s ij = 2ηD ij σ 11 +σ 22 +σ 33 σ 22 +σ 33 +σ 11

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1 Plasticity Z Suo NONLINEAR VISCOSITY Linear isotropic incompressible viscous fluid A fluid deforms in a homogeneous state with stress σ ij and rate of deformation The mean stress is σ m = σ kk / 3 and the deviatoric stress is = σ ij σ m δ ij isotropic incompressible viscous fluid we mean a model specified by D kk = 0 = 2η By a linear where η is the viscosity The model assumes that η is independent of the rate of deformation so that the deviatoric stress is linear in the rate of deformation The expression D kk = 0 ensures incompressibility and represents a single equation: D 11 + D 22 + D 33 = 0 The expression = 2η ensures isotropy and represents six equations: σ 12 = 2ηD 12 σ 23 = 2ηD 23 σ 31 = 2ηD 31 σ 11 σ 11 +σ 22 +σ 33 3 σ 22 σ 22 +σ 33 +σ 11 3 = 2ηD 11 = 2ηD 22 σ 33 σ +σ +σ = 2ηD 3 33 The sum of the last three equations gives an identity 0 = 0 Thus the last three equations consist of only two independent equations The two types of relations D kk = 0 and = 2η consist of a total of six independent linear equations between the twelve components of stress and rate of deformation Thermodynamic state and thermodynamic property A fluid can be in many thermodynamic states and has many thermodynamic properties Examples of thermodynamic properties include temperature pressure volume energy entropy smell color electric field and polarization When the fluid is in a particular thermodynamic state the value of every thermodynamic property is fixed We specify a thermodynamic state of a fluid by values of its thermodynamic properties October Nonlinear viscosity 1

2 Plasticity Z Suo How many thermodynamic properties do we need to differentiate thermodynamic states of a fluid? The answer is two How do we know? The answer depends on the evidence from our experience and on the limit of our attention One property is too few: if we hold the value of the temperature we can still change the thermodynamic state of the fluid by changing the pressure Three properties are too many: once the temperature and pressure are fixed the thermodynamic state of the fluid is fixed if we neglect minor effects of say the electric field and magnetic field As a model we specify the thermodynamic state using two thermodynamic properties for example temperature and pressure Once the values of the two thermodynamic properties are fixed the fluid is in a fixed thermodynamic state and values of all other thermodynamic properties are fixed In this model the deviatoric stress is not a thermodynamic property and does not affect thermodynamic state of the fluid Viscosity is a thermodynamic property The viscosity of a fluid is a function of temperature and pressure η( T p) The experimental data are sometimes fit to an expression ( ) =η 0 exp q +V a p η T p Here kt is the temperature in the unit of energy; the pre-factor η 0 the activation energy q and the activation volume V a are parameters used to fit the experimental data The model also requires that the mean stress equal the thermodynamic pressure: σ 11 +σ 22 +σ 33 = p 3 The pressure-dependent viscosity makes the model of viscosity = 2η nonlinear Is this nonlinear effect significant? kt = J/K! " kt $ & % At room temperature ( )( 300K ) = J Assume that the activation volume is comparable to the volume per molecule V a m 3 and assume a value of pressure p = 10 6 Pa We find that V a p << kt Thus in many applications the effect of pressure on viscosity is small so that the viscosity is assumed to be independent of the pressure Often the activation energy is large q >> kt and the viscosity is sensitive to temperature At the atmospheric pressure the viscosity of water is around 10-3 Pa s at room temperature Pa s at the freezing point and Pa s October Nonlinear viscosity 2

3 Plasticity Z Suo at the boiling point We will usually consider isothermal flows in which the temperature is held fixed Thermodynamic inequality A fluid is subject to an applied force and is in thermal contact with a heat bath held at a fixed temperature We assume that the fluid and the heat bath are in thermal equilibrium so that the temperature of the fluid is fixed at that of the heat bath We further assume that the fluid is incompressible Once the values of the two thermodynamic properties temperature and volume are fixed the fluid is in a fixed thermodynamic state with all its thermodynamic properties fixed In particular the Helmholtz free energy of the fluid is constant even when the fluid flows The potential energy of the applied force changes at the rate σ ij V where V is the volume of the fluid For an incompressible fluid D kk = 0 we can confirm an identity σ ij = The fluid and the applied force together constitute a composite thermodynamic system at a fixed temperature The Helmholtz free energy of the composite system sums over its parts: 0 σ ij V Thermodynamics requires that the Helmholtz free energy of the composite system should never increase: 0 The decrease in the potential entirely dissipates into the heat bath That is in an isothermal flow the incompressible viscous fluid does not change its own Helmholtz free energy but converts the potential energy of the applied force into energy in the heat bath This thermodynamic condition applies when an incompressible viscous fluid flows at a fixed temperature The condition does not require linearity and isotropy For a linear isotropic incompressible viscous fluid = 2η Note that 0 for any rate of deformation The model satisfies the thermodynamic inequality for arbitrary rate of deformation if and only if the viscosity is non-negative: η 0 For many materials stress is nonlinear in rate of deformation Metals creep at elevated temperatures but the stress is often nonlinear in the rate of deformation (Frost and Ashby 1982) Nonlinear viscosity is also observed in many other materials such as ice ice cream skin cream toothpaste and chocolate We can test a material in shear and measure the relation between the stress and the rate of deformation We often fit the experimental data to a power law: τ = A γ N October Nonlinear viscosity 3

4 Plasticity Z Suo where A and N are parameters used to fit the experimentally measured curve between the stress and the rate of deformation The power law is also written as! γ = τ $ & " A % where n = 1/N For example a representative values for ice is n = 3 (Glen 1955) The creep of ice contributes to the dynamics of glaciers We plot the experimental data as a curve in the plane with the axes of τ and γ The curve is often monotonic We represent the curve as a function: τ = g ( γ ) The same relation can also be written in another form: γ = h ( τ ) n stress τ τ = g ( γ ) γ = h ( τ ) Nonlinear isotropic incompressible viscous fluid We have tested a material under shear and measured the curve between the stress and the rate of deformation τ = g ( γ ) What can we do with the curve? We can compare the curve with that of another material We can study the microscopic origins for the values of A and N (Ashby and Frost 1982) We can even use the data to solve some boundary-value problems But to solve boundary-value problems in general we will need to have relation between stress and rate of deformation in arbitrary state of stress not just shear Isotropy of the material implies that the relation τ = g ( γ ) applies in all shearing directions: σ 12 = g( 2D 12 ) σ 23 = g( 2D 23 ) The experimental data τ = g γ rate of deformation γ between a tensile stress and the rate of extension σ 31 = g( 2D 31 ) ( ) in general do not allow us to predict the relation October Nonlinear viscosity 4

5 Plasticity Z Suo By viscosity here we mean that the deviatoric stress depends on the rate of deformation In general we need to determine functions of five independent variables: = D 11 D 22 D 23 D 31 D 12 ( ) Here we have dropped the dependence on D 33 ; due to incompressibility D 33 is not an independent quantity D 33 = D 11 D 22 A brute-force method to determine these functions of five variables is to measure them experimentally This method is impractical A function of one variable corresponds to a curve a function of two variables corresponds to many curves on a page a function of three variables corresponds to many pages of a book and a function of four variables corresponds to many books in a library A function of five variables will require many libraries We need to construct a model to reduce the number of experiments We wish to modify the multiaxial model of viscosity in a single aspect The new model will accommodate the nonlinear relation between stress and rate of deformation but will preserve incompressibility and isotropy The condition of incompressibility remains the same: D kk = 0 We need to learn how to preserve isotropy Invariant of a vector Let e 1 e 2 e 3 be an orthonormal basis of a Euclidean space Any vector u in the Euclidean space is a linear combination of the base vectors: u = u 1 e 1 + u 2 e 2 + u 3 e 3 We say that u 1 u 2 u 3 are the components of the vector u relative to the basis e 1 e 2 e 3 We are familiar with the geometric interpretations of these ideas The vector u is an arrow in the space The basis e 1 e 2 e 3 consists of three unit vectors normal to one another The components u 1 u 2 u 3 are the projection of the vector u on to the three unit vectors e 1 e 2 e 3 Once a vector u is given in the Euclidean space the vector itself does not change if we choose another basis However the components u 1 u 2 u 3 do change if we choose another basis We know the rule of the transformation of the components of the same vector relative to two bases The sum u i u i does not have any free index and is a scalar When a new basis is used the components u 1 u 2 u 3 change but u i u i remains invariant This invariant has a familiar geometric interpretation: u i u i is the length of the vector u The length of the vector is invariant when the basis changes October Nonlinear viscosity 5

6 Plasticity Z Suo Invariants of a tensor Let A be a second-rank tensor and A ij be the components of the tensor relative to an orthonormal basis e 1 e 2 e 3 The tensor is symmetric so that A ij = A ji The components of the tensor form three scalars: A ii A ij A ij A ij A jk A ki We form a scalar by combining the components of the tensor in a way that makes all indices dummy The three scalars are independent of the choice of the basis e 1 e 2 e 3 and are known as the invariants of the tensor A Invariants of stress tensor The state of stress is a physical fact independent of how we choose a basis Once we choose a basis e 1 e 2 e 3 we picture a unit cube in the fluid with the faces of the cube on the coordinate planes Forces acting on the faces of the cube define the components of the stress σ ij relative to the basis Whereas the state of stress does not depend on the choice of the basis the components of the stress do Some combinations of the components are invariants independent of the choice of basis For example the trace of the stress tensor σ kk is an invariant The flow of an incompressible fluid is unaffected when we superimpose any hydrostatic stress Define the deviatoric stress = σ ij σ kk δ ij / 3 and write the other two invariants of the stress tensor as s jk s ki Each invariant is a scalar measure of the state of stress The two invariants are sometimes designated as J 2 = /2 and J 3 = s jk s ki / 3 Invariants of rate-of-deformation tensor For an incompressible material the trace of the deformation gradient vanishes D kk = 0 We write the other two invariants of the rate of deformation as D jk D ki Each invariant is a scalar measure of the rate of deformation The two invariants are sometimes designated as I 2 = /2 and I 3 = D jk D ki / 3 The second-invariant fluid We have tested a fluid under shear and measured the curve between the stress and the rate of deformation τ = g γ ( ) Can we use this curve to predict the curve between the stress and the rate of deformation under all other states of stress? The answer is no However we October Nonlinear viscosity 6

7 Plasticity Z Suo fudge the prediction it will disagree with the experimental data for some states of stress But the desire to use the curve τ = g ( γ ) to predict the curves between stress and rate of deformation for all states of stress is so strong that we do it by making assumptions Fudge we do The most widely used model relies on the second invariant of the rate of deformation The model makes two assumptions First the relation between the stress and the rate of deformation takes the form: D kk = 0 = 2η Second the viscosity η varies with the rate of deformation in a particular way ( ) The use of the the viscosity is a function of the second invariant η invariant preserves isotropy The use of only one invariant simplifies the model The model ensures incompressibility Once we know the function η( ) the model calculates all components of the deviatoric stress for any given rate of deformation Fit the second-invariant model to experimental data Recall that γ 12 = 2D 12 When the fluid flows in shear the second-invariant model = 2η predicts that σ 12 =η γ 12 Comparing this prediction to the experimental data τ = g ( γ ) we obtain that η = g ( γ ) γ To make the model applicable to general state of deformation we need to convert the independent variable from the rate of shear to the second invariant Under the pure shear condition the second invariant is Define the equivalent rate of shear by = D 12 D 12 + D 21 D 21 = 1 2 γ 2 γ e = 2 The second invariant is positive-definite The equivalent rate of shear is just another way to write the second invariant Write ( ) = g ( γ e) η γ e γ e We have just described the most commonly used model of nonlinear isotropic incompressible viscosity We summarize the model as a recipe October Nonlinear viscosity 7

8 Plasticity Z Suo Experimentally measure the relation between the stress and the rate deformation under pure shear τ = g γ ( ) shear γ e = Given a general rate of deformation calculate the equivalent rate of 2 and calculate the deviatoric stress: ( ) = 2g γ e D γ ij e The second-invariant model uses the experimental data measured under shear to predict the relation between stress and rate of deformation for all types of flow The model achieves unusual economics: buy one and get everything else for free Power-law creep As an example suppose the experimental data under shear fit the power law τ = A γ N For the fluid at an arbitrary rate of deformation the second-invariant theory predicts the deviatoric stress as = 2A( γ e ) N 1 where the effective rate of shear is γ e = 2 Thermodynamics of the second-invariant fluid The viscosity is a function of the rate of deformation and is no longer a thermodynamic property The second invariant model satisfies the thermodynamic inequality 0 for arbitrary state of flow Note that =η( γ e ) The second invariant is positive definite Thus 0 provided g 0 The latter means that under shear the shearing stress is in the direction of the shearing deformation The statement is the thermodynamic condition under shear τ γ 0 Stress as independent variable We can also require that the viscosity be a function of the second invariant of deviatoric stress Multiply the equation = 2η by itself and we obtain that ( ) ( ) = 4! η D 2 " ij $ Dij D ij ( ) is monotonic the The right-hand side is! g γ " e $2 So long as the function g γe above equation is a one-to-one relation between the second invariant of stress and the second invariant of rate of deformation We can make the October Nonlinear viscosity 8

9 Plasticity Z Suo viscosity as a function of the second invariant of the deviatoric stress η( ) The second-invariant model is also known as the J 2 model Under the pure shear condition the deviatoric stress is! 0 τ 0! " sij $ = % & % τ 0 0 & % & " $ The second invariant is = s 12 s 12 + s 21 s 21 = 2τ 2 Define the equivalent shear by τ e = 1 2 The equivalent shear stress is just another way to write the second invariant of deviatoric stress and reproduces the applied shear stress under the pure shear condition The second-invariant model gives the following recipe Test a material in shear and measure the relation between the stress and the rate of deformation γ = h ( τ ) Given an arbitrary state of stress σ ij calculate the deviatoric stress = σ ij σ kk δ ij / 3 calculate the equivalent shear stress τ e = calculate the rate of deformation: = h ( τ e) s 2τ ij e /2 and We have presented the model using the second invariant of stress and using the second invariant of rate of deformation The two methods give the same model The two methods use the same experimental data and give identical predictions Indeed the equivalent stress relates to the equivalent rate of deformation: τ e = g γ e ( ) ( ) γ e = h τ e Yield stress Viscoplasticity Bingham flow Many materials creeps negligibly when the stress is small and creeps appreciably when the stress is large Bingham (1916) proposed an idealized model: " 0 for τ < τ Y $ γ = $ % $ τ τ Y η for τ > τ Y October Nonlinear viscosity 9

10 Plasticity Z Suo The model characterizes a fluid using two parameters: the yield stress τ Y and the viscosity η The combination of yield stress and viscous flow is called viscoplasticity The Bingham model provides a specific relation between the stress and the rate of deformation under shear The relation can be generalized to multiaxial loading using the second-invariant model Given an arbitrary state of stress σ ij calculate the deviatoric stress = σ ij σ kk δ ij / 3 calculate the equivalent shear stress τ e = calculate the rate of deformation: ( 0 for τ eq < τ Y * = ) 1 " τ eq τ %" Y s % ij * 2$ η ' $ & τ ' for τ > τ eq Y + * eq & /2 and The viscoplastic model satisfies the thermodynamic condition 0 for arbitrary state of stress stress τ η τ Y rate of deformation γ von Mises yield condition The equation τ e = τ Y or 1 2 s s = τ ij ij ( Y ) 2 is due to von Mises (1913) This equation uses the yield stress measured under shear to predict the yield condition under any type of stress The prediction uses the second invariant of deviatoric stress We will study the yield condition in more detail later Fit the second-invariant model to the experimental data measured under uniaxial tension Often it is more convenient to test a material under uniaxial tension We represent the experimentally measured curve of the tensile stress σ and the rate of extension ε as a function: ε = f σ ( ) October Nonlinear viscosity 10

11 Plasticity Z Suo We next use the data measured under uniaxial tension to fix the second-invariant model Under the uniaxial tension the stress tensor is! σ 0 0! " σ ij $ = % & % & % & " $ The mean stress is σ m = σ / 3 The deviatoric stress is! & 2σ / 3 0 0! " sij $ = & 0 σ / 3 0 & 0 0 σ / 3 "& The second invariant of the deviatoric stress is = 2 3 σ 2 Define the equivalent stress (or the von Mises equivalent stress) by σ e = 3 2 The equivalent stress is just another way to write the second invariant Under uniaxial tension the equivalent stress coincides with the applied stress Under uniaxial tension the second-invariant model = 2η predicts that σ / 3 =η ε Comparing this prediction with the experimental data ε = f σ we obtain that σ η = 3 f σ ( ) ' ' ' $ ' ( ) To make the model applicable to general state of deformation we write the viscosity as a function of the equivalent stress: η( σ e ) = 3 f ( σ e ) Thus the second-invariant model leads to the following recipe Conduct a uniaxial tensile test to measure the curve ε = f ( σ ) Given a state of stress σ ij calculate the deviatoric stress = σ ij σ kk δ ij / 3 the equivalent stress σ e σ e = 3 /2 and the rate of deformation ( ) = 3 f σ e s 2σ ij e October Nonlinear viscosity 11

12 Plasticity Z Suo This expression makes the viscosity as a function of the second invariant and reproduces the experimental data ε = f σ ( ) under uniaxial tension Inhomogeneous flow For a fluid deforming in inhomogeneous state we regard the body of fluid as a sum of many small pieces Each small piece underdoes homogeneous flow and obeys the relation between the stress and rate of deformation: D kk = 0 = 2η where the viscosity η is a known function of the second invariant of rate of deformation Different pieces in the body communicate through compatibility of deformation and balance of forces Let x be the coordinate of a place in space t be the time and v i ( xt) be the velocity of a small piece of fluid at the place x and time t Compatibility requites that the rate of deformation relates to the field of velocity: = 1 " v i + v % j 2$ x j x ' i & The balance of forces includes the inertial force: σ " ij +b x i = ρ v i j t + v v % i $ j x ' j & where b i is the body force and ρ is the density For an incompressible material the density is a constant Familiar boundary conditions include prescribed velocity and prescribed traction This boundary-value problem governs the inhomogeneous flow Creeping flow For highly viscous material we often neglect the effect of inertia and the balance of forces takes the form σ ij x j +b i = 0 Flow in a pipe Subject to a pressure gradient G a viscous fluid flows in a pipe of a circular radius of radius a The flow is of cylindrical symmetry and is invariant along the length of the pipe Balance of forces Consider part of the fluid in the pipe a cylinder of fluid of radius r and length l as a free body On the cylindrical surface of the cylinder the shearing stress has a constant magnitude τ The shearing stress gives a force in the axial direction 2πrlτ Acting on the two ends of the cylinder are pressures whose magnitudes differ by Gl This difference in pressures gives October Nonlinear viscosity 12

13 Plasticity Z Suo another force in the axial direction πr 2 Gl The balance of the two axial forces 2πrlτ = πr 2 Gl gives that τ = 1 2 Gr Thus the shearing stress is linear in the radial distance Compatibility of deformation The velocity field of the fluid particles directs in the axial direction and varies with the radial distance v r assume fluid at the surface of the pipe does not slip v a ( ) We ( ) = 0 The gradient of velocity gives the rate of shear γ = dv / dr Integrating we obtain that The rate of discharge is r ( ) = γ ( r ) dr v r a a ( ) 2πrdr Q = v r 0 Rheology of fluid We model the fluid by a nonlinear relation between rate of shear γ and shearing stress γ = h ( τ ) Mixing the three ingredients The three ingredients relate the rate of discharge Q to the gradient of pressure G First consider a linearly viscous fluid Tested under shear the rate of deformation relates to the stress as γ = τ /η with η being a constant Because the shearing stress is linear in radial distance τ = Gr /2 the rate of shear is also linear in radial distance γ = Gr 2η The velocity is quadratic in the radial distance: The rate of discharge is ( ) = G ( 4η r2 a 2 ) v r Q = πa4 G 8η We have modified the sign Next consider a power-law fluid Tested under shear the rate of deformation relates to the stress as γ = τ / A in radial distance τ = Gr /2 The rate of shear is ( ) n The shearing stress is still linear October Nonlinear viscosity 13

14 Plasticity Z Suo The velocity is The rate of discharge is ( ) = G v r n! γ = Gr $ & " 2A % n! $ r n+1 a n+1 & " 2A % n +1 n πa 3! Q = Ga $ & " 2A % n + 3 Growth of a cavity An incompressible material in a state of hydrostatic stress σ appl does not deform However the material will deform if it contains a cavity We neglect the surface energy of the cavity and assume that the cavity is traction-free Consequently near the cavity the material is not in a state of hydrostatic stress Far away from the cavity the material is still in the state of hydrostatic stress σ appl This applied stress will cause the cavity to enlarge We assume the cavity to be spherical so that the non-vanishing fields are the radial component of stress σ r circumferential components of stress σ θ and the radial component of the velocity v For the cavity at a given radius a all these fields are functions of radial coordinate r Balance of forces Consider a hemispherical shell inner radius r and outer radius r + dr cut from a spherical shell with a plane normal to the z axis The radial stress on the outer surface of the shell is σ r r + dr in the positive z direction π r + dr ( ) resulting in a force ( ) 2 σ r ( r + dr) The radial stress on the inner ( r ) resulting in a force in the negative z direction πa 2 σ r ( r ) The surface is σ r base of the hemispherical shell is a circular annulus and the circumferential stress gives a force in the negative z direction ( 2πrdr)σ θ ( r ) Balancing these forces acting on the hemispherical shell we obtain that dσ r dr + 2 ( r σ σ r θ ) = 0 Compatibility of deformation Denote the radius of cavity at time t by a t ( ) At a given time the velocity of the surface of the cavity is v ( a ) = da / dt and the volume of material per unit time crossing the surface of the cavity is 4πa 2 da / dt ( ) The volume of material per unit time crossing the spherical October Nonlinear viscosity 14

15 Plasticity Z Suo surface of radius r is 4πr 2 v r the two flows be equal 4πr 2 v r ( ) Incompressibility of the material requires that ( ) = 4πa 2 ( da / dt) giving the velocity field: v ( r ) = a2 da r 2 dt The velocity decays as a function of the distance r The rate of deformation in the radial direction is D r = dv / dr or D r = 2 a2 r 3 da dt The rates of deformation in the circumferential directions are D θ = v /r The material contacts in the radial direction and extends in the circumferential directions Rheology of material Consider a nonlinear viscous material Tested under uniaxial tensile stress the stress relates to the rate of deformation as σ = f ( ε ) Each small piece of material around the cavity is in a state of triaxial stress ( σ r σ θ σ θ ) and a state of triaxial rate of deformation ( D r D /2 r D r /2) Superimposing a state of hydrostatic stress does not affect the flow Superposing a state of hydrostatic stress σ θ σ θ σ θ ( ) on the original state ( σ r σ θ σ θ ) we ( ) and keep the state of ( ) We assume that the rheological obtain a state of uniaxial compressive stress σ r σ θ 00 deformation the same D r D /2 r D r /2 model is symmetric with respect to tension and compression In the relation σ = f ε ( ) we replace σ with ( σ r σ θ ) and ε with D r so that ( σ r σ θ ) = f ( D r ) Mixing the three ingredients Consider a power-law material σ = K ε Replacing σ with σ r σ θ ( ) and ε with D r we obtain that ( ) N " σ θ σ r = K 2a2 da % $ r 3 dt ' & Inserting this expression into the equation for the balance of force we obtain that dσ r dr = 2K r Integrating over the interval r a σ r! 2a 2 " r 3 ( ) and using the boundary conditions ( a ) = 0 and σ r ( ) = σ appl we obtain that da dt $ & % N N October Nonlinear viscosity 15

16 Plasticity Z Suo σ appl = 2K 3N This expression relates the rate of expansion da / dt to the applied stress σ appl The cavity expands in time exponentially ' ) a ( t ) = a 0 ) ) ( ( ) exp t 2! 2 " a! " da dt $ & % N 3Nσ appl 2K Ilyushin theorem (1946) For highly viscous material we often neglect the effect of inertia We also often neglect body force Under these conditions the balance of forces takes the form σ ij = 0 x j Compatibility of geometry requites that = 1 " v i + v % j 2$ x j x ' i & The incompressibility condition requires that D kk = 0 These partial differential equations are linear The uniaxial stress-strain curve is fit to the power law: ( ) n γ = τ / A Write the stress-strain relation in a generic form:! s = f 11 ij A s 22 A s 33 A s 23 A s 31 A s $ 12 " A & % The stress-strain relation developed in the previous paragraph has a useful property When all components of the stress tensor are multiplied by a factor λ n the components of the strain tensor are multiplied by a factor λ That is! λs f 11 ij A λs 22 A λs 33 A λs 23 A λs 31 A λs $! 12 " A & = λ s n f 11 ij % A s 22 A s 33 A s 23 A s 31 A s $ 12 " A & % Consider a boundary value problem characterized by a length a and load σ appl The Ilyushin theorem says that the fields of stress rate of deformation and velocity take the form! x σ ij = σ appl ˆσ ij " a n $ & % $ & % 1 N * + October Nonlinear viscosity 16

17 Plasticity Z Suo Here! = σ $ appl " A & % n! v i = a σ $ appl " A & % n ˆ! x " a n $ & %! x ˆv i " a n $ & % σˆ ˆDij and ˆv ij i stand for dimensionless functions The proof of this theorem is simple: the above form satisfies all the governing equations Thus when a body is subject to an applied stress σ appl the field of stress is linear in σ and the field of rate of deformation the field of velocity are appl proportional to ( σ appl ) n The Ilyushin theorem determines the dependence of the fields on the applied stress The distribution of the fields still need be determined by solving the boundary-value problem The Ilyushin theorem generalizes the consideration of linearity in the linear elastic theory into a scaling relation for the power-law material The second-invariant model sometimes fails The prediction of the second-invariant model often disagrees with experimental data For example once we fit the model to the experimental data measuring under shear we can use the model to predict the relation between stress and rate of deformation in other types flow It is not surprising that the prediction disagrees with experimental data Often we disregard the discrepancy but sometimes we cannot Let us look at a particular set of experimental observations Rate of shear can cause normal stresses When a nonlinear isotropic incompressible viscous fluid is subject to a rate of shear in the absence of all other components of the rate of deformation the second-invariant model predicts that the fluid only develops shearing stress This prediction is wrong Experiments show that the fluid also develops normal stresses Let the rate of shear be D 12 = D 21 = γ /2 ; all other components of the rate of deformation vanish We can determine the shearing stress as a function of the rate of shear: σ 12 = g ( γ ) Reversing the direction of the shear will also reverse the shearing stress so that g( γ ) = g ( γ ) ; it is an odd function For an isotropic fluid the symmetry precludes shearing stresses in other directions σ 23 = σ 31 = 0 The isotropy however does not preclude normal stresses The flow of an incompressible fluid is unaffected by the superimposition of a state of hydrostatic stress We subtract the normal stress in every direction by σ 22 the fluid is now October Nonlinear viscosity 17

18 Plasticity Z Suo subject to normal stress σ 11 σ 22 in direction 1 and normal stress σ 33 σ 22 in direction 3 The two normal stresses are also functions of the rate of shear: σ 11 σ 22 = N 1 γ ( ) ( γ ) σ 22 σ 33 = N 2 Reversing the direction of the shear will not affect the normal stresses so that γ N 1 ( ) = N 1 ( γ ) and N 2 ( γ ) = N 2 ( γ ) ; they are even functions Experiments show that the normal stress effects are pronounced in viscoelastic liquids; the first normal stress difference is larger in magnitude than the second normal stress difference (Barnes Hutton Walters 1989) Reiner-Rivlin fluid In the second invariant model = 2η the viscosity is taken to be a function of the second invariant We can of course assume that the viscosity is a function of second and third invariants η( D jk D ki ) This model is more general but requires more experimental data to fit the function Besides the two-invariant model still predicts that a rate of shear generates no normal stresses A fluid deforms in a homogeneous state with stress σ ij and rate of deformation Assume that the fluid is isotropic and that the state of stress is a function of the state of rate of deformation The most general form of the function is σ ij = aδ ij +b + cd ik D kj where a b and c are functions of the three invariants D ii D jk D ki This result is due to Reiner (1945) and Rivlin (1948) We further assume that the fluid is incompressible: D kk = 0 Superimposing a hydrostatic stress does not affect the deformation Let the mean stress be σ m = σ kk / 3 and the deviatoric stress be = σ ij σ m δ ij Modify the above relation between the stress and the rate of deformation as = 2η + 4ξ D ik D kj D mk D km δ ij ( ) where η and ξ are functions of the two remaining invariants D jk D ki Now consider a fluid subject to a rate of shear D 12 = D 21 = γ /2 with all other components of the rate of deformation being zero The model predicts that October Nonlinear viscosity 18

19 Plasticity Z Suo σ 12 =η γ σ 23 = 0 σ 31 = 0 σ 11 σ 11 +σ 22 +σ 33 3 σ 22 σ 11 +σ 22 +σ 33 3 = ξ γ 2 = ξ γ 2 σ 33 σ +σ +σ = 2ξ γ 2 3 The model predicts normal stresses: σ 11 σ 22 = 0 σ 22 σ 33 = 3ξ γ 2 This form of normal stresses disagrees with experimental observations The Reiner-Rivlin model is a general mathematical model of viscosity but the model does not predict experimentally observed form of normal stresses Rather the normal stress effect is pronounced in viscoelastic liquids Reiner-Rivlin model and the thermodynamic inequality Let us return to the Reiner-Rivlin model: σ ij = aδ ij +b + cd ik D kj The external force does work at the rate σ ij = ad kk +b + cd ik D kj D ji For an incompressible fluid D kk = 0 The second invariant is nonnegative for all rates of deformation 0 The third invariants however can be either positive or negative If we choose c as a constant the model will violate the thermodynamic inequality σ ij 0 for some rates of deformation If we choose the c as c =ζ D jk D ki with ζ as a positive constant then the Reiner-Rivlin model satisfies the thermodynamic inequality for all rates of deformation Rayleigh s dissipation function Rayleigh (1871) introduced the idea of dissipation function in a paper on the vibration of a viscous system The system has n generalized coordinates The time derivative of each generalized coordinate defines the associated generalized velocity Associated with this generalized velocity is a generalized viscous force Thus the system has n generalized velocities v 1 v n and n generalized viscous forces f 1 f n A general kinetic model is to write each force as a function of the n velocities: October Nonlinear viscosity 19

20 Plasticity Z Suo f 1 ( v 1 v n ) f n ( v 1 v n ) These functions are nonlinear if the viscous behavior is nonlinear Rayleigh introduced a dissipation function a scalar function of the generalized velocities Q v 1 v n form: ( ) He assumed that the viscous forces take the f i = Q ( v v 1 n) v i Rayleigh s kinetic model is more restricted than the general model but is also simpler His model only requires a single scalar function of n variables Q( v 1 v n ) Linear viscosity Rayleigh assumed that the dissipation function is quadratic in the generalized velocities: Q = 1 2 H ij v i v j where the matrix H ij represents generalized viscosities The value of F is unchanged by the anti-symmetric part of the matrix We assume that the matrix is symmetric H ij = H ji The viscous forces are linear in the velocities: f i = H ij v j We assume that the viscosity of the system dissipates energy f i v i 0 for all velocities The equality holds only when all velocities vanish In the case of linear viscosity f i v i = H ij v i v j The viscosity dissipates energy at all velocities if and only if the matrix of the viscosity matrix H is positive-definite When H is nonsingular we can invert the matrix and obtain v j = G ij f i The matrix G represents the generalized fluidities Introduce a scalar A = 1 2 G ij f i f j This scalar is a function of the viscous forces A( f 1 f n ) and is called the creep potential We confirm that v i = A ( f f 1 n) f i October Nonlinear viscosity 20

21 Plasticity Z Suo Graphic interpretation of Rayleigh s kinetic model We now ( ) generalized Rayleigh s dissipation function to nonlinear viscosity Let Q v 1 v n be a scalar function of the generalized viscosities but the function need not be a quadratic in the generalized velocities Still assume that the viscous forces are given by f i = Q ( v v 1 n) v i In a linear space of n dimensions the set of n forces are the components of a vector and the set of n velocities are the components of another vector A kinetic model maps one vector to another vector For a general nonlinear kinetic model the map is nonlinear The function Q( v 1 v n ) maps a vector to a scalar The gradient of the function maps a vector to a vector The equation Q( v 1 v n ) = constant represents a surface in the linear space The constant is independent of the velocity At any point on the surface the gradient of the function is a vector normal to the surface Rayleigh s kinetic model specifies that gradient of the function Q( v 1 v n ) gives the force vector Thus the force vector is normal to the surface Q( v 1 v n ) = constant The inner product of the two vectors f i v i is power dissipated For the power dissipated to be positive definite the angle between the two vectors must be acute The shape of the surface is restricted to ensure that the force vector and the velocity vector form an acute angle Forces as independent variables Rayleigh s kinetic model uses the n velocities as independent variables Alternatively we can use the n forces as independent variables Start with a scalar function of n generalized forces A( f 1 f n ) For want of a name different from the dissipation function let s call this function of forces the creep potential Assume that each velocity is the partial derivative of the creep potential with respect to the force associated with the velocity: ( ) v i = A f f 1 n f i Legendre transformation Now we have two kinetic models one based on a function of velocities and the other based on a function of forces Are the two kinetic models the same? Here is the power of mathematics We can answer this question without worrying whether either kinetic model is useful in the real world The mathematical idea is as follows October Nonlinear viscosity 21

22 Plasticity Z Suo Start with the model based on a function of n velocities Q( v 1 v n ) The forces are given by partial derivatives ( ) Recall an identify in calculus: ( ) f i = Q v v 1 n v i dv 1 ++ Q ( v v 1 n) dv v n n dq = Q v v 1 n v 1 A comparison of the two expressions gives that dq = f 1 dv 1 + f n dv n Next define a new scalar by A = v 1 f 1 ++ v n f n Q According to calculus this definition leads to da = v 1 df 1 ++ v n df n Q v 1 v n Thus start with the kinetic model based on the function of n velocities ( ) we obtain n functions of n variables f i ( v 1 v n ) The kinetic model ( v 1 v n ) are one-to-one map between is said to be invertible if the functions f i the set of velocities and the set of forces For the invertible kinetic model we can invert the functions to express velocities in terms of forces v i ( f 1 f n ) If the kinetic model is invertible the definition of the scalar A ensures that we can express A as a function of forces A f 1 f n ( ) The expression da = v 1 df 1 ++ v n df n tells us that each velocity is a partial derivative of the function with respected to the force associated with the velocity: ( ) v i = A f f 1 n f i This mathematical procedure known as the Legendre transformation starts from a function of one set of variables and reaches another function of another set of variables Apply Rayleigh s model to construct a nonlinear creep model under multiaxial stress state Start with a scalar function of stress F σ 11 σ 12 ( ) The function maps a tensor to a scalar Call the function the creep potential Assumes that each component of the rate-of-deformation tensor is a partial derivative of the creep potential with respect to the component of stress associated with the component of the rate of deformation: October Nonlinear viscosity 22

23 Plasticity Z Suo = F ( σ σ 11 12) σ ij For an incompressible isotropic material the creep potential is a function ( ) For simplicity we drop of the two invariants of the deviatoric stress F J 2 J 3 the dependence on J 3 and assume that the creep potential is a function of a single variable F J 3 deformation is ( ) According to Rayleigh s kinetic model the rate of = F ( J 2) ( ) = df J 2 J 2 σ ij dj 2 σ ij Recall that J 2 = /2 so that J 2 / σ ij = The kinetic model becomes that = 2η where η is a function of the second invariant This procedure recovers the J 2 model References HA Barnes The yield stress Journal of Non-Newtonian Fluid Mechanics (1999) HA Barnes JF Hutton K Walters An Introduction to Rheology Elsevier 1989 EC Bingham Fluidity and Plasticity McGraw-Hill 1922 DV Boger and K Walters Rheological Phenomena in Focus Elsevier 1993 BD Coleman H Markovitz W Noll Viscometric Flows of Non-Newtonian Fluids Springer 1966 HJ Frost and MF Ashby Deformation-Mechanism Maps: The Plasticity and Creep of Metals and Ceramics Pergamon Press 1982 JW Glen The creep of polycrystalline ice Proceedings of the Royal Society of London (1955) F Irgens Rheology and Non-Newtonian Fluids Springer 2014 AA Ilyushin The theory of small elastic-plastic deformations Prikadnaia Matematika i Mekhanika PMM (1946) H Markovitz The emergence of rheology Physics Today (1968) EM Purcell Life at low Reynolds number American Journal of Physics (1977) JW Strutt (Lord Rayleigh) Some general theorems relating to vibration Proceedings of London Mathematical Society s (1871) M Reiner A Mathematical Theory of Dilatancy American Journal of Mathematics (1945) October Nonlinear viscosity 23

24 Plasticity Z Suo RS Rivlin The hydrodynamics of non-newtonian fluids Proceedings of the Royal Society of London (1948) October Nonlinear viscosity 24

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