The OpenSees Community

Hi,

I have analyzed the model using the OpenSeesSP on Ranger (Supercomputer at TACC).

Since the computer has decommissioned, I have the new allocation on the Stampede (the newest computer).

Do you have any plan to make a compiler for the OpenSeesSP on that machine?


All the best,
Kyungtae Kim

Hello everyone,

Is there anyone who is familiar with creating a fiber section for an octagonal concrete cross section?

I would like to use a factor to account for confined concrete (core) in defining the core concrete.

The factor will be used to increase the concrete strength.

I am looking for any reference showing the factor or how to define the increase to make the fiber section.

I will appreciate any help on this.


All the best,
Kyungtae Kim

Thank you, Frank

However, I need to constrain only rotation with respect to the local x axis NOT global X.

In this case, do I need to provide the material for this rotation?

Otherwise, how can I constrain the rotation only in the local axis?

#..........o node 2 ..............^ Y
#........ /.........................|
#......./...........................|
#......o node 1................... -------> X (Global coordinates, Z is a normal direction of X-Y plane)

I have two nodes at the same location using the zerolength element.

In the local coordinates, the direction from node 1 to node 2 is defined as a longitudinal direction (which is local x).

Also, local z axis is a normal direction of x-y plane.

I want to constrain z-translation in local and theta-x in local.

To do it, I think that the only way is ...
Define the uniaxial material in the local degrees of freedom of 1, 2, 4, 5, 6. Then, Use equalDOF 1 2 3

Otherwise, is there any way NOT defining rotational material in 4, 5, 6 dofs?

Hi All,

I am trying to create 4 elements of standard brick elements (stdBrick) with uniform size of 20 m.

However, I got the message of ...
WARNING - Node::createDisp() ran out of memory for array of size -6
FATAL Node::getTrialDisp() -- ran out of memory

Is there anyone who knows this problem??

My script is shown below.




--------------------------------------------------------------------------------
wipe

model BasicBuilder -ndm 3 -ndf -3

set origin {}; # coordinates of node 1 in the 1st element
lappend origin 0. 0. 0.

set meshGrid 1; # size of brick element
set nEleX 2; # no. of elements in X direction
set nEleY 2; # no. of elements in Y direction
set nEleZ 1; # no. of elements in Z direction

# Node
set nodeCount 0
for {set k 1} {$k <= [expr $nEleZ+1]} {incr k} {
for {set j 1} {$j <= [expr $nEleY+1]} {incr j} {
for {set i 1} {$i <= [expr $nEleX+1]} {incr i} {

incr nodeCount
set x [expr [lindex $origin 0] + $meshGrid*($i-1)]
set y [expr [lindex $origin 1] + $meshGrid*($j-1)]
set z [expr [lindex $origin 2] + $meshGrid*($k-1)]

node $nodeCount $x $y $z
puts "node $nodeCount $x $y $z"

}
}
}

# Element
set nEle [expr $nEleX * $nEleY * $nEleZ]
puts "Total number of elements = $nEle "

nDMaterial ElasticIsotropic 1 994202.574 0.425;# 1.5

set eleCount 0
for {set k 1} {$k <= $nEleZ} {incr k} {
for {set j 1} {$j <= $nEleY} {incr j} {
for {set i 1} {$i <= $nEleX} {incr i} {

set node1 [expr $i + ($j-1)*($nEleX+1) + ($k-1)*($nEleX+1)*($nEleY+1)]
set node2 [expr $node1+1]
set node3 [expr $node2 + ($nEleX+1)]
set node4 [expr $node3 - 1]

set node5 [expr $node1 + ($nEleX+1)*($nEleY+1)]
set node6 [expr $node2 + ($nEleX+1)*($nEleY+1)]
set node7 [expr $node3 + ($nEleX+1)*($nEleY+1)]
set node8 [expr $node4 + ($nEleX+1)*($nEleY+1)]

incr eleCount
puts "$eleCount"
element stdBrick $eleCount $node1 $node2 $node3 $node4 $node5 $node6 $node7 $node8 1
puts "$eleCount $node1 $node2 $node3 $node4 $node5 $node6 $node7 $node8 1"


}
}
}

Hi, all

I am working on a bridge modeling.

For a 5% damping ratio, I used 'region' command with parameters following.. for all elements.

# RAYLEIGH damping parameters,
# D=$alphaM*M + $betaKcurr*Kcurrent + $betaKcomm*KlastCommit + $beatKinit*$Kinitial
set xDamp 0.05; # damping ratio
set MpropSwitch 1.0;
set KcurrSwitch 0.0;
set KcommSwitch 1.0;
set KinitSwitch 0.0;
set nEigenI 1; # mode 1
set nEigenJ 3; # mode 3
set lambdaN [eigen [expr $nEigenJ]]; # eigenvalue analysis for nEigenJ modes
set lambdaI [lindex $lambdaN [expr $nEigenI-1]]; # eigenvalue mode i
set lambdaJ [lindex $lambdaN [expr $nEigenJ-1]]; # eigenvalue mode j
set omegaI [expr pow($lambdaI,0.5)];
set omegaJ [expr pow($lambdaJ,0.5)];
set alphaM [expr $MpropSwitch*$xDamp*(2*$omegaI*$omegaJ)/($omegaI+$omegaJ)]; # M-prop. damping; D = alphaM*M
set betaKcurr [expr $KcurrSwitch*2.*$xDamp/($omegaI+$omegaJ)]; # current-K; +beatKcurr*KCurrent
set betaKcomm [expr $KcommSwitch*2.*$xDamp/($omegaI+$omegaJ)]; # last-committed K; +betaKcomm*KlastCommitt
set betaKinit [expr $KinitSwitch*2.*$xDamp/($omegaI+$omegaJ)]; # initial-K; +beatKinit*Kini
rayleigh $alphaM $betaKcurr $betaKinit $betaKcomm; # RAYLEIGH damping


However, I would like to assign a different damping ratio, for example, of 2% to some of zerolength elements with keeping the 5% damping for other elements.

For this control, do I need to change only 'set xDamp 0.05' into 'set xDamp 0.02'?
Then, assign new parameters obtained the above into zerolength elements using 'region' command?

Is it a right way?

Otherwise, I will appreciate any suggestion on this work.

All the best,

Hi, all

I am looking up an eigenvalue at a certain step in dynamic analysis.

For example, I am using an uniform excitation load pattern which has 878 time steps.

I am interested in the eigenvalue exactly at a 478 step.

I write down following..

...

pattern UniformExcitation 2 1 -accel $Gaccel1
pattern UniformExcitation 3 2 -accel $Gaccel2
pattern UniformExcitation 4 3 -accel $Gaccel3

#wipe analysis
constraints Penalty 1.0e+18 1.0e+18
numberer RCM
system UmfPack
test EnergyIncr 1.0e-6 20 0
algorithm KrylovNewton
integrator Newmark 0.5 0.25
analysis VariableTransient

set ok [analyze 477 0.02]
set controlTime [getTime]
puts "$controlTime"

# recorder
eval "recorder Node -file eigenvector1.out -node all -dof 1 2 3 \"eigen 1\" "
eval "recorder Node -file eigenvector2.out -node all -dof 1 2 3 \"eigen 2\" "
eval "recorder Node -file eigenvector3.out -node all -dof 1 2 3 \"eigen 3\" "
eval "recorder Node -file eigenvector4.out -node all -dof 1 2 3 \"eigen 4\" "
eval "recorder Node -file eigenvector5.out -node all -dof 1 2 3 \"eigen 5\" "
eval "recorder Node -file eigenvector6.out -node all -dof 1 2 3 \"eigen 6\" "

set ok [analyze 399 0.02]
set controlTime [getTime]
puts "$controlTime"

---------------------------------------------------------------

Is it a right way?

Otherwise, I will appreciate any suggestion on this.


All the best,

Hi, all

I am modeling a simple bridge with a intermediate hinge.

The hinge skews with a 45 degree with respect to a global coordinate.

To connect node 1 and node 2 at the hinge, I used the zerolength element.

< Case 1 >

uniaxialMaterial ElasticPPGap 302 4.529e+004 [expr 3478.5/3.] 0.0127
uniaxialMaterial ElasticPPGap 304 4.529e+006 [expr -3478.5/3.] 0.
uniaxialMaterial Parallel 307 302 304
uniaxialMaterial Elastic 308 4.529e+010
element zeroLength 1 1 2 -mat 307 308 -dir 1 2 -orient 0.707107 0.707107 0 -0.707107 0.707107 0
equalDOF 12 19 3 4 5 6
...................................................
It's because the vector of local x of zerolength element is defined with a 45 degree.

Moreover, I tried a different way where

< Case 2 >

element zeroLength 1 1 2 -mat 307 -dir 1
equalDOF 1 2 2 3 4 5 6

.............................................
Compared with the case 1, case 2 had the same results.

Here is my question.

If I did not assign the local x coordinate of the zerolength element in Case 2,
what is the vector of the local x of the element?

I think it might be the same as a vector of global X because two nodes locate at the same coordinates and the material direction is defined with 1st degree of freedom.

However, responses of the element are same...That's not my expect..

Could you clarify me?

Hi, all

I am just checking the elastoplastic material as a hook for a expansion joint by using the zero-length element.

A model is very simple following :

wipe
model BasicBuilder -ndm 2 -ndf 3

# Column
set Ac 3.3428e+1
set Ic 2.195e+1
set Ec 20.0e+10

# Shear keys
uniaxialMaterial ElasticPP 13 1.08e+6 3.4259e-3

node 1 0. 0.
node 2 0. 10.
node 3 0. 10.

fix 1 1 1 1

set transfColumn 2
geomTransf Linear $transfColumn

element elasticBeamColumn 1 1 2 $Ac $Ec $Ic $transfColumn
element zeroLength 2 2 3 -mat 13 -dir 1
equalDOF 2 3 2 3

pattern Plain 100 Linear {
load 3 100 0. 0.
}

recorder Node -file disp.txt -node 1 2 3 -dof 1 2 3 disp
recorder Element -file force.txt -ele 2 force

constraints Penalty 1.0e+18 1.0e+18
numberer Plain
system BandGeneral
integrator DisplacementControl 3 1 0.0001
#integrator LoadControl 0000.1

test EnergyIncr 1.e-6 10 0
algorithm Newton
analysis Static

set ok [analyze 1000]

------------------------------------------
The output just comes out only with a elastic range.
To look at a plastic part, how can I do?

I will appreciate any comments on that.

Thank you for your help.

I solved this problem.

But, I have a question on your second suggestion.

After fixing the dof, how can I remove or change the fixity?

Unit : meter
dt : 0.02

numbers are quite small...


from 0.02 sec.

0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000001
0.000001
0.000001
0.000001
0.000000
0.000000
0.000000
-0.000001
-0.000002
-0.000003
-0.000003
-0.000004
-0.000005
-0.000006
-0.000006
-0.000007
-0.000007
-0.000007
-0.000007
-0.000008
-0.000008
-0.000008
-0.000008
-0.000008
-0.000008
-0.000009
-0.000009
-0.000009
-0.000009
-0.000010
-0.000010
-0.000011

....

I am sorry to bother you again.

Would you check it out?

Since I applied an input to the base in the X direction.

I released the corresponding dof and used constraints of Penalty.

I did not scale the input.

If I changed the constraints into Transformation, I got almost zero displacements...

Why..?

#---------------------------------------------------------------------------------------

wipe
model BasicBuilder -ndm 3 -ndf 6

node 1 0. 0. 0. -mass 0. 0. 0. 0. 0. 0.
node 2 0. 0. 10. -mass 180.617 180.617 180.617 0. 0. 0.
node 3 0. 0. 20. -mass 180.617 180.617 180.617 0. 0. 0.


fix 1 0 1 1 1 1 1

geomTransf Linear 1 0. 1. 0.
element elasticBeamColumn 1 1 2 3.669 33094834.944 13789514.56 2.305 1.778 0.872 1
element elasticBeamColumn 2 2 3 3.669 33094834.944 13789514.56 2.305 1.778 0.872 1


set nEigenI 1
set nEigenJ 1
set lambdaN [eigen [expr $nEigenJ]]
set lambdaI [lindex $lambdaN [expr $nEigenI-1]]
set lambdaJ [lindex $lambdaN [expr $nEigenJ-1]]
set omegaI [expr pow($lambdaI, 0.5)]
set omegaJ [expr pow($lambdaJ,0.5)]
set lambda1 [lindex $lambdaN 0]
set omega1 [expr pow($lambda1,0.5)]
set PI 3.141592654
set freq1 [expr $omega1/2/$PI]
puts "Natural frequency = $freq1"
set xDamp 0.05
set MpropSwitch 1.0
set KcurrSwitch 0.0
set KcommSwitch 0.0
set KinitSwitch 1.0
set alphaM [expr $MpropSwitch*$xDamp*(2*$omegaI*$omegaJ)/($omegaI+$omegaJ)]
set betaKcurr [expr $KcurrSwitch*2.*$xDamp/($omegaI+$omegaJ)]
set betaKcomm [expr $KcommSwitch*2.*$xDamp/($omegaI+$omegaJ)]
set betaKinit [expr $KinitSwitch*2.*$xDamp/($omegaI+$omegaJ)]
rayleigh $alphaM $betaKcurr $betaKinit $betaKcomm
puts "rayleigh $alphaM $betaKcurr $betaKinit $betaKcomm"

# recorder
recorder Node -file topNode_multi.acc -time -node 3 -dof 1 2 3 accel
recorder Node -file topNode_multi.txt -time -node 3 -dof 1 2 3 disp

# Multiple Support Excitation
set GMfact [expr 1.]
pattern MultipleSupport 1 {
set DispSeries "Series -dt 0.02 -filePath xdisp_center.out -factor $GMfact"
groundMotion 1 Plain -disp $DispSeries
imposedMotion 1 1 1
}

#constraints Penalty 1.0e+18 1.0e+18
constraints Transformation
numberer RCM
system BandGeneral
test NormDispIncr 1.0e-3 20 1
algorithm ModifiedNewton
integrator Newmark 0.5 0.25
analysis VariableTransient

analyze 877 0.02 0.00015 0.02 20

Thank you.

When I released the direction and used constraints of penalty.

Displacements of the base node look good.

However, in terms of acceleration, responses of a top node became quite different. (even looked like wrong).

How can I impose the full displacements?

Just use a factor of 2?

Otherwise, do I need to change the constraints or release the dof?

Exactly, I fixed that direction of dofs at the base node and used penalty constraints.

If I released the direction at the node, the base displacement became the input but responses of the top node were different.

Is that why??