Jim Lux wrote:
>> 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).
I can't quite get my head around this. Without knowing how
much the top end moves, or the middle bows, how can you
know the bending moment? If the top does not move at all,
and the tower is perfectly rigid - is there still a base
bending moment? I would think there is only horizontal
shear force in that case (and straight downward force).
>> 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.
> 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.
Sure. Easy to see. I'll bet though that even the thin rod, in any
height which self-supports, can benefit from guy wires.
Steve K8LX
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