The "few" is actually the "many"!
13 ft.lbs times inertial load factor (3.8g)* = 593 ins.lbs
Times 1.5* = 889 ins.lbs ultimate
Times 1.5* = 1334 ins.lbs ultimate with nominal factor for non-interference
fit/potential for light hammering.
2no. 1/4" pins bearing on 0.063" wall of 1.5" dia tube generates a
bearing pressure of 27,587 psi.
This should be well within the bearing strength of 4130 N.
Hence, in my very first contribution to this thread I caveated my response
that I could not get the tube to fail by means of calculation.
Nevertheless, they do fail and, as Bob Harrison has pointed out, it is not
the 4130 tube that fails. Rather, the stainless tube instead. All we need
to do now is find out what grade it is.
(* Ref. Bruhn "The Analysis of Flight Vehicle Structures", JAR-VLA and
others).
Duncan mcFadyean
On Saturday, January 19, 2002 4:30 AM, Fred Fillinger
[SMTP:fillinger@ameritech.net] wrote:
> Indeed true, and I see where I made it sound like the mass balance has
> no effect, but I measured it. It takes only about 13 foot-pounds to
> lift the balance weight. That's about 1/20th of that required to
> elongate
> two 1/4" holes in 4130 at least, to the point of noticeable play, in
> my test. Maybe 13, reflecting inflation, is still a "few?" :-)
>
> Best,
> Fred F.
>
> > The "few pounds of rotational force" is the weight of the
counterbalance
> > times the length of the arm it sits on. Same difference in trimmed
flight
> > where any air-load eccentricity is reacted by the effect of the trim
tabs.
|