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: CT-1

1. 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=700mm

x方向桩中心距A=2000mm

y方向桩中心距B=2000mm

承台边缘至边桩中心距 C=500mm

2. 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/m

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=70mm

4. 作用在承台顶部荷载基本组合值

F=2495.000kN

Mx=0.000kN*m

My=45.000kN*m

Vx=32.000kN

Vy=0.000kN

四、计算参数

1. 承台总长 Bx=C+A+C=0.500+

2.000+0.500=

3.000m

2. 承台总宽 By=C+B+C=0.500+2.000+0.500=

3.000m

3. 承台根部截面有效高度 ho=H-as=0.700-0.070=0.630m

4. 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)=1.047

θ2=2*arcsin(0.5*A/B)=1.047

1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

=797.967kN

N2=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

=865.367kN

N3=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】①

1. ∑Ni=0=0.000kN

ho1=h-as=0.700-0.070=0.630m

2. αox=A/2-bc/2-bp/2=2.000/2-1/2*0.750-1/2*0.400=0.425m

αoy12=y2-hc/2-bp/2=0.577-0.750/2-0.400/2=0.002m

αoy3=y3-hc/2-bp/2=1.155-0.750/2-0.400/2=0.580m

3. λox=αox/ho1=0.425/0.630=0.675

λoy12=αoy12/ho1=0.126/0.630=0.200

λoy3=αoy3/ho1=0.580/0.630=0.920

4. αox=0.84/(λox+0.2)=0.84/(0.675+0.2)=0.960

αoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100

αoy3=0.84/(λoy3+0.2)=0.84/(0.920+0.2)=0.750

6. 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))=1.866m

CD=AD*tan(θ1)=1.866*tan(1.047)=3.232m

AE=C/tan(0.5*θ1)=0.500/tan(0.5*1.047)=0.866m

6.1 计算Umx1

Umx1=bc+αox=0.750+0.425=1.175m

6.2 计算Umx2

Umx2=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=1.175m

Umy=hc+αoy12+αoy3=0.750+0.126+0.580=1.456m

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

=(0.500*tan(1.047)/tan(0.5*1.047))-0.500-0.5*0.400

=0.800m

7. 计算冲切抗力

因 H=0.700m 所以βhp=1.0

γo*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】①

1. Nl=max(N1,N2)=865.367kN

ho1=h-as=0.700-0.070=0.630m

2. a11=(A-bc-bp)/2=(2.000-0.750-0.400)/2=0.425m

a12=(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=0.675

β11=0.56/(λ11+0.2)=0.56/(0.675+0.2))=0.640

C1=(C/tan(0.5*θ1))+0.5*bp=(C/tan(0.5*1.047))+0.5*0.400=1.066m

λ12=a12/ho=0.502/0.630=0.797

β12=0.56/(λ12+0.2)=0.56/(0.797+0.2))=0.562

C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*θ

2)=(3.232-0.500-|-0.577|-1.155+0.5*1.047)*cos(0.5*0.400)=1.039m

3. 因 h=0.700m 所以βhp=1.0

γo*Nl=1.0*865.367=865.367kN

β11*(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=831.667kN

β12*(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