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From: (John Bercovitz)
Subject: Pressure signs (was: Ti gun parts and physics)
Organization: Lawrence Berkeley Laboratory, California

In article <> infmx!
(Greg Mitchell) writes:

#This is quite interesting.  Is there some basis here by which to assess
#(quantify) (over-)pressure in a fired cartridge by measuring primer "dimple"
#depth and/or cratering and plugging those dimensions into some equation ?
#Given of course that we know enough about the gun the cartridge was fired in
#( firing pin parameters, firing pin hole size, primer hardness, etc. ).
#Is there a "safe" cratering envelope ?

I've always thought of pressure judgement by primer cratering as being
too tricky by half.  There are just too many variables to make the signs
reliable.  You've pointed out some of these variables.  If the primer
dent isn't a good hemisphere anymore, you've generally overdone it, I'm
afraid, at least if firing pin and hole are proper.

I prefer to check the pressure of a bolt gun by bolt handle lift and
extraction difficulty.  In a strong gun, you run into difficulty in
case extraction long before you're endangering the integrity of the
gun.  Some time back I posted an article which I think outlined some
of the calculations showing why this is true.  Did you catch that post?
Let me see if I can go find it and append it.

John Bercovitz     (

I see that that article was sort of long.  Let me preface it by giving
you a bare bones synopsis.  A brass cartridge case expands past its yield
point during firing, at least for the first firing when the case doesn't
fit the chamber so well.  That is to say, it stretches plastically to some
new size and then springs back from that new size to its new relaxed size.
If that springback is greater than the springback of the steel chamber,
then the case will extract easily.  If not, the case will want to be larger
than the sprung back chamber and so it will be a "press fit" in the chamber
and will be difficult to extract.  This extraction difficulty appears
approximately when chamber pressures exceed 70,000 psi.


Here's the article:

A friend asked why steel cases aren't more common since they would
allow higher chamber pressures.  I thought that as long as I had
written something up for him, I might as well post it here:

                   Material Properties
CDA 260 cartridge brass:          barrel steels:
Young's modulus = 16*10^6 psi     Young's modulus = 29*10^6 psi
Yield stress = 63,000 psi min.    Yield stress: usually > 100,000 psi

I was going to get back to you and explain further why brass is a better
cartridge case material than steel or aluminum.  Sorry I took so long.  I
left you with the nebulous comment that brass was "stretchier" and would
spring back more so it was easier to extract from the chamber after firing.
Now I'll attempt to show why this is true given the basic material properties
listed above.

A synopsis would be that the propellant pressure expands the diameter of
the thin wall of the cartridge case until it contacts the interior wall
of the chamber and thereafter it expands the case and the chamber
together.  The expansion of the cartridge case, however, is not elastic.
The case is enough smaller in diameter than the chamber that it has to
_yield_ to expand to chamber diameter.  After the pressure is relieved by
the departure of the bullet, both the chamber and the cartridge case
contract elastically.  It is highly desirable that the cartridge case
contract more than the chamber so that the case may be extracted with a
minimum of effort.

A quick review of the Young's modulus: this is sort of the "spring
constant" of a material; it is the inverse of how much a unit chunk of
material stretches under a unit load.  Its units are stress / strain =
psi/(inch/inch).  Here's a basic example of its use:  If you have a 2
inch by 2 inch square bar of steel which is 10 inches long and you put a
10,000 pound load on it, how much does it stretch?  First of all, the
stress on the steel is 10,000/(2*2) = 2500 psi.  The strain per inch will
be 2500 psi/29*10^6 = 0.000086 inches/inch.  So the stretch of a 10 inch
long bar under this load will be 10 * 0.000086 = 0.00086 inches or a
little less than 1/1000 inch.

Yield stress (aka yield strength) is the load per unit area at which a
material starts to yield or take a permanent set (git bint).  It's not
an exact number because materials often start to yield slightly and then
go gradually into full-scale yield.  But the transition is fast enough
to give us a useful number.

So how far can you stretch CDA 260 cartridge brass before it takes a
permanent set?  That would be yield stress divided by Young's modulus:
63,000 psi/16*10^6 psi/(inch/inch) = .004 inches/inch.

How far can you stretch cheap steel?  Try A36 structural steel:
36,000 psi/29*10^6 psi/(inch/inch) = .001 inches/inch.
How about good steel of modest cost such as C1118?
77,000 psi/29*10^6 psi/(inch/inch) = .003 inches/inch.
(Note that C1118 doesn't have anywhere near the formability of CDA 260.
Brass cases are made by the cheap forming process called "drawing"
while C1118 is a machinable steel, suitable for the more expensive machining
processes such as turning and milling.)

What about something that's expensive such as CDA 172 beryllium copper?
175,000 psi/19*10^6 psi/(inch/inch) = .009 inches/inch.
(This isn't serious because CDA 172 is pretty brittle when it's _this_
hard.  Not to mention that beryllium is poisonous even in an alloyed state.)

Note: I was corrected on this.  It seems that very low alloys of beryllium
are _not_ considered poisonous (except here at this Lab.  8-))

Titanium Ti-6AL-4V
150,000 psi/16.5*10^6 psi/(inch/inch) = .009 inches/inch
(This is an excellent material though expensive and hard to work with.)

Really expensive aluminum, 7075-T6
73,000 psi/10.4*10^6 psi/(inch/inch) = .007 inches/inch
Cheap aluminum, 3003 H18
29,000 psi/10*10^6 psi/(inch/inch) = .003 inches/inch
(Aluminum isn't a really good material because it isn't strong and cheap
at the same time, it hasn't much fatigue strength, and it won't go over
its yield stress very often without breaking.  So you can't reload it.
It makes a "one-shot" case at best.  Also, 7075 is a machinable rather
than a formable aluminum, primarily.)

Magnesium, AZ80A-T5
50,000/6.5*10^6 = .0077
(Impact strength and ductility are low.  Corrodes easily.)

+Here's the important part: Even if you stretch something until it
+yields, it still springs back some distance.  In fact, the springback
+amount is the same as if you had just barely taken the thing up to its
+yield stress.  This is because when you stretch it, you establish a new
length for it, and since you are holding it at the yield stress (at
least until you release the load) it will spring back the distance
associated with that yield stress.  So the figures given above such as
.004 inches/inch are the figures that tell us how much a case springs
back after firing.

Changing subjects for a moment:  How much does the steel chamber expand
and contract during a firing?  Naturally this amount is partially
determined by the chamber wall's thickness.  The outside diameter of a
rifle chamber is about 2 1/2 times the maximum inside diameter,
typically.  The inside diameter is around .48 inches at its largest.
Actual chamber pressures of high pressure rounds will run 60,000 psi or
even 70,000 psi range if you're not careful.

One of the best reference books on the subject is "Formulas for Stress
and Strain" by Roark and Young, published by MacGraw-Hill.  Everyone
just calls it "Roark's".  In the 5th edition, example numbers 1a & 1b,
page 504, I find the following:

For an uncapped vessel:
Delta b = (q*b/E)*{[(a^2+b^2)/(a^2-b^2)] + Nu}

For a capped vessel:
Delta b = (q*b/E)*{[a^2(1+Nu)+b^2(1-2Nu)]/(a^2-b^2)}

  a = the external radius of the vessel = 0.6 inch
  b = the internal radius of the vessel = .24 inch
  q = internal pressure of fluid in vessel = 70,000 psi
  E = Young's modulus = 29 * 10^6 psi for barrel steel
  Nu = Poisson's ratio = 0.3 for steel (and most other materials)

A rifle's chamber is capped at one end and open at the other but really
it's not too open at the other end because the case is usually bottle-
necked.  You'd have to go back to basics instead of using cookbook
formulae if you wanted the exact picture, but if we compute the results
of both formulas, the truth must lie between them but closer to the
capped vessel.

For an uncapped vessel:
D b = (70000*.24/29*10^6)*{[(.6^2+.24^2)/(.6^2-.24^2)] + .3} = .00097

For a capped vessel:
D b = (70000*.24/29*10^6)*{[.6^2(1.3)+.24^2(.4)]/(.6^2-.24^2)} = .00094

There's not a whole heck of a lot of difference between the two results
so let's just say that the chamber's expansion is .001 inch radial or
.002 inch diametral.

The cartridge case's outside diameter is equal to about .48 inch after
the cartridge has been fired.  So its springback, if made from CDA 260,
is .004 inches/inch (from above) * .48 inch = .002 inches diametral
which of course is just the amount the chamber contracted so we've just
barely got an extractable case when chamber pressures hit 70,000 psi in
this barrel.  This is why the ease with which a case can be extracted
from a chamber is such a good clue as to when you are reaching maximum
allowable pressures.  By the same token, you can see that if a chamber's
walls are particularly thin, it will be hard to extract cases (regardless
of whether or not these thin chamber walls are within their stress limits).
A really good illustration of this can be found when comparing the S&W
model 19 to the S&W model 27.  Both guns are 357 magnum caliber and both
can take full-pressure loads without bursting.  The model 27 has thick
chamber walls and the model 19 has thin chamber walls.  Cartridge cases
which contained full-pressure loads are easily extracted from a model 27
but they have to be pounded out of a model 19.  So manufacturers don't
manufacture full-pressure loads for the 357 magnums anymore.  8-(

We can see from the above calculations that a steel case wouldn't be a
good idea for a gun operating at 70,000 psi with the given 2.5:1 OD/ID
chamber wall ratio if reasonable extraction force is a criterion.  Lower
pressures and/or thicker chamber walls could allow the use of steel cases.


From: (John Bercovitz)
Subject: Re: Brass Expansion in a .357
Organization: Lawrence Berkeley Laboratory, California

In article <>
(Mike Janeczko) writes:

#A friend and I both have .357's.  I have a 6" stainless(S&W 686) and he has a
#4" blue (S&W Model 19, I think).  Anway, at the range the other day we both
#were firing maximum loads.  After shooting 6 rounds, the brass smoothly came
#out of my cylinder.  My friend's rounds didn't come out so smoothly.  He had
#to bang the ejector on the bench for a couple of minutes before they finally
#came out.  My question is: does the length of the barrel have somehting to
#do with the brass getting stuck or is it the bluing?  Anyone have any answers?

This used to be a really well-known phenomenon when there were a lot of
model 19s around. It's real simple - the model 19 is a K frame and has
thin chamber walls.  In fact, the wall expands more than the springback
of the cartridge case with a near-max pressure round.  So the cartridge
case gets stuck.  This won't happen with an L frame or an N frame because
they both have thick chamber walls.

Steel has a spring constant called the "Young's modulus" which governs
how much a given piece of it stretches under a given stress.  The strength
of the particular type of steel doesn't matter, this spring constant is
about the same for all types.  Even though the type and amount of steel in
the K frame's cylinder is strong enough to take full-pressure 357 loads, it
stretches a lot because it's so thin that the stress is very high.  Before the
model 19 came out, manufacturers commonly loaded factory 357 ammo a lot hotter
than they do now.  But one caveat here, even though the cylinder will take
hot loads (even if you have to pound 'em out of the chambers), the rest
of the gun will get pretty beat up in very short order.  Particularly, the
barrel/cylinder gap will be altered quickly.  Whom to blame?  Well, it
was Jordan's idea, he pushed S&W into it.  But they should have been smart
enough to resist.  Colt always built 357s on a frame equivalent to the L frame.
And no, I'm a Smith man, not a Colt man when it comes to revolvers.  8-)

My model 27 (N frame) has had thousands of rounds through it each of which
would jam a K frame.  I used to have a super-accurate match load using
Sierra bullets which was near max hence the "thousands of rounds".
The gun has never had any problems as a result of this.

Another thing to watch for on the 19 is locking notch dents in the
chamber.  If you raise the pressure high enough, the weakest part of the
cylinder, the part where the exterior locking notch is, gives way a little.
(plastic deformation).  Then you'll always have problems with stuck cases.
Best thing to do is send it back to Smith for a new cylinder.  You see
this especially with +P+ in K frame 38's.  The problem is that the
notch is right in the thinnest part of the chamber in the case of a S&W.
Not shifted off to the side as in so many other revolvers.

John Bercovitz     (

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