{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 3 solution

# 0443 040 4 0118 sn 424 8000 lb single 15000

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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) ►...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online