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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: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=
2、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
3、调整系数r1.=1.00Fk=FRk+Fqk=16.480+100.CIOQ=I16.48OkNMxk=MgxkPgk*(2-A1.)24WqxkFqk*(2-A1.)2=84.0006.480*(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)=55.20
4、OkNVyk=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
5、.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.*pk=1.00*76.1
6、78=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=(FkGk)+6*Mdyk(B2*By)-(116.480M32.000)/7.200*6*127.600/(2.4003.000)=85.761kPaPkmin_x=(Fk*GkA-6*IMdykI/(BxBy)=(116.480+432.000)/7.200-6*27.600
7、/(2.4002*3.000)=66.594kPaeyk=Mdxk(Fk+-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.O09-1=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
8、础底面反力设计值1 .1计%X方向基础底面反力设计fex=W6=0.500y方向小信心1.ax-y=(FG)A6*!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=149.36Ik
9、Pa4 .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.
10、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)*(YB2*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