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 (1010930013)

DHRUV GUPTA (1010930017)

GAURAB PAUL (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 Dr. Vineet Bajaj Mr. Pradeep Kumar

(Director) (Head of Department) (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

ABSTRACT

In this growing world, 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. As

a sequel to this an attempt has been made to learn the process of analysis and design of a multi-

storey Building using Limit State Method (IS 456:2000).

The project focuses on „Reinforced Concrete‟ buildings. The design using Limit State Method

(of collapse and serviceability) is taken up. In the limit state of collapse, the strength and stability

of structure is ensured. The guidelines being followed are as per IS 456:2000 and IS 13920 :

1993.

The structural components in a typical multi storey building, consists of floor system which

transfers the floor loads to a set of plane frames in one or both directions. The design study

comprises of the footing, columns, beams and slabs.

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, NCR Campus. The loadings are applied

and the design for beams, columns, slabs and footings is obtained.

STAAD Pro with its new features surpassed its predecessors and compotators with its data

sharing capabilities with other major software like AutoCAD, and MS Excel.

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.

CONTENTS

List of Tables i

List of Figures ii-iii

Assumptions and Notations iv-v

Symbols vi-vii

CHAPTER – 1 INTRODUCTION 1-2

CHAPTER – 2 LITERATURE SURVEY 3-12

2.1 Elements of Structural Design 4

2.2 Design Philosophies 7-9

2.3 Multi-Storey Building 9-11

2.4 Structural Planning 12

CHAPTER – 3 COMPUTER AIDED ANALYSIS & DESIGN 13-17

3.1 Staad Pro V8i 14

3.2 Alternatives for Staad Pro 15

3.3 Staad Editor 15

3.4 Staad Foundation V8i 16

3.5 Auto Cad 17

CHAPTER – 4 PLAN & ELEVATION 18-20

4.1 Plan 19

4.2 Elevation 20

CHAPTER – 5 LOADS 21-38

5.1 Load Conditions and Structural System Response 22

5.2 Building Loads Categorized by Orientation 22-23

5.3 Design Load for the Residential Building 24-30

5.4 Design Imposed Loads for Earthquake forces Calculation 31-35

5.4.1 Seismic Loading in Staad Pro V8i 32-33

5.5 Load Combinations 35-36

5.6 Inputs to Staad Editor for Loadings 37-38

CHAPTER – 6 ANALYSIS 39-54

6.1 Methods of Analysis 40-42

6.2 Seismic Analysis Procedure 43

6.3 Analysis using Staad Pro V8i 43

6.4 Analysis Results for Load Cases 1 to 4 44-47

6.5 Analysis Results for Support Reactions 48-54

CHAPTER – 7 DESIGN 55-105

Input to Staad Editor for Design 56

7.1 Beams 57-63

7.2 Columns 64-71

7.3 Slabs 72-86

7.4 Foundation 87-105

CONCLUSION 106-108

APPENDICES

APPENDIX A 109

APPENDIX B 110

REFERENCES 111

LIST OF TABLES

Table No. Title … Page No.

5.1 Zone Factor … 30

7.1 Dimensions of Continuous Strip Footing … 92

7.2 Design Results of Foundation … 93

7.3 Applied Loads-Allowable Stress Level … 95

7.4 Calculated Pressure at Four Corners … 96

7.5 Check for Stability against Overturning … 96

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LIST OF FIGURES

Figure No. Title … Page No.

5.1 Dead Load on the Structure … 25

5.2 Live Load on the Structure … 27

5.3 Seismic Parameters … 33

5.4 Seismic Load in X direction (SLX) … 34

5.5 Seismic Load in Z direction (SLZ) … 35

7.1 Location of Beam No. 1 in the Structure … 59

7.2 Beam Reinforcement … 60

7.3 Beam Web Reinforcement … 61

7.4 Skeleton Structure showing Column No. 1539 … 68

7.5 Shear Bending for Column No. 1539 … 70

7.6 One Way Slab … 72

7.7 Load Distribution in a One Way Slab … 73

7.8 Two Way Slab … 73

7.9 Load Distribution in a Two Way Slab … 74

7.10 Load Distribution showing One Way and Two Way … 74

7.11 Monolithic connection between Slab, Beam & Column … 75

7.12 Plan showing Slabs … 76

7.13 Detailing of Slabs … 86

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7.14 Staad Foundation Page showing Foundation … 89

7.15 Zoom View of Foundation … 89

7.16 Concrete and Rebar Parameters … 90

7.17 Cover and Soil Parameters … 90

7.18 Footings Dimensions … 91

7.19 Plan of Footings … 102

7.20 Elevation of Footings … 102

7.21 Strip Footing, FC1 … 103





A-1 Plan of the Multistorey SRM Hostel Building … 109

A-2 Elevation of the Multistorey SRM Hostel Building … 110

iii

ASSUMPTIONS AND NOTATIONS

The notations adopted throughout the work are same IS-456-2000.

Assumptions in Design:

1.Using partial safety factor for loads in accordance with clause 36.4 of IS-456-2000 as ϒt=1.5

2.Partial safety factor for material in accordance with clause 36.4.2 is IS-456-2000 is taken as 1.5

for concrete and 1.15 for steel.

3.Using partial safety factors in accordance with clause 36.4 of IS-456-2000 combination of

load.

D.L+L.L. 1.5

D.L+L.L+E.L 1.2

Density of materials used:

MATERIAL: DENSITY

i) Plain concrete 24.0KN/m3

ii) Reinforced 25.0KN/m3

iii) Flooring material (c.m) 20.0KN/m3

iv) Brick masonry 19.0KN/m3

v) Fly ash 5.0KN/m3

4.LIVE LOADS: In accordance with IS. 875-86

i) Live load on slabs 20.0KN/m2

ii) Live load on passage 4.0KN/m2

iii)Live load on stairs 4.0KN/m2

DESIGN CONSTANTS:

Using M30 and Fe 415 grade of concrete and steel for beams, slabs, footings, columns.

Therefore:-

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fck Characteristic strength for M30-30N/mm2

fy 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.

ii) Beams are assumed to be continuous over interior support and they frame in to the column at

ends.

Assumptions on design:-

1) M20 grade is used in designing unless specified.

2) For steel Fe 415 is used for the main reinforcement.

3) For steel Fe 415 and steel is used for the distribution reinforcement.

4) Mild steel Fe 230 is used for shear reinforcement.

v

SYMBOLS

The following symbols have been used in our project and its meaning is clearly mentioned

respective to it:

A Area

Ast Area of steel

b Breadth of beam or shorter dimension of rectangular column

D Overall depth of beam or slab

DL Dead load

d1 Effective depth of slab or beam

D Overall depth of beam or slab

Mu,max Moment of resistance factor

Fck Characters tic compressive strength

Fy Characteristic strength of of steel

Ld Devlopment length

LL Live load

Lx Length of shorter side of slab

Ly Length of longer side of slab

B.M. Bending moment

Mu Factored bending moment

Md Design moment

Mf Modification factor

Mx Mid span bending moment along short span

My Mid span bending moment along longer span

Mx Support bending moment along short span

My support bending moment along longer span

pt Percentage of steel

W Total design load

Wd Factored load

Tc max Maximum shear stress in concrete with shear

Tv Shear stress in concrete

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Tv Nominal shear stress

ɸ Diameter of bar

Pu Factored axial load

Mu,lim Limiting moment of resistance of a section without compression reinforcement

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 moment X

and Y axis respectively

Ac Area of concrete &

Asc Area of longitudinal reinforcement for column

SLX Seismic Load in X direction

SLZ Seismic Load in Z direction

vii

CHAPTER 1

INTRODUCTION

1

Building construction is the engineering deals with the construction of building such as

residential houses. In a simple building can be define as an enclose space by walls with roof,

food, cloth and the basic needs of human beings. In the early ancient times humans lived in

caves, over trees or under trees, to protect themselves from wild animals, rain, sun, etc. as the

times passed as humans being started living in huts made of timber branches. The shelters of

those old have been developed nowadays into beautiful houses. Rich people live in sophisticated

condition houses.

Buildings are the important indicator of social progress of the county. Every human has

desire to own comfortable homes on an average generally one spends his two-third life times in

the houses. These are the few reasons which are responsible that the person do utmost effort and

spend hard earned saving in owning houses. Nowadays the house building is major work of the

social progress of the county. Daily new techniques are being developed for the construction of

houses economically, quickly and fulfilling the requirements of the community engineers and

architects do the design work, planning and layout, etc. of the buildings. Draughtsman is

responsible for doing the drawing works of building as for the direction of engineers and

architects. 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, site plans and

layout plans etc., as for the requirements.

A building frame consists of number of bays and storey. A multi-storey, multi-paneled

frame is a complicated statically intermediate structure. A design of R.C building of G+9 storey

frame work is taken up. The building in plan consists of columns built monolithically forming a

network. It is residential complex. The design is made using software on structural analysis

design (STAAD PRO V8i). The building subjected to both the vertical loads as well as

horizontal loads. The vertical load consists of dead load of structural components such as beams,

columns, slabs etc. and live loads. The horizontal load consists of the wind forces thus building

is designed for dead load, live load and wind load as per IS 875. 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. The help is taken by software

available in institute and the computations of loads, moments and shear forces and obtained from

this software.

2

CHAPTER 2

LITERATURE SURVEY

3

BACKGROUND WORK (LITERATURE SURVEY)

2.1 Elements of Structural Design

Structures in concrete have become very common in civil engineering construction.

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. Its low tensile strength is

compensated by the use of steel reinforcement. Thus, the concrete is strengthened(i.e. reinforced)

by steel and the resultant composite mass is known as Reinforced Cement Concrete (R.C.C.) It

is this combination which allows almost unlimited use of reinforced concrete in construction of

buildings, bridges, tanks, dams etc., with the result that almost every civil engineer is intimately

concerned with reinforced concrete (R.C.) structures. It is therefore, necessary that every civil

engineer knows the basic principles involved in design of R.C. structures. So, it will be

approximate to begin by reviewing the basic principles of structural design in general and then

its application to reinforced concrete structures.

2.1.1. 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, space, an enclose and/or cover to serve the desired

function.

Structural design is a science and art of designing, 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.

2.1.2. 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. The basic requirements of the structural

design are as follows:

i. 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. Safety of structure is achieved by adequate

4

ii. strength and stability. Besides strength, ductility of structure is also nowadays considered

to be an additional desired quality from a view point that if at all failure occurs, 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. Ductility is

thus obtained by providing steel of such quality that it would yield prior to crushing of

concrete.

iii. Serviceability: The structure shall efficiently serve the intended function and also shall

give a satisfactory performance throughout the life span. The performance is rated buy

the fitness of the structure to maintain deflections, deformations, cracking and vibration

effects within acceptable limits. It is achieved by providing adequate stiffness and

cracking resistance.

iv. Durability: The structure shall resist effectively environmental action during its

anticipated exposure conditions, such as rain, alternate wetting and drying or freezing,

climatic variations in temperature and humidity, chemical actions of salt, abrasion action

etc.

v. 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.

vi. Aesthetics: The structure should be so designed that it should not only be safe,

serviceable and durable but should also give a pleasing appearance without affecting the

economy to a great extent.

vii. Feasibility, Practicability and Acceptability: The structure has to be so designed that

the proposed solution is feasible, practicable an acceptable.

2.1.3. The Design Process:

The entire process of design requires conceptual thinking, sound knowledge of engineering,

relevant design codes and byelaws, backed up by experience, imagination and judgment. The

codes of practice are compendia of good practice drawn by experienced and competent

engineers. They are intended to guide the engineers and should not be allowed to replace their

conscience and competence.

5

The design process commences with the planning of the structure primarily to meet its functional

requirement and then designed for safety and serviceability. Thus, 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. This

includes proper arrangement of rooms, halls, good ventilation, and acoustics,

unobstructed view in cinema theatre / community halls, proper water supply and

drainage arrangements etc.

2) Structural Design: As mentioned earlier Structural design is a science and art of

designing, 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.

It consists of the following steps:

a) Structural Planning

b) Determination of Loads

c) Analysis

d) Member Design

e) Drawing, Detailing and Preparation of Schedule.

2.1.4. Elements of a R.C. Building Frame

The principle elements of a R.C. building frame are slab, beam, column and footing.

a) Slab: It is two-dimensional or a planar member supporting a transverse load and

providing a working floor or a covering shelter. The loads are transferred to supporting

beams or walls in one or both directions.

b) Beam: A Beam is a one-dimensional (normally horizontal) flexural member which

provides support to the slab and the vertical walls.

c) Column: It is one dimensional vertical member providing a support to beam. Load is

transferred primarily by axial compression accompanied by bending and shear.

6

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

A wide range of approaches have been used for buildings of varying heights and

importance, from simple approximate methods which can be carried out manually, or with the

aid of a pocket calculator, to more refined techniques involving computer solutions. Till a few

years ago most of the multistory buildings were analyzed by approximate methods such as

substitute frame, moment distribution, portal and cantilever methods.

The recent advancement of abundance of ready-made computer package programs has

reduced the use of approximation methods. This has been induces from analysis to design, to

plotting, to detaining, to specification writing, to cost estimating, etc.

2.3.1. Structural Systems

A building is subjected to several loads which are transferred to ground through a system

of interconnected structural members.

In tall buildings, the biggest challenge comes from controlling lateral displacements

within the serviceability limit state.

The lateral stiffness may be achieved through a permutation and combination of

placement of columns and walls in plan.

A structural system may be classified as follows:-

1. Load Bearing wall system: -

Walls provide support for all gravity loads as well as resistance to lateral loads.

No columns.

The Walls and partition wall supply in-plane lateral stiffness and stability to resist

wind and earthquake loads.

Clause 8.2.1 and 8.4.8 of IS: 4326-1993 restricts the use of such system to 3

storey in seismic zone V and 4 storey in other zone.

2. Building with flexural (shear) wall system: -

Gravity load is carried by frame supported on columns rather than on bearing

walls.

The frame provides vertical stability to the building and prevents collapse after

damage to flexural wall or braced frames.

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3. Moment resisting frame system: -

Members and joints are capable of resisting vertical and lateral loads primarily by

flexure.

Relative stiffness of girders and columns is very important.

A frame can be designed using weak column-strong girder proportions or strong

column-weak girder proportions.

4. 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.

The America IBC 2000 permits use of flexural (shear) wall system up to 45m

high. However it can be used up to 70m; if and only if, shear walls in any plane

do not resist more than 33% of earthquake design force including torsional

effects.

5. Dual frame system: -

Moment resisting frame providing support for gravity loads.

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.

Flexural walls i.e. shear walls or braced frames must resist total required

lateral loads.

6. Space frame: -

3-Dimensional structural system without shear or bearing walls composed of

interconnected members laterally supported

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2.4. Structural Planning

Salient features:

Utility of building Residential Hostel Building

No of stories G+9

No of staircases 1

No. of Rooms/floor 22 rooms on each floor with attached washroom.

No of lifts 1

Type of construction R.C.C framed structure

Types of walls Brick wall

Ventilation Ventilated rooms with window in each room.

Geometric details:

Ground floor 2m

Floor to floor height 3.65m.

Height of plinth 2m

Depth of foundation 2m

Materials:

Concrete grade M35 (for footing) & M25 (for all other elements)

All steel grades Fe415 grade

Bearing capacity of soil: 175KN/m2

Depth of Water Table 4m.

12

CHAPTER 3

COMPUTER AIDED

ANALYSIS & DESIGN

13

COMPUTER AIDED ANALYSIS AND DESIGN

This project is mostly based on software and it is essential to know the details about these

software‟s.

List of software‟s used

1. Staad Pro (V8i)

2. Staad foundations 5(V8i)

3. Auto Cad 2010

STAAD PRO V8i STAAD FOUNDATIONV8i AUTOCAD 2010

3.1. 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. So first

find the outline of the structure, 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. Design phase is designing the type of materials and its dimensions to resist the load. This

we do after the analysis.

To calculate S.F.D and B.M.D of a complex loading beam it takes about an hour. So

when it comes into the building with several members it will take a week. 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. This software can be used to carry

RCC, steel, bridge, truss etc. according to various country codes.

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3.2. Alternatives for Staad Pro V8i:

STRUDS, ETAB, ROBOT, SAP which gives details very clearly regarding reinforcement and

manual calculations. But these software‟s are restricted to some designs only whereas Staad can

deal with several types of structure.

3.3. Staad Editor:

Staad has very great advantage to other software‟s i.e., Staad editor. Staad editor is the

programming

For the structure we created and loads we taken all details are presented in programming format

in Staad editor. This program can be used to analyze other structures also by just making some

modifications, but this require some programming skills. So load cases created for a structure can

be used for another structure using Staad editor.

Limitations of Staad Pro V8i:

1. Huge output data

2. Even analysis of a small beam creates large output.

3. Unable to show plinth beams.

3.4. Staad foundation:

Staad foundation is a powerful tool used to calculate different types of foundations. 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.

Analysis and design carried in Staad and post processing in Staad gives the load at various

supports. These supports are to be imported into this software to calculate the footing details i.e.,

regarding the geometry and reinforcement details.

This software can deal different types of foundations

SHALLOW (D<B)

� 1. Isolated (Spread) Footing

� 2.Combined (Strip) Footing

� 3.Mat (Raft) Foundation

15

DEEP (D>B)

� 1.Pile Cap

� 2. Driller Pier

1. Isolated footing is spread footing which is common type of footing.

2. Combined Footing or Strap footing is generally laid when two columns are very near to each

other.

3. Mat foundation is generally laid at places where soil has less soil bearing capacity.

4. Pile foundation is laid at places with very loose soils and where deep excavations are required.

So depending on the soil at type we have to decide the type of foundation required.

Also lot of input data is required regarding safety factors, soil, materials used should be given in

respective units.

After input data is give software design the details for each and every footing and gives the

details regarding

1. Geometry of footing

2. Reinforcement

3. Column layout

4. Graphs

5. Manual calculations

These details will be given in detail for each and every column.

Another advantage of foundations is even after the design; properties of the members can be

updated if required.

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.

16

3.5. AutoCAD:

AutoCAD is powerful software licensed by auto desk. The word auto came from auto Desk

Company and cad stands for computer aided design. AutoCAD is used for drawing different

layouts, details, plans, elevations, sections and different sections can be shown in auto cad. It is

very useful software for civil, mechanical and also electrical engineer. The importance of this

software makes every engineer a compulsion to learn this software‟s.

We used AutoCAD for drawing the plan, elevation of a residential building. We also used

AutoCAD to show the reinforcement details and design details of a stair case.

AutoCAD is a very easy software to learn and much user friendly for anyone to handle

and can be learn quickly. Learning of certain commands is required to draw in AutoCAD.

17

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

LOADS

5.1. 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.C.C structures. As shown in Table, building loads can be

divided into types based on the orientation of the structural action or forces that they induce:

vertical and horizontal (i.e. lateral) loads. Classification of loads is described in the following

sections.

5.2. Building Loads Categorized by Orientation:

Types of loads on a hypothetical building are as follows.

� Vertical Loads

� Dead Load (gravity)

� Live (gravity)

� Snow (gravity)

� Wind (uplift on roof)

� Seismic and wind (overturning)

� Seismic (vertical ground motion)

5.2.1. Horizontal (Lateral) Loads:

Direction of loads is horizontal w.r.t to the building.

� Wind

� Seismic (horizontal ground motion)

� Flood (static and dynamic hydraulic forces

� Soil (active lateral pressure)

5.2.2. Vertical Loads:

Gravity loads act in the same direction as gravity (i.e., downward or vertically) and include dead,

live, and snow loads. They are generally static in nature and usually considered a uniformly

distributed or concentrated load. Thus, 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, including the dead load (i.e., weight of the construction) and any applied loads(i.e.,

22

live load). For example, 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.

The structural designer then selects a standard beam or column model to analyze bearing

connection forces (i.e., reactions) internal stresses (i.e., bending stresses, shear stresses, and axial

stresses) and stability of the structural member or system a for beam equations. The selection of

an appropriate analytic model is, however no trivial matter, 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, but to varying degrees.

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.

As with gravity loads, the influence of wind uplift pressures on a structure or assembly (i.e. roof)

are analyzed by using the concept of tributary areas and uniformly distributed 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, particularly proximity to changes in geometry (e.g., eaves, corners, and ridges).Even

though the wind loads are dynamic and highly variable, the design approach is based on a

maximum static load (i.e., pressure) equivalent. 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, Earthquakes also produce vertical ground motions or accelerations which

increase the effect of gravity loads. However, Vertical earthquake loads are usually considered to

be implicitly addressed in the gravity load analysis of a light-frame building.

5.2.3. Lateral Loads:

The primary loads that produce lateral forces on buildings are attributable to forces associated

with wind, seismic ground motion, floods, and soil. Wind and seismic lateral loads apply to the

entire building. 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,

creating a combined push and-pull effect.

Seismic lateral forces are generated by a structure‟s dynamic inertial response to cyclic

ground movement. The magnitude of the seismic shear (i.e., lateral) load depends on the

23

magnitude of the ground motion, the buildings mass, and the dynamic structural response

characteristics (i.e., dampening, ductility, natural period of vibration, etc.). For houses and other

similar low rise structures, a simplified seismic load analysis employs equivalent static forces

based on fundamental Newtonian mechanics (F=ma) with somewhat subjective (i.e., experience-

based) adjustments to account for inelastic, ductile response characteristics of various building

systems.

Flood loads are generally minimized by elevating the structure on a properly designed

foundation or avoided by not building in a flood plain. Lateral loads from moving flood waters

and static hydraulic pressure are substantial. Soil lateral loads apply specifically to foundation

wall design, mainly as an “out-of-plane” bending load on the wall. Lateral loads also produce an

overturning moment that must be offset by the dead load and connections of the building.

Therefore, overturning forces on connections designed to restrain components from rotating or

the building from overturning must be considered. Since wind is capable of the generating

simultaneous roof uplift and lateral loads, the uplift component of the wind load exacerbates the

overturning tension forces due to the lateral component of the wind load. 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.

5.3. 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.e., safety and serviceability) throughout the structure‟s useful life. The anticipated

loads are influenced by a building‟s intended use (occupancy and function); configuration (size

and shape) and location (climate and site conditions).Ultimately, the type and magnitude of

design loads affect critical decisions such as material collection, construction details and

architectural configuration.

Since building codes tend to vary in their treatment of design loads the designer should, as a

matter of due diligence, identify variances from both local accepted practice and the applicable

24

code relative to design loads as presented in this guide, even though the variances may be

considered technically sound.

5.3.1. Dead Loads:

Dead loads consist of the permanent construction material loads compressing the roof, floor,

wall, and foundation systems, including claddings, finishes and fixed equipment. Dead load is

the total load of all of the components of the components of the building that generally do not

change over time, such as the steel columns, concrete floors, bricks, roofing material etc. In staad

pro assignment of dead load is automatically done by giving the property of the member. In load

case we have option called self-weight which automatically calculates weights using the

properties of material i.e., density and after assignment of dead load the skeletal structure looks

red in color as shown in the figure.

Figure 5. 1

25

Example for calculation of dead load:

Dead load calculation

Weight=Volume x Density

Self-weight floor finish=0.12*25+1=3kn/m^2

The above example shows a sample calculation of dead load.

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 -20.063N/mm.

5.3.2. Imposed Loads

Live loads are produced by the use and occupancy of a building. Loads include those from

human occupants, furnishings, no fixed equipment, storage, and construction and maintenance

activities. As required to adequately define the loading condition, loads are presented in terms of

uniform area loads, concentrated loads, and uniform line loads. The uniform and concentrated

live loads should not be applied simultaneously n a structural evaluation. 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. For example, the stair

load of 300 pounds should be applied to the center of the stair tread between supports.

In staad we assign live load in terms of:

Floor load = 2.125KN/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. After the

assignment of the live load the structure appears as shown below.

26

Figure 5.2

Live loads are calculated as per IS 875 Part 2

5.3.3 Wind loads:

In the list of loads we can see wind load is present both in vertical and horizontal loads. 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 produces non static loads on a structure at highly

variable magnitudes. 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. Therefore, 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. The peak pressures in one zone for a given wind direction may not, However, occur

simultaneously in other zones. For some pressure zones, the peak pressure depends on an arrow

range of wind direction. Therefore, the wind directionality effect must also be factored into

determining risk consistent wind loads on buildings.

27

Assignment of wind speed is quite different compared to remaining loads.

We have to define a load case prior to assignment.

After designing wind load can be assigned in two ways

1. Collecting the standard values of load intensities for particular heights and assigning of the

loads for respective height.

2. Calculation of wind load as per IS 875 part 3.

We designed our structure using second method which involves the calculation of wind load

using wind speed.

In Delhi we have a wind speed of 47 kmph for 10 m height and this value is used in calculation.

Basic wind speed:

It gives the basic wind speed of India, as applicable to 1m height above means ground level for

different zones of the country. 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.

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.

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

5.3.4. Earthquake Loads

Earthquake or seismic load on a building depends upon its geographical location, lateral stiffness

and mass, and is reversible. 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. During an earthquake,

the mass is imparted by the building whereas the acceleration is imparted by ground

disturbances. In order to have a minimum force, the mass of the building should be as low as

possible. 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.

Once there is a force, there has to be an equal and opposite reaction to balance the force. 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.

There are two methods to determine the earthquake force in a building: -

a) Seismic coefficient method or static method.

b) Response spectrum method or modal analysis method or spectral acceleration method or

dynamic method.

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.

In the IS : 1893:2002 code, an elastic response spectrum has been proposed for the Maximum

Considered Earthquake (MCE) condition.

NOTE: - The wind loads and earthquake loads are assumed not to act simultaneously. A

building is designed for the worst of the two loads. The fact is that the design forces for

wind are greater than the seismic design forces (i.e. wind governs the design) does not

obviate the need for seismic detailing. While wind forces govern, the design must provide at

least the type of seismic detailing that corresponds to the seismic forces calculated for that

building.

But for this structure the seismic loads are predominant than that of the wind loads,

therefore, the seismic loads govern the design.

29

Design Spectrum

For the purpose of determining seismic forces, the country is classified into four seismic zones as

shown in Fig. 1. of IS 1893

The design horizontal seismic coefficient Ah for a structure shall be determined by the following

expression:

Ah= Z.I.Sa /2.R.g

Provided that for any structure with T <0.1 s, the value of Ah will not be taken less than Z/2

whatever be the value of I/R, where,

Z = Zone factor given in Table 2 of IS 1893, is for the Maximum Considered Earthquake (MCE)

and service life of structure in a zone. 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).

I = Importance factor, depending upon the functional use of the structures, characterized by

hazardous consequences of its failure, post-earthquake functional needs, historical value, or

economic importance (Table 6, IS 1893).

R= Response reduction factor, depending on the perceived seismic damage performance of the

structure, characterized by ductile or brittle deformations. However, the ratio (I/R) shall not be

greater than 1.0 (Table 7, IS 1893). The values of R for buildings are given in Table 7.

Sa/g = Average response acceleration coefficient for rock or soil sites as given by Fig. 2 and

Table 3 of IS 1893, based on appropriate natural periods and damping of the structure. These

curves represent free field ground motion.

Table 5.1 ZONE FACTOR (Z)

Seismic Zone II III IV V

Seismic Intensity LOW MODERATE SEVERE VERY SEVERE

Z 0.10 0.16 0.24 0.36

30

5.4. Design Imposed Loads for Earthquakes Force Calculation

For various loading classes as specified in IS 875(Part 2), the earthquake force shall be

calculated for the full dead load plus the percentage of imposed load as given in Table 8.

For calculating the design seismic forces of the structure, the imposed load on roof need not be

considered. The percentage of imposed loads should be 25% for floor loads up to and including

3KN/m2.

Seismic Weight of Floors

The seismic weight of each floor is its full dead load plus appropriate amount of imposed load.

While computing the seismic weight of each floor, the weight of columns and walls in any storey

shall be equally distributed to the floors above and below the storey.

Seismic Weight of Building

The seismic weight of the whole building is the sum of the seismic weights of all the floors. Any

weight supported in between storeys shall be distributed to the floors above and below in inverse

proportion to its distance from the floors.

Design Lateral Force

Buildings and portions thereof shall be designed and constructed, to resist the effects of design

lateral force.

The design lateral force shall first be computed for the building as a whole. This design lateral

force shall then be distributed to the various floor levels. The overall design seismic force thus

obtained at each floor level, shall then be distributed to individual lateral load resisting elements

depending on the floor diaphragm action.

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,

31

Ah = Design horizontal acceleration spectrum value, using the fundamental natural period

T, in the considered direction of vibration, and

W = Seismic weight of the building.

5.4.1. Seismic Loading in Staad Pro V8i:

Now since we know the basic criterion for earthquake loads, the seismic weights as assigned in

Staad Pro V8i software are as follows: -

Defining Seismic parameters, which includes the following:

Earthquake Zone for Delhi is Zone IV (i.e. Zone Factor = 0.24)

Response Reduction Factor = 5, (for Special RC moment-resisting frame (SMRF) as per

Table 7, IS 1893.)

Importance Factor = 1.0, ( for All Other Buildings other than Important service and

community buildings, such as hospitals; schools; monumental structures; emergency

buildings like telephone exchange, television stations, radio stations, railway stations,

fire station buildings; large community halls like cinemas, assembly halls and subway

stations, power stations for which I = 1.5.)

Response spectra for Rock and Soil Site Type (SS) = 2, (For Medium Type Soil at 5%

damping.)

Type of Structure = 1 (for Reinforced Concrete Framed Structure)

Damping Ratio = 5%

Depth of foundation = 2m

32

Figure5.3

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.

The figure of Staad Editor is shown as below:

33

Figure5.4 Seismic Load in X direction (SLX)

34

Figure5.5 Seismic Load in Z direction (SLZ)

5.5. LOAD COMBINATIONS

Load combinations as per IS 875 Part 5 are taken into consideration.

A judicious combination of the loads (specified in IS 875 Parts 1 to 4 of this standard and

earthquake), keeping in view the probability of:

a) Their acting together, 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.

Load Combinations - The various loads should, therefore, be combined in accordance with the

stipulations in the relevant design codes. In the absence of such recommendations, the following

35

loading combinations, whichever combination produces the most unfavorable effect in the

building, foundation or structural member concerned may be adopted ( as a general guidance ). It

should also be recognized in load combinations that the simultaneous occurrence of maximum

values of wind, earthquake, imposed and snow loads is not likely: -

Where, the numerals 1.5, 0.9, 1.2, 1.0 represents the load factors as per IS 875 Part 5.

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.

Earthquake is not likely to occur simultaneously with wind or maximum flood or maximum sea

waves. Since the wind velocity in the region is less and less dominant than the seismic zone

(Zone IV), therefore wind load is not considered for design.

1) DL + LL 2) DL + LL +SLX 3) DL + LL + SLZ 4) DL + LL –SLX 5) DL + LL –SLZ 6) 1.5 (DL + LL) 7) 1.5 (DL + SLX) 8) 1.5 (DL - SLX) 9) 1.5 (DL + SLZ) 10) 1.5 (DL -SLZ) 11) 0.9DL +1.5SLX 12) 0.9DL -1.5SLX 13) 0.9DL + 1.5SLZ 14) 0.9DL - 1.5SLZ 15) 1.2 (DL +LL +SLX 16) 1.2(DL +LL -SLX ) 17) 1.2(DL+LL+SLZ) 18) 1.2(DL + LL - SLZ)

36

5.7. INPUT TO STAAD EDITOR FOR LOADING:

37

CHAPTER 6

ANALYSIS

39

ANALYSIS

6.1 Method Of Analysis

The various methods of analysis of statistically indeterminate portal frames are :

1. Method of flexibility coefficients.

2. Slope displacements methods (iterative methods)

3. Moment distribution method

4. Kane‟s method

5. Cantilever method

6. Portal method

7. Matrix method

8. Using STAAD Pro V8i

6.1.1 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.

Limitations:

It is not applicable for degree of redundancy>3

6.1.2. Slope displacement equations:

It is advantageous when kinematic indeterminacy <static indeterminacy. This procedure was first

formulated by axle bender in 1914 based on the applications of compatibility and equilibrium

conditions.

The method derives its name from the fact that support slopes and displacements are explicitly

comported. Set up simultaneous equations is formed the solution of these parameters and the

joint moment in each element or computed from these values.

Limitations:

A solution of simultaneous equations makes methods tedious for manual computations. This

method is not recommended for frames larger than two bays and two storeys. .

40

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.

Limitations of hardy cross method:

It presents some difficulties when applied to rigid frame especially when the frame is susceptible

to side sway. The method cannot be applied to structures with intermediate hinges.

6.1.3 Kani’s method:

This method over comes some of the disadvantages of hardy cross method. Kani‟s approach is

similar to H.C.M to that extent it also involves repeated distribution of moments at successive

joints in frames and continues beams. However there is a major difference in distribution process

of two methods. H.C.M distributes only the total joint moment at any stage of iteration.

The most significant feature of Kani‟s method is that process of iteration is self-corrective. 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.

Advantages:

It is used for side way of frames.

Limitations:

The rotational of columns of any storey should be functioning a single rotation value of same

storey.

The beams of storey should not undergo rotation when the column undergoes translation. That is

the column should be parallel.

Frames with intermediate hinges cannot be analyzed.

6.1.4. Approximate method:

Approximate analysis of hyper static structure provides a simple means of obtaining a quick

solution for preliminary design. 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. it is possible to sketch the deflected

41

profile of the structure for the given loading and hence by locate the print inflection. Since each

point of inflection corresponds to the location of zero moment in the structures. The inflection

points can be visualized as hinges for the purpose of analysis. The solution of structures is

sundered simple once the inflection points are located. The loading cases are arising in

multistoried frames namely horizontal and vertical loading. The analysis carried out separately

for these two cases.

Horizontal cases:

The behavior of a structure subjected to horizontal forces depends upon its heights to width ratio

among their factor. It is necessary it differentiate between low rise and high rise frames in this

case.

Low rise structures:

Height < width

It is characterized predominately by shear deformation.

High rise buildings

Height > width

It is dominated by bending action

6.1.5. 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 .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. However stiffness method is used degree of freedom is greater than degree of

redundancy especially for computers.

42

6.2. 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. The design base shear shall first be computed as a

whole, and then be distributed along the height of buildings based on simple formulae

appropriate for buildings with regular distribution of mass and stiffness. The design lateral force

obtained at each floor level shall be distributed to individual lateral load resisting elements

depending upon floor diaphragm action.

The design lateral force or design base shear and the distribution are given by some empirical

formulae given in the I.S 1893.

Response Spectrum analysis:

This method is applicable for those structures where modes other than the fundamental

one affect significantly the response of the structure. In this method the response of Multi degree

of freedom system is expressed as the superposition of modal response, each modal response

being determined from the spectral analysis of Single–degree of freedom system, which is then

combined to compute the total response.

Elastic Time history analysis:

A linear analysis, time history analysis over comes all disadvantages of modal response spectrum

provided nonlinear behavior is not involved. The method requires greater computational efforts

for calculating the response at discrete times. One interesting advantage of this is that the relative

signs of response quantities are preserved in the response histories.

6.3. 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.

This is the main advantage of using the software or computer aided analysis.

43

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

**The above results are displayed from the Staad Output file.

**These reaction forces and moments are evaluated for the critical load combinations 5 to 9 as

shown above under load combinations.

**The joints 69 to 113 show the column positions the ultimate position of reaction supports for

the RC framed structure.

53

CHAPTER 7

DESIGN

54

INPUT TO STAAD EDITOR FOR DESIGN

55

7.1. BEAMS

Beams are the horizontal members of the RC framed structure. Generally, beam is of two types-

i) Singly Reinforced Beam and ii) Doubly Reinforced Beams. Design of beams is done as per

Limit State Design of collapse (IS 456: 2000).

Using Staad Pro software, the design of beam is simply done by assigning the parameters for the

structure which includes the clear cover, yield strength of steel, compressive strength of concrete,

maximum and minimum size of bars to be used, etc.

A reinforced concrete beam should be able to resist tensile, compressive and shear stress induced

in it by loads on the beam.

There are three types of reinforced concrete beams

1.) Single reinforced beams

2.) Double reinforced concrete

3.) Flanged beams

Beams transfer loads from slabs to columns and hence are designed for bending.

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. I cantilever beams

reinforcing bars placed near the top of the beam, for the same reason as in the case of simply

supported beam.

Doubly reinforced concrete beams:

It is reinforced under compression and tension regions. The necessity of steel of compression

region arises due to two reasons; when depth of beam is restricted, the strength availability singly

reinforced beam is in adequate. 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.

Figure shows the bottom and top reinforcement details at three different sections.

These calculations are interpreted manually.

Due to the huge output of Staad Pro V8i, here we only show the design result of a beam.

56

7.1.1. Design Result for Beam No. 1

57

FIGURE 7. 1 Location of Beam 1 in the structure

58

7.1.2. Detailing of Beam Reinforcement as per IS 13920 : 1993

FIGURE 7. 2 Beam Reinforcement

59

FIGURE 7. 3 Beam Web Reinforcement

7.1.3. Check for the design of a Beam No.1:

Given data:

Cross section of beam : b x d = 300mm x600 mm

Vertical shear force = Vu =145.93 KN

τc = 0.29 N/mm2 (from table 19 of IS 456 200)

Minimum Shear Reinforcement:

When τv is less than τc , given in Table 19, minimum shear reinforcement shall -be provided

Design of Shear Reinforcement:

When τv exceeds τc, given in Table 19, shear reinforcement shall be provided in any of the

following forms:

a) Vertical stirrups,

b) Bent-up bars along with stirrups, and

c) Inclined stirrups,

60

τv = Vu/(b x d) (As per clause 40.1 of IS 456-2000)

= 145.93 x 103/(550x300)

=1.216 N/mm2

τv ≥ τc

Design reinforcement

Vus = Vu- τc x b x d (As per clause 40.4 of IS 456-2000)

= 145.93 x103 -0.29x550x300

= 111100 N

Shear reinforcement shall be provided to carry a shear equal to Vu - τc bd

The strength of shear reinforcement Vus, shall be calculated as below:

For vertical stirrups:

Vus = 0.87 fyAsvd/Sv (As per clause 40.4 of IS 456-2000)

Asv = total cross-sectional area of stirrup legs or bent-up bars within a distance Sv.

Sv = spacing of the stirrups or bent-up bars along the length of the member,

τv = nominal shear stress

τc = design shear strength of the concrete,

b = breadth of the member which for flanged beams, shall be taken as the breadth of

the web bw,

fy = characteristic strength of the stirrup or bent-up reinforcement which shall not be

taken greater than 415 N/mm2,

α = angle between the inclined stirrup or bent- up bar and the axis of the member, not

less than 45”, and

d = effective depth.

111130 N= 0.87 x 415 x 2 x π x 82 x 550/Sv

Sv = 140 mm

61

Sv should not be more than the following

1. 0.75xd = 0.75 x 550 = 300 mm

2. 300 mm

3. Minimum shear reinforcement spacing = Sv,min

Minimum shear reinforcement:

Minimum shear reinforcement in the form of stirrups shall be provided such that:

Asv/bSv ≥ 0.4/ 0.87fy (As per clause 26.5.1.6 of IS 456-2000)

Asv = total cross-sectional area of stirrup legs effective in shear,

Sv = stirrup spacing along the length of the member,

b = breadth of the beam or breadth of the web of flanged beam, 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)x82x0.87x415/(0.4x300) = 605 mm.

Provided 2 legged 8mm @100 mm stirrups .

Hence matched with Staad output.

62

7.2. COLUMNS

A column or strut is a compression member, which is used primary to support axial compressive

loads and with a height of at least three it is least lateral dimension.

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.G 0f the column I the

longitudinal direction.

Depending upon the architectural requirements and loads to be supported, R.C columns may be

cast in various shapes i.e. square, rectangle, and hexagonal, octagonal, circular. Columns of L

shaped or T shaped are also sometimes used in multistoried buildings.

The longitudinal bars in columns help to bear the load in the combination with the concrete.

The longitudinal bars are held in position by transverse reinforcement, or lateral binders.

The binders prevent displacement of longitudinal bars during concreting operation and also

check the tendency of their buckling towards under loads.

7.2.1. 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, but for the columns on the property line as the following

requirements some area beyond the column, the column can be shifted inside along a

cross wall to provide the required area for the footing with in the property line.

alternatively a combined or a strap footing may be provided.

B) The spacing between the columns is governed by the lamination on spans of supported

beams, as the spanning of the column decides the span of the beam. As the span of the of

the beam increases, the depth of the beam, and hence the self-weight of the beam and the

total.

63

7.2.2. Effective length:

The effective length of the column is defined as the length between the points of contraflexure of

the buckled column. The code has given certain values of the effective length for normal usage

assuming idealized and conditions shown in appendix D of IS - 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.

Axially loaded columns:

All compression members are to be designed for a minimum eccentricity of load into principal

directions. In practice, a truly axially loaded column is rare ,if not nonexistent.

Therefore, every column should be designed for a minimum eccentricity .clause 22.4 of IS code

E min = (L/500) + (D/300), subjected to a minimum of 200 mm.

Where L is the unsupported length of the column (see 24.1.3 of the code for definition

unsupported length) and D is the lateral dimension of the column in the direction under the

consideration.

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 0.002

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 0.0035-0.75

times the strain at least compressed extreme fiber.

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.

64

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 38.1 and 38.2 with neutral axis so chosen as to satisfy the

equilibrium of load and moment about two weeks.

Alternatively such members may be designed by the following equation:

(Mux/ Muy)αn +(Muy/ Muy1)αn <= 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.

αn is related to Pu/puz

Puz = 0.45*fck*Ac+0.75*fy*Asc

For values of pu/Puz=0.2 to 0.8, the values of αn vary linearly from 1.0 to 2.0 for values less

than 0.2, αn is values greater than 0.8 , αn is 2.0

The main duty of column is to transfer the load to the soil safely. Columns are designed for

compression and moment. 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. Also the amount

if load depends on number of beams the columns is connected to. As beam transfer half of the

load to each column it is connected.

7.2.3. 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. The strength of column depends upon

the strength of materials, shape and size of cross section, length and degree of proportional and

dedicational restrains at its ends.

A column may be classify based on deferent criteria such as

65

1.) Shape of the section

2.) Slenderness ratio (A=L+D)

3.) Type of loading, land

4.) Pattern of lateral reinforcement.

The ratio of effective column length to least lateral dimension is released to as slenderness ratio.

In our structure we have 3 types of columns.

� 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. So create three types of column sections

and assign to the respective columns depending on the connection. But in these structure we

adopted same cross section throughout the structure with a rectangular cross section .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.

The column design is done by selecting the column and from geometry page assigns the

dimensions of the columns. 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. Minimum reinforcement, max, bar sizes, maximum and minimum spicing.

2. Select the appropriate design code and input design column command to all the column.

3. 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 figure represents details regarding

1. Transverse reinforcement

2. Longitudinal reinforcement

The type of bars to be used, amount of steel and loading on the column is represented in the

below figure.

66

Table 7. 4 Skeleton Structure Showing Column No. 1539

67

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 = 0.45fckAc + 0.75fyAsc

19732.59 = 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

7.3. SLABS

A slab is a flat two – dimensional, planar structural element having thickness small compared to

its other two dimensions. 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. Therefore volume of concrete and hence, dead load is large in the case of slab. A small

reduction in depth of slab therefore, leads to a considerable economy. But care has to be taken to

see that its performance (serviceability) is not affected due to excessive deflection and cracking.

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.

FIGURE 7.6 One Way Slab (lb/la > 2)

71

FIGURE 7.7 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. 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

FIGURE 7.8 Two Way Slab (lb/la > 2)

FIGURE 7.9 Load Distribution in a Two Way Slab

Checks:

There is no need to check serviceability conditions, because design satisfying the span for depth

ratio.

a.) Simply supported slab

b.) Continuous beam

Slabs are designed for deflection. Slabs are designed based on yield theory

This diagram shows the distribution of loads in two slabs.

FIGURE 7.10 Load Distribution showing One way & Two waySlabs

73

In order to design a slab we have to create plates by selecting the plate cursor. Now selecting the

members to form slab and use form slab button. Now give the thickness of plate as 0.125 m.

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. After analysis is carried out go to

advanced slab design page and collect the reinforcement details of the slab.

FIGURE 7. 11 Monolithic connection between Slab, Beam & Column.

7.3.1. Design detail and sample calculation of a typical slab:

6310mm

3584mm S1

74

FIGURE 7. 12 Plan showing slabs

i. DESIGN OF TWO WAY SLAB:-

Calculation of thickness of slab using l/D = 26

Therefore, an overall depth of slab is 140 mm.

Using 8mm dia bars and providing 20 mm clear cover,

dxx= 140-Ø/2-cover=140-8/2-20=116mm

dyy=140-Ø/2-cover-8=140-108mm

ii. CALCULATION OF EFFECTIVE SPAN

lx = 3.58+dxx=3.58+.116=3.696

ly = 6.32+.133=6.456

ly/lx = 6.456/3.696=1.76<2

Hence it is two-way slab

iii. LOAD CALCULATION

Considering width of slab 1m

75

Dead load=DL=1×25×.140 = 3.5kN/m2

Live load=LL = 2kN/m2

Floor finishing (25mm thick) = 0.040×24×1=1.0 kN/m2

Plaster (6mm thick) = 0.006×24×1=0.25 kN/m2

Total load = 3.5+2+1+0.25=6.75 kN/m2

Factor load = 6.75×factor of safety = 6.75×1.5=10.125kN/m2

Taking factor of safety 1.5

iv. CALCULATION FOR MOMENT

There will be negative moment at continuous edge and positive

Moment at mid span=

Mx = αx×Wu×lx2

My = αy×Wu×lx2

Where αx = short span coefficient

Where αY = long span coefficient

v. Calculation of coefficient according to IS 456,clauses D-1.1 and 24.4.1

Type of panel = Two adjacent edge continuous.

αx (-ve) at 1.76 = 0.084

αx(+ve) at 1.76 = 0.063

vi. Moment calculation

Mux(-ve) = 0.084×10.125×3.6962= 10.90 kNm

Mux(+ve) = 0.063×10.125×3.6962= 8.175 kNm

76

Muy(+ve) = 0.035×10.125×3.6962= 6.099 kNm

Muy(-ve) = 0.047×10.125×3.6962= 4.5418kNm

vii. CHECK FOR DEPTH

d= √(M/Rb)

R = 0.36 Xu max/d (1-0.42Xumax/d)×fck

R =0.36×0.48×(1-0.42×0.48)×25 = 3.45 kN/mm2

b = 1000 mm

M =Max (10.90, 8.175, 6.099 4.5418 ) = 10.90 kNm

dreq = √(10.90×106)/(3.45×10

3×100) = 72 mm <d available

Hence safe.

viii. CALCULATION OF AREA REQUIRED IN THE MID SPAN

Equation for finding Ast

Ast = (0.5 fck/fy){1-[√(1-(4.6Mu/fck.b.d2))]}b.d

Astxx= 273.439mm2

Check

Astmin=.12×b×D/100= .12×1000×160/100=192mm2

Hence it is ok.

77

Span Position Mu

(KNm)

d

(mm)

Req. Ast

(mm2)

Dia – Spacing

(mm)

Provided Ast

(mm2)

Short – at Support 10.90 115 274 #8 – 180 279

-At Midspan 8.175 115 202 #8 – 240 309

Long – at Support 4.5418 115 111.18 #8 – 300 168

-At Midspan 6.099 115 150.24 #8 – 300 168

ix. CHECK FOR DEFLECTION:-

Deflection=(Lx/d)×Mf ,

For safe, it should be less than 26.

Where, Mf is modification factor.

x. CHECK FOR SERVICEABILITY

Req. pt at Shorter Midspan = Ast*100/b.d = 202*100/100*115 = 0.17%

Since Req. pt < Assumed p t (0.30) Hence SAFE.

xi. CHECK FOR SHEAR

a) Long Edge Continuous :

Vu,max = 1.2 qu[Lx(e/2e+1)] {where e = Ly/Lx}

Vu,max = 1.2*10.125[3.58(1.76/2*1.76+1)]

= 16.93KN

Since, Ast1 = 279mm2; pt = 100*279/1000*115 = 0.24%

τc = from Table 19 of IS 456 = 0.35

Cl.40.2.1.1 IS 456, k=1.30 for D<150mm

Vu = k. τc.b.d = 1.3*0.35*1000*115 = 52.32KN > 16.93KN

Hence SAFE

Long Edge Discontinuous:

Vu,max = 0.9*(16.93/2) = 12.70KN

Therefore, Astx = 202mm2 at midspan.

Assuming 50% bent up to resist moment due to partial fixity.

78

Ast1 = 101mm2; pt = 101*100/1000*115 = 0.087%

τc = 0.218n/mm2

k=1.3

Vuc = 1.3*0.218*1000*115/1000 = 32.59 > 12.70 ; Hence OK.

b) Short Edge Continuous:

Vu,max = 1.2qu.(Lx/3) = 1.2*10.125*3.58/3 = 14.50KN

Ast1 = 168mm2

Vuc = 50.85 > 14.50 ; Hence OK.

Short Edge Discontinuous:

Vu,max = 0.9*(14.50/2) = 12.70KN

Therefore, Astx = 168mm2 at midspan.

Assuming 50% bent up to resist moment due to partial fixity.

Ast1 = 84mm2; pt = 84*100/1000*115 = 0.07%

τc = 0.22n/mm2

k=1.3

Vuc = 1.3*0.22*1000*115/1000 = 32.89 > 10.875 ; Hence OK.

xii. CHECK FOR DEVELOPMENT LENGTH

a) 1. Long Edge Continuous :

Req.

79

For Fe415, M25; Ld = 64.47*8 = 515.78mm

Ld (available) = L/4 = 3584/4 = 896mm; Hence OK.

2. Long Edge Discontinuous:

Ld = 64.47 * 8 = 515.78mm

Assuming 50% bars bent up , M1 = 8.175/2 = 4.08KNm

Vu,max = 12.70KN

Lex => (Ld-1.3M1/V) = 515.78 – 1.3*4.08/12.70 = 98.14mm

Lex => (Ld/3 – bs/2) = 98.14 + 300/2

Lex = 248.14mm from inner face of support.

Straight Length available inside inner support = B =bs-A

B = 300-(5*8+25) = 235mm

Using 90degree bend, available anchorage length = 8db + 235 = 64 + 235 = 299mm >

235mm

Hence OK.

b) 1)Short Edge Continuous:

Req. Ld = 64.47 * 8 = 515.78mm

Available Ld = L/4 = 896mm; Hence OK

2) Short Edge Discontinuous:

Ld = 64.47 * 8 = 515.78mm

Assuming 50% bars bent up , M1 = 6.099/2 = 3.049KNm

Vu,max = 10.875KN

80

Lex => (Ld-1.3M1/V) = 515.78 – 1.3*3.049/10.875 = 151.30mm

Lex => (Ld/3 – bs/2) = 151.30 + 300/2

Lex = 301.30mm from inner face of support.

Straight Length available inside inner support = B =bs-A

B = 300-(5*8+25) = 235mm

Using 90degree bend, available anchorage length = 8db + 235 = 64 + 235 = 299mm >

235mm

Hence OK.

xiii. TORSION STEEL

a) At corners near column C127 & C128,

Since slab is discontinuous over both edger,

Full Torsion Steel = 0.75 Astx = 0.75*202 = 150mm2 ;

will be required in both direction at right angles in each of the two meshes, One at the top

and the other at the bottom up to the length of:

Lx/5 = 3584/5 = 716.8mm

b) At corner near column C126,

Required area of torsion steel = 1/2(150) = 75mm2

81

7.3.2. STAAD OUTPUT for Element Design:

82

****************************************************************************

FIGURE 7. 13

85

7.4. FOUNDATION

Foundations are structural elements that transfer loads from the building or individual column to

the earth .If these loads are to be properly transmitted, foundations must be designed to prevent

excessive settlement or rotation, to minimize differential settlement and to provide adequate

safety against sliding and overturning.

7.4.1. General:

1) Footing shall be designed to sustain the applied loads, 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.

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.

7.4.2. Bearing Capacity of Soil:

The size foundation depends on permissible bearing capacity of soil. The total load per unit area

under the footing must be less than the permissible bearing capacity of soil to the excessive

settlements.

7.4.3. 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. 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.

1.) Bearing capacity of soil

2.) Type of structure

3.) Type of loads

4.) Permissible differential settlements

5.) Economy

86

A footing is the bottom most part of the structure and last member to transfer the load.

In order to design footings we used the software named STAAD FOUNDATION V8i.

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

7.4.4. 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. In such case, Continuous

Strip Footing is provided to support more than two columns in a row, instead of individual

footing.

Thus the continuous strip footing runs along the column row. The strip footings have T section

and the flange of T section faces downwards. The projection of T-section behaves as a

Cantilever.

The thickness of the flange is kept constant, when the cantilever projection is of small length.

Otherwise, the depth of flange is increased towards the rib.

The weight of the footing is not considered in structural design because it is assumed to be

carried by the subsoil.

It is similar to a floor resting on a system on a system of beams and columns.

7.4.5. Design using STAAD FOUNDATION V8i:

87

- 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, there is no need to specify the column

positions, 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.

FIGURE 7. 15 Zoom View of continuous strip Foundation & Columns

88

- The load combination or the load cases are generated (selected) for which the

foundation is to be designed. Assign Loading: - 1.5(DL + LL)

- The next step is to create the job for the footing (i.e. Combined Footing.)

- Now the design parameters are entered which includes: Concrete & Rebar, Cover

& Soil, Footing Geometry

FIGURE 7. 16 Concrete & Rebar

Parameters FIGURE 7. 17 Cover & Soil

Parameters

89

FIGURE 7. 18 Footing Dimensions

The following input data is required regarding materials, Soil type, Type of foundation, safety

factors.

� Type of foundation: Combined.

� Unit weight of concrete: 25KN/m^3

� Minimum bar spacing: 50mm

� Maximum bar spacing: 500mm

� Strength of concrete: 35N/mm^2

� Yield strength of steel: 415 n/mm^2

� Minimum bar size: 12mm

� Maximum bar size: 60mm

� Bottom clear cover: 50mm

� Unit weight of soil: 22 KN/m^3

� Soil bearing capacity: 175 KN/m^3

� Minimum length: 1000mm

90

� Minimum width: 3500mm

� Minimum thickness: 500mm

� Maximum length: 70000mm

� Maximum width: 40000mm

� Maximum thickness: 2000mm

� Plan dimension: 50mm

� Aspect ratio: 1

� Safety against friction, 0.5; overturning, 1.5; sliding,1.5

Now the last step is to click on DESIGN.

After the analysis, detailed calculation of each and every footing is given with plan and

elevation.

Table 7.1 Dimensions of the Continuous Strip Footings

Footing No. Left Overhang

(m)

Right Overhang

(m)

Length

(m)

Width

(m)

Thickness

(m)

1 3.875 3.875 23.040 9.25 0.700

2 4.975 4.975 62.790 11.450 1.100

3 2.775 2.775 20.840 7.050 0.700

4 6.475 6.475 65.760 14.450 1.300

5 8.225 8.225 55.210 17.950 1.250

91

Table 7.2. DESIGN RESULTS

Footing No. Footing Reinforcement

- Main Steel

Top

Main Steel

Bottom

Secondary Steel

Top

Secondary Steel

Bottom

1 #12 @ 125mm c/c #32 @ 75mm c/c #12 @ 125mm c/c #16 @50mm c/c

2 #12 @ 75mm c/c #40 @75mm c/c #12 @ 75mm c/c #16 @50mm c/c




7.4.6. Design Calculations for Combined Footing 1 (FC1)

Column Dimensions for Column No. 69, 103, 102 and 101 (Combined Footing No. FC1)

Column Shape: Rectangular

Column Length - X (Pl): 1000mm

Column Width - Z (Pw): 500mm

Length of left overhang : 1.00 m

Length of right overhang : 1.00 m

Is the length of left overhang fixed? No

Is the length of right overhang fixed? No

Minimum width of footing (Wb) : 3.50 m

Minimum Thickness of footing (Do) : 500.00 mm

Maximum Width of Footing (Wb) : 40000.00 mm

Maximum Thickness of Footing (Do) : 2000.00 mm

92

Maximum Length of Footing (Lo) : 70000.00 mm

Length Increment : 50.00 mm

Depth Increment : 50.00 mm

Cover and Soil Properties

Pedestal Clear Cover : 50.00 mm

Footing Clear Cover : 50.00 mm

Unit Weight of soil : 22.00 kN/m3

Soil Bearing Capacity : 175.00 kN/m2

Soil Surcharge : 44.00 kN/m2

Depth of Soil above Footing : 2.00 m

Depth of Water Table : -4000mm

Concrete and Rebar Properties

Unit Weight of Concrete 25.000 kN/m3

Compressive Strength of Concrete : 35.000 N/mm2

Yield Strength of Steel : 415.000 N/mm2

Minimum Bar Size : 12

Maximum Bar Size : 60

Minimum Bar Spacing : 50.00 mm

Maximum Bar Spacing : 400.00 mm

Design Calculations

93

Footing Size Calculations

Reduction of force due to buoyancy = -0.00 kN

Minimum area required from bearing pressure, Amin = Pcritical / qmax : 123.46 sq m

Area from initial length and width, Ao = L x W: 60.51 sq m

Therefore, Final footing dimensions are:

Length of footing, L : 23.04 m

Width of footing, W : 9.25 m

Depth of footing, Do : 0.70 m

Area, A : 213.12 sq m

Length of left overhang, Lleft_overhang : 3.88 m

Length of right overhang, Lright_overhang : 3.88 m

Table 7.3.

94

If Au is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for

uplift, areas of negative pressure will be set to zero and the pressure will be redistributed to

remaining corners.

Design for Flexure

Sagging moment along length

Effective Depth =

= 0.63 m

Governing moment (Mu)

=17882.520713 kNm

As Per IS 456 2000

ANNEX G G-1.1C

Limiting Factor1 (Kumax) =

= 0.479107

Limiting Factor2 (Rumax) =

= 4822.007604 kN/m^2

Limit Moment Of Resistance (Mumax)=

= 7928.346683 kNm

Table 7.4.

Table 7.5.

95

Mu <= Mumax hence, safe

Hogging moment along length

Effective Depth =

= 0.63 m

Governing moment (Mu)

= 3.771009 kNm

As Per IS 456 2000 ANNEX G G-1.1C


= 0.479107


= 4822.007604 kN/m^2

Limit Moment Of Resistance (Mumax) =

=18498.368019

kNm


Transverse direction

Effective Depth =

= 0.64 m

Governing moment (Mu) =

= 6300.321341 kNm

As Per IS 456 2000 ANNEX G G-

1.1C


= 0.479107


=4822.007604 kN/m^2

Limit Moment Of Resistance

(Mumax) =

=45790.132556

kNm

96


Check trial depth for one way shear(along length)

Shear Force(S)

= 3538.28 kN

Shear Stress(Tv)

= 0.000000 kN/m^2

Percentage Of Steel(Pt)

= 0.080

As Per IS 456 2000 Clause 40 Table 19

Shear Strength Of Concrete(Tc)

= 0.35 kN/m^2

Tv< Tc hence, safe

Check trial depth for two way shear

For Column 1

Shear Force(S)

= 2618.45 kN

Shear Stress(Tv)

= 729.18 kN/m^2

As Per IS 456 2000 Clause 31.6.3.1

Ks =

= 1.00

Shear Strength(Tc)=

= 1479.0199 kN/m^2

Ks X Tc

= 1479.0199 kN/m^2

Tv<= Ks X Tc hence, safe

For Column 2

Shear Force(S)

= 4890.83 kN

Shear Stress(Tv)

= 1361.99 kN/m^2

As Per IS 456 2000 Clause 31.6.3.1

Ks =

= 1.00

97

Shear Strength(Tc)=

= 1479.0199 kN/m^2

Ks X Tc

= 1479.0199 kN/m^2


For Column 3

Shear Force(S)

= 4996.80 kN

Shear Stress(Tv)

= 1391.50 kN/m^2

As Per IS 456 2000 Clause 31.6.3.1

Ks =

= 1.00

Shear Strength(Tc)=

= 1479.0199 kN/m^2

Ks X Tc

= 1479.0199 kN/m^2


For Column 4

Shear Force(S)

= 2639.60 kN

Shear Stress(Tv)

= 735.07 kN/m^2

As Per IS 456 2000 Clause 31.6.3.1

Ks =

= 1.00

Shear Strength(Tc)=

= 1479.0199 kN/m^2

Ks X Tc

= 1479.0199 kN/m^2


98

Selection of reinforcement

Top reinforcement along length

As Per IS 456 2000 Clause 26.5.2.1

Minimum Area of Steel (Astmin) = 7770.00 mm2

Calculated Area of Steel (Ast) = 7770.00 mm2

Provided Area of Steel (Ast,Provided) = 7770.00 mm2

Astmin<= Ast,Provided Steel area is accepted

Selected bar Dia = 32.000

Minimum spacing allowed (Smin) = = 50.00 mm

Selected spacing (S) = 134.38 mm

Smin <= S <= Smax and selected bar size < selected maximum bar

size...

The reinforcement is accepted.

Along width

As Per IS 456 2000 Clause 26.5.2.1

Provided Minimum Area of Steel (Astmin) = 19353.57 mm2

Selected bar Dia = 16.000 mm

Minimum spacing allowed (Smin) = 50.00 mm



size...


99

Bottom reinforcement along length

As Per IS 456 2000 Clause 26.5.2.1









size...


Along width

As Per IS 456 2000 Clause 26.5.2.1








Smin <= S <= Smax and selected bar size < selected maximum bar size. The reinforcement is accepted.

100

FIGURE 7. 19

FIGURE 7. 20

101

7.4.7. Detail Drawings

FIGURE 7. 21 - Strip Footing FC1


102



103


*************************************************************************************

104

CONCLUSION

105

CONCLUSION

STAAD PRO has the capability to calculate the reinforcement needed for any concrete section.

The program contains a number of parameters which are designed as per IS: 456 : 2000 and IS

13920 : 1993. Beams are designed for flexure, shear and torsion.

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. Each of these sections is designed to resist both of these critical sagging and

hogging moments. Where ever the rectangular section is inadequate as singly reinforced section,

doubly reinforced section is tried.

Design for Shear:

Shear reinforcement is calculated to resist both shear forces and torsional moments. Shear

capacity calculation at different sections without the shear reinforcement is based on the actual

tensile reinforcement provided by STAAD program. Two-legged stirrups are provided to take

care of the balance shear forces acting on these sections.

Beam Design Output:

The default design output of the beam contains flexural and shear reinforcement provided along

the length of the beam.

Column Design:

Columns are designed for axial forces and biaxial moments at the ends. All active load cases are

tested to calculate reinforcement. The loading which yield maximum reinforcement is called the

critical load. Column design is done for square section. 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. 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.

106

Slab Design:

Slabs are designed for the load combinations as specified in IS 456:2000. All active load cases

are tested to calculate reinforcement. The loading which yield maximum reinforcement is called

the critical load. Slabs are designed as two way and one way. This enables to understand the

detailing of reinforcement in the slabs.

Foundation Design:

Footing is decided on the soil type, loading conditions and area available. It is designed to carry

the load distributed by the structure through slabs to beams to columns to the footings.

Use of Software’s:

Use of Staad Pro V8i, Staad Foundation V8i And Auto Cad is well known after the completion

of the project. This enables to relate theoretical knowledge to real life practicalities.

107

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. S.R. Karve & Dr. V.L. Shah - “Illustrated design of Reinforced concrete

Buildings”, Structures Publications.

o Dr. Ram Chandra - “Limit State Design”, Standard Book House, New Delhi.

o Dr. Ashok K. Jain – “Reinforced Concrete Limit State Design”, New Chand & Bros,

Roorkee

o S. Ramamrutham, R. Narayana – “Theory of Structures”, Dhanpat Rai Publishing

Company

o “STAAD Pro V8i – Getting started & tutorials” - Published by: R .E. I.

o “STAAD Pro 2004 & STAAD FOUNDATION V8i – Technical reference manual” -

Published by: R.E.I.

CODE BOOKS:

o IS 875(Part 1,2,3,5) - Bureau Of Indian Standards, Manak Bhavan, 9 Bahadur Shah

Zafar Marg, New Delhi 110002

o IS 456 : 2000 - Bureau Of Indian Standards, Manak Bhavan, 9 Bahadur Shah Zafar

Marg, New Delhi 110002


Marg, New Delhi 110002.


Marg, New Delhi 110002.

111

analysis & design of multistorey building

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design of multistorey