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 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 timely 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 section
Ab = Equivalent area of helical reinforcement.
Ac = Equivalent area of section
Ah = Area of concrete core.
Am = Area of steel or iron core.
Asc = Area of longitudinal reinforcement (comp.)
Ast = Area of steel (tensile.)
Al = Area of longitudinal torsional reinforcement.
Asv= Total cross-sectional are of stirrup legs or bent up bars within distance Sv
Aw =Area of web reinforcement.
AФ= Area of cross –section 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 = effective depth
dc = cover to compressive steel
ds= depth of slab
dt= cover to tensile steel

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e = eccentricity.
dc/d = compressive steel depth factor
F =shear characteristic force.
Fd= design load
Fr = radial shear force.
f= stress (in general)
fck = characteristic compressive stress of concrete.
Fy = characteristic tensile strength of steel.
H = height.
I = moment of inertia.
Ie=equivalent moment of intertia.
j= lever arm factor.
Ka=coefficient of active earth pressure.
Kp =coefficient of passive earth pressure.
k = neutral axis depth factor (n/d).
L=length.
Ld =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.
Mt=torsional moment.
Mu=ultimate bending moment
Mθ=circumferential bending moment
m = modular ratio.
n = depth of neutral axis.

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nc=depth of critical neutral axis.
Pa=active earth pressure.
Pp= passive earth pressure.
Pu= ultimate axial load on the member(limit state design).
P = percentage steel.
P’= reinforcement ratio.
Pa=active 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=ultimate torsional moment.
ߪst or t= tensile stress in steel.
tc= compressive stress in compressive steel.
Vu=ultimate 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 = inclination.
β = surcharge angle.
γ = unit weight of soil
γf = partial safety factor appropriate to the loading.
γm = partial 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 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 the 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 class
Water Consumption for Various Purposes:
Types of Consumption Normal Range
(lit/capita/day)
Average %
1 Domestic 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 sometimes treated as a function ofpopulation and is worked out from
following empirical formulae:
Authority Formula (P in
thousand)
Q for 1 lakh
Population)
1 American
InsuranceAssociation
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 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 variation: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 these 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 the 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 stagging
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 M20 is not used. The
minimum quantity 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 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 water-
tight 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 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 partial contraction 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 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/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 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 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× 10ି଺
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 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
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 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 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 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 satisfy 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 restricted 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 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 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.

Intze tank
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 dome depends on the type of the curve
and the direction of the axis of 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 of shell.

Intze tank
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 the element is shown in fig (b).
Shell of Revolution.

Intze tank
two forces N߮(rd ߠ) have the resultant N߮(rd ߠ)d߮ as shown in Fig.(c) and the
resultant acts normal to the surface pointed towards the inner side. Forces Nߠ(r1d߮)
again have horizontal resultant of magnitude N߮(r1 d߮) dߠ as shown in Fig (d). It
has a component N߮(r1d߮)dߠsin߮ 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(rdߠ) 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 equation pr is considered
positive when acting towards the inner side and negative when acting towards the
outerside of the shell.Value
s and Nf 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 ߠ =
constant, are the linesof curvature of the surface.
Forces in shell Force Nf 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
2prNf sin f + W= 0,
Where W= total load in the vertical dirction on the surface of the shell above the
cut.
This gives,
Nf = -W/2prsinf
Eq. is readily solved for Nfand 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.

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 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 W2 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 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 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 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 resulting 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 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/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 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 of 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.

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 respectively, 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 mentioned above should not be exceeded. The

Intze tank
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 of 300×10-6
may be assumed.
When shrinkage stresses are considered, the permissible stresses may be
increasedby 33
ଵ
ଷ
%.
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 distinct advantages as they 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 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 direction.

Intze tank
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 ~0.585D3
for case (2), the proportion was suggested by Grey and Total Volume is given by

Intze tank
V1 = p݀
మ
ర × ‫ܪ‬ = 0.39‫ܦ‬ଷ
. for H = D/2.
V2 =
௣.௛
ଵଶ
×(‫ܦ‬ଶ
+ ݀ଶ
+ ݀) = 0.102D3
.
V3 =
௣௛భ
଺
(3‫ݎ‬ଶ
+ ℎଵ
ଶ
) = 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 ratio 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.
Solution: 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 =
ௐோ
௧
[cos ߠ −
ଵ
ଵିୡ୭ୱ ఏ
]
=
ହଶହ଴×଺∙ହଶହ
଴∙ଵହ
[0.72−
ଵ
ଵ.଻ଶ
]
f =0.0298 N/mm2
Hoop stress @ Crown:
ߠ=0°

Intze tank
f =
ସଽହ଴×଺.ହଶହ
଴.ଵହ
[1−
ଵ
ଶ
]
f =0.107 N/mm2
Meridional thrust @ level of sprining:
T =
ௐோ
ଵାୡ୭ୱ ఏ
=
ସଽହ଴×଺.ହଶହ
ଵା଴.଻ଶ
=18778.34 N/m
Compressive stress
=
ଵ଼଻଻଼.ଷସ
ଵହ଴×ଵ଴଴଴
=0.125 N/mm2
provoide 8mm
Ring beam @ top :
Horizontal component of T= Tcos ߠ
=13520.40 N/m
Hoop stress in the ring beam
=14339.82×
ଽ
ଶ
=60841.82
Area of steel required
=
଺ସହଶଽ.ଶ
଴.ଽଶଷ଴
=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 section = A mm2
Equivalent concrete area = ‫ܣ‬௖+(m-1)‫ܣ‬௦௧
=‫ܣ‬௖+(13.33-1)452.38

Intze tank
= ‫ܣ‬௖+5577.8454
Limiting 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×
ଽ
ଶ
= 176580 N
Ast=
ଵ଻଺ହ଴
ଶଷ଴×଴.ଽ
= 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.
Distribution steel =
଴∙ଶସ
ଵ଴଴
[200×1000]
= 480 mm2
Provide 8 mm diametre bars.
=
ସ଼଴
ଵ଺×ଷ.ଵଶ
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 ߠ
= 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 =
ଷ଼ଵ଼ସ.ଵଷ଻
ଶ଴଴×ଵ଴଴଴
= 0.19 N/mm2
Nominal vertical stress is equql to 0.24% of gross area.
Vertical steel =
଴.ଶସ
ଵ଴଴
× 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 junction B.
Resolving vertically at B
T2sin ߙ= V2
tan ߙ=
ଵ.ହ
ଵ.ହ
= 1
ߙ = 45°.
T2 =
௏మ
ୱ୧୬ఈ
=
ୱ୧୬ସହ°
= 54000.52 N/m.
Resolving horizontally at B
H2 =T2cos ߙ=54000.52× cos 45°
= 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×
ௗ
ଶ
=38184.137×
ଽ
ଶ
N
=171828.6165N
Let the rinmg beam be 500mm deep
Water pressure on the ringh beam
=9810× 4 ×
ହ଴଴
ଵ଴଴଴
=19620 N/m
Hoop tension due to water = 19620×
ଽ
ଶ
=88290 N
Total hoop tension = 88290 +171828.61
= 260118.61 N
Steel for hoop tension =
ଶ଺଴ଵଵ଼.଺ଵ
ଶଷ଴×଴.ଽ
= 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
Limiting the tensile stress on the equivalent concrete area to 2 N/mm2
ଶ଺଴ଵଵ଼.଺ଵ
஺ାଵହ଺଼଼.଴ଵସ
= 2
Ac =11233.688 mm2

Intze tank
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:
௪ೢା௪ೞ
ଶగ
+
௪ೢ
ଶగ
tan ߙ
Where
Ww= weight of water resting on the conical slab.
Ws = weight of the conical slab.
ߙ = inclination of the conical slab with the horizontal.
Area of water section standing on the conical slab
=
ହ.ହାସ
ଶ
× 1.5 = 7.125 m2
.
X =
଺ା[
మ.మఱ
య
]
଻.ଵଶହ
= 0.52 m.
Weight of water resting on the conical slab Ww = 9810×7.125×2ߨ[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 = 0.2×2.121×25000×2ߨ[
଻.ହ
ଶ
] = 249874.42 N.
Hoop tension =
ଵହସହ଼଼ଶ.ଶସାଶସଽ଼଻ସ.ସଶାଵହସହ଼଼ଶ
ଶగ
=531838.349 N.
Hoop steel on the entire section =
ହଷଵ଼ଷ଼.ସଽ
ଶଷ଴×଴.ଽ

Intze tank
= 2569.267 mm2
.
Provide 14 bars of 6 mm ∅
=14× ߨ ×64 = 2814.86 mm2
.
Design for bending moment:
Load per metre width of the conical slab =
ௐೢାௐೞ
ଶగ × ௠௘௔௡ ௥௔ௗ௜௨௦
=
ଵହସହ଼଼ଶ.ଶସାଶସଽ଼଻ସ.ସଶ
ଶగ ×ଷ.଻ହ
= 76214.279 N.
Maximum bending moment =
ௐ௟
଼
=
଻଺ଶଵସ.ଶ଻ଽ×ଵ.ହ
଼
= 14290.177 Nm.
Axial compression V2 = T2sin ߙ =
ୱ୧୬ ସହ°
= 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−
௧
ଶ
= 167−100
= 67 mm
Resultant bending moment = M+T2.x =14290.177× 10ଷ
+ 54000 × 67
= 17908212.15 Nmm.
Ast =
ଵ଻ଽ଴଼ଶଵଶ.ଵହ
ଵ଺଻×ଶଷ଴×଴.ଽ
= 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 2ߠ be the angle subtended by the dome.
sin ߠ =
ଷ
ସ∙ଷହ
= 43°36°°
cos ߠ = 0.68
Thickness of dome = 200 mm.
Loads:
Dead load = 25000×0.2 = 5000 N/mm2
.
Weight of water resting on the dome = ߛ௪[ߨ‫ݎ‬ଶ
h−
గ௛೎
ଷ
(3R−ℎ௖)]
=9810[155.508−17.869] = 1350234.872
Area of dome surface = 2ߨRh = 2ߨ × 4.315 × 1.2

Intze tank
= 32.79 m2
.
Load intensity due to weight of water =
ଵଷହ଴ଶଷସ.଼଻
ଷଶ.଻ଽ
= 41178.25 N/m2
.
Total load intensity = 5000+41178.25 = 46178.25 N/m2
.
Meridional thrust =
ௐோ
ଵାୡ୭ୱ ఏ
=
ସ଺ଵ଻଼.ଶହ×ସ.ଷହ
ଵା଴.଻ଶ
= 116788.016 N/m.
Meridional compressive stress =
ଵଵ଺଻଼଼.ଵ଺
ଶ଴଴×ଵ଴଴଴
= 0.583 N/mm2
.
Hoop stress =
ௐோ
௧
[cos ߠ −
ଵ
ଵାୡ୭ୱ ఏ
]
=
ସ଺ଵ଻଼.ଶହ×ସ.ଷହ
଴.ଶ଴଴
[0.72−
ଵ
ଵ.଻ଶ
]
= 0.139 N/mm2
.
Hoop stress at the crown ߠ = 0°.
Maximum hoop stres =
ௐோ
௧
[cos ߠ −
ଵ
ଵାୡ୭ୱ ఏ
] = 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 = 38184.137×2ߨ ×4.5
= 1079631.039 N.
Weight of ring beam at B W3 = 0.25×0.5×25000× 2ߨ ×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 =
ௐ
଼
=
ସହଽସହ଺ଵ.ସସ଼଼
଼
= 573878.39 N.
Shear force at the support =
ௐ
ଶ
, 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 effective depth = 600 mm.

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Equivalent shear force = V+1.6
்
௕
= 286939.199+1.6
்
௕
.
= 287160.093+(
ଵ.଺×଼ଶ଻଴.ଶଵ×ଵ଴଴଴
ସ଴଴
).
Vc = 319994.559.
Equivalent nominal shear stress ߬ve =
௏೐
௕ௗ
=
ଷଵ଺ହ଺ହ.ଶ଼
ସ଴଴×ହ଺଴
= 1.42 N/mm2
.
Maximum shear stress ߬max> ߬v.
߬max= 1.8 N/mm2
.
߬c< ߬v.
Provide longitudinal and transverse reinforcement according to B-6.4
Longitudinal reinforcement:
Me = M+Mt , Mt =
்(ଵା
ವ
್
)
ଵ.଻
=
଼ଶ଺ଷ.଼ସ[ଵା
లబబ
రబబ
]×ଵ଴଴଴
ଵ.଻
= 12152705.88 Nmm.
M = moment at crosssection.
Mer = 1000×114316.577+12152705.88 = 126469282.9 Nmm.
Ast=
ெ೐ೝ
ଶଷ଴×଴.ଽ×ହ଺଴
=
ଵଶହଶଵହଷଵଶ
ଶଷ଴×଴.ଽ×ହ଺଴
= 1080.187 mm2
.
Transverse reinforcement:
Asv=
்∙௦ೡ
௕భௗభఙೞೡ
+
௏∙௦ೡ
ଶ.ହௗభఙೞೡ
, b1 = 400−80 = 320 mm , d1 = 600−80 = 520 mm.
Asv = [
଼ଶ଺ଷ.଼ସ×ଵ଴଴଴
ଷଶ଴×ହଶ଴×ଶଷ଴
+
ଶ଼଺ଽ.ଵଽଽ
ଶ.ହ×ହଶ଴×ଶଷ଴
]Sv
Providing 4 legged 10 mm stirrups.
Asv= 315 mm2
, 315 = 1.175, Sv = 267.95 mm.
Take Sv as 250 mm.
[
ఛೡ೐ିఛ೎
ఙೞೡ
]b×Sv ,
ଵ.ସଶି଴.ଶ଼
ଶଷ଴
×400×Sv = 315 , Sv= 158.88 mm.

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Provide 150 mm spacing.
Steel for sagging moment =
ହ଻ଶଽ଺.଴ଵ×ଵ଴଴଴
ଶଷ଴×଴.ଽ×ହ଺଴
= 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 ߙ ×2ߨr = Ww+Ws+weight of cylindrical wall and upper dome.
Tcsin ߙ ×2ߨr = 154588.24+249874.42+1079631.039
Tcsin ߙ ×2ߨr = 2875387.699.
Tc =
ଶ଼଻ହଷ଼଻.଺ଽଽ
ଶగ×ଷ×ୱ୧୬ ସହ°
= 215729.87 N.
Horizontal component of Tc = 215729.87× cos 45°, 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 =
ଶ଴ଷଽ଴଼.ଷ଼
ସ଴଴×଺଴଴
= 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 =
ସହଽଵ଴ଶ଻.ଵଽ଻
଼
= 573878.399 N.
Let ߙ be the inclination of the column with the vertical.
tan ߙ=
ଵ
ଵ଴
, ߙ = 5°42′ , sin ߙ = 0.0995, cos ߙ =
ଵ଴
√ଵ଴ଵ
= 0.995.
Actual length of column = √10ଶ + 1ଶ = 10.05 m.
Providing 300 mm × 300 mm column.

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Wt. Of column =10×0.3×0.3×25000
= 22500 N
Total vertical load = 573878.399+22500 N
= 596378.399 N
Corresponding axil load =
ହଽ଺ଷ଻଼.ଷଽଽ
଴.ଽଽହ
= 59375.2754 N
When tank is full = 599375.2754 N
Wt. Of water in tank =
ଶ଼ଽ଺ଵଵ଻.ଵଵଶ
଼
=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=
ଶଶସ଻ଽ଼.ଶଶଶ
଴.ଽଽହ
= 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 − ‫ܣ‬sc] +190×Asc =599375.275
Asc =807.433 mm2
.
Min. Requirement of steel = 0.8% =
଴.଼
ଵ଴଴
[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.

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Analysis due to wind pressure:
Wind pr. =1200 mm2
.
Wind force on the top dome & cylindrical walls =(4+
ଵ.଼
ଶ
)×9.4×1200
@Ht=13.95 =55272 N
Wind force on the circular wall =
ଽ.ସା଺.ସ
ଶ
×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×
଺ା଼
ଶ
×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 =
ெ௫
∑ ௫^ଶ
∑ ‫ݔ‬2
=2×42
+4(
ସ
√ଶ
)2
=64m2
Max. Wind load force in the most leeward side &the most windward side.
=
ଵ଴ଽ଼ହ଴଴.ସ×ସ
଺ସ
=68656.275 N
Max. Wind force in columns marked 5
=
ଵ଴ଽ଼ହ଴଴.ସ
଺ସ
×
ସ
√ଶ
=48547.317 N
Consider the windword column 1
Vertical load due to dead +wind load
=596378.399 +68656.275 N

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=665034.674 N.
Corresponding axial load =
଺଺ହ଴ଷସ.଺଻ସ
଴.ଽଽହ଴
=668376.556 N
Horizontal comoponent of the axial forces caused by wind action
=2×68456.275×0.0995+4×48547.317× 0.0995 ×
ଵ
√ଶ
=27285.39 N.
Aactualhorizontal force @ base
= 55272+14220+4608+25560−27285.39 = 72374.61
Horizontal shear column =
଻ଶଷ଻ସ.଺ଵ
଼
= 9046.826 N.
Maximum bending moment for the column = 9046.826×
ଶ.ହ
ଶ
= 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.
Ast = 1884 mm2
,
Equivalent concrete area = Ac+(m-1)Ast = (300×300)+(12.33×1884)
= 113229.72 N
Polar moment of inertia of the equivalent concrete section,
=
௔ర
ଵଶ
+(mAst×effective depth fromcentre),
=
ଷ଴଴ర
଺
+1884×12.33[150-50]2
= 1.582×109
mm4
.
Equivalent moment of inertia about full section =
ଵ.ହ଼ଶ×ଵ଴వ
ଶ
= 791.14×106
mm4.

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Direct stress in concrete =
ௗ௜௥௘௖௧ ௟௢௔ௗ
௘௤௨௜௩௔௟௘௡௧ ௖௢௡௖௥௘௧௘ ௔௥௘௔
= 5.9 mm2
.
Bending stress in concrete =
ଵହ଴×ଵଵଷ଴଼.ହଷଶ×ଵ଴଴଴
଻ଽଵ.ଵସ×ଵ଴ల = 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 section and a doubly reinforced beam with equal steel at
top and bottom.
Ast = Asc =
ଷଵଽ଼ହ.ଷହ଼×ଵ଴଴଴
ଶଷ଴×ଶଶ଴×଴.ଽ
= 702.357 mm2
.
Provide 4 bars of 18 mmdiameter at top and equal amount at bottom.
Shear force for brace =
௕௘௡ௗ௜௡௚ ௠௢௠௘௡௧ ௙௢௥ ௕௥௔௖௘
భ
మ
௦௣௔௡ ௢௙ ௕௥௔௖௘
,
Span of brace =2 ×
଻
ଶ
× sin 22°30′ = 2.678 m.
Shesr force for brace =
ଷଵଽ଼ହ.ଷହ଼
భ
మ
×ଶ.଺଻଼
= 23887.49 N.
Nominal shear stress ߬v =
௏
௕ௗ
=
ଶଷ଼଼଻.ସଽ
ଷ଴଴×ଶ଺଴
= 0.30 N/mm2
.
Provide nomonal stirrups say 2 legged 10 mmdiameter stirrups at 200 mm clear
cover.

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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/m2
,
Area =
௟௢௔ௗ
ௌ஻஼
=
ହଶ଻ସହ଴ଶ.ସଶଷ
ଶ଴଴×ଵ଴య = 26.37 m2
.
Let us provide outer dia of 9.5 m and inner dia of 6.5 m.
=
గ
ସ
[9.52
−6.52
] = 37.69 m2
.
Net intensity =
ହଶ଻ସହ଴ଶ.ସଶଷ
ଷ଻.଺ଽ

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= 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 =
ସ଻ଽହ଴଴ଶ.ଶ଴ଷ
ଶ×଼
(from table 2)
= 299687.63 N.
Design at support section;
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+1.6
்
௕
= 299687.63+1.6
ଵଵହ଴଼.଴଴ହ×ଵ଴଴଴
ହ଴଴
,
= 336550.0176 N.
Equivalent nominal shear ߬v =
௏ೡ
௕ௗ
= 1.14 N/mm2
, but ߬c = 1.8N/mm2
,
Hence ߬c< ߬v .
Longitudinal reinforcement:
Mel = M+Mt , Mt =
்(ଵା
ವ
್
)
ଵ.଻
=
ଵଵହ଴଼.଴଴ହ[ଵା
లఱబ
ఱబబ
]×ଵ଴଴଴
ଵ.଻
= 15569653.82 N ,
Mel = 1000[159194.073+15569.653] = 174763.726×1000 N.
Ast =
ଵ଻ସ଻଺ଷ.଻ଶ଺×ଵ଴଴଴
ଶଷ଴×଴.ଽ×ହଽ଴
= 1430.964 mm2
,

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Provide 9 bars of 16 mm diameter bars.
Hence area os steel required is Ast = 1809.55 mm2
.
Transverse reinforcement:
Asv=
்∙௦ೡ
௕భௗభఙೞೡ
+
௏∙௦ೡ
ଶ.ହௗభఙೞೡ
, providing 4 legged 10 mm diamater of stirrups.
Asv = 4ߨ ×52
= 314 mm2
, b1 = 500-80 = 420 mm , d1 = 650-120 = 530 mm,
314 =
ଵଵହ଴଼.଴଴ହ×ଵ଴଴଴
ସଶ଴×ହଷ଴×ଶଷ଴
+
ଶଽଽ଺଼଻.଼ଷ
ଶ.ହ×ହଷ଴×ଶଷ଴
, 314 = Sr[0.224+0.983] , Sv = 260 mm.
Let us provide 200 mm clear cover spacing.
Steel for hogging mommentAst =
଻ଽ଻଼଼.଼ଷ଺×ଵ଴଴଴
ଶଷ଴×଴.ଽ×ହଽ଴
= 653.31 mm2
,
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×
଴.଺మ
ଶ
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 effective depth and 40 mm clear cover.
D = 210 mm. Ast =
ଶହଵ଼ଽ.ଽ଼×ଵ଴଴଴
ଶଷ଴×଴.ଽ×ଵ଻଴
= 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:

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Provide 0.12 % steel and the steel required is =
଴.ଵଶ×ଶଵ଴×ଵ଴଴଴
ଵ଴଴
= 252 mm2
.
Provide 6 bars of 8 mm diameter bars and spacing =
ଵ଴଴଴
଺
= 160 mm clear cover.
Check for sliding:
Total load on the foundation when tank is empty = 5274502.423-2896117.112
= 2378385.311 N
Horizantal force on the base = 72374.61 N.
Let coefficient of friction = 0.5
Fs=
଴.ହ×ଶଷ଻଼ଷ଼ହ.ଷଵଵ
଻ଶଷ଻ସ.଺ଵ
= 16.43.

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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
OS
L m B
m
A m2
Hor
D
(m)
QTY
m3
REMARKS
1 Earthwork in
excavation
1 73.89 1 73.89 ‫ܣ‬ = ߨ݀ଶ
/4
=ߨ × 9.72
/4
=73.89
Earthwork in
filling
1 64.316
2 a)R.C.C work in
foundation
b)steel in
foundation
i )Longitudanal
ii)Transverse
1
9
4
7.068
ߨ ×0.0082
0.0082
× ߨ
0.2 1.4136
0.045
0.02 = ߨ ×8

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3 R.C.C in columns
Steel in columns
8
8×
6
0.3 0.3 0.09
ߨ ×0.012
10.0
49
10.0
49
7.235
0.151
4 RCC in
Bracings@2.5
Steel in Bracings
@2.5m from G.L
8
8×
8
0.637
5
0.637
5
0.3 0.09
ߨ ×0.0092
0.3 =0.459
=0.01
A=ߨ ×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
ߨ ×0.0092
0.3 0.414
0.0093
6
A=ߨ ×0.0092
=0.000254
6 a)RCC in bracings
7.5m
b)steel
8
8×
8
0.45
0.45 ߨ ×0.0092
0.324
0.0072
A=ߨ ×0.0092
=0.000254
7 Top ring girder
a)R.C.C
b)steel
longitudinal
transeverse
1
5
125
ߨD
6ߨ
6ߨ
0.4 0.24
ߨ ×0.0092
0.6 4.52
0.02
0.066
8 Bottom dome
a)RCC in dome
b) steel
1
L=6.6
2
=22.619
0.067
0.2 4.523
0.443
A=2ߨrh
=2ߨ ×3×1.2
=22.619
‫ܮ‬ =
௫
ଷ଺଴°
× 2ߨr
=6.62
9 a)RCC conical
slab
b)steel
steel for B.M.
1
14
3
23.56
23.56
23.56
0.2
ߨ ×0.0082
2.12
1
9.994
0.066
0.014
ߨ[‫1ܦ‬ + ‫]2ܦ‬
2
= ߨ(9+6)/2
10 a)RCC ring beam
@ B
b)steel
1
6
28.27
0.25
ߨ ×0.0082
0.5 3.534
0.034
11 Cylindrical wall
a)Main steel
b)Distribution
steel
1
20
4
4.32
28.27
0.2
ߨ ×0.0062
ߨ ×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 9 ߨ 0.2 0.04 0.2 1.13 L=ߨD
=9 ߨ

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b)Steel 4 9 ߨ ߨ ×0.0062
0.012
13 Top dome R.C.C
a) concrete
b) Steel
1
100 9.93
2ߨrh
=50.89
ߨ ×0.0042
0.15
0
7.63
0.05
A=2ߨrh
=
‫ܮ‬ =
௫
ଷ଺଴°
× 2ߨr
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=2ߨrh
=50.89
15 Plastering in CM
(1:6) for outer
surface Of conical
dome (12mm)
1 55.135 9.92 A=2ߨrh
=55.135
16 Plastering in CM
(1:2) for Inner
surface Of
cylindrical wall
(12mm)
1 ߨ ×D 28.2 112.8 4 20.354
17 Plastering in CM
(1:6) for outer
surface Of
cylindrical wall
(12mm)
1 ߨ ×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=2ߨrh
=2ߨ ×3×1.2
=22.619
19 Plastering in CM
(1:6) for outer
surface Of domed
roof (12mm)
1 26.38 4.74 A=2ߨrh
=2ߨ ×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 9 ߨ 0.2
5.65
1.01
22 Plastering in CM
(1:2) for ring beam
at bottom (12mm)
1 1.27

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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 ߨ[‫1ܦ‬ + ‫]2ܦ‬
2
= ߨ(9+6)/2
27 Plastering in
CM(1:6) for outer
surface of conical
slab(12mm)
1 4.46
28 Total plastering 105.53
3
29 Thick water proof
cement painting
for tank portion
85.278
30 white washing for
columns
8 0.312 0.31
2
10.04 7.826
31 Total white
washing
93.104

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ABSTRACT
S.NO DESCRIPTION
OF WORK
QTY OR
NOS
RATE
RS PS
COST
RS PS
1 Earth work in
excavation
73.89
2 Beldars 13 250 3250
3 Mazdoors 11 250 2750
4 Total 6000
5 Earth work in
Filling In foundation
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

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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

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COLUMNS
RS
Sand Cum 3.078 375 1154.25
Cement Cum 2.052 1620 3324.24
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
Add MA 20% 2912.198
Add TOT 4% 582.43
RCC RING BEAM AT TOP
RS
Sand Cum 0.48 375 180
Cement Cum 0.32 1620 518.4
capacity Hour 0.26 250 65
Add MA 20% 747.73
Add TOT 4% 149.54

Intze tank
RCC DOMED ROOF 150mm THICK
RS
Sand Cum 3.24 375 1215
Cement Cum 2.16 1620 3499.2
2nd Class Mason Day 5.6 285 1596
capacity Hour 0.267 250 66.75
Add MA 20% 5558.33
Add TOT 4% 1111.61
CONICAL SLAB 200mm THICK
RS
Sand Cum 4.25 375 1593.75
Cement Cum 2.83 1620 4584.6
Add MA 20% 5555.328
Add TOT 4% 1111.06

Intze tank
RCC CYLINDRICAL WALL
RS
Sand Cum 9.62 375 3607.5
Cement Cum 6.41 1620 10384.2
Add MA 20% 10352.066
Add TOT 4% 2070.432
RCC RING BEAM AT BOTTOM OF
CYLINDRICAL WALL
RS
20mm HBG graded metal Cum Cum 3 1076 3228
Sand Cum 1.5 375 562.5
Cement Cum 1 1620 1620
Add MA 20% 1818.7
Add TOT 4% 363.74

Intze tank
RCC CIRCULAR GIRDER
RS
Sand Cum 1.92 375 720
Cement Cum 1.28 1620 2073.6
Add MA 20% 2262.58
Add TOT 4% 452.517
RCC BRACING AT 2.5m HT.
RS
Sand Cum 0.19 375 71.25
Cement Cum 0.13 1620 210.6
Add MA 20% 447.213
Add TOT 4% 89.44

Intze tank
RCC BRACING AT 5m HT.
RS
Sand Cum 0.17 375 63.75
Cement Cum 0.11 1620 178.2
Add MA 20% 426.32
Add TOT 4% 85.264
RCC BRACING 7.5m HT.
A. MATERIALS UNIT QTY RATE
RS
AMOUNT
RS
Sand Cum 0.13 375 48.75
Cement Cum 0.09 1620 145.8
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
Painting 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 coatings : 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
Mazdoor (unskilled) Day 2 250 500
Add MA 20% 3926
Add TOT 4% 786
Total cost 24342

Intze tank
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 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
2nd class painter Day 1 250 250
cost of water proof cement paint Cum 6 35 210
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 painting = 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 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.

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

Intze tank design

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