×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

spatially varying displacement boundary condition

spatially varying displacement boundary condition

spatially varying displacement boundary condition

(OP)
Hi

I would like to define displacement of nodes as BC or in other words I want to create spatially varying displacement boundary condition.

For this can I use analytical field? The probelm I see with analytical field is that the after I give the coordinates of node (X,Y,Z) I onylhave option to enter one feild value while the the displacement BC for any point (node) has 3 displacement values (U1,U2,U3).

Please can anyone guide to create spatially varying displacement boundary condition.

Thanks
Prab

RE: spatially varying displacement boundary condition

There is an example in the Analytical Dialog box (Expression field): 2.4*X+pow(Y,3)

This translates in to linear variation along X direction and cubic variation in Y direction, where X and Y are axes of a global or local coordinate system. The other option is to use the Mapped field, if you have the point cloud data.

http://www.eng-tips.com/faqs.cfm?fid=376
http://www.eng-tips.com/faqs.cfm?fid=1083

RE: spatially varying displacement boundary condition

(OP)
Thanks. Earlier I might not have explained very well.

I want to use discrete set of displacements for individual nodes, therefore I looked into mapped fields. The problem I see there is that the after I give the coordinates of node (X,Y,Z) I onyl have option to enter one feild value while the displacement BC for any point (node) has 3 displacement values (U1,U2,U3). How to input these values.

RE: spatially varying displacement boundary condition

Use the DISP subroutine, it's the easiest subroutine there is, so also a good intro into using subroutines :).

RE: spatially varying displacement boundary condition

(OP)
Thanks sdebock, I have never used user subroutine in ABAQUS before, am looking on the DISP subroutine example from verification manual.

I don't quite understand, how to use it for my case. I have, an array with original coordinated of the nodes (U1, U2, U3) and another array with corresponding displacements (delta_U1, delta_U2, delta_U3) that I want to apply as BC. I am not sure how to program these displacements into subroutine.

Any suggestions?

RE: spatially varying displacement boundary condition

lets look at the disp subroutine:

Variable to be defined

U(1)

(your delta)

Variables passed in for information
TIME(1)
Current value of step time.
JDOF
Degree of freedom.
COORDS
An array containing the current coordinates of this point.

So, you get the COORDS array from abaqus
you search in your list with coordinates the corresponding delta you want (just for loop over them and use if to get the right one)
use JDOF to check if you need delta1 2 or 3
You multiply your final delta with TIME to get the delta at that timestep.

set U(1) to this value

finished!

RE: spatially varying displacement boundary condition

(OP)
Thank you sdebock, I understand it much better. I think its just the syntax I have to stuggle with now.

This is what I have comeup as:

SUBROUTINE DISP(U,KSTEP,KINC,TIME,NODE,NOEL,JDOF,COORDS)
C
INCLUDE 'ABA_PARAM.INC'
C
DIMENSION U(3),TIME(1),COORDS(3)

C delta_U array stores the node numbers (21 nodes here) and the corresponding displacement required for the BC (X,Y,Z)

delta_U = [43 0.00 0.00 0.07; 46 0.00 0.00 0.06; 48 0.00 0.00 0.04;
49 0.00 0.00 0.06; 50 0.00 0.00 0.07; 52 0.00 0.00 0.04;
53 0.00 0.00 0.06; 54 0.00 0.00 0.04; 55 0.00 0.00 0.05;
57 0.00 0.00 0.04; 58 0.00 0.00 0.05; 59 0.00 0.00 0.07;
61 0.00 0.00 0.04; 62 0.00 0.00 0.06; 63 0.00 0.00 0.05;
68 0.00 0.00 0.05; 69 0.00 0.00 0.07; 72 0.00 0.00 0.06;
78 0.00 0.00 0.07; 85 0.00 0.00 0.07; 89 0.00 0.00 0.08]


FOR i =1:1:21
IF(NODE.EQ.delta_U[i].AND.JDOF.EQ.1) THEN
C THE FUNCTION IS DISPLACEMENT FOR NODE and X direction
U(1)=COORDS-delta_U[i,2]*TIME(1)
IF(NODE.EQ.delta_U[i].AND.JDOF.EQ.2) THEN
C THE FUNCTION IS DISPLACEMENT FOR NODE and Y direction
U(1)=COORDS-delta_U[i,3]*TIME(1)
IF(NODE.EQ.delta_U[i].AND.JDOF.EQ.3) THEN
C THE FUNCTION IS DISPLACEMENT FOR NODE and Z direction
U(1)=COORDS-delta_U[i,4]*TIME(1)
ENDIF
RETURN
END
END

What do you think?

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources