九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》 (JGJ 94－2008)③二、示意图三、计算信息承台类型: 九桩承台计算类型: 自动计算截面尺寸构件编号: CT-11. 几何参数矩形柱宽bc=4200mm 矩形柱高hc=200mm方桩边长ls=400mm承台根部高度H(自动计算)=1300mm承台端部高度h(自动计算)=1300mmx方向桩中心距A=1400mmy方向桩中心距B=1400mm承台边缘至边桩中心距 C=400mm2. 材料信息柱混凝土强度等级: C30 ft_c=1.43N/mm2, fc_c=14.3N/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=70mm4. 作用在承台顶部荷载标准值Fgk=11234.000kN Fqk=100.000kNMgxk=466.000kN*m Mqxk=0.000kN*mMgyk=17.000kN*m Mqyk=0.000kN*mVgxk=0.000kN Vqxk=0.000kNVgyk=0.000kN Vqyk=0.000kN永久荷载分项系数rg=1.20可变荷载分项系数rq=1.40Fk=Fgk+Fqk=11234.000+100.000=11334.000kNMxk=Mgxk+Fgk*(B2-B1)/2+Mqxk+Fqk*(B2-B1)/2=466.000+11234.000*(0.000-0.000)/2+(0.000)+100.000*(0.000-0.000)/2=466.000kN*mMyk=Mgyk+Fgk*(A2-A1)/2+Mqyk+Fqk*(A2-A1)/2=17.000+11234.000*(0.000-0.000)/2+(0.000)+100.000*(0.000-0.000)/2=17.000kN*mVxk=Vgxk+Vqxk=0.000+(0.000)=0.000kNVyk=Vgyk+Vqyk=0.000+(0.000)=0.000kNF1=rg*Fgk+rq*Fqk=1.20*11234.000+1.40*100.000=13620.800kNMx1=rg*(Mgxk+Fgk*(B2-B1)/2)+rq*(Mqxk+Fqk*(B2-B1)/2)=1.20*(466.000+11234.000*(0.000-0.000)/2)+1.40*(0.000+100.000*(0.000-0.000)/2) =559.200kN*mMy1=rg*(Mgyk+Fgk*(A2-A1)/2)+rq*(Mqyk+Fqk*(A2-A1)/2)=1.20*(17.000+11234.000*(0.000-0.000)/2)+1.40*(0.000+100.000*(0.000-0.000)/2)=20.400kN*mVx1=rg*Vgxk+rq*Vqxk=1.20*(0.000)+1.40*(0.000)=0.000kNVy1=rg*Vgyk+rq*Vqyk=1.20*(0.000)+1.40*(0.000)=0.000kNF2=1.35*Fk=1.35*11334.000=15300.900kNMx2=1.35*Mxk=1.35*466.000=629.100kN*mMy2=1.35*Myk=1.35*17.000=22.950kN*mVx2=1.35*Vxk=1.35*(0.000)=0.000kNVy2=1.35*Vyk=1.35*(0.000)=0.000kNF=max(|F1|,|F2|)=max(|13620.800|,|15300.900|)=15300.900kNMx=max(|Mx1|,|Mx2|)=max(|559.200|,|629.100|)=629.100kN*mMy=max(|My1|,|My2|)=max(|20.400|,|22.950|)=22.950kN*mVx=max(|Vx1|,|Vx2|)=max(|0.000|,|0.000|)=0.000kNVy=max(|Vy1|,|Vy2|)=max(|0.000|,|0.000|)=0.000kN四、计算参数1. 承台总长 Bx=C+2*A+C=0.400+2*1.400+0.400=3.600m2. 承台总宽 By=C+2*B+C=0.400+2*1.400+0.400=3.600m3. 承台根部截面有效高度 ho=H-as=1.300-0.070=1.230mho1=h-as=1.300-0.070=1.230mh2=H-h=1.300-1.300=0.000m4. 方桩换算截面宽度 bp=ls=0.400m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=-1.400m, y1=-B=-1.400m)2号桩 (x2=0, y2=-B=-1.400m)3号桩 (x3=A=1.400m, y3=-B=-1.400m)4号桩 (x4=A=1.400m, y4=0)5号桩 (x5=A=1.400m, y5=B=1.400m)6号桩 (x6=0, y6=B=1.400m)7号桩 (x7=-A=-1.400m, y7=B=1.400m)8号桩 (x8=-A=-1.400m, y8=0)9号桩 (x9=0, y9=0)2. 各桩净反力设计值, 计算公式:【8.5.3-2】①∑x i=x12*6=11.760m∑y i=y12*6=11.760mN i=F/n-Mx*y i/∑y i2+My*x i/∑x i2+Vx*H*x i/∑x i2-Vy*H*y1/∑y i2N1=15300.900/9-629.100*(-1.400)/11.760+22.950*(-1.400)/11.760+0.000*1.300*(-1.400)/11.760-0.000*1.300*(-1.400)/11.760=1772.261kNN2=15300.900/9-629.100*(-1.400)/11.760+22.950*0.000/11.760+0.000*1.300*0.000/11.760-0.000*1.300*(-1.400)/11.760=1774.993kNN3=15300.900/9-629.100*(-1.400)/11.760+22.950*1.400/11.760+0.000*1.300*1.400/11.760-0.000*1.300*(-1.400)/11.760=1777.725kNN4=15300.900/9-629.100*0.000/11.760+22.950*1.400/11.760+0.000*1.300*1.400/11.760-0.000*1.300*0.000/11.760=1702.832kNN5=15300.900/9-629.100*1.400/11.760+22.950*1.400/11.760+0.000*1.300*1.400/11.760-0.000*1.300*1.400/11.760=1627.939kNN6=15300.900/9-629.100*1.400/11.760+22.950*0.000/11.760+0.000*1.300*0.000/11.760-0.000*1.300*1.400/11.760=1625.207kNN7=15300.900/9-629.100*1.400/11.760+22.950*(-1.400)/11.760+0.000*1.300*(-1.400)/11.760-0.000*1.300*1.400/11.760=1622.475kNN8=15300.900/9-629.100*0.000/11.760+22.950*(-1.400)/11.760+0.000*1.300*(-1.400)/11.760-0.000*1.300*0.000/11.760=1697.368kNN9=15300.900/9-629.100*0.000/11.760+22.950*0.000/11.760+0.000*1.300*0.000/11.760-0.000*1.300*0.000/11.760=1700.100kN六、柱对承台的冲切验算【8.5.17-1】①1. ∑Ni=N9=1700.100kN因为αox<0.2*ho, 所以αox=0.2*ho=0.246mαoy=B-hc/2-bp/2=1.400-0.200/2-0.400/2=1.100m3. λox=αox/ho=0.246/1.230=0.200λoy=αoy/ho=1.100/1.230=0.8944. βox=0.84/(λox+0.2)=0.84/(0.200+0.2)=2.100βoy=0.84/(λoy+0.2)=0.84/(0.894+0.2)=0.7685. 因 H=1.300m 所以βhp=0.958γo*Fl=γo*(F-∑Ni)=1.0*(15300.900-1700.100)=13600.80kN2*[βox*(hc+αoy)+βoy*(bc+αox)]*βhp*ft_b*ho=2*[2.100*(200+1100)+0.768*(4200+246)]*0.958*1.43*1230=38675.88kN≥γo*Fl=13600.80kN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, N3, N5, N7)=1777.725kN2. a1x=A-bc/2-bp/2=1.400-4.200/2-0.400/2=-0.900m因为 a1x<0.2*ho1 所以 a1x=0.2*ho1=0.246ma1y=B-hc/2-bp/2=1.400-0.200/2-0.400/2=1.100m3. λ1x=a1x/ho1=0.246/1.230=0.200λ1y=a1y/ho1=1.100/1.230=0.8944. β1x=0.56/(λ1x+0.2)=0.56/(0.200+0.2)=1.400β1y=0.56/(λ1y+0.2)=0.56/(0.894+0.2)=0.512C1=C+1/2*bp=0.400+0.400/2=0.600mC2=C+1/2*bp=0.400+0.400/2=0.600m5. 因 h=1.300m 所以βhp=0.958γo*Nl=1.0*1777.725=1777.725kN[β1x*(C2+a1y/2.0)+β1y*(C1+a1x/2)]*βhp*ft_b*ho1=[1.400*(600+1100/2)+0.512*(600+246/2)]*0.958*1.43*1230 =3337.491kN≥γo*Nl=1777.725kN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+2*A+C=0.400+2*1.400+0.400=3.600mbx2=bc=4.200mbxo=[1-0.5*h2/ho*(1-bx2/bx1)]*bx1=[1-0.5*0.000/1.230*(1-4.200/3.600)]*3.600=3.600mby1=By=C+2*B+C=0.400+2*1.400+0.400=3.600mby2=hc=0.200mbyo=[1-0.5*h2/ho*(1-by2/by1)]*by1=[1-0.5*0.000/1.230*(1-0.200/3.600)]*3.600=3.600m2.计算剪切系数因 0.800ho=1.230m<2.000m,βhs=(0.800/1.230)1/4=0.898λx=ax/ho=-0.900/1.230=-0.732因为λx<0.3 所以λx=0.3βx=1.75/(λx+1.0)=1.75/(0.300+1.0)=1.346ay=B-hc/2-bp/2=1.400-0.200/2-0.400/2=1.100mλy=ay/ho=1.100/1.230=0.894βy=1.75/(λy+1.0)=1.75/(0.894+1.0)=0.9243. 计算承台底部最大剪力【8.5.18-1】①因为 N178=N1+N7+N8=1772.261+1622.475+1697.368=5092.104kNN345=N3+N4+N5=1777.725+1702.832+1627.939=5108.496kN所以 Vx=max(|N178|, |N345|)=5108.496kN因为 N123=N1+N2+N3=1772.261+1774.993+1777.725=5324.979kNN567=N5+N6+N7=1627.939+1625.207+1622.475=4875.621kN所以 Vy=max(|N123|, |N567|)=5324.979kNγo*Vx=1.0*5108.496=5108.496kNβhs*βx*ft_b*byo*ho=0.898*1.346*1.43*3600*1230=7654.813kN≥γo*Vx=5108.496kNγo*Vy=1.0*5324.979=5324.979kNβhs*βy*ft_b*bxo*ho=0.898*0.924*1.43*3600*1230=5253.238kN<γo*Vy=5324.979kN承台斜截面受剪不满足规范要求九、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因为 Mdx178=(N1+N7+N8)*(A-1/2*bc)=(1772.261+1622.475+1697.368)*(1.400-1/2*4.200)=-3564.47kN*mMdx345=(N3+N4+N5)*(A-1/2*bc)=(1777.725+1702.832+1627.939)*(1.400-1/2*4.200)=-3575.95kN*m所以 Mx=max(|Mdx178|, |Mdx345|)=max(|3564.47|,|3575.95|)=3575.95kN*m因为 Mdy123=(N1+N2+N3)*(B-1/2*hc)=(1772.261+1774.993+1777.725)*(1.400-1/2*0.200)=6922.47kN*mMdy567=(N5+N6+N7)*(B-1/2*hc)=(1627.939+1625.207+1622.475)*(1.400-1/2*0.200)=6338.31kN*m所以 My=max(|Mdy123|, |Mdy567|)=max(|6922.47|,|6338.31|)=6922.47kN*m2. 计算配筋面积Asx=γo*Mx/(0.9*ho*fy)=1.0*3575.95*106/(0.9*1230*360)=8973.1mm2Asx1=Asx/By=8973.1/4=2493mm2/mAsy=γo*My/(0.9*ho*fy)=1.0*6922.47*106/(0.9*1230.000*360)=17370.5mm2Asy1=Asy/Bx=17370.5/4=4825mm2/m3. 计算最小配筋率受弯最小配筋率为ρmin=0.150%4. 承台最小配筋面积As1min=ρmin*H*1000=0.150%*1300*1000=1950mm2因As1min≤Asx1 所以承台底面x方向配筋面积为 2493mm2/m选择钢筋f20@125, 实配面积为2513mm2/m。