某钢结构独立基础计算书J-1.doc

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)混凝土结构设计规范 (gb50010-2010)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=250mmb2=300mm 一阶宽度 a1=300mma2=300mm二阶长度 b3=250mmb4=300mm 二阶宽度 a3=300mma4=300mm2. 材料信息基础混凝土等级:c30ft_b=1.43n/mm2fc_b=14.3n/mm2柱混凝土等级:c30ft_c=1.43n/mm2fc_c=14.3n/mm2钢筋级别:hrb335 fy=300n/mm23. 计算信息结构重要性系数: o=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:=20.000kn/m3最小配筋率:min=0.100%fgk=176.780knfqk=0.000knmgxk=141.340kn*mmqxk=0.000kn*mmgyk=0.000kn*mmqyk=0.000kn*mvgxk=54.380knvqxk=0.000knvgyk=0.000knvqyk=0.000kn永久荷载分项系数rg=1.20可变荷载分项系数rq=1.40fk=fgk+fqk=176.780+(0.000)=176.780knmxk=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*mmyk=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*mvxk=vgxk+vqxk=54.380+(0.000)=54.380knvyk=vgyk+vqyk=0.000+(0.000)=0.000knf1=rg*fgk+rq*fqk=1.20*(176.780)+1.40*(0.000)=212.136knmx1=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*mmy1=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*mvx1=rg*vgxk+rq*vqxk=1.20*(54.380)+1.40*(0.000)=65.256knvy1=rg*vgyk+rq*vqyk=1.20*(0.000)+1.40*(0.000)=0.000knf2=1.35*fk=1.35*176.780=238.653knmx2=1.35*mxk=1.35*141.340=190.809kn*mmy2=1.35*myk=1.35*(0.000)=0.000kn*mvx2=1.35*vxk=1.35*54.380=73.413knvy2=1.35*vyk=1.35*(0.000)=0.000knf=max(|f1|,|f2|)=max(|212.136|,|238.653|)=238.653knmx=max(|mx1|,|mx2|)=max(|169.608|,|190.809|)=190.809kn*mmy=max(|my1|,|my2|)=max(|0.000|,|0.000|)=0.000kn*mvx=max(|vx1|,|vx2|)=max(|65.256|,|73.413|)=73.413knvy=max(|vy1|,|vy2|)=max(|0.000|,|0.000|)=0.000kn5. 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=1.800m2. 基础总宽 by=a1+a2+a3+a4+hc=0.300+0.300+0.300+0.300+0.900=2.100ma1=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=1.050mb1=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=0.900m3. 基础总高 h=h1+h2=0.300+0.400=0.700m4. 底板配筋计算高度 ho=h1+h2-as=0.300+0.400-0.040=0.660m5. 基础底面积 a=bx*by=1.800*2.100=3.780m26. gk=*bx*by*dh=20.000*1.800*2.100*1.500=113.400kng=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*mmdyk=myk+vxk*h=0.000+54.380*0.700=38.066kn*mmdx=mx-vy*h=190.809-0.000*0.700=190.809kn*mmdy=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【5.2.1-2】因o*pk=1.0*76.767=76.767kpafa=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=0.300mx方向小偏心,由公式【5.2.2-2】和【5.2.2-3】推导pkmax_x=(fk+gk)/a+6*|mdyk|/(bx2*by) =(176.780+113.400)/3.780+6*|38.066|/(1.8002*2.100) =110.335kpapkmin_x=(fk+gk)/a-6*|mdyk|/(bx2*by) =(176.780+113.400)/3.780-6*|38.066|/(1.8002*2.100) =43.199kpaeyk=mdxk/(fk+gk)=141.340/(176.780+113.400)=0.487m因 |eyk| by/6=0.350my方向大偏心, 由公式【8.2.2-2】推导ayk=by/2-|eyk|=2.100/2-|0.487|=0.563mpkmax_y=2*(fk+gk)/(3*bx*ayk) =2*(176.780+113.400)/(3*1.800*0.563) =190.921kpapkmin_y=(fk+gk)/a-6*|mdxk|/(by2*bx) =(176.780+113.400)/3.780-6*|141.340|/(2.1002*1.800) =-30.066kpa3. 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 =224.489kpao*pkmax=1.0*224.489=224.489kpa1.2*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=0.300mx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) =148.952kpapmin_x=(f+g)/a-6*|mdy|/(bx2*by) =(238.653+153.090)/3.780-6*|51.389|/(1.8002*2.100) =58.319kpa1.2 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=0.350y方向大偏心, 由公式【8.2.2-2】推导ay=by/2-|ey|=2.100/2-|0.487|=0.563mpmax_y=2*(f+g)/(3*bx*ay) =2*(238.653+153.090)/(3*1.800*0.563) =257.744kpapmin_y=01.3 因 mdx0 mdy0 pmax=pmax_x+pmax_y-(f+g)/a=148.952+257.744-(238.653+153.090)/3.780=303.061kpa1.4 计算地基净反力极值pjmax=pmax-g/a=303.061-153.090/3.780=262.561kpapjmax_x=pmax_x-g/a=148.952-153.090/3.780=108.452kpapjmax_y=pmax_y-g/a=257.744-153.090/3.780=217.244kpa2. 验算柱边冲切yh=h1+h2=0.700m, yb=bc=0.700m, yl=hc=0.900myb1=b1=0.900m, yb2=b2=0.900m, yl1=a1=1.050m, yl2=a2=1.050myho=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=0.400myb=bc+b2+b4=1.300myl=hc+a2+a4=1.500myb1=b1=0.900m, yb2=b2=0.900m, yl1=a1=1.050m, yl2=a2=1.050myho=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mdx0 , mdy0 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=0.550mpj1=(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=0.600may=by/2-ey=2.100/2-0.487=0.563mpj2=(pmax_y*(3*ay-a)/(3*ay)-g/a=(257.744*(3*0.563-0.600)/(3*0.563)-153.090/3.780=125.671kpax=1.062y=1.085mi_1=1/48*x*(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*mmii_1=1/48*y*(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*m3. 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=0.250mpj1=(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=0.300may=by/2-ey=2.100/2-0.487=0.563mpj2=(pmax_y*(3*ay-a)/(3*ay)-g/a=(257.744*(3*0.563-0.300)/(3*0.563)-153.090/3.780=171.457kpax=1.023y=1.035mi_2=1/48*x*(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*mmii_2=1/48*y*(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asxasx_1=o*mi_1/(0.9*(h-as)*fy) =1.0*25.83*106/(0.9*