Cie 633 Final Report1

7/28/2019 Cie 633 Final Report1

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CIE 633

FINITE ELEMENT ANALYSIS

PROJECT

BEAM ANALYSIS

BY

Kalpesh Parikh


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Acknowledgment

My deepest gratitude goes to Dr. Richard Wilson Perkins for his continuous and constructive advice and

follows up. His successive advisories and comments were the pillars in my every step during the Project.

I am thankful to him for the fact that he has inspired and helped me to know about the Finite Element

Analysis.


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I ndex

1.Introduction 042.Buckling Analysis 053.Modal Analysis 244.

Transient and Dynamic Analysis 29

5.Conclusion 35-Reference 36

- Appendix 37


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1 Introduction

The Project analysis is done in the software called ANSYS .Basically Software takes all the Inputs and it has in itsmemory all the solver which are required to be carried out. So I basically started using an W shape section (I

section) and here i took structural components as a Beams. Where I studied the various behavior of the structure in

order to understand that when we design some structural component we should be clear with the behavior of the

structure. So that we dont land up to an failure of the structure.

General Information

A wide flange structural steel beam W10 x 26 (10 means deep and 26 is weight of the section lb/ft)

Material properties

ASTM Designation: A992 Steel Section Modulus of Elasticity: E =30000 Ksi Poisson ratio =0.3 Density of Steel= 490 lb/ft3 Yield Strength Fy = 50 ksi Tensile Strength Fu = 65 ksi

Cross sectional Properties

Flange Thickness (tf) = 0.44 inch Web Thickness (tw) = 0.3 inch Flange width (bf) = 5.75 inch Depth of section (d) = 10.3 inch Moment of Inertia in z-z direction (Izz) = 14.1 inch 4 Moment of Inertia in y-y direction (Iyy) = 144 inch4


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2. Buckling Analysis

Buckling loads are critical loads where certain types of structures become unstable. Each load has an associated

buckled mode shape; this is the shape that the structure assumes in a buckled condition. There are two primary

means to perform a buckling analysis:

1. EigenvalueEigenvalue buckling analysis predicts the theoretical buckling strength of an ideal elastic structure. It

computes the structural eigenvalues for the given system loading and constraints. This is known as classical

Euler buckling analysis. Buckling loads for several configurations are readily available from tabulated

solutions. However, in real-life, structural imperfections and nonlinearities prevent most real-world

structures from reaching their eigenvalue predicted buckling strength; ie. it over-predicts the expected

buckling loads. This method is not recommended for accurate, real-world buckling prediction analysis.

2. NonlinearNonlinear buckling analysis is more accurate than eigenvalue analysis because it employs non-linear, large-

deflection; static analysis to predict buckling loads. Its mode of operation is very simple: it gradually

increases the applied load until a load level is found whereby the structure becomes unstable (ie. suddenly a

very small increase in the load will cause very large deflections). The true non-linear nature of this analysis

thus permits the modeling of geometric imperfections, load perterbations, material nonlinearities and gaps.

For this type of analysis, note that small off-axis loads are necessary to initiate the desired buckling mode.

Study of Buckling Analysis and its comparison with various boundary condition and loading condition:

1 Cross section used W10 x 26 as defined.

2 Elements used is Beam 188, Beam 189, Beam 4, and Shell element 63 and Shell element 93

3 Study of Eigenvalue Analysis.

Model Descriptions

1. Beam 4 Element was model using the cross sectional properties and followed and two key points were usedto define the length of the beam.. Boundary condition and loading condition was assigned. Analysis is

Buckling so general post processor was defined. As per the required analysis

2. Beam 188 element and Beam 189 was model using standard SECDATA and SECTYPE and then Iassigned the sectional properties to the command, four key points were used to define the length and

width of the beam.. Boundary condition and loading condition was assigned. Analysis is Buckling so

general post processor was defined. As per the required analysis

3. Shell 63 and Shell 93 element was modeled using real constant and modeling I beam is not so proper as itgoes into it as shown below (Area overlapped)


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These is only model imperfection involved in the shell element as it makes shell element more

stronger. Apart from that the main think is to define loading at various position in order to get proper result

so as it is made up of three part web, flange and Flange I applied loading at all three position and at the tip

of the flange at node 1 and node 2 so in order to get buckled the structure.

The Euler buckling load for a cantilever beam length L=120 inch, with the above material and cross sectional

properties is given by

When L = 120

Pcr=

() =72.47 kips

When L = 60

Pcr=

() = 289.91 kips

Case 1 Cantilever Beam Subjected Axial Load

When Length is 120 inch Using Various Element Beam 188, Beam 189, Beam 4, Shell Element 63 and Shell 93 Element. AnalysisEigenvalue AnalysisBuckling Loads obtained using Set list when simulated

Element

Theoretical Pcr

in Kips

Ansys Pcr in

Kips

% of Comparison of

Theoretical with

Ansys

Beam 4 72.47 72.48 100.0137988

Beam 188 72.47 71.635 98.84779909

Beam 189 72.47 71.635 98.84779909

Shell 63 72.47 71.994 99.34317649

Shell 93 72.47 71.523 98.69325238

Study of Variation in Length

When Length is 60 inch Using Various Element Beam 188, Beam 189, Beam 4, Shell Element 63 and Shell 93 Element. AnalysisEigenvalue Analysis


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Comparision Table showing Theoretical value and Ansys Eulers Buckling Load

Element

Theoretical Pcr

in Kips

Ansys Pcr in

Kips

% of Comparison of

Theoretical with

Ansys

Beam 4 289.91 289.93 100.0068987Beam 188 289.91 287.01 98.99968956

Beam 189 289.91 287.01 98.99968956

Shell 63 289.91 281.71 97.17153599

Shell 93 289.91 277.2 95.61588079

Interpretation

The results form ANSYS are almost the same as the theoretical buckling load obtained using Euler Formula.Below Figures shows the buckled shape of the cantilever beam for the BEAM4, BEAM188, BEAM189,

SHELL63 and SHELL93 models respectively.

The deformed shapes in all the cases (where the beam is modeled using the beam and shell elements), clearlyagree with the mode shape that would result if a first mode Euler buckling is to occur. It is also important to

note from the buckling shapes of BEAM188, BEAM189, SHELL63 and SHELL93 that the buckling occurs

about the weak axis.

Distribution of load helped to identify the correct buckling value and behavior of the buckling which wasrequired s it is important how we apply loads in the beam to get correct buckling.

As I Tried to study the variation of the length in the beam I reduced my length to 50% and noted the result theresult are not changing as it has nothing to do with geometric propertie s , apart from how you apply loads.

Figure Buckling for Beam 4


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Figure Showing Buckling for Beam 188 and Beam 189

Figure Showing Buckling for Shell 63 and Shell 93

Case 2 Simply Supported Beam Subjected to axial Load:The Modeling Procedure remains same I tried all 5 element that are described for cantilever beam to study the

Eigenvalue buckling the model built up procedure remains same only changes would be boundary conditions it was

really painful to understand the behavior when I inserted the boundary condition as a simply supported then I found

in the textbook Structural Stability by Dr. Eric M Lui and Chen in that boundary conditions I need to apply (ROTX)

need to be restrained at one end of the beam in order to give structural stability. So now the behavior of the structure


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were quiet reasonable. So I applied restrained on the translation (Ux, Uy, Uz in node 1 and Uy in node 2 ) and made

restrained in (Rotx) then simulate as using the post processor for Eigen value buckling.

The Euler buckling load for a simply supported

When L = 120

Pcr=

() = 289.91 kips

Where E = 30000 ksi

Izz = Moment of Inertia about weak axis =14.1 inch4

Study of Eigenvalue Using simply supported boundary condition

When Length is 120 inch Using Various Element Beam 188, Beam 189, Beam 4, Shell Element 63 and Shell 93 Element.

AnalysisEigenvalue Analysis


Element

Theoretical Pcr

in Kips

Ansys Pcr in

Kips

% of Comparison of

Theoretical with

Ansys

Beam 4 289.91 289.92 100.0034493

Beam 188 289.91 281.2 96.99561933

Beam 189 289.91 281.2 96.99561933Shell 63 289.91 405.88 140.0020696

Shell 93 289.91 585.63 202.0040702

Interpretation:

1. Beam 4 give same result as we obtained by theoretical value reason behind these is there is no localbuckling. Now the question comes what is local buckling generally it is defined as

> rLocal Buckling will occur in the beam

r = obtained from AISC manual depending on the beam is compact section or non compact section

= Width/ Thickness ratio (We can determine Flange value as well as web value)

2 Beam 188 gives bit off result reason is that there is a local buckling effect taking place in the flange portion

and the web portion of the beam. So we can interprete from that generally the local buckling takes place if the

section is assembled and it is general effect is under when we use high yield strength steel so these clear says

that it is not adequate using beam188 element and same apply to beam 189 element as these also shows the

Local buckling in the web and flange portion.


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3 Mode shape difference are found when I changes the number of element so it signifies the importance of the

element size if we want to study the behavior of mode shape.

4 As Shell 63 and Shell 93 value is to high as I signifies that shell element would not be an good idea to study

the exact behavior of the buckling as it has model imperfection as web goes into the flange portion so when

we study the Eigen value buckling it requires more load to buckle because of its strong action in the model.

As always buckling occurs about the weak axis where there is an Inertia value is lesser so in Shell63 andShell 93 in both Local buckling take place in the web and the flange portion.

5 When we calculate the stiffness of the beam for shell 93 an shell 63 we can find that the stiffness is better of

shell 93 and shell 63 apart from that when we study the effect of Stress in both the beam during the buckling

Mode we can see There are more stresses in the shell 63 compare to shell 93 element.

6. The behavior using simply supported boundary condition is very unusual as the constraining of model is an

big challenge to get the adequate results.

7. For Practical purpose when we do the execution of steel beam the behavior of shell element would more be

given focus as in actual high rise building we always use High strength steel where the yield strength and

local buckling is always there compare to low yield strength so we can say if you have more yield strength wehave more chances of local buckling taking place in the structure.

8. Study of variation in length was made only the percentage of accuracy of the ansys remains same so I did not

tabulated the value for it so there is no percentage changes in the ansys.

CASE 3: FIXEDPINED END CONDITIONS

The Modeling Procedure remains same I tried all 5 element that are described for cantilever beam to study the

Eigen value buckling the model built up procedure remains same only changes would be boundary conditions

become Translation in x,y and z were restrained and other end I made restrained to all the degree of freedom. And

following same repetitive steps for general post processors.

The Euler buckling load when boundary condition is fixed and Pinned end

Pcr=

() = 591.67 kips

Izz =Weak axis of moment of inertia (Why I can say that only z-z axis is weak in moment of inertia it is very much

clear from the concept of mechanics of materials that bending takes place in the z-z axis ).

Here study is made on the all the elements used to compare theoretical eigenvalue buckling with Ansys so that we

can understand that when boundary condition changes what are the changes.

Study of Eigenvalue Using Fixed and Pinned boundary condition

When Length is 120 inch Using Various Element Beam 188, Beam 189, Beam 4, Shell Element 63 and Shell 93 Element. AnalysisEigenvalue Analysis


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Element

Theoretical Pcr

in Kips

Ansys Pcr in

Kips

% of Comparison of

Theoretical with

Ansys

Beam 4 591.67 594 100.39Beam 188 591.67 574.51157 97.1

Beam 189 591.67 574.51157 97.1

Shell 63 591.67 829.048004 140.12

Shell 93 591.67 1192.451718 201.54

Interpretation

1 When Beam 4 Element used the results are perfect as per the theoretical value. I did not notice indication oflocal buckling in the flange and web portion of the beam.

2 Results are same as obtained in the simply supported beam there is a local buckling effect in the shell elementand beam element local buckling taking place in the flange an d web portion .same case as it makes sense

because I have modified boundary conditions and obtained the result.

3 Shell element showing higher value as compared to theoretical reason is same as mentioned in the simplysupported beam.

4 Variation of Length is being studied the percentage of the accuracy for ansys does not changes.

Study of Lateral Torsional Buckling of beam

Lateral torsional buckling occurs when the distance between the lateral brace points is large enough that the beam

fails by lateral , outward movement in combination with a twisting action ( and respectively. Generally beam

with wider flanges provide more resistance to lateral displacement . In general adequate restraint against lateral

torsional buckling is accomplished by the addition of the brace or similar restraint somewhere between the centroid

of the member and the compression flange.

The Euler Buckling Load formula obtained from Timeshenko and Gere

When Length = 120 inch

Pcr=

() =17.45 kips

Izz = Weak axis of moment of Inertia.

J= torsional value for the given section


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When Length =60 inch

Pcr=

() =69.82 kips

Case 4 : Study of Lateral torsional Buckling of Cantilever beam

Modeling Procedure

The Modeling Procedure remains same I tried all 4 element that are described for cantilever beam to study the

Eigen value buckling the model built up procedure remains same only changes would be boundary Now the

loading was applied laterally to beam and the Post Processor remains same and eigen value analysis is being

performed.

When Length is 120 inch Using Various Elements Beam 188, Beam 189, Beam 4, Shell Element 63 and Shell 93 Element. AnalysisEigen value Analysis. Load Applied Laterally to the beam.


Element

Theoretical Pcr

in Kips

Ansys Pcr in

Kips

% of Comparison of

Theoretical with

Ansys

Beam 188 17.45 17.101 98

Beam 189 17.45 17.101 98

Shell 63 17.45 25.4072 145.6

Shell 93 17.45 32.9805 189

Study of Variation in Length

When Length is 60 inch Using Various Element Beam 188, Beam 189, Beam 4, Shell Element 63 and Shell 93 Element. AnalysisEigenvalue AnalysisComparision Table showing Theoretical value and Ansys Eulers Buckling Load

Element

Theoretical Pcr

in Kips

Ansys Pcr in

Kips

% of Comparison of

Theoretical with

Ansys

Beam 188 69.82 68.4236 98

Beam 189 69.82 68.4236 98

Shell 63 69.82 108.221 155

Shell 93 69.82 131.9598 189


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Interpretation

1. Result obtained for Beam 188 and Beam 189 are near to the theoretical value which makes sense. And thebehavior shown in the ansys matches the theoretical behavior of the beam means lateral displacement in

combination with twisting show exact behavior of the lateral torsional buckling.

2. Shell 63 and Shell 93 elements are of concern because of the Model imperfection that is the Areaoverlapped changes the ansys behavior as the K matrix changes means when we calculate the stiffness of

the beam using rotation and torsion the overlapping of the elements add to an more result hence the results

are higher as we can get the same result as theoretical value required.

3. Study of length variation is being carried out and result are same as expected by theoretical nothingchanges so effect of length variation does not changes the behavior of the structure.

4. Below figure shows the behavior of the lateral torisonal buckling we can notice the lateral displacement incombination with twisting.

Figure Showing Buckling Using Lateral LoadBeam 188

Figure Showing Buckling Using Lateral LoadBeam 189


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Figure Showing Buckling Using Lateral LoadShell 63

Figure Showing Buckling Using Lateral LoadShell 93

Study of effect of holes in the cantilever beam

Modeling Procedure

The beam is modeled using shell 63 and shell 93 element and the six holes where made into the beam and then the

basic post processor would be Eigenvalue Analysis. Here two variation were made . a case of when subjected to

axial load and when subjected to lateral load.

Case 5 Effect of Holes in cantilever beam using shell 63 element and shell 93 element when subjected to axial

load

When Length is 120 inch Using Various Element Shell element 63 and Shell 93 element. AnalysisEigen value Analysis. Six hexagonal holes ( h= 4 inch)


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Comparison of Shell Element when subjected to axial load with and without effect of holes of ansys value

Element

Ansys Pcr

With Hole

Ansys Pcr

without Hole Comments

Shell 63 65.55 71.994 Results are Better

when used Shell 93Shell 93 69.051 71.523

Interpretation

1. In cantilever beam when holes are made then its eulers buckling load decreases for both the Shell element.2. Shell 93 elements results are better because we can see that effect of holes has not taken much effect in the

buckling load.

3. We can notice the behavior of the buckling for the shell 93 element as shown below

Figure showing buckling behavior of shell 93 element with and without hole when subjected to axial load

Case 6 Effect of Holes in cantilever beam using shell 63 element and shell 93 element when subjected to

Lateral Load

When Length is 120 inch Using Various Element Shell element 63 and Shell 93 element. AnalysisEigen value Analysis. Six hexagonal holes ( h= 4 inch) Loading is lateral load

Comparison of Shell Element when subjected to lateral load and study the effect of with and without effect of holes

of ansys value

Element

Ansys Pcr

With Hole

Ansys Pcr

without Hole Comments

Shell 63 25.41 25.4072 Results were suprising i

got better result for shell

63 but cannot interpreteShell 93 31.98 32.98

Interpretation

1. In cantilever beam when holes are made then its eulers buckling load Shell 63 elements I was not able tointerprete it

2. results are better because we can see that effect of holes has not taken much effect in the buckling load.


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3. We can notice the behavior of the buckling for the shell 63 element as shown below

Figure showing buckling behavior of shell 63 element with and without hole when subjected to lateral load

Case 7 Effect of Number of Modes

Model Description

Here i tried element Beam 189 and made a model of cantilever beam subjected to axial loading model was made in

same fashion as previously describe for the cantilever beam but now i made variation in the method as subspace

method and Block lancoz method and changing number of modes to 3,4 and 5 and try to study the behavior due tochange in the modes. Reason of trying with one element is to notice the difference is coming for one element then it

should come with other element also

When Length is 120 inch Using Element Beam 189 AnalysisEigen value Analysis. Loading is Axial load.

Comparison of number modes and two different method for eigenvalue analysis

Element No. of Modes

Ansys Pcr using Subspace

method

Ansys Pcr using Block

Lancoz method

Beam 189

3 71.635 71.635

4 71.635 71.635

5 71.635 71.635

Interpretation


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1 Compared and obtained result that there is no changes in the eigenvalue due to change in number of modes.

Behavior forpldisp,1 remains same if I simulate with 3 modes or 4 modes or 5 modes.

2 Even I tried to change the method to notice some difference but both the method gave same result so we can say

these the method which we solve will play importance in the modal analysis and that to we can add some equation

solver to see its effect to change in mode shape. So buckling there is no significance changing methods for analysis.

Figure showing buckling behavior of Beam 189 element when subjected to Axial load using both method

Case 8 Effect of Shift Variables

Model Description

Here i tried element Beam 188 and made a model of cantilever beam subjected to axial loading model was made in

same fashion as previously describe for the cantilever beam but now i made variation in the method as subspace

method and Block lancoz method and changing number of Shifts in the command to 0,0.5 and 1 and try to study the

behavior due to change in the modes. Reason of trying with one element is to notice the difference is coming for one

element then it should come with other element also

When Length is 120 inch Using Element Beam 188 AnalysisEigen value Analysis. And notice the difference in behavior due to shift variable Loading is axial load.

Comparison using Shift variable and two different method for eigenvalue analysis

Element

Shift

Variable

Ansys Pcr using Subspace

method

Ansys Pcr using Block

Lancoz method Comments

Beam 188

0 71.635 71.635

Effect of Shift

Variable is notice

in SubspaceMethod

0.5 218.34 71.635

1 218.34 71.635


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Interpretation

1. When I command I tried with changing in the shift variables using block lancoz method I did not notice anychanges in the buckling value.

2. But when I tried with subspace method I notice the difference when I put the shift variable as 0.5 theeigenvalue changes and it takes more load to buckle.

3.

What is shift variable when we plot a graph of load vs displacement shift variable is assigned using symbol() we can see below how it shifts the value.

4. Genrally effect of shift variable is due to complexity in the models.

Non Linear Analysis

Model description

Cantilever beam using all 5 element type was used to compute the eigenvalue using Non linear analysis for that

model remains same only the post processor changes. The difference is that we put the Eulers Buckling Load and

we factored it by two times and then we put that into the command and monitor the graph and notice the

displacement.

The main modeling is how we apply lateral load

Element Apply Load Weak axis Plot Displacement

Beam 4 Y direction small lateral

Load

Z-Z axis is weak axis Uy

Beam188 Z direction small lateral

load

Y-Y axis is weak axis Uz

Beam 189 Z direction small lateral

load


Shell 63 Z direction small lateral

load


Shell 93 Z direction small lateralload


Modeling of the model all different elements was challenge it all depends on the orientation how we place the beam

but from mechanics of material we know that the bending axis is z-z it is the weakest axis.

Method used here is Newton Raphson Method

Comparison of Eigenvalue obtained from Set list using non linear approach

Element

TheoreticalPcr

in Kips

Ansys Pcr in

Kips

% of Comparison ofTheoretical with

Ansys

Beam 4 72.47 71.35 98.45453291

Beam 188 72.47 70.6 97.41962191

Beam 189 72.47 70.6 97.41962191

Shell 63 72.47 72.44 99.95860356

Shell 93 72.47 72.18 99.59983441


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Figure Showing Plot of Load vs deflection for Beam 4

Figure Showing Plot of Load vs deflection for beam 189

Figure Showing Plot of Load vs deflection for beam 188


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Figure Showing Plot of Load vs deflection for Shell 63 Element

Figure Showing Plot of Load vs deflection for Shell 93 Element (Uz)

Interpretation

1. Nonlinear buckling analysis is more accurate than eigenvalue analysis because it employs non-linear, large-deflection; static analysis to predict buckling loads. Its mode of operation is very simple: it gradually

increases the applied load until a load level is found whereby the structure becomes unstable (ie. suddenly a

very small increase in the load will cause very large deflections). The true non-linear nature of this analysisthus permits the modeling of geometric imperfections, load perterbations, material nonlinearities and gaps.

For this type of analysis, note that small off-axis loads are necessary to initiate the desired buckling mode.

2. As we can see from the graph of shell element and beam element the value which I obtained theoreticallyfrom eulers buckling load as the same load we can see that there is an buckling indication (means

displacement is noticed in the graph).

3. The value which I obtained from eigenvalue for shell element where to high but it is figured out in the nonlinear analysis as it does not account much for model imperfection it gives better result compare to

eigenvalue. Almost wecan read from the graph the value at which the buckling starts and afterwards it

gives high displacement.


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Displacement figure during buckling

Figure showing Displacement at time Buckling Figure showing Displacement at time Takes PlaceForBeam 188 (Uz) Buckling takes place- For Beam189 (Uz)

Figure showing Displacement at time Buckling Figure showing Displacement at time Takes PlaceFor Shell63 (Uz) Buckling takes place- For Shell 93 (Uz)


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Stresses Study

It was interesting to study stresses when the Buckling takes place for that I used the same model cantilever beam

using all the 5 element type when we do nonlinear analysis model the point where the buckling starts I noted the

stresses at that point where I went through all the stresses all 14 type of stresses I found the maximum among which

gave the stresses for beam was Von Mises Stress.

Figure showing Stresses at time Buckling takes Figure showing Stresses at time Buckling takes

PlaceFor Shell 93 (von mises Stress) Place- For Shell 63 (Von mises Stress)

Figure showing Stresses at time Buckling takes Figure showing Stresses at time Buckling takes

PlaceFor Beam 188 (von mises Stress) Place- For Beam 189 (Von mises Stress)


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Interpretation

When Stresses study are made it is important to know which one gives maximum stresses in the case where I find

the stresses for the Beam element an d shell element what I observe that maximum stresses we can see from figure .

The Comparison between element we do then the Shell element gave more stresses in von mises and where as stress

by beam element was less on the comparison.


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3 Modal analysisModal Analysis is the study of dynamic properties of structures under vibrational excitation . The main purpose to

study modal analysis in the beam is to understand it behaviors.

We can determine Natural Frequency, Mode Shapes and Mode participation factors means how much a given mode

participates in a given direction.

Basically I followed Method and try to obtained result for both the method

1. Subspace Method2. Reduced MethodBasic Information about model which I used

Material properties

Modulus of Elasticity: E =30000 Ksi Poisson ratio =0.3 Density of Steel= 490 lb/ft3


Thickness in Y direction (ty) = 0.44 inch Thickness in Z direction (tz) = 0.3 inch Moment of Inertia in z-z direction (Izz) = 14.1 inch 4 Moment of Inertia in y-y direction (Iyy) = 144 inch4 Area of Cross section = 7.61 inch2

Study of Modal Analysis and its comparison with various Method

1 Cross section used as defined.

2 Elements used is Beam 4 and Mass 21

3 Study of Natural Frequency.

Model Descriptions

1. Beam 4 Element using Subspace method was model using the cross sectional properties and followed andtwo key points were used to define the length of the beam..which is L=120 inch.

2. I use a case of Beam as a Cantilever beam subjected with two concentrated masses in order to carry out themodal analysis it is important to find the weight of the beam and Moment of inertia for the masses. Below

show the calculation for the beam carried out which was given in the command

E= 30000 ksi

nu =0.3

Gfactor=1


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Gmod= Gfactor*E/(2*(1+nu)) = 1*30000/(2*(1+0.33)) =11278.196

g =386

A = 7.61 inch2, Iyy =144 inch2 Izz=14.1 inch2

Specific Gravity = Weight/ Force = Iron density/ Water Density = 7850/1000 =7.85

Wt density = 7.85 *62.4/1728 =0.28167

= wt.density/g = 0.28167/386 =7.29 *10-4

Wt. Beam = *A*g =7.29 * 10 -4 *7.61*386 = 2.143

Mass m1 = 0.25/g Assumed

Mass m2 = 0.25/g. Assumed

Calculation of Moment of inertia

Z1= m1 (x12 + y1

2)/12, Z2= m2 (x22 + y2

2)/12

Im1= Z1 +m1*(x1/2)2 Im2= Z2 +m2*(x2/2)

2

X1 = 2, Y1 = 1, X2 = 2, Y2= 1.Assumed

3. Element use to define mass was Mass 21 element and I assign the location at the tip and the mid node ofbeam 4 element. Mass 21 element was created using real constant

4. Now its time to add the post processor to obtained the desired result so I used subspace method to simulatemy first result to determine the natural frequency and obtain the mode shape.

Theoretical Calculation of Natural Frequency obtained from the Structural dynamics by Roy R craig.

n= n2(EI/mL4)

n= 1.875 ..To obtain mode 1

n= 4.694To obtain mode 2

n= 7.844To obtain mode 3

(Dependent on *Mass,*Length and *Moment Inertia)

Total Mass = (Mass of Beam + Mass of m1 and Mass of m2) = (2.643/g)

Length =120 inch

Izz= Weak axis of moment of inertia from mechanics of materials = 144.1 inch 4.


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Ansys Model

Figure Showing Beam 4 element

used and obtained-Mode 1 and Mode 2 subspace method

Table Showing Comparision of Natural Frequency of Theory with Ansys when use subspace method

Element Mode

Theory

Natural

Frequency

Ansys

Natural

Frequency % Ansys Accuracy

Beam 4

1 5.2683 5.1 96.81

2 35.08068 33.96 96.81

3 245.5623 238.41 97.09

Interpretation

1. Result obtained from the ansys for natural frequency is not so accurate may be I am missing someconcept.

2. But it is closer by from here onwards we can know add forces and notices the behavior and notice thetransient dynamic analysis for the same model.

Case 2 Using Reduced Method

1 Cross section used as defined.

2 Elements used is Beam 4 and Mass 21

3 Study of Natural Frequency


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Model Descriptions

1. Beam 4 Element using Reduce method was model using the cross sectional properties and followed andtwo key points were used to define the length of the beam. which is L=120 inch.

2. Only difference is that now we add Master degree of freedom to the beam and use the post processor forreduced method. . These are degrees of freedom that govern the dynamic characteristics of a structure.

3. Rest everything remain same as we have defined earlier. And let us find the result

Figure Showing Beam 4 element

used and obtained-Mode 1 and Mode 2 by Reduced method

Table Showing Comparision of Natural Frequency of Theory with Ansys when use Reduced method

Element Mode

Theory

Natural

Frequency

Ansys

Natural

Frequency % Ansys Accuracy

Beam 4

1 5.2683 5.1 96.81

2 35.08068 33.96 96.81

3 245.5623 238.41 97.09


28/87

28

Interpretation

1. Result obtained from the ansys for natural frequency is not so accurate may be I am missing someconcept.

2. But it is closer by from here onwards we can know add forces and notices the behavior and notice thetransient dynamic analysis for the same model.

3. Result are dot same for both the method reason there is not so complexity in the model the model is notso complex if model becomes then we can notice the difference in the reduced method.

4. It is better idea if in the execution we have complex structure it is better from the Ansys to calculateusing reduced method obtained the analysis.

5. Subspace method and reduced method are frontal solver. But Subspace method is much slower thenreduced method.


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29

4 Transient Dynamic Analysis

Transient dynamic analysis is a technique used to determine the dynamic response of a structure under a time-

varying load. The time frame for this type of analysis is such that inertia or damping effects of the structure are

considered to be important. Cases where such effects play a major role are understep orimpulse loading

conditions. For my case, we will impact the end of the beam with an impulse force and view the response at thelocation of impact.

Figure Showing Example how I will apply impulse force at certain time and release after certain time

Since an ideal impulse force excites all modes of a structure, the response of the beam should contain all mode

frequencies. However, we cannot produce an ideal impulse force numerically. We have to apply a load over a

discrete amount of time dt

After the application of the load, we track the response of the beam at discrete time points for as long . The smaller

the time step, the higher the mode frequency we will capture.

The main Fundamental is

Time = (1/No of discrete point * Highest mode of frequency)

Material properties

Modulus of Elasticity: E =30000 Ksi Poisson ratio =0.3 Density of Steel= 490 lb/ft3


30/87

30


Thickness in Y direction (ty) = 0.44 inch Thickness in Z direction (tz) = 0.3 inch Moment of Inertia in z-z direction (Izz) = 14.1 inch 4 Moment of Inertia in y-y direction (Iyy) = 144 inch4 Area of Cross section = 7.61 inch2

Method used

The Full Method

All types of non-linearity are allowed. It is dependent on spacing in the disk to use full system matrices. This is the

easiest method to use.

The Mode Superposition Method

This method requires a preliminary modal analysis, as factored mode shapes are summed to calculate the structure's

response. It is the quickest of all the methods.

Case1 Cantilever with two concentrated masses

Modeling Procedure

1 Same Model is used which was modeled for modal analysis as I am performing mode of superpositionmethod so I need the factored mode shape from the modal analysis.

2 Damping Method #2 is added zeta , alpha and beta was added as Beam 4 supports most of theapplication . Like just for discussion it has gyroscopic effect which can be added as a real constant.

3

Now here I applied Impulse force at the tip of beam at time =0.01 and widraw that force at time 0.02

4 Now in /post26 we can note the dynamic response at the time 0.005, we can note the displacement in ydirection at the tip and middle to effect of impulse force.

Figure Showing at the Substep 1 and 20


31/87

31

Table Showing Load step and Sub step and Time

Set Time

Load

Step Sub Step

1 4.42E-03 1 120 0.1001 1 20

Figure Showing Time vs Displacement at the tip and Middle

Interpretation

1 We can see the plot of Uy displacement at the position of the tip and Middle. As we can see that tipdisplacement is more at time 0.1 sec and mid displacement is less at same time.

2 Like if we imagine the beam when force is applied at the tip at one time we can see that the beam displacesmore at the tip and less at the middle portion

3 The Behavior of beam can be known clearly how beam behave under loading condition.

Case 2 A cantilever beam has attached masses. One mass is free to hop or Leave the surface

Case Description

Two concentrated masses are affixed to a cantilever beam. A third mass is resting on the top of the beam. A hammer

Strike is applied at the beam free end. The load is of the form of a triangle. The object is to investigate the possibility

of a hop of the third mass which is connected to the beam using contact12 elements. Springs elements, Combin14,

are used to simulate the attachment to beam of hopping mass.


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32

Modeling Procedure

1. Element chosen to built the Beam is Beam4 then i assign the geometric properties of the Cylinder whichare going to be attached to the beam.

2. Then I assigned the various elements like Mass21 for Masses, Contac12 to give contact surface to thecylinder and the Combin 14 in Uy dof and Ux dof that is Spring element

3.

These all element were generate by the real constant as shown below.

Real Constant Assign Properties Purpose

r1 Area, Izz, Tz,Tx,h Beam Model Creation

r2 mass1,Izz1 Accelerometer#1-Cylinder

r3 mass2,Izz2 Accelerometer#2-Cylinder

r4 mass3,Izz3 Free Cylinder

r5 Uy Dof Contact element

r6 Ux Dof Spring attached to accelerometer

to the Beam

r7 thk Spring for fastening free cylinder.

4 Accelerometer #1 & #2, Free cylinder, springs, Contact element were assigned with respect to the keypoint.5 After these is done assigning a boundary condition to its6 Now hammer Load is applied at the tip of the beam at time t=0.01 sec and removed at time t=0.02 and

effect are noted till time t=0.5 sec.

7 In the post processor Command was given to plot y displacement at 2 location.

Figure showing Front view of the model

Figure Showing Eplot of the Beam Figure showing at last sub step where can see Hop of Free mass


33/87

33

Figure Showing Uy Plot for beam displacement at the tip and mid accel

Interpretation

1 The Graph of Y displacement looks reasonable was lumpm,on ( Lumped mass option). Thedisplacement shows between time 0.01 sec to 0.075 sec after that there is a constant line of Uy these

signifies that as a hammer load was released at time 0.02 The load impact remains till 0.075 sec only

after that it stabilize.

2 These result may be very important design firm to note the response based on the loading which wasgiven and look at the behavior of Beam.

3 As we can see that the mass which was kept free on the third cylinder it hop out of the cylinder asshown above in one of the figure.

Case 3 A Block falls on the Cantilever Beam

Case description: A Blocks Falls on the cantilever beam and study of displacement is noted at each position of the

beam.

Modeling Procedure

1 Model was Prepared using Beam 4 element for Beam structure, Plane 42 element for the Block, Contac171was given between Beam and Block as Block strikes the beam so targe169 element was taken in to

consideration.2 With Real constant Block and Beam were made. Block was made at the centre of beam . At the centre of

the Beam in order to give stability to the structure the structure was Constrained at centre.

3 Method used for Transient Analysis is full method4 Modeling was very simple inafe here Block Falls after some time and effects is studied.


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Figure showing eplot Figure Showing Block Position Load Step3 and Sub Step10

Figure Showing Y displacement at Node at Various Nodes

Interpretation

1 The Behavior of the Beam can notice at various time when Block Falls on the beam we can see at time0.28 sec where all the nodes have almost same displacement even it is a peak point of the graph where n

maximum displacement take place.

2 At different time we noted different displacement of the beam . from all of the study main purpose wouldbe that we full fill our design criteria..


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Conclusion

1 I study the behavior of the beam using various analysis. The main aim of doing BucklingAnalysis, Modal analysis and Dynamic Analysis was to understand the behavior of theBeam under each analysis.

2 From The Buckling Analysis, I carried out Eigenvalue analysis we have seen thebuckling in beams takes place at the same load one which is given by Eulers Buckling

Load.3 Spotted that if complexity in the model is involved ansys solve easily using the shift

variables.

4 Non Linear Study was made and it gave better result than the Eigenvalue analysis as inthat geometric imperfection is not much of effect.

5 Modal analysis was carried out where I tried and spotted that the natural frequency givenby ansys where slightly lower than I calculate by the theoretical formula.

6 In Modal Analysis we came to know the natural frequency of the beam. I came to knoweven while comparing between two methods if the complexity involved in the model

ansys can solve complex structure using reduced method.

7 In Transient Dynamic Analysis, I several cases were tried applying an impulse force forcertain time and know the response of structure due to that impulse force.

Collapse Structure of WTC- we dont want to see these again

Being a Civil Engineers it is my duty to Analyze the structure Correctly and design it

correctly based on that analysis


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References

1. Concepts and applications of Finite Element Analysis 4th Edition by Robert D Cook,

David S. Malkus, Michael E. Plesha, Robert J. Witt.

2. Structural Analysis- 3rd Edition by Aslam Kassimali3. Structural Steel Design- 1st Edition by Abi Aghayere and Jason Vigil4. Design of Steel Structure- 3rd Edition by Edwin H Gaylord5. Stability of Structure- 2nd edition by Dr. Eric m Lui and Chen6. Theory of Structures By Timoshenko, Stephen.7. University Alberta Ansys help

http://www.mece.ualberta.ca/tutorials/ansys/IT/Transient/Transient.html


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Appendix

Beam 4 Eigenvalue buckling of a an I-shaped solid beam.

!

!Eigenvalue buckling of a an I-shaped solid beam.

!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam

/title,Buckling of solid cantilever beam- BEAM4

/prep7/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam4

!keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

!mp,gxy,1,1e5

!mp,gxz,1,1e5

! GEOMETRY FORMATION

! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120

a=7.61

Izz=14.1

Iyy=144

r,1,a,Izz,Iyy

k,1,

k,2,120

l,1,2


38/87

38

LESIZE,1,,,4

lmesh,1

! Boundary Conditions

d,1,all

!f,2,fy,1 !Apply a small lateral load

! Display Commands

/pbc,f,1

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static

pstres,on !Calculate prestress effects

outres,all,alld,1,all

f,2,fx,-1 !Initial small load

save

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,buckle !Perform eigenvalue buckling analysis

bucopt,subsp,3 !Use subspace method, extract 3 modes

savesolve

finish

/solu

expass,on

mxpand,3 !Expand 3 mod

outres,all,all

solve

finish

/post1

set,list

set,1,1 !Select 1 or 2 mode

/eshape,1

pldisp,1 !Plot mode shape


39/87

39

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!

!Eigenvalue buckling of a an I-shaped solid beam.!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam

/title,Buckling of solid Cantilever beam - BEAM188

/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3et,1,beam188

keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

!mp,gxy,1,1e5

!mp,gxz,1,1e5


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120

k,1,

k,2,120

k,3,0,4

k,4,0,-4

l,1,2

LESIZE,1,,,10SECTYPE,1,BEAM,I,W10x26

SECDATA,bf,bf,dg,tf,tf,tw

SECNUM,1

LATT,1,,1,,3,4,1

lmesh,1



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40

d,1,all


! Display Commands

/pbc,f,1

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

d,1,all


savesolve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu


bucopt,lanb,3,0.5 !Use subspace method, extract 3 modes

save

solve

finish

/soluexpass,on

mxpand,3,,,YES !Expand 3 mod

outres,all,all

solve

finish

/post1

set,list


/eshape,1



!


41/87

41


!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam


/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam189

keyopt,1,8,2

keyopt,1,9,2

mp,EX,1,exxmp,nuxy,1,0.3

!mp,gxy,1,1e5

!mp,gxz,1,1e5

! GEOMETRY INFORMATION

! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120k,1,

k,2,120

k,3,0,4

k,4,0,-4

l,1,2

LESIZE,1,,,10

SECTYPE,1,BEAM,I,W10x26


SECNUM,1

LATT,1,,1,,3,4,1

lmesh,1


d,1,all


! Display Commands

/pbc,f,1


42/87

42

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

d,1,all


save

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/soluantype,buckle !Perform eigenvalue buckling analysis

bucopt,subsp,3,0.5 !Use subspace method, extract 3 modes

outres,all,all

save

solve

finish

/solu

expass,on


outres,all,allsolve

finish

/post1

set,list


/eshape,1


Buckling of a cantilever beam using shell63

/filnam,cantileverbucklingshell63

/title,Buckling of solid cantilever beam - shell63

/prep7

/plopts,logo,0


43/87

43

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,Shell63

mp,ex,1,exx

mp,nuxy,1,nu


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

r,1,tw

r,2,tflb=120

k,1,,dg/2

k,2,lb,dg/2

k,3,lb,-dg/2

k,4,0,-dg/2

k,5,0,dg/2,-bf/2

k,6,0,dg/2,bf/2

k,7,lb,dg/2,bf/2

k,8,lb,dg/2,-bf/2

k,9,0,-dg/2,-bf/2k,10,0,-dg/2,bf/2

k,11,lb,-dg/2,bf/2

k,12,lb,-dg/2,-bf/2

a,1,2,3,4

a,1,2,7,6

a,1,2,8,5

a,3,4,10,11

a,3,4,9,12

k=12

f=4

w=8

lsel,s,Length,,lb

lesize,all,,,k

lsel,s,Length,,bf/2

lesize,all,,,f

lsel,s,Length,,dg


44/87

44

lesize,all,,,w

type,1

real,1

amesh,1

type,1

real,2

amesh,2,3

type,1

real,2

amesh,4,5


nsel,s,loc,x,0

d,all,all

/pbc,f,1

/pbc,u,1nsell,all

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

!nsel,s,loc,x,lb

!nsel,r,loc,z,0!f,all,fx,-1/(w+1) !Equal Initial small load at center line only

!f,all,fx,-1/(4*f+w+3) !Equal Initial small load at all tip nodes

!**LOADING APPLIED TO THE FLANGE AND THE WEB PROPORTIONAL TOCORRESPONDING AREA

nsel,s,loc,x,lb

nsel,r,loc,z,0

f,all,fx,-.3/(w+1) !WEB

nsel,s,loc,x,lb

nsel,r,loc,Y,DG/2

f,all,fx,-.35/(2*F+1) !UPPER FLANGEnsel,s,loc,x,lb

nsel,r,loc,Y,-DG/2

f,all,fx,-.35/(2*F+1) !LOWER FLANGE

nsell,all

NODE1=NODE(LB,-DG/2,0)

f,NODE1,fx,(-.35/(2*F+1))+(-.3/(w+1))


45/87

45

NODE2=NODE(LB,DG/2,0)

f,NODE2,fx,(-.35/(2*F+1))+(-.3/(w+1))

nsell,all

save

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu



save

solve

finish

/solu

expass,onmxpand,3 !Expand 3 mod

outres,all,all

solve

finish

/post1

set,list


/eshape,1pldisp,1 !Plot mode shape

!Eigenvalue BUCKLING PROBLEM

/filnam,SOLID BEAM

/title,Buckling of solid Cantilever Shell 93

/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3et,1,Shell93

mp,ex,1,exx

mp,nuxy,1,nu


! A W10x26

bf=5.75


46/87

46

dg=10.3

tf=0.44

tw=0.26

r,1,tw

r,2,tf

!keyopt,1,3,2

!e=156.25

!b=112.5

!ho=375

!dt=(dg-ho)/2

lb=120

k,1,,dg/2

k,2,lb,dg/2

k,3,lb,-dg/2

k,4,0,-dg/2k,5,0,dg/2,-bf/2

k,6,0,dg/2,bf/2

k,7,lb,dg/2,bf/2

k,8,lb,dg/2,-bf/2

k,9,0,-dg/2,-bf/2

k,10,0,-dg/2,bf/2

k,11,lb,-dg/2,bf/2

k,12,lb,-dg/2,-bf/2

a,1,2,3,4

a,1,2,7,6a,1,2,8,5

a,3,4,10,11

a,3,4,9,12

k=12

f=4

w=8

lsel,s,Length,,lb

lesize,all,,,k

lsel,s,Length,,bf/2

lesize,all,,,f

lsel,s,Length,,dg

lesize,all,,,w

type,1

real,1

amesh,1

type,1


47/87

47

real,2

amesh,2,3

type,1

real,2

amesh,4,5


nsel,s,loc,x,0

d,all,all

/pbc,f,1

/pbc,u,1

nsell,all

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/soluantype,static


outres,all,all

!nsel,s,loc,x,lb

!nsel,r,loc,z,0

!f,all,fx,-1/(w+1) !Equal Initial small load at center line only

!f,all,fx,-1/(4*f+w+3) !Equal Initial small load at all tip nodes

!**LOADING APPLIED TO THE FLANGE AND THE WEB PROPORTIONAL TOCORRESPONDING AREA

nsel,s,loc,x,lbnsel,r,loc,z,0

f,all,fx,-.3/(2*w+1) !WEB

nsel,s,loc,x,lb

nsel,r,loc,Y,DG/2

f,all,fx,-.35/(4*F+1) !UPPER FLANGE

nsel,s,loc,x,lb

nsel,r,loc,Y,-DG/2


nsell,all

NODE1=NODE(LB,-DG/2,0)f,NODE1,fx,(-.35/(4*F+1))+(-.3/(2*w+1))


f,NODE2,fx,(-.35/(4*F+1))+(-.3/(2*w+1))

nsell,all

save

solve


48/87

48

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu



save

solve

finish

/solu

expass,on

mxpand,3 !Expand 3 mod

outres,all,all

solve

finish

/post1set,list



/filnam,catellatedbeam

/title,Buckling of Castellated Beam

/prep7

/plopts,logo,0

/plopts,date,0

antype,static

antype,static

exx=30000

nu=0.3

et,1,Shell93

mp,ex,1,exx

mp,nuxy,1,nu


!###########################e=8

b=2

h=10

d=e

!########################

tf=0.44


49/87

49

tw=0.26

r,1,tw

r,2,tf

bf=5.75

dg=10.3+h/2

L=E+2*B+D

k,1,

k,2,L

k,3,L,dg

k,4,,dg

k,7,0,0,-bf/2

k,8,0,0,bf/2

k,9,L,0,bf/2

k,10,L,0,-bf/2

k,11,0,dg,-bf/2k,12,0,dg,bf/2

k,13,L,dg,bf/2

k,14,L,dg,-bf/2

a,1,2,3,4 ! WEB AREA

a,1,2,10,7

a,1,2,9,8

a,4,3,13,12

a,4,3,14,11

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@! CASTELLATION OF BEAM WEB

dt=(dg-h)/2

K,15,d/2+b,dt,

k,16,d/2,dg/2,

k,17,d/2+b,dg-dt,

k,18,d/2+e+b,dg-dt,

k,19,d/2+e+b+b,dg/2

k,20,d/2+e+b,dt

a,15,16,17,18,19,20

ASBA,1,6,,

AGEN,6,all,,,d+e+2*b

NUMMRG,KP

REAL,1

AMESH,ALL

ASEL,S,LOC,Y,DG

REAL,2


50/87

50

aCLEAR,ALL

REAL,2

AMESH,ALL

ASEL,S,LOC,Y,0

REAL,2

aCLEAR,ALL

REAL,2

AMESH,ALL

ASEL,ALL


nsel,s,loc,x,0

d,all,all

/pbc,f,1

/pbc,u,1

nsell,allsave

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

!**LOADING APPLIED TO THE FLANGE AND THE WEB PROPORTIONAL TO

CORRESPONDING AREAnsel,s,loc,x,6*L

nsel,r,loc,z,0

f,all,fx,-.3/23 !WEB

nsel,all

f,3485,fy,0

f,3486,fy,0

f,3537,fy,0

nsel,s,loc,x,6*L

nsel,r,loc,Y,DG

f,all,fx,-.35/9 !UPPER FLANGEnsel,s,loc,x,6*L

nsel,r,loc,Y,0

f,all,fx,-.35/9 !LOWER FLANGE

nsell,all

NODE1=NODE(6*L,0,0)

f,NODE1,fx,(-.35/9-.3/23)


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51

NODE2=NODE(6*L,DG,0)

f,NODE2,fx,(-.35/9-.3/23)

nsell,all

save

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu



save

solve

finish

/solu

expass,onmxpand,3 !Expand 3 mod

outres,all,all

solve

finish

/post1

set,list



Non Linear Analysis of the I beam- Beam 4

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!


!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam


/prep7/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam4


52/87

52

!keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

!mp,gxy,1,1e5

!mp,gxz,1,1e5

Pcr=72.47

Fmax=2*Pcr

L=120


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120a=7.61

Izz=14.1

Iyy=144

r,1,a,Izz,Iyy

k,1,

k,2,120

l,1,2

LESIZE,1,,,4

lmesh,1

! Boundary Conditionsd,1,all

f,2,fy,1 !Apply a small lateral load

! Display Commands

/pbc,f,1

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

solcontrol,on

monitor,1,2,uy

monitor,2,2,fy

monitor,3,2,fx

/gst,on

nropt,auto !Solution method is selected


53/87

53

nlgeom,on

sstif,on

autots,on

pred,on !Activates a predictor in nonlinear analysis

outres,all,all*do,i,1,10

time,i*Fmax/10

f,2,fx,-(i*Fmax/10)

lswrite,i

*enddo

save

lssolve,1,10,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@@@@@@@@@@@@@@@@@

!Postprocessing:

!

!In post1 use set,list to list loads sets.

!Use /dscale,window,multiplier to magnifiy displacement plot as

!desired. Eg., /dscale,1,100 will magnify the plot by 100.

!Use pldisp,1 to observe deflection shape for sequential load steps.

!

!Use post26 to get the time history plot. See below.

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@@@@@@@@@@@@@@@@@

/post26

nsol,2,2,u,y,uy2

prod,3,1,,,LOAD,,,Fmax

xvar,3

/axlab,x,Total Load, lbs

/axlab,y,Displacement, inch

plvar,2


!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@


54/87

54

!

!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam

/title,Non Linear Buckling of solid Cantilever beam - BEAM188

/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam188

keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3!mp,gxy,1,1e5

!mp,gxz,1,1e5

Pcr=72.47

Fmax=2*Pcr

L=120


! A W10x26

bf=5.75

dg=10.3

tf=0.44tw=0.26

lb=120

a=7.61

Izz=14.1

Iyy=144

r,1,a,Izz,Iyy

k,1,

k,2,120

k,3,0,4

k,4,0,-4

l,1,2

LESIZE,1,,,10



SECNUM,1

LATT,1,,1,,3,4,1


55/87

55

lmesh,1


d,1,all

f,2,fz,1 !Apply a small lateral load

! Display Commands

/pbc,f,1

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

solcontrol,on

monitor,1,2,uz

monitor,2,2,fz

monitor,3,2,fx/gst,on


nlgeom,on

sstif,on

autots,on


outres,all,all

*do,i,1,10

time,i*Fmax/10

f,2,fx,-(i*Fmax/10)lswrite,i

*enddo

save

lssolve,1,10,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!Postprocessing:

!





!


!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@


56/87

56

@@@@@@@@@@@@@@@@@@@@@@@@@@@

/post26

nsol,2,2,u,z,uz2


xvar,3



plvar,2


!

/filnam,Solid Ibeam

/title,Non Linear Buckling of solid cantilever beam- BEAM189

/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam189

keyopt,1,8,2

keyopt,1,9,2

mp,EX,1,exx

mp,nuxy,1,0.3!mp,gxy,1,1e5

!mp,gxz,1,1e5

Pcr=72.47

Fmax=2*Pcr

L=120

! GEOMETRY INFORMATION

! A W10x26

bf=5.75dg=10.3

tf=0.44

tw=0.26

lb=120

a=7.61

Izz=14.1


57/87

57

Iyy=144

r,1,a,Izz,Iyy

k,1,

k,2,120

k,3,0,4

k,4,0,-4

l,1,2

LESIZE,1,,,10



SECNUM,1

LATT,1,,1,,3,4,1

lmesh,1

! Boundary Conditionsd,1,all

f,2,fz,1 !Apply a small lateral load

! Display Commands

/pbc,f,1

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,staticpstres,on !Calculate prestress effects

outres,all,all

d,1,all


save

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

solcontrol,on

monitor,1,2,uz

monitor,2,2,fz

monitor,3,2,fx

/gst,on


nlgeom,on


58/87

58

sstif,on

autots,on


outres,all,all

*do,i,1,10

time,i*Fmax/10

f,2,fx,-(i*Fmax/10)

lswrite,i

*enddo

save

lssolve,1,10,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!Postprocessing:!





!


!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/post26nsol,2,2,u,z,uy2


xvar,3



plvar,2

!Non Linear Buckling of a cantilever beam using shell63


/title,Buckling of solid cantilever beam - shell63/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3


59/87

59

et,1,Shell63

mp,ex,1,exx

mp,nuxy,1,nu

Pcr=72.47

Fmax=2*Pcr

L=120


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

a=7.61

Izz=14.1

Iyy=144

lb=120

r,1,tw

r,2,tf

k,1,,dg/2

k,2,lb,dg/2

k,3,lb,-dg/2

k,4,0,-dg/2

k,5,0,dg/2,-bf/2

k,6,0,dg/2,bf/2

k,7,lb,dg/2,bf/2

k,8,lb,dg/2,-bf/2

k,9,0,-dg/2,-bf/2

k,10,0,-dg/2,bf/2

k,11,lb,-dg/2,bf/2

k,12,lb,-dg/2,-bf/2

a,1,2,3,4

a,1,2,7,6

a,1,2,8,5

a,3,4,10,11


60/87

60

a,3,4,9,12

k=12

f=4

w=8

lsel,s,Length,,lb

lesize,all,,,k

lsel,s,Length,,bf/2

lesize,all,,,f

lsel,s,Length,,dg

lesize,all,,,w

type,1

real,1

amesh,1

type,1

real,2amesh,2,3

type,1

real,2

amesh,4,5


nsel,s,loc,x,0

d,all,all

/pbc,f,1

/pbc,u,1nsell,all

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

!nsel,s,loc,x,lb

!nsel,r,loc,z,0

!f,all,fz,-1/(w+1) !Equal Initial small load at center line only

!f,all,fz-1/(4*f+w+3) !Equal Initial small load at all tip nodes!**LOADING APPLIED TO THE FLANGE AND THE WEB PROPORTIONAL TO

CORRESPONDING AREA

nsel,s,loc,x,lb

nsel,r,loc,z,0


61/87

61

f,all,fx,-.3/(w+1) !WEB

nsel,s,loc,x,lb

nsel,r,loc,Y,DG/2

f,all,fx,-.35/(2*F+1) !UPPER FLANGE

nsel,s,loc,x,lb

nsel,r,loc,Y,-DG/2


nsell,all


f,NODE1,fz,(.35/(2*F+1))+(.3/(w+1))


f,NODE2,fz,(.35/(2*F+1))+(.3/(w+1))

nsell,all

save

solvefinish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

solcontrol,on

monitor,1,2,uz

monitor,2,2,fz

monitor,3,2,fx

/gst,on

nropt,auto !Solution method is selectednlgeom,on

sstif,on

autots,on


outres,all,all

*do,i,1,10

time,i*Fmax/10

f,2,fx,-(i*Fmax/10)

lswrite,i

*enddo

save

lssolve,1,10,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!Postprocessing:


62/87

62

!





!


!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@@@@@@@@@@@@@@@@@@@

/post26

nsol,2,2,u,z,uy2


xvar,3



plvar,2

Lateral Buckling

!Lateral Buckling


!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam


/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam188

keyopt,1,3,2

mp,EX,1,exxmp,nuxy,1,0.3

!mp,gxy,1,1e5

!mp,gxz,1,1e5


! A W10x26

bf=5.75


63/87

63

dg=10.3

tf=0.44

tw=0.26

lb=120

k,1,

k,2,120

k,3,0,4

k,4,0,-4

l,1,2

LESIZE,1,,,10



SECNUM,1

LATT,1,,1,,3,4,1

lmesh,1! Boundary Conditions

d,1,all


! Display Commands

/pbc,f,1

/pbc,u,1

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/soluantype,static


outres,all,all

d,1,all

f,2,fy,-1 !Initial small load

save

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu



save

solve

finish

/solu


64/87

64

expass,on


outres,all,all

solve

finish

/post1

set,list


/eshape,1


!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

Effect of Shift Variables using Subspace Method


!!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam


/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,beam188keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

!mp,gxy,1,1e5

!mp,gxz,1,1e5


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120

k,1,

k,2,120

k,3,0,4

k,4,0,-4


65/87

65

l,1,2

LESIZE,1,,,10



SECNUM,1

LATT,1,,1,,3,4,1

lmesh,1


d,1,all


! Display Commands

/pbc,f,1

/pbc,u,1

save

finish!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

d,1,all


save

solve

finish!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu



save

solve

finish

/solu

expass,on


outres,all,all

solve

finish

/post1

set,list



66/87

66


Lateral Buckling Loading

!Buckling of a cantilever beam using shell63


/title,Buckling of solid cantilever beam - shell63

/prep7

/plopts,logo,0

/plopts,date,0

antype,static

exx=30000

nu=0.3

et,1,Shell63

mp,ex,1,exx

mp,nuxy,1,nu


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

r,1,tw

r,2,tf

lb=120k,1,,dg/2

k,2,lb,dg/2

k,3,lb,-dg/2

k,4,0,-dg/2

k,5,0,dg/2,-bf/2

k,6,0,dg/2,bf/2

k,7,lb,dg/2,bf/2

k,8,lb,dg/2,-bf/2

k,9,0,-dg/2,-bf/2

k,10,0,-dg/2,bf/2k,11,lb,-dg/2,bf/2

k,12,lb,-dg/2,-bf/2

a,1,2,3,4

a,1,2,7,6

a,1,2,8,5

a,3,4,10,11


67/87

67

a,3,4,9,12

k=12

f=4

w=8

lsel,s,Length,,lb

lesize,all,,,k

lsel,s,Length,,bf/2

lesize,all,,,f

lsel,s,Length,,dg

lesize,all,,,w

type,1

real,1

amesh,1

type,1

real,2amesh,2,3

type,1

real,2

amesh,4,5


nsel,s,loc,x,0

d,all,all

/pbc,f,1

/pbc,u,1

nsell,allsave

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu

antype,static


outres,all,all

!nsel,s,loc,x,lb

!nsel,r,loc,z,0

f,all,fx,-1/(w+1) !Equal Initial small load at center line only

f,all,fx,-1/(4*f+w+3) !Equal Initial small load at all tip nodes!**LOADING APPLIED TO THE FLANGE AND THE WEB PROPORTIONAL TO

CORRESPONDING AREA

nsel,s,loc,x,lb

nsel,r,loc,z,0

f,all,fy,-.3/(w+1) !WEB


68/87

68

nsel,s,loc,x,lb

nsel,r,loc,Y,DG/2

f,all,fy,-.35/(2*F+1) !UPPER FLANGE

nsel,s,loc,x,lb

nsel,r,loc,Y,-DG/2

f,all,fy,-.35/(2*F+1) !LOWER FLANGE

nsell,all


f,NODE1,fy,(-.35/(2*F+1))+(-.3/(w+1))


f,NODE2,fy,(-.35/(2*F+1))+(-.3/(w+1))

nsell,all

save

solve

finish!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/solu



save

solve

finish

/solu

expass,on

mxpand,3 !Expand 3 modoutres,all,all

solve

finish

/post1

set,list


/eshape,1



69/87

69

Modal Analysis

BEAM 4

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!

!Modal Analysisof a an I-shaped beam. Using Reduced Method

!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam

/title,Modal analysis of solid cantilever beam- BEAM4

/prep7

/plopts,logo,0

/plopts,date,0

exx=30000

nu=0.3

A=7.61

Gfactor=1

Gmod=Gfactor*exx/(2*(1+nu))

g=386

sg=7.8

wtden=sg*62.4/1728

rho=wtden/g

wtbeam=rho*A*g

L=120

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@

! Mass and moment of inertiam1=.25/g

m2=.25/g

hblock1=2

wblock1=1

hblock2=2


70/87

70

wblock2=1

Ic_block1=m1*(hblock1**2+wblock1**2)/12


I_block1=Ic_block1+m1*(hblock1/2)**2


et,1,beam4

et,2,mass21,,,3

!keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

mp,gxy,Gmod

mp,dens,1,0


! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120

a=7.61

Izz=14.1

Iyy=144

r,1,a,Izz,Iyy,tf,tw

r,2,m1,I_block1r,3,m2,I_block2

k,1,

k,2,120

l,1,2

LESIZE,1,,,4

lmesh,1

midnode=node(L/2,0,0)

type,2

real,2

e,midnode

type,2

real,3

tipnode=node(L,0,0)

e,tipnode

outpr,nsol,all


71/87

71

d,1,all

save

finish

! Subspace Method

/soluantype,modal

modopt,reduc,3

mxpand,3

m,4,uy

m,2,uy

total,2

solve

*get,ff1,mode,1,freq



finish

/post1

set,list


/eshape,1


!

!Modal Analysisof a an I-shaped beam. Using Subspace method

!!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

/filnam,Solid Ibeam

/title,Modal analysis of solid cantilever beam- BEAM4

/prep7

/plopts,logo,0

/plopts,date,0

exx=30000

nu=0.3

A=7.61

Gfactor=1


g=386

sg=7.8

wtden=sg*62.4/1728

rho=wtden/g

wtbeam=rho*A*g


72/87

72

L=120

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@

! Mass and moment of inertia

m1=.25/g

m2=.25/g

hblock1=2

wblock1=1

hblock2=2

wblock2=1





et,1,beam4

et,2,mass21,,,3!keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

mp,gxy,Gmod

mp,dens,1,0


! A W10x26

bf=5.75

dg=10.3

tf=0.44tw=0.26

lb=120

a=7.61

Izz=14.1

Iyy=144

r,1,a,Izz,Iyy,tf,tw

r,2,m1,I_block1

r,3,m2,I_block2

k,1,

k,2,120

l,1,2

LESIZE,1,,,4

lmesh,1


type,2

real,2


73/87

73

e,midnode

type,2

real,3

tipnode=node(L,0,0)

e,tipnode

outpr,nsol,all

d,1,all

save

finish

! Subspace Method

/solu

antype,modal

modopt,subsp,3

mxpand,3

solve*get,ff1,mode,1,freq


finish

/post1

set,list


/eshape,1


Transient Dynamic Analysis

BEAM 4

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

!

!Transient dynamic analysis

!

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@/filnam,Solid Ibeam

/title,Transient dynamic analysis solid cantilever beam- BEAM4

/prep7

/plopts,logo,0

/plopts,date,0

exx=30000


74/87

74

nu=0.3

A=7.61

Gfactor=1


g=386sg=7.8

wtden=sg*62.4/1728

rho=wtden/g

wtbeam=rho*A*g

L=120

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@

! Mass and moment of inertia

m1=.25/g

m2=.25/g

hblock1=2

wblock1=1

hblock2=2

wblock2=1





et,1,beam4et,2,mass21,,,3

!keyopt,1,3,2

mp,EX,1,exx

mp,nuxy,1,0.3

mp,gxy,Gmod

mp,dens,1,0


! A W10x26

bf=5.75dg=10.3

tf=0.44

tw=0.26

lb=120

a=7.61

Izz=14.1


75/87

75

Iyy=144

r,1,a,Izz,Iyy,tf,tw

r,2,m1,I_block1

r,3,m2,I_block2

k,1,k,2,120

l,1,2

LESIZE,1,,,4

lmesh,1


type,2

real,2

e,midnode

type,2

real,3

tipnode=node(L,0,0)

e,tipnode

outpr,nsol,all

d,1,all

save

finish

! Subspace Method

/solu

antype,modal

modopt,subsp,3

mxpand,3

solve



finish

/post1

set,listset,1,1 !Select 1 or 2 mode

/eshape,1


!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

! Two damping methods are provided below

!@@@@@@@@@@@@@@@@@@@@@@


76/87

76

/solu

antype,trans

trnopt,msup,3,,1,yes ! Output modal coordinates

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@! Modal coordinates are printed in file .mcf

! Import this data to excel and plot contributions

! associated with each mode. Observe different

! frequencies and magnitudes of each mode.

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@

! Damping method #2

! Calculation for proportional damping

f1=ff1

f2=ff2

zeta1=.1

zeta2=.2

PI=ACOS(-1)

*dim,coeff,,2,2

*dim,rhs,,2,1

*dim,alfbeta,,2,1

coeff(1,1)=1/(4*PI*f1),1/(4*PI*f2)

coeff(1,2)=PI*f1,PI*f2

rhs(1,1)=zeta1,zeta2

*moper,alfbeta(1,1),coeff(1,1),solve,rhs(1,1)

alpha=alfbeta(1)

beta=alfbeta(2)

alphad,alfbeta(1)

betad,alfbeta(2)

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@

f,3,fy,5

kbc,1

time,.01


77/87

77

ITS=1/(20*f2)

deltim,ITS

outpr,all,all

outres,all,all

lswrite,1time,.02

f,3,fy,0

lswrite,2

time,0.6

lswrite,3

/output,file,dat

lssolve,1,3,1

finish

/output

/post26

file,,rdsp

time,.005,.25

nsol,2,4,u,y,middisp

nsol,3,2,u,y,tipdisp

/axlab,y,Displacement, in

/axlab,x,Time, sec

plvar,2,3

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@@@@@@@@@@@@@

finish

/solu

expass,on

numexp,20,0.0,0.1

outres,all,all

solve

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@@@@@@@@@@@@@@

/post1

/eshape,1

set,list


78/87

78

set,1,1

pldisp,2

Hammer Strike Case

!@@@@@@@@@@@@@@@@@@@@@@

! A cantilever beam has attached masses. One mass is freee to "hop" or! leave the surface of the beam!! Two concentrated masses are affixed to a cantilever beam. A third! mass is resting on the top of the beam. A hammer strike is applied! at the beam free end. The load is of the form of a triangle. The! object is to investigate the possibility of a "hop" of the third! mass which is connected to the beam using contact12 elements.! Spring elements, combin14, are used to simulate the attachment to! beam of the "hopping" mass. This simulates something like an! accelerometer attached to the beam with magnets.!

!@@@@@@@@@@@@@

/title, Transient Dynamic Analysis of Cantilever Beam With Added Masses

/filnam,hopc

/prep7

/plopts,logo

/plopts,date

! System parameter definitions

acy1=.5 !Radius of cylinder 1



Lcy1=1 !Length of cylinder 1

Lcy2=1 !Length of cylinder 2

Lcy3=1.5 !Length of cylinder 3


79/87

79

A=7.61

! A W10x26

bf=5.75

dg=10.3

tf=0.44

tw=0.26

lb=120

a=7.61

Izz=14.1

Iyy=144

r,1,a,Izz,Iyy,tf,tw

rho=0.000728 !Beam material density

rho1=sqrt((3*(acy1**2)+(Lcy1**2))/12) !Mass radius of gyration 1



g=386

w=rho*A*g

L=120

hcy1=4.5 !Height of cylinder1 cg above beam

hcy2=4.5 !Height of cylinder2 cg above beam

hcy3=6 !Height of cylinder3 cg above beam

m1=.2/g !Mass of accelerometer#1

m2=.2/g !Mass of accelerometer#2

m3=.4/g !Mass of "free" object

Iz1=m1*(rho1**2)

Iz2=m2*(rho2**2)

Iz3=m3*(rho3**2)

c***Element Definitions

et,1,beam4

!keyopt,1,3,2

et,2,mass21,,,3

et,3,contac12

et,4,combin14,,2 !UY dof

et,5,combin14,,1 !UX dof

keyopt,3,7,1

mp,ex,1,30e6

mp,ex,2,100e6 !For very stiff mass structures


80/87

80

mp,dens,1,rho

mp,damp,1,0.005

r,1,a,Izz,Iyy,tf,tw

r,2,m1,Iz1 !Accelerometer#1

r,3,m2,Iz2 !Accelerometer#2r,4,m3,Iz3 !Free cylinder

r,5,,1e4 !Contact element

r,6,1e5 !Spring to attach accelerometers to beam

r,7,0.1 !Spring for fastening free cylinder to beam

k,1

k,2,120

l,1,2

c*** Beam Construction

!Set beam node locations at 1 inch intervals along the beam

lesize,1,,,20

type,1

mat,1

real,1

lmesh,1

c*** Construction of accelerometers and free cylinder

/com Locations are at positions determined by inch locations

/com following from the lmesh operation of the beam

!Accelerometer #1

type,1

mat,2

real,1

k,3,60

k,4,66

k,5,63,-hcy1

l,3,4

l,4,5

l,5,3

lsel,s,,,2,4,1

lesize,all,,,1

lmesh,2,4,1

!Accelerometer #2

k,6,108


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k,7,114

k,8,111,-hcy2

l,6,7

l,7,8

l,8,6lsel,s,,,5,7,1

lesize,all,,,1

lmesh,5,7,1

!Free cylinder

k,9,102

k,10,108

k,11,105,hcy3

l,9,10

l,10,11

l,11,9

lsel,s,,,8,10,1

lesize,all,,,1

lmesh,8,10,1

/pnum,node,1

!Accelerometers

!Accel Masses

type,2

real,2

e,24

real,3

e,27

!Accel springs

type,4 !Uy dof

real,6

e,12,22

e,13,23

e,20,25

e,21,26

type,5 !Ux dof

real,6

e,12,22

e,13,23


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82

e,20,25

e,21,26

!Free Cylinder

!Mass

type,2real,4

e,30

!Springs

type,4 !Uy dof

real,7

e,19,28

e,20,29

type,5 !Ux dof

real,7

e,19,28

e,20,29

!Contact elements

type,3

real,5

e,19,28

e,20,29

save

finish

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@

! Solution Commands

/solu

outres,all,all

outpr,nsol,1


ksel,s,,,1

nslk,s

d,all,all

c*** Analysis

allsel,all,all

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@@@@@@@


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antype,trans

lumpm,on !Lumped mass option

tintp,.05 !Transient integration parameters

nlgeom,on

pred,on!autots,on

kbc,1

timint,off

nsubst,50

time,.001

acel,,g !Apply acceleration

lswrite,1

timint,on !Beginning of loading commands

kbc,0

nsubst,40

time,0.01

f,2,fy,300 !Hammer load applied at beam tip

lswrite,2

time,0.02

f,2,fy,0 !Hammer load removed

lswrite,3

time,0.5

nsubst,100

lswrite,4

lssolve,1,4,1

finish

/output

!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

@@@@

! Note: Beamleft and Cyleft represent the left

! side of the hopping mass and beam.

! BeamRite and Clyrite represent the right

! hand side of the hopping mass.

/post26

nsol,2,24,u,y,MidAccel

nsol,3,27,u,y,TipAccel

plvar,2


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nsol,4,19,u,y,BeamLeft

nsol,5,28,u,y,Cyleft

nsol

Documents

Cie 633 Final Report1