雕塑风荷载及基础计算

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不锈钢作品风荷载及基础计算一、概况不锈钢雕塑外观为镂空三角形,高度8米,宽度16米,厚度0.75米,置于建筑之间的广场上。

拟在其底部设置独立基础。

三角形的断面为750mm ×1500mm ×5的不锈钢箱型截面。

详见建筑图纸。

立面图 剖面图二、设计依据及荷载取值1.依据规范《建筑结构荷载规范》 (GB50009-2001) 《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) 《钢结构设计规范》(GB50017-2003)《建筑结构可靠度统一设计标准》(GB50068-2001) 《冷弯薄壁型钢结构技术规范》(GB50018-2002) 甲方提供的系列工程技术资料2. 作用在基础顶部的竖向荷载(自重G 及活荷载Q ):取不锈钢密度为80kN/m 3,则G =(0.75+2)×2×0.005×27×80=59.4 kN Q = 0.5×16×0.75 = 6 kN3. 作用在基础顶部的水平荷载(风荷载F ):取风压值为2.0kN/m 2,迎风面积为AA =0.5×(16×8) – 10= 54 m 2F =2.0×54 = 108 kN4. 作用在基础顶部的弯矩M :M = F ×h=108×8×2÷3 =576 kNm三、基础设计基本参数1. 基础类型:锥型柱基 计算形式:验算截面尺寸迎风面基础宽度3600mm ,另一方向长度4400 mm 。

如下图|Å3600mm Æ||Å4400mm Æ|立面剖面平面示意图 剖面示意图2.几何参数: 已知尺寸: B 1 = 2200 mm, A 1 = 1800 mmH 1 = 500 mm,H 2 = 450 mmB =750 mm, A =750 mm 无偏心: B 2 = 2200 mm,A 2 = 1800 mm基础埋深d = 2.80 m钢筋合力重心到板底距离a s = 80 mm3.荷载计算: (1)作用在基础顶部的标准值荷载 F gk = 60.00 kN F qk = 6.00 kN M gxk = 0.00 kN·m M qxk = 0.00 kN·m M gyk = 0.00 kN·m M qyk = 576.00 kN·m V gxk = 108.00 kNV qxk = 0.00 kNV gyk = 0.00 kNV qyk = 0.00 kN(2)作用在基础底部的弯矩标准值M xk = M gxk +M qxk = 0.00+0.00 = 0.00 kN ·mM yk = M gyk +M qyk = 0.00+576.00 = 576.00 kN ·mV xk = V gxk+V qxk = 108.00+0.00 = 108.00 kN·mV yk = V gyk+V qyk = 0.00+0.00 = 0.00 kN·m绕X轴弯矩: M0xk = M xk-V yk·(H1+H2) = 0.00-0.00×0.95 = 0.00 kN·m绕Y轴弯矩: M0yk = M yk+V xk·(H1+H2) = 576.00+108.00×0.95 = 678.60 kN·m(3)作用在基础顶部的基本组合荷载不变荷载分项系数r g = 1.20 活荷载分项系数r q = 1.40F = r g·F gk+r q·F qk = 80.40 kNM x = r g·M gxk+r q·M qxk = 0.00 kN·mM y = r g·M gyk+r q·M qyk = 806.40 kN·mV x = r g·V gxk+r q·V qxk = 129.60 kNV y = r g·V gyk+r q·V qyk = 0.00 kN(4)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) = 0.00-0.00×0.95 = 0.00 kN·m绕Y轴弯矩: M0y = M y+V x·(H1+H2) = 806.40+129.60×0.95 = 929.52 kN·m4.材料信息:混凝土: C30 钢筋: HRB335(20MnSi)5.基础几何特性:底面积:S = (A1+A2)(B1+B2) = 3.60×4.40 = 15.84 m2绕X轴抵抗矩:Wx = (1/6)(B1+B2)(A1+A2)2 = (1/6)×4.40×3.602 = 9.50 m3绕Y轴抵抗矩:Wy = (1/6)(A1+A2)(B1+B2)2 = (1/6)×3.60×4.402 = 11.62 m3四、底板反力1.轴心荷载作用下底板反力计算公式:按《建筑地基基础设计规范》(GB 50007-2002)下列公式验算:(5.2.2-1)p k = (F k+G k)/AF k = F gk+F qk = 60.00+6.00 = 66.00 kNG k = 20S·d = 20×15.84×2.80 = 887.04 kNp k = (F k+G k)/S = (66.00+887.04)/15.84 = 60.17 kPa ≤f a,满足要求。

2.偏心荷载作用下底板反力计算公式: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)X方向:偏心距e xk = M0yk/(F k+G k) = 678.60/(66.00+887.04) = 0.71 me = e xk = 0.71 m ≤ (B1+B2)/6 = 4.40/6 = 0.73 mp kmaxX = (F k+G k)/S+M0yk/W y=(66.00+887.04)/15.84+678.60/11.62 = 118.59 kPap kminX = (F k+G k)/S - M0yk/W y=(66.00+887.04)/15.84 - 678.60/11.62 = 118.59 kPa≥0 kPa根据反力计算结构,基础对底板的平均压力为60.17 kN/m2,最大压力118.59kN/m2,最小压力1.77kN/m2。

最小压力为正,满足结构抗倾覆要求。

五、基础承载力验算1.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 = 80.40/15.84+929.52/11.62 = 85.10 kPap jmin,x = F/S-M0y/W y = 80.40/15.84-929.52/11.62≤0, p jmin.x = 0.00 kPap jmax,y = F/S+M0x/W x = 80.40/15.84+0.00/9.50 = 5.08 kPap jmin,y = F/S-M0x/W x = 80.40/15.84-0.00/9.50 = 5.08 kPap j = p jmax,x+p jmax,y-F/S = 85.10+5.08-5.08 = 85.10 kPa(1)柱对基础的冲切验算:H0 = H1+H2-a s = 0.50+0.45-0.08 = 0.87 mX方向:A lx = 1/2·(A1+A2)(B1+B2-B-2H0)-1/4·(A1+A2-A-2H0)2= (1/2)×3.60×(4.40-0.75-2×0.87)-(1/4)×(3.60-0.75-2×0.87)2 =3.13 m2F lx = p j·A lx = 85.10×3.13 = 266.35 kNa b = min{A+2H0, A1+A2} = min{0.75+2×0.87, 3.60} = 2.49 ma mx = (a t+a b)/2 = (A+a b)/2 = (0.75+2.49)/2 = 1.62 mFlx ≤ 0.7·βhp·f t·a mx·H0 = 0.7×0.99×1430.00×1.620×0.870=1393.17kN,满足要求。

Y方向:A ly = 1/4·(2B+2H0+A1+A2-A)(A1+A2-A-2H0)= (1/4)×(2×0.75+2×0.87+3.60-0.75)(3.60-0.75-2×0.87) =1.69 m2F ly = p j·A ly = 85.10×1.69 = 143.81 kNa b = min{B+2H0, B1+B2} = min{0.75+2×0.87, 4.40} = 2.49 ma my = (a t+a b)/2 = (B+a b)/2 = (0.75+2.49)/2 = 1.62 mFly ≤ 0.7·βhp·f t·a my·H0 = 0.7×0.99×1430.00×1.620×0.870=1393.17kN,满足要求。

2.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 = 80.40 kN混凝土局部受压面积:A ln = A l = B×A = 0.75×0.75 = 0.56 m2混凝土受压时计算底面积:A b = min{B+2A, B1+B2}×min{3A, A1+A2} = 5.06 m2混凝土受压时强度提高系数:βl = sq.(A b/A l) = sq.(5.06/0.56) = 3.001.35βc·βl·f c·A ln=1.35×1.00×3.00×14300.00×0.56= 32577.19 kN ≥ F l = 80.40 kN ,满足要求。