This document summarizes the design of a circular overhead water tank with the following key details: - The tank will be located in Panchampalli village and have a capacity of 750 cubic meters to serve a population of 1873 people. - The tank dimensions include a 15 meter height and 12.6 meter diameter. - The structural components including the dome, wall, ring beam, floor slab, columns, and footings will be designed using the Limit State method. - STAAD and AutoCAD software will be used to analyze and detail the structural design. Reinforcement will be designed to resist forces from water pressure and other loads.

1. DESIGN OF CIRCULAR OVERHEAD
WATER TANK
Submitted by GUIDED BY
S.SURECA (961119103005) ASST.PROFESSOR
S.VIMALA (961119103302) MISS.D.J.SUBHASHINI
LORD JEGANATH COLLEGE OF ENGINEERING AND TECHNOLOGY

2. Abstract
• Water tanks are the storage contains for storing water .
• In this project , we have planned , analyzed and designed an
overhead circular reinforced cement tank
• To clear to requirement of Panchampalli village Panchayat
• The population of the village panchayat is estimate as 1873
using population forecast method.
• The water tank is planned using the drafting software
AUTOCAD 2D
• It is further analysis using the premiere analysis software
STAADPRO
• The design and drawing are presented in this project work all
designing procedure by using LIMIT STATE METHOD

3. Objective
• To store potable drinking water within a
designation area or community.
• To distribute the water in the particular area.
• To study the analysis and design of the water
tank

4. Water Tank
A water tank is container for storing water and any other
liquid . Elevation water tank is a water storing container constructed
for the purpose of holding a water supply at a height sufficient to
Pressurize a water distribution system . In major cities the main
Supply scheme is augmented by individual supply system .
Classification of water tank :
water tank can be classified into two types
 based on location
• Tank resting on ground
• Tank under ground
• Elevated tank
Introduction

5.  Based on shapes
– Circular tanks
– Rectangular tanks
– Square tanks
– Spherical tanks
– Intze tanks
According to monolithic , a round structure is the strongest
Geometrical structure than the square ones. Therefore, a
cylindrical water tank will withstand all outward force such as
Strong wind or heavy rain in a better way than the square or a
rectangle tank . This way that tank will last longer as well.

6. DESIGN COMPONENTS OF TANK :
Dome
Tank wall
Ring beam
Floor slab
Column
Bracings beam
Stair case
Footings

7. Literature review
Name of literature Author name Remark
International Research
Journal of Engineering and
Technology(IRJET)
Dr. Vithalrao Vikhe Design of circular wall,
dome, column .
Design of RC and brick
masonry structures
Dr. P.Purushothamaraj
Dr. V. Ramasamy
Design of ring beam and
bottom slap
Reniforced concrete
structure
Dr. B.C. Punmai reniforced detaial of
structure

8. methodology
Site
investigation
Structural
analysis
Plan and elevation
design
Design of element
staircase dome Tank wall Floor slab
Base
slab
Column
footing
Brace
beam
STRADPRO
Drawing
Autocad

9. PLAN

10. Plan

11. ANALYSIS

12. Analysis
General
After clear analysis of water tank, the next step has the
effect on tank at different height of structure with the help of
STAAD PRO software. For this at-tempt, we have tried to
consider same loading condition, and earthquake zones
coming on the structure so that it will be easy for comparison.
The tank selected for study will be circular type. Tank is
analysis by using STAAD PRO analysis package and
performance with respect to displacement, base shear and
maximum force are present

13. • Material and property:

14. Self weight analysis:

15. • Column and braces analysis :

16. Assign load:

17. Analysis report:

18. Geometry structure:

19. DESIGN

20. Design of circular overhead water tank
Calculate the population :
Arithmetic increase method
Incremental increase method
a)Arithmetic increase method:
year population Increase
1971
1981
683
792
-
109
1991 836 44
2001 1586 750
2011 1873 287
Average 595

21. Arithmetic increase = PO +nx
= 1873+(1*595)
= 2468
b)Incremental increase method:
Year Population Increase Incremental
1971 683 - -
1981 792 109 -
1991 836 44 65
2001 1586 750 706
2011 1873 287 463
1234

22. Pn = Po +nx +n(n+1)/2*y
= 1893+(1*2468)+1(1+1)/2*1234
= 5595
Per capital demand = 135 Lpcd
Capacity = 5595*135
= 7,55,325 l
Volume = 755325/1000
= 750 m³
Height = 15 m

23. Design Procedure For Circular Overhead tank
Dimension of tank:
volume of tank = 750 m³
Height = 15 m
Free board = 200 mm
Permissible stress in concrete = 30 N/mm²
Yield stress in steel = 415 N/mm²
Soil bearing of capacity = 270 KN /m²

24. Design of Top Dome:
Live load = 1.5 KNm²
Dead load = 2.5 KNm²
Tank load = unit weight of water × height o walldepth of
water
= 10 × 6.2 6
= 10.33 KNm²
Total load = dead load + live load
= 2.5 +1.5
= 4 KNm²
top beam = r₁
r₁ =

25. Sinθ = D2 = 12.62
r₁ 9.14
= 0.689 ≈ 0.70
Sin θ = 0.70
θ = 0.70sin
= 44.42
Cos θ = 0.714 ≈ 0.71
Maximum meridonal thrust = T₁

26. T₁ = W × r₁ = 4 × 9.14
1+ cos θ 1 + 0.71
= 21.38 KNm
Meridonal stress (radial) = T₁area
= 21.38 = 1.42 × 10ˉ⁴ Nmm²
1000× 150
Dome thickness of the wall = 150 mm
Minimum Ast = 0.3 ٪bd
= ( 0.31000) ×1000×150
= 450 mm²
Spacing = 1000(π×8²) = 111.68 mm ≈ 150
450
provide 8mm Ф bars @ 150 mm

27. Design of Top Ring Beam:
Hoop tension = (T₁ cos θ ) × D2
= (21.38 × 0.71) × 12.62
= 95.63 KN
σst = 230 Nmm
Permissible compressive stress = 190 Nmm²
Steel required = 95.63 ×10³
230
= 415.78 mm²
At 8mm φ bars = 415.78
50
= 8.31 ≈ 8 bars
Provide 8mm φ @ 8 bars

28. Size of the ring beam = l×b
= 200 × 300
= 60,000 mm²
5.5 Design of Tank Wall:
Hoop tension due to hydraulic pressure (γ):
γ = p × hw × D2
= 10 × 6.2 × 12.62
= 390.6 Nmm
Ast = 390.6 × 10³
230
= 1698.26 mm²
spacing = (astAst) × 1000

29. Assume 16 mm φ bars
= (π × 16²)4 × 1000
1698.26
= 118.3 ≈ 150 mm
Provide 16mm φ 150 mm
Tensile stress in concrete = F
Ac + (m-1) Ast
= 370818
(130×100)+ (9.33-1)1698.26
= 1612.25
m = 280
3σcbc

30. = 280 = 9.33
3×10
Minimum rft = 0.24٪Bd
= (24100)×1000×200
= 480 mm²
Spacing 12 mm φ
= π×12²4 × 1000
480
= 235.6 ≈ 250 mm
Provide 12 mm dia bar hoop at 300 mm
Center form top 1m for the middle 1 m
Adopt a spacing = 225 mmcc

31. Distribution rft Ast = 0.2 ٪ bD
= (0.2100) × 1000×200
= 400 mm²
Provide 12 mm dia bar 300 mm centre in vertical direction
5.6 Design of the Bottom Slab:
Slab thickness = 150 mm
Weight of water = (3×1×1×9.8)
= 23.4 KNm²
Self weight of slab = 0.15 × 1 × 25
= 3.75 KNm²
Weight of roof slab = 0.3 × 1 × 25
= 7.5 KNm²

32. Weight of tank wall = 0.3 × 1 × 25
= 7.5 KNm²
P = 23.4 + 3.75 + 7.5 + 7.5
= 42.15 ≈ 43 KNm²
Circumferential moment = (m₀)c
= (116) × p ×a²
= (116) × 43 × (11.252)²
= 226.75 KNm
Radial moment = 226.75 KNm
Radial shear = (12)pa
= (12) 43 × (11.252)
= 120.93 KNm

33. (Mr)c = (216) × 43 × (11.252)²
= 170.06 KNm
Radial moment is zero at radius
Given by,
Mr = 0
= (116) × p ×(a²- 3r²)
r = a2√3
= 11.25 2×√3
= 3.24 m
d =
= 202.67 ≈ 200 mm
Using 25 mm clear cover
Effective depth = 200 + 25
= 225mm

34. Spacing
Ast = (Muσst×j×d)
= 170.06 × 10⁶
270 × 0.68 × 225
= 4116.67 mm²
Provide 25 mm dia rods
Spacing = (ast Ast ) × 1000
= (4904116.67) × 1000
= 119.02 mm ≈ 200 mm
Provide 20 mm dia bars @ 200 mm cc

35. Check for shear,
Pr = 170.06 KNm
Tv = prbd
= 170.06 × 10² = 1.82
1000 ×225
Tv < Tc
0.75 < 1.82
Hence it safe for shear section
Tank Floor Slab:
W = (weight of water) + (self weight of slab)
= (10 × 7.56) + ( 25 × 0.3)
= 83.1 KNm²

36. Average depth = 0.6 × D = 0.6 × 12.6
= 7.56 m
Positive moment at center
Mrp = (113) Wr²
= (113) ×83.1×6.3²
= 618.41 KNm
Negative moment at center
Mrp = (Wr²8)
= 83.1 × 6.3²
8
= 412.27 KNm
Circumferential moment,
Mc = Wr²16

37. = 83.1 × 6.3² = 206.13 KNm
16
Effective Depth of slab,
d =
=
= 273.28 mm
Adopt, d = 270 mm and over all
Depth = 270 + 30
= 300 mm
Reinforcement in circular slab ,

38. Ast (center span) = Mu
σst × j × d
= 618.41 ×10⁶
230×0.68×270
= 14.64×10⁴ mm²
M = 280 = 280
3σcbc 3× 10
= 9.33
n = Mσcbc = 618.41×10
Mσcbc+σst 618.41+230
= 0.96

39. J = 1- (n3)
= 1-(0.963)
= 0.68
Ast (circumferential) = Mu
σst×j×d
= 206.13×10⁶
230×0.68×270
= 4.88 × 10³ mm²
5.8 Design of Column Section:
Size = 300 × 300
Axial load = 1000KN
SBC of the soil = 270 KNm²
Material = M₃₀ & Fe ₄₁₅
Main column reniforcement,

40. Factored load = (1.5 × 1000)
= 1500 KN
P = 0.4 fck Ac + 0.67 fck Asc
Ag = 300×300 = 90000 m²
1×10³ = 0.4 × 30(9×10⁴-Asc)+0.67×415×Asc
Asc = 4055.62 mm²
Ac = Ag – Asc
= 90000 – 4055.62
= 85.94× 10³
Min reinforcement = 0.8 ٪ of gross area
= (0.8100)×85.94×10³
= 687.52 mm²

41. No of bars = 687.52 = 1.80 ≈ 2 bars
380.13
Provide 2 bars of 22 mm dia
Lateral ties,
1) 222 = 11
2) 5 mm
Adopt 12 mm dia ties
Pitch of ties is the least of,
1) Least lateral Dimension = 300 mm
2) 16 times of Longitudinal bars = 16×22 = 352 mm
Adopt 12 mm ties @ 300 mm cc

42. Design of Footing:
Load on column = 1000 × 1.5
= 1500 KN
Self weight of footing (10٪) = 150 KN
Total ultimate, Wu = 1650 KN
Footing area = 1650 = 4.70 m
1.5 × 270
Short side = 300 mm
Long side = 500 mm
Hence (3x × 5x) = 4.07
X = 0.52 m

43. Short side of footing = (3×0.52) = 1.56 m
Long side of footing = (5×0.52) = 2.6 m
Adopt mat foundation,
Upward soil pressure at service load,
= (1002) × 3
= 167 KNm³ < 270 KNm³
Hence it is safe
Factorial soil pressure, Pu = 1.5 × 167
= 250.5 KNm² (or) 0.2505 Nmm²
Factored Moment,
Cantilever projection from the = 0.5 (3- 0.5)
Short side face of the column = 1.25 m

44. Cantilever projection from the = 0.5 (5- 0.5)
Long side face of the column = 2.25 m
BM at the short side face of the column is = (0.5 ×Pu×L²)
= (0.5 × 250.50×1.25²)
= 195.7×10⁶ KNm
BM at the long side face of the column is = ( 0.5×Pu×L²)
= (0.5×250.5×2.25)
= 281.81KNm
Large Direction,
Depth of the Footing =
= 266.3 mm ≈ 270 mm

45. Mu = 0.87 fy Ast d (1- Ast fy)
bd fck
195.7×10⁶ = 0.87×415×270×Ast (1- Ast×415 )
1000×270×30
Ast = 2271.98 mm²
Spacing = (astAst) × 100
= (201.062271.98) × 100
= 88.53 ≈ 100 mm
Provide 16 mm dia bars at 100 mm cc
Shorter Direction,
(281.81×10⁶) = 0.87×Ast×415×270 × (1- Ast×415 )
1000×270×30
Ast = 3528.86 mm²

46. Ratio of long to short side = (53) = 1.66
Reinforcement in central band width of 2m,
= × Ast
= × 3528.86
= 2652 mm
Spacing,
= 113.09 × 1000
2652
= 42.64 mm ≈ 150 mm
Provide 12 mm dia bars @ 150 mm cc

48. Design of Braces :
Fig: 5.2 Wind pressure acting on Braces AB
For the condition of the maximum BM for the brace BC,
seismic should act normal to an adjoining brace AB .

49. Moment in Brace BC = Moment for column × sec 45 ֯
= 1× 10³×√2
= 141.21 Nm
Providing (300×500) mm section and designing as doubly
reinforcement beam with equal steel at the top and bottom.
Ast = 1414.2 × 1000
(300×500)
= 942 mm²
Provide 3 bars of 20 mm dia
No of bars = 942 = 3 bars
314.15
Shear force for brace = BM for brace
0.5 × span of brace

50. = 1414.21
0.5× 3.518
= 803.98 N
Nominal shear stress, tv = 803.98
1000×250
= 1.27 Nmm²
1.27Nmm < 1.8Nmm
Hence that is safe
Design of staircase :
Effective span,
Le = cc distance between support
Le = 3 + (2×1) + 0.3
= 5.3 m

51. Effective Depth,
Le d = 530025
= 212 mm
Clear cover = 20 mm
D = 212 + 20 = 232 mm
Loads,
Cos (ϴ) = 300 (300²+150²)⁰ ׄ ⁵
= 0.894
Self weight of slab = 25(0.23)cos (ϴ)
= 6.43
Self weight of steps = 24 = 1.8
(0.152)

52. Finishing load = 1 KNm²
Live load = 5 KNm²
Total load = 14.3 KNm²
Factored loads = 1.5×4.3 = 21.5 KNm²
Raise = 150 mm
Tread = 300 mm
Floor to floor height = 3000 mm
No of rise = 30001500
= 20 rise