J4双柱阶梯基础计算.docx
J4双柱阶梯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(GB50007-201.1)E混凝土结构设计规范(GB50010-2010)二、不意图三、计算信息龙形柱高矩形柱高hc-50O11mhc=50011m构件编号：JC-I1 .几何参数台阶数n=1.矩形柱宽矩形柱宽基础高度计算类型:bc-500三nbc=500amh1.=500(m验算嵌面尺寸一阶长度b1.=115Omnb2=115On一阶宽度a1.=I250mma2=125Chm2 .材料信息基础混凝十.等级：C30ft>=1.43Kmfc_b=14.3Nm柱混凝土等级：C30ft_c=1.43Nmfc_c=1.-1.3Nm钢筋级别：HRMOOfy=360Nm3 .计算信息结构重要性系数：丫。=1.0基础埋深：dh=2.500m纵筋合力点至近边距离：as=40m基础及其上覆土的平均容重：丫二20.OOOkN/最小配筋率:Pain=0.150%4 .作用在基础顶部荷我标准组合值F1.=1650.OOOkNF2=1400.OOkNMx1-40.OOOkNtmMyI=40.0()0kN*mVx1.-IO.OOOkNVy1=-40.OOOkNMx2-20.OOOkN*mMy2=10.OOOkN*mV2=10.OOOkNVy2=20.OOOkNks=1.25Fk=FIZksI+F2ks2=1650.000/1.25+1400.000/1.25=2440.OOOkNMx=Mx1.ks+F1.*(Bx2-B1.)ks+Mx2ks+F2*(B2-Bx2)ks-10.000/1.25H650.000*(4.500/21.400)/1.25<-20.000/1.25H100.000(1.400/21.500)/1.25=186.(>OOkX*mMy=My1.ks÷F1.*(A2-A1.)2ks+My2ks+F2*(A2-A1.)2ks-10.000/1.25H650.000*(1.5001.500)/2/1.2510.000/1.25ni00.000*(1.5001.500)/2/1.25=-10.000k*mVx=Vx1.ks+Vx2ks=40.000/1.25+10.000/1.25=40.OOOkNVy=Vy1.ks÷Vy2ks=-W.000/1.25+20.000/1.25=16.OOOkNMxk-Mx1.F1*(Bx2-B1.)Mx2*F2*(B2Bx2)FO.000+1650.000*(4.500/2-1.W0)-20.000÷14(W.00<)*(1.-JOO-4.500/2)=232.500kN*mMyk=My1+F1*(A2-A1)2+My2ks+F2*(A2-A1)2MO.OOO*1650.000*(1.5001.500)/2*10.OOOH400.OOO*(1.5001.500)/2=50.OOOkN*mVxk=Vx1.+Vx2=10.000+10.000=50.OOOkNVyk=VyHVy2=-40,Oo(H20.000=-20.OOOkN5.修正后的地基承段力特征值fa=2-IO.OOOkPa四、计算ft1 .基础总长Bx=B1.+B2+Bc=1.400+1.400+1.700=1.500m2 .基础总宽By=A1.+A2=1.500*1.500=3.OOOm1.=a1.+ax(he1.,hc2)/2=1.250+三ax(0.500,0.500)/2=1.50OnA2=a2+三ax(hc1.,hc2)2=1.250+三ax(0.500,O.500)2=1.5Q0nB1.-b1.+bc1.2=1.150*0.500/2-1.100mB2=b2+bc22=1.1500.500/2=1.100m3 .基础总高H=h1.=O.5=0.500a从底板鼠筋计豫i¾度ho=h1.-as=0.500-0.(H0=0.160m5 .基础底面积A=BX*By=4.500*3.000=13.5006 .Gk=*BxBy*dh-20.000*1.500*3.000*2.500=675.OOOkNG=1.35*Gk=1.35*675.OOo=9II.25OkN五、计算作用在若破底部穹矩值MdXk=MXkTXk*H=232.500-50.000*0.500=207.500kN*mMdyk-MykHyk*11-50.000*20.000*0.500-40.OOOkN-HnMdX=MX-YX*H=186.OOtt-W.()0()«().500=166.OOOkN*三Mdy=My÷Vy*H=10.OoOI6.000*0.500=32.OOOkBm六、MMX承M1. 5金券轴心荷栽作用下地基承载力Pk=(HkHJk)/A=(24W.OoO-675.000)/13.500=230.741kPa15.2.2-2因o*pk=1.0*230.711=230.711kPafa=210.OOOkPa轴心荷载作用下地培承我力湎足要求2. 5金算偏心荷就作用下的地基承载力exk=Mdyk/(Fk4k)=40.000/(2440.000+675.()00)=«.013m因IexkBx6=0.750mX方向小偏心，由公式【5.2.2-2和©5,2.2-3推导Pkmax,x=(Fk*Gk>A+6*Mdyk(B*By)=(2440.0C0÷675.000)/13.500+6«40.(X)O/(4.503.000)=234.69IkPaPkmin_x=(Fk+Gk)/A-6*Mdyk(B*By)-(2440.000*675.000)/13.5006*40.000/(4.503.000)=226.79OkPaeykMdxk(Fk+Gk)=207.500/(2140.OOo+675.000)=0.067m因IeykBy6-0.5O0my方向小信心PkmHXy=(Fk*Gk>+6*Mdxk(B*Bx)=(244O.OOO+675.000)/13.500+6«207.5001/(3.00*4.500)=261.18IkPaPkminy=(Fk-Gk)-6*Mdxk(B*Bx)-(2440.000*675.000)/13.5006*207.5001/(3.00*4.500)=200.OOOkPa3.确定法础底面反力设计伯Pkmax=(Pk三ax-pk)+(Pk»ax_y-pk)+pk-(234.691230.741)*(261.481230.741)*230.741=265.432kPayo*Pknax=1.0*265.132=265.432kPa1.2*fa=1.2*2W.000=288.OOOkPa偏心荷载作用下地基承栽力湎足要求七、基砒冲切险算1 .计。基础底面反力设计伯1.1 1计算X方向塞础底面反力设计值©x=Mdy/(F+G)=32.000/(2440.00011.250)=0.O1.Oio因exBx6.0=0.75O三X方向小信心Paaxx=(F+G)A+6*Mdy/(B*By)=(2440.000-911.250)/13.500+6*32.0001/(4.50+3.000)=251.40IkPaPin,x=(F+G)A-6Mdy/(B*By)=(2440.000-911.250)/13.500-6*32.0(K)/(4.50*3.000)=215.08OkPa1.2 计比、,方向基础底面反力设计值ey-Mdx(FfG)166.000/(2440.000*911.250)-0.05Om因eyBy6=0.500y方向小(ft心PBau=(F+G)A+6*Mdx(B*Bx)=(2440.000-911.250)/13.500+6*166.000(3.00*1,500)=272.833kPnPain_y=(F+G)A-6*IMdx/(B*Bx)=<2410.000-911.250)/13.500-6*166.000/(3.00»1.500)=223.648kPa1.3 因MdXWOMdy0Pax=Pmax-x÷Pmax-y-(F+G)=251.401+272.833-(2140.OOO+911.250)/13.500=275.99妹Pa1.4 计算地基净反力极值PjInaX=P三ax-GA=275.991-911.250/13.500=208.494kPaPjmax_x=Pmax_x-G/A=252.191-911.250/13.500=181.69IkPaPjmax_y=Pmax_y-G/A=278,981-911.250/13.500=211.48IkPa2 .验算柱边冲切YII=h1.=O.500a,YB=bc=2.HOOm1Y1.=hc=O.500三YBI=B1+Bc2=2.25011,YB2=B2*Bc2=2.250«,Y1.1.=A1.=I.500Y1.2=A2=I.500«YHo=YH-as=0.460m2.1因(YH800)hp=1.0X冲切位置斜械面上边长X冲切位置斜板面卜边长X冲切不利位置X冲切嵌面上的地基净反力设计优2.2 X方向柱对基础的冲切验算bt=YB=2.80Ombb=YB+2YHo=3.720mbn-(btbb)2(2.800+3.720)/2-3.260mF1.x=A1.x*Pjmax=0.000*208.-194=0.OOOkNY。，FIX=I.0*0.000=0.OOkNoF1.x0.7*Bhp*ftbb<n*YHo(6.5.51)=0.7*1.OoO1.43*3260*460=1501.IOkNX方向柱对基础的冲切湎足规范要求2.3 y方向柱对基础的冲切5金算y冲切位置斜敲面上边长at=Y1.=0.50Omy冲切位置斜能面下边长ab=Y1.÷2*YHo=1.120my冲切面枳1.y-max(YB1.YB/2YHo)*(YU2YHo)t(YB1YB/2Y1.1.ot(YB2YB2YHo)(Y1.t2*Y1.1.o)(YB2YB/2Y1.1.o)=max(250-2.800/2-0.160)»(0.500+0.460)+(2.250-2.800/2-0.160,(2.250-2.800/2-0.-160)*(0.500-0.160)+(2.250-2.800/2-0.460)-三x(0.706,0.706)=0.706y冲切收面上的地净反力设计值FIy=A1.yPjmax=O.706*208.494=147.176kNoF1.y-1.0*147.176=147.18kXYo*F1.y0.7*hp*ftb*am*Y1.1.o(6.5.5-1)=0.7*1.000*1.43*96O*-I6O=442.04kNy方向柱对基础的冲切满足规箍要求八、柱下基础的局部受压及重因为基础的混凝土覆度等级大于等于柱的混凝土强度等级，所以不用验算柱下扩展舰础顶面的局部受压承载力。九、基砒受穹计算1 .因Mdx>O.Mdy>O此基础为双向受弯2 .计算IT截面考矩因exBx6=O.750nX方向小偏心PjI=11ax(Bx-a1.)(Pxmax-Pminx)Bx)+Pmin_x-G/A,(Bx-a2)*(Pxmax-Pmin_x)/Bx)+Pmin_x-G/A)=max(4.500-1.250)*(252.191-244.290)/4.500)+244.290-911.250/13.500,(4.500-1.250)*(252.191-244.290)/4.500)*2M.290911.250/13.500)=182.497kPa因eyWBy6=O500my方向小儡心PJ2=max(By-a1)*(Pya-Py三in)/By)+Py三inHA,(Bya2)*(Pymax-Pymin)By)-Pynin-GA)=max(3.000-1.250)*(278.<)81-217.500)/3.000)+217.500-911.250/13.500,(3.000-1.250)*(27H.981-217.500)/3.000)+217.500-911.250/13.500)=185,864kPabbc1.2*bc22Bc-0.500/2*0.500/2H.700-2.200mh=maxhcI,hc2)=max(0.500,0.500)=0.50()b=1.021y=1.030MI-1=148x*(Bxb*(2By+h)CPj1+Pjmnx_x)=1.48*1.021*(4.500-2.200*(2*3.000-0.500)*(182.497+184.691)=268.59kN*mMIIJ=I/48*y*(By-h(2*Bx)*(Pj2+Pj«ax_y)-1481.030*(3.0000.500*(2*4.500*2.200)*(185.864*211.481)=596.91kN*m十、计算配解10.1计算AsxAsx_1.-Yo*MI_1./(0.9*(H-as)fy)=1.0*268.59*1/(0.9*(500.000-10.000)*360)=1802.ImAsx1.=Asx_1.=18O2.Imsx-sx1.By-1802.1/3.000-601.mmAsx=max(Asx,min*H*10()0)=max(601,0.150¾*500*1000)=750a三选择钢筋12150,实配面枳为754m11u10.2计算ASyAsy_I=Yo*M11-1.(0.9*(H-as)*fy)=1.0*596.91*1.(0,9(500.OOO-40.OOO)*360)二4005.OnAsy1.=Asy_1.=4005.OmASy=ASyI/Bx=1005.0/4.500=890m11Asy=11ax(Asy,pmin*H1000)-max(890.O.150%*500*1000)=S9O<b选拜钢筋121.25,实配面枳为905mm,
