Fyp.pdf

ACKNOWLEDGEMENT

The trend of getting structures analyzed scientifically for structural safety and economic reasons is getting more and more popular nowadays. The desire to learn about such an analysis and design has motivated us for this project, entitled “Detailed Structural Analysis and Design of Five and a half Storey Residential Building. We are highly indebted to the help and advice of our dedicated teachers and our helpful friends collaborating with in the preparation of the project report. Our advisor Asst. Prof. Ajay Khadka, Department of Civil Engineering, Nepal Engineering College, deserves our special acknowledgement for his valuable guidance, care, expertise and devotion of time in bringing out this report in this format in complete and concise volume. We benefited a lot in a great deal from his logical thoughts, experience and incisive comments We also sincerely thank all the friends directly and indirectly supporting us in bringing out this report to its completion.. Finally, we want to express our heartily thanks to all those who directly or indirectly provided us their cooperation in the completion of this project.

PREFACE

This project work is a mandatory part of B.E Civil final semester curriculum. Our project group has chosen to do analysis and design of Reinforced Concrete framed building under the guidance of our dedicated teacher Asst. Prof. Ajay Khadka. Among the two category of building structure, via, load bearing and framed structures, here, in the project work the building frame structure is practiced. Building frame is a three dimensional structure as space consist of rigidly interconnected beams, slabs and columns. It produces greater number of the redundancy thus reduces the moments and facilitates the even distribution of the load. The degree of sophistication to which structural analysis is carried out depends on the importance of structure and available resource. A highly redundant space structure is splited into different 3D frames and is then analyzed by using the help of STAAD Pro software. Though the vivid use of computer software has lessened the burden of repeated calculations on analysis, it is clear that for understanding the process physically and realizing the structural behavior, manual step by step procedure is necessary. However, due to time constrain and to be familiar with computer software, the structural analysis part is performed using computer software “STAAD Pro.” and design work is performed manually with reference to different codes of practice. Though every care has been taken to make the report free from errors, yet we shall be obliged, if errors present shall be brought to our notice. We will warmly welcome constructive criticism

List of Abbreviations Symbols Ac Ag Ast Asc Asv b bf bw D emin fck fsc fst fy Ld L Lo lo Leff Mu Mlim Mux Muy Muxl Muyl Mi M P Pu p pc pt Sv V Vu Xu Xumax Z abc sc st v bd cmax

Description : Areaof concrete : Gross sectional area : Areaof tensile steel : Area of compression steel : Area of vertical stirrups : width of beam or shorter dimensions of column : Effective width of flange in a T-beam : breadth of web in T-beam : effective depth of beam or slab : minimum eccentricity : characteristic compressive strength of concrete : compressive stress in steel corresponding to strain of 0.002 : tensile stress in steel : characteristic yield strength of steel : development length of bar : length of column or span of beam : anchorage length of bar : distance between points of zero moments : effective length of beam or column or slab : factored design moment : limiting moment of resistance : factored design moment along x-x axis : factored design moment along y-y axis : maximum uniaxial moment capacity of the section with axial load, bending about x – x axis : maximum uniaxial moment capacity of the section with axial load, bending about y –y axis : moment of inertia of the section : modular ratio : Axial load : factored design axial load : percentage of reinforcement : percentage of compression reinforcement : percentage of tension reinforcement : spacing of vertical stirrups : shear force : factored shear force : depth of neutral axis at the limit state of collapse : maximum depth of neutral axis : lever arm : permissible stress in concrete in bending compression : permissible stress in steel in compression : permissible stress in steel in tension : nominal shear stress in concrete with shear reinforcement : design bond stress : maximum shear stress in concrete with shear reinforcement

 Ast, req Asc, req Ast, pro Asc, pro Sv, sp16 Sv, DUC Sv, prov  p

: Diameter of tor steel bar : required area of tensile steel : required area of compressive steel : provided area of tensile steel : provided area of compressive steel : spacing of vertical stirrups from ductility consideration : spacing of vertical stirrups from ductility consideration : provided spacing of vertical stirrups : unit weight of soil at site : safe bearing capacity of soil at site

CHAPTER 1 INTRODUCTION 1.1 BACKGROUND The basics needs of human existences are food, clothing‟s and shelter. From times immemorial man has been making efforts in improving their standard of living. Development of economically and socially appropriate and environmentally comfortable shelter has been priority of the people in attempt to enhancing their quality of life. Appropriate shelter has been identified important element to lead productive and dignified life. Kathmandu being the capital city of Nepal is one of the densely populated cities. As it is rapidly developing, construction of residential house is becoming costly and unaffordable by large section of population and especially those belonging to lower economic strata. The increase in the cost of development of a house in the city core is due essentially to high cost of housing plots and higher labor cost. Contrary to urban areas, the development of residential housing in the peri urban areas is largely spontaneous and has been proceeding without much consideration of human comfort and environmental sanitation. In the developing countries like Nepal, present situation of water scarcity, energy crisis and challenges in waste management have becoming worse day by day and are being the headache for people especially in the capital city Kathmandu. Growing environmental issues have affected the construction process of residential building. So residential building might be the better choice if constructed on peri-urban areas of Kathmandu. This building is designed for the general case of Kathmandu valley which lies on earthquake zone V. It indicates higher value of earthquake. Hence the effect of earthquake is predominant than the wind load. So, the building is analyzed for earthquake as lateral load. The seismic coefficient design method as stipulated an IS 1893:2002 (Part 1) is applied to analyze the building for earthquake. Special reinforced concrete moment resisting frame is considered as the main in structural system of the building. This project report has been prepared in complete conformity with various stipulations in Indian Standards, Code of Practice for Plain and Reinforced Concrete IS 456-2000, Design Aids for Reinforced Concrete to IS 456-2000(SP-16), Criteria Earthquake Resistant Design Structures IS 1893-2000, Handbook on Concrete Reinforcement and Detailing SP-34. Use of these codes have emphasized on providing sufficient safety, economy, strength and ductility besides satisfactory serviceability requirements of cracking and deflection in concrete structures. These codes are based on principles of Limit State of Design. The earthquake resistant residential building possessing its own characteristics such as rain water harvesting located in the peri-urban areas is the main output expected in undertaking this project work. The project methodology briefly involves: Load calculation.  Preliminary design. • Slab • Beam 1

• Column  Modeling in STAAD  Detail design: • Slab, Column, Beam, Staircase, Foundation. • Rain water Harvesting  Design methods:  Limit state design.

1.2 OBJECTIVE Concerning with the building project, every client of a designer will obviously hope the safe and economic proposal from him. So the aim of the project is to design the safe and economical residential building especially focusing on the peri-urban areas of Kathmandu valley. The overall objective of this project is to propose appropriate design of an earthquake resistant residential building. The specific objectives of the project are as stated here under.     

To develop plan of five storied building in which ground floor is subscribed for mini mart & remaining floors for residential purpose. To analyze the static and dynamic loads on proposed building using STAADPRO. To perform structural analysis and design the components of the proposed building. To design the residential building in peri-urban areas to minimize the rising over crowded situation in Kathmandu city. To develop the self-confidence to attain the similar project in the near future as professional and to give the client full satisfaction.

1.3 RATIONALE OF THE PROPOSED PROJECT The proposed project has been chosen as a final year project in partial fulfillment of the academic requirement of the degree of Bachelor of Engineering (B.E.) in Civil and Rural Engineering. This project is focused on design of residential building keeping in mind the growing overcrowded issues in Kathmandu city and need for developing economical housing. This design is therefore expected to lower the dependence on the utility services and prove to be economical over a long run. The project is expected to enhance our knowledge and analytical skills relating to design of residential buildings and especially the design of residential facilities in the peri-urban areas. Upon completion of the project, this is expected to enhance our capacity to undertake similar project independently.

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1.4 DESCRIPTION OF THE PROPOSED PROJECT  Name of the Project : Commercial Building with Rental Housing  Location : Mulpani  Structure system : RCC framed structure  No. of storey : 4 & a half storied  Type of Slab : Two way Slab  Type of beam : Rectangular (400*300mm)  Type of column : Square (400*400mm)  Type of foundation : Raft Foundation  Type of staircase : Open well Staircase  Method of analysis : STAAD PRO  Design concept : Limit State Design  Concrete grade : M20  Reinforcement grade : Fe415  Dead load : As per materials usage in building  Live load : As per usage and as specified relevant code  Seismic load : As per usage and as specified relevant code  Topography : Plain terrain  Floor to Floor height : 2..74m  Plinth Area : 125.32m2  Occupancy of the building: Ground floor used as mini mart & other floors are residential. 1.5 BRIEF DESCRIPTION OF THE PLAN The grid plan given in the adjoining page has got the following features: Each floor having 

  

The different rooms having different dimensions are listed below:  One room of dimension: 4.19m2.92m  One room of dimension: 3.35m*3.20m  Two rooms of dimension: 3.048m3.20m  One room of dimension: 3.048m*4.31m Two Bathroom of size: 1.70m2.21m Six verandah of size: 1.18m4.878m The space for the purpose of staircase and for the storing purpose spaced with the area of size 2.8963m3.048m in each floor.

1.6 SCOPE OF THE PROJECT The project we selected is a part of academic course of B.E /Civil & Rural Fifth Year, First semester designed on how to acquaint the students with complete project work from the initial stage of analysis and design to the final working. It will provide us confidence on how to accomplish the structural design work of a building in its entirety by strictly adhering to the theoretical knowledge gained during academic year of B.E.

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It will improve our analytical technique to some extent that the design may become reality, by repeated systematic analysis until the design constraints are satisfied, we will not be benefited until our theoretical knowledge be correlated to the design works. By designing the structure we can develop judgment, perception, imagination and creativity, in short the ability to handle the works. 

Desk study and job formulation



Identification of different members of the building and computation of loads.



Project work is carried out for structural analysis, design and reinforcement detailing of RCC framed structures



Detailed design of each components or structural member with due consideration to economy in both size and use of materials as well as labor cost.



Detailing of the reinforcement of different elements of the system.

Besides, this project work that we have tried to perform does not deal with all the criteria of the requirements of the building construction. So there are some of the criteria within which the scope of the project work is limited which are as below: 

This project work is limited to the structural analysis and the design only.



Design and layout of the building services like pipeline, electrical appliances, sanitary and sewage system are not covered.



The project is not concerned with the existing soil condition of the locality.The bearing capacity of the soil is assumed.



The environmental, social and economic condition of that locality is not taken into consideration.



The project work is related only with the practical application of the studied courses in the field.

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CHAPTER 2

LITERATURE REVIEW

2.1 GENERAL Structurally a building may consist of a load bearing walls and floors. The floor slab may be supported on beams which in turn may be supported on walls or columns. But, for a multistoried structure a building frame is made either in steel or a reinforcement concrete. This frame is designed for all the vertical and horizontal loads transmitted through the frame. The openings between the columns where necessary will be filled with thin brick walls. A frame of this type will consist of columns and beams built monolithically forming a network. This provides rigidity to the connection members by this arrangement the bending moments for the members of the structures are reduced. Earthquake loads and other horizontal loads due to winds etc. are evenly distributed to the whole structures. This makes the structure not only safe but economical. The tentative size of structural elements are determined through the preliminary design so that after analysis the pre assumed dimensions might not deviate considerably, thus making the final design both safe and economical. 2.2 TYPES OF BUILDING Depending upon the character of occupancy or the type of use, buildings have been classified as: 2.2.1 Residential Buildings These building include any building in which sleeping accommodation provide for normal residential purposes, with or without cooking and dining facilities. It includes single or multi-family dwellings, apartment houses, lodgings or rooming houses, restaurants, hostels, dormitories and residential hostels. 2.2.2 Educational Buildings These include any building used for school, college or day-care purposes involving assembly for instruction, education or recreation and which is not covered by assembly buildings. 2.2.3 Institutional Buildings These buildings are used for different purposes, such as medical or other treatment or care of persons suffering from physical or mental illness, diseases or infirmity, care of infants, convalescents or aged persons and for penal detention in which the liberty of the inmates is restricted. Institutional buildings ordinarily provide sleeping accommodation for the occupants. 2.2.4 Assembly Buildings

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These are the buildings where groups of people meet or gather for amusement, recreation, social, religious, assembly halls, city halls, marriage halls, exhibition halls, museums, places of work ship etc.

2.2.5 Business Buildings These buildings are used for transaction of business, for keeping of accounts and records and for similar purposes, offices, banks, professional establishments, courts houses, libraries. The principal function of these buildings is transaction of public business and keeping of books and records.

2.2.6 Mercantile Buildings These buildings are used as shops, stores, market, for display and sale of merchandise either wholesale or retail, office, shops, and storage service facilities incidental to the sale of merchandise and located in the same building. 2.2.7 Industrial Buildings These are buildings where products or materials of all kinds and properties are fabrication, assembled, manufactured or processed, as assembly plant, laboratories, dry cleaning plants, power plants, pumping stations, smoke houses, laundries etc. 2.2.8 Storage Buildings These buildings are used primarily for the storage or sheltering of goods, wares or merchandise vehicles and animals, as warehouses, cold storage, garages, trucks. 2.2.9 Hazardous Buildings These buildings are used for the storage, handling, manufacture or processing of highly combustible or explosive materials or products which are liable to burn with extreme rapidly and/or which may produce poisonous elements for storage handling, acids or other liquids or chemicals producing flames, fumes and explosive, poisonous, irritant or corrosive gases processing of any material producing explosive mixtures of dust which result in the division of matter into fine particles subjected to spontaneous ignition. 2.3 COMPONENTS OF BUILDING A building can be broadly divided in two parts viz. (i) sub-structure and (ii) Superstructure. The portion of the building below the surrounding ground is known as substructure and the portion above the ground is termed as super-structure. The components of the building can be broadly summarized as under: i. ii. iii. iv.

Foundations Plinth Walls Columns 6

v. Beams vi. Slabs vii. Doors, windows and ventilations viii. Staircase ix. Roof x. Building finishes xi. Building services

CHAPTER 3 METHODOLOGY The project provided to us is completed performing each section works mentioned in the contents before .The following stages are involved in the analysis and design of four storied building. 3.1 LOAD CALCULATION Load calculation is done using the IS 875(Part I) -1987 as reference. The exact value of unit weights of the materials from the code is used in the calculation. The thickness of materials is taken as per design requirements. 3.2 PRELIMINARY DESIGN The tentative size of structural elements are determined through the preliminary design so that after analysis the pre assumed dimensions might not deviated considerably , thus making the final design both safe and economical . Tentative size of various elements has been determined as follows: 3.2.1 Slab For slab, preliminary design is done according to deflection criteria span /effective depth = 26*modification factor.( IS456-2000 Art 23.2.1) 3.2.2 Beam Thumb rule of 1‟-0”=0‟-1” (d=L/12 to L/15) basis is adopted to consider the preliminary design of the beam section. b/D=1/2 3.2.3 Column Preliminary design of column is done consideration and interior column. For the load acting in the column, live load is decreased according to IS 875-1978. Cross-section of the columns are adopted considering the economy. Square column section is adopted in this building project as per the internal aesthetic requirements. 3.2.4 Staircase

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Stairs is designed as per drawing. Column for stairs boxes is not included in the grid system but they are assumed to be simply tied with main frame with beam. 3.3 LOADING PATTERNS Loading pattern from slab to beam is obtained by drawing 450 offset lines from each corners then obtained trapezoidal as well as the triangular loading and is converted into the equivalent UDL as described in the respective sections .The loading from cantilever slab part is converted to UDL acting in beam by dividing the total load by beam. Load from all cantilever part is converted to UDL acting in beam by dividing total load (wall UDL*total wall length) by length of the beam. Self-weight of the projected beam part is assumed as point load in nearby column.

3.4 ANALYSIS There are three types of loads for which the provided proposed project is designed: 3.4.1 Dead Load Dead load consists of the load from each element of building i.e. weight of column, beam, slab, and wall. Dimensions of column, beam, slab are taken from preliminary design and corresponding density from code. For wall load thickness of wall is taken from plan. 3.4.2 Live Load Live load is taken from relevant code. In case of different live load in one panel of slab, highest value of load is taken for the panel. 3.4.3 Seismic Load For seismic load whole mass lump of building is calculated from which base shear is obtained according to code. 3.5 METHODS AND TOOLS FOR ANALYSIS For analysis, different software are available during these days. Concerning to our project we are using “STAAD PRO” for analysis. 3.6 DATA 3.6.1 Architectural Drawings The architectural drawing is prepared by our group members. 3.6.2 Geo-technical Data The project work is only for the purpose of learning. So since the soil testing is high costing technology, we could not afford this test. For the fulfillment of this we take an average value of the bearing capacity considering the overall/general value for it applied in the Kathmandu valley. So the bearing capacity is assumed 120.0 KN/m2 3.6.3 Load Data The load data is taken from the standard code of practices.

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3.7 DESIGN METHOD Design methods available (or in use) are: Working stress method Ultimate load method Limit state method 3.7.1 Limit State Method It uses the concept of probability and based on the application of method of statistics to the variation that occurs in practice in the loads acting on the structures or in the strength of material. The structures may reach a condition at which it becomes unfit for use for one of many reasons e.g. collapse, excessive deflection, cracking, etc and each of this condition is referred to a limit state condition. The aim of limit state design is to achieve an acceptable probability that a structure will not become unserviceable in its life time for the use for which it has been intended i. e it will not reach a limit state. It means structures should be able to withstand safely all loads that are liable to act on it throughout its life and it would satisfy the limitations of deflection and cracking. We adopt limit state method for design.

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CHAPTER 4 PRELIMINARY DESIGN Preliminary design is carried out to estimate approximate size of the structural members before analysis of structure. Grid diagram is the basic factor for analysis in both Approximate and Exact method and is presented below. 4.1 PRELIMINARY DESIGN OF SLAB

Let % of steel=0.1% to 0.4% =0.2% (assumed) 10

From Art 23.2.1,(IS 456-2000) We know, (Span/d) =αβγδλ Where, α=26 β=1, δ=1 λ =1 fs=N/mm2 γ=1.45

Now fot th modification factor γ for tension reinforcement Fs = 0.58*fy*area of x section of steel required area of x section of steel provided =0.58*415*1 240.7 N/mm2 Now from graph of modification factor of tension reinforcement γ =1.64 For,shortest span =2.8963 =2896.3mm span =26*1.64 =42.640 d Effective depth,d = span(2896) (26*1.64) = 67.92mm Provide 10mmΦ bar and 15mm clear cover Overall depth (D) =67.92+10/2+15 = 87.92mm Adopt (D) =120 mm 4.2. PRELIMINARY DESIGN OF BEAM The approximate size of the structural elements were determined through preliminary design so that after analysis the preassumed dimension might not deviated considered ,thus making the find design for safe and economic purpose.Approximate size of various element have been determined as follow: Input data; Grade of concrete (fck)=M20 (20N/mm2) Gradeof steel (Fy)=Fe 415 (415N/mm2)

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Depth of beam(d)=L/10 to L/15 1. Along alphabetical grid(longest span=4.87m Adopt length(Lc/c)=5m (Leff)=5 m Taking value of L/d=14 d =L/14 =5000/14 =357.14 Adopt (d)=365mm Diameter of bar (mm)=20mm Clear cover (mm)=25mm D=d + Clear cover + Ǿ/2 = 400mm 2. Along Numerical grid (Longest span = 4.87) Adopt length (Lc/c) = 5m Leff = 5m Take L/d = 14 d = L/14 = 357.14 mm Adopt (d) = 365 mm Diameter of bar (mm) = 20mm Clear cover (mm) = 25mm D=d + Clear Cover +Ǿ/2 = 400mm Width of beam: Take 1/2 to 2/3 of d 1. Along alphabetical grid (longest span = 4.87) D = 400mm Take b/D = 1/2 b =0.5*D = 0.5*400 = 200mm adopt (b) =300mm 3. Along Alphabetical grid (longest span =4.87) 4. Same as above [as symmetric] b = 300mm D = 400mm Hence,

Along Alphabetical grid , Adopt, B*D = 300*400mm 12

Adopt Numerical grid, Adopt B*D = 300*400mm

Preliminary design of Tie beam Size of tie beam along X axis =350mm * 350mm Size of tie beam along Y axis = 350mm * 350mm 4.3 LOAD DESCRIPTION0 4.3.1 DEAD LOAD (DL) IS 875 (Part 1)-1987, (Second Revision) Remark Thickness

Unit

Unit Weight

Cement Concrete, Reinforced with sand and gravel or crushed natural stone with 5 percent steel

KN/m3

24.8 26.5

2.

Screeding /cement plaster (1:4)

KN/m3 20.40

8-25

3.

Cement mortar (1:4)

KN/m3 20.40

8-24

4.

Marble

KN/m3 26.70

25-47

5.

Terrazzo flooring

KN/m2 0.24

26-51

6.

Common burnt clay bricks masonry

KN/m3 18.85

6-13

7.

100mm partition wall

KN/m3 1.910

31-9

S.N. Material 1.

10mm

100mm

to

Page/ S.N. 8-22

4.3.2 LIVE LOAD (LL) IS 875 (Part 2)-1987, (Reaffirmed 1992)(Second Revision) Remark S.N. Description

UDL

Unit Page/ S.N.

1.

Room without separate storage

4.0

KN/m2

9/v(b)

2.

Cafeterias & dining rooms

3.0

KN/m2

10/v(g)

3.

Corridors, passage and staircases

4.0

KN/m2

10/v(j)

4.

Toilets and bath rooms

2.0

KN/m2

10/v(k)

5.

Roof with access provided

1.5

KN/m2

8/11

13

6.

Roof access not provided

0.75

KN/m2

8/11

4.4 VERTICAL LOAD CALCULATION 4.4.1. LOAD CALCULATION FOR ROOF SLAB ABOVE STAIRCASE S.N 1. 2. 3. 4.

Dead Load Water tank (8000 liters) 25 mm thick screeding (1:4) 140 mm thick rcc slab 12.5mm thick cement plaster

Unit load(KN/m²) 80 kN/(3.04*2.89)=9.10 20.40*25/1000=0.51 25*140/1000=3.5 20.40*12.5/1000=0.255

Total Dead Load = 13.365 KN/M2 Live load (access provided) = 1.5 KN/ M2

4.4.2. LOAD CALCULATION FOR ROOF SLAB TERRACE S.N 1. 2. 3.

Dead Load 25 mm thick screeding (1:4) 140 mm thick rcc slab 12.5mm thick cement plaster

Unit load 20.40*25/1000=0.51 25*140/1000=3.5 20.40*12.5/1000=0.255

Total Dead Load = 4.265 KN/M2 Live load (access provided) = 1.5 KN/ M2

4.4.3. LOAD CALCULATION FOR INTERMEDIATE SLAB (1, 2 & 3) S.N 1. 2. 3. 4. 5.

Dead Load 110 mm thick partition wall 25 mm thick screeding (1:4) 140 mm thick rcc slab 12.5mm thick cement plaster 25mm thick marble

Unit load 1.910*110/1000=2.101 20.40*25/1000=0.51 25*140/1000=3.5 20.40*12.5/1000=0.255 26.7*25/1000=0.668

Total Dead Load = 6.876 KN/M2 Live load = 2 KN/M2

4.4.4. LOAD CALCULATION FOR SLAB S4 S.N 1. 2. 3.

Dead Load 110mm thick partition wall 25 mm thick screeding (1:4) 140 mm thick rcc slab

Unit load 1.910*110/1000=2.101 20.40*25/1000=0.51 25*140/1000=3.5 14

4. 5.

25mm thick marble 12.5mm thick cement plaster

26.7*25/1000= 0.668 20.40*12.5/1000=0.255

Total Dead Load = 6.932 KN/M2 Live load = 2 KN/M2

4.4.5. LOAD CALCULATION FOR PROJECTION SLAB S.N

Dead Load

Unit load

1.

3 mm thick cement punning

20.4*3/1000=0.061

2.

25 mm thick screeding (1:4)

20.40*25/1000=0.51

3.

140 mm thick rcc slab

25*140/1000=3.5

4.

12.5mm thick cement plaster

20.40*12.5/1000=0.255

Total Dead Load = 4.326 KN/M2 Live load =0.75 KN/ M2

4.4.6. LOAD CALCULATION FOR VERANDAH SLAB S.N

Dead Load

Unit load

1.

3 mm thick cement punning

20.4*3/1000=0.061

2.

25 mm thick screeding (1:4)

20.40*25/1000=0.51

3.

140 mm thick rcc slab

25*140/1000=3.5

4.

12.5mm thick cement plaster

20.40*12.5/1000=0.255

5.

110 mm parapet wall

18.85*110/1000=2.07

Total Dead Load = 6.427 KN/M2 Live load = 1.5 KN/ M2

Unit load of different slab for different floors

(1) For roof slab above staircase Overall depth = 140mm (a) Dead Load = 13.365 kN/m2 15

(b) Live Load = 1.5 KN/m2

(2) For roof floor slab Overall depth = 140mm (a) Dead Load = 4.265 kN/m2 (b) Live Load = 1.5 KN/m2

(3) For intermediate floor slab Overall depth = 140mm (a) Dead Load = 6.876 kN/m2 (b) Live Load 2 KN/m2

(4) For slab S3 Overall depth = 140mm (a) Dead Load = 6.932 kN/m2 (b) Live Load =2 KN/m2

(5) For projection slab Overall depth = 140mm

(6 )Roof slab above staircase (a) Dead Load = 4.326 kN/m2 (b) Live Load =0.75 KN/m2

(7) Roof slab (a) Dead Load = 6.427 kN/m2 (b) Live Load = 1.5 KN/m2

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4.5 PRELIMINARY DESIGN OF CRITICAL COLUMN

Column B3 as a critical column Column size = 0.4m * 0.4m Size of beam along x direction =0.3*0.4 Size of beam along y direction = 0.3*0.4 Floor to floor height = 2.74 m Ground floor height = 3.048 m Foundation depth = 2 m (assumed) Thickness of cement plaster = 0.0125m Height of column = (2.74-0.14)*5 + 2 + 0.457 =15.457m 4.5.1 DEAD LOAD CALCULATION CALCULATION OF SELF WEIGHT OF ELEMENTS TO DESIGN CRITICAL COLUMN (A) Loading due to beam 17

(a) Self-load of beam Along (B-B):(3.96-0.4)*(0.4-0.14)*0.3*25 = 6.942 KN Along (3-3):(3.89-0.4)*(0.4-0.14)*0.3*25 = 6.805 KN (b) Using 12.5 mm plaster Along (B-B):((3.96-0.4)*0.0125*((0.4-0.14)*2+ 0.3))*20.4 =0.744 KN Along (3-3):((3.89-0.4)*0.0125*(0.4-0.14)*2+ 0.3))*20.4 =0.729 KN Hence total load = (6.942+6.805+0.744+0.729) =15.22 KN (B) Loading due to wall Loading due to partition wall= area * 1 KN/m2 =15.4044 * 1 =15.4044KN. (b) Using 12.5 mm plaster in both sides Along (B-B):(((3.96-0.4)*0.0125*(2.74-0.4)*2)*20.4) =4.248 KN Along (3-3):(((3.89-0.4)*0.0125*(2.74-0.4)*2)*20.4) =4.165KN Hence total load of floor = 15.4044+4.248+4.165 = 233.817 KN (C) Loading due to slab Area * thickness *25 =15.4044*0.14*25=53.915 KN Floor finishing = area *1 =15.4044*1 = 15.4044 KN

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4.5.2 Load Calculation

floor

Beam

Wall

Slab

Floor finishin g

Column

Dead load

Live load

Staircas e cover

-

-

-

-

-

-

3.96*3.89*1.5 =23.1066

Terrace

15.22

-

53.915

15.404 4

2.60*25=65

149.539

3.96*3.89*1.5*0.6 =13.86

3rd floor

15.22

23.817

53.915

15.404 4

2.60*25=65

173.356

3.96*3.89*2*0.8 =24.64

2nd floor

15.22

23.817

53.915

2.60*25=65

173.356

1st floor

15.22

23.817

53.915

15.404 4 15.404 4

2.60*25=65

173.356

3.96*3.89*2*0.9 =27.72 3.96*3.89*2*1 =30.80

Ground floor

15.22

-

-

-

2.457*25 =61.425

76.645

Total dead load=746.252 KN Total live load=120.1266 KN Hence grand total load = 866.3786 KN Load factor = 1.5 Factored load = 1.5*866.3786 = 1299.567 KN Increasing by 30% for EQ consideration, = 1.3*1299.567 = 1689.43827 KN Assuming % of steel (0.8% - 4 %) Now from Sp 16 chart 25 using M20 concrete and Fe 415 Pu/Ag =13 N/mm² Ag

= 1689.43827 * 1000/13 = 129956.79 mm2

For square column One side of column = 360.49 mm Hence, adopt size of column 400mm * 400mm 19

-

CHAPTER 5

LATERAL ANALYSIS

5.1 INTRODUCTION TO LATERAL LOAD Seismic weight is the total dead load plus appropriate amount of specified imposed load. While computing the seismic load weight of each floor, the weight of columns and walls in any story shall be equally distributed to the floors above and below the storey. The seismic weight of the whole building is the sum of the seismic weights of all the floors. It has been calculated according to IS: 1893(Part I) – 2002. IS: 1893(Part I) – 2002 states that for the calculation of the design seismic forces of the structure the imposed load on roof need not be considered. 5.2 CALCULATION OF SELF-WEIGHT OF ELEMENTS 5.2.1 SELF-WEIGHT OF BEAM (a) Along X-X direction Size of beam = 300*400 Slab thickness = 140mm Self-weight of beam = (0.4-0.14)*0.3*25 =1.95KN/M (b) Along Y-Y direction = 1.95 KN/M 5.2.2 SELF-WEIGHT OF TIE BEAM For 1m span tie beam = (0.4*0.3) * 25 = 3 KN/M 5.2.3 SELF-WEIGHT OF COLUMN Size of column = 0.4 *0.4 Let us take unit length of column 1m Dead load on column = 0.4*0.4*25 = 4 KN/M 5.2.4 SELF-WEIGHT OF BRICK WALL (a) Partition wall (110mm thick) Along X-X Height = (2.74-0.4) = 2.34 m Dead load of wall = 0.11*2.34*18.85 = 4.851 KN/M Using 12.5 mm plaster on both sides, = 2*2.34*0.0125*20.4 = 1.1934 KN/M Total dead load = (4.815+ 1.193) = 6.04 KN/M Assuming 30 % openings, =6.04*0.7 = 4.23 KN/M (b) Parapet walls (110 mm thick & 600mm height) Dead load of brick wall = 0.11 *0.6*18.85 = 1.244 KN/M 12.5mm thick 2 side plaster = 0.0125*20.40*2*0.6 =0.30KN/M 20

12.5 mm thick plaster at coping = (0.23*0.0125+0.075*0.0125*2)+(0.06*0.0125*2)*20.4 = 0.127KN/M Total load = 1.244+0.3+0.1275 = 1.6715 KN/M (c) Main outside brick wall (230mm thick) (i) Along X-X direction = 2.74-0.4 = 2.34m Dead load of wall = 2.34*0.23*18.85 = 10.14 KN/M Using 12.5mm plaster on both sides of wall = 0.0125*20.4*2*2.34 = 1.1934KN/M Total dead load = (10.14+1.1934)KN/M=11.3334KN/m Assuming 30 % openings Then total dead load =0.7* 11.3334 = 7.93338 KN/M (ii) Along Y-Y direction Dead load of wall = 2.34*0.23*18.85 = 10.145KN/M Using 12.5mm plaster on both sides of wall = 0.0125*20.4*2*2.34 = 1.1934KN/M Total dead load = (10.145+1.1934)KN/M=11.3334KN/M Assuming 30 % openings Then total dead load =0.7* 11.3334 = 7.93338 KN/M 5.3 LUMP MASS CALCULATION 5.3.1 LUMPED LOAD OF SLAB As per Art. IS 1893-1984, Clause 4.1 The earthquake force shall be calculated for the full dead load plus the percentage of imposed Load as given below.

SN 1 2

Imposed Uniformly Distributed Floor loads (KN/m2) Up to and including 3.0 Above 3.0

21

% of load 25 50

imposed

For calculating the design seismic force of the structure, the imposed load on the roof need not To be considered.

Floor

Above staircase

Load to

due Area (m2)

Cover slab Projection slab(x) Projection slab(y) Slab S1 Slab S2 Slab S3

Roof floor

Slab S4 Projection slab(X) Projection Slab(Y) Slab S1 Slab S2

Slab S3 Slab S4 1st,2nd, Verandah 3rd & (X) 4th floor Verandah (Y) Projection Slab(X) Projection slab(Y)

No of Sla b

U.L.(KN/m2) Load in KN DL 13.3.6 5

DL+LL (KN)

LL 1.5 0 1.5 0

DL

LL

118.011

13.244

131.255

5.7258

0.992

6.7178

8.8299

1

1.3236

1

4.326

1.8123

1

4.326

1.5

7.843

1.359

9.202

8.8299 14.128 14.87 23.794 8 13.186 4 10.915 4 8.8299 14.428 2 14.871 5 23.794

3 2 2

4.265 4.265 4.265

1.5 1.5 1.5

112.97 120.513 126.854

39.734 42.384 44.614

152.704 162.894 171.468

1

4.265

1.5

101.485

35.69

137.175

1

4.326

1.5

57.044

9.889

66.933

1

4.326

1.5

47.22

8.186

55.4065

3

6.876

2

182.14

52.979

2

6.876

2

194.291

56.512

235.122 250.803 8

2

6.876

2

1

6.932

2

204.513 5 164.946

59.486 2 47.589

6.568

1

6.427

1.5

42.21

9.852

52.062

2.7102

1

6.427

1.5

17.418

4.0653

29.4833

6.6184

1

4.326

1.5

28.6311

4.9638

33.5949

8.2052

1

4.326

1.5

35.495

6.1583

41.648

22

Total load of slab (KN)

147.174 8

746.581 3

263.999 212.536 1111.25 06

5.3.2 LUMPED LOAD OF BEAM Self-weight = 1.95KN/M Self-weight of tie beam = 3KN/M

S.N

floor

Span beam

1.

staircase

4.878(X)

of No.of beam

2

Load of Total load beam

9.5121

Total load in a floor (KN)

19.0242 30.9122

2.

Roof floor

3.048(Y)

2

5.944

11.888

2.8963(X)

8

5.6477

45.1816

8

5.944

47.552

3.0487(Y)

168.830 4

9.5121

38.0484

4

9.5121

38.0484

8

5.6477

45.1816

8

5.944

47.552

4.878(X) 4.878(Y) 3.

Intermediate- 2.8963(X) floors (1st, nd rd 2 &3 ) 3.0487(Y)

168.8304

4.

Tie beam

4.878(X)

4

9.5121

38.0484

4.878(Y)

4

9.5121

38.0484

2.8963(X)

8

8.688

69.5112

3.0487(Y)

8

9.1461

73.1688

4.878(X)

4

14.634

58.536

4.878(Y)

4

14.634

58.536

23

259.3752

5.3.3 LUMPED LOAD OF COLUMN S.N 1 2 3 4 5 6

Floor Stair cover Roof cover 3rd cover 2nd cover 1st cover Ground floor

Weight of column Above the slab Below the slab 0 41.6 124.8 124.8 124.8 124.8 124.8 124.8 124.8 124.8 29.248

Total load (KN) 41.6 166.4 249.6 249.6 249.6 154.048

5.3.4 LUMPED LOAD OF WALL a) On above staircase wall: no parapet wall b) Roof floor i. parapet wall Along long wall (X-X direction) =[ (2.8963*3)+(4.878*2)]*1.67 = 30.8029KN Along short wall (Y-Y direction) = [(3.0487*30 +(4.878*2)*1.67= 31.5665KN Total =62.3694KN ii. Wall (230 mm thick) Along long wall (X-X direction)= [(2.8963-0.4)*2*7.93338]= 39.6 KN Along short wall (Y-Y direction) = [(3.0487-0.4)*2*7.93338=42.0 KN Total = 81.6 KN c) Intermediate floor i) Outside wall (230 mm thick) Along X-X direction = [(2.8963*3-0.4*3)+(4.878*2-0.4*2)]*7.93338+10.067 = 247.4567 KN Along Y-Y direction = [(3.0487*4-0.4*4)+(4.878*2-0.4*2)]*7.93338= 155.1039KN ii) Inside wall (partition wall 0.11m) Along X-X direction = [(2.8963-0.4)*4+(4.878-0.4)*2+2.2865+1.70223]*4.23 = 96.9936 KN Along Y-Y direction = [(3.0487-0.4)*2+(3.1087*2)+(4.8780-0.4)+1.5243 = 74.0973 KN

iii) Verandah (parapet wall) Along X-X direction= 6.2245*1.67=10.395 KN 24

Along Y-Y direction= 3.633*1.67= 6.067 KN Therefore, Grand total =590.09 KN d) Ground floor Outside wall (230 mm) Along X-X direction = [(11.0264-4*0.4)*2]*7.93338= 149.566KN Along Y-Y direction = [(11.0264-4*0.4)*2]*7.93338= 154.4042 KN e)Inside wall Along X-X(0.11m) =[(2.8963-0.4)+ 3.766]*4.23=26.49 Along X-X(0.23m) =[0.366+0.366+2.8963-0.4]*7.93338=25.61 Along Y-Y = [(3.0487 + 0.4)*2*4.568=29.176 KN 5.3.5 LUMP LOAD OF STAIRCASE (i) Staircase inclined slab = 0.31 *1.06* 0.15 * 16 * 25 = 19.716 KN (ii) For steps = ½ * 0.17 * 0.26* 1.06* 25 * 16 = 10.8 KN (iii) For landing = 2*1.06*1.06*0.15*25 = 8.427 KN (iv)For the slab between step to column = 3.048* 0.7077 * 0.15 *25 = 5.8043 KN Therefore, total lump mass = 19.716+9.37+8.427+5.8043 = 43.3173 KN

5.4 DETERMINATION OF BASE SHEAR According to IS 1893 (Part I): 2002 Cl. No. 6.4.2 the design horizontal seismic coefficient Ah for a structure shall be determined by the following expression:

25

Z I Sa 2R g

Ah 

Where, Z = Zone factor given by IS 1893 (Part I): 2002 Table 2, Here for Zone V, Z = 0.36 I = Importance Factor, I = 1.5 (being similar to community building) R = Response reduction factor given by IS 1893 (Part I): 2002 Table 7, R = 5.0 Sa/g = Average response acceleration coefficient which depends on Fundamental natural period of vibration (T a). According to IS 1893 (Part I): 2002 Cl. No. 7.4.2

Ta 

0.09 h d

Where, h = height of building in m, h = 14.157 m d = Base dimension of the building at the plinth level in m along the considered direction of the lateral force. Along X-axis; d = 10.67m

Ta x 

0.09 x 14.157 10.67

 0.39 sec

For Ta = 0.39 sec and medium type of soil; Sa/g = 2.5 Now, Ah x 

0.36 x 1 x 2.5 2 x5

 0.09

Along Y-axis; d = 10.97 m

Ta y 

0.09 x 14.157 10.97 26

 0.384 sec

Now, Ah y 

0.36 x 1x 2.5 2 x5

 0.09

According to IS 1893 (Part I) : 2002 Cl. No. 7.5.3 the total design lateral force or design seismic base shear (VB) along any principle direction is given by VB = Ah x W Where, W = Seismic weight of the building = 9059.2498 KN The total design base shear along X- direction; (VB)X = 0.09*9059.2498 = 815.332 KN The total design base shear along Y- direction; (VB)Y = 0.09*9053.2498 = 815.332 KN

According to IS 1893 (Part I): 2002 Cl. No. 7.7.1 the design base shear (VB) computed above shall be distributed along the height of the building as per the following expression: Qi  VB

Wi h i2 n

 Wj h 2j

j1

Where, Qi = Design lateral force at floor i Wi = Seismic weight of floor i hi = Height of floor I measured from base n = No. of storeys in the building

27

5.4.1 CALCULATION OF LATERAL FORCES AND SHEAR AT STOREY (VB)X = 815.2498 KN (VB)Y = 815.2498 KN SN

STOREY

1 2 3 4 5 6

Staircase cover Roof Floor 3rd Floor 2nd Floor 1st Floor Tie Beam

hi (m)

Wi (KN)

Wi*hi2 (KNm2)

14.157 11.417 8.677 5.937 3.197 0.457

242.07 1554.27 2221.2317 2221.2317 2221.2317 599.214

48515.9668 202595.8134 167237.2655 78293.9061 22702.78 125.1452

Qi (KN)

Cumulative Qi (KN)

76.14 317.982 262.4861 122.8856 35.63299 0.196

76.14 394.122 656.6081 779.4937 815.1266 815.332

Total Wi = 9059.2498 KN Total (Wi*hi2) =519470 KN

5.4.2 CALCULATION OF NODAL MASS APPLIED FORCE

SN

FLOOR

1 2 3 4 5 6

Staircase cover Roof Floor 3rd Floor 2nd Floor 1st Floor Tie Beam

LATERAL FORCE (KN) 76.14 317.982 262.4861 122.8856 33.63299 0.196

ALONG XDIRECTION 38.07 79.4955 65.621 30.7214 8.9082 0.049

28

ALONG YDIRECTION 55.093 79.4955 65.621 30.7214 8.9082 0.049

29

5.5 LOAD COMBINATION Different load cases and load combination cases are considered to obtain most critical element stresses in the structure in the course of analysis. There are together four load cases considered for the structural analysis and are mentioned as below: i.) ii.) iii.) iv.)

Dead Load (D.L.) Live Load (L.L) Earthquake load in X-direction (EQx) Earthquake load in Y-direction (EQy)

Following Load Combination are adopted as per IS 1893 (Part I): 2002 Cl. No. 6.3.1.2 i.) ii.) iii.) iv.) v.) vi.) vii.) viii.) ix.) x.) xi.) xii.) xiii.)

1.5 (D.L + L.L) 1.5 (D.L + EQx) 1.5 (D.L - EQx) 1.5 (D.L + EQz) 1.5 (D.L - EQz) 1.2 (D.L + L.L + EQx) 1.2 (D.L + L.L - EQx) 1.2 (D.L + L.L + EQz) 1.2 (D.L + L.L - EQz) 0.9 D.L + 1.5 E.Qx 0.9 D.L -1.5 EQx 0.9 D.L + 1.5 EQz 0.9 D.L -1.5 EQz

30

CHAPTER 6 STRUCTURAL ANALYSIS

6.1 SALIENT FEATURE OF STAAD PRO STAAD Pro is comprehensive structural engineering software that addresses all aspects of structural engineering including model development, verification, analysis, design and review of results. It includes advanced dynamic analysis and push over analysis for wind load and earthquake load. Design in STAAD supports more than 10 codes. INDIAN standard codes (Concrete and Steel) are also supported; Limit state method is used for the design. 6.2INPUT AND OUTPUT The design of earthquake resistant structure should aim at providing appropriate dynamic and structural characteristics so that acceptable response level results under the design earthquake. The aim of design is the achievement of an acceptable probability that structures being designed will perform satisfactorily during their intended life. With an appropriate degree of safety, they should sustain all the loads and deformations of normal construction and use and have adequate durability and adequate resistance to the effects of misuse and fire. For the purpose of seismic analysis of our building we used the structural analysis program STAAD PRO 2007. It helps in creating the real model and analyzes all the forces acting on it. This type of modeling is very useful in the lateral dynamic analysis of building. The base shear and earthquake lateral forces are calculated as per code IS 1893(part-I) 2002 and are applied at each master joint located on every storey of the building. The sample input and output are shown in annex.

6.3METHODOLOGY OF ANALYSIS       

Creating model Assigning material properties Assigning section properties Creating different load cases Creating load combination Analysis (pre analysis and post analysis) Design of column and beam

31

Fig: Isometric View

Fig: Plan view

32

CHAPTER 7 STRUCTURAL DESIGN 7.1 LIMIT STATE METHOD In the method if design based on limit state concept, the structure shall be designed to withstand safely all loads liable to act on it throughout its life; it shall also satisfy the serviceability requirements, such as limitations on deflection and cracking. The acceptable limit for the safety and serviceability requirements before failure occurs is called a „limit state‟. The aim of design is to achieve acceptable probabilistic that the structure will not become unfit for the use for which it is intended, that is, that it will not reach a limit state. Assumptions for flexural member i)

Plane sections normal to the axis of the member remain plane after bending.

ii)

The maximum strain in concrete at the outermost compression fiber is 0.0035.

iii)

The relationship between the compressive stress distribution in concrete and the strain in concrete may be assumed to be rectangle, trapezoidal, parabola or any other shape which results in prediction of strength in substantial agreement with the result of test. For design purposes, the compressive strength of concrete in the structure shall be assumed to be 0.67 times the characteristic strength. The partial safety factor γm = 1.5 shall be applied in addition to this.

iv)

The tensile strength of concrete is ignored.

v)

The design stresses in reinforcement are derived from representative stressstrain curve for the type of steel used. For the design purposes the partial safety factor γm = 1.15 shall be applied.

vi)

The maximum strain in the tension reinforcement in the section at failure shall not be less than:

fy 1.15E s

 0.002

Where, fy= characteristic strength of steel Es = modulus of elasticity of steel Assumption for compression members In addition to the assumptions given above from i) to v), the following shall be assumed: i.)

The maximum compressive strain in concrete in axial compression is taken as 0.002. 33

ii.)

The maximum compressive strain at highly compressed extreme fiber in concrete subjected to axial compressive and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fiber.

The limiting values of the depth of neutral axis for different grades of steel based on the assumptions are as follows:

Fy

xu,max/d

250

0.53

415

0.48

500

0.46

7.2 SLAB Slabs are the most widely used structural elements forming floors and roofs of building for supporting load normal to its surface. Slab may be simply supported or continuous over one or more supports and is classified according to the manner of support as: oneway slab spanning in one direction, two-way slab spanning in directions, circular slabs and grid floor slabs resting directly on columns with no beams and grid floor and ribbed slab. The beams supporting the slabs are considered stiff with direction relatively very small as compared to that of slabs. Slabs supported directly on columns without beam are known as flat slabs. Slabs are designed using the same theory of bending and shear as are used for beams. The following methods of analysis are available: a) Elastic analysis-idealization into strips or beams. b) Semi empirical coefficients as given in the code, and c) Yield line theory. Slabs are analyzed and designed as having a unit width that is one meter wide strips. Compression reinforcement is used only in exceptional cases in the slab. Shear stresses are usually very low and shear reinforcement is never provided in slabs. It is preferred to increase the depth of a slab and hence reduce the shear stress rather than provide shear reinforcement. Temperature reinforcement is invariably provided at the right angles to the longitudinal reinforcement in a slab. 7.2.1 One way slab One way slabs are supported continuously on the two opposite sides so that the loads are carried along one direction only. The direction in which the load is carried in one way slab is called the span. One way slabs are usually made to span in shorter direction since the corresponding bending moment and shear forces are the least. One way slab are those in which the length is more than twice the breadth. One way slab may be simply 34

supported can be analyzed in a manner similar to that of continuous beam. The clear cover, minimum spacing of the reinforcement and minimum amount of reinforcement are provided as per IS 456: 2000 CL: 26.3-26.5. Similarly curtailments of bars are provided as per 456: 2000 CL: 26.2.3. Shear stress in slabs are generally not critical under normal loads but should be checked in accordance with requirements set out in IS 456:2000 CL40.2. the development length is checked at same critical points as for the beams. The check for deflection is a very important consideration in slab design. The strip of slab may check against span to effective depth ratio as discussed in IS456:200 CL23.2.1 7.2.2 Two way slabs When slabs are supported on four sides, two ways spanning action occurs. Such slabs may be simply supported or continuous or any other types. The deflection and bending moments in a two way slabs are considerably reduced as compared to those in one way slab. In a square slab, two way actions is equal in each direction. In long narrow slabs, when the length is greater than twice the breadth, the way action effectively reduces to one way action In the direction of short span although the end beam do carry some slab loads. A two-way slab is considered to consist of a series of inter connected beams of unit width in either direction which transfers the loads to the respective supports. The vertical deflection of the center point, common to both beams will be same. Elastic analysis of two way slabs may be based on these assumptions. The exact analysis of stress in such slabs is quiet complex. These are generally designed using approximate theories, which are known to yield satisfactory results on experimental data. The most popular method of analysis of two way slab is „yield line theory‟. In this theory, the strength of slab is assumed to be governed by flexure alone. The effects of shear and deflection are to be considered separately. It is assumed that mechanism is formed in the slab at failure. The reinforcing steel is assumed to have fully yielded along the yield line or cracks at failure. A simplified design can be carried out in accordance with IS 456:2000 CL 24.4 or by using Annex D. i.

Simply supported slabs:

A two slab, which is simply supported at its edges, tends to lift off if supports near the corner when loaded. Such slab is only truly simply supported slab. The values of bending moments used for design of such slabs can be obtained as follows: Mx = αxwlx2 My = αywlx2 Where, Mx, My = Maximum moments at mid span on strip of unit width Lx = Length of the shorter side αx, αy = Moment Coefficients can be obtained from IS 456:2000 table 27. ii. Restrained slabs

35

A slab may have its few or all edges restrained. The degree of restraints may vary depending whether it is continuous over its supports or cast monolithically with its supporting beams. A hogging or negative moment will develop in the top face of the slab at the supported sides. In these slabs the corners are prevented from lifting and provision is made for torsion. The maximum moments Mx and My at mid span on strips of unit width for spans lx, ly are given by: Mx= βxwlx2 My = βyw lx2 Where, βx ,βy= moment coefficients can be obtained from table 26 of IS 456:2000 for different arrangements of slabs.

7.3DESIGN OF SLAB (1st, 2nd, 3rd& 4th floor slab) SLAB PANEL DESCRIPTION S1

Two Adjacent Edges Discontinuous

S2

One Long Edge Discontinuous

S3

One Long Edge Discontinuous

S4

Interior panels

S5 (roof cover slab) Staircase cover slab

One Short Edge Discontinuous four Edges Discontinuous

As per IS 456-2000, ANNEX D, Clause D-1-1 The maximum bending moment per unit width in a slab are given by the following equations: Mx = xw lx2 My

= yw lx2

Value of x and yare takenfrom Table 26, IS 456-2000

36

Bending co-efficient for rectangular panels supported on four sides with provision for torsion at corners. Types of panel and moments

w

Cases

lx

Mx

My

ly

considered

Remark (KN/m2) KN-m KN-m

S1 1

Negative moments at continuous edge 2.6963 2.848

5.384 5.061 14.814

Positive moments at mid span

2.6963 2.848

4.56

3.76

7.606

3.98

6.03

3.01

S2 2

Negative moments at continuous edge 2.6963

4.678 14.814

Positive moments at mid span

2.6963

4.678

S3 3

Negative moments at continuous edge 2.848 4.678

8.65 4.445 14.814

Positive moments at mid span

2.848 4.678

6.6

4.36

S4 4

Negative moments at continuous edge 4.678

4.678

9.708 9.708 13.4

Positive moments at mid span

4.678

4.678

2.6963

2.848

7.28

7.28

7.71

7.2

Staircase cover slab 5 Positive moments at mid span

6

Roof slab S5

37

17.61

Negative moments at continuous edge 2.848

4.678

4.98

3.26

17.61 Positive moments at mid span

2.848

4.678

4.11

2.47

Design of Slab Slab S1 (Two adjacent edges discontinuous) (2.8963*3.048) Clear shorter span = 2.8963-0.3=2.5963m Effective span (lx) = 2.8963-0.3/2-0.3/2+0.1/2+0.1/2=2.6963m Effective span(ly) =3.048-0.3/2-0.3/2+0.1/2+0.1/2=2.848m As per IS-456:2000 ANNEX D, the maximum bending moment per unit width in slab are given by the following equations: Mx=αxωlx2 My=αyωlx2 Value of αx and αy are taken from table 26 IS-456:2000, bending coefficient of rectangular panels supported on four sides S.N.

1.

2.

Type of lx ly panel and moment considered Negative moment at continuous 2.6963 2.848 edge Positive moment at 2.6963 2.848 mid span

ly /lx

αx

αy

0.05

0.047

1.05

Factored Mx Load, ω

5.384

5.061

4.56

3.76

14.814 0.0377 0.035

Determination of slab thickness BM= 0.138fck bd2 5.384*106 =0.138*20*1000*d2 or, d =44.125mm Adopt effective depth as 100mm Overall depth = 120mm Calculation of steel along short span At support, BM = 0.87*415*Ast{d-415Ast/(20*1000)} 5.384*106 = 0.87*415*Ast{100-415Ast/(20*1000)} Ast = 154mm2 Hence provide 8mm υ bar @ 200mm c/c having area 251 mm2> 144mm2 (OK)

38

My

At the mid span, BM = 0.87*415*Ast{d-415Ast/(20*1000)} 4.56*106 = 0.87*415*Ast{85-415Ast/(20*1000)} Ast = 166mm2

1

Hence provide 8mm υ bar @ 200 mm c/c having area 251 mm2 Calculation of steel along longer span d‟

At support, BM 5.061*106 Ast

= d-υ/2- υ/2 =100- 4-4 = 92mm = 0.87*415*Ast{d‟-415Ast/(20*1000)} = 0.87*415*Ast{92-415Ast/(20*1000)} = 158mm2

Hence provide 8mm υ bar @ 200mm c/c having area 251mm2 At the mid span, BM = 0.87*415*Ast{d‟-415Ast/(20*1000)} 3.76*106 = 0.87*415*Ast{92-415Ast/(20*1000)} Ast =118mm2 Hence provide 8mm υ bar @ 200mm c/c having area 251mm2 Check for minimum reinforcement According to clause 26.5.2.1 of IS-456:2000 Ast = 0.12% of the total X-section area = 0.12*100*1000/100 = 120mm2 < actual reinforcement provided

Hence Ok

Check for spacing As per IS-456:2000 clause 26.3.3(b)1

spacing < 3 deff or 300mm < 3*100 or 300mm < 360 or 300mm

Check for shear Along short span at support edge, Maximum shear force at face of support (Vu)=ωlx/2 = 14.814*2.6813/2 = 17.86 KN Nominal shear stress, τv = Vu/bd = (17.86*103)/(1000*100) = 0.223 N/mm2 Percentage of steel = 100Ast/bd‟ = 100*(251/2)/(1000*92) = 0.136% 39

τc= 0.28 N/mm2 τc‟ = k τc = 1.3*0.28 = 0.364 N/ mm2 >τc>τv

Hence, ok.

Check for deflection From clause 23.2.1, IS-456:2000 L/d= αβγδλ α β δ λ

= 26 =1 =1 =1

for modification factor fs

= = =

100Ast/bd γ= d

0.58fyAst required/Astprovided 0.58*415*166/251 174N/mm2 = 0.24% 2

= Lx /αβγδλ = (2.6813*1000)/26*1*2*1*1 =55.69mm

Hence, ok.

Check for development length M1

= 0.87*415*Ast(82-415Ast/(20*1000) = 0.87*415*125.5(82-(415*125.5)/(20*1000) = 4.05KN-m According to IS- 456:2000 clause 26.2.1.1 For M20 concrete, bond stress τbd

= =

1.2*1.6 (60% is increased for deformed bar) 1.92 N/mm2

Anchorage length (Lo) Now, Ld (fs*υ)/4*τbd (0.87*415* υ)/4*1.92 υ

=

16υ

≤ ≤ ≤ ≤

1.3*(M1/Vu)+Lo 1.3*(M1/Vu)+Lo 1.3*(4.05*106/17.86*103)+16* υ 9.50 >8mm Hence

Torsional Reinforcement According to IS: 456-2000 ANNEX D(1.8 & 1.9 ) Torsional reinforcement area = 0.75*Astmaxm at midspan 40

ok.

= 0.75*251 = 188.25 mm2 Provide 8mm- υ bars – 5nos. Ast provided = 251mm2 at one direction as torsional reinforcement. Slab S2 (One long edge discontinuous) 2.8963*4.878 Clear shorter span = 2.8963-0.3=2.6963 Effective span (lx) = 2.8963-0.3+0.1=2.6963m Effective span(ly) =4.878-0.3+0.1=4.678m As per IS-456:2000 ANNEX D, the maximum bending moment per unit width in slab are given by the following equations: Mx=αxωlx2 My=αyωlx2 Value of αx and αy are taken from table 26 IS-456:2000, bending coefficient of rectangular panels supported on four sides S.N.

1.

2.

Type of lx ly panel and moment considered Negative moment at continuous 2.6963 4.678 edge Positive moment at 2.6963 4.678 mid span

ly /lx

αx

αy

Factored Mx Load, ω

0.07412 0.037 1. 78

7.606

3.98

6.03

3.01

14.814 0.056

0.028

Determination of slab thickness BM= 0.138fck bd2 7.606*106 =0.138*20*1000*d2 or, d = 52.49 mm Adopt effective depth as 100mm Overall depth = 120mm Calculation of steel along short span At support, BM = 0.87fy*415*Ast{d-415Ast/(20*1000)} 7.606*106 = 0.87fy*415*Ast{100-415Ast/(20*1000)} Ast = 220.84 mm2 Hence provide 8mm υ bar @ 150mm c/c having area 335 mm2> 144mm2 (OK)

41

My

At the mid span, BM = 0.87fy*415*Ast{d-415Ast/(20*1000)} 6.03*106 = 0.87fy*415*Ast{100-415Ast/(20*1000)} Ast = 173.24mm2 Hence provide 8mm υ bar @ 150 mm c/c having area 335 mm2 > 144mm2 (OK) Calculation of steel along longer span d‟

At support, BM 3.98*106 Ast

= d-υ/2- υ/2 = 100- 4-4 = 92mm = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} = 0.87fy*415*Ast{92-415Ast/(20*1000)} = 123.22mm2

Hence provide 8mm υ bar @ 160mm c/c having area 314 mm2> 144mm2 (OK) At the mid span, BM = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} 3.01*106 = 0.87fy*415*Ast{92-415Ast/(20*1000)} Ast =92.54mm2 Hence provide 8mm υ bar @ 160 mm c/c having area 314mm2> 144mm2 (OK) Check for minimum reinforcement According to clause 26.5.2.1 of IS-456:2000 Ast = 0.12% of the total X-section area = 0.12*120*1000/100 = 144mm2 < actual reinforcement provided

Hence ok.

Check for spacing As per IS-456:2000 clause 26.3.3(b) 1

spacing < 3 deff or 300mm < 3*120 or 300mm < 360 or 300mm

Check for shear Along short span at support edge, Maximum shear force at face of support (Vu)=ωlx/2 = 14.814*2.6963/2 = 19.97 KN Nominal shear stress, τv = Vu/bd = (19.97*103)/(1000*100) = 0.20 N/mm2 Percentage of steel = 100Ast/bd‟ = 100*157/(1000*92) = 0.17% 2 τc = 0.28 N/mm τc‟= k τc 42

= 1.3*0.28 = 0.364 N/ mm2 >τc>τv

Hence ok.

Check for deflection From clause 23.2.1, IS-456:2000 L/d= αβγδλ α β δ λ

= 26 =1 =1 =1

for modification factor fs

= = =

100Ast/bd γ=

0.58fyAst required/Astprovided 0.58*415*151/335 185 N/mm2 = 0.335% 2

d = Lx /αβγδλ = (2.696*1000)/26*1*2*1*1 =57.61 mm <100 mm (effective depth)

Hence ok.

Check for development length M1

= 0.87*415*Ast(92-415Ast/(20*1000) = 0.87*415*157(92-(415*157)/ (20*1000) = 5.03 KN-m According to IS- 456:2000 clause 26.2.1.1 For M20 concrete, bond stress τbd

= =

1.2*1.6 (60% is increased for deformed bar) 1.92 N/mm2

Anchorage length (Lo) Now, Ld (fs*υ)/4*τbd (0.87*415* υ)/4*1.92

=

16υ

Φ≤

≤ 1.3*(M1/Vu)+Lo ≤ 1.3*(M1/Vu)+Lo ≤ 1.3*(6.16*106/26.76*103)+16* υ 10.56>8mm Hence ok.

Torsional Reinforcement According to IS: 456-2000 ANNEX D(1.8 & 1.9 ) Torsional reinforcement area = 0.75*Astmaxm at midspan = 0.75*335 = 251.25 mm2 43

Provide 8mm- υ bars – 6 nos. Ast provided = 301 mm2 at one direction as torsional reinforcement.

Slab S3 (One long edge discontinuous) 3.048*4.878 Clear shorter span = 3.048-0.3=2.748 Effective span (lx) = 3.048-0.3+0.1=2.848m Effective span(ly) =4.878-0.3+0.1=4.678m As per IS-456:2000 ANNEX D, the maximum bending moment per unit width in slab are given by the following equations: Mx=αxωlx2 My=αyωlx2 Value of αx and αy are taken from table 26 IS-456:2000, bending coefficient of rectangular panels supported on four sides S.N.

1.

2.

Type of lx panel and moment considered Negative moment at continuous 2.848 edge Positive moment at 2.848 mid span

ly

4.678

ly /lx

1.625

αx

0.072

αy

Factored Mx Load, ω

My

8.65

4.445

6.6

3.36

0.037 14.814

4.678

0.055

0.028

Determination of slab thickness BM= 0.138fck bd2 8.65*106 =0.138*20*1000*d2 or, d = 55.98 mm Adopt effective depth as 100mm Overall depth = 120mm Calculation of steel along short span At support, BM = 0.87fy*415*Ast{d-415Ast/(20*1000)} 8.65*106 = 0.87fy*415*Ast{100-415Ast/(20*1000)} Ast = 252.84 mm2 Hence provide 8mm υ bar @ 150mm c/c having area 335 mm2> 144mm2 (OK)

At the mid span, 44

BM 6.6*106 Ast

= 0.87fy*415*Ast{d-415Ast/(20*1000)} = 0.87fy*415*Ast{100-415Ast/(20*1000)} = 190.31mm2

Hence provide 8mm υ bar @ 150 mm c/c having area 335 mm2 > 144mm2 (OK) Calculation of steel along longer span d‟

At support, BM 4.445 *106 Ast

= d-υ/2- υ/2 = 100-4-4 = 92mm = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} = 0.87fy*415*Ast{92-415Ast/(20*1000)} = 138.12mm2

Hence provide 8mm υ bar @ 150mm c/c having area 314 mm2> 144mm2 (OK) At the mid span, BM = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} 3.36*106 = 0.87fy*415*Ast{92-415Ast/(20*1000)} Ast =118.3 mm2 Hence provide 8mm υ bar @ 150 mm c/c having area 314mm2> 144mm2 (OK) Check for minimum reinforcement According to clause 26.5.2.1 of IS-456:2000 Ast = 0.12% of the total X-section area = 0.12*120*1000/100 = 144mm2 < actual reinforcement provided

Hence ok.

Check for spacing As per IS-456:2000 clause 26.3.3(b)1

spacing < 3 deff or 300mm < 3*120 or 300mm < 360 or 300mm

Check for shear Along short span at support edge, Maximum shear force at face of support (Vu)= ωlx/2 = 14.814*2.848/2 = 21.095 KN Nominal shear stress, τv = Vu/bd = (21.095*103)/ (1000*100) = 0.21 N/mm2 Percentage of steel = 100Ast/bd‟ = 100*157/ (1000*92) = 0.17% 2 τc = 0.28 N/mm 45

τc‟

= k τc = 1.3*0.28 = 0.364 N/ mm2 >τc>τv

Hence ok

Check for deflection From clause 23.2.1, IS-456:2000 L/d= αβγδλ α β δ λ

= 26 =1 =1 =1

For modification factor fs

= = =

0.58fyAst required/Astprovided 0.58*415*190/335 156.5 N/mm2

100Ast/bd= 0.28% γ= 2 d = Lx /αβγδλ = (2.848*1000)/26*1*2*1*1 =54.76 mm <120 mm (effective depth) Hence ok.

Check for development length M1

= 0.87*415*Ast(92-415Ast/(20*1000) = 0.87*415*157(112-(415*157)/(20*1000) = 5.03 KN-m According to IS- 456:2000 clause 26.2.1.1 For M20 concrete, bond stress τbd

= =

1.2*1.6 (60% is increased for deformed bar) 1.92 N/mm2

Anchorage length (Lo) Now, Ld (fs*υ)/4*τbd (0.87*415* υ)/4*1.92 υ Ok.

=

16υ

≤ ≤ ≤ ≤

1.3*(M1/Vu)+Lo 1.3*(M1/Vu)+Lo 1.3*(5.03*106/21.095*103)+16* υ 9.60>8mm

Torsional Reinforcement According to IS: 456-2000 ANNEX D(1.8 & 1.9 ) Torsional reinforcement area = 0.75*Astmaxm at midspan 46

Hence

= 0.75*335 = 251.25 mm2 Provide 8mm- υ bars – 6 nos. Ast provided = 301 mm2 at one direction as torsional reinforcement.

Slab S4 (Interior panel) 4.878*4.878 Clear shorter span = 4.878-0.3=4.578 Effective span (lx) = 4.878-0.3+0.1=4.678m Effective span(ly) =4.878-0.3+0.1=4.678m As per IS-456:2000 ANNEX D, the maximum bending moment per unit width in slab are given by the following equations: Mx=αxωlx2 My=αyωlx2 Value of αx and αy are taken from table 26 IS-456:2000, bending coefficient of rectangular panels supported on four sides S.N.

1.

2.

Type of lx panel and moment considered Negative moment at continuous 4.678 edge Positive moment at 4.678 mid span

ly

4.678

ly /lx

αx

αy

Factored Mx Load, ω

My

0.032

0.032

14.814

9.708

9.708

0.024

0.024

7.28

7.28

1

4.678

Determination of slab thickness BM= 0.138fck bd2 9.708*106 =0.138*20*1000*d2 or, d = 59.3 mm Adopt effective depth as 100mm Overall depth = 120mm Calculation of steel along short span At support, BM = 0.87fy*415*Ast{d-415Ast/(20*1000)} 9.708*106 = 0.87fy*415*Ast{100-415Ast/(20*1000)} Ast = 285.83 mm2 Hence provide 8mm υ bar @ 150mm c/c having area 335 mm2> 144mm2 (OK)

47

At the mid span, BM = 0.87fy*415*Ast{d-415Ast/(20*1000)} 7.28*106 = 0.87fy*415*Ast{100-415Ast/(20*1000)} Ast = 210.85mm2 Hence provide 8mm υ bar @ 150 mm c/c having area 335 mm2 > 144mm2 (OK) Calculation of steel along longer span d‟

At support, BM 9.708*106 Ast

= d-υ/2- υ/2 = 100-4-4 = 92mm = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} = 0.87fy*415*Ast{92-415Ast/(20*1000)} = 314.58mm2

Hence provide 8mm υ bar @ 160mm c/c having area 314 mm2> 144mm2 (OK) At the mid span, BM = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} 7.28*106 = 0.87fy*415*Ast{92-415Ast/(20*1000)} Ast =231.22 mm2 Hence provide 8mm υ bar @ 150 mm c/c having area 335mm2> 144mm2 (OK) Check for minimum reinforcement According to clause 26.5.2.1 of IS-456:2000 Ast = 0.12% of the total X-section area = 0.12*120*1000/100 = 144mm2 < actual reinforcement provided

Hence Ok

Check for spacing As per IS-456:2000 clause 26.3.3(b)1

spacing < 3 deff or 300mm < 3*120 or 300mm < 360 or 300mm

Check for shear Along short span at support edge, Maximum shear force at face of support (Vu) = ωlx/2 = 14.814*4.678/2 = 34.64 KN Nominal shear stress, τv = Vu/bd = (34.64*103)/(1000*100) = 0.34 N/mm2 Percentage of steel = 100Ast/bd‟ = 100*157/(1000*92) = 0.18% 48

= 0.28 N/mm2 = k τc = 1.3*0.28 = 0.364 N/ mm2 >τc>τv

τc τc‟

Hence ok

Check for deflection From clause 23.2.1, IS-456:2000 L/d= αβγδλ α β δ λ

= 26 =1 =1 =1

for modification factor fs

= = =

100Ast/bd γ=

0.58fyAst required/Astprovided 0.58*415*210.85/335 178.49 N/mm2 = 0.28% 1.95

d = Lx /αβγδλ = (4.678 *1000)/26*1*1.95*1*1 =92.26 mm <120 mm (effective depth) Hence ok

Check for development length M1

= 0.87*415*Ast(92-415Ast/(20*1000) = 0.87*415*157(92-(415*157)/(20*1000) = 5.85 KN-m According to IS- 456:2000 clause 26.2.1.1 For M20 concrete, bond stress τbd

= =

1.2*1.6 (60% is increased for deformed bar) 1.92 N/mm2

Anchorage length (Lo) Now, Ld (fs*υ)/4*τbd (0.87*415* υ)/4*1.92 υ Ok.

=

16υ

≤ ≤ ≤ ≤

1.3*(M1/Vu)+Lo 1.3*(M1/Vu)+Lo 1.3*(5.85*106/34.64*103)+16* υ 8.35>8mm

49

Hence

Torsional Reinforcement According to IS: 456-2000 ANNEX D(1.8 & 1.9 ) Torsional reinforcement area = 0.75*Astmaxm at midspan = 0.75*335 = 251.25 mm2 Provide 8mm- υ bars – 6 nos. Ast provided = 301 mm2 at one direction as torsional reinforcement.

Staircase cover slab (four edge discontinuous) 3.048 * 2.8963 Clear shorter span = 2.8963-0.3=2.5963 Effective span (lx) = 2.8963-0.3+0.1=2.6963m Effective span(ly) =3.048-0.3+0.1=2.848m As per IS-456:2000 ANNEX D, the maximum bending moment per unit width in slab are given by the following equations: Mx=αxωlx2 My=αyωlx2 Value of αx and αy are taken from table 26 IS-456:2000, bending coefficient of rectangular panels supported on four sides S.N.

1

Type of lx ly panel and moment considered Positive moment at 2.6963 2.848 mid span

ly /lx

αx

αy

Factored Mx Load, ω

My

1.05

0.06

0.056

17.61

7.2

Determination of slab thickness BM= 0.138fck bd2 7.71*106 =0.138*20*1000*d2 or, d = 52.85 mm Adopt effective depth as 100mm Overall depth = 120mm Calculation of steel along short span At the mid span, BM = 0.87fy*415*Ast{d-415Ast/(20*1000)} 7.71*106 = 0.87fy*415*Ast{100-415Ast/(20*1000)} Ast = 223.95mm2 Hence provide 8mm υ bar @ 150 mm c/c having area 335 mm2 > 144mm2 (OK)

50

7.71

Calculation of steel along longer span d‟

= d-υ/2- υ/2 = 100-4-4 = 92mm

At the mid span, BM = 0.87fy*415*Ast{d‟-415Ast/(20*1000)} 6 7.2*10 = 0.87fy*415*Ast{92-415Ast/(20*1000)} Ast =228.53 mm2 Hence provide 8mm υ bar @ 160 mm c/c having area 314mm2> 144mm2 (OK) Check for minimum reinforcement According to clause 26.5.2.1 of IS-456:2000 Ast = 0.12% of the total X-section area = 0.12*120*1000/100 = 144mm2 < actual reinforcement provided

Hence Ok

Check for spacing As per IS-456:2000 clause 26.3.3(b)1

spacing < 3 deff or 300mm < 3*120 or 300mm < 360 or 300mm

Check for shear Along short span at support edge, Maximum shear force at face of support (Vu) = ωlx/2 = 17.61*2.6963/2 = 23.74 KN Nominal shear stress, τv = Vu/bd = (23.74*103)/(1000*100) = 0.237 N/mm2 Percentage of steel = 100Ast/bd‟ = 100*157/(1000*92) = 0.17% τc = 0.28 N/mm2 τc‟ = k τc = 1.3*0.28 = 0.364 N/ mm2 >τc>τv

Check for deflection From clause 23.2.1, IS-456:2000 L/d= αβγδλ 51

Hence ok

α β δ λ

= 26 =1 =1 =1

For modification factor fs

= = =

100Ast/bd γ=

0.58fyAst required/Astprovided 0.58*415*223.95/335 170.9 N/mm2 = 0.335% 1.82

d = Lx /αβγδλ = (2.6963*1000)/26*1*1.82*1*1 =56.98 mm <100 mm (effective depth) Hence ok

Check for development length M1

= 0.87*415*Ast(112-415Ast/(20*1000) = 0.87*415*157(92-(415*157)/ (20*1000) = 5.03 KN-m According to IS- 456:2000 clause 26.2.1.1 For M20 concrete, bond stress τbd

= =

1.2*1.6 (60% is increased for deformed bar) 1.92 N/mm2

Anchorage length (Lo) Now, Ld (fs*υ)/4*τbd (0.87*415* υ)/4*1.92 υ Ok.

=

16υ

≤ ≤ ≤ ≤

1.3*(M1/Vu)+Lo 1.3*(M1/Vu)+Lo 1.3*(5.03*106/23.74*103)+16* υ 8.88>8mm

Torsional Reinforcement According to IS: 456-2000 ANNEX D(1.8 & 1.9 ) Torsional reinforcement area = 0.75*Astmaxm at midspan = 0.75*335 = 251.25 mm2 Provide 8mm- υ bars – 6 nos. Ast provided = 301 mm2 at one direction as torsional reinforcement.

Roof cover slab (S5) (One long edge discontinuous) 4.878* 3.048 Clear shorter span = 3.048-0.3=2.748 52

Hence

Effective span (lx) = 3.048-0.3/2-0.3/2+0.1/2+0.1/2=2.848m Effective span(ly) =4.878-0.3/2-0.3/2+0.1/2+0.1/2=4.678m As per IS-456:2000 ANNEX D, the maximum bending moment per unit width in slab are given by the following equations: Mx=αxωlx2 My=αyωlx2 Value of αx and αy are taken from table 26 IS-456:2000, bending coefficient of rectangular panels supported on four sides S.N.

1.

2.

Type of lx panel and moment considered Negative moment at continuous 2.848 edge Positive moment at 2.848 mid span

ly

ly /lx

4.678

αx

αy

Factored Mx Load, ω

My

0.037

10.89

4.98

3.26

4.11

2.47

0.07 1.64

4.678

0.055

0.028

Determination of slab thickness BM= 0.138fck bd2 4.98*106 =0.138*20*1000*d2 or, d = 42.47mm Adopt effective depth as 100mm Overall depth = 120mm Calculation of steel along short span At the mid span, BM = 0.87fy*415*Ast {d-415Ast/(20*1000)} 4.98*106 = 0.87fy*415*Ast {100-415Ast/(20*1000)} Ast = 155mm2 Hence provide 8mm υ bar @ 150 mm c/c having area 335 mm2 > 144mm2 (OK) At the mid span, BM 4.11*106 Ast

= 0.87fy*415*Ast {d-415Ast/(20*1000)} = 0.87fy*415*Ast {100-415Ast/(20*1000)} = 136.28mm2

Hence provide 8mm υ bar @ 150 mm c/c having area 335 mm2 > 144mm2 (OK)

Calculation of steel along longer span 53

d‟

At support, BM 3.26*106 Ast

= d-υ/2- υ/2 = 100-4-4 = 92mm = 0.87fy*415*Ast {d‟-415Ast/(20*1000)} = 0.87fy*415*Ast {92-415Ast/(20*1000)} = 100mm2

Hence provide 8mm υ bar @ 160mm c/c having area 314 mm2> 144mm2 (OK) At the mid span, BM = 0.87fy*415*Ast {d‟-415Ast/(20*1000)} 2.47*106 = 0.87fy*415*Ast {110-415Ast/(20*1000)} Ast =95 mm2 Hence provide 8mm υ bar @ 160 mm c/c having area 314mm2> 144mm2 (OK) Check for minimum reinforcement According to clause 26.5.2.1 of IS-456:2000 Ast = 0.12% of the total X-section area = 0.12*120*1000/100 = 144mm2 < actual reinforcement provided

Hence Ok

Check for spacing As per IS-456:2000 clause 26.3.3(b)1

spacing < 3 deff or 300mm < 3*120 or 300mm < 360 or 300mm

Check for shear Along short span at support edge, Maximum shear force at face of support (Vu) = ωlx/2 = 10.89*2.848/2 = 15.5KN Nominal shear stress, τv = Vu/bd = (15.5*103)/(1000*100) = 0.168 N/mm2 Percentage of steel = 100Ast/bd‟ = 100*157/ (1000*92) = 0.17% τc = 0.28 N/mm2 τc‟ = k τc = 1.3*0.28 = 0.364 N/ mm2 >τc>τv Check for deflection From clause 23.2.1, IS-456:2000 L/d= αβγδλ 54

Hence ok

α = 26 δ β =1 λ For modification factor fs

=1 =1

=0.58fyAst required/Astprovided =0.58*415*300/335 =123.32 N/mm2

100Ast/bd= 0.28% γ=2 d = Lx /αβγδλ = (2.848*1000)/26*1*2*1*1 =54.76 mm <120 mm (effectivedepth)

Hence ok.

Check for development length M1

= 0.87*415*Ast(112-415Ast/(20*1000) = 0.87*415*157(112-(415*157)/ (20*1000) = 5.0316 KN-m According to IS- 456:2000 clause 26.2.1.1 For M20 concrete, bond stress τbd

= =

1.2*1.6 (60% is increased for deformed bar) 1.92 N/mm2

Anchorage length (Lo) Now, Ld (fs*υ)/4*τbd (0.87*415* υ)/4*1.92 υ

=

16υ

≤ ≤ ≤ ≤

1.3*(M1/Vu) +Lo 1.3*(M1/Vu) +Lo 1.3*(5.0316*106/15.5*103) +16* υ 13.91>8mm Hence ok.

Torsional Reinforcement According to IS: 456-2000 ANNEX D (1.8 &1.9) Torsional reinforcement area = 0.75*Astmaxm at midspan = 0.75*335 = 251.25 mm2 Provide 8mm- υ bars – 6 nos. Ast provided = 301 mm2 at one direction as torsional reinforcement.

55

7.4DESIGN OF BEAM Design of moment for design is taken as maximum value of B.M. obtained from analysis using software STAAD. For design purpose, maximum moment of support or mid span moment is taken in consideration for a bay for a single floor or more than one floor according to quantity of variation. The mid-span of beam is designed as flanged beam and end section are designed as rectangular section.

DESIGN CONSTANT: Concrete

=M20

Steel Grade

=Fe 415

Beam size

= 400*300 mm 56

Effective cover of tensile reinforcement (d‟)

= 35 mm

Beam No. 75

Mz(kip-in) 1482

1500 1000 500 37 500 1000 1500 -1255

1

2

Fig: bending moment diagram (beam 75) Fy(kip) 28.6

30 20 10 37 10 20 -19.1 30

30 20 10 38 9.48 10 20 30 -29.3 18.7

5

Fig: shear force diagram (beam no75)

57

1420 1500 1000 500 38 500 2.89 1000 -1133 1500

END SECTION Design value: i) ii) iii) iv)

Mu:167.466KN-m Vu:130.35 KN Clear cover:25 mm Assumed diameter of bar:25 mm

Therefore, effective depth =400-25-0.5*25 =362.5 mm Ultimate moment of resistance, Mu,lim= 0.36fck b xm× 𝑑 − 0.42 ∗ 𝑥𝑚 =0.36*20*300*(0.48*362.5)(362.5-0.42*0.48*362.5) =108.775 KN-M < Mu Therefore,the section must be designed as doubly reinforced section. Calculation of reinforcement Area of tension steel corresponds to Mu,lim Mu,lim = 0.87*fy*Ast,1*(d-0.42*xm) Or, 108.775*106 =0.87*415*Ast1*(362.5-0.42*0.48*362.5) Thus, Ast1=1040.95 mm2 The remaining moment has to be resisted by a couple consisting compression steel and corresponding to the tensile steel.

𝑑 ′ /d = 37.5/362.5 =0.1 The corresponding stress in compression steel can be obtained from SP-16 (table: F),page no,13. Therefore, fsc =353 N/mm2 Now, Mu-Mu-lim= (fsc-fcc)*Asc*(d-𝑑 ′ ) (167.466-108.775)*106= (353-0.446*20) Asc (362.5-37.5) Or, Asc=524.84mm2 For equilibrium, the corresponding tension steel 58

Mu-Mu-lim=0.87fy*Ast2*(d-𝑑 ′ ) Or,58.691*106=0.87*415* Ast2*(362.5-37.5) Or, Ast2=500.17 mm2 Total tension steel (Ast) =Ast1+ Ast2=1040.95+500.17=1541.12mm2 Provide 4-25 mm diameter rods. Check for minimum reinforcement As per IS: 456-2000, clause 26.5.1.1 For minimum reinforcement 𝐴0 𝐵∗𝑑

= 0.85/fy 0.85∗300 ∗362 .5

Or, A0=

415

=222.74
Check for maximum reinforcement Ast,max = 0.04*b*D =0.04*300*400 = 4800 mm2 (ok) Curtailment of Bars Simplified rule for curtailment of bars in continuous beam according to Art. 3.12.10.2 of BS 8110 – 1985 is adopted. These rule applied to continuous beam of nearly equal spans which are designed for predominantly uniformly distributed load. First 40% of steel bars at support is curtailed at 0.25l from the center of the support and 70% of steel at mid span is curtailed at 0.15l. Check for development length As per IS: 456-2000, clause 26.2.1 Ld= 47 Φ 1.3  M1 And Ld= +Lo V Where, M1

f y  Ast    = 0.87fyAst  d  f  b ck   415 ∗1963 = 0.874151963(362.5- 20∗300 )

= 160.68 KN-m

 47 

V =130.35KN Lo = 8  + 100 = 825 +100 = 300mm  1.3  160.68  10 3     300  130.35   59

 40.47mm



OK.

Check for shear reinforcement Maximum shear occur at the face of the support. Vu = 130.35 KN Nominal shear stress, τv % of tension steel

Vu bd A 100 = st bd

=

130.35  10 3 =1.19 N/mm 2 300  362.5 1963  100 = = 1.8% 300  362.5 =

From IS: 456-2000, Table -19 Shear strength for M20 concrete, τc =0.758 N/mm 2 <τ v From IS: 456-2000, Table -20 τc max = 2.8 N/mm 2  τc max rel="nofollow">τ v >τc Hence shear reinforcement must be designed for shear value (τv-τc) bd Provide 2-legged stirrup of 8mm  bar having area 100 mm 2 . The shear that has to be contributed by the vertical stirrup Vus=(τv-τc)*b*d=(1.19-0.758)*300*362.5=46.98KN Now, spacing of vertical stirrup  0.87  f y  Asv  d   Sv =  Vus    0.87  415  100.53 * 362.5  =  =248.957 mm. 46.98 *1000  

But, as per IS: 456-2000, clause 26.5.1.5 Sv < 300 mm < 0.75d = 0.75362.5 =271.875mm >=100 mm Check for stability: Maximum spacing for vertical stirrups at support Sv,max=0.25*d=0.25*362.5=90.63 mm C/C Therefore, provide 2-legged 8mm  vertical stirrup @ 100 mm C/C up to 2d (i.e. 2362.5) from vertical face of the support

MID SECTION FLANGED BEAM (T-BEAM) Design Value: Mu

= 7.5 KN/m

Clear cover = 25 mm Assumed diameter of bar

= 25 mm

60

Now, the effective depth D

=120 mm

b

=300 mm

= 400-25-25/2

=362.5mm

Effective width of the flange As per IS: 456-2000, clause 23.1.2

For intermediate slab L bf=

=

0.7*l = 0.7*2896 =2027.2 mm

2027.2  300  6  100 = 1237.86 mm 6

Let the neutral axis lies in the flange of the beam Ultimate moment of resistance, Mu,lim = 0.36*fck*b*xm*(d - 0.42*xm) =0.36*20*300*0.48*362.5*(362.5-0.42*0.48*362.5) = 108.775 KN-m

>Mu

Calculation of reinforcement 𝑓𝑦 ∗𝐴𝑠𝑡

Mu

= 0.87*fy*Ast(d- 𝑓𝑐𝑘 ∗𝑏 )

7.5*106

=0.87*415*Ast*(362.5 – 20∗300 )

415 ∗𝐴𝑠𝑡

Therefore, Ast=57.944 mm² Provide 2– 12 mm Φ in tension having area 226 mm² Check for minimum reinforcement As per IS: 456-2000, Clause 26.5.1.1 For minimum reinforcement Ao(b*d)

=0.85/fy

Thus,Ao

=222.74


OK

Check for maximum reinforcement Ast, max

= 0.04*b*D = 0.04*300*400 =4800 mm² rel="nofollow">Ast

61

(ok)

Check for ductility: Sv,max=0.5*362.5=181.25mm Therefore provide 2-legged 8 mm diameter vertical stirrups @195mm c-c Check for deflection: l/d

=        

where,

 =26

 =1(for L<10)

Now, Fs=0.58*fy*(required steel cross section area/provided steel cross section area) =0.58*415*(57.944/226) =61.71 N/mm2 %p=100*Ast provided/(b*d) =100*226/(300*362.5) =0.207 So,  =2 (from table)

(L/d) allowable =26*1*1*2*0.8 =41.6 (L/d)actual =2896/362.5 =7.98 (OK)

62

DESIGN CONSTANT: Concrete

=M20

Steel Grade

=Fe 415

Beam size

= 400*300 mm

Effective cover of tensile reinforcement (d‟)

= 35 mm

63

Beam No. 76 Fig: bending moment diagram (beam 76) Mz(kip-in) 1500 1390 1000 500 38 500 1000 -508 1500

1390 1500 1000 500 39 15 16 500 -507 1000 1500

5.34 5 -524

10

Fy(kip) 30 24 20 10 38 10 20 30

5

30 20 10 39 15 16 10 20 -24 30

10

Fig: shear force diagram (beam no76)

END SECTION Design value: i) ii) iii) iv)

Mu:160.46KN-m Vu:106.776 KN Clear cover:25 mm Assumed diameter of bar:25 mm

Therefore, effective depth =400-25-0.5*25 =362.5 mm Ultimate moment of resistance, Mu,lim= 0.36fck b xm× 𝑑 − 0.42 ∗ 𝑥𝑚 =0.36*20*300*(0.48*362.5)(362.5-0.42*0.48*362.5)

64

=108.775 KN-M < Mu Therefore,the section must be designed as doubly reinforced section. Calculation of reinforcement Area of tension steel corresponds to Mu,lim Mu,lim = 0.87*fy*Ast,1*(d-0.42*xm) Or, 108.775*106 =0.87*415*Ast1*(362.5-0.42*0.48*362.5) Thus, Ast1=1040.95 mm2 The remaining moment has to be resisted by a couple consisting compression steel and corresponding to the tensile steel.

𝑑 ′ /d = 37.5/362.5 =0.1 The corresponding stress in compression steel can be obtained from SP-16 (table: F),page no,13. Therefore, fsc =353 N/mm2 Now, Mu-Mu-lim= (fsc-fcc)*Asc*(d-𝑑 ′ ) (160.46-108.775)*106= (353-0.446*20) Asc (362.5-37.5) Or, Asc=462.19mm2 For equilibrium, the corresponding tension steel Mu-Mu,lim=0.87fy*Ast2*(d-𝑑 ′ ) Or,51.685*106=0.87*415* Ast2*(362.5-37.5) Or, Ast2=440.467 mm2 Total tension steel (Ast) =Ast1+ Ast2=1040.95+440.467=1481.417mm2 Provide 4-25 mm diameter rods. Check for minimum reinforcement As per IS: 456-2000, clause 26.5.1.1 For minimum reinforcement 𝐴0 𝐵∗𝑑

= 0.85/fy 65

0.85∗300 ∗362 .5

Or, A0=

415

=222.74
Check for maximum reinforcement Ast,max = 0.04*b*D =0.04*300*400 = 4800 mm2 (ok) Curtailment of Bars Simplified rule for curtailment of bars in continuous beam according to Art. 3.12.10.2 of BS 8110 – 1985 is adopted. These rule applied to continuous beam of nearly equal spans which are designed for predominantly uniformly distributed load. First 40% of steel bars at support is curtailed at 0.25l from the center of the support and 70% of steel at mid span is curtailed at 0.15l. Check for development length As per IS: 456-2000, clause 26.2.1 Ld= 47 Φ 1.3  M1 And Ld= +Lo V Where, M1

 47  

f  Ast    = 0.87fyAst  d  y f ck  b   415 ∗1963 = 0.874151963(362.5- 20∗300 ) = 160.68 KN-m V =106.776KN Lo = 8  + 100 = 825 +100 = 300mm  1.3  160.68  10 3     300  106.776    48mm OK.

Check for shear reinforcement Maximum shear occur at the face of the support. Vu = 106.776 KN Nominal shear stress, τv % of tension steel

Vu bd Ast 100 = bd

=

106.776  10 3 =0.98 N/mm 2 300  362.5 1963  100 = = 1.8% 300  362.5 =

From IS: 456-2000, Table -19 Shear strength for M20 concrete, τc =0.758 N/mm 2 From IS: 456-2000, Table -20 τc max = 2.8 N/mm 2  τc max rel="nofollow">τ v >τc 66

<τ v

Hence shear reinforcement must be designed for shear value (τv-τc) bd Provide 2-legged stirrup of 8mm  bar having area 100 mm 2 . The shear that has to be contributed by the vertical stirrup Vus=(τv-τc)*b*d=(0.98-0.758)*300*362.5=24.142KN Now, spacing of vertical stirrup  0.87  f y  Asv  d   Sv =  Vus    0.87  415  100.53 * 362.5  =  =545mm. 24.142 *1000  

But, as per IS: 456-2000, clause 26.5.1.5 Sv < 300 mm < 0.75d = 0.75362.5 =271.875mm >=100 mm Spaccing of the stirrups shouldn‟t be greater than 0.75d i.e.271.875.hence, spacing is adopted as 250mm among the stirrups. Check for stability: Maximum spacing for vertical stirrups at support Sv,max=0.25*d=0.25*362.5=90.63 mm C/C Therefore, provide 2-legged 8mm  vertical stirrup @ 100 mm C/C up to 2d (i.e. 2362.5) from vertical face of the support

MID SECTION FLANGED BEAM (T-BEAM) Design Value: Mu

=59.212KN/m

Clear cover = 25 mm Assumed diameter of bar Now, the effective depth D

=120 mm

b

=300 mm

= 25 mm = 400-25-25/2

=362.5mm

Effective width of the flange As per IS: 456-2000, clause 23.1.2

For intermediate slab L

=

0.7*l = 0.7*4878 =3414.6 mm

67

bf=

3414.6  300  6  100 = 1469.1 mm 6

Let the neutral axis lies in the flange of the beam Ultimate moment of resistance, Mu,lim = 0.36*fck*b*xm*(d - 0.42*xm) =0.36*20*300*0.48*362.5*(362.5-0.42*0.48*362.5) = 108.775 KN-m

>Mu

Calculation of reinforcement 𝑓𝑦 ∗𝐴𝑠𝑡

Mu

= 0.87*fy*Ast(d- 𝑓𝑐𝑘 ∗𝑏 )

59.212*106

=0.87*415*Ast*(362.5 – 20∗300 )

415 ∗𝐴𝑠𝑡

Therefore, Ast=500.14 mm² Provide 3– 16 mm Φ in tension having area603 mm² Check for minimum reinforcement As per IS: 456-2000, Clause 26.5.1.1 For minimum reinforcement Ao(b*d)

=0.85/fy

Thus,Ao

=222.74


OK

Check for maximum reinforcement Ast, max

= 0.04*b*D = 0.04*300*400 =4800 mm² rel="nofollow">Ast

Check for ductility: Sv,max=0.5*362.5=181.25mm Therefore provide 2-legged 8 mm diameter vertical stirrups @195mm c-c Check for deflection: l/d

=        

where,

 =26

 =1(for L<10) 68

(ok)

Now, Fs=0.58*fy*(required steel cross section area/provided steel cross section area) =0.58*415*(500.14/603) =200 N/mm2 %p=100*Ast provided/(b*d) =100*603/(300*362.5) =0.554 So,  =2 (from table)

(L/d) allowable =26*1*1*1.55*0.8 =32.24 (L/d)actual =4878/362.5 =13.456 (OK)

69

DESIGN CONSTANT: Concrete

=M20

Steel Grade

=Fe 415

Beam size

= 400*300 mm

Effective cover of tensile reinforcement (d‟)

= 35 mm

70

Beam No.77 Fig: bending moment diagram (beam 77) Mz(kip-in) 1500 1417

1000 500 39 500 1000 1500 -1127

5

Fy(kip) 29.1

30 20 10 39 10 20 -18.5 30

30 20 10 40 9.51 10 20 30 -28.5 18.9

5

Fig: shear force diagram (beam no 77)

END SECTION Design value: i) ii)

Mu:166.675KN-m Vu:129.465 KN 71

1475

1500 1000 500 40 500 9.51 1000 -1248 1500

iii) iv)

Clear cover:25 mm Assumed diameter of bar:25 mm

Therefore, effective depth =400-25-0.5*25 =362.5 mm Ultimate moment of resistance, Mu,lim= 0.36fck b xm× 𝑑 − 0.42 ∗ 𝑥𝑚 =0.36*20*300*(0.48*362.5)(362.5-0.42*0.48*362.5) =108.775 KN-M < Mu Therefore,the section must be designed as doubly reinforced section. Calculation of reinforcement Area of tension steel corresponds to Mu,lim Mu,lim = 0.87*fy*Ast,1*(d-0.42*xm) Or, 108.775*106 =0.87*415*Ast1*(362.5-0.42*0.48*362.5) Thus, Ast1=1040.95 mm2 The remaining moment has to be resisted by a couple consisting compression steel and corresponding to the tensile steel.

𝑑 ′ /d = 37.5/362.5 =0.1 The corresponding stress in compression steel can be obtained from SP-16 (table: F),page no,13. Therefore, fsc =353 N/mm2 Now, Mu-Mu,lim= (fsc-fcc)*Asc*(d-𝑑 ′ ) (166.675-108.775)*106= (353-0.446*20) Asc (362.5-37.5) Or, Asc=517.768mm2 For equilibrium, the corresponding tension steel Mu-Mu,lim=0.87fy*Ast 2*(d-𝑑 ′ ) Or,57.9*106=0.87*415* Ast2*(362.5-37.5) Or, Ast2=493.43 mm2

72

Total tension steel (Ast) =Ast1+ Ast2=1040.95+493.43=1534.38mm2 Provide 4-25 mm diameter rods. Check for minimum reinforcement As per IS: 456-2000, clause 26.5.1.1 For minimum reinforcement 𝐴0 𝐵∗𝑑

= 0.85/fy 0.85∗300 ∗362 .5

Or, A0=

415

=222.74
Check for maximum reinforcement Ast,max = 0.04*b*D =0.04*300*400 = 4800 mm2 (ok) Curtailment of Bars Simplified rule for curtailment of bars in continuous beam according to Art. 3.12.10.2 of BS 8110 – 1985 is adopted. These rule applied to continuous beam of nearly equal spans which are designed for predominantly uniformly distributed load. First 40% of steel bars at support is curtailed at 0.25l from the center of the support and 70% of steel at mid span is curtailed at 0.15l. Check for development length As per IS: 456-2000, clause 26.2.1 Ld= 47 Φ 1.3  M1 And Ld= +Lo V Where, M1

 47  

f y  Ast    = 0.87fyAst  d  f  b ck   415 ∗1963 = 0.874151963(362.5- 20∗300 ) = 160.68KN-m V =129.465KN Lo = 8  + 100 = 825 +100 = 300mm  1.3  160.68  10 3     300  129.465    40.71mm OK.

Check for shear reinforcement Maximum shear occur at the face of the support. 73

Vu

= 129.465 KN

Nominal shear stress, τv

=

% of tension steel

129.465  10 3 =1.19 N/mm 2 300  362.5 1963  100 = = 1.8% 300  362.5

Vu bd Ast 100 = bd

=

From IS: 456-2000, Table -19 Shear strength for M20 concrete, τc =0.758 N/mm 2 <τ v From IS: 456-2000, Table -20 τc max = 2.8 N/mm 2  τc max rel="nofollow">τ v >τc Hence shear reinforcement must be designed for shear value (τv-τc) bd Provide 2-legged stirrup of 8mm  bar having area 100 mm 2 . The shear that has to be contributed by the vertical stirrup Vus=(τv-τc)*b*d=(1.19-0.758)*300*362.5=46.98KN Now, spacing of vertical stirrup  0.87  f y  Asv  d   Sv =  Vus    0.87  415  100.53 * 362.5  =  =227.44 mm. 46.98 *1000  

But, as per IS: 456-2000, clause 26.5.1.5 Sv < 300 mm < 0.75d = 0.75362.5 =271.875mm >=100 mm Check for stability: Maximum spacing for vertical stirrups at support Sv,max=0.25*d=0.25*362.5=90.63 mm C/C Therefore, provide 2-legged 8mm  vertical stirrup @ 100 mm C/C up to 2d (i.e. 2362.5) from vertical face of the support

MID SECTION FLANGED BEAM (T-BEAM) Design Value: Mu

= 7.5 KN/m

Clear cover = 25 mm Assumed diameter of bar Now, the effective depth D

=120 mm

b

=300 mm

= 25 mm = 400-25-25/2

74

=362.5mm

Effective width of the flange As per IS: 456-2000, clause 23.1.2

For intermediate slab L bf=

=

0.7*l = 0.7*2896 =2027.2 mm

2027.2  300  6  100 = 1237.86 mm 6

Let the neutral axis lies in the flange of the beam Ultimate moment of resistance, Mu,lim = 0.36*fck*b*xm*(d - 0.42*xm) =0.36*20*300*0.48*362.5*(362.5-0.42*0.48*362.5) = 108.775 KN-m

>Mu

Calculation of reinforcement 𝑓𝑦 ∗𝐴𝑠𝑡

Mu

= 0.87*fy*Ast(d- 𝑓𝑐𝑘 ∗𝑏 )

7.5*106

=0.87*415*Ast*(362.5 – 20∗300 )

415 ∗𝐴𝑠𝑡

Therefore, Ast=57.944 mm² Provide 2– 12 mm Φ in tension having area 226 mm² Check for minimum reinforcement As per IS: 456-2000, Clause 26.5.1.1 For minimum reinforcement Ao(b*d)

=0.85/fy

Thus,Ao

=222.74


OK

Check for maximum reinforcement Ast, max

= 0.04*b*D = 0.04*300*400 =4800 mm² rel="nofollow">Ast

Check for ductility: Sv,max=0.5*362.5=181.25mm Therefore provide 2-legged 8 mm diameter vertical stirrups @195mm c-c 75

(ok)

Check for deflection: l/d

=        

where,

 =26

 =1(for L<10)

Now, Fs=0.58*fy*(required steel cross section area/provided steel cross section area) =0.58*415*(57.944/226) =61.71 N/mm2 %p=100*Ast provided/(b*d) =100*226/(300*362.5) =0.207 So,  =2 (from table)

(L/d) allowable =26*1*1*2*0.8 =41.6 (L/d) actual =2896/362.5 =7.98 (OK)

76

77

7.5 DESIGN OF RAFT FOUNDATION If the loads transmitted by the columns in a structure are so heavy or the allowable soil bearing pressure so small that individual footing would cover more than about one half of the area, it may be better to provide a continuous footing under all the columns and walls. Such a footing is called a raft foundation. The raft is divided into series of continuous strips centered on the appropriate column rows in the both directions as shown in figure below.The shear and bending moment diagrams may be drawn using continuous beam analysis or coefficients for each strip. The depth is selected to satisfy shear requirements. The steel requirements will vary from strip. This method generally gives a conservative design since the interaction of adjacent strips is neglected. Design of Raft Foundation Total load = 11749.66 KN Load area = 11749.66/120 = 97.91 m2 Total plinth area = 125.32m2 Footing area > 50 % of plinth area provide raft foundation. Safe bearing capacity of soil, = 120 KN/m2 Total vertical column load, P = 11749.66KN Eccentricity along X direction about grid 1-1 ={2.8963(688.577+1075.388+1110.059+725.082)+7.774(726.918+1111.13+1110.735+7 26.064) +10.67(417.84+722.969+722.98+417.488)}/11749.66 = 4.8 m Therefore, ex= 10.67/2 – 4.8 = 0.535 m Eccentricity along Y direction about grid D-D ={3.048(684.582+1075.388+1111.13+722.969) +7.927(722.972+1110.059+1110.735+722.98)+10.9757(416.89+725.082+726.064+417.4 88)}/11749.66 = 5.54 m Therefore, ey=5.54- 5.487= 0.052 m Ix = 11.47 *11.77573/12

=1560.78 m4

Iy= 11.7757 * 11.473/12 = 1480.79 m4 Area (A) = 11.47*11.7757 = 135.067 m2 Mx= P * ey = 11749.66*0.052 = 610.98 KN-M My = P * ex = 11749.66*0.535 = 6286.068 KN-M P/A=11749.66/135.067=87 Soil pressure at different points is as follows 76

σ=

P MY MX  x y A IY IX

Corner A4, σ A-4 = 87+(6286.068/1480.79)*5.735+(610.98/1560.78)*5.887 = 113.64< 120 KN/M2 OK. Corner D4,0. .σ. .D.-.4. .=. 87+4.245*5.735-0.39*5.887= 87+24.345-2.295=109.05 KN/M2 Corner A1, σ A-1 =87-24.345+2.295 = 64.95 KN/M2 Corner D1, σ D-1 = 87-24.345-2.295 = 60.36 KN/M2 Grid B4, σ B-4 = 87+24.345+0.951= 112.296 KN/M2 Grid B1, σB-1 = 87-24.345+0.951= 63.606 KN/M2 Grid C4, σc-4 = 87+24.345-0.951 = 110.394 KN/M2 Grid C1, σc-1= 87-24.345-0.951 = 61.704 KN/M2 In X – direction, the raft is divided into 4 strips, i.e. 3 equivalent beams: i) ii) iii) iv)

Beam A-A soil pressure, σ = 113.64 KN/m2 Beam B-B soil pressure, σ = 0.5*( 113.64+112.296) = 112.818 KN/m2 Beam C-C soil pressure, σ = 0.5(110.394+112.296) = 111.345 KN/m2 Beam D-D soil pressure, σ = 0.5*(110.394+109.05) = 109.722 KN/m2

-m /For X – direction, the bending moment is obtained by using a coefficient of 1/10 and L as Centre of column distance, w L2 +M = -M = 10

For strip A-A, maximum bending moment = For strip B-B, maximum bending moment =

113.64∗4.878 2 10

= 270.405 KN-m /m

112.818∗4.878 2

77

10

= 268.449 KNm

For strip C-C, maximum bending moment = For strip D-D, maximum bending moment =

111.345∗4.878 2 10 109.722∗4.878 2 10

= 264.944 KN-m /m

=261.082KN-m /m

In Y direction, Grid A3, σA-3 = 87+(4.245)*(5.735-0.4-2.8963)+(0.39)*5.8878 = 99.64 KN/M2 Grid A2, σA-2 = 87-4.245*2.4387+0.39*5.8878 = 78.946 KN/M2 In Y – direction, soil pressure (i) Beam 4-4, σ = 113.64 KN/m2 (ii) Beam3-3,σ = 0.5*(113.64+99. 64) = 106.64 KN/m2 (iii) Beam 2-2, σ = 0.5*(99.64+78.946) = 89.293 KN/m2 (iv) Beam 1-1, σ = 0.5*(78.946+64.95) = 71.948 KN/m2 For Y – direction, the bending moment is obtained by using a coefficient of 1/10 and L as centre of column distance, w L2 +M = -M = 10

For strip 4-4, maximum bending moment = For strip 3-3, maximum bending moment = For strip 2-2, maximum bending moment = For strip 1-1, maximum bending moment =

113.64∗4.878 2 10 106.64∗4.878 2 10 89.293∗4.878 2 10 71.948∗4.878 2 10

78

= 270.405 KN-m /m =253.748 KN-m /m.

= 212.471 KN-m /m = 171.199 KN-m /m

The depth of the raft will be governed by two way shear at one of the exterior columns. In case location of critical shear is not obvious, it may be necessary to check all possible locations. Shear strength of concrete,′c = c =0.25

𝑓𝑐𝑘 = 0.25 20 = 1.11 N/mm2

For a corner column D1;

Perimeter b0 =2(0.5d+600) = d+1200 mm 1.5∗369.981∗1000 v = Vu/ (b0d) = 𝑑+1200 𝑑 1.5∗369.981∗1000 Therefore, 1.11 = d+1200 d

79

On solving we get, d = 327.34 mm. For an exterior column say A2,

Perimeter, b0 = (d+400) + 2(0.5d+600) = 2d +1600 Therefore, 1.11 =

1.5∗725.082∗1000 2d+1600 d

On solving we get, d = 406.176 mm So, adopt effective depth, d = 500 mm and overall depth, D = 500 +50 + 10/2 =555 mm Reinforcement calculation: Reinforcement along long direction A-A

80

Maximum bending moment in long direction is given by A-A 𝑓𝑦 ∗𝐴𝑠𝑡 BM = 0.87 fyAst (d ) 𝑓𝑐𝑘𝑏

270.405 * 106 = 0.87 * 415 * Ast(500 -

415∗𝐴𝑠𝑡

20∗1000 2 Therefore, we get Ast =1604.75 mm /m

)

Minimum reinforcement = 0.12 % of b D = 0.12/100 * 555 * 1000 = 666 mm2 /m < 1604.75 mm2 /m (okay) Minimum steel will govern in the remaining shaft So, provided 16 mm dia bar @ 120 mm c/c Ast,provided = 1675 mm2 In both direction at top and bottom. Reinforcement along short direction 4-4 Maximum bending moment in short direction 𝑓𝑦 ∗𝐴𝑠𝑡 BM = 0.87 fyAst (d ) 𝑓𝑐𝑘𝑏

270.405 * 106 = 0.87 * 415 * Ast(500 -

415∗𝐴𝑠𝑡

20∗1000 2 Therefore, we get Ast =1604.75 mm /m

So, provided 16 mm dia bar @ 120 mm c/c Ast,provided = 1675 mm2 In both directions at top and bottom.

Check for development length, Required development length for tension Ld Ld= (0.87*fy*𝜑)/4𝜏bd =(0.87*415*16)/4*1.2*1.6 = 752.187 mm The available vertical length L1 for anchorage is L1 = (555-8-50-16-8) = 473. < 905.11 mm OK.

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7.6 Column 3 PAGES

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7.7DESIGN OF STAIRCASE Staircase is an inclined structural system for the movement from one level to another. Since it is stepped, it is called staircase. A staircase behaves like an ordinary slab. It may span either in the direction of the steps or in the direction of going. Structurally staircase may be classified largely into two categories, depending on the predominant direction in which the slab staircase component of the stair undergoes flexure- stair slab spanning transversely and stair slab spanning longitudinally. The design of staircase requires proportioning of its different components and determination of reinforcement and its detailing to satisfy both the serviceability and strength requirements. The design of staircase is made for serviceability requirements of deflection and cracks. The serviceability requirement of deflection is controlled by the effective span to effective depth ratio. The design of reinforcement is made to satisfy the strength requirements for moments and shear. The design for moment is made for maximum moments either by working stress method or by the limit state method. The area of steel is expressed as diameter and spacing of bars. It is provided along the span of staircase and necessary curtailment is made wherever it is not required as in the case of edge supported slab. Generally the shear reinforcement is not required in the staircase as the shear strength of concrete is much greater than the nominal shear stress. The shear strength of concrete in staircase is determined as in the case of edge supported slabs. The detailing of reinforcement in staircase shall be similar to that of the edge supported slab except at the junction of landing and flight of staircase where it should ensure that the reinforcement bars in tension tending to straighten out do not cause cracking in concrete. General rules 

Between consecutive floors there should be equal rise for every parallel steps. Similarly there should be equal going.



There should be at least 2m headroom measured vertically above any steps.



The sum of going of a single step plus the twice the rise should be between 550mm and 700mm



The rise of steps should not be more than about 200mm and the going not less than 240mm



The slope of the staircase should be not more than 38 degrees.



Width of staircase depends upon the usage. The width required in residential building differs from other public building.

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7.7.1DESIGN OF STAIRCASE Size of staircase Total floor to floor height=2.74m 2R+T=600 R=170mm No. of riser=2.74/0.17=16no.s No. of tread = 16-1=15no.s Width of staircase= 1.06m Along Y-Y axis, 3.048+0.4=3.448m 3.448-0.3-0.3=2.848m 2.848-1.3-1.060=0.488m Beam c/c distance 0.3/2+1.060+1.3+0.488+0.3/2 = 3.148m Leff= 3.148m Deff=Leff/35 = 3148/35 = 110.514m Overall depth of waist slab (D) =99.94+20= 119.20mm Let us take D=150mm

A) Inclined flight section a.Dead load Step Section =1/2*0.26*0.17 =0.0221m2 Inclined Slab= (0.262+0.172) *0.15 =0.04659m2 Now total X-section area of slab only=0.0221+0.04659 =0.0686m2 Density of concrete=25KN/m3

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Load on RCC slab=25*0.0686 =1.717KN/m3

Step finishing Density of screeding=20.4KN/m3 Density of marble =26.7KN/m3 Marble 20mm thick= (0.26+0.17)*0.02*26.7 =0.22962KN/m3 Plastering 12.5 mm thick= (0.262+0.172) *0.0125*20.4=0.0792 KN/m3 Screeding 25mm thick= (0.26+0.17)*0.025*20.4=0.2193KN/m3 Total dead load of step=2.24512KN/m Now dead load per m2 on plan=2.24512/0.26=8.635KN/m2

b) Live load Live load per m2 on plan =5KN/m2 Total load = Dead load + live load=8.635+5=13.635KN/m2 Factored load =1.5*13.635=20.4525 KN/m2 Taking 1.06m width of slab, total load=1.06*20.4525 = 21.67965 KN/m

B) Landing a. Dead load Waist slab (W)=0.15*25=3.75KN/m2 Marble finishing 20mm thick=0.020*26.7=0.534KN/m2 Thick plastering 12.5mm=0.0125*20.4KN/m2=0.2448KN/m2 Screeding 25mm thick=0.025*20.4=0.51 KN/m2 Total dead load=3.75+0.534+0.2448+0.51=5.038 KN/m2 b. live load=5KN/m2 Total load=5+5.038 =10.038 KN/m2 Factored load=1.5*10.038 = 15.057 KN/m2

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Taking 1.06m width of slab=15.057*1.06=15.96 KN/m 1) Calculation of bending moment Ra+Rb=15.96*0.5013+21.679*1.3+15.96*1.095 Ra+Rb= 53.6562 KN/m2

…………………..(1)

21.679 KN/m 15.96 KN/m

15.96 KN/m

0.5013m

1.3 m

1.095m

Fig: Loading diagram of Staircase Moment at A 15.96*0.5013*(0.5013/2) +21.679*1.3*(1.3/2+0.5013)+15.96*1.095 ((1.095/2) +1.3+0.5013) =Rb*(0.5013+1.3+1.095) Rb= 26.06 KN, Ra= 53.599-26.06 = 27.539KN If the point of zero SF at a distance „x‟ from A 27.539=15.96*0.5013+21.679*(x-0.5013) X= 1.4m Maximum bending moment occurs at a distance „x‟ from A(i.e.1.4 m from the support A) Mx=Ra*1.4-15.96*0.5013(0.5013/2+1.4-0.5013)+21.679 (1.4-0.5013/2)2 =27.539*1.4-15.96*0.5013(0.5013/2+1.4-0.5013)+21.679 (1.4-0.5013/2)2

= 58 KN-m II) Calculation of depth Minimum depth can be calculated considering absolute maximum bending moment We know,

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Effective depth (d) = BMmaximum/0.138*b*fck (d)= 58*106/0.138*20*1060 =140 D= 140 + 20 + 10/2 ` Adopt D=165mm Available effective depth (d) =165-20-10/2 = 140mm So, D=165mm And d=140mm III) Area of tension steel We know,` Mu=0.87*fy*Ast(d-(fy*Ast)/(fck*b)) 58*106=0.87*415*Ast*(140-(415*Ast)/ (20*1060)) Ast=1435.63 mm2 From code,Ast=1470 mm2 𝜑=12mm No. of bars = 13+ 1 = 14 nos. And, spacing of the 12mm𝜑 reinforcement =1470 / 14 = 105 mm = 105mm IV) Temperature reinforcement/Distribution bar Provide 1 nos-10mm 𝜑 bar as temperature reinforcement in each riser In waist slab,(provide minimum) 0.12%of bD So, Ast minimum=0.12*1060*165/100 = 209.88 mm2 Using 8mm 𝜑bar,@220mm c/c. Ast provided=228 mm2 Check for shear, Nominal shear stress,𝜏v=vu/bd=28.8409*103/1060*145 =0.185 N / mm2 Percent of tension steel= (100*Ast) / bd = (100*1470)/ (1060*140) =0.99 88

From clause 40.2 .1 IS 456: 2000 table 19 Interpolating Tc = 0.617 N/mm2 Shear strength for slab 𝜏c‟=k𝜏c From code, IS456:2000 for overall depth of slab (mm) K=1.25 𝜏c‟= 1.25*0.6 N/mm2 =0.77N/mm2>𝜏v(ok) Deflection 𝑙

d=𝛼∗𝛽 ∗𝛾𝑥 ∗𝛿∗𝜆 α = 26, β =1, δ =1, λ=1 Now, for modification factor „γ‟ for tension reinforcement fs = 0.58*fy*(Area of x-section of steel required) /(Area of x-section of steel provided) =0.58*415 *1435.63/1470 = 235.07 N/mm2 Therefore, γ = 1.18 Therefore, d =99.34 mm ≤ 140 mm (ok) Check for development length Development length for 12mm 𝜑bar Ld=47𝜑 =47*12=564mm =600mm Ld=0.87*fy*𝜑/4𝜏bd For M20,𝜏bd=1.2N/mm2 Ld=0.87*415*12/4*1.2*1.6 =564.14mm=600mm Moment of resistance M1=0.87*fy*Ast*(d-(fy*Ast)/(fck*b)) =0.87*415*1470*(140-(415*1470)/ (20*1060)) = 59.03*106 89

Hence, Ld< 1.3M1/v + LO 0.87∗415 ∗𝜙 4∗1.2∗1.6

59.03∗10 6

=1.3 *27.539∗10 3 + 16 *𝜙

Therefore, 𝜙=89.85 mm Since bar diameter provided is 12mm ok. Along x-x axis, Leff = 1.134/2+0.78+1.134/2 = 1.914m 1) Calculation of bending moment Ra+Rb=15.96*0.567+21.679*0.78+15.96*0.567 Ra+Rb= 35 KN/m2

…………………..(1)

21.679 KN/m 15.96 KN/m

15.96 KN/m

0.567 m

0.78m

0.567 m

Fig: Loading diagram of Staircase (landing to landing) Moment at A 15.96*0.567*(0.567/2) +21.679*0.78*(0.78/2+0.567)+ 15.96*0.567 ((0.567/2) +0.78+0.567) =Rb*(1.914) Rb= 17.50 KN Ra= 35-17.50=17.50 KN If the point of zero SF at a distance „x‟ from A 17.5=15.96*0.567+21.679*(x-0.567) X= 0.956 m Maximum bending moment occurs at a distance „x‟ from A Mx=Ra*0.956-15.96*0.567(0.567/2+0.956-0.567)+21.679 (0.956-0.567/2)2 =17.5*0.956-15.96*0.567(0.567/2+0.956-0.567)+21.679 (0.956-0.567/2)2 90

= 20.449 KN-m

II) Calculation of depth Minimum depth can be calculated considering absolute maximum bending moment We know, Effective depth (d) = BMmaximum/0.138*b*fck (d)= 20.449*106/0.138*20*1060 =83.60 mm D= 83.60 + 20 + 10/2 Adopt D=110mm Available effective depth (d) =110-20-10/2 = 85mm So, D= 110mm And d= 85mm III) Area of tension steel We know, Mu=0.87*fy*Ast(d-(fy*Ast)/(fck*b)) 20.449*106=0.87*415*Ast*(85-(415*Ast)/ (20*1060)) Ast=821.89 mm2 From code,Ast=904 mm2𝜑=12mm No. of bars = 8+ 1 = 9 nos. And, spacing of the 12mm𝜑 reinforcement =821 / 6 = 90mm IV) Temperature reinforcement/Distribution bar Provide 1 nos-10mm 𝜑 bar as temperature reinforcement in each riser In waist slab,(provide minimum) 0.12%of bD So, Ast minimum=0.12*1060*110/100 = 139.92 mm2 Hence for the staircase design take the data along y-y axis for design.

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CHAPTER 8 CONCLUSION The project work entitled “DETAILED STRUCTURAL ANALYSIS AND DESIGN OF FIVE STOREY RESIDENTIAL BUILDING” was carried out by the group of four students. The analysis of the building was done using appropriate method of analysis. The design of elements was done using limit state design philosophy which is economic, safe and reliable. The detailing of structure was done as per IS and IS seismic codes. The Software used in this project is STAAD Prov8i. Project has indeed widened our perspective and acquainted us on how to perceive and counteract the worst possible difficulties regarding the analysis and design of five storied structures. This project work has mainly focused towards the structural analysis and design only. Nevertheless, the attempts have been made in the architectural planning and for the presentation of the analysis and design results in the tabular form with necessary drawing and details. The principle and methodology applied while analyzing and designing the five storey frame structure in this project is universally valid for any type of the framed structures of five storied in peri – urban areas of Kathmandu valley. A constant painstaking study and devotion to the work by the project group couple with the valuable guidance of the advisor made it possible in bringing up the project work to this level. The purpose of the project, through purely academic oriented, we had made every effort to make it feasible for the real construction of the site. Finally, we will consider our project report as successful if it process to be useful to the junior students or other designers. We hope that project work will prove much useful to us in our career.

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BIBLIOGRAPHY AND REFERENCES 1. Jain, A.K, "Reinforced concrete (Limit State Design) ", Nemchand and Bros, 5th Edition 1990 2. Ramamrutham, S., “Design of Reinforced Concrete Structure”, DhanpatRai Publishing Company, 11th Edition 1989 3. Varghese, P, C. "Limit State design of reinforced concrete", Princeton Hall of India 1996 4. Sinha, S. N., "Reinforced Concrete design", Tata McGraw - Hill, 2nd Edition 1996 5. Reddy, C S, “Basic Structural Analysis”, Tata McGraw – Hill, 3rd Edition 2011 6. I. S 875 (part - I) 1987, Code of practice for design loads (other than Earthquake) for building and structures, dead loads 7. I.S 875 (part 2nd) 1987, code of practice for design loads (other than Earthquake) for the building and structures, Dead loads. 8. I.S 1893- 1975 Criteria for Earthquake Resistant Design of structures. 9. I.S: 1893-1975 and IS 4326-1976, Explanatory Handbook on codes for Earthquake Engineering. 10. Design Aids for RCC to I.S 456-1978, SP 16:1980 11. I.S. 456-2000 Indian Standard plain and RC code of practice (fourth revision) 12. I.S. 1893 (part I):2002.

APPENDICES

Contents CHAPTER 1

INTRODUCTION ................................ Error! Bookmark not defined.

1.1 BACKGROUND ................................................... Error! Bookmark not defined. 1.2 OBJECTIVE.......................................................... Error! Bookmark not defined. 1.3 RATIONALE OF THE PROPOSED PROJECT .... Error! Bookmark not defined. 1.4 DESCRIPTION OF THE PROPOSED PROJECT . Error! Bookmark not defined. 1.5 BRIEF DESCRIPTION OF THE PLAN ................ Error! Bookmark not defined. 1.6 SCOPE OF THE PROJECT................................... Error! Bookmark not defined. CHAPTER 2

LITERATURE REVIEW ..................... Error! Bookmark not defined.

2.1 GENERAL ............................................................ Error! Bookmark not defined. 2.2 TYPES OF BUILDING ......................................... Error! Bookmark not defined. 2.3 COMPONENTS OF BUILDING ........................... Error! Bookmark not defined. CHAPTER 3

METHODOLOGY ............................... Error! Bookmark not defined.

3.1 LOAD CALCULATION ....................................... Error! Bookmark not defined. 3.2 PRELIMINARY DESIGN ..................................... Error! Bookmark not defined. 3.3 LOADING PATTERNS ........................................ Error! Bookmark not defined. 3.4 ANALYSIS ........................................................... Error! Bookmark not defined. 3.5 METHODS AND TOOLS FOR ANALYSIS ........ Error! Bookmark not defined. 3.6 DATA ................................................................... Error! Bookmark not defined. 3.7 DESIGN METHOD............................................... Error! Bookmark not defined. CHAPTER 4

PRELIMINARY DESIGN .................... Error! Bookmark not defined.

4.1 PRELIMINARY DESIGN OF SLAB .................... Error! Bookmark not defined. 4.2. PRELIMINARY DESIGN OF BEAM .................. Error! Bookmark not defined. 4.3 LOAD DESCRIPTION ......................................... Error! Bookmark not defined. 4.3.1 Dead load (DL) ............................................... Error! Bookmark not defined. 4.3.2 Live Load (LL) ............................................... Error! Bookmark not defined. 4.4 VERTICAL LOAD CALCULATION ................... Error! Bookmark not defined. 4.4.1. Load calculation for roof slab above staircase Error! Bookmark not defined. 4.4.2. Load calculation for roof slab terrace ............. Error! Bookmark not defined. 4.4.3. Load calculation for intermediate slab (1, 2, 3 & 4)Error! defined.

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4.4.4. Load calculation for slab S3 ........................... Error! Bookmark not defined. 4.4.5. Load calculation for projection slab................ Error! Bookmark not defined. 4.4.6. Load calculation for verandah slab ................. Error! Bookmark not defined. 4.5 PRELIMINARY DESIGN OF CRITICAL COLUMNError! defined.

Bookmark

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4.5.1 Dead load calculation ...................................... Error! Bookmark not defined.

4.5.2 Load Calculation ............................................. Error! Bookmark not defined. CHAPTER 5

LATERAL ANALYSIS ........................ Error! Bookmark not defined.

5.1 INTRODUCTION TO LATERAL LOAD ............. Error! Bookmark not defined. 5.2 CALCULATION OF SELF-WEIGHT OF ELEMENTSError! defined.

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5.2.1 Self-weight of beam ........................................ Error! Bookmark not defined. 5.2.2 Self-weight of tie beam ................................... Error! Bookmark not defined. 5.2.3 Self-weight of column ..................................... Error! Bookmark not defined. 5.2.4 Self-weight of brick wall ................................. Error! Bookmark not defined. 5.3 LUMP MASS CALCULATION ............................ Error! Bookmark not defined. 5.3.1 Lumped load of slab ........................................ Error! Bookmark not defined. 5.3.2 Lumped load of beam ................................................................................... 23 5.3.3 Lumped load of column ................................................................................ 24 5.3.4 Lumped load of wall ..................................................................................... 24 5.3.5 Lump load of staircase .................................................................................. 25 5.4 DETERMINATION OF BASE SHEAR .............................................................. 25 5.4.1 Calculation of lateral forces and shear at storey ............................................. 28 5.4.2 Calculation of nodal mass applied force ........................................................ 28 5.5 LOAD COMBINATION ..................................................................................... 30 CHAPTER 6

STRUCTURAL ANALYSIS .............................................................. 31

6.1 SALIENT FEATURE OF STAAD PRO .............................................................. 31 6.2 INPUT AND OUTPUT ....................................................................................... 31 6.3 METHODOLOGY OF ANALYSIS .................................................................... 31 CHAPTER 7

STRUCTURAL DESIGN ................................................................... 33

7.1 LIMIT STATE METHOD ................................................................................... 33 7.2 SLAB .................................................................................................................. 34 7.3 DESIGN OF SLAB (1st, 2nd, 3rd& 4th floor slab) .................................................. 36 7.4 DESIGN OF BEAM ............................................................................................ 56 7.5 DESIGN OF RAFT FOUNDATION ................................................................... 76 7.6 Column.................................................................................................................... 82 7.7 DESIGN OF STAIRCASE .................................................................................. 84 CHAPTER 8

CONCLUSION .................................................................................. 92

BIBLIOGRAPHY AND REFERENCES APPENDICES