1.15 Fluids

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Solids and fluids are two major classes of materials. Fluids include both liquids and gases.

There are several ways in which a fluid can be described. Assume that there is no gravity.

Then, the stress in a liquid at rest is zero. If you stir the liquid and wait until there is

no motion the stress will again be zero. If you take a container filled with gas at a given

pressure, stir the gas without increasing the external pressure and wait till there is no motion

the stress reduces to the hydrostatic pressure before stirring. Consequently, the deformation

of liquid materials does not induce stress as long as the liquid is at rest and the density is

unchanged. Accordingly, the deformation gradient for quasistatic deformations leaving the

density unchanged reduces to the shift operator (Eringen 1980):

xk

,K

= gk

K. (1.511)

In a similar way, one arrives at the following simplifications:

CKL = xk

,Kxl

,Lgkl = GKL (1.512)

˙C

KL = 2dklxk

,Kxl

,L

= 2dklgk

Kgl

L (1.513)

θ,K = θ,kgk

K. (1.514)

Just as for elastic materials, we start from the material formulation of mechanical grade 1

and thermal grade 1, but now we keep the first time derivatives of the mechanical quantities

as well:

S(X, t) = S(C, ˙C , ρ

−1, ρ˙, θ,∇0θ,X). (1.515)

Now, Equation (1.269) yields

σ kl = J

−1xk

,KSKLxl

,L

= ρ

ρ0

SKLgk

Kgl

L. (1.516)

60 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

Since (see Equation (1.178))

ρ˙ =

˙

_ρ0

J

  = −ρ0

J 2

˙ J = −ρdk

k (1.517)

and

∂

∂XK

= ∂

∂xk

gk

K (1.518)

the Cauchy stress takes the form

σ(X, t) = σ(d, ρ

−1, θ,∇θ,X). (1.519)

Since any configuration leaving the density unchanged is undeformed, X can be replaced

by x:

σ(x, t) = σ(d, ρ

−1, θ,∇θ, x). (1.520)

The principle of objectivity requires that Equation (1.520) does not change its form after

applying an arbitrary time-dependent translation, for example, x(t):

σ(d, ρ

−1, θ,∇θ, x) = σ (d, ρ

−1, θ,∇θ, 0) (1.521)

and the explicit dependence on x drops out:

σ = σ(d, ρ

−1, θ,∇θ) (1.522)

and similar expressions for q, ε and η. Just as in the derivation of the constitutive laws

for elastic materials the entropy inequality plays a major role. The spatial formulation of

Equation (1.389) reads

−ρ

θ

( ˙ψ + ˙ θη) + 1

θ

d : σ − 1

θ2 q · ∇θ ≥ 0 (1.523)

where ψ = ε − θη, Equation (1.387), and

ψ = ψ(d, ρ

−1, θ,∇θ) (1.524)

because of similar dependencies of ε and η. The time derivative of ψ reads

˙ψ

= ∂ψ

∂d

: ˙d + ∂ψ

∂ρ−1

·

˙

ρ−1 + ∂ψ

∂θ

˙ θ + ∂ψ

∂∇θ

· ∇˙θ. (1.525)

Substituting Equation (1.525) into Equation (1.523) yields

− ρ

θ

∂ψ

∂d

: ˙d − ρ

θ

∂ψ

∂ρ−1

˙

ρ−1 − ρ

θ

_∂ψ

∂θ

+ η

_ ˙ θ

− ρ

θ

∂ψ

∂∇θ

· ∇˙θ + 1

θ

d : σ − 1

θ2 q · ∇θ ≥ 0. (1.526)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 61

Since (see Equation (1.517))

˙

ρ−1 = 1

ρ

d : g (1.527)

this is equivalent to

− ρ

θ

∂ψ

∂d

: ˙d + 1

θ

d : _

σ − ∂ψ

∂ρ−1 g

_ − ρ

θ

_∂ψ

∂θ

+ η

_ ˙ θ

− ρ

θ

∂ψ

∂∇θ

· ∇˙θ − 1

θ2 q · ∇θ ≥ 0. (1.528)

Since this equation is linear in the time derivatives, it can only be satisfied if the corresponding

coefficients reduce to zero:

∂ψ

∂d

= 0 (1.529)

η = −∂ψ

∂θ

(1.530)

∂ψ

∂∇θ

= 0. (1.531)

Hence,

ψ = ψ(ρ

−1, θ) (1.532)

and Equation (1.528) reduces to

1

θ

d : _

σ − ∂ψ

∂ρ−1 g

_ − 1

θ2 q · ∇θ ≥ 0. (1.533)

Defining the pressure p by

p = − ∂ψ

∂ρ−1 (1.534)

and the dissipative stress by

t := σ + pg (1.535)

we finally arrive at the following equations:

p = − ∂ψ

∂ρ−1 (ρ

−1, θ) (1.536)

η = −∂ψ

∂θ

(ρ

−1, θ) (1.537)

t = t(d, ρ

−1, θ,∇θ) (1.538)

q = q(d, ρ

−1, θ,∇θ) (1.539)

ε = ψ(ρ

−1, θ) − θ

∂ψ

∂θ

(ρ

−1, θ) (1.540)

σ = −pg + t (1.541)

62 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

subject to

1

θ

d : t − 1

θ2 q · ∇θ ≥ 0. (1.542)

Equation (1.542) implies that t and q must be at least linear in d and ∇θ respectively.

Equation (1.538) and Equation (1.539) can be replaced by

t = tL(d, ρ

−1, θ,∇θ) : d (1.543)

q = −κL(d, ρ

−1, θ,∇θ) · ∇θ (1.544)

where tL is a fourth-order tensor, κL is a second-order tensor. Notice that the dissipative

stress cannot be derived from a potential function, only the hydrostatic part p can. This is

a major difference compared to elastic materials. Equation (1.542) is the fluid equivalent

of Equation (1.396) for elastic materials.

Because of the principle of objectivity, Equation (1.543) can be further reduced to

t = α0g_ + α1d_ + α2(d2)_ (1.545)

where

αK(ρ

−1, θ,∇θ, I1d, I2d, I3d ) (1.546)

Linearization yields

α0 = λv(ρ

−1, θ,∇θ)I1d (1.547)

α1 = 2μv(ρ

−1, θ,∇θ) (1.548)

α2 = 0 (1.549)

and one arrives at the well-known stress expressions for linear Stokesian fluids:

σ = (−p + λvg : d)g + 2μvd. (1.550)

For details, the reader is referred to (Eringen 1980).

The energy equation, Equation (1.355), reads in spatial coordinates:

ρ˙ε = d : σ −∇ · q + ρh. (1.551)

Substitution of Equation (1.540) yields

ρ

_−_

p + θ

∂2ψ

∂ρ−1∂θ

_ ˙

ρ−1 − θ

∂2ψ

∂θ2

˙ θ

_ = d : σ −∇ · q + ρh, (1.552)

which reads by the use of Equation (1.527) and Equation (1.534):

ρθ

∂2ψ

∂θ2

˙ θ + θ

∂2ψ

∂ρ−1∂θ

d : g + d : t −∇ · q + ρh = 0. (1.553)

For most gases, t = 0 is assumed (no stress dissipation) and Equation (1.553) reduces to

ρθ

∂2ψ

∂θ2

˙ θ + θ

∂2ψ

∂ρ−1∂θ

d : g −∇ · q + ρh = 0. (1.554)

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