Continuum Mechanics Fundamentals

James R. Rice, notes for ES 220, 12 November 2009; corrections 9 December 2009

Conserved Quantities

Let a conseved quantity have amount F per unit volume. Examples are as follows:

For conservation of mass, we take F = ρ , where ρ is mass density.

For conservation of linear momentum, F = ρu, where u is the velocity vector (relative tosome Newtonian reference frame).

For conservation of angular momentum, F = ρx × u, where x is the position vector(relative to a fixed point in that frame).

For conservation of energy (first law of thermodynamics), F = ρ (e+ 12|u|2) where e is the

internal energy per unit mass and 12|u|2 is the kinetic energy per unit mass.

There is also the non-conservation law of entropy (second law of thermodynamics) for whichthe same framework here will be used, replacing an “ = ” with a “ ≥ ”, and in which F = ρswhere s is entropy per unit mass.

We assume that at the continuum scales of interest here, the mass density ρ as well as thethe momentum per unit mass ρu can be interpreted as well-defined averages of microscopicmasses and momenta. That is, they are averages over minute spatial regions and timeintervals, which can nevertheless be chosen large enough that there are negligible statisticalfluctuations about those averages at a given x,t in different macroscopic realizations of thesame deformation or flow process. Accordingly, u must then be interpreted as the massaveraged velocity. When studying the motion of a surface S in the continuum which issaid to move with the material, we will interpret that as moving with the local velocity u.Correspondingly, then, in writing the law of conservation of, e.g., linear momentum for aregion of material momentarily occupying volume V , the surface flux rate of momentum T(where T is the stress vector along the bounding surface S of V ), is understood to includecontributions not just from forces acting on S but also from transfers of momentum acrossS associated with microscopic scale motions relative to the mass-averaged motion u.

We consider a single-component continuum. More complicated expressions than given herewould be required to describe multi-component continua in which one component movesmacroscopically relative to the other(s), like for fluids entraining solid particles or dropletsof other fluids, or for deformable porous solids with fluid infiltration, etc.

Representation of Accumulation Rates in Conservation Laws; Reynolds Trans-port Theorem

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We use Eulerian coordinates (fixed in space) so, focusing attention on an arbitrarily selectedvolume V of space, with bounding surface S. Conservation laws have the form:

Rate of accumulation of quantity F within the matter which, at time t, occupies region V

= Rate of generation of that quantity within V and and along its surface S

Because of the Eulerian coordinates, to properly judge the rate of accumulation of F , wemust account for the change within V as well as for the fact that some of the matter withinV at time t will, at an infinitesimally later time t + δt, have exited V across S at locationswhere n · u > 0 (here n is the outer unit normal to S), and that other matter will haveentered V through S at places where n · u < 0. (Had we used Lagrangian coordinates (i.e.,fixed along material points), the terms crossing S would not appear and, rather, we’d saythat the boundary of V moves with velocity u.) Hence, with the Eulerian description,

Accumulation of quantity in V within time δt

=

∫V

[F (x, t+ δt)− F (x, t)]dV +

∫S

F (x, t)(n · uδt) dS

to first order in δt. Dividing by δt and letting δt→ 0, this shows that

Rate of accumulation of quantity in V =

∫V

∂F

∂tdV +

∫S

n · uFdS =

∫V

(∂F

∂t+ ∇·(uF )

)dV

where the Divergence Theorem has been used in the last step. So we simply choose F = ρ,or ρu, or ρx×u, or ρ (e+ 1

2|u|2) to get the left side of the conservation law for the respective

cases of mass, linear momentum, angular momentum, and energy. (The result of the abovecalculation, e.g., when rewritten as

d

dt

∫V (t)

F dV =

∫V

(∂F

∂t+ ∇·(uF )

)dV ,

is sometimes referred to as the Reynolds transport theorem.)

Representation of Generation Rates in Conservation Laws

The form taken by the generation rate will, if non-zero, be of the form

Generation rate =

∫V

H dV +

∫S

hn dS

Thus, in integral form, each conservation law will read∫V

(∂F

∂t+ ∇·(uF )

)dV =

∫V

H dV +

∫S

hn dS .

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In thise we note that H must be of the same physical dimensions, and same tensorial rank,as is F , whereas hn must have physical dimensions with length elevated by a power of onerelative to F and H (a surface density rather than a volume density) but must share thesame tensorial rank as F and H.

Tensors Associated with Surface Fluxes

An important conclusion may now be drawn, from the conservation law itself, about theallowable form of any surface generation term hn: Divide both sides of the last equation bythe total surface area S of the region, and then let S → 0 and V → 0 in such a way that thegreatest linear dimension of V also→ 0 (i.e., V and S shrink onto a point). In that limit thevolume intergrals scale with V and do not contribute in the statement of the conservationlaw because V/S → 0. Thus

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S

∫S

hn dS → 0 as S → 0 .

Now, in this limit, as V and S are shrunk onto a point, the quantity hn can only vary withthe orientation n of the surface element at that point. The Cauchy tetrahedron argumentthen requires that for general n (= n1, n2, n3), the value hn on a surface element of thatorientation has the form

hn = n1h(1) + n2h

(2) + n3h(3) ≡ nkh

(k)

Here h(1), h(2) and h(3) are the values of hn for surface elements with normals n pointing,respectively, in the positive 1, 2, and 3 directions. This result suffices for us to conclude thatthe quantity h defined by

h = e1h(1) + e2h

(2) + e3h(3) ≡ ekh

(k) (with hn = n · h)

is a dyad (hence has components following tensor transformation laws) having a tensorialrank which is higher by unity than the rank of F , H, and hn.

[For example, in the case of conservation of linear momentum, F , H, and hn are of rank1, i.e., are vectors, so that h (the stress tensor) is of rank 2. In the case of conservation ofenergy, F , H, and hn are of rank 0, i.e., they are scalars, so that h (one term of which is theheat flux vector) is of rank 1.]

Local Form of a Conservation Law

Thus in the above integral form of a conservation law, the term∫

ShndS may now be rewritten

as∫

Sn · h dS =

∫V

∇· h dV by the divergence theorem. Therefore, if the conservation lawapplies for any choice of V and if the terms within it are assumed to be locally continuous,then the local (pde) form of the conservation law is [with notation (...), t ≡ ∂(...)/∂t]

F, t + ∇·(uF ) = ∇· h+H

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The conservation laws now follow.

Mass:

In the case of mass, F = ρ and there will be no generation rate, H = hn = 0. Thus

ρ, t + ∇ · (ρu) = 0

This may be rearranged to

ρ, t + u ·∇ρ+ ρ∇ · u = 0 orDρ

Dt+ ρ∇ · u = 0

where D(...)/Dt ≡ (...), t + u ·∇(...) is the short notation for a rate of change following amaterial particle moving with velocity u.

Linear Momentum:

Linear momentum, for which F = ρu, is generated by force, which may be divided, ac-cording to the Cauchy stress hypothesis, into a force ρg per unit volume within V and bya force per unit area acting along S. The sum of that force per unit area on S plus thecontribution already noted from momentum flux per unit area associated with microscopicmotions relative to the mass-averaged velocity u defines the surface stress vector T. ThusH = ρg (here g is body force per unit mass) and hn = T.

Accordingly, the conservation law requires the existence of a second rank tensor, the stresstensor σ (= σijeiej), satisfying

T = n · σ (Tj = niσij)

Evidently, σijej is the stress vector T associated with a surface element having the unitoutward normal n = ei at the point of interest.

Thus, with F = ρu, H = ρg and hn = n · σ, the local form is

(ρu), t + ∇ · (ρuu) = ∇ · σ + ρ g

and simplifying with use of conservation of mass, this becomes

ρ (u, t + u ·∇u) = ∇ · σ + ρ g or ρDu

Dt= ∇ · σ + ρ g

Angular Momentum:

In this case, F = ρ x×u, H = ρ x× g, and hn = x×T = x× (n · σ) so that the associated2nd rank tensor h, associated with the vector hn by hn = n · h, is h = −σ × x. The localexpression is then

(ρ x× u), t + ∇·(u(ρ x× u)) = ∇· (−σ × x) + ρ x× g

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That can be greatly simplified by use of conservation of mass and of linear momentum sothat there results an expression which is given most simply in component form as

σij ei × ej = 0 (implying that σij = σji)

That is, σ is symmetric; σ = σT .

Energy:

Now F = ρ (e+ 12|u|2), and the energy generation terms are H = ρ g · u + r in the volume,

where r is the rate of radiant energy supply per unit volume, and hn = T · u − qn on thesurface, where qn is the rate of heat outflow per unit area across S. As before, hn can berepresented as n·h where in this case, for which hn is a scalar, h will be a vector. In particular,h = σ · u− q where the heat flow vector q, the existence of which is a consequence of theconservation law, satisfies qn = n · q on S.

The resulting local form is

[ρ (e+1

2|u|2)], t + ∇·[ρu (e+

1

2|u|2)] = ∇ · (σ · u) + ρ g · u−∇ · q + r

and simplifying with the use of conservation of mass and linear and angular momentum, theenergy equation is

ρ (e, t + u ·∇e) = σ : D−∇ · q + r or ρDe

Dt= σ : D−∇ · q + r

where

D = sym(∇u) ≡ 1

2[ (∇u) + (∇u)T ]

is the rate of deformation tensor.

It may be observed that σ : D (= σijDij) is the rate of stress working per unit volume, andtherfore that w, the rate of stress working per unit mass, is given by

w ≡ σ : D

ρ

In terms of w, the last form of the energy equation reads De/Dt = w + (r−∇ · q)/ρ, whichis the form expected from elementary considerations when it is recognized that (r−∇ · q)/ρis the rate of heat absorption per unit mass.

For example, in the case that the stress state is just a pure hydrostatic pressure p, i.e. whenσij = −p δij, like for inviscid fluid models, it is seen that σ : D ≡ σijDij = −p Djj =−p ∇ · u = p (Dρ/Dt)/ρ, where the last form given above for conservation of mass hasbeen used. Thus w = p (Dρ/Dt)/ρ2 = −p D(1/ρ)/Dt, which is the expected result in thiscase.

[Aside on elasticity (not required for ES 220): In the case of elastic materials, energye and other thermodynamic functions of state would depend on strain E from a reference

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configuration as well as on another state variable such as temperatuure or entropy s perunit mass. Several appropriate definitions of E can be made. For example, letting ρo andX denote ρ and x in the reference configuration, one commonly adopted definition of E isas the Green strain E, which is based on the change in metric tensor under deformation. Itis defined by requiring that dx · dx = dX · dX + 2 dX · E · dX for arbitrary choices of dX.Here dx denotes the material line element which was described by dX when in the referenceconfiguration. Thinking of x as a function of X and t, and writing dxi = (∂xi/∂Xk)dXk thisthen defines the components of E as

Ekl =1

2

(∂xi

∂Xk

∂xi

∂Xl

− δkl

)=

1

2

(∂ξk∂Xl

+∂ξl∂Xk

+∂ξi∂Xk

∂ξi∂Xk

)where ξ = x −X is the displacement of the considered material particle from its referenceposition. A work-conjugate symmetric stress tensor S, called the second Piola-Kirchhoffstress, may then be defined with the property

w =σ : D

ρ=

S : DE/Dt

ρo

which implies that S = det[F] F−1 · σ · FT−1

Here F (not to be confused with use of F above as a general symbol for density of a con-served quantity) is defined by dx = F·dX, so that it has components Fij = ∂xi/∂Xj, FT

is its transpose (F Tij = Fji), F−1 is its inverse (dX = F−1·dx), and det[F] (= ρo/ρ) is its

determinant.]

Entropy:

Entropy is not a conserved quantity and, rather, can only be constrained to be non-decreasingin time (second law of thermodynamics). In this case the general integral and local forms ofa conservation law above are replaced by the inequalities∫

V

(∂F

∂t+ ∇·(uF )

)dV ≥

∫V

H dV +

∫S

hn dS (with hn = n·h)

and F, t + ∇·(uF ) ≥ ∇· h+H

In these expressions we take F = ρs, H = r/T , and h = −q/T to form what is called theClausius-Duhem inequality, of which the local form is

(ρs), t + ∇·(ρus) ≥ −∇· ( q

T) +

r

T

and which becomes, after simplification with conservation of mass,

ρ (s, t + u ·∇s) ≥ −∇· ( q

T) +

r

Tor ρ

Ds

Dt≥ q ·∇T

T 2− ∇ · q

T+r

T

The last expression can be combined with the energy equation to rewrite the Clausius-Duheminequality as

De

Dt+

q ·∇T

ρT≤ σ : D

ρ+ T

Ds

Dt

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Thermodynamic Constraints on Constitutive Laws, Simple Viscous Fluids

For simple viscous fluids, which we think of as compressible in general, we assume that thestress tensor σ is a function of ρ, T and the instantaneous rate of deformation tensor D (butnot of its history, like would be the case for complex fluids).

When D = 0, σ must reduce to its equilibrium value σeq where σeqij = −p(ρ, T )δij Here

p(ρ, T ) is the equilibrium pressure given by the thermal equation of state.

Then when D 6= 0, we write

σij = τij + σeqij ≡ τij − p(ρ, T )δij

where p(ρ, T ) continues to be given by the thermal equation of state and the tensor τ maybe called the viscous part of the stress tensor σ.

Other thermodynamic properties such as e and s relate to ρ, T and p just like at equilibium,i.e., in a manner constraind by continuing validity of the perfect differential form

−p d(1/ρ) + T ds = de ,

that is, of

−p D(1/ρ)

Dt+ T

Ds

Dt=De

Dt.

We now observe that because Dkk = (D(1/ρ)/Dt) /(1/ρ),

−p D(1/ρ)

Dt= −pDkk

ρ≡ σeq : D

ρ≡ (σ − τ ) : D

ρ

and the equation before this last one then becomes an expression for De/Dt in the form

De

Dt= T

Ds

Dt+

(σ − τ ) : D

ρ

Substituting this last expression into the thermodynamic inequality given just before thestart of this section, and concelling like terms on each side, we get (after multiplying throughby −ρ)

τ : D− q · (∇T )/T ≥ 0 .

This must hold for all possible deformation rates and temperature gradients, and for all ρand T .

Now, by our assumptions about the viscous stress, τ is a function only of ρ, T and D, butnot of ∇T . Thus, the inequality still gives a valid general constraint at points in space-time

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for which ∇T = 0, and such points could in principle be encountered for any values of ρ, Tand D. Hence we must require that

τ : D ≥ 0

for all ρ, T and D. In a related manner, when D = 0 it is clear that the inequality

−q ·∇T ≥ 0

It must hold for all D if we assume that q is a function only of ρ, T and ∇T , but not ofD.

Linear viscous resistance and heat conduction relations: We here make the conven-tional additional assumptions that τ is linear in D, and that q is linear in ∇T , in ways thatrecognize the isotropy of fluid properties. The most general* laws of that type are

τ ′ = 2µD′ and tr(τ )/3 = κ tr(D) (Newtonian viscosity)

[where τ ′ ≡ τ − tr(τ )I/3 and tr(τ ) ≡ τkk; similarly D′ ≡ D − tr(D)I/3 and tr(D) ≡ Dkk;here “tr” stands for “trace”, or sum of diagonal elements] and

q = −K∇T (Fourier law for heat conduction) ,

The factors in those equations are the shear viscosity µ, the bulk viscosity κ, and the thermalconductivity K. In general, all are functions of ρ and T .

The “primed” quantities τ ′, D′, and σ′ are called the deviatoric parts of τ , D and σ,respectively, and we note that tr(τ ′) = 0, tr(D′) = 0 and tr(σ′) = 0. Also, the relationσ = τ − pI shows that the stress tensor satisfies

σ′ ≡ τ ′ = 2µD′ and tr(σ)/3 ≡ tr(τ )/3− p = κ tr(D)− p .

That may be rewritten as an equivalent single expression for σ (rather than separately forits deviatoric part and its trace) as

σ = τ − pI = 2µD + (κ− 2µ/3)tr(D)I− pI ,

and the expressions may be inverted to represent D as

D =1

2µτ −

(1

6µ− 1

9κ

)tr(τ )I =

1

2µσ −

(1

6µ− 1

9κ

)tr(σ)I +

1

3κpI .

In these equations, we recall that p is the function p = p(ρ, T ) characterizing the thermalequation of state, and is not to be equated to −tr(σ)/3 (unless κ = 0 and/or tr(D) = 0).For example, if an “isotropic” stress state σ11 = σ22 = σ33 = −C (the notation C is usedto denote a compressive stress), with off-diagonal σij = 0, is applied to a fluid element, thecorresponding deformation rates will satisfy D11 = D22 = D33, and off-diagonal Dij = 0,where 3κD11 = p(ρ, T ) − C. Recognizing that 3D11 = −(1/ρ)Dρ/Dt in this circumstance,

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it is seen that ρ would evolve according to the first order differential equation κDρ/Dt =ρ [C − p(ρ, T )]. To characterize the solution to that differential equation, we would have tosay something about the thermal conditions on the fluid element. In the simplest case, whenwe assume T is held fixed by appropriate heat extraction from the element, it is easy tosee that for fixed C, ρ will evolve monotonically, at a finite rate, towards the value whichmakes p(ρ, T ) = C, provided that κ > 0 (see next sub-section) and that the isothermalcompressibility is positive (assured by ∂p(ρ, T )/∂ρ > 0).

[* At the start of this sub-section expressions were given for τ ′, tr(τ ) and q (or, equivaletly,for τ and q) which were claimed to provide the most general possible linear relations to D and∇T for an isotropic material. It is easy to understand that the result stated for q satisfiesthat claim, but the relation given between τ and D may merit more discussion. Such isprovided here, making use of the symmetry of τ (which holds because of symmetry of σ andI). That symmetry means that at any given x and t, there exist three mutually orthogonalprincipal directions so that τ is diagonal relative to those axes. Let those directions bechosen as the 1, 2 and 3 directions, so that the τij vanish if i 6= j. Now we make use oflinearity and isotropy. By linearity the instantaneous respose D to τ is the sum of theseparate responses to τ11, τ22 and τ33. By isotropy the response to τ11 alone must be of theform

D11 = A τ11, D22 = D33 = −B τ11, Dij = 0 for i 6= j

where A and B are scalar fluid properties. The response to τ22 alone, and to τ33 alone, may bewritten similarly, changing 1,2,3 to 2,3,1, and to 3,1,2, respectively, keeping the same factorsA and B. Thus, summing these separate linear responses, we conclude that, e.g.,

D11 = A τ11 −B τ22 −B τ33 = (A+B) τ11 −B (τ11 + τ22 + τ33)

with similar expressions for D22 and D33, and with the off-diagonal Dij = 0. Thus, in thisspecial, local principal axes coordinate system,

Dij = (A+B) τij −B δij τkk

holds for all i, j, i.e., for all deformation and stress components. However, Dij, τij and δij aresecond rank tensors, whereas τkk is invariant under coordinate transformation. Thus if wechoose any other orthogonal coordinate directions which are rotated from the principal axes,the same relation between components of D and τ must hold in that new system. RenamingA and B in terms of µ and κ, that justifies the form asserted for relations between D and τtowards the start of this sub-section.]

Requirement of non-negative viscosities and conductivity: It is easy to show thatthe viscous dissipation rate τ : D satisfies

τ : D = τ ′ : D′ + tr(τ ) tr(D)/3 = 2µ (D′ : D′) + κ [tr(D)]2 .

Thus the thermodynamic inequality τ : D− q · (∇T )/T ≥ 0 becomes

2µ (D′ : D′) + κ [tr(D)]2 +K |∇T |2/T ≥ 0 .

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That can be satisfied for all possible D [meaning all possible D′ and tr(D)] and ∇T as mayarise in flow processes only if all of µ, κ and K are non-negative.

Energy Equation

We now recall the energy conservation equation ρDe/Dt = σ : D − ∇ · q + r as wellas the above transformation of the thermodynamic identity −p d(1/ρ) + T ds = de intoρDe/Dt = ρTDs/Dt + (σ − τ ) : D. Eliminating De/Dt from the expressions, and usingq = −K∇T , allows the energy equation to be rewritten as

∇ · (K ∇T ) + τ : D + r = ρ TDs

Dt

In this expression, we recall that s is related to p, ρ and T (themselves related by p = p(ρ, T ))just as for equilibrium states. Thus we may write s = s(T, ρ) or s = s(T, p), according topreference, and write

Ds

Dt=∂s(T, ρ)

∂T

DT

Dt+∂s(T, ρ)

∂ρ

Dρ

Dtor

Ds

Dt=∂s(T, p)

∂T

DT

Dt+∂s(T, p)

∂p

Dp

Dt,

Two of the above partial derivatives of s are given in terms of specific heats per unit mass,cv or cp, at constant ρ or p, respectively, as ∂s(T, ρ)/∂T = cv/T and ∂s(T, p)/∂T = cp/T .Maxwell reciprocal relations associated with recognition that −p d(1/ρ)− s dT and dp/ρ−s dT are perfect differentials enable the other partial derivatives to be computed from thethermal equation of state.

The result may be written as

∇ · (K ∇T ) + τ : D + r = ρcvDT

Dt− β̂ p

ρ

Dρ

Dt,

or as ∇ · (K ∇T ) + τ : D + r = ρcpDT

Dt− α̂ Dp

Dt,

where β̂ =T

p

∂p(ρ, T )

∂Tand α̂ = −T

ρ

∂ρ(p, T )

∂T.

It may be noted that α̂ > 0 when the thermal expansion under fixed pressure is positive, andthen β̂ > 0 too, so long as p > 0 and the isothermal compressibility of the fluid is postive(as we would always assume on grounds of thermodynamic stability). Also, for an ideal gas,β̂ = α̂ = 1.

Simplified energy equation: In simplified analyses of some flow and heat transfer prob-lems, distinctions between cv abd cp are ignored (with notation c used instead), and the

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Dρ/Dt and Dp/Dt terms are ignored (they are often small compared to other terms), sothat the energy equation is written approximately as

∇ · (K ∇T ) + τ : D + r = ρcDT

Dt.

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