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BINDS
Description
These elements are used to impose a linear relationship between the DOF of some nodes
They must refer to the material law BINDS (ITYPE= 30)
Element type: 99
Implemented by: Bertarini I. (1986) & revised by Pascon F. (1999)
Example of relationship:
$\displaystyle\sum_{i=1}^{m} \alpha_i . \delta_i = \alpha_0$
With:
$\delta_i$: the degree of freedom (translation or rotation: 1, 2 …)
$\alpha_i$ and $\alpha_0$: data specified in the section “COLAW” for the material: LMATE and specified in each relation
$m$: number of terms in the linear relationship
See example below
Files
Prepro: BINDSA.F
Lagamine: BINDSB.F
Input file
| Title (A5) | |
|---|---|
| TITLE | “BINDS” |
| Control data (I5) | |
| NELEM | number of elements |
| Definition of the elements (3I5/3*NNODE I5) | |
| LMATE | material number (type 30) |
| NNODE | number of terms in the relation (max: 10) (see note 1) |
| INDISO | position of $\alpha_0$ in the list of the $\alpha$ coefficients of the law LMATE |
| NODES(I) | node number in the relation |
| IDOF(I) | componant of the node used in the relation (1: first DOF, 2: second DOF …) (see note 2) |
| INDIS(I) | position of the $\alpha_i$ coefficient in the law LMATE |
| Note: These 3 consecutive terms are written for I=1:NNODE on the same line | |
Notes
1: if NNODE >10, the subroutines: BINDSA and BINDSB must be adapted
2: Number of DOF in global axes
1= displacement in X direction 2= displacement in Y direction 3= displacement in Z direction 4= rotation around the axis X 5= rotation around the axis Y 6= rotation around the axis Z
3: The $\alpha_0$ coefficient is multiplied by DMULT
Example
If you have the 2 following relations:
$4.U_1 + 2. V_2 - U_3 = 0$ (element n° 1)
$2.V_1 + 2. U_2 + 4.U_3 = 4$ (element n° 2)
where, for example, $U_3$ is related to the first DOF (U) of the node 3
You have to describe 1 law with all the $\alpha_i$ coefficients:
$\alpha_1$= 4
$\alpha_2$= 2
$\alpha_3$= -1
$\alpha_4$= 0
And you have to define 1 element for each relation (here: 2 elements)
After the title, you have 5 lines:
| Line 1 |
|---|
| NELEM = 2 |
First element:
| Line 2 | ||
|---|---|---|
| LMATE = law nb. | NNODE = 3 | INDISO = 4 |
| Line 3 | ||||||||
|---|---|---|---|---|---|---|---|---|
| NODES(1) = 1 | IDOF(1) = 1 | INDIS(1) = 1 | NODES(2) = 2 | IDOF(2) = 2 | INDIS(2) = 2 | NODES(3) = 3 | IDOF(3) = 1 | INDIS(3) = 3 |
| for $4.U_1$ | for $2.V_2$ | for $-U_3$ | ||||||
Second element:
| Line 4 | ||
|---|---|---|
| LMATE = law nb. | NNODE = 3 | INDISO = 1 |
| Line 5 | ||||||||
|---|---|---|---|---|---|---|---|---|
| NODES(1) = 1 | IDOF(1) = 2 | INDIS(1) = 2 | NODES(2) = 2 | IDOF(2) = 1 | INDIS(2) = 2 | NODES(3) = 3 | IDOF(3) = 1 | INDIS(3) = 1 |
| for $2.V_1$ | for $2.U_2$ | for $4.U_3$ | ||||||
Results
Effort in the spring
