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HW 3 solution

HW 3 solution - CEE 362 Homework#3 Solution Manual...

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CEE 362 - Homework #3 Solution Manual 1 Problem #1: Given: = 0.5 Δ PSI = 2.0 TSI = 2.5 = 0.9 = 0.9 = 0.8 = 5000 lb/ Required: number of 25-kip single-axle SN = + using table: SN: (.44)(4) + 0.18 (7) *0.9 + 0.11(10) * 0.8 SN = 3.774 4 inch HMA 7 inch crushed stone 10 inch crushed stone GO ON TO THE NEXT PAGE

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CEE 362 - Homework #3 Solution Manual 2 = 5.9742 18 W = = 941890 Finding ESAL Axle load equivalency using table Axle load (kip) SN = 3 SN = 4 24 3.09 2.89 26 4.31 3.91 25 3.7 3.4 by interpolation : = 3.468 = 271597 25-kip single axle loads 07 . 8 log 32 . 2 1 SN 1094 40 . 0 7 . 2 / PSI log 0.20 - 1 SN log 9.36 = log 10 19 . 5 10 10 o 18 10 R R M S Z W 07 . 8 5000 log 32 . 2 1 4.744 1094 40 . 0 7 . 2 / 2 log 0.20 - 1 4.744 log 9.36 .4 0 * .9 0 = log 10 19 . 5 10 10 18 10 W
CEE 362 - Homework #3 Solution Manual 3 Problem # 2: Given: 1300 single 8000lb 900 Tandem 15000lb 20 single 40000lb 200 tandem 40000lb 4 inch HMA 4 inch HMA 8 inch crushed stone = 70% = 0.5 Δ PSI = 2.0 W = 1 N = 12 SN = + SN: (.044)(3) + 0.40 (4) + 0.11(8) SN = 4.24 8000 lb single: 15000 lb tandem: 40000 lb single: 40000 lb tandem: GO ON TO THE NEXT PAGE

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CEE 362 - Homework #3 Solution Manual 4 = (1300 * 0.03932)+(900*0.04308)+(20*22.164) +(200 * 2.042) Solving equation for = 5250.31 psi 07 . 8 log 32 . 2 1 SN 1094 40 . 0 7 . 2 / PSI log 0.20 - 1 SN log 9.36 = log 10 19 . 5 10 10 o 18 10 R R M S Z W
CEE 362 - Homework #3 Solution Manual 5 Problem # 3: 200 single 20 200 tandem 40 80 single 22 5 inch HMA 6 inch HMA 10 inch crushed stone = 3000 lb / = 0.6 Δ PSI = 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)

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