Subject: Re: [TowerTalk] Tower wind loads
Date: Wed, 03 Sep 2003 16:20:39 -0400
At 02:59 PM 2003-09-03, Bob Gates wrote:
If you have a tower rated to hold 30sf of antenna at 70 mph, and 10sf
at 90mph, what correlation, if any, can you draw about the load handling
ability at 80mph. Is there a linear relationship, or does it require a
whole new set of calcs?
The relationship is NOT linear, although in Bob's
particular example, a more accurate answer isn't very far from the number
he would get if he did assume it was linear.
The charts in the Rohn catalogs for decades showed the
pressure on a tower and beam to be related to the SQUARE of the wind
velocity. Thus, the ratio of allowable antenna square footage at 90 mph to
that at 70 mph would be 4900 / 8100, or 60%, if the tower itself had no
surface area. So if you ignored the wind load on the tower, you would
conclude that the allowable antenna surface area at 90 mph was 60% of that
at 70 mph, or 18 square feet. Clearly, the tower has surface area, and
it's pretty likely the manufacturer has included it in the 90 mph
calculation, which came out at 10 square feet for allowable antenna area,
A VERY ROUGH APPROXIMATION for finding the allowable
antenna area at other wind speeds is to assume that the entire tower can
be represented by an effective wind loading of "T" square feet
applied at the top. (We know that's a gross over-simplification, but since
we don't know anything about the tower at this point, it's the best we can
do.) Then we surmise from the manufacturer's 70 mph and 90 mph ratings the
(30 + T) x 70 x 70 x K = (10 + T) x 90 x 90 x K.
The equality is based on the assumption that the tower
manufacturer uses the SAME total force from the wind loading on the tower
plus antenna in his calculations at various different wind speeds. K is an
"arbitrary" constant that goes away, so we're not going to worry
about its dimensions or meaning.
Dividing both sides by K and collecting all terms with T
in them on the right hand side, we get
(147,000 - 81,000) = (8100 - 4900) x T
or: T = 20.63 square feet (effective tower wind loading
applied at the top).
Thus, at 70 mph, the total force on the tower and antenna
is K x (30 + 20.63) x 70 x 70, or K x 248,000. Similarly, at 90 mph, we
get K x (10 + 20.63) x 90 x 90, or K x 248,000.
Now we can calculate the allowable antenna square footage
(call it "F") at 80 mph:
(F + 20.63) x 80 x 80 x K = 248,000 x K
or: F = 18 square feet of allowable antenna area for 80
mph winds, versus the 20 square feet Bob would calculate by assuming it
was a linear curve between the figures for 70 and 90 mph.