阶梯柱基计算书项目名称_____________日期_____________设计者_____________校对者_____________一、示意图基础类型：阶梯柱基计算形式：验算截面尺寸平面:剖面:二、基本参数1．依据规范《建筑地基基础设计规范》（GB 50007－2002）《混凝土结构设计规范》（GB 50010－2002）《简明高层钢筋混凝土结构设计手册（第二版）》2．几何参数：已知尺寸：B1 = 1400 mm, A1 = 700 mmH1 = 300 mm, H2 = 300 mmB = 800 mm, A = 500 mmB3 = 1400 mm, A3 = 700 mm无偏心：B2 = 1400 mm, A2 = 700 mm基础埋深d = 1.50 m钢筋合力重心到板底距离a s = 80 mm3．荷载值：(1)作用在基础顶部的基本组合荷载F = 146.15 kNM x = 0.00 kN·mM y = 105.38 kN·mV x = 25.37 kNV y = 0.00 kN折减系数K s = 1.35(2)作用在基础底部的弯矩设计值绕X轴弯矩: M0x = M x－V y·(H1＋H2) = 0.00－0.00×0.60 = 0.00 kN·m绕Y轴弯矩: M0y = M y＋V x·(H1＋H2) = 105.38＋25.37×0.60 = 120.60 kN·m(3)作用在基础底部的弯矩标准值绕X轴弯矩: M0xk = M0x/K s = 0.00/1.35 = 0.00 kN·m绕Y轴弯矩: M0yk = M0y/K s = 120.60/1.35 = 89.33 kN·m4．材料信息：混凝土：C30 钢筋：HRB400(20MnSiV、20MnSiNb、20MnTi) 5．基础几何特性：底面积：S = (A1＋A2)(B1＋B2) = 1.40×2.80 = 3.92 m2绕X轴抵抗矩：Wx = (1/6)(B1+B2)(A1+A2)2 = (1/6)×2.80×1.402 = 0.91 m3绕Y轴抵抗矩：Wy = (1/6)(A1+A2)(B1+B2)2 = (1/6)×1.40×2.802 = 1.83 m3三、计算过程1．修正地基承载力修正后的地基承载力特征值f a = 110.00 kPa2．轴心荷载作用下地基承载力验算计算公式：按《建筑地基基础设计规范》（GB 50007－2002）下列公式验算：p k = (F k＋G k)/A (5.2.2－1)F k = F/K s = 146.15/1.35 = 108.26 kNG k = 20S·d = 20×3.92×1.50 = 117.60 kNp k = (F k＋G k)/S = (108.26＋117.60)/3.92 = 57.62 kPa ≤f a，满足要求。

3．偏心荷载作用下地基承载力验算计算公式：按《建筑地基基础设计规范》（GB 50007－2002）下列公式验算：当e≤b/6时，p kmax = (F k＋G k)/A＋M k/W (5.2.2－2)p kmin = (F k＋G k)/A－M k/W (5.2.2－3) 当e＞b/6时，p kmax = 2(F k＋G k)/3la (5.2.2－4) X方向:偏心距e xk = M0yk/(F k＋G k) = 89.33/(108.26＋117.60) = 0.40 me = e xk = 0.40 m ≤(B1＋B2)/6 = 2.80/6 = 0.47 mp kmaxX = (F k＋G k)/S＋M0yk/W y= (108.26＋117.60)/3.92＋89.33/1.83 = 106.45 kPa≤1.2×f a = 1.2×110.00 = 132.00 kPa，满足要求。

4．基础抗冲切验算计算公式：按《建筑地基基础设计规范》（GB 50007－2002）下列公式验算：F l≤0.7·βhp·f t·a m·h0(8.2.7－1)F l = p j·A l(8.2.7－3)a m = (a t＋a b)/2 (8.2.7－2)p jmax,x = F/S＋M0y/W y = 146.15/3.92＋120.60/1.83 = 103.21 kPap jmin,x = F/S－M0y/W y = 146.15/3.92－120.60/1.83≤0, p jmin.x = 0.00 kPap jmax,y = F/S＋M0x/W x = 146.15/3.92＋0.00/0.91 = 37.28 kPap jmin,y = F/S－M0x/W x = 146.15/3.92－0.00/0.91 = 37.28 kPap j = p jmax,x＋p jmax,y－F/S = 103.21＋37.28－37.28 = 103.21 kPa(1)柱对基础的冲切验算：H0 = H1＋H2－a s = 0.30＋0.30－0.08 = 0.52 mX方向：A lx = 1/2·(B1＋B2－B－2H0)(A1＋A2)= (1/2)×(2.80－0.80－2×0.52)×1.40= 0.67 m2F lx = p j·A lx = 103.21×0.67 = 69.36 kNa b = min{A＋2H0, A1＋A2} = min{0.50＋2×0.52, 1.40} = 1.40 ma mx = (a t＋a b)/2 = (A＋a b)/2 = (0.50＋1.40)/2 = 0.95 mFlx ≤0.7·βhp·f t·a mx·H0 = 0.7×1.00×1430.00×0.950×0.520= 494.49 kN，满足要求。

(2)变阶处基础的冲切验算：X方向：A lx = 1/2·(A1＋A2)(B1＋B2－B3－2H0)－1/4·(A1＋A2－A3－2H0)2= (1/2)×1.40×(2.80－1.40－2×0.22)－(1/4)×(1.40－0.70－2×0.22)2= 0.66 m2F lx = p j·A lx = 103.21×0.66 = 67.61 kNa b = min{A3＋2H0, A1＋A2} = min{0.70＋2×0.22, 1.40} = 1.14 ma mx = (a t＋a b)/2 = (A3＋a b)/2 = (0.70＋1.14)/2 = 0.92 mFlx ≤0.7·βhp·f t·a mx·H0 = 0.7×1.00×1430.00×0.920×0.220= 202.60 kN，满足要求。

Y方向：A ly = 1/4·(2B3＋2H0＋A1＋A2－A3)(A1＋A2－A3－2H0)= (1/4)×(2×1.40＋2×0.22＋1.40－0.70)(1.40－0.70－2×0.22)= 0.26 m2F ly = p j·A ly = 103.21×0.26 = 26.43 kNa b = min{B3＋2H0, B1＋B2} = min{1.40＋2×0.22, 2.80} = 1.84 ma my = (a t＋a b)/2 = (B3＋a b)/2 = (1.40＋1.84)/2 = 1.62 mFly ≤0.7·βhp·f t·a my·H0 = 0.7×1.00×1430.00×1.620×0.220= 356.76 kN，满足要求。

5．基础受压验算计算公式：《混凝土结构设计规范》（GB 50010－2002）F l≤1.35·βc·βl·f c·A ln(7.8.1-1)局部荷载设计值：F l = 146.15 kN混凝土局部受压面积：A ln = A l = B×A = 0.80×0.50 = 0.40 m2混凝土受压时计算底面积：A b = min{B＋2A, B1＋B2}×min{3A, A1＋A2} = 2.10 m2混凝土受压时强度提高系数：βl = sq.(A b/A l) = sq.(2.10/0.40) = 2.291.35βc·βl·f c·A ln= 1.35×1.00×2.29×14300.00×0.40= 17693.32 kN ≥F l = 146.15 kN，满足要求。

6．基础受弯计算计算公式：按《建筑地基基础设计规范》(GB 50007－2002)下列公式验算：MⅠ=a12[(2l＋a')(p max＋p－2G/A)＋(p max－p)·l]/12 (8.2.7－4)MⅡ=(l－a')2(2b＋b')(p max＋p min－2G/A)/48 (8.2.7－5)(1)柱根部受弯计算：G = 1.35G k = 1.35×117.60 = 158.76kNX方向受弯截面基底反力设计值：p minx = (F＋G)/S－M0y/W y = (146.15＋158.76)/3.92－120.60/1.83 = 11.86 kPap maxx = (F＋G)/S＋M0y/W y = (146.15＋158.76)/3.92＋120.60/1.83 = 143.71 kPap nx = p minx＋(p maxx－p minx)(2B1＋B)/[2(B1＋B2)]= 11.86＋(143.71－11.86)×3.60/(2×2.80)= 96.62 kPaⅠ-Ⅰ截面处弯矩设计值：MⅠ= [(B1＋B2)/2－B/2]2{[2(A1＋A2)＋A](p maxx＋p nx－2G/S)＋(p maxx－p nx)(A1＋A2)}/12= (2.80/2－0.80/2)2((2×1.40＋0.50)(143.71＋96.62－2×158.76/3.92)＋(143.71－96.62)×1.40)/12= 49.31 kN.mⅡ-Ⅱ截面处弯矩设计值：MⅡ= (A1＋A2－A)2[2(B1＋B2)＋B](p maxx＋p minx－2G/S)/48= (1.40－0.50)2(2×2.80＋0.80)(143.71＋11.86－2×158.76/3.92)/48= 8.05 kN.mⅠ-Ⅰ截面受弯计算：相对受压区高度：ζ= 0.009151 配筋率：ρ= 0.000363ρ < ρmin = 0.001500 ρ = ρmin = 0.001500计算面积：900.00 mm2/mⅡ-Ⅱ截面受弯计算：相对受压区高度：ζ= 0.000744 配筋率：ρ= 0.000030ρ < ρmin = 0.001500 ρ = ρmin = 0.001500计算面积：900.00 mm2/m(2)变阶处受弯计算：X方向受弯截面基底反力设计值：p minx = (F＋G)/S－M0y/W y = (146.15＋158.76)/3.92－120.60/1.83 = 11.86 kPap maxx = (F＋G)/S＋M0y/W y = (146.15＋158.76)/3.92＋120.60/1.83 = 143.71 kPap nx = p minx＋(p maxx－p minx)(2B1＋B3)/[2(B1＋B2)]= 11.86＋(143.71－11.86)×4.20/(2×2.80)= 110.75 kPaⅠ-Ⅰ截面处弯矩设计值：MⅠ= [(B1＋B2)/2－B3/2]2{[2(A1＋A2)＋A3](p maxx＋p nx－2G/S)＋(p maxx－p nx)(A1＋A2)}/12= (2.80/2－1.40/2)2((2×1.40＋0.70)(143.71＋110.75－2×158.76/3.92)＋(143.71－110.75)×1.40)/12= 26.67 kN.mⅡ-Ⅱ截面处弯矩设计值：MⅡ= (A1＋A2－A3)2[2(B1＋B2)＋B3](p maxx＋p minx－2G/S)/48= (1.40－0.70)2(2×2.80＋1.40)(143.71＋11.86－2×158.76/3.92)/48= 5.33 kN.mⅠ-Ⅰ截面受弯计算：相对受压区高度：ζ= 0.027918 配筋率：ρ= 0.001109ρ < ρmin = 0.001500 ρ = ρmin = 0.001500计算面积：450.00 mm2/mⅡ-Ⅱ截面受弯计算：相对受压区高度：ζ= 0.002753 配筋率：ρ= 0.000109ρ < ρmin = 0.001500 ρ = ρmin = 0.001500计算面积：450.00 mm2/m四、计算结果1．X方向弯矩验算结果：计算面积：900.00 mm2/m采用方案：f14@150实配面积：1026.25 mm2/m2．Y方向弯矩验算结果：计算面积：900.00 mm2/m采用方案：f14@150实配面积：1026.25 mm2/m。