DESIGN OF RECTANGULAR UNDERGROUND WATER TANK

Name of work : HOMING OF ELECTRIC LOCO at DLS, KGP

Tank capacity size : 10.00 x 5.00 x 2.00 m
Depth of water : 2.00 m
Tank capacity : 100000 ltr
Free Board : 300 mm
Tank dimension 10.00 x 5.00 x 2.30 m
Soil unit wt, we = 18.60 kN/m3 = 18600 N/m3
Conrete M 20
Steel fy = 415 N/mm2
scbc = 7 N/mm2
Nominal cover ,c= 50 mm
unit weight = 25 KN/m 3

Tensile stess = 230 N/mm2

m= 13
unit wt. of water,w = 10 KN/m3
SBC = 54 T/m2 as per soil report
Angle of internal friction = 36 O as per soil report

1. Design Constants :
For HYSD bars , Concrete M = 20
sst = 230 N/mm2 sin 36 = 0.587
wt. of concrete = 25 KN/m3
scbc = 7 N/mm2
m*c 13 x 7 =
k= = 0.283
m*c+sst 13 x 7 + 230
j= 1-k/3 = 1 - 0.283 / 3 = 0.906
R= 0.5.c. j .k = 3.5 x 0.906 x 0.283 = 0.897

2. Design of long wall :

Case 1 Tank full and dry earth outside
Max Water pressure Pw =w (H ) = 20.00 KN/m2
Max Earth pressure Pe =Ka we (H )= we (H ) (1 -sinQ/1+sinQ) = 37.20 KN/m2
Net pressure P = 17.20 KN/m2
Max negative BM producing tension on inner face of wall= P*H2/15 = 4.59 KN-m
Max positive BM producing tension on outer face of wall= P*H2/33.5 = 2.05 KN-m
Case 2 Tank empty and dry earth outside
Max Earth pressure Pe =Ka we (H )= we (H ) (1 -sinQ/1+sinQ) = 37.20 KN/m2
Max negative BM producing tension on outer face of wall= P*H2/15 = 9.92 KN-m
Max positive BM producing tension on inner face of wall= P*H2/33.5 = 4.44 KN-m
Therefore, Max. design B.M. = 9.92 KN-m

2.1 Design of section : Considring bending effect alone,

9920.00 x 1000
Required depth = = 105 mm = 120 mm
0.897 x 1000
Provide total depth D = 120 + 50 = 170 mm
so that available d = 120 mm
2.2 Reinforcement details for long walls :
2.2.1. Vertical Reinforcemnet
a. Ast for outer face of wall ( Max B.M. on outer face )
Mf 9.92 x 1000000
Ast for B.M. = = = 397 mm2
sst.j.d 230 x 0.906 x 120
Using 12 mm bars , A = 3.14xdia 2
3.14 x 144
= 113 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 113 / 397 = 285 = 250 mm
Actual Ast for B.M. = 452.2 mm2
Hence Provided 12 mm F bar, @ 250 mm c/c.
b. Ast for inner face of wall ( Max B.M. on inner face )
Mf 4.59 x 1000000
Ast for B.M. = = = 183 mm2
sst.j.d 230 x 0.906 x 120
Using 12 mm bars , A = 3.14xdia 2
3.14 x 144
= 113 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 113 / 183 = 618 = 300 mm
(3d or 300 mm whichever is lesser)
Actual Ast for B.M. = 376.8 mm2
Hence Provided 12 mm F bar, @ 300 mm c/c.
2.2.2. Horizontal Reinforcemnet
c. Ast for Secondary reinforcement (min)
0.20 % of nominal reinforcement
0.20 x 1000 x 170
Area of steel = = 340 mm2
100
Using 10 mm bars , A = 3.14xdia2 3.14 x 100
= 79 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 79 / 340 = 231 = 220 mm
Actual Ast for B.M. = 356.8 mm2
Hence Provided 10 mm F bar, @ 220 mm c/c in both direction

3. Reinforcement details for short walls. :

Case 1 Tank full and dry earth outside
Net pressure P = 17.20 KN/m2
Max B.M at support (causing tension at the inner side )= Ph 2/12 = 5.73 KN-m
Max B.M at centre (causing tension at the outer side )= Ph 2/24 = 2.87 KN-m
Case 2 Tank empty and dry earth outside
Net pressure P = 37.20 KN/m2
Max B.M at support (causing tension at the outer side )= Ph 2/12 = M= 12.40 KN-m
Max B.M at centre (causing tension at the inner side )= Ph 2/8 - M = 6.20 KN-m
Therefore, Max. design B.M. = 12.40 KN-m

3.1 Design of section : Considring bending effect alone,

12400.00 x 1000
Required depth = = 118 mm < 120 mm OK
0.897 x 1000
Provide total depth D = 120 + 50 = 170 mm
 so that available d = 120 mm
3.2 Reinforcement details for short walls :
3.2.1. Vertical Reinforcemnet
a. Ast for outer face of wall ( Max B.M. on outer face )
Mf 12.40 x 1000000
Ast for B.M. = = = 496 mm2
sst.j.d 230 x 0.906 x 120
Using 12 mm bars , A = 3.14xdia2 3.14 x 144
= 113 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 113 / 496 = 228 = 220 mm
Actual Ast for B.M. = 513.8 mm2
Hence Provided 12 mm F bar, @ 220 mm c/c.
b. Ast for inner face of wall ( Max B.M. on inner face )
Mf 6.20 x 1000000
Ast for B.M. = = = 248 mm2
sst.j.d 230 x 0.906 x 120
Using 12 mm bars , A = 3.14xdia2 3.14 x 144
= 113 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 113 / 248 = 456 = 300 mm
(3d or 300 mm whichever is lesser)
Actual Ast for B.M. = 376.8 mm2
Hence Provided 12 mm F bar, @ 300 mm c/c.
3.2.2. Horizontal Reinforcemnet
c. Ast for Secondary reinforcement (min)
0.20 % of nominal reinforcement
0.20 x 1000 x 170
Area of steel = = 340 mm2
100
Using 10 mm bars , A = 3.14xdia2 3.14 x 100
= 79 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 79 / 340 = 231 = 220 mm
Actual Ast for B.M. = 356.8 mm2
Hence Provided 10 mm F bar, @ 220 mm c/c in both direction

4. Design of base slab :

In this case the water pressure on base slab will be concentrated by the soil pressure below it
Assume projection on Z = 1.377 m
Base slab dimension , m= 8.094 m x 13.09 m
Weight of two walls= 19.55 KN
Soil upward reaction below 2.415369 KN/m
base = per each load =
Reaction 9.775 KN
BM at center of the slab= 9.775 x ( 5.17 ) /2) - ( 2.41537 x ( 4.047 ) x ( 2.024
= 5.488663 KN-m
BM at support= 2.415369 x 1.377 2 / 2 = 2.29 KN-m
Case 1 Tank full and dry earth outside
Net max BM at centre (produces tension on inner side) = 5.489 + 4.59 = 10.08 KN-m
Net max BM at support(produces tension on inner side) = 4.59 - 2.29 = 2.30 KN-m
Case 2 Tank empty and dry earth outside
Net max BM at center (produces tension on outer side) = 12.40 - 5.49 = 6.91 KN-m
Net max BM at support(produces tension on outer side) = 12.40 + 2.29 = 14.69 KN-m
Therefore, Max. design B.M. = 14.69 KN-m
Thickness of the slab on cracking stress consideration
14.69 x 1000
Required depth = = 128 mm = 130 mm
0.897 x 1000
Overall depth ,d = 180 mm
Since tank on underground, provide a 180 mm thick base slab.
Check for projection on "Z"
Weight of soil on projection Z = 2Zx 18600 x 2.00 = 74400 Z N
Uplift Pressure, pgh = 1000*9.81*(2.0+0.18) = 21386 N/m 2

Upward force = 21385.8 x ( 5.00 + 0.34 + 2Z)x1

= 114200 + 42771.6 Z N
Upward force = downward force (weight of base slab = 36.42 + 9 Z)
114.2002 + 42.7716 Z = 21.85 + (Weight of base slab) + 74.4 Z
114.2002 + 42.7716 Z = 21.85 + 36.423 + 83.4 Z
40.6284 Z = 55.93
Z = 1.377 m = 1.377 m (proved)
Bearing Pressure
Total vertical load , V= Load above footing + weight of slab footing
Load above footing = weight of water tank+ 2*(weight of long wall ) + 2*(wight of short wall) =
1000 + 195.5 + 97.75 = 1293 KN
weight of slab footing = 476.9228 KN
V= 1770 KN
Area of footing , A = 105.983 m2
Zx = b x d2 / 6 = 231.29 m3
Zz = b x d / 6 =
2
142.97 m3
Gross bearing pressure, s = V/A ± Mx/Zx
smax = 16.77 KN/m2 s1 = 1.677 T/m2 < 54 T/m2 OK
smin = 16.64 T/m2 s2 = 1.664 T/m2
a. Ast for outer face of wall ( Max B.M. on outer face )
Mf 14.69 x 1000000
Ast for B.M. = = = 542 mm2
sst.j.d 230 x 0.906 x 130
Using 12 mm bars , A = 3.14xdia2 3.14 x 144
= 113 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 113 / 542 = 209 = 200 mm
Actual Ast for B.M. = 565.2 mm2
Hence Provided 12 mm F bar, @ 200 mm c/c.
b. Ast for inner face of wall ( Max B.M. on inner face )
Mf 10.08 x 1000000
Ast for B.M. = = = 372 mm2
sst.j.d 230 x 0.906 x 130
Using 12 mm bars , A = 3.14xdia 2
3.14 x 144
= 113 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 113 / 372 = 304 = 300 mm
(3d or 300 mm whichever is lesser)
Actual Ast for B.M. = 376.8 mm2
Hence Provided 12 mm F bar, @ 300 mm c/c.
3.2.2. Horizontal Reinforcemnet
c. Ast for Secondary reinforcement (min)
0.20 % of nominal reinforcement
0.20 x 1000 x 180
Area of steel = = 360 mm2
Area of steel = = 360 mm2
100
Using 10 mm bars , A = 3.14xdia2 3.14 x 100
= 79 mm2
4 x100 4 x 100
Spacing of Bars = 1000 x 79 / 360 = 218 = 210 mm
Actual Ast for B.M. = 373.8 mm2
Hence Provided 10 mm F bar, @ 210 mm c/c in both direction
)