displacement protocol for pushover

[mokiao]

Hi, there
can anyone help me with this question:

I have the displacement protocol from the experiment data,
how can I make the pushover follow the same random cyclic displacement track?

thank you so much!!!

[mokiao]

I have figured this out, the pushover displacement is the same as the protocol displacement, but the weird thing is at the beginning of the pushover, the force is huge.
does it means the model is unstable or what?

Thank you for helping in advance.

source Wall01and00_beamcolumn.tcl;

test EnergyIncr 1.0e-2 300 0
algorithm KrylovNewton
system UmfPack -lvalueFact 15
numberer RCM
constraints Plain
set fileID [open protocol.out r];
set i 2;
set Dmax(1) 0;
foreach vari [split [read $fileID] \n] {
regexp {([^\s]+)} $vari match sub1
set Dmax($i) $sub1;
set j [expr $i-1];
set Dincr($j) [expr $Dmax($i)-$Dmax($j)];

set dU $Dincr($j)
#puts $dU
integrator DisplacementControl 5 1 $dU 1 $dU $dU
analysis Static
analyze 1
set i [expr $i+1]
}

[linguan118]

The problem seems to stem from your model. Check such as material parameters. Compare the elastic region first.

[mokiao]

linguan118 wrote:
> The problem seems to stem from your model. Check such as material
> parameters. Compare the elastic region first.

Thank you for your answer. Yes, the material I used is highly nonlinear, because I was modeling the wood shear wall, so I used the zerolength element to include the SAWS material in Opensees. I was just modeling a simple wall, I really don't see the problem, is that ok for you to take a look at my model. Thank you so much in advance.

wipe;
model BasicBuilder -ndm 3 -ndf 6;
set rigid 10e6
set nov 10e-7;
set A 1000;
node 1 0. 0. 0.;
node 2 766.66 0. 0.;
node 3 1533.33 0. 0.;
node 4 2300 0. 0.;
node 5 0. 0. 2300.;
node 6 766.66 0. 2300.;
node 7 1533.33 0. 2300.;
node 8 2300 0. 2300.;
node 11 0. 0. 0.;
node 21 766.66 0. 0.;
node 31 1533.33 0. 0.;
node 41 2300 0. 0.;

geomTransf PDelta 1 0 -1 0;
geomTransf Linear 2 0 -1 0;

element elasticBeamColumn 1 1 5 $A $rigid $rigid $rigid $rigid $rigid 2;
element elasticBeamColumn 2 2 6 $A $rigid $rigid $rigid $rigid $rigid 2;
element elasticBeamColumn 3 3 7 $A $rigid $rigid $rigid $rigid $rigid 2;
element elasticBeamColumn 4 4 8 $A $rigid $rigid $rigid $rigid $rigid 2;

element elasticBeamColumn 5 5 6 $A $rigid $rigid $rigid $rigid $rigid 2;
element elasticBeamColumn 6 6 7 $A $rigid $rigid $rigid $rigid $rigid 2;
element elasticBeamColumn 7 7 8 $A $rigid $rigid $rigid $rigid $rigid 2;


fix 11 1 1 1 1 1 1;
fix 21 1 1 1 1 1 1;
fix 31 1 1 1 1 1 1;
fix 41 1 1 1 1 1 1;



set F0 [expr 800.*18/1000*4.448] ;# kN
set FI [expr 40.*18/1000*4.448] ;# kN
set DU [expr 2*25.4] ;# mm
set S0 [expr 800*18/1000*4.448/20] ;# kN/mm
set R1 [expr 0.005]
set R2 [expr -0.4];
set R3 [expr 1.00]
set R4 [expr 0.01]
set alph [expr 0.5];
set bet [expr 1.10];

uniaxialMaterial SAWS 2 $F0 $FI $DU $S0 $R1 $R2 $R3 $R4 $alph $bet;
uniaxialMaterial Elastic 8 $nov;
uniaxialMaterial Elastic 9 $rigid;
# set rigid1 1000;
# uniaxialMaterial Elastic 10 $rigid1;
uniaxialMaterial ENT 10 $rigid;
uniaxialMaterial Parallel 11 10 2;
uniaxialMaterial Elastic 12 1000000;

element zeroLength 10 11 1 -mat 12 9 11 8 8 8 -dir 1 2 3 4 5 6;
element zeroLength 11 21 2 -mat 12 9 11 8 8 8 -dir 1 2 3 4 5 6;
element zeroLength 12 31 3 -mat 12 9 11 8 8 8 -dir 1 2 3 4 5 6;
element zeroLength 13 41 4 -mat 12 9 11 8 8 8 -dir 1 2 3 4 5 6;



recorder Node -file wall01node.dat -time -node 5 -dof 1 disp;

## Apply the nodal Load
pattern Plain 1 Linear { load 5 1.0 0.0 0.0 0.0 0.0 0.0}

##apply gravity load
set vertical [expr 10./4]; #kN/mm
pattern Plain 2 Constant {
load 5 0.0 0.0 -$vertical 0.0 0.0 0.0
load 6 0.0 0.0 -$vertical 0.0 0.0 0.0
load 7 0.0 0.0 -$vertical 0.0 0.0 0.0
load 8 0.0 0.0 -$vertical 0.0 0.0 0.0
}

test EnergyIncr 1.0e-5 500 0
#test NormDispIncr 1e-3 200 0
algorithm KrylovNewton
system UmfPack -lvalueFact 15
#system BandGeneral
numberer RCM
constraints Plain

set IDctrlNode 5;
set IDctrlDOF 1;
set fileID [open protocol01.out r];
set i 2;
set Dmax(1) 0;
foreach vari [split [read $fileID] \n] {
regexp {([^\s]+)} $vari match sub1
set Dmax($i) $sub1;
set j [expr $i-1];
set Dincr($j) [expr $Dmax($i)-$Dmax($j)];

set dU $Dincr($j)
#puts $dU
integrator DisplacementControl 5 1 $dU 1 $dU $dU
analysis Static
analyze 1
set i [expr $i+1]
}

[linguan118]

It seems like a 2D model. Why you modeled it in 3D?
You used rigid for in plane rotation dof for the zero length element, there will be no rotation at the bottom.
The only free DOF is 3, while you use Parallel material of rigid + your SAWS model. So I guess the stiffness is huge.
I don't know if it is right to model a wall just using elastic beam elements and some zerolength elements at the bottom.

[vesna]

The easiest way to define displacement protocol to match the experimental one is to use Path time series (http://opensees.berkeley.edu/wiki/index ... TimeSeries) and assign it to a plain load pattern. Use "sp" command within "load pattern" command along with Penalty constraints.

[juanolarte]

Hi everyone,

I am trying to define the following displacement control:

time {0.0 0.2 0.4 1.0}
values {0.0 12 -12 23} (Displacement)

How can I insert the displacement given into sp command?

For instance:

pattern Plain 4 "Path 100 -time {0.0 0.2 0.4 1.0} -values {0.0 12 -12 23}" {
sp 13 2 ???; # impose displacement ??? at node 13. DOF 2
}

Thanks

Juan Olarte

[fmk]

the ??? would be a 1. .. bye-the-way you are not using displacement control but imposed displacements.