Post-Buckling Analysis and Design of 3D Trusses

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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




TRUSSES

By



Faculty of Engineering, Cairo UniversityIn Partial Fulfillment of the

Requirements for the Degree of

MASTER OF SCIENCEIn


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


GIZA, EGYPT

2016




TRUSSES

By



Faculty of Engineering, Cairo University

In Partial Fulfillment of the

Requirements for the Degree of

MASTER OF SCIENCEIn


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


GIZA, EGYPT2016



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

mailto:[email protected]

mailto:[email protected]



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.



ii

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





iv

B.1.5. RectangleDatabase.dat ....................................................................... 115

B.2. NFA Inputs for problem in 4.1.1 ............................................................... 115




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



v

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



vi

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



vii

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 5-6 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



viii

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.





2

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.



3

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.



4

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]



5

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



6

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]



7

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



8

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.



9

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].



10

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.



11

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.



12

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 _à 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 [\]^ àbc0

defghi jklmn opq

0 rstu vwx

y z{| )

Eq. 2-3

} = ~ |•|

! Eq. 2-4



13

"#$ =

%

)* 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^_ ( à 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




14

~• =

"#

!

"0 0 ( #$

%0 0

0 &' )*

+,(-./)

01 34

56(789) 0

:;< >?

@A(BCD)

EF HI

JK(LMN)

0 OP RSTU(VWX)

(YZ[) ]^_(àb)

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[\)

(]^_) àb(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[#

(\]^)_

àb 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 [ \ ] ^#

(_à)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.



15

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



16

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].



17

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);



18

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



19

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



20

2.2.2.3. Elastic element


f i = [ N M1 M2 ] Eq. 2-42

Where

y = z{| } ~ Eq. 2-43

• = ! "#$ ( 2%& +'() Eq. 2-44

)* = + ,-. ( /0 + 212) Eq. 2-45


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

[ = \]^_à = 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



21

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.


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


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 _

àbcd Eq. 2-60



22

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

]

^_ à

=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.


f i = [ N M1 M2 ] Eq. 2-69

Where



23

N = 0.5 ( sda1 + sda2 ) Eq. 2-70

M1 =HI J

KLMNO ( P

QR STU [VWXY + Z[\]] +! _à bcdef +

gh ! ij

klmno Eq. 2-71

M2 = pq r

stuvw ( x

yz { =|} [~•!" + #$%&] + ! '()* +,-./ +

01 ! 23 45678 Eq. 2-72


K l = 9:;<= >?@A BCDEFGHI JKLM NOPQRSTU VWXY Z[\]

^ Eq. 2-73

Where

_àb =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



24

<=>? =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_àbcd

( 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.



25

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



26

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



27

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.



28

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 àbc def ghi 1 (4 jklm nop qrstu vwx yz {| }~• !

Eq. 2-98



29

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]



30

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.



31

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.



32

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



33

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



34

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.



35

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



36

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.



37

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.



38

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]



39

Figure 3-8 Arc-length method [36]

Figure 3-9 Types of arc-length method [36]



40

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]



41

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.



42

Figure 3-11 Parameters of Arc-length control method [36]



43

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.



44

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



45

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.



46

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 (Linear strain)

NFA (Arch strain)

(Battini 2002)



47

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.



48

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 (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 (Linear strain)

NFA (Arch strain)

(Battini 2002)



49

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.



50

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


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



51

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



52

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)



53

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.



54

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.)



55

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.



56

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 )



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 )



NTA (Advanced)NTA (E. large disp.)

Liew 1997 (E. large disp.)



57

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.



58

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



59

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 )


Liew 1997 ,CS-2

Liew 1997 ,CS-1

NTA ,CS-1

NTA ,CS-2



60

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


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



61

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



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


displacement u at point B ( mm )

NTA,Mixed CS-1 & CS-2

NTA ,CS-1



62

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


displacement w at point B ( mm )

NTA,Mixed CS-1 & CS-2NTA,CS-1



63

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



64

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 #



65

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



66



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



67

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



68

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 )



69

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).



70

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



71

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.





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,



72

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 )


slope 7.76slope 6.86slope 5.96slope 4.66 (Liew)slope 3.76

Code limit

x107



73

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 )



x107

Where

δ1-6 = L / 326

δ7-12

= L / 326δ13-24 = L / 337



74

Figure 5-12 Star dome: effect of roof rise using DA500


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 )



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 )


slope 7.76slope 6.86

slope 5.96slope 4.66 (Liew)slope 3.76

x107



75


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 )


slope 7.76slope 6.86slope 5.96

slope 4.66 (Liew)slope 3.76

x107



76

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)



77

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)



78


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)



79


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)



80

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)



81

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



82

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)



83

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.



84

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



85

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


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



86

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


LM+LG

LM+NG

DA-Liew((=L/204 : L/150)

DA-500

See detailed curve



87

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


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



88

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 )


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 )


All Pipes #48 DA-500

All Pipes #48 DA-Liew



89

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.



90

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


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


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



91

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


LM+LG

LM+NG

DA-500

DA-Liew

See detailed



92

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


LM+LG

LM+NG

DA-500

DA-500 with #DA-Liew1.005

(CL)

0.826

1.07861.03

(CL)0.92



93

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


LM+LG

LM+NG

DA-1500DA-1000

DA-500

DA-500 with #DA-Liew



94

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


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


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



95

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.



96

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



97

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.



98

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[3] K. M. Hsiao, H. J. Horng and Y. R. Chen, "A Corotational procedure that handles

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769-781, 1987.

[4] J. S. Sandhu, K. Stevens and G. A. Davies, "A 3D Co-rotational, Curved and

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Methods Appl. Mech. Engrg. 144, pp. 163-197, 1997.[6] C. Rankin and F. Brogan, "An element-independent co-rotational procedure for

the treatment of large rotations," ASME J. Pressure Vessel Technology, vol. 108,

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[7] J. Teigen, "Nonlinear analysis of concrete structures based on a 3D shear-beam

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[8] B. Haugen and C. Felippa, "A unified formulation of co-rotational finite elements:

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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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[12] C. Pacoste, "Co-rotational flat facet triangular elements for shell instability

analysis," Computer Methods in Applied Mechanics and Engineering, vol. 156,

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[13] J. M. Battini and C. Pacoste, "Co-rotational beam elements with warping effects,"

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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

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Engineering, vol. 3, pp. 3181-3187, 1977.



99

[17] M. Nakashima and M. Wakabayashi, "Analysis and design of steel braces and

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[25] K. Bathe and A. Cimento, "Some practical procedures for the solution of nonlinear

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Engineering , vol. 22, pp. 59-85, 1980.

[26] M. Clarke and G. Hancock, "A Study of incremental-iterative strategies for

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vol. 29, pp. 1365-1391, 1990.

[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.

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1979.

[29] G. Wempner, "Quadratic convergence of crisfield's method," Computers &

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through," Computers & Structures, vol. 13, pp. 55-62, 1981.

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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,"

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[33] J. Meek and H. Tan, "Geometrically nonlinear analysis of space frames by an

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[34] J. Meek and S. Loganathan, "Geometrically nonlinear behavior of space-frame

structures," Computers & Structures, vol. 31, no. 1, pp. 35-45, 1989.



100

[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

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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.

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101

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



102

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



103

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.



104

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.



105

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]



106

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

… … …



107

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



108

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].



109

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.



110

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




111

A.7.11. RectangleDatabase.dat file




112

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



113

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



114

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



115

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



116

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


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



117

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 ,


1e-5 40 190 2 3 3.0 0 20


1 1 0

1 2 0

21 1 021 2 0

For elastic large displacement according to method in 2.2.2.1



1e-5 40 190 2 1 3.0 0 20


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



118

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


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



119

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



1e-6 80 390 2 2 0.145 0 20


1 1 0

1 2 0

21 2 0


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



120

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



121

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


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



122

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



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



123

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



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




0.000001 40 100 2 2 70.0 0 20


For inelastic large displacement according to method in 2.3.2 & load case 1



1 1

1 2

displayelement.txt



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


convergence tol,max iteration,# of Load step, ControlType , MatrixType ,CtrlFactor,CtrlPara,Splitter



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




0.000001 40 100 2 2 0.16 0 20

For inelastic large displacement according to method in 2.3.2 & load case 2


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



0.000001 40 100 2 3 0.16 0 20

For elastic large displacement according to method in 2.3.1 & load case 2




126

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



0.000001 40 100 2 2 0.3 0 20


For inelastic large displacement according to method in 2.3.2 & CS-1 sections



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


2 0 -1500000 0

5 0 -1000000 0

8 0 -1500000 0

gcoord.txt



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



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



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



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



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



0.000001 50 100 2 2 25.6 0 20



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



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


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.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



4856.09 C 0.00 0.00 0.00

13 ST PIPE (BRITISH SECTIONS)


16429.04 C 0.00 0.00 3157.72



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.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

=======================================================================



14526.82 C 0.00 0.00 2511.74



4615.59 C 0.00 0.00 0.00

13 ST PIPE (BRITISH SECTIONS)


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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