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Calculating wind turbine tower loads - Page 3

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Posted by Curbie on February 24, 2009, 5:14 pm
Point well taken.

I have lived in south Florida for the past 25 years (moving soon) and
have personal experience with hurricanes, building on swap land, and
ground water 10' below the land/fill's surface. Must have been a
pretty challenging job. ("fun with pillions")

I'm not an mechanical engineer, but if it doesn't make sense in a
spreadsheet it probably won't function well, and if it does it's dumb
luck. (I'm not very lucky)



On Tue, 24 Feb 2009 16:35:20 GMT, "vaughn"

Posted by Ron Rosenfeld on February 24, 2009, 1:11 am
On Mon, 23 Feb 2009 18:29:22 GMT, "vaughn"

My wind turbine tower has that system, too.

Posted by bw on February 25, 2009, 1:48 am

That looks right. Torsional forces MUST balance of course. I wouldn't let a
novice do it, there can be resonances if not careful.
"Marks" handbook has a good readable section on tower loads. I think that
HAWT are approximated by using a drag coefficient equal to that of a flat
disk of the same diameter. I think that the ARRL antenna handbook also has a
section on towers.

Posted by Curbie on February 25, 2009, 5:09 am

Excellent explanation!

Basically, if tower weighs 500lbs and turbine weight 500lbs then there
is 1000lbs of static (or gravitational) force on the tower base. If a
moment of wind exerts a horizontal force 500lbs on the turbine the
guys exert an additional 500lbs on the tower base for a total equal to
1500lbs of static force. Both guys and anchors will receive 500 x
1.414 = 707lbs diagonal tension.

For the home built user/builder there are a lot of good (hopefully)
manuals out there. They are detailed in saying do this, that, and the
other thing based on the author's practical building experience. Bless
them for the help. Checking or double-checking their math doesn't cost
much, crashing a turbine will. Cross-checking their advise will give
me less to worry about in a pretty important investment.

Thank you so much for your time and patients with me.
On to Critical Buckling Load of a columns.

CurbieOn Tue, 24 Feb 2009 20:04:22 -0500, "daestrom"

Posted by Jim Wilkins on February 25, 2009, 12:41 pm
The math there is strange and non-intuitive (except to Euler) but it
really works. I made some 2x4x12' shear legs to lift logs onto a
sawmill and overloaded one side right to the critical point. As
predicted the leg bent and was stable at any degree of curvature. I
could bend or straighten the center of it with one finger even though
it was supporting a 350 Lb log.

Jim Wilkins
keeping l/d<50

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