186_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

186_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - =...

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SL12 Equalizing the Component Factors of Safety - c/, soil o Taylor 2 chart provides N s required if FS~ = 1, i.e. fully mobilized friction component (tancl» o remember FS =..!.= c+O'tancjl FS =(~) FS =( tancjl) 'td Cd + 0' tan cjld C Cd ~ tancjld o to mobilize cohesion component first: calculate N s = (;) read'" (developed) in Taylor 2 chart for the given slope angle, and calculate FS, (Le. use chart in reverse order) o desirable that cohesion strength (c) and friction strength (tan,) components be mobilized at the same rate Le. FS = FS c = FS, o to obtain this overall FS equilibrium must use chart to iteratively calculate either FS c or FS~ until FS c = FSq o simple to plot the iterative results and read plot for answer (see example) o example: Find FS if c = SOO pst, H = 2S', 'Y= 12S pet, cI>
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Unformatted text preview: = So, i = 30 assume FS~ = 1.0 =( tancl :. cl>d = tan-1 (tanS O ) = So ~ N s = 0.110 trom Taylor 2 tancl>d 1.0 and FS = _c _ = SOOpsf = 1.4S eNs'YH 0.11(12Spcf )(2S') assume FS~ = 1.2 =( tancl :. cl>d = tan-1 (tanS O ) = 4.2 ~ N s = 0.118 trom Taylor 2 tancl>d 1.2 and FS = _c _ = soopst = 1.36 c N s 'Y H 0.118(12Spct)(2S') ( tan, ) _(tanS O ) assume FS~ = 1.S =--:. cl>d = tan 1--= 3.3 ~ N s = 0.12S from Taylor 2 tancl>d 1.S and FS = _c _ = soopst = 1.28 c N s yH 0.12S(12Spcf)(2S') ,. FS c = FS~ , , , , , , , , , , , ~~~c":::~ FS = 1.32 answer 1.2 1.4 1.0 ~-r-"'--r--r--1.0 1.2 1.4 1.6 FS~ FS c 1.6 plot results:...
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