¬‚uid domain: for all relevant values of x, the velocity pro¬le is the same. For

z = 0 and z = h the ¬‚uid adheres to the bottom and the top plate respectively

(˜no slip™ boundary conditions, see previous section). It is simple to verify that the

Solution strategies for solid and fluid mechanics problems

224

z

h vx

x

y

Figure 13.5

Flow between two stationary parallel plates.

solution for (the components of) the velocity ¬eld v( x) and the pressure ¬eld p( x)

according to:

1 ‚p

vx = ’ z( h ’ z) , vy = 0, vz = 0 (13.37)

2· ‚x

‚p ‚p ‚p

= 0, = 0,

constant, (13.38)

‚x ‚y ‚z

satis¬es exactly the continuity equation, the Navier“Stokes equation (with q = 0)

and the prescribed boundary conditions. It can be observed that the inertia term

(the right-hand side of the Navier“Stokes equation) for the de¬ned problem van-

ishes. This is quite obvious as material particles of the ¬‚uid move with constant

velocity (no acceleration) in the x-direction.

The situation described above is called the ˜plane Poiseuille ¬‚ow™. The ¬‚ow is

characterized by a parabolic (in the z-coordinate) pro¬le for the velocity in the

x-direction, coupled to a constant pressure gradient in the x-direction. Note, that

the pressure itself cannot be determined. In order to do that the pressure should

be prescribed for a certain value of x, in combination with the pressure gradient

in the x-direction or in combination with for example the total mass ¬‚ux (per unit

time) in the x-direction.

In Fig. 13.6 the stationary ¬‚ow is depicted of a ¬‚uid between two parallel ¬‚at

plates (distance h) in the case where the bottom plate (z = 0) is spatially ¬xed and

the top plate (z = h) translates in the x-direction with a constant velocity V. There

is no pressure gradient. In this example a fully developed ¬‚ow is the starting point

(plates are assumed to be ˜in¬nitely wide™ and peripheral phenomena at the in¬‚ow

and out¬‚ow are left out of consideration). Also in this case for z = 0 and z = h a

perfect adhesion between the ¬‚uid and plates occurs. Again, it is simple to verify

that for the velocity components the following solution holds:

z

vx = vy = 0, vz = 0.

V, (13.39)

h

225 13.4 Diffusion and filtration

V

z

vx

h

x

y

Figure 13.6

Flow between two mutually translating parallel plates.

Thus, for the velocity in x-direction a linear pro¬le with respect to the z-coordinate

is found. The ¬‚ow in this example is known as the ˜Couette ¬‚ow™.

In this section two very special cases (yet both of practical interest) of ¬‚uid

¬‚ows are treated that permit an analytical solution. For very few practical prob-

lems this is possible. The Finite Element Method as discussed in Chapters 14 to

18 enables us to construct approximate solutions for very complex ¬‚ows, however

the speci¬c algorithms to do that for viscous ¬‚ows are beyond the scope of the

contents of this book.

13.4 Diffusion and filtration

In Section 12.8 the constitutive equations for diffusion and ¬ltration to describe

transport of material (˜¬‚uid™) through a stationary porous medium, have been

discussed. By adding the relevant balance laws (see Chapter 11) the problem

description can be further elaborated. This will be the subject of the present

section.

For diffusion Fick™s law can be applied:

ρv = ’D∇ρ, (13.40)

with D the diffusion coef¬cient.

The mass balance, see Section 11.2, can be written as

δρ

+ ∇ · ( ρv) = 0. (13.41)

δt

Elimination of the velocity v from the two equations above, leads to a linear partial

differential equation for the density ρ = ρ( x, t) according to:

δρ

’ D∇ · ( ∇ρ) = 0. (13.42)

δt

Solution strategies for solid and fluid mechanics problems

226

This differential equation should be satis¬ed for all x in the considered domain V

and for all times t. By specifying the initial conditions (prescribed ρ for all x in

V at t = 0) and with one single boundary condition for each boundary point of

V (either the density ρ or the outward mass ¬‚ux ρv · n = ’D( ∇ρ) ·n should be

prescribed), in principle a solution for ρ( x, t) can be calculated.

To illustrate some of the problems that arise, we con¬ne ourselves to an attempt

to solve a simple one-dimensional problem. A domain is given by 0 ¤ x ¤ L.

Diffusion of a certain material (diffusion coef¬cient D) in the x-direction can take

place. For the density ρ = ρ( x, t) the following partial differential equation holds:

‚ 2ρ

‚ρ

’ D 2 = 0, (13.43)

‚t ‚x

emphasizing that the spatial derivative δρ/δt is written here as the partial deriva-

tive of ρ to the time t (with constant x). Misunderstandings because of this will

not be introduced, because exclusively an Eulerian description will be used. The

differential equation in this example is completed with:

• the initial condition:

ρ = 0 for 0 ¤ x ¤ L and t = 0,

• the boundary conditions:

ρ = ρ0 (with ρ0 a constant) for x = 0 and t > 0

‚ρ

= 0 (no out¬‚ow of material) for x = L and t > 0.

‚x

Even for this very simple situation an exact solution is very dif¬cult to determine.

A numerical approach (for example by means of the Finite Element Method) can

lead to a solution in a simple way. This is the topic of Chapter 14. Here, it can be

stated that the solution for ρ( x, t) at t ’ ∞ has to satisfy ρ( x, t) = ρ0 for all x. A

large number of closed form solutions can be found in [4].

For ¬ltration problems Darcy™s law can be applied:

ρv = ’κ ∇p, (13.44)

with κ the permeability. The mass balance, see Section 11.2, can be written as

follows:

δρ

+ ∇ · ( ρv) = 0. (13.45)

δt

Further elaboration is limited to stationary ¬ltration (time t does not play a role).

In that case

δρ

= 0 and so ∇ · ( ρv) = 0.

δt

227 Exercises

Combination of this equation with Darcy™s constitutive law leads to:

∇ · ∇p = 0. (13.46)

This equation for the pressure p can formally be solved when for every boundary

point of the volume V one single condition is speci¬ed. This can either be formu-

lated in the pressure p, or in the outward mass ¬‚ux ρv · n = ’κ( ∇p) ·n. When the

solution for p is determined it is easy to calculate directly the mass ¬‚ux ρv with

Darcy™s law.

In the one-dimensional case (with x as the only relevant independent variable)

the differential equation for p reduces to

d2 p

= 0. (13.47)

dx2

In this case p will be a linear function of x.

Exercises

13.1 Consider a material element with the shape of a cube (length of the

edges ). The cube is placed in a Cartesian xyz-coordinate system, see

¬gure.

z

y

x

All displacements from the bottom face of the element (coinciding with the

xy-plane) are suppressed. The top face has a prescribed displacement in the

y-direction, which is small with respect to . The side faces are unloaded.

Assume that a homogeneous stress state occurs with σyz = σzy , the only

components of the stress matrix σ unequal to zero.

Why can this assumption not be correct?

13.2 Consider a thin rectangular piece of material (constant thickness h). The

midplane of the material coincides with the xy-plane of a Cartesian xyz-

coordinate system. The material behaviour is described by means of

Hooke™s law (Young™s modulus E and Poisson™s ratio ν). The plate is

Solution strategies for solid and fluid mechanics problems

228

statically loaded with pure shear (plane stress state). The accompanying

boundary conditions are shown in the ¬gure below.

For an arbitrary point of the midplane, with the coordinates ( x0 , y0 ) in

the unloaded reference con¬guration, the displacements in the x- and y-

direction, as a result of the stress „ , are indicated with ux ( x0 , y0 ), uy ( x0 , y0 ),

respectively.

Why does no unique solution exist for the ¬elds ux ( x0 , y0 ), uy ( x0 , y0 ), based

on the information that is given above?

y

σyy = 0

σxy = „

σxx = 0

σyx = „

σxx = 0

σyx = „

x

σyy = 0

σxy = „

13.3 A thin trapezium shaped plate is clamped on the left side and statically

loaded on the right side with an extensional load (local extensional stress

p), as given in the ¬gure below.

y

p

x

2d d

z

In the plate plane stress conditions occur (σxz = σyz = σzz = 0).

With respect to the stress components still relevant for the plane stress

conditions, the following assumptions are made:

x+

σxx = pσyy = 0, σxy = 0.

,

2

Give two reasons why this assumed stress ¬eld cannot be correct.

13.4 Consider a rectangular plate of some material ( — b). The thickness h is

small with respect to and b, see ¬gure.

229 Exercises

y

thickness h

P

x

b

z

The left side ( x = 0) is clamped. The right side (x = ) is loaded with a

distributed tangential load (the resultant force P is known) in the negative

y-direction. The top and bottom surface of the plate are unloaded. A plane

stress condition is supposed. With respect to the stress ¬eld the following

assumption is proposed:

σxx = c1 ( ’ x) y