intze tank design

Intze tank

MAIN PROJECT REPORT

ON

DESIGN AND ESTIMATION OF INTZE TANK

Submitted in partial fulfillment of the Requirements for the award of the

degree of Bachelor of Technology in Civil Engineering

By

M.LOKESH 09241A0175

K.NAGA RAJU 09241A0178

R.RAJASHEKAR 09241A0188

J.RAJEEV 09241A0190

Under the esteemed guidance of

G.V.V SATYA NARAYANA

(Associate professor of Civil Engineering Department)

DEPARTMENT OF CIVIL ENGINEERING

GOKARAJU RANGARAJU INSTITUTE OF ENGINEERING AND

TECHNOLOGY

(Affiliated to JNTU)

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ABSTRACT

Due to enormous need by the public, water has to be stored and supplied according

to their needs. Water demand is not constant throughout the day. It fluctuates hour

to hour. In order to supply constant amount of water, we need to store water. So to meet the public water demand, water tank need to be constructed.

Storage reservoirs and overhead tanks are used to store water, liquid petroleum,

petroleum products and similar liquids. The force analysis of the reservoirs or tanks

is about the same irrespect ive of the chemical nature of the product. All tanks are designed as crack free structures to eliminate any leakage.

This project gives in brief, the theory behind the design of liquid retaining structure

(Elevated circular water tank with domed roof and conical base) using working stress method. Elements are design in working stress method.

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ACKNOWLEDGEMENT

We would like to express our gratitude to all the people behind the screen who

helped us to transform an idea into a real applicat ion. We would like to express our

heart-felt grat itude to our parents without whom we would not have been privileged

to achieve and fulfill our dreams. We are grateful to our principal

Dr.JandyalaN.Murthi who most ably run the inst itution and has had the major hand in enabling us to do our project.

We profoundly thank Dr. G.Venkataramana, Head of the Department of CIVIL

ENGINEERING who has been an excellent guide and also a great source of inspirat ion to our work.

We would like to thank our internal guide Sri. G.V.V.Satyanarayana Associate

Professor for his technical guidance, constant encouragement and support in carrying out our project at college.

The sat isfact ion and euphoria that accompany the successful complet ion of the task

would be great but incomplete without the ment ion of the people who made it

possible with their constant guidance and encouragement crowns all the efforts with

success. In this context, We would like thank all the other staff members, both teaching and non-teaching, who have extended their t imely help and eased our task.

M.LOKESH 09241A0175

K.NAGA RAJU 09241A0178

R.RAJASHEKAR 09241A0188

J.RAJEEV 09241A0190

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INDEX

CONTENTS: PAGE NO.

1 SYMBOLS 1

2 INTRODUCTION 5

2.1 SOURCE OF WATER 6

3 WATER DEMAND 7

3.1 WATER QUANTITY ESTIMATION 7

3.2 WATER CONSUMPTION RATE 7

3.3 FIRE FIGHTING DEMAND 7

3.4 FACTORS EFFECTING PER CAPITA DEMAND 8

3.5 FLUCTUATION IN RATE OF DEMAND 8

4 POPULATION FORECASTING 10

4.1 DESIGN PERIOD OF POPULATION 10

4.2 POPULATION FORECASTING METHODS 10

5 WATER TANKS 11

5.1 CLASSIFICATION OF WATER TANKS 11

6 DESIGN REQUIREMENTS OF CONCRETE 12

6.1 JOINTS IN LIQUID RETAINING STRUCTURES 12

6.1.1 MOVEMENT JOINTS 13

6.1.2 CONTRACTION JOINTS 14

6.1.3 TEMPORARY JOINTS 15

7 GENERAL DESIGN REQUIREMENTS 16

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7.1 PLAIN CONCRETE STRUCTURES 16

7.2 PERMISSIBLE STRESSES IN CONCRETE 16

7.3 PERMISSIBLE STRESSES IN STEEL 16

7.4 STRESSES DUE TO DRYING SHRINKAGE OR

TEMPERATURE CHANGE 17

7.5 FLOORS 17

7.6 WALLS 19

7.7 ROOFS 20

7.8 MINIMUM REINFORCEMENT 21

7.9 MINIMUM COVER TO REINFORCEMENT 21

8 DOMES 22

9 MEMBERANE THEORY OF SHELLS OF REVOLUTION 23

10 WATER TANK WITH SPHERICAL DOME 25

11 DESIGN OF RCC DOME 26

12 OVER HEAD WATER TANK AND TOWERS 29

13 DESIGN 32

13.1 DETAILS OF DESIGN 32

14 ESTIMATION 53

14.1 DETAILED ESTIMATION 53

14.2 DATA SHEET 58

15 CONCLUSION 66

16 REFERENCES 67

17 REFERENCE BOOKS 69

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1. SYMBOLS

A = Total area of sect ion

Ab = Equivalent area of helical reinforcement.

Ac = Equivalent area of sect ion

Ah = Area of concrete core.

Am = Area of steel or iron core.

Asc = Area of longitudinal reinforcement (comp.)

Ast = Area of steel (tensile.)

A l = Area of longitudinal torsional reinforcement.

Asv= Total cross-sect ional are of st irrup legs or bent up bars within distance Sv

Aw =Area of web reinforcement.

AФ= Area of cross –sect ion of one bars.

a = lever arm.

ac = Area of concrete.

B =flange width of T-beam.

b = width.

br =width of rib.

C =compressive force.

c = compressive stress in concrete.

c’= stress in concrete surrounding compressive steel.

D = depth

d = effect ive depth

dc = cover to compressive steel

ds= depth of slab

d t= cover to tensile steel

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e = eccentricit y.

dc/d = compressive steel depth factor

F =shear characterist ic force.

Fd= design load

Fr = radial shear force.

f= stress (in general)

fck = characterist ic compressive stress of concrete.

Fy = characterist ic tensile strength of steel.

H = height.

I = moment of inertia.

Ie=equivalent moment of intert ia.

j= lever arm factor.

Ka=coefficient of act ive earth pressure.

Kp =coefficient of passive earth pressure.

k = neutral axis depth factor (n/d).

L=length.

Ld =devolopment length.

l = effect ive length of column or length or bond length.

M = bending moment or moment.

Mr=moment of resistance or radial bending moment.

Mt=torsional moment.

Mu=ult imate bending moment

Mθ=circumferent ial bending moment

m = modular rat io.

n = depth of neutral axis.

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nc=depth of crit ical neutral axis.

Pa=act ive earth pressure.

Pp= passive earth pressure.

Pu= ult imate axial load on the member(limit state design).

P = percentage steel.

P’= reinforcement ratio.

Pa=act ive earth pressure indencity.

Pe=net upward soil pressure.

Q= shear resistance.

� = shear stress.

q’=shear stress due to torsion

R= radius.

s= spacing of bars.

sa= average bond stress.

sb= local bond stress.

T=tensile force.

Tu=ult imate torsional moment.

�st or t= tensile stress in steel.

tc= compressive stress in compressive steel.

Vu=ult imate shear force due or design load.

Vus=shear carried by shear reinforcement.

W= point load.

X= coordinate.

xu= depth of neutral axis.

Z= distance.

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α = inclinat ion.

β = surcharge angle.

γ = unit weight of soil

γf = part ial safety factor appropriate to the loading.

γm = part ial safety factor appropriate to the material.

σcc = permissible stress in concrete.

σcbc = permissible compressive stress in concrete due to bending.

σsc = permissible compressive stress in bars.

σst = permissible stress in steel in tension.

σst = permissible tensile strss in shear reinforcement.

σsy = yield point compressive stress in steel.

µ = co efficient of frict ion.

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2. INTRODUCTION

A water tank is used to store water to tide over the daily requirement. In the

construction of concrete structure for the storage of water and other liquids the

imperviousness of concrete is most essent ial .The permeabilit y of any uniform and

thoroughly compacted concrete of given mix proportions is mainly dependent on

water cement rat io .The increase in water cement rat io results in increase in the

permeability .The decrease in water cement rat io will therefore be desirable to

decrease the permeability, but very much reduced water cement rat io may cause

compact ion difficult ies and prove to be harmful also. Design of liquid retaining

structure has to be based on the avoidance of cracking in the concrete having regard

to its tensile strength.Cracks can be prevented by avoiding the use of thick t imber

shuttering which prevent the easy escape of heat of hydrat ion from the concrete

mass the risk of cracking can also be minimized by reducing the restraints onfree expansion or contraction of the structure.

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1. Objective:

1. To make a study about the analysis and design of water tanks.

2. To make a study about the guidelines for the design of liquid retaining Structure according to is code.

3. To know about the design philosophy for the safe and economical design of water tank.

4. To develop programs for the design of water tank of flexible base and rigid base and the under ground tank to avoid the tedious calculat ions.

5. In the end, the programs are validated with the results of manual calculat ion given in concrete Structure.

2.1 Sources of water supply:

The various sources of water can be classified into two categories:

Surface sources, such as

1. Ponds and lakes,

2. Streams and rivers,

3. Storage reservoirs, and

4. Oceans, generally not used for water supplies, at present.

Sub-surface sources or underground sources, such as

1. Springs,

2. Infiltrat ion wells, and

3. Wells and Tube-wells.

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3. WATER DEMAND

3.1 Water Quantity Estimation: The quantity of water required for municipal uses

for which the water supply scheme hasto be designed requires following data:Water

consumption rate (Per Capita Demand in litres per day per head)Populat ion to be served.

Quant ity= Per demand x Populat ion

3.2 Water Consumption Rate: It is very difficult to precisely assess the quant it y

of water demanded by the public, sincethere are many variable factors affect ing

water consumption. The various types of waterdemands, which a cit y may have, may be broken into following class

Water Consumption for Various Purposes:

Types of Consumption Normal Range (lit/capita/day)

Average %

1 Domest ic Consumption 65-300 160 35

2 Industrial and Commercial Demand

45-450

135 30

3 Public including Fire Demand Uses

20-90

45 10

4 Losses and Waste 45-150

62 25

3.3 Fire Fighting Demand:The per capita fire demand is very less on an

average basis but the rate at which the wateris required is very large. The rate of

fire demand is somet imes treated as a function ofpopulat ion and is worked out from following empirical formulae:

Authority Formula (P in thousand)

Q for 1 lakh Populat ion)

1

American InsuranceAssociat ion

Q(L/min)=4637P(1-0.01 ÖP)

41760

2 Kuchling'sFormula Q(L/min)=3182 P 31800 3 Freeman'sFormula Q(L/min)=1136.5(P/5+10) 35050 4 Ministry

ofUrbanDevelopmentManual Formula

Q(kilo liters/d)=100P for P>50000

31623

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3.4 Factors affecting per capita demand:

• Size of the city: Per capita demand for big cit ies is generally large as compared tothat for smaller towns as big cit ies have sewered houses.

• Presence of industries.

• Climatic conditions.

• Habits of economic status.

• Quality of water: If water is aesthet ically $ people and their

. Medically safe, the consumption will increase as people will not resort to privatewells, etc.

• Pressure in the distribution system.

• Efficiency of water works administration: Leaks in water mains and services;and un authorised use of water can be kept to a minimum by surveys.

• Cost of water.

• Policy of metering and charging method: Water tax is charged in two

different ways on the basis of meter reading and on the basis of certain fixed monthly rate.

3.5 Fluctuations in Rate of Demand:

Average Daily Per Capita Demand

= Quant ity Required in 12 Months/ (365 x Populat ion)

If this average demand is supplied at all the times, it will not be sufficient to meet thefluctuations.

•Seasonal variation:The demand peaks during summer.Firebreak outs are generally more in summer, increasing demand. So,there is seasonal variat ion

.• Daily variation depends on the act ivity. People draw out more water on

Sundaysand Fest ival days, thus increasing demand on these days.

• Hourly variations are very important as they have a wide range. During

act ivehousehold working hours i.e. from six to ten in the morning and four to eight

inthe evening, the bulk of the daily requirement is taken. During other hours

therequirement is negligible. Moreover, if a fire breaks out, a huge quant ity of

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wateris required to be supplied during short duration, necessitat ing the need for

amaximum rate of hourly supply.So, an adequate quant ity of water must be

available to meet the peak demand. To meet allthe fluctuations, the supply pipes,

service reservoirs and distribut ion pipes must beproperly proportioned. The water is

supplied by pumping direct ly and the pumps anddistribut ion system must be

designed to meet the peak demand. The effect of monthlyvariat ion influences the

design of storage reservoirs and the hourly variat ions influencesthe design of pumps and service reservoirs. As the populat ion decreases, the fluctuationrate increases.

Maximum daily demand = 1.8 x average daily demand

Maximum hourly demand of maximum day i.e. Peak demand

= 1.5 x average hourly demand

= 1.5 x Maximum daily demand/24

= 1.5 x (1.8 x average daily demand)/24

= 2.7 x average daily demand/24

= 2.7 x annual average hourly demand

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4. POPULATION FORECAST

4.1 Design Periods & Population Forecast:

This quant ity should be worked out with due provision for the est imated

requirements ofthe future. The future period for which a provision is made in the water supply scheme isknown as the design period.

Design period is est imated based on the following:

• Useful life of the component , considering obsolescence, wear, tear, etc.

• Expandability aspect.

• Ant icipated rate of growth of populat ion, including industrial, commercial

developments& migrat ion-immigrat ion.

• Available resources.

• Performance of the system during init ial period.

4.2 Population Forecasting Methods:

The various methods adopted for estimat ing future populat ions are given below.

Theparticular method to be adopted for a particular case or for a particular cit y

dependslargely on the factors discussed in the methods, and the select ion is left to the discrect ionand intelligence of the designer.

1. Incremental Increase Method

2. Decreasing Rate of Growth Method

3. Simple Graphical Method

4. Comparat ive Graphical Method

5. Ratio Method

6. Logist ic Curve Method

7. Arithmet ic Increase Method

8. Geometric Increase Method

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5. WATER TANKS

5.1 CLASSIFICATIONS:

Classificat ion based on under three heads:

1. Tanks rest ing on ground

2. Elevated tanks supported on stagging

3. Underground tanks.

Classificat ion based on shapes

1. Circular tanks

2. Rectangular tanks

3. Spherical tanks

4. Intze tanks

5. Circular tanks with conical bottom

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6. DESIGN REQUIREMENT OF CONCRETE (I. S. I)

In water retaining structure a dense impermeable concrete

is requiredtherefore,proportion of fine and course aggregates to cement should

besuch as to give highqualityconcrete.Concrete mix lesser than M20 is not used. The

minimum quant ity ofcement in the concrete mix shall be not less than 30 kN/m3.The

design of the concretemix shall be such that the resultant concrete issu efficient ly

impervious. Efficientcompact ion preferably by vibrat ion is essent ial. The

permeability of the thoroughlycompacted concrete is dependent on water cement

ratio. Increase in water cement ratioincreases permeability, while concrete with low

water cement ratio is difficult to compact.Other causes of leakage in concrete are

defects such as segregat ion and honey combing.All joints should be made water-

tight as these are potent ial sources of leakage. Design ofliquid retaining structure is

different from ordinary R.C.C. structures as it requires thatconcrete should not

crack and hence tensile stresses in concrete should be withinpermissible limits. A

reinforced concrete member of liquid retaining structure is designedon the usual

principles ignoring tensile resistance of concrete in bending. Addit ionally itshould

be ensured that tensile stress on the liquid retaining ace of the equivalent

concretesect ion does not exceed the permissible tensile strength of concrete as

given in table 1. For calculat ion purposes the cover is also taken into concrete area.

Cracking may be caused due to restraint to shrinkage, expansion and contract ion of

concrete due to temperature or shrinkage and swelling due to moisture effects. Such restraint may be caused by .

(i) The interact ion between reinforcement and concrete during shrinkage due to drying.

(ii) The boundary condit ions.

(iii) The different ial condit ions prevailing through the large thickness of massive

concrete Use of small size bars placed properly, leads to closer cracks but of

smaller width. The risk of cracking due to temperature and shrinkage effects may

be minimized by limit ing the changes in moisture content and temperature to which

the structure as a whole is subjected. The risk of cracking can also be minimized by

reducing the restraint on the free expansion of the structure with long walls or slab

founded at or below ground level, restraint can be minimized by the provision of a

sliding layer. This can be provided by founding the structure on a flat layer

ofconcrete with interposit ion of some material to break the bond and facilitate

movement.Incaselength of structure is large it should be subdivided into suitable

lengths separated by movement joints, especially where sect ions are changed the

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movement joints should be provided.Where structures have to store hot liquids,

stresses caused by difference in temperature between insideand outside of the

reservoir should be taken into account.The coefficient of expansion due to

temperature change is taken as 11 x 10 -6 /° C and coefficient of shrinkage may be taken as 450 x 10 -6 for init ial shrinkage and 200 x 10 -6 for drying shrinkage.

6.1 JOINTS IN LIQUID RETAINING STRUCTURES:

6.1.1 MOVEMENT JOINTS. There are three types of movement joints.

(i)Contraction Joint. It is a movement joint with deliberate discont inuity without

init ial gap between the concrete on either side of the joint. The purpose of this joint is to accommodate contraction of the concrete. The joint is shown in Fig. (a)

Fig (a)

A contraction joint may be either complete contraction joint or partial contraction

joint. A complete contraction joint is one in which both steel and concrete

areinterrupted and a part ial contract ion joint is one in which only the concrete is interrupted, the reinforcing steel running through as shown in Fig.(b)

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Fig (b)

(ii)Expansion Joint. It is a joint with complete discont inuity in both reinforcing

steel and concrete and it is to accommodate either expansion or contraction of the structure. A typical expansion joint is shown in Fig.(c)

Fig(c)

This type of joint is provided between wall and floor in some cylindrical tank designs.

6.1.2 CONTRACTION JOINTS:

This type of joint is provided for convenience in construct ion. This type of joint

requires the provision of an init ial gap between thead joining parts of a structure

which by closing or opening accommodates the expansion or contraction of the structure.

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Fig (d)

(iii) Sliding Joint. It is a joint with complete discont inuity in both reinforcement

and concrete and with special provision to facilitate movement in plane of the joint.

A typical joint is shown in Fig. This type of joint is provided between wall and floor in some cylindrical tank designs.

Fig (e)

6.1.3 TEMPORARY JOINTS:

A gap is somet imes left temporarily between the concrete of adjoining parts of a

structurewhich after a suitable interval and before the structure is put to use, is

filled with mortaror concrete completely with suitable joint ing materials. In the

first case width of the gap should be sufficient to allow the sidesto be prepared before filling.Figure (g)

Fig (g)

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7. GENERAL DESIGN REQUIREMENTS (I.S.I)

7.1 Plain Concrete Structures:

Plain concrete member of reinforced concrete liquid retaining structure may be

designed against structural failure by allowing tension in plain concrete as per the

permissible limits for tension in bending. This will automat ically take care of

failure due to cracking. However, nominal reinforcement shall be provided, for plain concrete structural members.

7.2. Permissible Stresses in Concrete:

(a) For resistance to cracking: For calculat ions relat ing to the resistance of

members to cracking, the permissible stresses in tension (direct and due to bending)

and shear shall confirm to the values specified in Table 1.The permissible tensile

stresses due to bending apply to the face of the member in contact with the liquid.

In members less than 225mm ∅ thick and in contact with liquid on one side these permissible stresses in bending apply also to the face remote from the liquid.

(b) For strength calculations: In strength calculat ions the permissible concrete

stresses shall be in accordance with Table 1. Where the calculated shear stress in

concrete alone exceeds the permissible value, reinforcement act ing in conjunct ion

with diagonal compression in the concrete shall be provided to take the whole of the shear.

7.3 Permissible Stresses in Steel:

(a) For resistance to cracking. When steel and concrete are assumed to act

together for checking the tensile stress in concrete for avoidance of crack, the

tensile stress in steel will be limited by the requirement that the permissible tensile

stress in the concrete is not exceeded so the tensile stress in steel shall be equal to

the product of modular rat io of steel and concrete, and the corresponding allowable tensile stress in concrete.

(b) For strength calculations:

In strength calculat ions the permissible stress shall be as follows:

a) Tensile stress in member in direct tension 1000 kg/cm2.

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b) Tensile stress in member in bending on liquid retaining face of members or face away from liquid for members less than 225mm thick 1000 kg/cm2.

c) On face away from liquid for members 225mm or more in thickness 1250 kg/cm2.

d) Tensile stress in shear reinforcement For members less than 225mm thickness 1000 kg/cm2 for members 225mm or more in thickness 1250 kg/cm2.

(v)Compressive stress in columns subjected to direct load 1250 kg/cm2.

7.4 Stresses due to drying Shrinkage or Temperature Change:

( i)Stresses due to drying shrinkage or temperature change may be ignored provided that .

(a) The permissible stresses specified above in (ii) and (iii) are not otherwise

exceeded.

(b) Adequate precaut ions are taken to avoid cracking of concrete during the construction period and unt il the reservoir is put into use.

(c) Recommendat ion regarding joints given in art icle 8.3 and for suitable sliding

layer beneath the reservoir are complied with, or the reservoir is to be used only for

the storageof water or aqueous liquids at or near ambient temperature and the

circumstances aresuch that the concrete will never dry out.

(ii)Shrinkage stresses may however be required to be calculated in special cases,

when ashrinkage co-efficient of 300× 10�� may be assumed.

(iii) When the shrinkage stresses are allowed, the permissible stresses,tensile

stresses to concrete (direct and bending) as given in Table 1 may be increased by 33.33 per cent.

7.5 Floors:

(i) Provision of movement joints.

Movement joints should be provided as discussed in art icle 3.

(ii) Floors of tanks resting on ground.

If the tank is rest ing direct ly over ground, floor may be constructed of concrete

with nominal percentage of reinforcement provided that it is certain that the ground

will carry the load without appreciable subsidence in any part and that the concrete

floor is cast inpanels with sides not more than 4.5m.with contract ion or expansion

joints between. Insuch cases a screed or concrete layer less than 75mm thick shall

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first be placed on theground and covered with a sliding layer of bitumen paper or

other suitable material todestroy the bond between the screed and floor concrete. In

normal circumstances the screed layer shall be of grade not weaker than M10,where

injurious soils or aggressivewater are expected, the screed layer shall be of grade

not weaker than M15 and if necessary a sulphate resist ing or other special cement

should be used.

(iii) Floor of tanks resting on supports

(a) If the tank is supported on walls or other similar supports the floor slab shall

bedesigned as floor in buildings for bending moments due to water load and self weight.

(b)When the floor is rigidly connected to the walls (as is generally the case) the

bending moments at the junct ion between the walls and floors shall be taken into

account in the design of floor to gether with any direct forces transferred to the

floor from the walls orfrom the floor to the wall due to suspension of the floor from

the wall.If the walls are non-monolithic with the floor slab, such as in cases, where

movement joints have been provided between the floor slabs and walls, the floor shall be designed only for the vert ical loads on the floor.

(c) In cont inuous T-beams and L-beams with ribs on the side remote from the

liquid, the tension in concrete on the liquid side at the face of the supports shall not

exceed the permissible stresses for controlling cracks in concrete. The width of the

slab shall be determined in usual manner for calculat ion of the resistance to cracking of T-beam, L beam sect ions at supports.

(d)The floor slab may be suitably t ied to the walls by rods properly embedded in

both the slab and the walls. In such cases no separate beam (curved or straight) is

necessary under the wall, provided the wall of the tank itself is designed to act as a beam over the supports under it .

(e)Sometimes it may be economical to provide the floors of circular tanks,in the

shape of dome. In such cases the dome shall be designed for the vert ical loads ofthe

liquid over it and the rat io of it s rise to its diameter shall be so adjusted that the

stresses in the dome are, as far as possible, wholly compressive. The dome shall be

supported at its bottom on the ring beam which shall be designed for resultan tcircumferent ial tension in addit ion to vertical loads.

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7.6 Walls:

(i)Provision of joints

(a)Where it is desired to allow the walls to expand or contract separately from the

floor, or to prevent moments at the base of the wall owing to fixity to the floor, sliding joints may be employed.

(b)The spacing of vert ical movement joints should be as discussed in article 3.3

while the majority of these joints may be of the part ial or complete contraction

type, sufficient joints of the expansion type should be provided to sat isfy the requirements given in art icle

(ii) Pressure on Walls.

(a) In liquid retaining structures with fixed or float ing covers the gas pressure developed above liquid surface shall be added to the liquid pressure.

(b)When the wall of liquid retaining structure is built in ground, or has earth

embanked against it , the effect of earth pressure shall be taken into account.

(iii) Walls or Tanks Rectangular or Polygonal in Plan.

While designing the walls of rectangular or polygonal concrete tanks, the following points should be borne in mind.

(a) In plane walls, the liquid pressure is resisted by both vert ical and horizontal

bendingmoments. An est imate should be made of the proportion of the pressure

resisted by bending moments in the vertical and horizontal planes. The direct

horizontal tension caused by the direct pull due to water pressure on the end walls,

should be added to that result ing from horizontal bending moments. On liquid

retaining faces, the tensile stressesdue to the combinat ion of direct horizontal tension and bending act ion shall sat isfy the following condit ion:

(t./t )+ ( óc t . /óct ) ≤ 1

t . = calculated direct tensile stress in concrete

t = permissible direct tensile stress in concrete (Table 1)

óc t = calculated tensile stress due to bending in concrete.

óc t = permissible tensile stress due to bending in concrete.

(d)At the vert ical edges where the walls of a reservoir are rigidly joined, horizontal

reinforcement and haunch bars should be provided to resist the horizontal bending

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moments even if the walls are designed to withstand the whole load as vert ical beams or cant ilever without lateral supports.

(c) In the case of rectangular or polygonal tanks, the side walls act as two way

slabs,where by the wall is cont inued or restrained in the horizontal direct ion, fixed

or hinged atthe bottom and hinged or free at the top. The walls thus act as thin

plates subjected triangular loading and with boundary condit ions varying between

full restraint and freeedge. The analysis of moment and forces may be made on the

basis of any recognizedmethod.

(iv) Walls of Cylindrical Tanks.

While designing walls of cylindrical tanks the following points should be borne in mind:

(a)Walls of cylindrical tanks are either cast monolithically with the base or are set

in grooves and key ways (movement joints). In either case deformat ion of wall

under influence of liquid pressure is restricted at and above the base. Consequent ly,

only part ofthe triangular hydrostatic load will be carried by ring tension and part of the load at bottom will be supported by cantilever act ion.

(b)It is difficult to restrict rotation or settlement of the base slab and it is advisable

toprovide vertical reinforcement as if the walls were fully fixed at the base, in

addit ion to the reinforcement required to resist horizontal ring tension for hinged at

base, condit ions of walls, unless the appropriate amount of fixity at the base is

established by analysis with due consideration to the dimensions of the base slab

the type of joint between the wall and slab, and , where applicable, the type of soil supporting the base slab.

7.7 Roofs;

(i) Provision of Movement joints:

To avoid the possibility of sympathet ic cracking it is important to ensure that

movement joints in the roof correspond with those in the walls, if roof and walls are

monolithic. It , however, provision is made by means of a sliding joint for movement between the roof and the wall correspondence of joints is not so important.

(ii) Loading:

Field covers of liquid retaining structures should be designed for gravity loads,

such asthe weight of roof slab, earth cover if any, live loads and mechanical

equipment. They should also be designed for upward load if the liquid retaining

structure is subjected to internal gas pressure. A superficial load sufficient to

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ensure safety with the unequalintensity of loading which occurs during the placing

of the earth cover should be allowedfor in designing roo fs. The engineer should

specify a loading under these temporarycondit ions which should not be exceeded. In

designing the roof, allowance should bemade for the temporary condit ion of some

spans loaded and other spans unloaded, eventhough in the final state the load may

be small and evenly distributed.

(iii) Water tightness: In case of tanks intended for the storage of water for

domest ic purpose, the roof must be made water-tight. This may be achieved by

limit ing the stresses as for the rest of the tank, or by the use of the covering of the water proof membrane or by providing slopes to ensure adequate drainage.

(iv) Protection against corrosion: Protection measure shall be provided to the underside of the roof to prevent it from corrosion due to condensat ion.

7.8 Minimum Reinforcement:

(a)The minimum reinforcement in walls, floors and roofs in each of two direct ions

atright angles shall have an area of 0.3 per cent of the concrete sect ion in that

direct ion for sect ions up to 100mm, thickness. For sect ions of thickness greater than

100mm, and lessthan 450mm the minimum reinforcement in each of the two

direct ions shall be linearly reduced from 0.3 percent for 100mm thick sect ion to 0.2

percent for 450mm, thicksect ions. For sections of thickness greater than 450mm,

minimum reinforcement in eachof the two direct ions shall be kept at 0.2 per cent. In

concrete sect ions of thickness225mm or greater, two layers of reinforcement steel

shall be placed one near each faceof the sect ion to make up the minimum reinforcement.

(b)In special circumstances floor slabs may be constructed with percentage of

reinforcement less than specified above. In no case the percentage of reinforcement inany member be less than 0.15% of gross sect ional area of the member.

7.9 Minimum Cover to Reinforcement:

(a)For liquid faces of parts of members either in contact with the liquid (such as

innerfaces or roof slab) the minimum cover to all reinforcement should be 25mm or

the diameter of the main bar whichever is grater. In the presence of the sea water

and soil sand water of corrosive characters the cover should be increased by 12mm but thisaddit ional cover shall not be taken into account for design calculat ions.

(b)For faces away from liquid and for parts of the structure neither in contact with

theliquid on any face, nor enclosing the space above the liquid, the cover shall be as forordinary concrete member.

Intze tank

8. DOMES

A dome may be defined as a thin shell generated by the revolut ion of a regular

curve about one of it s axes. The shape of the dome depends on the type of the curve

and the direct ion of the axis of revolut ion. In spherical and conoidal domes, surface

is described by revolving an arc of a circle. The centre of the circle may be on the

axis of rotation (spherical dome) or outside the axis (conoidal dome). Both types

may or may not have assymmetrical lantern opening through the top. The edge of

the shell around its base isusually provided with edge member cast integrally with the shell.

Domes are used in variety of structures, as in the roof of circular areas, in circular

tanks, in hangers, exhibit ion halls, auditoriums, planetorium and bottom of tanks,

bins andbunkers. Domes may be constructed of masonry, steel, t imber and

reinforced concrete.However, reinforced domes are more common nowadays since

they can be constructed over large spans membrane theory for analysis of shells of

revolut ion can be developed neglect ing effectof bending moment, twist ing moment

and shear and assuming that the loads are carriedwholly by axial stresses. This

however applies at points of shell which are removed somedistance away from the

discont inuous edge. At the edges, the results thus obtained maybe indicated but are not accurate.

The edge member and the adjacent hoop of the shells must have very nearly the

same strain when they are cast integrally. The significance of this fact is usually

ignored and the forces thus computed are, therefore, subject to certain

modificat ions.Stresses in shells are usually kept fairly low, as effect of the edge

disturbance, as ment iioned above is usually neglected. The shell must be thick

enough to allow space and protection for two layers of reinforcement. From this

point of view 80 mm is considered as the minimum thickness of shell.

Intze tank

9. MEMBRANE THEORY OF SHELLS OF

REVOLUTION

Fig shows a typical shell of revolut ion, on which equilibrium of an element,

obtained by intersect ion of meridian and lat itude, is indicated. Forces along the

circumference are denotted by Nf and are called meridian stresses and forces at

right angles to the meridian plane and along the lat itude are horizontal and called

the hoop stresses, denoted by N .Neglect ing variat ions in the magnitudes of Nf and N , since they are very small.the state of stress in the element is shown in fig (b).

Shell of Revolut ion.

Intze tank

two forces N(rd�) have the resultant N(rd�)d as shown in Fig.(c) and the

resultant acts normal to the surface pointed towards the inner side. Forces N�(r1d)

again have horizontal resultant of magnitude N(r1 d) d� as shown in Fig (d). It

has a component N(r1d)d�sin directed normally to the shell and point ing

towards the inner side. These two forces and the external force normal to the

surface and a magnitude Pr(rd�) must bein equilibrium.

Thus,Nf (rd)df++N (r1df)dsinf+Pr(rd)(r1d )= 0

Combining and as r = r2 sinf from Fig. ((a)

Nf /r1+N/r2 = -Pr = pressure normal to the surface In this equat ion pr is considered

posit ive when act ing towards the inner side and negat ive when act ing towards the outerside of the shell.Value

s and N f and N will be posit ive when tensile andnegat ive compressive.

The equat ion is valid not only for shells in thform of a surface of revolut ion, but

may be apped to allshells, when the coordinate lines for = constant and � =

constant, are the linesof curvature of the surface.

Forces in shell Force N f act tangent ially to the surface aall around the

circumference. Considering thequilibrium of a segment of shell cut along the parallel to lat itude defined by the angle as shown in Fig

2prNf sin f + W= 0,

Where W= total load in the vert ical dirct ion on the surface of the shell above the cut.

This gives,

Nf = -W/2prsinf

Eq. is readily solved for N fand N may then be detrminedby Eq. This theory is

applicable to a shell of any material as only the condit iions of equilibrium have

been applied and no compat ibility relat ionsships in terms of deformat ion have been introduced. It is, therefore, immaterial whetherr Hooke's law is applicable or not.

Intze tank

10. WATER TANK WITH SPHERICAL BOTTOM

Referring to the tank in Fig.(a),supported along the circumference as shown,the

magnitude of Na may be obtained from considerat ion of equilibrium. If it is

required to obtain Na at sect ion 1 - 1 from calculat ion of the total downward load,

there are two possibilit ies. The downward load may be taken to be the weight of water and tank of the annular part i.e. W1 shown in Fig.(b)

Fig (a) Fig (b)

Fig. Water tank with spherical bottom.

Alterrnat ively, the downward load may be calculated from the weight of water and

tank bottom of the part i.e W2 less upward react ion of the support as shown in Fig.

For sect ion which cuts the tank bottom inside the support, the react ion has to be

considered with the weight of water and tank of the annular part. Simillar is the

case with Intze reservoir as in Fig. (a), which combines a truncated dome with a

spherical segment. Pattern of the two forces Nf1and Nf2 at point A are shown in Fig (b). To eliminate horizontal forces on the supporting ring girder, it is necessary that

Nf1cos a1 = Nf2cos a2.

Intze tank

11. DESIGN OF REINFORCED CONCRETE DOMES

The requirements of thickness of dome and reinforcement from the point of view o f

induced stresses are usually very small. However, a minimum of 80 mm is provided

so as to accommodate two layers of steel with adequate cover. Similarly a minimu

m of steel provided is 0.15% of the sectional area in each direct ion along the

meridians as well as along the lat itudes. This reinforcement will be in addit ion to the requirements for hoop tensile stresses.

The reinforcement is provided in the middle of the thickness of the dome shell Near

the edges usually some ring beam is provided for taking the horizontal component

of the meridian stress. Some bending moment develops in the shell near the edges.

As shown in Fig. it is normal to thicken the shell near the edges and provide

increased curvature. Reinforcements near the top as well as near the bottom face of

the shell are also provided. The size of the ring beam is obtained on basis of the

hoop tension developed in the ring due to the horizontal component of the meridian

stress. The concrete area is obtained so that the result ing tensile stress when

concrete alone is considered does not exceed 1.1N/mm2 to 1.70 N/mm2 for direct

tension and 1.5 N/mm2 to 2.40 N/mm2 for tension due to bending in liquid resist ing structure depending on the grade of concrete.

Reinforcement for the hoop stress is also provided with the allowable stress in steel

as 115 N/mm2 (or 150N/mm2) in case of liquid retaining structures and 140 N/mm2

(or190 N/ mm2) in other cases. The ring should be provided so that the central line

of the shell passes through the centroid of the ring beam. Reinforcement has to be

provided in both the direct ions. If the reinforcement along the meridians is

cont inued upto the crown, there will be congest ion of steel there. Hence, from

pract ical considerat ions, the reinforcement along the meridian is stopped below the

crown and a separate mesh, as shown in Fig (a), is provided. Alternat ively, the arrangement of the bars may be made as shown in plan in Fig (b)

In case of domes with lantern opening with concentrated load act ing there, ring

beam has to be provided at the periphery of the opening. The edge beam there will, however, be subjected to hoop compression in place of hoop tension.

Openings may be provided in the dome as required from other funct ional or

architectural requirements. However, reinforcement has to be provided all around

theopening as shown in Fig. (c). The meridian and hoop reinforcement reaching the opening should be well anchored to such reinforcement.

Intze tank

The allowable stresss specified in IS 3370 for such tanks are as follows:

Type of stresses: Permissible stress in N/mm2 High yield strength Plain bars

confirming to deformed bars as per Grade-I of IS 432-1966. IS 1786-1966 or is

1139-1966. Tensile stress in members under no table of contents entries found direct load.

Direct tensile stress in concrete a may be taken as 1.1 N/mm2, 1.2. N/mm2,1.32

N/mm2, 1.5 N/mm2, 1.6N/mm2 and 1.7 N/mm2 for M15, M20, MM25, M30, M35and

M40 respect ively, the value in tension due to bending i.e.,being1.5N/mm2,1.7N/mm2,1.82N/mm2,2.0 N/mm2,2.2 N/mm2 and 2.4 N/mm2.

When steel and concrete are assumed to act together for checking thetensile stress

in concrete for avoidance of cracks, the tensile streess in the steel will be limited by

the requirements that the stress as ment ioned above should not be exceeded. The

Intze tank

tensill stress in steel will be modular ratio mult iplied by the corresponding allowable tensile stress in concrete.

Stresses due to shrinkage or temperature change may be ignored if the permissible

stresses in concrete and steel are not exceeded and adequate precaut ions are taken

to avoid cracking of concrete during construction period, unt il the reservoir is put

into use and if it is assured that the concrete will never dry out. If it is required to

calculate shrinkage stresses, a shrinkage strain of 300×10-6 may be assumed.

When shrinkage stresses are considered, the permissible stresses may be

increasedby 33�

%.

When shrinkage stresses are considered it is necessary to check the thickness for no crack.

Minimum reinforcement of each of two directions at right angles shall have an areof

0.3% for 100 mm thick concrete to 0.2% for 450 mm thick concrete wall. In floor

slabs, minimum reinforcement to be provided is 0.15%. The minimum

reinforcement as specified above may be decreased by 20%), if high strength deformed bars are used.

Minimum cover to reinforcement on the liquid face is 25 mm or diameter of the bar,

whichever is larger and should be increased by 12 mm for tanks for sea water or liquid of corrosive character.

Intze tank

12. OVERHEAD WATER TANKS AND TOWERS

Overhead water tanks of various shapes can be used as service reservoirs, as a

balancing tank in water supply schemes and for replenishing the tanks for various

purposes. Reinforced concrete water towers have dist inct advantages as they are not

affected by climat ic changes, are leak proof, provide greater rigidity and are adoptable for all shapes.

Components of a water tower consists of-

(a) Tank portion with

(1) Roof and roof beams (if any) (2) sidewalls

(3) Floor or bottom slab (4) floor beams, including circular girder

(b)Staging portion, consist ing of

(5) Columns (6) Bracings and

(7)Foundat ions

Types of water Tanks may be

(a) Square open or with cover at top (b) Rectangular open or with cover at top (c) Circular open or with cover at which may be flat or domed.

Among these the circular types are proposed for large capacit ies. Such circular

tanks may have flat floors or domical floors and these are supported on circular girder.

The most common type of circular tank is the one which is called an Intze Tank. In

such cases, a domed cover is provided at top with a cylindrical and conical wall at

bottom. A ring beam will be required to support the domed roof.A ring beam is also

provided at the junct ion of the cylindrical and conical walls.The conical wall and

the tank floor are supported on a ring girder which is supported on a number of columns.

Usually a domed floor is shown in fig a result of which the ring girder supported on

the columns will be relieved from the horizontal thrusts as the horizonal thrusts of

the conical wall and the domed floor act in opposite direct ion.

Intze tank

Sometimes, a vert ical hollow shaft may be provided which may be supported on the domed floor.

The design of the tank will involve the following.

(1) The dome: at top usually 100 mm to 150 mm thick with reinforcement along themeridians and lat itudes. The rise is usually l/5th of the span.

(2) Ring beam supporting the dome: The ring beam is necessary to resist

thehorizontal component of the thrust of the dome. The ring beam will bedesigned for the hoop tension induced.

(3) Cylindrical walls: This has to be designed for hoop tension caused due tohorizontal water pressure.

(4) Ring beam at the junction of the cylindrical walls and the conical wall:This

ring beam is provided to resist the horizontal component of the react ion of the

conical wall on the cylindrical wall.The ring beam will be designed for theinducedhoop tension.

(5) Conical slab: This will be designed for hoop tension due to water pressure.The

slab will also be designed as a slab spanning between the ring beam at top and the ring girder at bottom.

(6)Floor of the tank.The floor may be circular or domed. This slab is supportedon the ring girder.

(7) The ring girder: This will be designed to support the tank and its

contents.Thegirder will be supported on columns and should be designed for result ing bending moment and Torsion.

(8) Columns: These are to be designed for the total load transferred to them. The

columns will bebraced at intervels and have to be designed for wind pressure or seismic loads whichever govern.

(9)Foundations: A combined footing is usuals provided for all supporting columns.

When this is done it is usual to make the foundat ion consist ing of a ring girder and acircular slab.

Suitable proportions for the Intze.

for case(1) suggested by Reynolds. Total volume ~0.585D3

for case (2), the proportion was suggested by Grey and Total Volume is given by

Intze tank

V1 = p�� × � = 0.39� . for H = D/2.

V2 = �.�

��×(�� + �� + �) = 0.102D3.

V3 = ��

�(3�� + ℎ�

�) = 0.017D3.

With h1 = 3/25D and r = 0.0179D3.

Volume V = 0.4693D3.

With h1 = D/6 and r = 3/10D.

Volume V = 0.493D3.

Intze tank

13. DESIGN

13. DETAILS OF DESIGN:

Design of tank:

Design of an intze tank for a capacity of 300000 lts .

Assuming height of tank floor above the ground level is 17.3m.

Safe bearing capacity of soil 200kn/m2

Wind pressure as per IS875 1200N/m2

Assuming M20 concrete

For which σcbe = 7N/mm2, σcc = 5N/mm2

Direct tension σt = 5N/mm2

Tension in bending = 1.70 N/mm2

Modular rat io m = 13

For Steel stress,

Tensile stress in direct tension =115 N/mm2

Tensile stress in bending on liquid face =115 N/mm2 for t < 225 mm

and 125 N/mm2 for > 225 mm.

Solut ion: Taking the volume as 0.585 D3 for proportion given in Fig.

D = 9.0 m. The dimension of the Tank is shown in fig.

Intze tank

Design of Roof Dome:

Considering a rise of 1.80 m, radius of the roof dome is given from

1.80(2R-1.80) = (4.75)2

R = 6.525m.

Sin φ = (4.5)/6.525= 0.7241

and φ= 43.36< 51.8°

Hence no tension

Assuming t = 100mm.

Hoop stress @ level of springing:

f =��

�[cos� −

�

��"#$%]

=&�&'×�∙&�&

'∙�&[0.72−

�

�.)�]

f =0.0298 N/mm2

Hoop stress @ Crown:

�=0°

Intze tank

f =+,&'×�.&�&

'.�&[1−

�

�]

f =0.107 N/mm2

Meridional thrust @ level of sprining:

T =��

�-"#$%

=+,&'×�.&�&

�-'.)�

=18778.34 N/m

Compressive stress

=�.)).. +

�&'×�'''

=0.125 N/mm2 provoide 8mm

Ring beam @ top :

Horizontal component of T= Tcos �

=13520.40 N/m

Hoop stress in the ring beam

=14339.82×,

�

=60841.82

Area of steel required

=�+&�,.�

'.,� '

=311.73 mm2

We have to provide 12 mm ⱷ,4 bars of 452.38 mm

Size of the ring beam:

Let the area of the ring beam sect ion = A mm2

Equivalent concrete area = /0+(m-1)/1�

=/0+(13.33-1)452.38

Intze tank

= /0+5577.8454

Limit ing tensile stress on the eqvivalent concrete area to 2 N/mm2

Cylindrical wall:

Pressure intensity at the bottom of cylindrical wall = 4×9810

=39240 N/mm2

Consider bottom strip of the wall as 1 mm.

Hoop tension = 39240×,

�

= 176580 N

Ast= �)�&'

� '×'.,

= 853.04 mm2

Provide 8 bars of 12 mm diametre of 142.85 mm distance.

Thickness of the wall may be kept as 200 mm.

Distribut ion steel = '∙�+

�''[200×1000]

= 480 mm2

Provide 8 mm diametre bars.

= +.'

��× .��

Provide 10 mm diametre bars of spacing 100 mm between them.

Check for compressive stress at the bottom of the cylindrical wall.

Vertical component of T1 = V1 = T1sin �

= 24917 × 0.68

= 17184.137 N/m.

Weight of the wall = 0.2×4×25000

= 20000 N/m.

Intze tank

Weight of ring beam = 0.2×0.2×25000

= 1000 N/m.

Total load V2 =38184.137 N/m.

Compressive stress = .�.+.� )

�''×�'''

= 0.19 N/mm2

Nominal vert ical stress is equql to 0.24% of gross area.

Vertical steel = '.�+

�''× 200 × 100

= 480 mm2

Provide 10 bars of 8 mm diametre of spacing 100 mm.

Ring beam at B :

Let T2 be the thrust /m run exerted by the conical wall at the junct ion B.

Resolving vert ically at B

T2sin 6= V2

tan 6= �.&

�.&

= 1

6 = 45°.

T2 = 9�

$:;<

= .�.+.� )

$:;+&°

= 54000.52 N/m.

Resolving horizontally at B

H2 =T2cos6=54000.52× cos45°

= 38184.137 N/m

Intze tank

This horizontal load H2 will produce a hoop tension in ring beam B

Hoop tension due to H2 =H2×?

�

=38184.137×,

�N

=171828.6165N

Let the rinmg beam be 500mm deep

Water pressure on the ringh beam

=9810× 4 ×&''

�'''

=19620 N/m

Hoop tension due to water = 19620×,

�

=88290 N

Total hoop tension = 88290 +171828.61

= 260118.61 N

Steel for hoop tension =��'��..��

� '×'.,

= 1256.611mm2

Provide 6 bars 18 mm ∅

Ast = 1526.81 mm2.

Let ‘A’ be the area of ring beam

Equivalent concrete area = A+(m-1)Ast

= A+(13.33-1)× 1526.81

= A+18825.61

Limit ing the tensile stress on the equivalent concrete area to 2 N/mm2

��'��..��

B-�&�...'�+ = 2

Ac =11233.688 mm2

Intze tank

Provide 250× 500mm size

Design of conical slab:

Conical slab should be designed for

a) Hoop tension

b) Bending as it spans on a sloping slab from the ring beam @ B at the ring girder @ ‘c’

Design for hoop tension:

CD-CE

�F +

CD

�Ftan 6

Where

Ww= weight of water resting on the conical slab.

Ws = weight of the conical slab.

6 = inclinat ion of the conical slab with the horizontal.

Area of water sect ion standing on the conical slab

= &.&-+

�× 1.5 = 7.125 m2.

X = �-[

�.�GH

]

).��& = 0.52 m.

Weight of water resting on the conical slab Ww = 9810×7.125×2J[3.52]

= 1545882.24 N

Length of conical slab = 2.121 m.

Take thickness of the slab as 200 mm.

Weight of the conical slab Ws = 0.2×2.121×25000×2J[).&

�] = 249874.42 N.

Hoop tension = �&+&..�.�+-�+,.)+.+�-�&+&..�

�F

=531838.349 N.

Hoop steel on the ent ire sect ion = & �. ..+,

� '×'.,

Intze tank

= 2569.267 mm2.

Provide 14 bars of 6 mm ∅

=14× J ×64 = 2814.86 mm2.

Design for bending moment:

Load per metre width of the conical slab = �D-�E

�F×KLMNOM?PQ1

= �&+&..�.�+-�+,.)+.+�

�F× .)& = 76214.279 N.

Maximum bending moment = �R

. =

)��+.�),×�.&

.

= 14290.177 Nm.

Axial compression V2 = T2sin 6 = .�.+.� )

$:; +&°

= 54000.52 N.

Providing 16 mm diametre bar at clear covers of spacing 25 mm.

Effect ive depth = 200−25 − 8 = 167 mm.

Distance between centre of sect ion and centre of steel x = d−�

� = 167−100

= 67 mm

Resultant bending moment = M+T2.x =14290.177× 10 + 54000 × 67

= 17908212.15 Nmm.

Ast = �),'.��.�&

��)×� '×'., = 518.04 mm2

Spacing of 16 mm diameter bars = 333.33 mm and provide 3 bars.

Intze tank

The bottom dome:

Let R be the radius of the dome,then 32 = 1.2(2R−1.2)

= 4.35 m.

Let 2� be the angle subtended by the dome.

sin � =

+∙ & = 43°36°°

cos� = 0.68

Thickness of dome = 200 mm.

Loads:

Dead load = 25000×0.2 = 5000 N/mm2.

Weight of water resting on the dome = VC[J��h−F�W

(3R−ℎ0)]

=9810[155.508−17.869] = 1350234.872

Area of dome surface = 2JRh = 2J × 4.315 × 1.2

Intze tank

= 32.79 m2.

Load intensity due to weight of water = � &'� +..)

�.), = 41178.25 N/m2.

Total load intensity = 5000+41178.25 = 46178.25 N/m2.

Meridional thrust = ��

�-"#$% =

+��)..�&×+. &

�-'.)�

= 116788.016 N/m.

Meridional compressive stress = ��)...��

�''×�''' = 0.583 N/mm2.

Hoop stress = ��

�[cos� −

�

�-"#$ %]

= +��)..�&×+. &

'.�''[0.72−

�

�.)�]

= 0.139 N/mm2.

Hoop stress at the crown � = 0°.

Maximum hoop stres = ��

�[cos� −

�

�-"#$ %] = 502188.46

= 0.502 N/mm2.

These stresses are low and hence provide nominal 0.3% steel.

Provide 8 mm ∅ bars @100 mm spacing.

Circular girder:

The total load on the circular girder consists of the following;

Total weight of water W1 = weight of water on conical slab + weight of water on dome.

= 1545882.24+1350234.872 = 2896117.112.

Weight of dome + cylindrical wall + ring beam at A W2 = 38184.137×2J ×4.5

= 1079631.039 N.

Weight of ring beam at B W3 = 0.25×0.5×25000× 2J ×4.5

= 88357.29 N.

Weight of conical wall W4 = 249874.42 N.

Intze tank

Weight of lower dome W5 = 5000×32.79 = 163950 N.

= 0.0075×4560396.668×3 = 102608.925 N.

Torsion = 0.0015×W×r = 20521.785 N. (from table 2)

Angular distance for maximum torsion = 12°44°°.

Let us provide 8 coloumns.

Bending momment at the support = 0.0083×W×r = 0.0083×4591027.197×3

= 114316.577 Nm.

Bending moment at centre = 0.00416×W×r = 0.00416×4591027.197×3

=57296.01 Nm.

Torsion = 0.0006×4591027.197×3 = 8263.84 Nm.

Angular distance for maximum torsion = 9°33°°.

Load at each support = �

. =

+&,+&��.++..

.

= 573878.39 N.

Shear force at the support = �

� , V = 286939.199.

Design at support section:

Equating moment of resistance to the bending moment at support

0.913bd2 = 114316.577×1000,

0.913×400×d2 = 114316.577×1000,

Then d2 = 278458.26, d =560 mm.

Let the clear cover be 40 mm.

Over all depth of beam = 600 mm.

Actual effect ive depth = 600 mm.

Intze tank

Equivalent shear force = V+1.6X

Y = 286939.199+1.6

X

Y.

= 287160.093+(�.�×.�)'.��×�'''

+'').

Vc = 319994.559.

Equivalent nominal shear stress �ve = 9ZY?

= ��&�&.�.

+''×&�' = 1.42 N/mm2.

Maximum shear stress �ma x> �v.

�ma x= 1.8 N/mm2.

�c< �v.

Provide longitudinal and transverse reinforcement according to B-6.4

Longitudinal reinforcement:

Me = M+Mt , Mt = X(�-

^_)

�.) =

.�� ..+[�-`aa�aa

]×�'''

�.)

= 12152705.88 Nmm.

M = moment at crosssect ion.

Mer = 1000×114316.577+12152705.88 = 126469282.9 Nmm.

Ast= bZc

� '×'.,×&�' =

��&��& ��

� '×'.,×&�'

= 1080.187 mm2.

Transverse reinforcement:

Asv= X∙1d

Y�?�eEd+

9∙1d�.&?�eEd

, b1 = 400−80 = 320 mm , d1 = 600−80 = 520 mm.

Asv = [ .�� ..+×�'''

�'×&�'×� '+

�.�,.�,,

�.&×&�'×� ']Sv

Providing 4 legged 10 mm st irrups.

Asv= 315 mm2 , 315 = 1.175, Sv = 267.95 mm.

Take Sv as 250 mm.

[ fdZ�fWeEd

]b×Sv , �.+��'.�.

� '×400×Sv = 315 , Sv= 158.88 mm.

Intze tank

Provide 150 mm spacing.

Steel for sagging moment = &)�,�.'�×�'''

� '×'.,×&�'

= 494.27 mm2.

Provide 5 bars of 12 mm diameter.

Ast = 565.48 mm2.

Hoop stress:

Tc = thrust exerted by the conical slab on the girder.

Tcsin 6 ×2Jr = Ww+Ws+weight of cylindrical wall and upper dome.

Tcsin 6 ×2Jr = 154588.24+249874.42+1079631.039

Tcsin 6 ×2Jr = 2875387.699.

Tc = �.)& .).�,,

�F× ×$:; +&° = 215729.87 N.

Horizontal component of Tc = 215729.87× cos45°, H1 = 152544.055 N.

Horizontal component due to dome = 11678.016× cos 43°36′, H2 = 84574.59,

H1−H2 = Net,Net = 67969.46 N. Hoop stress = 67969.46×3 = 203908.38 N.

Hoop compressive stress = �' ,'.. .

+''×�'' = 0.849 N/mm2.

Coloumns:

Coloumns should be designed for direct loads coming upon them and for the bending moments caused by wind load.

Vertical load on one column at top = +&,�'�).�,)

.

= 573878.399 N.

Let 6 be the inclinat ion of the column with the vert ical.

tan 6= �

�' , 6= 5°42′ , sin 6 = 0.0995, cos 6 =

�'

√�'� = 0.995.

Actual length of column = √10� + 1� = 10.05 m.

Providing 300 mm × 300 mm column.

Intze tank

Wt. Of column =10×0.3×0.3×25000

= 22500 N

Total vert ical load = 573878.399+22500 N

= 596378.399 N

Corresponding axil load =&,� ).. ,,

'.,,&

= 59375.2754 N

When tank is full = 599375.2754 N

Wt. Of water in tank =�.,��).��

.

=3620124.639 N on each column

Vertical load on each column when tank is empty

= 596378.399−362014.239

= 237361.036 N

Corresponding axial load= ��+),..��

'.,,&

= 238553.805 N

Ignoring wind load effect if the steel requirement is Asc

Then cAc + tAsc =599375.275 N

5×Ac + 190 ×Asc =599375.275

5[400× 400 − /sc] +190×Asc =599375.275

Asc =807.433 mm2.

Min. Requirement of steel = 0.8% ='..

�''[300×300]

=720 mm2

Provide 6 bars of 20mm dia. =1884 mm2

More steel has been subjected since column is subjected to B.M caused by wind load.

Intze tank

Analysis due to wind pressure:

Wind pr. =1200 mm2.

Wind force on the top dome & cylindrical walls =(4+�..

�)×9.4×1200

@Ht=13.95 =55272 N

Wind force on the circular wall =,.+-�.+

�×1.5×01200

=14220 N

Wind force on circular girder =0.6×6.4×1200

=4608 N

Wind force on column & braces =5×0.3×10×1200+3×�-.

�×0.3×1200

=25560 N

Total moment of wind pr. About the base

=55272×13.95+14220×0.8+4608×10+25560×5

=10982500Nm.

Vetrical load on any column due to wind load =bj

∑j^�

∑m2 =2×42 +4(+

√�)2 =64m2

Max. Wind load force in the most leeward side &the most windward side.

= �',.&''.+×+

�+ =68656.275 N

Max. Wind force in columns marked 5

=�',.&''.+

�+×

+

√� =48547.317 N

Consider the windword column 1

Vertical load due to dead +wind load

=596378.399 +68656.275 N

Intze tank

=665034.674 N.

Corresponding axial load =��&' +.�)+

'.,,&'

=668376.556 N

Horizontal comoponent of the axial forces caused by wind act ion

=2×68456.275×0.0995+4×48547.317× 0.0995 ×�

√�

=27285.39 N.

Aactualhorizontal force @ base

= 55272+14220+4608+25560−27285.39 = 72374.61

Horizontal shear column = )� )+.��

. = 9046.826 N.

Maximum bending moment for the column = 9046.826×�.&

� = 11308.532 N.

Analysis of column section:

Direct load = 668376.556 N.

Bending moment = 11308.532 Nm.

Provide 300×300 column.

Provide 6 bars of 20 mm diameter at effect ive cover of 50 mm.

Ast = 1884 mm2,

Equivalent concrete area = Ac+(m-1)Ast = (300×300)+(12.33×1884)

= 113229.72 N

Polar moment of inert ia of the equivalent concrete sect ion,

= M�

��+(mAst×effect ive depth fromcentre),

= ''�

�+1884×12.33[150-50]2 = 1.582×109 mm4.

Equivalent moment of inertia about full section = �.&.�×�'n

� = 791.14×106mm4.

Intze tank

Direct stress in concrete = ?POL0�RoM?

LpQPqMRLN�0oN0OL�LMOLM = 5.9 mm2.

Bending stress in concrete = �&'×�� '..& �×�'''

),�.�+×�'` = 2.14 N/mm2 .

Maximum stress = 5.9+2.14 = 8.04 N/mm2.

Design of braces:

Moment in brace BC = 2×moment for the column× sec 45°,

= 2×11308.532× √2 = 31985.358 Nm.

Provide 300×300 mm bar sect ion and a doubly reinforced beam with equal steel at

top and bottom.

Ast = Asc = �,.&. &.×�'''

� '×��'×'., = 702.357 mm2.

Provide 4 bars of 18 mmdiameter at top and equal amount at bottom.

Shear force for brace = YLN?PNsKoKLN�toOYOM0L

��1�MNotYOM0L

,

Span of brace =2 ×)

�× sin 22°30 ′ = 2.678 m.

Shesr force for brace = �,.&. &.��×�.�).

= 23887.49 N.

Nominal shear stress �v = 9

Y? =

� ..).+,

''×��'

= 0.30 N/mm2.

Provide nomonal st irrups say 2 legged 10 mmdiameter st irrups at 200 mm clear cover.

Intze tank

Design of foundation:

Total load on the column = 599375.2754×8 = 4795002.203 N.

Approximate weight of foundat ion is 10% of column loads.

= 479500.22 N.

Then total load is equal to 5274502.22 N.

Safe bearing capacity of 200 KN/m2 ,

Area = RoM?

uvw =

&�)+&'�.+�

�''×�'H = 26.37 m2.

Let us provide outer dia of 9.5 m and inner dia of 6.5 m.

= F

+[9.52−6.52] = 37.69 m2.

Net intensity = &�)+&'�.+�

).�,

Intze tank

= 139.9 KN/m2.

139.9 KN/m2< 200 KN/m2.

Design of circular girder:

Maximum bending moment occurs at support = 0.00416×W×r = 11508.005 Nm.

Maximum bending moment occurs at support = 0.0083× 4795002.203×4

= 159194.073 Nm.

Maximum torsion = 0.0006×W×r = 11508.005 Nm.

Maximum shear force at support = +),&''�.�'

�×. (from table 2)

= 299687.63 N.

Design at support sect ion;

Moment of resistence = maximum bending moment at support.

0.913bd2 = 159194.073×1000 , bd2 = 174363716.30 ,

d = 590 mm ,clear cover = 60 mm , D = 650 mm.

Equivalent shear stress Vv = V+1.6X

Y = 299687.63+1.6

��&'..''&×�'''

&'' ,

= 336550.0176 N.

Equivalent nominal shear �v = 9dY?

= 1.14 N/mm2 , but �c = 1.8N/mm2 ,

Hence �c< �v .

Longitudinal reinforcement:

Mel = M+Mt , Mt = X(�-

^_)

�.) =

��&'..''&[�-`GaGaa

]×�'''

�.)

= 15569653.82 N ,

Mel = 1000[159194.073+15569.653] = 174763.726×1000 N.

Ast = �)+)� .)��×�'''

� '×'.,×&,'

= 1430.964 mm2 ,

Intze tank

Provide 9 bars of 16 mm diameter bars.

Hence area os steel required is Ast = 1809.55 mm2 .

Transverse reinforcement:

Asv= X∙1d

Y�?�eEd+

9∙1d�.&?�eEd

, providing 4 legged 10 mm diamater of st irrups.

Asv = 4J ×52= 314 mm2, b1 = 500-80 = 420 mm , d1 = 650-120 = 530 mm,

314 = ��&'..''&×�'''

+�'×& '×� '+

�,,�.)..

�.&×& '×� ' , 314 = Sr[0.224+0.983] , Sv = 260 mm.

Let us provide 200 mm clear cover spacing.

Steel for hogging mommentAst = ),).... �×�'''

� '×'.,×&,'

= 653.31 mm2 ,

Provide 4 bars of 16 mm diameter.

Design of bottom slab:

Provide a cant ilever project ion beyond the face of the beam = 0.6 m.

Maximum bending moment for 1 m wide st irup = 139944.346×'.��

�Nm ,

= 2518.98 Nm.

Equating moment of resistence to bending moment ,

0.913×bd2 = 25189.98×1000 , b = 1000 mm.

Then d2 = 27590.339 , d = 166.1 mm.

Let us provide 170 mm effect ive depth and 40 mm clear cover.

D = 210 mm. Ast = �&�.,.,.×�'''

� '×'.,×�)' = 715.82 mm2.

Provide 4 bars of 18 mm diameter. Ast = 1017.87 mm2,and spacing of the bars is 250 mm clear cover.

Distribution steel:

Intze tank

Provide 0.12 % steel and the steel required is = '.��×��'×�'''

�''

= 252 mm2.

Provide 6 bars of 8 mm diameter bars and spacing = �'''

� = 160 mm clear cover.

Check for sliding:

Total load on the foundat ion when tank is empty = 5274502.423-2896117.112

= 2378385.311 N

Horizantal force on the base = 72374.61 N.

Let coefficient of frict ion = 0.5

Fs= '.&×� ). .&. ��

)� )+.�� = 16.43.

Intze tank

14. ESTIMATION

14.1 Detailed estimation:

Detailed est imate is an accurate est imate and consists of working out the quant it ies

of each item of works, and working the cost. The dimensions, length, breadth and

height of each item are taken out correctly from drawing and quant it ies of each item are calculated, and abstract ing and billing are done.

The detailed est imate is prepared in two stages:

Details of measurement and calculat ion of quant it ies.

The details of measurements of each item of work are taken out correctly from plan

and drawing and quant it ies under each item are calculated in a tabular form named as details of measurement form.

Abstract of est imated cost:

The cost of each item of work is calculated in a tabular form the quant it ies already

computed and total cost is worked out in abstract est imate form. The rates of

different items of work are taken as per schedule of rates or current workable rates for finished item of work.

Detailed estimation:

S.No

DECRIPTION OF WORK

NOS

L m B m

A m2 HorD (m)

QTY m3

REMARKS

1

Earthwork in excavat ion

1 73.89 1 73.89 / = J��/4 =J × 9.72/4 =73.89

Earthwork in filling

1 64.316

2 a)R.C.C work in foundat ion b)steel in foundat ion i )Longitudanal ii)Transverse

1 9 4

7.068 J ×0.0082 0.0082× J

0.2 1.4136 0.045 0.02

= J ×8

Intze tank

3 R.C.C in columns Steel in columns

8 8×6

0.3 0.3 0.09 J ×0.012

10.049 10.049

7.235 0.151

4 RCC in [email protected] Steel in Bracings @2.5m from G.L

8 8×8

0.6375 0.6375

0.3 0.09 J ×0.0092

0.3

=0.459 =0.01

A=J ×0.0092 =0.000254

5 a) R.C.C in bracings @5m from G.L. b) Steel

8 8×8

0.575 0.575

0.3 0.09 J ×0.0092

0.3 0.414 0.00936

A=J ×0.0092 =0.000254

6 a)RCC in bracings 7.5m b)steel

8 8×8

0.45 0.45

J ×0.0092

0.324 0.0072

A=J ×0.0092 =0.000254

7 Top ring girder a)R.C.C b)steel longitudinal transeverse

1 5 125

JD 6J 6J

0.4 0.24 J ×0.0092

0.6 4.52 0.02 0.066

8 Bottom dome a)RCC in dome b) steel

1

L=6.62

=22.619 0.067

0.2

4.523 0.443

A=2Jrh =2J ×3×1.2 =22.619

y =j

�'°× 2Jr

=6.62

9 a)RCC conical slab b)steel steel for B.M.

1 14 3

23.56 23.56 23.56

0.2 J ×0.0082

2.121

9.994 0.066 0.014

J[�1 + �2]2

= J(9+6)/2

10 a)RCC ring beam @ B b)steel

1 6

28.27

0.25 J ×0.0082

0.5

3.534 0.034

11 Cylindrical wall a)Main steel b)Distribut ion steel

1 20 4

4.32 28.27

0.2 J ×0.0062

J ×0.0042

4 22.619 0.098 0.0056

L=4+16d =4+16×0.012 =4.32

12 Ring beam @ A a)concrete

1

9J

0.2

0.04

0.2

1.13

L=JD =9J

Intze tank

b)Steel 4 9J J ×0.0062 0.012 13 Top dome R.C.C

a) concrete b) Steel

1 100

9.93

2Jrh =50.89 J ×0.0042

0.150

7.63 0.05

A=2Jrh = y =

j

�'°× 2Jr

12 Total RCC work 63.795

6

13 Total steel 1.017 14 Plastering in CM

(1:2) for Inner surface Of conical dome (12mm)

1 50.89 9.15 A=2Jrh =50.89

15 Plastering in CM (1:6) for outer surface Of conical dome (12mm)

1 55.135 9.92 A=2Jrh =55.135

16 Plastering in CM (1:2) for Inner surface Of cylindrical wall (12mm)

1 J ×D 28.2 112.8 4 20.354

17 Plastering in CM (1:6) for outer surface Of cylindrical wall (12mm)

1 J ×D 29.5 118.82 4 28.349

18 Plastering in CM (1:2) for Inner surface Of domed roof (12mm)

1 22.619 4.07 A=2Jrh =2J ×3×1.2 =22.619

19 Plastering in CM (1:6) for outer surface Of domed roof (12mm)

1 26.38 4.74 A=2Jrh =2J ×3×1.4

20 Plastering in CM (1:6) for columns (12mm)

8 0.3 0.3 0.09 17.28

21

Plastering in CM (1:2) for ring beam at top (12mm)

1

9J

0.2

5.65 1.01

22 Plastering in CM (1:2) for ring beam at bottom (12mm)

1 1.27

Intze tank

23 Plastering in CM (1:6) for bracings at 2.5m ht.(12mm)

1 0.27

24 Plastering in CM (1:6) for bracings at 5m ht.(12mm)

1 0.24

25 Plastering in CM (1:6) for bracings at 7.5m ht.(12mm)

1 0.19

26 Plastering in CM(1:2) for inner surface of conical slab(12mm)

1 4.239 J[�1 + �2]2

= J(9+6)/2

27 Plastering in CM(1:6) for outer surface of conical slab(12mm)

1 4.46

28 Total plastering 105.533

29 Thick water proof cement paint ing for tank portion

85.278

30 white washing for columns

8 0.312 0.312

10.04 7.826

31 Total white washing

93.104

Intze tank

ABSTRACT

S.NO

DESCRIPTION OF WORK

QTY OR NOS

RATE RS PS

COST RS PS

1

Earth work in excavat ion

73.89

2 Beldars 13 250 3250

3 Mazdoors 11 250 2750 4 Total 6000 5 Earth work in

Filling In foundat ion 64.316

6 Beldar 7 250 1750 7 Bhisthi 2 285 570

8 Mazdoors 5 250 1250 9 Total 3570 10 Total earth work in

Filling

11 Disposal of surplus earth in a lead 30m

9.574

12 Mazdoor 4 250 1000 13 Total 1000 Total cost of earth

work

10,570

Intze tank

14.2 DATA SHEET:

RCC M- 20 Nominal mix (Cement:fine aggregate: coarse aggregate) corresponding to Table 9 of IS 456 using 20mm size graded machine crushed hard granite metal (coarse aggregate) from approved quarry including cost and conveyance of all materials likecement FOUNDATION

A. MATERIALS UNIT QTY RATERS

AMOUNT RS

20mm HBG graded metal Cum Cum 0.601 1076 646.676 Sand Cum 1.2 375 450 Cement Cum 0.4 1620 648 1st Class Mason Day 0.38 285 108.3 2nd Class Mason Day 1.03 285 293.55 Mazdoor (Both Men and Women) Day 2.33 250 582.5 Concrete Mixer 10/7 cf (0.2/0.8cum)capacity

Hour

1

250

250

Cost of Diesel for Miller Liter 0.52 45 23.4 Cost of Petrol for Vibrator Liter 0.75 68 51 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 629.16

Add MA 20% 629.16

Add TOT 4% 176.166

BASIC COST per 1 cum 4580.31

Intze tank

COLUMNS


AMOUNT RS

20mm HBG graded metal Cum Cum 6.156 1076 6623.85 Sand Cum 3.078 375 1154.25 Cement Cum 2.052 1620 3324.24 1st Class Mason Day 1.99 285 567.15 2nd Class Mason Day 5.26 285 1499.1 Mazdoor (Both Men and Women) Day 11.96 250 2990 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

1

250

250

Labour centering Cum 1 971 971 Material hire charges for centering Cum 1 89 89 Water (including for curing) Ki 1.2 77.0 92.4 Add 20% in Labour (1st Floor) 2912.198

Add MA 20% 2912.198

Add TOT 4% 582.43


RCC RING BEAM AT TOP


AMOUNT RS

20mm HBG graded metal Cum Cum 0.96 1076 1032.96 Sand Cum 0.48 375 180 Cement Cum 0.32 1620 518.4 1st Class Mason Day 0.31 285 88.35 2nd Class Mason Day 0.83 285 236.55 Mazdoor (Both Men and Women) Day 1.86 250 465 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

0.26

250

65


Add MA 20% 747.73

Add TOT 4% 149.54


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RCC DOMED ROOF 150mm THICK


AMOUNT RS

20mm HBG graded metal Cum Cum 6.48 1076 6972.48 Sand Cum 3.24 375 1215 Cement Cum 2.16 1620 3499.2 1st Class Mason Day 2.1 285 598.5 2nd Class Mason Day 5.6 285 1596 Mazdoor (Both Men and Women) Day 12.6 250 3150 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

0.267

250

66.75


Add MA 20% 5558.33

Add TOT 4% 1111.61


CONICAL SLAB 200mm THICK


AMOUNT RS

20mm HBG graded metal Cum Cum 8.49 1076 9135.24 Sand Cum 4.25 375 1593.75 Cement Cum 2.83 1620 4584.6 1st Class Mason Day 2.75 285 783.75 2nd Class Mason Day 7.34 285 2091.9 Mazdoor (Both Men and Women) Day 16.52 250 4130 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

0.26

250

65


Add MA 20% 5555.328

Add TOT 4% 1111.06


Intze tank

RCC CYLINDRICAL WALL


AMOUNT RS

20mm HBG graded metal Cum Cum 19.23 1076 20691.48 Sand Cum 9.62 375 3607.5 Cement Cum 6.41 1620 10384.2 1st Class Mason Day 6.23 285 1775.55 2nd Class Mason Day 16.62 285 4736.7 Mazdoor (Both Men and Women) Day 37.39 250 9347.5 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

0.26

250

65


Add MA 20% 10352.066

Add TOT 4% 2070.432


RCC RING BEAM AT BOTTOM OF

CYLINDRICAL WALL


AMOUNT RS

20mm HBG graded metal Cum Cum 3 1076 3228 Sand Cum 1.5 375 562.5 Cement Cum 1 1620 1620 1st Class Mason Day 0.97 285 267.45 2nd Class Mason Day 2.59 285 738.15 Mazdoor (Both Men and Women) Day 5.84 250 1460 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

0.26

250

65


Add MA 20% 1818.7

Add TOT 4% 363.74


Intze tank

RCC CIRCULAR GIRDER


AMOUNT RS

20mm HBG graded metal Cum Cum 3.84 1076 4131.84 Sand Cum 1.92 375 720 Cement Cum 1.28 1620 2073.6 1st Class Mason Day 1.24 285 353.4 2nd Class Mason Day 3.32 285 946.2 Mazdoor (Both Men and Women) Day 7.47 250 1867.5 Concrete Mixer 10/7 cf (0.2/0.8cum) capacity

Hour

0.26

250

65


Add MA 20% 2262.58

Add TOT 4% 452.517


RCC BRACING AT 2.5m HT.


AMOUNT RS


Hour

0.26

250

65


Add MA 20% 447.213

Add TOT 4% 89.44


Intze tank

RCC BRACING AT 5m HT.


AMOUNT RS


Hour

0.26

250

65


Add MA 20% 426.32

Add TOT 4% 85.264


RCC BRACING 7.5m HT.

A. MATERIALS UNIT QTY RATE RS

AMOUNT RS


Hour

0.26

250

65


Add MA 20% 384.91

Add TOT 4% 76.98


Intze tank

Plastering with

CM(1:6)&(1:2),12 mm thick

Cement Mortor 1:6 1:2

cum cum

105.533 65.44 40.09

552 780

36165 31673

Mason 1st class day 39 285 11115

Bhisthi day 14 285 3990 Mazdoor (unskilled) day 39 250 9750 Add MA 20% 18539 Add TOT 4% 3719 Grand Total 114951

Paint ing to new walls of tank portion with 2 coats of water proof cement paint of approved brand and shade over a base coat of approved cement primer grade I making making 3 coats in all to give an even shade after thourughly brushing the surface to remove all dirt and remains of loose powdered materials, including cost and conveyance of all materials to work site and all operational, incidental, labour charges etc. complete for finished item of work as per SS 912 for walls Epoxy primer for Hibond floor & protective coat ings : Procoat SNP2 or Zoriprime EFC 2

Pack 26 548 14250

1st class painter Day 4 355 1420 Mazdoor Day 4 250 1000 cost of water proof cement paint Cum 50 35 1750 1st class painter Day 2 355 710 Mazdoor (unskilled) Day 2 250 500

Add MA 20% 3926

Add TOT 4% 786

Total cost 24342

Intze tank

Paint ing to new columns of tank portion with 2 coats of water proof cement paint of approved brand and shade over a base coat of approved cement primer grade I making making 3 coats in all to give an even shade after thourughly brushing the surface to remove all dirt and remains of loose powdered materials, including cost and conveyance of all materials to work site and all operational, incidental, labour charges etc. complete for finished item of work as per SS 912 for walls

Cost of cement primer Pack 18 100 1800 1st class painter Day 1 355 355 2nd class painter Day 1 250 250 cost of water proof cement paint Cum 6 35 210 1st class painter Day 1 355 355 Mazdoor (unskilled) Day 1 250 250

Add MA 20% 644

Add TOT 4% 129 Total cost 3993

Total cost of project:

Total cost of R.C.C = 2,23,930

Total cost of steel = 5,18,924

Total cost of plastering = 1,14,951

Total cost of paint ing = 28,335

Total cost of earthwork = 10,570

8,96,710

Intze tank

15. CONCLUSION

Storage of water in the form of tanks for drinking and washing purposes, swimming pools for exercise and enjoyment, and sewage sedimentat ion tanks are gaining increasing importance in the present day life. For small capacit ies we go for rectangular water tanks while for bigger capacit ies we provide circular water tanks. Design of water tank is a very tedious method. With out power also we can consume water by gravitat ional force.

Intze tank is constructed to minimize the project cost why because lower dome in

this construct ion resists the horizontal thrust.

Intze tank

16. REFERENCES

Table 16.2. Coefficients for moment in cylindrical wall fixed at base (As Per IS3370)

Moment = Coefficient (wH3) Nm/m

H2 Co efficient at points

DT 0.1 H0.2 H 0.3 H0.4 H 0.5 H 0.6 H 0.7 H 0.8H

0.4 + 0.0005 + 0.0014 + 0.0021 + 0.0007 - 0.0042 -0.0150 -0.0302-0.0529

0.8 + 0.0011 + 0.0037 + 0.0063 + 0.0080 + 0.0070 + 0.0023 + 0.0068 -0.0024

1.2 + 0.0012 + 0.0042 + 0.0077 + 0.0103 + 00112 + 0.0090 + 0.0022 -0.0108

1.6+ 0.0011 + 0.0041 + 0.0075 + 0.0107 + 0.0121 + 0.0111 + 0.0058 -0.0051

2.0+ 0.0010 + 0.0035 + 0.0068 + 0.0099 + 0.0120 + 0.0115 + 0.0075 -0.0021

3.0 + 0.0006 + 0.0024 + 0.0047 + 0.0071 + 0.0090 + 0.0097 + 0.0077 +0.0012

4.0 + 0.0003 + 0.0015 + 0.0028 + 0.0047 + 0.0066 + 0.0077 + 0.0069 +0.0023

5.0 + 0.0002 + 0.0008 + 0.0016 + 0.0029 + 0.0046 + 0.0059 + 0.0059 +0.0028

6.0 + 0.0001 + 0.0003 + 0.0008 + 0.0019 + 0.0032 + 0.0046 + 0.0051 +0.0029

8.0 0.0000 + 0.0001 + 0.0002 + 0.0008 + 0.0016 + 0.0028 + 0.0038 +0.0029

10.0 0.0000 + 0.0000 + 0 0001 + 0.0004 + 0.0007 + 0.0019 + 0.0029 +0.0028

12.0 0.0000 + 0.0000 + 0.0001 + 0.0002 + 0.0003 + 0.0013 + 0.0023 +0.0026

14.0 0.0000 0.0000 0.0000 0.0000 + 0.0001 + 0.0008 + 0.0019 +0.0023

16.0 0.0000 0.0000 -0.0001 - 0.0002 -0.0001 + 0.0004 + 0.0013 +0.0019

Table 1:

Intze tank

Permissible stresses in concrete

All values in N/mm2

Grade permissible stresses in compression permissible stress in bond

Of concrete for plain bars in tension Bending Direct (average)

zcbc zcc {bd

M 10 3.0 2.5 _

M 15 5.0 4.0 0.6

M 20 7.0 5.0 0.8

M 25 8.5 6.0 0.9

M 30 10.0 8.0 1.0

M 35 11.5 9.0 1.1

M 40 13.0 10.0 1.2

M 45 14.5 11.0 1.3

M 50 16.0 12.0 1.4

Table 1.1:

Grade of M10 M15 M20 M25 M30 M35 M40 M45 M50

Concrete

Tensile 1.2 2.0 2.8 3.2 3.6 4.0 4.4 4.8 5.2

Stress(N/mm2)

Table 2:

Moments for circular girders

For 8 columns B.M@ B.M@ Torsion

Support centre

0.0083Wr 0.00416Wr 0.0006Wr

Intze tank

17. REFERENCE BOOKS

• I.S 456:2000 for RCC.

• I.S 800:1984 for STEEL.

• I.S 872 Part I and Part II.

• I.S 3373 (Part IV-1967).

• Reinforced concrete structures (M.Ramamrutham).

• Element of environmental engineering (BIRIDI).

. Estimat ing, costing and evaluat ion (B.N.Datta).

. Standard schedule of rates (SSR)

Intze tank

intze tank design

Design

source of water

public water demand

classification of water

water quantity estimation

water consumption rate

estimation of intze

intze tank abstract

intze tank acknowledgement