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Calculating wind turbine tower loads
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Calculating wind turbine tower loads Curbie 02-23-2009
Posted by Jim Wilkins on February 25, 2009, 1:20 pm
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> ...but couldn't find my problem.
>
> Cylinder Outside Diameter (do) =A01.5 =A0 =A0 Inches =A0 =A0 =A0 =A0 =A0
> Cylinder Inside Diameter (di) =A0 1.4375 =A0Inches =A0 =A0 =A0 =A0 =A0
> Cylinder Wall Thickness (Tc) =A0 =A00.0625 =A0Inches =A0 =A0 =A0 =A0 =A0


Posted by Curbie on February 25, 2009, 4:56 pm
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 Jim,

Well, Cylinder Outside Diameter (do) & Cylinder Wall Thickness (Tc)
are just input to feed radius calculations, which in turn feed the
Moment of Inertia (I) calculation =PI() / 4 * ((ro ^ 4) - (ri ^ 4))

Since the results from the Moment of Inertia (I) calculation matches
the results given in the example I presume that's not my problem. The
only calculation which is not producing the same results as the
examples is for Critical Buckling Load Pcr=PI() ^ 2 * E * I / L ^ 2

I don't expect you find my problem, thanks for trying though, I
thought something might jump out at you (mathematically or
syntactically) from the Critical Buckling Load (Pcr) expression.

I'll just keep hammering away at it, it's what I do, learn the
properties of thing long enough to define them in computer equation
then move on and let the programs do their work.

Thank for your help.

Curbie
On Wed, 25 Feb 2009 10:20:40 -0800 (PST), Jim Wilkins

>> ...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          


Posted by Jim Wilkins on February 25, 2009, 5:41 pm
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> Jim,
>
> Well, Cylinder Outside Diameter (do) & Cylinder Wall Thickness (Tc)
> are just input to feed =A0radius calculations, which in turn feed the
> Moment of Inertia (I) calculation =A0 =3DPI() / 4 =A0* ((ro ^ 4) - (ri ^ =
4))

If the wall thickness is 0.062, the radii are 0.750 and 0.6875. 1/2 of
1.437 is 0.71875.

Posted by daestrom on February 25, 2009, 4:45 pm
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> 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

Wouldn't that be .03125?? (1.5-1.4375)/2 There are two walls

I run into this a lot with heat exchanger fluid calcs. I get the ID and the
wall thickness and have to calculate the OD to figure out the volume of
water displaced in the shell side.

daestrom


Posted by vaughn on February 25, 2009, 9:33 am
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> 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.

More that that! Don't forget the down thrust from the tension on the guy
wires, (and I suppose, even the weight of the guy wires). If you insist on
the guys being banjo-string tight, you could add thousands of pounds of
stress.

For my personal ham radio towers, I always went for a bit more sway, and
a lot less guy tension. With the big communications towers that I managed
at work, I listened to the engineers and insured the guy wires were tuned to
the specified stress.

Vaughn



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