阶梯基础计算DJJ01.docx

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1、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（GB50007-20U）E混凝土结构设计规范GB50010-2010（2015年版）简明高层钢筋混凝土结构设计手册?李国胜（第二版）二、示意图三、计算信息构件编号：JCT计算类型：验算板面尺寸1 .几何参数台阶数n=1.矩形柱宽bc=500nm矩形柱高hc=75O三n基础高度hIMOOrai一阶长度b1.=500mnb2=500三m一阶宽度a1.=525mna2=525m2 .材料信息基础混凝土等级：C30ft.b=1.43Nmn,fc_b=14.3N11三1柱混凝土等级：C30ft.c=1.13Nrnfc_c=11.3N

2、m1钢筋级别：IIRMOOfy=36（Wm三13 .计算信息结构重要性系数：Yo=1.OO基础埋深：dh=3.OOOni纵筋合力点至近边距离：as=40mn基础及其上覆土的平均容重：y=20.OOOkNZm最小配筋率：P三in=O.150%1 .作用在茶础顶部荷我设计伯Fgk=120,350kNFqk=100.OOOkNMgxk-84.000k!*mMqxk-O.OOOkVmMgyk=O.000kN*wMqyk=O.OOOkMnVgxk=33.690kNVqxk=O.OOOkNVgyk=O.OOOkNVqyk=O.OOOkN永久荷载分项系数rg=1.35可变荷我分项系数rq=50可变荷我调整系

3、数r1.=1.00Fk=FkFqk=120.350+100.000=220.350kNMxk=MgxkPgk*(2-A1.)24WqxkFqk*(2-A1.)2=84.000+120.350*(0.900-0.900)/2+(0.000)+100.000(0.900-0.900)/2=&1.OoOkMff1.Myk=Mgyk+Fk*(B2-B1.)2+Mqyk+Fqk*(B2-BI)/2-0.000-120.350*(0.7500.750)/2-(0.000)*100.000*(0.7500.750)/2=0.OOOkN*mVXk=VgXk+Yqxk=33.690+(0.000)=33.69O

4、kNVyk=Vgyk+Yqyk=O.000+(0.000=0.OOOkNF-rg*Fgk*rqr1.*Fqk1.35*120.3501.50*1.00*100.000-312.472kNMx=rg*(Mgxk-tFgk*(A2-I)2)-rq*r1.*(MqXk+Fqk*(2-A1)/2)=1.35*(84.000+120.350*(0.900-0.900)/2)+1.50*1.00*(0.000+100.000*(0.900-0.900)/2)=113,400kN*mMy-rg(MgykFgk*(B2B1.)/2)rq*r1.*(MqykFqk*(B2B1.)/2)=1.35*(0.00(H

5、120.350*(0.75Q-0.750)/2)+1.50*1.00(0.000100.000*(0.750-0.750)/2)=0.OOOkN*mVx=rVxk+rq*r1.*Vqk=1.35(33.690)+1.50*1.00*(0.000)=45.48IkNVy-rgVgyk*rq*r1.*Vqyk1.35(0.000)H.501.00*(0.000)=0.OOOkN5.修正后的地基承我力特征值fa=230.OOOkPa四、计算参数1 .基础总长Bxbb2bc=0.500*0.500*0.500-1.500m2 .基础总宽By=a1.+a2+hc=0.5250.525+0.750=1.8

6、00mA1.=a1.hc2=0.525+0.750/2=0.900mA2=a2+hc2=O.525+0.750/2=0.90OmB1.=b1.+bc2=0,500+0.500/2=0.750mB2=b2+bc2=O.500-0.500/2=0.7503 .基础总高H=h14.4009.100n4 .底板配筋计算岛度ho=h1.-ns=0.W0-0.(M0=0.360m5 .葩础底面积A=Bx*By=1.500*1.800=2.700n:6 .Gk=y*BxBy*dh=20.0001.5001.800*3.000=162.OOOkNG=1.35Gk-1.35*162.000-218.70OkX五

7、、计算作用在旃底部弯矩值Mdxk=Mxk-Vyk*H=84.0000.000*0.100=84.OOOkN*Mdyk=Myk+Yk*H=O000+33.690*0.100=13.476kN*Udx=Mx-VytH=I13.4000.000*0.400=113.400kNmMdy=My+Yx*H=O.00Q45.481*0.400=18.193kNtm六、设算地基承或力1 .脸算轴心荷我作用下地基承载力pk(Fk*Gk)=(220.350162.000)/2.700-141.61IkPa【5.2.1-2】因*pk=1.00*141.611=141.611.kPafa=230.OOOkPa轴心荷我

8、作用下地茶承找力满足要求2 .般用偏心荷栽作用下的地基承载力exk-Mdyk/(FkGk)13.476/(220.350H62.000)-0.035m因IeXkBx6=0.250mX方向小偏心，由公式【5.2.2-2】和【5.2.2-3】推好Pkmuxx=(FkGk)+6*Mdyk(B2*By)-(220.350*162.000)/2.700*6*13.476/(1.5001.800)=161.576kPaPkmin_x=(Fk*GkA-6*IMdyk|/(Bx-By)=(220.350+162.000)/2.700-6*113.476/(1.5001.800)=121.64TkPaCyk=M

9、dXk/(FkWk)=84.(X)0-6*Mdxk(By*Bx)=(220.350+162.000)/2.700-6*184.000/(1.8001.500)=37.907kPa3 .确定基础底面反力设计值Pkmax=(Pkax_x-pk)+(Pkax_y-pk)+pk=(161.576-141.611)+(245.315141.611)+141.611=265.279kPayo*1.,kmax=1.00*265.279=265.279kPa1.2*a=1.2*230.000=276.OOOkPa偏心荷我作用下地范承驶力满足要求七、基冲切险算1 .计以基础底面反力设计值1.1 计%X方向基础底

10、面反力设计fex=Mdy(FG)=18.193/(312.172+218.700)=0.03111因CXWBX/6.0=0.25OaX方向小信心PaaX_X=(F+G)A*6*Mdy/(BxBy)=(312.472+218.700)/2.70(H6*1H.193(1.5001.800)=223.683kPa1.in-x=(FG)A-6*1.Mdy/(BxBy)=(312.472218.700)/2.7006*18.1931(1.500;1.800)=169.779kPa1.2计算y方向法础底面反力设计他ey=Mdx(FG)=113.100/(312.472+218.700)=0.213m因ey

11、By6=0.300y方向小信心1.ax-y=(FG)A6*!Mdx(By*Hx)=(312.172+218.700)/2.700+6*113.WO1./(1.800:*1,500)=336.73IkPain-y=(F+G)A6*MdxI/(Byz*Bx)=(312.472+218.700)/2.70tt-6*113.WO/(1.8001.500)=56.73IkPa1.3因MdxOMdy0htax-Pmax_x，PmaXy-(F+G)/A=223.683+336.731-(312.472+21.700)/2.700=363.683kPa1.4计算地基净反力极值PjmaX-PBaXG/A-363

12、.683218.700/2.700-282.683kPaPjmaxX=MaX_x-G/A=223.683-218.700/2.700=142.683kPnPjmax_y=P(nax_y-G/A=336.731-218.700/2.700=255.73IkPa2.般算柱边冲切YII=h1.O.100n.YBbc=O.500m.Y1.=hc=O.75O三YB1.=B1.=O.750m,YB2=B2=0.750m,Y1.1.=A1.=O.900m,YI2=A2=0.9(X)mYHO=YH-as=0.360m因YH800.取BhP=1.O2.1 X方向柱对基础的冲切险竟X冲切位置斜敲面上边长at=Y1

13、.=O.75OmX冲切位置斜能面下边长ab=Y1.2Y1.1.o=1.170mX冲切面积AIX=max(YB1.-YB/2-YHu)*(Y1.+2*YHu)+(YB1.-YB/2-YHu)：(YB2-YB/2-YHo)*(YI2*Y11o)(YB2YB2-YHo)i)=三ax(0.750-0.500/2-0.360)*0.75OO.360)+(0.750-0.500/2-0.360)s,(0.750-0.500/2-0.360)*(0.750+。.360)+(0.750-0.500/2-0.360):)=三ax(0.225,0.225)=0.225mjX冲切极面上的地基净反力设计值F1.X=A

14、IX*Pjmax=Q.225*282.683=63.717kNyo11x=1.00*63.717=63.72kNYoF1.x0.7*hp*ftb*am*Y1.1.o(6.5.5-1)=0.7*1.000*1.43*1110*360=400.OOkNX方向柱对域础的冲切满足规范要求2.2 y方向柱对基础的冲切5金舞y冲切位置斜蔽面上边长bt=YB=O.500my冲切位置斜豉面下边长bb=YB*2Y1.1.o=1.220my冲切不利位置b三=(bt+bb)2=(0.5O01.220)/2=0.860my冲切面积1.y=max(Y1.1.Y1./2YHo)*(YB1.YB2)-(YB1YB/2Y1.1.o)72-(YB2-YB/2YHo)72.(YI.2-YI./2-YIIO)*(YBI+YB2)-(YB2-YB2-YHo)72-(YB2-YB2-YHo)72=ax(0.900-0.750/2-0.360)*(0.750+0.750)-(