Design Intze Water tank mazor project Report

MAJOR PROJECT REPORT
ON
DESIGN AND ESTIMATION OF INTZE TANK
Submitted in the partial fulfilment of
the Requirement for the award of the degree of
Bachelor of Technology
In Civil Engineering and Rural Engineering.
By
Guidance By Ravi Singh 124843
Er. Ramakant Tripathi Mayank Prajapati 124829
Amit singh 124811
Batch 2013-14
Department of Civil Engineering and Rural Engineering
Mahatma Gandhi Chitrakoot Gramodaya Vishwavidhalaya Satna (M.P)
2017

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 and limit state method. Elements are design in limit state
method.

ACKNOWLEDGEMENTS
I would like to express my gratitude to all the people behind the screen who helped me to
transform an idea into a real application.
I would like to express my heart-felt gratitude to my parents without whom I would not
have been privileged to achieve and fulfill my dreams. I am grateful to our Dean
I profoundly thank Dr. A.K Pandey, and Head of the Department of CIVIL
And Rural Engineering who most ably run the institution and support in carrying out my
project at college.
I would like to thank my guide Er. Ramakant Tripathi
Asst.Professor for his technical guidance, constant encouragement and who has
been an excellent guide and also a great source of inspiration to my work.
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, I would like thank all the other staff members, both teaching and non-
teaching, who have extended their timely help and eased my task.
Ravi Singh 124843
Mayank Prajapati 124829
Amit singh 124811
B.Tech.(CARE) VIIITh
Semester
Batch-2013 -14

DECLARATION
It is to declared that the major project report titled “DESIGN AND ESTIMATION OF
INTZE TANK” submitted to the Faculty of Engineering and Technology, Mahatma Gandhi
Chitrakoot Gramodaya Vishwavidyalaya is a bonafide record of major project work carried
out by me. No part of this report has been reproduced in any manner or whatsoever, in any
published article or book.
Ravi Singh 124843
Mayank Prajapati 124829
Amit singh 124811
B.Tech. (CARE) VIIITh
Semester
Batch- 2013 -14

FACTULTY OF ENGINEERING AND TECHNOLOGY
(Institute of People’s Science and Technology)
Mahatma Gandhi Chitrakoot Gramodaya Vishwavidyalaya
Chitrakoot, Satna (M.P.) – 485334
Ref.: E&T/16 Date: …. /…… /20….
This is certificate that Ravi Singh , Mayank Prajapati , Amit Singh students of IV year
Bachelor of Technology VIII semester of Civil and Rural Engineering , faculty of
Engineering and Technology “Mahatma Gandhi Chitrakoot Gramodya
vishvidyalaya Satna” have complete their major project entitled.
Design And Estimation of Over Head Intze Tank.
They have submitted their project for the partial fulfilment of the curriculum of the Degree of
Bachelor of Technology (civil and Rural engineering).
Thanking you
Dr. A.K Pandey Guided By
Dean And Head of Department Er. Ramakant Tripathi
Faculty of Engineering & Technology
Phone Dean Office -07670-265682, Registrar -07670-265411, Fax No. -07070-265411
Email – deanfetmgcgv@gmail.com, Website: www.mgcgvchitrakoot.com

SELF ATTESTATION
Certified that I have worked on the project title “DESIGN AND ESTIMATION OF INTZE
TANK”. The results mentioned in the project report have been generated during the work and are genuine.
Data /information obtained from other agencies have been duly acknowledged. None of the finding
/information pertaining to the work has been concealed. The results embodied in this project report have not
been submitted to any other University or Institute for the award of any degree .
RAVI SINGH
B.Tech.(CARE.) VIIIIIh
Semester
Batch- 2013- 14

SELF ATTESTATION
AMIT SINGH
Semester
Batch- 2013- 14

SELF ATTESTATION
MAYANK PRAJAPATI
Semester
Batch- 2013- 14

INDEX
8. WATER TANK
9. CLASSIFICATION
10. DESIGN REQUEMENT OF CONCRETE
11. OVER HEAD WATER TANKS AND TOWERS
12. SITE SELECTION
13. SOIL SAMPLING (TRACTOR DIGGER )
14. SOIL TESTING (C B R TEST)
15. LOAD CALCULATION
S.No Content Page
1. SYMBOL
2. INTRODUCTION
3. OBJECTIVE
4. SCOPE
5. SOURCE OF WATER SUPPLY
6. POPULATION FOR -CASTING
7. WATER QUANITY ESTIMATION

24 SITE LAYOUT
16 DESIGN
17 DESIGN OF INTZE TANK
18 ESTIMATION
19 DETAILED ESTIMATION
20 DATA SHEET
21 ABSTRACT
22 REFERENCE
23 CONCULATION

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
e = eccentricity. = compressive steel depth factor (=dc/d).

F =shear force characteristic load.
Fd= design load
Fr = radial shear force.
f= stress (in general)
fck= characteristic compressive stress.
Fy= characteristic strength of steel.
H = height.
I = moment of inertia.
Ie=equivalent moment of inertia of stress.
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=development length.
l = effective length of column; length; bond length.
M = bending moment; moment.
Mr=moment of resistance; radial bending moment.
Mt=torsional moment.
Mu= bending moment (limit state design)
Mθ=circumferential bending moment
m = modular ratio.
n = depth of neutral axis.
nc=depth of critical neutral axis.
Pa=active earth pressure.
Pp= passive earth pressure.
Pu= axial load on the member(limit state design).

P = percentage steel.
P’= reinforcement ratio.
Pa=active earth pressure intensity.
Pe=net upward soil pressure.
Q= shear resistance.
q = shear stress due to bending.
q’=shear stress due to torsion
R= radius.
s= spacing of bars.
sa= average bond stress.
sb= local bond stress.
T=tensile force.
Tu= torsional moment.
t= tensile stress in steel.
tc= compressive stress in compressive steel.
Vu= shear force due to design load.
Vus=strength of shear reinforcement.
W= point load.
X= coordinate.
xu= depth of neutral axis.
Z= distance.
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 stress in shear reinforcement.
σsy= yield point compressive stress in steel.
μ = co efficient of friction

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
on free expansion or contraction of the structure.

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 underground tank to avoid the tedious calculations.
5. In the end, the programs are validated with the results of manual
calculation given in .Concrete Structure.
SCOPE
1.1 This standard lays down guidelines for layout for overhead water
tanks and criteria for analysis for RCC staging both for steel and concrete
tanks.
1.2 While some of the provisions of this standard in the case of RCC staging
for elevated tanks, though refer to the storage of water, the recommendations
are equally applicable to other materials stored.
1.3 The requirements given in this code applicable for column type staging
and circular and polygonal shaft staging for overhead water tanks.

SOURCES OF WATER SUPPLY:
The various sources of water can be classified into two categories:
Surface sources, such as
1. Streams and rivers (At future).
2. generally, not used for water supplies, at present.
Sub-surface sources or underground sources, such as
1. Wells and Tube-wells.
Population fore-casting
Present population
s.no Department No of
population
1. Teachers 20
2. Non-Teaching staff 20
3. Staff living room 10
4. Students 530
5. Formers 150
Total 730

Future development
s.no Department No of
population
1. Girls Hostel 200
2. Boys Hostel 500
3. Staff Resistance
3.1 class “A” 50
3.2 Class “B” 60
3.3 Class “C” 70
4. Canteen 1200
5. Hospital (bed) 20
Total 2100
No of total present population = 2830 person

Population fore-casting
The various methods adopted for estimating future populations are given below. The
particular method to be adopted for a particular case or for a particular city depends
largely on the factors discussed in the methods, and the selection is left to the discretion
and 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.
“Population fore-casting according to census of chitrakoot population
data”
S.No Year No . of population Increase in
population
1. 1981 9113
2887
6981
4320
2. 1991 12000
3. 2001 18981
4. 2011 23301
5. 2021
6. 2031
7. 2041
8. 2051
Total 14188/3=4729

Using Arithmetic Increase Method
Pn =Po + nx
Pn = total population
Po = present population
n = no. of decay
x = increase in population
P2021 = 23301 + 1*4729
= 28030
P2031 = P2021 + 2*4729
= 28030 + 2*4729
= 37488
P2041= 37488 + 3* 4729
= 51663
P2051= 51663 + 4* 4729
= 70579
According to this calculation of the increase population at rate 33%
Hence design population is 3764 person

Water Quantity Estimation
The quantity of water required for municipal uses for which the water supply scheme
has to 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
Water Consumption Rate
It is very difficult to precisely assess the quantity of water demanded by the public,
since there are many variable factors affecting water consumption. The various types
of water demands, which a city may have, may be broken into following class
Water Consumption for Various Purposes:
s/n Use Consumption in l/h/d
1 Drinking 5
2 Cooking 5
3 Bathing 55
4 Bathing of utensils 10
5 Washing and clinging of houses and
residences
10
7 Flushing of water closet etc 30
8 Washing of clothes 20
Total 135 l/h/d

Water consumption
Particular students Teachers Staff
(family)
Non-
teaching
Former Hostel Canteen Staff
hostel
Hospital
Flushing
Of W.C.
30 30 30 30 30 30 X 30 200
Washing
Of house
x X 10 X X 10 X 10 X
Washing
Of utensils
x X 10 X 10 10 10 10 10
Washing of cloth x X 20 X 20 20 X 20 250
Bathing x X 55 X 55 55 x 55 55
Cooking x X 5 X 5 5 5 X 5
Drinking 5 5 5 5 5 5 3 5 20
Water Consumption
(l/h/d)
35 35 135 35 125 135 18 130 630
No .of population 530 20 10 20 150 700 1200 180 20
Total (l/d) 18550 700 1350 700 18750 94500 21600 23400 12600

Total average water consumption = 200000 L/D
Maximum water consumption = 1.8 * average water consumption
Maximum water consumption = 1.8* 200000 = 360000 L/D
Design of maximum water demand = 400000 L/D
Factors affecting per capita demand:
• Size of the city: Per capita demand for big cities is generally large as compared
to that 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 private
wells, 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.
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 the
fluctuations.
• 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
Sundays and Festival days, thus increasing demand on these days.
• Hourly variations are very important as they have a wide range. During active
household working hours i.e. from six to ten in the morning and four to eight in

the evening, the bulk of the daily requirement is taken. During other hours the
requirement is negligible. Moreover, if a fire breaks out, a huge quantity of
water is required to be supplied during short duration, necessitating the need
for a maximum rate of hourly supply.
So, an adequate quantity of water must be available to meet the peak demand. To
meet all the fluctuations, the supply pipes, service reservoirs and distribution pipes
must be properly proportioned. The water is supplied by pumping directly and the
pumps and distribution system must be designed to meet the peak demand. The effect
of monthly variation influences the design of storage reservoirs and the hourly
variations influences the design of pumps and service reservoirs. As the population
decreases, the fluctuation rate 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

WATER TANKS
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
1 DESIGN REQUIREMENT OF CONCRETE (I. S. I)
In water retaining structure a dense impermeable concrete is required
therefore, proportion of fine and course aggregates to cement should be such as to give
high quality concrete. Concrete mix weaker than M20 is not used. The minimum
quantity of cement in the concrete mix shall be not less than 30 k N/m3.The design of
the concrete mix shall be such that the resultant concrete issue efficiently impervious.
Efficient compaction preferably by vibration is essential. The permeability of the
thoroughly compacted concrete is dependent on water cement ratio. Increase in water
cement ratio increases 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 of liquid retaining structure is different from
ordinary R.C.C, structures as it requires that concrete should not crack and hence
tensile stresses in concrete should be within permissible limits. A reinforced concrete
member of liquid retaining structure is designed on the usual principles ignoring tensile
resistance of concrete in bending. Additionally, it should be ensured that tensile stress

on the liquid retaining ace of the equivalent concrete section 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 of concrete with
interposition of some material to break the bond and facilitate movement. In case
length of structure is large it should be subdivided into suitable lengths separated by
movement joints, especially where sections are changed the movement joints
should be provided. Where structures have to store hot liquids, stresses caused by
difference in temperature between inside and 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.

GENERAL DESIGN REQUIREMENTS (I.S.I)
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.
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.
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:
(i) Tensile stress in member in direct tension 1000 kg/cm2
(ii) 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
(iii)On face away from liquid for members 225mm or more in thickness
1250 kg/cm2
(iv) 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
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
storage of water or aqueous liquids at or near ambient temperature and the
circumstances are such that the concrete will never dry out.
(ii)Shrinkage stresses may however be required to be calculated in special cases, when a
shrinkage co-efficient of 300 x 10-6may 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.
5 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-
(i) Tank portion with
 Roof and roof beams (if any)
 sidewalls
 Floor or bottom slab
 floor beams ,including circular girder
(ii) Staging portion, consisting of
 Columns
 Bracings
 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.
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 the
meridians and latitudes. The rise is usually l/5th of the span.
(2) Ring beam supporting the dome. The ring beam is necessary to resist the
horizontal component of the thrust of the dome. The ring beam will be designed
for the hoop tension induced.
(3) Cylindrical walls : This has to be designed for hoop tension caused due to
horizontal 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 the
induced hoop 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 supported on
the ring girder.

(7) The ring girder: This will be designed to support the tank and its contents. The
girder 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 be braced at intervals and have to be designed for wind pressure or
seismic loads whichever govern.
(9) Foundations: A combined footing is usual provided for all supporting columns.
When this is done it is usual to make the foundation consisting of a ring girder
and a circular slab.

Site selection
Criteria
1. Soil testing
2. elevation of site
3. according To height of building
4. area of site
5. according Water supply system
6. according To source of water

Soil testing (C.B.R. test)
Processer
1. soil sampling from tractor digger.
Fig: - tractor digger

2. California bearing ratio test
Fig :- CBR with apparatus
CBR value =
Test load
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑙𝑜𝑎𝑑
x 100

Standard loads at specified penetrations
S.No Penetration
depth (mm)
Unit standard
load kgf/cm2
Total standard
load (kgf)
1. 2.50 70 1370
2. 5.00 105 2055
3. 7.50 134 2630
4. 10.00 162 3180
5. 12.50 183 3600
Report
 Report the CBR value to the nearest second decimal.
 Take the average of 3 test specimens as the CBR value of the
test.
 Generally CBR value at 2.5 mm penetration will be greater
then that 5.0 mm penetration in such case take the value at
2.50 mm as the CBR value .
If the CBR value corresponding to a penetration of 5.0mm exceed that
of 2.50 mm repeat the test .
SAFE BEARING CAPACITY = 110 KN /M2

LOADS
Dead Loads - Dead loads shall be calculated on the basis of unit weights taken in
accordance with IS : 1911-1967*. Unless more accurate calculations arc warranted, the unit
weight of reinforced concrete made with sand and gravel or crushed natural stone aggregate
may be taken as 2 500 kg/ms. Loads due to pipings and stair cases should also be
considered.
Dead load = 2500 N/ m2
Imposed Loads - Imposed loads like live loads, snow loads and wind loads shall be in
accordance with IS : 875-1964t. Weight of the water may be taken as live load for members
directly containing the same. The weight of water shall be considered as dead load in the
design of staging.
Weight of water = 1500 N/ m2
Wind Load - Wind load shall be applied in accordance with IS : 875-19641 while
analyzing the stresses the combination shall be as follows:
a) wind load with tank empty; and
b) wind load with tank full. The worst combination of the stress on account of the
above shall be considered while working out the permissible stresses.
Wind load = 1230.58 N/m2
Seismic Forces - When seismic loading is considered, following two cases may be
considered :
a) tank empty = 150.53 KN
b) tank full. = 404.3 KN
In addition, wherever required the effect of surge due to wave for- motion of the water may
be considered. The seismic force acting on the support for the tank and its analysis shall be
in accordance with IS : 1893-1975.
Vibration Forces - Vibration forces such as due to blast forces ( see IS : 6922-1973s ) as
experienced in mines, collieries and in the close proximity of railway tracks shall be
considered in the design.

Design of intze tank
Design of an intze tank for a capacity of = 400000 litre
Assuming height of tank floor above ground level---( Gl + 3) =16 m
Safe bearing capacity of soil = 110 kn / m 2
.
wind pressure as per is 875 = 1230 .8 n/m2
.
assuming M20 concrete [ IS code 456-2000].
σcbe = 7N/mm2
, σcc = 5N/mm2
permissible stress in concrete [Is cod 456-1964] for M20
direct tension (σt) = 1.2 N/mm2
Tension in bending = 1.70 N/mm2
Modular ratio m = 13
Permissible stresses in steel reinforcement according to Is 482 (part -1)
Type Plain
round mild
steel
HYSD
Tensile stress in direct
tension
115
N/mm2
150
N/mm2
Tensile stress in bending on
liquid face for t < 225 mm
and 125 N/mm2
for / > 225
mm.
115
N/mm2
150
N/mm2

1. Diameter of tank
Volume of water stored in tank = 0.585 D3.
D= internal diameter of intze tank
Assume D=8m v=0.585x 83=300.0 m2. [fail]
V original =400 m2.
D =8.8 m and 9 m2 are also fail.
So, for D = 10 m
H =2/3D = 6.67 m
Do =5/8D = 6.25 m
H0 =3/16D =1.87m
H2 =1/8D = 1.25 m
H1 = 5 m (assume)
H2( 2R2 – H2) =(Do/Ho)2
1.25(2R2-1.25) =3.332
R2 = 5.069m
Sin ф2 = 3.33 / 5.069
Sin ф2 = 0.6569
ф2 = 41.060
capacity of tank: -
V= (π D2 H)/ 4 + [π H0(D2+ D0
2+DD0)]/4 –[π H2
2 (3R2-H2)
400=(πx102H)/4 + [π x 1.875(102+ 6.252+ 10x6.25)] /12-[πX1.252 (3x5.069
-1.25)] /3
H=4.12 m
We add free board height in height of dome
So, total height = height + free board height.
Let assume value of free board according to IS provision,
Free board = 0.8 m in case water storage tank.
So ,Total height = 4.12 + 0.8
= 4.92 m = 5.0 m
Hence height of dome is H=5.0

2.DESIGN OF TOP DOME
FOR top dome, the radius R1 is
1.50(2R1-1.50) = (5)2
R1 = 9.08 m
Sin ф1 =5/ 9.08
ф1 =sin -1
0.5504
ф1 = 33.390
let thickness of top dome is t1=100 mm =0.1 m
live load on tank is 1500 n/m2
according to [ IS code 875 - 1964] provision.
total pressure (P) per square meter of dome = .1x 25000 +1500
= 4000 n/m2
MERIDONIAL THRUST at edge of dome
T1= PR/(1+cosφ)
= (4000 x 9.08)/1+cos (33.390
)
= 19793.52 N/m
Meridional stress = 19793.52 / 1000 x 100
= 0.1979 N / mm2
Maximum hoop stress at crown (at middle) =PR1 / 2T1
= 4000 x 9.08 / 2 x 0.1
= 181600.0 N / m2
= 0.1816 N / mm2
(SAFE)
The stress are within the safe limit provide normal reinforcement @ 0.3%
As = 0.3 x 100 x 1000 /1000
= 300 mm2
Using 8mm ф bars,
Aф = π x 82
/ 4
= 50 mm2

SO,
SPACING= 1000 x 50 / 300
= 160 mm
Hence , provide 8mm ф bars @ 160 mm c/c in both the direction
3. DESIGN OF TOP RING BEAM B1
Horizontal component of T1 is given by
P1 = T1 cos ф1
= 19793.52 cos (33.39)
= 16526 N/ m
Total tension bending to rupture the beam = P1 x D /2
= 16526x 10/2
= 82632 N
Permissible stress in high yield strength deformed bars (HYSD bar)
=150 N / mm2
(according to IS code)
Ash = 82632 /150
= 550 mm2
Since, number of 14 mm ф bars Aф = 153.93 mm2
=550/ 153.93
= 3.57 ≅4 nos
Actual Ash provided = 153.93 x 4 = 579 mm2
Hence safe 550 mm2
< 579 mm2
.
The area of cross section of rings beam is given by
= 82632/ 1.2
= 68860.0 mm2
Using ring beam 230 x 300 mm
Area provided are = 69000 mm2
> 68860 mm2
(SAFE)
So, provided ring beam of 230 mm depth and 300 mm width .
Use 6mm diameter nominal reinforcement stirrup @ 150 mm c/c .

4.DESIGN OF CYLINDRICAL WALL
Height of the wall = 5.0 m
W =11000 N/mm2
P = WHD/2
P = 11000 x 5 x10 /2
= 275000 N/m height
Area of steel Ash= 27500 / 150
= 1833.33 mm 2
per meter height
Providing rings on both the faces Ash on each face
= 1834 /2
= 917 mm2
Spacing of 12 mm ф rings = 1000 x 113 / 917
= 123.2 mm
Use 12 mm bar @ 120 mm c/c both face.
Actual Ash provide = 1000 x 113 /120
= 941.66 mm2
Permitting 1.2 N / mm2
stress on composite section
= 275000/(1000T + 12 x 941.66 X 2)
T = 206.56 mm
Minimum thickness = 3H +5
= 3 x 5 + 5
= 20 cm
T =250 mm at bottom, tapped 200 mm at top
Average T = 250 + 200 /2 = 225 mm
Minimum Ash = 0.3 [250 -100 / 450 -100] x0.1
= 0.24
Maximum Ash = (.24 x 250 x 1000) / 100
= 650 mm2
Area of steel of each face = 325 mm2
Spacing of 8mm ф bars = 1000 x 50.3 / 325
= 155 mm
Hence provide 8mm ф bars @ 150 mm c/c on both face and a clear cover of 25 mm.
To resist the hoop tension at 2m below top .
Ash = 2 x 1833.33 / 5
= 733.33 mm2
since Spacing of 12 mm ф rings = 1000 x 113 / (733.33/2)
= 308 mm
Hence provide the rings @ 300 mm c/c in the top 2m height.

At 3m blow the top
Ash =3 x 1833.3 / 5
= 1099.98 mm2
=1100 mm2
Spacing of 12 mm ф rings = 1000 x 113 / (1100 /2)
= 241.81 mm
Hence provide rings @ 240 mm c/c in the next 1m height .
At 4m below the top
Ash = 4x 1833 /5
= 1466.4 mm2
Spacing of 12mm ф rings = 1000 x 113 / 1466.412
= 154.11 mm
Hence provide rings @ 150 mm c/c for the next 1m height.
At 5 m below the top
Ash = 5x 1833 /5
= 1833 mm2
Spacing of 12mm ф rings = 1000 x 113 / 1833
= 61.64 mm
Hence provide rings @ 60 mm c/c .
5 DESIGN OF RINGS BEAM B3
Top of conical dome consist of the following.
1. Load of top dome= T1 sinф = 19793.52 sin(33.39)
= 10893 N/m
2. Load due to the ring beam B1
0.36(.4 - .2 ) x 1x 25000 =1800 N/m
3. load use to tank wall = 5( 0.2+0.3 /2)x 25000
= 31250 N/m

4. Self-load of beam B3 (1m x 0.6 m )
= (1-0.3) x 0.6 x 25000
= 10500 N/m
Therefore, total W = 10893 + 1800 + 31250 + 10500
= 54443.0 N/m
Inclination of conical dome wall with vertical =ф0 = 450
sin ф0 = cos ф0 =0.7071 =1/√2
tan ф0 = 1
PW = W tan ф0
= 54442 x1
=544443 N/m
Pw = w Hd3
= 11000 x 5 x 0.6
= 33000 n/m
Hence hoop tension in the ring beam is given by
P3 = (PW + Pw) D/2
= (54443 + 33000) X10 /2
= 437215 N
This to be resisted entirely by steel hoops, the area of which is
Ash = 437215 /150 2914.76 N/mm2
Number of 25 mm ф bars = 2914.76 / 490.87
= 5.9 =6 nos
Hence provided 6 rings of 25 mm ф bars.
Ash =490.87 x 6
=2945.2 mm2
Stress in equivalent section = 437215 / (1000 X 600)+ (12x29454)
= 0.69 N/mm2
< 1.2 N/mm2
Hence, safe
The 8 mm ф distribution bar provided in the wall @ 150 mm c/c.

6. Design of conical dome
A. Meridional thrust
Ww weight of water
Ww =π [102
– 6.252
]x 5x11000 +{ (π x 2 x11000/12) x(102
+ 6.252
+10x 6.25)} – (π x 6.252
x 2 x 11000)
= 2632311.032 +1160916.660 – 674951.5467
= 3118276.145 N
Let the thickness of conical slab be 400 mm.
Total self weight Ws
Ws = {π[(D+D0)/2] x L T0} ϒC
Slanting length L and thickness T0
Ws =25000 π(10 +6.25 /2)x 2√2 x 0.4
= 721968.4775N
Weight W at B3 = 54443 N /m
Hence vertical load W2 per meter run
W2= π x DW + W w + Ws/ πD0
W2 =(π x 10 x 54443 + 3118276.145 + 721968.4775)/π x 6.25
= 282690.85 N
Hence meridional thrust T0 in the conical dome is
T0= w2/(1+cosφ0)
= 282690.85 √2
T0 = 399785.23 N/m
Meridional stress = 399785.23/ 1000 x 400
= 0.99 N/ mm2
(safe)
B. Hoop tension
Diameter of conical dome at any height h‘ above base is
D’ =6.25 +(14-6.25)h’ /2

= 6.25 + 1.87 h’
Intensity of water pressure P= (5+2-h’)x 11000
= (7-h’)11000 N/ m2
Self weight q = 0.4 x 1 x1 x 25000
= 10000 N/ m2
Hence hoop tension P0
‘
is given
P0
‘
=[(𝑝 ÷ cosφ0 ) + qtanφ0] D’/2
= [(7-h’)11000 x √2 + (1000 x tan 45 )] x (6.25 + 1.87 h’) /2
= 479844.5 + 46509.6 h’ – 14467.3 h’ 2
The value of P0
‘
at h’ =0 , h’ =1 , and h’ =2
For maximum d P0
‘
/ dh’ = 0
0 = 46509.6 -2x14467.3 h’
H’= 1.60 m
Maximum P0
‘
= 479844.5 + 46509.6 x 1.60 – 14467.3 X 1.62
= 513912 N
c. Design of walls
meridional stress = 0.99 N/ mm2
max hoop stress = 513912
h’ Hoop tension (N)
0 479844.5
1 511886.8
2 512994.5

whole of which is to be resisted by steel As
As = 513912 /150 = 3426 mm2
Area of each face = 1713 mm2
Spacing of 16mm ф bar Aф =201 mm2
Ash = 1000 x 201 / 1713
= 117.3 mm
Hence provide 16 mm ф hoop @ 110 mm c/c on each face
Actual AS = 1000 x 201 / 110
= 1827.27 mm2
Max. tensile stress in composite section
=513912/ (400 x 1000 )+(12 x 1827 x 2)
= 1.157 N/ mm2
Hence this is less tan the permissible value of 1.2 N/ mm2
In the meridional direction ,provide reinforcement @
={0.3-[400-100 / 400 -100] 0.1 } %
Asd = 0.21 x 3654
= 767.34 mm2
Or 384 mm2
on each face
Spacing of 10 mm ф bars Aф =78.5mm2
= 1000 x 78.5 /384
= 204 mm
Hence provided 10 mm bars ф @ 200 mm c/c on each face .
provide a clear cover of 25 mm.

7. Design of bottom dome
R2 = 5.069 m
Sin ф2 = 0.6569 cos ф2=0.7539
ф2 = SIN-1
(0.6569) ф2=cos(0.7539)
ф2 = 41.060
ф2=41.060
weight of water W0 on the dome
W0 ={(π x D0
2
x H0 )4 – π D0
2
x(3x R2 – H2)/3 } x W
W0 ={(π x 6.252
x 6.87 )4 – π x 1.252
(3x 5.065 – 1.25 )/3 } x 11000
= [207.76- 22.81]X11000
= 2067548.56 N
Let the thickness of bottom dome be 250 mm .
self weight = 2 x π R2 h2 T2 x 25000
= 2 x π x 5.069 x 1.25 x 0.25 x 25000
= 248823.95 N
Total weight WT =2067548.56 + 248823.95
= 2316372.513 N
Meridional thrust T2 =WT/π D0 Sin ф2
= 2316372.513 / π x 6.25 x 0.6569
= 179588.80 N/m
Meridional stress = 179588.8 /250 x 1000
= 0.72 N/ mm2
Intensity of load per unit area P2
P2 = Wt /2 x π R2 h2
= 2316372.513/2 x π x 5.069 x 1.25
= 58183.01 N/m2
Max hoop stress at Centre of dome
= P2 R2 / 2 T2

= 58183.01 x 5.069 / 2x 0.2 = 0.58 N/ mm2
(safe)
Area of minimum stress = {0.3-[250-100 / 450 -100] 0.1 } %
= 0.26%
As = 0.26 x 2500 = 650 mm2
in each direction
Spacing of 10 mm ф bars= 100 x78.5 /65 = 121mm
Hence provide 10 mm ф bars 120 mm c/c in both direction .
also provide 16 mm ф meridional @ 110 mm c/c .
8. Design of bottom circular beam B2
Inward thrust from conical dome = T0 Sin ф0
= (39985.23 x 1)/√2
= 282690.6845 N/m
Outward thrust from bottom = T2 cos ф2
= 279588.8 x 1.7539
= 265391.9 N/m
Net inward trust = 282690 – 265391
= 17299.0 N/m
Hoop compression in beam = 17299x 6.25 /2
= 54059.37 N
Assuming the size of the beam to be 500 x 1000 mm
Hoop stress = 54059 / 500 x 1000
= 0.108 N/mm 2
Vertical load on beam , per meter run = T0 cos ф0 + T2 sin ф2
= 399785.23 /√2 + 279588.9 x 0.6569
= 466352.5 N/m.

Self weight = 0.5 x 1.0 x 1x 25000
= 12500 N/m
Load on beam = W = 466352.5 + 12500
= 478852.5 N/m
Let us support the beam on 8 equally spaced columns at a mean diameter of 10
m . Mean radius of curved beam is R = 5 m.
2 Ѳ = 450
= π/4 ;
Ѳ = 22.50
= π /8 radians
C1 = 0.066 ;
C2 = 0.03;
C3 = 0.005 ;
Фm = (19/2)0
WR2
(2Ѳ) = 478852.5 x 52
X π/ 4
= 9402246.851 N/m
Maximum negative bending moment at support M0
M0 =C1 x WR2
(2Ѳ)
= 0.066 x 9402246.85
= 620548 .29 N/ m
Maximum bending moment at support Mc
Mc = C3 x WR2
(2Ѳ)
= 0.03 x 9402246.85
= 282067.40 N/m
Maximum torsional moment = Mt
m = C3 x WR2
(2Ѳ)
= 0.005 x 9402246.85
= 47011.23 N-m
For M20 concrete (σcbe =7 N/mm2
) and HYSD bars (σst = 150 N/mm2
)
[IS Code].

We have k = 0.378 ; J = 0.874 ; and
R = 1.156 ;
Therefore required effective depth =
√620548.29𝑥1000
500 𝑥 1.156
= 1036 mm
However , keep total depth = 1200 mm from shear point of view
Let D = 1160 mm
Maximum shear force at support F0 = WR Ѳ
= 478852.5 x 5x π/8
= 940224.68 N
Shear force any point is given by
F =WR (Ѳ-ф)
At ф = Фm = 478852.5 x 5 x[22.50
- 9.50
] x π / 180
= 543240.92 N
Bending moment at the point of maximum torsional moment (ф = Фm= (19/2)0
)
Mф = WR2
(Ѳ sin ф + Ѳ cotф-1)
=478852.5 x 52
[(π sin 9.5)/8 + (π cot 22.5 cos 9.5 )/8-1]
= - 1612 N/m (sagging )
= 1621 N/m (hogging )
The torsional moment at point is
Mф’ = WR2
[Ѳ Cos ф – Ѳ cot Ѳ sinф –(Ѳ-ф)]
At the support ф =0 Mф’ = WR2
(Ѳ-ф)=0
At mid span ф = Ѳ = 22.50
=π/8 radians
Mt
ф = WR2
[ Ѳ cosѲ – Ѳ( cos ф / sin ф )sinф ] =0
Hence we have the following combination of bending moment and torsional moment .
(a) At the support
M0 = 620548.29 N/m [Hogging or negative ]
M0
T
= zero

(b) At mid span
Mc = 282067 N/m (sagging or positive )
M0
T
= zero
(c) at the point of maximum tension ( ф = Ѳ = 19/20
)
Mф =1612 N/m (hogging or negative )
Mt
m =47011.23 N/m
Main and longitudinal reinforcement
(A) Section at point of maximum torsion
T = Mt
MAX = 47011.23 N/m
Mф =m =1612
Me1 = M + Mt
Mt =T [1+(D/6)/ 1.7 ]
= 47011.23 [1+(1.2/0.5)/1.7]
= 94022.46 N/m
Therefore Me1 = 1612 + 94022
= 95634.46 N/m
Ast1 = Me1 / σst + J x d
= 95634 x 1000 / 150 x 0.874 x 1160
= 628.85 mm2
Therefore no of 25 mm ф bars = 1
However provide a minimum of 2 bars thus at the point of maximum torsion provide 2- 25
mm ф bars each at top and bottom .
(b) section at maximum hogging bending moment (support)
M0 = 620548.29 N/m = Mmax
M0
T
= zero
Ast = 620548 x 1000 / (150 x 0.874 x 1160)
Ast = 4080.5 mm2
Therefore no of 25 mm ф bars = 4080/491

= 8.3 ~ 8
Hence provide 6nos of 25 mm ф bar in one layer and 2 bar in the second layer .
(c) Section at maximum sagging (bending moment ) (mid span)
Mc = 282067 N/m M’c =0
Therefore for positive bending moment steel will be to the other face where stress σst can
be taken as 190 N/mm2
.the constant for M20 concrete
Having C =7 N/mm2
and m=13 will be
K =0.324 J =0.892 AND R=1.011
Ast = 282067 x 100 /(190 x 0.892 x 1160)
= 1434.7 mm2
No of 25 mm ф bars =1434.7 / 490
= 2.92
At the support provide 6- 25 mm ф bar at top layer and 2-25 mm ф bar in second layer.
Continuous these upto the section of maximum tension (Фm = 9.50
= 0.166 rad)
At a distance = R Фm = 5 x 0.166 = 0.83 m
Equal to Ld = 52 ф = 1300 mm from support provide 4 bars of 25 mm ф at the bottom .
Transverse reinforcement
(a) at point of maximum torsional moment at the point of maximum torsion
V= 543240.92 N
VE = V + 1.6 T /B
V = shear force
VE = equivalent shear
Where T = Mt
m = 47011.23 N/m
B= 500 mm = 0.5 m
Ve = 543240.9+ (1.6 x 47011.23) / 0.6
Ve = 668604. 2 N
Therefore Ꚍve = Ve / b d

= 668604.2 /(500 x 1160)
= 1.15 N/mm2
This is less than Ꚍc max = 1.8 N/mm2
for M20 concrete ; hence ok
100 As / bd =100 x 4x 491 / 500 x 1160
= 0.338
Ꚍc = 0.23 N/mm2
Since Ꚍve > Ꚍc shear reinforcement is necessary the area of cross section Asv of the stirrups .
Asv = T . Sv /b1 .d1 σsv + V.Sv /2.5 d1 σsv
Where , b1 = Centre of Centre distance between corner bars in the direction of the width .
B1 = 500 –(40 x 2) -25 = 395 mm
D1 = 1000 – (40x2) – 25 = 895 mm
Therefore Asv /Sv=47011 x1000 /(395 x 895 x 150 ) + 543240.3 /2.5x 895 x 150
= 0.886 + 1.61
= 2.50
Minimum transverse reinforcement is governed by
Asv /Sv > ((Ꚍve - Ꚍc)/ σsv )b
Asv /Sv = ((1.15 – 0.23 )/150 )x 600 )
= 3.68
Hence depth Asv /Sv = 3.68
Using 14 mm ф 4 lagged stirrups Asv = 4 x 153.9
= 615.75 mm2
Sv = 615.75 / 3.68= 168 mm
However , the spacing should not exceed the least of X1 , X1 + Y1 /4 and 300mm
Where X1= 395+ 25+12 = 432 mm
Y1 = 895 + 25+ 12 = 932 mm

X1 + Y1 /4 = 432+932 /4
= 341 mm
Hence provide 14 mm ф 4 lagged stirrups @ 170 mm c/c.
(b) at the point of maximum hear (support)
At support F0=940224.685 N
Ꚍv =940224.6 /(500 x 1160 )
= 1.62 N/mm 2
At support 100 As / bd =100 x 8x 491 / 500 x 1160
= 0.677
Ꚍc 0.31 N/mm 2
. hence shear reinforcement is necessary
Vc = 0.31 x 500 x 1160
= 179800
Therefore V s = F0 – Vc
= 940224.6- 179800
= 760424.6
The spacing of 12 mm ф 4 lagged stirrups having Asv = 314 mm2
sv = Asv σsv d / Vs
= 150 x 314 x 1160 /760424.6 = 71.84 mm (fail )
This is small, hence aare 12 mm ф 4 lagged stirrups
Asv = 4x πx 12 2
/4 = 452.3 mm 2
Spacing
Sv = 150x 452.3 x 1160 / 760424.6
= 103.5 = 110 mm
(c) At mid span : at the mid span shear force is zero hence provide minimum nominal
shear reinforcement
Asv / b.Sv = > 0.4 /Fy
For HYSD bars ,Fy = 415 N/mm 2
Asv /Sv = 0.4 x 500 / 415

= 0.48
Choosing 10 mm ф 4 lagged stirrups Asv = 314 mm2
Sv =314/0.48
= 615.5 mm
Maximum permissible spacing =0.75 d
= 0.75 x 1160
= 870 or 300 mm
Which ever is less hence provide 10 mm ф 4 lagged stirrups @ 300 mm c/c.
Side face reinforcement
Since the depth is more than 450mm provide side face reinforcement @ 0.1%
Al =0.1(500 x 1000 ) /100 = 500 mm2
Provide 3- 16 mm ф bar on each face having
total Al = b x 201
= 1206 mm2
9 . Design of columns
The tank is supported on columns.
Placed on a circle of 10m mean diameter.
Height of stagging above ground level is
Let us divide this height into four panels each of 4m height .
Let the column be connected to raft foundation by mean of a ring beam.
The tope of which is provide of 10m below the ground level so that actual height of bottom
penal is 5m.
(a) vertical load on columns.
(1) weight of water = ww + w0

=3118276.145 + 2067548.56
=5185824 N
(2) weight of tank
(i) weight of top dome + cylindrical well = w
W= 54443 x r x 14 =2394528.204 N
(ii) weight of conical dome ws =721968.47 N
(iii) weight of bottom =248823.95 N
(iv) weight of bottom ring beam
18000 x r x 10 = 565487 N
(v) total weight of tank = 3930807.5 N
Total superimposed load = 518524 + 3930807
= 9116631.2 N
Load per column = 9116631.2 /.8
=1139578.9 N
Let the column be of 600 mm diameter .
Weight of column per meter height
=π (.60)2
x1 x 25000
=7068.58 N
Let the brace be of 250 x 500 mm
Length of each brace =L =(R sin 2π/R) /cosπ/n
=5 x sin π/4 /cos π/8
=3.83 m
Clear length of each brace
= 3.83 – 0.6 = 3.25 m
Weight of each braces
= 0.25 x 0.5 x 3.26m x 25000
= 10187.5 N

Hence total weight of column just above each is tabulted below .
Brace GH
W = 1200000 + 4 x 7068.5
= 1228274 N
Brace EF
W = 1200000 + 8 x 7068 + 10187.5
= 1266731.5 N
Brace CD
W = 1200000 + 12 x 7068 + 2 x 101875
= 1200000 + 84816 + 20375
= 1305191 N
Bottom of column
W = 1200000 + 17 x 7068 + 3 x 10187.5
= 1200000 + 120156 + 30562.5
= 1350718.5 N
(b) Wind load
Intensity of wind pressure = 1500 N/m2
Let us take a shape factor of 0.7 for circular section in plan.
Wind load on tank , dome and ring beam
= {( 5 x 10.4 ) + ( 10.2 x 1.5 x 2/3 ) +(2 x8.2 ) + (6.85 x 1.2 )x 1500 x 0.7
= 115720 N
This may be assumed to at about 5.57m above the bottom of ring beam
Wind load on each panel (braces) of 4m
Height of column = (4 x 0.7 x 8 ) 1500
= 29685.0 N
Wind load at top and of top panel
= 1/2 ( 4 x 0.7 x 8 x 1500 x 0.7 )

= 1/2 x 23520
= 11760 N
The point of contra flexure O1 , O2 , O3 and O4 are assumed to be at the mid height of each
panel .The shear force Qw and moment Mw due to wind at these planes are
Level
Qw
Mw (N/m)
O4 146480 114720 x 7.57 + 111760 X 2 = 891950 N
O3 179540 114720 x11.57 + 111760 X 6 + 29685 x 2 =1457240 N
O2 212600 114720 x15.57 + 111760 X 10 + 29685 x 8 =2141270N
O1 245660 114720 x20.07 + 111760 X 14.5+ 29685 x 19.5 =
3051807N
The axial thrust ϒmax =4 Mw / n D0
= 4 Mw /8 x6.25
= 0.08 Mw
In the farthest leeward column
The shear force Smax = 2Qw /n
= 0.25 Qw
In the column on the bending axis at each of the above level and
The bending moment M =Smax x H/2 in the column.
Level Vmax =0.08 Mw Smax(N) = 0.25
Qw
M
=Smax
x H/2
O4 0.08 x 891950 = 71356 0.25 x 146480 =
53040
73240
O3 0.08 x 1457249=116579.2 0.25x 179540 = 44890 89780
O2 0.08x2141270 = 171301.6 0.25x 212600= 53150 106300
O1 0.08 x 3051807 =244144 0.25 x 245660 =
61420
153550

Panel Farthest leeward column Column on bending axis
Axial load Vmax Axial load M (N/m)
O4 1228274 71356 1228274 73240
O3 1266731 116579.2 1266731 89780
O2 1305191 171301.6 1305191 106300
O1 1350718.5 244144 1350718.5 153550
Use M20 concrete for which and σcbe = 7N/mm2
and
σcc = 5N/mm2
for steel σst = 230 N/mm2
Dimeter of column = 600 mm use 12 bars of 30 mm diameter at an effective cover of 40 mm
Asc =π x 30 2
x 12 /4
= 8482 mm2
equivalent area of column
= π x 6002
/4 + (13-1) 8482
= 384527 mm2
equivalent moment of inertia
Lc= π x d4
/64 + (n-1) Asc d’2
/8
. d= 600 mm ;
d’ = 600- 2 x40 = 520 mm
Lc= π x 6004
/64 + (13-1) 8482 x 5202
/8
= 9.8 x 109
mm4
= 1.0 x 1010
mm4
Direct stress in column = σcc’
= 1941795 / 384527
= 5.04 N/mm

Bending stress in column = σcbc’
= (153550 x 1000 x 350) / (1.0 x 1010
)
= 5.37 N/mm2
For the safety of the column , we have the condition
σcc'/ σcc + σcbc’/σcc >= 1
5.04/1.33x 5 + 5.37/ 1.33 x 5 >1
0.757 + 0.577 > 1
1.334 >1 (hence un safe )
Hence increase diameter from 600 mm to 700 mm
So ,
3.99/ 1.33x 5 + 3.22 /1.33 x 7 >1
0.95<1 Hence safe
Use 10 mm ф wire rings of 250 mm c/c to tie up the main reinforcement .
Since the column are of 700 mm diameter , increase the width of curved beam B2 from
600 mm to 700 mm.
10. Design of bracing
Ꜫ m at joint above = 20587.64 + 9167 (2)
= 38921.64 N-m
Moment of bracing = 38921.64 √2
Hence providing 300 mm x 300 mm section and design as doubly
Reinforced beam with equal steel of p and bottom
Ast = Ast =[( 38921.64 (1.2( 10000 ] / 0.87 9415 ) x260
= 497.54 mm2
Use 4 Nos 16 mm 60 at top and bottom
Use 2 legged 12 mm a stirrups @ 200 c/c

11. Design of raft foundation
Vertical load from filled tank and column
= 5185824 x 8 = 41486592 N
Weight of water = 5185824
Vertical load of empty tank and column
= 6390733 N
v max due to wind load = 244144 x 8
= 1953152
Which is less than 100/3 % of super imposed load. Assume self-weight of foundation
as 10 %
= 4148659.2 N
Total load = 41486592 + 4148659.2

= 45635251.20 N
Area of foundation required
= 45635251.2 / 150000
= 304.23 m2
Circumfrom of column circle = r x 10 = 31.42 m
Width of foundation = 304.2 / 31.42
= 4.67 m
Hence inner diameter = 10 + 4.67 = 5.33 m
Outer diameter = 10 + 4.67 = 14.67 m
Area of annular raft = π /4 ( 14.672
_ 5.332
)
= 114.35 m2
Moment of inertia of slab about a diametrical axis
= π / 64 [ 14.674
– 5.334
]
= 2216.76 m4
Total load tank empty
= 6390733 + 1553436
= 47877325
Stabilizing moment
= 47877325 x 14.67 /2
= 351180178.9 N-m
Let the base of be 2 m below ground level
Mw at base
= ( 114720 x 23.7 ) + ( 11760 x 18 ) + 33060 ( 14 + 1o + 6 )
= 3591744.00 N-m
Hence the soil pressure at the edge along a diameter area

(a) Tank full = 45635251/114.35 + 4396344/ 1618.8 x 13.64/ 2
167956 N-m 2
(b) tank empty
= 47877325 / 114.35 + (4396344 / 1618.8 ) x 13.64/2
= 87994 N/m2
Hence provide 10 mm ф bar @ 110 mm c/c at the slab
Main or longitudinal reinforcement
In provide 4-25 mm ф bar at top these bars will take care of both the maximum positive
bending moment .
And Transverse reinforcement
10 mm ф 4 lagged stirrup ,hence provide 10 mm ф 4 lagged stirrup @ 300 mm c/c.
Side face reinforcement
Since depth is more than 450 mm provide side face reinforcement @ 0.1 %
Al= 0.1(700 x 1200) /100
= 840 mm2
Provide 3-16 mm ф bars on each face having total Al = 6 x201 = 1206 mm2
.

ESTIMATION
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 NOS L( m) B (m ) A(
m2
)
D(m) QTY
M
3
REMARKS
1 EARTH WORK IN
EXCUVATION
1 64.32 2 128.64 L =2πR = 2π*2.55 =
16.022m, R =5.1/2 =
2.55m
2 EARTH WORK
IN FILLING
1 100.198 L =2πR = 2π*3.75 =
23.56m,R =7.5/2 = 3.75m
3 RCC WORK IN
FOUNDATION (1:1.5:3)
1 64.32 0.4 25.728 L =2πR = 2π*3.75 =
23.56m,R =7.5/2 = 3.75m
4 RCC WORK IN
COLOUMNS
BELOW G.L (1:1.5:3)
8 0.282 1.6 2.714 Sa = 2πhRc =π (h2 + r2)
=π (1.5^2+5.4375^2)
=99.95m^2,h =1.5m,r =
5.4375
5 RCC WORK IN
COLOUMNS ABOVE
G.L UPTO 4M HT (1:1.5:3)
8 0.282 4 6.785 Davg =(7.5+5.1)/2
=6.3m,R=6.3/2= 3.15m,
Sa =πr(r+h) =
π*3.15(3.15+1.6) =
47.006m^2
6 RCC WORK IN
COLOUMNS FROM
4M TO 10M HT(1:1.5:3)
8 0.282 4 6.785 R = 3.3816m,Sa = 2πhRc
=π (h2 + r2) =π
(0.950^2+3.3816^2) =
38.760m^2
7 RCC WORK IN
COLOUMNS FROM
10M TO 16M HT (1:1.5:3)
8 0.282 4 6.785 D = (0.23+0.2) =.215m,Sa
=2πR h= 2π*3.75*5 =
117.80m
8 TOTAL RCC WORK
IN COLOUMNS (1:1.5:3)
23.069 QTY = 2*6*0.3*0.3*0.6
=0.648m^3
9 RCC WORK IN
BRACING AT 4m HT (1:1.5:3)
1 18.535 0.3 0.3 1.668 QTY = 23.609 ‐ 0.648 =
22.961m^3
10 RCC WORK IN
BRACING AT 8m HT
(1:1.5:3)
1 17.278 0.3 0.3 1.555 QTY
=25.728+2.714+3*6.785+
22.961+1.668+1.555+3.8
45+3.675+0.848+9.995+1
1.751+7.752+25.327=138
.174m^3

11 RCC WORK IN
CIRCULAR GIRDER
(1:1.5:3)
1 16.022 0.4 0.6 3.845 ,R=6.3/2= 3.15m,Sa
=πr(r+h) =
π*3.15(3.15+1.6) =
47.006m^2
12 RCC WORK IN
RING BEAM AT
BOTTOM OF THE CL
WALL (1:1.5:3)
1 23.56 0.3 0.52 2.675 R=6.3/2+0.5= 3.65m,Sa
=πr(r+h) =
π*3.65(3.65+1.6)
= 60.2m^2
13 RCC WORK IN
RING BEAM AT TOP
OF THE CL WALL (1:1.5:3)
1 23.56 0.16 99.95 0.225 0.848 R = 3.3816m,Sa = 2πhRc
=π (h2 + r2) =π
(0.950^2+3.3816^2) =
38.760m^2
14
RCC WORK IN
DOMED ROOF(1:1.5:3)
1 0.1 9.995 Sa = 2πhRc =π (h2 + r2)
=π (1.5^2+5.4375^2)
=99.95m^2,h =1.5m,r =
5.4375
15
RCC WORK IN
CONICAL SLAB (1:1.5:3)
1
47.06 0.25
11.751
Davg =(7.5+5.1)/2 =
6.3m,R=6.3/2= 3.15m,Sa
=πr(r+h) =
π*3.15(3.15+1.6) =
47.006m^2
16 RCC WORK IN
CONICAL DOME (1:1.5:3)
1 38.76 0.2 7.752 R = 3.3816m,Sa = 2πhRc
=π (h2 + r2) =π
(0.950^2+3.3816^2) =
38.760m^2
17 RCC WORK IN
CYLINDRICAL
WALL (1:1.5:3)
1 0.215 117.8 5 126.35 D = (0.23+0.2) =.215m,Sa
=2πR h= 2π*3.75*5 =
117.80m
18 DEDUCTIONS IN
RCC WORK IN
BRACINGS IN COLOUMNS
2*6 0.3 0.3 0.6 0.648 QTY = 2*6*0.3*0.3*0.6
=0.648m^3
19 T0TAL RCC WORK
IN COLOUMNS
AFTER DEDUCTIONS
22.901 QTY = 23.609 ‐ 0.648 =
22.961m^3
20 TOTAL RCC WORK
(1:1.5:3)
138.174 QTY
=25.728+2.714+3*6.785+
22.961+1.668+1.555+3.8
45+3.675+0.848+9.995+1
1.751+7.752+25.327=138

.174m^3
21 PLASTERING IN
C M (1:2) FOR INNER
SURFACE OF CONIVAL
SLAB (12MM)
1 47.06 47.006 ,R=6.3/2= 3.15m,Sa
=πr(r+h) =
π*3.15(3.15+1.6) =
47.006m^2
22 PLASTERING IN
C M (1:6) FOR OUTER
SURFACE OF CONICAL
SLAB (12MM)
60.2 60.2 R=6.3/2+0.5= 3.65m,Sa
=πr(r+h) =
π*3.65(3.65+1.6) =
60.2m^2
23 PLASTERING IN
C M (1:2) FOR INNER
SURFACE OF CONICAL
DOME (12MM)
1 38.76 38.76 R = 3.3816m,Sa = 2πhRc
=π (h2 + r2) =π
(0.950^2+3.3816^2) =
38.760m^2
24 PLASTERING IN
C M (1:6) FOR OUTER
SURFACE OF CONICAL
DOME (12MM)
43.13
5
43.135 R = 3.3816+0.2m =
3.5816,Sa = 2πhRc =π (h2
+ r2) =π
(0.950^2+3.3.5816^2) =
43.135m^2
25 PLASTERING IN
C M (1:2) FOR INNER
SURFACE OF
CYLINDRICAL
WALL (12MM)
117.8 117.8 D = (0.23+0.2) =.215m,Sa
=2πR h= 2π*3.75*5 =
117.80m
26 PLASTERING IN
C M (1:6) FOR OUTER
SURFACE OF
CYLINDRICAL WALL
(12MM)
125.0
3
125.03 D = (0.23+0.2)
=.215m,R=3.75+.23
=3.98m,Sa =2πR h=
2π*3.98*5 = 125.03m
27 PLASTERING IN
C M (1:2) FOR INNER
SURFACE OF DOMED
ROOF (12MM)
96.5 96.556 Sa = 2πhRc =π (h2 + r2)
=π (1.5^2+5.3375^2)
=96.56m^2,h =1.5m,r =
5.3375
28 PLASTERING IN
C M (1:6) FOR OUTER
SURFACE OF DOMED ROOF
(12MM)
99.95 99.95 Sa = 2πhRc =π (h2 + r2)
=π (1.5^2+5.4375^2)
=99.95m^2,h =1.5m,r =
5.4375
29 PLASTERING IN
C M (1:6) FOR COLUMNS
(12MM)
8 45.23 271.433 P =2πRh =2π*.6*12 =
45.23m^2
30 PLASTERING IN
C M (1:6) FOR
CIRCULAR GIRDER (12MM)
1 16.022 0.6 91.732 L =2πR = 2π*2.55 =
16.022m,R =5.1/2 =
2.55m

31 PLASTERING IN
C M (1:2) FOR RING
BEAM AT TOP (12MM)
23.56 0.16 18.213 Sa
=2*23.56*0.225+2*0.225
*0,16+2*0.16*23.56 =
18.213m^2
32 PLASTERING IN
C M (1:2) FOR RING
BEAM AT BOTTOM (12MM)
23.56 0.3 0.225 38.95 Sa
=2*23.56*0.52+2*0.52*0
.3+2*0.3*23.56 =
38.950m^2
33 PLASTERING IN
C M (1:6) FOR
BRACING AT 4M HT
(12MM)
18.535 0.3 0.52 22.422 Sa
=2*18.535*0.3+2*0.3*0.
3+2*0.3*18.535
= 22.422m^2
34 PLASTERING IN
C M (1:6) FOR
BRACING AT 8M HT
(12MM)
17.278 0.3 0.3 20.936 Sa
=2*17.278*0.3+2*0.3*0.
3+2*0.3*17.278 =
20.936m^2
35 TOTAL PLASTERING IN
CM (1:2) 12MM THICK
0.3 357.289 QTY =
47.006+38.76+117.8+96.
56+18.213+38.95 =
357.289m^2
36 TOTAL PLASTERING IN
CM (1:6) 12MM
652.838 QTY =
60.2+43.135+125.03+99.
95+271.433+9.732+22.42
2+20.936 = 652.838m^2
37 THICK WATER PROOF
CEMENT PAINTING
FOR TANK PORTION
647.174 QTY
=47.006+60.2+38.76+43.
135+117.8+125.03+96.56
+99.95+18.213+0.52=647
.174m^2
38 WHITE WASHING
FOR COLUMNS
6 45.23 271.433 P =2πRh =2π*.6*12 =
45.23m^2
39 TOTAL WHITE
WASHING
918.607 QTY =647.174+271.433 =
918.607m^2

ABSTRACT
S.NO DESCRIPTION
OF WORK
QTY OR NOS RATE
RS PS
COST
RS PS
1 Earth work in
excuvation
28.30cumec
2 Beldars 4 nos 180.00 720.00
3 Mazdoors 3 nos 160.00 480.00
4 Total 1200.00
5 Total earth work
in Excavation for
128.64cumec
128.64/28.30
=4.6*1200
=5520
5520.00
6 Earth work in filling
In foundation
28.30
7 Beldar 2 180.00 360.00
8 Bhisthi ½ 240.00 120.00
9 Total 480.00
10 Total earth work in
Filling 100.198 cumec
100.198/28.30
=3.6*480
=1728
1728.00
11 Disposal of surplus
earth in a lead 30m
12 Mazdoor 2 160.00 320.00
13 Total 9248.00

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 like cement
FOUNDATION
A. MATERIALS: UNIT QTY RATE RS AMOUNT RS
20mm HBG graded metal Cum 0.7 1200.04 840.54
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
1st Class Mason Day 0.133 250 33.25
2nd Class Mason Day 0.267 230 61.41
Mazdoor (Both Men and Women) Day 3.6 160 576.00
Concrete Mixer 10 / 7 cft (0.2 / 0.8 cum) capacity
Hour 1 205.40 205.40
Cost of Diesel for Miller Liters 0.133 45 5.99
Cost of Petrol for Vibrator Liters 0.667 68 45.36
Water (including for curing) Kl 1.2 60.00 72.00
Add 20% in Labour (1st Floor) 176.27
Add MA 20% 211.52
Add TOT 4% 212.99
BASIC COST per 1 cum 3726.09
Description Unit Quantity
Rate Rs.
Amount Rs.
COLUMNS
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Labour for centering Cum 1 712 712.00
Material hire charges for centering Cum 1 59 59.00
Hour 1 205.40 205.40
Add 20% in Labour 232.304
Add MA 20% 300.96
Add TOT 4% 183.96

RCC RING BEAM AT TOP UNIT QTY RATE RS COST
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Concrete Mixer 10 / 7 cft (0.2 / 0.8 cum) capacity Hour
0.267 205.40
66.32
Water (including for curing) kl 1.2 60.00 72.00
Add MA 20% 258.08
Add TOT 4% 186.56
RCC Domed roof 100
mm thick
20mm HBG graded metal cum 0.7 1200.04 840.54
Sand cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
1st Class Mason day 0.067 250 16.75
2nd Class Mason day 0.133 230 34.58
Mazdoor (Both Men and
Women)
day 2.5 160 400.0
Labour for centering Sqm 10 1543 15430.00
Material hire charges for
centering
Sqm 10 1615 16150.0
Concrete Mixer 10 / 7 cft (0.2 /
0.8 cum) capacity
hour
0.267 205.40
66.32
Add MA 20% 5161.09
Add TOT 4% 1351.56
BASIC COST per 1 cum day 45440.95

C0NE SHAPED DOMICALSLAB AND INCLIND SLAB 200 mm thick
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Labour for centering Sqm 5 1243 6215.00
Material hire charges for centering Sqm 5 1615 8075.0
Concrete Mixer 10 / 7 cft (0.2 / 0.8 cum) capacity Hour 0.267 205.40 66.32
Add MA 20% 1239.48
Add TOT 4%
BASIC COST per 1 cum Day 19260.72
RCC RING BEAM AT BOTTOM OF CYLINDRICAL WALL
Sand cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Mazdoor (Both Men and Women) day 2.5 160 400.0
Concrete Mixer 10 / 7 cft (0.2 / 0.8 cum) capacity hour 0.267 205.40 66.32
Add 20% in Labour 241
Add MA 20% 245.036
Add TOT 4% 182.10

Description Unit Quantity
Rate Rs.
Amount Rs.
RCC CYLINDRICAL WALL
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Mazdoor (Both Men and Women) Day 3.7 160 592
Hour 0.267 205.40
66.32
Add MA 20% 279.40
Add TOT 4% 168.79
RCC CIRCULAR GIRDER
Sand cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Mazdoor (Both Men and Women) day 2.5 160 400.0
Add MA 20% 222.12
Add TOT 4% 165.92

RCC BRACING AT 5M HEIGHT
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Add MA 20% 251.88
Add TOT 4% 188.35
RCC BRACING AT 8M HEIGHT
Sand Cum 0.40 450.92 180.36
Cement Kgs 4.42 250 1105.00
Concrete Mixer 10 / 7 cft (0.2 / 0.8 cum) capacity Hour
0.267 205.40
66.32
Add MA 20% 240.84
Add TOT 4% 192.93

Plastering
Plastering with CM (1:3), 12 mm thick - 10 Sqm
Cement Mortor (1:3) cum 0.12 2391.00 286.92
Mason 1st class day 0.6 250.00 150.00
Mazdoor (unskilled) day 0.96 160 153.60
Add MA 20% 75.48
Add TOT 4% 37.26
Grand Total 702.66
Painting
Plastering with CM (1:6), 12 mm thick - 10 Sqm
Cement Mortor (1:6) cum 0.13 1489.00 193.57
Mason 1st class day 0.6 250.00 150
Add MA 20% 75.48
Add TOT 4% 29.45
Grand Total 601.50
Painting to new walls of tank portion with 2 coats of water proof cement paint of apporved 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 1
525.00 525
1st class painter day 0.21 260.00 54.60
2nd class painter day 0.49 240 117.60
cost of water proof cement paint
1.00 cum
(35.28 cft)
3.50
35 122.50
Add MA 20% 128.70
Add TOT 4% 39.79
Total cost/ 10 sqm 1250.89

Cement Primer
Painting to new columns with 2 coats of water proof cement paint of apporved 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 kg 1.00 100 100.00
cost of water proof cement paint
1.00 cum
(35.28 cft)
3.50
50.0 122.50
Add MA 20% 128.70
Add TOT 4% 39.79
Total cost/ 10 sqm 826.19

S.no Quantity Description per rate Amount
1
V.R.C.C (1:1 1/2 :3) 20mm size HBG, machine
crushed chips including cost, seignorage and
conveyance of all materials and labour charges
such as
25.728
Cum
Machine mixing, vibrating, curing etc., -Foundation - SF 1
cum 3726.09 95864.84
2
V.R.C.C (1:1 1/2 :3) 20mm size HBG, machine crushed
chips including cost,
23.069
seignorage and conveyance of all materials and labour
charges such as
1
cum 4531.06 104527.0
Cum
Machine mixing, vibrating, curing etc., -columns - SF
3
0.848
cumec
such as
Machine mixing, vibrating, curing etc., -Ring beam at
top - SF
1
cum
4950.43
4183.11
4
conveyance of all materials and
labour charges such as
9.995
Cum
Machine mixing, vibrating, curing etc., -domical roof -
SF
1
cum 45440.95 454182.3
5
such as
7.752
Cum
Machine mixing, vibrating, curing etc., -conical dome
base slab - SF
1
cum 19260.48 149307.3

6
126.635
cum
seignorage and conveyance of all materials and labour
charges such as
Machine mixing, vibrating, curing etc., -cylindrical wall
- SF
1
cum 4414.07 558975.8
7
3.675 seignorage and conveyance of all materials and labour
charges such as
Machine mixing, vibrating, curing etc., -ring beam at
bottom of cylindrical wall - SF
Cum
1
cum 5338.686 19619.67
8 3.845 cum
chips including cost, seignorage and conveyance of all
materials and labour charges such as Machine mixing,
vibrating, curing etc., -circular girder - SF
1
cum 4501.03 17306.46
9
11.751
cum
crushed chips including cost seignorage and
such as
Machine mixing, vibrating, curing etc., - inclined cone
shaped slab - SF
1
cum 11225.3 131908.5
10
1.668 cum
such as
Machine mixing, vibrating, curing etc., - Bracing at 5m
height – S Ft
1
cum 5012.01 8460.27
S.no
Quantity Description per rate Amount
11 1.555 cum
conveyance of all materials and labour charges such
as Machine mixing, vibrating, curing etc., - Bracing at
8m height – S Ft
1
cum 5057.35 7864.17

12 41.45 mt
Supplying, placing and fitting of HYSD bars
reinforcement, complete as per drawings and technical
specifications for bars below 36 mm diameter including
over laps and wastage, where they are not welded-SF
1 mt
30419.00
1260867.5
13 357.29 sqm
Plastering inside 12mm thick in single coat in cm (1:3)
with finishing including of cost of conveyance of all
materials and water to work site and all operational
incidental labour charges such as scaffolding. Mixing
mortar ,curing etc., complete for finished item of work.-
SF
10
sqm 702.66 25105
14
652.84 sqm
Plastering outside 12mm thick in single coat in cm (1:6)
with finishing including of cost of conveyance of all
materials and water to work site and all operational
incidental labour charges such as scaffolding. Mixing
mortar ,curing etc., complete for finished item of work.-
SF
10
sqm
601.5 39268.32
15 Painting to new outer walls with 2 coats of
647.174
sqm
columns 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
10
sqm 1250
80896.75
operational, incidental, labour charges etc. complete for
finished item of work as per SS 912 for walls
16
271.433
sqm
10
sqm 826.19 22425.52
Total 29,80762
Eath work 9248.00
Over all cost 29,90010
add 10% contractors profit 29,9001
Total cost 3289011
Total cost = Rs. 32,89,011 /

REFERENCE
• I.S 496:2000 for RCC.
• I.S 800:1984 for STEEL.
• I.S 872 Part I and Part II.
• I.S 3373 (Part IV-1967).
• I.S 1911 : 1967 (dead load ).
• I.S 875 : 1964 Part I , II and III (loading).
• I.S 1893 : 1975 (seismic force).
• I.S 6922 : 1973 (vibration force).
• I.S 465: 1978 (Permissible stress).
• I.S 11682 :1985 (Design for RCC Staging of over head tank)
• Theory of structure ( Dr .R. S. RAMANUTHAUM) .
• Rein force concrete structures (Dr B.C PUNMIA).
• Element of environmental engineering (BIRIDI).

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. Without power also we can consume water
by gravitational force.

Design Intze Water tank mazor project Report

Design Intze Water tank mazor project Report

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Design Intze Water tank mazor project Report

Similar to Design Intze Water tank mazor project Report (20)

Recently uploaded

Recently uploaded (20)

Design Intze Water tank mazor project Report