Hi gents,
My barbershop comments... :)
I would think that if there were say, 3 guys running to the same
anchor, that one would want to line up the anchor with
the "middle" guy wire. If there are two guys then I
would think the anchor would line up between the two
guys. This is because all the guy wires should be
tensioned the same, no one guy wire pulling more than
the others during low wind conditions, thus averaging
out the load.
It would be nearly impossible to predict dynamic wind
load on any one guy wire because wind speed is
not always the same at all levels of elevation. Thus, there
is a logical reason (other than mechanical "neatness") why
here are equalizer plates installed at the anchors.
BTW, quite a few years ago there was a QST article named
"Guys for Guys that have to Guy"...or something to that
effect...might be of interest to some of you.
73,
Charlie, N0TT
On Wed, 05 Sep 2001 22:25:30 -0400 "Michael Rauh"
<michaelrauh@hotmail.com> writes:
>
>Jerry, Stan, Dave, Tom,
>
>I agree with Jerry and Stan - the angle is probably not critical.
>Rohn
>presents the angle to 0.1 degree in their guyed tower drawings. This
>seems
>impossibly precise to me - there is probably way more sag or bow in
>the guy
>anchor than that.
>
>I also suspect that Dave is correct when he suggests taking the vector
>sum
>of the loads on the guys when the tower is at its rated wind load to
>find
>the guy rod angle.
>
>In thinking about this problem I realized I don't know how to
>calculate the
>individual loads on each guy wire. I use a rigid body model for the
>tower
>that will give the total load on the guys, but won't tell me how the
>guys
>share the load, at least as far as I can see. To figure out the load
>on
>each individual guy, I think I will need to consider how the tower
>deflects
>under load.
>
>Consider a tower with two sets of guys and an antenna on top. I could
>start
>by assuming the tower is fixed at the top and the base. I calculate
>the
>wind load and how the tower will deflect under that load. Draw a
>curve,
>deflection vs. load. Then assume the lower guy is attached at
>mid-point,
>draw a curve of elongation vs. load. The load on the second guy is
>where
>the curves meet. The reminder of the load on the tower, including the
>
>antenna load, is absorbed by the top guy.
>
>But wait! The top guy elongates under load too, relieving the load on
>the
>bottom guy. So I will have to set up a system of simultaneous
>equations in
>order to get the loads on each guy. I will need equations for the
>elongation of the top guy, the bottom guy, the deflection (bowing) and
>the
>pivoting of the tower. If the tower has a pivot base that is all I
>need,
>but if the base is cast in concrete I might need an equation to
>account for
>the force needed to tilt the tower as the guys elongate.
>
>Does that sound right? Do you know of any references where this kind
>of
>problem has been worked out? I know Kurt Andress has found guy loads
>using
>a finite element program. I need to understand the principles
>involved
>before I resort to a program!
>
>I think it would be very cool to know how to find the loads on the
>individual guy wires in a tower system.
>
>Best Regards,
>
>Mike Rauh, NV7X
>
>Reply to mikerauh@ic.org
>
>
>
>
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