吉林多层现代感市级中学各教学含体育馆建筑施工图多张图

结构构件计算书PAGE第6页，共6页锥形基础计算项目名称_____________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 《简明高层钢筋混凝土结构设计手册》李国胜二、示意图三、计算信息 构件编号:JC-4 计算类型:验算截面尺寸 1.几何参数 矩形柱宽 bc=400mm 矩形柱高 hc=400mm 基础端部高度 h1=200mm 基础根部高度 h2=50mm 基础长度 B1=750mm B2=750mm 基础宽度 A1=400mm A2=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: HRB400 fy=360N/mm2 3.计算信息 结构重要性系数: γo=1.0 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值 Fgk=116.700kN Fqk=0.000kN Mgxk=-1.900kN*m Mqxk=0.000kN*m Mgyk=-4.100kN*m Mqyk=0.000kN*m Vgxk=-1.400kN Vqxk=0.000kN Vgyk=0.600kN Vqyk=0.000kN 永久荷载分项系数rg=1.20 可变荷载分项系数rq=1.40 Fk=Fgk+Fqk=116.700+(0.000)=116.700kN Mxk=Mgxk+Fgk*(A2-A1)/2+Mqxk+Fqk*(A2-A1)/2 =-1.900+116.700*(0.400-0.400)/2+(0.000)+0.000*(0.400-0.400)/2 =-1.900kN*m Myk=Mgyk+Fgk*(B2-B1)/2+Mqyk+Fqk*(B2-B1)/2 =-4.100+116.700*(0.750-0.750)/2+(0.000)+0.000*(0.750-0.750)/2 =-4.100kN*m Vxk=Vgxk+Vqxk=-1.400+(0.000)=-1.400kN Vyk=Vgyk+Vqyk=0.600+(0.000)=0.600kN F1=rg*Fgk+rq*Fqk=1.20*(116.700)+1.40*(0.000)=140.040kN Mx1=rg*(Mgxk+Fgk*(A2-A1)/2)+rq*(Mqxk+Fqk*(A2-A1)/2) =1.20*(-1.900+116.700*(0.400-0.400)/2)+1.40*(0.000+0.000*(0.400-0.400)/2) =-2.280kN*m My1=rg*(Mgyk+Fgk*(B2-B1)/2)+rq*(Mqyk+Fqk*(B2-B1)/2) =1.20*(-4.100+116.700*(0.750-0.750)/2)+1.40*(0.000+0.000*(0.750-0.750)/2) =-4.920kN*m Vx1=rg*Vgxk+rq*Vqxk=1.20*(-1.400)+1.40*(0.000)=-1.680kN Vy1=rg*Vgyk+rq*Vqyk=1.20*(0.600)+1.40*(0.000)=0.720kN F2=1.35*Fk=1.35*116.700=157.545kN Mx2=1.35*Mxk=1.35*(-1.900)=-2.565kN*m My2=1.35*Myk=1.35*(-4.100)=-5.535kN*m Vx2=1.35*Vxk=1.35*(-1.400)=-1.890kN Vy2=1.35*Vyk=1.35*0.600=0.810kN F=max(|F1|,|F2|)=max(|140.040|,|157.545|)=157.545kN Mx=max(|Mx1|,|Mx2|)=max(|-2.280|,|-2.565|)=-2.565kN*m My=max(|My1|,|My2|)=max(|-4.920|,|-5.535|)=-5.535kN*m Vx=max(|Vx1|,|Vx2|)=max(|-1.680|,|-1.890|)=-1.890kN Vy=max(|Vy1|,|Vy2|)=max(|0.720|,|0.810|)=0.810kN 5.修正后的地基承载力特征值 fa=180.000kPa四、计算参数 1.基础总长Bx=B1+B2=0.750+0.750=1.500m 2.基础总宽By=A1+A2=0.400+0.400=0.800m 3.基础总高H=h1+h2=0.200+0.050=0.250m 4.底板配筋计算高度ho=h1+h2-as=0.200+0.050-0.040=0.210m 5.基础底面积A=Bx*By=1.500*0.800=1.200m2 6.Gk=γ*Bx*By*dh=20.000*1.500*0.800*1.500=36.000kN G=1.35*Gk=1.35*36.000=48.600kN五、计算作用在基础底部弯矩值 Mdxk=Mxk-Vyk*H=-1.900-0.600*0.250=-2.050kN*m Mdyk=Myk+Vxk*H=-4.100+(-1.400)*0.250=-4.450kN*m Mdx=Mx-Vy*H=-2.565-0.810*0.250=-2.768kN*m Mdy=My+Vx*H=-5.535+(-1.890)*0.250=-6.008kN*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=(116.700+36.000)/1.200=127.250kPa 【①5.2.2-2】 因γo*pk=1.0*127.250=127.250kPa≤fa=180.000kPa 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)=-4.450/(116.700+36.000)=-0.029m 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) =(116.700+36.000)/1.200+6*|-4.450|/(1.5002*0.800) =142.083kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(116.700+36.000)/1.200-6*|-4.450|/(1.5002*0.800) =112.417kPa eyk=Mdxk/(Fk+Gk)=-2.050/(116.700+36.000)=-0.013m 因|eyk|≤By/6=0.133m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(116.700+36.000)/1.200+6*|-2.050|/(0.8002*1.500) =140.063kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(116.700+36.000)/1.200-6*|-2.050|/(0.8002*1.500) =114.438kPa 3.确定基础底面反力设计值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(142.083-127.250)+(140.063-127.250)+127.250 =154.896kPa γo*Pkmax=1.0*154.896=154.896kPa≤1.2*fa=1.2*180.000=216.000kPa 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)=-6.008/(157.545+48.600)=-0.029m 因ex≤Bx/6.0=0.250m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(157.545+48.600)/1.200+6*|-6.008|/(1.5002*0.800) =191.813kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(157.545+48.600)/1.200-6*|-6.008|/(1.5002*0.800) =151.762kPa 1.2计算y方向基础底面反力设计值 ey=Mdx/(F+G)=-2.768/(157.545+48.600)=-0.013m 因ey≤By/6=0.133 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(157.545+48.600)/1.200+6*|-2.768|/(0.8002*1.500) =189.084kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(157.545+48.600)/1.200-6*|-2.768|/(0.8002*1.500) =154.491kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =191.813+189.084-(157.545+48.600)/1.200 =209.109kPa 1.4计算地基净反力极值 Pjmax=Pmax-G/A=209.109-48.600/1.200=168.609kPa Pjmax_x=Pmax_x-G/A=191.812-48.600/1.200=151.312kPa Pjmax_y=Pmax_y-G/A=189.084-48.600/1.200=148.584kPa 2.柱对基础的冲切验算 2.1因(H≤800)βhp=1.0 2.2x方向柱对基础的冲切验算 因hc/2+ho>=A1和hc/2+ho>=A2 x方向基础底面外边缘位于冲切破坏锥体以内,不用计算x方向的柱对基础的冲切验算 2.3y方向柱对基础的冲切验算 y冲切面积Aly=max((B1-bc/2-ho)*(A1+A2),(B2-bc/2-ho)*(A1+A2)) =max((0.750-0.400/2-0.210)*(0.400+0.400),(0.750-0.400/2-0.210)*(0.400+0.400)) =max(0.272,0.272) =0.272m2 y冲切截面上的地基净反力设计值 Fly=Aly*Pjmax=0.272*168.609=45.862kN γo*Fly=1.0*45.862=45.86kN 因γo*Fly≤0.7*βhp*ft_b*am*ho(6.5.5-1) =0.7*1.000*1.43*610.000*210 =128.23kN 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 pk=(Fk+Gk)/A=(116.7+36.0)/1.2=127.3kPa A'=(A1+A2)*(max(B1,B2)-0.5*bc) =(400+400)*(max(750,750)-0.5*400) =0.44m2 Vs=A'*pk=0.44*127.3=56.0kN 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=(A1+A2)*h1+0.5*(bo+100+(A1+A2))*h2 =(400+400)*200+0.5*(400+100+(400+400))*50 =192500.00mm2 γo*Vs=1.0*56.0=56.0kN ≤0.7βhsftAo=0.7*1.0*1.43*192500.0=192.7kN 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 双向受弯 因ex≤Bx/6=0.250mx方向小偏心 a=(Bx-bc)/2=(1.500-0.400)/2=0.550m Pj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx+Pmin_x-G/A =(1.500-0.550)*(191.812-151.762)/1.500+151.762-48.600/1.200 =136.627kPa 因ey≤By/6=0.133my方向小偏心 a=(By-hc)/2=(0.800-0.400)/2=0.200m Pj2=(By-a)*(Pmax_y-Pmin_y)/By+Pmin_y-G/A =(0.800-0.200)*(189.084-154.491)/0.800+154.491-48.600/1.200 =139.936kPa βx=1.023 βy=1.021 MI=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pj