178_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

178_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - cos...

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SL5 Infinite Slope Examples o Sand slope, (\) = 35°, f3 = 30°, 1581 = 120 pet Find FS tor dry slope, submerged slope and slope with steady state seepage. Dry Slope: FS=(tan35) =1.21 ,,( ~~ tan30 ~ Slope with seepage: FS = i tan cj) = 120pcf - 62.4pcf (tan 35) = 0.58 unstable! Ysat tanp 120pcf tan 30 Note: above FS are independent at slope height and soil weight. o Sandy Clay slope, c= 600 pst, (\) = 10°, f3 = 30°, 1m = 1881 = 120 pct (y = 120-62.4 = 57.6 pst) Find FS tor dry slope, submerged slope and slope with steady state seepage, h = 12'. Dry Slope: FS = c + tane!» = 600psf + tan10 = 1.27 yHcos 2 ptanp tanp 120pcf(12')cos 2 30tan30 tan30 H c _ 600psf = 16.6' er = Y
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Unformatted text preview: cos 2 p(tanp - tan cj» 120pcfcos 2 (30)(tan 30- tan1 0) Submerged Slope: FS = c + tane!» = 600psf + tan10 = 2.31 y'Hcos 2 ptanp tanp 57.6pcf(12')cos 2 30tan30 tan30 Her = C _ 600psf = 34.6' y' cos 2 p(tanp - tan cj» 57.6pcf cos 2 (30)(tan 30 - tan1 0) Slope with seepage: FS= C +i tancj) YsatHCOS2 ptanp Ysat tanp = 600psf + 57.6pcf tan10 = 1.11 120pcf(12')cos 2 30tan30 120pcf tan30 c Her =---:---------cos 2 P(y sat tanp -y'tancj» 600psf-135' (I I) = 2-• C osee COS (30)[120psf(tan30) -57 .6psf(tan10)]...
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