Civil Engineering Reference
In-Depth Information
L
E=10
6
kN/m
2
υ
= 0.3
rigid footing displaced
vertically by -1 x 10
-5
m
1
4
7
10
10 m
2
5
8
11
3
6
9
12
30 m
type_2d
'plane'
element nod dir
'quadrilateral' 4 'y'
nxe nye nip np_types
3 2 4 1
prop(e,v)
1.0e6 0.3
etype(not needed)
x_coords, y_coords
0.0 10.0 20.0 30.0
0.0 -5.0 -10.0
nr,(k,nf(:,k),i=1,nr)
8
1 0 1 2 0 1 3 0 0 6 0 0 9 0 0 10 0 1 11 0 1 12 0 0
loaded_nodes
0
fixed_freedoms,(node(i),sense(i),value(i),i=1,fixed_freedoms)
2
1 2 -1.0e-5 4 2 -1.0e-5
Figure 5.11 Mesh and data for third Program 5.1 example
use of “reduced” integration, by putting
nip
equal to 4, improves the performance of this
element. This is found to be particularly true of the plasticity applications described in
Chapter 6.
The simple mesh in Figure 5.15 is to be analysed and the consistent nodal loads
(Appendix A) necessary to reproduce a uniform stress field of 1 kN/m
2
should be noted
in the data. The computed results given in Figure 5.16, indicate a vertical displacement at
node 1 of
10
−
5
m and a vertical centroid stress in the element under the load
−
0
.
5311
×
0
.
9003 kN/m
2
.
of
σ
y
=−
