I'm modeling a simple 3D frame using 2 different approaches for slab or diaphragm. In Rigid Diaphragm case time period is 0.07073167481663893 where as in Flexible Diaphragm case time period is 0.05693672810516015. My under standing is that for Rigid Diaphragm I should gate lower time period compare to Flexible Diaphragm. Can anyone kindly look into the code and point out the issue in modeling.
Copied below is the code.
Thank You.
wipe
wipe all
model BasicBuilder -ndm 3 -ndf 6;
set Rigid_Slab True;
# Create Nodes
# Command: node $nodeTag $X $Y $Z
node 1 0 0 0;
node 2 5 0 0;
node 3 0 0 5;
node 4 5 0 5;
node 5 0 4 0;
node 6 5 4 0;
node 7 0 4 5;
node 8 5 4 5;
# Fixing base nodes
fix 1 1 1 1 1 1 1;
fix 2 1 1 1 1 1 1;
fix 3 1 1 1 1 1 1;
fix 4 1 1 1 1 1 1;
# Define mass to nodes
set m 1000;
mass 5 $m $m $m $m $m $m;
mass 6 $m $m $m $m $m $m;
mass 7 $m $m $m $m $m $m;
mass 8 $m $m $m $m $m $m;
if {$Rigid_Slab == "True"} {
# Assigning Rigid Diaphragm
set PerpDirn 2;
node 7634 2.5 4 2.5;
fix 7634 0 1 0 1 0 1;
rigidDiaphragm $PerpDirn 7634 5 6 7 8;
} else {
# Define quadrilateral element for plate element
set matTag 99;
set E 13.0e9; # Modulus of elasticity of Mixed Southern Pine (According to AWC-NDS 2012, Table-4B)
set v 0.20;
# Define Material
nDMaterial ElasticIsotropic $matTag $E $v;
# Define section for slab
set secID 999;
set thick 0.150;
set elemID 888;
section PlateFiber $secID $matTag $thick;
element ShellMITC4 $elemID 5 6 8 7 $secID;
}
# Define materials for force columns
# CONCRETE tag f'c ec0 f'cu ecu
# Core concrete (confined)
uniaxialMaterial Concrete01 1 -44.0e6 -0.004 -36.0e6 -0.014
# Cover concrete (unconfined)
uniaxialMaterial Concrete01 2 -44.0e6 -0.004 1.0 -0.006
# STEEL
# Reinforcing steel
set fy 500e6; # Yield stress
set Es 200e9; # Young's modulus
set dia 25e-3; # Diameter of the Reinforcing bar
set steelTagCol 3;
# tag fy E0 b
uniaxialMaterial Steel01 $steelTagCol $fy $Es $dia
# Define cross-section for nonlinear columns
# set some paramaters
set Width 0.3
set Depth 0.3
set cover 0.04;
set As 3927e-6; # area of no. 8 bars
set secID 1; # section tag for section fibre
# some variables derived from the parameters
set y1 [expr $Depth/2.0]
set z1 [expr $Width/2.0]
section Fiber $secID {
# Create the concrete core fibers
patch rect 1 10 1 [expr $cover-$y1] [expr $cover-$z1] [expr $y1-$cover] [expr $z1-$cover]
# Create the concrete cover fibers (top, bottom, left, right)
patch rect 2 10 1 [expr -$y1] [expr $z1-$cover] $y1 $z1
patch rect 2 10 1 [expr -$y1] [expr -$z1] $y1 [expr $cover-$z1]
patch rect 2 2 1 [expr -$y1] [expr $cover-$z1] [expr $cover-$y1] [expr $z1-$cover]
patch rect 2 2 1 [expr $y1-$cover] [expr $cover-$z1] $y1 [expr $z1-$cover]
# Create the reinforcing fibers (left, middle, right)
layer straight 3 3 $As [expr $y1-$cover] [expr $z1-$cover] [expr $y1-$cover] [expr $cover-$z1]
layer straight 3 2 $As 0.0 [expr $z1-$cover] 0.0 [expr $cover-$z1]
layer straight 3 3 $As [expr $cover-$y1] [expr $z1-$cover] [expr $cover-$y1] [expr $cover-$z1]
}
# set up geometric transformation of element
set TransfTag 1;
geomTransf Linear $TransfTag 0 0 -1; # PDelta Transformation
set NumIntgrPts 5; # Number of Integration Points
# Define Column elements
element forceBeamColumn 1 1 5 $NumIntgrPts $secID $TransfTag;
element forceBeamColumn 2 2 6 $NumIntgrPts $secID $TransfTag;
element forceBeamColumn 3 3 7 $NumIntgrPts $secID $TransfTag;
element forceBeamColumn 4 4 8 $NumIntgrPts $secID $TransfTag;
# set up geometric transformation of beam element
set TransfTag 2;
geomTransf Linear $TransfTag 0 1 0; # PDelta Transformation
set NumIntgrPts 5; # Number of Integration Points
# Define beam elements
element forceBeamColumn 5 5 6 $NumIntgrPts $secID $TransfTag;
element forceBeamColumn 6 6 8 $NumIntgrPts $secID $TransfTag;
element forceBeamColumn 7 7 8 $NumIntgrPts $secID $TransfTag;
element forceBeamColumn 8 5 7 $NumIntgrPts $secID $TransfTag; # columns
# Eiven Analysis
set numModes 3
set lambda [eigen $numModes];
# calculate frequencies and periods of the structure ---------------------------------------------------
set omega {}
set f {}
set T {}
set pi 3.141593
foreach lam $lambda {
lappend omega [expr sqrt($lam)]
lappend f [expr sqrt($lam)/(2*$pi)]
lappend T [expr (2*$pi)/sqrt($lam)]
}
puts "periods are $T"
Rigid Diaphragm has lesser stiffness than Flexible Diaphragm
Moderators: silvia, selimgunay, Moderators
Re: Rigid Diaphragm has lesser stiffness than Flexible Diaph
the rigid diaphragm does not add any additional bending stiffness at the nodes .. the shell does .. a more comparable element in terms of the stiffness, or more correctly lack there of provided viz-a-vie these stiffness components, is a quad element.
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- Posts: 52
- Joined: Fri Mar 17, 2017 11:01 am
- Location: IIT Patna
Re: Rigid Diaphragm has lesser stiffness than Flexible Diaph
fmk wrote:
> the rigid diaphragm does not add any additional bending stiffness at the
> nodes .. the shell does .. a more comparable element in terms of the
> stiffness, or more correctly lack there of provided viz-a-vie these
> stiffness components, is a quad element.
HI,
I am working RC 3D buildings. I have to find the time periods. In opensees, when am performing eigen analysis, it is considering slab stiffness I guess. I have checked with as small model. In fact we are providing any commands for slab.
So how to include the stiffness of slab also, so that my Time period matches excatly?
Thanks for your replies. Your suggestions are always helpful to me in learning Opensees .
> the rigid diaphragm does not add any additional bending stiffness at the
> nodes .. the shell does .. a more comparable element in terms of the
> stiffness, or more correctly lack there of provided viz-a-vie these
> stiffness components, is a quad element.
HI,
I am working RC 3D buildings. I have to find the time periods. In opensees, when am performing eigen analysis, it is considering slab stiffness I guess. I have checked with as small model. In fact we are providing any commands for slab.
So how to include the stiffness of slab also, so that my Time period matches excatly?
Thanks for your replies. Your suggestions are always helpful to me in learning Opensees .
-
- Posts: 52
- Joined: Fri Mar 17, 2017 11:01 am
- Location: IIT Patna
Re: Rigid Diaphragm has lesser stiffness than Flexible Diaph
fmk wrote:
> the rigid diaphragm does not add any additional bending stiffness at the
> nodes .. the shell does .. a more comparable element in terms of the
> stiffness, or more correctly lack there of provided viz-a-vie these
> stiffness components, is a quad element.
Hi fmk,
As you said, If one uses the quad element then the model dimensions also need to be changed. For 3d, model defining is done as ndm-3 ndf-6.
But by using this , am getting an error model dimensions and/or nodal DOF not compatible with quad element. When changing it to ndm-2 ndf-2, though it woks , the elaticbeamcolumn cannot be defined properly.
Can you help me in this?
> the rigid diaphragm does not add any additional bending stiffness at the
> nodes .. the shell does .. a more comparable element in terms of the
> stiffness, or more correctly lack there of provided viz-a-vie these
> stiffness components, is a quad element.
Hi fmk,
As you said, If one uses the quad element then the model dimensions also need to be changed. For 3d, model defining is done as ndm-3 ndf-6.
But by using this , am getting an error model dimensions and/or nodal DOF not compatible with quad element. When changing it to ndm-2 ndf-2, though it woks , the elaticbeamcolumn cannot be defined properly.
Can you help me in this?