Brainery Byte

</font> How To Release Element Ends Aligned Along Arbitraty Directions in OpenSees

R210821
[email protected]

</font> The objective of this workbook is to demonstrate the effectivenes of differe methods of managing Multi-point constraints in OpenSees.
For simplication, we are only looking at managing the axial deformation, but the concept can be extended to other DOFs
This issue is important when we are dealing with elements that are not aligned with the global axes.
There are different ways of imposing this type of constraint in OpenSees:

1. ZeroLength Element
2. EqualDOF Command
3. Rigid Truss Element
Each of these methods manage the constraint differently, and are more or less effective under different conditions

We will first test our model using a horizontal beam. We will then test the different options using an inclined beam

### Part 1: (Horizontal Beam)¶

• Two Head Fixed Beam with one central node
• Two Head Fixed Beam with two central EqualDOF nodes
• Two Head Fixed Beam with two central Axial EqualDOF nodes
• Altenatives:
1. ZeroLength Element
3. Stiff Truss Element
• Final Results of Part 1

### Part 2: (Inclined Beam)¶

1. ZeroLength Element
2. Stiff Truss Element
• Final Reults of Part 2
In [55]:
# set up the environment:
import openseespy.opensees as ops
import eSEESminiPy as slv
import matplotlib.pyplot as plt
#import markdown
%matplotlib notebook
import numpy as np
import math
import pandas as pd

def PerformAnalysis():
'''
This Function is set to run analysis
'''

ops.wipeAnalysis()
ops.system("BandSPD")                   # create SOE
ops.numberer("RCM")                     # create DOF number
ops.constraints("Plain")                # create constraint handler
ops.algorithm("Linear")                 # create algorithm
ops.analysis("Static")                  # create analysis object
ops.analyze(1)                          # perform the analysis


# Part 1 : Horizontal Beam¶

## Two Head Fixed Beam With Concentrated Force at the middle¶

Consider the simple 2D Below model:

In [56]:
Rs=[] #Results

#Build the 2D model
ops.wipe()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,0]);ops.fix(1,1,1,1)
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,0]);ops.fix(3,1,1,1)

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
A=100.0
E=2.1e6
I= 10.0*10.0*10.0*10.0/12.0
G=2.0e5
transfTag=1
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,2], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

tsTag=1
ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

f=ops.eigen(1)
f=f[0]**0.5

PerformAnalysis()
Rs.append(['Two Head Fixed Beam, node 2',ops.nodeDisp(2),f])
#Rs.append(['Two Head Fixed Beam, node 21',ops.nodeDisp(21),f])

slv.drawModel('2D','Two Head Fixed Beam With Concentrated Force at the middle')
slv.drawDeformedShape(50)


For the first Case Displacement at $node2=${{ops.nodeDisp(2)}}

and eigen value is equal={{f}}

## Two Head Fixed Beam with two central EqualDOF nodes¶

Now we want to make a release in node 2 how only axial force can be transfer. For this purpose we use $equalDOF$ command and we consider two cases. 1st the nodes are without any release (So the results should be exactly the same in previous step), then in 2nd case we release the rotations and only put axial DOF remains:

In [57]:
transfTag=1
tsTag=1

ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,0]);ops.fix(1,1,1,1)
ops.node(21, *[150,0]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,0]);ops.fix(3,1,1,1)

ops.equalDOF(21,2,*[1,2,3]) #Without Release

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()
Rs.append(['Two Head Fixed Beam with Central EqualDOF, node 2',ops.nodeDisp(2),f])
Rs.append(['Two Head Fixed Beam with Central EqualDOF, node 21',ops.nodeDisp(21),f])

slv.drawDeformedShape(50)


For the first Case Displacement at $node2=${{ops.nodeDisp(2)}}

and eigen value is equal={{f}}

As it is seen the results are completely Compatible with the model with 1 node at the middle and it shows EqualDOF connects degrees of freedom together correctly.

# Two Head Fixed Beam with two central Axial EqualDOF nodes¶

## Now Reseasing All Degree Freedoms Except Axial¶

Because All degree of freedoms are in global Direction, We can using equalDOF command release all Degree of Freedom Except axial that is our purpose of this study and consider the results as exact solution:

In [58]:
transfTag=1
tsTag=1

Case=2

ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,0]);ops.fix(1,1,1,1)
ops.node(21, *[150,0]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,0]);ops.fix(3,1,1,1)

ops.equalDOF(21,2,*[1])     #  With Release

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)

ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()

Rs.append(['EqualDOF nodes at the Middle With Release (Node2)',ops.nodeDisp(2),f])
Rs.append(['EqualDOF nodes at the Middle With Release (Node21)',ops.nodeDisp(21),f])
slv.drawDeformedShape(5)


Displacement Results at $node2$={{ops.nodeDisp(2)}}

and for $node21$={{ops.nodeDisp(21)}}

and eigen value is $f=${{f}}

So we consider these results as correct results and now search for alternative options with different tools. Alternative tools that will be checked are:

1. Zero Length Element
3. Stiff Truss Element

## 1. Zero Length Element¶

Now we want to check the Zero Length Element and check the results. We expect that we get same results that we got for previouse case:

In [59]:
ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,-10*0]);ops.fix(1,1,1,1)
ops.node(21, *[150,0]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,10*0]);ops.fix(3,1,1,1)

#element('zeroLength', eleTag, *eleNodes, '-mat', *matTags, '-dir', *dirs, <'-doRayleigh', rFlag=0>, <'-orient', *vecx, *vecyp>)
matTag=1
ops.uniaxialMaterial('Elastic', matTag, E*1e9)
ops.element('zeroLength', 3, *[21,2], '-mat', *[matTag],'-dir', *[1])#, '-orient', *[300,20,0], *[0,1,0])

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()

Rs.append(['ZeroLength Element nodes 2',ops.nodeDisp(2),f])
Rs.append(['ZeroLength Element nodes 21',ops.nodeDisp(21),f])

slv.drawDeformedShape(5)


Displacement Results at $node2$={{ops.nodeDisp(2)}}

and for $node21$={{ops.nodeDisp(21)}}

and eigen value is $f=${{f}}

In this Step we use RigidLink to see the results of this element.

In [60]:
ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,-0]);ops.fix(1,1,1,1)
ops.node(21, *[150,0]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,0]);ops.fix(3,1,1,1)

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()

slv.drawDeformedShape(5)


Displacement Results at $node2$={{ops.nodeDisp(2)}}

and for $node21$={{ops.nodeDisp(21)}}

and eigen value is $f=${{f}}

So, Seems that RigidLink element is not a proper choice and we can say that:

if we use rigidLink element for two nodes that are compatible on each other with 'bar' is equal internal pin (Look at the Slope) and if we use rigidLink element for two nodes that are compatible on each other with 'beam' is equal continiues beam

## 3. Really Stiff Truss Element¶

In [61]:
ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,-10*0]);ops.fix(1,1,1,1)
ops.node(21, *[149.9,-0.1*10/150*0]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,10*0]);ops.fix(3,1,1,1)

#element('Truss', eleTag, *eleNodes, A, matTag, <'-rho', rho>, <'-cMass', cFlag>, <'-doRayleigh', rFlag>)
matTag=1
ops.uniaxialMaterial('Elastic', matTag, E*1e6)
ops.element('Truss', 3, *[21,2], A, matTag)

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()

Rs.append(['Truss Element nodes 2',ops.nodeDisp(2),f])
Rs.append(['Truss Element nodes 21',ops.nodeDisp(21),f])

slv.drawDeformedShape(5)


Displacement Results at $node2$={{ops.nodeDisp(2)}}

and for $node21$={{ops.nodeDisp(21)}}

and eigen value is $f=${{f}}

# Final Results of Part 1:¶

In [62]:
print('Results:')
df = pd.DataFrame([[i[1][0],i[1][1],i[1][2],i[2]] for i in Rs],index=[i[0] for i in Rs], columns=['Ux', 'Uy', 'Theta','Frequency'])
df

Results:

Out[62]:
Ux Uy Theta Frequency
Two Head Fixed Beam, node 2 0.357143 -0.803571 0.000000 11.155467
Two Head Fixed Beam with Central EqualDOF, node 2 0.357143 -0.803571 0.000000 11.155467
Two Head Fixed Beam with Central EqualDOF, node 21 0.357143 -0.803571 0.000000 11.155467
EqualDOF nodes at the Middle With Release (Node2) 0.357143 -6.428571 0.064286 3.944053
EqualDOF nodes at the Middle With Release (Node21) 0.357143 0.000000 0.000000 3.944053
ZeroLength Element nodes 2 0.357143 -6.428571 0.064286 3.944053
ZeroLength Element nodes 21 0.357143 0.000000 0.000000 3.944053
RigidLink nodes 2 0.357143 -3.214286 0.032143 5.577734
RigidLink nodes 21 0.357143 -3.214286 -0.032143 5.577734
Truss Element nodes 2 0.357024 -6.428571 0.064286 3.944053
Truss Element nodes 21 0.357024 0.000000 0.000000 3.944053

Results: As it is seen in above table, ZeroLength Element Gives Exact Results and Truss Element also Gives proper Results But Rigid Link in not a good choice for this purpose! This is because the rigidLink command only forces the two nodes to have equal translational displacements, which is not what we want here!

# Part 2 : Inclined Beam¶

## Two Head Fixed Beam With Concentrated Force at the middle¶

In this part Because the EqualDOF command is not working on local directions, Using ZeroLength Element and Truss Element, we will model an Inclined Beam and check the Deformation Shape and Results.

## 1. Zero Length Element¶

The same Beam is considered with slope equal to $m=20/300=${{round(20/300,4)}}

In [63]:
Rss=[]
ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,-10]);ops.fix(1,1,1,1)
ops.node(21, *[150,0]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,10]);ops.fix(3,1,1,1)

#element('zeroLength', eleTag, *eleNodes, '-mat', *matTags, '-dir', *dirs, <'-doRayleigh', rFlag=0>, <'-orient', *vecx, *vecyp>)
matTag=1
ops.uniaxialMaterial('Elastic', matTag, E*1e9)
ops.element('zeroLength', 3, *[21,2], '-mat', *[matTag],'-dir', *[1], '-orient', *[300,20,0], *[0,1,0])

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()

slv.drawDeformedShape(1)


Displacement Results at $node2$={{ops.nodeDisp(2)}}

and for $node21$={{ops.nodeDisp(21)}}

and eigen value is $f=${{f}}

As Expected the node 21 don't has any rotation and the it's displcament vector ratio $(Ratio=V_y/V_x=${{ops.nodeDisp(21)[1]}}$/${{ops.nodeDisp(21)[0]}}$=${{round(ops.nodeDisp(21)[1]/ops.nodeDisp(21)[0],4)}}$)$ is exactly equal to the slope of the beam $(${{round(20/300,4)}}$)$ that means only axial deformation as we need is transforming.

## 2. Really Stiff Truss Element¶

In [64]:
ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,-10]);ops.fix(1,1,1,1)
ops.node(21, *[149.9,-0.1*10/150]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,10]);ops.fix(3,1,1,1)

#element('Truss', eleTag, *eleNodes, A, matTag, <'-rho', rho>, <'-cMass', cFlag>, <'-doRayleigh', rFlag>)
matTag=1
ops.uniaxialMaterial('Elastic', matTag, E*1e6)
ops.element('Truss', 3, *[21,2], A, matTag)

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

ops.wipeAnalysis()
ops.system("BandSPD")                   # create SOE
ops.numberer("RCM")
ops.test('NormDispIncr',1e-6,10)
ops.constraints("Plain")                # create constraint handler
ops.algorithm("Newton")                 # create algorithm
ops.analysis("Static")                  # create analysis object
ops.analyze(1)                          # perform the analysis

slv.drawDeformedShape(1)


Displacement Results at $node2$={{ops.nodeDisp(2)}}

and for $node21$={{ops.nodeDisp(21)}}

and eigen value is $f=${{f}}

As Expected the node 21 don't has any rotation and the it's displcament vector ratio $(Ratio=V_y/V_x=${{ops.nodeDisp(21)[1]}}$/${{ops.nodeDisp(21)[0]}}$=${{round(ops.nodeDisp(21)[1]/ops.nodeDisp(21)[0],4)}}$)$ is Again Exactly equal to the slope of the beam (With Newton Algorithm) $(${{round(20/300,4)}}$)$ that means only axial deformation as we need is transforming.

In this Step we use RigidLink to see the results of this element. You will see that the two nodes are forced to have the same displacement, hence we cannot release the axial deformation!

In [65]:
ops.wipe()
ops.wipeAnalysis()
ops.model('basic', '-ndm',2)

#Nodes
ops.node(1, *[0,-10]);ops.fix(1,1,1,1)
ops.node(21, *[149.9,-0.1*10/150]);
ops.node(2, *[150,0]); ops.mass(2,100,100,0)
ops.node(3, *[300,10]);ops.fix(3,1,1,1)

#Elements: element('elasticBeamColumn', eleTag, *eleNodes, Area, E_mod, Iz, transfTag, <'-mass', mass>, <'-cMass'>, <'-release', releaseCode>)
ops.geomTransf('Linear', transfTag)
ops.element('elasticBeamColumn', 1, *[1,21], A, E, I, transfTag)
ops.element('elasticBeamColumn', 2, *[2,3], A, E, I, transfTag)

f=ops.eigen(1)
f=f[0]**0.5

ops.timeSeries('Linear', tsTag)
ops.pattern('Plain', 1, tsTag)

#Predefine Function to perform Analysis
PerformAnalysis()

slv.drawDeformedShape(1)


# Final Results of Part 2:¶

In [66]:
print('Results:')
df = pd.DataFrame([[i[1][0],i[1][1],i[1][2],i[2]] for i in Rss],index=[i[0] for i in Rss], columns=['Ux', 'Uy', 'Theta','Frequency'])
df

Results:

Out[66]:
Ux Uy Theta Frequency
ZeroLength nodes 2 3.649125 -49.371425 4.939517e-01 3.930958
ZeroLength nodes 21 0.356114 0.023744 3.402922e-08 3.930958
Truss Element nodes 2 3.649003 -49.371384 4.939512e-01 3.930955
Truss Element nodes 21 0.355996 0.023733 1.213810e-14 3.930955
RigidLink nodes 2 2.000852 -24.649120 2.467285e-01 5.561996
RigidLink nodes 21 2.000852 -24.649120 -2.468931e-01 5.561996

Final Result: For Axial transformation of Inclined Elements, ZeroLength Element and Truss Element according above results are proper choices. This results as seen are reached according static analysis and user should check the numeric problems that may arise in other analysis methods and different algorithms.