某钢结构独立基础计算书J-1

阶梯基础计算（J-1）

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》 (GB50007-2002)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=700mm 矩形柱高 hc=900mm 基础高度 h1=300mm 基础高度 h2=400mm

一阶长度 b1=250mm b2=300mm 一阶宽度 a1=300mm a2=300mm 二阶长度 b3=250mm b4=300mm 二阶宽度 a3=300mm a4=300mm 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: HRB335 fy=300N/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=0.100% Fgk=176.780kN Fqk=0.000kN

Mgxk=141.340kN*m Mqxk=0.000kN*m Mgyk=0.000kN*m Mqyk=0.000kN*m Vgxk=54.380kN Vqxk=0.000kN Vgyk=0.000kN 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=176.780+(0.000)=176.780kN

Mxk=Mgxk+Fgk*(B2-B1)/2+Mqxk+Fqk*(B2-B1)/2

=141.340+176.780*(0.900-0.900)/2+(0.000)+0.000*(0.900-0.900)/2 =141.340kN*m

Myk=Mgyk+Fgk*(A2-A1)/2+Mqyk+Fqk*(A2-A1)/2

=0.000+176.780*(1.050-1.050)/2+(0.000)+0.000*(1.050-1.050)/2 =0.000kN*m

Vxk=Vgxk+Vqxk=54.380+(0.000)=54.380kN Vyk=Vgyk+Vqyk=0.000+(0.000)=0.000kN

F1=rg*Fgk+rq*Fqk=1.20*(176.780)+1.40*(0.000)=212.136kN Mx1=rg*(Mgxk+Fgk*(B2-B1)/2)+rq*(Mqxk+Fqk*(B2-B1)/2)

=1.20*(141.340+176.780*(0.900-0.900)/2)+1.40*(0.000+0.000*(0.900-0.900)/2) =169.608kN*m

My1=rg*(Mgyk+Fgk*(A2-A1)/2)+rq*(Mqyk+Fqk*(A2-A1)/2)

=1.20*(0.000+176.780*(1.050-1.050)/2)+1.40*(0.000+0.000*(1.050-1.050)/2) =0.000kN*m

Vx1=rg*Vgxk+rq*Vqxk=1.20*(54.380)+1.40*(0.000)=65.256kN Vy1=rg*Vgyk+rq*Vqyk=1.20*(0.000)+1.40*(0.000)=0.000kN F2=1.35*Fk=1.35*176.780=238.653kN Mx2=1.35*Mxk=1.35*141.340=190.809kN*m My2=1.35*Myk=1.35*(0.000)=0.000kN*m Vx2=1.35*Vxk=1.35*54.380=73.413kN Vy2=1.35*Vyk=1.35*(0.000)=0.000kN

F=max(|F1|,|F2|)=max(|212.136|,|238.653|)=238.653kN

Mx=max(|Mx1|,|Mx2|)=max(|169.608|,|190.809|)=190.809kN*m My=max(|My1|,|My2|)=max(|0.000|,|0.000|)=0.000kN*m Vx=max(|Vx1|,|Vx2|)=max(|65.256|,|73.413|)=73.413kN Vy=max(|Vy1|,|Vy2|)=max(|0.000|,|0.000|)=0.000kN 5. 修正后的地基承载力特征值 fa=200.000kPa 四、计算参数

1. 基础总长 Bx=b1+b2+b3+b4+bc=0.250+0.300+0.250+0.300+0.700=1.800m 2. 基础总宽 By=a1+a2+a3+a4+hc=0.300+0.300+0.300+0.300+0.900=2.100m

A1=a1+a2+hc/2=0.300+0.300+0.900/2=1.050m A2=a3+a4+hc/2=0.300+0.300+0.900/2=1.050m B1=b1+b2+bc/2=0.250+0.300+0.700/2=0.900m B2=b3+b4+bc/2=0.250+0.300+0.700/2=0.900m 3. 基础总高 H=h1+h2=0.300+0.400=0.700m

4. 底板配筋计算高度 ho=h1+h2-as=0.300+0.400-0.040=0.660m 5. 基础底面积 A=Bx*By=1.800*2.100=3.780m2

6. Gk=γ*Bx*By*dh=20.000*1.800*2.100*1.500=113.400kN

G=1.35*Gk=1.35*113.400=153.090kN 五、计算作用在基础底部弯矩值

Mdxk=Mxk-Vyk*H=141.340-0.000*0.700=141.340kN*m Mdyk=Myk+Vxk*H=0.000+54.380*0.700=38.066kN*m Mdx=Mx-Vy*H=190.809-0.000*0.700=190.809kN*m Mdy=My+Vx*H=0.000+73.413*0.700=51.389kN*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=(176.780+113.400)/3.780=76.767kPa 【①5.2.1-2】 因γo*pk=1.0*76.767=76.767kPa≤fa=200.000kPa 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)=38.066/(176.780+113.400)=0.131m 因 |exk| ≤Bx/6=0.300m 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)

=(176.780+113.400)/3.780+6*|38.066|/(1.8002*2.100) =110.335kPa

Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By)

=(176.780+113.400)/3.780-6*|38.066|/(1.8002*2.100) =43.199kPa

eyk=Mdxk/(Fk+Gk)=141.340/(176.780+113.400)=0.487m

因 |eyk| >By/6=0.350m y方向大偏心, 由公式【①8.2.2-2】推导 ayk=By/2-|eyk|=2.100/2-|0.487|=0.563m Pkmax_y=2*(Fk+Gk)/(3*Bx*ayk)

=2*(176.780+113.400)/(3*1.800*0.563) =190.921kPa

Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx)

=(176.780+113.400)/3.780-6*|141.340|/(2.1002*1.800) =-30.066kPa 3. 确定基础底面反力设计值

Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk

=(110.335-76.767)+(190.921-76.767)+76.767 =224.489kPa

γo*Pkmax=1.0*224.489=224.489kPa≤1.2*fa=1.2*200.000=240.000kPa 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)=51.389/(238.653+153.090)=0.131m 因 ex≤ Bx/6.0=0.300m x方向小偏心 Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By)

=(238.653+153.090)/3.780+6*|51.389|/(1.8002*2.100) =148.952kPa

Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By)

=(238.653+153.090)/3.780-6*|51.389|/(1.8002*2.100) =58.319kPa

1.2 计算y方向基础底面反力设计值

ey=Mdx/(F+G)=190.809/(238.653+153.090)=0.487m

因 ey >By/6=0.350 y方向大偏心, 由公式【①8.2.2-2】推导 ay=By/2-|ey|=2.100/2-|0.487|=0.563m Pmax_y=2*(F+G)/(3*Bx*ay)

=2*(238.653+153.090)/(3*1.800*0.563) =257.744kPa Pmin_y=0

1.3 因 Mdx≠0 Mdy≠0

Pmax=Pmax_x+Pmax_y-(F+G)/A

=148.952+257.744-(238.653+153.090)/3.780 =303.061kPa 1.4 计算地基净反力极值

Pjmax=Pmax-G/A=303.061-153.090/3.780=262.561kPa Pjmax_x=Pmax_x-G/A=148.952-153.090/3.780=108.452kPa Pjmax_y=Pmax_y-G/A=257.744-153.090/3.780=217.244kPa 2. 验算柱边冲切

YH=h1+h2=0.700m, YB=bc=0.700m, YL=hc=0.900m

YB1=B1=0.900m, YB2=B2=0.900m, YL1=A1=1.050m, YL2=A2=1.050m YHo=YH-as=0.660m

因 ((YB+2*YHo)≥Bx) 并且 (YL+2*YHo)≥By)

基础底面处边缘均位于冲切锥体以内, 不用验算柱对基础的冲切 3. 验算h2处冲切 YH=h2=0.400m

YB=bc+b2+b4=1.300m YL=hc+a2+a4=1.500m

YB1=B1=0.900m, YB2=B2=0.900m, YL1=A1=1.050m, YL2=A2=1.050m YHo=YH-as=0.360m

因 ((YB+2*YHo)≥Bx) 并且 (YL+2*YHo)≥By)

基础底面处边缘均位于冲切锥体以内, 不用验算柱对基础的冲切 八、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。 九、基础受弯计算

1. 因Mdx>0 , Mdy>0 此基础为双向受弯 2. 计算I-I截面弯矩

因 ex ≤Bx/6=0.300m x方向小偏心 a=(Bx-bc)/2=(1.800-0.700)/2=0.550m

Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A

=((1.800-0.550)*(148.952-58.319)/1.800)+58.319-153.090/3.780 =80.759kPa

因 ex >By/6=0.350m y方向大偏心 a=(By-hc)/2=(2.100-0.900)/2=0.600m ay=By/2-ey=2.100/2-0.487=0.563m Pj2=(Pmax_y*(3*ay-a)/(3*ay))-G/A

=(257.744*(3*0.563-0.600)/(3*0.563))-153.090/3.780 =125.671kPa

βx=1.062 βy=1.085

MI_1=1/48*βx*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x)

=1/48*1.062*(1.800-0.700)2*(2*2.100+0.900)*(80.759+108.452) =25.83kN*m

MII_1=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y)

=1/48*1.085*(2.100-0.900)2*(2*1.800+0.700)*(125.671+217.244) =47.98kN*m 3. 计算II-II截面弯矩 因 x方向小偏心

a=(Bx-bc-b2-b4)/2=(1.800-0.700-0.300-0.300)/2=0.250m Pj1=((Bx-a)*(Pmax_x-Pmin_x)/Bx)+Pmin_x-G/A

=((1.800-0.250)*(148.952-58.319)/1.800)+58.319-153.090/3.780 =95.864kPa 因 y方向大偏心

a=(By-hc-a2-a4)/2=(2.100-0.900-0.300-0.300)/2=0.300m ay=By/2-ey=2.100/2-0.487=0.563m Pj2=(Pmax_y*(3*ay-a)/(3*ay))-G/A

=(257.744*(3*0.563-0.300)/(3*0.563))-153.090/3.780 =171.457kPa βx=1.023 βy=1.035

MI_2=1/48*βx*(Bx-bc-b2-b4)2*(2*By+hc+a2+a4)*(Pj1+Pjmax_x)

=1/48*1.023*(1.800-0.700-0.300-0.300)2*(2*2.100+0.900+0.300+0.300)*(95.864+108.452)

=6.20kN*m

MII_2=1/48*βy*(By-hc-a2-a4)2*(2*Bx+bc+b2+b4)*(Pj2+Pjmax_y)

=1/48*1.035*(2.100-0.900-0.300-0.300)2*(2*1.800+0.700+0.300+0.300)*(171.457+108.452)

=14.78kN*m 十、计算配筋 10.1 计算Asx

Asx_1=γo*MI_1/(0.9*(H-as)*fy)

=1.0*25.83*106/(0.9*(700.000-40.000)*300) =144.9mm2

Asx_2=γo*MI_2/(0.9*(H-h2-as)*fy)

=1.0*6.20*106/(0.9*(700.000-400.000-40.000)*300) =88.4mm2

Asx1=max(Asx_1, Asx_2) =max(144.9, 88.4) =144.9mm2

Asx=Asx1/By=144.9/2.100=69mm2/m Asx=max(Asx, ρmin*H*1000)

=max(69, 0.100%*700*1000) =700mm2/m

选择钢筋?14@150, 实配面积为1026mm2/m。

10.2 计算Asy

Asy_1=γo*MII_1/(0.9*(H-as)*fy)

=1.0*47.98*106/(0.9*(700.000-40.000)*300) =269.3mm2

Asy_2=γo*MII_2/(0.9*(H-h2-as)*fy)

=1.0*14.78*106/(0.9*(700.000-400.000-40.000)*300) =152.1mm2

Asy1=max(Asy_1, Asy_2) =max(269.3, 152.050) =269.3mm2

Asy=Asy1/Bx=269.3/1.800=150mm2/m Asy=max(Asy, ρmin*H*1000)

=max(150, 0.100%*700*1000) =700mm2/m

选择钢筋?14@150, 实配面积为1026mm2/m。

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