Design and analysis of Reinforced Concrete Multistory commercial Building using aci-318 metric manually and extensive design by robot analysis,

September 4, 2017 | Author: Abdul Azeem Baig | Category: Beam (Structure), Reinforced Concrete, Concrete, Strength Of Materials, Bending
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Descripción: Design of multistory building by solving a sample manually ans rest of the building by solving on autodesk ...

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DESIGN & ANALYSIS OF REINFORCED CONCRETE MULTISTORY 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

iv

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.

vi

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 Research Background 1.2 Problem statement 1.3 Objectives 1.4 Scope of Study 1.5 Outline of Thesis CHAPTER 2 2.LITERATURE REVIEW 2.1 Introduction 2.2 Building Codes & Standards 2.3 Optimum Cost of Reinforced Concrete Building 2.4 Factors Contributing To the Cost of Building Construction 2.5 Construction Cost 2.6 Research Methodology 2.7 Model OF Design 2.8 Design Specifications And properties of the structure 2.8.1 Materials and Design Parameters 2.9 Design Loading 2.9. Partial Safety Factors 2.9.3 Factors of Safety Loads And Strength Of Section By Strength 2.10 Design Methods 2.10.1 Object of Structural Design 2.10.2 Philosophy of Limit State Design 2.11 Project Flow Chart 2.11. Expected Results CHAPTER 3 3. Design Of Slabs 3.1.1 Definition 3.1.2 Introduction 3.1.3 DesignConcepts 3.1.4 Types of slabs 3.1.5 One & two way slabs outlined 3.1.6 Econlomical Choice According to size and loading 3.1.7 Calculation of thickness for one way slab 3.1.8 Design procedure for one way slab 3.1.9 ACI Code specified method for two-way slabs 3.1.10 Two-way slab design procedure 3.1.11 Classification of slabs 3.1.12 Purpose of main steel in slabs 3.1.13 Analysis methods for slabs 3.1.14 Slabs direction in ribbed slab 3.1.15 Design concept

1 1 2 3 3 3 4 4 4 8 9 9 10 11 14 14 15 16 17 18 18 19 21 22 23 23 23 23 24 25 26 27 28 29 29 29 29 30 30 31

viii

3.1.16 Maximum Reinforcement Ratio 3.1.17 Shrinkage Reinforcement Ratio 3.1.18 Loads assigned to slabs 3.2.1 Elevation Plans of slabs 3.2.2 Design procedure for one-way slab 3.2.3 Design procedure for two-way slab 3.2.4 Data for Design 3.3 Design of two-way slab 3.5 Design of one -way slab CHAPTER 4 4. Design of Beams 4.1 Introduction 4.2 Struvtural theory of beams 4.2.1 Types of a beam 4.2.2 Scope of usage of beam 4.2.3 Relation of reinforcement with section 4.3 Assumptions 4.4 Loading data 4.5 Design Procedure 4.6 Flow charts 4.7 Information about Sample of design 4.7. Desing assumptions 4.7.2 Check Deflection 4.7.3 Sizing the cross-section 4.8 Design of Flexure 4.8.1 Actual Depth 4.8.2 Minimum Ratio of steel required 4.8.3 Design Reinforcement for every moment in beam 4.9 Design of shear 4.10 Development length CHAPTER 5 5. Design of Stairs 5.1 Geometrical design of stairs 5.1.2 Check for reliabilty 5.1.3 Check for angle 5.2 Detailed design of stair 5.3 No. of steps in each flight 5.4 Structural design of stairs 5.5 Design for flight no.1 & 3 5.6 Data for design for flight no 2 5.6.2 Design for flexure for flight no 2 5.7 Reinforcement details

31 31 32 33 35 37 38 39 45 50 50 50 50 51 51 54 54 55 56 58 59 59 59 60 60 60 60-68 69 75 78 78 78 79 81 82 83 86 87 88 71

ix

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 7.12 Development length in dowels CHAPTER 8

141 142 -

8.

143 143 144

CONCLUSION AND FUTURE WORK 8.1 Conclusion 6.2 Suggestion of Further Works 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 2 .3: 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

A

TITLE

Charts

PAGE

145

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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.

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Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 ACI318-08.

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.

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‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

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‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪+‬‬ ‫‪Fig 1.3 Right Side view‬‬

‫‪Fig1.4 perspective view‬‬

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Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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.

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Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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.

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Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 (ACI31802), (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

8

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

9

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

concerning the size of the buildings and decisions concerning material specifications and building configuration (Figure 2.1).

Cost

Area

Specificatio n and shape

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.

10

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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.

11

2016™ Area = 405.5 m²

1.20m

‫ﻋﺑ د اﻟﻌظﯾ م ﻧﻌﯾ م ﺑﯾ ك‬

MENS TOILET

3.60m

3.50m

1.5

2.40m

6.20m

3.50m

Toilet

Shop

3.50m

7.00m

‫ﻣﺷ روع ﺗﺻ‬

21.20m

Architectural Layout of Elevation plan

‫ﻣﯾم ﻣﺑ ﻧﻲ ﺗﺟ ﺎري‬

4.40m

Date : 2/8/2016

‫ﻛﻠﯾ ﺔ اﻟﮭﻧدﺳ ﺔ ﺟﺎﻣﻌﺔ ﺑ ﻧﻲ وﻟﯾ د‬

19.60m 4.20m

1.40m

1.00m

6.20m

1.70m WOMENS TOILET

1.40

‫اﻟﻣﺳ ﻘط اﻻﻓﻘ‬

6.00m

‫ﻲ ﻟ دور اﻻرﺿ ﻲ ﻟﻠﻣﺑ‬

2.40m

‫ﻧﻰ‬

CLEANERS ROOM

‫ﻛﻠﯾ ﺔ اﻟﮭﻧدﺳ ﺔ ﺟﺎﻣﻌﺔ ﺑ ﻧﻲ وﻟﯾ د‬

6.00m

Shop

Prayer Room

2.30m

Shop Shop ‫اﻋداد‬ ‫اﻟرﺳ م‬

0.30m

3.00m 7.00m 7.00m

2016™ Area = 405.5 m²

1.20m

2.40m

‫ﻧﻰ‬

1.5

CLEANERS ROOM

6.20m

3.50m

3.50m

‫ﻣﯾم ﻣﺑ ﻧﻲ ﺗﺟ ﺎري‬

Toilet

Shop Shop

2.40m

‫اﻟﻣﺳ ﻘط اﻻﻓﻘ‬

7.00m

‫ﻣﺷ روع ﺗﺻ‬

21.20m

Architectural Layout of Elevation plan

4.40m

:

‫ﻛﻠﯾ ﺔ اﻟﮭﻧدﺳ ﺔ ﺟﺎﻣﻌﺔ ﺑ ﻧﻲ وﻟﯾ د‬

19.60m 4.20m

6.00m

‫ﻲ ﻟ دور اﻻول و اﻟﺛ‬

1.40m

‫ﻛﻠﯾ ﺔ اﻟﮭﻧدﺳ ﺔ ﺟﺎﻣﻌﺔ ﺑ ﻧﻲ وﻟﯾ د‬

‫ﺎﻧﻲ ﻟﻠﻣﺑ‬

2.30

1.00m

6.20m

1.70m WOMENS TOILET

MENS TOILET

3.60m

3.50m

‫ﻋﺑ د اﻟﻌظﯾ م ﻧﻌﯾ م ﺑﯾ ك‬.‫م‬

‫اﻋداد‬ ‫اﻟرﺳ م‬

0.30m

6.00m

Shop

Prayer Room 2.30m

Shop Shop

3.00m 7.00m 7.00m

2.80m

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

Building usage

Shops

Story height

Ground floor

4m

1th & 2th

4m

Length of building

21.00 m

Width of building

20.00 m

Height of building

13.8 m 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 Columns

Beams

Dimensions(exterior)

Dimensions(Interior)

Ground floor

500x250 mm

600x250 mm

1th to 2th

400x250 mm

500x250 mm

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

14

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

24 kN/m 21.718 KN/mm

Modulus of elasticity; E

420 N/mm

Yield stress fy

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

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Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬



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

Area pressure

4.2

KN/m2

200 mm thickness (without finishes) 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.

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Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

Office use

2.4

Computer use

4.79

Access floor system

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

4.79

anticipated occupancy Lobbies and first-floor corridors 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

17

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

18

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

19

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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).

20

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

2.11 Project flow chart.

START

Generate Plan in AutoCAD Software

Import Structure Plan into (Autodesk Robot Structure Analysis)

Assigning Of Elements Based On assumed Structure Plan

Import the 2d plan to 3d software to process architectural visualization and establish ensuring it compliance with structural midel

General a 3D Model

Assign Loads and Load Cases Acting On the Structure

Correct any errors, Recheck Properties of Elements

Run Analysis

Change the dimensions of section

Ρmax< ρ< ρmin

Error

Decrease Spacing between members or add new members(such as column )

Run R.C member Required Reinforcement Calculations

Run Provided Reinforcement wizard & Check The Steel Ratio ρ.

Ρmax< ρ< ρmin

Report Design Results Flow chart 1 : Design procedures

Architectural output Drawings

21

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪2.10.1 EXPECTED RESULT.‬‬

‫‪22‬‬

‫‪The most economical alternative solution would be identified.‬‬

‫‪1-‬‬

‫‪The required quantity of material would be evaluated.‬‬

‫‪2-‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 twoway 3.1.2 Introduction The slab provides a horizontal surface and is usually supported by columns, beams or walls. One-wa y 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 validit y 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 capacit y 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 deformatio ns 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 carr y 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-wa y slabs.

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬23 Design Of Slabs

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ma ybe solid of uniform thickness or ribbed with ribs running in one or two directions. Slabs with var ying depth are generally not used. Slab are horizontal plate elements forming floor and roof in building and normally carr y 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 b y 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 b y 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 capacit y. 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 d esigned As one-way slab. 2. One-way joist floor system: This t yp e of slab, also called ribbed slab, is supported by reinforced Concrete ribs 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 ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬24 Design Of Slabs

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪transferred to the Supporting column, with punching shear resisted by drop‬‬ ‫‪panels.‬‬ ‫‪3. Two-way waffle slab: This t ype of slab consists of a floor slab with a‬‬ ‫‪length-to-width ratio less Than 2, supported b y waffles in two directions.‬‬

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

‫‪ 25‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 3.1.6 ECONOMICAL CHOICE OF CONCRETE FLOOR SYSTEMS ACCORDING TO SIZE, DIMENS IONS 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 t ype of building, architectural la yout, 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 capacit y 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-wa y 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-wa y 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.

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬26 Design Of Slabs

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪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

‫‪ 27‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 3.1.8 Design Procedure: 

One-wa y 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 distributed loads and supported on equally or nearly equally sp aced 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 limitatio ns 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.

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬28 Design Of Slabs

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 distribution 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 distributed 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 distribution 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

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬29 Design Of Slabs

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 3.1.13 Types of analysis-methods for slabs: 

Elastic analysis co vers 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 wa y slab : In one-way ribbed slabs ribs ma y 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.

`

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬30 Design Of Slabs

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ribs 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-wa y 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 p er 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 ma y be 0.12 percent where high yield strength deformed bars .

3.1.16 Maximum Reinforcement Ratio: One-wa y 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=



bh;



shrinkage = 0.0018.

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

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬31 Design Of Slabs

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 densit y of the reinforced concrete. (2) Weight of slab covering materials: This weight per unit area depends on the t ype 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 la yer 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. ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬32 Design Of Slabs

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪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

‫‪ 33‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Fig 3.1.3 Elevation plan for slabs on second floor‬‬

‫‪ 34‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪ 35‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪ 36‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

3.2 Steps for design of two way solid slab:  Find the moment coefficients in each slab:  For continuous edges ( -Ve moments ): ( M - ve )a  ( Ca ) neg . Wut . L2a ( M - ve ) b  ( Ca ) neg . Wut . L2b

 Span moments ( +Ve moments ): ( M  ve )a  [ (Ca )dL . Wud  ( Ca )LL . WuL ] L2a ( M  ve ) b  [ (Cb )dL . Wud  ( Cb ) LL . WuL ] L2b



For discontinuous edges ( -Ve moments ): ( M - ve )a  ( M  ve )a / 3 ( M - ve ) b  ( M  ve )b / 3

M(-Ve) -

+ M(+Ve)

 Effective Depth (d):

d  h  c.c  d u st  Percentage of steel (ρ): ρ

Mu

840  f y  d 2 h ρ min  0.002  s d fy Check : ρ  '  0.113

OK

fc

S  Spacing fy

For ρ  ρ  ω

f c' f

' c

fy

between

bars 

(A s )one ρd

bar

 2h s    450 mm

 0.113  ρ min

As  ρ  b  d

,

b  1000 mm

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬37 Design Of Slabs

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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-wa y 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 ma y be 0.12 percent where high yield strength deformed bars .

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬38 Design Of Slabs

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪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 :‬‬

‫‪lb‬‬

‫‪hs ‬‬

‫‪ 100mm‬‬ ‫‪6‬‬ ‫‪( 34 ‬‬ ‫)‬ ‫‪M‬‬ ‫‪La‬‬ ‫‪if‬‬ ‫‪m‬‬ ‫‪ 0.5‬‬ ‫)‪then (one - way slab‬‬ ‫‪Lb‬‬ ‫‪6.50‬‬ ‫‪else if m ‬‬ ‫)‪ 0.85  0.5 (two - way solid slab‬‬ ‫‪7.50‬‬ ‫‪7.50  1000‬‬ ‫‪hu ‬‬ ‫‪ 183.03mm‬‬ ‫‪6‬‬ ‫‪( 34 ‬‬ ‫)‬ ‫‪0.86‬‬

‫‪USE hs =200mm _________________________eq 3.1‬‬ ‫‪d  hu  c.c  d st‬‬ ‫‪12‬‬ ‫‪d  200  25   169‬‬ ‫‪2‬‬ ‫‪3.3.1.2 Calculation of loads on slabs S2:‬‬ ‫‪hs‬‬ ‫‪  c  flooring‬‬ ‫‪1000‬‬

‫‪WD ‬‬

‫‪200‬‬ ‫‪ 24  2.97  7.77KN/m 2‬‬ ‫‪1000‬‬ ‫‪Wu  1.4 DL  1.7 L L‬‬ ‫‪W  4 KN / m 2  14777  1.7  4‬‬ ‫‪WD ‬‬

‫‪LL‬‬

‫‪Wu 17.67 KN/m 2‬‬ ‫‪ 39‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 3.3.2 Moments at short direction:  For Discontinuous edge M

1 3  3 ( M ve )a 24.7 M   8.23KN .m 1 3 M  1



2

For mid-span M  [(Ca) WUD  ( Ca ) W ]  La 2 2 Dl ll ul 2  [ 0.029 10.78  0.040  6.8 ] 6.50  24.7 KN .m

 For Continuous Edge M  ( Ca )neg Wut  L2a 3 M  ( 0.049 ) 17.67  6.50 2  36.58KN .m 3

Slab

Case

m

S2

8

0.85

DL

LL

-Ve

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 4  ( Cb )neg Wut  L2a M 4  ( 0.046 )17.67  7.502  45.72 KN .m  For Mid-span M 5  [(Cb)  W  ( Cb )  W ]  La 2 uD Dl ll ul M  [ 0.017  10.78  0.022  6.8 ] 7.502  18.72KN .m 5

 For continuous edge M M 6 4

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬40 Design Of Slabs

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪3.4 Design for flexure:‬‬‫‪3.4.1 Reinforcement At Short direction:‬‬ ‫‪A . (La-discontinuous): (-ve) M1‬‬

‫‪8.23106‬‬

‫‪M u 106‬‬

‫‪‬‬

‫‪ 0.0008‬‬ ‫‪840  Fy  d 2 840  420 169 2‬‬ ‫‪h‬‬ ‫‪200‬‬ ‫‪ min  0.002  s  0.002 ‬‬ ‫‪ 0.0023  control‬‬ ‫‪d‬‬ ‫‪169‬‬ ‫‪‬‬

‫‪ min   req‬‬ ‫‪Use  min‬‬

‫‪check :‬‬ ‫‪ 0.113‬‬ ‫‪Fc‬‬ ‫‪  min‬‬ ‫‪Fy‬‬

‫‪fy‬‬ ‫‪fc‬‬

‫‪for  ‬‬

‫‪then use    ‬‬

‫‪420‬‬ ‫‪ 0.03  0.113 O.K (NO need) for new value‬‬ ‫‪25‬‬

‫‪0.0023‬‬

‫‪Ast    b  d  0.00231000130  299mm 2‬‬ ‫‪A‬‬ ‫‪‬‬ ‫‪sb  2hs‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ d‬‬ ‫‪450mm‬‬

‫‪ D 2  12 2‬‬ ‫‪‬‬ ‫‪ 113‬‬ ‫‪4‬‬ ‫‪4‬‬

‫‪S‬‬

‫‪A ‬‬ ‫‪sb‬‬

‫‪113‬‬ ‫‪ 377.9mm  2hs‬‬ ‫‪0.0024 130‬‬

‫‪S‬‬

‫‪Use S  320mm‬‬ ‫'‪Use  12/320mm/m‬‬

‫‪ 41‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪B. At Short direction (La-middle) : (+ve) M2‬‬ ‫‪24.7  10 6‬‬

‫‪ req ‬‬

‫‪ 0.0024‬‬ ‫‪840  420 1692‬‬ ‫‪h‬‬ ‫‪200‬‬ ‫‪ min  0.002  s  0.002 ‬‬ ‫‪ 0.0023‬‬ ‫‪d‬‬ ‫‪169‬‬ ‫‪ req   min , use  req‬‬

‫‪check :‬‬ ‫‪ 0.113‬‬

‫‪fy‬‬ ‫‪fc‬‬

‫‪‬‬

‫‪420‬‬ ‫‪ 0.04  0.113 O.K‬‬ ‫‪25‬‬ ‫‪A‬‬ ‫‪2hs‬‬ ‫‪‬‬ ‫‪S  s one bar  ‬‬ ‫‪‬‬ ‫‪ d‬‬ ‫‪450mm‬‬

‫‪0.0024 ‬‬

‫‪ d2 / 4‬‬ ‫‪113‬‬ ‫‪ 278.59mm‬‬ ‫‪  d 0.0024 169‬‬ ‫‪Use 270mm‬‬ ‫‪Use  12/270mm‬‬ ‫‪S‬‬

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

‫‪ 0.0036‬‬

‫‪36.58  10 6‬‬

‫‪ req ‬‬

‫‪840  420  1692‬‬ ‫‪ 0.0023‬‬

‫‪, use  req‬‬

‫‪min‬‬

‫‪‬‬

‫‪min‬‬ ‫‪ req  ‬‬

‫‪check :‬‬ ‫‪420‬‬ ‫‪ 0.060  0.113‬‬ ‫‪25‬‬ ‫‪113‬‬ ‫‪S‬‬ ‫‪ 185.7mm‬‬ ‫‪0.0036  169‬‬

‫‪0.0036 ‬‬

‫‪Use  12/180mm‬‬

‫‪ 42‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪3.4.2 Reinforcement At long direction‬‬ ‫‪A. (Lb-edge) continuous: M6-ve‬‬

‫‪45.72  106‬‬

‫‪ req ‬‬

‫‪ 0.005‬‬ ‫‪840  420  157 2‬‬ ‫‪0.002hs 0.002  200‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ 0.0025‬‬ ‫‪min‬‬ ‫‪d‬‬ ‫‪157‬‬ ‫‪check :‬‬ ‫‪420‬‬ ‫‪ 0.084  0.113 O.K‬‬ ‫‪25‬‬ ‫‪113‬‬ ‫‪S‬‬ ‫‪ 102.82mm‬‬ ‫‪0.005 157‬‬

‫‪0.005 ‬‬

‫‪Use  12/100mm‬‬ ‫‪B. At long direction (Lb-mid span): M5+ve‬‬

‫‪ 0.0021‬‬

‫‪18.72  106‬‬ ‫‪840  420  157 2‬‬

‫‪ req ‬‬ ‫‪‬‬

‫‪ 0.0025,‬‬ ‫‪min‬‬ ‫‪ req  ‬‬ ‫‪min‬‬ ‫‪use ‬‬ ‫‪ 0.0025‬‬ ‫‪min‬‬ ‫‪113‬‬ ‫‪S‬‬ ‫‪ 287mm‬‬ ‫‪0.0025  157‬‬ ‫‪Use  12/280mm‬‬

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

‫‪M M‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪Hence use‬‬ ‫‪Use  12/100mm‬‬

‫‪ 43‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫)‪S(req‬‬

‫‪use‬‬

‫‪req‬‬

‫‪320‬‬

‫‪377‬‬

‫‪0.0023‬‬

‫‪0.0008‬‬

‫‪270‬‬ ‫‪180‬‬ ‫‪100‬‬ ‫‪280‬‬ ‫‪100‬‬

‫‪278.5‬‬ ‫‪185.7‬‬ ‫‪102.8‬‬ ‫‪287‬‬ ‫‪102.8‬‬

‫‪0.0024‬‬ ‫‪0.0036‬‬ ‫‪0.005‬‬ ‫‪0.0025‬‬ ‫‪0.005‬‬

‫‪0.0024‬‬ ‫‪0.0036‬‬ ‫‪0.005‬‬ ‫‪0.0021‬‬ ‫‪0.005‬‬

‫‪used‬‬

‫‪S‬‬

‫‪min‬‬

‫‪‬‬

‫‪0.0023‬‬ ‫‪0.0023‬‬ ‫‪0.0023‬‬ ‫‪0.0025‬‬ ‫‪0.0025‬‬ ‫‪0.0025‬‬

‫)‪d (mm‬‬ ‫‪169‬‬

‫‪157‬‬

‫)‪M ( KN / m‬‬ ‫‪u‬‬

‫‪sec‬‬

‫‪8.23‬‬ ‫‪24.7‬‬ ‫‪36.58‬‬ ‫‪45.72‬‬ ‫‪18.72‬‬ ‫‪45.72‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬

‫‪Direction‬‬ ‫‪short‬‬

‫‪long‬‬

‫‪Table (3-4) two- way slabs reinforcement‬‬

‫‪ 44‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ap art than 45 cm, center-tocenter. 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 : L b hs   100mm 6 (34  ) m La m  0.5 L b 1.90m m  0.301  0.5( one way solid slab) 6.30m 6.30 1000 hs   116.8mm 6 ( 34  ) 0.301

 Use slab thickness from largest span hs= 200mm ( Eq 3.1 ) ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬45 Design Of Slabs

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪3.5.2 Calculation of loads on slabs S8:‬‬ ‫‪h‬‬ ‫‪WD  s   c  floowing‬‬ ‫‪1000‬‬

‫‪200‬‬ ‫‪ 24  2.97  7.77KN/ \ m 2‬‬ ‫‪1000‬‬ ‫‪Wu  1.4 DL  1.7 L L‬‬ ‫‪W  4KN / m 2‬‬ ‫‪WD ‬‬

‫‪LL‬‬

‫‪Wu 14177  1.7  4‬‬ ‫‪Wu 17.67 KN/m 2‬‬ ‫ ‪3.5.3 Moments at short direction:‬‬‫‪ For continuous edge‬‬ ‫‪W L2 17.67 1.92‬‬ ‫‪M  u n‬‬ ‫‪ 7.08KN .m‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪2‬‬

‫‪ For Mid-span‬‬ ‫‪Wu  L2n 17.67 1.9 2‬‬ ‫‪‬‬ ‫‪ 4.5 KN .m‬‬ ‫‪8 14‬‬ ‫‪14‬‬

‫‪M‬‬

‫‪ For continuous edge‬‬ ‫‪Wu  L2n 17.67  1.9 2‬‬ ‫‪‬‬ ‫‪ 2.65KN .m‬‬ ‫‪9‬‬ ‫‪24‬‬ ‫‪24‬‬

‫‪M‬‬

‫‪3.5.4 Minimum Steel Reinforcement:‬‬

‫‪10‬‬ ‫‪ 170mm‬‬ ‫‪2‬‬ ‫‪h‬‬ ‫) ‪min  0.002( s‬‬ ‫‪d‬‬ ‫‪200‬‬ ‫( ‪min  0.002 ‬‬ ‫)‬ ‫‪170‬‬ ‫‪min  2.37 10 3‬‬

‫‪d  200  25 ‬‬

‫‪min  0.00235‬‬

‫‪ 46‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪A . Reinforcement For edge-moment M7:‬‬ ‫‪M  7.08KN .m‬‬ ‫‪7‬‬ ‫‪M‬‬ ‫‪u‬‬ ‫‪Ku ‬‬ ‫'‬ ‫‪2‬‬ ‫‪ f c d  1000‬‬ ‫‪7.08 106‬‬

‫‪Ku ‬‬

‫‪ 0.0108‬‬ ‫‪0.9  25 1000 170 2‬‬ ‫) ‪1  1  2.36( 0.0108‬‬ ‫‪‬‬ ‫‪ 0.0109‬‬ ‫‪1.18‬‬ ‫'‪f c‬‬ ‫‪25‬‬ ‫‪f     0.0108 ‬‬ ‫‪ 0.00064   min‬‬ ‫‪fy‬‬ ‫‪420‬‬ ‫‪Use ‬‬ ‫‪min  0.00235‬‬

‫‪A   b d‬‬ ‫‪S‬‬ ‫‪2‬‬ ‫‪A  0.00235 170 1000  399.5 mm m‬‬ ‫‪S‬‬ ‫‪A b 1000‬‬ ‫‪S S‬‬ ‫‪A‬‬ ‫‪S‬‬ ‫‪ D 2  10 2‬‬ ‫‪A ‬‬ ‫‪‬‬ ‫‪ 78.5mm 2‬‬ ‫‪S‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪78.5 1000‬‬ ‫‪S req ‬‬ ‫‪ 196.4mm  190mm‬‬ ‫‪4‬‬ ‫'‪Ues 5 / 190/m‬‬ ‫‪B. Reinforcement for Mid-span:‬‬

‫‪M  4.5 kN.m‬‬ ‫‪8‬‬ ‫‪1.45  10 6‬‬ ‫‪Ku ‬‬ ‫‪ 0.0089‬‬ ‫‪0.9  20  1000  952‬‬ ‫) ‪1  1  2.36( 0.0069‬‬ ‫‪ω ‬‬ ‫‪ 0.0070‬‬ ‫‪1.18‬‬ ‫'‪f c‬‬ ‫‪25‬‬ ‫‪ρ ‬‬ ‫‪ 0.0070 ‬‬ ‫‪ 0.00041   min‬‬ ‫‪fy‬‬ ‫‪420‬‬ ‫‪Ues‬‬ ‫‪min  0.00235‬‬ ‫‪ 47‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪As    b  d‬‬ ‫‪2‬‬

‫'‪m‬‬

‫‪AS  0.00235  170  1000  399.5 mm‬‬ ‫‪AS b  1000‬‬ ‫‪AS‬‬

‫‪S‬‬

‫‪78.5  1000‬‬ ‫‪ 196.4mm  190mm‬‬ ‫‪399.5‬‬ ‫'‪Use 5 10/190/m‬‬ ‫‪S‬‬

‫‪C . Reinforcement For edge-moment M9 :‬‬

‫‪* M  2.65KN .m‬‬ ‫‪9‬‬ ‫‪2.65  10 6‬‬

‫‪Ku ‬‬

‫‪ 0.0040‬‬ ‫‪0.9  25  1000  170 2‬‬ ‫) ‪1  1  2.36( 0.0040‬‬ ‫‪‬‬ ‫‪ 0.0041‬‬ ‫‪1.18‬‬ ‫'‪f‬‬ ‫‪25‬‬ ‫‪    c  0.0041 ‬‬ ‫‪ 0.0002   min‬‬ ‫‪fy‬‬ ‫‪420‬‬ ‫‪Use  min  0.00235‬‬ ‫‪2‬‬ ‫'‪A    b  d  0.00235  170  1000  399.5 mm m‬‬ ‫‪S‬‬ ‫‪S  196.4mm  200mm‬‬ ‫'‪Use 5 10/200/m‬‬

‫‪ 48‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪3.5.5 Secondary reinforcement:‬‬

‫‪A s t  0.002 bh s‬‬ ‫‪2‬‬ ‫'‪A s t  0.002  1000  200  400 mm m‬‬ ‫‪A b  1000‬‬ ‫‪S s‬‬ ‫‪At‬‬ ‫‪s‬‬

‫‪ D 2 3.14  8 2‬‬ ‫‪As b ‬‬ ‫‪‬‬ ‫‪ 50.27mm 2‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪50.27 1000‬‬ ‫‪S‬‬ ‫‪ 125.67mm‬‬ ‫‪400‬‬ ‫‪Use S  120 mm‬‬ ‫'‪Use S10 / 120 mm/m‬‬

‫‪3‬‬

‫‪2‬‬

‫‪1‬‬

‫‪Section‬‬

‫‪2.65‬‬

‫‪4.5‬‬

‫‪7.08‬‬

‫‪0.0040‬‬

‫‪0.0069‬‬

‫‪0.0108‬‬

‫‪Mu‬‬ ‫‪Ku‬‬

‫‪0.0049‬‬

‫‪0.0070‬‬

‫‪0.0109‬‬

‫‪0.00235‬‬

‫‪0.00235‬‬

‫‪0.00235‬‬

‫‪399.5‬‬

‫‪399.5‬‬

‫‪399.5‬‬

‫‪196.4‬‬

‫‪196.4‬‬

‫‪196.4‬‬

‫‪190‬‬

‫‪190‬‬

‫‪190‬‬

‫‪ω‬‬ ‫‪ρ‬‬ ‫‪As /m‬‬ ‫)‪S(req‬‬

‫)‪(used‬‬

‫‪S‬‬

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

‫‪ 49‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Slabs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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. ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬50 Design Of Beams

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ( ) and strain ( ). While, concrete at compression zone has not yet reached to the ultimate strain ( =0.003).



In case of compression failure, provided (ρ) at tension zone is greater than ρbal . Concrete at compression zone will reach ultimate stress (

ult)

and strain (

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.

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬51 Design Of Beams

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 or Φ Mn ≥ Mu” Where Mu is the moment due to the factored loads, which the ACI code refer to as the required ultimate moment. This is a load effect computed by structural analysis from the governing combination of factored loads given in ACI section9.2. The term Mn refers to the 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 ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬52 Design Of Beams

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ account for possible variations in dimensions and material strengths and possible inaccuracies in the strength equations. In ACI318 ultimate moment required as flowing:-

= 0.85

′ . . ( − )

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: Φ Vn ≥ 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.

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬53 Design Of Beams

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 4.3 Design Assumptions :

fc' = 25 MPa

f y = 420 MPa

Cover= 25 mm

d = 16 mm b

d st = 8 mm 4.4 Loading Data: The beam carries load from following: 1- Load from wall

W  γ b  height of wall thickness of wall1m' 2- Concentrated load from other beams & columns ( kN )

3 - Own weight of beam

W  γc  h  b (kN/m) 4 - Loads from one way solid slab

W  Wu (kN/m2 ) 

span of slab  kN/m 2

5 - Load from two way solid slab

 Short Beam  W 

Wu  S 3

Wu  S  3  m2    2  3   Short Span m Long Span

 Long Beam  W 

S  Short Span (C.L to C.L)

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬54 Design Of Beams

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪4.5 Design procedure:‬‬

‫‪1- Design of flexure:‬‬

‫‪db‬‬ ‫‪2‬‬ ‫‪,   0 .9‬‬

‫‪d  h  cover  d s t ‬‬ ‫‪M u 106‬‬ ‫‪  f c'  b  d‬‬ ‫‪1  1  2.36 ku‬‬ ‫‪ω‬‬ ‫‪1.18‬‬ ‫'‬ ‫‪f‬‬ ‫‪ρ  ω c‬‬ ‫‪fy‬‬ ‫‪ku ‬‬

‫‪if min    max‬‬ ‫‪also  max  0.7 b‬‬ ‫‪ ‬‬ ‫‪‬‬ ‫‪ ‬‬ ‫‪ ‬‬

‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬

‫'‪f c‬‬

‫‪‬‬ ‫‪ 600‬‬ ‫‪β ‬‬ ‫‪1‬‬ ‫‪f‬‬ ‫‪ 600  f y‬‬ ‫‪y‬‬ ‫‪‬‬

‫‪‬‬ ‫‪‬‬ ‫‪ 0.85 ‬‬ ‫‪‬‬ ‫‪‬‬

‫‪‬‬ ‫‪b‬‬

‫‪‬‬ ‫' ‪fc‬‬ ‫‪1.4‬‬ ‫‪also ‬‬ ‫‪ the max of ‬‬ ‫‪or‬‬ ‫‪min‬‬ ‫‪ fy‬‬ ‫‪4  fy‬‬ ‫‪‬‬ ‫)‪  0.85 - 0.007 (fc ' - 28‬‬ ‫‪where 0.85    0.65 .65‬‬ ‫‪As  ρ b  d‬‬

‫‪‬‬

‫‪‬‬

‫‪‬‬

‫‪‬‬

‫‪fc ' ‬‬ ‫‪‬‬ ‫‪ the max of 1.4 or‬‬ ‫‪min‬‬ ‫‪4  fy ‬‬ ‫‪ fy‬‬

‫‪also ‬‬

‫‪OR‬‬

‫‪Mu‬‬

‫‪‬‬

‫‪fc ' ‬‬

‫‪/ 0 . 85‬‬

‫‪1  ( 2 Ru‬‬

‫‪b d 2‬‬ ‫' ‪0 . 85 fc‬‬ ‫‪1 ‬‬ ‫‪Fy‬‬

‫‪‬‬

‫‪Ru‬‬

‫‪‬‬ ‫‪‬‬

‫‪‬‬

‫‪ 55‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪4.6.1 Flow-chart for design of flexure for singly reinforced beam:‬‬

‫‪Flow-Chart: 1‬‬

‫‪ 56‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪4.6.2 Flow-chart for design of flexure for doubly reinforced beam:‬‬

‫‪4.7 Sample of design:‬‬‫‪ 57‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

‫‪ 58‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Fig 4.3. Spans and sections for beam 59‬‬ ‫‪4.7.1 Design Assumptions‬‬

‫‪f c' = 25 MPa‬‬ ‫‪f y = 420 MPa‬‬

‫‪d = 16 mm‬‬ ‫‪b‬‬ ‫‪d st = 8 mm‬‬

‫‪Cover = 25 mm‬‬

‫‪b  250 mm‬‬ ‫‪H  400 mm‬‬ ‫‪4.7.2Check deflection:‬‬

‫‪Ln  6.75 0.25/2- 0.25/2 6.5m‬‬ ‫‪Ln  6.30- 0.25/2- 0.25/2 6.30m‬‬ ‫‪Ln  6.75 0.25 / 2  0.25 / 2  6.50m‬‬ ‫‪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

‫‪h min  6500  351.35mm‬‬ ‫‪18.5‬‬

‫‪hmin < h assume‬‬ ‫)‪351.35 < 400 mm _______________(OK‬‬ ‫)‪(1‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ‪Use hmin = 400 mm‬‬ ‫‪Also,‬‬ ‫‪b=d/2‬‬ ‫‪b=400/2‬‬ ‫‪b=200mm‬‬ ‫‪ 59‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪4.8 Design of flexure:‬‬

‫‪4.8.1 Actual depth :‬‬

‫‪d‬‬

‫‪d  h  c.c  dst  b‬‬ ‫‪2‬‬ ‫‪16‬‬ ‫‪d  400 258   359 mm‬‬ ‫‪2‬‬ ‫)‪(2‬ــــــــــــــــــــــــــــــــــــــــــــــــــ‪d  359mm‬‬ ‫‪4.8.2 Minimum Ratio of Steel Required for section:‬‬

‫) ‪  0.85  0.007( fc'  28‬‬ ‫) ‪ 0.85  0.007( 25  28‬‬ ‫‪ 0.871‬‬ ‫‪0.65    0.85‬‬ ‫‪use   0.85‬‬ ‫)‪- - - - - - - - - - - - - - - - - - - - - - - - - - (3‬‬ ‫‪25‬‬ ‫‪600‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ b   0.85‬‬ ‫( ‪ 0.85‬‬ ‫‪)‬‬ ‫‪420‬‬ ‫‪600  420 ‬‬ ‫‪‬‬

‫‪  0.0252 mm‬‬ ‫‪d‬‬

‫‪‬‬

‫‪ 0.75  b  0.75  0.0252  0.0189‬‬ ‫‪max‬‬

‫)‪- - - - - - - - - - - - - - - - - - - - - - (4‬‬

‫‪  0.0189‬‬ ‫‪max‬‬

‫‪25 ‬‬ ‫‪‬‬ ‫‪4  420 ‬‬

‫‪0r‬‬

‫‪ 1.4‬‬

‫‪‬‬

‫‪ 420‬‬

‫‪min‬‬

‫‪‬‬

‫]‪ max of [ 0.0033 or 0.0029‬‬ ‫‪ min  0.0033‬‬ ‫)‪- - - - - - - - - - - - - - - - - - - - - - (5‬‬

‫‪ 60‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪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

‫‪(Mu)- ve  33.03 KN .m‬‬ ‫‪33.03  10 6‬‬ ‫‪K ‬‬ ‫‪ 0.0455  Ku  0.0455‬‬ ‫‪u 0.9  25  250  (359) 2‬‬ ‫)‪1  1  2.36  ( 0.0455‬‬ ‫‪ 0.046‬‬ ‫‪1.18‬‬ ‫‪fc‬‬ ‫‪25‬‬ ‫‪ρ     0.046 ‬‬ ‫‪ 0.0027‬‬ ‫‪420‬‬ ‫‪fy‬‬

‫‪ω‬‬

‫)‪ρ min  0.0033 - - - - - - - - - - - - - - - - - - - - - -(From eq 5‬‬ ‫)‪ρ max  0.0189 - - - - - - - - - - - - - - - - - - - - - -(From eq 4‬‬ ‫‪ρ‬‬

‫)‪ 0.0027 - - - - - - - - - - - - - - - - - - - - - -(6‬‬ ‫‪act‬‬ ‫‪ρ min  ρ  ρ max‬‬

‫‪Hence use ,‬‬

‫‪ 0.003‬‬

‫‪min‬‬

‫‪ρ‬‬

‫‪As  ρ  b  d  0.0033  250  359  296.175‬‬ ‫‪As‬‬ ‫‪296.175 296.175‬‬ ‫‪‬‬ ‫‪‬‬ ‫)‪As(bar‬‬ ‫‪d 2‬‬ ‫‪  16 2‬‬ ‫‪4‬‬

‫‪Number of bars ‬‬

‫‪4‬‬ ‫‪ 1.47  2bar‬‬ ‫‪Use 2 16‬‬

‫‪ 61‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪2 .Check Nominal moment capacity of section‬‬

‫∗‬

‫∗‬

‫ ‪= 31.77‬‬

‫∗‬

‫∗‬

‫∗‬

‫‪.‬‬

‫‪) ∗ 10‬‬

‫∗`‬

‫‪.‬‬

‫‪−‬‬

‫‪2‬‬ ‫‪.‬‬

‫=‬

‫∗‬

‫=‪a‬‬

‫=‬

‫‪=401.9 * 420*(359‬‬‫‪=57.91 KN.m‬‬ ‫∅ = ‬

‫‪= 0.9 *57.91= 52.12 KN.m‬‬ ‫)‬

‫( ‪> 33.03‬‬

‫ ‪= 52.12‬‬

‫‪.‬‬

‫‪3. Spacing:‬‬ ‫‪ 25mm‬‬ ‫‪= 16‬‬ ‫‪≈ 25‬‬

‫ ‬

‫ ‬

‫ ‬

‫‪max‬‬

‫‪( .‬‬

‫)‬ ‫)‬

‫=‪S‬‬

‫(‬

‫=‪S‬‬

‫‪S= 92 mm c/c‬‬

‫‪4. Check cracking:‬‬‫ ‪= 252‬‬ ‫ ‬ ‫‪∗ = 2*41*250 = 20500‬‬

‫‪=0.6‬‬ ‫ ‪= 41‬‬ ‫‪=2‬‬

‫‪Number of Bars =2‬‬ ‫=‬

‫ ‪= 10250‬‬

‫=‪A‬‬

‫‪5. Check for Exposure:‬‬ ‫‪= 252 ∗ √41 ∗ 10250 ∗ 10‬‬

‫∗‬

‫∗‬

‫=‪Z‬‬

‫‪ 62‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

Z=18.87 ‬

‫=‬

‫∅‬

‫‪ Case2 :- 3‬‬

‫‪74.79 > 37.64 > 37.39‬‬

‫‪= 180‬‬

‫‪600‬‬ ‫ ‪= 179.5 ≈ 180‬‬ ‫‪= 422‬‬

‫∗ ‪.‬‬

‫ ≤‬ ‫∗‪3‬‬ ‫‪S‬‬

‫‪=S‬‬

‫‪Use S= 180mm‬‬ ‫‪2. Zone = 2 :‬‬‫∗`‬

‫‪1‬‬ ‫∗ =‬ ‫‪6‬‬

‫‪= ∗ √25 ∗ 250 ∗ 359 = 74.79 KN‬‬ ‫‪= 74.79 ∗ 5 = 373.95‬‬

‫‪5‬‬

‫ ‪= 74.79 ∗ 3 = 223.47‬‬

‫‪3‬‬

‫‪74.79‬‬ ‫ ‪= 37.39‬‬ ‫‪2‬‬

‫=‬

‫‪2‬‬

‫‪ 72‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪section (2) from R.A = 4.91‬‬ ‫‪4.91‬‬ ‫‪5.77‬‬ ‫‪0.85‬‬

‫=‬

‫‪‬‬

‫‪ at‬‬

‫ ‬

‫ ‬

‫∅‬ ‫‪ Case 4 :-‬‬

‫∅‬

‫=‬

‫‪2‬‬

‫‪= 37.39 > 5.77‬‬ ‫‪Note: NO stirrups are required by code‬‬ ‫‪3. Zone= 3 :‬‬‫‪section 3.3 = 41.56‬‬ ‫‪41.56‬‬ ‫‪= 48.89‬‬ ‫‪0.85‬‬

‫‪ at‬‬

‫ =‬

‫ ‬

‫ ‬

‫∅‬

‫‪ Case 3:‬‬‫ ‬

‫ ‬

‫‪2‬‬

‫ >‬

‫ ‬

‫ ‬

‫∅‬

‫>‬

‫‪74.79 > 48.89 > 37.39‬‬ ‫‪=S‬‬ ‫)‪(From zone 3‬‬

‫‪S‬‬

‫‪S = 180 mm‬‬ ‫‪4. Zone = 4:-‬‬

‫‪section 5.5 = 12.80‬‬ ‫‪= 15.05‬‬

‫‪.‬‬ ‫‪.‬‬

‫‪ at‬‬

‫= ‬

‫ ‬

‫ ‬

‫∅‬

‫‪ Case 4 :‬‬ ‫ ‬ ‫∅‬

‫ >‬

‫ ‬

‫ ‬

‫‪2‬‬

‫‪Hence, NO need of Stirrups‬‬

‫‪ 73‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪5. Zone 5:‬‬ ‫‪41.96 from R.S.A value of Shear resembles to Zone 3,‬‬

‫= ‬

‫‪Hence we can use S = 180 mm‬‬ ‫‪6. Zone 6:‬‬ ‫ ‬

‫ ‪ . .‬‬

‫=‬

‫‪.‬‬

‫‪46.08‬‬ ‫ ‪= 54.21‬‬ ‫‪0.85‬‬

‫ ‬

‫ ‬

‫ ‬

‫ =‬

‫∅‬

‫‪ Case 4:‬‬ ‫ ‬

‫ ‬

‫ ‬

‫>‬

‫‪2‬‬

‫ ‬

‫ >‬

‫∅‬

‫ =‬

‫ ‬

‫) ‪( From Zone 2‬‬

‫‪Use‬‬

‫‪S= 180 mm‬‬ ‫‪7. Zone 7:‬‬ ‫ ‬

‫‪ . .‬‬

‫ ‬

‫=‬

‫‪.‬‬

‫‪19.30‬‬ ‫ ‪= 22.7‬‬ ‫ ‪0.85‬‬

‫ ‬

‫ =‬

‫ ‬

‫ ‬

‫∅‬

‫‪ Case 4 :‬‬‫ ‬

‫ ‬

‫∅‬

‫>‬

‫ ‬

‫ ‬

‫‪2‬‬

‫‪37.39 > 22.7‬‬ ‫‪Note: No stirrups are required by code.‬‬ ‫‪8. Zone 8:‬‬ ‫‪ . .‬‬ ‫ ‬

‫ ‪= 48.71‬‬

‫‪48.71‬‬ ‫‪= 58.30‬‬ ‫‪0.85‬‬

‫=‬

‫ ‬

‫ ‬

‫ ‬

‫∅‬

‫ ‬

‫‪ 74‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 

Case 3:

>









>

74.8 > 58.3 > 37.39 =



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]

C K k4= confining reinforcement factor =



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=

Ld=

tr  2.5



db

A sreq Aprovided

0.9  Fy  d b  (k 1  k 2  k 3 )  300 mm k 4  Fc'

‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬75 Design Of Beams

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪296 .17‬‬ ‫‪ 0 .73‬‬ ‫‪401 .92‬‬

‫‪‬‬

‫‪AS req‬‬ ‫‪AS prov‬‬

‫‪k 1  1 .3 , k 2  0 .8 , k 3 ‬‬

‫‪C  K tr‬‬ ‫‪ 2 .5‬‬ ‫‪db‬‬ ‫)) ‪c  25 mm , k tr  ( Atr F y / 10( s )( n‬‬ ‫‪k4 ‬‬

‫‪ 100 .5  350 / 10  180  2‬‬ ‫‪k tr  9 .77‬‬ ‫‪25  9 .77‬‬ ‫‪ 2 .17  2 .5 O.K‬‬ ‫‪16‬‬ ‫‪use K 4  2 .17‬‬ ‫‪k4 ‬‬

‫) ‪0 .9  350  16  ( 1 .3 )( 0 .8 )( 0 .73‬‬

‫‪L‬‬

‫‪2 .17  25‬‬ ‫‪ 352 .66 mm  300 mm‬‬ ‫‪Thats O.K‬‬

‫‪ 76‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪ 77‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

‫‪ 78‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Beams‬‬

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

DESIGN BEAM NO SECTION POSITION MOMENT 1

25X50 SPAN 1 SPAN2

2

SPAN 3 25X50 SPAN 1 SPAN2

3

SPAN 3 25X50 SPAN 1 SPAN2

4

SPAN 3 25X50 SPAN 1 SPAN2

5

SPAN 3 25X50 SPAN 1 SPAN2 SPAN 3

DESIGN FORCE N (KN)

1/0.40 1/5.39 1/10.37 1/15.36 1/20.35

-23.43 8.35 43.35 -30.47 -14.73

10.33 5.85 3.46 3.81 7.69

2/0.40 2/4.96 2/9.52 2/14.09 2/18.65

-27.06 -1.14 39.19 1.6 -30.61

3/0.40 3/5.39 3/10.37 3/15.36 3/20.35

LONGITUDINAL REINFORCEMENT BOTTOM TOP SUPPORT (+) (-) (-) 4Ø14 2Ø14 6Ø14 4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

6Ø14

15.64 1.86 3.62 4.09 14.92

4Ø14

2Ø14

6Ø14

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

6Ø14

-26.41 9.41 34.73 -23.36 -16.95

10.11 7.1 2.91 4.63 6.04

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

4/0.40 4/4.96 4/9.52 4/14.09 4/18.65

-29.71 1.54 33.42 -1 -26.93

15.35 4.32 4.16 1.72 16.13

4Ø14

2Ø14

6Ø14

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

6Ø14

5/0.40 5/5.39 5/10.38 5/15.36

-13.42 -19.11 29.73 9.39

3.95 2.89 2.84 5.95

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

8Ø14

DESIGN FOR Qz REQUIRED FOR STIRRUP REINFORCEMENT% SHEAR SPACING

TRANSVERSAL REINFORCEMENT TYPE DISTRIBUTION

0.27 0.27 0.27 0.27 0.27

48.97 -43.97 4.28 51.82 -42.48

22.5 22.5 22.5 22.5

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

0.27 0.53 0.27 0.53 0.27

43.75 -17.58 0.54 18.22 -45.51

22.5 22.5 22.5 22.5

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

0.27 0.27 0.27 0.27 0.27

46.38 -17.82 2.77 43.59 -39.43

22.5 22.5 22.5 22.5

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

0.27 0.53 0.27 0.53 0.27

44.59 -17.82 0.98 16.75 -42.66

22.5 22.5 22.5

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

0.27 0.27 0.27 0.27

20.48 -20.86 -2.03 15.03

22.5 22.5 -

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

6

25X50 SPAN 1 SPAN2

7

SPAN 3 25X50 SPAN 1 SPAN2

8

SPAN 3 25X50 SPAN 1 SPAN2

9

SPAN 3 25X50 SPAN 1 SPAN2 SPAN 3

5/20.35

-22.85

8.49

4Ø14

2Ø14

4Ø14

0.27

-42.41

22.5

6/0.40 6/5.39 6/10.37 6/15.36 6/20.35

-20.83 8.38 34.5 -25.98 -12.74

9.66 6.57 4.35 4.32 7.01

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

0.27 0.27 0.27 0.27 0.27

45.93 -16.6 1.85 45.65 -39.27

22.5 22.5 22.5 22.5

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

7/0.40 7/4.96 7/9.52 7/14.09 7/18.65

1.11 2.39 50.41 -3.72 -32.91

-0.36 -1.63 10.35 8.35 14.42

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

-

4Ø14

2Ø14

6Ø14

0.27 0.53 0.27 0.27 0.27

7.83 -6.97 1.01 49.65 -53.2

22.5 22.5

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

8/0.40 8/4.96 8/9.52 8/14.09 8/18.65

-29.63 -1.81 49.87 2.61 0.58

13.68 1.53 10.09 -1.35 0.01

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

-

4Ø14

2Ø14

6Ø14

0.27 0.53 0.27 0.53 0.27

49.12 -44.85 -0.94 6.8 -7.97

22.5 22.5 -

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

9/0.40 9/2.01 9/3.63 9/5.24 9/6.85

-38.71 11.43 25.71 14.16 -33.09

5.18 5.18 4.92 4.92 4.92

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

0.27 0.27 0.27 0.27 0.27

42.17 8.7 7.87 -7.95 -40.38

22.5 22.5

2T11 [email protected][email protected][email protected]

Table 2 : Sections and reinforcement for ground level beams

BEAM NO SECTION 10

25X40

DESIGN FORCE DESIGN POSITION N (KN) MOMENT

LONGITUDINAL REINFORCEMENT BOTTOM TOP(assembly) (+) (-)

SUPPORT (-)

REQUIRED REINFORCEMENT%

DESIGN FOR Qz FOR STIRRUP SHEAR SPACING

TRANSVERSAL REINFORCEMENT TYPE DISTRIBUTION

SPAN 1 SPAN2

11

SPAN 3 25X40 SPAN 1 SPAN2

12

SPAN 3 25X40 SPAN 1 SPAN2

13

SPAN 3 25X50 SPAN 1 SPAN2

14

SPAN 3 25X50 SPAN 1 SPAN2

15

SPAN 3 25X50 SPAN 1

-14.29 9.59 21.71 -14 -9.28

-0.96 -0.22 0.14 -0.15 -0.62

-45.36 18.82 58.63 -39.72 -29.75

4Ø14

2Ø14

8Ø14

0.29 0.29 0.29 0.29 0.29

35.77 -11.51 -2.54 32.24 -30.55

17.5 17.5 17.5

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

4Ø14

-19.58 3.57 20.32 2.63 -17.28

-1.02 -0.38 -0.75 -0.3 -1.34

-52.58 3.84 55.62 3.46 -50.77

4Ø14

2Ø14

8Ø14

0.29 0.58 0.29 0.92 0.29

34.93 -12.32 5.39 22.89 -33.51

17.5 10.8 17.5

4Ø14

2Ø14

-

4Ø14

2Ø14

8Ø14

-6.87 -10.72 17.29 10 -12.9

-0.07 0.13 0.11 -0.24 -1.06

-23.95 -35.1 53.41 18.6 -42.1

4Ø14

2Ø14

8Ø14

0.29 0.29 0.29 0.29 0.29

16.8 -15.97 3.03 9.42 -33.21

17.5 17.5 17.5

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

4Ø14

-16.54 2.47 23.85 3.37 -21.32

-0.83 -0.22 -0.3 -0.38 -1.21

-50.69 3.05 61.65 3.82 -56.82

4Ø14

2Ø14

8Ø14

0.29 0.92 0.29 0.58 0.29

33.78 -24.13 -4.45 13.06 -36.69

17.5 10.8 17.5

4Ø14

2Ø14

-

4Ø14

2Ø14

8Ø14

-22.3 4.27 38.04 2.69 0.6

-1.47 -2.05 2.06 -0.06 -0.05

-61.79 -2.8 94.28 -3.03 8.25

4Ø14

2Ø14

8Ø14

4Ø14

0.27 0.27 0.62 0.83 0.53

51.41 -49.24 -8.61 13.53 -8.98

22.5 22.5 13.3 13.3 -

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

-0.22

-0.29

2.92

4Ø14

2Ø14

8Ø14

0.53

9.39

-

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

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

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

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

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

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

SPAN2

16

SPAN 3 25X50 SPAN 1 SPAN2

17

SPAN 3 25X50 SPAN 1 SPAN2

18

SPAN 3 25X50 SPAN 1 SPAN2 SPAN 3

2.5 36.06 4.32 -20.8

-0.25 2.14 -1.63 -1.34

2.81 72.76 -2.79 -59.4

-30.2 12.98 20.44 9.14 -37.68

0.03 0.03 0.03 -0.26 -0.26

-11.25 45.18 65.9 30.77 -37.9

-12.3 20.01 32.14 -18.02 -6.62

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

4Ø14

-68.65 36.95 83.14 27.11 -88.27

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

-1.2 -1.2 -1.2 -1.85 -1.85

-42.56 93.37 169.88 66.24 -91.62

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

-0.95 -1.21 0.34 -1.2 -0.76

-42.8 34.43 60.22 -50.08 -26.39

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

0.85 0.27 0.27 0.27

-13.87 -0.91 45.42 -48.63

13.3 22.5 22.5

0.27 0.27 0.27 0.53 0.27

22.53 6.71 -9.1 -7.7 -40.11

22.5 22.5

0.27 0.27 0.29 0.27 0.27

46.07 9.51 -6.31 -12.91 -53.66

22.5 22.5

0.27 0.27 0.6 0.27

49.09 -43.5 19.74 50.64 -42.87

2T10 22.5 [email protected][email protected][email protected] 22.5 13.3 22.5 22.5

0.27

2T10 [email protected][email protected]

2T10 [email protected][email protected]

Table 3 : Sections and reinforcement for first floor beams

BEA M NO

SECTIO N

POSITIO N

DESIGN MOMEN T

DESIG N FORCE N (KN)

LONGITUDINAL REINFORCEMENT

REQUIRE D REINFOR

DESIG N FOR Qz

STIRRUP SPACIN G

TRANSVERSAL REINFORCEMENT TYPE DISTRIBUTION

CEMENT %

19

25X40 SPAN 1 SPAN2 SPAN 3

20

25X40 SPAN 1 SPAN2 SPAN 3

21

25X40 SPAN 1 SPAN2 SPAN 3

22

25X50 SPAN 1 SPAN2 SPAN 3

23

25X50

-14.29 9.59 21.71 -14 -9.28

-16.48 10.58 22.06 -13.18 -10.3

0.17 -0.08 -0.25 -0.1 0.06

BOTTO M (+) 4Ø14

TOP(asse mbly) (-) 2Ø14

SUPPOR T (-)

4Ø14

2Ø14

-

4Ø14

2Ø14

8Ø14

8Ø14

FOR SHEAR 2T10 [email protected][email protected]+49 @14.0

0.29 0.29 0.29 0.29 0.29

36.44 -11.2 -2.51 31.86 -30.96

17.5 17.5 17.5 2T10 [email protected][email protected]+43 @16.0

-19.58 3.57 20.32 2.63 -17.28

-21.66 4.28 20.94 3.24 -19.27

-0.34 -0.22 -0.2 -0.16 -0.45

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

-

4Ø14

2Ø14

8Ø14

0.29 0.58 0.29 0.92 0.29

35.71 -12.12 5.34 22.58 -34.34

17.5 10.8 17.5 2T10 [email protected][email protected]+63 @10.0

-6.87 -10.72 17.29 10 -12.9

-7.26 -10.68 17.64 10.1 -13.47

0.18 -0.07 0.09 0.1 0.5

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

4Ø14

0.29 0.29 0.29 0.29 0.29

16.85 -15.94 3.03 9.41 -33.42

17.5 17.5 17.5 2T10 [email protected][email protected]+43 @16.0

-16.54 2.47 23.85 3.37 -21.32

-18.71 3.17 24.37 4.54 -21.81

-0.77 -0.23 -0.36 -0.24 -0.38

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

8Ø14

0.29 0.93 0.29 0.58 0.29

34.6 -23.64 -4.44 12.26 -36.04

17.5 10.8 17.5 2T10 [email protected][email protected]+39 @16.0

SPAN 1 SPAN2 SPAN 3

24

25X50 SPAN 1 SPAN2 SPAN 3

25

25X50 SPAN 1 SPAN2 SPAN 3

26

25X50 SPAN 1 SPAN2 SPAN 3

27

25X50 SPAN 1 SPAN2 SPAN 3

-22.3 4.27 38.04 2.69 0.6

-14.07 20.87 32.89 -17.47 -7.48

0.21 0.26 -0.41 0.36 0.18

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

0.27 0.27 0.6 0.27 0.27

49.2 -42.93 19.66 50.37 -43.22

22.5 22.5 13.3 22.5 22.5 2T10 [email protected][email protected]+49 @14.0

-0.22 2.5 36.06 4.32 -20.8

-11.82 45.05 66.22 31.07 -37.53

0.33 0.33 0.33 0.57 0.57

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

0.27 0.27 0.29 0.27 0.27

46.35 9.57 -6.25 -12.89 -53.61

22.5 22.5 2T10 [email protected][email protected]

-30.2 12.98 20.44 9.14 -37.68

-25.52 5.1 39.05 2.24 -1.62

0.23 0.46 -0.76 -0.13 -0.24

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

4Ø14

0.27 0.27 0.27 0.83 0.27

53.53 -49.85 0.06 14.91 -10.08

22.5 22.5 13.3 2T10 [email protected][email protected]+52 @12.0

-11.25 45.18 65.9 30.77 -37.9

-2.54 2.89 37.44 4.69 -22.84

-0.1 -0.12 -1.25 0.51 0.24

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

-

4Ø14

2Ø14

8Ø14

0.27 0.85 0.27 0.27 0.27

9.78 -12.69 -1.38 45.04 -49.36

13.3 22.5 22.5 2T10 [email protected][email protected]+28 @22.0

-12.3 20.01 32.14 -18.02 -6.62

-30.05 13.19 20.71 9.72 -36.46

-0.05 -0.05 -0.05 0.09 0.09

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

0.27 0.27 0.27 0.27 0.27

22.54 6.73 -9.09 -7.52 -39.71

22.5 22.5

28

25X50 SPAN 1 SPAN2 SPAN 3

2T10 [email protected][email protected] -12.3 20.01 32.14 -18.02 -6.62

-23.36 26.62 51.1 35.77 -6.31

0.33 0.33 0.33 0.16 0.16

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

0.27 0.27 0.27 0.27 0.27

46.04 11.15 -2.21 -8.52 -40.24

22.5 22.5

Table 4 : Sections and reinforcement for second floor beams

BEA M NO 29

SECTIO N 25X40 SPAN 1 SPAN2

30

SPAN 3 25X40 SPAN 1 SPAN2

31

SPAN 3 25X40 SPAN 1 SPAN2

32

SPAN 3 25X50

SPAN 1 SPAN2

33

SPAN 3 25X50 SPAN 1

POSITIO N

DESIGN MOMEN T

DESIG N FORCE N (KN)

LONGITUDINAL REINFORCEMENT BOTTO M (+) 4Ø14

TOP(assembl y) (-) 2Ø14

SUPPOR T (-)

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

-14.29 9.59 21.71 -14 -9.28

-6.4 5.24 19.16 -10.64 -4.46

0.71 0.6 0.13 0.24 0.18

4Ø14

-19.58 3.57 20.32 2.63 -17.28

-10.96 1.54 19.67 0.91 -9.94

4.02 1.71 1.38 0.83 3.9

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

-

4Ø14

2Ø14

-

-6.87 -10.72 17.29 10 -12.9

-2.87 -7.69 12.7 5.74 -5.6

-0.3 0.14 -0.08 0.47 0.25

4Ø14

2Ø14

4Ø14

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

-16.54 2.47 23.85 3.37 -21.32

-9.48 0.93 22.97 1.47 -12.25

4.1 0.86 1.37 1.71 4.08

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

-

4Ø14

2Ø14

8Ø14

-22.3 4.27

-1.88 2.8

1.62 1.17

4Ø14

2Ø14

4Ø14

REQUIRED REINFORCEMENT %

DESIG N FOR Qz FOR SHEAR

STIRRU P SPACIN G

0.29 0.29 0.29 0.29 0.29

9.75 -7.91 2.73 9.58 -8.43

-

0.29 0.58 0.29 0.58 0.29

9.65 -7.49 -1.3 7.1 -9.34

-

2T10 [email protected][email protected][email protected] .0

0.29 0.29 0.29 0.29 0.29

7.5 -8.15 -1.54 6.43 -9

-

2T10 [email protected][email protected][email protected] .0

0.29 0.58 0.29 0.58 0.29

9.48 -7.49 2.12 8.24 -10.3

-

0.27 0.27

8.63 -14.47

22.5

TRANSVERSAL REINFORCEMENT TYPE DISTRIBUTION

2T10 [email protected][email protected][email protected] .0

2T10 [email protected][email protected][email protected] .0

SPAN2

34

SPAN 3 25X50 SPAN 1 SPAN2

35

SPAN 3 25X50 SPAN 1 SPAN2

36

SPAN 3 25X50 SPAN 1 SPAN2

37

SPAN 3 25X50 SPAN 1 SPAN2

38

SPAN 3 25X50

SPAN 1 SPAN2

38.04 2.69 0.6

41.45 -25.67 -0.6

0.51 0.79 1.23

4Ø14

2Ø14

8Ø14

4Ø14

2Ø14

4Ø14

-0.22 2.5 36.06 4.32 -20.8

-1.4 30.44 62.29 10.47 -40.12

1.45 1.45 1.45 1.18 1.18

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

-30.2 12.98 20.44 9.14 -37.68

-8.42 -5.24 54.38 -11.1 0.24

3.07 2.09 1.06 0.38 0.24

4Ø14

2Ø14

4Ø14

4Ø14

4Ø14

4Ø14

4Ø14

2Ø14

-

-11.25 45.18 65.9 30.77 -37.9

-0.72 -8.53 50.81 -5.25 -7.78

0.66 0.49 1.25 1.9 2.88

4Ø14

2Ø14

4Ø14

4Ø14

4Ø14

4Ø14

4Ø14

2Ø14

-

-12.3 20.01 32.14 -18.02 -6.62

-21.36 3.36 28.08 0.14 -27.35

0.53 0.53 0.53 0.11 0.55

4Ø14

-

-

-

2Ø14

-

-

-

8Ø14

-12.3 20.01 32.14

-30.25 8.33 47.75

0.84 0.84 1.14

4Ø14

-

-

-

2Ø14

-

0.27 0.27 0.27

12.75 13.81 -7.68

22.5 -

2T10 [email protected][email protected][email protected] .0

0.27 0.27 0.27 0.27 0.27

7.72 7.72 7.72 -12.33 -12.33

-

2T10 [email protected][email protected][email protected] .0

0.27 0.27 0.27 0.27 0.53

9.96 -39.33 -11.34 4.74 -1.85

22.5 -

2T10 [email protected][email protected][email protected] .0

0.27 0.27 0.27 0.27 0.27

1.46 -3.51 10.12 14.13 -9.11

22.5 -

2T10 [email protected][email protected][email protected] .0

0.27 0.27 0.27 0.53

6.1 6.1 6.1 -6.79 -6.79 -

0.27

0.27 0.27 0.27

11.12 11.12 -6.88

-

-

2T10 [email protected][email protected][email protected] .0 2T10 [email protected][email protected][email protected] .0

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪5. Design Of Stairs:‬‬ ‫‪5.1 Geometrical Design of Stairs:‬‬‫‪Given:‬‬‫‪Floor height = 4.2m‬‬

‫‪‬‬

‫‪‬‬ ‫‪ Assume:-‬‬

‫)‪(O.K‬‬ ‫)‪(O.K‬‬

‫‪Rise …. R = 150 mm‬‬

‫‪R ≤ 190 mm‬‬

‫‪Going … G = 300 mm‬‬

‫‪G ≥ 220 mm‬‬

‫‪N R  4.20  28‬‬

‫‪- Number of riser's‬‬

‫‪0.15‬‬

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

‫‪5.1.2 Check:‬‬‫‪570  2R  G  630‬‬ ‫‪570  [2(150)  300]  630‬‬ ‫‪570  600  630‬‬ ‫‪OK‬‬

‫‪5.1.3 Check for angle:-‬‬

‫‪   26.5‬‬ ‫‪OK‬‬

‫‪R 150‬‬ ‫‪‬‬ ‫‪ 0.5‬‬ ‫‪G 300‬‬ ‫‪ α  40‬‬ ‫‪ 26.5  40‬‬

‫‪tan α ‬‬ ‫‪25‬‬ ‫‪25‬‬

‫‪Figure (5 - 1) Dimensions of stair‬‬

‫‪ 78‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪5.2 - Detailed design of stair:‬‬

‫‪Assumption and Requirement of design:‬‬

‫‪Table (5-1) Data of design‬‬

‫‪WL.L = 4 kN/m 2‬‬

‫‪Live Load‬‬

‫‪H = 4.20 m‬‬

‫‪Height of story‬‬

‫‪f y = 420 MPa‬‬

‫‪Yield stress‬‬

‫‪f c' = 25 MPa‬‬

‫‪Compressive strength of‬‬ ‫‪concrete‬‬

‫‪C.C = 25 mm‬‬

‫‪Concrete Cover‬‬

‫‪d b = 14 mm‬‬

‫‪Diameter of main steel‬‬

‫‪d st = 12 mm‬‬

‫‪Diameter of secondary steel‬‬

‫‪2.5 kN/m 2‬‬

‫‪Flooring‬‬

‫‪ 79‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪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.‬‬

‫‪ 80‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪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) = 1500  10 rise‬‬ ‫‪150‬‬

‫‪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 = 1500  10‬‬ ‫‪150‬‬

‫‪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 ‫‪That’s O.K‬‬

‫‪4.20 => 4.20‬‬

‫‪ 81‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪5.4 Structural Design of Stairs‬‬ ‫‪5.4.1 Steps of Design :‬‬‫‪1. - Design for flexure:-‬‬

‫‪‬‬ ‫‪2‬‬ ‫‪M‬‬ ‫‪u‬‬

‫‪,   0.9‬‬

‫‪dh-c‬‬‫‪Ku ‬‬

‫‪  f c'  b  d 2‬‬ ‫‪1- 1- 2.36 K u‬‬ ‫‪1.18‬‬ ‫'‬ ‫‪c‬‬

‫‪f‬‬

‫‪fy‬‬

‫‪ω‬‬

‫‪ρ  ω‬‬

‫‪‬‬ ‫‪‬‬ ‫‪h‬‬ ‫‪1.4 ‬‬ ‫‪‬‬ ‫‪s‬‬ ‫‪ρ‬‬ ‫‪ 0.002  or‬‬ ‫‪min ‬‬ ‫‪d‬‬ ‫‪F ‬‬ ‫‪‬‬ ‫‪y‬‬

‫‪min‬‬

‫‪ρ  ρ‬‬

‫‪2. Main Reinforcement:-‬‬

‫‪As  ρ  b  d‬‬ ‫‪A‬‬

‫‪s‬‬ ‫‪(As ) one bar‬‬ ‫) ‪(A‬‬ ‫‪ b  450 mm‬‬ ‫‪s one bar‬‬ ‫‪‬‬ ‫‪As‬‬ ‫‪ 5h s mm‬‬

‫‪N‬‬

‫‪Sreq ‬‬

‫‪3. Secondary Reinforcement:‬‬

‫‪A s  0.002  1000  h s‬‬ ‫‪ 450 mm‬‬

‫‪ ‬‬

‫‪ 5h s mm‬‬

‫‪(As ) one bar  b‬‬ ‫‪As‬‬

‫‪S‬‬

‫‪ 82‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪5.5 Design for flight No 1 & 3 :‬‬

‫‪L‬‬ ‫‪6000‬‬ ‫‪hs  n ‬‬ ‫‪ 300 mm‬‬ ‫‪20‬‬ ‫‪20‬‬ ‫‪14‬‬ ‫ ‪d  300 - 25‬‬‫‪ 268 mm‬‬ ‫‪2‬‬ ‫‪WLanding  1.4( 0.324)2.51.73 20.38 KN/m‬‬ ‫‪0.3‬‬ ‫‪‬‬ ‫‪‬‬ ‫(‪WFlight  1.4 ‬‬ ‫‪ 0.5( 0.30 ))  24  2.5  1.7  4‬‬ ‫‪COS‬‬ ‫‪26.5‬‬ ‫‪‬‬ ‫‪‬‬

‫‪WFlight 25.6 KN/m ‬‬

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

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

‫‪ 83‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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

‫‪L‬‬ ‫)‪ n (for Fy  420 Mpa‬‬ ‫‪min 20‬‬ ‫‪3000‬‬ ‫) ‪(h s‬‬ ‫‪‬‬ ‫‪ 150 mm‬‬ ‫‪min‬‬ ‫‪20‬‬

‫) ‪(h s‬‬

‫)‪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) :-‬‬

‫‪29 10 6‬‬ ‫‪ 0.011‬‬ ‫‪0.9  25 1500  2682‬‬ ‫‪1  1  2.36  0.011‬‬ ‫‪ω‬‬ ‫‪ 0.012‬‬ ‫‪1.18‬‬ ‫‪25‬‬ ‫‪ρ  0.012 ‬‬ ‫‪ 0.0007‬‬ ‫‪420‬‬ ‫‪Ku ‬‬

‫‪1.4‬‬ ‫‪ 0.003‬‬ ‫‪420‬‬

‫‪min‬‬

‫‪‬‬

‫‪min‬‬

‫‪ρ‬‬

‫‪ρ req  ρ‬‬

‫‪ 84‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪2‬‬

‫‪2‬‬

‫‪ 0.003  1500  268  692 mm  1206 mm‬‬

‫‪st‬‬

‫‪A‬‬

‫‪A‬‬

‫‪st  1206 7.88  8‬‬ ‫‪A‬‬ ‫)‪(π  14 /4‬‬ ‫‪sb‬‬

‫‪N‬‬

‫‪Use 8  14‬‬

‫‪A‬‬ ‫‪bar  b 153.8  1500‬‬ ‫‪S  sone‬‬ ‫‪‬‬ ‫‪ 191.29mm‬‬ ‫‪1206‬‬ ‫‪As‬‬ ‫‪Use  14 / 190 mm‬‬

‫‪5.5.5 Secondary reinforcement (temp + shrinkage):‬‬

‫‪ 12mm‬‬

‫‪st‬‬

‫‪Use d‬‬

‫‪A  0.002  1000  268‬‬ ‫‪s‬‬ ‫‪2‬‬

‫‪max‬‬

‫‪A  568 mm‬‬ ‫‪s‬‬ ‫‪113  1000‬‬ ‫‪S‬‬ ‫‪ 199.01 mm  S‬‬

‫‪568‬‬

‫‪ 450 mm‬‬ ‫‪Use S max  ‬‬ ‫‪ 3h s  3  150  450 mm‬‬ ‫‪Us e  12 / 190 mm‬‬

‫‪A‬‬

‫‪568‬‬ ‫‪s total‬‬ ‫‪‬‬ ‫‪ 5.02  6‬‬ ‫‪2‬‬ ‫‪A‬‬ ‫)‪(π  12 /4‬‬ ‫‪s one bar‬‬

‫‪N‬‬

‫‪Us e 6  12‬‬

‫‪ 85‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪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

‫‪ 86‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪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

‫‪22.55  106‬‬ ‫‪ 0.0089‬‬ ‫‪0.9  24  1500  268‬‬

‫‪Ku ‬‬

‫‪1  1  2.36  0.089‬‬ ‫‪ 0.0089‬‬ ‫‪1.18‬‬ ‫‪24‬‬ ‫‪ρ  0.0089 ‬‬ ‫‪ 0.0005‬‬ ‫‪420‬‬ ‫‪ 1.4‬‬ ‫‪‬‬ ‫‪ 0.00333‬‬ ‫‪‬‬ ‫‪420‬‬ ‫‪‬‬ ‫‪ρmin  ‬‬ ‫‪‬‬ ‫'‪Fc‬‬ ‫‪‬‬ ‫‪ 0.003 ‬‬ ‫‪ 4 Fy‬‬ ‫‪‬‬

‫‪ω‬‬

‫)‪ρ  ρmin (control‬‬ ‫‪use ρmin‬‬

‫‪As  0.003  1500  268  1206mm 2‬‬ ‫‪AST‬‬ ‫‪1206‬‬ ‫‪N‬‬ ‫‪‬‬ ‫‪ 7.83  8‬‬ ‫‪2‬‬ ‫‪As‬‬ ‫‪(π‬‬ ‫‪‬‬ ‫‪14‬‬ ‫)‪/4‬‬ ‫‪one bar‬‬ ‫‪Us e 8  14‬‬ ‫‪ Spacing :‬‬

‫‪A‬‬ ‫‪bar  b 154 1500‬‬ ‫‪ sone‬‬ ‫‪‬‬ ‫‪ 191.54mm‬‬ ‫‪As‬‬ ‫‪1206‬‬

‫‪req‬‬

‫‪S‬‬

‫‪Use  14 / 190 mm‬‬

‫‪ 87‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪5.6.4 Secondary Reinforcement:-‬‬

‫‪As  0.002  1000  268‬‬ ‫‪As  568 mm2‬‬ ‫‪ 199.01 mm‬‬

‫‪113 1000‬‬ ‫‪568‬‬

‫‪S‬‬

‫‪Us e  12 / 190 mm‬‬

‫‪5.7 Reinforcement details:‬‬

‫‪Figure (5 -7 ) Reinforcement of stair‬‬

‫‪ 88‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Figure (5 - 8) Connection between stair and beam‬‬

‫‪ 89‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Stairs‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

6. Design of columns Introduction: This chapter presents an introductory discussion of reinforced concrete

6.1.

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 ,

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 90

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 91

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 crosssectional 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.

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 92

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 93

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 94

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 95

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 6.1.6 Effective Length: The effective length of a column is defined as the length between the points of contraflexure 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,

Where

= area of concrete and,

is given by the equation ,

= 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.

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 96

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 crosssectional 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.

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 97

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ (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 onefourth 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 ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 98

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 99

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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.

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 100

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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)

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 101

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 102

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Fig 6.5. Elevation plan showing Columns at first floor‬‬

‫‪Fig6.6 . Elevation plan showing beams and Columns for second floor‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪103‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪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 ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪104‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪6.3 Design of column in detail:‬‬ ‫‪493.58 KN‬‬

‫‪pu‬‬

‫‪18.33 KN.m‬‬

‫‪Mux top‬‬

‫‪-15.51 KN.m‬‬

‫‪Mux bottom‬‬

‫‪27.70 KN.m‬‬

‫‪Muy top‬‬

‫‪-21.64 KN.m‬‬

‫‪Muy bottom‬‬

‫‪Table 6.2 Design values obtained from R.S.A‬‬ ‫‪6.3.1Design moments:‬‬ ‫‪6.3.1.1 Inertia At X - Direction:‬‬

‫‪= 0.0045mm‬‬

‫‪.‬‬

‫‪× .‬‬

‫ ‬

‫=‬

‫)‪(Ig‬‬

‫=‬

‫‪‬‬

‫‪Ic = 0.7 × 0.0045 = 0.00315m‬‬

‫‪= 0.0026‬‬

‫‪.‬‬

‫‪× .‬‬

‫ ‬

‫=‬

‫)‪(Ig‬‬

‫=‬

‫‪‬‬

‫‪Ic = 0.7 × 0.0026 = 0.00182m‬‬ ‫‪× .‬‬

‫‪= 0.0026m‬‬

‫‪.‬‬

‫ ‬

‫=‬

‫= )‪(Ig‬‬

‫‪‬‬

‫‪Ib =× 0.35 × 0.0026 = 0.00091m‬‬

‫‪.‬‬

‫ ‬

‫‪.‬‬

‫ ‪= 1.02‬‬

‫‪.‬‬

‫=‬

‫ ‬

‫=‬

‫ ‬

‫‪G‬‬

‫‪‬‬

‫‪.‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪105‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪.‬‬

‫ ‪= 18.88‬‬

‫ ‬

‫‪.‬‬

‫‪.‬‬

‫ ‬

‫=‬

‫‪.‬‬

‫=‬

‫ ‬

‫‪G‬‬

‫ ‬

‫‪‬‬

‫‪.‬‬

‫=‬

‫‪, whereas plinth level column height is 0.7m‬‬

‫>= ‪Note‬‬

‫‪6.3.1.2 Now, take K the smaller of:‬‬‫‪≤1‬‬

‫) ‬

‫ ‪+‬‬

‫ ‬

‫( ‪K = 0.7 +0.05‬‬

‫)‪= 0.7 +0.05 (1.02 +18.88‬‬ ‫‪= 1.695 > 1‬‬ ‫‪) ≤1‬‬

‫( ‪K = 0.85 +0.05‬‬

‫‪K = 0.85 + 0.05(1.02) = 0.901 ≤ 1‬‬ ‫ ‪K = 0.9‬‬ ‫‪Using K = 0.9‬‬ ‫‪0.9 × 4‬‬ ‫‪= 20‬‬ ‫‪0.3 × 0.6‬‬ ‫‪M1‬‬ ‫‪≤ 40‬‬ ‫‪M2‬‬

‫=‬

‫‪K L‬‬ ‫‪r‬‬ ‫‪K L‬‬ ‫‪r‬‬

‫‪= 34 − 12‬‬

‫‪Whereas:‬‬‫‪value of smaller factored end moment at X-direction‬‬

‫‪M‬‬

‫= ‬

‫‪= value of larger factored end moment at X-direction‬‬

‫ ‬

‫‪M‬‬

‫‪6.3.1.3 Check for Short or long column :‬‬ ‫)‪(−15.512‬‬ ‫‪= 44.14‬‬ ‫‪18.324‬‬

‫‪= 34 − 12‬‬

‫ ‪ Hence , its a Short Column‬‬

‫‪K L‬‬ ‫‪r‬‬

‫ 7.40 KN.m (Thats O.K.‬‬

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

‫‪.‬‬

‫ ‪= 18.324‬‬

‫‪6.3.2 At Y - Direction:‬‬

‫‪= 0.00078‬‬

‫‪. × .‬‬

‫=‬

‫ ‬

‫=‬

‫)‪(Ig‬‬

‫‪‬‬

‫‪= 0.7 × 0.00078 = 0.00054‬‬

‫‪= 0.00065‬‬

‫‪. × .‬‬

‫=‬

‫ ‬

‫=‬

‫)‪(Ig‬‬

‫‪‬‬

‫‪= 0.7 × 0.00065 = 0.00045‬‬

‫‪= 0.00065‬‬

‫‪. × .‬‬

‫=‬

‫ ‬

‫= )‪(Ig‬‬

‫‪‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪107‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Ib = 0.35 × 0.00065 = 0.00022‬‬ ‫‪E I‬‬ ‫‪Lu‬‬ ‫=‬ ‫‪E I‬‬ ‫‪Ln‬‬

‫‪0.00054 0.00045‬‬ ‫‪+‬‬ ‫‪0.00024‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫=‬ ‫=‬ ‫‪= 8.1‬‬ ‫‪0.00022‬‬ ‫‪0.000030‬‬ ‫‪7.3‬‬ ‫‪0.00054 0.00045‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫⎛‬ ‫⎞‬ ‫‪0.00022‬‬ ‫⎜ =‬ ‫‪⎟ = 30.07‬‬ ‫‪7.3‬‬ ‫ ‬ ‫⎠‬

‫‪E I‬‬ ‫‪Lu‬‬ ‫ =‬ ‫‪E I‬‬ ‫‪Ln‬‬

‫⎝‬

‫‪G‬‬

‫ ‬

‫‪G‬‬

‫ ‬

‫‪6.3.2.1 Now, take K the smaller of:‬‬‫‪≤1‬‬

‫ ‬

‫ ‪+‬‬

‫ ‪K = 0.7 + 0.05‬‬

‫‪K = 0.7 + 0.05(8.1 + 30.07) ≤ 1‬‬ ‫‪2.6 ≠ 1‬‬ ‫‪) ≤ 1‬‬

‫( ‪K = 0.85 + 0.05‬‬ ‫)‪= 0.85 0.05 (8‬‬ ‫‪=1.25 ≠ 1‬‬

‫‪Hence use K=1‬‬ ‫‪6.3.2.2 Check for short or long:‬‬‫‪1∗4‬‬ ‫‪= 53.33‬‬ ‫‪0.3 × 0.25‬‬

‫=‬

‫‪K L‬‬ ‫‪r‬‬

‫‪M1‬‬ ‫‪M2‬‬

‫‪= 34 − 12‬‬

‫‪K L‬‬ ‫‪r‬‬

‫)‪(−21.64‬‬ ‫‪= 43.37‬‬ ‫‪27.7‬‬

‫‪= 34 − 12‬‬

‫‪K L‬‬ ‫‪r‬‬

‫‪= 43.37‬‬

‫‪K L‬‬ ‫‪r‬‬

‫ ‪=> Hence, its a Long Column‬‬

‫‪K L‬‬ ‫‪r‬‬

‫>‬

‫‪K L‬‬ ‫‪r‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪108‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

Note: The effect if the slenderness ratio

may be ignored if



0.1 ∗ 2830‬‬ ‫)‬

‫ ‬

‫‪493 > 283 => ( ℎ‬‬ ‫‪1‬‬

‫‪ = 0.0015‬‬ ‫‪- - - - OK‬‬

‫‪−‬‬

‫‪−‬‬

‫‪1‬‬

‫‪+‬‬

‫‪+‬‬

‫‪1‬‬

‫≥‬

‫‪1‬‬

‫ ‬

‫ =‬

‫‪= 630.5 KN >493 KN‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪113‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪6.5.4 Number of Main Steel bars:‬‬ ‫‪ℎ‬‬

‫ =‬

‫ ‬

‫‪= 0.01*250*600‬‬ ‫‪=1500mm2‬‬ ‫ ∅ ‪Use 8‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪114‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪6.6 Spacing for ties :‬‬

‫‪ Least dimension of column  250 mm‬‬ ‫‪‬‬ ‫‪S   48  (tie diameter)  48  8  384 mm‬‬ ‫‪ 16  (mainbars)  16  16  256 mm‬‬ ‫‪‬‬

‫‪use S  250 mm‬‬

‫‪Fig 6.8 shows Sec A-A of column 59‬‬ ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪115‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪6.6.1 Splices for Column :‬‬ ‫‪Compression deformed bars:‬‬

‫‪For f c'  20 MPa‬‬ ‫‪L  0.073 f  d  300 mm‬‬ ‫‪s‬‬ ‫‪y b‬‬ ‫‪Ls  0.073 420 16  490.56 mm  300‬‬ ‫‪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.‬‬ ‫‪Force in ties  1.5/6  A  f y  (1.5/6)  402  42042.21 kΝ‬‬ ‫'‪s‬‬ ‫‪Extra Αv  42.21 10 3  / 420100.5 mm 2‬‬

‫‪Fig.6.9 shows minimum requirements for splices for columns under ACI 318-08‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪116‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪π d2‬‬ ‫‪ 100.5 mm  d  11.32  use d  12 mm‬‬ ‫‪4‬‬ ‫‪Development of deformed bars in compression :‬‬ ‫‪fy d‬‬ ‫‪b‬‬ ‫‪L ‬‬ ‫‪ 0.044  d  f y‬‬ ‫' ‪db 4 f‬‬ ‫‪b‬‬ ‫‪c‬‬ ‫‪420 16‬‬ ‫‪L ‬‬ ‫‪ 336  0.044 16  420  295.68 mm‬‬ ‫‪db 4  25‬‬ ‫‪Av ‬‬

‫‪ 340 mm‬‬

‫‪db‬‬

‫‪Use L‬‬

‫‪L  L  K1'  K '2  200 mm‬‬ ‫‪d‬‬ ‫‪db‬‬ ‫‪A‬‬ ‫‪s req‬‬ ‫‪1500‬‬ ‫‪K1' ‬‬ ‫‪‬‬ ‫‪ 0.933‬‬ ‫‪2‬‬ ‫‪A‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪6‬‬ ‫‪s prov 8 ‬‬ ‫‪4‬‬ ‫‪K '  0.75‬‬ ‫‪2‬‬

‫‪L  340  0.933  0.75  237.92 mm‬‬ ‫‪d‬‬ ‫‪(5) =325 mm‬‬ ‫‪(6) = 75 mm‬‬ ‫‪(7) = 500 – db/2 – 50 – 8 – db‬‬ ‫‪= 500 – 8–60–8–16=408mm‬‬ ‫‪(8) = maximum slope = 1: 6‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪117‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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)

must be  Ld

(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.

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ Design Of Columns 118

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Figure (6.11 ) Reinforcement details for columns‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫‪Design Of Columns‬‬ ‫‪119‬‬

Fig1. Naming of columns from R.A

Section

DESIGN CASE

Column

DESING MOMENT

DESIGN MOMENT

Design force N

(KN.M)

Mz(KN)

(KN)

Required Reinforcement Ratio %

(cm)

1 R 25x60 2 R 25x60

3 3

-45.5 11.22

-55.64 -6.93

764.87 1000.96

1.9 1

3 R 25x60 4 R 25x60

3 3

-0.51 31.04

-6.14 -49.76

921.3 665.98

1 1

5 R 25x60 6 R 25x60

3 3

44.44 51.19

8.19 -8.83

1323.49 1361.61

1.83 2.33

7 R 25x60 8 R 25x60

3 3

37.96 0.93

53.7 88.7

699.44 1497.77

1.2 2.34

9 R 25x60 10 R 25x60

3 3

10.93 -51.86

99.9 59.38

1643.74 798.17

3.55 2.61

11 R 25x60 12 R 25x60

3 3

-70.65 -63.88

-7.48 10.62

1538.06 1497.37

3.9 3.5

13 14 15 16 17 18

3 3 3 3 3 3

1.28 8.53 19.57 -2.13 0.65 16.79

1.51 0.74 -84.43 -78.25 -11.39 -13.89

1409.86 1304.92 2203.57 1983.62 2726.58 2976.53

1.33 1 1.56 1 2.89 3.53

R 25x60 R 25x60 R 25x80 R 25x80 R 25x80 R 25x80

Reinforcments

stirrup spacing

8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 14Ø16 14Ø16 14Ø16 14Ø16

Design force Qz for Shear(KN)

Transver-sal Reinforc Ement Distribution & Spacing

10 10

-13 7.23

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

10 10

1.64 8.8

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

10 10

33.71 17.77

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

10 10

10.83 1.55

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

10 10

18.21 -14.78

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

10 10

-24.96 -22.95

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

10 10 10 10 10 10

-2.83 10.24 8.09 -1.9 0.31 13.07

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

Required Section

DESIGN CASE

Column

Transver-sal Reinforc

DESIGN MOMENT

DESING MOMENT

Design force N

Reinforment

Mz

(KN.M)

(KN)

Raitio

(cm)

19 R 25x40 20 R 25x40

3 3

34.78 49.92

-21.2 60.93

585.93 1206.36

1 3.53

21 R 25x40 22 R 25x40

3 3

45.6 31.86

58.5 13.56

1098.5 512.67

2.97 1

23 R 25x60 24 R 25x60

3 3

-1.08 -3.18

69.63 67.91

1001.98 976.16

2.28 2.17

25 R 25x60 26 R 25x60

3 3

-28.15 -7

10.14 -15.84

486.41 681.32

1 1

27 R 25x60 28 R 25x60

3 3

-7.39 -31.39

20.03 -16.62

742.24 559.38

1 1

29 R 25x60 30 R 25x60

3 3

-4.2 -0.08

-74.13 -77.82

1102.5 1131.71

2.84 3.02

31 32 33 34 35 36

3 3 3 3 3 3

-1.46 -1.25 21.58 -35.09 -32.1 20.22

70.15 -54.32 101.33 122.26 -116.4 -60.09

1016.83 942.92 1512.02 2099.59 1917.21 1339.12

2.35 1.2 2.34 4.23 3.66 1

R 25x60 R 25x60 R 25x80 R 25x80 R 25x80 R 25x80

Reinforcments

stirrup spacing

6Ø16 6Ø16 6Ø16 6Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 8Ø16 10Ø16 10Ø16 10Ø16 10Ø16

Design force Qz for Shear(KN)

ement Distribution Spacing

10 10

-5.75 0.89

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

10 10

0.26 4.2

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

10 10

4.78 4.18

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

10 10

3.34 0.09

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

10 10

0.97 -4.85

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

10 10

-5.86 -6.51

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

10 10 10 10 10 10

-0.05 0.89 4.49 2.94 -1.27 -2.74

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

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)

Required Section

DESIGN CASE

Column

DESIGN MOMENT

DESING MOMENT

Design force N

Mz

(KN.M)

(KN)

Reinforment Reinforcments Raitio

(cm)

37 R 25x40 38 R 25x40

3 3

20.22 42.72

17.97 -21.72

372.58 762.3

1 1

39 R 25x40 40 R 25x40

3 3

59.76 -54.08

16.49 12.32

692.15 325.22

1 1

41 R 25x40 42 R 25x40

3 3

-38.26 1.28

23.62 20.09

629.31 614.16

1 1

43 R 25x40 44 R 25x40

3 3

-2.76 -32.6

8.86 6.33

307.53 429.24

1 1

45 R 25x40 46 R 25x40

3 3

-10.58 -11.25

-7.42 -14.44

471.96 355.07

1 1

47 R 25x40 48 R 25x40

3 3

35.73 -4.51

33.88 39.46

694.65 713.63

1 1

49 50 51 52 53 54

3 3 3 3 3 3

3.19 -2.72 -4.37 -16.11 47.73 42.85

-12.98 11.51 33.61 -74.27 48.76 41.69

627.47 594.03 1033.12 1393.03 1278.86 899.9

1 1 1 1.19 1 1

R 25x40 R 25x40 R 25x60 R 25x60 R 25x40 R 25x40

stirrup spacing

8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø16 8Ø16 8Ø16 8Ø14

Design

Transver-sal Reinforc

force Qz for Shear(KN)

ement Distribution Spacing

10 10

-7.44 0.89

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

10 10

0.39 5.39

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

10 10

5.81 5.1

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

10 10

4.07 -0.12

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

10 10

1.11 -6.23

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

10 10

-7.28 -8.07

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

10 10 10 10 10 10

0 0.67 7.7 4.52 -2.71 -3.81

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

Section

DESIGN MOMENT Mz

DESIGN CASE

Column

DESING MOMENT

Design force N

Required Reinforment

My (KN.M)

(KN)

Raitio

stirrup Reinforcments spacing (cm)

55 R 25x40 56 R 25x40

3 3

46.31 -57.52

13.68 -4.02

155.12 316.71

1 1

57 R 25x40 58 R 25x40

3 3

-52 41.53

3.5 -10.08

286.25 133.92

1 1

59 R 25x40 60 R 25x40

3 3

-0.94 1.22

-12.91 -11.3

258.13 251.64

1 1

61 R 25x40 62 R 25x40

3 3

-36.47 -14.36

-7.41 -1.86

126.76 173.11

1 1

63 R 25x40 64 R 25x40

3 3

-15.6 -41.06

3.2 10.79

192.57 147.77

1 1

65 R 25x40 66 R 25x40

3 3

2.03 -1.68

15.82 18.11

288.84 297.72

1 1

67 R 25x40 68 R 25x40

3 3

-5.88 -8.29

3.02 -2.81

255.9 242.44

1 1

69 R 25x40 70 R 25x40

3 3

-18.01 36.22

-13.64 -10.85

542.32 693.26

1 1

71 R 25x40 72 R 25x40

3 3

28.41 -18.15

10.68 7.5

639.47 473.66

1 1

8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14 8Ø14

Design force Qz for Shear(KN)

Transver-sal Reinforc ement Distribution Spacing

10 10

-6.83 0.73

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

10 10

0.6 5.13

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

10 10

5.03 4.38

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

10 10

3.86 -0.13

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

10 10

1.32 -5.47

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

10 10

-5.9 -6.77

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

10 10

0.44 0.45

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

10 10

3.52 2.86

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

10 10

-1.07 -2.15

2T10 [email protected] 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)

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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( )

____________________________________________________________________________________________ 120

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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

____________________________________________________________________________________________ 121

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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 :

A

W  

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:

____________________________________________________________________________________________ 122

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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.

____________________________________________________________________________________________ 123

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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

____________________________________________________________________________________________ 124

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.5 Data for Design:

Fig 7.2: showing Foundation Plan for the building while displaying symbols depicting initially assumed sections

____________________________________________________________________________________________ 125

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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

____________________________________________________________________________________________ 126

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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 > S q r t ( A r e q ) For rectangular footings choose B

X

L > A

____________________________________________________________________________________________ 127

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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: (∅)P n = (∅ 0.85f' c A x where (∅ = 0.65 and A1 is the area of the column (Where With confinement: ∅ P n = ∅ 0.85 f' c A 1

X

A2 A1

cannot exceed 2)

A2 A1

Note: IF the column concrete strength is lower than the footing, calculate ∅ Pn for the column too. b. Find the load transferred by dowels ∅

Do we ls =

If ∅ Pn >Pu

P u - ∅P n 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

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____________________________________________________________________________________________ 129

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.7 Design of Foundation Type Reaction

footing

D.L

Ps

Ms

Vs

868.67

6.45

-5.11

412.39

3.10

-2.61

F L.L

Table (7.7-1): serviceability load (SLS) of column 59 on Foundation 33

7.7.1 Area of footing: Case : (D+L)

Assume L  2.6 m P  PD.L  PL.L  868.67  412  1280.67______________ (1) M  M D.L  M L.L  (Vd  VL )  h  M  6.45 3.10 (6.45 3.10) 0.5  14.325 M 14.325 L 2.6  0.0111   0.43 _________(2) P 1280.67 6 6 (qnet )all  qall   s ( D f - h f )c h f (q )  250  16 (1.6 -0.5)- 240.5  222 kN/m2 e

net all

(qnet )all  222 kN/m2 ________________________________( 3 )

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Design Of Foundations

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qnet 

Ps  e  6  1 B  L  L 

1280.67  0.0111 6  1 222  2.6  2.6  B  2.27m B

useB  2.40m

qnet 

L  2.6m

( 3b )

1280.67  0.0111 6   210.49  22 _________( 4 ) (O.K) 1  2.6  2.40  2.6 

7.7.2 Footing Stability:

Pu  1.4 D.L + 1.7 L.L  1.4 868.67  1.7  412.39  Pu  1917.201KN ______________________________________( 5) M u = 1.4  Mu  1.7MuL.L  1.4  6.45  1.7  3.10 d.L

 1.4  6.45  1.7  3.10  14.3KN.m Vu = 1.4 D.L  1.7 L.L  1.4  5.11  1.7  2.61  11.591KN M  V  h 14.3  ( 11.59  0.5 ) e u u  Pu 1917.201 e  0.0104 _____________________________________________( 6 ) L e (The eccentricity should be lesser than sixth part of long direction) 6 2.6 0.0104   0.43,0.0104  0.43 ( O. K ) 6 Stability Moment F .O.S   Over turning moment

2.6 2  1 (Thats OK) ..Stable.. ( 6.45  3.10 )  0.5

(868.67  412.39)

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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.7.3 Strength Design:

P e6   qmax  u  1  ,min B  L  L  1917.201  0.0104  6   1  qmax    314.6kN/m 2 _______________(7) 2.6  2.4  2.6  1917.201 

1  qmin  2.6  2.4 

0.0104  6    299.86 kN/m 2 ______________( 8 ) 2.6 

7.7.4 Check one way shear:

d  h  c.c  d /2 b d  500  75  16/2  417 mm (L  X ) d (q (q )  q   q ) u d max min min L L C1 2.6 0.6 X   d  - 0.417  0.583 d 2 2 2 2

7.7.5 Actual Shear Stress

( 2.6  0.583)

 (314.6 - 299.86)  311.29kN/m2 2.6  (q )  (q u )  d (Vu )  B ( X )  u max  d d 2 (q u )  299.86  d



  314.6  311.29  (Vu )  2.4  0.583     437.87 kN d 2  



Allowable Shear Stress

f c'  Vc  0.85   B d 6 0.85  25  2400  417  Vc   708.9 kN 6  103  Vc  708.9  (Vu )  437.87 OK d Hence , Allowable Shear > Actual Shear (That’s Ok.) _______________________________________________________________________________________ 133

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.7.6 Check two way shear: d=h-C.C-db-500-75-16=409 mm =C1+d=600+409=1009mm b1 b 2=C2+d=250+409=659mm b 0=2(b1+b 2)=2(1009+659) =3336 mm

A c  b 0  d  3336  409  1 364424 mm 2 J

(b 13 ) d 6



b1 ( d 3 ) 6

b   2b 2  d   1   2   

2

2

(1009 ) 3 (409) 1009 ( 409) 3  1009    2 (659) ( 409 )    2 . 187  10 11 mm 4 6 6  2  b C a  1  5 04 .5 mm 2 P 1917.201 q av  u   307 .24 kN/m 2 B  L 2 .6  2 . 4 

(V u )

bo

 Pu  q av (b 1  b 2 )

 1 917  307 .24 (1.009  0.659)  1 712 .90 kN/m 2 1 1 γ 1 1  0.491 2 500 2 C1 1 1 3 250 3 C 2

Vu 

(Vu )b 0 γ  M u  Ca  Ac J

0.49  14.3  106  504.5 1364424 2.187 1011  1.255  0.0161  1.271 MPa 

1712.90 103



_______________________________________________________________________________________ 134

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

2 2  1  ( β )  1  ( 600 /250 )  1.833control  c  (40  409) K  1  (α s  d)/2b 0  1   3.45 2  3336  2  

useK  1.83 0.85  K  f c' 0.85  1.83  25  Vc    1.3 MPa 6 6  Vc  Vu OK IF  Vc  Vu Increase H Increase column size Increase



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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.8 Design of Flexure in Long direction :

q

q



L / 2  C1 / 2

 (q max  q ) uf min min L From eq 3b, 7,8 2.6 0.60 (  ) 2  (314.6  299.86)  299  2 2.60  299  9.07  308 kN/m 2 B  (X ) 2 f [q  2(q ) M  u max ] uf uf 6 2.4  .(1) 2  [308  2(314.6)]  374.88 kN.m 6 16 d  500  75   417 mm 2 (M u ) 374.88  106 f Ku    0.0399 2 0.9  f c'  B  d 2 0.9  25  2400  417 1  1  2.36  K u ω 1.18 1  1  2.36  (0.0399) ω 0.0408 1.18

f c' ρ  ω fy ρ  0.0408 

25  0.0024 420

use  req  0.0024

 req   min o.oo24  0.002

As  ρbd  0.0024  2400  417  2401.4 mm2

number of bars ( N ) 

(As )total 2401.4   11.86  12 (As )one bar π 16 2 /4

use 1216in long direction

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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

L  2  C.C 2400  2  75   204.5  204 mm N 1 12  1 Use 12  16 / 204 mm (secondry reinforcem ent) N 12 m'    516 / m' L 2 .4 Spacing between bars ( S ) 

7.9 Design of flexure in Short direction:

d  h  c.c  1.5d b  500  75  1.5  16  401 mm P 1280.67 qav  u   205.23 kN/m 2 B  L 2.4  2.6 ( (B  C2 ) / 2 )2 M  qav  uf 2  (2.4  0.25) / 2 2 M  205.23  118.85 kN.m uf 2 M u f 106 Ku  0.9  f c'  L  d 2 118.85  106 Ku   0.0126 0.9  25  2600  4012 ω

1  1  2.36  (0.0126)  0.0126 1.18

ρ  ω

f c' fy

 0.0126 

25  0.0008   min 420

use  min  0.002

As  ρ  b  d  0.002  2600  401  2085 mm 2

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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

number of bars ( N ) 

(As ) total  2085.2  10.37  11 2 (As ) one bar π 16  /4

use 1116 Spacing between bars ( S )  2600 275  250 mm 101 11  16 / 245mm (main reinforcement)

_______________________________________________________________________________________ 138

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

Fig 7.5 Detailed Reinforced for foundation _______________________________________________________________________________________ 139

Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.10 Development length in footing :

L C L provided  - - C.C d 2 2 0.9  f y  d  K1  K 2  K 3 b L required  d K  f c' 4 where : d = 16 mm f y = 420 MPa b K1 = 1.0 for bottom bars K1  1.3 for top bars (A s ) req 2401 K 2 = 0.8 for d < 19 mm K3 =  = 0.995 b (A s ) prov 2411.52 C K4   2.5 d b L  C1 2600  600 L   C.C   75  925 mm d provided in long 2 2 B  C2 2400  250 L   C.C   75  1000 mm d provided in short 2 2 0.9  420  16  1 0.8  0.995 L required   385.13 mm d 2.5  25 L provided  L required OK d d

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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.11 Bearing strength 

Bearing strength of column:

 Puo   ( 0.85  f c'  C1  C2 ) column  Puo  0.7( 0.85  25  250  600 )  10 -3  2231.25 kN  Pu ( 1280.67 ) A1  C1  C2  0.6  0.25  0.15 m2 A2  L  B  2.6  2.4  6.24 m 2 where



A2 6.24   6.44  2 A1 0.15

(code clause)

Bearing strength of Footing:

 Puo   ( 0.85  f c'

column

 C1  C 2 )

A2 A1

 Puo  0.7( 0.85  25  250  600  2 )  4462.5

use

A2 A1

2

(ACI 318 - 02 Clause)

We can use min (As) dowels:

( A ) dowel  0.005  ( A ) col s s  0.005  ( 250  600 )  750 mm 2  4  14

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Design Of Foundations

DESIGN OF REINFORCED CONCRETE MULTI STORY BUILDING

7.12 Development of dowels :

  (L ) d com   L 1       

L1

L

2

f y .d b . K 1. K ' 4 fc  296.94 mm 

2



 h - c.c - 2  d

b  500 - 75 - 2  16  393 mm  296.14  1.3  (L ) com d  1.3  296 . 94  386 . 022 available

420  16  1  0.8 4  25

mm

O.K

use 450mm

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

_______________________________________________________________________________________ 142

Design Of Foundations

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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. ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬143

Conclusions & Recommendations

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬144

Conclusions & Recommendations

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪APPENDICES‬‬

‫‪Fig 8.1 Column interaction diagram‬‬ ‫‪ 145‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Conclusions & Recommendations‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Table (8-2) Coefficients for live- load positive moments in slabs‬‬ ‫‪ 146‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Conclusions & Recommendations‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Table (8-3) Coefficients for dead- load positive moments in slabs‬‬

‫‪ 147‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Conclusions & Recommendations‬‬

‫‪Design of Reinforced Concrete Multi-story Commercial building‬‬ ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Table (8-4) Coefficients for negative moments in slabs‬‬

‫‪ 148‬ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪Conclusions & Recommendations‬‬

Design of Reinforced Concrete Multi-story Commercial building ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

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 … ‫ ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬149

Conclusions & Recommendations

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