At 12:45 PM 4/11/2005, Steve Maki wrote:
>Jim,
>
>Thanks for going through those calcs, it's much appreciated.
>
>Just a couple of questions:
>
>Why did you choose 60% guy anchors instead of the commonly
>recommended 80%? I agree though that 60% is probably in that
>"sensible" range, but just barely.
To make the math easy. 60% out is actually about 31 degrees.. I just
rounded to 30 degrees, so sin(30) = 0.5, cos(30) = 0.866
If you want to go out to 80%, the angle becomes 38.6 degrees
sin(38.6) = .62
cos(38.6) = .781
Revise all the numbers accordingly, if you wish... It will reduce the
download from the guys somewhat..
>I'm not quite sure how you so quickly came up with a number
>for base bending moment in the guyed configuration. It seems
>unlikely to me that it can be even 1/4 of the unguyed number,
>unless the guys were left slack.
I looked up a uniformly loaded beam with fixed end on one end and pin joint
support on the other. the moment is 1/8 w*l^2 compared to 1/2 w*l^2 for
fixed/free. FWIW, the same 1/8 w * l^2 applies for a pin/pin
support. (because it's really like two half length beams.. and the moment
goes as the square of the length for uniform loading).
>But even so, 5700 lbs compression vs. 19,000 compression
>is quite an improvement, don't you think?
Yes, but...
I'd still worry about the distribution of loads.
>I do note that you're still worried about increased stress on
>on the middle of the tower (column buckling). Are you speaking
>of straight downward force collapsing the middle of the tower?
Bending moment and downforce combined.
But this is just a static rigid body analysis. Haven't even looked at the
classic column buckling analysis (and a tapered column with an average
width of 1.5 ft, 64 ft high has a slenderness of 85... that's quite
slender). You'll note that stiffness never entered into any of my
calculations (or those of Rohn, except where they used the standard column
calculation for the short 1 ft structural segments)). You'd need stiffness
to calculate buckling load:
Pcrit = pi*E*I/l^2 where E is modulus, I is moment of inertia of the
section, l is the length.
>Or a bend, where the downward force helps it along once it gets
>started?
That's the classic buckle... The column deflects in the middle, which
causes the force at the end points to point somewhere other than towards
the anchor, reducing the force resisting the buckle. If the column is
springy, then the force at the ends is a combination of the compressive
force (along the column) as well as a force at right angles to the column
(the bending moment springiness). The vector combination of those two
forces has to be bigger than the applied load, or it will collapse.
Think about this example:
Consider a 3" diameter, 6" long aluminum tube, 1 mil wall thickness. (A
extra heavy wall coke can)... It can support a 200 pound load axially
applied, without too much trouble. (the stress is about 22 kpsi)
Now take the same amount of aluminum ( about 0.009 square inches cross
section) and turn it into a single rod about a tenth of an inch in
diameter. Try and support that same 200 pound load. It will fail.
What's the difference? The compression load on the aluminum is exactly the
same in both cases. The difference is that the can is a LOT stiffer than
the rod.
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