Book Nonlinear Models

7/31/2019 Book Nonlinear Models

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Peter Hansbo

Applied Partial Differential

Equations for Computational

Science and Engineering

Part II: Nonlinear models in continuum

mechanics

March 10, 2011


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Contents

1. Material Models in Small Deformation Solid Mechanics . . 11.1 Viscous stressstrain relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Maxwell materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Kelvin materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Generalization to several dimensions . . . . . . . . . . . . . . . . 3

1.2 Elastoplastic stressstrain relations . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 An elastoviscoplastic model . . . . . . . . . . . . . . . . . . . . . . . 9

2. General Principles in Continuum Mechanics . . . . . . . . . . . . . . 112.1 Velocity of a continuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Flux through a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Reynolds transport theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Newtons equations of motion in a continuum setting . . . . . . . 17

3. Large Deformation Solid Mechanics . . . . . . . . . . . . . . . . . . . . . . . 193.1 Strain measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.1.1 The deformation gradient and a general approach tostrain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.2 The boundary value problem on the deformed domain 233.1.3 Stressstrain relations in large deformation theory . . . . 23

4. Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.1 Energy balance and the first law of thermodynamics . . . . . . . . 274.2 Entropy and the second law of thermodynamics . . . . . . . . . . . . 29

5. Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.1 Inviscid compressible fluids and conservation laws . . . . . . . . . . 335.2 Nonlinear conservation laws and shocks . . . . . . . . . . . . . . . . . . . 375.3 Linearization of the Euler equations and acoustics . . . . . . . . . . 395.4 Inviscid incompressible fluids. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.5 Viscous incompressible fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.6 Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45


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4 Contents

6. Stability problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.1 A simple example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2 The beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Instability and the minimum of potential energy . . . . . . . . . . . . 54


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1. Material Models in Small Deformation Solid

Mechanics

In this Chapter we extend the simple Hookes law to linear viscous modelsand nonlinear elastoplastic modes.

1.1 Viscous stressstrain relations

We recall the simple Hookes law in one dimension:

= E = Edu

dx

where is the stress (force per unit area) and is the strain (increase in lengthdivided by initial length), and u is local displacement. We can view this lawas a generalization of an elastic spring and visualize the model as in Fig.1.1 (left). In a viscous material, like a Newtonian fluid mentioned in AppliedPartial Differential Equations for Computational Science and Engineering,

Part I: Linear models in continuum mechanics (henceforth called Part I), itis instead the strain velocity that is proportional to the stress, so that

d

dt=

1

, or =

1

, or

du

dx=

1

,

where (in solid mechanics) is known as the modulus of viscosity. This ma-terial can be represented as a damper, Fig 1.1 (right). These two models cannow be combined in various ways to create ever more complex material mod-els in order to try to predict realistic material behavior. We shall only givethe two simplest possible models.

1.1.1 Maxwell materials

The Maxwell material can be viewed as a damper and a spring coupled inseries, cf. Fig 1.2 (left). The the stress in the spring part, s, and the damperpart d, must be equal due to equilibrium,

= s = d, (1.1)

and the total strain is given by


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2 1. Material Models in Small Deformation Solid Mechanics

Fig. 1.1. Spring model (left) end damper model (right).

= s + d. (1.2)

The constitutive relations are

s = Es, (1.3)

and

d = 1

d. (1.4)

Thus,

=1

E +

1

. (1.5)

In this material, a loadingunloading cycle leaves a residual deformation (inthe damper part) that does not go away, only the deformation in the springpart returns to its initial state.

1.1.2 Kelvin materials

Many viscous materials return to their original state after deformation (albeit

slowly). In order to model such materials, one can couple a spring and adamper in parallel, see Fig. 1.2 (right). This is the Kelvin model, in whichthe strain in the damper and the spring are equal,

= s = d, (1.6)

and the stress is the sum of stresses in the spring and in the damper,

= s + d. (1.7)

The constitutive relations are the same as for the Maxwell material, and thus

+ E = . (1.8)

In this model, a loadingunloading cycle leaves a residual stress in the springwhich slowly takes the deformation back to its initial state.


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1.1 Viscous stressstrain relations 3

E

E

Fig. 1.2. Maxwell model (left) and Kelvin model (right).

1.1.3 Generalization to several dimensions

The generalization of Kelvins onedimensional model to threedimensions isstraightforward. Hookes law for an isotropic material states that (cf. Part I)

= 2(u) + uI= 2(u) + tr (u) I,

where we recall that the trace of the strain matrix, tr , means the sum ofthe diagonal elements of. A generalization to a Kelvin model could thus be

(t) = (u) + 2 (u) + tr (u) I. (1.9)

This constitutive relation must then be combined with the equilibrium equa-tions

(t) = f(t),and the problem is that of an initialboundarvalue problem in u. The cor-

responding extension of a Maxwell material to get an initialboundaryvalueproblem for u is more involved because of the presence of in the constitutivelaw. We then first need to invert Hookes law to obtain

(u) =1

2

2(3 + 2)tr, (1.10)

and a Maxwell constitutive model could be

(u) =1

+

1

2

2(3 + 2)tr . (1.11)

Viscoelastic models are thus linear, but time dependent. We next turn toa model which is nonlinear.


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1.2 Elastoplastic stressstrain relations

Elastic materials exhibit a linear relation between stress and strain. There is,however, a limit to the load a material can carry before it breaks. In metals,

the elastic regime ends when atomic planes starts sliding along each other,and irreversible strains occur. In Fig. 1.3 the principle is illustrated. Oncethe stressstrain curve passes point 2, irreversible strains will occur. Uponunloading, the stress-strain curve will return to = 0 along a straight line,leaving residual strains. The actual graph of the function = () is complexand differs from material to material, and there is a need to make simplifyingassumptions. A simple model for plasticity is to postulate that the material

Fig. 1.3. Stressstrain relations in metals.

can only carry a stress below a fixed value y, the yield stress, and assume theidealized stressstrain relation shown in Fig. 1.4. The extension of this simplemodel to three dimensions can be done in various ways. Experimentally, onefinds that the material does not yield (the atomic planes do not slip) incase of pure compression (pure volume change). It is therefore sometimesassumed that the shear forces alone are responsible for plastic yield. It turnsout however, that a mathematically more pleasing theory can be developedby splitting the stress matrix into the part responsible for volumetric changeand the part responsible for change of form. By Hookes law we have

= 2(u) + uI,


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1.2 Elastoplastic stressstrain relations 5

Fig. 1.4. Idealized plasticity model in one dimension.

and we know that

u can be seen as the volume change (cf. Part I). Taking

the trace of the left hand side and of the right hand side, followed by divisionby 3, we obtain the mean (or volumetric) stress

V :=1

3(11 + 22 + 33) =

2

3 u + u := K u

where K = + 2/3 is called the bulk modulus. What remains, the stressdeviator

D = VI (= 2D from Hookes law),which has zero trace, is thus responsible for change of shape. Standard plas-ticity theory thus typically posits that there is a limit to the maximum sizeof the stress deviator. Richard von Mises suggested that we take as criterion

|D| = D : D y

23

,

where y is, as above, the yield stress in uniaxial tension. Note that theleft-hand side is an invariant and does not depend on the chosen coordinatesystem.

It is possible to visualize von Mises yield criterion if we use the principalstresses (the eigenvalues of the stress matrix) as coordinate axes. We are thuslooking at the case when the physical coordinates are chosen so that

=

1 0 00 2 00 0 3

.

If all stresses are equal, 1 = 2 = 3 =: , then the volumetric stress V = and, since


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D =

0 00 0

0 0

I=

0 0 00 0 0

0 0 0

,

there is no yield whatever the size of . Otherwise we find

D =

1 0 00 2 0

0 0 3

1

3

1 + 2 + 3 0 00 1 + 2 + 3 0

0 0 1 + 2 + 3

or

D =1

3

21 2 3 0 00 22 1 3 0

0 0 23 1 2

and thus

D : D =1

9

(21 2 3)2 + (22 1 3)2 + (23 1 2)2

.

A closer inspection shows that the equation for the yield surface,

1

9

(21 2 3)2 + (22 1 3)2 + (23 1 2)2

=

22y3

,

describes a slanted cylinder in the space of principal stresses shown in Fig.1.5. Thus, in von Mises model, the principal stresses are confined to staywithin this cylinder.

In many texts on plasticity, one employs the yield function, in our casedefined as

() :=D : D y

2

3,

and then the stress must reside in the set E of admissible stresses defined by

E =

{ : ()

0

},

i.e., inside the cylinder of Fig. 1.5. We note that this is a convex set and thustools of convex analysis can be used.

The simplest model for plasticity is the Hencky (or total deformation)model. In standard engineering notation this model may be formulated asfollows: for a given body load f (and given boundary conditions), find thedisplacement field u such that

= f (equilibrium) = 2e + tr e I (constitutive equation for stress)

()ij =1

2

uixj

+ujxi

(small strain)

= e + p (additive decomposition, cf. (1.2))

p = g() (constitutive equation for plastic strain)


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Fig. 1.5. Yield surface in the space of principal stresses.

where e is the elastic strain, p is the plastic strain, and g is the over-stress function (to be defined). The Hencky model is suitable for proportionalloading only; for general loading/unloading, these equations have to be timedependent, replacing u by u, strains by strain rates, etc.

Now, for a convex set E, there must hold, for E, where E is thesurface of E,

( ) : q 0 Eif q is an outward-pointing normal to E, cf. Fig 1.6. This simple factof convex sets has been given a mechanical interpretation as the MaximumPlastic Dissipation (MPD) property: the actual stress in a plastic state, E, is the one, among all admissible stresses E, leading to maximumplastic dissipation D, where

D : p 0.

Clearly, D 0 is implied by thermodynamics since it physically means adissipation of energy (the stresses must perform work when rupturing thematerial and this work cannot be regained). The maximization of D, however,is a physical assumption that may or may not hold for a given material. Wemay write MPD as

: p

: p

Eor


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( ) : p 0 E,and thus we can identify q = p; in other word: p must by necessity benormal to the yield surface. This means that

p = ()|=0 (or g() = ()|=0),where is called the plastic multiplier. (The reason that is called a mul-tiplier lies in the fact that we have a minimization problem (of potentialenergy) with a side condition () 0, which can be solved using Lagrangemultiplier techniques).

Fig. 1.6. Convex set of admissible stresses.

A very simple way of approaching the problem of obtaining a constitutivelaw for the plasticity problem, without the use of multipliers, is to introducethe projection onto the yield surface, i.e.,

(D) =

D if () 0,

D

|D|y

2

3if () > 0.

This corresponds to a projection () of as () = (D) + VIwhichwe note is always in E. Alternatively, we can view the projection as an elasto-plastic constitutive relation and formulate our original plasticity problem as

that of finding the displacement field u such that


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= f (equilibrium)

= (2D(u)) +3 + 2

3tr (u) I (constitutive equation for stress)

()ij =

1

2 ui

xj +

ujxi

(small strain)

and need not concern ourselves with plastic strains at all. All that needs tobe done is to project the deviatoric part of the elastic stress onto the yieldsurface. This approach is known as the radial return method (we scale theelastic stress radially back to the yield surface) and can only be used forplasticity models where the yield surface is convex and where MPD holds.

We remark that plasticity models with non-convex yield surfaces as wellas models violating MPD are in use, in particular for friction materials suchas sand. The basic idea of projecting back the elastic stress to a yield surfaceis often used in these cases as well, albeit in a more complicated fashion.

The elastic strains can be recovered from (1.10) and the plastic strainsfrom the difference between total strain (computed from u) and elastic strain.

1.2.1 An elastoviscoplastic model

The projection operator onto the yield surface can also be used in a penaltymethod. Starting from a time-dependent setting and using the additive split(u) = e + vp, where vp is a viscoplastic strain, we can argue as in thecase of a Maxwell material and define

(u) =1

2

2(3 + 2)tr +

1

( ()) .

This is known as the DuvautLions model of elasto-viscoplasticity. Note that

vp = 1 ( ()) = 1 D (D) , the mean stress cancels out

which takes the place of the viscous strain rate / in Maxwells model,forces onto the yield surface when 0 in order to stay finite. In thissense, the viscoplastic model is close to the von Mises plasticity model. Theconstitutive parameter is known as the relaxation time.

The main difference between this viscoplastic model and the plasticitymodel from which it is derived lies in the fact that in the former we have anapproximation of the gradient of the yield function (in the time dependentplasticity model) so that

p = ( ()) / = vp.

This approximation makes the problem more regular than the plasticity prob-lem, and thus easier to solve numerically.


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2. General Principles in Continuum Mechanics

In this Chapter we introduce the general conservation and balance laws thathold in all continuum mechanical theories. They thus form the basis for fluidand solid mechanics in general.

2.1 Velocity of a continuum

Two types of description of motion are mainly used: referential descriptionand spatial description. The referential description uses a coordinate systemof material coordinates X referring typically to the configuration occupiedby the continuum in question at a reference time, often t = 0. A referentialcoordinate of a particle thus remains constant during motion through space.If we use the coordinates of space itself, through which the object moves,then we talk of a spatial description with coordinates x. So we can say thatx is the place occupied by the particle which was originally at X, and wemay write

x = x(X, t).

Note that the material coordinates X do not depend on time! X and x mayor may not be measured with respect to the same coordinate axes.

The spatial description is very important in fluid mechanics, where oneusually looks at fixed volumes in space through which the fluid flows. Oftenthe velocity in a fluid is viewed a a primal variable, and as a spatial quantity,

v = v(x, t),

but if we want to definevelocity, we must look at particles and their velocities.Thus the velocity is defined as a function of material coordinates,

v = v(X, t) :=x(X, t)

t.

Clearlyv(X, t)

v(x(X, t), t),

so by the chain rule


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12 2. General Principles in Continuum Mechanics

v(X, t)

t=

v(x, t)

t+

x(X, t)

t v(x, t) = v

t+ v v =: Dv

Dt,

here we reintroduced v = v(x, t) and introduced the notation D/Dt forthe material time derivative obtained when following the motion (the time

derivative obtained when keeping the reference coordinate fixed).

2.2 Flux through a surface

Fig. 2.1. Flux through an infinitesimal surface element dS during an infinitesimaltime interval dt.

In Fig. 2.1 we show a small element dS of a fictitious surface S, fixed inspace, with a fluid medium flowing through it. A normal to the surface isdenoted n, and the fluid velocity is denoted v. By definition of velocity, wesee that during a time interval dt, particles initially on dS will move out toform a slanted cylinder with side length |v|dt. The height of the cylinder isimmediately found as v ndt, since v n is the length ofv in the direction n.Hence the volume of material flowing through dS in time dt equals v n dtdS.The amount of volume per unit time flowing through dS, v n dS, in knownas the flux of volume through dS. The total flux of volume through S is thusgiven by

Volume flux = Sv n dS.Similarly, we can compute the total mass per unit time flowing through thesurface. Since mass equals density times volume, we have


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2.3 Conservation of mass 13

Mass flux =

S

v n dS,

where is the density of the fluid. In general, any quantity carried by theflow (energy, heat, momentum, etc.), has a corresponding flux, so that

Flux of =

S

v ndS.

Here is not necessarily a scalar, momentum for example is a vectorialquantity. Typically we are interested in quantities per unit mass, and thenthe flux is instead defined as

Flux of =

S

v ndS.

If the surface is closed and the normal points outwards from S, then theintegral denotes the total net outflow ofx through S. In Part I we addressedheat conduction. There we had two fluxes, diffusive flux and convective flux.

The convective flux was of the type discussed here, i.e., heat carried by fluidflow. The diffusive flux, on the other hand, is a nonmechanical flux, whichof course also can occur; another such example is electric current flow. Thus,nonmechanical fluxes do not require motion of the medium in order to occur.In some respects they can still be treated as if they were fluid flows; indeed,heat was originally perceived as a fluid, caloric.

2.3 Conservation of mass

Consider an arbitrary volume V, fixed in space, with surface S. The totalmass M contained in the volume is given by

M =V

dV.

If the density in the volume changes with time, then so does the mass, andwe have

dM

dt=

V

tdV.

If mass is not created or destroyed inside V, then the rate of change of massmust also equal the total flux of mass through S. If there is an increase ofmass, so that

tis a positive quantity, then the rate of inflow of mass must

equal the rate of change of mass, so

Sv n dS = V

t

dV,

or, by use of Gauss divergence theorem,


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V

(v) dV =V

tdV,

and, since V is completely arbitrary, we obtain the continuity equation

t

+ (v) = 0. (2.1)

If we carry out the spatial differentiation,

xi(vi) =

xivi +

vixi

,

this can be written

t+ v + v = 0. (2.2)

Recalling the material time derivative,

D

Dt

:=

t

+ v

,

we can write the continuity equation as

D

Dt+ v = 0. (2.3)

If the fluid is incompressible, so that the density of an infinitesimal volumeis constant as it moves, then the continuity equation takes the simpler form

v = 0. (2.4)

This is also known as the incompressibility condition.

2.4 Reynolds transport theorem

In solid mechanics, the material is moving through space, so an importantquestion is how to evaluate the material time derivative of integrals overmoving volumes V(t), volumes that contain a given mass system (think ofV(t) as the volume of an object). For example, by conservation of mass

D

Dt

V(t)

dV = 0,

since the total mass of a body does not change as we follow its motion.In Reynolds theorem we think of a particular time t when V(t) is sur-

rounded by a surface S. If this S is fixed in time, then S is the boundary ofV(t) only at the particular time t when the volume of mass passes though it.In that case we have, for a quantity per unit mass (scalar, vector, tensor)


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2.4 Reynolds transport theorem 15V(t)

t( ) dV =

D

Dt

V(t)

dV S

v n dS (2.5)

or: the rate of increase of inside the fixed surface S equals the rate ofincrease of at this moment inside S minus the flux of through S.

The derivation of this theorem differs between fluid and solid mechanics.In fluid mechanics, one first defines a control volume, fixed in space, throughwhich the material flows. In solid mechanics, this control volume is simplythe object (or a part of an object) at a particular time t, in which case thesurface S is simply the boundary of V(t), S = V(t). Here, we will give thestandard fluid mechanics explanation for the transport theorem.

Fig. 2.2. A set of material points, a system, passing through a control volume(between the dotted lines) at time t and at time t + t.

In Figure 2.2 we have a system occupying a volume V(t) going througha control volume VCV. The total amount of in the system is denoted byBsys, so that


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Bsys(t) :=

V(t)

dV,

and the amount of within the control volume is denoted by

BCV(t) :=VCV

dV.

Note that BCV is also a time dependent quantity since the amount of insidethe control volume may change in time. The objective is to compute

D

DtBsys(t) = lim

t0

Bsys(t + t) Bsys(t)t

.

Now, obviously be have that

Bsys(t) = BCV(t) + Bin,

and

Bsys(t + t) = BCV(t + t) + Bout,where Bout is the amount of that is leaving the control volume and Binis the amount of that going in to the control volume. Thus

D

DtBsys(t) = lim

t0

BCV(t + t) + Bout BCV(t) Bint

or

D

DtBsys(t) = lim

t0

BCV(t + t) BCV(t)

t+

Boutt

Bint

(2.6)

Now, since the control volume is fixed in space, we have

limt0

BCV(t + t) BCV(t)t =VCV

t ( ) dV,

the local rate of change of inside the control volume. We recall from Chapter2.2 that the amount of that flows through a surface S over time t can befound as

B = t

S

v ndS

The different signs on Bin and Boutin (2.6) reflects the different signs onv n, and we arrive finally at

D

Dt

V(t)

dV =

VCV

t( ) dV +

S

v n dS

which, if we consider the instant in time when VCV = V(t), gives Reynoldstransport theorem (2.5).


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2.5 Newtons equations of motion in a continuum setting 17

By Gauss theorem, we can rewrite (2.5) as

D

Dt

V(t)

dV =

V(t)

t( ) + ( v)

dV, (2.7)

where we interpret

( v) :=j

xj( vj).

We can exemplify by setting = 1, in which case

0 =D

Dt

V(t)

dV =

V(t)

t+ (v)

dV,

which again gives the continuity equation (2.1).By rearranging terms, we find that the integral on the right in (2.7) can

be written

V(t)

t

+ v + t

+ (v) dV.By the continuity equation, the second expression equals zero, and we obtainReynolds transport theorem in its simplest form

D

Dt

V(t)

dV =

V(t)

D

DtdV, (2.8)

which is the most important form of Reynolds transport theorem in solidmechanics.

2.5 Newtons equations of motion in a continuumsetting

Newtons law of motion states that the (time) rate of change of linear momen-tum must equal external forces. For a body occupying V(t) moving throughspace, we have the rate of linear momentum as

D

Dt

V(t)

v dV,

where v is the velocity of the material points in the body. Newtons law ofmotion states that the time rate of change of linear momentum must equalthe applied force, so we have that

D

Dt

V(t)

v dV =

V(t)

fdV +

S

t dS,


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where f is a volume force per unit mass and t is a surface traction. In Part Iwe derived the relation between the surface traction and the stress tensor as

t = n.

By Gauss theorem and the transport theorem we thus haveV(t)

fdV =

V(t)

Dv

DtdV

S

n dS =V(t)

Dv

DtdV

V(t)

dV.

Since the volume is arbitrary, we find the equations of motion for a continuumto be

Dv

Dt = f, (2.9)

or

v

t+ v v = f, (2.10)

which must be supplied with initial and boundary conditions. The equations

(2.10) are valid in the spatial coordinate system, i.e., a coordinate systemfixed with respect to the motion. Next, we shall look at some models forcomputing the stress matrix in large deformations.


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3. Large Deformation Solid Mechanics

In this Chapter we derive the basic relations in large deformation solid me-chanics. We extend the notion of strain from small to large straining anddiscuss some basic stressstrain relations in the large deformation setting.

3.1 Strain measures

In small deformation theory we have, in the two dimensional case, defines thestrain by

xx =uxx

, yy =uyy

, xy =1

2

uyx

+uxy

.

These definitions require the deformations to be small, so that there is nopractical difference between initial and deformed state. Even if only the dis-placements are large, and the material remains stress free, this definition fails.Consider the case of a wheel undergoing a 90 rotation around its medianaxis. The corresponding displacements from P to P is given by

ux =

Y

X, uy = X

Y.

and if we extend the small deformation strain by siumply replacing derivativeswith respect to (x, y) with derivatives with respect to the initial coordinates(X, Y) we obtain

xx = yy = 1, xy = 0.Thus in small strain theory, a rigid body rotation will give rise to strainingwhich is clearly not desirable. We need a better definition of strain so thatrigid body rotations do not give rise to straining. Consider an infinitesimalsegment dX that undergoes deformation to length ds, see Fig. 3.2. The lengthafter deformation can be evaluated as

ds2 = dX+uxX

dX2

+ uyX

dX2

.

Green introduced the following definition of large deformation strain:


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20 3. Large Deformation Solid Mechanics

P

P' (X,Y)

(-Y,X)

Fig. 3.1. Counterclockwise rotation 90 from (X, Y) to (Y, X).

Exx =1

2

ds2 dX2dX2

=1

2

1 +

uxX

2+

uyX

2 1

=uxX

+1

2

uxX

2+

uyX

2,

(3.1)

and similarly

Eyy =uyY

+1

2

uxY

2+

uyY

2,

and (less straightforwardly)

Exy = 12

uxY

+ uy

X

+ 1

2

uxX

uxY

+ uy

XuyY

.

Taking again the case of the rigid body motion discussed earlier, we find

Exx = 1 + 12

(1 + 1) = 0, Eyy = 1 + 12

(1 + 1) = 0,

Exy =1

2(1 + 1) + 1

2(1 1) = 0,

which is as expected. We note that the quadratic terms can be neglectedif the displacements are small, which justifies the small deformation strainmatrix. Next, we shall take a more general approach to arrive at the Green

stain.


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3.1 Strain measures 21

X,x

Y,y

dX

ds

u(X,Y)

u(X+dX,Y)

(uy/X)dX

(1+ux/X)dX

Fig. 3.2. Large deformations in two dimensions.

3.1.1 The deformation gradient and a general approach to strain

The most important quantity in large deformation analysis is the deformationgradient F. Consider a material point P located initially at X and connectedto a point Q by the vector dX. After deformation during a time interval t,P has moved to p at x(X, t) and Q has moved to q at x(X+ dX, t). Thevector connecting p and q is denoted dx, and can be found by


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dx = FdX, where F =x

X=

x

X

x

Y

x

Zy

X

y

Y

y

Z

zX

zY

zZ

.

Note that F is in fact the transpose of J which we have used for mappingreference elements to physical elements in the finite element method. This isbecause J transforms gradients whereas F transforms infinitesimals.

Objects transforming material quantities such as dX to spatial quantitiessuch as dx will de denoted using capital boldface letters (such as F), thereverse transformation by lowercase bold letters.

Now, the length of ds a segment dx in the deformend configuration canbe found by ds2 = dx dx, and similarly, its original length dS2 = dX dX.We also see that

dx

dx = dX

(FTFdX) = dX

CdX,

where C = FTF is called the right CauchyGreen deformation tensor (be-cause F is on the right). In differential geometry, where ds is typically aphysical length and dS is the length in a parameter space, C is known as themetric tensor, because it it related to the measuring of distances.

If we interchange the order ofF and its transpose we obtain b = FFT,consequently called the left CauchyGreen deformation tensor, and we have

dX dX= dx b1dxThe Green strain tensor E is defined, in analogy with (3.1), by the relation

1

2(dx dx dX dX) = dX EdX, (3.2)

and thus

E=1

2(C I). (3.3)

Alternatively, we can express the strain in spatial coordinates through therelation

1

2(dx dx dX dX) = dx edx,

where

e =1

2(I b1)

is called the Almansi strain tensor. We have here used uppercase lettersto denote quantities associated with the reference coordinate system andlowercase for quantities associated with the spatial coordinate system. Thedeformation gradient is the link between the two coordinate systems since itis used as a map F : X x.


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3.1.2 The boundary value problem on the deformed domain

We also need to define stress in large deformation theory. The good newsis that the stress tensor in the deformed configuration must fulfill the same

equilibrium equations as in small deformation theory. The bad news is thatwe then have to deal with a domain that keeps changing with time. If we wantto differentiate on a fixed domain, this has to be the material domain wherewe use X as coordinates. The problem is that we then have to transformback the physical (Cauchy) stress from x to X.

Starting with the equilibrium on the deforming domain (t), from Chap-ter 2.5, we have that the external equilibrium, in the absence of acceleration,requires that volume forces and surface tractions have to be in balance, thus

= f in (t). (3.4)The gradient is here taken with respect to x, and we can also view the primaryunknown as x since the displacement u = xXwith X known.

With natural boundary conditions n = t on N(t), we have the

nonlinear boundary value problem: Find and x such that

= f in (t), n = t on N(t),

x = g on D(t),

where g represents a known motion of the D(t) part of the boundary. Wenow need to relate to x, or rather F, which is the objective of the nextSection.

3.1.3 Stressstrain relations in large deformation theory

A common assumption in large deformation theory is that there exists ascalar strain energy density W from which the stress tensor can be derived.In small deformation elasticity, this strain energy density takes the form

W =

0

() : d =

0

(2 + tr I) : d =1

2

2||2 + (tr)2 . (3.5)

The point of defining this quantity is that the stress can be directly derivedfrom it through

=W

, which means ij =

W

ij.

If the stress can be derived from such a strain energy density, we say thatthe material is hyperelastic. In large deformations, the assumption is that Wis a function of the deformation gradient or quantities associated with it, sothat for example


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Jx q(x) = X q = X (JF1q(x(X))).

We have that

JF1 =

y

Y x

Y

yX

x

X

, JF1q = y

Y qx x

Y qy

yX

qx +x

Xqy

.

Thus,

X (JF1q) = X

y

Yqx x

Yqy

+

Y

y

Xqx +

x

Xqy

= qx

2y

XY

2y

Y X

+ qy

2x

Y X

2x

XY

+y

Y

qxX

xY

qyX

yX

qxY

+x

X

qyY

.

Since qxX

=qxx

x

X+

qxy

y

X,

etc., we find that

X (JF1q) = yY

qxx

x

X+

qxy

y

X

xY

qyx

x

X+

qyy

y

X

yX

qxx

x

Y+

qxy

y

Y

+x

Xqyx

x

Y+

qyy

y

Y=

qxx

+ qy

y

yY

xX

yX

xY

= Jx q.

The corresponding transform in the case of stress tensors is the rule

=1

JFFT =

1

JF (2E+ trE I)FT,

which defines the Cauchy stress for the St. Venant-Kirchhoff model. Notethat a simple way to define a tensor (called a dyadic tensor) is to take theouter product of a vector by itself:

B=aaT

,where a is a column vector. Ifa = Fa, then


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B = FaaTFT = FBFT,

where we can think ofB as being defined in the reference coordinate system.The inclusion of 1/J is again done to counter volume change between referen-

tial and spatial coordinates. This allows for a traction to be defined in relationto the PiolaKirchhoff stress in the referential domain, and a Gaussian the-orem for stresses to hold in the referential domain. A further examination ofthese relations are however beyond the scope of this course.

To return to the stress definitions, it is generally agreed that the St.VenantKirchhoff model is not very accurate outside the small deformationtheory; it works well with large rotations but not give a physically realisticdeformation under large strains. A better model is given by the NeoHookestress definition

=

J(b I) + ln J

JI, (3.7)

which is here directly defined in the spatial coordinate system, but is actuallyderived from the referential system, and a PiolaKirchhoff stress tensor

= IC1+ C1 ln J.

We note that

u = xX, F = uX

+ I

b = FFT =u

X

u

X

T

+u

X+

u

X

T

+ I,

so that, in small deformations, b 2(u)+I. Further (two-dimensional case,but three dimensions is similar),

J =

uxX

+ 1uxY

uyX

uyY

+ 1

=ux

X

ux

Y +

ux

X +

ux

Y + 1 ux

Y

ux

X

so that J 1 + u 1 and ln J u, since ln(1 + x) x for x small.Thus, the Neo-Hooke model is also a generalization of small deformationelasticity.


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4. Thermodynamics

In this Chapter we give some fundamental facts from thermodynamics. I par-ticular, we derive the first and second laws of thermodynamics in a continuummechanics setting.

4.1 Energy balance and the first law of thermodynamics

Apart from mechanical work, which is a form of energy, we must also takeinto account heat energy. In fact, straining a body generates heat, and heatleads to strains and, in general, to stresses in a body. Energy transfer mayalso involve other non-mechanical entities such as electricity, but these effectswill not be taken into account here.

The power input Pin resulting from forces acting on a body occupying thevolume V(t) moving wiht velocity v is given by

Pin =

V(t)

t v ds +V(t)

f v dV,

where t is a surface traction and f is a volume force per unit mass. Thetraction is connected to the Cauchy stress via t =

n, so by Gauss theorem

and the symmetry of we can write

Pin =

V(t)

(( + f) v + : (v)) dV

By Newtons law of motion (2.9), we can replace the first two terms by densitytimes acceleration, and thus

Pin =

V(t)

Dv

Dt v + : (v)

dV.

SinceDv

Dt v =

1

2

D

Dt |v

|2,

we have, using Reynolds transport theorem on the form (2.8),


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28 4. Thermodynamics

Pin =1

2

D

Dt

V(t)

|v|2 dV +V(t)

: (v) dV. (4.1)

The first term is the material derivative of the kinetic energy of the volume,and the second is called stress power. Note that the stress power is that partof the external power input which does not contribute to change of kineticenergy.

We must also consider heat input rate Qin into the volume in order toget at a total energy balance. In a moving body, this input consists solely ofconducted heat through the surface V(t) (convective heat flux only occursthrough a fixed control volume). Thus

Qin = V(t)

q n ds +V(t)

rdV,

where q is the heat flux vector, q n is the outward heat flux, and r is aheat source per unit mass inside the body, resulting, e.g., from radiation. Itis assumed here that Qin and Pin are expressed in the same units, so that the

total input of power is the straight sum Pin +Qin (which requires a conversionconstant if Qin is expressed in thermal units).

It is well known from experiments that if a system goes through a cycleback to its initial state, then

Pindt = 0,

Qindt = 0,

where

denotes the sum of contributions during the cycle. This means thatPindt and Qindt are not differentials of any scalar function (in which casethe integrals would be zero). In thermodynamics, a scalar function belongingto the system, such as density (an intensive quantity) or mass (an extensivequantity), is called a state function. The quality of being a state function lies

in the fact that we may talk of the system content of it at any given moment;this is thus not true for the power and heat input. On the other hand, it hasbeen experimentally verified that

(Pin + Qin) dt = 0,

meaning that there is a scalar function such that (Pin + Qin) dt is a differentialthereof. This function is known as the total energy of the system, Etot, andwe have

D

DtEtot = Pin + Qin,

D

DtEtot dt = 0, (4.2)

total energy is conserved in a closed cycle. This is known as the first law ofthermodynamics. (Often this law is stated as the total energy of a closedsystem is constant; in a closed system there is no work or heat input andDDt

Etot = 0.)


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4.2 Entropy and the second law of thermodynamics 29

Usually, the total energy of a system is formally divided into two parts,its kinetic energy EK and its internal energy EI (which may actually also beof kinetic nature locally, as heat motion), Etot = EK + EI, where

EK =1

2V(t)

|v|2 dV.

Let us now define the internal energy density per unit mass, u, such that

EI =

V(t)

udV

The first law of thermodynamics (4.2) then gives

D

Dt

V(t)

2|v|2 + u

dV =

1

2

D

Dt

V(t)

|v|2 dV +V(t)

: (v) dV

V(t)q

nds +

V(t)

rdV.

Using Gauss divergence theorem on the heat flux term together with (2.8),we find that

V(t)

q + Du

Dt : (v) r

dV = 0.

This leads to the energy equation

Du

Dt= : (v) + r q, (4.3)

alternatively, with e the total energy density,

DeDt

= 12

DDt

|v|2 + : (v) + r q, (4.4)

which are the differential equations expressing conservation of energy.

4.2 Entropy and the second law of thermodynamics

The main use of the second law of thermodynamics in continuum mechanicsis for setting limits to constitutive laws. It deals with the inevitable lossof energy to heat dissipation, and it does so by introducing a new statefunction called entropy. Entropy is one of those concepts that is easy inprinciple to understand but very difficult to nail down. Indeed, Wikipedia has

a disambiguation page spanning twelve entries for thermodynamical entropy,six for information theory, and three for computer science (other uses not


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4.2 Entropy and the second law of thermodynamics 31

mobile. In an irreversible cyclic process thermodynamics still postulates thatthere exists a state function s such that

ds = 0,but on the other hand it is invariably found that

q

0.

If we now consider an irreversible process from state a to state b and itscorresponding imaginary reversible process, then

s =

ba

q

rev

,

(the entropy increase equals the entropy input in a reversible process) andwe have that for a full cycle (comprising one reversible and one irreversiblebranch)

0

q

irrev

=

ba

q

irrev

+

ab

q

rev

=

ba

q

irrev

s.

This means that the entropy increase in the irreversible process from a to bis larger that the entropy input by heating,

s ba

q

irrev

.

We shall now translate this inequality into a continuum mechanical setting.Recall that

Qin =

V(t)

qdV =

V(t)

rdV V(t)

q nds,

so the entropy input rate equalsV(t)

r

dV

V(t)

q

nds

and we must have

D

Dt

V(t)

sdV V(t)

r

dV

V(t)

q

n ds

or: the rate of entropy increase is larger than the rate of entropy input. Onlocal form this becomes


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32 4. Thermodynamics

Ds

Dt r

q

, (4.5)

known as the ClausiusDuhem inequality. This inequality is rarely used forcomputing anything in practice in continuum mechanics (except in isentropic

processes); its importance lies in the limits in puts on modeling materiallaws. Being able to show that a material law of your invention obeys (4.5)is generally viewed as an important step. If your material law on the otherhand violates (4.5), you probably had better modify it.


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5. Fluids

This Chapter concerns the modeling of compressible and incompressibleflows.

5.1 Inviscid compressible fluids and conservation laws

In a fluid at rest, shear stresses cannot be sustained. This means that the

only stress can be pressure, so that

= p0I,

where p0 is called the static pressure. The static pressure is assumed, fromthermodynamics, to be functionally related to the density and (absolute)temperature of the fluid, so that there exists an equation f(p0, , ) = 0. Thethermodynamic pressure p is defined as the quantity that fulfills the samerelation also when the fluid is moving, i.e.,

f(p, , ) = 0,

called the equation of state. A classical example of such an equation is the

perfect gas law which states p = R,

where R is called the gas constant. In basic books on thermodynamics thislaw is stated as

pV = nR0 =m

MR0

where V is the volume of gas and n is the amount of gas (in moles), m isthe mass of the gas, and M is its molar mass (the mass of a given amount ofsubstance). We can rewrite this as

pV

m=

R0M

= R,

giving the relation between R and R0.There is also a caloric equation of state which relates the internal energy

to temperature and density,


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34 5. Fluids

u = u(, ).

In an ideal gas, the internal energy is simply proportional to temperature, sothat

u = cv,where cv is the specific heat at constant volume. If energy is added to a gaswhile its volume is kept fixed, then the energy change is related to tempera-ture change by du = cvd. But if the gas expands and instead the pressure inthe gas is held fixed, then some energy is used to expand the specific volumev = 1/, so that

du +pdv = cpd, or d(u +pv) = cpd,

where cp is the specific heat at constant pressure. The quantity

h := u +pv = u +p

is known as the enthalpy of the gas. We note that in an ideal gas

p

= R = cp cv,

so also cp is constant. The ration of specific heats is denoted by ,

:=cpcv

,

and can be computed theoretically for different gases. For air we have = 1.4.For an ideal gas we can then deduce

u = cv =cvR

p

=

cvcp

cv

p

=

p

(

1)

,

and we obtain the most common form of the caloric equation of state:

e =p

1 +1

2|v|2 (5.1)

where e is the total energy density of the gas.Another important property of a gas is its speed of sound c. As we saw

in Part I, speed of waves in general can be found as

c =

stiffness

density.

In a gas, the stiffness related to sound speed is defined as the bulk modulusK when no heat is added or removed from the fluid (an adiabatic process).Then the bulk modulus has been theoretically found to be approximately


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5.1 Inviscid compressible fluids and conservation laws 35

K = p,

so

c =p

.

An ideal frictionless fluid is defined as a fluid that cannot support shearstresses even in motion. No such fluid exists, but it is a convenient approxi-mation for a large class of flows. Then

= pI,

where p is the thermodynamic pressure. These fluids are said to obey theEuler equations, which are characterized by the conservation of mass (2.1),

t+ (v) = 0.

energy (4.4),

De

Dt=

1

2

D

Dt|v|2 + : (v) + r q,

and linear momentum (2.9),

Dv

Dt = f.

We want to write these on conservation form

u

t+i

qi(u)

xi= 0, (5.2)

where the qi are known as the flux vectors. The reason for this will be eluci-dated below. We have for our ideal gas that

: (v) = pI : (v) = p v.

If we further assume no heat flow or heat source and no volume forces, wehave from transport of linear momentum

0 = Dv

Dt = v

t+ v v + p. (5.3)

From the conservation of mass we find

(v)

t

=

t

v + v

t

= v

t (v)v

so that (5.3) can be written


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5.2 Nonlinear conservation laws and shocks 37

also called the quasilinear form.Similarly in three dimensions, we can derive (5.2) with

u =

v1v2v3e

, qi = viu +p

0

1i2i3ivi

,

where ij is the Kronecker delta. Again we must add the caloric equation ofstate (5.1),

p = ( 1)

e 12

|v|2

.

5.2 Nonlinear conservation laws and shocks

To illustrate the fundamental problem of nonlinear conservation laws, weshall consider a scalar problem of the type

u

t+

q(u)

x= 0,

more specifically Burgers equation where q(u) = u2/2. On quasilinear form,this problem is

u

t+ u

u

x= 0,

and we note that this equation says that the material derivative following theparticle paths

x

t

= u(x(t), t)

is zero. This has the consequence of not allowing for a unique function u asa solution since if u varies, the characteristics for high u will overtake thoseof low u giving several values of u for a given x, see Fig. 5.1. This cannot bethe physically correct solution to this problem, and the explanation to whathappens is that we have completely neglected dissipation. Consider insteadthe viscous Burgers equation

u

t+

1

2

x(u2)

2u

x2= 0,

with a very small viscous parameter. Now, however small is there areinstances when the effect of this term cannot be neglected. As u becomessteeper the second derivatives will increase in size and counter the overtakingcharacteristics. We will obtain an almost vertical front, a shock, travelingwith a characteristic speed s.


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38 5. Fluids

Fig. 5.1. Physically impossible solutions to Burgers equation for smooth initialdata.

Using the conservation property of our equation, we have that for a solu-tion that takes on the constant value ul on the left of the shock and ur on

the right of the shock, called a Riemann problem,aa

u

tdx = q(ul) q(ur) = 1

2u2l

1

2u2r =

1

2(ul + ur)(ul ur)

for Burgersequation (assuming a is away from the shock). By definition ofs, we have

u(x, t) =

ul x < stur x > st,

and then aa

u(x, t) dx = (a + st)ul + (a st)ur

so that a

a

ut

dx = s(ul ur)

which gives the shock speed s = (ul + ur)/2. More generally, this sameargument gives the RankineHugoniot jump condition

s =q(ul) q(ur)

ul ur ,

which for systems of conservation laws becones

q(ul) q(ur) = s (ul ur).

We note that if we have a linear system

q = Au,


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5.3 Linearization of the Euler equations and acoustics 39

where A is a constant matrix, then A(ul ur) = s (ul ur), and s is aneigenvalue to A.

Now, conservation laws are very sensitive to manipulations. Consider forexample multiplying Burgers equation by 2u; then we get, if the derivatives

exist,2u

u

t+ 2u2

u

x= 0 =

tu2 +

2

3

xu3 = 0.

For smooth solutions this gives the same solution, but ifu is not differentiablethe implication is not true and we note that the shock speed in the modifiedequation is

smod =2

3

u3r u3lu2r u2l

and it is a simple exercise to show that

smod s = 16

(ul ur)2ul + ur

,

i.e., the manipulation resulted in shock waves moving with the wrong speed.This is the crucial fact concerning conservation laws: they have to be onconservation form, and not only that, the right physical variables have to beconserved.

5.3 Linearization of the Euler equations and acoustics

If we consider an on average still inviscid liquid of constant density 0 subjectto small oscillatory movements of its constituents particles, we can introducethe fluid displacement u = u(x, t) around the average state of rest. Then theincrease in volumetric stress (over the already existing ambient pressure in

the fluid) fulfills

v = K u, or p = K u,

where K is the bulk modulus and p now denotes the deviation from theambient pressure, known as the acoustic pressure. Introducing the soundspeed

c2 =K

0,

this can be writtenp + 0c

2 u = 0. (5.4)The velocity of the particles is given by

v = ut

,


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40 5. Fluids

and the conservation of linear momentum can be written

02u

t2+ p = 0. (5.5)

We can now eliminate the fluid displacement by taking the divergence of (5.5)and two time derivatives of (5.4), leading to

2p

t2 c22p = 0, (5.6)

the acoustic wave equation, or by taking the gradient of (5.4), leading to

2u

t2 c2 ( u) = 0. (5.7)

The formulation (5.6) is the most common for computations in acousticssince it only involves the scalar pressure. It describes the propagation of anacoustic pressure wave in the fluid, moving with the speed of sound c, but its

derivation involves taking an additional spatial derivative of the pressure notmotivated by physical reasoning (to put it another way: the pressure in (5.6)has more regularity than can be motivated from the original equations). Inthe weak form of (5.6) this leads to natural boundary conditions involvingnormal derivatives of the pressure whereas in a stress computation we havethe pressure itself in the natural boundary condition. This leads to difficultiesif we want to couple acoustics in a room with the sound propagation in thewalls of the room (for example modeled as an elastic wave equation), sincethe transmission conditions involve continuity of displacements and of stress,not continuity of derivatives of stress.

The second form (5.7) does not suffer from this increase in regularitysince the gradient of the pressure is already present in (5.5). Equations of thetype (5.7) are called graddiv equations and similar problems are commonin electrodynamical wave propagation problems (where artificially increasingregularity as in (5.6) typically does not work).

5.4 Inviscid incompressible fluids

For compressible fluids, we the density is an unknown and part of the problem.In an incompressible fluid on the other hand, the density does not changefollowing the motion we have v = 0 as derived in (2.4). For these types offlows there can be no shocks since a shock wave requires compression. Anotherassumption often used for incompressible flows is that the equation of staterelating pressure, density, and temperature is independent of temperature.

Then the equation of state takes the simple form = constant, and we candrop the energy equation (unless we are interested in the transport of heat in


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5.4 Inviscid incompressible fluids 41

the fluid). Note that the energy equation cannot be dropped for compressibleflows since we need the energy in order to compute the pressure, as in (5.1).

We still have the relation = pI, and the conservation of momentumcan then be written (and used) on nonconservative form as

v

t+ v v + p = f. (5.8)

Together with the continuity equation v = 0 this gives a system of fourequations for the three components of velocity and for the pressure. Theseare known as the incompressible Euler equations. Note that we do not hereneed an extra equation to compute the pressure and that the pressure in facttakes the role of a Lagrange multiplier enforcing divergence zero.

A further simplification of inviscid flows can be derived under the as-sumption that the flow is irrotational, that is v = 0, where we recallthat

v = vzy

vyz

,vxz

vzx

,vyx

vxy

is a measure of the rotation of the fluid. This means that we can assume thatthe velocity field is the gradient of a potential function

v = (x,y,z),

since the curl of a gradient is always zero. Applying the continuity equationto this velocity, we obtain

2 = 0,and we do not even need to invoke the conservation of momentum in orderto find the velocity field! The boundary conditions for this equation are ofNeumann type if the boundaries do not allow for flow going through, sincethen

0 = n v = n ,and in case we know that part of the domain is irrotational, we can couplethis equation to the full Euler equations via an interface by the couplingcondition

n = n vEwhere vE is the velocity in the rotational part of the flow.

In two space dimensions, we can get rid of the pressure for more generalflows. Introducing the vorticity through

:= v = vyx

vxy

,

and subtracting the y

derivative of the first equation of (5.8) from thexderivative of the second, we obtain (under the assumption that f is con-servative, i.e., a gradient of a potential field)


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5.5 Viscous incompressible fluids 43

5.5 Viscous incompressible fluids

In Part I we gave the Stokes equations of creeping incompressible flow. Weshall now extend these equations to include the effect of the material deriva-

tive.Stokes assumption was more generally that the difference between the

stress in a static fluid and a moving fluid is a function of the symmetricvelocity gradient, so that such a fluid carries stresses on the form

= p I+ F((v)),

where p is the thermodynamic pressure. F is called the viscous stress, and ifF is linear the fluid is called Newtonian. Such a linear relation can in generalbe written, componentwise,

ij = pij +k

l

cijklkl(v),

where cijkl are the components of a fourth order tensor. Now, it turns outthat if the fluid is isotropic and the stress symmetric, it is possible to provemathematically that the general form of this relation must be on the form

= p I+ 2(v) + (tr(v)) I, (5.10)

where and are material parameters. This is known as the NavierPoissonlaw of a Newtonian fluid.

Denoting, as above, the volumetric stress by v = (tr)/3, the deviatoricstress by D = v I, and the deviatoric strain rate by D = (tr)I/3,we can rewrite NavierPoisson as

D = (

v

p)I+ + 2

3

tr (v) I+ 2D(v).Taking the trace of this equation gives us

v p +

+2

3

tr(v) = 0,

so NavierPoisson can be written as

D = 2D(v) and v = p + K v = p K

D

Dt, (5.11)

where K is the bulk modulus, K = + 2/3. In the last step conserva-tion of mass was used. This means that the negative volumetric stress equals

the thermodynamic pressure in two situations: if the flow is incompressible,D/Dt = 0, or if Stokes hypothesis: K = 0 is fulfilled. Since typically only


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44 5. Fluids

one material parameter, the viscosity, is considered, Stokes hypothesis is usu-ally assumed to hold for compressible flows. Thus the NavierPoisson law isusually written

= p I+ 2D(v), (5.12)valid both for compressible and incompressible flows. Introducing this consti-tutive relation into the conservation of momentum equation (2.10), writtenin spatial coordinates, we obtain the incompressible NavierStokes equations

v

t+ v v 2 D(v) + p = f,

v = 0.(5.13)

As in Part I, we can rewrite these equations using the incompressibility con-dition to obtain

v

t+ v v 2v + p = f,

v = 0.(5.14)

Dividing by the constant density, we obtain the variant

v

t+ v v 2v + p = f,

v = 0,(5.15)

where = / is known as the kinematic viscosity and p = p/. As remarkedin Part I, the natural boundary conditions are different for (5.13) and (5.14)in a weak formulation. The formulation (5.13) gives the traction as a naturalboundary condition which is physically correct and must be used in case of afree boundary flow. On the other hand, (5.14) gives a good artificial boundarycondition for pipe flows. If no natural boundary conditions are present, bothformulations give the same weak solution.

Also for the NavierStokes equations, we can eliminate the pressure intwo dimensional flows. Following the derivation of (5.9), we find that we canseek (, ) such that

t+

y,

x

2 = 0,

2 = ,(5.16)

but now the boundary conditions for must be taken into account andrepresent an additional difficulty. The question of course is how the weak

form is derived. For stationary Stokes, where 2

= 0 and 2

= , thefunctional that takes on a stationary value is


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5.6 Turbulence 45

F(, ) =

1

22

dxdy,

Assuming is fixed on this leads to

v dxdy = 0, v H10 ()

( w w)dxdy = 0, w H1(),

and we see that the second equation is in fact the equation for vorticity! Thusno boundary condition for vorticity is usually assumed in the weak form of(5.16).

5.6 Turbulence

Laminar flowis when there is (almost) no mixing between adjacent layersof fluid moving in parallel. In such situations, the viscosity can be measured

and the NavierPoisson law can be evaluated. It is then found to accuratelydescribe Newtonian fluids.

In a turbulent flow the velocity, and hence stress, fluctuates more or lessrandomly in time at any given point. It is then not possible to experimentallyverify the NavierPoisson law, which nevertheless is still assumed to hold.Vortices in the flow can then occur at all scales down to the a limit set by theviscosity, where finally a vortex would dissipate. In order to analyze turbulentflows, it is usually assumed that there exists an average flow v from whichthe actual flow is perturbed, so that v = v+v. Thus NavierStokes, withoutforce terms, takes the form

t (v + v

) + (v + v

) (v + v) = (p +p) + 2(v + v),and we can rewrite the nonlinear term as

(vj + v

j)

xj(vi + v

i) =

xj

(vi + v

i)(vj + v

j) (vi + vi) xj (vj + vj)

where the last term vanishes when summed on j because of the divergencezero condition. Thus we have component wise

t(vi + v

i) +j

xj((vi + v

i)(vj + v

j))

=

xi(p+p)+ 2(vi+vi).

Finally, this equation is averaged. Under the rules of ensemble averaging, wehave


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6. Stability problems

In this Chapter we introduce the general conservation and balance laws thathold in all continuum mechanical theories. They thus form the basis for fluidand solid mechanics in general.

6.1 A simple example

Consider the situation depicted in Figure 6.1, consisting of two rigid beamsconnected by a rotation spring fulfilling the linear material law M = k,where M is the rotating moment, kis a material constant, and is the angleof rotation.

Fig. 6.1. A simple mechanical model.

The equilibrium, geometry, and constitutive law thus gives the system ofequations

M P L

2sin = 0,

= 2,

M = k,

leading to the relation

sin =4k

P L. (6.1)

This is a nonlinear equation since we are taking into account the change inloading on the spring due to deformation.


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6.1 A simple example 49

2k =P L

2sin P L

2 =

4k

P L 1

= 0,

so we have two possibilities: either = 0 or we have a critical load

Pcr = 4kL

in which case is indeterminate (unlike the nonlinear case where we cansolve the nonlinear equation to get = 1). This is the simplest form of aneigenvalue problem. To see what happens in a slightly more complicated sit-uation, consider the system in Figure 6.3, where the geometrical linearizationhas been performed. We have

Fig. 6.3. A problem with two rotational springs.

B = + = 2 , A = ( ) = 2 .from equailibrium: MB = P L, MA = P L, and from the constitutive law:MB = kB, MA = kA. Together, this gives

2 = P Lk

,

2 = P Lk

,

or 2 1

1 2

1 0

0 1

=

0

0

so that := PL/k is an eigenvalue in a generalized eigenvalue problem. Thecorresponding eigenvalues are 1 = 1, 2 = 3. For 1 we find the eigenvectorby solving

1 1

1 1

=

0

0

giving = , a symmetric deformation mode, and for 2 we find


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50 6. Stability problems1 11 1

=

0

0

giving =

, an antisymmetric deformation mode. The lowest load

Pcr = k/L is associated with the symmetric deformation. We realize that asthe number of springs increases, so do the number of possible modes and as-sociated eigenvalues. This is a simplified version of a reduced elasticity model,the beam. The beam is a long thin structure which makes possible somesimplifying assumptions in the elasticity equations.

6.2 The beam

Already in the 1490:s Leonardo da Vinci figured out (see Fig. 6.4) that abeam elongates on one side and compresses on the other if it is bent. Thusthere is an axis, going through the centroid of the cross section of the beam,

that remains unstretched.

Fig. 6.4. Leonardo draws a beam.

If we put the beam in a coordinate system with the xaxis along thecentroid of the cross section, and bending is assumed to take place aroundthe yaxis, then the deflection in the zdirection is w(x) := uz, see Fig. 6.5.We can also see that ux = z, where = dw/dx is the (small) angle ofrotation. We thus have

xx =uxx

= z d2w

dx2,

If we assume that the only stress is xx along the beam, Hookes law gives

xx =

zEd2w

dx2.

Summing up the stresses on the cross section surface, Fig. 6.6, we find that


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6.2 The beam 51

Fig. 6.5. Deformation of a beam.

A

xxz dydz = M,

so that

M = A Ed2w

dx2z2 dydz = E

d2w

dx2 A z2dzdy = EI

d2w

dx2

where I is known as the second moment of area.

Fig. 6.6. Forces at the cross section of a beam.

Inside the beam (assuming no axial forces) acts the moment M(x) andthe shear force T(x), which together balance the external distributed load q,see Fig. 6.7. From equilibrium we have

T(x + x) T(x) = qx, dTdx

= q,

and

M(x + x) M(x) T(x + x)x + qxx/2 = 0, dMdx

= T.


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52 6. Stability problems

Thusd2M

dx2= q,

and

d2

dx2

EId2

wdx2

= q. (6.2)

This is known as the beam equation and can (in special cases) be integratedto give the deflections of the beam. Boundary conditions for the beam includew = 0, a support; dw/dx = 0, no rotation; d2w/dx2, no moment (a frictionfree joint).

Fig. 6.7. Equilibrium in a beam.

We are now interested in the stability problem of the beam. In order tofind this, we have to study equilibrium in a deformed setting, Fig. 6.8.

Fig. 6.8. Equilibrium in a beam including the effect of an axial force.


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6.2 The beam 53

The axial force N acts along the centroid line, and w = w(x+x)w(x).For the equilibrium of bending moments we now get

M(x + x) M(x) T(x + x)x N(x + x)w + qxx/2 = 0,

leading to (N is constant, since N(x + x) N(x) = 0)d2M

dx2= q+ N

d2w

dx2.

ord2

dx2

EI

d2w

dx2

Nd

2w

dx2= q.

Now, if q = 0 and N = P, we have, with constant EI, thatd4w

dx4+

P

EI

d2w

dx2= 0.

This is a continuous eigenvalue problem with := P/EI, which can be solvedusing the characteristic equation

r4 +P

EIr2 = 0 = r1 = r2 = 0, r3 = i

P

EI, r4 = i

P

EI.

and the exact solution is on the form

w(x) = A sin

P

EIx + B cos

P

EIx + Cx + D.

The constants are decided from the boundary conditions. For example, asimply supported beam has boundary conditions w(0) = w(0) = w(L) =w(L) = 0, so

w(0) = 0 B + D = 0,

w(L) = 0 A sin

P L2

EI+ B cos

P L2

EI+ CL + D = 0,

w(0) = 0 PEI

B = 0,

w(L) = 0 A PEI

sin

P L2

EI B P

EIcos

P L2

EI= 0.

which means B = D = 0 and

A sinP L2EI

= 0,

leading to C = 0. Thus, either w = 0 or


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54 6. Stability problems

sin

P L2

EI= 0 =

P L2

EI= n.

The lowest nontrivial root to this equation is n = 1, and thus

Pcr =

2

EIL2 .

This is known as the buckling load for which a catastrophic instability occurs.Again, large deformation analysis may be used to predict the deflection afterbuckling, which is unknown in this linearized problem.

In 1991, a similar phenomenon led to the collapse of the Sleipner A oilplatform, evidently due to the analysts missing to take into account the sta-bility problem; the cost of this little oversight: $700 million.

6.3 Instability and the minimum of potential energy

Returning to our original simple problem in Fig 6.1, we can set up the poten-

tial energy . It consists of the elastic energy stored in the spring, minus thepotential decrease of the external load. From Fig. 6.1 we see that the loadhas moved the distance

= L L2

cos L2

cos = (1 cos )L.Thus the potential energy of the system is

() =1

2k2 P L(1 cos ) = 2k2 P L(1 cos ).

To find the rotation angle, we may use the principle of minimum potentialenergy to find

0 =d

d= 4k P L sin ,

which we already deduced. We may now also see if this is a true minimum ofthe potential energy. Thus we compute

d2

d2= 4k P L cos .

Now, since | cos | 1 we have d2/d2 > 0 as long as 4k/L > P. On theother hand, if P > 4k/L, d2/d2 < 0 at = 0, and the equilibrium isunstable. The situation is depicted in Fig. 6.9. Note that the local minimumchanges to a local maximum as P L/k > 4.

For a continuous problem, the energy is a functional, and we must usevariational derivatives to deduce the sign of the second derivative of potentialenergy. The change of sign of second derivatives of potential energy is echoedin the finite element approximation of large deformation problems in which

the linearized system becomes singular. (The system itself being the firstderivative of potential energy.)


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6.3 Instability and the minimum of potential energy 55

4 3 2 1 0 1 2 3 45

0

5

10

15

20

q

W

PL/k = 1

PL/k = 4

PL/k = 6

Fig. 6.9. Potential energy function ().


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