solution6 - log D H = -16.213 + 1.532 * 7.4 1.406 log...

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CE 728 GEOTECHNICAL EARTHQUAKE ENGINEERING ASSIGNMENT VI POST LIQUEFACTION GROUND DEFORMATIONS a) Hamada et al. (1986) D = 0.75 * H 0.75 * θ 0.33 H = 8.2 – 0.8 =7.4 m D = 0.75 * 7.4 0.75 * 3.0 0.33 = 2.93 m b) Youd et al. (2002) log D H = -16.213 + 1.532M – 1.406 logR* - 0.012R + 0.338 logS + 0.54 logT 15 + 3.413 log (100-F 15 ) – 0.795 log (D 50,15 + 0.1 mm) R* = R + R 0 R 0 = 10 (0.89M-5.64) Average values of fines content and D 50 value is used without considering the N < 15 criteria as the whole layer is liquefied according to the results of liquefaction analysis.
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Unformatted text preview: log D H = -16.213 + 1.532 * 7.4 1.406 log (9.83) - 0.012 *1 + 0.338 log(3.0) + 0.54 log (7.4) + 3.413 log (100-15.57) 0.795 log (0.21 + 0.1 mm) = 1.326 D H = 21.2 m c) Ishihara (1997) Average values of (F.S) and (N 1,45 ) are calculated for given soil layer as 0.33 and 19.11 respectively. Using the chart given by Ishihara (1997) max = 10 % D = 0.1 * 7.4 = 0.74 m...
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This note was uploaded on 05/09/2008 for the course CIVIL ENGI CE 728 taught by Professor Cetin during the Fall '06 term at Middle East Technical University.

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