阶梯基础计算四

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》(GB50007-2002)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=2 矩形柱宽 bc=600mm 矩形柱高 hc=800mm 基础高度 h1=300mm 基础高度 h2=300mm 一阶长度b1=300mm b2=300mm 一阶宽度a1=400mm a2=400mm 二阶长度b3=300mm b4=300mm 二阶宽度a3=400mm a4=400mm 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: C30 ft_c=1.43N/mm2 fc_c=14.3N/mm2 钢筋级别: HPB300 fy=270N/mm2 3.计算信息 结构重要性系数: γo=1.0 基础埋深: dh=2.500m 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=0.100% 4.作用在基础顶部荷载标准组合值 F=980.000kN Mx=1.000kN*m My=70.000kN*m Vx=15.000kN Vy=1.000kN ks=1.35 Fk=F/ks=980.000/1.35=725.926kN Mxk=Mx/ks=1.000/1.35=0.741kN*m Myk=My/ks=70.000/1.35=51.852kN*m Vxk=Vx/ks=15.000/1.35=11.111kN Vyk=Vy/ks=1.000/1.35=0.741kN 5.修正后的地基承载力特征值 fa=236.000kPa四、计算参数 1.基础总长Bx=b1+b2+b3+b4+bc=0.300+0.300+0.300+0.300+0.600=1.800m 2.基础总宽By=a1+a2+a3+a4+hc=0.400+0.400+0.400+0.400+0.800=2.400mA1=a1+a2+hc/2=0.400+0.400+0.800/2=1.200mA2=a3+a4+hc/2=0.400+0.400+0.800/2=1.200mB1=b1+b2+bc/2=0.300+0.300+0.600/2=0.900mB2=b3+b4+bc/2=0.300+0.300+0.600/2=0.900m 3.基础总高H=h1+h2=0.300+0.300=0.600m 4.底板配筋计算高度ho=h1+h2-as=0.300+0.300-0.040=0.560m 5.基础底面积A=Bx*By=1.800*2.400=4.320m2 6.Gk=γ*Bx*By*dh=20.000*1.800*2.400*2.500=216.000kN G=1.35*Gk=1.35*216.000=291.600kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=0.741-0.741*0.600=0.296kN*m Mdyk=Myk+Vxk*H=51.852+11.111*0.600=58.519kN*m Mdx=Mx-Vy*H=1.000-1.000*0.600=0.400kN*m Mdy=My+Vx*H=70.000+15.000*0.600=79.000kN*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=(725.926+216.000)/4.320=218.038kPa 【①5.2.1-2】 因γo*pk=1.0*218.038=218.038kPa≤fa=236.000kPa 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)=58.519/(725.926+216.000)=0.062m 因|exk|≤Bx/6=0.300m 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) =(725.926+216.000)/4.320+6*|58.519|/(1.8002*2.400) =263.192kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(725.926+216.000)/4.320-6*|58.519|/(1.8002*2.400) =172.885kPa eyk=Mdxk/(Fk+Gk)=0.296/(725.926+216.000)=0.000m 因|eyk|≤By/6=0.400m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(725.926+216.000)/4.320+6*|0.296|/(2.4002*1.800) =218.210kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(725.926+216.000)/4.320-6*|0.296|/(2.4002*1.800) =217.867kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(263.192-218.038)+(218.210-218.038)+218.038 =263.363kPaγo*Pkmax=1.0*263.363=263.363kPa≤1.2*fa=1.2*236.000=283.200kPa 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)=79.000/(980.000+291.600)=0.062m 因ex≤Bx/6.0=0.300m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(980.000+291.600)/4.320+6*|79.000|/(1.8002*2.400) =355.309kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(980.000+291.600)/4.320-6*|79.000|/(1.8002*2.400) =233.395kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=0.400/(980.000+291.600)=0.000m 因ey≤By/6=0.400 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(980.000+291.600)/4.320+6*|0.400|/(2.4002*1.800) =294.583kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(980.000+291.600)/4.320-6*|0.400|/(2.4002*1.800) =294.120kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =355.309+294.583-(980.000+291.600)/4.320 =355.540kPa 1.4计算地基净反力极值 2.验算柱边冲切 YH=h1+h2=0.600m,YB=bc=0.600m,YL=hc=0.800m YB1=B1=0.900m,YB2=B2=0.900m,YL1=A1=1.200m,YL2=A2=1.200m YHo=YH-as=0.560m 2.1因(YH≤800)βhp=1.0 2.2x方向向柱对基础础的冲切验验算 x冲切位置斜截面面上边长 bt=YYB=0..600mm x冲切位置斜截面面下边长 bb=YYB+2**YHo==1.7220m x冲切不利位置 bm=((bt+bbb)/22=(0..600++1.7220)/22=1.1160m x冲切面积AAlx=mmax(((YL1--YL/22-ho))*(YBB1+YBB2)+((YB1--YB/22-ho))2/2-((YB2--YB/22-ho))2/2,((YL2--YL/22-ho))*(YBB1+YBB2)+((YB2--YB/22-ho))2/2-((YB2--YB/22-ho))2/2 =max((11.2000-0.8800/22-0.5560)**(0.9900+00.9000)+(00.9000-0.6600/22-0.5560)22/2-((0.9000-0..600//2-0..560))2/2,((1.2000-0..800//2-0..560))*(0..900++0.9000)+((0.9000-0..600//2-0..560))2/2-((0.9000-0..600//2-0..560))2/2) =max(0..433,,0.4333) =0.432mm2 x冲切截面上的地地基净反力力设计值 Flx==Alx**Pjmaax=0..432**288..040==124..433kkNγo*Flx=11.0*1124.4433=1124.443kNγo*Flx≤00.7*ββhp*fft_b**bm*YYHo((6.5..5-1)) =0..7*1..000**1.433*11660*5660 =650.255kN x方向柱对基础的的冲切满足足规范要求求 2.3y方向向柱对基础础的冲切验验算 y冲切位置斜截面面上边长 at=YYL=0..800mm y冲切位置斜截面面下边长 ab=YYL+2**YHo==1.9220m y冲切面积AAly=mmax(((YB1--YB/22-ho))*(YLL+2*hho)+((YB1--YB/22-ho))2,(YBB2-YBB/2-hho)*((YL+22*ho))+(YBB2-YBB/2-hho)2) =max((00.9000-0.6600/22-0.5560)**(0.8800+00.5600)+(00.9000-0.6600/22-0.5560)22,(0..900--0.6000/2--0.5660)*((0.8000+0..560))+(0..900--0.6000/2--0.5660)2) =max(0..078,,0.0778) =0.078mm2 y冲切截面上的地地基净反力力设计值 Fly==Aly**Pjmaax=0..078**288..040==22.5582kNNγo*Fly=11.0*222.5882=222.58kkNγo*Fly≤00.7*ββhp*fft_b**am*YYHo((6.5..5-1)) =0.7*1..000**1.433*13660*5660 =762.366kN y方向柱对基础的的冲切满足足规范要求求 3.验算h22处冲切 YH=h2=00.3000m YB=bc+bb2+b44=1.2200m YL=hc+aa2+a44=1.6600m YB1=B1==0.9000m,YB2==B2=00.9000m,YYL1=AA1=1..200mm,YLL2=A22=1.2200m YHo=YH--as=00.2600m 3.1因((YH≤800))βhp=11.0 3.2x方向向变阶处对对基础的冲冲切验算 因YL/2+hho>=YYL1和YL/22+h0>>=YL22 x方向基础底面外外边缘位于于冲切破坏坏锥体以内内,不用计算x方向的柱柱对基础的的冲切验算算 3.3y方向向变阶处对对基础的冲冲切验算 因YB/2+hho>=YYB1和YB/22+ho>>=YB22 y方向基础底面外外边缘位于于冲切破坏坏锥体以内内,不用计算y方向的柱柱对基础的的冲切验算算八、柱下基础的的局部受压压验算 因为基础的混凝凝土强度等等级大于等等于柱的混混凝土强度度等级，所所以不用验验算柱下扩扩展基础顶顶面的局部部受压承载载力。九、基础受弯计计算 1.因Mdxx>0,,Mdyy>0此基础为为双向受弯弯 2.计算I--I截面弯矩矩 因ex≤Bxx/6=00.3000mx方向小偏偏心 a=(Bx-bbc)/22=(1..800--0.6000)/22=0.6600m Pj1=((BBx-a))*(Pmmax_xx-Pmiin_x))/Bx))+Pmiin_x--G/A =((11.8000-0.6600)**(3555.3099-2333.3955)/1..800))+2333.3955-2911.6000/4.3320 =2477.1711kPa 因ey≤Byy/6=00.4000my方向小偏偏心 a=(By-hhc)/22=(2..400--0.8000)/22=0.8800m Pj2=((BBy-a))*(Pmmax_yy-Pmiin_y))/By))+Pmiin_y--G/A =((22.4000-0.8800)**(2944.5833-2944.1200)/2..400))+2944.1200-2911.6000/4.3320 =2266.9299kPaβx=1.0333βy=1.0233 MI_1=1//48*ββx*(BBx-bcc)2*(2**By+hhc)*((Pj1++Pjmaax_x)) =1/48*11.0333*(1..800--0.6000)2*(2**2.4000+0..800))*(2447.1771+2887.8009) =92.83kkN*m MII_1=11/48**βy*(BBy-hcc)2*(2**Bx+bbc)*((Pj2++Pjmaax_y)) =1/48*11.0233*(2..400--0.8000)2*(2**1.8000+0..600))*(2226.9229+2227.0883) =104.000kN*mm 3.计算III-II截面弯矩矩 因x方向小偏心 a=(Bx-bbc-b22-b4))/2=((1.8000-0..600--0.3000-0..300))/2=00.3000m Pj1=((BBx-a))*(Pmmax_xx-Pmiin_x))/Bx))+Pmiin_x--G/A =((1.8000-0..300))*(3555.3009-2333.3995)/11.8000)+2333.3995-2991.6000/4..320 =267.4990kPaa 因y方向小偏心 a=(By-hhc-a22-a4))/2=((2.4000-0..800--0.4000-0..400))/2=00.4000m Pj2=((BBy-a))*(Pmmax_yy-Pmiin_y))/By))+Pmiin_y--G/A =((22.4000-0.4400)**(2944.5833-2944.1200)/2..400))+2944.1200-2911.6000/4.3320 =2277.0066kPaβx=1.0200βy=1.0200 MI_2=1//48*ββx*(BBx-bcc-b2--b4)22*(2**By+hhc+a22+a4))*(Pjj1+Pjjmax__x) =1/48*11.0200*(1..800--0.6000-0..300--0.3000)2*(2**2.4000+0..800++0.4000+0..400))*(2667.4990+2887.8009) =27.18kkN*m MII_2=11/48**βy*(BBy-hcc-a2--a4)22*(2**Bx+bbc+b22+b4))*(Pjj2+Pjjmax__y) =1/48*11.0200*(2..400--0.8000-0..400--0.4000)2*(2**1.8000+0..600++0.3000+0..300))*(2227.0006+2887.8009) =29.63kkN*m十、计算配筋 10.1计算算Asx Asx_1=γγo*MII_1/((0.9**(H-aas)*ffy) =1.0*992.833*1066/(0..9*(6600.0000-440.0000)*2270) =682.22mm2 Asx_2=