eindhoven university of technology master engineering

Eindhoven University of Technology

MASTER

Engineering model for coupled thermomechanical behaviour of steel elements under fireconditions

Titulaer, R.H.A.

Award date:2016

Link to publication

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https://research.tue.nl/en/studentTheses/d8b5455d-850c-4d41-a129-5081045d1e18

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


R.H.A. TITULAER II

ACKNOWLEDGEMENTS

This report comprises the final result of my master’s 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


R.H.A. TITULAER III

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.


R.H.A. TITULAER IV

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

�̇�𝒆 Rate of change of the elastic strain

Ε𝑖𝑗 Strain tensor

Ε𝑖𝑗𝑒 Elastic strain

Ε𝑖𝑗𝑡ℎ Thermal strain

Gr Grashof number

In Intensity of the thermal radiation in the normal direction to the emitting surface

Iθ Intensity of thermal radiation in the direction at an angle θ

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

𝑓𝑖 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

𝑟 Body heat source per unit volume

t Time

𝑢 Displacement x direction

w Displacement y direction

x,y,z Directions


R.H.A. TITULAER V

GREEK UPPER CASE SYMBOLS

Φ Entropy flux

𝚺 Stress

GREEK LOWER CASE SYMBOLS

𝛼 Thermal diffusivity

𝛼 Linear coefficient of thermal expansion

𝛽 Coefficient of thermal expansion of a fluid

𝛽 Parameter in the relation for relative buckling resistance

γ Modification factor

𝛿 Variation of variable

ε Emissivity

𝜖̇ Rate of change of the internal energy density

𝜂 Entropy

𝜂𝑒 Elastic entropy

𝜂𝑡ℎ Thermal entropy

𝜃 Temperature

𝜃𝑎 Steel temperature

𝜃0 Ambient temperature

λ Thermal conductivity

λ Lamé constant

μ Absolute viscosity

μ Lamé constant

𝜌 Density

𝜌0 Referential mass density

σ Stefan-Boltzmann constant

τ Transmissivity

𝜙 View factor

ψ Helmholtz energy density


R.H.A. TITULAER 0

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


R.H.A. TITULAER 1

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


R.H.A. TITULAER 2

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 Young’s 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


R.H.A. TITULAER 3


R.H.A. TITULAER 4

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.


R.H.A. TITULAER 5


R.H.A. TITULAER 6

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.


R.H.A. TITULAER 7

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:

‘What is the coupled thermomechanical behaviour of steel structural elements under fire conditions?’

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)


R.H.A. TITULAER 8

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.


R.H.A. TITULAER 9


R.H.A. TITULAER 10

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 room’s 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 fire’s 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)


R.H.A. TITULAER 11

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 350°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 600°C)

Fully developed fire After flashover everything is burning and the fire is

fully developed (900-1200°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)


R.H.A. TITULAER 12

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 plume’s 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 plume’s

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)


R.H.A. TITULAER 13

2.1.2. TEMPERATURE-TIME CURVES

A fire can thus be described by a fire curve, which represents the fire’s 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 𝜃𝑔 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

𝜃𝑔= 293 + 345 · 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

𝜃𝑔= 293 + 1280 (1 – 0,325 · e-0,167 t – 0,675 · 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

𝜃𝑔 = 293 + 660 (1 – 0,687 · e-0,32 t – 0,313 · 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

𝜃𝑔= 293 + 154 t0,25 for 0 < t ≤ 21, [2-4]

𝜃𝑔= 293 + 345 log(8(t – 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 (Sundström &

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

𝜃

[K]

Time [min]

Hydrocarbon curve

ISO-834

Slow heating curve

External fire exposure curve


R.H.A. TITULAER 14

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 – 3,33 · 10-2 𝜃𝑎 for 293 K < 𝜃𝑎 ≤ 1073 K, [2-6]

k = 27,3 for 𝜃𝑎 > 1073 K. [2-7]

where 𝜃𝑎 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 ∙10-1 𝜃𝑎 - 1,69∙ 10-2 𝜃𝑎2 + 2,22 ∙ 10-6 𝜃𝑎

3 for 293 K ≤ 𝜃𝑎 ≤ 873 K, [2-8]

CE = 666 + (13002

738−𝜃𝑎

) for 873 K < 𝜃𝑎≤ 1008 K, [2-9]

CE = 545 + (17820

𝜃𝑎−731) for 1008 K < 𝜃𝑎 ≤ 1173 K, [2-10]

CE = 650 for 𝜃𝑎> 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 𝜃a [K]

Figure 2-5. Thermal conductivity of carbon steel as a function of the temperature

k = 54 – 3,33 · 10-2 𝜃𝑎

k = 27,3


R.H.A. TITULAER 15

The thermal elongation (Δl/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

Δl/l = 1,2 ∙ 10-5 𝜃𝑎 + 0,4 ∙ 10-8 𝜃𝑎2 – 2,416 ∙ 10-4 for 273 K ≤ 𝜃𝑎 < 1023 K, [2-12]

Δl/l = 1,1 ∙ 10-2 for 1023 K ≤ 𝜃𝑎 ≤ 1137 K, [2-13]

Δl/l = 2,0 ∙ 10-5 𝜃𝑎 – 6,2 ∙ 10-3 for 1137 K < 𝜃𝑎 ≤ 1473 K. [2-14]

where l is the length at 293 K and Δl is the temperature induced elongation.

The thermal diffusivity (α) 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 (ε) is a value of a surface’s 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 700-750°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 723°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/Δ

l]


0

1000

2000

3000

4000

5000

200 400 600 800 1000 1200 1400

Specific

heat

[J/k

gK

]


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 ∙10-1 𝜃𝑎 –

1,69∙ 10-2 𝜃𝑎2 + 2,22 ∙ 10-6 𝜃𝑎

3

CE = 666 + (13002

738−𝜃𝑎)

CE = 545

+ (17820

𝜃𝑎−731)

CE = 650

Δl/l = 1,2 ∙ 10-5 𝜃𝑎 +

0,4 ∙ 10-8 𝜃𝑎2 –

2,416 ∙ 10-4

Δl/l = 1,1 ∙ 10-2

Δl/l = 2,0 ∙ 10-5 𝜃𝑎 – 6,2 ∙ 10-3


R.H.A. TITULAER 16

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 Young’s modulus

which can be seen in Figure 2-8. The Young’s modulus is taken into account in this thesis and can be given as

𝐸𝑎,𝜃 = 𝑘𝐸,𝜃𝐸𝑎, [2-15]

where

𝐸𝑎 is the Young’s modulus of steel at ambient temperature

𝑘𝐸,𝜃 is the reduction factor at an elevated temperature

𝐸𝑎,𝜃 is the Young’s 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

,θ


Figure 2-8. Reduction factor of the Young’s modulus at elevated

temperatures

Young’s modulus Ea,θ


R.H.A. TITULAER 17

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 Fourier’s law. The negative sign indicates that the heat flows from the higher

temperature side to the lower temperature side. Fourier’s law is prescribed by

𝑞𝑖 = −𝑘𝑖𝑗𝜃,𝑗 , [2-16]

where 𝑘𝑖𝑗 = 𝜅 𝛿𝑖𝑗 is the thermal conductivity tensor and 𝜃,𝑗 is the temperature gradient. The material thermal

conductivity 𝑘𝑖𝑗 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

𝑞𝑖 = ℎ𝑓𝑖(𝜃𝑓𝑖 − 𝜃𝑠) . [2-17]

And on the ambient temperature air side:

𝑞𝑖 = ℎ𝑎(𝜃𝑠 − 𝜃𝑎) , [2-18]

where 𝜃fi and 𝜃a are the fire and air temperatures respectively. Quantities ℎ𝑓𝑖 and ℎ𝑎 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


R.H.A. TITULAER 18

limited number of small-scale experimental studies is extended by these dimensionless numbers. The convective

heat transfer coefficient (ℎ𝑐) is related to the Nusselt (𝑁𝑢) number as can be seen in the following equation

𝑁𝑢 =ℎ𝑐 𝐿

𝑘, giving ℎ𝑐 =

𝑁𝑢 ∙𝑘

𝐿, [2-19]

where L is the characteristic length of the solid surface, ℎ𝑐 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, 𝑅𝑒 and 𝑃𝑟 are the Reynolds and Prandtl numbers respectively. The Reynolds number is

defined by

𝑅𝑒 =𝜌 𝐿 𝑈0

𝜇 =

𝐿 𝑈0

𝑣 , [2-20]

where ρ is the fluid density, the flow velocity is 𝑈0, and μ 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

𝑃𝑟 =𝜇 𝑐

𝑘 , [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

𝑁𝑢 = 𝐵 ∙ 𝑅𝑎𝑚 , [2-22]

where B and m are given in the Table 2-3. 𝑅𝑎 is the Raleigh number and is defined as

𝑅𝑎 = 𝐺𝑟 ∙ 𝑃𝑟 , [2-23]

where 𝐺𝑟 is the Grashof number, which is given by

𝐺𝑟 =𝑔 𝐿3 𝛽 ∆𝜃

𝑣2 , [2-24]

Table 2-2. Nusselt number relations for forced convection (Wang, 2002)


R.H.A. TITULAER 19

where g is the gravity acceleration, 𝛽 the coefficient of thermal expansion of the fluid, and Δ𝜃 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

ℎ𝑐 = 𝐵[(𝑔 ∙𝑃𝑟

𝜃 ∙ 𝑣2 )]1/3 𝑘(∆𝜃)1/3 = 𝛼(∆𝜃)1/3 . [2-25]

By substituting values of B (= 0,14), g (= 9,81 m/s2), Pr, 𝜃, v, and k into this equation the value of 𝛼 can be found

and varies between 1,0 and 0,6 within realistic fire temperatures of 600 – 1300 K. So for conservative calculations of

temperatures of the fire exposed surface, the value of α 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.

ℎ𝑐 = 𝛼 (∆𝜃)1/4 , [2-26]

where 𝛼 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 ℎ𝑐 is suggested as 25 W/m2,

and on the air side the ℎ𝑐 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 𝛼 for the absorptivity, 𝜌 for the

reflectivity, and τ for the transmissivity. The representation of the fractions of incident thermal radiation that a body

absorbs, reflects, and transmits, respectively, is (Wang, 2002):

𝛼 + 𝜌 + τ = 1. [2-27]

Table 2-3. Different variables for the Nusselt number equation for natural convection (Wang, 2002)


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. 𝛼 = 1,0. This

ideal body is then called a blackbody. The total amount of thermal radiation (𝐸𝑏) by a blackbody surface is a function

of its temperature only and is given by the Stefan-Boltzmann law

𝐸𝑏 = 𝜎𝜃4 , [2-28]

where σ is the Stefan-Boltzmann constant which is equal to 5,67 × 10-8 W/(m2K4) and 𝜃 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

𝐼𝜃 = 𝐼𝑛𝑐𝑜𝑠𝜑, [2-29]

where 𝐼𝑛 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 𝜑 to the normal direction of the emitting surface is given by 𝐼𝜃. 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 𝑑𝐴 on the hemispherical

enclosure gives the subtended solid angle to the point of thermal radiation of 𝑑𝐴/𝑟2. If the normal direction of this

incident surface area makes an angle 𝜑 to the normal direction of the emitting blackbody surface, the incident

thermal radiation on this area 𝑑𝐴 is

𝑑𝑞𝜃 = 𝐼𝜃𝑑𝐴

𝑟2= 𝐼𝑛

𝑑𝐴

𝑟2𝑐𝑜𝑠𝜑 . [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

𝐸𝑏 = ∯ 𝑑 𝑞 = ∫ 𝐼𝑛 𝑐𝑜𝑠 𝜑𝜋/2

0

2𝜋 ∙ 𝑟 sin𝜑 ∙ r d𝜑

𝑟2= 𝜋 𝐼𝑛 . [2-31]

Consequently, the intensity of the thermal radiation in the normal direction to the blackbody emitting surface is

given by

𝐼𝑛 =𝐸𝑏

𝜋 . [2-32]

Figure 2-9. Directional intensity of radiant heat (Wang, 2002)


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 𝑑𝐴1 incident on 𝑑𝐴2 is

𝑑𝑞𝑑𝐴1→𝑑𝐴2= 𝐸𝑏1

𝑐𝑜𝑠𝜑1𝑐𝑜𝑠𝜑2

𝜋 𝑟2 𝑑𝐴1 𝑑𝐴2 . [2-33]

The total thermal radiation from 𝐴1 incident on 𝑑𝐴2 is

𝑞𝐴1→𝑑𝐴2 = ∫ 𝐸𝑏1𝐴1

𝑐𝑜𝑠𝜑1𝑐𝑜𝑠𝜑2

𝜋 𝑟2 𝑑𝐴1 𝑑𝐴2 = 𝛷𝐸𝑏1𝑑𝐴2 . [2-34]

𝐸𝑏1𝑑𝐴2 is the maximum incident thermal radiation on 𝑑𝐴2 and this occurs when 𝑑𝐴2 is entirely enclosed by 𝐴1. Φ

is often called the configuration or view factor and only depends on the spatial configuration between 𝐴1 and 𝑑𝐴2.

This factor represents the fraction of thermal radiation from 𝐴1 incident on 𝑑𝐴2, since in most cases the incident

thermal radiation on 𝑑𝐴2 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 ε. The

total radiant energy emitted by a general surface is then

𝐸 = 휀𝜎𝜃4 . [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)


R.H.A. TITULAER 22

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

𝜌0𝜖̇ + (div 𝒒 − 𝑟) − 𝚺 · �̇�𝑒 = 0 , [2-36]

where 𝜌0 is the referential mass density, 𝜖̇ is the rate of change of the internal energy density per unit mass, div is

the divergence, 𝒒 is the thermal flux, 𝑟 is the body heat source per unit volume, 𝚺 is the stress, and �̇�𝑒 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

𝜓 = 𝜖 − 𝜃𝜂 , [2-37]

where 𝜓 is the Helmholtz energy density, 𝜖 is the internal energy, 𝜃 is the absolute temperature, and 𝜂 is the entropy

per unit mass. For the Legendre transformation the 𝚬𝑒 and 𝜃 are used as independent variables for the Helmholtz

energy. This will give

𝜓(𝚬𝑒 , 𝜃) = 𝜖(𝚬𝑒 , 𝜂(𝚬𝑒 , 𝜃)) − 𝜃𝜂(𝚬𝑒 , 𝜃) . [2-38]

Combining equations [2-36] and [2-37] results in

𝜌0(�̇� + �̇�𝜂 + 𝜃�̇�) + (div 𝒒 − 𝑟) − 𝚺 · �̇�𝑒 = 0 , [2-39]

which from Eq. [2-38] can also be written as

𝜌0 (𝜕𝜓

𝜕𝚬𝑒

· �̇�𝑒 +𝜕𝜓

𝜕𝜃�̇� + �̇�𝜂 + 𝜃�̇�) + (div 𝒒 − 𝑟) − 𝚺 · �̇�𝑒 = 0 . [2-40]

This equation consequently leads to

𝜌0 (𝜕𝜓

𝜕𝚬𝑒− 𝚺) · �̇�𝑒 + 𝜌0 (

𝜕𝜓

𝜕𝜃+ 𝜂) �̇� + 𝜌0𝜃�̇� + (div 𝒒 − 𝑟) = 0 . [2-41]

The procedure of Coleman and Noll (1963) prescribes that the terms in Eq. [2-41] that are multiplied by the rates �̇�𝑒

and �̇� 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

𝜌0𝜃�̇� + (div 𝒒 − 𝑟) = 0 . [2-42]

The thermal flux 𝒒 is given by the Fourier’s heat conduction law

𝒒 = −𝐤∇𝜽 , [2-43]

where 𝐤 = 𝜅 𝛿𝑖𝑗 is the thermal conductivity tensor, 𝜅 is the thermal conductivity, and ∇𝜽 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

𝑞(0, 𝑡) = 𝑞(𝐿, 𝑡) = 0 . [2-44]


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

𝑟 = 𝑟𝑟 + 𝑟𝑐 , [2-45]

where 𝑟𝑟 is the radiant heat source and 𝑟𝑐 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

𝑟 = 휀𝜎(𝜃𝐴𝑆𝑇(𝑡)4 − 𝜃𝑠(𝑡)4)+ ℎ𝑐(𝜃𝐴𝑆𝑇(𝑡) − 𝜃𝑠(𝑡)) , [2-46]

where 휀 is the emissivity, 𝜎 is the Stefan-Boltzmann constant, 𝜃𝑠 is the steel’s surface temperature, and ℎ𝑐 is the

convective heat transfer coefficient. 𝜃𝐴𝑆𝑇 is the adiabatic surface temperature, which is a useful conceptual term for

linking gas temperatures with structural surface temperatures (Wickström, Duthinh, & McGrattan, 2007). The

radiant body heat source term can be rewritten, giving

𝑟 = ℎ𝑟(𝜃𝐴𝑆𝑇(𝑡) − 𝜃𝑠(𝑡)) + ℎ𝑐(𝜃𝐴𝑆𝑇(𝑡) − 𝜃𝑠(𝑡)) , [2-47]

where the radiant heat transfer parameter ℎ𝑟 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)

ℎ𝑟 = 휀𝜎(𝜃𝐴𝑆𝑇 (𝑡)2 + 𝜃𝑠(𝑡)2)(𝜃𝐴𝑆𝑇(𝑡) + 𝜃𝑠(𝑡)) . [2-48]

The second equation used for the coupled analysis is the equation of motion which in local form can be written as

div 𝚺 + 𝐟 = 𝜌0�̈� , [2-49]

where div 𝚺 is the divergence of the stress tensor, 𝐟 is the body force, and the acceleration �̈� 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

Ε𝑖𝑗 =1

2(𝑢𝑖,𝑗 + 𝑢𝑗,𝑖) , [2-50]

where Ε𝑖𝑗 is the strain tensor and 𝑢𝑖 is the displacement.

2.3.3. CONSTITUTIVE EQUATIONS

The Cauchy stress tensor Σ𝑖𝑗 is conjugated to the strain tensor Ε𝑖𝑗 as well as that the entropy is conjugated to the

temperature. These terms can be described by using the Helmholtz free energy (𝜓(Ε𝑖𝑗,𝜃) ) equations following

from Eq. [2-41] giving the Cauchy stress tensor

Σ𝑖𝑗 =𝜕𝜓

𝜕Ε𝑖𝑗 , [2-51]

and the thermal part of the reversible entropy density as

𝜂𝑒 = −𝜕𝜓

𝜕𝜃 . [2-52]

The constitutive relation for the Cauchy stress tensor in terms of the strain and temperature can be given with

Σ𝑖𝑗 = 𝐶𝑖𝑗𝑘𝑙Ε𝑘𝑙 − 𝛽𝑖𝑗𝜃 , [2-53]

where 𝐶𝑖𝑗𝑘𝑙 is the elasticity tensor and 𝛽𝑖𝑗 is the coupling coefficient. The strain decomposition is given with


R.H.A. TITULAER 24

Ε𝑖𝑗 = Ε𝑖𝑗𝑒 + Ε𝑖𝑗

𝑡ℎ , [2-54]

where Ε𝑖𝑗𝑒 is the elastic strain and Ε𝑖𝑗

𝑡ℎ is the thermal strain. The following parameters 𝐶𝑖𝑗𝑘𝑙 and 𝛽𝑖𝑗 from an isotropic

body can be used for obtaining the entropy density and Cauchy stress tensor

𝐶𝑖𝑗𝑘𝑙 = 𝜆𝛿𝑖𝑗𝛿𝑘𝑙 + 𝜇(𝛿𝑖𝑘𝛿𝑗𝑙 + 𝛿𝑖𝑙𝛿𝑗𝑘) , [2-55]

𝛽𝑖𝑗 = 𝛼(𝜆 + 2𝜇)𝛿𝑖𝑗 , [2-56]

where 𝜆 and 𝜇 are Lamé constants. In the x direction the strain decomposition is Ε𝑥𝑥𝑒 = Ε𝑥𝑥 − 𝛼𝜃 where 𝛼 is the

coefficient of thermal expansion. The stress tensor can then be derived from the constitutive relation (Hooke’s law)

Σ𝑥𝑥 = 𝜆Ε𝑒 + 2𝜇Ε𝑥𝑥𝑒 = 𝜆(Ε𝑥𝑥 − 3𝛼𝜃) + 2𝜇(Ε𝑥𝑥 − 𝛼𝜃)

= (𝜆 + 2𝜇)Ε𝑥𝑥 − 𝛼𝜃(3𝜆 + 2𝜇) . [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)

𝜂 = 𝜂𝑒 + 𝜂𝑡ℎ , [2-58]

where 𝜂𝑒 is referred to as the thermal part of the reversible entropy density and 𝜂𝑡ℎ 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

Φ𝑖 =𝑞𝑖

𝜃 , [2-59]

where 𝑞𝑖 is the heat flux per unit area and 𝜃 is the absolute temperature. The constitutive relation between 𝜃 and

𝜂𝑒 is given by (Turteltaub & Suiker, 2006)

𝜂𝑒 = 𝑐𝐸 ln𝜃

𝜃0

+ 𝜂𝑟 , [2-60]

where 𝜂𝑟 is the common value of 𝜂𝑒 at the reference temperature 𝜃0 . Including the reversible entropy density term

𝜂𝑡ℎ and linearizing this equation for small changes of temperature, the entropy equation becomes (Biot, 1956)

𝜂 =𝑐𝐸

𝜃0

𝜃 + 𝛽𝑖𝑗Ε𝑖𝑗 , [2-61]

where 𝑐𝐸 is the specific heat for the unit volume without deformation, 𝜃0 is the reference temperature. 𝑐𝐸𝜃

𝜃0

is the

thermal part of the reversible entropy density and 𝛽𝑖𝑗Ε𝑖𝑗 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 𝛽𝑖𝑗.

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 𝑢 and 𝜃. A variation of these state variables (𝛿𝑢 and 𝛿𝜃) is considered around

their equilibrium value. By multiplying with 𝛿𝜃 and integrating over the region of the element the local balance

energy conservation equation described in Eq. [2-42] becomes

∫ (𝜌0𝜃0 �̇� + 𝑞𝑖,𝑖 − 𝑟)𝛿𝜃𝑑𝑉𝑉 = 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

∫ 𝜌0𝜃0 �̇�𝛿𝜃𝑑𝑥𝑉 − ∫ 𝑞𝑖𝛿𝜃,𝑖𝑑𝑥 + ∫ 𝑞𝑖𝑛𝑖𝛿𝜃 d𝐴 − ∫ 𝑟𝛿𝜃𝑑𝑥𝑉𝐴 = 0𝑉 . [2-63]


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 𝛿𝑢 and integrated over the element, giving

∫ (Σ𝑖𝑗,𝑗 + 𝑓𝑖)𝛿𝑢𝑖 𝑑𝑉𝑉 = 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

− ∫ Σ𝑖𝑗𝛿휀𝑖𝑗 𝑑𝑉 + ∫ 𝑓𝑖𝛿𝑢𝑖 𝑑𝑉𝑉 + ∫ Σ𝑖𝑗𝑛𝑗𝛿𝑢𝑖𝑑𝐴𝐴𝑉 = 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

∫ 𝜌0𝜃0 �̇�𝛿𝜃𝑑𝑉𝑉 − ∫ 𝑞𝑖𝛿𝜃,𝑖𝑑𝑉 − ∫ 𝑟𝛿𝜃𝑑𝑥𝑉 + ∫ Σ𝑖𝑗𝛿휀𝑖𝑗 𝑑𝑉 − ∫ 𝑓𝑖𝛿𝑢𝑖 𝑑𝑉𝑉𝑉𝑉

= − ∫ 𝑞𝑖𝑛𝑖𝛿𝜃 𝑑𝐴𝐴 + ∫ 𝑡𝑖𝛿𝑢𝑖𝑑𝐴𝐴 . [2-66]

The right hand side of this weak formulation give the appropriate boundary conditions. The displacements and the

temperatures are prescribed by 𝑢𝑖 = �̂� on 𝐴𝑢 and 𝜃𝑖 = �̂� on 𝐴𝜃 where �̂� and �̂� are the prescribed values. On the

other hand, the traction and flux are given by 𝑡𝑖 = Σ𝑖𝑗𝑛𝑗 = 𝑡̂ on 𝐴𝜎 and 𝑞𝑖 = 𝑞𝑖𝑛𝑖 = 𝑞 on 𝐴𝑞 where 𝑡̂ and 𝑞 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,

𝐍𝑢 and 𝐍𝜃, which relates the displacement component vector 𝒖 and the temperature vector 𝜽 to the corresponding

nodal element values 𝒖(𝑒) and 𝜽(𝑒). This can then be expressed as

𝒖 = 𝐍𝑢𝒖(𝑒) , 𝜽 = 𝐍𝜃𝜽(𝑒) . [2-67]

By using this equation the strain 𝚬 and the temperature gradient ∇𝜽 = 𝜃,𝑖 can be expressed in terms of nodal

quantities

𝚬 = 𝐁𝑢𝒖(𝑒), ∇𝜽 = 𝐁𝜃𝜽(𝑒) , [2-68]

where 𝐁𝑢 and 𝐁𝜃 relate respectively to the first order derivative of components 𝐍𝑢 and 𝐍𝑢. The shape functions for

the variational forms (𝛿𝜽 and 𝛿𝒖) are similarly expressed (𝛿𝜽 = 𝐍𝜃𝛿𝜽(𝑒)). 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

�̇� =𝐶𝐸

𝜃0

𝐍𝜃�̇�(𝑒) + 𝛽𝐁𝑢�̇�(𝑒) . [2-69]

The Fourier’s law from Eq. [2-43] in matrix notion is

𝒒 = −𝐤∇𝜽 = −𝐤𝐁𝜃𝜽(𝑒) . [2-70]

Figure 2-11. Body V with mixed boundary conditions for prescribed displacements,

tractions, temperatures, and fluxes


R.H.A. TITULAER 26

The stress constitutive equation from Eq. [2-53] becomes

𝚺 = 𝐂𝑢𝐁𝑢𝒖(𝑒) − 𝛽𝐍𝜃𝜽(𝑒) . [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 𝑑𝑉 becomes 𝐴𝑑𝑥, and 𝑑𝐴 becomes 𝑃|0𝑙 , where 𝑃 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

∑ (∫ 𝜌0𝜃0𝐍𝜃𝛿𝜽(𝑒) (𝛽𝐁𝑢�̇�(𝑒) +𝐶𝐸

𝜃0𝐍𝜃�̇�(𝑒))𝐴𝑑𝑥

𝑙𝑒

0 − ∫ 𝐁𝜃T𝛿𝜽(𝑒)T

𝐤𝐁𝜃𝜽(𝑒)𝐴𝑑𝑥𝑙𝑒

0 𝑒

− ∫ ℎ𝑟(𝜃𝐴𝑆𝑇(𝑡) − 𝐍𝜃𝜽𝑠(𝑡)(𝑒))𝛿𝜽(𝑒)𝑃𝑑𝑥

𝑙𝑒

0 − ∫ ℎ𝑐(𝜃𝐴𝑆𝑇(𝑡) − 𝐍𝜃𝜽𝑠(𝑡)(𝑒))𝛿𝜽(𝑒)𝑃𝑑𝑥

𝑙𝑒

0

+ ∫ 𝐁𝑢T𝛿𝒖(𝑒)T( 𝐂𝑢𝐁𝑢𝒖(𝑒) − 𝛽𝐍𝜃𝜽(𝑒)) 𝐴𝑑𝑥

𝑙𝑒

0

= −�̂�𝐍𝜃T𝛿𝜽(𝑒)T𝑃|

𝑙𝑒

0+ �̂�𝐍𝑢

T𝛿𝒖(𝑒)T𝑃|

𝑙𝑒

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

∫ 𝜌0𝜃0𝐍𝜃𝛿𝜽(𝑒) (𝛽𝐁𝑢�̇�(𝑒) +𝐶𝐸

𝜃0𝐍𝜃�̇�(𝑒))𝐴𝑑𝑥

𝑙𝑒

0

= 𝛿𝜽(𝑔)T 𝜌0𝜃0 ∑ ∫ 𝐍𝜃

T𝛽𝐁𝑢�̇�(𝑔)𝐴𝑑𝑥𝑙𝑒

0𝑒

+𝛿𝜽(𝑔)T∑ ∫ 𝐍𝜃

T𝐶𝐸𝐍𝜃�̇�(𝑔)𝐴𝑑𝑥𝑙𝑒

0𝑒 . [2-73]

The global system components from Eq. [2-73] can be summarized for the capacity matrix components as

𝐂𝜃,𝑢 = 𝜌0𝜃0 ∑ ∫ 𝐍𝜃T𝛽𝐁𝑢𝐴𝑑𝑥

𝑙𝑒

0𝑒 , [2-74]

𝐂𝜃,𝜃 = ∑ ∫ 𝐍𝜃T𝐶𝐸𝐍𝜃𝐴𝑑𝑥

𝑙𝑒

0𝑒 , [2-75]

so that

∫ 𝜌0𝜃0𝐍𝜃𝛿𝜽(𝑒) (𝛽𝐁𝑢�̇�(𝑒) +𝐶𝐸

𝜃0

𝐍𝜃�̇�(𝑒))𝐴𝑑𝑥𝑙𝑒

0

= 𝛿𝜽(𝑔)T𝐂𝜃,𝑢�̇�(𝑔) + 𝛿𝜽(𝑔)T

𝐂𝜃,𝜃�̇�(𝑔) . [2-76]

The second term on the left hand side becomes

∫ 𝐁𝜃T𝛿𝜽(𝑒)T𝐤𝐁𝜃𝜽(𝑒)𝐴𝑑𝑥

𝑙𝑒

0

= ∑ ∫ 𝐁𝜃T𝛿𝜽(𝑔)T

𝐤𝐁𝜃𝜽(𝑔)𝐴𝑑𝑥𝑙𝑒

0𝑒 , [2-77]

in which the stiffness matrix component 𝐊𝜃,𝜃 can be found as

𝐊𝜃,𝜃 = ∑ ∫ 𝐁𝜃T𝐤𝐁𝜃𝐴𝑑𝑥

𝑙𝑒

0𝑒 , [2-78]

giving

∫ 𝐁𝜃T𝛿𝜽(𝑒)T𝐤𝐁𝜃𝜽(𝑒)𝐴𝑑𝑥

𝑙𝑒

0 = 𝛿𝜽(𝑔)T 𝐊𝜃,𝜃 𝜽(𝑔) . [2-79]


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


𝑙𝑒

0

= − ∑ ∫ 𝛿𝜽(𝑔)T 𝐍𝜃

Tℎ𝑟(𝜃𝐴𝑆𝑇(𝑡) − 𝐍𝜃𝜽𝑠(𝑡)(𝑔)})𝑃𝑑𝑥

𝑙𝑒

0𝑒

= − ∑ ∫ 𝛿𝜽(𝑔)T 𝐍𝜃

Tℎ𝑟(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥𝑙𝑒

0𝑒

+ ∑ ∫ 𝛿𝜽(𝑔)T 𝐍𝜃

Tℎ𝑟𝐍𝜃𝜽𝑠(𝑔)

𝑃𝑑𝑥𝑙𝑒

0𝑒 , [2-80]

which can be evaluated as

𝐊𝜃,𝜃(𝑟) = ∑ ∫ 𝐍𝜃Tℎ𝑟𝐍𝜃𝑃𝑑𝑥

𝑙𝑒

0𝑒 , [2-81]

𝐟𝜃(𝑟) = ∑ ∫ 𝐍𝜃Tℎ𝑟(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥

𝑙𝑒

0𝑒 , [2-82]

so that


𝑙𝑒

0

= −𝛿𝜽(𝑔)T𝐟𝜃(𝑟) + 𝛿𝜽(𝑔)T

𝐊𝜃,𝜃(𝑟) 𝜽𝑠(𝑔)

. [2-83]

The convection part of the body heat source term in Eq. [2-73] can be divided into

− ∫ ℎ𝑐(𝜃𝐴𝑆𝑇(𝑡) − 𝐍𝜃𝜽𝑠(𝑡)(𝑒))𝛿𝜽(𝑒)𝑃𝑑𝑥

𝑙𝑒

0

= − ∑ ∫ 𝛿𝜽(𝑔)T 𝐍𝜃

Tℎ𝑐(𝜃𝐴𝑆𝑇(𝑡) − 𝐍𝜃𝜽𝑠(𝑡)(𝑔))𝑃𝑑𝑥

𝑙𝑒

0𝑒

= − ∑ ∫ 𝛿𝜽(𝑔)T 𝐍𝜃

Tℎ𝑐(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥𝑙𝑒

0𝑒

+ ∑ ∫ 𝛿𝜽(𝑔)T 𝐍𝜃

Tℎ𝑐𝐍𝜃𝜽𝑠(𝑔)

𝑃𝑑𝑥𝑙𝑒

0𝑒 , [2-84]

in which the 𝐊𝜃,𝜃(𝑐) and 𝐟𝜃(𝑐) can be found as

𝐊𝜃,𝜃(𝑐) = ∑ ∫ (𝐍𝜃Tℎ𝑐𝐍𝜃𝑃𝑑𝑥

𝑙𝑒

0𝑒 , [2-85]

𝐟𝜃(𝑐) = ∑ ∫ 𝐍𝜃Tℎ𝑐(𝜃𝐴𝑆𝑇 (𝑡))𝑃𝑑𝑥

𝑙𝑒

0𝑒 , [2-86]

giving that

− ∫ ℎ𝑐(𝜃𝐴𝑆𝑇(𝑡) − 𝐍𝜃𝜽𝑠(𝑡)(𝑒))𝛿𝜽(𝑒)𝑃𝑑𝑥

𝑙𝑒

0

= −𝛿𝜽(𝑔)T𝐟𝜃(𝑐) + 𝛿𝜽(𝑔)T

𝐊𝜃,𝜃(𝑐)𝜽𝑠(𝑔)

. [2-87]

The last term on the right hand side can be written in global system form as

∫ 𝐁𝑢T𝛿𝒖(𝑒) T( 𝐂𝑢𝐁𝑢𝒖(𝑒) − 𝛽𝐍𝜃𝜽(𝑒)) 𝐴𝑑𝑥

𝑙𝑒

0

= ∑ ∫ 𝐁𝑢T𝛿𝒖(𝑔)T

( 𝐂𝑢𝐁𝑢𝒖(𝑔) − 𝛽𝐍𝜃𝜽(𝑔))𝐴𝑑𝑥𝑙𝑒

0𝑒 . [2-88]

In this equation the stiffness components 𝐊𝑢,𝑢 and 𝐊𝑢,𝜃 can be found as

𝐊𝑢,𝑢 = ∑ ∫ 𝐁𝑢T𝐂𝑢𝐁𝑢𝐴𝑑𝑥

𝑙𝑒

0𝑒 , [2-89]

𝐊𝑢,𝜃 = ∑ ∫ 𝐁𝑢T𝛽𝐍𝜃𝐴𝑑𝑥

𝑙𝑒

0𝑒 , [2-90]

which results in

∫ 𝐁𝑢T𝛿𝒖(𝑒) T( 𝐂𝑢𝐁𝑢𝒖(𝑒) − 𝛽𝐍𝜃𝜽(𝑒)) 𝐴𝑑𝑥

𝑙𝑒

0

= 𝛿𝒖(𝑔)T 𝐊𝑢,𝑢 𝛿𝒖(𝑔) − 𝛿𝒖(𝑔)T 𝐊𝑢,𝜃 𝜽(𝑔) . [2-91]


R.H.A. TITULAER 28

The first term on the right hand side in Eq. [2-73] gives in global system form

−�̂�𝐍𝜃T𝛿𝜽(𝑒)T

𝑃|𝑙𝑒

0= − ∑ 𝐍𝜃

T𝛿𝜽(𝑔)T�̂�𝑃|

𝑙𝑒

0𝑒 , [2-92]

in which the thermal force vector can be seen as

𝐟𝜃 = − ∑ 𝐍𝜃T

𝑒 �̂�𝑃 |𝑙𝑒

0 . [2-93]

This term then becomes

−�̂�𝐍𝜃T𝛿𝜽(𝑒)T

𝑃|𝑙𝑒

0= 𝛿𝜽(𝑔)T

𝐟𝜃 . [2-94]

The second term on the right hand side in Eq. [2-73] is derived as

�̂�𝐍𝑢T𝛿𝒖(𝑒)T

𝑃 |𝑙𝑒

0= ∑ 𝐍𝑢

T𝛿𝒖(𝑔)T�̂�𝑃 |

𝑙𝑒

0𝑒 , [2-95]

in which the mechanical force vector can be seen as

𝐟𝑢 = ∑ 𝐍𝑢T


0 . [2-96]

The last term on the right hand side then becomes

�̂�𝐍𝑢T𝛿𝒖(𝑒)T

𝑃 |𝑙𝑒

0= 𝛿𝒖(𝑔)T

𝐟𝑢 . [2-97]

The total weak formulation for the coupled thermoelasticity in FEM formulation then becomes

𝛿𝜽(𝑔)T𝐂𝜃,𝑢�̇�(𝑔) + 𝛿𝜽(𝑔)T 𝐂𝜃,𝜃�̇�(𝑔) + 𝛿𝜽(𝑔)T

𝐊𝜃,𝜃 𝜽(𝑔)

+ 𝛿𝜽(𝑔)T 𝐊𝜃,𝜃(𝑟) 𝜽𝑠(𝑔)

+ 𝛿𝜽(𝑔)T 𝐊𝜃,𝜃(𝑐) 𝜽𝑠

(𝑔)

+𝛿𝒖(𝑔)T 𝐊𝑢,𝑢 𝒖(𝑔) − 𝛿𝒖(𝑔)T

𝐊𝑢,𝜃 𝜽(𝑔)

= 𝛿𝜽(𝑔)T𝐟𝜃 + 𝛿𝒖(𝑔)T

𝐟𝑢+𝛿𝜽(𝑔)T𝐟𝜃(𝑟)+𝛿𝜽(𝑔)T

𝐟𝜃(𝑐) . [2-98]

This equation should hold for arbitrary variations 𝛿𝜽(𝑔) and 𝛿𝒖(𝑔). This gives

[0 0

𝐂𝜃,𝑢 𝐂𝜃,𝜃] [

�̇��̇�] + [

𝐊𝑢,𝑢 𝐊𝑢,𝜃

0 𝐊𝜃,𝜃] [

𝒖𝜽] = [

𝐟𝑢𝐟𝜃

] , [2-99]

where

𝐂𝜃,𝑢 = 𝜌0𝜃0 ∑ ∫ 𝐍𝜃T𝛽𝐁𝑢𝐴𝑑𝑥

𝑙𝑒

0𝑒 ,

𝐂𝜃,𝜃 = ∑ ∫ 𝐍𝜃T𝐶𝐸𝐍𝜃𝐴𝑑𝑥

𝑙𝑒

0 , 𝑒

𝐊𝑢,𝑢 = ∑ ∫ 𝐁𝑢T𝐂𝑢𝐁𝑢𝐴𝑑𝑥

𝑙𝑒

0𝑒 ,

𝐊𝑢,𝜃 = − ∑ ∫ 𝐁𝑢T𝛽𝐍𝜃𝐴𝑑𝑥

𝑙𝑒

0𝑒 ,

𝐊𝜃,𝜃 = ∑ ∫ 𝐁𝜃T𝐤𝐁𝜃𝐴𝑑𝑥

𝑙𝑒

0𝑒 + ∑ ∫ 𝐍𝜃Tℎ𝑟𝐍𝜃𝑃𝑑𝑥

𝑙𝑒

0𝑒 + ∑ ∫ 𝐍𝜃Tℎ𝑐𝐍𝜃𝑃𝑑𝑥

𝑙𝑒

0𝑒 ,

𝐟𝑢 = ∑ 𝐍𝑢T


0,

𝐟𝜃 = − ∑ 𝐍𝜃T


0+ ∑ ∫ 𝐍𝜃

Tℎ𝑟(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥𝑙𝑒

0𝑒 + ∑ ∫ 𝐍𝜃Tℎ𝑐(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥

𝑙𝑒

0𝑒 . [2-100]


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

𝐂�̇� + 𝐊𝐝 = 𝐟 . [2-101]

This linear algebraic equation has to be solved for every time step. By introducing the parameter 𝛾, different

transient schemes can be used, which are shown in Table 2-4. The following equation is used for introducing this

parameter

𝐝𝑛+ 𝛾 = 𝛾𝐝𝑛+1 + (1 − 𝛾)𝐝𝑛 . [2-102]

Table 2-4. Time-stepping schemes for transient analysis and accompanying parameter 𝛾

Eq. [2-101] can be rewritten in order to introduce the time-stepping schemes, which results in

𝐂{𝐝𝑛+1−𝐝𝑛

∆𝑡} + 𝐊𝐝𝑛+𝛾 = 𝐟𝑛+𝛾 . [2-103]

Rearranging this equation using Eq. [2-102] gives

(𝐂 + γ∆𝑡𝐊)𝐝𝑛+1 = (𝐂 − (1 − γ)∆𝑡𝐊)𝐝𝑛 + ∆𝑡(γ𝐟𝑛+1 + (1 − γ)𝐟𝑛) , [2-104]

which can be rewritten for the fully implicit method as

(𝐂 + ∆𝑡𝐊)𝐝𝑛+1 = 𝐂𝐝𝑛 + ∆𝑡𝐟𝑛+1 . [2-105]

Rewriting this equation for the unknown values gives

𝐝𝑛+1 = (𝐂 + ∆𝑡𝐊)−1(𝐂𝐝𝑛 + ∆𝑡𝐟𝑛+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.

γ 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


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

∆𝑡𝐾11 · 𝑢𝑖𝑛+1 + ∆𝑡𝐾12 · 𝑢𝑗

𝑛+1 + ∆𝑡𝐾13 · 𝑢𝑘𝑛+1 + ∆𝑡𝐾14 · 𝜃𝑖

𝑛+1 + ∆𝑡𝐾16 · 𝜃𝑘𝑛+1 = −∆𝑡

𝑃𝐿𝑡𝑖

2

𝑛+1 , [2-109]

∆𝑡𝐾21 · 𝑢𝑖𝑛+1 + ∆𝑡𝐾22 · 𝑢𝑗

𝑛+1 + ∆𝑡𝐾23 · 𝑢𝑘𝑛+1 + ∆𝑡𝐾24 · 𝜃𝑖

𝑛+1 + ∆𝑡𝐾26 · 𝜃𝑘𝑛+1 = 0 , [2-110]

∆𝑡𝐾31 · 𝑢𝑖𝑛+1 + ∆𝑡𝐾32 · 𝑢𝑗

𝑛+1 + ∆𝑡𝐾33 · 𝑢𝑘𝑛+1 + ∆𝑡𝐾34 · 𝜃𝑖

𝑛+1 + ∆𝑡𝐾36 · 𝜃𝑘𝑛+1 = ∆𝑡

𝑃𝐿𝑡𝑘

2

𝑛+1 , [2-111]

𝐶41 · 𝑢𝑖𝑛+1 + 𝐶42 · 𝑢𝑗

𝑛+1 + 𝐶43 · 𝑢𝑘𝑛+1 + (𝐶44 + ∆𝑡𝐾44) · 𝜃𝑖

𝑛+1 + (𝐶46 + ∆𝑡𝐾46) · 𝜃𝑘𝑛+1 = 𝐶41 · 𝑢𝑖

𝑛 + 𝐶42 · 𝑢𝑗𝑛 +

𝐶43 · 𝑢𝑘𝑛 + 𝐶44 · 𝜃𝑖

𝑛 + 𝐶46 · 𝜃𝑖𝑛 + ∆𝑡

𝑃𝐿𝑞𝑖

2

𝑛+1 , [2-112]

𝐶61 · 𝑢𝑖𝑛+1 + 𝐶62 · 𝑢𝑗

𝑛+1 + 𝐶63 · 𝑢𝑘𝑛+1 + (𝐶64 + ∆𝑡𝐾64) · 𝜃𝑖

𝑛+1 + (𝐶66 + ∆𝑡𝐾66) · 𝜃𝑘𝑛+1 = 𝐶61 · 𝑢𝑖

𝑛 + 𝐶62 · 𝑢𝑗𝑛 +

𝐶63 · 𝑢𝑘𝑛 + 𝐶64 · 𝜃𝑖

𝑛 + 𝐶66 · 𝜃𝑖𝑛 + ∆𝑡

𝑃𝐿𝑞𝑘

2

𝑛+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.


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

3634333231

2624232221

16143634131211333231

2624232221

16143634131211333231

2624232221

1614131211

6664636261

46446664434241636261

46446664434241636261

4644434241

o

i

o

i

o

m

k

i

o

n

m

l

k

j

i

o

m

k

i

o

n

m

l

k

j

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


R.H.A. TITULAER 32

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

. [2-115]

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.


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

∆𝑡𝐾12 · 𝑢𝑗𝑛+1 + ∆𝑡𝐾14 · 𝜃𝑖

𝑛+1 + ∆𝑡𝐾16 · 𝜃𝑘𝑛+1 + ∆𝑡

𝑃𝐿𝑡𝑖

2

𝑛+1= −∆𝑡𝐾11 · 𝑢𝑖

𝑛+1 − ∆𝑡𝐾13 · 𝑢𝑘𝑛+1 , [2-117]

∆𝑡𝐾22 · 𝑢𝑗𝑛+1 + ∆𝑡𝐾24 · 𝜃𝑖

𝑛+1 + ∆𝑡𝐾26 · 𝜃𝑘𝑛+1 = −∆𝑡𝐾21 · 𝑢𝑖

𝑛+1 − ∆𝑡𝐾23 · 𝑢𝑘𝑛+1 , [2-118]

∆𝑡𝐾32 · 𝑢𝑗𝑛+1 + ∆𝑡𝐾34 · 𝜃𝑖

𝑛+1 + ∆𝑡𝐾36 · 𝜃𝑘𝑛+1 − ∆𝑡

𝑃𝐿𝑡𝑘

2

𝑛+1= −∆𝑡𝐾31 · 𝑢𝑖

𝑛+1 − ∆𝑡𝐾33 · 𝑢𝑘𝑛+1 , [2-119]

𝐶42 · 𝑢𝑗𝑛+1 + (𝐶44 + ∆𝑡𝐾44) · 𝜃𝑖

𝑛+1 + (𝐶46 + ∆𝑡𝐾46) · 𝜃𝑘𝑛+1 = −𝐶41 · 𝑢𝑖

𝑛+1 − 𝐶43 · 𝑢𝑘𝑛+1 + 𝐶41 · 𝑢𝑖

𝑛 + 𝐶42 · 𝑢𝑗𝑛 +

𝐶43 · 𝑢𝑘𝑛 + 𝐶44 · 𝜃𝑖

𝑛 + 𝐶46 · 𝜃𝑖𝑛 + ∆𝑡

𝑃𝐿𝑞𝑖

2

𝑛+1 , [2-120]

𝐶62 · 𝑢𝑗𝑛+1 + (𝐶64 + ∆𝑡𝐾64) · 𝜃𝑖

𝑛+1 + (𝐶66 + ∆𝑡𝐾66) · 𝜃𝑘𝑛+1 = −𝐶61 · 𝑢𝑖

𝑛+1 − 𝐶63 · 𝑢𝑘𝑛+1 + 𝐶61 · 𝑢𝑖

𝑛 + 𝐶62 · 𝑢𝑗𝑛 +

𝐶63 · 𝑢𝑘𝑛 + 𝐶64 · 𝜃𝑖

𝑛 + 𝐶66 · 𝜃𝑖𝑛 + ∆𝑡

𝑃𝐿𝑞𝑘

2

𝑛+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]


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 �̃� in the general equation [2-101], where �̃� is the moved stiffness matrix the general

equation becomes

𝐂�̇� + 𝐊𝐝 = �̃�𝐟 , [2-124]

where �̇� is the time derivative from the components 𝒖(𝑒) and 𝜽(𝑒), and 𝐝 are the components 𝒖(𝑒) and 𝜽(𝑒). Solving

this equation can again be executed with the fully implicit method giving

(𝐂 + 𝜃∆𝑡𝐊)𝐝𝑛+1 = (𝐂 − (1 − 𝜃)∆𝑡𝐊)𝐝𝑛 + ∆𝑡(𝜃�̃�𝐟𝑛+1 + (1 − 𝜃)�̃�𝐟𝑛) , [2-125]

(𝐂 + ∆𝑡𝐊)𝐝𝑛+1 = 𝐂𝐝𝑛 + ∆𝑡�̃�𝐟𝑛+1 , [2-126]

𝐝𝑛+1 = (𝐂 + ∆𝑡𝐊)−1(𝐂𝐝𝑛 + ∆𝑡�̃�𝐟𝑛+1) . [2-127]


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.


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

𝐂�̇� + 𝐊𝐝 = 𝐟 . [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

𝐂�̇� + 𝐊𝐝 = �̃�𝐟 , [2-130]

where �̃� 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 𝑁 × 𝑁 identity matrix with

in the 𝑖’th and 𝑗’th row the one included for the prescribed displacements and the other rows are zero. Call this

matrix 𝐃𝑖𝑗. 𝑁 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 𝑖 and 𝑗 are the columns where the displacement is known. The

desired vector will be the result of the matrix with vector multiplication 𝐃𝑖𝑗𝐝 = 𝐝𝑑𝑒𝑠𝑖𝑟𝑒𝑑. The same procedure holds

for the force vector 𝐟 where the tractions want to be excluded. Multiplying the 𝑁 × 𝑁 identity matrix with in the

𝑖’th and 𝑗’th row the one included for the prescribed displacements and the other rows are zero. Call this matrix

𝐅𝑖𝑗. The desired vector will be the result of the matrix with vector multiplication 𝐅𝑖𝑗𝐟 = 𝐟𝑑𝑒𝑠𝑖𝑟𝑒𝑑. 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 𝑁 × 𝑁 identity matrix where the 𝑖’th and 𝑗’th row are all zeroes. Call

this matrix 𝐃𝑘𝑙. The adjusted vector can be obtained by the multiplication 𝐃𝑘𝑙𝐝 = 𝐝𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑. The same procedure

can be used for the force vector, which means that the adjusted vector can be obtained by 𝐅𝑘𝑙𝐟 = 𝐟𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑. Then the

desired vectors have to be added to the adjusted vector from the other side. This means that the addition 𝐝𝑑𝑒𝑠𝑖𝑟𝑒𝑑 +

𝐟𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑 gives the correct force vector where the displacements appear and the addition 𝐟𝑑𝑒𝑠𝑖𝑟𝑒𝑑. +𝐝𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑 gives

the correct 𝐝 vector with the tractions in it. For example move the displacement term to the right hand side.

𝐝 =

[ 𝑢𝑖

𝑢𝑗

𝑢𝑘

𝜃𝑖

𝜃𝑗𝜃𝑘 ]

, 𝐃𝑖𝑗 =

000000000000000000000100000000000001

, 𝐟 =

[ 𝑡𝑖0𝑡𝑘𝑞𝑖

0𝑞𝑘]

, 𝐅𝑘𝑙 =

100000010000001000000000000010000000

, [2-131]

𝐝𝑑𝑒𝑠𝑖𝑟𝑒𝑑 =

000000000000000000000100000000000001

[ 𝑢𝑖

𝑢𝑗

𝑢𝑘

𝜃𝑖

𝜃𝑗𝜃𝑘 ]

=

[ 𝑢𝑖

0𝑢𝑘

000 ]

, 𝐟𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑 =

100000010000001000000000000010000000

[ 𝑡𝑖0𝑡𝑘𝑞𝑖

0𝑞𝑘]

=

[ 000𝑞𝑖

0𝑞𝑘]

, [2-132]

𝐝𝑑𝑒𝑠𝑖𝑟𝑒𝑑 + 𝐟𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑 =

[ 𝑢𝑖

0𝑢𝑘

000 ]

+

[ 000𝑞𝑖

0𝑞𝑖]

=

[ 𝑢𝑖

0𝑢𝑘

𝑞𝑖

0𝑞𝑖 ]

. [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


R.H.A. TITULAER 37

this can be performed by multiplying matrix 𝐊 with a 𝑁 × 1 vector where the 𝑖’th and 𝑗’th row containing a one

belonging to the position of the displacement. Call these vectors 𝐊𝑖 and 𝐊𝑗. The desired vectors 𝐊𝑑𝑒𝑠𝑖𝑟𝑒𝑑 will contain

the columns with the values which will be transferred to the right hand side with a negative sign.

𝐊 =

6664

4644

3634333231

2624232221

1614131211

0000000000

0000

0

0

0

KK

KK

KKKKK

KKKKK

KKKKK

,𝐊𝑖 =

[ 100000]

, 𝐊𝑗 =

[ 001000]

, [2-134]

𝐊𝑑𝑒𝑠𝑖𝑟𝑒𝑑 =

6664

4644

3634333231

2624232221

1614131211

0000000000

0000

0

0

0

KK

KK

KKKKK

KKKKK

KKKKK

[ 100000]

=

[ 𝐾11

𝐾21

𝐾31

000 ]

, [2-135]

𝐊𝑑𝑒𝑠𝑖𝑟𝑒𝑑 =

6664

4644

3634333231

2624232221

1614131211

0000000000

0000

0

0

0

KK

KK

KKKKK

KKKKK

KKKKK

[ 001000]

=

[ 𝐾13

𝐾23

𝐾33

000 ]

. [2-136]

In order to add the negative values of the matrix 𝐊 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 𝑖’th and 𝑗’th column from the matrix �̃�. Write down the 𝑁 × 𝑁 identity matrix and replace the

𝑖’th and 𝑗’th column with a column of all zeroes. Call this matrix 𝐊𝑖𝑗. The desired matrix will be the result of the

matrix multiplication �̃�𝐊𝑖𝑗. For example you want the third column to be zero of matrix �̃�

�̃� =

2/00000000000002/0000002/00000000000002/

PL

PLPL

PL

, 𝐊𝑖𝑗 =

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:

𝐊𝑑𝑒𝑠𝑖𝑟𝑒𝑑 = −1 ∗

[ 𝐾11

𝐾21

𝐾31

000 ]

=

[ −𝐾11

−𝐾21

−𝐾31

000 ]

, [2-139]


R.H.A. TITULAER 38

𝐊𝑑𝑒𝑠𝑖𝑟𝑒𝑑 = −1 ∗

[ 𝐾13

𝐾23

𝐾33

000 ]

=

[ −𝐾13

−𝐾23

−𝐾33

000 ]

. [2-140]

If you want to add this vector into matrix �̃� 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 𝑁 × 1 with a one on the 𝑖’th

row where the zero column is positioned in the matrix �̃�. So

[ −𝐾11

−𝐾21

−𝐾31

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 �̃�

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

[ −𝐾13

−𝐾23

−𝐾33

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.


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

�̃� (1)(𝑥) = 𝐶1 + 𝐶2𝑥 + 𝐶3𝑥2 + 𝐶4𝑥

3 for 0 ≤ 𝑥 ≤ 𝑙 . [2-145]

Figure 2-12. Stability of equilibrium (Yoo & Lee, 2005)

Figure 2-13. Simply supported one-dimensional element


R.H.A. TITULAER 40

The corresponding rotation field can be obtained by taking the spatial derivative of the displacement field

�̃�(1)(𝑥) = −𝑑�̃�(1) (𝑥)

𝑑𝑥= −𝐶2 − 2𝐶3𝑥 − 3𝐶4𝑥

2 for 0 ≤ 𝑥 ≤ 𝑙 . [2-146]

These relations can be rewritten in matrix-vector format

[�̃�(1)(𝑥)

�̃� (1)(𝑥)] = [1 𝑥

0 −1 𝑥2 𝑥3

−2𝑥 −3𝑥2] [

𝐶1

𝐶2

𝐶3

𝐶4

] for 0 ≤ 𝑥 ≤ 𝑙 . [2-147]

The 4 unknowns 𝐶𝑖 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

𝑑𝑖 and a rotation 𝑟𝑖 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

𝑑1 = �̃�(1)(0) = 𝐶1, [2-148]

𝑟1 = �̃�(1)(0) = −𝐶2, [2-149]

𝑑2 = �̃� (1) (𝑙

2) = 𝐶1 + 𝐶2𝑙 + 𝐶3𝑙

2 + 𝐶4𝑙3 , [2-150]

𝑟2 = �̃�(1) (𝑙

2) = −𝐶2 − 2𝐶3𝑙 − 3𝐶4𝑙

2 . [2-151]

This can again be rewritten in matrix-vector form

[

𝑑1

𝑟1𝑑2

𝑟2

] =

2

32

32101

00100001

lllll

[

𝐶1

𝐶2

𝐶3

𝐶4

] . [2-152]

Inverting these relations results in

[

𝐶1

𝐶2

𝐶3

𝐶4

] =

2323

22

1212

132300100001

llll

llll[

𝑑1

𝑟1𝑑2

𝑟2

] . [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

�̃� (1)(𝑥) = [1 −3𝑥2

𝑙2+

2𝑥3

𝑙3, −𝑥 +

2𝑥2

𝑙−

𝑥3

𝑙2,3𝑥2

𝑙2−

2𝑥3

𝑙3,𝑥2

𝑙−

𝑥3

𝑙2] [

𝑑1

𝑟1𝑑2

𝑟2

] for 0 ≤ 𝑥 ≤ 𝑙 . [2-154]

This relation can be formulated as

�̃� (𝑚)(𝑥) = 𝐍(m)𝐝(m) . [2-155]

The relation between th element rotation field and the displacement and rotations can be formulated as

�̃�(𝑚)(𝑥) = −𝑑�̃�(𝑚) (𝑥)

𝑑𝑥= 𝐆(m)𝐝(m) , [2-156]


R.H.A. TITULAER 41

�̃�(1)(𝑥) = [6𝑥

𝑙2−

6𝑥2

𝑙3, 1 −

4𝑥

𝑙+

3𝑥2

𝑙2, −

6𝑥

𝑙2+

6𝑥2

𝑙3, −

2𝑥

𝑙+

3𝑥2

𝑙2] [

𝑑1

𝑟1𝑑2

𝑟2

] for 0 ≤ 𝑥 ≤ 𝑙 . [2-157]

The element shape functions of this method can be found in 𝐍(m), which thus

𝐍(m) = [1 −3𝑥2

𝑙2+

2𝑥3

𝑙3, −𝑥 +

2𝑥2

𝑙−

𝑥3

𝑙2,3𝑥2

𝑙2−

2𝑥3

𝑙3,𝑥2

𝑙−

−𝑥3

𝑙2] . [2-158]

The relations for the rotation field are expressed with 𝐆(m), which is the first derivative of the shape functions

multiplied with minus one. This reads thus

𝐆(m) = −𝑑

𝑑𝑥𝐍(m) = [

6𝑥

𝑙2−

6𝑥2

𝑙3, 1 −

4𝑥

𝑙+

3𝑥2

𝑙2,−

6𝑥

𝑙2+

6𝑥2

𝑙3,−

2𝑥

𝑙+

3𝑥2

𝑙2] . [2-159]

Now also the curvature field in an element can be computed with

�̃� (𝑚)(𝑥) = −𝑑2�̃�(𝑚) (𝑥)

𝑑𝑥2= 𝐁(m)𝐝(m) , [2-160]

where 𝐁(m) can be given with

𝐁(m) = −𝑑2

𝑑𝑥2𝐍(m) = [

6

𝑙2−

12𝑥

𝑙3, −

4

𝑙+

6𝑥

𝑙2, −

6

𝑙2+

12𝑥

𝑙3,−

2

𝑙+

6𝑥

𝑙2] . [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

𝑉 = ∫1

2𝐸𝐼 (

𝑑2𝑤

𝑑𝑥2)2

𝑑𝑥 + ∫ 𝑁01

2(𝑑𝑤

𝑑𝑥)

2

𝑑𝑥𝑙

0

𝑙

0 [2-162]

Using the axial equilibrium condition 𝑁0 = −𝐹 and 𝐹 = λ�̂�, with λ as an arbitrary multiplier and �̂� = 1,0 as a

reference magnitude. Using �̃�(𝑚)(𝑥) = −𝑑2�̃� (𝑚) (𝑥)

𝑑𝑥2 and �̃� (𝑚)(𝑥) = −

𝑑�̃�(𝑚) (𝑥)

𝑑𝑥 the total potential energy can be written

as

�̃� = �̃�(𝐝) =1

2∑ 𝐝T𝐊(m)𝐝 − λ

1

2∑ 𝐝T𝐊𝝈

(𝐦)𝐝 𝑀

𝑚=1 𝑀𝑚=1 . [2-163]

In this equation, the total nodal displacement and rotation vector gives

𝐝T = [𝑑1 𝑟1 𝑑2 𝑟2 𝑑3 𝑟3] , [2-164]

and the element stiffness matrices are formulated by

𝐊(m) = ∫ 𝐁(m)T𝐸𝐼𝐁(m) 𝑑𝑥ℎ(𝑚) , [2-165]

𝐊𝝈(𝐦)

= ∫ 𝐆(m) T�̂�𝐆(m) 𝑑𝑥ℎ(𝑚) , [2-166]

where 𝐊(m) is the conventional stiffness matrix and 𝐊𝝈(𝐦)

is the stress stiffness matrix. According to Appendix C the

conventional stiffness matrix can be computed as

𝐊(m) =𝐸𝐼

𝑙3

22

22

4626

6126122646

612612

llll

llllll

ll

. [2-167]


R.H.A. TITULAER 42

The stress stiffness matrix is computed as found in Appendix C

𝐊𝝈(𝐦)

=�̂�

30𝑙

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]

𝛿�̃� =𝜕𝑉

𝜕𝐝 ∙ 𝛿𝐝 = 0

𝜕𝑉

𝜕𝐝= 0 ∑ (𝑀

𝑚=1 𝐊(m) − λ𝐊𝝈(𝐦)

)𝐝 = 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. 𝑑1 = 0 and 𝑑2 = 0. The first and third columns and rows of the stiffness matrix

disappear. This gives

. [2-171]

The buckling loads 𝐹 = λ�̂� can be found by solving the above set of equations for λ. �̂� = 1,0 is used as a reference

intensity. Two eigenvalues λ are found by equating the determinant of the reduced stiffness matrix to zero. This

results in

λ1 = F1 = F𝑐𝑟 = 12𝐸𝐼

𝑙2 , λ2 = F2 = 60

𝐸𝐼

𝑙2 . [2-172]

The exact solution for this buckling phenomenon is given by

F𝑛 =𝑛2𝜋2𝐸𝐼

𝑙2 , with n=1,2,3,… [2-173]

The exact buckling loads are then

F1 = F𝑐𝑟 = 9,8696𝐸𝐼

𝑙2 , F2 = 39,4784

𝐸𝐼

𝑙2 . [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


R.H.A. TITULAER 43

2.4.3. GALERKIN APPROXIMATION FEM STABILITY

The approximate displacement �̃�(𝑥) may also be directly substituted in the variational equilibrium equation. The

potential energy for the example in Figure 2-13 is described with

𝑉 = ∫1

2𝐸𝐼 (

𝑑2𝑤

𝑑𝑥2)2

𝑑𝑥 + ∫ 𝑁01

2(𝑑𝑤

𝑑𝑥)

2

𝑑𝑥𝑙

0

𝑙

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

𝛿𝑉 = ∫ 𝐸𝐼𝑤′′𝛿𝑤′′𝑑𝑥 + ∫ 𝑁0𝑤′𝛿𝑤′𝑑𝑥 = 0

𝑙

0

𝑙

0 , [2-176]

where 𝑤′′ =𝑑2𝑤

𝑑𝑥2 , 𝑤′ =

𝑑𝑤

𝑑𝑥. [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

𝛿𝑉 = ∫ {((𝐸𝐼𝑤′′)′′− (𝑁0𝑤′)′)𝛿𝑤}𝑑𝑥 + (𝐸𝐼𝑤′′)𝛿𝑤′|0

𝑙 − ((𝐸𝐼𝑤′′)′ − 𝑁0𝑤′)

𝑙

0 𝛿𝑤|0𝑙 = 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 �̃�(𝑥) 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

𝑀(0) = 𝐸𝐼𝜅(0) = 0 �̃� (1)(0) = −𝑑2�̃� (1)(0)

𝑑𝑥2= 𝐁(1)(0)𝐝(1) = 0 , [2-179]

𝑀(𝑙) = 𝐸𝐼𝜅(𝑙) = 0 �̃� (2)(𝑙) = −𝑑2�̃� (2)(𝑙)

𝑑𝑥2= 𝐁(2)(𝑙)𝐝(2) = 0 , [2-180]

Using the equation [2-161] for 𝐁(m)(𝑥), the two natural boundary conditions can be elaborated as

𝑓1(𝐝) =24

𝑙2𝑑1 −

8

𝑙𝑟1 −

24

𝑙2𝑑2 −

4

𝑙𝑟2 = 0 , [2-181]

𝑓2(𝐝) =−24

𝑙2𝑑2 +

4

𝑙𝑟2 +

24

𝑙2𝑑3 +

8

𝑙𝑟3 = 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 Φ∗.

This potential can be described with

Φ∗(𝐝,𝛬1 , 𝛬2) = �̃�(𝐝) + 𝛬1𝑓1(𝐝) + 𝛬2𝑓2(𝐝) , [2-183]

where [𝛬1, 𝛬2] are the Lagrange multipliers. This equation can be rewritten as

Φ∗(𝐝,𝚲) = �̃�(𝐝) + 𝚲 ∙ 𝐟(𝐝) , [2-184]

where 𝚲T is the vector with Lagrange multipliers, and 𝐟(𝐝)T is the vector containing the natural boundary

conditions. For the equilibrium condition the stationarity of the auxiliary potential can be given with

δΦ∗ = (𝜕𝑉

𝜕𝐝+

𝜕(𝚲∙𝐟(𝐝))

𝜕𝐝)∙ 𝛿𝐝 +

𝜕(𝚲∙𝐟(𝐝))

𝜕𝚲∙ 𝛿𝚲 = 0 . [2-185]

For any variations 𝛿𝐝 and 𝛿𝚲 this gives the following set of coupled equations

𝜕𝑉

𝜕𝐝+

𝜕(𝚲∙𝐟(𝐝))

𝜕𝐝= 0 ∑ (𝑀


)𝐝 +𝜕(𝚲∙𝐟(𝐝))

𝜕𝐝= 0 , [2-186]


R.H.A. TITULAER 44

𝜕(𝚲∙𝐟(𝐝))

𝜕𝚲= 0 𝐟(𝐝) = 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


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)


R.H.A. TITULAER 46

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)


R.H.A. TITULAER 47

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

𝐊𝐝 = 𝐟 , [2-189]

where 𝐊 is the stiffness matrix, 𝐝 is the displacement vector, and 𝐟 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 ‘guess’ 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 𝐊. 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)

𝛙(𝐝) = 𝐟 − 𝐏(𝐝) = 0 , [2-190]

where 𝐝 is the set of discretisation parameters, 𝐟 a vector which is independent of the parameters and 𝐏 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

𝛙𝑛+1 = 𝛙(𝐝𝑛+1) = 𝐟𝑛+1 − 𝐏(𝐝𝑛+1) = 0 , [2-191]

which starts from a nearby solution at

𝐝 = 𝐝𝑛, 𝛙𝑛 = 0, 𝐟 = 𝐟𝑛 . [2-192]

The forcing function can be written as

𝐟𝑛+1 = 𝐟𝑛 + ∆𝐟𝑛 . [2-193]

Figure 2-18. Schematic representation of the Newton-Raphson iterative method (Becker, 2001)


R.H.A. TITULAER 48

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

𝛙(𝐝𝑛+1𝑖+1 ) ≈ 𝛙(𝐝𝑛+1

𝑖 )+ (𝜕ψ

𝜕a)𝑛+1

𝑖

𝑑𝐝𝒏𝒊 = 0 . [2-194]

The iteration of this equation starts by assuming

𝐝𝑛+1𝟎 = 𝐝𝑛 , [2-195]

in which 𝐝𝑛 is a converged solution at a previous load level or time step. Following from Eq. [2-194] the iterative

correction can be written as

𝐊T𝑖 𝑑𝐝𝑛

𝑖 = 𝛙𝑛+1𝑖 , [2-196]

in which 𝐊T𝑖 is the tangent direction given by

𝐊T =𝜕𝐏

𝜕𝐚= −

𝜕𝛙

𝜕𝐚 . [2-197]

So the set of discretization parameters can be obtained by

𝑑𝐝𝑛𝑖 = (𝐊T

𝑖 )−1

𝛙𝑛+1𝑖 . [2-198]

Successive approximations can be obtained by

𝐝𝑛+1𝑖+1 = 𝐝𝑛+1

𝑖 + 𝑑𝐝𝑛+1𝑖

= 𝐝𝑛 + ∆𝐝𝑛𝑖 . [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 𝐊 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)


R.H.A. TITULAER 49

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 λ is considered as an additional unknown. The equilibrium equation of

the non-linear system can be written as

𝛙(𝐝𝑛+1, λ) = (λ𝑛 + 𝑑λ)𝐟0 − 𝐏(𝐝𝑛+1) = 0 , [2-200]

where 𝐝 is the set of discretisation parameters, 𝐟 a vector which is independent of the parameters and 𝐏 a vector

dependent on the parameters. λ𝑛+1 is the loading factor, such that 𝐟𝑛+1 = λ𝑛+1 𝐟0 = (λ𝑛 + 𝑑λ)𝐟0, with 𝐟0 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)

𝑔(∆𝐝,∆λ) = 0 . [2-201]

Now Eq. [2-200] can be used to formulate the set of discretization parameters

𝑑𝐝𝒏𝒊 = (𝐊T

𝒊 )−𝟏

𝛙𝑛+1𝒊 , [2-202]

𝑑𝐝𝒏𝒊 = (𝐊T

𝒊 )−𝟏

{(λ𝑛 + 𝑑λ)𝐟𝟎 − 𝐏(𝐚𝒏+𝟏)} , [2-203]

𝑑𝐝𝒏𝒊 = {(𝐊T

𝒊 )−𝟏

(λ𝑛𝐟𝟎 − 𝐏(𝐚𝒏+𝟏))} + 𝑑λ{(𝐊T𝒊 )

−𝟏𝐟𝟎} , [2-204]

𝑑𝐝𝒏𝒊 = 𝑑𝑰𝐝𝒏

𝒊 + 𝑑λ𝑑𝑰𝑰𝐝𝒏𝒊 , [2-205]

where λ𝑛𝐟0 is the vector of external forces 𝐟. The path-following constraint 𝑔 is suggested by de Borst et al. (2012a)

which is the spherical arc-length constraint

𝑔(∆𝐝,∆λ) = (∆𝐝𝑖)T∆𝐝𝑖 + 𝛽2(∆λ𝑖)2𝐟𝟎T𝐟𝟎 − ∆𝑙2 = 0 , [2-206]

where

∆𝐝𝑖 = ∆𝐝𝑖−1 + 𝑑𝐝 , [2-207]

∆λ𝑖 = ∆λ𝑖−1 + 𝑑λ , [2-208]

where ∆𝐝𝑖 and ∆λ𝑖 are the sought increments between the next state 𝑖 and the last known equilibrium state 𝑛 and

where ∆𝐝𝑖−1 and ∆λ𝑖−1 are the known increments between the current state 𝑖 − 1 and the last known equilibrium

state 𝑛 (Danielsson, 2013). The value 𝛽 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 𝒅𝐝𝑛𝑖 and of the load (λ𝑛 + 𝑑λ)𝐟0 should be properly balanced. The last term in this equation ∆𝑙 is the

path step length. Setting this equation to zero, the following solutions can be obtained

𝑐1𝑑λ2 + 𝑐2𝑑λ + 𝑐3 = 0 , [2-209]

Figure 2-20. Path-following technique (de Borst et al., 2012)


R.H.A. TITULAER 50

where

𝑐1 = 𝑑𝑰𝑰𝐝T𝑑𝑰𝑰𝐝+ 𝛽2𝐟0T𝐟0 , [2-210]

𝑐2 = 2𝑑𝑰𝑰𝐝T(∆𝐝𝑖−1 + 𝑑𝑰𝐝)+ 2𝛽2∆λ𝑖−1𝐟0T𝐟0 , [2-211]

𝑐3 = (∆𝐝𝑖−1 + 𝑑𝑰𝐝)T(∆𝐝𝑖−1 + 𝑑𝑰𝐝)+ 𝛽2(∆λ𝑖−1)2𝐟0

T𝐟0 − ∆𝑙2 . [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 ∆𝐝𝑛𝑖 so that

(∆𝐝𝑖)T∆𝐝𝑖−1 > 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]

𝑔(∆𝐝,∆λ) = (∆𝐝𝑖)T∆𝐝𝑖 + 𝛽2∆λ𝑖−1∆λ𝑖𝐟0T𝐟0 − ∆𝑙2 = 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

𝑑λ =∆𝑙2−(∆𝐝𝑖−1)T∆𝐝𝑖−1−(∆𝐝𝑖−1)T𝑑𝑰𝐝−𝛽2(∆λ𝑖−1)2𝐟𝟎

T𝐟𝟎

(∆𝐝𝑖−1)T𝑑𝑰𝑰𝐝+𝛽2∆λ𝑖−1𝐟𝟎T𝐟𝟎

. [2-215]

Noting that ∆𝑙2 ≈ (∆𝐝𝑖−1)T∆𝐝𝑖−1, the previous equation can be approximated by

𝑑λ = −(∆𝐝𝑖−1)T𝑑𝑰𝐝+𝛽2(∆λ𝑖−1)2𝐟𝟎

T𝐟𝟎

(∆𝐝𝑖−1)T𝑑𝑰𝑰𝐝+𝛽2∆λ𝑖−1𝐟𝟎T𝐟𝟎

. [2-216]

This equation can also be simplified by keeping the direction of the tangent normal to the hyperplane constant,

which results in

𝑔(∆𝐝,∆λ) = (∆𝐝1)T∆𝐝𝑖 + 𝛽2∆λ1∆λ𝑖𝐟𝟎T𝐟𝟎 − ∆𝑙2 = 0 . [2-217]

This normal path method is very simple and results in the following increment of the loading factor

𝑑λ = −(∆𝐝𝑖−1)T𝑑𝑰𝐝

(∆𝐝𝑖−1)T𝑑𝑰𝑰𝐝+𝛽2∆λ1𝐟𝟎T𝐟𝟎

. [2-218]

This equation resembles the constraint equation originally introduced by Riks (1972). The following figure describes

the normal path method. Using 𝛽 = 0 seems to work for many engineering cases, so this is used in this analysis.

𝑑λ = −(∆𝐝𝑖−1)T𝑑𝑰𝐝

(∆𝐝𝑖−1)T𝑑𝑰𝑰𝐝 . [2-219]

Figure 2-21. Updated hyperplanes for the geometrical interpretation of alternative constraint equation g (de Borst et al., 2012)


R.H.A. TITULAER 51

2.5.3. THERMAL NON-LINEARITY

Eq. [2-104] can also be rewritten for �̇�𝑛+1 to incorporate the Newton-Raphson method for non-linear calculations.

For the first-order equation only the value of 𝐝𝑛 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

𝐝 = 𝐝𝑛 + ∆𝑡�̇� . [2-220]

The approximation to the sets of ordinary differential equations is then

𝐂�̇� + 𝐊(𝐝𝑛 + 𝛾∆𝑡�̇�) = 𝐟 . [2-221]

Rewriting this equation gives a solution for �̇� as

(𝐂 + 𝛾∆𝑡𝐊)�̇� = (𝐟 − 𝐊𝐝𝑛) . [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 𝑡𝑛+1 the equilibrium equation may be written in a residual form as

𝛙(�̇�𝑛+1) = 𝐟 − 𝐏(𝐝𝑛 + 𝛾∆𝑡�̇�𝑛+1) − 𝐂�̇�𝑛+1 = 0 , [2-223]

where 𝐏 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

(𝐂 + 𝜃∆𝑡𝐊T𝑖 )∆�̇�𝑛+1

𝑖 = 𝛙(�̇�𝑛+1𝑖 ) . [2-224]

Eq. [2-224] can now be solved and this gives the successive approximations as

�̇�𝑛+1𝑖+1 = �̇�𝑛+1

(𝑖)+ 𝜏∆�̇�𝑛+1

(𝒊) , [2-225]

where 𝜏 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)


R.H.A. TITULAER 52

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.


R.H.A. TITULAER 53


R.H.A. TITULAER 54

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 Young’s

modulus is taken as 2,1×108 kN/m2 and the Poisson’s ratio is 0,3. The expansion coefficient of steel is 11×10-6. 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


R.H.A. TITULAER 55

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 Young’s modulus is taken as 2,1×108 kN/m2 and the Poisson’s ratio is

0,3. The expansion coefficient of steel is 11×10-6. 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

1×10-6 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


R.H.A. TITULAER 56

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


R.H.A. TITULAER 57

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 Young’s 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 = F𝑐𝑟 = 9,8696𝐸𝐼

𝑙2 , F2 = 39,4784

𝐸𝐼

𝑙2 . [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


R.H.A. TITULAER 58

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

𝜃. For a thermomechanical problem, the total axial strain 휀0 is decomposed into a mechanical (elastic) part 휀0𝑒 and

a thermal part 휀0𝑡ℎ (Suiker, 2014a)

휀0 = 휀0𝑒 + 휀0

𝑡ℎ . [3-2]

Using its constitutive relation, the axial force in the column can be decomposed

𝑁0 = 𝐸𝐴휀0𝑒 = 𝐸𝐴(휀0 – 휀0

𝑡ℎ) . [3-3]

The thermal strain is defined as

휀0𝑡ℎ = 𝛼∆𝜃 , [3-4]

where ∆𝜃 = 𝜃 − 𝜃0 , and 𝛼 is the linear coefficient of thermal expansion, for steel 11×10-6. 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 𝜃

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


R.H.A. TITULAER 59

𝑁0 = 𝐸𝐴휀0𝑒 = 𝐸𝐴(– 휀0

𝑡ℎ) = −𝐸𝐴𝛼∆𝜃 . [3-5]

An increase in temperature by ∆𝜃 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

𝑑2

𝑑𝑥2 (𝐸𝐼

𝑑2𝑤

𝑑𝑥2 ) –

𝑑

𝑑𝑥 (𝑁0

𝑑𝑤

𝑑𝑥 ) =𝑞𝑧 . [3-6]

Since 𝐸𝐼 is taken as a constant and 𝑞𝑧 = 0 this results in

𝐸𝐼 𝑑4𝑤

𝑑𝑥4 + 𝐸𝐴𝛼∆𝜃

𝑑2𝑤

𝑑𝑥2 = 0 , [3-7]

which can be rewritten as

𝑑4𝑤

𝑑𝑥4 + β2

𝑑2𝑤

𝑑𝑥2 = 0 , with β2 =

𝐴 𝛼𝐿 ∆𝜃

𝐼 . [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; 𝑑𝑤

𝑑𝑥 |x=0 = 0 ,

x = l: M(0) = - 𝐸𝐼 𝑑2𝑤

𝑑𝑥2|x=0 = 0; M(l) = - 𝐸𝐼

𝑑2𝑤

𝑑𝑥2|x=l = 0 . [3-9]

Substituting these boundary conditions into the fourth-order differential equation gives the following solution

sin(𝛽𝑙) = 0 𝛽𝑛𝑙 = 𝑛𝜋 with n = 1,2,3,… [3-10]

since

𝛽2 =𝐴 𝛼𝐿 ∆𝜃𝑛

𝐼 ∆𝜃𝑛 =

𝑛2𝜋2𝐼

𝐴𝛼𝐿𝑙2 . [3-11]

The critical temperature increment can then be found for n=1, which gives

∆𝜃𝑐𝑟 =𝜋2𝐼

𝐴𝛼𝐿𝑙2 . [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 ×10-5 m4 (about the weak axis), 𝐴 = 5,38 ×10-3 m2,

𝐿 = 3 m ,

𝛼= 11 ×10-6 /K.

The critical temperature increment and the critical temperature for this example are

Δ𝜃𝑐𝑟 = 248 K 𝜃𝑐𝑟= 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

𝑉 = ∫1

2𝐸𝐼 (

𝑑2𝑤

𝑑𝑥2)2

𝑑𝑥 + ∫ 𝑁01

2(𝑑𝑤

𝑑𝑥)

2

𝑑𝑥𝑙

0

𝑙

0 . [3-14]

Using the axial equilibrium condition 𝑁0 = −𝐸𝐴𝛼∆𝜃 and 𝐸𝐴𝛼∆𝜃 = λ�̂�, with λ as an arbitrary multiplier and �̂� = 1,0

as a reference magnitude. Equilibrium is then described by stationarity of the potential energy

𝛿�̃� =𝜕𝑉

𝜕𝐝 ∙ 𝛿𝐝 = 0

𝜕𝑉

𝜕𝐝= 0 ∑ (𝑀


)𝐝 = 0, [3-15]


R.H.A. TITULAER 60

The buckling loads 𝐸𝐴𝛼∆𝜃 = λ�̂� can be found by solving the above set of equations for λ. �̂� = 1,0 is used as a

reference intensity. Four eigenvalues λ are found by equating the determinant of the reduced stiffness matrix to

zero. This results in

λ1 = EA𝛼𝐿 ∆𝑇1 = EA𝛼𝐿 ∆𝑇𝑐𝑟 = 9,9438𝐸𝐼

𝑙2,

λ2 = EA𝛼𝐿 ∆𝑇2 = 48,0000𝐸𝐼

𝑙2,

λ3 = EA𝛼𝐿 ∆𝑇3 = 128,7228𝐸𝐼

𝑙2,

λ4 = EA𝛼𝐿 ∆𝑇4 = 240,0000𝐸𝐼

𝑙2. [3-16]

which gives

∆𝜃𝑐𝑟 = 9,9438𝐼

𝐴𝛼𝐿𝑙2 ,

∆𝜃2 = 48,0000𝐼

𝐴𝛼𝐿𝑙2 ,

∆𝜃3 = 128,7228𝐼

𝐴𝛼𝐿 𝑙2 ,

∆𝜃4 = 240,0000𝐼

𝐴𝛼𝐿 𝑙2. [3-17]

This result and procedure is thus very similar to a purely mechanical problem. If now for �̂� a value is used which

is applied on the structure, the stability of the structure can be checked. If λ < 1,0 buckling occurs and if λ > 1,0 the

structure remains stable. The �̂� can now vary over the length of the structure, instead of using a reference intensity.

The values of �̂� 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 Δ𝜃 = 260 K is applied on the

structure, the Matlab code calculates the eigenvalues of this problem. The solutions of the eigenvalues are

λ1 = 0,9522 , λ2 = 3.8087, λ3 = 8.5695, λ4 = 15.2347. [3-19]

This means that the structure is buckled at the temperature of 553 K, since λ < 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 λ < 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

λ


R.H.A. TITULAER 61

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 λ < 1,0 then the structure is buckled. From the Matlab

code it follows that λ = 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 𝐊𝝈(𝐦)

. 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.

∑ (𝑀𝑚=1 𝐊(m) − λ𝐊𝝈

(𝐦))𝐝 = 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 Λ1

0 -1.70662 0.853311 1.706621 0 0 Λ2

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]


R.H.A. TITULAER 62

6762r1 + 7133d2 + 3937r2 – 1,71 Λ1 = 0, [3-21]

7133r1 + 15014d2 – 117r2 – 7250r3 – 1,71Λ1 – 1,71Λ2= 0, [3-22]

3937r1 – 117d2 + 14759r2 + 3879r3 – 0,85Λ1 + 0,85Λ2 = 0, [3-23]

-7250d2 + 3879r2 + 6996r3 + 1,71Λ2= 0, [3-24]

-1,71r1 – 1,71d2 – 0,85r2 = 0, [3-25]

-1,71d3 +0,85r2 + 1,71r3 = 0. [3-26]

Solving these equations for arbitrary constant r1 it follows that

Λ1 = -223r1 , Λ2 = -141r1 , d2 = -0,99r1 , r2 = -0,02r1 , r3 = -0,98r1 .

Inserting these values back into the �̃� (𝑚)(𝑥) = 𝐍(m)𝐝(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

�̃� (1)(𝑥) = (−𝑥 − 0,1518𝑥3)𝑟1 for 0 ≤ 𝑥 ≤ 𝑙

2 �̃� (1)(0) = 0, [3-27]

�̃� (2)(𝑥) = (−0,98766 + 0,02468𝑥 − 0,012𝑥2 + 0,03795𝑥3)𝑟1 for 𝑙

2 ≤ 𝑥 ≤ 𝑙 �̃� (2)(𝑙) = 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


R.H.A. TITULAER 63

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 – 1,89Λ1 = 0, [3-29]

16389,34r1 + 58043,22d2 – 198,26r2 – 30211,20d3 – 16587,6r3 -2,84Λ1= 0, [3-30]

5792,89r1 – 198,26d2 +21457,25r2 + 16587,6d3 + 5726,80r3 – 0,95Λ1= 0, [3-31]

-30211,2d2 + 16587,60r2 + 62793,54d3 – 197,6r3 – 16785,2r4 – 2,84Λ2= 0, [3-32]

-16587,6d2 + 5726,80r2 – 197,60d3 + 21985,06r3 + 5660,93r4 + 0,95Λ2=0, [3-33]

-16785,2d3 + 5660,93r3 + 11124,26r4 + 1,89Λ2= 0, [3-34]

-1,89r1 – 2,84d2 – 0,95r2 = 0, [3-35]

-2,84d3 +0,95r3 + 1,89r4 = 0. [3-36]

Solving these equations for arbitrary constant r1 it follows that

Λ1 = -10,7677r1 , Λ2= -103,884r1 , d2 = -0,8011r1 , r2 = 0,4033r1 , d3 = -0,7119r1 , r3 = -0,5055r1 , r4 = -0,8152r1 .

Inserting these values back into the �̃� (𝑚)(𝑥) = [𝐍(m)][𝐝(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

�̃� (1)(𝑥) = (−𝑥 − 0,1989𝑥3)𝑟1 for 0 ≤ 𝑥 ≤ 𝑙

3 �̃� (1)(0) = 0, [3-37]

�̃� (2)(𝑥) = (−0,8011 − 0,4033𝑥 + 0,5686𝑥2 + 0,0760𝑥3)𝑟1 for 𝑙

3 ≤ 𝑥 ≤

2𝑙

3 , [3-38]

�̃� (3)(𝑥) = (−0,7119 + 0,5055𝑥 + 0,3096𝑥2 − 0,1032𝑥3)𝑟1 for 2𝑙

3 ≤ 𝑥 ≤ 𝑙 �̃� (3)(𝑙) = 0. [3-39]

And for 6 one 3rd-order elements it gives

�̃� (1)(𝑥) = (−𝑥 + 0,2451𝑥3)𝑟1 for 0 ≤ 𝑥 ≤ 𝑙

6 �̃� (1)(0) = 0, [3-40]

�̃� (2)(𝑥) = (−0,4693 − 0,8162𝑥 + 0,3699𝑥2 + 0,0978𝑥3)𝑟1 for 𝑙

6 ≤ 𝑥 ≤

𝑙

3, [3-41]

�̃� (3)(𝑥) = (−0,7727 − 0,3728𝑥 + 0,5143𝑥2 − 0,0351𝑥3)𝑟1 for 𝑙

3 ≤ 𝑥 ≤

𝑙

2, [3-42]

�̃� (4)(𝑥) = (−0,8349 + 0,1152𝑥 + 0,45666𝑥2 − 0,1060𝑥3)𝑟1 for 𝑙

2 ≤ 𝑥 ≤

2𝑙

3, [3-43]

�̃� (5)(𝑥) = (−0,6764 + 0,4923𝑥 + 0,2936𝑥2 − 0,1230𝑥3)𝑟1 for 2𝑙

3 ≤ 𝑥 ≤

5𝑙

6 , [3-44]

�̃� (6)(𝑥) = (−0,3709 + 0,7012𝑥 + 0,1223𝑥2 − 0,0815𝑥3)𝑟1 for 5𝑙

6 ≤ 𝑥 ≤ 𝑙 �̃� (6)(𝑙) = 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

Λ1

Λ2

0 0.5 1 1.5 2 2.5 3

Length [m]

Figure 3-16. Buckling mode using 3 elements assembled

First buckling mode


R.H.A. TITULAER 64

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


R.H.A. TITULAER 65

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 �̂�

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

𝐹𝑟𝑜𝑜𝑓 = 6 ∙ 6 ∙ 1,1 ∙ 12 = 475,2 𝑘𝑁 , [3-46]

𝐹𝑓𝑙𝑜𝑜𝑟𝑠 = 2 ∙ 6 ∙ 6 ∙ 1,1 ∙ 15 = 1188,0 𝑘𝑁 , [3-47]

𝐹𝑡𝑜𝑡𝑎𝑙 = 1663,2 𝑘𝑁 . [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 λ < 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


R.H.A. TITULAER 66

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

λ

Thermal

Mechanical

Thermal and mechanical

0

4000

8000

12000

16000

20000


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


R.H.A. TITULAER 67

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 Euler’s formulation for the critical

buckling load. The Matlab solution is provided by inserting the Young’s 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 Young’s

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


R.H.A. TITULAER 68

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 Young’s 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 YOUNG’S MODULUS

The reduction of the Young’s 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

Young’s Modulus can decrease more until buckling occurs.

Table 3-3. Eigenvalues thermal problem

𝛌

Analytical -

Matlab 0,1732

ABAQUS 0,1732

Table 3-4. Eigenvalues coupled thermomechanical problem

𝛌

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


λ

10 minutes

10 minutes with E-reduction

30 minutes


60 minutes


Figure 3-27. Eigenvalues for thermomechanical buckling comparing time and E-reduction


R.H.A. TITULAER 69

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

𝐊′𝐝′ = 𝐟′ , [3-49]

where 𝐊′ is the member stiffness matrix, 𝐝′ is the local displacement vector, and 𝐟′ 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

𝐝′ =

[

'

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

𝐝′ = 𝐓𝐝, [3-52]

where 𝐓 is the transformation matrix and 𝒅 is the global displacement vector. The transformation matrix can be

derived from looking at a deformed element in Figure 3-28, which rotates clockwise.

𝑢1′ = 𝑢1 cos(𝜗) + 𝑤1sin (𝜗) , [3-53]

𝑤1′ = −𝑢1 sin(𝜗) + 𝑤1cos (𝜗) , [3-54]

𝜑1′ = 𝜑1 , [3-55]

𝑢2′ = 𝑢2 cos(𝜗) + 𝑤2sin (𝜗) , [3-56]

𝑤2′ = −𝑢2 sin(𝜗) + 𝑤2cos (𝜗) , [3-57]

𝜑2′ = 𝜑2 . [3-58]

Figure 3-28. Example non-linear calculation two elements (left), element 1 highlighted in local system (right)


R.H.A. TITULAER 70

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 𝑐 = cos (𝜗) and 𝑠 = sin (𝜗). The local force vector can also be rewritten into the global form. The local force

vector is given by

𝐟′ =

[

'

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

𝐟 = 𝐓−𝟏𝐟′ or 𝐟 = 𝐓𝐓𝐟′ , [3-61]

where 𝐓−𝟏 is the inverse of the transformation matrix and 𝐟 is the global force vector. The inverse of the

transformation matrix is also the transpose of the transformation matrix (𝐓𝐓) (de Borst, Crisfield, Remmers, &

Verhoosel, 2012b). The transpose transformation matrix can be derived by again looking at a deformed element in

Figure 3-28

𝑁1 = 𝑁1′ cos(𝜗) − 𝑉1

′sin (𝜗) , [3-62]

𝑉1 = 𝑁1′ sin(𝜗) + 𝑉1

′cos (𝜗) , [3-63]

𝑀1 = 𝑀1′ , [3-64]

𝑁2 = 𝑁2′ cos(𝜗) − 𝑉2

′sin (𝜗) , [3-65]

𝑉2 = 𝑁2′ sin(𝜗) + 𝑉2

′cos (𝜗) , [3-66]

𝑀2 = 𝑀2′ . [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 𝑐 = cos (𝜗) and 𝑠 = sin (𝜗).


R.H.A. TITULAER 71

To formulate the global load-displacement relation Eq. [3-49] is inserted in Eq. [3-61]. This gives

𝐟 = 𝐓𝐓𝐊′𝒅′. [3-69]

Then Eq. [3-52] is inserted in the equation above, which results in

𝐟 = 𝐓𝐓𝐊′𝐓𝒅 , [3-70]

which can be rewritten for the global coordinate system as

𝐟 = 𝐊𝒅 , [3-71]

in which 𝐊 represents the global stiffness matrix

𝐊 = 𝐓𝐓𝐊′𝐓 . [3-72]

First an initial imperfection of 𝑙

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 𝜗 can be calculated. The angle 𝜗 = 1,9997e−2 𝑟𝑎𝑑 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


R.H.A. TITULAER 72

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 3×10-4 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 3×10-4 m

Figure 3-31. Zoomed in on the non-linear load-displacement relation for initial imperfection of 3×10-4 m

Non-linear load-displ. relation: ABAQUS

2 elements Matlab

4 elements Matlab

6 elements Matlab

Non-linear load-displ. relation: 8 elements Matlab


Non-linear load-displ. relation 8 elements ABAQUS


Non-linear load-displ. relation 8 elements Matlab


R.H.A. TITULAER 73

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 ‘chopped off’ (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


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)


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)


R.H.A. TITULAER 74

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


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


R.H.A. TITULAER 75

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

𝑃/𝑃𝑐𝑟 = 1,885

𝑃/𝑃𝑐𝑟 = 1,518

𝑃/𝑃𝑐𝑟 = 1,393

𝑃/𝑃𝑐𝑟 = 1,294

𝑃/𝑃𝑐𝑟 = 1,215

𝑃/𝑃𝑐𝑟 = 1,102

𝑃/𝑃𝑐𝑟 = 1,018

𝑃/𝑃𝑐𝑟 = 1,035

𝑃/𝑃𝑐𝑟 = 1,000

𝑃/𝑃𝑐𝑟 = 1,215

𝑃/𝑃𝑐𝑟 = 1,000

Figure 3-37. Buckling modes extensible elements (column) geometric non-linearity

𝑃/𝑃𝑐𝑟 = 0,834

𝑃/𝑃𝑐𝑟 = 0,625

𝑃/𝑃𝑐𝑟 = 0,391

𝑃/𝑃𝑐𝑟 = 0,208


R.H.A. TITULAER 76

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 100×100 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 11×10 -6 (CEN, 2011). The Young’s 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

5𝐹𝑐𝑟 ≈ 400 𝑘𝑁. 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 𝐿/250. Using this initial imperfection the non-linear mechanical response is

investigated and the transverse displacement for the one-dimensional element is found as 3,2×10-3 m. This value is

added up to the initial imperfection giving a total imperfection of 1,52×10-2 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

θ

[K]

x/L

Temp. distribution: t = 200 s

t = 150 s

t = 100 s

t = 50 s


R.H.A. TITULAER 77

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


R.H.A. TITULAER 78

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 ‘domains’. 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

𝑡𝑓𝑖,𝑑 ≥ 𝑡𝑓𝑖,𝑟𝑒𝑞𝑢, [3-73]

- in the strength domain as

𝑅𝑓𝑖,𝑑,𝑡 ≥ 𝐸𝑓𝑖,𝑑,𝑡 , [3-74]

- and in the temperature domain as

𝜃𝑑 ≤ 𝜃𝑐𝑟 ,𝑑, [3-75]

where

𝑡𝑓𝑖,𝑑 is the design value of the fire resistance;

𝑡𝑓𝑖,𝑟𝑒𝑞𝑢 is the required fire resistance time;

𝑅𝑓𝑖,𝑑,𝑡 is the design value of the resistance of the member in the fire situation;

𝐸𝑓𝑖,𝑑,𝑡 is the design value of the relevant effects of actions in the fire situation;

𝜃𝑑 is the design value of material temperature;

𝜃𝑐𝑟,𝑑 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

𝐸𝑓𝑖,𝑑,𝑡 = ∑ 𝐺𝑘,𝑗 + (𝜓1,1𝑜𝑟 𝜓2,1)𝑄𝑘,1 +𝑖≥1 ∑ 𝜓2,𝑖𝑄𝑘,𝑖𝑖≥1 , [3-76]


R.H.A. TITULAER 79

where

𝐺𝑘,𝑗 are the characteristic values of the permanent actions;

𝑄𝑘,1 is the characteristic leading variable action;

𝑄𝑘,𝑖 are the characteristic values of the accompanying variable actions;

𝜓1,1 is the factor for the frequent value of a variable action;

𝜓2,𝑖 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

𝐸𝑓𝑖,𝑑,𝑡 = 𝜂𝑓𝑖𝐸𝑑 , [3-77]

where

𝐸𝑑 is the design value of the corresponding force at ambient temperature (CEN, 2012a);

𝜂𝑓𝑖 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 휀

is adopted to take account for temperature influences. This value 휀 is here given as

휀 = 0,85√235

𝑓𝑦

. [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 𝜇0. For steel members subjected to instability problems, such as columns under

axial compressive forces, the degree of utilisation can be found with

𝜇0 =𝐸𝑓𝑖 ,𝑑,𝑡

𝑁𝑝𝑙 ,𝑓𝑖,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

𝜃𝑐𝑟 = 39,19 ln [1

0,9674𝜇03,833 − 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 𝐴𝑚/𝑉. 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


R.H.A. TITULAER 80

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

∆𝜃𝑎,𝑡 =𝑘𝑠ℎ

𝑐𝑎𝜌𝑎

𝐴𝑚

𝑉ℎ∆𝑡 , [3-81]

where

𝑘𝑠ℎ is the correction factor for the shadow effect;

𝑐𝑎 is the specific heat;

𝜌𝑎 is the density; 𝐴𝑚

𝑉 is the section factor;

∆𝑡 is the time increment (should be ≤ 5 seconds);

ℎ 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

ℎ = ℎ𝑟 + ℎ𝑐 . [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 𝑘𝑠ℎ. 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 𝐴𝑚/𝑉

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

θ

[K]

Time [min]

ISO-834

170 m-1

120 m-1

70 m-1

Am/V = 20 m-1


R.H.A. TITULAER 81

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 room’s 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 𝑙𝑓𝑖 = 0,7𝑙


R.H.A. TITULAER 82

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 Young’s 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 11×10-6 (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

θ

[K]

Time [min]

ISO-834

Matlab

CTM


R.H.A. TITULAER 83

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

θ[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

θ

[K]

𝜇0

CTM

Matlab

Figure 3-49. Influence degree of utilization on the critical temperature using the CTM and Matlab


R.H.A. TITULAER 84

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 𝐴𝑝/𝑉 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

]

𝜇0

Requirement

CTM

Matlab


R.H.A. TITULAER 85


R.H.A. TITULAER 86

4. CONCLUSIONS AND RECOMMENDATIONS

Finally, this thesis ends with the conclusions and recommendations.


R.H.A. TITULAER 87


R.H.A. TITULAER 88

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.


R.H.A. TITULAER 89

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 Young’s

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.


R.H.A. TITULAER 90

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ENGINEERING MODEL FOR COUPLED THERMOMECHANICAL BEHAVIOUR OF STEEL ELEMENTS UNDER FIRE CONDITIONS

R.H.A. TITULAER

26 JANUARY 2016

APPENDIX


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


R.H.A. TITULAER 1


R.H.A. TITULAER 2

A. CONFIGURATION FACTOR

For the total radiation the equation was given in Eq. [2-34]. In this equation the corresponding configuration factor

𝜙𝑗 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

𝑑𝐹𝑑1−𝑑2 =𝑐𝑜𝑠𝜃1𝑐𝑜𝑠𝜃2

𝜋𝑆1−22

𝑑𝐴2 . [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

𝜙 =1

2𝜋[

𝑎

(1+𝑎2)0.5tan−1 (

𝑏

(1+𝑎2)0.5) +

𝑏

(1+𝑏2)0.5tan−1 (

𝑎

(1+𝑏2)0.5)] , [A-2]

where

𝑎 =ℎ

𝑠 ,

𝑏 =𝑤

𝑠 ,

𝑠 is the distance from P to X,

ℎ is the height of the zone on the radiating surface,

𝑤 is the width of the zone on the radiating surface.

Figure A-1. Radiative heat transfer between two infinitesimal surface areas (CEN, 2011b)


R.H.A. TITULAER 3

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 𝜃 to the radiating surface. For the receiving surface perpendicular to the radiating surface the

configuration factor is given by

𝜙 =1

2𝜋[tan−1(𝑎) −

1

(1+𝑏2)0.5tan−1 (

𝑎

(1+𝑏2)0.5)] , [A-3]

and for the receiving surface in a plane at an angle 𝜃 to the radiating surface the configuration factor is written as

𝜙 =1

2𝜋[tan−1(𝑎) −

(1−𝑏𝑐𝑜𝑠𝜃)

(1+𝑏2−2𝑏𝑐𝑜𝑠𝜃)0.5tan−1 (

𝑎

(1+𝑏2−2𝑏𝑐𝑜𝑠𝜃)0.5) +

𝑎𝑐𝑜𝑠𝜃

(𝑎2+sin2 𝜃)0.5[tan−1 (

𝑏−𝑐𝑜𝑠𝜃

(𝑎2+sin2𝜃)0.5) +

tan−1 (𝑐𝑜𝑠𝜃

(𝑎2+sin2 𝜃)0.5)]] . [A-4]

A.2. CYLINDER AS RADIATIOR

The configuration factor for a cylinder is written as (Hamilton & Morgan, 1952)

𝜙 =1

𝜋𝐷tan−1 (

𝐿

√𝐷2−1) +

𝐿

𝜋[𝐴−2𝐷

𝐷√𝐴𝐵tan−1 √

𝐴(𝐷−1)

𝐵(𝐷+1)−

1

𝐷tan−1 √

𝐷−1

𝐷+1] , [A-5]

where

𝐷 =𝑑

𝑟 ,

𝐿 =𝑙

𝑟 ,

𝐴 = (𝐷 + 1)2 + 𝐿2 ,

𝐵 = (𝐷 − 1)2 + 𝐿2 .

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)


R.H.A. TITULAER 4

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

𝜙 =𝑎𝑏

√(𝑠2+𝑎2)(𝑠2+𝑏2) , [A-6]

where 𝑎, 𝑏, and 𝑠 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)


R.H.A. TITULAER 5


R.H.A. TITULAER 6

B. DERIVATION SHAPE FUNCTIONS AND FINAL FEM FORMULATION

The linear shape functions for the local system can be defined with the following figure.

𝑁1 = 𝑎1 + 𝑎2𝜉 ,

𝑁1(−1) = 𝑎1 − 𝑎2 = 1 ,

𝑁1(1) = 𝑎1 + 𝑎2 = 0 ,

𝑎2 = −1

2 ,

𝑎1 =1

2 ,

𝑁1 =1

2−

1

2𝜉 =

1

2(1 − 𝜉) . [B-1]

𝑁2 = 𝑎1 + 𝑎2𝜉 ,

𝑁1(−1) = 𝑎1 − 𝑎2 = 0 ,

𝑁1(1) = 𝑎1 + 𝑎2 = 1 ,

𝑎2 =1

2 ,

𝑎1 =1

2 ,

𝑁1 =1

2+

1

2𝜉 =

1

2(1 + 𝜉) . [B-2]

Combining these shape functions results in the shape function for the temperature. It has to be noted that 𝑑𝑥 =𝐿

2𝑑𝜉.

The shape functions and the first order derivative can then be formulated as

𝐍𝜃(𝜉) = [1

2(1 − 𝜉)

1

2(1 + 𝜉)] , [B-3]

𝐁𝜃 = 𝑑𝐍

𝑑𝑥=

𝑑𝐍

𝑑𝜉

𝑑𝜉

𝑑𝑥= [−

1

2

1

2]

2

𝐿= [−1 1]

1

𝐿 . [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.

𝑁1 = 𝑎1 + 𝑎2𝜉 + 𝑎3𝜉2 ,

𝑁1(−1) = 𝑎1 − 𝑎2 + 𝑎3 = 1 ,

𝑁1(0) = 𝑎1 = 0 ,

𝑁1(1) = 𝑎1 + 𝑎2 + 𝑎3 = 0 ,

𝑎2 = −1

2 ,

𝑎3 =1

2 ,

𝑁1 = −1

2𝜉 +

1

2𝜉2 = −

1

2𝜉(1 − 𝜉) . [B-5]

𝑁2 = 𝑎1 + 𝑎2𝜉 + 𝑎3𝜉2 ,

𝑁2(−1) = 𝑎1 − 𝑎2 + 𝑎3 = 0 ,

𝑁2(0) = 𝑎1 = 1 ,

𝑁2(1) = 𝑎1 + 𝑎2 + 𝑎3 = 0 ,

𝑎3 = −1 ,

𝑎2 = 0 ,

𝑁2 = 1 − 𝜉2 . [B-6]

Figure B-1. Global and local coordinate systems linear and quadratic (Eslami et al., 2013)


R.H.A. TITULAER 7

𝑁3 = 𝑎1 + 𝑎2𝜉 + 𝑎3𝜉2 ,

𝑁3(−1) = 𝑎1 − 𝑎2 + 𝑎3 = 0 ,

𝑁3(0) = 𝑎1 = 0 ,

𝑁3(1) = 𝑎1 + 𝑎2 + 𝑎3 = 1 ,

𝑎2 =1

2 ,

𝑎3 =1

2 ,

𝑁3 =1

2𝜉 +

1

2𝜉2 =

1

2𝜉(1 + 𝜉) . [B-7]

Again it has to be noted that 𝑑𝑥 =𝐿

2𝑑𝜉. The combined shape function for the displacement becomes

𝐍𝑢(𝜉) = [−1

2𝜉(1 − 𝜉) 1 − 𝜉2

1

2𝜉(1 + 𝜉)] , [B-8]

𝐁𝑢 = 𝑑𝐍

𝑑𝑥=

𝑑𝐍

𝑑𝜉

𝑑𝜉

𝑑𝑥= [−

1

2+ 𝜉 − 2𝜉

1

2+ 𝜉 ]

2

𝐿

= [−1 + 2𝜉 −4𝜉 1 + 2𝜉]1

𝐿 . [B-9]

Now the matrices can be solved using these shape functions. This gives

𝐂𝜃,𝑢(𝒆) = 𝜃0 ∫ 𝐍𝜃

T𝛽𝐁𝑢𝐴𝑑𝑥𝐿𝑒

0 =𝐿

2𝜃0 ∫ 𝐍𝜃

T𝛽𝐁𝑢𝐴𝑑𝜉1

−1 ,

𝐂𝜃,𝑢(𝒆) =

𝜃0𝛽

2∫ [

1

2(1 − 𝜉)

01

2(1 + 𝜉)

] [−1 + 2𝜉 −4𝜉 1 + 2𝜉]𝐴𝑑𝜉 =1

−1

𝜃0𝛽

2∫ [

−1

2+ 1

1

2𝜉 − 𝜉2 −2𝜉 + 2𝜉2

0 0

1

2+

1

2𝜉 − 𝜉2

0

−1

2+

1

2𝜉 + 𝜉2 −2𝜉 − 2𝜉2

1

2+ 1

1

2𝜉 + 𝜉2

] 𝐴𝑑𝜉1

−1 =𝜃0𝛽𝐴

6[−5 4 10 0 0

−1 −4 5] . [B-10]

𝐂𝜃,𝜃(𝒆) = ∫ 𝐍𝜃

T𝐶𝐸𝐍𝜃𝐴𝑑𝑥𝐿𝑒

0 =𝐿

2∫ 𝐍𝜃

T𝐶𝐸𝐍𝜃𝐴𝑑𝜉1

−1 ,

𝐂𝜃,𝜃(𝒆) =

𝐿𝐶𝐸

2∫ [

1

2(1 − 𝜉)

1

2(1 + 𝜉)

] [1

2(1 − 𝜉)

1

2(1 + 𝜉)] 𝐴𝑑𝜉 =

1

−1

𝐿𝐶𝐸

2∫ [

1

4−

1

2𝜉 +

1

4𝜉2 1

4−

1

4𝜉2

1

4−

1

4𝜉2 1

4+

1

2𝜉 +

1

4𝜉2

] 𝐴𝑑𝜉 =1

−1

𝐿𝐶𝐸𝐴

6[2 11 2

] =𝐿𝐶𝐸𝐴

6[20

00

10

1 0 2] . [B-11]

𝐊𝑢,𝑢(𝒆) = ∫ 𝐁𝑢

TC𝑢𝐁𝑢𝐴𝑑𝑥𝐿𝑒

0 =𝐿

2∫ 𝐁𝑢

TC𝑢𝐁𝑢𝐴𝑑𝜉1

−1 ,

𝐊𝑢,𝑢(𝒆) =

𝐶𝑈

2𝐿∫ [

−1 + 2𝜉−4𝜉

1 + 2𝜉] [−1 + 2𝜉 −4𝜉 1 + 2𝜉]𝐴𝑑𝜉

1

−1 =

𝐶𝑈

2𝐿∫ [

1 − 4𝜉 + 4𝜉2 4𝜉 − 8𝜉2 −1 + 4𝜉2

4𝜉 − 8𝜉2 16𝜉2 −4𝜉 − 8𝜉2

−1 + 4𝜉2 −4𝜉 − 8𝜉2 1 − 4𝜉 + 4𝜉2

] A 𝑑𝜉1

−1 =𝐶𝑈𝐴

6𝐿[

14−16

−1632

2−16

2 −16 14] . [B-12]

𝐊𝑢,𝜃(𝒆) = ∫ 𝐁𝑢

T𝛽𝐍𝜃𝐴𝑑𝑥𝐿𝑒

0 =𝐿

2∫ 𝐁𝑢

T𝛽𝐍𝜃𝐴𝑑𝜉1

−1 ,

𝐊𝑢,𝜃(𝒆) =

𝛽

2∫ [

−1 + 2𝜉−4𝜉

1 + 2𝜉] [

1

2(1 − 𝜉) 0

1

2(1 + 𝜉)] 𝐴𝑑𝜉 =

1

−1

𝛽

2∫

[ −

1

2+ 1

1

2𝜉 − 𝜉2 0

−2𝜉 + 2𝜉2 0

−1

2+

1

2𝜉 + 𝜉2

−2𝜉 − 2𝜉2

1

2+

1

2𝜉 − 𝜉2 0

1

2+ 1

1

2𝜉 + 𝜉2

]

𝐴𝑑𝜉 =𝛽𝐴

6

1

−1 [−5 0 −14 0 −41 0 5

] . [B-13]


R.H.A. TITULAER 8

𝐊𝜃,𝜃(𝒆) = ∫ 𝐁𝜃

TK𝜃𝐁𝜃𝐴𝑑𝑥𝐿𝑒

0 =𝐿

2∫ 𝐁𝜃

TK𝜃𝐁𝜃𝐴𝑑𝜉1

−1 ,

𝐊𝜃,𝜃(𝒆) =

K𝜃

2𝐿∫ [

−11

] [−1 1]𝐴𝑑𝜉1

−1 =K𝜃

2𝐿∫ [

1 −1−1 1

] 𝐴𝑑𝜉1

−1 =K𝜃𝐴

𝐿[

1 −1−1 1

]

=K𝜃𝐴

𝐿[

10

00

−10

−1 0 1] . [B-14]

𝐊𝜃,𝜃(𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛)(𝒆) = ∫ (𝐍𝜃

Tℎ𝑟𝐍𝜃𝑃𝑑𝑥

𝐿𝑒

0 =𝐿

2∫ (𝐍𝜃

Tℎ𝑟𝐍𝜃𝑃𝑑𝜉

1

−1 ,

𝐊𝜃,𝜃(𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛)(𝒆) =

𝐿ℎ𝑟

2∫ [

1

2(1 − 𝜉)

1

2(1 + 𝜉)

] [1

2(1 − 𝜉)

1

2(1 + 𝜉)] 𝑃𝑑𝜉 =

1

−1

𝐿ℎ𝑟

2∫ [

1

4−

1

2𝜉 +

1

4𝜉2 1

4−

1

4𝜉2

1

4−

1

4𝜉2 1

4+

1

2𝜉 +

1

4𝜉2

]𝑃𝑑𝜉 =1

−1

𝐿ℎ𝑟𝑃

6[2 11 2

] =𝐿ℎ𝑟𝑃

6[20

00

10

1 0 2] . [B-15]

𝐊𝜃,𝜃(𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛)(𝒆) = ∫ (𝐍𝜃

Tℎ𝑐𝐍𝜃𝑃𝑑𝑥

𝐿𝑒

0 =𝐿

2∫ (𝐍𝜃

Tℎ𝑐𝐍𝜃𝑃𝑑𝜉

1

−1 ,

𝐊𝜃,𝜃(𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛)(𝒆) =

𝐿ℎ𝑐

2∫ [

1

2(1 − 𝜉)

1

2(1 + 𝜉)

] [1

2(1 − 𝜉)

1

2(1 + 𝜉)]𝑃𝑑𝜉 =

1

−1

𝐿ℎ𝑐

2∫ [

1

4−

1

2𝜉 +

1

4𝜉2 1

4−

1

4𝜉2

1

4−

1

4𝜉2 1

4+

1

2𝜉 +

1

4𝜉2

] 𝑃𝑑𝜉 =1

−1

𝐿ℎ𝑐𝑃

6[2 11 2

] =𝐿ℎ𝑐𝑃

6[20

00

10

1 0 2] . [B-16]

𝐟𝑢(𝒆) = 𝐍𝑢

T�̂�𝑃|0𝐿𝑒

=𝐿

2𝐍𝑢

T�̂�𝑃|−11 =

𝐿

2�̂�𝑃

[ −

1

2𝜉(1 − 𝜉)

1 − 𝜉2

1

2𝜉(1 + 𝜉) ]

|−11

=𝐿

2�̂�𝑃 {

−101

} . [B-17]

𝐟𝜃(𝒆) = −𝐍𝜃

T�̂�𝑃|0𝐿𝑒

= −𝐿

2𝐍𝜃

T�̂�𝑃|−1

1 = −𝐿

2�̂�𝑃 [

1

2(1 − 𝜉)

1

2(1 + 𝜉)

] |−11 = −

𝐿

2�̂�𝑃 {

−11

}

= −𝐿

2𝑃�̂�{

−11

} . [B-18]

𝐟𝜃(𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛)(𝒆) = ∫ 𝐍𝜃

Tℎ𝑟(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥𝐿𝑒

0 = 𝐿ℎ𝑟𝜃𝐴𝑆𝑇

2∫ [

1

2(1 − 𝜉)

1

2(1 + 𝜉)

] 𝑃𝑑𝑥𝐿𝑒

0

=𝐿ℎ𝑟𝜃𝐴𝑆𝑇𝑃

2{11} . [B-19]

𝐟𝜃(𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛)(𝒆) = ∫ 𝐍𝜃

Tℎ𝑐(𝜃𝐴𝑆𝑇(𝑡))𝑃𝑑𝑥𝐿𝑒

0 = 𝐿ℎ𝑐𝜃𝐴𝑆𝑇

2∫ [

1

2(1 − 𝜉)

1

2(1 + 𝜉)

]𝑃𝑑𝑥𝐿𝑒

0

=𝐿ℎ𝑐𝜃𝐴𝑆𝑇𝑃

2{11} . [B-20]


R.H.A. TITULAER 9


R.H.A. TITULAER 10

C. DERIVATION EULER-BERNOULLI STIFFNESS MATRICES

The conventional stiffness matrix can be found with

𝐊(m) = ∫ 𝐁(m) T𝐸𝐼𝐁(m) 𝑑𝑥ℎ(𝑚) =

44434241

34333231

24232221

14131211

kkkk

kkkk

kkkk

kkkk

, [C-1]

𝑘11 → 𝐸𝐼 (36

𝑙4−

144𝑥

𝑙5+

144𝑥2

𝑙6) → 𝐸𝐼 [

36𝑥

𝑙4−

72𝑥2

𝑙5+

48𝑥3

𝑙6]0

𝑙

=12𝐸𝐼

𝑙3 , [C-2]

𝑘12 → 𝐸𝐼 (−24

𝑙3+

36𝑥

𝑙4+

48𝑥

𝑙4−

72𝑥2

𝑙5) → 𝐸𝐼 [

−24𝑥

𝑙3+

18𝑥2

𝑙4+

24𝑥2

𝑙4−

24𝑥3

𝑙5]0

𝑙

=−6𝐸𝐼

𝑙2 , [C-3]

𝑘13 → 𝐸𝐼 (−36

𝑙4+

144𝑥

𝑙5−

144𝑥2

𝑙6) → 𝐸𝐼 [

−36𝑥

𝑙4+

72𝑥2

𝑙5−

48𝑥3

𝑙6]0

𝑙

=−12𝐸𝐼

𝑙3 , [C-4]

𝑘14 → 𝐸𝐼 (−12

𝑙3+

36𝑥

𝑙4+

24𝑥

𝑙4−

72𝑥2

𝑙5) → 𝐸𝐼 [

−12𝑥

𝑙3+

18𝑥2

𝑙4+

12𝑥2

𝑙4−

24𝑥3

𝑙5]0

𝑙

=−6𝐸𝐼

𝑙2 , [C-5]

𝑘21 → 𝐸𝐼 (−24

𝑙3+

36𝑥

𝑙4+

48𝑥

𝑙4−

72𝑥2

𝑙5) → 𝐸𝐼 [

−24𝑥

𝑙3+

18𝑥2

𝑙4+

24𝑥2

𝑙4−

24𝑥3

𝑙5]0

𝑙

=−6𝐸𝐼

𝑙2 , [C-6]

𝑘22 → 𝐸𝐼 (16

𝑙2−

48𝑥

𝑙3+

36𝑥2

𝑙4) → 𝐸𝐼 [

16𝑥

𝑙2−

24𝑥2

𝑙3+

12𝑥3

𝑙4]0

𝑙

=4𝐸𝐼

𝑙 , [C-7]

𝑘23 → 𝐸𝐼 (24

𝑙3−

36𝑥

𝑙4−

48𝑥

𝑙4+

72𝑥2

𝑙5) → 𝐸𝐼 [

24𝑥

𝑙3−

18𝑥2

𝑙4−

24𝑥2

𝑙4+

24𝑥3

𝑙5]0

𝑙

=6𝐸𝐼

𝑙2 , [C-8]

𝑘24 → 𝐸𝐼 (8

𝑙2−

36𝑥

𝑙3+

36𝑥2

𝑙4) → 𝐸𝐼 [

8𝑥

𝑙2−

18𝑥2

𝑙3+

12𝑥3

𝑙4]0

𝑙

=2𝐸𝐼

𝑙 , [C-9]

𝑘31 → 𝐸𝐼 (−36

𝑙4+

144𝑥

𝑙5−

144𝑥2

𝑙6) → 𝐸𝐼 [

−36𝑥

𝑙4+

72𝑥2

𝑙5−

48𝑥3

𝑙6]0

𝑙

=−12𝐸𝐼

𝑙3 , [C-10]

𝑘32 → 𝐸𝐼 (24

𝑙3−

36𝑥

𝑙4−

48𝑥

𝑙4+

72𝑥2

𝑙5) → 𝐸𝐼 [

24𝑥

𝑙3−

18𝑥2

𝑙4−

24𝑥2

𝑙4+

24𝑥3

𝑙5]0

𝑙

=6𝐸𝐼

𝑙2 , [C-11]

𝑘33 → 𝐸𝐼 (36

𝑙4−

144𝑥

𝑙5+

144𝑥2

𝑙6) → 𝐸𝐼 [

36𝑥

𝑙4−

72𝑥2

𝑙5+

48𝑥3

𝑙6]0

𝑙

=12𝐸𝐼

𝑙3 , [C-12]

𝑘34 → 𝐸𝐼 (12

𝑙3−

36𝑥

𝑙4−

24𝑥

𝑙4+

72𝑥2

𝑙5) → 𝐸𝐼 [

12𝑥

𝑙3−

18𝑥2

𝑙4−

12𝑥2

𝑙4+

24𝑥3

𝑙5]0

𝑙

=6𝐸𝐼

𝑙2 , [C-13]

𝑘41 → 𝐸𝐼 (−12

𝑙3+

36𝑥

𝑙4+

24𝑥

𝑙4−

72𝑥2

𝑙5) → 𝐸𝐼 [

−12𝑥

𝑙3+

18𝑥2

𝑙4+

12𝑥2

𝑙4−

24𝑥3

𝑙5]0

𝑙

=−6𝐸𝐼

𝑙2 , [C-14]

𝑘42 → 𝐸𝐼 (8

𝑙2−

36𝑥

𝑙3+

36𝑥2

𝑙4) → 𝐸𝐼 [

8𝑥

𝑙2−

18𝑥2

𝑙3+

12𝑥3

𝑙4]0

𝑙

=2𝐸𝐼

𝑙 , [C-15]

𝑘43 → 𝐸𝐼 (12

𝑙3−

36𝑥

𝑙4−

24𝑥

𝑙4+

72𝑥2

𝑙5) → 𝐸𝐼 [

12𝑥

𝑙3−

18𝑥2

𝑙4−

12𝑥2

𝑙4+

24𝑥3

𝑙5]0

𝑙

=6𝐸𝐼

𝑙2 , [C-16]

𝑘44 → 𝐸𝐼 (4

𝑙2−

24𝑥

𝑙3+

36𝑥2

𝑙4) → 𝐸𝐼 [

4𝑥

𝑙2−

12𝑥2

𝑙3+

12𝑥3

𝑙4]0

𝑙

=4𝐸𝐼

𝑙 , [C-17]

𝐊(m) =𝐸𝐼

𝑙3

22

22

4626

6126122646

612612

llll

llllll

ll

. [C-18]


R.H.A. TITULAER 11

The stress stiffness matrix can be found with

𝐊𝝈(𝐦)

= ∫ 𝐆(m) T�̂�𝐆(m)𝑑𝑥ℎ(𝑚) =

44434241

34333231

24232221

14131211

ksksksks

ksksksks

ksksksks

ksksksks

, [C-19]

𝑘𝑠11 → �̂� (36𝑥2

𝑙4−

72𝑥3

𝑙5+

36𝑥4

𝑙6) → �̂� [

12𝑥3

𝑙4−

18𝑥4

𝑙5+

36𝑥5

5𝑙6]0

𝑙

=36�̂�

30𝑙, [C-20]

𝑘𝑠12 → �̂� (6𝑥

𝑙2−

30𝑥2

𝑙3+

42𝑥3

𝑙4−

18𝑥4

𝑙5) → �̂� [

3𝑥2

𝑙2−

10𝑥3

𝑙3+

42𝑥4

4𝑙4−

18𝑥5

5𝑙5]0

𝑙

=−3�̂�

30 , [C-21]

𝑘𝑠13 → �̂� (−36𝑥2

𝑙4+

72𝑥3

𝑙5−

36𝑥4

𝑙6) → �̂� [

−12𝑥3

𝑙4+

18𝑥4

𝑙5−

36𝑥5

5𝑙6]0

𝑙

=−36�̂�

30𝑙 , [C-22]

𝑘𝑠14 → �̂� (−12𝑥2

𝑙3+

30𝑥3

𝑙4−

18𝑥4

𝑙5) → �̂� [

−4𝑥3

𝑙3+

30𝑥4

4𝑙4−

18𝑥5

5𝑙5]0

𝑙

=−3�̂�

30 , [C-23]

𝑘𝑠21 → �̂� (6𝑥

𝑙2−

30𝑥2

𝑙3+

42𝑥3

𝑙4−

18𝑥4

𝑙5) → �̂� [

3𝑥2

𝑙2−

10𝑥3

𝑙3+

42𝑥4

4𝑙4−

18𝑥5

5𝑙5]0

𝑙

=−3�̂�

30, [C-24]

𝑘𝑠22 → �̂� (1 −8𝑥

𝑙+

22𝑥2

𝑙2−

24𝑥3

𝑙3+

9𝑥4

𝑙4) → �̂� [𝑥 −

4𝑥2

𝑙+

22𝑥3

3𝑙2−

6𝑥4

𝑙3+

9𝑥5

5𝑙4]0

𝑙

=4�̂�𝑙

30 , [C-25]

𝑘𝑠23 → �̂� (−6𝑥

𝑙2+

30𝑥2

𝑙3−

42𝑥3

𝑙4+

18𝑥4

𝑙5) → �̂� [

−3𝑥2

𝑙2+

10𝑥3

𝑙3−

42𝑥4

4𝑙4+

18𝑥5

5𝑙5]0

𝑙

=3�̂�

30 , [C-26]

𝑘𝑠24 → �̂� (−2𝑥

𝑙+

11𝑥2

𝑙2−

18𝑥3

𝑙3+

9𝑥4

𝑙4) → �̂� [−

𝑥2

𝑙+

11𝑥3

3𝑙2−

18𝑥4

4𝑙3+

9𝑥5

5𝑙4]0

𝑙

=−�̂�𝑙

30 , [C-27]

𝑘𝑠31 → �̂� (−36𝑥2

𝑙4+

72𝑥3

𝑙5−

36𝑥4

𝑙6) → �̂� [

−12𝑥3

𝑙4+

18𝑥4

𝑙5−

36𝑥5

5𝑙6]0

𝑙

=−36�̂�

30𝑙 , [C-28]

𝑘𝑠32 → �̂� (−6𝑥

𝑙2+

30𝑥2

𝑙3−

42𝑥3

𝑙4+

18𝑥4

𝑙5) → �̂� [

−3𝑥2

𝑙2+

10𝑥3

𝑙3−

42𝑥4

4𝑙4+

18𝑥5

5𝑙5]0

𝑙

=3�̂�

30 , [C-29]

𝑘𝑠33 → �̂� (36𝑥2

𝑙4−

72𝑥3

𝑙5+

36𝑥4

𝑙6) → �̂� [

12𝑥3

𝑙4−

18𝑥4

𝑙5+

36𝑥5

5𝑙6]0

𝑙

=36�̂�

30𝑙 , [C-30]

𝑘𝑠34 → �̂� (12𝑥2

𝑙3−

30𝑥3

𝑙4+

18𝑥4

𝑙5) → �̂� [

4𝑥3

𝑙3−

30𝑥4

4𝑙4+

18𝑥5

5𝑙5]0

𝑙

=3�̂�

30 , [C-31]

𝑘𝑠41 → �̂� (−12𝑥2

𝑙3+

30𝑥3

𝑙4−

18𝑥4

𝑙5) → �̂� [

−4𝑥3

𝑙3+

30𝑥4

4𝑙4−

18𝑥5

5𝑙5]0

𝑙

=−3�̂�

30 , [C-32]

𝑘𝑠42 → �̂� (−2𝑥

𝑙+

11𝑥2

𝑙2−

18𝑥3

𝑙3+

9𝑥4

𝑙4) → �̂� [−

𝑥2

𝑙+

11𝑥3

3𝑙2−

18𝑥4

4𝑙3+

9𝑥5

5𝑙4]0

𝑙

=−�̂�𝑙

30 , [C-33]

𝑘𝑠43 → �̂� (12𝑥2

𝑙3−

30𝑥3

𝑙4+

18𝑥4

𝑙5) → �̂� [

4𝑥3

𝑙3−

30𝑥4

4𝑙4+

18𝑥5

5𝑙5]0

𝑙

=3�̂�

30 , [C-34]

𝑘𝑠44 → �̂� (4𝑥2

𝑙2−

12𝑥3

𝑙3+

9𝑥4

𝑙4) → �̂� [

4𝑥3

3𝑙2−

3𝑥4

𝑙3+

9𝑥5

5𝑙4]0

𝑙

=4�̂�𝑙

30 , [C-35]

𝐊𝝈(𝐦)

=�̂�

30𝑙

22

22

433336336

343336336

llllllllllll

. [C-36]


R.H.A. TITULAER 12

D. ELEMENT MATRICES THERMAL NON-LINEARITY

For one quadratic element the formulation can be written as

𝐂�̇� + 𝐊𝐝 = 𝐟 , [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 𝑢, 𝑤, and 𝜑 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

𝒅′ = 𝐓𝒅 , [D-3]

where 𝐓 is the transformation matrix and 𝒅 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]


R.H.A. TITULAER 13


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).


R.H.A. TITULAER 15


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 (Δx).

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 Young’s 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


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


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


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


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 �̃� 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


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


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


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


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


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 Young’s 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 Young’s

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. Young’s modulus reduction


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


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


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


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


R.H.A. TITULAER 30

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


R.H.A. TITULAER 31

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 lamdba’s 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


R.H.A. TITULAER 32

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


R.H.A. TITULAER 33

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

Young’s 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

eindhoven university of technology master engineering

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