Design and analysis of reinforced concrete multistory commercial building using ACI 318-08 by robot structural analysis and by manual method

DESIGN & ANALYSIS OF REINFORCED CONCRETE MULTI-

STORY COMMERCIAL BUILDING USING ACI-318

MUHAMMAD ABDUL AZEEM BAIG

IMBIA ABD-EL-SALAM IMBIA AMMAR

A Thesis submitted in

Partial Fulfilment of the requirement for the award

Of the Degree of Bachelors of Civil Engineering

Faculty of Civil Engineering

University Of Bani-Walid Libya

SEPT 2016

ii

I hereby declare that the work in this project report is my own except for quotations

and summaries, which have been duly acknowledged

Student : MUHAMMAD ABDUL AZEEM BAIG

: IMBIA ABD-EL-SALAM IMBIA AMMAR

Date : SEPTEMBER 2016

Supervisor : Prof Dr Ibrahim Mohamed Elhaj

iii

For my beloved mother and Father

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ACKNOWLEDGEMENT

First, I would like to thank Almighty Allah for giving me faith, health and

intellectual capacity to carry out this project work. Then, I fully appreciate the moral

support and encouragement from my parents and other family members towards the

course of this study, thank you and may ALLAH (S.W.A) bless you with his infinite

mercy.

It has been a good fortune for me to have Dr Ibrahim Mohamed Elhaj as my

research supervisor, thank you sir; actually, there is no amount of words that i could

use to describe my profound gratitude to you.

I am also grateful to all the teaching staff who offered their contribution

during the conduct of this project. Finally, I am grateful to all my friends and

colleagues, those that were at University of Bani walid & at University of sirte,

students and others that were schooling at other universities.

v

ABSTRACT

All the building structures have to design based on the relevant code of practice of standard.

The choice of the standard code to be applied varies and sometimes depends on the

requirement of the local authority or familiarity of the designers. Standard code is essential in

the reinforced concrete structures design to provide a safety and economic design. Currently,

BS 8110 and ACI-318 are the most widely used standards in designing reinforced concrete

structures based on limit state principle. However, some of the design requirements such as

partial safety factors, material properties, load combinations, etc. Are made to be different

between BS 8110 and ACI-318. This may affect the cost of building structures that were

designed using these two standards. The aim of this study is to design the reinforced concrete

structures for a three-storey commercial building, which will be designed using ACI-318. The

material properties such as characteristics strength of reinforcements and concrete, and

dimensions of the structure elements are fixed. Autodesk Robot Structural Analysis is the

reinforced concrete structure design package that will be used to design and produce the

structural detailing for the three-storey building based on ACI-318. So then, generally, the

study found out that the correctly designed structure may result in economical output while

ensuring safety.

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CONTENTS

DESIGN & ANALYSIS OF REINFORCED CONCRETE MULTI-STORY COMMERCIAL BUILDING USING ACI-318 i

OATH ii

DEDICATION iii

ACKNOWLEDGEMENT iv

ABSTRACT v

CONTENTS vi

LIST OF TABLES xi

LIST OF FIGURES xv

LIST OF APPENDICES xi v

vii

CHAPTER 1 - INTRODUCTION 1

1.1 Research Background 1 1.2 Problem statement 2 1.3 Objectives 3 1.4 Scope of Study 3 1.5 Outline of Thesis 3

CHAPTER 2 - 2.LITERATURE REVIEW 4

2.1 Introduction 4 2.2 Building Codes & Standards 4

2.3 Optimum Cost of Reinforced Concrete Building 8 2.4 Factors Contributing To the Cost of Building Construction 9 2.5 Construction Cost 9

2.6 Research Methodology 10 2.7 Model OF Design 11 2.8 Design Specifications And properties of the structure 14

2.8.1 Materials and Design Parameters 14 2.9 Design Loading 15 2.9. Partial Safety Factors 16 2.9.3 Factors of Safety Loads And Strength Of Section By Strength 17 2.10 Design Methods 18 2.10.1 Object of Structural Design 18 2.10.2 Philosophy of Limit State Design 19 2.11 Project Flow Chart 21 2.11. Expected Results 22

CHAPTER 3 - 3. Design Of Slabs 23

3.1.1 Definition 23 3.1.2 Introduction 23 3.1.3 DesignConcepts 23 3.1.4 Types of slabs 24 3.1.5 One & two way slabs outlined 25 3.1.6 Econlomical Choice According to size and loading 26 3.1.7 Calculation of thickness for one way slab 27 3.1.8 Design procedure for one way slab 28 3.1.9 ACI Code specified method for two-way slabs 29 3.1.10 Two-way slab design procedure 29 3.1.11 Classification of slabs 29 3.1.12 Purpose of main steel in slabs 29 3.1.13 Analysis methods for slabs 30 3.1.14 Slabs direction in ribbed slab 30 3.1.15 Design concept 31

viii

3.1.16 Maximum Reinforcement Ratio 31 3.1.17 Shrinkage Reinforcement Ratio 31 3.1.18 Loads assigned to slabs 32 3.2.1 Elevation Plans of slabs 33 3.2.2 Design procedure for one-way slab 35 3.2.3 Design procedure for two-way slab 37 3.2.4 Data for Design 38

3.3 Design of two-way slab 39 3.5 Design of one -way slab 45 CHAPTER 4 - 4. Design of Beams 50

4.1 Introduction 50 4.2 Struvtural theory of beams 50 4.2.1 Types of a beam 50

4.2.2 Scope of usage of beam 51 4.2.3 Relation of reinforcement with section 51 4.3 Assumptions 54 4.4 Loading data 54 4.5 Design Procedure 55 4.6 Flow charts 56 4.7 Information about Sample of design 58 4.7. Desing assumptions 59 4.7.2 Check Deflection 59 4.7.3 Sizing the cross-section 59 4.8 Design of Flexure 60 4.8.1 Actual Depth 60 4.8.2 Minimum Ratio of steel required 60 4.8.3 Design Reinforcement for every moment in beam 60-68 4.9 Design of shear 69 4.10 Development length 75

CHAPTER 5 - 5. Design of Stairs 78

5.1 Geometrical design of stairs 78 5.1.2 Check for reliabilty 78 5.1.3 Check for angle 79 5.2 Detailed design of stair 81 5.3 No. of steps in each flight 82 5.4 Structural design of stairs 83 5.5 Design for flight no.1 & 3 86 5.6 Data for design for flight no 2 87 5.6.2 Design for flexure for flight no 2 88 5.7 Reinforcement details 71

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CHAPTER 6 - 6 . Design of Columns 90

6.1 Introduction 90 6.1.1 Types of reinforced concrete columns 90 6.1.2 Axial Load capacity of column 91 6.1.3 ACI code requirements for cast in place columns 92 6.1.4 General Configuration of moments with in columns 93 6.1.5 Classification of columns 94 6.1.6 Effective length 95 6.1.7 Design of axially loaded column 96 6.1.8 Types of reinforcements and their use 96 6.1.9 Safety provisions for columns 99 6.1.10 Design formula 101 6.2 Sample for Design 102 6.3 Design in detail 105 6.3.1 Design of moments 105 6.4 Buckling analysis for long column by moment magnification factor 109 6.6 Splices for columns 116 6.7 Usage of dowels 118

CHAPTER 7 - 7. Design of Foundations 120

7.1 Foundation design parameters 120 7.1.2 Allowable Settelment 121 7.2.1 General 122 7.2.2 Area of the footing 122 7.2.3 Depth of the footing 122 7.2.4 Depth from punching and shear consideration 122 7.3 General procedure of design of footing 122 7.4 Steps for structural Design 124 7.5 Data for design 125 7.6 Detailed Steps & formulas for design 127 7.7 Design of sample foundation 131 7.7.1 Area of footing 131 7.7.2 Footing Stability 132 7.7.3 Stregth of design 133 7.7.4 Check one way shear 133 7.7.5 Actual & allowable shear stress 133 7.7.6 Check two way shear 134 7.7.4 Check one way shear 133 7.8 Desing of flexure in long direction 136 7.9 Desing of flexure in short direction 137 7.10 Development length in footing 140

x

7.11 Bearing Stregth of column and footing 141 7.12 Development length in dowels 142

CHAPTER 8 - 8. CONCLUSION AND FUTURE WORK 143

8.1 Conclusion 143 6.2 Suggestion of Further Works 144

REFERENCES

xi

LIST OF TABLES

Table 2.1: Design input detail of building 13

Table 2.2: Initial Sizes and Specification of building 14

Table 2.3: Areas of groups of bars 15

Table 2.4:Detail Dead load 15

Table 2.5: Detail of Self weight of slab 16

Table 2.6: partial Safety factors according to ACI 318-02 17

Table 2.7: Live Loads from ASCE 17

Table 3.1 : Minimum thickness of beam & slabs 27

Table 3.2: Reinforcement of one-way slab 49

Table 4.1: Design for Shear by stirrups under ACI 318-08 69

Table 4.2: Beam Reinforcement tables 77

Table 5.1: Data for design of stairs 79

Table 6.1 : Preliminary assumed sections of columns 104

Table 6.2: Design Value Obtained from Robot -Analysis 105

Table 6.3: Columns Reinforcements -

Table 7.1: Servicbilty load for foundations 131

Table 7.2: Reinforcement table for foundations 142

xv

LIST OF FIGURES & FLOW –CHARTS

Figure 1.1: Front view of building 3

Figure 1.2: Side view –A of Buiding 4

Figure 1.3: Side view –B 5

Figure 1.4: perespective view 5

Figure 2.1:Design /Cost Relation Ship 10

Figure 2.2: Model of R.A for design 11

Figure 3.2 : Axis –Plan 11

Figure 2.4: First and Second floor plan 12

Figure 2.5: Ground Floor plan 13

Flow-chart1 : Desing Procedure 21

Figure 3.1 : Typical types of slabs 25

Figure 3.1.1:Elevation plans for ground floor showing assigned slab names 33

Figure 3.1.2:Elevation plans for first floor showing assigned slab names 33

Figure 3.1.3 :Elevation plans for Second floor showing assigned slab names 34

Figure 3.3: Slab S2 , two way slab as design sample 39

Figure 3.4: One way slab S8 for design 45

Figure 4.1: Types of beams 50

Flow-chart4.1 : Design procedure for singly reinforced rectangular section 56

Flow-chart4.2 : Design procedure for Doubly reinforced rectangular section 57

Figure 4.2: B.M.D for beam 59 from robot structural analysis 58

Figure 4.3: Spans and sections of beam 59 for design 59

Flow-chart4.3 : Design procedure for Shear of beams 70

Figure 4.4: S.F.D for beam 59 from robot structural analysis 71

Figure 4.5 : Reinforced concrete beams reinforcement model 77

Figure 5.1 : Dimensions of stairs 78

Figure 5.3 :plan for stairs 80

Figure 5.4 :vertical cut section of stairs 80

Figure 5.5 :Loading diagram for flight 1 & 3 83

Figure 5.6 :B.M.D for flight 1 & 3 83

Figure 5.8-5.10 :Loading -B.M.D and S.F.D or flight 2 86

Figure 5.7: Reinforcment for stairs 88

xv

Figure 6.1: Shows interior and exterior columns 94

Figure 6.3: Different kinds of column reinforcements 99

Figure 6.4: Elevation plans showing assigned names to coloumns of different

stories of the building 88

Figure 6.7: Shows governing case of column 59 with axial load and moments 99

Figure 6.8: shows section of column 59 115

Figure 6.9: shows minimum requirements for splices 116

Figure 6.10: Reinforcment detail of columns 118

Figure 6.11: Naming of columns from R.S.A 118

Figure 7.2: Foundation plans showing assined names to foundations 125

Figure 7.3: Shows load on foundation by 3d structural model on R.S.A 126

Figure 7.4: Showing dead and live load on foundation 33 under column 59 126

Figure 7.5: Typical reinforcment of foundation 139

xiv

LIST OF APPENDICES

APPENDIX TITLE PAGE

A Charts 145

Design of Reinforced Concrete Multi-story Commercial building ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

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INTRODUCTION To Design & Analysis of Reinforced Concrete Multi-story

Building Under ACI Code .

1.1 Research Background

Structural design is a process of selecting the material type and conducting in-depth calculation of

a structure to fulfill its construction requirements. The main purpose of structural design is to

produce a safe, economic and functional building. Structural design should also be an integration

of art and science. It is a process of converting an architectural perspective into a practical and

reasonable entity at construction site. (Chan Chee, 2007)

One of the important things to be considered in any construction is the cost effectiveness

(i.e. how economical the construction will be at the end of construction). Often a times,

constructions become uneconomical (too expensive) when too much emphasis is laid on the quality

alone. Therefore there should be a balance between quality control and cost effectiveness.

The codes and standards that impact modern building construction are constantly in flux and

changing, and it is difficult to keep up with copious changes and how they will impact building

design. In the structural design of concrete structures, Refereeing to standard code is essential. A

standard code serves as a reference document with important guidance. The contents of the

standard code generally cover comprehensive details of a design. These details include the basis

and concept of design, specification to be followed, design methods, safety factors, loading values

and etc. These codes and standards define the parameters in the reinforced concrete design process

that affect the cost of materials. This would include the dimensions(X, Y, Z) of the different

reinforced concrete elements, the area of reinforcements and ratio of reinforcement limit values.

1.2 Problem statement.

Accurately Analyzed structures are important during the design phase to minimize the construction

cost. Excellent designers must have the ability to organize and manage the process of design so

with special consideration to cost effectiveness during the design process.


2

In today’s construction industry, the commonest codes of practice used are the ACI and BS

codes. However the problem of cost ineffectiveness is becoming so rampant. Although lack of

experience from the engineers also affects the design which eventually affect the cost. For this

reason this research is dedicated to find out the process of assembling different building

components under strictly followed recommendations of one of the aforementioned code i.e ACI-

318-08.

1.3 Objectives.

The main objectives of this study are:

1- To make analysis by ACI code in order to obtain the most safe and sound solution.

2- To ascertain the accuracy of the analysis and the design using software (Robot Analysis)

3- To achieve an ultimate design in terms of quality at minimal cost.

1.4 Scope of study.

The project focuses mainly design of concrete and reinforcement, the structure is a three storey

building. This structure is intended to serve as a commercial building. The main reason why a

three-storey structure is being adopted is that it does not involve calculation for the wind load, The

code used is ACI 318-08 And the selected software to used is Robot Analysis.


3

1.5 Architectural model of Building : ALL of the Architectural work is done by author,


4


5

+

Fig 1.3 Right Side view

Fig1.4 perspective view


6

1.6 Outline of Thesis

The thesis is organized into five chapters. Each chapter begins with a brief introduction of what

to be encountered.

Chapter 1 is a brief overview of the research background and the objectives of the study

followed by the outline of thesis.

Chapter 2, which discusses the research methodology that was adopted for the research.

The chapter deals with the definition of model for designing multi stories reinforced concrete

multi-purpose building, which had built and consists of three floors. The properties of design

model are shown in the first part of its chapter such as the dimensions, the properties of materials

(concrete, steel), the unit weight of concrete and blocks, and the values of loads (dead load and

live load) which depends on the type of building.

Chapter 3 presents the general literature about slabs and proceeds with results of analysis

of slabs by designing a sample element.

Chapter 4 presents the general literature about beams and proceeds with results of analysis

of beams by designing a sample.

Chapter 5 presents the general literature about Stairs and proceeds with results of analysis

of stairs by designing in detail.

Chapter 6 presents the general literature about Columns and proceeds with results of

analysis of Columns by designing a sample element.

Chapter 7 presents the general literature about Foundations and proceeds with results of

analysis of foundations by designing a sample.

Chapter 8 summarizes the project results that have been carried out. The finding of the

study is described. A future recommendation to extend the study is also proposed.


7

2. LITERATURE REVIEW.

2.1 Introduction:

The term “Design of reinforced concrete building” consists of two main elements, which includes the concrete design and the design of reinforcement.

2.2.1 BUILDING CODES AND STANDARDS.

The codes and standards that impact modern building construction are constantly in flux, and it is

difficult at best to keep up with copious changes and how they will impact building design. For

engineers and architects who is working with structural design.

2.2.2 BS 8110 BUILDING CODE: PART 1:1997.

BS 8110 part 1 gives recommendations for the structural use of concrete building and structures,

excluding bridges and structural concrete made with high alumina cement. The aim of design is

the achievements of an acceptable probability that structures being design will perform satisfactory

during their intended life. With an appropriate degree of safety, they should sustain all the loads

and deformation of normal construction and use and have adequate durability and resistance to the

effects of misuse and fire. The structure should be so designed that adequate means exist to

transmit the design ultimate dead, wind and imposed loads safely from the highest supported level

to the foundations (British code, 1997).

The design strengths of materials and design loads should be based on the loads and material

properties as in the BS 8110 and as appropriate for the serviceability limit state (SLS). The design

should satisfy the requirement that no SLS is reached by rupture of any section, by overturning or

by buckling under the worst combination of ultimate loads.


8

2.2.3 ACI 318 BUILDING CODE (ACI 318-02).

The American concrete institute standard 318, building code requirements for reinforced concrete,

has permitted the design of reinforced concrete structure in accordance with limit state principles

using load and resistance factors since1963. A probabilistic assessment of these factors and

implied safety levels is made, along with consideration of alternate factors values and formats. (A

discussion of issues related to construction safety of existing structure is included). Working stress

principles and linear elastic theory formed the basis for reinforced concrete design prior to 1983,

when the concept of ultimate strength design was incorporated in the ACI building code (ACI318-

02), (Edward cohen, 1971). Because of the highly nonlinear nature of reinforced concrete behavior,

the linear approach was unable to provide a realistic assessment of true safety levels (Andrew

Scanlon, 1992).

The developers of ACI 318-02, who introduced the idea of load and resistance factors to

account for uncertainties in both load and resistance .Probabilistic methods were developed and

refined during the late 1960s in response to the need to consider variability and uncertainty,

explicitly and rationally. Proposed formulations include code incorporation of explicit second

moment probabilistic procedures. In such an approach, the designer would select a desired safety

index “B” and carry out the design utilizing the means standard deviations of the load and

resistance variables. The safety index positions the mean load effect to ensure attainment of the

target reliability (American code). The explicit second moment approach was not considered by

ACI38 or other major code writing organizations. (Edward Cohen, 1971).

2.3 OPTIMUM COST OF REINFORCED CONCRETE BUILDING.

The meaning of the optimum cost of reinforced concrete building with some studies, which it is

minimum quantity of concrete and steel in any construction or it is the minimum cost of the

construction but the most studies explains the optimum cost by minimum quantity of concrete and

steel in any construction.

Hence, the primary objective of economic analysis is to secure cost-effectiveness for the

client. In order to achieve this, it is necessary to identify and to evaluate the probable economic


9

outcome of a proposed construction project. An analysis is required from the viewpoint of the

owner of the project when doing the proposal, the analysis can be evaluated the followings

(Ashworth A., 1994) to achieve maximum profitability from the project concerned, to minimize

construction costs within the criteria set for design, quality and space, to maximize any social

benefits, to minimize risk and uncertainty and to maximize safety, quality and public image.

Cost and safety are one of the important factors that will affect method of construction,

quality of work, period of the construction and most of all, the success of a project. It seeks to

ensure the efficient use of all available sources to construction. Client’s requirements, possible

effect on the surrounding areas, relationship of space and shape, assessment of the initial cost, the

reason for, and method of, controlling costs, the estimation of the life of buildings and material

need to be studied so as to improve the efficiency of control in construction (Flanagan R. and Tate

B., 1997).

2.4 FACTORS CONTRIBUTING TO THE DESIGN OF BUILDING CONSTRUCTION.

Implementation of a construction projects is a complicated and complex process (Neap H.S and

Celik T., 2001). Phases of construction are divided into categories such as material, labor, plant,

supervision, All disturbances regarding the cost must be detected periodically (Popescu, 1977).

The collection, analysis, publication and retrieval of designed information are very important to

the construction industry. Contractors and surveyors will tend, wherever possible, to use their own

generated data in preference to commercially published data, since the former incorporate those

factors which are relevant to them. Published data will therefore be used for backup purpose. The

existence of a wide variety of published data leads one to suppose, that it is much more greatly

relied on than is sometimes admitted (Ashworth A., 1994)

2.5 BASIC PRINCIPLES OF COST

Most decision makers recognize that there are only a few variables that have a large influence on

a building’s costs. Brandon has classified these variables into two categories decisions


10

concerning the size of the buildings and decisions concerning material specifications and building

configuration (Figure 2.1).

Figure 2.1 Design / Costs relationships

2.6 RESEARCH METHODOLOGY.

The proposed methodology is based on designing the building by software program (Robot

Structural Analysis) with ACI Code, each code has different properties of concrete and steel ,such

as the concrete compressive strength (fc), the yield strength of steel (fy) ,the various combinations

of the load, the allowable ratio for minimum and maximum reinforcement and other properties ,

in practice ,design of the elements are governed by various architectural requirements. If the height

and width of the beam are located ,the designs allocates the right amount of steel but, in this

study ,we assumed that the dimension of the beams and columns are not given .hence ,during the

design by R.S.A software, we will start with small dimensions ,in this case the program will check

if the dimensions were acceptable or not ,here if the dimensions are small the message from

program report will come out “please note: max/min reinforcement sizes do not permit acceptable

bar spacing ,increase member size” .so, we will increase the member size till we get the first

acceptable dimensions that have the first acceptable amount of steel.

Specification and shape

Area

Cost


11

2.7 MODEL OF DESIGN.

The model that will be designed is a multi-stories reinforced concrete commercial building which

has length of 20.00 m x 21 m width and the building consists of three stories, two stories upon the

ground with height 4m. Figure 2.1, 2.2, 2.3, 2.4 shows the plan of the building.

Figure 2.2

Figure 2.3 Axis Plan.

PrayerRoom

6.00m

3.50m

4.40m7.00m

7.00m

6.00m

6.20m

19.60m

21.20m

7.00m

7.00m

4.20m

1.40m

2.30m

3.50m

0.30m

6.20m

3.50m

1.5

3.60m

2.40m2.40m

CLEANERS ROOM

Shop Shop

Shop

Toilet

WO

MEN

S TOILET

MEN

S TOILET

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نى ي للمب

ضي لدور االر

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د ني ولی جامعة ب

Area = 405.5 m

² ك م بی

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عباعداد م

الرس

2016™د

ني ولیة جامعة ب

ة الھندسكلی

Date : 2/8/2016

3.00m

1.00m

1.20m

1.70m

اري ني تج

میم مبص

روع ت مش

Architectural Layout of E

levation plan

1.40

Shop

ة ة الھندس كلی

نى اني للمب

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المس

د ني ولی جامعة ب

ك م بی

م نعید العظی

م.عباعداد م

الرس

2016™:

Area = 405.5 m

²PrayerRoom

6.00m

3.50m

4.40m7.00m

7.00m

6.00m

6.20m

19.60m

21.20m

7.00m

7.00m

2.30m

3.50m

0.30m

6.20m

3.50m

1.5

3.60m

2.40m2.40m

CLEANERS ROOM

Shop Shop

Shop

Toilet

WO

MEN

S TOILET

MEN

S TOILET

3.00m

1.00m

1.20m

1.70m

Shop

6.20m

4.20m

1.40m

1.5

Shop

دني ولی

ة جامعة بة الھندس

كلی

اري ني تج

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روع ت مش

Architectural Layout of E

levation plan

2.30

2.80m


14

Table 2.1 Design input detail of the building

2.8 DESIGN SPECIFICATIONS AND PROPERTIES OF THE STRUCTURE

The initial sizing of members and specifications of the frame building are shown in Table 3.2. The

initial sizes of member were checked against the conditions according to serviceability limit state

and ultimate limit state. The sizes were adjusted until the conditions of serviceability limit state

and ultimate limit state stated in ACI318-08 were satisfied.

Structural Elements Dimensions(exterior) Dimensions(Interior)

Columns Ground floor 500x250 mm 600x250 mm

1th to 2th 400x250 mm 500x250 mm

Beams Tie Beam(plinth) 250x600 mm 250x600 mm

1th to 2th 250x400 mm 250x500 mm

Slab 200 mm THK.

No. of stories 3 stories

Beam to column connection = fixed

Column to base connection = fixed Table 2.2 Initial sizes and specification of the building according to ACI318 code

2.8.2 MATERIAL PROPERTIES.

Every material has different properties that are simply of their own. Similarly, the material used in

the design of the structure in this research also has different properties and strength. Table 3.4, 3.5

lists the material properties applied in the preliminary analysis of the design of the structural

Building usage Shops

Story height Ground floor 4 m

1th & 2th 4 m

Length of building 21.00 m

Width of building 20.00 m

Height of building 13.8 m


15

members (beams, slabs and columns, etc.) The values of compressive strength of concrete, yield

stress of reinforcement, concrete density and modulus of elasticity conforms to ACI 318.

Structure Elements Parameters

Compressive strength; fcu. Beams, Slabs, Columns 25N/mm

Density of concrete 24kN/m

Modulus of elasticity; E 21.718KN/mm

Yield stress fy 420N/mm Table 2.3 Material Properties conform to ACI318 code

2.9 PROCEDURES OF DETERMINATION OF LOADING.

The simulation of load determination on members of the structure on three dimensional structural

frames was used; the procedure utilizes load analysis to find the dimension of members to be used

later on finding the optimal design. Dead load and live load were applied to the structure.

2.9.1 Determination Dead Load.

All of the dead loads are according to the (ACI318) Code. It is defined as the sum of all constant

and continuous loads occurred on the building

Which represents:

Own weight of structure

Floor covering

Wall loads

Flooring cover

Flooring cover represents the weight of finishing materials on floor, such as sand, bitumen, mortar

and marble. Table 3.8 shows the details of dead load on floor and surface slabs.

DEAD LOAD (FLOORING COVERING )

FROM TYPE MAGNITUDE UNIT

floor slab Area pressure 1.00 KN/m2

Surface slab( Roof) Area pressure 2 KN/m2 Table 2.4 Details of dead load on surfaces as component of concrete slab


16

Own weight of structure

Own weight of structure represents the weight of the main elements of the building,

such as slabs, beams and columns. Table 3.3 shows the details of slabs weight according to

ACI318-08 codes.

DEAD LOAD

FROM TYPE MAGNITUDE UNIT

Slab self weight of

200 mm thickness

(without finishes)

Area pressure 4.2 KN/m2

Table 2.5 Details of slab self-weight according to ACI318 Code Wall loads:

The wall in the building is from concrete blocks, the thickness of wall is 0.25m for

exterior wall and 0.2m for interior wall, therefore, the load of wall on beam will be:

For exterior walls:

H = 4m

W= 0.25 X 4 X18 + 0.02 X 4 X 24 =19.92 KN/m

For interior walls:

H = 4m

W= 0.2 X 4 X18+ 0.02 X 4 X 24 = 16.32 KN/m

2.9.2 Partial Safety Factors According To ACI 318-02 CODE

The strength reduction factors ,φ,are applied to specified strength to obtain the design strength provided by a member .the φ factors for flexure ,shear ,and torsion are as shown in Table 2.6.


17

Φ=0.9 for flexure (tension controlled)

Φ=0.75 For shear and torsion.

Φ=0.65 For axial compression (columns)

Table2.6: Partial safety factor according to ACI318-02

2.9.2.2 Determination Live Load

It is defined as the sum of all variable movable loads occurring in the building.

This represents:

Human weights

Furniture weights

Type of building (office building) LOAD

(KN/m2 )

Catwalks for maintenance access 1.92

Access floor system

Office use

2.4

Computer use 4.79

File and computer rooms shall be designed for heavier loads based on

anticipated occupancy

Lobbies and first-floor corridors

4.79

Offices 2.40

corridors above first floor 3.83

Balconies (exterior) 4.79

Catwalks for maintenance access 1.92

Private rooms and corridors serving them 1.92

Public rooms and corridors serving them 4.79 Table 2.7 American standard Design Minimum Loads for Building.

Note: Refer to ASCE 7-05 Section 4.9 (pg 12), Table 4.1


18

2.9.3 Design load combinations:-

Load factors according to ACI Code.

The design load combinations are the various combination of the load cases for which the structure

needs to be designed. For ACI 318-08 if a structure is subjected to dead load (D), live load (L), the

required strength U to resist dead load D and live load L shall not be less than Combination factors

(LRFD):

U = 1.4D

U = 1.2D + 1.6L

Where:

o D = dead load;

o L = live lopad.

2.10 Design Methods

There are two acceptable methods to design concrete: the working stress method and the

ultimate strength method (Wight & jamesr, 2005) (Mehdi & Robert, 2007). The ultimate

stress method is the one most commonly used. The reasons for this are the ultimate strength

method will require substantially less concrete and rebar, and the design calculations are

easier. Working stress design model assumes that as the concrete beam bends due to induced

moments the strain relationship between the rebar in tension and the concrete in compression

remain constant. Ultimate strength design places the rebar in full yield so the strain

relationship between reinforcement and concrete is ignored and a rectangular concrete

compression block stressed at design strength is formed.

2.10.1 Object of Structural Design

The permissible stress and ultimate strength methods have served their purpose over the

years. However, the engineers have always realized the shortcutting of these methods and

been on the outlook for improvements in the process of design.

The purpose of design may be stated the provision of a safe and economical structure

complying with the clients’ requirements (Rowe et al., 1995). In other words, the process


19

of design should ensure a balance between total cost of the structure and an acceptable

probability of the structure becoming unserviceable during its life. Limit state design is

based on this philosophy. It recognizes the need to provide a safe and efficient structure at

an economical price. Simultaneously, it gives clear idea of actual factors of safety used to

take into account elements of uncertainty and ignorance.

2.10.2 Philosophy of Limit State Design

Limit state design takes account of the variations and uncertainties that may occur in

the design and construction of structures. Different safety factors are provided for those

variations in design and construction. Safety and serviceability are expressed in terms of

the probability that the structure will not beware unfit for its intended pur-pose during its

life. Limit state for use may arise in various ways, the principal ones being as (Mehdi &

Robert, 2007) (Wight & jamesr, 2005):

(1) Ultimate limit states: the usual collapse limit. States including collapse due to fire,

explosive pressure etc.

(2) Serviceability limit state: focal damage and deflection limit states, durability, vibration,

ere penetration and heat trans-mission etc.

Limits states of collapse may be defined as occurring when a part or the whole of the

structure fails under extreme loads. It may be due to rupture of one or more critical sections, loss

of overall stability or buck-ling owing to elastic or plastic instability.

Limit states due to local damage may occur, when cracking or spalling of concrete impairs the

appearance or usefulness of the structure or adversely affects finishes, partitions etc. For example,

a check on the limit state of crack width may be necessary in water retaining structures or structures

situated in severe environments. Similarly, it may be necessary to check the limit state of crack

formation in compression to ensure that no initial microcracking, which could be harmful to the

durability of the member, is produced at any stage of construction in zones subject to high

compressive stresses.

Limit states of deflection or deformation may be defined as occurring when it becomes

excessive to impair the appearuruce or usefulness of the structure and may cause discomfort to

users. In certain cases limit states of other effects such as vibration, fatigue, impact, durability of


20

fire damage may also have to be considered: For example, the limit states design of bridges

requires the investigation of limit states of vibration and fatigue in addition to collapse, cracking

and deflection (Mehdi & Robert, 2007). Similarly, the consideration of limit states of impact

resistance is essential for structures, which may be subjected to impact, explosions or earthquakes.

The usual approach is to design the structure because of limit states for collapse and then check

that the criteria governing remaining limit states are satisfied. The limit state of collapse under

extreme loads is investigated by ultimate strength theory of reinforced concrete, while the limit

states of deflection and local damage both utilize the elastic theory (Mehdi & Robert, 2007).


21

2.11 Project flow chart.

Error

Flow chart 1 : Design procedures

START

Generate Plan in AutoCAD Software

Assigning Of Elements Based On assumed Structure Plan

General a 3D Model

Correct any errors, Recheck Properties of Elements

Assign Loads and Load Cases Acting On the Structure

Decrease Spacing between members or

add new members(such as column )

Run Analysis

Run R.C member Required Reinforcement

Calculations

Run Provided Reinforcement wizard & Check The Steel Ratio ρ.

Change the dimensions of section

Ρmax< ρ< ρmin Ρmax< ρ< ρmin

Report Design Results

Import the 2d plan to 3d software to process

architectural visualization and

establish ensuring it compliance with structural midel

Import Structure Plan into (Autodesk Robot Structure Analysis)

Architectural output Drawings


22

2.10.1 EXPECTED RESULT.

1- The most economical alternative solution would be identified.

2- The required quantity of material would be evaluated.

Design of Reinforced Concrete Multi-story Commercial building ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

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3. Design of Slabs

3.1.1 Definition: - A slab is structural element whose thickness is small compared to its own length and width. Slabs are usually used in floor and roof construction. According to the way loads are transferred to supporting beams and columns, slabs are classified into two types; one-way and two-way

3.1.2 Introduction

The slab provides a horizontal surface and is usually supported by columns, beams or walls. One-way slab is the most basic and common type of slab. One-way slabs are supported by two opposite sides and bending occurs in one direction only. Two-way slabs are supported on four sides and bending occurs in two directions. One-way slabs are designed as rectangular beams placed side by side.

3.1.3 DESIGN CONCEPTS:

An exact analysis of forces and displacements in a two-way slab is complex, due to its highly indeterminate nature; this is true even when the effects of creep and nonlinear behavior of the concrete are neglected. Numerical methods such as finite elements can be used, but simplified methods such as those presented by the ACI Code are more suitable for practical design. The ACI Code, Chapter 8, assumes that the slabs behave as wide, shallow beams that form, with the columns above and below them, a rigid frame. The validity of this assumption of dividing the structure into equivalent frames has been verified by analytical and experimental research. It is also established that factored load capacity of two-way slabs with restrained boundaries is about twice that calculated by theoretical analysis because a great deal of moment redistribution occurs in the slab before failure. At high loads, large deformations and deflections are expected; thus, a minimum slab thickness is required to maintain adequate deflection and cracking conditions under service loads.

However, slabs supported by four sides may be assumed as two-way slab when the ratio of lengths to width of two perpendicular sides exceeds 2. Although, while such slabs transfer their loading in four directions, nearly all load is transferred in the short direction. Two-way slabs carry the load to two directions, and the bending moment in each direction is less than the bending moment of one-way slabs. Also two-way slabs have less deflection than one-way slabs.



Compared to one-way slabs, Calculation of two-way slabs is more complex. Methods for two-way slab design include Direct Design Method (DDM), Equivalent frame method (EFM), Finite element approach, and Yield line theory. However, the ACI Code specifies two simplified methods, DDM and EFM.

Slabs maybe solid of uniform thickness or ribbed with r ibs running in one or two directions. Slabs with varying depth are generally not used. Slab are horizontal plate elements forming floor and roof in building and normally carry lateral actions.

3.1.4 Types of Slabs

Ribbed slabs: Slab cast integrally with a series of closely spaced joist which in turn are supported by a set of beams. Designed as a series of parallel T-beams and economical for medium spans with light to medium live loads.

Waffle slabs: A two-way slab reinforced by ribs in two-dimensions. Able to carry heavier loads and span longer than ribbed slabs.

Flat slabs: Slabs of uniform thickness bending and reinforced in two directions and supported directly by columns without beams.

Flat slabs with drop panel: Flat slab thickness at its column supports with column capitals or drop panels to increase strength and moment-resisting capacity. Suitable for heavily loaded span

3.1.5 One & two way slabs outlined:

One-way slabs

1. One-way Beam and slab / One-way flat slab: These slabs are supported on two opposite sides and all bending moment And deflections are resisted in the short direction. A slab supported on Two sides with length to width ratio greater than two, should be designed As one-way slab.

2. One-way joist floor system:

This type of slab, also called ribbed slab, is supported by reinforced Concrete r ibs or joists. The ribs are usually tapered and uniformly spaced And supported on girders that rest on columns.

Two-way slab

1. Two-way beam and slab: If the slab is supported by beams on all four sides, the loads are transferred to all four beams, assuming rebar in both directions.

2. Two-way flat slab: A flat slab usually does not have beams or girders but is

supported by Drop panels or column capitals directly. All loads are



transferred to the Supporting column, with punching shear resisted by drop panels.

3. Two-way waffle slab: This type of slab consists of a floor slab with a length-to-width ratio less Than 2, supported by waffles in two directions.

Fig. 3-1: Typical type of slabs (ACI,1994)



3.1.6 ECONOMICAL CHOICE OF CONCRETE FLOOR SYSTEMS

ACCORDING TO SIZE, DIMENSIONS AND LOADINGS REQUIRED:

Various types of floor systems can be used for general buildings, such as residential, office, and in institutional buildings. The choice of an adequate and economic floor system depends on the type of building, architectural layout, aesthetic features, and the span length between columns. In general, the superimposed live load on buildings varies between 5 and 10 KN/m. A general guide for the economical use of floor systems can be summarized as follows:

1. Flat plates: Flat plates are most suitable for spans of 6m to 7.5m and live loads between 4 and 6.5 KN/m. The advantages of adopting flat plates include low-cost formwork, exposed flat ceilings, and fast construction. Flat plates have low shear capacity and relatively low stiffness, which may cause noticeable deflection. Flat plates are widely used in buildings either as reinforced or prestressed concrete slabs.

2. Flat slabs: Flat slabs are most suitable for spans of 6mto 9m and for live loads of 5.5 to 10 KN/m they need more formwork than flat plates, especially for column capitals. In most cases, only drop panels witho ut column capitals are used.

3. Waffle slabs: Waffle slabs are suitable for spans of 9m to 14.5m and live loads of 5.5 to 10 KN/they carry, heavier loads than flat plates and have attractive exposed ceilings. Formwork, including the use of pans, is quite expensive.

4 . Slabs on beams: Slabs on beams are suitable for spans between 6m and 9m and live loads of 4 to 8 KN/m. The beams increase the stiffness of the slabs, producing relatively low deflection. Additional formwork for the beams is needed.

5. One-way slabs on beams: One-way slabs on beams are most suitable for spans of 0.9 to 1.8m and a live load of 4 to 7KN/m. They can be used for larger spans with relatively higher cost and higher slab deflection. Additional formwork for the beams is needed.

6. One-way joist floor system: A one-way joist floor system is most suitable for spans of 6 to 9 m and live loads of 5.5 to 8.2 KN/m, Because of the deep ribs, the concrete and steel quantities are relatively low, but expensive formwork is expected. The exposed ceiling of the slabs may look attractive.



3.1.7 Calculation of thickness for one way slab:

Table 3.1 Minimum thickness of beams

Table 3.2 Minimum thickness of beams for exterior panels



3.1.8 Design Procedure:

One-way slab design

1. Decide the type of slab according to aspect ratio of long and short side

Lengths.

2. Compute the minimum thickness based on ACI Code.

3. Compute the slab self-weight and total design load.

4. Compute factored loads (1.4 DL + 1.7 LL).

5. Compute the design moment.

6. Assume the effective slab depth.

7. Check the shear.

8. Find or compute the required steel ratio.

9. Compute the required steel area.

10. Design the reinforcement (main and temperature steel).

11. Check the deflection.

3.1.9 The ACI Code specifies two methods for the design of two-way slabs:

1 . The direct design method, DDM (ACI Code, Section 8.10), is an approximate procedure for the analysis and design of two-way slabs. It is limited to slab systems subjected to uniformly distr ibuted loads and supported on equally or nearly equally spaced columns. The method uses a set of coefficients to determine the design moments at critical sections. Two-way slab systems that do not meet the limitations of the ACI Code, Section 8.10.1.1, must be analyzed by more accurate procedures.

2 . The equivalent frame method, EFM (ACI Code, Section 8.11), is one in which three-dimensional building is divided into a series of two dimensional equivalent frames by cutting the building along lines midway between columns. The resulting frames are considered separately in the longitudinal and transverse directions of the building and treated floor by floor.



3.1.10 Two-way slab design procedure by the Direct Design Method

1. Decide the type of slab according to aspect ratio of long and short side Lengths.

2. Check the limitation to use the DDM in ACI Code. If limitations are not

met, the DDM cannot be used.

3. Determine and assume the thickness of slab to control deflection.

4. Compute the slab self-weight and total design load.

5. Compute factored loads (1.4 DL + 1.7 LL).

6. Check the slab thickness against one-way shear and two-way shear.

7. Compute the design moment.

8. Determine the distr ibution factor for the positive and negative moments using ACI Code.

9. Determine the steel reinforcement of the column and middle strips.

3.1.11Classification of slabs:

Slabs are plate elements forming floors and roofs in buildings which normally carry uniformly distr ibuted loads.

Slabs may be simply supported or continuous over one or more supports and are classified according to the method of support as follows:

One-end continuous

Both-End continuous

3.1.12 Purpose of main and secondary steel: The distr ibution steel should be tied

above the main steel, otherwise the lever arm which is measure up to the center of the main steel shall be reduced resulting in the reduction of the moment of the resistance

Purpose of Main steel:

It takes up all the tensile stresses developed in the structure

It increase the strength of concrete sections

Purpose of distribution steel:

It distribute the concentrated load on the slab

It guards against shrinkage and temperature stress

It also keeps the main reinforcement in the position



3 .1.13 Types of analysis-methods for slabs:

Elastic analysis covers three techniques:

(a) Idealization in to strips or beams spanning one way or a grid with the strips spanning two ways

(b) Elastic plate analysis

(c) Finite element analysis: (Used By the software Robot analysis in this project)

The best method for irregularly shaped slabs or slabs with non-uniform loads

Method of design coefficients use is made of the moment and shear coefficients given in the code, which have been obtained from yield line analysis.

The yield line and Hillerborg strip methods are limit design or collapse loads methods

3.1.14 Slabs direction In Ribbed Slab

Direction of one way slab : In one-way ribbed slabs ribs may be arranged in any of the two principal directions. Two options are possible; the first is by providing ribs in the shorter direction as shown in Figure a, which leads to smaller amounts of reinforcement in the ribs, while large amounts of reinforcement are required in the supporting beams, associated with large deflections.

`



The second option is by providing ribs in the longer direction as shown in Figure b, which leads to larger amount of reinforcement in the ribs, while smaller amounts of reinforcement are required in the supporting beams associated with smaller deflections compared to the first option. The designer has to make up his mind regarding the option he prefers. Some designers opt to run the r ibs in a direction that leads to smaller moments and shears in the supporting beams which means much more reinforcement in the ribs. Other designers opt to run the ribs in the shorter direction which leads to much more reinforcement in the supporting beams. The later option leads to more economical design.

3.1.15 Design Concept:

One-way solid slabs are designed as a number of independent 1 m wide strips which span in the short direction and supported on crossing beams.

Practical rules:

THE overall thickness of a slab shall not be less than 7.5 cm, the top surface of centering shall be given a camber of 7mm per meter span subject to maximum of 4.5 cm.

Reinforcements: the minimum reinforcement in slabs in either direction shall be not less than 0.15 percent of the gross sectional area of the concrete and which may be 0.12 percent where high yield strength deformed bars .

3.1.16 Maximum Reinforcement Ratio:

One-way solid slabs are designed as rectangular sections subjected to shear and moment. Thus, the maximum reinforcement ratio corresponds to a net stain in the reinforcement, e of 0.004.

3.2.17 Shrinkage Reinforcement Ratio

According to ACI Code 7.12.2.1 and for steels yielding at f 4200 kg / cm2 y = ,the

Shrinkage reinforcement is taken not less than 0.0018 of the gross concrete area, or

A= b h ; shrinkage = 0.0018.

Where, b = width of strip, and h = slab thickness.



3.2.18 Loads Assigned to Slabs

(1) Own weight of slab:

The weight of the slab per unit area is estimated by multiplying the thickness of the slab h by the density of the reinforced concrete.

(2) Weight of slab covering materials:

This weight per unit area depends on the type of finishing which is usually made of

- Sand fill with a thickness of about 5 cm, 0.05 × 1.80 t/m2

- Cement mortar, 2.5 cm thick. 0.025 × 2.10 t/m2

- Tiling 0.025 × 2.30 t/m2

- A layer of plaster about 2 cm in thickness. 0.02 × 2.10 t/m2

(3) Live Load:

It depends on the purpose for which the floor is constructed. Shows typical values used by the Uniform Building Code (UBC).

Note: During the analysis of the 3d frame of the building in this project, we assumed a uniformly distributed planar live load of 5kN per meter square (as the building falls in the whole sale stores category.



3.2.1 Plans showing the assigned slab names and direction for different stories:-

Fig 3.1.1 Elevation plan for slab of ground floor

Fig 3.1.2 Elevation plan for slab on first floor



Fig 3.1.3 Elevation plan for slabs on second floor







3.2 Steps for design of two way solid slab: Find the moment coefficients in each slab:

For continuous edges ( -Ve moments ):

2

2

bL . ut W. ) aC ( b) ve- M (

aL . ut W. ) aC ( a) ve-M (

neg

neg

Span moments ( +Ve moments ):

2

2

bL ] u W. )bC ( W. )b(C [ b) ve M (

aL ] u W. ) aC ( W. )a(C [ a) ve M (

LLLuddL

LLLuddL

For discontinuous edges ( -Ve moments ):

3 / b) ve M ( b) ve- M ( 3 / a) ve M ( a) ve- M (

Effective Depth (d):

stdc.cuhd

Percentage of steel (ρ):

mm 1000b ,d b ρ sA

minρyf

ω ρ

0.113yfρ For

mm 450 s2h

dρ)s(A

bars between SpacingS

0.113yfρ Check

dsh

0.002minρ

dyf840uM

ρ

'c

'c

'c

f

f

f

bar one

2

OK :

M(-Ve)

+

- -

M(+Ve)



3.2 Data for design:- Firstly: Defining the sample slab for design illustration Secondly: structural analysis design of the Unit 3.2.1 The slabs S2 and S8 are taken as design samples which are assumed to be solid slab , as

shown in fig given below:

3.2.1.2 Design Concept:

One-way solid slabs are designed as a number of independent 1 m wide strips which span in the short direction and supported on crossing beams.

Practical rules:

THE overall thickness of a slab shall not be less than 7.5 cm, the top surface of centering shall be given a camber of 7mm per meter span subject to maximum of 4.5 cm.

Reinforcements: the minimum reinforcement in slabs in either direction shall be not less than 0.15 percent of the gross sectional area of the concrete and which may be 0.12 percent where high yield strength deformed bars .



3.3 Firstly, let’s consider Slab S2 (Two-Way Slab) :

Fig 3.3 shows Slab S2 which is a two-way slab 3.3.1 Determine the thickness of the solid slab S2 :

mm.)

.(

.uh

slab)way solid-(two ..7.506.50m if else

way slab)-(one then .LbLam if

mm)M(

lbsh

03183

860634

1000507

50850

50

100634

USE hs =200mm _________________________eq 3.1

1692

1225200

d

stdc.cuhd

3.3.1.2 Calculation of loads on slabs S2:

2KN/m 17.67uW

.77741m/KNW

L L 1.7DL .uW

27.77KN/m2.97 24 W

LL

D

flooring h

W cs

D

47124

411000200

1000



3.3.2 Moments at short direction: For Discontinuous edge

m.KN..M

a)veM(M

M

2383

7241

31

32

1

For mid-span

m.KN.. ]....[

La]ulWll)Ca(UDWDl[(Ca) 2M

724250686040078100290

2

For Continuous Edge

m.KN...).(M

aLutWneg)Ca(3M

58362506671704903

2

Slab Case m DL LL -Ve

S2 8 0.85 Ca=0.029 Ca=0.040 Ca=0.049

Cb=0.017 Cb=0.022 Cb=0.046

Table (3-4) moment coefficients

3.3.3 Moments at Long direction: -

d=200-25-1.5*12=157mm

For continuous edge

m.KN...).(M

aLutWneg)Cb(M 4

7245250767170460

2

4

For Mid-span

m.KN.. ]....[M

La]ulWll)Cb(uDWDl[(Cb) M

5

5

7218250786022078100170

2

For continuous edge

46 MM



3.4 Design for flexure:-

3.4.1 Reinforcement At Short direction:

A . (La-discontinuous): (-ve) M1

'12/320mm/m Use

320mm SUse

hsmm..

S

2D sbA

450mm2hs d

sbAS

mm.dbstA

value new for need) (NO O.K ...

minFyFc use then

.fcyf

for

min Use

reqmin

control..dsh.min

..dFy

uM

:check

2937713000240

113

1134

2124

2299130100000230

11300302542000230

1130

0023016920000200020

000802169420840

6102382840

610



B. At Short direction (La-middle) : (+ve) M2

12/270mm Use270mm Use

mm..d/d S

450mm2hs

dbar one sA

S

O.K ...

.fcyf

req use ,minreq

..dsh.min

.req

2

:check

.

5927816900240

1134

11300402542000240

1130

0023016920000200020

002402169420840

610724

C. At short direction (La right-edge continuous): M3 (-ve)

12/180mm Use

7.1851690036.0

113

113.0060.0254200036.0

use ,min

0023.0min

0036.02169420840

:

61058.36

mmS

reqreq

req

check



3.4.2 Reinforcement At long direction

A. (Lb-edge) continuous: M6-ve

12/100mm Use

82.102157005.0

113

O.K 113.0084.025

420005.0

:

0025.0157

200002.0002.0min

005.02157420840

61072.45

mmS

check

dhs

req

B. At long direction (Lb-mid span): M5+ve

12/280mm Use

2871570025.0

113

0025.0min use min

,0025.0min

0021.02157420840

61072.18

mmS

req

req

C. At long direction (Lb-cont edge): M4-ve

12/100mm Use

use HenceMM

64



Direction sec )/(uM mKN d(mm) min req use (req)S usedS

short

1 8.23 169

0.0023 0.0008 0.0023 377 320 2 24.7 0.0023 0.0024 0.0024 278.5 270 3 36.58 0.0023 0.0036 0.0036 185.7 180

long 4 45.72

157 0.0025 0.005 0.005 102.8 100

5 18.72 0.0025 0.0021 0.0025 287 280 6 45.72 0.0025 0.005 0.005 102.8 100

Table (3-4) two- way slabs reinforcement



3.5 Design of one-way slab : 3 .5.1.1 Minimum Reinforcement Ratio

According to ACI Code 10.5.4, the minimum flexural reinforcement is not to be less than the shrinkage reinforcement, or A b h s 0.0018 min ³ .

3.5.1.2 Spacing of Flexural Reinforcement Bars

Based on ACI 10.5.4, flexural reinforcement is to be spaced not farther than three times the slab thickness, nor farther apart than 45 cm, center-to-center.

3.5.1.3 Spacing of Shrinkage Reinforcement Bars

Based on ACI 7.12.2.2, shrinkage reinforcement is to be spaced not farther than five times the slab thickness, nor farther apart than 45 cm, center-to-center.

3.5.1.4 Now Let’s consider Slab S8 (One-Way Slab)

Fig 3.5 shows one –way slab named S8

Determine the thickness of the solid slab :

mm.)

.(

.sh

slab) way solidone(..m.m.m

.bLaL

m

mm)

m6 (34

bLsh

8116

3010634

1000306

503010306901

50

100

Use slab thickness from largest span hs= 200mm ( Eq 3.1 )



3.5.2 Calculation of loads on slabs S8:

2KN/m 17.67uW

.77141uWm/KNW

L L 1.7DL .uW

2m\7.77KN/2.97 24 W

LL

D

floowing h

W cs

D

471

24

411000200

1000

3.5.3 Moments at short direction: - For continuous edge

m.KN...

29nL uWM 087

929167172

7

For Mid-span

m.KN...nLuWM 5414

291671714

28

For continuous edge

m.KN...nLuWM 652

242916717

24

29

Reinforcement: 3.5.4 Minimum Steel

002350

3103721702000020

0020

1702

1025200

.min

.min

)(.min

)dsh

(.min

mmd



A . Reinforcement For edge-moment M7:

002350

0006404202501080

01090181

0108036211

01080217010002590

610087

0877

1000

.min

min..fy

'cff

..

).(.

..

.uK

d 'cf

uMuK

m.KN.M

Use

2

190/m' / 5 Ues

mmmm..reqS

mm.2D

SA

SAbAS

mmm..SA

dbSA

S

190419641000578

25784

2104

1000

253991000170002350

B. Reinforcement for Mid-span:

0.00235 min

min0.00041.yf

'cfρ

0.00701.18

1 ω

0.0089951000200.9

101.45uK

kN.m.8M

Ues

).(.

2

6

4202500700

54

006903621



dbsA

10/190/m' 5 Use

mmmm..

.S

AbA

S

'mmm..A

S

S

S

19041965399

1000578

1000

539910001700023502

C . Reinforcement For edge-moment M9 :

10/200/m' 5 Use

mmmm.S

'mmm..dbSA

.min Use

min..fy

'cf

..

).(.

..

.uK

m.KN.9* M

2004196

253991000170002350

002350

000204202500410

00410181

0040036211

00400217010002590

610652

652



3.5.5 Secondary reinforcement:

mm/m'120/10S Usemm120S Use

67.125400

100027.50S

27.504

814.34D

1000S

m'2mm4002001000 0.002sA

sbh 0.002sA

222

mm

mmbsA

tsA

bsAt

t

Section 1 2 3

uM 7.08 4.5 2.65

uK 0.0108 0.0069 0.0040 ω 0.0109 0.0070 0.0049 ρ 0.00235 0.00235 0.00235 /msA 399.5 399.5 399.5

(req)S 196.4 196.4 196.4

(used)S 190 190 190

Table (3-5) reinforcement of one-way solid slab

Design of Reinforced Concrete Multi-story Commercial building ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ 50 Design Of Beams

4. Design of Beams

4.1 Introduction

The beams are a basic component of reinforced concrete structures , the beams

carries and transfers the loads from the slabs and walls to the columns and then to the

foundations. The beams should be correctly restrained and appropriate studies and analysis

should be done to overcome and resist the moments and shrinkage and other deformations

resulted upon loading. The explanation of design is shown firstly through formulas and

then a sample (continuous beam) is taken and is designed, the related moments and shears

forces acting upon it is calculated through Autodesk Robot structural analysis

4.2 Structural Theory of beams:

4.2.1 Types of beams:

There are many ways in which the beams may be supported, some of the more

common methods are given below ,

Fig. 4.1

The first beam in Fig is called a simply supported, or simple beam. It has Supports near its

ends, which restrain it only against vertical movement. The ends of the beam are free to

rotate. When the loads have a horizontal component, or when change in length of the beam

due to temperature may be important, the Supports may also have to prevent horizontal

motion. In that case, horizontal restraint at one support is generally sufficient. The distance

between the supports is called the span. The load carried by each support is called a reaction.



The beam which is a cantilever. It has only one support, which restrains it from rotating or

moving horizontally or vertically at that end. Such a support is called a fixed end.

When a beam extends over several supports, it is called a continuous beam

For flexural design of R/C rectangular section beams, there is a number of steps procedure

and equations provided by ultimate strength design method according to ACI-code. The large

number of equations and fork of solution steps causes a lot of confusion and boredom for

student or designer

4.2.2 Scope of Usage:

Concrete beams are widely used as a primary members to con-struct buildings.

Robot Structural analysis is widely used as a structural analysis and design program.

Sometimes, to give design more confidently, we need to compare the results of program

with the results of manual calculations. Most common beams, which are rectangular

section and T-section, apply vertical loads (dead and live loads).

Beams, like that, will be subjected to the bending moment, shear force and torsional

moment. This study focuses on beams of rectangular section considering bending

moment only. Flexural created by bending moment, makes the beam in case of tension

or com-precision failure.

4.2.3 Relation of Reinforcement place with section properties on loading:

In case of tension failure, provided reinforcement ratio (ρ) at tension zone is

less than balanced reinforcement ratio ρbal . So that, steel will reach to the yield

stress (훿) and strain (휀). While, concrete at compression zone has not yet

reached to the ultimate strain (휀=0.003).

In case of compression failure, provided (ρ) at tension zone is greater than ρbal .

Concrete at compression zone will reach ultimate stress ( 훿ult) and strain (휀 ult),

While steel has not yet reached . For each case, there is a number of design

equations derived. To identify those equations, textbooks can be reviewed for

the principles of the design of reinforced concrete.



Usually,

” longitudinal reinforcement is used for flexural strength in addition to flexural

resist by concrete in compression zone.”

Whereas,

“ Secondary reinforcements (Stirrups ) are used to accompany the concrete ability to

resist shear force , mostly intense at ends of beams and at meeting point of spans or

sometimes at mid span of very long beams “

4.2.4 Design of Beams

In the beginning, unification of the dimensions of the sections in the beams and the

parameters such as density of concrete, specified compressive strength of concrete , specified

yield strength of steel is done .we used ultimate strength theory as basic of design. According

to the following information, we must design beam firstly for flexure and then for shear.

4.2.5 Reinforced Concrete Flexure

The theory of flexure for reinforced concrete is based on three basic assumptions. Which are

sufficient to allow a person to calculate the moment resistance of a beam. These are presented

first and used to illustrate the behaviour of a beam cross section under increasing moment.

Following this, four additional simplifying assumptions from the ACI code are presented to

simplify the analysis for practical application (Mehdi & Robert, 2007).

4.2.6 Required Strength and Design Strength

The basic safety equation for flexure is “Factored resistance ≥ factored load effects

”u≥ M n or Φ M

is the moment due to the factored loads, which the ACI code refer to as the uWhere M

required ultimate moment. This is a load effect computed by structural analysis from the

he refers to t ngoverning combination of factored loads given in ACI section9.2. The term M

nominal moment capacity of a cross section computed from the nominal dimensions and

specified material strengths. The factor Φ is a strength reduction factor (ACI section 9.3) to



account for possible variations in dimensions and material strengths and possible inaccuracies

in the strength equations.

In ACI318 ultimate moment required as flowing:-

훷푀푛 = 0.85푓푐′. 푎. 푏(푑 −)

4.2.7 Shear in Beams

When loads applied to beams produce not only bending moment but also internal shear

forces. In the reinforced concrete beams, the primary longitudinal bending reinforcement is

usually considered first. This leads to the size of the section and the arrangement of the

reinforcement to provide the necessary moment resistance. Limits are placed on the amount

of bending reinforcement to ensure that if failure were ever to occur, it would gradually,

giving warning to the occupants (Mehdi & Robert, 2007).

Once the primary longitudinal reinforcement has been determined, then the reinforced

concrete beams are designed to resist the shear forces resulting from the various combinations

of ultimate loads. Most of shear failure is frequently sudden and brittle, hence the design for

shear must ensure that the shear strength equals or exceeds the flexural strength at all points in

the beam. The manner in which shear failure can occur varies widely depending on the

dimensions, geometry, loading and properties of the members (Mehdi & Robert, 2007).

4.2.8 Design of Reinforced Concrete Beams for Shear

In the ACI code, the basic design equation for the shear capacity of slender concrete beams is:

ΦVn ≥ Vu

Where, Vu is the shear force due the factored loads; Φ is a strength-reduction factor. The

nominal shear resistance is

Vn = Vc + Vs

Where Vc is the shear carried by the concrete and Vs is the shear carried by the stirrups.



4.3 Design Assumptions :

mm 8 = std

mm 16 = bd mm 25 = Cover

MPa420 = yf MPa25 = 'cf

4.4 Loading Data:

The beam carries load from following:

1- Load from wall

1m' wallof thickness wallofheight γWb

2- Concentrated load from other beams & columns ( kN )

3 - Own weight of beam

4 - Loads from one way solid slab

kN/m 2

slabofspan)(kN/muWW 2

5 - Load from two way solid slab

C.L)to(C.LSpanShort S

Span LongSpanShort m

2m3

3SuWWBeam Long

3SuWWBeamShort

2

(kN/m) bhγWc



4.5 Design procedure:

1- Design of flexure:

maxmin ifyf'cfωρ

1.18uk2.3611

ω

. , db'cf

uMuk

2

db tsd cover h d

90610

yf600600β

yf

'cf

0.85b

b.max also

1

70

dbρsA

fy

'fcorfy

1.4 of max themin also

0.65 .65 0.85 where28) - '(fc 0.007-0.85

4

fy

fcor 4'

fy1.4 ofmax themin also

OR

'fc./Ru(Fy

'fc.db

MuRu

850211850

2



4.6.1 Flow-chart for design of flexure for singly reinforced beam:

Flow-Chart: 1



4.6.2 Flow-chart for design of flexure for doubly reinforced beam:

4.7 Sample of design:-



Fig 4.2 . B.M.D for Beam 59 from Robot structural analysis



Fig 4.3. Spans and sections for beam 59

4.7.1 Design Assumptions

mm 400 Hmm 250 bmm 8 = std

mm 16 = bd

mm 25 =Cover

MPa 420 = yf MPa 25 = '

cf

4.7.2Check deflection:

m././.. L6.30m0.25/2-0.25/2-6.30 L

m6.50.25/2-0.25/26.75nL

nn

50622502250756

4.7.2 Sizing the cross-section:

Per ACI Table 9.5(a), minimum thickness = L/18.5 (For Fy= 420Mpa)

Note: We can use hmin from considering longest span=6.50m

mmh 35.3515.18

6500min

hmin < h assume

351.35 < 400 mm _______________(OK)

Use hmin = 400 mm ـــــــــــــــــــــــــــــــــــ (1)ــــــــــــــــــــــــ

Also,

b=d/2

b=400/2

b=200mm



4.8 Design of flexure:

4.8.1 Actual depth :

359mmd ــــــــــــــــــــــــــــــــــــــــــــــــــ(2)

mm 3592

16825400d

2bd

d.chd st

c

4.8.2 Minimum Ratio of Steel Required for section:

mm .0 d

b )(.42025 0.85

(3) - ------------------------- 0.85 use..

.)(..

)fc(.. '

0252420600

600850

8506508710

28250070850280070850

(5) ----------------------

r4201.4

0.0033 min

0.0029] or .[ of max

(4) ---------------------- .0.01890.02520.75 b .

min

max

max

00330

01890750

4204250



4.8.3 Design of Reinforcement for every moment at different places in beam :-

1) Section 1-1 from Robot Analysis , (Mu) –ve =-33.03 KN.m

16 2

24714

216

175296

4

2175296

17596

0030

0027001890

00330

045500455033

0333

, use Hence

Use

bar.

.d.

As(bar)As bars of Number

.23592500.0033dbρsA

.ρ

maxρρminρ-(6)--------------------- .actρ

4) eq -(From--------------------- .maxρ5) eq -(From--------------------- . minρ

0.0027420250.046

fyfcρ

0.0461.180.0455)(2.3611ω

.Ku.0(359)250250.9

10.3uK

m.KN .ve-(Mu)

min

2

6



2 .Check Nominal moment capacity of section

a= . ∗ `∗

=∗ ∗ ∗

. ∗ ∗= 31.77푚푚

푀 = 퐴 퐹 푑 −푎2

=401.9 * 420*(359- . ) ∗ 10

=57.91 KN.m

푀 = ∅푀

= 0.9 *57.91= 52.12 KN.m

푀 = 52.12퐾푁.푚 > 33.03(푂퐾)

3. Spacing:

25mm

푆 푑 = 16

max 푠푖푧푒표푓퐴푔푔 ≈ 25푚푚

S= ( . )

S= ( )

S= 92 mm c/c

4. Check cracking:-

푓 =0.6푓 = 252

푑 = 41푚푚

퐴 = 2푑 ∗ 푏= 2*41*250 = 20500 푚푚

Number of Bars =2

A= = = 10250푚푚

5. Check for Exposure:

Z= 푓 ∗ 푑 ∗ 퐴 = 252 ∗ √41 ∗ 10250 ∗ 10



Z=18.87 <25KN for exterior exposure

Z=18.87 <30KN for interior exposure (O.K)

2) From Robot Analysis Section 2-2, (Mu) +ve =75.34 KN.m

16 3 Use

mm .53592500.0066

maxρreqρminρ

0.0066420250.1111

fy

'fcρ

0.11111.18

0.1039)(2.3611ω

0.1039Ku(359)250250.9

10.7uK

..AAs bars of NO

bdA

. use0.0189 0.0066 0033.0

bar

S

req

total

2

2

6

3942

41635592

00660

2

3592

345

3 ) From Robot Analysis Section 3-3, (Mu) -ve =-96.61 KN.m

bars 6Use

mm .73592500.0086db sA

...maxρreqρminρ

0.0086420250.1457

fy

'fcρ

..

).)(.(1.18

)Ku(2.3611ω

0.1332(359)250250.9

10.9uK

..d.

AA

bar of No

0.0086 Use

bar

st

req

2

2

6

1 4

8571

018900086000330

14570181

1332036211

616

4843

41685771

4

8577122



4) From Robot Analysis Section 4-4, (Mu) -ve =-40.30 KN.m

162Use

mm .33592500.0034dbsΑ

...maxreqminρ

0.00341420250.0574

fy

'fcreqρ

0.05741.18

0.05552.3611ω

0.0555(359)250250.9

10.4uK

.)(

.AA bar of no

use

bar

S

req

2

2

6

25181

416

153052

1505

0189000341000330

300

5) From Robot Analysis at Section 5-5, (Mu) -ve =-84.46 KN.m

144

31272

4303

4

31672

00740

464

2

Use

mm .63592500.0074dbsΑreq use

maxρreqρminρ

0.0074250.1258 fy

'fcreqρ

0.12581.18

0.11652.3611ω

0.1165(359)250250.9

108uK

.d.

AA bar of no

.

420

.

bar S

S

2

2

6



1. Check Nominal moment capacity:

O.K ..).(.Mn.Mu.)/.(.)/ad(AsfMn

mm..

.

b'fc0.85

AsFya

y46844299471109090

411025563359420848032

556325025850

42084803

2. Check Cracking

251254

205004

205002504122

412

16825

2

2524206060

mmN

AA

NbdcA

mm

dbdC.Cd

.f.f

act

act

stc

ys

3. Check exposure:

exposure exterior for m.KN.Zexposure interior for m.KN.Z

. 512541 dcAf Z 3s

309814259814

65149812523

(OK)

4. Spacing :

bars) between space(mm.S

)(1-nstirrup)2(c.c-bw S

25mm Agg of max34

16mmdb25mm

S* req

336114

8252250



6) From Robot Analysis at Section 6-6, (Mu) +ve =-61.40 KN.m:

bar .)(

.AA bar of no

420

bar

s

mm2

2

6

.43592500.0053dbsΑ

.reqρ use maxreqρminρ

0.0053250.0894fy

'fcρ

0.08941.18

0.08462.3611ω

0.0846(359)250250.9

10.6K

33662

416

674752

6775

00530

401

2. Check Nominal moment capacity :

406138768684908684

102664735942088602

2

664725025850

4204163

850

6

2

.m.KN...MnMum.KN.Mn

).(.)ad(AsFyMn

mm..

)(

bfc.fA

a 'y S

3. Spacing

mmS

)(1-nstirriup)2(c.c-bw S

25mm Agg of sizemax34

16mmdb25mm

S* req

9214

8252250



4. Check Cracking :

3368333

205003

20500250412241

25260

2

.N

AA

NmmbdcA

mmdcfy.f

act

act

s

ok exposure interior for 30 16.48 Zok exporsure exterior for m.KN.Z

106833.3341 dcA fsZ 3 3-3

254816

252

7) From Robot Analysis at Section 7-7, (Mu) -ve =-24.8 KN.m

1.

162 use

..AA

bar of no

..dbAs. useHenu

..fyfc

..

).(.

..

.k

BAR

S

min

maxreqmin

'

req

u

2471

416175296

1752963592500033000330

00200420

2503400340

181034036211

03403592502590

10824

2

2

6



2. Check Nominal moment capacity :

(Mu)maxMu

m.KN.Mu

Mu...

MnMu

).(.)ad(AsFyMn

mm..b'fc0.8AsFya

269725752890

610263313594201200

2

633125025850

44202162

7528

3. Spacing:

mm)(n

c.c)2(stirrup-bwS

18412

25822501



4.9 Design for shear:-

Vu= Factored shear force at section considered.

Φ = shear strength reduction factor = 0.85

Vc = nominal shear strength provided by concrete

Fc'= 25Mpa, bw= 250mm, d=359mm.

Fyt= yield strength of stirrups (Mpa ) = 350Mpa

S = spacing between stirrups in direction parallel to longitudinal stirrups (mm)

Case1

c3VuVc5V

w/bytfv3Ad/4

mm300

maxS maxS)cV/u(VdytfvA

reqS

Case 2

2 cVuVc3V

w/bytfv3Ad/2

mm600

maxS maxS)cV/u(VdytfvA

reqS

Case 3

3 2cVuV

cV

w/bytfv3Ad/2

mm600

maxS maxSreqS

Case 4

4 uV

2cV No stirrups are required by the code

If S ≤ 75 mm increase ( vA ) and recalculate (S)

Table (4-1) Design for shear by using vertical stirrups



4.9.1 Flow-chart for design of shear for beam :



4.9.2 Data for design of shear for Beam: -

Fig 4.4 S.F.D for Beam 59 from Robot structural analysis



4.9.3 Design of Spacing between Stirrups for resisting shear in beam :-

1. Zone = 1

푉 = 16 ∗ 푓 ` ∗ 푏푤 ∗ 푑

= ∗ √25 ∗ 359 = 74791.66푁 =74.79 KN

5푉 = 74.79 ∗ 5 = 373.95퐾푁

3푉 = 74.79 ∗ 3 = 223.47퐾푁

푉2 =

74.792 = 37.39퐾푁

Vu from RA, use S.F of section 1.1 = 32KN

푉∅ =

320.85 = 37.64퐾푁

Case2 :- 3푉 > ∅>

74.79 > 37.64 > 37.39

푆 ≤

600푚푚= 179.5 ≈ 180

3 ∗ . ∗ = 422푆 = 180푚푚

S = S

Use S= 180mm

2. Zone = 2 :-

푉 =16 ∗ 푓 ` ∗ 푏 푑

= ∗ √25 ∗ 250 ∗ 359 = 74.79 KN

5푉 = 74.79 ∗ 5 = 373.95퐾푁

3푉 = 74.79 ∗ 3 = 223.47퐾푁

푉2 =

74.792 = 37.39퐾푁



퐕퐮at section (2) from R.A = 4.91

푉

∅ =4.910.855.77

Case 4 :-

푉2 =

푉∅

= 37.39 > 5.77

Note: NO stirrups are required by code

3. Zone= 3 :-

푽풖at section 3.3 = 41.56

푉

∅ = 41.560.85 = 48.89퐾푁

Case 3:-

푉 >푉

∅ >푉

2

74.79 > 48.89 > 37.39

S = S

S = 180 mm (From zone 3)

4. Zone = 4:-

퐕퐮at section 5.5 = 12.80

∅= .

.= 15.05

Case 4 :

푉

2 >푐∅

Hence, NO need of Stirrups



5. Zone 5:

푽풖= 41.96 from R.S.A value of Shear resembles to Zone 3,

Hence we can use S = 180 mm

6. Zone 6:

푽풖 = ퟒퟔ. ퟎퟖ풇풓풐풎푹. 푺. 푨

푉

∅ =46.080.85 = 54.21퐾푁

Case 4:

푉 > 푉

∅ >푉

2

Use 푆 = 푆

S= 180 mm ( From Zone 2 )

7. Zone 7:

푽풖 = ퟏퟗ. ퟑퟎ푲푵푭풓풐풎푹. 푺. 푨

푉

∅ = 19.300.85 = 22.7퐾푁

Case 4 :-

푉

2 >푉

∅

37.39 > 22.7

Note: No stirrups are required by code.

8. Zone 8:

푉 = 48.71푓푟표푚푅. 푆. 퐴

푉

∅ =48.710.85 = 58.30퐾푁



Case 3:

푉 > ∅

>

74.8 > 58.3 > 37.39

푆 = 푆

S = 180 mm

4.10 Development Length:

La= development length, fy = yield strength(Mpa)

Fc'= compressive strength of concrete (Mpa)

Db= diameter of bar (mm)

K1= reinforcement location factor= 1.3 for top bar

For bottom bar

k2= reinforcement size factor= 0.8 for db < 19

for db > 20

k3= excess reinforcement factor = [(As) required /As provided]

k4= confining reinforcement factor = 2.5db

trK

C

c = spacing or cover (mm)

Atr= Area of transverse reinforcement (ties of stirrup)

S= spacing of transverse reinforcement within Ld

h= number of bar being developed

Fyt= yield strength of transverse reinforcement

K1= 1.3, K2=0.8, K3= Aprovided

sreqA

Ld= mm'Fck

)kk(kdFy0.94

1b 30032



O.K Thats mmmm..

).)(.)(.(.L

.K use

O.K ...k

.k/.

))n)(s(/FA(k ,mmc

.db

KCk

...

AA

k ,.k ,.k

4

tr

ytrtr

tr

prov S

req S31

3006635225172

73080311635090172

5217216

77925779

21801035051001025

52

73092401172968031

4

4

2





Fig . 4.10 Shows Beam names assigned by R.S.A and can be used to evaluate

reinforcement table

Table 1: Sections and reinforcement for plinth level beams (tie beams)

BEAM NO SECTION POSITION DESIGN

MOMENT

DESIGN FORCE N (KN)

LONGITUDINAL REINFORCEMENT

REQUIRED REINFORCEMENT%

DESIGN FOR Qz

FOR SHEAR

STIRRUP SPACING

TRANSVERSAL REINFORCEMENT TYPE DISTRIBUTION

1 25X50 BOTTOM

(+) TOP (-)

SUPPORT (-)

SPAN 1 1/0.40 -23.43 10.33 4Ø14 2Ø14 6Ø14 0.27 48.97 22.5

2T10 [email protected][email protected][email protected]

1/5.39 8.35 5.85 0.27 -43.97 22.5 SPAN2 1/10.37 43.35 3.46 4Ø14 2Ø14 4Ø14 0.27 4.28 -

SPAN 3 1/15.36 -30.47 3.81 0.27 51.82 22.5

1/20.35 -14.73 7.69 4Ø14 2Ø14 6Ø14 0.27 -42.48 22.5 2 25X50 SPAN 1 2/0.40 -27.06 15.64 4Ø14 2Ø14 6Ø14 0.27 43.75 22.5


2/4.96 -1.14 1.86 0.53 -17.58 22.5 SPAN2 2/9.52 39.19 3.62 4Ø14 2Ø14 4Ø14 0.27 0.54 -

SPAN 3 2/14.09 1.6 4.09 0.53 18.22 22.5

2/18.65 -30.61 14.92 4Ø14 2Ø14 6Ø14 0.27 -45.51 22.5 3 25X50 SPAN 1 3/0.40 -26.41 10.11 4Ø14 2Ø14 4Ø14 0.27 46.38 22.5


3/5.39 9.41 7.1 0.27 -17.82 22.5 SPAN2 3/10.37 34.73 2.91 4Ø14 2Ø14 8Ø14 0.27 2.77 -

SPAN 3 3/15.36 -23.36 4.63 0.27 43.59 22.5

3/20.35 -16.95 6.04 4Ø14 2Ø14 4Ø14 0.27 -39.43 22.5 4 25X50 SPAN 1 4/0.40 -29.71 15.35 4Ø14 2Ø14 6Ø14 0.27 44.59 22.5


4/4.96 1.54 4.32 0.53 -17.82 22.5 SPAN2 4/9.52 33.42 4.16 4Ø14 2Ø14 4Ø14 0.27 0.98 -

SPAN 3 4/14.09 -1 1.72 0.53 16.75 -

4/18.65 -26.93 16.13 4Ø14 2Ø14 6Ø14 0.27 -42.66 22.5 5 25X50 SPAN 1 5/0.40 -13.42 3.95 4Ø14 2Ø14 4Ø14 0.27 20.48 22.5


5/5.39 -19.11 2.89 0.27 -20.86 22.5 SPAN2 5/10.38 29.73 2.84 4Ø14 2Ø14 8Ø14 0.27 -2.03 - SPAN 3 5/15.36 9.39 5.95 0.27 15.03 -

5/20.35 -22.85 8.49 4Ø14 2Ø14 4Ø14 0.27 -42.41 22.5 6 25X50 SPAN 1 6/0.40 -20.83 9.66 4Ø14 2Ø14 4Ø14 0.27 45.93 22.5


6/5.39 8.38 6.57 0.27 -16.6 22.5 SPAN2 6/10.37 34.5 4.35 4Ø14 2Ø14 8Ø14 0.27 1.85 -

SPAN 3 6/15.36 -25.98 4.32 0.27 45.65 22.5

6/20.35 -12.74 7.01 4Ø14 2Ø14 4Ø14 0.27 -39.27 22.5 7 25X50 SPAN 1 7/0.40 1.11 -0.36 4Ø14 2Ø14 8Ø14 0.27 7.83 -


7/4.96 2.39 -1.63 0.53 -6.97 - SPAN2 7/9.52 50.41 10.35 4Ø14 2Ø14 - 0.27 1.01 -

SPAN 3 7/14.09 -3.72 8.35 0.27 49.65 22.5

7/18.65 -32.91 14.42 4Ø14 2Ø14 6Ø14 0.27 -53.2 22.5 8 25X50 SPAN 1 8/0.40 -29.63 13.68 4Ø14 2Ø14 8Ø14 0.27 49.12 22.5


8/4.96 -1.81 1.53 0.53 -44.85 22.5 SPAN2 8/9.52 49.87 10.09 4Ø14 2Ø14 - 0.27 -0.94 -

SPAN 3 8/14.09 2.61 -1.35 0.53 6.8 -

8/18.65 0.58 0.01 4Ø14 2Ø14 6Ø14 0.27 -7.97 - 9 25X50 SPAN 1 9/0.40 -38.71 5.18 4Ø14 - - 0.27 42.17 22.5


9/2.01 11.43 5.18 0.27 8.7 - SPAN2 9/3.63 25.71 4.92 - 2Ø14 - 0.27 7.87 -

SPAN 3 9/5.24 14.16 4.92 0.27 -7.95 -

9/6.85 -33.09 4.92 - - 8Ø14 0.27 -40.38 22.5

Table 2 : Sections and reinforcement for ground level beams

BEAM NO SECTION POSITION

DESIGN FORCE N (KN)

DESIGN MOMENT


REQUIRED REINFORCEMENT%

DESIGN FOR Qz

FOR SHEAR

STIRRUP SPACING

TRANSVERSAL REINFORCEMENT TYPE DISTRIBUTION

10 25X40 BOTTOM

(+) TOP(assembly)

(-) SUPPORT

(-)

SPAN 1 -14.29 -0.96 -45.36 4Ø14 2Ø14 8Ø14 0.29 35.77 17.5 2T10 [email protected][email protected][email protected]

9.59 -0.22 18.82 0.29 -11.51 - SPAN2 21.71 0.14 58.63 4Ø14 2Ø14 4Ø14 0.29 -2.54 -

SPAN 3 -14 -0.15 -39.72 0.29 32.24 17.5

-9.28 -0.62 -29.75 4Ø14 2Ø14 4Ø14 0.29 -30.55 17.5 11 25X40


3.57 -0.38 3.84 0.58 -12.32 - SPAN2 20.32 -0.75 55.62 4Ø14 2Ø14 - 0.29 5.39 -

SPAN 3 2.63 -0.3 3.46 0.92 22.89 10.8

-17.28 -1.34 -50.77 4Ø14 2Ø14 8Ø14 0.29 -33.51 17.5 12 25X40


-10.72 0.13 -35.1 0.29 -15.97 17.5 SPAN2 17.29 0.11 53.41 4Ø14 2Ø14 4Ø14 0.29 3.03 -

SPAN 3 10 -0.24 18.6 0.29 9.42 -

-12.9 -1.06 -42.1 4Ø14 2Ø14 4Ø14 0.29 -33.21 17.5 13 25X50


2.47 -0.22 3.05 0.92 -24.13 10.8 SPAN2 23.85 -0.3 61.65 4Ø14 2Ø14 - 0.29 -4.45 -

SPAN 3 3.37 -0.38 3.82 0.58 13.06 -

-21.32 -1.21 -56.82 4Ø14 2Ø14 8Ø14 0.29 -36.69 17.5 14 25X50


4.27 -2.05 -2.8 0.27 -49.24 22.5 SPAN2 38.04 2.06 94.28 4Ø14 2Ø14 4Ø14 0.62 -8.61 13.3

SPAN 3 2.69 -0.06 -3.03 0.83 13.53 13.3

0.6 -0.05 8.25 4Ø14 2Ø14 4Ø14 0.53 -8.98 - 15 25X50

SPAN 1 -0.22 -0.29 2.92 4Ø14 2Ø14 8Ø14 0.53 9.39 - 2T10 [email protected][email protected][email protected]

2.5 -0.25 2.81 0.85 -13.87 13.3 SPAN2 36.06 2.14 72.76 4Ø14 2Ø14 4Ø14 0.27 -0.91 -

SPAN 3 4.32 -1.63 -2.79 0.27 45.42 22.5

-20.8 -1.34 -59.4 4Ø14 2Ø14 4Ø14 0.27 -48.63 22.5 16 25X50

SPAN 1 -30.2 0.03 -68.65 4Ø14 - - 0.27 22.53 22.5 2T10 [email protected][email protected]

12.98 0.03 36.95 0.27 6.71 - SPAN2 20.44 0.03 83.14 - 2Ø14 - 0.27 -9.1 -

SPAN 3 9.14 -0.26 27.11 0.53 -7.7 -

-37.68 -0.26 -88.27 - - 8Ø14 0.27 -40.11 22.5 17 25X50

SPAN 1 -11.25 -1.2 -42.56 4Ø14 - - 0.27 46.07 22.5 2T10 [email protected][email protected]

45.18 -1.2 93.37 0.27 9.51 - SPAN2 65.9 -1.2 169.88 - 2Ø14 - 0.29 -6.31 -

SPAN 3 30.77 -1.85 66.24 0.27 -12.91 -

-37.9 -1.85 -91.62 - - 8Ø14 0.27 -53.66 22.5 18 25X50

SPAN 1 -12.3 -0.95 -42.8 4Ø14 - - 0.27 49.09 22.5 2T10 [email protected][email protected][email protected]

20.01 -1.21 34.43 0.27 -43.5 22.5 SPAN2 32.14 0.34 60.22 - 2Ø14 - 0.6 19.74 13.3

SPAN 3 -18.02 -1.2 -50.08 0.27 50.64 22.5

-6.62 -0.76 -26.39 - - 8Ø14 0.27 -42.87 22.5

Table 3 : Sections and reinforcement for first floor beams

BEAM

NO SECTIO

N POSITIO

N

DESIGN MOMEN

T

DESIGN

FORCE N (KN)


REQUIRED

REINFOR

DESIGN FOR

Qz

STIRRUP SPACIN

G

TRANSVERSAL REINFORCEMENT TYPE

DISTRIBUTION

CEMENT%

FOR SHEAR

19 25X40 BOTTOM (+)

TOP(assembly) (-)

SUPPORT (-)

2T10 [email protected][email protected]+49

@14.0 SPAN 1 -14.29 -16.48 0.17 4Ø14 2Ø14 8Ø14 0.29 36.44 17.5 9.59 10.58 -0.08 0.29 -11.2 - SPAN2 21.71 22.06 -0.25 4Ø14 2Ø14 - 0.29 -2.51 -

SPAN 3 -14 -13.18 -0.1 0.29 31.86 17.5

-9.28 -10.3 0.06 4Ø14 2Ø14 8Ø14 0.29 -30.96 17.5

20 25X40


@16.0 SPAN 1 -19.58 -21.66 -0.34 4Ø14 2Ø14 4Ø14 0.29 35.71 17.5 3.57 4.28 -0.22 0.58 -12.12 - SPAN2 20.32 20.94 -0.2 4Ø14 2Ø14 - 0.29 5.34 -

SPAN 3 2.63 3.24 -0.16 0.92 22.58 10.8

-17.28 -19.27 -0.45 4Ø14 2Ø14 8Ø14 0.29 -34.34 17.5

21 25X40


@10.0 SPAN 1 -6.87 -7.26 0.18 4Ø14 2Ø14 8Ø14 0.29 16.85 17.5 -10.72 -10.68 -0.07 0.29 -15.94 17.5 SPAN2 17.29 17.64 0.09 4Ø14 2Ø14 4Ø14 0.29 3.03 -

SPAN 3 10 10.1 0.1 0.29 9.41 -

-12.9 -13.47 0.5 4Ø14 2Ø14 4Ø14 0.29 -33.42 17.5

22 25X50


@16.0 SPAN 1 -16.54 -18.71 -0.77 4Ø14 2Ø14 4Ø14 0.29 34.6 17.5 2.47 3.17 -0.23 0.93 -23.64 10.8 SPAN2 23.85 24.37 -0.36 4Ø14 2Ø14 4Ø14 0.29 -4.44 -

SPAN 3 3.37 4.54 -0.24 0.58 12.26 -

-21.32 -21.81 -0.38 4Ø14 2Ø14 8Ø14 0.29 -36.04 17.5

23 25X50


@16.0

SPAN 1 -22.3 -14.07 0.21 4Ø14 2Ø14 4Ø14 0.27 49.2 22.5 4.27 20.87 0.26 0.27 -42.93 22.5 SPAN2 38.04 32.89 -0.41 4Ø14 2Ø14 8Ø14 0.6 19.66 13.3

SPAN 3 2.69 -17.47 0.36 0.27 50.37 22.5

0.6 -7.48 0.18 4Ø14 2Ø14 4Ø14 0.27 -43.22 22.5

24 25X50


@14.0 SPAN 1 -0.22 -11.82 0.33 4Ø14 - - 0.27 46.35 22.5 2.5 45.05 0.33 0.27 9.57 - SPAN2 36.06 66.22 0.33 - 2Ø14 - 0.29 -6.25 -

SPAN 3 4.32 31.07 0.57 0.27 -12.89 -

-20.8 -37.53 0.57 - - 8Ø14 0.27 -53.61 22.5

25 25X50 2T10

[email protected][email protected] SPAN 1 -30.2 -25.52 0.23 4Ø14 2Ø14 8Ø14 0.27 53.53 22.5 12.98 5.1 0.46 0.27 -49.85 22.5 SPAN2 20.44 39.05 -0.76 4Ø14 2Ø14 4Ø14 0.27 0.06 -

SPAN 3 9.14 2.24 -0.13 0.83 14.91 13.3

-37.68 -1.62 -0.24 4Ø14 2Ø14 4Ø14 0.27 -10.08 -

26 25X50


@12.0 SPAN 1 -11.25 -2.54 -0.1 4Ø14 2Ø14 8Ø14 0.27 9.78 - 45.18 2.89 -0.12 0.85 -12.69 13.3 SPAN2 65.9 37.44 -1.25 4Ø14 2Ø14 - 0.27 -1.38 -

SPAN 3 30.77 4.69 0.51 0.27 45.04 22.5

-37.9 -22.84 0.24 4Ø14 2Ø14 8Ø14 0.27 -49.36 22.5

27 25X50


@22.0 SPAN 1 -12.3 -30.05 -0.05 4Ø14 - - 0.27 22.54 22.5 20.01 13.19 -0.05 0.27 6.73 - SPAN2 32.14 20.71 -0.05 - 2Ø14 - 0.27 -9.09 -

SPAN 3 -18.02 9.72 0.09 0.27 -7.52 -

-6.62 -36.46 0.09 - - 8Ø14 0.27 -39.71 22.5

28 25X50 2T10

[email protected][email protected] SPAN 1 -12.3 -23.36 0.33 4Ø14 - - 0.27 46.04 22.5 20.01 26.62 0.33 0.27 11.15 - SPAN2 32.14 51.1 0.33 - 2Ø14 - 0.27 -2.21 -

SPAN 3 -18.02 35.77 0.16 0.27 -8.52 -

-6.62 -6.31 0.16 - - 8Ø14 0.27 -40.24 22.5

Table 4 : Sections and reinforcement for second floor beams

BEAM

NO SECTIO

N POSITIO

N

DESIGN MOMEN

T

DESIGN

FORCE N (KN)


REQUIRED REINFORCEMENT

%

DESIGN FOR

Qz FOR

SHEAR

STIRRUP

SPACING

TRANSVERSAL REINFORCEMENT TYPE

DISTRIBUTION

29 25X40 BOTTOM (+)

TOP(assembly) (-)

SUPPORT (-)

SPAN 1 -14.29 -6.4 0.71 4Ø14 2Ø14 4Ø14 0.29 9.75 - 2T10

[email protected][email protected][email protected]

9.59 5.24 0.6 0.29 -7.91 - SPAN2 21.71 19.16 0.13 4Ø14 2Ø14 8Ø14 0.29 2.73 -

SPAN 3 -14 -10.64 0.24 0.29 9.58 -

-9.28 -4.46 0.18 4Ø14 2Ø14 4Ø14 0.29 -8.43 - 30 25X40 SPAN 1 -19.58 -10.96 4.02 4Ø14 2Ø14 8Ø14 0.29 9.65 -

2T10 [email protected][email protected]+63@10

.0

3.57 1.54 1.71 0.58 -7.49 - SPAN2 20.32 19.67 1.38 4Ø14 2Ø14 - 0.29 -1.3 -

SPAN 3 2.63 0.91 0.83 0.58 7.1 -

-17.28 -9.94 3.9 4Ø14 2Ø14 - 0.29 -9.34 - 31 25X40 SPAN 1 -6.87 -2.87 -0.3 4Ø14 2Ø14 4Ø14 0.29 7.5 -


.0

-10.72 -7.69 0.14 0.29 -8.15 - SPAN2 17.29 12.7 -0.08 4Ø14 2Ø14 8Ø14 0.29 -1.54 -

SPAN 3 10 5.74 0.47 0.29 6.43 -

-12.9 -5.6 0.25 4Ø14 2Ø14 4Ø14 0.29 -9 - 32 25X50

SPAN 1 -16.54 -9.48 4.1 4Ø14 2Ø14 8Ø14 0.29 9.48 -


2.47 0.93 0.86 0.58 -7.49 - SPAN2 23.85 22.97 1.37 4Ø14 2Ø14 - 0.29 2.12 -

SPAN 3 3.37 1.47 1.71 0.58 8.24 -

-21.32 -12.25 4.08 4Ø14 2Ø14 8Ø14 0.29 -10.3 - 33 25X50

SPAN 1 -22.3 -1.88 1.62 4Ø14 2Ø14 4Ø14 0.27 8.63 - 4.27 2.8 1.17 0.27 -14.47 22.5

SPAN2 38.04 41.45 0.51 4Ø14 2Ø14 8Ø14 0.27 12.75 - 2T10 [email protected][email protected]+39@16

.0

SPAN 3 2.69 -25.67 0.79 0.27 13.81 22.5

0.6 -0.6 1.23 4Ø14 2Ø14 4Ø14 0.27 -7.68 - 34 25X50 SPAN 1 -0.22 -1.4 1.45 4Ø14 - - 0.27 7.72 -


.0

2.5 30.44 1.45 0.27 7.72 - SPAN2 36.06 62.29 1.45 - 2Ø14 - 0.27 7.72 -

SPAN 3 4.32 10.47 1.18 0.27 -12.33 -

-20.8 -40.12 1.18 - - 8Ø14 0.27 -12.33 - 35 25X50 SPAN 1 -30.2 -8.42 3.07 4Ø14 2Ø14 4Ø14 0.27 9.96 -


.0

12.98 -5.24 2.09 0.27 -39.33 22.5 SPAN2 20.44 54.38 1.06 4Ø14 4Ø14 4Ø14 0.27 -11.34 -

SPAN 3 9.14 -11.1 0.38 0.27 4.74 -

-37.68 0.24 0.24 4Ø14 2Ø14 - 0.53 -1.85 - 36 25X50 SPAN 1 -11.25 -0.72 0.66 4Ø14 2Ø14 4Ø14 0.27 1.46 -


.0

45.18 -8.53 0.49 0.27 -3.51 - SPAN2 65.9 50.81 1.25 4Ø14 4Ø14 4Ø14 0.27 10.12 -

SPAN 3 30.77 -5.25 1.9 0.27 14.13 22.5

-37.9 -7.78 2.88 4Ø14 2Ø14 - 0.27 -9.11 - 37 25X50 SPAN 1 -12.3 -21.36 0.53 4Ø14 - - 0.27 6.1 -


.0

20.01 3.36 0.53 0.27 6.1 - SPAN2 32.14 28.08 0.53 - 2Ø14 - 0.27 6.1 -

SPAN 3 -18.02 0.14 0.11 0.53 -6.79 -

-6.62 -27.35 0.55 - - 8Ø14 0.27 -6.79 - 38 25X50

SPAN 1 -12.3 -30.25 0.84 4Ø14 - - 0.27 11.12 -


20.01 8.33 0.84 0.27 11.12 - SPAN2 32.14 47.75 1.14 - 2Ø14 - 0.27 -6.88 -


ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ 78 Design Of Stairs

5. Design Of Stairs: 5.1 Geometrical Design of Stairs:-

Given:- Floor height = 4.2m

Assume:-

Rise …. R = 150 mm R ≤ 190 mm (O.K)

Going … G = 300 mm G ≥ 220 mm (O.K)

- Number of riser's 8 R 0.15.N 2204

- Number of going's NG = NR-1 = 28-1= 27

5.1.2 Check:-

OK 630 600 570630 300] [2(150) 570

630 G 2R 570

5.1.3 Check for angle:-

OK 40 26.52540α25

26.5 0.5300150

GR αtan

Figure (5 - 1) Dimensions of stair



:Detailed design of stair - 5.2

Assumption and Requirement of design:

Table (5-1) Data of design

2kN/m 4 =WL.L

Live Load

m 4.20 = H Height of story

MPa 420 = yf Yield stress

MPa 25 ='cf

Compressive strength of

concrete

mm 25 = C.C Concrete Cover

mm 14 = bd Diameter of main steel

mm 12 = std Diameter of secondary steel

2kN/m 2.5 Flooring



Figure (5-3). Plan for stairs

Figure (5- 4) Vertical section for stairs

Note: The stairs were directed at horizontal planar direction to compensate limitation of

other side (5m) with respect to height.

After the distribution of steps (goings) we, now know there are three flights.



5.3 No. of Steps for each flight:

1- Flight (No.1):

Height = height of one riser x NO.OF goings

= 0.15* 10= 1.50m

No rise in flight (1) = rise 0 150

11500

No of going in flight (1) = 10-1= 9 treads

2- Flight (No.2):

H2=1.2m

G = 7

R= 7+1=8

3- Flight (No.3):

H3 = HTotal- (H2+H3)

H3= 4.20 – (8* 0.15 + 10*0.15) = 4.20 – 2.70

H3= 1.50m

No of Rise = 10 150

1500

No of goings = 10 -1 = 9 treads

Check: sum of risers height = HTotal (Floor Height)

(Sum of risers in all flights) * 0.15= 4.20

(10+8+10)*0.15= 4.20m

4.20 => 4.20 That’s O.K



5.4 Structural Design of Stairs

5.4.1 Steps of Design :-

1. - Design for flexure:-

minρ ρ yF

1.4or dsh

0.002 minρ

yf

ω ρ

1.18uK 2.36-1-1

ω

0.9 , db

uM uK

2 - c -h d

'c

'c

f

f 2

2. Main Reinforcement:-

mms5h mm 450

sAb)s(A

reqS

)s(AsA

N

dbρsA

bar one

bar one

3. Secondary Reinforcement:

mm s5h mm 450

sA

b)s(AS

sh10000.002sA

bar one



5.5 Design for flight No 1 & 3 :

KN/m 25.6 W

KN/m 20.38 31.72.524)0.3(1.4 W

mm 268 2

14 - 25 - 300 d

mm 300 20

6000 20nL

sh

Flight

Flight

Landing

..)).(.26.5 COS

0.3( .W 47152243005041

Figure (5-5): shows loading diagram for flight 1 & 3

Figure (5-6): Shows bending moment diagram for flight 1 & 3



Figure (5-7) Shows Shear diagram for flight 1 & 3

mm 15020

3000 min)s(h

Mpa) 420 yF(for 20

nLmin)s(h

5.5.2 Design for flexure: (For Flight No .1 & No.3)

Given :

Mmax(–ve) = 29 KN.m

5.5.3 Design for - ve moment (Main Reinforcement) :-

minρ reqρ

0.003 minρ

0.0007420250.012ρ

0.0121.180.0112.3611ω

0.0112681500250.9

1092uK

4201.4

2

6



mm 190 / 14 Use

191.29mm 1206

15008.153bbar

14 8 Use

8 7.88/4)14(πb

N

mm 692 26815000.003

sAS

1206

sAstA

mm 1206 stA 22

soneA

5.5.5 Secondary reinforcement (temp + shrinkage):

12 6 e Us

602.5/4)12(π

sA

s

AN

mm 190 / 12 e Us

mm 450 0513 s

3h

mm 450 maxS Use

max

S mm 01.9911000113

S

mm 685s

A

268 1000 0.002 s

A

2

2

568

568

12std Use

bar one

total

mm



5.6 Data for Design of Flight No 2:-

Figure (5.8) Loading diagram for flight No 2

Figure (5.9) Bending Moment diagram for flight No 2

Figure (5.10) Shear diagram for flight No 2



5.6.2 Design for flexure: (For Flight No .2) :-

5.6.3 Design - ve moment (Main Reinforcement) :-

Given:

Mmax = M(–ve) = 22.55 KN.m

minρ use(control) minρ ρ

.Fy4

Fc'

0.003334201.4

minρ

0.0005420240.0089ρ

0.00891.18

0.0892.3611ω

0.00892681500240.9

10.uK

6

0030

5522

14 8 eUs

./4)14(πsA

AN

mm 268 1500 0.003 sA

21206

bar one

ST

8837

21206

Spacing :

mm 190 / 14 Use

191.54mm 1206

b bar soneAreqS

sA

1500154



5.6.4 Secondary Reinforcement:-

mm 190 / 12 eUs

mm .11000S

mm 5sA

268 1000 0.002 sA

568

2

019911368

5.7 Reinforcement details:

Figure (5 -7 ) Reinforcement of stair



Figure (5 - 8) Connection between stair and beam


ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــDesign Of Columns 90

6. Design of columns

6.1. Introduction: This chapter presents an introductory discussion of reinforced concrete

columns, with particular emphasis on short, stocky columns subjected to small bending

moments. Such columns are often said to be “axially loaded.

6.1.1 Concrete columns can be roughly divided into the following three categories:

Short compression blocks or pedestals—if the height of an upright compression

member is less than three times its least lateral dimensions, it may be considered to be a

pedestal(support). The ACI (2.2 and 10.14) states that a pedestal may be designed with

unreinforced or plain concrete with a maximum design compressive stress equal to

0.85φf c’, where φ is 0.65. Should the total load applied to the member be larger than

0.85φf c’ Ag, it will be necessary either to enlarge the cross-sectional area of the

pedestal or to design it as a reinforced concrete column.

Short reinforced concrete columns—should a reinforced concrete column fail due to

initial material failure, it is classified as a short column. The load that it can support is

controlled by the dimensions of the cross section and the strength of the materials of

which it is constructed. We think of a short column as being a rather stocky member

with little flexibility.

Long or slender reinforced concrete columns—As columns become more slender,

bending deformations will increase, as will the resulting secondary moments. If these

moments are of such magnitude as to significantly reduce the axial load capacities of

columns, those columns are referred to as being long or slender.

Fig 6 Cropped image from ACI 318-05 ,



6.1.2 Axial Load Capacity of Columns:

In actual practice, there are no perfect axially loaded columns, but a discussion of such

members provides an excellent starting point for explaining the theory involved in

designing real columns with their eccentric loads. Several basic ideas can be explained for

purely axially loaded columns, and the strengths obtained provide upper theoretical limits

that can be clearly verified with actual tests. It has been known for several decades that

the stresses in the concrete and the reinforcing bars of a column supporting a long-term

load cannot be calculated with any degree of accuracy. You might think that such stresses

could be determined by multiplying the strains by the appropriate moduli of elasticity. But

this idea does not work too well practically because the modulus of elasticity of the

concrete is changing during loading due to creep and shrinkage. Thus, the parts of the

load carried by the concrete and the steel vary with the magnitude and duration of the

loads. For instance, the larger the percentage of dead loads and the longer they are applied,

the greater the creep in the concrete and the larger the percentage of load carried by the

reinforcement. Though stresses cannot be predicted in columns in the elastic range with

any degree of accuracy, several decades of testing have shown that the ultimate strength of

columns can be estimated very well. Furthermore, it has been shown that the proportions

of live and dead loads, the length of loading, and other such factors have little effect on the

ultimate strength. It does not even matter whether the concrete or the steel approaches its

ultimate strength first. If one of the two materials is stressed close to its ultimate strength,

its large deformations will cause the stress to increase quicker in the other material. For

these reasons, only the ultimate strength of columns is considered here. At failure, the

theoretical ultimate strength or nominal strength of a short axially loaded column is quite

accurately determined by the expression that follows, in which Ag is the gross concrete

area and Ast is the total cross-sectional area of longitudinal reinforcement, including bars

and shapes:

Pn = 0.85f c’(Ag − Ast ) + fyAst



6.1.3 ACI Code Requirements for Cast-in-Place Columns: The ACI Code specifies quite a

few limitations on the dimensions, reinforcing, lateral restraint, and other items pertaining

to concrete columns. Some of the most important limitations are as follows.

1. The percentage of longitudinal reinforcement may not be less than 1% of the gross

cross-sectional area of a column (ACI Code 10.9.1). It is felt that if the amount of steel

is less than 1%, there is a distinct possibility of a sudden non-ductile failure, as might

occur in a plain concrete column. The 1% minimum steel value will also lessen creep

and shrinkage and provide some bending strength for the column. Actually, the code

(10.8.4) does permit the use of less than 1% steel if the column has been made larger

than is necessary to carry the loads because of architectural or other reasons. In other

words, a column can be designed with 1% longitudinal steel to support the factored

load, and then more concrete can be added with no increase in reinforcing and no

increase in calculated load-carrying capacity. In actual practice, the steel percentage

for such members is kept to an absolute minimum of 0.005.

2. The maximum percentage of steel may not be greater than 8% of the gross cross-

sectional area of the column (ACI Code 10.9.1). This maximum value is given to

prevent too much crowding of the bars. Practically, it is rather difficult to fit more

than 4% or 5% steel into the forms and still get the concrete down into the forms and

around the bars. When the percentage of steel is high, the chances of having

honeycomb in the concrete is decidedly increased. If this happens, there can be a

substantial reduction in the column’s load-carrying capacity. Usually the percentage

of reinforcement should not exceed 4% when the bars are to be lap spliced. It is to be

remembered that if the percentage of steel is very high, the bars may be bundled.

3. The minimum numbers of longitudinal bars permissible for compression members

(ACI Code 10.9.2) are as follows: four for bars within rectangular or circular ties,

three for bars within triangular-shaped ties, and six for bars enclosed within spirals.

Should there be fewer than eight bars in a circular arrangement, the orientation of the

bars will affect the moment strength of eccentrically loaded columns.



4. The code does not directly provide a minimum column cross-sectional area, but it is

obvious that minimum widths or diameters of about 2 cm to 5 cm are necessary to

provide the necessary cover outside of ties or spirals and to provide the necessary

clearance between longitudinal bars from one face of the column to the other. To use

as little rentable floor space as possible, small columns are frequently desirable. In

fact, thin columns may often be enclosed or “hidden” in walls.

5. When tied columns are used, the ties shall not be less than #3, provided that the

longitudinal bars are #10 or smaller. The minimum size is #4 for longitudinal bars

larger than #10 and for bundled bars. Deformed wire or welded wire fabric with an

equivalent area may also be used (ACI 7.10.5.1).

6. The center-to-center spacing of ties shall not be more than 16 times the diameter of

the longitudinal bars, 48 times the diameter of the ties, or the least lateral dimension

of the column.

6.1.4 General Configurations of moments with in columns:

When a column is subjected to primary moments (those moments caused by applied

loads, joint rotations, etc.), the axis of the member will deflect laterally, with the result that

additional moments equal to the column load times the lateral deflection will be applied to the

column. These latter column that has large secondary moments is said to be a slender column,

and it is necessary to size its cross section for the sum of both the primary and secondary

moments. The ACI’s intent is to permit columns to be designed as short columns if the

secondary or P∆ effect does not reduce their strength by more than 5%.moments are called

secondary moments.

Fig 6. Cropped image from ACI Code 318-08 metric



Fig 6.1 shows interior & exterior columns

The effects of slenderness can be neglected in about 40% of all unbraced columns and

about 90% of those braced against sidesway. These percentages are probably decreasing year by

year, however, due to the increasing use of slenderer columns designed by the strength method,

using stronger materials and with a better understanding of column buckling behavior.

6.1.5 Classification of Columns:

A plain concrete column can support very little load, but its load-carrying capacity will be

greatly increased if longitudinal bars are added. Further substantial strength increases may

be made by providing lateral restraint for these longitudinal bars. Under compressive

loads, columns tend not only to shorten lengthwise but also to expand laterally due to the



Poisson effect. The capacity of such members can be greatly increased by providing lateral

restraint in the form of closely spaced closed ties or helical spirals wrapped around the

longitudinal reinforcing.

Fig6.2. Shows ACI 315-08 regarding Requirements for distance between supports

Reinforced concrete columns are referred to as tied or spiral columns, depending on the

method used for laterally bracing or holding the bars in place. If the column has a series

of closed ties, as shown in Figure 9.2(a), it is referred to as a tied column. These ties are

effective in increasing the column strength. They prevent the longitudinal bars from

being displaced during construction, and they resist the tendency of the same bars to

buckle outward under load, which would cause the outer concrete cover to break or spall

off. Tied columns are ordinarily square or rectangular, but they can be octagonal, round,

L shaped, and so forth. The square and rectangular shapes are commonly used because

of the simplicity



6.1.6 Effective Length:

The effective length of a column is defined as the length between the points of contra-

flexure of the buckled column. The code has given two charts to calculate the effective

length of columns in a framed structure.

6.1.7 DESIGN OF AXIALLY LOADED COLUMN

1. SHORT COLUMN UNDER AXIAL COMPRESSION

The factored axial load, is given by the equation ,

Where = area of concrete and, = area of longitudinal reinforcement of columns.

This equation can be recast as:

Where P = percentage of reinforcement. Design charts are prepared based on this equation.

6.1.8 REINFOCEMENTs:

There are two kinds of reinforcement in a column, longitudinal and transverse

reinforcement. The purpose of transverse reinforcement is to hold the vertical bars in

position, providing lateral support so that individual bars cannot buckle outward and split

the concrete.

6.1.8.1. Longitudinal Reinforcement in columns

a) The cross-sectional area of longitudinal reinforcement shall be not less than 0.8 percent

nor more than 6 percent of the gross cross-sectional area of the column.

Note: the use of 6 percent reinforcement may involve practical difficulties in placing and compacting of

concrete, hence lower percentage is recommended. Where bars from the columns below have to be

lapped with those in the column under consideration, the percentage of reinforcement steel shall usually

not exceed 4 percent.



b) In any column that has a larger cross-sectional area than that required to support the

load, the minimum percentage of steel shall be based upon the area of concrete required to

resist the direct stress and not upon the actual area.

c) The minimum number of longitudinal bars provided in a column shall be four in

rectangular columns and six in circular columns.

d) The bars shall not be less than 12mm in diameter.

e) A reinforced concrete column having helical reinforcement shall have at least six bars

of longitudinal reinforcement within the helical reinforcement.

f) In a helically reinforced columns, the longitudinal bars shall be in contact with the

helical reinforcement and equidistant around its inner circumference.

g) Spacing of longitudinal bars measured along the periphery of the column shall not

exceed 300mm.

h) In case of pedestals in which the longitudinal reinforcement is not taken into account in

strength calculations, nominal reinforcement not less than 0.15 percent of the cross-

sectional area shall be provided.

Note: Pedestal is a compression member, the effective length of which does not exceed three

times the least lateral dimension.

6.1.8.2 Transverse Reinforcement in columns:

(a) A reinforced compression member shall have transverse reinforcement or helical

reinforcement so disposed that every longitudinal bar nearest to the compression face has

effective lateral support against buckling subject to provisions. The effective lateral

support is given by transverse reinforcement either in the form of circular rings capable

of taking up circumferential tension or by polygonal links (lateral ties) with internal

angles not exceeding . The ends of the transverse reinforcement shall be properly

anchored.



(b) Arrangement of transverse reinforcement:

If the longitudinal bars are not spaced more than 75mm on either side, transverse

reinforcement need only to go round corner and alternate bars for the purpose of

providing effective lateral supports. If the longitudinal bars spaced at a distance of not

exceeding 48 times the diameter of the tie are effectively tied in two directions, additional

longitudinal bars in between these bars need to be tied in one direction, by open ties .

Where the longitudinal reinforcing bars in a compression member are placed in more

than one row, effective lateral support to the longitudinal bars at the inner rows may be

assumed to have been provided, if Transverse reinforcement is provided for the outer row

and No bar of the inner row is closer to the nearest compression face than three times the

diameter of the largest bar in the inner row. Where the longitudinal bars in a compression

member are grouped (not in contact) and each group adequately tied with transverse

reinforcement, the transverse reinforcement for the compression member as a whole may

be provided on the assumption that each group is a single longitudinal bar for purpose of

determining the pitch and diameter of the transverse reinforcement. The diameter of such

transverse reinforcement need not however exceed 20mm .

(c) Pitch and diameter of lateral ties:

Pitch – The pitch of transverse reinforcement shall be not more than the least of the

following distances:

a. The least lateral dimension of the compression member

b. Sixteen time the smallest diameter of the longitudinal reinforcement bar to be tied

c. Forty-eight times the diameter of the transverse reinforcement.

Diameter – The diameter of the polygonal links or lateral ties shall be not less than one-

fourth of the diameter of the largest –longitudinal bar, and in no case less than 5mm.

(d) Helical Reinforcement:

Pitch – Helical reinforcement shall be of regular formation with the turns of the helix

spaced evenly and its ends shall be anchored properly by providing one and a half extra



turns of the spiral bar. Where an increased load on the column on the strength of helical

reinforcement is allowed for, the pitch of helical turns shall be not more than 77 mm nor

more than one-sixth of the core diameter of the column, nor less than 25mm, nor less than

3 times the diameter of the steel bar forming the helix. In other cases, the requirements of

transverse reinforcement shall be complied with.

Diameter – The diameter of the helical reinforcement shall be in accordance with para (c)

above.

Fig 6.3 showing different kinds of columns reinforcement

6.1.9 Safety Provisions for Columns:

The values of φ to be used for columns as specified in Section 9.3.2 of the code are well

below those used for flexure and shear (0.90 and 0.75, respectively). A value of 0.65 is

specified for tied columns and 0.75 for spiral columns. A slightly larger φ is specified for

spiral columns because of their greater toughness.

The failure of a column is generally a more severe matter than is the failure of a beam,

because a column generally supports a larger part of a structure than does a beam. In other


ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــDesign Of Columns

100

words, if a column fails in a building, a larger part of the building will fall down than if a

beam fails. This is particularly true for a lower-level column in a multistory building. As a

result, lower φ values are desirable for columns. There are other reasons for using lower φ

values in columns. As an example, it is more difficult to do as good a job in placing the

concrete for a column than it is for a beam. The reader can readily see the difficulty of

getting concrete down into narrow column forms and between the longitudinal and lateral

reinforcing. As a result, the quality of the resulting concrete columns is probably not as

good as that of beams and slabs

The failure strength of a beam is normally dependent on the yield stress of the tensile

steel—a property that is quite accurately controlled in the steel mills. The failure strength

of a column is closely related to the concrete’s ultimate strength, a value that is quite

variable. The length factors also drastically affect the strength of columns and thus make

the use of lower φ factors necessary. It seems impossible for a column to be perfectly

axially loaded. Even if loads could be perfectly centered at one time, they would not stay

in place. Furthermore, columns may be initially crooked or have other flaws, with the

result that lateral bending will occur. Wind and other lateral loads cause columns to bend,

and the columns in rigid-frame buildings are subjected to moments when the frame is

supporting gravity loads alone.



101

6.1.10 Design Formulas :

In the pages that follow, the letter e is used to represent the eccentricity of the load. The

reader may not understand this term because he or she has analyzed a structure and has computed

an axial load, Pu, and a bending moment, Mu, but no specific eccentricity, e, for a particular

column. The term e represents the distance the axial load, Pu, would have to be off center of the

column to produce Mu. Thus,

Pu ×e = Mu

Or

e = Mu/ Pu

Nonetheless, there are many situations where there are no calculated moments for the

columns of a structure. For many years, the code specified that such columns had to be designed

for certain minimum moments even though no calculated moments were present. This was

accomplished by requiring designers to assume certain minimum eccentricities for their column

loads. These minimum values were 1 in. or 0.05h, whichever was larger, for spiral columns and

1 in. or 0.10h for tied columns. (The term h represents the outside diameter of round columns or

the total depth of square or rectangular columns.) A moment equal to the axial load times the

minimum eccentricity was used for design.

In today’s code, minimum eccentricities are not specified, but the same objective is

accomplished by requiring that theoretical axial load capacities be multiplied by a factor

sometimes called α, which is equal to 0.85 for spiral columns and 0.80 for tied columns. Thus,

as shown in Section 10.3.6 of the code, the axial load capacity of columns may not be greater

than the following values:

For tied columns (φ = 0.65)

φPn (max) = 0.80φ[0.85f’c’(Ag − Ast ) + fy Ast ] (ACI Equation 10-2)



102

6.2 Sample for Design :-

We will design column 28, which is at edge of the building (corner). Also this column

is located at ground floor. The assumed size of the column primarily was 25*40cm which

resulted in excessive reinforcement i.e 12∅16; hence considering the size of the building

and its huge live load due to its commercial nature. The new size assumed is 25*60cm;

while keeping in mind the size will decrease for every new floor. For example the assumed

size of exterior column for first floor will be 25*50cm. Also the beams are connected to the

respected column are of dimension 25*50cm.

Fig 6.4 Elevation plan showing Columns originating from foundations

while exterior columns shortens after first and second floor



103

Fig 6.5. Elevation plan showing Columns at first floor

Fig6.6 . Elevation plan showing beams and Columns for second floor



104

Table 6.1 : shows initial preliminary assumed sections

of columns for different stories and locations

Fig.6.7 Shows Governing case of column 59 with axial load & moments



105

6.3 Design of column in detail:

Table 6.2 Design values obtained from R.S.A

6.3.1Design moments:

6.3.1.1 Inertia At X - Direction:

(Ig) = = . × . = 0.0045mm

Ic = 0.7 × 0.0045 = 0.00315m

(Ig) = = . × . = 0.0026

Ic = 0.7 × 0.0026 = 0.00182m

(Ig) = = . × . = 0.0026m

Ib =× 0.35 × 0.0026 = 0.00091m

G =

=. .

..

= 1.02

pu

493.58 KN

Mux top

18.33 KN.m

Mux bottom -15.51 KN.m

Muy top 27.70 KN.m

Muy bottom -21.64 KN.m



106

G =

=. .

...

= 18.88

Note => 퐼 = 퐼 , whereas plinth level column height is 0.7m

6.3.1.2 Now, take K the smaller of:-

K = 0.7 +0.05 (퐺 +퐺 ) ≤ 1

= 0.7 +0.05 (1.02 +18.88)

= 1.695 > 1

K = 0.85 +0.05 (퐺 ) ≤1

K = 0.85 + 0.05(1.02) = 0.901 ≤ 1

K = 0.9

Using K = 0.9

KLr =

0.9 × 40.3 × 0.6 = 20

KLr = 34 − 12

M1M2 ≤ 40

Whereas:-

M = value of smaller factored end moment at X-direction

M = value of larger factored end moment at X-direction

6.3.1.3 Check for Short or long column :

KLr = 34 − 12

(−15.512)18.324 = 44.14

KLr <

KLr Hence, itsaShortColumn

Now, the loading on the column can be axial or eccentric;

We can check minimum eccentricity



107

e = . = . × . = 0.015 m

푀 ≥푀 ___________푒푞 (1.2)

Where :-

Mc = M = value of larger factor end moment

And Mmin= e × Pu___________________ (moment due to eccentricity)

Then eq. 1.2 becomes

푀 ≥ 푒 ∗ 푃

18.324 ≥ 0.015 × 493.58

18.324 > 7.40 KN.m (Thats O.K. )

Hence the design moment, will be critical moment, which is the ultimate moment Mu

푀 = 18.324퐾푁.푀

6.3.2 At Y - Direction:

(Ig) = = . × . = 0.00078

퐼 = 0.7 × 0.00078 = 0.00054

(Ig) = = . × . = 0.00065

퐼 = 0.7 × 0.00065 = 0.00045

(Ig) = = . × . = 0.00065



108

Ib = 0.35 × 0.00065 = 0.00022

G =

EILuEILn

=0.00054

4 + 0.000454

0.000227.3

=0.000240.000030 = 8.1

G =

EILuEILn

=

⎝

⎜⎛

0.000544 + 0.00045

40.000227.3

⎠

⎟⎞= 30.07

6.3.2.1 Now, take K the smaller of:-

K = 0.7 + 0.05 퐺 +퐺 ≤ 1

K = 0.7 + 0.05(8.1 + 30.07) ≤ 1

2.6 ≠ 1

K = 0.85 + 0.05 (퐺 ) ≤ 1

= 0.85 0.05 (8)

=1.25 ≠ 1

Hence use K=1

6.3.2.2 Check for short or long:- KLr =

1 ∗ 40.3 × 0.25 = 53.33

KLr = 34 − 12

M1M2

KLr = 34 − 12

(−21.64)27.7 = 43.37

KLr = 43.37

KLr >

KLr => Hence, itsaLongColumn



109

Note: The effect if the slenderness ratio may be ignored if <40, but if its greater then 40

(as in this case), slenderness effect must be considered

6.4 The design of the long column includes buckling analysis, and which includes moment

magnification factor , according to ACI code section 6.64 :-

1- Compute the curved shape factor Cm

4.04.06.02

1 ns

ns

MM

Cm

퐶 = 0.6 + 0.421.6427.70 = 0.912(푇ℎ푎푡푠표. 푘)

Where the is positive if column is bent in single curvature and is negative if the

column is bent in double curvature.

2- Determine if the frame is braced against sideway and find unsupported length Lu

and effective length factor K (may be assumed 1)

3- Calculate the member stiffness , EI using

EI = .

_____________________ (3.1)

Where, 퐸 =4700 푓푐` _____________________ (3.2)

I = grossmomentofinertiaofsectionabouttheaxis, neglectingA

퐵 =

< 1

= . ( ) . . . . .

< 1

=1.2 × 279.6

(1.2 × 279.6 + 1.6 × 98.78)

= 0.68 < 1(푡ℎ푎푡`푠표푘)



110

From (eq 3.2)

퐸 = 4700 ∗ √25

= 23500 Mpa

Then, substituting above (eq) in eq.(3.1)

EI = . ∗ ∗ . ∗ .

(푚푚)

=3.03*10 푁푚푚

4. Determine the Euler buckling load, 푷풄 :-

푃 = ∗( )∗

= ∗ . ∗( ∗ )

= 1862.88 KN

5. Calculate moment magnifying factor:-

훿 = 퐶푚

1 − 푃0.75푃

≥ 1.0

0.912

1 − ( 493.581862.88)= 1.24

훿 = 1.24

6. Design the compression member, using the axial factored load, the design values of

(푴풄&푷풖) for long column

푀 = 훿푀 ≥ 훿푝 ∗ 푒 ____________________(6)

Where :-

푒 = (15 + 0.03 ∗ ℎ)/100

푒 = ∗( . ∗ )=0.0225 m



111

By substituting the calculated values above in eq. (6), We have:

1.24*27.7 ≥ 1.24 ∗ 493 ∗ 0.022

34.34 ≥ 13.75 ____________________ (that’s Ok )

Hence,

푀 = 푀 = 34.34 Kn.m

7. Calculation of reinforcements in column by using Equivalent moment (푴풄) ∶ −

푒 = = . ∗ = 69.65푚푚

푒 = 푀푃 =

18.324 ∗ 10493 = 37.16푚푚

푒푒 =

69.6537.16 = 1.874;

푏푏 =

250600 = 0.417

If, >

Then,

(푀 ) = 푀 + 0.8 (푀 )

=34.34 + 0.8×0.417×18.324

(푀 ) = 40.45 KN.m

Also, the above calculated moment act along y-axis ,



112

6.5 Data required for determining steel reinforcement ratio from graph :

푃퐴 =

493 ∗ 10250 ∗ 600 = 3.28푀푃푎

∗= . ∗

∗ ∗= 0.449 Mpa

d`= cover+푑 + = 40 + 8 + = 56푚푚

훾= ` = ∗ = 0.81

When,

훾 = 0.81,휌 = 1.0% (From Chart)

6.5.1 Compute the axial load capacity (푷풖풙) when only eccentricity (풆풚) is present :

1- (ex =0)

h=푏 = 600푚푚휌 = 1.0%

푒ℎ =

푒푏 =

37.16600 = 0.061

From chart

푃퐴 = 11푀푝푎

푃 =퐴 ∗ 11 = (250 ∗ 600 ∗ 11) ∗ 10

= 1650 KN



113

6.5.2 Compute the axial load capacity (퐏퐮퐲)퐰퐡퐞퐧퐨퐧퐥퐲퐞퐜퐜퐞퐧퐭퐫퐢퐜퐢퐭퐲(퐞퐱)퐢퐬퐩퐫퐞퐬퐞퐧퐭:

ퟐ.(푒 = 0)

E=푒 = 69.65푚푚

h=푏 = 250푚푚

푒ℎ =

69.65250 = 0.2786

훾 =푏 − 2푑`

푏 =250 − 2(56)

250 = 0.552

푃퐴 = 5푀푝푎(푓푟표푚퐶ℎ푎푟푡)

푃 = 퐴 ∗ 5 = 5 ∗ 600 ∗ 250 ∗ 10 =750 KN

6.5.3 Check :

푃 > 0.1푃

493 > 0.1 ∗ 2830

493 > 283 => (푡ℎ푎푡푠푂푘)

1푃 ≥

1푃 +

1푃 −

1푃

= + − = 0.0015

푃 = 630.5 KN >493 KN - - - - OK



114

6.5.4 Number of Main Steel bars:

퐴 = 휌푏ℎ

= 0.01*250*600

=1500mm2

Use 8∅ퟏퟔ



115

6.6 Spacing for ties :

mm 2 Suse

mm 2116(mainbars) 16 mm 384848diameter) (tie 48

mm 250column of dimension Least S

50

566

Fig 6.8 shows Sec A-A of column 59



116

6.6.1 Splices for Column :

Compression deformed bars:

300 mm 56.904160240.073 sL

mm 300 bd yf 0.073 sLMPa 20 '

cfFor

6.6.2 Column details:

Interior column (using off – set bars):-

(1) Lap splices length.

(2) Equal (S/2) = 250/2 = 125 mm, where S = tie spacing.

(3) Terminated not more than 75 mm below the main reinforcement.

(4) Extra ties = 150 mm.

23 mm .1420 / 10.4vΑ Extra

kΝ .4044(1.5/6) yf s'A 1.5/6 ties in Force

500212

212202

Fig.6.9 shows minimum requirements for splices for columns under ACI 318-08



117

mm 237.92 0.75933.0340 dL0.75 'K

0.933

461π8

sAsA

'K

mm 200 'K'KdbL dL

mm 340 dbL Use

mm 295.68 024160.044 336 25416420 dbL

yfbd0.044 4

bd yf dbL

:n compressioin bars deformed oft Developmen

mm 12 d use 11.32 d mm 5.0014d π vA

2

prov

req1

21

2

2

1500

f 'c

(5) =325 mm

(6) = 75 mm

(7) = 500 – db/2 – 50 – 8 – db

= 500 – 8–60–8–16=408mm

(8) = maximum slope = 1: 6



118

Interior column Exterior column

Figure (6.10 ) Reinforcement details for columns

6.7 Exterior column (using dowels):-

(1) Where face of column above is off-set 75mm or more from the column below.

(2) Cut-off column verticals stop 75mm below finished floor, i.e. length (2)=h= 75.

(3) Length of dowels equals (two lap length + 75mm).

(4) This length must equal lap length.

(5) dL be must

(6) First tie must be located no more than S/2 above floor.

(7) Where beams frame from four sides (direction) into a column, tie may be terminated

not more than 75 mm below the main reinforcement of such beams.



119

Figure (6.11 ) Reinforcement details for columns

Fig1. Naming of columns from R.A

Column

DESIGN CASE

DESING MOMENT

Design force N

stirrup spacing

Section DESIGN

MOMENT

Required Reinforcement

Ratio % Reinforcments Design force

Transver-sal Reinforc Ement

Distribution

(KN.M) Mz(KN) (KN) (cm) Qz for Shear(KN) & Spacing

1 R 25x60 3 -45.5 -55.64 764.87 1.9 8Ø16 10 -13 2T10 [email protected] 2 R 25x60 3 11.22 -6.93 1000.96 1 8Ø16 10 7.23 2T10 [email protected]

3 R 25x60 3 -0.51 -6.14 921.3 1 8Ø16 10 1.64 2T10 [email protected] 4 R 25x60 3 31.04 -49.76 665.98 1 8Ø16 10 8.8 2T10 [email protected]

5 R 25x60 3 44.44 8.19 1323.49 1.83 8Ø16 10 33.71 2T10 [email protected] 6 R 25x60 3 51.19 -8.83 1361.61 2.33 8Ø16 10 17.77 2T10 [email protected]

7 R 25x60 3 37.96 53.7 699.44 1.2 8Ø16 10 10.83 2T10 [email protected] 8 R 25x60 3 0.93 88.7 1497.77 2.34 8Ø16 10 1.55 2T10 [email protected]

9 R 25x60 3 10.93 99.9 1643.74 3.55 8Ø16 10 18.21 2T10 [email protected] 10 R 25x60 3 -51.86 59.38 798.17 2.61 8Ø16 10 -14.78 2T10 [email protected]

11 R 25x60 3 -70.65 -7.48 1538.06 3.9 8Ø16 10 -24.96 2T10 [email protected] 12 R 25x60 3 -63.88 10.62 1497.37 3.5 8Ø16 10 -22.95 2T10 [email protected]

13 R 25x60 3 1.28 1.51 1409.86 1.33 8Ø16 10 -2.83 2T10 [email protected] 14 R 25x60 3 8.53 0.74 1304.92 1 8Ø16 10 10.24 2T10 [email protected] 15 R 25x80 3 19.57 -84.43 2203.57 1.56 14Ø16 10 8.09 2T10 [email protected] 16 R 25x80 3 -2.13 -78.25 1983.62 1 14Ø16 10 -1.9 2T10 [email protected] 17 R 25x80 3 0.65 -11.39 2726.58 2.89 14Ø16 10 0.31 2T10 [email protected] 18 R 25x80 3 16.79 -13.89 2976.53 3.53 14Ø16 10 13.07 2T10 [email protected]

Table 1 : Columns under plinth level originating directly from foundations (bold font columns have their initial sections changed to resist loads appropriately with excess reinforcements on demand by Robot analysis)

Table 2 : Columns above plinth level (Ground floor columns ) (bold font columns have their initial sections changed to resist loads appropriately with excess reinforcements on demand by Robot analysis)

Column

DESIGN

CASE

DESING

MOMENT Design force N

Required stirrup spacing

Transver-sal Reinforc

Section DESIGN

MOMENT Reinforment Reinforcments Design force ement Distribution

Mz (KN.M) (KN) Raitio (cm) Qz for Shear(KN) Spacing

19 R 25x40 3 34.78 -21.2 585.93 1 6Ø16 10 -5.75 2T10 [email protected] 20 R 25x40 3 49.92 60.93 1206.36 3.53 6Ø16 10 0.89 2T10 [email protected]

21 R 25x40 3 45.6 58.5 1098.5 2.97 6Ø16 10 0.26 2T10 [email protected] 22 R 25x40 3 31.86 13.56 512.67 1 6Ø16 10 4.2 2T10 [email protected]

23 R 25x60 3 -1.08 69.63 1001.98 2.28 8Ø16 10 4.78 2T10 [email protected] 24 R 25x60 3 -3.18 67.91 976.16 2.17 8Ø16 10 4.18 2T10 [email protected]

25 R 25x60 3 -28.15 10.14 486.41 1 8Ø16 10 3.34 2T10 [email protected] 26 R 25x60 3 -7 -15.84 681.32 1 8Ø16 10 0.09 2T10 [email protected]

27 R 25x60 3 -7.39 20.03 742.24 1 8Ø16 10 0.97 2T10 [email protected] 28 R 25x60 3 -31.39 -16.62 559.38 1 8Ø16 10 -4.85 2T10 [email protected]

29 R 25x60 3 -4.2 -74.13 1102.5 2.84 8Ø16 10 -5.86 2T10 [email protected] 30 R 25x60 3 -0.08 -77.82 1131.71 3.02 8Ø16 10 -6.51 2T10 [email protected]

31 R 25x60 3 -1.46 70.15 1016.83 2.35 8Ø16 10 -0.05 2T10 [email protected] 32 R 25x60 3 -1.25 -54.32 942.92 1.2 8Ø16 10 0.89 2T10 [email protected] 33 R 25x80 3 21.58 101.33 1512.02 2.34 10Ø16 10 4.49 2T10 [email protected] 34 R 25x80 3 -35.09 122.26 2099.59 4.23 10Ø16 10 2.94 2T10 [email protected] 35 R 25x80 3 -32.1 -116.4 1917.21 3.66 10Ø16 10 -1.27 2T10 [email protected] 36 R 25x80 3 20.22 -60.09 1339.12 1 10Ø16 10 -2.74 2T10 [email protected]

Table 3 : First story column (bold font columns have their initial sections changed to resist loads appropriately with excess reinforcements on demand by Robot analysis)

Column

DESIGN

CASE

DESING

MOMENT Design force

N

Required stirrup spacing

Design Transver-sal Reinforc

Section DESIGN

MOMENT Reinforment Reinforcments force ement Distribution

Mz (KN.M) (KN) Raitio (cm) Qz for Shear(KN) Spacing

37 R 25x40 3 20.22 17.97 372.58 1 8Ø14 10 -7.44 2T10 [email protected] 38 R 25x40 3 42.72 -21.72 762.3 1 8Ø14 10 0.89 2T10 [email protected]

39 R 25x40 3 59.76 16.49 692.15 1 8Ø14 10 0.39 2T10 [email protected] 40 R 25x40 3 -54.08 12.32 325.22 1 8Ø14 10 5.39 2T10 [email protected]

41 R 25x40 3 -38.26 23.62 629.31 1 8Ø14 10 5.81 2T10 [email protected] 42 R 25x40 3 1.28 20.09 614.16 1 8Ø14 10 5.1 2T10 [email protected]

43 R 25x40 3 -2.76 8.86 307.53 1 8Ø14 10 4.07 2T10 [email protected] 44 R 25x40 3 -32.6 6.33 429.24 1 8Ø14 10 -0.12 2T10 [email protected]

45 R 25x40 3 -10.58 -7.42 471.96 1 8Ø14 10 1.11 2T10 [email protected] 46 R 25x40 3 -11.25 -14.44 355.07 1 8Ø14 10 -6.23 2T10 [email protected]

47 R 25x40 3 35.73 33.88 694.65 1 8Ø14 10 -7.28 2T10 [email protected] 48 R 25x40 3 -4.51 39.46 713.63 1 8Ø14 10 -8.07 2T10 [email protected]

49 R 25x40 3 3.19 -12.98 627.47 1 8Ø14 10 0 2T10 [email protected] 50 R 25x40 3 -2.72 11.51 594.03 1 8Ø14 10 0.67 2T10 [email protected] 51 R 25x60 3 -4.37 33.61 1033.12 1 8Ø16 10 7.7 2T10 [email protected] 52 R 25x60 3 -16.11 -74.27 1393.03 1.19 8Ø16 10 4.52 2T10 [email protected] 53 R 25x40 3 47.73 48.76 1278.86 1 8Ø16 10 -2.71 2T10 [email protected] 54 R 25x40 3 42.85 41.69 899.9 1 8Ø14 10 -3.81 2T10 [email protected]

Table 4 : Second story column (bold font columns have their initial sections changed to resist loads appropriately with excess reinforcements on demand by Robot analysis)

Column

DESIGN CASE

DESIGN MOMENT DESING

MOMENT Design force

N Required stirrup

spacing Design Transver-sal Reinforc

Section Mz Reinforment Reinforcments force ement Distribution

My (KN.M) (KN) Raitio (cm) Qz for Shear(KN) Spacing

55 R 25x40 3 46.31 13.68 155.12 1 8Ø14 10 -6.83 2T10 [email protected] 56 R 25x40 3 -57.52 -4.02 316.71 1 8Ø14 10 0.73 2T10 [email protected]

57 R 25x40 3 -52 3.5 286.25 1 8Ø14 10 0.6 2T10 [email protected] 58 R 25x40 3 41.53 -10.08 133.92 1 8Ø14 10 5.13 2T10 [email protected]


61 R 25x40 3 -36.47 -7.41 126.76 1 8Ø14 10 3.86 2T10 [email protected] 62 R 25x40 3 -14.36 -1.86 173.11 1 8Ø14 10 -0.13 2T10 [email protected]

63 R 25x40 3 -15.6 3.2 192.57 1 8Ø14 10 1.32 2T10 [email protected] 64 R 25x40 3 -41.06 10.79 147.77 1 8Ø14 10 -5.47 2T10 [email protected]


67 R 25x40 3 -5.88 3.02 255.9 1 8Ø14 10 0.44 2T10 [email protected] 68 R 25x40 3 -8.29 -2.81 242.44 1 8Ø14 10 0.45 2T10 [email protected]



DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

120 ____________________________________________________________________________________________

Design Of Foundations

7. Design OF Foundations

7.1.1 Foundation Design Parameters

Determining the settlement of the structure is one of the primary obligations of the geotechnical

engineer. In general, three parameters are required: maximum total settlement ( max),

maximum differential settlement (), and rate of settlement. Another parameter that may be

useful in the design of the foundation is the maximum angular distortion ( /L), defined as the

differential settlement between two points divided by the distance between them. Figure given

below illustrates the maximum total settlement (( max), maximum differential settlement (),

and maximum angular distortion ( /L), of a foundation. Note in Fig. that the maximum angular

distortion ( /L),does not necessarily occur at the location of maximum differential settlement()


121 ____________________________________________________________________________________________


7.1.2 Allowable Settlement

The allowable settlement is defined as the acceptable amount of settlement of the structure and it

usually includes a factor of safety. The allowable settlement depends on many factors,

The Use of the Structure: Even small cracks in a house might be considered

unacceptable, whereas much larger cracks in an industrial building might not even be noticed.

The Presence of Sensitive Finishes: Tile or other sensitive finishes are much less

tolerant of movements.

The Rigidity of the Structure: If a footing beneath part of a very rigid structure

settles more than the others, the structure will transfer some of the load away from the footing.

However, footings beneath flexible structures must settle much more before any significant

load transfer occurs. Therefore, a rigid structure will have less differential settlement than a

flexible one.

Aesthetic and Serviceability Requirements: The allowable settlement for most

structures, especially buildings, will be governed by aesthetic and serviceability

requirements, not structural requirements. Unsightly cracks, jamming doors and windows,

and other similar problems will develop long before the integrity of the structure is in danger.

Another example of allowable settlements for buildings is Table ,where the allowable ––

foundation displacement has been divided into three categories: total settlement, tilting, and

differential movement


122 ____________________________________________________________________________________________


7.2.1 General:

The load from an isolated column may be distributed on the bearing strata, by providing square,

rectangular, and circular footings. The footing may be in form of a flat slab, it may be stepped or

sloped at the edges. the stepping or slopping of foundations is done to save the concrete and thus the

effect of economy in the cost of the footing .

7.2.2 Area of the footing: To determine the area of the footing, total load on the base of the footing

plus the self-weight of the footing is divided by the safe bearing capacity of the soil. Thus if W is the

load from the column and the is the bearing capacity of the soil, then the area of the footing

of the footing is given by :

WA

7.2.3 Depth of the footing : The depth of the footing is fixed from consideration of punching shear

and max. Bending moment in the footing. The shear and bending moments are caused on account of

net upward pressure of the soil below. Since the weight of the footing acts downward, while the net

upward pressure acts upward, the self-weight of the footing is excluded while computing the net

upward pressure due to the soil.

7.2.4 Depth from punching and shear considerations: the depth of the footing slab must be

sufficient to resist the tendency of the column to penetrate or punch through it.

7.3 - Design of a footing typically consists of the following steps:

1. Determine the requirements for the footing, including the loading and the nature of the

supported structure.

2. Select options for the footing and determine the necessary soils parameters. This step is often

completed by consulting with a Geotechnical Engineer.

3. The geometry of the foundation is selected so that any minimum requirements based on soils

parameters are met. Following are typical requirements:


123 ____________________________________________________________________________________________


a. The calculated bearing pressures need to be less than the allowable bearing pressures.

Bearing pressures are the pressures that the footing exerts on the supporting soil.

Bearing pressures are measured in units of force per unit area, such as Kilo Newton

per meter Area.

b. The calculated settlement of the footing, due to applied loads, needs to be less than

the allowable settlement.

c. The footing needs to have sufficient capacity to resist sliding caused by any horizontal

loads.

d. The footing needs to be sufficiently stable to resist overturning loads. Overturning

loads are commonly caused by horizontal loads applied above the base of the footing.

e. Local conditions.

f. Building code requirements.

4. Structural design of the footing is completed, including selection and spacing of reinforcing steel

in accordance to the structural design requirements specific to foundations, as defined in ACI

318-05 Chapter 15.


124 ____________________________________________________________________________________________


7.4 - Structural Design

The following steps are typically followed for completing the structural design of footings ,

based on ACI 318-05:

1. Determine footing plan dimensions by comparing the gross soil bearing pressure and the

allowable soil bearing pressure.

2. Apply load factors in accordance with Chapter 9 of ACI 318-05.

3. Determine whether the footing will be considered as spanning one-way or two-ways.

4. Confirm the thickness of the footing by comparing the shear capacity of the concrete

section to the factored shear load. ACI 318-05 Chapter 15 provides guidance on

selecting the location for the critical cross-section for one-way shear. ACI 318-05

Chapter 11 provides guidance on selecting the location for the critical cross-section for

two-way shear.

6. Structural design of the footing is completed, including selection and spacing of

reinforcing steel in accordance with ACI 318 and any applicable building code. During

this step, the previously selected geometry may need to be revised to accommodate the

strength requirements of the reinforced concrete sections. Integral to the structural design

are the requirements specific to foundations, as defined in ACI 318-05 Chapter 15.

5. Determine reinforcing bar requirements for the concrete section based on the flexural

capacity along with the following requirements in ACI 318-05.

Requirements specific to footings

Temperature and shrinkage reinforcing requirements

Bar spacing requirements

Development and splicing requirements

Seismic Design provisions

Other standards of design and construction, as required


125 ____________________________________________________________________________________________


7.5 Data for Design:

Fig 7.2: showing Foundation Plan for the building while displaying symbols

depicting initially assumed sections


126 ____________________________________________________________________________________________


Fig 7.3: Showing Loads on Foundations by 3d Structural Model built on R.S.A

Fig 7.4: Showing dead & Live Loads on Foundation 33 under column 59


127 ____________________________________________________________________________________________


7.6 Detailed Steps Of Design:

1. Find service dead and live column loads:

PD = Service dead load from column

PL = Service live load from column

P = Pd + PL (typically - see ACI 9.2)

2. Find design (factored) column load, Pu:

PU = 1.2PD + 1.6PL

3. Find an approximate footing depth, hf

h f = d + 10cm and is usually in multiples of 5, 10 or 15 cm.

For Rectangular Column:

4. Find net allowable soil pressure, qnet:

By neglecting the weight of any additional top soil added, the net allowable soil

pressure takes into account the change in weight when soil is removed

where y c is the unit weight of concrete and y s is the unit weight of the displaced

soil and replaced by concrete:

5. Find required area of footing base and establish length and width:

For square footings choose B > Sq r t (A r e q )

For rectangular footings choose B X L > A


128 ____________________________________________________________________________________________


6. Check transfer of load from column to footing: ACI 15.8

a. Find load transferred by bearing on concrete in column: ACI 10.17 basic:

(∅)Pn = (∅ 0.85f'cAx where (∅ = 0.65 and A1 is the area of the column

(Where 1A2A cannot exceed 2)

Note: IF the column concrete strength is lower than the footing, calculate ∅ Pn for the column too.

b. Find the load transferred by dowels

∅푃Do we ls = Pu- ∅Pn

I f ∅ Pn >Pu only nominal dowels are required

c. Find required area of dowels and choose bars

Choose dowels to satisfy the required area and nominal requirements

1. Minimum of 4 bars

2. Minimum As = 0.005Ag ACI 15.8.2.1 ( where Ag is the gross column area)

3. 4 ∅16mm bar

d. Find length of lapped splices of dowels with column bars: ACI 12.16 Ls is the largest

of:

1. larger of ldc or 0.0005 f yd b ( Fy of grade 60 or less)

2. l dc of larger bar

3. not less than 30cm

With confinement: ∅ Pn = ∅ 0.85 f 'cA1 X1A2A


129 ____________________________________________________________________________________________



130 ____________________________________________________________________________________________



131 _______________________________________________________________________________________


7.7 Design of Foundation

sV sM sP footing Type Reaction

5.11- 6.45 868.67 F

D.L

2.61- 3.10 412.39 L.L

Table (7.7-1): serviceability load (SLS) of column 59 on Foundation 33

7.7.1 Area of footing:

Case : (D+L)

)__(______________________________2kN/m 222)(q

2kN/m 222 .24-0.5)-(1.6 6 250 )(q

fh c)fh -fD ( s q )(q

(2)_________ 0.436

2.66L 0.0111

1280.6714.325

PM e

14.325 0.5 3.10) (6.45 3.10 6.45 Mh ) V (V M M M

(1)_ __________1280.67___ 412 868.67 P P P m 2.6 L Assume

allnet

allnet

allallnet

LdL.LD.L

L.LD.L

3

501


132 _______________________________________________________________________________________


7.7.2 Footing Stability:

)5________(______________________________KN201.1917uP

412.391.7868.671.4 L.L 1.7 + D.L 1.4uP

)K .O( ..,...

Le

)_____(________________________________________.euP

huVuM e

= uV

14.3KN.m3.101.76.451.4 3.101.76.451.4 1.7Mu Mu1.4 = uM

direction) long of part sixththan lesser be ty shouldeccentrici (The

1917.201)..(.

11.591KN 2.611.75.111.4 1.7 1.4 L.L D.L

L.Ld.L

4300104043066201040

6

601040

505911314

OK) (Thats .)..(

2.6412.39)(868.67

moment turning Over MomentStabilityS.O.F 1

501034562

..Stable..

(O.K) )_________(2.2.6

60.011111280.67netq

m.B2.6

60.011112.6222

1280.67 B

L6e1

LBsP

netq

..

)b( m.L m.useB

4249210

272

40262

362402


133 _______________________________________________________________________________________


7.7.3 Strength Design:

)____(__________kN/m .22.6

60.010412.42.6minq

_____(7)__________ kN/m.32.6

60.010412.42.6maxq

L6e1

LBuP

min,maxq

2

2

1917.201

1917.201

88699

614

7.7.4 Check one way shear:

7.7.5 Actual Shear Stress

kN 87.4372

29.1136.1430.5832.4d)u(V

2d)u(qmax)u(q

)dX ( B d)u(V

311.29kN/m 299.86) -(314.62.6

299.86d)u(q 2)583.06.2(

Allowable Shear Stress

OK 87.437d)u(V 9.708Vc

kN 708.9106

4172400250.85Vc

dB6

'cf

0.85 Vc

3

Hence , Allowable Shear > Actual Shear (That’s Ok.)

0.5830.417--Ld

)minq maxq(L

dX L minq d)u(q

mm 417 16/2 75 500 d

/2bd c.c h d

20.6

22.6

2C1

2

)(

dX


134 _______________________________________________________________________________________


7.7.6 Check two way shear:

d=h-C.C-db-500-75-16=409 mm

=C1+d=600+409=1009mm b1

b2=C2+d=250+409=659mm

b0=2(b1+b2)=2(1009+659) =3336 mm

0.491

250500

321

11

CC

321

11γ

2kN/m 90.71210.659)(1.00924.3079171

)b(bavquPbo)u(V

kN/m 24.307LB

uPavq

mm .50452b

aC

mm 10187.22

1009 )409( 659)( 26409)( 1009

6409)()1009(

2b

d2b6

)d (b6

d )(bJ

mm 36442414093336dbcA

2

1

21

1

12

11

0

2

411233

233

2

4.26.21917.201

MPa 1.2710161.0255.110187.2

.5045103.140.491090.7121

JaCuMγ

cA)bu(V

uV

11

63

1364424

0


135 _______________________________________________________________________________________


Vu Vc IF

OKVu Vc

MPa 1.3 6

2583.10.85 6

'cfK0.85

Vc

83.1

2

3.4533362409)(401d)/2bs(α1

control833.1)(1)cβ

(1

K0

/25060022

useK

Increase H

Increase column size

Increase 푓 ′


136 _______________________________________________________________________________________


7.8 Design of Flexure in Long direction :

0.04081.18

0.0399)(2.3611ω

1.18uK2.3611

ω

0399.04172400250.9

1088.374dB'

cf0.9f

)u(MuK

mm 4172

1675500d

kN.m 88.374)]6.14(32[3086

)1.(4.2

]max)u2(quf[q6)fX(B

ufM

kN/m 30807.9299

)86.2996.314(60.2

)260.0

26.2(

299

7,8 3b, eq From

)minqmax(q2/C2/L

minqufq

2

6

2

2

2

2

L1

2mm .4170020.0024 ρbdsA

0.002 o.oo24 .req useminreq 0.00240.0408 ρ

yf

'cfωρ

42025

424014

00240

direction long in

2401.4

12 use

12 11.86 /416 π

)s(A)s(A

) N ( bars of number 2bar one

total

16


137 _______________________________________________________________________________________


'm/.L

N'm

ent)reinforcem(secondry mm 204 / 16 12 Use

mm 204204.5 1

752 2400 1NC.C2 L ) S( bars between Spacing

16542

12

12

7.9 Design of flexure in Short direction:

01260

85118

610

8511823205

2305642

.1.18

(0.0126)2.3611ω

0.01264012600250.9

10.uK

d L 'cf 0.9fuM

uK

kN.m .2

2 / 0.25) (2.4.ufM

2)2 / )C(B (

avqufM

kN/m .2.2.LB

uPavq

mm 401 161.5 75 500 b1.5d c.c h d

2

6

2

2

2

2

2

1280.67

2mm 240126000.002 dbρsA

min.420250.0126

yf

'cfωρ

. use min

085

00080

0020


138 _______________________________________________________________________________________


ent)reinforcem (main 245mm / 16 11

mm 250 110

752 2600 ) S( bars between Spacing 11 use

1110.37 /4216 π

2085.2 bar one)s(A

total)s(A ) N ( bars of number

16


139 _______________________________________________________________________________________


Fig 7.5 Detailed Reinforced for foundation


140 _______________________________________________________________________________________


7.10 Development length in footing :

OK required dL provided dL

mm 385.13 252.5

0.9950.81160240.9required dL

mm 0001752

2502400C.C 2CB

shortin provided dL

mm 92575 2

6002600 C.C2CL

longin provided dL

2.5 bd

C K

0.995 = 52.2411

)s(A)s(A

= K mm 19 < bdfor 0.8 = K

bars for top 1.3 K bars bottomfor 1.0 = K

MPa 420 = yf mm 16 = bd : where

'cf4K

KK Kbdyf0.9required dL

C.C - 2C - 2

L provided dL

2

1

4

prov

req32

11

321

2401


141 _______________________________________________________________________________________


7.11 Bearing strength

Bearing strength of column:

clause) (code 2 .60.15AA

where

m 6.24 2.4 2.6 BL A

m 0.15 0.25 0.6 C C A

).(uP kN 2231.25 10 ) 600 250 25 0.85 0.7( uoP

) C C 'cf 0.85 ( uoP

6.24

1

2

2

211

21column

2

2

-3

44

671280

Bearing strength of Footing:

4462.5 ) 2 600 250 25 0.85 0.7( uoP

AA

) C C 'cf 0.85 ( uoP

1

221column

use Clause) 02-318 (ACI 2AA

1

2

We can use min (As) dowels:

14 4 mm 750) 600250 ( 0.005

) sA ( 0.005 ) sA (2

coldowel


142 _______________________________________________________________________________________


7.12 Development of dowels :

022.38694.2961.3 com)d(L 1.3 L

O.K mm 296.14 mm 393 16 2 - 75 - 500

bd2 - c.c -h L

mm 296.94

0.8125416420 K .K .

'cf4

b.dyf )d(L

L

450mm use

2

available1

21com

1

Footing Long direction Short direction

Symbol Length(m) Width(m) As Spacing As Spacing

F1 2.5 2.3 10Ø16 250 9Ø16 250

F2 3.1 3.1 12Ø16 250 12Ø16 250

F3 2.6 2.4 12Ø16 225 11Ø16 250

Table7.10 shows dimension and reinforcement of foundation


ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ 143Conclusions & Recommendations

8. CONCLUSION, RECOMMENDATION AND FUTURE WORK

8.1 Conclusion & Recommendations

Accurate Loads & analysis are the key to correct design also during the design phase it

plays important role to minimize the construction cost. Excellent designers must have

the capacity for organization and management to conduct the process of design so that

it includes cost consideration during the design process.

This research presented a model for design of reinforced concrete elements

since they represent the high value of the total cost of the constructed facility.

In the study, it was found out that, in beams, the ACI 318 allows designer to

use sections more than required. Hence, care should be taken while making

preliminary assumptions for sections as the minimum reinforcement is directly

connected to the gross area (area of section),therefore if too large section is supposed

it may be safe but may have more than enough required reinforcement and hence may

increase in overall cost.

For columns, ACI 318 gives very few limitations for columns , it is stated that

percentage of steel should be in between 1% - 8 %, where as it is felt that amount of

steel less than 1% has a distinct possibility of non-ductile failure as may occur in plain

concrete column. It should be interesting to know that actually , the code (10.8.4)

does permit the use of less than 1% steel if the column is larger than necessary to

carry the loads required . Practically it is rather difficult to fit more than 4% - 5% steel

into formworks and still get concrete down into forms .

The code used was ACI318. The calculations were done on the design of three story

structures elements, which are beams, columns, slabs and foundation. A specific load

was applied and designs were carried by Robot analysis software using ACI code to

find the minimum cost and maximum safety of design according to the code.


ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ 144Conclusions & Recommendations

From the study, it was concluded that all design done on software require sound

knowledge about design methods and philosophies,

The Recommendation I would like to make is that all Civil Engineers should

understand the theory of structures before diving in to lengthy calculations and

mathematics of design and analysis.

In addition, it is worth mentioning that an engineer should not entirely rely on

computerized results as many errors and mistakes are usually resulted from fresh

Computer output results. Hence, it is recommended to design a sample with known

results to compare it with the newly generated ones.

8.2 Suggestion for Further Works

After designing the building with ACI code and explaining the optimum design ,

which is the economical option, the design engineer must improve this method in

order to obtain the optimum design more accurately and easily.

In future, I suggest to any designer to use more than one software of design

especially when to design a large building in which the difference in quantities of

materials is high and that makes the building successful and economic.

By using software programs for design any construction quickly and accurate,

designer engineer can choose and compare more than code to get the minimum cost.

An addition, for more accuracy in design, designer must evaluate the labour costs,

time of construction and finishing costs for the building to obtain the optimum design

of building.

Scope for future works:

1. The above Study can be repeated with different types of steels with

different yield strength and different kinds of concrete with different

compressive strengths

2. The Work can be extended for different kinds of supports other than

fixed

3. The work can be extended and compared by designing by another code

4. The design could be extended by using different load case

Design of Reinforced Concrete Multi-story Commercial building ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ 145 ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــConclusions & Recommendations

APPENDICES

Fig 8.1 Column interaction diagram



Table (8-2) Coefficients for live- load positive moments in slabs



Table (8-3) Coefficients for dead- load positive moments in slabs



Table (8-4) Coefficients for negative moments in slabs



4.0 REFERENCES.

1- ACI, 2008. Building Code Requirements for Structural Concrete (Aci318-08) and

Commentary (ACI318R-08), American Concrete Institute.

2- Design of RC ACI-14-Dr. Nadim , 6th ed

3- Structural Concrete, Theory and Design,4th ed by M.Nadim hassoun

4- IBC Code 2007 Edition

5- Civil-Handbook-by-p-n-Khanna

6- Hibbeler structural analysis 8th Edition

7- Structural Design Guide to the ACI Building Code, 4th ed, 1998_2

8- ACI, Practitioner’s Guide for Slabs on Ground, American Concrete Institute,

Farmington Hills, MI, 1998..

9- Reinforced concrete Design theory and examples by T,J MACGinley and BS

CHOO

10- Other final year projects

11- Autodesk Robot Analysis, 2014. structural analysis, design and detailing

software. user manual window version 7.

12- And many other random informative website and resources …