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Momentum Heat Mass Transfer. MHMT3. Kinematics and dynamics. Constitutive equations. Kinematics of deformation , stresses, invariants and r heological constitutive equations . Fluids, solids and viscoelastic materials . Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010.

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momentum heat mass transfer
Momentum Heat Mass Transfer

MHMT3

Kinematics and dynamics. Constitutive equations

Kinematics of deformation, stresses, invariants and rheological constitutive equations. Fluids, solids and viscoelastic materials.

Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

constitutive equations
Constitutive equations

MHMT3

Material (fluid, solid) reacts by inner forces only if the body is deformed or in nonhomogeneous flows.

deformation

Stresses in elastic solids depend only upon the stretching of a short material „fiber“. Stretching of fibers depends upon the initial fiber orientation. Stresses are independent of the rate of stretching and of the rigid body motion (translation and rotation). Deformation is reversible, after removal of external forces, original configuration is restored.

l0

l0

Material with memory

Stresses in viscous fluids depend only upon the rate of changes of distance between neighboring molecules. The distance changes are caused by nonuniform velocity field (by nonzero velocity gradient). Stresses are again independent of the rigid body motion (translation and rotation). Process is irreversible and molecules have no memory of an initial configuration (stresses are caused by relative motion of instantaneous neighbors).

Material without memory

Many materials are something between – viscoelastic fluids have partial memory (fading memory). Example: polymers, food products…

constitutive equations1
Constitutive equations

MHMT3

Kinematic variables in solids and fluids

Reference and deformed configuration is distinguished. Motion is described by displacement of material particles.

reference configuration

displacement

l0

l0

stretch

Solids

Only the current configuration is considered (changes of configuration during infinitely short time interval dt). Motion is described by velocities of material particles.

velocity

time t+dt

time t

Fluids

Velocity is the time derivative of displacement

slide4

Constitutive equations FLUID

MHMT3

Motion of viscous fluid is a fully irreversible process, mechanical energy is converted to heat.

Fluid has no memory to previous spatial configuration of fluid particle, there is no recoil after unloading external stress.

Newton’s law:

Macke

fluids kinematics of flow
Fluids: Kinematics of flow

MHMT3

In case of fluids without a long time memory the role of displacements (differences between the current and the reference position of material particles) is taken over by the fluid velocities at near points. Viscous stresses are response to changing distances, for example between the near points A,B during the time dt.

B’

Gradient of velocity

A’

Arbitrary tensor can be decomposed to the sum of symmetric and antisymmetric part

B

A

Spin (antisymmetric) Rate of deformation

slide6

Fluids: Kinematics of flow

B’

A’

B

A

MHMT3

Result can be expressed as Helmholtz kinematic theorem, stating that any motion of fluid can be decomposed to translation + rotation + deformation

Vorticity tensor

Rate of deformation tensor

Viscous stresses are not affected by translation nor by rotation (tensor of spin, vorticity), because these modes of motion preserve distance between the nearest fluid particles.

slide7

Fluids: stresses

MHMT3

  • Tensor of total stresses can be decomposed to pressure and viscous stresses
  • Hydrostatic pressure (isotropic, independent of relative motion of fluid particles). For pure fluids the pressure can be derived from thermodynamics relationships,
  • e.g. where u is internal energy and s – entropy.
  • Viscous stresses are fully described by a symmetric tensor independent of the rigid body motion. Viscous stress is in fact the momentum flux due to molecular diffusion.
slide8

Fluids: Constitutive equation

MHMT3

Constitutive equations represent a relationship between

Kinematics(characterised by the rate of deformation for fluids)

Viscous stress(dynamic response to deformation)

The simplest constitutive equation for purely viscous (Newtonian) fluids is linear relationship between the tensor of viscous stresses and the tensor of rate of deformation (both tensors are symmetric)

Second (volumetric) viscosity [Pa.s]

Dynamic viscosity [Pa.s]

Index notation

In terms of velocities

slide9

Fluids: Constitutive equation

MHMT3

The coefficient of second viscosity represents resistance of fluid to volumetric expansion or compression. According to Lamb’s hypothesis the second (volumetric) viscosity can be expressed in terms of dynamic viscosity 

This follows from the requirement that the mean normal stresses are zero (this mean value is absorbed in the pressure term)

Constitutive equation for Newtonian fluids (water, air, oils) is therefore characterized by only one parameter, dynamic viscosity

slide10

Fluids: Constitutive equation

MHMT3

Viscosity  is a scalar dependent on temperature.

Viscosity of gases can be calculated by kinetic theory as =(lmv)/3 (as a function of density, mean free path and the mean velocity of random molecular motion, see the previous lecture) and these parameters depend on temperature. This analysis was performed 150 years ago by Maxwell, giving temperature dependence of viscosity of gases

Viscosity of liquids is much more difficult. According to Eyring the liquid molecules vibrate in a “cage” of closely packed neighbors and move out only if an energy barrier is overcame. This energy level depends upon temperature by the Arrhenius term

Viscosity of gases therefore increases and viscosity of liquids decreases with temperature. Typical viscosities at room temperature

water=0.001 Pa.s

air=0.00005 Pa.s

slide11

Fluids: Constitutive equation

MHMT3

  • Viscosity  dependent on the rate of deformation and stress.
  • There exist many liquids with viscosity dependent upon the intensity of deformation rate (“apparent” viscosity usually decreases with the increasing shear rate), these liquids are called
  • generalized newtonian fluids(viscosity depends only upon the actual deformation rate, examples are food liquids, polymers…)
  • There are also materials which flow like liquids only as soon as the intensity of stress exceeds some threshold (and below this threshold the material behaves like solid, or an elastic solid)
  • yield stress (viscoplastic) fluids(example is toothpaste, paints, foods like ketchup)

And there exist also liquids with viscosity dependent upon the whole history of previous deformation, changing an inner structure of liquid in time

  • thixotropic fluids(examples are thixotropic paints, plasters, yoghurt).

Viscosity (T,t, rate of deformation, stress) is a scalar, so the intensity of deformation and the characteristic stress should be also scalars. However the rate of deformation and the stress are tensors

fluids invariants
Fluids: Invariants

MHMT3

How large is a tensor? Magnitude of a stress tensor or intensity of the deformation rate are important characteristics of stress and kinematic state at a point x,y,z, information necessary for constitutive equations but also for decision whether a strength of material was exhausted (do you remember HMH criterion used in the structural analysis?) and many others.

Easy answer to this question is for vectors, it is simply the length of an arrow.

Magnitude of a tensor should be independent of the coordinate system, it should be INVARIANT. We will show, that there are just 3 invariants (3 characteristic numbers) in the case of second order tensors, telling us whether the material is compressed/expanded, what is the average value of the rate of deformation, density of deformation energy and so on (it depends upon the nature of tensor).

fluids invariants1
Fluids: Invariants

MHMT3

Any tensor of the second order is defined by 9 numbers arranged in a matrix. However these numbers depend upon rotation of the coordinate system. For the symmetric tensors (like stress, or deformation tensors) the rotation of coordinate system can be selected in such a way that the matrix representation will be a diagonal matrix ([[]], see the first lecture)

In view of orthogonality of [[R]] we obtain by multiplying the equation by [[R]]T

This is so called eigenvalue problem: given the matrix [[]]3x3 calculate three eigenvectors (columns of the matrix [[R]]T=[[n1],[n2],[n3]]) and corresponding eigenvalues 1, 2, 3, that satisfy the previous equation.

fluids invariants2
Fluids: Invariants

y

x

ux(y)

x

MHMT3

Physical interpretation (stresses):

The product represents the vector of forces acting at the cross section perpendicular to . In the case, that the normal is an eigenvector, the vector of normal will have the same direction as the vector of forces, and the corresponding eigenvalue is the value of the normal force. Eigenvectors are principal directions of the cross sections, where only normal stresses (and not shear stresses) act.

Shear deformation of an elastic block. Principal direction is at angle 450

Physical interpretation (rate of deformations):

The product represents the vector of velocity differences at near points . (only velocities after the rigid body rotation removal are considered). As soon as the vector of distance is an eigenvector, the velocity difference vector will have the same direction.

y

Drag flow between parallel plates. Principal direction is at angle 450

fluids invariants3
Fluids: Invariants

MHMT3

The eigenvalue problem can be reformulated to a system of linear algebraic equations for components of the eigenvector

This system is homogeneous (trivial solution n1=n2=n3=0) and non-trivial solution exists only if the matrix of system is singular, therefore if

Expanding this determinant gives a cubic algebraic equation for eigenvalues 

fluids invariants4
Fluids: Invariants

MHMT3

The values I , II , III are three principal invariants of tensor . Eigenvalues  are also invariants and generally speaking any combination of principal invariants forms an invariant too. For example the second invariant of the deformation rate is frequently expressed in the following form and called intensity of the strain rate (it has a right unit 1/s).

Example: at simple shear flow u1(x2)0, u2=u3=0 holds

The first and third invariants of rate of deformation are no so important. For example the first invariant of incompressible liquid is identically zero and brings no information neither on shear nor elongational flows.

shear rate

This is also the explanation why the constant 2 was introduced in the previous definition

slide17

Fluids: GNF (power law)

n=1 Newtonian

n<1 pseudoplastic fluid

ux(y)

x

MHMT3

Generalised Newtonian Fluids are characterised by viscosity function dependent upon the second invariant of the deformation rate tensor

The most frequently used constitutive equation of GNF is the power law model (Ostwald de Waele fluid)

K-coefficient of consistency

n-flow behavior index

Example: for the simple shear flow and incompressible liquid the power law reduces to

Graph representing relationship between the shear rate and the shear stress is called rheogram

slide18

Fluids: Yield stress (Bingham)

for

while for lower stresses the rate of deformation is zero.

MHMT3

Even for fluids exhibitinga yield stress it is possible to preserve the linear relationship between the stress and the rate of deformation tensor (Bingham fluid).

Constitutive equation for Bingham liquid (incompressible) can be expressed as

(this is von Mises criterion of plasticity)

p-plastic viscosity y=yield stress

Example: for the simple shear flow and incompressible liquid the Bingham model reduces to

y=0 Newtonian

Bingham fluid

This is orginal 1D model suggested by Bingham. The 3D tensorial extension was suggested by Oldroyd

slide19

Fluids: Thixotropic(HZS)

MHMT3

Thixotropic fluids have inner structure, characterised by a scalar structural parameter  (=1 fully restored gel-like structure, while =0 completely destroyed fluid-like structure). Evolution of the structural parameter (identified with the flowing particle) can be described by a kinetic equation, depending upon the history of rate of deformation.

a-restoration parameter, b-decay parameter, m-decay index

The structural parameter  increases at rest exponentially with the time constant 1/a. Rate of structure decay increases with the rate of deformation.

Value of the structural parameter  determines the actual viscosity or the consistency coefficient of the power law model and the yield stress (Herschel Bulkley)

see Sestak J., Zitny R., Houska M.: Dynamika tixoropnich kapalin. Rozpravy CSAV Praha 1990

slide20

Fluids: Rheometry

Rotating cylinder

Plate-plate, or cone-plate

MHMT3

Experimental identification of constitutive models

-Rotational rheometers use different configurations of cylinders, plates, and cones. Rheograms are evaluated from measured torque (stress) and frequency of rotation (shear rate).

-Capillary rheometers evaluate rheological equations from the experimentally determined relationship between flowrate and pressure drop. Theory of capillary viscometers, Rabinowitch equation, Bagley correction.

slide21

Fluids: Summary

MHMT3

Constitutive equations for fluids assume that the stress tensor depends only upon the state of fluid (characterized by the rate of deformation and by temperature) at a given place x,y,z. It is also assumed that the same coefficient of proportionality  holds for all components of the viscous stress tensor and the deformation rate tensor, therefore

It does not mean that the constitutive equations are always linear because the viscosity  can depend upon the deformation rate itself (and there exist plenty of models for viscosity as a function of the second invariants of deformation rate and stresses: power law, Bingham, Herschel Bulkley, Carreau model, naming just a few). Nevertheless, some features are common, for example the absence of normal stresses (rr, zz,…) as soon as the corresponding component of the deformation rate tensor is zero. The exception are rheological models of the second order, for example (Rivlin), exhibiting features typical for viscoelastic fluids, like normal stresses, secondary flows in channels, nevertheless these models are not so important for engineering practice.

solids
Solids:

MHMT3

Reversible accumulation of external loads to internal deformation energy

Time is of no importance, unloaded material recoils immediately to initial configuration.

Hook’s law

Macke

solids kinematics deformation
Solids: Kinematics-deformation

MHMT3

In case of solids the deformation means that an infinitely short material fiber is stretched.

Kinematics of motion can be decomposed to stretching followed by a rotation:

(the same decomposition was done with fluids, but in solids the decomposition cannot be expressed in terms of velocities because time is not considered)

b

Right stretch tensor

Rotation tensor

Deformation gradient

a

B

A

Deformed (current) configuration (loaded body)

Reference configuration (unloaded body)

Reference ( ) and deformed ( ) configurations are distinguished.

slide24

Solids: Kinematics-deformation

MHMT3

F-tensor (deformation gradient) transfers a short material line (vector dX) into new position (vector dx). Written in the index notation

Deformation gradient can be derived from relationship between coordinates of material points xi(X1,X2,X3) as partial derivatives

Deformation gradient can be decomposed to stretch tensor U (symmetric and positive definite) and the rotation tensor R (orthogonal)

Stretching first

Followed by rotation

slide25

Solids: Kinematics-deformation

MHMT3

The stretch tensor U transforms a material fiber dX to the vector UijdXj which is extended or compressed (stretched). Length of the stretched fiber can be derived as

Cjk is the Green tensorof deformation (more precisely right Cauchy-Green deformation tensor), reflecting only the true deformation when the rigid body motion is eliminated

The Cauchy-Green tensor C gives us the square of local change in distances due to deformation and equals the square of the stretch tensor C=U2. The C tensor can be calculated directly from the deformation gradient by matrix multiplication C=FTF. This is much easier than the calculation of stretch tensors U by decomposition.

slide26

Solids: Kinematics-deformation

MHMT3

There exists a confusion in definitions of deformation tensors (there is no generally accepted unique nomenclature and naming the tensors), differences are caused by the fact that the decomposition of F tensor is not unique (F=RU stretching followed by rotation is not the same decomposition as the F=VR rotation followed by stretch). Tensors can be related to the unloaded reference configuration (denoted by capitals X) or to the deformed configuration (small letters x are usually used). In solid mechanics where the Lagrangian approach (tracking material particles) is preferred it is assumed that the reference configuration (X) is the starting one. This is not true In fluid mechanics, when the Eulerian approach is usually used, tracking history of motion starting from the current configuration. Therefore in the solid mechanics the Green’s deformation is preferred (given X(t=0) the x(t) position of particle is calculated) while its role is substituted by the Finger tensor or Cauchy deformation tensor in fluid mechanics (given the current position of material particle x(t) the values at previous times X(t’<t) are calculated).

Remark: I am confused not only by the nomenclature of kinematic tensors, but also by the conjugated stress tensors. Graphical representation of deformations and nine different stress tensors is available as a seminal work of Abbasi (student in UCI)

slide27

Solids: Kinematics-deformation

MHMT3

Interpretation: There are obviously four possible combinations

example elongation i
Example:elongation I.

x1

X3

x3

X2

x2

MHMT3

Simple extension of an elastic rod from an incompressible material

X1

Incompressibility (constant volume)

Cauchy deformation tensor B-1 equals Finger deformation tensor C-1

Left Cauchy Green deformation tensor C equals right Cauchy Green tensor B

solids kinematics strains
Solids: Kinematics - strains

MHMT3

Without deformation the C-tensor is simply the identity tensor (all stretches are 1). Strain tensors are measures of deviation between the stretch tensors (C or U) and the unit tensor (strains are zero for rigid body motion). The most frequently used measure of strain is the Green-Lagrange strain tensor, defined as

or the Hencky strain (called also logarithmic stretches or true strain)

Remark: There are many other definitions of strains, Biot, Seyth… but the Green-Lagrange one is probably the most important.

Remark: The words “strain” and “deformation” are more or less equivalent. I used the “strain tensor” and “deformation tensor” to distinguish between the kinematic tensors that are reduced to the zero or to the identity tensor at rigid body motion.

slide30

Solids: Kinematics - strains

Remark: symbol is usually used in literature for displacement. The reason why the symbol is used here is to avoid conflict with velocities .

b

a

B

A

MHMT3

The Green Lagrange strains can be expressed also in terms of displacements of material points

Extension of material fibre (difference of squares of final and initial lengths) is

giving the final expression of the Green Lagrange strain tensor in terms displacements

this is the same as

slide31

Solids: Kinematics - strains

MHMT3

In relatively stiff solids (for example steel) the deformations, displacements and gradient of displacement are small quantities, therefore the last term in the Green Lagrange tensor (product of two small quantities) can be neglected, giving the tensor of small deformations

For one dimensional case|dx|=l, |dX|=l0

Small deformation

example elongation ii
Example:elongation II.

x1

X3

x3

X2

x2

MHMT3

Simple extension of an elastic rod made from an incompressible material

displacements

X1

incompressibility constrain

Green Lagrange strain (1 is the principal stretch in axial direction)

Small deformations

example small deformation
Example – small deformation

Let us consider a solid rod rotated along its axis by the angle 

X2

Reference configuration Current configuration (after rotation) Displacements

X1

Determine xx component of the small deformation tensor

This term should be zero because it is a rigid body motion (deformation is small only for small rotation angle )

Determine xx component of the Green Lagrange tensor

This term is zero for arbitrary large rotation

MHMT3

This example demonstrates limited applicability of small deformation tensors

solids stresses
Solids: Stresses

MHMT3

Tensor of stresses [N/m2]describes a vector of force F [N] acting to a small surface ([m2] cross-section of material). But which cross-section? The cross-section defined in the reference configuration (A) or in the deformed body (a)?

x2

X3

X2

V-volume in deformed configuration

V0-volume in reference configuration

F

A

x3

F

A0

V

V0

X1

x1

The stress tensor reflecting the actual load of material is related to the deformed cross section A (true stress, Cauchy stress, denoted by ).

Other stress tensors are more or less artificial constructs related to the reference configuration (cross section A0), for example the Kirchhoff tensor S

ratio of volumes (I am not quite sure, Valenta reports reciprocal value)

The reason why and when the Kirchhoff stress can be useful is associated with energetic considerations, see next slide…

solids deformation energy
Solids: Deformation energy

MHMT3

The deformation tensors enable calculation of deformation energy (reversibly accumulated in a deformed body) according to different constitutive models. Knowing the deformation energy W as a function of stretches (or deformations) it is possible to calculate the stress tensors as partial derivatives of W, e.g.

[stress]  [displacement] = [deformation energy change], e.g.

W [J/m3] is deformation energy related to unit volumeof sample in the reference configuration, Sij are components of Kirchhoff stresses and Eij are components of Green Lagrange strain. Kirchhoff stresses and Green Lagrange strains are the so called conjugated tensors (they are both related to the reference configuration).

Proof will be presented on the next slide…

solids deformation energy1
Solids: Deformation energy

MHMT3

It will be later derived that the rate of dissipation of mechanical energy to heat in a unit volume [W/m3] is the scalar product of Cauchy stress tensor and the rate of deformation

In case of solids the mechanical work is not dissipated to heat, but to the internal (deformation) energy increase. However the energy conversion term remains.

The rate of deformation can be expressed in terms of material derivative of the Green Lagrange strain

2kl

Substituting for the rate of deformation and the Cauchy stress

qn

pm

…..completes the proof that

solids deformation energy2
Solids: Deformation energy

MHMT3

Let us consider the simplest linear elastic and isotropic material described by the famous Hook’s law, when the deformation energy can be expressed in terms of the Green Lagrange tensor

This is the Hook’s law for perfectly elastic linear material

wherefrom the component of Kirchhoff stresses can be calculated by partial derivatives

E-Young’s modulus of elasticity [Pa] -Poisson’s constant

slide38

Solids: Deformation energy

MHMT3

The Hook’s law is an example of

CONSTITUTIVE EQUATION (relation between kinematics and stresses)

which is a linear one (in 1D loads and for small deformations the stress is proportional to deformation =E).

More complicated nonlinear or anisotropic materials are characterized by more complicated expression of deformation energy and by using different deformation tensors and their invariants (look at the frequently used Mooney Rivlin model as an example in next slides).

Kirchhoff stresses for isotropic materials can be expressed in rather general form

where I,II,III are three invariants of the Cauchy Green tensor

slide39

Solids: Deformation energy

MHMT3

There is a large group of simple constitutive equations based upon assumption that the deformation energy depends only upon the first invariant of the Cauchy Green tensor

Then the Cauchy stresses can be expressed as

this is Finger deformation tensor

Neo-Hook (linear model similar to Hook, but not the same, has only 1 parameter). Suitable only for description at small stretches (for rubber < 50%)

Yeoh model (third order polynomial)

Gent model characterised by a limited extensibility of material fibers (Imax is the first invariant corresponding to stretches giving infinite deformation energy)

Arruda Boyce model having 7 parameters (max is a limiting stretch). This is the “8 springs” model based upon idea of 8 springs connected in a cube.

example hook 1 3
Example:Hook (1/3)

x1

X3

x3

X2

x2

MHMT3

Homogeneous extension of an elastic rod

X1

Hook’s law

verify that without deformation (i=1) all stresses are zero

example hook 2 3
Example:Hook (2/3)

MHMT3

The previous result holds for compressible Hook material and large deformations. The stretches are not related by any constraint (for example by the incompressibility constraint).

Let us consider the isotropic loading (uniform expansion/compression) with stretches

Kirchhoff stresses are in this case

Cauchy stresses are in this case

because

The result indicates that the compressibility of a Hook material is determined by the ratio

…and it can be seen that =1/2 represents an incompressible material

example hook 3 3
Example:Hook (3/3)

MHMT3

Hook’s law is usually presented in the simple form

This statement is correct for the case of small deformations and unidirectional load in the x-direction (S22=S33=0) , when

this follows from requirements S22=S33=0

slide43

Example: Mooney Rivlin (1/3)

33

11

X3

X1

x3

x1

22

1

1

3

1

X2

x2

2

MHMT3

In many cases the principal directions of deformation and stress tensors are known, examples are a pressurized thick wall tube with axial load and uniform elongation, stretched rod or plate….

In this case the tensors written in the coordinate system of principal axes are represented by diagonal matrices,

Green deformation Green Lagrange strains

Stretches and the Green Lagrange strains are obviously related by

slide44

Example: Mooney Rivlin (2/3)

MHMT3

Mooney Rivlin model defines the deformation energy in terms of two invariants of the Green deformation tensor, or in terms of principal stretches as

Kirchhoff stresses corresponding to stretches are

Cauchy stresses follow from definition

Ou…!!!! non zero stresses are predicted at the reference configuration. Where is the error?

Evaluating partial derivatives of the deformation energy gives

….

The next slide demonstrates how to obtain the same result in a simpler way…

slide45

Example: Mooney Rivlin (3/3)

11

d1

MHMT3

Work done by Cauchy stresses can be expressed directly by stretches

Consider unit cube (dimensions 1 x 1 x 1 in the reference configuration) deformed to the brick with sides 1,2,3. These sides are principal directions and only the normal (Cauchy) stresses 1=11, 2=22, 3=33 act on the faces. Infinitesimally small motion of faces d1,d2,d3 requires the work dW (calculated as the sum of works done by forces 1 23, 2 13, 3 21).

dW is total differential and by comparing coefficients of d the Cauchy stresses can be expressed

slide46

Solids: Experimental methods

MHMT3

  • Biaxial testers
  • Sample in form of a plate, clamped at 4 sides to actuators and stretched
  • Static test
  • Creep test
  • Relaxation test
slide47

Solids: Experimental methods

MHMT3

  • Inflation tests
  • Tubular samples inflated by inner overpressure.
  • Internal pressure load
  • Axial load
  • Torsion

CCD cameras of correlation system Q-450

Pressure transducer

Pressurized sample (latex tube)

Laser scanner

Axial loading (weight)

Confocal probe

elastic solids summary
Elastic Solids (summary)

MHMT3

  • Relationships between coordinates of material points at reference (X) and loaded (x) configuration must be defined. For example in the finite element method the reference body is a cube and the loaded body is a deformed hexagonal element with sides defined by an isoparametric transformation.
  • Function x(X) enables to calculate components of the deformation gradient F and the Cauchy Green deformation tensor (by multiplication C=FTF) at arbitrary point x,y,z.
  • In terms of the Cauchy Green deformation or strain tensor the density of deformation energy W(C) can be expressed.
  • Components of stress tensors are evaluated as partial derivatives of deformation energy with respect to corresponding components of strain tensor.

Anyway, what is typical for solid mechanics, the constitutive equations are expressed in terms of deformation energy W. This is possible because the deformation of elastic solids is a reversible process, time has absolutely no effect upon the stress reactions, and it has a sense to speak about potential of stresses. Nothing like this can be said about fluids, where viscous stresses depend upon the rate of deformation, and viscous friction makes the process irreversible. Therefore the constitutive equations of purely viscous fluids cannot be based upon the deformation energy.

slide49

Viscoelastic FLUIDs

MHMT3

Exhibit features of fluids and solids simultaneously. Deformation energy is partly dissipated to heat (irreversibly) and partly stored (reversibly). Both the deformation and the rate of deformation should be considered.

Macke

slide50

Viscoelastic FLUIDs

MHMT3

The simplest idea of viscoelastic fluids is based upon the spring+dashpot models

Maxwell fluids are represented by a serial connection of spring and dashpot

These materials are more like fluids, because at constant force F no finite (equilibrium) deformation is achieved. Only during the time changes a part of mechanical work is converted to deformation energy, later on all mechanical work is irreversibly degraded to heat.

Voight elastoviscous materials are represented by a parallel connection of spring and dashpot

These materials are more like elastic solids. At constant force F finite (equilibrium) deformation is achieved but not immediately. A sudden change of deformation results to infinite force response.

slide51

k

x1

z

x

Viscoelastic FLUIDsMaxwell

MHMT3

Viscoelastic fluids of the Maxwell body (serially connected spring and dashpot) can be described by ordinary differential equations (k-stiffness of spring, z attenuation)

giving after elimination of x1

F

This 1D mechanical model is the basis of the Maxwell model

where  is a relaxation time, the time required for the stress to relax to (1/e) value of the initial stress jump (due to sudden extension).

slide52

k

x1

z

x

F

Viscoelastic FLUIDsMaxwell

MHMT3

There exist many objections about the previous generalisation of the Maxwell model, first of all against the way how the time derivative of the stress tensor was evaluated.

Time derivative does not fulfill the principle of material objectivity, which means that the time derivative depends upon to motion of observer (material objectivity requires, that even the time derivatives should satisfy the tensor transformation rules[[A’]]=[[R]].[[A]].[[R]]T).

The material objectivity is satisfied either by the model

or by the upper convected Maxwell (UCM)

Remark: Both lower and upper convective Maxwell models are acceptable from the point of view of mathematical requirements, but they are not the same. The UCM model prevails in practice, at least in papers on polymer melts rheology.

slide53

Viscoelastic FLUIDsMaxwell

MHMT3

Example: Fiber spinning (steady elongation flow of a polymer extruded from a dye). Axial velocity w is determined by the radius of fiber R

This is continuity equation relating radial u and axial w velocity (cylindrical coordinate system)

R

and this is a simplified solution based upon assumption of linear radial velocity profile

L

z

Normal stresses for the upper convected Maxwell

F

This is the system of two ODE for the two unknown normal stresses zz and rr. Given axial force F it is possible to calculate axial profile of thickness R(z)

see also the paper Tembely M.et al, Journal of Rheology vol.56 (2012),pp.159-183, fiber spinning using Oldroyd-B and the structural FENE CR rheological model (you will see close similarity of equations presented here).

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k

z1

x1

z2

x

F

Viscoelastic FLUIDs Oldroyd

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Viscoelastic fluids of the Voigt body combined with a serial dashpot can be described by ordinary differential equations (k-stiffness of spring, z1,z2-attenuations)

giving after elimination of x1

or

for new coefficients

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k

z1

x1

z2

x

Viscoelastic FLUIDs Oldroyd

MHMT3

Replacing viscous stress for F and the rate of deformation for dx/dt

This equation reduces to Newtonian fluid as soon as the time constants 1=2=0 (1 , relaxation time, describes exponential decrease of stress after the flow is stopped).

Oldroyd suggested generalization of this model by replacement the common time derivative by the convected derivative in the same way like in the previously discussed Maxwell model

Material derivative D/Dt

covariant convected time derivative

Remark: Introducing convected derivatives makes the Oldroyd’s model nonlinear (due to products of velocity times gradient of stresses and stressed times velocity gradients). Theory is rather complicated, the original Oldroyds model suggested even more complex Jaumann derivatives with vorticity terms, reflecting not only a parallel convection of a moving fluid particle, but also its rotation.

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x3

particle at time t

particle at a past time t’

x2

x1

Viscoelastic FLUIDintegral models

MHMT3

Let us assume a fluid particle moving along a streamline:

rate of strain (e.g. rate of elongation in 1D case)

Viscous fluids: stress depends only upon the rate of deformation at the current time t

Thixotropic fluids: stress depends also only upon the current rate of deformation but viscosity is affected by the deformation history

Viscoelastic fluids: stress at time t is a weighted sum of stresses corresponding to the history of deformation rates. Stress tensors calculated at previous positions of particle must be recalculated to the present position.

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Viscoelastic FLUID Boltzmann

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Integral viscoelastic models are based upon the Boltzmann superposition principle.

The idea can be expressed for 1D case (a dashpot, taking into account only scalar stress and rate of strain) in form of integral

strain rate

memory function

The monotonically decreasing memory function describes the rate of relaxation (rearrangement of fibers, entanglements…) and can be identified from the relaxation unidirectional experiments.

The unidirectional model can be extended to 3D for example as

i-relaxation times, ai-relaxation moduli

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Viscoelastic SOLIDBKZ

MHMT3

Bernstein, Kearsley, Zapas

Nowadays the more common integral methods are oriented to the hyperelastic rather than the fluid-like constitutive equations.

The Kaye-BKZ equation introduces the potential function  which is an analogy of elastic potential (deformation energy W) depending upon the invariants of the Finger tensor C-1 expressing deformation at present time t relative to the configuration at past time t’

Bernstein, B., Kearsley, E.A., Zapas, L.J.: Trans. Soc. Rheol. 7, p.391/410 (1963)

where IC-1 and IC are first invariants (traces) of the Finger and Cauchy tensors respectively.

There exist several modifications of the BKZ model, for example with the extracted memory function or just simply the model without the second term (Cauchy Green tensor), preserving only the Finger tensor in the integrand (Wagner model)

Wagner, M.H.: Rheol. Acta 15, p.136/142, (1976)

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Viscoelastic SOLID Haslach

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Differential models (Maxwell, Oldroyd) are suitable for viscoelastic liquids, integral models for liquids as well as for viscoelastic solids (BKZ). However the integral models are rather complicated (need many experimentally determined parameters).

Haslach suggested a method based upon a perfect knowledge of equilibrium state (for example a relationship between stress and strain at a steady state). Time evolution of stresses from a non-equilibrium state towards the equilibrium in selected in such a way that the dissipated energy is maximised .

Do you remember the Lagrange method, looking for the minimum of total energy?

()

Trajectory on the * surface starting from a non-equilibrium state (,) and maximising dissipated energy

*(,)

Deformation energy

*(,)= -

Curve of equilibrium states



()

Equilibrium deformation  corresponds to the minimum of total energy.

It is quite difficult for me to understand the statement about the maximisation of dissipated energy. There are many other trajectories between the starting and ending points when the dissipated energy will be greater

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Viscoelastic SOLID Haslach

MHMT3

Principle of the suggested method is based upon definition of affinity X(,)

which represents a measure of distance between the non-equilibrium point (,) and the equilibrium curve, when

The evolution proces (trajectory of (t),(t)) is described as the motion on the surface * towards equilibrium in the direction of gradient * (gradient relaxation method)

….giving the final result

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Viscoelastic SOLID Haslach

MHMT3

The nonlinear maximum dissipation evolution is described by the system of first-order nonlinear ordinary differential equations

corresponding for example to a creeping test, when stresses are prescribed and deformations are calculated. Deformation energy  is expressed in terms of deformations and the parameter k is relaxation modulus.

The role of stresses and deformations can be exchanged (describing a relaxation test) and the Haslach principle can be applied in fact for almost any pair of conjugated variables (pressure/volume, Kirchhoff stress/Green deformation,…). The deformation energy can be described by any hyperelastic model.

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Viscoelastic SOLID Haslach

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Haslach presents application of the Mooney Rivlin model (energy expressed by principal stretches 1, 2) for description of biaxial creeping test of a rubber sheet

The values of the Mooney-Rivlin constants, C1 = 2.240×105 Pa and C2 = 8.512× 103. Biaxial creep test with unequal loads, an initially unloaded rubber sheet is stressed by s1 = 9.456 × 105 Pa and s2 = 7.359 ×105 Pa with various relaxation moduli, k = 5.0 ×107, 2.0 ×107, and 1.0×107 Pa/s respectively, for the initial conditions λ1 = λ2 = 1.0 and s1 = s2 = 0 at t = 0. Fig shows that the principal stretches relax asymptotically to the equilibrium values of λ1 = 2.0 and λ2 = 1.5, corresponding to the perturbed stresses as calculated from the Mooney-Rivlin equation. Furthermore, the larger the relaxation modulus, k, the faster the relaxation to equilibrium.

This example is presented without permission-it is a copy from the book

Henry W. Haslach Jr.: Maximum Dissipation Non-Equilibrium Thermodynamics and its Geometric Structure. Springer N.Y. 2011

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Point near the

Rotating

rotating disk

disk

w

Stretched filament

H

Filament pushed

towards the center by

R

2

the stretched filament

R

1

Weissenberg effect (material climbing up on the rotating rod)

Point near

the fixed disk

Barus effect (die swell)

Kaye effect

Viscoelastic EFFECTs

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Polymer melt flows against the centrifugal forces towards the rotation axis

so that you will have a better idea about what is going on, imagine that instead of a homogeneous fluid there are entangled spaghetti fibers

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Viscous liquid- zero stress corresponds to zero strain rate (maximum ) =900

Hookean solid-stress is in phase with strain (phase shift =0)

Viscoelastic material – phase shift 0<<90

Experiments: solid and fluids

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Characteristic features of elastic, viscoelastic and viscous liquids are best seen using oscillating rheometer (usually cone and plate configuration)

Sinusoidaly applied stress  and measured strain (not the rate of strain!)

Weissenberg rheogoniometer from Wikipedia

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EXAM

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Constitutive equations

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What is important (at least for exam)

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Fluids

You should know Helmholtz decomposition of the velocity gradient tensor into spin and rate of deformation tensors

Decomposition of stress tensor

Newtonian fluid

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What is important (at least for exam)

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Non Newtonian Fluids

Invariants

Power law liquids

Bingham liquids

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What is important (at least for exam)

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Solids

Deformation gradient

(what is it x and X?)

Decomposition of deformation to stretch and rotation tensors

(what is it R and U?)

Cauchy Green deformation tensor

Hook’s law in terms of Kirchhoff stresses and Cauchy Green deformation tensor

(what is it E and ?)

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What is important (at least for exam)

MHMT3

Viscoelastic fluids

Maxwell model

(draw a combination of dashpots and springs corresponding to this model)