design of water tank-311207

7/30/2019 Design of Water Tank-311207

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1

For proper assessment of load acting on Overhead Water Tank we have followed the appropriate

clauses of DBR , specification no PE-DC-250-600-C001.We have taken the basic wind speed

& factors K1,K2,& K3 as per DBR and IS:875(part -III) for wind load assessment (as shown in detail in

load calculation ). For seismic load calculation we have considered data for zone-II and importance factor

(I=1.5) according to IS:1893 , part-IV.The whole structure has been modeled and analyzed with the help of STAAD Pro. Each component of the

structure has been designed to withstand forces and moments as obtained from the Staad output.The tank

slabs and beams have been designed as uncracked sections in accodance with IS:3370 and other parts of the

structure have been designed as per IS 456 :2000.

Total factored load = 283.76 T ==>

Unfactored load = 189.17 T

Total load = 189.17 T for other load cases )

Area of raft = 1.6x3.5 = 5.60 m2

Base Pressure = 33.8 T/m2

< 30x1.25 = 37.5 T/ m2

Hence OK.

Design base pressure = 33.78 T/m2

Column A

GROUND LEVEL

DIA = 7650 mm

2000

3500

A

33.78 T/mm2

b = 1000 mm ; d = 600 - 50 - 8 = 542 mm

Moment at section A-A = 4.22 T-m

Mu/bd2

= 0.22 N/mm2

Minimum reinforcement required @ 0.2 % Fe415

Ast= 1168.6 mm M25

Provide 16 mm diameter bar @ 150 mm c/c (1340 mm2

) as main reinforcement.

Calculation of top reinforcement :

Soil pressure at top of footing = 3.4x1.8 = 6.12 T/m2

Bending moment at section A-A = 0.765 T-m

Mu/bd2

= 0.04 N/mm2

Minimum reinforcement required @ 0.2 %

DESIGN PHILOSOPHY

DESIGN OF FOUNDATION

FOOTING SLAB

( For Node No. 89 & Load Case 104 in STAAD Output )

4000

2500

600

( As this reaction is maximum among all load combination so ,there is no need of checking bearing capacity


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2

Astmin = 1084 mm

Provide 16 mm diameter bar @ 150 mm c/c at top as main reinforcement.

Distribution reinforcement: = 0.12 % in two faces

Astmin / face / m = 325 mm2

Provide 12 mm diameter @ 200 mm c/c top & bottom (565 mm2) .

Additional Overturning Moment Check For The Structure :

( For Cross check Through Hand Calculation )

Total overturning moment due to wind about base bottom = 0.77x6x7.25x32.5 + 19.597x30 +

14.37x25 + 9.17x20 + 8.39x15 +12.997x10

= 2475.0 T-m

Back fill Soil load = 237.81 T

Empty Water tank Wt. = 935 T ( From STAAD Output )

Total load = 1172.808 T

Restoring moment = 5620.7 T-m

Overturning Check With Staad Output :

( For Wnid load )

Horizontal Vertical Horizonta Moment

Node L/C Fx Mton Fy Mton Fz Mton Mx MTon- My MTon- Mz MTon-m

89 145 0.285 57.405 30.604 0 -3.7 0

90 145 13.29 173.91 44.048 0 -3.9 0

91 145 -13.576 -60.07 44.532 0 -3.9 0

139 145 0 102.51 0 0 0 0

140 145 0 141.9 0 0 0 0

141 145 0 170.59 0 0 0 0

142 145 0 183.03 0 0 0 0143 145 0 136.89 0 0 0 0

144 145 0 84.049 0 0 0 0

145 145 0 26.368 0 0 0 0

146 145 0 -24.93 0 0 0 0

147 145 0 -70.689 0 0 0 0

148 145 0 -59.457 0 0 0 0

149 145 0 -30.522 0 0 0 0

150 145 0 10.51 0 0 0 0

Per node dead load = 62.33 T-m

Overturnig moment about A-A Section using Staad output = 1274.3507 T-m

( Algebric sum of all moment )


Staad Output for Seismic Load

A Horizontal Vertical Horizonta Moment Moment Moment

Node L/C Fx Mton Fy Mton Fz Mton Mx kNm My kNm Mz kNm

89 149 49.502 -67.986 -0.011 0 -1.5 0

Overturning moment

Restoring moment= 2.3 > 1.5

Restoring moment

Overturning moment= 4.4 > 1.5 Hence OK.

Hence OK.


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3

90 149 31.355 119.81 9.953 0 -0.7 0

91 149 31.479 119.42 -9.942 0 -2.3 0

139 149 0 -54.762 0 0 0 0

140 149 0 -21.268 0 0 0 0

141 149 0 25.013 0 0 0 0

142 149 0 75.092 0 0 0 0

143 149 0 148.75 0 0 0 0

144 149 0 163.1 0 0 0 0145 149 0 163 0 0 0 0

146 149 0 148.49 0 0 0 0

147 149 0 74.672 0 0 0 0

148 149 0 24.632 0 0 0 0

149 149 0 -21.554 0 0 0 0

150 149 0 -54.915 0 0 0 0

Per node dead load = 62.33 T-m

Overturnig moment about A-A Section using Staad output = 563.74333 T-m

( Algebric sum of all moment )


A A

PLAN SHOWING SPRING SUPPORT

STAAD OUTPUT FOR FOUNDATION CIRCULAR BEAM

Beam L/C Node Fx Mton Fy Mton Fz Mton Mx kNm Mz kNm

DESIGN OF FOUNDATION RING BEAM (SECTION 2000mmx2500mm)

DETAIL OF FORCES & MOMENTS AS OBTAINED FROM STAAD OUTPUT

Hence OK.Restoring moment

= 10.0 > 1.5

Overturning moment


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4

Max Fx 174 101 139 0 142.805 0 399.7 -152.5

Min Fx 174 101 139 0 142.805 0 399.7 -152.5

Max Fy 184 105 147 0 496.683 0 4123.9 -42.4

Min Fy 178 125 142 0 -498.214 0 -4131.2 -46.9

Max Fz 174 101 139 0 142.805 0 399.7 -152.5

Min Fz 174 101 139 0 142.805 0 399.7 -152.5

Max Mx 184 105 147 0 496.683 0 4123.9 -42.4

Min Mx 178 125 90 0 -464.864 0 -4131.2 -7165.1Max My 174 101 139 0 142.805 0 399.7 -152.5

Min My 174 101 139 0 142.805 0 399.7 -152.5

Max Mz 181 124 145 0 16.872 0 3.3 6512.4

Min Mz 188 104 89 0 -386.57 0 -3280.8 -7542.7

Bending moment , Muz = 7542.7 kN-m

Torsion Tu = 4131.2 kN-m Ref. STAAD Output

Shear Vu = 4982.1 kN Mem - 188 & 178 And

L/C - 104 & 125

b = 2000 mm D = 2500 mm

b1 = 1875 mm d1 = 2375 mm

Cover = 50 mm Bar dia = 25 mm

25 mmd = 2412.5 mm

(Cl. 41.4 of IS : 456 - 2000 )

Mu = 7542.7 kN-m

Mt = Tu (1+D/b)/1.7

= 5467.8 kN-m

Me1 = 13010 kN-m

Me1/bd= 1.12 N/mm ; ptreqd = 0.322 %

( From Table -3 of SP-16)

Astreqd = 15536.5 mm

Shear :

Equivalent shear , Ve = Vu + 1.6 Tu/b = 8287.1 kN

Tve = 1.72 N/mm < Tcmax = 3.1 N/mm Hence OK.

( From Table - 20 of IS: 456 -2000)

Percentage of steel = 0.33

Design shear strength Tc = 0.4 N/mm (From Table -19 of IS:456 - 2000)

Shear capacity = 1930 kN

For ten legged 12 mm diameter stirrup, spacing = 154.97 mmProvide spacing sv = 150 mm

Minimum Transverse Reinforcement :

(Tve -Tc)b*Sv / (0.87 fy) = 1094.8 mm

Provided 10 Legged - 12 mm diameter stirrup @ 150 mm c/c (1130 mm2) Hence OK.

For side face reinforcement :

Provided 6 nos. 20 diameter bars at each face @ 300 c/c .

20 - 25 mm dia

2000

DESIGN:-

Spacer bar dia. =

Equivalent bending moment Me1 = Mu + Mt

Provide (20+12 ) nos. 25 mm diameter(15707 mm2) bars as reinforcement at bottom & top .


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5

6 - 20 dia

12 - 25 mm dia

6 - 20 dia

12 dia @ 600 c/c

10 Legged 12 dia

4-12 dia at each row stirrup @ 150 c/c

16 dia @ 150 c/c

12 dia @ 200 c/c

12 dia @ 200 c/c

12 mm dia @ 200 c/c 12 dia @ 200 c/c

12-25 dia 16 mm dia @ 150 mm c/c

20-25 dia

Loading :

DL

Weight of 150 thk landing slab = 0.150x2.5 = 0.375 T/m2

LL Effective depth = 125 mm

Live load on landing = 0.25 T/m2

Total DL+LL = 0.625 T/m2

Load for ladder = 6x0.03 + 2x0.0304 = 0.241 T/m

Reaction on landing due to ladder = 0.241x5.112/2 = 0.616 T

Case 1 :

Considering simply supported max. span of landing = 1.965 m

For conservative design reaction due to ladder is considered at mid point of simply supported span

Max. moment due to simply supported = 0.60 T-m

Mu/bd2 = 0.58

Ast reqd = 2.3 cm2(Table 2 of sp - 16 )

Provide 10 mm dia @ 300 c/c (2.62 cm2 ) top & bottom along simply supported span direction

Case 2 :

Considering cantilever max. span of landing = 1.181 m

For conservative design reaction due to ladder is considered at end of cantilever span

Max. moment due to cantilever = 1.163 T-m

Mu/bd2 = 1.117

Ast reqd = 4.5 cm2

(Table 2 of sp - 16 )

Provide 10 mm dia @ 150 c/c (5.24 cm2 ) top along cantilever span direction

Provide 10 mm dia @ 300 c/c (2.62 cm2 ) bottom along cantilever span direction

1000

Group - 1 & 2 20 - 32 mm Dia 600

600 x 1000

Group - 1

600

3500

DESIGN OF COLUMNS

2

500

DESIGN OF LANDING SLAB


6/32

6

600 x1150

Group - 2

5 Sets - 12 dia @

300 c/c

600x 1350

Group -3

Group - 3

600

32 - 36 mm

Dia bars 1400Group -1 STAAD OUTPUT FOR COLUMNS

Beam L/C Node Fx Mton Fy Mton Fz Mton Mx MTon- My MTon-m

Max Fx 36 116 71 285.746 -0.105 -42.873 -0.1 53.2

Min Fx 222 122 48 1.649 -0.115 0.194 0 -0.2

Max Fy 36 125 71 142.69 21.457 -3.435 -1.6 6.8

Min Fy 36 105 71 143.289 -21.47 -1.925 1.4 3.8

Max Fz 29 125 72 245.292 10.412 49.844 -1.7 -63.2

Min Fz 36 104 71 249.287 0.257 -51.893 -0.1 64.5

Max Mx 29 105 72 40.628 -10.328 -44.511 1.5 52.5

Min Mx 29 125 72 245.292 10.412 49.844 -1.7 -63.2

Max My 29 125 7 233.288 10.412 49.844 -1.7 178.6

Min My 36 104 2 237.284 0.257 -51.893 -0.1 -187.2

Max Mz 36 105 2 131.286 -21.47 -1.925 1.4 -5.6

Min Mz 36 125 2 130.686 21.457 -3.435 -1.6 -9.8

Group -2 Beam L/C Node Fx Mton Fy Mton Fz Mton Mx MTon- My MTon-m

Max Fx 33 104 80 661.862 -0.454 -77.176 -1 201.3

Min Fx 33 124 77 -219.73 -0.451 73.606 -1 175.7

Max Fy 33 125 80 228.096 28.114 -1.782 -8.2 4.5

Min Fy 33 105 80 228.103 -29.02 -1.787 6.2 4.5

Max Fz 33 124 80 -205.66 -0.451 73.606 -1 -192.4

Min Fz 33 104 80 661.862 -0.454 -77.176 -1 201.3

Max Mx 33 105 80 228.103 -29.02 -1.787 6.2 4.5

Min Mx 33 125 80 228.096 28.114 -1.782 -8.2 4.5

Max My 33 104 80 661.862 -0.454 -77.176 -1 201.3

Min My 33 124 80 -205.66 -0.451 73.606 -1 -192.4

Max Mz 33 125 80 228.096 28.114 -1.782 -8.2 4.5

Min Mz 33 105 80 228.103 -29.02 -1.787 6.2 4.5

Group-3

Beam L/C Node Fx Mton Fy Mton Fz Mton Mx MTon- My MTon-mMax Fx 30 104 89 1057.43 -0.019 -73.356 -0.2 335.5

Min Fx 30 124 86 -458.74 -0.019 74.3 -0.2 -162.8

Max Fy 30 125 89 303.247 45.904 0.475 -5.6 -5.6

Min Fy 30 105 89 303.243 -45.942 0.469 5.1 -5.6

Max Fz 31 124 86 -402 -0.256 87.26 -0.6 -335.4

Min Fz 31 104 86 971.702 -0.257 -91.02 -0.6 344.7

Max Mx 31 105 86 284.853 -28.941 -1.875 7.7 4.6

Min Mx 32 125 83 256.421 29.572 -1.81 -9.4 4.5

Max My 31 104 86 971.702 -0.257 -91.02 -0.6 344.7

Min My 30 124 89 -450.94 -0.019 74.3 -0.2 -346.7

-1.1

-74.7

-35.6

-0.7

75.8

-74.7

-1.1

72.5

-28.4

14.9

0.5

-14.6

-75.7

Mz MTon-m

-1.1

1.2

14.9

Mz MTon-m

-0.1

0.1

28.3

0

-92.6

128.8

82.6

-128.8

0.1

0.1

0

72.5

0.1

-74.7

Mz MTon-m0

-1.1

-1.1

72.5

8 sets - 12 mm dia @ 300 c/c


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7

Max Mz 30 125 89 303.247 45.904 0.475 -5.6 -5.6

Min Mz 30 105 89 303.243 -45.942 0.469 5.1 -5.6

DESIGN OF COLUMN ( Group - 1 )

Member no 29,39,40,22,37,38,36,41,42

b = 600 mm

D = 1000 mm

Factored axial load Pu = 233.29 T Member - 29

Factored moment Muy = 178.60 T-m Load Case - 125 ZFactored moment Muz = 35.600 T-m M 25

Clear Cover = 40 mm Fe 415

Dia of bar = 25 mm

Assume percentage of steel 2.5

600

P/fck = 2.5/25 = 0.1

Y Y

Pu

fckbD

Uniaxial moment capacity about Z-Z axis

1000

d/D = 0.09 Z

Refer chart no - 44Mu

fckbD2

Muz1 = 139.5 T-m

Uniaxial moment capacity about Y-Y axis

d/D = 0.05

Refer chart no - 43

Mu

fckbD2

Muy1 = 247.5 T-m

Puz

Ag

Puz = 1140 T

PuPuz

n nMuy Muz

Muy1 Muz1

1 1

178.60 35.600

247.5 139.5

Provide 20 - 32 Dia bars. 161 Cm2

% of reinforcement = 2.68

Design of ties:

According to cl.26.5.3.2 c of IS 456 - 2000

The diameter of the ties should not be less than 1/4 th the diameter of the largest longitudinal bar

subject to a minimum = 8

The c/c spacing of the ties should be least of the following :

i) Least lateral dimension of column = 600 mm

ii) 16 x dia. Of longitudinal bar = 512 mmiii) 48 x dia.of tie = 384 mm


Member no 26,27,28,19,20,21,33,34,35

b = 600 mm Y

D = 1150 mm 600


Factored moment Muy = 201.200 T-m Load Case - 104

Factored moment Muz = 1.100 T-m M 25


= 0.156

= 0.155

= 0.165

= 19 (From Table 63 )

= 0.20 = => n =1 1

= = 0.98 < 1 Hence O.K

-128.8

128.8


8/32

8

Dia of bar = 25 mm

Assume percentage of steel 2.2 Z Z

P/fck = 2.2/25 = 0.088

Pu

fckbD Y

Uniaxial moment capacity about Z-Z axisd/D = 0.09

Refer chart no - 44

Mu

fckbD

Muz1 = 124.2 T-m

Uniaxial moment capacity about Y-Y axis

d/D = 0.046

Refer chart no - 43

Mu

fckbD2

Muy1 = 258 T-m

From Chart - 63

PuzAg

Puz = 1207.5 T

Pu

Puz

n = 1.4

n nMuy Muz

Muy1 Muz1

1.4 1.4

201.200 1.100

258 124.2

Provide 20 - 32 Dia bars. 161 Cm2


Design of ties:

According to cl.26.5.3.2 c of IS 456 - 2000The diameter of the ties should not be less than 1/4 th the diameter of the largest longitudinal bar

subject to a minimum = 8

The c/c spacing of the ties should be least of the following :

i) Least lateral dimension of column = 600 mm

ii) 16 x dia. Of longitudinal bar = 512 mm

iii) 48 x dia.of tie = 384 mm


Member no 23,24,25,16,17,18,30,31,32

b = 600 mm

D = 1350 mm


Factored moment Muy = 344.700 T-m Load Case - 104

Factored moment Muz = 0.100 T-m M 25


Dia of bar = 25 mmAssume percentage of steel 4 % Z

600

P/fck = 4/25 = 0.16

Pu

fckbD

Y Y

Uniaxial moment capacity about Z-Z axis

1350

= 0.38

= 0.12

= 0.13

= 17.5

= 0.55

= = 0.71 For d'/d = 0.1,

required pt = % => Astreqd =

required pc = % => Ascreqd =

mm diameter bars as tension reinforcement at support.

Therefore, provided pt = % for which, Tc = N/mm2

[Ref. Table 61 of SP16]


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14

312620 N = 312.62 kN

267.38 kN

3.2929

roviding 4L 8 220 mm[Ref. Table 62 of SP16]

232 kN-m

0.7037 N/mm

2

0.201 816 mm2

=> 1.6617949 nos. 25 mm bars

0.05 %

16 1.0092

Provide 2 no. 16

Provide 9 no. 32

Provide 5 no. 32

Provide 4L 8

9-32 mm Diameter Bars


500 500

804.2 kN

2412.09 kN-m

500 mm

900 mm

40 mm

32 mm

Separator bar dia. ' = 32 mmTherefore, 812 mm

7.317 N/mm2

2.35 9541 mm2

=> 11.86 nos. 32 mm bars

1.22 4953 mm2

=> 6.156321 nos. 32 mm bars

Provide 12 no. 32

2.38 0.86

349160 N = 349.16 kN

455.04 kN

5.6039

Providin 4L 12 250 mm[Ref. Table 62 of SP16]

291 kN-m

0.8827 N/mm2

[Refer "4th layer from bottom" in

Details of Forces & Moments Acting

on Tie Beams]

Shear capacity of the section = Tcbd =

Vus =Vu Tcbd =

Vus /d = kN/cm

mm diameter bars, required spacing =

At span:- Factored Bending Moment Mu =

Therefore, Mu/bd

2

=From Table 3 of SP16 ==>


For side face reinforcement required percentage of steel per face =

Therefore number mm diameter bars required per face =

mm diameter bars at each face for side face reinforcement.

mm diameter bars as top & bottom reinforcement at support.

mm diameter bars as top & bottom reinforcement at span.

mm diameter bars @ 200 c/c as stirrup.

4L-8 mm Dia.Bar @ 200 c/c


900

At Supports At Span

DESIGN OF BEAMS IN 2ND, 4TH AND 5TH LAYER FROM BOTTOM (TB2 ,TB4 & TB5)

At face of column:

Maximum factored shear force Vu =

Maximum factored bending moment Mu =

Assume section of beam ==> 500mm x 900mm

Here, width of the section, b=

overall depth of the section, D=

Assuming, clear cover, c =

Diameter of reinforcement, =

effective depth, d = D - c - - ('/2) =

Mu/bd2

=From Table 51 of SP16 ==> For d'/d = 0.1,




Therefore, provided pt = % for which, Tc = N/mm2 [Ref. Table 61 of SP16]


Vus =Vu Tcbd =

Vus /d = kN/cm



Therefore, Mu/bd2



15/32

15

0.261 1060 mm2

=> 1.317049 nos. 32 mm bars

0.05 %

16 mm diameter bars required per face = 1.00923295

Provide 2 no. 16

Provide 12 no. 32

Provide 6 no. 32

Provide 4L 12

500

826.2 kN

2481.36 kn-m

500 mm

900 mm

30 mm

32 mm

Separator bar dia. = 32 mm

Therefore, 822 mm

7.345 N/mm2

2.38 9782 mm2

=> 11.79 nos. 32 mm bars

1.46 nos. 16 mm bars

1.254 5154 mm2

=> 6.4058203 nos. 32 mm bars

Provide 12 no. 32

2 no. 16

2.45 0.85

349350 N = 349.35 kN

476.85 kN

5.8011

roviding 4L 12 250 mm[Ref. Table 62 of SP16]

296 kN-m0.8761 N/mm

2

0.261 1073 mm2

=> 1.3332688 nos. 32 mm bars

0.05 %


Provide 2 no. 16

Therefore number of





Therefore number of



12 32mm Diameter Bars 6 32mm Diameter Bars

At Span

DESIGN OF BEAMS IN 3RD LAYER FROM BOTTOM (TB3)

2 16mm Diameter Bars900

4L 12 mm bars @ 250c/c

At face of column:

Maximum factored shear force Vu =

Maximum factored bending moment Mu =

500

At Supports






effective depth, d = D - c - -('/2) =

Mu/bd2

=From Table 51 of SP16 ==> For d'/d = 0.1,



mm diameter bars and


Therefore, provided pt = % for which, Tc = N/mm2

[Ref. Table 61 of SP16]


Vus =Vu Tcbd =

Vus /d = kN/cm



Therefore, Mu/bd2





[Refer "3rd layer from bottom" in

Details of Forces & Moments Acting

on Tie Beams]


16/32

16

Provide 12 no. 32

2 no. 16

Provide 6 no. 32

Provide 4L 12

626.19 kN1033.11 kN-m

500 mm

1000 mm

40 mm

32 mm

Separator bar dia ' = 32 mmTherefore, 912 mm

8.5 N/mm

10.980392

150 N/mm2

0.3835616

0.8721461

1.4217176

688.74 kN-m

591.3 kN-m

4967.841 mm2

97.5 kN-m

819.11205

5786.953 mm2

1052.38802

639 kN-m

425.67333 kN-m

3567.8049 N/mm2

216mm DiameterBars

2 16mm Diameter Bars900


mm diameter bars and




6 32mm Diameter Bars1232mm Diameter Bars

500

At Supports At Span

500

DESIGN OF BEAMS SUPPORTING THE TANK BOTTOM SLAB(B1)

At face of column:

Maximum factored shear force Vu = [Refer Details of Forces & MomentsActing on Beams Supporting the

Tank Bottom Slab]Maximum factored bending moment Mu =






effective depth, d = D - c - - ('/2) =

Allowable Stress in steel tension for HYSD bars, st =

Therefore k = m.cbc/(m.cbc + st) =

Therefore j = 1 k / 3 =

Designing by working stress method with the following parameters:

Allowable Stress for M25 concrete in bending compression, cbc

=

Modular Ratio, m = 280/(3cbc) =

[Ref. Table 2 of IS:3370II ]

[Ref. Table 21 of IS: 456 ]

Therefore R = (1/2).cbc.jk =

Design Moment M = Mu / 1.5 =

Moment of resistance of the section under balanced condition Mbal =

Balance tensile steel Ast1 = Mbalx106/(stx0.87d) =

Remaining Moment Mrem = M Mbal =

Ast2 = Mremx106/(stx0.87d) =

Hence total tensile steel Ast = Ast1 + Ast2 =

Area of compression steel Asc =mm

2


Design Moment M = Mu / 1.5 =

Therefore, required area of steel Ast = M / (st.j.d) =

xjdmkd

ckdx

xM

cbc

rem

)]15.1()(

[

10 6


17/32

17

Therefor 4.43 no. 32

Provide 6+2= 8 no. 32

& bottom reinforcement at support.

Provide 6 no. 32

1.05 N/mm2

< 1.9

6436.571 mm2

1.41 %

0.45 N/mm2

212260 N = 212.26 kN

[where V = Vu / 1.5]

Allowable Tensile Stress in shear reinforcement, sv = 150 N/mm2

12 mm diameter bars, spacing Sv = sv.Asv.d/Vs = 291.68 mmProvide 4L 12

203.729 kN

135.81933 kN

0.2407122

Allowable Stress in direct tension, cc = 1.3 N/mm2

0.05 %


Provide 2 no. 16

[where V = Vu / 1.5]

[Ref. Table 23 of IS:456]



mm diameter bars are required as top & bottom reinforcement at span.

Assuming 4L

mm diameter bars as top


Shear Check:-Now actual shear stress Tv = V / bjd =

N/mm2 as per Table 1 of IS:3370II

Area of tensile steel provided at support Astprovd =

Percentage of tensile steel provided at support p t =

for which, permissible shear stress Tc =

Excess shear force, Vs = V Tcbd =


Check for direct tension :-

> than tensile stress. Hence O.K.


Therefore number of

Maximum axial tension acting on the beam,Tu =

Design axial tension acting on the beam,T = Tu /1.5 =

Therefore, tensile stress, ct = T / [Ac + (m 1)Astprovd ] =



18/32

18

6+2=

500

Beam L/C Node Fx kN Fy kN Fz kN Mx kNm My kNm Mz kNm

Max Fx 10 105 11 429.894 -165.307 61.131 -82.5 -28.3 -46.7

Min Fx 10 125 11 -316.56 82.08 -61.522 101.5 31.8 -16.1

Max Fy 11 140 12 275.072 217.439 -18.551 22.7 25.8 234.4

Min Fy 10 117 12 364.911 -210.218 50.129 -63.9 55.2 232.2

Max Fz 1 124 2 -211.71 0.552 80.897 -96.1 -81.5 -41.7Min Fz 15 124 16 -223.77 27.051 -82.006 96.9 39.8 -22.6

Max Mx 10 125 11 -316.56 82.08 -61.522 101.5 31.8 -16.1

Min Mx 11 125 12 -155 54.44 65.46 -105.1 -67.4 17.4

Max My 11 105 12 268.991 97.78 -63.931 85.5 68.8 97.4

Min My 15 124 2 -223.77 -6.209 -82.006 96.9 -83.8 -38.3

Max Mz 11 140 12 275.072 217.439 -18.551 22.7 25.8 234.4

Min Mz 10 125 12 -316.56 48.82 -61.522 101.5 -61 -115

sv = 150 N/mm2

M 25

cbc = 8.5 N/mm Fe 415

st = 190 N/mm2

m = 10.98

k = 0.329412

j = 0.89

R = 1.248 N/mm2Bending moment at support , 234.4 kN-m

Bending moment at span = 75.5 kN-m Load Case : 140 & 125

Torsional moment ,Tu = 105.1 kN-m

= 217.44 kN

Considering increase in material stress 33.33 %

Design Bending moment , M = 234.44/1.33 = 176.24 kN-m

Design torsion, T = 105.1/1.33 = 79.02 kN-m

Design shear , V = 217.44/1.33 = 163.49 kN

b = 1000 mm D = 600 mm


b1 = 925 mm d1 = 575 mm

d = 562.5 mm

Calculation of longitudinal reinforcement :

(Cl. B-6.2 of IS : 456 - 2000 )

M = 176.24 kN-mMt = T (1+D/b)/1.7

= 74.37 kN-m

Me1 = 250.61 kN-m

Moment of resistance of a balanced section = 394.875 kN-m > 248.32 kN-m

Astreqd = 2634.18 mm

Provide 4 nos. 28 mm diameter bars + 3 nos. 25 mm diameter bars (3935mm2) at top .

Provide 7 nos. 20 mm diameter bars (2198mm2) at bottom.

Provide 1 no. 16 diameter bars as side reinforcement on each side.

Shear :

1000

6 32mm Diameter Bars



DESIGN:-

At Supports

(From STAAD OUTPUT )


Me1/(st*j*d) =

500

DESIGN OF RING BEAM AT BOTTOM OF TANK (SECTION 600mmx1000mm)


Shear force , Vu

Equivalent bending moment Me1 = M+ Mt

At Span


19/32

19

Equivalent shear , Ve = V + 1.6 T/b = 289.9 kN

Tve = 0.52 N/mm < Tcmax = 1.9 N/mm2

Hence OK.

( From Table - 24 of IS: 456 -2000)

Percentage of steel = 0.79


Transverse Reinforcement :

Two legged 10 mm diameter stirrup Asv = 157.08 mm2

Asv = ((T*Sv)/(b1*d1*sv))+((V*Sv)/(2.5*d1*sv))

Sv = 89.83 mm

Asv = (Tve - Tc)*b*Sv/sv

Sv = 162.03 mm

Provide 4 legged 10 diameter stirrup @ 200 c/c .

DIA = 6250

WATER TANK

4 nos.28 dia. Bars

3 nos 25 dia Bars.

7 nos. 20 dia Bars.

Tank bottom slab

1000

4 Legged 10 mm diameter stirrup @ 200 mm c/c

Beam L/C Node Fx kN Fy kN Fz kN Mx kNm My kNm

Max Fx 48 139 50 76.957 3.462 -0.072 0.1 -0.4

Min Fx 56 136 58 -66.708 5.096 -2.358 -0.6 1.9

Max Fy 46 139 48 14.464 6.719 -0.717 1.7 0.9

Min Fy 55 124 58 -28.118 -7.374 -0.087 -0.6 -0.5

Max Fz 60 116 62 7.454 3.657 2.921 -0.4 -1.5

Min Fz 51 136 53 22.82 3.568 -2.67 1 2.7

Max Mx 56 120 58 -24.976 6.565 -0.428 2.1 0.1

Min Mx 50 118 52 35.145 0.677 0.468 -2.2 -0.1

Max My 60 116 48 7.454 -3.616 2.921 -0.4 2.9

Min My 56 136 59 -66.708 -2.177 -2.358 -0.6 -1.7

Max Mz 55 140 58 -42.458 -7.361 0.633 -1 0.2

Min Mz 49 118 51 68.176 2.445 -0.769 -1.6 0.1

sv = 150 N/mm2

cbc = 8.5 N/mm

st = 190 N/mm2

m = 10.98k = 0.329412

j = 0.89

R = 1.248

Bending moment at support , M u = 4.6 kN-m Bending moment at span = 2.243 kN-m

Torsional moment ,Tu = 2.2 kN-m (From STAAD Output ; Load case-140 & 118)

= 7.4 kN

Design Bending moment , M = 4.60 kN-m

Design torsion, T = 2.20 kN-m

Shear force , Vu

600

3.1

3.4

4.6

4

1.1

0.7

0.9

-1.8

DESIGN OF RING BEAM AT TOP OF TANK (SECTION 400mmx400mm)


DESIGN:-

1 no. 16 dia Bar

TANK WALL

1 no. 16 dia Bar

-1.5

1.1

Mz kNm

-1.7

3.9


20/32

20

Design shear , V = 7.40 kN

b = 400 mm D = 400 mm


b1 = 330 mm d1 = 370 mm

d = 365 mm

Calculation of longitudinal reinforcement :

(Cl. B-6.2 of IS : 456 - 2000 )

M = 4.60 kN-m

Mt = T (1+D/b)/1.7

= 2.59 kN-m

Me1 = 7.19 kN-m

Moment of resistance of a balanced section = 66.50592 kN-m >7.19 kN-m

Astreqd = 116.44 mm

Minimum % of reinforcement is required

Astmin = 292 mm

Provide 2 nos. 16 mm diameter bars ( 402 mm2

) at top & bottom althrough .

Shear :

Equivalent shear , Ve = V + 1.6 T/b = 16.2 kN

= 0.11 N/mm < Tcmax = 1.9 N/mm2

( From Table - 24 of IS: 456 -2000)

Hence OK.

Shear stress Tv = 0.05 N/mm

0.25


Transverse Reinforcement :

Provide 2 legged 8 diameter stirrup @ 200 mm c/c .

2 - 16 mm dia

TANK ROOF SLAB

2 - 16 mm dia

2 Legged 8 mm diameter stirrup @ 200 mm c/c

WATER TANK

Me1/(st*j*d) =

400

Percentage of steel =

Equivalent bending moment Me1 = M+ Mt

400

TANK WALL

DESIGN OF SLABS

DIA = 6250


21/32

21

8.5 N/mm [Ref. Table 21 of IS: 456 ]

10.98039

150 N/mm2 [Ref. Table 2 of IS:3370II ]

0.383562

0.872146

1.421718

1.088 kN-m /m

0.83 kN-m /m

0.032 N/mm2

0.021 N/mm2

1.088 kN-m /m

0.83 kN-m /m

0.032 N/mm2

0.021 N/mm2

1000 mm

150 mm

25 mm

10 mm

120 mm

27.66354 mm

Provide 10 mm dia. bars @ 250 c/c Hence, Astprovd = 314.29 mm2

0.26 %

0.26 0.235 N/mm2

O.K

69.305 mm2

O.K

Reinforcement in circumferential direction:-

52.87 mm2

Provide 10 250 c/c in radial and circumferential directions

at top and bottom .

DESIGN OF TANK BOTTOM SLAB ( 300 mm thk. )

From

STAAD

Output

mm diameter bars @

Tank slab designed as uncracked section as per IS-3370.



DESIGN OF TANK TOP SLAB ( 150 mm thk. )

Designed as per working stress method.

Allowable Stress in steel tension for Fe 415 bars, st =

cbc =


Therefore j = 1 k / 3 =Therefore R = (1/2).cbc.jk =

X ==> Radial direction.

Y ==> Circumferential direction.

Moment Mxu = Plate No.107; L/C 139Moment Myu = Plate No.107; L/C 139

Shear Stress SQXu = Plate No.100; L/C 121Shear Stress SQYu = Plate No.94; L/C 118

Design Moment Mx = Mxu =

Design Moment My = Myu =

Design Shear Stress SQX = SQXu =

Design Shear Stress SQY = SQYu =

Width of slab is b=

Overall depth of slab provided, D =Assuming, clear cover, c =


Effective depth, d = D - c - (/2) =

Required overall depth of slab = [Mmax/(R.b)] =< than provided overall depth. Hence O.K.

and pprovd =

%, is given by Tc=

> than maximum shear stress. Hence O.K.

Reinforcement in radial direction:-

Allowable shear stress, for pprvd =

Required area of steel is Astx = Mx/(jdst) =< than provided reinforcement. Hence O.K.

Required area of steel is Asty = My/(jdst) =< than provided reinforcement. Hence O.K.


22/32

22

8.5 N/mm [Ref. Table 21 of IS: 456 ]

10.98039


0.383562

0.872146

10.13 kN-m /m

11.24 kN-m /m

0.214 N/mm2

0.151 N/mm2

10.13 kN-m /m

11.24 kN-m /m

0.214 N/mm2

0.151 N/mm2

1000 mm

300 mm

30 mm

12 mm

264 mm

0.88

R = 0.3193 N/mm2

Provide 12 mm diameter bars @ 150 c/c Hence, Astprovd = 754.3 mm2

0.29 %

0.29 0.255 N/mm2

O.K

From

STAAD

Output

cbc =



X ==> Radial direction.



Y ==> Circumferential direction.

Moment Mxu = Plate No.88; L/C 125Moment Myu = Plate No.87; L/C 125

Shear Stress SQXu = Plate No.82; L/C 104Shear Stress SQYu = Plate No.87; L/C 105

Design Moment Mx = Mxu/ 1.2 =

Design Moment My = Myu/ 1.2 =

Design Shear Stress SQX = SQXu / 1.2 =

Design Shear Stress SQY = SQYu / 1.2 =

Width of slab is b=

Overall depth of slab provided, D =Assuming, clear cover, c =


Effective depth, d = D - c - (/2) =Hence, d/D =

For uncracked section ==>

and pprovd =

%, is given by Tc=Allowable shear stress, for pprvd =

> than maximum shear stress. Hence O.K.



23/32

23

187.622 mm

293.3087 mm2

O.K

Reinforcement in circumferential direction:-325.4482 mm

2

Provide 12 150 mm c/c in radial and circumferential

directions at top and bottom.

DESIGN OF INNER WALL( 175 mm thk. )

8.5 N/mm2 [Ref. Table 21 of IS: 456 ]

Allowable stress in bending tension = 1.8 N/mm2

Allowable stress for direct tension = 1.3 N/mm2

10.98039

150 N/mm2 [Ref. Table 2 of IS:3370II]0.383562

0.872146

X ==> Vertical direction

Y ==> Circumferential direction

3.652 kN-m /m

8.635 kN-m /m

0.399 N/mm2

0.262 N/mm2

1000 mm

175 mm

40 mm

Diameter of reinforcement, = 12 mm

Effective depth, d = D - c - (/ 129 mm0.737143

R = 0.3111 N/mm2

166.6024 mm

216.4016 mm2

Min. percentage of reinforcement = 0.25

437.5 mm in two faces

Astreqd./face = 219mm

511.6726 mm2 < 754 mm

2

Provide 12 mm diameter bars @ 250 mm c/c at inner face as vertical reinforcement

Provide 16 mm diameter bars @ 120 mm c/c up to 2.0 m from tank bottom and

12 mm diameter @ 120 mm c/c from 2 m to 4.450 m at outer face as vertical reinforcement.

Provide 12 mm diameter bars @ 150 mm c/c at both faces as circumferential reinforcement

Allowable stress for concrete in bending compression cbc =

Required overall depth of slab = [Mmax/(R.b)] =< than provided overall depth. Hence O.K.

Reinforcement in radial direction:-

Required area of steel is Astx = Mx/(jdst) =< than provided reinforcement. Hence O.K.


Required area of steel is Asty = My/(jdst) =< than provided reinforcement. Hence O.K.

Wall is be designed as uncracked section as per IS : 3370.

mm diameter bars @




Moment Mxu =From

STAAD

Output


Width of wall b=

Overall depth of wall provided, D =

Shear stress SQYu = Plate No.273; L/C 125

Plate No.283; L/C 117Moment Myu = Plate No.299; L/C 121

Shear stress SQXu = Plate No.281; L/C 117

< Provided O/A depth . Hence O.K.

Hence, d/D =

Required area of steel in Vertical

direction is Astx = Mx/(jdst) =

Astreqd =

Required area of steel in

circumferential direction is

For uncracked section,

Required O/A depth of wall = [Mmax/(R.b)] =


24/32

24

Shear :


Permissible shear stress Tc = 0.45 N/mm2

> 0.399 N/mm2

( From Table 23 pf IS 456 : 2000 )

Hence OK.

175

Roof

12 mm dia @ 250 mm c/c

TANK

16 mm dia

@ 120 mm c/c2.0m

up to 2 m

4.4

50m

12mm

dia@1

50mmc/c

12mm

dia@1

50mmc/c

12 mm dia

@ 120 mm c/c

from 2 m to 4.45

m


25/32

25

DESIGN OF RCC PARTITION WALL( 200 mm thk. ) INSIDE TANK

8.5 N/mm2 [Ref. Table 21 of IS: 456 ]

10.98039


Allowable stress for concrete in bending tension = 1.8 N/mm2

1.3 N/mm2

0.383562

0.872146

X ==> Vertical direction M 25

Y ==> Fe 415

1.038 kN-m /m

1.469 kN-m /m

0.01 N/mm2

0.016 N/mm2

1000 mm

200 mm

30 mm

12 mm

164 mm

0.82

R = 0.3141 N/mm2

68.38753 mm

48.4 mm2

Min. percentage of reinforcement = 0.22

440 mm in two faces

Astreqd./face = 220 mm

68.46954 mm2

Provide 12 mm diameter bars @ 200 mm c/c ( 565 mm2) at both faces as vertical reinforcement

Provide 12 mm diameter bars @ 200 mm c/c (565 mm2) at both faces as horizontal reinforcement

Shear :


Permissible shear stress Tc 0.25 N/mm2

> 0.016 N/mm2

( From Table 23 pf IS 456 : 2000 )

Hence OK.

200

Required O/A depth of wall = [Mmax/(R.b)] =

Effective depth, d = D - c - (/2) =

Horizontal direction


Allowable stress for concrete in direct tension =

Allowable stress for concrete in bending compression cbc =


Partition wall shall be designed as uncracked section as per IS : 3370.

From

STAAD

Output

Moment Mxu = Plate No.319;L/C 139Moment Myu = Plate No.316; L/C 140Shear SQXu = Plate No.320; L/C 121Shear SQYu = Plate No.317; L/C 118

Hence, d/D =


< than provided O/A depth. Hence O.K.



Astreqd =

Required area of steel in horizontal

direction is Asty =


Width of wall b=

Overall depth of wall provided, D =Assuming, clear cover, c =




26/32

26

12 mm dia

@ 200 mm c/cTANK

TANK

12 mm dia @ 200 mm c/c

WATER TANK WALL

DESIGN OF OUTER WALL

Allowable stress for concrete in bending compres 8.5 N/mm2 [Ref. Table 21 of IS: 456 ]

Allowable stress in bending tension = 1.8 N/mm2

Allowable stress for direct tension = 1.3 N/mm2

10.98039


0.383562

0.872146 M - 25

Fe - 415

X ==> Vertical direction

Y ==> Circumferential direction

8.607 kN-m /m

4.87 kN-m /m

0.099 N/mm2

0.085 N/mm2

1000 mm

250 mm at base

150 mm at top

40 mm

12 mm

204 mm at bottom104 mm at top

0.816

R = 0.3141 N/mm2

165.5358 mm

Hence O/A depth of wall 250 mm at bottom and 150 mm at top is O.K.


Required overall depth of wall = [Mmax/(R.b)] =

Hence, d/D =

Effective depth, d = D - c - (/2) =Effective depth, d = D1 - c - (/2) =


Overall depth of wall provided,D1 =


Shear stress SQYu = Plate No.220; L/C 141

width of wall b =

Overall depth of wall provided, D =

Shear stress SQXu =

Plate No.206; L/C 125

Tank wall shall be designed as uncracked section as per IS : 3370.


From

STAAD

Output



Moment Mxu =

Moment Myu =



12mmdia@2

00m

mc/c

12mmdia@2

00mmc/c




27/32

27

322.5085 mm2

182.4813 mm2

0.886 N/mm2

Plate no.-220, L/C -117

221500 N

Astreqd due to direct tension = 1476.7 mm

920.8146 mm

Total Astprovided = 2 x754 = 1508.0 mm

Equivalent area of steel =2x(m-1)xAst = 30100.9 mm2

Total area of concrete & steel = mm2

Direct Stress due to membrane force = 0.791 N/mm2

Equivalent moment of inertia of steel I1 =2x(m-1)xAstxd2

= mm

[ where, d = ( 250 -40x2-12)/2 =79 ]

Momen of inertia of the secion I2 = mm4

Total moment of inertia = I = 1.4E+09 mm4

Stress = M.y/I = 0.43 N/mm2

Total Stress = 1.22 , Hence O.K.

Provide 12 mm diameter bars @ 150 mm c/c at inner face as vertical reinforcement

Provide 12 mm diameter bars @ 250 mm c/c at outer face as vertical reinforcementProvide 12 mm diameter bars @ 150 mm c/c at both faces as circumferential reinforcement

Shear :


0.27 N/mm2

> 0.099 N/mm2

Table - 23 of IS 456 : 2000

Hence OK.

1.1E+08

Max. membrane tensile stress in wall sy =

1.3E+09

N/mm2

< 1.8

Total Astreqd due to direct tension & bending =

Max. Membrane tensile force in wall /m =


Required area of steel in

circumferential direction is Asty =

280100.9

Permissible shear stress Tc =


28/32

28

150

TANK

12 mm dia. bars @ 12 mm dia. bars @

250 mm c/c 150 mm c/c

250

TANK WALL

12mmdiabars@1

50c/c

12mmdiabars@1

50c/c


29/32

LOAD CALCULATION

DEAD LOAD:Weight of 150 thk landing slab = 0.150x2.5x1.796x1.25 = 0.421 T

Load for ladder = 6x0.03 + 2x0.0304 = 0.241 T/m

Reaction at column due to ladder = 0.241x5.112/2 = 0.616 T

Total load at column due to ladder & landing 1.037 TResultant load at column = 2.07 T

Moment at column due to ladder & landing = 3.28 T-m

Load at column due to circular slab at RL (+) 436.5 m = 3.14/4*6.25 2*0.150*2.5/3 = 4.42T

CALCULATION OF LIVE LOAD :

Live load on tank top slab = 0.150 T/m2

;

Live load for ladder & landing = 0.25 T/m2

Reaction at column due to LL = 0.639 T

Resultant load at column = 1.278 T

Moment at column due to LL on landing & ladder = 2.02 T-m

Load at column due to circular slab at RL (+) 436.5 m = 3.14/4*6.25^2*0.250/3 2.56 T

CALCULATION OF WIND LOAD WITH GUST FACTOR METHOD :

Along wind load on a structure on a strip area (Ae) at any height (z) is given by :

Fz = Cf.Ae.pz.G

Where ,

Fz = along wind load on the structure at any height z corresponding to strip area Ae,

Cf = force coefficient for the building , (Ref.Table - 28 of IS 875 part -3)

Ae = effective frontal area considered for the structure at height z,

Vz = Vb.k1.k2.k3

pz = design pressure at height z due to hourly mean wind obtained as 0.6Vz(N/m )

Vb = 47.3 m/sec

k1 = 1.07 ( As per Cl. 5.3.1 Of 875 ,Part-3 )

k2 = 0.67 (From Table -33 of IS 875 , Part - 3 ) (For 10 m height )k3 = 1.0 ( As per Cl. 5.3.3 Of 875 ,Part-3 )

Vh = 33.9 m/sec

Pz = 689.9072 N/m2

k1 = 1.07 ( As per Cl. 5.3.1 Of 875 ,Part-3 )

k2 = 0.72 (From Table -33 of IS 875 , Part - 3 ) (For 15 m height )

k3 = 1.0 ( As per Cl. 5.3.3 Of 875 ,Part-3 )

Vh = 36.4 m/sec

Pz = 796.7207 N/m2

k1 = 1.07 ( As per Cl. 5.3.1 Of 875 ,Part-3 )


k3 = 1.0 ( As per Cl. 5.3.3 Of 875 ,Part-3 )Vh = 38.0 m/sec

pz = 864.4972 N/m2

k1 = 1.07 ( As per Cl. 5.3.1 Of 875 ,Part-3 )


k3 = 1.0 ( As per Cl. 5.3.3 Of 875 ,Part-3 )

Vh = 40.0 m/sec

pz = 959.1693 N/m2


30/32

k1 = 1.07 ( As per Cl. 5.3.1 Of 875 ,Part-3 )


k3 = 1.0 ( As per Cl. 5.3.3 Of 875 ,Part-3 )

Vh= 43.0 m/sec

pz = 1110.399 N/m2

G = gust factor and is given by :

G = 1 + gfr ( B ( 1 + )2 + SE/ ))L (h) = 1250 From fig.8

gfr = 1.25 From fig.9

Cy = 10 From code

b = 6.9 m

Cz = 12 From code

h = 35.5 m

0.16

0.3408

B = 0.8 (From Fig.9)

fo = 0.501 ( From STAAD output )

5.416106 for 10 m ht. ( where Vz=Vh)

4.424241 for 15 m ht.4.269907 for 20 m ht.

4.125978 for 30 m ht.

3.898217 for 37 m ht.

18.46835 for 10 m ht.

17.18582 for 15 m ht.

13.47458 for 20 m ht.

13.47458 for 30 m ht.

12.52343 for 37 m ht.

S = 0.35 for 10 m ht.

0.38 for 15 m ht.

0.39 for 20 m ht. From Fig.10

0.40 for 30 m ht.

0.41 for 37 m ht.

E = 0.08 for 10 m ht.

0.10 for 15 m ht.

0.101 for 20 m ht. From Fig.11

0.102 for 30 m ht.

0.103 for 37 m ht.

= 0.016

gfr B

4

Gust Factor , G = 3.186505 for 10 m ht.3.399451 for 15 m ht.

3.427572 for 20 m ht.

3.455769 for 30 m ht.

3.484039 for 37 m ht.

Cf = 1.6 (Solidity ratio =0.52) From Table - 28

Cf = 2 From Table - 28

=Cy.b/Cz.h =Cz.h/L (h) =

Fo=Cz.fo.h/ Vh =

Fo=Cz.fo.h/ Vh =Fo=Cz.fo.h/ Vh =

Fo=Cz.fo.h/ Vh =

Fo=Cz.fo.h/ Vh =

fo.L (h)./Vh =

fo.L (h)./Vh =

fo.L (h)./Vh =

fo.L (h)./Vh =

fo.L (h)./Vh =

= = 0.279508


31/32

Ae = 36.9516 m2 for 10 m ht.

19.3508 m2 for 15 m ht.

19.3508 m2 for 20 m ht.

36.9516 m2 for 30 m ht.

43.2 m2 for 37 m ht.

Fz = 12.99746 T for 10 m ht.

8.385593 T for 15 m ht.

9.17422 T for 20 m ht.

14.37 T for 25 m ht.

19.59716 T for 30 m ht.

33.42533 T for 37 m ht.

FOR STAAD INPUT

Wind load per node for 10 m height = 3.2 T





Wind load at tank for 37 m height = 0.77 T/m2

CALCULATION OF THE WIND LOAD AS PER NORMAL METHOD :

Basic wind speed Vb = 47.3 m/sec ( As per DBR )

Terrain category - 2

Class - B

Design wind speed Vz = Vb x k1 x k2 x k3 m/sec

Factors for various heights as per IS 875 Part - III :

For 37 m height

For 30 m height

For 20 m height

For 15 m height

For 10 m height

k1 = 1.07

k2 = 1.15

k3 = 1.0

k1 = 1.07

k2 = 1.1

k3 = 1.0

k1 = 1.07

k2 = 1.05

k3 = 1.0

k2 = 0.98

k3 = 1.0

Wind pressure Pz = 0.6 x Vz2

k1 = 1.07

k2 = 1.02

k3 = 1.0

k1 = 1.07


32/32

Wind Pressure at 37 m height = 200 kg/m2





Wind load for staad input on tie beam at 10 m height from GL = 0.148x0.9 = 0.133 T/m





Average wind force on columns between ht 10 m to 20 m = ( 0.148 + 0.183 ) / 2 x 1.3 = 0.215 T/m

Average wind force on columns between ht 20 m to 30 m = ( 0.183 + .186 ) / 2 x 1.3 = 0.24 T/m

Wind pressure on tank = 0.200 T/m2

The value of wind load in gust factor method is higher than the normal wind load

calculation, so the value of wind load in gust factor method is applied in STAAD input.

CALCULATION OF THE SEISMIC LOAD :

Here Zone factor Z = 0.1

Importance factor = 1.5

Response reduction factor = 3Soil type = 2

Response sectrum method of analysis carried out using STAAD , considering

X = 0.025

Y = 0.005

Z = 0.0075

design of water tank-311207

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