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noticeable anymore from the detailed conditions near the in¬‚ow or out¬‚ow of the
¬‚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

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