JC-2锥形基础计算.doc

结构构件计算书锥形基础计算项目名称_日 期_设 计 者_校 对 者_一、设计依据建筑地基基础设计规范 (GB50007-2002)混凝土结构设计规范 (GB50010-2010)简明高层钢筋混凝土结构设计手册李国胜二、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. 几何参数矩形柱宽bc=600mm矩形柱高hc=600mm基础端部高度h1=350mm基础根部高度h2=100mm基础长度B1=950mmB2=950mm基础宽度A1=950mmA2=950mm2. 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=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:=20.000kN/m3最小配筋率:min=0.150%4. 作用在基础顶部荷载标准组合值F=190.100kNMx=-58.880kN*mMy=-58.880kN*mVx=-31.250kNVy=-31.250kNks=1.35Fk=F/ks=190.100/1.35=140.815kNMxk=Mx/ks=-58.880/1.35=-43.615kN*mMyk=My/ks=-58.880/1.35=-43.615kN*mVxk=Vx/ks=-31.250/1.35=-23.148kNVyk=Vy/ks=-31.250/1.35=-23.148kN5. 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=0.950+0.950=1.900m2. 基础总宽 By=A1+A2=0.950+0.950=1.900m3. 基础总高 H=h1+h2=0.350+0.100=0.450m4. 底板配筋计算高度 ho=h1+h2-as=0.350+0.100-0.040=0.410m5. 基础底面积 A=Bx*By=1.900*1.900=3.610m26. Gk=*Bx*By*dh=20.000*1.900*1.900*2.000=144.400kNG=1.35*Gk=1.35*144.400=194.940kN五、计算作用在基础底部弯矩值Mdxk=Mxk-Vyk*H=-43.615-(-23.148)*0.450=-33.198kN*mMdyk=Myk+Vxk*H=-43.615+(-23.148)*0.450=-54.031kN*mMdx=Mx-Vy*H=-58.880-(-31.250)*0.450=-44.818kN*mMdy=My+Vx*H=-58.880+(-31.250)*0.450=-72.942kN*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=(140.815+144.400)/3.610=79.007kPa【5.2.1-2】因o*pk=1.0*79.007=79.007kPafa=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)=-54.031/(140.815+144.400)=-0.189m因 |exk|Bx/6=0.317mx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) =(140.815+144.400)/3.610+6*|-54.031|/(1.9002*1.900) =126.272kPaPkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(140.815+144.400)/3.610-6*|-54.031|/(1.9002*1.900) =31.742kPaeyk=Mdxk/(Fk+Gk)=-33.198/(140.815+144.400)=-0.116m因 |eyk|By/6=0.317my方向小偏心Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(140.815+144.400)/3.610+6*|-33.198|/(1.9002*1.900) =108.047kPaPkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(140.815+144.400)/3.610-6*|-33.198|/(1.9002*1.900) =49.966kPa3. 确定基础底面反力设计值Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(126.272-79.007)+(108.047-79.007)+79.007 =155.312kPao*Pkmax=1.0*155.312=155.312kPa1.2*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)=-72.942/(190.100+194.940)=-0.189m因 exBx/6.0=0.317mx方向小偏心Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(190.100+194.940)/3.610+6*|-72.942|/(1.9002*1.900) =170.467kPaPmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(190.100+194.940)/3.610-6*|-72.942|/(1.9002*1.900) =42.852kPa1.2 计算y方向基础底面反力设计值ey=Mdx/(F+G)=-44.818/(190.100+194.940)=-0.116m因 eyBy/6=0.317y方向小偏心Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(190.100+194.940)/3.610+6*|-44.818|/(1.9002*1.900) =145.864kPaPmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(190.100+194.940)/3.610-6*|-44.818|/(1.9002*1.900) =67.455kPa1.3 因 Mdx0 Mdy0 Pmax=Pmax_x+Pmax_y-(F+G)/A=170.467+145.864-(190.100+194.940)/3.610=209.671kPa1.4 计算地基净反力极值Pjmax=Pmax-G/A=209.671-194.940/3.610=155.671kPaPjmax_x=Pmax_x-G/A=170.467-194.940/3.610=116.467kPaPjmax_y=Pmax_y-G/A=145.864-194.940/3.610=91.864kPa2. 柱对基础的冲切验算2.1 因 (H800) hp=1.02.2 x方向柱对基础的冲切验算x冲切面积 Alx=max(A1-hc/2-ho)*(bc+2*ho)+(A1-hc/2-ho)2,(A2-hc/2-ho)*(bc+2*ho)+(A2-hc/2-ho)2=max(0.950-0.600/2-0.410)*(0.600+2*0.410)+(0.950-0.600/2-0.410)2,(0.950-0.600/2-0.410)*(0.600+2*0.410)+(0.950-0.600/2-0.410)2)=max(0.398,0.398)=0.398m2x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=0.398*155.671=62.019kNo*Flx=1.0*62.019=62.02kN因 o*Flx0.7*hp*ft_b*bm*ho =0.7*1.000*1.43*1010*410=414.51kNx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冲切面积 Aly=max(B1-bc/2-ho)*(hc+2*ho)+(B1-bc/2-ho)2,(B2-bc/2-ho)*(hc+2*ho)+(B2-bc/2-ho)2)=max(0.950-0.600/2-0.410)*(0.600+2*0.410)+(0.950-0.600-0.410)2/2,(0.950-0.600/2-0.410)*(0.600+2*0.410)+(0.950-0.600-0.410)2/2)=max(0.398,0.398)=0.398m2y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=0.398*155.671=62.019kNo*Fly=1.0*62.019=62.02kN因o*Fly0.7*hp*ft_b*am*ho =0.7*1.000*1.43*1010.000*410 =414.51kNy方向柱对基础的冲切满足规范要求八、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、基础受弯计算双向受弯因 ex Bx/6=0.317m x方向小偏心a=(Bx-bc)/2=(1.900-0.600)/2=0.650mPj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx+Pmin_x-G/A =(1.900-0.650)*(170.467-42.852)/1.900+42.852-194.940/3.610 =72.809kPa因ey By/6=0.317m y方向小偏心a=(By-hc)/2=(1.900-0.600)/2=0.650mPj2=(By-a)*(Pmax_y-Pmin_y)/By+Pmin_y-G/A =(1.900-0.650)*(145.864-67.455)/1.900+67.455-194.940/3.610 =65.040kPax=1.100y=1.074MI=1/48*x*(Bx-bc)2*(2*By+hc)*(Pj1+Pjmax_x) =1/48*1.100*(1.900-0.600)2*(2*1.900+0.600)*(72.809+116.467) =32.26kN*mMII=1/48*y*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y) =1/48*1.074*(1.900-0.600)2*(2*1.900+0.600)*(65.040+91.864) =26.11kN*m十、计算配筋1. 计算基础底板x方向钢筋Asx=o*MI/(0.9*ho*fy) =1.0*32.26*106/(0.9*410.000*300) =291.4mm2Asx1=Asx/By=291.4/1.900=153mm2/mAsx1=max(Asx1, min*H*1000)=max(153, 0.150%*450*1000)