柱下独立基础结构计算书

柱下独立基础结构设计1、设计题目：基础埋深按2.0m计算地基承载力fak=180kPa1）混凝土标号，钢筋级别自己确定2）柱尺寸为400×6003）计算出地基承载力4）验算基础抗冲切选用C20混凝土ft=1.1×103kN/mm2,钢筋采用HRB400，fy=360kN/mm2,地基土质按粉土计算，b=0.3，d=1.5，钢筋混凝土重度c=25kN/m,土的重度m=20kN/m.1、地基承载力验算：1）荷载计算：标准值设计值Nk=2000kNN=1.35Nk=2700kNMk=60kN.mM=1.35Mk=81kN.mVk=12.5kNV=1.35Vk=16.88kNfa=fak+br(b-3)+d(d-0.5)m假定基础宽度为3~6m则：fa=180+1.5×1.5×2.0=225kPaA’≥dfNmaK=2202252000=10.8m2选用l=b=3.6mA=l×b=12.96m210.08m2W=62bl=7.82)基底净反力平均值：Ps=AN=26.32700=208.33kPa柱截面尺寸：400×600系数：C1=sttPfbb/6.0122=33.208/11006.016.06.322=3.02C1=sttPfbb/6.0122=3.07基础有效高度：h0=21(-bt+Cbt2)解得：h01=0.62h02=0.7h=h0+保护层厚度故去h=800mm3)采用两级台阶，每阶高400mm,台阶宽度分别为800、750mm则剪力产生弯矩MV=Vh=10kN.m∑Mk=70kN.m∑M=94.5kN.mGk=mAd=518.4kNG=1.35Gk=699.84kNe=kkFM=0.0356b故Pmax=WMAGNkkk=203.3kPa,Pmin=WMAGNkkk=185.4kPaPmaxfa=225kPa,满足地基承载力要求2、抗冲切验算：1)x方向：柱与基础交界处（第二级台阶处）h0=h-40=760mmAl=3.6×(0.8-0.7)+(3.6+2.2)×0.7/2=2.39m2地基净反力：Pmax=WMAN=8.75.946.327002=220.5kPaFl=PmaxAL=527kNam=(at+ab)/2=(0.6+0.6+2×0.76)/2=1.360.70hafmthp=795.82kNFl满足抗冲切要求2)Y方向柱与基础交接处（第二级台阶处）h0=800-40=760mmAl=(bt+2h0)(02hLLt)+(02hLLt)2=1.89m2am=(at+ab)/2=(0.4+0.4+2×0.76)÷2=1.160.70hafmthp=0.7×1100×1.16×0.76=678.32kNFl=PmaxAL=416.75kN0.70hafmthp=678.32kN满足抗冲切要求3）x方向：第一级台阶处h0=400-40=360mmam=(at+ab)/2=(2.1+2.1+2×0.36)÷2=2.46Al=1.31m2Fl=PmaxAL=228.9kN0.70hafmthp=0.7×1×1100×2.46×0.36=681.912kNFl=228.9kN故满足抗冲切要求。4）y方向：第一级台阶处h0=400-40=360mmAl=1.103m2am=(at+ab)/2=(2+2+2×0.36)÷2=2.36Fl=PmaxAL=243.2kN0.70hafmthp=0.7×1×1100×2.36×0.36=654.2kNFl=243.2kN满足抗冲切要求。3、配筋计算：Pmax=AGN+WM=96.1284.6992700+8.75.94=274.45kPaPmin=AGN-WM=96.1284.6992700-8.75.94=250.22kPa1)第一级台阶:当弯矩和剪力作用在x方向时:P1=Pmin+(bb8.0)(Pmax-Pmin)=266.37kPaMI=])()2)('2[(1211max1max21lpPAGppalaMI=]6.3)37.26645.274()96.1284.699237.26645.274)(88.26.32[(8.01212=234.35kN.mMII=)2)('2()'(481minmax2AGPPbbalMII=)96.1284.699222.25045.274)(8.26.32()88.26.3(4812=45kN.mX方向：AS=087.0hfMyI=36.036087.035.234=2079mm2Y方向：AS=087.0hfMyII=36.036087.045=399mm2=bhAs=40036002079=0.14%min=0.15%故两个方向按最小配筋率计算，最小配筋面积As=bhmin=3600×400×0.15%=2160mm2当弯矩和剪力作用在Y方向：P1=Pmin+(bb75.0)(Pmax-Pmin)=250.22+(6.375.06.3)(274.45-250.22)=269.4kPaMI=])()2)('2[(1211max1max21lpPAGppalaMI=]6.3)4.26945.274()96.1284.69924.26945.274)(8.26.32[(0751212=205.16kN.mMII=)2)('2()'(481minmax2AGPPbbalMII=)96.1284.699222.25045.274)(88.26.32()8.26.3(4812=56kN.mY方向：AS=087.0hfMyI=36.036087.016.205=1820mm2X方向：AS=087.0hfMyI=36.036087.056=497mm2=bhAs=40036001820=0.13%min=0.15%故两个方向按最小配筋率计算，最小配筋面积As=bhmin=3600×400×0.15%=2160mm22)第二级台阶:当弯矩和剪力作用在X方向：P2=Pmin+(bb6.1)(Pmax-Pmin)=250.22+(6.36.16.3)(274.45-250.22)=263.68kPaMI=])()2)('2[(1211max1max21lpPAGppalaMI=]6.3)68.26345.274()96.1284.699268.26345.274)(2.26.32[(6.11212=870.83kN.mMII=)2)('2()'(481minmax2AGPPbbalMII=)96.1284.699222.25045.274)(26.32()2.26.3(4812=156.53kN.mX方向:AS=087.0hfMyI=76.036087.083.870=7723mm2Y方向：AS=087.0hfMyI=76.036087.053.156=658mm2最小配筋面积As=bhmin=3600×800×0.15%=4320mm2故y方向按最小配筋率配当弯矩和剪力作用在y方向：P2=Pmin+(bb5.1)(Pmax-Pmin)=250.22+(6.35.16.3)(274.45-250.22)=264.35kPaMI=])()2)('2[(1211max1max21lpPAGppalaMI=]6.3)35.26445.274()96.1284.699268.26345.274)(26.32[(5.11212=725.95kN.mMII=)2)('2()'(481minmax2AGPPbbalMII=)96.1284.699222.25045.274)(2.26.32()26.3(4812=208.89kN.mX方向：AS=087.0hfMyI=76.036087.089.208=878mm2Y方向：AS=087.0hfMyI=76.036087.095.725=3050mm2最小配筋面积As=bhmin=3600×800×0.15%=4320mm2故x方向按最小配筋率配