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Posted by Ron Rosenfeld on February 23, 2009, 8:11 pm
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On Mon, 23 Feb 2009 18:29:22 GMT, "vaughn"
>Many guyed
>radio towers actually come to a point where their base contacts the
>foundation, and are only secured there by a pin which merely serves to keep
>the base from slipping off the foundation.
My wind turbine tower has that system, too.
--ron
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Posted by bw on February 24, 2009, 8:48 pm
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>
> At a 45 degree guy wire angle, it just happens to work out that whatever
> the side load of the wind is, the additional downward force caused by
> acting through the guy wire is the same amount. 100 lbf wind load
> horizontal gets turned into 100 lbf *additional* down force on the base.
>
> And the wire tie down has a similar force. The 141 lbf diagonal force in
> the guy wire will pull *up* on the base with 100 lbf and *sideways* on the
> guy (toward the tower of course) of 100 lbf.
>
> daestrom
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.
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Posted by Curbie on February 25, 2009, 12:09 am
Please log in for more thread options daestrom,
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"
>daestrom
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Posted by Jim Wilkins on February 25, 2009, 7:41 am
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> Thank you so much for your time and patients with me.
> On to Critical Buckling Load of a columns.
>
> Curbie
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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Posted by Curbie on February 25, 2009, 9:14 am
Please log in for more thread options Jim,
Sounds like you're where I want to be, metaphorically, spend a little
time on Euler, spend a little time on logs. Bliss.
I found the kind of site I like (almost) yesterday, its has an example
with not only the explanations to the math, but answers to check the
equations.
https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec=09.1&page=case_sol
I felt the whole thing was making perfect mathematical sense until the
last equation (the "almost" above), I was getting 7,955 for Pcr and
the example said I should get 1,593. I tore my spreadsheet expression
apart Pcr =PI() ^ 2 * E * I / L ^ 2 but couldn't find my problem.
Cylinder Outside Diameter (do) 1.5 Inches
Cylinder Inside Diameter (di) 1.4375 Inches
Cylinder Wall Thickness (Tc) 0.0625 Inches
Cylinder Radius (Radius) 0.75 Inches
Cylinder Outside Radius (ro) 0.75 Inches
Cylinder Inside Radius (ri) 0.6875 Inches
Moment of Inertia (I) 0.073043851
Length of Mast (Lm) 4 Feet
Length of Member in Feet (Lf) 4.272 Feet
Length of Member in Inches (L) 51.26402247 Inches
Young's modulus (E) 29,000,000 PSI
Factor of Safety (Fs) 2.00
Critical Buckling Load (Pcr) 7955 Lbs.
Will hit it again today.
Thanks for help and advise on "statics", I read the site and made
notes, haven't quite figured out where it fit in, but I'm sure to run
into "it's place" in the scheme of things.
Curbie
On Wed, 25 Feb 2009 04:41:48 -0800 (PST), Jim Wilkins
>>
>> Thank you so much for your time and patients with me.
>> On to Critical Buckling Load of a columns.
>>
>> Curbie
>
>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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