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1、CT5桩基承台十算项目名称日 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 (GB50010-2010建筑桩基技术规范(JGJ 942008二、示意图TW2嘶ri jL 一 L十一-HJ- -L-H三、计算信息承台类型：三桩承台构件编号：CT-1计算类型：验算截面尺寸1.几何参数矩形柱宽bc=450mm矩形柱高圆桩直径d=500mm承台根部高度H=1200mmx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hc=600mm柱混凝土强度等级：C30ft _c=1.43N/mm2,fc_c=14
2、.3N/mm2承台混凝土强度等级:C30ft _b=1.43N/mm2,fc_b=14.3N/mm2桩混凝土强度等级：C80ft _p=2.22N/mm2,fc_p=35.9N/mm2承台钢筋级别：HRB400fy=360N/mm23.计算信息结构重要性系数:Yo=l.l纵筋合力点至近边距离: as=150mm4. 作用在承台顶部荷载基本组合值F=3244.000kNMx=57.700kN*mMy=103.500kN*mVx=69.800kNVy=-38.800kN四、计算参数1. 承台总长 Bx=C+A+C=0.500+1.750+0.500=2.750m2. 承台总宽 By=C+B+C=0
3、.500+1.750+0.500=2.750m3. 承台根部截面有效高度 ho=H-as=1.200-0.150=1.050m4. 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:9 1二arccos(0.5*A/B)=1.047G2=2*arcsin(0.5*A/B)=1.0471 号桩(x1=-A/2=-0.875m, y1=-B*cos(0.5* 9 2)/3=-0.505m)2号桩 (x2=A/2=0.875m, y2=-B*cos(0.5*92)/3=-0.505m)3 号桩(x3=0, y3=B*cos(0.5*9
4、 2)*2/3=1.010m)2. 各桩净反力设计值，计算公式：【8. 5.3-2Ex =x 2*2=1.531mi1Eyi=y12*2+y3 2= 1.531mN=F/n-Mx*y/Ey 2+My*x /E x 2+Vx*H*x /E x 2-Vy*H*y / E y 2ii ii ii i1 iN1=3244.000/3-57.700*(-0.505)/1.531+103.500*(-0.875)/1.531 +69.800*1.200*(-0.875)/1.531-38.800*1.200*(-0.505)/1.531 =1008.724kNN2=3244.000/3-57.700*(-
5、0.505)/1.531+103.500*0.875/1.531 +69.800*1.200*0.875/1.531-38.800*1.200*(-0.505)/1.531 =1222.736kNN3=3244.000/3-57.700*1.010/1.531+103.500*0.000/1.531 +69.800*1.200*0.000/1.531-38.800*1.200*1.010/1.531 =1012.540kN六、柱对承台的冲切验算【8. 5.17-1 1. ENi=0=0.000kNho1=h-as=1.200-0.150=1.050m2. aox=A/2-bc/2-bp/2=1
6、.750/2-1/2*0.450-1/2*0.400=0.450maoy12二y2-hc/2-bp/2=0.505-0.600/2-0.400/2=0.005m aoy3=y3-hc/2-bp/2=1.010-0.600/2-0.400/2=0.510m3. 入o x=aox/ho1=0.450/1.050=0.429入 oy12=aoy12/ho1=0.210/1.050=0.200入 oy3二 aoy3/ho1=0.510/1.050=0.4864. Box=0.84/ (入 ox+0.2)=0.84/(0.429+0.2)=1.336Boy12=0.84/ (入 oy12+0.2)=0.
7、84/(0.200+0.2)=2.100Boy3=0.84/ (入 oy3+0.2)=0.84/(0.486+0.2)=1.2246. 计算冲切临界截面周长AD=0.5*A+C/ tan(0.5*ei)=0.5*1.750+0.500/tan(0.5*1.047)=1.741mCD二AD* tan(ei)=1.741* tan(1.047)=3.016mAE=C/ tan(0.5*ei)=0.500/tan(0.5*1.047)=0.866m6.1 计算 UmxlUmxl二bc+aox=0.450+0.450=0.900m6.2 计算 Umx2Umx2=2*AD*(CD-C-|y1|-|y3|
8、+0.5*bp)/CD =2*1.741*(3.016-0.500-|-0.505|-|1.010|+0.5*0.400)/3.016=1.386m因 Umx2Umxl,取 Umx2=Umx1=0.900mUmy二hc+aoyl2+aoy3=0.600+0.210+0.510=1.320m因 Umy(C*tan(ei)/tan(0.5*Gl)-C-0.5*bpUmy=(C* tan(ei)/ tan(0.5*Gl)-C-0.5*bp =(0.500*tan(1.047)/tan(0.5*1.047)-0.500-0.5*0.400=0.800m7. 计算冲切抗力因 H=1.200m 所以 Bh
9、p=0.967Yo*Fl=Yo*(F-YNi)=l.l*(3244.000-0.000)=3568.40kNBox*2*Umy+Boyl2*Umxl+Boy3*Umx2*Bhp*f t_b*ho=1.336*2*0.800+2.100*0.900+1.224*0.900*0.967*1.43*1.050*1000=7446. 122kN三yo*F1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)=1222.736kNho1=h-as=1.200-0.150=1.050m2. a11=(A-bc-bp)/2=(
10、1.750-0.450-0.400)/2=0.450m a12=(y3-(hc+d)*O.5)*cos(O.5*G2) = (1.O10-(O.6OO-O.4OO)*O.5)*cos(O.5*1.O47)=O.442m 入 11=a11/ho=0.450/1.050=0.429B11=0.56/ (入 11+0.2)=0.56/(0.429+0.2)=0.891C1=(C/ tan(0.5*G1)+0.5*bp=(C/ tan(0.5*1.047)+0.5*0.400=1.066m入 12=a12/ho=0.442/1.050=0.421B12=0.56/ (入 12+0.2)=0.56/(0
11、.421+0.2)=0.902C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*e 2) = (3.016-0.500-|-0.505|-1.010+0.5*1.047)*cos(0.5*0.400)=1.039m3. 因 h=1.200m 所以 Bhp=0.967Yo*Nl=1.1*1222.736=1345.010kNB11*(2*C1+a11)*( tan(0.5*G1)*Bhp*f t_b*ho =0.891*(2*1066.025+450.000)*(tan(0.5*1.047)*0.967*1.43*1050.000=1927.701kN三Yo*Nl=1345.010k
12、N底部角桩对承台的冲切满足规范要求Yo*N3=1.1*1012.540=1113.794kNB12*(2*C2+al2)*( tan(0.5*G2)*Bhp*f t_b*ho=0.902*(2*1039.230+441.987)*(tan(0.5*1.047)*0.967*1.43*1050.000*1000 =1904.835kN三Yo*N3=1113.794kN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=1.050m2.000m,Bhs=(0.800/1.050)i/4=0.934ay=|
13、y3|-0.5*hc-0.5*bp=|1.010|-0.5*0.600-0.5*0.400=0.510入 y=ay/ho=0.510/l. 050=0.486By=1.75/(入 y+1.0)=1.75/(0.486+1.0)=1.1783. 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 =1.750*(2/3+0.600/2/sqrt(1.7502-(1.750/2)2)+2*0.500 =2.513mYo*Vy=1.1*2231.460=2454.606kNBhs*By*f t_b*bxo*ho=0.934*1.178*1
14、.43*2513.077*1050.000=4151.528kN三Yo*Vy=2454.606kN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)=1222.736kN2. 承台底部弯矩最大值【8. 5.16-1】【8. 5.16-2M=Nmax*(A-(sqrt(3)/4)*bc)/3 =1222.736*(1.750-(sqrt(3)/4)*0.450)/3 =633.844kN*m3. 计算系数C30 混凝土 a1=1.0as二M/(a1*fc_b*By*
15、ho*ho) =633.844/(1.0*14.3*2.750*1.050*1.050*1000)=0.0154. 相对界限受压区高度Eb二B1/(1+fy/Es/&cu)=0.518E=1-sqr t(1-2as)=0.015WEb=0.5185. 纵向受拉钢筋Asx=Asy=a1*fc_b*By*ho*E/fy=1.0*14.3*2750.000*1050.000*0.015/360=1689mm2最小配筋面积：B=|y1|+C=|-505.2|+500=1005.2mmAsxmin二Asymin二Pmin*B*H=0.200%*1005.2*1200=2412mm2AsxAsxmin,取 Asx二Asxmin=2412mm2AsyAsymin,取 Asx=
