JC-1階梯基礎(chǔ)計(jì)算

結(jié)構(gòu)構(gòu)件計(jì)算書(shū)第1頁(yè)，共1頁(yè)P(yáng)AGEJC-1階梯基礎(chǔ)計(jì)算項(xiàng)目名稱_____________日期_____________設(shè)計(jì)者_(dá)____________校對(duì)者_(dá)____________一、設(shè)計(jì)依據(jù) 《建筑地基基礎(chǔ)設(shè)計(jì)規(guī)范》(GB50007-2011)① 《混凝土結(jié)構(gòu)設(shè)計(jì)規(guī)范》GB50010-2010(2015年版)② 《簡(jiǎn)明高層鋼筋混凝土結(jié)構(gòu)設(shè)計(jì)手冊(cè)》李國(guó)勝(第三版)③二、示意圖三、計(jì)算信息 構(gòu)件編號(hào):JC-1 計(jì)算類型:驗(yàn)算截面尺寸 1.幾何參數(shù) 臺(tái)階數(shù) n=1 圓柱直徑 dc=400mm 截面換算 矩形柱寬 bc=0.8*dc=320mm 矩形柱高 hc=0.8*dc=320mm 基礎(chǔ)高度 h1=500mm 一階長(zhǎng)度b1=800mm b2=800mm 一階寬度a1=800mm a2=800mm 2.材料信息 基礎(chǔ)混凝土等級(jí): C30 ft_b=1.43N/m fc_b=14.3N/m 柱混凝土等級(jí): C30 ft_c=1.43N/m fc_c=14.3N/m 鋼筋級(jí)別: HRB400 fy=360N/m 3.計(jì)算信息 結(jié)構(gòu)重要性系數(shù): γo=1.00 基礎(chǔ)埋深: dh=1.500m 縱筋合力點(diǎn)至近邊距離: as=40mm 基礎(chǔ)及其上覆土的平均容重: γ=20.000kN/ 最小配筋率: ρmin=0.150% 4.作用在基礎(chǔ)頂部荷載設(shè)計(jì)值 Fgk=350.000kN Fqk=0.000kN Mgxk=50.000kN*m Mqxk=0.000kN*m Mgyk=50.000kN*m Mqyk=0.000kN*m Vgxk=50.000kN Vqxk=0.000kN Vgyk=50.000kN Vqyk=0.000kN 永久荷載分項(xiàng)系數(shù)rg=1.30 可變荷載分項(xiàng)系數(shù)rq=1.50 可變荷載調(diào)整系數(shù)rl=1.00 Fk=Fgk+Fqk=350.000+(0.000)=350.000kN Mxk=Mgxk+Fgk*(A2-A1)/2+Mqxk+Fqk*(A2-A1)/2 =50.000+350.000*(0.960-0.960)/2+(0.000)+0.000*(0.960-0.960)/2 =50.000kN*m Myk=Mgyk+Fgk*(B2-B1)/2+Mqyk+Fqk*(B2-B1)/2 =50.000+350.000*(0.960-0.960)/2+(0.000)+0.000*(0.960-0.960)/2 =50.000kN*m Vxk=Vgxk+Vqxk=50.000+(0.000)=50.000kN Vyk=Vgyk+Vqyk=50.000+(0.000)=50.000kN F=rg*Fgk+rq*rl*Fqk=1.30*(350.000)+1.50*1.00*(0.000)=455.000kN Mx=rg*(Mgxk+Fgk*(A2-A1)/2)+rq*rl*(Mqxk+Fqk*(A2-A1)/2) =1.30*(50.000+350.000*(0.960-0.960)/2)+1.50*1.00*(0.000+0.000*(0.960-0.960)/2) =65.000kN*m My=rg*(Mgyk+Fgk*(B2-B1)/2)+rq*rl*(Mqyk+Fqk*(B2-B1)/2) =1.30*(50.000+350.000*(0.960-0.960)/2)+1.50*1.00*(0.000+0.000*(0.960-0.960)/2) =65.000kN*m Vx=rg*Vgxk+rq*rl*Vqxk=1.30*(50.000)+1.50*1.00*(0.000)=65.000kN Vy=rg*Vgyk+rq*rl*Vqyk=1.30*(50.000)+1.50*1.00*(0.000)=65.000kN 5.修正后的地基承載力特征值 fa=180.000kPa四、計(jì)算參數(shù) 1.基礎(chǔ)總長(zhǎng)Bx=b1+b2+bc=0.800+0.800+0.320=1.920m 2.基礎(chǔ)總寬By=a1+a2+hc=0.800+0.800+0.320=1.920m A1=a1+hc/2=0.800+0.320/2=0.960mA2=a2+hc/2=0.800+0.320/2=0.960m B1=b1+bc/2=0.800+0.320/2=0.960mB2=b2+bc/2=0.800+0.320/2=0.960m 3.基礎(chǔ)總高H=h1=0.500=0.500m 4.底板配筋計(jì)算高度ho=h1-as=0.500-0.040=0.460m 5.基礎(chǔ)底面積A=Bx*By=1.920*1.920=3.686 6.Gk=γ*Bx*By*dh=20.000*1.920*1.920*1.500=110.592kN G=1.35*Gk=1.35*110.592=149.299kN五、計(jì)算作用在基礎(chǔ)底部彎矩值 Mdxk=Mxk-Vyk*H=50.000-50.000*0.500=25.000kN*m Mdyk=Myk+Vxk*H=50.000+50.000*0.500=75.000kN*m Mdx=Mx-Vy*H=67.500-67.500*0.500=33.750kN*m Mdy=My+Vx*H=67.500+67.500*0.500=101.250kN*m六、驗(yàn)算地基承載力 1.驗(yàn)算軸心荷載作用下地基承載力 pk=(Fk+Gk)/A=(350.000+110.592)/3.686=124.944kPa 【①5.2.1-2】 因γo*pk=1.00*124.944=124.944kPa≤fa=180.000kPa 軸心荷載作用下地基承載力滿足要求 2.驗(yàn)算偏心荷載作用下的地基承載力 exk=Mdyk/(Fk+Gk)=75.000/(350.000+110.592)=0.163m 因|exk|≤Bx/6=0.320m 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|/(B*By) =(350.000+110.592)/3.686+6*|75.000|/(1.92*1.920) =188.522kPa Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(B*By) =(350.000+110.592)/3.686-6*|75.000|/(1.92*1.920) =61.365kPa eyk=Mdxk/(Fk+Gk)=25.000/(350.000+110.592)=0.054m 因|eyk|≤By/6=0.320m y方向小偏心 Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(B*Bx) =(350.000+110.592)/3.686+6*|25.000|/(1.92*1.920) =146.136kPa Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(B*Bx) =(350.000+110.592)/3.686-6*|25.000|/(1.92*1.920) =103.751kPa 3.確定基礎(chǔ)底面反力設(shè)計(jì)值 Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(188.522-124.944)+(146.136-124.944)+124.944 =209.715kPa γo*Pkmax=1.00*209.715=209.715kPa≤1.2*fa=1.2*180.000=216.000kPa 偏心荷載作用下地基承載力滿足要求七、基礎(chǔ)沖切驗(yàn)算 1.計(jì)算基礎(chǔ)底面反力設(shè)計(jì)值 1.1計(jì)算x方向基礎(chǔ)底面反力設(shè)計(jì)值 ex=Mdy/(F+G)=101.250/(472.500+149.299)=0.163m 因ex≤Bx/6.0=0.320m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(B*By) =(472.500+149.299)/3.686+6*|101.250|/(1.92*1.920) =254.505kPa Pmin_x=(F+G)/A-6*|Mdy|/(B*By) =(472.500+149.299)/3.686-6*|101.250|/(1.92*1.920) =82.843kPa 1.2計(jì)算y方向基礎(chǔ)底面反力設(shè)計(jì)值 ey=Mdx/(F+G)=33.750/(472.500+149.299)=0.054m 因ey≤By/6=0.320 y方向小偏心 Pmax_y=(F+G)/A+6*|Mdx|/(B*Bx) =(472.500+149.299)/3.686+6*|33.750|/(1.92*1.920) =197.284kPa Pmin_y=(F+G)/A-6*|Mdx|/(B*Bx) =(472.500+149.299)/3.686-6*|33.750|/(1.92*1.920) =140.064kPa 1.3因Mdx≠0Mdy≠0 Pmax=Pmax_x+Pmax_y-(F+G)/A =254.505+197.284-(472.500+149.299)/3.686 =283.115kPa 1.4計(jì)算地基凈反力極值 Pjmax=Pmax-G/A=283.115-149.299/3.686=242.615kPa Pjmax_x=Pmax_x-G/A=254.505-149.299/3.686=214.005kPa Pjmax_y=Pmax_y-G/A=197.284-149.299/3.686=156.784kPa 2.驗(yàn)算柱邊沖切 YH=h1=0.500m,YB=bc=0.320m,YL=hc=0.320m YB1=B1=0.960m,YB2=B2=0.960m,YL1=A1=0.960m,YL2=A2=0.960m YHo=YH-as=0.460m 因YH≤800，取βhp=1.0 2.1x方向柱對(duì)基礎(chǔ)的沖切驗(yàn)算 x沖切位置斜截面上邊長(zhǎng) at=YL=0.320m x沖切位置斜截面下邊長(zhǎng) ab=YL+2*YHo=1.240m x沖切面積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((0.960-0.320/2-0.460)*(0.320+0.460)+(0.960-0.320/2-0.460,(0.960-0.320/2-0.460)*(0.320+0.460)+(0.960-0.320/2-0.460) =max(0.537,0.537) =0.537 x沖切截面上的地基凈反力設(shè)計(jì)值 Flx=Alx*Pjmax=0.537*242.615=130.333kN γo*Flx=1.00*130.333=130.33kN γo*Flx≤0.7*βhp*ft_b*am*YHo(6.5.5-1) =0.7*1.000*1.43*780*460 =359.16kN x方向柱對(duì)基礎(chǔ)的沖切滿足規(guī)范要求 2.2y方向柱對(duì)基礎(chǔ)的沖切驗(yàn)算 y沖切位置斜截面上邊長(zhǎng) bt=YB=0.320m y沖切位置斜截面下邊長(zhǎng) bb=YB+2*YHo=1.240m y沖切不利位置 bm=(bt+bb)/2=(0.320+1.240)/2=0.780m y沖切面積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((0.960-0.320/2-0.460)*(0.320+2*0.460)+(0.960-0.320/2-0.460,(0.960-0.320/2-0.460)*(0.320+2*0.460)+(0.960-0.320/2-0.460) =max(0.537,0.537) =0.537 y沖切截面上的地基凈反力設(shè)計(jì)值 Fly=Aly*Pjmax=0.537*242.615=130.333kN γo*Fly=1.00*130.333=130.33kN γo*Fly≤0.7*βhp*ft_b*bm*YHo(6.5.5-1) =0.7*1.000*1.43*780*460 =359.16kN y方向柱對(duì)基礎(chǔ)的沖切滿足規(guī)范要求八、基礎(chǔ)受剪承載力驗(yàn)算 1.計(jì)算剪力 Az=a1+a2+hc =800+800+320 =1920mm Bz=b1+b2+bc =800+800+320 =1920mm A'=Az*max(b1,b2) =1920.0*max(800.0,800.0) =1.54m2 Vs=A'*p=1.5*128.2=196.9kN 基礎(chǔ)底面短邊尺寸大于柱寬加兩倍基礎(chǔ)有效高度，不需驗(yàn)算受剪承載力！九、柱下基礎(chǔ)的局部受壓驗(yàn)算 因?yàn)榛A(chǔ)的混凝土強(qiáng)度等級(jí)大于等于柱的混凝土強(qiáng)度等級(jí)，所以不用驗(yàn)算柱下擴(kuò)展基礎(chǔ)頂面的局部受壓承載力。十、基礎(chǔ)受彎計(jì)算 1.因Mdx>0,Mdy>0此基礎(chǔ)為雙向受彎，根據(jù)③中751頁(yè)條目4（3）中公式 2.計(jì)算I-I截面彎矩 因ex≤Bx/6=0.320mx方向小偏心 a=(Bx-bc)/2=(1.920-0.320)/2=0.800m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A =((1.920-0.800)*(254.505-82.843)/1.920)+82.843-149.299/3.686 =142.479kPa 因ey≤By/6=0.320my方向小偏心 a=(By-hc)/2=(1.920-0.320)/2=0.800m Pj2=((By-a)*(Pmax_y-Pmin_y)/By)+Pmin_y-G/A =((1.920-0.800)*(197.284-140.064)/1.920)+140.064-149.299/3.686 =132.942kPa βx=1.093 βy=1.039 MI_1=1/48*βx*(Bx-bc*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.093*(1.920-0.320*(2*1