The document provides design details for a 350KL overhead water tank at a university campus. Key points include: - The tank will be an Intze tank with a column and brace staging 25m high to hold 350KL of water. - Water demand was estimated at 120KL for the college campus and 216KL for hostels, totaling 346KL. - Design requirements include using M-25 concrete and Fe-415 steel, with minimum reinforcement. - The height of the staging was calculated as 25m based on pipe diameter, flow rate and head loss calculations. - Dimensions of the tank include a 12m diameter cylindrical portion with 1m and 1.5m domes at

1. 1
STRUCTURAL DESIGN OF 350KL
OVERHEAD WATER TANK AT INDIRA
GANDHI NATIONAL OPEN
UNIVERSITY, TELIBAGH LUCKNOW

2. 2
DATA
1. Type of Tank: Intze Tank
2. Capacityof the tank: 350KL
3. Type of staging: Column& Brace type
4. Depthof foundation: 2.5m
5. Safe BearingCapacityof Soil: 100KN/m2
6. Type of foundation: CircularRing&Raft foundation
7. Grade of Concrete: M-25
8. Grade of Steel: Fe-415
9. Heightof staging: 25m
10. Type of soil: SoftClay
11. Heightof BuildinguptoTerrace: 15.6m
12. No.of floorsinBuilding: G+3
13. Basic WindPressure: 1500N/m2
14. SesmicZone of Lucknow: Zone 3
15. No.of studentinCollege: 2000
16. Water consumptionrate
(Percapitademandinlitresperdayper head): 45
17. Designperiodfortank: 30 years
18. No.of studentinhostels: 1600

3. 3
OBJECTIVE
1:- To make a studyaboutthe analysisanddesignof watertank
2:- To make a studyaboutthe guidelinesforthe designof liquidretainingstructure accordingto
IS Code
IS: 3370 part 2-2009
IS: 456:2000
3:- To knowabout the designphilosophyforthe safe andeconomical designof watertank
4:- To estimate the overall costformakingthe Intze Tank

4. 4
WATER QUANTITY ESTIMATION IN COLLEGE CAMPUS
Populationorthe numberof studentstobe servedin2014 = 2000
Let populationtobe increasedatrate of 10% per decade
Numberof students(2014) = 2000
Numberof studentsin2024 = 2200
Numberof studentsin2034 = 2420
Numberof studentsin2044 = 2662
Quantity = per capitademand× Population
= 45 × 2662
= 1,19,790 litres
= 120 KL (assume)

5. 5
FLUCTUATION IN RATE OF DEMAND
Average dailypercapitademandincollege campus = 45 lpcd
If this average suppliedatall the timesitwill notbe sufficienttomeetthe fluctuation.
HOURLY VARIATION
(1) Duringthe entryof college from8to 9 inthe morning.
(2) Duringthe lunchfrom12 to 1 in the afternoon.

6. 6
WATER CONSUMPTION IN HOSTEL
Average dailypercapitademandinhostels=135 lpcd.
Quantity = 136 × 1600
= 216 KL
Total quantity = 216 + 130
= 346 KL
͌ 350 KL

7. 7
DESIGN REQUIREMENTOFTANK
* Concrete mix weakerthanM-20 isnot usedbecause of highergrade lesserporosityof
concrete.
* Minimumquantityof cementinconcrete shall be notlessthan30 KN/m3
.
* Use of small size bars.
* Coefficientof expansiondue totemperature=11×10-6
/˚C
* Coefficientof shrinkage maybe taken= 450 × 10-6
forinitial and200 × 10-6
fordrying
shrinkage.
* Minimumcovertoall reinforcementshouldbe 20 mmor the diameterof mainbarwhichever
isgreater.
* Anoverheadliquidretainingstructure isdesignusingworkingstressmethodavoidingthe
cracking inthe tank and to preventthe leakage andthe componentof tankcanbe designusing
LIMIT STATE METHOD
(example:-column,foundation,bracing,stairsetc.).
* Code usingIS:3370-PART 2-2009
IS: 456:2000
* The leakage ismore withhigherliquidheadandithas beenobservedthadwaterheadupto
15m doesnotcause leakage problem.
* Inorder to minimizecrackingdue toshrinkage andtemperature,minimumreinforcementis
recommendedas-
(i) For thickness≤100 mm = 0.3%
(ii) Forthickness≥450 mm = 0.2%
For thicknessbetween100mm to 450 mm= varieslinearlyfrom0.3% to0.2%
* For concrete thickness≥225 mm, twolayerof reinforcementbe placedone nearwaterface
and otherawayfrom waterface.

8. 8
FROM IS -3370
(i) For loadcombinationwaterloadtreatedasdeadload.
(ii) Cracking– The maximumcalculatedsurface widthof concrete fordirecttensionandflexure
or restrainedtemperatureandmoisture effectshall notexceed0.2mmwithspecified cover.
(iii) Shrinkagecoefficientmaybe assumed= 300 × 10-6
.
(iv) Bar spacingshouldgenerallynotexceedthan300 mm or the thicknessof the section
whicheverisless.

9. 9
DETERMINATION OFFOUNDATION
ℎ =
ℎ
ℎ
(
1 − ℎℎℎℎ
1 + ℎℎℎℎ
)
=
100
17
(
1−ℎℎℎ12
1+ℎℎℎ12
)
-Fromtestingof soil sample
For clay ∅=12˚
r = densityof soil
= 1.76 gm/cm3
= 17.6 KN/m3
p = 100 KN/m3
ℎ = 2.52 ℎ

10. 10
9. DETERMINATION OFHEIGHTOFSTAGGING
We knowturbulentflow occursina pipe
So Re > 4000
f =
0.079
ℎℎ
1/4
L = lengthof pipe,
v = meanvelocityinpipe of flow
d = diameterof pipe
ℎd=
4ℎℎℎ2
2ℎℎ
ℎ =
ℎℎ
2
The kinematicviscosityof water(ℎ) =0.01×10-4
m2
/s
Assume diameterof pipe = 15 cm
ℎ = ℎℎ
ℎ =
ℎ
4
× 0.15
2
= 0..176ℎ2
=
Volume (V) =350 m3
Onlyforone hour maximumvelocityoccursinthe pipe sothe discharge duringthatperiod
ℎ =
ℎ
ℎ
=
350
60×60
= 0.097
ℎ3
ℎ
ℎ = ℎℎ
0.097= 0.0176 × ℎ
ℎ = 5.52 m/sec.
Maximumvelocity=5.52 m/sec.
ℎℎ =
ℎℎ
ℎ
=
5.52×0.15
0.01×10−4 = 8.2 × 105
(O.K.)
ℎ =
0.079
(8.2 × 105
)
1
4
= 2.61 × 10−3

11. 11
Minimumlengthof pipe requirement
= 2 × heightof buildingupto3 storeysfromthe level +lateral distance uptothe centre of tank
= 2 × 15.6 + 18
= 49.2 m
≈ 50 m
Headloss ℎℎ =
4×2.61×10−3
×50×5.522
2×9.81×0.15
= 5.40 m
HEIGHT OF STAGGING
Total hydrostaticpressure ontank P = ρgh
Total head=
ℎ
ℎ
+
ℎ2
2ℎ
+ ℎ + ℎℎ+ ℎℎℎℎℎ ℎℎℎℎℎℎ
Minor loss(assume) =1 m.
=
ℎℎ
ℎ
+
ℎ2
2ℎ
+ ℎ+ ℎℎ+ 1
= 4.5 +
5.522
2×9.81
+ 15.6+ 5.4 + 1
= 28.08 ℎ
Usingtotal head= 29.5
Heightof stagging= 29.5 – 4.5
= 25 m

12. 12
DESIGN OF TOP DOME
Assume thicknessof topdome =100 mm.
Meridional thrustatedges ℎ1 =
ℎℎ1
1+ℎℎℎℎ1
Deadload of top dome = 0.100 × 25 = 2.5 KN/m2
Live loadon topdome = 0.75 KN/m2
(assume)
Total load P = 3.25 KN/m2
ℎ1 =
3.25 × 103
× 18.5
1 + ℎℎℎ 18.92
= 30897.15 N/m
Meridional stress=
30897.15
100×100
= 0.308MPa < 5 MPa (OK)
Maximumhoopstressoccurs at the centre and itsmagnitude
ℎℎ1
2ℎ1
=
3.25×103
×18.5
2×0.100
=0.30 N/mm2
=0.3 MPa < 5MPa (OK)
Provide nominal reinforcementof 0.24%.
ℎℎℎ =
0.24×100×1000
100
= 240ℎℎ2
Use 8 mmbars.
ℎℎ = 50 ℎℎ2
Spacing =
1000×50
240
= 208.33
= 205 mm c/c.
Provide 8 mmbars @ 205 mm c/c radiallyandcircumtentiallyasshowninfigure.
The 205 mm c/c for radial bar isprovidedatthe springingof the dome.
At the crown the spacingreducestozero.
Hence the curtailmentof radial barsmay be carriedout at the appropriate distance.

13. 13

14. 14
DIMENSION OF TANK
Innerdiameterof cylindrical portion D= 12 m
Rise of top dome h1 = 1 m
Rise of bottomdome h2 = D/8 = 1.5 m (centre)
Free board= 0.15 m
Diameterof ringbeamDo = 5/8 D = 7.5 = 8 m
Rise of bottomdome (side) ho = 3/16 × D
= 2.25 m
= 2.5 m
Capacityof tank:-
ℎ =
ℎℎ2
ℎ
4
+
ℎℎℎ
12
(ℎ2
+ ℎℎ
2
+ ℎℎℎ)−
ℎℎ2
2
(3ℎ2−ℎ2)
3
Radiusof bottomcircular dome:-
1.5 × (2R2 – 1.5) = 42
2R2 – 1.5 = 10.67
R2 =6 m
SinƟ2 =
4
6
Ɵ2 = 41.8o
ℎ =
ℎℎ2
ℎ
4
+
ℎℎℎ
12
(ℎ2
+ ℎℎ
2
+ ℎℎℎ) −
ℎℎ2
2
(3ℎ2−ℎ2)
3
350 =
ℎ×122
×ℎ
4
+
ℎ×2
12
(122
+ 82
+ 12 × 8) −
ℎ×1.52
(3×6−1.5)
3
350 = 113ℎ + 160− 38.87
ℎ = 2 ℎ
Radiusof top circulardome:-
1 × (2R1-1) = 6 × 6
R1 = 18.5 m

15. 15
SinƟ1 = 6/18.5
Ɵ1 = 18.92o
Designof top ringbeam:-
A ringbeamis providedatthe junctionof topdome and the vertical wall toresisthooptension
inducedbythe top dome.
Horizontal componentof meridional thrust P1 = T1 cos Ɵ1
= 30897.15 cos 18.92o
= 29227.8 N/m.
Total hoop tension tending to rupture of beam =
ℎ1×ℎ
2
=
29227.8×12
2
= 175366.8ℎ
Permissible stress in HYSD bars = 150 N/m2
Ash = 175366.8/150 = 1170 mm2
Provide 20 mm bars (314.15) as hoop.
Number of 12 mm bars = 1170 / 314.15
= 3.72
= 4
Actual Ash = 4 × ℎ/4 × 202
= 1256.63 mm2
= 1257 mm2
Provide 4-20 mm ø hoop and 8 mm bars tie @ 205 mm c/c.
Hence the cross sectional area of concrete
1.3=
175366.8
ℎ+1257×8
Ac = 124841.53
Provide ring beam of 320 mm × 400 mm.

16. 16
Designof cylindrical wall:-
In the membrane analysisthe tankwall isassumedtobe free attop andbottom maximumhoop
tensionoccursat the base of the wall and itsmagnitude:-
=
ℎℎℎℎ
2
=
9800×ℎ×12
2
= 58800 ℎ
Hoop tensionatanydepthx fromthe top
X (m) Hoop tension(N/m)
0 0
1 58800
2 117600
Minimumthicknessof cylindrical wall =3 H + 5
= 3 × 2 + 5
= 11 cm.
Provide 20 cm at the bottomand taperit to12 cm at top.
At x = 1 m.
Areaof steel Ash = 58800/150
= 392 mm2
Provide 8 mmbars.
Aø = 50.26 mm2
Spacing= (1000 × 50.26) / 392
= 130 mm c/c.
At x = 2 m.
Areaof steel Ash = 117600/150
= 784 mm2
Provide 10 mm bars.
Aø = 78.53 mm2
Spacing= (1000 × 78.53) / 784
= 100 mm c/c.

17. 17
The hoop steel maybe curtailedaccordingtohooptensionatdifferentheightalongthe wall
provided0.24%of minimumvertical reinforcement.
Average thicknessof wall =(120+200) / 2 = 160 mm.
Ash =
0.24×160×1000
100
= 384 mm2
Provide 8 mmø.
Aø = 50.26 mm2
Spacing=
50.26×1000
384
= 130mm c/c.
Designof ringbeamB3:-
Thickness=100 mm
Rise = 1.5 m (centre)
Base dia.= 8 m
Raidusof curvature = 6 m
Cos 41.8o
= 0.745
The ring beamconnectthe tank wall withinconical dome.The vertical loadatthe junctionof the
wall withconical dome istransferredtothe ringbeamB3 by horizontal thrust.Inthe conical dome
the horizontal componentof thrustcauseshooptensionatthe junction.
W = Load transferredthroughthe tankwall atthe topof conical dome /unitlength.
Øo = Inclinationof conical dome.
T = Meridional thrustinconical dome at the junction.
tan Øo = 2/2.5

18. 18
Øo = 38.65

19. 19
= 30897.15 sin18.92
= 10018.32 N/m
(ii) Loaddue to ringbeamB1 = 320 mmdepth
= 400 mm width
= 0.32×(0.4-0.1)×1×25000
= 2400 N/m
(iii) Loaddue totank wall = 2 × (
0.12+0.2
2
)× 1 × 25000
= 8000 N/m
(iv) Seif loadof beamB3 (1m× 0.6m say)
=(1-0.3) ×0.6×25000
= 10500 N/m
Total W = 10018.32 + 2400 + 8000 + 10500
= 30918.32 N/m
SinØo = sin38.65 = 0.62 , cos 38.65 = 0.78
Force Pw due to load Pw1 = W tan Øo
= 30918.32 tan 38.65
= 24725.97 N/m
Force Pw causeddue to waterpressure attop of conical dome
Pw2 = rw × hd3
h = depthof wateruptocentre of ringbeam
d3 = depthof ringbeam
Pw2 = 9800 × 2 × 0.6
= 11760 N/m
Hence hooptensioninthe ringbeamis givenby:-
P = (
ℎℎ1+ℎℎ2
2
) × ℎ
=(
24725.97+11760
2
)× 12 = 218915.82 ℎ
Thisis to be resistedbysteel hoopsthe areaof whichis

20. 20
Ash =
218915.82
150
= 1460 mm2
Use 20 mm bars = 314.15
Numberof 20 mm bars =
1460
314.15
= 4.64
= 5 bars
Hence provide 5 ringof 20 mm diabars.
Actual area As= π/4 × 20 × 5
= 1570 mm2
Stressinequivalentsection=
218915.82
(1000×600)+10×1570
= 0.35 N/mm2
<1.2 N/mm2
(SAFE) (OK)
The 10 mmdiameterdistributionbara(vertical bars) providedinthe wall@100 mmc/c shouldbe
takenroundthe above ring to act as stirrups.
Designof conical dome :-
(a)Meridionalthrust:-

21. 21
Ww = Total weightof wateron the conical dome
W = Weightof top dom,cylindrical wall etc.
Ws = Self weightof conical dome
Ww = ℎℎ[
ℎ
4
(ℎ2
+ ℎℎ
2
)+
ℎ
12
(ℎ2
+ ℎℎℎ
2
+ ℎℎℎ)−
ℎ
4
ℎℎ
2
× ℎℎ]
=9800[
ℎ
4
(122
+ 82
) +
ℎ
12
× 2.5(122
+ 82
+ 12× 8) −
ℎ
4
82
× 2.5]
=9800[326.72+ 198.96− 125.67]
=3920098 N
Let the thicknessof conical slab= 400 mm
Ww = [ℎ(
ℎ+ℎℎ
2
) × ℎ× ℎ0] × ℎℎ
l = √2.5
2
+ 22
= 3.2 m
Ws = 25000ℎ(
12+8
2
)× 3.2 × 0.4
= 1005309.649 N
WeightW at B3 = 30918.32 N/m.
Hence vertical loadW2 per metre runis givenby
W2 =
ℎℎℎ+ℎℎ+ℎℎ
ℎℎℎ
=
ℎ×12×30918.32+3920098 +1005309.64
ℎ×8
=242353.22 N/m
Meridional thrustTo in the conical dome
To = W2 / cos Øo
= 242353.22 / cos 38.65
= 310321.06 N/m.
Meridional stress =310321.06 / (1000×400)
= 0.775 N/mm2
< 5 N/mm2
(Safe).
(b) Hoop stress

22. 22
Diameterof conical dome at anyheighth’above base is
D’ = 8 + (12-8)/2 × h’= 8 + 2h’
Intensityof waterpressure P = [(4+2)-h’]×9800
= [6-h’]9800 N/mm2
Self weight q = 0.4 × 25000
= 10000 N/mm2
Hoop tension ℎℎ
′
= (
ℎ
ℎℎℎØℎ
+ ℎ ℎℎℎØℎ )
ℎ′
2
= (
(6−ℎ′)9800
ℎℎℎ38.65
+ 1000 ℎℎℎ38.65 )
(8+2ℎ′)
2
Po’= 333150.48 + 12548.4 h’2
+ 12548.4h’2
h’ Hoop Tension
0 333150.48 N
1 353695.41 N
2 349144.86 N
2.5 337457.95 N
For maximum
ℎℎℎ′
ℎℎ′ = 0
33093.99 – 25096.8 h’= 0
h’= 1.31 m.
Maximum Po’= 354969.2977 N
(c) Designof walls:-

23. 23
Meridional stress=0.775 N/mm2
Maximumhoopstress= 354969.29 N
Whichis to be resistedbysteel
As = 354969.29/150
= 2366.46 mm2
Areaof eachface = 1183.23 mm2
Spacingof 16 mmdia bars = (1000 × 201)/1183.23
= 170 mm c/c
Hence provide 16 mm diahoops@170 mm c/c on eachface.
Actual As = (1000 × 201)/170
= 1182 mm2
Maximumtensile stressincomposite section=
354969.29
(400×1000)+(2×1182×10)
=0.83 N/mm2
< 1.2 N/mm2
(Safe) (OK)
In the meridional directionthe reinforcement=
.21×400×1000
100
= 840 mm2
Or 420 mm2 on eachface
Use 10mm diameterbarsAℎ = 78.53 mm2
Spacing=
1000×78.53
420
= 180 mmc/c
Hence provide 10 mm bars @ 180 mm c/c on eachface witha clearcover20 mm
DESIGN OFBOTTOM DOME
R2= 6 m
Ɵ2= 41.8
Weightof wateron w0 onthe dome isgivenby
W0=[
ℎ
4
× ℎ0
2
ℎ0 –
ℎℎ2
2
3
(3ℎ2 − ℎ2)]ϒw
D0=8,H0 =2.5 ,R2=6,h2=1.5,
W0=850471.35N
Let the thicknessof bottomdome =250mm
Self WeightWs =2πR2h2t2×25000

24. 24
R2=6, h2=1.5, t2=0.25
Ws=353429.17N
Total Weight=1203900.52N
Meridional thurst=T2=
1203900.52
ℎℎℎℎℎℎ41.8
D0=8
T2=71866.98N/m
Meridional Stress=
718866.98
250
× 1000
=0.287N/mm2
<5N/mm2
safe o.k.
Intensityof loadperunitarea
P2=
1203900.52
2ℎℎ2ℎ2
R2 =6,h2=1.5
P2=21289.63N/m2
Max hoopstressat centre of dome
ℎ2ℎ2
2ℎ2
=
(21289.63×6)
2×.25
=255475.625N/mm2
=0.25N/mm2
<5MPa O.K.
Areaof minsteel=0.26× 250 ×
1000
100
=650 mm2
ineachdirection
Use 10 mm Ɵbars
Spacing=1000×
78.5
650
=120mm
Hence provide 10mm ᶲ bars @ 120mm c/c inbothdirectionalsoprovide 16mmø meridoinal bars
@170mm c/c nearwaterface.
Designof bottomcircularBeam B2
Inwardthrustfrom conical dome =T0Sinøo
=310321.06 Sin38.65
=193814.5 N/m

25. 25
Outwardthrustfrombottomdam=T2cosƟ2
=7866.98cos41.8
=56126.36N/m
Netinwardthrust=137688.14N/m
Hoop compressioninbeam=137688.14×8/2
=550752.56N
Vertical loadonbeam,permeterrun=T0Cosø0+T2SinƟ2
=310321.06 Cos38.6 +71866.98 Sin41.8
=290423.93 N/m
Self weightof beam=0.6×1.2×25000
=18000N/m
Total loadonbeam=290423.93+18000=308423.90N/m
Let ussupportthe beamon 8 equallyspacedcolumnata meandia 8m
Mean radiusof curvedbeamR=4m
For support8:-coefficientof B.H.&twistingmomentincircularbeam
2Ɵ=45o
C1=0.066,C2=0.030,C3=0.005
Øm=9.5o
,Ɵ=π/4=22.5=π/8 Radius
wR2
Ɵ=308423.9×42
×π/4
=3875769.4
Max –ve B.M.at supportM0=C1wR2
2Ɵ
=255800.78N.m
Max +ve B.M. at supportMc=C2wR2
2Ɵ
=116273.08N.m
Max torsional moment ℎℎ
ℎ
=C3 wR2
2Ɵ
=19378.84Nm
For σcbc=8.5

26. 26
Hysd bars σst=150 N/mm2
Neuteral axisdepthfactor(K)
K=
ℎℎℎℎℎ
ℎℎℎℎℎ+ℎℎℎ
m=
280
3ℎℎℎℎ
=
280
3×8.5
=10.98
=10.98 ×
8.5
10.98×8.5+150
=0.383
LeverArm
J=1-K/3=0.872
R=1/2×σcbc×J×k=1/2×8.5×0.872×0.383
1.41
Mr=Rbd2
Reqeff.Depth(d)-
255800.78=1.41×600×d2
d=550mm
Howeverkeeptotal depth=700mm fromshearpointof view.
Max shearforce at support Fo=WRƟ
=308423.9×4×π/8
=484471.12N
S.F.at any pointF=WR(Ɵ-φ)
=308423.9×4×(22.5-9.5) ×π/180
=279916.6N
B.M. at the pointyof max torssional momentφm=9.50
Mφ=WR2
(ƟSinφ+ƟCosƟCosφ-1) sagging
=308423.9×42
×(π/8×sin9.5+π/8×cot22.5×cos9.5-1)
=4934.78Nm sagging
The torsionmomentat any point-
Mpt
=WR2
[Ɵcosφ-Ɵcosφsinφ-(Ɵ-φ)]

27. 27
At the support φ=0 M0
t
=WR2
(Ɵ-φ)=0
At the midspan φ=Ɵ=22.5=π/8 radian
Mφ
t
= WR2
[ƟcosƟ]-[
Ɵℎℎℎøℎℎℎø
ℎℎℎø
]=0
Hence we have the followingcombinationof B.M.& torsional moment:-
(a)atthe support
M0 =255800.78 NM(hoggingornegative)
M0
t
=0