阶梯基础计算(新规范)

1、- -阶梯根底计算一、设计依据"建筑地基根底设计标准" (GB50007-2021)"混凝土构造设计标准" (GB50010-2021)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根底高度h1=300mm根底高度h2=300mm根底高度h3=350mm一阶长度 b1=450mmb2=450mm 一阶宽度 a1=450mma2=450mm二阶长度 b3=450mmb4=450mm 二阶宽度 a3=450mma4=450mm三阶长度 b5=500mmb6=500m

2、m 三阶宽度 a5=500mma6=500mm2. 材料信息根底混凝土等级: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:HRB335 fy=300N/mm23. 计算信息构造重要性系数: o=1.0 根底埋深: dh=3.000m纵筋合力点至近边距离:as=40mm根底及其上覆土的平均容重:=19.000kN/m3最小配筋率:min=0.150%4. 作用在根底顶部荷载标准组合值F=1300.000kNMx=-20.000kN*mMy=21.000kN*mVx=-15.000kNVy=16.

3、000kNks=1.25Fk=F/ks=1300.000/1.25=1040.000kNMxk=Mx/ks=-20.000/1.25=-16.000kN*mMyk=My/ks=21.000/1.25=16.800kN*mVxk=Vx/ks=-15.000/1.25=-12.000kNVyk=Vy/ks=16.000/1.25=12.800kN5. 修正后的地基承载力特征值fa=160.000kPa 四、计算参数1. 根底总长 Bx=b1+b2+b3+b4+b5+b6+bc=0.450+0.450+0.450+0.450+0.500+0.500+0.400=3.200m2. 根底总宽 By=a1

4、+a2+a3+a4+a5+a6+hc=0.450+0.450+0.450+0.450+0.500+0.500+0.400=3.200mA1=a1+a2+a3+hc/2=0.450+0.450+0.450+0.400/2=1.550m A2=a4+a5+a6+hc/2=0.450+0.500+0.500+0.400/2=1.650mB1=b1+b2+b3+bc/2=0.450+0.450+0.450+0.400/2=1.550m B2=b4+b5+b6+bc/2=0.450+0.500+0.500+0.400/2=1.600m3. 根底总高 H=h1+h2+h3=0.300+0.300+0.35

5、0=0.950m4. 底板配筋计算高度 ho=h1+h2+h3-as=0.300+0.300+0.350-0.040=0.910m5. 根底底面积 A=Bx*By=3.200*3.200=10.240m26. Gk=*Bx*By*dh=19.000*3.200*3.200*3.000=583.680kNG=1.35*Gk=1.35*583.680=787.968kN五、计算作用在根底底部弯矩值Mdxk=Mxk-Vyk*H=-16.000-12.800*0.950=-28.160kN*mMdyk=Myk+Vxk*H=16.800+(-12.000)*0.950=5.400kN*mMdx=Mx-V

6、y*H=-20.000-16.000*0.950=-35.200kN*mMdy=My+Vx*H=21.000+(-15.000)*0.950=6.750kN*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=(1040.000+583.680)/10.240=158.563kPa【5.2.1-2】因o*pk=1.0*158.563=158.563kPafa=160.000kPa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)=5.400/(1040.000+583.680)=0.003m因 |exk| Bx/6=

7、0.533mx方向小偏心,由公式【5.2.2-2】和【5.2.2-3】推导Pkmax_x=(Fk+Gk)/A+6*|Mdyk|/(Bx2*By) =(1040.000+583.680)/10.240+6*|5.400|/(3.2002*3.200) =159.551kPaPkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(1040.000+583.680)/10.240-6*|5.400|/(3.2002*3.200) =157.574kPaeyk=Mdxk/(Fk+Gk)=-28.160/(1040.000+583.680)=-0.017m因 |eyk| By/6=0

8、.533my方向小偏心Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(1040.000+583.680)/10.240+6*|-28.160|/(3.2002*3.200) =163.719kPaPkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(1040.000+583.680)/10.240-6*|-28.160|/(3.2002*3.200) =153.406kPa3. 确定根底底面反力设计值Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(159.551-158.563)+(163.719-158.563)+1

9、58.563 =164.708kPao*Pkmax=1.0*164.708=164.708kPa1.2*fa=1.2*160.000=192.000kPa偏心荷载作用下地基承载力满足要求七、根底冲切验算1. 计算根底底面反力设计值1.1 计算x方向根底底面反力设计值ex=Mdy/(F+G)=6.750/(1300.000+787.968)=0.003m因 ex Bx/6.0=0.533mx方向小偏心Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(1300.000+787.968)/10.240+6*|6.750|/(3.2002*3.200) =205.139kPaPmin

10、_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(1300.000+787.968)/10.240-6*|6.750|/(3.2002*3.200) =202.667kPa1.2 计算y方向根底底面反力设计值ey=Mdx/(F+G)=-35.200/(1300.000+787.968)=-0.017m因 ey By/6=0.533y方向小偏心Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(1300.000+787.968)/10.240+6*|-35.200|/(3.2002*3.200) =210.348kPaPmin_y=(F+G)/A-6*|Mdx|/(By

11、2*Bx) =(1300.000+787.968)/10.240-6*|-35.200|/(3.2002*3.200) =197.458kPa1.3 因 Mdx0 Mdy0 Pmax=Pmax_x+Pmax_y-(F+G)/A=205.139+210.348-(1300.000+787.968)/10.240=211.584kPa1.4 计算地基净反力极值Pjmax=Pmax-G/A=211.584-787.968/10.240=134.634kPaPjmax_x=Pmax_x-G/A=205.139-787.968/10.240=128.189kPaPjmax_y=Pmax_y-G/A=21

12、0.348-787.968/10.240=133.398kPa2. 验算柱边冲切YH=h1+h2+h3=0.950m, YB=bc=0.400m, YL=hc=0.400mYB1=B1=1.550m, YB2=B2=1.600m, YL1=A1=1.550m, YL2=A2=1.650mYHo=YH-as=0.910m2.1 因 800<YH<2000 hp=0.9882.2 x方向柱对根底的冲切验算x冲切位置斜截面上边长bt=YB=0.400mx冲切位置斜截面下边长bb=YB+2*YHo=2.220mx冲切不利位置bm=(bt+bb)/2=(0.400+2.220)/2=1.31

13、0mx冲切面积 Alx=max(YL1-YL/2-ho)*(YB+2*ho)+(YL1-YL/2-ho)2,(YL2-YL/2-ho)*(YB+2*ho)+(YL2-YL/2-ho)2=max(1.550-0.400/2-0.910)*(0.400+2*0.910)+(1.550-0.400/2-0.910)2,(1.650-0.400/2-0.910)*(0.400+2*0.910)+(1.650-0.400/2-0.910)2)=max(1.170,1.490)=1.490m2x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=1.490*134.634=200.659kNo*Flx

14、=1.0*200.659=200.66kNo*Flx0.7*hp*ft_b*bm*YHo (6.5.5-1) =0.7*0.988*1.43*1310*910=1178.38kNx方向柱对根底的冲切满足标准要求2.3 y方向柱对根底的冲切验算y冲切位置斜截面上边长at=YL=0.400my冲切位置斜截面下边长ab=YL+2*YHo=2.220my冲切面积 Aly=max(YB1-YB/2-ho)*(YL+2*ho)+(YB1-YB/2-ho)2,(YB2-YB/2-ho)*(YL+2*ho)+(YB2-YB/2-ho)2)=max(1.550-0.400/2-0.910)*(0.400+0.9

15、10)+(1.550-0.400/2-0.910)2,(1.600-0.400/2-0.910)*(0.400+0.910)+(1.600-0.400/2-0.910)2)=max(1.170,1.328)=1.328m2y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=1.328*134.634=178.781kNo*Fly=1.0*178.781=178.78kNo*Fly0.7*hp*ft_b*am*YHo (6.5.5-1)=0.7*0.988*1.43*1310*910=1178.38kNy方向柱对根底的冲切满足标准要求3. 验算h2处冲切YH=h2+h3=0.650mYB=

16、bc+b3+b6=1.350mYL=hc+a3+a6=1.350mYB1=B1=1.550m, YB2=B2=1.600m, YL1=A1=1.550m, YL2=A2=1.650mYHo=YH-as=0.610m3.1 因 (YH800) hp=1.03.2 x方向变阶处对根底的冲切验算x冲切位置斜截面上边长bt=YB=1.350mx冲切位置斜截面下边长bb=YB+2*YHo=2.570mx冲切不利位置bm=(bt+bb)/2=(1.350+2.570)/2=1.960mx冲切面积 Alx=max(YL1-YL/2-ho)*(YB1+YB/2+ho)+(YL1-YL/2-ho)2/2,(YL

17、2-YL/2-ho)*(YB1+YB/2+ho)+(YL2-YL/2-ho)2/2=max(1.550-1.350/2-0.910)*(1.550+1.350/2+0.910)+(1.550-1.350/2-0.910)2/2,(1.650-1.350/2-0.910)*(1.550+1.350/2+0.910)+(1.650-1.350/2-0.910)2/2)=max(-0.109,0.206)=0.206m2x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=0.206*134.634=27.720kNo*Flx=1.0*27.720=27.72kNo*Flx0.7*hp*ft_b

18、*bm*YHo (6.5.5-1) =0.7*1.000*1.43*1960*610=1196.80kNx方向变阶处对根底的冲切满足标准要求3.3 y方向变阶处对根底的冲切验算y冲切位置斜截面上边长at=YL=1.350my冲切位置斜截面下边长ab=YL+2*YHo=2.570my冲切面积 Aly=max(YB1-YB/2-ho)*(YL1+YL/2+ho)+(YB1-YB/2-ho)2/2,(YB2-YB/2-ho)*(YL1+YL/2+ho)+(YB2-YB/2-ho)2/2)=max(1.550-1.350/2-0.910)*(1.550+1.350/2+0.910)+(1.550-1.

19、350/2-0.910)2/2,(1.600-1.350/2-0.910)*(1.550+1.350/2+0.910)+(1.600-1.350/2-0.910)2/2)=max(-0.109,0.047)=0.047m2y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=0.047*134.634=6.346kNo*Fly=1.0*6.346=6.35kNo*Fly0.7*hp*ft_b*am*YHo (6.5.5-1)=0.7*1.000*1.43*1960*610=1196.80kNy方向变阶处对根底的冲切满足标准要求4. 验算h3处冲切YH=h3=0.350mYB=bc+b2+b

20、3+b5+b6=2.300mYL=hc+a2+a3+a5+a6=2.300mYB1=B1=1.550m, YB2=B2=1.600m, YL1=A1=1.550m, YL2=A2=1.650mYHo=YH-as=0.310m4.1 因 (YH800) hp=1.04.2 x方向变阶处对根底的冲切验算因 YL/2+ho>=YL1和YL/2+h0>=YL2x方向根底底面外边缘位于冲切破坏锥体以内, 不用计算x方向的柱对根底的冲切验算4.3 y方向变阶处对根底的冲切验算因 YB/2+ho>=YB1和YB/2+ho>=YB2y方向根底底面外边缘位于冲切破坏锥体以内, 不用计算y

21、方向的柱对根底的冲切验算八、根底受剪承载力验算根底底面短边尺寸大于柱宽加两倍根底有效高度，不需验算受剪承载力！九、柱下根底的局部受压验算因为根底的混凝土强度等级大于等于柱的混凝土强度等级，所以不用验算柱下扩展根底顶面的局部受压承载力。十、根底受弯计算1. 因Mdx>0 , Mdy>0 此根底为双向受弯2. 计算I-I截面弯矩因 ex Bx/6=0.533m x方向小偏心a=(Bx-bc)/2=(3.200-0.400)/2=1.400mPj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =(3.200-1.400)*(205.139-202.667)

22、/3.200)+202.667-787.968/10.240 =127.108kPa因 ey By/6=0.533m y方向小偏心a=(By-hc)/2=(3.200-0.400)/2=1.400mPj2=(By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =(3.200-1.400)*(210.348-197.458)/3.200)+197.458-787.968/10.240 =127.759kPax=1.025y=1.025MI_1=1/48*x*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x)=1/48*1.025*(3.200-0.400)2*

23、(2*3.200+0.400)*(127.108+128.189)=290.57kN*mMII_1=1/48*y*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y)=1/48*1.025*(3.200-0.400)2*(2*3.200+0.400)*(127.759+133.398)=297.24kN*m3. 计算II-II截面弯矩因 x方向小偏心a=(Bx-bc-b3-b6)/2=(3.200-0.400-0.450-0.500)/2=0.925mPj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A=(3.200-0.925)*(205.139-2

24、02.667)/3.200)+202.667-787.968/10.240=127.475kPa因 y方向小偏心a=(By-hc-a3-a6)/2=(3.200-0.400-0.450-0.500)/2=0.925mPj2=(By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =(3.200-0.925)*(210.348-197.458)/3.200)+197.458-787.968/10.240 =129.672kPax=1.022y=1.022MI_2=1/48*x*(Bx-bc-b3-b6)2*(2*By+hc+a3+a6)*(Pj1+Pjmax_x)=1/48*

25、1.022*(3.200-0.400-0.450-0.500)2*(2*3.200+0.400+0.450+0.500)*(127.475+128.189)=144.35kN*mMII_2=1/48*y*(By-hc-a3-a6)2*(2*Bx+bc+b3+b6)*(Pj2+Pjmax_y)=1/48*1.022*(3.200-0.400-0.450-0.500)2*(2*3.200+0.400+0.450+0.500)*(129.672+133.398)=148.54kN*m4. 计算III-III截面弯矩因 x方向小偏心a=(Bx-bc-b2-b4-b3-b6)/2=(3.200-0.40

26、0-0.450-0.450-0.450-0.500)/2=0.475mPj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A=(3.200-0.475)*(205.139-202.667)/3.200)+202.667-787.968/10.240=127.822kPa因 y方向小偏心a=(By-hc-a2-a4-a3-a6)/2=(3.200-0.400-0.450-0.450-0.450-0.500)/2=0.475mPj2=(By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =(3.200-0.475)*(210.348-197.458

27、)/3.200)+197.458-787.968/10.240 =131.485kPax=1.019y=1.019MI_3=1/48.0*x*(Bx-bc-b3-b6-b2-b4)2*(2*By+hc+a3+a6+a2+a4)*(Pj1+Pjmax_x)=1/48.0*1.019*(3.200-0.400-0.450-0.500-0.450-0.450)2*(2*3.200+0.400+0.450+0.500+0.450+0.450)*(127.822+128.189)=42.45kN*mMII_3=1/48.0*y*(By-hc-a3-a6-a2-a4)2*(2*Bx+bc+b3+b6+b2+b4)*(Pj2+Pjmax_y)=1/48.0*1.019*(3.200-0.400-0.450-0.500-0.450-0.450)2*(2*3.200+0.400+0.450+0.500+0.450+0.450)*(131.485+133.398)=43.92kN*m十一、计算配筋10.1 计算AsxAsx_1=o*MI_1/(0.9*(H-as)*fy)