杯口基础计算.doc

结构构件计算书杯口基础计算项目名称_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 (GB50010-2002)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=400mm矩形柱高hc=600mm基础端部高度h1=250mm基础根部高度h2=350mm基础杯口高度h3=300mm基础杯底厚度a1=200mm基础长度B1=1500mmB2=1500mm基础宽度A1=1250mmA2=1250mm杯壁厚t=200mm2. 材料信息基础混凝土等级:C20ft_b=1.10N/mm2fc_b=9.6N/mm2柱混凝土等级:C15ft_c=0.91N/mm2fc_c=7.2N/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=0.150%4. 作用在基础顶部荷载标准值Fgk=920.000kNFqk=0.000kNMgxk=197.000kN*mMqxk=0.000kN*mMgyk=0.000kN*mMqyk=0.000kN*mVgxk=18.000kNVqxk=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=920.000+(0.000)=920.000kNMxk=Mgxk+Fgk*(B2-B1)/2+Mqxk+Fqk*(B2-B1)/2 =197.000+920.000*(1.500-1.500)/2+(0.000)+0.000*(1.500-1.500)/2 =197.000kN*mMyk=Mgyk+Fgk*(A2-A1)/2+Mqyk+Fqk*(A2-A1)/2 =0.000+920.000*(1.250-1.250)/2+(0.000)+0.000*(1.250-1.250)/2 =0.000kN*mVxk=Vgxk+Vqxk=18.000+(0.000)=18.000kNVyk=Vgyk+Vqyk=0.000+(0.000)=0.000kNF1=rg*Fgk+rq*Fqk=1.20*(920.000)+1.40*(0.000)=1104.000kNMx1=rg*(Mgxk+Fgk*(B2-B1)/2)+rq*(Mqxk+Fqk*(B2-B1)/2) =1.20*(197.000+920.000*(1.500-1.500)/2)+1.40*(0.000+0.000*(1.500-1.500)/2) =236.400kN*mMy1=rg*(Mgyk+Fgk*(A2-A1)/2)+rq*(Mqyk+Fqk*(A2-A1)/2) =1.20*(0.000+920.000*(1.250-1.250)/2)+1.40*(0.000+0.000*(1.250-1.250)/2) =0.000kN*mVx1=rg*Vgxk+rq*Vqxk=1.20*(18.000)+1.40*(0.000)=21.600kNVy1=rg*Vgyk+rq*Vqyk=1.20*(0.000)+1.40*(0.000)=0.000kNF2=1.35*Fk=1.35*920.000=1242.000kNMx2=1.35*Mxk=1.35*197.000=265.950kN*mMy2=1.35*Myk=1.35*(0.000)=0.000kN*mVx2=1.35*Vxk=1.35*18.000=24.300kNVy2=1.35*Vyk=1.35*(0.000)=0.000kNF=max(|F1|,|F2|)=max(|1104.000|,|1242.000|)=1242.000kNMx=max(|Mx1|,|Mx2|)=max(|236.400|,|265.950|)=265.950kN*mMy=max(|My1|,|My2|)=max(|0.000|,|0.000|)=0.000kN*mVx=max(|Vx1|,|Vx2|)=max(|21.600|,|24.300|)=24.300kNVy=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=1.500+1.500=3.000m2. 基础总宽 By=A1+A2=1.250+1.250=2.500m3. 基础总高 H=h1+h2=0.250+0.350=0.600m4. 底板配筋计算高度 ho=h1+h2-as=0.250+0.350-0.040=0.560m5. 基础底面积 A=Bx*By=3.000*2.500=7.500m26. Gk=*Bx*By*dh=20.000*3.000*2.500*1.500=225.000kNG=1.35*Gk=1.35*225.000=303.750kN五、计算作用在基础底部弯矩值Mdxk=Mxk-Vyk*H=197.000-0.000*0.600=197.000kN*mMdyk=Myk+Vxk*H=0.000+18.000*0.600=10.800kN*mMdx=Mx-Vy*H=265.950-0.000*0.600=265.950kN*mMdy=My+Vx*H=0.000+24.300*0.600=14.580kN*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=(920.000+225.000)/7.500=152.667kPa【5.2.1-2】因o*pk=1.0*152.667=152.667kPafa=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)=10.800/(920.000+225.000)=0.009m因 |exk|Bx/6=0.500mx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) =(920.000+225.000)/7.500+6*|10.800|/(3.0002*2.500) =155.547kPaPkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By) =(920.000+225.000)/7.500-6*|10.800|/(3.0002*2.500) =149.787kPaeyk=Mdxk/(Fk+Gk)=197.000/(920.000+225.000)=0.172m因 |eyk|By/6=0.417my方向小偏心Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx) =(920.000+225.000)/7.500+6*|197.000|/(2.5002*3.000) =215.707kPaPkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx) =(920.000+225.000)/7.500-6*|197.000|/(2.5002*3.000) =89.627kPa3. 确定基础底面反力设计值Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk =(155.547-152.667)+(215.707-152.667)+152.667 =218.587kPao*Pkmax=1.0*218.587=218.587kPa1.2*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)=14.580/(1242.000+303.750)=0.009m因 exBx/6.0=0.500mx方向小偏心Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By) =(1242.000+303.750)/7.500+6*|14.580|/(3.0002*2.500) =209.988kPaPmin_x=(F+G)/A-6*|Mdy|/(Bx2*By) =(1242.000+303.750)/7.500-6*|14.580|/(3.0002*2.500) =202.212kPa1.2 计算y方向基础底面反力设计值ey=Mdx/(F+G)=265.950/(1242.000+303.750)=0.172m因 eyBy/6=0.417y方向小偏心Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx) =(1242.000+303.750)/7.500+6*|265.950|/(2.5002*3.000) =291.204kPaPmin_y=(F+G)/A-6*|Mdx|/(By2*Bx) =(1242.000+303.750)/7.500-6*|265.950|/(2.5002*3.000) =120.996kPa1.3 因 Mdx0 Mdy0 Pmax=Pmax_x+Pmax_y-(F+G)/A=209.988+291.204-(1242.000+303.750)/7.500=295.092kPa1.4 计算地基净反力极值Pjmax=Pmax-G/A=295.092-303.750/7.500=254.592kPaPjmax_x=Pmax_x-G/A=209.988-303.750/7.500=169.488kPaPjmax_y=Pmax_y-G/A=291.204-303.750/7.500=250.704kPa2. 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b=bc+2*t+2*0.075=0.400+2*0.200+2*0.075=0.950mh=hc+2*t+2*0.075=0.600+2*0.200+2*0.075=1.150mx冲切面积 Alx=max(A1-h/2-ho)*(b+2*ho)-(A1-h/2-ho)2,(A2-h/2-ho)*(b+2*ho)-(A2-h/2-ho)2=max(1.250-1.150/2-0.560)*(0.950+2*0.560)-(1.250-1.150/2-0.560)2,(1.250-1.150/2-0.560)*(0.950+2*0.560)+(1.250-1.150/2-0.560)2)=max(0.225,0.225)=0.225m2x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=0.225*254.592=57.239kNo*Flx=1.0*57.239=57.24kNo*Flx0.7*hp*ft_b*bm*ho=0.7*1.000*1.10*1510*559=651.11kNx方向柱对基础的冲切满足规范要求2.3 验算y方向柱对基础的冲切验算y冲切面积 Aly=max(B1-b/2-ho)*(A1+A2)-(A1-h/2-ho)2/2-(A2-h/2-ho)2/2,(B2-b/2-ho)*(A1+A2)-(A1-h/2-ho)2/2-(A2-h/2-ho)2/2)=max(1.500-0.950/2-0.560)*(1.250+1.250)-(1.250-1.150-0.560)2/2-(1.250-1.150-0.560)2/2,(1.500-0.950/2-0.560)*(1.250+1.250)-(1.250-1.150-0.560)2/2-(1.250-1.150-0.560)2/2)=max(1.149,1.149)=1.149m2y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=1.149*254.592=292.596kNo*Fly=1.0*292.596=292.60kNo*Fly0.7*hp*ft_b*am*ho=0.7*1.000*1.10*1709*559=737.35kNy方向柱对基础的冲切满足规范要求八、基础受弯计算双向受弯因 ex Bx/6=0.500m x方向小偏心a=(Bx-bc-2*t-2*0.075)/2=(3.000-0.400-2*0.200-2*0.075)/2=1.025mPj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx+Pmin_x-G/A =(3.000-1.025)*(209.988-202.212)/3.000+202.212-303.750/7.500 =166.831kPa因 ey By/6=0.417m y方向小偏心a=(By-hc-2*t-2*0.075)/2=(2.500-0.600-2*0.200-2*0.075)/2=0.675mPj2=(By-a)*(Pmax_y-Pmin_y)/By+Pmin_y-G/A =(2.500-0.675)*(291.204-120.996)/2.500+120.996-303.750/7.500 =204.748kPax=1.025y=1.067MI=1/48*x*(Bx-bc-2*t-2*0.075)2*(2*By+hc+2*t+2*0.075)*(Pj1+Pjmax_x) =1/48*1.025*(3.000-0.400-2*0.200-2*0.075)2*(2*2.500+0.600+2*0.20