pipeline Highway Crossing Design spreadsheet

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simple spreadsheet for calcuting stresses on pipeline crossing a highway.

Transcript of pipeline Highway Crossing Design spreadsheet

initial informationPipe and operational characteristicsvaluesunitsOutside diameter D12.75inOperating Pressurep1000psiSteel GradeX42Specific Minimum Yield Strength SMYS42000psiDesign Factor F0.72Longitudinal Joint Factor E1Installationn Temperature T0Maximum or minimum operating Temperature T0Temperature derating Factor T0Wall Thickness tw0.25inInstallation and Site CharacteristicsDepth H6ftBored Diameter Bd14.8inSoil Typeloose sandModulos of Soil Reaction E'0.5ksiResilient Modulos Er10ksiUnit Weight 120lb/ft30.0694444444lb/in3type of longitudinal wielderwDesign Wheel Load from single axle Ps12kipsDesign Wheel Load from tandem axle Pt10kipsPavement TypeflexibleOther Steel PropertiesYoungs Modulus Es30000ksiPoissons Ratio vs0.3Coeffiecient of thermal expansion T0.0000065/F

calculationsStep 1Check allowable Barlow StresspD/2twF*E*T*SMYSpD/2tw=25500FETSmys30240okStep 2 Circumferential Stress Due to Earth Loadusing tables to obtainKHe3024figure 3 usetw/D=0.02E=0.5Bc1.09figure 4 useH/Bd =4.9soil typeloose sandEe1.11figure 5 useBd/D=1.16She =KHeBeEeD=3239.50725psiStep 3 Impact Factor, Fi, and Applied Design Surface Pressure, w

Fi1.47Use Figure 7 for highwaysH=6refer to section 4.7.2.2.1w=P/Ap69.4444444444psiAp= contact area over load is applied =144inSttep 4 Cyclic Stress

Cyclic Circumferential StressSHh = KHhGHhRLFiw1445.19375psi

KHh14.3figure 14 usetw/D0.02Er10GHh0.99figure 15 useD12.75H6Cyclic Longitudinal StressSLh = KLhGLhRLFiw1020.73125

KLh9.9Figure 16 usetw/D0.02Er10GLh1.01figure 17 useD12.75H6R1L1Table 2 with flexible pavement and TANDEM AXLEH=6D=12.75Step 5Circumferential Stress Due to Internal Pressurization, SHiSHi=p(D-tw)/2tw24500Step 6Principal Stresses, S1,S2,S3 T0 T0 Es30000 vs0.3 T0.0000065S1=SHe+SH+SHi28760.2385S2=SLH-EsT(T2-T1)+vs(SHe+SHi)9342.583425S3=-p-1000Seff=[0,5((S1-S2)2+(S2-S3)2+(S3-S1)2]0,526169.5066450596Allowable Effective Stress = F x SMYS30240Step 7Check FatigueGirth WeldsFrom table 3 useSLh =1020.73125SFG=12000F=0.72SFG x F=8640Longitudinal WeldsFrom Table 3SHh=1445.19375SFL=21000F=0.72SFL x F=15120

tablesCRITICAL AXLE CONFIGURATIONS FOR DESIGN WHELL LOADS OF PS=12Kips (53.4kN) and Pt= 10 Kips(44.5kN)

Depth of burial, H