3.4 BOUNDARY CONDITIONS Z88I2.TXT
Mind the following formats:
[Long] = 4 bytes or 8 bytes integer number
[Double] = 8 bytes floating point number, alternatively with or without
point
1st input group, i. e. first line, contains:
Number of the boundary conditions: loads and constraints [Long]
2nd input group, starting in line 2, contains:
Boundary conditions and loads. For every boundary condition and for
every
load respectively one line.
node number with boundary condition: load or constraint [Long]
Respective degree of freedom (1,2,3,4,5,6) [Long]
Header flag: 1 = force [Long] or 2 = displacement [Long]
Value of the load or displacement [Double]
Example: The node 1 shall be fixed respectively at his 3 degrees of freedom: support. Node 3 gets a load of -1,648 N in Y direction (i.e. DOF 2), the degrees of freedom 2 and 3 is supposed to be fixed for the node 5. Resulting in 6 boundary conditions. Thus :
6 |
|
|
|
1 |
1 |
2 |
0 |
1 |
2 |
2 |
0 |
1 |
3 |
2 |
0 |
3 |
2 |
1 |
-1648 |
5 |
2 |
2 |
0 |
5 |
3 |
2 |
0 |
For edge loads and surface loads pay attention to:
It is a good idea to define surface and pressure loads in the file Z88I5.TXT. However, for plates no.18, no.19 and no.20 you may define the surface
load directly in the material information lines of Z88I1.TXT –
which is
much more convenient than via Z88I5.TXT!
Only forces and constraints should
entered here into Z88I2.TXT.
Of course, it is
possible, too,
to convert surface loads into single forces and to write these forces
into
Z88I2.TXT (which is the classical way but somewhat cumbersome).
For the elements with linear shape function, e.g. Hexahedrons No.1 and Torus No.6, edge loads and surface loads are distributed to the elements simply and straightly onto the respective nodes.
However, for elements with higher shape functions, i. e. square (Plane Stress No.3, No.7, Torus No.8, Hexahedron No.10) or cubic (Plane StressNo.11 and Torus No.12) edge and surface loads have to be put onto the elements according to fixed rules which are not always physically obvious. Really funny, some load components can have negative values. Though these facts are not obvious, nethertheless they lead to correct results which is not the case for intuitive distribution of loads to the respective nodes.
An example shall clarify the facts:
A FE structure consists of three plane stress elements No.7 with the load of 1,000 N distributed on the upper edge in Y direction. Above incorrect, below correct load sharing:
Incorrect: 1,000N/7=142.86 N per node. Not correct for elements with square shape function.
Correct: 2 * 1/6 + 2 * (1/6+1/6) + 3 * 2/3 = 18/6 = 3, corresponds to 1,000 N
"1/6 points" = 1,000/18*1 = 55.55
"2/6 points" = 1,000/18*2 = 111.11
"2/3 points" = 1,000/18*4 = 222.22
Control: 2*55.55 + 2*111.11 +3*222.22 = 1,000 N, o.k. Here is why: