M1 Garand and operating rod bending investigation


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GarandOwner
June 5, 2008, 02:48 PM
Like many others I have a love for the M1 Garand, It is pretty much common knowledge (at least among Garand shooters) that the operating rod is sensitive to buckling due to its length. I have always tried to find what pressure this occurs at but have never found it.

So the engineer in me took over and I decided to calculate it myself. After some research through the multiple books I have on the m1 Garand I have found that the operating rod is made out of 1050 steel:

According to the MIL-HDBK (version 5j) 1050 steel has the following properties:

Elastic modulus: 2.75572x10^7 PSI

Using Euler's theory, a rod under compression will buckle when its axial load reaches:

(pi^2EI)/L^2

where E is the elastic modulus, I is the area moment of inertia of the rod's cross section, and L is the length of the member

This is then divided by the area to get the normal stress required to buckle the rod. The area used to compute is (2*pi*r*t) where r is the outer radius of the rod, and t is the thickness of the cylinder wall. So we get:

(pi^2*2.75572x10^7*(pi/64*(.506^4-.401^4)))/(15.83"^2)
(2*pi*.253"*((.506"-.401")/2))

This gives a value of: 25342.3 PSI

The diameter of end of the op rod where the gas piston is, is .5255"

using Sigma = P/A where sigma is the normal stress (pressure on the end of the rod) P is the axial force and A is the area of the end of the op rod, we get:


(pi^2*2.75572x10^7*(pi/64*(.506^4-.401^4)))/(15.83"^2)
(pi*(.5255"/2)^2)

= 9751.46 PSI

So the max port pressure the Garand can handle before op rod to buckle is about 9752 PSI
(of course you would want to use less than this to be safe)

Things that can cause this number to fluctuate are cyclic fatigue from multiple cycling through the rifles life, variations in the strength of the steel used in each op rod due to imperfections, and the fact that since the op rod is bent (for design purposes) the strength is lower since the rod has already yielded.

Some may find this information interesting or useful, others may find it dry and boring.

So where am I going with this?

Well I came across the difference in gas cylinder locks between the early garands and the late type. And I noticed that the Late type has a valve stem in it to act as a pressure relief valve, so I am curious, how many people have either had or heard of an operating rod being bent by a Garand that had the late type gas cylinder lock on it?

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Rugerlvr
June 5, 2008, 02:52 PM
Are you considering that the op-rod is not straight? Or are you only considering failures that happen in the straight and tubular portion? What about failures at the cutout, or at the bolt cam?

GarandOwner
June 5, 2008, 02:58 PM
The only failure I was interested in in this example was failure due to buckling, not failure due to other yielding or fracture. This would only occur in the rod due to its length. Since this part has already reached its yield strength (from the rod being bent in production) this would be ONE of the likely places failure would occur. Of course it could happen at other places, but due to the time it would take to go through all of those calculations, I only chose the tube portion of the operating rod in this example. Sheer force at the bend in the rod is another likely place for failure, but due to steels high yield stress (sheer would only occur perpendicular to the rod) buckling is a more immediate threat to failure. When I get home tonight I will go through the sheer calculations to see what the force would be to cause that yielding due to sheer stress (though it is almost certain to be above the force necessary to cause buckling)

Also since buckling is known to occur before other failures in Garand operating rods, they can be neglected

jlbraun
June 5, 2008, 03:18 PM
Yay for engineering!

LB7_Driver
June 5, 2008, 03:33 PM
Are the formulas used for a straight section of rod, or a bent section?

A section of tubing that is not straight will bend under a much lower load than a section that is straight. You mention this in your write-up, though I wonder if the formula(s) are valid for a curved section of tubing.

30Cal
June 5, 2008, 04:12 PM
My guess would be it'd be significantly lower due to the bends pre-existing in the rod.

Also, the poppet isn't a relief. It opens inward (as in when you install the grenade launcher). Pressure inside the gas cyl seals the poppet firmly against it's seat. Scroll down a this link a couple pages to see the grenade launching action.

http://www.chestnutridge.com/gchart.asp

10X
June 5, 2008, 05:03 PM
I think you may have missed something regarding the late plug with the valve.

That valve is there so a rifle grenade launcher pin presses open the valve to allow gas to escape when a rifle grenade is fired.

In my young M1 days in the 1960s I bent a rod. I was firing USGI ammo. What I didn't know was that WWI ammo was not compatible with the M1. It had to be WWII ammo. The ammo wasn't over (chamber) pressure, it had too much port pressure. The late gas plug mattered not.

GarandOwner
June 5, 2008, 05:28 PM
Also, the poppet isn't a relief. It opens inward (as in when you install the grenade launcher). Pressure inside the gas cyl seals the poppet firmly against it's seat. Scroll down a this link a couple pages to see the grenade launching action.

.30 cal Thanks for the information about the Gas plug, I was wondering about how it worked.

The late gas plug mattered not.

Thanks I didn't know how the "relief valve" worked on it so I was wondering if it made a difference,

To all:

Since pressure is uniform, it acts at the centriod of the operating rod, for this reason there is no eccentricity in the force applied, so Euler's theory can be applied. Buckling occurs because of NORMAL stress, that is stress that occurs along the length of the rod. The offset will cause SHEER stress, but in order for it to fail due to sheer yielding, the sheer stress would need to be significantly larger than the normal stress. For this reason, the bar will fail due to buckling before it will fail by sheer yielding. What will cause the value to be lower than the calculated value is the fact that the bar has been yielded before, however it is nearly impossible to calculate the strength of a member that has already yielded.

This was the understanding that I took from my Solid mechanics/ statics and structural design courses, however I am an aeronautical engineer not a civil engineer. If there is anyone else will a higher understanding of these types of loadings I would appreciate any corrections if my assumptions are wrong Thanks!

MutinousDoug
June 5, 2008, 07:33 PM
GarandOwner;
I assume you are describing the rod OD as .506" and it's ID is .401"? If that is the case then the wall would be (.506-.401)/2.
The button on the end of the rod is nominally .526" dia. if memory serves so the area of that diameter is the pressure area =.2169in^2.

Please correct me if I am misunderstanding.

Doug

SlamFire1
June 5, 2008, 08:48 PM
Is the bending you calculated a permanent deformation or just a flexing?

I believe the permanent bend in the operating rod occurs in the saddle section. And that happens when the operating rod stops in front of the receiver.

I could be wrong, as I have never "bent" a M1 Garand operating rod.

However I have "bent" a M14 operating rod. The M14 operating rod is much stiffer. I had an earlier Super Match. The operating rod guide was not a GI configuration. The operating rod guide was a piece that fit into a cut dovetail on the bottom of the barrel. The dovetail was cut off center. I assume that is why the saddle section bent, an offcenter unbalanced load on impact with the front of the receiver.

So I assume that is what happens with a M1 operating rod.

Could be wrong.

bpm990d
June 5, 2008, 08:52 PM
In Hatcher's Book of the Garand he states that the M1 Rifle was developed with the intent that it would use the M1 ammunition. The development of M2 ball was so that the ammunition would stay within the safety fan of most post rifle ranges.

FM 23-5 dated Nov. 14 1941 lists M1 Ball ammunition as an acceptable cartridge for the M1 Rifle.

B

GarandOwner
June 6, 2008, 04:15 AM
Is the bending you calculated a permanent deformation or just a flexing?

buckling as calculated in this example is permanent deformation

GarandOwner
June 6, 2008, 01:42 PM
Also I left out the step on how to get from normal stress in the rod to port pressure, this step has been added to make it easier to follow the calculations

Art Eatman
June 6, 2008, 01:45 PM
eclancy oughta have some data about the design pressure at the gas port. I've read--albeit decades ago--that it was around 2,000 psi, but don't take my memory as Gospel.

The muzzle velocity is 32,400 INCHES per second, and the bullet's exit means the pressure instantly falls to zero. Thus the time of application of pressure is in milliseconds. The diameter of the gas port won't affect the pressure.

Dunno if all this matters, but like I said, I'd ask eclancy for input.

Art

GarandOwner
June 6, 2008, 01:50 PM
Thanks for the info Art, 10000 PSI is what I got, there is a range since 1050 steel has a range of Elastic Modulus since it can vary due to how it is heat treated, I took an average of the values for my example, I was hoping eclancy would have chimed in, he always has great information and insight into the Garand's history

strat81
June 6, 2008, 02:16 PM
Let's say you can say determine what pressure will bend the op rod. How do you then determine the port pressure of a given load with X type of powder and Y type of bullet?

Art Eatman
June 6, 2008, 02:40 PM
You'd have to have a graph of the pressure curve of the powder. Such a graph shows the change in pressure during its burning time as the bullet travels down the barrel.

Slower powders have a peak pressure farther down the barrel than faster powders, although the curves pretty much always have quite similar shapes.

Folks get into trouble with the Garand by using slower powders and heavier bullets. They read the reloading manuals, see the nice velocities available, and don't realize that at the gas port the pressure is above the design level--even though the maximum chamber pressures are the same as the GI load.

Only the powder manufacturers have the graphs. I don't know about their availability to the public.

From the FWIW department: Back around 1951, a buddy had a Garand. I had a ton of GI ammo, and also was reloading for the '06. We shot a lot of GI stuff, and not many of the handloads except at a few jackrabbits and skunks in our nighttime forays. Kids do stuff like that. (I know a grownup who still does.) Anyhow, the limited amount of "hot load" shots didn't hurt the op-rod. IIRC, 53.5 grains of 3031 behind Mr. Hornady's 110-grain spire points.

But if you stick with the GI load of 4895 and any good 150-grain bullet, you'll never harm the op-rod--and that load will kill any critter in the lower 48.

MutinousDoug
June 6, 2008, 02:45 PM
GarandOwner;
Your formulas are still all messed up. Grinding them out does not yield the numbers you have given.
And the formula for area of the rod piston is: (pi*.5255^2)/4 where you have: (2*pi*(.5255"/2) ?

MutinousDoug
June 6, 2008, 03:31 PM
Strat81;
You could generate the graphs Art E is talking about by mounting a pressure transducer just above your Garand's chamber and recording the P/T trace. Or, more simply by mounting one in the gas plug (old style: un-vented, drilled and tapped). That would only give you the trace once the bullet had passed the port for the port pressure but that's what you were interested in anyway. Pretty inexpensive (after you buy an oscilloscope) and you could move it from Garand to Garand!
Then, when you were done, you could report here and tell us what port pressure begins to bend your op rods.

GarandOwner
June 6, 2008, 04:16 PM
(pi*.5255^2)/4 where you have: (2*pi*(.5255"/2)

I forgot to change it from the area of a cylinder which is 2*pi*r*t :banghead: thats what happens when I tried to rush through it and not recheck my answers. I reworked the equations and used a different formula for the area moment of inertia. An old professor of mine gave us the equation Pi*r^3*t, but I have not seen it anywhere else, so I used the more common form which is pi/64*(Do^4-Di^4)


Thanks for double checking my work Doug, sometimes it takes another person to go through and catch "simple" mistakes

hps1
June 6, 2008, 05:16 PM
As 30Cal and 10X pointed out, the "late style" gas cyl. lock plug valve serves only as a grenade launching valve and does nothing to reduce gas port pressure.

The valve described in the following link can be adjusted to vent excess pressure and should elliminate op=rod damage due to slow powder/heavy bullet use. Can't comment on it, however, as I have never used one.

http://www.adcofirearms.com/gasnuts.cfm

Regards,
hps

peyton
June 6, 2008, 11:56 PM
And who says John Garand was not a genius?? Do you all think that the design team used this math to create the rifle? I mean does the math we use now (the formula's above) exist back then or did they just wing it? It is my favorite rifle I own.

MutinousDoug
June 7, 2008, 11:20 PM
Peyton;
My copy of Timoshenko's: Elements of Strength of Materials is the 2nd edition from 1943 but the 1st edition was published in 1935. I'm sure it's precursors were available to Springfield Armory and the Dept of the Army for John Garand's people to peruse. It is not rocket science; it is static mechanics as applied 100 years before then and now.
John Garand and John Browning were not college educated. Their application of mechanics was certainly inspired. I suspect the secondary schools they attended were more rigorous than ours now.

GarandOwner
June 8, 2008, 12:37 AM
Euler's equation has been around since 1757, most solid mechanics have been around for hundreds of years

Art Eatman
June 8, 2008, 12:42 AM
FWIW, Euler is pronounced oiler. :)

Funny-odd: I busted my brain trying to pass advanced calculus in college. After I got out, I only used integral calculus when driving a race car deep into a corner or shooting at a running buck. :D:D:D

Swampy
June 8, 2008, 08:36 AM
eclancy oughta have some data about the design pressure at the gas port. I've read--albeit decades ago--that it was around 2,000 psi, but don't take my memory as Gospel.

Art, IIRC, and I could be wrong here too.... the gas port pressure the M1 is supposed to have for correct function is in the 4000 to 6000 psi range.

Best,
Swampy

Garands forever
2007 NRA Missouri State 600 yard Service Rifle Champion.
Score 774-29X.... with an M1.

GarandOwner
June 8, 2008, 10:34 AM
Thanks for the information Swampy, If 6000 psi is the max design pressure, then it would make sence for the rod to be made to withstand a force of about 1.5 times that. Alot of times in engineering (at least in Aeronautical applications) we will use a factor of saftey when designing things of around 1.5 (more for critical components).........although other engineering disciplines think we are crazy for using such low factors of saftey ;)

Art Eatman
June 8, 2008, 11:02 AM
:D Swampy, as I said, I'm nowhere near positive about my 2,000 number.

It would be nice to have the pressure curve of the 4895 burn, with the horizontal axis being length in inches. That would give a direct plot for the pressure at the gas port.

Again, relying on an imperfect memory, most of the published curves are pressure vs. time.

Grumble, gritch...

redneck2
June 8, 2008, 05:03 PM
...did you take into account the inertia it takes to overcome the mass and get the rod moving? The rod's trying to stay still and you're trying to move it to hyper velocity in a millisecond.

I just looked at the stuff briefly, as I'm off to do something more exciting...

like watching paint dry....

:neener:

Swampy
June 8, 2008, 05:57 PM
...did you take into account the inertia it takes to overcome the mass and get the rod moving? The rod's trying to stay still and you're trying to move it to hyper velocity in a millisecond.

Nahh.... not really.

While the accelleration rate may be pretty impressive, the actual top speed of the op-rod is only something on the order of 20 mph (29.333 fps).

Best to all,
Swampy

Garands forever
2007 NRA Missouri State 600 yard Service Rifle Champion.
Score 774-29X... with an M1

GarandOwner
June 8, 2008, 07:04 PM
...did you take into account the inertia it takes to overcome the mass and get the rod moving? The rod's trying to stay still and you're trying to move it to hyper velocity in a millisecond.

actually yes thats what the area moment of inertia is for (how the member deals with inertia due to its cross sectional area). The inertial force IS P, because all an inertial force is, is the objects mass times its acceleration

GarandOwner
June 9, 2008, 06:27 PM
If you really wanted to, you can solve for the maximum acceleration. Either by Newton's method where the sum of the forces = ma or by D'alembert's method where there is an inertial force, and the sum of the forces = 0 (Both will give you the same result). I will show it using D'alembert's method:

Sum of Forces along the axis of the length of the operating rod = 0

so from the Free Body Diagram of the Operating rod:

http://www.thehighroad.org/attachment.php?attachmentid=79688&d=1213050439



- Pcr +FI = 0

where FI is the inertial force = (w/g)a that acts opposite acceleration


Solving for a we get:

aMAX = ((pi^2EI)g)/(wL^2)

(I cant get a numerical answer for this as I do not know how much the operating rod weighs.)

GarandOwner
June 9, 2008, 06:35 PM
I just looked at the stuff briefly, as I'm off to do something more exciting...

like watching paint dry....

hehe as I said in the OP some may find this information dry, but being an engineer, when I think of a problem, I can not rest until I solve it. I know math is not everyone's cup of tea. We engineers are a strange breed ;)

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