Analysis and design of multi storey building

Analysis and Design of Multi-Story Building

Project Submitted as Part Requirement for the Degree of Bachelor of Science in Civil Engineering

By:

Hawdin Ismael Karokh Hassan

Supervisor:

Prof. Dr. Mohamad Raouf

The University of Sulaimani Department of Civil Engineering 2015

DECLARATION STATEMENT We certify that, to the best of our knowledge, our dissertation does not infringe upon anyone’s copyright not violate any proprietary rights and that any ideas, techniques, quotations, or any other material from the work of other people included in our dissertation, published or otherwise, are fully acknowledged in accordance with the standard referencing practices.

Hawdin ismail Karox Hassan

In my capacity as supervisor of the group’s dissertation, I certify that the above statements are true to the best of my knowledge …

Dr. Mohamed Raouf (PhD)

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ACKNOWLEDGEMENTS Foremost, we would like to thank Allah then express our sincere gratitude to our supervisor prof. Dr. Mohamed Raouf for the continuous support of our dissertation, for this patience, motivation, enthusiasm, and immense knowledge. His guidance helped us in all the time of research and writing of this dissertation. We could not have imagined having a better supervisor and for this project.

Our sincere thanks also go to instructor Brwa Hama Saeed, who showed effort during the project preparation.

Last but not least, we would like to thank our parent for their continuous support and their firm love as always.

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ABSTRACT Our project deal with the analysis and design of multi-story commercial building locate in sulaimani city, our goal is to re-analysis and re-design the building from start to finish using completely different method from what was previously used. The hardships we incomer on the way with reach us as students and benefit us as designers in the future.

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TABLE OF CONTENTS

Page No.

1 General Information in Reinforced Concrete Building ………….………… 1 1.1 Introduction ……………………………………………………..…….. 1 1.2 Concrete structural system ………………………………………… 2 1.3 Concrete and property……………………………………………… 3 1.4 Type of loading subjected to the building …………………………. 4 1.5 Aim and objective...………………………………………………… 5 1.5 Out Line…………..………………………………………………… 6

2 Slab Analysis and Design ………………………………………………..…….…..7 2.1 Introduction …………………………………………………..…… 7 2.1.1 Type of slabs ……………………………………………..…… 7 2.2 Methodology……………………………………………….……… 9 2.2.1 Analyze and Design Procedure……………………….…….... 10 2.2.2 ACI-Code provision for slab design and detail …………...… 11 2.3 Calculation ……………………...……………………….……….. 19 2.3.1 Loading on the project (ASCE-2012)………………………….. 19 2.3.2 Analysis and Design the slab …………..……………….…… 20 2.4 Detail of slab ………………………..…….…….………………… 37 2.5 Conclusion ……………………………………………..…………...... 39

3 Beam Analysis and Design ……………………………………….…….………. 40 3.1 Introduction …………………………………………………………….. 40 3.1.1 Types of Beam …………………………………………………….. 41 3.2 Methodology…………………………………………………………..... 42 3.3 Calculation ……………………………………………………………… 43 3.3.1 Analysis of wind …………………………………………………… 43 3.3.2 Analyze and design of beam by X-direction….……………………. 45 3.3.3 Analyze and design of beam by Y-direction….……………………. 50 3.3.4 Check for stirrup spacing at critical beam on the frame …………… 56 3.4 Detail of beam…………………………………………………………… 58 3.5 Conclusion…………………………………………………………... 64

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4 Column Analysis and Design …………………………………………………… 65 4.1 Introduction ………………………………………………………………. 65 4.2 Types of Columns ………………………………………………………... 66 4.3 Methodology ……………………………………………………………... 68 4.3.1 Short Columns with small eccentricities …………………………... 68 4.3.2 Short Columns with large eccentricities ……………………………. 68 4.4 Calculation ………………………………………………………...……… 69 4.4.1 Analyze of column ………………………………………………….. 69 4.4.1.1 Calculate Axial Load on Column …….………………………. 69 4.4.1.2 Calculate Moment due to Wu ………………………….……... 70 4. 4.1.3 4.4.1.3 Calculate Moment due to Wind load ………….……….……... 73 4.4.1.4 Load combination according to ACI-code 1995 ………...….. 74 4.4.2 Design of Column ……………………………………………………. 75 4.4.2.1 Check for Slender Column ……………………………………. 75 4.4.2.2 Uniaxial Column Design ……………………….…………...…. 76 4.4.2.3 Biaxial Column Design …………………………………...…… 78

4.5 Detail of column reinforcement …………………………………..… 85 4.6 Conclusion ………………………………………………………..… 89

5 Foundation Analysis and Design ……………………………………………..… 90 5.1 Introduction ………………………………………………………… 90 5.1.1 Types of Footings ………………………………………………..…... 91

5.2 Methodology ……………………………………………………………... 93 5.3 Calculation ………………………………………………….……………. 94 5.3.1 Calculate Load on the Foundation ………………………….……….. 94 5.3.2 Find dimension of mat foundation ………………………………….. 96 5.3.3 Analysis of the mat …………………………………………….…….. 99 5.3.4 Design of the mat …………………………………………………… 103 5.4 Detail reinforcement in mat foundation …………………………………. 104 5.5 Conclusion ………………………………………………………………. 105

6 Analysis and Design of Stair and Concrete Wall ……….………………..… 106 6.1 Introduction ……………………………………………………….. 106 6.2 Calculation ……………………………………………………………… 107 6.2.1 Analysis and design of stair ……………………………………….. 107 6.2.2 Design of concrete wall …………………………………………... 110

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Conclusion ……………………………………………………………………………111 Appendix ……………………………………………………………………………..112 Appendix A: Architectural plan and section …………………………… 112 Appendix B: Factor moment according to ACI-Code …..………………… 114 Appendix C: Reinforcement detail at slab according to ACI-Code 318-11… 115 Appendix D: Interaction chart (all side steel)……...……………………… 116 Appendix E: Table 9.5(a): minimum thickness of beam and one way slab … 120 Suggestion and Recommendation …………………………………………………. 121 References …………………………………………………………………………… 122

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LIST OF FIGURES

Page No.

Figure 1-1; 3D Frame of the Building ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ 1 Figure 1-2; Typical elements of the building ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 2 Figure 2-1; Type of Structural Slab ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ 8 Figure 2-2; Moment distribution on slab in (DDM) according to ACI-Code ‌‌‌‌‌‌ 10 Figure 2-3; Slab system divided into X (a) and Y (b) direction strips ‌‌‌‌‌‌‌‌‌ 12 Figure 2-4; Longitudinal moment distribution ‌‌‌‌‌‌‌‌‌‌‌‌‌.‌‌‌‌.. 14 Figure 2-5; Cross-section of torsional member (edge beam) for calculation of βt ‌‌‌.‌ 15 Figure 2-6; Minimum length of slab reinforcement ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌... 18 Figure 2-7; Detail of span length ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌.... 20 Figure 2-8; Detail of slab analysis ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌.. 21 Figure 2-9; Strip by X-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 22 Figure 2-10; � calculation ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌.‌. 23 Figure 2-11; Bending moment diagram by X-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌.‌ 25 Figure 2-12; Strip by Y-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌... 28 Figure 2-13; Bending moment diagram by Y-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌..... 31 Figure 3-1; Reinforced rectangular beam ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ 40 Figure 3-2; Common shapes of concrete beam ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌..‌‌ 41 Figure 3-3; Analyze wind load by Y-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ 43 Figure 3-4; Moment acting on the beam ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 44 Figure 3-5; Moment acting on the foundation ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ 44 Figure 3-6; Shear Diagram in roof slab by X-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌... 46 Figure 3-7; Detail typical beam by X-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 49 Figure 3-8; Shear Diagram in roof slab by Y-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌.. 52 Figure 3-9; Detail typical beam by Y-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 55 Figure 3-10; Shear Diagram in office slab by X-direction ‌‌‌‌‌‌‌‌‌‌‌‌‌‌.. 56 Figure 3.11; Calculated moment and steel at roof slab for beam ‌‌‌‌‌‌‌‌..‌‌ 58 Figure 3.12; Calculated moment and steel at 3- 4 floor slabs for beam ‌‌‌‌‌‌‌.... 60 Figure 3.13; Calculated moment and steel at 1- 2 floor slabs for beam ‌‌‌‌‌‌‌‌ 62 Figure 4.1; Column shear failure at Tohoku – Japan ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ 65 Figure 4.2; Use column to Support slab and appearance ‌‌‌‌‌‌‌‌‌‌‌‌.‌. 65 Figure 4-3; Column types ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 66 Figure 4-4; The column types depending on applied load ‌‌‌‌‌‌‌‌‌‌‌‌‌ 67 Figure 4-5; Eccentric loaded conditions (Spiegel, 1998) ‌‌‌‌‌‌‌‌‌‌‌‌‌.. 67 Figure 4-6; Column Load Area (Top View) ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. 69 Figure 4-7; Ultimate axial load and moment due to (Wu) for strip 3.5 m width ‌‌‌‌.. 72 Figure 4-8; Ultimate axial load and moment due to (Wu) for strip 7 m width ‌‌‌‌..‌72 Figure 4-9; Calculated moment due to (Wu) about Y-axis ‌‌‌‌‌‌‌‌‌‌‌‌‌ 73 Figure 4-10; Calculated Axial load and moment due to wind load parallel to Y-direction ‌.. 73 Figure 4-11; Load combination according to ACI-Code 1995 for strip 3.5 m ‌‌‌‌‌.. 74 vii

Figure 4-12; Load combination according to ACI-Code 1995 for strip 7 m ………………. 74 Figure 4.13; Equivalent eccentricity of column load ……………………………………… 76 Figure 4.14; Detail of column 6 ……………………………………………………………. 77 Figure 4-15; Interaction diagram for compression plus biaxial bending ………………….. 78 Figure 4-16; Detail of column 2 ……………………………………………………………. 81 Figure 4-17; Column plan detail …………………………………………………………… 82 Figure 4-18; Detail of column reinforcement on floor 4 and floor 3 ………………………. 85 Figure 4-19; Detail of column reinforcement on floor 2 …………………………………… 86 Figure 4-20; Detail of column reinforcement on floor 1 …………………………………… 87 Figure 4-21; Detail of column reinforcement on ground floor …………………………….. 88 Figure 5-1; Mat foundation on the pile Dubai- Burj Khalifa ………………………………. 90 Figure 5-2; Footing Type ………………………………………………………………….. 92 Figure 5-3; Factor and unfactor load on the soil foundation ………………………………. 94 Figure 5-4; Factor load on the strip of the mat foundation ………………………………… 95 Figure 5-5; Detail of Strip by X & Y direction …………………………….………………. 99 Figure 5-6; Draw shear and bending moment diagram by X-direction ……………………. 100 Figure 5-7; Draw shear and bending moment diagram by Y-direction ……………………. 102 Figure 6-1; Architectural detail stair ………………………………….………………..…. 106 Figure 6-2; Architectural design of stair …………………………………………………. 107 Figure 6-3; Shear and bending moment diagram …………………………………………. 108 Figure 6-4; Reinforcement detail of Stair …………………………….…………………… 109 Figure 6-5; Reinforcement detail of concrete wall ……………………….………………. 110

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LIST OF TABELS

Page No.

Table 2-1; Total static moment according to ACI-Code ……………………………………... 13 Table 2-2; Moment distribution for strip from ACI-Code ……………………………………. 16 Table 2-3; Moment on roof slab strip 6.6 m by X-direction …………………………………. 27 Table 2-4; Moment on roof slab strip 7 m by Y-direction …………………………………… 32 Table 2-5; Moment on office slab strip 6.6 m by X-direction ……………………………….. 35 Table 2-6; Moment on office slab strip 7 m by Y-direction …………………………………. 36 Table 4-1; Design of column in all floors with biaxial moment (using excel program) ……... 83 Table 4-2; Design of column in all floors with uniaxial moment (using excel program) ……. 84 Table 5-1; Estimate quantity of steel in foundation with bend bar diameter ………………… 104

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Ch. -1- INTRODUCTION

Chapter One General Information in Reinforced Concrete Building 1.1 Introduction: Structure is defined as system of interconnected members assembles in a stable configuration, the main purpose of a structure is to transfer load from one point to another: bridge deck to pier; slab to beam; beam to girder; girder to column; column to foundation; foundation to soil.

The structural components in a typical multi-story building, consists of a floor system which transfers the floor loads to a set of plane frames in one or both directions. The floor system also acts as a diaphragm to transfer lateral loads from wind or earthquakes. The frames consist of beams and columns and in some cases braces or even reinforced concrete shear walls. As the height of the building increases beyond ten stories (tall building), it becomes necessary to reduce the weight of the structure for both functionality and economy.

Figure 1-1: 3D Frame of the Building

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Ch. -1- INTRODUCTION

1.2 Concrete structural system: Any structure is made up of structural element (load-caring such as beams, columns,…) and non- structural element (such as partitions, false ceilings, doors, windows,…) structural element, put together constitute the structural system its function is to resist effectively the action of gravitational and environmental loads, and to transmit the resulting forces to the supporting ground, without significantly disturbing the geometry, integrity and serviceability of the structure.

Such structural concrete element are composed of variety of concrete structural elements, its component can be broadly classified in to: 

Floor slabs



Beams



Columns



Walls



Foundations

Floor slabs:

Figure 1-2: Typical elements of the building

Floor slabs are the main horizontal elements that transmit the moving live loads as well as the stationary dead loads to the vertical framing supports of a structure.

Beams: Beams are structural elements that transmit the tributary loads from floor slabs to vertical supporting columns.

Columns: The vertical elements support the structural floor, system, they are compression members subjected in most cases to both bending and axial load are of the major importance in the safety considerations of any structure.

Walls: Walls are the vertical enclosures for building frames .they are not usually or necessarily made of concrete but of any material that esthetically fulfills the form and function needs of structural system.

Foundations: Foundations are the structural concrete elements that transmit the weight of super structure to the supporting soil.

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Ch. -1- INTRODUCTION

1.3 Concrete and property: Stone like martial obtained by permitting a carefully proportioned mixture of cement, sand, gravel, and water in forms of the shape and dimensions of desired structure. Main properties of an acceptable concrete can be described as follow: 

Compactness

The space occupied by concrete should as much as possible be filled with solid aggregate and cement get. 

Strength

Concrete is desired material in building because the concrete is a very strong in compression, and should always have sufficient strength and internal resistance to the various types of failure, according to ACI 318M-08 minimum compressive strength not to be less than 17.5Mpa, but concrete in tension is a very weak which 10% of compression strength. 

Water/cement ratio

The water/cement ratio should be suitably controlled to give required design strength and workability. 

Texture

Exposed concrete surface should have a dense and hard texture that can withstand adverse weather condition.

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Ch. -1- INTRODUCTION

1.4 Type of loading subjected to the building: Dead load: Dead loads consist of the weights of the various structural members and the weights of any objects that are permanently attached to the structure. For a building, the dead loads include the weights of the columns, beams, and girders, the floor slab, roofing, walls, windows; plumbing, electrical fixtures .Dead loads are these that are constant in magnitude and fixed in location throughout the time life of structure. Usually the major part of the dead loads is the weight of the structure itself.

Live load: Live loads are those loads produced by the use and occupancy of the building or other structure and do not include construction or environmental loads such as wind load, snow load, rain load, earthquake load, flood load, or dead load.

Wind load: This is very important design factor, especially for tall building, or building with large surface area. Building are designed not to resist the everyday wind condition, but extreme condition that may occur once every 100 years or so. These are called design wind speed, and are specified in building codes.

Earthquake Loads: Earthquakes produce loadings on a structure through its interaction with the ground and its response characteristics. These loadings result from the structure’s distortion caused by the ground’s motion and the lateral resistance of the structure. The specified earthquake loads are based upon post-elastic energy dissipation in the structure, and because of this fact, the requirements for design, detailing, and construction shall be satisfied.

Soil pressure: Lateral earth pressure is the pressure that soil exerts against a structure in a sideways, mainly horizontal direction. The common applications of lateral earth pressure theory are for the design of ground engineering structures such as retaining walls, basements, tunnels, and to determine the friction on the sides of deep foundations.

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Ch. -1- INTRODUCTION

1.5 Aim and objective: The aim of this project is to finish the project by analyzing and designing the commercial building using different system. Learn to work as a group. Learn what problems a designer might face. And potentially have a good background for designing.

Commercial building usually consists of more than one floor to decrease the land use.

Our project which is structural design of a commercial multi-story building and it is re-design of building in Sulaymaniyah. But with difference methodology and reduce number of floor on the land area of (780m2) the designed frame system involves slabs with beams and columns.

In conclusion, with this project we can achieve idea and information related to designing of a concrete frame structure, which aids to design more complex structure after graduation or at postgraduate studies.

For a structure, two factor influence it, first is construction cost second is it’s aesthetic, civil engineer can keep a balance between the cost, time and it’s quality, when this factors are controlled, then a great outcome will be achieved.

It is essential to choose a suitable plan, structure type and method of design for a research, we choose a plan with the aid of our professor that has been built in Sulaymaniyah in one of the most famous real estate project, We did some modifications to the structure in order not to be as same as the built one, the frame type which we designed is slabs with columns and beams.

There are several ways to design frame from finite element to approximate design. The method which converges with our knowledge and suitable for our frame is direct design method for slabs and for find the moment and reinforcement in beam and column by hand-calculation.

Several trial and error calculations performed before any decision, and finally we chose the most suitable result, the design is very logical.

This project was a good reason to force us learn engineering programs perceptibly, We drew the 3D view of the building with Google sketch up program. Then we increased our knowledge about AutoCAD with drawing a very detailed plan. Excel forms prepared for moment.

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Ch. -1- INTRODUCTION

1.6 Out line:

Chapter One: “Introduction” Introduce the project goal, description and general information.

Chapter Two: “Slab Analysis and Design” Involve the total slab design and details.

Chapter Three: “Beam Analysis and Design” Involve analysis of wind load on beams and design of beams.

Chapter Four: “Column Analysis and Design” Involve analysis of wind load on columns and design of columns.

Chapter Five: “Foundation Analysis and Design” Involve analysis of wind load on foundation and design by mat foundation type.

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Ch. -2- Analysis and Design of Slab

Chapter two Analysis and Design of Slab 2.1 Introduction: The slab provides a horizontal surface and is usually supported by columns, beams or walls.

Slabs can be categorized into two main types: one-way slabs and two-way slabs.

One-way slab is the most basic and common type of slab. One-way slabs are supported by two opposite sides and bending occurs in one direction only. Two-way slabs are supported on four sides and bending occurs in two directions. One-way slabs are designed as rectangular beams placed side by side. However, slabs supported by four sides may be assumed as one-way slab when the ratio of lengths to width of two perpendicular sides exceeds 2.

2.1.1 Type of slabs: a) One-way slab b) Two-way slab c) Flat plate slab d) Flat slab with drop panel e) Gird or waffle slab The slab may be supported in two opposite sides only as shown in figure (a) at which the load will be transferred in the direction of beam only. The slab may be supported in its four sides as shown in figure (b) at which the load will be transferred into two directions. Sometimes the slab will be provided with intermediate beam as shown in figure (c).

In some cases the concrete slab is carried directly by the column without drop panel as shown in figure (d).in this type a special care must be taken to check the punching shear of column, therefore if the punching shear exceeds the shear resistance of the column, a column capital or drop panel will be provided as shown in figure (e). There is another type of slab called gird slab consists of joists in both directions which provide a rigid slab with minimum dead load that has a minimum deflection amount compare to the other types of slab as shown in figure (f).

Note: In this project the system was column beam with slab system consist of two way slab.

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Ch. -2- Analysis and Design of Slab

Figure 2-1: Type of Structural Slab (at Design of concrete structure by Arthur H. Nilson – pa.425) -8-

Ch. -2- Analysis and Design of Slab

2.2 Methodology: The methods for analysis slab are:  Direct Design Method (DDM)  Equivalent Frame Method (EFM)  Yield Line Method (YLM)  Finite Element Method (FEM)  Coefficient method Each of these methods has its limitation, and direct design method DDM was chosen because all limitation in our project is ok, and limitations are from ACI-Code (13.6.1):  There must be three or more spans in each direction.  Panels should be rectangular and the long span is no more than twice the short span.  Successive span lengths center-to-center of supports in each direction shall not differ by more than 1/3 of the longer span.  Columns must be near the corners of each panel with an offset from the general column line of no more 10% of the span in each direction as shown below:

 The live load should not exceed 3 times the dead load in each direction. All loads shall be due gravity only and uniformly distributed over an entire panel.  If there are beams, there must be beams in both directions, and the relative stiffness of the beam in the two directions must be related as follows:

0.2 >

> 2.0

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Ch. -2- Analysis and Design of Slab

2.2.1 Analyze and Design Procedure: The basic design procedure of a two-way slab system has four steps: 1. Determine moments at critical sections in each direction, normally the negative moments at supports and positive moment near mid-span.

2. Distribute moment’s transverse at critical sections to column and middle-strip and if beams are used in the column strip, distribute column strip moments between slabs and beams.

3. Determine the area of steel required in the slab at critical sections for column and middle strips.

4. Detail steel reinforcement.

Figure 2-2: Moment distribution on slab in (DDM) according to ACI-Code

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Ch. -2- Analysis and Design of Slab

2.2.2 ACI-Code provision for slab design and detail: Definitions related to Direct Design Method:1.

Strips: Slab is considered to be a series of strips in two directions.

2.

Panel (ACI 13.2.3): A panel is bounded by column, beam, or wall centerlines on all sides. A panel includes all flexural elements between column centerlines.

3.

Column Strip (ACI 13.2.1): Column strip is a design strip with a width on each side of a column centerline equal to 0.25l2, Column strip includes beams, and the

4.

l1: is the length of span in direction that moments are being determined, measured centerto-center of supports.

5.

l2: is the length of span transverse to l1, measured center-to-center of supports.

6.

Middle strips (ACI 13.2.2): Middle strip is a design strip bounded by two column strips.

7.

Distribution of Moments:-

a) Total static Moment: Mo (ACI 13.6.2): the total static moment for a span length ln and width l2 of a given frame is given by ACI equation 13-3 as: Mo=

ACI-Eq (13-3)

Where: Wu= Factored load per unit area. ln = length of clear span in direction that moments are being determined, measured faceto-face of supports.

ACI 13.6.2.4 — when the span adjacent and parallel to an edge is being considered, the distance from edge to panel centerline shall be substituted for l2 in Eq. (13-3).

ACI 13.6.2.5 — clear span ln shall extend from face to face of columns, capitals, brackets, or walls. Value of ln used in Eq. (13-3) shall not be less than 0.65 l1. Circular or regular polygon shaped supports shall be treated as square supports with the same area.

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Ch. -2- Analysis and Design of Slab

Detail:

Figure 2-3: Slab system divided into X (a) and Y (b) direction strips

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Ch. -2- Analysis and Design of Slab

b) Longitudinal Distribution of Static Moment (ACI 13.6.3): For a typical interior panel, the total static moment is divided into positive moment 0.35Mo and negative moment of 0.65Mo. For an exterior panel, the total static moment is dependent on the type of restraint at the outside edge.

ACI table 13.6.3.3 (table 13.3 Nilson 13th Ed) as shown in table 2-1 and figure 2-4 of this document can be used for longitudinal distribution:

Table 2-1: Total static moment according to ACI-Code:

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Ch. -2- Analysis and Design of Slab

Figure 2-4: Longitudinal moment distribution:

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Ch. -2- Analysis and Design of Slab

c) Transverse or Lateral distribution of Longitudinal Moments:Tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI on the next page are used to assign moments to column strip. The remaining moments are assigned to middle strip in accordance with ACI 13.6.3. Beams between supports shall be proportioned to resist 85 percent of column strip moments if Îą1l2/l1 {Where l2 shall be taken as full span length irrespective of frame location (exterior or interior)} is equal to or greater than 1.0 (ACI 13.6.5.1).

Transverse distribution of the longitudinal moments to middle and column strips is a function of the ratio of length (l2/l1), ι1, and βt.

Where: βt = the ratio of torsional stiffness of edge beam section to flexural stiffness of a width of slab equal to span length of beam, center-to-center of supports.

If there is no edge beam, βt is taken equal to zero. If there is edge beam, βt is calculated as follows: đ?›˝t= Where, C is the torsional constant of the edge beam. This is roughly equal to the polar moment of inertia of edge beam and is given as: C=∑ Where, “xâ€? is the shorter side of the rectangle and “yâ€? is the longer one.

Figure 2-5: Cross-section of torsional member (edge beam) for calculation of βt.

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Ch. -2- Analysis and Design of Slab Table 2-2: Moment distribution for strip from ACI-Code:

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Ch. -2- Analysis and Design of Slab

8. a)

Minimum Slab Thickness for two-way construction (ACI 9.5.3):-

h=

If αm > 2

But not less than 9 cm. fy in Mpa

b)

h=

If 0.2 < αm < 2

But not less than 13 cm. fy in Mpa Where: The definitions of the terms are:h = Minimum slab thickness without interior beams. ln = length of clear span in direction that moments are being determined, measured faceto-face of supports. β = ratio of clear spans in long to short direction of two-way slabs. αm = average value of α for all beams on edges of a panel.

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Ch. -2- Analysis and Design of Slab

9.

Max Spacing and Min Reinforcement requirement of the ACI code:

• One way slab (ACI 7.6.5 & 7.12.2.2):  Main Reinforcement = 3 hf, or 18 in whichever is less (hf = slab thickness)  Temp reinforcement = 5 hf or 18 in whichever is less.

• Two way slab (ACI 13.3.2):  2 hf in each direction.

• Min Reinforcement in all cases (ACI 7.12.2.1):  0.0018 b hf for grade 60.  0.002 b hf for grade 40 and 50.

10. Reinforcement detail at slab according to ACI-Code 318-11 article 13-3-8

Figure 2-6: Minimum length of slab reinforcement

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Ch. -2- Analysis and Design of Slab

2.3 Calculation: 2.3.1 Loading on the project (Commercial building): according to ASCE-2012 Table 4-1 Roof slab: L.L= 1.5 kN/m2 D.L (Tiling) = 2 kN/m2 Office slab: L.L= 2.5 kN/m2 D.L (Tiling) = 1.5kN/m2 Light weight of block = 0.8 kN/m3 Stair: L.L= 4 kN/m2 D.L (Tiling) = 1.5kN/m2 Available data: fc’ = (21- 26) Mpa fy = 400 Mpa Height of building= 3.4 m for all slab except ground floor = 4.7 m γ (concrete) = 24.5 kN/m3 γ (light weight block) = 8 kN/m3

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Ch. -2- Analysis and Design of Slab

2.3.2 Analysis and Design: Analysis and Design the slab by using D.D.M (Direct Design Method) according to ACI-Code 318-11 chapter 13 article 13-6.

Size for beams, slab and column: ď&#x201A;ˇ

Beams: let assume all beam dimensions = 40 x 60 cm

ď&#x201A;ˇ

Column: let the column dimension = 40 x 40 cm

ď&#x201A;ˇ

Slab thickness: Determine the slab thickness according to ACI-Code (9.5.3):

h=

If Îąm > 2

h=

If 0.2 < Îąm < 2

Îą=

*2 for internal beam

Îą=

*1.5 for external beam

Îą1=

*2= 4.23 = Îą3

Îą2=

*1.5= 6.00 = Îą4

Îąm=

= 5.115 > 2

ln= 6.6 m ;

sn= 6.2 m

Figure 2-7: detail of span length

đ?&#x203A;˝= â&#x20AC;ŤŘ&#x17E;â&#x20AC;Źh=

= 15.57 cm â&#x2030;&#x2C6; 16 cm

Use slab thickness 18 cm

- 20 -

0.5 mid. strip

Col.strip

0.5 mid. strip

4.50 m

Y

X

6.60 m

0.5 mid. strip

Col.strip

0.5 mid. strip 6.60 m

All beam ( 40 X 60 ) cm All col.( 40 X 40 ) cm

1.65 m

7.00 m

7.00 m

7.00 m

7.00 m

7.00 m

Figure 2-8: Detail of slab analysis

- 21 -

Ch. -2- Analysis and Design of Slab

Roof slab: Determination of Wu (Ultimate load): Wu= 1.4 D.L + 1.7 L.L D.L (slab) =0.18 * 24.5 = 4.41 kN/m2 Total D.L = 4.41 + 2 = 6.41 kN/m2 L.L= 1.5 kN/m2 Wu= 1.4 (6.41) + 1.7 (1.5) = 11.524 kN/m2 ≈ 11.53 kN/m2 Analysis the strip by X direction and Y direction: X-direction:

Figure 2-9: Strip by X-direction

Mo=

…. From ACI-Code (13.6.2.2)

ln = 7.0 – 0.4 = 6.6 m ; l2= 6.6 m Mo=

= 414.35 kN.m/6.6m

For external panel: -Min= 0.7* Mo= 0.7*414.35 = 290.1

kN.m/6.6m

+M= 0.57* Mo= 0.57*414.35 = 236.2 kN.m/6.6m

for external panel ACI (13.6.3.3)

-Mex= 0.16* Mo= 0.16*414.35 = 66.3 kN.m/6.6m - 22 -

Ch. -2- Analysis and Design of Slab

From table of ACI-Code (13.6.4.1) find the ratio of column strip by interpolation: â &#x201E; =

â &#x201E; =0.943

Îą1*

â &#x201E; â&#x2030;Ľ1

4.23*0.943 â&#x2030;Ľ 1 only for â&#x20AC;&#x201C;Min & +M

The ratio of column strip = 0.7671 The ratio of middle strip = 0.2329 -Min (290.1) =

+M (236.2) =

-M (column) = 0.7671*290.1= 222.5

kN.m/3.3m

-M (middle) = 0.2329*290.1= 67.6

kN.m/3.3m

+M (column) = 0.7671*236.2 = 181.2 kN.m/3.3m +M (middle) = 0.2329*236.2 = 55

(-Mex) depend on degree of rotation (đ?&#x203A;˝t, Îą1* đ?&#x203A;˝t=

â &#x201E;

kN.m/3.3m

â &#x201E;

:

from ACI-Code (13.6.4.2) and equation (13.5) & (13.6)

C=â&#x2C6;&#x2018;

Figure 2-10: đ?&#x203A;˝ calculation

Option 1: C= [

]+[

]

]+[

]

C= 8.02 * 109 Option 2: C= [ C= 4.95* 109 â&#x20AC;Ť Ř&#x17E;â&#x20AC;ŹC= 8.02 * 109

use greater of them ;

Is=

= 3.21 * 109

- 23 -

Ch. -2- Analysis and Design of Slab

â&#x20AC;Ťđ?&#x203A;˝ Ř&#x17E;â&#x20AC;Źt=

= 1.25

Use đ?&#x203A;˝t &

â &#x201E;

:

â &#x201E; =

â &#x201E; =0.943

:

â &#x201E; â&#x2030;Ľ1

to find (-Mex for column strip) from ACI-Code (13.6.4.2): only for â&#x20AC;&#x201C;Mex

The ratio of column strip = 0.8835 The ratio of middle strip = 0.1165

-Mex (66.3) =

-M (column) = 0.8835*66.3 = 58.6 kN.m/3.3m -M (middle) = 0.1165*66.3 = 7.75 kN.m/3.3m

For internal panel: -Min= 0.65* Mo= 0.65*414.35 = 270 kN.m/6.6m +M = 0.35* Mo = 0.35*414.35 = 145 kN.m/6.6m

for internal panel ACI (13.6.3.3)

-Mex= 0.65* Mo= 0.65*414.35 = 270 kN.m/6.6m

From table of ACI-Code (13.6.4.1) find the ratio of column strip by interpolation: â &#x201E; =

â &#x201E; =0.943

Îą1*

The ratio of column strip = 0.7671

â &#x201E; â&#x2030;Ľ1

4.23*0.943 â&#x2030;Ľ 1 for â&#x20AC;&#x201C;Min & +M

The ratio of middle strip = 0.2329

-Min (270) =

+M (145) =

-M (column) = 0.7671*270 = 207.5 kN.m/3.3m -M (middle) = 0.2329*270 = 63

kN.m/3.3m

+M (column) = 0.7671*145 = 112

kN.m/3.3m

+M (middle) = 0.2329*145 = 34

kN.m/3.3m

- 24 -

Ch. -2- Analysis and Design of Slab

Bending moment diagram in slab:

According to ACI-Code (13.6.5) the moment on column strip transfer to the beam: If (α1* l 2/l1 > 1) the moment transfer to the beam by 85 % of column strip moment and 15% remain on column slab strip but if (α1* l 2/l1 < 1) determine the moment on beam by interpolation between (0-85)% . α1* l 2/l1 > 1

4.23*.943 > 1

‫ ؞‬the moment on column slab strip remain by 15 % of column strip.

Figure 2-11: Bending moment diagram by X-direction

- 25 -

Ch. -2- Analysis and Design of Slab

Find As of roof slab: Using Ă&#x2DC; 12 mm (113 mm2) ; fcâ&#x20AC;&#x2122;=21 Mpa ; fy=400 Mpa M= Ă&#x2DC; As fy ( d- ) Ă&#x2DC; = 0.9

Reduction factor

d= đ?&#x2019;˝ - Ă&#x2DC; -

ACI -11 (9-3-2-1)

â&#x20AC;&#x201C; cover = 0.0068 As â&#x20AC;&#x201C; 20 = 142 mm

d= 180 -12-

-M= 67.6 kN.m/3.3m = 67.6 * 106 N.mm/3.3m 67.6 * 106 = 0.9 * As * 400 (142 â&#x20AC;&#x201C; 0.0034 As) Find As =?

As = 1368 mm2 As (min) = 0.0018* h *â&#x201E;&#x201C;strip

ACI-Code (7.12.2.1) but for more safety we can use 0.002

As (min) = 0.002*180*3300 As (min) =1188 mm2 Check for d considering max moment (67.6 kN.m/3.3m): 2

M= Ă&#x2DC; Ď b d fy (1-

)

Let Ď = Ď max = 0.75 Ď balance = 0.75 [0.85 * đ?&#x203A;˝1*

â&#x20AC;Ť Ř&#x17E;â&#x20AC;ŹĎ = 0.75 [0.85 * 0.85*

]

] = 0.017

â&#x20AC;Ť Ř&#x17E;â&#x20AC;Ź67.6 * 106 = 0.9 * 0.017 * 3300 * d2 * 400 * (1-

)

d = 65 mm < 142 mm d provide

- 26 -

Ch. -2- Analysis and Design of Slab

Table 2-3: Moment on slab strip 6.6 m by X-direction:

Middle strip

Column strip

strip

Section -ve ex-in -ve in -ve ex-ex +ve ex +ve in -ve ex-in -ve in -ve ex-ex +ve ex +ve in

Moment(kN.m/3.3m) 67.6 63 7.75 55 34 33.4 31.2 8.8 27.2 16.8

As required (mm2) 1368 1270 157 1113 688 676 632 178 550 340

As design (mm2) 1368 1270 1188 1188 1188 1188 1188 1188 1188 1188

Determination spacing: If As = 1368 mm2/3.3m ; Area of one bar (12mm) = 113 mm2 No.of bars/3.3m = Spacing =

≈ 13 bars ; No.of spaces = 13-1= 12 space

= 0.275m = 27.5 cm

Use Ø 12 mm @ 25 cm ⁄

If As = 1188 mm2/3.3m No.of bars/3.3m = Spacing =

≈ 11 bars ; No.of spaces = 11-1= 10 space

= 0.33m = 33 cm

Use Ø 12 mm @ 30 cm ⁄

- 27 -

Ch. -2- Analysis and Design of Slab

Y-direction:

Figure 2-12: Strip by Y-direction For external panel: ln = 4.5 - 0.4 = 4.1 m ; l2= 7m Mo=

= 169.6 kN.m/7m

-Min= 0.7* Mo= 0.7*169.6 = 118.72 kN.m/7m +M = 0.57* Mo = 0.57*169.6 = 96.67 kN.m/7m

for external panel ACI (13.6.3.3)

-Mex= 0.16* Mo= 0.16*169.6 = 27.14 kN.m/7m From table of ACI-Code (13.6.4.1) find the ratio of column strip by interpolation: ⁄ = ⁄

=1.56

α1*

The ratio of column strip = 0.582

⁄ ≥ 1 ; 4.23*1.56 ≥ 1 only for –Min & +M

The ratio of middle strip = 0.418

-Min (118.72) =

-M (column) = 0.582*118.72= 69.1 -M (middle) = 0.418*118.72 = 49.63

+M (96.67) =

kN.m/3.5m kN.m/3.5m

+M (column) = 0.582*96.67 = 56.27

kN.m/3.5m

+M (middle) = 0.418*96.67 = 40.41

kN.m/3.5m - 28 -

Ch. -2- Analysis and Design of Slab

(-Mex) depend on degree of rotation (đ?&#x203A;˝t), (Îą1*

â &#x201E;

â &#x201E;

):

đ?&#x203A;˝t= C= 8.02 * 109 Is=

â&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Ś from page (23)

= 3.40 * 109

â&#x20AC;Ťđ?&#x203A;˝ Ř&#x17E;â&#x20AC;Źt =

= 1.178

Use đ?&#x203A;˝t &

â &#x201E;

:

â &#x201E; = â &#x201E;

=1.56

to find (-Mex for column strip) by interpolation:

The ratio of column strip = 0.803

only for â&#x20AC;&#x201C;Mex

The ratio of middle strip = 0.197

-Mex (27.14) =

-M (column) = 0.803*27.14 = 21.8 kN.m/3.5m -M (middle) = 0.197*27.14 = 5.35

kN.m/3.5m

- 29 -

Ch. -2- Analysis and Design of Slab

For Internal panel: ln = 6.6 – 0.4 = 6.2 m ; l2= 7m Mo=

= 387.81 kN.m/7m

-Min= 0.65* Mo= 0.65*387.81 = 252.1 kN.m/ 7m

for internal panel

+M= 0.35* Mo= 0.35*387.81 = 135.75 kN.m/ 7m From table of ACI-Code (13.6.4.1) find the ratio of column strip by interpolation: ⁄ = ⁄

= 1.06

α1*

⁄ ≥ 1 ; 4.23*1.04 ≥ 1

The ratio of column strip = 0.732

only for –Min & +M

The ratio of middle strip = 0.268

-Min (252.1) =

+M (135.75) =

-M (column) = 0.732*252.1= 184.54

kN.m/3.5m

-M (middle) = 0.268*252.1= 67.57

kN.m/3.5m

+M (column) = 0.732*135.75 = 99.37 kN.m/3.5m +M (middle) = 0.268*135.75 = 36.40

kN.m/3.5m

Moment distribution at cantilever: Length of cantilever = 1.65 m -Mc= -Mc (

)=

-M (column) = 0.732*

= 80.43

kN.m/3.5m

-M (middle) = 0.268*

= 29.45

kN.m/3.5m

- 30 -

Ch. -2- Analysis and Design of Slab

Bending moment diagram:

If (α1* l 2/l1 > 1) the moment transfer to the beam by 85 % of column strip moment and 15% remain on column slab strip. α1* l 2/l1 > 1 4.23*0.643 > 1 ‫ ؞‬the moment on column slab strip remain by 15 % of column strip.

Figure 2-13: Bending moment diagram by Y-direction

- 31 -

Ch. -2- Analysis and Design of Slab

Find As of roof slab by Y direction: Using Ă&#x2DC; 12 mm (113 mm2) ; fcâ&#x20AC;&#x2122;=21 Mpa ; fy=400 Mpa d= 142 mm â&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Ś.. From page (26) = 0.0064 As -M= 67.57 kN.m/3.5m = 67.57 * 106 N.mm/3.5m 67.57 * 106 = 0.9 * As * 400 (142 â&#x20AC;&#x201C; 0.0032 As) Find As =?

As = 1364 mm2 As (min) = 0.002*đ?&#x2019;˝*â&#x201E;&#x201C;strip As (min) = 0.002*180*3500 As (min) =1260 mm2

Table 2-4: Moment on slab strip 7 m by Y-direction:

Middle strip

Column strip

strip

Section -ve in -ve ex-in -ve ex-ex -ve cantilever +ve ex +ve in -ve in -ve ex-in -ve ex-ex -ve cantilever +ve ex +ve in

Moment(kN.m/3.5m) 67.57 49.63 5.35 29.45 40.41 36.40 27.67 10.37 3.27 12.1 8.45 14.91

As required (mm2) 1364 1002 108 595 816 735 560 210 67 245 171 301

As design (mm2) 1364 1260 1260 1260 1260 1260 1260 1260 1260 1260 1260 1260

- 32 -

Ch. -2- Analysis and Design of Slab

Determination spacing: If As = 1364 mm2/3.5m ; Area of one bar (12mm) = 113 mm2 No.of bars/3.5m = Spacing =

≈ 13 bars ; No.of spaces = 13-1= 12 space

= 0.29m = 29 cm

Use Ø 12 mm @ 25 cm ⁄

If As = 1260 mm2/3.5m No.of bars/3.5m = Spacing =

≈ 12 bars ; No.of spaces = 12-1= 11 space

= 0.31m = 31 cm

Use Ø 12 mm @ 30 cm ⁄

- 33 -

Ch. -2- Analysis and Design of Slab

Office slab: The procedure of all slabs is same as roof slab with some changes such as: Determination of ultimate load on office slab: Slab thickness = 0.18 m

………………. from page (19)

L.L = 2.5 kN/m2 D.L (slab) =h* γ (concrete) D.L (slab) =0.18* 24.5= 4.41 kN/m2 D.L (Tile) =1.5 kN/m2 D.L (partition wall) = T.D.L (partition wall) = Volume * γ (light weight block) Volume = h * t * l l= 272 m

:

h = 3.5 m

:

t = 25 cm

Volume = 272 * 3.5 * 0.25 = 238 m3 Total weight on slab = 238.8 * 8 ≈ 1910 kN Office slab area = L * W = 35.4 * 19.55 = 692 m2 D.L (partition wall) =

kN/m2

Wu = 1.4(1.5+4.41+2.76) + 1.7(2.5) ≈ 16.4 kN/m2 Check for d considering max moment (96.08 kN.m/3.3m): 2

M= Ø ρ b d fy (1-

)

‫ ؞‬96.08 * 106 = 0.9 * 0.017 * 3300 * d2 * 400 * (1-

)

d = 77 mm < 142 mm d provide - 34 -

Ch. -2- Analysis and Design of Slab

For X-direction: Width of strip = 6.6 m Table 2-5: Moment on slab strip 6.6 m by X-direction:

strips

section Moment (kN.m) 96.08 -ve( ) -ve(in) 89.22 Middle strip 10.97 -ve( ) (3.3m) +ve(ex) 78.24 +ve(in) 48.04 47.47 ) column -ve( -ve(in) 44.08 strip (slab 12.49 -ve( ) moment) +ve(ex) 38.65 (3.3m) +ve(in) 23.74

As required (mm2) 1972 1832 226 1606 986 975 905 257 793 488

As design (mm2) 1972 1832 1188 1606 1188 1188 1188 1188 1188 1188

Determination spacing: If As = 1972 mm2/3.3m ; Area of one bar (12mm) = 113 mm2 No.of bars/3.3m = Spacing =

≈ 18 bars

= 0.19m = 19 cm

;

No.of spaces = 18-1= 17 space ‫ ؞‬Use Ø 12 mm @ 18 cm ⁄

If As = 1832 mm2/3.3 m Spacing =

= 0.20 m = 20 cm

‫ ؞‬Use Ø 12 mm @ 18 cm ⁄

If As = 1606 mm2/3.3 m Spacing =

= 0.23 m = 23 cm

‫ ؞‬Use Ø 12 mm @ 18 cm ⁄

If As = 1188 mm2/3.3 m Spacing =

= 0.33 m = 33 cm

‫ ؞‬Use Ø 12 mm @ 30 cm ⁄ - 35 -

Ch. -2- Analysis and Design of Slab

For Y-direction: Width of strip = 7.0m Table 2-6: Moment on slab strip 7 m by Y-direction:

Strips

section Moment (kN.m) 70.58 -ve( ) -ve(in) 96.09 Middle 7.6 -ve( ) strip -ve (canti.)

29.45 (3.5m) +ve(ex) 57.47 +ve(in) 51.74 14.74 -ve( ) column -ve(in) 39.37 strip 4.65 -ve( ) (slab -ve (canti.)

12.07 moment) +ve(ex) 12.01 (3.5m) +ve(in) 21.20

As required (mm2) 1427 1967 154 1586 1162 1058 298 805 94 231 243 434

As design (mm2) 1427 1967 1260 1586 1260 1260 1260 1260 1260 1260 1260 1260

Determination spacing: If As = 1967 mm2/3.5m Spacing =

= 0.2m = 20 cm

‫ ؞‬Use Ø 12 mm @ 20 cm ⁄

If As = 1586 mm2/3.5 m Spacing =

= 0.25 m = 25 cm

‫ ؞‬Use Ø 12 mm @ 20 cm ⁄

If As = 1427mm2/3.5 m Spacing =

= 0.29 m = 29 cm

‫ ؞‬Use Ø 12 mm @ 20 cm ⁄

If As = 1260mm2/3.5 m Spacing =

= 0.31 m = 31 cm

‫ ؞‬Use Ø 12 mm @ 30 cm ⁄ - 36 -

2.4 Detail of slab:

Y

Roof Slab B

X

18 cm

1.60

1.6

4.50

6.60

0.40 4.40

4.40

Section ( B-B)

3.6

3.6

6.60

3.85

3.85

1.65

A

A 4.40

2.30

4.40

4.40

4.40

2.30

B Continuous 2 bar to support the bar 2.30

both direction-cut bar 4.40

4.40

4.40

2.30 18 cm 0.40

.

7.00

7.00

7.00

Section ( A-A)

7.00

7.00

Note:

fy = 400 mpa fcu = 25 mpa

Splice minimum 20 cm

- 37 -

Y

Office Slab B

X

1.60

1.6

4.50

3.6

3.6

4.40

Section ( B-B)

6.60

4.40

6.60

A

A

3.85

3.85

1.65

5.25 4.40

2.30

4.40

4.40

4.40

2.30

B

2.30

4.40

4.40

4.40

4.40

by x direction-addtional bar at exterior panel only

2.30

.

7.00

7.00

Splice minimum 20 cm

7.00

Section ( A-A)

7.00

7.00

Note:

fy = 400 mpa fcu = 25 mpa - 38 -

Ch. -2- Analysis and Design of Slab

2.5 Conclusion: Most of the slabs were designed with the accordance to Amin because there was no heavy applied weight on the slab. Since the applied live load were equal to 2.5 kPa, and slab thickness was assumed 18 cm (with the accordance to serviceability limit of deflection was requirement), this loads can be considered as light weight.

The using of direct design method lead us to have a safe design, also this method provided us a good time to design the other part of the structure, especially in design beam we can use 85% of moment of column strip to design beam according ACI. After the slab design the second steps is to transfer slab weight to the beam then using the available method to analysis the critical shears and moment applied those beams.

- 39 -

Ch. -3- Analysis and Design of Beam

Chapter Three Analysis and Design of Beam 3.1 Introduction: Beams can be described as members that are mainly subjected to flexure and it is essential to focus on the analysis of bending moment, shear, and deflection.

When the bending moment acts on the beam, bending strain is produced. The resisting moment is developed by internal stresses. Under positive moment, compressive strains are produced in the top of beam and tensile strains in the bottom.

Concrete is a poor material for tensile strength and it is not suitable for flexure member by itself.

The tension side of the beam would fail before compression side failure when beam is subjected a bending moment without the reinforcement. For this reason, steel reinforcement is placed on the tension side. The steel reinforcement resists all tensile bending stress because tensile strength of concrete is zero when cracks develop. In the Ultimate Strength Design (USD), a rectangular stress block is assumed as shown in figure below:

Figure 3-1: Reinforced rectangular beam

- 40 -

Ch. -3- Analysis and Design of Beam

3.1.1 Types of Beam: Figure (3-2) shows the most common shapes of concrete beams: single reinforced rectangular beams, doubly reinforced rectangular beams, T-shape beams, spandrel beams, and joists.

Figure 3-2: Common shapes of concrete beam

In castâ&#x20AC;&#x201C;in-place construction, the single reinforced rectangular beam is uncommon. The T-shape and L-shape beams are typical types of beam because the beams are built monolithically with the slab. When slab and beams are poured together, the slab on the beam serves as the flange of a T-beam and the supporting beam below slab is the stem or web. For positive applied bending moment, the bottom of section produces the tension and the slab acts as compression flange. But negative bending on a rectangular beam puts the stem in compression and the flange is ineffective in tension. Joists consist of spaced ribs and a top flange.

- 41 -

Ch. -3- Analysis and Design of Beam

3.2 Methodology: 1. Assume the depth of beam using the ACI Code reference, minimum thickness unless consideration the deflection.

2. Assume beam width (ratio of width and depth is about 1:2).

3. Compute self-weight of beam and design load.

4. Compute factored load.

5. Compute design moment (Mu).

6. Compute maximum possible nominal moment for singly reinforced beam (Ď&#x2020;Mn).

7. Decide reinforcement type by Comparing the design moment (Mu) and the maximum possible moment for singly reinforced beam (Ď&#x2020;Mn). If Ď&#x2020;Mn is less than Mu, the beam is designed as a doubly reinforced beam else the beam can be designed with tension steel only.

8. Determine the moment capacity of the singly reinforced section.(concrete-steel couple) 9. Compute the required steel area for the singly reinforced section.

10. Find necessary residual moment, subtracting the total design moment and the moment capacity of singly reinforced section.

11. Compute the additional steel area from necessary residual moment.

12. Compute total tension and compressive steel area.

13. Design the reinforcement by selecting the steel.

14. Check the actual beam depth and assumed beam depth.

- 42 -

4.365kn.m

9.94kn.m

8.21kn.m

2.62kn.m

14.78 kN

1.32 kN

0.388 kN

17.292kn.m

0.55 kN

39.7kn.m

1.164 kN

32.94kn.m

10.476kn.m

29.56 kN

6.56 kN

1.94 kN

35.24kn.m

2.68 kN

80.1kn.m

67.2kn.m

5.1 kN

7.1 kN

5.82 kN 21.33kn.m

29.56 kN

17.3kN 51.249kn.m

117.68kn.m

15.3 kN

97.3kn.m

31.1kn.m

29.56 kN

32.786 kN 35.72 kN

9.7 kN

86.9kn.m

199.5kn.m

59.1 kN

59.1 kN

52.54kn.m

24.12 kN

6.60m

17.48 kN

29.1 kN

165kn.m

17.48 kN

6.60m

20.43 kN

13.386 kN

52.46 kN

4.50m

24.12 kN

52.46 kN

Figure 3-3: Analyze wind load by Y-direction - 43 -

Figure 3-4: Moment acting on the beam by (kN.m):

5.584

4.356

2.62

14.78 kN 2.62

5.584

4.356

3.40m

22.408

17.292

10.476

29.56 kN 10.476

17.292

22.408 3.40m

44.86

35.24

21.33

29.56 kN 21.33

35.24

44.86 3.40m

66.431

51.249

31.1

29.56 kN 31.1

51.249

66.431

3.40m

112.6

86.9

52.54

35.72 kN 52.54

86.9

112.6 4.70m

20.43 kN 6.60m

6.60m

4.50m

Figure 3-5: Moment acting on Foundation:

57.96 kN.m 59.1 kN

132.2 kN.m 17.5 kN

109.1 kN.m 24.12 kN

34.9 kN.m 52.46 kN

- 44 -

Ch. -3- Analysis and Design of Beam

3.3.2 Analyze and design of beam at roof slab: Design one interior span of beams by X-direction: Moment: Find max moment for positive and negative due to 85% of column strip moment plus self-weight beam moment and plus moment due to portion wall on the beam if available and neglect effect of wind load by this direction.

1- 85% of column strip from due to DL and LL on slab and obtain in (page 25):

2- Moment due to weight of beam: Minimum h (beam) =

to control deflection according to ACI-Code table (9-5) chapter9 ‫ ؞‬Assume beam dim (40X60) cm

Minimum h (beam) = Wu = 1.4(0.4 * 0.6 * 24.5) = 8.232 kN/m Mo =

=

;

ln = 6.6 m

= 44.82 kN

3- Moment due to portion wall load: the moment is equal to zero because no wall available in roof slab.

Total moment on this beam:

Max negative moment (-M) = 220.5 kN.m

;

Max positive moment (+M) = 179.56 kN.m

- 45 -

Ch. -3- Analysis and Design of Beam

Shear: Find shear due to Wu (slab) plus weight of beam per meter: Wu (slab) = 11.53 kN/m A slab on beam = (3.32 * 2) + (0.4*6.6) = 24.42 m2 Wu = 11.53*

+ (1.4*0.4*0.6*24.5) = 48.45 kN/m

∑MB = (48.45 * 7 * 7/2) + 57 – 220.5 – RA * 7 RA = 146.2 kN RB = 192.93 kN Maximum shear occur at distance d from face of column according to ACI-Code (11.1.3.1) d = h - concrete cover – Ø stirrup d = 0.6 – 0.04 – 0.01 -0.02/2 = 0.54 m Vd (A) = Vd (B) left =

= 119.88 kN = 166.88 kN

Vd (B) Right =

= 143.4 kN

Figure 3-6: Shear Diagram in roof slab by X-direction.

- 46 -

Ch. -3- Analysis and Design of Beam

Design of beam at roof by X-direction: Max. Moment (-M) = 220.5 kN.m Assume beam dimension (40X60) cm Use Ø 20 mm for main bar and Ø 10 mm for stirrup And fc’=21 Mpa & fy = 400 Mpa ⁄ = 540 mm

d = 600 - 40 - 10 a=

=

= 0.056 As

Find As: MD = Ø As fy (d -

)

220.5*106 = 0.9*400*As (540 – 0.028As) As = 1210 mm2

Solve for As =?

No. of bars =

= 3.85 ≈ 4 bar

:

As provide = 4*314 = 1256 mm2

Check for ρ: ρ min = ρ act =

= =

= 0.0035 = 0.00581

ρ max = 0.75 * ρb =0.75 [0.852 *

‫ ؞‬ρ min ˂ ρ act ˂ ρ max

*

] = 0.017

O.K

Max. Moment (+M) = 179.56 kN.m As = 986 mm2 No. of bars =

= 3.14 ≈ 4 bar

:

As provide = 4*314 = 1256 mm2

ρ act = 0.00581

‫ ؞‬ρ min ˂ ρ act ˂ ρ max

O.K - 47 -

Ch. -3- Analysis and Design of Beam

Note: For negative moment (205 kN.m) between 220.5 and 179 because this use 4 bar Ø 20 mm.

For +M = 110.7 kN.m As = 608 mm2 = 1.93 ≈ 2 bar

No. of bars = ρ act =

:

As provide = 2*314 = 628 mm2

‫ ؞‬use ρ min

0.00029 < ρ min

As = 0.0035 *540 *400 = 756 mm2 = 2.4 ≈ 3 bar

No. of bars =

:

As provide = 3*314 = 942 mm2

Note: Use 3 bars for moment 57 kN.m because less than 110.7 kN.m Design for shear: Design the beam according to ACI-Code (11.4.5 & 11.4.6 & 11.4.7): ØVc = Ø 0.17 * √fc’ * b*d ØVc = 0.85*0.17* √21 * 400 * 540/1000 = 143 kN ØVc < Vd(B) Left (166.88 kN) Vs = Vd - ØVc = 166.88 – 143 = 23.88 kN But Vs < ØVc

because this (Spacing * 0.5) not need

Use Ø 10 mm for stirrup (A= 78.5 mm2) Spacing(S) =

=

= 1420 mm =1.42 m

Use minimum value of them according to ACI-Code 2011 (11-4-5 & 11-4-6): Smax = 61 cm Smax = ⁄ = Smax=

⁄ =

‫ ؞‬Use Ø 10 mm @ 20 cm

⁄

for all length

- 48 -

A

B

A

A

C

A

2.15 m 2.35 m

2.35 m

2.35

0.85 m

150 mm

5.25 m

A

A

B

A

7.00 m

18 cm

60 cm

40 cm

Sec (A-A)

A

C

7.00 m

18 cm

60 cm

40 cm

Sec (B-B)

18 cm

60 cm

40 cm

Sec (C-C)

Figure 3-7: Detail typical beam by X-direction

- 49 -

Ch. -3- Analysis and Design of Beam

3.3.3 Analysis and design of beams at roof slab by Y-direction: Moment: Find max moment for positive and negative due to 85% of column strip moment plus self-weight beam moment and plus moment due to portion wall on the beam if available and moment due to win load. 

85% of column strip from due to DL and LL on slab and obtain in (page 31):



Moment due to weight of beam: Assume beam dim (40X60) cm Wu = 1.4(0.4 * 0.6 * 24.5) = 8.232 kN/m ;



Mo (ln = 6.6m) =

=

= 39.56 kN.m

Mo (ln = 4.1m) =

=

= 17.3 kN.m

Mo (Cantilever) =

=

ln = 6.2m ; ln = 4.1 ; l cantilever = 1.65 m

= 11.2 kN.m

Moment due to portion wall load: the moment is equal to zero because no wall available in roof slab.

A. Total moment due to Wu without wind load:

B. Moment due to wind load from (page 44):

- 50 -

Ch. -3- Analysis and Design of Beam

Total moment on this beam: to find total moment in the beam we can load combination with wind load according to ACI- Code 318M-1995 (9.2.4): 1. U= 1.4DL + 1.7LL

(due to Wu)

2. U= 0.9DL + 1.3WL 3. U= 0.75 * (1.4DL + 1.7WL +1.7LL) 1.4 DL = 1.4 * 6.41 + 1.4 (13.6 * 0.2 * 0.6 * 24.5) / (7*6.6) =10.185 kN/m 1.7 LL = 2.55 kN/m Wu = 12.735 kN/m Moment due to DL= (

u)

M due to DL for (–Mex= 70.81 kN.m) = ( Moment due to LL= ( M due to LL for (–Mex= 70.81 kN.m) = (

/ 1.4 = kN.m ) / 1.4 = 40.45 kN.m

u)

/ 1.7 = kN.m ) / 1.7 = 8.34 kN.m

For –M at external span: U = 1.4DL + 1.7LL = 1.4(40.45) + 1.7(8.34) = 70.81 KN.m

it is critical

U = 0.9DL + 1.3WL = 0.9(40.45) + 1.3(2.62) = 39.81 KN.m U = 0.75(1.4DL + 1.7WL +1.7LL) = 0.75[1.4(40.45) + 1.7(2.62) + 1.7(8.34)] = 56.45 KN.m Note: For all moment as the same procedure.

‫ ؞‬Factor U= 1.4DL + 1.7LL is critical for beam at roof slab.

Diagram for moment due to U= 1.4DL + 1.7LL

Max negative moment (-M) = 182.5 kN.m and Max positive moment (+M) = 98.3 kN.m - 51 -

Ch. -3- Analysis and Design of Beam

Shear: Find shear due to Wu (slab) plus weight of beam per meter: Wu (slab) = 11.53 kN/m A slab on beam for long span = (3.32 * 2) = 21.78 m2 Wu = 11.53*

+ (1.4*0.4*0.6*24.5) = 46.28 kN/m

∑MB = (46.28 * 6.6 * 6.6/2) + 182.56 – 182.56 – RA * 6.6 RA = 152.73 kN = RB Maximum shear occur at distance (d) from face of column according to ACI-Code (11.1.3.1) d = 0.54 m Vd (A & B) =

= 129.2 kN

Figure 3-8: Shear Diagram in roof slab by Y-direction.

- 52 -

Ch. -3- Analysis and Design of Beam

Design of beam at roof by Y-direction: Max. Moment (-M) = 182.5 kN.m Assume beam dimension (40X60) cm Use Ø 20 mm for main bar and Ø 10 mm for stirrup And fc’=21 Mpa & fy = 400 Mpa d = 540 mm

:

a = 0.056 As

Find As: MD = Ø As fy (d -

)

182.5 *106 = 0.9*400*As (540 – 0.028As) As = 990 mm2

Solve for As =?

No. of bars =

= 3.15 ≈ 4 bar

:

As provide = 4*314 = 1256 mm2

Check for ρ: ρ min = ρ act =

= =

= 0.0035 = 0.00581

ρ max = 0.75 * ρb =0.75 [0.852 *

‫ ؞‬ρ min ˂ ρ act ˂ ρ max

*

] = 0.017

O.K

Max. Moment (+M) = 98.3 kN.m Use ρ min because this moment less than 110.7 kN.m ρ min = 0.0035 As = 0.0035 *540 *400 = 756 mm2 No. of bars =

= 2.4 ≈ 3 bar

:

As provide = 3*314 = 942 mm2

Note: Use 3 bars for all other moment because less than 98.3 kN.m

- 53 -

Ch. -3- Analysis and Design of Beam

Design for shear: ØVc = Ø 0.17 * √fc’ * b*d ØVc = 0.85*0.17* √21 * 400 * 540/1000 = 143 kN ØVc > Vd (A & B) (129.2 kN) But ØVc < 4 * Vd (A & B) (129.2 kN) because this minimum reinforcement must be provide.

Provide minimum reinforcement, use max value of them: Smax = 61 cm Smax = ⁄ = Smax=

⁄ =

‫ ؞‬Use Ø 10 mm @ 20 cm

⁄

for all length

- 54 -

B

C

A

A

C

A

1.55 m 1.5 m

2.3 m

2.3 m

0.5 m

150 mm

5.00 m

B

A

C

A

4.5 m

18 cm

60 cm

40 cm

Sec (A-A)

A

C

6.6 m

18 cm

60 cm

40 cm

Sec (B-B)

18 cm

60 cm

40 cm

Sec (C-C)

Figure 3-9: Detail typical beam by Y-direction

- 55 -

Ch. -3- Analysis and Design of Beam

3.3.4 Check for stirrup spacing at critical beam on the frame: Shear: Find shear due to Wu (slab) plus weight of beam per meter plus wall if available: Wu (slab) = 16.4 kN/m A slab on beam = (3.32 * 2) + (0.4*6.6) = 24.42 m2 Wu = 16.4*

+ (1.4*0.4*0.6*24.5) = 65.45 kN/m

∑MB = (65.45 * 7 * 7/2) + 57 – 220.5 – RA * 7 RA = 197.3 kN RB = 260.9 kN Maximum shear occur at distance d from face of column according to ACI-Code (11.1.3.1) d = h - concrete cover – Ø stirrup d = 0.6 – 0.04 – 0.01 -0.02/2 = 0.54 m Vd (A) = Vd (B) left = Vd (B) Right =

= 161.8 kN = 225.7 kN = 193.8 kN

Figure 3-10: Shear Diagram in office slab by X-direction.

- 56 -

Ch. -3- Analysis and Design of Beam

Design for maximum shear in the beam ØVc = Ø 0.17 * √fc’ * b*d ØVc = 0.85*0.17* √21 * 400 * 540/1000 = 143 kN ØVc < Vd (B left) (255.7 kN) Vs = Vd - ØVc = 255.7 – 143 = 112.7 kN But Vs < ØVc use Ø 10 mm for stirrup (A= 78.5 mm2) Spacing(S) =

=

= 300 mm = 30 cm

Use minimum value of them according to ACI-Code 2011 (11-4-5 & 11-4-6): Smax = 61 cm Smax = ⁄ = Smax=

⁄ =

‫ ؞‬Use Ø 10 mm @ 20 cm

⁄

for all length and for any beam in this building

- 57 -

3.4 Detail of beam:

71

71

71

42

71

33 104 58

98 183

221

57 180

58

98 45 142

45

80

80

88

175

104

183 162

104

183

162 88

49 155

183

183

162 80

80

88

98

98

162

111

183

111

205

183

183

45

142

205

183

98

58 104

175

45

98

205 111

183

180

95

183

183

183 205

78

177 191

183

98

98

58 104

221

57

95

25

58

177

183

95

183

104

155

177

96

22

191 177

48 58

58

58

33 42

49

48

89

22

48

89

12

89 22

78

89

22

12

96

25

Figure 3.11: Calculated moment at roof slab for beam at critical (1.4 DL + 1.7 LL) OR 0.75(1.4DL + 1.7LL +1.7 WL) - 58 -

Beam Detail in Roof Slab

Beam Detail

Longitudinal Reinforcement

Beam Number

Size (bXh) by cm

Ty p Sp e of an

Ty Se pe o cti f on

B1

40 x 60

1,2

1,2

B2

40 x 60

1

2

191

155

B3

40 x 60

2

2

205

B4

40 x 60

1

2

221

Max. Moments(kN.m)

Bottom Reinforcement

-M

-M +M interior exterior 96 78 25 88 162 162

a

20

20

20

49

20

20

20

111

205

20

20

20

180

57

20

20

20

BEAM PLAN AT ROOF SLAB

B1 B1

B1 B1

B1 B1

Top Reinforcement Spacing (cm) d Left hand Right hand zone -1-zone -2-zone -3-

c

b

B3

B1

B1

B1 B3

B1

B1

GREATER OF

GREATER OF

(L1/4 or L2/4)

(L1/3 or L2/3)

c

d

d

B3

B3

B3

B3

B2

b

a

L1/8

ZONE-2-

ZONE-3-

L1

B1

B3

h

150 mm

ZONE-1-

B1

TYPE OF BEAM SECTION

TYPE OF BEAM SPANS

150 mm

B2

Stirrup

b

L2

TYPE -1-

TYPE -1-

B4

B3

B3

B3

GREATER OF

GREATER OF

(L1/3 or L2/3)

(L1/3 or L3/3)

B4

c

d

d

h

150 mm

B1

B3

B3

B3

B1

B3

L1/8

ZONE-1-

L2

B2

B1

B1

B1

B2

b

a

ZONE-2-

L1

L1/8

ZONE-3-

L3

b TYPE -2-

TYPE -2-

Figure 5.20 at Design Concrete Structures Fourteenth Edition by Arthur H. Nilson at page 197

- 59 -

41.2

38.63 126.23 53.7

92

76.7

113 74.1

289.6

179.4

235.8

285.6

88 78

167.2

244.6

285.6

88

154 61 191.5

71.2

120

120

117.6

235.2

167.2

285.6 218.4

179.4

154 300.4

285.6

154 218.4 117.6

155.1

52

154 150.2

285.6

218.4 120

120

117.6

279

285.6

150.2

154 285.6

285.6

71.2

191.5

279

285.6

154

88 167.2

235.2 218.4

61

144.8

285.6

279 150.2

268.9

285.6

154

279

285.6

167.2

244.6

144.8

285.6

285.6 300.4

268.9

285.6

154

88 179.4

78

77.6 92

268.9 144.8

144.1

113

268.9

285.6

179.4

235.8

113

113 289.6

77.6 92

92

53.7 76.7

74.1

144.1 52

77.6

52

126.23

144.1

144.1

155.1 52

41.2

38.63

Figure 3.12: Calculated moment at 3- 4 floor slab for beam at critical (1.4 DL + 1.7 LL) OR 0.75(1.4DL + 1.7LL +1.7 WL) - 60 -

Beam Detail in Floor 3 and 4

Beam Detail

Longitudinal Reinforcement

Beam Number

Size (bXh) by cm

Ty p Sp e of an

Ty Se pe o cti f on

B1

40 x 60

1,2

1,2

B2

40 x 60

1

2

B3

40 x 60

2

2

B4

40 x 60

1

2

B5

40 x 60

2

1,2

B6

40 x 60

2

2

Max. Moments(kN.m)

Bottom Reinforcement

-M

-M +M interior exterior 155.1 126.3 38.63 167.2 87.8 167.2

a

20

20

20

78

20

20

20

279 150.2 279

20

20

20

20

20

20

218.4 117.6 218.4

20

20

20

285.6 154 285.6

20

20

20

300.4 244.6

235.2 191.5

61

BEAM PLAN AT ROOF SLAB

B1 B1

B1 B1

B1 B1

Top Reinforcement Spacing (cm) d Left hand Right hand zone -1-zone -2-zone -3-

c

b

B3

B1

B1

B1 B3

B1

B1

GREATER OF

GREATER OF

(L1/4 or L2/4)

(L1/3 or L2/3)

c

d

d

B3

B6

B6

B6

B2

b

a

L1/8

ZONE-2-

ZONE-3-

L1

B5

B6

h

150 mm

ZONE-1-

B5

TYPE OF BEAM SECTION

TYPE OF BEAM SPANS

150 mm

B2

Stirrup

b

L2

TYPE -1-

TYPE -1-

B2

B3

B3

B3

GREATER OF

GREATER OF

(L1/3 or L2/3)

(L1/3 or L3/3)

B2

c

d

d

h

150 mm

B1

B6

B6

B6

B1

B6

L1/8

ZONE-1-

L2

B4

B5

B5

B5

B4

b

a

ZONE-2-

L1

L1/8

ZONE-3-

L3

b TYPE -2-

TYPE -2-

Figure 5.20 at Design Concrete Structures Fourteenth Edition by Arthur H. Nilson at page 197

- 61 -

38.63

81.1

126.23 53.7

92

265.8

235.8

88

300.4

78

233.1

325

244.6

88 61 71.2

120

191.5

233.1

154 235.2

325

120

117.6

265.8

154 357.7

154 325

218.4

116.4

74.1

289.6

357.7

150.2

218.4 117.6

38.63

152 279

325

120

120

117.6

155.1

91.8

154 357.7

218.4

218.4

144.8

279

154 325

325

71.2

191.5

150.2

325

150.2

268.9

357.7

154 279

154

88 233.1

235.2

61

144.8

357.7

279

325

233.1

244.6

268.9

357.7

144.8

357.7 300.4

92

268.9

154

88 265.8

78

152

268.9

357.7

265.8

235.8

152

152 289.6

144.1 77.6

91.8

92

92

53.7 116.4

74.1

144.1 77.6

91.8

77.6

91.8

81.1

126.23

144.1

144.1

155.1

Figure 3.13: Calculated moment at 1-2 floor slab for beam at critical (1.4 DL + 1.7 LL) OR 0.75(1.4DL + 1.7LL +1.7 WL)

- 62 -

Beam Detail in Floor 1 and 2

Beam Detail

Longitudinal Reinforcement

Beam Number

Size (bXh) by cm

Ty p Sp e of an

Ty Se pe o cti f on

B1

40 x 60

1,2

1,2

B2

40 x 60

1

2

B3

40 x 60

2

2

B4

40 x 60

1

2

B5

40 x 60

2

1,2

B6

40 x 60

2

2

Max. Moments(kN.m)

Bottom Reinforcement

-M

-M +M interior exterior 155.1 126.3 38.63 167.2 87.8 167.2

a

20

20

20

78

20

20

20

279 150.2 279

20

20

20

20

20

20

218.4 117.6 218.4

20

20

20

285.6 154 285.6

20

20

20

300.4 244.6

235.2 191.5

61

BEAM PLAN AT ROOF SLAB

B1 B1

B1 B1

B1 B1

Top Reinforcement Spacing (cm) d Left hand Right hand zone -1-zone -2-zone -3-

c

b

B3

B1

B1

B1 B3

B1

B1

GREATER OF

GREATER OF

(L1/4 or L2/4)

(L1/3 or L2/3)

c

d

d

B3

B6

B6

B6

B2

b

a

L1/8

ZONE-2-

ZONE-3-

L1

B3

B6

h

150 mm

ZONE-1-

B3

TYPE OF BEAM SECTION

TYPE OF BEAM SPANS

150 mm

B2

Stirrup

b

L2

TYPE -1-

TYPE -1-

B2

B3

B3

B3

GREATER OF

GREATER OF

(L1/3 or L2/3)

(L1/3 or L3/3)

B2

c

d

d

h

150 mm

B6

B3

B6

B6

B6

B3

L1/8

ZONE-1-

L2

B4

B5

B5

B5

B4

b

a

ZONE-2-

L1

L1/8

ZONE-3-

L3

b TYPE -2-

TYPE -2-

Figure 5.20 at Design Concrete Structures Fourteenth Edition by Arthur H. Nilson at page 197

- 63 -

Ch. -3- Analysis and Design of Beam

3.5 Conclusion: In previous chapter we try to design slab by DDM, and in this chapter we try to analysis and design the beam. The beams have a good reinforcement because we chose a suitable section according to control deflection, Beams on the same line of action is designed is each of its maximum negative and positive moments achieved from different loading condition. The shear designed is similar for all the beams. For the maximum value of shear minimum spacing is required.

Wind load controls the combination only for the beams from ground to third floor and for other floor combination of dead load and live load is the critical case.

In this next chapter, columns are designed. Columns are short columns; point load is achieved from rectangular area on the column and moments from strip.

- 64 -

Ch. -4- Analysis and Design of Column

Chapter Four Analysis and Design of Column 4.1 Introduction: Columns support primarily axial load but usually also some bending moments. The combination of axial load and bending moment defines the characteristic of column and calculation method.

A column subjected to large axial force and minor moment is design mainly for axial load and the moment has little effect. A column subjected to significant bending moment is designed for the combined effect. The ACI Code assumes a minimal bending moment in its design procedure, although the column is subjected to compression force only. Compression force may cause lateral bursting because of the low-tension stress resistance. To resist shear, ties or spirals are used as column reinforcement to confine vertical bars.

The complexity and many variables make hand calculations tedious which makes the computeraided design very useful.

Figure 4.1: Column shear failure at Tohoku â&#x20AC;&#x201C; Japan

Figure 4.2: Use column to Support slab and appearance - 65 -

Ch. -4- Analysis and Design of Column

4.2 Types of Columns Reinforced concrete columns are categorized into two main types; rectangular tied column, round tied column or round spiral column, and columns of other geometry (Hexagonal, Lshaped, T-Shaped, etc).

Figure 4-3: Column types

Fig. 4-3 shows the rectangular tied and round spiral concrete column. Tied columns have horizontal ties to enclose and hold in place longitudinal bars. Ties are commonly 8mm or 10mm steel bars. Tie spacing should be calculated with ACI Code.

Spiral columns have reinforced longitudinal bars that are enclosed by continuous steel spiral.

The spiral is made up of either large diameter steel wire or steel rod and formed in the shape of helix. The spiral columns are slightly stronger than tied columns.

- 66 -

Ch. -4- Analysis and Design of Column

The columns are also categorized into three types by the applied load types; the column with small eccentricity, the column with large eccentricity (also called eccentric column) and biaxial bending column. Fig 4-4 shows the different column types depending on applied load.

Figure 4-4: The column types depending on applied load.

Eccentricity is usually defined by location: • Interior columns usually have small eccentricity • Exterior columns usually have large eccentricity • Corner column usually has biaxial eccentricity.

But eccentricity is not always decided by location of columns. Even interior columns can be subjected by biaxial bending moment under some load conditions Fig. 4-5 shows some examples of eccentric load conditions.

Figure 4-5: Eccentric loaded conditions (Spiegel, 1998) - 67 -

Ch. -4- Analysis and Design of Column

4.3 Methodology: 4.3.1 Short Columns with small eccentricities 1. Establish the material strength and steel area.

2. Compute the factored axial load.

3. Compute the required gross column area.

4. Establish the column dimensions.

5. Compute the load on the concrete area.

6. Compute the load to be carried by the steel.

7. Compute the required steel area.

8. Design the lateral reinforcing (ties or spiral).

9. Sketch the design.

4.3.2 Short Columns with large eccentricities 1. Establish the material strength and steel area.

2. Compute the factored axial load (Pu) and moment (Mu).

3. Determine the eccentricity (e).

4. Estimate the required column size based on the axial load and 10% eccentricity.

5. Compute the required gross column area.

6. Establish the column dimensions.

7. Compute the ratio of eccentricity to column dim. perpendicular to the bending axis.

8. Compute the ratio of a factored axial load to gross column area.

9. Compute the ratio of distance between centroid of outer rows of bars to thickness of the cross section, in the direction of bending.

10. Find the required steel area using the ACI chart.

11. Design the lateral reinforcing (ties or spiral).

12. Sketch the design.

- 68 -

Ch. -4- Analysis and Design of Column

4.4 Calculation: 4.4.1 Analyze of column 4.4.1.1 Calculate Axial Load on Column: Total weight on the column = weight slab + self weight beam + column weight if available Calculate colomn load at roof slab: For corner column find area = 3.7*2.45 = 9.065 m2 Wu = 11.53 kN/m2 Slab load on this column = Wu * Area = 11.53 * 9.065 = 104.52 kN Beam weight = (1.4 * 0.4 * 0.6 * 24.5) * (3.7+2.45 â&#x20AC;&#x201C; 0.4)= 47.34 kN Column weight equal to zero because at roof slab not column Total weight on this column = 104.52 + 47.34 + 0 = 151.86 kN = C1 Note: Calculate load for all other column remain by same procedure by excel program and for all other roof : C1= 151.86 kN : C2= 309.7 kN : C5= 272.3 kN

:

C6= 548 kN

C3= 363.1 kN : C4= 279.1 kN

: C7= 641.4 kN

:

C8= 494.6 kN

Figure 4-6: Column Load Area (Top View) - 69 -

Ch. -4- Analysis and Design of Column

4.4.1.2 Calculate Moment due to Wu: Find moment at column three (C3) at roof slab: About Y-axis: Moment on this column equal to moment due to slab plus beam and wall if available: Moment due to slab = 66.3 kN.m

…… total moment of this strip from page 25

Moment due to beam weight = 7.2 kN.m

from page 45

Moment due to wall weight equal to zero because not wall available ‫ ؞‬Total moment about Y-axis = 66.3 + 7.2 = 73.5 kN.m About X-axis: Moment due to slab = 252.1 – 252.1= 0.00 kN

…… total moment of this strip from page31

Moment due to beam weight = 25.7 – 25.7 = 0.00 kN.m

from page 50

Moment due to wall weight equal to zero ‫ ؞‬Total moment about X-axis = 0.00 kN.m Note: For column (C7) by (X & Y) direction, moment equal to zero because it is load at the center of the column, but we must find moment due to alternative live load by eq(13.7) ACI-Code 2011 chapter 13 article 13.6.9.

Mu = 0.07[(qDu + 0.5*qLu) l2 * ln2 - qDu’ l2’ * l’n2]

eq (13.7) From ACI-Code

Mu = 0.07[(1.4(6.41) + 0.5*(1.7*1.5)) 7 * 6.62 – 1.4* 6.41* 6.62* 7] Mu = 27.2 kN.m Note: we can neglect this value because is very small value.

- 70 -

Ch. -4- Analysis and Design of Column

Find moment at column six (C6) at roof slab: Note: moment about Y-axis equal to zero because balanced by slab weight.

Moment about X-axis due to slab = (-Mo in (6.6 m span) ) - (-M

in (4.5 m span) )

= 252.1 - 118.75 = 133.35 kN.m Moment about X-axis due to beam = 25.7 – 12.11 = 13.6 kN.m Total moment = 133.4 + 13.6 = 147 kN.m about X-axis Find moment at column two (C2) at roof slab: About Y-axis: Moment due to slab = 55.75 kN.m Moment due to beam weight = 7.2 kN.m

from excel program from page 45

Moment due to wall weight equal to zero ‫ ؞‬Total moment About Y-axis = 55.75 + 7.2 = 63 kN.m About X-axis: Moment due to slab = 126 – 59.35= 66.74 kN.m Moment due to beam weight = 25.7 – 12.11 = 13.6 kN.m

from excel program from page 50

Moment due to wall weight equal to zero ‫ ؞‬Total moment about X-axis =66.74 + 13.6 = 80.3 kN.m Note: for all other column as the same procedure and we can use excel program, then analyze column moment by wind load at X-direction, finally find ultimate strength by load combination according to ACI-Code-318M-1995 at article (9.2.1 & 9.2.2): 1. U= 1.4DL + 1.7 LL 2. U= 0.75 (1.4 DL +1.7LL + 1.7 WL) 3. U= 0.9 DL + 1.3 WL Note: Neglect equation (3) because when it is occurring the building is not complete and the frame is open for wind load.

- 71 -

Figure 4-7: Ultimate axial load and moment due to (1.4 DL + 1.7LL) for strip 3.5 m width : Zero

85.61 kN.m

279 kN

80.3 kN.m

363 kN

310 kN

16.4 kN.m

152 kN

Y Zero

63.22 kN.m

706 kN

57.12 kN.m

917 kN Zero

63.22 kN.m

1132 kN

783 kN 57.12 kN.m

1471 kN Zero

63.22 kN.m

1558 kN

1255 kN 57.12 kN.m

2025 kN Zero

63.22 kN.m

1985 kN

1728 kN 57.12 kN.m

2579 kN

2200 kN

6.6 m

6.6 m

31.61 kN.m

1985 kN

2579 kN

413 kN 11.63 kN.m

674 kN 11.63 kN.m

935 kN 11.63 kN.m

1195 kN

4.5 m

28.56 kN.m Zero

11.63 kN.m

2200 kN

5.82 kN.m

1195 kN

Figure 4-8: Ultimate axial load and moment due to (1.4 DL + 1.7LL) for strip 7 m width : Zero

143.12 kN.m

494.6 kN Y

641.4 kN

1213 kN

548 kN Zero

113.8 kN.m

1580 kN

1931 kN

2518 kN

2649 kN

3456 kN

3367 kN

4395 kN

104.6 kN.m

3971 kN

56.9 kN.m

3367 kN

104.6 kN.m

3115 kN Zero

113.8 kN.m

104.6 kN.m

2260 kN Zero

113.8 kN.m

104.6 kN.m

1404 kN Zero

113.8 kN.m

147 kN.m

52.3 kN.m

4395 kN

Zero

3971 kN

29.9 kN.m

273 kN 21.3 kN.m

721 kN 21.3 kN.m

1169 kN 21.3 kN.m

1617 kN 21.3 kN.m

2066 kN

10.65 kN.m

2066 kN

- 72 -

Figure 4-9: Calculated moment due to (1.4 DL+1.7LL) about Y-axis:

29.8 63 k

kN.m

N.m

19.8 73.5

Y

57.4

kN.m

kN.m

43.3

kN.m 19.8

50.8

kN.m

43.3

39.4

39.4

39.4

19.8 kN.m

50.8

43.3

kN.m

50.8

50.8

kN.m

43.3

kN.m

kN.m 19.8

kN.m

kN.m

kN.m

kN.m 19.8

kN.m

43.3

50.8

39.4

kN.m

kN.m

kN.m

X 39.4

kN.m

kN.m

kN.m

kN.m

kN.m

4.4.1.3 Calculate moment due wind load: Figure 4-10: Calculated Axial load and moment due to effect of wind load parallel to Y-direction: 4.37 kN.m Y

9.94 kN.m

8.2 kN.m

1.32 kN

0.34 kN

0.55 kN

12.93 kN.m

29.76 kN.m

24.7 kN.m

6.56 kN

1.94 kN

22.31 kN.m

50.34 kN.m

17.3 kN 28.94 kN.m 32.79 kN

5.1 kN 67.34 kN.m 9.7 kN

2.68 kN

2.62 kN.m 1.164 kN 7.86 kN.m 5.82 kN

41.5 kN.m

13.5 kN.m

7.1 kN

15.3 kN

56.1 kN.m

17.63 kN.m

13.39 kN

29.1 kN

109.1 kN.m

34.9 kN.m

57.96 kN.m

132.2 kN.m

59.1 kN

17.5 kN

24.12 kN

52.46 kN

57.96 kN.m

132.2 kN.m

109.1 kN.m

34.9 kN.m

59.1 kN

17.5 kN

24.12 kN

52.46 kN

- 73 -

4.4.1.4 Load combination According to ACI-Code 1995: Figure 4-11: Load combination according to ACI-Code 1995 [0.75(1.4 DL +1.7 LL+ 1.7 WL)] for strip 3.5 m: 12.68 kN.m 85.61 kN.m 279 kN

80.3 kN.m 363 kN

63.9 kN.m Y

16.4 kN.m

310 kN

152 kN

74.34 kN.m

18.75 kN.m

783 kN

413 kN

95.75 kN.m

25.94 kN.m

37.95 kN.m 706 kN

917 kN 75.86 kN.m 64.2 kN.m

1132 kN

1471 kN

1255 kN

84.33 kN.m

114.37 kN.m

674 kN 31.2 kN.m

85.86 kN.m 1558 kN

2025 kN 121.32 kN.m

1728 kN

935 kN

181.95 kN.m

53.23 kN.m

168.6 kN.m 1985 kN

2579 kN

168.6 kN.m

97.6 kN.m 1985 kN

2200 kN

160.54 kN.m 2200 kN

2579 kN

1195 kN

48.88 kN.m 1195 kN

Figure 4-12: Load combination according to ACI-Code 1995 [0.75(1.4 DL +1.7 LL+ 1.7 WL)] for strip 7 m: 12.68 kN.m 143.12 kN.m 494.6 kN

147 kN.m 641.4 kN

548 kN

29.9 kN.m 273 kN

Y 113.8 kN.m

109.95kN.m

26 kN.m

37.95 kN.m 1213 kN

1580 kN

1404 kN

113.8 kN.m

131.4 kN.m

721 kN 33.2 kN.m

64.2 kN.m 1931 kN

2518 kN

2260 kN

122.3 kN.m

150 kN.m

1169 kN 38.45 kN.m

85.86 kN.m 2649 kN

3456 kN

3115 kN

159.3 kN.m

217.5 kN.m

1617 kN 60.5 kN.m

168.6 kN.m 3367 kN

4395 kN

168.6 kN.m 178.33 kN.m

116.6 kN.m 3367 kN

3971 kN

4395 kN

3971 kN

2066 kN

52.5 kN.m 2066 kN

- 74 -

Ch. -4- Analysis and Design of Column

4.4.2 Design of Column 4.4.2.1 Check for Slender Column: ≤ 34 – 12

ACI-Code-article (10-10) eq (10-7)

K = 0.5 because end point in any column is fix.

L= 3.4 m with section (40X40) cm : L (at Ground) = 4.7 m with section (50X50) cm (r) Radius of gyration= √ r for (40X40)cm = 0.1155 and M1 = 113.8 kN.m and M2 = 122.3 kN.m from Page…. r for (50X50)cm = 0.1444 and M1 = 52.5 kN.m and M2 = 60.5 kN.m from Page….

But critical for

=1

For column (40X40) cm and h max = 3.4 m:

14.72 ≤ 22 ‫ ؞‬this column is short column For column (50X50) cm h max = 4.7 m:

16.3 ≤ 22 Therefor this column is short column and all columns can be design as short column.

- 75 -

Ch. -4- Analysis and Design of Column

4.4.2.2 Uniaxial Column Design: In this type of column axial load and one moment available, and it can be design by using interaction chart and find minimum steel require (ρg) of the column by knowing ⁄ and Kn. Where: e = eccentricity =

≥ 0.1 h

if not e = 0.1 h

Kn = M = Maximum moment on the column.

Pu = Maximum factor load on the column. h = Column dimension parallel to moment direction. Ø = 0.65 for tie stirrup, 0.75 for spiral. fc’ = Ultimate strength of concrete.

Ag = Gross area of cross-section of column. ρg = Minimum steel required to resist axial and moment, calculate from interaction chart.

(a) Figure 4.13: (a) Equivalent eccentricity of column load

(b)

(b) Stress and force at nominal strength

- 76 -

Ch. -4- Analysis and Design of Column

Note: C6 and C7 at floor 5 can be design as uniaxial column because on the column (C6 & C7) only about X-axis moment is available.

Design (C6): Pu = 548 kN

from page 69

M max = 147 kN.m fc’ = 26 Mpa :

fy = 414 Mpa

Assume column dim = (40X40) cm

γ=

=

Find ⁄ and Kn: e=

= 0.268 m > (0.1 * 0.4)

⁄ =

⁄

Kn =

= 0.203

By using ( ⁄ , Kn & γ) in interaction chart we can find minimum steel required ( ρg ): ρg = 0.012 > 0.01(ρg) As = ρg * b * h As = 0.012 * 400 * 400 = 1920 mm2 ; Using Ø 25 mm (490 mm2) No. of bars =

≈ 3.9 = 4 bars

‫ ؞‬Use 4 Ø25 mm Design for tie according to ACI-Code (7.10.5): Use Ø10 mm

Figure 4.14: Detail of column 6

Minimum spacing equal to shall not exceed of the following:  16 Ø longitudinal bar or 48 Ø tie bar or least dimension of column  Spacing for tie = 16 * 2.5 = 40 cm or 48 * 1.0 = 48 cm or 40 cm

‫ ؞‬Use Ø10 mm @ 25 cm ⁄ - 77 -

Ch. -4- Analysis and Design of Column

4.4.2.3 Biaxial Column Design: In this type of column axial load and two moments available, and it can be design by using interaction chart and find ultimate capacity of the column by knowing ( ⁄ x and ( ⁄ y and assume (ρg).

Where: ex =

≥ 0.1 hx

if not e = 0.1 hx

ey =

≥ 0.1 hy

if not e = 0.1 hy

Rn = Mx = Maximum moment on the column about X-Direction.

My = Maximum moment on the column about Y-Direction.

Pu = Maximum factor load on the column. h = Column dimension parallel to moment direction. Ø = 0.65 for tie stirrup, 0.75 for spiral.

Ag = Gross area of cross-section of column. ρg = Minimum steel required to resist axial and moment, Frist assume then checked.

Figure 4-15: (at Design Concrete Structures by Arthur H. Nilson) Interaction diagram for compression plus biaxial bending: (a) Uniaxial bending about Y-axis, (b) Uniaxial bending about X-axil, (c) biaxial bending about diagonal axis. (d) Interaction surface.

(At Design Concrete Structures -14th Edition - by A\rthur H. Nilson)

- 78 -

Ch. -4- Analysis and Design of Column

Note: There are two simple approximate methods used widely in order to design or check column subjected to biaxial bending with axial load: 1- Load Contour Method:

+

≤1

Where: Mnx = Maximum moment on the column about X-Direction.

Mny = Maximum moment on the column about Y-Direction.

Mnxo = Resisting moment of the column when ex = zero from chart.

Mnyo = Resisting moment of the column when ey = zero from chart. α1 and α2 constant varying from ( 1.15 to 1.55) 2- Reciprocal Method:

Where: Pn = Approximate value of ultimate capacity of the column when ex & ey is available Pnxo = Ultimate load when ex = zero Pnyo = Ultimate load when ey =zero Po = Ultimate load when ex & ey equal to zero If Pn ≥ Pu (Ultimate load on the column) the column is safe.

Note: Use Contour Method in this Project.

- 79 -

Ch. -4- Analysis and Design of Column

Now Design C2 at floor 5: Mny = 63 kN.m Mnx = 80.3 kN.m Pu = 310 kN fc’ = 26 Mpa :

fy = 414 Mpa

Assume column dim = (40X40) cm

γ= Find ⁄ and assume (ρg = 0.01) then find Rn:

ey = ex = ( ⁄ ) = ( ⁄ ) = By using interaction chart use ( ⁄ ) & ρg = 0.01 then find Rnx = 0.115 By using interaction chart use ( ⁄ ) & ρg = 0.01 then find Rny = 0.13 Mnxo = Pu*e =

= 0.115 * 0.65 * 26 * 4002 * 400/106 = 124.4 kN.m

Mnyo = Pu*e =

= 0.13 * 0.65 * 26 * 4002 * 400/106 = 140.61 kN.m

+

≤1

0.6045 + 0.3972 = 1.002 > 1 the column is not safe

Assume (ρg = 0.015) then : Rnx = 0.14 & Rny = 0.15 Mnxo = 151.4 kN.m : Mnyo = 162.3 kN.m

+

≤1

0.483 + 0.337 = 0.82 < 1 the column is safe.

- 80 -

Ch. -4- Analysis and Design of Column

‫ ؞‬As = 0.015 * 400 * 400 = 2400 mm2 Using Ø 20 mm (314 mm2) No. of bars =

≈ 7.65 = 8 bars

‫ ؞‬Use 8 Ø20 mm Design for Tie according to ACI-Code (7.10.5): Use Ø10 mm Minimum spacing equal to shall not exceed of the following:  16 Ø longitudinal bar (32 cm).  48 Ø tie bar (48 cm).  Least dimension of column (40 cm).

‫ ؞‬Use Ø10 mm @ 25 cm ⁄ Check for longitudinal bar is support by lateral reinforcement: h = 400 mm Cover = 40 mm Space between two longitudinal bars must be less than 150 mm (ACI-Code (7.10.5.3).

Space = 400 – 2(40+10) – (3*20) = 240 mm for 3 bar Spacing = 240/2 = 120 mm < 150 mm ‫ ؞‬Other lateral reinforcement no need.

Note: And overlap at top of column = 55 Ø main bar Overlap = 55 * 20 =1100 mm = 1.1 m

Figure 4-16: Detail of column 2:

- 81 -

Figure 4-17: Column plan detail: 7.0 m

7.0 m

7.0 m

7.0 m

C1

C5

C5

C5

C5

C1

C2

C6

C6

C6

C6

C2

C3

C7

C7

C7

C7

C3

C4

C8

C8

C8

C8

C4

6.60 m

6.60 m

4.50 m

7.0 m

1.55 m

- 82 -

Table 3-1: Design of column in all floor with biaxial moment (using excel program):

C1

slab Mnx Mny pu 4 16.6 29.8 152 3 18.75 19.8 413 2 25.94 19.8 674 1 31.2 19.8 935 G 53.23 19.8 1195

ey=mx/p ex= my/p

0.10921 0.0454 0.03849 0.03337 0.04454

0.19605 0.04794 0.02938 0.02118 0.01657

h 0.1*h (e or 0.1h)y (e or 0.1h)x 0.4 0.04 0.10921 0.19605 0.4 0.04 0.0454 0.04794 0.4 0.04 0.04 0.04 0.5 0.05 0.05 0.05 0.5 0.05 0.05 0.05

(e/h)y (e/h)x 0.273 0.49 0.113 0.12 0.1 0.1 0.1 0.1 0.1 0.1

pg 0.01 0.01 0.01 0.01 0.01

Rnx 0.12 0.085 0.08 0.08 0.08

M /Mno^ alpha

Rny 0.13 0.09 0.08 0.08 0.08

Mnxo Mnyo Mx/Mnx My/Mnyo alpph Mx/Mnx 129.79 140.61 0.127897 0.211937 1.15 0.09395 91.936 97.344 0.203946 0.203402 1.15 0.16067 86.528 86.528 0.299787 0.228828 1.15 0.25023 169 169 0.184615 0.11716 1.15 0.14329 0.31497 0.11716 1.15 0.26486 169 169

C2

slab Mnx Mny pu ey=mx/p 4 80.3 63 310 0.25903 3 74.34 43.3 783 0.09494 2 95.75 43.3 1255 0.07629 1 114.4 43.3 1728 0.06619 G 182 43.3 2200 0.0827

h 0.1 * h (e or 0.1h)y (e or 0.1h)x 0.20323 0.4 0.04 0.25903 0.20323 0.0553 0.4 0.04 0.09494 0.0553 0.0345 0.4 0.04 0.07629 0.04 0.02506 0.5 0.05 0.06619 0.05 0.01968 0.5 0.05 0.0827 0.05

ex= my/p

(e/h)y (e/h)x pg 0.648 0.508 0.015 0.237 0.138 0.01 0.191 0.1 0.0225 0.132 0.1 0.01 0.165 0.1 0.01

Rnx 0.14 0.125 0.14 0.095 0.11

Mnx 12.68 37.95 64.2 85.86 168.6

Mny 73.5 50.8 50.8 50.8 50.8

pu 363 917 1471 2025 2579

h 0.1 * h (e or 0.1h)y (e or 0.1h)x 0.03493 0.20248 0.4 0.04 0.04 0.20248 0.04138 0.0554 0.4 0.04 0.04138 0.0554 0.04364 0.03453 0.4 0.04 0.04364 0.04 0.02509 0.0424 0.5 0.05 0.05 0.05 0.06537 0.0197 0.5 0.05 0.06537 0.05

ey=mx/p ex= my/p

(e/h)y (e/h)x 0.1 0.506 0.103 0.138 0.109 0.1 0.1 0.1 0.131 0.1

pg 0.01 0.01 0.02 0.01 0.015

Rnx 0.08 0.08 0.1 0.08 0.105

Rny 0.15 0.1 0.09 0.08 0.08

Mnx 85.61 63.9 75.86 84.33 121.3

Mny 57.4 39.4 39.4 39.4 39.4

pu 279 706 1132 1558 1985

ey=mx/p ex= my/p

0.30685 0.09051 0.06701 0.05413 0.06112

0.20573 0.05581 0.03481 0.02529 0.01985

h 0.1 * h (e or 0.1h)y (e or 0.1h)x 0.4 0.04 0.30685 0.20573 0.4 0.04 0.09051 0.05581 0.4 0.04 0.06701 0.04 0.5 0.05 0.05413 0.05 0.5 0.05 0.06112 0.05

(e/h)y (e/h)x 0.767 0.514 0.226 0.14 0.168 0.1 0.108 0.1 0.122 0.1

pg 0.015 0.01 0.015 0.01 0.01

Rnx 0.135 0.095 0.12 0.085 0.09

4 4 4 8 8

Mnxo Mnyo Mx/Mnx My/Mnyo alpph Mx/Mnx 151.42 162.24 0.530299 0.388314 1.15 0.48217 135.2 108.16 0.549852 0.400333 1.15 0.50267 151.42 97.344 0.63233 0.444814 1.15 0.59032 200.69 169 0.569891 0.256213 1.15 0.52379 232.38 169 0.783002 0.256213 1.15 0.75479

My/Mny 0.3369 0.349 0.3939 0.2089 0.2089

t result as 0.819 Ok 2400 0.852 Ok 1600 0.984 Ok 3600 0.733 Ok 2500 0.964 Ok 2500

Detail 8 20 mm 8 20 mm 8 25 mm 8 20 mm 8 20 mm

t result as 0.584 Ok 1600 0.807 Ok 1600 0.994 Ok 3200 0.71 Ok 2500 0.964 Ok 3750

Detail 4 25 mm 4 25 mm 8 25 mm 8 20 mm 8 25 mm

t result as 0.844 Ok 2400 0.892 Ok 1600 0.917 Ok 2400 0.607 Ok 2500 0.784 Ok 2500

Detail 8 20 mm 8 20 mm 8 20 mm 8 20 mm 8 20 mm

M /Mno^ alpha

Rny 0.13 0.1 0.095 0.08 0.085

C4

slab 4 3 2 1 G

Detail 25 mm 25 mm 25 mm 20 mm 20 mm

t result as 0.262 Ok 1600 0.321 Ok 1600 0.434 Ok 1600 0.228 Ok 2500 0.35 Ok 2500

M /Mno^ alpha

C3

slab 4 3 2 1 G

My/Mny 0.1679 0.1602 0.1834 0.0849 0.0849

Mnxo 86.528 86.528 108.16 169 221.81

Mnyo 140.61 108.16 102.75 169 179.56

Mx/Mnx 0.146542 0.438586 0.593565 0.508047 0.760101

My/Mnyo 0.52273 0.469675 0.494394 0.300592 0.28291

alpph 1.15 1.15 1.15 1.15 1.15

Mx/Mnx 0.10986 0.38758 0.54889 0.45898 0.72946

My/Mny 0.4743 0.4193 0.4448 0.251 0.2341

M /Mno^ alpha

Rny 0.15 0.1 0.085 0.08 0.08

Mnxo Mnyo Mx/Mnx My/Mnyo alpph Mx/Mnx 146.02 162.24 0.586306 0.353797 1.15 0.54118 102.75 108.16 0.621886 0.364275 1.15 0.57912 129.79 91.936 0.584474 0.428559 1.15 0.53924 179.56 169 0.469641 0.233136 1.15 0.41931 190.13 169 0.638107 0.233136 1.15 0.59652

My/Mny 0.3027 0.3131 0.3774 0.1874 0.1874

Note: Use interaction chart to find Rnx and Rny in Appendix D

-83

Table 3-2: Design of column in all floor with uniaxial moment (using excel program):

slab 4 3 2 1 G

slab 4 3 2 1 G

slab 4 3 2 1 G

slab 4 3 2 1 G

m 29.9 26 33.2 38.45 60.5

m 147 110 131.2 150 217.5

m 12.68 37.95 64.2 85.86 168.6

m 143.12 113.8 113.8 122.3 159.3

pu 273 721 1169 1617 2066

pu 548 1404 2260 3115 3971

pu 641.4 1580 2518 3456 4395

pu 494.6 1213 1931 2649 3367

e=m/p 0.10952 0.03606 0.0284 0.02378 0.02928

e=m/p 0.26825 0.07835 0.05805 0.04815 0.05477

e=m/p 0.01977 0.02402 0.0255 0.02484 0.03836

e=m/p 0.28937 0.09382 0.05893 0.04617 0.04731

h (m) 0.4 0.4 0.4 0.5 0.5

h (m) 0.4 0.4 0.4 0.5 0.5

h (m) 0.4 0.4 0.4 0.5 0.5

h (m) 0.4 0.4 0.4 0.5 0.5

0.1 * h 0.04 0.04 0.04 0.05 0.05

0.1 * h 0.04 0.04 0.04 0.05 0.05

0.1 * h 0.04 0.04 0.04 0.05 0.05

0.1 * h 0.04 0.04 0.04 0.05 0.05

e or 0.1h e/h 0.10952 0.27381 0.04 0.1 0.04 0.1 0.05 0.1 0.05 0.1

C5 fc'(mpa) 26 26 26 26 26

b(mm) 400 400 400 500 500

h(mm) 400 400 400 500 500

kn 0.10096 0.26664 0.43232 0.38272 0.48899

Pg(av) 0.01 0.01 0.01 0.01 0.01

As 1600 1600 1600 2500 2500

4 4 4 8 8

e or 0.1h 0.26825 0.07835 0.05805 0.05 0.05477

e/h 0.67062 0.19587 0.14513 0.1 0.10954

C6 fc'(mpa) 26 26 26 26 26

b(mm) 400 400 400 500 500

h(mm) 400 400 400 500 500

kn 0.20266 0.51923 0.8358 0.73728 0.93988

Pg(av) 0.012 0.01 0.024 0.01 0.028

As 1920 1600 3840 2500 7000

Detail 4 25 mm 4 25 mm 8 25 mm 8 20 mm 16 25 mm

e/h 0.1 0.1 0.1 0.1 0.1

C7 fc'(mpa) 26 26 26 26 26

b(mm) 400 400 400 500 500

h(mm) 400 400 400 500 500

kn 0.2372 0.58432 0.93121 0.81799 1.04024

Pg(av) 0.01 0.01 0.023 0.013 0.031

As 1600 1600 3680 3250 7750

Detail 4 25 mm 4 25 mm 8 25 mm 8 25 mm 16 25 mm

C8 fc'(mpa) 26 26 26 26 26

b(mm) 400 400 400 500 500

h(mm) 400 400 400 500 500

kn 0.18291 0.44859 0.71413 0.62698 0.79692

Pg(av) 0.012 0.01 0.012 0.01 0.011

As 1920 1600 1920 2500 2750

4 4 4 8 12

e or 0.1h 0.04 0.04 0.04 0.05 0.05

e or 0.1h e/h 0.28937 0.72341 0.09382 0.23454 0.05893 0.14733 0.05 0.1 0.05 0.1

Detail 25 mm 25 mm 25 mm 20 mm 20 mm

Detail 25 mm 25 mm 25 mm 20 mm 20 mm

Note: Use interaction chart to find Ď g (av) in Appendix D

-84

Ch. -4- Analysis and Design of Column

4.6 Conclusion: Columns have a different reinforcement from one floor to others due to same suitable cross section of the columns that we chose for this project and according to the loading condition, at top floor some of columns controlled by As (min) and from bottom most of columns designed.

Wind load combination controlled some of columns in all floors and some of columns controlled by dead load and live load combination.

All the load that were transmitted to the columns must now is transferred to a stronger structure base known as a foundation.

The upcoming chapter will analysis and design of foundation...

- 89 -

Ch. -5- Analysis and Design of Foundation

Chapter Five Analysis and Design of Foundation 5.1 Introduction: The foundation of a building is the part of a structure that transmits the load to ground to support the superstructure and it is usually the last element of a building to pass the load into soil, rock or piles. The primary purpose of the footing is to spread the loads into supporting materials so the footing has to be designed not to be exceeded the load capacity of the soil or foundation bed.

The footing compresses the soil and causes settlement. The amount of settlement depends on many factors.

Excessive and differential settlement can damage structural and nonstructural elements.

Therefore, it is important to avoid or reduce differential settlement. To reduce differential settlement, it is necessary to transmit load of the structure uniformly. Usually footings support vertical loads that should be applied concentrically for avoid unequal settlement. Also the depth of footings is an important factor to decide the capacity of footings. Footings must be deep enough to reach the required soil capacity.

Figure 5-1: Mat foundation on the pile Dubai- Burj Khalifa - 90 -

Ch. -5- Analysis and Design of Foundation

5.1.2 Types of Footings: The most common types of footing are strip footings under walls and single footings under columns.

Common footings can be categorized as follow: 1. Individual column footing (Fig 5-2-a) This footing is also called isolated or single footing. It can be square, rectangular or circular of uniform thickness, stepped, or sloped top. This is one of the most economical types of footing.

The most common type of individual column footing is square of rectangular with uniform thickness.

2. Wall footing (Fig 5-2-b) Wall footings support structural or nonstructural walls. This footing has limited width and a continuous length under the wall.

3. Combined footing (Fig 5-2-e) They usually support two or three columns not in a row and may be either rectangular or trapezoidal in shape depending on column. If a strap joins two isolated footings, the footing is called a cantilever footing.

4. Mat foundation (Fig 5-2-f) Mats are large continuous footings, usually placed under the entire building area to support all columns and walls. Mats are used when the soil-bearing capacity is low, column loads are heavy, single footings cannot be used, piles are not used, or differential settlement must be reduced through the entire footing system.

5. Pile footing (Fig 5-2-g) Pile footings are thick pads used to tie a group of piles together and to support and transmit column loads to the piles.

Note: The Figure 5-2 shown in net page.

- 91 -

Ch. -5- Analysis and Design of Foundation

Figure 5-2: Footing Type

- 92 -

Ch. -5- Analysis and Design of Foundation

5.2 Methodology: Procedure for Analyze and Design of Foundation: 1. Compute the unfactor loads.

2. Compute the factored loads.

3. Assume the total footing thickness.

4. Compute the effective allowable soil pressure for superimposed service loads.

5. Compute required footing area.

6. Compute the factored soil pressure from superimposed loads.

7. Assume the effective depth for the footing.

8. Check the punching shear and beam shear.

9. Compute the design moment at the critical section.

10. Compute the required steel area.

11. Check the ACI Code minimum reinforcement requirement.

12. Check the development length.

13. Check the concrete bearing strength at the base of the column

- 93 -

Ch. -5- Analysis and Design of Foundation

5.3 Calculation 5.3.1 Calculate Load on the Foundation: Allowable soil pressure (Bearing Capacity) = 100 kN/m2 ď&#x201A;ˇ

Calculate Unfactor Load on the Pedestal on the Foundation ( using excel program):

ď&#x201A;ˇ

Calculate factored Load on the Pedestal on the Foundation ( using excel program):

Note: Moment on the foundation can be take in page 74.

Figure 5-3: Factor and unfactor load on the soil foundation - 94 -

Ch. -5- Analysis and Design of Foundation

Try to design the foundation by single (isolate) footing: B.C soil = 100 kN/m2 Total unfactor load of the building on the soil = 49057 kN Required area for single footing =

m2

Approximate minimum area required for mat foundation (20 X 36) m equal (720 m2) Ratio provide area by single footing to mat =

* 100 = 68 % > 50 %

‫ ؞‬Use mat foundation:

Figure 5-4: Factor load on the strip of the mat foundation

- 95 -

Ch. -5- Analysis and Design of Foundation

5.3.2 Find dimension of mat foundation first find length of foundation: Total factor load of the building = ∑ column load = 71105 kN P1 = 3367 + 4395 + 3971 + 2065 = 13798 kN P2 = 1985 + 2578 + 2200 + 1195 = 7958 kN M1 & M2 = 19.8 + 43.3 + 50.8 + 39.4 = 153.3 kN.m 71105 * X = 13798 * 7 + 13798 * 14 +13798 * 21 +13798 * 28 +7958 * 35 + 153.3 – 153.3 = 1244390 kN.m X = 17.5 m ‫ ؞‬Length of mat = (17.5 + 0.25 + 0.75) * 2 = 37 m Find width of foundation: P2 = (2 * 2200) + (4 * 3971) = 20284 kN P3 = (2 * 2478) + (4 * 4395) = 22536 kN P4 = (2 * 1985) + (4 * 3367) = 17438 kN M1 = (2 * 48.88) + (4 * 52.5) = 308 kN.m M2 = (2 * 160.54) + (4 * 178.33) = 1035 kN.m M3 = (2 * 168.6) + (4 * 168.6) = 1012 kN.m M4 = (2 * 97.6) + (4 * 116.6) = 662 kN.m 71105 * Y = 20284 * 4.5 + 22536 * 11.1 +17438 * 17.7 + 308 + 1035 + 1012 - 662 = 651773 kN.m Y = 9.17 ≈ 9.2 m ‫ ؞‬Width of mat = (9.2 + 0.25 + 0.8) * 2 = 20.5 m Check for bearing capacity of soil: Area of mat foundation = 20.5 * 37 = 758.5 m2 Total unfactor load = 49057 kN Applied soil pressure =

64.676 kN/m2 < 100 kN/m2 - 96 -

Ch. -5- Analysis and Design of Foundation

Find or check effective depth (d): Assume effective depth (d) = 70 cm Check due to punching shear: For column load (C7) = 4395 kN Critical sections for punching shear occur at distance (d/2): Area for punching shear = 4 * (0.5+0.7) * 0.7 = 3.36 m2 Concrete resistance for punching shear = ØVc * A + Soil reaction ØVc = Ø * 0.34 * √fc’ ØVc = 0.85 * 0.34 * √21 = 1324 kpa q Contact =

= 93.75 kN/m2

Soil reaction = Area * q contact = 1.2 * 1.2 * 93.75 = 135 kN Concrete resistance for punching shear = (1324*3.36) + 128 = 4583 > Column load (4395) O.K!

- 97 -

Ch. -5- Analysis and Design of Foundation

Check effective depth by one-way shear: For critical section (A-A): Find net applied shear force: Net applied shear force = Soil reaction – Column load Area soil reaction = (8.1 – d) * L = (8.1- 0.7) * 37 = 273.8 m2 Soil reaction = q contact * Area = 93.75 * 273.8 = 25668 kN Column load = (2 * 1985) + (4 * 3367) = 17438 kN Net applied shear force = 25668 – 17438 = 8230 kN Concrete resistance for one-way shear = ØVc * Area of concrete for resisting shear ØVc = Ø * 0.17 * √fc’

From ACI-Code 2011

ØVc = 0.85 * 0.17 * √21 = 662 kpa Assume d = 0.7 m Area = 37 * 0.7 = 25.9 m2 Concrete resistance for one-way shear = 662 * 25.9 = 17146 kN > applied shear (8230 kN) O.K For critical section (B-B): Area soil reaction = (5.3 – d) * L = (5.3 - 0.7) * 37 = 170.2 m2 Soil reaction = q contact * Area = 93.75 * 170.2= 15956 kN Column load = (2 * 1195) + (4 * 2065) = 10650 kN Net applied shear force = 15956 – 10650 = 5315 kN Concrete resistance for one-way shear = 17146 kN > applied shear (5315 kN) O.K For critical section (C-C): Area of soil reaction = (7.25 – d) * L = (7.75 - 0.7) * 20.5 = 144.52 m2 Soil reaction = q contact * Area = 88.937 * 145.5 = 13549.2 kN Column load = 1985 + 2578 + 2200 + 1195 = 7958 kN Net applied shear force = 12938 – 7958 = 5591.2 kN Area = 20.5 * 0.7 = 14.35 m2

;

ØVc = 662 kpa

Concrete resistance for one-way shear = 9500 kN > applied shear (5591.2 kN) O.K - 98 -

Ch. -5- Analysis and Design of Foundation

5.3.3 Analysis of the mat: Note: check static equilibrium of each strip and then draw shear and bending moment diagrams and considering each strip as a continuous beam.

Figure 5-5: Detail of Strip by X & Y direction

Note:

q Contact = 93.75 kN/m2 for all strip uniformly because eccentricity (ex & ey) = 0

Strip A: Column load = (2 * 1985) + (4 * 3367) = 17438 kN Soil reaction = Area of strip * q Contact = (5.05 * 37) * 93.75 = 17517.2 kN ∑ Fy = 0 ?

Fy = 17517.2 – 17438 = 79.2 kN

≠0

‫ ؞‬Average of column load with soil reaction =

kN

Decrease q Contact and increase column load until static equilibrium satisfied. q Contact =

* 93.75 = 93.538 kN/m2

For column (C4) =

* 1985 = 1989.51 kN

For column (C8) =

* 3367 = 3374.646 kN

Check: Total column load on this strip = (2 * 1989.51) + (4 * 3374.646) = 17477.6 kN Soil reaction = (5.05 * 37) * 93.538 = 17477.6 kN ‫ ؞‬Now ∑ Fy = 0 and static equilibrium is satisfied.

- 99 -

Figure 5-6: Draw shear and bending moment diagram by X-direction: Now draw shear and bending moment diagram for strip (A): 1989.51 kN

3374.646 kN

3374.646 kN

3374.646 kN

3374.646 kN

1989.51 kN

39.4 kN.m

39.4 kN.m X 1.00 m 7.00 m

7.00 m

7.00 m

1789.44 kN

1721.36 kN

1.00 m 7.00 m

7.00 m

1653.3 kN

1585.2 kN

5.05 * 93.538 = 472.367 kN/m

1517.13 kN

472.4 kN 472.4 kN 1585.2 kN

1517.13 kN

1228.6 kN.m

1653.3 kN

1721.36 kN

1705.2 kN.m

1705.2 kN.m

1789.44 kN

1228.6 kN.m

236.2 kN.m

236.2 kN.m

1188 kN.m 1428.12 kN.m

1428.12 kN.m

2156 kN.m

2156 kN.m

Width of the strip = 5.05 m (-M max = 2156 kN.m/strip = 2156/5.05 = 427 kN.m) & (+M max = 1705 kN.m/strip = 1705/5.05 = 338 kN.m)

Draw shear and bending moment diagram for strip (B): 2586.94 kN

4410.24 kN

4410.24 kN

4410.24 kN

4410.24 kN

1989.51 kN 50.8 kN.m

50.8 kN.m

1.00 m 7.00 m

7.00 m

7.00 m

2345.9 kN

2252 kN

1.00 m 7.00 m

7.00 m

2158.1 kN

2064.2 kN

2252 kN

2345.9 kN

6.6 * 93.427 = 616.6182 kN/m

1970.2 kN

616.6 kN 616.6 kN 2064.17 kN

1970.2 kN

1674 kN.m

2158.1 kN

2331.4 kN.m

2331.4 kN.m

1674 kN.m

308.3 kN.m

308.3 kN.m

1445.2 kN.m 1776.4 kN.m 2783.8 kN.m

1776.4 kN.m 2783.8 kN.m

Width of the strip = 6.6 m (-M max = 2783.8 kN.m/strip = 2783.8/6.6 =327 kN.m) & (+M max = 2331.4 kN.m/strip = 2331.4/6.6 = 354 kN.m)

- 100 -

Draw shear and bending moment diagram for strip (C):

2144 kN

3869.9 kN

3869.9 kN

3869.9 kN

3869.9 kN

2144 kN 43.3 kN.m

43.3 kN.m

1.00 m 7.00 m

7.00 m

7.00 m

2130.2 kN

2000 kN

1.00 m 7.00 m

7.00 m

1869.8 kN

1739.6 kN

5.55 * 96.26 = 534.243 kN/m

1609.4 kN

534.5 kN 534.5 kN 1739.6 kN

1609.4 kN

2000 kN

1869.8 kN

2133.5 kN.m

3044.9 kN.m

2130.2 kN

3044.9 kN.m

2133.5 kN.m

267.4 kN.m

267.4 kN.m

227.3 kN.m 692.3 kN.m

692.3 kN.m

2112.7 kN.m

2112.7 kN.m

Width of the strip = 5.55 m (-M max = 2112.7 kN.m/strip = 2112.7/5.55 =381 kN.m) &(+M max = 3044.9 kN.m/strip = 3044.9/5.55=549 kN.m)

Draw shear and bending moment diagram for strip (D): 1239.7 kN

2142.2 kN

2142.2 kN

2142.2 kN

2142.2 kN

1239.7 kN 19.8 kN.m

19.8 kN.m

1.00 m 7.00 m

7.00 m

7.00 m

1149.2 kN

1097.1 kN

1.00 m 7.00 m

7.00 m

1045 kN

992.9 kN

1097.1 kN

1149.2 kN

3.3 * 90.49 = 298.584 kN/m

940.8 kN

298.8 kN 298.8 kN 992.9 kN

940.8 kN

1045 kN

898.9 kN.m

1263.7 kN.m

1263.7 kN.m

898.9 kN.m

149.5 kN.m

149.5 kN.m

565.1 kN.m 748.9 kN.m 1311.8 kN.m

748.9 kN.m 1311.8 kN.m

Width of the strip = 3.3 m (-M max = 1311.8 kN.m/strip = 1311.8/3.3 =398 kN.m) & (+M max = 1263.7 kN.m/strip = 1263.7/3.3 = 383 kN.m)

- 101 -

Figure 5-7: Draw shear and bending moment diagram by Y-direction: Draw shear and bending moment diagram for strip (1): 2071.1 kN

2689.8 kN

2295.43 kN

1246.8 kN

168.6 kN.m 160.54 kN.m

97.6 kN.m

Draw shear and bending moment diagram for strip (2): 3324.9 kN

48.8 kN.m

4340 kN

3921.3 kN

2039.2 kN

168.6 kN.m 178.33 kN.m

116.6 kN.m

52.5 kN.m

Y 1.05 m

1.75 m 6.60 m

6.60 m

1294 kN 821 kN

709 kN

1.05 m

1.75 m

4.50 m

6.60 m

1311 kN

4.5 * 90 = 405 kN/m

4.50 m

6.60 m

2225 kN

2272 kN 1341 kN

1163 kN

426 kN 1379 kN

1362 kN

718 kN.m

550 kN.m

1002 kN

469 kN.m

698 kN 2162 kN

223 kN.m

1134 kN.m

2115 kN

1345 kN.m

1649 kN

1696 kN.m

682 kN.m 1572 kN.m

1966 kN.m

7 * 94.95 = 664.65 kN/m

824 kN.m

530 kN.m 2382 kN.m

2189 kN.m

Width of the strip = 4.5 m

Width of the strip = 7 m

(-M max = 1553 kN.m/strip = 1966/4.5 = 437 kN.m)

(-M max = 2382 kN.m/strip = 2382/7 = 340.3 kN.m)

(+M max = 718 kN.m/strip = 718/4.5 = 160 kN.m)

(+M max = 1696 kN.m/strip = 1696/7 = 242.3 kN.m)

Maximum moment by X-direction: -M = 427 kN.m +M max = 549 kN.m

Maximum moment by Y-direction: -M = 437 kN.m +M max = 242.3 kN.m

- 102 -

Ch. -5- Analysis and Design of Foundation

Note: use +M to find steel reinforcement at the bottom of the mat for all strips. use –M to find steel reinforcement at the top of the mat for all strips.

5.3.4 Design of the mat: Find steel at bottom of mat by X-direction: d= 700 mm ; b= 1000 mm a=

;

fc’= 21 Mpa ;

=

fy= 400 Mpa

As

Mu = Ø As fy ( d549 * 106 = 0.9 * As * 400 (700 -

As = 2260 mm2 /m

Solve for As ? ρ act = ρ min =

)

0.003228 = 0.0035

ρ min > ρ act

‫ ؞‬use ρ min

As = 0.0035 * 1000 * 700 = 2450 mm2 /m Use Ø 25 mm, Area= 490 mm2 No of bars = Spacing =

5 bars/m = 4 space / m = 25 cm c/c

Find steel at top of mat by X-direction: As = 1812 mm2 < As min (2450 mm2) ‫ ؞‬Use Ø 25 mm @ 20 cm c/c Note: provide same steel reinforcement for bottom and top of mat by Y-direction because +M & -M on this direction less than +M & -M by -direction: ‫ ؞‬Use Ø 25 mm @ 20 cm c/c at top and bottom of mat foundation.

- 103 -

Ch. -5- Analysis and Design of Foundation

5.5 Conclusion: Foundation is a structural base used to transfer the load of the building to the soil. The load must be transferred in such a way that soil failure will be prevented with limited settlement. In this project the require area for single footing was more than 50% of total area of project therefore we use mat foundation, it Couse the differential settlement in this building become minimize.

In this project mat foundation controlled by Asmin because of its large effective depth of foundation to control one way shear and two ways shear.

- 105 -

Ch. -1- Analysis and Design of Stair and Concrete Wall

Chapter Six Analysis and Design of Stair and Concrete Wall 6.1 Introduction: Staircase is an inclined structural system for movement from one level to another. Since it is stepped, it is called staircase. Stairs are constructed to provide access to the different floor levels, within buildings. They consist of a number of steps arranged in series. Most of stairs are designed as supported one-way slabs.

Definitions: 1. Going (G): The horizontal distance of the upper surface of a step.

2. Rise (R): The vertical distance between horizontal of two consecutive Steps.

3. Waist (h): The least thickness of the stairs slab.

4. Landing: The horizontal platform which is usually provided at the beginning and the end of series of steps.

5. Flight: Comprised of a number of steps provided between two consecutive landings.

Figure 6-1: Architectural detail stair

- 106 -

Ch. -1- Analysis and Design of Stair and Concrete Wall

6.2 Calculation: 6.2.1 Analysis and design of stair: Architectural design of stair: According to (Neufert-Architectsâ&#x20AC;&#x2122; Data- Fourth Edition- Pa. 120) standard for tread and riser is: Riser = 17 cm Tread = 29 cm

Figure 6-2: Architectural design of stair

fcâ&#x20AC;&#x2122; = 21 Mpa fy = 400 Mpa LL stair = 4.5 kN/m2 Weight Tile = 22 kN/m3 Assume thickness for mortar = 10 cm

- 107 -

Ch. -1- Analysis and Design of Stair and Concrete Wall

Structural Design of stair: The stair design as one-way slab and assume at ends of stair not fixed and not simple support because this use the following limit as shown below to find thickness to control deflection: h=

for simple support

h=

for both end continuous

at ACI-Code- table 9.5(a) and shown in Appendix E.

Length of span = 4.5 m center to center of beam h=

= 22.5 cm

h=

= 16 cm

‫ ؞‬Use h = 20 cm

DL (Stair) = = 11.25 kN/m2

DL (Stair) = wu (Stair) = 1.4 (11.25) + 1.7(4.5) = 23.4 kN/m2 = 18.2 * 1 = 23.4 kN/m length DL (Landing) = 0.2 * 24.5 + 0.1 * 22 = 7.1 kN/m2 wu (Landing) = 1.4(7.1) + 1.7(4.5) = 18.2 kN/m2 = 17.6 * 1 = 17.6 kN/m length d = h – cover -

= 200 – 20 -

= 174 mm = 17.4 cm

.

- 108 -

Ch. -1- Analysis and Design of Stair and Concrete Wall

Stair design: Check for shear: Vd (Applied) = 44 kN ØVc = Ø * 0.17 * √fc’ * b * d ØVc = 0.85 * 0.17 * √21 * 1000 * 174 /1000 = 115 > 44 OK! Find As: Use Ø 12 mm (113 mm2) AS =

= 956 mm2 /meter length

=

AS min = 0.002 * 1000 * 200 = 400 mm2 / meter length < AS AS (Total) = width of stair * AS = 1.3 * 956 = 1242 mm2 No. of bar =

= 10.9 ≈ 11 bars = 10 spaces

Spacing between bar = No. of bar =

= 13 cm c/c for longitudinal bar

≈ 4 bars = 3 spaces

Spacing between horizontal bar =

= 33 ≈ 30 cm c/c for lateral bar

- 109 -

Ch. -1- Analysis and Design of Stair and Concrete Wall

6.2.2 Design of concrete wall: Note: the wall in the building in elevator place is concrete wall not shear wall because not lateral force such as wind load because the beam carry 100 % effect of wind load, because this provide AS

min

for horizontal and vertical bar according to ACI- Code article (14.3), and with minimum

thickness according to ACI-Code article (14.5 & 14.6): Height of one floor (h) = 3.4 m Thickness of concrete wall =

h=

= 11.3 ≈ 12 cm > minimum thickness (10 cm)

Use thickness of wall = 15 cm Find steel reinforcement: Vertical reinforcement: Minimum ratio of vertical reinforcement area to gross concrete area ρl = 0.0012 for deformed bar and not longer than No. 16 with fy not less than 420 Mpa. Horizontal reinforcement: Minimum ratio of horizontal reinforcement area to gross concrete area ρt = 0.002 for deformed bar and not longer than No. 16 with fy not less than 420 Mpa. AS vertical = 0.0012 * 150 * 1000 = 180 mm2/m Use Ø 8 mm ; No. of bars =

= 4 bar = 3 space

AS Horizontal = 0.002 * 150 * 1000 = 300 mm2 Use Ø 10 mm ; No. of bars =

= 4 bar = 3 space

- 110 -

Conclusion

Conclusion: This project involves the analysis and design of a multi-story commercial building. The American institute code (ACI) 318 was used in this project. It is most commonly practiced code

by Iraqi engineer syndicate. The concept of ultimate design strength method was used to increase the applied and reduce the ultimate strength of the material.

The assumed dimension of the building has been checked for serviceability limits as well as all the used reinforcements. The dimensions were satisfactory to resist the deflection and the minimum reinforcement was used to prevent shrinkage and expansion due to temperature. The maximum reinforcement limit which prevents sudden failure was not exceeded in our design.

Live load and dead load value is taken from (ASCE), resulted some of slabs controlled by minimum reinforcement and governed by shrinkage and temperature limit, and some designed by bending moment that is larger than minimum reinforcement limit.

In design beams, all of beams same cross section, some of beams at top floor obtain the maximum positive and negative moment from dead load and live load combination, and beams in first, second and third floor maximum positive and negative moment obtained from wind load combination.

Also in design column most of columns controlled by wind load combination and some by dead and live load combination.

Soil reaction and the column loads were the main factors in selecting foundation type. And this foundation was mat foundation.

Concrete wall designed by minimum thickness and minimum reinforcement to prevent shrinkage and temperature because there is no lateral load on wall.

- 111 -

Appendix-A: Architectural plan (plan of the building): A 7.00

7.00

2.00

2.00

3.00

4.40

2.00

3.50

2.40

Manager room

4.70

Manager room

Staff room

7.00

2.00 3.60

2.40

3.10

3.50

4.70

Staff room

7.00

3.50 3.00 2.40

3.00

4.30

4.30

Manager room

Staff room

7.00

2.00

3.10

2.00

2.40 4.70

4.30

4.70

Manager room

Manager room

Staff room

Staff room

6.60 2.00

2.50

2.00 1.80

2.50

2.00 1.80

0.66

0.65

2.20 3.60

2.60

2.00

3.60

Reciption 1.60

Reciption

2.00 1.80

2.00 1.80

1.60

2.00 1.60

Reciption

2.00

2.00

4.50

2.60

2.50

2.10

3.60

Reciption

2.00 1.60

2.00

4.60

2.00

Reciption 2.00

4.60

1.89

4.50

6.60

3.40 1.60

4.60 2.20

2.00

2.00

2.00 2.203.60

2.00 3.60

Reciption 2.00 1.80

2.00

3.20

4.50

4.40

2.00 1.60

3.50 2.00

2.00

2.20

2.50

2.70

2.10

Staff room

4.40

4.40

Manager room

Store 3.40 2.00

2.00 1.80

Staff room

4.40

Manager room 2.00

2.60

Staff room

4.40

3.00

2.00 1.80

1.40

1.40

1.60

0.66

2.00

Reciption 3.60

2.00

2.50

Staff room

2.00

Reciption

3.60

Reciption

2.00

1.60 1.60

3.60

2.00 2.00

2.00

2.00

4.40

4.40

Manager room

Manager room

3.70

3.50

2.00

2.00

A

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+22.40m +21.90m

+18.50m

Office +15.10m

Office +11.70m

Office +8.30m

Office +4.90m

Shop +0.20m +0.00m

base ment -3.20m

Section A-A

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Appendix

Appendix-B: Factor moment according to ACI-Code at interior and exterior panel:

- 114 -

Appendix

Appendix-C: Reinforcement detail at slab according to ACI-Code 318-11

- 115 -

Appendix

Appendix-D: Interaction chart (all side steel):

- 116 -

Appendix

- 117 -

Appendix

- 118 -

Appendix

- 119 -

Appendix

Appendix-E: Table 9.5(a): minimum thickness of beam and one way slab:

- 120 -

SUGGESTION AND RECOMMENDATION Suggesting and recommending several points for this study:  Suggesting that, this project could be designed by computer program and make comparisons between both results.  Recommending that, this project might be analysis by another method such as: Coefficient method, Yield line theory, and finite element method…etc.  This research might be taken for an estimation project in the future.

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References:  Nilson, A. H & winter, G., 1995. Design of concrete structure.14th ed. Boston McGraw Hill  Nawy, E. G., 2004. Reinforced concrete. 5th ed. Upper Saddle River: Prentice Hall.  ACI Committee 318. Building Code Requirements for Structural Concrete (ACI 318M-11 & 95).American concrete institute 38800 Country Club Drive Farmington Hills.

 Neufert, 2012. Architects’ Data. 4th ed. Blackwell publishing Ltd.

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