廊架基础计算书

结构构件计算书第1页，共1页PAGE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 《简明高层钢筋混凝土结构设计手册》李国胜(第三版)③二、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=1 矩形柱宽 bc=500mm 矩形柱高 hc=500mm 基础高度 h1=300mm 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/m fc_b=14.3N/m 柱混凝土等级: C30 ft_c=1.43N/m fc_c=14.3N/m 钢筋级别: HRB400 fy=360N/m 3.计算信息 结构重要性系数: γo=1.00 基础埋深: dh=1.000m 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/ 最小配筋率: ρmin=0.150% 4.作用在基础顶部荷载设计值 4.作用在基础顶部荷载标准组合值 F=11.800kN Mx=3.300kN*m My=0.100kN*m Vx=0.100kN Vy=1.700kN ks=1.35 Fk=F/ks=11.800/1.35=8.741kN Mxk=Mx/ks=3.300/1.35=2.444kN*m Myk=My/ks=0.100/1.35=0.074kN*m Vxk=Vx/ks=0.100/1.35=0.074kN Vyk=Vy/ks=1.700/1.35=1.259kN 5.修正后的地基承载力特征值 fa=159.000kPa四、计算参数 1.基础总长Bx=b1+b2+bc=0.350+0.350+0.500=1.200m 2.基础总宽By=a1+a2+hc=0.350+0.350+0.500=1.200m A1=a1+hc/2=0.350+0.500/2=0.600mA2=a2+hc/2=0.350+0.500/2=0.600m B1=b1+bc/2=0.350+0.500/2=0.600mB2=b2+bc/2=0.350+0.500/2=0.600m 3.基础总高H=h1=0.300=0.300m 4.底板配筋计算高度ho=h1-as=0.300-0.040=0.260m 5.基础底面积A=Bx*By=1.200*1.200=1.440 6.Gk=γ*Bx*By*dh=20.000*1.200*1.200*1.000=28.800kN G=1.35*Gk=1.35*28.800=38.880kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=2.444-1.259*0.300=2.067kN*m Mdyk=Myk+Vxk*H=0.074+0.074*0.300=0.096kN*m Mdx=Mx-Vy*H=3.300-1.700*0.300=2.790kN*m Mdy=My+Vx*H=0.100+0.100*0.300=0.130kN*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=(8.741+28.800)/1.440=26.070kPa 【①5.2.1-2】 因γo*pk=1.00*26.070=26.070kPa≤fa=159.000kPa 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)=0.096/(8.741+28.800)=0.003m 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|/(B*By) =(8.741+28.800)/1.440+6*|0.096|/(1.20*1.200) =26.404kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(B*By) =(8.741+28.800)/1.440-6*|0.096|/(1.20*1.200) =25.736kPa eyk=Mdxk/(Fk+Gk)=2.067/(8.741+28.800)=0.055m 因|eyk|≤By/6=0.200m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(B*Bx) =(8.741+28.800)/1.440+6*|2.067|/(1.20*1.200) =33.246kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(B*Bx) =(8.741+28.800)/1.440-6*|2.067|/(1.20*1.200) =18.894kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(26.404-26.070)+(33.246-26.070)+26.070 =33.580kPa γo*Pkmax=1.00*33.580=33.580kPa≤1.2*fa=1.2*159.000=190.800kPa 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)=0.130/(11.800+38.880)=0.003m 因ex≤Bx/6.0=0.200m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(B*By) =(11.800+38.880)/1.440+6*|0.130|/(1.20*1.200) =35.646kPa Pmin_x=(F+G)/A-6*|Mdy|/(B*By) =(11.800+38.880)/1.440-6*|0.130|/(1.20*1.200) =34.743kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=2.790/(11.800+38.880)=0.055m 因ey≤By/6=0.200 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(B*Bx) =(11.800+38.880)/1.440+6*|2.790|/(1.20*1.200) =44.882kPa Pmin_y=(F+G)/A-6*|Mdx|/(B*Bx) =(11.800+38.880)/1.440-6*|2.790|/(1.20*1.200) =25.507kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =35.646+44.882-(11.800+38.880)/1.440 =45.333kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=45.333-38.880/1.440=18.333kPa Pjmax_x=Pmax_x-G/A=35.646-38.880/1.440=8.646kPa Pjmax_y=Pmax_y-G/A=44.882-38.880/1.440=17.882kPa 2.验算柱边冲切 YH=h1=0.300m,YB=bc=0.500m,YL=hc=0.500m YB1=B1=0.600m,YB2=B2=0.600m,YL1=A1=0.600m,YL2=A2=0.600m YHo=YH-as=0.260m 因YH≤800，取βhp=1.0 2.1x方向柱对基础的冲切验算 x冲切位置斜截面上边长 at=YL=0.500m x冲切位置斜截面下边长 ab=YL+2*YHo=1.020m x冲切面积Alx=max((YB1-YB/2-YHo)*(YL+2*YHo)+(YB1-YB/2-YHo,(YB2-YB/2-YHo)*(YL+2*YHo)+(YB2-YB/2-YHo) =max((0.600-0.500/2-0.260)*(0.500+0.260)+(0.600-0.500/2-0.260,(0.600-0.500/2-0.260)*(0.500+0.260)+(0.600-0.500/2-0.260) =max(0.100,0.100) =0.100 x冲切截面上的地基净反力设计值 Flx=Alx*Pjmax=0.100*18.333=1.831kN γo*Flx=1.00*1.831=1.83kN γo*Flx≤0.7*βhp*ft_b*am*YHo(6.5.5-1) =0.7*1.000*1.43*760*260 =197.80kN x方向柱对基础的冲切满足规范要求 2.2y方向柱对基础的冲切验算 y冲切位置斜截面上边长 bt=YB=0.500m y冲切位置斜截面下边长 bb=YB+2*YHo=1.020m y冲切不利位置 bm=(bt+bb)/2=(0.500+1.020)/2=0.760m y冲切面积Aly=max((YL1-YL/2-YHo)*(YB+2*YHo)+(YL1-YL/2-YHo,(YL2-YL/2-YHo)*(YB+2*YHo)+(YL2-YL/2-YHo =max((0.600-0.500/2-0.260)*(0.500+2*0.260)+(0.600-0.500/2-0.260,(0.600-0.500/2-0.260)*(0.500+2*0.260)+(0.600-0.500/2-0.260) =max(0.100,0.100) =0.100 y冲切截面上的地基净反力设计值 Fly=Aly*Pjmax=0.100*18.333=1.831kN γo*Fly=1.00*1.831=1.83kN γo*Fly≤0.7*βhp*ft_b*bm*YHo(6.5.5-1) =0.7*1.000*1.43*760*260 =197.80kN y方向柱对基础的冲切满足规范要求八、基础受剪承载力验算 1.计算剪力 Az=a1+a2+hc =350+350+500 =1200mm Bz=b1+b2+bc =350+350+500 =1200mm A'=Az*max(b1,b2) =1200.0*max(350.0,350.0) =0.42m2 Vs=A'*p=0.4*8.2=3.4kN 基础底面短边尺寸大于柱宽加两倍基础有效高度，不需验算受剪承载力！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，根据③中751页条目4（3）中公式 2.计算I-I截面弯矩 因ex≤Bx/6=0.200mx方向小偏心 a=(Bx-bc)/2=(1.200-0.500)/2=0.350m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.200-0.350)*(35.646-34.743)/1.200)+34.743-38.880/1.440 =8.383kPa 因ey≤By/6=0.200my方向小偏心 a=(By-hc)/2=(1.200-0.500)/2=0.350m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.200-0.350)*(44.882-25.507)/1.200)+25.507-38.880/1.440 =12.231kPa βx=1.022 βy=1.078 MI_1=1/48*βx*(Bx-bc*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.022*(1.200-0.500*(2*1.200+0.500)*(8.383+8.646) =0.52kN*m MII_1=1/48*βy*(By-hc*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.078*(1.200-0.500*(2*1.200+0.500)*(12.231+17.882) =0.96kN*m十一、计算配筋 10.1计算Asx Asx_1=γo*MI_1/(0.9*(H-as)*fy) =1.00*0.52*1/(0.9*(300.000-40.000)*360) =6.115m Asx1=Asx_1=6.115m Asx=Asx1/By=6.115/1.200=5.096m/m Asx=max(Asx,ρmin*H*1000) =max(5.096,0.150%*300*1000)