8/10/2019 Circular Water Tank (Rigid Joi

1/6

Nameof work:-

Tank capacity 400 K ltr mm

Depth of water 4.00 m mm

Free board 0.20 m mm

Conrete M 20

Steel fy 415 N/mm2

= 150 N/mm2

cbc 7 N/mm2

m = 13.3

Nominal cover 35 mm Effective cover = 40 mm

Design Cons tants:-For HYSD Bars = 20

sst = 150 N/mm2

= #### N/mm2

scbc = 7 N/mm2

m = 13.3x

13.3 x 7 + 150

j=1-k/3 = 1 - 0.383 / 3 =R=1/2xc x j x k = 0.5 x 7 x 0.872 x 0.383 =

Dimention of tank:- = 4.00 - 0.20 =

x

x

400 x 4

3.143 x 3.80

= m

Determinat ion of bending moment and h oop tens ion: -

Thickness of wall from empirical formula = 3H+5 = 3 x 4.00 + 5 = 17 cm

H2 = 4.00 x 4.00

DT 11.60 x 0.17

The height above base, upto which cantilever action will be there is given by H/3 or1 m which e

D

2

11.60

2

Maximum prerssure at bottom = wH = 9800 x 4.00 = N

1 1.33

2 3

Design of set ion for can ti lever act ion:-

if d is the effective thickness of tank wall,

Mf x

1000 R x

= 100 + 35 = 135 =However, provide a minimum thickness equal to the greater of following;

(I) 150mm (II) 30H+ 50 = 30 x 4.00 + 50 = 170 Hence provide T=

35 mm cover to the center of reinforcement, availble d =

x ###

150 x 0.872 x 135

3.14xdia2

3.14 x 12 x 12

4 x100 4 x

Spacing of hoop Bars = 1000 x 113 / 658 = 172 say = 170 m

Hence Provided 12 mm F bar, @ 170 mm c/c upto height of 1.40 mAbove this height, curtail half bars and continue the other half upto top. Let us test this for devlopment leng

0

=

D =

Mf =\maximum cantilever B.M. =

=

pressure at h/3

=

=

Maximum ring tension at this level, per meter height

Provide a diameter of

1.3339800 x ( 4.00

x D2

4

say

3.80x

m

=400000

1000

where w =

= 11.60 m

= 11615

200

Effective depth of tank

7

wt. of concrete

11.57

If D is the inside diameter of tank, we have =

from which D

1.169

Tensile stess

m*c+sst0.383

0.872

400000

4000

k=m*c

=13.3

DESIGN OF CIRCULAR WATER TANK (Rigid joint)

=

Cocrete M

3 =

11.60

=

h =

9800

H/38 /= 4.00

w (H-h)=

1000

1000

1.17

151573

11615

39200

-

xx 39200 x 1.33

)x = N

100.00 mm

Area of steel for cantilever bending is given by= Ast =Mf

sst . J .d=

11615

=

# mm bars A

say 140

Providing 135

=

\ Total thickness =d +cover

100==using

8/10/2019 Circular Water Tank (Rigid Joi

2/6

f .sst 12 x 1504. tbd 4 x 0.8

Hence half bars may be curtailed at 1.40 m height above the base. These vertical bars are t

provided inner face . Keep a clear cover of = 25 mm

Design o f sect ion for hoops act ion;

N at 1.33 m above th

/ 150 = mm

Let us provide rings atboth faseHence of area of ring in each face= mm2

3.14xdia2

3.14 x 12 x 12

4 x

Spacing of hoop Bars = 1000 x 113 / 505 = 224 say = 220 m

Hence Provided # mm F bar, @ 220 mm c/c upto height of 1.40 m from baseand above this, the spacing may be increased

1000 x 113

1000 x 220 +( 13.3 - 1 )x 1028

Hoop tension at 2.00 m below top = 9800 x 2.00 x 11.60 / 2.00 =

/ 150 = mm

Let us provide rings atboth faseHence of area of ring in each face= mm2

3.14xdia2

3.14 x 12 x 12

4 x100 4 x

Spacing of hoop Bars = 1000 x 113 / 379 = 298 say = 290 m

Hence Provided # mm F bar, @ 290 mm c/c upto height of 2.00 m from base0.3

100

= 1000 x 113 / 255 = 443 say = 440 m

Distr ibut ion reinforcement:-

170 - 100

450 - 100

0.280100

However , no additional reinforcement will be provided at inner face since the vertical steel for cantilever act

will serve this purpose. Hence provided Ast= 238 mm2in vertical direction at the outer face.

3.14xdia2

3.14 x 8 x 8

4 x100 4 x

Spacing of hoop Bars = 1000 x 50 / 238 = 211 say = 200 m

8 mm F bar, 200 mm c/c

Prov is ion for haunches: -

It is customary to provide150 mmx150mm haunchesat the junction of wall and base.

200 mm c/c may be provided.

Design of tank f loor. :-

0.30

100

= 225 mm2

3.14xdia2

3.14 x 8 x 8

4 x100 4 x

Spacing of hoop Bars = 1000 x 50 / 225 = 223 say = 200 m

However provide 8 mmfbars @ 200 mm c/c in both direction , at top and bottom of floor slab.The floor slab will rest on 75 mm thick layer of lean concrete covered with a layer of tar felt.

= =100

Provide half the reinfocement near each face, Ast

using 8 mm bars

x 150 x 1000 = 450 mm2in each direc

100

A =

Since the tank floor is resting on ground throughout , provide a minimum thickness of

Minimum Ast =

Hence Provided these

A haunches reinforcement of 8 mmf@

\ Ash = x 1000 x 170 =

Percentage area of distribution reinforcement is x

510 2 = 255=

Spacing of hoop Bars

0.30

=

At the top, minimum Ash x 1000 x 170 = /

\ sct

Actual , Ast 2 x= = 1028

0.652