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Calculating wind turbine tower loads - Page 7

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Posted by Ron Rosenfeld on February 25, 2009, 8:45 pm
 
On Wed, 25 Feb 2009 11:36:39 -0600, david.williams@bayman.org (David
Williams) wrote:


Well, that depends on the sizes.  Although I agree that the large
commercial turbines, that I've seen, are all on unguyed towers.

My Bergey-XL is on a guyed lattice tower, though.  It has a 7m diameter
rotor; it's on a 30m tower, with two sets of three guys -- one at 27 m, and
the second at 14.8m.
--ron

Posted by Curbie on February 25, 2009, 8:52 pm
 
Oh crap, I thought I had it until reading your reply!

I thought (from previous examples), but expanding using your example:

"For example, suppose there are three guys, each at 45 degrees from
horizontal, and at compass bearings 0, 120, and 240 degrees from the
tower. Suppose the wind is blowing from bearing 60 degrees, exerting
500 lbs force on the turbine. The additional downward force produced
by the tensions in the two guys at 0 and 120 degrees will be
500 x sqrt(2), or about 707 lbs, making a total downward force on
the tower base of 1707 lbs. The tensions in those two guys will each
be 500 lbs. The third guy, at 240 degrees bearing, will be slack."

That the additional vertical (downward) force on the mast from a 60
degree windward breeze placing a 500lb horizontal load on the turbine
would add 500lb to the base loading (through the guys and mast) and
707lbs of additional (45 degree) guy tension divided equally between
guys at compass points 0 & 120?

How can guys at compass points 0 & 120 have 707lbs each of tension
without exerting an additional 707 x 2 vertical load on the base?

Thanks for the reply, I'm not trying to argumentive, I'm just confused
now?

Curbie

On Wed, 25 Feb 2009 10:13:40 -0600, david.williams@bayman.org (David
Williams) wrote:



Posted by daestrom on February 25, 2009, 10:48 pm
 

I'm not sure David's numbers are right.

First, looking down from above we have a force acting at 0 degrees and a
force acting at 120 degrees that have to add up to a net force of 500 lbf
acting at 60 degrees (opposite the direction the wind force is acting).
This forms a nice neat equalateral triangle with each leg equal to 500 lbf.

Now, for a guy wire that is 45 from the horizontal to generate a reaction
force of 500 lbf horizontally, it must have 500 lbf / sin(45) of tension 707
lbf.  The third guy will be slack.

A tension of 707 lbf in the guy will create 500 lbf in the horizontal
direction from the tower and also 500 lbf downward.  With two guys, the
total downward force is double that or 1000 lbf.  Add to that the static
load mentioned and you have 2000 lbf total downward.

The issue is that the two guys are pulling 'against' each other horizontally
if you look at it from above, but pulling 'together' downward if you look at
it from the side.

daestrom
P.S.  Hopefully I got it right this time, I've worked it out three times and
gotten three different answers :-/


Posted by Curbie on March 2, 2009, 10:37 pm
 On Sun, 01 Mar 2009 20:51:31 -0600, david.williams@bayman.org (David
Williams) wrote:


Oh please do!

I'm finally coming to realization that the use of angles in the guys
create new forces which add together on the design elements of tower.
The sharper the angle, the greater the newly created force. I'm still
chasing around the forces in the previous round of examples, but would
like to see an example of wind blowing from bearing 90 degrees.

Thanks.

Curbie

Posted by ralleyrat on March 2, 2009, 11:51 pm
 
The guy wires add a compression, (or 'down') load to the mast.
Their tension takes side loads to the ground.
Stress on the anchor points is a resultant vector.
Draw the triangles and figure it out.

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