Modeling Super-Tall Buildings Shear Walls

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imurdereri
Posts: 28
Joined: Wed Feb 18, 2015 6:58 pm
Location: Universidad Técnica Federico Santa María

Re: Modeling Super-Tall Buildings Shear Walls

Post by imurdereri »

That works, thanks!. the problem is when i use the command that you said for the layered shell:

"recorder Element -file stressH.txt -time -ele 1 material 1 fiber 1 stress"

The xml file contain this:


<?xml version="1.0" encoding="UTF-8"?>
<OpenSees
xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance"
xsi:noNamespaceSchemaLocation = "http://OpenSees.berkeley.edu/xml-schema ... enSees.xsd">
<TimeOutput>
<ResponseType>time</ResponseType>
</TimeOutput>
<ElementOutput eleType="ShellMITC4" eleTag="1" node1="1" node2="2" node3="7" node4="6">
<GaussPoint number="1" eta="-0.57735" neta="-0.57735">
<SectionOutput secType="LayeredShellFiberSection" secTag="1">
<FiberOutput number="1" zLoc="-0.0562499" thickness="0.0125">
<NdMaterialOutput matType="UnknownMovableObject" matTag="4">
<ResponseType>UnknownStress</ResponseType>
<ResponseType>UnknownStress</ResponseType>
<ResponseType>UnknownStress</ResponseType>
<ResponseType>UnknownStress</ResponseType>
<ResponseType>UnknownStress</ResponseType>
</NdMaterialOutput>
</FiberOutput>
</SectionOutput>
</GaussPoint>
</ElementOutput>
<Data>
0.1 -10.397 -182894 175.511 1.84701e-014 -2.73496e-013
0.2 -20.5169 -365778 346.467 3.9806e-013 -3.39094e-013
0.3 -30.7486 -548587 519.896 -4.24484e-013 -1.56135e-012
0.4 -41.1621 -731317 695.871 -1.17091e-012 -7.1719e-012
.
.
.

so OpenSees understands that i am asking for the $gp=1 and the fiber 1 is the first layer of the material (with thickness 0.0125 with matTag=4). That's perfect, but why the recorder return this?

<ResponseType>UnknownStress</ResponseType>
fmk
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Posts: 5884
Joined: Fri Jun 11, 2004 2:33 pm
Location: UC Berkeley
Contact:

Re: Modeling Super-Tall Buildings Shear Walls

Post by fmk »

beacuse the person who wrote it did not provide the info as needed.

the material is using a 3d material, but sets the 33 strain to 0 and ignores the 33 stress from the material
so the values output are 11, 22, 12, 23, and 31 stresses,
imurdereri
Posts: 28
Joined: Wed Feb 18, 2015 6:58 pm
Location: Universidad Técnica Federico Santa María

Re: Modeling Super-Tall Buildings Shear Walls

Post by imurdereri »

Hi again dear fmk.

i'm having trouble running pushover analysis using ShellMITC4 or ShellNLDKGQ. I was working on comparing these elements with cyclic tests available in the data base "ACI445B Structural Wall DataBase" avaible on NEEShub and works really fine. But when i run pushover analysis the stress-strain relationship makes no sense.

If you can comunicate with me for e-mail (felipe.guerreroa@alumnos.usm.cl) would be great to send you my opensees and matlab scripts for post-procesing the stress-strain response of the wall or send you the plot results directly.

It's weird that the elements works fine for cyclic pattern and not for incremental pushover analysis.
fmk
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Posts: 5884
Joined: Fri Jun 11, 2004 2:33 pm
Location: UC Berkeley
Contact:

Re: Modeling Super-Tall Buildings Shear Walls

Post by fmk »

do the stress-strain relationships look o.k. for cyclic cases?
imurdereri
Posts: 28
Joined: Wed Feb 18, 2015 6:58 pm
Location: Universidad Técnica Federico Santa María

Re: Modeling Super-Tall Buildings Shear Walls

Post by imurdereri »

Yes, they look ok for cyclic cases and not ok for pushover analysis. At the stress-strain relationships for the pushover analysis i obtain a higher strength for the concrete as the one that is defined using the PlaneStressUserMaterial specially for the ShellMITC4.
imurdereri
Posts: 28
Joined: Wed Feb 18, 2015 6:58 pm
Location: Universidad Técnica Federico Santa María

Re: Modeling Super-Tall Buildings Shear Walls

Post by imurdereri »

This is the code for pushover analysis using ShellMITC4 or ShellNLDKGQ with layered shell section. Look at the stress-strain (plot third column = sigma11 ). Something is not working with the element or material (althrough that works fine for cyclic analysis). If you can help me i would be really grateful. I just need to know if is a bug for this element, material or section.

wipe;
model basic -ndm 3 -ndf 6;

#define nodes
node 1 0.0 0 0
node 2 0.2 0 0
node 3 0.5 0 0
node 4 0.8 0 0
node 5 1.0 0 0

node 6 0.0 0.2 0
node 7 0.2 0.2 0
node 8 0.5 0.2 0
node 9 0.8 0.2 0
node 10 1.0 0.2 0

node 11 0.0 0.4 0
node 12 0.2 0.4 0
node 13 0.5 0.4 0
node 14 0.8 0.4 0
node 15 1.0 0.4 0

node 16 0.0 0.6 0
node 17 0.2 0.6 0
node 18 0.5 0.6 0
node 19 0.8 0.6 0
node 20 1.0 0.6 0

node 21 0.0 0.8 0
node 22 0.2 0.8 0
node 23 0.5 0.8 0
node 24 0.8 0.8 0
node 25 1.0 0.8 0

node 26 0.0 1.0 0
node 27 0.2 1.0 0
node 28 0.5 1.0 0
node 29 0.8 1.0 0
node 30 1.0 1.0 0

node 31 0.0 1.2 0
node 32 0.2 1.2 0
node 33 0.5 1.2 0
node 34 0.8 1.2 0
node 35 1.0 1.2 0

node 36 0.0 1.4 0
node 37 0.2 1.4 0
node 38 0.5 1.4 0
node 39 0.8 1.4 0
node 40 1.0 1.4 0

node 41 0.0 1.6 0
node 42 0.2 1.6 0
node 43 0.5 1.6 0
node 44 0.8 1.6 0
node 45 1.0 1.6 0

node 46 0.0 1.8 0
node 47 0.2 1.8 0
node 48 0.5 1.8 0
node 49 0.8 1.8 0
node 50 1.0 1.8 0

node 51 0.0 2.0 0
node 52 0.2 2.0 0
node 53 0.5 2.0 0
node 54 0.8 2.0 0
node 55 1.0 2.0 0

set N 246000
set OutPut SW1-1

nDMaterial PlaneStressUserMaterial 1 40 7 20.7e6 2.07e6 -4.14e6 -0.002 -0.007 0.001 0.07
nDMaterial PlaneStressUserMaterial 2 40 7 20.7e6 2.07e6 0 -0.002 -0.004 0.001 0.07
nDMaterial PlateFromPlaneStress 4 1 1.2e10
nDMaterial PlateFromPlaneStress 5 2 1.2e10

uniaxialMaterial Steel02 7 379e6 202.7e9 0.01 18.5 0.925 0.15
uniaxialMaterial Steel02 8 392e6 200.6e9 0.01 18.5 0.925 0.15
nDMaterial PlateRebar 9 7 90
nDMaterial PlateRebar 10 8 90
nDMaterial PlateRebar 11 8 0

section LayeredShell 1 10 5 0.0125 11 0.0002855 11 0.0004367 9 0.001118 4 0.04816 4 0.04816 9 0.001118 11 0.0004367 11 0.0002855 5 0.0125
section LayeredShell 2 8 5 0.0125 11 0.0002855 10 0.0002799 4 0.049435 4 0.049435 10 0.0002799 11 0.0002855 5 0.0125

#define element
element ShellMITC4 1 1 2 7 6 1
element ShellMITC4 2 2 3 8 7 2
element ShellMITC4 3 3 4 9 8 2
element ShellMITC4 4 4 5 10 9 1

element ShellMITC4 5 6 7 12 11 1
element ShellMITC4 6 7 8 13 12 2
element ShellMITC4 7 8 9 14 13 2
element ShellMITC4 8 9 10 15 14 1

element ShellMITC4 9 11 12 17 16 1
element ShellMITC4 10 12 13 18 17 2
element ShellMITC4 11 13 14 19 18 2
element ShellMITC4 12 14 15 20 19 1

element ShellMITC4 13 16 17 22 21 1
element ShellMITC4 14 17 18 23 22 2
element ShellMITC4 15 18 19 24 23 2
element ShellMITC4 16 19 20 25 24 1

element ShellMITC4 17 21 22 27 26 1
element ShellMITC4 18 22 23 28 27 2
element ShellMITC4 19 23 24 29 28 2
element ShellMITC4 20 24 25 30 29 1

element ShellMITC4 21 26 27 32 31 1
element ShellMITC4 22 27 28 33 32 2
element ShellMITC4 23 28 29 34 33 2
element ShellMITC4 24 29 30 35 34 1

element ShellMITC4 25 31 32 37 36 1
element ShellMITC4 26 32 33 38 37 2
element ShellMITC4 27 33 34 39 38 2
element ShellMITC4 28 34 35 40 39 1

element ShellMITC4 29 36 37 42 41 1
element ShellMITC4 30 37 38 43 42 2
element ShellMITC4 31 38 39 44 43 2
element ShellMITC4 32 39 40 45 44 1

element ShellMITC4 33 41 42 47 46 1
element ShellMITC4 34 42 43 48 47 2
element ShellMITC4 35 43 44 49 48 2
element ShellMITC4 36 44 45 50 49 1

element ShellMITC4 37 46 47 52 51 1
element ShellMITC4 38 47 48 53 52 2
element ShellMITC4 39 48 49 54 53 2
element ShellMITC4 40 49 50 55 54 1


fixY 0.0 1 1 1 1 1 1; # Se aplica empotramiento en los nodos de la base

file mkdir $OutPut;
recorder Node -file $OutPut/VMITC4.txt -time -node 1 2 3 4 5 -dof 1 reaction
recorder Node -file $OutPut/DMITC4.txt -time -node 53 -dof 1 disp

file mkdir Strain
file mkdir Stress

foreach ele {4} {
foreach gp {1 2 3 4} {
recorder Element -file Strain/strainH$ele$gp.txt -time -ele $ele material $gp fiber 5 strains
recorder Element -file Strain/strainA$ele$gp.txt -time -ele $ele material $gp fiber 4 strains
recorder Element -file Stress/stressH$ele$gp.txt -time -ele $ele material $gp fiber 5 stresses
recorder Element -file Stress/stressA$ele$gp.txt -time -ele $ele material $gp fiber 4 stresses
}
}


pattern Plain 1 Linear {
#load 51 0 [expr -$N/4] 0 0 0 0
load 52 0 [expr -$N/2] 0 0 0 0
#load 53 0 [expr -$N] 0 0 0 0
load 54 0 [expr -$N/2] 0 0 0 0
#load 55 0 [expr -$N/4] 0 0 0 0
}


constraints Plain
numberer RCM
system BandGeneral
test NormDispIncr 1.0e-6 200 ;
algorithm BFGS -count 100
integrator LoadControl 0.1;
analysis Static
analyze 10;
puts "gravity analyze ok..."
loadConst -time 0.0;


#PUSHOVER
pattern Plain 2 Linear {
load 51 1000 0 0 0 0 0
}

set DIncr 0.0001;
set DMax 0.02;

# OPCIONES DE ANALISIS PARA CARGA CICLICA
constraints Plain;
numberer RCM;
system BandGeneral;
test NormDispIncr 1.0e-5 100;
algorithm KrylovNewton;
integrator DisplacementControl 51 1 $DIncr;
analysis Static ;

set Nsteps [expr int($DMax/$DIncr)];
set ok [analyze $Nsteps];
dgale
Posts: 57
Joined: Fri Jan 15, 2016 4:01 am
Location: UPM

Re: Modeling Super-Tall Buildings Shear Walls

Post by dgale »

Hello,
I have the same problem.

I define the plate material (element ShellMITC4) like this:
# Materiales para las placas
# Hormigón fc ft fcu epsc0 epscu epstu stc
nDMaterial PlaneStressUserMaterial 3 40 7 43e6 4e6 -30e6 -0.0025 -0.0035 0.1 0.08;
# nDMaterial PlateFromPlaneStress $matTag $PlaneStressMatTag $OutOfPlaneShearModulus
nDMaterial PlateFromPlaneStress 4 3 14.33e9; #
# Armadura
# Armadura en dirección Y
nDMaterial PlateRebar 11 $Acero 90
# Armadura en dirección X
nDMaterial PlateRebar 12 $Acero 0

section LayeredShell 1001 11 4 0.0175 11 0.0025 4 0.0175 11 0.0025 4 0.0175 4 0.0175 4 0.0175 4 0.0175 4 0.0175 11 0.005 4 0.0175

And when I get the results from plates, the stresses are larger than 43e6 defined above.

recorder Element -file StressPlaca/stress11$ele$gp.txt -time -ele $ele material $gp fiber 11 stresses
Results:
0.026 6.51388e+006 -6.57496e+007>43e6 -4.47308e+007 -3.05238e+006 -1.70626e+007

What could be the reasons for that?

Thanks,
samsonama
Posts: 10
Joined: Wed Mar 02, 2016 12:42 am

Re: Modeling Super-Tall Buildings Shear Walls

Post by samsonama »

Hei,
I am trying to use the same model, but I couldn't understand how the section layer is implemented. Especially, the absolute thickness of the reinforcement (rebars). Is it the diameter of the rebar or what? I would appreciate for any suggestion.

section LayeredShell 2 8 4 0.0125 11 0.0002403 10 0.0002356 4 0.0495241 4 0.0495241 10 0.0002356 11 0.0002403 4 0.0125

Thanks.
dgale
Posts: 57
Joined: Fri Jan 15, 2016 4:01 am
Location: UPM

Re: Modeling Super-Tall Buildings Shear Walls

Post by dgale »

Hello samsonama,
you must keep the initial area of the all rebars, so the idea is to make a layer whose thinkness multiply by width is the total area of the rebar layer.

I hope this answer helps to clear up your question.
imurdereri
Posts: 28
Joined: Wed Feb 18, 2015 6:58 pm
Location: Universidad Técnica Federico Santa María

Re: Modeling Super-Tall Buildings Shear Walls

Post by imurdereri »

Dr McKenna:

Have someone look at the error for the ShellMITC4 problem using layered shell element for pushover analysis?
Have someone look at the stress-strain relationship for this case of analysis?

I don't understand why the element works for cyclic analysis and for pushover gives results with no sense (which is more simple)

Please, I' am working with this element and I need to validate the test results with the model

I really appreciate any help

Thanks in advance
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