Aluminum Compression Longitudial Deflection 7

[jimmyjimmyv]

I regularly perform compression tests on aluminum tubing. The tubing must meet a requirement of a minimum length deflection. Whoever drew the print up is using the length (before compression), multiplied by the yield offset value of .002 (0.2%). I'm wondering if anyone knows why an engineer would use this calculation to come up with a max value for the length deflection? My values are always higher as I use a calculator that utilizes "δ = (Pl)/(AE)" which takes into account; the load, modulus of elasticity, length and cross-section area.

 

[jimmyjimmyv]

Would going to a T6 temper be acceptable, if not would going to T4511 or T4510 yield more consistent test results?

These parts are T6. I'm not familiar with the others you reference, I'll have to research them.


Edit, I see, they are stress relieved through stretching. This wouldn't work well for the trouble parts, as the amount of stretch needed would pull the dimensions out of tolerance.

 

[IFRs]

My mistake on the T4 - "extruded as T4, artificially aged to T6". Sorry to have wasted your time.

After a brief reading of AA specs I respectfully withdraw the suggestion of going to T62, T6510 or T6511 - I don't see any benefit and T62 is clearly not applicable.

Too bad you can't go to 6005A - the air vs water quenching has several advantages.

Your original question was how to predict failure. Using force / area = stress and comparing this stress to the specified minimum yield stress is one answer, as the learned responders have elucidated above. Because if the applied stress is less than the yield stress, the part should not have any permenant deformation (ignoring the Bauschinger effect if it is relevent to your parts).

I think "buckling" is not quite the appropriate word but who cares - it's what the customer uses. Same for referring to yield offset value.

It's goofy that they draw a part, hold you to those dimensions and alloy specifications and then require you to test past yield strength while requiring a maximum permenant deformation. Leave it to the auto industry!!! Or, more likely a carry over from years ago now used by a new PIC and/or SME who have too much on their plate and depend on outside suppliers like you for technical expertise and suggestions on how to make it work in real life.

 

[IFRs]

Also, I'd be careful about putting your customer's drawing on the web - they tend to be very picky about their secret stuff...

 

[jimmyjimmyv]

Well I figure there isn't anything that could point to them. From what rb1957 had taught me above, I think I can calculate the stress in the part at a given load during compression, compare that to the yield we typically see during tension and make a judgement based from that. The mechanical properties we measure during tensile testing is fairly consistent across our alloys and tempers.

 

[Settingsun]

In that print quote, they're asking for material with 0.2% proof stress (nominal yield stress) of 272MPa or more, higher than the usual minimum for 6061-T6. So you are correct that you'll need to look at your history and process to check that you can achieve this reliably, or query whether that is in fact their intention.

The customer expects the part to yield or they would not be asking you to measure permenant deformation

With aluminium, permanent deformation begins at less than nominal yield stress. By definition, the nominal yield stress is the point of 0.2% permanent deformation.

Jimmy, reading up on Ramberg-Osgood might help with your review of historic test results and general knowledge of strain/deformation at stresses near nominal yield.

PS: kN, ksi, inches, mm - that's a winning combination...