阶梯基础计算计算书

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)(2015年版)②二、示意图三、计算信息 构件编号:DJa 计算类型:验算截面尺寸 1.几何参数 台阶数 n=1 矩形柱宽 bc=400mm 矩形柱高 hc=400mm 基础高度 h1=600mm 一阶长度b1=200mm b2=200mm 一阶宽度a1=200mm a2=200mm 2.材料信息 基础混凝土等级: C40 ft_b=1.71N/mm2 fc_b=19.1N/mm2 柱混凝土等级: C40 ft_c=1.71N/mm2 fc_c=19.1N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.00 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组合值 F=35.600kN Mx=0.000kN*m My=0.000kN*m Vx=0.000kN Vy=0.000kN ks=1.35 Fk=F/ks=35.600/1.35=26.370kN Mxk=Mx/ks=0.000/1.35=0.000kN*m Myk=My/ks=0.000/1.35=0.000kN*m Vxk=Vx/ks=0.000/1.35=0.000kN Vyk=Vy/ks=0.000/1.35=0.000kN 5.修正后的地基承载力特征值 fa=1626.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.200+0.200+0.400=0.800m 2.基础总宽By=a1+a2+hc=0.200+0.200+0.400=0.800mA1=a1+hc/2=0.200+0.400/2=0.400mA2=a2+hc/2=0.200+0.400/2=0.400mB1=b1+bc/2=0.200+0.400/2=0.400mB2=b2+bc/2=0.200+0.400/2=0.400m 3.基础总高H=h1=0.600=0.600m 4.底板配筋计算高度ho=h1-as=0.600-0.040=0.560m 5.基础底面积A=Bx*By=0.800*0.800=0.640m2 6.Gk=γ*Bx*By*dh=20.000*0.800*0.800*1.500=19.200kN G=1.35*Gk=1.35*19.200=25.920kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=0.000-0.000*0.600=0.000kN*m Mdyk=Myk+Vxk*H=0.000+0.000*0.600=0.000kN*m Mdx=Mx-Vy*H=0.000-0.000*0.600=0.000kN*m Mdy=My+Vx*H=0.000+0.000*0.600=0.000kN*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=(26.370+19.200)/0.640=71.204kPa 【①5.2.1-2】 因γo*pk=1.00*71.204=71.204kPa≤fa=1626.000kPa 轴心荷载作用下地基承载力满足要求 因Mdyk=0,Mdxk=0 Pkmax=(Fk+Gk)/A=(26.370+19.200)/0.640=71.204kPa七、基础冲切验算 1.计算基础底面反力设计值 因Mdx=0并且Mdy=0 Pmax=Pmin=(F+G)/A=(35.600+25.920)/0.640=96.125kPa Pjmax=Pmax-G/A=96.125-25.920/0.640=55.625kPa 2.验算柱边冲切 YH=h1=0.600m,YB=bc=0.400m,YL=hc=0.400m YB1=B1=0.400m,YB2=B2=0.400m,YL1=A1=0.400m,YL2=A2=0.400m YHo=YH-as=0.560m 因((YB+2*YHo)≥Bx)并且(YL+2*YHo)≥By) 基础底面处边缘均位于冲切锥体以内,不用验算柱对基础的冲切八、基础受剪承载力验算 1.计算剪力 Az=a1+a2+hc =200+200+400 =800mm Bz=b1+b2+bc =200+200+400 =800mm A'=Az*max(b1,b2) =800.0*max(200.0,200.0) =0.16m2 Vs=A'*p=0.2*55.6=8.9kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/800.0)1/4=1.0 3.剪切承载力验算 Ao=Az*h1 =800*600 =480000.00mm2 γo*Vs=1.0*8.9=8.9kN ≤0.7βhsftAo=0.7*1.0*1.71*480000.0=574.6kN 受剪承载力验算满足要求！九、柱下基础的局部受压验算 因为基础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基础顶面的局部受压承载力。十、基础受弯计算 因Mdx=0Mdy=0基础轴心受压,根据公式【①8.2.7-4】【①8.2.7-5】推导: MI_1=1/24*(Bx-bc)2*(2*By+hc)*Pjmax =1/24*(0.800-0.400)2*(2*0.800+0.400)*55.625 =0.74kN*m MII_1=1/24*(By-hc)2*(2*Bx+bc)*Pjmax =1/24*(0.800-0.400)2*(2*0.800+0.400)*55.625 =0.74kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.00*0.74*106/(0.9*(600.000-40.000)*360) =4.1mm2 Asx1=Asx_1=4.1mm2 Asx=Asx1/By=4.1/0.800=5mm2/m Asx=max(Asx,ρmin*H*1000) =max(5,0.150%*600*1000) =900mm2/m 选择钢筋12@125,实配面积为905mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.00*0.74*106/(0.9*(600.000-40.000)*360) =4.1mm2 Asy1=Asy_1=4.1mm2 Asy=Asy1/Bx=4.1/0.800=5mm2/m Asy=max(Asy,ρmin*H*1000) =max(5,0.150%*600*1000) =900mm2/m 选择钢筋12@125,实配面积为905mm2/m。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)(2015年版)②二、示意图三、计算信息 构件编号:DJb 计算类型:验算截面尺寸 1.几何参数 台阶数 n=1 矩形柱宽 bc=600mm 矩形柱高 hc=600mm 基础高度 h1=600mm 一阶长度b1=250mm b2=250mm 一阶宽度a1=250mm a2=250mm 2.材料信息 基础混凝土等级: C40 ft_b=1.71N/mm2 fc_b=19.1N/mm2 柱混凝土等级: C40 ft_c=1.71N/mm2 fc_c=19.1N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.00 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组合值 F=256.400kN Mx=45.800kN*m My=-8.000kN*m Vx=-11.500kN Vy=-27.700kN ks=1.35 Fk=F/ks=256.400/1.35=189.926kN Mxk=Mx/ks=45.800/1.35=33.926kN*m Myk=My/ks=-8.000/1.35=-5.926kN*m Vxk=Vx/ks=-11.500/1.35=-8.519kN Vyk=Vy/ks=-27.700/1.35=-20.519kN 5.修正后的地基承载力特征值 fa=1626.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.250+0.250+0.600=1.100m 2.基础总宽By=a1+a2+hc=0.250+0.250+0.600=1.100mA1=a1+hc/2=0.250+0.600/2=0.550mA2=a2+hc/2=0.250+0.600/2=0.550mB1=b1+bc/2=0.250+0.600/2=0.550mB2=b2+bc/2=0.250+0.600/2=0.550m 3.基础总高H=h1=0.600=0.600m 4.底板配筋计算高度ho=h1-as=0.600-0.040=0.560m 5.基础底面积A=Bx*By=1.100*1.100=1.210m2 6.Gk=γ*Bx*By*dh=20.000*1.100*1.100*1.500=36.300kN G=1.35*Gk=1.35*36.300=49.005kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=33.926-(-20.519)*0.600=46.237kN*m Mdyk=Myk+Vxk*H=-5.926+(-8.519)*0.600=-11.037kN*m Mdx=Mx-Vy*H=45.800-(-27.700)*0.600=62.420kN*m Mdy=My+Vx*H=-8.000+(-11.500)*0.600=-14.900kN*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=(189.926+36.300)/1.210=186.964kPa 【①5.2.1-2】 因γo*pk=1.00*186.964=186.964kPa≤fa=1626.000kPa 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)=-11.037/(189.926+36.300)=-0.049m 因|exk|≤Bx/6=0.183m x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) =(189.926+36.300)/1.210+6*|-11.037|/(1.1002*1.100) =236.717kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(189.926+36.300)/1.210-6*|-11.037|/(1.1002*1.100) =137.210kPa eyk=Mdxk/(Fk+Gk)=46.237/(189.926+36.300)=0.204m 因|eyk|>By/6=0.183m y方向大偏心,由公式【①8.2.2-2】推导 ayk=By/2-|eyk|=1.100/2-|0.204|=0.346m Pkmax_y=2*(Fk+Gk)/(3*Bx*ayk) =2*(189.926+36.300)/(3*1.100*0.346) =396.703kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(189.926+36.300)/1.210-6*|46.237|/(1.1002*1.100) =-21.468kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(236.717-186.964)+(396.703-186.964)+186.964 =446.456kPa γo*Pkmax=1.00*446.456=446.456kPa≤1.2*fa=1.2*1626.000=1951.200kPa 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)=-14.900/(256.400+49.005)=-0.049m 因ex≤Bx/6.0=0.183m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(256.400+49.005)/1.210+6*|-14.900|/(1.1002*1.100) =319.568kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(256.400+49.005)/1.210-6*|-14.900|/(1.1002*1.100) =185.233kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=62.420/(256.400+49.005)=0.204m 因ey>By/6=0.183 y方向大偏心,由公式【①8.2.2-2】推导 ay=By/2-|ey|=1.100/2-|0.204|=0.346m Pmax_y=2*(F+G)/(3*Bx*ay) =2*(256.400+49.005)/(3*1.100*0.346) =535.548kPa Pmin_y=0 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =319.568+535.548-(256.400+49.005)/1.210 =602.716kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=602.716-49.005/1.210=562.216kPa Pjmax_x=Pmax_x-G/A=319.568-49.005/1.210=279.068kPa Pjmax_y=Pmax_y-G/A=535.548-49.005/1.210=495.048kPa 2.验算柱边冲切 YH=h1=0.600m,YB=bc=0.600m,YL=hc=0.600m YB1=B1=0.550m,YB2=B2=0.550m,YL1=A1=0.550m,YL2=A2=0.550m YHo=YH-as=0.560m 因((YB+2*YHo)≥Bx)并且(YL+2*YHo)≥By) 基础底面处边缘均位于冲切锥体以内,不用验算柱对基础的冲切八、基础受剪承载力验算 1.计算剪力 Az=a1+a2+hc =250+250+600 =1100mm Bz=b1+b2+bc =250+250+600 =1100mm A'=Az*max(b1,b2) =1100.0*max(250.0,250.0) =0.28m2 Vs=A'*p=0.3*211.9=58.3kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/800.0)1/4=1.0 3.剪切承载力验算 Ao=Az*h1 =1100*600 =660000.00mm2 γo*Vs=1.0*58.3=58.3kN ≤0.7βhsftAo=0.7*1.0*1.71*660000.0=790.0kN 受剪承载力验算满足要求！九、柱下基础的局部受压验算 因为基础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基础顶面的局部受压承载力。十、基础受弯计算 1.因Mdx>0,Mdy>0此基础为双向受弯 2.计算I-I截面弯矩 因ex≤Bx/6=0.183mx方向小偏心 a=(Bx-bc)/2=(1.100-0.600)/2=0.250m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.100-0.250)*(319.568-185.233)/1.100)+185.233-49.005/1.210 =248.538kPa 因ex>By/6=0.183my方向大偏心 a=(By-hc)/2=(1.100-0.600)/2=0.250m ay=By/2-ey=1.100/2-0.204=0.346m Pj2=(Pmax_y*(3*ay-a)/(3*ay))-G/A =(535.548*(3*0.346-0.250)/(3*0.346))-49.005/1.210 =365.919kPa βx=1.023 βy=1.056 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.023*(1.100-0.600)2*(2*1.100+0.600)*(248.538+279.068) =7.87kN*m MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.056*(1.100-0.600)2*(2*1.100+0.600)*(365.919+495.048) =13.25kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.00*7.87*106/(0.9*(600.000-40.000)*360) =43.4mm2 Asx1=Asx_1=43.4mm2 Asx=Asx1/By=43.4/1.100=39mm2/m Asx=max(Asx,ρmin*H*1000) =max(39,0.150%*600*1000) =900mm2/m 选择钢筋12@125,实配面积为905mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.00*13.25*106/(0.9*(600.000-40.000)*360) =73.0mm2 Asy1=Asy_1=73.0mm2 Asy=Asy1/Bx=73.0/1.100=66mm2/m Asy=max(Asy,ρmin*H*1000) =max(66,0.150%*600*1000) =900mm2/m 选择钢筋12@125,实配面积为905mm2/m。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)(2015年版)②二、示意图三、计算信息 构件编号:DJc 计算类型:验算截面尺寸 1.几何参数 台阶数 n=1 矩形柱宽 bc=600mm 矩形柱高 hc=600mm 基础高度 h1=700mm 一阶长度b1=350mm b2=350mm 一阶宽度a1=350mm a2=350mm 2.材料信息 基础混凝土等级: C40 ft_b=1.71N/mm2 fc_b=19.1N/mm2 柱混凝土等级: C40 ft_c=1.71N/mm2 fc_c=19.1N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.00 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组合值 F=2982.600kN Mx=-20.600kN*m My=-201.000kN*m Vx=-1.100kN Vy=10.900kN ks=1.35 Fk=F/ks=2982.600/1.35=2209.333kN Mxk=Mx/ks=-20.600/1.35=-15.259kN*m Myk=My/ks=-201.000/1.35=-148.889kN*m Vxk=Vx/ks=-1.100/1.35=-0.815kN Vyk=Vy/ks=10.900/1.35=8.074kN 5.修正后的地基承载力特征值 fa=1626.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.350+0.350+0.600=1.300m 2.基础总宽By=a1+a2+hc=0.350+0.350+0.600=1.300mA1=a1+hc/2=0.350+0.600/2=0.650mA2=a2+hc/2=0.350+0.600/2=0.650mB1=b1+bc/2=0.350+0.600/2=0.650mB2=b2+bc/2=0.350+0.600/2=0.650m 3.基础总高H=h1=0.700=0.700m 4.底板配筋计算高度ho=h1-as=0.700-0.040=0.660m 5.基础底面积A=Bx*By=1.300*1.300=1.690m2 6.Gk=γ*Bx*By*dh=20.000*1.300*1.300*1.500=50.700kN G=1.35*Gk=1.35*50.700=68.445kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=-15.259-8.074*0.700=-20.911kN*m Mdyk=Myk+Vxk*H=-148.889+(-0.815)*0.700=-149.459kN*m Mdx=Mx-Vy*H=-20.600-10.900*0.700=-28.230kN*m Mdy=My+Vx*H=-201.000+(-1.100)*0.700=-201.770kN*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=(2209.333+50.700)/1.690=1337.298kPa 【①5.2.1-2】 因γo*pk=1.00*1337.298=1337.298kPa≤fa=1626.000kPa 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)=-149.459/(2209.333+50.700)=-0.066m 因|exk|≤Bx/6=0.217m x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) =(2209.333+50.700)/1.690+6*|-149.459|/(1.3002*1.300) =1745.471kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(2209.333+50.700)/1.690-6*|-149.459|/(1.3002*1.300) =929.125kPa eyk=Mdxk/(Fk+Gk)=-20.911/(2209.333+50.700)=-0.009m 因|eyk|≤By/6=0.217m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(2209.333+50.700)/1.690+6*|-20.911|/(1.3002*1.300) =1394.406kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(2209.333+50.700)/1.690-6*|-20.911|/(1.3002*1.300) =1280.190kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(1745.471-1337.298)+(1394.406-1337.298)+1337.298 =1802.579kPa γo*Pkmax=1.00*1802.579=1802.579kPa≤1.2*fa=1.2*1626.000=1951.200kPa 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)=-201.770/(2982.600+68.445)=-0.066m 因ex≤Bx/6.0=0.217m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(2982.600+68.445)/1.690+6*|-201.770|/(1.3002*1.300) =2356.385kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(2982.600+68.445)/1.690-6*|-201.770|/(1.3002*1.300) =1254.319kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=-28.230/(2982.600+68.445)=-0.009m 因ey≤By/6=0.217 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(2982.600+68.445)/1.690+6*|-28.230|/(1.3002*1.300) =1882.448kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(2982.600+68.445)/1.690-6*|-28.230|/(1.3002*1.300) =1728.256kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =2356.385+1882.448-(2982.600+68.445)/1.690 =2433.481kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=2433.481-68.445/1.690=2392.981kPa Pjmax_x=Pmax_x-G/A=2356.385-68.445/1.690=2315.885kPa Pjmax_y=Pmax_y-G/A=1882.448-68.445/1.690=1841.948kPa 2.验算柱边冲切 YH=h1=0.700m,YB=bc=0.600m,YL=hc=0.600m YB1=B1=0.650m,YB2=B2=0.650m,YL1=A1=0.650m,YL2=A2=0.650m YHo=YH-as=0.660m 因((YB+2*YHo)≥Bx)并且(YL+2*YHo)≥By) 基础底面处边缘均位于冲切锥体以内,不用验算柱对基础的冲切八、基础受剪承载力验算 1.计算剪力 Az=a1+a2+hc =350+350+600 =1300mm Bz=b1+b2+bc =350+350+600 =1300mm A'=Az*max(b1,b2) =1300.0*max(350.0,350.0) =0.46m2 Vs=A'*p=0.5*1764.9=803.0kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/800.0)1/4=1.0 3.剪切承载力验算 Ao=Az*h1 =1300*700 =910000.00mm2 γo*Vs=1.0*803.0=803.0kN ≤0.7βhsftAo=0.7*1.0*1.71*910000.0=1089.3kN 受剪承载力验算满足要求！九、柱下基础的局部受压验算 因为基础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基础顶面的局部受压承载力。十、基础受弯计算 1.因Mdx>0,Mdy>0此基础为双向受弯 2.计算I-I截面弯矩 因ex≤Bx/6=0.217mx方向小偏心 a=(Bx-bc)/2=(1.300-0.600)/2=0.350m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.300-0.350)*(2356.385-1254.319)/1.300)+1254.319-68.445/1.690 =2019.175kPa 因ey≤By/6=0.217my方向小偏心 a=(By-hc)/2=(1.300-0.600)/2=0.350m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.300-0.350)*(1882.448-1728.256)/1.300)+1728.256-68.445/1.690 =1800.435kPa βx=1.028 βy=1.021 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.028*(1.300-0.600)2*(2*1.300+0.600)*(2019.175+2315.885) =145.57kN*m MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.021*(1.300-0.600)2*(2*1.300+0.600)*(1800.435+1841.948) =121.53kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.00*145.57*106/(0.9*(700.000-40.000)*360) =680.7mm2 Asx1=Asx_1=680.7mm2 Asx=Asx1/By=680.7/1.300=524mm2/m Asx=max(Asx,ρmin*H*1000) =max(524,0.150%*700*1000) =1050mm2/m 选择钢筋16@190,实配面积为1058mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.00*121.53*106/(0.9*(700.000-40.000)*360) =568.3mm2 Asy1=Asy_1=568.3mm2 Asy=Asy1/Bx=568.3/1.300=437mm2/m Asy=max(Asy,ρmin*H*1000) =max(437,0.150%*700*1000) =1050mm2/m 选择钢筋16@190,实配面积为1058mm2/m。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:DJd 计算类型:验算截面尺寸 1.几何参数 台阶数 n=1 矩形柱宽 bc=600mm 矩形柱高 hc=600mm 基础高度 h1=800mm 一阶长度b1=450mm b2=450mm 一阶宽度a1=450mm a2=450mm 2.材料信息 基础混凝土等级: C40 ft_b=1.71N/mm2 fc_b=19.1N/mm2 柱混凝土等级: C40 ft_c=1.71N/mm2 fc_c=19.1N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.00 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组合值 F=3536.400kN Mx=-26.600kN*m My=-53.500kN*m Vx=-28.500kN Vy=14.200kN ks=1.35 Fk=F/ks=3536.400/1.35=2619.556kN Mxk=Mx/ks=-26.600/1.35=-19.704kN*m Myk=My/ks=-53.500/1.35=-39.630kN*m Vxk=Vx/ks=-28.500/1.35=-21.111kN Vyk=Vy/ks=14.200/1.35=10.519kN 5.修正后的地基承载力特征值 fa=1626.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.450+0.450+0.600=1.500m 2.基础总宽By=a1+a2+hc=0.450+0.450+0.600=1.500mA1=a1+hc/2=0.450+0.600/2=0.750mA2=a2+hc/2=0.450+0.600/2=0.750mB1=b1+bc/2=0.450+0.600/2=0.750mB2=b2+bc/2=0.450+0.600/2=0.750m 3.基础总高H=h1=0.800=0.800m 4.底板配筋计算高度ho=h1-as=0.800-0.040=0.760m 5.基础底面积A=Bx*By=1.500*1.500=2.250m2 6.Gk=γ*Bx*By*dh=20.000*1.500*1.500*1.500=67.500kN G=1.35*Gk=1.35*67.500=91.125kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=-19.704-10.519*0.800=-28.119kN*m Mdyk=Myk+Vxk*H=-39.630+(-21.111)*0.800=-56.519kN*m Mdx=Mx-Vy*H=-26.600-14.200*0.800=-37.960kN*m Mdy=My+Vx*H=-53.500+(-28.500)*0.800=-76.300kN*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=(2619.556+67.500)/2.250=1194.247kPa 【①5.2.1-2】 因γo*pk=1.00*1194.247=1194.247kPa≤fa=1626.000kPa 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)=-56.519/(2619.556+67.500)=-0.021m 因|exk|≤Bx/6=0.250m x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) =(2619.556+67.500)/2.250+6*|-56.519|/(1.5002*1.500) =1294.724kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(2619.556+67.500)/2.250-6*|-56.519|/(1.5002*1.500) =1093.770kPa eyk=Mdxk/(Fk+Gk)=-28.119/(2619.556+67.500)=-0.010m 因|eyk|≤By/6=0.250m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(2619.556+67.500)/2.250+6*|-28.119|/(1.5002*1.500) =1244.235kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(2619.556+67.500)/2.250-6*|-28.119|/(1.5002*1.500) =1144.258kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(1294.724-1194.247)+(1244.235-1194.247)+1194.247 =1344.713kPa γo*Pkmax=1.00*1344.713=1344.713kPa≤1.2*fa=1.2*1626.000=1951.200kPa 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)=-76.300/(3536.400+91.125)=-0.021m 因ex≤Bx/6.0=0.250m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(3536.400+91.125)/2.250+6*|-76.300|/(1.5002*1.500) =1747.878kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(3536.400+91.125)/2.250-6*|-76.300|/(1.5002*1.500) =1476.589kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=-37.960/(3536.400+91.125)=-0.010m 因ey≤By/6=0.250 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(3536.400+91.125)/2.250+6*|-37.960|/(1.5002*1.500) =1679.718kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(3536.400+91.125)/2.250-6*|-37.960|/(1.5002*1.500) =1544.749kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =1747.878+1679.718-(3536.400+91.125)/2.250 =1815.362kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=1815.362-91.125/2.250=1774.862kPa Pjmax_x=Pmax_x-G/A=1747.878-91.125/2.250=1707.378kPa Pjmax_y=Pmax_y-G/A=1679.718-91.125/2.250=1639.218kPa 2.验算柱边冲切 YH=h1=0.800m,YB=bc=0.600m,YL=hc=0.600m YB1=B1=0.750m,YB2=B2=0.750m,YL1=A1=0.750m,YL2=A2=0.750m YHo=YH-as=0.760m 因((YB+2*YHo)≥Bx)并且(YL+2*YHo)≥By) 基础底面处边缘均位于冲切锥体以内,不用验算柱对基础的冲切八、基础受剪承载力验算 1.计算剪力 Az=a1+a2+hc =450+450+600 =1500mm Bz=b1+b2+bc =450+450+600 =1500mm A'=Az*max(b1,b2) =1500.0*max(450.0,450.0) =0.68m2 Vs=A'*p=0.7*1571.7=1060.9kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/800.0)1/4=1.0 3.剪切承载力验算 Ao=Az*h1 =1500*800 =1200000.00mm2 γo*Vs=1.0*1060.9=1060.9kN ≤0.7βhsftAo=0.7*1.0*1.71*1200000.0=1436.4kN 受剪承载力验算满足要求！九、柱下基础的局部受压验算 因为基础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基础顶面的局部受压承载力。十、基础受弯计算 1.因Mdx>0,Mdy>0此基础为双向受弯 2.计算I-I截面弯矩 因ex≤Bx/6=0.250mx方向小偏心 a=(Bx-bc)/2=(1.500-0.600)/2=0.450m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.500-0.450)*(1747.878-1476.589)/1.500)+1476.589-91.125/2.250 =1625.991kPa 因ey≤By/6=0.250my方向小偏心 a=(By-hc)/2=(1.500-0.600)/2=0.450m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.500-0.450)*(1679.718-1544.749)/1.500)+1544.749-91.125/2.250 =1598.727kPa βx=1.022 βy=1.022 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.022*(1.500-0.600)2*(2*1.500+0.600)*(1625.991+1707.378) =206.96kN*m MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.022*(1.500-0.600)2*(2*1.500+0.600)*(1598.727+1639.218) =201.03kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.00*206.96*106/(0.9*(800.000-40.000)*360) =840.5mm2 Asx1=Asx_1=840.5mm2 Asx=Asx1/By=840.5/1.500=560mm2/m Asx=max(Asx,ρmin*H*1000) =max(560,0.150%*800*1000) =1200mm2/m 选择钢筋14@125,实配面积为1232mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.00*201.03*106/(0.9*(800.000-40.000)*360) =816.4mm2 Asy1=Asy_1=816.4mm2 Asy=Asy1/Bx=816.4/1.500=544mm2/m Asy=max(Asy,ρmin*H*1000) =max(544,0.150%*800*1000) =1200mm2/m 选择钢筋14@125,实配面积为1232mm2/m。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)(2015年版)②二、示意图三、计算信息 构件编号:DJe 计算类型:验算截面尺寸 1.几何参数 台阶数 n=1 矩形柱宽 bc=600mm 矩形柱高 hc=600mm 基础高度 h1=800mm 一阶长度b1=550mm b2=550mm 一阶宽度a1=550mm a2=550mm 2.材料信息 基础混凝土等级: C40 ft_b=1.71N/mm2 fc_b=19.1N/mm2 柱混凝土等级: C40 ft_c=1.71N/mm2 fc_c=19.1N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.00 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组合值 F=4041.200kN Mx=14.800kN*m My=-78.400kN*m Vx=-41.900kN Vy=-8.000kN ks=1.35 Fk=F/ks=4041.200/1.35=2993.481kN Mxk=Mx/ks=14.800/1.35=10.963kN*m Myk=My/ks=-78.400/1.35=-58.074kN*m Vxk=Vx/ks=-41.900/1.35=-31.037kN Vyk=Vy/ks=-8.000/1.35=-5.926kN 5.修正后的地基承载力特征值 fa=1626.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.550+0.550+0.600=1.700m 2.基础总宽By=a1+a2+hc=0.550+0.550+0.600=1.700mA1=a1+hc/2=0.550+0.600/2=0.850mA2=a2+hc/2=0.550+0.600/2=0.850mB1=b1+bc/2=0.550+0.600/2=0.850mB2=b2+bc/2=0.550+0.600/2=0.850m 3.基础总高H=h1=0.800=0.800m 4.底板配筋计算高度ho=h1-as=0.800-0.040=0.760m 5.基础底面积A=Bx*By=1.700*1.700=2.890m2 6.Gk=γ*Bx*By*dh=20.000*1.700*1.700*1.500=86.700kN G=1.35*Gk=1.35*86.700=117.045kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=10.963-(-5.926)*0.800=15.704kN*m Mdyk=Myk+Vxk*H=-58.074+(-31.037)*0.800=-82.904kN*m Mdx=Mx-Vy*H=14.800-(-8.000)*0.800=21.200kN*m Mdy=My+Vx*H=-78.400+(-41.900)*0.800=-111.920kN*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=(2993.481+86.700)/2.890=1065.807kPa 【①5.2.1-2】 因γo*pk=1.00*1065.807=1065.807kPa≤fa=1626.000kPa 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)=-82.904/(2993.481+86.700)=-0.027m 因|exk|≤Bx/6=0.283m x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) =(2993.481+86.700)/2.890+6*|-82.904|/(1.7002*1.700) =1167.053kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(2993.481+86.700)/2.890-6*|-82.904|/(1.7002*1.700) =964.561kPa eyk=Mdxk/(Fk+Gk)=15.704/(2993.481+86.700)=0.005m 因|eyk|≤By/6=0.283m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(2993.481+86.700)/2.890+6*|15.704|/(1.7002*1.700) =1084.985kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(2993.481+86.700)/2.890-6*|15.704|/(1.7002*1.700) =1046.629kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(1167.053-1065.807)+(1084.985-1065.807)+1065.807 =1186.231kPa γo*Pkmax=1.00*1186.231=1186.231kPa≤1.2*fa=1.2*1626.000=1951.200kPa 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)=-111.920/(4041.200+117.045)=-0.027m 因ex≤Bx/6.0=0.283m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(4041.200+117.045)/2.890+6*|-111.920|/(1.7002*1.700) =1575.521kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(4041.200+117.045)/2.890-6*|-111.920|/(1.7002*1.700) =1302.157kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=21.200/(4041.200+11