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I got it, they are the ratio between the area of the element for shear (y and z in local direction, respectively) to area of the element for axial loading

I don't think the following description about the rigid link command is correct. is there any idea?


http://opensees.berkeley.edu/wiki/index ... nk_command
beam both the translational and rotational degrees of freedom are constrained.

would you please explain more about these parameters and its application?

Should the response of UniformExcitation and MultipleSupport be the same?
I understand that UniformExcitation uses Trapezoidal as the only option and MultipleSupport as the default option.
I expected their results to be the same. Therefore, I provided an example. According to the acceleration values and Trapezoidal method I expected 3.35 as the final displacement for both of them; however, this value for UniformExcitation is 3.35 while this value for MultipleSupport is 3.333. Where did this difference come from?

the acceleration serries is as the following with the time step of .1 sec

0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00

Let me ask you my question in other way.
I understand that UniformExcitation uses Trapezoidal as the only option and MultipleSupport as the default option.
I expected their results to be the same. Therefore, I provided an example. According to the acceleration values and Trapezoidal method I expected 3.35 as the final displacement for both of them; however, this value for UniformExcitation is 3.35 while this value for MultipleSupport is 3.333. Where did this difference come from?

What are the parameters alphaY and alphaZ in the following command??

section Elastic $secTag $E $A $Iz <$G $alphaY>
section Elastic $secTag $E $A $Iz $Iy $G $J <$alphaY $alphaZ>

http://opensees.berkeley.edu/wiki/index ... ic_Section


In the manual it is just stated that they are shape factor.

Would you please provide more description about it?

Thanks for your consideration.
In this regard, the only option for UniformExcitation and default option for MultipleSupport is Trapezoidal. With this in mind, why the final results, as mentioned earlier (3.35 and 3.33), are not the same?

Thanks, thats right.
actually, when I use disp rather than accel it works. However, if I use accel, I encounter a warning which states that integration is required. I thought it assumed something (Trapezoidal) as default. Would you please let me know what should we add to the mentioned commands?

I am trying to use UniformExcitation command and use displacement record rather than acceleration record.
There are two ways for implementing that:

1
timeSeries Path 1 -dt .1 -filePath record/recdis.txt -factor 1
pattern UniformExcitation 1 1 -disp 1

2
set Disp "Series Path 1 -dt .1 -filePath record/recdis.txt -factor 1"
pattern UniformExcitation 1 1 -disp $Disp


However, it turns out that this command just works with -accel. Actually, I did not encounter any errors; nevertheless the results are not valid.
In this regard, is my conclusion correct? or is there something that I neglect in my command and model?
_____________________________________________________________________________
another question is that when I use UniformExcitation with dt=0.1 and the following values as acceleration (a=20t):
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00

the final displacement should be 3.33 (20t^3/3) but I get 3.35. Actually, it turns out that OpenSees uses Trapezoidal (I am not sure about Trapezoidal correct me if I am wrong*) method. In this regard, is there any way to change the default integration type in this command (UniformExcitation )?

* It is stated that MultipleSupport uses Trapezoidal as default, and if MultipleSupport is used rather than UniformExcitation, the final displacement is 3.33. Thus, it seems one of these commands do not use Trapezoidal as default.

Thanks for your response.
In order to sum up for others:
The earlier example is analyzed by constraints Transformation. In other words, if one uses constraints Transformation should not define the node A in a command as the master node and in another comment as the slave node.

However, we do not encounter this problem if we use all other constraints handler:
constraints Plain
constraints Lagrange
constraints Penalty 1.0e14 1.0e14

As we can see in the Manual, there are three commands for Multi-Point Constraints.
1 rigidDiaphragm
2 equalDOF
3 rigidLink
3-1 rigidLink bar
3-2 rigidLink beam

In this topic I want to shed light on the difference between the second and third ones:
if you see this topic: http://opensees.berkeley.edu/community/ ... =2&t=63430
I stated that rigidLink bar A B and equalDOF A B 1 2 3 are the same. As you can see in the manual, it is asserted that only the translational degree-of-freedom will be constrained to be exactly the same as those at the master node by using rigidLink bar and here nodes A and B are constrained on transitional directions, x y and z.

In the following I provide a 3D example:
4 cantilever columns are modelled and then a force is exerted in node 5.

node 1 0. 0. 0.
node 2 4. 0. 0.
node 3 0. 4. 0.
node 4 4. 4. 0.
node 5 0. 0. 5.
node 6 4. 0. 5.
node 7 0. 4. 5.
node 8 4. 4. 5.
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

pattern Plain 1 Linear {
load 5 100. 0. 0. 0. 0. 0.
}


when we use
rigidLink bar 5 6 or equalDOF 5 6 1 2 3 (the resualts are exatly the same)
as we expected, the force is distributed between the columns, which are involved, equally.
reaction for nodes 1 to 4
-50 0 0 0 -250 0
-50 0 0 0 -250 0
0 0 0 0 0 0
0 0 0 0 0 0
displacement for nodes 5 to 8
0.0013 0 0 0 0.0004 0
0.0013 0 0 0 0.0004 0
0 0 0 0 0 0
0 0 0 0 0 0
_____________________________________________________________
To recap of this part:
rigidLink bar A B = equalDOF A B 1 2 3
1 2 3 means transitionally constrained

_____________________________________________________________
The problem is here:
According to manual http://opensees.berkeley.edu/wiki/index ... nk_command
using command, rigidLink beam, both the translational and rotational degrees of freedom are constrained. Therefore, based on which is provided earlier, we expect that:
rigidLink beam 5 7 and equalDOF 5 7 1 2 3 4 5 6 to be the same. On the following some examples are provided seperately:


Using rigidLink beam command
rigidLink beam 5 7
reaction for nodes 1 to 4
-99.7013 0 0 0 -374.2530 -0.5974
0 0 0 0 0 0
-0.2987 0 0 0 -125.7470 -0.5974
0 0 0 0 0 0
displacement for nodes 5 to 8
0.0016 0 0 0 0.0004 0.0002
0 0 0 0 0 0
0.0009 0 0 0 0.0004 0.0002
0 0 0 0 0 0

Rotationally Constrained
equalDOF 5 7 4 5 6
reaction for nodes 1 to 4
-100 0 0 0 -375 0
0 0 0 0 0 0
0 0 0 0 -125 0
0 0 0 0 0 0
displacement for nodes 5 to 8
0.0016 0 0 0 0.0004 0
0 0 0 0 0 0
0.0009 0 0 0 0.0004 0
0 0 0 0 0 0

Transitionally and Rotationally Constrained
equalDOF 5 7 1 2 3 4 5 6
reaction for nodes 1 to 4
-50 0 0 0 -250 0
0 0 0 0 0 0
-50 0 0 0 -250 0
0 0 0 0 0 0
displacement for nodes 5 to 8
0.0013 0 0 0 0.0004 0
0 0 0 0 0 0
0.0013 0 0 0 0.0004 0
0 0 0 0 0 0
____________________________________________________
As it can be seen, rigidLink beam 5 7 provide almost the same results as equalDOF 5 7 4 5 6 rather than equalDOF 5 7 1 2 3 4 5 6 (this negligble difference ie because of the fact that the constraint object constructed for the beam option assumes small rotations). Thus, we can conclude that rigidLink beam just constrains rotational degree of freedom rather than both transitional and rotational.

To recap, if my examples and description are clear, please let me know why the manual expresses that "beam: both the translational and rotational degrees of freedom are constrained."

I modeled 4 cantilever columns. Then, push them with a force in node 5.

node 1 0. 0. 0.
node 2 4. 0. 0.
node 3 0. 4. 0.
node 4 4. 4. 0.
node 5 0. 0. 5.
node 6 4. 0. 5.
node 7 0. 4. 5.
node 8 4. 4. 5.
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

pattern Plain 1 Linear {
load 5 100. 0. 0. 0. 0. 0.
}

when we use
rigidLink bar 5 6 or equalDOF 5 6 1 2 3
as we expected, the force is distributed between them equally.
reaction for nodes 1 to 4
-50 0 0 0 -250 0
-50 0 0 0 -250 0
0 0 0 0 0 0
0 0 0 0 0 0
displacement for nodes 5 to 8
0.0013 0 0 0 0.0004 0
0.0013 0 0 0 0.0004 0
0 0 0 0 0 0
0 0 0 0 0 0


_________________________________________________________________
And, if we use (5 is the master for both commands)
rigidLink bar 5 6 or equalDOF 5 6 1 2 3
rigidLink bar 5 7 or equalDOF 5 6 1 2 3
as we expected, the force is distributed between all these three nodes, 5 6 7, equally.
reaction for nodes 1 to 4
-33.3333 0 0 0 -166.6670 0
-33.3333 0 0 0 -166.6670 0
-33.3333 0 0 0 -166.6670 0
0 0 0 0 0 0
displacement for nodes 5 to 8
1.0e-03 *
0.8333 0 0 0 0.2500 0
0.8333 0 0 0 0.2500 0
0.8333 0 0 0 0.2500 0
0 0 0 0 0 0
_________________________________________________________________
After all, the question is why do we meet the following results when we change the arrangement of nodes as the following (first 6 is the master and then 5 is the master node):
rigidLink bar 6 5 or equalDOF 6 5 1 2 3
rigidLink bar 5 7 or equalDOF 5 6 1 2 3

reaction for nodes 1 to 4
-52.0930 0 0 0 -246.5120 0
-47.9070 0 0 0 -239.5350 0
0 -13.9535 0 69.7674 0 0
0 0 0 0 0 0
displacement for nodes 5 to 8
0.0012 0 0 0 0.0003 0
0.0012 0 0 0 0.0004 0
0 0.0003 0 -0.0001 0 0
0 0 0 0 0 0


As we can see, the results are completely changed. And I cannot understand where these results come from.

proc Isection { secID matID d bf tw tf Nfx Nfy Nwx Nwy} {
# ###################################################################
# Isection $secID $matID $d $bf $tf $tw $nfdw $nftw $nfbf $nftf
# _________________________________________________________________ #
# Because the shape may be distorted here
# You can download the code here: https://www.zeta-uploader.com/1864663169
# _________________________________________________________________ #

# This Procedure creat an stimated I-sectio with nominal properties using section Fiber command
# create a standard W section given the nominal section properties
# written: Mohsen Vazirizade s.m.vazirizade@gmail.com
# date: 05/2016
# INPUT Parameters
# secID - section ID number
# matID - material ID number
# d = nominal depth dw=d-2tf
# tw = web thickness
# bf = flange width
# tf = flange thickness
# Nfx = number of fibers along flange width (X-direction or JK)
# Nfy = number of fibers along flange thickness (Y-direction or IJ)
# Nwx = Number of fibers along Web thickness (X-direction or IJ)
# Nwy = Number of fibers along Web depth (Y-direction or JK)
#Isection 7 1 .2 .1 .01 .02 2 10 1 20
#Isection { secID matID d bf tw tf Nfx Nfy Nwx Nwy}
# y
# ^
# |
# |
# |
# |
# ------->x
#
# | bf |
# _ _____________
# |_____ tw_____| |||||| Nfx = Nfy
# | |
# | | =
# | | =
# d | |tf |||| Nwx = Nwy
# | | =
# | | =
# _____| |_____
# _ |_____________|
set dw [expr $d - 2 * $tf]
set x1 [expr -$bf/2]
set x2 [expr -$tw/2]
set x3 [expr $tw/2]
set x4 [expr $bf/2]
set y1 [expr -$d/2]
set y2 [expr -$dw/2]
set y3 [expr $dw/2]
set y4 [expr $d/2]
section fiberSec $secID {
# NIJ NJK xI yI xJ yJ xK yK xL yL
patch quad $matID $Nfx $Nfy $x1 $y1 $x4 $y1 $x4 $y2 $x1 $y2
patch quad $matID $Nwx $Nwy $x2 $y2 $x3 $y2 $x3 $y3 $x2 $y3
patch quad $matID $Nfx $Nfy $x1 $y3 $x4 $y3 $x4 $y4 $x1 $y4
}

Would your please let me know why I encounter the followoing error when I run this commands?

set a {1 2}
recorder Node -file $direction/Reigenvector-Mode$ModeNumber-Node$NodeNumber-x.out -precision 10 -node 1 -dof $a "eigen 1 "


NodeRecorder::NodeRecorder - dataToStore 1 2not recognized (disp, vel, accel, incrDisp, incrDeltaDisp)

I think the problem will be solved if you try the following command:

set accelSeries "Series -dt $dt -filePath $GMfile -factor [expr $Scalefact*$g]";

Actually timeSeries is replaced by Series.