阶梯基础计算DJJ03.docx
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：JC-3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=50011m矩形柱高Hc=I1.OOmB基础高度h1.=500rm一阶长度b1.=950mnb2=950三一阶宽度a1.=950mnu2=950am2 .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=11.3Nm1钢筋级别：IIRMOOfy=36（Wm三13 .计算信息结构重要性系数：Y0=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Fk=16.480kNFqk=100.OOOkNMgxk-84.000k!*mMqxk-O.OOOkVmMgyk=O.000kN*wMqyk=O.OOOkMnVgxk=55.200kNVqxk=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可变荷我调整系数r1.=1.00Fk=FRk+Fqk=16.480+100.CIOQ=I16.48OkNMxk=Mgxk÷Pgk*(2-A1.)24Wqxk÷Fqk*(2-A1.)2=84.000÷6.480*(1.500-1.500)/2+(0.000)+100.000*(1.500-1.500)/2=&1.OoOkMff1.Myk=Mgyk+Fk*(B2-B1.)2+Mqyk+Fqk*(B2-BI)/2-0.000*16.480*(1.2001.200)/2+(0.000)H00.000(1.200-1.200)/2=0.OOOkN*mVXk=VgXk+Yqxk=55.200+(0.000)=55.20OkNVyk=Vgyk+Yqyk=O.000+(0.000>=0.OOOkNF-rg*Pgk*rqr1.*Fqk=1.35*16.480*1.501.00*100.000-172.248kNMx=rg*(Mgxk-tFgk*(A2-I)2)-rq*r1.*(MqXk+Fqk*(2-A1)/2)=1.35*(84.000+16.180*(1.SO(T500)/2)+1.50*1.00*(0.000+100.000*(1.500-1.500)/2)=113,400kN*mMy-rg(MgykFgk*(B2B1.)/2)rq*r1.*(MqykFqk*(B2B1.)/2)=1.35*(0.000+16.480*(1.200-1.200)/2)+1.50*1.00*(0.000+100.000*(1.20Q-1.200)/2)=0.OOOkN*mVx=rVgk+rq*r1.*Vqxk=1.35»(55.200)+1.50*1.00*(0.000>=74,520kNVy-rgVgyk*rq*r1.*Vqyk1.35(0.000)*1.501.00*(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=0.950*0.950*0.5002.100m2 .基础总宽By=a1.+a2+hc=0.950+0.<)50+1.100=3.OOOmA1.=a1.÷hc2=0.950+1.100/2=1.500mA2=a2+hc2=O.950+1.100/2=1.500mB1.=b1.+bc2=0.950÷0.500/2=1.200mB2=b2+bc2=O.950+0.500/2=1.20Om3.基础总高H=h1.).500=0.500a1.底板配筋计算离度ho=h1.-as=0.500-0.(M0=0.160m5 .葩础底面积A=Bx*By=2.100*3.000=7.200«:6 .Gk=y*BxBy*dh=20.0002.4003.000*3.Ooo=432.OOOkNG=1.35Gk-1.35*432.000-583.20OkX五、计算作用在旃底部弯矩值Mdxk=Mxk-Vyk*H=84.0000.000*0.500=84.OOOkN*Mdyk=Myk+Yxk*H=0.000+55.200*0.500=27.600WweUdx=Mx-VytH=I13.4000.000*0.500-113.400kNmMdy=My+Vx*H=O.000+74.520*0.500=37.26OkN六、设算地基承或力1 .脸算轴心荷我作用下地基承载力Pk=(FkHJk)/R=6.480+432.000)/7.20076.178kPa5.2.1-2因f>*pk=1.00*76.178=76.178kPafa=230.OOOkPa轴心荷我作用下地菸承找力满足要求2 .般用偏心荷栽作用下的地基承载力exk-Mdyk/(Fk*Gk)=27.600/(116.480*432.000)-0.050m因IeXkBx6=0.100mX方向小倡心，由公式【5.2.2-2】和【5.2.2-3】推好Pkmuxx=(Fk÷Gk)+6*Mdyk(B2*By)-(116.480M32.000)/7.200*6*127.600/(2.4003.000)=85.761kPaPkmin_x=(Fk*Gk>A-6*IMdykI/(BxBy)=(116.480+432.000)/7.200-6*27.600/(2.4002*3.000)=66.594kPaeyk=Mdxk(Fk+<Jk)=84.«)0/(116.180÷432.000)=0.153m因IeykBy=O.5001.ny方向小假心Pkmuxy=(Fk-Gk)+6*Mdxk(ByBx)-(116.180*432.000)/7.200*6*84.000/(3.000*2.400)=99.51IkPaPkmin_y=(Fk*Gk>-6*Mdxk(By*Bx)=(116.180+432.000)/7.200-6*184.000/(3.000*2.100)=52.844kPa3.确定基础底面反力设计值Pkmax=(PkBaX_X-Pk)+(PkBaj1.y-Pk)+pk=(85,761-76.178)+(99,511-76.178)+76.178三109.091kPayo*1.,kmax=1.(X)*1.O<09-1=109.(XMkPaW1.2*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方向基础底面反力设计fex=W<F+G)=37.260/(172.248+583.200)=0.049m因CXWBx6.0=0.100nX方向小信心PaaX_X=(F+G)A*6*Mdy/(BxBy)=(172.248+583.200)/7.20(H6*37.2601/(2.400X1.000)=117.86IkPa1.in-x=(F÷G)A-6*1.Mdy/(BxBy)=(172.248+583.200)/7.2006*37.2601(2.4003.000)=91.986kPa2 .2计算y方向法础底面反力设计他ey=Mdx(F÷G)=113.100/(172.248+583.200)=0.150m因eyB>6=0.500y方向小信心1.ax-y=(F÷G)A÷6*!Mdx(By*Hx)=(172.248+583.200)/7.20(H6*113.WO1./(3.000s*2.400)=136.423kPain-y=(F*G)-6*!Mdx(By2*Bx)=(172.248+583.200)/7.200-6*113.WO/(3.000;*2.-100)=73.123kPa3 .3因MdxOMdy0htax-Pmax_x，PmaXy-(F+G)/A=117.861+136.123-(172.248+583.200)/7.200=149.36IkPa4 .4计算地基净反力极值Pjmax-PaaxG-149.361583.200/7.200-68.36IkPaPjmaxx=Pinax_x-G/A=117.861-583.200/7.200=36.«6IkPaPjmax_y=P(nax_y-G/A=136.423-583.200/7.200=55.123k1.2 .般算柱边冲切YII=h1.O.500a.YBbc=O.500m.Y1.=hc=1.100nYB1.=B1.=I.2()0m,YB2=B2=1.200m,Y1.1.=A1.=I.500m,YI2=A2=1.5(X)mYHO=YH-as=0.160m因YH800.取BhP=1.O2.1 X方向柱对基础的冲切险竟X冲切位置斜截面上边长at=Y1.=1.100mX冲切位置斜裁面下边长a>=Y1.+2YHo=2.020mX冲切面积AIX=max(YB1.-YB2-YHu)*(Y1.+2*YHu)+(丫BI-YB/2-YH。)：(YB2-YB/2-YHo)*(YI2*Y11o)(YB2YB2-YHo)i)=三ax(1.200-0.500/2-0.460)1.1000.-160)+(1.200-0.52-0.460)s,(1.200-0.500/2-0.460)*<1.,100÷0.460)+(1.200-0.500/2-0.460):)=三ax(1.230,1.230)=1.230n*X冲切极面上的地基净反力设计值F1.X=AIX*Pjmax=1.230*68.361=84.077kNyo11x=1.00*84.077=84.08kNyoF1.x0.7*hp*ftb*am*Y1.1.o(6.5.5-1)=0.7*I.Ooo*1.43*1560*460=718.32kNX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=1.120my冲切不利位置b三=(bt+bb)/2=(0.5O0÷1.420)/2=0.960my冲切面积AIyHnaX(YuY1.2-YHo>*(YB÷2*YHo)(Y1.1.-Y1.2-YHo)j,(Y1.2-Y1./2-YHo)*(YB+2*YHo)+(Y1.2-Y1.2-YHo)=vax(1.500-1.100/2-0.-160)*(0.50(K2*0.460)+(1.500-1.100/2-0.460)z,(1.500-1.100/2-0.160)»(0.500+2*0,460)+(1.500-1.10
