《J3双柱阶梯基础计算.docx》由会员分享，可在线阅读，更多相关《J3双柱阶梯基础计算.docx（7页珍藏版）》请在第壹文秘上搜索。
1、J3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(GB50007-201.1)混凝土结构设计规范(GB50010-2010)二、不意图三、计算侑息构件编号：JC-I1.几何参数台阶数n=1.矩形柱宽矩形柱宽求础高度计算类型:验算嵌面尺寸bc-500三nbc=500amh1.=500(m龙形柱高矩形柱高hc-50O11mhc=50011m一阶长度b1.=900mn2 .材料信息基础混凝十.等级：柱混凝土等级：钢筋级别：3 .计算信息结构重要性系数：基础埋深：b2=900m一阶宽度C30ft_b=1.43NmC30ft_c=1.13NmHRIHOOfy=360No=1
2、.0dh=2.500ma1.=900nma2=900nnfc.b=1.4.3Nmfc-c=1.1.3Nm纵筋合力点至近边距离：as=40m基础及其上覆土的平均容重：丫二20.OOOkN/最小配筋率:Pain=0.150%4 .作用在基础顶部荷我标准组合伯F1=1230.OOOkNF2=IO5O.OOkNMx1.-30.OOOkN*mMx2=20.OOOkN*nMy1=10.OOOkN*nMy2=-10.OOOk.X*OOkX*mMy=My1.ksF1.*(A2-A1.)2ks+My2ks+F2*(A2-A1.)2ks-10.000/1.25H230.000*(1.1501.150)/2/1.2
3、5-10.000/1.25H050.000(1.1501.150)/2/1.25=24.000k*mVx=Vx1.ks+Vx2ks=40.000/1.25+-10.000/1.25=24.OOOkNVy=Vy1.ksVy2ks=30.000/1.25+10.000/1.25=32.OOOkNMxk-Mx1.F1*(Bx2-B1.)Mx2*F2*(B2Bx2)=-30.OOOT230.00()(4.300/2-1.150)+-20.O(X)+1050.000*(1.150-4.300/2)=130.OOOkN*mMyk=My1+F1*(A2-A1)2+My2ks+F2*(A2-A1)210.00
4、0*1230.000*(1.1501.150)/210.0001050.000(1.1501.150)/2=30.OOOkN*mVxk=Vx1.+Vx2=10.000+-10.000=30.OOOkNVyk=VyI-Vy2=30.000+10.000=40.OOOkN5 .修正后的地基承段力特征值fa=2-IO.OOOkPa四、计算1 .基础总长Bx=B1.+B2+Bc=1.150+1.150+2.000=1.300m2 .基础总宽By=1.2=1.150*1.1502.300m1.=a1.+ax(he1.,hc2)/2=0.9()0+三ax(O.500,O.500)/2=1.I50bA2=a
5、2+三ax(hc1.,hc2)2=0.900+三ax(O.500,O.500)2=1.150B1.-b1.+be1/2=0.900*0.500/2-1.150mB2=b2+bc220.900*0.500/2=1.150m3 .基础总高H=h1.=O.5=0.500a从底板鼠筋计豫i度ho=h1.-as=0.500-0.(H0=0.160m5 .基础底面积A=BX*By=4.300*2.300=9.8906 .Gk=*BxBy*dh-20.000*1.3002.300*2.500M94.50OkNG=1.35*Gk=1.35*494.500=667.575k五、计算作用在IM1.底部学矩值Mdx
6、k=Mxk-Vxk*!=130.000-30.000*0.500=115.OOOkN+mMdyk-MykHyk*11-30.000*40.000*0.500-50.OOOkX*mMdX=MX-YX*H=104.Om-24.000*0.500=92.O(WkwmMdy=MyVy*H=21.Ooo+32.000*0.500=40.OOOkN*三六、MMX承M1. 5金券轴心荷栽作用下地基承载力Pk=(HkHJk)/A=(1824.000491.500)/9.890=234.429kPa15.2.2-2因o*pk=1.0*23-1.429=234.429kPafa=210.OOOkPa轴心荷载作用下
7、地培承我力湎足要求2. 5金算偏心荷就作用下的地基承载力exk=Mdyk/(Fk)=50.000/(1H24.000+491.500)=0.022m因IexkBx6=0.717mX方向小偏心，由公式【5.2.2-2和5,2.2-3推导Pkmax,x=(Fk*GkA+6*Mdyk(B*By)=(1821.OOO+494.500)/9.890+050.000/(4.30*2.300)=211.183kPaPkmin_x=(Fk+Gk)/A-6*Mdyk(B*By)-(1824.000494.500)/9.8906*50.OOO/(4.30*2.300)=227.374kPaeyk=Mdxk(Fk+
8、Gk)=115.000(1821.000191.500)=0.050m因IeykBy6-0.383my方向小信心PkmHXy=(Fk*Gk+6*Mdxk(B*Bx)=(1824.OOO+494.500)/9.890+6*115.()00：/(2.30*1.300)=264.762kPaPkminy=(Fk-Gk)-6*Mdxk(B*Bx)-(1824.000*491.500)/9.8906*115.000/(2.30*-1.300)=204.095kPa3确定法础底面反力设计伯Pkmax=(Pk三ax-pk)+(Pkax_y-pk)+pk=(241.483-234.429)+(264.762-
9、234.429)*234.429=271.817kPao*Pknax=1.0*271.817=271.817kPa1.2*fa=1.2*2W.000=288.OOOkPa偏心荷载作用下地基承栽力湎足要求七、基砒冲切险算1 .计。基础底面反力设计伯1.1 1计算X方向塞础底面反力设计值ex=M1.y(F+G)=40.000/(1824.Ooo珀67,575)=0.016m因exBx6.0=0.717bX方向小信心Paaxx=(F+G)A+6*Mdy/(B*By)=(1824.000667.575)/9.8906140.0001/(1.30*2.300)=257.572kPaPBi1.X-(FM)
10、A-6Mdy/(B*By)=(1824.000-667.575)/9.890-640.000/(4.30*2.300)=246.285kPa1.2 计比、,方向基础底面反力设计值ey-Mdx(FtG)=92.000/(1824.000*667.575)-0.03711)ISeyBy6=0.383y方向小信心PBau=(F+G)A+6*Mdx(B*Bx)=1824.000667.575)/9.890-6*92.0001/(2.30*4.300)=276.196kPaPain_y=(F+G)A-6*IMdx/(B*Bx)=(1824.000-667.575)/9.890-6192.0001/(2.
11、30*4.300)=227.662kPa1.3 因MdXWOMdy0Pax=Pmax-xPmax-y-(F+G)=257.572+276.196-(1821.000+667.575)/9.890=281.839kPa1.4 计算地基净反力极值Pjmax=fax-GA=28I.839-667.575/9.890=214.339kPnPjmax_x=Pmax_x-G/A=258.983-667.575/9.890=191.483kPaPjmax_y=Pmax_y-G/A=282.262-667.575/9.890=214.762kPa2 .验算柱边冲切YII=h1.=O.500a,YB=bc=3.
12、250m.Y1.=hc=O.500三YBI=B1+Bc2=2.15011,YB2=B2*Bc2=2.150Y1.1.=A1.=I.150Y1.2=A2=I.150三YHo=YH-as=0.460m2.1因(YH800)hp=1.02.2 X方向柱对基础的冲切验算X冲切位置斜械面上边长X冲切位置斜板面卜边长X冲切不利位置X冲切嵌面上的地基净反力设计优bt=YB=3.250mbb=YB+2YHo=4.170mbn-(bttbb)/2-(3.250*4.170)/2-3.71OmF1.x=A1.x*Pjmax=0.000*211.339=0.OOOkNYFIX=I.0*0.000=0.OOkNoF1
13、.x0.7*Bhp*ftbbn*YHo(6.5.51)=0.7*1.000*1.13*3710*160=1708.3IkNX方向柱对基础的冲切湎足规范要求2.3 y方向柱对基础的冲切5金算y冲切位置斜敲面上边长at=Y1.=0.50Omy冲切位置斜能面下边长ab=Y1.2*YHo=1.120my冲切面枳1.y-max(YB1.YB/2YHo)*(YU2YHo)t(YB1YB/2Y1.1.ot(YB2YB2YHo)(Y1.t2*Y1.1.o)(YB2YB/2Y1.1.o)=max(150-3.250/2-0.-160)(0.500+0.460)+2.15(113.250/2-0.160,(2.1
14、50-3.250/2-0.-160)*(0.500-0.160)+(2.150-3.250/2-0.460)Fax097,0.097)=0.097y冲切收面上的地净反力设计值F1.y=A1.ytPjmax=O.097*214.339=20.689kNoF1.y-1.0*20.689=20.69kNYo*F1.y0.7*hp*ftb*am*Y1.1.o(6.5.5-1)=0.7*1.000*1.43*96O*-I6O=442.04kNy方向柱对基础的冲切满足规箍要求八、柱下基础的局部受压及重因为基础的混凝土覆度等级大于等于柱的混凝土强度等级，所以不用验算柱下扩展舰础顶面的局部受压承载力。九、基砒
