Civil Engineering Reference
In-Depth Information
similar to Program 5.4 for general elastic analysis of two- or three-dimensional solids,
and can use any of the two- or three-dimensional elements referred to in this topic (see
Appendix B). The main difference from the programs described previously in this chapter, is
that this program includes no “geometry” subroutine, so all nodal coordinates
g coords
and element node numbers
g num
must be provided as data. In addition, some of the
variables that were previously fixed in the declaration statements, must now be read as data
in order to identify the dimensionality of the problem and the type of element required.
There are no variables required by this program that have not already been encountered in
earlier programs of this chapter.
A three-dimensional seepage example is shown in Figure 7.20. The model represents
one-eighth of a symmetrical cube with a point source of 100 units at its centroid with all
outside faces maintained at a total head of zero. Referring to the figure, node numbers
are indicated in circles and some of the element numbers have also been included. The
example has 125 nodes and 64 elements.
The first line of data identifies the element type
element
which in this case is a
'
hexahedron
', the number of nodes on each element
nod
, the number of elements
nels
, the number of nodes in the mesh
nn
, the number of integrating points
nip
,the
number of dimensions of the problem
ndim
, and the number of property types
np types
.
It may be noted that numerical integration of an 8-node hexahedral element usually requires
8 Gauss points, (2 in each of the three coordinate directions), so
nip
is read as 8.
The problem includes 2 property types, so the next two lines of data provide the property
values for each of the
np types
groups. A 3D problem (
ndim=3
) such as this, requires
3 permeabilities terms (
k
x
,
k
y
and
k
z
) for each property group. In this example, the first
group is applied to elements 1 to 32, which are isotropic with
k
x
=
k
y
=
k
z
=
2, and the
y
fixed potential
f
=0
plane of symmetry
z
101
105
elements 1-32
kx = 2,ky = 2,kz =2
elements 33-64
kx = 1,ky = 1,kz = 5
49
52
fixed
potential
36
33
f
=0
17
20
26
30
x
1
2
4
3
5
1
2
3
4
24
6
5
8
28
64
4.0
11
125
48
12
9
32
4.0
16
13
16
50
21
25
4.0
Q
21
= 100
(source)
planes of symmetry
Figure 7.20 Mesh and data for Program 7.4 example (
Continued on page 343
)
