msc dissertation 1

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Reliability Analysis of Reinforced Concrete Beams

[Student name]

Khaled Mohammad A Emhamed

[Student Registration Number] 091586571

A dissertation submitted in partial fulfilment of the requirements for the degree of Master of Engineering in

Structural Engineering

[Course title] Structural and Foundation Engineering

Dissertation Supervisor: [Supervisor’s name]

Dr Dmitry Val

Heriot-Watt University School of the Built Environment

[Date]

23 / 08 / 2010

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STATEMENT OF AUTHORSHIP

I, Khaled Mohammad A Emhamed, confirm that this work submitted for assessment in

partial fulfilment of requirements for the degree of Master of Science in Structural

and Foundation Engineering, at Heriot -Watt University (Edinburgh), is my own and

is expressed in my own words. Any uses made within of the works of other authors in

any form (e.g. ideas, equations, figures, texts, tables, programs) are properly

acknowledged at their point of use and have been referenced. A full list of the references

employed has been included and may be reviewed at the back of this text.

………………………… ….………………..

Khaled Emhamed Date

Reg. No: 091586571

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“Reliability of Reinforced Concrete Beams” MSc Dissertation

TABLE OF CONTENTS

STATEMENT OF AUTHORSHIP......................................................................i

TABLE OF CONTENTS………………………………………………….….…ii

ACKNOLOWLEDGEMENT………………………………………………......v ABSTRACT…………………………………………………………….……..…vi CHAPTER 1 INTRODUCTION …………………………………….…....1 CHAPTER 2 AIM AND OBJECTIVES………………………....…….….3 CHAPTER3 LITRATURE REVIEW 3-1 Reliability Developments ……………………………………………….……4

3-2 Reliability Analysis of Reinforced concrete beam philosophy……….……….5

3-3 Reliability Assessment of Reinforced Concrete Beams…………….…….…...6

CHAPTER 4 REINFORCED CONCRETE BEAMS

4-1 Introduction ………………………………………………………………......8

4-2 Brittle and Ductile Failure …………………………………………………....8

4-3 Brittle Failure of Reinforced Concrete Beams ……………………………….9

4-3-1 Flexural Failure………………………………………………………….....11

4-3-2 Shear Failure…………………………………………………………….....13

4-4 Effective of Shear Links on Failure Mechanism……………………………. 13

4-5 Designing of Reinforced Concrete Beams…………………………………...17

CHAPTER 5 DISCRIPTION OF THE METHOD

5-1 Reliability Engineering Method…………………………………………...….18

5-2 Selection of the Beams………………………………………………………..19

5-3 Load Combination and Patterns for the Limit States…………………………21

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5-4 Reliability Analysis Methods ………………………………………………...22

5-4-1 First Order Reliability Method (FORM)……………………………….…...22

5-4-2 Monte Carlo Simulation Method (MCS)…………………………………...25

5-5 Reliability Analysis of Reinforced Concrete Beams and Results..............…...27

5-5-1 Partial of Safety Factors……………………………………………….…....27

5-5-2 Limit State Functions………………………………………………….…....27

CHAPTER 6 APPLICATIONS OF RELIABILITY ANALYSIS

6-1 First Order Reliability Method………………………………………………..29

6-1-1 Limit State Function of Flexural Failure B1…………………………….....29

6-1-2 Limit State Function of Crushing Concrete Failure B1……………………32

6-1-3 Limit State Function of Shear Failure B1……………………………….....33

6-1-4 Limit State Function of Flexural Failure B2……………………………….34

6-1-5 Limit State Function of Crushing Concrete Failure B2 …………………...36

6-1-6 Limit State Function of Shear Failure B2………………………………….36

6-1-7 Limit State Function of Flexural Failure B3……………………………….38

6-1-8 Limit State Function of Crushing Concrete Failure B3 …………………...40

6-1-9 Limit State Function of Shear Failure B3……………………………….....41

6-2 Monte Carlo Simulation Method …………………………………………….42

6-2-1 Monte Carlo Applications B1……………………………………………....42

CHAPTER 7 RESULTS AND DISCUSSION

7-1 The Results…………………………………………………………………...43

7-1-1 Results of the Reliability Analysis B1……………………………………..43



7-2 Discussion…………………………………………………………………….45

7-2-1 Discussion Results of Reliability Analysis B1…………………………….45

7-2-2 Discussion Results of Reliability Analysis B2…………………….………46

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7-2-3 Discussion Results of Reliability Analysis B3……………….……………46

CHAPTER 8 CONCLUSION………………………………………………47

BIBLIUGRAPHY………………………………………………………………..48

APPINDECES

APPENDIX A

Design of Reinforced Concrete Beams…………………………….……….…..…51

APPENDIX B

Output Excel Result Using First Order Reliability Method……………………....56

APPENDIX C

Calculation Carried Out Using Monte Carlo Simulation………………….……....66

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ACKNOWLEDGEMENTS The author of this report would like to thank my supervisor Dr.Dimitry Val for his help

in producing this dissertation, and I would like to thank the staff at Heriot-Watt

University, School of Built Environment, Department of Structural and Foundation

Engineering and for help to achieve this project work.

Also I would like to extend my regards to all my colleagues, for supporting me in my

project and for providing the journal papers to use them in my dissertation.

Finally, I would like also to thank all my family and my Friend Mohammed who

supported me all along my period of study, in the UK.

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ABSTRACT Reliability of structural engineering is the key to estimate the probability of failure for

the majority of structures this time .In this project the main points are to set up to

calculate the probability of brittle failure in reinforced concrete beams , rather than

ductile failure , as we know there are two types of concrete failure , the first type is

ductile failure , this type of failure occurs gradually , and there will be enough time to

keep this failure under control , and the maintenance will coast low price , the second

type is called brittle failure , and this type of failure is more dangerous than the first

type, the brittle failure occurs suddenly and without any warning , also there is no

enough time to prevent this type , and if it is occur it may cause a big disaster for the

structure .

In this project, I am going to estimate the probability of failure, for three different beams

with a various dimensions and loads, for these three beams it has been mentioned to the

three types of failure, flexural due to yielding of reinforcement (ductile failure) , flexural

due to crushing of concrete (brittle failure) , and shear failure, for each failure a

probability of failure will be calculated, and according to this value of probability of

failure, we can decide how to reduce this failure before it occurs.

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CHAPTER 1 INTRODUCTION

Reliability analysis of reinforced concrete beams with respect to the different failures

according to Euro Code 2 design [1] Structural reliability theory has been developed in

order to provide a balanced treatment to understand the uncertainties that exist in

structures and apply various methodologies available in order to asses the safety and

serviceability of such structures .

In general there are two types of failure can occurs in reinforced concrete beams brittle

and ductile failure where the brittle failure is more dangerous than the ductile failure.

According to the ductile for the reinforced concrete structures there is a lot of time to

maintenance the structure and to recover the structure to his original situation before it

collapse but the brittle failure occurs suddenly and the structure will collapse without

any chance to repair or maintained the structure, by using the reliability analysis

methods to calculate the probability of brittle failure to prevent this failure for the

reinforced concrete beams.

The response of structures can be considered satisfactory if the requirements such as

safety against brittle failure and limitation damage or deflection are satisfied each of

these requirements may be treated as (Limit State). The violation of limit state can thus

be treated as failure for the structures. In this project the study of structural reliability is

concerned with preventing the brittle failure of reinforced concrete beams such as

flexural and shears failures and to achieve these results. I have designed various number

of reinforced concrete beams have various failure modes and different dimensions

according to Euro Code 2.

In the reliability analysis of structure engineering a very large number of the system

parameters, can be considered to be random variables. The difficulty in calculating the

probability of failure increase rapidly with the number of variables one of the aims of

this project is to compute the probability of failure of the reinforced concrete beams with

different failure modes and from that values it should be different number of probability

of failure for each failure, the probability of failure for the flexural in tension case more

likely than the compression and shear failure.

Theory and methods for structural reliability [2] have been developed substantially in

recent time and they are very useful tool for computing the probability of failure and

evaluating the safety of complex structures such as high buildings and bridges.

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There are many reliability methods such as Monte Carlo Simulation (MCS) First Order

Reliability Method (FORM) and First Order Second Moment (FOSM). In this project I

have used the First Order Reliability Method to estimate the probability of failure for the

reinforced concrete beams have different dimensions according to Euro Code 2; There

are certain limitations that restricted the enhancement of the scope of this project these

are mentioned as follows:-

� In this project I have designed a three different reinforced concrete beams having

different dimensions according to EC 2 and the analysis by using the reliability

methods. Hence estimate the probability of failure for each beam.

� This project is completely based on the concept of reliability of structural

engineering and hence would be more convenient for some are who have prior ideas

and knowledge of reliability or reinforced concrete structures.

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CHAPTER 2 AIM AND OBJECTIVES The aim of this project is to investigate the reliability of reinforced concrete beams. In

order to do those probabilities of failure determine the reliability of failure in different

failure modes, for several reinforced concrete beams with different dimensions using

FORM and Monte Carlo simulation.

The objectives of this project can be summarised by the following set of bullet points:

� The study is carried out to design several number of reinforced concrete beams , with

different dimension and reinforcement ratio , according to the Eurocode 2 [1] , with

respect to the three types of failure :-

• Beams – flexure – ductile (due to yielding of tensile reinforcement)

• Beams – flexure – brittle (due to crushing of compressed concrete)

• Beams – shear failure.

� To carry out the reliability analysis of the beams using FORM and Monte Carlo

simulation, and calculate the safety index (probability of failure), for three different

failure modes, flexural due to yielding of reinforcement (ductile), flexural due to

crushing of concrete (brittle), and shear failure.

� To compare the results of the safety index and probability of failure, for three

different beams, with different failure modes, using the results obtained using FORM

and Monte Carlo simulation.

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CHAPTER 3 LITERATURE REVIEW 3-1 Reliability Developments In 1986 Arafa [14] has presented a reliability analysis of reinforced concrete beam

section at the flexural limit state failure situation. The mode of the bending failure of the

beam section was modelled placement the main constitutive laws for concrete and steel

structures he used a Monte Carlo Simulation method to compute the action of the beam

section at their flexural limit state function the results of the reliability based analysis

were computed and presented by Arafa in condition and steps of the reliability index of

reinforced concrete beams at a different levels of reinforcement at the reinforcement

about 0.4 of the maximum allowable reinforcement ratio the reliability index was about

4.0 this value decreased to about 2.4 where the maximum allowable steel reinforcement

was used. Those results indicate that the reliability of reinforced concrete beams section

was highly critical to changeable in the compression and the design safety factors were

kept same and constant.

In his study Arafa [14] recommended the results shows that the reliability of reinforced

concrete beams was highly critical to variation in the compression and tension steel

reinforcement. Also when the factors of safety included in the design were kept constant.

High reliability values were noticed for less value of tension steel reinforcement ratio. In

opposite the compression steel reinforcement rise the section ductility and reliability and

finally found that the compression reinforcement was larger and recommended in beam

section.

Nearly during 2006 [8] at the University federal de Santa Catarina SC-Brazil structural

reliability is developed and linked to the ANSYS finite element program and they have

been suggested that the definition of random variable parameters distribution and by

definition a limit state function based on finite element analysis to get the results of the

probability of failure for many types of complex structures.

Reliability analysis also used [18] to design the simply supported beams too, in

accordance with BS5950 (1990), was carried using First Order Reliability Method

(FORM), at ultimate and serviceability limit states. Design variable such as load ratio,

span and intensity of imposed loading were considered random and stochastic. It was

shown among other findings that, when the span (L) of the beam and the load ratio ( o� )

are kept constant, as the magnitude of imposed load ( kQ ) increased by 200 %, they

found that the safety of the designed section decreased by 66% considering bending,

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84% considering shear and by 82 % when deflection criterion was considered. Also, the

weighted average safety indices are respectively 2.03, 5.69 and 1.72 for bending, shear

and deflection failure criteria of the BS5950 (1990). Therefore, BS 5950 design results

seems conservative with respect to shear, unsafe with respect to bending (under low

shear load ), and satisfactory with respect to deflection.

3-2 Reliability Analysis of Reinforced concrete beam philosophy The design of structures may be regarded as the process of selecting proper materials

and proportioned [8] elements of the structures according to the engineering science and

technology. In order to fulfil its purpose a structural must meet its requirements of safety

serviceability economy and functionality.

In the last decades the structural reliability have been widely developed and the

structural designers used it to predict the safety index since the huge number of

constructors were collapsed during the natural disasters or by human errors.

The concept of structural reliability is to extreme importance due to the fact that all

structural members should be able to resist or hold the external acting such as live loads

(snow wind) …etc .The various acting loads may cause different types of failure in the

structures members, such as flexural and shear failure in the reinforced concrete beams

structural engineers involved in the constructions of any structure play, an important part

in design process against brittle failure where structural consistency of the design item is

a matter of safety and reliability. Furthermore design codes play an important role in

verifying the safety of important design solution to prevent any kind of failure in the

structures. From the prescription of structural engineers it is essential to ensure that these

design rules can achieve the required safety level of the structure and to prevent the

flexural and shear failure on that structure. Most a concrete structures have a difference

two types of failure brittle and ductile failure for the building fails under a ductile failure

the owner of the building has a chance to repair the damage of that structure or to leave

the building before the final collapse .On the other hand if the structure has a brittle

failure (sudden failure) that mean the structure is over loaded and the most probable the

structure has a final collapse without any warning to the owner of that building. And to

prevent this type if failure the structural engineers have used the reliability engineering

to estimate the probability of failure � �fP by computing the safety index � �� according

to the designing code and to estimate the different values for � �� to decide which is the

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most failure will occur and the ways to prevent this kind of failure . This types of

failure are occurs in different dimension of structures such as reinforced concrete beams

and the diagram below shows the structural reliability analysis and potential way to

measure of structural safety the failure probability.

3-3 Reliability Assessment of Reinforced Concrete Beams

Recent reinforced concrete design code [3] as Eurocode 2, which I have used in this

project or ACI 318 is based on limit state theory.

The following figure (3-1) show the example of three points flexural failure in

reinforced concrete beam loaded by force P designed by Euro Code 2 allow to determine

the acceptable value of P such as fixedP Pf f� .

Figure 3-1 steps computing for bending beam

Figure (3-1) Reliability analysis of a structure with random parameters

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The majority of reinforced concrete beams failures cause by bending failure the case of

high loaded and less depth of beam can be divided in two points. Firstly the residual

reliability of bending reinforced concrete beam ( bendingPf ) provide an data about the

safety reduction of the beam .The next step is the expecting of the remaining loading

capacity of the crushing beam ( bendingP ) that ensure the same probability of failure as

initially designed

( bending fixedffP P� ).

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CHAPTER 4 REINFORCED CONCRETE BEAMS 4-1 Introduction Nowadays[10] reinforced concrete beams are designed traditional methods such as BS

8110 Eurocode 2 and ACI 318 based on truss action a unified design approach has been

developed using a flexural - shear interaction that is ignored in the traditional methods.

The flexural -Shear interaction reduces the compressive concrete strength and hence the

ultimate flexural and shear failure capacity of a section the reinforced concrete beams

consist of concrete and steel each one of theme has his function when both of them

combined make high strength concrete beam when they are combined the steel is

working to increase the tensile strength and probably some of the shear strength wile the

concrete strong in compression protects the steel to give durability and fire resistance

.Figure (4-1) below shows how the steel and the concrete are working.

` 4-2 Brittle and Ductile Failure

In this project the main concept is about computing the probability of brittle failure for

the reinforced concrete beams and how to prevent this type of concrete failure because

this type of failure is the most dangerous in concrete structures and when this failure is

happen there is no time to make the maintenance for that structure. On the[4] other hand

for the ductile failure when this type of failure appears on the concrete structure there is

Figure 4-1 Steel reinforcement and concrete behaviour

16

a chance and enough time to recover the structure the figure below shows the different

between the brittle and ductile failure.

4-3 Brittle Failure of Reinforced Concrete Beams Preventing brittle failure [5] of reinforced concrete beams is being developed by

using many strength materials such as steel reinforcement to reduce of ductility in a

brittle reinforced concrete beam means that there is very little potential for stress

redistribution. For example the lower bond thermo of plasticity (Which permit many of

assumption made in steel reinforce concrete analysis) can not be applied.

Analysis of brittle failure of reinforced concrete beams should be on detailed tests of

compatibility needed within a beam of the phenomenon crack occurs and this is the

Figure (4-2) Brittle and Ductile Failure Behaviour

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important point. A crack based model is designed in this chapter based on compatibility

requirements, where reinforcement crosses a crack and compatibility in the compression

zone concrete. The analysis combines dowel rupture of the reinforced concrete and for

the shear at the corner will reduce by the stirrup see figure (4-2).It highlights the

requirements for this project into preventing the flexural and shear failure by according

to reliability analysis and the figure below shows how the stirrups preventing the brittle

shear failure.

In this project I have mentioned to the most important of failure types occurred in the

reinforced concrete beams and these failures are:

� Flexural failure (Yielding of steel reinforcement).

� Flexural failure (Crushing of concrete).

� Shear failure.

Figure (4-3) Prevent Shear Failure by steel stirrup reinforcement

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4-3-1 Flexural failure (Tension & Compression )

The compression response of reinforced concrete beams as the figure (4-3) shows below

the ascending branch describes the concrete while it is safe and there is no failure and

like any other elastic material can be described by strain and stress –response. However

as the load increase the width of the cracks increase and the probability of failure will be

increase too and the descending branch is dominated by the manner in which combined

and grow and can not be described in terms simply strain. The descending branch is not

a material property it depends on the place and the quantity of the steel reinforcement .In

this work the design for the flexural is taken according to Euro Code 2.

An extreme amount of reinforcement sometimes indicates that a member is undersized

and it may also cause a brittle failure and many problems such as difficulty fixing the

steel bars and pouring the concrete there fore the Euro Code stipulates that.

100 / 4%S CA A � Except at laps

In opposite too use the reinforcement less than the required to use therefore

m100 / 26 %cS t

yk

fA b d

f� And not less than 0.13%

Design for flexural of rectangular section with moment redistribution according to

EC2:-

The steps for design [1]:-

1- Calculate � �0.4balX d��

2- Calculate 2/ ckK M bd f�

Figure (4-3) Cross section of a beam subjected to the stress and strain diagrams

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3- Take balK from table (4-1) appendix A or alternatively calculate

� � � �20.454 0.4 0.182 0.4 50bal ckK For f C� ��

balIf K K� Compression steel is required.

4- Calculate the Area of compression steel from

� ��

2bal ck

ssc

K K f bdA

f d d

��

��

scf is the stress in the compression steel.

/ 0.38If d x� � The compression steel has yielded and 0.87sc ykf f�

/ 0.38If d x� � Then the strain sc� in the compressive steel must be calculated from

the proportion of the strain diagram and 3200 10sc s sc scf E � ��

5- Calculate the area of tension steel from

Where 0.8 / 2balZ d x� �

5- Check the areas of steel required and the areas provided that

, , ,,( ) ( )s prov s prov s reqs reqA A A A� ��

This is ensure that the depth of the neutral axis has not exceeded the maximum value of

balX by providing an over–exceeds of tensile reinforcement. For more details illustrate

the dimensions see figure (4-4).

Figure (4-4) Rectangular reinforced concrete beam dimensions

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4-3-2 Shear Failure

The shear failure is caused by the compressive and tensile stresses the direction of the

compressive stresses takes the form of an arch while the tensile stresses have an curve of

a catenary or suspended chain at the mid span of the beam where the shear is lower than

the external sides of the beam and the opposite for the bending at the mid span of the

beam is greater than the sides of the beam where the shear forces are greater and the

principle stresses become included and as the shear force is large the angle of inclination

will be large the tensile strength due to shear are liable to cause the diagonal cracking of

the beam near to the supports so that shear reinforcement at the side should be grater

than the middle of the beam. Mechanical overlap occurs between the shear

reinforcement for example the pull-out model would consider the transfer of load from

reinforcement to concrete across the bend [6] Moreover the shear reinforcement is

often failed at the bend that depending on the way used to form the corner and it is

diameter . Corner strength less than 50 % of the straight reinforcement has been reported

.Figure (4-4) shows a crack – span of a reinforced concrete beam angle between 22�

and 45� to the beam axis [6]. This angle is the shear failure angle for the cracks which

occurs in the reinforced concrete beams the reinforced steel stirrups cross the crack and

these stirrups will prevent the beam from the shear failure and it must be these

reinforcement stirrups used according to the design rules and the required diameters and

dimensions.

4-4 Effects of shear links on failure Mechanism

Before the cracks occurs in reinforced concrete beams the shear links do not carry any

stress .On the other hand when the cracks appear the links which prevent the cracks

carry a proportion of the shear force V . Furthermore the links that do not intercept the

Figure (4-5) shear links on failure Mechanism

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diagonal cracks the figure (4-6) shows how the load effects on the beam to cause the

shear failure and the cracks along the beam .Where: a d czV V V V� � �

According to Euro Code design approximately between of (35 % to 50 %) of shear

force is resisted by the aggregate interlock ( aV ) about 15 % to 25 % [7] of the shear

force is resisted by the dowel action of longitudinal reinforcement ( dV ) and

approximately between of 20% to 40 % is resisted by the un-cracked compression

concrete zone czV .

The important resistance comes from the shear links sV then

a d cz sV V V V V��

Finally as I mentioned before there are too many conditions and rules for using the shear

links according to Euro Code 2 the table (4-2) below illustrate the spacing of links [7].

Figure 4-6 Effects of shear links on failure Mechanism

Table (4-1) the spacing of links [7]

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Design for Shear.

Designing reinforced concrete beams [1] against the shear failure it depends on the

designing code in this project I have mentioned to Euro Code 2 and I have used it in this

work to design the reinforced concrete beams against the shear failure.

The shear reinforcement will usually take the form of vertical links or combination

of links and bent up bars to make a high reinforcement and should be use in the required

dimensions beams and to design reinforced concrete beams according to Euro Code 2

should be follow the next steps:

1- Calculate the ultimate design shear forces edV along the beam’s span.

2- Check the crushing strength ,Rd MaxV of the concrete diagonal strut at the section of

maximum shear usually at the face of the beam support.

For most cases the angle of inclination of the strut is 22� � � with 2.5Cot� � and

tan 0.4� � so that from the equation

� �, 0.124 1 / 250Rd Max ckw ck

V b d f f� �

And if ,Rd Max EdV V� then go to step (3) with 22� � � and 2.5Cot� �

But if ,Rd Max EdV V� then � must be calculated from the equation

Table (4-1) the spacing of links

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� �10.5Sin 45

0.18 1 / 250Ed

ck ck

V

b df fw� �

� ��

3- The shear links required can be calculated from the equation :

0.78 cot

sw Ed

yk

A V

S df ��

Where swA is the cross –sectional area of the legs of the stirrups ( 22 / 4�� for

single stirrups).

For a predominately uniformly distributed load the shear EdV should be taken at a

distance d from the face of the support and the shear reinforcement should continue

to the face of the support.

4- Calculate the minimum links required by Euro Code 2 from

0.5

,min 0.08sw ck w

yk

A f b

S f�

And the shear resistance for the links actually specified

min

0.78SW

yk

AV df Cot

S��

This value should be marked on the shear force envelop to show the extent of these

links.

5- Calculate the additional longitudinal tensile force caused by the shear force

0.5td EdF V Cot��

These additional tensile forces increase the curtailment length of the tension bars.

The minimum spacing of the links is governed by the requirements of placing and

computing the concrete and should not normally be less than 80 mm. Euro Code 2

gives the following guidance on the maximum link spacing:

(a) – Maximum longitudinal spacing between shear links in a series of links

1,max 0.75 (1 )S d Cot��

Where � the inclination of the shear reinforcement for the longitudinal axis of the

beam.

(b) –The maximum transverse spacing between legs in a series of shear links.

,max 0.75 ( 600)bS d� � .

24

4-5 DESIGNING OF REINFORCED CONCRETE BEAMS

In this chapter I have used to design the reinforced concrete beams EC2 for various

number of different dimension beams against the flexural failure (tension compression)

and shear failure.

The table (4-3) below shows the beams in different dimensions:-

The applications of designing reinforced concrete beams see Appendix (B).

Beam No. Length (L)

mm

Width (b)

mm

Depth (d)

mm

Cover (d � ) mm

B1 6000 300 450 50

B2 6000 300 450 50

B3 6000 300 550 50

Table (4-2) Beams dimensions

25

CHAPTER 5 Description of the Method

5-1 Reliability Engineering Method:-

The reliability analysis is one of the method for designing building [17] and other

structures and nowadays the structural engineers used this method to estimate and to

help them to compute the probability of failure for the structures.

The aim of the structural engineers is to produce the design for safe and economical

structures that fulfils it is intended purpose.

Structural design is achieve by computing the internal forces and bending moment

of the structure (beam column ........etc ) in this project reinforced concrete beams

after calculating the forces and moments it will followed by the selection of

appropriate cross section for a given grade of concrete and steel . The carrying load

capacity (C) of the reinforced concrete beam and this load is resisted by the moment

called demand (D) and according to the Euro code 2 using the limit state function

for each failure flexural crushing due to concrete and shear failure and for each

failure limit state from this limit state and by using the one of the reliability method

First Order Reliability Method, or Monte Carlo method it can be easy to calculate

the probability of failure .Furthermore the risk of failure of the reinforced concrete

beams is often measured in terms of the probability of failure, fp .Alternatively the

probability of survival is measured in terms of the reliability index � Both

quantities are related as will be explained later in this chapter.

The safety of a structural component depends on its resistance (Capacity) which can

be expressed in terms of a limit state function G. The limit state function can be as

simple as the difference between the random resistance of the reinforced concrete

beams (Demand) and the random load affect acting on the member.

G=C – D (5 - 1)

In this case If G > 0 the structure is safe [17] otherwise it fails. The condition failure

does not essential mean the catastrophic failure but is used to indicate that the

structure does not meet the intended performance. Thus the reliability of structure is

equal the probability that a structure will not fail to perform its intended function.

The probability of failure kkyc QGdff ,,,, (Eq. 5-2)

)0(Pr)0(Pr �� GobDCobp f (5 - 2)

26

Figure (5-1) Beam 1 Compression Failure

Since C and D are treated as random variables the outcome G will also be a random

variable. In general the limit state function can be a function of many variables

which are ( kkyc QGdff ,,,, ) representing the depth of the beam characteristics

material strength loads and other factors such as analysis method. Accordingly G

becomes.

G(x) = g( nXXX ............, ,21 ) ( 5 – 3)

A direct calculation of the probability of failure may be very difficult for complex

limit state functions and therefore it is convenient to measure structural safety in

terms of the reliability index� the relationship between the reliability index� and

the probability of failure fp is )( ��fp

5-2 Selection of the Beams

In this project I have chose three different reinforced concrete beams each beam

has been selected according to the mode of failure the table (4-1) illustrate the

dimensions for the beams.

In the first case beam 1 the concrete in the ultimate compression failure and reaches

a 0.003 of the assumed crushing strain and ( balx x� ) as the figure (5-1) shows that

before the compression steel starts yield at ( 0.45x d� ). [1].

In this case the mode failure called compression failure with yielding of

reinforcement steel, and this beam may failed suddenly in a brittle failure without

any warning to the building owner and the consequences may have a serious

problem the figure (5-4) B1 shows the moment curvature diagram and the behaviour

of the compression failure [16] .In this case we have to calculate the probability of

flexural failure with yielding of reinforcement steel before it occurs.

27

Figure (5-2) Beam 2 Crushing Concrete Failure

In the second case for the beam 2 the load has been increased by the same ratio

2k

k

g

q� the figure (5-2) shows the strain distribution of failure includes simultaneous

crushing and yielding of the steel and s y� �� where

10.0035

baly

dx x

��

�and in this

calculations 2500 /yf N mm� then 0.00217y� � this gives 0.617balx d� this mean

the ductile failure due to tension and brittle due compression failure expected and

most probably as the figure (5-4) B2 shows that . [1]. In this case we have to

calculate the probability of crushing brittle failure before it occurs.

In the last case for beam 3 where an incense in load, (at the same ratio 2k

k

g

q� ), led to

an increase in the depth. The reinforcement in this beam has yielded as shown in the

figure (5-3) in this type the tension failure is most probably to occurs with a large

centrality and a small depth of neutral axis (x) and the failure starts with a yielding of

the tensile steel and has a ductile moment curvature response as shown in figure (5-

4) B3 [16]. In this case we have to calculate the probability of shear failure before it

occurs.

Figure (5-3) Beam 3 Tension Failures

28

5-3 Load Combination and Patterns for the Limit States

The load applied to design the different three beams according to the EC2 had

different values but the same ratio between characteristic values of the permanent

loads KG and variable loads kQ , of 2 (i.e. 2K

k

G

Q� ). It was done in order to exclude

the influence of load uncertainties on the relation, between the probabilities of failure

calculated for the three beams.

The total design load is calculated as :-

. .u G k Q kW G Q� ��

Where

G Qand� � are the partial safety factors equal to 1.35 and 1.5, respectively.

Figure (5-4) Moment – Curvature diagram

29

5-4 Reliability Analysis Methods:-

5 – 4-1 First Order Reliability Method

Probabilistic design is concerned [18] with the probability that the structure will

realise the function assigned to it. In this work, the reliability method is a briefly

reviewed. If C is the strength capacity and R the loading affect(s) of a structural

system which are random variables. The solution in this method set of (might

correlated) random variables x, is converted into a set on unaccompanied slandered

Gaussian variables G [8], by the relation:-

( )G x C D� �

Where G is the function in different number of random variables and the output

results are different and the probability of failure will be considered according to this

three situations for more illustration see figure( 5-1 ) .

G(x) < 0 failure in the structure.

G(x) >0 safe structure

G(x) =0 the structure is in the critical situation

Figure (5 – 5) First order reliability approximation in standard normal space

30

The procedure for computing the safety index by using First Order Reliability

Method is summarised in this steps [12]:-

1-Select the starting values for ( 1 2, ............. nx x x ) which are always in this project as

the mean values and with different distribution types (Normal Lognormal and

Gumbel). The table (5-1) below shows all the mean values COV and standard

deviation for different three beams (B1 B2 B3) and the same mean values for yf cf .

Variable Distribution

type

Mean COV Standard

Deviation

B1 Normal 52.5 KN/m 0.10 5.25 KN/m

B2 Normal 69.3 KN/m 0.1 6.93 KN/m

G

B3 Normal 94.5 KN/m 0.1 9.45 KN/m

B1 Gumbel 25 KN/m 0.25 6.25 KN/m

B2 Gumbel 33 KN/m 0.25 8.25 KN/m

Q

B3 Gumbel 45 KN/m 0.25 11.25 KN/m

yf Lognormal 500 MPa 0.08 40 MPa

cf Lognormal 38 MPa 0.158 6 MPa

B1 Normal 450 mm 0.02 9 mm


d


2-Calculate the limit state function ( )( )mG x and find the first derivative� �mi x

G

x

� ��

and

for Gumbel distribution� �

� �1 ( )

( )

( )i

i

mx i

mx i

F x

f x

��

� � ��

( ) 1 ( )( )i

m mi i x ix F x� � � � �� (i = 1 2 3 ……. n). (5-4)

Table (5-1) Basic Random Variable

31

3- Calculate the sum of � �, ,x � �

Where

( )

1

nm

ii i x

Gx x

x ��

� ��

� � (5-5)

1

n

ii i x

G

x� �

��

� ��

� � (5-6)

� �2

2

1

n

ii i x

G

x� �

��

� ��

� � (5-7)

4- Calculate i and� �

Where

� �2( 1, 2,3,............, )

ii x

i

G

xi n

�

��

�

� ��

(5-8)

� ��

( )

2

mx G x��

�

� ��

� ��

(5-9)

5-Calculate the new iteration values of ( 1)mx �

Where

( 1) ( 1, 2,3,................., )mi i ix i n� � � �� (5-10)

6-Conteinue produces [9] a new Iteration on steps 2 and 5 until convergence on �

and ( ) 0G x � or close to 0.

32

5-4-2 Monte Carlo Simulation Method (MCS)

The Monte Carlo Simulation is a numerical method to calculate the probability of

failure and was introduced in the 1940s by physicists [12] (Stanislaw Ulam, Enrico

Ferm, John von Neumann and others) working in nuclear weapon projects in the Los

Alamos National Laboratory, and there is no method is easy to understand and ready

compatible to engineering solution as Monte Carlo method [10]. The procedure for

computing the safety index by using Monte Carlo Method is summarised in this

steps [12]:-

1- Generate a number of different random u uniformly distributed between 0 and 1.

2- Find values for a basic random variable ( iX ) with the generated number u.

3- Produce the limit state function and repeat the last steps for all random variable

which appear in the limit state function G(x).

4- Substitute the output values for ( )jX of the basic random variables into the

limit state function G(x) and then calculate its value.

5- Repeat the last four steps for number of trials and compute a new value of G(x) at

each cycle.

5- Calculate the number of trials fn for ( )( ) 0jG x �

( )

1

( )N

jf

j

n I x�

��

Where

� �( )

( )

( )

1, ( ) 0

0, ( ) 0

jj

j

G xI x

G x

� ��

(5-11)

6- Calculate the estimate fP of the probability of failure using the equation

ff

nP

N� (5-12)

The next steps for generating normal, lognormal and Gumbel distribution random

variables [12].

1- Normal random variable –

2 ln( ) (2 )1 2g u Cos ug g� �� (5-13)

33

1 22ln( ) (2 )d dd u Sin u� �� (5-14)

2- Lognormal random variable –

� �3 4exp 2ln( ) (2 )y yy f ff u Cos u� � �� (5-15)

� �3 4exp 2 ln( ) (2 )cc f fcf u Sin u� � �� (5-16)

Where

21ln

2x x x� �� (5-17)

2ln(1 ( ))x xCov� � � (5-18)

3-Gumbel random variable

� �n5

1log( log( ))

n

q u u�

� � � � (5-19)

Where

16n q

��

0.5772157n q

n

u�

� � �

The last steps are used in a visual basic simulation program in excel spreadsheet to

estimate the probability of failure in different cycles.

34

5-5 Reliability analysis of Reinforced Concrete Beams and Results

This part of this chapter is contain a various number of reliability analysis for a

different three beams each of these beams have a different loads and different

dimension. Furthermore a different amount of reinforced steel obtained by using EC

2 designs sees Appendix (A). Also for each failure for these beams has limit state

function according to the capacity and demand for each beam.

5-5-1 Partial Safety Factors

In the design procedure for any structure the partial safety factors must be taken into

account and in this project for designing reinforced concrete beams the partial factors

were taken into account. The partial safety factors are not taken into account in

reliability analysis since uncertainties associated with loads and resistance are

considered explicitly in this kind of analysis.

5-5-2 Limit State Functions

The limit state functions [1] in this project have been considered corresponding to

the obtained of the ultimate bending moment (Flexural Failure) and for tension and

compression shear failure at the middle section of the reinforced concrete beam the

steps below illustrate the different limit state functions for each type of failure

according to EC 2.

1-Limit State Function of Flexural Failure:-

The limit state function due to brittle flexural failure can be written as

( )G x C D� � (5-20)

Where C the capacity or resistance effect (Bending Moment) and D is the demand or

load affect.

35

Where

� �0.87 ( )

0.87 ( ) 0.871.134

yk S Syk S S yk S

ck

f A AC f A A d f A d d

f b

��

� � (5-21)

and

8uW l

D�

� (1.05 )u k kW G Q� � (5-22)

So that

(5-23)

2-Limit State Function due Crushing of Concrete:-

The limit state function for the collapse due to crushing of concrete can be written

as-

( )G x C D� �

Where C the capacity or resistance effect (Bending Moment) and D is the demand or

load affect.

21

3 w ckC b d f� ��

(5-24)

( )

2k kG Q l

D�

� (5-25)

� �21( )

3 2k k

w ck

G Q lG x b d f

��

[9] (5-26)

2-Limit State Function of Shear Failure:-

The limit state function for shear failure of the concrete can be written as –

( )G x C D� �

Where C the capacity or resistance effect (Maximum Shear) and D is the demand or

load affect.

0.124 1 0.78250

ck swck yk

f AC bd f df

S� ��

(5-27)

( )

2k kG Q l

D�

� (5-28)

� �( ) 0.124 1

250 2K Kck sw

ck yk

G Q lf AG x bd f df

S

��

(5-29)

S t a r t s i m u l a t i o n

36

Chapter 6 Applications of Reliability Analysis

Reliability Analysis of Reinforced Concrete Beams

6-1- First Order Reliability Method (FORM)

Beam 1 –

6-1-1-Limit state function of flexural failure: B1

8

uW lD

�� (1.05 )u k kW G Q� �

� �0.87 ( )

0.87 ( ) 0.871.134

yk S Syk S S yk S

ck


f b

��

� �

� �2

( )

0.87 ( ) ( )( ) 0.87 ( ) 0.87

1.134 8yk S S K K

yk S S yk Sck

G x C D

f A A G Q lG x f A A d f A d d

f b

� �

��

Variable Distribution type Mean COV Standard

Deviation

G Normal 1.05 50 /KN m� 0.10 5.25 /KN m

Q Gumbel 25 /KN m 0.25 6.25 /KN m


cf Lognormal 38MPa 0.158 6 MPa

d Normal 450mm 0.02 9mm

6000L mm� 50d mm� �

2 , 22860 964S SA mm A mm��

� � � �4.85

1649.52 838.68 50 4500000yk

yk yk k k

ck

f

f d f d g qf

� ��

� � � � � � � � ��


37

1-Normal distribution types:

1.05 50 52.5 / , 5.25 /k kg gN mm N mm��

450 , 9d dmm mm��

2- Lognormal distribution of ckf :-

2 238 / , 6 /ck ckf fN mm N mm��

2ln 1ckf COV� � �� = 2ln 1 0.158 0.157� ��

� ��

� �21ln 1

2ck ckf fck fCOV� � � � �

� �21ln 38 ln 1 0.157 3.625

2ckf� � � � �

3-Lognormal distribution of ykf :-

2 2500 / , 40 /

yk ykf fN mm N mm��

2ln 1yk ykf fCOV� � ��

2ln 1 0.08 0.08� ��

� �21ln ln 1

2yk yk ykf f fCOV� � � � � � �21ln 500 ln 1 0.08 6.211

2� � � �

4-Parameters of the Gumbel distribution:-

25 / , 6.25 /kq qkN mm N mm��

1

6n

qk

��

��

10.205

6.256

��

0.5772157qk qk

n

u�

� � �

0.577215725 22.18

0.205� � �

38

Denote 1 2 3 4 5, , , ,yk ck k kx f x d x f x g x q� � � � �

� � � �4.85

1649.52 838.68 50 4500000yk

yk yk k k

ck

f

f d f d g qf

� ��

� � � � � � � � ��

� � � �11 2 1 2 4 5

3

4.851649.52 838.68 50 4500000

xx x x x x x

x

� ��

� �

� �12 2

1 3

16000.341649.52 838.68 50

xGx x

x x

� ��

1 1 12

1649.52 838.68 2488.2G

x x xx

��

�

21

23 3

8000.2xG

x x

��

�

4

4500000G

x

��

�

5

4500000G

x

��

�

Parameters of equivalent normal distribution

lognormalykf �

* 1 10.08x� �

� �1 1 11 ln( )ykfx x ��

� � � �1 1 1 11 ln( ) 6.211 7.211 ln( )x x x x� � � � �

d=Normal

* 2 29 , 450 mm� � � �

lognormalckf �

* 3 30.158x� �

� �3 3 31 ln( )ckfx x ��

� � � �3 3 3 31 ln( ) 3.25 4.625 ln( )x x x x� � � � �

Normalg �

* 4 45.25 / , 52.5 /N mm N mm� � � �

39

Gumbelq �

* � �5 50.205 22.18par x� � � �

� ��

15

55 5

(exp( exp( )))

0.205exp exp( )

par

par par

� ��

� ��

�

� �15 5 5 5exp( exp( ))x par� ��

� � � �11 2 1 2 4 5

3

4.85( ) 1649.52 838.68 50 4500000

xG x x x x x x x

x

� ��

� �

6-1-2 Limit state function for Crushing of the Concrete B1 [9]

( )G x C D� � 6000L mm� 300b mm�

21


( )

2k kG Q l

D�

�

� �21( )

3 2K K

w ck

G Q lG x b d f

��

� �2 60001( ) 300

3 2K K

ck

G QG x d f

��

� � � �2( ) 100 3000ck K KG x d f G Q� � � �

� � � �2( ) 100 3000ckd f gk qkG x � � � ��

Denote:

1 2 , 3 4, ,ck k kx d x f x g x q� � � �

� � � �21 2 3 4( ) 100 3000G x x x x x� � �

1 21

200G

x xx

��

�

21

2

100G

xx

��

�

3

3000G

x

��

�

4

3000G

x

��

�

40

6-1-3- Limit state function for Shear failure B1

( )G x C D� �

( )

2k kG Q l

D�

� 6000L mm� 300b mm�

0.124 1 0.78250

ck swck yk

f AC bd f df

S� ��

� �( ) 0.124 1

250 2K Kck sw

ck yk

G Q lf AG x bd f df

S

��

� � 60000.124 300 1 0.287 0.78 2.5

250 2ck

ck yk

fd f df g q� ��

� �37.2 1 0.56 3000250

ck

ck yk k k

fd f d f g q� �

��

� �

Denote:-

1 2 , 3 4 5, , ,ck yk k kx d x f x f x g x q� � � � �

� �21 2 1 3 4 5( ) 37.2 1 0.56 3000 3000

250

xG x x x x x x x

� ��

22 3

1

37.2 1 0.56250

xGx x

x

� � ��

21

2

37.2 1125

xGx

x

� � ��

13

0.56G

xx

��

�

4

3000G

x

��

�

5

3000G

x

��

�

41

Beam 2 –

6-1-4 Limit state function for flexural failure B2:

8

W luD�

� (1.05 )u k kW G Q� �

� �0.87 ( )

0.87 ( ) 0.871.134

yk S Syk S S yk S

ck


f b

��

� �

� �2

( )

0.87 ( ) ( )( ) 0.87 ( ) 0.87

1.134 8yk S S K K

yk S S yk Sck

G x C D


f b

� �

��



type Mean COV

Standard

Deviation

G Normal 1.05 66 /KN m� 0.10 6.93 /KN m

Q Gumbel 33 /KN m 0.25 8.25 /KN m




6000L mm� 50d mm� �

21833SA mm� � 23729sA mm�

� � � �4.85

1896 1833 50 4500000yk

yk yk k k

ck

f

f d f d g qf

� ��

� � � � � � � � ��


1.05 66 69.3 / , 6.93 /k kg gN mm N mm��

450 , 9d dmm mm��


33 / , 8.25 /kq qkN mm N mm��

1

6n

qk

��

��

42

10.155

8.256

��

0.5772157qk qk

n

u�

� � �

0.577215733 29.27

0.155� � �


� � � �11 2 1 2 4 5

3

4.851896 1833 50 4500000

xx x x x x x

x

� ��

� �

12 2

1 3

18391.21896 1833( 50)

xGx x

x x

� ��

1 1 12

1896 1833 3729G

x x xx

��

�

21

23 3

9195.6xG

x x

��

�

4

4500000G

x

��

�

5

4500000G

x

��

�

Normalg �

* 4 46.93 / , 69.3 /N mm N mm� � � �

Gumbelq �

* � �5 50.155 29.27par x� � � �

� ��

15

55 5

(exp( exp( )))

0.155exp exp( )

par

par par

� ��

� ��

�

� �15 5 5 5exp( exp( ))x par� ��

43

6-1-5 Limit state function for Crushing of the Concrete [9]

( )G x C D� � 6000L mm� 300b mm�

21


( )

2k kG Q l

D�

�

� �21( )

3 2K K

w ck

G Q lG x b d f

��

� �2 60001( ) 300

3 2K K

ck

G QG x d f

��

� � � �2( ) 100 3000ck K KG x d f G Q� � � �

� � � �2( ) 100 3000ckd f gk qkG x � � � ��

Denote:

1 2 , 3 4, ,ck k kx d x f x g x q� � � �

� � � �21 2 3 4( ) 100 3000G x x x x x� � �

1 21

200G

x xx

��

�

21

2

100G

xx

��

�

3

3000G

x

��

�

4

3000G

x

��

�

6-1-6- Limit state function for Shear failure

( )G x C D� �

( )

2k kG Q l

D�

� 6000L mm� 300b mm�

0.124 1 0.78250

ck swck yk

f AC bd f df

S� ��

� �( ) 0.124 1

250 2K Kck sw

ck yk

G Q lf AG x bd f df

S

��

44

� � 60000.124 300 1 0.287 0.78 2.5

250 2ck

ck yk


� �37.2 1 0.56 3000250

ck

ck yk k k

fd f d f g q� �

��

� �

Denote:-


� �21 2 1 3 4 5( ) 37.2 1 0.56 3000 3000

250

xG x x x x x x x

� ��

22 3

1

37.2 1 0.56250

xGx x

x

� � ��

21

2

37.2 1125

xGx

x

� � ��

13

0.56G

xx

��

�

4

3000G

x

��

�

5

3000G

x

��

�

45

Beam 3 –

6-1-7 Limit state function for flexural failure:

8

uW lD

�� (1.05 )u k kW G Q� �

� �0.87 ( )

0.87 ( ) 0.871.134

yk S Syk S S yk S

ck


f b

��

� �

� �2

( )

0.87 ( ) ( )( ) 0.87 ( ) 0.87

1.134 8yk S S K K

yk S S yk Sck

G x C D


f b

� �

��



type Mean COV

Standard

Deviation

G Normal 1.05 90 /KN m� 0.10 9.45 /KN m

Q Gumbel 45 /KN m 0.25 11.25 /KN m




6000L mm� 50d mm� �

2 , 24132.5 1815S SA mm A mm��

� � � �6

2016 1579 50 4500000yk

yk yk k k

ck

f

f d f d g qf

� ��

� � � � � � � � ��


1.05 90 94.5 / , 9.45 /k kg gN mm N mm��

550 , 11d dmm mm��


45 / , 11.25 /kq qkN mm N mm��

1

6n

qk

��

��

46

10.114

11.256

��

0.5772157qk qk

n

u�

� � �

0.577215745 39.93

0.114� � �


� � � �6

2016 1579 50 4500000yk

yk yk k k

ck

f

f d f d g qf

� ��

� � � � � � � � ��

� � � �11 2 1 2 4 5

3

62016 1579 50 4500000

xx x x x x x

x

� ��

� �

� �12 2

1 3

241922016 1579 50

xGx x

x x

� ��

1 1 12

2016 1579 3595G

x x xx

��

�

21

23 3

12096xG

x x

��

�

4

4500000G

x

��

�

5

4500000G

x

��

�

Parameters of equivalent normal distribution

d=Normal

* 2 211 , 550 mm� � � �

Normalg �

* 4 49.45 / , 94.5 /N mm N mm� � � �

Gumbelq �

* � �5 50.114 39.93par x� � � �

� ��

15

55 5

(exp( exp( )))

0.114exp exp( )

par

par par

� ��

� ��

�

47

� �15 5 5 5exp( exp( ))x par� ��

6-1-8 Limit state function for Crushing of the Concrete [9].

( )G x C D� � 6000L mm� 300b mm�

21


( )

2k kG Q l

D�

�

� �21( )

3 2K K

w ck

G Q lG x b d f

��

� �2 60001( ) 300

3 2K K

ck

G QG x d f

��

� � � �2( ) 100 3000ck K KG x d f G Q� � � �

� � � �2( ) 100 3000ckd f gk qkG x � � � ��

Denote:

1 2 , 3 4, ,ck k kx d x f x g x q� � � �

� � � �21 2 3 4( ) 100 3000G x x x x x� � �

1 21

200G

x xx

��

�

21

2

100G

xx

��

�

3

3000G

x

��

�

4

3000G

x

��

�

6-1-9 Limit state function for Shear failure

( )G x C D� �

48

( )

2k kG Q l

D�

� 6000L mm� 300b mm�

0.124 1 0.78250

ck swck yk

f AC bd f df

S� ��

� �( ) 0.124 1

250 2K Kck sw

ck yk

G Q lf AG x bd f df

S

��

� � 60000.124 300 1 0.287 0.78 2.5

250 2ck

ck yk


� �37.2 1 0.56 3000250

ck

ck yk k k

fd f d f g q� �

��

� �

Denote:-


� �21 2 1 3 4 5( ) 37.2 1 0.56 3000 3000

250

xG x x x x x x x

� ��

22 3

1

37.2 1 0.56250

xGx x

x

� � ��

21

2

37.2 1125

xGx

x

� � ��

13

0.56G

xx

��

�

4

3000G

x

��

�

5

3000G

x

��

�

6-2 Monte Carlo simulation (MCS)

6-2-1 Beam 1 –

49

1-Normal distribution

2 ln( ) (2 )1 2g u Cos ug g� �� 52.5 5.25 2 ln( ) (2 )Rnd Cos Rnd��

1 22ln( ) (2 )d dd u Sin u� �� 450 9 2ln( ) (2 )Rnd Sin Rnd��

2-Lognormal distribution

2 2ln(1 ( )) ln(1 (0.08 )) 0.08fy fyCov� � � � � �

2 21 1ln ln 500 0.08 6.211

2 2fy fy� ��

� �3 4exp 2ln( ) (2 )y yy f ff u Cos u� � �� exp 6.211 0.08 2 ln( ) (2 )Rnd Cos Rnd��

2 2ln(1 ( )) ln(1 (0.158 )) 0.158c cf fCov� � � � � �

2 21 1ln ln 38 0.158 3.625

2 2c cf f� ��

� �3 4exp 2 ln( ) (2 )cc f fcf u Sin u� � ��

� �exp 3.625 0.158 2ln( ) (2 )Rnd Sin Rnd��

3-Gumbel distribution

16n q��

10.205

6.256

��

0.5772157n q

n

u�

� � �

0.577215725 22.2

0.205� � �

� �1log( log( ))n

5q u u

n�

� � � � � �122.2 log( log( ))0.205 Rnd� � � �

CHAPTER 7 Results and Discussion 7-1 The Results

50

Method

Failure

Failure

With respect to the reliability analysis of reinforced concrete beams by using first

Order Reliability Method and Monte Carlo Simulation to check the results of the safety

index and the probability of failure for these methods the results about the same for

both methods and the table (7 -1) below illustrates the computed values for � and fP by

using FORM and MCS method. Also for more information about the output data see the

Appendix (B) and (C).

7-1-1 Results of the Reliability Analysis Beam 1 –

The safety index and probability of failure for the first beam where d=450 mm and with

load uw = 105 KN /m.

Table (7-1)

Bending

Yielding of

reinforcement

Crushing of

concrete

Shear

� fP � fP � fP

FORM 2.525 0.005785 3.990 53.29 10�� 6.072 106.288 10��

Monte Carlo 2.456 0.00700 3.926 54.3 10�� 4.991 7103 ��


The safety index and probability of failure for the first beam where d=450 mm and with

load uw = 138.6 KN /m. The difference between this beam and the previous one is the

load in this beam is greater than the first beam.

Table (7 -2)

Bending

Yielding of

reinforcement

Crushing of

concrete

Shear


FORM 3.340 0.000148 2.649 0.004026 4.666 61.529 10��

Monte Carlo 3.300 0.000483 2.593 0.00474 4.602 62.08 10��


Method

51

Failure

The safety index and probability of failure for the first beam where d=550mm and with

load uw = 189 KN /m. The differences between this beam and the first two beams, in

this beam the load and the depth of the beam was increased (increase in shear force), and

the results obtained see table (7-3) below:-

Table (7-3)

Bending

Yielding of

reinforcement

Crushing of

concrete

Shear


FORM 2.478 0.0066 3.101 0.000962 4.099 52.07 10��

Monte Carlo 2.418 0.0077 3.052 0.00113 4.075 52.29 10��

7-2 Discussion

Method

52

From the previous chapter where steps of calculation of the safety index by using First

Order Reliability Method and Monte Carlo Simulation according to the limit state

functions of the flexural crushing concrete and shear failure and the output results from

the excel spreadsheet give the estimate value of the safety index and probability of

failure for different beams have a various number of failure and dimensions. In the next

steps I am going to discuss the values of safety index � and probability of failure fP for

the different three reinforced concrete beams.

7-2-1 Results of Reliability Analysis Beam 1 –

According to the output results obtained for the first beam using FORM and Monte

Carlo simulation, the results were satisfied. The third value for shear failure, the result

obtained using Monte Carlo simulation was inaccurate, and to get the accurate result it

needs a huge number of cycles, and to do this it needs a long time.

The lowest result was for � =2.525, and high probability of failure fP =0.005785 with a

ductile failure (yielding of reinforcement steel).

The value for 0.1018fc� � , this value is not important, and there is no a significant

increase in� , and for 0.55fy� � , this value is very important in this case.

For fc� the sensitivity factor =1

1.0051 0.1018

��

1.005 2.525 2.537� � � �

We can notice that � is increased by a factor 2

1

1 �� when the corresponding random

variable cf is considered as deterministic parameters equal to their mean value without

any significant increase or effect on the estimated safety index � Thus neglecting

uncertainty of a random variable cf with 0.1018fc� � resulted in an increase of � by

about 2 % .For more results about the output excel data see Appendix B Table (B-1).


53

In this case there is an increase in the reinforcement ratio , and the lowest results in this

beam of safety index with probability of failure, � and fP obtained by FORM were

about 2.649 and 0.004026 respectively, with a brittle failure (Crushing of concrete), and

this is not good, the beam will failed suddenly without any warning. In this case the

fc� is very important, and fy� is not important.

For more details about the results obtained using FORM and Monte Carlo Simulation

see appendices B and C.


In this beam the lowest value is still for the bending failure due to yielding of

reinforcement, and the results obtained using FORM and Monte Carlo simulation for the

shear failure were the same, and for this beam � is decreased with high probability of

failure with respect to the shear failure, were 4.099� � and 52.07 10fP �� , this mean

the beam will failed due to shear failure, and this result as expected, because in this

beam of design, the shear force increased, and this cause the failed due to shear. For

more results calculated by FORM and Monte Carlo simulation, see appendices B and C.

54

CHAPTER 8 CONCLUSION

The objective of this dissertation was to study and understand the reliability of

reinforced concrete beams and using reliability structural to estimate the probability of

different types of concrete structural failure and the main points for the conclusion of

this work are mentioned by the following bullets points:

� The main conclusion of this study was that to calculate the probability of failure

of the reinforced concrete beams with a various number of beams with a various

loads and the results were as expected according to the limit state functions using

EC 2 approach.

� The results obtained using First Order Reliability Method (FORM) and Monte

Carlo simulation (MCS) were the same, just for the second beam with respect to

the shear failure the result was slightly difference using these methods and in this

case to reach the same result obtained by FORM was more accurate, and the

result obtained using Monte Carlo simulation was inaccurate, to obtain the

accurate result using Monte Carlo simulation, it need a huge number of

simulation trials and in this case we need to wait for a long time.

55

BIBLIOGRAPHY

1. Bill M.John B & Ray H. (2007). Reinforced Concrete Design. Eurocode 2 . 6th Ed.

Basingstoke Hampshire: Palgrave Macmillan.

2. C. O. Arun & Dr. T M M Pillai.(2006). Reliability studies of deteriorating

reinforced concrete flexural members. Structural Safety. Vol 86 P. 139-143.

3. Bruno. C Oliver. B & Bruno. G (2003). Reliability Assessment of Reinforced

Concrete Beam subjected to corrosion 17-22 2003. Czech Republic.

Available at: http://www.iasmirt.org/SMiRT17/M04-3.pdf 4. Leoind. Z. (2001). Failure How do materials break? P. 3-4.

Available at: http://people.virginia.edu/~lz2n/mse209/ 5. Chris. B & Tim. S. (2001). Shear analysis of concrete with brittle reinforcement

Vol.7 No.4 P.323-329.

Available at: http://www-civ.eng.cam.ac.uk/cjb/papers

6. C.J Burgoyne & T.J Stratford. (2002). Crack – based analysis of concrete with

brittle reinforcement, Magazine of concrete research. Vol 45 No. 5. P. 321-332.

7. Gummaa. H. (2008). Design of reinforced concrete beams Scotland. UK MSc:

Heriot-Watt University.

8. Ander. T & Edison da Rosa. (2006). Structural reliability analysis using

deterministic finite element program. Latin American Journal of Solids and Structures

3(2006) P. 197-202.

Available at: www.lajss.org

9. R. Lu Y. Luo & J. P. Conte. (1994). Reliability evaluation of reinforced concrete

beams. Structural Safety, Vol 17 (1995) P. 117-127.

56

10. Tirthankar. B. (2006) .Reliability assessment of plane from offshore structures

Scotland. UK MSc: Heriot-Watt University.

11. Schneider. J. (1997). Introduction to safety and reliability of structures. 5th Ed.

Zurich 1997.

12. Safety Risk and Reliability Engineering Lecture Notes delivered in class during the

study course 2 2010.

13. Prakash Desayi K. Balajirao. (1989). Reliability of reinforced concrete beams in

limit state of cracking – failure rate analysis approach. Materials and Structures.

Vol. 22. P. 269 – 279.

14. M. Arafa. (1986). Reliability of reinforced concrete beam section as affected by

their reinforcement ratio P. 1-6.

15. Safety Risk and Reliability Engineering Mini- Project Reliability based design of

glued laminated (glulam) timber beam was achieved during 2nd the second course

(2010).

16. Mac Gregor & James. G. (1997). Reinforced concrete mechanics and design. 3rd

Ed United States of America.

17. Sujata Kulkarni. (2006). Calibration of flexural design of concrete members

reinforced with FRP bars 2002 India. MSc dissertation Civil Engineering, Louisiana

State University. P. 89-97.

18. Abubakar Idris & Pius Edache. (2007). Reliability analysis of simply supported

steel beams, Australian Journal of Basic and Applied Sciences , P. 20-29, 2007.

57

APPENDICES (A B C)

58

Appendix A

Design of Reinforced Concrete Beams (B1 B2 B3)

59

DESIGN BEAMS (B1)

(a)- Design Reinforcement

Use:-

230 / mmfck � �

2500 / mmf yk � �

� Ultimate loading and max moment

Actions -:

Permanent load 50 / mg kk� �

Variable load 25 / mq kk � �

Ultimate load � �1.35 1.5W g qu k k� �

� �1.35 50 1.5 25 105 / mK� � � � � �

Maximum design moment 2

8

wl� �

2105 6472.5 m

8K

��

Bending Reinforcement 2

Kbd fck

��

6

2

472.5 100.259

300 450 30

��

� �

0.259 0.167balK K� � �

Therefore compression reinforced required

60

� Compression Steel sA � :-

50

0.111 0.171450

d

d

��

Therefore 0.87f fsc yk� and Compression steel will have yielded.

20.87 500 435 / mmfsc� � � � �

Compression steel � � 2

( )

bdbal ckK K fAs f d dsc

��

��

� ��

220.259 0.167 30 300 450

964mm435 450 50

� � ��

�

4 16 mmUse �

� Tension steel sA :-

0.82z d�

2

0.87bal ck

s syk

k f bdA A

f z��

� �2

20.167 30 300 450964 2860 mm

0.87 500 0.82 450

� � ��

� � �

� �0.87

0.567yk S s

ck

f A AS

f b

��

� �0.87 500 2860 964161.6

0.567 30 300mm

� � ��

� �

161.6202

0.8 0.8

sx mm� � �

161.60.448 0.617

450

x

d� � � Tension steel will have yielded.

(b)– Design for Shear.

� Check maximum shear at face of support

Maximum design shear 6

105 3152 2

Lw Ku� � � � � �

� Shear EFV at face of support

,max

0.15EF EdV V Wu� � �

315 105 0.15 299.25 KN� � � �

61

� Shear EdV at a distance d from face of supports

0.45Ed EF uV V w� � � 299.25 105 0.45 252 KN� � � �

� Check the crushing strength ,maxRdV with 22� � �

,max 0.124 (1 / 250)Rd w ck ckV b d f f� �

3300.124 1 30 10 442

250KN��

� �

,max 442 299.25Rd EFV V� � � Therefore 22� � � .

� Shear resistance of the links -:

0.9

VAsw EdS df Cotyk �

�

3252 10

0.574

0.78 450 500 2.5

��

� � �

Provide 10@ 275mm� then 0.571sA

S�

� Minimum Links

0.50.08,min ckA f bs

S f yk

� 0.50.08 300

0.263500

� �� Go to table (A-1)

Use stirrup diameter 8 mm spacing 350 mm 0.287sA

S�

0.87sRds yk

AV f dCot

S��

30.78 450 500 2.5 10 250.5KN��

Rds EdV V�

Therefore shear resistance of links =250.5 KN Therefore the beam can support in

shear the ultimate load of 103.5 KN/m [1].

62

Shear Reinforcement (EC 2) [1]

Table (A-1)

Note:

The design steps for the second and third beam the same procedure only in this

case the loads are difference for these beams and the depth for the third beam will be

550 mm.

Stirrup Spacing (mm) Stirrup

Diameter

(mm) 85 90 100 125 150 175 200 225 250 275 300

8 1.183 1.118 1.006 0.805 0.671 0.575 0.503 0.447 0.402 0.366 0.335

10 1.847 1.744 1.57 1.256 1.047 0.897 0.785 0.698 0.628 0.571 0.523

12 2.659 2.511 2.26 1.808 1.507 1.291 1.13 1.004 0.904 0.822 0.753

16 4.729 40467 4.02 3.216 2.68 2.297 2.01 1.787 1.608 1.462 1.34

63

Appendix B

Output Excel Result Using First Order Reliability Method

64

Variable Iteration 1 Iteration 2 Iteration 3 Iteration 4 Iteration 5 Iteration 6 Iteration 7 Iteration 8 Iteration 9

x_1 (f_y) 500.0000 425.8192 436.6795 443.4960 445.3921 445.7574 445.8104 445.8158 445.8159

x_2 (d) 450.0000 445.6861 446.0483 446.5100 446.6378 446.6629 446.6667 446.6670 446.6671

x_3 (f_c) 38.0000 35.4113 35.8177 35.9941 36.0341 36.0405 36.0412 36.0412 36.0412

x_4 (G_k) 50.0000 57.8096 58.2111 57.4184 57.1655 57.1109 57.1020 57.1009 57.1008

x_5 (Q_k) 25.0000 30.8362 36.3727 38.4941 38.9660 39.0390 39.0470 39.0473 39.0472

G(x(m)) 148745289.4737 14487925.7765 -1878749.9401 -1193725.3965 -336238.2389 -65016.1174 -9445.4216 -959.6148 -26.4740

(dG/dx_1)x(m) 1056698.8947 1047778.1969 1048412.4787 1049353.7830 1049609.4643 1049659.3253 1049666.5980 1049667.3304 1049667.3483

(dG/dx_2)x(m) 1244100.0000 1059523.2937 1086545.9514 1103506.8652 1108224.5155 1109133.4781 1109265.4369 1109278.8420 1109279.2118

(dG/dx_3)x(m) 1385070.9834 1156819.1970 1189133.2344 1214548.7677 1222239.9301 1223813.2197 1224057.8280 1224085.8608 1224087.4426

(dG/dx_4)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

(dG/dx_5)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

sigma_1 40.0000 34.0655 34.9344 35.4797 35.6314 35.6606 35.6648 35.6653 35.6653

mu_1 498.1960 492.6665 494.2341 495.0796 495.2960 495.3368 495.3427 495.3433 495.3433

sigma_2 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000

mu_2 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000

sigma_3 5.9660 5.5596 5.6234 5.6511 5.6574 5.6584 5.6585 5.6585 5.6585

mu_3 37.5217 37.4641 37.4853 37.4931 37.4947 37.4950 37.4950 37.4950 37.4950

sigma_4 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 mu_4 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 par_5 -0.5740 -1.7704 -2.9054 -3.3403 -3.4370 -3.4520 -3.4536 -3.4537 -3.4537

sigma_5 5.9763 8.1477 10.2095 10.9732 11.1404 11.1662 11.1690 11.1691 11.1691 mu_5 23.9559 22.6175 19.8938 18.5841 18.2771 18.2291 18.2239 18.2237 18.2238

Sum_x 803327144.7368 560437278.7394 559437154.7149 570221396.9718 573913225.8981 574734392.0413 574870407.9897 574887177.8371 574888367.6161 Sum_mu 794206876.6153 698300974.3491 725909664.2824 741750461.6963 745898991.7220 746650691.6500 746752563.3464 746761640.7888 746761578.3530

Sum_sigma 57110722.5795 57521459.1461 64425649.0672 67292935.9541 67950932.1412 68056653.9457 68068880.6410 68069574.9037 68069459.0203 alpha_1 -0.7401 -0.6205 -0.5685 -0.5533 -0.5504 -0.5500 -0.5500 -0.5500 -0.5500 alpha_2 -0.1961 -0.1658 -0.1518 -0.1476 -0.1468 -0.1467 -0.1467 -0.1467 -0.1467 alpha_3 -0.1447 -0.1118 -0.1038 -0.1020 -0.1018 -0.1018 -0.1018 -0.1018 -0.1018 alpha_4 0.4137 0.4107 0.3667 0.3511 0.3477 0.3471 0.3471 0.3471 0.3471 alpha_5 0.4709 0.6374 0.7131 0.7338 0.7378 0.7383 0.7384 0.7384 0.7384

beta 2.4448 2.6486 2.5548 2.5313 2.5261 2.5251 2.5250 2.5250 2.5250 P_f 0.007246369 0.004041237 0.005312652 0.005682817 0.005767152 0.005782924 0.005785208 0.00578544 0.005785446

Table B-1 FORM Calculations of the Safety Index Beam 1 Flexural Failure

57

65

Variable

Iteration 1 Iteration 2 Iteration 3 Iteration 4 Iteration 5 Iteration 6 Iteration 7 Iteration 8 Iteration 9

x_1 (d) 450.0000 443.1542 442.1311 442.6668 442.9882 443.1194 443.1678 443.1847 443.1906

x_2 (f_c) 38.0000 19.7101 21.9429 22.8916 23.3943 23.5988 23.6734 23.6993 23.7083

x_3(g) 52.5000 55.5651 59.3974 58.2871 57.7978 57.5832 57.5023 57.4736 57.4637

x_4 (q) 25.0000 27.9277 35.9671 41.3851 44.2182 45.3882 45.8175 45.9676 46.0191

G(x(m)) 420750000.0000 11361108.8235 -200181.5476 45599.3380 14544.3749 2418.5093 325.0677 39.7269 4.6882

(dG/dx_1)x(m) 3420000.0000 1746926.0038 1940331.4009 2026673.5348 2072679.0446 2091415.9610 2098254.1615 2100637.8947 2101455.7887

(dG/dx_2)x(m) 20250000.0000 19638567.1702 19547988.4219 19595389.9851 19623854.6630 19635484.5404 19639769.4017 19641271.9025 19641788.7103

(dG/dx_3)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

(dG/dx_4)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

sigma_1 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000

mu_1 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000

sigma_2 5.9660 3.0945 3.4450 3.5940 3.6729 3.7050 3.7167 3.7208 3.7222

mu_2 37.5217 32.4008 33.7165 34.2053 34.4483 34.5440 34.5785 34.5904 34.5945

sigma_3 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500

mu_3 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000

par_4 -0.5740 -1.1742 -2.8223 -3.9329 -4.5137 -4.7536 -4.8416 -4.8724 -4.8829

sigma_4 5.9763 7.0535 10.0614 11.9811 12.9289 13.3086 13.4463 13.4942 13.5106

mu_4 23.9559 23.5168 20.1299 16.6275 14.5604 13.6723 13.3421 13.2261 13.1862

Sum_x 1959750000.0000 785518757.1736 857680619.4226 897186696.8341 918186906.5541 926749505.1931 929878993.9837 930970711.3759 931345407.1794

Sum_mu 1954763462.3124 1080347102.5099 1205404979.6595 1271196402.7289 1306941849.3615 1321650127.0899 1327038485.4487 1328919758.4475 1329565602.7082

Sum_sigma 129708242.7190 74202249.8849 86303090.5279 93580982.8305 97396419.2116 98952664.8629 99520644.8270 99718719.0506 99786690.2110

alpha_1 -0.2373 -0.2119 -0.2023 -0.1949 -0.1915 -0.1902 -0.1898 -0.1896 -0.1895 alpha_2 -0.9314 -0.8190 -0.7803 -0.7526 -0.7400 -0.7352 -0.7335 -0.7329 -0.7327

alpha_3 0.1821 0.3184 0.2737 0.2525 0.2426 0.2388 0.2374 0.2369 0.2368 alpha_4 0.2073 0.4278 0.5246 0.5761 0.5974 0.6052 0.6080 0.6090 0.6093

beta 3.2054 4.1264 4.0268 3.9971 3.9916 3.9908 3.9907 3.9907 3.9907 P_f 0.000674435 1.84229E-05 2.82721E-05 3.20576E-05 3.28117E-05 3.29215E-05 3.29354E-05 3.29371E-05 3.29373E-05

Table B-2 FORM Calculations of the Safety Index Beam 1 Crushing Concrete Failure

58

66


x_1 (d) 450.0000 440.7590 443.8177 443.6913 444.3577 444.4603 444.4749 444.4768 444.4770

x_2 (f_c) 36.0000 7.0393 16.0209 20.8755 22.5339 22.7347 22.7606 22.7637 22.7640

x_3(f_y) 500.0000 465.9462 446.3454 464.8456 472.4332 473.8567 474.0030 474.0230 474.0253

x_4(g) 52.5000 59.1137 64.7447 60.4731 58.4255 58.0050 57.9538 57.9474 57.9466

x_5 (q) 25.0000 32.5256 56.2171 83.8140 93.7808 95.2131 95.3974 95.4201 95.4230

G(x(m)) 409359.8400 -47742.9845 -4396.6114 -1577.1254 -144.7808 -1.9556 -0.0316 -0.0004 0.0000

(dG/dx_1)x(m) 1426.3552 515.4176 807.7388 972.0365 1027.2673 1034.1806 1035.0499 1035.1551 1035.1684

(dG/dx_2)x(m) 11918.8800 15472.8950 14393.9739 13748.8644 13550.2008 13526.7749 13523.7950 13523.4446 13523.3996

(dG/dx_3)x(m) 252.0000 246.8251 248.5379 248.4671 248.8403 248.8978 248.9059 248.9070 248.9071

(dG/dx_4)x(m) -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000

(dG/dx_5)x(m) -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000

sigma_1 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000

mu_1 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000

sigma_2 5.6520 1.1052 2.5153 3.2775 3.5378 3.5693 3.5734 3.5739 3.5740 mu_2 37.4933 18.8195 29.6564 33.1174 34.0257 34.1272 34.1402 34.1417 34.1419

sigma_3 40.0000 37.2757 35.7076 37.1876 37.7947 37.9085 37.9202 37.9218 37.9220

mu_3 498.1960 497.1319 495.4018 497.0569 497.5211 497.5946 497.6019 497.6029 497.6030

sigma_4 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 mu_4 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 52.5000 par_5 -0.5781 -2.1208 -6.9776 -12.6350 -14.6782 -14.9718 -15.0096 -15.0142 -15.0148

sigma_5 5.9835 8.7913 16.5272 22.9902 24.9456 25.2149 25.2494 25.2536 25.2542 mu_5 23.9346 21.8826 4.8002 -19.8496 -29.0735 -30.4076 -30.5795 -30.6006 -30.6034

Sum_x 964439.5200 176182.1564 337141.6800 400935.8841 422754.8842 425466.9880 425791.3413 425830.6668 425835.6241 Sum_mu 984979.7356 422686.3835 741581.4204 918293.9020 976849.7540 984585.6215 985571.7998 985692.0400 985707.6073

Sum_sigma 73313.0242 36636.3559 64411.1178 84837.6330 91217.4120 92070.5215 92179.8626 92193.1905 92194.9405 alpha_1 -0.1751 -0.1266 -0.1129 -0.1031 -0.1014 -0.1011 -0.1011 -0.1011 -0.1011 alpha_2 -0.9189 -0.4668 -0.5621 -0.5311 -0.5255 -0.5244 -0.5243 -0.5242 -0.5242 alpha_3 -0.1375 -0.2511 -0.1378 -0.1089 -0.1031 -0.1025 -0.1024 -0.1024 -0.1024

alpha_4 0.2148 0.4299 0.2445 0.1856 0.1727 0.1711 0.1709 0.1708 0.1708 alpha_5 0.2448 0.7199 0.7698 0.8130 0.8204 0.8216 0.8217 0.8218 0.8218

beta 5.8639 5.4252 6.2108 6.0796 6.0729 6.0727 6.0727 6.0727 6.0727 P_f 2.26063E-09 2.89374E-08 2.63615E-10 6.02324E-10 6.28281E-10 6.28886E-10 6.28896E-10 6.28896E-10 6.28896E-10

Table B-3 FORM Calculations of the Safety Index Beam 1- Shear Failure

59

67


x_1 (f_y) 500.0000 393.3091 426.9848 438.9885 442.0432 442.6594 442.7793 442.8026 442.8071

x_2 (d) 450.0000 442.6360 444.6921 445.5600 445.7896 445.8372 445.8466 445.8484 445.8488 x_3 (f_c) 38.0000 33.1403 34.7738 35.1654 35.2495 35.2646 35.2675 35.2680 35.2681 x_4 (G_k) 66.0000 79.8378 78.9559 76.7401 76.1624 76.0378 76.0133 76.0085 76.0076

x_5 (Q_k) 33.0000 45.3328 59.2086 65.2694 66.8179 67.1357 67.1980 67.2101 67.2125

G(x(m)) 287202631.5789 6954442.2806 -1035004.5818 -292710.9753 -69064.2245 -12747.4383 -2395.1487 -460.5661 -89.1705 (dG/dx_1)x(m) 1344410.5263 1340672.5753 1340782.2527 1340255.6424 1340066.2521 1340021.4901 1340012.3802 1340010.5889 1340010.2400

(dG/dx_2)x(m) 1864500.0000 1466649.5525 1592226.4510 1636988.2778 1648379.0548 1650676.7560 1651123.9351 1651210.7278 1651227.5712 (dG/dx_3)x(m) 2524536.3573 2053829.5009 2198520.7011 2272400.2706 2293146.8073 2297572.3300 2298449.0880 2298620.0048 2298653.2092

(dG/dx_4)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 (dG/dx_5)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

sigma_1 40.0000 31.4647 34.1588 35.1191 35.3635 35.4127 35.4223 35.4242 35.4246 mu_1 498.1960 486.2890 492.8479 494.5323 494.9082 494.9815 494.9956 494.9984 494.9989

sigma_2 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 mu_2 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000

sigma_3 5.9660 5.2030 5.4595 5.5210 5.5342 5.5365 5.5370 5.5371 5.5371 mu_3 37.5217 37.2580 37.4213 37.4489 37.4543 37.4553 37.4554 37.4555 37.4555

sigma_4 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 mu_4 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 par_5 -0.5782 -2.4897 -4.6405 -5.5799 -5.8199 -5.8692 -5.8788 -5.8807 -5.8811

sigma_5 7.9138 12.5161 17.3661 19.2646 19.7287 19.8230 19.8414 19.8450 19.8457 mu_5 31.5906 27.7119 18.5482 13.7872 12.5284 12.2683 12.2173 12.2073 12.2054

Sum_x 1161662644.7368 681287574.4160 735254468.5861 758600376.9756 764618564.3141 765848279.4289 766087287.4930 766133613.7294 766142600.4551 Sum_mu 1149522146.3900 951914759.7903 1064258281.1344 1110651221.3017 1122640573.8731 1125089100.0625 1125567702.6467 1125660686.7032 1125678735.4276

Sum_sigma 75107062.8367 78820292.3279 97603929.9573 105250563.5626 107150073.9816 107536609.0165 107612188.0471 107626873.9442 107629724.6656 alpha_1 -0.7160 -0.5352 -0.4692 -0.4472 -0.4423 -0.4413 -0.4411 -0.4410 -0.4410 alpha_2 -0.2234 -0.1675 -0.1468 -0.1400 -0.1385 -0.1381 -0.1381 -0.1381 -0.1381 alpha_3 -0.2005 -0.1356 -0.1230 -0.1192 -0.1184 -0.1183 -0.1183 -0.1183 -0.1183 alpha_4 0.4152 0.3956 0.3195 0.2963 0.2910 0.2900 0.2898 0.2898 0.2897

alpha_5 0.4742 0.7146 0.8007 0.8237 0.8285 0.8295 0.8297 0.8297 0.8297 beta 3.6623 3.5217 3.3602 3.3421 3.3407 3.3405 3.3405 3.3405 3.3405

P_f 0.000124996 0.000214392 0.000389429 0.000415732 0.000417884 0.00041811 0.000418145 0.000418152 0.000418153

Table B-4 FORM Calculations of the Safety Index Beam 2- Flexural Failure

60

68


x_1 (d) 450.0000 445.2363 444.9421 445.0996 445.1654 445.1860 445.1923 445.1942 445.1948

x_2 (f_c) 38.0000 25.1275 26.5947 26.9281 27.0468 27.0836 27.0948 27.0982 27.0992

x_3(g) 69.3000 73.0163 75.3311 74.8245 74.6810 74.6338 74.6193 74.6149 74.6136

x_4 (q) 33.0000 36.4370 41.6784 43.7259 44.4284 44.6485 44.7158 44.7361 44.7423

G(x(m)) 309150000.0000 5575844.6837 -38216.0944 4740.7646 706.6569 68.5122 6.4130 0.5877 0.0536

(dG/dx_1)x(m) 3420000.0000 2237532.4628 2366620.8177 2397133.7094 2408063.7758 2411443.9044 2412476.0135 2412788.2351 2412882.4976

(dG/dx_2)x(m) 20250000.0000 19823540.4779 19797345.9004 19811369.4061 19817220.3526 19819055.3159 19819617.8827 19819788.3099 19819839.7853

(dG/dx_3)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

(dG/dx_4)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

sigma_1 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000

mu_1 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000

sigma_2 5.9660 3.9450 4.1754 4.2277 4.2464 4.2521 4.2539 4.2544 4.2546

mu_2 37.5217 35.2045 35.7509 35.8636 35.9028 35.9148 35.9184 35.9195 35.9199

sigma_3 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300

mu_3 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000

par_4 -0.5782 -1.1109 -1.9233 -2.2407 -2.3496 -2.3837 -2.3941 -2.3972 -2.3982

sigma_4 7.9138 9.1764 11.1478 11.9170 12.1795 12.2616 12.2866 12.2942 12.2965

mu_4 31.5906 31.1036 29.4303 28.5162 28.1735 28.0632 28.0292 28.0189 28.0158

Sum_x 1848150000.0000 1001806624.9583 1052970984.9181 1066968104.0462 1071987299.9730 1073541076.4815 1074015737.6141 1074159348.6766 1074202708.2078

Sum_mu 1844807254.7706 1252952242.4144 1328466624.9622 1349044735.3079 1356491479.3589 1358812416.4903 1359522233.5378 1359737092.6048 1359801971.3213

Sum_sigma 133355029.5021 95911902.8736 103805490.0493 106437352.6079 107366114.7095 107656752.0455 107745599.3922 107772496.7791 107780618.6853

alpha_1 -0.2308 -0.2100 -0.2052 -0.2027 -0.2019 -0.2016 -0.2015 -0.2015 -0.2015

alpha_2 -0.9059 -0.8154 -0.7963 -0.7869 -0.7838 -0.7828 -0.7825 -0.7824 -0.7824

alpha_3 0.2338 0.3251 0.3004 0.2930 0.2905 0.2897 0.2894 0.2894 0.2893

alpha_4 0.2670 0.4305 0.4833 0.5038 0.5105 0.5125 0.5132 0.5133 0.5134

beta 2.2932 2.6766 2.6536 2.6502 2.6499 2.6498 2.6498 2.6498 2.6498

P_f 0.010918786 0.003718245 0.003982002 0.004022084 0.004026288 0.00402669 0.004026727 0.00402673 0.004026731

Table B-5 FORM Calculations of the Safety Index Beam 2- Crushing Concrete Failure

61

69


x_1 (d) 450.0000 443.0025 444.6157 445.1252 445.5193 445.6086 445.6265 445.6299 445.6306

x_2 (f_c) 38.0000 13.2949 21.2797 23.9888 24.9396 25.1355 25.1731 25.1803 25.1816

x_3(f_y) 500.0000 474.6408 465.2428 476.0050 479.0677 479.8225 479.9659 479.9932 479.9984

x_4(g) 69.3000 77.7169 82.3467 78.1051 76.7254 76.3785 76.3100 76.2969 76.2945

x_5 (q) 33.0000 42.5670 64.8662 81.2952 87.1646 88.4575 88.7097 88.7578 88.7669

G(x(m)) 358529.7600 -35657.7363 -3798.4075 -440.4252 -47.9915 -1.9868 -0.0722 -0.0026 -0.0001

(dG/dx_1)x(m) 1478.7328 734.0681 984.7609 1073.3170 1103.4804 1109.7317 1110.9284 1111.1556 1111.1987

(dG/dx_2)x(m) 11651.0400 14726.9274 13724.0208 13380.8788 13266.6609 13243.3367 13238.8854 13238.0412 13237.8812

(dG/dx_3)x(m) 252.0000 248.0814 248.9848 249.2701 249.4908 249.5408 249.5508 249.5527 249.5531

(dG/dx_4)x(m) -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000

(dG/dx_5)x(m) -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000

sigma_1 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000 9.0000

mu_1 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000 450.0000

sigma_2 5.9660 2.0873 3.3409 3.7662 3.9155 3.9463 3.9522 3.9533 3.9535

mu_2 37.5217 27.0899 33.3505 34.7217 35.1285 35.2078 35.2228 35.2257 35.2262

sigma_3 40.0000 37.9713 37.2194 38.0804 38.3254 38.3858 38.3973 38.3995 38.3999

mu_3 498.1960 497.6332 497.0843 497.6973 497.8271 497.8560 497.8614 497.8624 497.8626

sigma_4 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300 6.9300

mu_4 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000 69.3000

par_5 -0.5782 -2.0610 -5.5174 -8.0639 -8.9737 -9.1741 -9.2132 -9.2206 -9.2220 sigma_5 7.9138 11.4822 19.1424 23.7118 25.1721 25.4837 25.5441 25.5556 25.5578

mu_5 31.5906 29.0497 14.1125 0.2326 -4.9352 -6.0834 -6.3078 -6.3506 -6.3587 Sum_x 927269.2800 277884.8817 404083.2341 439204.4746 450340.4696 452611.8062 453039.7988 453120.8257 453136.1764

Sum_mu 925470.4724 557686.7891 774374.8719 863062.6845 893713.0163 900233.5410 901493.7470 901733.5483 901779.0260 Sum_sigma 78142.5802 51923.8951 77439.1497 90639.8826 94991.3387 95915.7041 96094.5926 96128.6354 96135.0917

alpha_1 -0.1703 -0.1272 -0.1144 -0.1066 -0.1045 -0.1041 -0.1040 -0.1040 -0.1040 alpha_2 -0.8895 -0.5920 -0.5921 -0.5560 -0.5468 -0.5449 -0.5445 -0.5444 -0.5444 alpha_3 -0.1290 -0.1814 -0.1197 -0.1047 -0.1007 -0.0999 -0.0997 -0.0997 -0.0997 alpha_4 0.2661 0.4004 0.2685 0.2294 0.2189 0.2168 0.2163 0.2163 0.2163 alpha_5 0.3038 0.6634 0.7416 0.7848 0.7950 0.7971 0.7975 0.7975 0.7976


62

Table B-6 FORM Calculations of the Safety Index Beam 2- Shear Failure

70

Variable Iteration 1 Iteration 2 Iteration 3 Iteration 4 Iteration 5 Iteration 6 Iteration 7 Iteration 8

x_1 (f_y) 500.0000 426.1349 446.0745 450.9513 452.2578 452.5649 452.6362 452.6527

x_2 (d) 550.0000 543.8167 545.3182 545.7342 545.8533 545.8817 545.8883 545.8898

x_3 (f_c) 38.0000 35.4309 36.1160 36.2111 36.2368 36.2428 36.2441 36.2444

x_4 (G_k) 94.5000 105.9246 104.6497 103.3345 102.9950 102.9126 102.8934 102.8889

x_5 (Q_k) 45.0000 57.8955 67.2299 70.4651 71.3191 71.5225 71.5698 71.5808

G(x(m)) 242882105.2632 1101326.6566 62393.7429 2775.2388 239.4051 13.3875 0.7291 0.0390

(dG/dx_1)x(m) 1584228.4211 1588988.5608 1586653.5874 1585708.5576 1585487.4787 1585435.7792 1585423.8823 1585421.1410

(dG/dx_2)x(m) 1797500.0000 1531954.8149 1603637.8245 1621170.0286 1625866.6924 1626970.8162 1627227.1731 1627286.3725

(dG/dx_3)x(m) 2066260.3878 1726400.6969 1820658.3208 1850924.8713 1859018.6288 1860936.5629 1861381.9093 1861484.7586

(dG/dx_4)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

(dG/dx_5)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

sigma_1 40.0000 34.0908 35.6860 36.0761 36.1806 36.2052 36.2109 36.2122

mu_1 498.1960 492.7159 495.3720 495.8844 496.0127 496.0423 496.0491 496.0507

sigma_2 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000

mu_2 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000

sigma_3 5.9660 5.5627 5.6702 5.6851 5.6892 5.6901 5.6903 5.6904

mu_3 37.5217 37.4652 37.4979 37.5015 37.5024 37.5026 37.5026 37.5026 sigma_4 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500

mu_4 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000

par_5 -0.5780 -2.0481 -3.1122 -3.4810 -3.5784 -3.6016 -3.6069 -3.6082 sigma_5 10.7595 15.5690 19.0165 20.1693 20.4692 20.5403 20.5568 20.5606

mu_5 43.0853 39.6807 34.6901 32.6217 32.0527 31.9159 31.8840 31.8767 Sum_x 1231507105.2632 834204003.8539 874555363.9058 884730715.5065 887484937.1851 888133532.5635 888284217.6136 888319020.1325

Sum_mu 1236277008.3285 1086361892.1332 1154900185.3420 1175336393.7127 1180878838.5592 1182195446.0657 1182501644.9604 1182572393.1537 Sum_sigma 93333896.4840 100136896.2610 112938117.1292 117249157.1480 118385748.1666 118655906.7506 118718690.6960 118733196.8607

alpha_1 -0.6790 -0.5410 -0.5013 -0.4879 -0.4846 -0.4838 -0.4836 -0.4835 alpha_2 -0.2118 -0.1683 -0.1562 -0.1521 -0.1511 -0.1508 -0.1508 -0.1508 alpha_3 -0.1321 -0.0959 -0.0914 -0.0897 -0.0893 -0.0892 -0.0892 -0.0892

alpha_4 0.4556 0.4247 0.3765 0.3627 0.3592 0.3584 0.3582 0.3582 alpha_5 0.5188 0.6996 0.7577 0.7741 0.7781 0.7790 0.7792 0.7793

beta 2.6534 2.5291 2.4828 2.4786 2.4783 2.4783 2.4783 2.4783 P_f 0.003984288 0.005717285 0.00651699 0.006595793 0.006600702 0.006600974 0.006600988 0.006600989

Table B-7 FORM Calculations of the Safety Index Beam 3- Flexural Failure

63

71


x_1 (d) 550.0000 543.2813 542.6993 543.0094 543.1610 543.2156 543.2345 543.2409 543.2431

x_2 (f_c) 38.0000 23.2192 24.9845 25.4924 25.7057 25.7815 25.8076 25.8164 25.8194

x_3(g) 94.5000 99.8382 104.1107 103.0614 102.7042 102.5689 102.5219 102.5060 102.5006

x_4 (q) 45.0000 50.0055 59.4357 63.9719 65.8237 66.4909 66.7206 66.7983 66.8245

G(x(m)) 521750000.0000 11028602.9229 -110800.9727 17346.4865 3569.8511 457.7380 54.3667 6.2232 0.7041

(dG/dx_1)x(m) 4180000.0000 2522911.2185 2711808.8086 2768522.2893 2792463.8978 2800985.6107 2803914.6807 2804904.5603 2805237.3998

(dG/dx_2)x(m) 30250000.0000 29515461.4945 29452248.1996 29485925.4319 29502388.8754 29508322.3167 29510372.9909 29511067.6226 29511301.3777

(dG/dx_3)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

(dG/dx_4)x(m) -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000 -4500000.0000

sigma_1 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000

mu_1 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000

sigma_2 5.9660 3.6454 3.9226 4.0023 4.0358 4.0477 4.0518 4.0532 4.0536

mu_2 37.5217 34.3649 35.1468 35.3483 35.4298 35.4584 35.4682 35.4715 35.4726

sigma_3 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500

mu_3 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000

par_4 -0.5780 -1.1486 -2.2237 -2.7408 -2.9519 -3.0279 -3.0541 -3.0630 -3.0660

sigma_4 10.7595 12.6001 16.1470 17.8309 18.5076 18.7496 18.8327 18.8608 18.8702

mu_4 43.0853 42.3496 38.9762 36.6147 35.5418 35.1415 35.0021 34.9547 34.9388

Sum_x 2820750000.0000 1381679191.6972 1471585820.6995 1503351090.5314 1516761096.3241 1521539627.2999 1523183264.3890 1523738886.0167 1523925727.9564

Sum_mu 2814898377.2576 1786073339.9836 1926003838.0325 1974947275.5527 1995931450.3489 2003473653.2512 2006073052.5062 2006952445.2025 2007248234.1710

Sum_sigma 197070487.7091 131796912.0442 146029675.7615 151989650.4335 154482172.2442 155379031.4426 155687844.1978 155792306.4120 155827440.5440

alpha_1 -0.2333 -0.2106 -0.2043 -0.2004 -0.1988 -0.1983 -0.1981 -0.1980 -0.1980

alpha_2 -0.9158 -0.8164 -0.7911 -0.7764 -0.7707 -0.7687 -0.7680 -0.7678 -0.7677

alpha_3 0.2158 0.3227 0.2912 0.2798 0.2753 0.2737 0.2731 0.2730 0.2729

alpha_4 0.2457 0.4302 0.4976 0.5279 0.5391 0.5430 0.5443 0.5448 0.5449

beta 2.6178 3.1520 3.1111 3.1029 3.1018 3.1017 3.1017 3.1017 3.1017

P_f 0.004424456 0.000810805 0.000932082 0.000958069 0.000961716 0.000962162 0.000962213 0.000962219 0.00096222

Table B-8 FORM Calculations of the Safety Index Beam3 Crushing Concrete Failure

64

72


x_1 (d) 550.0000 542.5788 544.0981 544.6868 545.0459 545.1397 545.1618 545.1668 545.1679

x_2 (f_c) 38.0000 16.4994 23.2368 25.1879 25.9044 26.0812 26.1218 26.1310 26.1330

x_3(f_y) 500.0000 477.7564 472.0255 479.3014 481.3089 481.8708 482.0002 482.0296 482.0362

x_4(g) 94.5000 105.6118 110.0415 105.7210 104.3731 103.9953 103.9063 103.8861 103.8816

x_5 (q) 45.0000 57.4900 81.2188 96.0844 101.5432 102.9496 103.2765 103.3507 103.3674

G(x(m)) 394803.0400 -33098.6221 -3349.4185 -271.3690 -33.5913 -2.0164 -0.1057 -0.0054 -0.0003

(dG/dx_1)x(m) 1478.7328 840.8124 1048.3980 1110.9959 1133.3272 1138.8507 1140.1177 1140.4041 1140.4686

(dG/dx_2)x(m) 14240.1600 17519.7541 16477.8676 16179.4173 16073.8610 16047.9479 16042.0124 16040.6721 16040.3700

(dG/dx_3)x(m) 308.0000 303.8441 304.6949 305.0246 305.2257 305.2782 305.2906 305.2934 305.2940

(dG/dx_4)x(m) -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000

(dG/dx_5)x(m) -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000 -3000.0000

sigma_1 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000 11.0000

mu_1 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000 550.0000

sigma_2 5.9660 2.5904 3.6482 3.9545 4.0670 4.0948 4.1011 4.1026 4.1029

mu_2 37.5217 30.0565 34.3733 35.2287 35.5043 35.5692 35.5839 35.5872 35.5880

sigma_3 40.0000 38.2205 37.7620 38.3441 38.5047 38.5497 38.5600 38.5624 38.5629

mu_3 498.1960 497.7739 497.4993 497.8362 497.9096 497.9286 497.9329 497.9339 497.9341

sigma_4 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500 9.4500

mu_4 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000 94.5000

par_5 -0.5780 -2.0018 -4.7069 -6.4016 -7.0239 -7.1842 -7.2215 -7.2300 -7.2319

sigma_5 10.7595 15.4165 23.8004 28.3086 29.8310 30.2129 30.3011 30.3211 30.3256

mu_5 43.0853 39.8571 24.9083 12.9448 8.3305 7.1281 6.8479 6.7843 6.7700 Sum_x 1089929.1200 401129.7065 523366.8144 553453.1695 563258.5285 565652.6239 566196.2918 566318.8466 566346.4467

Sum_mu 1088306.8855 737204.1788 936378.6276 1010545.8456 1037503.5373 1044302.2385 1045872.4722 1046227.9941 1046308.1401 Sum_sigma 97363.4852 72268.7202 98898.5517 111336.8172 115670.9901 116758.8869 117010.0767 117066.9462 117079.7661

alpha_1 -0.1671 -0.1280 -0.1166 -0.1098 -0.1078 -0.1073 -0.1072 -0.1072 -0.1072 alpha_2 -0.8726 -0.6280 -0.6078 -0.5747 -0.5652 -0.5628 -0.5623 -0.5621 -0.5621 alpha_3 -0.1265 -0.1607 -0.1163 -0.1050 -0.1016 -0.1008 -0.1006 -0.1006 -0.1006 alpha_4 0.2912 0.3923 0.2867 0.2546 0.2451 0.2428 0.2423 0.2422 0.2421 alpha_5 0.3315 0.6400 0.7220 0.7628 0.7737 0.7763 0.7769 0.7770 0.7770


Table B-9 FORM Calculations of the Safety Index Beam 3 - Shear Failure

65

73

Appendix C

Calculations are carried out in Excel using a Visual Basic Subroutine

74

Calculations are carried out in Excel using a Visual Basic Subroutine.

Beam 1

Table C-1

Reinforced concrete beam 1 Flexural Failure

Number of simulation trials 10000000 Number of times G(x) < 0 70085 Probability of failure 0.0070085 COV_Pf 0.00376409 beta 2.45682743

Private Sub CommandButton1_Click ()

Numfails = 0

Ntrials = 10000000

Range ("C8").Value = Ntrials

Pi_2 = 2 * 3.14159265358979

Randomize

For i = 1 to Ntrials

u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

End If

fy = Exp(6.211 + 0.08 * Sqr(-2 * Log(u1)) * Cos(Pi_2 * u2))

fc = Exp(3.625 + 0.157* Sqr(-2 * Log(u1)) * Sin(Pi_2 * u2))

g = 52.5 + 5.25 * Sqr (-2 * Log (u3)) * Cos (Pi_2 * u4)

d = 450 + 9 * Sqr (-2 * Log (u3)) * Sin (Pi_2 * u4)

q = 22.2 - 1 / 0.205 * (Log (-Log (u5)))

Start simulation

75

Gx = 1649.52 * fy * (d - 4.85 * fy * fc ^ -1) + 838.68 * fy * (d - 50) - 4500000 * (g + q)

If Gx < 0! Then

Numfails = Numfails + 1

End If

Next i

Pf = Numfails / Ntrials

beta = -Application.WorksheetFunction.NormSInv (Pf)

COV_Pf = Sqr ((1# - Pf) / (Ntrials - 1) / Pf)

Range ("C10").Value = Numfails

Range ("C12").Value = Pf

Range ("C14").Value = COV_Pf

Range ("C16").Value = beta

End Sub

Table C-2

Reinforced concrete beam 1 Crushing concrete failure


Private Sub CommandButton1_Click()

Numfails = 0

Ntrials = 1000000


Pi_2 = 2 * 3.14159265358979

Randomize

For i = 1 to Ntrials

u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

Start simulation

76

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

fc = Exp(3.625 + 0.157 * Sqr(-2 * Log(u1)) * Sin(Pi_2 * u2))

g = 52.5 + 5.25 * Sqr(-2 * Log(u3)) * Cos(Pi_2 * u4)

d = 450 + 9 * Sqr(-2 * Log(u3)) * Sin(Pi_2 * u4)

q = 22.2 - 1 / 0.205 * (Log(-Log(u5)))

Gx = 100 * d ^ 2 * fc - 4500000 * (g + q) If Gx < 0! Then


End If

Next i


beta = -Application.WorksheetFunction.NormSInv(Pf)






End Sub

Table C-3

Reinforced concrete beam 1

Shear failure Number of simulation trials 10000000 Number of times G(x) < 0 3

Probability of failure 0.0000003

COV_Pf 0.57735021

beta 4.9912173Private Sub CommandButton1_Click ()

Numfails = 0

Ntrials = 10000000


Pi_2 = 2 * 3.14159265358979

Randomize

Start simulation

77

For i = 1 To Ntrials

u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

u5 = Rnd

End If



g = 52.5 + 5.25 * Sqr(-2 * Log(u3)) * Sin(Pi_2 * u4)

d = 450 + 9 * Sqr(-2 * Log(u3)) * Cos(Pi_2 * u4)

q = 22.2 - 1 / 0.205 * (Log(-Log(u5)))

Gx = 37.2 * d * (1 - fc / 250) * fc + 0.56 * fy * d - 3000 * (g + q)

If Gx < 0! Then


End If

Next i


Beta = -Application.WorksheetFunction.NormSInv (Pf)

COV_Pf = Sqr((1# - Pf) / (Ntrials - 1) / Pf)





End Sub

78

Table C-4


Beam 2

Reinforced Concrete Beam 2

Bending Failure Number of simulation trials 10000000 Number of times G(x) < 0 4831 Probability of failure 0.0004831 COV_Pf 0.0143839 beta 3.30018823


Numfails = 0

Ntrials = 10000000


Pi_2 = 2 * 3.14159265358979

Randomize


u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

End If


fc = Exp(3.625 + 0.157* Sqr(-2 * Log(u1)) * Sin(Pi_2 * u2))

g = 69.3 + 6.93 * Sqr(-2 * Log(u3)) * Cos(Pi_2 * u4)

d = 450 + 9 * Sqr (-2 * Log(u3)) * Sin(Pi_2 * u4)

Start simulation

79

q = 29.27 - 1 / 0.155 * (Log (-Log(u5)))

Gx = 1896 * fy * (d - 4.85 * fy * fc ^ -1) + 1833 * fy * (d - 50) - 4500000 * (g + q)

If Gx < 0! Then


End If

Next i








End Sub

Table C-5

Reinforced Concrete Beam 2 Crushing Concrete failure



Numfails = 0

Ntrials = 10000000


Pi_2 = 2 * 3.14159265358979

Randomize


u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

Start simulation

80

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

u5 = Rnd

End If


g = 69.3 + 6.93 * Sqr (-2 * Log (u3)) * Cos (Pi_2 * u4)

d = 450 + 9 * Sqr (-2 * Log (u3)) * Sin (Pi_2 * u4)

q = 29.27 - 1 / 0.155 * (Log (-Log (u5)))

Gx = 100 * d ^ 2 * fc - 4500000 * (g + q)

If Gx < 0! Then


End If

Next i


Beta = -Application.WorksheetFunction.NormSInv (Pf)






End Sub

Table C-6

Reinforced Concrete Beam 2 Shear Failure

Number of simulation trials 1000000000 Number of times G(x) < 0 2086 Probability of failure 2.086E-06 COV_Pf 0.02189487 beta 4.60262451


Numfails = 0

Ntrials = 1000000000

Start simulation

81


Pi_2 = 2 * 3.14159265358979

Randomize


u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

u5 = Rnd

End If



g = 69.3 + 6.93 * Sqr (-2 * Log (u3)) * Cos (Pi_2 * u4)

d = 450 + 9 * Sqr (-2 * Log (u3)) * Sin (Pi_2 * u4)

q = 29.27 - 1 / 0.155 * (Log(-Log(u5)))

Gx = 37.2 * d * (1 - fc / 250) * fc + 0.56 * fy * d - 3000 * (g + q)

If Gx < 0! Then


End If

Next i








End Sub

82

Table C-7


Beam 3

Reinforced Concrete Beam 3 Bending Failure



Numfails = 0

Ntrials = 10000000


Pi_2 = 2 * 3.14159265358979

Randomize


u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

End If



g = 94.5 + 9.45 * Sqr (-2 * Log (u3)) * Cos (Pi_2 * u4)

d = 550 + 11 * Sqr (-2 * Log (u3)) * Sin (Pi_2 * u4)

q = 39.93 - 1 / 0.114 * (Log (-Log (u5)))

Start simulation

83

Gx = 2016 * fy * (d - 5.92 * fy * fc ^ -1) + 1579 * fy * (d - 50) - 4500000 * (g + q)

If Gx < 0! Then


End If

Next i








End Sub

Table C-8

Reinforced concrete beam 3 Crushing concrete failure



Numfails = 0

Ntrials = 10000000


Pi_2 = 2 * 3.14159265358979

Randomize


u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

Start simulation

84

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

u5 = Rnd

End If


g = 94.5 + 9.45 * Sqr(-2 * Log(u3)) * Cos(Pi_2 * u4)

d = 550 + 11 * Sqr(-2 * Log(u3)) * Sin(Pi_2 * u4)

q = 39.93 - 1 / 0.114 * (Log(-Log(u5)))

Gx = 100 * d ^ 2 * fc - 4500000 * (g + q)

If Gx < 0! Then


End If

Next i



COV_Pf = Sqr((1# - Pf) / (Ntrials - 1) / Pf)





End Sub

Table C-9

Reinforced concrete beam 3 Shear failure



Numfails = 0

Ntrials = 100000000

Start simulation

85


Pi_2 = 2 * 3.14159265358979

Randomize


u1 = Rnd

u2 = Rnd

u3 = Rnd

u4 = Rnd

u5 = Rnd

If u1 = 0 Then

u1 = Rnd

End If

If u3 = 0 Then

u3 = Rnd

End If

If u5 = 0 Then

u5 = Rnd

End If



g = 94.5 + 9.45 * Sqr(-2 * Log(u3)) * Cos(Pi_2 * u4)

d = 550 + 11 * Sqr(-2 * Log(u3)) * Sin(Pi_2 * u4)

q = 39.93 - 1 / 0.114 * (Log(-Log(u5)))

Gx = 37.2 * d * (1 - fc / 250) * fc + 0.56 * fy * d - 3000 * (g + q) If Gx < 0! Then


End If

Next i








End Sub

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