eindhoven university of technology master engineering
TRANSCRIPT
Eindhoven University of Technology
MASTER
Engineering model for coupled thermomechanical behaviour of steel elements under fireconditions
Titulaer, R.H.A.
Award date:2016
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ENGINEERING MODEL FOR COUPLED THERMOMECHANICAL BEHAVIOUR OF STEEL ELEMENTS UNDER FIRE CONDITIONS
R.H.A. TITULAER
26 JANUARI 2016
GRADUATION THESIS
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER I
A thesis submitted for the degree of Master of Science at the University of Technology Eindhoven
Publication:
Version Final graduation thesis
Date 26-01-2016
Place Eindhoven
Author:
Name R.H.A. (Rick) Titulaer
Student number 0739564
Address Heezerweg 52
5614 HE Eindhoven
Phone number +31625241550
University of Technology Eindhoven:
Education Architecture, Building and Planning
Faculty Structural Design
Chair Applied Mechanics and Design
Mail address [email protected]
Supervisors:
1st supervisor prof.dr.ir. A.S.J. (Akke) Suiker
2nd supervisor dr.ir. H. (Herm) Hofmeyer
3th supervisor prof.dr.ir. J. (Johan) Maljaars
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ACKNOWLEDGEMENTS
This report comprises the final result of my masterβs thesis at the University of Technology Eindhoven. It consists
of several theoretical aspects belonging to the coupled thermomechanical behaviour of elements under fire
conditions and an engineering model describing this behaviour. The model was used to investigate several coupled
thermomechanical examples and increase insight in the coupled behaviour.
During my graduation project I read numerous papers, articles, and books to increase my knowledge in this
difficult topic. I also managed to increase my Matlab skills significantly, since the coupled behaviour has been
elaborated in this high-level numerical computing environment. I am now able to configure and write simple
Matlab codes in just a second. This code writing did not only improve my program skills, but also increased insight
in every step made in the coupled thermomechanical analysis.
This graduation project has become a success due to the support and encouragement of my environment. Therefore
I am very thankful for the excellent guidance and clear explanations from my first two supervisors, Akke Suiker
and Herm Hofmeyer. Their encouragement, enthusiasm, and motivation were incredibly helpful during my
graduation project. I would also like to thank Johan Maljaars for the interesting meetings and his support in this
project.
Finally, I would like to thank my family and friends for their support and interest throughout this graduation
project. A special thanks goes to my dad, who has always been my biggest supporter.
Rick Titulaer
Eindhoven, February 2016
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SUMMARY
The field of steel structures exposed to fire has developed as an important area of research over the years. During
a fire, structural stability needs to be achieved so that people can escape the building safely, the property is
protected, and firefighters can safely enter the building. The aim of this thesis is to increase insight into the coupled
thermomechanical behaviour of structural columns and to investigate their behaviour under fire conditions.
Firstly the theory is described for fire, in which the general development of a fire is explained. Then the most
common temperature-time curves and the steel material properties at elevated temperatures are given. The heat
transfer mechanisms are also described in the theory. The variational framework of the coupled thermoelasticity
equations is derived, and subsequently discretised to obtain the Finite Element Method (FEM) formulation.
Furthermore, the linear buckling and non-linear buckling theory is explained, followed by the coupled formulation,
in which the linear and non-linear theory are used to write a Matlab code which represents the coupled
thermomechanical engineering model.
By using the Matlab model, some examples were extensively investigated. Firstly the simply supported one-
dimensional beam element restrained at both sides and subjected to a sudden temperature increase was
investigated and verified with ABAQUS. Secondly the linear buckling theory was used to perform simple analyses
on simply supported one-dimensional elements. This was also verified with ABAQUS. Then the non-linear
buckling analysis was performed on a simple column subjected purely to a mechanical load. Finally a restrained
column in a structural system was studied and compared with the Eurocode. An analysis on a simple steel member
of a braced system showed similar results as methods described in the Eurocode.
From the results it is concluded that the model presented in this thesis is able to perform several coupled
thermomechanical analyses, linear buckling analyses, and non-linear buckling analyses on structural columns. This
model provides insight into the general theory of the fully coupled thermomechanical behaviour.
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ABBREVIATIONS, SYMBOLS, AND NOTATION
ABBREVIATIONS
AST Adiabatic Surface Temperature
FE Finite Element
FEM Finite Element Method
CFD Computational Fluid Dynamics
SYMBOLS
LATIN UPPER CASE SYMBOLS
A Cross-section
Bi Biot number
Ca Specific heat
E Modulus of elasticity
οΏ½ΜοΏ½π Rate of change of the elastic strain
Ξππ Strain tensor
Ξπππ Elastic strain
Ξπππ‘β Thermal strain
Gr Grashof number
In Intensity of the thermal radiation in the normal direction to the emitting surface
IΞΈ Intensity of thermal radiation in the direction at an angle ΞΈ
L Length
N Shape function
Nu Nusselt number
P Perimeter
Pr Prandtl number
Ra Raleigh number
Re Reynolds
LATIN LOWER CASE SYMBOLS
c Specific heat
ππ Body force
ha Heat exchange coefficient air
hc Convective heat transfer coefficient
hr Radiant heat transfer coefficient
hfi Heat exchange coefficient fire
k Thermal conductivity
q Heat flux
qc Convective heat flux
qr Radiative heat flux
qtot Net total heat flux
π Body heat source per unit volume
t Time
π’ Displacement x direction
w Displacement y direction
x,y,z Directions
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GREEK UPPER CASE SYMBOLS
Ξ¦ Entropy flux
πΊ Stress
GREEK LOWER CASE SYMBOLS
πΌ Thermal diffusivity
πΌ Linear coefficient of thermal expansion
π½ Coefficient of thermal expansion of a fluid
π½ Parameter in the relation for relative buckling resistance
Ξ³ Modification factor
πΏ Variation of variable
Ξ΅ Emissivity
πΜ Rate of change of the internal energy density
π Entropy
ππ Elastic entropy
ππ‘β Thermal entropy
π Temperature
ππ Steel temperature
π0 Ambient temperature
Ξ» Thermal conductivity
Ξ» LamΓ© constant
ΞΌ Absolute viscosity
ΞΌ LamΓ© constant
π Density
π0 Referential mass density
Ο Stefan-Boltzmann constant
Ο Transmissivity
π View factor
Ο Helmholtz energy density
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TABLE OF CONTENTS
Acknowledgements ........................................................................................................................................................ II
Summary ........................................................................................................................................................................ III
Abbreviations, symbols, and notation .......................................................................................................................... IV
Abbreviations ............................................................................................................................................................ IV
Symbols...................................................................................................................................................................... IV
Latin upper case symbols...................................................................................................................................... IV
Latin lower case symbols ...................................................................................................................................... IV
Greek upper case symbols ..................................................................................................................................... V
Greek lower case symbols ...................................................................................................................................... V
1. Introduction ............................................................................................................................................................. 4
1.1. Motivation ..................................................................................................................................................... 6
1.1.1. Aim ............................................................................................................................................................ 6
1.1.2. Objectives .................................................................................................................................................. 6
1.2. Literature review ........................................................................................................................................... 6
1.3. Research questions ........................................................................................................................................ 7
2. Theory ...................................................................................................................................................................... 8
2.1. Fire ............................................................................................................................................................... 10
2.1.1. Compartment fire model ........................................................................................................................ 12
2.1.2. Temperature-time curves ....................................................................................................................... 13
2.1.3. Material properties at elevated temperatures ....................................................................................... 14
2.1.3.1. Thermal material properties ......................................................................................................... 14
2.1.3.2. Mechanical material properties..................................................................................................... 16
2.2. Heat transfer mechanisms........................................................................................................................... 17
2.2.1. Conductive heat transfer ........................................................................................................................ 17
2.2.2. Convective heat transfer......................................................................................................................... 17
2.2.2.1. Forced convection .......................................................................................................................... 18
2.2.2.2. Natural convection ........................................................................................................................ 18
2.2.2.3. Boundary condition on the fire side ............................................................................................. 19
2.2.2.4. Boundary condition on the air side .............................................................................................. 19
2.2.3. Radiant heat transfer .............................................................................................................................. 19
2.2.3.1. Blackbody radiation....................................................................................................................... 20
2.2.3.2. Radiant heat transfer of greybody surfaces.................................................................................. 21
2.3. Coupled thermomechanical analysis ......................................................................................................... 22
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2.3.1. Balance equations ................................................................................................................................... 22
2.3.2. Kinematic equation................................................................................................................................. 23
2.3.3. Constitutive equations............................................................................................................................ 23
2.3.4. Weak formulation ................................................................................................................................... 24
2.3.5. Finite Element formulation .................................................................................................................... 25
2.3.6. Boundary conditions .............................................................................................................................. 29
2.3.6.1. Natural boundary conditions prescribed ..................................................................................... 29
2.3.6.2. Essential boundary conditions prescribed ................................................................................... 33
2.3.6.3. Mathermatical description rewriting boundary conditions ........................................................ 36
2.4. Linear buckling............................................................................................................................................ 39
2.4.1. Introduction ............................................................................................................................................ 39
2.4.2. Ritz approximation FEM stability.......................................................................................................... 39
2.4.2.1. One 3rd-order element ................................................................................................................... 39
2.4.3. Galerkin approximation FEM stability .................................................................................................. 43
2.5. Non-linear buckling .................................................................................................................................... 45
2.5.1. Introduction ............................................................................................................................................ 45
2.5.1.1. Non-linear classifications .............................................................................................................. 45
2.5.1.2. Material non-linearity.................................................................................................................... 45
2.5.1.3. Geometric non-linearity ................................................................................................................ 45
2.5.1.4. Boundary non-linearity ................................................................................................................. 46
2.5.2. Non-linear Finite Element procedures .................................................................................................. 47
2.5.2.1. Newton-Raphson iterative method .............................................................................................. 47
2.5.2.2. Load incrementation procedure.................................................................................................... 48
2.5.2.3. Arc-length method ........................................................................................................................ 49
2.5.3. Thermal non-linearity............................................................................................................................. 51
3. Results and discussion .......................................................................................................................................... 52
3.1. Coupled thermomechanical analysis ......................................................................................................... 54
3.1.1. Sudden boundary temperature increase Matlab code .......................................................................... 54
3.1.2. Sudden boundary temperature increase ABAQUS .............................................................................. 55
3.2. Linear buckling analysis ............................................................................................................................. 57
3.2.1. Ritz approximation FEM stability.......................................................................................................... 57
3.2.2. Galerkin approximation FEM stability .................................................................................................. 57
3.2.3. Matlab code............................................................................................................................................. 58
3.2.3.1. Example coupled thermomechanical problem............................................................................. 60
3.2.3.2. Example buckling mode 2 elements ............................................................................................. 61
3.2.3.3. Example buckling mode 3 and 6 elements ................................................................................... 62
3.2.3.4. Optimal number of elements ........................................................................................................ 64
3.2.3.5. Example coupled thermomechanical problem continuation....................................................... 65
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3.2.4. Abaqus verification ................................................................................................................................ 67
3.2.4.1. Mechanical verification ................................................................................................................. 67
3.2.4.2. Thermal verification ...................................................................................................................... 67
3.2.4.3. Coupled thermomechanical verifcation ....................................................................................... 68
3.2.5. Reduction Youngβs modulus.................................................................................................................. 68
3.3. Non-linear buckling analysis ...................................................................................................................... 69
3.3.1. Example geometric non-linear mechanical behaviour.......................................................................... 69
3.3.2. Thermal non-linearity............................................................................................................................. 76
3.4. Comparison with the Eurocode .................................................................................................................. 78
3.4.1. CTM......................................................................................................................................................... 78
3.4.1.1. Step 1: Determination of applied design load steel member in fire ............................................ 78
3.4.1.2. Step 2: Classification of the steel member under fire conditions................................................. 79
3.4.1.3. Step 3: Determination of the design load-bearing capacity steel member .................................. 79
3.4.1.4. Step 4: Determination of the degree of utilisation steel member ................................................ 79
3.4.1.5. Step 5: Determination of the critical temperature steel member ................................................. 79
3.4.1.6. Step 6: Determination of the section factor and correction factor for the shadow effect ........... 79
3.4.2. Case study ............................................................................................................................................... 81
4. Conclusions and recommendations ..................................................................................................................... 86
4.1. Conclusions ................................................................................................................................................. 88
4.2. Recommendations ....................................................................................................................................... 89
5. References .............................................................................................................................................................. 90
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1. INTRODUCTION
The first chapter is an introduction to the topic and starts with the motivation for the field of study. Then it continues with
giving the aim and objectives of this research, followed by a literature review. Finally, this chapter ends with the presentation
of the research question, divided into three sub questions.
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1.1. MOTIVATION
During recent years the field of steel structures exposed to fire has developed as an important area of research. Fire
resistance of buildings and building parts is an important factor for the building safety, since fires could occur at
any moment. The building safety has become one of the most essential aspects in the built environment. In this
built environment, thin-walled steel structures are commonly being used in industrial and commercial buildings
all over the world, since they have a very high strength to weight ratio and are easy to construct (Gunalan, Kolarkar,
& Mahendran, 2013). The thin-walled steel is vulnerable to various buckling modes when exposed to fire conditions
and therefore very complex (Ranawaka & Mahendran, 2004). Consequently, the fire resistance of these structures
has an enormous social relevance and therefore increasing insight in the coupled thermomechanical behaviour of
steel structures subjected to fire is of great importance.
1.1.1. AIM
The purpose of this graduation thesis is to study the fully coupled thermomechanical behaviour of one-dimensional
elements under fire conditions and increase insight into the general theory of this behaviour. In order to investigate
this behaviour a model was developed by using the governing equations of the coupled thermoelasticity. The
analysis was performed by modelling the coupled behaviour into the software Matlab, which is a tool for numerical
computation and visualization. After modelling this behaviour, the Matlab solutions were verified with commercial
FE software ABAQUS. Although some simple coupled calculations could be performed in ABAQUS, the more
difficult analyses, taking into account extensible elements and three unknowns (temperature and displacement in
two directions) instead of two, could be accomplished by using the Matlab code.
1.1.2. OBJECTIVES
The main objective of this study was to increase insight in the fully coupled thermomechanical behaviour of a one-
dimensional element under fire conditions. Another objective was to gain experience in analytical techniques,
numerical methods, Matlab, and finite element software. In the end, the model was able to predict at what
temperature an element would buckle, how long it would take for this instability to occur, what the influence was
of the material degradation, and what the influence was of applying the appropriate boundary conditions.
1.2. LITERATURE REVIEW
General requirements for the design of steel building components subjected to fire can be found in the Eurocode
for steel structures (CEN, 2011). The scope of this code is that it is applicable to buildings, where the fire load can
be related to the occupancy of the building. The Eurocode also describes the material properties at elevated
temperatures, which is of great importance during a structural stability analysis. Although the Eurocode can
prescribe the critical temperature and time for specific problems, the general background should be understood.
Therefore Eslami et al. (2013) described a finite element method (FEM) formulation for coupled thermoelasticity by
using the Galerkin formulation for the equation of motion and the conservation of energy. However, this
formulation neglects the Babuska-Brezzi condition for coupled formulations, in which oscillations should be
prevented by applying the appropriate polynomial degrees for the approximation functions (Mijuca, 2008). In this
thesis, this condition has been taken into account.
The structural response of restrained systems under fire conditions has been well investigated during recent years.
Sanad et. al. (2000) studied the structural action of composite beams in large buildings using numerical models, in
which they showed that the restrained thermal expansion dominates the response of the structural system over the
degradation of material properties at elevated temperatures. Usmani et. al. (2001) also verified the influence of the
restrained thermal expansion on the structural response by theoretical models. This was confirmed by Tan et. al.
(2007) using experimental models. They presented the structural response of restrained steel columns at elevated
temperatures and concluded that the restrains have a significant influence on the failure time. Although all these
models prescribe structural behaviour of restrained systems under fire conditions, hardly any research has been
carried out on the fully coupled thermomechanical behaviour of restrained structural systems.
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Kumar et al. (2005) and Welch et. al. (2008) investigated the opportunity of coupling the Computational Fluid
Dynamics (CFD) with the Finite Element (FE) software in the European research project FIRESTRUC. They studied
the simultaneous modelling of FE simulations and CFD giving a two-way interaction, shown in Figure 1-1. Using
common interchange file formats, information could be transferred from one program to the other to take into
account the two-way coupling. However, this European research project focussed mainly on the interaction
between CFD and FE software, whereas this thesis investigates and shows the general theory behind the
thermomechanical coupling and presents the influence of the coupling components.
1.3. RESEARCH QUESTIONS
The main research question of this thesis is:
βWhat is the coupled thermomechanical behaviour of steel structural elements under fire conditions?β
The main research question of this graduation project has been elaborated into three sub questions. These sub
questions are as follows:
- What is the coupled thermomechanical behaviour of a structural one-dimensional element under simple
conditions?
- At what temperature will this one-dimensional element buckle under these conditions and how long will
it take for this instability to occur? And what are the influences of the degradation of the properties on the
instability time?
- What is the influence of the appropriate fire boundary conditions? And what is the non-linear coupled
thermomechanical behaviour of steel?
These research questions have been answered during this project by investigating and modelling the coupled
thermomechanical behaviour into Matlab and investigating simple problems.
Figure 1-1. Coupling the CFD with FEM software for simulations of structures under fire conditions (Welch et al., 2008)
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2. THEORY
In the second chapter the general theory is explained. Firstly the fire development is explained, followed by some information
about fire models, temperature-time curves, and the degradation of the material properties at elevated temperatures. In the
second paragraph this chapter continues with general information about the heat transfer mechanisms. In this paragraph the
conductive, convective, and radiant heat transfer is explained. Next the governing equations for the coupled thermomechanical
analysis are derived and written in finite element (FE) formulation. In this paragraph also the boundary conditions are
explained and elaborated in more detail. The following paragraph describes the linear buckling theory in which the Ritz
approximation and Galerkin approximation methods are presented. Finally, the non-linear buckling theory is described giving
solution strategies which can be used to solve non-linear problems.
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2.1. FIRE
This paragraph is an introduction to the topic fire. In order to analyse steel structures in case of fire, a deeper
understanding of how a fire behaves during enclosure fires is required. A fire can start and grow in many different
ways. The fire development is very random and can vary for many specific situations. Despite this randomness,
the development of a compartment fire can be generally explained and understood. The main components that
affect the fire development are the roomβs geometry, the quantity of the combustible material, the type of
combustible material, its arrangement in the compartment, and the oxygen supply. The thermal properties of the
surfaces, such as thermal conductivity and heat capacity, are also contributory factors. The fire development can
be described by using Figure 2-1. The horizontal axis describes the time and the vertical axis the temperature.
The figure above describes the possible paths for the fireβs development. The early stage of the fire development is
defined as the period from ignition to flashover. In this early fire development stage, the temperature will gradually
increase if there is an opening, such as a door or window. Then the fire can advance to flashover, which mea ns that
any combustible surface in the compartment will emit pyrolysis products. Pyrolysis can be described as when a
solid material heats up and starts to emit gases, so the organic material starts to change the chemical composition
and physical phase. These gases start to burn when they are mixed with oxygen. The entire room or compartment
will be filled with the flames produced by this flashover. These flames will generate very high levels of radiation.
After the event of a flashover, the access to oxygen mainly controls the heat release rate. This period is known as a
fully developed fire. This stage is important for the bearing capacity of the building. Lastly, when all the material
in the compartment has been burning for a while, the heat release rate decreases because the mass loss rate of the
fuel decreases. This period is the decay period. A fire can be ventilation controlled and fuel controlled. The
ventilation controlled fire means that the magnitude of the fire depends on the amount of oxygen. Consequently,
the fuel controlled fire is dependent on the amount of fuel. If a fire is ventilation controlled, the fire can beha ve in
several ways illustrated in Figure 2-2. In a few percentage of all fires, air will rush into the compartment when the
firefighters open the door, which could result that the smoke gases in the room ignite. This type is represented by
line 3 in Figure 2-2. Line 1 represents a backdraft. This phenomenon describes that the smoke gases may ignite very
quickly when oxygen suddenly can enter the compartment, which results in flames shooting out of the room. Lines
1 and 2 describe the scenario that the fire spontaneously diminishes due to a lack of oxygen. The difference between
these lines is that line 2 is ventilation controlled, and line 1 is fuel controlled. In this last case (line 4), the air is easily
available and the fire is controlled by the amount of fuel.
Figure 2-1. Fire growth curve featuring different types of fire behaviour (Bengtsson, 1999)
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Table 2-1. Phenomena fire development
The difference between a flashover and backdraft is that a flashover is caused by thermal change, since the ignition
is caused by heat attaining the auto ignition temperature of the combustible material and gases. Backdrafts on the
other hand are caused by chemical change, since the re-introduction of oxygen may lead to ignition of the already
heated enclosure.
Phenomenon Description
Early stage of fire development Period from ignition to flashover, the start of the fire
(windows break around 350Β°C)
Flashover Pyrolysis occurs within the compartment, gases start
burning when mixed with oxygen, flashover occurs
when all the combustible materials in the compartment
reach their ignition temperature at the same time
(around 600Β°C)
Fully developed fire After flashover everything is burning and the fire is
fully developed (900-1200Β°C)
Decay When all the material in the compartment has been
burning, the heat release rate decreases because the
mass loss rate of the fuel decreases
Backdraft When the compartment is out of oxygen and suddenly
the oxygen can enter again, resulting in combustion of
the gases still present in the compartment
Figure 2-2. Typical fire behaviour (Bengtsson, 1999)
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2.1.1. COMPARTMENT FIRE MODEL
The temperatures generated in a compartment fire can be calculated or predicted by a description or model of the
fire. Therefore, such a model is an idealization of the compartment fire phenomena. The fire development was
described in Figure 2-1 and changes with time. As the fire plume rises, it draws in cool air from within the
compartment. This decreases the plumeβs temperature and increases the volume flow rate. A hot gas layer is formed
when the plume reaches the ceiling, however, the temperature of the gas layer decreases with time as the plumeβs
gases continue to flow into it. The interface between the hot upper layer and the air in the lower part of the
compartment is relatively sharp, which means that the compartment can be divided into two zones. This is called
the two-zone model. The description of the compartment fire can thus be defined with the two-zone model, which
can be seen in the figure below. It is assumed that the temperature and other properties are the same throughout
each layer. The temperature of the upper layer will remain superior even though the temperature in the lower layer
will rise during the course of the fire. The upper layer is therefore the most important factor in compartment fires.
Another model which is commonly used for compartment fires is the one-zone model. The one-zone model is based
on the fundamental hypothesis that, during the fire, the gas temperature is uniform in the compartment.
Enclosure gas temperatures can vary significantly depending on the position in the enclosure. The two-zone model
gives a simplified description of the compartment fire. This model is used for the pre-flashover stage which is the
early stage of the fire development. After flashover (post-flashover stage), the structural stability needs to be
achieved so that the property is protected and the firefighters can safely enter the building without the risk of
structural collapse. The fire is assumed to have caused flashover at a very early stage and is fully developed. For
this case the one-zone model is the most commonly used assumption, where the entire compartment is assumed to
be filled with fire gases of uniform temperature. The objective of calculating the temperatures is to cover the whole
process of the fire development. The design of structural components under fire conditions must be based on
knowledge of the thermal exposure to which the structural component is subjected. This is usually computed by
temperature-time curves. The gas temperature comprises thus the total temperature of the gas in the compartment
originated from convection and radiation, and is used as the applied temperature on the surface of the structural
elements. The temperature-time curves are then used to test these elements.
Figure 2-3. Two-zone model compartment fire (Walton & Thomas, 2002)
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2.1.2. TEMPERATURE-TIME CURVES
A fire can thus be described by a fire curve, which represents the fireβs development in time. The fire development
of any compartment will vary depending on different conditions as described earlier in this chapter. Several
national and international fire curves have been developed to simulate fires. The gas temperature in the
compartment is represented by ππ and the time is represented by t. All curves presented in this paragraph are post-
flashover.
- Standard ISO Cellulosic curve (ISO-834): This curve is used in standards all over the world. It represents
a model of a ventilation controlled natural fire in a normal building. The equation for this curve is
ππ= 293 + 345 Β· log(8t + 1). [2-1]
- Hydrocarbon curve: This curve represents a ventilated oil fire and is relevant for compartments where
petroleum fires might occur. The equation for this curve is
ππ= 293 + 1280 (1 β 0,325 Β· e-0,167 t β 0,675 Β· e-2,5 t). [2-2]
- External fire exposure curve: This curve is used for structural members in a façade external to the main
structure. The external fire curve is given by
ππ = 293 + 660 (1 β 0,687 Β· e-0,32 t β 0,313 Β· e-3,8 t). [2-3]
- Slow heating curve: The fire development may also be slow growing. Another name for this curve is the
smouldering curve. It is described by
ππ= 293 + 154 t0,25 for 0 < t β€ 21, [2-4]
ππ= 293 + 345 log(8(t β 20) +1) for t > 21. [2-5]
The ISO-834 fire curve is commonly used for fire resistance calculations and is independent of fuel, fuel load,
compartment geometry, opening size and the thermal properties of the surrounding structure (SundstrΓΆm &
Samuelsson, 2014). The cooling phase can be very important with regard to structural performance. In order to take
these factors into account the parametric fire curve can be used (CEN, 2011). The parametric time temperature curve
would be the most optimal curve to use for representations of a fire. This curve includes the cooling phase, which
can be an important phase in a fire. Despite the most significant representation of a fire, the parametric curve has
some limitations and uses several equations and parameters. Therefore, the ISO curve is easier to implement and
is used in this thesis.
Figure 2-4. Temperature-time curves
0
200
400
600
800
1000
1200
1400
1600
1800
0 20 40 60 80 100 120
Tem
pera
ture
π
[K]
Time [min]
Hydrocarbon curve
ISO-834
Slow heating curve
External fire exposure curve
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2.1.3. MATERIAL PROPERTIES AT ELEVATED TEMPERATURES
The next important aspect is the degradation of material properties at elevated temperatures. The material
properties of steel at elevated temperatures can be divided into two groups, the thermal material properties and
the mechanical material properties.
2.1.3.1. THERMAL MATERIAL PROPERTIES
The thermal material properties are described as follows. The density () is a physical property. It is defined as the
weight of objects with a constant volume. The unit used for the density is kg/m3. The density of structural steel
proposed by NEN-EN 1993-1-2 is 7850 kg/m3. The density is assumed to be constant under an increasing
temperature (CEN, 2011). The thermal conductivity (k) is defined as the amount of heat flux that would pass
through a certain material depending on the temperature gradient over the material. Commonly the units are
W/mK. For the thermal conductivity a standard value is suggested by Eurocode 3, Part 1.2, namely 45 W/mK.
However, the thermal conductivity of steel varies with the change in temperature based on the following equations
k = 54 β 3,33 Β· 10-2 ππ for 293 K < ππ β€ 1073 K, [2-6]
k = 27,3 for ππ > 1073 K. [2-7]
where ππ is the steel temperature. These relations can be seen in the following picture.
The specific heat (CE) is defined as the amount of heat per unit mass required to raise the temperature by one degree
Celsius. The Specific Heat is denoted as J/kgK. The following equations are suggested by Eurocode 3, Part 1.2 for
the change of specific heat of steel. These equations are graphically presented in Figure 2-6.
CE = 425 + 7,73 β10-1 ππ - 1,69β 10-2 ππ2 + 2,22 β 10-6 ππ
3 for 293 K β€ ππ β€ 873 K, [2-8]
CE = 666 + (13002
738βππ
) for 873 K < ππβ€ 1008 K, [2-9]
CE = 545 + (17820
ππβ731) for 1008 K < ππ β€ 1173 K, [2-10]
CE = 650 for ππ> 1173 K. [2-11]
0
10
20
30
40
50
60
200 400 600 800 1000 1200 1400
Therm
al
conductivi
ty [
W/m
K]
Steel temperature πa [K]
Figure 2-5. Thermal conductivity of carbon steel as a function of the temperature
k = 54 β 3,33 Β· 10-2 ππ
k = 27,3
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R.H.A. TITULAER 15
The thermal elongation (Ξl/l) is the change in length occurring in a member by a change in temperature. The units
are usually m/m/K. The relative thermal elongation of steel should be determined from the following
Ξl/l = 1,2 β 10-5 ππ + 0,4 β 10-8 ππ2 β 2,416 β 10-4 for 273 K β€ ππ < 1023 K, [2-12]
Ξl/l = 1,1 β 10-2 for 1023 K β€ ππ β€ 1137 K, [2-13]
Ξl/l = 2,0 β 10-5 ππ β 6,2 β 10-3 for 1137 K < ππ β€ 1473 K. [2-14]
where l is the length at 293 K and Ξl is the temperature induced elongation.
The thermal diffusivity (Ξ±) is defined as the thermal conductivity divided by the density and the specific heat at a
constant pressure. The thermal diffusivity is used in the heat equation, which the distribution of heat in a given
region over time. Common units are mm2/s. The emissivity (Ξ΅) is a value of a surfaceβs efficiency as a source of
radiation ranging from 0 to 1, where 0 is a low emissivity (reflective) and 1 is a high emissivity (black body).
Eurocode 3 recommends a constant value of 0,625 for steel (CEN, 2011). As can be seen in Figure 2-6, the specific
heat at around 700-750Β°C increases to infinity. This can be described by latent heat, which means that the energy is
absorbed by the material. An example of the latent heat is phase transformation. The phase transformation for steel
is called eutectoid transformation, which occurs for all iron-carbon alloys at around 723Β°C. Hence, this eutectoid
transformation causes the changes in the graphs in the previous pictures.
0
2
4
6
8
10
12
14
16
18
20
200 400 600 800 1000 1200 1400
Therm
al
elo
ngatio
n x
10
-3[l/Ξ
l]
Steel temperature πa [K]
0
1000
2000
3000
4000
5000
200 400 600 800 1000 1200 1400
Specific
heat
[J/k
gK
]
Steel temperature πa [K]
Figure 2-7. Relative thermal elongation of carbon steel as a function of the temperature
Figure 2-6. Specific heat of carbon steel as a function of the temperature
CE = 425 + 7,73 β10-1 ππ β
1,69β 10-2 ππ2 + 2,22 β 10-6 ππ
3
CE = 666 + (13002
738βππ)
CE = 545
+ (17820
ππβ731)
CE = 650
Ξl/l = 1,2 β 10-5 ππ +
0,4 β 10-8 ππ2 β
2,416 β 10-4
Ξl/l = 1,1 β 10-2
Ξl/l = 2,0 β 10-5 ππ β 6,2 β 10-3
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2.1.3.2. MECHANICAL MATERIAL PROPERTIES
The mechanical properties of steel also degrade at elevated temperatures. The effective yield strength degrades
significantly at temperatures larger than 700 K (CEN, 2011). The other mechanical property is the Youngβs modulus
which can be seen in Figure 2-8. The Youngβs modulus is taken into account in this thesis and can be given as
πΈπ,π = ππΈ,ππΈπ, [2-15]
where
πΈπ is the Youngβs modulus of steel at ambient temperature
ππΈ,π is the reduction factor at an elevated temperature
πΈπ,π is the Youngβs modulus of steel at an elevated temperature
0
0.2
0.4
0.6
0.8
1
273 473 673 873 1073 1273 1473
Reductio
n f
acto
r kE
,ΞΈ
Steel temperature πa [K]
Figure 2-8. Reduction factor of the Youngβs modulus at elevated
temperatures
Youngβs modulus Ea,ΞΈ
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2.2. HEAT TRANSFER MECHANISMS
Heat transfer is also an important aspect in the investigation of the structural performance during a fire event.
Mathematical equations can describe the temperature distribution through a structure or material. These equations
can be used in several numerical heat transfer analyses. The three basic mechanisms that can be presented are
Conduction,
Convection,
Radiation.
Understanding these three heat transfer modes is necessary for performing numerical heat transfer analyses.
2.2.1. CONDUCTIVE HEAT TRANSFER
Conduction is the transmission of internal energy through microscopic diffusion and collisions between
neighbouring particles due to a temperature gradient without any motion of solid matter relative to one another.
Heat conduction can be described by Fourierβs law. The negative sign indicates that the heat flows from the higher
temperature side to the lower temperature side. Fourierβs law is prescribed by
ππ = βππππ,π , [2-16]
where πππ = π πΏππ is the thermal conductivity tensor and π,π is the temperature gradient. The material thermal
conductivity πππ within the appropriate temperature range may be estimated as a constant. In order to determine
the heat flux through the material, the exterior temperatures of the element are required. These exterior
temperatures are unknown variables, however, the surfaces of the element are in contact with the fluids of known
temperatures. To define the temperature distribution in the construction element, these fluid temperatures are used
as boundary conditions. When applying these thermal boundary conditions, it is often assumed that the heat
exchange between the fluid and the element surface is related to the temperature difference at the interface.
Consequently, the rate of heat transfer on the fire side is
ππ = βππ(πππ β ππ ) . [2-17]
And on the ambient temperature air side:
ππ = βπ(ππ β ππ) , [2-18]
where πfi and πa are the fire and air temperatures respectively. Quantities βππ and βπ are the overall heat exchange
coefficients on the fire and air side respectively. These values depend on the convective and radiant heat transfer
(h = hc + hr). The main issue of heat transfer in case of fire is to define appropriate heat transfer coefficients at the
fluid/solid interface. These heat transfer coefficients consist of two parts: the convective part and the radiant part.
Only when the fluid is in contact with the solid surface convective heat transfer may be applied. Radiant heat
transfer will always occur whether the fluid is in contact with the solid surface or not (Wang, 2002).
2.2.2. CONVECTIVE HEAT TRANSFER
Convection is defined as the transfer of heat by the collective movement of groups of molecules within fluids.
Convective heat transfer involves the combined processes of advection or diffusion or as a combination of both of
them. Convection is usually the dominant form of heat transfer in liquids and gases. Convective heat transfer is
difficult to study because it is highly unpredictable. Certain parameters have to be estimated to succeed the goal of
safety in case of flame spread. Fluid movement passing a solid surface can be separated into two categories, namely
forced convection or natural convection. Natural convection can be described by warmer air rising up by buoyancy,
which means that it has a natural cause. If the movement of the fluid or gas is controlled, then it is called forced
convection. The flow can be categorised into either laminar or turbulent. The laminar flow arises when a fluid flows
in parallel layers, without disruption between the layers. In the turbulent flow, vortices, eddies, and wakes cause
the flow to be highly unpredictable. The turbulent flow is therefore less arranged. In case of fire, the heat transfer
process occurs through the medium of air.
The convective heat transfer coefficients are necessary for modelling temperature distributions in structures in case
of fire. Dimensionless numbers are usually introduced for heat transfer analysis. The range of applicability of the
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R.H.A. TITULAER 18
limited number of small-scale experimental studies is extended by these dimensionless numbers. The convective
heat transfer coefficient (βπ) is related to the Nusselt (ππ’) number as can be seen in the following equation
ππ’ =βπ πΏ
π, giving βπ =
ππ’ βπ
πΏ, [2-19]
where L is the characteristic length of the solid surface, βπ is the convective heat transfer coefficient, and k is the
thermal conductivity of the fluid. The thermal conductivity of air varies at different temperatures.
2.2.2.1. FORCED CONVECTION
Table 2-2 presents the relations between the Nusselt number and additional dimensionless numbers for forced
convection. In this table, π π and ππ are the Reynolds and Prandtl numbers respectively. The Reynolds number is
defined by
π π =π πΏ π0
π =
πΏ π0
π£ , [2-20]
where Ο is the fluid density, the flow velocity is π0, and ΞΌ is the absolute viscosity of the fluid. The density of the
fluid and the absolute viscosity for air vary at different temperatures. The relative viscosity is given by v, which is
the absolute viscosity divided by the density. A fast velocity of the fluid results in a high Reynolds number and
therefore a high convective heat transfer.
The Prandtl number is defined as
ππ =π π
π , [2-21]
where k is the thermal conductivity and c is the specific heat of air, which both vary at different temperatures. The
Prandtl number for air is close to 0,7 (Wang, 2002).
2.2.2.2. NATURAL CONVECTION
Due to density differences arising from temperature gradients in the fluid the natural convection occurs. When the
temperature increases, the density change in the boundary layer will cause the fluid to rise and be replaced by a
cooler fluid. In case of natural convection, heat exchange between the fluid and the solid surface not only depends
on the fluid properties, but also on how the surface is placed in relation to the fluid (perpendicular/parallel and
above/below the fluid). The general equation for the Nusselt number for natural convection is given by
ππ’ = π΅ β π ππ , [2-22]
where B and m are given in the Table 2-3. π π is the Raleigh number and is defined as
π π = πΊπ β ππ , [2-23]
where πΊπ is the Grashof number, which is given by
πΊπ =π πΏ3 π½ βπ
π£2 , [2-24]
Table 2-2. Nusselt number relations for forced convection (Wang, 2002)
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R.H.A. TITULAER 19
where g is the gravity acceleration, π½ the coefficient of thermal expansion of the fluid, and Ξπ the temperature
difference between the fluid and the solid surface. The coefficient of thermal expansion of air at different
temperatures is 1 divided by the absolute temperature of air, according to the ideal gas law .
Unfortunately, an iterative process is necessary in order to calculate the convective heat transfer coefficients, since
most variables in the equations are temperature dependent and the surface temperatures are unknown variables in
a fire. However, radiation is the dominant mode of heat transfer in most cases of heat transfer analysis under fire
conditions, and therefore temperature calculations, will not be very sensitive to immense variations of the
convective heat transfer coefficient. Consequently simplified methods can be used.
2.2.2.3. BOUNDARY CONDITION ON THE FIRE SIDE
Natural convection is the main part of the convective heat transfer. The convective heat transfer is usually turbulent
at the fire/solid interface. The convective heat transfer coefficient can now be generalised by
βπ = π΅[(π βππ
π β π£2 )]1/3 π(βπ)1/3 = πΌ(βπ)1/3 . [2-25]
By substituting values of B (= 0,14), g (= 9,81 m/s2), Pr, π, v, and k into this equation the value of πΌ can be found
and varies between 1,0 and 0,6 within realistic fire temperatures of 600 β 1300 K. So for conservative calculations of
temperatures of the fire exposed surface, the value of Ξ± may be taken as 1,0 (Wang, 2002).
2.2.2.4. BOUNDARY CONDITION ON THE AIR SIDE
A similar exercise as for the fire side gives the following equation for the convective heat transfer coefficient
assuming a laminar flow on the ambient temperature air side.
βπ = πΌ (βπ)1/4 , [2-26]
where πΌ can be taken as approximately 2,2. Eurocode 3, Part 1.2 simplifies the coefficient calculations even further
and recommends a constant convective heat transfer coefficient. On the fire side the βπ is suggested as 25 W/m2,
and on the air side the βπ is proposed as 10 W/m2 (CEN, 2011).
2.2.3. RADIANT HEAT TRANSFER
Radiation is the emission or transmission of energy through space or a material medium in the form of waves or
particles. When radiant thermal energy passes a medium, any object within the path can absorb, reflect, and
transmit the incident thermal radiation. These three phenomena can be denoted as πΌ for the absorptivity, π for the
reflectivity, and Ο for the transmissivity. The representation of the fractions of incident thermal radiation that a body
absorbs, reflects, and transmits, respectively, is (Wang, 2002):
πΌ + π + Ο = 1. [2-27]
Table 2-3. Different variables for the Nusselt number equation for natural convection (Wang, 2002)
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R.H.A. TITULAER 20
2.2.3.1. BLACKBODY RADIATION
The factors in Eq. [2-27] are functions of the temperature, the electromagnetic wave length, and the surface
properties of the incident body. When all the incident thermal radiation is absorbed by the body, i.e. πΌ = 1,0. This
ideal body is then called a blackbody. The total amount of thermal radiation (πΈπ) by a blackbody surface is a function
of its temperature only and is given by the Stefan-Boltzmann law
πΈπ = ππ4 , [2-28]
where Ο is the Stefan-Boltzmann constant which is equal to 5,67 Γ 10-8 W/(m2K4) and π the absolute temperature in
K. This thermal radiation is not uniformly distributed in space, it is directionally dependent and can be expressed
by the Lambert Law
πΌπ = πΌππππ π, [2-29]
where πΌπ is the intensity of the thermal radiation in the normal direction to the emitting surface. The intensity of
thermal radiation in the direction at an angle π to the normal direction of the emitting surface is given by πΌπ. These
terms can be seen in the following figure.
The intensity of thermal radiation is defined as the radiant heat flux per unit area of the emitting surface per unit
subtended solid angle. The intensity of the directional thermal radiation can be derived by using this definition and
Eq. [2-29]. The total thermal radiation of a unit blackbody surface is covered by a hemispherical enclosure of radius
r, which can be seen in the following Figure 2-10 . The entire thermal radiation has to go through the hemispherical
enclosure, which means that the total incident thermal radiation on the hemispherical enclosure is equal to the total
thermal radiation emitted by the blackbody surface. An infinitesimally small surface area ππ΄ on the hemispherical
enclosure gives the subtended solid angle to the point of thermal radiation of ππ΄/π2. If the normal direction of this
incident surface area makes an angle π to the normal direction of the emitting blackbody surface, the incident
thermal radiation on this area ππ΄ is
πππ = πΌπππ΄
π2= πΌπ
ππ΄
π2πππ π . [2-30]
By integrating this equation over the entire hemisphere surface, the total thermal radiation incident on the
hemisphere can be found. Equate this to the total thermal radiation of the unit area of emitting surface gives
πΈπ = β― π π = β« πΌπ πππ ππ/2
0
2π β π sinπ β r dπ
π2= π πΌπ . [2-31]
Consequently, the intensity of the thermal radiation in the normal direction to the blackbody emitting surface is
given by
πΌπ =πΈπ
π . [2-32]
Figure 2-9. Directional intensity of radiant heat (Wang, 2002)
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R.H.A. TITULAER 21
Using the directional intensity of thermal radiation, the radiant thermal exchange between two blackbody
surfaces can be calculated. The thermal radiation from ππ΄1 incident on ππ΄2 is
ππππ΄1βππ΄2= πΈπ1
πππ π1πππ π2
π π2 ππ΄1 ππ΄2 . [2-33]
The total thermal radiation from π΄1 incident on ππ΄2 is
ππ΄1βππ΄2 = β« πΈπ1π΄1
πππ π1πππ π2
π π2 ππ΄1 ππ΄2 = π·πΈπ1ππ΄2 . [2-34]
πΈπ1ππ΄2 is the maximum incident thermal radiation on ππ΄2 and this occurs when ππ΄2 is entirely enclosed by π΄1. Ξ¦
is often called the configuration or view factor and only depends on the spatial configuration between π΄1 and ππ΄2.
This factor represents the fraction of thermal radiation from π΄1 incident on ππ΄2, since in most cases the incident
thermal radiation on ππ΄2 is much less. The configuration factor can be found by using tables and equations found
in the book from (Wang, 2002). In the Eurocode the configuration factor is also described and can be found in
Appendix A.
2.2.3.2. RADIANT HEAT TRANSFER OF GREYBODY SURFACES
Emitting and absorbing radiation according to laws of the blackbody occurs for no real material. Generally, in order
to define the radiant energy of an emitting surface, an additional term is required. This term is the emissivity Ξ΅. The
total radiant energy emitted by a general surface is then
πΈ = νππ4 . [2-35]
This emissivity of a surface in the total radiant energy equation is dependent on the wavelength of radiant energy,
the temperature of the surface, and the angle of radiation. A resultant emissivity is introduced when two surfaces
radiate heat. Eurocode 1, Part 1.2 presents the emissivity of fire and a general construction element surface as 0,8
and 0,7 respectively (CEN, 2011).
Figure 2-10. Determination of the intensity of thermal radiation (Wang, 2002)
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2.3. COUPLED THERMOMECHANICAL ANALYSIS
2.3.1. BALANCE EQUATIONS
The governing equations of generalized thermoelasticity are given in this section. The first equation used in the
coupled analysis follows from the first law of thermodynamics which is also known as the conservation of energy.
The first law of thermodynamics can be written as
π0πΜ + (div π β π) β πΊ Β· οΏ½ΜοΏ½π = 0 , [2-36]
where π0 is the referential mass density, πΜ is the rate of change of the internal energy density per unit mass, div is
the divergence, π is the thermal flux, π is the body heat source per unit volume, πΊ is the stress, and οΏ½ΜοΏ½π is the rate of
change of the elastic strain. The centre dot indicates a double contraction and the formulation used in this paper is
based on small strains. By using the relation between the Helmholtz energy density and the internal energy density
the temperature can be used as a state variable. This relation can be obtained from a Legendre transformation
(Yadegari, Turteltaub, & Suiker, 2012). The Helmholtz free energy is used in thermodynamics as a potential that
measures the valuable work obtainable from a closed thermodynamic system at a constant temperature. This
Helmholtz energy density is defined as
π = π β ππ , [2-37]
where π is the Helmholtz energy density, π is the internal energy, π is the absolute temperature, and π is the entropy
per unit mass. For the Legendre transformation the π¬π and π are used as independent variables for the Helmholtz
energy. This will give
π(π¬π , π) = π(π¬π , π(π¬π , π)) β ππ(π¬π , π) . [2-38]
Combining equations [2-36] and [2-37] results in
π0(οΏ½ΜοΏ½ + οΏ½ΜοΏ½π + ποΏ½ΜοΏ½) + (div π β π) β πΊ Β· οΏ½ΜοΏ½π = 0 , [2-39]
which from Eq. [2-38] can also be written as
π0 (ππ
ππ¬π
Β· οΏ½ΜοΏ½π +ππ
πποΏ½ΜοΏ½ + οΏ½ΜοΏ½π + ποΏ½ΜοΏ½) + (div π β π) β πΊ Β· οΏ½ΜοΏ½π = 0 . [2-40]
This equation consequently leads to
π0 (ππ
ππ¬πβ πΊ) Β· οΏ½ΜοΏ½π + π0 (
ππ
ππ+ π) οΏ½ΜοΏ½ + π0ποΏ½ΜοΏ½ + (div π β π) = 0 . [2-41]
The procedure of Coleman and Noll (1963) prescribes that the terms in Eq. [2-41] that are multiplied by the rates οΏ½ΜοΏ½π
and οΏ½ΜοΏ½ must vanish since a) an assumption can be made that these terms do not depend on the corresponding rates
and b) if these terms were non-zero, the dissipation could be negative, which is impossible (Turteltaub & Suiker,
2006). This will give the first equation for the coupled thermoelasticity
π0ποΏ½ΜοΏ½ + (div π β π) = 0 . [2-42]
The thermal flux π is given by the Fourierβs heat conduction law
π = βπ€βπ½ , [2-43]
where π€ = π πΏππ is the thermal conductivity tensor, π is the thermal conductivity, and βπ½ is the temperature gradient.
The body heat source term from Eq. [2-42] can be used for introducing heat from the sides of a one-dimensional
element while insulating the ends (Ericksen, 1991). This means that the boundary conditions for the thermal flux at
the ends of the element are zero. This is given by
π(0, π‘) = π(πΏ, π‘) = 0 . [2-44]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 23
The body heat source term can then be divided into radiant and convective heat source terms. These terms can be
summed up to obtain the total heat source in the element. This is described by
π = ππ + ππ , [2-45]
where ππ is the radiant heat source and ππ is the convective heat source, respectively. According to Ericksen (1991),
these terms are prescribed by the heat flows into the bar, which is radiation and convection. The body heat source
can then be rewritten as
π = νπ(ππ΄ππ(π‘)4 β ππ (π‘)4)+ βπ(ππ΄ππ(π‘) β ππ (π‘)) , [2-46]
where ν is the emissivity, π is the Stefan-Boltzmann constant, ππ is the steelβs surface temperature, and βπ is the
convective heat transfer coefficient. ππ΄ππ is the adiabatic surface temperature, which is a useful conceptual term for
linking gas temperatures with structural surface temperatures (WickstrΓΆm, Duthinh, & McGrattan, 2007). The
radiant body heat source term can be rewritten, giving
π = βπ(ππ΄ππ(π‘) β ππ (π‘)) + βπ(ππ΄ππ(π‘) β ππ (π‘)) , [2-47]
where the radiant heat transfer parameter βπ is defined to simplify the finite element formulation of the body heat
source, so that both the radiant and convective term in Eq. [2-47] have the same form. The radiant heat transfer
parameter can then be given as (Wang, 2002)
βπ = νπ(ππ΄ππ (π‘)2 + ππ (π‘)2)(ππ΄ππ(π‘) + ππ (π‘)) . [2-48]
The second equation used for the coupled analysis is the equation of motion which in local form can be written as
div πΊ + π = π0οΏ½ΜοΏ½ , [2-49]
where div πΊ is the divergence of the stress tensor, π is the body force, and the acceleration οΏ½ΜοΏ½ is neglected for the
formulation of thermomechanical coupling in this paper.
2.3.2. KINEMATIC EQUATION
For convenience, tensor component notation will be used from now on. The kinematic equation for small strains
and displacements is the strain tensor
Ξππ =1
2(π’π,π + π’π,π) , [2-50]
where Ξππ is the strain tensor and π’π is the displacement.
2.3.3. CONSTITUTIVE EQUATIONS
The Cauchy stress tensor Ξ£ππ is conjugated to the strain tensor Ξππ as well as that the entropy is conjugated to the
temperature. These terms can be described by using the Helmholtz free energy (π(Ξππ,π) ) equations following
from Eq. [2-41] giving the Cauchy stress tensor
Ξ£ππ =ππ
πΞππ , [2-51]
and the thermal part of the reversible entropy density as
ππ = βππ
ππ . [2-52]
The constitutive relation for the Cauchy stress tensor in terms of the strain and temperature can be given with
Ξ£ππ = πΆππππΞππ β π½πππ , [2-53]
where πΆππππ is the elasticity tensor and π½ππ is the coupling coefficient. The strain decomposition is given with
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 24
Ξππ = Ξπππ + Ξππ
π‘β , [2-54]
where Ξπππ is the elastic strain and Ξππ
π‘β is the thermal strain. The following parameters πΆππππ and π½ππ from an isotropic
body can be used for obtaining the entropy density and Cauchy stress tensor
πΆππππ = ππΏπππΏππ + π(πΏπππΏππ + πΏπππΏππ) , [2-55]
π½ππ = πΌ(π + 2π)πΏππ , [2-56]
where π and π are LamΓ© constants. In the x direction the strain decomposition is Ξπ₯π₯π = Ξπ₯π₯ β πΌπ where πΌ is the
coefficient of thermal expansion. The stress tensor can then be derived from the constitutive relation (Hookeβs law)
Ξ£π₯π₯ = πΞπ + 2πΞπ₯π₯π = π(Ξπ₯π₯ β 3πΌπ) + 2π(Ξπ₯π₯ β πΌπ)
= (π + 2π)Ξπ₯π₯ β πΌπ(3π + 2π) . [2-57]
Similarly the decomposition of the total entropy is used to find the equation for the total entropy rate (Yadegari et
al., 2012; Biot, 1956)
π = ππ + ππ‘β , [2-58]
where ππ is referred to as the thermal part of the reversible entropy density and ππ‘β is the reversible entropy density
that accounts for the coupling between the mechanical and thermal fields. Both terms follow from the linearization
of the entropy expression. The entropy flux is given by
Ξ¦π =ππ
π , [2-59]
where ππ is the heat flux per unit area and π is the absolute temperature. The constitutive relation between π and
ππ is given by (Turteltaub & Suiker, 2006)
ππ = ππΈ lnπ
π0
+ ππ , [2-60]
where ππ is the common value of ππ at the reference temperature π0 . Including the reversible entropy density term
ππ‘β and linearizing this equation for small changes of temperature, the entropy equation becomes (Biot, 1956)
π =ππΈ
π0
π + π½ππΞππ , [2-61]
where ππΈ is the specific heat for the unit volume without deformation, π0 is the reference temperature. ππΈπ
π0
is the
thermal part of the reversible entropy density and π½ππΞππ is the reversible entropy density that accounts for the
coupling between the mechanical and thermal fields. Note that the strong formulation equations (7) and (14) are
coupled through the Cauchy stress tensor in Eq. (18) and the entropy in Eq. (26) by the coupling coefficient π½ππ.
2.3.4. WEAK FORMULATION
Matrix equations for FEM implementation can be derived by using the weak formulation of equilibrium. The state
variables for thermoelasticity are π’ and π. A variation of these state variables (πΏπ’ and πΏπ) is considered around
their equilibrium value. By multiplying with πΏπ and integrating over the region of the element the local balance
energy conservation equation described in Eq. [2-42] becomes
β« (π0π0 οΏ½ΜοΏ½ + ππ,π β π)πΏππππ = 0 . [2-62]
The second term in Eq. [2-62] can be extended by using the divergence theorem and the total weak formulation of
the heat balance equation becomes
β« π0π0 οΏ½ΜοΏ½πΏπππ₯π β β« πππΏπ,πππ₯ + β« πππππΏπ dπ΄ β β« ππΏπππ₯ππ΄ = 0π . [2-63]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 25
Similarly the weak formulation for the equation of motion described in Eq. [2-49] is formulated. The equation is
multiplied by πΏπ’ and integrated over the element, giving
β« (Ξ£ππ,π + ππ)πΏπ’π πππ = 0 . [2-64]
Now the first term Eq. [2-64] can be extended by using the divergence theorem and the weak formulation for the
equation of motion therefore becomes
β β« Ξ£πππΏνππ ππ + β« πππΏπ’π πππ + β« Ξ£πππππΏπ’πππ΄π΄π = 0 . [2-65]
The already coupled weak formulations for the heat balance and the equation of motion can be obtained by
multiplying Eq. [2-65] with minus one and summing this equation up with Eq. [2-63] to find the total weak
formulation for the coupled thermoelasticity
β« π0π0 οΏ½ΜοΏ½πΏππππ β β« πππΏπ,πππ β β« ππΏπππ₯π + β« Ξ£πππΏνππ ππ β β« πππΏπ’π πππππ
= β β« πππππΏπ ππ΄π΄ + β« π‘ππΏπ’πππ΄π΄ . [2-66]
The right hand side of this weak formulation give the appropriate boundary conditions. The displacements and the
temperatures are prescribed by π’π = οΏ½ΜοΏ½ on π΄π’ and ππ = οΏ½ΜοΏ½ on π΄π where οΏ½ΜοΏ½ and οΏ½ΜοΏ½ are the prescribed values. On the
other hand, the traction and flux are given by π‘π = Ξ£ππππ = π‘Μ on π΄π and ππ = ππππ = π on π΄π where π‘Μ and π are the
prescribed traction and flux, respectively. These boundary conditions can be seen in Figure 2-11.
2.3.5. FINITE ELEMENT FORMULATION
In order to find the numerical solution the weak formulation of equilibrium or the variational equation must be
rewritten in matrix notation. Body forces are from now on neglected. Two sets of shape functions are introduced,
ππ’ and ππ, which relates the displacement component vector π and the temperature vector π½ to the corresponding
nodal element values π(π) and π½(π). This can then be expressed as
π = ππ’π(π) , π½ = πππ½(π) . [2-67]
By using this equation the strain π¬ and the temperature gradient βπ½ = π,π can be expressed in terms of nodal
quantities
π¬ = ππ’π(π), βπ½ = πππ½(π) , [2-68]
where ππ’ and ππ relate respectively to the first order derivative of components ππ’ and ππ’. The shape functions for
the variational forms (πΏπ½ and πΏπ) are similarly expressed (πΏπ½ = πππΏπ½(π)). Now the constitutive equations need to
be written in matrix form. Using the temperature component from Eq. [2-67] and the strain from Eq. [2-68] the
entropy constitutive law from Eq. [2-61] with respect to time becomes
οΏ½ΜοΏ½ =πΆπΈ
π0
πποΏ½ΜοΏ½(π) + π½ππ’οΏ½ΜοΏ½(π) . [2-69]
The Fourierβs law from Eq. [2-43] in matrix notion is
π = βπ€βπ½ = βπ€πππ½(π) . [2-70]
Figure 2-11. Body V with mixed boundary conditions for prescribed displacements,
tractions, temperatures, and fluxes
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 26
The stress constitutive equation from Eq. [2-53] becomes
πΊ = ππ’ππ’π(π) β π½πππ½(π) . [2-71]
By using the above equations the coupled FEM notation can be formulated. From now on the problem formulation
will be one-dimensional, which causes that ππ becomes π΄ππ₯, and ππ΄ becomes π|0π , where π is the perimeter. Since
the radiation and convection is applied as a boundary condition on a one-dimensional element, it is applied on the
perimeter to ensure a realistic outcome. The total weak formulation derived in Eq. [2-66] can written in matrix form.
Using equations [2-67]-[2-71] the FEM formulation for the total weak formulation can be written as
β (β« π0π0πππΏπ½(π) (π½ππ’οΏ½ΜοΏ½(π) +πΆπΈ
π0πποΏ½ΜοΏ½(π))π΄ππ₯
ππ
0 β β« ππTπΏπ½(π)T
π€πππ½(π)π΄ππ₯ππ
0 π
β β« βπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πΏπ½(π)πππ₯
ππ
0 β β« βπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πΏπ½(π)πππ₯
ππ
0
+ β« ππ’TπΏπ(π)T( ππ’ππ’π(π) β π½πππ½(π)) π΄ππ₯
ππ
0
= βοΏ½ΜοΏ½ππTπΏπ½(π)Tπ|
ππ
0+ οΏ½ΜοΏ½ππ’
TπΏπ(π)Tπ|
ππ
0) . [2-72]
Since in this problem the elements are of the same type and the DOFs at each node are the same, the assemblage
can be easily written by firstly describing the global size of the matrices and vectors followed by summing up the
element matrices and vectors accompanying the correct DOFs. Assembling the global matrices (g) can be written as
follows. The first term on the left hand side for the global system becomes
β« π0π0πππΏπ½(π) (π½ππ’οΏ½ΜοΏ½(π) +πΆπΈ
π0πποΏ½ΜοΏ½(π))π΄ππ₯
ππ
0
= πΏπ½(π)T π0π0 β β« ππ
Tπ½ππ’οΏ½ΜοΏ½(π)π΄ππ₯ππ
0π
+πΏπ½(π)Tβ β« ππ
TπΆπΈπποΏ½ΜοΏ½(π)π΄ππ₯ππ
0π . [2-73]
The global system components from Eq. [2-73] can be summarized for the capacity matrix components as
ππ,π’ = π0π0 β β« ππTπ½ππ’π΄ππ₯
ππ
0π , [2-74]
ππ,π = β β« ππTπΆπΈπππ΄ππ₯
ππ
0π , [2-75]
so that
β« π0π0πππΏπ½(π) (π½ππ’οΏ½ΜοΏ½(π) +πΆπΈ
π0
πποΏ½ΜοΏ½(π))π΄ππ₯ππ
0
= πΏπ½(π)Tππ,π’οΏ½ΜοΏ½(π) + πΏπ½(π)T
ππ,ποΏ½ΜοΏ½(π) . [2-76]
The second term on the left hand side becomes
β« ππTπΏπ½(π)Tπ€πππ½(π)π΄ππ₯
ππ
0
= β β« ππTπΏπ½(π)T
π€πππ½(π)π΄ππ₯ππ
0π , [2-77]
in which the stiffness matrix component ππ,π can be found as
ππ,π = β β« ππTπ€πππ΄ππ₯
ππ
0π , [2-78]
giving
β« ππTπΏπ½(π)Tπ€πππ½(π)π΄ππ₯
ππ
0 = πΏπ½(π)T ππ,π π½(π) . [2-79]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 27
The body heat source term on the left hand side of Eq. [2-73] can for the radiation part be described by
β β« βπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πΏπ½(π)πππ₯
ππ
0
= β β β« πΏπ½(π)T ππ
Tβπ(ππ΄ππ(π‘) β πππ½π (π‘)(π)})πππ₯
ππ
0π
= β β β« πΏπ½(π)T ππ
Tβπ(ππ΄ππ(π‘))πππ₯ππ
0π
+ β β« πΏπ½(π)T ππ
Tβππππ½π (π)
πππ₯ππ
0π , [2-80]
which can be evaluated as
ππ,π(π) = β β« ππTβππππππ₯
ππ
0π , [2-81]
ππ(π) = β β« ππTβπ(ππ΄ππ(π‘))πππ₯
ππ
0π , [2-82]
so that
β β« βπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πΏπ½(π)πππ₯
ππ
0
= βπΏπ½(π)Tππ(π) + πΏπ½(π)T
ππ,π(π) π½π (π)
. [2-83]
The convection part of the body heat source term in Eq. [2-73] can be divided into
β β« βπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πΏπ½(π)πππ₯
ππ
0
= β β β« πΏπ½(π)T ππ
Tβπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πππ₯
ππ
0π
= β β β« πΏπ½(π)T ππ
Tβπ(ππ΄ππ(π‘))πππ₯ππ
0π
+ β β« πΏπ½(π)T ππ
Tβππππ½π (π)
πππ₯ππ
0π , [2-84]
in which the ππ,π(π) and ππ(π) can be found as
ππ,π(π) = β β« (ππTβππππππ₯
ππ
0π , [2-85]
ππ(π) = β β« ππTβπ(ππ΄ππ (π‘))πππ₯
ππ
0π , [2-86]
giving that
β β« βπ(ππ΄ππ(π‘) β πππ½π (π‘)(π))πΏπ½(π)πππ₯
ππ
0
= βπΏπ½(π)Tππ(π) + πΏπ½(π)T
ππ,π(π)π½π (π)
. [2-87]
The last term on the right hand side can be written in global system form as
β« ππ’TπΏπ(π) T( ππ’ππ’π(π) β π½πππ½(π)) π΄ππ₯
ππ
0
= β β« ππ’TπΏπ(π)T
( ππ’ππ’π(π) β π½πππ½(π))π΄ππ₯ππ
0π . [2-88]
In this equation the stiffness components ππ’,π’ and ππ’,π can be found as
ππ’,π’ = β β« ππ’Tππ’ππ’π΄ππ₯
ππ
0π , [2-89]
ππ’,π = β β« ππ’Tπ½πππ΄ππ₯
ππ
0π , [2-90]
which results in
β« ππ’TπΏπ(π) T( ππ’ππ’π(π) β π½πππ½(π)) π΄ππ₯
ππ
0
= πΏπ(π)T ππ’,π’ πΏπ(π) β πΏπ(π)T ππ’,π π½(π) . [2-91]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 28
The first term on the right hand side in Eq. [2-73] gives in global system form
βοΏ½ΜοΏ½ππTπΏπ½(π)T
π|ππ
0= β β ππ
TπΏπ½(π)TοΏ½ΜοΏ½π|
ππ
0π , [2-92]
in which the thermal force vector can be seen as
ππ = β β ππT
π οΏ½ΜοΏ½π |ππ
0 . [2-93]
This term then becomes
βοΏ½ΜοΏ½ππTπΏπ½(π)T
π|ππ
0= πΏπ½(π)T
ππ . [2-94]
The second term on the right hand side in Eq. [2-73] is derived as
οΏ½ΜοΏ½ππ’TπΏπ(π)T
π |ππ
0= β ππ’
TπΏπ(π)TοΏ½ΜοΏ½π |
ππ
0π , [2-95]
in which the mechanical force vector can be seen as
ππ’ = β ππ’T
π οΏ½ΜοΏ½π |ππ
0 . [2-96]
The last term on the right hand side then becomes
οΏ½ΜοΏ½ππ’TπΏπ(π)T
π |ππ
0= πΏπ(π)T
ππ’ . [2-97]
The total weak formulation for the coupled thermoelasticity in FEM formulation then becomes
πΏπ½(π)Tππ,π’οΏ½ΜοΏ½(π) + πΏπ½(π)T ππ,ποΏ½ΜοΏ½(π) + πΏπ½(π)T
ππ,π π½(π)
+ πΏπ½(π)T ππ,π(π) π½π (π)
+ πΏπ½(π)T ππ,π(π) π½π
(π)
+πΏπ(π)T ππ’,π’ π(π) β πΏπ(π)T
ππ’,π π½(π)
= πΏπ½(π)Tππ + πΏπ(π)T
ππ’+πΏπ½(π)Tππ(π)+πΏπ½(π)T
ππ(π) . [2-98]
This equation should hold for arbitrary variations πΏπ½(π) and πΏπ(π). This gives
[0 0
ππ,π’ ππ,π] [
οΏ½ΜοΏ½οΏ½ΜοΏ½] + [
ππ’,π’ ππ’,π
0 ππ,π] [
ππ½] = [
ππ’ππ
] , [2-99]
where
ππ,π’ = π0π0 β β« ππTπ½ππ’π΄ππ₯
ππ
0π ,
ππ,π = β β« ππTπΆπΈπππ΄ππ₯
ππ
0 , π
ππ’,π’ = β β« ππ’Tππ’ππ’π΄ππ₯
ππ
0π ,
ππ’,π = β β β« ππ’Tπ½πππ΄ππ₯
ππ
0π ,
ππ,π = β β« ππTπ€πππ΄ππ₯
ππ
0π + β β« ππTβππππππ₯
ππ
0π + β β« ππTβππππππ₯
ππ
0π ,
ππ’ = β ππ’T
π οΏ½ΜοΏ½π |ππ
0,
ππ = β β ππT
π οΏ½ΜοΏ½π |ππ
0+ β β« ππ
Tβπ(ππ΄ππ(π‘))πππ₯ππ
0π + β β« ππTβπ(ππ΄ππ(π‘))πππ₯
ππ
0π . [2-100]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 29
Now the shape functions can be implemented, which can be found in Appendix B. It should be noted that the
polynomial degree of approximation functions used for displacements should be one order higher than for the
temperatures to overcome oscillations according to the Babuska-Brezzi condition (Mijuca, 2008). For the transverse
displacement components the Euler-Bernoulli beam theory (Cirak, 2015) is used for the implementation of the
stiffness matrix. The finite element formulation can then easily be shortened by
ποΏ½ΜοΏ½ + ππ = π . [2-101]
This linear algebraic equation has to be solved for every time step. By introducing the parameter πΎ, different
transient schemes can be used, which are shown in Table 2-4. The following equation is used for introducing this
parameter
ππ+ πΎ = πΎππ+1 + (1 β πΎ)ππ . [2-102]
Table 2-4. Time-stepping schemes for transient analysis and accompanying parameter πΎ
Eq. [2-101] can be rewritten in order to introduce the time-stepping schemes, which results in
π{ππ+1βππ
βπ‘} + πππ+πΎ = ππ+πΎ . [2-103]
Rearranging this equation using Eq. [2-102] gives
(π + Ξ³βπ‘π)ππ+1 = (π β (1 β Ξ³)βπ‘π)ππ + βπ‘(Ξ³ππ+1 + (1 β Ξ³)ππ) , [2-104]
which can be rewritten for the fully implicit method as
(π + βπ‘π)ππ+1 = πππ + βπ‘ππ+1 . [2-105]
Rewriting this equation for the unknown values gives
ππ+1 = (π + βπ‘π)β1(πππ + βπ‘ππ+1) . [2-106]
2.3.6. BOUNDARY CONDITIONS
Classical boundary conditions are considered for the coupled thermoelasticity. The Dirichlet type or the Neumann
type can be used. The Dirichlet type of boundary conditions are the essential boundary conditions, which includes
the displacements or the temperatures as given data. For the Neumann type the given boundary conditions are the
normal stresses or the heat flux, which are also called the natural boundary conditions.
2.3.6.1. NATURAL BOUNDARY CONDITIONS PRESCRIBED
First the general equation needs to be discretised in order to solve the system of equations. This procedure to solve
this system has been elaborated in the previous paragraph. If the natural boundary conditions are prescribed the
general equation [2-106] can be used to solve the linear system. Initial boundary conditions are needed to solve the
system for the first time step. It is also possible that essential boundary conditions are given. This means that the
known values should be rewritten to the right hand side of the equation.
Ξ³ Name of the scheme Comments
0.0 Fully explicit scheme Forward difference method
1.0 Fully implicit scheme Backward difference method
0.5 Semi-implicit scheme Crank-Nicolson method
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 30
Equation [2-101] for one element looks like
2/0
2/
2/0
2/
0000000000
0000
0
0
0
0000000
0000000000000000000
6664
4644
3634333231
2624232221
1614131211
6664636261
4644434241
k
i
k
i
k
j
i
k
j
i
k
j
i
k
j
i
PLq
PLq
PLt
PLt
u
u
u
KK
KK
KKKKK
KKKKK
KKKKK
u
u
u
CCCCC
CCCCC
. [2-107]
If you use the discretised equation for one element, this will give the following equation in which the red box is
indicated as the known values, and the blue box is indicated as the unknown values
1
6664636261
4644434241
1
6664
4644
3634333231
2624232221
1614131211
6664636261
4644434241
2/0
2/
2/0
2/
0000000
0000000000000000000
0000000000
0000
0
0
0
0000000
0000000000000000000
n
k
i
k
i
n
k
j
i
k
j
i
n
k
j
i
k
j
i
PLq
PLq
PLt
PLt
tu
u
u
CCCCC
CCCCC
u
u
u
KK
KK
KKKKK
KKKKK
KKKKK
t
CCCCC
CCCCC
. [2-108]
Now the six equations will be given separately
βπ‘πΎ11 Β· π’ππ+1 + βπ‘πΎ12 Β· π’π
π+1 + βπ‘πΎ13 Β· π’ππ+1 + βπ‘πΎ14 Β· ππ
π+1 + βπ‘πΎ16 Β· πππ+1 = ββπ‘
ππΏπ‘π
2
π+1 , [2-109]
βπ‘πΎ21 Β· π’ππ+1 + βπ‘πΎ22 Β· π’π
π+1 + βπ‘πΎ23 Β· π’ππ+1 + βπ‘πΎ24 Β· ππ
π+1 + βπ‘πΎ26 Β· πππ+1 = 0 , [2-110]
βπ‘πΎ31 Β· π’ππ+1 + βπ‘πΎ32 Β· π’π
π+1 + βπ‘πΎ33 Β· π’ππ+1 + βπ‘πΎ34 Β· ππ
π+1 + βπ‘πΎ36 Β· πππ+1 = βπ‘
ππΏπ‘π
2
π+1 , [2-111]
πΆ41 Β· π’ππ+1 + πΆ42 Β· π’π
π+1 + πΆ43 Β· π’ππ+1 + (πΆ44 + βπ‘πΎ44) Β· ππ
π+1 + (πΆ46 + βπ‘πΎ46) Β· πππ+1 = πΆ41 Β· π’π
π + πΆ42 Β· π’ππ +
πΆ43 Β· π’ππ + πΆ44 Β· ππ
π + πΆ46 Β· πππ + βπ‘
ππΏππ
2
π+1 , [2-112]
πΆ61 Β· π’ππ+1 + πΆ62 Β· π’π
π+1 + πΆ63 Β· π’ππ+1 + (πΆ64 + βπ‘πΎ64) Β· ππ
π+1 + (πΆ66 + βπ‘πΎ66) Β· πππ+1 = πΆ61 Β· π’π
π + πΆ62 Β· π’ππ +
πΆ63 Β· π’ππ + πΆ64 Β· ππ
π + πΆ66 Β· πππ + βπ‘
ππΏππ
2
π+1 . [2-113]
As can be seen in this equation, if the natural boundary conditions are prescribed and initial boundary conditions
are used, all the known values are positioned in the right hand side of the equation. The unknown values are in the
left hand side of the equation. The prescribed boundary conditions are given in the red box as described earlier,
and the initial boundary conditions can be implemented in the blue box. In the next time steps the previous found
vector is used to calculate the following step. The fully implicit method can then be used to solve the equation in
time. The same procedure can be executed for three elements.
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 31
Equation [2-101] for three elements looks like
2/00000
2/
2/00000
2/
0
0
0
00000000000000000000000000
0000000000000000000000000
0000000000000000000000000
000000000000
000000000
000000000
000000
000000000
000000
000000000
000000000
0
0
0
00000000000000000000000
00000000000000000000
00000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
6664
46446664
46446664
4644
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16143634131211333231
2624232221
16143634131211333231
2624232221
1614131211
6664636261
46446664434241636261
46446664434241636261
4644434241
o
i
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i
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m
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n
m
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k
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o
n
m
l
k
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i
PLq
PLq
PLt
PLt
u
u
u
u
u
u
u
KK
KKKK
KKKK
KK
KKKKK
KKKKK
KKKKKKKKKK
KKKKK
KKKKKKKKKK
KKKKK
KKKKK
u
u
u
u
u
u
u
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
. [2-114]
This equation needs then to be discretised in order to solve this system. If you use the discretised equation for three
elements, this will give
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 32
1
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4644434241
1
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2624232221
16143634131211333231
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46446664434241636261
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2/00000
2/
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0
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0
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00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0
0
0
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000000000000
000000000
000000000
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00000000000000
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00000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
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i
PLq
PLq
PLt
PLt
tu
u
u
u
u
u
u
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
u
u
u
u
u
u
u
KK
KKKK
KKKK
KK
KKKKK
KKKKK
KKKKKKKKKK
KKKKK
KKKKKKKKKK
KKKKK
KKKKK
t
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
. [2-115]
Again if the natural boundary conditions are prescribed and initial boundary conditions are given, all the known
values can be found in the right hand side of the equation. The unknown essential values are in the left hand side
of the equation. The prescribed boundary conditions are again given in red box, and the initial boundary conditions
can again be implemented in the blue box, which also gives the unknown vector after time step 1 . In the next time
steps the previous found essential vector is used to calculate the following step where the vector is used as input in
the blue part.
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 33
2.3.6.2. ESSENTIAL BOUNDARY CONDITIONS PRESCRIBED
If the general equation is discretised and the essential boundary conditions are prescribed, the known values can
be moved to the right hand side of the equation. For example if the displacement and the flux is given at the
boundary nodes, the general equation can be rewritten. The tractions will move to the left hand side of the equation
and the known displacements to the right hand side.
The discretised equation for one element is given with equation [2-109]
1
6664636261
4644434241
1
6664
4644
3634333231
2624232221
1614131211
6664636261
4644434241
2/0
2/
2/0
2/
0000000
0000000000000000000
0000000000
0000
0
0
0
0000000
0000000000000000000
n
k
i
k
i
n
k
j
i
k
j
i
n
k
j
i
k
j
i
PLq
PLq
PLt
PLt
tu
u
u
CCCCC
CCCCC
u
u
u
KK
KK
KKKKK
KKKKK
KKKKK
t
CCCCC
CCCCC
. [2-116]
Now the six equations will be given separately
βπ‘πΎ12 Β· π’ππ+1 + βπ‘πΎ14 Β· ππ
π+1 + βπ‘πΎ16 Β· πππ+1 + βπ‘
ππΏπ‘π
2
π+1= ββπ‘πΎ11 Β· π’π
π+1 β βπ‘πΎ13 Β· π’ππ+1 , [2-117]
βπ‘πΎ22 Β· π’ππ+1 + βπ‘πΎ24 Β· ππ
π+1 + βπ‘πΎ26 Β· πππ+1 = ββπ‘πΎ21 Β· π’π
π+1 β βπ‘πΎ23 Β· π’ππ+1 , [2-118]
βπ‘πΎ32 Β· π’ππ+1 + βπ‘πΎ34 Β· ππ
π+1 + βπ‘πΎ36 Β· πππ+1 β βπ‘
ππΏπ‘π
2
π+1= ββπ‘πΎ31 Β· π’π
π+1 β βπ‘πΎ33 Β· π’ππ+1 , [2-119]
πΆ42 Β· π’ππ+1 + (πΆ44 + βπ‘πΎ44) Β· ππ
π+1 + (πΆ46 + βπ‘πΎ46) Β· πππ+1 = βπΆ41 Β· π’π
π+1 β πΆ43 Β· π’ππ+1 + πΆ41 Β· π’π
π + πΆ42 Β· π’ππ +
πΆ43 Β· π’ππ + πΆ44 Β· ππ
π + πΆ46 Β· πππ + βπ‘
ππΏππ
2
π+1 , [2-120]
πΆ62 Β· π’ππ+1 + (πΆ64 + βπ‘πΎ64) Β· ππ
π+1 + (πΆ66 + βπ‘πΎ66) Β· πππ+1 = βπΆ61 Β· π’π
π+1 β πΆ63 Β· π’ππ+1 + πΆ61 Β· π’π
π + πΆ62 Β· π’ππ +
πΆ63 Β· π’ππ + πΆ64 Β· ππ
π + πΆ66 Β· πππ + βπ‘
ππΏππ
2
π+1 . [2-121]
The matrices can then be rewritten as
1
6664636261
4644434241
1
6664
4644
363432
262422
161412
6664636261
4644434241
0
0
2/00000000000002/000000330310002302100013011
0000000
0000000000000000000
0000000000
0000
02/0
000
002/
0000000
0000000000000000000
n
k
i
k
i
n
k
j
i
k
j
i
n
k
j
i
k
j
i
q
q
u
u
PL
PLKKKKKK
tu
u
u
CCCCC
CCCCC
t
u
t
KK
KK
KKPLK
KKK
KKKPL
t
CCCCC
CCCCC
. [2-122]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 34
For three elements the same procedure can be executed. If the displacements and fluxes are again given on the
boundary nodes, the discretised equation first looks like
1
6664636261
46446664434241636261
46446664434241636261
4644434241
1
6664
46446664
46446664
4644
3634333231
2624232221
16143634131211333231
2624232221
16143634131211333231
2624232221
1614131211
6664636261
46446664434241636261
46446664434241636261
4644434241
2/00000
2/
2/00000
2/
0
0
0
00000000000000000000000
00000000000000000000
00000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0
0
0
00000000000000000000000000
0000000000000000000000000
0000000000000000000000000
000000000000
000000000
000000000
000000
000000000
000000
000000000
000000000000000000
00000000000000
00000000000000000000
00000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
n
o
i
o
i
n
o
m
k
i
o
n
m
l
k
j
i
n
o
m
k
i
o
n
m
l
k
j
i
PLq
PLq
PLt
PLt
tu
u
u
u
u
u
u
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
u
u
u
u
u
u
u
KK
KKKK
KKKK
KK
KKKKK
KKKKK
KKKKKKKKKK
KKKKK
KKKKKKKKKK
KKKKK
KKKKK
t
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
. [2-123]
If we introduce a matrix οΏ½ΜοΏ½ in the general equation [2-101], where οΏ½ΜοΏ½ is the moved stiffness matrix the general
equation becomes
ποΏ½ΜοΏ½ + ππ = οΏ½ΜοΏ½π , [2-124]
where οΏ½ΜοΏ½ is the time derivative from the components π(π) and π½(π), and π are the components π(π) and π½(π). Solving
this equation can again be executed with the fully implicit method giving
(π + πβπ‘π)ππ+1 = (π β (1 β π)βπ‘π)ππ + βπ‘(ποΏ½ΜοΏ½ππ+1 + (1 β π)οΏ½ΜοΏ½ππ) , [2-125]
(π + βπ‘π)ππ+1 = πππ + βπ‘οΏ½ΜοΏ½ππ+1 , [2-126]
ππ+1 = (π + βπ‘π)β1(πππ + βπ‘οΏ½ΜοΏ½ππ+1) . [2-127]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 35
The tractions will move to the left hand side of the equation and the known displacements to the right hand side.
1
33
23
13
31
21
11
6664636261
46446664434241636261
46446664434241636261
4644434241
1
6664
46446664
46446664
4644
36343231
26242221
161436341211333231
2624232221
161436341312113332
26242322
16141312
6664636261
46446664434241636261
46446664434241636261
4644434241
00000
00000
2/000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002/0000000
0000000000000
0000000000000
000000000000000000000000000
0000000000000
0000000000000
0000000000000
0
0
0
00000000000000000000000
00000000000000000000
00000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0
0
0
00000000000000000000000000
0000000000000000000000000
0000000000000000000000000
000000000000
000002/0000
0000000000
0000000
000000000
0000000
0000000000
0000000002/000000000
00000000000000
00000000000000000000
00000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
n
o
i
o
i
n
o
m
k
i
o
n
m
l
k
j
i
n
o
m
k
i
o
n
m
l
k
j
i
q
q
u
u
PL
PL
K
K
K
K
K
K
t
u
u
u
u
u
u
u
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
t
u
u
u
u
u
t
KK
KKKK
KKKK
KK
KKPLKK
KKKK
KKKKKKKKK
KKKKK
KKKKKKKKK
KKKK
KKKKPL
t
CCCCC
CCCCCCCCCC
CCCCCCCCCC
CCCCC
. [2-128]
The same procedure can be used if the temperatures are prescribed instead of the fluxes.
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 36
2.3.6.3. MATHERMATICAL DESCRIPTION REWRITING BOUNDARY CONDITIONS
Boundary conditions can thus be given by either the natural or essential boundary conditions. If the natural
boundary conditions are given, no transformations are needed. The original equation can then be used, namely
ποΏ½ΜοΏ½ + ππ = π . [2-129]
However, if the essential boundary conditions are prescribed, these known values are transported to the right hand
side of the equation. The equation used for solving the system is then given with
ποΏ½ΜοΏ½ + ππ = οΏ½ΜοΏ½π , [2-130]
where οΏ½ΜοΏ½ is the moved stiffness matrix.
Moving the known values to the right hand side of the equation results in changed stiffness matrices. The capacity
matrix stays the same. If for example the displacements are known, they move to the right hand side of the equation
and the traction moves to the left hand side. This can be performed by multiplying the π Γ π identity matrix with
in the πβth and πβth row the one included for the prescribed displacements and the other rows are zero. Call this
matrix πππ. π is the number of nodes multiplied by two and minus the number of elements (since the zero rows
and columns for the temperatures disappear) and π and π are the columns where the displacement is known. The
desired vector will be the result of the matrix with vector multiplication ππππ = ππππ ππππ. The same procedure holds
for the force vector π where the tractions want to be excluded. Multiplying the π Γ π identity matrix with in the
πβth and πβth row the one included for the prescribed displacements and the other rows are zero. Call this matrix
π ππ. The desired vector will be the result of the matrix with vector multiplication π πππ = ππππ ππππ. Then the desired
vectors need to be added to the original vectors and replace the old values. Therefor the old values have to be firstly
deleted by multiplying the matrix with the π Γ π identity matrix where the πβth and πβth row are all zeroes. Call
this matrix πππ. The adjusted vector can be obtained by the multiplication ππππ = πππππ’π π‘ππ. The same procedure
can be used for the force vector, which means that the adjusted vector can be obtained by π πππ = πππππ’π π‘ππ. Then the
desired vectors have to be added to the adjusted vector from the other side. This means that the addition ππππ ππππ +
πππππ’π π‘ππ gives the correct force vector where the displacements appear and the addition ππππ ππππ. +πππππ’π π‘ππ gives
the correct π vector with the tractions in it. For example move the displacement term to the right hand side.
π =
[ π’π
π’π
π’π
ππ
ππππ ]
, πππ =
000000000000000000000100000000000001
, π =
[ π‘π0π‘πππ
0ππ]
, π ππ =
100000010000001000000000000010000000
, [2-131]
ππππ ππππ =
000000000000000000000100000000000001
[ π’π
π’π
π’π
ππ
ππππ ]
=
[ π’π
0π’π
000 ]
, πππππ’π π‘ππ =
100000010000001000000000000010000000
[ π‘π0π‘πππ
0ππ]
=
[ 000ππ
0ππ]
, [2-132]
ππππ ππππ + πππππ’π π‘ππ =
[ π’π
0π’π
000 ]
+
[ 000ππ
0ππ]
=
[ π’π
0π’π
ππ
0ππ ]
. [2-133]
The values belonging to these variables will transfer with them. The stiffness values of the known displacements
will move to the right hand side, which means that the column numbers of the nodes where the displacements are
known will be negative when transferred to the right hand side in the same location in the matrix. The values
belonging to the traction will be moved instead of the stiffness values but also with a negative sign. Mathematically
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 37
this can be performed by multiplying matrix π with a π Γ 1 vector where the πβth and πβth row containing a one
belonging to the position of the displacement. Call these vectors ππ and ππ. The desired vectors ππππ ππππ will contain
the columns with the values which will be transferred to the right hand side with a negative sign.
π =
6664
4644
3634333231
2624232221
1614131211
0000000000
0000
0
0
0
KK
KK
KKKKK
KKKKK
KKKKK
,ππ =
[ 100000]
, ππ =
[ 001000]
, [2-134]
ππππ ππππ =
6664
4644
3634333231
2624232221
1614131211
0000000000
0000
0
0
0
KK
KK
KKKKK
KKKKK
KKKKK
[ 100000]
=
[ πΎ11
πΎ21
πΎ31
000 ]
, [2-135]
ππππ ππππ =
6664
4644
3634333231
2624232221
1614131211
0000000000
0000
0
0
0
KK
KK
KKKKK
KKKKK
KKKKK
[ 001000]
=
[ πΎ13
πΎ23
πΎ33
000 ]
. [2-136]
In order to add the negative values of the matrix π to the matrix on the other side, this matrix should first make the
columns zero for which the vectors will be replaced later on. So if you know which columns should be zero, you
want to remove the πβth and πβth column from the matrix οΏ½ΜοΏ½. Write down the π Γ π identity matrix and replace the
πβth and πβth column with a column of all zeroes. Call this matrix πππ. The desired matrix will be the result of the
matrix multiplication οΏ½ΜοΏ½πππ. For example you want the third column to be zero of matrix οΏ½ΜοΏ½
οΏ½ΜοΏ½ =
2/00000000000002/0000002/00000000000002/
PL
PLPL
PL
, πππ =
100000010000001000000000000010000000
, [2-137]
2/00000000000002/0000002/00000000000002/
PL
PLPL
PL
100000010000001000000000000010000000
=
2/00000000000002/000000000000000000000
PL
PL . [2-138]
Multiply the vectors you want to add with minus one:
ππππ ππππ = β1 β
[ πΎ11
πΎ21
πΎ31
000 ]
=
[ βπΎ11
βπΎ21
βπΎ31
000 ]
, [2-139]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 38
ππππ ππππ = β1 β
[ πΎ13
πΎ23
πΎ33
000 ]
=
[ βπΎ13
βπΎ23
βπΎ33
000 ]
. [2-140]
If you want to add this vector into matrix οΏ½ΜοΏ½ instead of the zero column, two steps have to be executed. The first
step is to multiply the vector you want to move with the transpose of a column vector π Γ 1 with a one on the πβth
row where the zero column is positioned in the matrix οΏ½ΜοΏ½. So
[ βπΎ11
βπΎ21
βπΎ31
000 ]
[1 0 0 0 0 0] =
000000000000000000
00000
00000
00000
31
21
11
K
K
K
[2-141]
This matrix can then be added to the matrix οΏ½ΜοΏ½
000000000000000000
00000
00000
00000
31
21
11
K
K
K
+
2/00000000000002/000000000000000000000
PL
PL =
2/00000000000002/000
00000
00000
00000
31
21
11
PL
PL
K
K
K
[2-142]
And
[ βπΎ13
βπΎ23
βπΎ33
000 ]
[0 0 1 0 0 0] =
000000000000000000
00000
00000
00000
33
23
13
K
K
K
[2-143]
000000000000000000
00000
00000
00000
33
23
13
K
K
K
+
2/00000000000002/000
00000
00000
00000
31
21
11
PL
PL
K
K
K
=
2/00000000000002/000000330310002302100013011
PL
PLKKKKKK
[2-144]
The same procedure can be executed if the temperatures are known.
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 39
2.4. LINEAR BUCKLING
2.4.1. INTRODUCTION
The stability in a bar or beam can be described with an initial situation of the structure. These bars and beams are
connected through nodes grids, where forces and moments are introduced into each single element. The element
reacts linear elastic as long as the loads applied on the element are lower than the critical load. However, if the load
is larger than the critical value, buckling occurs. So if the centrally applied compressive force is smaller than its
critical value, the compressed element remains straight. This straight form of equilibrium is stable. If the load is
slightly increased above its critical value, two theoretical forms of equilibrium can occur. On one hand the element
remains straight and on the other hand the element buckles sideways. In case of a compressed element it can be
said that the column may be stable or unstable depending on the magnitude of the axial load. The concept of the
stability of various forms of equilibrium can be illustrated by considering the equilibrium of a ball in several
positions (Timoshenko, 1961) .
In Figure 2-12 it can be seen that the ball is in equilibrium in each state shown, however, important differences
among the three cases can be distinguished. If the ball in state a) is moved slightly from its original position of
equilibrium, it will return to that position after removing he disturbing force. This is called stable equilibrium. The
ball in state b) does not return to its original equilibrium position if it is disturbed slightly from its position of
equilibrium. This equilibrium is called unstable equilibrium. The ball in state c) neither returns to its original
position nor continues to move away after removing the disturbing force. This is called neutral equilibrium. These
equilibrium states can also be explained by the potential energy of the ball. For the ball in state a), a certain amount
of work is required to move the ball. Therefore, the potential energy of this situation is increased. In state b), any
slight displacement from the position of equilibrium will decrease the potential energy of the ball. Consequently,
in the case of stable equilibrium the energy of the system is a minimum, and in case of unstable equilibrium the
energy of the system is a maximum (Yoo & Lee, 2005).
2.4.2. RITZ APPROXIMATION FEM STABILITY
First a simple buckling problem is elaborated by using two different methods for the Finite Element Method. The
Ritz method assumes a single polynomial solution for spanning the entire structure. The FEM uses several
polynomial solutions which results in reflection a piece-wise polynomial interpolation, since each solution is
defined over only a portion of the structure. Giving a column with length l the stiffness matrices can be found
(Suiker, 2014b).
2.4.2.1. ONE 3RD-ORDER ELEMENT
Using a third-order polynomial function, the displacement field in the element is approximated with
οΏ½ΜοΏ½ (1)(π₯) = πΆ1 + πΆ2π₯ + πΆ3π₯2 + πΆ4π₯
3 for 0 β€ π₯ β€ π . [2-145]
Figure 2-12. Stability of equilibrium (Yoo & Lee, 2005)
Figure 2-13. Simply supported one-dimensional element
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 40
The corresponding rotation field can be obtained by taking the spatial derivative of the displacement field
οΏ½ΜοΏ½(1)(π₯) = βποΏ½ΜοΏ½(1) (π₯)
ππ₯= βπΆ2 β 2πΆ3π₯ β 3πΆ4π₯
2 for 0 β€ π₯ β€ π . [2-146]
These relations can be rewritten in matrix-vector format
[οΏ½ΜοΏ½(1)(π₯)
οΏ½ΜοΏ½ (1)(π₯)] = [1 π₯
0 β1 π₯2 π₯3
β2π₯ β3π₯2] [
πΆ1
πΆ2
πΆ3
πΆ4
] for 0 β€ π₯ β€ π . [2-147]
The 4 unknowns πΆπ of each element need to be replaced by 4 unknown displacements and rotations in the element
nodes in order to formulate a beam element. This means that for each node at the two element ends a displacement
ππ and a rotation ππ is needed. These nodal displacements and rotations represent the generalised coordinates, often
also called degrees of freedom. For the element depicted in Figure 2-13, the nodal displacements and rotations need
to match the corresponding displacement and rotation field. This gives
π1 = οΏ½ΜοΏ½(1)(0) = πΆ1, [2-148]
π1 = οΏ½ΜοΏ½(1)(0) = βπΆ2, [2-149]
π2 = οΏ½ΜοΏ½ (1) (π
2) = πΆ1 + πΆ2π + πΆ3π
2 + πΆ4π3 , [2-150]
π2 = οΏ½ΜοΏ½(1) (π
2) = βπΆ2 β 2πΆ3π β 3πΆ4π
2 . [2-151]
This can again be rewritten in matrix-vector form
[
π1
π1π2
π2
] =
2
32
32101
00100001
lllll
[
πΆ1
πΆ2
πΆ3
πΆ4
] . [2-152]
Inverting these relations results in
[
πΆ1
πΆ2
πΆ3
πΆ4
] =
2323
22
1212
132300100001
llll
llll[
π1
π1π2
π2
] . [2-153]
Substituting equations [2-153] back into [2-147] gives the relations between the element displacement fields and
the displacement and rotation in the element nodes
οΏ½ΜοΏ½ (1)(π₯) = [1 β3π₯2
π2+
2π₯3
π3, βπ₯ +
2π₯2
πβ
π₯3
π2,3π₯2
π2β
2π₯3
π3,π₯2
πβ
π₯3
π2] [
π1
π1π2
π2
] for 0 β€ π₯ β€ π . [2-154]
This relation can be formulated as
οΏ½ΜοΏ½ (π)(π₯) = π(m)π(m) . [2-155]
The relation between th element rotation field and the displacement and rotations can be formulated as
οΏ½ΜοΏ½(π)(π₯) = βποΏ½ΜοΏ½(π) (π₯)
ππ₯= π(m)π(m) , [2-156]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 41
οΏ½ΜοΏ½(1)(π₯) = [6π₯
π2β
6π₯2
π3, 1 β
4π₯
π+
3π₯2
π2, β
6π₯
π2+
6π₯2
π3, β
2π₯
π+
3π₯2
π2] [
π1
π1π2
π2
] for 0 β€ π₯ β€ π . [2-157]
The element shape functions of this method can be found in π(m), which thus
π(m) = [1 β3π₯2
π2+
2π₯3
π3, βπ₯ +
2π₯2
πβ
π₯3
π2,3π₯2
π2β
2π₯3
π3,π₯2
πβ
βπ₯3
π2] . [2-158]
The relations for the rotation field are expressed with π(m), which is the first derivative of the shape functions
multiplied with minus one. This reads thus
π(m) = βπ
ππ₯π(m) = [
6π₯
π2β
6π₯2
π3, 1 β
4π₯
π+
3π₯2
π2,β
6π₯
π2+
6π₯2
π3,β
2π₯
π+
3π₯2
π2] . [2-159]
Now also the curvature field in an element can be computed with
οΏ½ΜοΏ½ (π)(π₯) = βπ2οΏ½ΜοΏ½(π) (π₯)
ππ₯2= π(m)π(m) , [2-160]
where π(m) can be given with
π(m) = βπ2
ππ₯2π(m) = [
6
π2β
12π₯
π3, β
4
π+
6π₯
π2, β
6
π2+
12π₯
π3,β
2
π+
6π₯
π2] . [2-161]
By substituting the displacement derivatives from [2-160] and [2-156] into the usual expression for the total
potential energy, the total potential energy for the FEM model can be found. The expression for the potential energy
for the example from Figure 2-13 is given with
π = β«1
2πΈπΌ (
π2π€
ππ₯2)2
ππ₯ + β« π01
2(ππ€
ππ₯)
2
ππ₯π
0
π
0 [2-162]
Using the axial equilibrium condition π0 = βπΉ and πΉ = Ξ»οΏ½ΜοΏ½, with Ξ» as an arbitrary multiplier and οΏ½ΜοΏ½ = 1,0 as a
reference magnitude. Using οΏ½ΜοΏ½(π)(π₯) = βπ2οΏ½ΜοΏ½ (π) (π₯)
ππ₯2 and οΏ½ΜοΏ½ (π)(π₯) = β
ποΏ½ΜοΏ½(π) (π₯)
ππ₯ the total potential energy can be written
as
οΏ½ΜοΏ½ = οΏ½ΜοΏ½(π) =1
2β πTπ(m)π β Ξ»
1
2β πTππ
(π¦)π π
π=1 ππ=1 . [2-163]
In this equation, the total nodal displacement and rotation vector gives
πT = [π1 π1 π2 π2 π3 π3] , [2-164]
and the element stiffness matrices are formulated by
π(m) = β« π(m)TπΈπΌπ(m) ππ₯β(π) , [2-165]
ππ(π¦)
= β« π(m) TοΏ½ΜοΏ½π(m) ππ₯β(π) , [2-166]
where π(m) is the conventional stiffness matrix and ππ(π¦)
is the stress stiffness matrix. According to Appendix C the
conventional stiffness matrix can be computed as
π(m) =πΈπΌ
π3
22
22
4626
6126122646
612612
llll
llllll
ll
. [2-167]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 42
The stress stiffness matrix is computed as found in Appendix C
ππ(π¦)
=οΏ½ΜοΏ½
30π
22
22
433336336
343336336
llllllllllll
. [2-168]
It can be noted that both the element stiffness matrices [2-167] and [2-168] are symmetric. Equilibrium for this
problem is described by stationarity of the potential energy from equation [2-163]
πΏοΏ½ΜοΏ½ =ππ
ππ β πΏπ = 0
ππ
ππ= 0 β (π
π=1 π(m) β Ξ»ππ(π¦)
)π = 0 . [2-169]
The global stiffness matrix for one element can now be straightforwardly assembled from the element stiffness
matrices by using equation [2-169]
. [2-170]
This element stiffness matrix can be used to assemble more than one element. Then the essential boundary
conditions can be applied, e.g. π1 = 0 and π2 = 0. The first and third columns and rows of the stiffness matrix
disappear. This gives
. [2-171]
The buckling loads πΉ = Ξ»οΏ½ΜοΏ½ can be found by solving the above set of equations for Ξ». οΏ½ΜοΏ½ = 1,0 is used as a reference
intensity. Two eigenvalues Ξ» are found by equating the determinant of the reduced stiffness matrix to zero. This
results in
Ξ»1 = F1 = Fππ = 12πΈπΌ
π2 , Ξ»2 = F2 = 60
πΈπΌ
π2 . [2-172]
The exact solution for this buckling phenomenon is given by
Fπ =π2π2πΈπΌ
π2 , with n=1,2,3,β¦ [2-173]
The exact buckling loads are then
F1 = Fππ = 9,8696πΈπΌ
π2 , F2 = 39,4784
πΈπΌ
π2 . [2-174]
It can be noted that the relative error of the FEM buckling loads are 22% and 52%. These errors will reduce when
the number of elements is increased. This is called h-refinement. The FEM solution also converges to the exact
solution when the order of the polynomial shape function is increased. This is called p-refinement.
0000
30
~44
30
~36
30
~2
30
~36
30
~36
30
~3612
30
~36
30
~3612
30
~2
30
~36
30
~44
30
~36
30
~36
30
~3612
30
~36
30
~3612
2
2
1
1
22
2323
22
2323
r
d
r
d
lF
l
EIF
l
EIlF
l
EIF
l
EI
F
l
EI
l
F
l
EIF
l
EI
l
F
l
EI
lF
l
EIF
l
EIlF
l
EIF
l
EI
F
l
EI
l
F
l
EIF
l
EI
l
F
l
EI
00
30
~44
30
~2
30
~2
30
~44
2
1
r
r
lF
l
EIlF
l
EI
lF
l
EIlF
l
EI
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 43
2.4.3. GALERKIN APPROXIMATION FEM STABILITY
The approximate displacement οΏ½ΜοΏ½(π₯) may also be directly substituted in the variational equilibrium equation. The
potential energy for the example in Figure 2-13 is described with
π = β«1
2πΈπΌ (
π2π€
ππ₯2)2
ππ₯ + β« π01
2(ππ€
ππ₯)
2
ππ₯π
0
π
0 , [2-175]
The equilibrium conditions can be found by requiring the potential energy to become stationary. This means that
the variation of the potential energy needs to be zero. This gives
πΏπ = β« πΈπΌπ€β²β²πΏπ€β²β²ππ₯ + β« π0π€β²πΏπ€β²ππ₯ = 0
π
0
π
0 , [2-176]
where π€β²β² =π2π€
ππ₯2 , π€β² =
ππ€
ππ₯. [2-177]
Using integration by parts gives the final equation where three terms are required to be zero in order to satisfy the
variation of the potential energy to be zero
πΏπ = β« {((πΈπΌπ€β²β²)β²β²β (π0π€β²)β²)πΏπ€}ππ₯ + (πΈπΌπ€β²β²)πΏπ€β²|0
π β ((πΈπΌπ€β²β²)β² β π0π€β²)
π
0 πΏπ€|0π = 0 . [2-178]
In this equation the buckling equation can be seen and also the combined natural and essential boundary
conditions. All terms need to be equal to zero separately. Substituting the displacement function οΏ½ΜοΏ½(π₯) into this
equation, the critical buckling loads (eigenvalues) can be found. The Galerkin approximation leads to the same set
of FEM equations as obtained with the Ritz method. However, the Galerkin approximation includes the satisfaction
of the natural boundary conditions. This means that both the essential and natural boundary conditions need to be
satisfied.
The two natural boundary conditions for the two-element model of the simply-supported column can be given by
π(0) = πΈπΌπ (0) = 0 οΏ½ΜοΏ½ (1)(0) = βπ2οΏ½ΜοΏ½ (1)(0)
ππ₯2= π(1)(0)π(1) = 0 , [2-179]
π(π) = πΈπΌπ (π) = 0 οΏ½ΜοΏ½ (2)(π) = βπ2οΏ½ΜοΏ½ (2)(π)
ππ₯2= π(2)(π)π(2) = 0 , [2-180]
Using the equation [2-161] for π(m)(π₯), the two natural boundary conditions can be elaborated as
π1(π) =24
π2π1 β
8
ππ1 β
24
π2π2 β
4
ππ2 = 0 , [2-181]
π2(π) =β24
π2π2 +
4
ππ2 +
24
π2π3 +
8
ππ3 = 0 . [2-182]
Combining the potential energy derived from the Ritz method with the natural boundary conditions gives the
corresponding equilibrium solution. This combination can be established by defining an auxiliary potential Ξ¦β.
This potential can be described with
Ξ¦β(π,π¬1 , π¬2) = οΏ½ΜοΏ½(π) + π¬1π1(π) + π¬2π2(π) , [2-183]
where [π¬1, π¬2] are the Lagrange multipliers. This equation can be rewritten as
Ξ¦β(π,π²) = οΏ½ΜοΏ½(π) + π² β π(π) , [2-184]
where π²T is the vector with Lagrange multipliers, and π(π)T is the vector containing the natural boundary
conditions. For the equilibrium condition the stationarity of the auxiliary potential can be given with
δΦβ = (ππ
ππ+
π(π²βπ(π))
ππ)β πΏπ +
π(π²βπ(π))
ππ²β πΏπ² = 0 . [2-185]
For any variations πΏπ and πΏπ² this gives the following set of coupled equations
ππ
ππ+
π(π²βπ(π))
ππ= 0 β (π
π=1 π(m) β Ξ»ππ(π¦)
)π +π(π²βπ(π))
ππ= 0 , [2-186]
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R.H.A. TITULAER 44
π(π²βπ(π))
ππ²= 0 π(π) = 0 . [2-187]
The first part of equation [2-186] gives the stiffness response of the structure, which was already derived in the Ritz
method, and the coupling between the natural boundary conditions and the response of the structure. The second
part of the equation characterises the satisfaction of the two natural boundary conditions.
An example can be given for a two-element model of a simply-supported beam shown in Figure 2-14. This example
results in 8 homogeneous, linear algebraic equations by substituting all the values into equation [2-186]. The result
is the following matrix-vector form
00000000
00824424
00
0000424824
80
15
~8
10
~24
60
~4
10
~24
00
240
10
~24
5
~1296
10
~24
5
~1296
00
44
60
~4
10
~24
15
~216
060
~4
10
~24
2424
10
~24
5
~1296
05
~24192
10
~24
5
~1296
08
0060
~4
10
~24
15
~8
10
~24
024
0010
~24
5
~1296
10
~24
5
~1296
1
1
3
3
2
2
1
1
22
22
22
22323
22
2223323
22
22323
r
d
r
d
r
d
llll
llll
l
lF
l
EIF
l
EIlF
l
EIF
l
EIl
F
l
EI
l
F
l
EIF
l
EI
l
F
l
EIll
lF
l
EIF
l
EIlF
l
EIlF
l
EIF
l
EIll
F
l
EI
l
F
l
EI
l
F
l
EIF
l
EI
l
F
l
EIl
lF
l
EIF
l
EIlF
l
EIF
l
EIl
F
l
EI
l
F
l
EIF
l
EI
l
F
l
EI
[2-188]
This global stiffness matrix is again symmetric and by applying the 2 essential boundary conditions, the eigenvalues
can be calculated, since the first and fifth columns and rows vanish. By equating the determinant of the reduced
stiffness matrix to zero, the eigenvalues are found. Now the critical buckling load is overestimated by 1,3%. This is
slightly larger than for the Ritz method, however now the natural and essential boundary conditions are satisfied.
The buckling modes (eigenmodes) are found when inserting the eigenvalues back into the original equations.
Figure 2-14. Simply supported one-dimensional element divided into two elements
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R.H.A. TITULAER 45
2.5. NON-LINEAR BUCKLING
2.5.1. INTRODUCTION
The displacements in a linear analysis are assumed to be linearly dependent on the applied loads and the behaviour
of the structure is assumed to be completely reversible. However, many structural applications consist of a load
history and thus the behaviour may depend on this history. Also after the elastic limit, large deformation could
occur. Linear analysis can therefore be summarized by the fact that all loads can be applied instantaneously and
the loading history is irrelevant, which means that the displacements are linearly dependent on the loads. In a non-
linear analysis the loading history is usually important and the material properties and loads must be specified.
The non-linear FEM is used to solve a large system of non-linear equations.
2.5.1.1. NON-LINEAR CLASSIFICATIONS
In structural analysis usually three main types of non-linear classifications are used: a) Material non-linearity, b)
Geometric non-linearity, and c) Boundary non-linearity. These classifications are now explained.
2.5.1.2. MATERIAL NON-LINEARITY
The constitutive stress-strain relationship is an example of non-linearity. Material non-linearities are normally
further classified into three different categories:
I) Time-independent behaviour such as the elastic-plastic behaviour of metals in which the structure is
loaded past the yield point.
II) Time-dependent behaviour such as creep of metals at high temperatures in which the effect of
variation of stress/strain with time is of interest and a power law stress-strain relationship is often
used.
III) Viscoelastic/viscoplastic behaviour in which both the effects of plasticity and creep are presented.
An example of elastic-plastic behaviour is shown in the following figure where a uniaxial test specimen is loaded
under tension. Linear assumption can be used till the yield point is reached. However, after the yield point a
significant inaccuracy will occur if linear theory is applied, since the stress-strain relationship becomes non-linear.
2.5.1.3. GEOMETRIC NON-LINEARITY
When changes in the geometry of a structure are taken into account in analysing its behaviour geometric non -
linearity occurs. A change in geometry affects both equilibrium and kinematic relationships. In linear analysis the
equilibrium equations are always based on the original (undeformed) shape, whereas in geometric non-linearity,
the equilibrium equations take into account the deformed shape. Subsequently the strain-displacement
relationships may have to be redefined in order to take into account the current (updated) deformed shape. In
geometric non-linearity it should be emphasised that large and small displacements, strains, or rotations are
possible.
Figure 2-16 shows two examples of geometric non-linearity. In a) an example of a cantilever beam under a
transverse load is given. The linear assumption of this example at large loads is that the displacement is
proportional to the loads. However, in real-life applications, the cantilever begins to stiffen after a certain tip
displacement. The linear assumption is therefore inadequate. Example b) shows a two-bar system in which the
structure softens, which is an example of geometric non-linearity. The vertical displacement increases up to a limit
point after which the load has to drop to maintain equilibrium.
Figure 2-15. Example of material non-linearity problem (Becker, 2001)
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2.5.1.4. BOUNDARY NON-LINEARITY
The last main type of non-linear classification is the boundary non-linearity, which occurs in most contact problems.
The displacements and stresses of the contacting bodies are usually not linearly dependent on the applied loads.
Even if the material behaviour is assumed to be linear and the displacements are infinitesimally small, this type of
non-linearity may occur.
In the following figure two examples of boundary non-linearity are shown. Example a) represents a typical contact
problem of a cylindrical roller on a flat plane. The contact starts at a single point and then gradually spreads as the
load is increased. The increase in the contact area and the change in the contact pressure are not linearly
proportional to the applied load. The other example b) shows the non-linear contact behaviour of a cantilever
coming in contact with a rigid surface. As can be seen in the graph, at a certain point the load is still increasing but
the displacement of the tip remains at the same position. This can be described as the non-linear behaviour. When
no-contact is made, the problem becomes a geometric non-linearity problem. A combination of the above non-
linearities may occur in actual structural problems, for example material and geometric non-linearities.
Figure 2-16. Examples of geometric non-linearity problems (Becker, 2001)
Figure 2-17. Examples of boundary non-linearity problems (Becker, 2001)
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2.5.2. NON-LINEAR FINITE ELEMENT PROCEDURES
The main strategy of non-linear Finite Element formulations is to break down the loading history into a series of
simpler piecewise-linear or weakly non-linear steps. Commercial FE codes use a combination of load increments
and iteration procedures to find the final solution. The reliability of the FE solutions are significantly influenced by
the solution procedures. Linear elastic problems contain linear algebraic equations which can be solved by
commercial software such as Matlab. These equations can be written as
ππ = π , [2-189]
where π is the stiffness matrix, π is the displacement vector, and π is the vector of applied external forces.
2.5.2.1. NEWTON-RAPHSON ITERATIVE METHOD
One effective numerical method used for solving non-linear equations is the Newton-Raphson iterative method.
This method guesses a trial solution and then gradually improves the βguessβ by using the slope of the load-
displacement curve. This curve is consequently approximated by a series of appropriate tangents. The following
figure schematically represents the Newton-Raphson method. Two conditions need to be fulfilled, which is that the
initial guess should not be very far from the exact solution and the slope of the non-linear load-displacement curve
does not change its sign. If these conditions are not met, the solution may not converge.
The Newton-Raphson method has a high computation time if the slope is updated after each iteration, since this
method needs to calculate the slope of the load-displacement curve, i.e. the stiffness matrix π. However, the slope
can also be kept constant during the iterations, which usually results in a slower convergence, but at a lower cost
per iteration. If the slope is kept constant for sequential iterations, this is generally called the modified Newton-
Raphson method. In general the set of algebraic non-linear equations are indicated as (Zienkiewicz & Taylor, 2000b)
π(π) = π β π(π) = 0 , [2-190]
where π is the set of discretisation parameters, π a vector which is independent of the parameters and π a vector
dependent on the parameters. This set of equations may have multiple solutions. In order to obtain realistic answers
small-step incremental approaches should be used. The general solution should always be obtained by using
ππ+1 = π(ππ+1) = ππ+1 β π(ππ+1) = 0 , [2-191]
which starts from a nearby solution at
π = ππ, ππ = 0, π = ππ . [2-192]
The forcing function can be written as
ππ+1 = ππ + βππ . [2-193]
Figure 2-18. Schematic representation of the Newton-Raphson iterative method (Becker, 2001)
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The Newton-Raphson method is the process in which the convergence is quadratic. This method is the most rapid
process for non-linear problems. In this iterative method, Eq. [2-191] can be approximated as
π(ππ+1π+1 ) β π(ππ+1
π )+ (πΟ
πa)π+1
π
ππππ = 0 . [2-194]
The iteration of this equation starts by assuming
ππ+1π = ππ , [2-195]
in which ππ is a converged solution at a previous load level or time step. Following from Eq. [2-194] the iterative
correction can be written as
πTπ πππ
π = ππ+1π , [2-196]
in which πTπ is the tangent direction given by
πT =ππ
ππ= β
ππ
ππ . [2-197]
So the set of discretization parameters can be obtained by
ππππ = (πT
π )β1
ππ+1π . [2-198]
Successive approximations can be obtained by
ππ+1π+1 = ππ+1
π + πππ+1π
= ππ + βπππ . [2-199]
2.5.2.2. LOAD INCREMENTATION PROCEDURE
The load incrementation procedure divides the total applied load into small increments and each increment is
applied individually. The material behaviour may be assumed linear during the load increment giving that the
increments are small. In each increment a new stiffness matrix π can be used. Therefore the load history is treated
as piecewise-linear. In Figure 2-19 a schematic representation of the load incrementation procedure can be seen. It
shows that the solution tends to drift away from the exact solution. Consequently, iterations should be performed
within each load increment to ensure that the solution remains below a specified tolerance. In some cases
displacement control should be used where the displacement of a specified node or a set of nodes is limited to a
small values. The load path is then followed correctly.
Figure 2-19. Schematic representation of the load incrementation procedure (Becker, 2001)
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2.5.2.3. ARC-LENGTH METHOD
During a structural stability analysis snap-through and snap-back buckling can occur. Since these behaviours
reduce the force in the force-displacement relation, the Newton-Raphson method can never converge at a certain
force increment. In order to be able to trace the force-displacement relation correctly, the arc-length method can be
used. This method was originally developed by Riks (1972). The essential idea of the arc-length method or the path-
following method is that the loading factor Ξ» is considered as an additional unknown. The equilibrium equation of
the non-linear system can be written as
π(ππ+1, Ξ») = (Ξ»π + πΞ»)π0 β π(ππ+1) = 0 , [2-200]
where π is the set of discretisation parameters, π a vector which is independent of the parameters and π a vector
dependent on the parameters. Ξ»π+1 is the loading factor, such that ππ+1 = Ξ»π+1 π0 = (Ξ»π + πΞ»)π0, with π0 as the
normalised external load vector (Suiker, Askes, & Sluys, 2000). In order to complete the definition of the problem
an additional equation is introduced, depending on the solution strategy adopted (Suiker et al., 2000)
π(βπ,βΞ») = 0 . [2-201]
Now Eq. [2-200] can be used to formulate the set of discretization parameters
ππππ = (πT
π )βπ
ππ+1π , [2-202]
ππππ = (πT
π )βπ
{(Ξ»π + πΞ»)ππ β π(ππ+π)} , [2-203]
ππππ = {(πT
π )βπ
(Ξ»πππ β π(ππ+π))} + πΞ»{(πTπ )
βπππ} , [2-204]
ππππ = ππ°ππ
π + πΞ»ππ°π°πππ , [2-205]
where Ξ»ππ0 is the vector of external forces π. The path-following constraint π is suggested by de Borst et al. (2012a)
which is the spherical arc-length constraint
π(βπ,βΞ») = (βππ)Tβππ + π½2(βΞ»π)2ππTππ β βπ2 = 0 , [2-206]
where
βππ = βππβ1 + ππ , [2-207]
βΞ»π = βΞ»πβ1 + πΞ» , [2-208]
where βππ and βΞ»π are the sought increments between the next state π and the last known equilibrium state π and
where βππβ1 and βΞ»πβ1 are the known increments between the current state π β 1 and the last known equilibrium
state π (Danielsson, 2013). The value π½ as a user-specified value that weighs the importance of the contributions of
the displacement degrees of freedom and the load increment. The different magnitudes of the displacement
increments π πππ and of the load (Ξ»π + πΞ»)π0 should be properly balanced. The last term in this equation βπ is the
path step length. Setting this equation to zero, the following solutions can be obtained
π1πΞ»2 + π2πΞ» + π3 = 0 , [2-209]
Figure 2-20. Path-following technique (de Borst et al., 2012)
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where
π1 = ππ°π°πTππ°π°π+ π½2π0Tπ0 , [2-210]
π2 = 2ππ°π°πT(βππβ1 + ππ°π)+ 2π½2βΞ»πβ1π0Tπ0 , [2-211]
π3 = (βππβ1 + ππ°π)T(βππβ1 + ππ°π)+ π½2(βΞ»πβ1)2π0
Tπ0 β βπ2 . [2-212]
Usually the root of the above quadratic equation is chosen, since this results in the incremental displacement vector
that consists of the same direction as that which was obtained in the previous loading step βπππ so that
(βππ)Tβππβ1 > 0 . [2-213]
However, this method does not always work well, especially with snap-back behaviour and at strongly curved
parts of the equilibrium path. It is possible that two imaginary roots are computed. To prevent this non-physical
result the increment size can be decreased. An alternative formulation was given by Ramm (1981), which linearizes
Eq. [2-206]
π(βπ,βΞ») = (βππ)Tβππ + π½2βΞ»πβ1βΞ»ππ0Tπ0 β βπ2 = 0 . [2-214]
This linearization results in the updated normal path method. This can be seen in the following figure.
The increment of the loading factor can then be determined with
πΞ» =βπ2β(βππβ1)Tβππβ1β(βππβ1)Tππ°πβπ½2(βΞ»πβ1)2ππ
Tππ
(βππβ1)Tππ°π°π+π½2βΞ»πβ1ππTππ
. [2-215]
Noting that βπ2 β (βππβ1)Tβππβ1, the previous equation can be approximated by
πΞ» = β(βππβ1)Tππ°π+π½2(βΞ»πβ1)2ππ
Tππ
(βππβ1)Tππ°π°π+π½2βΞ»πβ1ππTππ
. [2-216]
This equation can also be simplified by keeping the direction of the tangent normal to the hyperplane constant,
which results in
π(βπ,βΞ») = (βπ1)Tβππ + π½2βΞ»1βΞ»πππTππ β βπ2 = 0 . [2-217]
This normal path method is very simple and results in the following increment of the loading factor
πΞ» = β(βππβ1)Tππ°π
(βππβ1)Tππ°π°π+π½2βΞ»1ππTππ
. [2-218]
This equation resembles the constraint equation originally introduced by Riks (1972). The following figure describes
the normal path method. Using π½ = 0 seems to work for many engineering cases, so this is used in this analysis.
πΞ» = β(βππβ1)Tππ°π
(βππβ1)Tππ°π°π . [2-219]
Figure 2-21. Updated hyperplanes for the geometrical interpretation of alternative constraint equation g (de Borst et al., 2012)
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R.H.A. TITULAER 51
2.5.3. THERMAL NON-LINEARITY
Eq. [2-104] can also be rewritten for οΏ½ΜοΏ½π+1 to incorporate the Newton-Raphson method for non-linear calculations.
For the first-order equation only the value of ππ needs to be specified as an initial value for any computation. The
SS11 algorithm (Zienkiewicz & Taylor, 2000a) is most often used for this calculation. The approximation leads to
π = ππ + βπ‘οΏ½ΜοΏ½ . [2-220]
The approximation to the sets of ordinary differential equations is then
ποΏ½ΜοΏ½ + π(ππ + πΎβπ‘οΏ½ΜοΏ½) = π . [2-221]
Rewriting this equation gives a solution for οΏ½ΜοΏ½ as
(π + πΎβπ‘π)οΏ½ΜοΏ½ = (π β πππ) . [2-222]
This equation can easily be used to formulate the non-linear problem. Non-linear transient problems can be solved
by using a discrete approximation in time in order to formulate the set of algebraic equations (Zienkiewicz & Taylor,
2000b). For each discrete time π‘π+1 the equilibrium equation may be written in a residual form as
π(οΏ½ΜοΏ½π+1) = π β π(ππ + πΎβπ‘οΏ½ΜοΏ½π+1) β ποΏ½ΜοΏ½π+1 = 0 , [2-223]
where π is the vector of non-linear internal forces. This set of equations may have multiple solutions. In order to
obtain realistic answers small-step incremental approaches should be used. Using the formulation as used in Eq.
[2-223] the iterative correction for the non-linear transient problem can be written as
(π + πβπ‘πTπ )βοΏ½ΜοΏ½π+1
π = π(οΏ½ΜοΏ½π+1π ) . [2-224]
Eq. [2-224] can now be solved and this gives the successive approximations as
οΏ½ΜοΏ½π+1π+1 = οΏ½ΜοΏ½π+1
(π)+ πβοΏ½ΜοΏ½π+1
(π) , [2-225]
where π is the step size for non-linear problems as described in by Zienkiewicz & Taylor (2000b, Chapter 13). After
each iteration the residue is checked for convergence by using the Euclidean norm. If convergence is obtained, the
iteration process is terminated and the next time step is calculated. If convergence is not obtained, a new iteration
is performed. The element matrices including transverse displacement can be seen in Appendix D.
Figure 2-22. Fixed hyperplane for the geometrical interpretation of alternative constraint equation g (de Borst et al., 2012)
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3. RESULTS AND DISCUSSION
Since now the general theory of the fire development, heat transfer mechanisms, coupled thermomechanical analysis, linear
buckling, and non-linear buckling are explained, several analyses can be performed. In the first paragraph a simple one-
dimensional element is subjected to a sudden temperature increase at one boundary. This analysis gives temperatures and
displacements in x direction and these results are then verified. Next the linear buckling analysis is performed. Firstly on purely
mechanical problems for both the Ritz approximation and Galerkin approximation, followed by the implementation in Matlab
now also taken into account thermal and thermomechanical problems, and then the verification in ABAQUS. After linear
analyses most often non-linear analyses can be performed. This is described in the third paragraph, in which a simple non-
linear mechanical problem is extensively studied followed by the thermal and thermomechanical implementation. Finally, this
chapter ends with a comparison with the Eurocode.
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3.1. COUPLED THERMOMECHANICAL ANALYSIS
The coupled thermomechanical behaviour is then programmed into Matlab. The accuracy of the code is checked
for several errors from top to bottom. The next step is to investigate if the parameters were all correctly entered into
the code. Then some literature is reviewed in order to check the equation used for the calculation of the
displacements and temperatures and the coupling coefficient. Since the code solves the temperatures and
displacements simultaneously, an analysis is performed on a simple problem.
3.1.1. SUDDEN BOUNDARY TEMPERATURE INCREASE MATLAB CODE
The first analysis is performed on a beam with a length of 3 meters. The density is 7850 kg/m3. The Youngβs
modulus is taken as 2,1Γ108 kN/m2 and the Poissonβs ratio is 0,3. The expansion coefficient of steel is 11Γ10-6. The
thermal conductivity is equal to 43 W/mK and the specific heat is 0,5 kJ/kg K. Applying a sudden boundary
temperature of 1200 K gives the temperature and displacement distributions shown in Figure 3-1 and 3-2. It can be
seen that the longer the element is heated, the further the heat reaches into the element. At some point in time the
steady-state distribution can be found for both components. The shape for the displacement distribution can be
explained by the fact that the left boundary is heated and pushes the nodal points to the right, since the elements
want to expend. The total displacement increases in time similar to the temperature distribution.
Parameter Value
L 3
nr_elements 60
kxx 43
cp 500
T left 1200
T right 293
t_limit 3600
Figure 3-1. Simply supported one-dimensional beam element with a sudden temperature
increase on the left boundary and the accompanying input
200
300
400
500
600
700
800
900
1000
1100
1200
0 0.5 1 1.5 2 2.5 3
Tem
pera
ture
[K
]
Length [m]
Figure 3-2. Temperature distribution for the coupled thermomechanical analysis of Fig 3-1
Temp. distribution: t = 600 s
t = 1200 s
t = 1800 s
t = 2400 s
t = 3000 s
t = 3600 s
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3.1.2. SUDDEN BOUNDARY TEMPERATURE INCREASE ABAQUS
These temperature and displacement distributions are then checked with ABAQUS. In ABAQUS the 2D Planar
modelling space is chosen. The type of the model is deformable and is modelled as a wire. Then the steel properties
are added. The density is 7850 kg/m3. The Youngβs modulus is taken as 2,1Γ108 kN/m2 and the Poissonβs ratio is
0,3. The expansion coefficient of steel is 11Γ10-6. The thermal conductivity is equal to 43 W/mK and the specific
heat is 0,5 kJ/kg K. The section chosen is a truss type, since a truss element type can be oriented anywhere in a 3D
space. They transmit forces axially only and often used for linear elastic structural analysis. The truss elements are
3 DOF elements which allow translation only and not rotation. Beam elements are 6 DOF elements which also allow
translation and rotation. These elements will be used when buckling is studied. The cross-sectional area is taken
1Γ10-6 in order to predict the 1D behaviour. Applying a sudden boundary temperature of 1200 K gives for 600
seconds and 60 elements the following figures in ABAQUS. The NT11 are the nodal temperatures and U is the
displacement.
Figure 3-4. Nodal temperatures and displacements for time is 600 seconds and 60 elements ABAQUS
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0.0018
0.002
0 0.5 1 1.5 2 2.5 3
Axi
al
dis
pla
cem
ent
[m]
Length [m]
Figure 3-3. Displacement distribution for the coupled thermomechanical analysis of Fig 3 -1
Displ. distribution: t = 600 s
t = 1200 s
t = 1800 s
t = 2400 s
t = 3000 s
t = 3600 s
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Plotting these results from Figure 3-4 gives the following graphs comparing the ABAQUS solution with the solution
obtained from the Matlab code.
In Figure 3-5 and 3-6 can be seen that the results from the Matlab code coincide with the results from the ABAQUS
analysis. It can thus be concluded that the Matlab code gives the correct answers for a coupled thermomechanical
behaviour analysis.
200
300
400
500
600
700
800
900
1000
1100
1200
0 0.5 1 1.5 2 2.5 3
Tem
pera
ture
[K
]
Length [m]
Figure 3-5. Temperature distribution Matlab code and ABAQUS at 600 seconds
ABAQUS solution: t = 600 s
Matlab code: t = 600 s
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0 0.5 1 1.5 2 2.5 3
Dis
pla
cem
ent
[m]
Length [m]
Figure 3-6. Displacement distribution Matlab code and ABAQUS at 600 seconds
ABAQUS solution: t = 600 s
Matlab code: t = 600 s
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3.2. LINEAR BUCKLING ANALYSIS
3.2.1. RITZ APPROXIMATION FEM STABILITY
By applying the Ritz approximation in Matlab the optimal number of elements can be calculated for the critical
buckling load and second eigenvalue for the problem in Fig 3-7. Using the Youngβs Modulus of 2,1e8 kN/m2 and
a moment of inertia of a 100x100 steel profile, the critical buckling load can be calculated for a beam with a length
of 3 meters.
The exact buckling loads are as given before
F1 = Fππ = 9,8696πΈπΌ
π2 , F2 = 39,4784
πΈπΌ
π2 . [3-1]
By calculating the critical buckling loads in Matlab it can be seen that the critical buckling load can be calculated
almost exactly with 2 elements. The second eigenvalue (buckling load) can be predicted by using at least 3 elements.
3.2.2. GALERKIN APPROXIMATION FEM STABILITY
For the Galerkin method the same procedure is performed. The approximation is implemented in Matlab and the
optimal number of elements is calculated for the critical buckling load and for the second eigenvalue. The exact
values are again obtained from Eq. [3-1] and the model in Fig 3-7 is used for the calculation. In Fig 3-9 the results
of the Galerkin approximation can be seen. The graph shows that the critical buckling load is obtained using 2
elements. The accurate second eigenvalue can be found using 4 elements.
0
2000
4000
6000
8000
10000
12000
1 2 3 4 5 6
Eig
enva
lue [k
N]
Number of elements
Critical buckling load Matlab
Second eigenvalue Matlab
Figure 3-8. Critical buckling load and second eigenvalue calculated in Matlab according to the Ritz approximation for the problem depicted in Fig 3-7
Figure 3-7. Simply supported one-dimensional element
Second eigenvalue exact
Critical buckling load exact
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3.2.3. MATLAB CODE
The methods described in the previous paragraphs have been implemented in Matlab in order to find some
solutions for purely mechanical problems. In order to couple the heat behaviour with the mechanical behaviour,
the latter is explained with an example. In Fig 3-10, a simple 1D column is subjected to a uniform thermal loading
π. For a thermomechanical problem, the total axial strain ν0 is decomposed into a mechanical (elastic) part ν0π and
a thermal part ν0π‘β (Suiker, 2014a)
ν0 = ν0π + ν0
π‘β . [3-2]
Using its constitutive relation, the axial force in the column can be decomposed
π0 = πΈπ΄ν0π = πΈπ΄(ν0 β ν0
π‘β) . [3-3]
The thermal strain is defined as
ν0π‘β = πΌβπ , [3-4]
where βπ = π β π0 , and πΌ is the linear coefficient of thermal expansion, for steel 11Γ10-6. No external mechanical loading (or deformation) results in the total strain equals to zero. This results in
Figure 3-10. One-dimensional beam element subjected to uniform thermal loading π
0
2000
4000
6000
8000
10000
12000
2 3 4 5 6
Eig
enva
lue [k
N]
Number of elements
Second eigenvalue Matlab
Figure 3-9. Critical buckling load and second eigenvalue calculated in Matlab according to the Galerkin approximation for the problem depicted in Fig 3-7
Second eigenvalue exact
Critical buckling load Matlab
Critical buckling load exact
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π0 = πΈπ΄ν0π = πΈπ΄(β ν0
π‘β) = βπΈπ΄πΌβπ . [3-5]
An increase in temperature by βπ induces thus a compressive normal force in the column. This can be used in the
general fourth-order differential equation for the curved system, resulting in
π2
ππ₯2 (πΈπΌ
π2π€
ππ₯2 ) β
π
ππ₯ (π0
ππ€
ππ₯ ) =ππ§ . [3-6]
Since πΈπΌ is taken as a constant and ππ§ = 0 this results in
πΈπΌ π4π€
ππ₯4 + πΈπ΄πΌβπ
π2π€
ππ₯2 = 0 , [3-7]
which can be rewritten as
π4π€
ππ₯4 + Ξ²2
π2π€
ππ₯2 = 0 , with Ξ²2 =
π΄ πΌπΏ βπ
πΌ . [3-8]
This structure of the differential equation is exactly the same as for a column subjected to a point load. Since the
structures are exactly the same, the procedures for finding a solution for these types of problems are identical. The
boundary conditions for this specific problem are
x = 0: w(0) = 0; ππ€
ππ₯ |x=0 = 0 ,
x = l: M(0) = - πΈπΌ π2π€
ππ₯2|x=0 = 0; M(l) = - πΈπΌ
π2π€
ππ₯2|x=l = 0 . [3-9]
Substituting these boundary conditions into the fourth-order differential equation gives the following solution
sin(π½π) = 0 π½ππ = ππ with n = 1,2,3,β¦ [3-10]
since
π½2 =π΄ πΌπΏ βππ
πΌ βππ =
π2π2πΌ
π΄πΌπΏπ2 . [3-11]
The critical temperature increment can then be found for n=1, which gives
βπππ =π2πΌ
π΄πΌπΏπ2 . [3-12]
For example a practical application, considering a steel profile HEA-200, with the following parameters (Suiker,
2014a)(Suiker, 2014a) the critical temperature increment for buckling can be calculated with the three cases. The critical temperature for buckling can then be calculated by adding the initial temperature (293 K)
Iyy = 1,336 Γ10-5 m4 (about the weak axis), π΄ = 5,38 Γ10-3 m2,
πΏ = 3 m ,
πΌ= 11 Γ10-6 /K.
The critical temperature increment and the critical temperature for this example are
Ξπππ = 248 K πππ= 541 K. [3-13]
In the FEM this value can be calculated with calculating the eigenvalues with the potential energy equation for a
simply supported beam shown in Figure 3-10. The expression for this potential energy is given with
π = β«1
2πΈπΌ (
π2π€
ππ₯2)2
ππ₯ + β« π01
2(ππ€
ππ₯)
2
ππ₯π
0
π
0 . [3-14]
Using the axial equilibrium condition π0 = βπΈπ΄πΌβπ and πΈπ΄πΌβπ = Ξ»οΏ½ΜοΏ½, with Ξ» as an arbitrary multiplier and οΏ½ΜοΏ½ = 1,0
as a reference magnitude. Equilibrium is then described by stationarity of the potential energy
πΏοΏ½ΜοΏ½ =ππ
ππ β πΏπ = 0
ππ
ππ= 0 β (π
π=1 π(m) β Ξ»ππ(π¦)
)π = 0, [3-15]
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The buckling loads πΈπ΄πΌβπ = Ξ»οΏ½ΜοΏ½ can be found by solving the above set of equations for Ξ». οΏ½ΜοΏ½ = 1,0 is used as a
reference intensity. Four eigenvalues Ξ» are found by equating the determinant of the reduced stiffness matrix to
zero. This results in
Ξ»1 = EAπΌπΏ βπ1 = EAπΌπΏ βπππ = 9,9438πΈπΌ
π2,
Ξ»2 = EAπΌπΏ βπ2 = 48,0000πΈπΌ
π2,
Ξ»3 = EAπΌπΏ βπ3 = 128,7228πΈπΌ
π2,
Ξ»4 = EAπΌπΏ βπ4 = 240,0000πΈπΌ
π2. [3-16]
which gives
βπππ = 9,9438πΌ
π΄πΌπΏπ2 ,
βπ2 = 48,0000πΌ
π΄πΌπΏπ2 ,
βπ3 = 128,7228πΌ
π΄πΌπΏ π2 ,
βπ4 = 240,0000πΌ
π΄πΌπΏ π2. [3-17]
This result and procedure is thus very similar to a purely mechanical problem. If now for οΏ½ΜοΏ½ a value is used which
is applied on the structure, the stability of the structure can be checked. If Ξ» < 1,0 buckling occurs and if Ξ» > 1,0 the
structure remains stable. The οΏ½ΜοΏ½ can now vary over the length of the structure, instead of using a reference intensity.
The values of οΏ½ΜοΏ½ can be exported from the coupled thermomechanical analysis. The same example for the stability
analysis is used as before in this paragraph. If now a temperature increment of Ξπ = 260 K is applied on the
structure, the Matlab code calculates the eigenvalues of this problem. The solutions of the eigenvalues are
Ξ»1 = 0,9522 , Ξ»2 = 3.8087, Ξ»3 = 8.5695, Ξ»4 = 15.2347. [3-19]
This means that the structure is buckled at the temperature of 553 K, since Ξ» < 1,0. This can be verified by the fact
that the critical temperature for this problem is 541 K.
3.2.3.1. EXAMPLE COUPLED THERMOMECHANICAL PROBLEM
For the example used in the coupled thermomechanical analysis, the eigenvalues can be calculated. This example
was shown in Figure 3-1. The temperature and displacement distributions accompanying this example were shown
in Figure 3-2 and 3-3. These distributions are used to formulate the internal forces. Then a buckling analysis is
performed. The eigenvalues are plotted at 10 minutes of coupled behaviour for various steel profiles. As can be
seen in Figure 3-11, only the HEA100 and HEA120 are buckled after 10 minutes, since Ξ» < 1,0. The other profiles
have eigenvalues larger than 1,0, which means that these profiles are still stable after 10 minutes of coupled
thermomechanical behaviour.
Figure 3-11. Eigenvalues at 10 minutes coupled behaviour for various steel profiles
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
HEA100 HEA120 HEA140 HEA160
Ξ»
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In the next figure the critical times are plotted for four steel profiles under the conditions of the example above. It
can be seen that a HEA100 profile buckles just after 170 seconds. The next profile, which is the HEA120, becomes
unstable after 10 minutes. The other two profiles buckle after 23 minutes and after 45 minutes. Using this code the
critical time for a steel profile can thus be calculated for a certain thermomechanical problem.
3.2.3.2. EXAMPLE BUCKLING MODE 2 ELEMENTS
For this example 2 elements are used in order to check the buckling modes. Firstly this is performed analytically.
Using the Matlab code, the example is tested on stability. If Ξ» < 1,0 then the structure is buckled. From the Matlab
code it follows that Ξ» = 0,36 for this temperature increment, which means that this structure is unstable.
Then the equilibrium is described by the stationarity of the potential energy for the total structure. This formulation
can be exported from the Matlab code, in which the temperature increments are implemented as internal forces in
the ππ(π¦)
. The eigenvalues are substituted back into the equilibrium equation, which allows then solving the system
up to an arbitrary constant. The total assembled matrix can then be separated into eight separate equations in which
the boundary conditions are applied and deleted.
β (ππ=1 π(m) β Ξ»ππ
(π¦))π = 0 . [3-20]
6762.1744 7133.087 3937.456 0 -1.70662 0 r1
7133.0872 15013.86 -117.225 -7250.31 -1.70662 -1.70662 d2
3937.4564 -117.225 13758.8 3878.844 -0.85331 0.853311 r2
0 -7250.31 3878.844 6996.624 0 1.706621 r3
-1.706621 -1.70662 -0.85331 0 0 0 Ξ1
0 -1.70662 0.853311 1.706621 0 0 Ξ2
Parameter Value
L 3
nr_elements 2
kxx 43
cp 500
T left 1200
T right 730
t_limit -
Figure 3-12. Critical times until buckling for various steel profiles
Figure 3-13. Temperature increment in the beam divided into two elements and input
0
500
1000
1500
2000
2500
3000
HEA100 HEA120 HEA140 HEA160
Cri
tical t
ime [
s]
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6762r1 + 7133d2 + 3937r2 β 1,71 Ξ1 = 0, [3-21]
7133r1 + 15014d2 β 117r2 β 7250r3 β 1,71Ξ1 β 1,71Ξ2= 0, [3-22]
3937r1 β 117d2 + 14759r2 + 3879r3 β 0,85Ξ1 + 0,85Ξ2 = 0, [3-23]
-7250d2 + 3879r2 + 6996r3 + 1,71Ξ2= 0, [3-24]
-1,71r1 β 1,71d2 β 0,85r2 = 0, [3-25]
-1,71d3 +0,85r2 + 1,71r3 = 0. [3-26]
Solving these equations for arbitrary constant r1 it follows that
Ξ1 = -223r1 , Ξ2 = -141r1 , d2 = -0,99r1 , r2 = -0,02r1 , r3 = -0,98r1 .
Inserting these values back into the οΏ½ΜοΏ½ (π)(π₯) = π(m)π(m)from Eq. [2-156] together with the essential boundary
conditions d1 = 0 and d3 = 0 gives the eigen mode for 2 one 3rd-order elements it gives
οΏ½ΜοΏ½ (1)(π₯) = (βπ₯ β 0,1518π₯3)π1 for 0 β€ π₯ β€ π
2 οΏ½ΜοΏ½ (1)(0) = 0, [3-27]
οΏ½ΜοΏ½ (2)(π₯) = (β0,98766 + 0,02468π₯ β 0,012π₯2 + 0,03795π₯3)π1 for π
2 β€ π₯ β€ π οΏ½ΜοΏ½ (2)(π) = 0. [3-28]
This gives the following buckling mode
As can be seen in the graph in Figure 3-14, the minimum of the buckled structure is moved from the middle of the
structure (1,50 m) to the left of the middle (1,48 m). This can be explained that a higher internal force gives a more
exaggerated buckling mode. If the structure is divided into more elements this movement will probably be more
clear. The more elements, the higher the accuracy of the buckling mode.
3.2.3.3. EXAMPLE BUCKLING MODE 3 AND 6 ELEMENTS
In the following example 3 elements are investigated.
Parameter Value
L 3
nr_elements 3
kxx 43
cp 500
T left 1200
T right 293
t_limit -
0 0.5 1 1.5 2 2.5 3
Length [m]
Figure 3-14 Buckling mode using 2 elements assembled
First buckling mode
Figure 3-15. Temperature increment in the beam divided into three elements and input
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The equilibrium equation now becomes
10596.45 16389.34 5792.887 0 0 0 -1.89403 0
16389.34 58043.22 -198.257 -30211.2 -16587.6 0 -2.84105 0 5792.887 -198.257 21457.25 16587.6 5726.801 0 -0.94702 0
0 -30211.2 16587.6 62793.54 -197.603 -16785.2 0 -2.84105
0 -16587.6 5726.801 -197.603 21985.06 5660.934 0 0.947017 0 0 0 -16785.2 5660.934 11124.26 0 1.894033
-1.89403 -2.84105 -0.94702 0 0 0 0 0
0 0 0 -2.84105 0.947017 1.894033 0 0
10596,45r1 + 16389,34d2 + 5792,89r2 β 1,89Ξ1 = 0, [3-29]
16389,34r1 + 58043,22d2 β 198,26r2 β 30211,20d3 β 16587,6r3 -2,84Ξ1= 0, [3-30]
5792,89r1 β 198,26d2 +21457,25r2 + 16587,6d3 + 5726,80r3 β 0,95Ξ1= 0, [3-31]
-30211,2d2 + 16587,60r2 + 62793,54d3 β 197,6r3 β 16785,2r4 β 2,84Ξ2= 0, [3-32]
-16587,6d2 + 5726,80r2 β 197,60d3 + 21985,06r3 + 5660,93r4 + 0,95Ξ2=0, [3-33]
-16785,2d3 + 5660,93r3 + 11124,26r4 + 1,89Ξ2= 0, [3-34]
-1,89r1 β 2,84d2 β 0,95r2 = 0, [3-35]
-2,84d3 +0,95r3 + 1,89r4 = 0. [3-36]
Solving these equations for arbitrary constant r1 it follows that
Ξ1 = -10,7677r1 , Ξ2= -103,884r1 , d2 = -0,8011r1 , r2 = 0,4033r1 , d3 = -0,7119r1 , r3 = -0,5055r1 , r4 = -0,8152r1 .
Inserting these values back into the οΏ½ΜοΏ½ (π)(π₯) = [π(m)][π(m)] from Eq. [2-156] together with the essential boundary
conditions d1 = 0 and d4 = 0 gives the eigen mode for 3 one 3rd-order elements
οΏ½ΜοΏ½ (1)(π₯) = (βπ₯ β 0,1989π₯3)π1 for 0 β€ π₯ β€ π
3 οΏ½ΜοΏ½ (1)(0) = 0, [3-37]
οΏ½ΜοΏ½ (2)(π₯) = (β0,8011 β 0,4033π₯ + 0,5686π₯2 + 0,0760π₯3)π1 for π
3 β€ π₯ β€
2π
3 , [3-38]
οΏ½ΜοΏ½ (3)(π₯) = (β0,7119 + 0,5055π₯ + 0,3096π₯2 β 0,1032π₯3)π1 for 2π
3 β€ π₯ β€ π οΏ½ΜοΏ½ (3)(π) = 0. [3-39]
And for 6 one 3rd-order elements it gives
οΏ½ΜοΏ½ (1)(π₯) = (βπ₯ + 0,2451π₯3)π1 for 0 β€ π₯ β€ π
6 οΏ½ΜοΏ½ (1)(0) = 0, [3-40]
οΏ½ΜοΏ½ (2)(π₯) = (β0,4693 β 0,8162π₯ + 0,3699π₯2 + 0,0978π₯3)π1 for π
6 β€ π₯ β€
π
3, [3-41]
οΏ½ΜοΏ½ (3)(π₯) = (β0,7727 β 0,3728π₯ + 0,5143π₯2 β 0,0351π₯3)π1 for π
3 β€ π₯ β€
π
2, [3-42]
οΏ½ΜοΏ½ (4)(π₯) = (β0,8349 + 0,1152π₯ + 0,45666π₯2 β 0,1060π₯3)π1 for π
2 β€ π₯ β€
2π
3, [3-43]
οΏ½ΜοΏ½ (5)(π₯) = (β0,6764 + 0,4923π₯ + 0,2936π₯2 β 0,1230π₯3)π1 for 2π
3 β€ π₯ β€
5π
6 , [3-44]
οΏ½ΜοΏ½ (6)(π₯) = (β0,3709 + 0,7012π₯ + 0,1223π₯2 β 0,0815π₯3)π1 for 5π
6 β€ π₯ β€ π οΏ½ΜοΏ½ (6)(π) = 0. [3-45]
The eigen mode plotted using 3 one 3rd-order elements gives the following graph, in which again can be seen that
the top is slightly moved to the left. This is caused by the heat on the left boundary.
r1
d2
r2
d3
r3
r4
Ξ1
Ξ2
0 0.5 1 1.5 2 2.5 3
Length [m]
Figure 3-16. Buckling mode using 3 elements assembled
First buckling mode
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The eigen mode using 6 elements gives the buckling mode shown in Figure 3-17.
3.2.3.4. OPTIMAL NUMBER OF ELEMENTS
In order to decrease the calculation time of the buckling analysis, the optimal number of elements is investigated.
The coupled thermomechanical analysis is used to define the optimal number of elements, since this analysis was
benchmarked with ABAQUS. For this specific problem the optimal number of elements is 30 elements. This can be
seen in the following graph. The length is plotted up until 0,8 meters of the total length of 3 meters, since this range
is the most interesting. As can be seen, the temperature distributions of 120 and 60 elements are positioned directly
underneath the ABAQUS distribution. The first number of elements which slightly differs is 30 elements, however,
the differences are significantly low. On the other hand, when 15 elements are used the temperature distribution
clearly differs from the ABAQUS distribution. Using this information it can be said that 30 elements is the most
optimal number of elements to use for this problem.
If then the example from Figure 3-1 is investigated for a HEA100 profile, the buckling mode can be plotted for 30
elements. From Figure 3-11 could be seen that the HEA100 profile is buckled after 170 seconds. The buckling mode
belonging to this instability can be seen in Figure 3-19. The graph shows that the buckling mode is slightly tilted to
the left side. This can again be explained by the fact that the sudden temperature increase is positioned at the left
boundary.
200
300
400
500
600
700
800
900
1000
1100
1200
0 0.2 0.4 0.6 0.8
Tem
pera
ture
[K
]
Length [m]
Figure 3-17. Buckling mode using 6 elements assembled
0 0.5 1 1.5 2 2.5 3
Length [m]
Figure 3-18. Temperature distribution at 10 minutes for different number of elements compared with exact ABAQUS solution
First buckling mode
Temp. distribution ABAQUS
120 elements Matlab
60 elements Matlab
30 elements Matlab
15 elements Matlab
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3.2.3.5. EXAMPLE COUPLED THERMOMECHANICAL PROBLEM CONTINUATION
In paragraph 3.2.3.1 the eigenvalues of the coupled thermomechanical problem were calculated for the induced
internal forces caused by the thermal expansion of the element. Various steel profiles were investigated for their
eigenvalues for a specific coupled thermomechanical problem. Now a column is investigated in a 3-storey building
which is depicted in the figure below. For simplicity reasons the column is hinged at both sides. In the Matlab code
the number of storeys can be adjusted. The total internal force per element can be calculated by summing up the
thermally induced internal force and the mechanical force. This internal force is then again implemented into the οΏ½ΜοΏ½
of the stress stiffness matrix. The thermally induced internal force is computed with the coupled thermomechanical
analysis. The mechanical force is computed by multiplying the grid distance with a factor for the number of
supports and with an estimated load. The estimated load for a roof is taken as 12 kN/m2 and for a floor 15 kN/m2
(concrete slabs) taking into account all safety factors. Then the total force can be calculated. The grid distance for
this example is 6 meters and the number of supports is 4 which means that the factor for the reaction force in the
middle is 1,1 according to simple mechanics rules. For this example the total force is equal to
πΉππππ = 6 β 6 β 1,1 β 12 = 475,2 ππ , [3-46]
πΉππππππ = 2 β 6 β 6 β 1,1 β 15 = 1188,0 ππ , [3-47]
πΉπ‘ππ‘ππ = 1663,2 ππ . [3-48]
In the following graph the differences in eigenvalues can be seen by comparing the purely thermal buckling, purely
mechanical buckling, and the thermal and mechanical buckling combined. The graph shows that the first four steel
profiles are already buckled at a mechanical load of 1663,2 kN since they have a Ξ» < 1,0. It also shows that only the
first two steel profiles buckle within the first 10 minutes of thermal exposure. It can also be seen that the influence
of the thermal exposure on eigenvalues of the combined analysis is significantly low at the first three or four steel
profiles and will gradually increase for the larger steel profiles.
Figure 3-19. Buckling mode for the thermomechanical problem from Figure 3-1 for 30 elements
Figure 3-20. Fire in a building with structural stability analysis for the coupled thermomechanical problem
0 0.5 1 1.5 2 2.5 3Length [m]
First buckling mode
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In Figure 3-22 the critical time is plotted for various steel profiles investigating the thermal exposure and thermal
and mechanical exposure. The purely mechanical part has no critical time since it has no transient component. It
can be seen that the critical time for purely thermal exposure gradually increases for larger steel profiles. The first
four steel profiles are not combined since they are already buckled at 1663,2 kN, which can be seen in the previous
graph. The last three profiles show the time they can resist the additional thermal forces.
0
0.5
1
1.5
2
2.5
3
3.5
HEA100 HEA120 HEA140 HEA160 HEA180 HEA200 HEA220
Ξ»
Thermal
Mechanical
Thermal and mechanical
0
4000
8000
12000
16000
20000
HEA100 HEA120 HEA140 HEA160 HEA180 HEA200 HEA220
Cri
tical t
ime [
s]
Thermal
Thermal and mechanical
Figure 3-21. Eigenvalues at 10 minutes and/or 1663,2 kN thermal and/or mechanical exposure
Figure 3-22. Critical time for various steel profiles subjected to 1663,2 kN and thermal exposure
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3.2.4. ABAQUS VERIFICATION
In this paragraph the previous results are verified with ABAQUS. Firstly the mechanical results will be verified.
Afterwards the purely thermal results will be reviewed and lastly the combined thermal and mechanical problem.
3.2.4.1. MECHANICAL VERIFICATION
The first verification is the purely mechanical problem. The critical buckling load is for a beam with a length of 3
meters is investigated. The analytical solution could be obtained by using the Eulerβs formulation for the critical
buckling load. The Matlab solution is provided by inserting the Youngβs modulus, the moment of inertia, and the
length. The solution is plotted for 6 3rd-order elements. In ABAQUS the solution is obtained by using 6 B23 elements.
In the following table the results for the three different methods are given for a HEA200 profile.
Table 3-1. Critical buckling loads
3.2.4.2. THERMAL VERIFICATION
The second verification is the thermal problem with a constant temperature increment over the length of the beam.
The critical buckling temperature increment is investigated for a beam with a length of 3 meters. The analytical
solution is obtained by using the formulas in paragraph 3.2.3. The Matlab solution is provided by inserting the
moment of inertia, the area, the thermal expansion coefficient, and the length. The solution is now given for 10 3 rd-
order elements. In ABAQUS the solution is found by using 10 B23 elements. For this analysis a simple square profile
is used with the dimensions 100 x 100 mm2. The results of the critical buckling temperature increment is given in
the following table.
Table 3-2. Critical buckling temperature increment
The third verification is the thermal problem with a varying temperature. This time the eigenvalues are plotted
instead of the critical values, since the internal forces are now inserted into the model. A simple square profile is
again used with the dimensions 100 x 100 mm2. The results of eigenvalues are given in the Table 3-3. Since this
problem has a varying temperature over the length of the beam, no analytical solution can be given. In the Matlab
code the temperature differences are averaged in order to find the internal force for the element. The same method
is then used in ABAQUS in order to check this. 6 B23 elements are used in ABAQUS and the simple square profile
is investigated. The internal forces are found by multiplying the temperature increments with the Youngβs
modulus, area, and thermal expansion coefficient. In Table 3-3 can be seen that the Matlab solution is similar to the
ABAQUS solution. It can thus be concluded that the Matlab code is correct, and the method to implement this in
ABAQUS should be to insert the internal forces instead of the temperature increments.
Strong axis Weak axis
Analytical 8502,34 kN 3076,68 kN
Matlab 8503,23 kN 3077,02 kN
ABAQUS 8503,20 kN 3077,10 kN
Strong/weak axis
Analytical 83,08 K
Matlab 83,08 K
ABAQUS 83,08 K
Figure 3-23. Mechanical problem with HEA cross section
Figure 3-24. Thermal problem with square cross section
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3.2.4.3. COUPLED THERMOMECHANICAL VERIFCATION
The example of the coupled thermomechanical behaviour of a beam with a length of 3 meters can also be verified
(paragraph 3.1). Using the method from the previous verification of the thermal problem with internal forces the
coupled problem is investigated. For this problem the simple square profile is used and the beam is divided into
30 elements, which was the most optimal described in paragraph 3.2.3.4. Then the internal forces are applied on the
beam, by multiplying the temperature increments at 600 seconds with the Youngβs modulus, the area, and the
expansion coefficient. In the following table the results can be seen. It shows that the Matlab solution is similar to
the ABAQUS solution. This proves that the Matlab code works correctly. Again no analytical solution can be found.
Since the method used in the Matlab code is correct, it can be assumed that the eigenvalues of the coupled
thermomechanical behaviour combined with mechanical forces will be correct.
3.2.5. REDUCTION YOUNGβS MODULUS
The reduction of the Youngβs Modulus was described in paragraph 2.1.3.2. For the example in Figure 3-20 the
influence of this reduction was investigated. In the figure below it can be seen that the influence of the degradation
of the mechanical property is significantly small. This can be explained by the fact that the heat does not reach deep
into the beam, which results in degradation of only the first few elements where the temperature increment is large.
However, Figure 3-27 shows that the larger the profile, the larger the influence of the degradation. This is caused
by the fact that the internal force is calculated with the area in it, which results in a larger internal force, so the
Youngβs Modulus can decrease more until buckling occurs.
Table 3-3. Eigenvalues thermal problem
π
Analytical -
Matlab 0,1732
ABAQUS 0,1732
Table 3-4. Eigenvalues coupled thermomechanical problem
π
30 elements
Analytical -
Matlab 0,9035
ABAQUS 0,9034
Figure 3-25. Thermal problem with square cross section
Figure 3-26. Coupled thermomechanical problem from Figure 3-1
with square cross section
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
HEA100 HEA120 HEA140 HEA160 HEA180 HEA200 HEA220
Ξ»
10 minutes
10 minutes with E-reduction
30 minutes
30 minutes with E-reduction
60 minutes
60 minutes with E-reduction
Figure 3-27. Eigenvalues for thermomechanical buckling comparing time and E-reduction
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3.3. NON-LINEAR BUCKLING ANALYSIS
3.3.1. EXAMPLE GEOMETRIC NON-LINEAR MECHANICAL BEHAVIOUR
In order to understand the non-linear calculation an example is executed for mechanical geometric non-linearity.
In the following figure this example can be seen. A initial imperfection is applied on the beam, and the beam is
divided into two elements. Each element has a stiffness matrix belonging to the local coordinate system. The red
numbers describe the number of elements, and the blue numbers describe the number of nodes.
The load-displacement relation in the local coordinate system can be written as
πβ²πβ² = πβ² , [3-49]
where πβ² is the member stiffness matrix, πβ² is the local displacement vector, and πβ² is the local force vector. In matrix
format this relation is represented by according to the Euler-Bernoulli stiffness matrix as
'
2
'
2
'
2
'
1
'
1
'
1
'
2
'
2
'
2
'
1
'
1
'
1
22
22
22
22
3
46026061206120
0000
26046061206120
0000
M
V
N
M
V
N
w
u
w
u
llllll
I
Al
I
Alllll
llI
Al
I
Al
l
EI
. [3-50]
The local displacement vector is thus given with
πβ² =
[
'
2
'
2
'
2
'
1
'
1
'
1
w
u
w
u
]
, [3-51]
where the apostrophes indicate the local values. The local displacement vector can be rewritten to the global
displacement vector by performing the following multiplication
πβ² = ππ, [3-52]
where π is the transformation matrix and π is the global displacement vector. The transformation matrix can be
derived from looking at a deformed element in Figure 3-28, which rotates clockwise.
π’1β² = π’1 cos(π) + π€1sin (π) , [3-53]
π€1β² = βπ’1 sin(π) + π€1cos (π) , [3-54]
π1β² = π1 , [3-55]
π’2β² = π’2 cos(π) + π€2sin (π) , [3-56]
π€2β² = βπ’2 sin(π) + π€2cos (π) , [3-57]
π2β² = π2 . [3-58]
Figure 3-28. Example non-linear calculation two elements (left), element 1 highlighted in local system (right)
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These equations can be written in matrix format
2
2
2
1
1
1
'
2
'
2
'
2
'
1
'
1
'
1
1000000000000000010000000000
w
u
w
u
cssc
cssc
w
u
w
u
, [3-59]
where π = cos (π) and π = sin (π). The local force vector can also be rewritten into the global form. The local force
vector is given by
πβ² =
[
'
2
'
2
'
2
'
1
'
1
'
1
M
V
N
M
V
N
]
, [3-60]
where the apostrophes again indicate the local values. The local displacement vector can be rewritten to the global
displacement vector by performing the following multiplication
π = πβππβ² or π = πππβ² , [3-61]
where πβπ is the inverse of the transformation matrix and π is the global force vector. The inverse of the
transformation matrix is also the transpose of the transformation matrix (ππ) (de Borst, Crisfield, Remmers, &
Verhoosel, 2012b). The transpose transformation matrix can be derived by again looking at a deformed element in
Figure 3-28
π1 = π1β² cos(π) β π1
β²sin (π) , [3-62]
π1 = π1β² sin(π) + π1
β²cos (π) , [3-63]
π1 = π1β² , [3-64]
π2 = π2β² cos(π) β π2
β²sin (π) , [3-65]
π2 = π2β² sin(π) + π2
β²cos (π) , [3-66]
π2 = π2β² . [3-67]
These equations are then written in matrix format
'
2
'
2
'
2
'
1
'
1
'
1
2
2
2
1
1
1
1000000000000000010000000000
M
V
N
M
V
N
cssc
cssc
M
V
N
M
V
N
, [3-68]
where π = cos (π) and π = sin (π).
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To formulate the global load-displacement relation Eq. [3-49] is inserted in Eq. [3-61]. This gives
π = πππβ²π β². [3-69]
Then Eq. [3-52] is inserted in the equation above, which results in
π = πππβ²ππ , [3-70]
which can be rewritten for the global coordinate system as
π = ππ , [3-71]
in which π represents the global stiffness matrix
π = πππβ²π . [3-72]
First an initial imperfection of π
100 is taken for a beam with the length of 3 meters using both the Newton-Raphson
method and the Arc-length method. Both methods gave the same solution during this analysis, however the
Newton-Raphson method had significantly smaller calculation time. Therefore, the Newton-Raphson method was
used for this analysis. The beam is a 100 x 100 mm2 steel profile. This means that the initial imperfection is 0,03
meters. Using this imperfection, the angle π can be calculated. The angle π = 1,9997eβ2 πππ for this initial
imperfection. The angle is then updated per load step using the new coordinates of each node. For this imperfection
the load-displacement relation is plotted for the vertical displacement of the middle node in Figure 3-29. As can be
seen in this graph, increasing the number of elements results in a more accurate solution for the load-displacement
relation for this problem. The difference between 8 and 10 elements is almost negligible, which means that 8
elements should be used to plot the load-displacement relation for this problem. It can also be seen that the Matlab
solution starts to diverge from the ABAQUS solution around 0.15 meters vertical displacement. Numerous checks
have been performed to investigate this divergence.
The first check was the Matlab code. It has been examined extensively in order to find formulation errors. Also
using the load controlled formulation of the Newton-Raphson method was used to check if the results were the
same. Both formulations gave the same results. The transformation of the stiffness matrix has been tested in several
ways by implementing the adjusted angle in four different ways. By examining these implementations, the correct
implementation was found, which is the adjustment of each coordinate. The stiffness matrix was compared for the
Matlab and ABAQUS solution without and with transformation for one element. For one straight element the
Figure 3-29. Load-displacement relation middle of the beam for initial imperfection of 0,03m
0
400
800
1200
1600
2000
2400
0 0.2 0.4 0.6 0.8 1 1.2
Forc
e [kN
]
Displacement [m]
Non-linear load-displ. relation: ABAQUS
2 elements Matlab
4 elements Matlab
6 elements Matlab
Non-linear load-displ. relation: 8 elements Matlab
Linear critical buckling load
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stiffness matrix was correct except a small difference in the stiffness in the normal direction. Then one element was
rotated and gave the same results for Matlab as for ABAQUS. This concludes that the transformation is applied
correctly. For two elements a strange solution was obtained in the stiffness matrix from ABAQUS. Some values in
the stiffness matrix were completely different comparing Matlab with ABAQUS. This means that ABAQUS used
another stiffness matrix than the Matlab code, which means another element type. Then the number of elements
used for this problem was studied. By increasing the number of elements in Matlab the result was a more accurate
solution. However for ABAQUS no difference occurred when the number of elements was increased. This was
strange since finite element methods are approximation methods and, in general, the accuracy of the approximation
increases with the number of elements used. Therefore the initial imperfection was reduced in order to study if the
behaviour would follow the linear behaviour. This can be seen in the following graph in Figure 3-30. Increasing the
number of elements causes the load-displacement relation to follow the linear critical buckling load, as expected.
However, the Matlab solution is still different when the vertical displacement increases. As can be seen in Figure
3-31, the non-linear load-displacement relation for an initial imperfection of 3Γ10-4 meters is practically the same
when the graph is enlarged from 0 to 0,14 meters. This could mean that the Matlab code resembles the correct
behaviour.
0
400
800
1200
1600
2000
2400
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
Forc
e [kN
]
Length [m]
0
400
800
1200
1600
2000
2400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Forc
e [kN
]
Displacement [m]
Figure 3-30. Load-displacement relation middle of the beam for initial imperfection of 3Γ10-4 m
Figure 3-31. Zoomed in on the non-linear load-displacement relation for initial imperfection of 3Γ10-4 m
Non-linear load-displ. relation: ABAQUS
2 elements Matlab
4 elements Matlab
6 elements Matlab
Non-linear load-displ. relation: 8 elements Matlab
Linear critical buckling load
Non-linear load-displ. relation 8 elements ABAQUS
Linear critical buckling load
Non-linear load-displ. relation 8 elements Matlab
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In Figure 3-32 the initial imperfection is varied between L/100, L/1000, and L/1000 for the Matlab and ABAQUS
solution for 8 elements. The graph shows that the difference between the Matlab code and the ABAQUS solution is
significantly small. However, again from 0,2 meters a variation starts to occur which is shown in Figure 3-33.
The cause of these differences could be a numerical error. In numerical calculations two major sources of errors can
be present, namely, truncation errors and round-off errors. A truncation error is the difference between the
truncated value and the actual value. The truncated value is represented by a numeral consisting of a fixed number
of allowed digits, with any excess digits βchopped offβ (Zienkiewicz & Taylor, 2000a). The difference between the
Matlab and ABAQUS solution could be the allowed number of digits. This could eventually increase the variation
between both solutions. A round-off error is the difference between a rounded-off numerical value and the actual
value. The rounded value is given by a numeral with a fixed number of allowed digits. In this value the last digit
is set to the value that produces the smallest difference between the actual and rounded quality (Zienkiewicz &
Taylor, 2000a). This could also be the cause of the difference between the Matlab and ABAQUS solution. However,
since the vertical displacement up to 0,2 meters is significantly accurate, the Matlab solution is assumed to be
correct.
0
400
800
1200
1600
2000
2400
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
Forc
e [kN
]
Displacement [m]
0
400
800
1200
1600
2000
2400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Forc
e [kN
]
Displacement [m]
Figure 3-32. Load-displacement relation middle of the beam for three different initial imperfections
Figure 3-33. Load-displacement relation middle of the beam for three different initial imperfections
Linear critical buckling load
Matlab (L/10000)
ABAQUS (L/10000)
Matlab (L/1000)
ABAQUS (L/1000)
Non-linear load-displ. relation 8 elements ABAQUS (L/100)
Matlab (L/100)
Linear critical buckling load
ABAQUS (L/10000)
Matlab (L/10000)
ABAQUS (L/1000)
Matlab (L/1000)
ABAQUS (L/100)
Non-linear load-displ. relation 8 elements Matlab (L/100)
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The next step is to investigate the exact solution for the secondary equilibrium path for post buckling behaviour.
This secondary path can be found in numerous literature (Godoy, 1999; Ramachandra & Roy, 2001) Inserting this
secondary path into Figure 3-33 shows that the ABAQUS solution follows the secondary path. This implies that the
ABAQUS solution follows the exact solution. However, in the exact secondary path following the ABAQUS
solution, it is assumed that the elements are inextensible, which means that the axial strains are neglected. The red
line in the Figure 3-34 is the secondary equilibrium path for the example in Figure 3-28 if the influence of the axial
strains in the total potential energy is neglected. This is called a strut element. The column element on the other
hand takes into account the influence of the axial strain on the potential energy and can be called extensible. The
extensible elements are used in the Matlab code, since these elements are more realistic. This is depicted in Figure
3-34 as the blue line.
If then the inextensibility is used in the Matlab code to confirm the behaviour with the ABAQUS solution it can be
seen that the inextensible solution clearly follows the ABAQUS solution. This is shown in Figure 3-35. The ABAQUS
element type B23 therefore neglects the axial strain. The small difference still visible in the graph could be explained
by numerical errors. However, this dissimilarity is significantly small. Although the inextensibility for this example
was confirmed with ABAQUS, it was not possible to plot the extensible solution in ABAQUS, since no element type
belongs to this behaviour. However, since the extensibility was benchmarked, it could be concluded that if then
extensible elements were taken into account in Matlab, the solution should be correct.
0
400
800
1200
1600
2000
2400
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
Forc
e [kN
]
Displacement [m]
Figure 3-34. Load-displacement relation middle of the beam for three different initial imperfections and the exact secondary paths where the red line is the exact secondary equilibrium path for the inextensible elements (strut) and the blue line is the exa ct
secondary equilibrium path for the extensible elements (column)
Figure 3-35. Load-displacement relation middle of the beam for extensible and inextensible elements comparing the Matlab solution with ABAQUS
Linear critical buckling load
Matlab (L/100) inextensible
Non-linear load-displ. relation 8 elements ABAQUS (L/100) inextensible
Non-linear load-displ. relation 8 elements Matlab (L/100) extensible
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The differences between the strut and column can also be seen in the buckling modes below. In Figure 3-36 the
buckling modes are depicted for the strut, where the axial strain is neglected. It can be seen that this element type
shows a significantly larger deformation compared to the buckling modes of the column, which are shown in Figure
3-37. The column depicted in Figure 3-37 also takes into account geometric non-linearity, which is shown by the
deflection of the column at a load smaller than the critical load. The buckling modes in Figure 3-36 are derived from
post-buckling, which in this figure indicates a linear behaviour up until the critical load.
The Arc-length method gives the same results as the Newton-Raphson method.
-0.1
0
0.1
0.2
0.3
0.4
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
w/L
x/L
Figure 3-36. Buckling modes inextensible elements (strut) post-buckling
π/πππ = 1,885
π/πππ = 1,518
π/πππ = 1,393
π/πππ = 1,294
π/πππ = 1,215
π/πππ = 1,102
π/πππ = 1,018
π/πππ = 1,035
π/πππ = 1,000
π/πππ = 1,215
π/πππ = 1,000
Figure 3-37. Buckling modes extensible elements (column) geometric non-linearity
π/πππ = 0,834
π/πππ = 0,625
π/πππ = 0,391
π/πππ = 0,208
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3.3.2. THERMAL NON-LINEARITY
For the thermal non-linear analysis the example from Figure 3-1 is again used. Since a square steel profile with the
dimensions 100Γ100 mm2 is easy to investigate, this profile was again used to calculate the critical time. The length
of the beam is 3 meters and for this simple calculation the thermal properties are taken as constants to reduce
calculation time. The specific heat is taken as 600 J/kgK and the thermal conductivity is 45 W/mK. The density of
steel is 7850 kg/m3 and the thermal expansion of steel is taken as 11Γ10 -6 (CEN, 2011). The Youngβs modulus is
taken as temperature dependent as described in paragraph 2.1.3.2. The thermal load on the beam element is given
by a sudden temperature increase on the left boundary. This temperature increase is taken as 1200 K whereas both
ends are restrained. Performing a coupled thermomechanical analysis gives the temperature and displacement
distribution for the one-dimensional beam element. These distributions can be seen in Figures 3-38, 3-39, and 3-40.
The mechanical load is taken as 1
5πΉππ β 400 ππ. This is a design load which is normally taken as the maximum for
the applied load on a column in practice. The linear critical buckling time for this profile subjected to a sudden
temperature increase on the left boundary as can be seen in Figure 3-1 is around 230 seconds. This number is found
by using the linear implementation of coupled thermoelasticity formulation in combination with the Galerkin
method. The next step is to check the non-linear load-displacement relation for the mechanical problem which
results in the accompanying initial imperfection applied in the non-linear analysis. The initial imperfection for a
column in practice is taken as πΏ/250. Using this initial imperfection the non-linear mechanical response is
investigated and the transverse displacement for the one-dimensional element is found as 3,2Γ10-3 m. This value is
added up to the initial imperfection giving a total imperfection of 1,52Γ10-2 m. Consequently, this means that the
load is applied on the element, the response is calculated and then restrained at the point of total imperfection.
Now the non-linear thermomechanical analysis can be performed restraining both ends of the element. In Figure
3-41 this response is plotted over time and compared with the linear critical buckling time. As one can see in non-
linear mechanical analyses, the non-linear curve approaches the linear critical buckling load when applying an
imperfection. Figure 3-41 shows that the non-linear thermomechanical response approaches the linear critical
buckling time.
Figure 3-38. Temperature distribution for the coupled thermomechanical analysis of the problem in Figure 3-1
200
300
400
500
600
700
800
900
1000
1100
1200
0 0.02 0.04 0.06 0.08 0.1
Tem
pera
ture
ΞΈ
[K]
x/L
Temp. distribution: t = 200 s
t = 150 s
t = 100 s
t = 50 s
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Figure 3-41: Comparison between the linear critical buckling time and the non-linear thermomechanical transverse displacement of the middle of the element from Figure 3-1 with an initial imperfection of 1,52e-2 m
Figure 3-39: Displacement distribution in x direction using coupled thermomechanical analysis on the problem
depicted in Figure 3-1
Figure 3-40: Displacement distribution in y direction using coupled thermomechanical analysis on the problem
depicted in Figure 3-1
0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0 0.2 0.4 0.6 0.8 1
Axi
al
dis
pla
cem
ent
[m]
x/L
Displ. u: t = 200 s
t = 150 s
t = 100 s
t = 50 s
0
0.01
0.02
0.03
0.04
0.05
0.06
0 0.2 0.4 0.6 0.8 1
Tra
nsve
rse dis
pla
cem
ent
[m]
x/L
t = 150 s
t = 100 s
t = 50 s
Displ. w: t = 200 s
0
50
100
150
200
250
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
Tim
e [
s]
Transverse displacement w [m]
Linear buckling analysis
Non-linear coupled
thermomechanical analysis
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3.4. COMPARISON WITH THE EUROCODE
The coupled thermomechanical analysis can now be used to assess the stability of steel structural elements
subjected to a fire. A parametric study is carried out on the structural columns subjected to a fire in comparison
with the Eurocode. This Eurocode defines the critical temperature as the temperature at which the structural
member fails. The time until this failure is called the critical time. The fire resistance of steel structures can be
evaluated by means of the three βdomainsβ. The first domain is the time domain, which is usually only found by
using advanced models. However, the code written in for this paper prescribes the critical temperature and time
simultaneously. The second domain is the strength domain. This domain can easily be calculated by hand by
finding the reduced resistance at a required resistance time. The last domain is the temperature domain, which is
mostly used for simple fire resistance calculations in the Eurocode. The three domains are given
- in the time domain as
π‘ππ,π β₯ π‘ππ,ππππ’, [3-73]
- in the strength domain as
π ππ,π,π‘ β₯ πΈππ,π,π‘ , [3-74]
- and in the temperature domain as
ππ β€ πππ ,π, [3-75]
where
π‘ππ,π is the design value of the fire resistance;
π‘ππ,ππππ’ is the required fire resistance time;
π ππ,π,π‘ is the design value of the resistance of the member in the fire situation;
πΈππ,π,π‘ is the design value of the relevant effects of actions in the fire situation;
ππ is the design value of material temperature;
πππ,π is the design value of the critical material temperature (CEN, 2011).
The Eurocode also describes three types of design methods which can be used to assess the mechanical behaviour
of steel structures exposed to fire. The first method is the simple calculation method (SCM), which is practically
limited for member analysis. This method uses a simple equation to calculate the critical temperature belonging to
a specific degree of utilisation. The second method is the critical temperature method (CTM), which is most
commonly used for designing steel structures exposed to fire conditions. The third method uses the advanced
calculation model (ACM). These models are increasingly being used in the modern fire safety engineering due to
the numerous advantages it can provide, however, they cost a significant amount of time. The main focus of this
paper is on the CTM, since this method is the most convenient to use.
3.4.1. CTM
A step by step procedure has to be conducted in order to take all necessary features of the Eurocode for fire design
of steel structures into account. These steps are extensively elaborated in the background and applications report
of the Eurocode for structural fire design (Vassart et al., 2014).
3.4.1.1. STEP 1: DETERMINATION OF APPLIED DESIGN LOAD STEEL MEMBER IN FIRE
The applied loads to a steel member exposed to fire can be obtained according to relation 6.11b of NEN-EN 1990.
This equation is given by
πΈππ,π,π‘ = β πΊπ,π + (π1,1ππ π2,1)ππ,1 +πβ₯1 β π2,πππ,ππβ₯1 , [3-76]
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where
πΊπ,π are the characteristic values of the permanent actions;
ππ,1 is the characteristic leading variable action;
ππ,π are the characteristic values of the accompanying variable actions;
π1,1 is the factor for the frequent value of a variable action;
π2,π is the factor for the quasi-permanent values of the variable actions.
A simplified calculation for the applied loads in structural fire design can be obtained with
πΈππ,π,π‘ = ππππΈπ , [3-77]
where
πΈπ is the design value of the corresponding force at ambient temperature (CEN, 2012a);
πππ is the reduction factor for design loads in fire situations.
3.4.1.2. STEP 2: CLASSIFICATION OF THE STEEL MEMBER UNDER FIRE CONDITIONS
In order to classify steel members, 4 classes are prescribed depending on their levels of slenderness of the cross-
section and on their stress distribution. A detailed description of this classification is given in NEN-EN 1993-1-1
(CEN, 2012b). The classification has the same procedure as at ambient temperature, however, a different value of ν
is adopted to take account for temperature influences. This value ν is here given as
ν = 0,85β235
ππ¦
. [3-78]
3.4.1.3. STEP 3: DETERMINATION OF THE DESIGN LOAD-BEARING CAPACITY STEEL MEMBER
This step is performed at time equals zero, which is at ambient temperature. The load-bearing capacity in this case
should either be the simple plastic or the elastic resistance of the cross-section. In this step also the non-dimensional
slenderness is calculated.
3.4.1.4. STEP 4: DETERMINATION OF THE DEGREE OF UTILISATION STEEL MEMBER
A parameter which relates the design load of a steel member in the ire situation to its design load-bearing capacity
is given by the degree of utilisation π0. For steel members subjected to instability problems, such as columns under
axial compressive forces, the degree of utilisation can be found with
π0 =πΈππ ,π,π‘
πππ ,ππ,0 . [3-79]
3.4.1.5. STEP 5: DETERMINATION OF THE CRITICAL TEMPERATURE STEEL MEMBER
Using the non-dimensional slenderness calculated in step 3 and the degree of utilisation from step 4 the critical
temperature can easily be obtained. For members without any instability the critical temperature can simply be
calculated with
πππ = 39,19 ln [1
0,9674π03,833 β 1] + 755 . [3-80]
The critical temperature for steel members subjected to instability phenomena can be calculated according to the
tables given by Vassart et al. (2014) shown in Appendix E.
3.4.1.6. STEP 6: DETERMINATION OF THE SECTION FACTOR AND CORRECTION FACTOR FOR
THE SHADOW EFFECT
Lastly, the section factor is of great importance in the calculation of the steel temperatures during a fire event. The
section factor is a ratio of the heated perimeter to the area of the cross-section. In the Eurocode this factor is defined
as π΄π/π. The rate of temperature increase of a steel element depends on this section factor. According to the
Eurocode the temperature of a steel element can be described with Figure 3-42 in which several section factors are
depicted with their belonging temperature-time curve. It can be seen that the larger the section factor, the faster the
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temperature increase. This can be explained by the fact that the section factor consists of the heated perimeter in
the numerator. This means that if the perimeter is large, the section factor is large, and therefore more heat can
enter the profile. The steel temperature increase can be calculated according to the following equation
βππ,π‘ =ππ β
ππππ
π΄π
πββπ‘ , [3-81]
where
ππ β is the correction factor for the shadow effect;
ππ is the specific heat;
ππ is the density; π΄π
π is the section factor;
βπ‘ is the time increment (should be β€ 5 seconds);
β is the design value of the net heat flux per area.
This net heat flux consists of two parts, the radiation part and the convection part and is described with
β = βπ + βπ . [3-82]
The NEN-EN 1993-1-2 also takes into account the so called shadow effect (SE). This is the reduction on the ratio
between the perimeter through which heat is transferred into the steel over the steel volume, which is ππ β. This
reduction has to be applied on I-profiles, since the radiation will not be able to reach the full perimeter which can
be seen in Figure 3-43.
Figure 3-42. Temperature-time curves for the ISO-834 and several sections factors π΄π/π
Figure 3-43. Reduced radiation giving the shadow effect for a steel I-profile
0
200
400
600
800
1000
1200
1400
0 20 40 60 80 100 120
Tem
pera
ture
ΞΈ
[K]
Time [min]
ISO-834
170 m-1
120 m-1
70 m-1
Am/V = 20 m-1
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3.4.2. CASE STUDY
The next step is to compare the results obtained from the Matlab code with the calculation methods used in practice
by investigating a simple example. In Figure 3-43 the simply supported one-dimensional beam element can be seen
with a compartment fire as the boundary condition. This fire gives convective and radiant heat transfer as boundary
conditions on the element, which was prescribed in equations [2-42]-[2-46] The appropriate fire boundary
conditions can be found by using the body heat source term to introduce heat from the sides of a one-dimensional
element. A fire can start and grow in many different ways. The fire development is very random and can vary for
many specific situations. Despite this randomness, the development of a compartment fire can be generally
explained and understood. The main components that affect the fire development are the roomβs geometry, the
quantity of the combustible material, the type of combustible material, its arrangement in the compartment, and
the oxygen supply. The thermal properties of the surfaces, such as thermal conductivity and heat capacity, are also
contributory factors.
The fire resistance of steel structures can be studied according to three different approaches. The first approach is
to analyse a steel member. Every member of the total structure can then be evaluated by separating each member
from the other members. The second approach is the analysis of parts of the structure, in which the link between
several parts of the structure is taken into account. The last approach is the global structural analysis, in which the
total structure is assessed for its fire resistance (CEN, 2011). This paper uses the first approach for the assessment
of the fire resistance of the steel structure in a compartment fire. The temperatures generated in a compartment fire
can be calculated or predicted by a specific fire description or a model of the fire. A model most often used in order
to investigate the structural stability is the one-zone model. The one-zone model is based on the fundamental
hypothesis that, during the fire, the gas temperature is uniform in the compartment. One-zone models are valid for
post-flashover conditions, in which the structural stability needs to be achieved. The gas temperature in the
compartment can be described by temperature-time curves, also known as fire curves. Several national and
international fire curves have been developed to simulate fires. The ISO-834 fire curve is most often and was shown
in Eq. [2-1]. The gas temperatures are then used as input data for the compartment fire boundary conditions on the
element from Figure 3-44. A six-storey braced building is used in this example to investigate a steel column
subjected to a compartment fire on the ground floor. In Figure 3-45 (a) this building structure can be seen. The
column which is subjected to the compartment fire is highlighted in Figure 3-45 (b) with the appropriate mechanical
boundary conditions. This example is chosen since the influence of the mechanical load can also be investigated by
varying the number of floors applied for this building.
Figure 3-44: Simply supported one-dimensional beam element with
body heat source representing the fire boundary conditions
(a) (b) Figure 3-45. (a) Braced frame building subjected to a compartment fire on the ground level, (b) highlighted steel column subjected to the compartment fire hinged on the ground floor and clamped at the first floor giving a buckling length of πππ = 0,7π
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The fire resistance requirement of the example described in Figure 3-46 should be 60 minutes according to the
norms for office buildings, which can be seen in Fig. 14. The height of each floor is 3,6 meters. The grid used for this
office building is 6m x 7m. The floors are composite slabs and the columns at the ground floor are HEB300 with the
steel grade S275. The Youngβs modulus is described as temperature dependent, the specific heat is taken as 600
J/kgK, and the thermal conductivity is 45 W/mK. The density of steel is 7850 kg/m3 and the thermal expansion of
steel is taken as 11Γ10-6 (CEN, 2011). According to the Eurocode the convective heat transfer coefficient should be
taken as 25 W/m2. The radiant heat transfer coefficient should be temperature dependent according to Eq. [2 -46].
The steel temperature obtained from Matlab is uniformly throughout the cross-section, since the heat is applied as
an internal body heat source. This body heat source is for one-zoned models uniform over the length of the column,
however, this term could also be used non-uniformly. In Figure 3-47 the temperature-time curves for the CTM and
Matlab can be seen. The section factor of 116 m-1 for a HEB300 is taken into account for the CTM calculation, and
the correct perimeter is used in Matlab. It can be seen that the temperature development for both methods results
in the same temperature-time curves. The graph also shows that after 30-35 minutes the steel temperature reaches
the compartment fire gas temperature. This can also be seen in Figure 3-48, which shows the comparison between
the temperature-time curves for the CTM and Matlab solution with and without the shadow effect (SE) taken into
account. This graph shows that the shadow effect has an influence on the temperature-time curve which is caused
by the reduced radiation explained in Figure 3-43. The graph in Figure 3-48 also shows that the temperature
developments are again the same for both methods.
Figure 3-47: ISO-834 fire curve and the steel temperature development according to the CTM and Matlab
Figure 3-46. Fire resistance requirements office buildings
0
200
400
600
800
1000
1200
1400
0 10 20 30 40 50 60
Tem
pera
ture
ΞΈ
[K]
Time [min]
ISO-834
Matlab
CTM
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Figure 3-49 shows the critical temperatures for the HEB300 column on the ground floor at a varying degree of
utilisation calculated with the CTM and the Matlab code. The range of the calculations is taken according to the
tables obtained from the Vassart et. al. (2014). It can be seen that the critical temperature obtained with Matlab at a
small degree of utilisation starts almost at the same value (a difference of 2,56%). This means that purely thermal
calculations result in almost the same critical temperatures. However, increasing the degree of utilisation, and
therefore the applied load on the column, the critical temperature obtained in Matlab is slightly higher than for the
other method. At a degree of utilization of 0,5 the Matlab solution is 12,57% higher than the CTM method. This
indicates that the degree of utilisation has a slightly higher influence when using the CTM. This can be explained
by the fact that the tabulated data used for the CTM calculation also takes into account the degradation of the yield
strength at elevated temperatures. This degradation of the yield strength was neglected in the Matlab calculations.
The columns calculated according to the CTM will fail earlier due to this degradation of the yield strength,
prescribed in section 3 of the NEN-EN 1993-1-2.
0
200
400
600
800
1000
1200
1400
0 10 20 30 40 50 60
Tem
pera
ture
ΞΈ[K
]
Time [min]
CTM with SE
CTM without
SE Matlab without
SE
Matlab with SE
Figure 3-48. CTM and Matlab steel temperature development without and with the shadow effect (SE)
600
700
800
900
1000
1100
1200
0 0.1 0.2 0.3 0.4 0.5 0.6
Cri
tical t
em
pera
ture
ΞΈ
[K]
π0
CTM
Matlab
Figure 3-49. Influence degree of utilization on the critical temperature using the CTM and Matlab
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The critical time is also calculated with the Matlab code. This can be seen in Figure 3-50, in which the critical time
from Matlab is compared with the CTM depending on the degree of utilisation. The graph shows that in both
calculations the critical time is smaller than the requirement of 60 minutes fire resistance. It can also be seen that a
slight difference is visible between the critical time calculated with the CTM and Matlab. At the smallest degree of
utilization plotted in this figure, the CTM calculated critical time is 14,64% higher than the Matlab solution. On the
other hand, at the largest degree of utilization plotted, the Matlab solution is 17,91% higher than the CTM solution.
This difference was also noticeable in Fig. 15 where the critical temperatures were plotted. It can again be noted
that the column calculated with the CTM will fail earlier than the column calculated with Matlab, which can be
explained by the degradation of the yield strength in the CTM.
In order to increase the critical time, insulation can be applied on the steel members. This insulation has to be taken
into acccount in the calculation of the net heat flux. Common fire insulation systems for steel members are sprays,
boards, and coatings. The insulation will have a significant effect on the steel temperature. Two parameters have a
great influence on the temperature development of the protected steel elements. The first is the section factor, which
is given by π΄π/π according to the Eurocode. The second parameter is the insulation itself, where the thickness,
thermal conductivity , density , and specific heat are important. The temperature development including the
insulation effect can be calculated according to relation (4.27) described in the NEN-EN 1993-1-2.
Figure 3-50: Influence degree of utilization on the critical time using the CTM and Matlab
0
10
20
30
40
50
60
70
0 0.1 0.2 0.3 0.4 0.5 0.6
Cri
tical t
ime [
min
]
π0
Requirement
CTM
Matlab
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4. CONCLUSIONS AND RECOMMENDATIONS
Finally, this thesis ends with the conclusions and recommendations.
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4.1. CONCLUSIONS
An engineering model describing the coupled thermomechanical behaviour of steel elements under fire conditions
has been developed in order to increase insight into the general thermomechanical behaviour and to investigate
steel members exposed to fire. After deriving the governing equations of coupled thermoelasticity, the Matlab code
has been written in FE formulation and has been verified with ABAQUS. Although some simple coupled
calculations could be performed using ABAQUS, the more difficult calculations were accomplished by using the
Matlab code.
It could be seen that the coupling effect of mechanical behaviour on the thermal part was significantly small in
comparison with the thermal effect on the mechanical behaviour. The model written in Matlab was used to
investigate the coupled thermomechanical behaviour of steel elements under fire conditions. Linear and non-linear
thermomechanical problems were solved using the Matlab model. The results were presented as the critical
temperature or time for the linear problems and the nodal displacements and temperatures changing in time were
given for the non-linear problems.
In the first example from Chapter 3 a simple coupled thermomechanical analysis was performed with a sudden
temperature increase on the left boundary. It was clearly visible that the temperatures and displacements gradually
increased over time. After the verification in ABAQUS it could be concluded that the model written in Matlab is
able to perform simple coupled thermomechanical analyses. In the second paragraph from Chapter 3 the linear
buckling analysis was performed mechanically, thermally, and thermomechanically. By using the Matlab code,
simple linear buckling analyses could be performed in order to obtain the critical temperature or time for specific
steel profiles subjected to certain boundary conditions. By verifying these results it could be concluded that also a
buckling analysis could be performed for the coupled thermomechanical problems. The third paragraph showed
the non-linear buckling analysis results, in which the mechanical case was extensively investigated, since the
solution obtained from the Matlab code was significantly different compared to the ABAQUS solution. However,
after some intense programming the cause of the difference was found in the extensibility of the elements chosen
in the calculation. ABAQUS used strut elements, which are inextensible and therefore neglect axial strain. In the
more practical solution obtained from the Matlab code, the column elements were used, which are extensible and
take into account axial strain. After the verification of the results it could be concluded that the Matlab solution
would be more realistic. Subsequently using the non-linear calculation for the thermal problems, it was clearly
visible that the non-linear solution approached the linear solution, as expected. In the last paragraph in which the
comparison with the Eurocode was made, the perimeter or section factor of the column had a significant influence
on the steel temperature development, and therefore also the critical time. Consequently, it should be noted that
using the correct perimeter or section factor is of great importance during a fire resistance calculation. This example
also showed that the temperature development calculated with the Matlab code resembled the solution obtained
from the Eurocode. However, the critical temperature or time slightly differed when increasing the degree of
utilisation. For larger degrees of utilisation it could be seen that the Matlab solution was slightly higher, which was
explained by the fact that the CTM also took into account the degradation of the yield strength at elevated
temperatures. Although some differences occur between the critical temperature or time calculated with Matlab or
the Eurocode, the steel temperature development remains the same and the differences remain relatively small.
It may be concluded that using the engineering model presented in this thesis, quick assessments can be performed
on the fire resistance of steel members. Furthermore, the engineering model gives a complete understanding of the
fully coupled thermomechanical behaviour and can be used in the future for more difficult calculations.
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4.2. RECOMMENDATIONS
Although this thesis presents a fully coupled thermomechanical model which can perform one-dimensional
calculations, this subject is still in its infancy. This graduation thesis can be used to increase the knowledge in
coupled thermomechanical behaviour and to understand every step made in the procedure. This thesis also
presents some examples performed with the code written in Matlab.
However, further research can be performed on many aspects of the coupled thermomechanical analysis to increase
the knowledge on these models and even write more codes which can be applied in practice. First of all, during this
thesis only some non-linear aspects were taken into account in the calculation, such as the non-linear Youngβs
modulus, radiant heat transfer coefficient, and geometry. The non-linear stress strain relationship was not taken
into account, since this would take considerably more time to understand. Taking this non-linearity into account
would provide more realistic solutions. Plasticity is an aspect in non-linear calculations which should be considered
in the future. The one-dimensional model can then be extended to full non-linear coupled thermomechanical
behaviour.
The next step in the process of developing a fully coupled thermomechanical model should be to extend the one-
dimensional model calculations to two dimensions, for example portal structures. The portal structures have the
same governing equations, however, the assemblage of the FEM formulation differs. Consequently plate structures
could be analysed. The general equations of thermoelasticity are already given for three dimensions, so two
dimensional shape functions should be used for the implementation in the FE formulation. After finishing and
verifying the two-dimensional elements, the three-dimensional elements can be modelled. When the three-
dimensional model performs correctly, this model could be used in two-way coupled calculations, in which the
mechanical part could have a more significant influence on the thermal part. For example, if a thin-walled steel
façade buckles or fails, this could lead to the introduction of more oxygen into the compartment, or change the
velocity and paths of the fluid. Therefore, fully coupled thermomechanical analysis can be used to investigate the
two-way coupled behaviour of the failure of the thin-walled structural system.
Another important aspect would then be the boundary conditions. The fire development is very random and can
vary for many situations. Therefore applying the appropriate boundary conditions is very difficult. Investigating
the boundary conditions could be of great importance for the research on thin-walled steel structures and how to
apply these boundary conditions. The behaviour of the full system of thin-walled structures could then be
investigated using both the coupled formulation and the appropriate boundary conditions. Furthermore, insulation
could be applied on the structural elements or in between structural elements to investigate the complete behaviour.
In the end a model could be developed which uses both the mechanical and thermal aspects in a fully coupled
manner for members in a structure or for the whole structural system. Using this model, design rules could be
developed, since the current design rules do not take into account the system effects.
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 90
5. REFERENCES
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Bengtsson, L. G. (1999). Enclosure fires. Huskvarna: Swedish Rescue Services Agency.
Biot, M. A. (1956). Thermoelasticity and irreversible thermodynamics. Journal of Applied Physics, 27(3), 240β253.
CEN. (2011). Eurocode 3: Design of steel structures - Part 1-2: General rules - Structural fire design. EN1993-1-2:2011.
Brussels: European Standard.
CEN. (2012a). Eurocode 0: Basis of structural design. EN1990:2012. Brussels: European Standard.
CEN. (2012b). Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings. EN1993-1-1:2012. Brussels: European Standard.
Cirak, F. (2015). Finite Element Formulation for Beams Review of Euler-Bernoulli Beam. University of Cambridge.
Coleman, B. D., & Noll, W. (1963). The thermodynamics of elastic materials with heat conduction and viscosity. Archive for Rational Mechanics and Analysis, 13(1), 167β178. doi:10.1007/BF01262690
Danielsson, H. (2013). Path following solution approaches and integration of constitutive relations for a 3d wood cohesive zone model.
De Borst, R., Crisfield, M. a., Remmers, J. J. C., & Verhoosel, C. V. (2012a). Geometrically Non-linear Analysis. In Non-Linear
Finite Element Analysis of Solids and Structures.
De Borst, R., Crisfield, M. a., Remmers, J. J. C., & Verhoosel, C. V. (2012b). Non-linear Finite Element Analysis. In Non-Linear
Finite Element Analysis of Solids and Structures (Second Edi). Chichester: John Wiley & Sons.
Ericksen, J. (1991). Introduction to the thermodynamics of solids. New York: Springer.
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ENGINEERING MODEL FOR COUPLED THERMOMECHANICAL BEHAVIOUR OF STEEL ELEMENTS UNDER FIRE CONDITIONS
R.H.A. TITULAER
26 JANUARY 2016
APPENDIX
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R.H.A. TITULAER 0
TABLE OF CONTENTS
A. Configuration Factor .............................................................................................................................................. 2
A.1. Rectangle as radiator ....................................................................................................................................... 2
A.2. Cylinder as radiatior ........................................................................................................................................ 3
A.3. Ellipse as radiator ............................................................................................................................................ 4
B. Derivation shape functions and final FEM formulation........................................................................................ 6
C. Derivation Euler-Bernoulli stiffness matrices...................................................................................................... 10
D. Element matrices thermal non-linearity .............................................................................................................. 12
E. Critical temperatures of steel members with steel grade S275............................................................................ 14
F. Matlab tutorial ....................................................................................................................................................... 16
F.1. Coupled thermomechanical analysis............................................................................................................. 16
F.2. Linear buckling ............................................................................................................................................... 23
F.3. Non-linear buckling........................................................................................................................................ 26
F.4. Code description............................................................................................................................................. 33
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A. CONFIGURATION FACTOR
For the total radiation the equation was given in Eq. [2-34]. In this equation the corresponding configuration factor
ππ is of great importance. The configuration factor determines the influence of the radiation source on the receiving
surface. In fire safety engineering projects it can be important to make calculations of the radiation intensity. One
example is the fire spread from one building to another by radiation. These calculations require the use of the
configuration factor. Fire models can be used for the most accurate calculations of the incident radiation. The
configuration factor calculation can be very difficult and tedious.
In Eurocode 1 (NEN-EN 1991-1-2 Annex G) the configuration factor is explained. The general formulation of this
configuration factor in mathematical form is written as
ππΉπ1βπ2 =πππ π1πππ π2
ππ1β22
ππ΄2 . [A-1]
The total radiative heat that arrives at a given receiving surface emitted from a radiating surface is measured by
the configuration factor. In Figure A-1 it can be seen that this factor depends on the size of the radiating surface, on
the distance from the radiating surface to the receiving surface, and on their relative orientation. The contribution
of the heat can be calculated according to three shapes, the rectangle, the cylinder, and the ellipse. In Eurocode 1
the formulas for the rectangular calculations are given.
A.1. RECTANGLE AS RADIATOR
The configuration factor according to the rectangular calculation should be determined as the sum of the
contributions from each of the zones on the radiating surface (usually four) that are visible from point P on the
receiving surface. The equation for the receiving surface configuration factor parallel to the radiating surface is
given by
π =1
2π[
π
(1+π2)0.5tanβ1 (
π
(1+π2)0.5) +
π
(1+π2)0.5tanβ1 (
π
(1+π2)0.5)] , [A-2]
where
π =β
π ,
π =π€
π ,
π is the distance from P to X,
β is the height of the zone on the radiating surface,
π€ is the width of the zone on the radiating surface.
Figure A-1. Radiative heat transfer between two infinitesimal surface areas (CEN, 2011b)
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This configuration factor in Eq. [2-34] is most often used for rectangular calculations. The factor can also be
calculated when the receiving surface is perpendicular to the radiating surface and when the receiving surface in a
plane at an angle π to the radiating surface. For the receiving surface perpendicular to the radiating surface the
configuration factor is given by
π =1
2π[tanβ1(π) β
1
(1+π2)0.5tanβ1 (
π
(1+π2)0.5)] , [A-3]
and for the receiving surface in a plane at an angle π to the radiating surface the configuration factor is written as
π =1
2π[tanβ1(π) β
(1βππππ π)
(1+π2β2ππππ π)0.5tanβ1 (
π
(1+π2β2ππππ π)0.5) +
ππππ π
(π2+sin2 π)0.5[tanβ1 (
πβπππ π
(π2+sin2π)0.5) +
tanβ1 (πππ π
(π2+sin2 π)0.5)]] . [A-4]
A.2. CYLINDER AS RADIATIOR
The configuration factor for a cylinder is written as (Hamilton & Morgan, 1952)
π =1
ππ·tanβ1 (
πΏ
βπ·2β1) +
πΏ
π[π΄β2π·
π·βπ΄π΅tanβ1 β
π΄(π·β1)
π΅(π·+1)β
1
π·tanβ1 β
π·β1
π·+1] , [A-5]
where
π· =π
π ,
πΏ =π
π ,
π΄ = (π· + 1)2 + πΏ2 ,
π΅ = (π· β 1)2 + πΏ2 .
Figure A-2. Receiving surface b in a plane parallel to that of the radiating surface a (CEN, 2011b)
Figure A-3. Configuration parameters for a cylindrical radiator with a parallel receiver (Hamilton & Morgan, 1952)
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A.3. ELLIPSE AS RADIATOR
For many practical fire engineering calculations the configuration factor for an ellipse can be used (Tanaka, 1999).
The equation for the configuration factor is written as
π =ππ
β(π 2+π2)(π 2+π2) , [A-6]
where π, π, and π can be found in the following figure. For the receiver at point A the equation above can be used.
If the heat is received at point B the configuration factor is halved.
Figure A-4. Configuration parameters for an elliptical radiator with a parallel receiver at A or B (Tanaka, 1999)
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B. DERIVATION SHAPE FUNCTIONS AND FINAL FEM FORMULATION
The linear shape functions for the local system can be defined with the following figure.
π1 = π1 + π2π ,
π1(β1) = π1 β π2 = 1 ,
π1(1) = π1 + π2 = 0 ,
π2 = β1
2 ,
π1 =1
2 ,
π1 =1
2β
1
2π =
1
2(1 β π) . [B-1]
π2 = π1 + π2π ,
π1(β1) = π1 β π2 = 0 ,
π1(1) = π1 + π2 = 1 ,
π2 =1
2 ,
π1 =1
2 ,
π1 =1
2+
1
2π =
1
2(1 + π) . [B-2]
Combining these shape functions results in the shape function for the temperature. It has to be noted that ππ₯ =πΏ
2ππ.
The shape functions and the first order derivative can then be formulated as
ππ(π) = [1
2(1 β π)
1
2(1 + π)] , [B-3]
ππ = ππ
ππ₯=
ππ
ππ
ππ
ππ₯= [β
1
2
1
2]
2
πΏ= [β1 1]
1
πΏ . [B-4]
The same method is used for the quadratic shape functions. Again Figure A-1 is used to come up with the local
shape functions.
π1 = π1 + π2π + π3π2 ,
π1(β1) = π1 β π2 + π3 = 1 ,
π1(0) = π1 = 0 ,
π1(1) = π1 + π2 + π3 = 0 ,
π2 = β1
2 ,
π3 =1
2 ,
π1 = β1
2π +
1
2π2 = β
1
2π(1 β π) . [B-5]
π2 = π1 + π2π + π3π2 ,
π2(β1) = π1 β π2 + π3 = 0 ,
π2(0) = π1 = 1 ,
π2(1) = π1 + π2 + π3 = 0 ,
π3 = β1 ,
π2 = 0 ,
π2 = 1 β π2 . [B-6]
Figure B-1. Global and local coordinate systems linear and quadratic (Eslami et al., 2013)
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 7
π3 = π1 + π2π + π3π2 ,
π3(β1) = π1 β π2 + π3 = 0 ,
π3(0) = π1 = 0 ,
π3(1) = π1 + π2 + π3 = 1 ,
π2 =1
2 ,
π3 =1
2 ,
π3 =1
2π +
1
2π2 =
1
2π(1 + π) . [B-7]
Again it has to be noted that ππ₯ =πΏ
2ππ. The combined shape function for the displacement becomes
ππ’(π) = [β1
2π(1 β π) 1 β π2
1
2π(1 + π)] , [B-8]
ππ’ = ππ
ππ₯=
ππ
ππ
ππ
ππ₯= [β
1
2+ π β 2π
1
2+ π ]
2
πΏ
= [β1 + 2π β4π 1 + 2π]1
πΏ . [B-9]
Now the matrices can be solved using these shape functions. This gives
ππ,π’(π) = π0 β« ππ
Tπ½ππ’π΄ππ₯πΏπ
0 =πΏ
2π0 β« ππ
Tπ½ππ’π΄ππ1
β1 ,
ππ,π’(π) =
π0π½
2β« [
1
2(1 β π)
01
2(1 + π)
] [β1 + 2π β4π 1 + 2π]π΄ππ =1
β1
π0π½
2β« [
β1
2+ 1
1
2π β π2 β2π + 2π2
0 0
1
2+
1
2π β π2
0
β1
2+
1
2π + π2 β2π β 2π2
1
2+ 1
1
2π + π2
] π΄ππ1
β1 =π0π½π΄
6[β5 4 10 0 0
β1 β4 5] . [B-10]
ππ,π(π) = β« ππ
TπΆπΈπππ΄ππ₯πΏπ
0 =πΏ
2β« ππ
TπΆπΈπππ΄ππ1
β1 ,
ππ,π(π) =
πΏπΆπΈ
2β« [
1
2(1 β π)
1
2(1 + π)
] [1
2(1 β π)
1
2(1 + π)] π΄ππ =
1
β1
πΏπΆπΈ
2β« [
1
4β
1
2π +
1
4π2 1
4β
1
4π2
1
4β
1
4π2 1
4+
1
2π +
1
4π2
] π΄ππ =1
β1
πΏπΆπΈπ΄
6[2 11 2
] =πΏπΆπΈπ΄
6[20
00
10
1 0 2] . [B-11]
ππ’,π’(π) = β« ππ’
TCπ’ππ’π΄ππ₯πΏπ
0 =πΏ
2β« ππ’
TCπ’ππ’π΄ππ1
β1 ,
ππ’,π’(π) =
πΆπ
2πΏβ« [
β1 + 2πβ4π
1 + 2π] [β1 + 2π β4π 1 + 2π]π΄ππ
1
β1 =
πΆπ
2πΏβ« [
1 β 4π + 4π2 4π β 8π2 β1 + 4π2
4π β 8π2 16π2 β4π β 8π2
β1 + 4π2 β4π β 8π2 1 β 4π + 4π2
] A ππ1
β1 =πΆππ΄
6πΏ[
14β16
β1632
2β16
2 β16 14] . [B-12]
ππ’,π(π) = β« ππ’
Tπ½πππ΄ππ₯πΏπ
0 =πΏ
2β« ππ’
Tπ½πππ΄ππ1
β1 ,
ππ’,π(π) =
π½
2β« [
β1 + 2πβ4π
1 + 2π] [
1
2(1 β π) 0
1
2(1 + π)] π΄ππ =
1
β1
π½
2β«
[ β
1
2+ 1
1
2π β π2 0
β2π + 2π2 0
β1
2+
1
2π + π2
β2π β 2π2
1
2+
1
2π β π2 0
1
2+ 1
1
2π + π2
]
π΄ππ =π½π΄
6
1
β1 [β5 0 β14 0 β41 0 5
] . [B-13]
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R.H.A. TITULAER 8
ππ,π(π) = β« ππ
TKππππ΄ππ₯πΏπ
0 =πΏ
2β« ππ
TKππππ΄ππ1
β1 ,
ππ,π(π) =
Kπ
2πΏβ« [
β11
] [β1 1]π΄ππ1
β1 =Kπ
2πΏβ« [
1 β1β1 1
] π΄ππ1
β1 =Kππ΄
πΏ[
1 β1β1 1
]
=Kππ΄
πΏ[
10
00
β10
β1 0 1] . [B-14]
ππ,π(ππππππ‘πππ)(π) = β« (ππ
Tβππππππ₯
πΏπ
0 =πΏ
2β« (ππ
Tβππππππ
1
β1 ,
ππ,π(ππππππ‘πππ)(π) =
πΏβπ
2β« [
1
2(1 β π)
1
2(1 + π)
] [1
2(1 β π)
1
2(1 + π)] πππ =
1
β1
πΏβπ
2β« [
1
4β
1
2π +
1
4π2 1
4β
1
4π2
1
4β
1
4π2 1
4+
1
2π +
1
4π2
]πππ =1
β1
πΏβππ
6[2 11 2
] =πΏβππ
6[20
00
10
1 0 2] . [B-15]
ππ,π(ππππ£πππ‘πππ)(π) = β« (ππ
Tβππππππ₯
πΏπ
0 =πΏ
2β« (ππ
Tβππππππ
1
β1 ,
ππ,π(ππππ£πππ‘πππ)(π) =
πΏβπ
2β« [
1
2(1 β π)
1
2(1 + π)
] [1
2(1 β π)
1
2(1 + π)]πππ =
1
β1
πΏβπ
2β« [
1
4β
1
2π +
1
4π2 1
4β
1
4π2
1
4β
1
4π2 1
4+
1
2π +
1
4π2
] πππ =1
β1
πΏβππ
6[2 11 2
] =πΏβππ
6[20
00
10
1 0 2] . [B-16]
ππ’(π) = ππ’
TοΏ½ΜοΏ½π|0πΏπ
=πΏ
2ππ’
TοΏ½ΜοΏ½π|β11 =
πΏ
2οΏ½ΜοΏ½π
[ β
1
2π(1 β π)
1 β π2
1
2π(1 + π) ]
|β11
=πΏ
2οΏ½ΜοΏ½π {
β101
} . [B-17]
ππ(π) = βππ
TοΏ½ΜοΏ½π|0πΏπ
= βπΏ
2ππ
TοΏ½ΜοΏ½π|β1
1 = βπΏ
2οΏ½ΜοΏ½π [
1
2(1 β π)
1
2(1 + π)
] |β11 = β
πΏ
2οΏ½ΜοΏ½π {
β11
}
= βπΏ
2ποΏ½ΜοΏ½{
β11
} . [B-18]
ππ(ππππππ‘πππ)(π) = β« ππ
Tβπ(ππ΄ππ(π‘))πππ₯πΏπ
0 = πΏβπππ΄ππ
2β« [
1
2(1 β π)
1
2(1 + π)
] πππ₯πΏπ
0
=πΏβπππ΄πππ
2{11} . [B-19]
ππ(ππππ£πππ‘πππ)(π) = β« ππ
Tβπ(ππ΄ππ(π‘))πππ₯πΏπ
0 = πΏβπππ΄ππ
2β« [
1
2(1 β π)
1
2(1 + π)
]πππ₯πΏπ
0
=πΏβπππ΄πππ
2{11} . [B-20]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 9
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 10
C. DERIVATION EULER-BERNOULLI STIFFNESS MATRICES
The conventional stiffness matrix can be found with
π(m) = β« π(m) TπΈπΌπ(m) ππ₯β(π) =
44434241
34333231
24232221
14131211
kkkk
kkkk
kkkk
kkkk
, [C-1]
π11 β πΈπΌ (36
π4β
144π₯
π5+
144π₯2
π6) β πΈπΌ [
36π₯
π4β
72π₯2
π5+
48π₯3
π6]0
π
=12πΈπΌ
π3 , [C-2]
π12 β πΈπΌ (β24
π3+
36π₯
π4+
48π₯
π4β
72π₯2
π5) β πΈπΌ [
β24π₯
π3+
18π₯2
π4+
24π₯2
π4β
24π₯3
π5]0
π
=β6πΈπΌ
π2 , [C-3]
π13 β πΈπΌ (β36
π4+
144π₯
π5β
144π₯2
π6) β πΈπΌ [
β36π₯
π4+
72π₯2
π5β
48π₯3
π6]0
π
=β12πΈπΌ
π3 , [C-4]
π14 β πΈπΌ (β12
π3+
36π₯
π4+
24π₯
π4β
72π₯2
π5) β πΈπΌ [
β12π₯
π3+
18π₯2
π4+
12π₯2
π4β
24π₯3
π5]0
π
=β6πΈπΌ
π2 , [C-5]
π21 β πΈπΌ (β24
π3+
36π₯
π4+
48π₯
π4β
72π₯2
π5) β πΈπΌ [
β24π₯
π3+
18π₯2
π4+
24π₯2
π4β
24π₯3
π5]0
π
=β6πΈπΌ
π2 , [C-6]
π22 β πΈπΌ (16
π2β
48π₯
π3+
36π₯2
π4) β πΈπΌ [
16π₯
π2β
24π₯2
π3+
12π₯3
π4]0
π
=4πΈπΌ
π , [C-7]
π23 β πΈπΌ (24
π3β
36π₯
π4β
48π₯
π4+
72π₯2
π5) β πΈπΌ [
24π₯
π3β
18π₯2
π4β
24π₯2
π4+
24π₯3
π5]0
π
=6πΈπΌ
π2 , [C-8]
π24 β πΈπΌ (8
π2β
36π₯
π3+
36π₯2
π4) β πΈπΌ [
8π₯
π2β
18π₯2
π3+
12π₯3
π4]0
π
=2πΈπΌ
π , [C-9]
π31 β πΈπΌ (β36
π4+
144π₯
π5β
144π₯2
π6) β πΈπΌ [
β36π₯
π4+
72π₯2
π5β
48π₯3
π6]0
π
=β12πΈπΌ
π3 , [C-10]
π32 β πΈπΌ (24
π3β
36π₯
π4β
48π₯
π4+
72π₯2
π5) β πΈπΌ [
24π₯
π3β
18π₯2
π4β
24π₯2
π4+
24π₯3
π5]0
π
=6πΈπΌ
π2 , [C-11]
π33 β πΈπΌ (36
π4β
144π₯
π5+
144π₯2
π6) β πΈπΌ [
36π₯
π4β
72π₯2
π5+
48π₯3
π6]0
π
=12πΈπΌ
π3 , [C-12]
π34 β πΈπΌ (12
π3β
36π₯
π4β
24π₯
π4+
72π₯2
π5) β πΈπΌ [
12π₯
π3β
18π₯2
π4β
12π₯2
π4+
24π₯3
π5]0
π
=6πΈπΌ
π2 , [C-13]
π41 β πΈπΌ (β12
π3+
36π₯
π4+
24π₯
π4β
72π₯2
π5) β πΈπΌ [
β12π₯
π3+
18π₯2
π4+
12π₯2
π4β
24π₯3
π5]0
π
=β6πΈπΌ
π2 , [C-14]
π42 β πΈπΌ (8
π2β
36π₯
π3+
36π₯2
π4) β πΈπΌ [
8π₯
π2β
18π₯2
π3+
12π₯3
π4]0
π
=2πΈπΌ
π , [C-15]
π43 β πΈπΌ (12
π3β
36π₯
π4β
24π₯
π4+
72π₯2
π5) β πΈπΌ [
12π₯
π3β
18π₯2
π4β
12π₯2
π4+
24π₯3
π5]0
π
=6πΈπΌ
π2 , [C-16]
π44 β πΈπΌ (4
π2β
24π₯
π3+
36π₯2
π4) β πΈπΌ [
4π₯
π2β
12π₯2
π3+
12π₯3
π4]0
π
=4πΈπΌ
π , [C-17]
π(m) =πΈπΌ
π3
22
22
4626
6126122646
612612
llll
llllll
ll
. [C-18]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 11
The stress stiffness matrix can be found with
ππ(π¦)
= β« π(m) TοΏ½ΜοΏ½π(m)ππ₯β(π) =
44434241
34333231
24232221
14131211
ksksksks
ksksksks
ksksksks
ksksksks
, [C-19]
ππ 11 β οΏ½ΜοΏ½ (36π₯2
π4β
72π₯3
π5+
36π₯4
π6) β οΏ½ΜοΏ½ [
12π₯3
π4β
18π₯4
π5+
36π₯5
5π6]0
π
=36οΏ½ΜοΏ½
30π, [C-20]
ππ 12 β οΏ½ΜοΏ½ (6π₯
π2β
30π₯2
π3+
42π₯3
π4β
18π₯4
π5) β οΏ½ΜοΏ½ [
3π₯2
π2β
10π₯3
π3+
42π₯4
4π4β
18π₯5
5π5]0
π
=β3οΏ½ΜοΏ½
30 , [C-21]
ππ 13 β οΏ½ΜοΏ½ (β36π₯2
π4+
72π₯3
π5β
36π₯4
π6) β οΏ½ΜοΏ½ [
β12π₯3
π4+
18π₯4
π5β
36π₯5
5π6]0
π
=β36οΏ½ΜοΏ½
30π , [C-22]
ππ 14 β οΏ½ΜοΏ½ (β12π₯2
π3+
30π₯3
π4β
18π₯4
π5) β οΏ½ΜοΏ½ [
β4π₯3
π3+
30π₯4
4π4β
18π₯5
5π5]0
π
=β3οΏ½ΜοΏ½
30 , [C-23]
ππ 21 β οΏ½ΜοΏ½ (6π₯
π2β
30π₯2
π3+
42π₯3
π4β
18π₯4
π5) β οΏ½ΜοΏ½ [
3π₯2
π2β
10π₯3
π3+
42π₯4
4π4β
18π₯5
5π5]0
π
=β3οΏ½ΜοΏ½
30, [C-24]
ππ 22 β οΏ½ΜοΏ½ (1 β8π₯
π+
22π₯2
π2β
24π₯3
π3+
9π₯4
π4) β οΏ½ΜοΏ½ [π₯ β
4π₯2
π+
22π₯3
3π2β
6π₯4
π3+
9π₯5
5π4]0
π
=4οΏ½ΜοΏ½π
30 , [C-25]
ππ 23 β οΏ½ΜοΏ½ (β6π₯
π2+
30π₯2
π3β
42π₯3
π4+
18π₯4
π5) β οΏ½ΜοΏ½ [
β3π₯2
π2+
10π₯3
π3β
42π₯4
4π4+
18π₯5
5π5]0
π
=3οΏ½ΜοΏ½
30 , [C-26]
ππ 24 β οΏ½ΜοΏ½ (β2π₯
π+
11π₯2
π2β
18π₯3
π3+
9π₯4
π4) β οΏ½ΜοΏ½ [β
π₯2
π+
11π₯3
3π2β
18π₯4
4π3+
9π₯5
5π4]0
π
=βοΏ½ΜοΏ½π
30 , [C-27]
ππ 31 β οΏ½ΜοΏ½ (β36π₯2
π4+
72π₯3
π5β
36π₯4
π6) β οΏ½ΜοΏ½ [
β12π₯3
π4+
18π₯4
π5β
36π₯5
5π6]0
π
=β36οΏ½ΜοΏ½
30π , [C-28]
ππ 32 β οΏ½ΜοΏ½ (β6π₯
π2+
30π₯2
π3β
42π₯3
π4+
18π₯4
π5) β οΏ½ΜοΏ½ [
β3π₯2
π2+
10π₯3
π3β
42π₯4
4π4+
18π₯5
5π5]0
π
=3οΏ½ΜοΏ½
30 , [C-29]
ππ 33 β οΏ½ΜοΏ½ (36π₯2
π4β
72π₯3
π5+
36π₯4
π6) β οΏ½ΜοΏ½ [
12π₯3
π4β
18π₯4
π5+
36π₯5
5π6]0
π
=36οΏ½ΜοΏ½
30π , [C-30]
ππ 34 β οΏ½ΜοΏ½ (12π₯2
π3β
30π₯3
π4+
18π₯4
π5) β οΏ½ΜοΏ½ [
4π₯3
π3β
30π₯4
4π4+
18π₯5
5π5]0
π
=3οΏ½ΜοΏ½
30 , [C-31]
ππ 41 β οΏ½ΜοΏ½ (β12π₯2
π3+
30π₯3
π4β
18π₯4
π5) β οΏ½ΜοΏ½ [
β4π₯3
π3+
30π₯4
4π4β
18π₯5
5π5]0
π
=β3οΏ½ΜοΏ½
30 , [C-32]
ππ 42 β οΏ½ΜοΏ½ (β2π₯
π+
11π₯2
π2β
18π₯3
π3+
9π₯4
π4) β οΏ½ΜοΏ½ [β
π₯2
π+
11π₯3
3π2β
18π₯4
4π3+
9π₯5
5π4]0
π
=βοΏ½ΜοΏ½π
30 , [C-33]
ππ 43 β οΏ½ΜοΏ½ (12π₯2
π3β
30π₯3
π4+
18π₯4
π5) β οΏ½ΜοΏ½ [
4π₯3
π3β
30π₯4
4π4+
18π₯5
5π5]0
π
=3οΏ½ΜοΏ½
30 , [C-34]
ππ 44 β οΏ½ΜοΏ½ (4π₯2
π2β
12π₯3
π3+
9π₯4
π4) β οΏ½ΜοΏ½ [
4π₯3
3π2β
3π₯4
π3+
9π₯5
5π4]0
π
=4οΏ½ΜοΏ½π
30 , [C-35]
ππ(π¦)
=οΏ½ΜοΏ½
30π
22
22
433336336
343336336
llllllllllll
. [C-36]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 12
D. ELEMENT MATRICES THERMAL NON-LINEARITY
For one quadratic element the formulation can be written as
ποΏ½ΜοΏ½ + ππ = π , [D-1]
k
k
k
k
j
j
j
j
i
i
i
i
EE
EE
w
u
w
u
w
u
ALcATATALcAT
ALcATATALcAT
300
6
5000
6
4
600
6
0000000000000000000000000000000000000000000000000000000000000000000000000000000000006
006
0006
4
300
6
5000000000000000000000000000000000000
000
000
2
2
0000
2
2
0000000000
0824
00424
00000
02496
002496
00000
6
500
6
14000
6
16
600
6
2000000000000
0424
0016
000424
0
02496
000192
002496
0
6
400
6
16000
6
32
6
400
6
16
0000000000
00000424
00824
0
000002496
002496
0
600
6
2000
6
16
6
500
6
14
22
2323
22
23323
22
2323
i
k
k
i
i
i
i
i
k
k
k
k
j
j
j
j
i
i
i
i
QPL
M
V
tPL
QPL
M
V
tPL
w
u
w
u
w
u
l
AK
l
AKl
EI
l
EI
l
EI
l
EIl
EI
l
EI
l
EI
l
EI
A
l
CuA
l
CuAA
l
CuA
l
EI
l
EI
l
EI
l
EI
l
EIl
EI
l
EI
l
EI
l
EI
l
EI
A
l
CuA
l
CuAA
l
CuAl
AK
l
AKl
EI
l
EI
l
EI
l
EIl
EI
l
EI
l
EI
l
EI
A
l
CuA
l
CuAA
l
CuA
. [D-2]
However, the terms π’, π€, and π have to be transformed to the local coordinate axis. The local displacement vector
can be rewritten to the global displacement vector by performing the following multiplication
π β² = ππ , [D-3]
where π is the transformation matrix and π is the global displacement vector without the thermal components. The
transformation matrix is given by
3
3
3
2
2
2
1
1
1
'
3
'
3
'
3
'
2
'
2
'
2
'
1
'
1
'
1
100000000000000000000000001000000000000000000000000010000000000000000
w
u
w
u
w
u
cssc
cssc
cssc
w
u
w
u
w
u
. [D-4]
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 13
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 14
E. CRITICAL TEMPERATURES OF STEEL MEMBERS WITH STEEL GRADE S275
Table E-1 Critical temperatures of steel members with steel grade S275 based on the non-dimensional slenderness in the fire situation and degree of utilisation (Vassart et al., 2014).
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 15
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 16
F. MATLAB TUTORIAL
Matlab is a numerical computing environment developed by MathWorks. This software is used to analyse the
coupled thermomechanical behaviour. In this chapter a small tutorial is given for the Matlab code written for this
graduation project.
F.1. COUPLED THERMOMECHANICAL ANALYSIS
The first part of the Matlab code can be seen in Figure F-1. Lines 1-11 prescribe some general information about the
code, the date it was firstly adjusted, and the date it was lastly revised. Line 13 removes all items from the workspace
in Matlab, freeing up system memory. It also closes all windows that are still open, such as figures and tables, and
clears the command window.
Part 2 prescribes the mesh generation of the element. In line 16 the length of the linear bar element can be prescribed.
Line 17 gives the number of nodes, which divides the linear bar element into smaller elements. Then lines 18 -20
create the element matrix, for this graduation project a quadratic element matrix. This element matrix is then
included into the database in line 21. Line 22 calculates the number of elements, and for this code you cannot use
less than 3 elements, or you have to change other factors in the code. Line 23 gives the nodal coordinates of the
whole element, and line 24 gives the distances between the nodes (Ξx).
The next part describes the input for all the parameters. The most important parameters for the coupled
thermomechanical behaviour can be found in line 27, the thermal conductivity, line 28, the Youngβs Modulus, line
29, the Poisson ratio, line 33, the thermal linear expansion coefficient, line 35, the specific heat, line 43, the time step,
and in line 44 the time limit. These parameters have a significant influence on the coupled behaviour.
Figure F-2. Mesh generation
Figure F-1. General information
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 17
Then the boundary conditions can be prescribed. Either the traction and flux are given, the displacement and flux,
the displacement and temperature, the traction and temperature, or the force and displacement.
In line 48 the boundary conditions can be set. Lines 50-54 describe the boundary conditions which can be
implemented in line 48. Lines 58-78 can be used to apply the boundary values.
Figure F-3. Input parameters
Figure F-4. Boundary conditions
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 18
The next part of the code produces empty global matrices, which will be filled with values in the following parts of
the code. Lines 81-91 are to form the matrices for every single part of the total equation.
Figure F-6. Element stiffness matrices
Then the global matrices can be filled with the element matrices. The first part of this part 6 of the Matlab code
determines the element matrices, which can be seen in lines 95-101. Lines 103-109 assemble these element matrices.
The function LinearBarElementX() and LinearBarAssembleX() are prescribed in sub codes. Two sub codes will be
explained. In Figure F-7 the first sub code can be seen, which is called in the main code in line 95. This sub code
forms the element matrix.
Figure F-5. Global stiffness matrices
Figure F-7. Linear bar element stiffness
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 19
Line 103 then assembles these element matrices into the global matrix. This sub code can be seen in Figure F-8.
When all the matrices are assembled, the total matrix can be formed by using the following part of the code. Since
the mathematical formulation of the matrices consist of four parts, the following part of the Matlab code adds all
the four parts into one matrix. This means that lines 112-114 add the displacement part (K_UU) on the top left side
of the total stiffness matrix, the thermal part (K_TT) on the bottom right side of the total stiffness matrix, and the
thermomechanical part (K_UT) on the top right side of the total stiffness matrix.
Figure F-8. Linear bar assemblage
Figure F-9. Assemblage matrices
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 20
Since the temperature is prescribed linearly and the displacement is prescribed quadratic, the rows and columns
which are zero can be deleted for the thermal part. The same procedure can be executed for the capacity matrix
from lines 121-127, the force vector from lines 129-135, and for the added matrix οΏ½ΜοΏ½ from lines 137-145.
The following part describes the boundary conditions from lines 48-54. If the traction and flux are prescribed, the
number 1 can be added in line 48 and then lines 148-162 describe the applied boundary conditions. Lines 149-150
create the empty component values UT, which contain the unknown values. Then the traction and flux can be
applied.
The next part describes the option when the displacement and flux are prescribed. This can be seen in Figure A-11.
Lines 166-167 create again the empty component values UT, which contain again the unknown values. Then the
displacement and flux can be applied and a trick is used in lines 170-171 described in paragraph 6.8.
The following boundary condition is the displacement and temperature. This boundary condition is used in the
analysis of the Matlab code and the benchmark with ABAQUS in paragraph 6.9. Lines 189 -192 create the empty
component values UT, which contain the unknown values (the displacements and temperatures for the nodes in
between the boundaries). This can be seen in Figure F-12.
Figure F-10. Applying boundary condition traction/flux
Figure F-11. Applying boundary condition displacement/flux
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 21
Figure F-12. Applying boundary condition displacement/temperature
The next boundary condition is the traction and temperature, where lines 215-218 create the empty component
values UT, which contain again the unknown values. This boundary condition is number 4 and can again be added
into line 48. Lines 220-221 prescribe the traction and lines 225-231 prescribe the temperature. This boundary
condition can be seen in Figure F-13.
Finally, the last boundary condition is the force and displacement, which is a purely mechanical problem. This
boundary condition was added to check the code for errors. The temperature distribution is taken as the ambient
temperature, so the temperature has no influence on the displacements. Therefore, a force can be applied on one
boundary side, and the other side can be constrained. The result will give a linear elastic expansion, which can be
mathematically calculated.
Figure F-13. Applying boundary condition traction/temperature
Figure F-14. Applying boundary condition force/displacement
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 22
In Figure F-15 the data container is formed. Line 261 creates the force data, when for example a changing boundary
condition is used over time. Line 262 creates the unknown components data, which is used as a result for plotting
graphs. Then line 263 and line 264 are used to calculate the thermal stresses, where line 263 is the delta temperature,
and line 264 are the thermal stresses. Line 265 is used for the calculation of the unknown components. Lines 266 -
268 create data containers for the applied boundary conditions. The conditions are prescribed in lines 271-312.
However, these containers are not of importance for the calculation.
The last part of the code contains the calculation and time step loop. Lines 314-315 and 337 create the loop, where
every new unknown values are used in line 315 for the following time step, which is created in line 337. The most
important line of this code is line 325, which is the backward difference method calculation of the coupled
thermomechanical behaviour. All the other equations can also be used, but they use different time stepping
schemes. Finally, lines 331-335 calculate the thermal stresses.
Code: LinearBarHeatSteadyExample_Transient_ThermomechanicalCoupling_3.m
Figure F-15. Data container
Figure F-16. Calculation and plotting
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 23
F.2. LINEAR BUCKLING
For the linear buckling analysis information and mesh generation is the same as for the coupled thermomechanical
analysis found in paragraph F.1. The next part is the input parameters for the linear buckling analysis. This can be
seen in Figure F-17. In this figure also the boundary conditions are explained.
Then the global and element stiffness matrices are determined. First the global conventional and stress stiffness
matrices in lines 44 and 45, followed by the element stiffness matrices and the assemblage in lines 48-56.
The conventional and stress stiffness matrices are found in Figure F-19.
Figure F-17. Input parameters and boundary conditions
Figure F-18. Global stiffness matrix and element stiffness matrices with assemblage
Figure F-19. Conventional and stress stiffness matrix
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 24
The conventional and stress stiffness matrices are assembled using the linear bar assemblage from Figure F-20.
The next step is to calculate the eigenvalues and buckling loads, which is described in lines 59-80.
Codes: LinearBuckling_SimplysupportedRitz_1.m
LinearBuckling_SimplysupportedGalerkin_1.m
Figure F-20. Assemblage stiffness matrices
Figure F-21. Eigenvalues calculation
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 25
The linear buckling calculation can then be implemented into the Matlab code for the coupled thermomechanical
analysis. The forces implemented into the buckling calculation can be calculated according to lines 307 -315.
For each element now a different force can be implemented. This is shown in line 340, in which the force is visible
for each element in the data container FORCE. Now again the linear buckling procedure can be performed in order
to calculate the eigenvalues.
By using lines 409-580 the buckling modes can be calculated.
Code: Transient_ThermomechanicalCoupling_GALERKIN_3_THERMALMECHANICAL.m
The reduced Youngβs modulus is defined in the following figure. It can be seen that for each temperature found in
the nodes of the model, the temperature is averaged over the elements and then the accompanying Youngβs
modulus is calculated in lines 315-343.
Code: Transient_ThermomechanicalCoupling_GALERKIN_4_THERMECH_Ereduc.m
Figure F-22. Force calculation for buckling analysis
Figure F-23. Determining element matrices and assemblage
Figure F-24. Youngβs modulus reduction
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 26
F.3. NON-LINEAR BUCKLING
The non-linear buckling analysis is described for the Newton-Raphson method and Arc-length method.
F.3.1. NEWTON-RAPHSON METHOD
First the Newton-Raphson Method code is explained. Again the general information can be found in the first lines,
which is the same as for the other codes. In lines 15-31 the mesh is generated and the coordinates are described.
In the following lines 32-39 the input parameters are described. So in this section you can adjust the profiles you
want to investigate. The initial imperfection is the most important parameter in this code, which can be seen in line
37.
In Figure F-27 the geometry is defined in which the element lengths are defined in lines 43-45. Furthermore some
data containers are generated.
In the next section the angles are defined for each element. First a data container is defined in line 52 and 53. Then
the angles are calculated for each element. The angles are later on used for the transformation matrix.
Figure F-25. Mesh generation
Figure F-26. Input parameters
Figure F-27. Geometry and some data containers
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 27
Then the boundary conditions are described in lines 74-79.
In the following section the data containers are described. These containers are empty, however needed for the
calculations.
Figure F-28. Define angles
Figure F-29. Boundary conditions
Figure F-30. Data containers
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 28
Now the Newton-Raphson calculation can be performed. In lines 105-113 the mechanical force is applied on the
element and the tolerance and norm are set. These are needed for the iterations in the next step. Line 104 describes
the start of the load increments (10), the increment size (100), and the increment limit (1810).
Figure F-31. Increments Newton-Raphson procedure
The Newton-Raphson uses iterations to calculate the non-linear behaviour. In line 116 the convergence
requirements are set as while the tolerance is larger than the norm, the iteration procedure should continue. Also a
maximum number of iterations is implemented in order to overcome endless calculation time or errors. Lines 117 -
120 describe the settings for each iteration, which shows that every iteration uses the values calculated in the
previous iteration. Lines 122-132 determine the element matrices and global matrices. In line 125 the transformation
can be seen which is applied on the conventional stress stiffness matrix in line 129. The angles calculated in Figure
F-28 are used for the transformation in line 125. The element matrices and assemblage are calculated according to
the same procedures as used before.
Using the matrices and the residue, the displacement vector can be calculated. In lines 135-142 first the boundary
conditions are applied. Then the internal forces are calculated in line 146. The residue is calculated in line 149.
Using the residue and the conventional stiffness matrix, the displacement vector can be calculated using lines 152
and 153. The total displacement vector is calculated in lines 155-160.
Figure F-32. Iteration settings and determination of element matrices and assemblage
Graduation Thesis University of Technology Eindhoven
R.H.A. TITULAER 29
After the first iteration, the model consists of a new geometry, which is used for the second iteration. The updated
angles and element lengths are calculated in lines 163-180. These new values are used in the next iteration.
Then the convergence parameters are calculated in lines 187-196 and used for the next iteration. If the convergence
is visible, the calculation is stopped and the values are put into data containers in lines 200 and 201.
Code: NonLinear_18_NEWTON_RAPHSON_NEW_good.m
Figure F-33. Boundary conditions and calculation of displacement vector
Figure F-34. Preparing next iteration
Figure F-35. Convergence check
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F.3.2. ARC-LENGTH METHOD
Secondly the Arc-length method is described and explained. Up until line 110 the input is the same as for the
Newton-Raphson method. Then the first increment is described in order to get a first estimation of the load. From
line 110-201 the Newton-Raphson method is used to estimate the first solution. Using the first solution the Arc-
length method can be performed. In line 202-204 the important parameters for the Arc-length method are described.
Lines 208-217 describe some data.
The next step is to perform iterations. In line 220 the maximum number of iterations is implemented (100).
Furthermore lines 224 and 225 give the convergence parameters and line 227 gives the iteration values.
Then the iteration procedure is started. If k = 1 in line 231 indicates the second increment, which is needed to
perform the analysis. Lines 234-245 set some data and values used for new iterations.
Figure F-36. Important data Arc-length method
Figure F-37. Increments
Figure F-38. Iteration procedure 1 with data
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The element matrices and assemblage procedure is the same as for the Newton-Raphson method. This is shown in
Figure F-39.
The calculation procedure of the Arc-length method can be seen in the following figure. First the conventional
stiffness matrix is defined with the boundary conditions. Then the internal force is calculated and the displacement
vector components I and II, which are defined in paragraph 2.5.2.3. Lines 274-276 describe the additional equation
g. The most important part of the Arc-length method can be found in lines 278-280, in which the lambda is calculated
(Fafard, 1993). This value is then used to calculate the updated lamdbaβs and updated displacement vectors in lines
282-285 and 287-291 respectively.
Figure F-39. Determination of element matrices and assemblage
Figure F-40. Boundary conditions and calculation of lambda and displacement vector
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The third and following increments can be calculated with the same procedure and sections as described above.
However, the lambda is calculated differently as can be seen in line 355. Then the convergence is checked and the
calculation is terminated.
Code: NonLinear_19_ARCLENGTH_NEW.m
Figure F-41. Boundary conditions and calculation of lambda and displacement vector rest increments
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F.4. CODE DESCRIPTION
Every code should be used and opened in the folder it appears in, so the accompanying files can be applied in the
code. In the following table the used Matlab codes can be seen.
Table F-1. Matlab codes descriptions
CODE USED FOR
LINEARBARHEATSTEADYEXAMPLE_TRANSIENT_THERMOMECHANICALCOUPLING_3.M Coupled thermomechanical
analysis
LINEARBUCKLING_SIMPLYSUPPORTEDRITZ_1.M Linear buckling according to
Ritz approximation
LINEARBUCKLING_SIMPLYSUPPORTEDGALERKIN_1.M Linear buckling according to
Galerkin approximation
TRANSIENT_THERMOMECHANICALCOUPLING_GALERKIN_3_THERMALMECHANICAL.M Linear coupled
thermomechanical buckling
Galerkin approximation
TRANSIENT_THERMOMECHANICALCOUPLING_GALERKIN_4_THERMECH_EREDUC.M Linear coupled
thermomechanical buckling
Galerkin approximation with
Youngβs modulus reduction
NONLINEAR_18_NEWTON_RAPHSON_NEW_GOOD.M Non-linear buckling Newton-
Raphson method mechanically
NONLINEAR_19_ARCLENGTH_NEW.M Non-linear buckling Arc-length
method mechanically
TRANSIENT_THERMOMECHANICALCOUPLING_NONLINEAR_15_NR_WORKS.M Non-linear coupled
thermomechanical analysis
Newton-Raphson
EC.M Eurocode steel temperature
development
EC_INSULATION.M Eurocode steel temperature
development with insulation
TRANSIENT_THERMOMECHANICALCOUPLING_NEWBOUNDARY_EC_PAPERNLR_GOOD.M Coupled thermomechanical
behaviour comparing with
Eurocode