chris macosko department of chemical engineering and materials science nsf- mrsec (national science...
Post on 15-Jan-2016
231 views
TRANSCRIPT
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)
IMA Annual Program Year TutorialAn Introduction to Funny (Complex) Fluids: Rheology, Modeling and Theorems
September 12-13, 2009
Understanding silly putty, snail slime and other funny fluids
What is rheology? (Greek) =
=
rheology =
honey and mayonnaise
rate of deformation
stress= f/area
to flow
every thing flows
study of flow?, i.e. fluid mechanics?
honey and mayo
rate of deformation
viscosity = stress/rate
What is rheology? (Greek) =
=
rheology =
rubber band and silly putty
time of deformation
modulus = f/area
to flow
every thing flows
study of flow?, i.e. fluid mechanics?
honey and mayo
rate of deformation
viscosity
4 key rheological phenomena
fluid mechanician: simple fluids complex flows
rheologist: complex fluids simple flows
materials chemist: complex fluids complex flows
rheology = study of deformation of complex materials
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)
G dt
d
Simplest constitutive relations:Newton’s Law: Hooke’s Law:
• shear thinning (thickening) • time dependent modulus G(t)
• normal stresses in shear N1
• extensional > shear stress u>
Key Rheological Phenomena
1-8
ELASTIC SOLID1The power of any spring
is in the same proportion
with the tension thereof.
Robert Hooke (1678)
f L
L´
k1 k2
L´
k1
k2
ff kL
modulus
stress
Young (1805)
strain
G
1-9
Uniaxial Extension
Natural rubberG=3.9x105Pa
11 a
fT
L
L
extension
or strain 1L - L
L
a = area
natural rubber G = 400 kPa
1-10
Silicone rubber G = 160 kPaGoal: explain different results in extension and shear
obtain from Hooke’s Law in 3D If use stress and deformation tensors
= 0
= -0.4
= 0.4
Shear gives different stress response
1-11
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ
xx xy xz
yx yy yz
zx zy zz
x y z
T T T
T T T
T T T
T xx xy xz
yx yy yz
zx zy zz
T xt yt zt
3 3
1 1
ˆ ˆi j ij iji j
T x x T T
Stress Tensor - Notation
zzzyzx
yzyyyx
xzxyxx
TTT
TTT
TTT
T
ˆ ˆ xyTxydirection of stress on plane
plane stress acts on
Other notation besides Tij: ijorij
1 2 3ˆ ˆ ˆ ˆ ˆ ˆx y z x x x, , , ,
333231
232221
131211
TTT
TTT
TTT
ijT
dyad
1-12
1. Uniaxial Extension
Rheologists use very simple T
000
000
0011T
T T22 = T33 = 0
or
33
22
00
00
000
T
TT
T22 = T33
T22
T33
T11 -T22 causes deformation
T11 = 0
1-13
33
22
11
00
00
00
T
T
T
ijT
Consider only normal stress components
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1 0 0
0 1 0
0 0 1
ij
p
p p
p
p
T
T I I
T I T
Hydrostatic Pressure T11 = T22 = T33 = -p
If a liquid is incompressible
G ≠ f(p) ≠ f(p)
Then only the extra or viscous stresses cause deformation
T = -pI +
and only the normal stress differences cause deformation
T11 - T22 = 11 - 22 ≡ N1 (shear)
1-14
2. Simple Shear
But to balance angular momentum
000
000
00
000
00
000
12
21
T
TTT ijij
( )T Ti j ij i j jii j i j
T T T x x x x
Stress tensor for simple shear Only 3 components:
T12
T11 – T22 = 11 – 22 ≡ N1
T22 – T33 = 22 – 33 ≡ N2
ˆ ˆ
T T
n T T n
T
Tnt
in general
11 12
12 22
33
0
0
0 0ij
T T
T T T
T
2x̂
3x̂
1̂x
T21
T21
2x̂
3x̂
1̂x
T12
T12
21 12T T
Rheologists use very simple T
Hooke → Young → Cauchy → Gibbs Einstein
(1678) (~1801) (1830’s) (1880,~1905)
Stress Tensor Summary
T11T11
n
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 + .
5. τ causes deformation
6. normal stress differences cause deformation, 11-22 = T11-T22
7. symmetric T = TT i.e. T12=T21
1. stress at point on any plane
1-16
s'
P
Q
Motions
P
x3
x2
x1
Rest or past state at t' Deformed or present state at t
Q
ˆ
ˆ
ˆ
Deformation Gradient Tensor
( ) ( )
2
w x w x y s
xx y s Os
x y
= or iij
j
xF
x
F
s w - y F s
x
x
a new tensor !
x(y ) F s
y F s
s = w – yw = y + s
w
y
s
P
Q
s′ = w′ - y′w’ = y’ + s’
w′
y′
s′
P
Q
x = displacement functiondescribes how material points move
s’ is a vector connecting two very close points in the material, P and Q
1-17
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.
1
ij 2
3
0 0
F 0 0
0 0
1 1 1 1 2 1 3
2 1 2 2 2 2 3
3 1 3 2 3 3 3
0 0
0 0
0 0
ij i jF x x
x x x x x x
x x x x x x
x x x x x x
1-18
1
2
3
0 0
0 0
0 0
ij i jF x x
11 2
11 2
1
0 0
0 0
0 0ijF
Can we write Hooke’s Law as ? τ F - IG
Assume: 1) constant volume V′ = V
2) symmetric about the x1 axis
1-19
Can we write Hooke’s Law as ?
Solid Body Rotation – expect no stresses
For solid body rotation, expect F = I
= 0But
F ≠ IF ≠ FT
Need to get rid of rotation create a new tensor!
τ F - IG
1-20Bij gives relative local change in area within the sample.
Finger Tensor
2V
V·I·V
·VV·R·R
V·R·V·RF·F
T
TT
TT
Solid Body Rotation
cos sin 0 cos sin 0 1 0 0
sin cos 0 sin cos 0 0 1 0
0 0 1 0 0 1 0 0 1ijB
stretch
rotation
F V R
V
R
or jT iij ik jk
k k
xxB F F
x x
B F F
1-21
1. Uniaxial Extension
since T22 = 0
Neo-Hookean Solid BITI-Bτ GpG or
1-22
1 1 2 1 22
2 2
3 3
'
sx x x x x
x
x x
x x
2. Simple Shear
21
211 22
G
G
G
τ B - I
agrees with experiment
Silicone rubber G = 160 kPa
1 0
0 1 0
0 0 1ijF
21 0 1 0 0 1 0
0 1 0 1 0 1 0
0 0 1 0 0 1 0 0 1ijB
1. area change around a point on any plane
2. symmetric
3. eliminates rotation
4. gives Hooke’s Law in 3D
fits rubber data fairly well
predicts N1, shear normal stresses
Finger Deformation Tensor Summary
TB F F
Course Goal: Understand Principles of Rheology: (constitutive equations)
stress = f (deformation, time)
G dt
d
Simplest constitutive relations:Newton’s Law: Hooke’s Law:
• shear thinning (thickening) • time dependent modulus G(t)
• normal stresses in shear N1
• extensional > shear stress u>
Key Rheological Phenomena
Gτ B - I
VISCOUS LIQUID2The 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)
yx d x
dy
measured in shear1856 capillary (Poiseuille)1880’s concentric cylinders
(Perry, Mallock, Couette, Schwedoff)
dv
dy x
yx
dv
dy
Newton, 1687 Stokes-Navier, 1845
Bernoulli
Familiar materials have a wide range in viscosity
Adapted from Barnes et al. (1989).
measured in shear1856 capillary (Poiseuille)1880’s concentric cylinders
(Perry, Mallock, Couette, Schwedoff)
measured in extension1906 Trouton
u = 3
dv
dy x
yx
dv
dy
Newton, 1687 Stokes-Navier, 1845
Bernoulli
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.
“A variety of pitch which gave by the traction method = 4.3 x 1010 (poise) was found by the torsion method to have a viscosity = 1.4 x 1010 (poise).” F.T. Trouton (1906)
polystyrene 160°CMünstedt (1980)
Goal1.Put Newton’s Law in 3 dimensions
• rate of strain tensor 2D• show u = 3
Separation and displacement of point Q from P
( , t)
w y F s
s = w - y = F s
x
s = w - ys′ = w′ - y′
w′
y′
s′
P
Q
recall Deformation Gradient Tensor, F
s'
P
Q
Motions
P
x3
x2
x1
Rest or past state at t' Deformed or present state at t
Q
ˆ
ˆ
ˆ
w
y
s
P
Q
rate of separation
t
s F s
s F ss F
t t
Alternate notation:v
ˆ ˆ jTi j
i j ix
v L x x
dt
d
d d d
sv = F s F x
v F Lx x
is the velocity gradient tensor.
or
lim lim x x x xort t
d d d d
d d d
L
vv v L
v F L
F L F
F I F L
x xx
x x
Velocity Gradient Tensor
Viscosity is “proportional to the velocity by which the parts of the fluids are being separated from each other.” —Newton
θ
r z
v Ωr
v v 0
0 Ω 0
Ω 0 0
0 0 0
Lij
lim lim ( )
lim +
x x x x
x x
t t
• • •
• • •
F V R
F = V R + V R
V R I
F L V R
( )
is anti-symmetric
T T • • • •
•
L V R V R
R
Can we write Newton’s Law for viscosity as = L?
solid body rotation
2 ( ) + 2•
DV L L v vT T Rate of Deformation Tensor D
≠
2 •
D
Other notation:Vorticity Tensor W 2 2
+ ( )
•
R W L L
L D W v
T
T
t t lim 2
BShow D
d
dt
Example 2.2.4 Rate of Deformation Tensor is a Time Derivative of B.
t t
t t
lim +
recall that lim
lim + 2
Thus
B F F
F L
B L L D
T
x x
T
0 0 0 0 0 0
2 0 0 0 0 0 2 - 2 0 0
0 0 0 0 0 0 0 0 0
T TD L L W L L
Show that 2D = 0 for solid body rotation
t t t t
lim lim
BB F F F F F F
F F I
T T T
T
d
dt
Here planes of fluid slide over each other like cards in a deck.
Steady simple shear
Newtonian Liquid
1 1 2 2 2 3 3x x γx x = x x = x
= 2D or T = -pI + 2D
Time derivatives of the displacement functions for simple, shear
2 2
11 2
x
12 3
2
lim v 0 v
vv and v v 0 (2.2.10)
22
x
1 2 2
dx dγ dxx
dt dt dt
dx x
dx
12 12 21 T
0 0 0 0 0 0 0
0 0 0 0 0 2 0 0
0 0 0 0 0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0 0
ij ji ij
ij
L L D
T p
1 1
1 1 11 1 1 1
2 2 2 3 3 3
x x
α or
t
1 1 1
dx dα d x α x x v x x
d dt dt
v x v x
time derivatives of the displacement functionsat
Similarly
1
ij 2
3
0 0
L 0 0
0 0
1
1
1
0 0 0 0
L 0 2 0 0 2 0
0 0 2 0 0 2
2
2 (L L )
ij
ij ij ji
0 0
D 0 0
0 0
1 2 3
2 3
12 3 2 3
(1.7.9)
0
0
υ υ
2
incompressible fluid
or
symmetric, and thus
v
Steady Uniaxial Extension
Newtonian Liquid
Apply to Uniaxial Extension = 2D
11
22 33
2
From definition of extensional viscosity
11 22 3u
2
2 ij
0 0
D 0 0
0 0
Newton’s Law in 3 Dimensions•predicts 0 low shear rate•predicts u0 = 30
but many materials show large deviation
Newtonian Liquid
melt ePolystyren
T11T11
n1. stress at point on plane
Summary of Fundamentals
simple T - extension and shear
T = pressure + extra stress = -pI + .
symmetric T = TT i.e. T12=T21
2. area change around a point on planeTB F Fsymmetric, eliminates rotation
gives Hooke’s Law in 3D, E=3G
3. rate of separation of particles 2 ( ) + T T D L L v v
symmetric, eliminates rotation
gives Newton’s Law in 3D, 3u
Course Goal: Understand Principles of Rheology:
stress = f (deformation, time) NeoHookean: Newtonian:
• shear thinning (thickening) • time dependent modulus G(t)
• normal stresses in shear N1
• extensional > shear stress u>
Key Rheological Phenomena
Gτ B - I = 2D