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1、J8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)(混凝土结构设计规范(GB50010-2010)二、不意图三、计算信息构件编号：JC1.计算类型：验尊限而尺寸1 .几何参数台阶数n=1.hc=50Onn矩形柱宽bc=50Oa矩形柱高基础高度hI=100nm阶长度b1.=800mmb2=800三m-一阶宽度a1=800m112=800nm2 .材料信息域础混凝土等级：C30ft1.43Nnfc_b=14.3Nm柱混凝土等级：C30ft.c=1.43Nmfc.c=14.3Nm钢筋级别：IIKB-100fy=360Nm3 .计算信息结构重要性系
2、数：Y0=1.00基础埋深：dh=2.500m纵筋合力点至近边距离：as-Wrrm基础及其上班士的平均容H1.：Y=20.OOOkN/最小配筋率:P三in=O.15O4 .作用在基础顶部荷载标准组合值E=95().OOOkXMx=-30.OOOkN*11My=-IO.OOOkN*wVx=-10.OOOkNVy=3O.OOOkNks=1.25Fk=Fks=950.000/1.25=760.OOOkNMXk=MXks=-30.000/1.25:24.OOOkVmMyk=MyZks=-IO.000/1.25=-8.OOOk.X*mVxk=VxZks=-IO.000/1.25=-8.OOOkNVyk=
3、VyZkS=30.000/1.25=24.OOOkN5 .修正后的地基承我力特征值fa=240.OOOkPa四、计算1 .基础总长Bx=b1.+b2+bc=0.800+0.800+0.500=2.100m2 .基础总宽By=a1.+a24hc-0.800*0.800*0.500-2.100mA1.=a1.hc2=0.H00+0.500/2=1.O5O三2=a2+hc2=O.800+0.500/2=1.050mB1.=b1.+bc2=0.800+0.500/2=1.050三B2=b2+bc2=0.8000.500/2=1.050m3 .基础总离H=h1.=O.400=0.100n4 .底板配筋计
4、算高度ho-h1.as-0.4000.040=0.360m5 .基础底面积A=BX*By=2.100*2.100M.4106 .Gk=y*Bx*By*dh=20.0002.100*2.100*2.500=220.50OkNG=1.35*Gk=1.35*220.500=297.675kN五、计算作用在基础底部弯矩值MdXk=MXk-Yyk*H=-2(000-24.000*0.400=-33.60()k*mMdyk=Myk+Yk*H=-8.000(-8.000)*0.100=-H.200kN*11Mdx=M-Vy*H=-30.000-30.0000.100=-42.OOOkN*.Mdy-MyWxM
5、kTO.000*(-10.000)*0.400-14.OOOkNm六、改墓地茶承力5.2.211 .蕤算轴心荷载作用下地携承我力Pk=(FkX)/A=(760.000+220.500)/4.410=222.336kPa因yopk=1.00*222.336-222.336kPafa=240.OOOkPa轴心荷我作用下地基承致力满足要求2 .蕤算偏心荷载作用下的地携承效力Cxk-Mdyk/(FkGk)=-11.200/(760.000+220.500)=-0.O1.1.m因IexkBx6-0.350mX方向小ft心，由公式5.2.2-2和105.2.2-3】推导PkmaxX=(Fkyk)A+6*M
6、dyk(B*By)=(760.000+220.500)/4.410+6*1-11.2001/(2.10*2.100)=229.592kPaPkmin-X=(Fk+CkA-6*Mdyk(B*By)=(760.000+220.500)/4.410-6*-11.2001/(2.10*2.100)=215,079kPaeyk-Mdxk(Fk*Gk)-33.600/(760.000*220.500)0.034m因IeykBy6=0.350my方向小信心Pkmax-y=(Fk-Gk)A+6*MdxkI/(B*Bx)=(760.000+220.500)/4.410+6*I-33.6001/(2.10*2.1
7、00)=244.104kPa1.,kmin-y=(HkIk)-6*Mdxk(B*Bx)=(760.000+220.500)/4.410-6I-33.6001/(2.10*2.100)=200,567kPa3 .确定基础底面反力设计值Pkmax=(PkBaX_X-Pk)+(Pkaaxj-Pk)+pk=(229.592-222.336)+(244.104-222.336)+222.336=251.36IkPayo*Pkmnx1.00*251.361=251.36IkPaS1.2*fa1.2*240.000-288.OOOkPn偏心荷我作用下地基承毅力满足要求七、涮冲切殴算1 .计算基础底面反力设计
8、值1.I计算X方向基础底面反力设计值ex=Mdy(FG)=-14.000/(950.000+297.675)=-0.O1.1.m因exWBx60=0.350X方向小信心1.ax.x=(F+G)A+6SYdy/(B*By)=(950.000*297.675)/4.410*6*14.000/(2.102.100)=291.990kPnPein-X=(F+G)/A-6IJMyI/(B*By)=950,000+297.675)/4.410-6*-14.000/(2.102.100)=273.849kPa1.2计算y方向基础底面反力设计值ey=MJxFG)=-12.000/(950.000+297.67
9、5)=-0.031m因cyWBy/6=0.350y方向小偏心Pnax_y-FG)/A6Mdx/(B*Bx)=(950.000+297.675)4.410+6*-42.000/(2.10*2.100)=310.13OkPaPMin.y=(F+G)A-6*1.M1.x(B*Bx)=(950.000297.675)/4.4106*42.000/(2.102.100)=255.709kPa1. 3因YdK0Mdy0PBaX=PmaX_x+Ptnax_y-(F+A二291.990310.130-(950.000*297.675)/4.410=319.20IkPa1.1 计篇地艇净反力极Pjmax=P三-
10、GA=319.201-297.675/4.410=251.70IkPaPjmaxx-Pmax_x-G/A291.990-297.675/4.410二224.49OkPaPjmax_y=Pfcax_y-C/A=310.130-297.675/4.410=242.63OkPa2 .骁算柱边冲切YH=h1.=O.100n,YB=bc=O.5O0m,Y1.=hc=O.500bYB1.-B1.=I.050m,YB2-B2=1.050m,Y1.1=1=1.050m1Y1.2=2=J.050mYHo=YH-as=0.360m2. 1因(YH800)hp=1.03. 2X方向柱对基础的冲切脸算X冲切位比斜极面
11、上边长bt=YB=O.50OmX冲切位置斜蔽面下边Kbh=YB2*YHo=1.220mX冲切不利位置=bt+bb)2=(0.500+1.220)/2=0.860InX冲切面积1x-nax(Y1.1.-V1.2-YHo)*(YB2YHo)(Y1.1Y1./2YHo,(Y1.2-Y1.2-YHo)(YB*2*Y1.1.o)(Y1.2-Y1.2-Y1.1.o=max(1.050-0.500/2-0.360)(0.500+2*0.360)+(1.05Q-0.500/2-0.360,(1.050-0.500/2-0.360)*(0,500+2*0.360)+(1.050-0.500/2-0.360)Hn
12、ax(0.730,0.730)=0.730X冲切板面上的地展净反力设计值F1.x=A1.xPjmax=0.730*251.701=183.842kNoF1.x-1.00*183.842-183.84kNoF1.x0.7*hp*ftb*1.xn*YHo(6.5.5-1)=0.7*1.000*1.43*860*360=309.9IkNX方向柱对基础的冲切满足规范要求4. 3y方向柱对基础的冲切验算y冲切位置斜能面上边长at=Y1.=0.500my冲切位置斜裁面下边长ab=Y1.*2YHo=1.220my冲切而积A1.y=max(YB1-YB2-Y1.1.o)*(Y1.+2*YHo)+(YB1-YB
13、2-YHo,(YB2-YB2-YHo)(Y1.+2*YHc)+(YB2-YB2-YHo)=max(1.050-0.500/2-0.360)(0.500+0.36O)+(1.050-0.500/2-0.360,(1.050-0.500/2-0.360)*(0.5000.360)1.0500.500/2-0.360)=BaX(0.730,0.730)=0.730y冲切截面上的地第净反力设计值F1.y-1.y*Pjmax-0.730*251.701-183.842kN丫FIy=1.00*183.842=183.84kNo*F1.y0.7*Bhp*ft_bam*YHo(6.5.5-1)=0.7*1.0
14、001.43*860*360=309.9IkNy方向柱对基础的冲切满足规范要求八、拿H受剪承假力好1 .计算剪力z=a1.*a2hc=800+800+500=2100mnBz=b1.+b2+bc800*800*500=2100amA,=Az*ma(b1.,b2)=2100,0*maxO,MdyO此基础为双向受弯2 .计算IT截面弯矩因exBx6=O.350mX方向小儡心a=(B-bc)2=(2.100-0.500)/2=0.800Pj1.=(Bx-a)*(Pinax_x-Pmin_x)/Bx)+Pmin_x-G/A=(2.1000.800)*(291.990273.819)/2.100)273.8-19297.675/4.410=217.579kPa因eyWBy6=O.350my方向小儡心a=(Byhc)/2=(2.100-0,500)/2=0.800nPj2=(B-a)*(Pmax_y-Pminy)/By)*Pmin_y-C/=(2,10
