几何尺寸与公差论坛

 找回密码
 注册
查看: 695|回复: 0

wind force to velocity

[复制链接]
发表于 2009-9-16 21:38:16 | 显示全部楼层 |阅读模式
wind force to velocity
can anyone give me a few pointers.if i had a wind force acting on wall of given dimensions (say 3m h x 4m w), how easy is it to calculate the wind velocity?
check out our whitepaper library.
f = (0.5*q air)(v)^2
q air = 1.2kg/m^3
f = force per unit area
v = wind speed m/s
also
f =p*a
p = wind pressure
a = area
hope this helps!
force = pressure*area
pressure = 1/2*rho*v^2
rho = 0.00238
v in ft/sec
a in ft^2
force in lbs
also i forgot to mention f is in pascals it should have really been p.
hopefully i haven't confussed!

you would have to know also the gust effect factor, directionality factor, importance factor, topographic factor, exposure coefficients and force coefficients.  if the wall encloses a space you may also need to consider internal pressure coefficients.  these are terms used by asce 7, but if you are under a different code or use a different analysis method, you'll still have to consider these effects to some degree to have a meaningful answer.
it sounds like you're asking "my sign blew over, so how fast must the wind have been blowing?"  is that the case?
there are several issues to consider.  the two formulas above are for stagnation pressure, and will not necessarily give you the net force on a rectangular surface.  the factors suggested by ucfse would be used to get from a mapped wind velocity as specified by asce 7 or similar source to the design loading on a surface.  but the building codes aren't set up to be worked in reverse, and i'm not sure how meaningful the results would be when you do this.  probably the best you could do is to state that per a given building code, the structure should have been adequate for a specific wind speed.

to answer your question; it is not easy and it's impossible to do to any meaningful accuracy.
nicam and rb1957 gave you the correct answer to your question.  the others are just saying that your question is too simple.  
i beg to differ hokie.
the formulas presented by nicam and rb1957 have to be multiplied by the pressure coefficent appropriate for the structure in question to be able to calculate the force on that structure.
the pressure coefficients found in codes would be safe averages, but probably not accurate enough for the reverse calculation suggested.
i agree with comments made here, i just gave him a basic equ to answer his question. but i should have added that other factors are also required.

well, apsix, if he already knows the design force as he said, any pressure coefficients have been included in factoring the gust dynamic wind pressure to get the design wind pressure.  for instance, if he knows that the force on his 3m x 4m wall is 12 kn, the wind pressure is 1.0 kpa, and the wind speed which causes this force is 40.8 m/sec.  i agree that wind codes are complex, and differ in different countries, but this is a simple question which deserves a simple answer.
您需要登录后才可以回帖 登录 | 注册

本版积分规则

QQ|Archiver|小黑屋|几何尺寸与公差论坛

GMT+8, 2025-1-9 03:23 , Processed in 0.039306 second(s), 19 queries .

Powered by Discuz! X3.4 Licensed

© 2001-2023 Discuz! Team.

快速回复 返回顶部 返回列表