At 04:07 PM 4/11/2005, Steve Maki wrote:
>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).
Yes, there's still a moment. If it's perfectly rigid, there's no deflection.
A moment is a just a torque (i.e. a rotational force, as opposed to a
translational force, like a shear or compression or tension). In static
analysis (which this is), the torque must be resisted by an equal and
opposite reaction force from the the support (the ground, in this case). A
pin joint cannot have any moment, because it pivots.
> >> 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.
It's not entirely clear.. You'd have to compare the bending strength of the
aluminum against the column loading critical load. 6" long by 0.1" diameter
is a slenderness ratio of 6/0.05 = 120 (which is quite slender). There's a
shorthand calculation using buckling coefficients (DIN 4114).. I have a
table which gives the buckling coefficient for Aluminum (2017) as 7.57...
this means that the buckling stress is about 1/8th of the permissible
compressive stress. The compressive stress for that 0.1" diameter rod
corresponding to failure would be around 270 pounds (I used 30ksi for the
yield). If your guys exerted more than 35 pounds of downforce, you'd
buckle the column. (anyone want to try balancing a 35 pound weight on a 6"
length of 0.1" diameter rod and verify my little handbook table?)
Looking at bending stress.. the section moment of the solid rod is
pi*d^4/64 = 4.9E-6 in4
30ksi = M*(d/2) / I = M * 0.05/4.9E-6
30E3 lb/in2 * 4.9E-6 in4 / 0.05 in = 2.94 in lb for failure in bending
That's about a 6 pound lateral load at the top of your mighty 6" mast.
Say you use 80% base guys...(4.8" out) sin = .624, cos=.781 Assume no pre
tension, the same 6 pound load.. the guy tension to resist the 6 pounds
will be 6/.681= 8.81 lb.. the down force on the column will be 6.88
pounds... since we previously calculated that the column load should stay
less than 35 pounds, it looks like guying is a winner for this application.
Also bear in mind that solid aluminum rods aren't a very good model for
lattice work antenna structures (nobody is building 60 ft aluminum towers
using a 1 ft diameter bar...) There's a lot of scaling here that would
need to be taken into account (the whole 30 ft high ant would collapse
under its own weight problem because mass goes as the cube and strength
goes as the square problem)
Also, bear in mind that the idea of guying is to increase the load
capacity. In our example above, the whole idea would be to improve the
system with guys (i.e. allow it to withstand more than 6 pounds of lateral
load)... otherwise, why go to the trouble of guying.
Having looked at the numbers for BX, it kind of looks like the windload on
the structure is actually more important than the windload on the antenna.
(if only because there's a lot more square feet of structure than antenna).
There's also the issue of using multiple sets of guys, which breaks the
column up into a series of shorter columns. In our thin aluminum rod
example, if you hold the center of the rod from moving, then some of the
loads trying to bend it will be resisted (you've basically made it two
short columns).
This is where all those guying schedules with all the multiple tiers of
guys come from. If you put enough sets of guys in, then the column
doesn't have to be as stiff. Stiffness comes from adding material (i.e.
wall thickness for tubes), so you trade off weight of the tower (and cost
of materials) against complexity of guying. And, of course, the desire of
most manufacturers to not make an arbitrarily large number of different
sections.
Fascinating stuff.. but of little practical use...
Either you're subject to regulatory constraints, in which case you need to
get a "real Engineer" to do a real analysis, or you aren't, in which case
you can do pretty much whatever you're comfortable with, as far as
overloading the tower.
Now... back to my antenna project... does anyone have a good handle on
matrix math to deembed a series of T matrixes from measured data between
multiple ports?
Jim, W6RMK
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