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Hi Vesna,
Thanks for notifying me this. Is it because the zerolengthsection element doesn't output disp but curvature instead? I tried zerolengthsection since some of the researchers used zerolengthsection element to model Strain penetration effect. It actually makes sense that the a "section" can only do MC analysis. Instead, an "element" can do Moment Rotational analysis.

And here is my script:

# unit is in ksi kips

# SET UP ----------------------------------------------------------------------------
wipe; # clear memory of all past model definitions
model BasicBuilder -ndm 2 -ndf 3; # Define the model builder, ndm=#dimension, ndf=#dofs
set dataDir DataSPspr_st02_US_Test9Rhis_Axialload; # set up name of data directory (can remove this)
file mkdir $dataDir; # create data directory


# define GEOMETRY -------------------------------------------------------------
# define structure-geometry paramters

node 1 0 0
node 2 0 0


print node;


# BOUNDARY CONDITIONS
fix 1 1 1 1;
fix 2 0 1 0; # fix axial first



# Definition of parameters for fiber RC section

# MATERIAL parameters

set matTag 1;
set fc -7.; # CONCRETE Compressive Strength, ksi (+Tension, -Compression)
set Ec [expr 0.14504*5000*sqrt(48.2)]; # Concrete Elastic Modulus,ksi
# confined concrete
set Kfcb 1.29; # ratio of confined to unconfined concrete strength
set Kresb 0.6; # ratio of residual/ultimate to maximum stress
set fc1C [expr $Kfcb*$fc]; # CONFINED concrete (mander model), maximum stress;ksi # CONFINED concrete (mander model), maximum stress;
set eps0 -0.002; # UNCONFINED concrete strain
set fac [expr $fc1C/$fc];
set eps1C [expr $eps0*(1+5*($fac-1))]; # strain at confined concrete strength
set fc2C [expr $Kresb*$fc1C]; # crushing strength;ksi
set eps2C -0.05;
# unconfined concrete---------------------------------------------------------------------------------
set fc1U $fc; # UNCONFINED concrete (todeschini parabolic model), maximum stress
set eps1U $eps0; # strain at maximum strength of unconfined concrete
set fc2U -0.001; # crushing strength;ksi
set eps2U -0.0064; # strain at ultimate stress

set IDconcCore 1;
set IDconcCover 2;

uniaxialMaterial Concrete01 $IDconcCore $fc1C $eps1C $fc2C $eps2C ; # Column core concrete (confined)
uniaxialMaterial Concrete01 100 $fc1C $eps1C $fc2C $eps2C ; # Footing core concrete

uniaxialMaterial Concrete01 $IDconcCover $fc1U $eps1U $fc2U $eps2U; # Column cover concrete (unconfined)
uniaxialMaterial Concrete01 200 $fc1U $eps1U $fc2U $eps2U; # Footing cover concrete (unconfined)

set Fy [expr 0.14504*464.0]; # STEEL yield stress;MPa
set Fu [expr 0.14504*650];

# Strain penetration material-------------------------------------------------------------------------------------

set IDsp01 5;
set sy [expr 1.0*0.447/25.4]; # bar slip at member interface under yield stress (mm)
puts "sy = $sy";
set su [expr 35*$sy]; # bar slip at ultimate stress of rebar (in)
set bsp 0.4; # Initial hardening ratio in the monotonic slip
set rsp 0.8; # pinching factor for cyclic slip

uniaxialMaterial Bond_SP01 $IDsp01 $Fy $sy $Fu $su $bsp $rsp;

#uniaxialMaterial Elastic $matTag $E;

#uniaxialMaterial ReinforcingSteel $IDreinf $Fy $Fu $Es $Esh $esh $eult ;

set Es [expr 0.14504*200000]; # modulus of steel;MPa
set IDreinf 4;
## STEEL rebar
##STEEL02 tag $Fy $E $b $R0 $cR1 $cR2 $a1 $a2 $a3 $a4
uniaxialMaterial Steel02 $IDreinf [expr $Fy/2] $Es 0.01 18.5 0.925 0.15 0.01 0.5 0.05 1.2

set ColSecTag 1;

# section GEOMETRY -------------------------------------------------------------
set pi 3.1416;
set DSec 24; # Column Diameter
set Across [expr (3.14/4)*pow($DSec,2)];
set coverSec [expr 31.75/25.4]; # Column cover to reinforcing steel NA.
set numBarsSec 16; # number of longitudinal-reinforcement bars in column
set barAreaSec [expr 284.9/pow(25.4,2)]; # area of longitudinal-reinforcement bars
set ColSecTag 1; # set tag for symmetric section
set dbar [expr 19.05/25.4] ; # bar diameter

# Generate a circular reinforced concrete section
# with one layer of steel evenly distributed around the perimeter and a confined core.
# confined core.
#
#
# Notes
# The center of the reinforcing bars are placed at the inner radius
# The core concrete ends at the inner radius (same as reinforcing bars)
# The reinforcing bars are all the same size
# The center of the section is at (0,0) in the local axis system
# Zero degrees is along section y-axis
#
set ri 0.0; # inner radius of the section, only for hollow sections
set ro [expr $DSec/2]; # overall (outer) radius of the section
set nfCoreR 8; # number of radial divisions in the core (number of "rings")
set nfCoreT 24; # number of theta divisions in the core (number of "wedges")
set nfCoverR 2; # number of radial divisions in the cover
set nfCoverT 24; # number of theta divisions in the cover

set ZerSecTag 1;
source RCcircSection.tcl;
RCcircSection $ZerSecTag $ri $ro $coverSec 100 200 $IDreinf $numBarsSec $barAreaSec $nfCoreR $nfCoreT $nfCoverR $nfCoverT

set IDColTransf 1; # all columns
set ColTransfType Linear ; # options: Linear, PDelta, Corotational
# set ColTransfType PDelta ; # options: Linear, PDelta, Corotational
geomTransf $ColTransfType $IDColTransf; # only columns can have PDelta effects (gravity effects)

# Duplicate the spring in our tiger model
# tag ndI ndJ secID
element zeroLengthSection 1 1 2 $ZerSecTag

print element;

recorder Node -file $dataDir/disp.xls -node 2 -dof 3 disp; # check input dipl
recorder Element -file $dataDir/eleglobal_1.xls -time -ele 1 globalForce;
recorder Node -file $dataDir/sectionforce.xls -node 1 -dof 3 reaction;
#recorder Element -file $dataDir/eleSection_1.xls -time -ele 1 section 1 force;
recorder Element -file $dataDir/elelocal_1.out -time -ele 1 localForce ;

set IDloadTag 100; # load tag

set PCol [expr 0.000225*-756200]; # Axial load
# define GRAVITY -------------------------------------------------------------
pattern Plain 1 Linear {
load 2 $PCol 0 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 $Tol 1000 ; # determine if convergence has been achieved at the end of an iteration step
algorithm Newton; # use Newton's solution algorithm: updates tangent stiffness at every iteration
set NstepGravity 10; # apply gravity in 10 steps
set DGravity [expr 1./$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
# ------------------------------------------------- maintain constant gravity loads and reset time to zero

puts "Model Built"

loadConst -time 0.0

# -------------define computational parameters for cyclic test---------
source LibAnalysisStaticParameters2.tcl;

source Test9Rhis.tcl; # displacement history for test

# create load pattern for monotonic load

set Hload 100; # define the lateral load as a proportion of the weight so that the pseudo time equals the lateral-load coefficient when using linear load pattern
set iPushNode 2; # define nodes where lateral load is applied in static lateral analysis
pattern Plain 200 Linear { ; # define load pattern -- generalized
load $iPushNode 0.0 0.0 $Hload
};

# Set up parameters that are particular to the model for displacement control

set IDctrlNode 2; # node where displacement is read for displacement control
set IDctrlDOF 3; # degree of freedom of displacement read for displacement control

set Dpre 0;
set Dincr 0.0001;

foreach Dmax $iDmax {
set Ddiff [expr $Dmax-$Dpre];
set Nsteps [expr int(abs($Ddiff)/$Dincr)];
if {$Ddiff<0} {; # avoid the divide by zero
set dx [expr -1*$Dincr];
} else {
set dx $Dincr;
}
integrator DisplacementControl $IDctrlNode $IDctrlDOF $dx;
analyze $Nsteps;
set Dpre $Dmax;
}; # end of iDmaxCycl

puts "Spring Simulation Done";


vesna wrote:
> With zerolengthsection element you can only do moment-curvature analysis,
> not moment rotation analysis.
>
> If you post your example I can take a look.

Hi Vesna,
I observed an interesting result when recording global and local forces.
In a 2D cantilever model, I used one beam column element. Global force and local force are the same since I applied the default orientation of the section that local coordinate coincide with global coordinate.
When I did a moment rotational analysis with zerolengthsection element, recorded global moment is not the same as local moment. And the local force equated to $Hload times global force. $Hload is the lateral load I specified in the load pattern. (Section sill have the default orientation.) It seems that global force recording command is recording the lateral load coefficient.

Do you know why is this happening?

Yuhao Feng

Hi Hidemasa,
I am working on rotational spring, too. The element I used is zerolength element. Are you using this element also?

Thanks.

Yuhao

Hidemasa KAIHARA wrote:
> Hi,
>
> I have a question.
> Generally, the performance of rotational spring gives Moment-angle
> relation.
> According to the Command manual (page 127), stress-strain or
> force-deformation must be used in "uniaxialMaterial ElasticPP".
> There is a problem in the case of definition of tha material of Rotational
> spring.
> Although I verified Moment-angle relation can be used.
> I mean, stiffness of rotation(Nmm/rad) as $E instead of modulus of
> elastic(N/mm2).
> Above is no problem.
>
> I want to use "uniaxialMaterial SAWS" as rotational spring. But I
> can't understand how to define this material.
> This material is defined force-displacement relation according to the
> internet source. In the case of rotational spring, I must give parameters
> in M-θ.
> I try several case.
> For example, I give the angle at ultimate as $DU instead of the
> displacement at ultimate load.
> Other parameters also give in the M-θ　field.
>
> But the result of the analysis and M-θ behavior I assumed does not match.
> (Upside down "T" shape model with one rotational spring at 2
> element, and apply cyclic load. It is able to test rot spring, I think.)
>
> I can't understand this reason. Because I try the same way in 2case
> (elasticPP and SAWS),although one is sucssess and the other is not.
>
> If you know how to define SAWS material as rotational spring or rot spring
> material definition, teach me please.
>
> Thank you for reading.

Hi Vesna,
I wrote a static analysis script for rotational spring. Though I specified integrator to be displacement control, OpenSees kept showing me warning " no Integrator specified". I checked a lot of times and failed to find the problem. Can you take a look at it? Thanks

# unit is in ksi kips

# SET UP ----------------------------------------------------------------------------
wipe; # clear memory of all past model definitions
model BasicBuilder -ndm 2 -ndf 3; # Define the model builder, ndm=#dimension, ndf=#dofs
set dataDir DataTrial; # set up name of data directory (can remove this)
file mkdir $dataDir; # create data directory


# define GEOMETRY -------------------------------------------------------------
# define structure-geometry paramters

node 1 0 0
node 2 0 0


print node;


# BOUNDARY CONDITIONS
fix 1 1 1 1;
fix 2 1 1 0; # fix axial first



# Definition of parameters for fiber RC section

# MATERIAL parameters

set matTag 1;
set fc 50.; # MPa confined concrete strength
set E [expr 5000*pow($fc,0.5)]; # MPa young's modulous of confined concrete strength

uniaxialMaterial Elastic $matTag $E;

set ColSecTag 1;

# section GEOMETRY -------------------------------------------------------------
set DSec 609.; # Column Diameter
set Across [expr (3.14/4)*pow($DSec,2)];

set IDColTransf 1; # all columns
set ColTransfType Linear ; # options: Linear, PDelta, Corotational
# set ColTransfType PDelta ; # options: Linear, PDelta, Corotational
geomTransf $ColTransfType $IDColTransf; # only columns can have PDelta effects (gravity effects)

# Create the columns using displacement controlled beam-column elements
# tag ndI ndJ secID
element zeroLength 1 1 2 -mat $matTag -dir 6
#print element;

recorder Node -file $dataDir/disp.xls -node 2 -dof 3 disp; # check input dipl
recorder Element -file $dataDir/eleglobal_1.xls -time -ele 1 globalForce;

set IDloadTag 100; # load tag

# -------------define computational parameters for cyclic test---------
source LibAnalysisStaticParameters2.tcl;


source test9US.tcl; # displacement history for test

# create load pattern for monotonic load

set Hload 100; # define the lateral load as a proportion of the weight so that the pseudo time equals the lateral-load coefficient when using linear load pattern
set iPushNode 2; # define nodes where lateral load is applied in static lateral analysis
pattern Plain 200 Linear { ; # define load pattern -- generalized
load $iPushNode 0.0 0.0 $Hload
}

# Set up parameters that are particular to the model for displacement control

set IDctrlNode 2; # node where displacement is read for displacement control
set IDctrlDOF 3; # degree of freedom of displacement read for displacement control

set Dpre 0;
set Dincr 0.1;

foreach Dmax $iDmax {
set Ddiff [expr $Dmax-$Dpre];
set Ncycles [expr int(abs($Ddiff)/$Dincr)];
if {$Ddiff<0} { # avoid the divide by zero
set dx [expr -1*$Dincr];
} else {
set dx $Dincr;
# puts "Spring Simulation not Done";
}
integrator DisplacementControl $IDctrlNode $IDctrlDOF $dx;
analyze $Ncycles;
set Dpre $Dmax;
}; # end of iDmaxCycl

Vesna,
The way to assign moment-rotation ralation is to play with the material in the element, right?
Thank you.

Yuhao

vesna wrote:
> You can use zeroLength element and assign moment-rotation relation in direction 6.
> Do not forget to equlDOF all other dofs.

Hi all,
What is the best way to model a nonlinear rotational spring? The spring should have a flexible force displacement response (moment_rotation) that can be specified.
Does anyone have some experience on that?
Thank you.

Yuhao

Nice. Let me try your method and see what I get.

Thank you.

Yuhao

vesna wrote:
> What I do is the following:
> since I know position of the last two integration points (e.g., if #IP=5,
> one point is going to be at location 0, and another point is going to be at
> 0.173L, where L is the length of the element) I make sure to encompass the
> plastic hinge region with this two points (for this example the Lp~0.173L).

Vesna,
With one beam with hinge element, I imposed the calculated plastic hinge length to the weight of the base integration point which is at the fix end.
I also tried one nonlinear beam column element to mode the column. For this case, I used 5 integration points along the element that the weight of end integration point will represent the plastic hinge length.
Do you mean that you use 2 integration points to represent the plastic hinge length? Do you set plastic hinge length equal to the sum of weights of this two integration points?

Thank you

Yuhao

vesna wrote:
> Where are your integration points with respect to the plastic hinges? The
> way I usually decide about the number of integration points and and number
> of elements needed to model the column using force-based beam column
> element is to by imposing constraint that the two integration points
> encompass the plastic hinge region. Most of the time this gives good
> results.

Hi Vesna,
Yes, I recorded the strain at plastic hinge region. In the experiment, cracks were all over the length of the whole column and concentrates in plastic hinge region. We block out the cover concrete of our column and put a series of LED sensors along the bar to cover the whole plastic hinge region. We use camera to track the movement of the LEDs and calculate the engineering strain based on the elongation of distance between two LED sensors. Basicly, spacing(gage length) of the LED sensors are around 2". So we got the strain distribution along the plastic hinge region.
Strain is usually higher at the gage which have a crack acrossing it.
The strain in fiber model is even lower than the strain from a uncrack gage of experiment.

Yuhao


vesna wrote:
> Hi Yuhao,
>
> At what location along your cantilever beam-column element are you
> recording the strain? Is it in the plastic hinge region? How about your
> experimental element? Did you have any cracks in concrete at the location
> where you recorded strains in your bars?

Hi Vesna,
I am using a beamcolumn element to model a RC cantilever. I tried beam with hinge element, nonlinear beam column element and displacement based element. The material constitutive model is calibrated also based on material test result.
Meanwhile, we are doing some experimental tests in the lab. After calibration of the model, the force displacement response match with test result really good. But the strain predicted by the model is always lower than the test result.
It looks like that the steel material constitutive model dominates the strain we got from the model. In material test, we see a lot of strain hardening for the rebar. We match the constitutive model with the material test result by involving strain hardening in Steel02.
As soon as we involve the strain hardening in Steel02, the strain at a specific displacement for a cyclic analysis is much lower than the one without strain hardening that it is also lower than the experimental result.
Is there any way to increase the strain in the model with keeping the strain hardening in the material model?

Thank you

Yuhao Feng

Frank,
Yes. I am using force based Beam with hinge element. But acturally, I am reducing the time step and the result varies. Do you have any suggestion on selecting the time step?

Yuhao


[quote="fmk"]a problem with increasing time step is that the changes in the response get larger at each step (dU) .. certain elements (force Beam Column) and the solution algorithm (if discontinuities exist) can have problems depending on the state.[/quote]

Vesna,
Yes, my tolerance is 10^-8.

Yuhao

[quote="vesna"]Yuhao,
What is the tolerance for the test? For EnergyIncr convergence test (residual force)*(incr disp)<tol. This would mean that your tolerance is on the order of 10^4. Please check on this one.[/quote]

Vesna,
I have significant deviations. I reduced the time step because model cant converge at a specific time during dynamic analysis.
I also checked the norm unbalance force and incremental displacement at the point which fail to converge(i am using EnergyIncr convergence test). Interestingly, residual force is large and the incremental displacement is very small. ie. deltaX=0.0003, deltaR=1.26e+008.
It doesn't make sense having a large residual force. I think there must be an error in the numerical process. Also, the same error may lead to significant deviation on responses with different time steps.
Do you have any idea about this?

Thank you.

Yuhao

[quote="vesna"]Yuhao,

As you reduce the time step of analysis compared to ground motion time step the responses will not be exactly the same but they should be almost the same. Is this what you are getting, or you have significant deviations?[/quote]

Hi Vesna,
We said analytical time step is ok as long as it is less than 1/10-1/20 of natural period.
Most of the examples in OpenSee example manual is making DtAnalysis equate to ground motion time step. And myself used the time step of ground motion as the analytical time step also. It works pretty good.
However, when I reduced the analytical time step, result deviated. I didn't expect this except a quicker convergence. Do you have any idea on it?

Thank you.

Yuhao

Hi all,
Can the velocity time history be the input ground motion for dynamic analysis? Or can we use UniformExcitation Pattern with velocity or displacement?


Thank you.

Yuhao