185_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

# 185_CEG 4012 Notes - base failure NsyH 0.181(115pcf(40 4 What is the maximum slope angle i that will provide FS c = 2 for H = 42 c = 1000 psf 41 =

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SL11 Example Stability Problems with Taylor Charts: 1. Find FS c for an open slope with H = 30', i = 70°, c = 500 psf, 41 = 20° and 'Y = 120 pcf. in Taylor 2 chart read Ns = 0.122 (zone A- toe failure critical) then FS c = c _ 500psf =1.14 N s y H 0.122(120pcf)(30') 2. Find FS c for the side slope of a deep, triangular ditch excavation in a thick clay layer with H = 40', i = 40°, c = 800 psf, 41 = and 'Y = 115 pcf. assume a toe failure is forced and read Ns = 0.165 in Taylor 2 chart (zone 8) (or use Case 8 dashed line in Taylor 1 chart) then FS = c _ 800psf =1.05 c NsyH 0.165(115pcf)(40') 3. Same as 2. above but allow all failure circles to occur (the Panama canal at a wide spot). in Taylor 2 chart zone 8, read greatest possible (most critical) Ns = 0.181 (for D = 00) (or use Case A in Taylor 1 chart, as D -700 all lines converge to N s = 0.181 ) ~ then FS c = c _ 800 pSf = 0.96 < 1.0
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Unformatted text preview: base failure! NsyH 0.181(115pcf)(40') 4. What is the maximum slope angle i that will provide FS c = 2 for H = 42', c = 1000 psf, 41 = 1 0°, and 'Y = 105 pcf. N c 1000psf =0.113 calculate s = FScyH -2(105pcf)(42') in Taylor 2 chart using Ns = 0.113 and 41 = 10° read i = 47° (almost zone 8, toe failure) 5. Given slope with H = 15', i = 30°, c = 500 psf, 41 = 0°, 'Y= 100 pcf, and a hard layer at the toe (15' below crown), find FS c • D = 15'/15' = 1 in Taylor charts read Ns = 0.133 with , i = 30 ° (easier to read in Taylor 2, face failure) h FS-c 500 pSf- 2 5 t en---• c NsyH 0.133(100pcf)(15')...
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## This note was uploaded on 02/10/2012 for the course CEG 4012 taught by Professor Staff during the Fall '08 term at University of Florida.

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