pkpm桩基承台计算

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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》(GB50007-2023)①《混凝土结构设计规范》(GB50010-2023)②《建筑桩基技术规范》(JGJ94－2023)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-11.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=1200mmx方向桩中心距A=1750mmy方向桩中心距B=1750mm承台边缘至边桩中心距C=500mm2.材料信息

柱混凝土强度等级:C30ft_c=1.43N/mm2

,

fc_c=14.3N/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:C80ft_p=2.22N/mm2

,

fc_p=35.9N/mm2

承台钢筋级别:HRB400fy=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

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结构构件计算书

4.作用在承台顶部荷载基本组合值F=3244.000kNMx=57.700kN*mMy=103.500kN*mVx=69.800kNVy=-38.800kN四、计算参数

1.承台总长Bx=C+A+C=0.500+1.750+0.500=2.750m2.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=1.050m4.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)=1.047θ2=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:①

∑x=x2

i1*2=1.531m

∑y22

i=y1*2+y3=1.531m

N2222

i=F/n-Mx*yi/∑yi+My*xi/∑xi+Vx*H*xi/∑xi-Vy*H*y1/∑yi

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①1.∑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=0.450mαoy12=y2-hc/2-bp/2=0.505-0.600/2-0.400/2=0.005mαoy3=y3-hc/2-bp/2=1.010-0.600/2-0.400/2=0.510m3.λox=αox/ho1=0.450/1.050=0.429λoy12=αoy12/ho1=0.210/1.050=0.200λoy3=αoy3/ho1=0.510/1.050=0.486

4.βox=0.84/(λox+0.2)=0.84/(0.429+0.2)=1.336βoy12=0.84/(λoy12+0.2)=0.84/(0.200+0.2)=2.100βoy3=0.84/(λoy3+0.2)=0.84/(0.486+0.2)=1.2246.计算冲切临界截面周长

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结构构件计算书

AD=0.5*A+C/tan(0.5*θ1)=0.5*1.750+0.500/tan(0.5*1.047))=1.741mCD=AD*tan(θ1)=1.741*tan(1.047)=3.016m

AE=C/tan(0.5*θ1)=0.500/tan(0.5*1.047)=0.866m6.1计算Umx1

Umx1=bc+αox=0.450+0.450=0.900m6.2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*bpUmy=(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=0.800m7.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①计算公式:①1.Nl=max(N1,N2)=1222.736kNho1=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)=0.442mλ11=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=1.066mλ12=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=0.967γo*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

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

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结构构件计算书

=0.902*(2*1039.230+441.987)*(tan(0.5*1.047))*0.9