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POST-BUCKLING ANALYSIS AND DESIGN OF 3D
TRUSSES
By
Ahmed Shaban Mahmoud
A Thesis Submitted to the
Faculty of Engineering, Cairo University
In Partial Fulfillment of theRequirements for the Degree of
MASTER OF SCIENCE
In
Structural Engineering
FACULTY OF ENGINEERING, CAIRO UNIVERSITY
GIZA, EGYPT
2016
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POST-BUCKLING ANALYSIS AND DESIGN OF 3D
TRUSSES
By
Ahmed Shaban Mahmoud
A Thesis Submitted to the
Faculty of Engineering, Cairo UniversityIn Partial Fulfillment of the
Requirements for the Degree of
MASTER OF SCIENCEIn
Structural Engineering
Under the Supervision of
Dr. Sherif Ahmed Mourad Dr. Maheeb Abdel-Ghaffar
Professor of Steel Structures and Bridges
Dean, Faculty of Engineering,
Cairo University
Associate Professor
Structural Engineering Department
Faculty of Engineering,
Cairo University
FACULTY OF ENGINEERING, CAIRO UNIVERSITY
GIZA, EGYPT
2016
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POST-BUCKLING ANALYSIS AND DESIGN OF 3D
TRUSSES
By
Ahmed Shaban Mahmoud
A Thesis Submitted to the
Faculty of Engineering, Cairo University
In Partial Fulfillment of the
Requirements for the Degree of
MASTER OF SCIENCEIn
Structural Engineering
Approved by the
Examining Committee
____________________________
Prof. Dr. Sherif Ahmed Mourad, Thesis Main Advisor
____________________________
Assoc. Prof. Dr. Maheeb Abdel-Ghaffar, Member
____________________________
Prof. Dr. Ahmed Atef Rashed, Internal Examiner
____________________________Prof. Dr. Ahmed Abdelsalam El-Serwi, External Examiner
Faculty of Engineering, Ain Shams University
FACULTY OF ENGINEERING, CAIRO UNIVERSITY
GIZA, EGYPT2016
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Engineers Name: Ahmed Shaban Mahmoud
Date of Birth: 29/2/1984
Nationality: Egyptian
E-mail: [email protected]
Phone: 01002663205
Address: Pyramids Gardens,GizaRegistration Date: 1 / 10 / 2010
Awarding Date: …./…./2016
Degree: Master of Science
Department: Structural Engineering
Supervisors:Prof. Dr. Sherif Ahmed Mourad
Assoc. Prof. Dr. Maheeb Abdel-Ghaffar
Examiners:Prof. Dr. Ahmed Abdelsalam El-Serwi (External
examiner) Faculty of Engineering, Ain Shams
University
Prof. Dr. Ahmed Atef Rashed (Internal examiner)
Porf. Dr. Sherif Ahmed Mourad (Thesis main
advisor)
Assoc. Prof. Dr. Maheeb Abdel-Ghaffar (Member)
Title of Thesis:
POST-BUCKLING ANALYSIS AND DESIGN OF 3D TRUSSES
Key Words: Nonlinear, co-rotational, inelastic buckling, post buckling, advanced design, direct
analysis.
Summary:
A full nonlinear analysis is implemented using MATLAB to solve 3D trusses and2D frames. The proposed analysis considers nonlinear geometry, inelastic material and
initial out-of-straightness. This program is verified using common benchmark
problems. The analysis results using the proposed 3D truss element show more logic
results compared with the analysis using the equivalent out-of-straightness. This
advanced analysis provides some important results concerning the behavior of the post
buckling and shows major differences compared with typical designs using the linear
analysis.
Insert
photo h ere
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i
Acknowledgments
Firstly, I would like to express my sincere gratitude to my advisors Porf. Dr. Sherif
Ahmed Mourad and Assoc. Prof. Dr. Maheeb Abdel-Ghaffar for the continuous support
of my MSc study and related research, for their patience, motivation, and immense
knowledge. Their guidance helped me in all the time of research and writing of this thesis.
I could not have imagined having a better advisors and mentors for my MSc study.
Besides my advisors, I would like to thank the rest of my thesis committee: Prof.
Dr. Ahmed Abdelsalam El-Serwi and Prof. Dr. Ahmed Atef Rashed for their insightful
comments and encouragement, but also for the hard question that incented me to widen
my research from various perspectives.
Last but not the least; I would like to thank my family: my parents, my sisters and
special thanks to my wife for supporting me spiritually throughout writing this thesis andmy life in general.
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ContentsAcknowledgments ..................................................................................................... i
Contents .................................................................................................................... ii
List of Tables ............................................................................................................ v
List of Figures .......................................................................................................... vi Abstract ................................................................................................................. viii
Chapter 1 : Introduction ............................................................................................ 1
1.1. Introduction .................................................................................................... 1
1.2. Literature review ............................................................................................ 1
1.2.1. Frame models .......................................................................................... 1
1.2.2. Truss models ........................................................................................... 2
1.2.3. Solution of nonlinear system of equations .............................................. 7
1.3. Problem statement.......................................................................................... 9
1.4. Methodology .................................................................................................. 9
1.5. Thesis organization ...................................................................................... 10
Chapter 2 : Finite Elements Used ........................................................................... 11 2.1. Introduction .................................................................................................. 11
2.2. 2D frame element......................................................................................... 11
2.2.1. 2D frame element small-displacement ................................................. 11
2.2.2. 2D frame element large displacement .................................................. 15
2.3. 3D truss element .......................................................................................... 25
2.3.1. Elastic element large displacement ....................................................... 25
2.3.2. 3D inelastic element large displacement .............................................. 27
Chapter 3 : Nonlinear Solution Techniques ............................................................ 32
3.1. Critical points in nonlinear equilibrium paths ............................................. 32
3.1.1. Load limit points ................................................................................... 32
3.1.2. Displacement limit points ..................................................................... 32 3.2. Common methods for solving nonlinear problems ..................................... 33
3.2.1. Single step iterative method .................................................................. 33
3.2.2. Simple incremental method .................................................................. 34
3.2.3. Standard Newton Raphson method....................................................... 34
3.2.4. Modified Newton Raphson method ...................................................... 35
3.2.5. Controls used in the incremental-iterative methods ............................. 36
3.3. Unified approach to nonlinear solution ....................................................... 40
3.3.1. Base algorithm ...................................................................................... 40
3.3.2. Load control .......................................................................................... 41
3.3.3. Arc length control ................................................................................. 41
Chapter 4 : Verification Examples.......................................................................... 43
4.1. 2D frame element......................................................................................... 43
4.1.1. Example 1: Cantilever beam ................................................................. 43
4.1.2. Example 2: Lees frame ........................................................................ 44
4.1.3. Example 3: Euler beam ......................................................................... 49
4.1.4. Example 4: Rectangular frame with X-bracing .................................... 50
4.2. 3D truss element .......................................................................................... 53
4.2.1. Example 1: Two-bar truss system ......................................................... 53
4.2.2. Example 2: Space truss dome system ................................................... 54
4.2.3. Example 3: Twelve-bar space truss system .......................................... 56
Chapter 5 : Case-Studies ......................................................................................... 63 5.1. Effect of variables on local element curve................................................... 63
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B.1.5. RectangleDatabase.dat ....................................................................... 115
B.2. NFA Inputs for problem in 4.1.1 ............................................................... 115
B.3. NFA Inputs for problem in 4.1.2 ............................................................... 116
B.4. NFA Inputs for problem in 4.1.3 ............................................................... 117
B.5. NFA Inputs for problem in 4.1.4 ............................................................... 119
B.6. NTA Inputs for problem in 4.2.1 .............................................................. 122 B.7. NTA Inputs for problem in 4.2.2 .............................................................. 123
B.8. NTA Inputs for problem in 4.2.3 .............................................................. 126
Appendix C : STAAD PRO report ....................................................................... 129
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List of TablesTable 4-1 Dimension of sections used in the rectangular frame ........................... 51
Table 5-1 Summary of limit load for star dome ................................................... 70
Table 5-2 Summary of limit load for star dome (continued) ................................ 71
Table 5-3 Result of star dome optimum design model and upsizing model ......... 89 Table 5-4 Mild steel dome Vs. high tensile using NTA ....................................... 95
Table A-1 Description of files in NFA ............................................................... 101
Table A-2 Description of files in NTA ............................................................... 102
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List of FiguresFigure 1-1 Hysteresis Loops Method for Stee1 Braces [15] .................................. 4
Figure 1-2 Formulation of Hysteresis Behavior of Braces [16] ............................. 5
Figure 1-3 Normalized Compression Curves ......................................................... 5
Figure 1-4 Stress-strain relations [19] .................................................................... 6 Figure 1-5 Assumed Stress-Strain Behavior for the Brace Model [20] .................. 7
Figure 2-1 Deformed shape of the beam element [5] ........................................... 17
Figure 2-2 Co-rotational formulation [37] ............................................................ 18
Figure 2-3 Stress-Strain relation ........................................................................... 25
Figure 2-4 Axial load-axial displacement for strut model [22] ............................ 29
Figure 2-5 Strut in the elastic compression range [22] ......................................... 29
Figure 2-6 Strut in plastic compression range [22]............................................... 29
Figure 2-7 Axial load-axial displacement for tie model [23] ............................... 30
Figure 3-1 Limit points in nonlinear equilibrium paths ........................................ 32
Figure 3-2 Single step iterative method ................................................................ 33
Figure 3-3 Simple incremental method ................................................................ 34 Figure 3-4 Standard Newton Raphson method ..................................................... 35
Figure 3-5 Modified Newton Raphson method .................................................... 36
Figure 3-6 Snap through behavior ........................................................................ 37
Figure 3-7 Snap back behavior ............................................................................. 38
Figure 3-8 Arc-length method [36] ....................................................................... 39
Figure 3-9 Types of arc-length method [36] ......................................................... 39
Figure 3-10 Incremental-iterative procedure [36] ................................................ 40
Figure 3-11 Parameters of Arc-length control method [36] ................................. 42
Figure 4-1 Cantilever beam .................................................................................. 43
Figure 4-2 Cantilever beam: P-!v curve ............................................................... 44
Figure 4-3 Cantilever beam: deformations ........................................................... 44 Figure 4-4 Lee's frame .......................................................................................... 45
Figure 4-5 Lee's frame N.L.E: Load vs. displacement u ...................................... 46
Figure 4-6 Lee's frame N.L.E: Load vs. displacement v ...................................... 46
Figure 4-7 Lee's frame N.L.E: Deformed shapes ................................................. 47
Figure 4-8 Lee's frame E.P.A: Deformed shapes ................................................. 47
Figure 4-9 Lee's frame E.P.A: Load vs. displacement u ...................................... 48
Figure 4-10 Lee's frame E.P.A: Load vs. displacement v .................................... 48
Figure 4-11 Strut model for imperfect element .................................................... 49
Figure 4-12 Strut model: Load vs. axial shortening ............................................. 50
Figure 4-13 Strut model: Deformed shapes .......................................................... 50
Figure 4-14 Rectangular frame with X-bracing [22] ............................................ 51
Figure 4-15 X-bracing frame: Load vs. displacement u4 ..................................... 52
Figure 4-16 X-bracing frame: Load vs. displacement v4 ..................................... 52
Figure 4-17 Two-bar truss system ........................................................................ 53
Figure 4-18 two-bar truss: Load vs. displacement ................................................ 54
Figure 4-19 Space truss dome system................................................................... 55
Figure 4-20 Space truss dome: Load vs. displacement at load case 1 .................. 56
Figure 4-21 Space truss dome: Load vs. displacement at load case 2 .................. 56
Figure 4-22 Twelve-bar space truss system [22] .................................................. 57
Figure 4-23 Twelve-bar truss E.L.D. analysis: Load vs. disp. w at point A ......... 58
Figure 4-24 Twelve-bar truss E.L.D. analysis: Load vs. disp. u at point B .......... 58 Figure 4-25 Twelve-bar truss E.L.D. analysis: Load vs. disp. w at point B ......... 59
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Figure 4-26 Twelve-bar truss advanced analysis: Load vs. disp. w at point A .... 59
Figure 4-27 Twelve-bar truss advanced analysis: Load vs. disp. u at point B ..... 60
Figure 4-28 Twelve-bar truss advanced analysis: Load vs. disp. w at point B .... 60
Figure 4-29 Twelve-bar truss advanced analysis: Load vs. disp. w at point A .... 61
Figure 4-30 Twelve-bar truss advanced analysis: Load vs. disp. u at point B ..... 61
Figure 4-31 Twelve-bar truss advanced analysis: Load vs. disp. w at point B .... 62 Figure 5-1 Critical compressive stress .................................................................. 63
Figure 5-2 Allowable compression stress AISC 360-10 vs. DA500 .................... 64
Figure 5-3 Effect of Fy and initial imperfection at L/r =200 ................................ 65
Figure 5-4 Effect of Fy and initial imperfection at L/r =157 ................................ 66
Figure 5-5 Effect of Fy and initial imperfection at L/r =100 ................................ 66
Figure 5-6 Effect of initial imperfection at Fy = 248.2 N/mm2 and L/r =157 ...... 67
Figure 5-7 Effect of Fy at L/r =157 and initial imperfection L/1000 ................... 68
Figure 5-8 Effect of yield stress on model at L/r=157 and same code limit ........ 68
Figure 5-9 Space truss dome system..................................................................... 69
Figure 5-10 Star dome: effect of roof rise using LM+NG .................................... 72
Figure 5-11 Star dome: effect of roof rise using DA-Liew .................................. 73
Figure 5-12 Star dome: effect of roof rise using DA500 ...................................... 74
Figure 5-13 Star dome: effect of roof rise using DA1000 .................................... 74
Figure 5-14 Star dome: effect of roof rise using DA1500 .................................... 75
Figure 5-15 Star dome: AISC 360-10 (LRFD) design vs. DA-Liew.................... 76
Figure 5-16 Star dome: AISC 360-10 (LRFD) design vs. DA500 ....................... 77
Figure 5-17 Star dome: AISC 360-10 (LRFD) design vs. DA1000 ..................... 78
Figure 5-18 Star dome: AISC 360-10 (LRFD) design vs. DA1000 ..................... 79
Figure 5-19 Star dome: Effect of raise angle on analysis DA vs. LM+NG .......... 80
Figure 5-20 Star dome: Effect of raise angle on analysis DA vs. LM+LG .......... 81
Figure 5-21 Star dome: capacity Vs. slope ........................................................... 81
Figure 5-22 Ratio of different analysis to NG without code limit Vs. slope ........ 82
Figure 5-23 Space truss dome system for real design .......................................... 83
Figure 5-24 Loaded area of the space truss dome ................................................ 84
Figure 5-25 Analysis result of real dome (optimum design) using NTA ............. 85
Figure 5-26 Detailed curve of real dome (optimum design) analysis result ......... 85
Figure 5-27 Analysis result of real dome (all pipes"48) using NTA .................. 86
Figure 5-28 Detailed curve of real dome (all pipes "48) analysis result ............. 87
Figure 5-29 Internal force of elements No.1 to 6 in the two models .................... 87
Figure 5-30 Internal force of elements No.13 to 24 in optimum design model ... 88
Figure 5-31 Internal force of elements No.13 to 24 in All Pipes #48 model ....... 88
Figure 5-32 Analysis result of real dome (load case 1) using NTA ..................... 90 Figure 5-33 Detailed curve of real dome (load case 1) analysis result ................. 90
Figure 5-34 Analysis result of real dome (load case 2) using NTA ..................... 91
Figure 5-35 Detailed curve of real dome (load case 2) analysis result ................. 92
Figure 5-36 Analysis result of real dome (load case 2) with different imperf. ..... 93
Figure 5-37 Analysis result of dome with A570-50 using NTA .......................... 94
Figure 5-38 Detailed curve of dome with A570-50 analysis result ...................... 94
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Abstract
A full nonlinear analysis of structures requires the investigation of the entire
equilibrium path. This work considers available algorithms for tracing equilibrium paths
such as displacement control method and arc length method.
Two programs are developed by the author using MATLAB, which are NFA
(Nonlinear Frame Analysis), and NTA (Nonlinear Truss Analysis). Different types of
element matrices are used in both programs. NFA uses an element stiffness matrix
considering elastic material and nonlinear geometry (small or large displacement) and
uses another element stiffness matrix considering inelastic material and nonlinear
geometry assuming large displacement matrix. NTA uses an element stiffness matrix that
considers elastic material and linear or nonlinear large displacement and uses another
element stiffness matrix that considers inelastic material and nonlinear large
displacement matrix.
The accuracy of the programs is verified with some common benchmark problems;
then they are used to investigate the behavior of some structures. The structural capacity
is estimated and post-buckling curve of the structure is drawn. It is found that the
knowledge of post-buckling behavior is crucial for some specific structures such as
space-trusses and shallow arches.
Keywords: nonlinear, co-rotational, inelastic buckling, post buckling, arc length
control, advanced design, direct analysis.
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geometric nonlinearity caused by the large rigid-body motion, is integrated in the
transformation matrices. Assuming the pure deformation part to be small, a $linear theory can be used. As highlighted by Haugen and Felippa [8] , [9] in their review of the
topic, the main advantage of these assumption is the possibility to reuse existing high-
performance linear elements.
In order to fully utilize this possibility, Rankin and co-workers [6] [10] introduced
the $element independent co-rotational formulation. The definition of the element relates
to several changes of variables from the local frame to the global one. This is done using
a projector matrix that relates the variations of the local displacements to the variations
of the global displacements, by excluding the rigid body modes from the global
displacements. A very similar line of work was proposed by Pacoste and Eriksson [5]
[11] [12] [13].
The main interest of the co-rotational approach compared to the total Lagrangian is
its independence on the assumptions used to derive the internal forces and tangent
stiffness in local coordinates.
An elasto-plastic element based on co-rotational framework was presented by Pattini
[14].In the elasto-plastic context, one additional interest of the co-rotational approach is
the separation of material and geometrical nonlinearities. The geometrical nonlinearity
is included in the rigid-body motion of the element; however, the material nonlinearity
exists only in the local deformations. According to that, the internal forces and tangent
stiffness matrix in local coordinates is expressed in a simple manner. However, opposite
to the elastic case, analytical expressions cannot be derived. Due to the material
nonlinearity, numerical integration over the cross-section is required.
1.2.2. Truss models
Several models are used in the post-limit analysis of steel trusses and braced frames.
Kato and co-workers [15] developed a simple model to represent the hysteretic behavior
of braces. The basic features of the model are shown in Figure 1-1 ( a and b ) the load
capacity at B is equa1 to that at the previous unloading point A and the slope of line AB
is equal to the initial elastic stiffness at O. The same rules are applicable to point F and
point G in Figure 1-1 ( c ). The maximum load capacity was estimated by the buckling
strength according to the Japanese specification. The unloading curve and post buckling
capacity were calculated by an empirical equation depending on the cross sectional type
and the slenderness ratio.
Another model proposed by Wakabayashi and co-workers [16] is based on the
hysteresis properties noticed in tests, According to this model, curves BCD, EH,FG, ...
etc, in Figure 1-2 have a stated configuration. The slope of the unloading line AB which
starts from point A after yielding in tension is equal to the initial elastic slope, and the
slope of the unloading line DE from point D is stated using the property of ∆t / ∆c = ∆t
/ ∆c. The formulated curves are composed of four basic elements; tension and
compression mechanism curves, elastic unloading line and plastic yield line in axial
tension. The basic parameters of the formulated loops are described as a function of
slenderness ratio, a similar model was used by Nakashima and Wakabayashi [17] to
design the braced frames under cyclic and earthquake loading.
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For the limit states analysis of transmission towers, Pricken and co-workers [18]
modeled the post buckling behavior by a bilinear approximation as shown in Figure 1-3.
In this model, the limit loads for members are dimensionless to represent the post
buckling resistance of the member as a ratio of the buckling load based on the slenderness
ratio of the member. In this approximation, the member is assumed elastic until the
buckling load is reached. After that, the member stiffness is reduced linearly until itsaxial shortening reaches a certain displacement after which the buckled member will
oppose a constant force for any additional displacements in the post buckling range. The
tension behavior of all members was assumed elastic-perfectly plastic. This analysis did
not include unloading and strain hardening.
Papadrakakis [19] compared between several stress-strain relations shown in
Figure 1-4 for truss members in the post-critical load analysis of space trusses. The
trusses displayed great changes in its response.
Hill and co-workers [20] modeled the stress-strain behavior as shown in Figure 1-5.
Compression behavior until buckling is assumed linear elastic with the same slope as thematerial's elastic modulus. The post buckling behavior is modeled by an empirical
equation that depends on slenderness ratio and the asymptotic lower stress limit,
Unloading from the inelastic post buckling region is assumed to be a straight line from
point r on the post buckling curve to point a that corresponds to one half the material
yield stress Fy . After reaching point a in Figure 1-5, the member behavior will be
modeled as a tension member. Based on tests, which include a normal framing
eccentricity, Mueller and co-workers [21] developed a computer model to capture the
post-buckling behavior of angle struts with slenderness ratio greater than 120. Regression
model was used to obtain an empirical relation between axial load P and axial shortening
∆. The formula is P = Area *[ a * exp(b* L/r)] , where a and b are constants that depend
on the axial shortening ∆. Although the fit is fairly good for several cases as shown by
Abdel-Ghaffar [22], the model is complicated as a and b are indirect functions of ∆ but
series of 19 independent constants at values of ∆. In addition, the model is only valid for
L/r >120.
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Axial
Load,P
Axial
Shortening,O
P
_
Puy
Axial
Load,P
Axial
Shortening,
O
A
-Puy B
CD
Pyt
Puy
AxialLoad,P
Axial
Shortening,
O
Puyi
- Puyi
D
G
F
E
( a )
( c )
( b )
Figure 1-1 Hysteresis Loops Method for Stee1 Braces [15]
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AxialLoad,P
Axial
Shortening,
_ c
_'c
_ t
_'t
HG I
A
F
E
D
C
B
J
P
_
Figure 1-2 Formulation of Hysteresis Behavior of Braces [16]
PP
_
1.00
0.66
0.50
0.33
1.00 20 cr 0.500.250.10
0.05
r
L= 240
r
L= 180
r L = 80
r
L= 130
Pcr
Figure 1-3 Normalized Compression Curves
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y
y
y
y
E
y
E
y
E
y
E
y
( a ) ( b ) ( c ) ( d )
( e ) ( f ) ( g )
perfectly plastic elasto plastic elasto plastic
Figure 1-4 Stress-strain relations [19]
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cr
y
r
r
e
e
Figure 1-5 Assumed Stress-Strain Behavior for the Brace Model [20]
Abdel-Ghaffar [22] develop a simple model for the pre- and post-buckling behaviorof the bracing members under compression, tension, leading, unloading, etc. The model
is based on the usage of relationship between axial load and axial deformation to provide
the elastic and inelastic member behavior. Every bracing member is modelled as one
element in the finite element code. The change in stiffness throughout the analysis is
applied by updating the member stiffness according to the member's axial force-axial
deformation relationship. For tension members, one model is assumed that considers
both residual stress and strain hardening. For compression members, two models are
used; one for the members that do not show any torsional or local failure, the second is
for members that may suffer local and/or torsional failure. The first model is based on
closed form solution for different behavior regions; the second model is based on fitting
test results by two or more curves according to the cross-sectional shape andconfiguration.
Liew and co-workers [23] developed an advanced analytical technique for space
truss structures using strut model provided in Abdel-Ghaffar [22] and assuming that the
maximum strength of the strut is calculated based on a member with an equivalent out-
of-straightness to achieve the specification's strength for an axially loaded column.
1.2.3. Solution of nonlinear system of equations
Large deformation analysis of structures requires solution of a nonlinear system ofequations. Nonlinear systems of equations are commonly solved using iterative
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incremental techniques where small incremental changes in displacement are found by
applying small incremental changes in load on the structure. The resulting solutions are
used to plot a curve of the equilibrium path for the structure. Presentation for the most
popular solution techniques, are given by Crisfield [24]. The most common technique
for solving nonlinear finite element equations is the Newton-Raphson method. The
Newton-Raphson method is famous for its rapid convergence but is known to fail at points (limit points) on the equilibrium path where the tangent stiffness is singular or
nearly singular. Bathe and Cimento [25] showed the problems with the Newton-Raphson
method and presented many versions of the method that include accelerations or line
searches to ensure convergence during the solution process.
More recently, arc length methods have been used to overcome the problem of
tracing the equilibrium path in the limit points. The arc length methods are similar to the
Newton-Raphson method except that the applied load increment becomes variable. A
comparative study of arc length methods was presented by Clarke and Hancock [26]. The
original idea behind the arc length method was proposed by Riks [27] [28] and Wempner
[29]. The original method proposed by Riks and Wempner destroyed the symmetry ofthe finite element matrix and made the numerical solution wasting time. The Riks-
Wempner method modified by Crisfield [30] and Ramm [31] to maintain the symmetry
of the finite element matrix. Both researchers proposed two methods for modifying the
original method of Riks and Wempner. The first forced the iterative process to lie on a
plane normal to a tangent to the equilibrium path. The second, constrained iteration to
the surface of a sphere whose radius is a tangent to the equilibrium path. In both methods,
the user specifies the length of the tangent. Both methods are used widely in current finite
element work. Iteration on a normal plane is the easiest solution to implement, but
iteration on a sphere has proven to converge in more cases. Watson and Holzer [32]
presented a study of the convergence of iteration on a sphere. The method was found to
have quadratic convergence for a single degree-of-freedom system, and a slightly lower
average rate of convergence for a 21 degree-of-freedom system. The major problem with
iteration on a sphere is that the technique gives two solutions to the unknown load
increment and in some cases does not give a real solution at all. Crisfield [30] [24]
proposes a method for choosing the correct solution from the two solutions. Meek and
Tan [33] and Meek and Loganathan [34] [35] examined the problem of imaginary
solutions and found that this problem only occurred for certain types of structures and
made recommendations on how to solve the problem. They also studied the problem of
determining the correct sign of the load increment in the neighborhood of limit points.
Crisfield [24] has also proposed a version of the arc length method, which is known as
the cylindrical arc length method. Many of the same problems occurred with the sphericalarc length method also occur when using the cylindrical arc length method.
Other algorithms are also developed to solve the nonlinear problems as
displacement, work, generalized displacement and orthogonal residual control
algorithms. One single algorithm may not be capable of solving all nonlinear problems.
Leon [36] presented an algorithm that unified into a single framework all of these control
algorithms. Each of these solution methods differs in the use of a constraint equation for
the incremental-iterative process. The finite element equations and constraint equation
for each solution method are combined into a single matrix equation, which identifies the
unified approach. This concept leads to an effective implementation.
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1.3. Problem statement
Advanced design (Direct analysis) of structures requires carrying stability analysis
considering nonlinear material, nonlinear geometry, residual stress and the initial out-of-
straightness of individual members. In large space truss structures, it is not practical to
divide each truss element into many frame elements to account for the initial out-of-straightness. Thus, it is preferred to use a truss element after putting all these effects into
consideration.
After including all above effects, the structure can be analyzed under a certain
factored load. However, to have good evaluation of the structure near failure, it is also
necessary to investigate the full equilibrium path. The load controlled Newton-Raphson
method was the first attempt to obtain the equilibrium. However, the method diverges
after a limit point, and therefore, only part of the curve is usually obtained. The %collapse
loads& were then often associated with such limit points and with the failure to achieve
convergence with the iterative solution procedure. Nevertheless, as Crisfield [24] quoted,
many questions remained open:
- Was it really a limit point or was the iterative solution procedure that has
collapsed?
- Was it only a local maximum and the structure can be further loaded without
collapsing?
- How is the collapse process, ductile or brittle?
To overcome the difficulties with limit points, a technique is needed that is able to
draw the entire equilibrium path in the case of snap-through or snap-back structural
behaviors.
The method of estimating the initial out-of-straightness of elements to mimic the
allowable stress provided in design codes that was presented by Liew and co-workers
[23] has some illogical results in some cases. Another method is required to study the
effect of the shallowness of a star dome on its loading capacity.
1.4. Methodology
In order to perform the required analysis a program called NTA (Nonlinear Truss
Analysis) is programmed by the author using MATLAB. The program is able to analyze
3D truss structure considering nonlinear geometry, nonlinear material and initial out-of-
straightness of individual members. An element behavior proposed by Abdel-Ghaffar[22] is included to take these effects.
The Arc-Length control method is used to solve the system of nonlinear equations
and to trace the full equilibrium path. This method is able to overcome load and
displacement limit points. However, the method is implemented using the Unified
Approach cited in Leon [36] that enables the use of other control methods in an easy way.
The 3D truss element is compared with nonlinear frame analysis using the truss
element divided into many elements considering all above nonlinearities. For this scope,
a program called NFA (Nonlinear Frame Analysis) is programmed by the author in
MATLAB using elements introduced by Battini [37] and by Pacoste and Eriksson [5].
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Through some case studies, advantages and disadvantages of the method of
assuming an equivalent out-of-straightness to achieve the specification's strength that
was presented by Liew and co-workers [23] are highlighted. Moreover, an alternative
method is proposed to overcome the disadvantages.
1.5. Thesis organization
The body of this work is divided into several chapters in order to give a theoretical
background for the implemented programs, the element matrix formulation and the path
following algorithms.
In Chapter 2, the basic equations for different types of finite elements that are used
in the proposed programs are shown.
In Chapter 3, the major critical points during tracing the equilibrium path are
discussed. Some of the common path following techniques and the %Unified approach to
nonlinear solution schemes& algorithm are presented.
In Chapter 4, the programs NFA & NTA are used to solve several problems from
the textbooks and results are verified with others.
In Chapter 5, some case-studies using the implemented programs are shown. The
effects of changing yield stress, slenderness ratio and initial out-of-straightness on the
performance of a truss element are discussed. The effect of changing rise height on the
behavior of the star dome is shown using NTA program. In addition, a star dome with
real dimensions is analyzed for different type of loading and different type of materials
using the available elements in NTA program.
In Chapter 6, a summary of all chapters is presented. Conclusions from the study
of problems are described, and suggestions for future research are given.
Appendix A, flow charts and job descriptions for main routine files of the proposed
programs NFA & NTA are shown. Clarifications about the input files required by NFA
& NTA are provided.
In appendix B, all input files of verification examples that were given in Chapter 4
are attached.
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Chapter 2 : Finite Elements Used
2.1. Introduction
Finite elements used in NFA to analyze 2D frames and finite elements used in NTA
to analyze 3D trusses are presented in this chapter. The presented elements contain both
elastic elements and inelastic elements. The following six finite elements are stated for
analyzing 2D frame. First two elements are based on assumptions of elastic materials
with small displacement and ignoring shear deformation in the first element while the
second considers it. The next two elements are based on assumptions of elastic materials
with large displacement. Using total formulation the third element is able to handle large
displacement while the fourth element is based on Co-Rotational method. The last two
elements are based on assumptions of inelastic materials with large displacement. The
fifth element is based on the classical linear beam theory using a linear interpolation for
the axial displacement. However, the sixth element uses a shallow arch definition for the
local strain. The initial out-of-straightness can be considered indirectly in the 2D frame
elements through dividing the element into many elements and modifying the joint
coordinates of these subdivides according to the value and shape of the out-of-
straightness.
For using in the analysis of 3D trusses, two elements are stated. The first element is
based on assumptions of elastic materials with large displacements while the other
element considers inelastic materials, large displacements and the initial out-of-
straightness. The initial out-of-straightness can be considered directly only in the second
3D truss element.
2.2. 2D frame element
2D frame elements are classified here into small and large displacement elements.
The small displacement elements assume that the inclination of the deformed frame
element is approximately the same as the inclination of the element before deformation.
However, the large displacement element assumes that the structure is under large
deformation, which requires changing the inclination of the elements during loading and
checking the equilibrium of the internal forces in the structure with the external applied
loads at deformed shape.
2.2.1. 2D frame element small-displacement
Two elements based on small displacement and elastic material assumptions are
shown here, the first element cited in [38] is based on stability function, ignoring the
shear deformation while the second element cited in [39] is derived taking into account
the shear deformation.
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2.2.1.1. Chen element
This element was cited in [38]; both P-! and P-( can be incorporated directly in the
analysis procedure. The major weakness of this method is assuming small displacement.
Local internal force vectorf i = k T u Eq. 2-1
Where:
f i local internal force vector for the element ,size [6,1]
k T local tangent stiffness matrix for the element will be stated in the next, size [6,6]
u local nodal displacement vector for each element node, size [6,1]
Local tangent stiffness matrix
! =
"
%&' 0 0 ( )*+
0 0
0 ,- /0
1234 56
78 0 9:; <=
>?@A BC
DE
0 FG HI
JKL MN
O 0
P QRST
U VWX
( YZ[
0 0 \]
^ 0 0
0 _`a bc
def gh
ij 0 kl mn
opq rs
tu
0 vw xy
z{| }~
• 0
!"#$
% &'( )
Eq. 2-2
,- =
./0
"
1/ 2 0 0 ( 3/ 4 0 0
0 56789: <=>?@ (AB)C
DEFGHIJK MNOP
Q 0 RSTUVWX Z[\]^ (_`)a
bcdefghi klmn
o
0 pqrstu wxyz
{ |}~ 0 •!" $%&' ()*
( +/ , 0 0 -/ . 0 0
0 /012345 789:; (<=)>
?@ ABCD FGHI 0 JKLMNO QRSTU (VW)X
YZ [\]^ `abc0
defghi jklmn opq
0 rstu vwx
y z{| )
Eq. 2-3
} = ~ |•|
! Eq. 2-4
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"#$ =
%
)* sin+, ( (-.)/ cos012 ( 2cos23 ( 45 sin67 , 89: ; < 0
4.0 , <=> @ = 0
(AB)C coshDE ( FG sinhHI2 ( 2coshJK +LM sinhNO , PQR S > 0
Eq. 2-5
TUV =
%
(WX)Y ( Z[ sin\]2 ( 2 cos^_ ( `a sinbc , def g < 0
2.0 , hij l = 0
mn sinhop ( (qr)s2 ( 2cosh
tu +
vwsinh
xy ,
z{| } > 0
Eq. 2-6
k T = k E + k G Eq. 2-7
Where:
k E local elastic stiffness matrix, size [6,6]
k G local geometric stiffness matrix, size [6,6]
k T local tangent stiffness matrix, size [6,6]
k is a factor defined in Eq. 2-4.
E elastic modulus for element material.
A cross section area of the element.
I moment of inertia of the element.
L unreformed length of the element.
N axial force in the element, positive for tension.
2.2.1.2. Gavin element
This element was derived in [39]; same as in Chen element both P-! p-δ can be
incorporated directly in the analysis procedure. The advantage of this method compared
to Chen element is considering the shear deformation into account. However, this
advantage does not affect the accuracy of the results in majority of cases.
Local internal force vector
f i = k T u Eq. 2-8
Local tangent stiffness matrix
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~• =
"#
!
"0 0 ( #$
%0 0
0 &' )*
+,(-./)
01 34
56(789) 0
:;< >?
@A(BCD)
EF HI
JK(LMN)
0 OP RSTU(VWX)
(YZ[) ]^_(`ab)
0 c efgh(ijk)
(lmn) pqr(stu)
(vw
x0 0 yz
{0 0
0 |}~ !"#($%&)
' )*+,(-./)
0 01 3456(789)
: <=>?(@AB)
0 CD EFGH(IJK)
(LMN) OPQ(RST)
0 U VWXY(Z[\)
(]^_) `ab(cde) )
*
Eq. 2-9
Where
f = gh ijk lm no
Eq. 2-10
pq =
r
s
"
0 0 0 0 0 0
0t
uvwxyz
{
(|}~)• !"#
($%&)' 0
()
* +,-./0
(123)4
5 67#
(89:);
0 < =>#
(?@A)B
CDE FGHIJK LMNOPQ RS###
(TUV)W 0
XY Z[#
(\]^)_
`ab cdefgh ijklmn op###
(qrs)t
0 0 0 0 0 0
0uv
w xyz{|}
(~•!)"
# $ % &#
('())* 0
+
,-./01
2
(345)6
7 8 9 :#
(;<=)>
0 ? @A#
(BCD)E
FGH
IJKLM
N O PQR
ST
UV###
(WXY)Z 0 [ \ ] ^#
(_`a)b
cde
fghij
k l mno
pq
rs###
(tuv)w )
Eq. 2-11
k T = k E + k G Eq. 2-12
Where:
A cross section area of the element.
As shear area of the element.
E elastic modulus for element material.G shear modulus for element material.
I moment of inertia of the element.
k E local elastic stiffness matrix, size [6,6]
k G local geometric stiffness matrix, size [6,6]
k T local tangent stiffness matrix, size [6,6]
L unreformed length of the element.
N axial force in the element, positive for tension.
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2.2.1.3. Transformation from local to global coordinates
The transformation process aims to convert tangent stiffness matrix and the internal
force vector from local coordinate system into global one. A transformation matrix
required to do this transformation is called T. This matrix is identical to the one used for
the elastic linear element. The transformation matrix, T, is
x =
"
cosy sin z 0 0 0 0( sin { cos| 0 0 0 0
0 0 1 0 0 00 0 0 cos} sin ~ 00 0 0 ( sin • cos 00 0 0 0 0 1)
Eq. 2-13
Fi = TT f i Eq. 2-14
K T = TT k T T Eq. 2-15
Where:
) the counter-clockwise angle between the global X-axis and the element axial axis.
Fi global internal force vector for the element, size [6, 1]
K T global tangent stiffness matrix for the element, size [6, 6]
2.2.2. 2D frame element large displacement
Based on assumptions of large displacement, four elements are introduced here. The
first element is based on the total formulation approach derived by Pacoste and Eriksson
[5] while the other three elements are based on the co-rotational approach. First and
second element assume elastic material, while third and fourth elements assume inelastic
material based on the Bernoulli strain assumption. Third and fourth elements use a linear
and a shallow arch strain definition, respectively.
2.2.2.1. Total formulation
This element was derived in [5]; element deformed configuration (Figure 2-1) isdescribed by a regular curve defined by the position vector:
r (x) = [ x + u(x) ] i + w (x) j Eq. 2-16
Where the abscissa x is measured on the straight reference configuration of the
beam, u(x), w(x) represent the axial and transversal displacement components and i and
j are unit axis vectors. In addition, each point on the beam axis has an associated cross
section (S); the angle )(x) defines the rotation of the cross sections in the deformed
configuration. Deformation measures *, , ! are defined, according to
r ,x = ( 1+ * ) a + b
" = ),x Eq. 2-17
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Where a(x) = cos ) i + sin ) j and b(x) = - sin ) i + cos ) j are the unit vectors,
orthogonal and parallel to the cross section. Using Eq. 2-16 the definition in Eq. 2-17 can
be reformulated to give:
* = (1+u,x) cos ) + w,x sin ) -1 Eq. 2-18
= w,x cos ) - (1+u,x) sin ) Eq. 2-19
# = ),x Eq. 2-20
The constitutive relations are taken as linear, according to
N=EA * T=GA M =EI $ Eq. 2-21
With these assumptions, the strain energy can be written as
%(&) = 12 '[ ()*+ +,-./ +0123 ] 456
7 Eq. 2-22
One-point Gaussian quadrature is used to perform the integral in Eq. 2-22.The one-
point quadrature has the advantage of avoiding locking problems. Moreover, in contrast
to shells, the beam elements do not exhibit spurious energy modes due to reduced
integration. Finally, the expressions of the internal force vector T = [Ni Ti Mi N j T j M j ]T
and tangent stiffness matrix K t are obtained through successive differentiation.
The MAPLE inputs that perform the above-mentioned operations and produce the
necessary MATLAB code are listed in following section. The finite element defined inthis way is denoted $ b2dtt by [5].
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y,w
x,u
l
i
j
S
x
S '
r(x)
u(x)
w ( x )
b(x)
a(x)
r 'x
(x)
Figure 2-1 Deformed shape of the beam element [5]
Local internal force vector and tangent stiffness matrix
The following is the local internal force vector and local tangent stiffness matrixrelated to the local axis of the unreformed element, to convert to the global coordinates
a transformation matrix similar to the one used in the ordinary linear analysis is to be
used. Using Maple software V. 11 to perform needed mathematical operations, we enter
the following code and the result will be the local internal force vector and local tangent
stiffness matrix.
with(linalg);
ux:=(uj-ui)/L; wx:=(wj-wi)/L;
t:=(ti+tj)/2; tx:=(tj-ti)/L;
e:=(1+ux)*cos(t)+wx*sin(t)-1;g:=wx*cos(t)-(1+ux)*sin(t);
k:=tx;
Fie:=1/2*L*EA*e^2;
Fig:=1/2*L*GA*g^2;Fik:=1/2*L*EI*k^2;
Fi:=Fie+Fig+Fik;
T:=grad(Fi,[ui,wi,ti,uj,wj,tj]);
Kt:=hessian(Fi,[ui,wi,ti,uj,wj,tj]);
Matlab(T,optimize);Matlab(Kt,optimize);
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2.2.2.2. Co-rotational framework
The purpose of this section is to state the relations between the local and global
expressions of the internal force vector and tangent stiffness matrix.
Figure 2-2 Co-rotational formulation [37]
The notations used in this section are defined in Figure 2-2. The coordinates for the
nodes 1 and 2 in the global coordinate system are (x, z) are (x1, z1) and (x2, z2). The
vector of global displacements is defined by
Pg = [ u1 w1 )1 u2 w2 )2 ]T Eq. 2-23
The vector of local displacements is defined by
89 = [ : ; < => ? =@ ]A Eq. 2-24
The components of pl can be computed according to
B C = DE ( FG Eq. 2-25
H =I = JK ( L Eq. 2-26
M =N = OP ( Q Eq. 2-27
In the above equations lo and ln denote the initial and current lengths of the element
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lo = [ ( x2 + x1 )2 + ( z2 + z1)2 ]1/2 Eq. 2-28
Ln = [ ( x2 + u2 + x1 , u1)2 + ( z2 + w2 + z1 , w1)2 ]1/2 Eq. 2-29
and
denotes the rigid rotation which can be computed as
sin - = co s + so c Eq. 2-30
cos - = co c + so s Eq. 2-31
co = cos .o = ( x2 , x1 ) / lo Eq. 2-32
so = sin .o = ( z2 , z1 ) / lo Eq. 2-33
c = cos . = ( x2 + u2 , x1 , u1 ) / ln Eq. 2-34
s = sin . = ( z2 + w2 , z1 , w1 ) / ln Eq. 2-35
Then, provided that |-| < /, - is given by
= sin! 1 (sin )
= cos! 1 (cos )
= sin! 1 (sin )
= ! cos! 1 (cos )
if sin - 0 0 and cos - 0 0
if sin - 0 0 and cos - < 0
if sin - < 0 and cos - 0 0
if sin - < 0 and cos - < 0
Eq. 2-36
The transformation matrix B is given by
B = R (S (T 0(U/ VW X/ YZ 1
[ \ 0]/ ^_ (`/ ab 0(c/ de f/ gh 0 i/ jk (l/ mn 1
o Eq. 2-37
The following notations are introduced
r = [ -c -s 0 c s 0 ]
T
Eq. 2-38z = [ s -c 0 s c 0 ]T Eq. 2-39
The global tangent stiffness matrix becomes
Kg = BT K l B +p qrst N +
uvwx ( r zT + z r T) (M1 + M2) Eq. 2-40
The global internal force vector becomes
Fi = BT f i Eq. 2-41
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2.2.2.3. Elastic element
Local internal force vector
f i = [ N M1 M2 ] Eq. 2-42
Where
y = z{| } ~ Eq. 2-43
• = ! "#$ ( 2%& +'() Eq. 2-44
)* = + ,-. ( /0 + 212) Eq. 2-45
Local tangent stiffness matrix
K 3 =
"#
456 0 0
0 478
92:;<
0 2=>
?4@AB )* Eq. 2-46
2.2.2.4. Inelastic linear strain Bernoulli element
This element is based on the classical linear beam theory, using a linear interpolation
for the axial displacement u and a cubic one for the vertical displacement w.
C = DE FG Eq. 2-47
H = I J1 (KLM
N O =P +
QRS T
UV ( 1W
X Y =Z Eq. 2-48
The curvature at any section located at distance x from element start
[ = \]^_`a = b (
4
c + 6 def gh =i + j (
2
k + 6 lmn op =q Eq. 2-49
Strain at any fiber
r = stsu ( vw =
xyz ( {| Eq. 2-50
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Where: z is the distance from center of fiber to the natural axis.
By using Gauss integration based on two points per element
}~ = •2 1 ( 1! 3" #$ = %2 &1 + 1! 3' Eq. 2-51
At each point of the two Gauss points per element, the following is to be calculated
sda = () *+
szda = (,- ./
Eda = (01 34
Ezda =
(567
89
Ez2da = (:;<= >?
Eq. 2-52
Where @ A is the material modulus of elasticity E or tangent modulus of elasticity Et
according to reached strain in the fibers.
Local internal force vector
f i = [ N M1 M2 ] Eq. 2-53
Where
N = 0.5 ( sda1 + sda2 ) Eq. 2-54
M1 =! BCDE FGHIJ + KL ! MN OPQRS
Eq. 2-55
M2 =! TUVW XYZ[\ + ]^ ! _` abcde
Eq. 2-56
Local tangent stiffness matrix
fg = hijkl mnop qrstuvwx yz{| }~•!"#$% &'() *+,-. Eq. 2-57
Where
/012 =s3s45 =
1
2 6 [ 789: + ;<=> ] Eq. 2-58
?@AB = sCsD =E =
! 3 + 1
2 F GHIJK + 1 ( ! 3
2 L MNOPQ Eq. 2-59
RSTU =
s
VsW =X =
! 3 ( 1
2 Y
Z[\]^ (
1 +
! 3
2 _
`abcd Eq. 2-60
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efgh =sijskl =
m! 3 + 1no2 p qr2stu +
v! 3 ( 1wx2 y z{2|}~ Eq. 2-61
•!" =s#$s% =& =
1
' [ ()2*+, + -.2/01 ] Eq. 2-62
2345 = s67s89 = :! 3 ( 1;<2 = >?2@AB + C! 3 + 1DE
2 F GH2IJK Eq. 2-63
2.2.2.5. Inelastic shallow arch Bernoulli element
This element uses a shallow arch definition for the local strain
Curvature at any section located at distance x from element start
L = MNO
PQR = S (
4
T + 6
U
VW XY
=Z + [ (
2
\ + 6
]
^_ `a
=b Eq. 2-64
rc =1
de[ sfsg +
1
2 hsisjk
l]
mno
= pqr +
1
15 s =tu (
1
30 v =wx =y +
1
15 z ={|
Eq. 2-65
Strain at any fiber
r = r} ( ~• Eq. 2-66
Where: z is the distance from center of fiber to the natural axis.
By using Gauss integration based on two points per element
! ="# $1 ( %! &' () =
*+ ,1 +
-! ./ Eq. 2-67
At each point of the two Gauss points per element calculate the following
sda =
(0 12
szda = (34 56
Eda = (78 :;
Ezda = (<=> ?@
Ez2da = (ABCD EF
Eq. 2-68
Where G A is calculated according to 2.2.2.6.
Local internal force vector
f i = [ N M1 M2 ] Eq. 2-69
Where
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N = 0.5 ( sda1 + sda2 ) Eq. 2-70
M1 =HI J
KLMNO ( P
QR STU [VWXY + Z[\]] +! _`a bcdef +
gh ! ij
klmno Eq. 2-71
M2 = pq r
stuvw ( x
yz { =|} [~•!" + #$%&] + ! '()* +,-./ +
01 ! 23 45678 Eq. 2-72
Local tangent stiffness matrix
K l = 9:;<= >?@A BCDEFGHI JKLM NOPQRSTU VWXY Z[\]
^ Eq. 2-73
Where
_`ab =scsde =
1
2 f [ ghij + klmn ] Eq. 2-74
opqr = ssst =u =
1
2v 2
15w =x (
1
30 y =z{ [|}~• +!"#]
+ ! 3 + 1
2 $ %&'() + 1 ( ! 3
2 * +,-./ Eq. 2-75
0123 = s4s5 =6 = 127 2
1589 ( 130 : =;< [=>?@ +ABCD]
+ ! 3 ( 1
2 E FGHIJ (1 + ! 3
2 K LMNOP Eq. 2-76
QRST =sUVsW =X =
Y2Z 2
15[ =\ (
1
30 ] = _` [abcd +efgh]
+ j 2
15k =l (
1
30 n =op qr! 3 + 1stuvwx
+y1 ( ! 3z{|}~• + !! 3 + 1
"#
2 % '(2)*++
,! 3 ( 1-.2 / 012234 +
515
[6781 +9:;2] Eq. 2-77
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<=>? =s@AsB =C =
D2E 2
15F =G (
1
30 H =IJ K 2
15L =M (
1
30 N =OP [QRST
+UVWX]
+
Z[! 3 ( 1\
2 ] 2
15 =_ (
1
30
a =bc+ d! 3 + 1e2
f 2
15g =h (
1
30 j =klmnopqr
( tu! 3 + 1v2
w 2
15x =y (
1
30 { =|}
+ ~! 3 ( 1•
2 2
15! =" (
1
30 $ =%&'()*+,
+1
- [./2011 +232452] (6
60[7891 +:;<2] Eq. 2-78
=>?@ = sABsC =D = E
2F 2
15G =H (
1
30 IJKL [MNOP +QRST]
+ V 2
15W =X (
1
30 Z =[\ ]^! 3 ( 1_`abcd
( e1 +! 3fghijkl +m! 3 ( 1no
2 q st2uvw+ x! 3 + 1yz
2 { |}2~• + !
15["#$1 +%&'2] Eq. 2-79
2.2.2.6. Stress-Strain curve for inelastic elements
A bilinear strain , stress relation, see Figure 2-3, with an isotropic hardening is
adopted. This model presents a constant modulus E t in the plastic range and E for
unloading. If unloading happens before reaching yield stress then, unloading will be on
the same loading curve. However, if unloading at stress > yielding stress then, unloading
will be on a new curve with reduced opposite yield stress. The reduced opposite yield
stress assumed to be Fs , 2 Fy.
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Strain
S t r e s s
y
L o a d i n g
U n L
o a d i
n g
E
Et
Ey-
-Fy
Fy
Figure 2-3 Stress-Strain relation
2.3. 3D truss elementTruss elements assume that all connections are pinned type and carry only axial
loads. Due to this fact, the stiffness matrix becomes simpler than the one of frame
elements and the analysis becomes faster. This is the reason why only truss elements with
large displacement will be presented in below sections.
2.3.1. Elastic element large displacement
This element assumes elastic material and large displacement of the structure.
2.3.1.1. Global tangent stiffness matrix
The tangent stiffness matrix for an elastic truss element
K T = K E + K G Eq. 2-80
Local elastic stiffness matrix for 3D truss element
() =
*+,- "
1 0 0 (1 0 00 0 0 0 0 00 0 0 0 0 0
(1 0 0 1 0 00 0 0 0 0 00 0 0 0 0 0
)
Eq. 2-81
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Global elastic stiffness matrix for 3D truss element
K E = TT k E T Eq. 2-82
Transformation matrix for 3D truss element T can be calculated as following [40, p.
cluse 8.1]
. =
"
c c cz 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 c c cz0 0 0 0 0 00 0 0 0 0 0)
Eq. 2-83
Where:
cx = ( X2 , X1 ) / Ln Eq. 2-84
cy = ( Y2 , Y1 ) / Ln Eq. 2-85
cz = ( Z2 , Z1 ) / Ln Eq. 2-86
X1, Y1, Z1 is the coordinate of first point of the element based on deformed shapeX2, Y2, Z2 is the coordinate of second point of the element based on deformed shape
Ln is the length of the element based on deformed shape.
[41, p. Cluse 9.1] Stated the geometric matrix for 2D elastic truss element
/0 =12 3
1 0 (1 00 1 0 (1
(1 0 1 00 (1 0 1
4 Eq. 2-87
Where
N axial force in the element, positive for tension calculated in 2.3.1.2.
[36] Stated the geometric matrix for 3D elastic truss element
56 = 789
"
1 0 0 (1 0 00 1 0 0 (1 00 0 1 0 0 (1
(1 0 0 1 0 00 (1 0 0 1 0
0 0 (1 0 0 1
)
Eq. 2-88
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K G is the geometric matrix for the truss element and doesnt need any transformation
to be used either in local or global coordinate system because multiply transformation
matrix T * T will result in a unitary matrix.
2.3.1.2. Global internal force vector
The internal force vector can be calculated in the global coordinates
Fi = TT f i Eq. 2-89
The local internal force vector :; ="
(<0
0=00
)
Eq. 2-90
Where
N axial force in the element, negative for tension.
Axial force in the element N => ?@A B Eq. 2-91
! = Ln , L0 Eq. 2-92
Where
! axial deformation of the element based on local coordinate system, negative for
stretching.
Ln the new length of the element based on its deformed configuration.
L0 the initial length of the element.
When the difference between Ln and L0 is too small, the error in calculation of !
may be large due to round off error. Crisfield [24] recommended to avoid subtraction
operation between Ln and L0 by multiply it by CDE
GHIJK MN .O = PQRS TUVWXY Z[ Eq. 2-93
2.3.2. 3D inelastic element large displacement
The behavior of a truss element is modeled by a strut under compression force and
by a tie under tension force. Strut element follows the elastic curve (point A to B in
Figure 2-4) and then follows the plastic curve (point B to C in Figure 2-4). Figure 2-5
shows the deformation of a strut in the elastic region while Figure 2-6 shows the
deformation in the plastic region. Elastic and plastic curves for tie element are shown inFigure 2-7.
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The axial force-shortening relationship of a strut in the compression elastic range
can be computed as presented in [22]
\] = (
^_
`bcd
+ 1 (1
1 +
8
3e fghi(1 ( j/kl)mno Eq. 2-94
In the compression plastic range
qr = ( st u vwx + 1 ( y 1 ( z2{|}~•!"#$% &'( Eq. 2-95
)* = +,-./ & 01 =
23456 Eq. 2-96
In order to perform above differentia the MAPLE software is used to do and the
results shown below.
1 78# = 9:; = +
16 >?@3 A1 +
83 BCDE0F G1 ( HIJ KLM
N OP Q1 ( RSTUV WX
Eq. 2-97
1 YZ# = [\] _ ( 4 `abc def ghi 1 (4 jklm nop qrstu vwx yz {| }~• !
Eq. 2-98
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Axial
Load,P
Axial
Shortening,A
B
C
Elastic Curve
Plastic Curve
P
Post-Buckling
Strength
_
D
Elastic Unloading Curve
Pmax
Figure 2-4 Axial load-axial displacement for strut model [22]
P
_
L0
Arc Length=L0
Ln
Figure 2-5 Strut in the elastic compression range [22]
P
_ Ln
0 .5 L 0 0. 5 L 0
Figure 2-6 Strut in plastic compression range [22]
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In th tension elastic rang N = (" #$% |&| Eq. 2-99
In th tension plastic rang N
= (' ()*
+,-. (/0 123
( |
4| (
5678 )
Eq. 2-100
Axial
Load,P
Axial
Elongation,-
P l a s t i c C u r
v e
Py
y
E A/L
1
E A/Lt
E A/L
1
E l a s t i c
C u
r v e
E l a s t i c
U n l o
a d i n
g
1
Figure 2-7 Axial load-axial displacement for tie model [23]
9: =
%
&;<=, 0 < |>| < ?@ABCDEFG, |H| I JKLMNOPQ RST U < 0
%
VWXY
, |
Z| <
[\]^_ abc , |d| I efghi mno q I 0
Eq. 2-101
Where
K a The axial stiffness of the element.
! The axial shorting of the element calculated in 2.3.2.2, negative for
stretching.
!Pmax The axial shorting of the element corresponding to N = Pmax.
Pmax The axial load capacity based on the code.E Elastic modules of the elements material.
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2.3.2.1. Global tangent stiffness matrix
Same as 2.3.1.1 except the elastic stiffness matrix shall be as follows
rs =tu"
1 0 0 (1 0 00 0 0 0 0 00 0 0 0 0 0
(1 0 0 1 0 00 0 0 0 0 00 0 0 0 0 0
)
Eq. 2-102
In addition, the axial compression force N used in K G to be calculated according
to 2.3.2.2.
2.3.2.2. Global internal force vector
Axial shortening ! to be calculated according to 2.3.1.2 and solving equation
Eq. 2-94 or Eq. 2-95 to get axial compression force N in the element.
Construct global internal force vector same as 2.3.1.2 using N calculated from
above.
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Chapter 3 : Nonlinear Solution Techniques
Nonlinear problems are common in structural engineering to solve material,
geometry and contact nonlinear problems. Many algorithms are developed to solve these
problems. No single algorithm is capable of solving all nonlinear problems; dependingon the system and the degree of nonlinearity, one solution scheme may be preferred overanother. In this Chapter, a brief review of common nonlinear problems and the most
common algorithms are summarized.
Afterwards a description of the algorithm used to trace nonlinear problems in this
research is presented in detail.
3.1. Critical points in nonlinear equilibrium paths
Tracing an equilibrium path beyond the simple linear region and into a nonlinear
region is a complicated task in structural analysis. In fact, in many cases, it may seem
unnecessary to trace a path beyond the first load limit point. However, the full
equilibrium path, including critical points and regions of instability, gives more
information about the structural behavior than a simpler analysis [36].
Generally, there are two types of critical points based on direction of the tangent at
that point: load limit point and displacement limit point.
Figure 3-1 Limit points in nonlinear equilibrium paths
3.1.1. Load limit points
Load limit points occur when a local maximum or minimum load is reached on the
load versus displacement curve, as shown at points A and D in Figure 3-1. A horizontal
tangent is present at load limit points.
3.1.2. Displacement limit points
Displacement limit points are shown at points B and C in Figure 3-1, they occur atvertical tangents on the solution curve. Displacement limit points are also commonly
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referred to as snap-back points or turning points in the literature. Methods capable of
passing displacement limit points are said to capture snap-back behavior.
3.2. Common methods for solving nonlinear problems
3.2.1. Single step iterative method
The most common procedure to solve a nonlinear problem is iterative procedure. It
is used to solve a problem for a given load in case of no need to trace the full Load-
Displacement curve.
Steps:
a. Construct the stiffness matrix for the un-deformed shape.
b. Solve the stiffness matrix with full load getting the nodal deformations.
c. Add nodal displacements to the nodal coordinates to get deformed shape.d. Construct the internal forces in members at the deformed shape.
e. Calculate the unbalance load equal to the difference between externally applied
load and internally member forces at deformed shape.
f. Reconstruct the stiffness matrix at the deformed shape.
g. Solve the new stiffness matrix with the unbalance load to update the nodal
deformation.
h. Repeat steps from c to f until convergence occurs.
Figure 3-2 Single step iterative method
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3.2.2. Simple incremental method
The simple incremental method is the simplest method in concept and the easiest in
implementation.Steps:
a. Determine number of increments or steps based on desired accuracy, large
number mean higher accuracy.
b. Calculate step load equal to total load divided by number of steps.
c. Construct the stiffness matrix for the un-deformed shape.
d. Solve the stiffness matrix with step load getting the nodal deformation.
e. Add nodal displacement to the nodal coordinates to get deformed shape.
f. Reconstruct the stiffness matrix at the deformed shape.
g. Solve the new stiffness matrix with the next step load to update the nodal
deformation.
h. Repeat steps from e to g till complete all step loads.
Displacement
Load
s t e p 1
Applied Load
s t e p 2
s t e p 3
Figure 3-3 Simple incremental method
3.2.3. Standard Newton Raphson method
The standard Newton Raphson method mixed the simple iterative method with the
incremental method. The load is divided into some equal steps and iterations are made at
each step to guarantee balanced structure at deformed shape. The description of
%Standard& is related to update the stiffness matrix at each iteration in each step.
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Steps:
a. Determine number of increments or steps based on desired accuracy, large
number mean higher accuracy.
b. Calculate step load equal to total load divided by number of steps.
c. Construct the stiffness matrix for the un-deformed shape.
d. Solve the stiffness matrix with first step load getting the nodal deformation.e. Until convergence occurs at current step load repeat the following :
1. Add nodal displacement to the nodal coordinates to get deformed shape.
2. Reconstruct the stiffness matrix for the deformed shape.
3. Construct the internal forces in members at the deformed shape.
4. Calculate the unbalance load equal to the difference between externally
applied load and internally member forces at deformed shape.
5. Solve the updated stiffness matrix with the unbalance load to update the
nodal deformation.
f. Solve the new stiffness matrix with the next step load and update the nodal
deformation.
g. Repeat steps from e to f until complete all step loads.
Displacement
Load
s t e p i
1 s t i t e r .
2 n d i t e r .
Figure 3-4 Standard Newton Raphson method
3.2.4. Modified Newton Raphson method
The modified Newton Raphson method is similar to the standard one except that theupdating stiffness matrix is performed once at the start of each load step. During the
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iteration in each load step, same stiffness matrix is used instead of constructing a new
one. This modification will save the time of constructing new stiffness matrix, but
unfortunately, many iterations may be needed and may lead to expensive working time.
Displacement
Load
s t e p i
iterations
Figure 3-5 Modified Newton Raphson method
3.2.5. Controls used in the incremental-iterative methods
A number of nonlinear solution procedures are proposed to trace equilibrium paths,
such as load control, displacement control, work control, arc-length and generalized
displacement control. Load control and arc-length are chosen to be explained in the
following.
3.2.5.1. Load Control Method (LCM)
The standard Newton Raphson method and the modified one can be classified as
load control type. This method is very widely used although it is not the most powerful.
Since the externally applied loads are kept constant in each step, this method has
difficulties near load limit points. If LCM is chosen to trace the equilibrium path in
Figure 3-6, the structure will snap through from point A to point E directly.
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Figure 3-6 Snap through behavior
3.2.5.2. Displacement Control Method (DCM)
The displacement control method is very effective in tracing the post-critical
response and for controlling the step size in the incremental procedure.
While the load control methods iterate at a constant load, this method iterates at a
constant displacement in the solution procedure, a particular nodal displacement isselected and incremented. This means that the external load is not kept constant during
the iterations. However, the displacement for a selected degree-of-freedom should be
kept constant during iterations. Since the displacement is kept constant in each step, this
method has difficulties near displacement limit points. If DCM is chosen to trace the
equilibrium path in Figure 3-7, the structure will break down from point B to point B1
directly without tracing the snap back.
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Load
Displacement
B
E
B1
Figure 3-7 Snap back behavior
3.2.5.3. Arc-Length Control Method (ALCM)
In order to trace the equilibrium path beyond critical points, a more general
incremental control strategy is needed, in which displacement and load increments are
controlled simultaneously. Such methods are known as %arc-length methods& in which
the $arc length of the combined displacement-load increment is controlled during
equilibrium iterations. The basic idea behind arc length methods is that instead of keeping
the load or the displacement fixed during an incremental step, both the load and
displacement increments are modified during iterations, see Figure 3-8. Either load limit
points or displacement limit points can be passed to this method. There are several
versions of the arc-length method, including cylindrical, spherical, elliptical andlinearized, see Figure 3-9. [36]
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Figure 3-8 Arc-length method [36]
Figure 3-9 Types of arc-length method [36]
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3.3. Unified approach to nonlinear solution
Many methods and algorithms were developed to solve such problems. One of the
most powerful algorithms is called %Unified approach to nonlinear solution schemes&
cited in [36]. This method unified into one algorithm all of the following methods: Load
control, displacement control, arc-length control, work control, generalized displacementcontrol, and orthogonal residual. One algorithm is used for all these control methods
while the difference is through calculation of the factor ! at each step or iteration. If load
control method is the desired method to solve a problem, then using algorithm 3.3.1 and
calculate 1 based on 3.3.2 . However, 1 is calculated based on 3.3.3 if the arc-length
method is the desired method.
3.3.1. Base algorithm
A flow chart describes the main process of the method is shown in Figure 3-10.
Step i
K K(u )i
o
i-1
j 1
j==1 or Standard K
Update
Yes
Yes
Yes
No
No
j ==1
Compute K j-1
Compute q(u )i
i 1+1
j j+1
Nou
i
r j( K ) r
i
j-1
-1 i
j-1
Compute i
j
ui
r j0
ui
p j( K ) p
i
j-1
-1
i i i
j+
r i
j p - q (u )i i
pi
p +i i
j p
ui
u + (i i
ju + u )i
p j
i
r j
r i
j
p
< TOL
i
Figure 3-10 Incremental-iterative procedure [36]
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3.3.2. Load control
For load control, the value of ! is assumed as one / Number of load steps for the first
iteration. The value of ! for the following iterations is zero.
3.3.3. Arc length control
The concept of this method is to force the solution path to an arc-length δ si j .
The following general constraint equation can govern different versions of arc-length as
spherical, cylindrical and elliptical:
δ ui j . δ ui
j + η (δ λi j )
2 = (δ si j )
2 Eq. 3-1
Where η is a non-negative real parameter and is a unique for each version of arc-
length method. Constant η is equal to one in the spherical arc-length, equal to zero in the
cylindrical arc-length and equal to a value greater than zero (not equal to one) in the
elliptical arc-length. Figure 3-11 describes the parameters used in the method.
The constraint equation with respect to iterations instead of increments can be
rewritten as follows:
δ ui1 . δ ui
j + η δ λi1 δ λi
j = (δ si j )
2 Eq. 3-2
δ vwx = y(∆ z{ ) 2, = 10, I 2
Eq. 3-3
Where ∆ "# is the prescribed arc-length to be assigned at the first iteration. The load
factor is then given by.
δ λ'( =
± ∆ )*
. δ
/012
.δ
3456
+ η , 7 = 1
δ 9:; .δ =>?@A δ BCD .δ EFGH + η δ λIJ
, K I 2
Eq. 3-4
The sign of the load factor on the first iteration is positive for structure is being
loaded and negative for unloading. The ability of changing the sign of the load factor
enables it to capture complex nonlinearities at load and displacement limit points.
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Figure 3-11 Parameters of Arc-length control method [36]
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Chapter 4 : Verification Examples
In this chapter, the proposed programs are verified by comparing the results of the
programs with those available in textbooks. Two main objectives for this comparison,
the first one is to verify the ability of the used nonlinear algorithms to capture the fullequilibrium path in problems with different critical points. The second objective is toassure the implementation the finite elements and its capability in handling problems
with large displacements. Four examples are used to test the 2D frame module and three
examples are used to test the 3D truss module.
4.1. 2D frame element
4.1.1. Example 1: Cantilever beam
This example is solved in [37] by the Elasto-plasticity analysis. The structure shownin Figure 4-1
Figure 4-1 Cantilever beam
This example is modelled with co-rotational linear strain Bernoulli element. The
beam is divided into 10 elements while the cross section had been sliced into 20 slices
for the purpose of numerical integration on cross section.
Figure 4-2 shows that NFA program succeeded to solve inelastic cantilever problem
with same results stated in [37]. Figure 4-3 shows the deformation at different load levels.
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Figure 4-2 Cantilever beam: P-!v curve
Figure 4-3 Cantilever beam: deformations
4.1.2. Example 2: Lees frame
This example is solved in [37] by the elastic large displacement analysis and Elasto-
plasticity. The structure shown in Figure 4-4 exhibits snap-through behavior and a limit
point in the load , deflection diagram. For comparison purposes in the analysis of this
problem, the following methods are used:
Elastic large displacement analysis:
• Total formulation [5]
• Co-rotational linear Bernoulli element [37]• Co-rotational shallow arch element [37]
0
20
40
60
80
100
120
140
160
0.0 0.5 1.0 1.5 2.0 2.5 3.0
L o a d P
Displacement v
NFA (Linear strain)
(Battini 2002)
P = 0
P = 200
P = 98
P = 68
P = 48
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Elasto- plasticity analysis:
• Co-rotational linear Bernoulli element [37]
• Co-rotational shallow arch element [37]
Figure 4-4 Lee's frame
In the Co-rotational linear Bernoulli element, we use Et = E to get an elastic analysis
for comparative purpose with the total formulation method.
Figure 4-5 and Figure 4-6 shows that NFA program is able to trace the full
equilibrium path of the Lees frame with an elastic material and overcomes the load limit
point. In addition, its result is identical to the one shown in [14] for either total
formulation method or co-rotational framework using linear/Arch strain methods.
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Figure 4-5 Lee's frame N.L.E: Load vs. displacement u
Figure 4-6 Lee's frame N.L.E: Load vs. displacement v
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
0 20 40 60 80 100
L o a d P
Displacement u
NFA (total formulation)
NFA (Linear strain)
NFA (Arch strain)
(Battini 2002)
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
0 20 40 60 80 100
L o a d P
Displacement v
NFA (total formulation)
NFA (Linear strain)
NFA (Arch strain)
(Battini 2002)
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The NFA program is able to trace the equilibrium path at high large deformation as
shown in Figure 4-7 and in Figure 4-8.
Figure 4-7 Lee's frame N.L.E: Deformed shapes
Figure 4-8 Lee's frame E.P.A: Deformed shapes
Figure 4-9 and Figure 4-10 shows that load-displacement curves from NFA
program is identical to the one shown in [14] for either elastic material or inelasticmaterial.
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Figure 4-9 Lee's frame E.P.A: Load vs. displacement u
Figure 4-10 Lee's frame E.P.A: Load vs. displacement v
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
0 20 40 60 80 100 120
L o a d P
Displacement u
NFA (total formulation)
NFA (Linear strain)
NFA (Arch strain)
(Battini 2002)
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
0 20 40 60 80 100
L o a d P
Displacement v
NFA (total formulation)
NFA (Linear strain)
NFA (Arch strain)
(Battini 2002)
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4.1.3. Example 3: Euler beam
The purpose of this example to make a comparison between the results of strut
formula proposed in section 2.3.2 with result of frame analysis considering both
geometry and material nonlinearities according to section 2.2.2.4. Figure 4-11 shows the
geometry of the model. The material parameters are assumed as following:E = 203400, Et =203.424 and Fy = 400.
L= 251.17
? 0 . 7
7 0 1 4
5 . 4
8
0 . 5
0 2
P
Figure 4-11 Strut model for imperfect element
Figure 4-12 proves that the result of using one strut element by NFA is almost the
same result from analysis using 20 frame elements by NFA. In addition, its clear that
analysis using elastic frame elements (NFA total formulation) with initial out-of-
straightness will reach Euler limit load at higher axial shortening. In Figure 4-13 thedeformed shape of the strut is shown against the strut with its initial out-of-straightness
at different load levels.
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Figure 4-12 Strut model: Load vs. axial shortening
Figure 4-13 Strut model: Deformed shapes
4.1.4. Example 4: Rectangular frame with X-bracing
A rectangular frame with x bracing as shown in Figure 4-14 is studied under lateral
load at point No.4. All members are assumed to be rigidly connected together and with
square tube sections. The dimension of the chosen tubes and member imperfection are
shown in Table 4-1.
Imperfection is assumed to be half sine wave with the maximum in the middle of an
element.
0
200
400
600
800
1000
1200
1400
0 0.5 1 1.5 2 2.5 3 3.5 4
A x i a l C o m p r e s s i o n L o a d
( N )
Axial Shortening ( mm )
NFA (Linear strain)
Abdel-Ghaffar Formula
NFA (total formulation)
Perfect Euler
0.0
1.0
2.0
3.0
4.0
5.0
0 50 100 150 200 250 300
Load P = 0
Load P = 770
Load P = 882
Load P = 986Load P = 1,096
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Figure 4-14 Rectangular frame with X-bracing [22]
Table 4-1 Dimension of sections used in the rectangular frame
Member
No.
Width
in
Thickness
in
Imperfection
in
1 3.01960 0.08134 0
2 3.31058 0.39831 0.317261542
3 3.99968 0.08135 0
4 2.95322 0.57128 0.262966982
5 2.95322 0.57128 0.262966982
6 2.67020 0.23170 0
7 2.67020 0.23170 0
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Figure 4-15 X-bracing frame: Load vs. displacement u4
Figure 4-15 shows that the frame can carry some loads after the failure of the
member number 5. The lateral stiffness of the structure becomes softer after that failure.
The structure loses its vertical stiffness at node number four after failure of member
number two. Figure 4-16 shows that the frame losses its vertical stiffness at point 4
suddenly (after failure of member number two) and the frame becomes only able to carry
about 20% of its previous loading.
Figure 4-16 X-bracing frame: Load vs. displacement v4
0
20
40
60
80
100
120
140
0 10 20 30 40
H o r i z o n t a l L o a
d k i p s
Displacement u at point 4
NFA (Inelastic
linearstrain)
0
20
40
60
80
100
120
140
0 10 20 30 40 50 60
H o r i z o n t a l L o a d k i p s
Displacement v at point 4
NFA (Inelastic
linearstrain)
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4.2. 3D truss element
4.2.1. Example 1: Two-bar truss system
This example is solved by Liew and co-workers [23] using elastic largedisplacement analysis and advanced analysis according to clause 2.3.2.
The structure shown in Figure 4-17 consists of two square bars of 0.254m x 0.254m
size with member slenderness ratio L/r of 150. The modulus of elasticity of the material
is E = 2.06 2 105 N/mm2 and yield strength, 3y = 235 N/mm2. The truss supports are
restrained against translations. The structure is subjected to a concentrated downward
load, P, applied at the crown joint resulting in compressive forces in both members.
For comparative purposes in the analysis of this problem, the following methods are
used:
• Elastic large displacement analysis (discussed in clause 2.3.1).• Advanced analysis (discussed in clause 2.3.2).
For the advanced analysis, the strut capacity, Pmax is computed as 4618 KN using
the column curve 'b' of [42].
Figure 4-17 Two-bar truss system
Figure 4-18 shows that NTA program is able to trace the full equilibrium path of thetwo bar truss system and overcomes the load limit point. In addition, its result is identical
to the one shown by Liew and co-workers [23] either for elastic large displacement or
for advanced analysis.
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Figure 4-18 two-bar truss: Load vs. displacement
4.2.2. Example 2: Space truss dome system
This example is solved by Liew and co-workers [23] using elastic large
displacement analysis and advanced analysis. The structure shown in Figure 4-19consists of 24 members to form a star-shaped structure. All members are square hollow
section 5.48mm width and 0.50223mm thickness. Cross-sectional properties are: A = 10
mm2; I = 41.7 mm4; E = 2.034 2 105 N/mm2, 3y= 400 N/mm2 and E p = E/1000.
All the supports are restrained against translations while the remaining nodes are
free to translate in the space. The slenderness ratios (L/r) for members 1-6 and 13-24 are
123 and 155, respectively.
-1500
-1000
-500
0
500
1000
1500
2000
0 200 400 600 800 1000 1200 1400 1600
V e r t i c a l l o a d a t c r
o w n ( N )
Vertical displacement at crown ( mm )
Liew 1997 (Advanced)
Liew 1997 (Elastic large disp.)
NTA (Advanced)
NTA (Elastic large disp.)
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1
23
4
5 6
7
8
9
10
11
12
15
16
1718
19
20
21
22
23 24
13
14
62.16mm20mm
433 mm 433 mm
Plan
ElevationSupports
Figure 4-19 Space truss dome system
The space truss is analyzed for two load cases:
1. The single downward concentrated load applied at the crown joint.
2. Downward concentrated loads applied at all the seven unrestrained joints.
For comparative purposes, the following analysis methods are used:
• Elastic large displacement analysis (discussed in clause 2.3.1).
• Advanced analysis (discussed in clause 2.3.12.3.2).
For the advanced analysis, the strut capacity, Pmax is computed as 1129 N,1135 N
and 745 N for members 1 to 6 , 7 to 12 and 13 to 24 respectively using the column curve
'b' of [42].
Figure 4-20 and Figure 4-21 show that NTA program is able to trace the full
equilibrium path of the star dome and overcome the load limit point. In addition, its result
is identical to the one shown by Liew and co-workers [23] either by using elastic large
displacement or by using advanced analysis.
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Figure 4-20 Space truss dome: Load vs. displacement at load case 1
Figure 4-21 Space truss dome: Load vs. displacement at load case 2
4.2.3. Example 3: Twelve-bar space truss system
This example is solved in Liew and co-workers [23] using elastic large displacement
analysis and advanced analysis.
The structure shown in Figure 4-22 consists of 12-truss element with circular
section. For CS-1, the pipe assumed to be 193.7mm out diameter and 10mm thickness
while for CS-2 the pipe assumed to be 168.3mm out diameter and 5mm thickness.
0
100
200
300
400
500
600
700
0 5 10 15 20
V e r t i c a l l o a d a t c r o
w n ( N )
Vertical displacement at crown ( mm )
Liew 1997 (Advanced)
NTA (Advanced)
NTA (E. large disp.)
Liew 1997 (E. large disp.)
0
200
400
600
800
1000
1200
1400
1600
1800
0 2 4 6 8 10 12
V e r t i c a l l o a d
a t c r o w n ( N )
Vertical displacement at crown ( mm )
Liew 1997 (Advanced)
NTA (Advanced)NTA (E. large disp.)
Liew 1997 (E. large disp.)
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Figure 4-22 Twelve-bar space truss system [22]
Using NTA software with elastic large displacement analysis, the load-displacement
curve is shown in Figure 4-23 , Figure 4-24 and Figure 4-25 for displacement w at point
A, displacement u at point B, displacement w at point B respectively. These three curvesshow that the maximum vertical load P that cause global elastic buckling without limiting
internal forces to any limit is equal to 811120 KN and 36298 KN for structure with pipes
CS-1 and CS-2 respectively. The structure perform unloading after this load limit.
Results from NTA software with elastic large displacement analysis are identical to the
results shown in Liew [23] for the twelve-bar space truss system either using cross
section CS-1 or CS-2.
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Figure 4-23 Twelve-bar truss E.L.D. analysis: Load vs. disp. w at point A
Figure 4-24 Twelve-bar truss E.L.D. analysis: Load vs. disp. u at point B
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
-150 -100 -50 0 50 100 150 200
V e r t i c a l l o a d a t P o i n t A ( K N )
displacement W at point A ( mm )
Liew 1997 ,CS-2
Liew 1997 ,CS-1
NTA ,CS-1
NTA ,CS-2
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
0 200 400 600 800 1000
V e r t i c a l l o a d a t P o i n t A (
K N )
displacement U at point B ( mm )
Liew 1997 ,CS-2
Liew 1997 ,CS-1
NTA ,CS-1
NTA ,CS-2
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Figure 4-25 Twelve-bar truss E.L.D. analysis: Load vs. disp. w at point B
Using NTA software with advanced analysis, the load-displacement curve is shown
in Figure 4-26 , Figure 4-27 and Figure 4-28 for displacement w at point A, displacement
u at point B and displacement w at point B respectively. Results of all curves are similar
to the results shown in [23] for the twelve-bar space truss system either using cross
section CS-1 or CS-2.
Figure 4-26 Twelve-bar truss advanced analysis: Load vs. disp. w at point A
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
0 200 400 600 800 1000 1200 1400 1600
V e r t i c a l l o a d a t P o i n
t A ( K N )
displacement W at point B ( mm )
Liew 1997 ,CS-2
Liew 1997 ,CS-1
NTA ,CS-1
NTA ,CS-2
0
200
400
600
800
1000
1200
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50
V
e r t i c a l l o a d a t P o i n t A ( K N )
displacement W at point A ( mm )
Liew 1997 ,CS-2
Liew 1997 ,CS-1
NTA ,CS-1
NTA ,CS-2
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Figure 4-27 Twelve-bar truss advanced analysis: Load vs. disp. u at point B
Figure 4-28 Twelve-bar truss advanced analysis: Load vs. disp. w at point B
Above truss with pipes CS-1 has a limit load equal to 1095 KN while truss with
pipes CS-2 has 418 KN limit load. The truss with pipes CS-2 will be analyzed after
upsizing elements No. 1, 2, 11 & 12 to be pipes CS-1. Figure 4-29 shows that
displacement w at point A becomes more ductile and Figure 4-30 shows that the load
limit of the new truss will be 972 KN. however, Figure 4-31 shows that displacement wat point B become more brittle.
0
200
400
600
800
1000
1200
0 0.5 1 1.5 2 2.5
V e r t i c a l l o a d a t P o i n t A ( K N )
displacement U at point B ( mm )
Liew 1997 ,CS-2
Liew 1997 ,CS-1
NTA ,CS-1
NTA ,CS-2
0
200
400
600
800
1000
1200
0 20 40 60 80 100
V e r t i c a l l o a d a t P o i n t A ( K N
)
displacement W at point B ( mm )
Liew 1997 ,CS-2
Liew 1997 ,CS-1
NTA ,CS-1
NTA ,CS-2
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Figure 4-29 Twelve-bar truss advanced analysis: Load vs. disp. w at point A
Figure 4-30 Twelve-bar truss advanced analysis: Load vs. disp. u at point B
0
200
400
600
800
1000
1200
0 20 40 60 80 100 120
V e r t i c a l l o a d a t P o i n t A ( K N )
displacement W at point A ( mm )
NTA,Mixed CS-1 & CS-2
NTA ,CS-1
0
200
400
600
800
1000
1200
0.00 1.00 2.00 3.00 4.00 5.00
V e r t i c a l l o a d a t P o i n t A ( K N )
displacement u at point B ( mm )
NTA,Mixed CS-1 & CS-2
NTA ,CS-1
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Figure 4-31 Twelve-bar truss advanced analysis: Load vs. disp. w at point B
0
200
400
600
800
1000
1200
0 10 20 30 40 50 60 70
V e r t i c a l l o a d a t P o i n t A ( K N )
displacement w at point B ( mm )
NTA,Mixed CS-1 & CS-2NTA,CS-1
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Chapter 5 : Case-Studies
The behavior of a truss element under compression force depends on many
parameters. The most important three parameters are yield stress of the material,
slenderness ratio of the element, and initial out-of-straightness of the element. Throughthe examples and the figures in this chapter, this effect on element behavior is clarified.The problem of the star dome is sensitive to the shallowness of its rise. The bigger the
domes rise, the bigger the load it can carry. Using the NTA program, the load-
displacement curve is shown for the star dome at different values of the rise height. Due
to this sensitivity, the star dome is chosen to investigate the strength and weakness of
each analysis type available in NTA program. Using nonlinear analysis provided in
STAAD software, a star dome with real dimension be designed under certain loading.
NTA program is used to evaluate this design using different type of analysis that are
available in it.
5.1. Effect of variables on local element curve
The strut element according to 2.3.2 follows Eq. 2-94 in the elastic range and follows
Eq. 2-95 in the plastic range. The intersection of the two equations is the critical buckling
load as shown in Figure 5-1. Figure 5-2 shows that critical compressive stress calculated
based on this method at initial out-of-straightness 1/500 of strut length is almost the same
stress allowed by AISC [43] before applying the compression reduction factor φ.
The effect of changing the following variables will be discussed:
1. Initial out-of-straightness on the elastic equation.
2. Yielding stress on the plastic equation.3. Slenderness ratios.
Figure 5-1 Critical compressive stress
A x
i a l c o m p
r e s s i o n s t r e e
Axial strain
Elastic equation
Plastic equation
Critical compressive
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Figure 5-2 Allowable compression stress AISC 360-10 vs. DA500
In the following studies, a pipe section with diameter 48.3mm, thickness 3.2mm is
used and elastic modulus is 200 GPa.
Pipe section properties:
•
Area = 453.4 mm
2
• Inertia = 115856.5 mm4
• Plastic section modulus = 6519.7 mm3
• Radius of gyration = 15.985 mmThe axial capacity had been calculated according to [43] by following equations.
LM = NO PQRST UV
Eq. 5-1
WX =
%
YZ 0.658[\] _ a, bcd e 4.71 f ghijk _ 0.877 _ lm,
nop > 4.71 q Ers
Eq. 5-2
Where:
• 3c is the allowable compressive stress.
• 4c equal to 0.9
0
50
100
150
200
250
0 50 100 150 200
C o m p r e s s i v e s t r e s s
( M p a )
L/r
Elastic Euler
DA500
AISC LRFD w/o #
AISC LRFD with #
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5.1.1. Effect of yielding stress and initial imperfection at different L/r
The strut length is chosen 3197 mm to obtain slenderness ration equal to 200. The
elastic curve is shown at different initial imperfections values equal to L/1500, L/1000,
and L/500 & imperfection code.
Imperfection code is the initial imperfection that makes the intersection of Eq. 2-94
with Eq. 2-95 at the allowable compressive stress with the code [43].In this example;
imperfection code was 30.27mm that is equal to L/106.
Figure 5-3 shows the effect of changing initial out-of-straightness and material
yielding stress on a strut with a slenderness ratio equal to 200. The equation Eq. 2-94 is
plotted for the strut with different values of the initial out-of-straightness and the equation
Eq. 2-95 is plotted for different values of yielding stress. Increasing the yield stress shifts
up the plastic equation and thus increases the critical buckling load. However, increasing
the initial out-of-straightness shifts down the elastic equation and thus decreases the
critical buckling load. Figure 5-4 and Figure 5-5 show both elastic and plastic equationsfor strut with a slenderness ratio equal to 157 and 100 respectively.
Figure 5-3 Effect of Fy and initial imperfection at L/r =200
0
10
20
30
40
50
60
70
0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
A
x i a l c o m p r e s s i o n s t r e e ( N / m m 2 )
Axial strain
Elastic Imperf = L/1,500
Elastic Imperf = L/1,000
Elastic Imperf = L/500
Elastic Imperf code
Plastic Fy = 355Plastic Fy = 275
Plastic Fy = 241
Plastic Fy = 207
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Figure 5-4 Effect of Fy and initial imperfection at L/r =157
Figure 5-5 Effect of Fy and initial imperfection at L/r =100
0
20
40
60
80
100
120
0.000 0.001 0.002 0.003 0.004 0.005 0.006
A x
i a l c
o m p r e s s i o n s t r e e
( N / m m
2 )
Axial strain
Elastic Imperf = L/1,500Elastic Imperf = L/1,000Elastic Imperf = L/500Elastic Imperf Plastic Fy = 355
Plastic Fy = 275Plastic Fy = 241Plastic Fy = 207
0
50
100
150
200
250
300
0.000 0.001 0.001 0.002 0.002 0.003 0.003 0.004 0.004 0.005
A x
i a l c o m p r e s s i o
n s t r e e
( N / m m
2 )
Axial strain
Elastic Imperf = L/1,500Elastic Imperf = L/1,000Elastic Imperf = L/500
Elastic Imperf Plastic Fy = 355Plastic Fy = 275Plastic Fy = 241Plastic Fy = 207
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Figure 5-6 shows the behavior of a strut with yield stress 248.2 N/mm2 and
slenderness ratio equal to 157. The strut follows elastic equation before critical buckling
stress and then follows the plastic equation as explained before. It is clear from this figure
that the strut with higher value of the initial out-of-straightness has the lower axial
stiffness and the lower load limit imperfection.
Figure 5-6 Effect of initial imperfection at Fy = 248.2 N/mm2 and L/r =157
The critical load is located at the intersection of Eq. 2-94 with Eq. 2-95, the code
imperfection is the initial imperfection, which is used to make the critical load equal to
capacity specified by the code. In Figure 5-7 using higher tensile stress results in higher
capacity even at slenderness ratio 157.
According to [43] 1c equal to 1.6 > 1.5 at slenderness ratio 157.
tu = vw y z {|}
0
10
20
30
40
50
60
70
80
90
0.0000 0.0005 0.0010 0.0015 0.0020 0.0025
A x
i a l c o m p r e s s i o n s t r e e
( N / m m
2 )
Axial strain
Imperf. Code
Imperf. L/1500
Imperf. L/1000
Imperf. L/500
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Figure 5-7 Effect of Fy at L/r =157 and initial imperfection L/1000
The allowable compressive stress by the code at 1c > 1.5 becomes independent of
the yield stress, when the strut model is adjusted to get the load limit by the code; the
equivalent initial imperfection will be higher for high values of the yield stress with
slenderness ratio 157 Figure 5-8.
Figure 5-8 Effect of yield stress on model at L/r=157 and same code limit
0
10
20
30
40
50
60
70
80
90
0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030
A x
i a l c o m p r e s s i o n s t r e s s
( N / m m
2 )
Axial strain
Fy = 355.0
Fy = 275.8
Fy = 241.3
Fy = 206.8
0
10
20
30
40
50
60
70
0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030
A x
i a l c o m p r e s
s i v e s t r e s s
N / m m
2
Axial strain
Fy = 207 ( ( = L / 254 )Fy = 241 ( ( = L / 211 )
Fy = 275 ( ( = L / 182 )
Fy = 355 ( ( = L / 138 )
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It shall be noticed in Figure 5-8 that adjusting ( to get a code limit led to a non-logic
behavior because it gives a higher stiffness for a steel with lower yield stress.
5.2. Effect of changing rise height on star dome
Star dome solved in 4.2.2 is reanalyzed under different height of the crown (variable
X in Figure 5-9).The structure shown in Figure 4-19 consists of 24 members to form a
star-shaped structure. All members are square hollow section 5.48mm width and
0.50223mm thickness. Cross-sectional properties are: A = 10 mm2; I = 41.7 mm4; E =
2.034 2 105 N/mm2, 3y= 400 N/mm2 and E p = E/1000.
All the supports are restrained against translations and the remaining nodes are free
to translate in the space. The slenderness ratios (L/r) for members 1-6 and 13-24 are 123
and 155, respectively at X = 20mm.
1
23
4
5 6
7
8
9
10
11
12
15
16
1718
19
20
21
22
23 24
13
14
62.16mmX
433 mm 433 mm
Plan
ElevationSupports
X
E l e m e n t s 1 t o 6 S l o
p e
Roof slope
Figure 5-9 Space truss dome system
The space truss is analyzed for a single downward concentrated load applied at the
crown joint.
Capacity of the star dome is calculated by three methods:
• Elastic linear analysis
•
Elastic large displacement analysis (discussed in clause 2.3.1).• Advanced analysis (discussed in clause 2.3.12.3.2).
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All member capacities are checked according to BS [42] using the column curve 'b'.
The results of different analysis types for star dome with different rise height X are
tabulated in Table 5-1 and Table 5-2.in addition the ratios between different analysis
types are shown. These results are plotted in the next for more clearance.
Table 5-1 Summary of limit load for star dome
X ( m m )
S l o p e
( D e g .
)
D A - L i e w
L M + N G
( D A - L i e w ) /
( L M + N G )
( L M + N G + C L )
( D A - L i e w )
/ ( L M + N G + C L )
( L M + L G + C L )
( D A - L i e w )
/ ( L M + L G + C L )
10 2.3 71 77 0.923 77 0.923 271 0.2612 2.7 119 135 0.887 135 0.887 325 0.37
14 3.2 178 215 0.826 214 0.831 379 0.47
16 3.7 241 323 0.746 303 0.795 433 0.56
18 4.1 305 463 0.660 378 0.808 486 0.63
20 4.6 370 639 0.578 446 0.828 539 0.69
22 5.0 433 856 0.506 511 0.848 592 0.73
24 5.5 496 1118 0.443 573 0.865 645 0.77
26 5.9 557 1430 0.390 632 0.882 697 0.80
28 6.4 618 1797 0.344 690 0.896 749 0.83
30 6.8 679 2224 0.305 747 0.909 800 0.85
32 7.3 738 2716 0.272 803 0.920 851 0.87
34 7.7 797 3277 0.243 857 0.930 902 0.88
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Table 5-2 Summary of limit load for star dome (continued)
D A 5 0 0
D A 1 0 0 0
D A 1 5 0 0
( D A 5 0 0 )
/ ( L M + N G + C L )
( D A 1 0 0 0 )
/ ( L M + N G + C L )
( D A 1 5 0 0 )
/ ( L M + N G + C L )
( D A 5 0 0 )
/ ( L M + L G + C L )
( D A 1 0 0 0 )
/ ( L M + L G + C L )
( D A 1 5 0 0 )
/ ( L M + L G + C L )
71.41 71.41 71.41 0.92 0.92 0.92 0.26 0.26 0.26
126.67 132.21 133.45 0.94 0.98 0.99 0.39 0.41 0.41
191.82 205.51 209.79 0.90 0.96 0.98 0.51 0.54 0.55
262.19 285.83 295.02 0.87 0.94 0.97 0.61 0.66 0.68
332.42 363.66 376.41 0.88 0.96 1.00 0.68 0.75 0.77
401.06 437.83 453.00 0.90 0.98 1.02 0.74 0.81 0.84
468.05 509.04 525.97 0.92 1.00 1.03 0.79 0.86 0.89
533.57 577.95 596.28 0.93 1.01 1.04 0.83 0.90 0.93
597.84 645.02 664.50 0.95 1.02 1.05 0.86 0.93 0.95
660.99 710.56 731.02 0.96 1.03 1.06 0.88 0.95 0.98
723.14 774.80 796.10 0.97 1.04 1.07 0.90 0.97 0.99
784.37 837.90 859.91 0.98 1.04 1.07 0.92 0.98 1.01
844.75 899.92 922.60 0.99 1.05 1.08 0.94 1.00 1.02
Where:
Slope Angle between members 1 to 6 with the horizontal plane in
degree.
DA-Liew The load limit of the dome using advanced analysis using
adjusted imperfection to get a code limit [42].
DA-1500 The load limit of the dome using advanced analysis using
imperfection ( = L / 1500.
DA-1000 The load limit of the dome using advanced analysis using
imperfection ( = L / 1000.
DA-500 The load limit of the dome using advanced analysis using
imperfection ( = L / 500.
LM+NG The load limit of the dome using elastic large displacement
analysis
LM+NG+CL The load limit of the dome using elastic large displacement
analysis at which any member reached permissible force
according to the code.
LM+LG+CL The load limit of the dome using elastic linear analysis at
which any member reached permissible force according to
the code.
Figure 5-10 shows that this problem is very sensitive to the shallowness of the dome.The bigger the domes rise, the bigger the load it can carry. For slope with value 3.75,
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the capacity determined according to the code using nonlinear geometry is too close to
the load limit of the nonlinear analysis.
Figure 5-10 Star dome: effect of roof rise using LM+NG
Figure 5-11 shows that adjusting initial imperfection to get a code limit led to using(=L/326 in some members which is out of usual allowable imperfection in most of
international codes.
0
5
10
15
20
25
30
35
0 5 10 15 20 25
V e r t
i c a l l o a
d a
t c r o w n
( N )
Vertical displacement at crown ( mm )
slope 7.76slope 6.86slope 5.96slope 4.66 (Liew)slope 3.76
Code limit
x107
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Figure 5-11 Star dome: effect of roof rise using DA-Liew
Figure 5-12 shows the effect of changing slope values of elements 1 to 6 on the
behavior of the structure. The elements are assumed to have an initial out-of-straightness
equal to 1/500 of members length. Figure 5-13 and Figure 5-14 shows the same, but
with using a value of the initial out-of-straightness equal to 1/1000 and 1/1500 ofmembers length respectively. In these figures, a star dome with slope equal to 3.75 has
a capacity equal to one-third the capacity of the star dome with slope 7.75. Star dome
with higher initial out-of-straightness has the lowest load capacity.
0
1
2
3
4
5
6
7
8
9
0 2 4 6 8 10 12 14 16
V e r t i c a
l l o a
d a
t c r o w n (
N )
Vertical displacement at crown ( mm )
slope 7.76slope 6.86slope 5.96slope 4.66 (Liew)slope 3.76
x107
Where
δ1-6 = L / 326
δ7-12
= L / 326δ13-24 = L / 337
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Figure 5-12 Star dome: effect of roof rise using DA500
Figure 5-13 Star dome: effect of roof rise using DA1000
0
1
2
3
4
5
6
7
8
9
0 2 4 6 8 10 12 14
V e r t i c a
l l o a
d a
t c r o w n (
N )
Vertical displacement at crown ( mm )
slope 7.76slope 6.86slope 5.96slope 4.66 (Liew)slope 3.76
x107
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8
V e r t i c a
l l o a
d a
t c r o w n
( N )
Vertical displacement at crown ( mm )
slope 7.76slope 6.86
slope 5.96slope 4.66 (Liew)slope 3.76
x107
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Figure 5-14 Star dome: effect of roof rise using DA1500
In Figure 5-15, the maximum difference between analysis using DA-Liew and
design according to the code with nonlinear geometry occurs at slop 3.75. When the slope
of the star dome becomes lower than 3.75, then the structure becomes softer and the
structure fails without a local buckling in any elements. Therefore, for an extremelyshallow star dome, the difference between DA-Liew and LM+NG+CL became less than
the maximum that occurs at slope 3.75. In addition, as shown in Figure 5-16 the out-of-
straightness does not affect the star dome capacity at slope 2.35 due to structure overall
buckling.
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8
V e r t i c a
l l o a
d a
t c r o w n (
N )
Vertical displacement at crown ( mm )
slope 7.76slope 6.86slope 5.96
slope 4.66 (Liew)slope 3.76
x107
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Figure 5-15 Star dome: AISC 360-10 (LRFD) design vs. DA-Liew
Figure 5-16 shows that design using nonlinear geometrically overestimates the
capacity by about 15% than DA-500 at a value of slope 3.75. However, at slope 7.75 design using nonlinear geometry or DA-500 gives the same capacity for the star dome.
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
(DA-Liew) /(LM+NG+CL)
DA-Liew /(LM+LG+CL)
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Figure 5-16 Star dome: AISC 360-10 (LRFD) design vs. DA500
Figure 5-17 shows that applying the reduction factor φ with a value equal to 0.85 onthe result of analysis using DA500 will lead to 75% capacity of the one obtained usingdesign according to the code through nonlinear analysis.
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
DA-500 /(LM+NG+CL)
DA-1500 /(LM+NG+CL)
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Figure 5-17 Star dome: AISC 360-10 (LRFD) design vs. DA1000
Figure 5-18 shows that design using nonlinear geometry overestimates the capacity
by about 6% compared with DA-1000 at a value of slope 3.75 however; linear analysisgives large overestimation of capacity at that slope. At slope 7.75, design using nonlinear
geometry underestimates the capacity compared with DA-1000 by 5%.
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
0.85*DA-500 /(LM+NG+CL)
DA-500 /(LM+LG+CL)
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Figure 5-18 Star dome: AISC 360-10 (LRFD) design vs. DA1000
From Figure 5-19 it is clear that the structural capacity is independent of the initial
out-of-straightness for slope with a value equal to 2.35. In addition, it is clear that thehigh initial out-of-straightness is the low structural capacity because increasing initial
out-of-straightness will decrease stiffness, will increase the displacement, and therefore
will lead to lesser structure capacity.
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
DA-1000 /(LM+NG+CL)
DA-1000 /(LM+LG+CL)
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Figure 5-19 Star dome: Effect of raise angle on analysis DA vs. LM+NG
Figure 5-20 shows that star dome with the smallest slope will have the most
overestimation of load capacity if the linear analysis is used.
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
DA-1500 /(LM+NG+CL)
DA-1000 /(LM+NG+CL)
DA-500 /(LM+NG+CL)
(DA-Liew) /(LM+NG+CL)
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Figure 5-20 Star dome: Effect of raise angle on analysis DA vs. LM+LG
Figure 5-21 shows that the capacity of the star is reduced to the half by reducing theslope from 65 to be 35 if linear analysis is used. However, the capacity of the star is
reduced to the third if nonlinear analysis or DA-Liew is used.
Figure 5-21 Star dome: capacity Vs. slope
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
DA-1000 /(LM+LG+CL)
DA-500 /(LM+LG+CL)
DA-1500 /(LM+NG+CL)
DA-Liew /(LM+LG+CL)
0
100
200
300
400
500
600
700
800
900
1000
2.0 3.0 4.0 5.0 6.0 7.0 8.0
V e r t i c a
l L o a
d a
t c r o w n
P ( N )
Slope (Deg.)
(LM+LG+CL)
(LM+NG+CL)
DA-Liew
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In Figure 5-22, at high slope values, the capacity of the star dome becomes a small
ratio of global structure load limit, however, at low slope values the global structure load
limit governs the capacity. The design using linear analysis extremely overestimates the
capacity of the star dome at low values of slope and becomes quite acceptable at slope
75 and more.
Figure 5-22 Ratio of different analysis to NG without code limit Vs. slope
5.3. Star dome real design examples
Large scale of star dome example, as shown in Figure 5-23 is studied. Star dome is
assumed to carry the following loads:
• Cover load 20 kg/m2 at below shaded area.
• Live load 60 kg/m2 at below shaded area.
• Self-weight of all members
The large scale of star dome example will be analyzed for 4 types of analysis:
Analysis type 1 Linear elastic analysis, which limits the load ratio of the level
at which any element exceeds its allowed load by the code.
Analysis type 2 P-! large displacement elastic analysis, which limits the load
ratio of the level at which any element exceeds its allowed
load by the code.
Analysis type 3 P-! & p-( large displacement inelastic analysis, including
equivalent element imperfection effect that leads. Equivalent
imperfection will enforce any element not to carry loadsmore than capacity calculated by the code.
0.00
0.20
0.40
0.60
0.80
1.00
1.20
2.0 3.0 4.0 5.0 6.0 7.0 8.0
R a
t i o
Slope (Deg.)
DA-1500 /(LM+LG+CL)
DA-1500 / (LM+NG)
DA-1000 / (LM+NG)
DA-Liew / (LM+NG)
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Analysis type 4 P-! & p-( large displacement inelastic analysis, including
specified element imperfection (L/500) effect limits the load
ratio of the level at which any element exceeds its allowed
load by the code.
1
23
4
5 6
7
8
9
10
11
12
15
16
1718
19
20
21
22
23 24
13
14
621.6mm200mm
4330 mm 4330 mm
Plan
ElevationSupports
Figure 5-23 Space truss dome system for real design
Three types of load case are applied
Load case 1 Single downward concentrated load applied at the crown joint.
Its value will equal total factored load applied to the dome.
Load case 2 Downward equal concentrated loads applied at all the seven
unrestrained joints. Its value will equal total factored load
applied to the dome divided equally on seven joints.
Load case 3 Downward concentrated loads applied at all the seven
unrestrained joints. Load at crown joint not equal to the loadapplied at any other joint. This distribution is due to apply load
in shaded areas as shown in Figure 5-24.
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Figure 5-24 Loaded area of the space truss dome
5.3.1. Star dome real design (load case 3)
Dome members are designed according to AISC 360-10 (LRFD) using combination
1.2 Dead Load + 1.6 Live Load. Using STAAD PRO for analysis and design we got
members 1 to 12 to be pipe 42.4mm O.D. and 2.6mm thickness while members from 13
to 24 to be pipe 48.3mm O.D. and 3.2mm thickness. All pipes are assumed to be mild
steel A36 with yield stress = 248.2 N/mm2
and material modulus of elasticity 200000 N/mm2.All connections assumed in pinned connection. From STAAD PRO results we
found that maximum ratio of usage 0.961 using elastic linear analysis while the ratio of
usage 0.966 using elastic nonlinear analysis.This means that star dome with above
sections has load factor = 1/0.961 = 1.041 and 1/0.966 = 1.035 under elastic linear
analysis and elastic nonlinear analysis respectively.
The same problem is analyzed using NTA program and result shown in Figure 5-25
and in Figure 5-26.
For elastic linear analysis, Load factor = 1.042
For elastic nonlinear analysis, Load factor = 1.036
For DA-Liew, Load factor = 0.819
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Figure 5-25 Analysis result of real dome (optimum design) using NTA
Figure 5-26 Detailed curve of real dome (optimum design) analysis result
0
1
2
3
4
5
6
7
8
9
10
0 20 40 60 80 100 120
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-Liew((=L/170 : L/150)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0 20 40 60 80 100 120
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-500 with "=0.85
DA-Liew(#=L/170 : L/150)
LR=1.0791
LR=1.0089 at CL
LR=1.036
LR=0.818
LR=0.917
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Figure 5-27 and Figure 5-28 show the result of the star dome after upsizing members
1 to 12 to be pipe 48.3mm O.D. and 3.2mm thickness instead of 42.4mm O.D. and 2.6mm
thickness while members from 13 to 24 to remain 48.3mm O.D. and 3.2mm thickness.
Figure 5-29 to Figure 5-31 show the internal force in star dome elements.
One of the disadvantages of analysis type DA-Liew (imperfection calibration) thatit gives high value for imperfection that causes weakness in structural stiffness from
loading beginning.
Figure 5-27 Analysis result of real dome (all pipes"48) using NTA
0
2
4
6
8
10
12
0 20 40 60 80 100
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-Liew((=L/204 : L/150)
DA-500
See detailed curve
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Figure 5-28 Detailed curve of real dome (all pipes"48) analysis result
Figure 5-29 Internal force of elements No.1 to 6 in the two models
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
0 10 20 30 40 50
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-500 with #=0.85
DA-Liew((=L/204 : L/150)
LF=1.2172LF=1.0322 at CL
LF=1.036
LF=0.993
LF=1.035
0
2
4
6
8
10
12
14
16
18
20
0 1 2 3 4 5 6 7
A x
i a l C
o m p r e s s i o n
F o r c e
( K N )
Axial shortening (mm)
All Pipes #48 DA-500
All Pipes #48 DA-LiewOptimum design DA-Liew
Optimum design DA-500
Unloading Failure
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Figure 5-30 Internal force of elements No.13 to 24 in optimum design model
Figure 5-31 Internal force of elements No.13 to 24 in All Pipes#48 model
The results of the two design models of star dome are summarized in Table 5-3. Itis noticed that capacity is the same for analytical types LM+LG+CL & LM+NG+CL
0
2
4
6
8
10
12
14
16
18
20
0 0.5 1 1.5 2 2.5
A x
i a l C o m p r e s s i o n
F o r c e
( K
N )
Axial shortening (mm)
Optimum design DA-Liew
Optimum design DA-500
LoadingLoading
Unloading
Unloading
0
5
10
15
20
25
0 1 2 3 4 5 6 7 8
A x
i a l C
o m p r e s s i o n
F o r c e
( K N )
Axial shortening (mm)
All Pipes #48 DA-500
All Pipes #48 DA-Liew
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because that the dome capacity governed by elements 13 to 24 which have the same size
and capacity in the two models.
Table 5-3 Result of star dome optimum design model and upsizing model
Analysis type Optimum design All pipes "48.3 O.D Gain in L.F.
LM+LG+CL 1.042 1.042 0
LM+NG+CL 1.036 1.036 0
DA-Liew 0.819 0.993 21%
DA500 1.0089 1.032 2%
While in analysis DA-Liew the dome capacity governed by elements 1 to 6 that was
upsized in the second model, therefore capacity enhanced from 0.819 in optimum model
to 0.993 in the upsizing model.
In analysis type, DA500 elements 1 to 6 governed the dome capacity, but at this
moment members 13 to 24 have high ratio. Therefore, when upsizing elements 1 to 6,
the dome gains slightly higher capacity from 1.0089 to 0.032 as members 13 to 24
governed the capacity in the upsizing model.
5.3.2. Star dome large scale (load case 1)
Using load case one (described in 5.3) instead of load case 3 in the optimum model
in 5.3.1 and analyzed using NTA follow result had found. Optimum model has members
numbered 1 to 12 to be pipe 42.4mm O.D. and 2.6mm thickness while members from 13
to 24 to be pipe 48.3mm O.D. and 3.2mm thickness. The results from NTA program are
presented in Figure 5-32 and Figure 5-33.
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Figure 5-32 Analysis result of real dome (load case 1) using NTA
Figure 5-33 Detailed curve of real dome (load case 1) analysis result
0.0
2.0
4.0
6.0
8.0
10.0
12.0
0 20 40 60 80 100 120
L o a d R a t i o
Down displcaement at crown ( mm )
LM+LGLM+NG
DA-Liew
DA-500
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 20 40 60 80 100
L o a d R a t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-Liew
See detailed
LR=0.9911LR=0.989
7 at CL
LR=0.915
LR=1.036
at CL
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5.3.3. Star dome large scale (load case 2)
Using load case 2 (described in 5.3) instead of load case 3 in the optimal modelin 5.3.1 , analysis results using NTA are shown in Figure 5-34 and Figure 5-35.
Optimum model has members numbered 1 to 12 to be pipe 42.4mm O.D. and 2.6mm
thickness while members numbered 13 to 24 to be pipe 48.3mm O.D. and 3.2mm
thickness.
Figure 5-34 Analysis result of real dome (load case 2) using NTA
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
0 20 40 60 80 100 120
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-Liew
See detailed
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Figure 5-35 Detailed curve of real dome (load case 2) analysis result
5.3.4. Star dome large scale (load case 2) with different imperfection
Figure 5-36 show the result of reanalyzing problem 5.3.3 by new three types of
analysis.
DA-500 with # Same as analysis DA-500 (detailed in 5.3) but the load
ratio is multiplied with #=0.85.
DA-1000 Same as analysis DA-500 (detailed in 5.3) but setting
the imperfection as 1/1000 of member length instead
of 1/500.
DA-1500 Same as analysis DA-1500 (detailed in 5.3) but settingthe imperfection as 1/1500 of member length instead
of 1/500.
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 10 20 30 40 50 60 70
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-500 with #DA-Liew1.005
(CL)
0.826
1.07861.03
(CL)0.92
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Figure 5-36 Analysis result of real dome (load case 2) with different imperf.
5.3.5. Star dome large scale (load case 3) with steel A570-50
Example 5.3.1 (optimum section model) will be analyzed again, but with changing
material to be high, tensile with yielding strength 345 N/mm2 instead of 248.2 N/mm2
and the results are shown in Figure 5-37 and Figure 5-38.
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 10 20 30 40 50 60 70
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-1500DA-1000
DA-500
DA-500 with #DA-Liew
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Figure 5-37 Analysis result of dome with A570-50 using NTA
Figure 5-38 Detailed curve of dome with A570-50 analysis result
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0 20 40 60 80 100 120
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-Liew((=L/170 : L/150)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0 20 40 60 80 100 120
L o a
d R a
t i o
Down displcaement at crown ( mm )
LM+LG
LM+NG
DA-500
DA-500 with #=0.85
DA-Liew((=L/170 : L/150)LR=1.0791
LR=1.0089 at CL
LR=0.7
LR=1.036 at CL
LR=0.917
See detailed
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From Figure 5-26 of the dome with steel A36 and Figure 5-38 of the dome with steel
A570-50, the following results are summarized in Table 5-4.
Table 5-4 Mild steel dome Vs. high tensile using NTA
Analysis type Dome with A36 Dome with A570-50
LM+LG 1.064 1.064
LM+NG 1.036 1.036
DA-Liew 0.818 0.700
DA-500 1.0089 1.0089
Using analysis type LM+LG, LM+NG nor DA-500 will not affect the dome capacity
as same element stiffness and same element capacity are assumed while using type DA-Liew will reduce the dome capacity of high tensile steel as stiffness of elements reduced
(see Figure 5-8).
Although using high tensile steel will not reduce dome capacity in reality, but as the
AISC [43] assume same element capacity at slenderness ratio 157 for any yielding stress
so, more imperfection with analysis type DA-Liew needs to be assumed at high yield
stress that will lead to less element stiffness and less dome capacity.
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Chapter 6 : Summary and Conclusions
6.1. Summary
The unified approach to nonlinear solution with arc-length control method using
different finite elements was implemented into the proposed program. It is able to capture
the full equilibrium path in different problems. The result of the analysis was compared
with that available in textbooks. A proposed truss element is able to consider nonlinear
geometry, nonlinear material and initial out-of-straightness. It was compared with results
of analysis for strut member that consists of 20 frame elements, considering the inelastic
material, nonlinear geometry and out-of-straightness. This comparison showed that the
results are identical. By using this truss element, the effect of changing rise height of star
dome was investigated. The direct analysis of the shallow star dome shows some
differences compared with the results obtained from design using only nonlinear
geometry. The effect of changing values of out-of-straightness and yielding stress at
different slenderness rations was clarified. It was shown that using the method of an
equivalent out-of-straightness to mimic the design code equation is not a good idea for
star dome problem. Better results can be achieved by using the allowable tolerance for
values of out-of-straightness stated in the design code and then applying a global
reduction factor on the structure depending on brittle or ductile performance in the post-
buckling stage.
6.2. ConclusionsThe following conclusions drawn from the results of the study:
• The problem of star dome is very sensitive to the shallowness of its rise. The bigger the domes rise, the bigger the load it can carry.
• Adjusting initial out-of-straightness by some researchers to mimic code
limits led to using large out-of-straightness in some cases that is unjustifiably
out of usual out-of-straightness tolerance of international codes.
• Higher value of the initial out-of-straightness gives less axial stiffness and
load limit for compression elements.
• Adjusting the initial out-of-straightness to get a code limit led to an illogical behavior because it gives a lower stiffness to steel with higher yield stress.
• Traditional design using linear elastic or nonlinear elastic analysis and
member check formulas per codes gives same member capacity for λc > 1.5
whatever the value of yield stress is. On the contrary, higher yield stress in
advanced design (or Direct Analysis, DA) using full nonlinear techniques,
gives bigger ductility and at least equal failure load. DA using the equivalent
out-of-straightness method gave lower failure load for higher yield stress,
which does not make sense. DA using the equivalent out-of-straightness
method had to introduce bigger unreasonable δ (out-of-straightness) which
reached a value of L/175 for mild steel and L/110 for high-grade steel in
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order to mimic the code design equations. This led to losing three most useful
advantages of direct analysis:
1. Ductility (especially at post failure) assessment which controls
serviceability limits.
2. The high elastic stiffness at the start.3. Added member capacity (4% to 8%) at λc 0 1.5.
6.3. Suggestions for future research
Suggestions for future research and enhancements are discussed in the following:
1. Developing a truss element connected at one end or both ends through a
connection with an eccentricity.
2. Considering bolted connections with bolts, more than one bolt or using
welded connections with partially restrain for the truss element, which will
increase the critical buckling load. Therefore, some modifications are
required on element proposed in 2.3.2 to consider the gain in element
stiffness and buckling load from partially restrained connections.
3. There is a need to implement a truss element with three nodes similar to the
element proposed in 2.3.2 to be able to model continuous bracing intersected
in the internal node as x-bracing.
4. For the analysis by the program NFA using elements proposed in 2.2.2.4
and 2.2.2.5 there is a need to include the residual stresses and to limit the
stresses to the rupture stresses. Limiting stresses will need to use more
powerful control method in the nonlinear solver. In addition, an auto-stabilizing algorithm for the ruptured parts is required to be able to continue
the analysis up to the full collapse.
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[4] J. S. Sandhu, K. Stevens and G. A. Davies, "A 3D Co-rotational, Curved and
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[7] J. Teigen, "Nonlinear analysis of concrete structures based on a 3D shear-beam
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[10] B. Nour-Omid and C. Rankin, "Finite rotation analysis and consistentlinearization using projectors,," Computer Methods in Applied Mechanics
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[15] B. Kato and H. Akiyama, "Force characteristics of steel frames equipped with
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[16] M. Wakabayashi, C. Matsui and I. Mitani, "Cyclic behavior of Restrained steel
brace under axial loading," Proc. 6th World Conference on Earthquake
Engineering, vol. 3, pp. 3181-3187, 1977.
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[17] M. Nakashima and M. Wakabayashi, "Analysis and design of steel braces and
braced frames in building stsucturts," in Proc. U.S.-Japan Seminar on Cyclic
Behavior of Steel Structures, Osaka,Japan, 1991.
[18] S. Prickett, W. Mueller and F. Dewey, "Limit state analysis of lattice steel
transmission towers," ASCE Struct. Divi. Proceeding on Plastic and Other Limit
State Methods far Design Evaluation., 1984.
[19] M. Papadrakakis, "Inelastic post-buckling analysis of trusses," Structural
Engineering, ASCE,, vol. 109, no. 9, pp. 2129-2147, 1983.
[20] C. Hill, G. Blandford and S. Wang, "Post-Buckling Analysis of Steel space
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[21] W. H. Mueller and A. L. Wagner, "Plastic behavior of steel angle columns,"
School of Engineering and Applied Science,Portland State University, 1984.
[22] M. Abdel-Ghaffar, "Post-failure analysis for steel structure," 1992.
[23] J. Liew, N. Punniyakotty and N. Shanmugam, "Advanced Analysis and Design of
Spatial Structures," J. Construct. Steel Res., vol. Vol. 42, no. No. 1, p. pp. 2148,1997.
[24] M. A. Crisfield, Non-linear Finite Element Analysis of Solids and Structures, vol.
1, John Wiley & Sons Ltd., 1991.
[25] K. Bathe and A. Cimento, "Some practical procedures for the solution of nonlinear
finite element equations," Computer Methods in Applied Mechanics and
Engineering , vol. 22, pp. 59-85, 1980.
[26] M. Clarke and G. Hancock, "A Study of incremental-iterative strategies for
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[27] E. Riks, "The application of newton's methods to the problem of elastic stability," Journal of Applied Mechanics, vol. 39, pp. 1060-1065, 1972.
[28] E. Riks, "An incremental approach to the solution of snapping and nuckling
problems," International Journal of Solids and Structures, vol. 15, pp. 529-551,
1979.
[29] G. Wempner, "Quadratic convergence of crisfield's method," Computers &
Structures, vol. 17, pp. 69-72, 1971.
[30] M. Crisfield, "A fast incremental/iterative solution procedure that handles snap-
through," Computers & Structures, vol. 13, pp. 55-62, 1981.
[31] E. Ramm, "Strategies for tracing the nonlinear response near limit points,"
onlinear Finite Element Analysis in Structural Mechanics: Proceedings of the Europe-U.S. Workshop Ruhr-Universit " t Bochum, Germany, July 28% 31, 1980,
vol. Part II, pp. 63-89, 1980.
[32] L. Watson and S. Holzer, "Quadratic convergence of crisfield's method,"
Computers & Structures, vol. 17, pp. 69-72, 1983.
[33] J. Meek and H. Tan, "Geometrically nonlinear analysis of space frames by an
incremental iterative technique," Computer Methods in Applied Mechanics and
Engineering, vol. 47, pp. 261-282, 1984.
[34] J. Meek and S. Loganathan, "Geometrically nonlinear behavior of space-frame
structures," Computers & Structures, vol. 31, no. 1, pp. 35-45, 1989.
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[35] J. Meek and S. Loganathan, "Large displacement analysis of space-frame
structures," Computer Methods in Applied Mechanics and Engineering, vol. 72,
pp. 57-75, 1989.
[36] S. Leon, "A unified library of nonlinear solution schemes: An excursion into
nonlinear computational mechanics," Urbana-Champaign, 2010.
[37] J. Battini, "Co-rotational beam elements in instability problems.," Stockholm,
Sweden, 2002.
[38] W. Chen and E. M. Lui, Structural Stability: Theory and Implementation, Elsevier
Science Publishing Co.,Inc., 1987.
[39] H. Gavin, "Geometric Stiffness Effects in 2D and 3D Frames," 2012.
[40] A. Kassimali, Matrix Analysis of Structures, Second Edition ed., USA: Cengage
Learning, 2012.
[41] W. McGuire, R. H. Gallagher and R. D. Ziemian, Matrix sturctural analysis, 2nd
Edition ed., John Wiley & Sons,Inc., 2000.
[42] BS 5950, Structural use of steelwork in building, Corrigendum No. 1 ed., vol. Part1: Code of practice for design rolled and welded sections, 2000.
[43] AISC-360-10, Specification for Structural Steel Buildings, 2010, Ed., Chicago,
Illinois: American Institute of Steel Construction, Inc., June 22, 2010.
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Appendix A : NFA & NTA programs
A.1. NFA flow chart
A.2. NFA files descriptions
Table A-1 Description of files in NFA
NFA_Run Main function which NFA can run from.
1st read all files of the problem and read all sections database through
%readsectable&.Reformatting data in a way to be passed by NLSolver.
2nd
call NLSolver and passing all inputs and sections data base to it.3rd get problem solution from NLSolver and make all required
presentation and storing of the results.
readsectable A function which reads all data from sections database from hard files
and return it in a table format.
NLSolver The nonlinear solver will get all problem data and then assembling
structure stiffness matrix from element matrices (element matrices
constructed by %eleMatrix&). Assembling structure force vector and
solving both to get nodal displacements.
Solving the problem can be according to any control method as
described in 3.3.
boundAply Will apply boundary condition on both structure stiffness matrix andstructure force vector. The boundary condition can be used to define
plasticSecProelasticSecProp
NLSolver
eleMatrix qassemble feeldof boundApl
readsectabl
Results
NFA_Run
r
l
I
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any support or any support with settlement in degree of freedom
direction.
feeldof Returns a vector containing the degree of freedom numbers assigned
to the received node.
eleMatrix Returns global element stiffness matrix corresponding to the type of
analysis is chosen.qassemble Will assemble the internal force vector for the whole structure.
elasticSecProp Calculate section properties (area and second moment of area) for any
section in database.
plasticSecProp Make numerical integration on the section by slicing the section into
a lot of slices.
A.3. NTA flow chart
A.4. NTA files descriptions
Table A-2 Description of files in NTA
NFA_Run
Same descriptions as in A.2
readsectable
NLSolver
boundAply
feeldof
eleMatrix
qassemble
plasticSecProp elasticSecProp
NLSolver
eleMatrix qassemble feeldof boundAply
readsectable
Results
NFA_Run
P r o b l e m
I n u t s
ca acit ofcode
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elasticSecProp
plasticSecProp Calculate cross sectional area A, second moment of area I, plastic
section modulus Z, and reduced plastic section modulus Z pc at a
specified value of compression force.
capacityofcode Calculate axial compression capacity, according to code for any truss
element.
A.5. NFA & NTA MATLAB files
Attached with the thesis a CD contains all files for both programs.
A.6. NFA input manual
Nonlinear frame analysis (NFA) program requires the following files:
1. dispcurve.txt
2. DispDefomAtLoadSteps.txt
3. displayelement.txt
4. elprop.txt5. force.txt
6. gcoord.txt
7. general_input_data.txt
8. supports.txt
In addition to the following sections database files:
9. HSSDatabase.dat
10. IPEDatabase.dat11. IPNDatabase.dat
12. PipeDatabase.dat
13. RectangleDatabase.dat
A.6.1. Dispcurve.txt file
NFA program produces the P-! curve for any two degrees of freedom. The two
curves will be drawn on the screen and stored in the file will be asked for its name after
perform analysis.
In dispcurve.txt file first row is just a header without any effect; in second and third
rows first number represents the node number followed by a direction
header
1st node number Direction of D.O.F.
2nd node number Direction of D.O.F.
The only two rows after header must be filled with the above data.
Direction of D.O.F. = 1 for X-axis displacement, 2 for Y-axis displacement & 3 for
rotation about Z- axis.
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A.6.2. DispDefomAtLoadSteps.txt file
NFA program produces the structural deformed shape at a specified list of load
levels. The deformed shape of the structure will be drawn on the screen and stored in a
file called DefomAtLoadSteps.csv. To do enter the list of the percentage of required load
step to total number of load steps.I.e. if the total number of load steps = 300 and required to show structure
deformation at load step = 30,60&210 then DispDefomAtLoadSteps.txt will be as
follows:
header
10
20
70
NFA will automatically add percent at which maximum load factor occurs.
A.6.3. Displayelement.txt file
NFA program produces the stress-strain curve for top and bottom point of element
section. Each specified element has two sections at 0.211 L & 0.789 L. Therefore, four
stress-strain curves will be drawn on screen only for each element number in the
displayelement.txt
header
Element No.
Element No.
…….
Note: stress-strain curve will be drawn when the element matrix is chosen to be
based on [37]
A.6.4. Elprop.txt file
By this file, NFA program will get all element data. Each element is connected by
two nodes and has material and section dimension.
header
1st
node2nd node E Et G Fy Section type Section No. No. of slices
… … … … … … … … …
1st node number of the row in file gcoord.txt for the first node in the element.
2nd node number of the row in file gcoord.txt for the second node in the element.
E elastic modulus of element material.
Et tangent modulus of element material.
G shear modulus of element material.
Fy yielding stress of element material.Section type defines which database will be used. May be rec, HSS, IPN, IPE or pipe.
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Section No. number of the row of the section in database files. (Numbering after
header)
No. of slices number of section slicing for purpose of section properties by numerical
integration.20 slices will be usually enough.
A.6.5. Force.txt file
In this file the nodal forces will be assigned.
header
Node number Fx Fy Mz
… … … …
Node number number of the row in file gcoord.txt for node at which forces will be
applied.
Fx force in direction of global coordinate axis X.
Fy force in direction of global coordinate axis Y.Mz moment about global coordinate axis Z.
A.6.6. gcoord.txt file
By this file NFA program will get all node coordinates.no header row is allowed in
this file.
X-coordinate Y-coordinate
… …
A.6.7. general_input_data.txt file
header
e maxit ninc ControlType Matrixtype CtrlFactor
e Is the relative allowed tolerance used to check the convergence.
Value of 1e-6 may be usually good.
maxit Maximum number of iterations allowed in each load step. If the
number of iterations reached this value without achieving
convergence, then NTA program will stop the analysis process.
Value of 40 may be usually good.
ninc The number of load steps will be used. Value of 150 may be usually
good.
ControlType 0 for using standard Newton Raphson, 1 for displacement control
method while 2 for using arch length as a control method.
Matrixtype 0 for using element stiffness matrix based on 2.2.1.1 Chen element
by [38].
1 for using element stiffness matrix based on 2.2.2.1 Totalformulation by [5]
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2 for using element stiffness matrix based on 2.2.2.4 Inelastic linear
strain Bernoulli element by [37].
3 for using element stiffness matrix based on 2.2.2.5 Inelastic
shallow arch Bernoulli element by [37].
CtrlFactor Real number used for arch length control method. The high value
of CtrlFactor results long arc length and large load increment ateach step.
CtrlPara Integer number describes node number will be controlled during
displacement control method. While in arc length, control method
will be 0 for constant arc length and 1 for variable arc length.
Splitter Integer number used to be the maximum value of splitter counter.
When program couldnt achieve convergence then the load step
which didnt converged will restart using different CtrlFactor. A
counter will be grown at each splitting process and shrink at each
successive load increment. CtrlFactor will equal first inputted
CtrlFactor multiplied to 0.5 powered to splitter counter.
CtrlFactor~ = CtrlFactor• _ 0.5!"#$%&
A.6.8. supports.txt file
By this file NFA program will get all supports or node movement in the structure.
header
Node number Direction of D.O.F. Settlement value
… … …
Node number number of the row in file gcoord.txt for node at which settlement will be
applied.
Direction of D.O.F. = 1 for X-axis displacement, 2 for Y-axis displacement & 3 for
rotation about Z- axis.
Settlement value 0 for supports without any movements. If another value assigned,
then this value will be added incrementally with all other loads in a load step during the
analysis.
A.6.9. HSSDatabase.dat file
Hollow structural section is allowed to be defined as asection for any element.
header
b h t
… … …
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A.6.10. IPEDatabase.dat & IPNDatabase.dat file
IPE & IPN sections are allowed to be defined as a
section for any element.
header
h bf tf tw … … … …
A.6.11. PipeDatabase.dat file
Pipe section is allowed to be defined as a sectionfor any element, but restricted in this version to be not
used with inelastic analysis.
header
O.D. t
… …
A.6.12. RectangleDatabase.dat file
Rectangle section is allowed to be defined as a section for
any element.
header
b h
… …
A.7. NTA input files manual
Nonlinear frame analysis (NTA) program requires the following files:1. dispcurve.txt
2. displayelement.txt
3. elprop.txt
4. force.txt
5. gcoord.txt
6. general_input_data.txt
7. supports.txt
In addition to the following sections database files:
8. HSSDatabase.dat
9. IPEDatabase.dat
10. IPNDatabase.dat11. PipeDatabase.dat
t
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12. RectangleDatabase.dat
A.7.1. Dispcurve.txt file
NTA program produces the P-! curve for any two degrees of freedom. The two
curves will be drawn on the screen and stored in the file will be asked for its name after perform analysis.
In dispcurve.txt file first row is just a header without any effect; in second and third
rows first number represents the node number followed by a direction
header
1st node number Direction of D.O.F.
2nd node number Direction of D.O.F.
The only two rows after header must be filled with the above data.
Direction of D.O.F. = 1 for X-axis displacement, 2 for Y-axis displacement & 3 for Z-
axis displacement.
A.7.2. Displayelement.txt file
NTA program produces the P-! curve for any individual element.
header
Element No.
Element No.
…….
A.7.3. Elprop.txt file
By this file NTA program will get all element data. Each element is connected by
two nodes and has material and section dimension.
header
1st node 2nd node E Et G Fy Section type Section No.Code
capacity
… … … … … … … … …
1st node number of the row in file gcoord.txt for first node in the element.
2nd node number of the row in file gcoord.txt for first node in the element.
E elastic modulus of element material.
Et tangent modulus of element material.
G shear modulus of element material.
Fy yielding stress of element material.
Section type defines which database will be used. May be rec, HSS, IPN, IPE or pipe.
Section No. the number of the row of the section in database files. (Numbering after
header)
Code capacity positive number indicates entering the element axial compression
manually. While 0 for auto calculation according to [43] and -1 for auto calculatesaccording to [42].
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A.7.4. Force.txt file
In this file the nodal forces will be assigned.header
Node number Fx Fy Fz
… … … …
Node number number of the row in file gcoord.txt for node at which forces will be
applied.
Fx force in direction of global coordinate axis X.
Fy force in direction of global coordinate axis Y.
Fz force in direction of global coordinate axis Z.
A.7.5. gcoord.txt file
By this file NTA program will get all node coordinates.no header row is allowed in
this file.
X-
coordinate
Y-
coordinate
Z-
coordinate
… … …
A.7.6. general_input_data.txt file
header
e maxit ninc ControlType Matrixtype CtrlFactor CtrlPara Splitter
e Is the relative allowed tolerance used to check the convergence.
Value of 1e-6 may be usually good.
maxit Maximum number of iterations allowed in each load step. If the
number of iterations reached maxit without achieving
convergence, then NTA program will stop the analysis process.
Value of 40 may be usually good.
ninc Number of load steps will be used. Value of 150 may be usually
good.
ControlType 0 for using standard Newton Raphson, 1 for displacement control
method while 2 for using arch length as a control method.
Matrixtype 1 for using elastic element stiffness matrix based on [40].
2 for using element stiffness matrix based on 2.3.1 Elastic element
large displacement.
3 for using element stiffness matrix based on 2.3.2 3D inelastic
element large displacement limiting imperfection to a value that
limit capacity to code capacity in A.7.3.
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4 for using element stiffness matrix based on 2.3.2 3D inelastic
element large displacement limiting imperfection to a specified
value.
CtrlFactor Real number used for arch length control method. The high value
of CtrlFactor results long arc length and large load increment in
each step.CtrlPara Integer number describes node number will be controlled during
displacement control method. While in arc length control method
will be 0 for constant arc length and 1 for variable arc length.
Splitter Integer number used to be the maximum value for splitter counter.
When program couldnt achieve convergence then the load step
which didnt converged will restart using different CtrlFactor. A
counter will grow at each splitting process and shrink at each
successive load increment. CtrlFactor will equal first inputted
CtrlFactor multiplied to 0.5 powered to splitter counter.
CtrlFactor' = CtrlFactor(
_0.5)*+,-./
A.7.7. supports.txt file
By this file NTA program will get all supports or node movement in the structure.
header
Node number Direction of D.O.F. Settlement value
… … …
Node number Number of the row in file gcoord.txt for node at which settlement
will be applied.
Direction of D.O.F. Equal to 1 for X-axis displacement, 2 for Y-axis displacement &
3 for Z-axis displacement.
Settlement value 0 for supports without any movements. If another value assigned,
then this value will be added incrementally with all other loads in
a load step during the analysis.
A.7.8. HSSDatabase.dat fileSame as data base of NFA. Details in A.6.9
A.7.9. IPEDatabase.dat & IPNDatabase.dat file
Same as data base of NFA. Details in A.6.9
A.7.10. PipeDatabase.dat file
Same as data base of NFA. Details in A.6.9
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A.7.11. RectangleDatabase.dat file
Same as data base of NFA. Details in A.6.12
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Appendix B : NTA&NFA Input files for examples
This appendix contains the input files for solved examples in Chapter 4 :
B.1. NTA & NFA Database
B.1.1. HSSDatabase.dat
b,h,t
5.48 5.48 0.5022333
60 60 4.0
54.8 54.8 5.022
3.019595229 3.019595229 0.081340812
3.310583166 3.310583166 0.398314962
3.999677596 3.999677596 0.0813484492.953215349 2.953215349 0.571278881
2.670200436 2.670200436 0.23169975
100 100 8.0
0.0001 0.0001 0.00005
B.1.2. IPEDatabase.dat
h,bf,tf,tw (below is Standard IPE data base)
80 46 5.2 3.8
100 55 5.7 4.1
120 64 6.3 4.4140 73 6.9 4.7
160 82 7.4 5
180 91 8 5.3
200 100 8.5 5.6
220 110 9.2 5.9
240 120 9.8 6.2
270 135 10.2 6.6
300 150 10.7 7.1
330 160 11.5 7.5
360 170 12.7 8
400 180 13.5 8.6
450 190 14.6 9.4
500 200 16 10.2
550 210 17.2 11.1
600 220 19 12
B.1.3. IPNDatabase.dat
h,bf,tf,tw (below is Standard IPN data base)
80 42 5.9 3.9
100 50 6.8 4.5
120 58 7.7 5.1
140 66 8.6 5.7
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160 74 9.5 6.3
180 82 10.4 6.9
200 90 11.3 7.5
220 98 12.2 8.1
240 106 13.1 8.7
260 113 14.1 9.4280 119 15.2 10.1
300 125 16.2 10.8
320 131 17.3 11.5
340 137 18.3 12.2
360 143 19.5 13
380 149 20.5 13.7
400 155 21.6 14.4
450 170 24.3 16.2
500 185 27 18
550 200 30 19
600 215 32.4 21.6
B.1.4. PipeDatabase.dat
OD,t
6.3 0.554
6 4
168.3 5
193.7 10
21.3 3.2
26.9 3.2
33.7 2.633.7 3
33.7 3.2
42.4 2.6
33.7 3.6
48.3 2.5
42.4 3
33.7 4
42.4 3.2
48.3 3
42.4 3.6
48.3 3.2
60.3 2.5
42.4 4
48.3 3.6
60.3 3
48.3 4
60.3 3.2
76.1 2.5
60.3 3.6
88.9 2.5
48.3 576.1 3
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60.3 4
76.1 3.2
88.9 3
76.1 3.6
88.9 3.2
60.3 576.1 4
88.9 3.6
114.3 3
88.9 4
76.1 5
114.3 3.2
114.3 3.6
76.1 6
88.9 5
139.7 3.2
76.1 6.3114.3 4
139.7 3.6
88.9 6
88.9 6.3
168.3 3.2
139.7 4
114.3 5
168.3 3.6
114.3 6
168.3 4
139.7 5
114.3 6.3
139.7 6
168.3 5
139.7 6.3
193.7 5
168.3 6
168.3 6.3
139.7 8
219.1 5
193.7 6193.7 6.3
244.5 5
219.1 6
168.3 8
139.7 10
219.1 6.3
273 5
244.5 6
193.7 8
244.5 6.3
168.3 10323.9 5
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273 6
273 6.3
219.1 8
B.1.5. RectangleDatabase.dat
b,h
3 2
0.1 0.1
0.1 0.5
1.413 7.077140835
100.0 100.0
254.0 254.0
B.2. NFA Inputs for problem in 4.1.1
dispcurve.txt Node No. , direction 1 for x 2 y and 3 for rz ---this requires two rows for nodes that will
be presented in the P-delta curves
11 1
11 2
DispDefomAtLoadSteps.txt
Load level No. at which deformed structure will be shown in the text file and dialog
25
35
50
displayelement.txt
Element No. which stress -strain curve will be drawn to
1
elprop.txtist node,2nd node,E,Et,G,FY,Section Type , Sec ID ,Slice No
1 2 30000000 1000000 276.9230769 30000 rec 3 20
2 3 30000000 1000000 276.9230769 30000 rec 3 20
3 4 30000000 1000000 276.9230769 30000 rec 3 20
4 5 30000000 1000000 276.9230769 30000 rec 3 20
5 6 30000000 1000000 276.9230769 30000 rec 3 20
6 7 30000000 1000000 276.9230769 30000 rec 3 20
7 8 30000000 1000000 276.9230769 30000 rec 3 208 9 30000000 1000000 276.9230769 30000 rec 3 20
9 10 30000000 1000000 276.9230769 30000 rec 3 20
10 11 30000000 1000000 276.9230769 30000 rec 3 20
force.txt Node No. , Fx , Fy , Mz
11 0 1.0 0
gcoord.txt0 0
0.5 0
1 01.5 0
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2 0
2.5 0
3 0
3.5 0
4 0
4.5 05 0
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
1e-6 40 80 2 2 2.0 0 20 0 20
supports.txtSupport Node No. , direction 1 for x 2 y and 3 for rz , support settlement value
1 1 0
1 2 0
1 3 0
B.3. NFA Inputs for problem in 4.1.2
dispcurve.txt
Node No. , direction 1 for x 2 y and 3 for rz ---this require two rows for nodes which will
be represented in the P-delta curves
13 1
13 2
DispDefomAtLoadSteps.txt
Load level No. at which deformed structure will be shown in the text file and dialog
25
35
50
displayelement.txt
Element No. which stress -strain curve will be drawn to
5
elprop.txtist node,2nd node,E,Et,G,FY,Section Type , Sec ID,Slice No
1 2 720 72 276.9230769 10.44 rec 1 20
2 3 720 72 276.9230769 10.44 rec 1 20
3 4 720 72 276.9230769 10.44 rec 1 20
4 5 720 72 276.9230769 10.44 rec 1 205 6 720 72 276.9230769 10.44 rec 1 20
6 7 720 72 276.9230769 10.44 rec 1 20
7 8 720 72 276.9230769 10.44 rec 1 20
8 9 720 72 276.9230769 10.44 rec 1 20
9 10 720 72 276.9230769 10.44 rec 1 20
10 11 720 72 276.9230769 10.44 rec 1 20
11 12 720 72 276.9230769 10.44 rec 1 20
12 13 720 72 276.9230769 10.44 rec 1 20
13 14 720 72 276.9230769 10.44 rec 1 20
14 15 720 72 276.9230769 10.44 rec 1 20
15 16 720 72 276.9230769 10.44 rec 1 2016 17 720 72 276.9230769 10.44 rec 1 20
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17 18 720 72 276.9230769 10.44 rec 1 20
18 19 720 72 276.9230769 10.44 rec 1 20
19 20 720 72 276.9230769 10.44 rec 1 20
20 21 720 72 276.9230769 10.44 rec 1 20
force.txt
Node No. , Fx , Fy , Mz13 0 -1.0 0
gcoord.txt0 0
0 12
0 24
0 36
0 48
0 60
0 72
0 84
0 960 108
0 120
12 120
24 120
36 120
48 120
60 120
72 120
84 120
96 120
108 120
120 120
general_input_data.txt
convergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
1e-5 40 190 2 3 3.0 0 20
supports.txtSupport Node No. , direction 1 for x 2 y and 3 for rz , support settlement value
1 1 0
1 2 0
21 1 021 2 0
For elastic large displacement according to method in 2.2.2.1
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
1e-5 40 190 2 1 3.0 0 20
B.4. NFA Inputs for problem in 4.1.3
dispcurve.txt
Node No. , direction 1 for x 2 y and 3 for rz ---this require two rows for nodes that will be presented in the P-delta curves
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11 2
21 1
DispDefomAtLoadSteps.txtLoad level No. at which deformed structure will be shown in the text file and dialog
25
3550
displayelement.txtElement No. which stress -strain curve will be drawn to
10
11
elprop.txtist node,2nd node,E,Et,G,FY,Section Type , Sec ID ,No. of slices
1 2 203400 203.424 78230.8 400 HSS 1 20
2 3 203400 203.424 78230.8 400 HSS 1 20
3 4 203400 203.424 78230.8 400 HSS 1 20
4 5 203400 203.424 78230.8 400 HSS 1 205 6 203400 203.424 78230.8 400 HSS 1 20
6 7 203400 203.424 78230.8 400 HSS 1 20
7 8 203400 203.424 78230.8 400 HSS 1 20
8 9 203400 203.424 78230.8 400 HSS 1 20
9 10 203400 203.424 78230.8 400 HSS 1 20
10 11 203400 203.424 78230.8 400 HSS 1 20
11 12 203400 203.424 78230.8 400 HSS 1 20
12 13 203400 203.424 78230.8 400 HSS 1 20
13 14 203400 203.424 78230.8 400 HSS 1 20
14 15 203400 203.424 78230.8 400 HSS 1 20
15 16 203400 203.424 78230.8 400 HSS 1 20
16 17 203400 203.424 78230.8 400 HSS 1 20
17 18 203400 203.424 78230.8 400 HSS 1 20
18 19 203400 203.424 78230.8 400 HSS 1 20
19 20 203400 203.424 78230.8 400 HSS 1 20
20 21 203400 203.424 78230.8 400 HSS 1 20
force.txt Node No. , Fx , Fy , Mz
21 -1000 0 0
gcoord.txt
0 0.0000012.55865648 0.12048
25.11731297 0.23799
37.67596945 0.34963
50.23462594 0.45268
62.79328242 0.54457
75.35193891 0.62305
87.91059539 0.68620
100.4692519 0.73244
113.0279084 0.76066
125.5865648 0.77014
138.1452213 0.76066150.7038778 0.73244
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163.2625343 0.68620
175.8211908 0.62305
188.3798473 0.54457
200.9385037 0.45268
213.4971602 0.34963
226.0558167 0.23799238.6144732 0.12048
251.1731297 0.00000
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
1e-6 80 390 2 2 0.145 0 20
supports.txtSupport Node No. , direction 1 for x 2 y and 3 for rz , support settlement value
1 1 0
1 2 0
21 2 0
B.5. NFA Inputs for problem in 4.1.4
dispcurve.txt Node No. , direction 1 for x 2 y and 3 for rz ---this require two rows for nodes that will
be presented in the P-delta curves
4 1
4 2
DispDefomAtLoadSteps.txtLoad level No. at which deformed structure will be shown in the text file and dialog
25
35
50
displayelement.txtElement No. which stress -strain curve will be drawn to
15
46
elprop.txt
ist node,2nd node,E,Et,G,FY,Section Type , Sec ID ,No. of slices
1 6 29734 297.34 11436.15385 36 HSS 4 20
6 7 29734 297.34 11436.15385 36 HSS 4 207 8 29734 297.34 11436.15385 36 HSS 4 20
8 9 29734 297.34 11436.15385 36 HSS 4 20
9 10 29734 297.34 11436.15385 36 HSS 4 20
10 11 29734 297.34 11436.15385 36 HSS 4 20
11 12 29734 297.34 11436.15385 36 HSS 4 20
12 13 29734 297.34 11436.15385 36 HSS 4 20
13 14 29734 297.34 11436.15385 36 HSS 4 20
14 3 29734 297.34 11436.15385 36 HSS 4 20
2 15 29734 297.34 11436.15385 36 HSS 5 20
15 16 29734 297.34 11436.15385 36 HSS 5 20
16 17 29734 297.34 11436.15385 36 HSS 5 2017 18 29734 297.34 11436.15385 36 HSS 5 20
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18 19 29734 297.34 11436.15385 36 HSS 5 20
19 20 29734 297.34 11436.15385 36 HSS 5 20
20 21 29734 297.34 11436.15385 36 HSS 5 20
21 22 29734 297.34 11436.15385 36 HSS 5 20
22 23 29734 297.34 11436.15385 36 HSS 5 20
23 4 29734 297.34 11436.15385 36 HSS 5 203 24 29734 297.34 11436.15385 36 HSS 6 20
24 25 29734 297.34 11436.15385 36 HSS 6 20
25 26 29734 297.34 11436.15385 36 HSS 6 20
26 27 29734 297.34 11436.15385 36 HSS 6 20
27 28 29734 297.34 11436.15385 36 HSS 6 20
28 29 29734 297.34 11436.15385 36 HSS 6 20
29 30 29734 297.34 11436.15385 36 HSS 6 20
30 31 29734 297.34 11436.15385 36 HSS 6 20
31 32 29734 297.34 11436.15385 36 HSS 6 20
32 4 29734 297.34 11436.15385 36 HSS 6 20
3 33 29734 297.34 11436.15385 36 HSS 7 2033 34 29734 297.34 11436.15385 36 HSS 7 20
34 35 29734 297.34 11436.15385 36 HSS 7 20
35 36 29734 297.34 11436.15385 36 HSS 7 20
36 37 29734 297.34 11436.15385 36 HSS 7 20
37 38 29734 297.34 11436.15385 36 HSS 7 20
38 39 29734 297.34 11436.15385 36 HSS 7 20
39 40 29734 297.34 11436.15385 36 HSS 7 20
40 41 29734 297.34 11436.15385 36 HSS 7 20
41 5 29734 297.34 11436.15385 36 HSS 7 20
5 42 29734 297.34 11436.15385 36 HSS 7 20
42 43 29734 297.34 11436.15385 36 HSS 7 20
43 44 29734 297.34 11436.15385 36 HSS 7 20
44 45 29734 297.34 11436.15385 36 HSS 7 20
45 46 29734 297.34 11436.15385 36 HSS 7 20
46 47 29734 297.34 11436.15385 36 HSS 7 20
47 48 29734 297.34 11436.15385 36 HSS 7 20
48 49 29734 297.34 11436.15385 36 HSS 7 20
49 50 29734 297.34 11436.15385 36 HSS 7 20
50 2 29734 297.34 11436.15385 36 HSS 7 20
1 51 29734 297.34 11436.15385 36 HSS 8 20
51 52 29734 297.34 11436.15385 36 HSS 8 2052 53 29734 297.34 11436.15385 36 HSS 8 20
53 54 29734 297.34 11436.15385 36 HSS 8 20
54 55 29734 297.34 11436.15385 36 HSS 8 20
55 56 29734 297.34 11436.15385 36 HSS 8 20
56 57 29734 297.34 11436.15385 36 HSS 8 20
57 58 29734 297.34 11436.15385 36 HSS 8 20
58 59 29734 297.34 11436.15385 36 HSS 8 20
59 5 29734 297.34 11436.15385 36 HSS 8 20
5 60 29734 297.34 11436.15385 36 HSS 8 20
60 61 29734 297.34 11436.15385 36 HSS 8 20
61 62 29734 297.34 11436.15385 36 HSS 8 2062 63 29734 297.34 11436.15385 36 HSS 8 20
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63 64 29734 297.34 11436.15385 36 HSS 8 20
64 65 29734 297.34 11436.15385 36 HSS 8 20
65 66 29734 297.34 11436.15385 36 HSS 8 20
66 67 29734 297.34 11436.15385 36 HSS 8 20
67 68 29734 297.34 11436.15385 36 HSS 8 20
68 4 29734 297.34 11436.15385 36 HSS 8 20
force.txt Node No. , Fx , Fy , Mz
4 100 0 0
gcoord.txt0 0
240 0
0 180
240 180
120 90
0.000 18.0000.000 36.000
0.000 54.000
0.000 72.000
0.000 90.000
0.000 108.000
0.000 126.000
0.000 144.000
0.000 162.000
240.098 18.000
240.186 36.000
240.257 54.000
240.302 72.000
240.317 90.000
240.302 108.000
240.257 126.000
240.186 144.000
240.098 162.000
24.000 180.000
48.000 180.000
72.000 180.000
96.000 180.000120.000 180.000
144.000 180.000
168.000 180.000
192.000 180.000
216.000 180.000
11.951 171.065
23.907 162.124
35.872 153.170
47.850 144.200
59.842 135.210
71.850 126.20083.872 117.170
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95.907 108.124
107.951 99.065
131.951 81.065
143.907 72.124
155.872 63.170
167.850 54.200179.842 45.210
191.850 36.200
203.872 27.170
215.907 18.124
227.951 9.065
12.000 9.000
24.000 18.000
36.000 27.000
48.000 36.000
60.000 45.000
72.000 54.00084.000 63.000
96.000 72.000
108.000 81.000
132.000 99.000
144.000 108.000
156.000 117.000
168.000 126.000
180.000 135.000
192.000 144.000
204.000 153.000
216.000 162.000
228.000 171.000
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
1e-6 40 350 2 2 1.045 0 20
supports.txt
Support Node No. , direction 1 for x 2 y and 3 for rz , support settlement value
1 1 0
1 2 02 1 0
2 2 0
B.6. NTA Inputs for problem in 4.2.1
dispcurve.txt Node No. , direction 1 for x 2 y and 3 for z ---this require two rows for nodes that will be
presented in the P-delta curves
2 12 2
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displayelement.txt
Element No. which will be presented in the P-delta curves
1
elprop.txt
ist node,2nd node,E,Et,FY,Section Type , Sec ID , Pmax1 2 206000 206 235 rec 6 4618000
2 3 206000 206 235 rec 6 4618000
force.txt
Node No. , Fx , Fy , Fz
2 0 -1000 0
gcoord.txt-10977 0 0
0 695 0
10977 0 0
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 3 19.0 0 20
supports.txtSupport Node No. , direction 1 for x 2 for y and 3 for z , support settlement value
1 1 0
1 2 0
1 3 0
2 3 0
3 1 0
3 2 0
3 3 0
For elastic large displacement according to method in 2.3.1
general_input_data.txt
convergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 2 70.0 0 20
B.7. NTA Inputs for problem in 4.2.2
For inelastic large displacement according to method in 2.3.2 & load case 1
dispcurve.txt Node No. , direction 1 for x 2 y and 3 for z ---this require two rows for nodes that will be
presented in the P-delta curves
1 1
1 2
displayelement.txt
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124
Element No. which will be presented in the P-delta curves
1
7
13
elprop.txt
ist node,2nd node,E,Et,FY,Section Type , Sec ID , Pmax1 2 203400 203.4 400 HSS 1 1129
1 3 203400 203.4 400 HSS 1 1129
1 4 203400 203.4 400 HSS 1 1129
1 5 203400 203.4 400 HSS 1 1129
1 6 203400 203.4 400 HSS 1 1129
1 7 203400 203.4 400 HSS 1 1129
2 3 203400 203.4 400 HSS 1 1135
3 4 203400 203.4 400 HSS 1 1135
4 5 203400 203.4 400 HSS 1 1135
5 6 203400 203.4 400 HSS 1 1135
6 7 203400 203.4 400 HSS 1 11357 2 203400 203.4 400 HSS 1 1135
7 8 203400 203.4 400 HSS 1 745
2 8 203400 203.4 400 HSS 1 745
2 9 203400 203.4 400 HSS 1 745
3 9 203400 203.4 400 HSS 1 745
3 10 203400 203.4 400 HSS 1 745
4 10 203400 203.4 400 HSS 1 745
4 11 203400 203.4 400 HSS 1 745
5 11 203400 203.4 400 HSS 1 745
5 12 203400 203.4 400 HSS 1 745
6 12 203400 203.4 400 HSS 1 745
6 13 203400 203.4 400 HSS 1 745
7 13 203400 203.4 400 HSS 1 745
force.txt Node No. , Fx , Fy , Fz
1 0 -100 0
gcoord.txt0.0000 82.1600 0.0000
250.3756 62.1600 0.0000
125.1878 62.1600 -216.8316
-125.1878 62.1600 -216.8316-250.3756 62.1600 0.0000
-125.1878 62.1600 216.8316
125.1878 62.1600 216.8316
433.0000 0.0000 249.9927
433.0000 0.0000 -249.9927
0.0000 0.0000 -499.9853
-433.0000 0.0000 -249.9927
-433.0000 0.0000 249.9927
0.0000 0.0000 499.9853
general_input_data.txt
convergence tol,max iteration,# of Load step, ControlType , MatrixType ,CtrlFactor,CtrlPara,Splitter
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125
0.000001 40 100 2 3 0.16 0 20
supports.txt
Support Node No. , direction 1 for x 2 for y and 3 for z , support settlement value
8 1 0
8 2 0
8 3 09 1 0
9 2 0
9 3 0
10 1 0
10 2 0
10 3 0
11 1 0
11 2 0
11 3 0
12 1 0
12 2 012 3 0
13 1 0
13 2 0
13 3 0
For elastic large displacement according to method in 2.3.1 & load case 1
general_input_data.txt
convergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 2 0.16 0 20
For inelastic large displacement according to method in 2.3.2 & load case 2
force.txt Node No. , Fx , Fy , Fz
1 0 -100 0
2 0 -100 0
3 0 -100 0
4 0 -100 0
5 0 -100 06 0 -100 0
7 0 -100 0
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 3 0.16 0 20
For elastic large displacement according to method in 2.3.1 & load case 2
force.txt Node No. , Fx , Fy , Fz
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1 0 -100 0
2 0 -100 0
3 0 -100 0
4 0 -100 0
5 0 -100 0
6 0 -100 07 0 -100 0
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 2 0.3 0 20
B.8. NTA Inputs for problem in 4.2.3
For inelastic large displacement according to method in 2.3.2 & CS-1 sections
dispcurve.txt Node No. , direction 1 for x 2 y and 3 for z ---this require two rows for nodes that will be
presented in the P-delta curves
8 3
5 2
DispDefomAtLoadSteps.txtElement No. which will be presented in the P-delta curves
1
6
3
4
elprop.txt
ist node,2nd node,E,Et,FY,Section Type , Sec ID , Pmax
1 2 205000 2050 275 pipe 4 1193741.6
3 2 205000 2050 275 pipe 4 1193741.6
1 5 205000 2050 275 pipe 4 683681.1
2 5 205000 2050 275 pipe 4 1212371.4
3 5 205000 2050 275 pipe 4 683681.1
4 5 205000 2050 275 pipe 4 1212380.3
6 5 205000 2050 275 pipe 4 1212380.37 5 205000 2050 275 pipe 4 683681.1
8 5 205000 2050 275 pipe 4 1212371.4
9 5 205000 2050 275 pipe 4 683681.1
7 8 205000 2050 275 pipe 4 1193741.6
9 8 205000 2050 275 pipe 4 1193741.6
force.txt Node No. , Fx , Fy , Fz
2 0 -1500000 0
5 0 -1000000 0
8 0 -1500000 0
gcoord.txt
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127
-3535.5000 0.0000 6000.0000
0.0000 3535.5000 5000.0000
3535.5000 0.0000 6000.0000
-3535.5000 0.0000 0.0000
0.0000 3535.5000 0.0000
3535.5000 0.0000 0.0000-3535.5000 0.0000 -6000.0000
0.0000 3535.5000 -5000.0000
3535.5000 0.0000 -6000.0000
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 3 0.9 0 20
supports.txt
Support Node No. , direction 1 for x 2 for y and 3 for z , support settlement value1 1 0
1 2 0
1 3 0
3 1 0
3 2 0
3 3 0
4 1 0
4 2 0
4 3 0
6 1 0
6 2 0
6 3 0
7 1 0
7 2 0
7 3 0
9 1 0
9 2 0
9 3 0
For elastic large displacement according to method in 2.3.1 & CS-1 sections
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 2 25.6 0 20
For inelastic large displacement according to method in 2.3.2 & CS-2 sections
elprop.txtist node,2nd node,E,Et,FY,Section Type , Sec ID , Pmax
1 2 205000 2050 275 pipe 3 473093.1
3 2 205000 2050 275 pipe 3 473093.11 5 205000 2050 275 pipe 3 247617.3
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128
2 5 205000 2050 275 pipe 3 483352.6
3 5 205000 2050 275 pipe 3 247617.3
4 5 205000 2050 275 pipe 3 483357.6
6 5 205000 2050 275 pipe 3 483357.6
7 5 205000 2050 275 pipe 3 247617.3
8 5 205000 2050 275 pipe 3 483352.69 5 205000 2050 275 pipe 3 247617.3
7 8 205000 2050 275 pipe 3 473093.1
9 8 205000 2050 275 pipe 3 473093.1
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 40 100 2 3 1.115 0 20
For elastic large displacement according to method in 2.3.1 & CS-2 sections
elprop.txtist node,2nd node,E,Et,FY,Section Type , Sec ID , Pmax
1 2 205000 2050 275 pipe 3 473093.1
3 2 205000 2050 275 pipe 3 473093.1
1 5 205000 2050 275 pipe 3 247617.3
2 5 205000 2050 275 pipe 3 483352.6
3 5 205000 2050 275 pipe 3 247617.3
4 5 205000 2050 275 pipe 3 483357.6
6 5 205000 2050 275 pipe 3 483357.6
7 5 205000 2050 275 pipe 3 247617.3
8 5 205000 2050 275 pipe 3 483352.6
9 5 205000 2050 275 pipe 3 247617.3
7 8 205000 2050 275 pipe 3 473093.1
9 8 205000 2050 275 pipe 3 473093.1
general_input_data.txtconvergence tol,max iteration,# of Load step, ControlType , MatrixType ,
CtrlFactor,CtrlPara,Splitter
0.000001 50 100 2 2 25.6 0 20
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129
Appendix C : STAAD PRO report
This appendix contains the STAAD PRO report for solved example in 5.3.1.
****************************************************
* *
* STAAD.Pro *
* Version 2007 Build 04 *
* Proprietary Program of *
* Research Engineers, Intl. *
* Date= JUN 2, 2014 *
* Time= 17:20:32 *
* *
* USER ID: -- *
****************************************************
1. STAAD TRUSS
INPUT FILE: StarDome Real design.STD
2. START JOB INFORMATION
3. ENGINEER DATE 13-JAN-13
4. END JOB INFORMATION
5. INPUT WIDTH 79
6. UNIT MMS NEWTON
7. JOINT COORDINATES
8. 1 0 821.6 0; 2 2503.76 621.6 0; 3 1251.88 621.6 -2168.32
9. 4 -1251.88 621.6 -2168.32; 5 -2503.76 621.6 0; 6 -1251.88 621.6 2168.32
10. 7 1251.88 621.6 2168.32; 8 4330 0 2499.93; 9 4330 0 -2499.93; 10 0 0 -4999.85
11. 11 -4330 0 -2499.93; 12 -4330 0 2499.93; 13 0 0 4999.85
12. MEMBER INCIDENCES
13. 1 1 2; 2 1 3; 3 1 4; 4 1 5; 5 1 6; 6 1 7; 7 2 3; 8 3 4; 9 4 5; 10 5 6; 11 6 7
14. 12 7 2; 13 7 8; 14 2 8; 15 2 9; 16 3 9; 17 3 10; 18 4 10; 19 4 11; 20 5 11
15. 21 5 12; 22 6 12; 23 6 13; 24 7 13
16. DEFINE MATERIAL START17. ISOTROPIC STEEL
18. E 200000
19. POISSON 0.3
20. DENSITY 7.85E-005
21. ALPHA 1.2E-005
22. DAMP 0.03
23. END DEFINE MATERIAL
24. MEMBER PROPERTY BRITISH
25. 1 TO 6 TABLE ST PIP422.6
26. 7 TO 12 TABLE ST PIP422.6
27. 13 TO 24 TABLE ST PIPE OD 48.3 ID 41.9
28. CONSTANTS
29. MATERIAL STEEL ALL
30. SUPPORTS
31. 8 TO 13 PINNED
32. MEMBER TRUSS33. 1 TO 24
34. DEFINE REFERENCE LOADS
35. LOAD R1 SELF WEIGHT
36. SELFWEIGHT Y -1 LIST 1 TO 24
37. LOAD R2 COVER LOAD
38. JOINT LOAD
39. 1 FY -1090.4
40. 2 FY -847.352
STAAD TRUSS -- PAGE NO. 2
41. 3 FY -847.352
42. 4 FY -847.352
43. 5 FY -847.352
44. 6 FY -847.352
45. 7 FY -847.352
46. 8 FY -241.943
47. 9 FY -241.943
48. 10 FY -241.943
49. 11 FY -241.943
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130
50. 12 FY -241.943
51. 13 FY -241.943
52. LOAD R3 LIVE LOAD
53. JOINT LOAD
54. 1 FY -3271.19
55. 2 FY -2542.06
56. 3 FY -2542.06
57. 4 FY -2542.06
58. 5 FY -2542.0659. 6 FY -2542.06
60. 7 FY -2542.06
61. 8 FY -725.83
62. 9 FY -725.83
63. 10 FY -725.83
64. 11 FY -725.83
65. 12 FY -725.83
66. 13 FY -725.83
67. END DEFINE REFERENCE LOADS
68. LOAD 1 LINEAR ANALYSIS 1.2D.L.+1.6L.L.
69. REFERENCE LOAD
70. R1 1.2 R2 1.2 R3 1.6
71. PERFORM ANALYSIS
P R O B L E M S T A T I S T I C S-----------------------------------
NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 13/ 24/ 6
SOLVER USED IS THE IN-CORE ADVANCED SOLVER
TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 21
72. CHANGE
73. LOAD 2 NONLINEAR ANALYSIS 1.2D.L.+1.6L.L.
74. REFERENCE LOAD
75. R1 1.2 R2 1.2 R3 1.6
76. PDELTA KG 100 ANALYSIS
STAAD TRUSS -- PAGE NO. 3
PDELTA KG FOR CASE NO. 2
77. LOAD LIST 1
78. PARAMETER 1
79. CODE LRFD
80. FYLD 248.213 ALL
81. FU 413.688 ALL
82. CHECK CODE MEMB 1 7 13
STAAD TRUSS -- PAGE NO. 4
STAAD.Pro CODE CHECKING - (LRFD 3RD EDITION)
***********************
ALL UNITS ARE - NEWT MMS (UNLESS OTHERWISE NOTED)
MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/
FX MY MZ LOCATION
=======================================================================
1 ST PIP422.6 (BRITISH SECTIONS)
PASS COMPRESSION 0.941 1
14179.85 C 0.00 0.00 2511.74
7 ST PIP422.6 (BRITISH SECTIONS)
PASS COMPRESSION 0.320 1
4856.09 C 0.00 0.00 0.00
13 ST PIPE (BRITISH SECTIONS)
PASS COMPRESSION 0.961 1
16429.04 C 0.00 0.00 3157.72
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131
************** END OF TABULATED RESULT OF DESIGN **************
83. LOAD LIST 2
84. PARAMETER 2
85. CODE LRFD86. FYLD 248.213 ALL
87. FU 413.688 ALL
88. CHECK CODE MEMB 1 7 13
STAAD TRUSS -- PAGE NO. 5
STAAD.Pro CODE CHECKING - (LRFD 3RD EDITION)
***********************
ALL UNITS ARE - NEWT MMS (UNLESS OTHERWISE NOTED)
MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/
FX MY MZ LOCATION
=======================================================================
1 ST PIP422.6 (BRITISH SECTIONS)
PASS COMPRESSION 0.964 2
14526.82 C 0.00 0.00 2511.74
7 ST PIP422.6 (BRITISH SECTIONS)
PASS COMPRESSION 0.304 2
4615.59 C 0.00 0.00 0.00
13 ST PIPE (BRITISH SECTIONS)
PASS COMPRESSION 0.966 2
16517.42 C 0.00 0.00 3157.72
************** END OF TABULATED RESULT OF DESIGN **************
89. FINISH
*********** END OF THE STAAD.Pro RUN ***********
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