《DJ02阶梯基础计算.docx》由会员分享，可在线阅读，更多相关《DJ02阶梯基础计算.docx（7页珍藏版）》请在第壹文秘上搜索。
1、DJ02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(GB50010-20IO)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.=950rrta2=950mfc-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.=950mmb2=950nm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
2、=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=380.OOOkNMx=15.OOOkNnMy-15.OOOkNtmVx=50.OOOkNVy=50.OOOkNks=1.25Fk-F/ks-380.000/1.25-301.OOOkNMxk=MxZks=15.000/1.25=12.000k*mMyk=MyZks=15.000/1.25=12.OOOkN*mVxk=Vxks=50.000/1.25=40.OOOkNVyk=Vys=50.000/1.25
3、-40.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Bxb1.b2bc=0.950*0.950*0.400-2.300m2 .基础总宽By=a1.+a2+hc=0.950+0.95O+O.400=2.MOmA1.=a1.+hc2=0.950+0.100/2=1.150A2=a2+hc2=0.9500.1002=1.150mB1.=b1.+bc2=0.950+0.100/2=1.150B2=b2+bc2=0.950+0.400/2=1.150m3 .基础总高Ikh10.100=O-100a4 .底板毛筋计算高度ho=hI-HS=O.400-0.05
4、0=0.350m5 .粗础底面积A=BX*By=2.300*2.300=5.2906 .Gk=y*BxBy*dh=20.0002.3002.300*2.000=211.60OkNG1.35*Gk1.35*211.600-285.660kN五、计算作用在涮IjS部学矩值Mdxk=Mxk-Vyk*!=12.000-10.000*0.4OO=-4.OOOkBmMdyk=Myk+Yxk*H=12.Oo(H40.000*0.400=28.OOOkNnMdx-MxVyH-15.00050.000*0.400-5.000kN*三Mdy=My+VxH=1.5.000+50.000*0.4(K)=35.OOok
5、N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二(Fkgk)/A=(304.000+211.600)/5.290-97.467kPa因yo*pk=1.00*97.467=97.467kPafa=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exk=Ndyk/(FIdGk)=28.000/(304.OoO+211.600)=0.054m因IexkBx6=0.383mX方向小假心，由公式(5.2.2-2和5.2.2-3】推导PkmiixX=(FVGk)A+6*Mdyk(B*By)二(304.000+211.600)/5.29
6、0*6*28.0001/(2.30*2.300)=111.275kPaPkmiivx=(Fkyk)A-6*MdykI/(B*By)=(304.000+211.600)/5.29O-6*28.OOO/(2.30*2.300)=83.6S9kPaeyk=Mdxk(k+Gk)=-4.000/(304.0C0+2!1.600)=-0.008m因IeykBy6=0.383my方向小侑心Pkmaxy=(Fk-Gk)A+6*Mdxk(B*Bx)=(304.000+211.600)/5.290*6*1-4.000/(2.30*2.300)=99.439kPaPkmi1.Vy=(Fk-Gk)A-6*Mdxk|/
7、(B*Bx)=(304.000+211.600)/5.290-6*1-4.OOO/(2.30*2,300)=95.494kPa3 .确定基础底面反力设计伯Pkmax=(Pkeax_x-pk)+(Pkax-y-pk)+pk=(111.275-97.467)+(99.43-97.467)+97.467-113.247kPayo*Pkmax=1.00*113.247=113.247kPa1.2*fa=1.2*150.)0=180.OOCkPa偏心荷我作用下地葩承我力满足要求七、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=Miy(F+G)=35.000/380.OoO+
8、285.660)=0.053m因exBx6.0=0.383X方向小俏心P三ax-x=(F*G)*6Mdy(B*By)=(380.W)O+285.660)/5.290+6*35.0001/(2.30*2.300)=H3.093kPaPain.x=(F+G)/AfIMdy|/(B*By)=(380.000*285.660)/5.2906*35.0001/(2.30*2.300)=108.574kPa1.2 计算，方向基础底面反力设计他eyMdx(FG)=-5.000/(380.OoO+285.660)=-0.008mWeyBy6).383y方向小信心PBaK_y=(FM)A+6*,YB2=B2=1
9、.150m.Y1.1.=A1.=I.150m,YI2=A2=I.150mYHo=YH-as=0.350m2.2 1因(YH800)hp=1.02.3 X方向柱对基础的冲切5金算bt=YB=O.400mbb=YB2*YHo=1.100m1.三=(bt+bb)/2=(0.400+1.100)/2=0.750mX冲切位置斜椽面上边反X冲切位置斜豉面下边长X冲切不利位置X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(1.150-0.40020,350)(0.40
10、0+2*0,350)+(1.150-0.400/2-0,350,(1.150-0.100/2-0.350)*(0.100*2*0.350)1.1500.400/20.350)=max(1.020,1.020)=1.020X冲切截面上的地基净反力设计值F1.x-A1.xPjmax-1.020*91.559=93.390kN丫2FIX=1.00*93.390=93.39kNYo*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2.4 y方向柱对堪础的冲切验算y冲切位置斜低面上边长at=Y
11、1.=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)=11ax(1.150-0.400/2-0.350)(0.400W.350)+(1.150-0.400/2-0.350,(1.1500.400/20.350)*(0.400+0.350)+(1.150-0.400/2-0.350)=ax(1.020,1.020)=1.020y冲切破面上的地基净反力设计值E1.y=A1.y*Pjmax
12、=1.020*91.55tf=93.390kNo*F1.y=1.00*93.390=93.39kN5F1.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=950+950+100=23OOBmBz-b1.*b2+bc=950+950+400=2300wnA,=zna(b1.,b2)工2300.0*max950.0.950.0)=2.1O.Mdy0此基础为双向受弯2 .计算IT截面弯矩因exBx6=0.383mx方向小偏心a=(B-bc)2=300-0,100)/2=0.950PJ1-(Bxa)*(Pmax-xPmin_x)/Bx)Pmin_xG=(2.300-0.950)*(1.13.093-10.574)/2.300)+108.574-285.660/5.290=71.835kPa因cyBy6=0.383my方向小偏心a-(Byhe)/2-300-0.100)/2=0.95OaPj2=(By-b)*(Pmax_y-Pmin_y)/By)+Pmin_y-；/A=(2.300-0.950)*(128.299-123.368)/2.300
