How do I calculate thread length for fitting at end of pressure cyclinder

[perky416]

Evening all,

I am in the process of designing a fill valve and regulator that screw into either end of a cylinder that holds 250 bar of pressure. The thread at both ends of the cylinder is M25 x 1mm.

Does anybody have any idea how I would calculate the length of thread needed so the attachments dont blow out of the ends of the cylinder?

Regards
Lee

 

[3DDave]

All of them?

Typical thread deformation theory is that only the first 3 full threads take all the load, but since the material isn't specified it's not enough to determine if the parts are sturdy enough to resist the pressure.

If I had to this I'd start with the load F=PA for the section area of the fitting, and divide that by the Yield strength of the material, with a safety factor of 5, to get the shear area required. Divide by the thread pitch circumference to get the engagement. It should be less than 3 or 4 threads.

I would also contact the maker of the cylinder for any recommendations.

 

[perky416]

Hi dave

Thanks for the formula however I don’t seem to have got the answer I was looking for. I got an engagement length of 1.762 inches, the current fitting that came from the manufacturer only has an engagement length of 0.393 inches. The material being used for the fill valve is brass, material properties taken from

http://elginfasteners.com/resources/material-properties/brass-material-property-data-sheet/

Force = pressure / area
Force = 3625psi / 0.28 inches^2
Force = 12946.42 lbf

Area = Force/yield strength *5 safety factor
Area = 12946.42 lbf / 19600 psi * 5
0Area = 0.660 * 5 = 3.30

Engagement length = Area / Thread Pitch Circumference
Engagement length = 3.30 / 1.872
Engagement length = 1.762”

Contacting the manufacturer is out of the question, I have tried on many occasions about a different issue and when they do finally get back in touch after many attempts they just don’t want to speak to you.

I also found this formula on the internet to calculate pull out force which should help me determine engagement length however I seem to have messed this up as well. I used the engagement length of the original part to try and determine the pull out force.

Lbf = pi * pitch dia * shear strength * length of engagement


Pi = 3.1415
Pitch diameter of M25 x 1.0 thread = 0.596 inches
Shear strength of brass = 34100 psi
Engagement length = 0.393 inches

To work out pull out force:
Lbf = pi * pitch dia * shear strength * length of engagement

Lbf = 3.1415 * 0.596 * 34100 * 0.393

Lbf = 25091 lbs

To convert the force required into pressure:
Pressure = force / area
Pressure = 25091 / 0.28 inches^2
Pressure = 89610 psi or 6178.38 Bar.

As you can see I got an answer of over 6000 bar, when the original product is designed to operate at 232 bar this seems a little excessive.

Any ideas where I am going wrong?

Thanks.

 

[perky416]

Just gone over my pullout calcs again as i got the diameter wrong:

Calculate pullout force:
Pullout force = (pi * shear dia * length) * shear strength
Pullout force = (3.1415 * 24mm * 10mm) * shear strength
Pullout force = (3.1415 * 0.944" * 0.393") * 34100 psi
Pullout force = (1.165) * 34100 psi
Pullout force = 39726.5 lbs


Convert lbs to psi:
Psi = pullout force / area
Psi = 39726.5 / 0.699
Psi = 56833 psi

Convert to bar:
Bar = psi / 14.503
Bar = 56833 / 14.503
Bar = 3918 bar

Is 3918 bar an acceptable pullout pressure for something designed to operate at 232bar?

 

[3DDave]

Force = pressure times area.

 

[racookpe1978]

You're "thinking" too hard. Pipe threaded fittings (both metric and ANSI/standard) do not engage very far compared to a "nut and bolt/threaded rod/stud parallel mechanical thread.

Look up the required thread engagement for "metric pipe threads" for the 25 mm fitting you're working on at the design pressure. Are you designing both pressure vessel and regulator/valve, or the PV and threads for a commercial regulator/valve, or the regulator for an existing PV and threads?

 

[perky416]

What a dohnut regarding the formula, sorry I was at work at the time and rushing it quickly, will try it again a bit later.

racookpe the pressure cylinder is off of my air rifle, im designing a new valve & regulator to free up space inside the cylinder as well as increase the fill pressure from 232 bar to 250 bar. I've already run the calculations on the cylinder to make sure it can take the increase in pressure.

Standard fill valve is 41mm long, i've already saved 12mm by re designing the valve, if I can save another 3mm it will mean the cylinder capacity has increased by 10%. The stock valve has 10mm thread engagement length, im trying to calculate what the minimum engagement length I could get away with is whilst still being safe.

Regards

 

[perky416]

Just gone through the calcs again:

Force = pressure * area
Force = 3625psi * 0.699
Force = 2533.875

Shear area = force / yield strength
Shear area = 2533.875 / 19600psi
Shear area = 0.129
*5 safety factor = 0.645

Engagement = Shear Area / pitch circumference
Engagement = 0.645 / 0.956
Engagement = 0.674

That works out to be a little over 17mm, is a safety factor of 5 normal or can it be less as the engagement length on the original part is only 10mm? When calculating the pressure for the cylinder I was told to use a safety factor of 1.5.
Many thanks….getting there lol.

 

[3DDave]

Factor of safety is there to cover what is unknown, like someone abusing the fitting or the fitting being poorly made, et al. What value is ultimately assigned for personal use is more a matter of how safe you feel.

If the existing engagement is 10mm, it's about 3:1 with the original pressure.

I would look to test this at a higher pressure, using water, and under water to make sure that whatever engagement you settle on will not fail catastrophically.

You should also look at the way the item could catastrophically fail. For example, if it pushes the fill valve out, how fast will it be going based on the stored energy in the compressed air? If the cylinder splits is there a chance to create a small jet that produces an air-knife (I've nearly seen, but missed the failed demo, where a guy was injected with grease from a failed plastic tube at around 1000psi. I had asked at a previous meeting why they thought their 250 psi tubing was suitable for a 2000psi relief valve, but group-think determined it was 3000psi tubing somehow.) If it is contained, can the containment hold the volume of air without itself rupturing and also release or redirect it in a safe manner?

In regular product development one would cut the threads shorter and shorter, test each to failure and repeat until there was a good understanding of what variables needed to be controlled most closely, no calcs required, just spend time and money. If there won't be much money or time for testing, then more calculations and higher safety factors are substituted.

 

[tbuelna]

Does the valve use pipe threads or straight threads? Pipe threads are tapered in diameter along their length, and the engaged threads often do not load share well when mated. Plus, due to the taper there can be quite a bit of radial force produced when they are tightened together which must be accounted for.

 

[perky416]

The valve is using straight threads.

I have just been looking into what dave said about the first 3 threads taking all the load, I have found this site

http://www.gizmology.net/nutsbolts.htm

which says that it is the first 6 threads that take the entire load, with 2 extra threads as the first 2 are sometimes poorly formed.

So with an M25 x 1mm thread, 8 threads takes it to approximatley 10mm, as per the original valve.

Is it possible to use a stronger material such as 303 stainless steel to lower the thread count? If so how would I calculate this?

Thanks

 

[zeusfaber]

What value is ultimately assigned for personal use is more a matter of how safe you feel.

Unless you're expecting somebody else to fill the cylinder for you.

A.