1. Chris Macosko Department of Chemical Engineering and Materials Science NSF- MRSEC (National Science Foundation sponsored Materials Research Science and Engineering Center) IP RIME (Industrial Partnership for Research in Interfacial and Materials Engineering) 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

2. 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

3. 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. 4 key rheological phenomena

5. 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

6. from: Rheology: Principles, Measurement and Applications, VCH/Wiley (1994). ad majorem Dei gloriam

7. Goal: Understand Principles of Rheology: (stress, strain, constitutive equations) stress = f (deformation, time) Simplest constitutive relations: Newton’s Law:Hooke’s Law: shear thinning (thickening) time dependent modulus G(t) normal stresses in shear N 1 extensional > shear stress  u  >  Key Rheological Phenomena

8. 1-8 ELASTIC SOLID 1 The power of any spring is in the same proportion with the tension thereof. Robert Hooke (1678 ) f  L L´ k1k1 k2k2 k1k1 k2k2 f f  k  L modulus stress  Young (1805) strain

9. 1-9 Uniaxial Extension Natural rubber G=3.9x10 5 Pa a = area natural rubber G = 400 kPa

10. 1-10 Silicone rubber G = 160 kPa Goal: 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

11. 1-11 Stress Tensor - Notation direction of stress on plane plane stress acts on Other notation besides T ij :  ij  or  ij dyad

12. 1-12 1. Uniaxial Extension Rheologists use very simple T T 22 = T 33 = 0 or T 22 = T 33 T 22 T 33 T 11 -T 22 causes deformation T 11 = 0

13. 1-13 Consider only normal stress components Hydrostatic Pressure T 11 = T 22 = T 33 = -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 T 11 - T 22 =  11 -  22 ≡ N 1 (shear)

14. 1-14 2. Simple Shear But to balance angular momentum Stress tensor for simple shear Only 3 components: T 12 T 11 – T 22 =  11 –  22 ≡ N 1 T 22 – T 33 =  22 –  33 ≡ N 2 in general T 21 T 12 Rheologists use very simple T

15. Hooke→Young→Cauchy→Gibbs Einstein (1678)(~1801)(1830’s)(1880,~1905) Stress Tensor Summary T 11 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 = T 11 -T 22 7. symmetric T = T T i.e. T 12 =T 21 1. stress at point on any plane

16. 1-16 Deformation Gradient Tensor a new tensor ! s = w – y w = y + s w y s P Q s′ = w′ - y′ w’ = y’ + s’ w′ y′ s′ P Q x =displacement function describes how material points move s’ is a vector connecting two very close points in the material, P and Q

17. 1-17 Apply F to Uniaxial Extension Displacement functions describe how coordinates of P in undeformed state, x i ‘ have been displaced to coordinates of P in deformed state, x i.

18. 1-18 Can we write Hooke’s Law as ? Assume: 1)constant volume V′ = V 2) symmetric about the x 1 axis

19. 1-19 Can we write Hooke’s Law as ? Solid Body Rotation – expect no stresses For solid body rotation, expect F = I  = 0 But F ≠ I F ≠ F T Need to get rid of rotation create a new tensor!

20. 1-20 B ij gives relative local change in area within the sample. Finger Tensor Solid Body Rotation

21. 1-21 1. Uniaxial Extension since T 22 = 0 Neo-Hookean Solid

22. 1-22 2. Simple Shear agrees with experiment Silicone rubber G = 160 kPa

23. 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 N 1, shear normal stresses Finger Deformation Tensor Summary

24. Course Goal: Understand Principles of Rheology: (constitutive equations) stress = f (deformation, time) Simplest constitutive relations: Newton’s Law:Hooke’s Law: shear thinning (thickening) time dependent modulus G(t) normal stresses in shear N 1 extensional > shear stress  u  >  Key Rheological Phenomena

25. VISCOUS LIQUID 2 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)

26. measured  in shear 1856 capillary (Poiseuille) 1880’s concentric cylinders (Perry, Mallock, Couette, Schwedoff) Newton, 1687 Stokes-Navier, 1845 Bernoulli Familiar materials have a wide range in viscosity Adapted from Barnes et al. (1989).

27. measured  in shear 1856 capillary (Poiseuille) 1880’s concentric cylinders (Perry, Mallock, Couette, Schwedoff) measured in extension 1906Trouton  u = 3  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 10 10 (poise) was found by the torsion method to have a viscosity  = 1.4 x 10 10 (poise).” F.T. Trouton (1906)

28. polystyrene 160°C Münstedt (1980) Goal 1.Put Newton’s Law in 3 dimensions rate of strain tensor 2D show  u = 3 

29. Separation and displacement of point Q from P s = w - y s′ = w′ - y′ w′ y′ s′ P Q recall Deformation Gradient Tensor, F w y s P Q

30. Alternate notation : Velocity Gradient Tensor Viscosity is “proportional to the velocity by which the parts of the fluids are being separated from each other.” —Newton

31. Can we write Newton’s Law for viscosity as  =  L ? solid body rotation Rate of Deformation Tensor D   ≠   Other notation: Vorticity Tensor W

32. Example 2.2.4 Rate of Deformation Tensor is a Time Derivative of B. Show that 2D = 0 for solid body rotation

33. Here planes of fluid slide over each other like cards in a deck. Steady simple shear Newtonian Liquid  =  2D or T = -pI +  2D Time derivatives of the displacement functions for simple, shear

34. Steady Uniaxial Extension Newtonian Liquid

35. Apply to Uniaxial Extension  =  2D From definition of extensional viscosity Newton’s Law in 3 Dimensions predicts  0 low shear rate predicts  u0 = 3  0 but many materials show large deviation Newtonian Liquid

36. T 11 n 1. stress at point on plane Summary of Fundamentals simple T - extension and shear T = pressure + extra stress = -pI + . symmetric T = T T i.e. T 12 =T 21 2. area change around a point on plane symmetric, eliminates rotation gives Hooke’s Law in 3D, E=3G 3. rate of separation of particles symmetric, eliminates rotation gives Newton’s Law in 3D,

37. Course Goal: Understand Principles of Rheology: stress = f (deformation, time) NeoHookean: Newtonian: shear thinning (thickening) time dependent modulus G(t) normal stresses in shear N 1 extensional > shear stress  u  >  Key Rheological Phenomena  =  2D