A Design Example for a Circular Concrete Tank 美国规范实例.pdf

A Design Example for a Circular Concrete Tank 美国规范实例.pdf

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时间:2020-03-06

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1、ADesignExampleforaCircularConcreteTankPCADesignMethodCVEN4830/4434UniversityofColorado,BoulderSpringSemester2008PreparedbyBenBlackardCircularTankExampleD=90fttH=16ftgrade6ftgroundwatertablefluiddensityinsidetank=65pcff′c=4,000psify=60,000psisoilbearingcapacity=2,400psfWallsabovethegroundwat

2、ertableshouldbedesignedusingalateralearthpressureequivalenttothatdevelopedbyafluidweighing40pcf,belowthegroundwatertableuse90pcf.TensileHoopForcesAssumeawallthicknesst=12″2wu=1.65×1.7×65pcf=183pcfH=16ftR=45ftH/Dt=2.844≈3.0FromPCA-CAppendix:coeffromcoeffromlargerhooptensileforceAs=tableA-1ta

3、bleA-5coefTu=coef×wu×H×RTu/0.9fy2top0.1340.0740.13417,656lb0.33in20.1H0.2030.1790.20326,748lb0.50in20.2H0.2670.2810.28137,025lb0.69in20.3H0.3220.3750.37549,410lb0.92in20.4H0.3570.4490.44959,161lb1.10in20.5H0.3620.5060.50666,671lb1.24in20.6H0.3300.5190.51968,384lb1.27in20.7H0.2620.4790.47963

4、,114lb1.17in20.8H0.1570.3750.37549,410lb0.92in20.9H0.0520.2100.21027,670lb0.52inbottom0.0000.000Barswillbeinstalledonbothfacesasillustratedbelow:top″96′-0″A2@s=1.17in#60.375H″86′-8″A2@s=1.32in#60.80H″012As=1.05in3′-4″@#6bottomNotethattheabovesolutionisnotuniqueHoopTensioninConcreteC=0.00032

5、Ag=12″×12″=144in2As=1.32inEc=57,000f′=57,0004,000psi=3,605,000psicEs=29,000,000psin=Es/Ec≈8T=68,384lb/(1.65)(1.7)=24,380lb()()(2)C⋅E⋅A+T0.000329,000,000psi1.32in+24,380lbssf===233psic2()2A+n⋅A144in+81.32ingsfc/f′c=0.058=5.8%seemsreasonableCompressionHoopForcesThewallthicknesst=1′Conservativ

6、elyconsiderthesoiltobesaturated2wu=1.3×1.7×90pcf=199pcfH=6ftR=45ftH/Dt=0.4FromPCA-CAppendix:coeffromcoeffromlargerhoopcomp.forcetableA-1tableA-5coefCu=coef×wu×H×Rtop0.1490.4740.47425,469lb0.1H0.1340.4400.44023,642lb0.2H0.1200.3950.39521,224lb0.3H0.1010.3520.35218,913lb0.4H0.0820.3080.30816,

7、549lb0.5H0.0660.2640.26414,185lb0.6H0.0490.2150.21511,552lb0.7H0.0290.1650.1658,866lb0.8H0.0140.1110.1115,965lb0.9H0.0040.0570.0573,063lbbottom0.0000.000φP=0.55()0.75f′A=0.55()0.75(4,000psi)(12in)(12in)=237,600lbncgVerticalMomentandShearThewallthicknesst

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