41_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

41_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - at...

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p24 OC Settlement Example 1: Solution: calculate Pc for the clay layer. Use point Pat mid-height of layer to represent stress change in layer. Find Pc under center of mat foundation (Pc in sand = 0) Initial Stresses: avo Uo a'vo OC Clay (OCR = 1.4) 'Ysat = 18.6 kN/m 3 Cc = 0.28, eo = 0.90 CR:: 0.030 ..... : : : \ \/ \ 1 : ·: ',: :.,· :••·: ·:.: ·.\.·.\1.\,·.\.· :: :: 4m(15.7) + 13m(19.1) + 3m(18.6) = 366.9 kN/m 2 = (13m + 3m)(9.81 kN/m 3 ) :: 157.0 kN/m 2 :: av - u = 366.9 - 157.0 :: 209.9 kNlm 2 20m 15m~ ~--- 30 m lp I 40m Stress Change: the soil is excavated Dr (:: 2m) prior to footing placement q = net applied load:: (QlA) - (Dr 'Ym) = [300 MN(1000 kN/MN)/(30m x 40m)] - 2m(15.7 kN/m 3 ) = 218.6 kN/m 2 use Boussinesq "mn chart" for stress change chart gives tlav under corner of footing so use superposition at P
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Unformatted text preview: at point P m = 15 m 1(2 m + 13 m + 3 m) = 0.83 } I:: 0.166 n :: 20 m I (2 m + 13 m + 3 m) :: 1.11 tlav :: 4ql = 4(218.6 kN/m 2 )(0.166} = 145.2 kN/m 2 Consolidation: ClayisOC~alc:: (odl)cr~ :'1,lj(20".~"",/",,1.) ::. 'l'H .., w/,.,,'l. . .. (C1"~ol lo-v)~.(l(l".tt.,ll.r.1. .); 55>,1 i<Nj",l '/ a-~ :, LOL Pc =( CRH )loJ ~IC )+( CcH )IOg(CJlvo~aCJV) 1 + eo ~l CJ vo 1 + eo CJ c = (0.030(6 m) 10 (293.9) + (0.28(6 m) 10 (355.1) 1 + 0.90 g 209.9 1 + 0.90 9 293.9 = o. o/if + D.0""3 (tJ( IN"j lcp,. .- '. ) r ': (}. g:J "'l -: & ~ WI~ c. ,-Note: under corner of footing tlov:: ql :: (218.6 kN/m 2 )(0.232) :: 50.7 kN/m 2...
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