CIE 633
FINITE ELEMENT ANALYSIS
PROJECT
BEAM ANALYSIS
BY
Kalpesh Parikh
1
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.
2
Index
1. Introduction
04
2. Buckling Analysis
05
3. Modal Analysis
24
4. Transient and Dynamic Analysis
29
5. Conclusion
35
-Reference
36
- Appendix
37
3
1 Introduction
The Project analysis is done in the software called ANSYS .Basically Software takes all the Inputs and it has in its
memory 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 don’t 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 inch4
Moment of Inertia in y-y direction (Iyy) = 144 inch4
4
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.
Eigenvalue
Eigenvalue 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.
Nonlinear
Nonlinear buckling analysis is more accurate than eigenvalue analysis because it employs non-linear, largedeflection; 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.
2.
3.
Beam 4 Element was model using the cross sectional properties and followed and two key points were used
to 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
Beam 188 element and Beam 189 was model using standard SECDATA and SECTYPE and then I
assigned 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
Shell 63 and Shell 93 element was modeled using real constant and modeling I beam is not so proper as it
goes into it as shown below (Area overlapped)
5
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 =
𝜋2 ∗ 𝐸∗𝐼𝑧𝑧
(2𝐿)2
=72.47 kips
When L = 60
Pcr =
𝜋2 ∗ 𝐸∗𝐼𝑧𝑧
(2𝐿)2
= 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.
Analysis – Eigenvalue Analysis
Buckling Loads obtained using Set list when simulated
Element
Theoretical Pcr
in Kips
% of Comparison of
Theoretical with
Ansys
Ansys Pcr in
Kips
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.
Analysis – Eigenvalue Analysis
6
Comparision Table showing Theoretical value and Ansys Eulers Buckling Load
Element
Theoretical Pcr
in Kips
% of Comparison of
Theoretical with
Ansys
Ansys Pcr in
Kips
Beam 4
289.91
289.93
100.0068987
Beam 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), clearly
agree 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 was
required 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 the
result are not changing as it has nothing to do with geometric propertie s , apart from how you apply loads.
Figure Buckling for Beam 4
7
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
8
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 =
𝜋2 ∗ 𝐸∗𝐼𝑧𝑧
(𝐿)2
= 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.
Analysis – Eigenvalue Analysis
Comparision Table showing Theoretical value and Ansys Eulers Buckling Load
Element
Theoretical Pcr
in Kips
% of Comparison of
Theoretical with
Ansys
Ansys Pcr in
Kips
Beam 4
289.91
289.92
100.0034493
Beam 188
289.91
281.2
96.99561933
Beam 189
289.91
281.2
96.99561933
Shell 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 local
buckling. Now the question comes what is local buckling generally it is defined as
λ> λr Local Buckling will occur in the beam
λr = obtained from AISC manual depending on the beam is compact section or non compact section
λ = 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.
9
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 and
Shell 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 we
have 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: FIXED – PINED 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 =
𝜋2 ∗ 𝐸∗𝐼𝑧𝑧
(0.7∗𝐿)2
= 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.
Analysis – Eigenvalue Analysis
10
Comparision Table showing Theoretical value and Ansys Eulers Buckling Load
Theoretical Pcr
in Kips
Element
% of Comparison of
Theoretical with
Ansys
Ansys Pcr in
Kips
Beam 4
591.67
594
100.39
Beam 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 of
local 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 element
and 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 simply
supported 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 Ѳ 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 =
4.013∗𝐸
𝐿2
∗√
𝐼𝑧𝑧∗𝐽
2(1+𝑉)
=17.45 kips
Izz = Weak axis of moment of Inertia.
J= torsional value for the given section
11
When Length =60 inch
Pcr =
4.013∗𝐸
𝐿2
∗√
𝐼𝑧𝑧∗𝐽
2(1+𝑉)
=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.
Analysis – Eigen value Analysis.
Load Applied Laterally to the beam.
Comparision Table showing Theoretical value and Ansys Eulers Buckling Load
Element
Theoretical Pcr
in Kips
% of Comparison of
Theoretical with
Ansys
Ansys Pcr in
Kips
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.
Analysis – Eigenvalue Analysis
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 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
12
Interpretation
1.
Result obtained for Beam 188 and Beam 189 are near to the theoretical value which makes sense. And the
behavior 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 Area
overlapped 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 nothing
changes 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 in
combination with twisting.
Figure Showing Buckling Using Lateral Load –Beam 188
Figure Showing Buckling Using Lateral Load –Beam 189
13
Figure Showing Buckling Using Lateral Load –Shell 63
Figure Showing Buckling Using Lateral Load –Shell 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.
Analysis – Eigen value Analysis.
Six hexagonal holes ( h= 4 inch)
14
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
Shell 93
69.051
71.523
Results are Better
when used Shell 93
Interpretation
1.
2.
3.
In cantilever beam when holes are made then its euler’s buckling load decreases for both the Shell element.
Shell 93 elements results are better because we can see that effect of holes has not taken much effect in the
buckling load.
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.
Analysis – Eigen 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
Shell 63
25.41
25.4072
Shell 93
31.98
32.98
Comments
Results were suprising i
got better result for shell
63 but cannot interprete
Interpretation
1.
2.
In cantilever beam when holes are made then its euler’s buckling load Shell 63 elements I was not able to
interprete it
results are better because we can see that effect of holes has not taken much effect in the buckling load.
15
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 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 189
Analysis – Eigen value Analysis.
Loading is Axial load.
Comparison of number modes and two different method for eigenvalue analysis
Element
Beam 189
Ansys Pcr using Subspace
method
No. of Modes
Ansys Pcr using Block
Lancoz method
3
71.635
71.635
4
71.635
71.635
5
71.635
71.635
Interpretation
16
1 Compared and obtained result that there is no changes in the eigenvalue due to change in number of modes.
Behavior for pldisp,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
Analysis – Eigen 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
Beam 188
Shift
Variable
Ansys Pcr using Subspace
method
Ansys Pcr using Block
Lancoz method
0
71.635
71.635
0.5
218.34
71.635
1
218.34
71.635
17
Comments
Effect of Shift
Variable is notice
in Subspace
Method
Interpretation
1.
2.
3.
4.
When I command I tried with changing in the shift variables using block lancoz method I did not notice any
changes in the buckling value.
But when I tried with subspace method I notice the difference when I put the shift variable as 0.5 the
eigenvalue changes and it takes more load to buckle.
What is shift variable when we plot a graph of load vs displacement shift variable is assigned using symbol
(λ) we can see below how it shifts the value.
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 Euler’s 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
Y direction small lateral
Load
Z direction small lateral
load
Z direction small lateral
load
Z direction small lateral
load
Z direction small lateral
load
Beam 4
Beam188
Beam 189
Shell 63
Shell 93
Weak axis
Z-Z axis is weak axis
Uy
Plot Displacement
Y-Y axis is weak axis
Uz
Y-Y axis is weak axis
Uz
Y-Y axis is weak axis
Uz
Y-Y axis is weak axis
Uz
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
Theoretical
Pcr
in Kips
% of Comparison of
Theoretical with
Ansys
Ansys Pcr in
Kips
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
18
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
19
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.
2.
3.
Nonlinear buckling analysis is more accurate than eigenvalue analysis because it employs non-linear, largedeflection; 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.
As we can see from the graph of shell element and beam element the value which I obtained theoretically
from euler’s buckling load as the same load we can see that there is an buckling indication (means
displacement is noticed in the graph).
The value which I obtained from eigenvalue for shell element where to high but it is figured out in the non
linear 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.
20
Displacement figure during buckling
Figure showing Displacement at time Buckling Figure showing Displacement at time Takes Place – For
Beam 188 (Uz)
Buckling takes place- For Beam189 (Uz)
Figure showing Displacement at time Buckling Figure showing Displacement at time Takes Place – For Shell
63 (Uz)
Buckling takes place- For Shell 93 (Uz)
21
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
Place – For Shell 93 (von mises Stress)
Figure showing Stresses at time Buckling takes
Place- For Shell 63 (Von mises Stress)
Figure showing Stresses at time Buckling takes
Place – For Beam 188 (von mises Stress)
Figure showing Stresses at time Buckling takes
Place- For Beam 189 (Von mises Stress)
22
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.
23
3 Modal analysis
Modal 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.
2.
Subspace Method
Reduced Method
Basic Information about model which I used
Material properties
•
•
•
Modulus of Elasticity: E =30000 Ksi
Poisson ratio =0.3
Density of Steel= 490 lb/ft3
Cross sectional Properties





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 inch4
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.
2.
Beam 4 Element using Subspace method was model using the cross sectional properties and followed and
two key points were used to define the length of the beam..which is L=120 inch.
I use a case of Beam as a Cantilever beam subjected with two concentrated masses in order to carry out the
modal 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
24
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
ρ= wt.density/g = 0.28167/386 =7.29 *10 -4
Wt. Beam = ρ*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 + y12)/12,
Z2= m2 (x22 + y22)/12
Im1= Z1 +m1*(x1/2)2
Im2= Z2 +m2*(x2/2)2
X1 = 2, Y1 = 1, X2 = 2, Y2= 1………….Assumed
3.
4.
Element use to define mass was Mass 21 element and I assign the location at the tip and the mid node of
beam 4 element. Mass 21 element was created using real constant
Now its time to add the post processor to obtained the desired result so I used subspace method to simulate
my 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.
ώ n = αn2 √(EI/mL4)
αn = 1.875 ………..To obtain mode 1
αn = 4.694…………To obtain mode 2
αn = 7.844…………To 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.
25
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
Beam 4
Theory
Natural
Frequency
Ansys
Natural
Frequency
% Ansys Accuracy
1
5.2683
5.1
96.81
2
35.08068
33.96
96.81
3
245.5623
238.41
97.09
Interpretation
1.
2.
Result obtained from the ansys for natural frequency is not so accurate may be I am missing some
concept.
But it is closer by from here onwards we can know add forces and notices the behavior and notice the
transient 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
26
Model Descriptions
1.
2.
3.
Beam 4 Element using Reduce method was model using the cross sectional properties and followed and
two key points were used to define the length of the beam. which is L=120 inch.
Only difference is that now we add Master degree of freedom to the beam and use the post processor for
reduced method. . These are degrees of freedom that govern the dynamic characteristics of a structure.
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
Beam 4
Mode
Theory
Natural
Frequency
Ansys
Natural
Frequency
% Ansys Accuracy
1
5.2683
5.1
96.81
2
35.08068
33.96
96.81
3
245.5623
238.41
97.09
27
Interpretation
1.
2.
3.
4.
5.
Result obtained from the ansys for natural frequency is not so accurate may be I am missing some
concept.
But it is closer by from here onwards we can know add forces and notices the behavior and notice the
transient dynamic analysis for the same model.
Result are dot same for both the method reason there is not so complexity in the model the model is not
so complex if model becomes then we can notice the difference in the reduced method.
It is better idea if in the execution we have complex structure it is better from the Ansys to calculate
using reduced method obtained the analysis.
Subspace method and reduced method are frontal solver. But Subspace method is much slower then
reduced method.
28
4
Transient Dynamic Analysis
Transient dynamic analysis is a technique used to determine the dynamic response of a structure under a timevarying 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 under step or impulse loading
conditions. For my case, we will impact the end of the beam with an impulse force and view the response at the
location 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
29
Cross sectional Properties





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 inch4
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 superposition
method 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 the
application . 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 y
direction at the tip and middle to effect of impulse force.
Figure Showing at the Substep 1 and 20
30
Table Showing Load step and Sub step and Time
Set
Load
Step
Time
Sub Step
1
4.42E-03
1
1
20
0.1001
1
20
Figure Showing Time vs Displacement at the tip and Middle
Interpretation
1
2
3
We can see the plot of Uy displacement at the position of the tip and Middle. As we can see that tip
displacement is more at time 0.1 sec and mid displacement is less at same time.
Like if we imagine the beam when force is applied at the tip at one time we can see that the beam displaces
more at the tip and less at the middle portion
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.
31
Modeling Procedure
1.
2.
3.
4
5
6
7
Element chosen to built the Beam is Beam4 then i assign the geometric properties of the Cylinder which
are going to be attached to the beam.
Then I assigned the various elements like Mass21 for Masses, Contac12 to give contact surface to the
cylinder and the Combin 14 in Uy dof and Ux dof that is Spring element
These all element were generate by the real constant as shown below.
Real Constant
r1
r2
r3
r4
r5
r6
Assign Properties
Area, Izz, Tz,Tx,h
mass1,Izz1
mass2,Izz2
mass3,Izz3
Uy Dof
Ux Dof
r7
thk
Purpose
Beam Model Creation
Accelerometer#1-Cylinder
Accelerometer#2-Cylinder
Free Cylinder
Contact element
Spring attached to accelerometer
to the Beam
Spring for fastening free cylinder.
Accelerometer #1 & #2, Free cylinder, springs, Contact element were assigned with respect to the keypoint.
After these is done assigning a boundary condition to its
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.
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
32
Figure Showing Uy Plot for beam displacement at the tip and mid accel
Interpretation
1
2
3
The Graph of Y displacement looks reasonable was lumpm,on ( Lumped mass option). The
displacement 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.
These result may be very important design firm to note the response based on the loading which was
given and look at the behavior of Beam.
As we can see that the mass which was kept free on the third cylinder it hop out of the cylinder as
shown 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
2
3
4
Model was Prepared using Beam 4 element for Beam structure, Plane 42 element for the Block, Contac171
was given between Beam and Block as Block strikes the beam so targe169 element was taken in to
consideration.
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.
Method used for Transient Analysis is full method
Modeling was very simple inafe here Block Falls after some time and effects is studied.
33
Figure showing eplot
Figure Showing Block Position Load Step3 and Sub Step10
Figure Showing Y displacement at Node at Various Nodes
Interpretation
1
2
The Behavior of the Beam can notice at various time when Block Falls on the beam we can see at time
0.28 sec where all the nodes have almost same displacement even it is a peak point of the graph where n
maximum displacement take place.
At different time we noted different displacement of the beam . from all of the study main purpose would
be that we full fill our design criteria..
34
Conclusion
1 I study the behavior of the beam using various analysis. The main aim of doing Buckling
Analysis, Modal analysis and Dynamic Analysis was to understand the behavior of the
Beam under each analysis.
2 From The Buckling Analysis, I carried out Eigenvalue analysis we have seen the
buckling in beams takes place at the same load one which is given by Euler’s 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 in
that geometric imperfection is not much of effect.
5 Modal analysis was carried out where I tried and spotted that the natural frequency given
by 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 know
even 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 for
certain time and know the response of structure due to that impulse force.
“Collapse Structure of WTC- we don’t 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
35
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 Kassimali
3. “ Structural Steel Design”- 1st Edition by Abi Aghayere and Jason Vigil
4. “ Design of Steel Structure- 3rd Edition by Edwin H Gaylord
5. “ Stability of Structure”- 2nd edition by Dr. Eric m Lui and Chen
6.
“Theory of Structures” –By Timoshenko, Stephen.
7. “ University Alberta” – Ansys help
http://www.mece.ualberta.ca/tutorials/ansys/IT/Transient/Transient.html
36
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
37
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,all
d,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
save
solve
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
38
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
!
!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=0.3
et,1,beam188
keyopt,1,3,2
mp,EX,1,exx
mp,nuxy,1,0.3
!mp,gxy,1,1e5
!mp,gxz,1,1e5
! GEOMETRY FORMATION
! 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,,,10
SECTYPE,1,BEAM,I,W10x26
SECDATA,bf,bf,dg,tf,tf,tw
SECNUM,1
LATT,1,,1,,3,4,1
lmesh,1
! Boundary Conditions
39
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,all
d,1,all
f,2,fx,-1
!Initial small load
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,lanb,3,0.5
!Use subspace method, extract 3 modes
save
solve
finish
/solu
expass,on
mxpand,3,,,YES
!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
/title,Buckling of solid cantilever beam- BEAM189
!
40
!Eigenvalue buckling of a an I-shaped solid beam.
!
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/filnam,Solid Ibeam
/title,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
! GEOMETRY INFORMATION
! 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,,,10
SECTYPE,1,BEAM,I,W10x26
SECDATA,bf,bf,dg,tf,tf,tw
SECNUM,1
LATT,1,,1,,3,4,1
lmesh,1
! Boundary Conditions
d,1,all
!f,2,fy,1
!Apply a small lateral load
! Display Commands
/pbc,f,1
41
/pbc,u,1
save
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,static
pstres,on
!Calculate prestress effects
outres,all,all
d,1,all
f,2,fx,-1
!Initial small load
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,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
mxpand,3,,,YES
!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
Buckling of a cantilever beam using shell63
/filnam,cantileverbucklingshell63
/title,Buckling of solid cantilever beam - shell63
/prep7
/plopts,logo,0
42
/plopts,date,0
antype,static
exx=30000
nu=0.3
et,1,Shell63
mp,ex,1,exx
mp,nuxy,1,nu
! GEOMETRY FORMATION
! A W10x26
bf=5.75
dg=10.3
tf=0.44
tw=0.26
r,1,tw
r,2,tf
lb=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/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
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
43
lesize,all,,,w
type,1
real,1
amesh,1
type,1
real,2
amesh,2,3
type,1
real,2
amesh,4,5
! Boundary Conditions
nsel,s,loc,x,0
d,all,all
/pbc,f,1
/pbc,u,1
nsell,all
save
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,static
pstres,on
!Calculate prestress effects
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,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
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))
44
NODE2=NODE(LB,DG/2,0)
f,NODE2,fx,(-.35/(2*F+1))+(-.3/(w+1))
nsell,all
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,subsp,3
!Use subspace method, extract 3 modes
save
solve
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
!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=0.3
et,1,Shell93
mp,ex,1,exx
mp,nuxy,1,nu
! GEOMETRY FORMATION
! A W10x26
bf=5.75
45
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/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
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
46
real,2
amesh,2,3
type,1
real,2
amesh,4,5
! Boundary Conditions
nsel,s,loc,x,0
d,all,all
/pbc,f,1
/pbc,u,1
nsell,all
save
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,static
pstres,on
!Calculate prestress effects
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,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
f,all,fx,-.35/(4*F+1) !LOWER FLANGE
nsell,all
NODE1=NODE(LB,-DG/2,0)
f,NODE1,fx,(-.35/(4*F+1))+(-.3/(2*w+1))
NODE2=NODE(LB,DG/2,0)
f,NODE2,fx,(-.35/(4*F+1))+(-.3/(2*w+1))
nsell,all
save
solve
47
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,subsp,3
!Use subspace method, extract 3 modes
save
solve
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
/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
! GEOMETRY FORMATION
!###########################
e=8
b=2
h=10
d=e
!########################
tf=0.44
48
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/2
k,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
49
aCLEAR,ALL
REAL,2
AMESH,ALL
ASEL,S,LOC,Y,0
REAL,2
aCLEAR,ALL
REAL,2
AMESH,ALL
ASEL,ALL
! Boundary Conditions
nsel,s,loc,x,0
d,all,all
/pbc,f,1
/pbc,u,1
nsell,all
save
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,static
pstres,on
!Calculate prestress effects
outres,all,all
!**LOADING APPLIED TO THE FLANGE AND THE WEB PROPORTIONAL TO
CORRESPONDING AREA
nsel,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 FLANGE
nsel,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)
50
NODE2=NODE(6*L,DG,0)
f,NODE2,fx,(-.35/9-.3/23)
nsell,all
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,subsp,3
!Use subspace method, extract 3 modes
save
solve
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
Non Linear Analysis of the I beam- Beam 4
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
!
!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
51
!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
! 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
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
solcontrol,on
monitor,1,2,uy
monitor,2,2,fy
monitor,3,2,fx
/gst,on
nropt,auto
!Solution method is selected
52
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
Non Linear Analysis of the I beam- Beam 188
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
53
!
!
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/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
! 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
k,3,0,4
k,4,0,-4
l,1,2
LESIZE,1,,,10
SECTYPE,1,BEAM,I,W10x26
SECDATA,bf,bf,dg,tf,tf,tw
SECNUM,1
LATT,1,,1,,3,4,1
54
lmesh,1
! Boundary Conditions
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
nropt,auto !Solution method is selected
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.
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
55
@@@@@@@@@@@@@@@@@@@@@@@@@@@
/post26
nsol,2,2,u,z,uz2
prod,3,1,,,LOAD,,,Fmax
xvar,3
/axlab,x,Total Load, lbs
/axlab,y,Displacement, inch
plvar,2
Non Linear Analysis of the I beam- Beam 189
!
/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=5.75
dg=10.3
tf=0.44
tw=0.26
lb=120
a=7.61
Izz=14.1
56
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
SECTYPE,1,BEAM,I,W10x26
SECDATA,bf,bf,dg,tf,tf,tw
SECNUM,1
LATT,1,,1,,3,4,1
lmesh,1
! Boundary Conditions
d,1,all
f,2,fz,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,all
d,1,all
f,2,fx,-1
!Initial small load
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
solcontrol,on
monitor,1,2,uz
monitor,2,2,fz
monitor,3,2,fx
/gst,on
nropt,auto !Solution method is selected
nlgeom,on
57
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,z,uy2
prod,3,1,,,LOAD,,,Fmax
xvar,3
/axlab,x,Total Load, lbs
/axlab,y,Displacement, inch
plvar,2
!Non Linear Buckling of a cantilever beam using shell63
/filnam,cantileverbucklingshell63
/title,Buckling of solid cantilever beam - shell63
/prep7
/plopts,logo,0
/plopts,date,0
antype,static
exx=30000
nu=0.3
58
et,1,Shell63
mp,ex,1,exx
mp,nuxy,1,nu
Pcr=72.47
Fmax=2*Pcr
L=120
! GEOMETRY FORMATION
! 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
59
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,2
amesh,2,3
type,1
real,2
amesh,4,5
! Boundary Conditions
nsel,s,loc,x,0
d,all,all
/pbc,f,1
/pbc,u,1
nsell,all
save
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,static
pstres,on
!Calculate prestress effects
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
60
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
f,all,fx,-.35/(2*F+1) !LOWER FLANGE
nsell,all
NODE1=NODE(LB,-DG/2,0)
f,NODE1,fz,(.35/(2*F+1))+(.3/(w+1))
NODE2=NODE(LB,DG/2,0)
f,NODE2,fz,(.35/(2*F+1))+(.3/(w+1))
nsell,all
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
solcontrol,on
monitor,1,2,uz
monitor,2,2,fz
monitor,3,2,fx
/gst,on
nropt,auto !Solution method is selected
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:
61
!
!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,z,uy2
prod,3,1,,,LOAD,,,Fmax
xvar,3
/axlab,x,Total Load, lbs
/axlab,y,Displacement, inch
plvar,2
Lateral Buckling
!Lateral Buckling
!Eigenvalue buckling of a an I-shaped solid beam.
!
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/filnam,Solid Ibeam
/title,Buckling of solid Cantilever beam - BEAM189
/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
! GEOMETRY FORMATION
! A W10x26
bf=5.75
62
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
SECTYPE,1,BEAM,I,W10x26
SECDATA,bf,bf,dg,tf,tf,tw
SECNUM,1
LATT,1,,1,,3,4,1
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,all
d,1,all
f,2,fy,-1
!Initial small load
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,lanb,3,0.5
!Use subspace method, extract 3 modes
save
solve
finish
/solu
63
expass,on
mxpand,3,,,YES
!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
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
Effect of Shift Variables using Subspace Method
!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=0.3
et,1,beam188
keyopt,1,3,2
mp,EX,1,exx
mp,nuxy,1,0.3
!mp,gxy,1,1e5
!mp,gxz,1,1e5
! GEOMETRY FORMATION
! 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
64
l,1,2
LESIZE,1,,,10
SECTYPE,1,BEAM,I,W10x26
SECDATA,bf,bf,dg,tf,tf,tw
SECNUM,1
LATT,1,,1,,3,4,1
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,all
d,1,all
f,2,fx,-1
!Initial small load
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,lanb,3,0.5
!Use subspace method, extract 3 modes
save
solve
finish
/solu
expass,on
mxpand,3,,,YES
!Expand 3 mod
outres,all,all
solve
finish
/post1
set,list
set,1,1 !Select 1 or 2 mode
65
/eshape,1
pldisp,1 !Plot mode shape
Lateral Buckling Loading
!Buckling of a cantilever beam using shell63
/filnam,cantileverbucklingshell63
/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
! GEOMETRY FORMATION
! A W10x26
bf=5.75
dg=10.3
tf=0.44
tw=0.26
r,1,tw
r,2,tf
lb=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/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
66
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,2
amesh,2,3
type,1
real,2
amesh,4,5
! Boundary Conditions
nsel,s,loc,x,0
d,all,all
/pbc,f,1
/pbc,u,1
nsell,all
save
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,static
pstres,on
!Calculate prestress effects
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
67
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
NODE1=NODE(LB,-DG/2,0)
f,NODE1,fy,(-.35/(2*F+1))+(-.3/(w+1))
NODE2=NODE(LB,DG/2,0)
f,NODE2,fy,(-.35/(2*F+1))+(-.3/(w+1))
nsell,all
save
solve
finish
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
/solu
antype,buckle
!Perform eigenvalue buckling analysis
bucopt,subsp,3
!Use subspace method, extract 3 modes
save
solve
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
68
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 inertia
m1=.25/g
m2=.25/g
hblock1=2
wblock1=1
hblock2=2
69
wblock2=1
Ic_block1=m1*(hblock1**2+wblock1**2)/12
Ic_block2=m2*(hblock2**2+wblock2**2)/12
I_block1=Ic_block1+m1*(hblock1/2)**2
I_block2=Ic_block2+m1*(hblock2/2)**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
! 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,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
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
70
d,1,all
save
finish
! Subspace Method
/solu
antype,modal
modopt,reduc,3
mxpand,3
m,4,uy
m,2,uy
total,2
solve
*get,ff1,mode,1,freq
*get,ff2,mode,2,freq
*get,ff3,mode,3,freq
finish
/post1
set,list
set,1,1 !Select 1 or 2 mode
/eshape,1
pldisp,1 !Plot mode shape
!
!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
Gmod=Gfactor*exx/(2*(1+nu))
g=386
sg=7.8
wtden=sg*62.4/1728
rho=wtden/g
wtbeam=rho*A*g
71
L=120
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@
! Mass and moment of inertia
m1=.25/g
m2=.25/g
hblock1=2
wblock1=1
hblock2=2
wblock2=1
Ic_block1=m1*(hblock1**2+wblock1**2)/12
Ic_block2=m2*(hblock2**2+wblock2**2)/12
I_block1=Ic_block1+m1*(hblock1/2)**2
I_block2=Ic_block2+m1*(hblock2/2)**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
! 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,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
midnode=node(L/2,0,0)
type,2
real,2
72
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
*get,ff2,mode,2,freq
finish
/post1
set,list
set,1,1 !Select 1 or 2 mode
/eshape,1
pldisp,1 !Plot mode shape
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
73
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 inertia
m1=.25/g
m2=.25/g
hblock1=2
wblock1=1
hblock2=2
wblock2=1
Ic_block1=m1*(hblock1**2+wblock1**2)/12
Ic_block2=m2*(hblock2**2+wblock2**2)/12
I_block1=Ic_block1+m1*(hblock1/2)**2
I_block2=Ic_block2+m1*(hblock2/2)**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
! GEOMETRY FORMATION
! A W10x26
bf=5.75
dg=10.3
tf=0.44
tw=0.26
lb=120
a=7.61
Izz=14.1
74
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
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
d,1,all
save
finish
! Subspace Method
/solu
antype,modal
modopt,subsp,3
mxpand,3
solve
*get,ff1,mode,1,freq
*get,ff2,mode,2,freq
finish
/post1
set,list
set,1,1 !Select 1 or 2 mode
/eshape,1
pldisp,1 !Plot mode shape
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
! Two damping methods are provided below
!@@@@@@@@@@@@@@@@@@@@@@
75
/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
76
ITS=1/(20*f2)
deltim,ITS
outpr,all,all
outres,all,all
lswrite,1
time,.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
77
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
acy2=.5
!Radius of cylinder 2
acy3=.5
!Radius of cylinder 3
Lcy1=1
!Length of cylinder 1
Lcy2=1
!Length of cylinder 2
Lcy3=1.5
!Length of cylinder 3
78
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
rho2=sqrt((3*(acy2**2)+(Lcy2**2))/12) !Mass radius of gyration 2
rho3=sqrt((3*(acy3**2)+(Lcy3**2))/12) !Mass radius of gyration 3
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
79
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#2
r,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
80
k,7,114
k,8,111,-hcy2
l,6,7
l,7,8
l,8,6
lsel,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
81
e,20,25
e,21,26
!Free Cylinder
!Mass
type,2
real,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
! Boundary Conditions
ksel,s,,,1
nslk,s
d,all,all
c*** Analysis
allsel,all,all
!@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
@@@@@@@@@@
82
antype,trans
lumpm,on
tintp,.05
!Lumped mass option
!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
83
nsol,4,19,u,y,BeamLeft
nsol,5,28,u,y,Cyleft
nsol,6,20,u,y,BeamRite
nsol,7,29,u,y,Clyrite
plvar,2,3
!Block falls and strikes a beam. Based on WB11. Version 5.0.
!
/output,xbounce,txt
!@@@@@@@@@@@@@@@@@
/prep7
/plopts,logo
/plopts,date
et,1,beam4
et,2,plane42,,,3
et,3,conta171
et,4,targe169
!keyopt,1,3,2
keyopt,3,2,1
!@@@@@@@@@@
!Beam characteristics
! 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,tf,tw
L=120
!Block characteristics
blkthk=1
!Block thickness
84
blklen=2
!Block side length
hblock=1
!Block height
adist=L/2
!Distance to block corner
!Contact element properties
cstiff=.05 !Contact stiffness FKN Default 1, reduce for convergence
!@@@@@@@@@@@@@@@@@@@
!Reals
r,1,a,Izz,Iyy,tf,tw
r,2,blkthk
r,3,blklen,blklen,cstiff
!@@@@@@@@@@@@@@@@@@@@@@
!Material properties
mp,ex,1,30000
mp,nuxy,1,.3
mp,dens,1,.001
!Mass density
!@@@@@@@@@@@@@@@@@
!Model building
k,1
k,2,L
l,1,2
lesize,1,,,20
mat,1
lmesh,1
/pnum,elem,10
eplot
/eshape,1
eplot
rectang,adist,adist+blklen,hblock,hblock+blklen
esize,,1
type,2
real,2
mat,1
amesh,all
eplot
lsel,s,,,2
nsll,s,1
type,3
85
real,3
esurf
lsel,s,,,1
nsll,s,1
type,4
esurf
nsel,all
esel,all
finish
!@@@@@@@@@@@@@@
/solu
antype,trans
trnopt,full
nropt,full,,on
lnsrch,on
nlgeom,on
lumpm,on
fcr=0.1
!Force convergence value
ldcr=0.01
!Large displacement criteria
nsbstp=10000
neq=100
!Number of substeps
!Number of equilibrium iterations
!@@@@@@@@@@@@@@@@
ksel,s,kp,,1,2
nslk,s
d,all,all,0
nsel,all
esel,s,ename,,42
nsle,s
d,all,all,0
nsel,all
esel,all
acel,,386
time,.0002
deltim,.0001
kbc,1
betad,.0003
timint,off
86
outres,all,last
solve
esel,s,ename,,42
nsle,s
ddele,all,all
nsel,all
esel,all
tf=0.4
time,tf
cnvtol,f,,fcr
cnvtol,u,,ldcr
cnvtol,stat
!deltim,.0002,.00002,.02 !Initial, minimum and maximum delta times
neqit=neq
nsubst=nsbstp
solcontrol,on
autots,on
timint,on
pred,on
!Used to predict the start of the next substep
outres,all,all
solve
time,2*tf
solve
finish
/pnum,node
/pnum,elem
/output
!@@@@@@@@@@@@@@@@
/post26
nsol,2,12,u,y,uy12
nsol,3,14,u,y,uy14
nsol,4,7,u,y,uy7
nsol,5,8,u,y,uy8
plvar,2,3,4,5
87