Yield Strength Analysis Before Fatigue Analysis 2

[TomFin]

Hi,
I'm constructing an S&N Fatigue diagram for an Al 6063 T1 part. I performed a FEA analysis and found the largest VonMises stress 17ksi located where I expected it to be. I'm expecting the bending stress to fluctuate in tension (+), so a book I have instructs to designate (Sigma Alternating) as 1/2 the amplitude of the stress (8.5ksi), which will be what I use to intersect the S&N plot correct?
The S&N diagram starts at N=10^3 cycl, Sf = .9 UTS = 19.8ksi.
According to the yield strength of AL at 13 ksi this part has has plastically deformed at 10^3. It is hard to believe that this part will endure 1000 cycles at 19.8 ksi before "failing"
Does this sound right? By the way I thought the part "failed" as soon as you pass elastic deformation.
Comments and suggestions appreciated

 

[desertfox]

Hi FinTom

I think I agree with you normally fatigue failures occur at stresses way below the yield stress.
However you say your FEA analysis gives a result of 17ksi in bending which is also above the 13ksi yield stress you quote.

regards
desertfox

 

[corus]

A part doesn't fail in fatigue because it's passed the yield point. Take a paper clip and bend it back and forth. It'll take several bends before it cracks even though it had permanent deformation after the first bend.

corus

 

[desertfox]

Hi corus

I see what your saying, I interpreted the post working on the basis that when you design a component one usually works below yield stress with a margin of safety.
I think the original poster considers this also has is last line states that once you pass elastic deformation he considers the component failed.

regards
desertfox

 

[metengr]

I believe that there is one problem with this approach. You cannot directly obtain N cycles to failure on a typical SN fatigue curve unless the mean stress is 0. Instead, you need to use the Goodman equation to calculate a value of stress amplitude using your mean stress of (sigma max + sigma min)/2

The Goodman equation is as follows;

sigma (a) = sigma (N) - [ 1 - sigma(m)/sigma UTS]


sigma (a)= stress amplitude for N cycles to failure under a mean stress condition

sigma (N)= fatigue strength based on N cycles to failure for a 0 mean stress (this is a value found on a typical SN curve)

sigma (m) = mean stress

sigma UTS = ultimate tensile strength

 

[corus]

desertfox,
Even if you pass the yield point in a component it doesn't necessarily mean it has failed. If you take a notch in a component then the stress at the notch may exceed yield but stresses every where else may be ok and the component won't fail. It's how you classify the stresses at a location. Unfortunately many structural design codes don't go into much detail with regard to stress classification and hence the confusion. Equally I've seen welded parts designed with the highest strength steel they could use when the design itself is governed by fatigue. The high strength steel in that case will be of little benefit when fatigue damage occurs at the weld.

corus

 

[desertfox]

Hi corus

I don't disagree with your sentiments but your now talking about local yielding in a particular component ie: notch, keway etc. All I am saying in general when your designing a component you design it so that your below the yield stress with a margin of safety. If you have access to:- Mechanics of Materials volume 1 by E.J.Hearn page 402 it states:-

Maximum Principle stress theory:- This theory assumes that when the maximum principle stress in the complex stress system reaches the elastic limit stress in simple tension,failure occurs.

Whilst this theory is not reccomended for ductile materials I believe this is were the OP is coming from with regard to his last sentence.


The OP gives no detail on his component except the material and a tensile bending stress of 17ksi which can only be the stress range (which he is assuming to be)if the component is cycling between zero and 17ksi.
I would be concerned if I had a design that showed the component was above the yield stress and then was subjected to a fatigue situation in the ame area.

regards desertfox

 

[TomFin]

First of all, thanks for the responses.
This is an aluminum flange which is basically acting as a cantilever beam with a notch radius. FEA gives me 17ksi at this notch, but all other areas in the stress plot are safe (Blue). So yes this is local yielding.
I agree that by definition for most design requirements: a part has failed as soon as it passes the elastic region, but the part may still function for a long time. So I'd assume the fatigue diagram (S&N) is not based on the Yield Strength but the UTS. Joe Schmoe could see a bending paperclip and think your insane if you told him the paper clip has failed and is useless. From my understanding as we repeatedly bend the paperclip we are strain hardening it until we climb up the top of the stress/strain diagram and hit the UTS and observe the paperclip necking (breaking).
Say for example this stress concentration at the notch I brought up was 30 ksi. Half of that for Sigma"a" would be 15ksi, which is 2 ksi above the yield strength and 4.8ksi below Sf in the S&N diagram. So from what I brought up earlier this part fails locally after the first cycle at the notch but continues to function. Thousands of cycles later a crack will propagate and Joe Schmoe will tell you its broke.
Does this sound sensible? Comments Appreciated...

 

[corus]

FinTom,
The SN diagram is based upon tests carried out on a polished component and is then usually factored to allow for surface finish etc. A relationship has been found between the UTS and the endurance limit (Endurance limit = 0.5 UTS for 7.10^6 cycles) but the general SN diagram isn't related to either yield or UTS, as far as know. If you read structural design standards such as the german structural DIN standard or the EN pressure vessel standard or BS 5500 then the stresses at a notch aren't limited by yield as they're classed as a peak stress component plus a primary or secondary stress. Only the primary or secondary stress components are limited to some factor of yield. Basically the overall stress may be above yield at a notch but the stress is contained within an elastic region and hence doesn't fail. The limits on peak plus primary/secondary in these standards are governed by the number of cycles required by the design and hence the stress in the SN curve. If in doubt refer to a suitable design standard as a reputable source.

corus

 

[bvanhiel]

Aluminum doesn't have an endurance limit.

1000 cycles is not that many. Fatigue analysis may not be appropriate, hence the weird answers.

-b

 

[desertfox]

Hi FinTom

I think we need to clarify something about this flange which is in bending, firstly the tensile stress you quote presumbly is about its neutral axis, in which case half of the flange would be in compression so your stress amplitude would be +17.5 to -17.5 and the mean stress would be zero.
At the present situation your mean stress is not zero but 8.5 which suggests your component in bending is only seeing a tensile stress between zero and 17.5.
If the latter is correct then the cycle life of your component will be adveresly affected as the mean stress is
greater then zero and metengr in an earlier post as touched on this.

Fatigue curves are usually drawn up from experimetal data (as mentioned by corus)which you may have to do for your component and a guide to fatigue testing can be found in BS3518 parts 1 to 5 if your in the UK.

Finally as the last poster as stated Aluminium doesn't have an endurance limit so your design would have to be for a finite number of cycles which you might already be aware of.

At this stage it might be worth considering this:-
Assume your intial findings are correct then is a 1000 cycles before failure of the part sufficient for your requirements and how long will the part last before a 1000 cycles are reached. Might it be better to reduce the stresses to a lower value in the area of concern and in the meantime try to obtain a fatigue curve for the loading application you require from a supplier for the material your using.

 

[NickE]

Also as an additional piece of info:

Fatigue as evaluated by the S-N curve requires stresses below the yield point. Fatigue failure is falure occuring due to stresses that vary wrt time. These stresses can be quite small for some materials. Fatigue occurs entirely within the elastic region, once you plastically deform the part it has failed. It doesnt matter if it still works or not, it is no longer shaped to the B/P.

The bending of a paper clip is not fatigue if there is a permanent change in the shape of the paperclip.

We are not Joe Schmoe, We are engineers. Some things are way more detailed for us. Kind of like how scientific theroies are way different than everyday theroys.

Oh and aluminium alloys do have an endurance limit, the stress level is really quite small though.

Nick
I love materials science!

 

[cbrn]

Hi,
I think that:

1- a component must first of all serve to a Function: if you plastically "gross-deform" it, this Function will not be fulfilled any more. That's why we "engineeristically" design below yeld point. Anyway, local yelding, as far as it is REALLY "local" (I don't insist here on what "local" means) leads to stress redistribution, so that the "gross" stress is still below yeld limit; a local yelding doesn't lead to component's failure (see for ex. ASME code...)
2- Voeller curves do are drawn from experimental data. They are useful for long-life fatigue predictions/verifications, but almost meaningless for low-cycle fatigue: for this last purpose, fatigue "under deformation control" should be used instead (along with some crack-propagation theory). Moreover, Voeller curves are clipped to the rupture stress from 1 to 1000 cycles, usually
3- Design Fatigue Curves like those you find in ASME are drawn with a different concept in mind: the stress they relate to is a "fictitious" one: it only means that, if you calculate a LOCAL peak stress variation of, say, 700 MPa, your component may resist for, say, 100 cycles, regardless of the "real" yeld and rupture limits, because these last two relate to "gross" stress, not local peak
4- if you carry out a long-term fatigue analysis using Voeller curves, you do have to "scale" them using Goodman-Smith (or other) whenever you don't have R=-1 (or, the same, sigma-average=0)
5- Aluminium alloys may have an endurance limit, but as far as I know this is meaningless for common engineering purposes (which real-life part should undergo sigma-a of less than 5 MPa for more than 10E10 cycles ?)

Regards

 

[NickE]

I wholehartedly agree with you cbrn. I think your description of localized yielding is very good. However it may be usefull to define it wrt cycle count also. The first time a body is loaded then the localized yielding occurs, after that cycle no yielding should occur, since then the sresses have been re-distributed and now non-local stress conditions are defining the behavior of the body.

#5 is a great statement!

 

[cbrn]

Hi,
NickE, I believe you're right, one should pay very careful attention to local yeld progression wrt cycles count: it's the ratchetting phenomenon. All my description can be valid only in the case that ratchetting stops after some cycles (1, 2, usually not more than 10). I experienced it in a very sollicited component that we "wantedly" designed "at limit" both statically and for low-cycle fatigue: ratchetting stopped at the second cycle, that's why we accepted the local yelding (moreover, very limited because local plastic deformation was 0.5% of the elastic one - results were FEM-based with full non-linear analysis of course).

Regards

 

[corus]

For secondary stresses (strain dependent) shakedown will occur after a few cycles and ratchetting or progressive deformation will stop. The limit on the secondary stress range is twice yield. After 1 or 2 cycles the stresses will remain within yield. Again, refer to design standards such as those in pressure vessel or nuclear design codes which will explain these limits and how to classify stresses at a particular location.

corus

 

[cbrn]

Hi,
Corus, I can see your point and, if I refer to the definition of secondary stress, I find a perfect agreement in that "secondary stresses are self-limiting". But theoretically there may be cases in which progressive yelding will stop only when the "local" yelding has overcome a critical extent after which it can't be treated as local any more. I must admit, however, that I never encountered a geometry where this occurs. European PED however, in the Direct Route, wants the designer to check for ratchetting, or in better terms, the method in itself is something like "strain-based" (btw, it's a very powerful method which allows to push the limits of strength resistance far further than what ASME allows, but it's extremely resource-consuming... So stress categorization has still long happy days to live!)

Regards