Deformation of Polymers

Transcription

1 Department of Engineering Science University of Oxford Deformation of Polymers 4 Lectures for Paper C3 Prof. J.C. Tan Hilary Term

2 Contents of the lectures 1. Introduction 2. Preliminaries (B5 lectures): Large deformations & rotations, Stresses in the presence of finite deformations and rotations 3. Elasticity of polymers: Linear elasticity; Hyperelasticity; Phenomenological & physically-based models 4. Viscoelasticity: Linear viscoelasticity in 3D; Relaxation and retardation spectra; Thermal activation; Glass transition 2

3 1. Introduction The subject of these lectures is materials modelling applied to the deformation of polymers. What does this mean?... the business of describing quantitatively, via mathematics, the 3D deformation that occurs in polymer solids under the action of applied loads. 3

4 Why is this subject important in Engineering? Designers of manufacturing processes and of products need to be able to predict quantitatively in advance the deformation that will occur under applied loading. O-rings, gears, impellers, Materials engineers, developing and selecting materials for particular applications, need to predict how the (3D) deformation of a polymers depends on their composition and structure. Orthopaedic implants (UHMW-PE knee meniscus) 4

5 Computer-based engineering analyses Engineers nowadays use finite element (FE) numerical analysis to predict the complex patterns of stress and strain that usually arise in 3D... But as always, such computer-based analyses follow the general rule of computer programs: FE model of an artificial knee implant Rubbish input gives rubbish output So it is essential to feed FE analyses with good models of the material behaviour! 5

6 How do we create mathematical models of material behaviour? Where does the information come from? It comes to us from two sources at two different length-scales, which comprise the macro-world and the micro-world 6

7 Handouts P.4-5 (I) Macro-world > 1 mm few m This is the everyday world you are familiar with, visible with the naked human eye! (a) A few known physical laws applying at this length scale (e.g. Newton s laws, the laws of thermodynamics...) (b) Our idealisations of the material (a single continuum, assume some known symmetry...) (c) Experimental data of specific materials, specific test conditions & parameters, specific samples... Phenomenological Models 7

8 (II) Micro-world < 1 mm (a) Using microscopy and other advanced characterisation tools (eg. EMs, X-rays, neutrons) to gain insights into materials composition and microstructure TEM (b) Known physics and chemistry provide useful theories to devise more powerful models that capture more aspects of material behaviour (c) Guide us in making predictions of how the material is likely to behave in the macroworld Physically-based Models Synchrotron radiation (Diamond Light Source) 8

9 Handout P.5 What polymers are? Polymers are materials consisting of very long molecules, made up of many repeating chemical units (the monomer units), covalently bonded together. Amorphous (e.g. PS, PMMA, LDPE, atactic PVC) Semicrystalline (e.g. HDPE, Nylon, isotactic PVC) 9

10 linear Molecular topology e.g. linear PE ( high density polyethylene or HDPE) branched crosslinked e.g. branched PE with 6-10 branches per 100 carbon atoms ( low density polyethylene or LDPE) e.g. cross-linked polyethylene 10

11 What are the key features of the chemical architecture of a polymer? 1. Chemical composition(s) of its monomers 2. The number of monomers in each molecule (the degree of polymerisation = n) 3. The manner of covalent connection between the monomers (the molecular topology ) H H CH 3 * C H C H n * * C O C CH 3 O O * n Polyethylene (PE) (2 main-chain covalent bonds (n b0 ), 28 amu: molar mass M 0 = 28 g/mol) Polycarbonate (6 main-chain covalent bonds, M 0 = 254 g/mol) 11

12 Handout P.7 Molecule axis Stretched out lengths of polymers b Let the number of main chain bonds per monomer be n b0 b p a polyethylene (PE) molecule Then the total number of such bonds per molecule is n b = n n b0 n Let the mean main-chain bond length, as projected onto the axis of the molecule when stretched out straight, be b p Then the stretched out length of one monomer, L 0, and that of the whole molecule, L, are respectively L = n b ; L = nl = n b 0 b0 p 0 b p (1.1) 12

13 Structure of polymer crystals e.g. A lamella-shaped single crystal of polyethylene (PE) The crystals are tiny (10 50 nm thickness) looking down c-axis Its unit cell c the chains pack parallel along c-axis (254 pm) Thickness c-axis looking down c-axis The molecules are aligned normal to the lamella-shaped crystal 13

14 Handout P.7 Microstructures of spherulites in PE In a polymer crystallised from the melt, instead of lamellae forming independently, stacks of them grow outwards together from each nucleus, forming a spherulite. 200 μm Maltese crosses visible under cross-polarisers 14

15 Handout P.8 Sizes of typical structural features of polymers Micro-world vs. Typical size of polymer-based engineering components/products in the Macro-world : a few millimetres (mm) up to several metres (m) 15

16 Handout P.9 2. Preliminaries (B5 Revision) 2.1 Large deformations and rotations We must use a mathematical description that does not suffer from restriction to small strains Require a formulation that is valid for arbitrarily large deformations and rotations Assume it is a continuum: the new position x is a smooth function of the old position X The vector displacement: u = x X 16

17 Handout P.10 Deformation gradient F It follows that we can arbitrarily pick another point in the material, initially at position Q 0, close to P 0,... track it to its new position Q, by applying a transformation F to their initial vector separation dx to obtain their new vector separation dx dx = F d X; or in component terms dx = F dx F is called the Deformation Gradient It fully specifies the deformation and rotation of the sample of material as it transforms from B 0 to B F ij i ij j x = i = X j 0 x 17

18 Rigid body rotation R F can always be expressed as the product of an orthogonal matrix R and a symmetric matrix U or V: F = VR = RU R corresponds to a pure rigid body rotation U and V correspond to pure deformations The eigen values of U and V are equal (i.e. differ by only a rotation), and they are called the Principal Stretches Definition: stretch λ = new length original length λ1, λ2, λ3 18

19 Handout P.11 Left Cauchy Green tensor B Principle of Material Frame Indifference : Stresses that give rise to the deformation must be independent of any rigid body rotation R Consequently, to describe the deformation we do not use F itself, but instead we could use U or V... but actually it turns out to be more convenient to use instead the left Cauchy Green tensor B or right Cauchy Green tensor C defined Here we shall use: T B = FF ; C = F F T (3x3 Symm.) The eigen values of both B and C are , 2, 3 λ λ λ 19

20 Handout P.14 True stress in 3D Stresses in the presence of finite deformations and rotations t n n t Δa ΔP Assuming a continuum, we define traction vector t as the local force per unit area, in the limit t ΔP = Δ a Δ a 0 Cauchy Stress Tensor σ This is a measure of true stress; it relates the traction vector t to the direction n of the normal of the plane on which it is acting: = σ n or in component te t =σ rms i ij n j σ ij is the i-th component of the force per unit current area acting on a plane whose normal is currently in the j-th direction. 20

21 Handout P Elastic Deformation of Polymers 3.1 The general case Under some circumstances solid polymers are elastic in their mechanical response, particularly at very low temperatures, and at high temperatures (>T g ) if they are crosslinked or sufficiently entangled Polymer sample as a thermodynamic system with U and S at absolute T ΔQ ΔU, ΔS ΔW First and Second Laws of Thermodynamics: Δ U = Δ Q+ΔW; ΔS W U TS = A+ ST ΔQ T Helmholtz Free Energy (per unit initial volume) A U TS Rate of work done (power supplied) by applied loads: the equality applies to a reversible process = Elastic the inequality applies to an irreversible process 21

22 Elastic deformation of a polymer at constant temperature We define a material as elastic if deformation under applied loads is a reversible process (thus = sign). W = A + ST But at constant temperature T = 0 Power supplied to the system W = A Rate of change in the free energy To complete a model, we need to express in terms of σ, and A in terms of B. HOW to establish such relationships? W 22

23 Linking Cauchy stress σ to B... By reading handouts P.16 to P.20, we can show that: Cauchy stress tensor (symmetric, σ ij = σ ji ) σ ρ A = 2 B ρ B 0 (3.7) density ratio Note that we have not yet made any restriction to small deformations Eqn. (3.7) is valid for hyperelastic material (large elastic deformations) Isotropic invariants of B applicable to isotropic materials 1 ( ) 2 2 I1= tr B ; I2 = tr tr ; I3 = det 2 B B B (3.11) 23

24 Handout P Hyperelasticity: Phenomenological modelling A material that is elastic even at large deformations is called hyperelastic, i.e. rubbery What happens to the volume during deformation? Since mass is constant, the density ratio after deformation is ρ old volume = = = = ρ new volume λλ λ detb I (3.21) 3 new length rd isotropic invariant Principal λ = stretch: old length I3 = detb = λ1 λ2 λ3 24

25 Handout P.23 Hyperelastic rubbery polymers Experimentally, hyperelasticity is seen in polymers only under certain conditions: at temperatures above the glass transition, T >T g (rubbery) when the polymer is amorphous when there is connectivity between the molecules (chemical crosslinks or entanglements providing physical crosslinks ). - Under these conditions, the molecules move very freely past each other, but they are reluctant to move apart from each other. 25

26 Handout P.24 G << K and ν ~0.5 In hyperelastic polymers, the change of volume (K) meets much more resistance than change of shape (G), i.e. e.g. Typically G ~ 1 MPa, while K ~ 1 GPa Hence ν ~ 0.5, i.e. material is: G K INCOMPRESSIBLE (constant volume) ν = 2G 1 1 3K 1 2 G K (3.20) In light of these observations, there are two consequences for modelling hyperelasticity... 26

27 Handout P.24 Modelling rubbery polymers λλ λ = (a) I 3 =1 (see eqn (3.21)), hence A depends on only I 1 and I 2 and the density ratio is unity (b) The hydrostatic component of stress ( p) is indeterminate from the constitutive model alone, and must be found from the boundary conditions (BCs). Thus for a known deformation state, the Cauchy stress is obtained from either: (, ) A( λ, λ, λ ) σ AI1I = 2 p or = 2 B I B σ B B (3.22) pi 27

28 Handout P.25 Cauchy stress expressed in terms of the isotropic invariants For the models expressed in terms of the invariants, it is helpful to evaluate first their differentials with respect to B A A A 2 σ = 2 + I1 B B pi (3.25) I1 I2 I2 Hydrostatic stress -p (pressure) to be found from the boundary conditions (BCs) 28

29 Handouts P Example problems 29

30 Handout P.25 Phenomenological hyperelastic models The parameters C 0, C 1, C 2, C G, I m, μ n and α n are material parameters, to be found empirically by fitting the models to experimental data. 30

31 Handout P.28 neo-hookean model vs. experiments expt. λ ~ 1.2 Tension Compression Relatively simple to use as it involves only one material parameter C 0 ( ) A C0 I1 3 Predicts that the Cauchy stress is a linear function of B = Table P.25 σ = 2C0B pi (3.26) Limitations? - It cannot fit real data for rubbery polymers to within accuracy acceptable for many purposes, beyond a nominal strain of about 20% (λ ~ 1.2) 31

32 Handout P.29 neo-hookean vs. uniaxial tensile data Nominal stress s is found from true stress σ by exploiting the incompressibility: σ λ s = (3.30) There are two systematic errors: (1) over-estimate of stress at intermediate values of λ, and (2) under-estimate of stress at highest value of λ. Potential solutions...? 1) Mooney-Rivlin; 2) Gent Data are plotted as nominal stress (s) versus nominal strain (λ 1) nominal stress (MPa) Natural rubber Strain softening Strain stiffening Exp nominal strain λ 1 32

33 Handout P.30 Example problem 3 33

34 Handout P.31 Hyperelasticity: Physically-based model Chemical structures of some common polymers that are hyperelastic at room temperature (RT) H * C H CH 3 C H C H C H * n 1,4 polyisoprene ( natural rubber ) H H H H * C C C C H H 1,4 polybutadiene * n CH 3 * O Si * n CH 3 poly(dimethyl siloxane) (PDMS silicone rubber ) H * C H Cl C H C H C H * n 1,4 polychloroprene ( neoprene ) These are all flexible molecules with T g < RT They are liquids at RT until they are chemically crosslinked (e.g. using sulphur) at discrete points,... creating a 3D molecular network that cannot flow, and hence forms a soft solid. 34

35 Handout P.32 Entropic elasticity Rapid changes of shape But even though crosslinked, there is vigorous Brownian motion above T g, as they would in the liquid state Between crosslinks, the molecular chains move freely relative to each other Crosslink Some resistance to deformation comes entirely from the resistance of the system to reduction of its entropy The network changes shape with negligible cost in its internal energy U U The free energy of deformation A can be considered entirely entropic in origin (recall: A = U TS) A TS TS 35

36 Handout P.33 Physical model assumptions e 3 L K n K Kuhn segment Model derived following the approach first suggested by Kuhn a Kuhn chain r Imagine each long molecule has a set of rigid segments (n K, L K ) e 1 e 2 Crosslink (or entanglement) L K = length of each Kuhn segment n K = no. Kuhn segments per chain r = instantaneous end-to-end separation (a broad distribution); the directions of r will be random.... connected together by universal joints as if it were a freely-jointed chain Kuhn chain For such an imaginary chain molecule, the direction of each segment is, by definition, totally uncorrelated to the directions of the other segments. 36

37 Handout P.34 3D random walk In the absence of external forces (subscript 0 ) acting on the network, there will be equal numbers of bonds pointing in all directions The path r from one end of the chain to the other is a 3D random walk. This allows us to do some statistics on it, c.f. handout P NB. < > denotes avg. over the whole population of chains r 2 2 = nklk (3.34) 0 It is of interest to compare the unperturbed (i.e. coiled up & compact configuration) root-mean-square end-to-end distance r rms to the stretched-out length of the chain L: L r r = L n L = L n = n (3.35) 2 rms K K ; K K; hence K 0 rrms λ max 37

38 Handout P.35..with externally applied stresses.. e 3 i-th Kuhn segment r External forces f When stresses are applied to the polymer in the macro-world, in the micro-world this corresponds to microscopic forces being applied to each individual chain via the crosslinks. -f e 1 e 2 Crosslink -or- Entanglement The forces perturb the chain from its random shape, and thereby reduce its entropy Derivation of the relationships between magnitude of applied force f and end-to-end distance r, and the final link to continuum Cauchy stress σ... See full details in your Handouts P.36-40! 38

39 Handout P.41 Inverse Langevin Model λmax 1 λ λ λmax σ NkT = p 3 B I L (3.49) N chains per unit volume Hydrostatic stress -p (due to intermolecular potentials; vdw) to be found from BCs, as this model assumes incompressibility where L λ Maximum stretch ( x) is the rms stretch 1 max nk I 3 max λ = = 3x 1 x x I tr = = B is the Langevin function approximated using the Padé expression (3.46) (3.35) (3.48) 39

40 Handout P.43 Relating to real molecular dimensions To use the Inverse Langevin model to predict the behaviour of a particular rubbery polymer, we need to relate the parameters of the freely-jointed Kuhn chain to the real molecular dimensions... by equating the (a) unperturbed rms end-to-end distance, and (b) stretched-out length, respectively: Characteristic ratio Bond length Total # bonds between crosslinks rms = = K K= b ; = K K= b p 0 r r nl Cnb L nl nb (a) Kuhn Real (b) Kuhn Real If we call M c the molar mass of the chains between crosslinks, then from the definition of Avagadro s number N A, N and n b are both determined by M c thus # chains per unit volume ρ N = N n = n M c A ; b b0 Mc M0 crosslink monomer (3.53) (3.54) 40

41 Handout P.7 RECAP Molecule axis Stretched out lengths of polymers b Let the number of main chain bonds per monomer be n b0 b p a polyethylene (PE) molecule Then the total number of such bonds per molecule is n b = n n b0 n Let the mean main-chain bond length, as projected onto the axis of the molecule when stretched out straight, be b p Then the stretched out length of one monomer, L 0, and of the whole molecule, L, are respectively L = n b ; L = nl = n b 0 b0 p 0 b p (1.1) 41

42 Handout P.45 Example problem 4 42

43 Handout P Linear Viscoelasticity of Polymers Polymers frequently deviate from purely elastic behaviour. They are then said to be inelastic, in which deformation under applied loads is an irreversible process and hence the net work done is positive & W > Adt = 0 Net work done is dissipated as heat, or locked inside the material as a permanent change in structure (eg. plastic strain). One form of inelasticity is linear viscoelasticity, occuring at small strains of typically < ~0.005 (0.5%) (4.1) It behaves as a time-dependent, but linear system obeying the Boltzmann Superposition Principle (similar to an electric circuit containing only resistors, inductors and capacitors, see B5/B8). 43

44 Handout P RECAP (B5, A3) 3D linear elasticity ( A & dt = 0) Let the Infinitesimal strain tensor ε becomes vanishingly small relative to unity Dilation (vol. strain) Δ = tr = ε + ε + ε ε (3.16) Deviatoric strain tensor 3D equation of linear elasticity for modelling small strains σ Cauchy stress Δ e ε I 3 = KΔI + Bulk modulus 2Ge Shear modulus Mean strain (3.18) (3.19) 44

45 Handout P.47 3D equation of linear viscoelasticity For an isotropic linear viscoelastic polymer, the 3D equation of linear elasticity - eqn (3.19) is then replaced by the corresponding 3D equation of linear viscoelasticity Convolution integral: t σ = KΔI + dδ de σ() t = K( t u) dui + 2 G( t u) du (4.2) du du time-dependent Cauchy stress 0 0 K(x) time step 2Ge t G(x) (3.19) 3D stress response is fully determined by two material time-dependent functions: K(x) = Bulk stress relaxation modulus function G(x) = Shear stress relaxation modulus function 45

46 Handout P.47 Expressing σ(t) in terms of σ m & σ Using * to represent a convolution integral, we can express equation (4.2) in terms of the mean stress σ m and deviatoric stress σ : t dδ de σ() t = K( t u) dui + 2 G( t u) du (4.2) du du 0 0 σ = σ I + σ ; where σ = K Δ ; σ = 2G e m m t Hydrostatic response Deviatoric response Experimental results for polymers always show much less timedependence in the hydrostatic response than in the deviatoric response (e.g. a factor of ~1000 in G(x) relaxation vs factor ~2 in K(x) at T g ). (4.3) 46

47 Handout P Spring-dashpot analogy Zener model: Linear viscoelastic behaviour similar to that of a polymer is exhibited by Hookean springs (k) and Newtonian dashpots (c) coupled suitably in series or parallel.... Revise your B5 & B8 notes... Δk c 1 2 F k R y Zener model captures qualitatively all the key features typical of polymers, e.g. in tensile tests (creep, stress relaxation, dynamic tests/cyclic deformation...) The response of this model is defined by three equations: dy F1 1 d F F = F 1 1+ F2; = + ; F2 = kry dt c k dt Defining the two time constants of the system: Relaxation time c Retardation time ( k + k ) τ R y = ; F y k τ = k τ (input = y) (input = F) R ; τf > τy (4.8) (4.9) 47

48 Handout P.50 Zener model Combining equations (4.8), and employing time-constant definitions (4.9), gives a single differential equation governing the response of the Zener model: df F + τy = kr y + τ dt F dy dt (4.10) Relaxation time Retardation time Now we consider three example cases. 48

49 Handout P Case 1: Creep and recovery A step load F 0 is applied at time 0 and removed at time t u F = 1 N; y = 1 mm; t = 2 s. 0 0 u The resulting displacement is: t y = + 1 exp F 0 < t t ( k kr) kr ( k kr) + + τf ( ) ( ) 0 u 1 1 t tu t y = exp exp F0 ( t > tu). kr ( k kr) + τf τf (4.11) Let: force F (N) displacement y (mm) creep Response Input recovery time t (s) kr = 1 N/mm; k = 1 N/mm; c = 1 Ns/mm τy = 1 s; τf = 2 s. y 0 F 0 t u 49

50 Handout P.52 Case 2: Stress relaxation A step in displacement of magnitude y 0, is applied at time 0 and reversed at time t u. The required force is: t F = k + kexp y 0 < t t ( ) R 0 u τ y ( t t ) t u F = k exp exp y0 ( t > tu). τ y τ y (4.12) force F (N) displacement y (mm) y y 0 relaxation Input 0 u t u time t (s) = 1 mm; t = 2 s. Response recovery 50

51 Handout P.52 Case 3: Dynamic mechanical response A continuous oscillating displacement The required oscillating force is 2 2 kωτ y ωτy k F = kr + y*; F = y * ωτ y 1 ωτ + + y F (N) or y (mm) F F y y Response Input ω = 1 rad/s time t (s) Force F (N) y* = y expiωt displacement y (mm) is applied W = Fdy = area enclosed by the ellipse > 0 0 F* = F' + if" (4.13) where F* δ F' F" y* 51

52 Handout P Case 3b - Variation with frequency ω force F (N) τ y F' F" frequency ω (rad/s) (log scale) tan δ /τ F 1/τ y Loss tangent frequency ω (rad /s) (log scale) The tangent of the phase angle frequency ω = τ τ 1/ y F tan δ = F / F peaks at a The patterns of response seen here are similar to those of polymers in the linear viscoelastic regime (but only qualitatively) The one systematic discrepancy between the response of the particular spring-dashpot Zener model and the observed response of polymers is that it is too localised in the time/frequency domains. 52

53 Handout P.55 Generalised Maxwell model k 1 k 2 k j c 1 c 2 c j k R A truer representation of the viscoelasticity of a polymer would be given by a spring-dashpot model with multiple arms, such as N + 1 arms (j = 1..N) F y Replacing equation (4.8), the equations of the model are then: dy F 1 df F = F + F = + j = N F = k y N j j j N+ 1; ( 1.. ); N+ 1 R j = 1 dt cj kj dt The j-th relaxation time is now τ j = cj / kj The relaxation spectrum of the model: τ j, kj ( j = 1.. N) (4.14) 53

54 Handout P Shear relaxation spectrum The shear relaxation spectrum of such a generalised Maxwell model is given by the set of pairs of: τ j, Gj ( j = 1.. N) Example data: shear stress relaxation modulus G(t) for an amorphous polystyrene at 100 C, near, but just above its T g ~95 C G (GPa) The discrepancy is very clear... we need a shear relaxation spectrum G U single relaxation time approximation log 10 t (s) Zener model G j (GPa) G R ~1MPa polystyrene at 100 o C 13 elements log 10 τ j (s) 54

55 Handout P.56 Shear stress relaxation modulus, G(t) The equations governing the shear response would then be (by analogy with equations (4.14) and replacing y by a shear angle γ (=2 x tensorial shear strain e) and F by the corresponding shear stress σ ): N d j j d j G σ σ R where j ( 1.. ) d G γ σ = σ + γ t + τ = dt j = N j = 1 j (4.15) Applying the model to the case of shear stress relaxation (γ(t) is a step function at t = 0, of height γ 0 ), the shear stress relaxation modulus is then by analogy with eqn (4.12) σ N t Gt () = = GR + Gj exp γ0 j = 1 τ j Prony series (4.16) 55

56 Handout P.59 Dynamic shear modulus G* The shear relaxation spectrum given above for polystyrene at 100 C could be used to predict the dynamic shear modulus G* governing the oscillatory response of polystyrene in shear, via the analogue of equations (4.13) σ * G* = = G + ig γ * Complex shear modulus (GPa) Storage modulus Loss modulus log 10 ω (rad/s) by analogy with eqn (4.13) G' G" 2 2 N Gjωτ j N Gjωτ j R j= 1 = 1 where G = G + ; G = ( 1+ ωτ ) ( 1 ) j j + ωτ j (4.21) 56

57 Handout P.61 Example problem 5 57

58 Handout P.62 How are viscoelastic data obtained?... spanning such a broad range of times; there is no single technique capable of achieving this. We may exploit the phenomenon of time-temperature equivalence 4.2 The role of temperature Glassy << T g Rubbery ~T g Elasto-viscous fluid >> T g Some of the earliest data for an uncrosslinked sample of the amorphous polymer polyisobutylene (PIB): T g ~ -20 C H CH 3 * C Tensile stress relaxation modulus E(t) (in dynes/cm 2 = 0.1 Pa) vs time for a wide range of T (log scales on both axes) The data exhibit time-temperature superposition: with suitable horizontal shifts, they all superimpose to form a single master curve H C CH 3 * n 58

59 Handout P.63 Time-temperature superposition The master curve for PIB obtained by time-temperature superposition of these curves (in previous slide), onto the curve for 25 C (reference temperature T 0 ). Glassy << T g The factor by which the time scale for a particular curve has to be reduced/expanded to give a fit to the reference temperature curve (here T 0 Rubbery ~T g =25 C) is these days called a T (but called K 298 in the graph) a T Elasto-viscous fluid >> T g 59

60 Handout P.64 Linking E(t) to G(t) In and above the glass transition T g region, K >> G As such the Poisson s ratio is ν ~ 0.5 (i.e. incompressible) and hence E = 2 G(1 + ν ) 3G It follows that under these conditions, the tensile relaxation modulus function E(t) is related simply to the shear relaxation modulus function G(t): Et () 3 Gt () 60

61 Handout P.64 Time-temperature equivalence The observation of time-temperature superposition in the E(t) data shows, to a good approximation that,...not only E(t) but also the shear relaxation modulus G(t) at any temperature T is related to that at a reference temperature T 0 by a simple expansion or contraction of the time-scale. In terms of equations, this means: (4.23) t G t T G T G t T G t a T at (, ) =, or ( log, ) = ( log log, ) 0 T 0 Expansion / contraction factor Reference temperature Experimental data further suggest that the same a T function also applies to all τ j : Relaxation time at any temp T τ ( ) = τ ( ) T a T0 for all j j T j (4.24) 61

62 Handout P.65 Vogel-Tamman-Fulcher (VTF) equation lna T PIB calculated experiment /T e.g. the best-fit values of material parameters in PIB are: C = 1446 K; T = C This graph shows lna T plotted versus 1/T (absolute). Two points are notable: (a) The data do not give a straight line relation (i.e. do not fit the Arrhenius relation, when T > T g ). (b) The data are fitted by the alternative 2-parameter calculated line included in the graph. ln a T C = ( T T ) ( T T ) 0 C (4.25) Reference temperature 62

63 Handout P.67 Glass transition temperature T g Recall: the glass transition temperature T g of an amorphous material is the temperature below which it cannot maintain structural equilibrium, on a given experimental time-scale Cooling rates q 1 > q 2 As shown in plot, faster cooling leads to a higher measured T g. Because glass is not in thermodynamic equilibrium, its structure gradually evolves with time (e.g. polymer at B C). Fictive temperature T f = temperature at which the glass is believed to have the same structure in equilibrium Equilibrium line The value of T g is defined in terms of the volume V or enthalpy H measured during heating or cooling, and its values are rate dependent 63

64 Handout P.68 Physical ageing in glassy polymers The viscoelastic properties also vary with time in the glassy state (<T g ) Shown here are tensile creep curves for PVC (T g = 83 C) obtained after rapid cooling from 90 C to 20 C, and waiting ( ageing ) for varying times t e before testing. Note how the creep curves shift to the right with increasing ageing time...there appears to be time ageing-time (t e ) superposition going on. τ j ( ) = τ ( ) te ae j te0 for all j (4.27) Each relaxation time after ageing for a time t e is related by the same factor a e to its corresponding value after ageing for a time t e0 64

65 Handout P.69 Secondary relaxations of polymers The viscoelastic process visible in the glass transition (T g ) region is often labelled the primary relaxation of the polymer. But viscoelasticity is also prominent in other regions of the temperature scale. These are called the secondary relaxations They differ from the primary relaxation in that: (i) they are not usually associated with evidence of physical ageing (i.e. not linked to microstructural evolution). (ii) They are usually found, to reasonable accuracy, to show timetemperature superposition, with the relaxation times conforming to the Arrhenius relation as follows c c c τ exp hence lna T T = T T ΔH where 0 Activation energy (also denoted as Q) c = R 0 (4.28) 65

66 Handout P.70 Example problem 6 66

67 Handout P.71 Non-linear viscoelasticity Exp. data from tensile creep and shear creep tests on PMMA at 30 C With increase in stress, the viscoelasticity shown by a polymer becomes increasingly non-linear i.e. the relation between stress and strain is time-dependent, but the material ceases to behave as a linear system. It no longer obeys the Boltzmann Superposition Principle Many of the questions of (i) how to model them in the macroworld, and (ii) how to explain them in terms of what is going on in the micro-world form part of the ongoing research activities... 67

68 C3 Class: Week-7: Thursday 5th Venue: LR8 (IEB) 2 Slots: 2-3 pm / 3-4 pm PPT slides: 68