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》 (GB50007-2002)①《混凝土结构设计规范》 (GB50010-2002)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: 三桩承台计算类型: 验算截面尺寸构件编号: CT-11. 几何参数矩形柱宽bc=500mm 矩形柱高hc=500mm圆桩直径d=600mm承台根部高度H=1100mmx方向桩中心距A=2400mmy方向桩中心距B=2400mm承台边缘至边桩中心距 C=600mm2. 材料信息柱混凝土强度等级: C25 ft_c=1.27N/mm2, fc_c=11.9N/mm2承台混凝土强度等级: C25 ft_b=1.27N/mm2, fc_b=11.9N/mm2桩混凝土强度等级: C25 ft_p=1.27N/mm2, fc_p=11.9N/mm2承台钢筋级别: HRB335 fy=300N/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=40mm4. 作用在承台顶部荷载基本组合值F=2613.900kNMx=0.000kN*mMy=83.460kN*mVx=0.000kNVy=32.090kN四、计算参数1. 承台总长 Bx=C+A+C=0.600+2.400+0.600=3.600m2. 承台总宽 By=C+B+C=0.600+2.400+0.600=3.600m3. 承台根部截面有效高度 ho=H-as=1.100-0.040=1.060m4. 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:θ1=arccos(0.5*A/B)=1.047θ2=2*arcsin(0.5*A/B)=1.0471号桩 (x1=-A/2=-1.200m, y1=-B*cos(0.5*θ2)/3=-0.693m)2号桩 (x2=A/2=1.200m, y2=-B*cos(0.5*θ2)/3=-0.693m)3号桩(x3=0, y3=B*cos(0.5*θ2)*2/3=1.386m)2. 各桩净反力设计值, 计算公式:【8.5.3-2】①∑x i=x12*2=2.880m∑y i=y12*2+y32=2.880mN i=F/n-Mx*y i/∑y i2+My*x i/∑x i2+Vx*H*x i/∑x i2-Vy*H*y1/∑y i2N1=2613.900/3-0.000*(-0.693)/2.880+83.460*(-1.200)/2.880+0.000*1.100*(-1.200)/2.880-32.090*1.100*(-0.693)/2.880=828.033kNN2=2613.900/3-0.000*(-0.693)/2.880+83.460*1.200/2.880+0.000*1.100*1.200/2.880-32.090*1.100*(-0.693)/2.880=897.583kNN3=2613.900/3-0.000*1.386/2.880+83.460*0.000/2.880+0.000*1.100*0.000/2.880-32.090*1.100*1.386/2.880=888.283kN六、柱对承台的冲切验算【8.5.17-1】①1. ∑Ni=0=0.000kNho1=h-as=1.100-0.040=1.060m2. αox=A/2-bc/2-bp/2=2.400/2-1/2*0.500-1/2*0.480=0.710mαoy12=y2-hc/2-bp/2=0.693-0.500/2-0.480/2=0.203mαoy3=y3-hc/2-bp/2=1.386-0.500/2-0.480/2=0.896m3. λox=αox/ho1=0.710/1.060=0.670λoy12=αoy12/ho1=0.212/1.060=0.200λoy3=αoy3/ho1=0.896/1.060=0.8454. βox=0.84/(λox+0.2)=0.84/(0.670+0.2)=0.966βoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100βoy3=0.84/(λoy3+0.2)=0.84/(0.845+0.2)=0.8046. 计算冲切临界截面周长AD=0.5*A+C/tan(0.5*θ1)=0.5*2.400+0.600/tan(0.5*1.047))=2.239mCD=AD*tan(θ1)=2.239*tan(1.047)=3.878mAE=C/tan(0.5*θ1)=0.600/tan(0.5*1.047)=1.039m6.1 计算Umx1Umx1=bc+αox=0.500+0.710=1.210m6.2 计算Umx2Umx2=2*AD*(CD-C-|y1|-|y3|+0.5*bp)/CD=2*2.239*(3.878-0.600-|-0.693|-|1.386|+0.5*0.480)/3.878=1.663m因Umx2>Umx1,取Umx2=Umx1=1.210mUmy=hc+αoy12+αoy3=0.500+0.212+0.896=1.608m因Umy>(C*tan(θ1)/tan(0.5*θ1))-C-0.5*bpUmy=(C*tan(θ1)/tan(0.5*θ1))-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=1.100m 所以βhp=0.975γo*Fl=γo*(F-∑Ni)=1.0*(2613.900-0.000)=2613.90kN[βox*2*Umy+βoy12*Umx1+βoy3*Umx2]*βhp*ft_b*ho=[0.966*2*0.960+2.100*1.210+0.804*1.210]*0.975*1.27*1.060*1000=7045.583kN≥γo*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】①计算公式:【8.5.17-5】①1. Nl=max(N1,N2)=897.583kNho1=h-as=1.100-0.040=1.060m2. a11=(A-bc-bp)/2=(2.400-0.500-0.480)/2=0.710ma12=(y3-(hc＋d)*0.5)*cos(0.5*θ2)=(1.386-(0.500-0.480)*0.5)*cos(0.5*1.047)=0.776m λ11=a11/ho=0.710/1.060=0.670β11=0.56/(λ11+0.2)=0.56/(0.670+0.2))=0.644C1=(C/tan(0.5*θ1))+0.5*bp=(C/tan(0.5*1.047))+0.5*0.480=1.279mλ12=a12/ho=0.776/1.060=0.732β12=0.56/(λ12+0.2)=0.56/(0.732+0.2))=0.601C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*θ2)=(3.878-0.600-|-0.693|-1.386+0.5*1.047)*cos(0.5*0.480)=1. 247m3. 因 h=1.100m 所以βhp=0.975γo*Nl=1.0*897.583=897.583kNβ11*(2*C1+a11)*(tan(0.5*θ1))*βhp*ft_b*ho=0.644*(2*1279.230+710.000)*(tan(0.5*1.047))*0.975*1.27*1060.000=1594.630kN≥γo*Nl=897.583kN底部角桩对承台的冲切满足规范要求γo*N3=1.0*888.283=888.283kNβ12*(2*C2+a12)*(tan(0.5*θ2))*βhp*ft_b*ho=0.601*(2*1247.077+775.648)*(tan(0.5*1.047))*0.975*1.27*1060.000*1000=1489.247kN≥γo*N3=888.283kN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】①1. 计算承台计算截面处的计算宽度2.计算剪切系数因0.800ho=1.060m<2.000m,βhs=(0.800/1.060)1/4=0.932ay=|y3|-0.5*hc-0.5*bp=|1.386|-0.5*0.500-0.5*0.480=0.896 λy=ay/ho=0.896/1.060=0.845βy=1.75/(λy+1.0)=1.75/(0.845+1.0)=0.9493. 计算承台底部最大剪力【8.5.18-1】①bxo=A*(2/3+hc/2/sqrt(B2-(A/2)2))+2*C=2.400*(2/3+0.500/2/sqrt(2.4002-(2.400/2)2))+2*0.600=3.089mγo*Vy=1.0*1725.617=1725.617kNβhs*βy*ft_b*bxo*ho=0.932*0.949*1.27*3088.675*1060.000=3676.061kN≥γo*Vy=1725.617kN 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】①1. 确定单桩最大竖向力Nmax=max(N1, N2, N3)=897.583kN2. 承台底部弯矩最大值【8.5.16-1】【8.5.16-2】①M=Nmax*(A-(sqrt(3)/4)*bc)/3=897.583*(2.400-(sqrt(3)/4)*0.500)/3=653.289kN*m3. 计算系数C30混凝土α1=1.0αs=M/(α1*fc_b*By*ho*ho)=653.289/(1.0*11.9*3.600*1.060*1.060*1000)=0.0144. 相对界限受压区高度ξb=β1/(1+fy/Es/εcu)=0.550ξ=1-sqrt(1-2αs)=0.014≤ξb=0.5505. 纵向受拉钢筋Asx=Asy=α1*fc_b*By*ho*ξ/fy=1.0*11.9*3600.000*1060.000*0.014/300=2069mm2最小配筋面积:B=|y1|+C=|-692.8|+600=1292.8mmAsxmin=Asymin=ρmin*B*H=0.150%*1292.8*1100=2133mm2Asx<Asxmin,取Asx=Asxmin=2133mm2Asy<Asymin,取Asx=Asymin=2133mm26. 选择Asx钢筋选择钢筋7D20, 实配面积为2199mm2/m。