Static load equal 29.865

[tecpilot]

As known the JAR 29.865 describes external loads specification and design. I'm now in discussions how to interprete the static load in case of human external loads. A 3,5 load factor to the maximimum external load for which authorisation is requested. OK, but our CAA means that on a 1000kg approved and certified load hook, load attachment point or connection part produced by an external manufacturer our load is to cut to 285kg. Thats not my opinion because i think the external manufacturer was controlled to design and test the 1000kg hook or part to a load factor of at least 2,5 means 2500kg. Therefore our (human)load is right up to 714kg.
Additionally the new CAA inspector would love to add a 1,5 "general safety" factor. Means 3,5 + 1,5 = 5,0 and a limited human load from max. 200kg on a 1000kg certified part. The whole system is redundant of course and consists of two hooks and 2 lines. But the discussions turning around the max permitted human load to one system (in case of other system failure).
Any advices?

[Genghis the Engineer]

Tecpilot, you don't make your case all that clearly, but I'll have a stab - come back if I've answered the wrong question.

Lets start with the actual document, I've pasted this in below for reference.


JAR 29.865 External loads
(a) It must be shown by analysis, test, or both,
that the rotorcraft external load attaching means for
rotorcraft-load combinations to be used for nonhuman
external cargo applications can withstand a
limit static load equal to 2·5, or some lower load
factor approved under JAR 29.337 through 29.341,
multiplied by the maximum external load for which
authorisation is requested. It must be shown by
analysis, test, or both that the rotorcraft external
load attaching means and corresponding personnelcarrying
device system for rotorcraft-load
combinations to be used for human external cargo
applications can withstand a limit static load equal
to 3·5 or some lower load factor, not less than 2·5,
approved under JAR 29.337 through 29.341,
multiplied by the maximum external load for which
authorisation is requested. The load for any
rotorcraft-load combination class, for any external
cargo type, must be applied in the vertical direction.
For jettisonable rotorcraft-load combinations, for
any applicable external cargo type, the load must
also be applied in any direction making the
maximum angle with the vertical that can be
achieved in service but not less than 30º. However,
the 30º angle may be reduced to a lesser angle if—
(1) An operating limitation is established
limiting external load operations to such angles
for which compliance with this paragraph has
been shown; or
(2) It is shown that the lesser angle
cannot be exceeded in service.
(b) The external load attaching means, for
jettisonable rotorcraft-load combinations, must
include a quick-release system to enable the pilot to
release the external load quickly during flight. The
quick-release system must consist of a primary
quick-release subsystem and a backup quick-release
subsystem that are isolated from one another. The
quick-release system, and the means by which it is
controlled, must comply with the following:
<snip>.

The term used is "limit static load" of 3.5g, in other words, this is the minimum load which the system is designed to take in normal use - given snatch loads, etc. that seems not all that unreasonable.

It's important to appreciate what "limit static load means" in civil certification speak the system must take that load indefinitely. there are higher loads, commonly called "proof loads", which are those loads which the system must be able to take for at-least 3 seconds. The standard safety factor is given in 29.303...

JAR 29.303 Factor of safety
Unless otherwise provided, a factor of safety of
1.5 must be used. This factor applies to external
and inertia loads unless its application to the
resulting internal stresses is more conservative.

This means that any stress analysis, plus any load testing (it's fairly normal in such a case to do both) must show that for at-least three seconds the system will take 1.5 x 3.5 = 5.25g. This is mandatory, and means that for a 1000kg underslung load, you need at-least a 5250kg strong system.

However, it's not that simple; there are other factors to be considered. For example if your hook is manufactured from a casting, there will be an additional safety factor of somewhere between 1.25 and 2.0. There will also be a 1.15 fitting factor (29.625(a)) to cover, for-example, the junction of cable to hook - that brings you up to 5.25 x 1.15 = 6.04g.

JAR-29 also includes a 2.0 factor for control cables; now this isn't a control cable, but I'd check with your CAA whether they want this included, it wouldn't be unusual.


So, you are looking at a 5.25g+ system, but probably greater for some parts of it. My recommendation would be to (for analysis purposes) split the system up into component parts, and determine the applicable safety factors for each part - that way you can optimise the design and avoid an excessive safety factor for any particular part. Of-course if you do it that way, your sample load test will need to be compartmentalised.


With regard to you proprietary hook system, you really need to find out where in this process the item lies w.r.t the 1000kg limit, there is no guarantee about what approach was taken by the manufacturer. But if you know nothing to the contrary, you'll have to take your CAA's view.


Does that help at-all?

G

[tecpilot]

Thanks Genghis,

for your words. You are on the right way. I've splitted the system up into component parts some months before, as you advised.

To get the whole system breaking load Fb=[Cw+Hl] x Lf x Sf x Tf

Fb=Breaking Load

Cw=Components Weight
Hl=Human Load
together = 400kg / 3,92 kN

Lf = Load Factor = 3,5
Sf = Safety Factor = 1,5
Tf = Textile Factor = 2 (Aging and worn out factor of included textile lines and belts)

Fb =[3,92]x3,5x1,5 x 2 = 41,16kN

Ok up to this point only mathematics. All included metall parts must have breaking loads greater than 20,58kN, textile parts needs 41,16kN.

I splitted the parts and the system himself is safe, also in the view of our CAA. The problem is, the main system is hooked into the a/c air frame load hook, approved by STC and FM to 1000kg means 9,8 kN. Not enough to the CAA. Unfortunately no other papers to the hook available. The redundancy systems is attached on the airframe and ok to the CAA.
The problem is the hook. But as i wrote before, in my opinion, if the hook is approved to 1000kg on a FAR/JAR 29 helicopter, he must be certified with a load factor of at least 2,5 and could therefore count as a part with 24,5kN breaking load, greater than the needed 20,58kN. With the certification of the hook on a 29' helicopter the CAA should have accepted the hook with a breaking load of at least 24,5 kN and should therefore also in our case the hook count as described.

The system is designed to carry 3 to 4 persons. It makes no sense to cut it to only 1-2 persons because a 1000kg hook, usually used to transport 800-900 kg nonhuman load isn't safe enough.

[Genghis the Engineer]

If there's no supporting documentation on the hook, I can see that you've got a problem. Can you get hold of the original STC from whichever authority approved it.

If, for example, it was a UK-CAA approval (in which case the document is called an AAN - Airworthiness Approval Note) the approval documents are all available on their website at www.caa.co.uk

G