Simultaneous displacement control

[ccalong]

Hello~

I wana ask about 'Displacement Control'.

I have to use 'Displacement Control' to analyze a brige model which has three columns and four decks.

Now, I like to control the displacement of three nodes of the model, simultaneously.

The problem is that the displacement of each node must be increased respectively.

For instance, node 3 must be increased by 0.0001, node 5 must be increased by 0.0002, and node 7 must be increase by 0.0001.

If I use three 'displacement control integrator' as the following code, I cannot get the exact displacement at each node.

In the case of the following code, the displacement of node 5 is not 0.0002 but 0.00021129.

Is it possible to use 'Displacement Control' for control of each displacement?

I'm looking forward your mentions.
Thanks.

Code: Select all

model BasicBuilder -ndm 3 -ndf 6
set h1 14.0;       # height of column1 and 3
set h2 21.0;       # height of column2
set by 0.0;       # Bay width in Y-direction
set bx 50.0;      # Bay width in X-direction
# Create nodes
#    tag             X              Y           Z 
node  1 0 0	[expr  $h2]
node  2 [expr  $bx] 0 [expr  $h2-$h1]
node  3 [expr  $bx] 0 [expr  $h2] 
node  4 [expr  $bx*2] 0 0 
node  5 [expr  $bx*2] 0 [expr  $h2] 
node  6 [expr  $bx*3] 0 [expr  $h2-$h1] 
node  7 [expr  $bx*3] 0 [expr  $h2] 
node  8 [expr  $bx*4] 0 [expr  $h2]

....

pattern Plain 1 Linear {
   load  3 0 1 0 0 0 0
   }
pattern Plain 2 Linear {
   load  5 0 1 0 0 0 0
   }
pattern Plain 3 Linear {
   load  7 0 1 0 0 0 0
   }
   
   
constraints Transformation
numberer RCM
test NormDispIncr 1.0e-12 10 3
system BandGeneral
algorithm Newton
integrator DisplacementControl 3 2 0.0001
integrator DisplacementControl 5 2 0.0002
integrator DisplacementControl 7 2 0.0001
analysis Static
analyze 1

[fmk]

NO .. suggest you have a look at some papers or books on the theory behind it to see why not .. you might want imposed displacements instead; here you
specify the displacements at the nodes exactly, e.g. used in multi-support exciatations .. but again, you have to understand the theory before you use it to understand what results you are going to get.

[Boris]

Hello There,


Put three very stiff springs at those nodes, in directions you want to displace, then apply loads large enough to compress springs by the amount of displacements you want, the structure will follow (aka penalty method..., carefull with large spring stiffnesses, make it about million times stiffer that the stiffest axial member you have...).

Boris