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》 (GB50007-2011)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: CT2

1. 几何参数

矩形柱宽bc=500mm 矩形柱高hc=550mm

圆桩直径d=600mm

承台根部高度H=1250mm

x方向桩中心距A=1800mm

y方向桩中心距B=1800mm

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: C35 ft_c=1.57N/mm2, fc_c=16.7N/mm2

承台混凝土强度等级: C30 ft_b=1.43N/mm2, fc_b=14.3N/mm2

桩混凝土强度等级: C30 ft_p=1.43N/mm2, fc_p=14.3N/mm2

承台钢筋级别: HRB400 fy=360N/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=155mm

4. 作用在承台顶部荷载标准值

Fgk=4418.000kN Fqk=0.000kN

Mgxk=81.000kN*m Mqxk=0.000kN*m

Mgyk=6.000kN*m Mqyk=0.000kN*m

Vgxk=5.000kN Vqxk=0.000kN

Vgyk=57.000kN Vqyk=0.000kN

永久荷载分项系数rg=1.20

可变荷载分项系数rq=1.40

Fk=Fgk+Fqk=4418.000+(0.000)=4418.000kN

Mxk=Mgxk+Fgk*(B2-B1)/2+Mqxk+Fqk*(B2-B1)/2

=81.000+4418.000*(0.000-0.000)/2+(0.000)+0.000*(0.000-0.000)/2

=81.000kN*m

Myk=Mgyk+Fgk*(A2-A1)/2+Mqyk+Fqk*(A2-A1)/2

=6.000+4418.000*(0.000-0.000)/2+(0.000)+0.000*(0.000-0.000)/2

=6.000kN*m

Vxk=Vgxk+Vqxk=5.000+(0.000)=5.000kN

Vyk=Vgyk+Vqyk=57.000+(0.000)=57.000kN

F1=rg*Fgk+rq*Fqk=1.20*(4418.000)+1.40*(0.000)=5301.600kN

Mx1=rg*(Mgxk+Fgk*(B2-B1)/2)+rq*(Mqxk+Fqk*(B2-B1)/2)

=1.20*(81.000+4418.000*(0.000-0.000)/2)+1.40*(0.000+0.000*(0.000-0.000)/2) =97.200kN*m

My1=rg*(Mgyk+Fgk*(A2-A1)/2)+rq*(Mqyk+Fqk*(A2-A1)/2)

=1.20*(6.000+4418.000*(0.000-0.000)/2)+1.40*(0.000+0.000*(0.000-0.000)/2)

=7.200kN*m

Vx1=rg*Vgxk+rq*Vqxk=1.20*(5.000)+1.40*(0.000)=6.000kN

Vy1=rg*Vgyk+rq*Vqyk=1.20*(57.000)+1.40*(0.000)=68.400kN

F2=1.35*Fk=1.35*4418.000=5964.300kN

Mx2=1.35*Mxk=1.35*81.000=109.350kN*m

My2=1.35*Myk=1.35*6.000=8.100kN*m

Vx2=1.35*Vxk=1.35*5.000=6.750kN

Vy2=1.35*Vyk=1.35*57.000=76.950kN

F=max(|F1|,|F2|)=max(|5301.600|,|5964.300|)=5964.300kN

Mx=max(|Mx1|,|Mx2|)=max(|97.200|,|109.350|)=109.350kN*m

My=max(|My1|,|My2|)=max(|7.200|,|8.100|)=8.100kN*m

Vx=max(|Vx1|,|Vx2|)=max(|6.000|,|6.750|)=6.750kN

Vy=max(|Vy1|,|Vy2|)=max(|68.400|,|76.950|)=76.950kN

四、计算参数

1. 承台总长 Bx=C+A+C=0.600+1.800+0.600=3.000m

2. 承台总宽 By=C+B+C=0.600+1.800+0.600=

3.000m

3. 承台根部截面有效高度 ho=H-as=1.250-0.155=1.095m

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=-0.900m, y1=-B*cos(0.5*θ2)/3=-0.520m)

2号桩 (x2=A/2=0.900m, y2=-B*cos(0.5*θ2)/3=-0.520m)

3号桩 (x3=0, y3=B*cos(0.5*θ2)*2/3=1.039m)

2. 各桩净反力设计值, 计算公式:【8.5.3-2】①

∑x i=x12*2=1.620m

∑y i=y12*2+y32=1.620m

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=5964.300/3-109.350*(-0.520)/1.620+8.100*(-0.900)/1.620

+6.750*1.250*(-0.900)/1.620-76.950*1.250*(-0.520)/1.620

=1983.134kN

N2=5964.300/3-109.350*(-0.520)/1.620+8.100*0.900/1.620

+6.750*1.250*0.900/1.620-76.950*1.250*(-0.520)/1.620

=2001.509kN

N3=5964.300/3-109.350*1.039/1.620+8.100*0.000/1.620

+6.750*1.250*0.000/1.620-76.950*1.250*1.039/1.620

=1979.656kN

六、柱对承台的冲切验算【8.5.17-1】①

1. ∑Ni=0=0.000kN

ho1=h-as=1.250-0.155=1.095m

2. αox=A/2-bc/2-bp/2=1.800/2-1/2*0.500-1/2*0.480=0.410m

αoy12=y2-hc/2-bp/2=0.520-0.550/2-0.480/2=0.005m

αoy3=y3-hc/2-bp/2=1.039-0.550/2-0.480/2=0.524m

3. λox=αox/ho1=0.410/1.095=0.374

λoy12=αoy12/ho1=0.219/1.095=0.200

λoy3=αoy3/ho1=0.524/1.095=0.479

4. βox=0.84/(λox+0.2)=0.84/(0.374+0.2)=1.462

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

βoy3=0.84/(λoy3+0.2)=0.84/(0.479+0.2)=1.238

6. 计算冲切临界截面周长

AD=0.5*A+C/tan(0.5*θ1)=0.5*1.800+0.600/tan(0.5*1.047))=1.939m

CD=AD*tan(θ1)=1.939*tan(1.047)=3.359m

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+0.410=0.910m

6.2 计算Umx2

Umx2=2*AD*(CD-C-|y1|-|y3|+0.5*bp)/CD

=2*1.939*(3.359-0.600-|-0.520|-|1.039|+0.5*0.480)/3.359

=1.663m

因Umx2>Umx1,取Umx2=Umx1=0.910m

Umy=hc+αoy12+αoy3=0.550+0.219+0.524=1.293m

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