双柱阶梯基础计算.docx

1、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)混凝土结构设计规范 (GB50010- 2010)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h仁450mm基础高度h2=300mm基础高度h3=300mm一阶长度b1=600mmb3=500mm一阶宽度a仁600mma3=500mm二阶长度b2=600mmb4=500mm二阶宽度a2=600mma4=500mm

2、一阶长度b1=600mmb4=500mm一阶宽度a仁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=500mm2.材料信息基础混凝土等级 柱混凝土等级：C30 ft b=1.43N/mmC30 ft_c=1.43N/mmfc b=14.3N/mmfc_c=14.3N/mm钢筋级别：3.计算信息结构重要性系数基础埋深:HPB300fy=270N/mm丫 0=1.0dh=1.500m纵筋合力点至近边距离： as=40mm4.作用在基础顶部荷载标准值Fgk1=200.000

3、kN Fqk1=100.000kNFgk2=100.000kN Fqk2=50.000kNFk=Fgk1/rg1+Fqk1/rq1+Fgk2/rg2+Fqk2/rq2=200.000/1.200+100.000/1.400+100.000/1.200+50.000/1.400=357.143kNMxk=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*

4、(1.700-1.900)+0.000+50.000*(1.700-1.900)=70.000kN*mMyk=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*(4.400/2-1.90 0)+0.000+50.000*(4.400/2-1.900)=135.000kN*mVxk=Vgxk1+Vqxk1+Vgxk2+Vqx

5、k2=80.000+0.000+80.000+0.000=160.000kNVyk=Vgyk1+Vqyk1+Vgyk2+Vqyk2=90.000+0.000+90.000+0.000=180.000kNF1= 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*mMx1=rg*(Mgxk1+Fgk1*(A2-A1)+rq*(Mqxk1+Fqk1*(A2-A1)+rg*(Mgxk2+Fgk2*(A2-A1)+rq*(Mqxk2+Fq k2*(A2-A1)=1.20

6、*(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*mMy仁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*(0.000-100.0

7、00*(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*mVx仁 rg*Vgxk1+rq*Vqxk1+rg*Vgxk2+rq*Vqxk2 =1.40*80.000+1.40*0.000+1.20*80.000+1.40*0.000 =192.000kNVy仁 rg*Vgyk1+rq*Vqyk1+rg*Vgyk2+rq*Vqyk2 =1.40*90.000+1.40*0.000+1.20*90.000+1.40*0.000 =216.000kNF2=

8、1.35*Fk=1.35*357.143=482.143kNMx2=1.35*Mxk =1.35*177.000 =238.950kN*mMy2=1.35*Myk =1.35* 102.000 =137.700kN*mVx2=1.35*Vxk=1.35*160.000=216.000kNVy2=1.35*Vyk=1.35*180.000=243.000kNF=max(|F1|,|F2|)=max(|570.000|,|482.143|)=570.000kNMx=max(|Mx1|,|Mx2|)=max(|78.000|,|238.950|)=238.950kN*mMy=max(|My1|,|M

9、y2|)=max(|159.000|,|137.700|)=159.000kN*mVx=max(|Vx1|,|Vx2|)=max(|192.000|,|216.000|)=216.000kNVy=max(|Vy1|,|Vy2|)=max(|216.000|,|243.000|)=243.000kN5.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=4.400m2.基础总宽 By=A1+A2=1.900+1.700=3.600mA1= a1+a2+a3+max(hc1,hc2)/2=0.600+0.600+

10、0.500+max(0.400,0.400)/2=1.900mA2=a4+a5+a6+max(hc1,hc2)/2=0.500+0.500+0.500+max(0.400,0.400)/2=1.700mB仁 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=1.900m3.基础总高 H=h1+h2+h3=0.450+0.300+0.300=1.050m4.底板配筋计算高度 ho=h1+h2+h3-as=0.450+0.300+0.300-0.040=1.010m5

11、.基础底面积 A=Bx*By=4.400*3.600=15.840m6.Gk= y *Bx*By*dh=20.000*4.400*3.600*1.500=475.200kNG=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*mMdyk=Myk+Vyk*H=102.000+180.000*1.050=291.000kN*mMdx=Mx-Vx*H=238.950-216.000 *1.050=12.150kN*mMdy=My+Vy*H=59.000+243.000 *

12、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 【5.2.2 -2 】因 丫 o*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| w Bx/6=0.733m x 方向小偏心， 由公式【5.2.2 -2】和【5.2.2 -3】推导 Pkmax_x=(Fk+G

13、k)/A+6*|Mdyk|/(Bx 2*By)2=(357.143+475.200)/15.840+6*1 174.000 |/(4.400 *3.600)=67.526kPaPkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx 2*By)=(357.143+475.200)/15.840-6*1 174.000 |/(4.400 2*3.600)=37.568kPaeyk=Mdxk/(Fk+Gk)= -14.500 /(357.143+475.200)=-0.017m因|eyk| w By/6=0.600m y方向小偏心Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By 2

14、*Bx) =(357.143+475.200)/15.840+6*1 -14.500 |/(3.600 2*4.400)=54.073kPaPkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By 2*Bx)2 =(357.143+475.200)/15.840-6*1 -14.500 |/(3.600 *4.400)=51.021kPa3.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=69.052kPaY o*Pkmax=1.0*69.052=69.052kPa w

15、 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exw Bx/6.0=0.733m x方向小偏心Pmax_x=(F+G)/A+6*|Mdy|/(Bx 2*By) =(570.000+641.520)/15.840+6*| 385.800 |/(4.400 2*3.600)=109.698kPaPmin_x=(F+G)/A-6*|Mdy|/(Bx 2*By)2 =(570.000+

16、641.520)/15.840-6*| 385.800 |/(4.400 *3.600)=43.272kPa1.2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=0.600 y方向小偏心Pmax_y=(F+G)/A+6*|Mdx|/(By *Bx)2=(570.000+641.520)/15.840+6*| -16.200 |/(3.600 *4.400)=78.189kPaPmin_y=(F+G)/A-6*|Mdx|/(By 2*Bx)=(570.000+641.520)/15.840-6*| -1

17、6.200 |/(3.600 2*4.400)=74.780kPa1.3因 Md炉 0 Mdyz 0Pmax=Pmax_x+Pmax_y-(F+G)/A =109.698+78.189-(570.000+641.520)/15.840 =111.402kPa1.4计算地基净反力极值Pjmax=Pmax-G/A=111.402-641.520/15.840=70.902kPaPjmax_x=Pmax_x-G/A=101.537-641.520/15.840=61.037kPaPjmax_y=Pmax_y-G/A=77.432-641.520/15.840=36.932kPa2.验算柱边冲切YH=

18、h1+h2+h3=1.050m, YB=bc1/2+bc2/2+Bc=1.100m, YL=max(hc1,hc2)=0.400mYB1=B1+Bc/2=2.200m, YB2=B2+Bc/2=2.200m, YL1=A1=1.900m, YL2=A2=1.700mYHo=YH-as=1.010m2.1因 800YH2000 3 hp=0.9792.2x方向柱对基础的冲切验算x冲切位置斜截面上边长 x冲切位置斜截面下边长 x冲切不利位置 x冲切面积bt=YB=1.100mbb=YB+2*YHo=3.120mbm=(bt+bb)/2=(1.100+3.120)/2=2.110m2/2-(1.70

19、0-0.400/22/2-(1.700-0.400/=max(2.200-1.100/2-1.010)*(1.700+0.400/2+1.010)+(2.200-1.100/2-1.010)2-1.010) /2,(2.200-1.100/2-1.010)*(1.700+0.400/2+1.010)+(2.200-1.100/2-1.010)22-1.010) /2)=max(1.947,1.947)=1.947my冲切截面上的地基净反力设计值Fly=Aly*Pjmax=1.947*70.902=138.057kN丫 o*Fly=1.0*138.057=138.06kN丫 o*Fly 0.7*

20、 3 hp*ft_b*am*YHo (6.5.5 -1)=0.7*0.979*1.43*1410*1010=1395.83kNy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=0.750mYB=bc1/2+bc2/2+bc+b3+b6=2.200mYL=max(hc1,hc2)+a3+a6=1.400mYB1=B1+Bc/2=2.200m, YB2=B2+Bc/2=2.200m, YL1=A1=1.900m, YL2=A2=1.700mYHo=YH-as=0.710m2,(YL2-YL/2-YHo)*(YB+2*YHo)+(YL2-YL/2-YHo)=max(2.200-2.

21、200/2-0.710)*(1.700+1.400/2+0.710)+(2.200-2.200/2-0.710)2-0.710) /2,(2.200-2.200/2-0.710)*(1.700+1.400/2+0.710)+(2.200-2.200/2-0.710)22-0.710) /2)=max(1.247,1.247)=1.247my冲切截面上的地基净反力设计值Fly=Aly*Pjmax=1.247*70.902=88.408kN丫 o*Fly=1.0*88.408=88.41kN丫 o*Fly 0.7* 3 hp* ft_b*am*YHo (6.5.5-1)=0.7*1.000*1.4

22、3*2110*710=1499.60kNy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=0.300mYB=bc1/2+bc2/2+bc+b2+b3+b5+b6=3.300m2/2-(1.700-1.400/22/2-(1.700-1.400/YL=max(hc1,hc2)+a2+a3+a5+a6=2.500mYB1=B1+Bc/2=2.200m, YB2=B2+Bc/2=2.200m, YL1=A1=1.900m, YL2=A2=1.700mYHo=YH-as=0.260m=926.53kNx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

23、上边长 at=YL=2.500my冲切位置斜截面下边长 ab=YL+2*YHo=3.020my冲切面积2 2Aly=max(YB1-YB/2-YHo)*(YL2+YL/2+YHo)+(YB1-YB/2-YHo) /2-(YL2-YL/2-YHo) /2,(YB2-YB/2-YHo)*(YL22 22/2-(1.700-2.500/22/2-(1.700-2.500/+YL/2+YHo)+(YB2-YB/2-YHo) /2)-(YL2-YL/2-YHo) /2Fly=Aly*Pjmax=0.955*70.902=67.705kN丫 o*Fly=1.0*67.705=67.70kN丫 o*Fly

24、0 , Mdy0 此基础为双向受弯2.计算1-1截面弯矩因ex Bx/6=0.733m x 方向小偏心Pj仁m ax(Bx-a1)*(Pxmax-Pmin_x)/Bx)+Pmin_x-G/A,(Bx-a2)*(Pxmax-Pmin_x)/Bx)+Pmin_x-G/A)=max(4.400-0.600)*(101.537-51.433)/4.400)+51.433-641.520/15.840,(4.400-0.600)*(101.537-51 .433)/4.400)+51.433-641.520/15.840)=54.204kPa因ey w By/6=0.600m y 方向小偏心Pj2=ma

25、x(By-a1)*(Pymax-Pymi n)/By)+Pymi n-G/A,(By-a2)*(Pymax-Pymi n)/By)+Pymi n-G/A)=max(3.600-0.600)*(77.432-75.538)/3.600)+75.538-641.520/15.840,(3.600-0.600)*(77.432-75.5 38)/3.600)+75.538-641.520/15.840)=36.616kPab=bc1/2+bc2/2+Bc=0.400/2+0.800/2+0.600=1.200mh=max(hc1,hc2)=max(0.400,0.400)=0.400m3 x=1.0

26、773 y=1.0252MI_1=1/48* 3 x*(Bx -b) *(2*By+h)*(Pj1+Pjmax_x)=1/48*1.077*(4.400-1.200) 2*(2*3.600+0.400)*(54.204+61.037)=201.30kN*mMII_1=1/48* 3 y*(By -h) 2*(2*Bx+b)*(Pj2+Pjmax_y)=1/48*1.025*(3.600-0.400) *(2*4.400+1.200)*(36.616+36.932)=160.81kN*m3.计算ll-ll 截面弯矩因x方向小偏心Pj1= max(Bx-a1-a2)*(Pxmax-Pmin_x)/

27、Bx)+Pmin_x-G/A,(Bx-a3-a4)*(Pxmax-Pmin_x)/Bx)+Pmin_x- G/A)=max(4.400-0.600-0.600)*(101.537-51.433)/4.400)+51.433-641.520/15.840,(4.400-0.500-0.500) *(101.537-51.433)/4.400)+51.433-641.520/15.840)=49.649kPa因y方向小偏心Pj2=max(By-a1-a2)*(Pymax-Pymi n)/By)+Pymi n-G/A,(By-a3-a4)*(Pymax-Pymi n)/By)+Pymi n-G/A)

28、=max(3.600-0.600-0.600)*(77.432-75.538)/3.600)+75.538-641.520/15.840,(3.600-0.500-0.500)*(77.432-75.538)/3.600)+75.538-641.520/15.840)=36.406kPa3 x=1.0453 y=1.022MI_2=max(1/48* 3 x*(Bx -b-b3) *(2*By+h+a3)*(Pj1+Pjmax_x),1/48* 3 x*(Bx -b-b6) 2*(2*By+h+a6)*(Pj1+Pjmax_x)=max(1/48*1.045*(4.400-1.200-0.50

29、0) 2*(2*3.600+0.400+0.500)*(49.649+61.037),1/48*1.045*(4400-1.200-0.500) 2*(2*3.600+0.400+0.500)*(49.649+61.037)=142.35kN*m2 2MII_2=max(1/48* 3 y*(By -h-a3) *(2*Bx+b+b3)*(Pj2+Pjmax_y),1/4 8* 3 y*(By -h-a6) *(2*Bx+b+b6)*(Pj2+Pjmax_y)=max(1/48*1.022*(3.600-0.400-0.500) 2*(2*4.400+1.200+0.500)*(36.406+

30、61.037) ,1/48*1.022*(3.600-0.400-0.500) 2*(2*4.400+1.200+0.500)*(36.406+61.037)=119.54kN*m4.计算III-III 截面弯矩因x方向小偏心Pj仁 max(Bx-a1-a2-a3)*(Pxmax-Pmin_x)/Bx)+Pmin_x-G/A,(Bx-a4-a5-a6)*(Pxmax-Pmin_x)/Bx)+Pmin_x-G/A) =max(4.400-0.600-0.600-0.500)*(101.537-51.433)/4.400)+51.433-641.520/15.840,(4.400-0.500-0.500-0.500)*(101.537-51.433)/4.400)+51.433-641.520/15.840)=43.956kPa因y方向