Transcript IMA Annual Program Year Tutorial An Introduction to Funny (Complex) Fluids: Rheology, Modeling and Theorems September 12-13, 2009 Understanding silly putty, snail slime and.

IMA Annual Program Year Tutorial
An Introduction to Funny (Complex) Fluids: Rheology, Modeling and Theorems
September 12-13, 2009
Understanding silly putty, snail slime
and other funny fluids
Chris Macosko
Department of Chemical Engineering and Materials Science
NSF- MRSEC
(National Science Foundation sponsored Materials Research Science and Engineering Center)
IPRIME
(Industrial Partnership for Research in Interfacial and Materials Engineering)
What is rheology?
rein (Greek) = to flow
ta panta rei = every thing flows
rheology =
study of flow?, i.e. fluid mechanics?
honey and mayonnaise
stress=
f/area
honey and mayo
viscosity =
stress/rate
rate of deformation
rate of deformation
What is rheology?
rein (Greek) = to flow
ta panta rei = every thing flows
rheology =
study of flow?, i.e. fluid mechanics?
rubber band and silly putty
honey and mayo
modulus
= f/area
viscosity
rate of deformation
time of deformation
4 key rheological phenomena
rheology = study of deformation of complex materials
fluid mechanician: simple fluids
complex flows
materials chemist: complex fluids complex flows
rheologist:
complex fluids simple flows
rheologist fits data to constitutive equations which
- can be solved by fluid mechanician for complex flows
- have a microstructural basis
from: Rheology: Principles, Measurement and Applications, VCH/Wiley (1994).
ad majorem Dei gloriam
Goal: Understand Principles of Rheology:
(stress, strain, constitutive equations)
stress = f (deformation, time)
Simplest constitutive relations:
Newton’s Law:
Hooke’s Law:
d
t 
 
dt
t  G
Key Rheological Phenomena
•
•
•
•
shear thinning (thickening)
time dependent modulus
normal stresses in shear
extensional > shear stress
  
G(t)
N1
u > 
1
k1
ELASTIC SOLID
k2
L´
The power of any spring
is in the same proportion
with the tension thereof.
Robert Hooke (1678)
k1
f  L
f  kL
k2
f
L´
Young (1805)
modulus
stress
t
t  G
strain
1-8
Uniaxial Extension
Natural rubb
G=3.9x105Pa
T11 
f
a
extension a 
or strain e 
a = area
L
L
L - L
 a 1
L
1

T11  G  a 2  
a

natural rubber G = 400 kPa
1-9
Shear gives different stress response
=0
 = -0.4
T21  G
 = 0.4
Goal: explain different results in extension and shear
obtain from Hooke’s Law in 3D
If use stress and deformation tensors
T11  T22  G 2
Silicone rubber G = 160 kPa
1-10
Stress Tensor - Notation
Txx

T  Tyx
Tzx

ˆ ˆ Txx  xy
ˆ ˆ Txy  xz
ˆ ˆ Txz
T  xx
ˆ ˆ Tyx  yy
ˆ ˆ Tyy  yz
ˆ ˆ Tyz
 yx
ˆ ˆ Tzx  zy
ˆ ˆ Tzy  zz
ˆ ˆ Tzz
 zx
dyad
ˆ x  yt
ˆ y  zt
ˆ z
T  xt
ˆ ˆ Txy
xy
direction of stress on plane
plane stress acts on
Txz 

Tyz 
Tzz 
Txy
Tyy
Tzy
ˆ ,y
ˆ ,z
ˆ x
ˆ1 ,x
ˆ 2 ,x
ˆ3
x
T11
Tij  
T21

T31
3
T
i 1
T13 
T23 

T33 

T12
T22
T32
3
 xˆ xˆ T
j 1
i
j
ij
Other notation besides Tij: sij or Pij
 Tij
1-11
Rheologists use very simple T
1. Uniaxial Extension
T11 0 0
T   0 0 0
 0 0 0
T22 = T33 = 0
or
T22
T33
0
0
T  0  T22
0
0
0 
0  T11 = 0
 T33 
T22 = T33
T11 -T22 causes deformation
1-12
Consider only normal stress components
T11 0
Tij   0 T22
 0
0
Hydrostatic Pressure
 p
Tij   0

 0
T   pI
0
0 
T33 
T11 = T22 = T33 = -p
0 
1
 p 0    p 0


0  p 
 0
1 0 0
I  0 1 0


 0 0 1 
0
TI  T
0
1
0
0
0

1 
Then only t the extra or viscous stresses
cause deformation
T = -pI + t
If a liquid is
incompressible
and only the normal stress differences
cause deformation
G ≠ f(p)  ≠ f(p)
T11 - T22 = t11 - t22 ≡ N1 (shear)
1-13
Rheologists use very simple T
2. Simple Shear
x̂2
x̂2
T12
T21
x̂1
x̂3
x̂1
T21
 0 0 0
Tij  T21 0 0
 0 0 0
But to balance angular momentum
in general
x̂3
0 T12 0
Tij  0 0 0
0 0 0
T12
T21  T12
T  TT
T T   i  j (xi x jTij )T   i  j xi x jT ji
tn  nˆ T  TT nˆ
Stress tensor for simple shear
T11 T12 0 
Tij  T12 T22 0 
 0 0 T33 
Only 3 components:
T12
T11 – T22 = t11 – t22 ≡ N1
T22 – T33 = t22 – t33 ≡ N2
1-14
Stress Tensor Summary
1.
t n  nˆ  T stress at point
n
T11
T11
on any plane
2. in general T = f( time or rate, strain)
3. simple T for rheologically complex materials:
- extension and shear
4. T = pressure + extra stress = -pI + t.
5. τ causes deformation
6. normal stress differences cause deformation, t11-t22 = T11-T22
7. symmetric T = TT i.e. T12=T21
Hooke
(1678)
→
Young
(~1801)
→
Cauchy
(1830’s)
→
Gibbs
Einstein
(1880,~1905)
2
Deformation Gradient Tensor
Q
Q
s’ is a vector connecting
s
Motion
s'
two very close points in the
material, P and Q
xˆ 1
P
P
Rest or past state at t'
Deformed or present state at t
xˆ 3
Q
Q
s′
w′
y′
w
P
s′ = w′ - y′
w’ = y’ + s’
s
y
P
s=w–y
w=y+s
x = displacement function
describes how material points move
w  x( w)  x( y  s)
x
2
 s  Os
x y
 x(y)  F  s
 y  F  s
 x  y 
a new tensor !
x
F=
x
or Fij 
 xi
 xj
s  w - y  F  s
1-16
Apply F to Uniaxial Extension
Displacement functions describe how coordinates of P in undeformed state, xi‘
have been displaced to coordinates of P in deformed state, xi.
x1
x1 
x1  a1 x1
x1
x 2
x2 
x2  a 2 x2
x2
 x3
x3 
x3  a 3 x3
x3
Fij   xi  x j 
 x1  x1  a1  x1  x2  0  x1  x3  0 
  x  x  0  x  x  a  x  x  0 
2
2
2
2
3
 2 1

  x3  x1  0  x3  x2  0  x3  x3  a 3 
a1 0 0 
Fij   0 a 2 0 
 0 0 a 3 
1-17
a1 0
Fij   xi  xj   0 a
2

 0 0
0
0

a 3 
Assume:
1) constant volume V′ = V
x1x2 x3  x1x2 x3 or a1a 2a 3  1
x
x2
2
2) symmetric about the x1 axis

x3
 a2  a3
x3
0
0 
a1
a1a  1 a2  1 2
Fij   0 a11 2
0 
a1
 0
0
a11 2 
1-18
Can we write Hooke’s Law as τ  G  F - I  ?
2
2
1
Can we write Hooke’s Law as τ  G  F - I  ?
Solid Body Rotation – expect no stresses
x1  x1 cos θ  x2 sin θ
x2  x1 sin θ  x2 cos θ
x3  x3
Fij   xi  x j 
cos 
Fij   sin 

 0
 sin 
cos 
0
0
0

1 
For solid body rotation,
expect F = I
t=0
But
F≠I
F ≠ FT
Need to get rid of rotation
1-19
create a new tensor!
Finger Tensor
F·FT  V·R·V·R
T
FV R
 V·R·RT ·VT
V  stretch
 V·I·VT
R  rotation
BF F
T
 V2
 xi  x j
or Bij  Fik Fjk 
 xk  xk
Solid Body Rotation
cos 
Bij   sin 
 0
 sin 
cos 
0
0   cos 
0     sin 
1   0
sin 
cos 
0
0  1 0 0 
0   0 1 0 
1  0 0 1 
Bij gives relative local change in area within the sample.
1-20
Neo-Hookean Solid
τ  GB - I 
or
T   pI  GB
1. Uniaxial Extension
0
0 
a1
Fij   0 a11 2
0 


1 2
0
a1 
 0
0
0  a1
0
0  a12 0
0 
a1


Bij  Fik F jk   0 a11 2
0    0 a11 2
0    0 a11 0 

 

1 2
1 2
0
a1   0
0
a1   0
0 a11 
 0
t 11  G (a12  1)
t 22  G (a11  1)
t 11  t 22  G (a12 
t 22  t 33  0
1
a1
)  T11  T22  T11 
since
T22 = 0
f1
a1
1-21
2. Simple Shear
τ  G B - I
x1  x1 
x2  x2
x3  x3
s
x2  x1   x2
x '2
t 21  G
t 11  t 22  G 2
agrees with experiment
1  0 
Fij   0 1 0 
 0 0 1 
2  0
1  0   1 0 0  1  






Bij  0 1 0    1 0    
1 0

0 0 1   0 0 1   0
0
1


1-22
Silicone rubber G = 160 kPa
Finger Deformation Tensor Summary
1. B  F F area change around a point
on any plane
T
2. symmetric
3. eliminates rotation
4. gives Hooke’s Law in 3D
fits rubber data fairly well
predicts N1, shear normal stresses
for uniaxial tension B11  a1
for a1  1  e
f1
1
2
T11  T22  T11 
 G(a1  )
a1
a1
 (1  e )3  1 
T11  G 

1

e


for e  
T11
e
T11  3Ge
 tensile modulus  3G
Course Goal: Understand Principles of Rheology:
(constitutive equations)
stress = f (deformation, time)
Simplest constitutive relations:
Newton’s Law:
Hooke’s Law:
d
t 
 
dt
t  G
τ  G B - I 
Key Rheological Phenomena
•
•
•
•
shear thinning (thickening)
time dependent modulus
normal stresses in shear
extensional > shear stress
  
G(t)
N1
u > 
2
VISCOUS LIQUID
The resistance which arises
From the lack of slipperiness
Originating in a fluid, other
Things being equal, is
Proportional to the velocity
by which the parts of the
fluids are being separated
from each other.
Isaac S. Newton (1687)
t yx  
d x
dy
dv
t 
dy
t yx
Bernoulli
Newton, 1687
dvx

dy
Stokes-Navier, 1845
Material
Glass
Molten Glass (500C)
Asphalt
Molten polymers
Heavy syrup
Honey
Glycerin
Olive oil
Light oil
Water
Air
measured  in shear
1856 capillary (Poiseuille)
1880’s concentric cylinders
(Perry, Mallock,
Couette, Schwedoff)
Approximate
Viscosity
(Pa - s)
1040
1012
108
103
102
101
100
10-1
10-2
10-3
10-5
Familiar materials have a wide range in viscosity
Adapted from
Barnes et al.
(1989).
dv
t 
dy
Newton, 1687
t yx
Bernoulli
dvx

dy
Stokes-Navier, 1845
measured  in shear
1856 capillary (Poiseuille)
1880’s concentric cylinders
(Perry, Mallock,
Couette, Schwedoff)
measured in extension
1906 Trouton
u = 3
“A variety of pitch which gave by the traction method
l = 4.3 x 1010 (poise) was found by the torsion
method to have a viscosity m = 1.4 x 1010 (poise).”
F.T. Trouton (1906)
To hold his viscous pitch samples, Trouton forced a
thickened end into a small metal box. A hook was
attached to the box from which weights were hung.
polystyrene 160°C
Münstedt (1980)
Goal
1.Put Newton’s Law in 3 dimensions
• rate of strain tensor 2D
• show u = 3
recall Deformation Gradient Tensor, F
xˆ 2
Q
Q
s
Motion
s'
xˆ 1
P
P
Rest or past state at t'
Deformed or present state at t
xˆ 3
Separation and displacement
of point Q from P
Q
w′
y′
s′ = w′ - y′
s′
P
Q
w
w  x ( y, t)  F  s
s
y
P
s=w-y
s = w - y = F  s
Viscosity is “proportional to the velocity by which the parts of the
fluids are being separated from each other.” —Newton
Velocity Gradient Tensor
L is the velocity gradient tensor.
v
dv 
 dx or dv  L  dx
x
s  F  s
rate of separation

s F
s


s F
t t
t
s
 dv = F s  F dx
t



dv  F dx  L  dx
dv  F dx  L  dx

F  LF
lim F  I
x  x
or
t t

 lim F  L
x  x
Alternate notation:
v  LT    xˆ i xˆ j
i
j
 vj
 xi
Can we write Newton’s Law for viscosity as t = L?
F  VR
v θ  Ωr
•
vr  vz  0
•
•
F = V R+V R
solid body rotation
lim V  t    lim R (t )  I
x x
x x
•
•
•
lim F  L  V + R
 0 Ω 0
Lij   Ω 0 0


 0 0 0
x x
t2 ≠ t2
•
•
T
•
R is anti-symmetric
Rate of Deformation Tensor D
•
2 V  2D  L  LT  (v)T + v
Vorticity Tensor W
•
Other notation:
•
2 R  2W  L  LT
L  D+W  (v )
T
•
L  (V  R )  V  R
T
•
2D  
Example 2.2.4 Rate of Deformation Tensor is a Time Derivative of B.
Show lim
t  t
dB
 2D
dt

Thus


t  t

lim B  F+ FT

dB 
T
 B  F  F  F FT  F  FT
dt
lim F  lim FT  I

t  t

recall that lim F  L
t t
x  x

lim B  L + LT  2D
t  t
Show that 2D = 0 for solid body rotation
 0  0  0  0
2 D  L  LT    0 0   0 0  0

 

0 0  0
0 0
 0
 0  0
2W  L - LT  2   0 0


0 0
 0
Newtonian Liquid
t = 2D
or
T = -pI + 2D
Steady simple shear
Here planes of fluid slide over each other like cards in a deck.
x1  x1  γx2
x2 = x2
x3 = x3
Time derivatives of the displacement functions
for simple, shear
dx1 dγ
 x2  v 1
x2  x2 dt
dt
lim
v 1   x2 
0  0 
Lij  0 0 0 
0 0 0 
 0 0 0
L ji   0 0 
 0 0 0 
1 0 0 
 0  0
Tij   p 0 1 0     0 0 
0 0 1 
 0 0 0 
dv1
x2
dx2
dx2
 0  v2
dt
and
v2  v3  0
 0  0
2Dij   0 0 
 0 0 0 
T12  t12  t 21  
(2.2.10)
Newtonian Liquid
Steady Uniaxial Extension
time derivatives of the displacement functions at x1  x1
x1  α1 x1
dx1 dα1
dα

x1 or v  1 x1  a1 x1
dt
dt
dt
Similarly
v2  a 2 x2
v3  a 3 x3
a1 0 0 
Lij   0 a 2 0 


 0 0 a 3 
incompressible fluid (1.7.9)
v  0
or a1  a 2  a 3  0
symmetric, υ2  υ3 and thus
a 2  a3
a 2  a3  
a1
2
a1
Lij   0
 0
0
0
 e
 a1 2
0    0  e 2
0
0
 a1 2   0
0
e
0
0
 2e
2Dij  (Lij  L ji )   0
 0
0
e
0
0
0 
e 



2 
Newtonian Liquid
Apply to Uniaxial Extension
 2e
2Dij   0
 0
0
e
0
0
0 
e 
t = 2D
t 11  2e
t 22  t 33  e
From definition of extensional viscosity
t 11  t 22
u 
 3
e
Newton’s Law in 3 Dimensions
•predicts 0 low shear rate
•predicts u0 = 30
but many materials show large deviation
Polystyren
e melt
Summary of Fundamentals
1.
t n  nˆ  T stress at point on plane
n
T11
simple T - extension and shear
T = pressure + extra stress = -pI + t.
symmetric T = TT i.e. T12=T21
2. B  F F area change around a point on plane
symmetric, eliminates rotation
T
gives Hooke’s Law in 3D, E=3G
T
T
2
D

L

L

(

v
)
+ v rate of separation of particles
3.
symmetric, eliminates rotation
gives Newton’s Law in 3D,
u  3
T11
Course Goal: Understand Principles of Rheology:
stress = f (deformation, time)
NeoHookean:
τ  G B - I 
Newtonian:
t = 2D
Key Rheological Phenomena
•
•
•
•
shear thinning (thickening)
time dependent modulus
normal stresses in shear
extensional > shear stress
  
G(t)
N1
u > 