锥型柱基DJp1计算书.doc

锥型柱基DJp1计算书项目名称_日 期_设 计 者_校 对 者_一、示意图基础类型：锥型柱基计算形式：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 500072011）混凝土结构设计规范（GB 500102010）简明高层钢筋混凝土结构设计手册（第四版）2几何参数：已知尺寸：B1 = 2250 mm, A1 = 2250 mmH1 = 200 mm, H2 = 450 mmB = 500 mm, A = 500 mm无偏心：B2 = 2250 mm, A2 = 2250 mm基础埋深d = 2.00 m 钢筋合力重心到板底距离as = 50 mm 3荷载值：(1)作用在基础顶部的标准值荷载Fgk = 2200.00 kNFqk = 10.00 kNMgxk = 1.30 kNmMqxk = 0.00 kNmMgyk = 12.70 kNmMqyk = 0.00 kNmVgxk = 6.90 kNVqxk = 0.00 kNVgyk = -0.40 kNVqyk = 0.00 kN(2)作用在基础底部的弯矩标准值Mxk = MgxkMqxk = 1.300.00 = 1.30 kNmMyk = MgykMqyk = 12.700.00 = 12.70 kNmVxk = VgxkVqxk = 6.900.00 = 6.90 kNmVyk = VgykVqyk = -0.400.00 = -0.40 kNm绕X轴弯矩: M0xk = MxkVyk(H1H2) = 1.30-0.400.65 = 1.56 kNm绕Y轴弯矩: M0yk = MykVxk(H1H2) = 12.706.900.65 = 17.18 kNm(3)作用在基础顶部的基本组合荷载不变荷载分项系数rg = 1.25活荷载分项系数rq = 1.40F = rgFgkrqFqk = 2764.00 kNMx = rgMgxkrqMqxk = 1.63 kNmMy = rgMgykrqMqyk = 15.88 kNmVx = rgVgxkrqVqxk = 8.63 kNVy = rgVgykrqVqyk = -0.50 kN(4)作用在基础底部的弯矩设计值绕X轴弯矩: M0x = MxVy(H1H2) = 1.63-0.500.65 = 1.95 kNm绕Y轴弯矩: M0y = MyVx(H1H2) = 15.888.630.65 = 21.48 kNm4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 = (A1A2)(B1B2) = 4.504.50 = 20.25 m2绕X轴抵抗矩：Wx = (1/6)(B1+B2)(A1+A2)2 = (1/6)4.504.502 = 15.19 m3绕Y轴抵抗矩：Wy = (1/6)(A1+A2)(B1+B2)2 = (1/6)4.504.502 = 15.19 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 = 150.00 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 500072011）下列公式验算：pk = (FkGk)/A(5.2.21)Fk = FgkFqk = 2200.0010.00 = 2210.00 kN Gk = 20Sd = 2020.252.00 = 810.00 kN pk = (FkGk)/S = (2210.00810.00)/20.25 = 149.14 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 500072011）下列公式验算：当eb/6时，pkmax = (FkGk)/AMk/W(5.2.22) pkmin = (FkGk)/AMk/W(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) = 17.18/(2210.00810.00) = 0.01 me = exk = 0.01 m (B1B2)/6 = 4.50/6 = 0.75 mpkmaxX = (FkGk)/SM0yk/Wy = (2210.00810.00)/20.2517.18/15.19 = 150.27 kPa偏心距eyk = M0xk/(FkGk) = 1.56/(2210.00810.00) = 0.00 me = eyk = 0.00 m (A1A2)/6 = 4.50/6 = 0.75 mpkmaxY = (FkGk)/SM0xk/Wx = (2210.00810.00)/20.251.56/15.19 = 149.24 kPapkmax = pkmaxXpkmaxY(FkGk)/S = 150.27149.24149.14 = 150.37 kPa 1.2fa = 1.2150.00 = 180.00 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 500072011）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/Wy = 2764.00/20.2521.48/15.19 = 137.91 kPapjmin,x = F/SM0y/Wy = 2764.00/20.2521.48/15.19 = 135.08 kPapjmax,y = F/SM0x/Wx = 2764.00/20.251.95/15.19 = 136.62 kPapjmin,y = F/SM0x/Wx = 2764.00/20.251.95/15.19 = 136.37 kPapj = pjmax,xpjmax,yF/S = 137.91136.62136.49 = 138.04 kPa(1)柱对基础的冲切验算：H0 = H1H2as = 0.200.450.05 = 0.60 mX方向：Alx = 1/4(A2H0A1A2)(B1B2B2H0) = (1/4)(0.5020.604.50)(4.500.5020.60) = 4.34 m2Flx = pjAlx = 138.044.34 = 599.08 kNab = minA2H0, A1A2 = min0.5020.60, 4.50 = 1.70 mamx = (atab)/2 = (Aab)/2 = (0.501.70)/2 = 1.10 mFlx 0.7hpftamxH0 = 0.71.001430.001.1000.600= 660.66 kN，满足要求。Y方向：Aly = 1/4(B2H0B1B2)(A1A2A2H0) = (1/4)(0.5020.604.50)(4.500.5020.60) = 4.34 m2Fly = pjAly = 138.044.34 = 599.08 kNab = minB2H0, B1B2 = min0.5020.60, 4.50 = 1.70 mamy = (atab)/2 = (Bab)/2 = (0.501.70)/2 = 1.10 mFly 0.7hpftamyH0 = 0.71.001430.001.1000.600= 660.66 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 500102010）Fl 1.35clfcAln(7.8.1-1)局部荷载设计值：Fl = 2764.00 kN混凝土局部受压面积：Aln = Al = BA = 0.500.50 = 0.25 m2混凝土受压时计算底面积：Ab = minB2A, B1B2min3A, A1A2 = 2.25 m2混凝土受压时强度提高系数：l = sq.(Ab/Al) = sq.(2.25/0.25) = 3.00 1.35clfcAln = 1.351.003.0014300.000.25 = 14478.75 kN Fl = 2764.00 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（第四版）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 = 1.35810.00 = 1093.50kN-截面处弯矩设计值：pjnx = pjmin,x(pjmax,xpjmin,x)(B1B2B)/2/(B1B2) = 135.08(137.91135.08)(4.500.50)/2/4.50 = 136.65 kPa M = /48(B1B2B)22(A1A2)A(pjmax,xpjnx) = 1.0000/48(4.500.50)2(24.500.50)(137.91136.65) = 869.44 kNm-截面处弯矩设计值：pjny = pjmin,y(pjmax,ypjmin,y)(A1A2A)/2/(A1A2) = 136.37(136.62136.37)(4.500.50)/2/4.50 = 136.51 kPa M = /48(A1A2A)22(B1B2)B(pjmax,ypjny) = 1.0000/48(4.500.50)2(24.500.50)(136.62136.51) = 864.91 kNm-截面受弯计算：相对受压区高度： = 0.038263 配筋率