HW 3 solution

# 45 psi 2 y 20 years 12000 lb using table sn sn

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Unformatted text preview: SN = 3.36 single kip 18: = (400 * 1.49)+(900*1.24) log = 6.9739 GO ON TO THE NEXT PAGE 7 CEE 362 - Homework #3 log10W18 = Z R S o 9.36log10 SN 1 - 0.20 Solution Manual log10 PSI / 2.7 0.40 1094 SN 1 5.19 2.32 log10 M R 8.07 6.9739 = Z R 0.4 9.36log10 3.36 1 - 0.20 log10 2 / 2.7 5.19 0.40 1094 3.36 1 2.32 log1012000 8.07 6.9739 = Z R *0.4 +7.0384 Z R = -0.16125 8 CEE 362 - Homework #3 Solution Manual Problem # 5: Assume D=11" Axle loads: by interpolation we will have: Single 22,500 = 2.675 Tandem 25kip = 0.5295 Tandem 39kip = 3.55 Triple 48kip = 2.58 (available in the Table) 2.675*80 + 0.5295*570 + 3.55*50 + 2.58*80 = 899.715/day W18 = 899.715 *365 *20 = 6,567,920 Try D= 11 inch log10W18 Z R S o 7.35log10 D 1 - 0.06 log10 PSI / 3.0 8.46 1 1.624 107 D 1 ' S c Cd D 0.75 1.132 4.22 0.32TSI log10 0 .25 0.75 215.63J D 18.42 Ec k log10W18 1.645 * 0.4 7.35log10 11 1 - 0.06 log10 1.7 / 3.0 8.46 1 1.624 107 11 1 600 * 0.9 110.75 1.132 4.22 0.32 * 2log10 215.63 * 3.2 110.75 18.42 5000000 2000 .25 W18 = 6.710 W18 =...
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## This document was uploaded on 02/11/2014.

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