Depth of tank = 4.5 m,      Diameter of tank = 8 m

 Assume a free board of 200 mm

   Total height of tank H  = 4.5 + 0.2 = 4.7 m    

     Unit weight of water  = 9.8 kN/m3

 Assume M25 concrete and Fe415 steel

Permissible tensile stress in steel = 150 N/mm2 = st

Permissible tensile stress in concrete = 1.3 N/mm2 

  cbc = 8.5 N/mm2

   Modular ratio, m = = 11

Design constants :

  n = =  = 0.384

  j = 1 – =  1 – = 0.872

  k = cbc j  n =  8.5  0.872  0.384 = 1.423

Design for Cantilever Action :

• The height ‘h’ above base upto which cantilever action exist is give by h = or 1m whichever is more.

  h =  = 1.567 m

Cantilever moment  =  H h =  9.8  4.7  1.567 = 18.85 kN-m.

  Depth of balanced section d =   = = 115 mm

 Let d =  115 = 153.35 mm

 Let us provide d = 155 mm and total thickness = 190 mm

  Ast =  =   = 929.76 mm2

 Using 10 mm bars

  s =  = 80.11 mm

  = 84.47 mm

Provide 10 mm bars @ 80 mm c/c near inner face, keeping a clear cover of 30 mm.

Cultural alternate bars at a height of 1.56 m and spacing of 160 mm is available in top 3.14 m height.

Design for hoop action :

Maximum hoop tension is considered at 1.56 m height.

  T =  (H – h) = 9.8 (4.7 – 1.56) = 123.09 kN

   Ash =  = 820.58 mm2

 Using 10 mm bars,

  Spacing =  1000 = 95.71 mm

  Provide 10 mm bars @ 95 mm c/c.

Check for tensile stress in concrete :

   Ash =  1000 = 826.73 mm2

  ct =   = 

   = 0.61N/mm2< 1.3 N/mm2

 Hence OK.

Provide a spacing of 100 mm for 1.56 m height and 300 mm spacing above 1.56 m height .

Step 1 : Dimension of tank

 H =  5 – 0.3 = 4.7 and volume V = 700 m3

 A = 700/4.7 = 148.94 m2

 D =  √(4  148.94/) = 13.77 ≫14 m

Step 2 : Analysis for hoop tension and bending moment

 One meter width of the wall is considered and the thickness of the wall is estimated as t = 30H + 50 = 191 mm.

 The thickness of wall is assumed as 200 mm.

 Referring to table 9 of IS3370 (part IV), the maximum coefficient for hoop tension = 0.575

 Analysis for hoop tension and bending moment (Contd.)

  Tmax = 0.575  10  4.7  7 = 189.175 kN

 Referring to table 10 of IS3370 (part IV), the maximum coefficient for
bending moment = – 0.0146 (produces tension on water side)

  Mmax = 0.0146  10  4.73 = 15.15 kN-m

Step 3 : Design of section:

 For M20 concrete σcbc = 7, For Fe415 steel σst = 150 MPa and m=13.33 for M20 concrete and Fe415 steel

 The design constants are:

  k =  = 0.39

  j = 1-(k/3)=0.87

  Q = ½ σcbc x j x k = 1.19

 Effective depth is calculated as

  d =  =  = 112.94 mm

 Let over all thickness be 200 mm with effective cover 33 mm dprovided=167 mm

  Ast =  =  = 695.16 mm2

 Spacing of 16 mm diameter bar =  = 289.23 mmc/c

 (Max spacing 3d=501mm)

 Provide #16@275 c/c as vertical reinforcement on water face

Hoop steel :

  Ast1 =  =  = 1261 mm2

Spacing of 12 mm diameter bar =  = 89 mmc/c

Provide #12@80 c/c as hoop reinforcement on water face

Actual area of steel provided 

  Ast =  1412.5 mm2

Step 4 : Check for tensile stress:

  c = 

   =  = 0.87 N/mm2

Permissible stress = 0.27√fck=1.2 N/mm2>c Safe

Step 5 : Distribution Steel:

 Minimum area of steel is 0.24% of concrete area

  Ast = (0.24/100)  1000  200 = 480 mm2

 Spacing of 8 mm diameter bar = = 104.7 mmc/c 

Provide #8 @ 100 c/c as vertical and horizontal distribution on the outer face.

The thickness of base slab shall be 150 mm. The base slab rests on firm ground, hence only minimum reinforcement is provided.

  Ast = (0.24/100) 1000  150 = 360 mm2

 Reinforcement for each face = 180 mm2

 Spacing of 8 mm diameter bar =  = 279 mmc/c 

Provide #8 @ 250 c/c as vertical and horizontal distribution on the outer face.

The exact analysis of rectangular water tank is difficult and the design cab be done by approximate method as per is IS3370. Two cases are considered for rectangular water tank.

1. When <2 :

In this case the tank walls are assumed to cantilever continuous frame  subjected to hydrostatic pressure. The bottom  or 1 m (which is greater) is designed as cantilever and the top portion of a continuous frame in which moments are calculated by moment distribution. In addition to the walls are subjected to direct tension due to hydrostatic pressure.

The direct tension in long wall :

 Specific weight of water = w (H h) × B

The direct tension in short wall =  × L

Fig. 5.7.1

2. If >2 :

The longer wall are considered as vertical cantilever for the whole height from the base the short wall is to be designed as simply supported over long wall. The maximum BM in long wall at the base.

 P =   × wH × H

 BM=  × wH × H ×  =

The maximum BM at the junction of steel and long wall =  B2Fig. 5.7.2

Maximum positive BM at the centre of short wall = 

The maximum BM for the bottom of short wall =  or

Direct tension in long wall =  direct tension in short wall =

Given :

  Capacity = 45 kl

  1 kl = 1 m3 = 45 m3

 Depth of water = 3m

 Free board = 0.2 m

 Height of tank = 3 + 0.2 = 3.2 m

 Volume = L  B  H

 45 = L  B  3.2

 L  B = 14.06

  = 1.25 B

 L = 1.25 B

 1.25  B B = 14.06

 B = 3.35 m  3.5 m Fig.  P. 5.7.1

Intensity of pressure at 1 m from base

   = w  H = 10  2 = 20 kN/m2

  For 1 m depth of wall = 10  1 = 10 kN/m2

Fig . P. 4.7.1(A)

Moment distribution :

   =  =  = 32.26kN/m  w-at 1 m from base

   = –  =  = – 20.41

Maximum simply supported BM for longer span

   =  =  = 48.4 kN.m

  Net BM = 48.4 – 27.06

  At centre =  21.34kN.m

For shorter span :

  Maximum S. S. BM =  =  = 30.625 kN.m

  Net BM at centre = 30.625 – 27.06 = 3.57 kN.m

  The pull in longer wall = 

  TL =  =  = 35 kN

Fig. P. 5.7.1(B)

  Pull in short wall =  =  = 44 kN

Thickness of wall :

1. Minimum thickness = 100 mm

2. (30 mm per meter depth of water) + 50 = (30  3) + 50 = 140 mm

3. 60 mm per meter length of water = 60  4.4 = 264 mm

  Take greater value = 264 mm

  Assume t = 300 mm

Design of wall to direct tension and moment :

  x = Eccentricity of T  

  x = 150 – Efficiency cover

   = 150 – 40 = 110 mm = 0.11 m  

 Efficiency cover  = Clear cover +  = 30 +  = 40 mm

For longer wall  Net BM = M – TL x 

  (at mid span) = 21.35 – 35  0.11 = 17.50 kN.m

  And at junction = M – TL x = 27.06 – 35  0.11 = 23.2 kN.m

For shorter wall

  Net BM= m – TB x               (mid span) = 3.57 – 44  0.11 = – 1.27 kN.m

 Negative sign indicate tension on water face

 At junction = M – TB x =   27.06 – 44   0.11 = 22.22 kN.m

Check for depth :

  M = f   Z

  f = cbt for M20 cbt= 1.7 N/mm2

  Z =  

  b =1000 mm = 1 m

  Mr =   D = t = ?

Equating maximum BM to Mr

 Maximum BM = 23.2 106  N.mm

  23.2  106 = 

 D = 286 < 300 mm …OK

 Assuming    D = 300 mm 

  d = 300 – 30 –  = 260 mm

Area of steel at junction :

1. Ast in shorter wall at junction

  Ast =  +  =  +  = 1236.9 mm2

 For Fe 250st =  115 N/mm2

 M20 and Fe 255

  J = 0.87

Ast(read) in long wall at junction

  Ast =  +  =  +

    = 1207.74 mm2 (1211.5)

Ast (min) :

  = 0.2 +  (450 – 300)

 Ast min = 0.24 % of Ag

  =  1000  D =  1000  300

  = 720 mm2    Fig. P. 5.7.1(D)

 Assuming 16 mm  bars                                                                 

  Spacing =  = 162 = 160 mm c/c

Providing 16 mm  bar at the spacing 160 mm c/c horizontally. In the depth from 1 m up to 2 m above the base at the junction in the long and short wall for a distance 0.25 l from the support.

Since the water pressure decreasing towards the top area of steel above 2 m from base =  = 618 mm2 Ast (min). = 720 mm2.

Assuming 12 mm  bar

  Spacing=  150 mm c/c

Providing 12 mm  bar @ 150 mm c/c in the top 1.2 m height of the wall.

Ast in centre of span :

1. In shorter wall :

  Ast =  +

   =  +  = 431.62 < Ast (min)

  Provide Ast = 720 mm2

Providing 12 mm  bars

  Spacing =   = 150 mm c/c

Providing 12 mm  bars @ 150 mm c/c

2. Longer wall :

  Ast =  +

   =  +  = 977.08 < Ast (min)

Assuming 16 mm  bars  

  Spacing =  200 mm c/c

Distribution steel

  (Ast)min =  1000  D =  1000  300 = 720 mm2

 On each face =   = 360 mm2

Assuming 8 mm  bar  

  Spacing =   = 130 mm c/c

Design of bottom 1 m height :

 Pressure intensity at base = wH = 10  3 = 30 kN/m2

Cantilever moment at the base

   =  30  1  = 5 kN.m 

  Ast =  =

   = 192.21 < Ast (min) Fig. P. 5.7.1(E)

 Provide Ast  = 720 mm2

Steel on each face  =   = 360 mm2

 Providing 10 mm  @ 210 mm c/c on both faces.

Base slab :

 The water tank is resting on ground providing 150 mm thick slab.

  Ast = 0.3 % of Ag =  1000  150 = 450 mm2

 On each face =  = 225 mm2

Assume 8 mm  bars

  Spacing =  220 mm c/c

(1) Height to diameter ratio  = 0.4

(2) Around 10% of total capacity is because of bottom conical portion. Hence the capacity of vertical container can be equivated with 90% of capacity to have H and D.

(3) Diameter of top dome = diameter of container.

(4) Risk of top dome is 0.1 to 0.2 time dia.

(5) Height of conical dome = rise of top dome. It can be equal to

Bottom dome :

It is approximately 50% m diameter compare to diameter of top dome.  Rise

of bottom dame is kept somewhat greater approximately 20 to 25% of bottom dome diameter.

Volume of tank

 (1) Capacity of cylindrical portion =  D2H  

(2) Capacity of Bottom conical dome

   =  Fig. A Typical Intze tank

(3) Capacity of bottom dome =  (3r2 + h2) h 

(4) Total capacity = (1) + (2) – (3)

Capacity of tank=1000 k J = 1000  103liters = 1000 m3

  cbc = 8.5 N/mm2, st = 150 N/mm2 , m = 11

  ct = 1.3 N/mm2, cc = 6 n/mm2, n = 0.384, j = 0.872

  K = 1.423

Dimensions of tank :

 Let the diameter of cylindrical portion=D

 Diameter of bottom ring girder =0.6D

 Height of conical shell h3= 0.2D

 Height of bottom spherical dome=

The volume of tank

  V = Volume of cylindrical portion + Volume of conical portion

    – Volume of bottom spherical dome portion

   =  D2 h2 + h3  – (3R2 – h4)

  D1 = 0.6D, h2 = 0.4D, h3 = 0.2D, h4 =

 Radius of bottom dome R2 is

   = 0.36  = 0.09D2

   R2 = 0.386 D

   V = D2 0.4D + (D2 + 0.36D2 + 0.6D2)
    – 

   = 0.374 D3

    0.374 D3 = 1000

   D = 13.88 m

  Let  D = 14 m, D1 = 0.6D = 0.6  14 = 8.5 m

  h2 = 0.4D = 0.4  14 = 5.5 m

  h3 = 0.2D = 2.8 m

  h4 =  = 2 m

Actual volume of tank

 2 (2R2 – 2)  =   R2 = 5.52 m

  V =  142 5.5 +  2.8 (142 + 8.52 + 14  8.5)

    –   22 (3  5.52 – 2)

  = 1069 m3

Fig. Dimensions of Intize Tank

Fig. **

Design of Top dome :

  Diameter  =  13 m

  Let rise  =  1/7  13 = 1.86 m

 Radius of spherical dome – 2 (2R1 – 2) = 6.52 R1 = 11.56 m

 Semi-central angle    = cos– 1= 34.21

 Let the thickness of dome= 75 mm

 Self weight = 0.075  1  1  25 = 1.875 kN/m2

 Live load = 1.5 kN/m2

 Total = 3.375 kN/m2 4 kN/m2

 Meridoinal thrust    T1 = w R1= = 25.31 kN/m

 Meridoinal stress =  = 0.34 N/mm2< 6 N/mm2  O.K.

 Circumferential force = T2 = wR1

  = 4  11.56 = 12.93 kN/m

   Circumferential stress=  = 0.172 N/mm2 < 6 N/mm2  O.K.

 Provide nominal reinforcement of 0.3 % in circumferential and radial direction.

  Ast =  75  1000 = 225 mm2             

 Using 8mm bars, S =  1000 = 223 mm

Provide 8mm bars at 200 mm c/c in radial and circumferential directions.

Design of Top Ring (Beam A–A) :

 Hoop tension = T1 cos = 25.31 cos 34.21 = 136.05 kN

  Ast =  = 907mm2

Provide 5 bars of 16mm diameter

    Astprovided = 5  162 = 1005.31 mm2

Area of concrete :

  1.3 =  =

    Ac = 93595.44 mm2

  Assume b = 300mm

Provide 300  350 mm top ring beam with 5 bars of 16mm (2 at top and 3at bottom). Also provide nominal shear reinforcement of 6mm bars @ 300mm c/c.                           

Design of side wall :

Hoop tension at base of tank = 9.8  5.5 = 350.35 kN

  Ast =  = 2335.67 mm2

Use 16 mm bars on each face

  s =  1000= 172.16 mm

 Provide 16 mm bars on each face @ 170 mm c/c

 Ast provided = 2  162  = 2680 mm2/m height

Area of concrete :

  1.3 =  =

   Ac  = 240020 mm2 /m height

   t =  = 240.02 mm

  Provide t = 250 mm

Distribution steel :

  P = 0.3 – 0.1 = 0.25

 Area of distribution steel required =  275  1000 = 687.5 mm2

 Area of distribution steel required on each face = 687.5/2 = 343.75 mm2. Use 8 mm bars.

  S =  1000 = 146 mm

Provide 8 mm bars@ 140 mm c/c on each face.

 Hoop tension = 9.8  3.5  = 222.95 kN/m

 Ast =  = 1486.33 mm2

 Aston each face = 743.17 mm2

Use 12 mm bars

  S =  1000 = 152.18 mm

Use 12 mm bars @ 150 mm c/c on each face.