[TowerTalk] Fiberglass Insulator/Torque arms

[Kurt Andress]

K7LXC@aol.com wrote:

>       AM broadcast stations use them all the time but they're just part of the
> guying hardware - a big insulator. I think the smallest size they come in is
> around 15,000 pounds breaking strength. Should be plenty for typical ham use.
>
>       A 'torque arm' would have to be rigidly connected to the tower to do any
> good. These insulators just have shackles on the end. Like my grampappy used
> to say, "couldn't hurt".
>

Yup, better than wrapping guy wire around tower legs for torque resistance.

Because of the stiffness deficiency of the glass composite, one would need a much
larger insulator to equal the standard Rohn 45 torque bar.

If the insulators are made from approx 50% axially oriented glass fiber
construction, it would only take  a 3/8" - 7/16" diameter , at the narrow diameter
sections of the convolute insulator, to carry a 15000 Lb tension load. If the
insulator is compression molded from random fiber bulk molding compound, the cross
section would need to be approx. doubled.
Mark tells me that the glass insulators he was looking at are 1" diameter. If they
were G-10 or equivalent, the strength is about 35-45000 psi strength and a elastic
modulus of around 3.5 MSI (Millions of Psi). Good rolled mild steel has about the
same strength (usually the lower end) but has an elastic modulus of 29 MSI.

Overall effective stiffness of a component is defined by  E x I. E = the modulus
of elasticity (stiffness) of the material. I = the moment of inertia (or section
modulus) determined by the shape and ammount of material in the cross section.

First calculate the Moments of Inertia of the two sections:

Rohn 45 standard torque bar, in the horizontal plane (that's the one that combats
torque) :
Dim's: Thickness = .3125"  Width = 1.25"
Moment of Inertia for rectangular solid section = bd^3/12 = (.3125 x 1.25^3)/12 =
.05086 In^4    where d dimension is the axis of concern.

Solid round section, 1" diameter:
Moment of inertia in any direction (all the same) = (Pi x d^4)/64 = .049d^4 = .049
x 1^4 = .049 In^4

Next, we multipy the Moments of inertia by the elastic modulus of the materials:

Rohn 45 torque bar:    .05086 x 29 MSI = 1.4749

Fiberglass rod:            .049 x 3.5 MSI = .1715

Result:
The Rohn torque bar is:    1.4749/.1715 = 8.6 times stiffer than the fiberglass
rod.

Options:
#1    Go find a fiberglass insulator that is 8.6 times stiffer (in section
configuration) to make up for the material stiffness deficiency.

#2     Use the Rohn torque bar with an insulator near the torque bar connection,
like Rohn said (Prime Directive #1).


Like steve said, the insulator would need to be rigidly connected to the tower, to
even begin to work.
If it was my tower I'd put the torque bars on it!

73, Kurt

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