Dear colleagues,
Iam trying to simulate RC shear wall behavior under cyclic loading using force based element, and section aggregator to include shear behavior. first, without including bond-slip action, results were reasonable but with less pinching compared to experimental results. When i used Bond Sp01 material and zero length section element, results get highly underestimated and odd in pinching behavior. I would be grateful if you could help me. This is my script:
model basic -ndm 2 -ndf 3
#units N,mm
# Nodes Formation and Restrains
node 100 0 0
node 1 0 0
node 2 0 3660
fix 100 1 1 1
puts "Nodes Formation and Restrains Done"
#Material Definition
# Unconfined Concrete
set fc -42.8; set ec -0.0021; set Ec 31030;
set fcc2 -47.6; set ecc2 -0.0033; set xnc2 2; set rcc2 1.87;
set ft 2.028; set et 0.00008; set xp 100;
set xn 2; set rc 7;
uniaxialMaterial Concrete07 11 $fc $ec $Ec $ft $et $xp $xn $rc
uniaxialMaterial Concrete07 12 $fcc2 $ecc2 $Ec $ft $et $xp $xnc2 $rcc2
uniaxialMaterial MinMax 1 11 -min -0.005
uniaxialMaterial MinMax 2 12 -min -0.02
# Confined Concrete
#Steel
set Ev 200000; set R0 15
set cR1 0.925; set cR2 0.15
#Steel 2
set Fyp2 448; set Fyn2 448; set bp2 0.02; set bn2 0.02
#Steel 3
set Fyp3 434; set Fyn3 434;
set bp3 0.02; set bn3 0.02
uniaxialMaterial SteelMPF 13 $Fyp3 $Fyn3 $Ev $bp3 $bn3 $R0 $cR1 $cR2 0.01 7 0.01 7
uniaxialMaterial MinMax 3 13 -max 0.12
uniaxialMaterial SteelMPF 14 $Fyp2 $Fyn2 $Ev $bp2 $bn2 $R0 $cR1 $cR2 0.01 7 0.01 7
uniaxialMaterial MinMax 4 14 -max 0.1
#bar slip material
uniaxialMaterial Bond_SP01 300 448 0.36 670 11 0.5 0.5
uniaxialMaterial Bond_SP01 400 434 0.36 650 11 0.5 0.5
puts "Material Definition Done"
#Section Definition
geomTransf PDelta 1
section Fiber 1 {
patch quad 1 1 10 -610 -51 -596 -51 -596 51 -610 51
patch quad 1 119 1 -596 -51 596 -51 596 -37 -596 -37
patch quad 1 119 1 -596 37 596 37 596 51 -596 51
patch quad 2 16 8 -596 -37 -433 -37 -433 37 -596 37
patch quad 1 87 8 -433 -37 433 -37 433 37 -433 37
patch quad 2 16 8 433 -37 596 -37 596 37 433 37
patch quad 1 1 10 596 -51 610 -51 610 51 596 51
layer straight 3 4 71.33 -591 -32 -438 -32
layer straight 3 4 71.33 -591 32 -438 32
layer straight 3 4 71.33 438 -32 591 -32
layer straight 3 4 71.33 438 32 591 32
layer straight 4 4 31.67 -285 -32 -285 -32
layer straight 4 4 31.67 -285 32 285 32
}
#bar slip section
section Fiber 10 {
patch quad 1 1 10 -610 -51 -596 -51 -596 51 -610 51
patch quad 1 119 1 -596 -51 596 -51 596 -37 -596 -37
patch quad 1 119 1 -596 37 596 37 596 51 -596 51
patch quad 2 16 8 -596 -37 -433 -37 -433 37 -596 37
patch quad 1 87 8 -433 -37 433 -37 433 37 -433 37
patch quad 2 16 8 433 -37 596 -37 596 37 433 37
patch quad 1 1 10 596 -51 610 -51 610 51 596 51
layer straight 300 4 71.33 -591 -32 -438 -32
layer straight 300 4 71.33 -591 32 -438 32
layer straight 300 4 71.33 438 -32 591 -32
layer straight 300 4 71.33 438 32 591 32
layer straight 400 4 31.67 -285 -32 -285 -32
layer straight 400 4 31.67 -285 32 285 32
}
#Shear Model
uniaxialMaterial SelfCentering 5 1.3e09 5.1e07 112000 1
section Aggregator 2 5 Vy -section 1
#shear bar slip
section Aggregator 200 5 Vy -section 10
# Gaussian Points and Records
set N 7
set secTag "2 2 2 2 2 2 2"
set locations "0 0.44 0.44 0.54 0.72 0.9 1"
set weights "0.22 0.22 0.028 0.152 0.2 0.152 0.028"
set integration1 "UserDefined $N $secTag $locations $weights"
element forceBeamColumn 1 1 2 1 $integration1
element zeroLengthSection 2 100 1 200
# Apply Axial Load
set tolAx 1.0e-7
set iterAx 10
pattern Plain 1 Linear { load 2 0.0 -400000 0.0 }
integrator LoadControl 0.1;
system SparseGeneral -piv; test NormUnbalance $tolAx $iterAx 1;
numberer RCM; constraints Transformation;
algorithm KrylovNewton; analysis Static;
puts "Gravity Load Application Done"
# perform the gravity load analysis,
analyze 10
# Set the gravity loads to be constant & reset the time in the domain
loadConst -time 0.0
puts "Gravity load analysis completed"
# Set Lateral Loads
set tolLat 1.0e-12; set iterLat 100
recorder Node -file TopDisp2.out -time -node 2 -dof 1 disp
pattern Plain 2 Linear { load 2 1000.0 0.0 0.0 }
test NormDispIncr $tolLat $iterLat 1
puts "Lateral Load Application Done"
#Analyze lateral Loads
#first
#U=+2.3mm
integrator DisplacementControl 2 1 0.1
analyze 23
integrator DisplacementControl 2 1 -0.1
analyze 23
#U=-2.6mm
integrator DisplacementControl 2 1 -0.1
analyze 26
integrator DisplacementControl 2 1 0.1
analyze 26
#U=2.4mm
integrator DisplacementControl 2 1 0.1
analyze 24
integrator DisplacementControl 2 1 -0.1
analyze 48
integrator DisplacementControl 2 1 0.1
analyze 24
#second
#U=+5.9mm
integrator DisplacementControl 2 1 0.1
analyze 59
integrator DisplacementControl 2 1 -0.1
analyze 59
#U=-6.5mm
integrator DisplacementControl 2 1 -0.1
analyze 65
integrator DisplacementControl 2 1 0.1
analyze 65
#U=+6.1mm
integrator DisplacementControl 2 1 0.1
analyze 61
integrator DisplacementControl 2 1 -0.1
analyze 61
#U=-6.5mm
integrator DisplacementControl 2 1 -0.1
analyze 65
integrator DisplacementControl 2 1 0.1
analyze 65
#third
#U=+13mm
integrator DisplacementControl 2 1 0.1
analyze 130
integrator DisplacementControl 2 1 -0.1
analyze 130
#U=-13.7mm
integrator DisplacementControl 2 1 -0.1
analyze 137
integrator DisplacementControl 2 1 0.1
analyze 137
#U=13.8mm
integrator DisplacementControl 2 1 0.1
analyze 138
integrator DisplacementControl 2 1 -0.1
analyze 276
integrator DisplacementControl 2 1 0.1
analyze 138
#4th
#U=20.1mm
integrator DisplacementControl 2 1 0.1
analyze 201
integrator DisplacementControl 2 1 -0.1
analyze 201
#U=-21.2mm
integrator DisplacementControl 2 1 -0.1
analyze 212
integrator DisplacementControl 2 1 0.1
analyze 212
#U=20.5mm
integrator DisplacementControl 2 1 0.1
analyze 205
integrator DisplacementControl 2 1 -0.1
analyze 205
#U=-21.1mm
integrator DisplacementControl 2 1 -0.1
analyze 211
integrator DisplacementControl 2 1 0.1
analyze 211
#5th
#U=+28.5mm
integrator DisplacementControl 2 1 0.1
analyze 285
integrator DisplacementControl 2 1 -0.1
analyze 285
#U=-29.4mm
integrator DisplacementControl 2 1 -0.1
analyze 294
integrator DisplacementControl 2 1 0.1
analyze 294
#U=28.7mm
integrator DisplacementControl 2 1 0.1
analyze 287
integrator DisplacementControl 2 1 -0.1
analyze 287
#U=-29.7mm
integrator DisplacementControl 2 1 -0.1
analyze 297
integrator DisplacementControl 2 1 0.1
analyze 297
#6th
#U=+45.5mm
integrator DisplacementControl 2 1 0.1
analyze 455
integrator DisplacementControl 2 1 -0.1
analyze 455
#U=-46mm
integrator DisplacementControl 2 1 -0.1
analyze 460
integrator DisplacementControl 2 1 0.1
analyze 460
#U=+45.7mm
integrator DisplacementControl 2 1 0.1
analyze 457
integrator DisplacementControl 2 1 -0.1
analyze 457
#U=-46.1mm
integrator DisplacementControl 2 1 -0.1
analyze 461
integrator DisplacementControl 2 1 0.1
analyze 461
#7th
#U=62.1mm
integrator DisplacementControl 2 1 0.1
analyze 621
integrator DisplacementControl 2 1 -0.1
analyze 1242
integrator DisplacementControl 2 1 0.1
analyze 621
#U=+62.5mm
integrator DisplacementControl 2 1 0.1
analyze 625
integrator DisplacementControl 2 1 -0.1
analyze 625
#U=-62.3mm
integrator DisplacementControl 2 1 -0.1
analyze 623
integrator DisplacementControl 2 1 0.1
analyze 623
#8th
#U=79.8mm
integrator DisplacementControl 2 1 0.1
analyze 798
integrator DisplacementControl 2 1 -0.1
analyze 798
#U=-79.2mm
integrator DisplacementControl 2 1 -0.1
analyze 792
integrator DisplacementControl 2 1 0.1
analyze 792
#U=80.5mm
integrator DisplacementControl 2 1 0.1
analyze 805
integrator DisplacementControl 2 1 -0.1
analyze 805
#U=-79.6mm
integrator DisplacementControl 2 1 -0.1
analyze 796
integrator DisplacementControl 2 1 0.1
analyze 796
Zero length section element-FBE
Moderators: silvia, selimgunay, Moderators
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Zero length section element-FBE
Last edited by abdallahmontaser on Mon Jun 24, 2019 7:32 am, edited 1 time in total.
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- Posts: 917
- Joined: Mon Sep 09, 2013 8:50 pm
- Location: University of California, Berkeley
Re: Zero length section element-FBE (RC shear wall)
You may need to use an equaldof constraint for the lateral degree of freedom. To check whether the zerolength section does not result in any in compatibility issues in your current version, could you try defining your section as elastic with a large stiffness and compare it to the case without the zerolength section element?