三桩桩基承台计算.doc

结构构件计算书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)混凝土结构设计规范 (GB50010-2010)建筑桩基技术规范 (JGJ 942008)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: CT21. 几何参数矩形柱宽bc=500mm矩形柱高hc=550mm圆桩直径d=600mm承台根部高度H=1250mmx方向桩中心距A=1800mmy方向桩中心距B=1800mm承台边缘至边桩中心距 C=600mm2. 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: o=1.0 纵筋合力点至近边距离:as=155mm4. 作用在承台顶部荷载标准值Fgk=4418.000kNFqk=0.000kNMgxk=81.000kN*mMqxk=0.000kN*mMgyk=6.000kN*mMqyk=0.000kN*mVgxk=5.000kNVqxk=0.000kNVgyk=57.000kNVqyk=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=4418.000+(0.000)=4418.000kNMxk=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*mMyk=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*mVxk=Vgxk+Vqxk=5.000+(0.000)=5.000kNVyk=Vgyk+Vqyk=57.000+(0.000)=57.000kNF1=rg*Fgk+rq*Fqk=1.20*(4418.000)+1.40*(0.000)=5301.600kNMx1=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*mMy1=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*mVx1=rg*Vgxk+rq*Vqxk=1.20*(5.000)+1.40*(0.000)=6.000kNVy1=rg*Vgyk+rq*Vqyk=1.20*(57.000)+1.40*(0.000)=68.400kNF2=1.35*Fk=1.35*4418.000=5964.300kNMx2=1.35*Mxk=1.35*81.000=109.350kN*mMy2=1.35*Myk=1.35*6.000=8.100kN*mVx2=1.35*Vxk=1.35*5.000=6.750kNVy2=1.35*Vyk=1.35*57.000=76.950kNF=max(|F1|,|F2|)=max(|5301.600|,|5964.300|)=5964.300kNMx=max(|Mx1|,|Mx2|)=max(|97.200|,|109.350|)=109.350kN*mMy=max(|My1|,|My2|)=max(|7.200|,|8.100|)=8.100kN*mVx=max(|Vx1|,|Vx2|)=max(|6.000|,|6.750|)=6.750kNVy=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=3.000m2. 承台总宽 By=C+B+C=0.600+1.800+0.600=3.000m3. 承台根部截面有效高度 ho=H-as=1.250-0.155=1.095m4. 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:1=arccos(0.5*A/B)=1.0472=2*arcsin(0.5*A/B)=1.0471号桩 (x1=-A/2=-0.900m, y1=-B*cos(0.5*2)/3=-0.520m)2号桩 (x2=A/2=0.900m, y2=-B*cos(0.5*2)/3=-0.520m)3号桩 (x3=0, y3=B*cos(0.5*2)*2/3=1.039m)2. 各桩净反力设计值, 计算公式:【8.5.3-2】xi=x12*2=1.620myi=y12*2+y32=1.620mNi=F/n-Mx*yi/yi2+My*xi/xi2+Vx*H*xi/xi2-Vy*H*y1/yi2N1=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 =1983.134kNN2=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 =2001.509kNN3=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. Ni=0=0.000kN ho1=h-as=1.250-0.155=1.095m2. ox=A/2-bc/2-bp/2=1.800/2-1/2*0.500-1/2*0.480=0.410moy12=y2-hc/2-bp/2=0.520-0.550/2-0.480/2=0.005moy3=y3-hc/2-bp/2=1.039-0.550/2-0.480/2=0.524m3. ox=ox/ho1=0.410/1.095=0.374 oy12=oy12/ho1=0.219/1.095=0.200 oy3=oy3/ho1=0.524/1.095=0.4794. ox=0.84/(ox+0.2)=0.84/(0.374+0.2)=1.462 oy12=0.84/(oy12+0.2)=0.84/(0.200+0.2)=2.100 oy3=0.84/(oy3+0.2)=0.84/(0.479+0.2)=1.2386. 计算冲切临界截面周长AD=0.5*A+C/tan(0.5*1)=0.5*1.800+0.600/tan(0.5*1.047)=1.939mCD=AD*tan(1)=1.939*tan(1.047)=3.359mAE=C/tan(0.5*1)=0.600/tan(0.5*1.047)=1.039m6.1 计算Umx1Umx1=bc+ox=0.500+0.410=0.910m6.2 计算Umx2Umx2=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Umx2Umx1,取Umx2=Umx1=0.910mUmy=hc+oy12+oy3=0.550+0.219+0.524=1.293m因 Umy(C*tan(1)/tan(0.5*1)-C-0.5*bpUmy=(C*tan(1)/tan(0.5*1)-C-0.5*bp =(0.600*tan(1.047)/tan(0.5*1.047)-0.600-0.5*0.480 =0.960m7.计算冲切抗力 因 H=1.250m 所以 hp=0.963o*Fl=o*(F-Ni)=1.0*(5964.300-0.000)=5964.30kN ox*2*Umy+oy12*Umx1+oy3*Umx2*hp*ft_b*ho =1.462*2*0.960+2.100*0.910+1.238*0.910*0.963*1.43*1.095*1000 =8808.946kNo*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】计算公式:【8.5.17-5】1. Nl=max(N1,N2)=2001.509kN ho1=h-as=1.250-0.155=1.095m2. a11=(A-bc-bp)/2=(1.800-0.500-0.480)/2=0.410m a12=(y3-(hcd)*0.5)*cos(0.5*2)=(1.039-(0.550-0.480)*0.5)*cos(0.5*1.047)=0.454m 11=a11/ho=0.410/1.095=0.374 11=0.56/(11+0.2)=0.56/(0.374+0.2)=0.975 C1=(C/tan(0.5*1)+0.5*bp=(C/tan(0.5*1.047)+0.5*0.480=1.279m 12=a12/ho=0.454/1.095=0.415 12=0.56/(12+0.2)=0.56/(0.415+0.2)=0.911 C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*2)=(3.359-0.600-|-0.520|-1.039+0.5*1.047)*cos(0.5*0.480)=1.247m3. 因 h=1.250m 所以 hp=0.963 o*Nl=1.0*2001.509=2001.509kN 11*(2*C1+a11)*(tan(0.5*1)*hp*ft_b*ho =0.975*(2*1279.230+410.000)*(tan(0.5*1.047)*0.963*1.43*1095.000 =2518.101kNo*Nl=2001.509kN 底部角桩对承台的冲切满足规范要求 o*N3=1.0*1979.656=1979.656kN 12*(2*C2+a12)*(tan(0.5*2)*hp*ft_b*ho =0.911*(2*1247.077+453.997)*(tan(0.5*1.047)*0.963*1.43*1095.000*1000 =2337.378kNo*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. 计算承台计算截面处的计算宽度2.计算剪切系数因 0.800ho=1.095m2.000m,hs=(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 y=ay/ho=0.524/1.095=0.479 y=1.75/(y+1.0)=1.75/(0.479+1.0)=1.1833. 计算承台底部最大剪力【8.5.18-1】 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 o*Vy=1.0*3984.644=3984.644kN hs*y*ft_b*bxo*ho=0.925*1.183*1.43*2717.543*1095.000 =4655.743kNo*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. 确定单桩最大竖向力Nmax=max(N1, N2, N3)=2001.509kN2. 承台底部弯矩最大值 【8.5.16-1】【8.5.16-2】M=Nmax*(A-(sqrt(3)/4)*bc)/3 =2001.509*(1.800-(sqrt(3)/4)*0.500)/3 =1056.459kN*m3.计算系数C30混凝土 1=1.0 s=M/(1*fc_b*By*ho*ho) =1056.459/(1.0*14.3*3.000*1.095*1.095*1000) =0.0214.相对界限受压区高度 b=1/(1+fy/Es/cu)=0.518 =1-sqrt(1-2s)=0.021b=0.5185.纵向受拉钢筋 Asx=Asy=1*f