58147987 Design of Shopping Complex

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ABSTRACT

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 7! m has been designed. The limit state

method of analysis has been used to design the shopping complex. The design of

building is done using "S# $%&'!!!.

1



TABLE OF CONTENTS

Particulars Page No.

(ertificate

Acknowledgements

Abstract

Table of contents

Abbreviations

)ist of figures

)ist of tables

Chapter 1: General Consieration 1!"

* "ntroduction

+euirement

- rganisation of pro/ect work

Chapter ": Theor#$ Bac%groun an &ethoolog#

$ )imit State 0esign.*.* (haracteristic load

% 1orking Stress 2ethod

& 0esign (omponents of Shopping (omplex

* 3oundation

(olumn

- 4eam

$ Slab

% Staircase

& 1ater tank

7 Shell

Chapter ': Anal#sis$ (esign an Results

7 0esign of Slab 56round, 3irst and Top floor

8 0esign of 4eams 5Top floor

* 0esign of 4eam 9A4:

0esign of 4eam 9A;:

- 0esign of 4eam 9;2:

$ 0esign of 4eam 942:

< 0esign of (olumns 5Top floor

* 0esign of (olumn 0*

0esign of (olumn (*

2



- 0esign of (olumn (

$ 0esign of (olumn 0

*! 0esign of 4eams 53irst and 6round floor

* 0esign of 4eam 9A4:

0esign of 4eam 9A;:

- 0esign of 4eam 9;2:$ 0esign of 4eam 942:

** 0esign of (olumns 53irst floor


0esign of (olumn (*


$ 0esign of (olumn 0

* 0esign of (olumns 56round floor


0esign of (olumn (*


$ 0esign of (olumn 0*- 0esign of 3ootings

* 0esign of 3ooting for (olumn 0*

0esign of footing for (olumn (*

- 0esign of footing for (olumn (

$ 0esign of footing for (olumn 0

*$ 0esign of Staircase

*% 0esign of Tank

* 0esign of Slab for Tank

0esign of 4eam for Tank

*& 0esign of Shell

(hapter $# (onclusions

4ibliography

L)ST OF TABLES

Ta*les Page No

Table .* )imiting value of depth of neutral axis

Table . =alue of moment coefficientsTable .- =alue of bending moment coefficient

Table .$ =alue of shear force coefficients

Table -.* +esult of slab

Table -. +esult of beam 9A4: 5Top floor

Table -.- +esult of beam 9A;: 5Top floor

Table -.$ +esult of beam 9;2: 5Top floor

Table -.% +esult of beam 942: 5Top floor

Table -.& +esult of beam 9A4: 53irst and 6round floor

Table -.7 +esult of beam 9A;: 53irst and 6round floor

Table -.8 +esult of beam 942: 53irst and 6round floor

Table -.< 3inal moment calculation

3



L)ST OF F)G+RES

Figures Page No

3igure -.* 0ifferent beams

3igure -. 6rid of columns

3igure -.- 0istribution of pressure below the retaining wall

4



ABBRE,)AT)ONS

0) > 0ead load

)) > )ive load

wd > 0ead load intensity

w l > )ive load intensity

d'

> 0epth of compression reinforcement from the highly compressed face

Ast > Area of steel.

Asv > Total (ross'sectional area of stirrup legs or bent up bars within distance Sv .

A? > Area of cross'section of one bar.

b > 1idth of beam or shorter dimension of a rectangular column

0 > Total 0epth.

d > ;ffective depth.

e > ;ccentricity

f > Stress

f ck > (haracteristic compressive stress.

f y> (haracteristic strength of steel

f s > (haracteristic strength of steel

@ > Stiffness of member

" > 2ovement of inertia..

) > )ength.

)d > 0evelopment length.

2 > 2oment.

5



m > 2odular ratio.

S > Spacing of bars.

=u > Shear force due to design load 5limit state design.

=us > Strength of shear reinforcement 5limit state design.

1 > oint loadB Total load.

xu > 0epth of neutral axis.

C > Dnit weight of soil.

f cbc > ermissible stress in concrete due to bending.

f st > ermissible tensile stress in reinforcement.

? > 0iameter of bar, angle of repose

E bd > 0esign bond stress.

Ec > Shear stress in concrete.

Ecmax > 2aximum shear stress in concrete with shear reinforcement.

Ev > Fominal shear stress.

M x >3actored moment along G'axis

M y > 3actored moment along 'axis

M ux >2aximum moment capacity for bending along x'axis only at axial load p

M uy >2aximum 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.

2e > 4ending moment at the end of the beam framing in to the column assumingfixity at the connection.

6



2es > 2aximum 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

k) > Stiffness factor of lower column

kb* > Stiffness of beam on one of side of the column

kb > Stiffness of the beam on other side of column

M x , M y > 2aximum moments at mid span on strips of unit width and spans

l x ,l y B respectively.

l x > )ength of shorter side

l y > )ength of longer side

α x , α y> 2oment coefficients.

7



C-APTER!1

GENERAL CONS)(ERAT)ON

1.1 )NTRO(+CT)ON



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 I a modern, indoor version of the traditional

marketplace.

2odern Jcar'friendlyJ shopping malls corresponded with the rise of suburban living

in many parts of the 1orld. 3rom 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 JanchorJ or Jbig boxJ

8



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 7!m has been designed in the 1est campus of

H4T" @anpur. The limit state method of analysis has been used to design the

shopping complex. The design of building is done using "S# $%&'!!!.

1." RE+)RE&ENT

The 1est campus of H4T" houses about *!!! students and faculty members. To cater

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

design a shopping complex at a /udiciously 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 eually competitive pricing. Since the food Kone would be under

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

served.

1.' ORGAN)SAT)ON OF PRO/ECT 0OR

ur pro/ect report comprises of seven chapters.

(hapter *' "t involves brief introduction regarding the pro/ect

(hapter ' Theory, background and methodology has been discussed in this chapter.

(hapter -' This chapter includes the analysis, design and results.

(hapter $' 3inally (onclusions have been mentioned in this chapter.

9



C-APTER "

T-EOR2$ BACGRO+N( AN( &ET-O(OLOG2

".1 L)&)T STATE (ES)GN

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

safety and serviceability reuirement before failure is called the limit state ."n 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

reuirements 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 ensureadeuate degree of safety and serviceability. To ensure the above ob/ective the design

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

".1.1Characteristic loa: The term Lcharacteristic loadM means that value of load

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

"." 0OR)NG STRESS (ES)GN

10



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 sub/ected to the worst combination of working loads.

The working stress method is based on the following assumption#

*# A section which is plane before bending remains plane after bending.

# 4ond between steel and concrete is perfect within the elastic limit of steel.

-# The tensile strength of concrete is ignored.

$# (oncrete is elastic.

%# The modular ratio m has the value

280

3σ cbc

¿ where

σ cbc is the permissible

compressive stress in bending.

".' (ES)GN CO&PONENTS OF S-OPP)NG CO&PLE3

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

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

".'.1 Founation: 3oundations are structural elements that transfer loads from the

building or individual column to the earth. "f these loads are to be properly

transmitted, foundation must be designed to prevent excessive settlement or rotation,to minimiKe differential settlement and to provide adeuate safety against sliding and

overturning. 2ost foundations may be classified as follows#

5a "solated footing under individual columns. They may be suare, rectangular or

circular in plan.

5b Strip foundations and wall footings.

5c (ombined footing supporting two or more column loads. These may be

rectangular or trapeKoidal in plan or they may be isolated bases /oined by a

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

11



5d +aft 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.

5e "n 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,

5a Soil strata,

5b 4earing capacity and standard penetration test value F of soil.

5c Type of structure,

5d Type of loads,

5e ermissible differential settlement, and5f ;conomy.

The siKe 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. "n general, foundations have to resist vertical

load, horiKontal load and moments. 3or the purpose of design of foundation, soil

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

of internal friction etc of soil are reuired.

Two terms are used for bearing capacity# 6ross bearing capacity and net bearing

capacity. 6ross 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. Fet 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. 0epth of

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

0epth of foundation is governed by the following factors#

5a To secure safe bearing capacity,

5b To penetrate below the Kone where seasonal weather changes are likely to

cause significant movement due to swelling shrinkage of soils,

5c To penetrate below the Kone that may be affected by frost.

12



"S# *!8!'*<& reuires that in all soils a minimum depth of %!cm 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 +ankine formula, that is,

h > pγ [1−sinф1+sinф

]2

1here,

h> minimum depth of foundation

> gross bearing capacity

> density of soil

O> angle of repose of soil.

".'." Colu4n: 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 siKe 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#

5a Shape of the cross'section,

5b Slenderness ratio,

5c Type of loading,

5d attern 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

*. "ts design is based on the strength of the material and the applied loads. A long

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

of a column should not exceed &!. 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#

13



5a Axially loaded column,

5b A column sub/ected to axial load and uni'axial bending, and

5c A column sub/ected 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,

5a Tied column, and

5b Spiral column.

5c "n 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.

(lause %.$ of the code reuires that the minimum eccentricity should be as

follows#

emin ≥ l500

+ D300

¿20

1here,

)> Dnsupported length

0> )ateral dimension of column in the direction under consideration in mm.

2oreover 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. 6uidelines for longitudinal

reinforcement are#

5a The minimum area of cross'section of longitudinal bars must be at least !.8N

of the gross'sectional area of column.

5b The maximum area of cross'section of longitudinal bars must not exceed &N

of the gross'sectional area of the concrete.

5c The bars should not be less than *mm in diameter so that it is sufficiently to

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

14



5d The minimum number of longitudinal bars provided in a column must be four

in rectangular columns and six in circular bars.

5e Spacing of the longitudinal bars measured along the periphery of a column

should not exceed -!!mm.

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 *-%

° . The ends of the transverse reinforcement should be properly anchored. 1hile

providing lateral ties following points are worth note#

5a 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 &mm.5b The pitch of the lateral ties should not exceed following distances#

• The least lateral dimension of the compression member,

• Sixteen times the smaller diameter of the longitudinal

reinforcement bar to be tied, and

• -!!mm.

The general non dimensional euation for the load contour at constant 9: can be

expressed in the form#

5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P R*

1here,

M x >3actored moment along G'axis

M y > 3actored moment along 'axis

M ux >2aximum moment capacity for bending along x'axis only at axial load p

M uy >2aximum 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.

15



As per "S# $%&' !!!, 3rame Analysis method can be used to analyKe and design the

columns which are sub/ected to axial loading and bi'axial bending. The guidelines can

be summed up as follows#

External columns:

2oment for frames of one bay

2oment at foot of upper column > 2e 5ku

ku+kL+0.5 kb

2oment at head of lower column > 2e5kL

ku+kL+0.5 kb

2oment for frames of two or more bay

2oment at foot of upper column > 2e 5ku

ku+kL+kb

2oment at head of lower column > 2e 5kL

ku+kL+kb

Internal columns:

2oments for frame of two or more bays

2oment at foot of upper column > 2es 5ku

kL+ku+kb1+kb2

2oment at head of lower column > 2es 5kL

kL+ku+kb1+kb2

1here,

2e > 4ending moment at the end of the beam framing in to the column assuming

fixity at the connection.

16



2es > 2aximum 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

k) > Stiffness factor of lower column

kb* > Stiffness of beam on one of side of the column

kb > Stiffness of the beam on other side of column

".'.'!Bea4: 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 +(( beams#

5a Singly reinforced beams

5b 0oubly reinforced beams

5c Singly or doubly reinforced flanged beams

1hile designing the beams, following important rules must be kept in mind#

5a ;ffective 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.

• "n case of continuous beams, if the width of the supports is less than

** of the clear span, the effective span shall be taken as above. "f the

supports are wider than ** of the clear span or &!!mm whichever is

less, the effective span shall be taken as under#

i. 3or 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. 3or the end span with one end free and the other continuous,

the effective shall be eual 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.

17



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

span shall always be the distance between the centre of

bearings.

5b 3ollowing specifications can be used for providing reinforcement in beam#

• 2inimum area of tension reinforcement shall not be less than that

given by the following# A st

bd =

0.85

f y .

• 2aximum area of tension reinforcement shall not exceed !.!$b0.

5c 2aximum area of compression reinforcement shall not exceed !.!$b0

compression reinforcement in beams shall be enclosed by the stirrups for

effective lateral restraint.

5d 2aximum spacing of shear reinforcement measured along the axis of the

member shall not exceed !.7%d for vertical stirrups and d for inclined stirrups

at $% ° . in no case shall the spacing exceed -!!mm.

5e 2inimum shear reinforcement in the form of stirrups shall be provided such

that As

b !

= 0.4

0.87 f y

1here,

As > Total cross'sectional area of stirrup legs effective in shear

! >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 ad/acent table#

Ta*le ".1! Li4iting 5alue o6 epth o6 neutral a7is

f y xum"x

d

%! !.%-

$*% !.$8

%!! !.$&

18



".'.8 Sla*# 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#

5a ne 'way slab spanning in one direction,

5b Two'way slab spanning in both direction,

5c (ircular slabs,

5d 3lat slabs resting directly on columns with no beams, and

5e 6rid floor and ribbed slab.

ne 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

analyKed in a manner similar to that for a continuous beam. 1hen 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. "n a suare

slab, the two way action is eual in each direction. "n long narrow slabs where length

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 wl x2

1here,

M x , M y > 2aximum moments at mid span on strips of unit width and spans

l x ,l y B respectively.

19



l x > )ength of shorter side

l y > )ength of longer side

α x , α y > 2oment coefficients.

Table showing the values of moment coefficients#

Ta*le "."! ,alue o6 4o4ent coe66icients

l y

l x

*.! *.* *. *.- *.$ *.% *.7% .!

α x !.!& !.!7$ !.!8$ !.!<- !.!<< !.*!$ !.**- !.**8

α y !.!& !.!&* !.!%< !.!%% !.!%* !.!$& !.!-7 !.!<

"f 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 ;lastic theory analysis' idealiKation into strips or beams,

b Semi empirical coefficients as given in the code, and

c ield line theory.

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

Ta*le ".'! ,alue o6 *ening 4o4ent coe66icients

Type of load Fear middle of

end span

At middle of

interior span

At support next

to end support

At other interior

support

0ead load and

imposed

load5fixed

+112

+116

−1

10

−1

12

"mposed load

5not fixed

+110

+112

−1

9

−1

9

20



5a 1alls or beams spanning transversely to the flight, or

5b A landing slab spanning transversely to the flight, or

5c A landing slab spanning along the direction of the flight

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

5a 1here supported at bottom risers by beam or wall, the centre to centre

distance of beamsB

5b 1here spanning on to the edge of a landing slab which spans transversely to

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

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

5c 1here 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.

".'. 0ater tan%: A water tank is used to store water to tide over the daily

reuirements in general, water tanks can be classified under three heads#

5a Tanks resting on the ground,

5b ;levated tanks supported on staging, and

5c Dnderground tanks.

3rom shape point of view, water tanks may be of several types, such as#

5a (ircular tanks,

5b +ectangular tanks,

5c Spherical tanks,

5d "ntKe tanks, and

5e (ircular tanks with conical bottoms.

"n the construction of concrete structures for the storage of water and other liuids, the

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

less than --!kgm- of concrete. "t should also be less than %-!kgm- of concrete to

keep the shrinkage low. "t is usual to use rich mix like 2 -! grade in most of water

tanks.

1hen 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 /oint at the /unction of wall and bottom

slab. "f the /oint is flexible it will be free to move outward in a different position.

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

22



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

maximum increase in diameter there hence maximum movement there if /oint is

flexible. 1hen /oint is flexible, Hoop tension is developed everywhere in the wall.

2aximum Hoop tension at the bottom, per unit height of wall> wH D2

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

height at the base is given by,

s#=¿ w$D

2 f s

A¿

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 %mm if the thickness is

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

developed in the composite section is within safe limits. "ff ct is the permissible

tensile stress in the euivalent concrete section, and T is the thickness of the wall, wehave

f ct = w$D /2 $$

1000% +(m−1) As#

3or tanks of smaller capacity, the cost of shuttering for circular tank becomes high.

+ectangular tanks are therefore used in such circumstances. However, rectangular

tanks are not normally used in for large capacities since they are uneconomical andalso its exact analysis is difficult. "n such a tank wall of the tank is sub/ected 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 2! concrete may be taken as *.Fmm and *.7Fmm respectively for direct

tension and due to bending. 3or the design by approximate method, rectangular tanks

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

than and tanks in which ratio is more than .

23



5a "f the ratio )4 is less than , the tank walls are designed as continuous frame

sub/ected to a triangular load. Such a bending is assumed to take place from

top to a height h>H$ or *m above base whichever is more. 3or the bottom

height h, bending is assumed to act in vertical plane as cantilever, sub/ected to

triangular load, having Kero intensity at h and wH at base. 3or horiKontal

bending, the maximum horiKontal force per unit height is taken eual to p>

w5H'h per metre run. The value of bending moment and direct tension is

calculated as below#

4ending moment at centre of span> p &2

16 5producing tension on outer face

4ending moment at ends of span> p &

2

12 5producing tension on water face

0irect tension on long wall> w 5H'h 4

0irect tension on short wall> w 5H'h ).

5b "f ratio )4 is greater than , the long walls are assumed to bend vertically as

cantilever fixed at base and sub/ected to triangularly distributed load. The

short wall is assumed to bend horiKontally, 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 > w5H'h. The bottom portion of height h of the short wall is designed as

cantilever sub/ect to triangular load. Thus#

3or long wall, bending moment at the base, per unit length of the wall > wH-&

3or short wall, the maximum bending moment at ends and centre

> w5H'h4*&

2aximum cantilever bending moment> wHh&

The direct tension on long wall > w5H'h4

".'.; Shell: "t 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

24



construction. The use of curved space roofs reuires % N to $! N 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 reuired. Shells are broadly classified into two ma/or groups as,

5a Singly curved shells which are developable

5b 0oubly curved shells which are non developable

Dnder 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,

5a Shells of revolution

5b Shells of translation

5c +uled 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 !. "n general, most

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

guidelines followed for selecting the dimensions of the various structural components

of the shells are detailed below#

5a The overall thickness of a reinforced concrete shell should not be less than

%!mm for singly curved shells, $!mm for doubly curved shells and %mm for

precast shells. 6enerally the thickness is in the range of 8!mm to *!mm for

most of the shells based on practical considerations.

5b The span of reinforced concrete shells should not be greater than -!m to limit

the siKe and reinforcement within practicable limits in the edge beams. 3or

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

member is limited to to - times the thickness of the shell.

5c 3or large span shells, depth> *& to ** span, larger figures are applicable to

small spans. 3or shells without edge members, depth span*!. 3or shells

with chord width much larger than the span, depth U **! chord width.

5d The semi central angle should be in the range of -! to $% degrees. "f the angle

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

5e The diameter of reinforcement should not exceed *!mm for %!mm thick shellsand *mm for &%mm thick shells and *&mm for shells having thickness

25



greater than &%mm. "n the /unction Kones 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. 2inimum clear cover must be *mm or

the nominal siKe of reinforcing bars.

5f "n case of shells of long lengths exceeding $!m, expansion /oints have to be

provided. The construction /oints 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 4eam theory. "n the beam theory the shell is analyKed as a bean of

curved cross'section spanning between the end frames or transverses. "n 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 )+ ratio exceeding the value of V.

C-APTER!'

ANAL2S)S$ (ES)GN AN( RES+LTS

'.1 (ES)GN OF SLAB <TOP$ F)RST AN( GRO+N( FLOOR=

SiKe of slab> -.--mW8m

Assuming modification factor *.

sp"neffectiedept# > &W*.

;ffective depth > *!&.7mm

Taking total depth 0 > *$!mm

Hence effective depth d>**m

0ead load due to slab > %W!.*$

26



>-.%kFm

Surface finish load > *.%kFm

Hence dead load intensity,w d

> %kF m

)ive load intensity,w l > $kF m

l x > -.-- Q !.*8

> -.$%8 m

l y > 8 Q !.*8

> 8.*8 m

l y

l x > .- U

Hence one way slab.

Calculation of Bending Moment:

4ending moment near middle of end span

M 1 >

wd l2

12 Qwl l2

10 > <.7&-kF'm m run

4ending moment at support next to end support

&11 >

−wd l2

10 −w l l

2

9

> ' <.7&-kF'm m run

4ending moment at middle of interior span

27



M 2 >

wd l2

16 Qwl l2

12

> 7.7&kF'm m run

4ending moment at other interior support

&12 >

−wd l2

12 −w ll

2

9

> *!.<kF'm m run

2aximum bending moment > *!.<kF'm m run

!.*-8 f ck bd > *!.< W *.%W *!&

d > 7$.7mm *8mm ok.

Reinforcement for M

1 :

3actored 42 > *.%W<.7&-

>*$.&$kF'm m run

+einforcement Ast > --% mm

Spacing >1000(50.24

335 >*$<.<mm

Hence providing 8X Y *%!mm cc.

Actual reinforcement provided >1000(50.24

150 >--$.< mm

Hence N tension reinforcement >100(334.9

1000(121 > !.7&N

Reinforcement for &11 :

28



3actored 42 > *<.&kF'm m run

+einforcement Ast > -<* mm

Spacing >1000(50.24

391 >*-!mm

Hence providing 8X Y *-!mm cc.

Reinforcement for M

2 :

3actored 42 >**.%8<kF'm m run

+einforcement Ast > & mm

Spacing >

1000(50.24

262 > *<!mm

Hence providing 8X Y *<!mm cc.

Reinforcement for &12 :

3actored 42 > *%.$-%kF'm m run

+einforcement Ast > -%% mm

Spacing >

1000(50.24

355 > *$!mm

Hence providing 8X Y *$!mm cc

Codal specifications:

2aximum spacing > -d > -8$mm

2inimum area of reinforcement > !.* N of 40 >*&8mm

Check for shear force:

29



2aximum shear force occurs at first interior support&

11

) u > !.&W*.%W-.$%8W**.& > -&.*kFm

* >36.1

128 >!.8Fmm

pt >100(509.7

1000(28 > !.-8< N

* c > !.$Fmm, k > *.-

* c'

> *.-W!.$ > !.%$&Fmm

* c' >* , hence k.

Check for development length:

O +

4 * bd

0.87( f ck (

1.3 M 1

) + L0)

M 1 >

0.87( f y ( Ast (d−f y ( Ast

f ck ( b )

>7.-kF'm

Shear force = > !.$ w d+0.45w l

> *.< kF

O + 4(1.6

0.87(415(1.3(7.32

12.92 +300)

+9 mm.

Hence ok.

Check for deflection:

30



ercentage tension reinforcement > !.7&N

2odification factor *.%

d >3.333(1000

23(1.5 > <&.&mm

Hence safe.

A table showing results is as follows#

Ta*le '.1 > Results o6 sla*

0esignation +einforcement 0iameter Spacing5cc

Fear middle of end

span

--%mm 8mm *%!mm

At support next to

end support

-<* mm 8mm *-!mm

At middle of

interior span

& mm 8mm *<!mm

At other interior

support

-%% mm 8mm *$!mm

'." (ES)GN OF BEA& <TOP FLOOR=

'.".1 (esign o6 Bea4 ?AB@ O6 Span 14:

Taking width b > -!!mm

;ffective depth d > &!!mm

Total depth 0 > &%mm

)ive load of full slab > 4(10(8

>-!kF

31



0ead load > (8(10(0.14(25 )+(8(0.625( .300(25)

> -*7.%kF

A *! 4

8

; 2

Fig '.1! (i66erent *ea4s

)ive load for triangular portion >

16(320

80

> &$kF

)ive load for trapeKoidal portion >124(320

80

> <&kF

0ead load for triangular portion >16(317.5

80

> &-.%kF

0ead load for trapeKoidal portion >24 (317.5

80

> <%.%kF

0ead load of A4 >95.25

10 +.3( .625(10(25

10

wd > *$.kF m

)ive load of A4 >96

10

32



w l > <.&kF m

Calculation of bending moment and reinforcement:


M A& >wd l2

12 Q

wl l2

10

> *$.--kF'm

+einforcement Ast > *<!*.7 mm

Hence providing & ' ! X.


M & >wd l2

16 Qwl l2

12

> *&8.7%kF'm

+einforcement Ast > *-<.& mm

Hence providing %'!X


M & >−wd l2

10

−w ll2

9

> −248.66 kF'm

+einforcement Ast > -7!.- mm

Hence providing 8'!X.

Calculation of shear force:

33



> *!7.*kF

* >107100

300(600

> !.%<%Fmm

pt >100(8(314

300(600

> *.-< N

* c > !.&< Fmm

* c>* .

Safe hence providing minimum reinforcement

Dsing 8mm' legged vertical stirrups

As > *!!.% mm

Spacing >

0.87( f y ( As

0.4(b

> -!!mm

roviding 8mm' legged vertical stirrups Y -!!mm cc.


2aximum shear force > 7*.$kF

2aximum 4ending moment > $8.&&kF'm

L0 > &!!mm or * X

>$!mm

35



Ld + 1.3( M

) + L-

47

X +

1.3(248.66(1000

71.4

+240

O + $!mm

Hence ok.

Ta*le '."> Result o6 *ea4 ?AB@

0esignation +einforcement Fumber of bars 0iameter Fear middle of end

span

*<!*.7mm & !mm

At middle of interior

span

*-<.&mm5 !mm

At support next to

end support

-7!.-mm 8 !mm

'."." (esign o6 Bea4 ?AE@ o6 span 4:

0ead load > 0ead load of triangular portion Q dead load of beam

>63.5

10 +.3( .625(8(25

10

wd > *.&- kF m

)ive load >64

8

w l > 8kF m



36



M A >wd l2

12 Q

wl l2

10

> **8.%& kF'm

+einforcement Ast > <*8.& mm

Hence providing %'*&X.


M ./ >wd l

2

16 Qwl l

2

12

> <-.*8kF'm

+einforcement Ast > 7!.$ mm

Hence providing $'*&X.


M >−w d l2

10

−w l l2

9

> '*-7.7*kF'm

+einforcement Ast > *!<*.- mm

Hence providing &'*&X.


Shear force at end support 5A

! A >

0.4(wd (l

2 Q

0.45( wl (l

2

> -*.!* kF

37



Shear force at support next to end support 5;

! >

0.6(wd(l

2 Q

0.6(w l(l

2

> $$.7*kF

Check for shear force at ‘A’:

) u > *.% x -*.!*

> $&.%*% kF

* >46.515(1000

300(600

> !.-7Fmm

pt >100(6(200.96

300(600

> !.&& N

* c > !.%Fmm

* c>* .

Hence safe.

Check for shear force at ‘E’:

) u > *.% x $$.7*

> &7.!&8 kF

* > !.-7 Fmm

pt

> !.&7 N

38



* c > !.% Fmm

* c>* .



As > *!!.% mm

Spacing >

0.87( f y ( As

0.4(b

> -!!mm



2aximum shear force > $$.7* kF

2aximum 4ending moment > *-7.7* kF'm

L0 > &!!mm or * X

>$!mm

Ld + 1.3( M

) + L-

47 O +

1.3(137.71(1000

44.712 +240

O + $!mm. hence ok.

Ta*le '.'! Result o6 *ea4 ?AE@

0esignation +einforcement Fumber of bars 0iameter

Fear middle of end <*8.&mm % *&mm

39



span


span

7!.$mm4 *&mm

At support next to

end support

*!<*mm6 *&mm

'.".' (esign o6 Bea4 ?E&@ o6 span 14:

0ead load > 50ead load of trapeKoid Q dead load of beam

wd > -.7kF m

)ive load > 5)ive load of trapeKoid

w l > *<.kF m

Taking effective depth d > 7%!mm



M M >wd l2

12 Q

wl l2

10

> -8<.%kF'm

+einforcement Ast > <7$.7 mm

Hence providing <'!X.


M M0 >wd l

2

16 Qwl l

2

12

> -!8.*kF'm

+einforcement Ast > *-.- mm

40





M M > −wd l

2

10 −w ll

2

9

> '$%!.-- kF'm

+einforcement Ast > -888 mm

Hence providing *-'!X.


Shear force at end support 5;

! >

0.4(wd (l

2 Q

0.45( wl (l

2

> <!.& kF

Shear force at support next to end support 52

! M >

0.6( wd (l

2 Q

0.6( w l(l

2

> *8.7 kF


) u > *-%.<kF

* >135.9(1000

300(750

> !.&!Fmm

pt >100(9(314

300(750

41



2aximum shear force > *8.7kF

2aximum 4ending moment > $%!.--kF'm

L0

> &!!mm or * X

>$!mm

Ld + 1.3( M

) + L-

47 X +

1.3(450.33(1000

128.7 +240

O + $!mm

Hence ok.

Ta*le '.8 ! Result o6 *ea4 ?E&@


Fear middle of endspan

<7$.7mm

< !mm


span

*-.-mm7 !mm

At support next to

end support

-888mm 13 !mm

'.".8 (esign o6 Bea4 ?B&@ o6 Span 4:

0ead load > 50ead load of triangle Q dead load of beam

wd > *.%kF m

)ive load > 5)ive load of triangle

43



w l > *&kF m

Taking effective depth d > 7%!mm



M M& >wd l2

12 Q

wl l2

10

> *7.!&kF'm

+einforcement Ast > *-78 mm

Hence providing %'!X.


M M1 >wd l2

16 Qwl l2

12

>*<<.<<kF'm

+einforcement Ast > *%- mm

Hence providing $'!X.


M M >−wd l2

10

−w l l2

9

> ' %*.-7 kF'm

+einforcement Ast > *&$* mm

Hence providing &'!X.


44



Shear force at end support 54

!& >

0.4(w d (l

2 Q

0.45( wl (l

2

> &-.kF


! M >

0.6( wd (l

2 Q

0.6( w l (l

2

> <!kF

Check for shear force at ‘B’:

) u > <$.8kF

* >94.8(1000

300(750

> !.$Fmm

pt >100(5(314

300(750

> !.&< N

* c > !.% Fmm,* c>* .

Hence safe.

Check for shear force at ‘M’:

) u > *.% x <! > *-%kF

* > !.& Fmm, pt > !.8$N

* c > !.8 Fmm

45



* >* c .



As > *!!.% mm

Spacing >

0.87( f y ( As

0.4(b

> -!!mm



2aximum shear force > <!kF

2aximum 4ending moment > %*.-7kF'm

L0

> &!!mm or * X > >$!mm

Ld + 1.3( M

) + L-

47 X +

1.3(251.37(1000

90 +240

O+

$%mm. Hence O > !mm is ok.

Ta*le '.9 ! Result o6 *ea4 ?B&@


Fear middle of end

span

*-78mm % !mm


span

*%-mm4 !mm

At support next to *&$*mm 6 !mm

46



end support

****

Fig '."! Gri o6 Colu4ns

'.' (ES)GN OF COL+&NS <TOP FLOOR=

'.'.1 (esign o6 Colu4n (1:

0imension of beam > -!! ( &%

SiKe of column > -!! ( $!!

2oment of inertia of column "c > bd-* > -!! (400 -* > *& (10 8 mm$

2oment of inertia of beam "b > -!! (625 -* > &* (10 8 mm$

47

A3 A4A2

B1 B4B3B2

C4C3C2C1

D4D3D2

D1



Stiffness @ of column at /oint > $ (10 8

Stiffness kb of beam at /oint > &. 1(10 8

@u > !.&8

k) > !.&8

kb > !.$&$

Total dead load of slab Q beam Q intermediate beams in G'direction

> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm

Total live load> $kFm

Total load > 8.<<kFm

)oad intensity w in G'direction > 8.99(24 *!

> *.&kFm

2e > wl*

> *.& ( *!*

> *7%kF'm

2ux > !.&8 (175 kF'm

> $&.<kF'm

Total dead load due to slab Q beams in G'direction

> *! (8(0.14(25+0.625(0.3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm

Total load > 7.<7kFm

)oad intensity in 'direction > *7.<8kFm

2e > <%.8<kF'm

2uy > $.%%kF'm

3rom load distribution axial load u > *7<.8kF

)et assume N reinforcement provided > !.8N

48



pfck >

0.8

20 > !.!$

)et effective cover be %! mm on either side'

d D > 50

400 > !.*%

Dniaxial moment capacity of section

0ufck >

179.8(1.5(1000

20(300(400 > !.**

4y using p'2 interaction as given by S# *&

Mu

fc k . b . D2

> !.!<

2ux* > !.!< (fck(b( D2

>!.!< (20(300(4002

2ux* > 8&.$kF'm [ 2uy* > 8&.$kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *-78.8kFuK > !.*-

3rom graph P > *

The strength of the section can be checked using the interaction

5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 546.9

86.4 *Q 524.55

86.4 *

> !.8 R*

Hence design is safe. @ Thus the assumed value of reinforcement N p is @.

As > !.8N gross'area

> !.!!8 (300(400 > <&! mm

So provide $'*& diameters

Dse 8 mm lateral ties and should be at least of

*-!! mm

49



*& (12 > *<mm

-$8 (8 > -8$

rovide bars at *<! mm cc

'.'." (esign o6 Colu4n C1:




2oment of inertia of beam "b > -!! (625

-* > &* (10

8 mm$



@u > !.&8

k) > !.&8

kb > !.$&


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm



)oad intensity w in G'direction > $-.*%kFm

2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm

Total dead load due to slab Q beams 'direction

> *! (8(0.14(25+0.625(- .3(25( 8

50



> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm

3rom load distribution axial load u > *7<.8kF

)et assume N reinforcement p provided > !.8N

pfck >

0.820 > !.!$


d D >

50

400 > !.*%


0u

fck >

359.6(1.5(1000

20(300(400 > !.


Mu

fc k . b . D2 > !.*

2ux* > !.* (fck(b( D2

>!.* (20(300(4002

2ux* > **%.kF'm [ 2uy* > **%.kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *-78.8kF

uK > !.&

3rom graph P > *.!-7%


51



5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 596.37

115.2 *.!-7Q 532.98

115.2 *.!-7

> !.<- R*

Hence design is safe. @ Thus the assumed value of reinforcement N p is @.

As > !.8N gross'area

> !.!!8 (300(400 > <&! mm

So provide $'*& diameters


*-!! mm

*& (12 > *<mm

-$8 (8 > -8$


'.'.' (esign o6 Colu4n C ":



2oment of inertia of column "c > bd-* > -!! (400 -* > *&(10 8 mm$



Stiffness kb of beam at /oint > &.

1(10 8

@u > !.*!&

k) > !.*!&

@b > !.-<


> *! (

8 ( .*$ (

% Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm

52






2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm


> *! (8(0.14(25+0.625( - .3(25( 8

> -*7.%kF

0ead load> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78

kF'm

> -.<8kF'm

3rom load distribution axial load u > 7*<.kF

)et assume N reinforcement provided > !.8N

pfck >

0.7

20 > *.$N


d D >

50

400 > !.*%


0ufck >

719.5(1.5(1000

20(300(400 > !.$$


Mu

fc k . b . D2 > !.*

53



2ux* > !.* (fck(b( D2

>!.* (20(300(4002

2ux* > <&.-7kF'm [ 2uy* > <&.-7kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *-78.8kF

uK > !.-7

3rom graph P > *.!


5

Mux

Mux1¿ P

Q 5

Muy

Muy1¿ P

> 5

96.37

115.2

*.!-7

Q 5

32.98

115.2

*.!-7

> !.<- R*

Hence design is safe. k

Thus the assumed value of reinforcement N p is k.

As > *.$N gross'area

> !.!*$ (300(400 > *&8! mm

So provide 8'*& diameters


* -!! mm

*& (12 > *<mm

- $8 (8 > -8$


'.'.8 (esign o6 Colu4n (":



2oment of inertia of column "c > bd-* > -!! (400 -* > *&

(10 8 mm$



54




@u > !.*!&

k) > !.*!&

kb > !.-<


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm



)oad intensity w in G'direction > *.&kFm

2es > ( wl*

>(

*.& ( *!*

> -%!kF'm

2ux > -%! (0.106 kF'm

> -7.*kF'm


> *! (8(0.14(25+0.625(- .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm


)oad intensity in 'direction > *7.<8 (2 kFm

> -%.<&kF'm

2es > *<*.78kF'm

2uy > !.%& (191.78 kF'm

> $<.!<kF'm

3rom load distribution axial load u > -%<.&kF

)et assume N reinforcement p provided > !.$N

55



pfck >

0.4

20 > !.!


d D > 50

400 > !.*%


0ufck >

359.6(1.5(1000

20(300(400 > !.


Mu

fc k . b . D2

> !.!%

2ux* > !.!% (fck(b( D2

>!.!% (20(300(4002

2ux* > $<.<kF'm [ 2uy* > **%.kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *-78.8kFuK > !.&

3rom graph P > *.!-7%


5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 596.37

115.2 *.!-7Q 532.98

115.2 *.!-7

> !.<- R*

Hence design is safe. k Thus the assumed value of reinforcement N p is k.

As > !.$N gross'area

> !.!!$ (300(400 > $8! mm

So provide $'*&diameter


*-!! mm

56



*& (12 > *<mm

-$8 (8 > -8$


'.8 (ES)GN OF BEA& <F)RST GRO+N( FLOOR=

'.8.1 Bea4 ?AB@ o6 span 14

Taking width b > -!!mm

;ffective depth d > 7%!mm

Assuming thickness > *-!mm

Height of wall > $!!!' *$!

> -8&!mm

0ead load > dead load of trapeKoid Q dead load of beam Q self weight of wall

wd

> <%.% Q %W*!W!.-W!.7% Q *!W-.8&W!.*-W!

> %.kFm

)ive load > live load of trapeKoid

w l > <.&kFm



M A& >wd l2

12 Q

wl l2

10 > -!& kF'm

+einforcement Ast > *!$ mm


57




M & >wd l2

16 Q

wl l2

12

> -7.% kF'm

+einforcement Ast > *%-- mm



M & >−wd l

2

10

−w ll2

9

> '-%8.& kF'm

+einforcement Ast > &* mm

Hence providing <'!X.



! A > 7kF


!& > *!$.$kF


) u > *!8kF

* > !.$8Fmm

pt > !.<7 N

58



* c > !.& Fmm

* c>* .

Hence safe.


) u > *%&.&

* > !.&<& Fmm

pt > *.&N

* c > !.&8 Fmm

* >* c .

Hence shear reinforcement needed to be provided

) us > -&!! kF


As > *!!.% mm

Spacing > -!!mm



Ld + 1.3( M

) + L-

47

X

+ 1.3(358.66(1000

104.4

+240

59



O + $%mm. hence k.

Ta*le '. ! Results o6 *ea4 ?AB@

0esignation +einforcement Fumber of bars 0iameter Fear middle of end

span

*!$mm 7 !mm


span

*%--mm5 !mm

At support next to

end support

&*mm 9 !mm

'.8." Bea4 ?AE@ o6 span 4

0ead load > dead load of triangle Q dead load of beam Q self weight of wall

w d > -.%kFm

)ive load > live load of triangle

w l > 8kFm



M A >wd l

2

12 Qwl l

2

10 >*7&.%-kF'm

+einforcement Ast > *!87 mm

Hence providing $'!X.


M ./ >wd l2

16 Q

wl l2

12>*-&.&7kF'm

+einforcement Ast > 8*< mm

60



Hence providing -'!X.


M > −w d l

2

10 −w l l

2

9 > !7.-kF'm

+einforcement Ast > *-!& mm




! A >

0.4(wd (l

2 Q

0.45( wl (l

2

> %*.&kF

Shear force at support next to end support 5;

! >

0.6(wd(l

2 Q

0.6(w l(l

2 > 7%.&kF


) u > 77.$kF,* > !.-$Fmm

pt > !.%%N,* c > !.% Fmm

* c>* , hence safe.


) u > **-.$kF,* > !.% Fmm

pt > !.&<N,

* c > !.% Fmm

61



* c>* .

Hence minimum reinforcement reuired

Dsing 8mm' legged vertical stirrups, A s > *!!.% mm

Spacing >-!!mm



Ld +

1.3( M

) + L-

47 X +3804.68

O + $%mm, hence k.

Ta*le '.; ! Results o6 *ea4 ?AE@


Fear middle of end

span

*!87mm $ !mm


span

8*<mm3 !mm

At support next to

end support

*-!&mm5 !mm

'.8.' Bea4 ?E&@ o6 span 14

62



0ead load > 5dead load of trapeKoid Q dead load of beam Q self weight of wall

wd >-$.7kFm

)ive load > live load of trapeKoid

w l >*<.kFm



M M >

wd l2

12 Q

wl l2

10

> $8*.*&kF'm


M M0 >wd l2

16 Qwl l2

12

> -7&.8kF'm


M M >−wd l2

10

−w l l2

9 D !%&!.--kF'm

M lim ¿0.138 f

ck

bd2

.

>!.*-8W!W-!!W7%! > $&%.7%kF'm

lim ¿ M ̶ M ¿ > <$.%kF'm

xm > !.$8d

Area of tension steel corresponding to

63



!.87 f y At 1 > !.-&

f ck b xm

A t 1 > *%-mm

Takingd '

d >!.*

lim ¿ M ̶ M ¿ > ( f !, ( A!, −f cc(A sc )(d−d ' )

A!, > $!7.mm

(orresponding tension steel#

!.87 f y At 2 >

f ! ( A!

A t 2 > -<8.*$mm

A t-t"ltensi-n > A t 1+ At 2

> %%*$.*$mm.

2aximum area in tension > !.!$b0 > <-!!mm

A t-t"ltensi-n< Am"x , hence ok.

'.8.8 (esign o6 Bea4 ?B&@ o6 span 4

0ead load > 5dead load of triangle Q dead load of beam Q self weight of wall

wd > -*.%kFm

)ive load > live load of triangle

w l > *&kFm

64



! M >

0.6( wd (l

2 Q

0.6( w l (l

2 > **$kF


) u > **8.8kF,* > !.%-Fmm

pt > !.8-N,* c > !.%8 Fmm

* c>* , hence safe.

Check for shear force at ‘M’:

) u > *7*kF,* > !.7& Fmm

pt > !.<8N,* c > !.&* Fmm

* >* c .

) us > --7%!kF

Dsing 8mm' legged vertical stirrups, A s > *!!.% mm

Spacing >

0.87(f y( A s(d

) us > -!!mm



Ld + 1.3( M

) + L-

47 X +3596.32+240

O + $%mm, hence ok.

66



Ta*le '. ! Results o6 *ea4 ?B&@


Fear middle of end

span

*7<&mm & !mm


span

*--%.7mm5 !mm

At support next to

end support

*<!mm7 !mm

'.9 (ES)GN OF COL+&NS <F)RST FLOOR=

'.9.1(esign o6 Colu4n (1:







@u > !.&8

k) > !.&8

kb > !.$&$


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm



)oad intensity w in G'direction > 8.99(24 *!

> *.&kFm

2e > wl*

> *.& ( *!*

67



> *7%kF'm

2ux > !.&8 (175 kF'm

> $&.<kF'm


> *! (8(0.14(25+0.625(- .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2e > <%.8<kF'm

2uy > $.%%kF'm

3rom load distribution axial load u > -%<.!kF

)et assume N reinforcement p provided > N

pfck >

2

20 > !.*


d D >

50

400 > !.*%


0ufck >

359.6(1.5(1000

20(300(400 > !.


Mu

fc k . b . D2 > !.*8

2ux* > !.*8 (fck(b( D2

>!.*8 (20(300(4002

2ux* > *7.8kF'm [ 2uy* > *7.8kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *8kF

68



uK > !.*&

3rom graph P > !.<


5 Mux

Mux1 ¿ P Q 5 Muy

Muy 1 ¿ P > 546.9

172.8 !.<Q 524.55

172.8 !.<

> !.< R*

Hence design is safe. @

Thus the assumed value of reinforcement N p is @.

As > !.N gross'area

> !.! (300(400 > $!! mm

So provide 8'! diameters


*-!! mm

*& (12 > *<mm

-$8 (8 > -8$


'.9." (esign o6 Colu4nC1:

0imension of beam > -!! (

&%


2oment of inertia of column "c > bd-* > -!! (400 -* > *&(10 8 mm$

2oment of inertia of beam "b > -!! (625 -

* > &* (10 8

mm$



@u > !.&8

k) > !.&8

kb > !.$&


69



> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm




2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm


> *! (8(0.14(25+0.625(- .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm



pfck >

2

20 > !.*


d D >

50

400 > !.*%


0ufck >

719.2(1.5(1000

20(300(400 > !.$%

70




Mu

fc k . b . D2 > !.*$

2ux* > !.*$ (fck(b( D2

>!.*$ (20(300(4002

2ux* > *-$.$kF'm [ 2uy* > *-$.$kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *-78.8kF

uK > !.%<

3rom graph P > *.$


5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 596.37

134.4 *.$Q 532.98

134.4 *.$

> !.7& R*



As > !.N gross'area

> !.! (300(400 > $!! mm

So provide 8'! diameters


*-!! mm

*& (12 > *<mm

-$8 (8 > -8$


'.9.' (esign o6 Colu4n C":



71





Stiffness @ of column at /oint > $

(10 8


@u > !.*!&

k) > !.*!&

kb > !.-<


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87kF

> $.<<kFm




2es > wl*

> $-.*%

(*!

*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm


> *! (8(0.14(25+0.625(- .3(25( 8

> -*7.%kF> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm

72



3rom load distribution axial load u > *$-8.$kF

)et assume N reinforcement p provided > $N

pfck >

4

20 > !.N


d D >

50

400 > !.*%


0ufck >

1438.4(1.5(1000

20(300(400 > !.8<


Mu

fc k . b . D2 > !.!<

2ux* > !.!< (fck(b( D2

>!.!< (20(300(4002

2ux* > 8&.$kF'm [ 2uy* > 8&.$kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> %7$kF

uK > !.8

3rom graph P > *.8


5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 596.37

86.4 *.8Q 532.98

86.4 *.8

> !.< R*



As > $N gross'area

> !.!$ (300(400 > $8!! mm

So provide 8'% diameters


73



* -!! mm

*& (12 > *<mm

-$8 (8 > -8$


'.9.8 (esign o6 Colu4n (":







@u > !.&8

k) > !.&8

kb > !.$&


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm




2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm


74



> *! (8(0.14(25+0.625( - .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm



pfck >

2

20 > !.*


d D >

50

400 > !.*%


0ufck >

719.2(1.5(1000

20(300(400 > !.$%


Mu

fc k . b . D2 > !.*$

2ux* > !.*$ (fck(b( D2

>!.*$ (20(300(4002

2ux* > *-$.$kF'm [ 2uy* > *-$.$kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> *-78.8kF

uK > !.%<

3rom graph P > *.$


75



> $.<<kFm



)oad intensity w in G'direction > 8.99(24

*!

> *.&kFm

2e > wl*

> *.& ( *!*

> *7%kF'm

2ux > !.&8 (175 kF'm

> $&.<kF'mTotal dead load due to slab Q beams 'direction

> *! (8(0.14(25+0.625(- .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2e > <%.8<kF'm

2uy > $.%%kF'm

3rom load distribution axial load u > %-<.$kF

)et assume N reinforcement p provided > -N

pfck >

3

20 > !.*%


d D >

50

400 > !.*%


0ufck >

539.4(1.5(1000

20(300(400 > !.--


77



Mu

fc k . b . D2 > !.!7

2ux* > !.!7 (fck(b( D2

>!.!7 (20(300(4002

2ux* > &7.kF'm [ 2uy* > &7.kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> !!kF

uK > !.-&

3rom graph P > *.*8


5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 546.9

67.2 *.*8Q 524.55

67.2 *.*8

> !.< R*



As > -N gross'area

> !.!- (300(400 > -&!! mm

So provide *'! diameters


*-!! mm

*& (12 > *<mm

-$8(8

> -8$


'.." (esign o6 Colu4n C1:




78





Stiffness kb of beam at /oint > &.

1(10 8

@u > !.&8

k) > !.&8

kb > !.$&


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm




2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm


> *! (8(0.14(25+0.625( - .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm

3rom load distribution axial load u > *!78.8kF


79



*& (12 > *<mm

-$8 (8 > -8$


'..' (esign o6 Colu4n C":







@u > !.&8

k) > !.&8

kb > !.$&


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm




2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm


> *! (8(0.14(25+0.625(- .3(25( 8

> -*7.%kF

81



> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm

3rom load distribution axial load u > *!78.8kF


pfck >

3

20 > !.*%


d D >

50

400 > !.*%


0ufck >

1078.8(1.5(1000

20(300(400 > !.&7


Mu

fc k . b . D2 > !.*$

2ux* > !.*$ (fck(b( D2

>!.*$ (20(300(4002

2ux* > *-$.$kF'm [ 2uy* > *-$.$kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> !%kF

uK > !.7-

3rom graph P > *.&


5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 596.37

134.4 *.&Q 532.98

134.4 *.&

82



> !.&8R*



As > -N gross'area

> !.!- (300(400 > -&!! mm

So provide *'! diameters


*-!! mm

*& (12 > *<mm

-$8 (8 > -8$

rovide bars at *<! mm cc.

'..8 (esign o6 Colu4n (":







@u > !.*!&

k) > !.*!&kb > !.-<


> *! ( 8 ( .*$ ( % Q !.&% (25(10+2(.3(0.6(8(2 %

> -<8.87%kF

> $.<<kFm



83




2es > wl*

> $-.*% ( *!*

> -%<.&kF'm

2ux > -%<.& (0.268 kF'm

> <&.-7kF'm

Total dead load due to slab Q beam 'direction

> *! (8(0.14(25+0.625( - .3(25( 8

> -*7.%kF

> -.<&8kFm

)ive load > $kFm



2es > *<*.78kF'm

2uy > !.*7 (191.78 kF'm

> -.<8kF'm

3rom load distribution axial load u > *%7.&kF

)et assume N reinforcement p provided > $N

pfck >

4

20 > !.N


d D >

50

400 > !.*%


0ufck >

2157.6(1.5(1000

20(300(400 > !.8<


Mu

fc k . b . D2 > !.!8

2ux* > !.!8 (fck(b( D

2

84



>!.!8 (20(300(4002

2ux* > 7&.8kF'm [ 2uy* > 7&.8kF'm

K > !.$% (fck( Ac Q !.7%

( f y ( Asc

> %7$kF

uK > !.8

3rom graph P > *.8


5 Mux

Mux1¿ P Q 5

Muy Muy 1

¿ P > 596.37

76.8 *.8Q 532.98

76.8 *.8

> !.< R*Hence design is safe. @


As > $N gross'area

> !.!$ (300(400 > $8!! mm

So provide *&'! diameters


* -!! mm

*& (12 > *<mm

- $8 (8 > -8$


'.; (ES)GN OF FOOT)NG

'.;.1 For Colu4n (1:

pecification:

6rade of concrete > 2 %

4earing capacity > *!!kFm

)oad on column > %-<.$kFm

85



Area of footing >1.1(539.4

100 > %.<--$ m

Hence providing - (2 m rectangular footing

Bending moment:

Fet earth pressure acting upward due to load

>539.4

2(3 > 8<.<kF

Along long side'

2aximum bending moment > 8<.< ( *.-

> 7%.<kFm

Along short side Z

2aximum bending moment > 8<.< (0.85

> -.$8kFm

Hence design 4.2> 7%.<kFm

;ffective depth d > √ (1.5(75.9(106)/(0.138(25(1000)

> *8*.8 mm

Total depth 0 > *8*.8 Q %! Q20

2

> $*.8 mm

Assume 0 > $&! mm

d x > $&! '%!'20

2

86



pt >100(

1000(314

300

1000(400 > !.&

* c > !.-7 Fmm

* cm"x 3or 2 % > -.* Fmm

So,* < * c<* cm"x

Hence no reinforcement is reuired.

Check for #!o shear $%unching shear&:

(ombined effective depth d >

d x+d y

2

>400+382

2 > -<*mm

= > p \- (2 ̶0.9 ]

> 8<.< (5.19 > $&&.%8kF

* >1.5(466.58(1000

(791+691)(391 > !.!!- Fmm

* c >

k s .

* c

'

> * (1.25 > *.% Fmm

So

c<¿ * cm"x

<¿ * ¿* ¿

hence ok.

88



'.;." For Colu4n C1:

pecification:

6rade of concrete > 2%


)oad on column > *!78.8kFm


100 > **.8& m

Hence providing $

(3

m rectangular footing

Bending moment:


>1078.8

2(3 > 8<.<kF

Along long side'

2aximum bending moment > 8<.< (

*.8

> *$%.&$kFm

Along short side Z


> 8*.<kFm

Hence design 4.2. > *$%.&$kFm

;ffective depth d > √ (1.5(145.64(106) /(0.138(25(1000)

> %* mm

89



Total depth 0 > %* Q %! Q20

2

> -** mm

Assume 0 > $&! mm

d x > $&! '%!'20

2

> $!! mm

d y

> $&!'%!'! '

16

2 > -8 mm

Astx >*& mm

Asty > 88-.&8 mm

2inimum reinforcement >0.12(1000(460

100

> %% mm

Spacing of ! diameter bars >314 (1000

1622 > *<! mm cc

Spacing of *& diameter bars >201(1000

883.68 > ! mm cc

Hence provide * Z ! diameter bars Y *<! mm cc in G'direction

And *- Z *& diameter Y ! mm cc in ' direction

Check for one !a" shear:

Along long side, shear force 5= > p (1.4

> 8<.< (1.4 > *%.8&kF

90



So

c<¿ * cm"x

<¿ * ¿* ¿

. k.

'.;.' For Colu4n C":

pecification:



)oad on column > *%7.&kFm


100 > -.7- m

Hence providing & (4 m rectangular footing

Bending moment:


>2157.6

6(4 > 8<.<kF

Along long side'

2aximum bending moment > 8<.< ( .8

>-%.$kFm

Along short side Z


>*%-.8kFm

Hence design 4.2. > -%.$kFm

92



;ffective depth d > √ (1.5(352.4(106) /(0.138(25(1000)

> -<* mm

Total depth 0 > -<* Q %! Q20

2

> $%* mm

Assume 0 > 8!! mm

d x > 8!! '%!'25

2 > 7-7.% mm

d y > 8!!'%!'% '20

2 > &8% mm

Astx >!8$ mm, A sty > <&! mm


100

> <&! mm

Spacing of % diameter bar >490.625(1000

2084 > -! mm cc

Spacing of ! diameter bar >314(1000

960 > -!! mm cc

Hence provide & Z % diameters Y -! mm cc in G'direction

And *- Z ! diameter Y -!! mm cc in ' direction


Along long side, shear force 5= > p(2.0625

> 8<.< (2.0625 > *8%.$kF

93



Shear stress* >

) (1000

bd >185.4(1.5(1000

1000(737.5

> !.-7 Fmm

pt >100((

1000(490.625

230 )

1000(737.5 > !.<

* c > !.-< Fmm

* cm"x 3or 2 % > -.* Fmm

So,<¿ * c

* ¿ * cm"x


Check for t!o shear $%unching shear&:


d x+d y

2

>737.5+685

2 > 7**.% mm

= > p \& (4 ̶ 2.625 ]

> 8<.< (17.1

> *%-7 kF

* >1.5(1537(1000

(1111.25+1011.25)(711.25 > *.* Fmm

* c >k s . * c

'

> * (1.25 > *.% Fmm

94



So

c<¿ * cm"x

<¿ * ¿* ¿

. k.

'.;.8 For Colu4n ("#

pecification:



)oad on column > *!78.8kFm


100 > **.8& m

Hence providing $ (3 m rectangular footing

Bending moment:


>1078.8

2(3 > 8<.<kF

Along long side'

2aximum bending moment > 8<.< ( *.8

> *$%.&$kFm

Along short side Z


> 8*.<kFm

Hence design 4.2. > *$%.&$kFm

95



;ffective depth d > √ (1.5(145.64(106) /(0.138(25(1000)

> %* mm

Total depth 0 > %* Q %! Q20

2

> -** mm

Assume 0 > $&! mm

d x > $&! '%!'20

2

> $!! mm

d y > $&!'%!'! '16

2 > -8 mm

Astx > *& mm, A sty > 88-.&8 mm


100 > %% mm

Spacing of ! X bar >314(1000

1622 > *<! mm cc

Spacing of *& X bar >201(1000

883.68 > ! mm cc

Hence provide * Z ! XY *<! mm cc in G'direction

and*- Z *& XY ! mm cc in ' direction


Along long side, shear force 5= > p(1.4

> 8<.< (1.4 > *%.8&kF

96



Shear stress* >

) (1000

bd >125.86(1.5(1000

1000(400

> !.-& Fmm

pt >100((

1000(314

300 )

1000(400 > !.&

* c > !.-7 Fmm

* cm"x 3or 2 % > -.* Fmm

So,<¿ * c

* ¿ * cm"x


Check for t!o shear $%unching shear&:


d x+d y

2 >400+382

2 > -<*mm

= > p \$ (3−1.4 ]

> 8<.< (5.19 > <!.%<&kF

* >1.5(902.596(1000

(791

+691

)(391 > *.!7 Fmm

* c >k s . * c

'

> * (1.25 > *.% Fmm

So

c<¿ * cm"x

<¿ * ¿

* ¿. Hence k.

97



'. (ES)GN OF STA)R CASE:

Assuming thickness of waist slab !!mm.

'ead load of flight:

Step section >!.%W!.8W!.*%> !.!* mm

"nclined slab > !.-*8W!.> !.!&$!m

3inish > 5!.8Q!.*% W!.!->!.!*-!mm

Total area > !.!<8m

, 0ensity of concrete > %kFm-

0) of step section, *m in width and 8!mm in plan length >!.!<8W% > .$%kFm

0) per m on plan >2.45(1000

280 > 8.7%kFm

)) per m on plan >%kFm, Total load > *-.7%kFm, 3actored load > !.&kFm

Taking *.%m width of slab, load > *.%W!.&> -*kFm

(oading A:

Self weight of slab > !.W%> %kFm

3inish> !.!-x% > !.7%kFm

)ive load > %kFm

Total load > *!.7%kFm

3actored load > *&.*%kFm

Taking *.%m width of slab, load > *.%W*&.*%> -*kFm

(oading B:

"n a distance eual to *%!mm from the wall and a distance eual to 7%mm inside the

wall only dead load will be considered.

98



Total factored load > *.%W%.7%W*.% > *-kFm

'esign of stair flight:

+eaction at support 4, 2&

> &.%kF

2 A > *<.8 &.% > &7.-kF

Assuming point of Kero S3 occurs at distance 9x: from A

&7.- %W!.7% -*5x !.7% > ! or, x> .-m

Hence maximum 42 > &7.-W.- ̶

%W!.7%W*.<$% -8. > 8kF'm

d >82(10

6

0.138(20(1500 , d > *$*mm

Adopting effective depth as *&%mm and overall as *<!mm.

A t >82(10

6

0.695(415(165 > *7&%mm

Dsing *&'*mm bars eually spaced in *.%m width.

Check for shear:

* > !.8Fmm, pt > !.7% N

* c > !.%& Fmm

* c ' > !.&7 Fmm U

* ok.


Ld > $7X &!!mm

2oment of resistance M

1 >82(1810

1765 > 8$.*kF'm

99



= > &.%kF, Taking L

0 > !

Ld + 1.3( M

) + L-

O + *mm, hence ok.

Temperature reinforcement >0.12(19(100

100 >.8 cmm.

roviding *!mm bars Y-!!mm cc.

0esign of landing slab A#

;ffective span > *.%Q!.*%Q*.%Q!.*&% > -.-*%m

1idth > *.%m, 3actored load per m > *&.*%kFm

Total load > *&.*%x*.%x-.-*% > 8!kF

+eaction from one flight > ^-*W.%W.&8%Q%W*.W!.8%Q *-W!.%W 5!.%_

34.695

> %!kF

+eaction from two flight> *!!kF

2aximum 42 >

wu (l

8 > 7%kF'm, 2aximum S3 > <!kF

;ffective depth > *&%mm, A t > *$%!mm

Dsing *&'*mm bars in*.%m width

'. (ES)GN OF 0ATER TAN

(apacity > ! k), )ength >$m, 4readth >-m, Height > *.7m

100



l

b >*.-- , h > $

4 or *m whichever greater > *m.

Hence bottom *m will bend as vertical cantilever.

1ater pressure at point 0, p> w 5H'h > &8&! F'm

3;2 for long wall > pl2

12 >4 p

3 F'm

3;2 for short wall > p&2

12 > '3 p4 F'm

D/ A& > !.$-, D/ AD > !.%7

Moment calculation:

Ta*le '. > Final 4o4ent calculation

2ember A4 A003 !.$- !.%7

3;2 4 p

3

−3 p

4

4alancing moment −3.01 p

12

−3.99 p12

3inal moment 12.99 p12

−12.99 p

12

Hence moment at support M / > 7$%.<% F'm

4ending moment at centre of long span > p l

2

8 ̶ M / > &<$.!% F'm

4ending moment at centre of short span > p&2

8 ̶ M / > <*.%% F'm

101



2aximum bending moment > 7$%.<% F'm

'esign of section:

d >7425.95(1000

1.32(1000 , d > 7%mm

verall depth T> 7% Q -% !!mm

d > *&%mm

'etermination of pull:

0irect tension on long wall 0 L >

p&2 > *!<! F

0irect tension on short wall 0& >

pL

2 > *-7! F

Cantilever moment:

(antilever moment at base per unit height > wH #

2

6 > *&--.-- F'm

)einforcement at corner of long !all:

x > d %

2 > &%m

Ast 1 >

(7425.95(1000 ) 4 (10290(65)

115(0.853(165 > $*7mm

Ast 3or pull >10290

115 > 8<.$mm

Total Ast > %!&.8mm

Dsing 8mm bars A

ф > %!.$ mm

102



Spacing > *!!mm

Hence providing 8mm Y*!!mm cc at inner face near corners and at a height *m

above the base.

)einforcement at middle of long !all:

0esign bending moment > &<$.!% F'm

0irect pull 0 L > *!<! F

Ast 1 > -$7.%mm, Ast 2 > 8<.$7mm

Total Ast > $-&.<mm which is very near to the reinforcement provided at ends.

Hence providing 8X Y *!!mm cc, also providing additional reinforcement of 8X Y

!!mm cc at the outer face.

)einforcement for short !all:

4ending moment at ends > 7$%.<% F'm, 0irect pull, 0& > *-7! F

Ast 1 > $!-.7mm, Ast 2 > **<.-mm

Total Ast > %-mm, Spacing > *!!mm

Hence providing 8mm Y*!!mm cc at inner face near ends of short span.

)einforcement for cantilever moment:

2aximum cantilever moment >*&--.-- F'm

+einforcement > *$$.7mm


100 > &!!mm

roviding -!!mm vertically on inner face and remaining -!!mm vertically at outer

face with spacing > *&!mm.

103



'..1 (esign o6 sla* <eges si4pl# supporte= 6or tan%:

1 >*7.&kFm

Assume overall thickness as !! mm

;ffective depth > *8! mm

l x > -.$ Q !.*8 > -.%8 mm

l y > %Q !.*8 > %.*8mm

l y

l x > *.$$

∝ x > !.!<<,5 y > !.!%*

0ead load of slab >!. x % > %kFm

Total load >.&kFm

M x > !.!<<W.&W-.%8 >8.&kF m

M y > !.!%*W.&W-.%8 >*$.77kF m

2>!.*-8f ck b d

2

d >

6 28.6(10

6

0.138(20(1000

> *!*mm.@

Area of steel along short span

>

0.36( f ck b xm

0.87 f y

> 5!.-&W!W*!!!W!.$8W*8!5!.87x$*%

104



d > %! mm

Ast

>

1− 4.6(21.6

20(125(250(250

1−6 ¿

0.5( 20

415( ¿

(125(250

> <8 mm2

roviding# 7 −¿ 8 8 bars

2in steel >0.85b d

f y >0.85(125(250

415 > &$ mm2

Hence @.

Beam A':

M AD >7l

8 >99(5

8 > &*.87kFm

2>!.*-8f ck b d2

&*.87x 106

>!.*-8 W !Wd ( d

2

2

d > -&! mm

Ast >

1− 4.6(61.8720(125(250(250

1−6 ¿

0.5( 20

415( ¿

(180(360

> %8& mm2

rovide# &

−¿ *

8 bars

106



2inimum steel >

0.85b df y >

0.85(180(360

415 > *- mm2

Hence k.

'.1 (ES)GN OF S-ELL

+adius> *!.*! m, (entral rise> .-%m, (hord width> *-m, Span>-! m

Thickness of shell> 8!mm, Semi central angle> $! degree,

;dge beam siKe> !!W*88! mm, +einforcement in edge beam> *&'-X

1idth of edge beams> !!mm

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

about neutral axis, we have

9 1A > ∫0

α

2d:t ( 2c-s:− 2c-sα ) Q mW A t 5*.%8Q+cosP 7.7%

utting m>*-, +>*!.*, A t > 58x8.!$> &$.-cm> !.!!&$m

+ t 5sin P P cos P > m

A t 5+ cos P &.*7

"t gives P> & .25 ° , 9 1A > *.8-%m$

Self weight of shell > 5!.!8W%>*.<kFm

1ater proofing and live load > *.!kFm

Total load > .<kFm

Total weight per meter run> ∫0

40

2.92( 2d:) > $!kFm

1eight of edge beam > 5!.*x*.88x$ > <kFm

107



1eight of filling in the valley portion > *.%kFm

Total load > %!.%kFm

2aximum 42 >50.5

(30

2

8 > %7!!kF'm

2aximum S3 > !.%x%!.%x-! > 7%7.%kF

2aximum compressive stress at crown,f c >

5700(1330(106

1.835(1012 > -.%Fm

2aximum tensile stress at centre of gravity of steel, f t >

3.52(2600(23

1330 >

8<.%Fm

Shear stress E >757.5(10

3(435(106

1.835(2(80(1012

2aximum horiKontal shear stress at neutral axis >*.* Fmm

roviding 8mm diameter bars inclined at $% ° to the longitudinal axis of shell.

Spacing >√ 2(50(230

80(1.12 > *8!mm near supports.

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

%!mm centers.

C-APTER!9

CONCL+S)ONS



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

108



and designed to meet the optimum balance between economy and uality. Ample

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

cross ventilation. The guidelines of "S# $%& '!!! has been followed very strictly. Slab

of dimension 8mW-.--m has been designed with total depth *$!mm, live load $kFm

and surface finish load *.%kFm. 4eams have been designed with same value of live

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

77%mm at first and ground floor.

The shell has been designed using L4eam TheoryM. (olumns have been analyKed

using LSubstitute 3rame 2ethodM. The foundation depth has been kept m. "solated

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

B)BL)OGRAP-2

*. `ain, A.@., L0esign of +einforced (oncreteM, Fem (hand and 4ros

ublishers, +oorkee, !!7, Sixth reprint.

. @rishna +a/u, F., LAdvanced +einforce (oncrete 0esignM, (4S, ublishers [

0istributors, 0elhi, *<88, 3irst reprint.

109



-. illai, S.D., 2enon, 0., L+einforced (oncrete 0esignM, Tata 2c6raw'Hill

ublishing company )td., Few 0elhi, !!7, ;ight +eprint.

$. unmia, 4.(., `ain, A.@., `ain, A.@., L+.(.(. 0esigns 5+einforced (oncrete

StructuresM, )axmi ublication vt. )td., Few 0elhi, !!&, Third +eprint.

%. +eddy, (.S., L4asic Structural AnalysisM, Tata 2c6raw Hill ublishing

(ompany )td., Few 0elhi, !!&, 3ifteenth reprint.

&. "S $%&#!!! (ode of ractice lain and +einforced (oncrete.

7. "S# 87% 5art ' *<87 (ode of ractice for 0esign )oads 5ther than

;arthuake for 4uilding and Structures.

58147987 Design of Shopping Complex

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Transcript of 58147987 Design of Shopping Complex