PDA

View Full Version : How does an axial hole in a threaded piece affect its tensile strength?

Jimno2506
04-17-2013, 10:01 PM
Guys,

I machine parts for PCP airguns. I am making a coupler to add to this tubing by duplicating the factory end cap back-to-back. As of now I am using a small axial hole to connect the two chambers, I want to maximize air volume and minimize weight.

The tubing is 1.053ID x 1.25OD 4130. The endcap is 1.125-20tpi male made of 6061 aluminum. There is .537" of thread length.

I understand that the tensile strength of a threaded fastener is based on the area beneath the mean thread diameter, 1.096" dia. in my case. Area= .947in2 x 40KPSI = 37,909lbs could be held back by this end cap.

Keeping a 3:1 safety factor would mean I need enough area to hold back 13,333lbs of force trying to strip the threads.

Dividing 13,333 by 40KPSI = .333in2 Does that mean I would only need .333in2 of material between the mean thread diameter?

I just don't understand if the area of the hole in the center can be simply subtracted out of the Mean Thread Diameter or if there are other things to consider. IOW, would a 1/2-20 bolt with a hole removing half of its area be the same strength as a 1/4-20 bolt?

I found this on another site (my numbers in parentheses)

Using average pitch diameter method:
Thread width at the pitch line is p/2 = 0.0357 in. (.025)
Number of engaged threads is 14/0.5533 = 7.74 (10)
Thread shear area is pi*7.74*0.978*0.0357 = 0.849 in^2 (.860)
40,000 x .860 = 34,431lbs

The thread connection does not see pressure, as it is sealed by an o-ring. There is little torque on the system.

Am I doing that right?

Thanks,
Jim

Gary Paine
04-17-2013, 10:34 PM
I would have to do a lot more research to fully answer your question, but off the top:
Tensile strength is a function of the cross sectional area taking the load. The equation is F/A. To my thinking, that would be the root diameter of the threads squared minus the hole diameter squared and the product multiplied by PI /4 and multiplied by tensile force. The area in the threads will, IIRC, not improve the tensile strength, but would very possibly have the negative effect of a stress riser.
I think, though, that you really aren't trying to use a tensile load, but are more interested in rupture or burst strength, aren't you? If that is the case, take a look at this:
Barlow's formula can be used to estimate burst pressure of pipes or tubes.
P = 2 s t / (do SF) (1)
where
P = max. working pressure (psig)
s = material strength (psi)
t = wall thickness (in)
do = outside diameter (in)
SF = safety factor (in general 1.5 to 10)
The Barlow's estimate is based on ideal conditions at room temperature.
Material Strength
The strength of a material is determined by the tension test, which measure the tension force and the deformation of the test specimen.
 the stress which gives a permanent deformation of 0.2% is called the yield strength
 the stress which gives rupture is called the ultimate strength
Strength of some common materials:
Material Yield Strength
(psi) Ultimate Strength
(psi)
Stainless Steel, 304 30,000 75,000
6 Moly, S31254 45,000 98,000
Duplex, S31803 65,000 90,000
Nickel, N02200 15,000 55,000
 1 psi (lb/in2) = 6,894.8 Pa (N/m2) = 6.895x10-2 bar

J Tiers
04-17-2013, 10:45 PM
Mostly it IS area...... but the presence of stress-risers (scratches, gouges, tool marks) inside the hole is a possibility also, and might change the actual strength. Of course threads are a stress riser also, but they can be made with rounded minor diameter clearance.

Didn't completely follow the reasoning (quickly read) , but you need 3x the minimum area to have a 3:1 margin. The minimum being the deformation load in psi/kp, etc.

Then also, you may want to consider reducing it for aluminum, which has a tendency to fatigue with cycling at any load, let alone near maximum ones, unlike steel which essentially won't, below a certain load. That might suggest additional margin.

And you mentioned "stripping threads".... the part obviously cannot require a strength more than the threads can handle..... assuming they are the limiting factor (might or might not be the case, I don't know), in which case you need to use thread parameters, and not sectional area, to calculate allowable loads.

I know enough ME to be dangerous past simple structures.... The MEs can have that, and I'll stick to switching losses and gate drives..... You might consider that same general approach..........

wierdscience
04-17-2013, 10:50 PM
It's not as simple as subtracting area on a thin walled threaded tube connection.The reason being the elastic limit of the material can be exceeded allowing the male thread to compress and shrink in diameter and the female to expand.When that happens you lose thread engagement and the joint lets loose.Forgot to mention,a V-thread consists of two inclined planes acting against each other which exacerbates the problem in your example.A square thread or Buttress thread is optimal.

Black_Moons
04-18-2013, 08:01 AM
Why do you want to maximize air volume? that seems risky and I doubt it will improve proformance much if any, The gun should have its own rather metered amount of air used and the chamber it uses to meter it will refill in microseconds at the pressures those guns run at.

When I made a CO2 adapator for my gun, I only used a 1/8" hole in the neck of the adapator to minimize stress. I also used steel for extra safty margin.

Forestgnome
04-18-2013, 09:43 AM
"Dividing 13,333 by 40KPSI = .333in2 Does that mean I would only need .333in2 of material between the mean thread diameter?"

It's early in the morning, but didn't you just wipe out your 3:1 safety factor here? Also don't forget that shear strength is a portion of UTS. I don't remember but it's something like 80%.

becksmachine
04-18-2013, 10:46 AM
Also just a quick read, but the thing that jumps out at me is, " There is .537" of thread length."

This violates the "rule" of having a length of engagement at least equal to the diameter of the thread.

Dave

Forestgnome
04-18-2013, 03:22 PM
Here's the shear vs. UTS:
http://www.roymech.co.uk/Useful_Tables/Matter/shear_tensile.htm

J Tiers
04-18-2013, 08:44 PM
This violates the "rule" of having a length of engagement at least equal to the diameter of the thread.

Dave

Which isn't that great anyway in some cases.... with an elastic material, and a longer thread engagement you can stress the first threads to failure while the last ones are barely touched. Then you get "zipper" failure. It can be necessary to take the "stretch" of the material into account, or many of the threads may be worthless. That is not common with most materials and reasonable thread lengths, but serves to make the point.

This stuff looks simple, but unless you are familiar with the field, many important details may be overlooked.

Forestgnome
04-19-2013, 09:19 AM
Also just a quick read, but the thing that jumps out at me is, " There is .537" of thread length."

This violates the "rule" of having a length of engagement at least equal to the diameter of the thread.

Dave

That's just a rule of thumb when you're not going to the trouble of calculating thread strength. The calculation takes into account the number of threads engaged. The calculation gets less accurate with an increase in the number of threads due to what was already mentioned about threads having less contact as you increase the engagement length. So really a shorter engagement length can be more dependable in the sense that the calculation is more accurate.

Jimno2506
04-19-2013, 04:41 PM
Decided to go with 4140 instead of aluminum.