双柱阶梯基础计算.docx

双柱阶梯基础计算.H期.校对者.£建筑地基基础设计规范(GB50007-2011)£混凝土结构设计规范3(GB50010-2010)二、不意图三、计算侑息构件编号：JC-I1.几何参数台阶数n=1.矩形柱宽矩形柱宽求础高度计算类型:验算嵌面尺寸bc-60O三nbc=600三mh1.=500(m龙形柱高矩形柱高一阶长度b1.=700mm2 .材料信息基础混凝十.等级：柱混凝土等级：钢筋级别：3 .计算信息结构重要性系数：基础埋深：b2=70Omn一阶宽度C30ft_b=1.43NmC30ft_c=1.13NmHRIHOOfy=360Nhc-60011mhc=60011ma1.850mii2=850三nfc_b=14.3Nmfc_c=M.3Nmo=1.0dh=1.500m纵筋合力点至近边距离：as=40m基础及其上覆土的平均容重：丫二18.OOOkN/最小配筋率:P三in=0.15O4.作用在基础顶部荷我标准组合值F1=250.OOOkNF2=45.OOkNMx1=5.OOOkWmMx2=5.OOOkNMy1.=5.()OOkN>mMy2=5.OoOkN"Vx1=5.OOOkNV2=5.OOOkNVy1=5.OOOkNVy2=5.OOOkNks=1.30Fk=F1.ks1÷F2ks2=250.()00/1.30+45.OOO/1.30=226.923kNMx=Mx1ks+F1.*(Bx2-B1.)ks+Mx2ks+F2(B2-Bx2)ks=S.000/1.30+250.000(2.700/2-1.000)/1.30+5.000/1.30-45.000*(1.000/2-2.700)/1.30=62.885kNtmMy=My1.ks*F1.*(2-A1.)2ks+My2ks+2(A2-A1.)2ks=5.000/1.30+250.000»(1.150-1.150)/2/1.30+5.000/1.30*45.000*(1.i50-1.150)/2/1.30=7.692kN*mVx=Vx1.ks*Vx2ks=5.000/1.305000/1.30=7.692kNVy=VyIks+Vy2ks=5.000/1.30+5.000/1.30=7.692kNMxk=Mx1+F1<Bx2-B1.)+Mx2+F2*<B2-Bx2)=5.000*250.Oo0*700/2-1.000)+5,000-15.000*(1.000/2-2.700)=10.OOOkN*mMyk=My1.-E1.*(2-A1)2+Uy2ks+F2*(A2-AI)/2=5.000-250.000*(1.150-1.150)/2+5.OOo+45.000*(1.150-1.150)/2=61.625kN*mVXk=VXIVx2-5.000*5.000-10.OOOkNVyk=Vy1+Vy2=5.OOO+5.000=10.OOOkN5.修正后的地施承我力特征伯fa=90.OOOkPa四、计算割收1 .基础总长Bx=H1.÷B2+Bc=1.000+1.000+0.700=2.700m2 .茶础总宽By=A1.+A2=1.150*1.150=2.300mA1.=a1.+mx(hc1.,hc2)2=0.850÷三x(0.600,0.600)/2=1.150三2=a2*Bax(hc1.,hc2)2=0.850*box(0.600,0.600)/2=1.15O三B1.=b1.+bc1./2=0.700+0.600/2=1.OOOmB2=b2+bc22=O.7000.600/2=1.OOOm3 .基础总高H=h1.=O.500=0,500«4 .底板配筋计算高度ho=h1.as=0.500-0.040=0.460m5 .基础底面积A=BX*By=2.700*2.300=6.2106 .Gk=y*Bx*By*dh=18.OOo*2.700*2.300*1.500=167.67OkyG=1.35*Gk=1.35*167.670=226.355kN五、计算作用在基础底部弯矩值MdXk=MXkTXk*11=10.000-10.000*().500=5.OOOkN*aMdyk=Myk+Yyk*H=61.625+10.000*0.500=66.625kN*mMdX=MX-VX*H=62.885-7.6920.500=59.038kNmMdy-MyH'y1.k7.692*7.692*0.500=11.538kN*三六、设算地茶承力1 .骁算轴心荷我作用下地携承效力pk=(FkXIk)ZA=(226.923+167.670)/6.210=63.542kPa【5.2.2-2因yopk-1.0*63.542=63.542kPafa-90.OOOkPa轴心荷我作用下地基承毅力满足要求2 .骁算偏心荷载作用下的地及承我力Cxk=MdykZ(Fk-KJk)=S.769/(226.923+167.670>=0.015m因【exkBx0.150mX方向小信心，由公式【5.2.2-2和105.2.2-3】推导Pkmaxx=(Fk+Gk)+6*Mdyk(B*By)=(226.923+167.670)/6.210+6*15.7691/(2.70*2.300)=65.606kPaPkminX=(FkWk)A-6*Mdyk(B*By)=(226.923+167.670)/6.210-6*15.7691/(2.70*2.300)=61.477kPaeyk-Mdxk/(FHGk)=1.923/(226.923»167.670)-0.005m因ieykBy6=0.3S3my方向小偏心PkmaX_y=(Fk÷Gk)+6*MdxkI/(B*Bx)=(226.923+167.670)/6.210+6*11.9231/(2.30*2.700)=64.349kPPkmin_y=(FkYk)A-6*Mdxk/(B*Bx)=(226.923+167.670)/6.2106*1.9231/(2.30*2.700)=62.734kPa3 .确定基础底面反力设计值PkmaX=(PknaX_X-Pk)+(PkBaX_y-pk)+pk=(65,606-63.542)+(64.319-63.542)+63.542=66.4UkPaotPkmaxE.0*66.414-66.I1.4kPa1.2fa1.2*90.000-108.OOOkPa心荷我作用下地基承教力满足要求七、挈!肿切验算1 .计算基础底面反力设计值1.I计算X方向基础底面反力设计值ex=Mdy(F+G)=!1.5M(226.923+226.355)=0.025mISexBx/6.0=0.450«X方向小信心Peax_x=(F+G)A61.Mdy(B*By)=(226.923+226.355)/6.2106*11.5381/(2.70*2.300)=77.121kPaPin-X=(F+G)/A-6*1.Mdy/(B*By)=<226,923+226.355)/6.210-6*11.5381/(2.70*2.300)二68.863kPa1.2计算、,方向基础底面反力设计1ey=Mdx(F+G)=59.038/(226.923+226.355)=0.130m因CyBy6=0.383y方向小信心Paax_y-(FG)*6MdxI/(B*Bx)=(226.923+226.3S5)6.210+6*59.0381/(2.30*2.700)=97.792kPaPain-y=(F+G)A-6Mdx(B*Bx)=(226.923«226.355)/6.210-6*159.0381/(2.30*2.700)=48.19IkPa1. 3因Mdx#QJdyOPBaX=PmaX_x+Pmaxy-(F+)/A-77.121»97.792(226.923*226.355)/6.210=101.921kPa1.4计算地批净反力极值PJmax=PKix-GZA=IO1.921-226.355/6.210=65.47IkPaPjmnxX-Pmax,x-G-96.833226.355/6.210-60.383kPaPjmHXy=hnaxy-GA=75.092-226.355/6.210=38.642kPa2.验算柱边冲切YIhh1.=O.500YB=bc=1.350m,Y1.=hc=0.600«YB1.=B1.>Ik21.350m.YB2B2Bc2=1.35O三,Y1.1.=AI=1.I50b.Y1.2=A2=1.150三YIIo=YH-HS=O.160m2.1因(YHW800)Bhp=1.O2.2X方向柱对基础的冲切脸笄X冲切位置斜极面I.边长X冲切位置斜板面下边匕X冲切不利位置bt=YB=1.350mbb=YB÷2*Y1.1.o=2.270mb三=<bt+bb)/2=(1.350+2.270)/2=1.81OmX冲切成面上的地基净反力设计值F1.x-1.xPjmax-0.000*65.471-0.OOOkNo*F1.x=1.0*0.000=0.OOkSYo*F1.x0.7*Bhp*ft.b*tm*YHo(6.5.5-1)=0.7*1.0001.43*1810*460=833.43kNX方向柱对基础的冲切满足规范要求2.3y方向柱对暴础的冲切登算at=Y1.=O,600mabY1.2Mb-1.520my冲切位置斜鼓面上边长y冲切位置斜极面下边长y冲切面枳A1y=max(YBITB/2-YHO)»(Y1.+2*YHo)+(YB1-YB2-Y1.1.o,(YB2-YB2-YHo)»(Y1.+2*YHo)+(YB2-YB2-YHo)=InaX(1.350-1.350/20.460)(0.600*0.460)»(1.350-1.350/20.460.(1.3501.350/20.460)*9600+0.460)+(1.350-1.350/2-0.460)=ax(0.373,0.373)=0.373y冲切截面上的地基净反力设计值F1.y=A1.yPjmax=O.373*65.471=24.422kNYo»F1.y=1.0*24.422=24.42kNoF1.y0.7*1.*ft>am*YHo(6.5.5-1)-0.7*1.0001.43*1060*160F88.09kNy方向柱对基础的冲切满足规范要求八、柱下基础的局部受压险算因为基础的混凝土强度等级大干等干柱的混凝土强度等级.所以不用脸算柱下扩展基础段面的局部受压承我力.九、茶砒受穹计算1 .因Mdx>O,Mdy>O此基础为双向受弯2 .计算IT截面弯矩因exBx6=O.450mX方向小偏心Pj1.=max(<B-a1.)*(Pxma-Pmin-x)Bx)+Pmi11-<IA,(B-a2)*(Pxmax-Ptnin_x)/Bx)+Pmin-xGA)=max(2.700-0.850)*(96.833-49.150)/2.700)+49.150-226.355/6.210,(2.700-0.850)*(96.833-49.150)/2.700)+49.150-226.355/6.210)=45.372kPa因eySBy6=0.383my方向小偏心Pj2=max(By-a1.)(Pynax-PyBin)/By)+Py三in-CA,(Bv-h2)*(Pymax-Pymin)/By)-Pymin-GA)=max(«2.300-0.850)*(75.092-70,891)/2.300)+70.891-226.355/6.210,(2.300-0.850)*(75.092-70.891)/2.300)470.891-226.355/6.210)=37.089kPab=bc1.2+bc22+Bc=O.600/2*0.600/2+0.700=1.300mh=max(hc1.,hc2)=InaX(0.600,0.600)=0.600ax=1.046By=1.023MI1=1/48»Bx*(B-b*(2*By+h)*(Pj1.+Pjmax_x)=I481,016*700-1.300*(2*2.300'0,600)*(45.372-60.383)=23.48kN*mMII="48*By*(By-h(2*Bx-b)(Pj2+Pj三ax-y)=1/48*1.023*300-0.600*(2*2.700*1.300)*(37.089-38.612)=31,26kN*m十、计算配能10. 1计算AsxAsx-I=YoII-1.(0.9*(H-as)fy)=1,023.18*1.(0,9(500.000-40.000)360)=157.5«Asx1.=Asx_1.=157.5mAsx=Asx1.By=157.5/2.300=68*11sx=11ax(Asx,pmin*H1000)-max(68.0.150%5001000)=75O三,b选择钢筋121.50,实配面积为754m11u10.2计算ASyAsy-1.=oMI1.J(0.9(1.-as)*fy)=1.0*31.26*1.(0.9*(500.000-10.000)*360)=209.8nAsy1.=Asy_1.=209.8mAsy-Asy1./Bx-209.8/2.700二78三nAsy=mnx(sy,Pmin*H«1000)=max78,0.150S*500*1OOO)=75Ob选择钢筋12150,实配面枳为754mDu