[TowerTalk] Real numbers for Rohn BX-64 Re: guying

[Steve Maki]

Jim Lux wrote:

>>>> If I understand  him correctly, putting some side force on the
>>>> middle of my tripod will cause unequal scale readings, and I
>>>> don't understand why.

>>> ....because the tower can deflect and the guys are at the top.
>>> 
>>> That will lift some leg(s) and compress others more. My point is
>>> it never seems to get worse than without the guys.

>> Yes of course, but he came up with a number WITHOUT knowing the
>> deflection. That was my whole point. It seems to me that if you
>> hold the top still, and the middle does not bow, then you have pure
>> shear force at the bottom, and the three legs maintain equal
>> compression.

> Think of the problem as a single horizontal 10 ft long beam, rigidly
>  attached at both ends, for the mean time.
> 
> Let's just consider the forces from the weight in the middle
> 
> The beam's attached firmly to the wall at both ends
> 
> You've got a 10 pound weight at the midpoint
> 
> You've got a 5 pound shear force acting up at each end (so the
> overall ups and downs match).
> 
> But, there's also a moment (torque) trying to bend the beam down in
> the middle. If the beam is L feet long, and the weight (W) is at "a"
> feet from End A and "b" feet from the End B (i.e. L=a+b) then the
> moment at end A is MA = W * a * b^2/L^2, or 50/4 = 12.5 ft lb.  Since
> it's symmetric, the moment at End B is exactly the same, but in the
> reverse direction.
> 
> ---
> 
> In your case, you've got one end fixed to the wall (call that End A)
> and the other end supported by a free pivoting joint (pin joint)
> (call that End B).  Again, let's just consider the forces and moments
> from the weight in the middle.
> 
> The reaction force at the free end is 10*5/16 (=3.125) pounds, the
> reaction force at the fixed end is 10*11/16 = 6.875 pounds.
> 
> The moment at end A is W * (L+b)/2 * (a/L)*(b/L) = 10 * 
> (10+5)/2*(1/2)*(1/2) ft lb = 18.75 ft lb.  (you can get there by
> summing the two moments.. 3.125*10 = 31.25 ft lb from the end and
> 10*5= 50 ft lb in the middle acting in the opposite direction, for
> 18.75 ft lb net)
> 
> If the base is 3 feet wide, the moment of 18.75 ft lb will correspond
> to a force pulling away from the base of 6.25 pounds on one side, and
> pushing with 6.25 pounds on the other side.
> 
> --
> 
> The load at the top of the tower, resisted by the guy, doesn't add
> any net moment, since the up and down match (looking at the tower 
> horizontally).  However, since the guy now needs to provide the force
> to counter the 10 pounds at the top plus the 3.125 pounds from the
> load in the middle, there's now a 13.125 pound down force in the
> tower. (The guy tension is 18.6 lb).  The downforce will divide
> evenly between the two legs (Yeah, you said it's a tripod, but I'm
> calculating for a two legged thing, to keep things planar)
> 
> 6.5625 pounds in each leg
> 
> if you add in the forces from the moment at the fixed base,
> 
> The leg towards the guy will have a force of 6.5625-6.25 pounds
> (about 0.3 pounds).  The leg away from the guy will have a down force
> of  6.5625+6.25 pounds (about 12.8 pounds).
> 
> Both legs together have a shear load of 6.875 pounds.. That, plus the
>  13.125 horizontal force from the guy adds up to 20 pounds, which is
> the total load from your pair of 10 pound horizontal loads.  That
> balances.
> 
> It all balances.. the net of the downforces is equal to the total
> imposed by the guy (13 pounds).
> 
> Looking at the moments at the base.. 0.3125 pounds on a 1.5 ft lever
> arm (back foot), -12.8125 pounds on a 1.5 ft lever arm (front foot),
> 10 pounds on a 5 ft lever arm (middle load), and -3.125 pounds on 10
> ft lever arm (net of the horizontal guy load and the 10 pounds)...
> they all sum to zero
> 
> (of course, the back leg will tend to lift...)

Jim,

OK, I thought about that for quite a while (not all day
though - I actually worked today!). Went to a couple
of web sites that seemed to offer free physics lessons.
Drew some diagrams, got a headache but it started make
sense, then checked it with one of my down home
experiments using a step ladder for a one dimensional
tower. Of course the ladder flexes, so I couldn't really
test the concept of a perfectly rigid beam.

What it boils down to is that with just a top guy on
a relatively tall structure, the leg forces are not
as balanced as I expected - that's for sure. BUT, it
still seems to be a large improvement over the unguyed
state for  typical self supporters.

I'd still be interested in any other data points you
can think up related to middle tower components that
turn out to be more stressed in the guyed state. I can't
imagine what they could be - but I guess imagination
shouldn't be a part of this!

Steve K8LX