Introduction & under ground water tank problem

DESIGN OF REINFORCED CONCRETE STRUCTURES

1 SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II

UNIT-II

WATER TANKS


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


• What is design…????

• why is it????


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.


• 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.


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


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.


A special type of tank named Intze tank is used for

storing large amount of water for an area.

Intze tank 9 SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II

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


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.


Grade of

Concrete

Permissible Stresses Shear stress

N/mm2 Direct 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)


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.


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.


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.


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.


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


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


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

SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II 19

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


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.


• 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


• 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


• Step:6 Area of steel for inner face:


= (13.07x 10 6 / 115* 0.85* 280)

= 477.5 mm2


Spacing:

s = ( ast / Ast )*1000

= ( (π/4*(122)/477.5)*1000 = 236.85mm = 230mm

Provide 12mmΦ at 230 mm c/c along vertical direction


• 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.


• 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


Step-9 Area of steel for short wall:


= (36.48x 10 6 / 115* 0.85* 280)

= 1332.84 mm2


Spacing:

s = ( ast / Ast )*1000

= ( (π/4*(162)/1332.84)*1000 = 150mm

Provide 16mmΦ at 150 mm c/c on both faces.


• Step 10: Horizontal reinforcement for short wall:


= (0.3/100) (1000*320)

= 960 mm2


Spacing:

s = ( ast / Ast )*1000

= ( (π/4*(102)/960)*1000 = 81.87mm = 100mm

Provide 10mmΦ at 100 mm c/c.


• 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.


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


• Area of steel:


= (8.26x 10 6 / 115* 0.85* 125)

= 676.01 mm2


Spacing:

s = ( ast / Ast )*1000

= ( (π/4*(162)/761.12)*1000 = 103.198mm = 100mm

Provide 10mmΦ at 100 mm c/c.


• Distribution bar :


= (0.3/100) (1000*150)

= 450 mm2


Spacing:

s = ( ast / Ast )*1000

= ( (π/4*(82)/450)*1000 = 111.70mm = 110mm

Provide 8mmΦ at 110 mm c/c spacing.


SKCET / CIVIL / FINAL YEAR/ DRC-II / UNIT-II 33

Date post:	22-Jan-2017
Category:	Engineering
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Introduction & under ground water tank problem

Engineering