0443 040 4 0118 sn 424 8000 lb single 15000

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Unformatted text preview: 4.24 8000 lb single: 15000 lb tandem: 40000 lb single: 40000 lb tandem: GO ON TO THE NEXT PAGE 3 CEE 362 - Homework #3 Solution Manual = (1300 * 0.03932)+(900*0.04308)+(20*22.164) +(200 * 2.042) log10W18 = Z R So 9.36log10 SN 1 - 0.20 log10 PSI / 2.7 5.19 0.40 1094 SN 1 2.32 log10 M R 8.07 Solving equation for = 5250.31 psi 4 CEE 362 - Homework #3 Solution Manual Problem # 3: 200 200 80 single tandem single 20 40 22 5 inch HMA 6 inch HMA 10 inch crushed stone = 3000 lb / = 0.6 ΔPSI = 2.0 Y = 20 years SN = + SN: (.44)(5) + 0.40 (6) + 0.11(10) ► SN = 5.7 20 single: 40 tandem: 22 single: = (200 * 1.538)+(200*2.122)+(80*2.264) log10W18 = Z R S o 9.36log10 SN 1 - 0.20 log10 PSI / 2.7 0.40 1094 SN 1 5.19 2.32 log10 M R 8.07 GO ON TO THE NEXT PAGE 5 CEE 362 - Homework #3 log10W18 = Z R 0.6 9.36log10 5.7 1 - 0.20 Solution Manual log10 2.0 / 2.7 0.40 1094 5.7 1 5.19 2.32 log10 3000 8.07 = -0.6993 6 CEE 362 - Homework #3 Solution Manual Problem # 4: 1290 single 18 4 inch sand mixed asphalt 6 inch dense graded crushed stone 8 inch crushed stone = 0.45 ΔPSI = 2 Y = 20 years = 12000 lb / using table: SN = + SN: (.35)(4) + 0.18 (6)(1) + 0.11(8)(1) ►...
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