This document discusses the design of circular water tanks resting on ground. It describes tanks with both flexible and rigid bases. For flexible bases, the thickness is estimated as 30H+50 mm and hoop reinforcement is provided. For rigid bases, the bottom acts as a cantilever and the upper part experiences hoop tension. The IS code provides coefficients to calculate hoop tension, moment and shear at different heights to simplify the analysis. Reinforcement is designed based on bending moment and permissible tensile stress.

1. 1
WATER TANKS

2. 2
Learning out Come
• REVIEW
• TYPES OF TANKS
• DESIGN OF CIRCULAR WATER TANK
RESTING ON GROUND WITH
FLEXIBLE BASE
• DESIGN OF CIRCULAR WATER TANK
RESTING ON GROUND WITH RIGID
BASE

3. 3
INTRODUCTION
• Storage tanks are built for storing water, liquid
petroleum, petroleum products and similar liquids
• Designed as crack free structures to eliminate
any leakage
• Permeability of concrete is directly proportional
to water cement ratio.
• Cement content ranging from 330 Kg/m3 to 530
Kg/m3 is recommended in order to keep shrinkage
low.

4. 4
RESTING ON GROUND

5. 5
UNDERGROUND

6. 6
ELEVATED

7. 7
CIRCULAR

8. 8
RECTANGULAR

9. 9
SPHERICAL

10. 10
INTZ

11. 11
CONICAL BOTTOM

12. 12
Circular Tanks Resting On
Ground

13. 13
• The tank has tendency to increase in
diameter due to hydrostatic pressure
• This increase in diameter all along the
height of the tank depends on the
nature of joint at the junction of slab
and wall

14. 14
• When the joints at base are flexible,
hydrostatic pressure induces maximum
increase in diameter at base and no
increase in diameter at top
• When the joint at base is rigid, the base
does not move

15. 15
Design of Circular Tanks resting on
ground with flexible base

16. 16
• When the thickness ≤ 225 mm, the steel placed at
centre.
• When the thickness > 225mm, at each face Ast/2 of
steel as hoop reinforcement is provided
• The stress in concrete is computed as
If c  cat, where cat=0.27fck , then no crack appears
in concrete
st
st
c
c
A
)
1
m
(
t
1000
2
/
HD
A
)
1
m
(
A
T
−
+

=
−
+
=


17. 17
• While designing, the thickness of concrete wall
can be estimated as t=30H+50 mm, where H is in
meters
• Distribution steel in the form of vertical bars are
provided such that minimum steel area
requirement is satisfied
• As base slab is resting on ground and no
bending stresses are induced hence minimum
steel distributed at bottom and the top are
provided

18. 18
• While designing, the thickness of concrete wall
can be estimated as t=30H+50 mm, where H is in
meters
• Distribution steel in the form of vertical bars are
provided such that minimum steel area
requirement is satisfied
• As base slab is resting on ground and no
bending stresses are induced hence minimum
steel distributed at bottom and the top are
provided

19. 19
Problem on Circular Tanks resting
on ground with flexible base

20. 20
Design a circular water tank with flexible
connection at base for a capacity of 4,00,000
liters. The tank rests on a firm level ground.
The height of tank including a free board of
200 mm should not exceed 3.5m. The tank is
open at top. Use M 20 concrete and Fe 415
steel. Draw to a suitable scale:
Plan at base
Cross section through centre of tank.

21. 21
Step 1: Dimension of tank
Depth of water H=3.5 -0.2 = 3.3 m
Volume V = 4,00,000/1000 = 400 m3
Area of tank A = 400/3.3 = 121.2 m2
Diameter of tank 13 m
The thickness is assumed as
t = 30H+50=149 160 mm

22. 22
Step 2: Design of Vertical wall
Max hoop tension at bottom Area of steel Minimum
steel to be provided
Ast min=0.24%of area of concrete
= 0.24x 1000x160/100 = 384 mm2
The steel required is more than the minimum required
Let the diameter of the bar to be used be 16 mm, area
of each bar =201 mm2
Spacing of 16 mm diameter bar=1430x 1000/201= 140.6
mm c/c
Provide #16 @ 140 c/c as hoop tension steel

23. 23
Step 3: Check for tensile stress
Area of steel provided Ast
provided=201x1000/140 = 1436.16 mm2
Modular ratio m=Stress in concrete
Permissible stress cat=0.27fck= 1.2
N/mm2
Actual stress is equal to permissible
stress, hence safe.

24. 24
Step 4: Curtailment of hoop steel:
Quantity of steel required at 1m, 2m, and
at top are tabulated. In this table the
maximum spacing is taken an 3 x 160 =
480 mm

25. 25
Step 5: Vertical reinforcement:
For temperature and shrinkage
distribution steel in the form of vertical
reinforcement is provided @ 0.24 % ie.,
Ast=384 mm2.
Spacing of 10 mm diameter bar =
78.54x1000/384=204 mm c/c  200 mm
c/c

26. 26
Step 6: Tank floor:
As the slab rests on firm ground, minimum
steel @ 0.3 % is provided. Thickness of
slab is assumed as 150 mm.
8 mm diameter bars at 200 c/c is provided
in both directions at bottom and top of
the slab.

27. 27

28. 28

29. 29
Design of Circular Tanks resting on
ground with Rigid base

30. 30
• Due to fixity at base of wall, the upper
part of the wall will have hoop tension
and lower part bend like cantilever.
• For shallow tanks with large diameter, hoop
stresses are very small and the wall act more
like cantilever
• For deep tanks of small diameter the
cantilever action due to fixity at the base is
small and the hoop action is predominant

31. 31
• The exact analysis of the tank to determine
the portion of wall in which hoop tension is
predominant and the other portion in which
cantilever action is predominant, is difficult
1. Simplified methods of analysis are
2. Reissner’s method
3. Carpenter’s simplified method
4. Approximate method
5. IS code method

32. 32
IS code method
• Tables 9,10 and 11 of IS 3370 part IV gives
coefficients for computing hoop tension,
moment and shear for various values of
H2/Dt
• Hoop tension, moment and shear is
computed as
T= coefficient ( wHD/2)
M= coefficient (wH3)
V= coefficient (wH2)

33. 33
• Thickness of wall required is computed from
BM consideration
where,
Q= ½ cbcjk
j=1-(k/3)
b = 1000mm
Qb
M
d =
st
cbc
cbc
m
m
k

+


=

34. 34
IS code method
• Over all thickness is then computed as
t = d+cover.
• Area of reinforcement in the form of vertical
bars on water face is computed as
• Area of hoop steel in the form of rings is
computed as
jd
M
A
st
st

=
st
1
st
T
A

=

35. 35
IS code method
• Distribution steel and vertical steel for outer
face of wall is computed from minimum
steel consideration
• Tensile stress computed from the following
equation should be less than the
permissible stress for safe design
st
c
A
)
1
m
(
t
1000
T
−
+
=

the permissible stress is 0.27 fck

36. 36
IS code method
• Base slab thickness generally varies
from 150mm to 250 mm and minimum
steel is distributed to top and bottom
of slab.