Problem in modeling of eccentric bracing (bad result)

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peacemaker
Posts: 9
Joined: Thu Mar 29, 2007 2:24 pm
Location: IRAN

Problem in modeling of eccentric bracing (bad result)

Post by peacemaker » Tue Jun 24, 2008 8:25 am

Hi to all .
I want to model an 1story -1 bay frame with eccentric bracing .
Because the model has 1 story , the brace element is connected to ground as a support .
My applied latetal force to frame is in positive direction of X -axis . It is clear that reaction of supports in X-axis must be negetive . But when i checked the RBase.out file i saw the reaction of brace node is positive !
I think this result is incorrect .
I CAN NOT understand whats the problem of my model , Would u plz check my model and give mea reason of this fault ?

Thank u .

Code: Select all

#units: kg-cm-sec
#os :v2.0.0
#modeling of eccentric barcing  
wipe;
model basic  -ndm 2 -ndf 3;
file mkdir EBF;
#----------------------------------
node 1 0 0;
node 2 400 0;
node 3 0 300;
node 4 400 300;
node 5 100 0;
node 6 400 300;
#---------------------------------
fix 1	 1 1 1;
fix 2	 1 1 1;
fix 5  1 1 0;
#---------------------------------
mass 3 2000	0.0	0.0	 
mass 4 2000	0.0	0.0
#---------------------------------
equalDOF 4 6 1 2;
#---------------------------------
set Fy	2400 
set E		2100000 
set Bs	0.02 
set R0	18
set cR1 	0.925
set cR2 	0.15
uniaxialMaterial  Steel02 1	    $Fy $E $Bs $R0 $cR1 $cR2
#--------------------------------
geomTransf Linear 1
#--------------------------------
source Wsection.tcl
#--------------------------------
#Wsection     $SecTag    $matIDhard $d  $bf $tf $tw $nfdw $nftw $nfbf $nftf
Wsection      1		  1          23 24  1.2 1.0  16    4     16    4 ;
Wsection      2		  1          18 9   0.85 0.7 16    4     16    4 ;
#--------------------------------
#----------2unp 10-------------
section fiberSec 3 { 
  patch quad 1 10 5 -5.0 0.35 -5.0 4.75 -4.15 4.75 -4.15 0.35 
  patch quad 1 5 5 -5.0 4.75 -5 5.35 -4.15 5.35 -4.15 4.75 
  patch quad 1 20 5 -4.15 5.35 4.15 5.35 4.15 4.75 -4.15 4.75 
  patch quad 1 5 5 4.15 5.35 5.0 5.35 5.0 4.75 4.15 4.75 
  patch quad 1 10 5 4.15 0.35 4.15 4.75 5.0 4.75 5.0 0.35 
  patch quad 1 10 5 -5.0 -4.75 -5.0 -0.35 -4.15 -0.35 -4.15 -4.75 
  patch quad 1 5 5 -5.0 -5.35 -5 -4.75 -4.15 -4.75 -4.15 -5.35 
  patch quad 1 20 5 -4.15 -4.75 4.15 -4.75 4.15 -5.35 -4.15 -5.35 
  patch quad 1 5 5 4.15 -4.75 5.0 -4.75 5.0 -5.35 4.15 -5.35 
  patch quad 1 10 5 4.15 -4.75 4.15 -0.35 5.0 -0.35 5.0 -4.75 
  }
#----------------------------------
#element dispBeamColumn    $elemID $nodeI $nodeJ $np $SecTag $IDBeamTransf;
 element dispBeamColumn		1	1	 3	  5	1		1	;
 element dispBeamColumn		2	2	 4	  5	1		1	;
 element dispBeamColumn		3	3	 4	  5	2		1	;
 element dispBeamColumn		4	5	 6	  5	3		1	;
#--------------------------------
recorder Node -file EBF/pushoverXdisp.xls -time -node 4 -dof 1  disp ;
recorder Node -file  EBF/RBase.xls -time -node 1 2 5 -dof 1  reaction;
#--------------------------------
# Gravity Analasys 

pattern Plain 1 Linear {
   load 3 0. -2000. 0.
   load 4 0. -2000. 0.
}
#--------------------------------
# Gravity-analysis parameters -- load-controlled static analysis
set Tol 1.0e-8;				# convergence tolerance for test
constraints Plain;     			# how it handles boundary conditions
numberer Plain;				# renumber dof's to minimize band-width (optimization), if you want to
system BandGeneral;			# how to store and solve the system of equations in the analysis
test NormDispIncr $Tol 6 ; 		# determine if convergence has been achieved at the end of an iteration step
algorithm Newton;				# use Newton's solution algorithm: updates tangent stiffness at every iteration
set NstepGravity 10;  			# apply gravity in 10 steps
set DGravity [expr 1./$NstepGravity]; 	# first load increment;
integrator LoadControl $DGravity;	# determine the next time step for an analysis
analysis Static;				# define type of analysis static or transient
analyze $NstepGravity;			# apply gravity
 
loadConst -time 0.0 
puts "Model Built" ;
#--------------------------------
set IDctrlNode 4;	
set IDctrlDOF 1;
set Dmax 10;		
set Dincr 0.05;

	pattern Plain 2 Linear {			
	load 3 100 0.0 0.0;
}
constraints Plain;
numberer RCM;
system UmfPack;
set Tol 1.e-7;                        # Convergence Test: tolerance
set maxNumIter 10;                		# Convergence Test: maximum number of iterations that will be performed before "failure to converge" is returned
set printFlag 0;                # Convergence Test: flag used to print information on convergence (optional)        # 1: print information on each step; 
set TestType EnergyIncr ;	# Convergence-test type
test $TestType $Tol $maxNumIter $printFlag;
set algorithmType Linear
algorithm $algorithmType; 
integrator DisplacementControl 4 1 $Dincr 1 $Dincr  $Dincr
analysis Static


set Nsteps [expr int($Dmax/$Dincr)];        # number of pushover analysis steps
set ok [analyze $Nsteps];   

Last edited by peacemaker on Wed Jun 25, 2008 12:52 am, edited 1 time in total.

silvia
Posts: 3912
Joined: Tue Jan 11, 2005 7:44 am
Location: Degenkolb Engineers
Contact:

Post by silvia » Tue Jun 24, 2008 10:05 am

actually, the sum is positive. it seems that node 5 is picking up all the load.
Silvia Mazzoni, PhD
Structural Consultant
Degenkolb Engineers
235 Montgomery Street, Suite 500
San Francisco, CA. 94104

peacemaker
Posts: 9
Joined: Thu Mar 29, 2007 2:24 pm
Location: IRAN

Post by peacemaker » Wed Jun 25, 2008 12:45 am

I dont understand , what does it mean ?!
i checked this model with sap2000 , all support reactions is negetive .
u say that result is correct ?
I want to get story shear , i am confused !
i cannot understand what is problem of modeling ?

plz give me a help !

silvia
Posts: 3912
Joined: Tue Jan 11, 2005 7:44 am
Location: Degenkolb Engineers
Contact:

Post by silvia » Wed Jun 25, 2008 7:59 am

the truss element is taking all the story shear, as it is significantly stiffer than the rest in that direction.
check your model, please.
Silvia Mazzoni, PhD
Structural Consultant
Degenkolb Engineers
235 Montgomery Street, Suite 500
San Francisco, CA. 94104

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