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.

 

[rb1957]

are you measuring under load ? your deflection (PL/EA) is the deflection under load.

.2% permanent set (at yield) is the deformation post test. A much bigger strain happens under load.

is the load expected to yield the specimen ?

another day in paradise, or is paradise one day closer ?

 

[jimmyjimmyv]

rb1957,
I should state that these parts are extruded 6061 aluminum alloy tubes, with lengths not more than 3 inches. In most cases, the load required exceeds the min yield strength for 6061 T6, in others the load does not exceed the standard yield strength of the tube (ex: 198 kN); however, this material is subjected to tensile testing as well (in 12" lengths), with the yield strength and UTS results always above the ASTM min limits for this alloy/temper. These part/s were previously made from steel.

This is our process,
[ul]
[li]Measure the sample length[/li]
[li]Apply the load (Instron 60,000lb machine)[/li]
[li]Remove the load[/li]
[li]Measure the sample length[/li]
[li]Subtract the "after" length from the "before" length = Deformation[/li]
[/ul]

The engineer actually calls this "displacement", which I don't believe is the correct term. I really appreciate your help. I've been trying to figure this out for years, and I am only a shade-tree engineer. I've never been able to figure out why or where they got the formula of (yield offset X sample length), as this does not take cross-sectional area into account, and since this is a tube and not a solid rod or bar, I would think that would make a world of difference.

 

[jimmyjimmyv]

I'm also curious as to whether the Bauschinger effect comes into play, since these parts are extruded and then stretched. I've read that effect is when the yield strength is lowered when a load has been applied in the opposite direction.

 

[rb1957]

ok, so you're measuring the permanent set from the test. Your calc (PL/EA) is inappropriate (as that measures the deflection under load). I assume the point of the test is not to load to yield ? then using .2% offset is a limiting value. Do you know the yield load of the specimen ?

another day in paradise, or is paradise one day closer ?

 

[jimmyjimmyv]

rb1957, If you're meaning the yield strength, the minimum acceptable YS for 6061 T6 is 35 ksi. If you're meaning the total amount I am loading onto the sample, it varies, but a particularly troublesome part is 35 kN or 7900 lbs. And, yes we are not wanting the sample to yield, or yield only enough that the length deformation remains below the requirement, which on this particular sample is a maximum of .003" (.07 mm).

 

[dauwerda]

As rb1957 indicated, the PL/AE equation is for determining the elastic displacement under load. That is, when the specimen is loaded with a load, "P" it will displace the amount determined by the above equation.

What you are doing is loading then unloading the specimen and determining any permanent deformation. Technically speaking, as long as the specimen's stress remained less than it's yield stress it shouldn't have any permanent deformation. That is, the specimen should have remained fully elastic and recover all of the deformation that happened under load.
The deformation that you are seeing (difference between pre-load and post-load lengths) is due to plastic (non-recoverable) deformation.
Take a look at this page for more information on the offset yield point that is being described on the drawing:

https://en.wikipedia.org/wiki/Yield_(engineering)

 

[jimmyjimmyv]

Dauwerda, I realize that, theoretically, the length should not suffer permanent deformation at a load that is less than the the yield strength, the problem is that yield strength under compression is not the same as it is in tension. I know this because I tensile test 1,0000s of samples per month with no failures, but with samples from the same parts that the tensile samples were taken, they yield at a lower value, and allow the length deformation. I'll have to look up some historical compression and tensile tests to identify an average difference yield strengths between those two tests. Also, I want to compress some samples that have not been stretched, to see if there is some Bauschinger effect happening.

 

[rb1957]

ok, but whatever you're doing PL/EA is not correct to measure the post-test deformation.

if your test load is less than yield then the post-test deformation should be less than 0.2% permanent set.

another day in paradise, or is paradise one day closer ?

 

[jimmyjimmyv]

rb1957, sometimes. Unless the Bauschinger effect is interfering. But, yes I agree, my calculation is not the correct one to use, but I have to replace it with something.

 

[rb1957]

sure. you know the load applied, and so if tensile yield is happening. I thought you were looking for a pass/fail criteria, ie if permanent set > .2% for a load < yield then fail.

sure the may be some tension then compression effect happening, but the theory should be telling you what this is.

what is it you're trying to do ? validate a spec yield strength ? why test in tension then in compression ??

another day in paradise, or is paradise one day closer ?

 

[jimmyjimmyv]

rb1957, The material must meet the mechanical property specification for 6061 T6, which it always does, AND the amount of length deformation after compression cannot exceed .2% of the original length. We are required by ASTM to verify the material is within the limits for the alloy/temper, the customer requires the material to pass the compression tests; thus the tensile testing and compression testing. The tests are performed on different samples, meaning I'm not tensile testing and compressing the same sample. Both samples are taken from the same lot of material, meaning it was extruded, stretched/straightened, and aged together at the same time.

A five-year history shows that the material always passes tensile testing. A couple of time a year, I'll have samples that won't pass compression testing. If they fail, it is not by much. The requirement is a max of .07, my failures might be at .12mm, so very very small at .35% instead of .2% I have just recently discovered if I can maintain a wall thickness above a certain size during extrusion, I will have samples that consistently pass compression testing, but this is only for one part; we have 40 others and more are being requested monthly. The parts end up at all North American automotive OEM's.

Meeting the mechanical property requirements is easy. I need to be able to calculate a projected longitudinal deflection because I'm the one who accepts or rejects proposals for manufacturing parts that have prints with these compression testing requirements. I can't simply extrude the part and test it beforehand because I can't spend thousands of dollars for tooling to extrude a part that might not meet the print requirements, and I've now learned that I can't trust the customer's engineering department to place a valid value on the print for length displacement. They are simply saying, it's length X .002; I know that it's much more complicated than that.

 

[SWComposites]

No, you are wrong, and they are correct. The 0.2% is strain in inch/inch. Your allowed deformation after test is max strain*Length or 0.002*Length. Area is not relevant except to calculate the required load, which presumably comes from the customer.

 

[jimmyjimmyv]

SWComposites, ok, yeah I can almost agree with it stated that way, but that also means the customer has unrealistic load requirements, and mainly because I don't believe there's a precise equivalency between the offset yield in tension and compression. They definitely would never be able to require more than 156 kN, because that is the minimum yield strength for 6061 T6.

A question, if you don't believe area plays a part, why is it that I decrease the longitudinal deformation by increasing the wall thickness?

 

[jimmyjimmyv]

I've been reading about the bauschinger effect all night, and the more I read, the more I believe we may be inducing this in our stretch/straighten after extrusion. We have to stretch the metal past it's elastic limit in order to straighten it. The effect would reduce the elastic limit in the compressive direction, increasing the delta between the elastic limit and the offset yield, which is where the sample length is being shortened. At least I think, hehe.

I'm going to test this by comparing compression results from non-stretched parts, with results from parts that were stretched. I should know something within the next couple weeks.

Again, thanks everyone for helping me. Everything I know about material properties, I've learned in the last 30 years while working with aluminum and constantly perusing forums like this one, picking up bits and pieces that I can understand.

 

[rb1957]

sorry, but again, what exactly are you trying to do ? verify the yield load/stress ?? That sounds like a very difficult thing to do ... repeated load and unload until you see some permanent set. but why ... since most metal specimens are higher strength than book values ??

So you plastically deform the extrusion, and don't stress relieve, and then want to determine the compression strength of the final part. Yes, the manufacturing stresses will affect this but not, I think, in the way you're looking at it. I'm trying to understand your problem ...

I think you have (from your customer) an eng'g drwg that says "take this extrusion and form it this way" and also "required strength is ??" or "required deformation after forming under load of X is less than y = 0.2%*L".

another day in paradise, or is paradise one day closer ?

 

[dauwerda]

A question, if you don't believe area plays a part, why is it that I decrease the longitudinal deformation by increasing the wall thickness?

Area does play a part. You use the area to determine the stress based on the force or vice versa.
Anything with the units of kN is not a stress, it is a force. A stress is a force per area. The yield strength of a material is a stress, not a force.

If you increase the wall thickness of your part you are increasing the area. But if you load it with the same force that you loaded the previous part with a smaller side wall thickness (and therefore smaller area) you are not inducing the same amount of stress in the part.

I've now learned that I can't trust the customer's engineering department to place a valid value on the print for length displacement. They are simply saying, it's length X .002

They are not saying that that will be the deflection. They are saying that that is the max allowable deflection. If it is deflecting more than that, the compressive yield strength of the material isn't as high as it should be and is not acceptable (assuming the correct load is being applied to the part which would be determined by multiplying the cross sectional area of the part times the yield strength of the material). This test is for testing the material properties, so if the part is not meeting the requirements of the test it is the material properties that are failing.
There is a very real possibility that the material is failing to meet the requirements because of the forming process of the part that you are describing (which may affect parts with thin sidewalls more than parts with thicker sidewalls).

If you have certain parts that continuously fail to meet the deflection requirement (and therefore the yield requirement as that is what this test is actually testing) it is probably worth starting a discussion with the customer so that they can understand the issue and either accept it or start designing to a lower yield strength, or perhaps if it is determined that it is only an issue for parts with really thin sidewalls, start designing parts/tubes with a minimum thickness in mind that does not result in the material properties being so negatively affected by the forming/straitening process.

 

[jimmyjimmyv]

sorry, but again, what exactly are you trying to do ? verify the yield load/stress ??

We have to tensile test 1 sample for every 1,000 lbs that we produce to verify alloy/temper; it is an ASTM standard B221. In addition, the part print states "Buckling Strength: permanent deformation after 38kN loaded shall be 0.07mm max. (38mm X .002) in the axial dir."

And, no we don't relieve the stresses in this part. They are extruded as T4, then artificially aged to T6. We only do stress relief on O-tempered parts, and some that are partially annealed, like 3003 alloy/H24 or H27 Temper.

I'm going to print your post out and re-read it a couple of time, because there is some good stuff in there.

 

[jimmyjimmyv]

Area does play a part. You use the area to determine the stress based on the force or vice versa.
Anything with the units of kN is not a stress, it is a force. A stress is a force per area. The yield strength of a material is a stress, not a force.

If you increase the wall thickness of your part you are increasing the area. But if you load it with the same force that you loaded the previous part with a smaller side wall thickness (and therefore smaller area) you are not inducing the same amount of stress in the part.

Yeah the same for your posts, printing them out to read it a couple of times.

 

[jimmyjimmyv]

So, in your opinions, how should I determine whether we can, or cannot, meet these "Buckling Strength" compression test requirements going forward? Like I wrote earlier, I can't run a trial on a part that we don't yet make. And, I've already rejected a bunch of prints due to my use of the wrong formula, oops

I don't know if I'm going to use the right terminology to ask this question, but is there a formula to find the amount of stress that will be caused in a part at a certain load (kN)?