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)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-1

1. 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=1100mm

x方向桩中心距A=2400mm

y方向桩中心距B=2400mm

承台边缘至边桩中心距 C=600mm

2. 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/mm2

3. 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=40mm

4. 作用在承台顶部荷载基本组合值

F=2613.900kN

Mx=0.000kN*m

My=83.460kN*m

Vx=0.000kN

Vy=32.090kN

四、计算参数

1. 承台总长 Bx=C+A+C=0.600+

2.400+0.600=

3.600m

2. 承台总宽 By=C+B+C=0.600+2.400+0.600=

3.600m

3. 承台根部截面有效高度 ho=H-as=1.100-0.040=1.060m

4. 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)=1.047

θ2=2*arcsin(0.5*A/B)=1.047

1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=2.880m

N i=F/n-Mx*y i/∑y i2+My*x i/∑x i2+Vx*H*x i/∑x i2-Vy*H*y1/∑y i2

N1=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

=828.033kN

N2=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

=897.583kN

N3=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】①

1. ∑Ni=0=0.000kN

ho1=h-as=1.100-0.040=1.060m

2. αox=A/2-bc/2-bp/2=2.400/2-1/2*0.500-1/2*0.480=0.710m

αoy12=y2-hc/2-bp/2=0.693-0.500/2-0.480/2=0.203m

αoy3=y3-hc/2-bp/2=1.386-0.500/2-0.480/2=0.896m

3. λox=αox/ho1=0.710/1.060=0.670

λoy12=αoy12/ho1=0.212/1.060=0.200

λoy3=αoy3/ho1=0.896/1.060=0.845

4. βox=0.84/(λox+0.2)=0.84/(0.670+0.2)=0.966

βoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100

βoy3=0.84/(λoy3+0.2)=0.84/(0.845+0.2)=0.804

6. 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))=2.239m

CD=AD*tan(θ1)=2.239*tan(1.047)=3.878m

AE=C/tan(0.5*θ1)=0.600/tan(0.5*1.047)=1.039m

6.1 计算Umx1

Umx1=bc+αox=0.500+0.710=1.210m

6.2 计算Umx2

Umx2=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=1.210m

Umy=hc+αoy12+αoy3=0.500+0.212+0.896=1.608m

因Umy>(C*tan(θ1)/tan(0.5*θ1))-C-0.5*bp

Umy=(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

=0.960m

7. 计算冲切抗力

因 H=1.100m 所以βhp=0.975

γo*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】①

1. Nl=max(N1,N2)=897.583kN

ho1=h-as=1.100-0.040=1.060m

2. a11=(A-bc-bp)/2=(2.400-0.500-0.480)/2=0.710m

a12=(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=0.670

β11=0.56/(λ11+0.2)=0.56/(0.670+0.2))=0.644

C1=(C/tan(0.5*θ1))+0.5*bp=(C/tan(0.5*1.047))+0.5*0.480=1.279m

λ12=a12/ho=0.776/1.060=0.732

β12=0.56/(λ12+0.2)=0.56/(0.732+0.2))=0.601

C2=(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. 247m

3. 因 h=1.100m 所以βhp=0.975

γo*Nl=1.0*897.583=897.583kN

β11*(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=888.283kN

β12*(2*C2+a12)*(tan(0.5*θ2))*βhp*ft_b*ho