DESIGN OF REACTANGULAR GROUND WATER TANK Name of work :- pkn 1 Tank size L= 6.00 x B= 4.00 x H= 3.00 m 2 Tank capacity 72000 ltr 3 Saturated soil unit wt 17.00 17000 4 Conrete M 20 concrete unit 25000 7 m 13 5 Steel fy 415 Tensile stress 115 6 Nominal Cover 35 mm Water unit wt 9.80 7 Thickness Walls 220 mm 0.22 m Bottom slab 100 mm 10 cm 0.10 m 7 Reinforcement Long wall In side near corner Ho 20 110 mm c/c 1.00 m height above the base, near corner in side middle horizont 20 110 mm c/c upto top Out side middle horizon 20 220 mm c/c upto top Short wall In side near corner Ho 20 110 mm c/c upto top Out side middle horizon 20 220 mm c/c, upto top Distribution 10 230 mm c/c vertical 9 Base 8 250 mm c/c in both direction C 20 20 110 mm c/c 110 mm c/c 20 220 mm c/c 20 20 A 20 110 mm c/c 220 mmc/c 10 220 mmc/c 20 220 mm c/c 230 mmc/c 20 220 mm c/ (d) 10 230 mm c/c both side 10 20 10 20 110 mm c/ (d+e) 110 mmc/c 220 mmc/c 110 mmc/c B Bars(c) 20 250 mmc/c Section plan at depth of H/4 or 1 mt. Section on CD D Bar(a 20 220 mm c/c 20 440 mm c/c 10 230 mm c/c Bar(b 20 220 mm c/c 10 230 mm c/c Bar(c 20 220 mm c/c 20 220 mm c/c 8 250 mm c/c both way Bar(d 20 220 mm c/c Bar F Bar(e 10 230 mm c/c Section on AB kN/m 3 N/m 3 N/m 3 scbc N/mm 2 N/mm 2 kN/m 3 mm f bars mm f bars mm f bars mm f bars mm f bars mm f bars mm f bars mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f mm f
Transcript
DESIGN OF REACTANGULAR GROUND WATER TANK
Name of work :- pkn1 Tank size L= 6.00 x B= 4.00 x H= 3.00 m2 Tank capacity 72000 ltr
3 Saturated soil unit wt 17.00 17000
4 Conrete M 20 concrete unit wt. 25000
7 m 13
5 Steel fy 415 Tensile stress 1156 Nominal Cover 35 mm Water unit wt 9.807 Thickness Walls 220 mm 0.22 m
Bottom slab 100 mm 10 cm 0.10 m7 Reinforcement
Long wallIn side near corner Horizont 20 110 mm c/c 1.00 m height
above the base, near corners. in side middle horizontal 20 110 mm c/c upto top Out side middle horizontal 20 220 mm c/c upto top Short wallIn side near corner Horizont 20 110 mm c/c upto top Out side middle horizontal 20 220 mm c/c, upto top Distribution 10 230 mm c/c vertical
9 Base 8 250 mm c/c in both direction
C20 20
110 mm c/c 110 mm c/c
20 220 mm c/c
20 20
A 20 110 mm c/c 220 mmc/c 10 220 mmc/c
20 220 mm c/c 230 mmc/c
20 220 mm c/ (d)
10 230 mm c/c both side 10 20 10
20 110 mm c/ (d+e) 110 mmc/c 220 mmc/c 110 mmc/c
B
Bars(c) 20 250 mmc/c
Section plan at depth of H/4 or 1 mt. Section on CD
Name of work :- pkn1 Tank size L= 6.00 x B= 4.00 x H= 3.00 m2 Tank capacity 72000 ltr
3 Saturated soil unit wt 17.00 17000
4 Conrete M 20 unit weight 25000
7 m 13
5 Steel fy 415 Tensile stress 1156 Nominal Cover 35 mm Water unit wt 9.807 Thickness Walls 220 mm 0.22 m
Top roof 200 mm 20 cm 0.20 mBottom slab 100 mm 10 cm 0.10 m
7 Reinforcement Long wallIn side near corner Horizontal 20 110 mm c/c 1.00 m height
above the base, near corners. in side middle horizontal 20 110 mm c/c upto top Out side middle horizontal 20 220 mm c/c upto top Short wallIn side near corner Horizontal 20 110 mm c/c upto top Out side middle horizontal 20 220 mm c/c, upto top Distribution 10 230 mm c/c vertical
9 Base Main 8 250 mm c/c in both direction
kN/m3 N/m3
N/m3
scbc N/mm2
N/mm2
kN/m3
mm F
mm Fmm F
mm Fmm Fmm Fmm F
DESIGN OF REACTANGULAR GROUND WATER TANKName of work :- pknTank size 6.00 x 4.00 x 3.00 mTank capacity 72000 ltr
Saturated soil unit wt 17.00 17000
Conrete M 20 unit weight = ###
Steel fy 415 Tensile stess = 115
7 m = 13
Nominal cover 35 mm unit wt. of water = 9.8 9800
1 Design Constants:- For HYSD Bars Cocrete M = 20
= 115 wt. of concrete = 25000
= 7 m = 13m*c
=13 x 7
= 0.44213 x 7 + 115
= 1 - 0.442 / 3 = 0.853
= 0.5 x 7 x 0.853 x 0.442 = 1.320
3 Determination of B.M. for horizontal bending :--L / B = 6.00 / 4.00 = 1.50 < 2
Here h = H/4 or 1 m which ever is greater \ h = 1.00 mThus top 2.00 m height of walls will be bend horizontally while the bottom 1.00 m will bend as
vertical cantilever . The bending moments for horizontal bending may be determined by moment distribution by considering tank as continuos frame of unit height at level of D.
Water pressure p at point D is given by =p= w (H - h ) = 9800 ( 3.00 - 1.00 )= 19600 N-m
The Fixed end moments for long wall = =P x 6.00
3.00 P N-m12 12
Fixed end moments for short wall = =P x 4.00
1.33 P N-m12 12
+ 3.00 P while the fixed end
- 1.33 Pboth the walls to be the same, the stiffness of walls will be inversely proportional to these length.
Thus we have following table.Member Stiffness Relative stiffness Sum Distribution factor
AE1 1
x 6.00 = 25
2= 0.4
3 3 5
AF1 1
x 6.00 = 33
= 0.62 2 5
The moment distribution is carried out in the following table.Joint A
Member AE AFDistribution facvtor 0.4 0.6
Fixed end moments + 3.00 p - 1.33 p
Balancing moments - 0.667 p - 1 p
Final moments + 2.33 p - 2.33 p
kN/m3 N/m3
N/m3
N/mm2 N/mm2
scbc N/mm2
N/mm3
sst N/mm2 N/mm2
scbc N/mm2
k= m*c+sstj=1-k/3
R=1/2xc x j x k
Hence Both long and short walls will bend horizontally for upper portion, upto poin D, where horizontal water pressure is p=w(H-h).
PL2 2'=
PB2 2'=
Refer fig 1. Consider quarter frame FAE with joint A rigid. Taking clock wise moment as positive and anticlock
wise moment as negative, the fixed end moment MAF for long wall will be
end moments MAF for short wall will be Considreing Area A and moment of inertia l for
Hence moment at supports, Mf= 2.33 x 19600 = 45733 N-m/mThis support moment will cause tension at the water force.
B.M. at the center long span = - Mf "=### x 6.00
45733 = 42467 N-m/m8 8
This bending moment cause tension at outer face.
B.M. at the center short span = - Mf "=### x 4.00
45733 = -6533 N-m/m8 8
= 42467 N-m/m
3 Design of section :- Considring bending effect alone,
Required depth =42467 x 1000
= 179 mm1.320 x 1000
Provide total depth T= 179 + 35 = 220 mm so that available d = 185 mm
4 Determination pull :-
19600 x 4.00 / 2 = 39200 N
19600 x 6.00 / 2 = 58800 N
5 Cantilever Moment :- Cantilever moment atb the base, per unit length .
= w H xh2
=9800 x 4.00 x 1.00
6533 N-m6 6
This will cause tension at water face.6 Reinforcement at corners of long walls.:- The upper portion of long walls is subjected to both bending in
horizontal direction as well as pull. The reinforcement for both will be in horizontal direction. Hence
tension on water face). Similarly vertical section of unit height ( 1 m) of long wall, at its end, at the level of1.00 m above the base , where reinforcement is provided at the water face.
x = d -T
= 179 -220
= 69 mm2 2
=Mf - Pl x
=45733 x 1000 ) - ### x 69
= 2371115 x 0.853 x 185
=PL
=39200
= 341115
= 2371 + 341 = 2712 per meter height.
using ### mm bars A = =3.14 x 20 x 20
= 3144 x100 4 x 100Spacing of Bars = 1000 x 314 / 2712 = 116 say = 110 mm
Hence Provided 20 110 mm c/c. The above reinforcement is to be provided at
inner face, near the corners, and at a height 1.00 m above the base. For other height the above spacing may be varied, since bending moment will reduce.
7 Reinfocement at the middle of long wall. :-Tension occurs at outer face. However, since distance of corner of steel from
water face will be less than 225 mm, permissible stress will be 115
= 0.442 j = 0.853 R = 1.318Design B.M. = 42467 N-m per meter height. = 39200 N
= 3144 x100 4 x 100Spacing of Bars = 1000 x 314 / 2533 = 124 say = 120 mm
This is very near to the reinforcement provided at ends.Hence provided 20 110 mm
c/c. Bend half the bars provided at ends, outwards.at distance L/4 = 1.50 m form ends.
This reinforcement is to be provided at outer face. The additional 20 220 mm c/c.
are continued upto the end.
8 Reinforcement for shorts walls.:-
B.M. at ends=Mf = 45733 N-m 58800 N
= =45733 x 1000 ) - ### x 69
= 2297115 x 0.853 x 185
=PL
=58800
= 511115
= 2297 + 511 = 2808 per meter height.
using ### mm bars A = =3.14 x 20 x 20
= 3144 x100 4 x 100Spacing of Bars = 1000 x 314 / 2808 = 112 say = 110 mm
Hence provide 20 mm f bars @ 110 mm c/c at inner face near the ends of short span.The B.M. at the center of short walls cause tension at water face (unlikethat in the center of long walls where tension is produced at outer face ).since this B.M. is small, only nominal reinforcement is required. Similarlly, we have to provide nominal reinforcement at outer face,. Hence bend half bars outward at distance B/4= 1.00m from each end, and continue remaning half tjrought. Thus at the center of span, the reinforcement on each face will consist of 20 220 mm c/c.
9 Reinforcement for cantilever moment and distribution reinforcement.:-max. cantilever moment= 6533 N-m
=6533 x 1000
= 360115 x 0.853 x 185
But minimum reinforcementin vertical direction =0.3
x( 220 x 1000 )= 660100
Since half of this area of steel can reist cantilever momnt, we will provide = 330on the inner face and remaining area i.e.= 330
reinforcment. \ Area of steel on each face = 330 mm2.
using ### mm bars A = =3.14 x 10 x 10
= 78.54 x100 4 x 100Spacing of Bars = 1000 x 78.5 / 330 = 238 say = 230 mm
Hence Provided 10 230 mm c/c on out side face, at bottom of long wall
Since tank rest on ground, provide a 100 mm thick base slab.taking 1m length for calculation and 0.20 % of nominal reinforcement
area of steel =0.20 x 1000 x 100
= 200100
using 8 mm bars A = =3.14 x 8 x 8
= 50.24 x100 4 x 100Spacing of Bars = 1000 x 50.2 / 200 = 251 say = 250 mm
Hence Provided 8 250 mm c/c in both direction , at top and bottom of base slab.
shown in drawing.
3.14xdia2
mm2
mm f bars
mm f bars provided @
Direct pull pu =
Ast for B.M. M - PB x
mm2
sst.j.d
Ast for pull mm2
ss
Total Ast mm2
3.14xdia2
mm2
mm f bars @
Ast mm2
mm2
mm2 steel area vertically mm2 vertically at outer face to serve as distribution
