Design of Rectangular water tank CASE-1 ( L / B < 2 )
Capacity 80000 Litres
(given)
Material
M20 Grade Concrete
(given)
Fe 415 Grade HYSD reinforcement
(given)
Solution :Provide 6 m x 4 m x 3.5 m tank with free board of 150 mm.
Volume = 6 x 4 x 3.35 x 10 3 =
80400 Litres
6
/
4
=
1.5
<2.
L/B=
The top portion of side walls will be designed as a continuous frame.
bottom 1 m or H / 4 whichever is more is designed as cantilever.
H/4=
3.5 / 4 =
0.875 m
bottom 1 m will be designed as cantilever.
2.5 x YW = 2.5 x 9.8
Water pressure at 3.5 - 1 = 2.5 m height from top =
=
24.5
KN / m2
where Yw is unit weight of water = 9.8 KN / m3
To find moment in side walls, moment distribution or kani's method is used. As the
frame is symmetrical about both the axes, only one joint is solved
6m
A
F
2.5 m
3.5 m
6m
E
24.5 KN / m
2
1m
D
34.3 KN / m
2
Elevation
Plan
Fixed end moments :MAB = w x l2 / 12 =
MAD =
= 24.5 x 62 / 12 =
73.5
KNm
w x l2 / 12 =
= 24.5 x 42 / 12 =
-32.66
KNm
Kani's Method :-
0
0
D
- 32.66
-12.25
-12.25
0
-57.16
73.5
-3/10 40.84 -2/10 -8.17
A
-8.17
0
57.16
0
0
B
Rotation factor at Joint A
Joint
Member
Relative
Stiffness( K )
∑K
Rotation Factor
u =(-1/2) k / ∑ K
A
AB
AD
I/6
I/4
5 * I / 12
- 2 / 10
- 3 / 10
Sum of FEM
MAF = 73.5-32.66
40.84
KNm
MAB =
MABF + 2 MAB' + MBA'
= 73.5 + 2 x (- 8.17 ) + 0
= 57.16
MAD =
MADF + 2 MAD' + MDA'
= (- 32.66 ) + 2 x (- 12.25 ) + 0
= -57.16
B.M. at centre of long span =
w x l 2 / 8 - 57.16
= 24.5 x 62 / 8 - 57.16
53.09
KNm
B.M. at centre of short span =
w x l 2 / 8 - 57.16
= 24.5 x 42 / 8 - 57.16
-8.16
KNm
Direct tension in long wall =
=
Yw ( H - h ) x B / 2
24.5 x 4 / 2 =
49
KN
=
Yw ( H - h ) x L / 2
24.5 x 6 / 2 =
73.5
KN
Direct tension in short wall =
Design of Long Walls :At support
M=
57.16
T=
From Table 9-6
D=
KNm
49
KN
Tension on liquid face.
Q = 0.306 Assuming d / D = 0.9
√M / Q x b
= √57.16 x 10 / 0.306 x 1000
6
=
Take D = 450 mm
432.2
mm,
Assuming d / D = 0.9
d = 450 - 25 - 8
=
417 mm
From Table 9-5
Ast1 for moment = M / σst x j x d
=
=
57.16 x 10 6 / 150 x 0.872 x 417
1048
mm2
Ast2 for direct tension = T / σst
Table 9-5
=
=
Total Ast1 + Ast2 =
49 x 10 / 150
327
mm2
3
1048 + 327
= 1375
mm2
Provide 16 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 200.96 x 1000 / 1375
= 146.152727
mm
Provide 16 mm O bar @ 130 mm C/C…marked(a)
= 1546
mm2 / m.
Larger steel area is provided to match with the steel of short walls.
At centre
M=
53.09
KNm
T=
49
KN
tension on remote face
e=
M/T=
53.09 / 49
= 1.08 m
Line of action of forces lies outside the section
i.e.tension is small
E=
e+D/2-d
b
= 1080 + 450 / 2 - 417
=
888
mm
D
modified moment = 49 x 0.888
=
43.51
d
KNm
d'
Ast1 for moment = M / σst x j x d
=
=
Table 9-5
43.51 x 10 6 / 190 x 0.89 x 417
617
mm2
Ast2 for direct tension = T / σst
=
=
Total Ast1 + Ast2 =
Table 9-5
49 x 10 3 / 150
327
mm2
617 + 327
= 944
mm2
Provide 16 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
E=e+D/2-d
= 200.96 x 1000 / 944
= 212.881356
mm
Provide 16 mm O bar @ 200 mm C/C…marked(b)
= 1005
mm2
From Table 9-3
minimum reinforcement 0.16 %
Distribution steel = 0.16 / 100 x 1000 x 450
=
720
mm2
On each face =
360
mm2
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 /360
= 139.555556
mm
Provide 8 mm O bar @ 130 mm C/C…marked(d)
= 385
mm2
Vertical Steel ( c)
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 78.50 x 1000 /360
= 218.055556
mm
Provide 10 mm O bar @ 200 mm C/C…marked(c)
= 392
mm2
Horizontal steel :Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005
mm2
Liquid face ( d) - Provide 8 mm O @ 130 mm C/C =
385
mm2
Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392
Design of short walls :At support
M=
T=
From Table 9-5
Ast1 for moment = M / σst x j x d
=
=
57.16
73.5
KNm
KN
tension on liquid face
Table 9-5
57.16 x 10 6 / 150 x 0.872 x 417
1048
mm2
Ast2 for direct tension = T / σst
=
=
Total Ast1 + Ast2 =
Table 9-5
73.5 x 10 3 / 150
490
mm2
1048 + 490
= 1538
mm2
Provide 16 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 200.96 x 1000 / 1538
= 130.663199
mm
Provide 16 mm O bar @ 130 mm C/C…marked(b)
= 1546
mm2 / m.
At centre
mm2
M=
8.16
T=
73.5
From Table 9-5
Ast1 for moment = M / σst x j x d
=
=
KNm
KN
tension on liquid face
Table 9-5
8.16 x 10 6 / 150 x 0.872 x 417
150
mm2
Ast2 for direct tension = T / σst
=
=
Total Ast1 + Ast2 =
Table 9-5
73.5 x 10 3 / 150
490
mm2
150 + 490
= 640
mm2
Provide 12 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 113.04 x 1000 / 640
= 176.625
mm
Provide 12 mm O bar @ 130 mm C/C…marked(e)
= 869
mm2 / m.
From Table 9-3
minimum reinforcement 0.16 %
Distribution steel = 0.16 / 100 x 1000 x 450
=
720
mm2
On each face =
360
mm2
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 /360
= 139.555556
mm
Provide 8 mm O bar @ 130 mm C/C…marked(d)
= 385
mm2
Vertical Steel ( c)
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 78.50 x 1000 /360
= 218.055556
mm
Provide 10 mm O bar @ 200 mm C/C…marked(c)
= 392
mm2 .
Horizontal steel :Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869
mm2
Liquid face ( d) - Provide 8 mm O @ 130 mm C/C =
385
mm2
Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392
Bottom 1 m will be designed as cantilever
Cantilever moment : Yw x H x h 2 / 6
Yw x H / 6 , whichever is greater.
M=
OR
mm2
= 9.8 x 3.5 x 1 / 6
= 5.72
KNm
From Table 9-5
Ast for moment =
= 9.8 x 3.5 / 6
=
5.72
,tension on liquid face.
M / σst x j x d
Table 9-5
= 5.72 x 10 / 150 x 0.872 x 417
=
105
mm2
From Table 9-3
minimum reinforcement 0.16 %
Distribution steel = 0.16 / 100 x 1000 x 450
=
720
mm2
On each face =
360
mm2
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 /m
= 78.50 x 1000 /360
=
218
mm
Provide 10 mm O bar @ 200 mm C/C…marked(c)
= 392
mm2 .
each face
Base slab :Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab.
From table 9-3
Minimum steel =
0.229%
= 0.229 / 100 x 1000 x 150
=
344
mm2
,172 mm2 bothway
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 /172
=
292
mm
Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.
Ast =
346
mm2
Designed section,Elevation etc. are shown in fig.
Top slab may be designed as two-way slab as usual for a live load of 1.5 KN / m 2
Top slab : consider 1 m wide strip. Assume 150 mm thick slab.
lx =
4 + 0.15 = 4.15 say 4.5 m
6
ly =
6 + 0.15 = 6.15 say 6.5 m
Dead Load : self 0.15 x 25 =
3.75
KN / m2
floor finish =
1.0
KN / m2
Live load =
1.5
KN / m2
6.25
KN / m2
For 1 m wide strip
PU =
1.5 x 6.25
= 9.38
KN / m
ly / lx =
6.5 / 4.5
=
1.4
AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span.
αx =
Table 26
0.085
αy =
Mx =
αx x w x lx
0.056
My =
2
= 0.085 x 9.38 x 4.52
= 16.15
KNm
From Table 6-3 ,Q = 2.76
αy x w x lx2
= 0.056 x 9.38 x 4.52
= 10.64
KNm
√M / Q x b
= √16.15 x 10 / 2.76 x 1000
drequired =
6
=
76.50
mm
dshort = 150 - 15(cover) - 5
= 130 > 76.50
mm
…………(O.K.)
dlong = 130 - 10 =
120
mm
Larger depth is provided due to deflection check.
Mu / b x d2 (short) = 16.15 x 106 / 1000 x 130 x 130
=
0.96
Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)
fy / fck
=
50 1-√1-(4.6 / 20) x (0.96)
415 / 20
= 50 [(1-0.88) x 20 / 415 ]
= 0.29%
Ast (short) = 0.29 x 1000 x 130 / 100
=
377
mm2
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 /m
= 78.50 x 1000 /377
=
208
mm
Provide 10 mm O bar @ 210 mm c/c
= 374
mm2 .
Mu / b x d2 (long) = 10.64 x 106 / 1000 x 120 x 120
=
0.74
Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)
fy / fck
=
50 1-√1-(4.6 / 20) x (0.74)
415 / 20
= 50 [(1-0.91) x 20 / 415 ]
= 0.22%
Ast (long) = 0.22 x 1000 x 120 / 100
=
264
mm2
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 /264
=
190
mm
Provide 8 mm O bar @ 190 mm c/c
= 264
mm2 .
B
4m
C
TABLE 9-6
Balanced Design Factors for members in bending
For M20 Grade Concrete Mix
Mild steel
HYSD bars
d/D
Pt
Pt
Q = M / bD2
Q = M / bD2
0.75
0.3
0.4
0.295
0.289
0.8
0.305
0.37
0.299
0.272
0.85
0.9
0.31
0.314
0.355
0.335
0.302
0.306
0.258
0.246
TABLE 9-5
Members in bending ( Cracked condition )
Coefficients for balanced design
σst
Grade of
Grade of σcbc
2
concrete
steel
N / mm
N / mm2
k
j
For members less than 225mm thickness and tension on liquid face
M20
Fe250
7
115
0.445
0.851
Fe415
7
150
0.384
0.872
Q
1.33
1.17
For members more than 225mm thickness and tension away from liquid face
M20
Fe250
7
125
0.427
0.858
1.28
Fe415
7
190
0.329
0.89
1.03
s lies outside the section
D/2
e=M/T
TABLE 9-3
Minimum Reinforcement for Liquid Retaining Structures
Thickness, mm
100
150
200
250
300
% of reinforcement
Mild Steel
HYSD bars
0.3
0.24
0.286
0.271
0.257
0.243
0.229
0.217
0.206
0.194
350
400
450 or more
0.229
0.214
0.2
0.183
0.171
0.16
8 O @ 190 c/c
10 O @ 210 c/c
150
150 Free board
3500
A
10 O @ 200 c/c -
shape
10 O @ 200 c/c -
shape
1500
150
150
8 O @ 290 c/c both ways top and bottom
1500
A
1 : 4 : 8 P.C.C.
150 450
6000
450 150
Elevation
1500
450
1000
16 O @ 130 c/c (a)
16 O @ 200 c/c (b)
10 O @ 200 c/c both faces (c)
4000
8 O @ 130 c/c (d)
12 O @ 130 c/c (e)
1000
vv
(a)
(b)
(d)
(c)
(d)
( a ) 1000
vv
450
1500
1500
6000
450
450
Section A-A
Table 6-3
Limiting Moment of resistance factor Q lim, N / mm2
fck N / mm2
fy, N / mm2
15
20
25
250
2.22
2.96
3.70
415
2.07
2.76
3.45
500
2.00
2.66
3.33
550
1.94
2.58
3.23
30
4.44
4.14
3.99
3.87
Pt,bal
1.36
0.98
1.2
0.61
Design of Circular water tank with flexible base
capacity 600000 litres
material M20
grade concrete
( Given )
Fe415 HYSD reinforcement
Solution :Provide 5 m depth with 0.2 m free board . If D is the diameter of tank
( ¶ / 4 ) D2 x 4.8 = 600 m3
D = 12.62
m.
( 1 m3 = 1000 litres )
Provide 12.8 m diameter of the tank.
Maximum hoop tension at base
= ywHD / 2
= 9.8 x 5 x 12.8 / 2
= 313.6
KN.
Area of steel required
= 313.6 x 103 /150
Table 9-5
2
= 2091
mm /m.
Provide 16 mm O .
Area of one bar = 201
mm2.
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 201 x 1000 / 2091
= 96.1263
mm
Provide 16 mm O bar @ 190 mm c/c on both faces ( circular hoop steel ) .
Ast = 201 x 1000 x 2 / 190
= 2116
mm2.
Let t be the thickness of wall in mm.
Tensile stress σct = T / ( 1000 t + ( m - 1 ) Ast )
m = 13.33
σct = 1.2
1.2 = 313.6 x 103 / ( 1000 x t + 12.33 x 2116 )
t = 235.2
mm.
Provide 240 mm thickness.
Vertical reinforcement ( minimum steel )
= 0.208
%
As = ( 0.208 / 100 ) x 1000 x 240
Table 9-3
50 diff.
-0.011
10 diff.
?
-0.0022
= 499
mm2.
Provide 8 mm O .
Area of one bar = 50
mm2.
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 50 x 1000 / 499
= 100.2
mm
Provide 8 mm O bar @ 200 mm c/c on both faces .
Ast = 50 x 1000 x 2 / 200
= 500
mm2.
As the pressure decreases from bottom to top , the main reinforcement may be reduced.
At 4 m height from top
T = ywHD / 2
= 9.8 x 4 x 12.8 / 2
= 250.88 KN.
Ast = 250.88 x 103 /150
Table 9-5
= 1673
mm /m.
Provide 16 mm O .
Area of one bar = 201
mm2.
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 201 x 1000 / 1673
= 120.143
mm
Provide 16 mm O bar @ 240 mm c/c on both faces .
Ast = 201 x 1000 x 2 / 240
2
= 1675
mm2.
At 3 m height from top
T = ywHD / 2
= 9.8 x 3 x 12.8 / 2
= 188.16 KN.
Ast = 188.16 x 103 /150
Table 9-5
= 1254
mm /m.
Provide 12 mm O .
Area of one bar = 113
mm2.
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 113 x 1000 / 1254
= 90.1116
mm
Provide 12 mm O bar @ 180 mm c/c on both faces .
Ast = 113 x 1000 x 2 / 180
2
= 1256
mm2.
At 2 m height from top
T = ywHD / 2
= 9.8 x 2 x 12.8 / 2
= 125.44 KN.
Ast = 125.44 x 103 /150
Table 9-5
= 836
mm /m.
Provide 10 mm O .
Area of one bar = 79
mm2.
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 79 x 1000 / 836
= 94.4976
mm
Provide 10 mm O bar @ 180 mm c/c on both faces .
Ast = 79 x 1000 x 2 / 180
2
= 878
mm2.
Reinforcement will not be now reduced as it will be less than minimum.
Base :
For 5 m head of water , provide 150 mm thick base.
Minimum steel = 0.229 % ( table 9-3 ).
Ast = ( 0.229 / 100 ) x 1000 x 150 = 344 mm2 / m.
Provide 8 mm O .
Area of one bar = 50
mm2.
spacing of bar = Area of one bar x 1000 / required area in m 2 / m
= 50 x 1000 / 344
= 145.349
mm
Provide 8 mm O bar @ 290 mm c/c on both ways at top and bottom .
Ast = 50 x 1000 x 2 / 290
= 345
sketch :
1000
mm2.
240
10 O @ 180 c/c ( both faces ) horizontal steel
8 O @ 200 c/c ( both faces ) Vertical steel
1000
10 O @ 180 c/c ( both faces ) horizontal steel
1000
12 O @ 180 c/c ( both faces ) horizontal steel
1000
16 O @ 240 c/c ( both faces ) horizontal steel
1000
Prepared sliding surface or rubber pad
strip painting
joint sealing compound
16 O @ 190 c/c ( both faces ) horizontal steel
100
150
Elevation
v v v v v v v v v v v v v v v 150
v v v v v v v v v v v 150
Sliding layer of bitumen paper
M20
M7.5
8 O @ 290 c/c both ways
top and bottom
Lap in first ring
Lap in first ring
Lap in third ring
Minimum lap = 2 Ld
Four bar in each ring
Lap in third ring
Lap in second ring
Lap in second ring
Lap in first ring
Lap in first ring
Proposed laps in horizontal bars
y be reduced.
Circular water tanks are used for a large capacity in the fields of water
supply and sewage treatment pumping stations,chemical processing
and large underground sumps.
At base , maximum water pressure = yw H.
Hoop tension per meter = yw H D / 2.
where , yw = unit weight of water ( liquid)
= 9.8 KN / m3.
Tensile stress σct = T / ( 1000 t + ( m - 1 ) Ast )
where, m = 280 / 3 σcbc.
where , σcbc is permissible compressive stress due to
bending in concrete N / mm2.
Grade of concrete
σcbc N / mm2.
M10
3.0
M15
5.0
M20
7.0
M25
8.5
M30
10.0
M35
11.5
M40
13.0
where , σct is permissible stress in concrete in direct tension.
Table 9-2
Permissible concrete stresses in
calculations relating to resistance to
cracking
Grade of
concrete
Permissible stresses in N / mm2
Direct
tension σct
Tension
due to
bending
σcbt
Shear stress
‫ﺡ‬v = V / b j
d
M15
M20
1.1
1.2
1.5
1.7
1.5
1.7
M25
M30
M35
M40
1.3
1.5
1.6
1.7
1.8
1.9
2.0
2.2
2.2
2.5
2.4
2.7
TABLE 9-3
Minimum Reinforcement for Liquid Retaining Structures
Thickness, mm
100
150
% of reinforcement
Mild Steel
0.3
0.286
HYSD bars
0.24
0.229
200
0.271
0.217
250
300
350
0.257
0.243
0.229
0.206
0.194
0.183
400
450 or more
0.214
0.2
0.171
0.16
ty in the fields of water
,chemical processing
TABLE 9-5
Members in bending ( Cracked condition )
Coefficients for balanced design
σcbc
σst
Grade
Grade of
Pt,bal
concrete of steel N / mm2
N / mm2
k
j
Q
For members less than 225mm thickness and tension on liquid face
M20
Fe250
7
115
0.445
0.851
1.33
1.36
Fe415
7
150
0.384
0.872
1.17
0.98
For members more than 225mm thickness and tension away from liquid face
M20
Fe250
7
125
0.427
0.858
1.28
1.2
Fe415
7
190
0.329
0.89
1.03
0.61
irect tension.
Design of Rectangular water tank CASE-2 ( L / B ≥ 2 )
Size of tank : 3.6 m x 8.0 m x 3.0 m high
(given)
Material
M20
Grade Concrete
(given)
Fe 415
Grade HYSD reinforcement
(given)
Solution :Size of tank : 3.6 m x 8.0 m x 3.0 m high
Volume = 3.6 x 8 x 3.0 x 10 3 =
86400 Litres
8
/
3.6
=
2.22
>2.
L/B=
The long walls are designed as vertical cantilevers from the base.
The short walls are designed as supported on long walls.
If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m.
bottom 1 m or H / 4 whichever is more is designed as cantilever.
H/4=
3.0 / 4 =
0.75
m
bottom h = 1 m will be designed as cantilever.
Moments and tensions :
Maximum B.M. in long walls at the base
3
= (1 / 6 ) x Yw x H
= ( 1 / 6 ) x 9.8 x 33
= 44.1
KNm.
Maximum ( - ve ) B.M. in short walls at support
2
= Yw x ( H - h ) x B / 12
= 9.8 x ( 3 - 1 ) x 42 / 12
= 26.13
KNm.
Maximum ( + ve ) B.M. in short walls at centre
2
= Yw x ( H - h ) x B / 16
= 9.8 x ( 3 - 1 ) x 42 / 16
= 19.60
KNm.
For bottom portion
Yw x H x h 2 / 6
M=
= 9.8 x 3.0 x 1 / 6
=
4.90
KNm
Direct tension in long wall =
OR
Yw x H / 6 , whichever is greater
= 9.8 x 3.0 / 6
=
4.90
KNm
Yw x ( H - h ) x B / 2
= 9.8 x ( 3 - 1 ) x 3.6 / 2
=
35.28
KN
Yw ( H - h ) x 1
=
9.8 x ( 3 - 1 ) x 1
=
19.6
KN
It is assumed that end one metre width of long wall gives direct tension to short walls.
Direct tension in short wall=
Design of long walls : M(-)=
44.1
KNm
T=
35.28
KN
From Table 9-6
Assume d / D = 0.9
( water face )
( perpendicular to moment steel )
Q = 0.306
√
D= M/Qxb
=
=
Take D =
√44.1 x 10 / 0.306 x 1000
6
379.6
mm,
400 mm
d = 400 - 25 - 8
= 367 mm
Ast for Moment
From Table 9-5 ,
Ast = M / σst x j x d
Table 9-5
= 44.1 x 10 / 150 x 0.872 x 367
= 918.68
mm2
6
Provide 16 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 200.96 x 1000 / 918.68
= 218.749
mm
Provide 16 mm O bar @ 200 mm c/c = 1005
mm2 .
Note : The design is made at the base. The moment reduces from base to top.For economy, the
reinforcement can be curtailed or the thickness of wall can also be reduced as we have done for
cantilever retaining wall.
From Table 9-3
Distribution steel = 0.171 % for 400 mm depth
As = ( 0.171 / 100 ) x 1000 x 400
= 684
mm2 .
on each face = 342
Steel required for direct tension
= T / σst
mm2 .
…………………… ( 1 )
Table 9-5
= 35.28 x 10 / 150
= 235
mm2 . …………………… ( 2 )
From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.
Provide 8 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 / 342
= 146.901
mm
Provide 8 mm O bar @ 140 mm c/c on each face = 357
mm2
on each face
3
Design of short walls :-
At support
M=
T=
From Table 9-5
Ast1 for moment =
26.13
19.6
KNm
KN
M / σst x j x d
Table 9-5
= 26.13 x 10 6 / 150 x 0.872 x 367
= 544
mm2
Ast2 for direct tension = T / σst
=
=
Total Ast1 + Ast2 =
Table 9-5
19.6 x 10 3 / 150
131
mm2
544 + 131
=
675
mm2
Provide 12 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 113.04 x 1000 / 675
= 167.467
mm
Provide 12 mm O bar@160 mm c/c = 706
mm2.
1000
203.56
400 367
163.44
checking :
modular ratio m =
x=
280 / 3 x σcbc
= 13.33
b x D2 / 2 + Ast ( m - 1 ) x d
=
= 203.56
b x D + ( m - 1 ) x Ast
( 1000 x 4002 / 2 ) + ( 706 x ( 13.33 - 1 ) x 367 )
( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 )
mm
D-x=
196.44 mm
d-x=
163.44 mm
AT = b x D + ( m - 1 ) x Ast
= 1000 x 400 + (13.33 - 1 ) x 706
=
408705 mm2
Ixx = ( 1 / 3 ) x b x ( x3 + ( D - x )3 ) + ( m - 1 ) x Ast x ( d - x )2
= ( 1 / 3 ) x 1000 x ( 203.563 + 196.443 ) + ( 13.33 - 1 ) x 706 x 163.442
=
5.34E+09
+
2.33E+08
=
5.57E+09 mm4
fct = T / AT
= 19.6 x 10 3 / 408705
=
0.048 N / mm2
fcbt = M x ( d - x ) / Ixx
= 26.13 x 106 x 163.44 / 5.57 x 10 9
= 0.767
N / mm2
check :
( fct / σct ) + ( fcbt / σcbt )
≤1
( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) ≤ 1
From Table 9-2
0.04 + 0.4512 ≤ 1
0.4912 ≤ 1
………………….. ( O. K. )
At centre :
M=
T=
From Table 9-5
Ast1 for moment =
19.6
19.6
KNm
KN
M / σst x j x d
Table 9-5
= 19.6 x 10 / 150 x 0.872 x 367
= 408
mm2
6
Ast2 for direct tension = T / σst
=
=
Total Ast1 + Ast2 =
Table 9-5
19.6 x 10 / 150
131
mm2
3
408 + 131
=
539
mm2
Provide 12 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 113.04 x 1000 / 539
= 209.722
mm
Provide 12 mm O bar @ 200 mm c/c = 565
mm2.
From Table 9-3
Distribution steel = 0.171 % for 400 mm depth
As = ( 0.171 / 100 ) x 1000 x 400
= 684
mm2 .
on each face = 342
mm2 .
…………………… ( 1 )
Steel required for direct tension
= T / σst
= 19.6 x 103 / 150
Table 9-5
2
= 131
mm . …………………… ( 2 )
From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.
Provide 8 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 / 342
= 146.901
mm
Provide 8 mm O bar @ 140 mm c/c on each face = 357
mm2
on each face
Bottom cantilever
M = 4.9
KNm
From Table 9-5
Ast = M / σst x j x d
Table 9-5
= 4.9 x 10 6 / 150 x 0.872 x 367
= 102
mm2
Minimum steel = 342 mm2 on each face.
Provide 8 mm O bar @ 140 mm c/c on each faces = 357
mm2
on each face
Base slab :Base slab is resting on ground. For a water head 3 m, provide 150 mm thick slab.
From table 9-3
Minimum steel =
0.229%
= 0.229 / 100 x 1000 x 150
= 344
mm2
,172 mm2 bothway
Provide 8 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 50.24 x 1000 /172
= 292
mm
Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.
Ast =
346
mm2
Designed section,Elevation etc. are shown in fig.
Top slab may be designed as a one-way slab as usual for a live load of 1.5 KN / m 2
Top slab : consider 1 m wide strip. Assume 150 mm thick slab.
lx =
3.6 + 0.4 = 4 say 4 m
ly =
8 + 0.15 = 8.15 say 8.5 m
Dead Load : self 0.15 x 25 =
floor finish =
Live load =
3.75
1.0
1.5
KN / m2
KN / m2
KN / m2
6.25
KN / m2
1.5 x 6.25
=
9.38
KN / m
For 1 m wide strip
PU =
Maximum moment = 9.38 x 42 / 8
= 18.76
KNm
Maximum shear = 9.38 x 3.6 / 2
= 16.88
KN
From Table 6-3 ,Q = 2.76
√M / Q x b
= √18.76 x 10 / 2.76 x 1000
drequired =
6
=
82.44
mm
dprovided =
150 - 15(cover) - 6
= 129 > 82.44
…………(O.K.)
Larger depth is provided due to deflection check.
Design for flexure :
Mu / b x d2 = 18.76 x 106 / 1000 x 129 x 129
= 1.13
Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)
fy / fck
=
50 1-√1-(4.6 / 20) x (1.13)
415 / 20
= 50 [(1-0.86) x 20 / 415 ]
= 0.34%
Ast = 0.34 x 1000 x 129 / 100
= 439
mm2
Provide 8 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 /m
= 50.24 x 1000 /439
= 114
mm
Provide 8 mm O bar @ 110 mm c/c = 457
mm2 .
Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement
Distribution steel = ( 0.12 / 100 ) x 1000 x 150
= 180
mm2
Provide 6 mm O bar
spacing of bar =
Area of one bar x 1000 / required area in m 2 / m
= 28.26 x 1000 /180
= 157
mm
Provide 6 mm O bar @ 150 mm c/c = 188
mm2 .
4.0 m.
hort walls.
op.For economy, the
as we have done for
x
where, m = 280 / 3 σcbc.
where , σcbc is permissible compressive stress due to bending in concrete N /
mm2.
Grade of concrete
M10
M15
M20
M25
M30
M35
M40
σcbc N / mm2.
3.0
5.0
7.0
8.5
10.0
11.5
13.0
42
Table 9-2
Permissible concrete stresses in calculations relating to
resistance to cracking
Permissible stresses in N / mm2
Grade of
Direct
Tension due to Shear stress
concrete
‫ﺡ‬v = V / b j d
tension σct
bending σcbt
M15
1.1
1.5
1.5
M20
1.2
1.7
1.7
M25
M30
M35
M40
1.3
1.5
1.6
1.7
1.8
2.0
2.2
2.4
1.9
2.2
2.5
2.7
6 O @ 150 c/c
8 O @ 110 c/c
150
12 O @ 200 c/c
8 O @ 140 c/c -
shape
8 O @ 140 c/c
8 O @ 290 c/c both ways top and bottom
2000
3000
150
150
2000
150
1 : 4 : 8 P.C.C.
150 400
8000
400
Section A-A
B
2000
8 O @ 140 mm c/c
400
900
12 O @ 160 c/c(chipiya)
16 O @ 200 c/c ( chipiya )
8 O @ 140 c/c (chipiya)
3600
8 O @ 140 c/c
A
900
400
A
12 O @ 200 c/c
900
v
v
vv
2000
400
B
2000
8000
Sectional plan
8 O @ 110 c/c
150
16 O @ 200 c/c ( chipiya )
3000
8 O @ 140 c/c
400
6 O @ 150 c/c
900
900
Base details not shown
for clarity
150
150
150 400
3600
Section B- B
400 150
ue to bending in concrete N /