pkpm桩基承台计算

CT5 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》 (GB50010-2010)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=450mm 矩形柱高hc=600mm

圆桩直径d=500mm

承台根部高度H=1200mm

x方向桩中心距A=1750mm

y方向桩中心距B=1750mm

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/mm2, fc_c=14.3N/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: C80 ft_p=2.22N/mm2, fc_p=35.9N/mm2

承台钢筋级别: HRB400 fy=360N/mm2

3. 计算信息

结构重要性系数: γo=1.1

纵筋合力点至近边距离: as=150mm

4. 作用在承台顶部荷载基本组合值

F=3244.000kN

Mx=57.700kN*m

My=103.500kN*m

Vx=69.800kN

Vy=-38.800kN

四、计算参数

1. 承台总长 Bx=C+A+C=0.500+1.750+0.500=

2.750m

2. 承台总宽 By=C+B+C=0.500+1.750+0.500=2.750m

3. 承台根部截面有效高度 ho=H-as=1.200-0.150=1.050m

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=-0.875m, y1=-B*cos(0.5*θ2)/3=-0.505m)

2号桩 (x2=A/2=0.875m, y2=-B*cos(0.5*θ2)/3=-0.505m)

3号桩(x3=0, y3=B*cos(0.5*θ2)*2/3=1.010m)

2. 各桩净反力设计值, 计算公式:【8.5.3-2】①

∑x i=x12*2=1.531m

∑y i=y12*2+y32=1.531m

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=3244.000/3-57.700*(-0.505)/1.531+103.500*(-0.875)/1.531

+69.800*1.200*(-0.875)/1.531--38.800*1.200*(-0.505)/1.531

=1008.724kN

N2=3244.000/3-57.700*(-0.505)/1.531+103.500*0.875/1.531

+69.800*1.200*0.875/1.531--38.800*1.200*(-0.505)/1.531

=1222.736kN

N3=3244.000/3-57.700*1.010/1.531+103.500*0.000/1.531

+69.800*1.200*0.000/1.531--38.800*1.200*1.010/1.531

=1012.540kN

六、柱对承台的冲切验算【8.5.17-1】①

1. ∑Ni=0=0.000kN

ho1=h-as=1.200-0.150=1.050m

2. αox=A/2-bc/2-bp/2=1.750/2-1/2*0.450-1/2*0.400=0.450m

αoy12=y2-hc/2-bp/2=0.505-0.600/2-0.400/2=0.005m

αoy3=y3-hc/2-bp/2=1.010-0.600/2-0.400/2=0.510m

3. λox=αox/h o1=0.450/1.050=0.429

λoy12=αoy12/ho1=0.210/1.050=0.200

λoy3=αoy3/ho1=0.510/1.050=0.486

4. βox=0.84/(λox+0.2)=0.84/(0.429+0.2)=1.336

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

βoy3=0.84/(λoy3+0.2)=0.84/(0.486+0.2)=1.224

6. 计算冲切临界截面周长

AD=0.5*A+C/tan(0.5*θ1)=0.5*1.750+0.500/tan(0.5*1.047))=1.741m

CD=AD*tan(θ1)=1.741*tan(1.047)=3.016m

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

6.1 计算Umx1

Umx1=bc+αox=0.450+0.450=0.900m

6.2 计算Umx2

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

=2*1.741*(3.016-0.500-|-0.505|-|1.010|+0.5*0.400)/3.016

=1.386m

因Umx2>Umx1,取Umx2=Umx1=0.900m

Umy=hc+αoy12+αoy3=0.600+0.210+0.510=1.320m

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=1.200m 所以βhp=0.967

γo*Fl=γo*(F-∑Ni)=1.1*(3244.000-0.000)=3568.40kN

[βox*2*Umy+βoy12*Umx1+βoy3*Umx2]*βhp*ft_b*ho

=[1.336*2*0.800+2.100*0.900+1.224*0.900]*0.967*1.43*1.050*1000

=7446.122kN≥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)=122

2.736kN

ho1=h-as=1.200-0.150=1.050m

2. a11=(A-bc-bp)/2=(1.750-0.450-0.400)/2=0.450m

a12=(y3-(hc＋d)*0.5)*cos(0.5*θ2)=(1.010-(0.600-0.400)*0.5)*cos(0.5*1.047)=0.442m λ11=a11/ho=0.450/1.050=0.429

β11=0.56/(λ11+0.2)=0.56/(0.429+0.2))=0.891

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

λ12=a12/ho=0.442/1.050=0.421

β12=0.56/(λ12+0.2)=0.56/(0.421+0.2))=0.902

C2=(CD-C-|y1|-y3+0.5d)*cos(0.5*θ2)=(3.016-0.500-|-0.505|-1.010+0.5*1.047)*cos(0.5*0.400)=1. 039m

3. 因 h=1.200m 所以βhp=0.967

γo*Nl=1.1*1222.736=1345.010kN

β11*(2*C1+a11)*(tan(0.5*θ1))*βhp*ft_b*ho

=0.891*(2*1066.025+450.000)*(tan(0.5*1.047))*0.967*1.43*1050.000

=1927.701kN≥γo*Nl=1345.010kN

底部角桩对承台的冲切满足规范要求

γo*N3=1.1*1012.540=1113.794kN