Analysis & Design of Multistorey Building

“ANALYSIS AND DESIGN OF MULTISTOREY BUILDING”A PROJECT REPORT Submitted in partial fulfillment of the requirements for the award of the degree of BACHELOR OF TECHNOLOGY in CIVIL ENGINEERING By AVINASH SHARMA DHRUV GUPTA GAURAB PAUL (1010930013) (1010930017) (1010930018) Under the guidance of Mr. PRADEEP KUMAR DEPARTMENT OF CIVIL ENGINEERING SRM INSTITUTE OF MANAGEMENT AND TECHNOLOGY SRM UNIVERSITY – NCR CAMPUS, GHAZIABAD, U.P., INDIA May, 2013 SRM INSTITUTE OF MANAGEMENT AND TECHNOLOGY SRM UNIVERSITY – NCR CAMPUS, GHAZIABAD, U.P., INDIA DEPARTMENT OF CIVIL ENGINEERING CANDIDATE’S DECLARATION I hereby certify that the work which is being presented in the thesis entitled, “ANALYSIS AND DESIGN OF MULTISTOREY BUILDING” in partial fulfillment of the requirements for the award of the degree of Bachelor of Technology in Civil Engineering at SRM Institute of Management and Technology, NCR Campus, Ghaziabad is an authentic work carried out during a period from January, 2013 to May 2013 under the supervision of Mr. Pradeep Singh. The matter embodied in the thesis has not been submitted to any other University/Institute for the award of any Degree or Diploma. (Avinash Sharma) (Dhruv Gupta) (Gaurab Paul) Prof. (Dr.) Manoj Kumar Pandey (Director) Dr. Vineet Bajaj (Head of Department) Mr. Pradeep Kumar (Project Guide) (Project Co-ordinater) (External Examiner) ACKNOWLEDGEMENT I would like to express my gratitude to all the people behind the screen who helped me to transform an idea into a real application. I profoundly thank Dr. Vineet Bajaj, Head of the Department, Civil Engineering who has been an excellent guide and also a great source of inspiration to my work. I would like to thank my guide, Mr. Pradeep Kumar, Asst. Professor, for his technical guidance, constant encouragement and support in carrying out my project at college. I would like to thank Mr. Ashoka Kumar, Staad Pro Expert from Bentley, for his valuable guidance in whenever requirement for the successful fulfillment of my project needs. I wish to thank Er. Naveen Kumar Singh, Structural Consultant, for his valuable guidance in the practical aspects related to the project. The satisfaction and euphoria that accompany the successful completion of the task would be great but incomplete without the mention of the people who made it possible with their constant guidance and encouragement crowns all the efforts with success. In this context I would like to thank my friends who supported me in successfully completing this project. Thanking You. AVINASH SHARMA 1010930013 DHRUV GUPTA 1010930017 GAURAB PAUL 1010930018 columns. slabs and footings is obtained. In the limit state of collapse. The conclusion of this study is that the design parameters of a multi-storey building are successfully construed and Staad Pro is a very powerful tool which can save much time and is very accurate in Designs. STAAD Pro with its new features surpassed its predecessors and compotators with its data sharing capabilities with other major software like AutoCAD. The project focuses on „Reinforced Concrete‟ buildings. columns. as a Civil Engineering student one needs to be fully aware of the Structural elements and their safety parameters before and during the execution of the project. The design study comprises of the footing. consists of floor system which transfers the floor loads to a set of plane frames in one or both directions. As a sequel to this an attempt has been made to learn the process of analysis and design of a multistorey Building using Limit State Method (IS 456:2000). The guidelines being followed are as per IS 456:2000 and IS 13920 : 1993. beams and slabs. NCR Campus. the strength and stability of structure is ensured. The loadings are applied and the design for beams. The design using Limit State Method (of collapse and serviceability) is taken up. The present project deals with the analysis of a multi-storey residential hostel building of G+9 consisting of 22 rooms in each floor at SRM University. The structural components in a typical multi storey building. and MS Excel. .ABSTRACT In this growing world. 5 Staad Pro V8i Alternatives for Staad Pro Staad Editor Staad Foundation V8i Auto Cad PLAN & ELEVATION CHAPTER – 4 4.1 4.2 3.1 2.3 3.CONTENTS List of Tables List of Figures Assumptions and Notations Symbols CHAPTER – 1 CHAPTER – 2 2.2 Plan Elevation LOADS CHAPTER – 5 .2 2.4 3.1 3.4 i ii-iii iv-v vi-vii INTRODUCTION LITERATURE SURVEY 1-2 3-12 4 7-9 9-11 12 13-17 14 15 15 16 17 18-20 19 20 21-38 Elements of Structural Design Design Philosophies Multi-Storey Building Structural Planning COMPUTER AIDED ANALYSIS & DESIGN CHAPTER – 3 3.3 2. 5 5.4 6.3 7.1 Seismic Loading in Staad Pro V8i 22 22-23 24-30 31-35 32-33 35-36 37-38 39-54 40-42 43 43 44-47 48-54 55-105 56 57-63 64-71 72-86 87-105 106-108 5.4 Load Conditions and Structural System Response Building Loads Categorized by Orientation Design Load for the Residential Building Design Imposed Loads for Earthquake forces Calculation 5.4 Beams Columns Slabs Foundation CONCLUSION .1 5.6 Load Combinations Inputs to Staad Editor for Loadings ANALYSIS CHAPTER – 6 6.1 6.5 Methods of Analysis Seismic Analysis Procedure Analysis using Staad Pro V8i Analysis Results for Load Cases 1 to 4 Analysis Results for Support Reactions DESIGN CHAPTER – 7 Input to Staad Editor for Design 7.3 6.5.2 6.4.3 5.2 7.1 7.2 5. APPENDICES APPENDIX A APPENDIX B REFERENCES 109 110 111 . 5 Zone Factor Dimensions of Continuous Strip Footing Design Results of Foundation Applied Loads-Allowable Stress Level Calculated Pressure at Four Corners Check for Stability against Overturning Title … … … … … … … Page No.LIST OF TABLES Table No.1 7.2 7. 5.3 7.4 7. 30 92 93 95 96 96 i .1 7. 1 in the Structure Beam Reinforcement Beam Web Reinforcement Skeleton Structure showing Column No. Beam & Column Plan showing Slabs Detailing of Slabs … … … … … … … … … … … … … … … … … … … Page No.1 5.3 7.7 7.10 7.4 5.2 7. 25 27 33 34 35 59 60 61 68 70 72 73 73 74 74 75 76 86 ii .1 7.8 7.11 7.5 7. 5.13 Title Dead Load on the Structure Live Load on the Structure Seismic Parameters Seismic Load in X direction (SLX) Seismic Load in Z direction (SLZ) Location of Beam No.12 7.6 7.3 5.LIST OF FIGURES Figure No.5 7. 1539 One Way Slab Load Distribution in a One Way Slab Two Way Slab Load Distribution in a Two Way Slab Load Distribution showing One Way and Two Way Monolithic connection between Slab. 1539 Shear Bending for Column No.2 5.4 7.9 7. FC3 Strip Footing. FC4 Strip Footing.20 7.22 7.25 A-1 A-2 Staad Foundation Page showing Foundation Zoom View of Foundation Concrete and Rebar Parameters Cover and Soil Parameters Footings Dimensions Plan of Footings Elevation of Footings Strip Footing.7.21 7.14 7. FC5 Plan of the Multistorey SRM Hostel Building Elevation of the Multistorey SRM Hostel Building … … … … … … … … … … … … … … 89 89 90 90 91 102 102 103 103 104 104 105 109 110 iii . FC2 Strip Footing.16 7.23 7.19 7.15 7.24 7. FC1 Strip Footing.17 7.18 7. Therefore:- iv .0KN/m3 20.ASSUMPTIONS AND NOTATIONS The notations adopted throughout the work are same IS-456-2000.0KN/m3 5.0KN/m3 4.L+E. D.4.0KN/m2 4. columns. 875-86 i) Live load on slabs ii) Live load on passage iii)Live load on stairs 20.L 1.Partial safety factor for material in accordance with clause 36.15 for steel.2 is IS-456-2000 is taken as 1. 3.L+L.0KN/m3 19. Assumptions in Design: 1.5 2.4 of IS-456-2000 as ϒt=1.L+L.LIVE LOADS: In accordance with IS.5 for concrete and 1.Using partial safety factors in accordance with clause 36. slabs.2 Density of materials used: MATERIAL: DENSITY i) Plain concrete ii) Reinforced iii) Flooring material (c. D.m) iv) Brick masonry v) Fly ash 24.0KN/m3 25.4 of IS-456-2000 combination of load.Using partial safety factor for loads in accordance with clause 36.0KN/m2 DESIGN CONSTANTS: Using M30 and Fe 415 grade of concrete and steel for beams. footings.L.5 1.0KN/m2 4. fck fy Characteristic strength for M30-30N/mm2 Characteristic strength of steel-415N/mm2 Assumptions Regarding Design: i) Slab is assumed to be continuous over interior support and partially fixed on edges. due to monolithic construction and due to construction of walls over it. 3) For steel Fe 415 and steel is used for the distribution reinforcement. 2) For steel Fe 415 is used for the main reinforcement. v . Assumptions on design:1) M20 grade is used in designing unless specified. ii) Beams are assumed to be continuous over interior support and they frame in to the column at ends. 4) Mild steel Fe 230 is used for shear reinforcement. M. Mu Md Mf Mx My Mx My pt W Wd Tc max Tv Area Area of steel Breadth of beam or shorter dimension of rectangular column Overall depth of beam or slab Dead load Effective depth of slab or beam Overall depth of beam or slab Moment of resistance factor Characters tic compressive strength Characteristic strength of of steel Devlopment length Live load Length of shorter side of slab Length of longer side of slab Bending moment Factored bending moment Design moment Modification factor Mid span bending moment along short span Mid span bending moment along longer span Support bending moment along short span support bending moment along longer span Percentage of steel Total design load Factored load Maximum shear stress in concrete with shear Shear stress in concrete vi .SYMBOLS The following symbols have been used in our project and its meaning is clearly mentioned respective to it: A Ast b D DL d1 D Mu.max Fck Fy Ld LL Lx Ly B. Muy Nominal shear stress Diameter of bar Factored axial load Limiting moment of resistance of a section without compression reinforcement Moment about X and Y axis due to design loads Mux1. Muy1 Maximum uniaxial moment capacity for an axial load of pu.bending moment X and Y axis respectively Ac Asc SLX SLZ Area of concrete & Area of longitudinal reinforcement for column Seismic Load in X direction Seismic Load in Z direction vii .lim Mux.Tv ɸ Pu Mu. CHAPTER 1 INTRODUCTION 1 . over trees or under trees. 2 . These are the few reasons which are responsible that the person do utmost effort and spend hard earned saving in owning houses. The shelters of those old have been developed nowadays into beautiful houses.Building construction is the engineering deals with the construction of building such as residential houses. The vertical load consists of dead load of structural components such as beams. of the buildings. etc. It is residential complex. etc. and live loads. A multi-storey. The draughtsman must know his job and should be able to follow the instruction of the engineer and should be able to draw the required drawing of the building. The building is designed as two dimensional vertical frame and analyzed for the maximum and minimum bending moments and shear forces by trial and error methods as per IS 456-2000. live load and wind load as per IS 875. Daily new techniques are being developed for the construction of houses economically. Buildings are the important indicator of social progress of the county. quickly and fulfilling the requirements of the community engineers and architects do the design work. Draughtsman is responsible for doing the drawing works of building as for the direction of engineers and architects. as for the requirements. site plans and layout plans etc. Rich people live in sophisticated condition houses. moments and shear forces and obtained from this software. The building in plan consists of columns built monolithically forming a network. food. The design is made using software on structural analysis design (STAAD PRO V8i). columns. cloth and the basic needs of human beings. slabs etc. A building frame consists of number of bays and storey. sun. planning and layout. Every human has desire to own comfortable homes on an average generally one spends his two-third life times in the houses. The help is taken by software available in institute and the computations of loads. multi-paneled frame is a complicated statically intermediate structure. The building subjected to both the vertical loads as well as horizontal loads.. as the times passed as humans being started living in huts made of timber branches. Nowadays the house building is major work of the social progress of the county. to protect themselves from wild animals. A design of R.C building of G+9 storey frame work is taken up. The horizontal load consists of the wind forces thus building is designed for dead load. In the early ancient times humans lived in caves. rain. In a simple building can be define as an enclose space by walls with roof. CHAPTER 2 LITERATURE SURVEY 3 . C. Engineering Structure and Structural Design An engineering structure is an assembly of members or elements transferring load (or resisting the external actions) and providing a form. So. Safety: It has been the prime requirement of structural design right from the history of civilization and construction that a structure shall be so designed that it will not collapse in any way during its expected life span. the concrete is strengthened(i. reinforced) by steel and the resultant composite mass is known as Reinforced Cement Concrete (R.2.1 Elements of Structural Design Structures in concrete have become very common in civil engineering construction. 2. Objectives and Basic Requirements of Structural Design The objective of the structural design is to plan a structure which meets the basic requirements of structural science and those of the client or the user.C. The basic requirements of the structural design are as follows: i.C. structures.BACKGROUND WORK (LITERATURE SURVEY) 2.) structures.1.C.e.1. Concrete has established itself to be a universal building material because of its high compressive strength and its adaptability to take any form and shape. with the result that almost every civil engineer is intimately concerned with reinforced concrete (R.. bridges.1. it will be approximate to begin by reviewing the basic principles of structural design in general and then its application to reinforced concrete structures. 2. with economy and elegance. a durable structure which can safely carry the design forces and can serve the desired function satisfactorily in working environment during its intended service life span. Its low tensile strength is compensated by the use of steel reinforcement.) It is this combination which allows almost unlimited use of reinforced concrete in construction of buildings. Structural design is a science and art of designing. It is therefore. necessary that every civil engineer knows the basic principles involved in design of R. tanks. Safety of structure is achieved by adequate 4 . space. dams etc. an enclose and/or cover to serve the desired function. Thus. iv.ii. Ductility is thus obtained by providing steel of such quality that it would yield prior to crushing of concrete. Feasibility. The performance is rated buy the fitness of the structure to maintain deflections. imagination and judgment. Economy: The economy shall be of material by optimum utilization of its strength or it may be the economy of cost which includes cost of construction as well as cost of maintenance and repairs. v. ductility of structure is also nowadays considered to be an additional desired quality from a view point that if at all failure occurs. Durability: The structure shall resist effectively environmental action during its anticipated exposure conditions. The codes of practice are compendia of good practice drawn by experienced and competent engineers. Aesthetics: The structure should be so designed that it should not only be safe. The Design Process: The entire process of design requires conceptual thinking. vi. iii. alternate wetting and drying or freezing. sound knowledge of engineering. Serviceability: The structure shall efficiently serve the intended function and also shall give a satisfactory performance throughout the life span. such as rain. vii. abrasion action etc.1. relevant design codes and byelaws. it should not be sudden but should give prior warning of its probable occurrence so as to enable one to minimize the consequences of collapse and avoid loss of human life. chemical actions of salt. practicable an acceptable. serviceable and durable but should also give a pleasing appearance without affecting the economy to a great extent. cracking and vibration effects within acceptable limits. It is achieved by providing adequate stiffness and cracking resistance. climatic variations in temperature and humidity.3. 5 . Practicability and Acceptability: The structure has to be so designed that the proposed solution is feasible. 2. strength and stability. backed up by experience. Besides strength. deformations. They are intended to guide the engineers and should not be allowed to replace their conscience and competence. the design of any structure is categorized into the following two types: 1) Functional Design: The structure to be constructed must primarily serve the basic purpose for which it is to be constructed to satisfy the need of the user efficiently. Building Frame The principle elements of a R. beam. Load is transferred primarily by axial compression accompanied by bending and shear.C. This includes proper arrangement of rooms. It consists of the following steps: a) Structural Planning b) Determination of Loads c) Analysis d) Member Design e) Drawing. good ventilation. a) Slab: It is two-dimensional or a planar member supporting a transverse load and providing a working floor or a covering shelter. proper water supply and drainage arrangements etc. with economy and elegance.The design process commences with the planning of the structure primarily to meet its functional requirement and then designed for safety and serviceability. unobstructed view in cinema theatre / community halls. and acoustics. a durable structure which can safely carry the design forces and can serve the desired function satisfactorily in working environment during its intended service life span. building frame are slab.1. Detailing and Preparation of Schedule. Thus.4. 2. b) Beam: A Beam is a one-dimensional (normally horizontal) flexural member which provides support to the slab and the vertical walls. 6 . Elements of a R. halls. 2) Structural Design: As mentioned earlier Structural design is a science and art of designing. c) Column: It is one dimensional vertical member providing a support to beam. column and footing.C. The loads are transferred to supporting beams or walls in one or both directions. d) Footing: A footing can be considered as a horizontal two way cantilever slab providing a wide base to a column for distributing concentrated column load over a large area of supporting soil. Load transfer is affected partly by bending and partly by bearing. 2.1.5. Computer Programming It is important to emphasize that in every field the use of computer prevails. Access to personal computers, due to their affordable cost, has made it possible for almost every engineer and student to be equipped with such tools. The need is more apparent to utilize this powerful tool for simplifying engineering design works. It has now become practically obligatory for structural engineers or students to get conversant with the programming languages and techniques of computer aided design. 2.2. Design Philosophies Since the inception of the concept of reinforced concrete in the last twenties of the nineteenth century, the following design philosophies have been evolved for design of R.C. structures: a) Working Stress Method (WSM) b) Ultimate Load Method (ULM) c) Limit State Method (LSM) 2.2.1. Limit State Method (LSM) The limit state method ensures the safety at ultimate load and serviceability at working load rendering the structure fit for its intended use. Thus, it considers the fitness of the structure to perform its function satisfactorily during its life span. The salient features and the merits of the method are briefly given below: 1) It considers the actual behavior of the structure during the entire loading history up to collapse. 2) It adopts the concept of fitness of structure to serve the desired function during the service life span and defines the limiting state of fitness as the „limit state‟. 3) It attempts to define quantitatively the margins of safety or fitness on some scientific mathematical foundations rather than on adhoc basis of experience and judgment. 7 The mathematical basis is derived from classical reliability theory and statistical probability (e.g. the reliability of the fitness of the structure and the probability of attainment of a critical limit state). 4) The method, adopts the idea of probability of the structure becoming unfit, and attempts to achieve the minimum acceptable probability of failure. 5) The method is based on statistical probabilistic principles. The method examines the factors which can be quantified by statistical method (such as loads, material strength) and then they are accounted through characteristic loads and characteristic strength on the basis of statistical probabilistic principles and the others which are abstract (such as variation in dimensions, accuracy, variation in loads and material properties etc.) are taken into account through partial safety factors. In the limit state method, a structure is essentially designed for safety against collapse (i.e. for ultimate strength to resist ultimate load) and checked for its serviceability at working loads. The first part of design thus incorporates basic principles of ultimate load method. But at the same time, it eliminates the drawbacks of the ultimate load method by introducing the second part of check for serviceability. Since this second part relates to working loads at which the behavior of structure is elastic, the material uses the principles of working stress method to satisfy the requirements of serviceability. The limit state method, thus, makes a judicious combination of the ultimate load method and working stress philosophy avoiding the demerits of both. 2.2.2. Limit State of Collapse (Ultimate Limit State) It is the limit state on attainment of which the structure is likely to collapse. It relates to stability and ultimate strength of the structure. Design to this limit state ensures safety of structure from collapse. The structure failure can be any of the following types: i. Collapse of one or more members occurring as a result of force coming on the member exceeding its strength(Types (a) and (b) given below); 8 ii. Displacement of the structure bodily due to lack of equilibrium between the external forces and the resisting reactions (Types (c), (d), (e) given below). The various conditions leading to structural failure are as follows: a) Failure, breakage and hence division into segments of one or more members of the structure either due to material failure or on account of formation of mechanism by development of plastic hinges at one or more critical sections. b) Buckling; c) Sliding; d) Overturning; e) Sinking. This limit state is attended to by providing resistance greater than the force coming on it and keeping a margin of safety through safety factors. I.S. Code prescribes different safety factors for overturning and sliding without giving any special status to sinking or buckling. 2.2.3 Limit State of Serviceability Limit states of serviceability relate to performance or behavior of structure at working loads and are based on causes affecting serviceability of the structure. They are mainly subdivided into following categories: A. Limit State of Deflection, B. Limit State of Cracking, and C. Other Limit States. 2.3. MULTISTOREY BUILDINGS Reinforced concrete buildings consist of floor slabs, beams, girders and columns continuously placed to form a rigid monolithic system. This continuous system leads to greater redundancy, reduced moments and distributes the load more evenly. The floor slab may rest on a system of interconnected beams. A building frame is a three – dimensional structure or a space structure. 9 to more refined techniques involving computer solutions. Building with flexural (shear) wall system:   Gravity load is carried by frame supported on columns rather than on bearing walls.1.1 and 8. In tall buildings. Structural Systems A building is subjected to several loads which are transferred to ground through a system of interconnected structural members. No columns. The recent advancement of abundance of ready-made computer package programs has reduced the use of approximation methods.2. Clause 8. 2. portal and cantilever methods. to specification writing. 2. to plotting. to cost estimating. This has been induces from analysis to design.8 of IS: 4326-1993 restricts the use of such system to 3 storey in seismic zone V and 4 storey in other zone. moment distribution.A wide range of approaches have been used for buildings of varying heights and importance. The frame provides vertical stability to the building and prevents collapse after damage to flexural wall or braced frames. etc.4. from simple approximate methods which can be carried out manually. Till a few years ago most of the multistory buildings were analyzed by approximate methods such as substitute frame. Load Bearing wall system:     Walls provide support for all gravity loads as well as resistance to lateral loads.3. to detaining. 10 . The lateral stiffness may be achieved through a permutation and combination of placement of columns and walls in plan. or with the aid of a pocket calculator. A structural system may be classified as follows:1. The Walls and partition wall supply in-plane lateral stiffness and stability to resist wind and earthquake loads. the biggest challenge comes from controlling lateral displacements within the serviceability limit state. 6. Resistance to lateral loads by:  Special detailed moment resisting frame (concrete or steel) which is capable of resisting at least 25%of base shear including torsional effects. 5.3. Space frame:  3-Dimensional structural system without shear or bearing walls composed of interconnected members laterally supported 11 . shear walls in any plane do not resist more than 33% of earthquake design force including torsional effects. 4. if and only if. The America IBC 2000 permits use of flexural (shear) wall system up to 45m high. A frame can be designed using weak column-strong girder proportions or strong column-weak girder proportions. However it can be used up to 70m. Dual frame system:   Moment resisting frame providing support for gravity loads. shear walls or braced frames must resist total required lateral loads. Relative stiffness of girders and columns is very important. Flexural (shear) wall system:   Reinforced concrete wall designed to resist lateral forces parallel to the plane of the wall and detailed to provide ductility as per IS 13920-1993.e.  Flexural walls i. Moment resisting frame system:    Members and joints are capable of resisting vertical and lateral loads primarily by flexure. 1 R.C framed structure Brick wall Ventilated rooms with window in each room. Geometric details: Ground floor Floor to floor height Height of plinth Depth of foundation 2m 3.65m. 12 . of Rooms/floor No of lifts Type of construction Types of walls Ventilation Residential Hostel Building G+9 1 22 rooms on each floor with attached washroom. Structural Planning Salient features: Utility of building No of stories No of staircases No.2.4. 2m 2m Materials: Concrete grade All steel grades Bearing capacity of soil: Depth of Water Table M35 (for footing) & M25 (for all other elements) Fe415 grade 175KN/m2 4m.C. CHAPTER 3 COMPUTER AIDED ANALYSIS & DESIGN 13 . This software can be used to carry RCC. truss etc. Staad foundations 5(V8i) 3. So when it comes into the building with several members it will take a week.M. So first find the outline of the structure. This we do after the analysis. STAAD PRO V8i Staad Pro V8i is powerful design software licensed by Bentley .Staad stands for structural analysis and design Any object which is stable under a given loading can be considered as structure.D of a complex loading beam it takes about an hour.1. List of software‟s used 1. whereas analysis is the estimation of what are the type of loads that acts on the beam and calculation of shear force and bending moment comes under analysis stage. To calculate S. Staad Pro (V8i) 2. 14 .F. steel.COMPUTER AIDED ANALYSIS AND DESIGN This project is mostly based on software and it is essential to know the details about these software‟s. Design phase is designing the type of materials and its dimensions to resist the load. bridge. Staad pro is a very powerful tool which does this job in just an hour‟s staad is a best alternative for high rise buildings. Nowadays most of the high rise buildings are designed by staad which makes a compulsion for a civil engineer to know about this software. according to various country codes.D and B. Auto Cad 2010 STAAD PRO V8i STAAD FOUNDATIONV8i AUTOCAD 2010 3. Staad editor is the programming For the structure we created and loads we taken all details are presented in programming format in Staad editor. Even analysis of a small beam creates large output. Unable to show plinth beams. These supports are to be imported into this software to calculate the footing details i.e.. Limitations of Staad Pro V8i: 1.4. This software can deal different types of foundations SHALLOW (D<B) 1.e. ETAB. But these software‟s are restricted to some designs only whereas Staad can deal with several types of structure.3. Staad editor.2. Analysis and design carried in Staad and post processing in Staad gives the load at various supports. regarding the geometry and reinforcement details. Staad foundation: Staad foundation is a powerful tool used to calculate different types of foundations.Mat (Raft) Foundation 15 . 3.. So load cases created for a structure can be used for another structure using Staad editor. but this require some programming skills. Staad Editor: Staad has very great advantage to other software‟s i.Combined (Strip) Footing 3. SAP which gives details very clearly regarding reinforcement and manual calculations. Alternatives for Staad Pro V8i: STRUDS. Huge output data 2. This program can be used to analyze other structures also by just making some modifications. 3. It is also licensed by Bentley software‟s. All Bentley software‟s cost about 10 lakhs and so all engineers can‟t use it due to heavy cost. Isolated (Spread) Footing 2. ROBOT.3. 3. 4. Manual calculations These details will be given in detail for each and every column.Pile Cap 2. soil. Column layout 4. So depending on the soil at type we have to decide the type of foundation required. Mat foundation is generally laid at places where soil has less soil bearing capacity. Another advantage of foundations is even after the design. Combined Footing or Strap footing is generally laid when two columns are very near to each other. 2. properties of the members can be updated if required.DEEP (D>B) 1. Graphs 5. After input data is give software design the details for each and every footing and gives the details regarding 1. Reinforcement 3. The following properties can be updated Column Position Column Shape Column Size Load Cases Support List It is very easy deal with this software and we don‟t have any best alternative to this. Isolated footing is spread footing which is common type of footing. 3. materials used should be given in respective units. Also lot of input data is required regarding safety factors. 16 . Driller Pier 1. Pile foundation is laid at places with very loose soils and where deep excavations are required. Geometry of footing 2. mechanical and also electrical engineer. AutoCAD is used for drawing different layouts. elevations. elevation of a residential building. The word auto came from auto Desk Company and cad stands for computer aided design. 17 . It is very useful software for civil. We also used AutoCAD to show the reinforcement details and design details of a stair case. Learning of certain commands is required to draw in AutoCAD. The importance of this software makes every engineer a compulsion to learn this software‟s.5. We used AutoCAD for drawing the plan. AutoCAD: AutoCAD is powerful software licensed by auto desk. AutoCAD is a very easy software to learn and much user friendly for anyone to handle and can be learn quickly.3. sections and different sections can be shown in auto cad. details. plans. CHAPTER 4 PLAN & ELEVATION 18 4.1. PLAN The Annexure A represents the plan of a G+9 hostel building. The plan clearly shows that it is a combination of rooms and attached washrooms of the SRM Hostel building. The Hostel is located at SRM University, NCR Campus, Ghaziabad which is surrounded by other hostel blocks on the three sides except the backside. Every floor consists of 22 rooms along with attached bathroom. It represents a spacious surrounding with huge areas for each room. It is a G+9 proposed building, so for 9 floors we have 9*22=198 rooms. The plan shows the details of dimensions of each and every room. The entire plan area is about 810sq.m. The plan also gives the details of location of stair cases in different blocks. We have 2 stair cases for the building and designing of stair case is shown in AutoCAD plot no.3. At the left end of the building we have a small construction which consists of two lifts and those who want to fly through lift can use this facility and we know for a building with more than G+4 floors should compulsory have lift and the charges for the facilities is collected by all the members. So these represent the plan of our building and detailed explanation of remaining parts like elevations and designing is carried in the next sections. 19 4.2. Elevation The Annexure B represents the proposed elevation of building. It shows the elevation of the G+9 building representing the front view which gives the overview of a building block. Each floor consists of height 3m which is taken as per GHMC rules for residential buildings. The building is not designed for increasing the number of floors in future.so the number of floors is fixed for future also for this building due to unavailability of the permissions of respective authorities. Also special materials like fly ash and self-compacted concrete were also used in order to reduce the dead load and increase life of the structure and also improve economy. But these materials were not considered while designing in Staad to reduce the complexity and necessary corrections are made for considering the economy and safety of the structure as it is a very huge building. The construction is going to complete in the month of July 2013 and ready for the occupancy. This is regarding the elevation and details of the site and next section deals with the design part of the building under various loads for which the building is designed. 20 CHAPTER 5 LOADS 21 . Vertical Loads: Gravity loads act in the same direction as gravity (i.1. Vertical Loads Dead Load (gravity) Live (gravity) Snow (gravity) Wind (uplift on roof) Seismic and wind (overturning) Seismic (vertical ground motion) 5. building loads can be divided into types based on the orientation of the structural action or forces that they induce: vertical and horizontal (i. Building Loads Categorized by Orientation: Types of loads on a hypothetical building are as follows. and snow loads. 5. including the dead load (i.e.2. Wind Seismic (horizontal ground motion) Flood (static and dynamic hydraulic forces Soil (active lateral pressure) 5.. Horizontal (Lateral) Loads: Direction of loads is horizontal w.C structures.e.2. weight of the construction) and any applied loads(i.e.e.2. live.. As shown in Table. determining a gravity load on a beam or column is a relatively simple exercise that uses the concept of tributary areas to assign loads to structural elements. Load Conditions and Structural System Response: The concepts presented in this section provide an overview of building loads and their effect on the structural response of typical R. downward or vertically) and include dead. They are generally static in nature and usually considered a uniformly distributed or concentrated load.1. Thus.C. lateral) loads.t to the building. 22 . Classification of loads is described in the following sections.LOADS 5..r.2. bending stresses. Lateral Loads: The primary loads that produce lateral forces on buildings are attributable to forces associated with wind. The selection of an appropriate analytic model is. the tributary gravity load on a floor joist would include the uniform floor load (dead and live) applied to the area of floor supported by the individual joist.Even though the wind loads are dynamic and highly variable. The structural designer then selects a standard beam or column model to analyze bearing connection forces (i. Vertical forces are also created by overturning reactions due to wind and seismic lateral loads acting on the overall building and its lateral force resisting systems. but to varying degrees. creating a combined push and-pull effect. seismic ground motion. Wind uplift forces are generated by negative (suction) pressures acting in an outward direction from the surface of the roof in response to the aerodynamics of wind flowing over and around the building. and axial stresses) and stability of the structural member or system a for beam equations.e. roof) are analyzed by using the concept of tributary areas and uniformly distributed loads. the design approach is based on a maximum static load (i. and soil. Lateral forces from wind are generated by positive wind pressures on the windward face of the building and by negative pressures on the leeward face of the building.e.. especially if the structural system departs significantly from traditional engineering assumptions are particularly relevant to the structural systems that comprise many parts of a house.e.3.live load). the influence of wind uplift pressures on a structure or assembly (i. Earthquakes also produce vertical ground motions or accelerations which increase the effect of gravity loads. shear stresses. The magnitude of the seismic shear (i.2. However. eaves..g. lateral) load depends on the 23 . For example. corners. floods. Wind and seismic lateral loads apply to the entire building. 5.e. however no trivial matter. and ridges). Vertical earthquake loads are usually considered to be implicitly addressed in the gravity load analysis of a light-frame building.. reactions) internal stresses (i.e. particularly proximity to changes in geometry (e. Seismic lateral forces are generated by a structure‟s dynamic inertial response to cyclic ground movement. As with gravity loads.. The major difference is that wind pressures act perpendicular to the building surface (not in the direction of gravity) and that pressures vary according to the size of the tributary area and its location on the building.. pressure) equivalent. ). overturning forces on connections designed to restrain components from rotating or the building from overturning must be considered. identify variances from both local accepted practice and the applicable 24 . Flood loads are generally minimized by elevating the structure on a properly designed foundation or avoided by not building in a flood plain. Since wind is capable of the generating simultaneous roof uplift and lateral loads. the type and magnitude of design loads affect critical decisions such as material collection. Conversely the dead load may be sufficient to offset the overturning and uplift forces as is the case in lower design wind conditions and in many seismic design conditions.e. Lateral loads from moving flood waters and static hydraulic pressure are substantial. Soil lateral loads apply specifically to foundation wall design. as a matter of due diligence. and the dynamic structural response characteristics (i. the uplift component of the wind load exacerbates the overturning tension forces due to the lateral component of the wind load. ductility. a simplified seismic load analysis employs equivalent static forces based on fundamental Newtonian mechanics (F=ma) with somewhat subjective (i.magnitude of the ground motion. Since building codes tend to vary in their treatment of design loads the designer should. Lateral loads also produce an overturning moment that must be offset by the dead load and connections of the building. mainly as an “out-of-plane” bending load on the wall.. experiencebased) adjustments to account for inelastic. etc.3.. Design loads for the residential building: General Loads are a primary consideration in any building design because they define the nature and magnitude of hazards are external forces that a building must resist to provide a reasonable performance(i. For houses and other similar low rise structures. 5. construction details and architectural configuration. Therefore.. configuration (size and shape) and location (climate and site conditions). dampening.Ultimately. the buildings mass. ductile response characteristics of various building systems. The anticipated loads are influenced by a building‟s intended use (occupancy and function).e. safety and serviceability) throughout the structure‟s useful life.e. natural period of vibration. code relative to design loads as presented in this guide. concrete floors. In staad pro assignment of dead load is automatically done by giving the property of the member. and foundation systems. finishes and fixed equipment. Dead Loads: Dead loads consist of the permanent construction material loads compressing the roof. even though the variances may be considered technically sound..1. such as the steel columns. bricks. In load case we have option called self-weight which automatically calculates weights using the properties of material i. floor. 5.3. roofing material etc. including claddings. Dead load is the total load of all of the components of the components of the building that generally do not change over time. density and after assignment of dead load the skeletal structure looks red in color as shown in the figure. wall.e. 1 25 . Figure 5. Loads include those from human occupants. furnishings. and construction and maintenance activities. the stair load of 300 pounds should be applied to the center of the stair tread between supports. no fixed equipment. For example. The uniform and concentrated live loads should not be applied simultaneously n a structural evaluation. storage. Dead load is calculated as per IS 875 part 1 Here for the multistory building we need to define the loads distributed by the masonry brick wall which is shown in the above figure using UNI GY -20. After the assignment of the live load the structure appears as shown below. 26 .12*25+1=3kn/m^2 The above example shows a sample calculation of dead load.2.063N/mm. In staad we assign live load in terms of:   Floor load = 2. concentrated loads. and uniform line loads. Concentrated loads should be applied to a small area or surface consistent with the application and should be located or directed to give the maximum load effect possible in endues conditions. Imposed Loads Live loads are produced by the use and occupancy of a building.3. 5. As required to adequately define the loading condition.125KN/m2 (as per IS 875 Part 2) (for residential building including floor finish) Plate/Element Load = 2KN/m2 (Imposed/live load on slab) We have to create a load case for live load and select all the beams to carry such load. loads are presented in terms of uniform area loads.Example for calculation of dead load: Dead load calculation Weight=Volume x Density Self-weight floor finish=0. The peak pressures in one zone for a given wind direction may not. The variation in pressures at different locations on a building is complex to the point that pressures may become too analytically intensive for precise consideration in design. the peak pressure depends on an arrow range of wind direction.3 Wind loads: In the list of loads we can see wind load is present both in vertical and horizontal loads. occur simultaneously in other zones.3. Wind produces non static loads on a structure at highly variable magnitudes. For some pressure zones.Figure 5. However. the wind directionality effect must also be factored into determining risk consistent wind loads on buildings. Therefore. 27 . Therefore.2 Live loads are calculated as per IS 875 Part 2 5. This is because wind load causes uplift of the roof by creating a negative (suction) pressure on the top of the roof figure 3 a diagram of wind load. wind load specifications attempt to amplify the design problem by considering basic static pressure zones on a building representative of peak loads that are likely to be experienced. After designing wind load can be assigned in two ways 1. In Delhi we have a wind speed of 47 kmph for 10 m height and this value is used in calculation. We have to define a load case prior to assignment. a) Risk level b) Terrain roughness. height and size of the structure and c) Local topography It can be mathematically expressed as follows: Vs. =Vb* K1* K2* K3 Where Vz= design wind speed at any height Z in m/s K1= probability factor (risk coefficient) K2=terrain height and structure size factor and K3=topography factor 28 . Basic wind speed is based on peak just velocity averaged over a short time interval of about 3 seconds and corresponds to mean heights above ground level in an open terrain. Calculation of wind load as per IS 875 part 3.Assignment of wind speed is quite different compared to remaining loads. Design wind speed: The basic wind speed (Vb) for any site shall be obtained the following effects to get design wind velocity at any height (Vz) for the chosen structure. We designed our structure using second method which involves the calculation of wind load using wind speed. Collecting the standard values of load intensities for particular heights and assigning of the loads for respective height. as applicable to 1m height above means ground level for different zones of the country. Basic wind speed: It gives the basic wind speed of India. 2. 5. the mass of the building should be as low as possible.3. Its effect should be considered along both axes of a building taken one at a time. A force is defined as the product of mass and acceleration. Response Spectra: The representation of the maximum response of idealized single degree of freedom system having certain period of vibration and damping during a given earthquake is referred to as a response spectrum.e. But for this structure the seismic loads are predominant than that of the wind loads. therefore. There can be no control on the ground acceleration as it is an act of God! The point of application of this internal force is the center of gravity of the mass on each floor of the building. While wind forces govern. lateral stiffness and mass. 29 . The fact is that the design forces for wind are greater than the seismic design forces (i. and is reversible. Earthquake Loads Earthquake or seismic load on a building depends upon its geographical location. There are two methods to determine the earthquake force in a building: a) Seismic coefficient method or static method. During an earthquake. In order to have a minimum force. the mass is imparted by the building whereas the acceleration is imparted by ground disturbances. Once there is a force. an elastic response spectrum has been proposed for the Maximum Considered Earthquake (MCE) condition. b) Response spectrum method or modal analysis method or spectral acceleration method or dynamic method. The inertial force is resisted by the building and the resisting force acts at the center of rigidity at each floor of the building or shear center of the building at each storey. the seismic loads govern the design. the design must provide at least the type of seismic detailing that corresponds to the seismic forces calculated for that building.The wind loads and earthquake loads are assumed not to act simultaneously. wind governs the design) does not obviate the need for seismic detailing.4. there has to be an equal and opposite reaction to balance the force. In the IS : 1893:2002 code. NOTE: . A building is designed for the worst of the two loads. The values of R for buildings are given in Table 7. historical value.0 (Table 7.36 30 . the value of Ah will not be taken less than Z/2 whatever be the value of I/R.I.Sa /2. is for the Maximum Considered Earthquake (MCE) and service life of structure in a zone. R= Response reduction factor. of IS 1893 The design horizontal seismic coefficient Ah for a structure shall be determined by the following expression: Ah= Z. I = Importance factor. depending on the perceived seismic damage performance of the structure. or economic importance (Table 6. The factor 2 in the denominator of Z is used so as to reduce the Maximum Considered Earthquake (MCE) zone factor to the factor for Design Basis Earthquake (DBE). the country is classified into four seismic zones as shown in Fig. Z = Zone factor given in Table 2 of IS 1893. Sa/g = Average response acceleration coefficient for rock or soil sites as given by Fig. 1.R. where. depending upon the functional use of the structures. based on appropriate natural periods and damping of the structure.24 0.1 s. However.1 ZONE FACTOR (Z) II LOW III MODERATE IV SEVERE V VERY SEVERE Seismic Zone Seismic Intensity Z 0. characterized by hazardous consequences of its failure. These curves represent free field ground motion. 2 and Table 3 of IS 1893. IS 1893). post-earthquake functional needs. Table 5.Design Spectrum For the purpose of determining seismic forces. characterized by ductile or brittle deformations. IS 1893).16 0. the ratio (I/R ) shall not be greater than 1.g Provided that for any structure with T <0.10 0. The percentage of imposed loads should be 25% for floor loads up to and including 3KN/m2. the earthquake force shall be calculated for the full dead load plus the percentage of imposed load as given in Table 8. Design Lateral Force Buildings and portions thereof shall be designed and constructed. The overall design seismic force thus obtained at each floor level. For calculating the design seismic forces of the structure.4. to resist the effects of design lateral force. While computing the seismic weight of each floor. Any weight supported in between storeys shall be distributed to the floors above and below in inverse proportion to its distance from the floors. Seismic Weight of Building The seismic weight of the whole building is the sum of the seismic weights of all the floors. This design lateral force shall then be distributed to the various floor levels. Design Seismic Base Shear The total design lateral force or design seismic base shear ( Vb)along any principal direction shall be determined by the following expression: Vb= AhW Where. the weight of columns and walls in any storey shall be equally distributed to the floors above and below the storey. Seismic Weight of Floors The seismic weight of each floor is its full dead load plus appropriate amount of imposed load. The design lateral force shall first be computed for the building as a whole. Design Imposed Loads for Earthquakes Force Calculation For various loading classes as specified in IS 875(Part 2).5. shall then be distributed to individual lateral load resisting elements depending on the floor diaphragm action. 31 . the imposed load on roof need not be considered. large community halls like cinemas. Zone Factor = 0. Seismic Loading in Staad Pro V8i: Now since we know the basic criterion for earthquake loads.4. (for Special RC moment-resisting frame (SMRF) as per Table 7.e. (For Medium Type Soil at 5% damping. fire station buildings. schools. ( for All Other Buildings other than Important service and community buildings. and W = Seismic weight of the building. monumental structures. in the considered direction of vibration.) Type of Structure = 1 (for Reinforced Concrete Framed Structure) Damping Ratio = 5% Depth of foundation = 2m 32 . television stations. assembly halls and subway stations. power stations for which I = 1. radio stations.Ah = Design horizontal acceleration spectrum value.1. emergency buildings like telephone exchange. such as hospitals. railway stations.5. which includes the following: Earthquake Zone for Delhi is Zone IV (i.0. IS 1893.) Importance Factor = 1.)     Response spectra for Rock and Soil Site Type (SS) = 2. the seismic weights as assigned in Staad Pro V8i software are as follows:     Defining Seismic parameters.24) Response Reduction Factor = 5. 5. using the fundamental natural period T. 3 The weights are then defined for the structure which includes: SELFWEIGHT (represents the dead weight) FLOOR WEIGHT (represents the live load) PLATE WEIGHT (represents the live load on slab) MEMBER WEIGHT (masonry brick weight ) The load case for seismic loads is then defined in the two directions that are horizontally perpendicular (X and Z) directions.Figure5. The figure of Staad Editor is shown as below: 33 . 4 Seismic Load in X direction (SLX) 34 .Figure5. A judicious combination of the loads (specified in IS 875 Parts 1 to 4 of this standard and earthquake).Figure5. Load Combinations . and b) Their disposition in relation to other loads and severity of stresses or c) Deformations caused by combinations of the various loads are necessary to ensure the required safety and economy in the design of a structure. the following 35 . keeping in view the probability of: a) Their acting together.5. be combined in accordance with the stipulations in the relevant design codes. In the absence of such recommendations.The various loads should. LOAD COMBINATIONS Seismic Load in Z direction (SLZ) Load combinations as per IS 875 Part 5 are taken into consideration.5 5. therefore. 2(DL+LL+SLZ) 18) 1. earthquake. the numerals 1.0 represents the load factors as per IS 875 Part 5.1. It should also be recognized in load combinations that the simultaneous occurrence of maximum values of wind.9.9DL +1.5.2.9DL .loading combinations. DL = Dead Load LL = Live Load SLX = Seismic load in X direction SLZ = Seismic load in Z direction The negative sign in the above load combinations shows the directions opposite to the defined case.SLX) 9) 1.2(DL +LL -SLX ) 17) 1.9DL + 1. Earthquake is not likely to occur simultaneously with wind or maximum flood or maximum sea waves.5SLZ 15) 1.5SLX 12) 0.SLZ) Where.5 (DL + SLZ) 10) 1.5SLZ 14) 0.5 (DL + SLX) 8) 1.5SLX 13) 0. Since the wind velocity in the region is less and less dominant than the seismic zone (Zone IV).2 (DL +LL +SLX 16) 1. 1. 0. 36 . therefore wind load is not considered for design.5 (DL . imposed and snow loads is not likely: - 1) DL + LL 2) DL + LL +SLX 3) DL + LL + SLZ 4) DL + LL –SLX 5) DL + LL –SLZ 6) 1. foundation or structural member concerned may be adopted ( as a general guidance ).5 (DL + LL) 7) 1.9DL -1. whichever combination produces the most unfavorable effect in the building.5 (DL -SLZ) 11) 0.2(DL + LL . 1. 7.5. INPUT TO STAAD EDITOR FOR LOADING: 37 . 38 . CHAPTER 6 ANALYSIS 39 . . Limitations: It is not applicable for degree of redundancy>3 6. Method of flexibility coefficients. Moment distribution method 4.2. Slope displacements methods (iterative methods) 3. Using STAAD Pro V8i 6.ANALYSIS 6.1 Method Of Analysis The various methods of analysis of statistically indeterminate portal frames are : 1. Portal method 7.1. This procedure was first formulated by axle bender in 1914 based on the applications of compatibility and equilibrium conditions. Matrix method 8. Cantilever method 6. Set up simultaneous equations is formed the solution of these parameters and the joint moment in each element or computed from these values.1 Method of flexibility coefficients: The method of analysis is comprises reducing the hyper static structure to a determinate structure form by: Removing the redundant support (or) introducing adequate cuts (or) hinges. 2. Slope displacement equations: It is advantageous when kinematic indeterminacy <static indeterminacy. Limitations: A solution of simultaneous equations makes methods tedious for manual computations.1. 40 . The method derives its name from the fact that support slopes and displacements are explicitly comported. Kane‟s method 5. This method is not recommended for frames larger than two bays and two storeys. The method cannot be applied to structures with intermediate hinges. Advantages: It is used for side way of frames. That is the column should be parallel. 6. Frames with intermediate hinges cannot be analyzed.1.M to that extent it also involves repeated distribution of moments at successive joints in frames and continues beams. H. Limitations: The rotational of columns of any storey should be functioning a single rotation value of same storey.C. It makes Some simplifying assumptions regarding Structural behavior so to obtain a rapid solution to complex structures. The usual process comprises reducing the given indeterminate configuration to a determine structural system by introducing adequate no of hinges. Kani‟s approach is similar to H. The beams of storey should not undergo rotation when the column undergoes translation.Iterative methods: These methods involve distributing the known fixed and moments of the structural member to adjacent members at the joints in order satisfy the conditions of compatibility. Approximate method: Approximate analysis of hyper static structure provides a simple means of obtaining a quick solution for preliminary design.4.3 Kani’s method: This method over comes some of the disadvantages of hardy cross method.M distributes only the total joint moment at any stage of iteration. Limitations of hardy cross method: It presents some difficulties when applied to rigid frame especially when the frame is susceptible to side sway.1. However there is a major difference in distribution process of two methods. 6. Any error at any stage of iterations corrected in subsequent steps consequently skipping a few steps error at any stage of iteration is corrected in subsequent consequently skipping a few steps of iterations either by over sight of by intention does not lead to error in final end moments. The most significant feature of Kani‟s method is that process of iteration is self-corrective. it is possible to sketch the deflected 41 .C. The analysis carried out separately for these two cases. It is necessary it differentiate between low rise and high rise frames in this case. High rise buildings Height > width It is dominated by bending action 6. Horizontal cases: The behavior of a structure subjected to horizontal forces depends upon its heights to width ratio among their factor.5. Low rise structures: Height < width It is characterized predominately by shear deformation.profile of the structure for the given loading and hence by locate the print inflection. The inflection points can be visualized as hinges for the purpose of analysis.1.never the less the rudimentary flexibility and stiffness methods are applied to frames stiffness method is more useful because its adaptability to computer programming stiffness method is used when degree of redundancy is greater than degree of freedom. Since each point of inflection corresponds to the location of zero moment in the structures. However stiffness method is used degree of freedom is greater than degree of redundancy especially for computers. Matrix analysis of frames: The individual elements of frames are oriented in different directions unlike those of continues beams so their analysis is more complex . 42 . The loading cases are arising in multistoried frames namely horizontal and vertical loading. The solution of structures is sundered simple once the inflection points are located. The design base shear shall first be computed as a whole.S 1893. Seismic Analysis Procedures: Main features of seismic method of analysis based on Indian Standard 1893(part 1): 2002 are described as follows Equivalent lateral force method: The Equivalent lateral force method is the simplest method of analysis and requires less computational effort because the forces depend on the code based fundamental period of structures with some empirical modifier. time history analysis over comes all disadvantages of modal response spectrum provided nonlinear behavior is not involved. 6. and then be distributed along the height of buildings based on simple formulae appropriate for buildings with regular distribution of mass and stiffness.3. Elastic Time history analysis: A linear analysis. In this method the response of Multi degree of freedom system is expressed as the superposition of modal response.2. 43 . Analysis Using Staad Pro V8i: After assigning all the properties of a structural frame only a command is used to analyze the structure and the results are obtained within seconds of time. The method requires greater computational efforts for calculating the response at discrete times. This is the main advantage of using the software or computer aided analysis. One interesting advantage of this is that the relative signs of response quantities are preserved in the response histories. which is then combined to compute the total response. Response Spectrum analysis: This method is applicable for those structures where modes other than the fundamental one affect significantly the response of the structure. each modal response being determined from the spectral analysis of Single–degree of freedom system.6. The design lateral force or design base shear and the distribution are given by some empirical formulae given in the I. The design lateral force obtained at each floor level shall be distributed to individual lateral load resisting elements depending upon floor diaphragm action. 6.4. Analysis Result For Load Cases 1 To 4 For Load Case 1 (SLX) 44 . For Load Case 2 (SLZ) 45 . For Load Case 3 (Dead Load) 46 . For Load Case 4 (Live Load) 47 . 6.5. ANALYSIS RESULTS FOR SUPPORT REACTIONS 48 . 49 . 50 . 51 . 52 . **These reaction forces and moments are evaluated for the critical load combinations 5 to 9 as shown above under load combinations. 53 . **The joints 69 to 113 show the column positions the ultimate position of reaction supports for the RC framed structure.**The above results are displayed from the Staad Output file. CHAPTER 7 DESIGN 54 . INPUT TO STAAD EDITOR FOR DESIGN 55 . ) Flanged beams Beams transfer loads from slabs to columns and hence are designed for bending. The necessity of steel of compression region arises due to two reasons. Doubly reinforced concrete beams: It is reinforced under compression and tension regions. maximum and minimum size of bars to be used. for the same reason as in the case of simply supported beam. compressive and shear stress induced in it by loads on the beam. I cantilever beams reinforcing bars placed near the top of the beam. Figure shows the bottom and top reinforcement details at three different sections.1. the strength availability singly reinforced beam is in adequate. Using Staad Pro software.) Single reinforced beams 2. yield strength of steel. Due to the huge output of Staad Pro V8i. here we only show the design result of a beam. when depth of beam is restricted. These calculations are interpreted manually. Generally. beam is of two typesi) Singly Reinforced Beam and ii) Doubly Reinforced Beams.7. Singly reinforced beams: In singly reinforced simply supported beams steel bars are placed near the bottom of the beam where they are more effective in resisting in the tensile bending stress. the design of beam is simply done by assigning the parameters for the structure which includes the clear cover. A reinforced concrete beam should be able to resist tensile.) Double reinforced concrete 3. etc. At a support of continuous beam where bending moment changes sign such as situation may also arise in design of a beam circular in plan. BEAMS Beams are the horizontal members of the RC framed structure. Design of beams is done as per Limit State Design of collapse (IS 456: 2000). compressive strength of concrete. 56 . There are three types of reinforced concrete beams 1. 1. 1 57 .7. Design Result for Beam No.1. FIGURE 7. 1 Location of Beam 1 in the structure 58 . 1.7.2. 2 Beam Reinforcement 59 . Detailing of Beam Reinforcement as per IS 13920 : 1993 FIGURE 7. 3. minimum shear reinforcement shall -be provided Design of Shear Reinforcement: When τv exceeds τc.93 KN τc = 0. given in Table 19. shear reinforcement shall be provided in any of the following forms: a) Vertical stirrups. 60 . b) Bent-up bars along with stirrups.1: Given data: Cross section of beam : b x d = 300mm x600 mm Vertical shear force = Vu =145.29 N/mm2 (from table 19 of IS 456 200) Minimum Shear Reinforcement: When τv is less than τc . and c) Inclined stirrups.1. Check for the design of a Beam No. given in Table 19.FIGURE 7. 3 Beam Web Reinforcement 7. 111130 N= 0.4 of IS 456-2000) Asv Sv τv τc b = = = = = total cross-sectional area of stirrup legs or bent-up bars within a distance Sv. and d = effective depth. shall be calculated as below: For vertical stirrups: Vus = 0. breadth of the member which for flanged beams.216 N/mm2 τv ≥ τc Design reinforcement Vus = = = Vu.93 x 103/(550x300) =1.1 of IS 456-2000) 145.87 fyAsvd/Sv (As per clause 40.τv = = Vu/(b x d) (As per clause 40.87 x 415 x 2 x π x 82 x 550/Sv Sv = 140 mm 61 .29x550x300 111100 N Shear reinforcement shall be provided to carry a shear equal to Vu . spacing of the stirrups or bent-up bars along the length of the member.τc x b x d (As per clause 40.4 of IS 456-2000) 145. shall be taken as the breadth of the web bw.93 x103 -0. not less than 45”.τc bd The strength of shear reinforcement Vus.up bar and the axis of the member. = angle between the inclined stirrup or bent. nominal shear stress design shear strength of the concrete. fy α = characteristic strength of the stirrup or bent-up reinforcement which shall not be taken greater than 415 N/mm2. Minimum shear reinforcement spacing = Sv.87fy (As per clause 26. b = breadth of the beam or breadth of the web of flanged beam. 0. 605 mm.4x300) = Provided 2 legged 8mm @100 mm stirrups . Sv = stirrup spacing along the length of the member.1.4/ 0.75xd = 0.Sv should not be more than the following 1.min Minimum shear reinforcement: Minimum shear reinforcement in the form of stirrups shall be provided such that: Asv/bSv ≥ 0.87x415/(0. 62 . and fy = characteristic strength of the stirrup reinforcement in N/mm* which shall not be taken greater than 415 N/mn2 Sv=2x(π/4)x82x0. 300 mm 3.6 of IS 456-2000) Asv = total cross-sectional area of stirrup legs effective in shear. Hence matched with Staad output.75 x 550 = 300 mm 2.5. Positioning of columns: Some of the guiding principles which help the positioning of the columns are as follows:- A) Columns should be preferably located at or near the corners of the building and at the intersection of the wall.7. The binders prevent displacement of longitudinal bars during concreting operation and also check the tendency of their buckling towards under loads.C columns may be cast in various shapes i. and hence the self-weight of the beam and the total. B) The spacing between the columns is governed by the lamination on spans of supported beams. the column can be shifted inside along a cross wall to provide the required area for the footing with in the property line. Depending upon the architectural requirements and loads to be supported. alternatively a combined or a strap footing may be provided. octagonal.e.2. but for the columns on the property line as the following requirements some area beyond the column. R.2. which is used primary to support axial compressive loads and with a height of at least three it is least lateral dimension. circular. 7. Columns of L shaped or T shaped are also sometimes used in multistoried buildings. As the span of the of the beam increases. square. COLUMNS A column or strut is a compression member.G 0f the column I the longitudinal direction. as the spanning of the column decides the span of the beam. the depth of the beam. or lateral binders. The longitudinal bars in columns help to bear the load in the combination with the concrete. and hexagonal.1. A reinforced concrete column is said to be subjected to axially loaded when line of the resultant thrust of loads supported by column is coincident with the line of C. The longitudinal bars are held in position by transverse reinforcement. 63 . rectangle. clause 22.2. Design charts for combined axial compression and bending are in the form of intersection diagram in which curves for Pu/fck bD verses Mu/fck bD2 are plotted for different values of p/fck where p is reinforcement percentage. In practice.002 2) The maximum compressive strength at the highly compressed extreme fiber in concrete subjected to highly compression and when there is no tension on the section shall be 0.75 times the strain at least compressed extreme fiber. Axial load and uniaxial bending: A member subjected to axial force and bending shall be designed on the basis of 1) The maximum compressive strength in concrete in axial compression is taken as 0. subjected to a minimum of 200 mm.if not nonexistent.0035-0.456(Table 24) A column may be classified based as follows based on the type of loading: 1) Axially loaded column 2) A column subjected to axial load and uneasily bending 3) A column subjected to axial load and biaxial bending.7. Effective length: The effective length of the column is defined as the length between the points of contraflexure of the buckled column. Therefore. every column should be designed for a minimum eccentricity .3 of the code for definition unsupported length) and D is the lateral dimension of the column in the direction under the consideration. 64 . a truly axially loaded column is rare . The code has given certain values of the effective length for normal usage assuming idealized and conditions shown in appendix D of IS . Where L is the unsupported length of the column (see 24.4 of IS code E min = (L/500) + (D/300).2. Axially loaded columns: All compression members are to be designed for a minimum eccentricity of load into principal directions.1. As beam transfer half of the load to each column it is connected.45*fck*Ac+0.2. the values of αn vary linearly from 1.0 Mux&Muy=moment about x and Y axis due to design loads Mux1&Muy1=maximum uniaxial moment capacity for an axial load of Pu bending about x and y axis respectively.1 and 38. The cross section of the column generally increases from one floor to another floor due to the addition of both live and dead load from the top floors. αn is values greater than 0. Alternatively such members may be designed by the following equation: (Mux/ Muy)αn +(Muy/ Muy1)αn <= 1.2 with neutral axis so chosen as to satisfy the equilibrium of load and moment about two weeks.0 The main duty of column is to transfer the load to the soil safely. shape and size of cross section. length and degree of proportional and dedicational restrains at its ends. αn is 2.0 for values less than 0.2 to 0. Also the amount if load depends on number of beams the columns is connected to. The strength of column depends upon the strength of materials. A column may be classify based on deferent criteria such as 65 .Axial load and biaxial bending: The resistance of a member subjected to axial force and biaxial bending shall be obtained on the basis of assumptions given in 38.2. Columns are designed for compression and moment.3.8. Column design: A column may be defined as an element used primary to support axial compressive loads and with a height of a least three times its lateral dimension.8 .0 to 2.75*fy*Asc For values of pu/Puz=0. 7. αn is related to Pu/puz Puz = 0. bar sizes. 2. In our structure we have 3 types of columns. The figure represents details regarding 1. max. Now analyze the column for loads to see the reactions and total loads on the column by seeing the loads design column by giving appropriate parameters like 1.) Pattern of lateral reinforcement. land 4. amount of steel and loading on the column is represented in the below figure. But in these structure we adopted same cross section throughout the structure with a rectangular cross section . Column with beams on two sides Columns with beams on three sides Columns with beams on four sides So we require three types of column sections.) Type of loading.1. 66 . Select the appropriate design code and input design column command to all the column. So create three types of column sections and assign to the respective columns depending on the connection. Minimum reinforcement. 3.) Slenderness ratio (A=L+D) 3. Longitudinal reinforcement The type of bars to be used. The column design is done by selecting the column and from geometry page assigns the dimensions of the columns.In foundations we generally do not have circular columns if circular column is given it makes a circle by creating many lines to increase accuracy.) Shape of the section 2. Transverse reinforcement 2. maximum and minimum spicing. Now run analysis and select any column to collect the reinforcement details The following figure shows the reinforcement details of a beam in staad. The ratio of effective column length to least lateral dimension is released to as slenderness ratio. 4 Skeleton Structure Showing Column No.Table 7. 1539 67 . 68 Figure 7.5 - Shear Bending For Column No. 1539 7.2.4. Check for Column Design: Short axially Loaded columns: Given data fck = 25 N/mm2 fy = 415N/mm2 puz = 19732.59 N b = 450mm d = 900mm 69 Design of reinforcement Area: (As per clause 39.6 of IS 456 2000) Puz 19732.59 = = 0.45fckAc + 0.75fyAsc 0.45*25*(350*450-Asc) + 0.75*415*Asc On solving the above equation we get Asc = 3241.15 Sq.mm.((Matched with Output) Design of Main(Longitudinal) reinforcement: (As per clause 26.5.3.1 of IS 456-2000 ) 1. The cross sectional area of longitudinal reinforcement shall not be less 0.8% , not more than 6% of the gross cross sectional area of the column. 2. The bars shall not be less than 12 mm in diameter. 3. Spacing of longitudinal bars measured along the periphery of the column shall not exceed 300 mm. Provided main reinforcement : 32 – 12mm dia (0.89%, 3619.95 Sq.mm.) Check for Transverse reinforcement : (As per clause 26.5.3.2 of IS 456-2000 ) A) pitch : shall not be more than the least of the following 1) Least lateral dimension of the compression member (350mm). 2) 16 x diameter of longitudinal reinforcement bar = 16x 12 = 192 mm 3) 300 mm B) Diameter : 1) Shall not be less than one fourth of the diameter of main reinforcement. 2) Not less than 6 mm. Provided Tie Reinforcement: Provide 8 mm dia. rectangular ties @ 190 mm c/c. 70 6 One Way Slab (lb/la > 2) 71 . Classification of Slab on the basis of spanning direction: a) Spanning in one direction (One Way Slab)  One way slab are those in which the length is more than twice the breadth it can be simply supported beam or continuous beam. It provides a working flat surface or a covering shelter in buildings. It supports mainly transverse loads and transfers them to support primarily by bending action in one or more directions. Reinforced concrete slab covers relatively large are compared to beam or column. But care has to be taken to see that its performance (serviceability) is not affected due to excessive deflection and cracking. dead load is large in the case of slab. SLABS A slab is a flat two – dimensional.7. leads to a considerable economy. A small reduction in depth of slab therefore. planar structural element having thickness small compared to its other two dimensions. Therefore volume of concrete and hence. FIGURE 7.3. 7 Load Distribution in a One Way Slab b) Spanning in two orthogonal direction (Two Way Slab) When slabs are supported to four sides two ways spanning action occurs.FIGURE 7. Such as slab are simply supported on any or continuous or all sides the deflections and bending moments are considerably reduces as compared to those in one way slab. 72 . ) Continuous beam Slabs are designed for deflection.8 Two Way Slab (lb/la > 2) FIGURE 7.10 Load Distribution showing One way & Two waySlabs 73 . Slabs are designed based on yield theory This diagram shows the distribution of loads in two slabs. FIGURE 7.) Simply supported slab b.9 Load Distribution in a Two Way Slab Checks: There is no need to check serviceability conditions.FIGURE 7. a. because design satisfying the span for depth ratio. 125 m.1. Now similar to the above designs give the parameters based on code and assign design slab command and select the plates and assign commands to it. Beam & Column. Design detail and sample calculation of a typical slab: 6310mm S1 3584mm 74 . FIGURE 7. Now selecting the members to form slab and use form slab button. After analysis is carried out go to advanced slab design page and collect the reinforcement details of the slab.3. Now give the thickness of plate as 0. 11 Monolithic connection between Slab.In order to design a slab we have to create plates by selecting the plate cursor. 7. 58+dxx=3.58+.116=3.76<2 Hence it is two-way slab iii. LOAD CALCULATION Considering width of slab 1m 75 . an overall depth of slab is 140 mm.FIGURE 7.696=1.32+.696 6. dxx= 140-Ø/2-cover=140-8/2-20=116mm dyy=140-Ø/2-cover-8=140-108mm ii.456/3.133=6. CALCULATION OF EFFECTIVE SPAN lx ly ly/lx = = = 3. 12 Plan showing slabs i.456 6. Using 8mm dia bars and providing 20 mm clear cover. DESIGN OF TWO WAY SLAB:Calculation of thickness of slab using l/D = 26 Therefore. 1 Type of panel = Two adjacent edge continuous.175 kNm 76 . = Calculation of coefficient according to IS 456.6962= 10.1 and 24. αx (-ve) at 1.006×24×1=0. CALCULATION FOR MOMENT = = = = = = 3.040×24×1=1.5kN/m2 2kN/m2 0.084 αx(+ve) at 1.clauses D-1.6962= 0.25 kN/m2 3.125×3.0 kN/m2 0.5 iv.063 vi.063×10.75×factor of safety Taking factor of safety 1.125×3.75 kN/m2 6.76 = 0.25=6.90 kNm 8.5+2+1+0.5=10.125kN/m2 There will be negative moment at continuous edge and positive Moment at mid span= Mx My = = = αx×Wu×lx2 αy×Wu×lx2 short span coefficient long span coefficient Where αx Where αY v.Dead load=DL=1×25×.4. Moment calculation Mux(-ve) Mux(+ve) = = 0.75×1.140 Live load=LL Floor finishing (25mm thick) Plaster (6mm thick) Total load Factor load = 6.76 = 0.084×10. 5418 ) = 10. 77 .6962= 0.42×0.Muy(+ve) Muy(-ve) vii.48)×25 = 3.45 kN/mm2 b = 1000 mm M =Max (10.12×1000×160/100=192mm2 Hence it is ok. 8.d2))]}b.175.125×3. CALCULATION OF AREA REQUIRED IN THE MID SPAN Equation for finding Ast Ast = (0.035×10.099 4.42Xumax/d)×fck R =0.90.5418kNm CHECK FOR DEPTH d= √(M/Rb) R = 0.047×10.36 Xu max/d (1-0.d  Astxx= 273.b.36×0.099 kNm 4. viii.125×3.439mm2 Check Astmin=.12×b×D/100= . = = 0.45×103×100) = 72 mm <d available Hence safe. 6.90×106)/(3.90 kNm dreq = √(10.6Mu/fck.5 fck/fy){1-[√(1-(4.48×(1-0.6962= 6. 18 150.max = 1.2 qu[Lx(e/2e+1)] Vu.30 for D<150mm Vu = k.76+1)] = 16.b.58(1. pt at Shorter Midspan = Ast*100/b.3*0. k=1. pt < Assumed p t (0. it should be less than 26.max = 0.d = 1.35*1000*115 = 52. Astx = 202mm2 at midspan.76/2*1.17% xi.24% τc = from Table 19 of IS 456 = 0.24 Dia – Spacing Provided Ast (mm) #8 – 180 #8 – 240 #8 – 300 #8 – 300 279 309 168 168 (mm2) -At Midspan 8. pt = 100*279/1000*115 = 0.70KN Therefore.2. 78 .40. Ast1 = 279mm2.Span Position Short – at Support Long – at Support Mu (KNm) 10.5418 -At Midspan 6. Ast (mm2) 274 202 111.32KN > 16.1. CHECK FOR DEFLECTION:Deflection=(Lx/d)×Mf .93KN {where e = Ly/Lx} Hence SAFE Long Edge Discontinuous: Vu. CHECK FOR SHEAR a) Long Edge Continuous : Vu.d Since Req.30) = 202*100/100*115 Hence SAFE.93KN Since. CHECK FOR SERVICEABILITY Req.2*10.max = 1.175 4. Mf is modification factor. Where.93/2) = 12.1 IS 456.90 115 115 115 115 d (mm) Req.125[3.9*(16.35 Cl. For safe. = 0. τc. x. Assuming 50% bent up to resist moment due to partial fixity.099 ix. 07% xii.3*0.087% Vuc = 1. τc = 0.50 .70 . Long Edge Continuous : Req. CHECK FOR DEVELOPMENT LENGTH a) 1. Hence OK.50KN Short Edge Discontinuous: Vu.3*0. Astx = 168mm2 at midspan.218n/mm2 k=1. pt = 84*100/1000*115 = 0.59 > 12.Ast1 = 101mm2.85 > 14. Ast1 = 84mm2. Assuming 50% bent up to resist moment due to partial fixity.(Lx/3) Ast1 = 168mm2 Vuc = 50.3 pt = 101*100/1000*115 = 0.89 > 10. Hence OK.22*1000*115/1000 = 32. Hence OK.70KN Therefore.2qu.max = 0. 79 . = 1.2*10.50/2) = 12. τc = 0.875 .58/3 = 14.3 Vuc = 1.max = 1. b) Short Edge Continuous: Vu.9*(14.218*1000*115/1000 = 32.22n/mm2 k=1.125*3. 78 – 1.For Fe415.max = 10.47*8 = 515. Ld = 64.3M1/V) = 515.47 * 8 = 515.78mm Assuming 50% bars bent up .47 * 8 = 515.08/12.78mm Hence OK. M25.175/2 = 4. Hence OK 2) Short Edge Discontinuous: Ld = 64. M1 = 6.14mm from inner face of support.78mm Assuming 50% bars bent up .78mm Available Ld = L/4 = 896mm.max = 12.3*4. Long Edge Discontinuous: Ld = 64.08KNm Vu. b) 1)Short Edge Continuous: Req. available anchorage length = 8db + 235 = 64 + 235 = 299mm > 235mm Hence OK. Ld (available) = L/4 = 3584/4 = 896mm.099/2 = 3.049KNm Vu.875KN 80 . M1 = 8.14 + 300/2 Lex = 248.14mm Lex => (Ld/3 – bs/2) = 98. Ld = 64. Straight Length available inside inner support = B =bs-A B = 300-(5*8+25) = 235mm Using 90degree bend.70KN Lex => (Ld-1.47 * 8 = 515.70 = 98. 2. will be required in both direction at right angles in each of the two meshes.Lex => (Ld-1.78 – 1. Straight Length available inside inner support = B =bs-A B = 300-(5*8+25) = 235mm Using 90degree bend. Since slab is discontinuous over both edger. available anchorage length = 8db + 235 = 64 + 235 = 299mm > 235mm Hence OK. TORSION STEEL a) At corners near column C127 & C128.3*3. xiii.049/10. Required area of torsion steel = 1/2(150) = 75mm2 81 .875 = 151.75 Astx = 0. Full Torsion Steel = 0.8mm b) At corner near column C126.30mm from inner face of support.30mm Lex => (Ld/3 – bs/2) = 151. One at the top and the other at the bottom up to the length of: Lx/5 = 3584/5 = 716.3M1/V) = 515.75*202 = 150mm2 .30 + 300/2 Lex = 301. STAAD OUTPUT for Element Design: 82 .7.3.2. 83 . 84 . 13 85 .**************************************************************************** FIGURE 7. ) Permissible differential settlements 5. FOUNDATION Foundations are structural elements that transfer loads from the building or individual column to the earth . moments and forces and the induced reactions and to assure that any settlements which may occur will be as nearly uniform as possible and the safe bearing capacity of soil is not exceeded. 7.) Economy 86 . 7. 7. The total load per unit area under the footing must be less than the permissible bearing capacity of soil to the excessive settlements.4. to minimize differential settlement and to provide adequate safety against sliding and overturning. Bearing Capacity of Soil: The size foundation depends on permissible bearing capacity of soil.2.) Type of loads 4.1.4. 2) Thickness at the edge of the footing: in reinforced and plain concrete footing at the edge shall be not less than 150 mm for footing on the neither soil nor less than 300mm above the tops of the pile for footing on piles.4. General: 1) Footing shall be designed to sustain the applied loads.If these loads are to be properly transmitted.3. Foundation design: Foundations are structure elements that transfer loads from building or individual column to earth this loads are to be properly transmitted foundations must be designed to prevent excessive settlement are rotation to minimize differential settlements and to provide adequate safety isolated footings for multi storey buildings.7. foundations must be designed to prevent excessive settlement or rotation. These may be square rectangle are circular in plan that the choice of type of foundation to be used in a given situation depends on a number of factors.) Type of structure 3. 1.4.) Bearing capacity of soil 2. The strip footings have T section and the flange of T section faces downwards. the depth of flange is increased towards the rib. These are the types of foundations the software can deal.  Shallow (D<B)  Isolated (Spread) Footing Combined (Strip) Footing Mat (Raft) Foundation Deep (D>B) Pile Cap Driller Pier 7. In order to design footings we used the software named STAAD FOUNDATION V8i. The thickness of the flange is kept constant. Criterion for Combined Strip Footing: Heavily loaded column when these are supported on relatively weak or uneven soils having low bearing capacity (which is equal to 175KN/m2) need large bearing area.A footing is the bottom most part of the structure and last member to transfer the load. instead of individual footing. The weight of the footing is not considered in structural design because it is assumed to be carried by the subsoil. The projection of T-section behaves as a Cantilever. Design using STAAD FOUNDATION V8i: 87 .5.4. It is similar to a floor resting on a system on a system of beams and columns. Continuous Strip Footing is provided to support more than two columns in a row.4. when the cantilever projection is of small length. Thus the continuous strip footing runs along the column row. 7. Otherwise.4. In such case. as it is already specified in the Staad Pro file. The main advantage of this software is that it automatically generated the reaction and moment values at supports when the load cases are defined. there is no need to specify the column positions.- Import the Staad Pro V8i analyzed file into Staad Foundation V8i using the IMPORT option. FIGURE 7. 14 Staad Foundation Page Showing Continuous Strip Footing When the file is imported from the Staad Pro V8i. 15 Zoom View of continuous strip Foundation & Columns 88 . Figure 7. 5(DL + LL) - The next step is to create the job for the footing (i. Combined Footing.- The load combination or the load cases are generated (selected) for which the foundation is to be designed. Footing Geometry FIGURE 7. 17 Cover & Soil Parameters 89 .1. Cover & Soil. 16 Concrete & Rebar Parameters FIGURE 7.) Now the design parameters are entered which includes: Concrete & Rebar. Assign Loading: .e. Soil type. Type of foundation. Unit weight of concrete: Minimum bar spacing: Maximum bar spacing: Strength of concrete: Yield strength of steel: Minimum bar size: Maximum bar size: Bottom clear cover: Unit weight of soil: Soil bearing capacity: Minimum length: 25KN/m^3 50mm 500mm 35N/mm^2 415 n/mm^2 12mm 60mm 50mm 22 KN/m^3 175 KN/m^3 1000mm 90 . safety factors.FIGURE 7. 18 Footing Dimensions The following input data is required regarding materials. Type of foundation: Combined. 100 0.475 8.25 11.790 20.840 65. overturning.775 6.700 1.475 8.775 6.975 2. Table 7.225 Dimensions of the Continuous Strip Footings Right Overhang (m) 3.875 4.040 62.5.1 Footing No.Minimum width: Minimum thickness: Maximum length: Maximum width: Maximum thickness: Plan dimension: Aspect ratio: 3500mm 500mm 70000mm 40000mm 2000mm 50mm 1 Safety against friction.210 Width (m) 9. Left Overhang (m) 1 2 3 4 5 3.450 7.975 2. sliding.875 4.700 1.950 Thickness (m) 0. 1.760 55.300 1. 0. After the analysis.5 Now the last step is to click on DESIGN.225 Length (m) 23.5. detailed calculation of each and every footing is given with plan and elevation.450 17.050 14.250 91 .1. 69.00 mm 40000.00 mm 2000.4. Design Calculations for Combined Footing 1 (FC1) Column Dimensions for Column No. Main Steel Top 1 2 3 4 5 #12 @ 125mm c/c #12 @ 75mm c/c #12 @ 125mm c/c #12 @ 50mm c/c #12 @ 50mm c/c DESIGN RESULTS Footing Reinforcement Main Steel Bottom Secondary Steel Top #12 @ 125mm c/c #12 @ 75mm c/c #12 @ 125mm c/c #12 @ 50mm c/c #12 @ 75mm c/c Secondary Steel Bottom #16 @50mm c/c #16 @50mm c/c #12 @50mm c/c #20 @75mm c/c #25 @50mm c/c #32 @ 75mm c/c #40 @75mm c/c #20 @50mm c/c #40 @50mm c/c #40 @50mm c/c 7.Z (Pw): Rectangular 1000mm 500mm Length of left overhang : Length of right overhang : Is the length of left overhang fixed? Is the length of right overhang fixed? Minimum width of footing (Wb) : Minimum Thickness of footing (Do) : Maximum Width of Footing (Wb) : Maximum Thickness of Footing (Do) : 1.Table 7. Footing No.6.00 m 1.50 m 500.00 mm 92 . FC1) Column Shape: Column Length . 102 and 101 (Combined Footing No.X (Pl): Column Width . 103.00 m No No 3.2. Maximum Length of Footing (Lo) : Length Increment : Depth Increment : 70000.00 mm Design Calculations 93 .00 mm 400.000 kN/m3 35.00 kN/m2 2.00 kN/m2 44.00 m -4000mm Concrete and Rebar Properties Unit Weight of Concrete Compressive Strength of Concrete : Yield Strength of Steel : Minimum Bar Size : Maximum Bar Size : Minimum Bar Spacing : Maximum Bar Spacing : 25.00 mm Cover and Soil Properties Pedestal Clear Cover : Footing Clear Cover : Unit Weight of soil : Soil Bearing Capacity : Soil Surcharge : Depth of Soil above Footing : Depth of Water Table : 50.000 N/mm2 12 60 50.000 N/mm2 415.00 mm 50.00 mm 50.00 mm 22.00 mm 50.00 kN/m3 175. Lright_overhang : Table 7. L : Width of footing. Final footing dimensions are: Length of footing.46 sq m 60. Lleft_overhang : Length of right overhang.25 0.12 3.51 sq m 94 .00 kN 123.88 3.04 9. 23. Ao = L x W: Therefore.3. Do : Area. W : Depth of footing. Amin = Pcritical / qmax : Area from initial length and width.70 213.88 m m m sq m m m -0.Footing Size Calculations Reduction of force due to buoyancy = Minimum area required from bearing pressure. A : Length of left overhang. 007604 kN/m^2 = 7928.479107 = 4822. to account for uplift. Otherwise.5.520713 kNm 95 . Design for Flexure Sagging moment along length Effective Depth = Governing moment (Mu) As Per IS 456 2000 ANNEX G G-1. there is no uplift and no pressure adjustment is necessary.4.63 m =17882. Table 7.346683 kNm = 0.1C Limiting Factor1 (Kumax) = Limiting Factor2 (Rumax) = Limit Moment Of Resistance (Mumax)= = 0.Table 7. If Au is zero. areas of negative pressure will be set to zero and the pressure will be redistributed to remaining corners. 63 = 3.007604 kN/m^2 =45790.368019 kNm Mu <= Mumax Transverse direction hence.64 m = 6300.321341 kNm = 0.132556 kNm 96 . safe = 0.1C Limiting Factor1 (Kumax) = Limiting Factor2 (Rumax) = hence.771009 m kNm = 0.479107 =4822.007604 kN/m^2 Limit Moment Of Resistance (Mumax) = =18498.Mu <= Mumax Hogging moment along length Effective Depth = Governing moment (Mu) As Per IS 456 2000 ANNEX G G-1. safe Effective Depth = Governing moment (Mu) = As Per IS 456 2000 ANNEX G G1.479107 = 4822.1C Limiting Factor1 (Kumax) = Limiting Factor2 (Rumax) = Limit Moment Of Resistance (Mumax) = = 0. safe = 3538.18 kN kN/m^2 = 1.0199 kN/m^2 = 1479.00 97 .0199 kN/m^2 hence.1 Ks = Shear Strength(Tc)= Ks X Tc Tv<= Ks X Tc For Column 2 Shear Force(S) Shear Stress(Tv) As Per IS 456 2000 Clause 31.99 kN kN/m^2 = 1. safe = 4890.1 Ks = hence.6.Mu <= Mumax Check trial depth for one way shear(along length) Shear Force(S) Shear Stress(Tv) Percentage Of Steel(Pt) As Per IS 456 2000 Clause 40 Table 19 Shear Strength Of Concrete(Tc) Tv< Tc Check trial depth for two way shear For Column 1 Shear Force(S) Shear Stress(Tv) As Per IS 456 2000 Clause 31.000000 = 0.00 = 1479.28 = 0. safe kN/m^2 = 2618.3.6.45 = 729.35 hence.83 = 1361.080 kN kN/m^2 = 0.3. 6.0199 kN/m^2 hence.07 kN kN/m^2 = 1.1 Ks = Shear Strength(Tc)= Ks X Tc Tv<= Ks X Tc = 1479.80 = 1391.50 kN kN/m^2 = 1.00 = 1479.00 = 1479.0199 kN/m^2 hence. safe = 4996.Shear Strength(Tc)= Ks X Tc Tv<= Ks X Tc For Column 3 Shear Force(S) Shear Stress(Tv) As Per IS 456 2000 Clause 31.0199 kN/m^2 = 1479. safe = 2639.0199 kN/m^2 = 1479.3.3.6.0199 kN/m^2 = 1479.60 = 735.1 Ks = Shear Strength(Tc)= Ks X Tc Tv<= Ks X Tc For Column 4 Shear Force(S) Shear Stress(Tv) As Per IS 456 2000 Clause 31.0199 kN/m^2 hence. safe 98 . = 16.1 Minimum Area of Steel (Astmin) Calculated Area of Steel (Ast) Provided Area of Steel (Ast.38 mm mm The reinforcement is accepted.000 = 50.. = 19353... 99 .5.00 = 7770.2.00 = 7770.5..Selection of reinforcement Top reinforcement along length As Per IS 456 2000 Clause 26.00 = 134.2. Along width As Per IS 456 2000 Clause 26.000 = 50.08 mm mm mm The reinforcement is accepted.00 = 134.1 Provided Minimum Area of Steel (Astmin) = 32.00 Steel area is accepted mm2 mm2 mm2 Selected bar Dia Minimum spacing allowed (Smin) = Selected spacing (S) Smin <= S <= Smax and selected bar size < selected maximum bar size.57 mm2 Selected bar Dia Minimum spacing allowed (Smin) Selected spacing (S) Smin <= S <= Smax and selected bar size < selected maximum bar size.Provided) Astmin<= Ast.Provided = 7770. Provided) Astmin<= Ast.93 = 74809.5.00 = 61.73 = 97241.57 = 74809.5.Bottom reinforcement along length As Per IS 456 2000 Clause 26. = 19353..Provided = 7770.Provided Selected bar Dia Minimum spacing allowed (Smin) = Selected spacing (S) = 32.2. Along width As Per IS 456 2000 Clause 26.Provided) Astmin<= Ast.1 Minimum Area of Steel (Astmin) Calculated Area of Steel (Ast) Provided Area of Steel (Ast..00 = 75.00 = 97241.1 Minimum Area of Steel (Astmin) Calculated Area of Steel (Ast) Provided Area of Steel (Ast. 100 .000 = 50.62 mm2 mm2 mm2 mm mm mm Smin <= S <= Smax and selected bar size < selected maximum bar size.000 = 50.93 Steel area is accepted = 16.73 Steel area is accepted mm2 mm2 mm2 Selected bar Dia Minimum spacing allowed (Smin) = Selected spacing (S) Smin <= S <= Smax and selected bar size < selected maximum bar size.98 mm mm mm The reinforcement is accepted.2. The reinforcement is accepted. 20 101 .FIGURE 7. 19 FIGURE 7. 7. Detail Drawings FIGURE 7.4. 21 .7.Strip Footing FC2 102 . 22 .Strip Footing FC1 FIGURE 7. 24 - Strip Footing FC4 103 .Strip Footing FC3 FIGURE 7.FIGURE 7. 23 . Strip Footing FC5 ************************************************************************************* 104 . 25 .FIGURE 7. CONCLUSION 105 . Column Design: Columns are designed for axial forces and biaxial moments at the ends. The loading which yield maximum reinforcement is called the critical load. Beam Design Output: The default design output of the beam contains flexural and shear reinforcement provided along the length of the beam. Design for Flexure: Maximum sagging (creating tensile stress at the bottom face of the beam) and hogging (creating tensile stress at the top face) moments are calculated for all active load cases at each of the above mentioned sections.CONCLUSION STAAD PRO has the capability to calculate the reinforcement needed for any concrete section. Column design is done for square section. 106 . shear and torsion. Two-legged stirrups are provided to take care of the balance shear forces acting on these sections. Shear capacity calculation at different sections without the shear reinforcement is based on the actual tensile reinforcement provided by STAAD program. Beams are designed for flexure. Each of these sections is designed to resist both of these critical sagging and hogging moments. doubly reinforced section is tried. All active load cases are tested to calculate reinforcement. The program contains a number of parameters which are designed as per IS: 456 : 2000 and IS 13920 : 1993. All major criteria for selecting longitudinal and transverse reinforcement as stipulated by IS: 456 have been taken care of in the column design of STAAD. Design for Shear: Shear reinforcement is calculated to resist both shear forces and torsional moments. Square columns are designed with reinforcement distributed on each side equally for the sections under biaxial moments and with reinforcement distributed equally in two faces for sections under uni-axial moment. Where ever the rectangular section is inadequate as singly reinforced section. It is designed to carry the load distributed by the structure through slabs to beams to columns to the footings. Slabs are designed as two way and one way.Slab Design: Slabs are designed for the load combinations as specified in IS 456:2000. Staad Foundation V8i And Auto Cad is well known after the completion of the project. Foundation Design: Footing is decided on the soil type. 107 . All active load cases are tested to calculate reinforcement. This enables to relate theoretical knowledge to real life practicalities. The loading which yield maximum reinforcement is called the critical load. loading conditions and area available. This enables to understand the detailing of reinforcement in the slabs. Use of Software’s: Use of Staad Pro V8i. ANNEXURE A Plan of the Multi-storey Hostel Building at SRM University (Figure A-1): 108 . ANNEXURE B Elevation of the Multi-storey Hostel building at SRM University (Figure A-2): 110 . REFERENCES o Dr.Published by: R . o Dr.R. Manak Bhavan. o Dr. Ashok K.Bureau Of Indian Standards. Ram Chandra . Roorkee o S. 9 Bahadur Shah Zafar Marg.2. New Delhi 110002 o IS 1893 : 2002 . Jain – “Reinforced Concrete Limit State Design”.“Limit State Design”. CODE BOOKS: o IS 875(Part 1.5) . 9 Bahadur Shah Zafar Marg. New Chand & Bros. V.L. Manak Bhavan. o “STAAD Pro 2004 & STAAD FOUNDATION V8i – Technical reference manual” Published by: R. New Delhi. New Delhi 110002 o IS 456 : 2000 . New Delhi 110002. S. Manak Bhavan.Bureau Of Indian Standards.“Illustrated design of Reinforced concrete Buildings”.Bureau Of Indian Standards. Dhanpat Rai Publishing Company o “STAAD Pro V8i – Getting started & tutorials” .E.3. Manak Bhavan. Standard Book House. 9 Bahadur Shah Zafar Marg.E.I. Karve & Dr. R. 111 . Narayana – “Theory of Structures”.Bureau Of Indian Standards. o IS 13920 : 1993 . Structures Publications. 9 Bahadur Shah Zafar Marg. Shah . New Delhi 110002. Ramamrutham. I.