189_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

189_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 -...

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SL15 Location of Taylor Critical Circles Examples 1. In previous example 1 found FS c =1.14 for an open slope with H = 30', i = 70°, c = 500 psf, cI> = 20 ° and 1 = 120 pet. This is a toe failure. But center of circle and radius not specified. 2. In previous example 2. found FS c = 1.05 for the side slope of a deep, triangular ditch excavation in a thick clay layer with H = 40', i = 40°, c = 800 psf, cI> = 0° and 1= 115 pet. i = 40° < 530: This example forced a toe failure, Le. n = O. (see Taylor 1 and 2) with D == 1.07 ( can get D from Taylor 4 or from Taylor 1). Interpolate for a1 and a2 in table above: ~-~ ~-~ <X 1 = 28 + 33.68 _ 45 (26 - 28) = 27.1° <X 2 = 37 + 33.68 _ 45 (35 - 37) = 36.1° R = H sine (X2 + i) _ 40' sin(36.1° + 40°) = 62.0' sin i sine <X 1 + (X2 + i) sin 40° sin(27.1° + 36.1 ° + 40°) 3. In previous example 5. found FS c = 2.5 for slope with H = 15', i = 30°, c = 500 psf, cI> = 0°, and 1 = 100 pcf with a hard layer at the toe (15' below crown) so D = 15'/15' = 1. i = 30°
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Unformatted text preview: &lt; 53, D = 1: Slope failure. But center of circle and radius not specified. 4. If revise previous example 5. with deeper failure: H = 15', i = 30, c = 500 psf, cI&gt; = 0, and 'Y = 100 pcf with a hard layer 30' below crown, so D = 30'/15' = 2. i = 30 &lt; 53, D = 2: This is a midpoint circle with n =1.4 (from Taylor 4 or Taylor 1). Center of circle is above midpoint of slope tan(90 _ a) = nH + X = 1.4(15') + (1/2)(15' / tan 30) and a = 47.6 2 (0-1)H (2-1)15' R = nH + X = 1.4(15') + (1/2)(15' / tan30) = 46.0' sin a sin 47.6 5. If revise previous example 5. with steeper slope and no firm layer: H = 15', i = 60, c = 500 psf, cI&gt; = 0, and 1= 100 pet. i = 60 &gt; 53: This is a toe circle with a = 35 and e = 72.5 (from Taylor 3). R= H 2sinasin 9 2 15' = 22.1' 2 . 350 72.5 sin sin 2...
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