Thread design based on internal pressure 1

[mikerascally]

I am trying to calculate the necessary length of thread engagement for an ACME thread on a pipe (and pipe cap)based on the internal pressure.

I have calculated for the Stress Area At for ACME from the Machinery Handbook 26th, pg 1794. and then inputed it in the Length of engagement formula Le on pg 1490.

It gives me a value but isn't this based on the minimum length of engagement so a bolt breaks before the thread strips?

How do you correlate this to the internal pressure?

If I calculate the force on the pipe cap wall, F=PxA where the internal pressure is 50000psi and the area is .506in^2. This gives me 25300 lbf.

Now, if I use the Load to Break formula on pg 1491. P=S*At where my Ultimate Tensile Strength is 195000psi and At for my ACME is 1.266 it gives me 246903 lbf. Does this mean that it would take 246903 lbf to strip off the threads but because the force on the cap is 25300lbf I have a safety factor of 9.75?!

Help.

 

[MintJulep]

Don't forget to look at the hoop stress in the pipe to, as you will need that to figure the total stress.

There is also the stress concentration factor associated with the thread form.

 

[mikerascally]

MintJulep,
Calculated the hoop stress.... around 165600 psi. So my approach is correct?

 

[Cockroach]

You're on the right track, but got a little messed up with the logic towards the end.

Yes, you are indeed correct in using the loading of the end cap under internal pressure to determine the length of engagement to keep the piece in place. Afterall, the end reaction of the cap is to be forced away under internal pressure.

So this will provide you with a initial estimate of length of threads. Apply your factor of safety which will give you an added length.

You will need to check the normal stress UNDER the root of the external threads, the case where the male threads are literally stripped off the pipe by the end cap pulling away. If that stress is below material yield, then you will be okay, radial and hoop stress only add to locking the male/female threads together.

You never gave geometry, so it is hard to look at your specific case. Good luck with it.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[mikerascally]

I am using a 1.5-4 ACME-2G on a pipe with a .803 I.D.
My material yield is 150000psi.,
Ultimate Tensile is 195000psi,
Radial stress is 57400psi
and hoop is 165600psi.

Cockroach,when you say check the normal stress under the root of the external thread...do you mean by calculating the area of the of the crosssection of the pipe minus the threads A=pi(R^2-r^2)...pi(.598^2-.4025^2)=.618 where .598 is the radius of the Minimum Minor Diamter?

Then calculating stress=P/A where P is the 50000psi and A is .618, which gives stress=80912psi...and compare this with the yield of the material?

 

[Cockroach]

Yeah, I mean the wall between your bore, 0.803 inch and the minimum minor diameter (external) of the thread, 1.196 inches. You never stated the internal pressure of the system, so can't compute the factor of safety.

Also, I assume you are sealing the threads in some capacity, otherwise you would have a leak. This may proove interesting due to your geometry. I figure you're going with 025, 123, 217 or 321 o-ring. You would also need to look at the wall stress below the gland diameter if the design calls for a male.

Hope this helps.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[unclesyd]

I'm not that familiar with the API calculations but you should find either an example or some guidance
.

http://www.grantprideco.com/engineering/api/calculators.asp