J6双柱阶梯基础计算.docx

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1、J6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-70O11mbc=700nmh1=60Ofm1.龙形柱高矩形柱高hc-70O11mhc=70011m一阶长度b1.=220011nb2=2200nm一阶宽度a1.=2100mna2=21O0n2 .材料信息基础混凝十.等级：C35ft=1.57Nmfc_b=16.7Nm柱混凝土等级：C35ft_c=1.57Nmfc_c=16.

2、7Nm钢筋级别：HRMOOfy=360Nm3 .计算信息结构重要性系数：丫。=1.0基础埋深：dh=3.500m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值F1.=1700.OOOkNF2=1750.OOkNMxU150.000kX*mMx26O.OOOkN*mMy1.=IO.OOOkN*nMy2=-20.OOOk.X*nVx1.=IO.OOOkNV2=-20.OOOkNVy1.=-150.OOOkNVy2=-40.OOOkNks=1.00Fk=F1.ZksIF2ks2=1700.000/

3、1.00+1750.000/1.00=3450.OOOkNMx=Mx1ks+F1.*(Bx2-B1.)ks+Mx2ks+F2(B2-Bx2)ks-150.000/1.00-1700.000*(5.800/2-2.550)/1.00*60.000/1.00H750.OOO(2.550/25.800)/1.OO=192.500k*mMy=My1.ksF1.*(A2-A1.)2ks+My2ks+F2*(A2-A1.)2ks-10.000/1.00*1700.000*(2.1502.-150)/2/1.00-20.000/1.OO*-1750.000(2.450-2.450)/2/1.00=-10.(

4、)OOkX*mVx=VxIks+Vx2ks=10.000/1.Q0-20.000/1.OO=-IO.OOOkNVy=Vy1.ksVy2ks=-150.000/1.00-10.000/1.00=-190.OOOkNMxk-Mx1.F1*(Bx2-B1.)Mx2*F2*(B2Bx2)=150.OOOH700.000*(5.800/2-2.550)+60.000+1750.000(2.550-5.800/2)=192.500kN*mMyk=My1+F1*(A2-A1)2+My2ks+F2*(A2-A1)2-10.000*1700.OOO*(2.4502.450)/2,20.0001750.000(2

5、.450-2.450)/2=-10.000k*mVxk=Vx1+Vx2=10.000+-20.OOO=-IO.OOOkNVyk=Vy1.+Vy2=-150.00010.000=-190.OOOkN5.修正后的地基承段力特征值fa=194.OOOkPa四、计算1 .基础总长Bx=B1.+B2+Bc=2.550+2.550+0.700=5.800m2 .基础总宽By=1.*22.450*2.450-4.900m1.=a1.+三ax(hc1.,hc2)2=2.100+三ax(0.700,0.700)/2=2.450三A2=a2+三ax(hc1.,hc2)2=2.100+三ax(0.700,0.700

6、)/2=2.450B1.-b1.bc1/2=2.200*0.700/2=2.550mB2=b2+bc22=2.200*0.700/2=2.550m3 .基础总高H=h1.=O.600=0.600b从底板鼠筋计豫i度ho=h1.-as=0.600-0.(H0=0.560m5 .基础底面积A=BX*By=5.800*1.900=28.4206 .Gk-*BxBy*dh-20.000*5.800*1.900*3.500-1989.WOkNG=I.35*Gk=1.35*1989.400=2685.6A+6*Mdyk(B*By)=(3450.OoO+1989.400)/28.420+6*16.000/(

7、5.80*4.900)=191.976kPaPkmin_x=(Fk+Gk)/A-6*Mdyk(B*By)-(3450.0001989.400)/28.4206*16.OOO/(5.80*4.900)=190.81IkPaeyk=Mdxk(Fk+Gk)=2W.000/(3150.000+1989.400)=0.044a因IeykBy6-0.817my方向小信心PkmHXy=(Fk*Gk+6*Mdxk(B*Bx)=(3450.000+1989.400)/28.42(K6*240.000/4.90*5.800)=201.731kPaPkminy=(Fk-Gk)-6*Mdxk(B*Bx)-(3450.

8、000*1989.400)/28.4206*240.000/(4.9O5.800)=181.053kPa3.确定法础底面反力设计伯Pkmax=(Pk三ax-pk)+(Pkax_y-pk)+pk-(191.976-191.393)+(201.734191.393)*191.393=202.316kPao*Pknax=1.0*202.316=202.316kPa1.2*fa=1.2*194.OOO=232.80OkPa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=Mdy(F+G)=-16.000/(3450.W)O+2685

9、.690)=0.0O3NexBx6.00.967三X方向小信心Paaxx=(F+G)A+6*Mdy/(B*By)=(3450.000*2685.690)/28.420+6-16.0001/(5.80*1.900)=216.476kPaPBi1.X-(FM)A-6Mdy/(B*By)=(3450.000-2685.690)/2.420-6*-16.0(K)/(S.80*4.900)=215.31IkPa1.2 计比、,方向基础底面反力设计值ey-Mdx(FG)306.500/(3450.000+2685.690)-0.050aISeyBy6=O.81.7y方向小信心PBau=(F+G)A6*Md

10、x(B*Bx)=(3450.000-2685.690)/28.120+6*1306.500i(4.90*5.800)=229.099kPaP三in_y=(F+G)A-6*IMdx/(B*Bx)=1.82Omy冲切收面上的地净反力设计值FIy=A1.yPjmax=O.000*135.181=0.OOOkNoF1.y-1.0*0.000=0.OOkNYo*F1.y0.7*hp*ftb*am*Y1.1.o(6.5.5-1)=0.7*1.0001.57*1260*560=775.45kNy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

11、用验算柱下扩展舰础顶面的局部受压承载力。九、基砒受穹计算1 .因MdxO.MdyO此基础为双向受弯2 .计算IT截面考矩因CXBx6=0.967nX方向小偏心PjI=11ax(Bx-a1.)(Pxmax-Pminx)Bx)+Pmin_x-G/A,(Bx-a2)*(Pxmax-Pmin_x)/Bx)+Pmin_x-G/A)=max(5.800-2.100)*(220.107-211.380)/5.800)+211.380-2685.690/28.420,(8002.100)*(220.407211.380)/5.800)211.3802685.690/28.420)=122.639kPa因eyB

12、y6=0.817my方向小偏心PJ2=max(By-a1)*(Pya-Py三in)/By)+Py三inHA,(Bya2)*(Pymax-Pymin)By)-Pynin-GA)=max(4.900-2.100)*(224.446-207.341)/4.)00)+207.311-2685.690/28.420.(4.900-2.100)*(221.146-207.341)/4.900)+207.311-2685.690/28.420)=122,615kPabbc1.2*bc22Bc-0.700/2*0.700/2+0.700-1.400mh=maxhcI,hc2)=max(0.70(),O.700

13、)=0.700三X=1.024y=1.025MI-1=148x*(Bxb*(2By+h)CPj1+Pjmnx_x)=1.48*1.021*(5.800-1.400*(2*4.900-0.700)*(122.639+125.907)=1077.42kN*mMIIJ=I/48*y*(By-h(2*Bx)*(Pj2+Pjax_y)-1481.025*(4.9000.700*(2*5.800*1.100)*(122.615H29.946)=1236.26kN*m十、计算配解10 .1计算AsxAsx_1.-Yo*MI_1./(0.9*(H-as)fy)=1.O1O77.42*1.(0.9*(600.0

14、00-40.000)*360)=5938.ImAsx1.=Asx_1.=5938.ImASX=ASX1.By=5938.1.4.900;121211mAsx=max(Asx,min*H*10()0)=max(1212,0.150%*600*1000)=1212mm选择钢筋14包25,实配面枳为1232mm.11 .2计算ASyAsy_I=Yo*M11-1.(0.9*(H-as)*fy)=1.0*1236.26*1/(0.9*(600.000-40.000)*360)二6813.6nAsy1.=AsyJ=6813.6mASy=ASyI/Bx=6813.6/5.800=1175mmAsy=11ax(Asy,pmin*H1000)-max(1175,O.150*6001000)=II75mm选样钢筋18H70,实配面积为H83m