三桩桩基承台计算

结构构件计算书PAGE第1页，共1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:CT-1 1.几何参数 矩形柱宽 bc=1200mm 矩形柱高 hc=600mm 圆桩直径 d=600mm 承台根部高度 H=1000mm x方向桩中心距 A=1800mm y方向桩中心距 B=1800mm 承台边缘至边桩中心距C=600mm 2.材料信息 柱混凝土强度等级: C40ft_c=1.71N/m, fc_c=19.1N/m 承台混凝土强度等级: C40ft_b=1.71N/m, fc_b=19.1N/m 桩混凝土强度等级: C40ft_p=1.71N/m, fc_p=19.1N/m 承台钢筋级别: HRB400 fy=360N/m 3.计算信息 结构重要性系数: γo=1.0 纵筋合力点至近边距离: as=80mm 4.作用在承台顶部荷载基本组合值 F=5300.000kN Mx=-540.000kN*m My=-231.000kN*m Vx=38.000kN Vy=164.000kN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=1.000-0.080=0.920m 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 ∑*2=1.620m ∑*2=1.620m =F/n-Mx*/+My*/+Vx*H*/-Vy*H*/ N1=5300.000/3--540.000*(-0.520)/1.620+-231.000*(-0.900)/1.620 +38.000*1.000*(-0.900)/1.620-164.000*1.000*(-0.520)/1.620 =1753.287kN N2=5300.000/3--540.000*(-0.520)/1.620+-231.000*0.900/1.620 +38.000*1.000*0.900/1.620-164.000*1.000*(-0.520)/1.620 =1538.842kN N3=5300.000/3--540.000*1.039/1.620+-231.000*0.000/1.620 +38.000*1.000*0.000/1.620-164.000*1.000*1.039/1.620 =2007.871kN六、柱对承台的冲切验算【8.5.19-1】① 1.∑Ni=0=0.000kN ho1=h-as=1.000-0.080=0.920m 2.αox=A/2-bc/2-bp/2=1.800/2-1/2*1.200-1/2*0.480=0.060m 因αox<0.2*ho,所以αox=0.2*ho=0.184m αoy12=y2-hc/2-bp/2=0.520-0.600/2-0.480/2=-0.020m αoy3=y3-hc/2-bp/2=1.039-0.600/2-0.480/2=0.499m 3.λox=αox/ho1=0.184/0.920=0.200 λoy12=αoy12/ho1=0.184/0.920=0.200 λoy3=αoy3/ho1=0.499/0.920=0.543 4.αox=0.84/(λox+0.2)=0.84/(0.200+0.2)=2.100 αoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100 αoy3=0.84/(λoy3+0.2)=0.84/(0.543+0.2)=1.131 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=1.200+0.184=1.384m 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=1.384m Umy=hc+αoy12+αoy3=0.600+0.184+0.499=1.283m 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=1.000m所以βhp=0.983 γo*Fl=γo*(F-∑Ni)=1.0*(5300.000-0.000)=5300.00kN [αox*2*Umy+αoy12*Umx1+αoy3*Umx2]*βhp*ft_b*ho =[2.100*2*0.960+2.100*1.384+1.131*1.384]*0.983*1.71*0.920*1000 =13155.268kN≥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 【8.5.19-5】① 计算公式:【8.5.19-5】① 1.Nl=max(N1,N2)=1753.287kN ho1=h-as=1.000-0.080=0.920m 2.a11=(A-bc-bp)/2=(1.800-1.200-0.480)/2=0.060m a12=(y3-(hc＋d)*0.5)*cos(0.5*θ2)=(1.039-(0.600-0.480)*0.5)*cos(0.5*1.047)=0.432m λ11=a11/ho=0.060/0.920=0.065 β11=0.56/(λ11+0.2)=0.56/(0.065+0.2))=2.111 C1=(C/tan(0.5*θ1))+0.5*bp=(C/tan(0.5*1.047))+0.5*0.480=1.279m λ12=a12/ho=0.432/0.920=0.470 β12=0.56/(λ12+0.2)=0.56/(0.470+0.2))=0.836 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=1.000m所以βhp=0.983 γo*Nl=1.0*1753.287=1753.287kN β11*(2*C1+a11)*(tan(0.5*θ1))*βhp*ft_b*ho =2.111*(2*1279.230+60.000)*(tan(0.5*1.047))*0.983*1.71*920.000 =4938.058kN≥γo*Nl=1753.287kN 底部角桩对承台的冲切满足规范要求 γo*N3=1.0*2007.871=2007.871kN β12*(2*C2+a12)*(tan(0.5*θ2))*βhp*ft_b*ho =0.836*(2*1247.077+432.346)*(tan(0.5*1.047))*0.983*1.71*920.000*1000 =2184.860kN≥γo*N3=2007.871kN 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 【8.5.21-1】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=0.920m<2.000m,βhs=(0.800/0.9201/=0.966 ay=|y3|-0.5*hc-0.5*bp=|1.039|-0.5*0.600-0.5*0.480=0.499 λy=ay/ho=0.499/0.920=0.543 βy=1.75/(λy+1.0)=1.75/(0.543+1.0)=1.134 3.计算承台底部最大剪力【8.5.21-1】① bxo=A*(2/3+hc/2/sqrt(-(A/2))+2*C =1.800*(2/3+0.600/2/sqrt(1.80-(1.800/2))+2*0.600 =2.746m γo*Vy=1.0*3292.129=3292.129kN βhs*βy*ft_b*bxo*ho=0.966*1.134*1.71*2746.410*920.000 =4733.123kN≥γo*Vy=3292.129kN 承台斜截面受剪满足规范要求九、承台受弯计算 【8.5.21-1】【8.5.21-2】 计算公式:【8.5.21-1.2】① 1.确定单桩最大竖向力 Nmax=max(N1,N2,N3)=2007.871kN 2.承台底部弯矩最大值【8.5.21-1】【8.5.21-2】① M=Nmax*(A-(sqrt(3)/4)*bc)/3 =2007.871*(1.800-(sqrt(3)/4)*1.200)/3 =856.949kN*m 3. 计算系数 C30混凝土α1=1.0 αs=M/(α1*fc_b*By*ho*ho) =856.949/(1.0*19.1*3.000*0.920*0.920*1000) =0.018 4. 相对界限受压区高度 ξb=β1/(1+fy/Es/εcu)=0.800 ξ=1-sqrt(1-2αs)=0.018≤ξb=0.800 5. 纵向受拉钢筋 Asx=Asy=α1*fc_b*By*ho*ξ/fy =1.0*19.1*3000.000*920.000*0.018/360 =2611m 最小配筋面积: B=|y1|+C=|-519.6|+600=1119.6mm Asxmin=Asymin=ρmin*B*H =0.200%*1119.6*1000 =2239m