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土力学浅基础课程设计(Soil mechanics; shallow foundation; curriculum design)
土力学浅基础课程设计(Soil mechanics; shallow foundation; curriculum design) Independent foundation design under 1 columns Materials: No. 1 soil layer, miscellaneous fill soil, layer thickness 0.5m, containing some construction waste. Soil layer, silty clay, layer thickness 1.2m, soft plastic and humid, the bearing capacity characteristic value is kPafak130 =. Soil layer is clay, layer thickness 1.5m, plasticity, slightly wet, bearing capacity value kPafak180 = Soil layer, fine sand, layer thickness 2.7m, medium density, bearing capacity characteristic value kPafak240 =. The thickness of strongly weathered mudstone in uncovered soil layer is not uncovered, and the characteristic value of bearing capacity is kPafak300 = kF:1339kN, kM:284kN, kV:96kN F:1741kN M:369kN N:133kN The No. 3 soil layer is the bearing stratum. Design 2000 A Independent foundation under an axle column C25 concrete, grade HPB235 steel, assuming a base height of 0.8m. Groundwater is located below the ground surface 1.5m, in order to protect the base from human and other biological activities and so on, The foundation should be buried below the ground surface, and the minimum embedment depth is 0.5m. Take d=0.5m here. The height of outdoor to base is h=05+1.2+0.5=2.2m. One The eigenvalue of subgrade bearing capacity fa E=0.580.85, LI=0.780.85 Table b=0.3, ETA, average ETA d=1.6 base above soil gravity: y m=2.2 5.04.92.0) 1020 (1205.018 x + X + X * + *) =16.233/ MkN bearing capacity characteristic value FA, fa= () 15.224) 5.02.2 (23.166.11805.0= * * +=++dmd) AF gamma ETA kPa in D according to the outdoor ground up. 2: initial selection of base size, take column bottom load standard value, kF:1339kN, kM:284kN, kV:96kN calculation foundation and backfill weight Gk when the base depth is () md425.265.22.2 Two One =+= The basal area is Two hundred and thirty-three .1 725.1207.01015.224 13390= * *?? = ? = DG A KF F A Y M2 Because the bias is not large, the base area increased by 20%, or 8 .833.72.12.10= = Am2 *. P
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