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)①《混凝土结构设计规范》 (GB50010-2010)②《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=1170mm基础端部高度h1=200mm基础根部高度h2=150mm基础长度B1=1200mm B2=1200mm基础宽度A1=1800mm A2=1800mm2. 材料信息基础混凝土等级: C30 ft_b=1.43N/mm2fc_b=14.3N/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=1.800m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=18.000kN/m3最小配筋率: ρmin=0.150%4. 作用在基础顶部荷载标准值Fgk=201.000kN Fqk=0.000kNMgxk=234.000kN*m Mqxk=0.000kN*mMgyk=0.000kN*m Mqyk=0.000kN*mVgxk=59.000kN Vqxk=0.000kNVgyk=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=1.40Fk=Fgk+Fqk=201.000+(0.000)=201.000kNMxk=Mgxk+Fgk*(B2-B1)/2+Mqxk+Fqk*(B2-B1)/2=234.000+201.000*(1.200-1.200)/2+(0.000)+0.000*(1.200-1.200)/2=234.000kN*mMyk=Mgyk+Fgk*(A2-A1)/2+Mqyk+Fqk*(A2-A1)/2=0.000+201.000*(1.800-1.800)/2+(0.000)+0.000*(1.800-1.800)/2=0.000kN*mVxk=Vgxk+Vqxk=59.000+(0.000)=59.000kNVyk=Vgyk+Vqyk=0.000+(0.000)=0.000kNF1=rg*Fgk+rq*Fqk=1.20*(201.000)+1.40*(0.000)=241.200kNMx1=rg*(Mgxk+Fgk*(B2-B1)/2)+rq*(Mqxk+Fqk*(B2-B1)/2)=1.20*(234.000+201.000*(1.200-1.200)/2)+1.40*(0.000+0.000*(1.200-1.200)/2) =280.800kN*mMy1=rg*(Mgyk+Fgk*(A2-A1)/2)+rq*(Mqyk+Fqk*(A2-A1)/2)=1.20*(0.000+201.000*(1.800-1.800)/2)+1.40*(0.000+0.000*(1.800-1.800)/2)=0.000kN*mVx1=rg*Vgxk+rq*Vqxk=1.20*(59.000)+1.40*(0.000)=70.800kNVy1=rg*Vgyk+rq*Vqyk=1.20*(0.000)+1.40*(0.000)=0.000kNF2=1.35*Fk=1.35*201.000=271.350kNMx2=1.35*Mxk=1.35*234.000=315.900kN*mMy2=1.35*Myk=1.35*(0.000)=0.000kN*mVx2=1.35*Vxk=1.35*59.000=79.650kNVy2=1.35*Vyk=1.35*(0.000)=0.000kNF=max(|F1|,|F2|)=max(|241.200|,|271.350|)=271.350kNMx=max(|Mx1|,|Mx2|)=max(|280.800|,|315.900|)=315.900kN*mMy=max(|My1|,|My2|)=max(|0.000|,|0.000|)=0.000kN*mVx=max(|Vx1|,|Vx2|)=max(|70.800|,|79.650|)=79.650kNVy=max(|Vy1|,|Vy2|)=max(|0.000|,|0.000|)=0.000kN5. 修正后的地基承载力特征值fa=106.900kPa四、计算参数1. 基础总长 Bx=B1+B2=1.200+1.200=2.400m2. 基础总宽 By=A1+A2=1.800+1.800=3.600m3. 基础总高 H=h1+h2=0.200+0.150=0.350m4. 底板配筋计算高度 ho=h1+h2-as=0.200+0.150-0.040=0.310m5. 基础底面积 A=Bx*By=2.400*3.600=8.640m26. Gk=γ*Bx*By*dh=18.000*2.400*3.600*1.800=279.936kNG=1.35*Gk=1.35*279.936=377.914kN五、计算作用在基础底部弯矩值Mdxk=Mxk-Vyk*H=234.000-0.000*0.350=234.000kN*mMdyk=Myk+Vxk*H=0.000+59.000*0.350=20.650kN*mMdx=Mx-Vy*H=315.900-0.000*0.350=315.900kN*mMdy=My+Vx*H=0.000+79.650*0.350=27.878kN*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=(201.000+279.936)/8.640=55.664kPa 【①5.2.1-2】因γo*pk=1.0*55.664=55.664kPa≤fa=106.900kPa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)=20.650/(201.000+279.936)=0.043m因|exk|≤Bx/6=0.400m 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)=(201.000+279.936)/8.640+6*|20.650|/(2.4002*3.600)=61.639kPaPkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By)=(201.000+279.936)/8.640-6*|20.650|/(2.4002*3.600)=49.689kPaeyk=Mdxk/(Fk+Gk)=234.000/(201.000+279.936)=0.487m因|eyk|≤By/6=0.600m y方向小偏心Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx)=(201.000+279.936)/8.640+6*|234.000|/(3.6002*2.400)=100.803kPaPkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx)=(201.000+279.936)/8.640-6*|234.000|/(3.6002*2.400)=10.525kPa3. 确定基础底面反力设计值Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk=(61.639-55.664)+(100.803-55.664)+55.664=106.778kPaγo*Pkmax=1.0*106.778=106.778kPa≤1.2*fa=1.2*106.900=128.280kPa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)=27.878/(271.350+377.914)=0.043m因ex≤Bx/6.0=0.400m x方向小偏心Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By)=(271.350+377.914)/8.640+6*|27.878|/(2.4002*3.600)=83.213kPaPmin_x=(F+G)/A-6*|Mdy|/(Bx2*By)=(271.350+377.914)/8.640-6*|27.878|/(2.4002*3.600)=67.080kPa1.2 计算y方向基础底面反力设计值ey=Mdx/(F+G)=315.900/(271.350+377.914)=0.487m因ey≤By/6=0.600y方向小偏心Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx)=(271.350+377.914)/8.640+6*|315.900|/(3.6002*2.400)=136.084kPaPmin_y=(F+G)/A-6*|Mdx|/(By2*Bx)=(271.350+377.914)/8.640-6*|315.900|/(3.6002*2.400)=14.209kPa1.3 因Mdx≠0 Mdy≠0Pmax=Pmax_x+Pmax_y-(F+G)/A=83.213+136.084-(271.350+377.914)/8.640=144.150kPa1.4 计算地基净反力极值Pjmax=Pmax-G/A=144.150-377.914/8.640=100.410kPaPjmax_x=Pmax_x-G/A=83.213-377.914/8.640=39.473kPaPjmax_y=Pmax_y-G/A=136.084-377.914/8.640=92.344kPa2. 柱对基础的冲切验算2.1 因(H≤800) βhp=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)-(B1-hc/2-ho)2/2-(B2-bc/2-ho)2/2,(A2-hc/2-ho)*(bc+2*ho)-(B2-hc/ 2-ho)2/2-(B1-bc/2-ho)2/2=max((1.800-1.170/2-0.310)*(0.600+2*0.310)-(1.200-1.170/2-0.310)2/2-(1.200-0.600/2-0.310 )2/2,(1.800-1.170/2-0.310)*(0.600+2*0.310)-(1.200-1.170/2-0.310)2/2-(1.200-0.600/2-0.310)2/2) =max(0.884,0.884)=0.884m2x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=0.884*100.410=88.716kNγo*Flx=1.0*88.716=88.72kN因γo*Flx≤0.7*βhp*ft_b*bm*ho (6.5.5-1)=0.7*1.000*1.43*910*310=282.38kNx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((1.200-0.600/2-0.310)*(1.170+2*0.310)+(1.200-0.600-0.310)2/2,(1.200-0.600/2-0.310)* (1.170+2*0.310)+(1.200-0.600-0.310)2/2)=max(1.404,1.404)=1.404m2y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=1.404*100.410=140.996kNγo*Fly=1.0*140.996=141.00kN因γo*Fly≤0.7*βhp*ft_b*am*ho (6.5.5-1)=0.7*1.000*1.43*1480.000*310=459.26kNy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。