1

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

2

DATA 1. Type of Tank: Intze Tank

2. Capacity of the tank: 350KL

3. Type of staging: Column & Brace type

4. Depth of foundation: 2.5m

5. Safe Bearing Capacity of Soil: 100KN/m2

6. Type of foundation: Circular Ring &Raft foundation

7. Grade of Concrete: M-25

8. Grade of Steel: Fe-415

9. Height of staging: 25m

10. Type of soil: Soft Clay

11. Height of Building up to Terrace: 15.6m

12. No. of floors in Building: G+3

13. Basic Wind Pressure: 1500N/m2

14. Sesmic Zone of Lucknow: Zone 3

15. No.of student in College: 2000

16. Water consumption rate

(Per capita demand in litres per day per head): 45

17. Design period for tank: 30 years

18. No.of student in hostels: 1600

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OBJECTIVE 1:- To make a study about the analysis and design of water tank

2:- To make a study about the guidelines for the design of liquid retaining structure according to

IS Code

IS: 3370 part 2-2009

IS: 456:2000

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

4:- To estimate the overall cost for making the Intze Tank

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WATER QUANTITY ESTIMATION IN COLLEGE CAMPUS

Population or the number of students to be served in 2014 = 2000

Let population to be increased at rate of 10% per decade

Number of students (2014) = 2000

Number of students in 2024 = 2200

Number of students in 2034 = 2420

Number of students in 2044 = 2662

Quantity = per capita demand × Population

= 45 × 2662

= 1,19,790 litres

= 120 KL (assume)

5

FLUCTUATION IN RATE OF DEMAND

Average daily per capita demand in college campus = 45 lpcd

If this average supplied at all the times it will not be sufficient to meet the fluctuation.

HOURLY VARIATION

(1) During the entry of college from 8 to 9 in the morning.

(2) During the lunch from 12 to 1 in the afternoon.

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WATER CONSUMPTION IN HOSTEL

Average daily per capita demand in hostels = 135 lpcd.

Quantity = 136 × 1600

= 216 KL

Total quantity = 216 + 130

= 346 KL

350 KL

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DESIGN REQUIREMENT OF TANK * Concrete mix weaker than M-20 is not used because of higher grade lesser porosity of

concrete.

* Minimum quantity of cement in concrete shall be not less than 30 KN/m3.

* Use of small size bars.

* Coefficient of expansion due to temperature=11×10-6/˚C

* Coefficient of shrinkage may be taken = 450 × 10-6 for initial and 200 × 10-6 for drying

shrinkage.

* Minimum cover to all reinforcement should be 20 mm or the diameter of main bar whichever

is greater.

* An overhead liquid retaining structure is design using working stress method avoiding the

cracking in the tank and to prevent the leakage and the component of tank can be design using

LIMIT STATE METHOD

(example:-column ,foundation ,bracing ,stairs etc.).

* Code using IS: 3370-PART 2-2009

IS: 456:2000

* The leakage is more with higher liquid head and it has been observed thad water head up to

15m does not cause leakage problem.

* In order to minimize cracking due to shrinkage and temperature, minimum reinforcement is

recommended as-

(i) For thickness ≤ 100 mm = 0.3%

(ii) For thickness ≥ 450 mm = 0.2%

For thickness between 100 mm to 450 mm = varies linearly from 0.3% to 0.2%

* For concrete thickness ≥ 225 mm, two layer of reinforcement be placed one near water face

and other away from water face.

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FROM IS -3370

(i) For load combination water load treated as dead load.

(ii) Cracking – The maximum calculated surface width of concrete for direct tension and flexure

or restrained temperature and moisture effect shall not exceed 0.2 mm with specified cover.

(iii) Shrinkage coefficient may be assumed = 300 × 10-6.

(iv) Bar spacing should generally not exceed than 300 mm or the thickness of the section

whichever is less.

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DETERMINATION OF FOUNDATION

ℎ = ℎ

ℎ (1 − ℎℎℎℎ

1 + ℎℎℎℎ)

=100

17 (1−ℎℎℎ12

1+ℎℎℎ12)

-From testing of soil sample

For clay ∅=12˚

r = density of soil

= 1.76 gm/cm3

= 17.6 KN/m3

p = 100 KN/m3

ℎ = 2.52 ℎ

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9. DETERMINATION OF HEIGHT OF STAGGING

We know turbulent flow occurs in a pipe

So Re > 4000

f = 0 .079

ℎℎ1/4

L = length of pipe,

v = mean velocity in pipe of flow

d = diameter of pipe

ℎd=4ℎℎℎ2

2ℎℎ

ℎ =ℎℎ

2

The kinematic viscosity of water (ℎ) = 0.01×10-4 m2/s

Assume diameter of pipe = 15 cm

ℎ = ℎℎ

ℎ =ℎ

4× 0.15

2= 0. .176ℎ2 =

Volume (V) = 350 m3

Only for one hour maximum velocity occurs in the pipe so the discharge during that period

ℎ =ℎ

ℎ=

350

60×60= 0.097

ℎ3

ℎ

ℎ = ℎℎ

0.097= 0.0176 × ℎ

ℎ = 5.52 m/sec.

Maximum velocity =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)14

= 2.61 ×10−3

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Minimum length of pipe requirement

= 2 × height of building up to 3 storeys from the level + lateral distance up to the centre of tank

= 2 × 15.6 + 18

= 49.2 m

≈ 50 m

Head loss ℎℎ =4×2.61×10−3×50×5.522

2×9.81×0.15

= 5.40 m

HEIGHT OF STAGGING

Total hydrostatic pressure on tank 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

Using total head = 29.5

Height of stagging = 29.5 – 4.5

= 25 m

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DESIGN OF TOP DOME

Assume thickness of top dome = 100 mm.

Meridional thrust at edges ℎ1 =ℎℎ1

1+ℎℎℎℎ1

Dead load of top dome = 0.100 × 25 = 2.5 KN/m2

Live load on top dome = 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= 0.308 MPa < 5 MPa (OK)

Maximum hoop stress occurs at the centre and its magnitude

ℎℎ1

2ℎ1=

3.25×103×18.5

2×0.100

=0.30 N/mm2 =0.3 MPa < 5MPa (OK)

Provide nominal reinforcement of 0.24%.

ℎℎℎ =0.24×100×1000

100= 240ℎℎ2

Use 8 mm bars.

ℎℎ = 50 ℎℎ2

Spacing =1000×50

240= 208.33

= 205 mm c/c.

Provide 8 mm bars @ 205 mm c/c radially and circumtentially as shown in figure.

The 205 mm c/c for radial bar is provided at the springing of the dome.

At the crown the spacing reduces to zero.

Hence the curtailment of radial bars may be carried out at the appropriate distance.

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14

DIMENSION OF TANK

Inner diameter of cylindrical portion D = 12 m

Rise of top dome h1 = 1 m

Rise of bottom dome h2 = D/8 = 1.5 m (centre)

Free board = 0.15 m

Diameter of ring beam Do = 5/8 D = 7.5 = 8 m

Rise of bottom dome (side) ho = 3/16 × D

= 2.25 m

= 2.5 m

Capacity of tank:-

ℎ =ℎℎ2ℎ

4+

ℎℎℎ

12 (ℎ2 +ℎℎ

2 + ℎℎℎ) −ℎℎ2

2(3ℎ2−ℎ2)

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Radius of bottom circular dome:-

1.5 × (2R2 – 1.5) = 42

2R2 – 1.5 = 10.67

R2 =6 m

Sin Ɵ2 = 4

6

Ɵ2 = 41.8o

ℎ =ℎℎ2ℎ

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

Radius of top circular dome:-

1 × (2R1-1) = 6 × 6

R1 = 18.5 m

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Sin Ɵ1 = 6/18.5

Ɵ1 = 18.92o

Design of top ring beam:-

A ring beam is provided at the junction of top dome and the vertical wall to resist hoop tension

induced by the top dome.

Horizontal component of 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= 175366.8 ℎ

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.

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Design of cylindrical wall:-

In the membrane analysis the tank wall is assumed to be free at top and bottom maximum hoop

tension occurs at the base of the wall and its magnitude:-

=ℎℎℎℎ

2=

9800×ℎ×12

2= 58800 ℎ

Hoop tension at any depth x from the top

X (m) Hoop tension (N/m)

0 0 1 58800

2 117600

Minimum thickness of cylindrical wall = 3 H + 5

= 3 × 2 + 5

= 11 cm.

Provide 20 cm at the bottom and taper it to 12 cm at top.

At x = 1 m.

Area of steel Ash = 58800/150

= 392 mm2

Provide 8 mm bars.

Aø = 50.26 mm2

Spacing = (1000 × 50.26) / 392

= 130 mm c/c.

At x = 2 m.

Area of steel Ash = 117600/150

= 784 mm2

Provide 10 mm bars.

Aø = 78.53 mm2

Spacing = (1000 × 78.53) / 784

= 100 mm c/c.

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The hoop steel may be curtailed according to hoop tension at different height along the wall

provided 0.24% of minimum vertical reinforcement.

Average thickness of wall = (120+200) / 2 = 160 mm.

Ash = 0.24×160×1000

100= 384 mm2

Provide 8 mm ø.

Aø = 50.26 mm2

Spacing = 50.26×1000

384= 130 mm c/c.

Design of ring beam B3:-

Thickness =100 mm

Rise = 1.5 m (centre)

Base dia. = 8 m

Raidus of curvature = 6 m

Cos 41.8o = 0.745

The ring beam connect the tank wall within conical dome. The vertical load at the junction of the

wall with conical dome is transferred to the ring beam B3 by horizontal thrust. In the conical dome

the horizontal component of thrust causes hoop tension at the junction.

W = Load transferred through the tank wall at the top of conical dome / unit length.

Øo = Inclination of conical dome.

T = Meridional thrust in conical dome at the junction.

tan Øo = 2/2.5

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Øo = 38.65

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= 30897.15 sin 18.92

= 10018.32 N/m

(ii) Load due to ring beam B1 = 320 mm depth

= 400 mm width

= 0.32×(0.4-0.1)×1×25000

= 2400 N/m

(iii) Load due to tank wall = 2 × (0.12+0.2

2) × 1 × 25000

= 8000 N/m

(iv) Seif load of beam B3 (1m × 0.6m say)

=(1-0.3) ×0.6×25000

= 10500 N/m

Total W = 10018.32 + 2400 + 8000 + 10500

= 30918.32 N/m

Sin Øo = sin 38.65 = 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 caused due to water pressure at top of conical dome

Pw2 = rw × hd3

h = depth of water upto centre of ring beam

d3 = depth of ring beam

Pw2 = 9800 × 2 × 0.6

= 11760 N/m

Hence hoop tension in the ring beam is given by :-

P = (ℎℎ1+ℎℎ2

2) × ℎ

=(24725.97+11760

2) × 12 = 218915.82 ℎ

This is to be resisted by steel hoops the area of which is

20

Ash = 218915.82

150

= 1460 mm2

Use 20 mm bars = 314.15

Number of 20 mm bars = 1460

314.15= 4.64

= 5 bars

Hence provide 5 ring of 20 mm dia bars.

Actual area As = π/4 × 20 × 5

= 1570 mm2

Stress in equivalent section = 218915.82

(1000×600)+10×1570

= 0.35 N/mm2 < 1.2 N/mm2 (SAFE) (OK)

The 10 mm diameter distribution bara (vertical bars) provided in the wall@ 100 mm c/c should be

taken round the above ring to act as stirrups.

Design of conical dome :-

(a)Meridional thrust :-

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Ww = Total weight of water on the conical dome

W = Weight of top dom, cylindrical wall etc.

Ws = Self weight of conical dome

Ww = ℎℎ[ℎ

4(ℎ2 + ℎℎ

2) +ℎ

12(ℎ2 + ℎℎℎ

2 +ℎℎℎ) −ℎ

4ℎℎ

2 ×ℎℎ]

=9800[ℎ

4(122 + 82) +

ℎ

12× 2.5(122 + 82 + 12× 8) −

ℎ

482 × 2.5]

=9800[326.72+ 198.96− 125.67]

=3920098 N

Let the thickness of conical slab = 400 mm

Ww = [ℎ(ℎ+ℎℎ

2) × ℎ× ℎ0] ×ℎℎ

l = √2.52

+ 22

= 3.2 m

Ws = 25000ℎ(12+8

2) × 3.2 ×0.4

= 1005309.649 N

Weight W at B3 = 30918.32 N/m.

Hence vertical load W2 per metre run is given by

W2 = ℎℎℎ+ℎℎ+ℎℎ

ℎℎℎ

= ℎ×12×30918.32+3920098 +1005309.64

ℎ×8

=242353.22 N/m

Meridional thrust To 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

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Diameter of conical dome at any height h’ above base is

D’ = 8 + (12-8)/2 × h’ = 8 + 2h’

Intensity of water pressure 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) Design of walls:-

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Meridional stress = 0.775 N/mm2

Maximum hoop stress = 354969.29 N

Which is to be resisted by steel

As = 354969.29/150

= 2366.46 mm2

Area of each face = 1183.23 mm2

Spacing of 16 mm dia bars = (1000 × 201)/1183.23

= 170 mm c/c

Hence provide 16 mm dia hoops @170 mm c/c on each face.

Actual As = (1000 × 201)/170

= 1182 mm2

Maximum tensile stress in composite section = 354969.29

(400×1000)+(2×1182×10)

=0.83 N/mm2 < 1.2 N/mm2 (Safe) (OK)

In the meridional direction the reinforcement = .21×400×1000

100 = 840 mm2

Or 420 mm2 on each face

Use 10mm diameter bars Aℎ = 78.53 mm2

Spacing = 1000×78.53

420 = 180 mm c/c

Hence provide 10 mm bars @ 180 mm c/c on each face with a clear cover 20 mm

DESIGN OF BOTTOM DOME

R2= 6 m

Ɵ2= 41.8

Weight of water on w0 on the dome is given by

W0=[ℎ

4× ℎ0

2ℎ0 –ℎℎ2

2

3(3ℎ2 −ℎ2)]ϒw

D0=8,H0 =2.5 ,R2=6,h2=1.5,

W0=850471.35N

Let the thickness of bottom dome =250mm

Self Weight Ws =2πR2h2t2×25000

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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.

Intensity of load per unit area

P2=1203900.52

2ℎℎ2ℎ2

R2 =6,h2=1.5

P2=21289.63N/m2

Max hoop stress at centre of dome

ℎ2ℎ2

2ℎ2=

(21289.63×6)

2×.25

=255475.625N/mm2

=0.25N/mm2 <5MPa O.K.

Area of min steel=0.26× 250 ×1000

100

=650 mm2in each direction

Use 10 mm Ɵbars

Spacing=1000×78.5

650 =120mm

Hence provide 10mm ɸ bars @ 120mm c/c in both direction also provide 16mm ø meridoinal bars

@170mm c/c near water face.

Design of bottom circular Beam B2

Inward thrust from conical dome =T0Sinøo

=310321.06 Sin38.65

=193814.5 N/m

25

Outward thrust from bottom dam=T2cosƟ2

=7866.98cos41.8

=56126.36N/m

Net inward thrust=137688.14N/m

Hoop compression in beam =137688.14×8/2

=550752.56N

Vertical load on beam ,per meter run=T0Cosø0+T2SinƟ2

=310321.06 Cos 38.6 +71866.98 Sin 41.8

=290423.93 N/m

Self weight of beam=0.6×1.2×25000

=18000N/m

Total load on beam=290423.93+18000=308423.90N/m

Let us support the beam on 8 equally spaced column at a mean dia 8m

Mean radius of curved beam R=4m

For support 8:-coefficient of B.H.& twisting moment in circular beam

2Ɵ=45o C1=0.066,C2=0.030,C3=0.005

Øm=9.5o ,Ɵ=π/4=22.5=π/8 Radius

wR2Ɵ=308423.9×42×π/4

=3875769.4

Max –ve B.M.at support M0=C1wR22Ɵ

=255800.78N.m

Max +ve B.M. at support Mc=C2wR22Ɵ

=116273.08N.m

Max torsional moment ℎℎℎ=C3 wR22Ɵ

=19378.84Nm

For σ cbc=8.5

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Hysd bars σst=150 N/mm2

Neuteral axis depth factor (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

Lever Arm

J=1-K/3=0.872

R=1/2×σcbc×J×k=1/2×8.5×0.872×0.383

1.41

Mr=Rbd2

Req eff. Depth (d)-

255800.78=1.41×600×d2

d=550mm

However keep total depth =700mm from shear point of view.

Max shear force at support Fo=WRƟ

=308423.9×4×π/8

=484471.12N

S.F. at any point F=WR(Ɵ-φ)

=308423.9×4×(22.5-9.5) ×π/180

=279916.6N

B.M. at the pointy of max torssional moment φm=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 torsion moment at any point-

Mpt=WR2[Ɵcosφ-Ɵcosφsinφ-(Ɵ-φ)]

27

At the support φ=0 M0t=WR2 (Ɵ-φ)=0

At the mid span φ=Ɵ=22.5=π/8 radian

Mφt= WR2[ƟcosƟ]-[

Ɵℎℎℎøℎℎℎø

ℎℎℎø]= 0

Hence we have the following combination of B.M. & torsional moment:-

(a)at the support

M0 =255800.78 NM(hogging or negative)

M0t=0