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)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: 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=750mm 矩形柱高hc=750mm圆桩直径d=500mm承台根部高度H=700mmx方向桩中心距A=2000mmy方向桩中心距B=2000mm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/m, fc_c=14.3N/m承台混凝土强度等级: C30 ft_b=1.43N/m, fc_b=14.3N/m桩混凝土强度等级: C30 ft_p=1.43N/m, fc_p=14.3N/m承台钢筋级别: HRB400 fy=360N/m3. 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基本组合值F=2495.000kNMx=0.000kN*mMy=45.000kN*mVx=32.000kNVy=0.000kN四、计算参数1. 承台总长 Bx=C+A+C=0.500+2.000+0.500=3.000m2. 承台总宽 By=C+B+C=0.500+2.000+0.500=3.000m3. 承台根部截面有效高度 ho=H-as=0.700-0.070=0.630m4. 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:θ1=arccos(0.5*A/B)=1.047θ2=2*arcsin(0.5*A/B)=1.0471号桩 (x1=-A/2=-1.000m, y1=-B*cos(0.5*θ2)/3=-0.577m)2号桩 (x2=A/2=1.000m, y2=-B*cos(0.5*θ2)/3=-0.577m)3号桩 (x3=0, y3=B*cos(0.5*θ2)*2/3=1.155m)2. 各桩净反力设计值, 计算公式:【8.5.3-2】①∑*2=2.000m∑*2=2.000m=F/n-Mx*/+My*/+Vx*H*/-Vy*H*/N1=2495.000/3-0.000*(-0.577)/2.000+45.000*(-1.000)/2.000+32.000*0.700*(-1.000)/2.000-0.000*0.700*(-0.577)/2.000=797.967kNN2=2495.000/3-0.000*(-0.577)/2.000+45.000*1.000/2.000+32.000*0.700*1.000/2.000-0.000*0.700*(-0.577)/2.000=865.367kNN3=2495.000/3-0.000*1.155/2.000+45.000*0.000/2.000+32.000*0.700*0.000/2.000-0.000*0.700*1.155/2.000=831.667kN六、柱对承台的冲切验算【8.5.19-1】①1. ∑Ni=0=0.000kNho1=h-as=0.700-0.070=0.630m2. αox=A/2-bc/2-bp/2=2.000/2-1/2*0.750-1/2*0.400=0.425mαoy12=y2-hc/2-bp/2=0.577-0.750/2-0.400/2=0.002mαoy3=y3-hc/2-bp/2=1.155-0.750/2-0.400/2=0.580m3. λox=αox/ho1=0.425/0.630=0.675λoy12=αoy12/ho1=0.126/0.630=0.200λoy3=αoy3/ho1=0.580/0.630=0.9204. αox=0.84/(λox+0.2)=0.84/(0.675+0.2)=0.960αoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100αoy3=0.84/(λoy3+0.2)=0.84/(0.920+0.2)=0.7506. 计算冲切临界截面周长AD=0.5*A+C/tan(0.5*θ1)=0.5*2.000+0.500/tan(0.5*1.047))=1.866mCD=AD*tan(θ1)=1.866*tan(1.047)=3.232mAE=C/tan(0.5*θ1)=0.500/tan(0.5*1.047)=0.866m6.1 计算Umx1Umx1=bc+αox=0.750+0.425=1.175m6.2 计算Umx2Umx2=2*AD*(CD-C-|y1|-|y3|+0.5*bp)/CD=2*1.866*(3.232-0.500-|-0.577|-|1.155|+0.5*0.400)/3.232=1.386m因Umx2>Umx1,取Umx2=Umx1=1.175mUmy=hc+αoy12+αoy3=0.750+0.126+0.580=1.456m因 Umy>(C*tan(θ1)/tan(0.5*θ1))-C-0.5*bpUmy=(C*tan(θ1)/tan(0.5*θ1))-C-0.5*bp=(0.500*tan(1.047)/tan(0.5*1.047))-0.500-0.5*0.400=0.800m7. 计算冲切抗力因 H=0.700m 所以βhp=1.0γo*Fl=γo*(F-∑Ni)=1.0*(2495.000-0.000)=2495.00kN[αox*2*Umy+αoy12*Umx1+αoy3*Umx2]*βhp*ft_b*ho=[0.960*2*0.800+2.100*1.175+0.750*1.175]*1.000*1.43*0.630*1000=4401.186kN≥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【8.5.19-5】①计算公式:【8.5.19-5】①1. Nl=max(N1,N2)=865.367kNho1=h-as=0.700-0.070=0.630m2. a11=(A-bc-bp)/2=(2.000-0.750-0.400)/2=0.425ma12=(y3-(hc＋d)*0.5)*cos(0.5*θ2)=(1.155-(0.750-0.400)*0.5)*cos(0.5*1.047)=0.502m λ11=a11/ho=0.425/0.630=0.675β11=0.56/(λ11+0.2)=0.56/(0.675+0.2))=0.640C1=(C/tan(0.5*θ1))+0.5*bp=(C/tan(0.5*1.047))+0.5*0.400=1.066mλ12=a12/ho=0.502/0.630=0.797β12=0.56/(λ12+0.2)=0.56/(0.797+0.2))=0.562C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*θ2)=(3.232-0.500-|-0.577|-1.155+0.5*1.047)*cos(0.5*0.400)=1.039m3. 因 h=0.700m 所以βhp=1.0γo*Nl=1.0*865.367=865.367kNβ11*(2*C1+a11)*(tan(0.5*θ1))*βhp*ft_b*ho=0.640*(2*1066.025+425.000)*(tan(0.5*1.047))*1.000*1.43*630.000=851.593kN<γo*Nl=865.367kN底部角桩对承台的冲切不满足规范要求γo*N3=1.0*831.667=831.667kNβ12*(2*C2+a12)*(tan(0.5*θ2))*βhp*ft_b*ho=0.562*(2*1039.230+502.035)*(tan(0.5*1.047))*1.000*1.43*630.000*1000=753.987kN<γo*N3=831.667kN顶部角桩对承台的冲切不满足规范要求八、承台斜截面受剪验算【8.5.21-1】①1. 计算承台计算截面处的计算宽度2.计算剪切系数因 ho=0.630m≤0.800m,βhs=(0.800/0.8001/=1.0ay=|y3|-0.5*hc-0.5*bp=|1.155|-0.5*0.750-0.5*0.400=0.580 λy=ay/ho=0.580/0.630=0.920βy=1.75/(λy+1.0)=1.75/(0.920+1.0)=0.9113. 计算承台底部最大剪力【8.5.21-1】①bxo=A*(2/3+hc/2/sqrt(-(A/2))+2*C=2.000*(2/3+0.750/2/sqrt(2.00-(2.000/2))+2*0.500=2.766mγo*Vy=1.0*1663.333=1663.333kNβhs*βy*ft_b*bxo*ho=1.000*0.911*1.43*2766.346*630.000=2271.349kN≥γo*Vy=1663.333kN 承台斜截面受剪满足规范要求九、承台受弯计算【8.5.21-1】【8.5.21-2】计算公式:【8.5.21-1.2】①1. 确定单桩最大竖向力Nmax=max(N1, N2, N3)=865.367kN2. 承台底部弯矩最大值【8.5.21-1】【8.5.21-2】①M=Nmax*(A-(sqrt(3)/4)*bc)/3=865.367*(2.000-(sqrt(3)/4)*0.750)/3=483.232kN*m3. 计算系数C30混凝土α1=1.0αs=M/(α1*fc_b*By*ho*ho)=483.232/(1.0*14.3*3.000*0.630*0.630*1000)=0.0284. 相对界限受压区高度ξb=β1/(1+fy/Es/εcu)=0.518ξ=1-sqrt(1-2αs)=0.029≤ξb=0.5185. 纵向受拉钢筋Asx=Asy=α1*fc_b*By*ho*ξ/fy=1.0*14.3*3000.000*630.000*0.029/360=2162m最小配筋面积:B=|y1|+C=|-577.4|+500=1077.4mmAsxmin=Asymin=ρmin*B*H=0.200%*1077.4*700=1508mAsx≥Asxmin, 满足要求。