J3阶梯基础计算.docx
J3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、计算信息构件编号：JC1.计算类型：验尊限而尺寸1 .几何参数台阶数n=1.hc=70Onn矩形柱宽bc=700m矩形柱高基础高度h1.-60011m一阶长度b1.=1700mb2=170011m一阶宽度a1.=1.700mma2=170011m2 .材料信息域础混凝土等级：C35ft>1.57Nnfc_b=16.7Nm柱混凝土等级：C35ft.c=1.57Nmfc.c=16.7Nm钢筋级别：IIKB-100fy=360Nm3 .计算信息结构重要性系数：Y0=1.00基础埋深：dh=3.500m纵筋合力点至近边距离：as=W11m基础及其上覆土的平均容H1.：=20.OOOkN/最小配笳率：Pin=0.1504 .作用在基础顶部荷载标准组合值F=1950.OOOkNMx=80.OOOkHmMy=40.OOOkWmVx-10.OOOkNVy=80.OOOkNks=1.00Fk=FZks=1950.000/1.00=1950.OOOkNMxk=Mxks=80.000/1.00-80.OOOk*mMyk=M>ks=40.0001.00=40.(>00kX*mVxk=Vks=40.000/1.00=40.OOOkNVyk=Vyks=80.000/1.00=80.OOOkN5 .修正后的地基承我力特征值fa=194.OOOkPa四、计算1 .基础总长Bx=b1.+b2+bc=1.700+1.700).700=4.100m2 .基础总宽By=aHa2hc=1.700H.700*0.7001.100mA1.=a1.÷hc2=1.700+0.700/2=2.()5O三A2=a2+hc2=1.700+0.700/2=2.050mB1.=b1.+bc2=1.700+0.700/2=2.050bB2=b2÷2=1.700+0.700/2=2.050m3 .基础总离H=h1.=0.600=0.600n4 .底板配筋计算高度ho=h1.-as=0.6000.040=0.560m5 .基础底面积A=BX*By=4.100*1.100=16.8106 .Gk=y*Bx*By*dh=20.000»1.1OO-1.100*3.500=1176.700kNG=1.35*Gk=1.35*1176.700=1588.545kN五、计算作用在基础底部弯矩值MdXk=MXk-Yyk*H=80.000-80.000*0.600=32.OOOkNMiIMyk=Myk+yk*H=40.Oo(H40.000*0.GOo=64.OOokN加MdX=MX-Vy*H=80.000-80.000*0.600=32.OOOkN*Mdy-MyWxMkIO.000*40.000*0.600=64.OOOkN*三六、改墓地茶承力1 .蕤算轴心荷载作用下地携承我力5.2.1-2Pk=(FkM)/A=(1950.000+1176.700)/16.810=186.002kPayo*pk-1.00*186.002=186.002kPafa=194.OOOkPa轴心荷我作用下地基承我力满足要求2 .蕤算儡心荷载作用下的地携承致力xk=Mdyk/(Fk-KJk)=64.000/(1950.000+1176.700)=0.020n因IeXkBxfr-iO.683mX方向小信心，由公式(5.2.2-2和105.2.2-3】推导PkmaxX=(Fkyk)/A+6*Mdyk(B*By)=(1950.000÷1176.700)/16.810*64.0001/(4.10*1.100)=191.571.kPaPkmin-X=(Fk+Gk>A-6*Mdyk1./(B*By)=(1950.OoO+1176.700)/16.810-6*61.0001/(4.10*1.100)=180.43IkPaeyk-Mdxk(Fk4Gk)=32.000/(1950.0001176.700)-0.O1.Om因IeykBy6=0.683my方向小信心PkmaX-y=(Fk÷Gk)÷6*MdxkI/(B*Bx)=(1950.OOOH176,700)/16.810÷6*32.0001(4.10*4.100)-188.788kPa1.,kmin-y=(HkIk)-6*Mdxk(B*Bx)=(1950.000÷1176.700)/16.810-6*32.000(4.10*4.100)=183.217kPa3 .确定基础底面反力设计值Pkmax=(PkBaX_X-Pk)+(Pkaaxj-Pk)+pk=(191.574-186.002)+(188.788-186.002)÷186.002=194,360kPayo*Pkmnx=1.00*194.360-191.36OkPaS1.2*fa1.2*194.Ooo=232.80OkPa偏心荷我作用下地基承毅力满足要求七、涮冲切殴算1 .计算基础底面反力设计值1.I计算X方向基础底面反力设计值ex=Mdy(F÷G)=64.000<1950.000+1588.545)=0.018mHexBx/6.0=0.683X方向小心1.ax.x=(F+G)A+6SYdy/(B*By)=<1950.0001588.545)/16.8106)64.0001(4.10i.100)=216.074kPaPein-X=(F+G)/A-6»IJMyI/(B*By)=(1950.000-1588.545)/16.810-6»164.000/(4.104.100)=2M.931kPa1.2计算y方向基础底面反力设计值ey=M<Jx<F÷G)=32.000/(1950.000+1588.545)=0.009m因cyWBy/6=0.683y方向小偏心Pnax_y-FG)/A6Mdx/(B*Bx)=(1950.000-1588.515)/16.810-6«32.0001/(4.10*4.100)=213.288kPaPMin.y=(F+G)A-6*1.M<1.x(B*Bx)=(1950.000*1588.545)/16.810632.000(4.10M.100)=207.717kPa1. 3因YdK0Mdy0PBaX=PmaX_x+Ptnax_y-(F+A=216.074+213.288-(1950.OOoH588.S45)16.810=218.86OkPa1.1 计篇地艇净反力极fftPjnax=P三-GA=218,860-1588.545/16.810=124.360kPaPjmaxX-PInaX_x-G/A=216.074158S.545/16.810121.574kPaPjmaxy=hnax_y-G/A=2!3.288-1588.545/16.810=118.788kPa2 .骁算柱边冲切YH=h1=0.600n,YB=bc=O.7O0m,Y1.=hc=O.TOOaYB1.-B1.=2.050m,YB2-B2=2.050m,Y1.1=1=2.050m,Y1.2=2=2.050mY1.1.o=YH-HS=O.56Om2. 1因(YH800)hp=1.03. 2X方向柱对基础的冲切脸算X冲切位比斜极面上边长bt=YB=O.700mX冲切位置斜蔽面下边长bh=YB÷2*YHo=1.82OmX冲切不利位置=bt+bb)2=(0.700+1.820)/2=1.260mX冲切面积1x-<nax(Y1.1.-V1.2-YHo)*(YB2YHo)(Y1.1Y1./2YHo,(Y1.2-Y1.2-YHo)(YB*2*Y1.1.o)«(Y1.2-Y1.2-Y1.1.o=max(050-0.700/2-0.560)«(0.700+2*0.560)+(2.05(H0.700/2-0.560,(2.050-0.700/2-0.560)*(0.700+2*0.560)+(2.050-0.700/2-0.560)=uax(3.374.3.374)=3.374X冲切板而上的地展净反力设计值F1.x=A1.xPjmax=3.374*121.360=419.64OkNoF1.x-1.00*419.640-119.64kNo<F1.x0.7*hp*ftb*1.xn*YHo(6.5.5-1)=0.7*1.000*1.57*1260*560=775.45kNX方向柱对基础的冲切满足规范要求4. 3y方向柱对基础的冲切验算y冲切位置斜能面上边长at=Y1.=0.700my冲切位置斜裁面下边长ab=Y1.*2YHo=1.820my冲切而积A1.y=max(YB1.-YB/2-Y1.1.o)(Y1.+2*YHo)+(YB1-YB2-YHo,(YB2-YB2-YHo)(Y1.+2YHo)+(YB2-YB2-YHo)=max(050-0.700/2-0.560)(0.700+0.560)+050-0.700/2-0.560,(2.050-0.700/2-0.560)*(0.7000.560)+0500.700/2-0.560)=BaX(3.374,3.374)=3.371y冲切截面上的地第净反力设计值F1.y-1.y*Pjmax-3.374*124.360"19.610kNF1.y=1.00*4)9.640=119.64kNo*F1.y0.7*Bhp*ft_b»am*YHo(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范要求八、拿H受剪承假力好1.计算剪力z=a1.*a2hc=1700-1700+700=I1.OOnmBz=bi÷b2+bc=1700*1700*700=4100Hm*=xmaxb1.,b2)=4100.Otmax(1700.0>1700.0)=6.97«2Vs='*p=7.()116.0=808.5kN丛础底面短边尺寸大于柱宽加两倍基础有效高度，不需验算受剪承1«力！九、柱下脚的局部受压睑算因为基础的混凝土强度等级大于等于柱的混凝土强衣等级,所以不用脸舞柱下扩展基础蹊面的局部受压承教力.十、受穹计算1 .因Mdx>0,Mdy>O此基础为双向受弯2 .计算IT截面弯矩因exBx6=0.683mX方向小儡心a=(B-bc)2=(4.100-0.700)/2=1.700bPj1.=(Bx-a)*(Pinax_x-Pmin_x)/Bx)+Pmin_x-G/A=(4.100-1.700)*(216.074-20-1.931)/4.100)*204.9311588.545/16.810=116.954kPa因eyWBy6=0.683my方向小儡心a=(Byhc)/2=(4.100-0,700)/2=1.700nPj2=(B>-a)*(Pmax_y-Pminy)/By)*Pmin_y-C/=(01.100-1.700)*(213.288-207.717)/4.100)÷207.717-1588.545/16.810=116,478kPaBx=I.024By=1.024M1.1.=I/48*x*(Bx-bc*(2*By+hc>(PjI+Pjmaxx)=1/48*1.021*(4.100-0.700*(2*4.100+0.700)*(116.954+121.571)=523.68kN*mM11J=148*y*(By-hc*(2Btbc)*(PJ2+PjBaxy)=1.48*1.021*(4.100-0.700*(2*1.100-0.700)*(H6.478+118.78«)=516.52kN*m十一、计算配筋10.1计算sxAsx_1.=o*M1._1/(0.9*(H-as)*fy)=1.00*523.68*1/(0.9»(600.000-10.QOo)*360)=2886.3nsx1.-sx-1=2886.3mAsx=Asx1.Hy=2886.3/4.I00=704mmAsx=maxAsx,pmin*f1*1000)=max(704,0.150*600*1000)二900三选择钢筋12S125,实配面枳为905Wnu10.2计算AsyAsy_1.=oMII-1.(0.9(H-as)*fy)=1.00516.52*1/(0.9*(600.00040.000)*360)=2846.8aAsy1.=Asy_1.=2846.8mAsy=Asy1.Bx=2846.8/4.100=694mmAsy-max(sy,Pmin*H<,1000)=max(694,O.15(XVt600*1（三）O)=900a选择钢筋12W25,实配面枳为905mm,
