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97_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

# 97_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - 'f...

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BC12 Meyerhof BC Example 4: same as above, but foundation now eni1b!dtcleJ~I)po..-..ekfl" Bearing factors, shape factors and inclinations factors same as example 1 1. surcharge term = q = (~q) ('00 1',,) ~ S'" r'~ 2. Determine depth factors: D,/B = (8 ft / 30 ft) ~ 0.267 8ft Acd = 1+ 0.2(0.267)tan(45 + 3:0) = I, 10 ( 360) Aqd~A..,o =1+0.1(0.267)tan 45+ 2 = I,D5 q, = (200 psf)(50.59)(1.23)(1.1 0)(1) + (800 psf)(37.75)(1.12)(1.05)(1.00) + V, (100 pcf)(30 ft)(44.43)(1.12)(1.05)(1) = 11 J bqO~ 15"1 flS ! 1f,'n5'"= 1').1,)8 0 (l5t 0, = q, x area = ( '"1', 5S0 1'> c )(30 ft)(100 ft)(1 t / 2000 Ib) = I~ 1,"),0 ""J 0, = 1'/1",0 /3.0 = ('),l'[c> ~1>"5 .( ''''f »t it -'Uk
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Unformatted text preview: ''''f »t it-'Uk ttf.f) • Meyerhof BC Examp:l:e:.;5:.;:~=~~~~" same example 1 with d Bearing factors, shape factors and depth factors same as example 1 1. Determine inclination factors: ~i ~Aq' =[1-(:~:)r = O.W a = 20° 2. Calculate bearing: I q,~ cN,A,,;Acd~' + qNgAq,Aq,Aq, + v, :yBNb).~)J q, ~ (200 psf)(50.59)(1.23)(1 )(0.60) + 0 + V, (100 pcf)(30 ft)(44.43)(1.12)(1 )(0.20) = '1, 'I.) + (>. J'f,q?~ , 1l,H'i"f,r = q, x area = (H., 'I'" l,r)(30 ft)(100 ft)(1 ton / 2000 Ib) = 31 >13 tons ,...
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