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?

ρειν (Greek) = to flow τα παντα ρει = every thing flows rheology = study of flow?, i.e. fluid mechanics?

honey and mayonnaise honey and mayo

stress= viscosity = f/area stress/rate

rate of deformation rate of deformation

What is rheology?

ρειν (Greek) = to flow τα παντα ρει = every thing flows rheology = study of flow?, i.e. fluid mechanics?

honey and mayo rubber band and silly putty

viscosity modulus = f/area

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: dγ τ = η = ηγ τ = Gγ dt

Key Rheological Phenomena • shear thinning (thickening)     • time dependent modulus G(t)

• normal stresses in shear N1

η η • extensional > shear stress u > k k 1 ELASTIC SOLID 1 2

The power of any spring L´ is in the same proportion with the tension thereof. Robert Hooke (1678) k1 k f ∝ ∆L 2 f f = k∆L L´

Young (1805)

modulus

τ = Gγ stress τ

1-8 strain Uniaxial Extension

Natural rubber G=3.9x105Pa

f T = 11 a

L extension α = L′ L - L or strain     1 L a = area

natural rubber G = 400 kPa 1-9 Shear gives different stress response

γ = 0

γ = -0.4

γ = 0.4

Goal: explain different results in extension and shear Silicone rubber G = 160 kPa obtain from Hooke’s Law in 3D If use stress and deformation te nsors 1-10 Stress Tensor - Notation

T  xˆxˆTxx  xˆyˆTxy  xˆzˆTxz T T T   xx xy xz  T = T T T yˆxˆTyx  yˆyˆTyy  yˆzˆTyz  yx yy yz    Tzx Tzy Tzz  zˆxˆTzx  zˆyˆTzy  zˆzˆTzz

dyad xˆ ,yˆ ,zˆ  xˆ 1 ,xˆ 2 ,xˆ 3 T  xˆt x  yˆt y  zˆt z   T11 T12 T13 =  Tij T21 T22 T23    T31 T32 T33  xyˆ ˆT xy 3 3 direction of stress on plane T  xˆ ixˆ jTij  Tij plane stress acts on i1 j1

σ Π Other notation be sides Tij: ij or ij 1-11 Rheologists use very simple T 1. Uniaxial Extension

T 0 0  11  T =  0 0 0 T22 = T33 = 0  0 0 0

or

T 22 0 0 0    T = 0 −T 0  22  T11 = 0  −  T33 0 0 T33 

T22 = T33

1-12 T11 -T22 causes deformation Consider only normal stress components   T11 0 0 =   Tij  0 T22 0     0 0 T33 

Hydrostatic Pressure T11 = T22 = T33 = -p

p 0 0 1 0 0

Tij  0  p 0  p 0 1 0 0 0  p 0 0 1 1 0 0 T   pI I  0 1 0 Then only τ the extra or viscous stresses 0 0 1 cause deformation

T I  T T = -pI + τ

If a liquid is and only the normal stress differences incompressible cause deformation 1-13 G ≠ f(p) η ≠ f(p) τ τ T11 - T22 = 11 - 22 ≡ N1 (shear) Rheologists use very simple T

2. Simple Shear T xˆ2 12 xˆ2

T21

xˆ1 xˆ1 T21 xˆ3 xˆ3     T 0 0 0 0 T12 0 12 =   =   Tij T21 0 0 Tij 0 0 0  0 0 0 0 0 0

But to balance angular momentum T21  T12

T T in general T T  (x x T )  x x T T  T i j i j ij i j i j ji T tn  nˆ gT  T gnˆ

Stress tensor for simple shear Only 3 components:

T11 T12 0 T12 Tij  T12 T22 0 τ τ T11 – T22 = 11 – 22 ≡ N1 0 0 T 1-14 33 τ τ T22 – T33 = 22 – 33 ≡ N2 Stress Tensor Summary

1. stress at point n T T on any plane 11 11 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 + τ. 5. τ causes deformation τ τ 6. normal stress differences cause deformation, 11- 22 = T11-T22 T 7. symmetric T = T i.e. T12=T21

Hooke → Young → Cauchy → Gibbs Einstein

(1678) (~1801) (1830’s) (1880,~1905) ˆ x2 Q Deformation Gradient Tensor

Q s Motion s’ is a vector connecting s' ˆ two very close points in the x1 P P material, P and Q

Rest or past state at t' Deformed or present state at t xˆ 3 Q Q w′ w s′ s

y′ P y P s′ = w′ - y′ s = w – y w’ = y’ + s’ w = y + s a new tensor ! x = displacement function x x describes how material points move F= or F  i x ij x w  x(w )  x(y  s ) j x  x  y   s  Os 2 x y s  w - y  F s  x(y )  F s 1-16  y  F s 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.

Fij  xi x j 

x1 x1  1 x1 x2  0 x1 x3  0

x2 x1  0 x2 x2  2 x2 x3  0

x3 x1  0 x3 x2  0 x3 x3  3

1 0 0

Fij  0 2 0 1-17 0 0 3 1 0 0 Fij  xi x j  0 2 0

0 0 3

Assume: 2) constant volume V′ = V

2) symmetric about the x1 axis

1 0 0 1 2 Fij  0 1 0 1 2 0 0 1 Can we write Hooke’s Law as τ  G  F - I  ? 1-18 Can we write Hooke’s Law as τ  G  F - I  ?

Solid Body Rotation – expect no stresses

For solid body rotation, expect F = I τ = 0 But F ≠ I F ≠ FT

Need to get rid of rota1t-io19n create a new tensor! Finger Tensor T F·FT = ( V·R)·( V·R) F  VgR = V·R·RT ·VT V  stretch = V·I·VT R  rotation = V 2

T xi x j B  FgF or Bij  Fik Fjk  xk xk Solid Body Rotation cos sin 0 cos sin 0 1 0 0

Bij  sin cos 0 sin cos 0  0 1 0 0 0 1 0 0 1 0 0 1

Bij gives relative local change in are a within the sample. 1-20 Neo-Hookean Solid τ = G( B - I) or T = − pI + GB

1. Uniaxial Extension

since T 22 = 0 1-21 2. Simple Shear

τ  G  B - I    G s 21 2 x1  x1  x2  x1   x2 11  22  G x '2 x x 2  2 agrees with experiment x3  x3 1  0

Fij  0 1 0 0 0 1

2 1  0 1 0 0 1   0

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 g F T 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

Course Goal: Understand Principles of Rheology: (constitutive equations)

stress = f (deformation, time) Simplest constitutive relations: Newton’s Law: Hooke’s Law: dγ τ = η = ηγ τ = Gγ dt τ  G  B - I  Key Rheological Phenomena • shear thinning (thickening)     • time dependent modulus G(t)

• normal stresses in shear N1

η η • extensional > shear stress u > 2 VISCOUS LIQUID The resistance which arises

From the lack of slipperiness du x t yx=h Originating in a fluid, other dy 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)

dv dvx     yx   dy dy Bernoulli measured η in shear 1856 capillary (Poiseuille) Newton, 1687 Stokes-Navier, 1845 1880’s concentric cylinders (Perry, Mallock, Couette, Schwedoff)

Material Approximate Viscosity (Pa - s) Glass 1040 Molten Glass (500°C) 1012 Asphalt 108 Molten polymers 103 Heavy syrup 102 Honey 101 Glycerin 100 Olive oil 10-1 Light oil 10-2 Water 10-3 -5 Air 10 Adapted from Barnes et al. (1989). Familiar materials have a wide range in viscosity dv dvx     yx   dy dy Bernoulli measured η in shear 1856 capillary (Poiseuille) Newton, 1687 Stokes-Navier, 1845 1880’s concentric cylinders (Perry, Mallock, Couette, Schwedoff)

measured in extension 7 Trouton η η u = 3

“A variety of pitch which gave by the traction method λ = 4.3 x 1010 (poise) was found by the torsion method to have a viscosity µ = 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 2.Put Newton’s Law in 3 dimensions • rate of strain tensor 2D η η • show u = 3 recall Deformation Gradient Tensor, F ˆ x2

Q

Q s Motion s' ˆ x1 P P

Rest or past state at t' Deformed or present state at t ˆ x3

Separation and displacement of point Q from P Q Q w  x(y , t)  F s w′ w s′ s P y′ y P s = w - y = F s s′ = w′ - y′ s = w - y 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 dv  F dx  L dx s F s  s  F t t t F  L F

s lim F  I  lim F  L x x x x dv = F s  F dx or t t t Alternate notation: dv F dx L dx   T v j v  L  xˆ i xˆ j i j xi

Can we write Newton’s Law for viscosity as τ = ηL?

F  V R vΩθ r • • • v r  vz  0 F = VgR + VgR solid body rotation lim V t  lim R(t )  I x x   x x

• • • lim F  L  V+ R 0Ω 0 x x • • • • L Ω 0 0 τ ≠ τ T T ij   12 21 L  (V R )  V R 0 0 0 • R is anti-symmetric Rate of Deformation Tensor D • 2 V  2D  L  LT  ( v)T + v Other notation: • Vorticity Tensor W T 2 R  2W  L  L • 2D   L  D+W  ( v)T Example 2.2.4 Rate of Deformation Tensor is a Time Derivative of B.

dB Show lim  2D t t dt Thus

T dB T T T lim B  F+ F  B  F F  F F  F F t t dt

T recall that lim F  L lim F  lim F  I x x t t t t lim B  L + LT  2D t t

Show that 2D = 0 for solid body rotation

0  0 0  0 0  0 2D  L  LT   0 0   0 0  0 2W  L - LT  2  0 0 0 0 0 0 0 0 0 0 0

Newtonian Liquid τ = η2D or T = -pI + η2D Steady simple shear Here planes of fluid slide over each other like cards in a deck.

x1  x1γx 2 x = 2x 2 x = 3x 3 Time derivatives of the displacement functions for simple, shear

dx1 dγ dx2 lim  x2  v1  0  v2 x2 x2 dt dt dt

dv1 v1   x2  x2 and v2  v3  0 (2.2.10) dx2

0  0 0 0 0 0  0

Lij  0 0 0 Lji   0 0 2Dij   0 0 0 0 0 0 0 0 0 0 0

1 0 0 0  0

Tij   p 0 1 0   0 0 0 0 1 0 0 0 T12  12   21   Newtonian Liquid Steady Uniaxial Extension

time derivatives of the displacement functionsat x1 x1 dx dα dα xαx 1  x 1 ovr  x 1 x 1 1 1 dt dt 1 dt 1 1 1

Similarly v2   2 x2 v3   3x3

1 0 0

Lij  0  2 0

0 0  3 incompressible fluid(1.7.9)

v  0 1 0 0  0 0   or 1  2  3  0 Lij  0 1 2 0  0  2 0

0 0 1 2 0 0  2 symmetric, υ2  υ3 and thus

         1 2 3 2 3 2 2 0 0 2D  (L  L )  0  0 ij ij ji 0 0  Newtonian Liquid Apply to Uniaxial Extension τ = η2D

2 0 0 11  2 2Dij  0  0  0 0   22   33  

From definition of extensional viscosity     11 22  3 u 

Newton’s Law in 3 Dimensions Polystyrene melt η •predicts 0 low shear rate η η •predicts u0 = 3 0

but many materials show large deviation Summary of Fundamentals n 1. stress at point on plane T T11 11 simple T - extension and shear T = pressure + extra stress = -pI + τ. T symmetric T = T i.e. T12=T21

2. B  F g F T area change around a point on plane symmetric, eliminates rotation gives Hooke’s Law in 3D, E=3G

T T 3. 2D  L  L  ( v) + v rate of separation of particles symmetric, eliminates rotation

gives Newton’s Law in 3D, u  3 Course Goal: Understand Principles of Rheology:

NsetroeHsoso k=e af n(:d e f o r m Naetiwotno,n iatimn:e) τ  G  B - I  τ = η2D

Key Rheological Phenomena • shear thinning (thickening)     • time dependent modulus G(t)

• normal stresses in shear N1 η η • extensional > shear stress u >