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1、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：程序自动计算平面:剖面:二、基本参数1依据规范建筑地基基础设计规范（GB 500072002）混凝土结构设计规范（GB 500102002）简明高层钢筋混凝土结构设计手册（第二版）2几何参数：自动计算所得尺寸：B1 = 1450 mm, A1 = 1450 mmH1 = 250 mm, H2 = 250 mmB = 400 mm, A = 400 mmB3 = 1650 mm, A3 = 1650 mm无偏心：B2 = 1450 mm, A2 = 1450 mm基础埋深d = 1.50 m 钢筋合力重
2、心到板底距离as = 80 mm 3荷载值：(1)作用在基础顶部的基本组合荷载F = 500.00 kNMx = 390.00 kNmMy = 40.00 kNmVx = 40.00 kNVy = 100.00 kN折减系数Ks = 1.30(2)作用在基础底部的弯矩设计值绕X轴弯矩: M0x = MxVy(H1H2) = 390.00100.000.50 = 340.00 kNm绕Y轴弯矩: M0y = MyVx(H1H2) = 40.0040.000.50 = 60.00 kNm(3)作用在基础底部的弯矩标准值绕X轴弯矩: M0xk = M0x/Ks = 340.00/1.30 = 261
3、.54 kNm绕Y轴弯矩: M0yk = M0y/Ks = 60.00/1.30 = 46.15 kNm4材料信息：混凝土： C30钢筋： HPB235(Q235)5基础几何特性：底面积：S = (A1A2)(B1B2) = 2.902.90 = 8.41 m2绕X轴抵抗矩：Wx = (1/6)(B1+B2)(A1+A2)2 = (1/6)2.902.902 = 4.06 m3绕Y轴抵抗矩：Wy = (1/6)(A1+A2)(B1+B2)2 = (1/6)2.902.902 = 4.06 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 fa = 400.00 kPa2轴心荷载作用下地
4、基承载力验算计算公式：按建筑地基基础设计规范（GB 500072002）下列公式验算：pk = (FkGk)/A(5.2.21)Fk = F/Ks = 500.00/1.30 = 384.62 kN Gk = 20Sd = 208.411.50 = 252.30 kN pk = (FkGk)/S = (384.62252.30)/8.41 = 75.73 kPa fa，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 500072002）下列公式验算：当eb/6时，pkmax = (FkGk)/AMk/W(5.2.22) pkmin = (FkGk)/AMk/W(
5、5.2.23)当eb/6时，pkmax = 2(FkGk)/3la(5.2.24)X、Y方向同时受弯。偏心距exk = M0yk/(FkGk) = 46.15/(384.62252.30) = 0.07 me = exk = 0.07 m (B1B2)/6 = 2.90/6 = 0.48 mpkmaxX = (FkGk)/SM0yk/Wy = (384.62252.30)/8.4146.15/4.06 = 87.09 kPa偏心距eyk = M0xk/(FkGk) = 261.54/(384.62252.30) = 0.41 me = eyk = 0.41 m (A1A2)/6 = 2.90/
6、6 = 0.48 mpkmaxY = (FkGk)/SM0xk/Wx = (384.62252.30)/8.41261.54/4.06 = 140.07 kPapkmax = pkmaxXpkmaxY(FkGk)/S = 87.09140.0775.73 = 151.43 kPa 1.2fa = 1.2400.00 = 480.00 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 500072002）下列公式验算：Fl 0.7hpftamh0(8.2.71)Fl = pjAl(8.2.73)am = (atab)/2(8.2.72)pjmax,x = F/SM0y/W
7、y = 500.00/8.4160.00/4.06 = 74.21 kPapjmin,x = F/SM0y/Wy = 500.00/8.4160.00/4.06 = 44.69 kPapjmax,y = F/SM0x/Wx = 500.00/8.41340.00/4.06 = 143.10 kPapjmin,y = F/SM0x/Wx = 500.00/8.41340.00/4.060, pjmin.y = 0.00 kPapj = pjmax,xpjmax,yF/S = 74.21143.1059.45 = 157.86 kPa(1)柱对基础的冲切验算：H0 = H1H2as = 0.250
8、.250.08 = 0.42 mX方向：Alx = 1/4(A2H0A1A2)(B1B2B2H0) = (1/4)(0.4020.422.90)(2.900.4020.42) = 1.72 m2Flx = pjAlx = 157.861.72 = 271.22 kNab = minA2H0, A1A2 = min0.4020.42, 2.90 = 1.24 mamx = (atab)/2 = (Aab)/2 = (0.401.24)/2 = 0.82 mFlx 0.7hpftamxH0 = 0.71.001430.000.8200.420= 344.74 kN，满足要求。Y方向：Aly = 1
9、/4(B2H0B1B2)(A1A2A2H0) = (1/4)(0.4020.422.90)(2.900.4020.42) = 1.72 m2Fly = pjAly = 157.861.72 = 271.22 kNab = minB2H0, B1B2 = min0.4020.42, 2.90 = 1.24 mamy = (atab)/2 = (Bab)/2 = (0.401.24)/2 = 0.82 mFly 0.7hpftamyH0 = 0.71.001430.000.8200.420= 344.74 kN，满足要求。(2)变阶处基础的冲切验算：X方向：Alx = 1/4(A32H0A1A2)
10、(B1B2B32H0) = (1/4)(1.6520.172.90)(2.901.6520.17) = 1.11 m2Flx = pjAlx = 157.861.11 = 175.61 kNab = minA32H0, A1A2 = min1.6520.17, 2.90 = 1.99 mamx = (atab)/2 = (A3ab)/2 = (1.651.99)/2 = 1.82 mFlx 0.7hpftamxH0 = 0.71.001430.001.8200.170= 309.71 kN，满足要求。Y方向：Aly = 1/4(B32H0B1B2)(A1A2A32H0) = (1/4)(1.6
11、520.172.90)(2.901.6520.17) = 1.11 m2Fly = pjAly = 157.861.11 = 175.61 kNab = minB32H0, B1B2 = min1.6520.17, 2.90 = 1.99 mamy = (atab)/2 = (B3ab)/2 = (1.651.99)/2 = 1.82 mFly 0.7hpftamyH0 = 0.71.001430.001.8200.170= 309.71 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 500102002）Fl 1.35clfcAln(7.8.1-1)局部荷载设计值：Fl =
12、 500.00 kN混凝土局部受压面积：Aln = Al = BA = 0.400.40 = 0.16 m2混凝土受压时计算底面积：Ab = minB2A, B1B2min3A, A1A2 = 1.44 m2混凝土受压时强度提高系数：l = sq.(Ab/Al) = sq.(1.44/0.16) = 3.00 1.35clfcAln = 1.351.003.0014300.000.16 = 9266.40 kN Fl = 500.00 kN，满足要求。6基础受弯计算计算公式：按简明高层钢筋混凝土结构设计手册（第二版）中下列公式验算：M=/48(La)2(2Bb)(pjmaxpjnx)(11.47)M=/48(Bb)2(2La)(pjmaxpjny)(11.48)(1)柱根部受弯计算：G = 1.35Gk = 1.35252.30 = 340.61kN-截面处弯矩设计值：pjnx = pjmin,x(pjmax,xpjmin,x)(B1B2B)/2/(B1B2) = 44.69(74.2144.69)(2.900.40)/2/2.90 = 61.49 kPa M = /48(B1B2B)22(A1A2)A(pjmax,xpjnx) = 1.0444/48(2.900.40)2(22.900.40)(74.2161.49) = 114.41 kNm-截面处弯矩设计值：pj
