四桩桩基承台计算四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》 (GB50007-2002)①《混凝土结构设计规范》 (GB50010-2010)②《建筑桩基技术规范》 (JGJ 94－2008)③二、示意图三、计算信息承台类型: 四桩承台计算类型: 验算截面尺寸构件编号: CT-41. 几何参数矩形柱宽bc=600mm 矩形柱高hc=600mm圆桩直径d=400mm承台根部高度H=1000mm承台端部高度h=1000mmx方向桩中心距A=1600mmy方向桩中心距B=1600mm承台边缘至边桩中心距 C=400mm2. 材料信息柱混凝土强度等级: C35 ft_c=1.57N/mm2, fc_c=16.7N/mm2承台混凝土强度等级: C30 ft_b=1.43N/mm2, fc_b=14.3N/mm2桩混凝土强度等级: C30 ft_p=1.43N/mm2, fc_p=14.3N/mm2承台钢筋级别: HRB400 fy=360N/mm23. 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=100mm4. 作用在承台顶部荷载基本组合值F=4297.800kNMx=16.900kN*mMy=71.900kN*mVx=182.100kNVy=43.200kN四、计算参数1. 承台总长 Bx=C+A+C=0.400+1.600+0.400=2.400m2. 承台总宽 By=C+B+C=0.400+1.600+0.400=2.400m3. 承台根部截面有效高度 ho=H-as=1.000-0.100=0.900mho1=h-as=1.000-0.100=0.900mh2=H-h=1.000-1.000=0.000m4. 圆桩换算截面宽度 bp=0.8*d=0.8*0.400=0.320m五、内力计算1. 各桩编号及定位座标如上图所示:1号桩 (x1=-A/2=-0.800m, y1=-B/2=-0.800m)2号桩 (x2=A/2=0.800m, y2=-B/2=-0.800m)3号桩 (x3=A/2=0.800m, y3=B/2=0.800m)4号桩 (x4=-A/2=-0.800m, y4=B/2=0.800m)2. 各桩净反力设计值, 计算公式:【8.5.3-2】①∑xi =x12*4=2.560m∑yi =y12*4=2.560mN i =F/n-Mx*yi/∑yi2+My*xi/∑xi2+Vx*H*xi/∑xi2-Vy*H *y1/∑yi2N1=4297.800/4-16.900*(-0.800)/2.560+71.900*(-0.800)/2.560+182.100*1.000*(-0.800)/2.560-43.200*1.000*(-0.800)/2.560=986.856kNN2=4297.800/4-16.900*(-0.800)/2.560+71.900*0.800/2.560+182.100*1.000*0.800/2.560-43.200*1.000*(-0.800)/2.560=1145.606kNN3=4297.800/4-16.900*0.800/2.560+71.900*0.800/2.560+182.100*1.000*0.800/2.560-43.200*1.000*0.800/2.560=1162.044kNN4=4297.800/4-16.900*0.800/2.560+71.900*(-0.800)/2.560+182.100*1.000*(-0.800)/2.560-43.200*1.000*0.800/2.560=1003.294kN六、柱对承台的冲切验算【8.5.17-1】①1. ∑Ni=0=0.000kN2. αox=A/2-bc/2-bp/2=1.600/2-0.600/2-0.320/2=0.340mαoy=B/2-hc/2-bp/2=1.600-0.600/2-0.320/2=0.340m3. λox=αox/ho=0.340/0.900=0.378λoy=αoy/ho=0.340/0.900=0.3784. βox=0.84/(λox+0.2)=0.84/(0.378+0.2)=1.454βoy=0.84/(λoy+0.2)=0.84/(0.378+0.2)=1.4545. 因 H=1.000m 所以βhp=0.983γo*Fl=γo*(F-∑Ni)=1.0*(4297.800-0.000)=4297.80kN2*[βox*(hc+αoy)+βoy*(bc+αox)]*βhp*ft_b*ho=2*[1.454*(600+340)+1.454*(600+340)]*0.983*1.43*900=6918.08kN≥γo*Fl=4297.80kN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】①1. Nl=max(N1, N2, N3, N4)=1162.044kN2. a1x=(A-bc-bp)/2=(1.600-0.600-0.320)/2=0.340ma1y=(B-hc-bp)/2=(1.600-0.600-0.320)/2=0.340m3. λ1x=a1x/ho1=0.340/0.900=0.378λ1y=a1y/ho1=0.340/0.900=0.3784. β1x=0.56/(λ1x+0.2)=0.56/(0.378+0.2)=0.969β1y=0.56/(λ1y+0.2)=0.56/(0.378+0.2)=0.969C1=C+1/2*bp=0.400+0.320/2=0.560mC2=C+1/2*bp=0.400+0.320/2=0.560m5. 因 h=1.000m 所以βhp=0.983γo*Nl=1.0*1162.044=1162.044kN[β1x*(C2+a1y/2.0)+β1y*(C1+a1x/2)]*βhp*ft_b*ho1 =[0.969*(560+340/2)+0.969*(560+340/2)]*0.983*1.43*900 =1790.851kN≥γo*Nl=1162.044kN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】①1. 计算承台计算截面处的计算宽度bx1=Bx=C+A+C=0.400+1.600+0.400=2.400mbx2=bc=0.600mbxo=[1-0.5*h2/ho*(1-bx2/bx1)]*bx1=[1-0.5*0.000/0.900*(1-0.600/2.400)]*2.400=2.400mby1=By=C+B+C=0.400+1.600+0.400=2.400mby2=hc=0.600mbyo=[1-0.5*h2/ho*(1-by2/by1)]*by1=[1-0.5*0.000/0.900*(1-0.600/2.400)]*2.400=2.400m2.计算剪切系数因0.800ho=0.900m<2.000m,βhs=(0.800/0.900)1/4=0.971ax=1/2*(A-bc-bp)=1/2*(1.600-0.600-0.320)=0.340mλx=ax/ho=0.340/0.900=0.378βx=1.75/(λx+1.0)=1.75/(0.378+1.0)=1.270ay=1/2*(B-hc-bp)=1/2*(1.600-0.600-0.320)=0.340mλy=ay/ho=0.340/0.900=0.378βy=1.75/(λy+1.0)=1.75/(0.378+1.0)=1.2703. 计算承台底部最大剪力【8.5.18-1】①因为 N14=N1+N4=986.856+1003.294=1990.150kN因为 N23=N2+N3=1145.606+1162.044=2307.650kN所以 Vx=max(|N14|, |N23|)=max(1990.150,2307.650)=2307.650kN因 N12=N1+N2=986.856+1145.606=2132.463kNN34=N3+N4=1162.044+1003.294=2165.338kN所以 Vy=max(|N12|, |N34|)=max(2132.463,2165.338)=2165.338kNγo*Vx=1.0*2307.650=2307.650kNβhs*βx*ft_b*byo*ho=0.971*1.270*1.43*2400*900=3809.435kN≥γo*Vx=2307.650kNγo*Vy=1.0*2165.338=2165.338kNβhs*βy*ft_b*bxo*ho=0.971*1.270*1.43*2400*900=3809.435kN≥γo*Vy=2165.338kN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】1. 承台底部弯矩最大值【8.5.16-1】【8.5.16-2】①因 Mdx14=(N1+N4)*(A/2-1/2*bc)=(986.856+1003.294)*(1.600/2-1/2*0.600)=995.08kN*mMdx23=(N2+N3)*(A/2-1/2*bc)=(1145.606+1162.044)*(1.600/2-1/2*0.600)=1153.83kN*m所以 Mx=max(|Mdx14|, |Mdx23|)=max(|995.08|,|1153.83|)=1153.83kN*m因 Mdy12=(N1+N2)*(1/2*B-1/2*hc)=(986.856+1145.606)*(1/2*1.600-1/2*0.600)=1066.23kN*mMdy34=(N3+N4)*(1/2*B-1/2*hc)=(1162.044+1003.294)*(1/2*1.600-1/2*0.600)=1082.67kN*m所以 My=max((|Mdy12|, |Mdy34|)=max(|1066.23|,|1082.67|)=1082.67kN*m2. 计算配筋面积Asx=γo*Mx/(0.9*ho*fy)=1.0*1153.83*106/(0.9*900*360)=3956.9mm2Asx1=Asx/By=3956.9/2=1649mm2/mAsy=γo*My/(0.9*ho*fy)=1.0*1082.67*106/(0.9*900.000*360)=3712.9mm2Asy1=Asy/Bx=3712.9/2=1547mm2/m3. 计算最小配筋率受弯最小配筋率为ρmin=0.200%4. 承台最小配筋面积As1min=ρmin*H*1000=0.200%*1000*1000=2000mm2因 As1min>Asx1 所以承台底面x方向配筋面积为 2000mm2/m选择钢筋22@190, 实配面积为2001mm2/m。