[Clark]

357 mag max has gone below 38 Special:
"Speer 6" 1964 38 s&w special 160 gr. soft point 11 gr. 2400
"Speer 6" 1964 357 mag 160 gr. soft point 15 gr. 2400
Midway "Load map" 1999 357 mag Speer 160 gr. soft point 10.9 gr. 2400


What went wrong that Midway could get the max loads so far off and make
a useless load book?
They used an "Oehler System 83 and piezoelectric transducers, the latest
in industry standard equipment".

The way I understand this is from John Bercovitz's 1993 post when he was still at JPL:

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.)

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)}

Where:
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.
jb

What does it all mean?
1) The 357 mag max load started out at the threshold of sticky cases, and then backed off a safety margin of powder charge to avoid sickly loads.

2) Then they measured the pressure and registered that.

3) Then they made thin wall 357 mags that got sticky cases with less pressure. So they changed the pressure registration. Loads were reduced.

4) Then they came up with a new way of measuring pressure, and that inspired them to make loads reduced even further.