弧形廊架基础计算

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=350mm 矩形柱高 hc=350mm 基础高度 h1=400mm 一阶长度b1=575mm b2=575mm 一阶宽度a1=575mm a2=575mm 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=18.600kN Mx=2.800kN*m My=7.200kN*m Vx=3.800kN Vy=2.300kN ks=1.35 Fk=F/ks=18.600/1.35=13.778kN Mxk=Mx/ks=2.800/1.35=2.074kN*m Myk=My/ks=7.200/1.35=5.333kN*m Vxk=Vx/ks=3.800/1.35=2.815kN Vyk=Vy/ks=2.300/1.35=1.704kN 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=0.575+0.575+0.350=1.500m 2.基础总宽By=a1+a2+hc=0.575+0.575+0.350=1.500m A1=a1+hc/2=0.575+0.350/2=0.750mA2=a2+hc/2=0.575+0.350/2=0.750m B1=b1+bc/2=0.575+0.350/2=0.750mB2=b2+bc/2=0.575+0.350/2=0.750m 3.基础总高H=h1=0.400=0.400m 4.底板配筋计算高度ho=h1-as=0.400-0.040=0.360m 5.基础底面积A=Bx*By=1.500*1.500=2.250 6.Gk=γ*Bx*By*dh=20.000*1.500*1.500*1.000=45.000kN G=1.35*Gk=1.35*45.000=60.750kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=2.074-1.704*0.400=1.393kN*m Mdyk=Myk+Vxk*H=5.333+2.815*0.400=6.459kN*m Mdx=Mx-Vy*H=2.800-2.300*0.400=1.880kN*m Mdy=My+Vx*H=7.200+3.800*0.400=8.720kN*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=(13.778+45.000)/2.250=26.123kPa 【①5.2.1-2】 因γo*pk=1.00*26.123=26.123kPa≤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)=6.459/(13.778+45.000)=0.110m 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|/(B*By) =(13.778+45.000)/2.250+6*|6.459|/(1.50*1.500) =37.607kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(B*By) =(13.778+45.000)/2.250-6*|6.459|/(1.50*1.500) =14.640kPa eyk=Mdxk/(Fk+Gk)=1.393/(13.778+45.000)=0.024m 因|eyk|≤By/6=0.250m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(B*Bx) =(13.778+45.000)/2.250+6*|1.393|/(1.50*1.500) =28.599kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(B*Bx) =(13.778+45.000)/2.250-6*|1.393|/(1.50*1.500) =23.648kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(37.607-26.123)+(28.599-26.123)+26.123 =40.082kPa γo*Pkmax=1.00*40.082=40.082kPa≤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)=8.720/(18.600+60.750)=0.110m 因ex≤Bx/6.0=0.250m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(B*By) =(18.600+60.750)/2.250+6*|8.720|/(1.50*1.500) =50.769kPa Pmin_x=(F+G)/A-6*|Mdy|/(B*By) =(18.600+60.750)/2.250-6*|8.720|/(1.50*1.500) =19.764kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=1.880/(18.600+60.750)=0.024m 因ey≤By/6=0.250 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(B*Bx) =(18.600+60.750)/2.250+6*|1.880|/(1.50*1.500) =38.609kPa Pmin_y=(F+G)/A-6*|Mdx|/(B*Bx) =(18.600+60.750)/2.250-6*|1.880|/(1.50*1.500) =31.924kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =50.769+38.609-(18.600+60.750)/2.250 =54.111kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=54.111-60.750/2.250=27.111kPa Pjmax_x=Pmax_x-G/A=50.769-60.750/2.250=23.769kPa Pjmax_y=Pmax_y-G/A=38.609-60.750/2.250=11.609kPa 2.验算柱边冲切 YH=h1=0.400m,YB=bc=0.350m,YL=hc=0.350m YB1=B1=0.750m,YB2=B2=0.750m,YL1=A1=0.750m,YL2=A2=0.750m YHo=YH-as=0.360m 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=0.350m x冲切位置斜截面下边长 ab=YL+2*YHo=1.070m 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((0.750-0.350/2-0.360)*(0.350+0.360)+(0.750-0.350/2-0.360,(0.750-0.350/2-0.360)*(0.350+0.360)+(0.750-0.350/2-0.360) =max(0.276,0.276) =0.276 x冲切截面上的地基净反力设计值 Flx=Alx*Pjmax=0.276*27.111=7.490kN γo*Flx=1.00*7.490=7.49kN γo*Flx≤0.7*βhp*ft_b*am*YHo(6.5.5-1) =0.7*1.000*1.43*710*360 =255.86kN 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=0.350m y冲切位置斜截面下边长 bb=YB+2*YHo=1.070m y冲切不利位置 bm=(bt+bb)/2=(0.350+1.070)/2=0.710m 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((0.750-0.350/2-0.360)*(0.350+2*0.360)+(0.750-0.350/2-0.360,(0.750-0.350/2-0.360)*(0.350+2*0.360)+(0.750-0.350/2-0.360) =max(0.276,0.276) =0.276 y冲切截面上的地基净反力设计值 Fly=Aly*Pjmax=0.276*27.111=7.490kN γo*Fly=1.00*7.490=7.49kN γo*Fly≤0.7*βhp*ft_b*bm*YHo(6.5.5-1) =0.7*1.000*1.43*710*360 =255.86kN 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 =575+575+350 =1500mm Bz=b1+b2+bc =575+575+350 =1500mm A'=Az*max(b1,b2) =1500.0*max(575.0,575.0) =0.86m2 Vs=A'*p=0.9*8.3=7.1kN 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=0.250mx方向小偏心 a=(Bx-bc)/2=(1.500-0.350)/2=0.575m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.500-0.575)*(50.769-19.764)/1.500)+19.764-60.750/2.250 =11.884kPa 因ey≤By/6=0.250my方向小偏心 a=(By-hc)/2=(1.500-0.350)/2=0.575m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.500-0.575)*(38.609-31.924)/1.500)+31.924-60.750/2.250 =9.047kPa βx=1.150 βy=1.056 MI_1=1/48*βx*(Bx-bc*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.150*(1.500-0.350*(2*1.500+0.350)*(11.884+23.769) =3.78kN*m MII_1=1/48*βy*(By-hc*(2*Bx+bc)*(Pj2+Pj