Design Problem Dimensions of the tank

# Tank capacity = 1 million litre = 1000 m^3 Assuming D = Inside diameter of the tank = 12 m
H = average depth = 8 m
Height of supporting tower = 16 m ho = 2 m
Number of columns = 8 Do = 8 m
h1 = 2 m
Depth of foundations = 1.5 m below ground level
h2 = 1.6 m
Calculating from the property of circle,
R1= Top dome radius =10 m
R2 = Bottom dome radius = 5.8 m
tan 1 6 36.86o 36o 52
8
Capacity of the tank = 904.78 + 304 +42.38
= 1166.4 m^3 > 1000 m^3

Design of top dome

Assuming thickness of the top dome = t = 100mm
V = volume = V1 +V2 +V3
Self weight of dome = 0.1×25 = 2.5kN m 2

= D2 h + o
D 2 +D o2 +DD o Live load on dome = 1.5kN m 2
4 12 Total = 2

2 ×10
2 Meridional thrust = T1 = = = 22.22 kN m
- 3R 2 -h 2 1+cos
3
1
Circumferential force = ×10 0.8 - 1 =10 kN/m
1+cos 1.8

22.22×103
Meridional stress = = 0.22 N mm 2 < 6 N mm 2 for M25
1000×100

10×103
Hoop stress = = 0.10 N mm 2 < 6 N mm 2 for M25
1000×100
The stresses are within safe limits.
 Design of Cylindrical Wall
Providing nominal reinforcements of 0.3% both circumferentially
10×8×12
and meridionally Maximum hoop tension at the base of wall = Ft = = =480 kN m
2 2
0.3×100×1000
Ast =300mm 2 Ft 480×103
100 A st = = 3200 mm 2 m height
st
Hence, provide 8mm bars at160mm centre to centre
Provide 20mm at180 mm centres on each face (A st =3492mm 2 )
As we go up, the reinforcement is reduced
Provide 10mm at 180mm centres on each face
if t = thickness of side wall at bottom:
480×103
=1.3, t = 330mm
1000t+ 11×3492
adopt 400 mm thick walls at bottom gradually reducing to 200 mm at top.

Design of top ring beam Distribution Steel

T1cos
Hoop tension = Ft = = 22.22×0.8×12 =106.6kN
2 2
450
3
A st = 106.6×10 = 710mm 2
150
Provide 8 bars of 12mm A st =904mm 2
if A c =cross sectional area of ring
Ft
then, = 1.3 for M25
A c +mA st

106.6 103 = 1.3

Ac 11 904
which gives A c = 72056mm 2

Provide 300 mm by 300 mm size top ring beam, with 8 bars of

12 mm dia. as main reinforcement and 6 mm dia. stirrups at 200
mm centres.
The details of reinforcements provided in the cylindrical Hoop tension due to vertical load is given by
wall of the tank at different heights are as follows: Hoop Force.D 93.6×12
Hg = = 561.6 kN
2 2
Distance from Main hoop steel Vertical
Hoop tension due to water pressure is
top (each face) distribution of
steel (each face) 10×8×0.6×12
H = = = 288kN
2 2
0-2 m 10mm-180mmc/c 10mm-250mmc/c
Total hoop tension = H g +H = 561.6+288 = 849.6 kN
2m 4m 16mm-200mmc/c 10mm-250mmc/c
849.6×103
4m 8m 20mm-180mmc/c 10mm-250mmc/c A st = = 5564 mm 2 , Provide 18 bars of 20mm A st =5655 mm 2
150

Design of bottom ring beam

a Load due to top dome = meridional thrust × sin 849.6×103

Max. tensile stress = =1.08 N mm 2 <1.3 N mm 2
1200×600 + 11×5655
=22.22 x sin36.833 = 13.3kN m
Provide a ring beam of 1200 mm wide by 600mm deep with 18 bars of 20 mm
(b) load due to top ring beam = 0.3×0.3×25 = 2.25kN m
and distribution bars of 10mm from cylindrical wall taken round the main bars
0.4+0.2 as stirrups at 180mm centres.
(c) load due to cylindrical wall = ×8×25 = 60 kN m
2
(d) self weight of ring beam (assuming a section1.2 m by 0.6m)
= (1.2×0.6×25) = 18 kN m
Total vertical load = V1 = 93.6 kN m
Horizontal Hoop Force H = V1 cot × cot45o = 93.6 kN
Design of Conical Dome
Average diameter of conical dome = (12 + 8)/2 =10m o

Average depth of water=(8 + 2/2) = 9m D = 12m

Weight of water above conical dome = H = 80 cosec45o +15 cot450 12 2 = 768.82 kN
( x10 x 9 x 2 x 10) = 5655kN 768.82×103
A st = = 5125mm 2
Assuming 600 mm thick slab: 150
Self weight of slab = ( x 10 x 2.83 x 0.6 x 25) = 1333.6kN Provide 25mm bar at 180mm centres (A st = 5470mm 2 ) on both faces of the slab

Load from top dome, top ring beam , cylindrical wall and 0.2×600×1000
Distribution steel = =1200mm 2
bottom ring beam = ( x 12 x 93.6) = 3528.6kN 100
Provide 10mm bar at 130mm centres on both faces along the meridians.
Total load at the base of conical dome =
768.82×103
(5655 +1333.6 + 3528.6) = 10517.2 kN Maximum tensile stress = =1.16 N mm 2 <1.3 N mm 2
600×1000 + 11×5470
Load per unit length = V2 kN/m

Design of Bottom Spherical Dome

Meridional Thrust T = V2 .cosec × cosec45o = 591.8kN Assuming the thickness of the bottom dome = 300mm

591.8×103 Diameter at base = 8m

Meridional stress = = 0.986 N mm 2 < 5 N mm 2 Central rise =1/5 x 8 =1.6m
600×1000
Radius of the dome = 5.8m
Hoop tension in conical dome will be maximum at the top of Self weight of the dome slab = 2× ×5.8×1.6×0.3×25 = 437.31 kN
the conical dome slab since diameter D is maximum at this Volumeofwaterabovedome
section.
2 ×5.82 ×1.6 ×42
Hoop tension = H = p.cosec = ×42 8+2 - - 5.8-1.6 = 460.29 m 3
3 3
water pressure = p = 10×8 = 80 kN m 2 Weight of water = 460.29×10 = 4603 kN
2
weight of conical dome slab per m is computed as: Total load on dome = 437.31 + 4603 = 5040.3 kN
q= 0.6×25 =15 kN m 2 5040.3 2
load per unit area =
×42
 Design of Bottom Circular Girder
Thrust from conical dome = T1 = 413kN m
Meridional thrust = T1 =
1+cos acting at an angle of o
zontal
o
Calculated thrust from spherical dome = T2 = 357 kN m
100.27×5.8 acting at an angle of o
izontal
T1 = = 337.3kN m
1+0.724
Net force on the ring beam = T1cos 2
337.3×103 2
Meridional stress = = 1.124 N mm safe 413 0.7 357 0.713 38 kN m
300×1000
38 8
1 Hoop compression in the beam = 152kN
Circumferential force = 2
1+cos
Assuming the size of the ring girder at 600 mm wide by 1200 deep:
1
= 100.27×5.8 0.724 - = 83.72 kN m
1.724 152 103
Hoop stress = 0.21N mm 2 (safe)
600 1200

83.72×103 Vertical load on ring beam = T1sin 2

hoop stress = = 0.279 N mm 2 safe
300×1000 = 413×0.707+357×0.7 =542 kN m
Provide nominal reinforcement of 0.3% Self weight of beam = 0.6×1.2×25 = 17.28 kN m
0.3×300×1000 Ttotal load = 542+17.28 = 560 kN m
A st = = 900mm 2
100
Total design load on the ring girder = W = ×8×560 =14074 kN
Provide12mm at 120mm centres
the circular girder is supported on 8 columns.
circumferentially and along the meridians.
Using the moment coefficients given the table
Maximum negative B.M. at support section
= 0.0083W.R=0.0083×14074×4 = 467kNm
Maximum positive B.M. at mid span section = 0.0041W.R
= 0.0041×14074×4=231kN
Torsional moment = 0.0006WR = 0.0006×14074×4 = 34 kNm
 560×4×
shear force at support section is = V = 4 = 880kN
2
shear force at section of maximum torsion is
at anangle of 9.5o from the column support
560× ×4×9.5
V= 880- = 521kN
180
Design of support section:
M = 467 kNm
V = 880 KN
467×106
d= = 931 mm
0.897×600
Adopt effective depth d = 1150 mm and cover = 50 mm

467×106
A st = =3008mm 2
150×0.9×1150
Provide 6 bars of 25mm A st =2946mm 2
680×103 2
v
600×1150
100A st 100×2946 2
= = 0.246 c
bd 600×1150
Since c v nts are required.
0.27×600×1150
Shear taken by concrete = =186 kN
1000