双柱阶梯基础计算1.docx

《双柱阶梯基础计算1.docx》由会员分享，可在线阅读，更多相关《双柱阶梯基础计算1.docx（6页珍藏版）》请在优知文库上搜索。

1、双柱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)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=11.3Ni115fcc1.4.3Nn1

2、IIRB100fy=360N11三4钢筋级别：3.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=210.OOOkNFqk1-58.OOOkNFgk2=31.OOOkNFqk2=8.OOOkNMgXk1.=-6.000kN*11Mgxk2=13.OOOkNWiMgyk1-15.OOOkN*mMgyk2=18.OOOkN*mVgxk1.=4.OOOkNVgxk2=7.OOOkNVgyk1.=2.OOOkNVgyk2=-5.OOOkNMq

3、Nk1.=T.OOOkVmMqXk2=3.OOOkN*mMqyk1.=10.OOOkVmMqyk2=5.OOokN*mVqxk1=3.OOOkNVqxk2=2.OOOkNVqyk1.=I.OOOkNVqyk2=-1.OOOkS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=FgkbFqk1.+Fgk2+Fqk2=240.000+58.000*31.000+8.000=337.OOOkNMxk=MgxkPFgk1(A2-AD-Mqxk1.-

4、FqkI(A2-1)+Mgxk2+Fgk2*(A2-1)Mqxk2+Fqk2(A2-1)=-6.000+240.000*(1.150-1.15O)+-4.000-58.000*(1.150-1.150)13.000-31.000*(1.150-1.150) +3.OoO+8.000*(1.150-1.150)-6.OOokNomMyk=Mgyk1.-Fgk1(Bx2-B1.)-Mqyk1.-PqkI(Bx2-B1.)Mgyk2Pgk2*(Bx2-BI+Mqyk2+Fqk2*(Bx2-BD=15.000-210.000*(2.900/2-1.150+10.000-58.000*(2.900/2-

5、1. ISO)*18.00031.000*(2.900/2-1.150)+5.0008.000*(2.900/2-1.150)=-29.70OkwmVxk=Vgxk1+Vqxk1Vgk2+Vqxk2M.000+3.000-7.OoO+2.000=16.OOOkNVyk=Vgyk1.+Vqyk1.+Vgyk2+Vqyk2=2.000+1.000-5.OOO+-1.000=-3.OOOkNF-rg1.*Fgk1.rq1.*r1.1.*Fqk1.g2Fgk2rq2r1.2Fqk2=1.30*240.000+1.50*58.OOO1.30+1.00*31.000+1.50*1.00*8.000M51.

6、300kN*mMx=r1.*(Mxk1.+Fgk1.*(2-A1)+rq1.*r1.1(Mqxk1.+Fqk1.*(A2-A1)+rg2*(Mgxk2+Fgk2*(2-D)*rq2*r1.2*(Mqxk2*Fqk2*(A21)=1.30*(-6.000+240.000(1.150-1.150)+1.50*1.00*(-4.000+58.000*(1.150-1.150+1.30*(13.000+31.000*(1.150-1.150)+1.501.00*(3.000+8.000*(1.15Ch1.150)=7.600kNmMy-rg1.*(Mgyk1.Fgk1.*(Bx2B1.)+rq1.r1

7、.1(Mqyk1-Fqk1.*(Bx2-B1.)=1.30*(15.000-240.000*(2.900/2-1.150)+1.50*1.00*(10.QoO-58.000*(2.900/2-1. 150)+1.30*(18.000+31.000*(2.900/2-1.150)+1.501.00*(5.000+8.000*(2.900/2-1.150)=-38.610kN*mV=rg1*Vgk1+rq1*r11*Vqk1*rg2*Vgk2+r三-0,048n因IeXkBx60.183mX方向小信心，由公式(5.2.2-2和05.2.2-3】推导PkmaxX=(Fk+GkA+6*IMdyk(Bx

8、*By)=(337.000+146.740)/6.6706*-23.3001/(2.90012,300)=79.752kPaPkmin_x=(PkCkA-6*Mdyk/(Bx*By)=(337.000+1-16.710)/6.670-6*卜23.300(2.9002,300)=65.297kPaeyk-Mdxk(Fk*Gk)7.200/(337.000146.740)=0.015m因IeykBy=O.：183my方向小信心Pkmax,y=(FkGk)+6*MdxkI/(By*Bx)=(337.OOO+M6.740)/6.670+6*17.2001/(2.3002,900)=75.34IkPaP

9、kmin_y=(FkGk)A-6*;Mdxk/(By*Hx)=(337.OoO+146.710)/6.670-6*17.200/3002.900)=69.709kPa3 .确定基础底面反力设计值Pkmax=(PkIIaX_X-Pk)+(Pk三ax-y-pk)+pk=(79.752-72.525)+(75.341-72.525)+72.525=82.568kPao*Pkmax1.0*82.568-82.568kPa1.2*fa1.2*190.000-228.OOOkPa偏心荷我作用下地基承领力满足要求七、基地冲切验算1 .计算基础底面反力设计值1.1 计或X方向基础底面反力设计值ex=Mdy(F

10、+G)=-31.375(41.950+198.(W9)=-0.(MHm因exBx/6.0=0.483X方向小信心1.axx=(F+G)A+6*!M1.y(BxBy)=(454.950*198.099)/6.670*6*31.3751(2.900,*2.300)=107.64IkPaPain_x=(FG)A-6*Mdy1./(Bx*By)=(454.950+198.099)/6.670-6*-31.375(2.900s*2.300)=88.176kPa1.2 计算y方向基础底面反力设计1ey=Mdx(F+G)=9.720/(151.950+198.099)=0.015m因CyBy6=0.383y

11、方向小偏心P三ax-y-(F*G)/A6*YdXI/(By*Bx)=-154.950+198.099)/6.67(K6*9.720/(2.300*2.$)00)=IO1.T1.OkPaPin_y=(F+G)A-6*iMdx(By2*Bx)=(454.950+198.099)/6.6706*9.720(2.300z*2.900)=94.107kPa1. 3因Mdx#OMdyOPBaX=Pmax_x+Pmax_y-(FM)=107.6414101.710-(454.950198.099)/6.670=111.442kPa1 .4计算地法净反力极值PJmax=1.a-GA=1.11,442-198.099/6.670=81.742kPaPjmaxX-PtoaX_x-G/A=107.493198.099/6.670=77.793kPaPjmaxy=Pinax.y-GA=98.