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BHN vs. PSI - Mountain Molds

BHN vs. PSI

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mtngun
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BHN vs. PSI

Postby mtngun » Wed May 11, 2011 10:53 am

This has already been discussed elsewhere on my site, but I'll repeat it here because it never ceases to amaze me how many people believe in old wives tales.

Here is a chart I made showing the relationship between BHN and tensile strength for various lead alloys. Based on real measurements, not theory. Most of the data points came from Vulcan Lead and Alchemy Castings, just because they were nice enough to publish their data.
Image

In other words, even the hardest lead alloy at about 30 BHN will yield at about 11,000 psi.

There are various formulas floating around on the web that would have you believe that lead is stronger than steel. A common one is PSI = 1422 x BHN. If you believe this formula, then a 30 BHN lead bullet does not yield until 42,660 psi. NOT ! ! !

Grade A36 mild steel has a BHN of 149. It's yield strength is 36,000 psi and a tensile strength of 58,000. Yet if we apply the PSI = 1422 x BHN formula, we're supposed to believe that the yield strength of mild steel is 212,000 psi ! ! !

On the other hand, let's apply my formula for tensile (not yield) strength, PSI = 375 * BHN + 500. My formula says mild steel should have a tensile strength (not yield) of 56,375 psi, pretty close to the official spec of 58,000. I'm not advocating using my formula for steel, I'm merely pointing out that it's in the ball park, unlike the 1422 formula.

To clarify, yield strength is when the metal begins to deform. Tensile strength is when the metal actually breaks apart. There's a big difference between the yield strength and the tensile strength of steel, but not so much difference for lead.

The pressure that burning powder applies to the bullet will not be constant, but will vary from the base of the bullet to the tip, like this:
Image

What this means is that there are degrees of obturation. The bottom of the bullet may obturate, but the rest of the bullet may not obturate. Or the bullet may obturate when the chamber pressure peaks, but not before or after. That's life, we have to deal with it, and that' why I prefer to size bullets "big enough" so that they fit without obturating.

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Re: BHN vs. PSI

Postby jwp » Sun Jul 24, 2011 4:23 am

Brinell Hardness is measured in kilograms per square millimeter of the spherical surface area of the indentation made by the ball used for testing. One kilogram is a about 2.2046 pounds, and one square inch is about 645.16 square millimeters. Thus, the conversion factor between BHN and PSI is approximately 1422.32, and a BHN of 30 (kilograms per square millimeter) equates to about 42670 PSI.

In Modern Reloading - Second Edition, Chapter 10, Richard Lee discusses matching bullets to chamber pressure. This seems likely to be the original source of whatever information (probably now garbled and ill understood) might be floating around on the net. (Veral Smith may have also published something along these lines; I can't remember for sure.) Mr. Lee is very clear that he is concerned with compressive strength, not tensile strength. He is specifically concerned with the relationship between the ultimate compressive strength of the bullet and chamber pressure, and that relationship's effect on accuracy. Because BHN is a fair test of compressive strength, and because he wants to deal with both compressive strength and chamber pressure in PSI, he multiplies the BHN by 1422.

What Mr. Lee and his son seem to have found through testing was that the best accuracy occurred when the chamber pressure (in PSI) was around 90% of the bullet's ultimate compressive strength (as obtained by multiplying its BHN by 1422). It's strongly implied, but not explicitly stated, that the testing was done with gas checked bullets.

In this regard, your post of several years ago under the "Pressure Measurement" heading with subject "WW748 in the 30-06" is interesting. You didn't give the BHN of the bullets, but since they were heat treated wheel weight metal, I'd guess they were around 30 BHN. From the photograph, they also seem to have had copper gas checks. The bullets obviously didn't disintegrate even though the 53K to 61K PSI you reported far exceeds not only the tensile strength of lead, but also that of copper gas checks. More interesting, is that when you lowered the pressures, accuracy improved. From the chart in Mr. Lee's book, 30 BHN bullets should work best (for accuracy) at around 38K-40K PSI chamber pressure. Dropping the pressure that much might well result in too low a muzzle velocity (and therefore energy) for your hunting purposes - Accurate shows the max load for AA4064 for a 180gr Lyman bullet at 36K PSI and 2440 FPS - but it'd be interesting to know what it did for accuracy.

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Re: BHN vs. PSI

Postby mtngun » Sun Jul 24, 2011 5:52 am


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Re: BHN vs. PSI

Postby jwp » Sun Jul 24, 2011 9:35 am

Dan, here's the formula for BHN (the image is from ) :

image005.gif
image005.gif (9.63 KiB) Viewed 14097 times

where
BHN = the Brinell hardness number
F = the imposed load in kg
D = the diameter of the spherical indenter in mm
Di = diameter of the resulting indenter impression in mm

The denominator in the formula is the spherical area of the indentation in square millimeters. Hopefully that makes it clear that a material's BHN number is the number of kilograms per square millimeter (kg/mm^2) needed to make the test indentation with the test indenter in that material. If we want pounds per square inch (lb/in^2), we can use 1 kilogram and 1 millimeter and substitute the appropriate pound/inch values to get the conversion factor:

1 kilogram = 2.2046+ pounds , 1 millimeter = 0.03937+ inches therefore

1 kilogram per square millimeter = 2.2046 / (0.03937 * 0.03937) pounds per square inch
= 2.2046 / .0015499969
= 1422.3254 pounds per square inch (lb/in^2)

with a little round off error.

30 BHN means "30 kilograms per square millimeter", and that really does represent applying a pressure of something over 42K pounds per square inch (which is the same as 30 kg/mm^2) to the tested material. Here are some numbers.

First in kilograms and millimeters

F = 27.2 kg , D = 4 mm , Di = 1.067 mm

BHN = 27.2 / ( ( 3.14159 / 2 ) * 4 * ( 4 - sqrt(16 - 1.138489) )
= 27.2 / 1.570795 * 4 * (4 - 3.8555)
= 27.2 / 1.570795 * 4 * 0.1445
= 27.2 / 0.90792
= 29.96 kilograms per square millimeter

And now the same dimensions expressed in pounds and inches

F = 60 lb , D = 5/32 = 0.15625 in , Di = .042 in

BHN = 60 / ( ( 3.14159 / 2 ) * .15625 * ( .15625 - sqrt(.0244 - .001764) )
= 60 / 1.570795 * .15625 * ( .15625 - .15045 )
= 60 / 1.570795 * .15625 * .0058
= 60 / .0014235
= 42149.63 pounds per square inch

There's round off error in all of those numbers, but I didn't feel like typing all the needed decimal places. The total error over both calculations is about one percent in the ratio of lb/in^2 to kg/mm^2.

None of that says anything about tensile strength. There is a relationship between BHN and ultimate tensile strength, but that relationship is material-dependent and involves the exponent of some test-dependent value in SomebodyOrOther's Law that I no longer remember and probably never understood anyway.

The primary point of my original note was simply that 1422 has a basis in reality: It's what's needed to convert BHN, which is expressed as kg/mm^2, into a number expressed as PSI. Lee wanted to do that so he could compare it with chamber pressure. Was there any real value in doing that? I have no idea. He claims he found some relationship between BHN_as_PSI, chamber pressure, and accuracy. You and the CBA say you've never seen such a relationship, barring the generally accepted belief that lowering pressure (within reason) produces better accuracy. Personally, I think you're correct that the real answer is to shoot bullets that fit correctly.

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Re: BHN vs. PSI

Postby mtngun » Mon Jul 25, 2011 6:34 am

First, a correction: in my previous post, I stated that tensile strength can be assumed to be equal to compressive strength for most metals. That's a general rule of thumb, but there are exceptions. You can make a metal so hard that it shatters like glass if you drop in on a concrete floor. Such a brittle metal would be much stronger in compression than in tension. However, most common metal alloys are not brittle, and the tensile = compressive rule is used.

JWP, if you scroll down to the bottom of the webpage that you linked to, they offer a simple formula for converting BHN to PSI strength. PSI = 500 x BHN. That's ultimate (breaking) strength, not the more commonly used yield (begins to flex) strength.

For our 30 BHN bullet, that would be 15,000 psi, which seems a bit high to me, but still saner than the oft-quoted 1422 x BHN formula.

For a 21 BHN bullet, the strength would be 10,500 psi. That seems reasonable.

For a 5 BHN pure lead bullet, the strength would be 2500 psi. Again, quite reasonable.

For 146 BHN mild steel, their formula predicts 73,000 psi, which is a little high, but in the ballpark.

Here is a chart showing the 500 BHN formula (blue line) and comparing it to real world data. The blue line works well for everything except gray iron, probably because grey iron is somewhat brittle (compressive strength much higher than tensile strength).
Image

Getting back to bullets -- OK, so let's use the 500 x BHN formula and say we are using a 30 BHN bullet in a 38 special that produces 15,000 psi peak pressure. In theory, even the puny 38 special load could obturate the 30 BHN bullet, but it would only do so at peak pressure, not before or after, and it would only obturate the bottom band.

As I suggested before, there are DEGREES of obturation -- how far does the bullet have to travel before pressure is high enough to obturate ? Does only the bottom band obturate, or 1/3 or the bullet, or 2/3 of the bullet ?

If we are foolish enough to shoot undersize bullets, we generally wish that they would obturate in the throat. According to Quickload, rifle loads may not hit peak pressure until the bullet has traveled 1.5" (I'm going from memory, so forgive me if I mis-remember). Revolver loads may not hit peak pressure until the bullet has left the throat and entered the forcing cone. By then, our undersize bullet has wobbled around and lost accuracy. So we don't want to wait until peak pressure for obturation. We want our undersize bullet to obturate almost as soon as it begins to move, when chamber pressures are well below peak. That makes it next to impossible to come up with any simple formula for predicting the peak chamber pressure required to provide satisfactory obturation. Perhaps the 1422 x BHN rule is as good a rule of thumb as any for that purpose ?

My preferred solution is to size the bullet to fit the gun and then not have to rely on obturation, and not have to worry about BHN.

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Re: BHN vs. PSI

Postby jwp » Mon Jul 25, 2011 3:03 pm


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Re: BHN vs. PSI

Postby mtngun » Wed Jul 27, 2011 7:58 am


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Re: BHN vs. PSI

Postby atomictaco » Wed Aug 08, 2012 8:04 pm

OK, I'm just starting in the bullet casting hobby and trying to determine the proper BHN for 9 x 19. Based on this thread I've come up with a BHN of 84.3? Using PSI=375 x BHN +500 equation and the basic rule of "Multiply Divide Add Subtract" (Parenthesis and Exponents do not apply in this instance). In the Lee 2nd Edition reloading manual it lists a 125 gr lead bullet propelled by 4.7 gr of Bullseye at 32,100 PSI. My math shows 32100=375X+500 -- 31600=375X -- 64.2666=X -- X being the BHN. Is my math wrong? I know this is an old post, just trying to figure out my alloy for when I start pouring when my moulds arrive.

Ry

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Re: BHN vs. PSI

Postby mtngun » Thu Aug 09, 2012 6:24 am


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Re: BHN vs. PSI

Postby atomictaco » Thu Aug 09, 2012 4:39 pm

What would be a good BHN for the 9x19? Of have read anywhere from about 8 to about 24and just getting more confused. I've slugged all of my barrels which came out at, .3565 so was looking at 125 grey and 158 grain. .358 moulds which should cover both my 9mm and 38/357 pistols. I have wheel weights ans a little plumbers lead at the current time. Any suggiestions would be greatly appreciated.

Thanks,

Ry


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