9 major cracked slides? Read this.

[MikeBurgess]

10 hours ago, theWacoKid said:

 

It actually does change how the load is applied to the slide. A FBD diagram will explain how the accelerating bullet reduces the tensile load in the slide/barrel assembly as compared to treating it as a static pressure vessel. 

 

The extreme thought experiment is how firing a 9mm in a 40 barrel works. 

yep 9 in a 40 = very low case pressure,

I fail to see how that thought experiment is relevant or how it show the redirection of forces?

 

 

Edited September 4, 2019 by MikeBurgess

[theWacoKid]

2 hours ago, MikeBurgess said:

yep 9 in a 40 = very low case pressure,

I fail to see how that thought experiment is relevant or how it show the redirection of forces?

 

 

 

The magnitude of the pressures is irrelevant for the basis of how this works.  A 9mm in a 40 barrel will still fire and that's the concept that matters. 

 

It is relevant because a 9mm fired in a 40 barrel applies no static tensile load from the pressure build up to the barrel/slide assembly (because there no friction) yet there is still equal pressure acting against the breech face and bullet base.  Following your logic you would have to conclude the tensile load in the barrel/slide is still equal to the load that acts equal and opposite on the bullet base and breech face, but that is not the case. In this case all loads from pressure act to accelerate the bullet one way and the slide/barrel the other.  Therefore demonstrating how the equal and opposite forces from the internal pressure on the bullet and breech face only generate tensile load in the barrel/slide equivalent to the frictional force opposing the motion of the bullet down the barrel.

 

You previously said:

"If a load has a peak pressure of 65 KSI it will try to push the breach face away from he barrel with about 7,800 lbs of force."

 

All I'm saying is, yes, there is 7,800 lbf acting on the breech face but it's NOT 7,800 lbf pushing the breech face apart from the barrel (which puts the assembly in pure tension).  The amount of force pushing the breech face away from the barrel is only equal to the frictional force of the bullet traveling down the barrel.  The remaining force from the 7,800 lbf is unreacted on the slide/barrel assembly and thus becomes an external load acting on a free body causing the unit to accelerate.  

 

To cover all the bases, there will be significant tensile load in the slide due to its inertia and the sudden acceleration, but nobody seems to understand enough to get to this point.

 

I've heard your argument before and I'm just doing my best to share the basic correct information so that it exists.

 

I will reply again only to correct a FBD that you draw trying to defend your point.

  

 

 

 

[Racinready300ex]

How much effect does one of those stepped cases have on your pressure? What does that do to the gun?

[MikeBurgess]

46 minutes ago, theWacoKid said:

 

The magnitude of the pressures is irrelevant for the basis of how this works.  A 9mm in a 40 barrel will still fire and that's the concept that matters. 

 

It is relevant because a 9mm fired in a 40 barrel applies no static tensile load from the pressure build up to the barrel/slide assembly (because there no friction) yet there is still equal pressure acting against the breech face and bullet base.  Following your logic you would have to conclude the tensile load in the barrel/slide is still equal to the load that acts equal and opposite on the bullet base and breech face, but that is not the case. In this case all loads from pressure act to accelerate the bullet one way and the slide/barrel the other.  Therefore demonstrating how the equal and opposite forces from the internal pressure on the bullet and breech face only generate tensile load in the barrel/slide equivalent to the frictional force opposing the motion of the bullet down the barrel.

 

You previously said:

"If a load has a peak pressure of 65 KSI it will try to push the breach face away from he barrel with about 7,800 lbs of force."

 

All I'm saying is, yes, there is 7,800 lbf acting on the breech face but it's NOT 7,800 lbf pushing the breech face apart from the barrel (which puts the assembly in pure tension).  The amount of force pushing the breech face away from the barrel is only equal to the frictional force of the bullet traveling down the barrel.  The remaining force from the 7,800 lbf is unreacted on the slide/barrel assembly and thus becomes an external load acting on a free body causing the unit to accelerate.  

 

To cover all the bases, there will be significant tensile load in the slide due to its inertia and the sudden acceleration, but nobody seems to understand enough to get to this point.

 

I've heard your argument before and I'm just doing my best to share the basic correct information so that it exists.

 

I will reply again only to correct a FBD that you draw trying to defend your point.

  

 

 

 

Ok 

I agree my static model is too simplified, 

 

I see so many more open slides cracking at the ejection port than I do limited slides with both slides being milled similarly, it seems most likely to me that the portion of force that is forcing the barrel and breach apart is the likely culprit. 

[gianmarko]

On 9/4/2019 at 6:42 PM, theWacoKid said:

 

The magnitude of the pressures is irrelevant for the basis of how this works.  A 9mm in a 40 barrel will still fire and that's the concept that matters. 

 

It is relevant because a 9mm fired in a 40 barrel applies no static tensile load from the pressure build up to the barrel/slide assembly (because there no friction) yet there is still equal pressure acting against the breech face and bullet base.  Following your logic you would have to conclude the tensile load in the barrel/slide is still equal to the load that acts equal and opposite on the bullet base and breech face, but that is not the case. In this case all loads from pressure act to accelerate the bullet one way and the slide/barrel the other.  Therefore demonstrating how the equal and opposite forces from the internal pressure on the bullet and breech face only generate tensile load in the barrel/slide equivalent to the frictional force opposing the motion of the bullet down the barrel.

 

You previously said:

"If a load has a peak pressure of 65 KSI it will try to push the breach face away from he barrel with about 7,800 lbs of force."

 

All I'm saying is, yes, there is 7,800 lbf acting on the breech face but it's NOT 7,800 lbf pushing the breech face apart from the barrel (which puts the assembly in pure tension).  The amount of force pushing the breech face away from the barrel is only equal to the frictional force of the bullet traveling down the barrel.  The remaining force from the 7,800 lbf is unreacted on the slide/barrel assembly and thus becomes an external load acting on a free body causing the unit to accelerate.  

 

To cover all the bases, there will be significant tensile load in the slide due to its inertia and the sudden acceleration, but nobody seems to understand enough to get to this point.

 

I've heard your argument before and I'm just doing my best to share the basic correct information so that it exists.

 

I will reply again only to correct a FBD that you draw trying to defend your point.

  

 

 

 

i see your point but not sure i agree completely

at some stage there is indeed pressure pushing the case head against the breech face, else blowback would not work.
or blank firing guns. 

[Knguyen1904]

On 9/3/2019 at 8:07 AM, GetAwayDriva said:

I’ve cracked two slides in less than a year.  First crack was at the ejection port, second one on the replaced slide was at the extractor tunnel.  

 

Same ! According to my smith it’s bound to happen at some point . 

[echotango]

I cracked 2 last year within a month(2 diff guns). 1 on the left side, one on the right side.

Edited September 6, 2019 by echotango

[Ssanders224]

5 hours ago, gianmarko said:

i see your point but not sure i agree completely

at some stage there is indeed pressure pushing the case head against the breech face, else blowback would not work.
or blank firing guns. 

 

Wacos post isn't really something to agree or disagree with, as it is factually correct. 

He never said there was no pressure acting on the breech face.  He stated that it is equal to the frictional force of the bullet traveling down the barrel, which is physics 101. 

I'd agree with him that unless someone wants to sketch out an accurate FBD that shows otherwise (good luck), there isn't much more to discuss. 

[Yondering]

On 9/3/2019 at 1:59 PM, Whoops! said:

 

 

It’s going to get very complicated if we look at fatigue points of the slide with a greater pressure for a shorter period of time than a lesser pressure for a longer period of time.  But, I believe that once the calculations are done, you will see very little change in breech or slide fatigue from that initial force with fast versus slow powders operating at the same energy levels.

 

 

It's actually not complicated if we're looking at fatigue caused by case pressure - time doesn't contribute to fatigue. Fatigue is based on stress/strain level (resulting from pressure in this case) and number of cycles. We can load a steel part slowly or quickly and generate the same damage, as long as we're talking about fatigue and not elastic deformation or shock.

My professional career is in durability testing, mostly of metal parts and systems. When we generate cyclic loading in a part to test it, we can speed up the test by cycling faster at the same levels and still get the same result. 

 

If we have a fast powder and a slow powder producing the same peak pressure, for the same number of shots, the fatigue damage is the same from each. However the fast powder load will obviously be slower. If we're looking at a fast powder and a slow powder producing the same velocity, the fast powder will hit higher peak pressure, causing more fatigue damage. The area under the pressure curve affects bullet velocity, but does not affect fatigue on the steel slide (that is caused by pressure, recoil is a different thing). This is assuming that each pressure curve has only one peak, which is not always a correct assumption but works to understand this point. 

 

 

Edited September 6, 2019 by Yondering

[yigal]

the impact time between slide and frame is the most critical factor: f= mv/dt   and if u will work with higher rate RS this will make slide to live longer.

buffer can help to increase impact time and prevent slide cracking..

[gianmarko]

On 9/6/2019 at 5:12 PM, Ssanders224 said:

 

Wacos post isn't really something to agree or disagree with, as it is factually correct. 

He never said there was no pressure acting on the breech face.  He stated that it is equal to the frictional force of the bullet traveling down the barrel, which is physics 101. 

I'd agree with him that unless someone wants to sketch out an accurate FBD that shows otherwise (good luck), there isn't much more to discuss. 

perhaps i dont understand exactly what "frictional force" means

 

bullet intertia is no factor? 

[theWacoKid]

On 9/9/2019 at 8:40 AM, gianmarko said:

perhaps i dont understand exactly what "frictional force" means

 

bullet intertia is no factor? 

I never said there is no pressure against the breech face.  In fact, I said exactly the opposite.

 

Bullet inertia (mass) matters to the profile of the pressure curve which will change the equal and opposite forces from total chamber pressure acting on the bullet base and the breech face.  It's a smaller effect and not critical to the heart of this discussion.  *How* the pressure loads are carried and what they act upon is the item of discussion. 

 

The amount of static tensile load carried by the locked barrel/slide assembly (i.e. the force attempting to push the slide rearward and pull the barrel forward across the interaction of the locking lugs) is only equal to the resistance (friction) of the bullet traveling down the barrel.  The remaining equal and opposite forces not carried in tension by the slide/barrel assembly are unreacted at the bullet base and the breech face and thus accelerate the bullet forward and the slide rearward.

 

I'm not sure how many more ways I can continue to say the same thing.

 

[GOF]

On ‎9‎/‎3‎/‎2019 at 8:03 AM, Sarge said:

Sometimes you just get bad slides. A few years back STI made some defective slides. Matt used one on my build and it cracked. STI replaced it.

A Steel Challenge shooter at my club recently had a cracked slide on a STI comped gun, and he was shooting standard pressure loads. He sent it back. I do not know the outcome.

[Whoops!]

On 9/6/2019 at 5:19 PM, Yondering said:

 

It's actually not complicated if we're looking at fatigue caused by case pressure - time doesn't contribute to fatigue. Fatigue is based on stress/strain level (resulting from pressure in this case) and number of cycles. We can load a steel part slowly or quickly and generate the same damage, as long as we're talking about fatigue and not elastic deformation or shock.

My professional career is in durability testing, mostly of metal parts and systems. When we generate cyclic loading in a part to test it, we can speed up the test by cycling faster at the same levels and still get the same result. 

 

If we have a fast powder and a slow powder producing the same peak pressure, for the same number of shots, the fatigue damage is the same from each. However the fast powder load will obviously be slower. If we're looking at a fast powder and a slow powder producing the same velocity, the fast powder will hit higher peak pressure, causing more fatigue damage. The area under the pressure curve affects bullet velocity, but does not affect fatigue on the steel slide (that is caused by pressure, recoil is a different thing). This is assuming that each pressure curve has only one peak, which is not always a correct assumption but works to understand this point. 

 

 

 

 

If it’s not complicated, then get us the numbers so we know exactly what the difference is.  

 

My predication is there won’t be much difference, even though some difference is to be expected.

Edited September 10, 2019 by Whoops!

[Yondering]

1 hour ago, Whoops! said:

 

 

If it’s not complicated, then get us the numbers so we know exactly what the difference is.  

 

My predication is there won’t be much difference, even though some difference is to be expected.

 

Difference in what? 

 

I just said considering the amount oftime that pressure is applied doesn't make this complicated, because time doesn't matter for fatigue of metals. What did you think my post said? 

Edited September 10, 2019 by Yondering