Design Calculations for proposed 300 m3 R.C.
C Storage Tank for Construction Water at RSP Rourkella
Grade of concrete = M 20 Grade of steel = Fe 415 Ground water level : At ground level 10
5 Plan Of Walls
For 5 m Long Side walls :
2.5
2.5 Unit weight of water = w Lateral pressure due to water = 1 t/m 2.5 t/m 2a Free
b Fixed
2.5
5 2.5
From the figure we get, Therefore , Moody's tables give values for a/b = 1 Horizontal moment =
a= b= a/b =
2.5 m 2.5 m 1.0 m
P= 2.5 t/m Therefore, P x b = 2.5 x 2.5 x 2.5 = 15.625 t Again, Pxb= 2.5 x 2.5 = 6.25 t/m
Mx x P x b
My x P x b Vertical moment = Vertical Reinforcement : Using Figure 4 of Moody's Tables, Maximum vertical moment = 0.0845 x 15.625 (At the bottom of the wall) = 1.320 t-m/m st = Now, 230 cbc = 7 Modular ratio m = 13.33 (m x cbc)/st = k/(1-k) => k= 0.2886 Now, j= (1-k/3) = 0.90 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((1.3203125 x 10^7)/(0.91 x 1000))^0.5
N/mm N/mm
= Taking overall depth = Taking Effective depth = Ast = M / (st x j x d) = providing Ast provided 50
121
mm 230 mm 10 bar,
mm cover and 175 mm
2
364 mm /m 10 bars @ 150 mm c/c 2 524 mm^2/m > 364 mm /m
(O.K)
Horizontal Reinforcement : Maximum horizontal moment =
0.0644 x 15.625 (At the top of the wall) 1.01 t-m/m st = Now, 230 cbc = 7.0 Modular ratio m = 13.33 (m x cbc)/st = k/(1-k) => k= 0.2886 Now, j= (1-k/3) = 0.90 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((1.01 x 10^7)/(0.91 x 1000))^0.5 = 106 mm Taking overall depth = 180 mm Taking 50 mm cover and 10 bar, Effective depth = 112 mm Ast = M / (st x j x d) = = providing Ast provided
2 436 mm /m 10 bars @ 150 mm c/c 2 524 mm^2/m > 436 mm /m
N/mm N/mm
(O.K)
Maximum horizontal shear = = For Side Walls : 10 m long
0.2485 x 6.25 ton 1.553 ton
Condition 1 : No water inside. Earth outside with G.W at its highest level and surcharge outside. 0.18 WALL G.L F1 = Pressure due to water F1 = Pressure due to submerged earth F3 = Pressure due to surchage Length / Height = 4 0.5
1.5 F3 F1 F2
2.5
1.5
0.40 0.33 0.23
Fig: 1 - Pressure Diagram Dry unit weight of soil = Unit weight of water = Considering surcharge = Angle of repose () = Therefore Co-efficient of earth pressure (ca) = fck = fy = 1.8 1 1 30 0.33 25 415 N/mm N/mm t/m t/m t/m
Considering this is a free cantilever, Moment at the base of the wall due to force (F1 + F2), M1 = (0.5 x 1.9 x 1.5 x 1.5 / 3 ) = 0.71 t-m
Moment at the base due to force F3, M2 = (0.33 x 1.5 x 1.5 / 2 ) = 0.37 t-m Total moment at the base of the wall = Working on working stress method, For M 20 grade concrete and Fe 415 steel
1.08 t-m
st = cbc =
230 7
N/mm N/mm
Modular ratio m = 13.33 (m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.90 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((1.08 x 10^7)/(0.91 x 1000))^0.5 = 109 mm Taking overall depth = 230 mm Taking 50 mm cover and 10 Effective depth = 175 mm Ast = M / (st x j x d)
(O.K) bar,
2 = 297 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 10 bars @ 200 mm c/c 2 mm2/m Ast provided = 393 mm /m > 297
(O.K)
Condition 2 : Water upto full depth inside and no earth outside. Moment = (w x H) / 6 = = (1 x 15.625 / 6 ) 2.60 t-m st = cbc = Modular ratio m = 13.33 (m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.9 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((2.6 x 10^7)/(0.91 x 1000))^0.5 = 169 mm Taking overall depth = 230 mm Taking 50 mm cover and 12 Effective depth = 174 mm Ast = M / (st x j x d) 230 7 N/mm N/mm
Working on working stress method, For M 20 grade concrete and Fe 415 steel
(O.K) bar,
2 = 720 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 12 bars @ 150 mm c/c 2 mm2/m Ast provided = 754 mm /m > 720
(O.K)
For horizontal reinforcement : Again horizontal tension from 5 m long side walls due to water, = 1.553 ton Now, Steel required for tensile force, = (1.553 x 10000)/150 2 = 104 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 10 bars @ 200 mm c/c 2 2 mm /m Ast provided = 393 mm /m > 276 Check at 1 m above the base of the wall due to water pressure : Moment = (w x H) / 6 = = (1 x 3.375 / 6 ) 0.56 t-m st = cbc = Modular ratio m = 13.33 (m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.9 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((0.56 x 10^7)/(0.91 x 1000))^0.5 = 79 mm Taking overall depth = 180 mm Taking 50 mm cover and 12 Effective depth available = 124 mm For Intermediate Walls : 10 m long 0.18 230 7
(O.K)
Working on working stress method, For M 20 grade concrete and Fe 415 steel
N/mm N/mm
bar,
Length / Height = 4 2.5
0.5 2.5 0.23 Unit weight of water = w 1 t/m Lateral pressure at base of wall due to water = 2.5 x 1 = 2.5 t/m Considering this is a free cantilever, Total moment at the base = 0.5 x 2.5 x 2.5 x 2.5/3 = 2.60 t-m Working on working stress method, st = For M 20 grade concrete and Fe 415 steel cbc = Modular ratio m = 13.33 (m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.9 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((2.6 x 10^7)/(0.91 x 1000))^0.5 = 169 mm
230 7
N/mm N/mm
Taking overall depth = Taking Effective depth = Ast = M / (st x j x d)
230 mm 50 mm cover and 174 mm
2
12
(O.K) bar,
= 720 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 12 bars @ 150 mm c/c 2 2 mm /m Ast provided = 754 mm /m > 720 For horizontal reinforcement : Again horizontal tension from 5 m long side walls due to water, = 1.553 ton Now, Steel required for tensile force, = (1.553 x 10000)/150 2 = 104 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 10 bars @ 200 mm c/c 2 2 mm /m Ast provided = 393 mm /m > 276
(O.K)
(O.K)
Design of Base : 0.6
0.6
10 5 4.8 4.8
Plan
1 G.L
1.5
0.3 0.6 Elevation Total Load on Base Slab : Unit Weight Of concrete = 2.5 t/m Unit Weight Of soil = 1.8 t/m Weight of walls = 2 x (15.52 + 4 x 10) x 2.5 x 0.205 x 2.5 (Taking avg thickness of the wall) Base slab = 16.72 x 11.2 x 0.3 x 2.5 (Say 0.3 m thk) Soil = (2 x 16.72) + (2 x 11.20) x 1.5 x 0.6 x 1.8 Total load =
142.27 t 140.45 t 90.46 t 373.18 t
Check For Upift : D.L = 373.18 t Uplift force due to water upto G.L = F.O.S against uplift = 15.52 x 10.46 x (1.5 + 0.3) = 292.21 t 373.18 / 292.21 = 1.3 (O.K)
Base Pressure : Maximum load on base = base pressure = P/A = Calculation Of Reinforcement :
373.18 t 390.53/(16.72 x 11.2) = 1.99 t/m
Uniform Base pressure = 1.99 t/m Net pressure = 1.99 - 0.3 x 2.5 = 1.24 t/m Pressure per meter width of base = 1.24 t/m As Ly/Lx = 10/5 = 2, this is an one way slab Therefore, ultimate moment = Mu = (1.5 x 1.24 x 5^2 / 8) = 5.83 t-m d = effective depth (using 12f bar & 50 mm cover) Mu / bd = 1.0 From Table 2 of SP-16 we get, pt = Ast = (0.295 x 1000 x 244 / 100) =
= =
0.3 - 0.05 - 0.006 0.244 m =
m 244
mm
0.295 % 720 mm/m 140 mm c/c at both support and mid span along shorter span 807 mm 0.12 % of b x D at both support and mid-span along longer span
Provide 12 bars @ Area of reinforcement provided = Provide minimum steel i.e. Ast = Ast = (0.12 x 1000 x 300 / 100) = 360 mm Provide 10 bars @ Area of reinforcement provided =
150 mm c/c at both support and mid span along longer span 523 mm
Design Calculations for proposed 300 m3 R.C.C Storage Tank for Construction Water at RSP Rourkella
Grade of concrete = Grade of steel = M 20 Fe 415
10
5 For 5 m Long Side walls :
2.5
2.5 Unit weight of water = w Lateral pressure due to water = 1 t/m 2.5 t/m 2a Free
b Fixed
2.5
5 2.5
From the figure we get,
a=
2.5 m
P=
2.5
b= Therefore , a/b = Moody's tables give values for a/b = 1 Horizontal moment =
2.5 m 1.0 m
Therefore, P x b = = Again, Pxb= =
Mx x P x b
My x P x b Vertical moment = Vertical Reinforcement : Using Figure 4 of Moody's Tables, Maximum vertical moment = 0.0845 x 15.625 (At the bottom of the wall) = 1.320 t-m/m Now, st = 230 cbc = 7 Modular ratio m = 13.33 (m x cbc)/st = k/(1-k) => k= 0.2886 Now, j= (1-k/3) = 0.90 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((1.3203125 x 10^7)/(0.91 x 1000))^0.5 = 121 mm Taking overall depth = 230 mm Taking 50 mm cover and 10 bar, Effective depth = 175 mm Ast = M / (st x j x d) = providing Ast provided
2 364 mm /m 10 bars @ 150 mm c/c 2 524 mm^2/m > 364 mm /m
N/mm N/mm
(O.K)
Horizontal Reinforcement : Maximum horizontal moment = =
0.0644 x 15.625 (At the top of the wall) 1.01 t-m/m Now, st = cbc =
230 7.0
N/mm N/mm
Modular ratio m = 13.33 (m x cbc)/st = k/(1-k) => k= 0.2886 Now, j= (1-k/3) = 0.90 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((1.01 x 10^7)/(0.91 x 1000))^0.5 = 106 mm Taking overall depth = 180 mm Taking 50 mm cover and 10 Effective depth = 115 mm Ast = M / (st x j x d) = providing Ast provided
bar,
2 424 mm /m 10 bars @ 150 mm c/c 2 524 mm^2/m > 424 mm /m
(O.K)
Maximum horizontal shear = = For Side Walls : 10 m long
0.2485 x 6.25 ton 1.553 ton
Condition 1 : No water inside. Earth outside with G.W at its highest level and surcharge outside. 0.18 WALL G.L F1 = Pressure due to water F1 = Pressure due to submerged earth F3 = Pressure due to surchage
2.0 F3 F1 F2 0.5
2.5
2.0
0.53 0.33 0.23
Fig: 1 - Pressure Diagram Dry unit weight of soil = Unit weight of water = Considering surcharge = Angle of repose () = Therefore Co-efficient of earth pressure (ca) = fck = 1.8 1 1 30 0.33 25 N/mm t/m t/m t/m
fy = 415 N/mm Moment at the base of the wall due to force (F1 + F2), M1 = (0.5 x 2.53 x 2 x 2 / 3 ) = 1.69 t-m Moment at the base due to force F3, M2 = (0.33 x 2 x 2 / 2 ) = 0.66 t-m Total moment at the base of the wall = 2.35 t-m Working on working stress method, For M 20 grade concrete and Fe 415 steel Modular ratio m = (m x cbc)/st = k/(k-1) => k= j= (1-k/3) = Q = 0.5 x j x k x cbc = 0.91 13.33 0.2886 0.90
st = cbc =
230 7
N/mm N/mm
Now, Again,
Now,
dreq = =
Taking overall depth = Taking Effective depth = Ast = M / (st x j x d)
50
((2.35 x 10^7)/(0.91 x 1000))^0.5 161 mm 230 mm mm cover and 12 174 mm
(O.K) bar,
2 = 648 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 12 bars @ 150 mm c/c 2 mm2/m Ast provided = 754 mm /m > 648
(O.K)
Condition 2 : Water upto full depth inside and no earth outside. Moment = (w x H) / 6 = = (1 x 15.625 / 6 ) 2.60 t-m st = cbc = Modular ratio m = 13.33 (m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.9 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((2.6 x 10^7)/(0.91 x 1000))^0.5 = 169 mm Taking overall depth = 230 mm Taking 50 mm cover and 12 Effective depth = 174 mm Ast = M / (st x j x d) 230 7 N/mm N/mm
Working on working stress method, For M 20 grade concrete and Fe 415 steel
(O.K) bar,
2 = 720 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 12 bars @ 150 mm c/c 2 mm2/m Ast provided = 754 mm /m > 720
(O.K)
For horizontal reinforcement : Again horizontal tension from 5 m long side walls due to water, = 1.553 ton Now, Steel required for tensile force, = (1.553 x 10000)/150 2 = 104 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 10 bars @ 200 mm c/c 2 mm2/m Ast provided = 393 mm /m > 276 Check at 1 m above the base of the wall due to water pressure :
(O.K)
Moment =
(w x H) / 6
= =
(1 x 3.375 / 6 ) 0.56 t-m st = cbc = 230 7 N/mm N/mm
Working on working stress method, For M 20 grade concrete and Fe 415 steel
Modular ratio m = 13.33 (m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.9 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((0.56 x 10^7)/(0.91 x 1000))^0.5 = 79 mm Taking overall depth = 180 mm Taking 50 mm cover and 12 Effective depth = 124 mm Ast = M / (st x j x d)
(O.K) bar,
2 = 218 mm /m Minimum steel required = 0.12 x 180 x 1000 / 100 2 = 216 mm /m providing 12 bars @ 300 mm c/c 2 mm2/m Ast provided = 377 mm /m > 218
(O.K)
For Intermediate Walls : 10 m long 0.18
2.5
0.5 2.5 0.23 Unit weight of water = w 1 t/m Lateral pressure at base of wall due to water = 2.5 x 1 = 2.5 t/m Considering this is a free cantilever, Total moment at the base = 0.5 x 2.5 x 2.5 x 2.5/3 = 2.60 t-m Working on working stress method, st = For M 20 grade concrete and Fe 415 steel cbc = Modular ratio m = 13.33
230 7
N/mm N/mm
(m x cbc)/st = k/(k-1) => k= 0.2886 Now, j= (1-k/3) = 0.9 Q = 0.5 x j x k x cbc Again, = 0.91 dreq = Now, ((2.6 x 10^7)/(0.91 x 1000))^0.5 = 169 mm Taking overall depth = 230 mm Taking 50 mm cover and 12 Effective depth = 174 mm Ast = M / (st x j x d)
(O.K) bar,
2 = 720 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 12 bars @ 150 mm c/c 2 mm2/m Ast provided = 754 mm /m > 720
(O.K)
For horizontal reinforcement : Again horizontal tension from 5 m long side walls due to water, = 1.553 ton Now, Steel required for tensile force, = (1.553 x 10000)/150 2 = 104 mm /m Minimum steel required = 0.12 x 230 x 1000 / 100 2 = 276 mm /m providing 10 bars @ 200 mm c/c 2 mm2/m Ast provided = 393 mm /m > 276
(O.K)
Design of Base : 0.6
0.6
10 5 4.8 4.8
Plan
1 G.L
1.5
0.3 0.6 Elevation Total Load on Base Slab : Unit Weight Of concrete = 2.5 t/m Unit Weight Of soil = 1.8 t/m Weight of walls = 2 x (15.52 + 4 x 10) x 2.5 x 0.23 x 2.5 159.62 t Base slab = 16.72 x 11.2 x 0.3 x 2.5 140.45 t (Say 0.3 m thk) Soil = (2 x 16.72) + (2 x 11.20) x 1.5 x 0.6 x 1.8 90.46 t Total load = 390.53 t Check For Upift : D.L = 390.53 t Uplift force due to water upto G.L = 15.52 x 10.46 x (1.5 + 0.3) = 292.21 t F.O.S against uplift = 390.53 / 292.21 = 1.3 (O.K) Base Pressure : Maximum load on base = base pressure = P/A = Calculation Of Reinforcement : Uniform Base pressure = 2.09 t/m Net pressure = 2.09 - 0.3 x 2.5 = 1.34 t/m Pressure per meter width of base = 1.34 t/m As Ly/Lx = 10/5 = 2, this is an one way slab Therefore, ultimate moment = Mu = (1.5 x 1.34 x 5^2 / 8) = 6.26 t-m d = effective depth (using 12f bar & 50 mm cover)
390.53 t 390.53/(16.72 x 11.2) = 2.09 t/m
0.3 - 0.05 - 0.006
= Mu / bd = 1.05 From Table 2 of SP-16 we get, pt = Ast = (0.311 x 1000 x 244 / 100) = 0.311 % 759 mm/m
0.244
m =
Provide 12 bars @ Area of reinforcement provided = Provide minimum steel i.e. Ast = Ast = (0.12 x 1000 x 300 / 100) = 360 mm Provide 10 bars @ Area of reinforcement provided =
140 mm c/c at both support and mid span along shorter span 807 mm 0.12 % of b x D at both support and mid-span along longer span
150 mm c/c at both support and mid span along longer span 523 mm
Now, tensile force from 10 m long walls = 0.5 x 2.5 x 2.5 = 3.125 t/m Now, Steel required for tensile force, = (3.125 x 10000)/(2 x 150) 2 = 105 mm /m
for Construction
t/m
2.5 x 2.5 x 2.5 15.625 t 2.5 x 2.5 6.25 t/m
sure due to water sure due to submerged earth sure due to surchage
244
mm
an along shorter span
d-span along longer span
an along longer span
