三桩桩基承台计算演示教学

1、Good is good, but better carries it.精益求精，善益求善。三桩桩基承台计算-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)混凝土结构设计规范(GB50010-2010)建筑桩基技术规范(JGJ942008)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-31.几何参数矩形柱宽bc=600mm矩形柱高hc=600mm圆桩直径d=400mm承台根部高度H=1000mmx方向桩中心距A=1600mmy方向桩中心距B=1600mm承台边缘至边桩中心距C=400mm2.材料信息柱

2、混凝土强度等级:C35ft_c=1.57N/mm2,fc_c=16.7N/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:C30ft_p=1.43N/mm2,fc_p=14.3N/mm2承台钢筋级别:HRB400fy=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=100mm4.作用在承台顶部荷载基本组合值F=3881.200kNMx=42.200kN*mMy=4.500kN*mVx=2.300kNVy=-23.200kN四、计算参数1.承台总长Bx=C+A+C=0.400+1.600+0.400=2

3、.400m2.承台总宽By=C+B+C=0.400+1.600+0.400=2.400m3.承台根部截面有效高度ho=H-as=1.000-0.100=0.900m4.圆桩换算截面宽度bp=0.8*d=0.8*0.400=0.320m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=-0.800m,y1=-B*cos(0.5*2)/3=-0.462m)2号桩(x2=A/2=0.800m,y2=-B*cos(0.5*2)/3=-0.462m)3号桩(x3=0,y3=B*cos(0.

4、5*2)*2/3=0.924m)2.各桩净反力设计值,计算公式:【8.5.3-2】xi=x12*2=1.280myi=y12*2+y32=1.280mNi=F/n-Mx*yi/yi2+My*xi/xi2+Vx*H*xi/xi2-Vy*H*y1/yi2N1=3881.200/3-42.200*(-0.462)/1.280+4.500*(-0.800)/1.280+2.300*1.000*(-0.800)/1.280-23.200*1.000*(-0.462)/1.280=1313.083kNN2=3881.200/3-42.200*(-0.462)/1.280+4.500*0.800/1.280

5、+2.300*1.000*0.800/1.280-23.200*1.000*(-0.462)/1.280=1321.583kNN3=3881.200/3-42.200*0.924/1.280+4.500*0.000/1.280+2.300*1.000*0.000/1.280-23.200*1.000*0.924/1.280=1246.535kN六、柱对承台的冲切验算【8.5.17-1】1.Ni=0=0.000kNho1=h-as=1.000-0.100=0.900m2.ox=A/2-bc/2-bp/2=1.600/2-1/2*0.600-1/2*0.320=0.340moy12=y2-hc/2

6、-bp/2=0.462-0.600/2-0.320/2=0.002moy3=y3-hc/2-bp/2=0.924-0.600/2-0.320/2=0.464m3.ox=ox/ho1=0.340/0.900=0.378oy12=oy12/ho1=0.180/0.900=0.200oy3=oy3/ho1=0.464/0.900=0.5154.ox=0.84/(ox+0.2)=0.84/(0.378+0.2)=1.454oy12=0.84/(oy12+0.2)=0.84/(0.200+0.2)=2.100oy3=0.84/(oy3+0.2)=0.84/(0.515+0.2)=1.1746.计算冲切临

7、界截面周长AD=0.5*A+C/tan(0.5*1)=0.5*1.600+0.400/tan(0.5*1.047)=1.493mCD=AD*tan(1)=1.493*tan(1.047)=2.586mAE=C/tan(0.5*1)=0.400/tan(0.5*1.047)=0.693m6.1计算Umx1Umx1=bc+ox=0.600+0.340=0.940m6.2计算Umx2Umx2=2*AD*(CD-C-|y1|-|y3|+0.5*bp)/CD=2*1.493*(2.586-0.400-|-0.462|-|0.924|+0.5*0.320)/2.586=1.109m因Umx2Umx1,取U

8、mx2=Umx1=0.940mUmy=hc+oy12+oy3=0.600+0.180+0.464=1.244m因Umy(C*tan(1)/tan(0.5*1)-C-0.5*bpUmy=(C*tan(1)/tan(0.5*1)-C-0.5*bp=(0.400*tan(1.047)/tan(0.5*1.047)-0.400-0.5*0.320=0.640m7.计算冲切抗力因H=1.000m所以hp=0.983o*Fl=o*(F-Ni)=1.0*(3881.200-0.000)=3881.20kNox*2*Umy+oy12*Umx1+oy3*Umx2*hp*ft_b*ho=1.454*2*0.640

9、+2.100*0.940+1.174*0.940*0.983*1.43*0.900*1000=6250.314kNo*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)=1321.583kNho1=h-as=1.000-0.100=0.900m2.a11=(A-bc-bp)/2=(1.600-0.600-0.320)/2=0.340ma12=(y3-(hcd)*0.5)*cos(0.5*2)=(0.924-(0.600-0.320)*0.5)*cos(0.5*1.047)=0.402m11=a11/ho=0.3

10、40/0.900=0.37811=0.56/(11+0.2)=0.56/(0.378+0.2)=0.969C1=(C/tan(0.5*1)+0.5*bp=(C/tan(0.5*1.047)+0.5*0.320=0.853m12=a12/ho=0.402/0.900=0.44612=0.56/(12+0.2)=0.56/(0.446+0.2)=0.867C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*2)=(2.586-0.400-|-0.462|-0.924+0.5*1.047)*cos(0.5*0.320)=0.831m3.因h=1.000m所以hp=0.983o*Nl=1.0

11、*1321.583=1321.583kN11*(2*C1+a11)*(tan(0.5*1)*hp*ft_b*ho=0.969*(2*852.820+340.000)*(tan(0.5*1.047)*0.983*1.43*900.000=1448.689kNo*Nl=1321.583kN底部角桩对承台的冲切满足规范要求o*N3=1.0*1246.535=1246.535kN12*(2*C2+a12)*(tan(0.5*2)*hp*ft_b*ho=0.867*(2*831.384+401.628)*(tan(0.5*1.047)*0.983*1.43*900.000*1000=1307.064kNo*N3=1246.535kN