Rheology of Soft Materials Thomas G. Mason Department of Chemistry and Biochemistry Department of Physics and Astronomy California NanoSystems Institute Τ ω Τ ω γ τ γ γ (t) 0 0 ω τ (t) 2π ω γ ω t Δt © 2006 by Thomas G. Mason All rights reserved. Rheology Study of deformation and flow of materials Elastic Solids Viscous Liquids And everything in between… “Soft Materials” Emulsions, Polymers, Glassy Materials Thermally agitated colloidal structures (1 nm - 1 µm) Describe mechanical constitutive relationships: material response Focus on isotropic disordered materials (Math is more complicated for partially or fully ordered materials) Rheology of Soft Materials © 2006 Thomas G. Mason 1 Rheology - Outline Basic Definitions: Stress and Strain Common Shear Flows and Simple Examples Equations of Continuity and Momentum Linear Shear Viscoelastic Rheology Introduction to Non-linear Rheology Macroscopic Mechanical Shear Rheometry Example of a Soft Glassy Material: Concentrated Emulsions Rheology of Soft Materials © 2006 Thomas G. Mason Affine and Non-Affine Shear Deformations Affine Non-Affine All points in the material Different points in the material deform uniformly deform differently “Homogeneous flow” Apparent macroscopic strain is not the local strain everywhere Local non-affine motion of constituents in soft materials can have an important impact on their rheology For simplicity, we focus on affine rheology Rheology of Soft Materials © 2006 Thomas G. Mason 2 Incompressible Soft Materials Isothermal Compressibility 1 ∂V β = − V ∂p T Important for gases, but much smaller for condensed matter € Mass Conservation of Fluid Flow: Equation of Continuity ∂ρ r r ρ is density = −(∇ ⋅ ρv ) v is velocity ∂ t r Constant density implies: r (Incompressible) ( ∇ ⋅ v ) = 0 € Good approximation for most complex fluids, but not for foams € Rheology of Soft Materials © 2006 Thomas G. Mason Example of Shear Flow Shear Deformations: Force is applied along surface, not normal to it Unidirectional Simple Shear Flow Strain Tensor parallel plates (plane Couette) 0 γ 0 v y 0 0 γ = γ z x 0 0 0 fixed gap separation: d Δx(t) Rate of Strain Tensor Shear Strain: 0 ˙ 0 γ = Δx/d € γ r r † γ˙ = γ˙ 0 0 = [∇ vr + (∇ vr ) ] Shear Strain Rate: 0 0 0 γ˙ = v(t)/d = Δx˙ (t)/d velocity gradient tensor Convention: x is along shear direction (1-direction) y is normal to shearing surface (2-direction) € z is tangential€ to shearing surface (3-direction) Rheology of Soft Materials © 2006 Thomas G. Mason 3 Example of Extensional Flow Planar Extensional (Elongational) Flow “Pure Shear Flow” 4 roll mill y roller z x extension rate Rate of Strain Tensor ε ˙ 0 0 ˙ 0 ˙ 0 γ = −ε 0 0 0 Rheology of Soft Materials © 2006 Thomas G. Mason € Applying Shear Stresses and Pressures Fxy Area of Face: A Simple isotropic pressure: y − p 0 0 x p = 0 −p 0 = −pδ z Fxx F 0 0 −p τxy = Fxy/A, … xy Shear Stress Tensor Total Stress Tensor 0 τ xy τ xz € −p τ xy τ xz τ = τ yx 0 τ yz σ = τ − pδ = τ yx −p τ yz τ zx τ zy 0 τ zx τ zy −p Total description including shear-free flows is more complex (we’ve neglected normal stresses , , for now) € € τxx τyy τzz Rheology of Soft Materials © 2006 Thomas G. Mason 4 Momentum Conservation: Navier-Stokes Equation Newton’s Law for Fluid Elements Form for Simple Shear Flow ∂ r r ρvr = −[∇ ⋅ ρvr vr ]+[∇ ⋅σ ]+ ρgr ∂ t Assume viscous dissipation: σ = −pδ +ηγ˙ η is the viscosity neglect dilational viscosity € A Very Simple Rheological Equation: Navier-Stokes Equation € D r r ρ vr = −∇ p +η∇ 2vr + ρgr Dt r r where D = ∂ +v ⋅∇ is the convective or substantial derivative operator Dt ∂t (derivative operator in reference frame of moving liquid) € €We’ll simplify to scalar equations mostly from here on out Rheology of Soft Materials © 2006 Thomas G. Mason Simple Viscous Liquids and Elastic Solids Viscous Liquid . Stress: τ Strain Rate: γ(t) Viscosity resists motion proportional to rate: τ v = −ηγ˙ “Newtonian Liquid” Elastic Solid Stress: τ € Strain: γ(t) Elasticity resists motion proportional to strain: τ e = −Gγ “Hookean Solid” Both η and G are intensive thermodynamic properties Rheology of Soft Materials € © 2006 Thomas G. Mason 5 Thermodynamic Definition: Shear Modulus G Helmholtz Free Energy γ 2 F = F + VG + O[γ 4 ] 0 2 Intensive form of Hooke’s Law: differentiate once w.r.t. strain 1 ∂F € τ = − = −Gγ V ∂γ γ=0 γ = 0 is used to eliminate any nonlinear dependence Elastic Shear Modulus: differentiate again € 1 ∂ 2F φ ∂ 2F G = = related to curvature V 2 V 2 of energy well ∂γ γ=0 d ∂γ γ=0 Rheology of Soft Materials © 2006 Thomas G. Mason € € Linear Viscoelastic Response of Soft Materials Soft materials: colloidal components dispersed in a viscous liquid Examples: polymer chains, clay platelets, emulsion droplets… Strain response of material is dependent on strength of interactions between components (e.g. polymer chains, clay platelets, droplets) and relaxation time scales of microstructure Stress: τ Strain: γ(t) Response for small strains is neither perfectly viscous nor elastic Mechanical response is called “viscoelastic” Rheology of Soft Materials © 2006 Thomas G. Mason 6 Linear Shear Viscoelastic Rheology Study of small (near-equilibrium) deformation response of materials Key Idea There is only one real function of one real variable (e.g. time t) that describes the relaxation of shear stress Stress Relaxation Modulus: Gr(t) There are many equivalent ways of representing the information contained in this one real function Gr(t) Preference for a particular representation is based on expt. method This can be very confusing to people new to the field Rheology of Soft Materials © 2006 Thomas G. Mason Stress Response to a Step Strain Step the strain: γ(t) = γ0 u(t) Measure Shear Stress: τ(t) Strain: γ(t) = γ0 <<1 Define a function, the Stress Relaxation Modulus Gr(t): τ(t) Gr (t) ≡ γ0 This one function contains all of the information about the equilibrium (linear) stress-strain response of the soft material € Simple viscous liquid: Gr(t) = η δ(t) Simple elastic solid: Gr(t) = G u(t) Rheology of Soft Materials © 2006 Thomas G. Mason 7 Strain Response to a Step Stress Step the stress: τ = τ0 u(t) Measure Strain: γ(t) Define an Equivalent Function, the Creep Compliance, J(t): J(t) = γ(t)/τ0 In principle, if measured over a large enough dynamic range in J and t, we can perform mathematical manipulations to determine J(t) from Gr(t). This is cumbersome and is usually done numerically Simple viscous liquid: J(t) = t/η Simple elastic solid: J(t) = (1/G) u(t) Rheology of Soft Materials © 2006 Thomas G. Mason Modeling Viscoelastic Materials Spring and Dashpot Mechanical Models Are Useful For Schematically Representing Relaxation Modes in Soft Materials G Hookean Spring τ(t) = -G γ(t) Newtonian Dashpot τ(t) = −η γ˙ (t) η Maxwell combined€ the ideas behind these equations: η τ + τ˙ = −ηγ˙ G This is one of the first viscoelastic equations Rheology of Soft Materials € © 2006 Thomas G. Mason 8 General Linear Viscoelastic Model Use as many springs and dashpots in series or in parallel to model the response function of the material A solution of the set of coupled differential equations and crossover from discrete to continuum notation yields: (t) t G (t t )˙ (t )dt Check: step strain -> τ = − ∫−∞ r − ′ γ ′ ′ delta function strain rate The shear stress is related to the convolution integral of the stress relaxation modulus € with the strain rate history You have to know the history of applied shear! Equivalently: Memory function: t (t) M (t t ′) (t ′) dt ′ ∂Gr (t − t ′) τ = ∫−∞ − γ M (t − t ′) = ∂t ′ Rheology of Soft Materials © 2006 Thomas G. Mason € € Frequency Domain Representation: Complex Viscosity What is the Unilateral Fourier Transform of Gr(t)? Answer: the complex frequency-dependent viscosity ∞ η *(ω) = ∫0 Gr (t ′) exp(−iωt ′) dt ′ ∞ η *(ω) = ∫0 Gr (t ′)[cos( ωt ′) − isin(ωt ′ )]dt ′ ∞ ∞ η€* (ω) = [ ∫0 Gr (t ′) cos(ωt ′ )dt ′] − i[ ∫0 Gr (t ′) sin(ωt ′) dt ′ ] in-phase w.r.t. strain rate 90° out-of-phase w.r.t. strain rate € Complex Viscosity: η *(ω) = η ′( ω)− iη ′′ ( ω) € ∞ η ′( ω) = ∫0 Gr (t ′) cos(ωt ′ )dt ′ ∞ €η ′ ′ ( ω) = ∫0 Gr (t ′) sin(ωt ′) dt ′ Rheology of Soft Materials€ © 2006 Thomas G. Mason € 9 Frequency Domain: Complex Shear Modulus Another equivalent ω-domain representation of Gr(t) Complex Modulus: G*(ω) = iωη*(ω) pick up from time derivative G *(ω) = G ′( ω)+ iG ′′ ( ω) Real Part: G’(ω) “Storage Modulus” € Imaginary Part: G’’(ω) “Loss Modulus” Complex “phasor notation” is mathematically convenient But in reality, we measure real functions of real variables Rheology of Soft Materials © 2006 Thomas G. Mason Forced Parallel Plate Oscillatory Viscometry Controlled Strain Oscillations γ (t) γ 0 τ 0 γ(t) = γ0 sin(ωt) τ (t) γ˙ (t) = γ˙ 0 cos(ωt) 2π ω t Δt € Measure: Stress τ(t) phase lag: δ = ωΔt € Keep γ0 small enough that the stress response is also sinusoidal Measured Stress: τ(t) = τ 0(ω) sin [ωt + δ(ω)] τ 0 (ω) τ 0 (ω) τ(t) = cosδ(ω) γ0 sinωt + sinδ(ω) γ0 cosωt γ0 γ0 € In-phase w.r.t. strain 90° out-of-phase w.r.t. strain τ 0 (ω) τ 0 (ω) € G ′( ω) = cosδ(ω) G ′′ ( ω) = sinδ(ω) γ0 γ0 Rheology of Soft Materials © 2006 Thomas G. Mason € € 10 Forced Parallel Plate Oscillatory Viscometry Controlled Strain Oscillations γ (t) γ 0 τ 0 γ(t) = γ0 sin(ωt) τ (t) γ˙ (t) = γ˙ 0 cos(ωt) 2π ω t Δt € Measure: Stress τ(t) phase lag: ωΔt € (t) t G (t t )˙ (t )dt t G (t t )˙ cos( t )dt τ = − ∫−∞ r − ′ γ ′ ′ = − ∫−∞ r − ′ γ 0 ω ′ ′ ∞ τ(t) = −γ˙ 0 ∫0 Gr (s)cos[ω(t − s)]ds s = t − t ′ ∞ ∞ € τ(t) = −[ ∫0 Gr (s)cos(ωs)ds]γ˙ 0 cos(ωt)−[ ∫0 Gr (s)sin(ωs)ds]γ˙ 0 sin(ωt) € In-phase w.r.t strain: € G ′( ω) = ωη ′′ ( ω) 90° out-of-phase w.r.t strain: G ′′ ( ω) = ωη ′( ω) € Rheology of Soft Materials © 2006 Thomas G.