How To Calculate Average Yield Strength
How To Calculate Average Yield Strength
(OP)
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.
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





RE: How To Calculate Average Yield Strength
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
RE: How To Calculate Average Yield Strength
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
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
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:
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
RE: How To Calculate Average Yield Strength
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?
RE: How To Calculate Average Yield Strength
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
RE: How To Calculate Average Yield Strength
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.
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
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
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
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
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
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
RE: How To Calculate Average Yield Strength
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
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
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
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.
Cheers and thanks,
Tip
Western Design Services Co. Ltd.
Bangkok, Thailand
Western Design Services
Chicago, Illinois USA
RE: How To Calculate Average Yield Strength
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.
RE: How To Calculate Average Yield Strength
For your application(s), a material temper grade would be purchased. Each temper grade has minimal varience in hardness and tensile properties. You should be able to determine an average relatively easy; however, would you not like to know the maximum YS of the temper grade?