车库DJ计算书（改）

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:DJ1 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=1700mm 一阶长度b1=300mm b2=300mm 一阶宽度a1=300mm a2=300mm 2.材料信息 基础混凝土等级: C30 ft_b=1.43N/mm2 fc_b=14.3N/mm2 柱混凝土等级: C35 ft_c=1.57N/mm2 fc_c=16.7N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.10 基础埋深: dh=0.500m 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=6100.000kN Mx=20.000kN*m My=20.000kN*m Vx=10.000kN Vy=10.000kN ks=1.35 Fk=F/ks=6100.000/1.35=4518.519kN Mxk=Mx/ks=20.000/1.35=14.815kN*m Myk=My/ks=20.000/1.35=14.815kN*m Vxk=Vx/ks=10.000/1.35=7.407kN Vyk=Vy/ks=10.000/1.35=7.407kN 5.修正后的地基承载力特征值 fa=3630.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.300+0.300+0.600=1.200m 2.基础总宽By=a1+a2+hc=0.300+0.300+0.600=1.200mA1=a1+hc/2=0.300+0.600/2=0.600mA2=a2+hc/2=0.300+0.600/2=0.600mB1=b1+bc/2=0.300+0.600/2=0.600mB2=b2+bc/2=0.300+0.600/2=0.600m 3.基础总高H=h1=1.700=1.700m 4.底板配筋计算高度ho=h1-as=1.700-0.040=1.660m 5.基础底面积A=Bx*By=1.200*1.200=1.440m2 6.Gk=γ*Bx*By*dh=20.000*1.200*1.200*0.500=14.400kN G=1.35*Gk=1.35*14.400=19.440kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=14.815-7.407*1.700=2.222kN*m Mdyk=Myk+Vxk*H=14.815+7.407*1.700=27.407kN*m Mdx=Mx-Vy*H=20.000-10.000*1.700=3.000kN*m Mdy=My+Vx*H=20.000+10.000*1.700=37.000kN*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=(4518.519+14.400)/1.440=3147.860kPa 【①5.2.1-2】 因γo*pk=1.10*3147.860=3462.646kPa≤fa=3630.000kPa 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)=27.407/(4518.519+14.400)=0.006m 因|exk|≤Bx/6=0.200m 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) =(4518.519+14.400)/1.440+6*|27.407|/(1.2002*1.200) =3243.025kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(4518.519+14.400)/1.440-6*|27.407|/(1.2002*1.200) =3052.695kPa eyk=Mdxk/(Fk+Gk)=2.222/(4518.519+14.400)=0.000m 因|eyk|≤By/6=0.200m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(4518.519+14.400)/1.440+6*|2.222|/(1.2002*1.200) =3155.576kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(4518.519+14.400)/1.440-6*|2.222|/(1.2002*1.200) =3140.144kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(3243.025-3147.860)+(3155.576-3147.860)+3147.860 =3250.741kPa γo*Pkmax=1.10*3250.741=3575.815kPa≤1.2*fa=1.2*3630.000=4791.600kPa 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)=37.000/(6100.000+19.440)=0.006m 因ex≤Bx/6.0=0.200m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(6100.000+19.440)/1.440+6*|37.000|/(1.2002*1.200) =4378.083kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(6100.000+19.440)/1.440-6*|37.000|/(1.2002*1.200) =4121.139kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=3.000/(6100.000+19.440)=0.000m 因ey≤By/6=0.200 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(6100.000+19.440)/1.440+6*|3.000|/(1.2002*1.200) =4260.028kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(6100.000+19.440)/1.440-6*|3.000|/(1.2002*1.200) =4239.194kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =4378.083+4260.028-(6100.000+19.440)/1.440 =4388.500kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=4388.500-19.440/1.440=4375.000kPa Pjmax_x=Pmax_x-G/A=4378.083-19.440/1.440=4364.583kPa Pjmax_y=Pmax_y-G/A=4260.028-19.440/1.440=4246.528kPa 2.验算柱边冲切 YH=h1=1.700m,YB=bc=0.600m,YL=hc=0.600m YB1=B1=0.600m,YB2=B2=0.600m,YL1=A1=0.600m,YL2=A2=0.600m YHo=YH-as=1.660m 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 =300+300+600 =1200mm Bz=b1+b2+bc =300+300+600 =1200mm A'=Az*max(b1,b2) =1200.0*max(300.0,300.0) =0.36m2 Vs=A'*p=0.4*4236.1=1525.0kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/1700.0)1/4=0.8 3.剪切承载力验算 Ao=Az*h1 =1200*1700 =2040000.00mm2 γo*Vs=1.1*1525.0=1677.5kN ≤0.7βhsftAo=0.7*0.8*1.43*2040000.0=1691.3kN 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小于柱的混凝土强度等级，验算柱下扩展基础顶面的局部受压承载力。 1.Fl=F=6100.000kN 混凝土等级为C30,fcc=0.85*fc=0.85*14.3=12.155N/mm2,ω=1.0 2.计算混凝土局部受压面积 YB1=b1+bc/2=0.300+0.600/2=0.600m2 YB2=b2+bc/2=0.300+0.600/2=0.600m2 YA1=a1+hc/2=0.300+0.600/2=0.600m2 YA2=a2+hc/2=0.300+0.600/2=0.600m2 Al=bc*hc=0.600*0.600=0.360m2 Ab=bc*hc=0.600*0.600=0.360m2 Ab=(bc+2*min(bc,hc))*hc) =(0.600+2*min(0.600,0.600))*0.600) =1.080m2 Ab=(hc+2*min(bc,hc))*bc =(0.600+2*min(0.600,0.600))*0.600) =1.080m2 3.计算混凝土局部受压时的强度提高系数 【②附录D】 β1=sqrt(Ab/Al)=sqrt(1.080/0.360)=1.732 4.因 γo*Fl=1.10*6100.000=6710.00kN γo*Fl≤ω*β1*fcc*Al =1.350*1.732*12.155*360000.000/1000 =10231.80kN 柱下基础局部受压承载力满足规范要求十、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=0.200mx方向小偏心 a=(Bx-bc)/2=(1.200-0.600)/2=0.300m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.200-0.300)*(4378.083-4121.139)/1.200)+4121.139-19.440/1.440 =4300.347kPa 因ey≤By/6=0.200my方向小偏心 a=(By-hc)/2=(1.200-0.600)/2=0.300m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.200-0.300)*(4260.028-4239.194)/1.200)+4239.194-19.440/1.440 =4241.319kPa βx=1.021 βy=1.021 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.021*(1.200-0.600)2*(2*1.200+0.600)*(4300.347+4364.583) =199.06kN*m MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.021*(1.200-0.600)2*(2*1.200+0.600)*(4241.319+4246.528) =194.99kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.10*199.06*106/(0.9*(1700.000-40.000)*360) =407.1mm2 Asx1=Asx_1=407.1mm2 Asx=Asx1/By=407.1/1.200=339mm2/m Asx=max(Asx,ρmin*H*1000) =max(339,0.150%*1700*1000) =2550mm2/m 选择钢筋25@190,实配面积为2584mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.10*194.99*106/(0.9*(1700.000-40.000)*360) =398.8mm2 Asy1=Asy_1=398.8mm2 Asy=Asy1/Bx=398.8/1.200=332mm2/m Asy=max(Asy,ρmin*H*1000) =max(332,0.150%*1700*1000) =2550mm2/m 选择钢筋25@190,实配面积为2584mm2/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:DJ2 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=700mm 基础高度 h1=2000mm 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: C30 ft_b=1.43N/mm2 fc_b=14.3N/mm2 柱混凝土等级: C35 ft_c=1.57N/mm2 fc_c=16.7N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.10 基础埋深: dh=0.500m 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=7000.000kN Mx=20.000kN*m My=20.000kN*m Vx=10.000kN Vy=10.000kN ks=1.35 Fk=F/ks=7000.000/1.35=5185.185kN Mxk=Mx/ks=20.000/1.35=14.815kN*m Myk=My/ks=20.000/1.35=14.815kN*m Vxk=Vx/ks=10.000/1.35=7.407kN Vyk=Vy/ks=10.000/1.35=7.407kN 5.修正后的地基承载力特征值 fa=3630.000kPa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+0.700=1.400mA1=a1+hc/2=0.350+0.700/2=0.700mA2=a2+hc/2=0.350+0.700/2=0.700mB1=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=2.000=2.000m 4.底板配筋计算高度ho=h1-as=2.000-0.040=1.960m 5.基础底面积A=Bx*By=1.300*1.400=1.820m2 6.Gk=γ*Bx*By*dh=20.000*1.300*1.400*0.500=18.200kN G=1.35*Gk=1.35*18.200=24.570kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=14.815-7.407*2.000=0.000kN*m Mdyk=Myk+Vxk*H=14.815+7.407*2.000=29.630kN*m Mdx=Mx-Vy*H=20.000-10.000*2.000=0.000kN*m Mdy=My+Vx*H=20.000+10.000*2.000=40.000kN*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=(5185.185+18.200)/1.820=2859.003kPa 【①5.2.1-2】 因γo*pk=1.10*2859.003=3144.903kPa≤fa=3630.000kPa 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)=29.630/(5185.185+18.200)=0.006m 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) =(5185.185+18.200)/1.820+6*|29.630|/(1.3002*1.400) =2934.141kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(5185.185+18.200)/1.820-6*|29.630|/(1.3002*1.400) =2783.864kPa 因Mdxk=0Pkmax_y=Pkmin_y=(Fk+Gk)/A=(5185.185+18.200)/1.820=2859.003kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(2934.141-2859.003)+(2859.003-2859.003)+2859.003 =2934.141kPa γo*Pkmax=1.10*2934.141=3227.556kPa≤1.2*fa=1.2*3630.000=4791.600kPa 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)=40.000/(7000.000+24.570)=0.006m 因ex≤Bx/6.0=0.217m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(7000.000+24.570)/1.820+6*|40.000|/(1.3002*1.400) =3961.091kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(7000.000+24.570)/1.820-6*|40.000|/(1.3002*1.400) =3758.217kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=0.000/(7000.000+24.570)=0.000m 因ey≤By/6=0.233 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(7000.000+24.570)/1.820+6*|0.000|/(1.4002*1.300) =3859.654kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(7000.000+24.570)/1.820-6*|0.000|/(1.4002*1.300) =3859.654kPa 1.3因Mdx=0并且Mdy≠0 Pmax=Pmax_x=3961.091kPa Pmin=Pmin_x=3758.217kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=3961.091-24.570/1.820=3947.591kPa 2.验算柱边冲切 YH=h1=2.000m,YB=bc=0.600m,YL=hc=0.700m YB1=B1=0.650m,YB2=B2=0.650m,YL1=A1=0.700m,YL2=A2=0.700m YHo=YH-as=1.960m 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+700 =1400mm Bz=b1+b2+bc =350+350+600 =1300mm A'=Bz*max(a1,a2) =1300.0*max(350.0,350.0) =0.46m2 Vs=A'*p=0.5*3846.2=1750.0kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/2000.0)1/4=0.8 3.剪切承载力验算 Ao=Bz*h1 =1300*2000 =2600000.00mm2 γo*Vs=1.1*1750.0=1925.0kN ≤0.7βhsftAo=0.7*0.8*1.43*2600000.0=2069.8kN 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小于柱的混凝土强度等级，验算柱下扩展基础顶面的局部受压承载力。 1.Fl=F=7000.000kN 混凝土等级为C30,fcc=0.85*fc=0.85*14.3=12.155N/mm2,ω=1.0 2.计算混凝土局部受压面积 YB1=b1+bc/2=0.350+0.600/2=0.650m2 YB2=b2+bc/2=0.350+0.600/2=0.650m2 YA1=a1+hc/2=0.350+0.700/2=0.700m2 YA2=a2+hc/2=0.350+0.700/2=0.700m2 Al=bc*hc=0.600*0.700=0.420m2 Ab=bc*hc=0.600*0.700=0.420m2 Ab=(bc+2*min(bc,hc))*hc) =(0.600+2*min(0.600,0.700))*0.700) =1.260m2 Ab=(hc+2*min(bc,hc))*bc =(0.700+2*min(0.600,0.700))*0.600) =1.140m2 3.计算混凝土局部受压时的强度提高系数 【②附录D】 β1=sqrt(Ab/Al)=sqrt(1.140/0.420)=1.648 4.因 γo*Fl=1.10*7000.000=7700.00kN γo*Fl≤ω*β1*fcc*Al =1.350*1.648*12.155*420000.000/1000 =11354.44kN 柱下基础局部受压承载力满足规范要求十、基础受弯计算 因Mdx≠0Mdy=0并且ey<By/6=0.233my方向单向受压且小偏心 a=(By-bc)/2=(1.400-0.600)/2=0.400m P=((By-a)*(Pmax-Pmin)/By)+Pmin =((1.400-0.400)*(3961.091-3758.217)/1.400)+3758.217 =3903.127kPa MI_1=1/48*(Bx-bc)2*(2*By+hc)*(Pmax+Pmin-2*G/A) =1/48*(1.300-0.600)2*(2*1.400+0.700)*(3961.091+3758.217-2*24.570/1.820) =274.84kN*m MII_1=1/12*a2*((2*Bx+bc)*(Pmax+P-2*G/A)+(Pmax-P)*Bx) =1/12*0.4002*((2*1.300+0.600)*(3961.091+3903.127-2*24.570/1.820)+(3961.091-3903.127)*1.300) =335.39kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.10*274.84*106/(0.9*(2000.000-40.000)*360) =476.1mm2 Asx1=Asx_1=476.1mm2 Asx=Asx1/By=476.1/1.400=340mm2/m Asx=max(Asx,ρmin*H*1000) =max(340,0.150%*2000*1000) =3000mm2/m 选择钢筋22@125,实配面积为3041mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.10*335.39*106/(0.9*(2000.000-40.000)*360) =581.0mm2 Asy1=Asy_1=581.0mm2 Asy=Asy1/Bx=581.0/1.300=447mm2/m Asy=max(Asy,ρmin*H*1000) =max(447,0.150%*2000*1000) =3000mm2/m 选择钢筋22@125,实配面积为3041mm2/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:DJ3 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=500mm 矩形柱高 hc=500mm 基础高度 h1=1200mm 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: C30 ft_b=1.43N/mm2 fc_b=14.3N/mm2 柱混凝土等级: C35 ft_c=1.57N/mm2 fc_c=16.7N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.10 基础埋深: dh=0.500m 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=3600.000kN Mx=20.000kN*m My=20.000kN*m Vx=10.000kN Vy=10.000kN ks=1.35 Fk=F/ks=3600.000/1.35=2666.667kN Mxk=Mx/ks=20.000/1.35=14.815kN*m Myk=My/ks=20.000/1.35=14.815kN*m Vxk=Vx/ks=10.000/1.35=7.407kN Vyk=Vy/ks=10.000/1.35=7.407kN 5.修正后的地基承载力特征值 fa=3630.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.250+0.250+0.500=1.000m 2.基础总宽By=a1+a2+hc=0.250+0.250+0.500=1.000mA1=a1+hc/2=0.250+0.500/2=0.500mA2=a2+hc/2=0.250+0.500/2=0.500mB1=b1+bc/2=0.250+0.500/2=0.500mB2=b2+bc/2=0.250+0.500/2=0.500m 3.基础总高H=h1=1.200=1.200m 4.底板配筋计算高度ho=h1-as=1.200-0.040=1.160m 5.基础底面积A=Bx*By=1.000*1.000=1.000m2 6.Gk=γ*Bx*By*dh=20.000*1.000*1.000*0.500=10.000kN G=1.35*Gk=1.35*10.000=13.500kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=14.815-7.407*1.200=5.926kN*m Mdyk=Myk+Vxk*H=14.815+7.407*1.200=23.704kN*m Mdx=Mx-Vy*H=20.000-10.000*1.200=8.000kN*m Mdy=My+Vx*H=20.000+10.000*1.200=32.000kN*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=(2666.667+10.000)/1.000=2676.667kPa 【①5.2.1-2】 因γo*pk=1.10*2676.667=2944.333kPa≤fa=3630.000kPa 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)=23.704/(2666.667+10.000)=0.009m 因|exk|≤Bx/6=0.167m 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) =(2666.667+10.000)/1.000+6*|23.704|/(1.0002*1.000) =2818.889kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(2666.667+10.000)/1.000-6*|23.704|/(1.0002*1.000) =2534.444kPa eyk=Mdxk/(Fk+Gk)=5.926/(2666.667+10.000)=0.002m 因|eyk|≤By/6=0.167m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(2666.667+10.000)/1.000+6*|5.926|/(1.0002*1.000) =2712.222kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(2666.667+10.000)/1.000-6*|5.926|/(1.0002*1.000) =2641.111kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(2818.889-2676.667)+(2712.222-2676.667)+2676.667 =2854.444kPa γo*Pkmax=1.10*2854.444=3139.889kPa≤1.2*fa=1.2*3630.000=4791.600kPa 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)=32.000/(3600.000+13.500)=0.009m 因ex≤Bx/6.0=0.167m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(3600.000+13.500)/1.000+6*|32.000|/(1.0002*1.000) =3805.500kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(3600.000+13.500)/1.000-6*|32.000|/(1.0002*1.000) =3421.500kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=8.000/(3600.000+13.500)=0.002m 因ey≤By/6=0.167 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(3600.000+13.500)/1.000+6*|8.000|/(1.0002*1.000) =3661.500kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(3600.000+13.500)/1.000-6*|8.000|/(1.0002*1.000) =3565.500kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =3805.500+3661.500-(3600.000+13.500)/1.000 =3853.500kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=3853.500-13.500/1.000=3840.000kPa Pjmax_x=Pmax_x-G/A=3805.500-13.500/1.000=3792.000kPa Pjmax_y=Pmax_y-G/A=3661.500-13.500/1.000=3648.000kPa 2.验算柱边冲切 YH=h1=1.200m,YB=bc=0.500m,YL=hc=0.500m YB1=B1=0.500m,YB2=B2=0.500m,YL1=A1=0.500m,YL2=A2=0.500m YHo=YH-as=1.160m 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+500 =1000mm Bz=b1+b2+bc =250+250+500 =1000mm A'=Az*max(b1,b2) =1000.0*max(250.0,250.0) =0.25m2 Vs=A'*p=0.3*3600.0=900.0kN 2.计算截面高度影响系数βhs βhs=(800/h0)1/4=(800/1200.0)1/4=0.9 3.剪切承载力验算 Ao=Az*h1 =1000*1200 =1200000.00mm2 γo*Vs=1.1*900.0=990.0kN ≤0.7βhsftAo=0.7*0.9*1.43*1200000.0=1085.4kN 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小于柱的混凝土强度等级，验算柱下扩展基础顶面的局部受压承载力。 1.Fl=F=3600.000kN 混凝土等级为C30,fcc=0.85*fc=0.85*14.3=12.155N/mm2,ω=1.0 2.计算混凝土局部受压面积 YB1=b1+bc/2=0.250+0.500/2=0.500m2 YB2=b2+bc/2=0.250+0.500/2=0.500m2 YA1=a1+hc/2=0.250+0.500/2=0.500m2 YA2=a2+hc/2=0.250+0.500/2=0.500m2 Al=bc*hc=0.500*0.500=0.250m2 Ab=bc*hc=0.500*0.500=0.250m2 Ab=(bc+2*min(bc,hc))*hc) =(0.500+2*min(0.500,0.500))*0.500) =0.750m2 Ab=(hc+2*min(bc,hc))*bc =(0.500+2*min(0.500,0.500))*0.500) =0.750m2 3.计算混凝土局部受压时的强度提高系数 【②附录D】 β1=sqrt(Ab/Al)=sqrt(0.750/0.250)=1.732 4.因 γo*Fl=1.10*3600.000=3960.00kN γo*Fl≤ω*β1*fcc*Al =1.350*1.732*12.155*250000.000/1000 =7105.41kN 柱下基础局部受压承载力满足规范要求十、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=0.167mx方向小偏心 a=(Bx-bc)/2=(1.000-0.500)/2=0.250m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.000-0.250)*(3805.500-3421.500)/1.000)+3421.500-13.500/1.000 =3696.000kPa 因ey≤By/6=0.167my方向小偏心 a=(By-hc)/2=(1.000-0.500)/2=0.250m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.000-0.250)*(3661.500-3565.500)/1.000)+3565.500-13.500/1.000 =3624.000kPa βx=1.021 βy=1.021 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.021*(1.000-0.500)2*(2*1.000+0.500)*(3696.000+3792.000) =99.55kN*m MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.021*(1.000-0.500)2*(2*1.000+0.500)*(3624.000+3648.000) =96.68kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.10*99.55*106/(0.9*(1200.000-40.000)*360) =291.4mm2 Asx1=Asx_1=291.4mm2 Asx=Asx1/By=291.4/1.000=291mm2/m Asx=max(Asx,ρmin*H*1000) =max(291,0.150%*1200*1000) =1800mm2/m 选择钢筋18@140,实配面积为1818mm2/m。 10.2计算Asy Asy_1=γo*MII_1/(0.9*(H-as)*fy) =1.10*96.68*106/(0.9*(1200.000-40.000)*360) =282.9mm2 Asy1=Asy_1=282.9mm2 Asy=Asy1/Bx=282.9/1.000=283mm2/m Asy=max(Asy,ρmin*H*1000) =max(283,0.150%*1200*1000) =1800mm2/m 选择钢筋18@140,实配面积为1818mm2/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:DJ4 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=500mm 矩形柱高 hc=500mm 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: C30 ft_b=1.43N/mm2 fc_b=14.3N/mm2 柱混凝土等级: C35 ft_c=1.57N/mm2 fc_c=16.7N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.10 基础埋深: dh=0.500m 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=500.000kN Mx=20.000kN*m My=20.000kN*m Vx=10.000kN Vy=10.000kN ks=1.35 Fk=F/ks=500.000/1.35=370.370kN Mxk=Mx/ks=20.000/1.35=14.815kN*m Myk=My/ks=20.000/1.35=14.815kN*m Vxk=Vx/ks=10.000/1.35=7.407kN Vyk=Vy/ks=10.000/1.35=7.407kN 5.修正后的地基承载力特征值 fa=3630.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.200+0.200+0.500=0.900m 2.基础总宽By=a1+a2+hc=0.200+0.200+0.500=0.900mA1=a1+hc/2=0.200+0.500/2=0.450mA2=a2+hc/2=0.200+0.500/2=0.450mB1=b1+bc/2=0.200+0.500/2=0.450mB2=b2+bc/2=0.200+0.500/2=0.450m 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=0.900*0.900=0.810m2 6.Gk=γ*Bx*By*dh=20.000*0.900*0.900*0.500=8.100kN G=1.35*Gk=1.35*8.100=10.935kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=14.815-7.407*0.600=10.370kN*m Mdyk=Myk+Vxk*H=14.815+7.407*0.600=19.259kN*m Mdx=Mx-Vy*H=20.000-10.000*0.600=14.000kN*m Mdy=My+Vx*H=20.000+10.000*0.600=26.000kN*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=(370.370+8.100)/0.810=467.247kPa 【①5.2.1-2】 因γo*pk=1.10*467.247=513.972kPa≤fa=3630.000kPa 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)=19.259/(370.370+8.100)=0.051m 因|exk|≤Bx/6=0.150m 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) =(370.370+8.100)/0.810+6*|19.259|/(0.9002*0.900) =625.760kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(370.370+8.100)/0.810-6*|19.259|/(0.9002*0.900) =308.735kPa eyk=Mdxk/(Fk+Gk)=10.370/(370.370+8.100)=0.027m 因|eyk|≤By/6=0.150m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(370.370+8.100)/0.810+6*|10.370|/(0.9002*0.900) =552.600kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(370.370+8.100)/0.810-6*|10.370|/(0.9002*0.900) =381.895kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(625.760-467.247)+(552.600-467.247)+467.247 =711.113kPa γo*Pkmax=1.10*711.113=782.224kPa≤1.2*fa=1.2*3630.000=4791.600kPa 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)=26.000/(500.000+10.935)=0.051m 因ex≤Bx/6.0=0.150m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(500.000+10.935)/0.810+6*|26.000|/(0.9002*0.900) =844.776kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(500.000+10.935)/0.810-6*|26.000|/(0.9002*0.900) =416.792kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=14.000/(500.000+10.935)=0.027m 因ey≤By/6=0.150 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(500.000+10.935)/0.810+6*|14.000|/(0.9002*0.900) =746.010kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(500.000+10.935)/0.810-6*|14.000|/(0.9002*0.900) =515.558kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =844.776+746.010-(500.000+10.935)/0.810 =960.002kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=960.002-10.935/0.810=946.502kPa Pjmax_x=Pmax_x-G/A=844.776-10.935/0.810=831.276kPa Pjmax_y=Pmax_y-G/A=746.010-10.935/0.810=732.510kPa 2.验算柱边冲切 YH=h1=0.600m,YB=bc=0.500m,YL=hc=0.500m YB1=B1=0.450m,YB2=B2=0.450m,YL1=A1=0.450m,YL2=A2=0.450m 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+500 =900mm Bz=b1+b2+bc =200+200+500 =900mm A'=Az*max(b1,b2) =900.0*max(200.0,200.0) =0.18m2 Vs=A'*p=0.2*617.3=111.1kN 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 =900*600 =540000.00mm2 γo*Vs=1.1*111.1=122.2kN ≤0.7βhsftAo=0.7*1.0*1.43*540000.0=540.5kN 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小于柱的混凝土强度等级，验算柱下扩展基础顶面的局部受压承载力。 1.Fl=F=500.000kN 混凝土等级为C30,fcc=0.85*fc=0.85*14.3=12.155N/mm2,ω=1.0 2.计算混凝土局部受压面积 YB1=b1+bc/2=0.200+0.500/2=0.450m2 YB2=b2+bc/2=0.200+0.500/2=0.450m2 YA1=a1+hc/2=0.200+0.500/2=0.450m2 YA2=a2+hc/2=0.200+0.500/2=0.450m2 Al=bc*hc=0.500*0.500=0.250m2 Ab=bc*hc=0.500*0.500=0.250m2 Ab=(bc+2*min(bc,hc))*hc) =(0.500+2*min(0.500,0.500))*0.500) =0.750m2 Ab=(hc+2*min(bc,hc))*bc =(0.500+2*min(0.500,0.500))*0.500) =0.750m2 3.计算混凝土局部受压时的强度提高系数 【②附录D】 β1=sqrt(Ab/Al)=sqrt(0.750/0.250)=1.732 4.因 γo*Fl=1.10*500.000=550.00kN γo*Fl≤ω*β1*fcc*Al =1.350*1.732*12.155*250000.000/1000 =7105.41kN 柱下基础局部受压承载力满足规范要求十、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=0.150mx方向小偏心 a=(Bx-bc)/2=(0.900-0.500)/2=0.200m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((0.900-0.200)*(844.776-416.792)/0.900)+416.792-10.935/0.810 =736.168kPa 因ey≤By/6=0.150my方向小偏心 a=(By-hc)/2=(0.900-0.500)/2=0.200m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((0.900-0.200)*(746.010-515.558)/0.900)+515.558-10.935/0.810 =681.299kPa βx=1.024 βy=1.020 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.024*(0.900-0.500)2*(2*0.900+0.500)*(736.168+831.276) =12.30kN*m MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.020*(0.900-0.500)2*(2*0.900+0.500)*(681.299+732.510) =11.06kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.10*12.30*106/(0.9*(600.000-40.000)*360) =74.6mm2 Asx1=Asx_1=74.6mm2 Asx=Asx1/By=74.6/0.900=83mm2/m Asx=max(Asx,ρmin*H*1000) =max(83,0.150%*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) =1.10*11.06*106/(0.9*(600.000-40.000)*360) =67.1mm2 Asy1=Asy_1=67.1mm2 Asy=Asy1/Bx=67.1/0.900=75mm2/m Asy=max(Asy,ρmin*H*1000) =max(75,0.150%*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)②二、示意图三、计算信息 构件编号:DJ5 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=500mm 矩形柱高 hc=500mm 基础高度 h1=1300mm 一阶长度b1=300mm b2=300mm 一阶宽度a1=300mm a2=300mm 2.材料信息 基础混凝土等级: C30 ft_b=1.43N/mm2 fc_b=14.3N/mm2 柱混凝土等级: C35 ft_c=1.57N/mm2 fc_c=16.7N/mm2 钢筋级别: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.10 基础埋深: dh=0.500m 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=4000.000kN Mx=20.000kN*m My=20.000kN*m Vx=10.000kN Vy=10.000kN ks=1.35 Fk=F/ks=4000.000/1.35=2962.963kN Mxk=Mx/ks=20.000/1.35=14.815kN*m Myk=My/ks=20.000/1.35=14.815kN*m Vxk=Vx/ks=10.000/1.35=7.407kN Vyk=Vy/ks=10.000/1.35=7.407kN 5.修正后的地基承载力特征值 fa=3630.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.300+0.300+0.500=1.100m 2.基础总宽By=a1+a2+hc=0.300+0.300+0.500=1.100mA1=a1+hc/2=0.300+0.500/2=0.550mA2=a2+hc/2=0.300+0.500/2=0.550mB1=b1+bc/2=0.300+0.500/2=0.550mB2=b2+bc/2=0.300+0.500/2=0.550m 3.基础总高H=h1=1.300=1.300m 4.底板配筋计算高度ho=h1-as=1.300-0.040=1.260m 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*0.500=12.100kN G=1.35*Gk=1.35*12.100=16.335kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=14.815-7.407*1.300=5.185kN*m Mdyk=Myk+Vxk*H=14.815+7.407*1.300=24.444kN*m Mdx=Mx-Vy*H=20.000-10.000*1.300=7.000kN*m Mdy=My+Vx*H=20.000+10.000*1.300=33.000kN*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=(2962.963+12.100)/1.210=2458.730kPa 【①5.2.1-2】 因γo*pk=1.10*2458.730=2704.603kPa≤fa=3630.000kPa 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)=24.444/(2962.963+12.100)=0.008m 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) =(2962.963+12.100)/1.210+6*|24.444|/(1.1002*1.100) =2568.923kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(2962.963+12.100)/1.210-6*|24.444|/(1.1002*1.100) =2348.537kPa eyk=Mdxk/(Fk+Gk)=5.185/(2962.963+12.100)=0.002m 因|eyk|≤By/6=0.183m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(2962.963+12.100)/1.210+6*|5.185|/(1.1002*1.100) =2482.104kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(2962.963+12.100)/1.210-6*|5.185|/(1.1002*1.100) =2435.355kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(2568.923-2458.730)+(2482.104-2458.730)+2458.730 =2592.297kPa γo*Pkmax=1.10*2592.297=2851.526kPa≤1.2*fa=1.2*3630.000=4791.600kPa 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)=33.000/(4000.000+16.335)=0.008m 因ex≤Bx/6.0=0.183m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(4000.000+16.335)/1.210+6*|33.000|/(1.1002*1.100) =3468.045kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(4000.000+16.335)/1.210-6*|33.000|/(1.1002*1.100) =3170.525kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=7.000/(4000.000+16.335)=0.002m 因ey≤By/6=0.183 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(4000.000+16.335)/1.210+6*|7.000|/(1.1002*1.100) =3350.840kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2