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98_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

# 98_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - of 6...

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BC13 Meyerhof Be Example 6: same as example 4 with GWT changing, assume 'Ysat = 'Ym = 100 pcf 8ft 1. watN"tlfbllRft· aeptn at 20 ft below ground:surface -(d = 12 ftcbelow the-fo15ting '" wllic/; is30::ft) calculate y for the Ny term: d 12 ft 'Y ~sat 00 '~':" ?J~,6 f'lf.- \ /1. r ~ (fOI.) fC (. -1,.6 rd') t~. 6 pef d }bEl = flo calculate bearing: qu = (200 psf)(50.59)(1.23)(1.10)(1) + (800 psf)(37.75)(1.12)(1.05)(1.00) + \/2 (62.6 pcf)(30 ft)(44.43)(1.12)(1.05)(1) = 13,690 + 35,515 + 49,062 = 1 v, '). 61 fS r Q, =quxarea=( H, ;>.0r,r)(30ft)(100ft)(1 ton 12000 Ib)= 1"1," 0 ' '0"5 = 11.11/ 10 1/3.°= 1./'1, r3 1 ' jill"! (z'}";. Illls{ 41"t,,, c-)('. I) 2. ,water table aJ-a3tepth
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Unformatted text preview: of 6 ft below ground surface (noW"2 ft a60ve me footing) 6ft calculate y for the N, term: y = y" = 37.6pcf calculate q for N q term: d 2 ft = calculate bearing: qu = (200 psf)(50.59)(1.23)(1.1 0)(1) + (675 psf)(37.75)(1.12)(1.05)(1.00) + \/2 (37.6 pcf)(30 ft)(44.43)(1.12)(1.05)(1) = 13,690 + 29,966 + 29,469 = TI, I:>"psf = qu x area = (., 1, I~ s", )(30 ft)(1 00 ft)(1 ton 1 2000 Ib) = 1 u 'I, 6 ~~ I"J =lo'II.y~ 13.0=16s-o f.'! ("3". ','l ".",c,.v) , /...
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