Re: I-Beam deflection/failure

Be careful:
When working with alluminium alloys, plastics and some other materials
different to steels it is necessary to deal with very tight fatigue
constraints. The fact is that alluminium, will end up their life after
millions of load cycles for these material have not got resistance after
infinite number of load cycles. So will do some plastics just after a few
load cycles. So fatigue and TBF, not only ultimate yield, should be one of
the major design criterion from structural point of view when handling such
materials.
Best regards.
Ignacio Simón Yarza.
Mech&electronic-electrical eng.

"Jeff Finlayson" <finlayson@xxxxxxxxxxx> escribió en el mensaje
news:11gadqodbn5eab@xxxxxxxxxxxxxxxxxxxxx
> Jeff Finlayson wrote:
>> Brian Whatcott wrote:
>>> Jeff Finlayson wrote:
>>
>>>> Well, ultimate strength is the main criteria in aerospace. But
>>>> ultimate
>>>> plastic analysis is used sparely..
>>>
>>> Hmmm....0.1% yield is the usual aviation design criterion in a
>>> material with a soft knee like most aluminum alloys.
>>
>>> But there are composites with essentially no plastic range - and
>>> they are used with a generous safety factor from UTS
>>> I expect it is this class that Jeff has in mind?
>>
>> No I'm speaking of metallics. Not breaking is the main concern.
>> Functionality and fatigue concerns make no yielding important.
>> Also what's critical depends on the safety factors on yield and
>> ultimate. For aircraft with the ultimate SF = 1.5 and yield SF = 1,
>> most metals have a lower allowable at ultimate. For unmanned
>> flight hardware, with an ultimate SF of like 1.25 yield can be
>> critical. This is linear elastic.
>
> Here's a simple example to show how the safety factors come
> into play.
>
> Am aircraft part loaded in uniaxial tension is made from Aluminum
> 7075-T73 *** per AMS QQ-A-250/12. The max stress in the
> part is 35 ksi at 70 F.
>
> Stress, S = 35 ksi
> Ultimate tensile strength, Ftu = 67 ksi (A basis) at room temp
> Yield tensile strength, Fty = 56 ksi (A basis) at room temp
>
> At yield (safety factor = 1.0):
>
> Yield stress, Syd = 1*S = 35 ksi
> Margin of Safety = Fty/Syd - 1 = +0.60
>
>
> At ultimate (safety factor = 1.5):
>
> Ultimate stress, Sul = 1.5*S = 52.5 ksi
> Margin of Safety = Ftu/Sul - 1 = +0.27
>
> This shows that ultimate is critical.


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