Table of Contents

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) = 2NODES(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