As of the time of writing (early 2021), the safe handling of CO2 magazines for gas guns has become a bit of a topic of discussion.
Considering the current efforts to address environmental issues, it's easy to imagine that the number of CO2 gas gun users will increase in the future.
Since I have this opportunity, I'll try performing strength calculations from an engineer's perspective.
Still moreThis content is not intended to criticize CO2 gas guns or any specific company, organization, or system.
*This article aims to provide educational explanations from a mechanical engineering perspective regarding the safe hobby use of airsoft guns.
This does not promote acts of violence, the use of weapons, weapon modification, practical evaluation, or violation of laws and regulations.
Mechanism and operating environment
First, let's calculate the cost of what is called a sealed magazine.
The components related to strength are as follows:
Name the magazine body, valve, CO2 cylinder, sealing bolt (or cap bolt), and other parts.
Next, we'll consider the usage scenario using a flowchart.
When installing a carbon dioxide cylinder
When a CO2 magazine is used (when firing BBs)
When removing the carbon dioxide cylinder
This is roughly what the operating environment (lifecycle) of the magazine and carbon dioxide cylinder looks like.
There are a few parts that feel like leading questions, but let's try not to worry about them.
Next, let's check the dimensions around the magazine.
Survey of magazine dimensions
First, let's check the dimensions of a standard carbon dioxide cylinder. We'll measure a carbon dioxide cylinder from a certain domestic manufacturer (it should be almost the same as others).
All units are in millimeters (mm).
Next, we will measure the magazine, using a magazine from a Glock 17, a major domestic manufacturer.
The reason is that, although it's a simplified representation, it's a representative example of the 9mm, double-column type magazines that make up the majority of the world's ammunition, so I expect that there won't be any dimensional differences that would affect calculations for the same caliber, double-column type. (Except for .45 caliber and special ammunition.)
Since we are considering a hypothetical CO2 magazine here, the dimensions of the sealing bolts are unknown, but we can determine the range of possible dimensions for the sealing bolts from the dimensions we have researched.
First, the minimum diameter of the sealing bolt that can be used must be 18.6 mm or larger, as the carbon dioxide cylinder must be removable, according to the flowchart of the usage instructions.
Also, if the dimensions are too tight, it will be difficult to remove the carbon dioxide cylinder, so let's set the manufacturing tolerance to 0.5 in radius and a clearance of 1 for ease of use.
Then
φ21.6 < Sealing bolt diameter
Next, let's consider the maximum diameter from which a sealing bolt can be used.
Since the magazine's width is 23mm, it obviously cannot be made any larger.
Sealing bolt diameter <φ23
The available bolt sizes are determined by the standard, so you can't choose any bolt size other than M22. (Actually, if you were to set it to M22, the internal thread strength would likely be insufficient.)
Since it seals gas, the screw pitch is fine, and according to JIS it is 1.0, so M22x1.0 (considering the pilot hole diameter, the cylinder seems like it will be a tight fit).
Next, let's consider the length of the sealing bolt.
The length of the sealing bolts can be set to a fairly flexible size, so let's leave this as variable L and perform the strength calculation.
Image of the layout around the sealing bolts
Similarly, the available valve sizes are naturally determined by this.
This time, based on my observations and experience, I'll focus on the sealing bolt and the thickness of the magazine, as these are the points that catch my eye.
Investigating the force generated in a CO2 magazine
Since the working gas is carbon dioxide, you can quickly look up its physical properties online (you just need to look at the saturated vapor pressure curve).
・4.0 MPa at a temperature of 6.3℃
・4.5 MPa at a temperature of 10.9℃
・5.0 MPa at a temperature of 15.1℃
・5.5 MPa at a temperature of 19℃
・6.0 MPa at a temperature of 22.9℃
・6.5 MPa at a temperature of 26.1℃
• Critical point at 31.1°C and 7.38 MPa
This level of pressure should be sufficient for use in this environment. It's quite intense.
Atmospheric pressure is approximately 100 kPa, so this is 40 to 70 times greater.
For simplicity, we'll base our calculations on 5.5 MPa at 19°C. (If you use multivariate analysis, you can directly use the equation for the saturated vapor pressure curve, such as mode frontier analysis.)
In this case, since we are calculating for a sealed container, we assume that a uniform pressure of 5.5 MPa is applied inside the container. The force on the sealing bolt is pressure x area, so we find the area from the effective diameter of the screw.
The effective diameter of an M22x1.5 screw is φ21.0 mm according to the standard, and its area is 346.3 mm².
Therefore, the force acting on the sealing bolt is $ 5.5 × 10^6 × 346.3 × 10^{-6} = 1904 (N) $
It will be 1.9kN.
Yes, I can see that the screws are fine at this level. (M6x0.8 can handle up to about 10kN)
So the only remaining concern is the thickness of the magazine.
According to Pascal's principle, the pressure inside the magazine will be constant, so we can assume it is 5.5 MPa at 19°C.
Furthermore, the usage chart indicates that when a bullet is fired, gas is released, causing a decrease in pressure inside the magazine.
Therefore, since it will be subjected to repeated loads, it seems necessary to consider fatigue strength as well.
However, since the degree of decompression upon release is unknown, let's consider atmospheric pressure, half of the original pressure, and two-thirds of the original pressure.
- When pressure drops to atmospheric pressure: Initial pressure 5.5 MPa, average pressure 2.75 MPa, amplitude pressure 2.75 MPa
- When the pressure drops to half of its original level: Initial pressure 5.5 MPa, Average pressure 4.13 MPa, Amplitude pressure 1.38 MPa
- When the pressure drops to 2/3 of the original level: Initial pressure 5.5 MPa, average pressure 4.59 MPa, amplitude pressure 0.92 MPa
I think.
Next, let's consider the stress generated within the magazine.
Stress generated within the magazine
Next, we need to determine the stress generated inside the magazine. Since we don't have CAE or similar tools, we'll simplify the shape for our calculations.
We will consider this as a thin-walled cylindrical container.
For details on stress calculations for cylindrical pressure vessels, click here →Strength of Materials for Beginners 26: Stress in Thin-Walled Pressure Vessels (Cylindrical Pressure Vessels, Spherical Pressure Vessels, and Pressure Vessels in General)
First, let's imagine the magazine as a simple cylinder and let its wall thickness be t.
Furthermore, we will consider the stress-generating areas separately as the circumferential portion of the upper and lower lids and the cylindrical portion.
Since there is a formula, we can quickly calculate it by letting the pressure be P (MPa) and the wall thickness be t.
Stress at the bottom (axial direction): $ σz=\frac{P・22・10^{-6}}{4・t・10^{-6}}=\frac{5.5・P}{t} (MPa) $
The circumferential stress is given by $ σr = \frac{P・22・10^{-6}}{2・t・10{^-6}}=\frac{11・P}{t} (MPa) $
Next, we consider the range of values that P can take and the range of values that t can take.
P is 4-7.38 (MPa) based on material properties, and considering the wall thickness t, the magazine width is 23, the screw is M22x1.0, and the pilot hole is φ21 according to the standard, so if we draw a diagram...
The thinnest wall thickness is calculated using the diameter of the female thread root and the outer thickness t2.
The diameter of the female thread's valley is φ22 according to the standard, and the magazine width is 23, so it's 0.5 (this is scary).
Similarly, t4 is approximately 1 because the diameter of the threads of the female thread is φ20.917.
t2 has a female thread pilot hole diameter of φ21, so it's roughly 1
The remaining t3 just needs to avoid hitting the cylinder, so in an extreme case, considering the cylinder diameter φ18.6, it's 2.2
Therefore, the range of t is 0.5
The conditions are met. 4
I would actually like to create a 3D graph, but that's not possible with my current environment, so I'll just use a table.
Now we know the generated stress. Next, let's look at the material properties.
Magazine material
Considering cost, weight, and mass production feasibility, the only materials that can be chosen for the magazine are cast aluminum or zinc. (Casting aluminum is abbreviated as ADC, and zinc as ZDC.)
Looking at two physical properties at room temperature (there are various materials, but the table is created using approximate averages)
When it comes to aluminum, which I often use, considering casting quality, creep, and brittle fracture, I would like to consider the yield point of aluminum to be around 100 MPa. However, since this is a hobby in a different genre, I will use the values in the table.
For more detailed information on stress-strain diagrams, please click here.
Furthermore, the design of pressure vessels is specified in JIS as follows. Since the JIS document is quite long, here is an excerpt of the relevant parts:
If the generated stress is within 100 MPa, the value shall be less than or equal to the smaller of the following values.
(This time, there are areas exceeding 100 MPa, but we'll use them as a bonus. Actually, you shouldn't use them.)
1) $ \frac{1}{3}$ of the specified minimum tensile strength at room temperature
2) The tensile strength at each temperature $ \frac{1}{3}$
3) $ \frac{1}{1.5}$ of the specified minimum yield point or 0.2% proof stress at room temperature.
4) The yield strength or 0.2% proof stress at each temperature is $ \frac{1}{1.5}$ or 0.9
Therefore, if we adopt JIS B8266, the material properties will be as follows:
I don't understand why the yield point of zinc is higher than its tensile strength, but intuitively, it feels like "this is about right" for safe use, so I can accept the value.
Intensity determination
Next, we will determine the strength based on the generated stress and the material.
Since the generated stress is greater in the circumferential direction, we will examine it using the circumferential stress σr.
Furthermore, since amplitude stress is generated according to the usage flowchart, use within the elastic range is the principle, and therefore the evaluation is made within the yield point (0.2% proof stress).
First, in the case of aluminum ADCs considering JIS B8266, (the area in red is the damaged area)
Next, in the case of zinc (ZDC)
Therefore, if we were to adopt JIB 8266, it seems unlikely that the specifications would be met.
Furthermore, if we simply compare it to the strength of the material, there are areas where aluminum cannot withstand certain pressures.
Conclusion and Discussion
As conclusion,
The specifications do not apply to a Glock 17 size from a certain major domestic manufacturer with a magazine width of 23mm, a sealed type using standard CO2 cylinders, an M22x1.0 sealing bolt, and a material of aluminum die-cast or zinc die-cast, and there is a high possibility of breakage.
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The minimum wall thickness is 0.5 mm, but if casting quality issues or machining deviations occur, and the wall thickness decreases further, it is highly likely that the tensile strength will be exceeded and a single-shot fracture will occur.
The image of the destruction is as shown in the diagram.
As a countermeasure, considering casting quality, machining deviations, and other factors, it would be better to widen the magazine to achieve a minimum wall thickness of around 1.5, or even 2.
Alternatively, you could try reducing the gas pressure or using a relief valve.
If youIf you have a CO2 magazine of the size assumed in this strength calculation, we strongly recommend that you measure the threads of the sealing bolt and the magazine width to confirm that the wall thickness is at least 1 mm.
If the thickness is less than 1mm, I strongly recommend discarding it for safety reasons, even though it may seem wasteful.
If it breaks, a large pressure of 4 MPa to 7.38 MPa will be released, making it extremely dangerous.
Please refer to the diagram for instructions on how to measure.
If you can't accurately measure the diameter of the female thread's root, you can use the measured value to Google the screw specifications, identify the screw size, and then determine the diameter of the female thread's root.
You can definitely find the screw specifications right away.
またNever remove the sealing bolts before the carbon dioxide gas has been depleted.
If the gas pressure remains the same but the thread engagement of the sealing bolt decreases, the resulting stress increases, causing the sealing bolt to fly off at high speed.
At worst, you could get injured.
To ensure fun and safe play, please read the instructions carefully and handle the product with care.
Also, as I've written in the instruction manual, after you're finished playing, try to use up as much gas as possible, remove the gas canister from the magazine, and store it properly.
If high stress continues to occur for a long period of time, strain will gradually accumulate, and the material may break due to temperature changes or slight stimuli (creep phenomenon).
My initial impression was that fatigue failure was caused by gas pressure fluctuations inside the magazine during operation, but I was surprised to find that there was a very high possibility of sudden, one-shot failure even before that (which is impossible from a design perspective).
Everyone, please be careful when choosing products.
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