Fitting Factor and Fail Safe Conditions 1

[inertia4u]

Just a quick question.

Assume I have a fitting that is designed to carry both primary and fail safe loads.

Also assume that a 1.15 fitting factor is deemed appropriate for this fitting for ultimate analysis.

The question is: Is it necessary to use a fitting factor for the fitting when performing fail safe analysis?

If you could explain the answer to me I would greatly appreciate it.

Thanks,

Nert

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Nert

 

[Sparweb]

If this is a normal category aircraft, the answer is very clear-cut.

Quoting from FAR 23.572

[ul]
...2) A fail safe strength investigation in which it is shown by analysis, tests, or both, that catastrophic failure of the structure is not probable after fatigue failure, or obvious partial failure, of a principal structural element, and that the remaining structure is able to withstand a static ultimate load factor of 75 percent of the critical limit load at VC. (These loads must be multiplied by a factor of 1.15 unless the dynamic effects of failure under static load are otherwise considered....
[/ul]
I don't think there's any question.

If this is a transport category aircraft, then you'd better read .571 carefully, because fail-safe is only a part of a broader Damage Tolerance requirement. If this is going into a helicopter, inspection requirements are bundled up with it, too.

I would expect that in all cases, fitting factors apply to limit loads, by definition. FAR's 23/25/27/29 .625 all say the same thing. "For each fitting whose strength is not proven by limit and ultimate load tests..."

There's no avoiding it, unless you're prepared to test your joint.

Where did the figure "1.15" come from, you may ask? That, my friend, may be lost in the depths of time...


"Simplicate, and add more lightness" - Bill Stout
Steven Fahey, CET

 

[inertia4u]

Dear Sparweb,
Thank you for responding to my post.

In regards to Far 25 a/c:

I'm still not 100% convinced that it is true that the fitting factor has to be applied to a failsafe condition.

For the failsafe condition, it is assumed that the primary lug has failed, and the failsafe lug must carry "some" percentage of the limit load. Isn't that "percentage" somewhat of a "psuedo" load anyway?

I never did see the bolded text that you give in the FAR 23 quote...

Thanks for your help,

Nert

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Nert

 

[Sparweb]

A "pseudo" load is one way to call it: 75% of the ultimate load isn't limit, isn't necessarily yield of anything, and is probably an arbitrary number. I suspect that the 1.15 fitting factor is also arbritrary, too, to take care of "slip" in bolted joints.

This topic, though, is purely intellectual, and the bare fact is that these are the numbers that the FAA has specified. The original type design conforms. Why can't you?

To delve deeper into this topic, read FAR 25.571 (and 23.13 for that matter). All AC's are available off the FAA website.

BTW: FAR 23.571 was updated recently to include that line - it wasn't there before.


"Simplicate, and add more lightness" - Bill Stout
Steven Fahey, CET

 

[inertia4u]

Sparweb,

Thank you *very much* for your input.

Some structure that I work on assumes 100% limit for failsafe, hence the reason for the use of the "some" nomenclature.

Your posts were extremely helpful to me and my peers, and I really appreciate it.

Thank you and Godspeed,

Nert


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Nert

 

[Sparweb]

Glad to hear it!


"Simplicate, and add more lightness" - Bill Stout
Steven Fahey, CET

 

[aerohead56]

An additional question on fitting factors. Does the 1.15 fitting factor apply to ultimate loaded primary/secondary structure that has more than one fastener in the joint, that is not tested to ultimate load, for FAR 25 certified aircraft?

 

[inertia4u]

Just an FYI:
I talked to our DER, and the fitting factor is *not* required for the particular fail-safe condition on FAR 25 A/C.

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Nert

 

[aerohead56]

What about more than one fastener in a joint in non-failsafe structure? Does the fitting factor apply?

 

[Sparweb]

Read the text carefully on this one, AeroHead. Different joints behave differently if one fastener lets go, and you must make the analysis fit the intent of the rule. This is the factor that takes care of bumps, scratches, shakes, and inaccuracies in all of the other parts of the system that create the particular load at the particular joint of interest. In other industries, the factor is 5.0. Count yourself lucky (planes wouldn't fly otherwise).

Inertia, are you able to elaborate a little on why the FF doesn't apply to your case? I wouldn't like for your statement to be misinterpreted.



"Simplicate, and add more lightness" - Bill Stout
Steven Fahey, CET