Investigation of Ultimate... .pdf - SUST Repository

Sudan University of Science and Technology

College of Graduate Studies

Investigation of Ultimate Resisting Moments of R.C.

Slabs systems using Yield Line Theory, BS8110 and

Computer Software Programs

اخلرسانية العزم املقاوم الأقىص لأنظمة البالطات تقيص

س تخدام نظرية خط اخلضوع ، املدونة الربطانية املسلحة بإ

BS8110 وبرامج احلاسوب

By:

Izeldein Jadalla Izeldein Ahmed

Thesis Supervisor:

Dr.Fathelrahaman Mohammed Adam

A thesis submitted in partial fulfillment of the requirement for the degree

of master of sciences in civil engineering (structures) to Civil

Engineering Department, College of engineering

December 2015

I

ACKNOWLEDGEMENTS

First and foremost, I would like to thank God.

Secondly, I would like to express my utmost gratitude and thanks to my thesis

supervisor, Dr. Fathelrahaman, who gives me support, guidance, continuous

encouragement throughout my project thesis, with his patience.

Finally, I would like to give a special thanks to my parent, my family, my wife,

and my friends for their care, love, encouragement and support during my study

and given me hope that Nothing is impossible, everything can be achieved by

perseverance .

II

الخالصة

، بحاالث حثبيج هخخلفت وهن البالطاث الخشسانيت الوسلحت لعذة أنظوتإنشائي عول ححليل في هزه الذاسست حن

هن بالبالطاث ، العزوم الوخحصل عليها هقاوهتىع إليجاد أقصى عزوم بإسخخذام نظشيت خظ الخض ورلك

هج ابإسخخذام بشهن الخحليل حن الحصىل عليها الخي نخائج الحوج هقاسنخها ببنظشيت خظ الخضىع الخحليل

بذسجاث هخخلفت يخشاوح وأظهشث الوقاسنت حطابق ( ، الوذونت البشيطانيت ، بشوكن -إسخاد بشوالحاسىب )

%109719% إلى 090.0ن هابي

III

ABSTRACT

In this study, different types of reinforced concrete slabs system of different

support conditions have been analyzed using yield line theory to determine the

maximum resisting moment, the result of resisting moments obtained, were

compared with others those obtained by using software program (STAAD-

PRO,PROKON) and BS8110. The comparison revealed different conforming by

percentage range by 0.097 % and 17.81%.

IV

Table of Contents

CHAPTER ONE GENRAL INTRODUCTION 1

1.1 Introduction: 1

1.2 Objectives of the Research 2

1.3 Methodology of Research 2

1.4 Outlines of Research 3

CHAPTER TWO LITERATURE REVIE 4

2.1 Introduction 4

2.2 Historical back ground about yield line theory 7

2.3 Yield lines theory 10

2.4 The advantages and disadvantages of yield line theory 11

2.5 prediction of yield line pattern 13

CHAPTER THREE Methodology 19

3.1 Introduction 19

3.2 Method of solution by using yield line theory 19

3.3 The 10% rule 26

3.4 Serviceability and deflections 26

3.5. Analysis and design of R.C. slabs using BS8110 27

3.6 Software program 29

V

CHAPTER FOUR 30

Investigation of Ultimate Resisting Moment for the Reinforced Concrete

Slabs

4.1 Introduction 30

4.2 4.2 Analysis of flat Slab system 31

4.3. Analysis of beam Slab system 49

4.4 Special Beam slab system of different support conditions 67

4.5 Discussion of the results 75

CHAPTER FIVE 76

CONCLUSIONS AND RECOMMENDATION 76

5.1 Conclusions 76

5.2 Recommendations 77

REFERENCES 78

APPENDIX A 80

APPENDIX B 82

VI

LIST OF FIGURES

Figure NO. TITLE PAGE

Fig.2.1 Yield Line Pattern for the Two-Way Slab 13

Fig.2.2 Consisting of Yield Line Pattern 13

Fig.2.3 Yield line pattern for square slab 16

Fig.2.4 Notation used for yield line theory 16

Fig.2.5 Yield line pattern for the different slabs shape 18

Fig.3.1 Simply supported one-way slab 21

Fig.3.2 Deformed shape at failure 23

Fig.3.3 Principles of expenditure of external loads 24

Fig.3.4 Principles of dissipation of internal energy 25

Fig.4. 1 Plan of Flat Slab with the Expected Yield Line Pattern 31

VII

Fig.4.2 Expected yield line Pattern of external Corner Panel S1 33

Fig.4.3 Yield Pattern of Panel S1 36

Fig.4.4 Expected yield line Pattern of the Panel S2 37


Fig.4.6 Expected yield line Pattern of edge Panel S3 41


Fig.4.8 Expected yield line Pattern of interior Panel S4 45


Fig.4.1 Plan of beam Slab system with the Expected Yield Line Pattern 49

Fig.4.11 Expected yield line Pattern of the Corner Panel S5 50


Fig.4.13 Expected yield line Pattern of the edge Panel S6 54

Fig.4.14 Yield line Pattern of Panel S6 57

VIII

Fig.4.15 Expected Yield line Pattern of the edge Panel S7 58


Fig.4.17 Expected Yield line Pattern of the interior Panel S8 62


Fig.4.19 Expected Yield line Pattern of Corner Panel S9 67


Fig.4.21 Expected Yield line Pattern of Corner Panel S10 71


IX

LIST OF TABLES

TABLE NO. TITLE PAGE

Table.3.1 Bending moment coefficients for rectangular panels 28

Table.4.1 Analysis of Panel S1 by yield line theory 33




Table.4.5 The Ultimate Resisting Moment for flat slab system 48





Table.4.10 The Ultimate Resisting Moments for beam slab system 66

X



Table.4.13 The Ultimate Resisting Moments for Panels S9 &S10 76

XI

List of Symbols

Symbol Description Units

A Area cross-section m2

A, b, c Plan dimensions of slab supported on several sides m

D Dissipation of internal energy kNm

E Expenditure of energy by external loads kNm

g Ultimate distributed dead load kN/m

l Length of a yield line (projected onto a region’s axis of rotation) m

L Span (commonly edge to edge), distance m

m Positive moment, i.e the ultimate moment along the yield line kNm/m

m` Negative moment, i.e the ultimate moment along the yield line KNm/m

w Ultimate distributed load kN/m

x Distance to section of max. Positive moment from support m

∆δ max Deflection, maximum deflection (usually taken as unity) m

θ Angle of rotation …

CHAPTER ONE

GENRAL INTRODUCTION

1

CHAPTER ONE

GENRAL INTRODUCTION

1.1 Introduction

Reinforcement concrete slabs are among the most common structural element,

but despite the large number of slabs designer and built, the details of the elastic

and plastic behavior of slab are not always appreciated or properly taken into

account, this occurs at least partially because of the mathematical complexities

of dealing with elastic plate equation, especially for support condition, which

realistically approximate those in multi-panel building floor slabs.

Because the theoretical analysis of slab and plates is much less widely known

and practiced than is the analysis of element such as beam, the provisions in

building codes generally provide both design criteria and method of analysis for

slabs, whereas only criteria are provided for most other element, for example,

chapter 13 of the 1995 edition of the American Concrete Institute (ACI)

Building Code Requirements for structural Concrete one of the most widely

used codes for concrete design, is devoted largely to the determination of

moment in the slab structures, once the moments, shears, and torques are found,

section are proportioned to resist them using the criteria specified in other

sections of the same code.

Although the ACI code approach to slab design is basically one of using elastic

moment distributions, it is also possible to design slab using plastic analysis

(limit analysis) to provide the required moment.

Yield lines analysis is an analysis approach for determining the ultimate load

capacity of reinforced concrete slab and was pioneered by K.W.Johansen in the

2

1940s. It is closely related to the plastic collapse or limit analysis of steel frame,

and is an upper bound or mechanism approach.

Yield lines design is well founded method of designing reinforced concrete slabs

, and similar types of element , it uses the yield line theory to investigate failure

mechanism at limit state , the theory is based on the principle that :

Work done in yield line rotating = work done in load moving

Two of the most popular methods of application are working method and the

use of stander formula. Working method proposed to use in this research to

analysis and assessment different models of slabs (slab with beam and flat slab).

The proposed models consist of slab with beams with different edges conditions

and flat slab with different support conditions.

The suggested tools should be used is spread sheets and STAAD-Pro – Prokon

software and BS8110.

This research explains the use of practical and economic design of reinforced

concrete slabs.

1.2 Objectives of the Research

The objectives of this study are:

1- To identify yield line theory of reinforced concrete slabs.

2- To apply the yield line theory to obtain the ultimate resistance moments of

different types of slabs.

3- To compare the results obtained by yield line theory with that obtained by

software program STAAD-PRO, PROKON, and BS8110.

1.3 Methodology of Research

To achieve the objectives of this research; the following methods can be used:

1- A comprehensive reading and collecting data about yield line method must be

carried out.

3

2- Using three system of slabs (beam slab system flat slab system and special case

of support) with different of boundary condition as mathematical models in

order to apply the yield line theory for analysis.

3- Manual calculation is used for analyzing the different cases of slabs.

4- Structural Analysis Program STAAD-Pro, PROKON and BS8110 is used to

verify the results.

1.4 Outlines of Research

Chapter One contains general introduction.

Chapter Two contains literature review.

Chapter Three describes the methodology of analysis of reinforced concrete

slabs by using yield line theory.

Chapter Four (models) contains Analysis of the three systems of Slabs using

virtual work method.

Chapter Five contains conclusion and recommendation.

CHAPTER TWO

LITERATURE REVIEW

4

CHAPTER TWO

LITERATURE REVIEW

2.1 Introduction

There are a number of possible approaches to the analysis and design of the

reinforced concrete slab. The various approaches available are elastic theory,

limit analysis theory, and modifications to elastic theory and limit analysis

theory as in the ACI Code. Such methods can be used to analyze a given slab

system to determine rather the stresses in the slabs and the supporting system or

the load-carrying capacity. Alternatively, the methods can be used to determine

the distribution of moments and shears to allow the reinforcing steel and

concrete sections to be designed.

The bending and torsional moments, shear forces, and deflections of slab

systems, with given dimensions, steel content, and material properties, at any

stage of loading from zero to ultimate load, can be determined analytically using

conditions of static equilibrium and geometrical compatibility if the moment-

deformation relationships of the slab elements, and yield criteria for bending

and torsional moments and shear force when the strength of slab elements is

reached, are known. In such analysis of the complete behavior of slab system,

difficulties are caused by nonlinearity of the moment deformation relationships

of the slab elements at high levels of stress, and a step-by-step procedure with

load increased increment in generally necessary.

At low levels of load the slab elements are uncracked and the actions and

deformations can be computed from elastic theory using the uncracked flexural

rigidity of the slab elements. The slab elements are searched at each load

increment to ascertain whether cracking of the concrete has taken place. When

it is found that the cracking moment has been reached, the flexural rigidity of

5

the element is recomputed on the basis of cracked section value and the section

and deformations of the slab are recomputed. This procedure is repeated at the

same load level unit all the flexural rigidity values are correct. At higher load

increments, when the stresses in one or more elements begin to enter the

inelastic range, the flexural rigidities of those elements are reduced to that

corresponding to the particular point on the moment-deformation relationship

of the element. This requires the computation to be repeated at the load level

until the flexural rigidity of each element is correct. Eventually, with further

load increments, the strength of one or more elements is reached and if the

elements are sufficiently ductile, plasticity will spread through the slab system

with further loading. The ultimate load is reached when deflections occur

without further increase in load and hence further load cannot be carried.

Many finite element programs have been developed for the analysis of slabs,

and these programs have become more capable as computers have become more

powerful. The earlier efforts which included nonlinear effects generally were

for simple cases with idealized boundary conditions and were essentially

research tools.

Current commercial and semi commercial programs that can deal with slabs

with fairly general support conditions include the following non exhaustive list:

FIFNTE, SAP2000 , SAFE, RISA-3D, Lake Forest, ABAQUS, STAAD-Pro

and PCA-Mats is a specialized program adapted to the analysis of mat

foundations and slabs on grade. Some of these programs include nonlinear

effects such as cracking and yielding, and some include section of

reinforcement.

Code design procedures are usually based on elastic theory moments modified

in the light of some moment redistribution, which has been shown by tests to be

possible without excessive cracking or deflections at service loads.

6

Elastic theory moments without modification, and moments obtained from

methods of limit analysis, form alternative design approaches that are

recommended by some codes of practice.

Limit analysis recognizes that because of plasticity, redistribution of moments

and shears away from the elastic theory distribution can occur before the

ultimate load is reached. Such redistribution of moments occurs because for

typical reinforced concrete sections there is little change in moment with

curvature once the tension steel has reached the yield strength. Thus, when the

most highly stressed sections of slab reach the yield moment they tend to

maintain a moment capacity that is close to flexural strength with further

increase in curvature, while yielding of the slab reinforced spreads to other

sections of the slab with further increase in load. Limit analysis computes the

ultimate load of the slab, and the distribution of moments and shear at the

ultimate load, or the distribution of moments and shear at that load, of a given

slab system, either a lower bound or an upper bound method may be used.

The ultimate load is calculated from the equilibrium equation and the postulated

distribution of moments. For a given slab system the lower bound method gives

an ultimate load that is either correct or too low; that i.e., the ultimate load is

never overestimated.

The upper bound method postulates a collapse mechanism for the slab system

at the ultimate load

A collapse mechanism is composed of portions of the slab separated by lines of

plastic hinging, and the ultimate load is calculated from the postulated collapse

mechanism. The most commonly used lower bound approach is Hillerborgs

strip method. This method obtains the distribution of moments and shears by

replacing the slab by system of strip running in two directions. Which share the

load strip action is a valid lower bound procedure because if the loading is

carried entirely by flexure, and therefore satisfies the requirements of statics no

7

assistance from torsion is necessary .Wood has shown that alternative lower

bound solution, including torsion, are difficult to obtain in many cases, the upper

bound methods for slab is yield line theory, which was due primarily to

Johansen. If yield line theory is used, the designer must examine all possible

collapse mechanisms to ensure that the load-carrying capacity of the slab is not

overestimate, the correct collapse mechanisms in nearly all cases are well

known, however and therefore the designer is not often faced with the

uncertainty of whether further alternatives exist.

2.2 Historical back ground about yield line theory

The term "yield line" literally meaning is line of rupture was coined in 1921 by

Ingerslev [1] to describe lines in the slab along which the bending moment is

constant. In 1931 K W Johansen [2] gave the concept a geometrical meaning as

lines of relative rotation of rigid slab parts.

In 1938 Gvozdev [3] had already formulated the limit analysis theorems, but his

work was not widely known in the West until it was translated to English in

1960. Whereas the Pager school of plasticity was mainly concerned with

metallic structures, Gvozdev’s point of departure was reinforced concrete, in

particular slabs, [4].

Yield line analysis was adopted by the Danish concrete code [5], and introduced

into the curriculum at the Technical University of Denmark. There is anecdotic

evidence to the effect that the success of Danish engineers worldwide in the

decades immediately following the Second World War owed no small part to

their mastery of yield line analysis, allowing them to produce efficient designs

of reinforced concrete slabs of any shape and loading, [5].

In the 1960s yield line theory was the subject of considerable interest in the UK,

as evidenced by a flurry of papers and monographs, including a special

publication issued by Magazine of Concrete Research [6], including

contributions by L L Jones, K O Kemp, C T Morley, M P Nielsen and R H

8

Wood. A particular subject under debate was whether Johansen’s yield criterion

was compatible with limit analysis. Jones & Wood went so far as to state in 1967

[7] that such a criterion is useless within the strict framework of limit analysis,

which must develop its own idealized criteria of yield.

In 1970, however, Braestrup [9] showed that not only is the Johansen criterion

consistent with limit analysis, as evidenced by the work of Nielsen, it is indeed

the only possible yield condition for a slab that allows complete solutions

(coinciding upper and lower bounds) to be derived by yield line analysis. The

message was brought home in 1974 when Fox [10] determined the exact yield

load for the clamped, isotropic slab under uniform loading. This fairly simple

case had long defied attempts of solution, and this fact had been cited as

evidence of the incompatibility of yield line theory and limit analysis. Fox’s

analysis of the square, clamped slab is not a proper yield line solution, because

it includes finite regions with a negative Gaussian curvature [10]. However,

yield line analysis provides a close estimate, and by successively refining the

yield line pattern, the exact solution can be approximated to any desired degree,

which is the point. Slabs or plates obeying other yield conditions (egTresca or

v. Mises) can also be analyzed by yield lines, but except for trivial cases the

resulting upper bound will never approach the exact solution, however detailed

the yield line pattern. It is interesting to note that, unbeknownst to most

participants in the debate 40 years ago, limit analysis and yield line theory had

for many years peacefully coexisted in the Soviet Union.

In 1995 Gerg.E.Mertz mentions that: Yield line theory offers a simplified

nonlinear analytical method that can determine the ultimate bending capacity of

flat reinforced concrete plates subject to distributed and concentrated loads.

Alternately, yield line theory, combined with hinge rotation limits can determine

the energy absorption capacity of plates subject to impulsive and impact loads.

This method is especially useful in evaluating existing structures that cannot be

qualified using conservative simplifying analytical assumptions. Typical

9

components analyzed by yield line theory are basements, floor and roof slabs

subject to vertical loads along with walls subject to out of plane wall loads.

One practical limitation of yield line theory is that it is computationally difficult

to evaluate some mechanisms. This problem is aggravated by the complex

geometry and reinforcing layouts commonly found in practice. A yield line

evaluation methodology is proposed to solve computationally tedious yield line

mechanisms. This methodology is implemented in a small, PC based computer

program, which allows the engineer to quickly evaluate multiple yield line

mechanisms [11].

GregE.Mertz obtains, Yield line theory is capable of determining the ultimate

bending capacity of complex slabs, and when combined with rotation limits,

yield line theory can also be used to evaluate slabs for impact loads. [11]

In 2003 Tim Gudman-Hoyer papers treats the subject Yield line Theory for

Concrete Slabs Subjected to Axial Force. In order to calculate the load-carrying

capacity from an upper bound solution the dissipation has to be known.

For a slab without axial force the usual way of calculating this dissipation is by

using the normality condition of the theory of plasticity together with the yield

condition. This method is equivalent to the original proposal by K. W. Johansen.

This method has shown good agreement with experiments and has won general

acceptance.

The dissipation in a yield line is calculated on the basis of the Coulomb yield

condition for concrete in order to verify K. W. Johansen’s method. It is found

that the calculations lead to the same results if the axes of rotation are the same

for adjacent slab parts. However, this is only true if the slab is isotropic and not

subjected to axial load.

11

An evaluation of the error made using K. W. Johansen’s proposal for orthotropic

rectangular slabs is made and it is found that the method is sufficiently correct

for practical purposes.

For deflected slabs it is known that the load-carrying capacity is higher. If it is

assumed that the axis of rotation corresponds to the neutral axis of a slab part

and the dissipation is found from the moment capacities about these axes K. W.

Johansen’s proposal may be used to find the load- carrying capacity in these

cases too. In this paper this is verified by comparing the results with numerical

calculations of the dissipation. Also for deflected slabs it is found that the

simplified method is sufficiently correct for practical purposes.

The same assumptions are also used for rectangular slabs loaded with axial force

in both one and two directions and sufficiently good agreement is found by

comparing the methods.

Interaction diagrams between the axial load and the transverse load are

developed at the end of the paper for both methods. Different approaches are

discussed.

Only a few comparisons between experiments and theory are made. These

indicate that the theory may be used if a proper effectiveness factor is introduced

and the deflection at failure is known.

If the deflection is unknown an estimate of the deflection based on the yield

strains of the concrete and the reinforcement seems to lead to acceptable results

[12].

In this research an analysis of reinforced concrete slab was done by applying

yield line method with depending on virtual work method.

2.3 Yield lines theory

Yield lines analysis is an analysis approach for determining the ultimate load

capacity of reinforced concrete slab and was pioneered by K. W. Johansen in

11

the 1940s. It is closely related to the plastic collapse or limits analysis of steel

frame, and is an upper bound or mechanism approach.

Yield lines design is well-founded method of designing reinforcement concrete

slabs, and similar types of element. It uses the yield line theory to investigate

failure mechanism at limit state, the theory is based on the principle of virtual

work which state that:

Work done in yield lines rotating = work done in load moving (2.1)

One of the most popular method of applications is work method, this research

will explains how it may use in the practical of reinforced concrete slabs such

as flat slabs and slab with beam.

2.4 The advantages and disadvantages of yield line theory

Advantages of yield line theory over liner elastic analysis were summarized as

follows:

1. It is Simpler to use (computer not necessary).

2. Linear elastic only tell about slab first yield occurs, yield line theory give the

ultimate capacity of the slab what it takes to cause collapse.

3. It helps in understanding the ultimate behavior of the slabs.

4. Good for awkward shapes.

5. It more economical method.

6. It more Versatility.

Yield line design leads to slabs that are quick and easy to design , and are quick

and easy to construct , there is no need to resort to computer for analysis or

design , the resulting slab are thin and have very low amounts of reinforcement

in very regular arrangement .

The reinforcement therefore easy to detail and easy to fix and the slabs are very

quick to construct. Above all, yield line design generates very economic

concrete slabs, because it considers features at the ultimate limit state.

12

Disadvantages of yield line theory were summarized as follows:

1- Requires experience to know likely failure mechanism.

2- Dangerous design is possible with checking or experience.

3- Does not give an idea of slab behavior.

The yield line theory based on the following process:

1- Assume collapse mechanism by choosing a pattern of yield line.

2- Calculate the load factor corresponding the yield lines pattern.

3- Repeat for arrange of yield line pattern.

4- Actual failure occurs at the lowest collapse factor.

The yield line theory is based on the following assumptions and rules:

1- Yield line divides the slab in to rigid regions which remain plane through the

collapses.

2- Yield lines are straight.

3- Axes of rotation generally lie along line of support and pass over any columns.

4- Yield lines between adjacent regions must pass through the point of intersection

of the axes of rotation of these regions.

5- Yield lines must end at slab boundary.

6- Continuous supports repel and a simple attract a yield lines.

The mentioned above rules all are illustrated in Fig.2.1

13

Fig.2.1. Yield Line Pattern for the Two-Way Slab.

2.5 prediction of yield line pattern

When a simply supported, isotropically reinforced square slab is subjected to

uniformly distributed load of increasing intensity, initially it was observed that

the slab behaves elastically. As the load is gradually increased, cracking of

concrete on the tension side of the slab will cause the stiffness of the cracked

section to be reduced, and the distribution of moments in the slab to change

slightly.

Owing to this redistribution for equal increments of load, the increase in moment

at an un cracked section will be greater than before cracking occurred.

As the load is increased further the reinforcement will yield in the central area

of the slab, which is the region of highest moment.

Fig.2.2. Consisting of Yield Line Pattern.

14

Once the steel in an under-reinforced section has yielded, although the section

will continue to deform, its resistance moment will not increase by any

appreciable amount, and consequently an even greater redistribution of

moments takes place.

When even more load is applied, since an increased proportion of moment has

to be carried by the sections adjacent to the central area, this will cause the steel

in these sections to yield as well. In this manner, lines along which the steel has

yielded are propagated from the point at which yielding originally occurred. At

this stage of loading the yield lines might be as shown in Fig.2.2a. The

application of more load will cause the reinforcement in even more sections to

yield and further propagation of the yield lines, until eventually all the yield

lines reach the boundary of the slab. This is shown in Fig.2.2b. At this stage,

since the resistance moments along the yield lines are almost at their ultimate

values, and since the yield lines cannot propagate further, the slab is carrying

the maximum load possible.

Any slight increase in load will now cause a state of unstable equilibrium, and

the slab will continue to deflect under this load until the curvature at some

section along the yield lines is so great that the concrete will crush. This section

will then lose its capacity to carry any moment and this will increase the state of

unstable equilibrium yield lines. Thus the condition when the yield lines have

just reached the boundary may be regarded as the collapse criterion of the slab.

The system of yield lines or fracture lines such as that in Fig.2.2b is called a

yield-line pattern. From the description of the process of failure it will be

apparent how essential it is that the slab should have a long horizontal potion to

its moment-rotation curve, otherwise a section might fail and lose its moment

carrying capacity before the yield-line pattern reached the boundary of the slab.

If this occurred, it was not assume that the ultimate resistance moment of the

15

slab acted along the whole length of the yield line and analysis would be almost

impossible.

The first stage of the ultimate load analysis of any slab is to predict the yield-

line pattern at failure. For a given amount of reinforcement the ultimate

resistance moment along the yield lines can be calculated and hence by analysis

the slab at the failure condition, the value of the load which is in equilibrium

with these moments can be found.

As with most methods of analysis certain assumptions are made, which from

tests are known to be reasonably true. Since the steel has yielded along the yield

lines, the curvature of the slab in this region is larger than the curvature of the

parts of the slab between the yield lines, which are still behaving elastically.

Consequently it is quite reasonable to assume that the elastic deformations are

negligible in comparison with the plastic deformations, in other words the

assumption is made that the slab elements between the yield lines remain plane

and that all the deformations take place in the yield lines.

Thus in the deflected state, the plane elements of a slab such as A, B, C and D

in Fig.2.3 are inclined planes. Since the intersections between inclined planes

are straight lines, it follows that the yield lines, which are the intersection

between the planes elements, are also straight.

In order that the slab may deflect, the plane elements must rotate about certain

axes. Element A rotates about ab, and the element about bc, and it is a necessary

condition of deformation that the yield line, or intersection line between two

adjacent elements, passes through the inter section of the axes of rotation of

these

16

Fig.2.3.Yield line pattern for square slab.

Generally the axes of rotation will lie along the lines of support and pass over

any columns.

In order to represent in diagrammatic form the boundary conditions of any slab

and the sign of the yield lines, the notation given in Fig.2.4 will always be

adopted.

Column

Simple support

Continuous over support

Beam

Positive yield line

( tension in bottom face )

Negative yield line

Point load

Axis of rotation

Fig.2.4. Notation used for yield line theory.

17

Since the first step in any analysis is to postulate a failure mechanism or yield-

line pattern, the yield-line patterns of various slabs are shown in Fig.2.5(a-k) to

show how the confirm to the four conditions that have been given. Fig.2.5a

shows a possible yield-line pattern for a square slab subjected to uniformly

distributed load. The axes of rotation of element A is ab, the line of support,

while that of B is bc. The yield line between these element passes through the

point of intersection of these axes, which is the corner b. similarly yield lines

pass through the other corners. Since yielding starts in the center of the slab,

then the yield lines are straight lines between the center and the corners. Fig.2.5b

shows a yield-line pattern for the rectangular slab under uniform load. The yield

lines pass through the corner for the reasons given previously and yield line ef

is parallel to the longer sides- it intersects the parallel axes of rotation of adjacent

elements A and C at infinity. Initially it is only necessary to draw the general

yield-line pattern, the exact position of e and f can be found in the process of the

analysis.

For the continuous rectangular slab shown in Fig.2.5c, negative yield lines must

also form along the lines of support before they can become axes of rotation, in

Fig.2.5b which represents a trapezoidal slab, the yield line ef produced passes

through the point of intersection of the axes of rotation along the longer sides.

18

Fig.2.5. yield line pattern for the different slabs shape with different

support conditions.

The patterns other than those shown in Fig 3.5(e)-3.5(j) may be found by similar

reasoning.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j)

CHAPTER THREE

Methodology

19

CHAPTER THREE

Methodology

3.1 Introduction

According to Euro code 2, Yield Line Design is a perfectly valid method of

design. Section 5.6 of Euro code 2 states that plastic methods of analysis shall

only be used to check the ultimate limit state. Ductility is critical and sufficient

rotation capacity may be assumed provided x/d ≤ 0.25 for C50/60. Euro code 2

goes on to say that the method may be extended to flat slabs, ribbed, hollow or

waffle slabs and that corner tie down forces and torsion at free edges need to be

accounted for.

Section 5.11.1.1 of EC2 includes Yield Line as a valid method of analysis for

flat slabs. It is recommended that a variety of possible mechanisms are examined

and the ratios of the moments at support to the moment in the spans should lie

between 0.5 and 2.

3.2 Method of solution by using yield line theory

Once a failure pattern has been postulated two methods of solution are available

in order to find the relation between the ultimate resistance moments in the slab

and the ultimate load. Since the moment and the load are equilibrium when the

yield line pattern has formed, the slightest increment in load will cause the

structure to deflect. When this increase in load is infinitesimal, the work done

on the slab while the yield lines are rotating must be equal to the loss of work

due to the load deflecting. Thus, if a point on the slab is given a virtual deflection

take place along the yield lines. The internal work done on the slab will be the

sum of the rotations in the yield lines multiplied by the resisting ultimate

moments, while the external loss of work will be the sum of the loads multiplied

by their respective deflections. When the internal and external work is equated,

we have the relations between the ultimate resistance moments in the slab and

21

the ultimate load will be obtained. This method of solution is the well-known

principle of virtual work.

The second method of solution has been called the equilibrium method. When

using this method it is usually necessary to study the equilibrium of each of the

elements into which the slab is divided by the yield lines. Equilibrium equations

are obtained from each element, by equating internal and external moments and

by resolving vertical forces, and to find the desired solution the resulting

equations are solved simultaneously.

Both methods of the solution which are presented are upper bound solutions, to

find the most critical collapse mechanism several alternative yield-line patterns

may have to be studied.

Usually ultimate load solutions are obtained for slabs in which the percentage

of steel along any given line is constant. This implies that the same mesh is used

all over the slab. If the ultimate flexural strength along two orthogonal lines on

the slab surface is the same, the slab is called an isotropically reinforced slab; if

however the ultimate strength is different in the two directions the slab is called

an orthotropically reinforced slab.

The virtual work method of analysis is one-way, probably the most popular way

of applying yield line analysis to analyze slabs from first principles. It is

considered to be the quickest way of analyzing a slab using manual calculations

only. It can be applied and used on slabs of any configuration and loading

arrangement .

The only prerequisite is that the designer has a reasonably good idea of the

modes of failure and the likely shape of the crack pattern that will develop at

failure. This is not as difficult as it sounds. Having studied the basic failure

patterns that are formed by the majority of slab shapes encountered in practice,

the designer soon develops a feel for the way a slab is likely to fail and the

confidence to turn this feel into safe and practical designs. Provided the numeric

21

methods shown below are used, and, if necessary, iterations made, the virtual

work method gives solutions that are, almost always, within 10% of that attained

by an exact algebraic approach using a differentiation process. In recognition of

this possible inexactness, it is recommended that ‘the 10% rule.’

Before explaining how to apply the virtual work method of analysis it may help

to review the stages involved in the failure of a slab :

•Collapse occurs when yield lines form a mechanism .

•This mechanism divides the slab into rigid regions .

•Since elastic deformations are neglected these rigid regions remain as plane

areas .

•These plane areas rotate about their axes of rotation located at their supports.

•All deformation is concentrated within the yield lines: the yield lines act as

elongated plastic hinges.as shown in Fig.3.1.

Fig.3.1. Simply Supported One-way Slab.

22

The basis of the Work Method is simply that at failure the potential energy

expended by loads moving must equal the energy dissipated (or work done) in

yield lines rotating.

In other words:

External energy = Internal energy (3.1)

Expended by the displacement of loads = Dissipated by the yield lines rotating

Σ (Ν x δ) for all regions = Σ (m x l x θ) (3.2)

Where;

N : is the Load(s) acting within a particular region [kN].

δ : is the vertical displacement of the load(s) N on each region expressed as a

fraction of unity [m].

m : is the moment or moment of resistance of the slab per meter run represented

by the reinforcement crossing the yield line [kNm/m].

l : is the length of yield line or its projected length onto the axis of rotation for

that region [m].

θ : is the rotation of the region about its axis of rotation [m/m] for all regions.

Once a valid failure pattern (or mechanism) has been postulated, either the

moment, m, along the yield lines or the failure load of a slab, N (or indeed n

kN/m2), can be established by applying the above equation .

This, fundamentally, is the Work Method of analysis: it is a kinematic (or

energy) method of analysis. The deformed shape of slab at failure was shown in

Fig.3.2.

23

Fig.3.2. Deformed shape at failure.

Fig.3.3 shown the Principles of expenditure of external loads, The external

energy expended, E, is calculated by taking, in turn, the resultant of each load

type (i.e. uniformly distributed load, line load or point load) acting on a region

and multiplying it by its vertical displacement measured as a proportion of the

maximum deflection implicit in the proposed yield line pattern. For simplicity,

the maximum deflection is taken as unity, and the vertical displacement of each

load is usually expressed as a fraction of unity. The total energy expended for

the whole slab is the sum of the expended energies for all the regions.

24

Fig.3.3. Principles of expenditure of external loads.

Fig.3.4 shown the Principles of dissipation of internal energy, The internal

energy dissipated, D, is calculated by taking the projected length of each yield

line around a region onto the axis of rotation of that region, multiplying it by the

moment acting on it and by the angle of rotation attributable to that region. The

total energy dissipated for the whole slab is the sum of the dissipated energies

of all the regions.

Diagonal yield lines are assumed to be made up of small steps with sides parallel

to the axes of rotation of the two regions it divides. The 'length' of a diagonal or

inclined yield line is taken as the summation of the projected lengths of these

individual steps onto the relevant axes of rotation.

25

The angle of rotation of a region is assumed to be small and is expressed as

being δmax/length. The length is measured perpendicular to the axis of rotation

to the point of maximum deflection of that region.

Fig.3.4. Principles of dissipation of internal energy.

A fundamental principle of physics is that energy cannot be created or destroyed.

So in the yield line mechanism, External energy = Internal energy. By equating

these two energies the value of the unknown i.e. either the moment, m, or the

load, N, can then be established.

26

If deemed necessary, several iterations may be required to find the maximum

value of m (or the minimum value of load capacity) for each chosen failure

pattern.

3.3 The 10% rule

A 10% margin on the design moments should be added when using the virtual

work method or formulae for two-way slabs to allow for the method being upper

bound and to allow for the effects of corner levers [13].

The addition of 10% to the design moment in two-way slabs provides some

leeway where inexact yield line solutions have been used and some reassurance

against the effects of ignoring corner levers. At the relatively low stress levels

in slabs, a 10% increase in moment equates to a 10% increase in the

reinforcement design.

The designer may of course chase in search of a more exact solution but most

pragmatists are satisfied to know that by applying the 10% rule to a simple

analysis their design will be on the safe side without being unduly conservative

or uneconomic. The 10% rule can and usually is applied in other circumstances

where the designer wants to apply engineering judgment and err on the side of

caution.

The only situations where allowances under this ‘10% rule’ may be inadequate

relate to slabs with acute corners and certain configuration of slabs with

substantial point loads or line loads. In these cases guidance should be sought

from specialist literature.

3.4 Serviceability and deflections

Yield Line Theory concerns itself only with the ultimate limit state. The

designer must ensure that relevant serviceability requirements, particularly the

limit state of deflection, are satisfied.

Deflection of slabs should be considered on the basis of elastic design. This may

call for separate analysis but, more usually, deflection may be checked by using

27

span/effective depth ratios with ultimate (i.e. yield line) moments as the basis.

Such checks will be adequate in the vast majority of cases.

3.4.1 Deflection according to BS 8110

Deflection is usually checked by ensuring that the allowable span/effective

depth ratio is greater than the actual span/ effective depth ratio (or by checking

allowable span is greater than actual span). The basic span/depth ratio is

modified by factors for compression reinforcement (if any) and service stress in

the tension reinforcement. The latter can have a large effect when determining

the service stress, fs, to use in equation 8 in Table 3.10 of BS 8110. When

calculations are based on the ultimate yield line moments, one can,

conservatively, use βb values of 1.1 for end spans and 1.2 for internal spans.

Where estimates of actual deflections are required, other approaches, such as the

rigorous methods in BS 8110 Part 2, simplified analysis methods or finite

element methods should be investigated. These should be carried out as a

secondary check after the flexural design based on ultimate limit state principles

has been carried out.

In order to keep cracking to an acceptable level it is normal to comply (sensibly)

with the bar spacing requirements of BS 8110 Clauses 3.12.11.2.7 and 2.8.

3.4.2 Deflection according to Euro code 2

Euro code 2 treats deflection in a similar manner to BS 8110. Deemed-to-satisfy

span to depth ratios may be used to check deflection. Otherwise calculations,

which recognize that sections exist in a state between un cracked and fully

cracked, should be undertaken.

3.5. Analysis and design of R.C. slabs using BS8110

Any design process is governed by the recommendations of a specific code of

practice. In the UK, BS 8110 clause 3.5.2.1 says Alternatively Johansen’s Yield

Line method may be used for solid slabs. The proviso is that to provide against

serviceability requirements, the ratio of support and span moments should be

similar to those obtained by elastic theory. This sub-clause is referred to in

28

clauses 3.6.2 and 3.7.1.2 making the approach also acceptable for ribbed slabs

and flat slabs.

In slabs where the corners are prevented from lifting, and provision for torsion

is made, the maximum design moments per unit width are given by equations

3.3 and 3.4.

msx= βsxnlx2 (3.3)

msy= βsynlx2 (3.4)

Where design ultimate support moments are used which differ substantially

from those that would be assessed from Table 3.1.

Table 3.1. Bending moment coefficients for rectangular panels supported

on four sides with provision for torsion at corners.

29

3.6 Software program

3.6.1 STAAD-Pro

General purpose software suite for structural engineers involved in analysis and

design of structures. The structural analysis and design software, STAAD-Pro

was developed for practicing engineers. For static, pushover, dynamic, P-delta,

buckling or cable analysis, STAAD-Pro is the industry standard.

STAAD-Pro has design codes for most countries including US, BS, Canada,

Russia, Aus, France, India, China, Euro, Japan

3.6.2 PROKON

PROKON provides engineers with tools to streamline their Workflow in the

structural and geotechnical spheres. The tools are modular, but all are launched

from the Prokon Calcpad.

Analysis of Continuous Beam & Slab Design

The Continuous Beam and Slab Design module is used to design and detail

reinforced concrete beams and slabs as encountered in typical building projects.

The design incorporates automated pattern loading and moment redistribution.

Cross-sections can include a mixture of rectangular, I, T and L-sections. Spans

can have constant or tapered sections. Entered dead and live loads are

automatically applied as pattern loads during the analysis. At ultimate limit

state, moments and shears are redistributed to a user specified percentage. Both

short-term and long-term deflections are calculated. Complete bending

schedules can be generated for editing and printing using Pads. The

reinforcement details may be graphically edited by the designer, and is

presented in user friendly pages depicting entered, required and minimum

reinforcement (as specified by the applicable code of practice.

CHAPTER FOUR

Investigation of Ultimate

Resisting Moment for the

Reinforced Concrete Slabs

31

CHAPTER FOUR

Investigation of Ultimate Resisting Moment for the

Reinforced Concrete Slabs

4.1 Introduction

In this chapter, different types of reinforced concrete slab systems were analyzed

using yield line theory to determine ultimate resisting moment. Three systems

of reinforced concrete slabs were chosen. They were flat slab, beam slab

systems, have three panel in each direction. And special types of support of slabs

systems.

31

4.2 Analysis of flat Slab system

The Fig.4.1 shown below is the plan of flat slab have three equal spans at

direction X of 6 m length and three spans at direction Y of length 6 m for the

edges spans and 4 m length for middle span. , The slab is subjected to uniformly

distributed load of 20 kN/m2. By considering a reasonable pattern of positive

and negative yield lines is that shown in Fig.4.1, and with following the

procedure explained at previous Chapter, the ultimate resisting moment (MP)

can be obtained for each panel as named in Fig.4.1.

Fig.4.1. Plan of Flat Slab System with Expected of Yield Line Pattern.

32

4.2.1. Analysis of External Corner Panel S1

Panel S1 is the square panel have length of 6 m each with two adjacent edges

discontinuous and continuous in other tow edges, by considering a reasonable

pattern of positive and negative yield lines is that shown in Fig.4.2, we will

determine the Mp using work method.

Fig.4.2. Expected yield line Pattern of External Corner Panel S1.

For External work as explained in (3.2). = ∑N × δ

N = w x A, where A is the total area of panel

= ∑w × A × δ

= w (6*6) * 1

2

= 18w (4.1)

For Internal work =∑𝑚 × 𝑙 × 𝜃

Wint (total) = wint (A) +wint (B) +wint(C)+wint (D)

33

Table.4.1. Analysis of Panel S1 by yield line theory

Segment (A):

Length of (m) = √𝑥2 + 𝑦2

θB = 1

2𝑥 sin 𝛼

wint (A) = 𝑚 ∗ √𝑥2 + 𝑦2 ∗1

2𝑥 sin 𝛼

but ; sin 𝛼 = 𝑦

√𝑥2 + 𝑦2

wint (A) = (𝑚 ∗ √(6 − 𝑥)2 + 𝑦2 ∗ √(6−𝑥)2+𝑦2

2𝑦(6−𝑥)+ m` ∗

𝑦

2(6−𝑥))

= (𝑚 ∗(6−𝑥)2+𝑦2

2𝑦(6−𝑥)+ m` ∗

𝑦

2(6−𝑥))

Segment (B):

Length of (m) = √(6 − 𝑥)2 + 𝑦2

Length of (m`) = y ∗ sin 𝛽

Segment Length for

moment

Length for

moment

Component of

Rotation

Wint

Segment

(A)


√𝑥2 + 𝑦2

Segment

(B)

√(6 − 𝑥)2 + 𝑦2 y ∗ sin 𝛽 1

2(6 − 𝑥) sin 𝛽

but ; sin 𝛽

=𝑦

√(6 − 𝑥)2 + 𝑦2

𝑚 ∗(6 − 𝑥)2 + 𝑦2

2𝑦(6 − 𝑥)+ m`

∗𝑦

2(6 − 𝑥)

Segment

(C)

√(6 − 𝑥)2 + (6 − 𝑦)2 √(6 − 𝑥)2 + (6 − 𝑦)2

1

2(6 − 𝑦) sin 𝛾

but ; sin 𝛾

=6 − 𝑥

√(6 − 𝑥)2 + (6 − 𝑦)2

(𝑚 + m`) ∗(6 − 𝑥)2 + (6 − 𝑦)2

2(6 − 𝑦) (6 − 𝑥)

Segment

(D):

√(6 − 𝑦)2 + 𝑥2 x ∗ sin θ 1

2(6 − 𝑦) sin θ

but ; sin 𝛽

= 𝑥

√(6 − 𝑦)2 + 𝑥2

(𝑚 ∗(6 − 𝑦)2 + 𝑥2

2𝑥(6 − 𝑦)+ m`

∗𝑥

2(6 − 𝑦))

Total

wint 𝒎 ∗

(𝒙𝟐+𝒚𝟐)

𝟐𝒙𝒚+ (𝒎 ∗

(𝟔−𝒙)𝟐+𝒚𝟐

𝟐𝒚(𝟔−𝒙)+ 𝐦` ∗

𝒚

𝟐(𝟔−𝒙)) + (𝒎 + 𝐦`) ∗

(𝟔−𝒙)𝟐+(𝟔−𝒚)𝟐

𝟐(𝟔−𝒚) (𝟔−𝒙) + (𝒎 ∗

(𝟔−𝒚)𝟐+𝒙𝟐

𝟐𝒙(𝟔−𝒚)+ 𝐦` ∗

𝒙

𝟐(𝟔−𝒚))

√𝑥2 + 𝑦2 1

2𝑥 sin 𝛼 𝑚 ∗

(𝑥2 + 𝑦2)

2𝑥𝑦

34

θB = 1

2(6−𝑥) sin 𝛽

wint (B) = (𝑚 ∗ √(6 − 𝑥)2 + 𝑦2 + m` ∗ y ∗ sin 𝛽) ∗1


but ; sin 𝛽 = 𝑦

√(6 − 𝑥)2 + 𝑦2

wint (B) = (𝑚 ∗ √(6 − 𝑥)2 + 𝑦2 ∗ √(6−𝑥)2+𝑦2

2𝑦(6−𝑥)+ m` ∗

𝑦

2(6−𝑥))

= (𝑚 ∗(6−𝑥)2+𝑦2

2𝑦(6−𝑥)+ m` ∗

𝑦

2(6−𝑥))

Segment (C):

Length of (m) = √(6 − 𝑥)2 + (6 − 𝑦)2

Length of (m`) = √(6 − 𝑥)2 + (6 − 𝑦)2

θC = 1

2(6−𝑦) sin 𝛾

wint (C) = (𝑚 + m`) ∗ √(6 − 𝑥)2 + (6 − 𝑦)2 ∗1


but ; sin 𝛾 = 6 − 𝑥

√(6 − 𝑥)2 + (6 − 𝑦)2

wint (C) = (𝑚 + m`) ∗ √(6 − 𝑥)2 + (6 − 𝑦)2 ∗√(6−𝑥)2+(6−𝑦)2

2(6−𝑦) (6−𝑥)

= (𝑚 + m`) ∗(6−𝑥)2+(6−𝑦)2

2(6−𝑦) (6−𝑥)

Segment (D):

Length of (m) = √(6 − 𝑦)2 + 𝑥2

Length of (m`) = x ∗ sin θ

θD = 1

2(6−𝑦) sin θ

wint (D) = (𝑚 ∗ √(6 − 𝑦)2 + 𝑥2 + m` ∗ x ∗ sin θ) ∗1

2(6−𝑦) sin θ

but ; sin 𝛽 = 𝑥

√(6 − 𝑦)2 + 𝑥2

wint (D) = (𝑚 ∗ √(6 − 𝑦)2 + 𝑥2 ∗ √(6−𝑦)2+𝑥2

2𝑥(6−𝑦)+ m` ∗

𝑥

2(6−𝑥))

= (𝑚 ∗(6−𝑦)2+𝑥2

2𝑥(6−𝑥)+ m` ∗

𝑥

2(6−𝑥))

Total Internal work =

35

= 𝑚 ∗ (𝑥2+𝑦2)

2𝑥𝑦+ (𝑚 ∗

(6−𝑥)2+𝑦2

2𝑦(6−𝑥)+ m` ∗

𝑦

2(6−𝑥)) + (𝑚 + m`) ∗

(6−𝑥)2+(6−𝑦)2

2(6−𝑦) (6−𝑥) + (𝑚 ∗

(6−𝑦)2+𝑥2

2𝑥(6−𝑦)+ m` ∗

𝑥

2(6−𝑦)) (4.2)

Let m=m`

=3m [1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)] (4.3)

External work = Internal work

=3m [1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)] = 18 w (4.4)

𝑤

𝑚 = [

1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)]/ 6 (4.5)

Using the expression:

𝑑(𝑤

𝑚)

𝑑y = 0 ; when

𝑑

𝑑y[

1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)] = 0

i.e.

1

𝑦2 +

(−2)

(6−𝑦)2= 0

By using excel; substituting a range of values of (y) in the above equation

and solve the equation for (y),

Given; y = 2.485 m

That is (w/m) maximum when y = 2.485

For y = 2.485, the Eqn (4.1.5) reduces to

1

x +

2

(6−x)= 𝑚/𝑤 (4.6)

𝑑(𝑤

𝑚)

𝑑x = 0

i.e.

1

𝑥2 +

(−2)

(6−𝑥)2= 0

By using excel; substituting a range of values of (x) in the above equation

and solve the equation for (x),

gives x = 2.485 m

Substituting; x = 2.485 and y = 2.485 into Equ(4.1.5)

m = 3.09 w (4.7)

36

When; w = 20 kN.m2

Mmax =61.8 kN.m/m

By using “10% rule”; Mmax = 67.98kN.m/m

So; the actual yield lines pattern as per the calculation will be as shown in

figure below:

Fig.4.3.Yield Line Pattern of Panel S1.

37

4.2.2. Analysis of Edge Panel S2

Panel S2 is the square panel have length of 6m with one edge discontinuous and

three edges continuous, by considering a reasonable pattern of positive and

negative yield lines is that shown in Fig.4.4, we will determine the Mmax using

work method.

Fig.4.4.Expected yield line Pattern of the edge Panel S2.

Solution by using method work:


For External work = ∑N × δ

= ∑w × a × δ

= w (6*6) * 1

2

= 18w (4.8)



38

Table.4.3.Analysis of Panel S2 by yield line theory.

Segment (A):

Length of (m) = √9 + 𝑦2

Length of (m`) = y ∗ sin 𝛼

θA = 1

6 sin 𝛼

wint (A) =(𝑚 ∗ √9 + 𝑦2 + m` ∗ y ∗ sin 𝛼) ∗1

6 sin 𝛼


√9 + 𝑦2

wint (A) = (𝑚 ∗ √9 + 𝑦2 + m` ∗ y ∗ 𝑦

√9+𝑦2) ∗

√9+𝑦2

6𝑦

=(𝑚 ∗ (9 + 𝑦2) + m` ∗𝑦

6)

Segment (B):

Length of (m) = √9 + 𝑦2

Length of (m`) = y ∗ sin 𝛽

θB = 1

6 sin 𝛽

Segment Length for

moment

Length for

moment

Component of

Rotation

Wint

Segment

(A)

√9 + 𝑦2 y ∗ sin 𝛼 1

6 sin 𝛼 but ; sin 𝛼 =

𝑦

√9 + 𝑦2

𝑚 ∗ (9 + 𝑦2) + m` ∗𝑦

6

Segment

(B)

√9 + 𝑦2 y ∗ sin 𝛽 1

6 sin 𝛽

but ; sin 𝛽

=𝑦

√(6 − 𝑥)2 + 𝑦2

𝑚 ∗(9 + 𝑦2)

6𝑦+ m` ∗

𝑦

6

Segment

(C)

√9 + (6 − 𝑦)2 √9 + (6 − 𝑦)2 1

2(6 − 𝑦) sin 𝛾

𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝛾

= 3

√9 + (6 − 𝑦)2

(𝑚 + 𝑚`) ∗(9 + (6 − 𝑦)2)

6(6 − 𝑦)

Segment

(D):

√(6 − 𝑦)2 + 9 √(6 − 𝑦)2 + 9 1

2(6 − 𝑦) sin θ

𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝜃

= 3

√9 + (6 − 𝑦)2

(𝑚 + m`) ∗(9 + (6 − 𝑦)2)

6(6 − 𝑦)

Total wint 𝟐 (𝒎(𝟗+𝒚𝟐)

𝟔𝒚+ 𝒎`

𝒚

𝟔)+𝟐(𝒎 + 𝒎`)

(𝟗+(𝟔−𝒚)𝟐)

𝟔(𝟔−𝒚)

39

wint (B) = (𝑚 ∗ √9 + 𝑦2 + m` ∗ y ∗ sin 𝛽) ∗1

6 sin 𝛽

but ; sin 𝛽 = 𝑦

√9 + 𝑦2

wint (B) =(𝑚 ∗ √9 + 𝑦2 + m` ∗ y ∗ 𝑦

√9+𝑦2) ∗

√9+𝑦2

6𝑦

=(𝑚 ∗(9+𝑦2)

6𝑦+ m` ∗

𝑦

6)

Segment (C):

Length of (m) = √9 + (6 − 𝑦)2

Length of (m`) = √9 + (6 − 𝑦)2

θC = 1


wint (C) = (𝑚 + m`) ∗ √9 + (6 − 𝑦)2 ∗1


but ; sin 𝛾 = 3

√9 + (6 − 𝑦)2

wint (C) = (𝑚 + m`) ∗ √9 + (6 − 𝑦)2 ∗√9+(6−𝑦)2

6(6−𝑦)

= (𝑚 + m`) ∗(9+(6−𝑦)2)

6(6−𝑦)

Segment (D):

Length of (m) = √(6 − 𝑦)2 + 9

Length of (m`) = √(6 − 𝑦)2 + 9

θD = 1

2(6−𝑦) sin θ

wint (D) = (𝑚 + m`) ∗ √(6 − 𝑦)2 + 9 ∗1

2(6−𝑦) sin θ

but ; sin θ = 3

√9 + (6 − 𝑦)2

wint (D) = (𝑚 + m`) ∗ √9 + (6 − 𝑦)2 ∗√9+(6−𝑦)2

6(6−𝑦)

= (𝑚 + m`) ∗(9+(6−𝑦)2)

6(6−𝑦)


=𝟐 (𝑚(9+𝑦2)

6𝑦+ m`

𝑦

6)+2(𝑚 + m`)

(9+(6−𝑦)2)

6(6−𝑦) (4.9)

41

let m=m`

=𝑚 ((9+𝑦2)

3𝑦+

𝑦

3)+

2(9+(6−𝑦)2)

3(6−𝑦) ) (4.10)

= m [ 3

𝑦+

2𝑦

3 +

6

(6−𝑦) +

2(6−𝑦)

3] (4.11)


=m [ 3

𝑦+

2𝑦

3 +

6

(6−𝑦) +

2(6−𝑦)

3]= 18 w (4.12)

m = 18 w /[ 3

𝑦+

2𝑦

3 +

6

(6−𝑦) +

2(6−𝑦)

3] (4.13)


and finding the maximum Values of (m),

We find maximum values of (m) = 2.6w, coincide with y = 2.49m

That is (m) maximum when y = 2.49m

Substituting; x = 2.49m into Equ (4.13)

m = 2.6 w (4.14)

When; w = 20 kN.m2

Mmax =52.0kN.m/m

By using " 10% rule " ; Mmax = 57.2 kN.m/m


figure below:

Fig.4.5 Yield line Pattern of Panel S2.

41

4.2.3. Analysis of edge Panel S3

Panel S3 is the rectangular panel have length of 6m and 4m width with one edge

discontinuous and three edges continuous, by considering a reasonable pattern

of positive and negative yield lines is that shown in Fig.4.6, we will determine

the Mp using work method.

Fig.4.6.Expected yield line Pattern of the edge Panel S3.

Solution by using virtual work method:

External work = Internal work from Equ (3.1)


= ∑w × a × δ

= w (6*4) * 1

2

= 12w (4.15)



42

Table.4.3. Analysis of Panel S3 by yield line theory.

Segment (A):

Length of (m) = √𝑥2 + 4

Length of (m`) =2 sin 𝛼

θA = 1

2𝑥 sin 𝛼

wint (A) = 𝑚 ∗ √𝑥2 + 4 ∗1

2𝑥 sin 𝛼+ m` ∗ 2 sin 𝛼 ∗

1

2𝑥 sin 𝛼

but ; sin 𝛼 = 2

√𝑥2 + 4

wint (A) = 𝑚(𝑥2+4)

4𝑥+

m`

𝑥

Segment (B):

Length of (m) = √(6 − 𝑥)2 + 4

Length of (m`) =√(6 − 𝑥)2 + 4

θB = 1


Segment Length for

moment

Length

for

moment

Component

of Rotation

Wint

Segment

(A)

2 sin 𝛼 but ; sin 𝛼 =

2

√𝑥2 + 4

Segment

(B)

√(6 − 𝑥)2 + 4 √(6 − 𝑥)2 + 4 1

2(6 − 𝑥) 𝑠𝑖𝑛 𝛽

𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝛽

= 2

√(6 − 𝑥)2 + 4

(𝑚 + m`) ∗((6 − 𝑥)2 + 4)

4(6 − 𝑥)

Segment

(C)

√(6 − 𝑥)2 + 4 √(6 − 𝑥)2 + 4 1

2(6 − 𝑥) 𝑠𝑖𝑛 𝛽

𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝛽

= 2

√(6 − 𝑥)2 + 4

(𝑚 + m`) ∗((6 − 𝑥)2 + 4)

4(6 − 𝑥)

Segment

(D):

√4 + 𝑥2 𝑥 ∗ 𝑠𝑖𝑛 𝜃 1

4 𝑠𝑖𝑛 𝜃 𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝛽 =

𝑥

√4 + 𝑥2

𝑚 ∗(4 + 𝑥2)

4𝑥+ m` ∗

𝑥

4

Total wint 𝐦(𝐱𝟐+𝟒)

𝟒𝐱+

𝐦`

𝐱+ 𝟐 ∗ (𝐦 + 𝐦`) ∗

((𝟔−𝐱)𝟐+𝟒)

𝟒(𝟔−𝐱) + (𝐦 ∗

(𝟒+𝐱𝟐)

𝟒𝐱+ 𝐦` ∗

𝐱

𝟒)

√𝑥2 + 4 1

2𝑥 sin 𝛼 𝑚

(𝑥2 + 4)

4𝑥+

𝑚`

𝑥

43

wint (B) = (𝑚 + m`) ∗ √(6 − 𝑥)2 + 4 ∗1


but ; sin 𝛽 = 2

√(6 − 𝑥)2 + 4

wint (B) = (𝑚 + m`) ∗ √(6 − 𝑥)2 + 4 ∗√(6−𝑥)2+4

4(6−𝑥)

= (𝑚 + m`) ∗((6−𝑥)2+4)

4(6−𝑥)

Segment (C):

Length of (m) = √(6 − 𝑥)2 + 4

Length of (m`) =√(6 − 𝑥)2 + 4

θC = 1


wint (C) = (𝑚 + m`) ∗ √(6 − 𝑥)2 + 4 ∗1


but ; sin 𝛽 = 2

√(6 − 𝑥)2 + 4

wint (C) = (𝑚 + m`) ∗ √(6 − 𝑥)2 + 4 ∗√(6−𝑥)2+4

4(6−𝑥)

= (𝑚 + m`) ∗((6−𝑥)2+4)

4(6−𝑥)

Segment (D):

Length of (m) = √4 + 𝑥2

Length of (m`) = x ∗ sin θ

θD = 1

4 sin θ

wint (D) = (𝑚 ∗ √4 + 𝑥2 + m` ∗ x ∗ sin θ) ∗1

4 sin θ

but ; sin 𝛽 = 𝑥

√4 + 𝑥2

wint (D) = (𝑚 ∗ √4 + 𝑥2 ∗ √4+𝑥2

4𝑥+ m` ∗

𝑥

4)

= (𝑚 ∗(4+𝑥2)

4𝑥+ m` ∗

𝑥

4)


=𝑚(𝑥2+4)

4𝑥+

m`

𝑥+ 2 (𝑚 + m`) ∗

((6−𝑥)2+4)

4(6−𝑥) + (𝑚 ∗

(4+𝑥2)

4𝑥+ m` ∗

𝑥

4)

let m=m`

44

= m [ 2

𝑥+

3

4𝑥 +

(6−𝑥)

2 +

2

(6−𝑥)] (4.16)


= m [ 2

𝑥+

3

4𝑥 +

(6−𝑥)

2 +

2

(6−𝑥)]=12 w (4.17)

m = 12 w/ [ 2

𝑥+

3

4𝑥 +

(6−𝑥)

2 +

2

(6−𝑥)] (4.18)



We find maximum values of (m) = 2.41w , coincide with x = 2.257 m

That is (m) maximum when x = 2.257 m

Substituting; x = 2.257 m into Equ (4.18)

m = 2.41w (4.19)

when;

w = 20 kN.m2

Mmax =48.2kN.m/m



figure below:

Fig.4.7.Yield Line Pattern of Panel S3.

45

4.2.4. Analysis of Interior panel S4

Panel S4 is the rectangular panel have length of 6m and width 4m with four

edges continuous, by considering a reasonable pattern of positive and negative

yield lines is that shown in Fig.4.8, we will determine the Mp using virtual work

method.

Fig.4.8.Expected yield line Pattern for the interior Panel S4.




= ∑w × a × δ

= w (6*4) * 1

2

= 12w (4.20)



46

Table.4.4. Analysis of panel S4 by yield line theory.

Segment (A):

Length of (m) = √4 + 9 = 3.61 𝑚

Length of (m`) =√4 + 9 = 3.61 𝑚

θA = 1

6 sin 𝛼

wint (A) = (𝑚 + 𝑚`) ∗ 3.61 ∗1

6 sin 𝛼

but ; sin 𝛼 = 2

3.61

wint (A) = (𝑚 + 𝑚`) ∗ 3.61 ∗3.61

6𝑥2

=(𝑚 + 𝑚`) ∗ 1.09

Segment (B):

Length of (m) = √4 + 9 = 3.61 𝑚

Length of (m`) =√4 + 9 = 3.61 𝑚

θB = 1

6 sin 𝛽

wint (B) = (𝑚 + 𝑚`) ∗ 3.61 ∗1

6 sin 𝛽

Segment Length for +ve

moment

Length for

–ve

moment

Component

of Rotation

Wint

Segment

(A)

3.61 𝑚 but ; sin 𝛼 =

2

3.61

Segment

(B)

3.61 𝑚

3.61 𝑚 1

6 𝑠𝑖𝑛 𝛽 𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝛽 =

2

3.61

(𝑚 + 𝑚`) ∗ 1.09

Segment

(C)

3.61 𝑚 1

4 𝑠𝑖𝑛 𝛾 𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝛽 =

2

3.61

(𝑚 + 𝑚`) ∗ 1.09

Segment

(D):

3.61 𝑚

3.61 𝑚 1

4 𝑠𝑖𝑛 𝜃 𝑏𝑢𝑡 ; 𝑠𝑖𝑛 𝜃 =

3

3.61

(𝑚 + 𝑚`) ∗ 1.09

Total wint 𝟒 ∗ (𝒎 + 𝒎`) ∗ 𝟏. 𝟎𝟗

3.61 𝑚 1

6 sin 𝛼

(𝑚 + 𝑚`) ∗ 1.09

3.61 𝑚

47

but ; sin 𝛽 = 2

3.61

wint (B) = (𝑚 + 𝑚`) ∗ 3.61 ∗3.61

6𝑥2

=(𝑚 + 𝑚`) ∗ 1.09

Segment (C):

Length of (m) = √4 + 9 = 3.61 𝑚

Length of (m`) =√4 + 9 = 3.61 𝑚

θC = 1

4 sin 𝛾

wint (C) = (𝑚 + 𝑚`) ∗ 3.61 ∗1

4 sin 𝛾

but ; sin 𝛾 = 3

3.61

wint (C) = (𝑚 + 𝑚`) ∗ 3.61 ∗3.61

4𝑥3

=(𝑚 + 𝑚`) ∗ 1.09

Segment (D):

Length of (m) = √4 + 9 = 3.61 𝑚

Length of (m`) =√4 + 9 = 3.61 𝑚

θD = 1

4 sin θ

wint (D) = (𝑚 + 𝑚`) ∗ 3.61 ∗1

4 sin θ

but ; sin θ = 3

3.61

wint (D) = (𝑚 + 𝑚`) ∗ 3.61 ∗3.61

4𝑥3

=(𝑚 + 𝑚`) ∗ 1.09


=4 ∗ (𝑚 + 𝑚`) ∗ 1.09 (4.21)

let m=m`

=8.72 m


=8.72 m = 12 w (4.22)

m = 12w/ 8.72 (4.23)

48

m = 1.38 w (4.24)

When; w = 20 kN.m2

Mmax =27.6kN.m/m



figure below:

Fig.4.9. Yield Pattern of Panel S4.

The results obtained for the ultimate resisting moments for each panel of

reinforced concrete flat slabs were summarized at Table (4.5) and were

compared with value obtained by using STAAD-Pro Software.

Table .4.5. The Ultimate Resisting Moments for Flat Slab.

Panel Value of (MP) by

yield line theory

kN.m/m

Value of (MP) by

STAAD-pro

kN.m/m

Difference %

S1 61.80 61.74 0.097%

S2 52.00 52.96 -1.85%

S3 48.02 47.22 1.67%

S4 27.6 26.09 5.47%

49

4.3. Analysis of beam Slab system

The Fig.4.10. below is the plan of slab with beam have three equal spans at X

direction of 6 m length and three spans at Y direction of length 6 m for the edges

spans and 4 m length for middle span. , The slab is subjected to uniformly

distributed load of 20 kN/m2. By considering a reasonable pattern of positive

and negative yield lines is that shown in Fig.4.10, and with following the

procedure explained at previous Chapter, the ultimate moment (MP) can be

obtained for each panel as named in Fig.4.10.

Fig.4.10. Plan of beam Slab System with Expected of Yield Line Pattern.

51

4.3.1. Analysis of External Corner slab S5

Panel S5 is the square panel have length of 6 m each with two adjacent edges

discontinuous and continuous in other two sides, by considering a reasonable

pattern of positive and negative yield lines is that shown in Fig.4.11, we will

determine the MP using work method.

Fig.4.11. Expected Yield line Pattern of External Corner Panel S5.




= ∑w × a × δ

= w (6*6) * 1

3

= 12w (4.25)


Wint (total)= wint (A) +wint (B) +wint (D) +wint(C)

51


Segment (A):

Length of (m) = 6m

θA =1

𝑥

wint (A) = 𝟔𝒎

𝒙

Segment (B):

Length of (m) = 6m

θB = 1

𝑦

wint (B) = 𝟔𝒎

𝒚

Segment (C):

Length of (m) = 6m

Length of (m`) = 6m

θC = 1

(6−𝑥)

Segment Length for

moment

Length for

moment

Compone

nt of

Rotation

Wint

Segment

(A)

𝟔𝒎

𝒚

Segment

(B)

6m 1

𝑦

𝟔𝒎

𝒚

Segment

(C)

6m 6m 1

(6 − 𝑥)

𝟏𝟐𝒎

(𝟔 − 𝒙)

Segment

(D):

6m 6m 1

(6 − 𝑦)

𝟏𝟐𝒎

(𝟔 − 𝒚)

Total wint 𝟔𝐦

𝐱+

𝟔𝐦

𝐲 +

𝟏𝟐𝐦

(𝟔−𝐱) +

𝟏𝟐𝐦

(𝟔−𝐲)

6m 1

𝑥

52

wint (c) = 𝟔∗(𝒎+𝐦`)

(𝟔−𝒙)

Let; m`=m

wint (c) = 𝟔∗(𝒎+𝐦)

(𝟔−𝒙)=

𝟏𝟐𝒎

(𝟔−𝒙)

Segment (D):

Length of (m) = 6m

Length of (m`) = 6m

θC = 1

(6−𝑦)

wint (D) = 𝟔∗(𝒎+𝐦`)

(𝟔−𝒚)

Let; m`=m

wint (D) = 𝟔∗(𝒎+𝐦)

(𝟔−𝒚) =

𝟏𝟐𝒎

(𝟔−𝒚)


= 𝟔𝒎

𝒙+

𝟔𝒎

𝒚 +

𝟏𝟐𝒎

(𝟔−𝒙) +

𝟏𝟐𝒎

(𝟔−𝒚)

=6m [1

𝑥+

1

𝑦 +

2.33

(6−𝑥) +

2.33

(6−𝑦)] (4.26)


=6m [1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)]= 12w (4.27)

𝑤

𝑚 = [

1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)] / 2 (4.28)


𝑑(𝑤

𝑚)

𝑑y = 0 ; when

𝑑

𝑑y[

1

𝑥+

1

𝑦 +

2

(6−𝑥) +

2

(6−𝑦)] = 0

i.e.

1

𝑦2 +

(−2)

(6−𝑦)2= 0


to solve it for (y)

Given; y = 2.485 m

That is w/m maximum when y = 2.485

For y = 2.485, the Eqn (4.28) reduces to

53

1

x +

2

(6−x)= 𝑚/𝑤 (4.29)

𝑑(𝑤

𝑚)

𝑑x = 0 ; gives x = 2.485 m

Substituting; x = 2.485 and y = 2.485 into Equ (4.28)

m = 1. 03 w (4.30)

When; w = 20 kN.m2

Mmax =20.6 kN.m/m

By using “10% rule”; Mmax = 22.66 kN.m/m

From BS8110:

msx= βsxnlx2

0.034*20*36=24.48 kN.m/m


figure below:


54

4.3.2. Analysis of Edge Slab S6

Panel S6 is the square panel have length of 6m with one edge discontinuous and

three edges continuous, by considering a reasonable pattern of positive and

negative yield lines is that shown in Fig.4.13, we will determine the Mp using

work method.

Fig.4.13. Expected Yield Line Pattern of edge Panel S6.

Solution by using virtual method work:



= ∑w × a × δ

= w (6*6) * 1

3

= 12w (4.31)


Wint (total) = wint (A) +wint (B) +wint (D) +wint(C)

55


Segment (A):

Length of (m) = 6m

Length of (m`) = 6m

θA = 1

3

wint (A) = 𝟔 ∗(𝒎+m`)

𝟑

Let; m` = m

𝐰𝐢𝐧𝐭 (𝐀) = 𝟔 ∗(𝒎+m)

𝟑= 𝟒𝒎

Segment (B):

Length of (m) = 6m

θB = 1

𝑦

wint (B) = 𝟔𝒎

𝒚

Segment (C):

Length of (m) = 6m

Segment Length for

moment

Length for

moment

Compone

nt of

Rotation

Wint

Segment

(A)

6m

𝟒𝒎

Segment

(B)

6m 1

𝑦

𝟔𝒎

𝒚

Segment

(C)

6m 6m 1

3

𝟒𝒎

Segment

(D):

6m 6m 1

(6 − 𝑦)

𝟏𝟐𝒎

(𝟔 − 𝒚)

Total wint 𝟖𝐦+ 𝟔𝐦

𝐲 +

𝟏𝟐𝐦

(𝟔−𝐲)

6m 1

3

56

Length of (m`) = 6m

θc = 1

3

wint (c) = 𝟔 ∗(𝒎+m`)

𝟑

Let; m` = m

wint (c) = 𝟔 ∗(𝒎+m)

𝟑= 𝟒𝒎

Segment (D):

Length of (m) = 6m

Length of (m`) = 6m

θC = 1

(6−𝑦)

wint (D) = 𝟔∗(𝒎+𝐦`)

(𝟔−𝒚)

Let; m` = m

wint (D) = 𝟔∗(𝒎+𝐦)

(𝟔−𝒚)=

𝟏𝟐𝒎

(𝟔−𝒚)


= 8𝑚+ 𝟔𝒎

𝒚 +

𝟏𝟐𝒎

(𝟔−𝒚)

=2m[4+ 3

𝑦 +

6

(6−𝑦)] (4.32)


=2m[4+ 3

𝑦 +

6

(6−𝑦)]= 12w (4.33)

𝑚

𝑤 = 6/ [4+

3

𝑦 +

6

(6−𝑦)] (4.34)


and finding the maximum Values of (𝑚

𝑤),

We find maximum values of (𝑚

𝑤) = 0.87, coincide with y = 2.49 m

That is w/m maximum when y = 2.49

Substituting; y = 2.49 into Equ (4.34)

m = 0.87 w (4.35)

w = 20 kN.m2

57

Mmax =17.4 kN.m/m

By using "10% rule " ; Mmax = 19.14 kN.m/m.

msx= βsxnlx2

0.029*20*36=20.88 kN.m/m


figure below:


58

4.3.3. Analysis of Edge Slab S7

Panel S7 is the rectangular panel have length of 6m and width 4m with one edge

discontinuous and three edges continuous, by considering a reasonable pattern

of positive and negative yield lines is that shown in Fig.4.15, we will determine

the Mp using work method.

Fig.4.15. Expected yield line Pattern of edge Panel S7.

Solution by using virtual method work:



= ∑w × a × δ

= w [(4x+4y) * 1/3 +(6-x-y) (4) *1/2]

= 2

3w (18-x-y) (4.36)



59


Segment (A):

Length of (m) = 4m

θA = 1

𝑥

wint (A) = (0.5m)* (4)*1

𝑥=

2𝑚

𝑥

Segment (B):

Length of (m) = 6m

Length of (m`) = 6m

θB = 1

2

Let; m` = m

wint (B) = 6*(m+m`)*1

2

6*(m+m)*1

2 = 6 m

Segment Length for

moment

Length for

moment

Compone

nt of

Rotation

Wint

Segment

(A)

𝟐𝒎

𝒙

Segment

(B)

6m 6m 1

2

6m

Segment

(C)

4m 4m 1

𝑦

𝟒𝒎

𝒚

Segment

(D):

6m 6m 1

2

6m

Total wint 𝟏𝟐𝐦 +

𝟐𝒎

𝒙 +

𝟒𝒎

𝒚

4m 1

𝑥

61

Segment (C):

Length of (m) = 4m

Length of (m`) = 4m

θc = 1

𝑦

wint (c) = (0.5m+0.5m`) *4

𝑦

Let; m` = m

= (0.5m+0.5*m) *4

𝑦=

4𝑚

𝑦

Segment (D):

Length of (m) = 6m

Length of (m`) = 6m

θD = 1

2

wint (B) = 6*(m+m`)*1

2

Let; m` = m

6*(m+m)*1

2= 6m


=12m + 2𝑚

𝑥 +

4𝑚

𝑦

=2m(6 + 1

𝑥 +

2

𝑦 ) (4.37)


=2m(6 + 1

𝑥 +

2

𝑦 )=

2

3w (18-x-y) (4.38)

𝑤

𝑚 =

3 (6+ 1

𝑥 +

2

𝑦 )

(18−x−y) (4.39)


f1( 𝑥1, 𝑥2)

f2 ( 𝑥1, 𝑥2)=

𝑑

𝑑𝑥1f1( 𝑥1, 𝑥2)

𝑑

𝑑𝑥1f2 ( 𝑥1, 𝑥2)

By differentiate to x :

61

3 (6 + 1

𝑥 +

2

𝑦 )

(18 − x − y)=

3 ( − 1

𝑥2 )

−1

18-x-y = 6x2+2x +2𝑥2

y

(6 + 2

𝑦) 𝑥2+ 2x +(𝑦 − 18)= 0 (4.40)

Solve for "x" ;

x =

−2±√22−4(6+ 2

𝑦)(𝑦−18)

2∗(6+ 2

𝑦)

(4.41)


and finding the maximum Values of (x),

we find maximum values of (x) = 1.38 m , coincide with y = 1.95 m

m = 0.63 w (4.42)

w = 20 kN.m2

Mmax =12.6 kN.m/m

By using "10% rule " ; Mmax = 13.86kN.m/m

msx= βsxnlx2

0.043*20*16=13.76 kN.m/m


figure below:

Fig.4.16. Yield line Pattern of Panel S7.

62

4.3.4. Analysis of Interior Slab S8

Panel S8 is the rectangular panel have length of 6m and width 4m with four

edges continuous, by considering a reasonable pattern of positive and negative

yield lines is that shown in Fig.4.17, we will determine the Mp using work

method.

Fig4.17. Expected Yield Line Pattern of Interior Panel S8.




= ∑w × a × δ

= w [(4x+4y) *1

3+ (6-x-y) (4) *

1

2]

= 2

3w (18-x-y) (4.43)



63


Segment (A):

Length of (m) = 4m

Length of (m`) = 4m

θA = 1

𝑥

wint (A) = (0.5m+0.5m`)* 4

𝑥

Let; m` = m

wint (A) =(0.5m+0.5*m)* 4

𝑥 =

4𝑚

𝑥

Segment (B):

Length of (m) = 6m

Length of (m`) = 6m

θB = 1

2

wint (B) = 6*(m+m`)*1

2

Segment Length for

moment

Length for

moment

Compone

nt of

Rotation

Wint

Segment

(A)

4m

𝟒𝒎

𝒙

Segment

(B)

6m 6m 1

2

6m

Segment

(C)

4m 4m 1

𝑦

𝟒𝒎

𝒚

Segment

(D):

6m 6m 1

2

6m

Total wint 12m + 𝟒𝒎

𝒙 +

𝟒𝒎

𝒚

4m 1

𝑥

64

Let; m` = m

wint (B) =6*(m+m)*1

2 = 6m

Segment (C):

Length of (m) = 4m

Length of (m`) = 4m

θc = 1

𝑌

wint (c) = (0.5m+0.5m`) ∗ 4

𝑌

Let; m` = m

wint (c) = (0.5m+0.5*m) ∗ 4

𝑌=

4𝑚

𝑌

Segment (D):

Length of (m) = 6m

Length of (m`) = 6m

θD = 1

2

wint (D) = 6*(m+m`)*1

2

Let; m` = m

wint (D) = 6*(m+1.33m)*1

2= 6.99m


= 12m +4𝑚

𝑋 +

4𝑚

𝑌

= 4m (3 +1

𝑋+

1

𝑌) (4.44)


= 4m (3 +1

𝑋+

1

𝑌) =

2

3w (18-x-y) (4.45)

𝑤

𝑚=

6 (3 +1

𝑋+

1

𝑌 )

(18−x−y)


f1( 𝑥1, 𝑥2)

f2 ( 𝑥1, 𝑥2)=

𝑑

𝑑𝑥1f1( 𝑥1, 𝑥2)

𝑑

𝑑𝑥1f2 ( 𝑥1, 𝑥2)

By differentiate to x:

65

6(3 + 1

𝑥 +

1

𝑦 )

(18−x−y)=

6( − 1

𝑥2 )

−1 (4.46)

18-x-y = 3x2+x +𝑥2

y

(3 + 1

𝑦) 𝑥2+ 2x +(𝑦 − 18)= 0

Solve for x;

x = −2+√22−4(3+

1 𝑦

)(𝑦−18)

2(3+ 1 𝑦

) (4.47)



we find maximum values of (x) = 1.87 m , coincide with y = 1.87 m

m = 0.584 w (4.48)

w = 20 kN.m2

Mmax =11.68kN.m/m

By using " 10% rule " ; Mmax = 12.85kN.m/m

msx= βsxnlx2

0.04*20*16=12.8 kN.m/m


figure below:

Fig.4.18 Yield line Pattern of Panel S8.

66

The results obtained for the ultimate moments for each panel of reinforced

concrete slabs with beams are summarized as shown in Table. 4.10. and

compared with others obtained by using STAAD-Pro Software, and BS8110.

Table.4.10.The Ultimate Resisting Moments for Beam slab System.

Panel Value of

(MP) by

yield line

theory

KN.m/m

Value of

(MP) by BS

8110

KN.m/m

Difference

%

Value of (MP)

by STAAD-pro

KN.m/m

Difference

%

S5 22.65

24.48

8.08% 25.8 13.91%

S6 19.14

20.88

9.09% 20.02 4.6%

S7 13.86

13.76

0.72% 14.95 7.86%

S8 12.85

12.8

0.39% 12.76 0.70%

67

4.4 Special Beam slab system of different support conditions

4.4.1 Analysis of corner square slab S9

Panel S9 square slab length of 5m with edges continuous in two sides and simply

supported in other two sides, by considering a reasonable pattern of positive and

negative yield lines is that shown in Fig.4.19, the ultimate resisting moment Mp

can be determine using virtual work method

Fig.4.19. Expected yield line Pattern for the Corner Panel S9.

Solution by using virtual work method:



= ∑w × a × δ

=2 [(w*10∗(10−𝑥)

2*

1

3 )-(w* 5 *

5∗5(10−𝑥)

2𝑥 *

1

3 )]

5

𝑥+2 [(w *

(√2 ∗5)∗(√2 ∗(𝑥−5))

2∗

1

3 )] −(w*

((√2 ∗5))

2 *

(10𝑥−50)

(√2 ∗𝑥) ) ∗

(10𝑥−50)

√2 ∗√2 ∗(𝑥−5)

= 5𝑤

3 (10-x)

(2𝑥2−25)

𝑥2+

10𝑤

3 (x-5)

(𝑥2−25)

𝑥2

= 5𝑤

3𝑥2∗ {(10 − 𝑥)(2𝑥2 − 25) + 2 (𝑥 − 5)(𝑥2 − 25)}

=5𝑤

3𝑥 (10𝑥 − 25)(4.49)

68

Table.4.11 Analysis of Panel S9 by yield line theory.


Wint (total) = wint (A) +wint (B) +wint(C)

Segment (A):

Length of (m) =5m

Length of (m`) = 5m

θA = 1

(10−𝑥)

wint (A) = (m +m`)* (5)*1

(10−𝑥)

Let; m = m`

=10𝑚

(10 − 𝑥)

Segment (B):

Length of (m) = 5m

Length of (m`) = 5m

θB = 1

(10−𝑥)

wint (B) = (m +m`)* (5)*1

(10−𝑥)

let; m = m`

Segment Length for

moment

Length for

moment

Component

of Rotation Wint

Segment

(A)

5m 𝟏𝟎𝒎

(𝟏𝟎 − 𝒙)

Segment

(B)

6m 6m 1

2

𝟏𝟎𝒎

(𝟏𝟎 − 𝒙)

Segment

(C)

(10𝑥 − 50)

(√2 ∗ 𝑥)

1

√2(𝑥 − 5) 2𝑚

(10𝑥 − 50)

(√2 ∗ 𝑥)√2(𝑥 − 5)

Total wint 𝟏𝟎𝐦{

𝟐

(𝟏𝟎 − 𝒙) +

𝟏

𝒙}

5m 1

(10 − 𝑥)

69

=10𝑚

(10 − 𝑥)

Segment (C):

Length of (m) = (10𝑥−50)

(√2 ∗𝑥)m

θc = 1

√2(𝑥−5)

wint (c) = 2(m) *(10𝑥−50)

(√2 ∗𝑥)∗

1

√2(𝑥−5)

= 2𝑚(10𝑥 − 50)

(√2 ∗ 𝑥)√2(𝑥 − 5)


=10𝑚

(10−𝑥)+

10𝑚

(10−𝑥) + 2𝑚

(10𝑥−50)

(√2 ∗𝑥)√2(𝑥−5)

= 20𝑚

(10−𝑥) +

(10𝑥−50) 𝑚

𝑥(𝑥−5)

= 10m{ 2

(10−𝑥) +

1

𝑥} (4.50)


= 10𝑚 ( 2

(10−𝑥) +

1

𝑥 ) =

5𝑤

3𝑥 (10𝑥 − 25) (4.51)

= 2𝑚 ( 2

(10−𝑥) +

1

𝑥 ) =

(10𝑥−25) 𝑤

3𝑥

m = w(10𝑥−25)

3𝑥 ÷ (

4

(10−𝑥) +

2

𝑥 )(4.52)

Solve for m;



We find maximum values of (m) = 1.462w, coincide with x = 5.81 m

m = 1.462 w (4.53)

w = 19.9 kN.m2

Mmax =29.10kN.m/m

By using “10% rule”; Mmax = 32.00kN.m/m

By using prokon program = Mmax= 37.70kN.m/m

msx= βsxnlx2

71

0.062*20*25=31 kN.m/m


figure below:

Fig.4.20 Yield Pattern of Panel S9.

The results obtained for the ultimate moments for the panel S9 are summarized

at Table.4.12 and compared with others obtained by using Prokon Software,

and BS8110.

71

4.4.2 Analysis of rectangular slab S10

A rectangular slab of length 6m and width 4m with continuance edge in one side

and simply supported in two sides, and free edge in one, by considering a

reasonable pattern of positive and negative yield lines is that shown in Fig.4.12,

we will determine the Mmax using work method

Fig.4.21 Expected yield line Pattern of Corner Panel S10.




= ∑w × a × δ

=w[2*1

2 (x*y)*

1

3+2*

1

2 (x*(4-y))*

1

3+(6-x)*y)*

1

2+ (6 − x) ∗ (6 − y) ∗

1

2]

= w (12+4

3 x-2x) (4.54)

= w (12-2

3 x)



72

Table.4.12 Analysis of Panel S10 by yield line theory.

Segment (A):

Length of (m) =6m

θA = 1

𝑦

wint (A) = (m)* (6)*1

𝑦=

6𝑚

𝑦

Segment (B):

Length of (m) = 6m

Length of (m`) = 6m

θB = 1

(4−𝑦)

wint (B) = 6*(m+m`)*1

(4−𝑦)

Let; m`= m

wint (B) = 6*(m+m)*1

(4−𝑦) =

12𝑚

(4−𝑦)

Segment (C):

Length of (m) = 4m

θc = 1

𝑥

Segment Length for

moment

Length for

moment

Component

of Rotation Wint

Segment

(A)

𝟔𝒎

𝒚

Segment

(B)

6m 6m 1

(4 − 𝑦)

𝟏𝟐𝒎

(𝟒 − 𝒚)

Segment

(C)

4m 1

𝑥

𝟐𝒎

𝒙

Total wint 𝟏𝟐𝒎

(𝟒 − 𝒚)+

𝟐𝒎

𝒙 +

𝟔𝒎

𝒚

6m 1

𝑦

73

wint (c) = (0.5m) *4

𝑥=

2𝑚

𝑥


=12𝑚

(4−𝑦)+

2𝑚

𝑥 +

6𝑚

𝑦

= 2m (6

(4−𝑦)+

1

𝑥 +

3

𝑦 )


= 2m (6

(4−𝑦)+

1

𝑥 +

3

𝑦 )= w (12-

2

3 x)

= m (6

(4−𝑦)+

1

𝑥 +

3

𝑦 )=w (6-

x

3) (4.55)

𝑤

𝑚 =

(6

(4−𝑦)+

1

𝑥 +

3

𝑦 )

(6−x

3 )

(4.56)


f1( 𝑥1, 𝑥2)

f2 ( 𝑥1, 𝑥2)=

𝑑

𝑑𝑥1f1( 𝑥1, 𝑥2)

𝑑

𝑑𝑥1f2 ( 𝑥1, 𝑥2)

By differentiate to x:

(6

(4−𝑦)+

1

𝑥 +

3

𝑦 )

(6−x

3 )

= 3( −

1

𝑥2 )

−1 (4.57)

18-x = (6

(4−𝑦)+

3

𝑦 ) x2+x

(6

(4−𝑦)+

3

𝑦) 𝑥2+ 2x - (18)= 0 (4.58)

Solve for x;

x = −2±√22−4(

6(4−𝑦)

+3𝑦

)∗18

2(6

(4−𝑦)+

3𝑦

) (4.59)



we find maximum values of (x) = 1.8 m, coincide with y = 1.66 m

m = 1.1 w (4.60)

w = 20 kN.m2

Mmax =22kN.m/m

By using "10% rule"; Mmax = 24.2kN.m/m

74

By using prokon program, Mmax= 22.9kN.m/m.

msx= βsxnlx2

0.063*20*16=20.16 kN.m/m


figure below:

Fig.4.22 Yield Line Pattern for Panel S10.

The results obtained for the ultimate moments for the panel S10 are summarized

at Table (4.14) and compared with others obtained by using Prokon Software,

and BS8110.

Table.4.13 the Ultimate Resisting Moments for Panels S10.

panel Value of

(MP) by yield

line theory

kN.m/m

Value of

(MP) by

BS8110

Table 3.14

kN.m/m

Different

%

Value of

(MP) by

Prokon

kN.m/m

Different

%

S9 32.00 31.00 3.125% 37.7 17.%

S10 22.00 20.16 8.36% 22.9 4.09%

75

4.5 Discussion of the results

The discussion of the results may be summarized as follows:

1. From results of ultimate resisting moments for R.C. flat slab system it was

appeared that the difference about 0.097% to 5.47% in comparison of with yield

line theory.

2. The results of ultimate resisting moments obtained were shown in table (4.10)

for the R.C. beam slab system that were compared with others those obtained

by using STAAD-Pro software and BS8110. The comparison revealed different

conforming by percentages range by 0.70 % to 13.91% when comparing with

STAAD-PRO and range by 0.39 % to 9.09 % when comparing to BS8110.

3. The results of ultimate resisting moments obtained from table (4.13) for the R.C.

slabs that have special condition of supports that were compared with others

those obtained by using PROKON software and BS8110. The comparison

revealed different conforming by percentages range by 4.09 % to 17.81% when

comparing with PROKON and range by 3.125 % to 8.36 % when comparing

with BS8110.

CHAPTER FIVE

CONCLUSIONS AND

RECOMMENDATION

76

CHAPTER FIVE

CONCLUSIONS AND RECOMMENDATION

5.1 Conclusions

On Basis of this study, Conclusions that can be drawn are as follows:

1. By using yield line theory, different types of reinforced concrete slabs are used

to determine the ultimate resisting moments and their locations.

2. One of the most popular methods of application in yield line theory is the virtual

work method that was used in this research to analysis and assessment different

models of reinforced concrete slabs (beam slab and flat slab system) of different

shapes and different support conditions, In addition to slabs that have special

condition of supports.

3. The percentages range between 0.097 % to 5.47 % of the results of bending

moments for the reinforced concrete flat slab system that were compared with

others those obtained by using STAAD-Pro software. This results confirm to

the software program STAAD-Pro with manual calculations, the results were

classified as very good once.

4. The percentages range between 0.70 % and 13.91% of the results of bending

moments for the reinforced concrete beam slab system that were compared with

others those obtained by using STAAD-Pro software. This results confirm to

the software program STAAD-Pro with hand calculations, the results were

classified as good once. The same results from manual calculations were

compared with others those obtained by using BS8110 the percentages range of

difference between 0.39 %and 9.09 %. This results obtained by BS8110 and

manual calculations were closer than obtained by STAAD-Pro, the results were

classified good.

5. The percentages range between 4.09 % and 17.81% of the results of bending

moments for the reinforced concrete slabs that have special condition of

supports were compared with others those obtained by using PROKON

77

software. This results related to the software program PROKON with hand

calculations, the results were classified very good results. The same results from

hand calculations were compared with others those obtained by using BS8110

the percentages range of difference between 3.125 %and 8.36 %. This results

confirm to the BS8110 with hand calculations and it was closer than results from

PROKON, the results were classified very good results.

6. As the general, the results of this study were clearly demonstrate that acceptable

and close results can be achieved with analysis by using yield line theory.

5.2 Recommendations

Recommendations were summarized as follows:

1. Study the analysis by assuming the pattern of yield line bisecting the corners at

45 degrees "simple layout". It will make the calculations very quick and easy.

2. Check the analysis by assuming percentage for the distance between the yield

line and the edges of slab depend on support condition.

3. Irregular reinforced concrete slabs, raft foundations can be check by using yield

line theory.

78

REFERENCES

[1] Ingerslev A., (On a Simple Analysis of Two-Way Slabs)”, Ingeniren, 30,

69, 1921, pp 507-515. (See also: The Strength of Rectangular Slabs, Struct.

Eng., 1, 1, 1923, pp. 3-14.)

[2] Johansen, K.W. 3, 1, 1931, pp. 1-18 (German version, Memo. Int. Ass.

Bridge Struct. Eng. 1, 1932, pp. 277-296.).

[3] Gvozdev, A.A., (English translation: “The Determination of the Value of

the Collapse Load for Statically Indeterminate Systems Undergoing Plastic

Deformation”, International Journal of Mechanical Sciences, 1, 1960, pp. 322-

333.)

[4] Gvozdev, A.A, (Comments of the design standard for reinforced concrete

structures)”, 17, 3, 1939, pp. 51-58.

[5] Morley symposium on concrete plasticity and its application, university of

Cambridge 23th July, 2007, pp. 43.

[6] Recent Developments in Yield-Line Theory, MCR Special Publication,

Cement and Concrete Association, London, 1965.

[7] Jones, L.L. and Wood, R. H., Yield-Line Analysis for Slabs, Thames &

Hudson and Chatto & Windus, London, 1967, 405 pp.

[8] Wood, R. H., “A Partial Failure of Limit Analysis of Slabs”, Magazine of

Concrete Research, 21, 67, 1969, pp. 79-80 (Discussion, 22, 21, 1970, pp. 112-

113.).

79

[9] Braestrup, M.W., “Yield Line Theory and Limit Analysis of Plates and

Slabs”, Magazine of Concrete Research, 22, 71, 1970, pp. 99-106.

[10] Fox, E.N., “Limit Analysis for Plates: The Exact Solution for a Clamped

Square Plate of Isotropic Homogeneous Material Obeying the Square Yield

Criterion and Loaded by Uniformly Pressure”, Philosophical Transactions of

the Royal Society, London, 277, 1265, 1974, pp 121155.

[11] Ultimate capacity evaluation of reinforced concrete slabs using yield line

analyses, by GregE.Mertz, Structural Mechanics Section, Westinghouse

Savannah River Company, Aiken, South Carolina 29808, 1995 pp. 5.

[12] Yield line Theory for Concrete Slabs Subjected to Axial Force, by Tim

Gudmand-Høyer, Denmark's Tekniske University, December 2003, pp. 3.

[13] Practical yield line design, by Gerard Kennedy and Chales Good child,

Published by the British Cement Association, 2003 pp. 15.

[14] Reinforced Concrete Design, W.H.Mosley, J.H.Bungey&R.Hulse, Fifth

Edition.

[15] The plastic behavior and the calculation of plates subjected to bending, by

Prof. A.C.W.M. Vrouwenvelder, Prof. J. Witteveen, March 2003.

[16] Civil Engineering Design (1) Analysis and Design of Slabs, Dr. Colin

Caprani, Chartered Engineer, 2006.

81

APPENDIX A

The results of ultimate resisting moment for reinforced concrete flat slab system

by using STAAD-PRO.

The result of resisting moment at x direction.

81

The result of resisting moment at y direction.

82

APPENDIX B

The results of ultimate resisting moment for reinforced concrete beam slab

system by using STAAD-PRO.

The result of resisting moment at x direction.

83

The result of resisting moment at y direction.

Investigation of Ultimate... .pdf - SUST Repository

Documents

Transcript of Investigation of Ultimate... .pdf - SUST Repository