风荷载计算例题ppt课件.ppt

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1、例题1-1一高层钢筋混凝土结构，平面形状为正六边形，边长为20m。房屋共20层，除底层层高为5m外，其余层高为3.6m。该房屋的第一自振周期T1=1.2S，所在地区的基本风压20/7 . 0mkNw 地面粗糙度为C类。试计算各楼层处与风向一致方向总的风荷载标准值。34.64m风向解1 确定体形系数0-0.5+0.8-0.50-0.5 60该房屋共有6个面，查表1-11得到各个面的风荷载体形系数，如图所示，不为零的4个面分别用表示。20m2 计算各层的风压高度系数近似假定室内外地面相同，则二层楼面离室外地面高度为5m，查表1-10，对于C类地面粗糙度，74. 0z同理可求得其余各层楼面标高处的风

2、压高度系数。3 计算风振系数zzz101. 12 . 17 . 02210Tw查表1-12386. 1房屋高宽比H/B=(5+3.619)/34.64=2.1，查表1-13，485. 0对于质量和刚度沿高度比较均匀的房屋，结构振型系数可以取4 .73/zHzz对于C类地面，乘0.62根据0.621.01=0.634 计算各个面不同高度的分布风荷载0wwzsziz5 计算各个面各楼层处的集中风荷载顶层中间层底层izizizizjizwwwwhBW)2/6 . 3(206 . 3202/ )6 . 35(206 计算各楼层处总的风荷载60cos2321zzzzWWWWizwizW层数HI(m)zz

3、z面面面面面面zW150.740.071.060.440.280.2837.9023.6923.6985.2928.60.740.121.110.460.290.2933.0920.6820.6874.45312.20.740.171.150.480.300.3034.4421.5321.5377.50415.80.760.221.190.510.320.3236.6122.8822.8882.36519.40.840.261.210.570.360.3640.9225.5725.5792.066230.890.311.230.620.380.3844.2927.6827.6899.65726

4、.60.950.361.250.670.420.4248.0630.0430.04108.14830.210.411.280.710.450.4551.4332.1532.15115.72933.81.050.461.290.760.480.4854.8034.2534.25123.311037.41.10.511.310.810.500.5058.1736.3636.36130.8911411.140.561.330.850.530.5361.1438.2138.21137.571244.61.190.611.340.900.560.5664.5140.3240.32145.161348.2

5、1.230.661.360.940.590.5967.4842.1842.18151.831451.81.270.711.380.980.610.6170.4544.0344.03158.511555.41.30.751.391.010.630.6372.7445.4645.46163.6716591.340.81.401.050.660.6675.7147.3247.32170.351762.61.380.851.411.090.680.6878.6849.1749.17177.031866.21.410.91.431.130.710.7181.2450.7850.78182.801969.81.450.951.441.170.730.7384.2152.6352.63189.482073.41.4811.451.210.750.7543.3927.1227.1297.62风荷载计算结果