1阶梯基础计算

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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)②二、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=2矩形柱宽bc=500mm 矩形柱高hc=500mm基础高度h1=350mm基础高度h2=350mm一阶长度 b1=450mm b2=450mm 一阶宽度 a1=450mm a2=450mm二阶长度 b3=450mm b4=450mm 二阶宽度 a3=450mm a4=450mm2. 材料信息基础混凝土等级: C35 ft_b=1.57N/mm2fc_b=16.7N/mm2柱混凝土等级: C30 ft_c=1.43N/mm2fc_c=14.3N/mm2钢筋级别: HRB400 fy=360N/mm23. 计算信息结构重要性系数: γo=1.0基础埋深: dh=2.000m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/m3最小配筋率: ρmin=0.150%4. 作用在基础顶部荷载标准组合值F=695.000kNMx=0.000kN*mMy=51.900kN*mVx=17.000kNVy=0.000kNks=1.35Fk=F/ks=695.000/1.35=514.815kNMxk=Mx/ks=0.000/1.35=0.000kN*mMyk=My/ks=51.900/1.35=38.444kN*mVxk=Vx/ks=17.000/1.35=12.593kNVyk=Vy/ks=0.000/1.35=0.000kN5. 修正后的地基承载力特征值fa=150.000kPa四、计算参数1. 基础总长 Bx=b1+b2+b3+b4+bc=0.450+0.450+0.450+0.450+0.500=2.300m2. 基础总宽 By=a1+a2+a3+a4+hc=0.450+0.450+0.450+0.450+0.500=2.300mA1=a1+a2+hc/2=0.450+0.450+0.500/2=1.150m A2=a3+a4+hc/2=0.450+0.450+0.500/2=1.150m B1=b1+b2+bc/2=0.450+0.450+0.500/2=1.150m B2=b3+b4+bc/2=0.450+0.450+0.500/2=1.150m3. 基础总高 H=h1+h2=0.350+0.350=0.700m4. 底板配筋计算高度 ho=h1+h2-as=0.350+0.350-0.040=0.660m5. 基础底面积 A=Bx*By=2.300*2.300=5.290m26. Gk=γ*Bx*By*dh=20.000*2.300*2.300*2.000=211.600kNG=1.35*Gk=1.35*211.600=285.660kN五、计算作用在基础底部弯矩值Mdxk=Mxk-Vyk*H=0.000-0.000*0.700=0.000kN*mMdyk=Myk+Vxk*H=38.444+12.593*0.700=47.259kN*mMdx=Mx-Vy*H=0.000-0.000*0.700=0.000kN*mMdy=My+Vx*H=51.900+17.000*0.700=63.800kN*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=(514.815+211.600)/5.290=137.318kPa 【①5.2.1-2】因γo*pk=1.0*137.318=137.318kPa≤fa=150.000kPa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)=47.259/(514.815+211.600)=0.065m因|exk| ≤Bx/6=0.383m 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)=(514.815+211.600)/5.290+6*|47.259|/(2.3002*2.300)=160.624kPaPkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By)=(514.815+211.600)/5.290-6*|47.259|/(2.3002*2.300)=114.013kPa因 Mdxk=0 Pkmax_y=Pkmin_y=(Fk+Gk)/A=(514.815+211.600)/5.290=137.318kPa3. 确定基础底面反力设计值Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk=(160.624-137.318)+(137.318-137.318)+137.318=160.624kPaγo*Pkmax=1.0*160.624=160.624kPa≤1.2*fa=1.2*150.000=180.000kPa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)=63.800/(695.000+285.660)=0.065m因ex≤ Bx/6.0=0.383m x方向小偏心Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By)=(695.000+285.660)/5.290+6*|63.800|/(2.3002*2.300)=216.842kPaPmin_x=(F+G)/A-6*|Mdy|/(Bx2*By)=(695.000+285.660)/5.290-6*|63.800|/(2.3002*2.300)=153.918kPa1.2 计算y方向基础底面反力设计值ey=Mdx/(F+G)=0.000/(695.000+285.660)=0.000m因ey ≤By/6=0.383y方向小偏心Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx)=(695.000+285.660)/5.290+6*|0.000|/(2.3002*2.300)=185.380kPaPmin_y=(F+G)/A-6*|Mdx|/(By2*Bx)=(695.000+285.660)/5.290-6*|0.000|/(2.3002*2.300)=185.380kPa1.3 因 Mdx=0 并且Mdy≠0Pmax=Pmax_x=216.842kPaPmin=Pmin_x=153.918kPa1.4 计算地基净反力极值Pjmax=Pmax-G/A=216.842-285.660/5.290=162.842kPa2. 验算柱边冲切YH=h1+h2=0.700m, YB=bc=0.500m, YL=hc=0.500mYB1=B1=1.150m, YB2=B2=1.150m, YL1=A1=1.150m, YL2=A2=1.150mYHo=YH-as=0.660m2.1 因(YH≤800) βhp=1.02.2 x方向柱对基础的冲切验算x冲切位置斜截面上边长bt=YB=0.500mx冲切位置斜截面下边长bb=YB+2*YHo=1.820mx冲切不利位置bm=(bt+bb)/2=(0.500+1.820)/2=1.160mx冲切面积Alx=max((YL1-YL/2-ho)*(YB+2*ho)+(YL1-YL/2-ho)2,(YL2-YL/2-ho)*(YB+2*ho)+(YL2-YL/2-ho)2 =max((1.150-0.500/2-0.660)*(0.500+2*0.660)+(1.150-0.500/2-0.660)2,(1.150-0.500/2-0.660)*(0.500+2*0.660)+(1.150-0.500/2-0.660)2)=max(0.494,0.494)=0.494m2x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=0.494*162.842=80.509kNγo*Flx=1.0*80.509=80.51kNγo*Flx≤0.7*βhp*ft_b*bm*YHo (6.5.5-1)=0.7*1.000*1.57*1160*660=841.39kNx方向柱对基础的冲切满足规范要求2.3 y方向柱对基础的冲切验算y冲切位置斜截面上边长at=YL=0.500my冲切位置斜截面下边长ab=YL+2*YHo=1.820my冲切面积Aly=max((YB1-YB/2-ho)*(YL+2*ho)+(YB1-YB/2-ho)2,(YB2-YB/2-ho)*(YL+2*ho)+(YB2-YB/2-ho)2)=max((1.150-0.500/2-0.660)*(0.500+0.660)+(1.150-0.500/2-0.660)2,(1.150-0.500/2-0.660)*(0 .500+0.660)+(1.150-0.500/2-0.660)2)=max(0.494,0.494)=0.494m2y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=0.494*162.842=80.509kNγo*Fly=1.0*80.509=80.51kNγo*Fly≤0.7*βhp*ft_b*am*YHo (6.5.5-1)=0.7*1.000*1.57*1160*660=841.39kNy方向柱对基础的冲切满足规范要求3. 验算h2处冲切YH=h2=0.350mYB=bc+b2+b4=1.400mYL=hc+a2+a4=1.400mYB1=B1=1.150m, YB2=B2=1.150m, YL1=A1=1.150m, YL2=A2=1.150mYHo=YH-as=0.310m3.1 因(YH≤800) βhp=1.03.2 x方向变阶处对基础的冲切验算因 YL/2+ho>=YL1和YL/2+h0>=YL2x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3.3 y方向变阶处对基础的冲切验算因 YB/2+ho>=YB1和YB/2+ho>=YB2y方向基础底面外边缘位于冲切破坏锥体以内, 不用计算y方向的柱对基础的冲切验算八、基础受剪承载力验算1.柱与基础相接处Az=A1+A2+A3+A4=450+450=1800mmBz=B1+B2+B3+B4=450+450=1800mmA'=Az*(max(b1+b2,b3+b4)-0.5*bc)=1800.0*(max(450.0+450.0,450.0+450.0)-0.5*500.0)=1.17m2Vs=A'*pk=1.2*137.3=160.7kNβhs=(800/h0)1/4=(800/800.0)1/4=1.0Ao=Az*h1+(a2+a4+hc)*h2=1800*350+(450+450+500)*350=1120000mm2γo*Vs=1.0*160.7=160.7kN≤0.7βhs ftAo=0.7*1.0*1.57*1120000.0=1230.9kN受剪承载力验算满足要求!2.第一级台阶处A'=Az*max(b1,b3)=1800.0*max(450.0,450.0)=0.81m2Vs=A'*pk=0.8*137.3=111.2kNβhs=(800/h0)1/4=(800/800.0)1/4=1.0Ao=Az*h1=1800*350=630000mm2γo*Vs=1.0*111.2=111.2kN≤0.7βhs ftAo=0.7*1.0*1.57*630000.0=692.4kN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。