Design & Estimation of Intze Tanks-Major Project Report

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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 irrespective 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 application. We would like to express our heart-felt gratitude 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 institution 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 inspiration 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 satisfaction and euphoria that accompany the successful completion of the task would be great but incomplete without the mention of the people who made it possible with their constant guidance and encouragement crowns all the efforts with success. In this context, 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 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 3 WATER DEMAND

6 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

5.1 CLASSIFICATION OF WATER TANKS 6 DESIGN REQUIREMENTS OF CONCRETE

6.1 JOINTS IN LIQUID RETAINING STRUCTURES

11 11 12 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 14 ESTIMATION

32 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 section A b = Equivalent area of helical reinforcement. A c = Equivalent area of section A h = Area of concrete core. A m = Area of steel or iron core. A sc = Area o f longitudinal reinforcement (comp.) A st = Area of steel (tensile.) A l = Area of longitudinal torsional reinforcement. A sv = Total cross-sectional are of stirrup legs or bent up bars within distance Sv A w =Area of web reinforcement. A Ф = Area of cross –section of one bars. a = lever arm. a c = Area o f concrete. B =flange width of T-beam. b = width. b r =width of rib. C =compressive force. c = compressive stress in concrete. c’= stress in concrete surrounding compressive steel. D = depth d = effective depth d c = cover to compressive steel d s = depth of slab d t = cover to tensile steel

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e = eccentricity. d c /d = co mpress ive st ee l dept h fact or F =shear characteristic force. F d = design load F r = radial shear force. f= stress (in general) f ck = characteristic compressive stress of concrete. F y = characteristic tensile strength of steel. H = height. I = moment of inertia. I e =equivalent moment of intertia. j= le ver ar m fact or. Ka =coefficient of active earth pressure. K p =coefficient of passive earth pressure. k = neutral axis depth factor (n/d). L=length. L d =devolopment length. l = effective length of column or length or bond length. M = bending moment or moment. Mr=moment of resistance or radial bending moment. M t =torsional moment. M u =ultimate bending moment M θ =circumferential bending moment m = modular ratio. n = depth of neutral axis.

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n c =depth of critical neutral axis. P a =active earth pressure. P p = passive earth pressure. P u = ultimate axial load on the member(limit state design). P = percentage steel. P’= reinforcement ratio. P a =active earth pressure indencity. P e =net upward soil pressure. Q= shear resistance.

 = shear stress. q’=shear stress due to torsion R= radius. s= spacing of bars. s a = average bond stress. s b = local bond stress. T=tensile force. T u =ultimate torsional moment.

st or t= tensile stress in steel. tc= compressive stress in compressive steel. V u =ultimate shear force due or design load. V us =shear carried by shear reinforcement. W= point load. X= coordinate. x u = depth of neutral axis. Z= distance.

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α = inclination. β = surcharge angle. γ = unit weight of soil γ f = partial safety factor appropriate to the loading. γ m = partial safety factor appropriate to the material. σ cc = permissible stress in concrete. σ cbc = permissible compressive stress in concrete due to bending. σ sc = permissible compressive stress in bars. σ st = permissible stress in steel in tension. σ st = permissible tensile strss in shear reinforcement. σ sy = yield point compressive stress in steel. µ = co efficient of friction.

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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 essential .The permeability of any uniform and thoroughly compacted concrete of given mix proportions is mainly dependent on water cement ratio .The increase in water cement ratio results in increase in the permeability .The decrease in water cement ratio will therefore be desirable to decrease the permeability, but very much reduced water cement ratio may cause compaction difficulties 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 timber shuttering which prevent the easy escape of heat of hydration 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 t he design of water tank of flexible base and rigid base and the under ground tank to avoid the tedious calculations. 5. In the end, the programs are validated with the results of manual calculation 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. Infiltration 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)Population to be served. Quantity= Per demand x Population 3.2 Water Consumption Rate: It is very difficult to precisely assess the quantity

of water demanded by the public, sincethere are many variable factors affecting water consumption. The various types of waterdemands, which a city may have, may be broken into following c lass Water Consumption for Various Purposes:

Types of Consumption 1

Domestic Consumption

2

Industrial and Commercial Demand Public including Fire Demand Uses Losses and Waste

3 4

Normal Range (lit/capita/day) 65-300

Average

%

160

35

45-450

135

30

20-90

45

10

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 sometimes treated as a function ofpopulation and is worked out from following empirical formulae:

Authority 1 2 3 4

American InsuranceAssociation Kuchling'sFormula Freeman'sFormula Ministry ofUrbanDevelopmentManual Formula

Formula (P in thousand) Q(L/min)=4637P(1-0.01 ÖP) Q(L/min)=3182 P Q(L/min)=1136.5(P/5+10) Q(kilo liters/d)=100P for P>50000

Q for 1 lakh Population) 41760 31800 35050 31623

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3.4 Factors affecting per capita demand:

• Size of the city : Per capita demand for big cities is generally large as compared tothat for smaller towns as big cities have sewered houses. • Presence of industries. • Climatic conditions . • Habits of economic status . • Quality of water : If water is aesthetically $ 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 = Quantity Required in 12 Months/ (365 x Population) If this average demand is supplied at all the times, it will not be sufficient to meet thefluctuations. • Seasonal va riation :The demand peaks during summer.Firebreak outs are generally more in summer, increasing demand. So,there is seasonal variation .• Daily variation depends on the activity. People draw out more water on Sundaysand Festival days, thus increasing demand on t hese days. • Hourly variations are very important as they have a wide range. During activehousehold 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 quantity of

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wateris required to be supplied during short duration, necessitating the need for amaximum rate of hourly supply.So, an adequate quantity of water must be available to meet the peak demand. To meet allthe fluctuations, the supply pipes, service reservoirs and distribution pipes must beproperly proportioned. The water is supplied by pumping directly and the pumps anddistribution system must be designed to meet the peak demand. The effect of monthlyvariation influences the design of storage reservoirs and the hourly variations influencesthe design of pumps and service reservoirs. As t he population 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 quantity should be worked out with due provision for the estimated 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 estimated based on the following: • Useful life of the component , considering obsolescence, wear, tear, etc. • Expandability aspect. • Anticipated rate of growth of population, including industrial, commercial developments& migration-immigration. • Available resources. • Performance of the system during initial period. 4.2 Population Forecasting Methods:

The various methods adopted for estimating future populations are given below. Theparticular method to be adopted for a particular case or for a particular city dependslargely on the factors discussed in the methods, and the selection is left to the discrectionand intelligence of the designer. 1. Incremental Increase Method 2. Decreasing Rate of Growth Method 3. Simple Graphical Method 4. Comparative Graphical Method 5. Ratio Method 6. Logistic Curve Method 7. Arithmetic Increase Method 8. Geometric Increase Method

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5. WATER TANKS

5.1 CLASSIFICATIONS:

Classification based on under three heads: 1. Tanks resting on ground 2. Elevated tanks supported on st agging 3. Underground tanks. Classification 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 M 20 is not used. The minimum quantity ofcement in the co ncrete mix shall be not less than 30 kN/ m 3 .The design of the concretemix shall be such that the resultant concrete issu efficiently impervious. Efficientcompaction preferably by vibration is essential. 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 segregation and honey combing.All joints should be made watertight as these are potential 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. Additionally itshould be ensured that tensile stress on the liquid retaining ace of the equivalent concretesection does not exceed the permissible tensile strength of concrete as given in table 1. For calculation purposes the cover is also taken into concrete area. Cracking may be caused due to restraint to shrinkage, expansion and contraction of concrete due to temperature or shrinkage and swelling due to moisture effects. Such restraint may be caused by . (i) The interaction between reinforcement and concrete during shrinkage due to drying. (ii) The boundary conditions. (iii) The differential conditions 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 limiting 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 interposition 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 sections 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 initial 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 discontinuity without

initial gap between the concrete on either side o f 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 jo int. A co mplete co nt ra ct ion joint is one in which both st ee l and co ncret e areinterrupted and a partial contraction joint is one in which only the concrete is interrupted, the reinforcing steel running through as s hown in Fig.(b)

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Fig (b) (ii)Expansion Joint . It is a joint with complete discontinuity 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 construction. This type of joint requires the provision of an initial 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 discontinuity 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 sometimes 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 jointing 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 automatically 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 calculations relating 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 calculations the permissible concrete

stresses shall be in accordance with Table 1. Where the calculated shear stress in concrete alone exceeds the permissible value, reinforcement acting in conjunction 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 ratio of steel and concrete, and the corresponding allowable tensile stress in concrete. (b) For strength calculations:

In strength calculations the permissible stress shall be as follows: a) Tensile stress in member in direct tension 1000 kg/cm 2 .

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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/cm 2 . c) On face away from liquid for members 225mm or more in thickness 1250 kg/cm 2 . d) Tensile stress in shear reinforcement For members less than 225mm thickness 1000 kg/cm 2 for members 225mm or more in thickness 1250 kg/cm 2 . (v)Compressive stress in columns subjected to direct load 1250 kg/cm 2 . 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 precautions are taken to avoid cracking of concrete during the construction period and until the reservoir is put into use. (c) Recommendation regarding joints given in article 8.3 and for suitable sliding layer beneath the reservoir are complied with, or the reservoir is t o be used only for the storageof water or aqueous liquids at or near ambient temperature and the circumstances aresuch that the concrete w ill 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 d iscussed in article 3. (ii) Floors of tanks resting on ground.

If the tank is resting directly 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 contraction or expansion jo ints between. Insuc h cases a sc reed or co ncret e layer less tha n 75 mm thick sh all

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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 M 1 0 ,where injurious soils or aggressivewater are expected, the screed layer shall be of grade not weaker than M 1 5 and if necessary a sulphate resisting 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 junction 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 vertical loads on the floor. (c) In continuous 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 calculation of the resistance to cracking of T-beam, L beam sections at supports. (d)The floor slab may be suitably tied 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 vertical loads ofthe liquid over it and the ratio of its 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 tcircumferential tension in addition 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 vertical movement joints should be as discussed in article 3.3 while the majority of these joints may be of the partial or complete contraction type, sufficient joints of the expansion type should be provided to satisfy the requirements given in article (ii) Pressure on Walls.

(a) In liquid retaining structures with fixed or floating 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 t aken 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 vertical and horizontal bendingmoments. An estimate 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 resulting from horizontal bending moments. On liquid retaining faces, the tensile stressesdue to the combination of direct horizontal tension and bending action shall sat isfy the following condition: (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 vertical 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 vertical beams or cantilever 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 continued or restrained in the horizontal direction, 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 conditions 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 deformation of wall under influence of liquid pressure is restr icted at and above the base. Consequently, only part ofthe triangular hydrostatic load will be carried by ring tension and part of the load at bottom will be supported by cantilever action. (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 addition to the reinforcement required to resist horizontal ring tension for hinged at base, conditions 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 sympathetic cracking it is important to ensure that movement joints in the roof correspond with those in the wa lls, 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 temporaryconditions which should not be exceeded. In designing the roof, allowance should bemade for the temporary condition 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

domestic purpose, the roof must be made water-tight. This may be achieved by limiting 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 condensation. 7.8 Minimum Reinforcement:

(a)The minimum reinforcement in walls, floors and roofs in each of two directions atright angles shall have an area of 0.3 per cent of the concrete section in that direction for sections up to 100mm, thickness. For sections of thickness greater than 100mm, and lessthan 450mm the minimum reinforcement in each of the two directions shall be linearly reduced from 0.3 percent for 100mm thick section to 0.2 percent for 450mm, thicksections. For sections of thickness greater than 450mm, minimum reinforcement in eachof the two directions shall be kept at 0.2 per cent. In concrete sections of thickness225mm or greater, two layers of reinforcement steel shall be placed one near each faceof the section 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 sectional 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 thisadditional cover shall not be taken into account for design calculations. (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.

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8. DOMES A dome may be defined as a thin shell generated by the revolution of a regular curve about one of its axes. The shape of the do me depends on the type of the curve and the direction of the axis o f revolution. 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, exhibition halls, auditoriums, planetorium and bottom of tanks, bins andbunkers. Domes may be constructed of masonry, steel, timber 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 revolution can be developed neglecting effectof bending moment, twisting 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 discontinuous 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 modifications.Stresses in shells are usually kept fairly low, as effect of the edge disturbance, as mentiioned 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 o f shell.

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9. MEMBRANE THEORY OF SHELLS OF REVOLUTION Fig shows a typical shell of revolution, on which equilibrium of an element, obtained by intersection of meridian and latitude, 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 latitude are horizontal and called the hoop stresses, denoted by N .Neglecting variations in the magnitudes of Nf and N , since they are very small.the state of stress in t he element is shown in fig (b).

Shell of Revolution.

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two forces N (r d  ) have the resultant N (r d )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 pointing 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) N f /r1+N/r 2 = -Pr = pressure normal to the surface In this equation pr is considered positive when acting towards the inner side and negative when acting towards the outerside of the shell.Value s and N f and N will be positive when tensile andnegative compressive. The equation is valid not only for shells in thform of a surface of revolution, but may be apped to allshells, when the coordinate lines for = constant and  = constant, are the linesof curvature o f the surface. Forces in shell Force N f act tangentially to the surface aall around the circumference. Considering thequilibrium of a segment of shell cut along the parallel to latitude defined by the angle as shown in Fig 2prN f sin f + W= 0, Where W= total load in the vertical dirction on the surface of the shell above the cut. This gives, N f = -W/2prsinf Eq. is readily solved for N f a nd N may then be detrminedby Eq. This theory is applicable to a shell of any material as only the conditiions of equilibrium have been applied and no compatibility relationsships in terms of deformation have been introduced. It is, therefore, immaterial whetherr Hooke's law is applicable or not.

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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 consideration of equilibrium. If it is required to obtain Na at section 1 - 1 from calculation of the total downward load, there are two possibilities. 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. Alterrnatively, the downward load may be calculated from the weight of water and tank bottom of the part i.e W 2 less upward reaction of the support as shown in Fig. For section which cuts the tank bottom inside the support, the reaction 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 Nf 1 and Nf 2 at point A are shown in Fig (b). To eliminate horizontal forces on the supporting ring girder, it is necessary that Nf 1 cos a 1 = Nf 2 cos a 2 .

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11. DESIGN OF REINFORCED CONCRETE DOMES The requirements of thickness of dome and reinforcement from the point of view of 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 direction along the meridians as well as along the latitudes. This reinforcement will be in addition to the requirements for hoop tensile stresses. The reinforcement is provided in the middle o f 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 resulting tensile stress when concrete alone is considered does not exceed 1.1N/mm 2 to 1.70 N/mm 2 for direct tension and 1.5 N/mm 2 to 2.40 N/mm 2 for tension due to bending in liquid resisting 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/mm 2 ) in case of liquid retaining structures and 140 N/mm 2 (or190 N/ mm 2 ) 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 directions. If the reinforcement along the meridians is continued upto the crown, there will be congestion of steel there. Hence, from practical considerations, the reinforcement along the meridian is stopped below the crown and a separate mesh, as shown in Fig (a), is provided. Alternatively, 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 acting 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 o f hoop tension. Openings may be provided in the dome as required from other functional 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.

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The allowable stresss specified in IS 3370 for such tanks are as follows: Type of stresses: Permissible stress in N/mm 2 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/mm 2 , 1.2. N/mm2 ,1.32 N/mm 2 , 1.5 N/mm 2 , 1.6N/mm 2 and 1.7 N/mm 2 for M 1 5 , M 20 , MM 25 , M 3 0 , M 35 and M4 0 respectively, the value in tension due to bending 2 2 2 2 2 i.e.,being1.5N/mm ,1.7N/mm ,1.82N/mm ,2.0 N/mm ,2.2 N/mm and 2.4 N/mm 2 . 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 w ill be limited by the requirements that the stress as mentioned above should not be exceeded. The

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tensill stress in steel will be modular ratio multiplied 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 precautions are taken to avoid cracking of concrete during construction period, until 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 o f 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.

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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 distinct advantages as t hey are not affected by climatic 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, consisting of (5) Columns

(6) Bracings and

(7)Foundations

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 capacities. 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 junction 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 re sult of which the ring girder support ed 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 direction.

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Sometimes, a vertical 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 latitudes. 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 reaction 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 resulting 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 foundation consisting of a ring girder and acircular slab. Suitable proportions for the Intze. for case(1) suggested by Reynolds. Total volume ~0.585D 3 for case (2), the proportion was suggested by Grey and Total Volume is given by

�� 

V 1 = p   �  = 0.39  . for H = D/2. V2 =

. � (  

V3 =

 (3   

+   +  ) = 0.102D 3 .

+ ℎ  ) = 0.017D 3 .

With h 1 = 3/25D and r = 0.0179D 3 . Volume V = 0.4693D 3 . With h 1 = D/6 and r = 3/10D. Volume V = 0.493D 3 .

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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/m 2 Wind pressure as per IS875 1200N/m 2 Assuming M 20 concrete For which σ cbe = 7N/mm2 , σ cc = 5N/mm 2 Direct tension σ t = 5N/mm2 Tension in bending = 1.70 N/mm 2 Modular ratio m = 13 For Steel stress, Tensile stress in direct tension =115 N/mm 2 Tensile stress in bending on liquid face =115 N/mm2 for t < 225 mm and 125 N/mm 2 for > 225 mm. Solution: Taking the volume as 0.585 D3 for proportion given in Fig. D = 9.0 m. The dimension of the Tank is sho wn in fig.

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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/mm 2 Hoop stress @ Crown:

=0 �

��

f=

�.  [1 − ] . 

f =0.107 N/mm2 Meridional thrust @ level of sprining: 

T =  =

�. .

=18778.34 N/m Compressive stress .

= � =0.125 N/mm 2 provoide 8mm Ring beam @ top : Horizontal component of T= T cos  =13520.40 N/m Hoop stress in the ring beam 

=14339.82×  =60841.82 Area of steel required .

= .

=311.73 mm 2 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 section = A mm 2 Equivalent concrete area =  +(m-1)  =  +(13.33-1)452.38

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=  +5577.8454 Limiting tensile stress on the eqvivalent concrete area to 2 N/mm 2 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 A st =

 �.

= 853.04 mm 2 Provide 8 bars of 12 mm diametre of 142.85 mm distance. Thickness of the wall may be kept as 200 mm.

Distribution steel =

∙ [200 �1000] 

= 480 mm 2 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 T 1 = V 1 = T 1 sin = 24917 � 0.68 = 17184.137 N/m. Weight of the wall = 0.2 �4 �25000 = 20000 N/m.

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Weight of ring beam = 0.2 �0. 2 �25000 = 1000 N/m. Total load V 2 =38184.137 N/m. Compressive stress =

. �

= 0.19 N/mm 2 Nominal vertical stress is equql to 0.24% of gross area. Vertical steel =

. � 200 � 100 

= 480 mm 2 Provide 10 bars of 8 mm diametre of spacing 100 mm.

Ring beam at B :

Let T 2 be the thrust /m run exerted by the conical wall at the junction B. Resolving vertically at B T 2 sin= V 2

tan=

. .

=1

 = 45 �. T2 =

=

  . �

= 54000.52 N/m. Resolving horizontally at B H 2 =T 2 cos=54000.52 � cos 45� = 38184.137 N/m

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This horizontal load H 2 will produce a hoop tension in ring beam B 

=H 2 � 

Hoop tension due to H 2



=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 =

. �.

= 1256.611mm 2 Provide 6 bars 18 mm ∅ A st = 1526.81 mm 2 . Let ‘A’ be the area of ring beam Equivalent concrete area

= A+(m-1)A st = A+(13.33-1) � 1526.81 = A+18825.61

Limiting the tensile stress on the equivalent concrete area to 2 N/mm 2 . .

=2 A c =11233.688 mm 2

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Provide 250 � 500 mm 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:   



+  tan 

Where W w = weight of water resting on the conical slab. W s = weight of the conical slab.

 = inclination of the conical slab with the horizontal. Area of water section standing on the conical slab . �1.5 = 7.125 m2 . 

= X=

. ] 

[

.

= 0.52 m.

Weight of water resting on the conical slab W w = 9810 �7.125 �2 [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 W s = 0.2 �2.121 �25000 �2 [  ] = 249874.42 N. Hoop tension =

.. 

=531838.349 N. Hoop steel on the entire section =

. �.

��

= 2569.267 mm 2 . Provide 14 bars of 6 mm ∅ =14 �  �64 = 2814.86 mm 2 . Design for bending moment:

Load per metre width of the conical slab = =

.. =  �.

   �  

76214.279 N.

Maximum bending moment =

 

=

.�. 

= 14290.177 Nm. Axial compression V 2 = T 2 sin =

.  �

= 54000.52 N. Providing 16 mm diametre bar at clear covers of spacing 25 mm. Effective depth = 200 −25 − 8 = 167 mm. 

Distance between centre of section and centre of steel x = d −  = 167 −100 = 67 mm Resultant bending moment = M +T 2 .x =14290.177 � 10 +54000�67 = 17908212.15 Nmm. A st =

. = ��.

518.04 mm2

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

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The bottom dome:

Let R be the radius of the dome,then 3 2 = 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/mm 2 . Weight of water resting on the dome =  [  h − =9810[155.508 −17.869] = 1350234.872 Area of dome surface = 2 Rh = 2  � 4.315 � 1.2

 (3R −ℎ )] 

��

= 32.79 m2 . Load intensity due to weight of water =

. = .

41178.25 N/m 2 .

Total load intensity = 5000 +41178.25 = 46178.25 N/m 2 . Meridional thrust =

 

=

.�. .

= 116788.016 N/m. Meridional compressive stress = Hoop stress = =

 [ cos  

. = �

0.583 N/mm 2 .



−  ]

.�.  [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 he nce 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 W 1 = 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 W 2 = 38184.137 �2  �4.5 = 1079631.039 N. Weight of ring beam at B W 3 = 0.25 �0. 5 �25000 � 2  �4.5 = 88357.29 N. Weight of conical wall W 4 = 249874.42 N.

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Weight of lower dome W 5 = 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 0.913bd 2 = 114316.577 �1000, 0.913 �400 �d 2 = 114316.577 �1000, Then d 2 = 278458.26, d =560 mm. Let the clear cover be 40 mm. Over all depth of beam = 600 mm. Actual effective depth = 600 mm.

�� 



Equivalent shear force = V +1. 6  = 286939.199 +1. 6  . = 287160.093 +(

.�.� ). 

V c = 319994.559. Equivalent nominal shear stress ve =

 

=

. �

= 1.42 N/mm2 .

Maximum shear stress ma x >  v .

ma x = 1.8 N/mm 2 . c <  v. Provide longitudinal and transverse reinforcement according to B-6.4

Longitudinal reinforcement: 

M e = M +M t , Mt =

(  ) .



=

.[ ]� .

= 12152705.88 Nmm. M = moment at crosssection. M er = 1000 �114316.577 +12152705.88 = 126469282.9 Nmm. A st =

 �.�

=

 �.�

= 1080.187 mm 2 . Transverse reinforcement:

A sv =

∙   

A sv = [

+

∙ . 

, b1 = 400 −80 = 320 mm , d 1 = 600 −80 = 520 mm.

.� . ]S + �� .�� v

Providing 4 legged 10 mm stirrups. A sv = 315 mm 2 , 315 = 1.175, S v = 267.95 mm. Take S v as 250 mm. [

  ]b �S v 

,

.. 

�400 �S v = 315 , S v = 158.88 mm.

��

Provide 150 mm spacing. Steel for sagging moment =

.� �.�

= 494.27 mm 2 . Provide 5 bars of 12 mm diameter. A st = 565.48 mm 2 . Hoop stress:

T c = thrust exerted by the conical slab on the girder. T c sin  �2 r = W w +W s +weight of cylindrical wall and upper dome. T c sin  �2 r = 154588.24 +249874.42 +1079631.039 T c sin  �2 r = 2875387.699. Tc =

. = �� �

215729.87 N.

Horizontal component of T c = 215729.87 � cos45�, H1 = 152544.055 N. Horizontal component due to dome = 11678.016 � cos 43�36′, H 2 = 84574.59, H 1 −H 2 = Net,Net = 67969.46 N. Hoop stress = 67969.46 �3 = 203908.38 N. Hoop compressive stress =

. = �

0.849 N/mm 2 .

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  be the inclination of the column with the vertical.

tan=

 

,  = 5 �42 ′ , sin = 0.0995, cos =

 = √ 

Actual length of column = √ 10 + 1 = 10.05 m. Providing 300 mm �300 mm column.

0.995.

��

Wt. Of column =10 �0. 3 �0. 3 �25000 = 22500 N Total vertical 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 co lumn Vertical load on each co lumn 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 A sc Then cA c + tAsc =599375.275 N 5 �A c + 190 �A sc =599375.275 5[400 �400 −sc ] +190 �A sc =599375.275 A sc =807.433 mm 2 . .

Min. Requirement of steel = 0.8% = [300 �300] =720 mm2 Provide 6 bars of 20mm dia. =1884 mm 2 More steel has been subjected since column is subjected to B.M caused by wind load.

��

Analysis due to wind pressure:

Wind pr. =1200 mm 2 . Wind force on the top dome & cylindrical walls =(4+ @Ht=13.95

. ) �9. 4 �1200 

=55272 N

Wind force on the circular wa ll =

.. �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 = ∑ ^

∑ 2 =2 �4 2 +4( ) 2 =64m2 √ Max. Wind load force in the most leeward side &the most w indward side. =

.� =68656.275 

N

Max. Wind force in columns marked 5 =

. 

Consider the windword column 1 Vertical load due to dead +wind load =596378.399 +68656.275 N



� √  =48547.317 N

��

=665034.674 N. Corresponding axial load =

. .

=668376.556 N Horizontal comoponent of the axial forces caused by w ind action =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 effective cover of 50 mm. A st = 1884 mm 2 , Equivalent concrete area = A c +(m-1)A st = (300 �300) +(12.33 �1884) = 113229.72 N Polar moment of inertia of the equivalent concrete section, =

 +(mA st �effective 

=

depth fromcentre),

 +1884 �12.33[150-50] 2 = 

1.582 �10 9 mm 4 .

Equivalent moment of inertia about full section =

.�  = 

791.14 �10 6 mm4.

��

Direct stress in concrete =

  =   

Bending stress in concrete =

�.� = .�

5.9 mm2 .

2.14 N/mm 2 .

Maximum stress = 5.9+2.14 = 8.04 N/mm 2 . 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 section and a doubly reinforced beam with equal steel at top and bottom. A st = A sc =

.� = ��.

702.357 mm2 .

Provide 4 bars of 18 mmdiameter at top and equal amount at bottom. Shear force for brace =

        

,



Span of brace =2 �  � sin 22�30 ′ = 2.678 m. Shesr force for brace =

.  �. 

= 23887.49 N. Nominal shear stress v =

 

=

. �

= 0.30 N/mm 2 . Provide nomonal stirrups say 2 legged 10 mmdiameter stirrups at 200 mm clear cover.

��

Design of foundation:

Total load on the column = 599375.2754 �8 = 4795002.203 N. Approximate weight of foundation is 10% of column loads. = 479500.22 N. Then total load is equal to 5274502.22 N. Safe bearing capacity of 200 KN/m 2 , Area =

 

=

. = �

26.37 m2 .

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

=  [9.5 2 −6. 5 2 ] = 37.69 m2 . Net intensity =

. .

��

= 139.9 KN/m 2 . 139.9 KN/m 2 < 200 KN/m 2 . 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 section; Moment of resistence = maximum bending moment at support. 0.913bd 2 = 159194.073 �1000 , bd 2 = 174363716.30 , d = 590 mm ,clear cover = 60 mm , D = 650 mm. 

Equivalent shear stress V v = V +1. 6  = 299687.63+1.6

.� , 

= 336550.0176 N.

Equivalent nominal shear v =

 = 

1.14 N/mm 2 , but c = 1.8N/mm2 ,

Hence c <  v . Longitudinal reinforcement: 

M el = M+M t , Mt =

(  ) .



=

.[ ]� .

= 15569653.82 N , M el = 1000[159194.073 +15569.653] = 174763.726 �1000 N. A st =

.� �.�

= 1430.964 mm 2 ,

��

Provide 9 bars of 16 mm diameter bars. Hence area os steel required is A st = 1809.55 mm 2 . Transverse reinforcement:

A sv =

∙   

+

∙ , . 

providing 4 legged 10 mm diamater of stirrups.

A sv = 4  �5 2 = 314 mm2 , b1 = 500-80 = 420 mm , d 1 = 650-120 = 530 mm, 314 =

.� ��

.

+ .�� , 314 = S r [0.224+0.983] , S v = 260 mm.

Let us provide 200 mm clear cover spacing. Steel for hogging mommentA st =

.� �.�

= 653.31 mm 2 , Provide 4 bars of 16 mm diameter.

Design of bottom slab:

Provide a cantilever projection beyond the face of the beam = 0. 6 m. Maximum bending moment for 1 m wide stirup = 139944.346 �

. Nm 

,

= 2518.98 Nm. Equating moment of resistence to bending moment , 0.913 �bd 2 = 25189.98 �1000 , b = 1000 mm. Then d 2 = 27590.339 , d = 166.1 mm. Let us provide 170 mm effective depth and 40 mm clear cover. D = 210 mm. A st =

.� = �.�

715.82 mm 2 .

Provide 4 bars of 18 mm diameter. A st = 1017.87 mm 2 ,and spacing of the bars is 250 mm clear cover. Distribution steel:

��

Provide 0.12 % steel and the steel required is =

.�� 

= 252 mm 2 . Provide 6 bars of 8 mm diameter bars and spacing =

 = 

160 mm clear cover.

Check for sliding:

Total load on the foundation when t ank is empty = 5274502.423-2896117.112 = 2378385.311 N Horizantal force on the base = 72374.61 N. Let coefficient of friction = 0.5 F s=

.�. = .

16.43.

��

14. ESTIMATION

14.1 Detailed estimation :

Detailed estimate is an accurate estimate and consists of working out the quantities 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 quantities of each item are calculated, and abstracting and billing are done. The detailed estimate is prepared in two stages: Details of measurement and calculation of quantities. The details of measurements of each item of work are taken out correctly from plan and drawing and quantities under each item are calculated in a tabular form named as details of measurement form. Abstract of estimated cost: The cost of each item of work is calculated in a tabular form the quantities already computed and total cost is worked out in abstract estimate 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. N o

DECRIPTION OF WORK

N Lm OS

1

Earthwork in excavation

1

B m

A m2

Ho r QTY D m3 (m)

73.89

1

73.89

REMARKS  =   /4 = � 9.72 /4 =73.89

2

Earthwork in 1 filling a)R.C.C work in 1 foundation b)steel in foundation i )Longitudanal 9 ii)Transverse 4

64.316 7.068

 �0.008 2 0.008 2 � 

0.2

1.4136 0.045 0.02

=  �8

��

3

4

5

6

7

8

R.C.C in columns

8

Steel in columns

8� 6 8

RCC in [email protected] Steel in Bracings @2.5m from G.L a) R.C.C in bracings @5m from G.L. b) Steel

8� 8 8

8� 8 a)RCC in bracings 8 7.5m b)steel 8� 8 Top ring girder 1 a)R.C.C b)steel longitudinal 5 transeverse 125 Bottom dome a)RCC in dome 1

0.3

0.3

0.09

 �0.009 2

10.0 7.235 49 0.151 10.0 49 0.3 =0.459 A=  �0. 009 =0.000254 =0.01

0.09

0.3

 �0.01 2 0.637 5 0.637 5 0.575

0.3

0.3

0.09

 �0.009 2

0.575

D 6

 �0.009 2 0.4

0.24

0.6

 �0.009 2

6

=22.619

A=  �0. 009 =0.000254

0.0093 6 0.324 A=  �0. 009 =0.000254 0.0072

0.45 0.45

0.414

4.52 0.02 0.066

0.2

4.523

A= 2 rh =2  �3 �1.2 =22.619 

b) steel 9

a)RCC conical slab

1

L=6.6 2 23.56

b)steel

14

23.56

0.067 0.2

2.12 9.994 1

 �0.008 10 11

12

steel for B.M. a)RCC ring beam @B b)steel Cylindrical wall

3 1

a)Main steel

20

4.32

b)Distribution steel Ring beam @ A a)concrete

4

28.27

1

9 

0.443

28.27 6 1

 �0.008

2

0.2

0.5 4

 �0.006 2  �0.004 2

0.2

0.04

=6.62

[1 + 2] 2 =  (9+6)/2

0.066

2

23.56 0.25

 = � � 2 r

0.014 3.534 0.034 22.619 0.098 L=4+16d 0.0056 =4+16 �0.012 =4.32

0.2

1.13

L= D =9 

��

13

b)Steel Top dome R.C.C a) concrete b) Steel

12

Total RCC work

13 14

Total steel Plastering in CM (1:2) for Inner surface Of conical dome (12mm) Plastering in CM (1:6) for outer surface Of conical dome (12mm) Plastering in CM (1:2) for Inner surface Of cylindrical wall (12mm) Plastering in CM (1:6) for outer surface Of cylindrical wall (12mm) Plastering in CM (1:2) for Inner surface Of domed roof (12mm) Plastering in CM (1:6) for outer surface Of domed roof (12mm) Plastering in CM (1:6) for columns (12mm)

15

16

17

18

19

20

4

9 

 �0.006 2 rh =50.89

1 100 9.93

0.012 0.15 7.63 0

 �0.004 2

0.05

1

50.89

63.795 6 1.017 9.15

1

55.135

9.92

1

 �D

28.2 112.8

4

20.354

1

 �D

29.5 118.82

4

28.349

22



 = � � 2 r

A= 2 rh =50.89 A= 2 rh =55.135

1

22.619

4.07

A= 2 rh =2  �3 �1.2 =22.619

1

26.38

4.74

A= 2 rh =2  �3 �1.4

0.09

17.28

8

0.3

0.3

5.65 21

A= 2 rh =

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

9 

0.2

1.01 1.27

��

23 24 25 26

27

28 29 30 31

Plastering in CM (1:6) for bracings at 2.5m ht.(12mm) Plastering in CM (1:6) for bracings at 5m ht.(12mm) Plastering in CM (1:6) for bracings at 7.5m ht.(12mm) Plastering in CM(1:2) for inner surface of conical slab(12mm) Plastering in CM(1:6) for outer surface of conical slab(12mm) Total plastering

1

0.27

1

0.24

1

0.19

1

4.239

1

4.46

Thick water proof cement painting for tank portion white washing for 8 columns Total white washing

105.53 3 85.278 0.312

0.31 2

10.04

7.826 93.104

[1 + 2] 2 =  (9+6)/2

��

ABSTRACT

S . NO

DESCRIPTION OF WORK

QTY OR NOS

1

Earth work in excavation Beldars Mazdoors Total Earth work in Filling In foundation Beldar Bhisthi

73.89

2 3 4 5 6 7 8 9 10 11 12 13

Mazdoors Total Total earth work in Filling Disposal of surplus earth in a lead 30m Mazdoor Total Total cost of earth work

13 11

RATE RS PS

COST RS PS

250 250

3250 2750 6000

7 2

250 285

1750 570

5

250

1250 3570

250

1000 1000 10,570

64.316

9.574 4

��

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

20mm HBG graded metal Cum Sand Cement 1st Class Mason 2nd Class Mason Mazdoor (Both Men and Women) Concrete Mixer 10/7 cf (0.2/0.8cum)capacity Cost of Diesel for Miller Cost of Petrol for Vibrator Water (including for curing) Add 20% in Labour (1st Floor) Add MA 20% Add TOT 4% BASIC COST per 1 cum

Cum Cum Cum Day Day Day

0.601 1.2 0.4 0.38 1.03 2.33

1076 375 1620 285 285 250

Hour Liter Liter Ki

1 0.52 0.75 1.2

250 45 68 77.0

AMOUNT RS 646.676 450 648 108.3 293.55 582.5 250 23.4 51 92.4 629.16 629.16 176.166 4580.31

��

COLUMNS

A. MATERIALS

UNIT

QTY

RATERS

20mm HBG graded metal Cum Sand Cement 1st Class Mason 2nd Class Mason Mazdoor (Both Men and Women) Concrete Mixer 10/7 cf (0.2/0.8cum) capacity Labour centering Material hire charges for centering Water (including for curing) Add 20% in Labour (1st Floor) Add MA 20% Add TOT 4% BASIC COST per 1 cum


6.156 3.078 2.052 1.99 5.26 11.96

1076 375 1620 285 285 250

Hour Cum Cum Ki

1 1 1 1.2

250 971 89 77.0

AMOUNT RS 6623.85 1154.25 3324.24 567.15 1499.1 2990 250 971 89 92.4 2912.198 2912.198 582.43 20967.816

RCC RING BEAM AT TOP

A. MATERIALS

UNIT

QTY

RATERS



0.96 0.48 0.32 0.31 0.83 1.86

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 1 1 1.2

250 971 89 77.0

AMOUNT RS 1032.96 180 518.4 88.35 236.55 465 65 971 89 92.4 747.73 747.73 149.54 5383.66

��

RCC DOMED ROOF 150mm THICK

A. MATERIALS

UNIT

QTY

RATERS



6.48 3.24 2.16 2.1 5.6 12.6

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.267 10 10 1.2

250 971 89 77.0

AMOUNT RS 6972.48 1215 3499.2 598.5 1596 3150 66.75 9710 890 92.4 5558.33 5558.33 1111.61 40018.6

CONICAL SLAB 200mm THICK

A. MATERIALS

UNIT

QTY

RATERS



8.49 4.25 2.83 2.75 7.34 16.52

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 5 5 1.2

250 971 89 77.0

AMOUNT RS 9135.24 1593.75 4584.6 783.75 2091.9 4130 65 4855 445 92.4 5555.328 5555.328 1111.06 39998.35

��

RCC CYLINDRICAL WALL

A. MATERIALS

UNIT

QTY

RATERS



19.23 9.62 6.41 6.23 16.62 37.39

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 1 1 1.2

250 971 89 77.0

AMOUNT RS 20691.48 3607.5 10384.2 1775.55 4736.7 9347.5 65 971 89 92.4 10352.066 10352.066 2070.432 74534.89

RCC RING BEAM AT BOTTOM OF CYLINDRICAL WALL

A. MATERIALS

UNIT

QTY

RATERS



3 1.5 1 0.97 2.59 5.84

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 1 1 1.2

250 971 89 77.0

AMOUNT RS 3228 562.5 1620 267.45 738.15 1460 65 971 89 92.4 1818.7 1818.7 363.74 13094.64

��

RCC CIRCULAR GIRDER

A. MATERIALS

UNIT

QTY

RATERS



3.84 1.92 1.28 1.24 3.32 7.47

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 1 1 1.2

250 971 89 77.0

AMOUNT RS 4131.84 720 2073.6 353.4 946.2 1867.5 65 971 89 92.4 2262.588 2262.58 452.517 16290.61

RCC BRACING AT 2.5m HT.

A. MATERIALS

UNIT

QTY

RATERS



0.39 0.19 0.13 0.125 0.33 0.75

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 1 1 1.2

250 971 89 77.0

AMOUNT RS 419.64 71.25 210.6 35.625 94.05 187.5 65 971 89 92.4 447.213 447.213 89.44 3219.93

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RCC BRACING AT 5m HT.

A. MATERIALS

UNIT

QTY

RATERS



0.33 0.17 0.11 0.125 0.33 0.75

1076 375 1620 285 285 250

Hour Cum Cum Ki

0.26 1 1 1.2

250 971 89 77.0

AMOUNT RS 355.08 63.75 178.2 35.625 94.05 187.5 65 971 89 92.4 426.32 426.32 85.264 3069.50

RCC BRACING 7.5m HT.

A. MATERIALS

UNIT


QTY


RATE RS 0.27 1076 0.13 375 0.09 1620 0.08 285 0.23 285 0.535 250

AMOUNT RS 290.52 48.75 145.8 22.8 65.55 133.75

Hour Cum Cum Ki

0.26 1 1 1.2

65 971 89 92.4 384.91 384.91 76.98

250 971 89 77.0

2771.37

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

Cement Mortor 1:6 1:2 Mason 1st class Bhisthi Mazdoor (unskilled) Add MA 20% Add TOT 4%

cum cum day day day

105.533 65.44 40.09 39 14 39

552 780 285 285 250

36165 31673 11115 3990 9750 18539 3719

Grand Total

114951

Painting to new walls of tank portion with 2 coats of water proof cement paint o f 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 a fter 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 coatings : Procoat SNP2 or Zoriprime EFC 2 1st class painter Mazdoor cost of water proof cement paint 1st class painter Mazdoor (unskilled) Add MA 20% Add TOT 4% Total cost

Pack

26

548

14250

Day Day Cum Day Day

4 4 50 2 2

355 250 35 355 250

1420 1000 1750 710 500 3926 786

24342

��

Painting 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 t o give an even shade after t hourughly 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 wa lls Cost of cement primer 1st class painter 2nd class painter cost of water proof cement paint 1st class painter Mazdoor (unskilled) Add MA 20% Add TOT 4%

Pack Day Day Cum Day Day

18 1 1 6 1 1

Total cost

100 355 250 35 355 250

1800 355 250 210 355 250 644 129 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 painting

= 28,335

Total cost of earthwork

= 10,570 8,96,710

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15. CONCLUSION Storage of water in the form of tanks for drinking and washing purposes, swimming pools for exercise and enjoyment, and sewage sedimentation tanks are gaining increasing importance in the present day life. For small capacities we go for rectangular water tanks while for bigger capacities we provide circular water tanks. Design of water tank is a very tedious method. With out power also we can consume water by gravitational force. Intze tank is constructed to minimize the project cost why because lower dome in this construction resists the horizontal thrust.

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16. REFERENCES Table 16.2. Coefficients for moment in cylindrical wall fixed at base (As Per IS3370) 3

Moment = Coefficient (wH ) 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:

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Permissible stresses in concrete 2 All values in N/mm Grade Of concrete

permissible stresses in compression Bending

Direct

permissible stress in bond for plain bars in tension (average)

c bc

c c

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 Concrete

M10

M15

M20

M25

M30

M35

M40

Tensile 2 Stress(N/mm )

1.2

2.0

2.8

3.2

3.6

4.0

4.4

M45

4.8

Table 2: Moments for circular girders For 8 columns

B.M@ Support 0.0083Wr

B.M@ centre 0.00416Wr

Torsion

0.0006Wr

M50

5.2

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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). . Estimating, costing and evaluation (B.N.Datta). . Standard schedule of rates (SSR)

Design & Estimation of Intze Tanks-Major Project Report

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