Z十二桩桩基承台计算

1、十二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)混凝土结构设计规范 (GB50010-2002)建筑桩基技术规范 (JGJ94-94)二、示意图三、计算信息承台类型: 十二桩承台计算类型: 验算截面尺寸构件编号: Z31. 几何参数矩形柱宽bc=1300mm矩形柱高hc=1300mm圆桩直径d=500mm承台根部高度H=1500mm承台端部高度h=1000mmx方向桩中心距A=1750mmy方向桩中心距B=1750mm承台边缘至边桩中心距 C=500mm2. 材料信息柱混凝土强度等级: C30 ft_c=1.43N/mm

2、2,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: C40 ft_p=1.71N/mm2,fc_p=19.1N/mm2承台钢筋级别: HRB335 fy=300N/mm23. 计算信息结构重要性系数: o=1.0 纵筋合力点至近边距离:as=140mm4. 作用在承台顶部荷载基本组合值F=8486.000kNMx=1566.000kN*mMy=906.000kN*mVx=464.000kNVy=253.000kN四、计算参数1. 承台总长 Bx=C+3*A+C=0.500+3*1.750+0.500=6.25

3、0m2. 承台总宽 By=C+2*B+C = 0.500+2*1.750+0.500=4.500m3. 承台根部截面有效高度 ho=H-as=1.500-0.140=1.360mho1=h-as=1.000-0.140=0.860mh2=H-h=1.500-1.000=0.500m4. 圆桩换算截面宽度 bp=0.8*d=0.8*0.500=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=-1.5*A=-2.625m, y1=-B=-1.750m)2号桩 (x2=-0.5*A=-0.875m, y2=-B=-1.750m)3号桩 (x3=0.5*A=0.875m,

4、y3=-B=-1.750m)4号桩 (x4=1.5*A=2.625m, y4=-B=-1.750m)5号桩 (x5=1.5*A=2.625m, y5=0)6号桩 (x6=1.5*A=2.625m, y6=B=1.750m)7号桩 (x7=0.5*A=0.875m, y7=B=1.750m)8号桩 (x8=-0.5*A=-0.875m, y8=B=1.750m)9号桩 (x9=-1.5*A=-2.625m, y9=B=1.750m)10号桩 (x10=-1.5*A=-2.625m, y10=0)11号桩 (x11=-0.5*A=-0.875m, y11=0)12号桩 (x12=0.5*A=0.8

5、75m, y12=0)2. 各桩净反力设计值, 计算公式:【-2】xi=x12*6+(x22)*4=45.938myi=y12*8=24.500mNi=F/n-Mx*yi/yi2+My*xi/xi2+Vx*H*xi/xi2-Vy*H*y1/yi2N1=8486.000/12-1566.000*(-1.750)/24.500+906.000*(-2.625)/45.938 +464.000*1.500*(-2.625)/45.938-253.000*1.500*(-1.750)/24.500 =700.374kNN2=8486.000/12-1566.000*(-1.750)/24.500+90

6、6.000*(-0.875)/45.938 +464.000*1.500*(-0.875)/45.938-253.000*1.500*(-1.750)/24.500 =761.402kNN3=8486.000/12-1566.000*(-1.750)/24.500+906.000*0.875/45.938 +464.000*1.500*0.875/45.938-253.000*1.500*(-1.750)/24.500 =822.431kNN4=8486.000/12-1566.000*(-1.750)/24.500+906.000*2.625/45.938 +464.000*1.500*2.

7、625/45.938-253.000*1.500*(-1.750)/24.500 =883.460kNN5=8486.000/12-1566.000*0.000/24.500+906.000*2.625/45.938 +464.000*1.500*2.625/45.938-253.000*1.500*0.000/24.500 =798.710kNN6=8486.000/12-1566.000*1.750/24.500+906.000*2.625/45.938 +464.000*1.500*2.625/45.938-253.000*1.500*1.750/24.500 =713.960kNN7=

8、8486.000/12-1566.000*1.750/24.500+906.000*0.875/45.938 +464.000*1.500*0.875/45.938-253.000*1.500*1.750/24.500 =652.931kNN8=8486.000/12-1566.000*1.750/24.500+906.000*(-0.875)/45.938 +464.000*1.500*(-0.875)/45.938-253.000*1.500*1.750/24.500 =591.902kNN9=8486.000/12-1566.000*1.750/24.500+906.000*(-2.62

9、5)/45.938 +464.000*1.500*(-2.625)/45.938-253.000*1.500*1.750/24.500 =530.874kNN10=8486.000/12-1566.000*0.000/24.500+906.000*(-2.625)/45.938 +464.000*1.500*(-2.625)/45.938-253.000*1.500*0.000/24.500 =615.624kNN11=8486.000/12-1566.000*0.000/24.500+906.000*(-0.875)/45.938 +464.000*1.500*(-0.875)/45.938

10、-253.000*1.500*0.000/24.500 =676.652kNN12=8486.000/12-1566.000*0.000/24.500+906.000*0.875/45.938 +464.000*1.500*0.875/45.938-253.000*1.500*0.000/24.500 =737.681kN六、柱对承台的冲切验算 【-1】因最外侧桩距承台变阶处水平距离: (Bx-2*C-bc+bp-0.100)/2=(6.250-2*0.500-1.300+0.400-0.100)/2=2.125m>H 最外侧桩距承台变阶处垂直距离: (By-2*C-hc+bp-0.10

11、0)/2=(4.500-2*0.500-1.300+0.400-0.100)/2=1.250mH所以应该验算柱对承台的冲切.1. Ni=N1+N12=1438.055kN 当 (A/2+bp/2)<H 时2.650m因为 ox>ho, 所以 ox=ho=1.360m0.900m3. ox=ox/ho=1.360/1.360=1.000 oy=oy/ho=0.900/1.360=0.6624. ox=0.84/(ox+0.2)=0.84/(1.000+0.2)=0.700 oy=0.84/(oy+0.2)=0.84/(0.662+0.2)=0.9755. 因 H=1.500m 所以

12、hp=0.942o*Fl=o*(F-Ni)=1.0*(8486.000-1438.055)=7047.95kN2*ox*(hc+oy)+oy*(bc+ox)*hp*ft_b*ho=2*0.700*(1300+900)+0.975*(1300+1360)*0.9*1.43*1360=15137.30kNo*Fl=7047.95kN柱对承台的冲切满足规范要求七、角桩对承台的冲切验算 【-5】1. Nl=max(N1, N4, N6, N9)=883.460kN1.725m因为 a1x>ho1 所以 a1x=ho1=0.860m0.850m3. 1x=a1x/ho1=0.860/0.860=1

13、.000 1y=a1y/ho1=0.850/0.860=0.9884. 1x=0.56/(1x+0.2)=0.56/(1.000+0.2)=0.467 1y=0.56/(1y+0.2)=0.56/(0.988+0.2)=0.471 C1=C+1/2*bp=0.500+0.400/2=0.700m C2=C+1/2*bp=0.500+0.400/2=0.700m5. 因 h=1.000m 所以 hp=0.983 o*Nl=1.0*883.460=883.460kN 1x*(C2+a1y/2.0)+1y*(C1+a1x/2)*hp*ft_b*ho1 =0.467*(700+850/2)+0.471

14、*(700+860/2)*1.0*1.43*860 =1278.830kNo*Nl=883.460kN角桩对承台的冲切满足规范要求八、承台斜截面受剪验算 【-1】1. 计算承台计算截面处的计算宽度 bx1=Bx=C+3*A+C=0.500+3*1.750+0.500=6.250m bx2=bc+0.100=1.300+0.100=1.400m bxo=1-0.5*h2/ho*(1-bx2/bx1)*bx1 =1-0.5*0.500/1.360*(1-1.400/6.250)*6.250 =5.358m by1=By=C+2*B+C = 0.500+2*1.750+0.500=4.500m by

15、2=hc+0.100=1.300+0.100=1.400m byo=1-0.5*h2/ho*(1-by2/by1)*by1 =1-0.5*0.500/1.360*(1-1.400/4.500)*4.500 =3.930m2.计算剪切系数因 0.800ho=1.360m2.000m,hs=(0.800/1.360)1/4=0.8761.725m x1=ax/ho=1.725/1.360=1.268 x1=1.75/(x1+1.0)=1.75/(1.268+1.0)=0.771 ax2=1/2*(A-bc-bp)-0.050=1/2*(1.750-1.300-0.400)-0.050=-0.025

16、m x2=ax/ho=-0.025/1.360=-0.018 因为x2<0.3所以 x2=0.3 x2=1.75/(x2+1.0)=1.75/(0.300+1.0)=1.3460.850m y=ay/ho=0.850/1.360=0.625 y=1.75/(y+1.0)=1.75/(0.625+1.0)=1.0773. 计算承台底部最大剪力【-1】 因为 N1910=N1+N9+N10=700.374+530.874+615.624=1846.871kNN456=N4+N5+N6=883.460+798.710+713.960=2396.129kN 所以 Vx1=max(|N1910|,

17、 |N456|)=2396.129kN N12891011=N1+N2+N8+N9+N10+N11=3876.829kNN3456712=N3+N4+N5+N6+N7+N12=4609.171kN 所以 Vx2=max(|N12891011|, |N3456712|)=4609.171kN 因为 N1234=N1+N2+N3+N4=700.374+761.402+822.431+883.460=3167.667kNN6789=N6+N7+N8+N9=713.960+652.931+591.902+530.874=2489.667kN 所以 Vy=max(N1234, N6789)=3167.6

18、67kN 3. 计算承台底部最大剪力【-1】 4.1 验算x1截面承载力 o*Vx1=1.0*2396.129=2396.129kN hs*x1*ft_b*byo*ho=0.876*0.771*1.43*3930.147*1360.000=5164.084kNo*Vx1=2396.129kN 承台斜截面X1受剪满足规范要求 4.2 验算x2截面承载力 o*Vx2=1.0*4609.171=4609.171kN hs*x2*ft_b*byo*ho=0.876*1.346*1.43*3930.147*1360.000=9010.859kNo*Vx2=4609.171kN 承台斜截面X2受剪满足规范

19、要求 4.3 验算y截面承载力 o*Vy=1.0*3167.667=3167.667kN hs*y*ft_b*bxo*ho=0.876*1.077*1.43*5358.456*1360.000 =9828.496kNo*Vy=3167.667kN 承台斜截面Y受剪满足规范要求九、承台受弯计算 【-1】【8.5.16-2】1. 承台底部弯矩最大值 【-1】【8.5.16-2】因为 Mdx1910=(N1+N9+N10)*(1.5*A-1/2*bc) =(700.374+530.874+615.624)*(1.5*1.750-1/2*1.300) =3647.57kN*m Mdx2811=(N2+

20、N8+N11)*(0.5*A-1/2*bc) =(761.402+591.902+676.652)*(0.875-1/2*1.300) =456.74kN*m Mdx3712=(N3+N7+N12)*(0.5*A-1/2*bc) =(822.431+652.931+737.681)*(0.875-1/2*1.300) =497.93kN*m Mdx456=(N4+N5+N6)*(1.5*A-1/2*bc) =(883.460+798.710+713.960)*(2.625-1/2*1.300) =4732.35kN*m所以 Mx=max(|Mdx1910+Mdx2811|, |Mdx3712+

21、Mdx456|)=max(|4104.31|,|5230.29|)=5230.29kN*m因为 Mdy1234=(N1+N2+N3+N4)*(B-1/2*hc) =(700.374+761.402+822.431+883.460)*(1.750-1/2*1.300) =3484.43kN*m Mdy6789=(N6+N7+N8+N9)*(B-1/2*hc) =(713.960+652.931+591.902+530.874)*(1.750-1/2*1.300) =2738.63kN*m所以 My=max(|Mdy1234|, |Mdy6789|)=max(|3484.43|,|2738.63|)=3484.43kN*m2. 计算配筋面积Asx=o*Mx/(0.9*ho*fy)=1.0*5230.29*106/(0.9*1360*300)=14243.7mm2Asx1=Asx/By=142