DJ03阶梯基础计算.docx
DJ03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(GB50010-20IO)三、计算信息构件编号：JC-I计算类型：验算截面尺寸1.几何参数台阶数n=1.he-100nma1.=550rrta2=550wnfc-bM.3Nmfc_c=14.3Nm矩形柱宽bc-100三m走形柱高基础度h1.MOOmn一阶长度b1.=550mmb2=550三n一阶宽度2.材料信息基础混凝土等级：C30ft.b=1.43Nm柱混凝土等级：C30ft_c=1.43Nm钢筋级别：HRB400fy=360Nm3 .计竦信息结构重要性系数：YO=1.OO基础埋深：dh=2.OOOm板筋合力点至近边距离：as=50rm域础及其上覆土的平均容用：y=20.OOOkN/最小配筋率：P三in=0.1504 .作用在基础顶部荷技标准组合伯F=150.OOOkNMx=15.OOOkNnMy-15.OOOkNtmVx=20.OOOkNVy=20.OOOkNks=1.25Fk-F/ks，150.000/1.25-120.OOOkNMxk=MxZks=15.000/1.25=12.000k*mMyk=MyZks=15.000/1.25=12.OOOkN*mVxk=Vxks=20.000/1.25=16.OOOkNVyk=Vyks=20.000/1.25=16.OOOkN5 .修正后的地基承我力特征伯fa=150.OOOkPa四、计算参数1 .基础总长Bx=b11b2bc=0.550*0.550*0.400-1.500m2 .基础总宽By=a1.÷a2+hc=0.550+0.550+0.400=I.500mA1.=a1.+hc2=0.550+0.100/2=0.750«A2=a2+hc2=0.550÷0.1002=0.750mB1.=b1.+bc2=0.550+0.100/2=0.75OB2=b2+bc2=0.550+0.400/2=0.750m3 .基础总高Ikh10.100=O-100a4 .底板毛筋计算高度ho=hI-HS=O.400-0.050=0.350m5 .粗础底面积A=BX*By=1.500*1.500=2.2506 .Gk=y*BxBy*dh=20.00O1.5001.500*2.000=90.OOOkNG=1.35*Gk1.35*90.00"121.50OkN五、计算作用在涮!底部学矩值Mdxk=Mxk-Vyk*!=12.000-16.000»0.W0=5.600kN*Mdyk=Myk+Yxk*H=12.OOOH6.000*0.400=18.400kN>Mdx-MxVyH-15.00020.000*0.400-7.000kNmMdy=My+Yx*H=15.000+20.000*0.4(K)=23.OOOkN杭六、设算地基承假力15.2.1-2】1 .验葬柏心荷载作用下地基承武力Pk-(FktGk)AV(120.000*90.000)/2.250-93.333kPa因yo*pk=i.00*93.333=93.333k1.,afa=150.OOOkPa轴心荷我作用下地施承载力满足要求2 .粉总偏心荷战作用下的地基承载力exkMdyk/(Fk*Gk)=18.400/<120.000*90.000)-0.088m因IeXkBx6=0.250mX方向小偏心，由公式(5.2.2-2和5.2.2-3】推导PkmiixX=(FVGk)A+6*Mdyk(B*By):(120.000*90.000)/2.2506*118.400(1.5t>*1.500)=126.(M4kPaPkmiivx=(Fkyk)A-6*Mdyk(B*By)=(120.000+90.000)/2.250-6*i18.1001/(1.50*1.500)=60.622kPaeyk=Mdxk(k4Gk)=5.600/(120.000+90.000)=0.027m因IeykBy6=0.250my方向小侑心Pkmaxy=(Fk-Gk)A+6*Mdxk(B*Bx)二(120.000190.000)/2.250*6*5.600/(1.50*1.500)=103.289kPaPkmi1.Vy=(Fk-Gk)A-6*Mdxk|/(B*Bx)=(120.000÷90.000)/2.250-6*:5.600/(1,50*1.500)=83.378kPa3 .确定基础底面反力设计伯Pkmax=(Pkeax_x-pk)+(Pkax-y-pk)+pk=(126.044-93.333)+(103.289-93.333)+93.333-136.OOOkPayo*Pkmax=1.00*136.000=136.OOOkPa1.2*fa=1.2*150.«)0=180.OoCkPa偏心荷我作用下地茶承栽力满足要求七、基砒冲切段算1 .计以基础底面反力设计值I计算X方向基础底面反力设计值ex=M<iy(F+G)=23.000/(150.OoO+121.500)=0.085m因exBx6.0=0.250»X方向小俏心P三ax-x=(F*G)*6Mdy(B*By)=(150.W)O+121.500)/2.250+6*23.000(1.50*1.500)=161.556kPaPain.x=(F+G)/AfIMdy|/(B*By)=<150.OoO+121.500)/2.2506*23.000(1.50*1.500)=79.778kPa1.2 计算，方向基础底面反力设计他ey=Mdx<F+G)=7.000/(150.000+121.500)=0.026mWeyBy6).250y方向小信心PBaK_y=(FM)A+6*<Mdx(B*Bx)=(150.Oo(H421.500)/2.250+6*7.0001/(1.50*1.500)=133.HIkPaPain-y=(F÷G)A-6Mdx(B*Bx)=(150.OOOH21.500)/2.250-6*7.OOO/(1.50*1.500)=108.222kPa1.3 因YdXWOMdy0PBaX-Pmax_xtPmax_y-(F+G)/A=161.556÷133.I1.1-(150.000÷121.500)/2.250=174.OOOkPa1.4 计总地基净反力极值PjmiJX=PBaX-G/A=I74.000121.500/2.250=120.OOOkPaPjmaxX=Pmaxx-GA=161.556-121.500/2.250=107.556kPaPjmax_y=Pmax_y-G/A=133.111-121.500/2.250=79.11IkPa2 .3金算柱边冲切YH-h1-0.100a.YB=bcO.400m.Y1.hc=O.100aYB1.=B1.=O.750m,YB2=B2=0.750m,Y1.I=AI=O.750m.YI2=A2=0.750mYHo=YH-as=0.35Om2.1因(YH800)hp=1.02.2X方向柱对基础的冲切段舞X冲切位置斜戴面上边长X冲切位地斜豉面下边长X冲切不利位网bt=YB=O.400mbb=YB÷2*YHo=1.100m1.三=(bt+bb)/2=(0.400+1.100)/2=0.750mX冲切而积A1x=nax(Y1.1.-Y1.2-YHo)*(YB+2*YHo)+G1.1-Y1.2-YHo,(Y1.2-YI2-YHo)(YB+2*YHo)+(Y1.2-Y1.2-=max(0.750-0.400/2-0,350)(0.400+2*0.350)+(0.750-0.400/2-0,350,(0.750-0.100/2-0.350)*(0.100*2*0.350)0.7500.400/20.350)=BaX(0.260,0.260)=0.260X冲切截面上的地基净反力设计值F1.x-1.xPjmax-0.260*120.OoO=31.200kNo*F1.x=1.00*31.200=31.20kXYo*F1.x0.7*hp*ftb*1.xn*YHo(6.5.5-1)=0.7*1.000*1.43*750*350=262.76kNX方向柱对基础的冲切满足规范要求2.3y方向柱对堪础的冲切验算y冲切位置斜低面上边长at=Y1.=0.400my冲切位置斜极面下边长ab=Y1.2f1.1.o=1.100my冲切面枳A1y=max(YB1.-YB2-YHo)*(Y1.+2*Y1.1.o)+(YB1.-YB2-YHo,(YB2-YB2-YHo)»(Y1.÷2*YHo)+(YB2-YB2-YHo)=11>ax(0.750-0.400/2-0.350)(0.400W.350)+(0.750-0.400/2-0.350,(0.7500.400/20.350)*(0.400+0.350)+(0.750-0.400/2-0.350)=ax(0.260,0.260)=0.260y冲切破面上的地基净反力设计值E1.y=AIy*Pjmax=0.260*120.000=31.20OkNo*F1.y=1.00*31.200=31.20kN5F1.yW0.7*hp*ft>*am*YHo(6.5.5-1)=0.7*1.0001.43*750*350=262.76kNy方向柱对基础的冲切湎足规范要求八、IM1.受剪承也险算1,计算剪力Az=a1.*a2+hc=550+550+400=150OanBz-b1.*b2+bc=550+550+400=1500wnA,=zna(b1.,b2)=1500.0max(550.0,550.0)=0.82三2Vs=A'*p=0.8*66.7=55.OkN基础底面短边尺寸大于柱宽加两倍基础有效高度，不需脸算受剪承战力！九、柱下基砒的局部受压险售因为基础的混凝十.强度等级大于等于柱的混凝土强度等级,所以不用验翼柱下扩展基础顶面的局部受压承我力。十、基破受穹计算1 .WMdx>O.Mdy>0此基础为双向受弯2 .计算IT截面弯矩因exBx6=0.250mX方向小偏心a=(B-bc)2=(1.500-0,100)/2=0.550«PJ1-(Bxa)*(Pmax-xPmin_x)/Bx)»Pmin_xG=(<1.500-0.550)*(161.556-79.778)/1.500)+79.778-121.500/2.250=77.570kPa因cyBy6=0.250my方向小偏心a-(By-hc)2=(1.500-0.100)/2=0.550三Pj2=(By-b)*(Pmax_y-Pmin_y)/By)+Pmin_y-<；/A=(1.500-0.550)*(133.111-108.222)/1.500)+108.222-121.500/2.250=69.985kPax=1.072y=1.027MI_1.=1./48»x*(Bx-bc(2*By+hc>*(Pj1.+Pjmax_x)=1481.072*(1.500-0.400*(2*1.500-0,100)*(77.570+107,556)=17.01kN*mMI1.1.=1.48*y*(By-hc*(2*Bx+bc)*(Pj2+PjBax-y>=1/48*1.027*(1.500-0.400*(2*1.500+0.100)(69.985+79.I1.1.)=13,13kN*m十一、计算瓜B10.1计算sxAsx_1.=yo«MI_1./(0.9*(H-as)*fy)=1.00*17.01*1.(0.9*(400.000-50.000)*360)=150.OnAsx1.=Asx_1.=150.OmAsx=Asx1./By=150.0/1.500=100mmAsx=nax(Asx1pmin*H1000)-mnxdOO,0.150%*400*1000)=600a,a选择钢筋10R130,实配面积为601mm.10.2计算AsyAsy_1.=yo«III_1./(0.9(1.1.-as)*fy)=1.00*13.13*1(O.9*(1.(>0.000-50.000)*360)=115.8mAsy1.=Asy_1.=115.8msy-sy1.Bx-1.15.8/1.500=77bi11Asy=max(Asy,Pmin*1.1.*1000)=max<77,0.150*400»1000)=600三选择制前10130,实配面积为604mm.
