1、桩基竖向抗压承载力计算 桩身周长:u=πd=3.140.8=2.51m 桩端面积:Ap=πd2/4=3.140.82/4=0.5m2 承载力计算深度:min(b/2,5)=min(5/2,5)=2.5m fak=(2.5240)/2.5=600/2.5=240kPa 承台底净面积:Ac=(bl-nAp)/n=(55-40.5)/4=5.75m2 除以4是何意? 复合桩基竖向承载力特征值: Ra=ψuΣqsia?li+qpa?Ap+ηcfakAc=0.82.51(2.455+5.730+4.250+4.370)+20000.5+0.12405.75= 2779.89kN应该为 (2772.5)计算有误问题补充:帮忙解答:承台底净面积:Ac=(bl-nAp)/n=(5×5-4×0.5)/4=5.75m2 除以4是何意?2779.89kN应该为 (2772.5)计算有误
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