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原帖由 ashin 于 2008-1-9 10:43 发表
锥形的我不会算哦。lblqf大侠在哪?
发错版了
hehe, 我不是大虾,是外行,来这里学习的。 我按 ASCE 7-05 试一下吧。 虽然是锥形,我认为可以当 圆柱形来处理。
V=120km/h = 75 mph, h = 20m = ~66 feet
F = qz*G*Cf*Af [ASCE 7-05 Eq. 6-28]
qz = 0.00256*Kz*Kzt*Kd*V^2*I [ASCE 7-05 Eq. 6-15]
Kz = (0.85+0.89)/2= 0.87 [ASCE Table 6-3; h=66ft, Exposure B], (最不利场所 Kz=1.33 for Exp. D)
Kzt = 1.0 [ASCE 7-05 Section 6.5.7.2]
Kd = 0.95 [Table 6-4; Chinneys Tanks and similar structures/ Round]
I = 1.0 [Table 6-1; Category II]
=> qz = 0.00256*0.87*1.0*0.95*75^2*1.0 = 11.9 psf ( 1 psf = 0.0479kN/m^2)
G = 0.85 [Section 6.5.8.1; rigid]
Cf= 1.2 [Table 6-21; D/sqrt(qz)<=2.5, h/D>25]
=> F = qz*G*Cf*Af= 11.9*0.85*1.2 = 11.9*0.85*1.2*Af=12.14 psf * Af = 582 N/m^2 *Af (Af=投影面积)
(搞了半天,对 Exposure B, 跟直接按基本风压来算才不多 qs= 0.00256 * V^2 =0.00256 * 75^2 =14.4 psf (~690 N/m^2), 所以一般都只是直接用这个公式来算出来风压来算 wind load. )
锥形投影是梯形 Af=(d+D)/2*h = (0.08m+0.26m)/2*20m = 3.4 m^2
=> F = 582 N/m^2 *Af =582 *3.4 = 1979 N =~ 2000N
假设是 悬臂柱, 根部弯矩 M_max = F*(h/2)=2000*20/2=20000 N.m
I = pi*0.26^4/64-pi*0.25^4/64 = 3.2569971 × 10^-5 m^4
应力 fb= M*D/2/I = 80 MPa (普通结构钢管 fy >200 Mpa)
请高手指正. :) |
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