Question

Q5. A column foundation shown below is 3mx2m in plan. Given: D=2m, '=20°, and c'=45 kN/m2. Using Meyerhof's bearing capacity general equation and FS=3, determine the net allowable load the foundation could carry. Use bearing capacity, shape and depth factors. (15p) Y=16.8kN/m 2m Groundwater level_ -- Ysat=19.4kN/m 3m*2m

Please solve step by step and clearly thank you

Answer 1

N=16 kN/m Groundwateę 2M & sat = 19.4 kN/ 3 mx 2m O'= 20 c = u5kN/m² FS=3 According to Meyerhoff's general beering Capacity equation, ultimate bearing capacity can be determined as a qu = c'Ne fcs Fed & qNq Fas Fad + 0.588 No 1. Frsfod Ne, Nq, Nr > Bearing capacity of for d'= 20 Ne =. 14.83, Nq = 6.4, Ny=5.39 :وهياكها factors may of

14.83 fes, fqs, Frs > Shape factors fcs = 1+ B Nq = 1+ ? x 6.4 Fes = 1.288 fas= 1+ B Tan q = 1+ 2 Tanz . fqs = 1.243 Frs = 1-0.4 B = 1-0:4x? = 0.733 Fed, fqd, ford → depth factors fad = 1+ 27an e' (1-sino Of = 1+ 27an 20 (iSinzo) x ford = 1.158 Fed= fqd - (1-Fqd) - Ne Tan a fed = 1.150 - (1-1150) 14.83 Tanzo

1.187 Fed = frd = q= overbueden stress = roux Df 9= 16.8x1 + 1*(19.4-9.01) q= 26.39 kN/m² r= Submerged unit weight = Vsat - Nu g' = 19.4-9.81 = 9.59 Ianim! 1 Now qu = c'Nc Fcs fcd + q nq Fqs Fad t R 0.58B Ng Frs Frd qu = 45x14.83x1.288x 1.187 + 26.39x6.4x1.243 x 1.158 +0.5x9.59% 5.39 x2 0.733x1| que = 1020.20 +243.108 + 37.89

u = ( 3 2 8 IN / Net ultimate bearing capacity - quet = qu - Doux Df = 1301.28 - ( 16.8x1 + (19.4-9.01)x)) - quet = 1274.89 kN/m² Now not allowable bearing capacity quet call = quet - 1274.89 C = 424.96 KWE Now not allowable load - Onet (all) = quet (all) X BOL = 424.96x 2x3 = 2549.0 KN or Onet call) =2550 KN

Answer 2

wrong, recheck your formula