双柱阶梯基础2计算.docx

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1、双柱阶梯基础2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)混凝土结构设计规范GB50010-2010(2015年版)二、不意图M计算类型：验鳏被面尺寸三、计年信息构件编次JCT1.几何参数台阶数1hc=50(Hnhc=500mn矩形柱宽bc=500三m矩形柱高矩形柱宽bc=500三n矩形柱高基础高度h1.=400rm一阶长度b1.=900mm2.材料信息加础混凝土等级：柱混凝土等级：b290011mC30ft_b=1.13N/WC30ftc=1.43Nmni阶宽度a1.-9001111a2-90011mfc_b=11.3Ni115fcc1.4.3Nn

2、1钢筋级别：HRB-100fy=360N111153.计算信息结构虫要性系数：o=I.0基础埋深：dh=1.100b纵筋合力点至近边距离：as-40mn基础及其上检十的平均容或：=20.OOOkWm最小配筋率：p三in=O.150作用在基础顶部荷载设计值Fgk1=236.OOOkNFqk1-58.OOOkNFgk2=93.OOOkNFqk2=11.OOOkNMgxkI=-3.000kN*mMgxk2=-27.OOOkN*nMgyk1-15.OOOkN*mMgyk2=23.000k*11VgxkI=5.OOOkNVftXk2=8.OOOkNVgyk1.=1.OOOkNVgyk2=IO.OOOkN

3、Mqxk1.=I.OOOkN=HiiMqxk2=-7,OOOkN*mMqyk1=6.OOOkMmMqyk2=6.OOokN*mVqxk1=2.OOOkNVqxk2=2.OOOkNVqyk1.=I.OOOkNYqyk2=3.OOOkN永久荷我分项系数rg1.=1.30可变荷我分项系数rq1.=1.50可变荷载分项系数r1.1.=1.00永久荷战分攻系数rg2=1.30可变荷段分项系数rq2=1.5Q可变荷栽调整系数r12=1.00Fk=FgkHFqk1.+Fgk2+Fqk2=236.000*58.000*93.000+11.000-398.OOOkNMxk=MgxkPFgk1(A2-AD-Mqx

4、k1.-FqkI(A2-1)+Mgxk2+Fgk2*(A2-1)Mqxk2+Fqk2(A2-1)=-3.000+236.000*(1.150-1.150+1.000+58.000*(1.150-1.150)+-27.000-93.000*(1.150-1.150)+-7.000+11.000*(1.15Q-1.150)=-36.OOOkNXmMyk=Mgyk1.-Fgk1(Bx2-B1.)-Mqyk1.-PqkI(Bx2-B1.)+Mgyk2+k2(Bx2-BI+Mqyk2+Fqk2*(Bx2-BD=15.000-236.000*(2.900/2-1.150)+6.000-58.000*(2.

5、900/2-1.150)23.000*93.000*(2.900/21.150)*6.000H1.OOO(2.900/2-1.150)=-7.0()0kNVxk=Vgxk1+Vqxk1Vgk2+Vqxk2=5.000+2.OOo-8.Ooo+2.000=17.OOOkNVyk=Vgyk1.+Vqyk1.+Vgyk2+Vqyk2=1.000+1.000+10.000+3,000=15.OOOkNF-rg1.*Fgk1.rq1.*r1.1.*Fqk1.g2Fgk2rq2r1.2Fqk2=1.30*236.000+1.50*58.0001.：W+1.00*93.000+1.50*1.00*11.00

6、0=531.200kN*mMx=r1*(MRxk1+Fgk1*(2-A1)+rq1.*r1.1(Mqxk1.+Fqk1.*(A2-A1)+rg2*(Mgxk2+Fgk2*(2-D)*rq2*r1.2*(Mqxk2*Fqk2*(A21)=1.30*(-3.000+236.000(1.150-1.150)+1.50*1.00*(1.000+58.000*(1.150-1.150)+1.30*(-27.00093.000*(1.150-1.150)+1.50*1.00*(-7.000+11.000*(1.150-1.150)=-48.OOOkN*mMy-rg1.*(Mgyk1.Fgk1.*(Bx2B

7、1.)+rq1.r1.1(Mqyk1-Fqk1.*(Bx2-B1.)=1.30*(15.000-236.000(2.900/2-1.150)+1.50*1.00*(6.000-58.000(2.900/2-1.150)+1.30*(23.00093.000*(2.900/2-1.150)+1.501.00*(6.000+11.000*(2.900/2-1,150)=-9.520kNV=rg1*Vgk1+rq1*r11*Vqk1*rg2*Vgk2+rk1.+rq1.*r1.!*Vqyk1.+rg2*Vgyk2+r2q*r1.2*Vqyk2=1.30*1.000+1.50*1.00*1.000+1

8、.3010.OO0+1.50*1.00*3.000=20.300kN5.修IE后的地基承载力特征值fa=190.OOOkPa四、计算套敷1 .范础总长Bx=B1.B2+Bc=1.150+1.150+0.600=2.900112 .基础总宽By=A1.+A2=1.150-1.150=2.300mA1.-a1.*三ax(hc1.,hc2)20.900bhx(0.500,0.500)/2=1.150n2=a2+三ax(hc1.,hc2)2=0.!J()0+bhx(0.500.0.500)/2=1.150BB1.=b1.+bc1/2=0.900+0.500/2=1.150mB2=b2+bc22=O.9

9、000.500/2=1.150m3 .基础总高Ikh1.W_400=0.100a4 .底板花筋计算高度ho=h1.-HS=0.WO-O.0-10=0.360m5 .范础底面积A=BX*By=2.900*2.300=6.670:6 .Gk=*BxBy*dh=20.0002.9002.300*1,100=146.7WkNG=1.35*Gk-1.35*146.740=198.099kN五、计算作用在旃底部学矩值YdXk=MXk-Vyk*H=-36.000-15.000*0.100=-12.OOOkN*mMdyk=Myk+Vxk*i1.=-7,OOOH7,0000.4OO=-O.200kNmMdx-M

10、xy*1.1.-18.60020.300*0.100-56.720kN*三Ydy=My+,x*H=-9.520+22.950*0.100=-O.340k*a六、设算地基承假力1 .璇尊柏心荷俄作用下地基承武力Pk二(FkWk)/R=(398.000*146.710)/6.67(M1.670kPa【5.2.2-2因yo*pk=1.0*1.670=81.67OkpaWfa=I90.OOOkPa轴心荷敦作用下地范承毂力满足要求2 .麟尊偏心荷我作用下的地基承载力xk=Mdyk/(FkGk)=-O.200/(398.000H46.740)=-0.OOOm因IexkBx6=0.43mK方向小偏心，由公式

11、【旗.2.2-2】和【5.2.2-3】推导Pkmax.x=(Fk+GkA+6*MdykI/(BXxBy)-(398.OOOM6.740)/6.670*6*0.200/(2.900*2.300)=81.732kPaPkmiivx=(Fk+GkA-6*MdykI/(BX-By)=(398.000+M6.740)/6.670-6*-0.200/(2.9C02.300)=81.608kPaeyk=Mdxk(Fk4Gk)=-42.000/(398.000+146.740)=-0.077m因IeykBy6=0.383my方向小储心Pkmiixy=(Fk与kA+6*MdxkI/(ByBx)=(398.000

12、+146.740)/6.6706*-12.0001/(2.300,2.900)=98.097kPaPkmin_y=(FkGk)-6*iMdxk/(By,*Bx)=(398.000+146.740)/6.670-6*-12.0001/(2.30012,900)=65.244kPa3.确定基础杼面反力设计值PkmaX=(PkeaX_X-Pk)+(PkeaX_y-pk)+pk=(81,732-81.670)+(98.097-81.670)+81.670三98.159kPayo*1.,kmax=1.0*98.159=98.159kPa1.2*fa=1.2*190.000=228.(X)OkPa偏心荷我

13、作用下他菸承我力涵足要求七、基1冲切及算1.计算基础底面反力设计值1.1.计算X方向基础底面反力设计值ex=Mdy-537.300*198.099)/6.6706*0.3401/(2.90012.300)=110.!49kPa1.2计算y方向菸础底面反力设计俏ey=MdxF+G)=-56.720/(537.300+198.099)=-0.077m因eyBy60.383y方向小信心PeaX_产(F+C)A+6*Mdx(By-Bx)=(537.300+198.099)/6.67O6*-56.720(2.300:*2,900)=132.438kPaPin_y-FG)6*MdxI/(By*Bx)=(537.300+198.099)/6.670-6*-56.7201(2.300i*2.900)=88.07IkPa1.3因M1.xOMdy0Paax二PInnX_x*Pinaxy-(F*G)=110.360+132.438-(537.300+198.099)/6.670=132.544kPa1.4计算地基净反力极值PjmnX-P三a-GA=1.32.541198.099/6.670