双柱式标志结构设计计算书

1、双柱式标志结构设计计算书(1)基本资料5.6m4.55m铝合金矩形标志板, 立柱采用214a,立柱缀板606mm计算简图（略）(2)荷载计算1)永久荷载标志板单位重量为（5.4kg/m2 ）标志板重力为g1=5.64.555.49.8=1348.40（n）立柱采用214a,立柱缀板606mmg2=(487.34+94.92)9.8=5700.27（n）标志上部结构的总重量g按标志板和立柱重量的110计(考虑有关连接件及加劲肋等的重量),则g=( g1+ g2) 1.10=(1348.40+5700.27)1.10=7753.537（n）2)风荷载标志板fwb=0q(1/2cv2)(wb1hb1

2、)/1000=1.01.4(1/21.22581.2302) (2.84.55)/1000 =11.81(kn)立柱fwp=0q(1/2cv2)(wb1hb1)/1000=1.01.4(1/21.22581.4302) (0.1244.77)/1000 =0.64(kn)（3）强度验算 立柱的截面积为a=(0.0018512)m2半截面静矩s1 =0.060.006(0.07-0.0125)=2.0710-5 m3 s2 =(0.070.0060.035)2+(0.0520.0095)(0.07-0.00475)2=6.16310-5 m3 s= s1 + s2 =8.23310-5 m3 截

3、面惯性矩为i=0.060.006(0.07-0.0125)2 2+5.63710-6 2=13.65510-6 m4抗弯截面膜量为w=i/y=1.95110-4m3立柱根部所受到的由风载引起的弯矩为：m= fwbh2 + fwph1 =11.81(1.8+2.97+4.55/2)+0.644.77/2=84.73(kn.m)立柱根部所受到的由风载引起的剪力为：f= fwb + fwp=11.81+0.64=12.45(kn)1)最大正应力验算立柱根部横断面上由风荷载引起的最大正应力为:max =m/w=84.73103 /1.95110-4 =434.29(mpa) d =215(mpa),强

4、度不足2)最大剪应力验算max =fmax s/(ib)=12.45103 8.23310-5/(13.65510-60.012)=6.26(mpa)d=125mpa3)危险点应力验算危险点e所在位置为槽钢腹板与翼板交接处:s=(0.0580.0095)(0.07-0.0095/2)2=7.19110-5 (m3 )e =my/i=84.73103(0.07-0.0095/2)/(13.65510-6 )=404.88(mpa)e =fmaxs/(ib)=12.451037.19110-5/(13.65510-60.012)=5.46(mpa) d =215(mpa),强度不足（4）变形验算立

5、柱总高度为：l=4.55+1.8+2.97=9.32m标志板所受集中荷载标准值为：p= fwb /（0q ）=11.81/（1.01.4）=8.44（kn）该荷载与立柱根部之间的距离为：h2 =7.045m;由该荷载引起的立柱顶部挠度为：f2 =p h2 2 /(6ei) (3l- h2 )=8.44103 7.0457.045/(6206109 13.65510-6 ) (39.32-7.045)=0.519(m)标志板与基础之间的立柱所受均布荷载标准值为：q= fwp /（0q h）=0.64/（1.01.44.77）=0.096(kn/m)其顶部挠度为：f1 =q h1 4 /(8ei)

6、 =0.096103 4.774 /(8206109 13.65510-6 )=0.00221(m)转角为：= q h2 3 /(6ei)= 0.096103 4.773 /(6206109 13.65510-6 )=6.17310-4 (rad)立柱顶部的总变形为挠度为：f = f1 + f2 +(l-h) tan()=0.00221+0.519+4.55tan(6.17310-4 )=0.5213(m)f/l=0.5213/9.32=1/181/100,不足（5）基础验算b=0.8m l=1.2m h=1.5m 1基底所受外荷载为:n=g+v=7754+24(0.81.51.2)=42.31(kn)弯矩 m= fwb(h2+1.5)+fwp(h1 +1.5) =11.81(1.8+2.97+4.55/2+1.5)+0.64(4.77/2+1.5) =103.40(kn.m)基底应力最大值为:max =n/a+m/w=42.31/(1.20.8)+ 103.40/(1/60.81.22 )=582.61(kpa)=290(kpa)基底应力最小值为:min =n/a-m/w=42.31/(1.20.8)- 103.40/(1/60.81.22 )=-494.47(kpa)2基础抗倾覆稳定性验算抗倾覆稳定系数k0 =(l/2)/e0