How To Calculate Average Yield Strength 1

[Tippaporn]

I've searched hi and lo for a definition of "average yield strength" and have come up empty. I utilize a particular formula which determines the bottoming force of a forming operation in metal press stamping. Bottoming force is the force exerted when a forming operation is at the bottom of the press stroke, at which point the stackup of die, material, and punch is solid. Material specs include "yield strength" but do not include "average yield strength."

Is there another term for it? And how can I obtain a value for any given material? Not knowing what else to do I have been using a value betwixt ultimate tensile strenght and yield strength. Can't be right but I must input a value.

Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA

 

[metengr]

As mentioned in other posts, yield strength is a measured value based on an accepted definition, like 0.2% offset. With that said, yield strength values in most material specifications that are international standards (ASTM or ASME) are typically reported as minimum values.

 

[FennLane]

Yield strength is relatively difficult to define.

Some steels exhibit a 'yield discontinuity' and many International Standards call for the determination of both a Lower and Upper Yield Stress.

I have been concerned about 'Upper Yield' Stress for many years and there was a time when it vanished from the British Standard only to be re-introduced in EN10002-1.

Upper Yield points can be influenced by test piece alignment and the type of test machine being used as 'hard' machines will produce different results to 'soft' machines.

In general variations in Upper Yield can easily be between 10 and 15% for a given Lower Yield Point.

Materials which exhibit a continuous yield it can be even more difficult to define exactly what is meant.

As has been stated it is common practice to use a 0.2% offset proof stress and this value tends to be both measurable and repeatable.

There can still be issues of repeatability with this value depending on the class of extensometer being used and the 'n' value of the material.

If the n value is high then repeatability will be more difficult to determine as small errors in strain measurement may result in significant stress variation.

If you know the strain in the material when the punch bottoms and you have an idea of yield strength and the n value you should be able to estimate the 'flow stress' of the material at this point.

 

[OGMetEngr]

Just take a look at the MTR and use the yield strength given for the heat of material you are using. The yield strength will vary throughout the thickness and from different locations in the product, but the MTR value will get you in the ballpark. From there you should be able to "trial and error" your way to the correct settings.

 

[redpicker]

Just take a look at the MTR and use the yield strength given for the heat of material you are using. The yield strength will vary throughout the thickness and from different locations in the product, but the MTR value will get you in the ballpark. From there you should be able to "trial and error" your way to the correct settings.

No... just plain no.

Many people believe that the properties on the MTR represent the minimum, or maximum, or average, or is somehow representative of the heat or heat treat batch it is being used to certify. It is none of these. It is merely a demonstration that the material and processing meets the minimum requirements of the standard or specification it is being certified to meet.

The OP (Tippaporn) has not explained very well what he means by "average yield strength", or even what material he is working with, so we are left with wondering what to advise. With a low strain-hardening material, like a cold-worked low-carbon steel, the 0.2% offset yield strength may be appropriate, but with a high strain-hardening material, the ultimate tensile strength may be more accurate. Perhaps using 120% of the minimum specified yield strength (SMYS) might even make sense, but using the MTR values is just plain wrong. That is not what the values on the MTR mean.

rp

 

[OGMetEngr]

I completely disagree here. This is exactly what MTRs give, the actual results from mechanical testing to established standards. I warned that properties will vary at different locations of the material. Also take a look at E8 standard deviation for yield strength between labs and even at the same lab, it's surprisingly high. But your statement that the MTR is not "somehow representative of the heat or heat treat batch" is ridiculous. I see MTRs all day every day, along with additional testing for validation of those properties. Unless your MTR doesn't represent the heat treat condition or otherwise that the material you have is in, or a "separate QTC" was used in lieu of a prolongation for large diameter bar for instance, the MTR provides the results of the mechanical tests that were performed. Using your logic, the MTR doesn't represent your mechanical properties. So how would the additional mechanical testing you personally do represent those properties any better? Because you "know what you are doing" and the mill doesn't? Also, using 120% of SMYS as more suitable than the actual yield strength on the MTR? Are you kidding?

The original question was about a forming operation using an equation and "average yield strength". This is most certainly not a final or exact result he will get from his equation. As I said, it will get him in the ballpark. Trial and error will provide the correct end result. This is why forming operations using CNC machines (think brake press, punch press) often use an initial test piece to adjust the settings to correct values.

 

[Tippaporn]

OGMetEngr, I have not defined "average yield strength" as I have been searching for a long time for a definition and have not been able to find one. I'm assumming that "average" would be in the context of the mean yield lying between the low and high points of the yield curve. But since I'm not a metallurgist, nor possess any background in metallurgy, I understand that I risk sounding like a fool by that assumption. I would love to stand corrected.

As to the material I'm working with, well, that varies. I'll try to be brief in a fuller explaination of what I'm after and why.

My profession is that of a tooling designer. I take a metal product, be it for automotive, telcom, computing, appliance, or any other industry, and design a stamping tool (die) by which the product is mass produced. That tool is then placed in a stamping press; be it mechanical or hydraulic. A flat sheet of steel is fed into the tool as a blank or from a coil on a reel. In the case of a progressive die (a die with multiple stations for cutting and forming) the steel sheet is fed incrementally with each press stroke, each station performing either a cutting or forming operation, until the part is complete per the customer's part specifications.

Since I design for multiple industries with a wide range of applications the material varies; almost all CRS grades, various grades of high strength low alloy, stainless grades at various tempers, aluminum, copper, brass . . . etc.

One aspect of tool design is the need to calculate forces for cutting and forming. As an example, to bend a given material with a given thickness at a given angle I have to determine 1) the bending force and 2) the clamping force required so that the material does not pull while being forming. Calculations are made using industry standard formulae that take into account all of the particular variables.

One other calculation I need to make is "bottoming" force. Generally, all forming operations need to "bottom." In other words, at the bottom of the press stroke the stack-up of all plates top to bottom, including material, is solid. Bottoming produces added force and it is calculated using the following formulae:

Minimum bottoming force = Minimum yield strength x material thickness x length of bend / 2000​

Average bottoming force = Average yield strength x material thickness x length of bend / 2000​

Now, the stamper for whom I design the tool needs to know how much tonnage my design will produce (adding up all cutting, forming & bottoming pressures along with nitrogen cylinder/spring pressures used for clamping and stripping) in order for him to designate what size (tonnage) press to run the tool in.

I must make mention that the total force generated by any given tooling design need not be calculated to a degree of accuracy requiring one or more decimal places. The force value should be fairly accurate give or take a few tons. Press sizes range in tonnage. A 300 ton press, for instance, can run any tool that generates less than 300 tons. So often times a tooling design is well within the press tonnage capacity and the precise tonnage of the tool need not be calculated very accurately. On the other hand, if a stamper has an option of running a tool in a 150 ton press or a 200 ton press then the tonnage calculations I make need to be more accurate.

Hopefully, this explanation helps.

If calculating "minimum" or "average" yield strength is something that can fairly easily be done given general material specs supplied by a steel vendor then I would be happy. If it cannot be done from a spec sheet but only through analysis then I'm out of luck. I would then at best hope for a formula that approximates and which is not hugely off.

I appreciate the comments thus far and thankful for the education all of you are providing me.

Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA

 

[OGMetEngr]

For your first statement, google yield stress and you will see that it is as metengr described. You speak of "average yield strength", do you mean typical yield strength? Or specified minimum yield strength? If I was running a press feeding coil through it (and I have done this in the past), and I needed yield strength for a coil of material, the MTR actual results will give the best estimation. As for bending certain angles, now you'll have to account for springback and make adjustment along the way until your end result is what you were looking for.

I've worked with die designers, and even the most experienced individuals don't have their parts come out perfect immediately. It takes tweaking and adjustment to get it right. But searching for an elusive and completely correct "average yield strength" will get you nowhere.

One last comment, where did these equations containing "average yield stress" originate?

 

[Tippaporn]

To be honest, I'm not sure what the definition of "average yield strength" is. That was the term given in the formulae I posted. And those formulae apply only to the calculation of "bottoming" force. I'm assuming that there is a low and high point in the yield curve and the average would be the mean. Of course, mean and average are not the same thing. Average would be a number expressing the central or typical value in a data set.

I gleaned the formula from an article I had come across comparing ve and wipe bending to rotary bending. It was quite awhile ago and I no longer have the article but I may be able to find it again. I'll do a bit of searching.

Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA

 

[redpicker]

OGMetEngr
I have seen many engineers (even metallurgists) over the years make the same mistake you are making. The mechanical test results that appear on the MTR are only useful for demonstrating that the material meets the specification they were produced to meet, that is all. They do not represent anything other than the material that was destroyed during the test. To assume if the Y.S. results on the MTR exceed the minimum requirements of the original specification by 10,500 PSI means that all of the material will exceed the SMYS by 10,000 PSI is not only wrong, but dangerous.

Consider a processor heat treating some casing to P110 (110,000-140,000 PSI Y.S.). The metallurgist knows his equipment and knows the first joints through the furnace will have a higher YS than the rest of the run, but never by more than 10,000 PSI or so. Now, to be sure all of the product meets all the requirements (impact testing required), one of these first joints is tested (this is upset casing, so testing a joint causes it to be scrapped) and has a YS of 132,000 PSI and the impact tests are fine. So, the metallurgist releases the material, knowing that some joints may be as low as 122,000 (perhaps even a bit lower), but those lower strength joints still meet the specification and the impact test values would be even better than what he tested. The MTR reports 132,000 PSI yield for this lot.

Now, someone later wants some Q125 Casing, which has a Y.S. of 125,000 - 150,000 PSI. He sees the Y.S. for this lot is reported as 132,000 and the impacts meet the Q125 casing. Using your logic, this lot would be acceptable, which would be incorrect, since only the first few joints would have Y.S. this high.

So how would the additional mechanical testing you personally do represent those properties any better? Because you "know what you are doing" and the mill doesn't?

I never said any testing I would personally do would be any different. The issue is not that I "know what I am doing", rather that I don't know what the mill was doing. The only thing I can be sure of is that the mill certified the material met the specification it was produced to meet.

Now, to answer Tippaporn's question, it would appear that the value he is wanting for "Average Yield Strength" would be the average Y.S. for the material he is processing. In the absence of any additional information, using 1.20 times the minimum would seem to be a logical approach, as it is assuming the material YS can very by 20% or more. One could look at the average results from several MTRs and use that value, but for reasons described above, that approach also has its problems. Honestly, I believe the value in the two equations Tippaporn posted is that while the minimum bottoming force can be calculated as shown, the bottoming force is going to be proportional to material's Y.S. and a variation is to be expected to the same extent that the material's Y.S. will vary.

rp

 

[OGMetEngr]

I don't want to debate this back and forth any further, but a controlled process in a mill should provide somewhat consistent results throughout a heat. Anything less and the process, be it forming or heat heat treatment, is not in control. The company I work for performs technical audits of all heat treaters and for some mills. The situation you describe, which I've seen more than once, sounds like a continuous heat treat facility in which the process is not properly controlled.

The OP didn't imply that he was designing something utilizing 95% or 100% of actual yield, so a yield strength estimation or "average yield" would be the best that can be done for this situation. I don't disagree with your description of casing and variable yield throughout a heat having the potential to cause issues. As I stated in my first post, yield will vary through thickness and at different locations in the product. This is basic physical metallurgy knowledge and one of the reasons why companies employ metallurgists.

 

[redpicker]

[quote}I don't want to debate this back and forth any further, but...[/quote]
Just as long is you get in the last word, I guess.

Any facility I audit whose technical staff tells me their processes are always in control will get extra-special review of their controls and results. In my experience, it the guys who tell you nothing ever goes wrong who have something to hide.

rp

 

[OGMetEngr]

...says the guy who replies to get the last word.

All my arguments still stand valid for this situation. I think that it would serve the community best if we can stick to the technical debate.

 

[Tippaporn]

My goal was to determine a somewhat accurate value for "average yield strength" which would insure that my "bottoming" force calculations were not out in space somewhere. As long as the "bottoming" force formula results were within ± a few tons the numbers would be suitable from my practicle standpoint.

From the discussion thus far it's become evident that 1) "average yield strength" is a term which is nowhere defined well enough to allow for the determination of any specific value, and 2) since even a minimum yield strength can be a number with a degree of variation then an "average," assuming a value between the low and high values of the yield curve, may exhibit even greater variation.

Since I do have latitude in the results of my calculations then I thought I might run some example calculations to see how much variance is produced between the value I have been using and redpicker's suggested min. YS x 1.2.

HSLA 955X has a UTS of 70,000 psi and min. YS of 55,000 psi. Given .063 thick material, a 10" bend length, and an "average yield strength" per the value I have been using: (UTS-YS)/2+YS, I calculate a "bottoming" force of 62,500 x .063 x 10 / 2000 = 19.69 tons. Substitute with min. YS x 1.2 the formula works out to: 66,000 x .063 x 10 / 2000 = 20.79 tons. A difference of 1.1 tons. I can live with that.

AISI 1008 CRS has a UTS of 49,300 psi and min. YS of 41,300. Given material thickness and bend length as above, the current value I was using for "average yield strength" resulted in a "bottoming" force of 45,300 x .063 x 10 / 2000 = 14.27 tons. Substitution comes up with 49,560 x .063 x 10 / 2000 = 15.61 tons. Or a difference of 1.34 tons. Again, a variance I can live with.

What I note, though, is by using a 1.2 multiplier of min. YS in the second example it returns a value greater than the supposed UTS. That would appear to be illogical. Logic seems to tell me that 1) an "average" yield strength (whatever that might mean) must be greater than a min. yield strength, 2) the value must lie somewhere on the yield curve itself (centered?), and 3) it cannot be greater than UTS.

Any thoughts?

Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA

 

[OGMetEngr]

There's no such thing as an "average yield strength" aside from running multiple tensile tests and averaging the yield strength results. Also some refer to a "typical" yield strength for a given grade of steel in a given heat treatment condition. Search Google images for stress-strain curve and you'll see where a yield strength point is on this curve. UTS is simply the maximum stress on this stress-strain curve. Also note that this is engineering stress and strain, which uses the initial cross section of the tensile specimen, and shows stress tapering off after UTS. In actuality, if true stress and strain were used (using instantaneous cross-sectional area) the curve would continue going up until fracture.

As far as using the 1.2 multiplier for a typical yield, I'd consider which route would be more detrimental(i.e. - using too low a yield or too high a yield than actual) from a safety, cost, etc. standpoint. This should give you a starting point to work your way up (or down) as appropriate.

UTS will always be greater than yield strength. If actual yield is above the minimum specified yield (even if greater than minimum UTS), UTS will be greater than minimum UTS as well as yield. This is, of course, barring a brittle fracture with no plastic deformation where yield and UTS could be the same.

 

[mrfailure]

It sounds to me like you are looking in reality for typical yield strength. This is easily available for given materials - you can find published data (I first look in the ASME B&PV Code Materials volume which gives those values provided your material is covered under the code). You can also literally obtain an average value by running tensile tests on several samples of the same material and getting the average. Specifications are never going to give you an average - material will be specified for minimum and/or maximum yield strength.

 

[Tippaporn]

OGmetEng, I agree that there is no such thing as "average yield strength" . . . published. The fact that someone has taken the trouble to include it into their formula must mean, therefore, that an average yield strength value can be obtained. And I would, therefore, agree with mrfailure that the value can be obtained by running multiple tensile tests, recording all of the minimum yield strength values, each test obviously producing a varying value, and averaging these for an "average" minimum yield strength. There exists one other possibility as to what "average" could be in this "bottoming" force formula. Since there exists a minumum yield strength value there must, logically, also exist a maximum yield strength value. In which case an "average" value would lie between.

It's obvious that we'll never be able to ascertain what the formula creator's definition of "average" is, unless I were to hunt him down and ask him, which is not going to happen . Neither will I ever be in a position to request multiple tensile strength tests for a job merely so that I can plug in an accurate number in a formula whose result need not be that accurate since I can be off a few tons on my overall tonnage calculation. Hence, I will lay this thread to rest.

I do appreciate all of the input everyone has supplied. My understanding of this topic has been greatly increased and I feel very much satisfied.

Cheers,

Tip



Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA

 

[mrfailure]

Tippaporn,

I want to make sure to clarify that an individual tensile test provides a yield strength, not minimum yield strength. Yield strength is defined by test standards (such as ASTM E8 or B557). I proposed averaging the set of test values you obtain to get your average. Similarly, published typical yield strength values are really derived by averaging of a set of data.

 

[Tippaporn]

Thanks for that mrfailure. All along I was assuming that "minimum yield strength" meant the point at which the yield curve begins, "maximum" would be at the end of the curve, and "average" would then lie in between. It sounds like from what you're saying that "minimum" or "maximum" yield strength values do not exist, that it's simply "yield strength." Therefore published yield strength values are to certify that the yield strength of the material never falls below the published value, hence "minimum." Semantics can be a b*tch.

What, then, of steel specs in which a range is given for both yield strength and ultimate tensile strength? And quite a wide range at that.

Well, now, I hope I haven't sounded too foolish or thick-headed throughout this thread. I'm starting to empathize with what teachers must go through.

Cheers and thanks,

Tip

Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA

 

[mrfailure]

There are excellent reasons regarding material performance why some material specifications have a range for one or both properties and others just specify minimum, but that is not really germane to your question.

I would suggest you read a copy of ASTM E8 or an equivalent ISO or international standard which lays out in detail how yield strength and other properties are obtained in tensile testing of steels. That will help make this clearer.