huangyhg

wind load on a pipe & a pipe behind
i try to design a steel structure to support two big parallel pipes (about 2m diameter) having the same level. i work out the wind loading applied on a pipe. i wonder if the other pipe behind a few meter will take 100% of wind loadings or not. which australian or british or american standard show about that ?
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wind----->| pipe | |pipe |
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thanks in advance.
tngv752
check out our whitepaper library.
there is a small publication by the american society of civil engineers (asce) that deals with pipes and pipe racks. it is called "wind loads and anchor bolt design for petrochemical facilities" and is a good resource. for this case, according to the publication, you use an effective frontal area equal to the height of the largest pipe plus 10% of the distance parallel to the wind from the front pipe to the most leeward pipe. to put numbers to your example, if the pipes are 3m apart c-to-c, then the frontal area used for force calculation would be:
2m + 0.1 x (2m + 1m + 2m) = 2.5m^2 / meter length of rack.
note that this value is specifically designed to be used with the wind provisions of asce 7, "minimum design loads for buildings and other structures". i don't know of its applicability to other codes.
the british standard is bs 6399-2:1997. the standard that it replaced, cp3, chapter v, part 2, had an excellent solution for wind sheilding. i haven't seen the new standard.
we use the following formula for pipe racks, based on equating a pipe group to a single equivalent pipe. unfortunately i'm not sure of the original source/reference.
fw = cd*p*(dmax + dave*a)
where:
'fw' is the wind load
'cd' is the drag coefficient for the largest pipe in the group
'p' is the wind pressure
'dmax' is the largest diamter in the pipe group
'dave' is the average diamter of the pipe group
'a' is a factor for the number of pipes in the group
1 pipe, a = 0; 2, 0.70; 3, 1.19; 4, 1.53; 5, 1.77; 6, 1.94; 7, 2.06; 8, 2.14; 9, 2.20; 10, 2.24; >10, 2.30
thanks all for the replies.
dbuzz,
according to you, the first pipe will take
f1 = cd*p*diameter(1st pipe) (kn/m)
the 2nd pipe takes:
f2 = cd*p*(dmax + dave*0.7) (kn/m)
the 3rd pipe takes:
f3 = cd*p*(dmax + dave*1.19) (kn/m)
therefore, f3>f2>f1. that means the pipe behind the others takes more wind loading ???
regards,
tngv752
i see that the total wind load is fw = cd*p*(dmax + dave*1.19) (kn/m)(if there are 3 pipes).
so which pipe should the wind apply to ?

say all the pipes have the saem diameter. then,
fw = cd*p*dia* 2.19
fw = 2.19*fo (fo = cd*p*dia)
100%fo applied to the first pipe, 70% to 2nd & 49% to 3rd ?
don't forget.......the wind doesn't always blow in the same direction !! if you have shielding on one pipe in one wind direction you won't get it when the wind changes .
tngv752
i'm not sure you've undertstood my post.
the factors aren't applied to each pipe in the pipe group, rather it is an idealised single pipe representing the entire pipe group.
perhaps i misunderstood your initial query.
dbuzz
with vortex shedding, the shielded pipe can take a greater load. depends on the diameter, spacing and wind...
dik

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