscoelasticresponse that can qualitatively capture even the
most basic features of tissue; therefore, more complex network
models must be used
Three element models:
A Kelvin solid in series with a spring:
A Maxwell fluid in parallel with a spring:
1
ε
2
ε
1
ε
2
ε
η
η
2
E
1
E
1
E
2
E
()
εεηε
εεε
εεηετ
21121
21
22111
EEE
EE
=++⇒

+=
=+=

εεεη
εεε
εεηεετ
1111
21
21221
EE
EEE
=+⇒

+=
+=+=

3 parameters Network Models
Consider the response of the first three element model to creep
and stress relaxation:
Perfect creep:
A Kelvin solid in series with a spring:
Time
0
τ

䥮楴楡氠獴it攺
䥮獴φn瑡湥nu猠

䙩nal⁳tat攺
䭥汶楮⁥汥m敮e:
()()
(
)
η
τ
εεηετ
tE
e
E
tEtH
1
1
1
0
11110

−=⇒+=

Elastic element:
()()
(
)
2
0
2220
E
tH
tEtH
τ
εετ
=⇒=
(
)
tH
0
τ
τ
=
(
)
21
221110
εεε
ε
ε
η
ε
τ
τ
+=
=
+
=
=
EEtH

Total strain:
()

+

=

21
0
11
1
EE
e
t
tE
η
τε
η
1
E
2
E
Time
Strain
2
0
E
τ

+
21
0
11
EE
τ
13
3 parameter Network Models
Consider the response of the first three element model to creep
and stress relaxation:
Perfect stress relaxation:
A Kelvin solid in series with a spring:
Initial state:
Instantaneous
response:
Final state:
Time
0
ε

()
tH
0
ε
ε
=
η
1
E
2
E
(
)
(
)
tHEEEE
0221121
ε
ε
ε
η
ε
=
=
+
+

Kelvin element (can use Laplace transforms):
Elastic element:
Total stress:
()
()
[
]
η
ε
ε
tEE
e
EE
E
t
21
1
21
02
1
+−

+
=
()()
()
[
]
η
ε
εε
tEE
e
EE
E
tHt
21
1
21
02
02
+−

+
−=
()()
()
[]
η
ε
εετ
tEE
e
EE
E
EtEt
21
1
21
0
2
2
0222
+−

+
−==
Time
Stress
02
ε
E
21
021
EE
EE
+
ε
Creep and Relaxation Functions
The creep functionand relaxation functionfor a material model is
determined by setting the input (stress and strain, respectively)
to be unity
For the three component model considered previously:
Creep function:
Relaxation function:
()
()
11221
0
,
1111
1
EE
e
E
tJ
EE
e
t
t
tE
η
λτε
λ
η
=

+=⇒

+

=

()
()
[]
()
211
21
21
21
21
0
2
2
02
,1
21
EEE
eEE
EE
EE
tGe
EE
E
Et
t
tEE
+
=

+
+
=⇒−
+
−=

+−
η
λ
ε
ετ
λ
η
Standard Linear Solid model
Under a constant stress, the modeled material will instantaneously deform
to some strain, which is the elastic portion of the strain, and after that it
will continue to deform and asymptotically approach a steady-state strain.
This last portion is the viscous part of the strain. Although the Standard
Linear Solid Model is more accurate than the Maxwell and Kelvin-Voigt
models in predicting material responses, mathematically it returns
difficult to calculate.
The Standard Linear Solid Model
effectively combines the Maxwell Model
and a Hookeanspring in parallel. A
viscous material is modeled as a spring
and a dashpot in series with each other,
both of which are in parallel with a lone
spring.
Multi Parameter Model
Viscous flow
Elastic
deformation
τ=ηγ
τ=G*γ
14
Which model???
Kelvin
Voigt
Maxwell
RelaxationCreepModel
o
t
t
t
t
Creep and recovery experiment
γ
t
o
t
τ
τ
t
γ
Maxwell
Kelvin-Voigt
Creep-Recovery -PS
0100200300400500600
t [s]
0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
Â [-]
Creep-Recovery-Test
Creep1
Â = f (t)
Creep2
Â = f (t)
Creep3
Â = f (t)
Which model?
0100200300400500600
t [s]
0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
Â [-]
Creep-Recovery-Test
Creep1
Â = f (t)
Creep2
Â = f (t)
Creep3
Â = f (t)
γ
t
o
t
γ
Maxwell
Kelvin-Voigt
4 parameter model
15
Which model for Creep?
Kelvin
Voigt
Maxwell
CreepModel
o
t
t
Polymermelts
Polymer“solids”
Stress relaxation Maxwell
Which model???
Pure elasticneverwithviscoelasticity
Kelvin
Voigt
Maxwell
RelaxationCreepModel
o
t
t
t
t
Polymermelts
Relaxation
Polymer“solids”
And other stress patterns ??
τ
t
Δτ0
Δτ1
Δτ2
t0
t1
t2
16
Superposition principle
0<t<t1
For times larger then t
1
is applied
total response for times t > t
1
()
01
)(
τ
ε
tJt=
()
0112
)(
τ
ε
ttJtt−=−
()()
010
)(
τ
τ
ε
ttJtJt−+=
Suppose that the stress history input consist of a sequence of stress jumps τ0,
applied at successive times t
i
with t
0
=0
τ
t
Δτ
0
Δτ
1
Δτ2
t0
t1
t2
Then the resulting strain history is given by:
()()
....
21)(0
210
+
Δ
+
Δ
+
Δ
=
−−
τ
τ
τ
ε
tttttt
JJJ
()
()

=

Δ=
0j
jttt
j
J
τε
the response at time t on the first strain step starting at t0
is
ε
ε
00
00())t−=

=
G(t-0)(G(t-0)
Δ
the second step starting at t1:
ε
ε
ε
101
0())t−
=

=
G(t-t)(G(t-t)
11
Δ
τ
τ
17
ττ
t
Δτ
0
t0
t1
viscoelasticmaterial
OUTPUT?
Input:
1.Stress relaxationof Creep?
2.Kelvin-VoigtorMaxwell model?
3.Proportionaliaty?
4.Superposition?
5.Memorybeforet=0?
6.Whatis the elongationat t=300s?
MPa10
0
=
τ
t1= 100 s
λ= 200 s
Tmeas
= 300K, Tg= 373K
ρ= 1000kg/m
3
02.0
0
=
ε
τ
t
Δτ
0
t0
t1
1.Stress relaxationorCreep?
2.Kelvin-VoigtorMaxwell model?
3.Proportionaliaty?
4.Superposition?
=
+
t1
5. Memory before t<0?
Creep(stress inpossedand measuringthe strain)
Kelvin-Voigt:solidmaterial
We assume(smalldeformations)(?)
We assume not
(
)
(
)
(
)
(
)
1020
**tJttJtt
ε
ττ
=−+−−
()
)1(
1
1
λ
tt
eJttJ

−Δ=−
()
)1(
2
2
λ
tt
eJttJ

−Δ=−
0
0
τ
ε
=ΔJ
solving the equations:
6 Whatis the elongationat t=300s?
!!!
18
(
)
(
)
(
)
(
)
1020
**tJttJtt
ε
ττ
=−+−−
19
7
0
0
10*2
10*1
02.0
−−
===ΔPaJ
τ
ε
()
)1(
1
1
λ
tt
eJttJ

−Δ=−
()
)1(
2
2
λ
tt
eJttJ

−Δ=−
()
7
200
100300
97
200
0300
9
10*)1(10*210*)1(10*2300

−−−=ee
ε
300
300
0.00285 (theoretically)
0.00282 (measured)
ε
ε
=
=
ττ
t
Δτ0
t0
t1
OUTPUT?
Input:
ττ
t
t0
t1
Done it !
(
)
(
)
(
)
(
)
0010
*tJttJtt
ε
ττ
=−+−−
ττ
t
t0
t1
(
)
0
not constant = ft
τ
τ
τ
=≠
()
(')'(')'
'
dt
dtdttdt
dt
τ
ττ
==

From
-

()
()
11
()
i
totii
tt
ii
ttJ
ε
ετ

==
==Δ

()
'
0
()'
t
t
tJttd
ε
τ
=−

()
()
'
0
()''
t
t
tJttdt
ετ
=−

()
()
'
()''
t
t
tJttdt
ετ
−∞
=−

19
')'('
'
)(
)'(dttdt
dt
td
td
ε
ε
ε

==
From
-

()
()
11
()
i
totii
tt
ii
ttG
τ
τε

==
==Δ
∑∑
()
'
0
')(
t
t
dttGt
ετ

−=
()
()
'')(
'
0
dtttGt
t
t
ετ

−=
()
()
'')(
'
dtttGt
t
t
ετ

∞−
−=
τ
τ
t’
t0
t1
()
()
'
()''
t
t
tJttdt
ετ
−∞
=−

1.Stress relaxationof Creep?
2.Kelvin-VoigtorMaxwell model?
3.Proportionaliaty?
4.Superposition?
5.Memorybeforet=0s?
6.Whatis the elongationat t=300s?
912
5*10
J
Nm
−−
Δ=
λ= 200 s
Tmeas
= 300K,
Tg= 373K
ρ= 1000kg/m
3
τ
τ
t’
0
400
200
200 MPa
0
0
τ
ε
=ΔJ
1.Stress relaxationorCreep? Creep(stress inpossedand measuringthe strain)
2.Kelvin-VoigtorMaxwell model? Kelvin-Voigt:solidmaterial
3.Proportionaliaty? We assume(smalldeformations)(?)
4.Superposition?
5. Memory before t<0s? We assume not
()
()
'
()''
t
t
tJttdt
ετ
−∞
=−

20
()
()
'
()''
t
t
tJttdt
ετ
−∞
=−

()
()
()
()
200300
(300)
''
0200
''''
tt
J
ttdtJttdt
εττ
=−+−
∫∫

()
'
0
(1)
tt
JttJe
λ

−=Δ−
()
'
1
(1)
tt
JttJe
λ

−=Δ−
912
5*10
J
Nm
−−
Δ=
0
0
τ
ε
=ΔJ
()
()
200300
''
(300)
''
0200
1'1'
tttt
tt
J
edtJedt
λλ
εττ
−−
−−
⎛⎞⎛⎞
⎜⎟⎜⎟
=Δ−+Δ−
⎜⎟⎜⎟
⎜⎟⎜⎟
⎝⎠⎝⎠
∫∫

ττ
t’
0
400
200
200 MPa
λ= 200 s
912
5*10
J
Nm
−−
Δ=
τ
τ
t’
0
400
200
200 MPa
()
'
200
1 t<200
200
t
MPa
τ
==

()
'
200
1 200< t>400
200
t
MPa
τ

==−

()
()
200300
''
(300)
''
0200
1'1'
tttt
tt
J
edtJedt
λλ
εττ
−−
−−
⎛⎞⎛⎞
⎜⎟⎜⎟
=Δ−+Δ−
⎜⎟⎜⎟
⎜⎟⎜⎟
⎝⎠⎝⎠
∫∫

()
()
200300
''
(300)
''
0200
1'1'
tttt
tt
J
edtJedt
λλ
εττ
−−
−−
⎛⎞⎛⎞
⎜⎟⎜⎟
=Δ−+Δ−
⎜⎟⎜⎟
⎜⎟⎜⎟
⎝⎠⎝⎠
∫∫

200300
300'
(300)
0200
200300
300'
0200
*1*'*1*'
*(1)*'*(1)*'
t
t
J
dteedt
J
dteedt
λλ
λλ
ε
−−
−−
⎛⎞
⎜⎟
=Δ−
⎜⎟
⎜⎟
⎝⎠
⎛⎞
⎜⎟
+Δ−−−
⎜⎟
⎜⎟
⎝⎠
∫∫
∫∫
21
200
''
200
(300)
0
0
300
''
300
200
200
'
*1***1*
'
*1**(1)*
tt
tt
t
J
teed
t
Jteed
λ
λλ
λ
λ
λλ
λ
ελ
λ
λ
−−
−−
⎛⎞
⎜⎟
=Δ−
⎜⎟
⎜⎟
⎝⎠
⎛⎞
⎜⎟
+Δ−−−
⎜⎟
⎜⎟
⎝⎠

3002000
200
(300)
300300200
200
*200*200*1*
*1*100*200*(1)
Jeee
Jeee
λλ
λλ
ε

⎛⎞
⎜⎟
=Δ−−
⎜⎟
⎜⎟
⎝⎠
⎛⎞
⎜⎟
+
Δ−−−−
⎜⎟
⎜⎟
⎝⎠
()
300
0.00045
ε
=
Lab experiments in real world
the stress relaxation modulus G(t) and E(t)
or the creep compliance J(t) and D(t) the
mechanical response to an arbitrary time
22
linear region small strains!!
LinearVisco-
elasticity
Oscillatory Testing
Deformation and Flow
upper plate
‘moving’
lower plate
‘fixed’
sample
φ1
φ2
φ3
Dynamic Mechanical Thermal Analyzer
(DMTA)
Dynamic Mechanical Testing
Response for Classical Extremes
Stress
Strain
δ= 0°
δ= 90°
Purely Elastic Response
(HookeanSolid)
Purely Viscous
Response
(Newtonian Liquid)
Stress
Strain
23
Dynamic Mechanical Testing Viscoelastic
Material Response
Phase angle
0°< δ< 90°
Strain
Stress
Dynamic Flow Testing
Deformation
Response
Phase angle
δ

An oscillatory (sinusoidal)
deformation (stress or strain)
is applied to a sample.
The material response
(strain or stress) is measured.The phase angleδ, or phase
shift, between the control and
the response is measured.
Dynamic Experiments
()
(
)
γ
γ
ϖ
γ
t
t
=
0
sin: with amplitude
0
()
()
(
)
(
)
(
)
ttt
t
ω
δ
ω
δ
τ
δ
ω
τ
τ
cossinsincossin
00
+
=
+
=
INPUT:
OUTPUT:
24
δ
γ
τ
δ
γ
τ
sin='G'en cos'
0
0
0
0
=G
)cos()(")sin()(' = (t)
0 0
tGtG
ω
ω
γ
ω
ω
γ
τ

tan() =
G"()
G'()
δ
ω
ω
δ
δ
=
sin()
cos()
()
()
(
)()
tt
t
ω
δ
ω
δ
τ
τ
cossinsincos
0
+=
Oscillation -Input and response
•Input:
•Response:
•Separation in elastic and viscous components:
)(sin
0
tωτ=τ
)(sin
0
δ−ωγ=γt
[
]
δω−δωγ=δ−ωγ=γsin)(coscos)(sin)(sin
00
ttt
in Phase
(elastic component)
90°phase shift
(viscous component)
)cos()(")sin()(' = (t)
0 0
tGtG
ω
ω
γ
ω
ω
γ
τ

δ
γ
σ
δ
γ
σ
sin='G'en cos'
0
0
0
0
=G
G’: Storage modulus
G”: Loss modulus
δ
=AnglePhase
)('
)(''
tan
ω
ω
δ
G
G
=
Loss Tangent
LiquidViscous
MaterialicViscoelast
SolidElasticHookean
o
o
90
900
0
=
<<
=
δ
δ
δ
Viscoelastic Measurements
Torque bar
Sample
Cup
Bob
Strain γ
Stress σ
o
σ
o
γ
Oscillator
Phase Angle δ
0
cos
)('
γ
δσ
ω
o
G=
Storage Modulus
0
sin
)(''
γ
δσ
ω
o
G=
Loss Modulus
25
DMA Viscoelastic Parameters:
The Complex, Elastic, & Viscous Stress
The stress in a dynamic experiment is referred to as the
complex stressτ*
Phase angleδ
Complex Stress, σ*
Strain, ε
τ* = τ' + iτ"
The complex stress can be separated into two components:
1) An elastic stress in phase with the strain.τ' = τ*cosδ
τ
' is the degree to which material behaves like an elastic solid.
2) A viscous stress in phase with the strain rate.τ" = τ*sinδ
τ
" is the degree to which material behaves like an ideal liquid.
DMA Viscoelastic Parameters: Damping, tan
δ
Phase angle δ
G*
G'
G"
Dynamic measurement
represented as a vector
It can be seen here that
G* = (G’2
+G”2)1/2
The tangent of the phase angle is the ratio of the
loss modulus to the storage modulus.
tan δ
= G"/G'
"TAN DELTA" (tan δ)
is a measure of the damping
ability of the material.
Complex and Dynamic Viscosity
The viscosity measured in an oscillatory experiment is a
Complex Viscositymuch the way the modulus can be
expressed as the complex modulus. The complex viscosity
contains an elastic component and a term similar to the
The Complex viscosity is defined as:
η* = η’-i
η”
or
η* = G*/
ω
Cyclic Testing
Consider the response of the three element model to cyclic
deformation
(
)
(
)
tt
ω
τ
τ
sin
0
=

0
τ
1
ε
2
ε
η
1
E
2
E
(
)
()
220
1110
sin
sin
εωτ
ε
η
ε
ω
τ
Et
Et
=
+
=

26
Cyclic Testing
Response of the elastic
component:
()
1110
sin
ε
η
ε
ω
τ

+=
Et
Response of the Kelvin solid:
()()
()
t
E
tEt
ω
τ
εεωτ
sinsin
2
0
2220
=⇒=
Homogeneous solution:
()
η
ε
tE
cet
1
1

=
Particular solution: try
()()
(
)
tBtAt
ω
ω
ε
sincos
1
+
=
()
()()
[]
tEt
E
t
ωωωη
ηω
τ
ε
sincos
1
222
1
0
1
+−
+
=
Over time the homogeneous solution dies out and we are left
with the particular solution
Total strain:
()
(
)
()

++
+−
+
=t
E
EEE
t
E
ω
ηω
ωωη
ηω
τ
ε
sincos
2
22
211
222
1
0
Cyclic Testing
InputOutput
()
()
t
t
ω
τ
τ
sin
0
=
()
(
)
()

++
+−
+
=t
E
EEE
t
E
ω
ηω
ωωη
ηω
τ
ε
sincos
2
22
211
222
1
0
Cyclic Testing -General Formulation
To generalize the previous analysis, in which a particular forcing
mode was selected, consider the more general input of a
complex exponential, which can account for amplitude,
frequency, and phase:
()
ti
et
ω
ττ
0
=
Input:
Elastic response:
1110
εηετ
ω

+=Ee
ti
Response of the Kelvin solid (only considering the particular solution):
()
titi
e
E
tEe
ωω
τ
εετ
2
0
2220
=⇒=
Try
()
ti
Aet
ω
ε
=
1
ωη
τ
iE
A
+
=
1
0
Stress (input) amplitude:
Strain (output) amplitude:
0
τ
2
0
1
0
EiE
τ
ωη
τ
+
+
Complex modulus:
()
(
)
ωη
ωη
ω
iEE
iEE
iG
++
+
=
21
12
Cyclic Testing -General Formulation
The complex storage modulus is the ratio of the stress amplitudeto
the strain amplitude, and can be separated into a storage
modulus(representing the energy that is recovered during the
deformation) and a loss modulus(representing the energy that
is lost during deformation)
Storage modulus
()()
(
)
ω
ω
ω
GiGiG

+

=
Loss modulus
()
(
)
(
)
()
()
22
2
21
2
2
22
2
21
22
22121
21
12
ηω
ηω
ωη
ωη
ωη
ωη
ω
++
+
++
++
=
++
+
=
EE
E
i
EE
EEEEE
iEE
iEE
iG
The storage and loss moduliare dependent on the particular
viscoelasticmodel chosen; for the three element model
considered here,
27
1) Storage Modulus (E’) –elastic property2) Loss Modulus (E’’) –damping property
3) Tanδ–loss tangent or measure of the phase angle between E”and E’
Tanδ= E”/E’
Maxwell model
frequency dependence of G’and G”
poly(cyclohexylmethacrylate) (PCHMA)
Glass transition temperature: 104
oC.
Amorphous density at 25
oC: 1.10 g/cm
3.
Molecular weight of repeat unit: 168.24 g/mol.
28
poly(cyclohexylmethacrylate) (PCHMA)
Creep compliance of PS (M
w = 3.85 10
2
kg/mol)
Time dependence tensile modulus of PIB
reference temperature T = 298K
Time-Temperature Superposition
We can superimpose dynamic
oscillatory data :
)T ,(Gb)T,("G
)T ,(Gb)T,(G
refTT
refTT
ωα
′′

ω
α

=
ω

Where b
T
is a vertical shift factor and the temperature dependence of the
horizontal shift factor, αT
can be described by either the WLF equation or the
Arrhenius equation.
29
Example: TTS for LLDPE
The figure below shows data of elastic modulus obtained for LLDPE at various
temperatures
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
0.1110100
Frequency,
ω
Elastic modulus, G' (Pa)
190°C
180°C
170°C
160°C
150°C
140°C
135°C
130°C
125°C
123°C
123°C
190°C
Example: TTS for LLDPE
First we chose a reference temperature (T=123
°C in this example).The data at the
rest of the temperatures can be shifted horizontally by multiplying the frequency
ω
by
a horizontal shift factor
αT
. In addition sometimes a vertical shift of the data is also
required to obtain better superposition. This shift is obtained by multiplying the elastic
and loss moduliby a vertical shift factor b
T.
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
0.010.1110100
Frequency,
ωα
b
T
G' (Pa)
190°C
180°C
170°C
160°C
150°C
140°C
135°C
130°C
125°C
123°C
Temperature (°C)
αT
bT
12311
125
1
1.19
1300.8451.26
1350.731.21
1400.621.16
1500.511.16
1600.421.135
1700.331.035
1800.241.025
1900.221.08
3.43
Example: TTS for LLDPE
The values of the shift factor used are summarized in following Figure.
Usually αT
is follows an Arrhenius type or a WLF-type relationship
0
0.2
0.4
0.6
0.8
1
1.2
100120140160180200
Temperature (°C)
shift factor a
T

−=α
α
ref
T
T
1
T
1
R
E
ln
)TT(C
)TT(C
log
ref2
ref1
T10
−+

or
3.44
Example: TTS for LLDPE
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
0.010.1110100
Frequency,
b
T
G', b
T
G" (Pa)
G'
G"
The superposed G’and G”curves at T
ref
=123°C are shown below.
3.45
30
Shifting data
()
()
***
001101
(,),,
T
GTGTGTa
ω
ωω
==
ω
G*
ω0
ω1
T0
T1
()
()
1
00110
*,*,*,
T
t
GTtGTtGT
a
⎛⎞
==
⎜⎟
⎜⎟
⎝⎠
()()
***
001101
(,),,
T
GTGTGTa
ω
ωω
==
(
)
f
f
v
vvB
A

+=lnln
η
g
TT≈
g
TT<
No mobility
(
)
gfff
TTvv

+
=
α
Tg)(at
volumefree:
f
v
αf: expansioncoëfficiënt
()
g
f
g
TgT
TT
f
TT
f
B
−+

−=
α
ηη
0
0
3.2
loglog
0
0
v
v
f
f
=
Fraction free volume
(
)
ref
o
ref
o
ref
refref
T
TTc
TTc
T
T
a
−+
−−
==
2
1
loglog
ρη
ρη
31
g
g
gg
TTcc
c
cc
c
−+=
=
02
0
2
0
2
21
0
1
(
)
ref
o
ref
o
T
TTc
TTc
a
−+
−−
=
2
1
log
Kc
c
g
g
]7030[
]1814[
2
1
−≈
−≈
WLF parameters
Polymeer
T0
Tg
Polyisobuthyleen
298
8.61
200.4
205
Polyvinylacetaat
349
8.86
101.6
305
Polystyreen
373
13.7
50
373
Polymethylmethacrylaat
381
34
80
381
Polydimethylsiloxane
303
1.9
222
150
Buthylrubber
298
9.03
201.6
205
Polyurethaan
283
8.86
101.6
238
0
1
c
0
2
c
No WLF
No Mobility
E= 3-4 GPa
32
RT
E
AeN
*

=
During Shear
N

γ

RT
E
AeN
*

=
RT
Eact
Ae

=
0
η
Frequency Sweep: Material Response
Terminal
Region
Rubbery
Plateau
Region
Transition
Region
Glassy Region
1
2
Storage Modulus (E' or G')
Loss Modulus (E" or G")
l
o
g

G
'
a
n
d
G
"
33
Dynamic Temperature Ramp or Step and
Hold: Material Response
Temperature
Terminal Region
Rubbery Plateau
Region
Transition
Region
Glassy Region
1
2
Loss Modulus (E" or G")
Storage Modulus (E' or G')
L
o
g

G
'

a
n
d

G
"
Courtesy: TA Instruments
Relaxation processes in polymers
•local processes still active in the glassy state,
•co-operative processes involving longer chain sequences
•motions of the complete chain, related to flow behavior
•specific processes in partially crystalline polymers.
34
Relaxation processes in polymers
ATOMIC FORCE MICROSCOPY (AFM)
Nonlinear Viscoelasticity
The theory of linear viscoelasticityis only valid for very small or slow
deformations. Most practical applications however involve large
deformations. Important nonlinear effects manifest themselves even in
the simple flow situations such as steady simple shear. The predominant
nonlinear phenomena are the dependence of the viscosity on the shear
rate and the appearance of a nonzero first normal stress difference. The
following are topics of interest in the study of nonlinear phenomena:
–Normal stress differences, shear thinning and extensional
thickening
–Development and assessment of constitutive equations which can
describe realistically viscoelasticfluids and can model viscoelastic
flows
1. Normal Stress Differences in Shear
Consider two identical stationary cup-rotating rod arrangements, one
containing a Newtonian fluid and the other a polymer melt. The surface
profile in the polymer melt is exactly opposite than that of theNewtonian
profile. This phenomenon is called the rod-climbing
or Weisenberg
effect and it is due to the generation of Normal Stressesinside the
polymer melt.
2
22111
R
F2
N
π
=σ−σ=
35
Normal Stress Differences in Shear
We define the normal stress
coefficients:
2
2
2
2
1
1
N
N
γ

γ

Where N
1
and N2
are the first
and second normal stress
differences
33222
22111
N
N
σ−σ=
σ−σ=
3. Shear Thinning
4. Elongational(Extensional) Flow
Let’s consider the uniaxialstretching of a cylinder of fluid:
We define the Extensional Viscosity as:
1
1
11
11
e
x
V
and
A
F

=ε=σ
ε
σ

FF
A
Extensional Thickening
Although in a shear flow the viscosity of a polymeric fluid usually
decreases with increasing deformation rate, in an extensional flow the
viscosity frequently increases with increasing extension rate. That is the
fluid is extensional thickening. If a sample, initially at its rest state is
subjected to steady simple extension at a rate
ε
starting at a time=0, the
tensile stress growth coefficient is defined:
ε
ε
τ

ε
τ
≡εη+



),t(),t(
),t(
2211
E
.
36
Extensional Thickening
η
E
+
(t,ε)
.