双柱阶梯基础计算.docx
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双柱阶梯基础计算.docx
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双柱阶梯基础计算
双柱阶梯基础计算
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
二、示意图
三、计算信息
构件编号:
JC-1计算类型:
验算截面尺寸
1.几何参数
台阶数n=3
矩形柱宽bc=400mm矩形柱高hc=400mm
矩形柱宽bc=800mm矩形柱高hc=400mm
基础高度h1=450mm
基础高度h2=300mm
基础高度h3=300mm
一阶长度b1=600mmb3=500mm一阶宽度a1=600mma3=500mm
二阶长度b2=600mmb4=500mm二阶宽度a2=600mma4=500mm
一阶长度b1=600mmb4=500mm一阶宽度a1=600mma4=500mm
二阶长度b2=600mmb5=500mm二阶宽度a2=600mma5=500mm
三阶长度b3=500mmb6=500mm三阶宽度a3=500mma6=500mm
2.材料信息
基础混凝土等级:
C30ft_b=1.43N/mm2fc_b=14.3N/mm2
柱混凝土等级:
C30ft_c=1.43N/mm2fc_c=14.3N/mm2
钢筋级别:
HPB300fy=270N/mm2
3.计算信息
结构重要性系数:
γo=1.0
基础埋深:
dh=1.500m
纵筋合力点至近边距离:
as=40mm
基础及其上覆土的平均容重:
γ=20.000kN/m3
最小配筋率:
ρmin=0.150%
4.作用在基础顶部荷载标准值
Fgk1=200.000kNFqk1=100.000kN
Fgk2=100.000kNFqk2=50.000kN
Mgxk1=80.000kN*mMqxk1=0.000kN*m
Mgxk2=80.000kN*mMqxk2=0.000kN*m
Mgyk1=90.000kN*mMqyk1=0.000kN*m
Mgyk2=90.000kN*mMqyk2=0.000kN*m
Vgxk1=80.000kNVqxk1=0.000kN
Vgxk2=80.000kNVqxk2=0.000kN
Vgyk1=90.000kNVqyk1=0.000kN
Vgyk2=90.000kNVqyk2=0.000kN
永久荷载分项系数rg1=1.20
可变荷载分项系数rq1=1.40
永久荷载分项系数rg2=1.20
可变荷载分项系数rq2=1.40
Fk=Fgk1/rg1+Fqk1/rq1+Fgk2/rg2+Fqk2/rq2=200.000/1.200+100.000/1.400+100.000/1.200+50.000/1.400=357.143kN
Mxk=Mgxk1+Fgk1*(A2-A1)+Mqxk1+Fqk1*(A2-A1)+Mgxk2+Fgk2*(A2-A1)+Mqxk2+Fqk2*(A2-A1)
=80.000+200.000*(1.700-1.900)+0.000+100.000*(1.700-1.900)+80.000+100.000*(1.700-1.900)+0.000+50.000*(1.700-1.900)
=70.000kN*m
Myk=Mgyk1-Fgk1*(Bx/2-B1)+Mqyk1-Fqk1*(Bx/2-B1)+Mgyk2+Fgk2*(Bx/2-B1)+Mqyk2+Fqk2*(Bx/2-B1)
=90.000-200.000*(4.400/2-1.900)/2+0.000-100.000*(4.400/2-1.900)+90.000+100.000*(4.400/2-1.900)+0.000+50.000*(4.400/2-1.900)
=135.000kN*m
Vxk=Vgxk1+Vqxk1+Vgxk2+Vqxk2=80.000+0.000+80.000+0.000=160.000kN
Vyk=Vgyk1+Vqyk1+Vgyk2+Vqyk2=90.000+0.000+90.000+0.000=180.000kN
F1=rg*Fgk1+rq*Fqk1+rg*Fgk2+rq*Fqk2
=1.20*200.000+1.40*100.000+1.20*100.000+1.40*50.000
=570.000kN*m
Mx1=rg*(Mgxk1+Fgk1*(A2-A1))+rq*(Mqxk1+Fqk1*(A2-A1))+rg*(Mgxk2+Fgk2*(A2-A1))+rq*(Mqxk2+Fqk2*(A2-A1))
=1.20*(80.000+200.000*(1.700-1.900))+1.40*(0.000+100.000*(1.700-1.900))+1.20*(80.000+100.000*(1.700-1.900))+1.40*(0.000+50.000*(1.700-1.900))
=78.000kN*m
My1=rg*(Mgyk1-Fgk1*(Bx/2-B1))+rq*(Mqyk1-Fqk1*(Bx/2-B1))+rg*(Mgyk2+Fgk2*(Bx/2-B1))+rq*(Mqyk2+Fqk2*(Bx/2-B1)))
=1.20*(90.000-200.000*(4.400/2-1.900))+1.40*(0.000-100.000*(4.400/2-1.900))+1.20*(90.000+100.000*(4.400/2-1.900))+1.40*(0.000+50.000*(4.400/2-1.900))
=180.000kN*m
Vx1=rg*Vgxk1+rq*Vqxk1+rg*Vgxk2+rq*Vqxk2=1.40*80.000+1.40*0.000+1.20*80.000+1.40*0.000=192.000kN
Vy1=rg*Vgyk1+rq*Vqyk1+rg*Vgyk2+rq*Vqyk2=1.40*90.000+1.40*0.000+1.20*90.000+1.40*0.000=216.000kN
F2=1.35*Fk=1.35*357.143=482.143kN
Mx2=1.35*Mxk=1.35*177.000=238.950kN*m
My2=1.35*Myk=1.35*102.000=137.700kN*m
Vx2=1.35*Vxk=1.35*160.000=216.000kN
Vy2=1.35*Vyk=1.35*180.000=243.000kN
F=max(|F1|,|F2|)=max(|570.000|,|482.143|)=570.000kN
Mx=max(|Mx1|,|Mx2|)=max(|78.000|,|238.950|)=238.950kN*m
My=max(|My1|,|My2|)=max(|159.000|,|137.700|)=159.000kN*m
Vx=max(|Vx1|,|Vx2|)=max(|192.000|,|216.000|)=216.000kN
Vy=max(|Vy1|,|Vy2|)=max(|216.000|,|243.000|)=243.000kN
5.修正后的地基承载力特征值
fa=150.000kPa
四、计算参数
1.基础总长Bx=B1+B2+Bc=1.900+1.900+0.600=4.400m
2.基础总宽By=A1+A2=1.900+1.700=3.600m
A1=a1+a2+a3+max(hc1,hc2)/2=0.600+0.600+0.500+max(0.400,0.400)/2=1.900m
A2=a4+a5+a6+max(hc1,hc2)/2=0.500+0.500+0.500+max(0.400,0.400)/2=1.700m
B1=b1+b2+b3+bc1/2=0.600+0.600+0.500+0.400/2=1.900mB2=b4+b5+b6+bc2/2=0.500+0.500+0.500+0.800/2=1.900m
3.基础总高H=h1+h2+h3=0.450+0.300+0.300=1.050m
4.底板配筋计算高度ho=h1+h2+h3-as=0.450+0.300+0.300-0.040=1.010m
5.基础底面积A=Bx*By=4.400*3.600=15.840m2
6.Gk=γ*Bx*By*dh=20.000*4.400*3.600*1.500=475.200kN
G=1.35*Gk=1.35*475.200=641.520kN
五、计算作用在基础底部弯矩值
Mdxk=Mxk-Vxk*H=177.000-160.000*1.050=9.000kN*m
Mdyk=Myk+Vyk*H=102.000+180.000*1.050=291.000kN*m
Mdx=Mx-Vx*H=238.950-216.000*1.050=12.150kN*m
Mdy=My+Vy*H=159.000+243.000*1.050=414.150kN*m
六、验算地基承载力
1.验算轴心荷载作用下地基承载力
pk=(Fk+Gk)/A=(357.143+475.200)/15.840=52.547kPa【①5.2.2-2】
因γo*pk=1.0*52.547=52.547kPa≤fa=150.000kPa
轴心荷载作用下地基承载力满足要求
2.验算偏心荷载作用下的地基承载力
exk=Mdyk/(Fk+Gk)=174.000/(357.143+475.200)=0.209m
因|exk|≤Bx/6=0.733mx方向小偏心,
由公式【①5.2.2-2】和【①5.2.2-3】推导
Pkmax_x=(Fk+Gk)/A+6*|Mdyk|/(Bx2*By)
=(357.143+475.200)/15.840+6*|174.000|/(4.4002*3.600)
=67.526kPa
Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By)
=(357.143+475.200)/15.840-6*|174.000|/(4.4002*3.600)
=37.568kPa
eyk=Mdxk/(Fk+Gk)=-14.500/(357.143+475.200)=-0.017m
因|eyk|≤By/6=0.600my方向小偏心
Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx)
=(357.143+475.200)/15.840+6*|-14.500|/(3.6002*4.400)
=54.073kPa
Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx)
=(357.143+475.200)/15.840-6*|-14.500|/(3.6002*4.400)
=51.021kPa
3.确定基础底面反力设计值
Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk
=(67.526-52.547)+(54.073-52.547)+52.547
=69.052kPa
γo*Pkmax=1.0*69.052=69.052kPa≤1.2*fa=1.2*150.000=180.000kPa
偏心荷载作用下地基承载力满足要求
七、基础冲切验算
1.计算基础底面反力设计值
1.1计算x方向基础底面反力设计值
ex=Mdy/(F+G)=385.800/(570.000+641.520)=0.318m
因ex≤Bx/6.0=0.733mx方向小偏心
Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By)
=(570.000+641.520)/15.840+6*|385.800|/(4.4002*3.600)
=109.698kPa
Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By)
=(570.000+641.520)/15.840-6*|385.800|/(4.4002*3.600)
=43.272kPa
1.2计算y方向基础底面反力设计值
ey=Mdx/(F+G)=-16.200/(570.000+641.520)=-0.013m
因ey≤By/6=0.600y方向小偏心
Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx)
=(570.000+641.520)/15.840+6*|-16.200|/(3.6002*4.400)
=78.189kPa
Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx)
=(570.000+641.520)/15.840-6*|-16.200|/(3.6002*4.400)
=74.780kPa
1.3因Mdx≠0Mdy≠0
Pmax=Pmax_x+Pmax_y-(F+G)/A
=109.698+78.189-(570.000+641.520)/15.840
=111.402kPa
1.4计算地基净反力极值
Pjmax=Pmax-G/A=111.402-641.520/15.840=70.902kPa
Pjmax_x=Pmax_x-G/A=101.537-641.520/15.840=61.037kPa
Pjmax_y=Pmax_y-G/A=77.432-641.520/15.840=36.932kPa
2.验算柱边冲切
YH=h1+h2+h3=1.050m,YB=bc1/2+bc2/2+Bc=1.100m,YL=max(hc1,hc2)=0.400m
YB1=B1+Bc/2=2.200m,YB2=B2+Bc/2=2.200m,YL1=A1=1.900m,YL2=A2=1.700m
YHo=YH-as=1.010m
2.1因800 2.2x方向柱对基础的冲切验算 x冲切位置斜截面上边长bt=YB=1.100m x冲切位置斜截面下边长bb=YB+2*YHo=3.120m x冲切不利位置bm=(bt+bb)/2=(1.100+3.120)/2=2.110m x冲切面积Alx=max((YL1-YL/2-YHo)*(YB+2*YHo)+(YL1-YL/2-YHo)2,(YL2-YL/2-YHo)*(YB+2*YHo)+(YL2-YL/2-YHo)2 =max((1.900-0.400/2-1.010)*(1.100+2*1.010)+(1.900-0.400/2-1.010)2,(1.700-0.400/2-1.010)*(1.100+2*1.010)+(1.700-0.400/2-1.010)2) =max(2.629,1.769) =2.629m2 x冲切截面上的地基净反力设计值 Flx=Alx*Pjmax=2.629*70.902=186.395kN γo*Flx=1.0*186.395=186.39kN γo*Flx≤0.7*βhp*ft_b*bm*YHo(6.5.5-1) =0.7*0.979*1.43*2110*1010 =2088.79kN x方向柱对基础的冲切满足规范要求 2.3y方向柱对基础的冲切验算 y冲切位置斜截面上边长at=YL=0.400m y冲切位置斜截面下边长ab=YL+2*YHo=2.420m y冲切面积Aly=max((YB1-YB/2-YHo)*(YL2+YL/2+YHo)+(YB1-YB/2-YHo)2/2-(YL2-YL/2-YHo)2/2,(YB2-YB/2-YHo)*(YL2+YL/2+YHo)+(YB2-YB/2-YHo)2/2)-(YL2-YL/2-YHo)2/2 =max((2.200-1.100/2-1.010)*(1.700+0.400/2+1.010)+(2.200-1.100/2-1.010)2/2-(1.700-0.400/2-1.010)2/2,(2.200-1.100/2-1.010)*(1.700+0.400/2+1.010)+(2.200-1.100/2-1.010)2/2-(1.700-0.400/2-1.010)2/2) =max(1.947,1.947) =1.947m2 y冲切截面上的地基净反力设计值 Fly=Aly*Pjmax=1.947*70.902=138.057kN γo*Fly=1.0*138.057=138.06kN γo*Fly≤0.7*βhp*ft_b*am*YHo(6.5.5-1) =0.7*0.979*1.43*1410*1010 =1395.83kN y方向柱对基础的冲切满足规范要求 3.验算h2处冲切 YH=h1+h2=0.750m YB=bc1/2+bc2/2+bc+b3+b6=2.200m YL=max(hc1,hc2)+a3+a6=1.400m YB1=B1+Bc/2=2.200m,YB2=B2+Bc/2=2.200m,YL1=A1=1.900m,YL2=A2=1.700m YHo=YH-as=0.710m 3.1因(YH≤800)βhp=1.0 3.2x方向变阶处对基础的冲切验算 x冲切位置斜截面上边长bt=YB=2.200m x冲切位置斜截面下边长bb=YB+2*YHo=3.620m x冲切不利位置bm=(bt+bb)/2=(2.200+3.620)/2=2.910m x冲切面积Alx=max((YL1-YL/2-YHo)*(YB+2*YHo)+(YL1-YL/2-YHo)2,(YL2-YL/2-YHo)*(YB+2*YHo)+(YL2-YL/2-YHo)2 =max((1.900-1.400/2-0.710)*(2.200+2*0.710)+(1.900-1.400/2-0.710)2,(1.700-1.400/2-0.710)*(2.200+2*0.710)+(1.700-1.400/2-0.710)2) =max(2.014,1.134) =2.014m2 x冲切截面上的地基净反力设计值 Flx=Alx*Pjmax=2.014*70.902=142.790kN γo*Flx=1.0*142.790=142.79kN γo*Flx≤0.7*βhp*ft_b*bm*YHo(6.5.5-1) =0.7*1.000*1.43*2910*710 =2068.17kN x方向变阶处对基础的冲切满足规范要求 3.3y方向变阶处对基础的冲切验算 y冲切位置斜截面上边长at=YL=1.400m y冲切位置斜截面下边长ab=YL+2*YHo=2.820m y冲切面积Aly=max((YB1-YB/2-YHo)*(YL2+YL/2+YHo)+(YB1-YB/2-YHo)2/2-(YL2-YL/2-YHo)2/2,(YB2-YB/2-YHo)*(YL2+YL/2+YHo)+(YB2-YB/2-YHo)2/2)-(YL2-YL/2-YHo)2/2 =max((2.200-2.200/2-0.710)*(1.700+1.400/2+0.710)+(2.200-2.200/2-0.710)2/2-(1.700-1.400/2-0.710)2/2,(2.200-2.200/2-0.710)*(1.700+1.400/2+0.710)+(2.200-2.200/2-0.710)2/2-(1.700-1.400/2-0.710)2/2) =max(1.247,1.247) =1.247m2 y冲切截面上的地基净反力设计值 Fly=Aly*Pjmax=1.247*70.902=88.408kN γo*Fly=1.0*88.408=88.41kN γo*Fly≤0.7*βhp*ft_b*am*YHo(6.5.5-1) =0.7*1.000*1.43*2110*710 =1499.60kN y方向变阶处对基础的冲切满足规范要求 4.验算h3处冲切 YH=h3=0.300m YB=bc1/2+bc2/2+bc+b2+b3+b5+b6=3.300m YL=max(hc1,hc2)+a2+a3+a5+a6=2.500m YB1=B1+Bc/2=2.200m,YB2=B2+Bc/2=2.200m,YL1=A1=1.900m,YL2=A2=1.700m YHo=YH-as=0.260m 4.1因(YH≤800)βhp=1.0 4.2x方向变阶处对基础的冲切验算 x冲切位置斜截面上边长bt=YB=3.300m x冲切位置斜截面下边长bb=YB+2*YHo=3.820m x冲切不利位置bm=(bt+bb)/2=(3.300+3.820)/2=3.560m x冲切截面上的地基净反力设计值 Flx=Alx*Pjmax=0.000*70.902=0.000kN γo*Flx=1.0*0.000=0.00kN γo*Flx≤0.7*βhp*ft_b*bm*YHo(6.5.5-1) =0.7*1.000*1.43*3560*260 =926.53kN x方向变阶处对基础的冲切满足规范要求 4.3y方向变阶处对基础的冲切验算 y冲切位置斜截面上边长at=YL=2.500m y冲切位置斜截面下边长ab=YL+2*YHo=3.020m y冲切面积Aly=max((YB1-YB/2-YHo)*(YL2+YL/2+YHo)+(YB1-YB/2-YHo)2/2-(YL2-YL/2-YHo)2/2,(YB2-YB/2-YHo)*(YL2+YL/2+YHo)+(YB2-YB/2-YHo)2/2)-(YL2-YL/2-YHo)2/2 =max((2.200-3.3
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