DESIGN OF SHOPPING COMPLEX

DESIGN OF SHOPPING COMPLEX

By

Rohit Chaurasia

Under the guidance of

Dr. Sunil Kumar

Submitted

In partial fulfilment of the requirement for degree of

Bachelor of Technology in Civil Engineering

to the

Harcourt Butler Technological Institute, Kanpur

June, 2011

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CERTIFICATE

It is to certify that Mr. Rohit Chaurasia of 3rd year B. Tech Civil engineering (2010-

2011), had done their B. Tech project on “Design of Shopping Complex” under my

supervision and guidance. They took keen interest in all the activities regarding the

project. I appreciate their sincerity and efforts.

I wish them bright and prosperous future.

(Dr. Sunil Kumar)

Professor & Head

Deptt .of Civil Engineering.

H.B.T.I Kanpur

2

ACKNOWLEDGEMENTS

We are deeply indebted to all those without whose firm support, encouragement, and

guidance, this report would not have been at this stage.

We wish to express our sincere gratitude to Dr. Sunil Kumar, department of Civil

Engineering for invaluable suggestions and guidance throughout our work at HBTI

Kanpur. He always guided us through the difficulties and made us understand the

concepts needed for the project. His experimental and theoretical knowledge was

indeed very helpful.

Last but not the least we feel privileged in conveying our thanks to all faculty

members for their encouragement and moral support.

June , 2011

Rohit Chaurasia

(S.R. No. 71/08)

ABSTRACT

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Shopping complexes are imperative for catering the daily needs of people under one

roof. Their utility is of immense importance especially in institutions and residential

areas which are normally located away from main city. However, appropriate design

of such shopping complexes involve complex procedures, complex methods which

needs to be understood by designers, planners and architects. The present study has

been taken to address this aspect.

A shopping complex of built up area 720 m2 has been designed in the west campus of

HBTI Kanpur. The limit state method of analysis has been used to design the

shopping complex. The design of building is done using IS: 456-2000.

TABLE OF CONTENTS

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Particulars Page No.

Certificate

Acknowledgements

Abstract

Table of contents

Abbreviations

List of figures

List of tables

Chapter 1: General Consideration 1-2

1.1 Introduction

1.2 Requirement

1.3 Organisation of project work

Chapter 2: Theory, Background and Methodology

1.4 Limit State Design

2.1.1 Characteristic load

1.5 Working Stress Method

1.6 Design Components of Shopping Complex

1.6.1 Foundation

1.6.2 Column

1.6.3 Beam

1.6.4 Slab

1.6.5 Staircase

1.6.6 Water tank

1.6.7 Shell

Chapter 3: Analysis, Design and Results

1.7 Design of Slab (Ground, First and Top floor)

1.8 Design of Beams (Top floor)

1.8.1 Design of Beam ‘AB’

1.8.2 Design of Beam ‘AE’

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1.8.3 Design of Beam ‘EM’

1.8.4 Design of Beam ‘BM’

1.9 Design of Columns (Top floor)

1.9.1 Design of Column D1

1.9.2 Design of Column C1



1.10 Design of Beams (First and Ground floor)

1.10.1 Design of Beam ‘AB’

1.10.2 Design of Beam ‘AE’

1.10.3 Design of Beam ‘EM’

1.10.4 Design of Beam ‘BM’

1.11 Design of Columns (First floor)





1.12 Design of Columns (Ground floor)





1.13 Design of Footings

1.13.1 Design of Footing for Column D1

1.13.2 Design of footing for Column C1

1.13.3 Design of footing for Column C2

1.13.4 Design of footing for Column D2

1.14 Design of Staircase

1.15 Design of Tank

1.15.1 Design of Slab for Tank

1.15.2 Design of Beam for Tank

1.16 Design of Shell

Chapter 4: Conclusions

6

Bibliography

LIST OF TABLES

Tables Page No

Table 2.1 Limiting value of depth of neutral axis

Table 2.2 Value of moment coefficients

Table 2.3 Value of bending moment coefficient

Table 2.4 Value of shear force coefficients

Table 3.1 Result of slab

Table 3.2 Result of beam ‘AB’ (Top floor)

Table 3.3 Result of beam ‘AE’ (Top floor)

Table 3.4 Result of beam ‘EM’ (Top floor)

Table 3.5 Result of beam ‘BM’ (Top floor)

Table 3.6 Result of beam ‘AB’ (First and Ground floor)

Table 3.7 Result of beam ‘AE’ (First and Ground floor)

Table 3.8 Result of beam ‘BM’ (First and Ground floor)

Table 3.9 Final moment calculation

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LIST OF FIGURES

Figures Page No

Figure 3.1 Different beams

Figure 3.2 Grid of columns

Figure 3.3 Distribution of pressure below the retaining wall

ABBREVIATIONS

DL = Dead load

LL = Live load

wd= Dead load intensity

w l= Live load intensity

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d '= Depth of compression reinforcement from the highly compressed face

Ast = Area of steel.

Asv = Total Cross-sectional area of stirrup legs or bent up bars within distance Sv .

AΦ = Area of cross-section of one bar.

b = Width of beam or shorter dimension of a rectangular column

D = Total Depth.

d = Effective depth.

e = Eccentricity

f = Stress

fck= Characteristic compressive stress.

fy= Characteristic strength of steel

fs = Characteristic strength of steel

K = Stiffness of member

I = Movement of inertia..

L = Length.

Ld = Development length.

M = Moment.

m = Modular ratio.

S = Spacing of bars.

Vu = Shear force due to design load (limit state design).

Vus = Strength of shear reinforcement (limit state design).

W = Point load; Total load.

xu = Depth of neutral axis.

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γ = Unit weight of soil.

fcbc = Permissible stress in concrete due to bending.

fst = Permissible tensile stress in reinforcement.

Φ = Diameter of bar, angle of repose

τbd = Design bond stress.

τc = Shear stress in concrete.

τcmax = Maximum shear stress in concrete with shear reinforcement.

τv = Nominal shear stress.

M x=Factored moment along X-axis

M y= Factored moment along Y-axis

M ux=Maximum moment capacity for bending along x-axis only at axial load p

M uy=Maximum moment capacity for bending along y-axis only at axial load p

Αn=An exponent that depends on the dimension of the cross-section, the amount of

reinforcement, concrete strength and yield stress of steel.

Me = Bending moment at the end of the beam framing in to the column assuming

fixity at the connection.

Mes = Maximum difference between the moments at the ends of the beams framing in

to opposite sides of column each calculated on the assumption that the ends of the

beam are fixed and assuming one of beam unloaded.

ku = Stiffness factor of the upper column

kL = Stiffness factor of lower column

kb1 = Stiffness of beam on one of side of the column

kb2 = Stiffness of the beam on other side of column

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M x , M y= Maximum moments at mid span on strips of unit width and spanslx , l y;

respectively.

lx= Length of shorter side

l y= Length of longer side

α x , α y= Moment coefficients.

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CHAPTER-1

GENERAL CONSIDERATION

1.1 INTRODUCTION



areas which are normally located away from main city. However, appropriate design

of such shopping complexes involves complex procedures, complex methods which

needs to be understood by designers, planners and architects. The present study has

been taken to address this aspect. A shopping mall, shopping centre, shopping

precinct or simply mall is one or more buildings forming a complex of shops

representing merchandisers, with interconnecting walkways enabling visitors to easily

walk from unit to unit, along with a parking area — a modern, indoor version of the

traditional marketplace.

Modern "car-friendly" shopping malls corresponded with the rise of suburban living

in many parts of the World. From early on, the design tended to be inward-facing,

with malls following theories of how customers could best be enticed in a controlled

environment. Similar, the concept of a mall having one or more "anchor" or "big box"

stores was pioneered early, with individual stores or smaller-scale chain stores

intended to benefit from the shoppers attracted by the big stores.

A shopping complex of built up area 720m2 has been designed in the West campus of

HBTI Kanpur. The limit state method of analysis has been used to design the

shopping complex. The design of building is done using IS: 456-2000.

1.2 REQUIREMENT

The West campus of HBTI houses about 1000 students and faculty members. To cater

the everyday needs a weather proof destination was required. So, we decided to

design a shopping complex at a judiciously selected site so as to avail easy reach and

privacy to both students and faculty members. The complex would ensure that

students and faculties get hustle free shopping experience it would enable them to get

easy access to unadulterated milk and food stuff. Healthy competition among the

shops would help in equally competitive pricing. Since the food zone would be under

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straight monitoring by institution, healthy and delicious cuisines and snacks would be

served.

1.3 ORGANISATION OF PROJECT WORK

Our project report comprises of seven chapters.

Chapter 1- It involves brief introduction regarding the project

Chapter 2- Theory, background and methodology has been discussed in this chapter.

Chapter 3- This chapter includes the analysis, design and results.

Chapter 4- Finally Conclusions have been mentioned in this chapter.

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CHAPTER 2

THEORY, BACKGROUND AND METHODOLOGY

2.1 LIMIT STATE DESIGN

Limit state design is based on the concept of limit state. The acceptability for the

safety and serviceability requirement before failure is called the limit state .In the

limit state methods of design, the design is to withstand safely all the loads which act

on it during its life. Thus limit state design method certifies the serviceability

requirements also. The aim of design is to achieve acceptable probabilities that the

structure will not its limit state .All relevant limit states are considered to ensure

adequate degree of safety and serviceability. To ensure the above objective the design

is based on the characteristic values for material strength and in loads to be supported.

2.1.1Characteristic load: The term “characteristic load” means that value of load

which has 95% probability of not being exceeded during the life of structure.

2.2 WORKING STRESS DESIGN

This has been traditional method used for reinforced concrete where it is assumed that

concrete is elastic .steel and concrete act together elastically and the relationship

between loads and stresses is linear up to the collapse of structure .The basis of

method is that the permissible stress for concrete and steel are not exceeded anywhere

in the structure when it is subjected to the worst combination of working loads.

The working stress method is based on the following assumption:

1: A section which is plane before bending remains plane after bending.

2: Bond between steel and concrete is perfect within the elastic limit of steel.

3: The tensile strength of concrete is ignored.

4: Concrete is elastic.

5: The modular ratio m has the value¿ ) where σ cbc is the permissible compressive

stress in bending.

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2.3 DESIGN COMPONENTS OF SHOPPING COMPLEX

The design purpose of design elements of shopping complex is to provide adequate

stability and safety .There are following design elements of shopping complex:

2.3.1 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, foundation must be designed to prevent excessive settlement or rotation,

to minimize differential settlement and to provide adequate safety against sliding and

overturning. Most foundations may be classified as follows:

(a) Isolated footing under individual columns. They may be square, rectangular or

circular in plan.

(b) Strip foundations and wall footings.

(c) Combined footing supporting two or more column loads. These may be

rectangular or trapezoidal in plan or they may be isolated bases joined by a

beam. The latter case is referred to as strip footing.

(d) Raft or mat foundation is a large continuous foundation supporting all the

column of a structure. This is normally used when soil conditions are poor or

differential settlement is to be avoided.

(e) In pile foundation pile caps are used to tie a group of piles together. These

may be support isolated columns or group of several columns or load bearing

walls.

The choice of type of foundation to be used in a given situation depends on a number

of factors, for example,

(a) Soil strata,

(b) Bearing capacity and standard penetration test value N of soil.

(c) Type of structure,

(d) Type of loads,

(e) Permissible differential settlement, and

(f) Economy.

The size of foundation depends on permissible bearing capacity of the soil. Total load

per unit area under the footing must be less than the permissible bearing capacity of

the soil to prevent excessive settlement. In general, foundations have to resist vertical

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load, horizontal load and moments. For the purpose of design of foundation, soil

profile, depth of water table, values of density, bearing capacity, N value, coefficient

of internal friction etc of soil are required.

Two terms are used for bearing capacity: Gross bearing capacity and net bearing

capacity. Gross bearing capacity is the total safe bearing pressure at the bottom of the

footing including the load of the superstructure, weight of the footing and that of the

earth lying over the footing. Net safe bearing capacity is the safe bearing pressure at

the bottom of the footing additional to the weight of the earth which existed at that

level before the trench for footing was dug. Thus the net safe bearing capacity is the

gross bearing capacity minus the weight per unit area dug out of trench. Depth of

foundation plays a pivotal role in the load resisting capabilities of footing.

Depth of foundation is governed by the following factors:

(a) To secure safe bearing capacity,

(b) To penetrate below the zone where seasonal weather changes are likely to

cause significant movement due to swelling shrinkage of soils,

(c) To penetrate below the zone that may be affected by frost.

IS: 1080-1962 requires that in all soils a minimum depth of 50cm is necessary.

However, if good rock is met at smaller depth, only removal of top soil may be

sufficient. An estimate of depth of footing below ground level may be obtained by

using Rankine formula, that is,

h = pγ[ 1−sin ф

1+sin ф]2

Where,

h= minimum depth of foundation

P = gross bearing capacity

𝛾 = density of soil

Ф= angle of repose of soil.

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2.3.2 Column: A column may be defined as an element used primarily to support

axial compressive loads and with a height of at le.ast three times its least lateral

dimension. The strength of column depends on the strength of the materials, shape

and size of the cross-section, length and degree of positional and directional restraints

at its ends. A column may be classified based on different criteria such as:

(a) Shape of the cross-section,

(b) Slenderness ratio,

(c) Type of loading,

(d) Pattern of lateral reinforcement.

A column may be classified as short or long column depending on its effective

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

referred to as slenderness ratio. A short column has a maximum slenderness ratio of

12. Its design is based on the strength of the material and the applied loads. A long

column has a slenderness ratio greater than 12. However maximum slenderness ratio

of a column should not exceed 60. A long column is designed to resist the applied

load plus additional bending moment included due to its tendency to buckle. A

column may be classified as follows based on type of loading:

(a) Axially loaded column,

(b) A column subjected to axial load and uni-axial bending, and

(c) A column subjected to axial load and bi-axial bending.

A reinforced concrete column can also be classified according to the manner, in which

the longitudinal bars are laterally supported, that is,

(a) Tied column, and

(b) Spiral column.

(c) In practise a truly axially loaded column is rare, if not nonexistent. Therefore

every column should be designed for certain minimum eccentricity. This

accidental eccentricity may occur due to end conditions, inaccuracy during

construction or variation in materials even when the load is theoretically axial.

Clause 25.4 of the code requires that the minimum eccentricity should be as

follows:

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emin ≥l

500+ D

300

¿20

Where,

L= Unsupported length

D= Lateral dimension of column in the direction under consideration in mm.

Moreover there are two types of reinforcement in the column: longitudinal

reinforcement and transverse reinforcement. The purpose of transverse reinforcement

is to hold the vertical bars in position providing lateral support so that individual bars

cannot buckle outwards and split the concrete. Guidelines for longitudinal

reinforcement are:

(a) The minimum area of cross-section of longitudinal bars must be at least 0.8%

of the gross-sectional area of column.

(b) The maximum area of cross-section of longitudinal bars must not exceed 6%

of the gross-sectional area of the concrete.

(c) The bars should not be less than 12mm in diameter so that it is sufficiently to

stand up straight in the column forms during fixing and concreting.

(d) The minimum number of longitudinal bars provided in a column must be four

in rectangular columns and six in circular bars.

(e) Spacing of the longitudinal bars measured along the periphery of a column

should not exceed 300mm.

Transverse reinforcement may be provided in the form of lateral ties or spiral. The

lateral ties may be in the form of polygonal link with internal angle not exceeding 135

°. The ends of the transverse reinforcement should be properly anchored. While

providing lateral ties following points are worth note:

(a) The diameter of the polygonal link or lateral ties should not be less than one

fourth of the diameter of the longitudinal bars, and in no case less than 6mm.

(b) The pitch of the lateral ties should not exceed following distances:

The least lateral dimension of the compression member,

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Sixteen times the smaller diameter of the longitudinal

reinforcement bar to be tied, and

300mm.

The general non dimensional equation for the load contour at constant ‘P’ can be

expressed in the form:

(Mux

Mux1¿α + (

MuyMuy1

¿α ≤1

Where,

M x=Factored moment along X-axis

M y= Factored moment along Y-axis

M ux=Maximum moment capacity for bending along x-axis only at axial load p

M uy=Maximum moment capacity for bending along y-axis only at axial load p

Αn=An exponent that depends on the dimension of the cross-section, the amount of

reinforcement, concrete strength and yield stress of steel.

As per IS: 456- 2000, Frame Analysis method can be used to analyze and design the

columns which are subjected to axial loading and bi-axial bending. The guidelines can

be summed up as follows:

External columns:

Moment for frames of one bay

Moment at foot of upper column = Me (ku

ku+kL+0.5 kb)

Moment at head of lower column = Me(kL

ku+kL+0.5 kb)

Moment for frames of two or more bay

Moment at foot of upper column = Me (ku

ku+kL+kb)

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Moment at head of lower column = Me (kL

ku+kL+kb)

Internal columns:

Moments for frame of two or more bays

Moment at foot of upper column = Mes (ku

kL+ku+kb1+kb2)

Moment at head of lower column = Mes (kL

kL+ku+kb1+kb2)

Where,

Me = Bending moment at the end of the beam framing in to the column assuming

fixity at the connection.

Mes = Maximum difference between the moments at the ends of the beams framing in

to opposite sides of column each calculated on the assumption that the ends of the

beam are fixed and assuming one of beam unloaded.

ku = Stiffness factor of the upper column

kL = Stiffness factor of lower column

kb1 = Stiffness of beam on one of side of the column

kb2 = Stiffness of the beam on other side of column

2.3.3-Beam: A reinforced concrete flexure member should be able to resist tensile,

compressive and shear stresses induced in it by the loads acting on the member.

There are three types of RCC beams:

(a) Singly reinforced beams

(b) Doubly reinforced beams

(c) Singly or doubly reinforced flanged beams

While designing the beams, following important rules must be kept in mind:

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(a) Effective span of a member shall be as follows:

The effective span of a member that is not built integrally with its

supports shall be taken as clear span plus effective depth of beam or

centre to centre of supports, whichever is lesser.

In case of continuous beams, if the width of the supports is less than

1/12 of the clear span, the effective span shall be taken as above. If the

supports are wider than 1/12 of the clear span or 600mm whichever is

less, the effective span shall be taken as under:

i. For the end span with one end fixed and the other continuous

or for the intermediate spans, the effective span shall be clear

span between supports, and

ii. For the end span with one end free and the other continuous,

the effective shall be equal to the clear span plus half the

effective depth of the beam or the clear span plus half the

width of the discontinuous support, whichever is lesser.

iii. In case of spans with roller or rocker bearings, the effective

span shall always be the distance between the centre of

bearings.

(b) Following specifications can be used for providing reinforcement in beam:

Minimum area of tension reinforcement shall not be less than that

given by the following:

A st

bd=0.85

f y

.

Maximum area of tension reinforcement shall not exceed 0.04bD.

(c) Maximum area of compression reinforcement shall not exceed 0.04bD

compression reinforcement in beams shall be enclosed by the stirrups for

effective lateral restraint.

(d) Maximum spacing of shear reinforcement measured along the axis of the

member shall not exceed 0.75d for vertical stirrups and d for inclined stirrups

at 45° .in no case shall the spacing exceed 300mm.

(e) Minimum shear reinforcement in the form of stirrups shall be provided such

that

A sv

b Sv

= 0.40.87 f y

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Where,

A sv= Total cross-sectional area of stirrup legs effective in shear

Sv=Stirrup spacing along the length of the member.

The limiting values of depth of neutral axis for different grades of steel are shown in

the adjacent table:

Table 2.1- Limiting value of depth of neutral axis

f y xumax

d

250 0.53

415 0.48

500 0.46

2.3.4 Slab: Slabs are plate elements forming floors and roofs of buildings and

carrying distributed loads primarily by flexure. A slab may be simply supported or

continuous over one or more supports and is classified according to the manner of

support:

(a) One -way slab spanning in one direction,

(b) Two-way slab spanning in both direction,

(c) Circular slabs,

(d) Flat slabs resting directly on columns with no beams, and

(e) Grid floor and ribbed slab.

One way slab are those in which length is more than twice the breadth. A one way

slab can be simply supported or continuous. A continuous one way slab can be

analyzed in a manner similar to that for a continuous beam. When the slabs are

supported on four sides, two ways spanning occurs. Such slabs may be simply

supported or continuous on any or all sides. The deflection and bending moments in a

two way slab are considerably reduced as compared to those in one way slab. Thus a

thinner slab can carry the same load when supported on all the four edges. In a square

slab, the two way action is equal in each direction. In long narrow slabs where length

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is greater than twice breadth, the two way action effectively reduces to one way action

in the direction of short span although the end beams do carry some slab load.

A two way slab which is simply supported at its edges tends to lift off its supports

near the corners when loaded. Such a slab is the only truly simply supported slab. The

values of bending moments used for the design of such slabs can be obtained as

follows:

M x=α x w l x2

M y=α y w lx2

Where,

M x , M y= Maximum moments at mid span on strips of unit width and spanslx , l y;

respectively.

lx= Length of shorter side

l y= Length of longer side

α x , α y= Moment coefficients.

Table showing the values of moment coefficients:

Table 2.2- Value of moment coefficients

l y

lx

1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0

α x 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118

α y 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029

If the cross-sectional areas of the three basic structural elements: beam, slab and

column are related to the amount of steel reinforcement provided, it will be seen that

the percent steel is usually maximum in a column than a beam and the least in slab.

Slabs are designed by using the same theories and shear as are used for beams. The

following method of analysis is available:

a) Elastic theory analysis- idealization into strips or beams,

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b) Semi empirical coefficients as given in the code, and

c) Yield line theory.

Table showing the value of bending moment coefficients is as follows:

Table 2.3- Value of bending moment coefficients

Type of load Near middle of

end span

At middle of

interior span

At support next

to end support

At other interior

support

Dead load and

imposed

load(fixed)

+112

+116

−110

−112

Imposed load

(not fixed)

+110

+112

−19

−19

Table showing the value of shear force coefficients is as follows:

Table 2.4- Value of shear force coefficients

Type of load At end support At support next to

end support

At all other interior

supports

Dead load and

imposed load(fixed)

0.4 0.6 0.5

Imposed load (not

fixed)

0.45 0.6 0.6

2.3.5 Staircase: The purpose of a staircase is to provide pedestrian access to different

levels within buildings. The geometrical forms of staircases may be quite different

depending on the individual circumstances involved. The shape and structural

arrangement of a staircase generally depend on type of construction of the structure

around the staircase that is, loading bearing brick structure or reinforced concrete

frame structure and availability of space.

There are two main component of a staircase-

Stairs

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Landing slab

General rules to design staircase are as follows:

(a) Between consecutive floors there should be an equal rise for every parallel

step. Similarly, there should be an equal going.

(b) There should be at least 2m headroom measured vertically above any step.

(c) The sum of going of a single step plus twice the rise should be between 550m

and 700m.

(d) The rise of a step should not be more than about 200mm and the going not less

than240mm.

(e) The slope or pitch of stair case should not be more than about 38°.

(f) Width of the staircase will depend on the usage. The width required in a

residential building would be different from that for a cinema hall.

Generally a staircase consists of a sloping slab in the direction of flight and is

supported at each end. Steps are built over this slab. The support at either end may be

as follows:

(a) Walls or beams spanning transversely to the flight, or

(b) A landing slab spanning transversely to the flight, or

(c) A landing slab spanning along the direction of the flight

The effective span of stair slab should be taken as the following distances:

(a) Where supported at bottom risers by beam or wall, the centre to centre

distance of beams;

(b) Where spanning on to the edge of a landing slab which spans transversely to

the flight, a distance equal to the going of stairs plus at each end either half the

width of landing slab or one metre, whichever is smaller, and

(c) Where landing slab spans in the same direction as the stairs, they should be

considered as acting together to form a single slab and the span determined as

centre to centre distance of the supporting beams or walls.

2.3.6 Water tank: A water tank is used to store water to tide over the daily

requirements in general, water tanks can be classified under three heads:

(a) Tanks resting on the ground,

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(b) Elevated tanks supported on staging, and

(c) Underground tanks.

From shape point of view, water tanks may be of several types, such as:

(a) Circular tanks,

(b) Rectangular tanks,

(c) Spherical tanks,

(d) Intze tanks, and

(e) Circular tanks with conical bottoms.

In the construction of concrete structures for the storage of water and other liquids, the

imperviousness of concrete is most essential. The quantity of cement should not be

less than 330kg/m3 of concrete. It should also be less than 530kg/m3 of concrete to

keep the shrinkage low. It is usual to use rich mix like M 30 grade in most of water

tanks.

When water is filled in circular tank, the hydrostatic pressure will try to increase its

diameter at any section. However this increase in the diameter all along the height of

the tank will depend upon the nature of the joint at the junction of wall and bottom

slab. If the joint is flexible it will be free to move outward in a different position.

Hydrostatic pressure at top point is zero; hence there will be no change in the diameter

at the highest point. The hydrostatic pressure at bottom will be maximum, resulting in

maximum increase in diameter there hence maximum movement there if joint is

flexible. When joint is flexible, Hoop tension is developed everywhere in the wall.

Maximum Hoop tension at the bottom, per unit height of wall= wHD2

Taking permissible stress in steel in direct tension as f s, area of steel per metre height

at the base is given by,

Ash=¿ wHD

2 f s

¿

This area of steel may be provided at the centre of wall, if its thickness is small, or it

may be provided on each face, keeping a minimum cover of 25mm if the thickness is

more than 225mm. The thickness of the wall should be such that tensile stress

26

developed in the composite section is within safe limits. If f ct is the permissible

tensile stress in the equivalent concrete section, and T is the thickness of the wall, we

have

f ct=wHD /2 HH

1000 T +(m−1) A sh

For tanks of smaller capacity, the cost of shuttering for circular tank becomes high.

Rectangular tanks are therefore used in such circumstances. However, rectangular

tanks are not normally used in for large capacities since they are uneconomical and

also its exact analysis is difficult. In such a tank wall of the tank is subjected to both

bending moment as well as direct tension. Though reinforcement is provided for both

moments as well as direct tension, the maximum permissible value of tensile stresses

for M20 concrete may be taken as 1.2N/mm2 and 1.7N/mm2 respectively for direct

tension and due to bending. For the design by approximate method, rectangular tanks

may be divided into two categories, tanks in which ratio of length to breadth is less

than 2 and tanks in which ratio is more than 2.

(a) If the ratio L/B is less than 2, the tank walls are designed as continuous frame

subjected to a triangular load. Such a bending is assumed to take place from

top to a height h=H/4 or 1m above base whichever is more. For the bottom

height h, bending is assumed to act in vertical plane as cantilever, subjected to

triangular load, having zero intensity at h and wH at base. For horizontal

bending, the maximum horizontal force per unit height is taken equal to p=

w(H-h) per metre run. The value of bending moment and direct tension is

calculated as below:

Bending moment at centre of span= p B2

16 (producing tension on outer face)

Bending moment at ends of span= p B2

12 (producing tension on water face)

Direct tension on long wall= w (H-h) B/2

Direct tension on short wall= w (H-h) L/2.

27

(b) If ratio L/B is greater than 2, the long walls are assumed to bend vertically as

cantilever fixed at base and subjected to triangularly distributed load. The

short wall is assumed to bend horizontally, supported on long walls for the

portion from top to a height H-h. The load intensity for such a bending is taken

as p = w(H-h). The bottom portion of height h of the short wall is designed as

cantilever subject to triangular load. Thus:

For long wall, bending moment at the base, per unit length of the wall = wH3/6

For short wall, the maximum bending moment at ends and centre

= w(H-h)B2/16

Maximum cantilever bending moment= wHh2/6

The direct tension on long wall = w(H-h)B/2

2.3.7 Shell: It is basically a curved surface having small thickness compared to radius

and other dimensions. Shells or skin space roofs are preferable to plane roofs since

they can be used to cover large floor spaces with economical use of materials of

construction. The use of curved space roofs requires 25 % to 40 % less material than

that of plane elements. Shell roofs are architecturally very expressive. Shell roofs are

generally adopted for hangers, sports auditoriums, exhibition halls, industrial

buildings and a variety of other large span structures where uninterrupted floor space

is required. Shells are broadly classified into two major groups as,

(a) Singly curved shells which are developable

(b) Doubly curved shells which are non developable

Under the singly curved shells we have the conical and cylindrical shells. The

common example of doubly curved shells is the circular domes, paraboloid. These

shells are generally grouped under these categories designated as,

(a) Shells of revolution

(b) Shells of translation

(c) Ruled surfaces

Shells are also classified as thin and thick shells. A shell can be considered as thin if

the ratio of the radius to the thickness of the shell is greater than 20. In general, most

of the shells used in practise come under the category of thin shells. The general

28

guidelines followed for selecting the dimensions of the various structural components

of the shells are detailed below:

(a) The overall thickness of a reinforced concrete shell should not be less than

50mm for singly curved shells, 40mm for doubly curved shells and 25mm for

precast shells. Generally the thickness is in the range of 80mm to 120mm for

most of the shells based on practical considerations.

(b) The span of reinforced concrete shells should not be greater than 30m to limit

the size and reinforcement within practicable limits in the edge beams. For

longer spans, prestressed edge beams can be used. The width of the edge

member is limited to 2 to 3 times the thickness of the shell.

(c) For large span shells, depth= 1/6 to 1/12 span, larger figures are applicable to

small spans. For shells without edge members, depth < span/10. For shells

with chord width much larger than the span, depth > 1/10 chord width.

(d) The semi central angle should be in the range of 30 to 45 degrees. If the angle

is less than 45°, the effect of wind load may be ignored.

(e) The diameter of reinforcement should not exceed 10mm for 50mm thick shells

and 12mm for 65mm thick shells and 16mm for shells having thickness

greater than 65mm. In the junction zones where the shell is thickened, larger

diameter bars are permissible. The spacing of the bars should not be more than

five times the thickness of the shell. Minimum clear cover must be 12mm or

the nominal size of reinforcing bars.

(f) In case of shells of long lengths exceeding 40m, expansion joints have to be

provided. The construction joints are provided along the curved length of the

shells where the shear forces are minimum.

Though several methods have been discussed to design a shell but the most widely

used method is Beam theory. In the beam theory the shell is analyzed as a bean of

curved cross-section spanning between the end frames or transverses. In the case of

long shells, the longitudinal force components are predominant and hence the beam

theory is ideally suited for analysis. The beam theory is generally applicable for

cylindrical shells of L/R ratio exceeding the value of π.

29

CHAPTER-3

ANALYSIS, DESIGN AND RESULTS

3.1 DESIGN OF SLAB (TOP, FIRST AND GROUND FLOOR)

Size of slab= 3.33m×8m

Assuming modification factor ≈ 1.2

spaneffective depth

= 26×1.2

Effective depth = 106.7mm

Taking total depth D = 140mm

Hence effective depth d=121m

Dead load due to slab = 25×0.14

=3.5kN/m2

Surface finish load = 1.5kN/m2

Hence dead load intensity, wd = 5kN/ m2

Live load intensity, w l = 4kN/ m2

lx= 3.33 + 0.128

= 3.458 m

l y = 8 + 0.128

= 8.128 m

l y

lx

= 2.3 > 2

Hence one way slab.

30

Calculation of Bending Moment:

Bending moment near middle of end span

M 1= wdl2

12 +

wl l2

10 = 9.763kN-m/ m run

Bending moment at support next to end support

B11= −wd l2

10 −wl l

2

9

= - 9.763kN-m/ m run

Bending moment at middle of interior span

M 2= wdl2

16 + wl l

2

12

= 7.726kN-m/ m run

Bending moment at other interior support

B12= −wd l2

12 −wl l

2

9

= Z 10.29kN-m/ m run

Maximum bending moment = 10.29kN-m/ m run

0.138f ckbd2 = 10.29 × 1.5× 106

d = 74.7mm< 128mm ok.

Reinforcement forM 1:

Factored BM = 1.5×9.763

=14.64kN-m/ m run

Reinforcement A st = 335 mm2

Spacing = 1000× 50.24

335 =149.9mm

31

Hence providing 8ф @ 150mm c/c.

Actual reinforcement provided = 1000× 50.24

150 =334.9 mm2

Hence % tension reinforcement = 100× 334.91000× 121

= 0.276%

Reinforcement forB11:

Factored BM = 19.6kN-m/ m run


Spacing = 1000× 50.24391

=130mm


Reinforcement forM 2:

Factored BM =11.589kN-m/ m run


Spacing = 1000× 50.24262

= 190mm


Reinforcement forB12:

Factored BM = 15.435kN-m/ m run


Spacing = 1000× 50.24355

= 140mm

Hence providing 8ф @ 140mm c/c

Codal specifications:

Maximum spacing = 3d = 384mm

32

Minimum area of reinforcement = 0.12 % of BD =168mm2

Check for shear force:

Maximum shear force occurs at first interior support B11

V u = 0.6×1.5×3.458×11.6 = 36.1kN/m

τ v = 36.1128

=0.282N/mm2

pt= 100× 509.71000× 28

= 0.389 %

τ c= 0.42N/mm2, k = 1.3

τ c' = 1.3×0.42 = 0.546N/mm2

τ c' >τ v, hence Ok.

Check for development length:

Ф≤ 4 τbd

0.87 × f ck

(1.3 M 1

V+L0)

M 1= 0.87 × f y × A st(d−f y × Ast

f ck× b) =7.32kN-m

Shear force V = 0.4wd+0.45 wl

= 12.92 kN

Ф ≤ 4 × 1.6

0.87 × 415( 1.3 ×7.32

12.92+300)

≤ 9mm.

Hence ok.

Check for deflection:

Percentage tension reinforcement = 0.276%

Modification factor ≈1.5

33

d = 3.333× 1000

23× 1.5 = 96.6mm

Hence safe.

A table showing results is as follows:

Table 3.1 – Results of slab

Designation Reinforcement

Diameter

Spacing(c/c)

Near middle of end

span

335mm2 8mm 150mm

At support next to

end support

391 mm2 8mm 130mm

At middle of

interior span

262 mm2 8mm 190mm

At other interior

support

355 mm2 8mm 140mm

3.2 DESIGN OF BEAM (TOP FLOOR)

3.2.1 Design of Beam ‘AB’ Of Span 10m:

Taking width b = 300mm

Effective depth d = 600mm

Total depth D = 625mm

Live load of full slab = 4 ×10 ×8

=320kN

Dead load = (8×10 × 0.14 ×25 )+ (8× 0.625 × .300 ×25 )

= 317.5kN

A 10 B

34

8

E M

Fig 3.1- Different beams

Live load for triangular portion = 16 ×320

80

= 64kN

Live load for trapezoidal portion = 124 ×320

80

= 96kN

Dead load for triangular portion = 16 ×317.5

80

= 63.5kN

Dead load for trapezoidal portion = 24 ×317.5

80

= 95.25kN

Dead load of AB = 95.25

10 + .3× .625 ×10 ×25

10

wd = 14.2kN/ m

Live load of AB = 9610

w l = 9.6kN/ m

Calculation of bending moment and reinforcement:


35

M AB= wdl2

12 +

wl l2

10

= 214.33kN-m

Reinforcement A st = 1901.7 mm2

Hence providing 6 - 20 ф.


M BC= wdl2

16 +

wl l2

12

= 168.75kN-m


Hence providing 5-20ф


M B= −wd l2

10 −wl l

2

9

=−248.66 kN-m


Hence providing 8-20ф.

Calculation of shear force:

Shear force at end support (A)

SA= 0.4 × wd × l

2 +

0.45 ×w l× l

2

= 50kN

Shear force at support next to end support (B)

SB= 0.6 × wd ×l

2 +

0.6 × wl× l

2

36

= 71.4kN

Check for shear force at ‘A’:

V u = 1.5 x 50

= 75kN

τ v = 75000

300× 600

=0.416 N/mm2

pt= 100× 6 ×314

300 ×600

= 1.04 %

τ c= 0.64 N/mm2

τ c>τ v.

Hence safe.

Check for shear force at ‘B’:

V u = 1.5 x 71.4

= 107.1kN

τ v = 107100

300× 600

= 0.595N/mm2

pt= 100× 8 ×314

300 ×600

= 1.39 %

τ c= 0.69 N/mm2

τ c>τ v.

Safe hence providing minimum reinforcement

37

Using 8mm- 2 legged vertical stirrups

A sv= 100.5 mm2

Spacing = 0.87 × f y × A sv

0.4 × b

= 300mm

Providing 8mm-2 legged vertical stirrups @ 300mm c/c.


Maximum shear force = 71.4kN

Maximum Bending moment = 248.66kN-m

L0 = 600mm or 12 ф

=240mm

Ld ≤1.3 × M

V+Lo

47ф≤1.3 ×248.66 × 1000

71.4+240

Ф ≤ 40mm

Hence ok.

Table 3.2– Result of beam ‘AB’

Designation Reinforcement Number of bars Diameter

Near middle of end

span

1901.7mm2 6 20mm

At middle of

interior span

1392.6mm2 5 20mm

At support next to

end support

2370.3mm2 8 20mm

3.2.2 Design of Beam ‘AE’ of span 8m:

38

Dead load = Dead load of triangular portion + dead load of beam

= 63.510

+ .3× .625 ×8 ×25

10

wd = 12.63 kN/ m

Live load = 648

w l = 8kN/ m



M AE= wdl2

12 +

wl l2

10

= 118.56 kN-m




M EF= wdl2

16 +

wl l2

12

= 93.18kN-m




M E= −wd l2

10 −wl l

2

9

= -137.71kN-m


39




SA= 0.4 × wd × l

2 +

0.45 ×w l× l

2

= 31.01 kN

Shear force at support next to end support (E)

SE= 0.6 × wd ×l

2 +

0.6 × wl× l

2

= 44.712kN


V u = 1.5 x 31.01

= 46.515 kN

τ v = 46.515 ×1000

300 ×600

= 0.37N/mm2

pt= 100× 6 ×200.96

300×600

= 0.66 %

τ c= 0.5N/mm2

τ c>τ v.

Hence safe.

Check for shear force at ‘E’:

V u = 1.5 x 44.712

= 67.068 kN

40

τ v = 0.37 N/mm2

pt= 0.67 %

τ c= 0.52 N/mm2

τ c>τ v.



A sv= 100.5 mm2


0.4 × b

= 300mm



Maximum shear force = 44.712 kN

Maximum Bending moment = 137.71 kN-m

L0 = 600mm or 12 ф

=240mm

Ld ≤1.3 × M

V+Lo

47Ф≤1.3 ×137.71 ×1000

44.712+240

Ф ≤ 40mm. hence ok.

Table 3.3- Result of beam ‘AE’


41

Near middle of end

span

918.6mm2 5 16mm

At middle of

interior span

702.4mm2 4 16mm

At support next to

end support

1091mm2 6 16mm

3.2.3 Design of Beam ‘EM’ of span 10m:

Dead load = 2(Dead load of trapezoid) + dead load of beam

wd = 23.7kN/ m

Live load = 2(Live load of trapezoid)

w l = 19.2kN/ m

Taking effective depth d = 750mm



M EM= wdl2

12 +

wl l2

10

= 389.5kN-m




M MP= wdl2

16 +

wl l2

12

= 308.1kN-m


42



M M= −wd l2

10 −wl l

2

9

= -450.33 kN-m




Shear force at end support (E)

SE= 0.4 × wd × l

2 +

0.45 ×w l× l

2

= 90.6 kN

Shear force at support next to end support (M)

SM= 0.6 × wd ×l

2 +

0.6 × wl× l

2

= 128.7 kN


V u = 135.9kN

τ v = 135.9× 1000

300× 750

= 0.60N/mm2

pt= 100× 9 ×314

300 ×750

= 1.25 %

τ c= 0.67 N/mm2

43

τ c>τ v, hence safe.

Check for shear force at ‘M’:

V u = 1.5 x 128.7

= 193.05 kN

τ v = 0.85 N/mm2

pt= 1.8%

τ c= 0.76 N/mm2

τ v>τ c.

Hence shear reinforcement needed to be provided

V us= 193.05x1000 – 0.76x300x750

= 22050kN


A sv= 100.5 mm2

Spacing = 0.87 × f y × A sv × d

V us

= 300mm



Maximum shear force = 128.7kN


L0 = 600mm or 12 ф

=240mm

44

Ld ≤1.3 × M

V+Lo

47ф≤1.3 × 450.33 ×1000

128.7+240

Ф ≤ 40mm

Hence ok.

Table 3.4 - Result of beam ‘EM’


Near middle of end

span

2974.7mm2 9 20mm

At middle of

interior span

2123.3mm2 7 20mm

At support next to

end support

3888mm2 13 20mm

3.2.4 Design of Beam ‘BM’ of Span 8m:

Dead load = 2(Dead load of triangle) + dead load of beam

wd = 21.5kN/ m

Live load = 2(Live load of triangle)

w l = 16kN/ m

Taking effective depth d = 750mm



M MB= wdl2

12 +

wl l2

10

45

= 217.06kN-m

Reinforcement A st= 1378 mm2



M MN= wdl2

16 +

wl l2

12

=199.99kN-m




M M= −wd l2

10 −wl l

2

9

= - 251.37 kN-m




Shear force at end support (B)

SB= 0.4 × wd × l

2 +

0.45 ×w l× l

2

= 63.2kN


SM= 0.6 × wd ×l

2 +

0.6 × wl× l

2

= 90kN

46


V u= 94.8kN

τ v= 94.8 ×1000300 ×750

= 0.42N/mm2

pt= 100× 5 ×314

300 ×750

= 0.69 %

τ c= 0.5 N/mm2, τ c>τ v .

Hence safe.


V u= 1.5 x 90 = 135kN

τ v = 0.6 N/mm2,pt= 0.84%

τ c= 0.8 N/mm2

τ v>τ c.



A sv= 100.5 mm2


0.4 × b

= 300mm



Maximum shear force = 90kN

47


L0 = 600mm or 12 ф = =240mm

Ld ≤1.3 × M

V+Lo

47ф≤1.3 ×251.37 × 1000

90+240

Ф ≤ 45mm. Hence Ф = 20mm is ok.

Table 3.5 - Result of beam ‘BM’


Near middle of end

span

1378mm2 5 20mm

At middle of

interior span

1253mm2 4 20mm

At support next to

end support

1641mm2 6 20mm

1111

48

A3 A1 A2 A4

B1 B2 B3 B4

C1 C2 C3 C4

Fig 3.2- Grid of Columns

3.3 DESIGN OF COLUMNS (TOP FLOOR)

3.3.1 Design of Column D1:

Dimension of beam = 300×625

Size of column = 300×400

Moment of inertia of column Ic = bd3/12 = 300× 4003/12 = 16 ×108 mm4

Moment of inertia of beam Ib = 300×6253/12 = 61×108 mm4

Stiffness K of column at joint = 4×108

Stiffness kb of beam at joint = 6.1 ×108

Ku = 0.268

kL = 0.268

kb = 0.464

Total dead load of slab + beam + intermediate beams in X-direction

= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2

Total live load= 4kN/m2

Total load = 8.99kN/m2

Load intensity w in X-direction =8.99 ×24/10

= 21.6kN/m2

Me = wl2/12

= 21.6×102/12

= 175kN-m

Mux = 0.268×175 kN-m

= 46.9kN-m

Total dead load due to slab + beams in X-direction

= 10× 8× 0.14 ×25+0.625 ×0.3 ×25 ×8

49

D1 D2 D3 D4

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2


Load intensity in Y-direction = 17.98kN/m

Me = 95.89kN-m

Muy = 24.55kN-m

From load distribution axial load Pu = 179.8kN

Let assume % reinforcement provided = 0.8%

pfck

= 0.820

= 0.04

Let effective cover be 50 mm on either side-

dD

= 50

400 = 0.125

Uniaxial moment capacity of section

Pufck

= 179.8× 1.5 ×1000

20 × 300× 400 = 0.112

By using p-M interaction as given by SP: 16

Mu

fck .b . D2 = 0.09

Mux1 = 0.09× fck × b× D2

=0.09×20 × 300× 4002

Mux1 = 86.4kN-m & Muy1 = 86.4kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 1378.8kN

Pu/Pz = 0.13

From graph α = 1

The strength of the section can be checked using the interaction

(Mux

Mux1¿α + (

MuyMuy1

¿α = (46.986.4

) 1+ (24.5586.4

) 1

= 0.82 ≤1

Hence design is safe. OK

Thus the assumed value of reinforcement % p is OK.

A s = 0.8% gross-area

= 0.008×300× 400 = 960 mm2

50

So provide 4-16 diameters

Use 8 mm lateral ties and should be at least of

1)300 mm

2)16×12 = 192mm

3)48 × 8 = 384

Provide bars at 190 mm c/c

3.3.2 Design of Column C1:







Ku = 0.268

kL = 0.268

kb = 0.46


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2



Load intensity w in X-direction = 43.15kN/m2

Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m

Total dead load due to slab + beams Y-direction

= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2

51



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m


Let assume % reinforcement p provided = 0.8%

pfck

= 0.820

= 0.04


dD

= 50

400 = 0.125


Pufck

= 359.6 ×1.5 ×1000

20 × 300× 400 = 0.22


Mu

fck .b . D2 = 0.12

Mux1 = 0.12× fck × b× D2

=0.12×20 × 300× 4002

Mux1 = 115.2kN-m & Muy1 = 115.2kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 1378.8kN

Pu/Pz = 0.26

From graph α = 1.0375


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37115.2

) 1.037+ (32.98115.2

) 1.037

= 0.93 ≤1




= 0.008×300× 400 = 960 mm2



52

1)300 mm

2)16×12 = 192mm

3)48 × 8 = 384


3.3.3 Design of Column C 2:







Ku = 0.106

kL = 0.106

Kb = 0.39


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

Dead load= 3.968kN/m2

Live load = 4kN/m2



53

Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m


Let assume % reinforcement provided = 0.8%

pfck

= 0.720

= 1.4%


dD

= 50

400 = 0.125


Pufck

= 719.5× 1.5 ×1000

20 × 300× 400 = 0.44


Mu

fck .b . D2 = 0.1

Mux1 = 0.1× fck × b× D2

=0.1×20 × 300× 4002

Mux1 = 96.37kN-m & Muy1 = 96.37kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 1378.8kN

Pu/Pz = 0.37



(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37115.2

) 1.037+ (32.98115.2

) 1.037

= 0.93 ≤1

Hence design is safe. Ok

Thus the assumed value of reinforcement % p is Ok.


= 0.014×300× 400 = 1680 mm2



1) 300 mm

2) 16×12= 192mm

54

3) 48 × 8 = 384









Ku = 0.106

kL = 0.106

kb = 0.39


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes =2× wl2/12

=2 ×12.6×102/12

= 350kN-m

Mux = 350× 0.106 kN-m

= 37.1kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2


Load intensity in Y-direction = 17.98×2 kN/m

= 35.96kN-m

Mes = 191.78kN-m

55

Muy = 0.256×191.78 kN-m

= 49.09kN-m


Let assume % reinforcement p provided = 0.4%

pfck

= 0.420

= 0.02


dD

= 50

400 = 0.125


Pufck

= 359.6 ×1.5 ×1000

20 × 300× 400 = 0.22


Mu

fck .b . D2 = 0.052

Mux1 = 0.052× fck × b× D2

=0.052×20 × 300× 4002

Mux1 = 49.9kN-m & Muy1 = 115.2kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 1378.8kN

Pu/Pz = 0.26



(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37115.2

) 1.037+ (32.98115.2

) 1.037

= 0.93 ≤1

Hence design is safe. Ok

Thus the assumed value of reinforcement % p is Ok.


= 0.004×300× 400 = 480 mm2

So provide 4-16diameter


1)300 mm

2) 16×12 = 192mm

3)48 × 8 = 384

56


3.4 DESIGN OF BEAM (FIRST & GROUND FLOOR)

3.4.1 Beam ‘AB’ of span 10m

Taking width b = 300mm

Effective depth d = 750mm

Assuming thickness = 130mm

Height of wall = 4000- 140

= 3860mm

Dead load = dead load of trapezoid + dead load of beam + self weight of wall

wd = 95.25 + 25×10×0.3×0.75 + 10×3.86×0.13×20

= 25.2kN/m

Live load = live load of trapezoid

w l = 9.6kN/m



M AB= wdl2

12 +

wl l2

10 = 306 kN-m




M BC= wdl2

16 +

wl l2

12

= 237.5 kN-m

57




M B= −wd l2

10 −wl l

2

9

= -358.6 kN-m





SA = 72kN

Shear force at support next to end support (B)

SB = 104.4kN


V u = 108kN

τ v = 0.48N/mm2

pt= 0.97 %

τ c= 0.6 N/mm2

τ c>τ v.

Hence safe.


V u = 156.6

τ v = 0.696 N/mm2

58

pt= 1.26%

τ c= 0.68 N/mm2

τ v>τ c.

Hence shear reinforcement needed to be provided

V us = 3600 kN


A sv= 100.5 mm2

Spacing = 300mm



Ld ≤1.3 × M

V+Lo

47ф≤1.3 ×358.66 × 1000

104.4+240

Ф ≤ 45mm. hence Ok.

Table 3.6 - Results of beam ‘AB’


Near middle of end

span

2104mm2 7 20mm

At middle of

interior span

1533mm2 5 20mm

At support next to

end support

2621mm2 9 20mm

3.4.2 Beam ‘AE’ of span 8m

Dead load = dead load of triangle + dead load of beam + self weight of wall

wd = 23.5kN/m

59

Live load = live load of triangle

w l = 8kN/m



M AE= wdl2

12 +

wl l2

10 =176.53kN-m




M EF= wdl2

16 +

wl l2

12=136.67kN-m




M E= −wd l2

10 −wl l

2

9 = Z 207.3kN-m





SA= 0.4 × wd × l

2 +

0.45 ×w l× l

2

= 51.6kN

Shear force at support next to end support (E)

60

SE= 0.6 × wd ×l

2 +

0.6 × wl× l

2 = 75.6kN


V u = 77.4kN, τ v = 0.34N/mm2

pt= 0.55%, τ c= 0.5 N/mm2

τ c>τ v, hence safe.


V u = 113.4kN, τ v = 0.5 N/mm2

pt= 0.69%, τ c= 0.52 N/mm2

τ c>τ v.

Hence minimum reinforcement required

Using 8mm- 2 legged vertical stirrups, A sv= 100.5 mm2

Spacing =300mm



Ld ≤1.3 × M

V+Lo

47ф≤ 3804.68

Ф ≤ 45mm, hence Ok.

61

Table 3.7 - Results of beam ‘AE’


Near middle of end

span

1087mm2 4 20mm

At middle of

interior span

819mm2 3 20mm

At support next to

end support

1306mm2 5 20mm

3.4.3 Beam ‘EM’ of span 10m

Dead load = 2 (dead load of trapezoid) + dead load of beam + self weight of wall

wd =34.7kN/m

Live load = live load of trapezoid

w l =19.2kN/m



M EM= wdl2

12 +

wl l2

10

= 481.16kN-m


M MP= wdl2

16 +

wl l2

12

= 376.8kN-m


M M= −wd l2

10 −wl l

2

9 = -560.33kN-m

62

M lim ¿ 0.138 f ckbd 2.

=0.138×20×300×7502 = 465.75kN-m

M :̶ M lim ¿¿= 94.5kN-m

xm= 0.48d

Area of tension steel corresponding to

0.87f y A t 1= 0.36f ck b xm

At 1= 2153mm2

Taking d '

d =0.1

M :̶ M lim ¿¿= ( f SC × ASC−f cc× A sc ) (d−d' )

ASC = 407.22mm2

Corresponding tension steel:

0.87f y A t 2= f SC × ASC

At 2= 398.14mm2

Atotal tension = At 1+ A t 2

= 25514.14mm2.

Maximum area in tension = 0.04bD = 9300mm2

Atotal tension< Amax, hence ok.

3.4.4 Design of Beam ‘BM’ of span 8m

Dead load = 2(dead load of triangle) + dead load of beam + self weight of wall

wd = 31.5kN/m

Live load = live load of triangle

63

w l = 16kN/m



M BM= wdl2

12 +

wl l2

10 = 270.4kN-m




M MN= wdl2

16 +

wl l2

12 = 211.33kN-m




M M= −wd l2

10 −wl l

2

9 = Z 315.7kN-m




Shear force at end support (B)

SB= 0.4 × wd × l

2 +

0.45 ×w l× l

2

= 79.2kN


SM= 0.6 × wd ×l

2 +

0.6 × wl× l

2 = 114kN

64


V u = 118.8kN, τ v = 0.53N/mm2

pt= 0.83%, τ c= 0.58 N/mm2

τ c>τ v , hence safe.


V u = 171kN, τ v = 0.76 N/mm2

pt= 0.98%, τ c= 0.61 N/mm2

τ v>τ c.

V us = 33750kN

Using 8mm- 2 legged vertical stirrups, A sv= 100.5 mm2

Spacing = 0.87 × f y × A sv × d

V us = 300mm



Ld ≤1.3 × M

V+Lo

47ф≤ 3596.32+240

Ф ≤ 45mm, hence ok.

Table 3.8 - Results of beam ‘BM’


Near middle of end

span

1796mm2 6 20mm

At middle of

interior span

1335.7mm2 5 20mm

65

At support next to

end support

2190mm2 7 20mm

3.5 DESIGN OF COLUMNS (FIRST FLOOR)

3.5.1Design of Column D1:







Ku = 0.268

kL = 0.268

kb = 0.464


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




= 21.6kN/m2

Me = wl2/12

= 21.6×102/12

= 175kN-m

Mux = 0.268×175 kN-m

= 46.9kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2


66


Me = 95.89kN-m

Muy = 24.55kN-m


Let assume % reinforcement p provided = 2%

pfck

= 2

20 = 0.1


dD

= 50

400 = 0.125


Pufck

= 359.6 ×1.5 ×1000

20 × 300× 400 = 0.22


Mu

fck .b . D2 = 0.18

Mux1 = 0.18× fck × b× D2

=0.18×20 × 300× 4002

Mux1 = 172.8kN-m & Muy1 = 172.8kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 182kN

Pu/Pz = 0.16

From graph α = 0.9


(Mux

Mux1¿α + (

MuyMuy1

¿α = (46.9

172.8) 0.9+ (

24.55172.8

) 0.9

= 0.29 ≤1




= 0.02×300× 400 = 2400 mm2



1)300 mm

2)16×12= 192mm

67

3)48 × 8= 384


3.5.2 Design of ColumnC1:







Ku = 0.268

kL = 0.268

kb = 0.46


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

68

= 32.98kN-m



pfck

= 2

20 = 0.1


dD

= 50

400 = 0.125


Pufck

= 719.2× 1.5× 1000

20 ×300 × 400 = 0.45


Mu

fck .b . D2 = 0.14

Mux1 = 0.14× fck × b× D2

=0.14×20 × 300× 4002

Mux1 = 134.4kN-m & Muy1 = 134.4kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 1378.8kN

Pu/Pz = 0.59

From graph α = 1.4


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37134.4

) 1.4+ (32.98134.4

) 1.4

= 0.76 ≤1




= 0.02×300× 400 = 2400 mm2



1)300 mm

2)16×12= 192mm

3)48 × 8 = 384


69








Ku = 0.106

kL = 0.106

kb = 0.39


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.87kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m


70


pfck

= 4

20 = 0.2%


dD

= 50

400 = 0.125


Pufck

= 1438.4 ×1.5 ×1000

20× 300 ×400 = 0.89


Mu

fck .b . D2 = 0.09

Mux1 = 0.09× fck × b× D2

=0.09×20 × 300× 4002

Mux1 = 86.4kN-m & Muy1 = 86.4kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 2574kN

Pu/Pz = 0.8

From graph α = 1.8


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.3786.4

) 1.8+ (32.9886.4

) 1.8

= 0.9 ≤1



A s = 4% gross-area

= 0.04×300× 400= 4800 mm2



1) 300 mm

2)16×12= 192mm

3)48 × 8 = 384



71







Ku = 0.268

kL = 0.268

kb = 0.46


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m



72

pfck

= 2

20 = 0.1


dD

= 50

400 = 0.125


Pufck

= 719.2× 1.5× 1000

20 ×300 × 400 = 0.45


Mu

fck .b . D2 = 0.14

Mux1 = 0.14× fck × b× D2

=0.14×20 × 300× 4002

Mux1 = 134.4kN-m & Muy1 = 134.4kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 1378.8kN

Pu/Pz = 0.59

From graph α = 1.4


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37134.4

) 1.4+ (32.98134.4

) 1.4

= 0.76 ≤1




= 0.02×300× 400 = 2400 mm2



1)300 mm

2)16×12= 192mm

3)48 × 8 = 384


3.6 DESIGN OF COLUMNS (GROUND FLOOR)


73







Ku = 0.268

kL = 0.268

kb = 0.464


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




= 21.6kN/m2

Me = wl2/12

= 21.6×102/12

= 175kN-m

Mux = 0.268×175 kN-m

= 46.9kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Me = 95.89kN-m

Muy = 24.55kN-m



74

pfck

= 3

20 = 0.15


dD

= 50

400 = 0.125


Pufck

= 539.4 ×1.5 ×1000

20× 300 ×400 = 0.33


Mu

fck .b . D2 = 0.07

Mux1 = 0.07× fck × b× D2

=0.07×20 × 300× 4002

Mux1 = 67.2kN-m & Muy1 = 67.2kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 2200kN

Pu/Pz = 0.36



(Mux

Mux1¿α + (

MuyMuy1

¿α = (46.967.2

) 1.18+ (24.5567.2

) 1.18

= 0.9 ≤1



A s = 3% gross-area

= 0.03×300× 400 = 3600 mm2



1)300 mm

2)16×12= 192mm

3)48 × 8 = 384




75






Ku = 0.268

kL = 0.268

kb = 0.46


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m



pfck

= 3

20 = 0.15

76


dD

= 50

400 = 0.125


Pufck

= 1078.8× 1.5 ×1000

20 × 300× 400 = 0.67


Mu

fck .b . D2 = 0.14

Mux1 = 0.14× fck × b× D2

=0.14×20 × 300× 4002

Mux1 = 134.4kN-m & Muy1 = 134.4kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 2205kN

Pu/Pz = 0.73

From graph α = 1.6


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37134.4

) 1.6+ (32.98134.4

) 1.6

= 0.68≤1



A s = 3% gross-area

= 0.03×300× 400 = 3600 mm2



1)300 mm

2)16×12 = 192mm

3)48 × 8 = 384






77




Ku = 0.268

kL = 0.268

kb = 0.46


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m


= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m



pfck

= 3

20 = 0.15


78

dD

= 50

400 = 0.125


Pufck

= 1078.8× 1.5 ×1000

20 × 300× 400 = 0.67


Mu

fck .b . D2 = 0.14

Mux1 = 0.14× fck × b× D2

=0.14×20 × 300× 4002

Mux1 = 134.4kN-m & Muy1 = 134.4kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 2205kN

Pu/Pz = 0.73

From graph α = 1.6


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.37134.4

) 1.6+ (32.98134.4

) 1.6

= 0.68≤1



A s = 3% gross-area

= 0.03×300× 400 = 3600 mm2



1)300 mm

2) 16×12= 192mm

3)48 × 8 = 384

Provide bars at 190 mm c/c.





79




Ku = 0.106

kL = 0.106

kb = 0.39


= 10×8×.14×25 + 0.625×25 × 10+2× .3 ×0.6 × 8× 25

= 398.875kN

= 4.99kN/m2




Mes = wl2/12

= 43.152×102/12

= 359.6kN-m

Mux = 359.6× 0.268 kN-m

= 96.37kN-m

Total dead load due to slab + beam Y-direction

= 10× 8× 0.14 ×25+0.625 ×o .3× 25×8

= 317.5kN

= 3.968kN/m2

Live load = 4kN/m2



Mes = 191.78kN-m

Muy = 0.172×191.78 kN-m

= 32.98kN-m



pfck

= 4

20 = 0.2%


80

dD

= 50

400 = 0.125


Pufck

= 2157.6 ×1.5 ×1000

20 × 300× 400 = 0.89


Mu

fck .b . D2 = 0.08

Mux1 = 0.08× fck × b× D2

=0.08×20 × 300× 4002

Mux1 = 76.8kN-m & Muy1 = 76.8kN-m

Pz = 0.45× fck × Ac+ 0.75× f y × A sc

= 2574kN

Pu/Pz = 0.8

From graph α = 1.8


(Mux

Mux1¿α + (

MuyMuy1

¿α = (96.3776.8

) 1.8+ (32.9876.8

) 1.8

= 0.9 ≤1



A s = 4% gross-area

= 0.04×300× 400 = 4800 mm2



1) 300 mm

2) 16×12= 192mm

3) 48 × 8 = 384


3.7 DESIGN OF FOOTING

3.7.1 For Column D1:

Specification:

81

Grade of concrete = M 25

Bearing capacity = 100kN/m2

Load on column = 539.4kN/m2

Area of footing = 1.1× 539.4

100 = 5.9334 m2

Hence providing 3×2 m rectangular footing

Bending moment:

Net earth pressure acting upward due to load

P = 539.42× 3

= 89.9kN

Along long side-

Maximum bending moment = 89.9×1.32/2

= 75.9kN/m2

Along short side –

Maximum bending moment = 89.9 × 0.852/2

= 32.48kN/m2

Hence design B.M= 75.9kN/m2

Effective depth d = √(1.5 ×75.9 × 106) /(0.138 ×25 ×1000)

= 181.8 mm

Total depth D = 181.8 + 50 +202

= 241.8 mm

Assume D = 460 mm

d x= 460 -50-202

82

= 400 mm

d y= 460-50-20 -162

= 382 mm

A stx = 0.5 × f ck

f y[1- √1−(4.6 ×1.5 × B . M .×106¿)/¿¿2)]× b× d x

= 1029.92 mm2

A sty = 354 mm2

Minimum reinforcement = 0.12× 1000 ×460

100

= 552 mm2

Spacing of 20 diameter bar = 314 ×1000

1030 = 300 mm c/c

Spacing of 16 diameter bar = 201× 1000

600 = 300 mm c/c

Hence provide 10 – 20 diameter @ 300 mm c/c in X-direction

And 6 – 16 diameter @ 300 mm c/c in Y- direction

Check for one way shear:

Along long side, shear force (V) = p × 0.9

= 89.9× 0.9 = 80.91kN

Shear stress τ v = V × 1000

bd =

80.91× 1.5 ×10001000 × 400

= 0.34 N/mm2

pt = 100×

1000 × 314300

1000 × 400 = 0.26

τ c= 0.37 N/mm2

τ cmaxFor M 25 = 3.1 N/mm2

83

So, τ v<τ c<τ cmax

Hence no reinforcement is required.

Check for Two shear (Punching shear):

Combined effective depth d = dx+d y

2

= 400+382

2 = 391mm

V = p [3×2 :̶ 0.92]

= 89.9×5.19 = 466.58kN

τ v = 1.5× 466.58 ×1000(791+691)×391

= 0.003 N/mm2

τ c = k s.τ c'

= 1×1.25 = 1.25 N/mm2

So τ v<¿ τc<¿τcmax

¿¿ hence ok.

3.7.2 For Column C1:

Specification:

Grade of concrete = M25




100 = 11.86 m2


Bending moment:


84

P = 1078.82× 3

= 89.9kN

Along long side-


= 145.64kN/m2


Maximum bending moment = 89.9 ×1.352/2

= 81.92kN/m2

Hence design B.M. = 145.64kN/m2

Effective depth d = √(1.5 ×145.64 ×106)/(0.138 × 25 ×1000)

= 251 mm

Total depth D = 251 + 50 +202

= 311 mm

Assume D = 460 mm

d x= 460 -50-202

= 400 mm

d y= 460-50-20 -162

= 382 mm

A stx =1622 mm2

A sty = 883.68 mm2


100

= 552 mm2

85

Spacing of 20 diameter bars = 314 ×1000

1622 = 190 mm c/c

Spacing of 16 diameter bars = 201× 1000

883.68 = 220 mm c/c

Hence provide 21 – 20 diameter bars @ 190 mm c/c in X-direction



Along long side, shear force (V) = p ×1.4

= 89.9×1.4 = 125.86kN


bd =

125.86 ×1.5 ×10001000 × 400

= 0.36 N/mm2

pt = 100×

1000 × 314300

1000 × 400 = 0.26

τ c= 0.37 N/mm2

τ cmaxfor M 25 = 3.1 N/mm2

So, τ v<¿ τc ¿<τ cmax


Check for two shear (Punching shear):


2

= 400+382

2 = 391mm

V = p [4×3 :̶ 1.42]

= 89.9×5.19 = 902.596 kN

86

τ v = 1.5× 902.596 ×1000(791+691)×391

= 1.07 N/mm2

τ c = k s.τ c'

= 1×1.25 = 1.25 N/mm2


¿¿. Ok.

3.7.3 For Column C2:

Specification:





100 = 23.73 m2

Hence providing 6× 4m rectangular footing

Bending moment:


P = 2157.66× 4

= 89.9kN

Along long side-


=352.4kN/m2



=153.8kN/m2

87


Effective depth d = √(1.5 × 352.4 ×106)/(0.138× 25 ×1000)

= 391 mm

Total depth D = 391 + 50 +202

= 451 mm

Assume D = 800 mm

d x= 800 -50-252

= 737.5 mm

d y= 800-50-25 -202

= 685 mm

A stx =2084 mm2, A sty = 960 mm2


100

= 960 mm2

Spacing of 25 diameter bar = 490.625 ×1000

2084 = 230 mm c/c

Spacing of 20 diameter bar = 314 ×1000

960 = 300 mm c/c

Hence provide 26 – 25 diameters @ 230 mm c/c in X-direction




= 89.9×2.0625 = 185.4kN


bd =

185.4 ×1.5 ×10001000 ×737.5

88

= 0.37 N/mm2

pt = 100×( 1000 × 490.625

230)

1000 ×737.5 = 0.29

τ c= 0.39 N/mm2


So, τ v<¿ τc ¿<τ cmax




2

= 737.5+685

2 = 711.25 mm

V = p [6× 4 :̶ 2.6252]

= 89.9×17.1 = 1537 kN

τ v = 1.5 ×1537 × 1000

(1111.25+1011.25)× 711.25 = 1.12 N/mm2

τ c = k s.τ c'

= 1×1.25 = 1.25 N/mm2


¿¿. Ok.

3.7.4 For Column D2:

Specification:



89



100 = 11.86 m2


Bending moment:


P = 1078.82× 3

= 89.9kN

Along long side-


= 145.64kN/m2



= 81.92kN/m2


Effective depth d = √(1.5 ×145.64 ×106)/(0.138 × 25 ×1000)

= 251 mm

Total depth D = 251 + 50 +202

= 311 mm

Assume D = 460 mm

d x= 460 -50-202

= 400 mm

90

d y= 460-50-20 -162

= 382 mm

A stx = 1622 mm2, A sty = 883.68 mm2


100 = 552 mm2

Spacing of 20 ф bar = 314 ×1000

1622 = 190 mm c/c

Spacing of 16 ф bar = 201× 1000

883.68 = 220 mm c/c

Hence provide 21 – 20 ф@ 190 mm c/c in X-direction

and13 – 16 ф@ 220 mm c/c in Y- direction



= 89.9×1.4 = 125.86kN


bd =

125.86 ×1.5 ×10001000 × 400

= 0.36 N/mm2

pt = 100×( 1000 ×314

300)

1000 ×400 = 0.26

τ c= 0.37 N/mm2


So, τ v<¿ τc ¿ <τ cmax




2=

400+3822

= 391mm

91

V = p [4×3−1.42]

= 89.9×5.19 = 902.596kN

τ v = 1.5× 902.596 ×1000(791+691)×391

= 1.07 N/mm2

τ c = k s.τ c'

= 1×1.25 = 1.25 N/mm2


¿¿. Hence Ok.

3.8 DESIGN OF STAIR CASE:

Assuming thickness of waist slab 200mm.

Dead load of flight:

Step section =0.5×0.28×0.15= 0.021 mm2

Inclined slab = 0.318×0.2= 0.0640m2

Finish = (0.28+0.15) ×0.03=0.0130mm2

Total area = 0.098m2, Density of concrete = 25kN/m3

DL of step section, 1m in width and 280mm in plan length =0.098×25 = 2.45kN/m

DL per m2 on plan = 2.45× 1000

280 = 8.75kN/m2

LL per m2 on plan =5kN/m2, Total load = 13.75kN/m2, Factored load = 20.6kN/m2

Taking 1.5m width of slab, load = 1.5×20.6= 31kN/m

Loading A:

Self weight of slab = 0.2×25= 5kN/m2

Finish= 0.03x25 = 0.75kN/m2

Live load = 5kN/m2

92


Factored load = 16.125kN/m2

Taking 1.5m width of slab, load = 1.5×16.125= 31kN/m

Loading B:

In a distance equal to 150mm from the wall and a distance equal to 75mm inside the

wall only dead load will be considered.

Total factored load = 1.5×5.75×1.5 = 13kN/m

Design of stair flight:

Reaction at support B, RB = 62.5kN

RA = 129.8 Z 62.5 = 67.3kN

Assuming point of zero SF occurs at distance ‘x’ from A

67.3 Z 25×0.75 Z 31(x Z 0.75) = 0 or, x= 2.32m

Hence maximum BM = 67.3×2.32 :̶ 25×0.75×1.945 Z 38.2 = 82kN-m

d2 = 82 ×106

0.138 ×20 ×1500 , d = 141mm

Adopting effective depth as 165mm and overall as 190mm.

At= 82× 106

0.695 ×415 × 165 = 1765mm2

Using 16-12mm bars equally spaced in 1.5m width.

Check for shear:

τ v = 0.28N/mm2,pt= 0.75 %

τ c= 0.56 N/mm2

τ c ' = 0.67 N/mm2 > τ v ok.


93

Ld= 47ф ≈ 600mm

Moment of resistance M 1= 82× 1810

1765 = 84.1kN-m

V = 62.5kN, Taking L0= 0

Ld ≤1.3 × M

V+Lo

Ф ≤ 21mm, hence ok.

Temperature reinforcement = 0.12× 19× 100

100 =2.28 cm2/m.

Providing 10mm bars @300mm c/c.

Design of landing slab A:

Effective span = 1.5+0.15+1.5+0.165 = 3.315m

Width = 1.5m, Factored load per m2 = 16.125kN/m2

Total load = 16.125x1.5x3.315 = 80kN

Reaction from one flight = {31×2.52×2.685+25×1.2×0.825+ 13×0.225× (0.225/2)}

÷ 4.695

= 50kN

Reaction from two flight= 100kN

Maximum BM = wu ×l

8 = 75kN-m, Maximum SF = 90kN

Effective depth = 165mm, At= 1450mm2

Using 16-12mm bars in1.5m width

3.9 DESIGN OF WATER TANK

Capacity = 20 kL, Length =4m, Breadth =3m, Height = 1.7m

94

lb

=1.33 < 2, h = H4

or 1m whichever greater = 1m.

Hence bottom 1m will bend as vertical cantilever.

Water pressure at point D, p= w (H-h) = 6860 N-m

FEM for long wall = pl2

12 =

4 p3

N-m

FEM for short wall = pB2

12 = -

3 p4

N-m

DF AB= 0.43, DF AD= 0.57

Moment calculation:

Table 3.9 – Final moment calculation

Member AB AD

DF 0.43 0.57

FEM 4 p3

−3 p4

Balancing moment −3.01 p12

−3.99 p12

Final moment 12.99 p12

−12.99 p12

Hence moment at support M F= 7425.95 N-m

Bending moment at centre of long span = p l2

8:̶ M F = 6294.05 N-m

Bending moment at centre of short span = pB2

8̶: M F= 291.55 N-m

Maximum bending moment = 7425.95 N-m

Design of section:

95

d2 =7425.95× 1000

1.32× 1000 , d = 75mm

Overall depth T= 75 + 35 ≈200mm

d = 165mm

Determination of pull:

Direct tension on long wall PL= pB2

= 10290 N

Direct tension on short wall PB = pL2

= 13720 N

Cantilever moment:

Cantilever moment at base per unit height = wH h2

6 = 1633.33 N-m

Reinforcement at corner of long wall:

x = d Z T2

= 65m

A st 1= (7425.95× 1000 ) – (10290× 65)

115× 0.853×165 = 417mm2

A st For pull = 10290

115 = 89.4mm2

Total A st= 506.8mm2

Using 8mm bars Aф= 50.24 mm2

Spacing = 100mm

Hence providing 8mm @100mm c/c at inner face near corners and at a height 1m

above the base.

Reinforcement at middle of long wall:

Design bending moment = 6294.05 N-m

Direct pull PL= 10290 N

96

A st 1= 347.5mm2,A st 2= 89.47mm2

Total A st= 436.9mm2 which is very near to the reinforcement provided at ends.

Hence providing 8ф @ 100mm c/c, also providing additional reinforcement of 8ф @

200mm c/c at the outer face.

Reinforcement for short wall:

Bending moment at ends = 7425.95 N-m, Direct pull, PB= 13720 N

A st 1= 403.7mm2,A st 2= 119.3mm2

Total A st= 523mm2, Spacing = 100mm

Hence providing 8mm @100mm c/c at inner face near ends of short span.

Reinforcement for cantilever moment:

Maximum cantilever moment =1633.33 N-m

Reinforcement = 144.7mm2

Minimum reinforcement = 0.3 ×1000 ×200

100 = 600mm2

Providing 300mm2 vertically on inner face and remaining 300mm2 vertically at outer

face with spacing = 160mm.

3.9.1 Design of slab (edges simply supported) for tank:

W =17.6kN/m2

Assume overall thickness as 200 mm

Effective depth = 180 mm

lx = 3.4 + 0.18 = 3.58 mm

l y = 5+ 0.18 = 5.18mm

l y

lx

= 1.44

97

∝x = 0.099, ∝y = 0.051

Dead load of slab =0.2 x 25 = 5kN/m2

Total load =22.6kN/m2

M x = 0.099×22.6×3.582 =28.6kN m

M y = 0.051×22.6×3.5822 =14.77kN m

M=0.138 f ck bd2

d = √28.6 ×106

0.138 × 20× 1000

= 101mm.OK

Area of steel along short span

= 0.36 × f ck b xm

0.87 f y

= (0.36×20×1000×0.48×180)/(0.87x415)

= 1722 mm2

Provide: 25-10ф bars

Area of steel along long span

A sty =0.5 ×20415

×¿ ) ×1000 × 180

=233 mm2

Minimum steel = 0.12%

= 0.0012x180x1000

= 216 mm2

Hence OK.

3.9.2 Design of beam for tank

98

Dead Load = 200kN

Factored Load = 1.5 × 200 = 30kN

Load on AB = 51kN

Load on AD = 99kN

Beam AB:

M AB = Wl8

= 51× 3.4

8 = 21.6kNm

M = 0.138 f ck bd2

21.6x106 =0.138 x 20xd×d2

2

d = 250 mm

A st =0.5 ×20415

×¿ ) ×125× 250

= 298 mm2

Providing: 7−¿8Ф bars

Min steel = 0.85bd

f y =

0.85 ×125 ×250415

= 64 mm2

Hence OK.

Beam AD:

M AD = Wl8

= 99 ×5

8 = 61.87kNm

M=0.138 f ck bd2

61.87x106 =0.138 × 20×d × d2

2

d = 360 mm

99

A st =0.5 ×20415

×¿ ) ×180× 360

= 586 mm2

Provide: 6−¿ 12Ф bars

Minimum steel = 0.85bd

f y =

0.85 ×180 ×360415

= 132 mm2

Hence Ok.

3.10 DESIGN OF SHELL

Radius= 10.10 m, Central rise= 2.35m, Chord width= 13m, Span=30 m

Thickness of shell= 80mm, Semi central angle= 40 degree,

Edge beam size= 200×1880 mm, Reinforcement in edge beam= 16-32ф

Width of edge beams= 200mm

Assuming neutral axis cut the shell at an angleα . Taking moment of effective areas

about neutral axis, we have

I NA= 2∫0

α

Rdθt (Rcosθ−Rcosα ) + m×2At(1.58+Rcosα Z 7.75)

Putting m=13, R=10.1, At= (8x8.04)= 64.32cm2= 0.0064m2

R2 t (sin α Z α cos α) = m At(R cos α Z 6.17)

It gives α= 26.25 ° , I NA= 1.835m4

Self weight of shell = (0.08×25)=1.92kN/m2

Water proofing and live load = 1.0kN/m2


Total weight per meter run=2∫0

40

2.92(Rdθ)= 40kN/m

100

Weight of edge beam = 2(0.1x1.88x24) = 9kN/m

Weight of filling in the valley portion = 1.5kN/m

Total load = 50.5kN/m

Maximum BM =50.5× 302

8= 5700kN-m

Maximum SF = 0.5x50.5x30 = 757.5kN

Maximum compressive stress at crown, f c= 5700× 1330 ×106

1.835 × 1012 = 3.52N/m2

Maximum tensile stress at centre of gravity of steel, f t= 3.52× 2600× 23

1330= 89.5N/m2

Shear stress τ = 757.5× 103× 435 ×106

1.835 ×2×80 × 1012

Maximum horizontal shear stress at neutral axis =1.12 N/mm2

Providing 8mm diameter bars inclined at 45° to the longitudinal axis of shell.

Spacing = √2 ×50 × 23080 ×1.12

= 180mm near supports.

Towards the centre of span where shear stress is less, adopting 6mm diameter bars at

250mm centers.

CHAPTER-5

CONCLUSIONS



areas which are normally located away from main city. The complex has been

planned and designed to meet the optimum balance between economy and quality.

Ample connectivity has been provided. The complex has throughout natural lighting

101

and cross ventilation. The guidelines of IS: 456 -2000 has been followed very strictly.

Slab of dimension 8m×3.33m has been designed with total depth 140mm, live load

4kN/m2 and surface finish load 1.5kN/m2. Beams have been designed with same value

of live and surface finish load. Total depth of beam has been kept 625mm at top floor

and 775mm at first and ground floor.

The shell has been designed using “Beam Theory”. Columns have been analyzed

using “Substitute Frame Method”. The foundation depth has been kept 2m. Isolated

footings have been provided. The complex meets both design and aesthetic standards.

BIBLIOGRAPHY

1. Jain, A.K., “Design of Reinforced Concrete”, Nem Chand and Bros

Publishers, Roorkee, 2007, Sixth reprint.

2. Krishna Raju, N., “Advanced Reinforce Concrete Design”, CBS, Publishers &

Distributors, Delhi, 1988, First reprint.

3. Pillai, S.U., Menon, D., “Reinforced Concrete Design”, Tata McGraw-Hill

Publishing company Ltd., New Delhi, 2007, Eight Reprint.

102

4. Punmia, B.C., Jain, A.K., Jain, A.K., “R.C.C. Designs (Reinforced Concrete

Structures)”, Laxmi Publication Pvt. Ltd., New Delhi, 2006, Third Reprint.

5. Reddy, C.S., “Basic Structural Analysis”, Tata McGraw Hill Publishing

Company Ltd., New Delhi, 2006, Fifteenth reprint.

6. IS 456:2000 Code of Practice Plain and Reinforced Concrete.

7. IS: 875 (Part 2) - 1987 Code of Practice for Design Loads (Other than

Earthquake) for Building and Structures.

103

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