階梯基礎(chǔ)計算

結(jié)構(gòu)構(gòu)件計算書PAGE第1頁，共1頁階梯基礎(chǔ)計算DJ1一、設(shè)計依據(jù) 《建筑地基基礎(chǔ)設(shè)計規(guī)范》(GB50007-2011)① 《混凝土結(jié)構(gòu)設(shè)計規(guī)范》(GB50010-2010)②二、示意圖三、計算信息 構(gòu)件編號:JC-1 計算類型:驗算截面尺寸 1.幾何參數(shù) 臺階數(shù) n=1 矩形柱寬 bc=500mm 矩形柱高 hc=500mm 基礎(chǔ)高度 h1=600mm 一階長度b1=500mm b2=500mm 一階寬度a1=500mm a2=500mm 2.材料信息 基礎(chǔ)混凝土等級: C30 ft_b=1.43N/mm2 fc_b=14.3N/mm2 柱混凝土等級: C30 ft_c=1.43N/mm2 fc_c=14.3N/mm2 鋼筋級別: HRB400 fy=360N/mm2 3.計算信息 結(jié)構(gòu)重要性系數(shù): γo=1.1 基礎(chǔ)埋深: dh=0.800m 縱筋合力點至近邊距離: as=40mm 基礎(chǔ)及其上覆土的平均容重: γ=18.000kN/m3 最小配筋率: ρmin=0.150% Fgk=1200.000kN Fqk=0.000kN Mgxk=69.000kN*m Mqxk=0.000kN*m Mgyk=-19.000kN*m Mqyk=0.000kN*m Vgxk=-18.000kN Vqxk=0.000kN Vgyk=-38.000kN Vqyk=0.000kN 永久荷載分項系數(shù)rg=1.00 可變荷載分項系數(shù)rq=1.00 Fk=Fgk+Fqk=1200.000+(0.000)=1200.000kN Mxk=Mgxk+Fgk*(A2-A1)/2+Mqxk+Fqk*(A2-A1)/2 =69.000+1200.000*(0.750-0.750)/2+(0.000)+0.000*(0.750-0.750)/2 =69.000kN*m Myk=Mgyk+Fgk*(B2-B1)/2+Mqyk+Fqk*(B2-B1)/2 =-19.000+1200.000*(0.750-0.750)/2+(0.000)+0.000*(0.750-0.750)/2 =-19.000kN*m Vxk=Vgxk+Vqxk=-18.000+(0.000)=-18.000kN Vyk=Vgyk+Vqyk=-38.000+(0.000)=-38.000kN F1=rg*Fgk+rq*Fqk=1.00*(1200.000)+1.00*(0.000)=1200.000kN Mx1=rg*(Mgxk+Fgk*(A2-A1)/2)+rq*(Mqxk+Fqk*(A2-A1)/2) =1.00*(69.000+1200.000*(0.750-0.750)/2)+1.00*(0.000+0.000*(0.750-0.750)/2) =69.000kN*m My1=rg*(Mgyk+Fgk*(B2-B1)/2)+rq*(Mqyk+Fqk*(B2-B1)/2) =1.00*(-19.000+1200.000*(0.750-0.750)/2)+1.00*(0.000+0.000*(0.750-0.750)/2) =-19.000kN*m Vx1=rg*Vgxk+rq*Vqxk=1.00*(-18.000)+1.00*(0.000)=-18.000kN Vy1=rg*Vgyk+rq*Vqyk=1.00*(-38.000)+1.00*(0.000)=-38.000kN F2=1.35*Fk=1.35*1200.000=1620.000kN Mx2=1.35*Mxk=1.35*69.000=93.150kN*m My2=1.35*Myk=1.35*(-19.000)=-25.650kN*m Vx2=1.35*Vxk=1.35*(-18.000)=-24.300kN Vy2=1.35*Vyk=1.35*(-38.000)=-51.300kN F=max(|F1|,|F2|)=max(|1200.000|,|1620.000|)=1620.000kN Mx=max(|Mx1|,|Mx2|)=max(|69.000|,|93.150|)=93.150kN*m My=max(|My1|,|My2|)=max(|-19.000|,|-25.650|)=-25.650kN*m Vx=max(|Vx1|,|Vx2|)=max(|-18.000|,|-24.300|)=-24.300kN Vy=max(|Vy1|,|Vy2|)=max(|-38.000|,|-51.300|)=-51.300kN 5.修正后的地基承載力特征值 fa=2291.000kPa四、計算參數(shù) 1.基礎(chǔ)總長Bx=b1+b2+bc=0.500+0.500+0.500=1.500m 2.基礎(chǔ)總寬By=a1+a2+hc=0.500+0.500+0.500=1.500mA1=a1+hc/2=0.500+0.500/2=0.750mA2=a2+hc/2=0.500+0.500/2=0.750mB1=b1+bc/2=0.500+0.500/2=0.750mB2=b2+bc/2=0.500+0.500/2=0.750m 3.基礎(chǔ)總高H=h1=0.600=0.600m 4.底板配筋計算高度ho=h1-as=0.600-0.040=0.560m 5.基礎(chǔ)底面積A=Bx*By=1.500*1.500=2.250m2 6.Gk=γ*Bx*By*dh=18.000*1.500*1.500*0.800=32.400kN G=1.35*Gk=1.35*32.400=43.740kN五、計算作用在基礎(chǔ)底部彎矩值 Mdxk=Mxk-Vyk*H=69.000-(-38.000)*0.600=91.800kN*m Mdyk=Myk+Vxk*H=-19.000+(-18.000)*0.600=-29.800kN*m Mdx=Mx-Vy*H=93.150-(-51.300)*0.600=123.930kN*m Mdy=My+Vx*H=-25.650+(-24.300)*0.600=-40.230kN*m六、驗算地基承載力 1.驗算軸心荷載作用下地基承載力 pk=(Fk+Gk)/A=(1200.000+32.400)/2.250=547.733kPa 【①5.2.1-2】 因γo*pk=1.1*547.733=602.507kPa≤fa=2291.000kPa 軸心荷載作用下地基承載力滿足要求 2.驗算偏心荷載作用下的地基承載力 exk=Mdyk/(Fk+Gk)=-29.800/(1200.000+32.400)=-0.024m 因|exk|≤Bx/6=0.250m x方向小偏心, 由公式【①5.2.2-2】和【①5.2.2-3】推導(dǎo) Pkmax_x=(Fk+Gk)/A+6*|Mdyk|/(Bx2*By) =(1200.000+32.400)/2.250+6*|-29.800|/(1.5002*1.500) =600.711kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(1200.000+32.400)/2.250-6*|-29.800|/(1.5002*1.500) =494.756kPa eyk=Mdxk/(Fk+Gk)=91.800/(1200.000+32.400)=0.074m 因|eyk|≤By/6=0.250m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(1200.000+32.400)/2.250+6*|91.800|/(1.5002*1.500) =710.933kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(1200.000+32.400)/2.250-6*|91.800|/(1.5002*1.500) =384.533kPa 3.確定基礎(chǔ)底面反力設(shè)計值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(600.711-547.733)+(710.933-547.733)+547.733 =763.911kPa γo*Pkmax=1.1*763.911=840.302kPa≤1.2*fa=1.2*2291.000=3024.120kPa 偏心荷載作用下地基承載力滿足要求七、基礎(chǔ)沖切驗算 1.計算基礎(chǔ)底面反力設(shè)計值 1.1計算x方向基礎(chǔ)底面反力設(shè)計值 ex=Mdy/(F+G)=-40.230/(1620.000+43.740)=-0.024m 因ex≤Bx/6.0=0.250m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(1620.000+43.740)/2.250+6*|-40.230|/(1.5002*1.500) =810.960kPa Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(1620.000+43.740)/2.250-6*|-40.230|/(1.5002*1.500) =667.920kPa 1.2計算y方向基礎(chǔ)底面反力設(shè)計值 ey=Mdx/(F+G)=123.930/(1620.000+43.740)=0.074m 因ey≤By/6=0.250 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(1620.000+43.740)/2.250+6*|123.930|/(1.5002*1.500) =959.760kPa Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(1620.000+43.740)/2.250-6*|123.930|/(1.5002*1.500) =519.120kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =810.960+959.760-(1620.000+43.740)/2.250 =1031.280kPa 1.4計算地基凈反力極值 Pjmax=Pmax-G/A=1031.280-43.740/2.250=1011.840kPa Pjmax_x=Pmax_x-G/A=810.960-43.740/2.250=791.520kPa Pjmax_y=Pmax_y-G/A=959.760-43.740/2.250=940.320kPa 2.驗算柱邊沖切 YH=h1=0.600m,YB=bc=0.500m,YL=hc=0.500m YB1=B1=0.750m,YB2=B2=0.750m,YL1=A1=0.750m,YL2=A2=0.750m YHo=YH-as=0.560m 因((YB+2*YHo)≥Bx)并且(YL+2*YHo)≥By) 基礎(chǔ)底面處邊緣均位于沖切錐體以內(nèi),不用驗算柱對基礎(chǔ)的沖切八、基礎(chǔ)受剪承載力驗算 1.計算剪力 Az=A1+A2 =500+500 =1000mm Bz=B1+B2 =500+500 =1000mm A'=Az*(max(B1,B2)-0.5*bc) =1000.0*(max(500.0,500.0)-0.5*500.0) =0.25m2 Vs=A'*pk=0.3*547.7=136.9kN 2.計算截面高度影響系數(shù)βhs βhs=(800/h0)1/4=(800/800.0)1/4=1.0 3.剪切承載力驗算 Ao=Az*h1 =1000*600 =600000.00mm2 γo*Vs=1.1*136.9=150.6kN ≤0.7βhsftAo=0.7*1.0*1.43*600000.0=600.6kN 受剪承載力驗算滿足要求！九、柱下基礎(chǔ)的局部受壓驗算 因為基礎(chǔ)的混凝土強度等級大于等于柱的混凝土強度等級，所以不用驗算柱下擴展基礎(chǔ)頂面的局部受壓承載力。十、基礎(chǔ)受彎計算 1.因Mdx>0,Mdy>0此基礎(chǔ)為雙向受彎 2.計算I-I截面彎矩 因ex≤Bx/6=0.250mx方向小偏心 a=(Bx-bc)/2=(1.500-0.500)/2=0.500m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.500-0.500)*(810.960-667.920)/1.500)+667.920-43.740/2.250 =743.840kPa 因ey≤By/6=0.250my方向小偏心 a=(By-hc)/2=(1.500-0.500)/2=0.500m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.500-0.500)*(959.760-519.120)/1.500)+519.120-43.740/2.250 =793.440kPa βx=1.023 βy=1.037 MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.023*(1.500-0.500)2*(2*1.500+0.500)*(743.840+791.520) =114.49kN*m