Project : 8.2 m diameter Recycled water tank Project No. 22-17058Calculation for Shell Thickness of Tank

Tank diameter D 8.20 mTank Height H 9.50 mCalculated Tank Capacity 501.70 cu.mRequired Tank Capacity 425.00 cu.m OK

Roof Plate thickness 6.0 mm Corrosion Allowance for Roof plate 1.0 mm

Allowable stress in shell plate as per AWWA s = 103.4 MPaSpecific Gravity of Liquid G 1.00 Corrosion Allowance for shell CA 1.0 mm

Strake No. Width of Level from Design Design Adopted Weight of from each base Liquid shell Shell Shellbase of strake of tank Level thickness Plate Plate tank in m in m in m in mm thickness excludingupward hp t in mm CA

1 2.40 0.00 9.50 5.3 6 24.272 2.40 2.40 7.10 4.2 6 24.273 2.40 4.80 4.70 3.1 6 24.274 2.30 7.20 2.30 2.1 6 23.26

9.50

Sum 9.50Total Shell Plates Weight excluding CA in kN 96.06Roof Plate weight excluding CA in kN 20.73

Total Weight of Shell and Roof Plates excluding CA in kN 116.78Total Shell Plates gross weight in kN 115.27 kNRoof Plate gross weight in kN 24.87 kNC.G of Tank shell from base 1.19 m

Formula s -> Allowable Design stress in shell plate as per Table 5 , AWWA D100-05s = 103.4 MpaE -> Joint EfficiencyFrom table15, Joint Efficiency for single -groove butt joint in shell = 85 %Required shell thickness t = { 4.9 * hp * D * G / ( s *E) } + CA Eq. 3-40From table 16 , AWWA D100-05Minimum thickness of tank shell plate in contact with water = 4.76 mm

Required shell thickness including C.A = 5.76 mmMinimum adopted shell plate thickness = 6 mm

Hence OK

Project : 8.2 m diameter Recycled water tank Project No. 22-17058

Calculation for Intermediate Wind GirderTank diameter D = 8.20 mTank Height = 9.50 mWind Region A2Design Wind Speed ( service ) V = 141 km / h( 3-sec gust based on 50 year recurrence interval from AS 1170.2)Thickness of top shell course 6 mmThickness of top shell course - CA t uniform = 5 mm

Shell Level from Actual width Actual Actual shell TransposedLayers base of tank of shell shell thickness width offrom base in m in mm thickness minus corrosion each shell

adopted thickness in mmW t actual t actual - CA W tr

0.00 1 2.40 2400 6 5.0 2400.002 4.80 2400 6 5.0 2400.003 7.20 2400 6 5.0 2400.004 9.50 2300 6 5.0 2300.00

Sum 9500.00 mmHeight of Transformed Shell = 9.500 mMaximum Height of unstiffened shell H1 = 40.938 m

H1 greater than Height of Transformed ShellIntermediate Wind Girder NOT Required

Min. Section modulus of the Intermediate Wind Girder = 0.00 cm3

Half height of Transformed shell < H1Second Intermediate Girder Not Required

Formula Cl. 5.9.7 API 650Transposed width of each shell W tr = W ( t uniform / t actual )

5/2

Maximum Height of unstiffened Shell H1 = 9.47 * t * ( t / D ) 3/2 * (190 / V )2

't' is the actual thickness of top shell excluding CA in mmMin. Section modulus of the Intermediate Wind Girder = D2 H1 (V / 190)

2 / 17 If half the height of transformed shell > H1 , a second intermediate girder shall be used.

Corroded thickness of plates are used in calculation of intermediate wind girders.

Cl. 3.5 AWWA D100-05 Transposed width of each shell W tr = W ( t uniform / t actual )

5/2

Min.Section modulus of the Intermediate Wind Girder S= 0.06713HD2Paw

Project : 8.2 m diameter Recycled water tank Project No. 22-17058

Calculation for Bottom and Annulus Plates

Tank diameter D = 8.20 mTank Height = 9.50 m

Bottom Plate - API 650 clause 5.4Minimum Nominal thickness of Bottom plate 6 mmCorrosion Allowance for Bottom plate 2.0 mmMinimum Require thickness of Bottom plate 8.0 mm

Any plate in contact with water - AWWA section 3.10Minimum thickness of Bottom plate 6.35 mmCorrosion Allowance for Bottom plate 1.0 mmMinimum Require thickness of Bottom plate 7.4 mm

Adopted bottom plate thickness = 8 mmAdopted slope of bottom plate = 1 deg.

Annlular Bottom Plate - API 650 clause 5.5

Product stress of Bottom strake ( td * Sd / (t provided - CA) 132.41 MPaHydrostatic test stress of Bottom strake ( tt * St / t provided) 61.61 MPa

Effective Product Height H * G = 9.50 mRefer to Table 5-1 a for Annular Bottom-plate thicknessFrom Table 5-1a Annlus plate thickness 8 excluding CA

Minimum Required Annular Bottom Plate thickness = 10.0 mmProvided 10.0 mm OK

Minimum width of Annulus from inner face of shell L1 = 600 mmL1 = 215 * tb / (HG)0.5 or min 600 mmMinimum projection of Annulus from outer face of sheel L2= 50 mm cl. 5.4.2

Minimum required width of Annular plate L1 + L2 = 650 mmProvide 750 mm OK

Project : 8.2 m diameter Recycled water tank Project No. 22-17058

Calculation for Stability Check of Tank Tank diameter D 8.20 mTank Height H 9.50 mTank Capacity 425.00 cu.m

Design Wind Speed ( service ) V 141.00 km / hFrom API 650 clause 5.2.1 pshell =0.86(V/190)

2 0.47 kPaproof =1.44(V/190)

2 0.79 kPaAs per AS 1170.2 Design wind pressure on shell pshell = 0.78 kPaAs per AS 1170.2 Design wind pressure on roof proof = 0.73 kPa

Adopt the higher value for stability check.Total weight of shell excluding CA Wshell 96.06 kNWeight of roof plates excluding CA Wroof plate 20.73 kNWeigth of Roof Frame 11.10 kNTotal weight of Roof Wroof 31.83 kNDesign Internal pressure 0 kPa

Overturning moment due to wind Mw = 460.33 kNm(D.H.pshell.H/2)+(SD2.proof/4 * D/2)

Moment due to Internal pressure Mpi = 0 kNm

Resisting Moment due to Self Wt. of Tank MDL = 524.3305 kNm(Wshell.D/2) + (Wroof.D/2)Weight of band of liquid at the shell using wL = 23002.43 N/mG=0.7 and height of one-half of design liquid height

wL = 59 (tb - CA) (Fby. H)1/2

Moment about liquid weight wL MF 2429.525 kNmS*D*wL*D/2

Check 1 0.6Mw + Mpi < MDL / 1.5276.20 < 349.5537 OK

Check 2 Mw + 0.4 Mpi < (MDL + MF ) / 2

460.3293 < 1476.928 OK

Lateral force due to wind when tank is empty = D*H*pshell= 36.89 kNConsidering friction co-eff as P=0.4,Resisting force = P*Wtank= 51.15 kN Tank is stable for sliding against wind

Project : 8.2 m diameter Recycled water tank Project No. 22-17085

Calculation for Seismic Design of Tank as per AWWA D100

Tank diameter D = 8.20 mTank Height H = 9.50 mTank Capacity 425.00 cu.mTank Content Specific Gravity G 1.00

Peak Ground Acceleration Sp = 0.40Spectral Response Acceleration parameter Ss = 0.4 clause E4.3-1Acceleration parameter for 1 second S1 = 0.218 clause E4.3-2Acceleration parameter for 0 second S0 =Sp 0.40

Seismic User Group ISite Class DType of Tank Self Anchored / Mech Anchored Mechanically AnchoredFrom Table 26 Fa = 1.48From Table 27 Fv = 1.98 I = 1.00From Table 28 Ri = 3.00

Rc = 1.50Scaling Factor U= 0.67 Eqn. 13-8

Convective Sloshing Period Ks = 0.578 / [tanh (3.68 H/D)]0.5 Ks = 0.58Tc = 1.8 Ks D0.5 Tc = 2.98 sec

Regional dependent transition period for TL = 4 seclonger period ground motion (for outside USA )

Impulsive Spectral Acceleration parameter Ai = 0.329Ai=2.5UFaS0(I/Ri)

Check Ai > 0.007 Ai > 0.007 OK

Convective Spectral Acceleration parameter Ac = 0.121if Tc < TL Ac = 2.5 KUFaSo(Ts/Tc)(I/Ri)if Tc > TL , Ac = 2.5 KUFaSo(TsTL/Tc2)(I/Ri)where K = 1.50

Check Ai > 0.007 Ac < Ai OK

Effective Weight of product

Total Weight of Contents Wt =785.4 GHD2= Wt = 4,169,250.00 N Eqn. 3-27(calculated from working capacity of tank)

Effective Impulse Weight Wi is weight of effective mass of tank contents thatmoves in unison with tank shell.Effective Impulsive Weight Wi = 3,384,728.81 N if D/H > 1.33, Wi= tanh(0.866D/H) Wt / (0.866 D/H) Eqn. 13-24if D/H < 1.33, Wi= [ 1.0 - 0.218 D/H]Wt Eqn. 13-25

Effective Convective Weight Wc is weight of effective mass of the first modesloshing contents of the tank.Effective Convective Weight Wc = 827,370.32 N Wc = 0.23 * D/H * Wt * tanh(3.67 H/D) Eqn 13-26

Centre of Action for Overturning Moment across entire base of tank

Height from bottom of tank shell to centre Xi = 3.98 mof action of lateral seismic force applied tothe impulsive weight Wiif D/H > 1.33, Xi= 0.375 H Eqn. 13-28if D/H < 1.33, Xi= [0.5 - (0.094D/H)] H Eqn. 13-29

Height from bottom of tank shell to centre Xc = 7.33 mof action of lateral seismic force applied tothe effective convective weight WcXc = {1.0 - [(cosh(3.67H/D) -1) / (3.67H/D sinh(3.67H/D)]} H

Centre of Action for Overturning Moment at bottom of shell

Height from bottom of tank shell to centre Xis = 5.24 mof action of lateral seismic force related tothe impulsive weight Wi adjusted to include bottom pr.if D/H > 1.33, Xis = 0.375 H*{1.0 + [1.33(0.866D/H) / (tanh 0.866D/H) ]}if D/H < 1.33, Xis= [0.5 + (0.06D/H)] H

Height from bottom of tank shell to centre Xcs = 7.39 mof action of lateral seismic force related tothe convective weight Wc adjusted to include bottom pr.Xcs = {1.0 - [(cosh(3.67H/D) -1.937) / (3.67H/D sinh(3.67H/D)]} H

Total Weight of Tank Shell Ws = 115,267.86 NC.G of tank shell plates from base Xs = 1.19 m

Total weight of Roof Wr = 31,828.95 NC.G of tank Roof from base Xr = 9.84 m

Vf is the Design Shear at top of foundation due to horizontal design acceleration Vf = [Ai*(Wi + Ws + Wr +Wf ]2 + (Ac Wc )2}0.5 1264247.546 N Eqn 13-31 Design overturning Moment across the Mmf = 4,934,403.12 Nm Eqn 13-32entire base cross secrtion due to horizontal accelerationMmf={[Ai*(Wi Xi + Ws Xs+ Wr Xr)]2 + (Ac Wc Xc)2}0.5

Design overturning moment at the bottom Ms = 6,328,037.91 Nm Eqn 13-23of the shell caused by horizontal design accelerationMs={[Ai*(Wi Xis + Ws Xs+ Wr Xr)]2 + (Ac Wc Xc)2}0.5

Resistance to Design loads

Annulus thickness excluding CA ta 9 mmMax resisting weight of tank contents wa 41,457.13 N/mthat may be used to resist shell overturning momentwa = 99 ta (Fy H G)^

0.5 ( Fy = 250 Mpa) Max. wa = 15,665.69 N/m Av = 0.14 * 2.5 U Fa So = 0.138133Max wa = 201.1 H D Ge

Check wa < Max. wa wa > Max. wa . Adopt Max wa

Internal pressure Int.Pr 0 kPaInternal pressure along perimeter wint 0 N/m

Tank shell & roof weight acting at base wt 5,710.04 N/mwt = (Ws + Wr) / (S D)

Anchorage Ratio J = 4.47 ( Tank Not Stable ) J = Ms / [D2 (wt (1 - 0.4Av) + wa ] Eqn 13-36

( Provide Mechanical Anchors )

If J < 0.785 Tank is Self-anchored. No calculated uplift under seismic overturning moment.

If 0.785 < J < 1.54 Tank is self-anchored. Tank is stable provided shell compression requirements aresatisfied.

If J > 1.54 Tank is not self-anchored and not stable. Mechanical anchors to be provided.

Mechanical Anchors Design cl. E6.2.1.2 API 650

Minimum Anchorage Resistance per unit length of circumference of Tank wAB = 88024.478 N/mwAB = 1.273 Mmf / D

2 - wt (1-0.4Av)

In accordance with API, maximum spacing of anchors = 3 m

Minimum No. of Anchors required = 9

Provide 15 (nA) Anchors. Spacing of Anchors = 1.72 m

Anchor Seismic Design Load PAB 151.2 kNAdopting , Grade 4.6/S bolts, minimum yield strength = 240 N/mm2Allowable Design stress for Anchor bolts = 80% of yield stress = 192 N/mm2

Diameter of Anchor bolt = 36 mmAnchor capacity = 195.43 kN OK

Adopt 15 Nos. of 36 diameter Grade 4.6/S Anchor bolts.

Check for Sliding Resistance

Vf is the Design Shear at top of foundation due to horizontal design acceleration Vf = [Ai*(Wi + Ws + Wr +Wf ]2 + (Ac Wc )2}0.5 1264247.546 N Eqn 13-31Lateral Seismic force F = 1264.25 kN

Friction co-efficient of tank base P = 0.35Average bottom plate thickness tb = 8.0 mmFrictional Resistance to lateral seismic force Vs = 1438.22 kNVs = P (Ws +Wr + Wf + Wp)(1.0 - 0.4 Av)

Check F < Vs F < Vs OK ( Sliding check is OK )

Check for uplift force per mechanical anchors AWWA D-100

No. of mechanical anchors provided N= 15 Diameter of anchor circle Dac = 8.36 m

Design uplift force per mechanical anchor = Ps = ( 4 Ms / ( N Dac)) - (W / N) Eqn 3-41where W ->tank shell and roof weight = 147096.81 N Ms -> overturning moment at bottom of shell 6328037.91 Nm

Ps = 192044.8351 N

Calculating Overturning Stability Ratio E.6.2.3

The overturning stability ratio for mechanically-anchored tank system excluding vertical seismic effects shall be 2.0 or greater

0.5 D [Wp + Wf + WT + Wfd + Wg] / Ms > 2.0 Eqn. 6.2.3-1

where D --> Nominal diameter of Tank in m = 8.2Wp --> Weight of contents of the tank in N = 4,169,250 Wf --> Weight of Tank bottom in N = 33,165 WT --> Weight of Tank shell, roof =Ws+Wr in N = 147,097 Wfd --> Weight of tank foundationConcrete slab size B= 9000 Dp = 500 mm

Weight of foundation = pi()*D*B*Dp*25000 in N = 2,898,119

Wg --> Weight of soil directly over the foundation in N = - Ms --> Slab Moment ( as calculated by E6.1.5-2) in Nm = 8,563,722

0.5 D [Wp + Wf + WT + Wfd + Wg] = 29,715,286.37 NmMs = 6,328,038 Nm

0.5 D [Wp + Wf + WT + Wfd + Wg] / Ms = 4.7

0.5 D [Wp +Wf +WT +Wfd +Wg] / Ms > 2.0 Hence OK

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