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Class_25_Example_4.5(2)

# Class_25_Example_4.5(2) - 12-kip single = 18-kip tandem =...

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Rigid Pavement Design

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Given Service life = 20 years Initial PSI = 4.4; TSI = 2.5 Mod. Subgrade Reaction = 300lb./in 3 cars, pickups, light van volume = 20,000/day Truck traffic = 200/day single unit; 410/day tractor semi- trailer Axle weights: Cars et.al. = two 2000lb. Single axles SU Trucks = 10,000 steering axle (single) 20,000 drive axle (tandem) Tractor semi: 12,000 single (steering) 18,000 drive (tandem) 50,000 trailer (triple)
Given (continued) Reliability = 95% Overall standard deviation = 0.45 Concrete modulus = 4.5 million lb./in 2 Modulus of rupture = 900 lb./in 2 Load transfer Coefficient = 3.2 Drainage = 1.0 Determine required slab thickness

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Solution Assume D to start: 10 inches Given D = 10, 18 kip ESALs are: 2-kip single axle = 10-kip single axle (truck) = 22-kip tandem (truck) =

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Unformatted text preview: 12-kip single = 18-kip tandem = 50-kip triple = Computed 18-kip ESAL = 1449.27 Traffic = 1449.27 (365) 20 = 10,579,671 18-kip ESAL Solution (continued) Δ PSI = 4.4 – 2.5 = 1.9 Equation 4.19 gives D = 9.21 in. Are we done yet?? NO Different from assumed 10 in. Need to recalculate ESALs with D = 9 2-kip single axle = 10-kip single axle (truck) = 22-kip tandem (truck) = 12-kip single = 18-kip tandem = 50-kip triple = New 18-kip ESAL = 1418.09 [change from 1449.27] Solution (continued) Traffic over 20 years now: 1418.09 (365) (20) = 10, 352,057 18-kip ESALs Solving for D = 9.17 in. Close to assumed 9 in. and little change from 9.21 in in first iteration. Being conservative, round up to 9.5 in....
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Class_25_Example_4.5(2) - 12-kip single = 18-kip tandem =...

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