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Structural design of 350 kl overhead water tank at telibagh,lucknow

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Anchit Agrawal
Anchit Agrawal

The document provides design details for a 350KL overhead water tank at a university campus. Key points include: - The tank will be an Intze tank with a column and brace staging 25m high to hold 350KL of water. - Water demand was estimated at 120KL for the college campus and 216KL for hostels, totaling 346KL. - Design requirements include using M-25 concrete and Fe-415 steel, with minimum reinforcement. - The height of the staging was calculated as 25m based on pipe diameter, flow rate and head loss calculations. - Dimensions of the tank include a 12m diameter cylindrical portion with 1m and 1.5m domes at

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1 STRUCTURAL DESIGN OF 350KL OVERHEAD WATER TANK AT INDIRA GANDHI NATIONAL OPEN UNIVERSITY, TELIBAGH LUCKNOW
2 DATA 1. Type of Tank: Intze Tank 2. Capacityof the tank: 350KL 3. Type of staging: Column& Brace type 4. Depthof foundation: 2.5m 5. Safe BearingCapacityof Soil: 100KN/m2 6. Type of foundation: CircularRing&Raft foundation 7. Grade of Concrete: M-25 8. Grade of Steel: Fe-415 9. Heightof staging: 25m 10. Type of soil: SoftClay 11. Heightof BuildinguptoTerrace: 15.6m 12. No.of floorsinBuilding: G+3 13. Basic WindPressure: 1500N/m2 14. SesmicZone of Lucknow: Zone 3 15. No.of studentinCollege: 2000 16. Water consumptionrate (Percapitademandinlitresperdayper head): 45 17. Designperiodfortank: 30 years 18. No.of studentinhostels: 1600
3 OBJECTIVE 1:- To make a studyaboutthe analysisanddesignof watertank 2:- To make a studyaboutthe guidelinesforthe designof liquidretainingstructure accordingto IS Code IS: 3370 part 2-2009 IS: 456:2000 3:- To knowabout the designphilosophyforthe safe andeconomical designof watertank 4:- To estimate the overall costformakingthe Intze Tank
4 WATER QUANTITY ESTIMATION IN COLLEGE CAMPUS Populationorthe numberof studentstobe servedin2014 = 2000 Let populationtobe increasedatrate of 10% per decade Numberof students(2014) = 2000 Numberof studentsin2024 = 2200 Numberof studentsin2034 = 2420 Numberof studentsin2044 = 2662 Quantity = per capitademandร— Population = 45 ร— 2662 = 1,19,790 litres = 120 KL (assume)
5 FLUCTUATION IN RATE OF DEMAND Average dailypercapitademandincollege campus = 45 lpcd If this average suppliedatall the timesitwill notbe sufficienttomeetthe fluctuation. HOURLY VARIATION (1) Duringthe entryof college from8to 9 inthe morning. (2) Duringthe lunchfrom12 to 1 in the afternoon.
6 WATER CONSUMPTION IN HOSTEL Average dailypercapitademandinhostels=135 lpcd. Quantity = 136 ร— 1600 = 216 KL Total quantity = 216 + 130 = 346 KL อŒ 350 KL
7 DESIGN REQUIREMENTOFTANK * Concrete mix weakerthanM-20 isnot usedbecause of highergrade lesserporosityof concrete. * Minimumquantityof cementinconcrete shall be notlessthan30 KN/m3 . * Use of small size bars. * Coefficientof expansiondue totemperature=11ร—10-6 /หšC * Coefficientof shrinkage maybe taken= 450 ร— 10-6 forinitial and200 ร— 10-6 fordrying shrinkage. * Minimumcovertoall reinforcementshouldbe 20 mmor the diameterof mainbarwhichever isgreater. * Anoverheadliquidretainingstructure isdesignusingworkingstressmethodavoidingthe cracking inthe tank and to preventthe leakage andthe componentof tankcanbe designusing LIMIT STATE METHOD (example:-column,foundation,bracing,stairsetc.). * Code usingIS:3370-PART 2-2009 IS: 456:2000 * The leakage ismore withhigherliquidheadandithas beenobservedthadwaterheadupto 15m doesnotcause leakage problem. * Inorder to minimizecrackingdue toshrinkage andtemperature,minimumreinforcementis recommendedas- (i) For thicknessโ‰ค100 mm = 0.3% (ii) Forthicknessโ‰ฅ450 mm = 0.2% For thicknessbetween100mm to 450 mm= varieslinearlyfrom0.3% to0.2% * For concrete thicknessโ‰ฅ225 mm, twolayerof reinforcementbe placedone nearwaterface and otherawayfrom waterface.
8 FROM IS -3370 (i) For loadcombinationwaterloadtreatedasdeadload. (ii) Crackingโ€“ The maximumcalculatedsurface widthof concrete fordirecttensionandflexure or restrainedtemperatureandmoisture effectshall notexceed0.2mmwithspecified cover. (iii) Shrinkagecoefficientmaybe assumed= 300 ร— 10-6 . (iv) Bar spacingshouldgenerallynotexceedthan300 mm or the thicknessof the section whicheverisless.
9 DETERMINATION OFFOUNDATION โ„Ž = โ„Ž โ„Ž ( 1 โˆ’ โ„Žโ„Žโ„Žโ„Ž 1 + โ„Žโ„Žโ„Žโ„Ž ) = 100 17 ( 1โˆ’โ„Žโ„Žโ„Ž12 1+โ„Žโ„Žโ„Ž12 ) -Fromtestingof soil sample For clay โˆ…=12หš r = densityof soil = 1.76 gm/cm3 = 17.6 KN/m3 p = 100 KN/m3 โ„Ž = 2.52 โ„Ž
10 9. DETERMINATION OFHEIGHTOFSTAGGING We knowturbulentflow occursina pipe So Re > 4000 f = 0.079 โ„Žโ„Ž 1/4 L = lengthof pipe, v = meanvelocityinpipe of flow d = diameterof pipe โ„Žd= 4โ„Žโ„Žโ„Ž2 2โ„Žโ„Ž โ„Ž = โ„Žโ„Ž 2 The kinematicviscosityof water(โ„Ž) =0.01ร—10-4 m2 /s Assume diameterof pipe = 15 cm โ„Ž = โ„Žโ„Ž โ„Ž = โ„Ž 4 ร— 0.15 2 = 0..176โ„Ž2 = Volume (V) =350 m3 Onlyforone hour maximumvelocityoccursinthe pipe sothe discharge duringthatperiod โ„Ž = โ„Ž โ„Ž = 350 60ร—60 = 0.097 โ„Ž3 โ„Ž โ„Ž = โ„Žโ„Ž 0.097= 0.0176 ร— โ„Ž โ„Ž = 5.52 m/sec. Maximumvelocity=5.52 m/sec. โ„Žโ„Ž = โ„Žโ„Ž โ„Ž = 5.52ร—0.15 0.01ร—10โˆ’4 = 8.2 ร— 105 (O.K.) โ„Ž = 0.079 (8.2 ร— 105 ) 1 4 = 2.61 ร— 10โˆ’3
11 Minimumlengthof pipe requirement = 2 ร— heightof buildingupto3 storeysfromthe level +lateral distance uptothe centre of tank = 2 ร— 15.6 + 18 = 49.2 m โ‰ˆ 50 m Headloss โ„Žโ„Ž = 4ร—2.61ร—10โˆ’3 ร—50ร—5.522 2ร—9.81ร—0.15 = 5.40 m HEIGHT OF STAGGING Total hydrostaticpressure ontank P = ฯgh Total head= โ„Ž โ„Ž + โ„Ž2 2โ„Ž + โ„Ž + โ„Žโ„Ž+ โ„Žโ„Žโ„Žโ„Žโ„Ž โ„Žโ„Žโ„Žโ„Žโ„Žโ„Ž Minor loss(assume) =1 m. = โ„Žโ„Ž โ„Ž + โ„Ž2 2โ„Ž + โ„Ž+ โ„Žโ„Ž+ 1 = 4.5 + 5.522 2ร—9.81 + 15.6+ 5.4 + 1 = 28.08 โ„Ž Usingtotal head= 29.5 Heightof stagging= 29.5 โ€“ 4.5 = 25 m
12 DESIGN OF TOP DOME Assume thicknessof topdome =100 mm. Meridional thrustatedges โ„Ž1 = โ„Žโ„Ž1 1+โ„Žโ„Žโ„Žโ„Ž1 Deadload of top dome = 0.100 ร— 25 = 2.5 KN/m2 Live loadon topdome = 0.75 KN/m2 (assume) Total load P = 3.25 KN/m2 โ„Ž1 = 3.25 ร— 103 ร— 18.5 1 + โ„Žโ„Žโ„Ž 18.92 = 30897.15 N/m Meridional stress= 30897.15 100ร—100 = 0.308MPa < 5 MPa (OK) Maximumhoopstressoccurs at the centre and itsmagnitude โ„Žโ„Ž1 2โ„Ž1 = 3.25ร—103 ร—18.5 2ร—0.100 =0.30 N/mm2 =0.3 MPa < 5MPa (OK) Provide nominal reinforcementof 0.24%. โ„Žโ„Žโ„Ž = 0.24ร—100ร—1000 100 = 240โ„Žโ„Ž2 Use 8 mmbars. โ„Žโ„Ž = 50 โ„Žโ„Ž2 Spacing = 1000ร—50 240 = 208.33 = 205 mm c/c. Provide 8 mmbars @ 205 mm c/c radiallyandcircumtentiallyasshowninfigure. The 205 mm c/c for radial bar isprovidedatthe springingof the dome. At the crown the spacingreducestozero. Hence the curtailmentof radial barsmay be carriedout at the appropriate distance.
13
14 DIMENSION OF TANK Innerdiameterof cylindrical portion D= 12 m Rise of top dome h1 = 1 m Rise of bottomdome h2 = D/8 = 1.5 m (centre) Free board= 0.15 m Diameterof ringbeamDo = 5/8 D = 7.5 = 8 m Rise of bottomdome (side) ho = 3/16 ร— D = 2.25 m = 2.5 m Capacityof tank:- โ„Ž = โ„Žโ„Ž2 โ„Ž 4 + โ„Žโ„Žโ„Ž 12 (โ„Ž2 + โ„Žโ„Ž 2 + โ„Žโ„Žโ„Ž)โˆ’ โ„Žโ„Ž2 2 (3โ„Ž2โˆ’โ„Ž2) 3 Radiusof bottomcircular dome:- 1.5 ร— (2R2 โ€“ 1.5) = 42 2R2 โ€“ 1.5 = 10.67 R2 =6 m SinฦŸ2 = 4 6 ฦŸ2 = 41.8o โ„Ž = โ„Žโ„Ž2 โ„Ž 4 + โ„Žโ„Žโ„Ž 12 (โ„Ž2 + โ„Žโ„Ž 2 + โ„Žโ„Žโ„Ž) โˆ’ โ„Žโ„Ž2 2 (3โ„Ž2โˆ’โ„Ž2) 3 350 = โ„Žร—122 ร—โ„Ž 4 + โ„Žร—2 12 (122 + 82 + 12 ร— 8) โˆ’ โ„Žร—1.52 (3ร—6โˆ’1.5) 3 350 = 113โ„Ž + 160โˆ’ 38.87 โ„Ž = 2 โ„Ž Radiusof top circulardome:- 1 ร— (2R1-1) = 6 ร— 6 R1 = 18.5 m
15 SinฦŸ1 = 6/18.5 ฦŸ1 = 18.92o Designof top ringbeam:- A ringbeamis providedatthe junctionof topdome and the vertical wall toresisthooptension inducedbythe top dome. Horizontal componentof meridional thrust P1 = T1 cos ฦŸ1 = 30897.15 cos 18.92o = 29227.8 N/m. Total hoop tension tending to rupture of beam = โ„Ž1ร—โ„Ž 2 = 29227.8ร—12 2 = 175366.8โ„Ž Permissible stress in HYSD bars = 150 N/m2 Ash = 175366.8/150 = 1170 mm2 Provide 20 mm bars (314.15) as hoop. Number of 12 mm bars = 1170 / 314.15 = 3.72 = 4 Actual Ash = 4 ร— โ„Ž/4 ร— 202 = 1256.63 mm2 = 1257 mm2 Provide 4-20 mm รธ hoop and 8 mm bars tie @ 205 mm c/c. Hence the cross sectional area of concrete 1.3= 175366.8 โ„Ž+1257ร—8 Ac = 124841.53 Provide ring beam of 320 mm ร— 400 mm.
16 Designof cylindrical wall:- In the membrane analysisthe tankwall isassumedtobe free attop andbottom maximumhoop tensionoccursat the base of the wall and itsmagnitude:- = โ„Žโ„Žโ„Žโ„Ž 2 = 9800ร—โ„Žร—12 2 = 58800 โ„Ž Hoop tensionatanydepthx fromthe top X (m) Hoop tension(N/m) 0 0 1 58800 2 117600 Minimumthicknessof cylindrical wall =3 H + 5 = 3 ร— 2 + 5 = 11 cm. Provide 20 cm at the bottomand taperit to12 cm at top. At x = 1 m. Areaof steel Ash = 58800/150 = 392 mm2 Provide 8 mmbars. Aรธ = 50.26 mm2 Spacing= (1000 ร— 50.26) / 392 = 130 mm c/c. At x = 2 m. Areaof steel Ash = 117600/150 = 784 mm2 Provide 10 mm bars. Aรธ = 78.53 mm2 Spacing= (1000 ร— 78.53) / 784 = 100 mm c/c.
17 The hoop steel maybe curtailedaccordingtohooptensionatdifferentheightalongthe wall provided0.24%of minimumvertical reinforcement. Average thicknessof wall =(120+200) / 2 = 160 mm. Ash = 0.24ร—160ร—1000 100 = 384 mm2 Provide 8 mmรธ. Aรธ = 50.26 mm2 Spacing= 50.26ร—1000 384 = 130mm c/c. Designof ringbeamB3:- Thickness=100 mm Rise = 1.5 m (centre) Base dia.= 8 m Raidusof curvature = 6 m Cos 41.8o = 0.745 The ring beamconnectthe tank wall withinconical dome.The vertical loadatthe junctionof the wall withconical dome istransferredtothe ringbeamB3 by horizontal thrust.Inthe conical dome the horizontal componentof thrustcauseshooptensionatthe junction. W = Load transferredthroughthe tankwall atthe topof conical dome /unitlength. ร˜o = Inclinationof conical dome. T = Meridional thrustinconical dome at the junction. tan ร˜o = 2/2.5
18 ร˜o = 38.65
19 = 30897.15 sin18.92 = 10018.32 N/m (ii) Loaddue to ringbeamB1 = 320 mmdepth = 400 mm width = 0.32ร—(0.4-0.1)ร—1ร—25000 = 2400 N/m (iii) Loaddue totank wall = 2 ร— ( 0.12+0.2 2 )ร— 1 ร— 25000 = 8000 N/m (iv) Seif loadof beamB3 (1mร— 0.6m say) =(1-0.3) ร—0.6ร—25000 = 10500 N/m Total W = 10018.32 + 2400 + 8000 + 10500 = 30918.32 N/m Sinร˜o = sin38.65 = 0.62 , cos 38.65 = 0.78 Force Pw due to load Pw1 = W tan ร˜o = 30918.32 tan 38.65 = 24725.97 N/m Force Pw causeddue to waterpressure attop of conical dome Pw2 = rw ร— hd3 h = depthof wateruptocentre of ringbeam d3 = depthof ringbeam Pw2 = 9800 ร— 2 ร— 0.6 = 11760 N/m Hence hooptensioninthe ringbeamis givenby:- P = ( โ„Žโ„Ž1+โ„Žโ„Ž2 2 ) ร— โ„Ž =( 24725.97+11760 2 )ร— 12 = 218915.82 โ„Ž Thisis to be resistedbysteel hoopsthe areaof whichis
20 Ash = 218915.82 150 = 1460 mm2 Use 20 mm bars = 314.15 Numberof 20 mm bars = 1460 314.15 = 4.64 = 5 bars Hence provide 5 ringof 20 mm diabars. Actual area As= ฯ€/4 ร— 20 ร— 5 = 1570 mm2 Stressinequivalentsection= 218915.82 (1000ร—600)+10ร—1570 = 0.35 N/mm2 <1.2 N/mm2 (SAFE) (OK) The 10 mmdiameterdistributionbara(vertical bars) providedinthe wall@100 mmc/c shouldbe takenroundthe above ring to act as stirrups. Designof conical dome :- (a)Meridionalthrust:-
21 Ww = Total weightof wateron the conical dome W = Weightof top dom,cylindrical wall etc. Ws = Self weightof conical dome Ww = โ„Žโ„Ž[ โ„Ž 4 (โ„Ž2 + โ„Žโ„Ž 2 )+ โ„Ž 12 (โ„Ž2 + โ„Žโ„Žโ„Ž 2 + โ„Žโ„Žโ„Ž)โˆ’ โ„Ž 4 โ„Žโ„Ž 2 ร— โ„Žโ„Ž] =9800[ โ„Ž 4 (122 + 82 ) + โ„Ž 12 ร— 2.5(122 + 82 + 12ร— 8) โˆ’ โ„Ž 4 82 ร— 2.5] =9800[326.72+ 198.96โˆ’ 125.67] =3920098 N Let the thicknessof conical slab= 400 mm Ww = [โ„Ž( โ„Ž+โ„Žโ„Ž 2 ) ร— โ„Žร— โ„Ž0] ร— โ„Žโ„Ž l = โˆš2.5 2 + 22 = 3.2 m Ws = 25000โ„Ž( 12+8 2 )ร— 3.2 ร— 0.4 = 1005309.649 N WeightW at B3 = 30918.32 N/m. Hence vertical loadW2 per metre runis givenby W2 = โ„Žโ„Žโ„Ž+โ„Žโ„Ž+โ„Žโ„Ž โ„Žโ„Žโ„Ž = โ„Žร—12ร—30918.32+3920098 +1005309.64 โ„Žร—8 =242353.22 N/m Meridional thrustTo in the conical dome To = W2 / cos ร˜o = 242353.22 / cos 38.65 = 310321.06 N/m. Meridional stress =310321.06 / (1000ร—400) = 0.775 N/mm2 < 5 N/mm2 (Safe). (b) Hoop stress
22 Diameterof conical dome at anyheighthโ€™above base is Dโ€™ = 8 + (12-8)/2 ร— hโ€™= 8 + 2hโ€™ Intensityof waterpressure P = [(4+2)-hโ€™]ร—9800 = [6-hโ€™]9800 N/mm2 Self weight q = 0.4 ร— 25000 = 10000 N/mm2 Hoop tension โ„Žโ„Ž โ€ฒ = ( โ„Ž โ„Žโ„Žโ„Žร˜โ„Ž + โ„Ž โ„Žโ„Žโ„Žร˜โ„Ž ) โ„Žโ€ฒ 2 = ( (6โˆ’โ„Žโ€ฒ)9800 โ„Žโ„Žโ„Ž38.65 + 1000 โ„Žโ„Žโ„Ž38.65 ) (8+2โ„Žโ€ฒ) 2 Poโ€™= 333150.48 + 12548.4 hโ€™2 + 12548.4hโ€™2 hโ€™ Hoop Tension 0 333150.48 N 1 353695.41 N 2 349144.86 N 2.5 337457.95 N For maximum โ„Žโ„Žโ„Žโ€ฒ โ„Žโ„Žโ€ฒ = 0 33093.99 โ€“ 25096.8 hโ€™= 0 hโ€™= 1.31 m. Maximum Poโ€™= 354969.2977 N (c) Designof walls:-
23 Meridional stress=0.775 N/mm2 Maximumhoopstress= 354969.29 N Whichis to be resistedbysteel As = 354969.29/150 = 2366.46 mm2 Areaof eachface = 1183.23 mm2 Spacingof 16 mmdia bars = (1000 ร— 201)/1183.23 = 170 mm c/c Hence provide 16 mm diahoops@170 mm c/c on eachface. Actual As = (1000 ร— 201)/170 = 1182 mm2 Maximumtensile stressincomposite section= 354969.29 (400ร—1000)+(2ร—1182ร—10) =0.83 N/mm2 < 1.2 N/mm2 (Safe) (OK) In the meridional directionthe reinforcement= .21ร—400ร—1000 100 = 840 mm2 Or 420 mm2 on eachface Use 10mm diameterbarsAโ„Ž = 78.53 mm2 Spacing= 1000ร—78.53 420 = 180 mmc/c Hence provide 10 mm bars @ 180 mm c/c on eachface witha clearcover20 mm DESIGN OFBOTTOM DOME R2= 6 m ฦŸ2= 41.8 Weightof wateron w0 onthe dome isgivenby W0=[ โ„Ž 4 ร— โ„Ž0 2 โ„Ž0 โ€“ โ„Žโ„Ž2 2 3 (3โ„Ž2 โˆ’ โ„Ž2)]ฯ’w D0=8,H0 =2.5 ,R2=6,h2=1.5, W0=850471.35N Let the thicknessof bottomdome =250mm Self WeightWs =2ฯ€R2h2t2ร—25000
24 R2=6, h2=1.5, t2=0.25 Ws=353429.17N Total Weight=1203900.52N Meridional thurst=T2= 1203900.52 โ„Žโ„Žโ„Žโ„Žโ„Žโ„Ž41.8 D0=8 T2=71866.98N/m Meridional Stress= 718866.98 250 ร— 1000 =0.287N/mm2 <5N/mm2 safe o.k. Intensityof loadperunitarea P2= 1203900.52 2โ„Žโ„Ž2โ„Ž2 R2 =6,h2=1.5 P2=21289.63N/m2 Max hoopstressat centre of dome โ„Ž2โ„Ž2 2โ„Ž2 = (21289.63ร—6) 2ร—.25 =255475.625N/mm2 =0.25N/mm2 <5MPa O.K. Areaof minsteel=0.26ร— 250 ร— 1000 100 =650 mm2 ineachdirection Use 10 mm ฦŸbars Spacing=1000ร— 78.5 650 =120mm Hence provide 10mm แถฒ bars @ 120mm c/c inbothdirectionalsoprovide 16mmรธ meridoinal bars @170mm c/c nearwaterface. Designof bottomcircularBeam B2 Inwardthrustfrom conical dome =T0Sinรธo =310321.06 Sin38.65 =193814.5 N/m
25 Outwardthrustfrombottomdam=T2cosฦŸ2 =7866.98cos41.8 =56126.36N/m Netinwardthrust=137688.14N/m Hoop compressioninbeam=137688.14ร—8/2 =550752.56N Vertical loadonbeam,permeterrun=T0Cosรธ0+T2SinฦŸ2 =310321.06 Cos38.6 +71866.98 Sin41.8 =290423.93 N/m Self weightof beam=0.6ร—1.2ร—25000 =18000N/m Total loadonbeam=290423.93+18000=308423.90N/m Let ussupportthe beamon 8 equallyspacedcolumnata meandia 8m Mean radiusof curvedbeamR=4m For support8:-coefficientof B.H.&twistingmomentincircularbeam 2ฦŸ=45o C1=0.066,C2=0.030,C3=0.005 ร˜m=9.5o ,ฦŸ=ฯ€/4=22.5=ฯ€/8 Radius wR2 ฦŸ=308423.9ร—42 ร—ฯ€/4 =3875769.4 Max โ€“ve B.M.at supportM0=C1wR2 2ฦŸ =255800.78N.m Max +ve B.M. at supportMc=C2wR2 2ฦŸ =116273.08N.m Max torsional moment โ„Žโ„Ž โ„Ž =C3 wR2 2ฦŸ =19378.84Nm For ฯƒcbc=8.5
26 Hysd bars ฯƒst=150 N/mm2 Neuteral axisdepthfactor(K) K= โ„Žโ„Žโ„Žโ„Žโ„Ž โ„Žโ„Žโ„Žโ„Žโ„Ž+โ„Žโ„Žโ„Ž m= 280 3โ„Žโ„Žโ„Žโ„Ž = 280 3ร—8.5 =10.98 =10.98 ร— 8.5 10.98ร—8.5+150 =0.383 LeverArm J=1-K/3=0.872 R=1/2ร—ฯƒcbcร—Jร—k=1/2ร—8.5ร—0.872ร—0.383 1.41 Mr=Rbd2 Reqeff.Depth(d)- 255800.78=1.41ร—600ร—d2 d=550mm Howeverkeeptotal depth=700mm fromshearpointof view. Max shearforce at support Fo=WRฦŸ =308423.9ร—4ร—ฯ€/8 =484471.12N S.F.at any pointF=WR(ฦŸ-ฯ†) =308423.9ร—4ร—(22.5-9.5) ร—ฯ€/180 =279916.6N B.M. at the pointyof max torssional momentฯ†m=9.50 Mฯ†=WR2 (ฦŸSinฯ†+ฦŸCosฦŸCosฯ†-1) sagging =308423.9ร—42 ร—(ฯ€/8ร—sin9.5+ฯ€/8ร—cot22.5ร—cos9.5-1) =4934.78Nm sagging The torsionmomentat any point- Mpt =WR2 [ฦŸcosฯ†-ฦŸcosฯ†sinฯ†-(ฦŸ-ฯ†)]
27 At the support ฯ†=0 M0 t =WR2 (ฦŸ-ฯ†)=0 At the midspan ฯ†=ฦŸ=22.5=ฯ€/8 radian Mฯ† t = WR2 [ฦŸcosฦŸ]-[ ฦŸโ„Žโ„Žโ„Žรธโ„Žโ„Žโ„Žรธ โ„Žโ„Žโ„Žรธ ]=0 Hence we have the followingcombinationof B.M.& torsional moment:- (a)atthe support M0 =255800.78 NM(hoggingornegative) M0 t =0

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Structural design of 350 kl overhead water tank at telibagh,lucknow

  • 1. 1 STRUCTURAL DESIGN OF 350KL OVERHEAD WATER TANK AT INDIRA GANDHI NATIONAL OPEN UNIVERSITY, TELIBAGH LUCKNOW
  • 2. 2 DATA 1. Type of Tank: Intze Tank 2. Capacityof the tank: 350KL 3. Type of staging: Column& Brace type 4. Depthof foundation: 2.5m 5. Safe BearingCapacityof Soil: 100KN/m2 6. Type of foundation: CircularRing&Raft foundation 7. Grade of Concrete: M-25 8. Grade of Steel: Fe-415 9. Heightof staging: 25m 10. Type of soil: SoftClay 11. Heightof BuildinguptoTerrace: 15.6m 12. No.of floorsinBuilding: G+3 13. Basic WindPressure: 1500N/m2 14. SesmicZone of Lucknow: Zone 3 15. No.of studentinCollege: 2000 16. Water consumptionrate (Percapitademandinlitresperdayper head): 45 17. Designperiodfortank: 30 years 18. No.of studentinhostels: 1600
  • 3. 3 OBJECTIVE 1:- To make a studyaboutthe analysisanddesignof watertank 2:- To make a studyaboutthe guidelinesforthe designof liquidretainingstructure accordingto IS Code IS: 3370 part 2-2009 IS: 456:2000 3:- To knowabout the designphilosophyforthe safe andeconomical designof watertank 4:- To estimate the overall costformakingthe Intze Tank
  • 4. 4 WATER QUANTITY ESTIMATION IN COLLEGE CAMPUS Populationorthe numberof studentstobe servedin2014 = 2000 Let populationtobe increasedatrate of 10% per decade Numberof students(2014) = 2000 Numberof studentsin2024 = 2200 Numberof studentsin2034 = 2420 Numberof studentsin2044 = 2662 Quantity = per capitademandร— Population = 45 ร— 2662 = 1,19,790 litres = 120 KL (assume)
  • 5. 5 FLUCTUATION IN RATE OF DEMAND Average dailypercapitademandincollege campus = 45 lpcd If this average suppliedatall the timesitwill notbe sufficienttomeetthe fluctuation. HOURLY VARIATION (1) Duringthe entryof college from8to 9 inthe morning. (2) Duringthe lunchfrom12 to 1 in the afternoon.
  • 6. 6 WATER CONSUMPTION IN HOSTEL Average dailypercapitademandinhostels=135 lpcd. Quantity = 136 ร— 1600 = 216 KL Total quantity = 216 + 130 = 346 KL อŒ 350 KL
  • 7. 7 DESIGN REQUIREMENTOFTANK * Concrete mix weakerthanM-20 isnot usedbecause of highergrade lesserporosityof concrete. * Minimumquantityof cementinconcrete shall be notlessthan30 KN/m3 . * Use of small size bars. * Coefficientof expansiondue totemperature=11ร—10-6 /หšC * Coefficientof shrinkage maybe taken= 450 ร— 10-6 forinitial and200 ร— 10-6 fordrying shrinkage. * Minimumcovertoall reinforcementshouldbe 20 mmor the diameterof mainbarwhichever isgreater. * Anoverheadliquidretainingstructure isdesignusingworkingstressmethodavoidingthe cracking inthe tank and to preventthe leakage andthe componentof tankcanbe designusing LIMIT STATE METHOD (example:-column,foundation,bracing,stairsetc.). * Code usingIS:3370-PART 2-2009 IS: 456:2000 * The leakage ismore withhigherliquidheadandithas beenobservedthadwaterheadupto 15m doesnotcause leakage problem. * Inorder to minimizecrackingdue toshrinkage andtemperature,minimumreinforcementis recommendedas- (i) For thicknessโ‰ค100 mm = 0.3% (ii) Forthicknessโ‰ฅ450 mm = 0.2% For thicknessbetween100mm to 450 mm= varieslinearlyfrom0.3% to0.2% * For concrete thicknessโ‰ฅ225 mm, twolayerof reinforcementbe placedone nearwaterface and otherawayfrom waterface.
  • 8. 8 FROM IS -3370 (i) For loadcombinationwaterloadtreatedasdeadload. (ii) Crackingโ€“ The maximumcalculatedsurface widthof concrete fordirecttensionandflexure or restrainedtemperatureandmoisture effectshall notexceed0.2mmwithspecified cover. (iii) Shrinkagecoefficientmaybe assumed= 300 ร— 10-6 . (iv) Bar spacingshouldgenerallynotexceedthan300 mm or the thicknessof the section whicheverisless.
  • 9. 9 DETERMINATION OFFOUNDATION โ„Ž = โ„Ž โ„Ž ( 1 โˆ’ โ„Žโ„Žโ„Žโ„Ž 1 + โ„Žโ„Žโ„Žโ„Ž ) = 100 17 ( 1โˆ’โ„Žโ„Žโ„Ž12 1+โ„Žโ„Žโ„Ž12 ) -Fromtestingof soil sample For clay โˆ…=12หš r = densityof soil = 1.76 gm/cm3 = 17.6 KN/m3 p = 100 KN/m3 โ„Ž = 2.52 โ„Ž
  • 10. 10 9. DETERMINATION OFHEIGHTOFSTAGGING We knowturbulentflow occursina pipe So Re > 4000 f = 0.079 โ„Žโ„Ž 1/4 L = lengthof pipe, v = meanvelocityinpipe of flow d = diameterof pipe โ„Žd= 4โ„Žโ„Žโ„Ž2 2โ„Žโ„Ž โ„Ž = โ„Žโ„Ž 2 The kinematicviscosityof water(โ„Ž) =0.01ร—10-4 m2 /s Assume diameterof pipe = 15 cm โ„Ž = โ„Žโ„Ž โ„Ž = โ„Ž 4 ร— 0.15 2 = 0..176โ„Ž2 = Volume (V) =350 m3 Onlyforone hour maximumvelocityoccursinthe pipe sothe discharge duringthatperiod โ„Ž = โ„Ž โ„Ž = 350 60ร—60 = 0.097 โ„Ž3 โ„Ž โ„Ž = โ„Žโ„Ž 0.097= 0.0176 ร— โ„Ž โ„Ž = 5.52 m/sec. Maximumvelocity=5.52 m/sec. โ„Žโ„Ž = โ„Žโ„Ž โ„Ž = 5.52ร—0.15 0.01ร—10โˆ’4 = 8.2 ร— 105 (O.K.) โ„Ž = 0.079 (8.2 ร— 105 ) 1 4 = 2.61 ร— 10โˆ’3
  • 11. 11 Minimumlengthof pipe requirement = 2 ร— heightof buildingupto3 storeysfromthe level +lateral distance uptothe centre of tank = 2 ร— 15.6 + 18 = 49.2 m โ‰ˆ 50 m Headloss โ„Žโ„Ž = 4ร—2.61ร—10โˆ’3 ร—50ร—5.522 2ร—9.81ร—0.15 = 5.40 m HEIGHT OF STAGGING Total hydrostaticpressure ontank P = ฯgh Total head= โ„Ž โ„Ž + โ„Ž2 2โ„Ž + โ„Ž + โ„Žโ„Ž+ โ„Žโ„Žโ„Žโ„Žโ„Ž โ„Žโ„Žโ„Žโ„Žโ„Žโ„Ž Minor loss(assume) =1 m. = โ„Žโ„Ž โ„Ž + โ„Ž2 2โ„Ž + โ„Ž+ โ„Žโ„Ž+ 1 = 4.5 + 5.522 2ร—9.81 + 15.6+ 5.4 + 1 = 28.08 โ„Ž Usingtotal head= 29.5 Heightof stagging= 29.5 โ€“ 4.5 = 25 m
  • 12. 12 DESIGN OF TOP DOME Assume thicknessof topdome =100 mm. Meridional thrustatedges โ„Ž1 = โ„Žโ„Ž1 1+โ„Žโ„Žโ„Žโ„Ž1 Deadload of top dome = 0.100 ร— 25 = 2.5 KN/m2 Live loadon topdome = 0.75 KN/m2 (assume) Total load P = 3.25 KN/m2 โ„Ž1 = 3.25 ร— 103 ร— 18.5 1 + โ„Žโ„Žโ„Ž 18.92 = 30897.15 N/m Meridional stress= 30897.15 100ร—100 = 0.308MPa < 5 MPa (OK) Maximumhoopstressoccurs at the centre and itsmagnitude โ„Žโ„Ž1 2โ„Ž1 = 3.25ร—103 ร—18.5 2ร—0.100 =0.30 N/mm2 =0.3 MPa < 5MPa (OK) Provide nominal reinforcementof 0.24%. โ„Žโ„Žโ„Ž = 0.24ร—100ร—1000 100 = 240โ„Žโ„Ž2 Use 8 mmbars. โ„Žโ„Ž = 50 โ„Žโ„Ž2 Spacing = 1000ร—50 240 = 208.33 = 205 mm c/c. Provide 8 mmbars @ 205 mm c/c radiallyandcircumtentiallyasshowninfigure. The 205 mm c/c for radial bar isprovidedatthe springingof the dome. At the crown the spacingreducestozero. Hence the curtailmentof radial barsmay be carriedout at the appropriate distance.
  • 13. 13
  • 14. 14 DIMENSION OF TANK Innerdiameterof cylindrical portion D= 12 m Rise of top dome h1 = 1 m Rise of bottomdome h2 = D/8 = 1.5 m (centre) Free board= 0.15 m Diameterof ringbeamDo = 5/8 D = 7.5 = 8 m Rise of bottomdome (side) ho = 3/16 ร— D = 2.25 m = 2.5 m Capacityof tank:- โ„Ž = โ„Žโ„Ž2 โ„Ž 4 + โ„Žโ„Žโ„Ž 12 (โ„Ž2 + โ„Žโ„Ž 2 + โ„Žโ„Žโ„Ž)โˆ’ โ„Žโ„Ž2 2 (3โ„Ž2โˆ’โ„Ž2) 3 Radiusof bottomcircular dome:- 1.5 ร— (2R2 โ€“ 1.5) = 42 2R2 โ€“ 1.5 = 10.67 R2 =6 m SinฦŸ2 = 4 6 ฦŸ2 = 41.8o โ„Ž = โ„Žโ„Ž2 โ„Ž 4 + โ„Žโ„Žโ„Ž 12 (โ„Ž2 + โ„Žโ„Ž 2 + โ„Žโ„Žโ„Ž) โˆ’ โ„Žโ„Ž2 2 (3โ„Ž2โˆ’โ„Ž2) 3 350 = โ„Žร—122 ร—โ„Ž 4 + โ„Žร—2 12 (122 + 82 + 12 ร— 8) โˆ’ โ„Žร—1.52 (3ร—6โˆ’1.5) 3 350 = 113โ„Ž + 160โˆ’ 38.87 โ„Ž = 2 โ„Ž Radiusof top circulardome:- 1 ร— (2R1-1) = 6 ร— 6 R1 = 18.5 m
  • 15. 15 SinฦŸ1 = 6/18.5 ฦŸ1 = 18.92o Designof top ringbeam:- A ringbeamis providedatthe junctionof topdome and the vertical wall toresisthooptension inducedbythe top dome. Horizontal componentof meridional thrust P1 = T1 cos ฦŸ1 = 30897.15 cos 18.92o = 29227.8 N/m. Total hoop tension tending to rupture of beam = โ„Ž1ร—โ„Ž 2 = 29227.8ร—12 2 = 175366.8โ„Ž Permissible stress in HYSD bars = 150 N/m2 Ash = 175366.8/150 = 1170 mm2 Provide 20 mm bars (314.15) as hoop. Number of 12 mm bars = 1170 / 314.15 = 3.72 = 4 Actual Ash = 4 ร— โ„Ž/4 ร— 202 = 1256.63 mm2 = 1257 mm2 Provide 4-20 mm รธ hoop and 8 mm bars tie @ 205 mm c/c. Hence the cross sectional area of concrete 1.3= 175366.8 โ„Ž+1257ร—8 Ac = 124841.53 Provide ring beam of 320 mm ร— 400 mm.
  • 16. 16 Designof cylindrical wall:- In the membrane analysisthe tankwall isassumedtobe free attop andbottom maximumhoop tensionoccursat the base of the wall and itsmagnitude:- = โ„Žโ„Žโ„Žโ„Ž 2 = 9800ร—โ„Žร—12 2 = 58800 โ„Ž Hoop tensionatanydepthx fromthe top X (m) Hoop tension(N/m) 0 0 1 58800 2 117600 Minimumthicknessof cylindrical wall =3 H + 5 = 3 ร— 2 + 5 = 11 cm. Provide 20 cm at the bottomand taperit to12 cm at top. At x = 1 m. Areaof steel Ash = 58800/150 = 392 mm2 Provide 8 mmbars. Aรธ = 50.26 mm2 Spacing= (1000 ร— 50.26) / 392 = 130 mm c/c. At x = 2 m. Areaof steel Ash = 117600/150 = 784 mm2 Provide 10 mm bars. Aรธ = 78.53 mm2 Spacing= (1000 ร— 78.53) / 784 = 100 mm c/c.
  • 17. 17 The hoop steel maybe curtailedaccordingtohooptensionatdifferentheightalongthe wall provided0.24%of minimumvertical reinforcement. Average thicknessof wall =(120+200) / 2 = 160 mm. Ash = 0.24ร—160ร—1000 100 = 384 mm2 Provide 8 mmรธ. Aรธ = 50.26 mm2 Spacing= 50.26ร—1000 384 = 130mm c/c. Designof ringbeamB3:- Thickness=100 mm Rise = 1.5 m (centre) Base dia.= 8 m Raidusof curvature = 6 m Cos 41.8o = 0.745 The ring beamconnectthe tank wall withinconical dome.The vertical loadatthe junctionof the wall withconical dome istransferredtothe ringbeamB3 by horizontal thrust.Inthe conical dome the horizontal componentof thrustcauseshooptensionatthe junction. W = Load transferredthroughthe tankwall atthe topof conical dome /unitlength. ร˜o = Inclinationof conical dome. T = Meridional thrustinconical dome at the junction. tan ร˜o = 2/2.5
  • 19. 19 = 30897.15 sin18.92 = 10018.32 N/m (ii) Loaddue to ringbeamB1 = 320 mmdepth = 400 mm width = 0.32ร—(0.4-0.1)ร—1ร—25000 = 2400 N/m (iii) Loaddue totank wall = 2 ร— ( 0.12+0.2 2 )ร— 1 ร— 25000 = 8000 N/m (iv) Seif loadof beamB3 (1mร— 0.6m say) =(1-0.3) ร—0.6ร—25000 = 10500 N/m Total W = 10018.32 + 2400 + 8000 + 10500 = 30918.32 N/m Sinร˜o = sin38.65 = 0.62 , cos 38.65 = 0.78 Force Pw due to load Pw1 = W tan ร˜o = 30918.32 tan 38.65 = 24725.97 N/m Force Pw causeddue to waterpressure attop of conical dome Pw2 = rw ร— hd3 h = depthof wateruptocentre of ringbeam d3 = depthof ringbeam Pw2 = 9800 ร— 2 ร— 0.6 = 11760 N/m Hence hooptensioninthe ringbeamis givenby:- P = ( โ„Žโ„Ž1+โ„Žโ„Ž2 2 ) ร— โ„Ž =( 24725.97+11760 2 )ร— 12 = 218915.82 โ„Ž Thisis to be resistedbysteel hoopsthe areaof whichis
  • 20. 20 Ash = 218915.82 150 = 1460 mm2 Use 20 mm bars = 314.15 Numberof 20 mm bars = 1460 314.15 = 4.64 = 5 bars Hence provide 5 ringof 20 mm diabars. Actual area As= ฯ€/4 ร— 20 ร— 5 = 1570 mm2 Stressinequivalentsection= 218915.82 (1000ร—600)+10ร—1570 = 0.35 N/mm2 <1.2 N/mm2 (SAFE) (OK) The 10 mmdiameterdistributionbara(vertical bars) providedinthe wall@100 mmc/c shouldbe takenroundthe above ring to act as stirrups. Designof conical dome :- (a)Meridionalthrust:-
  • 21. 21 Ww = Total weightof wateron the conical dome W = Weightof top dom,cylindrical wall etc. Ws = Self weightof conical dome Ww = โ„Žโ„Ž[ โ„Ž 4 (โ„Ž2 + โ„Žโ„Ž 2 )+ โ„Ž 12 (โ„Ž2 + โ„Žโ„Žโ„Ž 2 + โ„Žโ„Žโ„Ž)โˆ’ โ„Ž 4 โ„Žโ„Ž 2 ร— โ„Žโ„Ž] =9800[ โ„Ž 4 (122 + 82 ) + โ„Ž 12 ร— 2.5(122 + 82 + 12ร— 8) โˆ’ โ„Ž 4 82 ร— 2.5] =9800[326.72+ 198.96โˆ’ 125.67] =3920098 N Let the thicknessof conical slab= 400 mm Ww = [โ„Ž( โ„Ž+โ„Žโ„Ž 2 ) ร— โ„Žร— โ„Ž0] ร— โ„Žโ„Ž l = โˆš2.5 2 + 22 = 3.2 m Ws = 25000โ„Ž( 12+8 2 )ร— 3.2 ร— 0.4 = 1005309.649 N WeightW at B3 = 30918.32 N/m. Hence vertical loadW2 per metre runis givenby W2 = โ„Žโ„Žโ„Ž+โ„Žโ„Ž+โ„Žโ„Ž โ„Žโ„Žโ„Ž = โ„Žร—12ร—30918.32+3920098 +1005309.64 โ„Žร—8 =242353.22 N/m Meridional thrustTo in the conical dome To = W2 / cos ร˜o = 242353.22 / cos 38.65 = 310321.06 N/m. Meridional stress =310321.06 / (1000ร—400) = 0.775 N/mm2 < 5 N/mm2 (Safe). (b) Hoop stress
  • 22. 22 Diameterof conical dome at anyheighthโ€™above base is Dโ€™ = 8 + (12-8)/2 ร— hโ€™= 8 + 2hโ€™ Intensityof waterpressure P = [(4+2)-hโ€™]ร—9800 = [6-hโ€™]9800 N/mm2 Self weight q = 0.4 ร— 25000 = 10000 N/mm2 Hoop tension โ„Žโ„Ž โ€ฒ = ( โ„Ž โ„Žโ„Žโ„Žร˜โ„Ž + โ„Ž โ„Žโ„Žโ„Žร˜โ„Ž ) โ„Žโ€ฒ 2 = ( (6โˆ’โ„Žโ€ฒ)9800 โ„Žโ„Žโ„Ž38.65 + 1000 โ„Žโ„Žโ„Ž38.65 ) (8+2โ„Žโ€ฒ) 2 Poโ€™= 333150.48 + 12548.4 hโ€™2 + 12548.4hโ€™2 hโ€™ Hoop Tension 0 333150.48 N 1 353695.41 N 2 349144.86 N 2.5 337457.95 N For maximum โ„Žโ„Žโ„Žโ€ฒ โ„Žโ„Žโ€ฒ = 0 33093.99 โ€“ 25096.8 hโ€™= 0 hโ€™= 1.31 m. Maximum Poโ€™= 354969.2977 N (c) Designof walls:-
  • 23. 23 Meridional stress=0.775 N/mm2 Maximumhoopstress= 354969.29 N Whichis to be resistedbysteel As = 354969.29/150 = 2366.46 mm2 Areaof eachface = 1183.23 mm2 Spacingof 16 mmdia bars = (1000 ร— 201)/1183.23 = 170 mm c/c Hence provide 16 mm diahoops@170 mm c/c on eachface. Actual As = (1000 ร— 201)/170 = 1182 mm2 Maximumtensile stressincomposite section= 354969.29 (400ร—1000)+(2ร—1182ร—10) =0.83 N/mm2 < 1.2 N/mm2 (Safe) (OK) In the meridional directionthe reinforcement= .21ร—400ร—1000 100 = 840 mm2 Or 420 mm2 on eachface Use 10mm diameterbarsAโ„Ž = 78.53 mm2 Spacing= 1000ร—78.53 420 = 180 mmc/c Hence provide 10 mm bars @ 180 mm c/c on eachface witha clearcover20 mm DESIGN OFBOTTOM DOME R2= 6 m ฦŸ2= 41.8 Weightof wateron w0 onthe dome isgivenby W0=[ โ„Ž 4 ร— โ„Ž0 2 โ„Ž0 โ€“ โ„Žโ„Ž2 2 3 (3โ„Ž2 โˆ’ โ„Ž2)]ฯ’w D0=8,H0 =2.5 ,R2=6,h2=1.5, W0=850471.35N Let the thicknessof bottomdome =250mm Self WeightWs =2ฯ€R2h2t2ร—25000
  • 24. 24 R2=6, h2=1.5, t2=0.25 Ws=353429.17N Total Weight=1203900.52N Meridional thurst=T2= 1203900.52 โ„Žโ„Žโ„Žโ„Žโ„Žโ„Ž41.8 D0=8 T2=71866.98N/m Meridional Stress= 718866.98 250 ร— 1000 =0.287N/mm2 <5N/mm2 safe o.k. Intensityof loadperunitarea P2= 1203900.52 2โ„Žโ„Ž2โ„Ž2 R2 =6,h2=1.5 P2=21289.63N/m2 Max hoopstressat centre of dome โ„Ž2โ„Ž2 2โ„Ž2 = (21289.63ร—6) 2ร—.25 =255475.625N/mm2 =0.25N/mm2 <5MPa O.K. Areaof minsteel=0.26ร— 250 ร— 1000 100 =650 mm2 ineachdirection Use 10 mm ฦŸbars Spacing=1000ร— 78.5 650 =120mm Hence provide 10mm แถฒ bars @ 120mm c/c inbothdirectionalsoprovide 16mmรธ meridoinal bars @170mm c/c nearwaterface. Designof bottomcircularBeam B2 Inwardthrustfrom conical dome =T0Sinรธo =310321.06 Sin38.65 =193814.5 N/m
  • 25. 25 Outwardthrustfrombottomdam=T2cosฦŸ2 =7866.98cos41.8 =56126.36N/m Netinwardthrust=137688.14N/m Hoop compressioninbeam=137688.14ร—8/2 =550752.56N Vertical loadonbeam,permeterrun=T0Cosรธ0+T2SinฦŸ2 =310321.06 Cos38.6 +71866.98 Sin41.8 =290423.93 N/m Self weightof beam=0.6ร—1.2ร—25000 =18000N/m Total loadonbeam=290423.93+18000=308423.90N/m Let ussupportthe beamon 8 equallyspacedcolumnata meandia 8m Mean radiusof curvedbeamR=4m For support8:-coefficientof B.H.&twistingmomentincircularbeam 2ฦŸ=45o C1=0.066,C2=0.030,C3=0.005 ร˜m=9.5o ,ฦŸ=ฯ€/4=22.5=ฯ€/8 Radius wR2 ฦŸ=308423.9ร—42 ร—ฯ€/4 =3875769.4 Max โ€“ve B.M.at supportM0=C1wR2 2ฦŸ =255800.78N.m Max +ve B.M. at supportMc=C2wR2 2ฦŸ =116273.08N.m Max torsional moment โ„Žโ„Ž โ„Ž =C3 wR2 2ฦŸ =19378.84Nm For ฯƒcbc=8.5
  • 26. 26 Hysd bars ฯƒst=150 N/mm2 Neuteral axisdepthfactor(K) K= โ„Žโ„Žโ„Žโ„Žโ„Ž โ„Žโ„Žโ„Žโ„Žโ„Ž+โ„Žโ„Žโ„Ž m= 280 3โ„Žโ„Žโ„Žโ„Ž = 280 3ร—8.5 =10.98 =10.98 ร— 8.5 10.98ร—8.5+150 =0.383 LeverArm J=1-K/3=0.872 R=1/2ร—ฯƒcbcร—Jร—k=1/2ร—8.5ร—0.872ร—0.383 1.41 Mr=Rbd2 Reqeff.Depth(d)- 255800.78=1.41ร—600ร—d2 d=550mm Howeverkeeptotal depth=700mm fromshearpointof view. Max shearforce at support Fo=WRฦŸ =308423.9ร—4ร—ฯ€/8 =484471.12N S.F.at any pointF=WR(ฦŸ-ฯ†) =308423.9ร—4ร—(22.5-9.5) ร—ฯ€/180 =279916.6N B.M. at the pointyof max torssional momentฯ†m=9.50 Mฯ†=WR2 (ฦŸSinฯ†+ฦŸCosฦŸCosฯ†-1) sagging =308423.9ร—42 ร—(ฯ€/8ร—sin9.5+ฯ€/8ร—cot22.5ร—cos9.5-1) =4934.78Nm sagging The torsionmomentat any point- Mpt =WR2 [ฦŸcosฯ†-ฦŸcosฯ†sinฯ†-(ฦŸ-ฯ†)]
  • 27. 27 At the support ฯ†=0 M0 t =WR2 (ฦŸ-ฯ†)=0 At the midspan ฯ†=ฦŸ=22.5=ฯ€/8 radian Mฯ† t = WR2 [ฦŸcosฦŸ]-[ ฦŸโ„Žโ„Žโ„Žรธโ„Žโ„Žโ„Žรธ โ„Žโ„Žโ„Žรธ ]=0 Hence we have the followingcombinationof B.M.& torsional moment:- (a)atthe support M0 =255800.78 NM(hoggingornegative) M0 t =0
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