Clear introduction and a problem of under ground water tank.

1. DESIGN OF REINFORCED CONCRETE
STRUCTURES
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2. UNIT-II
WATER TANKS
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3. Needs (or) uses for water tanks
• Water tank is a container for storing liquids
Needs :
Drinking purpose
Irrigation process
Fire suspension
Chemical manufacture
 preparation for well, etc
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4. • What is design…????
• why is it????
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5. RCC WATER TANK DESIGN BASIS
• RCC Water tank design is based on IS 3370: 2009
(Parts I – IV).
• The design depends on the location of tanks, i.e.
overhead, on ground or underground water tanks.
• The tanks can be made in different shapes usually
circular and rectangular shapes are mostly used.
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6. • The tanks can be made of RCC or even of
steel.
• The overhead tanks are usually elevated
from the roof top through column.
• In the other hand the underground tanks are
rested on the foundation.
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7. TYPES OF WATER TANKS
• Based on the location of the tank in a building s
tanks can be classified into three categories. Those
are:
 Underground tanks
 Tank resting on grounds
 Overhead tanks
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8.  In most cases the underground and on ground tanks are
circular or rectangular is shape.
 The shape of the overhead tanks are influenced by the
aesthetical view of the surroundings and as well as the design
of the construction.
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9.  A special type of tank named Intze tank is used for
storing large amount of water for an area.
Intze tank 9SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II

10. Spherical
Steel tanks are also used specially in railway yards.
Basing on the shape the tanks can be circular, rectangular,
square, polygonal, spherical and conical.
Conical
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11. PERMISSIBLE STRESSES IN CONCRETE
• To ensure impervious concrete mixture linear than M20
grade is not normally recommended to make the walls
leak proof the concretes near the water face need to such
that no crack occurs.
• To ensure this member thicknesses are so designed that
stress in the concrete is lesser then the permissible as
given in table 1.
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12. Grade of
Concrete
Permissible Stresses Shear stress
N/mm2Direct Tension
N/mm2
Tension due to
bending
N/mm2
M15 1.1 1.5 1.5
M20 1.2 1.7 1.7
M25 1.3 1.8 1.9
M30 1.5 2.0 2.2
M35 1.6 2.2 2.7
M40 1.7 2.4 2.7
Table 1 Permissible Stresses In Concrete
(For calculations relating to resistance to concrete)
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13. THE PERMISSIBLE STRESS IN STEEL
• The stress in steel must not be allowed to exceed the following values
under different positions to prevent cracking of concrete.
• When steel is placed near the face of the members in contact with liquid
 115 N/mm2 for ms Bars and
 150 N/mm2 for HYSD bars.
• When steel is placed on the face away from the liquid for members 225
mm or more in thickness:
 125 N/mm2 for M.S. bars and
 190 N/mm2 for HYSD bars.
 HYSD bars– High Yielding Strength Deformed bars.
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14. BASE FOR FLOOR SLAB
• The floor slab should be strong enough to transmit the load
from the liquid and the structure itself to the ground without
subsidence. The floor slab is usually 150 to 200 mm thick.
• Before laying the slab the bed has to be rammed and
leveled then a 75 mm thick layer of lean concrete of M100
grade should be laid and cured.
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15. MINIMUM REINFORCEMENT FOR WATER
TANK
• Minimum reinforcement required for 200mm thick sections is
0.3 % of the area of concrete section which reduced linearly to
0.2% for 450 mm thick sections.
• In case of floor slab for tank resting on ground the minimum
reinforcement from practical consideration should not be less
than 0.3% of the gross sectional area of the floor slab.
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16. UNDER GROUND WATER TANKS
EXAMPLE : 1
To Design an underground RC rectangular water tank of 10m x 3m x
3m. The soil surrounding the tank is in dry states & sometimes wet
state. Angle of repose of soil in dry state is 30° and in wet state 6°.
Density of soil is 20kN/m3. Adopt M20 grade of concrete & Fe415
steel.
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17. DESIGN PROCEDURE
• Given data:
Size = 10m x 3m x 3m
Angle of repose @ dry soil = 30o
Angle of repose @ wet soil = 6o
Density (or) unit wt of soil= 20 kN/m3
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18.  Step:1 permissible & design values:
Tension due to bending ( concrete),
σct = 1.7 N/mm2 (Pg.no.:2,table- 1, IS 3370;part-II)
Compression due to bending ( concrete),
σcbc = 7 N/mm2 (Pg.no.:3, table – 2, IS 3370;part-II)
or (IS 456, page no.81)
Steel reinforcement for strength,
σst = 115 N/mm2 (Pg.no.:3, table- 4)
Design constant, m = 280 / 3* σcbc
= 280 / 3*7
= 13.33
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19. j =1 –(n/3)
n = (1/( 1+ (σst /m* σcbc)))
= (1/(1+(115/13.33*7)))
=0.447
j = 1-(0.447/3)
= 0.85
Q = (1/2)* σcbc * n * j
= (1/2) * 7 * 0.447 * 0.85
= 1.32
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20.  Step : 2 design of long walls:
Ratio for long wall = L/H = 10 / 3 = 3.33 >2
Ratio for short wall = B/H = 3/3 = 1 < 2
Long walls are designed as vertical cantilevers and
Short walls are designed as continuous slab.
pressure intensity @ wet soil,
P = w*h* (1-sinφ/1+sin φ)
= 20 *3*(1-sin 6o
/1+sin6o
)
= 48.64 kN/m2
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21. Step:3 Design of long wall
By considering 1m run of wall
BM calculation for vertical reinforcement:
Max B.M at top (tension near water face )
Max BM @ top = pH2/33.5
= (48.65 * 3 * 3 ) / 33.5
= 13.06kNm
Max B.M @ bottom = pH2/15
= 48.64 * 3 * 3 / 15
= 29.18kNm
Walls of underground water tank must also be designed cracking stress
consideration.
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22. • Step 4 Thickness of wall:
Max BM, M = (σct * bD2) / 6
29.18x 106 = (1.7 * 1000 * D2 ) / 6
29.18x 106 = 283.33 * D2
D2 = 29.18x 106 / 283.33
D = 323mm
provide overall depth, D = 320mm
cover, d’ = 40mm
Effective depth, d eff = 320-40
= 280 mm
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23. • step-5 Area of steel required for long wall:
for outer face:
Ast = (max BM / σst * j *d)
= (29.18x 10 6 / 115* 0.85*280 )
= 1066.13 mm2
provide 16mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(162)/1066.13)*1000
= 189.15 mm = 180mm
Provide 16mmΦ at 180 mm c/c along vertical direction
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24. • Step:6 Area of steel for inner face:
Ast = (max BM / σst * j *d)
= (13.07x 10 6 / 115* 0.85* 280)
= 477.5 mm2
provide 12mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(122)/477.5)*1000
= 236.85mm = 230mm
Provide 12mmΦ at 230 mm c/c along vertical direction
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25. • Step 7: Horizontal reinforcement for long wall:
Area of distribution bar = 0.3% (b*D)
= (0.3/100) (1000*320)
= 960 mm2
provide 10 mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(102)/960)*1000
= 81.87mm = 100mm
Provide 10mmΦ at 100 mm c/c along horizontal direction
on both faces.
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26. • step-8 Design Of Short Wall
intensity of earth pressure, p = 48.64 kN/m2
Max BM = pH2/12
= 48.64 * 3*3/ 12
= 36.48 kNm
Effective span = clear span +( thickness/2)
= 3 + (0.32/2)+(0.32/2)
= 3.32 m
Effective depth, d = √M/ Q.b
= √(36.48x106/1.33*1000)
= 165.61 < 280 mm
adopt, d = 280 mm
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27. Step-9 Area of steel for short wall:
Ast = (max BM / σst * j *d)
= (36.48x 10 6 / 115* 0.85* 280)
= 1332.84 mm2
provide 16mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(162)/1332.84)*1000
= 150mm
Provide 16mmΦ at 150 mm c/c on both faces.
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28. • Step 10: Horizontal reinforcement for short wall:
Area of distribution bar = 0.3% (b*D)
= (0.3/100) (1000*320)
= 960 mm2
provide 10 mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(102)/960)*1000
= 81.87mm = 100mm
Provide 10mmΦ at 100 mm c/c.
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29. • Step-11 Design Of Roof Slab :
Let us assume the overall thickness as slab , t = 150mm
Assume cover = 25mm
Eff.depth provided = 150 -((25/2)+(25/2))
= 125mm
• Load calculation
self wt of slab/ m2 = (1 * 0.15* 1 * 20) = 3 kN/m2
Live load/ m2 = 2.5 kN/m2
Floor finish = 0.5 kN/m2
Total load = 6 kN/m2
Ratio of Ly/Lx = 10.35 / 3.35 = 3.09 > 2
Hence one way slab Max.
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30. Moment calculation:
BM (M) = W* D2/ 8
= 6 * 3.32* 3.32/ 8
= 8.26 kNm
check for depth :
d = √M/ Q.b
= √(8.26x106/1.33*1000)
= 78.80 < 100 mm
adopt, d= 100 mm
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31. • Area of steel:
Ast = (max BM / σst * j *d)
= (8.26x 10 6 / 115* 0.85* 125)
= 676.01 mm2
provide 10mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(162)/761.12)*1000
= 103.198mm = 100mm
Provide 10mmΦ at 100 mm c/c.
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32. • Distribution bar :
Area of distribution bar = 0.3% (b*D)
= (0.3/100) (1000*150)
= 450 mm2
provide 8 mm dia bar
Spacing:
s = ( ast / Ast )*1000
= ( (π/4*(82)/450)*1000
= 111.70mm = 110mm
Provide 8mmΦ at 110 mm c/c spacing.
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