Search found 2 matches

by xujin
Tue Nov 17, 2020 11:36 pm
Forum: OpenSees.exe Users
Topic: How to apply acceleration to a certain node
Replies: 0
Views: 4977

How to apply acceleration to a certain node

My model is as follow. I want apply acceleration to node 5 .But i don't kown my way is right or wrong.

wipe
model BasicBuilder -ndm 2 -ndf 3
set PI [expr 2.0*asin(1.0)]
#设置单位
set N 1.0
set m 1.0
set kg 1.0
set sec 1.0

set mm [expr $m*0.001]
set cm [expr $m*0.01]
set kN [expr $N*1000]
set Pa [expr $N/($m*$m)]
set kPa [expr $Pa*1000]
set MPa [expr $Pa*1e6]
set m2 [expr $m*$m]
set cm2 [expr $cm*$cm]
set mm2 [expr $mm*$mm]
set m4 [expr $m2*$m2]
set cm4 [expr $cm2*$cm2]
set mm4 [expr $mm2*$mm2]
set m3 [expr $m*$m*$m]
set m4 [expr $m2*$m2]
set p [expr $kg/$m3] ;# density
set a [expr $m/($sec*$sec)] ;#accelerated speed

set A [expr 130*$cm2]
set Iz [expr 4267*$cm4]
set E [expr 2100*$kN/$cm2]
set m [expr 0.0001067*$kN/$cm]


node 1 0 0
node 2 1.2 0
node 3 2.4 0
node 4 3.6 0
node 5 4.8 0

if { [file exists kangzhen] ==0} {
file mkdir kangzhen;
}

recorder Node -file kangzhen/disp_1.txt -time -node 1 -dof 2 disp

fix 1 1 1 1

mass 2 0 $m 0
mass 3 0 $m 0
mass 4 0 $m 0
mass 5 0 $m 0

geomTransf Linear 1
#element element elasticBeamColumn eleTag iNode jNode A E Iz transfTag <-mass massDens> <-cMass>
element elasticBeamColumn 1 1 2 $A $E $Iz 1
element elasticBeamColumn 2 2 3 $A $E $Iz 1
element elasticBeamColumn 3 3 4 $A $E $Iz 1
element elasticBeamColumn 4 4 5 $A $E $Iz 1


set ACC "Series -dt 0.0001 -filePath acc.txt -factor 1"
pattern MultiSupport 1 {
groundMotion 1 Plain -accel $ACC
imposedMotion 5 2 1
}
constraints Transformation
numberer RCM
system UmfPack
test EnergyIncr 1.0e-6 200
algorithm KrylovNewton
integrator Newmark 0.5 0.25
analysis Transient
analyze 3503 0.0001
puts "succeed"
by xujin
Mon Jul 27, 2020 9:00 pm
Forum: OpenSees.exe Users
Topic: Why can the model only add 3mm displacement?
Replies: 6
Views: 5903

Why can the model only add 3mm displacement?

My model can only be loaded with a displacement of 3mm under the required direction load, otherwise it will not converge.
Can you help me find the reason?
This is a simple reinforced concrete column model with 3 nodes and 2 elements


wipe
model BasicBuilder -ndm 3 -ndf 6
set PI [expr 2.0*asin(1.0)]
#设置单位
set N 1.0
set m 1.0
set kg 1.0
set sec 1.0
set mm [expr $m*0.001]
set kN [expr $N*1000]
set Pa [expr $N/$m]
set kPa [expr $Pa*1000]
set MPa [expr $Pa*1e6]
set m2 [expr $m*$m]
set mm2 [expr $mm*$mm]
set mm4 [expr $mm*$mm*$mm*$mm]
set m3 [expr $m*$m*$m]
set m4 [expr $m*$m*$m*$m]
set p [expr $kg/$m3] ;# density
set a [expr $m/($sec*$sec)] ;#accelerated speed



set ps 0.018 ;#钢筋体积配箍率
set fyh [expr 241*$MPa] ;#箍筋的屈服强度
set fc [expr 42.4*$MPa] ;#混凝土抗压强度
set h [expr 0.426*$m] ;#箍筋肢距
set sh [expr 0.1*$m] ;#箍筋间距
set K0 [expr 1.0+$ps*$fyh/$fc] ;#箍筋对混凝土强度的提高系数
set fpc0 [expr $K0*$fc] ;#混凝土28天抗压强度
set epsc0 [expr 0.002*$K0] ;#约束混凝土极限抗压强度对应的应变值
set E0 [expr 2*$fpc0/$epsc0]
set fpcu0 [expr 0.2*$fpc0] ;#混凝土退化强度
set Z0 [expr 0.5/((3+0.29*$fc)/(145*$fc-1000)+0.75*$ps*sqrt($h/$sh)-0.002*$K0)] ;#应变软化斜率系数
#set epsU0 [expr (1.0-$fpcu0/($K0*$fc))+$epsc0]
set epsU0 [expr (-0.004-0.9*($ps*$fyh)/300)]
set ft0 [expr 0.01*$fpc0]
set Ets0 [expr $ft0/0.002]
set lambda0 0.1

#定义核心混凝土
#Concrete
#uniaxialMaterial Concrete02 $matTag $fpc $epsc0 $fpcu $epsU $lambda $ft $Ets
uniaxialMaterial Concrete02 1 [expr -$fpc0] [expr -$epsc0] [expr -$fpcu0] [expr -$epsU0] $lambda0 $ft0 $Ets0

set K1 1 ;#箍筋对混凝土强度的提高系数
set fpc1 [expr $K0*$fc] ;#混凝土28天抗压强度
set epsc1 [expr 0.002*$K1] ;#约束混凝土极限抗压强度对应的应变值
set E1 [expr 2*$fpc1/$epsc1]
set fpcu1 [expr 0.2*$fpc1] ;#混凝土退化强度
set Z1 [expr 0.5/((3+0.29*$fc)/(145*$fc-1000))] ;#应变软化斜率系数
set epsU1 -0.004
set ft1 [expr 0.01*$fpc1]
set Ets1 [expr $ft1/0.002]
set lambda1 0.1
#定义无约束混凝土
#Concrete
#uniaxialMaterial Concrete02 $matTag $fpc $epsc0 $fpcu $epsU $lambda $ft $Ets
uniaxialMaterial Concrete02 2 [expr -$fpc1] [expr -$epsc1] [expr -$fpcu1] [expr -$epsU1] $lambda1 $ft1 $Ets1



#定义钢筋+钢筋的直径和面
set Fy [expr 412*$MPa]
set E [expr 2.0e11*$Pa]
#uniaxialMaterial Steel02 $matTag $Fy $E $b $R0 $cR1 $cR2 <$a1 $a2 $a3 $a4 $sigInit>
uniaxialMaterial Steel02 3 $Fy $E 0.0006 18 0.925 0.15
set d [expr 0.018*$m]
set A [expr $d*$d/4*$PI]

puts "material finish"

set Lcolumn1 [expr 0.54*$m]
set Lcolumn2 [expr 3.24*$m]


node 1 0.0 0.0 0.0
node 2 0.0 0.0 $Lcolumn1
node 3 0.0 0.0 $Lcolumn2



fix 1 1 1 1 1 1 1
fix 3 0 0 0 0 0 0

#墩的截面
set b [expr 0.5*$m]
set h [expr 0.34*$m]
set A1 [expr $b*$h]
set cover 0.037
set y1 [expr $b/2.0-$cover]
set y2 [expr $b/2.0-$cover-1.42]
set z1 [expr $h/2.0-$cover]

section Fiber 1 {
patch rect 1 10 10 [expr $b/2.0-$cover] [expr $h/2.0-$cover] [expr $cover-$b/2.0] [expr $cover-$h/2.0]
patch rect 2 10 1 [expr $b/2.0] [expr $cover-$h/2.0] [expr -$b/2.0] [expr -$h/2.0]
patch rect 2 2 1 [expr $b/2.0] [expr $h/2.0-$cover] [expr $b/2.0-$cover] [expr $cover-$h/2.0]
patch rect 2 2 1 [expr $cover-$b/2.0] [expr $h/2.0-$cover] [expr -$b/2.0] [expr $cover-$h/2.0]
patch rect 2 10 1 [expr $b/2.0] [expr $h/2.0] [expr -$b/2.0] [expr $h/2.0-$cover]
layer straight 3 3 $A $y1 $z1 $y1 [expr -$z1]
layer straight 3 2 $A $y2 $z1 $y2 [expr -$z1]
layer straight 3 3 $A [expr -$y1] $z1 [expr -$y1] [expr -$z1]
layer straight 3 2 $A [expr -$y2] $z1 [expr -$y2] [expr -$z1]
}

section Fiber 2 {
patch rect 2 10 10 [expr $b/2.0] [expr $h/2.0] [expr -$b/2.0] [expr -$h/2.0]
layer straight 3 3 $A $y1 $z1 $y1 [expr -$z1]
layer straight 3 2 $A $y2 $z1 $y2 [expr -$z1]
layer straight 3 3 $A [expr -$y1] $z1 [expr -$y1] [expr -$z1]
layer straight 3 2 $A [expr -$y2] $z1 [expr -$y2] [expr -$z1]
}
puts "section finish"

geomTransf Linear 1 0 -1 0
#define elements
set nP 5
#element dispBeamColumn eleTag iNode jNode numIntgrPts secTag transfTag <-mass massDens> <-cMass> <-integration intType>
element nonlinearBeamColumn 1 1 2 $nP 2 1
element nonlinearBeamColumn 2 2 3 $nP 1 1
puts "element finish"

set N 0.1
set fcd 18.4e6
set P [expr $N*$fcd*$A1]
pattern Plain 1 Linear {
load 3 0.0 0.0 [expr -$P] 0.0 0.0 0.0
}
puts "Gravity finish"

constraints Plain
numberer RCM
system BandGeneral
test EnergyIncr 1.0e-6 200
algorithm Newton
integrator LoadControl 0.01
analysis Static
analyze 100

loadConst -time 0.0
pattern Plain 2 Linear {
load 3 1 1 0.0 0.0 0.0 0.0

}
integrator DisplacementControl 3 1 0.004
analyze 1