3交A锥形基础计算.docx
3交A锥形基础计算项目名尊日M设计*校对*一、设计依据E建筑地基基础设计规他（GB50007-20U）£混凝土结构设计规范GB50010-2010（2015年版）简明高层钢筋混凝土结构设计手册?李国胜（第二版）二、示意图三、计算信息构件编号：JCT计算类型：验算截面尺寸1 .几何参数定形柱宽be-900Hm城形柱高基础端部高度hi=350nf11基础根部高度h2=150wn基础长度B1.=950mB2950n基础宽度A1.-95011m2-950mnhc-90011mfc_b=1.1.9N'11三:fcc=1.1.W/mn:2 .MMfSB加础混凝土等级：C25ft.b=1.27N三1柱混凝土等级：C25ftc=1.27N11tn1钢筋级别：HRB-100fy=360N11三,3 .计算信息结构虫要性系数：丫o=I.0基础埋深：dh=1.600纵筋合力点至近边距肉：as40mn基础及其上用土的平均容色：Y=20.OOCkNZm'最小配筋率：p三in=O.15O4 .作用在基础顶部荷载标准组合值F-882.OOOkNMx=67.OOOkN*mMy=10.OOOkN*mVx=7.OOOkNVy=-17.OOOkNks=1.35Fk=Fks=882.000/1.35=653.333kNMxk=Mxks=67.000/1.35=49.630kN*mMyk=MyZks=IO.000/1.35=7.407kNtaVxk=Vxks=7.000/1.35=5.185kNVyk=VyZks=-I7.000/1.35=-12.593kN5 .脩正后的地基承载力特征假fa=244.400kPa四、计算敷1 .茶础总长Bx=B1.+B2=O.95OP.95O=1.9OOm2 .荔础总宽By=A1.+2=0.9500.950=1.900m3 .基础总高Ihh1.*h20.350*0.150-0.500m4 .底板配筋计算高度ho=h1.+h2-as=0.350+0.150-0.(M0=0.WOm5 .施础底面积A=BX*By=1.900*1.900=3.610/6 .Gk=*BxBy*dh=20.0001.900*1.900*1,600=115.520kNG=1.35*Gk=1.35*115.52k1.55.952kN五、计算作用在旃底部弯矩值MdXk=XXk-Vyk*H=49.630-(-12.593)*0.500=55.926kN*三Mdyk=Myk+Vxk*H=7.407+5.I85O.500=10.OOOkWmMdx-MxVy*1.1.-67.OOO(-17.OOO)0.50(k75.50OkN*mMdy=My÷Vx*H=10.000+7.000*0.500=13.5(K)k.X*<n六、Ift算地基承敏力1 .脸尊轴心荷载作用下地域承栽力Pk=(FkWk)/R=(653.333+115.520)/3.610=212.979kPa【5.2.2-2因o*pk=1.0*212.979=212.979k1.,afa=244.400k1.,a轴心前我作用下地菸承载力in足要求2 .脸笄偏心荷栽作用下的地基承载力exk-Mdyk(Fk*Gk)10.000/(653.333+115.520)=0.013m因exkBx6=0.31711>X方向小信心，由公式05,2.2-2和【®5.2.2-31推导PkmiI1.X=(FkGk>A+6*!MdykI/(BxBy)-(653.333115.520)/3.610*6*10.000/(1.9001.900)=221.726kPaPkmin_x=(Fk+Gk>A-6*Mdyk/(Bx-By)=(653.333+115.520)/3.610-6*110.000/(1.9002*1.900)=204.23IkPaeyk=Mdxk(Fk+<)=55.926/(653.333+115.520)=0.073m因IeykBy6=0.317my方向小偏心PkmiIX_y=(FhGk)A+6*iMdxk(ByBx)-(653.333+115.520)/3.610*6*155.926/(1.900*1.900)=261.901kPaPkmin_y=(Fk+Gk)A-6*MdxkI/(By-Bx)=(653.333+115.520)/3.610-6*155.926/(1.9001.900)=164.057kPa3.确定基础底面反力设计值PkmaX=(Pk三ax-pk)+(Pk三ax-y-pk)+pk=(221.726-212.979)+(261.901-212.979)÷212,979=270.648kPayo*Pkmax=1.()*270.(W8=270.618k1.,aI.2<fa=1.2*244.400=293.28OkPa偏心荷我作用下地就承领力涵足要求七、基砒冲切破算1 .计灯基础底面反力设计值1.1计算X方向基础底面反力设计值ex=M<1.y(F+G)=13.500/(882.00(K155.952)=0.013m因cxBx6.0=0.317mX方向小俏心PaaxX-(FG)/A6*MdyI/(Bz*By)=(882.OOoT55.952)/3.61.(H6*13.5001/(1.<)00*1.900)=299.33IkPaIin.x=(F+G)A-6*Mdy/(BXXBy)=(882.000H55.952)/3.6106*13.500(1.900,1.900)=275.712kPa1.1 2计算y方向基础成面反力设计他ey=Mdx(F+G)=75.500/(882.OOOH55.952)=0.073meyBy60.317y方向小信心Paxc=(F÷G)A+6*Mdx(By-*Hx)=(882.OOO+155.952)/3.61/6*75.500(1.900,1.900)=353.566kPaP三in-y=(F+G)A6*MdxI/(Byz*Bx)=(882.OOCH-155.952)/3.610-6*75.500(1.900：*1.900)=221.477kPa1.3 因MdKWoMdy01.ax-Pmax_x*Pmaxy-(F*G)/A=299.331+353.566-(882.000+155.952)/3.610=365.375kPa1.4 计算地基净反力极值PjmiJX-PBaX-GA-365.375155.952/3.610322.175kPnPjmaxx=PmaxX-GA=299.331-155.952/3.610=256.K1.1.kPnPjmax_y=Pmax_y-G/A=353.566-155.952/3.610=310.366kPa2 .柱对基础的冲切脸算因1000,取BhP=1.o2 .1X方向柱对基础的冲切验算X冲切面积A1.x=ax(B1.-bc2-ho)*(hc+2*ho)÷(B1.-bc2-ho)(B2-bc2-ho)*(hc÷2*ho)+(B2-bc2-ho),>-max(0.9500.90020.460)*(0.900*2*0.460)(0.950-0.9000.460)72,(0.950-0.900/2-0.460)*(0.900+2*0.160)+(0.950-0.900-0.460)72)=三ax(0.074,0.074)=0.074m:X冲切破面上的地基净反力设计值F1.x-1.xPjmax-0.074*322.17523.970kNyo<F1.x=1.0*23.970=23.97k因o*F1.x0.70hp*ft-b*am*ho(6.5.5-1)=0.7*1.0001.27*1360.000*160=556.16kNX方向柱对基础的冲切满足规范要求3 .2y方向柱对艇础的冲切除算y冲切面枳A1y=max(A1-hc2-h)(bc+2*ho)+(A1-hc2-ho)*(A1-hc2-ho),(A2-hc2-ho)*(bc+2*ho)(2hc2*ho)*(B2hc/2ho)=JnaX(0.950-0.900/2-0.460)(0.900+2*0.460)+(0.950-0.900/2-0.460)*(0.950-0.900/2-0.-160),<0.950-0.900/2-0.160>*(0.900-2*0.160)+(0.950-0.900/2-0.-160)(0.950-0.900/2-0.460)-max(0.074,0.074)=0.074n2y冲切板面上的地鹤净反力设计值F1.y=A1.y*Pjmax=0.071*322.175=23.970kNyoF1.y=1.0*23.970=23.97kN因o*F1.y0.7*hpft-b*bn*ho(6.5.5-1)=0.7*1.000*1.27*1360*460=556.16kNy方向柱对基础的冲切满足规范要求八、基11受剪承武力险第I,计算剪力Pk=(FkXJk)/A=(653.3+115.5)/3.6=213.OkPaA,=(A1.+A2)*<max(B1.,B2)-0.51.>c)-(950*950)*(max(950,950)0.5*900)=0.95a2Vs=A'*pkO.95*213.0=202.3kN2 .计算豉面高度影响系数.=(800h0)1-(800/800.0)"=1.03 .剪切承统力验算Ao=(A1.+A2)*h1.0.5*(bo+100+(A1+A2)*112=(950-950)*35O÷O.5*(900+100+(950950)*150-882500.O0三2yo*Vs=1.0*202.3=202.3kN0.7BMtAo=O.7*1.0*1.27*882500.0=784.5kN受剪承载力验算满足要求！九、柱下基础的局部受压险第因为基础的混凝土强度等级大于等于柱的混凝十强度等级,所以不用验算柱下扩展基础丁负面的局部受压承我力。十、JM1.爻穹计售双向受理.根据中751页条目4(3)中公式因exBx6=0.317mX方向小偏心a=(B-bc)2=(1.900-0.900)/2=0.500mPj1.=(BX-a)*(PBa1.X-Pmin_x)/B+Pmin_x-G/A=(1.9000.500)(299.331-275.712)/1.900*275.712-155.952/3.610=249.915kPa因eyBy6=0.317ny方向小偏心a=(By-hc)2=(1.900-0.900)/2=0.500nPj2=(Bya)*(Paax_yPmin_y)/By»Pmin_y-G=(1.900-0.500)*(353.566-221.477)/1.900+221.177-155.952/3.610=275.606kPa6x=1.021
