三桩桩基承台计算

三桩桩基承台计算结构构件计算书PAGE第4页，共16页三桩桩基承台计算项目名称_____________日期_____________设计者_____________校对者_____________一、设计依据 《建筑地基基础设计规范》(GB50007-2011)① 《混凝土结构设计规范》(GB50010-2010)② 《建筑桩基技术规范》 (JGJ94－2008)③二、示意图三、计算信息 承台类型:三桩承台 计算类型:验算截面尺寸 构件编号:CT2 1.几何参数 矩形柱宽 bc=500mm 矩形柱高 hc=550mm 永久荷载分项系数rg=1.20 可变荷载分项系数rq=1.40 Fk=Fgk+Fqk=4418.000+(0.000)=4418.000kN Mxk=Mgxk+Fgk*(B2-B1)/2+Mqxk+Fqk*(B2-B1)/2 =81.000+4418.000*(0.000-0.000)/2+(0.000)+0.000*(0.000-0.000)/2 =81.000kN*m Myk=Mgyk+Fgk*(A2-A1)/2+Mqyk+Fqk*(A2-A1)/2 =6.000+4418.000*(0.000-0.000)/2+(0.000)+0.000*(0.000-0.000)/2 =6.000kN*m Vxk=Vgxk+Vqxk=5.000+(0.000)=5.000kN Vyk=Vgyk+Vqyk=57.000+(0.000)=57.000kN F1=rg*Fgk+rq*Fqk=1.20*(4418.000)+1.40*(0.000)=5301.600kN Mx1=rg*(Mgxk+Fgk*(B2-B1)/2)+rq*(Mqxk+Fqk*(B2-B1)/2) =1.20*(81.000+4418.000*(0.000-0.000)/2)+1.40*(0.000+0.000*(0.000-0.000)/2) =97.200kN*m My1=rg*(Mgyk+Fgk*(A2-A1)/2)+rq*(Mqyk+Fqk*(A2-A1)/2) =1.20*(6.000+4418.000*(0.000-0.000)/2)+1.40*(0.000+0.000*(0.000-0.000)/2) =7.200kN*m Vx1=rg*Vgxk+rq*Vqxk=1.20*(5.000)+1.40*(0.000)=6.000kN Vy1=rg*Vgyk+rq*Vqyk=1.20*(57.000)+1.40*(0.000)=68.400kN F2=1.35*Fk=1.35*4418.000=5964.300kN Mx2=1.35*Mxk=1.35*81.000=109.350kN*m My2=1.35*Myk=1.35*6.000=8.100kN*m Vx2=1.35*Vxk=1.35*5.000=6.750kN Vy2=1.35*Vyk=1.35*57.000=76.950kN F=max(|F1|,|F2|)=max(|5301.600|,|5964.300|)=5964.300kN Mx=max(|Mx1|,|Mx2|)=max(|97.200|,|109.350|)=109.350kN*m My=max(|My1|,|My2|)=max(|7.200|,|8.100|)=8.100kN*m Vx=max(|Vx1|,|Vx2|)=max(|6.000|,|6.750|)=6.750kN Vy=max(|Vy1|,|Vy2|)=max(|68.400|,|76.950|)=76.950kN四、计算参数 1.承台总长Bx=C+A+C=0.600+1.800+0.600=3.000m 2.承台总宽By=C+B+C=0.600+1.800+0.600=3.000m 3.承台根部截面有效高度ho=H-as=1.250-0.155=1.095m 4.圆桩换算截面宽度bp=0.8*d=0.8*0.600=0.480m五、内力计算 1.各桩编号及定位座标如上图所示: θ1=arccos(0.5*A/B)=1.047 θ2=2*arcsin(0.5*A/B)=1.047 1号桩(x1=-A/2=-0.900m,y1=-B*cos(0.5*θ2)/3=-0.520m) 2号桩(x2=A/2=0.900m,y2=-B*cos(0.5*θ2)/3=-0.520m) 3号桩(x3=0,y3=B*cos(0.5*θ2)*2/3=1.039m) 2.各桩净反力设计值,计算公式:【8.5.3-2】① ∑xi=x12*2=1.620m ∑yi=y12*2+y32=1.620m Ni=F/n-Mx*yi/∑yi2+My*xi/∑xi2+Vx*H*xi/∑xi2-Vy*H*y1/∑yi2 N1=5964.300/3-109.350*(-0.520)/1.620+8.100*(-0.900)/1.620 +6.750*1.250*(-0.900)/1.620-76.950*1.250*(-0.520)/1.620 =1983.134kN N2=5964.300/3-109.350*(-0.520)/1.620+8.100*0.900/1.620 +6.750*1.250*0.900/1.620-76.950*1.250*(-0.520)/1.620 =2001.509kN N3=5964.300/3-109.350*1.039/1.620+8.100*0.000/1.620 +6.750*1.250*0.000/1.620-76.950*1.250*1.039/1.620 =1979.656kN六、柱对承台的冲切验算【8.5.17-1】① 1.∑Ni=0=0.000kN ho1=h-as=1.250-0.155=1.095m 2.αox=A/2-bc/2-bp/2=1.800/2-1/2*0.500-1/2*0.480=0.410m αoy12=y2-hc/2-bp/2=0.520-0.550/2-0.480/2=0.005m αoy3=y3-hc/2-bp/2=1.039-0.550/2-0.480/2=0.524m 3.λox=αox/ho1=0.410/1.095=0.374 λoy12=αoy12/ho1=0.219/1.095=0.200 λoy3=αoy3/ho1=0.524/1.095=0.479 4.βox=0.84/(λox+0.2)=0.84/(0.374+0.2)=1.462 βoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100 βoy3=0.84/(λoy3+0.2)=0.84/(0.479+0.2)=1.238 6.计算冲切临界截面周长 AD=0.5*A+C/tan(0.5*θ1)=0.5*1.800+0.600/tan(0.5*1.047))=1.939m CD=AD*tan(θ1)=1.939*tan(1.047)=3.359m AE=C/tan(0.5*θ1)=0.600/tan(0.5*1.047)=1.039m 6.1计算Umx1 Umx1=bc+αox=0.500+0.410=0.910m 6.2计算Umx2 Umx2=2*AD*(CD-C-|y1|-|y3|+0.5*bp)/CD =2*1.939*(3.359-0.600-|-0.520|-|1.039|+0.5*0.480)/3.359 =1.663m 因Umx2>Umx1,取Umx2=Umx1=0.910m Umy=hc+αoy12+αoy3=0.550+0.219+0.524=1.293m 因Umy>(C*tan(θ1)/tan(0.5*θ1))-C-0.5*bp Umy=(C*tan(θ1)/tan(0.5*θ1))-C-0.5*bp =(0.600*tan(1.047)/tan(0.5*1.047))-0.600-0.5*0.480 =0.960m 7. 计算冲切抗力 因H=1.250m所以βhp=0.963 γo*Fl=γo*(F-∑Ni)=1.0*(5964.300-0.000)=5964.30kN [βox*2*Umy+βoy12*Umx1+βoy3*Umx2]*βhp*ft_b*ho =[1.462*2*0.960+2.100*0.910+1.238*0.910]*0.963*1.43*1.095*1000 =8808.946kN≥γo*Fl 柱对承台的冲切满足规范要求七、角桩对承台的冲切验算 【8.5.17-5】① 计算公式:【8.5.17-5】① 1.Nl=max(N1,N2)=2001.509kN ho1=h-as=1.250-0.155=1.095m 2.a11=(A-bc-bp)/2=(1.800-0.500-0.480)/2=0.410m a12=(y3-(hc＋d)*0.5)*cos(0.5*θ2)=(1.039-(0.550-0.480)*0.5)*cos(0.5*1.047)=0.454m λ11=a11/ho=0.410/1.095=0.374 β11=0.56/(λ11+0.2)=0.56/(0.374+0.2))=0.975 C1=(C/tan(0.5*θ1))+0.5*bp=(C/tan(0.5*1.047))+0.5*0.480=1.279m λ12=a12/ho=0.454/1.095=0.415 β12=0.56/(λ12+0.2)=0.56/(0.415+0.2))=0.911 C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*θ2)=(3.359-0.600-|-0.520|-1.039+0.5*1.047)*cos(0.5*0.480)=1.247m 3.因h=1.250m所以βhp=0.963 γo*Nl=1.0*2001.509=2001.509kN β11*(2*C1+a11)*(tan(0.5*θ1))*βhp*ft_b*ho =0.975*(2*1279.230+410.000)*(tan(0.5*1.047))*0.963*1.43*1095.000 =2518.101kN≥γo*Nl=2001.509kN 底部角桩对承台的冲切满足规范要求 γo*N3=1.0*1979.656=1979.656kN β12*(2*C2+a12)*(tan(0.5*θ2))*βhp*ft_b*ho =0.911*(2*1247.077+453.997)*(tan(0.5*1.047))*0.963*1.43*1095.000*1000 =2337.378kN≥γo*N3=1979.656kN 顶部角桩对承台的冲切满足规范要求八、承台斜截面受剪验算 【8.5.18-1】① 1.计算承台计算截面处的计算宽度 2.计算剪切系数 因0.800ho=1.095m<2.000m,βhs=(0.800/1.095)1/4=0.925 ay=|y3|-0.5*hc-0.5*bp=|1.039|-0.5*0.550-0.5*0.480=0.524 λy=ay/ho=0.524/1.095=0.479 βy=1.75/(λy+1.0)=1.75/(0.479+1.0)=1.183 3.计算承台底部最大剪力【8.5.18-1】① bxo=A*(2/3+hc/2/sqrt(B2-(A/2)2))+2*C =1.800*(2/3+0.550/2/sqrt(1.8002-(1.800/2)2))+2*0.600 =2.718m γo*Vy=1.0*3984.644=3984.644kN βhs*βy*ft_b*bxo*ho=0.925*1.183*1.43*2717.543*1095.000 =4655.743kN≥γo*Vy=3984.644kN 承台斜截面受剪满足规范要求九、承台受弯计算 【8.5.16-1】【8.5.16-2】 计算公式:【8.5.16-1.2】① 1.确定单桩最大竖向力 Nmax=max(N1,N2,N3)=2001.509kN 2.承台底部弯矩最大值【8.5.16-1】【8.5.16-2】① M=Nmax*(A-(sqrt(3)/4)*bc)/3 =2001.509*(1.800-(sqrt(3)/4)*0.500)/3 =1056.459kN*m 3. 计算系数 C30混凝土α1=1.0 αs=M/(α1*fc_b*By*ho*ho) =1056.459/(1.0*14.3*3.000*1.095*1.095*1000) =0.021 4. 相对界限受压区高度 ξb=β1/(1+fy/Es/εcu)=0.518 ξ=1-sqrt(1-2αs)=0.021≤ξb=0.518 5. 纵向受拉钢筋 Asx=Asy=α1*fc_b*By*ho*ξ/fy =1.0*14.3*3000.000*1095.000*0.021/360 =2708mm2 最小配筋面积: B=|y1|+C=|-519.6|+600=1119.6mm Asxmin=Asymin=ρmin*B*H =0.200%*1119.