How to calculate thread pull out in 6061 aluminum

[atomarc]

I have studied the on-line engineering calculators and they may as well be in Chinese for all the good they do me!

I have a 1/4" X 20 SHCS that is threaded 5/8" into a piece of 6061 aluminum...how much pull out force will this arrangement give me. The actual assembly will have multiple fasteners so I'll assume the load would be divided by the number of fasteners..right?


Stuart

 

[split tenth]

Drill and tap a 5/8 piece of aluminum. Thread the screw in and then try to push it out in a hydraulic press with a pressure gage. Yes this is old school but it will give some kind of idea of what you're looking for.

 

[muckalee]

destructive testing is the only way to determine near absolute. The stress may or may not equally distributed.

 

[johnoder]

P/2 X PD X "pi" X number of engaged threads = area working for you in shear

About 0.017 in ^2 per engaged thread

If 12.5 threads engaged, about 0.212 in ^2

If failure assumed at yield in shear of 30,000 psi, about 6300 lbs, but multiple destructive testing is the only way to give a real idea of actual

A reason for this is the use of the PD - which gives an average area. The alloy steel SHCH is certainly NOT going to fail at this load, meaning the 6061 T651 is going to fail nearer major diameter. Other things less obvious going on - like the aluminum having to get out the way

ERROR! - the alloy steel SHCH very likely WILL fail with that number of threads engaged

 

[Clive603]

Stress distribution is very non linear. Major stress concentration is at the entrance so only the first few threads do any real work. Typically thread depths between 3/4 and 1.5 times bolt diameter are sufficient to generate a stripping force equivalent to the ultimate tensile strength of the bolt. Failure is by bolt core shearing at the top of the hole.

See the Important Note in the blue box on this page http://www.roymech.co.uk/Useful_Tables/Screws/Thread_Calcs.html for a bit more background.

Some interesting graphs here showing stripping force against depth of tapped hole in various materials up to the ultimate tensile strength of high tensile cap-head screws:-

http://viewmold.com/Products/Techni...ation/STRIPPING STRENGTH OF TAPPED HOLES.html

Need to compare with standard maximum torque values for for ordinary nuts and bolts to establish a safe load. Which will be lots less than stripping force.


Clive

 

[CarbideBob]

..... The actual assembly will have multiple fasteners so I'll assume the load would be divided by the number of fasteners..right?
Stuart

Yes, No, and Maybe. Real life is not this simple.

 

[atomarc]

Well hell..it was pretty simple after all! I didn't understand most of the replies anymore than the engineering voodoo I found online. I guess I missed school the day they taught basic arithmetic.

I didn't miss any shop classes though and I'll bet "split tenth" was one of my instructors. I used his suggestion and did as he advised. The 1/4" X 20 SHCS mushroomed in on itself at 8900 lbs of ram force with nary a scratch to the threads in the aluminum...in fact I could remove the fastener with my fingers.

That's the kind of practical stuff I can understand.

Stuart

 

[EPAIII]

Johnoder and Clive 603 give two different approaches to this question. One assumes that the bolt will not distort so each thread that is engaged will provide the same amount of holding power. The other assumes that the bolt will stretch and the first few threads (from the head) will assume most of the hold while each successive one from there will have less effect. One of these assumes that the material with the female threads is very yielding so the load is evened out while the other assumes it has no yield. The four or five thread rule comes from a analysis that assumes similar materials for both male and female threads.

But in this case, the bolt is hardened steel (SHCS) while the female threads are in 6061 aluminum which is apparently not tempered. So if is very soft compared to the hardened steel of the bolt. What is going to happen, in reality is something between the two assumptions of the methods above. The aluminum will yield far more than the hardened steel of the SHCS so the load on the successive threads will be a lot more uniform than the rule of thumb shows. So, it is entirely possible for a lot more than four or five threads to be effectively engaged. The aluminum will quickly compensate for the slight (by comparison) yield of the steel. The loading on the threads will be very close to uniform. The first few threads in the aluminum will be slightly distorted and therefore SLIGHTLY less strong, but this will be a small factor, not a large one.

I think that adding additional threads that are engaged will be a very effective way of increasing the pull out resistance of this kind of joint. As many as ten or twelve threads would clearly be useful and perhaps even more could be added before the joint fails from the bolt breaking instead of the female threads pulling out. Could be as many as 10 or even more. If there isn't enough room for more threads, some steel threaded inserts could be used. These would increase the area of the steel to aluminum juncture and therefore the strength of the overall joint.

A lot of math could be generated, but I have to agree with Muckalee that the only way of knowing will be to do some destructive testing. I just reread Clive603's post and see he did give a reference to some tables showing the strength of various combinations of metals. Perhaps they will be of some help here. They seem to show fairly low numbers for aluminum but do not give any specific aluminum alloy. 6061 is fairly soft so those tables may not apply.

 

[Halcohead]

5/8" thread engagement is 2.5 diameters of 1/4" threads. In aluminum, the SHCS will pull apart before the thread strips. If you're really worried, add a helicoil. Guaranteed the fastener will fail first.

The hydraulic ram test is interesting, but unfortunately doesn't tell you when the joint will fail under tension. I've never seen a 1/4-20 SHCS rated for >7000 pounds tension.

Yes load is divided among all fasteners, assuming the joint is stiff, they're located properly, and well-tightened.

From memory, if your SHCS is steel and well-torqued, it should be good for ~5300 lbf of tension.

 

[jim rozen]

5/8" thread engagement is 2.5 diameters of 1/4" threads. In aluminum, the SHCS will pull apart before the thread strips. If you're really worried, add a helicoil. Guaranteed the fastener will fail first.

The hydraulic ram test is interesting, but unfortunately doesn't tell you when the joint will fail under tension. I've never seen a 1/4-20 SHCS rated for >7000 pounds tension.

Yes load is divided among all fasteners, assuming the joint is stiff, they're located properly, and well-tightened.

From memory, if your SHCS is steel and well-torqued, it should be good for ~5300 lbf of tension.

+1

The *real* question is, how small a thread engagement will cause the aluminum female threads to fail before the
cap screw snaps.

My SWAG is about 3/8 inch. Below that, the aluminum will strip out. Who knows? Only the shadow, er, instron, knows!

 

[atomarc]

5/8" thread engagement is 2.5 diameters of 1/4" threads. In aluminum, the SHCS will pull apart before the thread strips. If you're really worried, add a helicoil. Guaranteed the fastener will fail first.

The hydraulic ram test is interesting, but unfortunately doesn't tell you when the joint will fail under tension. I've never seen a 1/4-20 SHCS rated for >7000 pounds tension.

Yes load is divided among all fasteners, assuming the joint is stiff, they're located properly, and well-tightened.

From memory, if your SHCS is steel and well-torqued, it should be good for ~5300 lbf of tension.

Would it not be logical that the junction between threads and threaded hole would see no difference between a application of pressure versus and application of tension? The strength of the fastener is a different story.

Stuart

 

[Forrest Addy]

Consider a thread as a long loaded ledge where the length of the ledge is the pitch circumference times the number of threads engaged times the thickness about 1/3 (fudge factor) the pitch. This gives you the total shear area of the engaged thread. In your example: 0.210 x Pi x 13 engaged threads = 8.6 x 0.50/3 = .142 square inches. Multiply that by the shear strength of the weaker material - 6061 aluminum - that's roughly 1/3 the tensile strength or 40,000 PSI/3 or 13,333 PSI. .142 square in x 13,333 PSI = 1906 lb

Don't be misled by the 4 significant digit math: this is a ROUGH guess suited only for initial assumptions. It does not take into account the many variable involved in a competent analysis. In any case, actual pull-out strength of a threaded fastened in the base metal is best determined by actual tests.

I'm sure there are thread pull-out tables somewhere but remember this: thread pull-out is a catastrophic mode of failure. Threaded fastener connection strength and reliability figures are usually taken from fatigue or yield data of the connected elements.

 

[E.F. Thumann]

Am I reading this correctly? John calculates it at 6300, and Forrest at not even 20...............Btw I've read that the first two threads bear approx 70+% of the load due to bolt stretch FWIW.

 

[Forrest Addy]

Am I reading this correctly? John calculates it at 6300, and Forrest at not even 20...............Btw I've read that the first two threads bear approx 70+% of the load due to bolt stretch FWIW.

True when the fastener and the base metal are the same material but don' forget to crank in Young's modulus. Aluminum is three time as "stretchy" as steel.

 

[atomarc]

I noted this in a post above.. I drilled and tapped a 1/4" x 20 hole in a piece of 3/4" 6061 aluminum and threaded a SHCS through the full 3/4". I then pushed on the head of the SHCS with the ram in my press. I stopped at 8900 lbs of ram force and the fastener didn't move or bind in the threaded hole. The portion of the SHCS not threaded into the hole collapsed onto itself but never actually broke.

How can the pull out figures calculated above be so skimpy when the practical test in the press says otherwise?

Stuart

 

[Halcohead]

How can the pull out figures calculated above be so skimpy when the practical test in the press says otherwise?

Stuart

Threaded joints are surprisingly complicated. It is wise to just use numbers provided by reputable vendors, rather than get into the detail engineering. See page 39 of this pdf: http://www.holo-krome.com/HK_Consolidated_Tech_Manual_FULL_2015.pdf

With that said:

In compressive loading, the load is applied to the top thread first, which as it deflects, allows the next thread to take more load, and so on. Thus the pressure can be uniformly-ish distributed throughout the entire threaded hole, even if the top threads are starting to break.

As others have mentioned, in tensile loading, the top thread takes most of the tensile stress, with subsequent threads taking less and less. But there's nothing above the top thread to take this stress when it starts to fail. So as the top thread rips out it just allows the next thread to take full load and rip out until everything is zippered out.

Even if you somehow do get 9000 lb strength out of the female threads, your screw is only rated for 5720 lb tensile strength (180000 psi screw strength * .0318 in^2 tensile area). Why different for tensile and compression? Because steel's compressive strength is greater than its tensile strength.

If you want a stronger joint, use 1/4-28. Each screw has ~6550 lbf breaking strength (14% stronger).

 

[atomarc]

Threaded joints are surprisingly complicated. It is wise to just use numbers provided by reputable vendors, rather than get into the detail engineering. See page 39 of this pdf: http://www.holo-krome.com/HK_Consolidated_Tech_Manual_FULL_2015.pdf

With that said:

In compressive loading, the load is applied to the top thread first, which as it deflects, allows the next thread to take more load, and so on. Thus the pressure can be uniformly-ish distributed throughout the entire threaded hole, even if the top threads are starting to break.

As others have mentioned, in tensile loading, the top thread takes most of the tensile stress, with subsequent threads taking less and less. But there's nothing above the top thread to take this stress when it starts to fail. So as the top thread rips out it just allows the next thread to take full load and rip out until everything is zippered out.

Even if you somehow do get 9000 lb strength out of the female threads, your screw is only rated for 5720 lb tensile strength (180000 psi screw strength * .0318 in^2 tensile area). Why different for tensile and compression? Because steel's compressive strength is greater than its tensile strength.

If you want a stronger joint, use 1/4-28. Each screw has ~6550 lbf breaking strength (14% stronger).

Thank you..a very clear and understandable explanation. I was advised to use a NF fastener but thought the NC offered more strength.

Stuart

 

[dian]

looking at spec. torque for 8mm waterpump screws in an a356 engine block (30 ksi yield fro memory) its 30 nm. thats between the values for a 8.8 and a 10.9 screw. if there is only 12% of tolerance in this, the pullout strength equals at least the clamping force of a 10.9 screw being 27 kn. deductive logic tells me that would apply to any size screw. no idea if this is true.

"steel's compressive strength is greater than its tensile strength": do tell more about it.

 

[dian]

and this is what bossard has to say concerning thread depth (hopefully google will translate, values determined on experimental basis, gg25 is g35 in the states i believe):

Einschraubtiefe

 

[JST]

One of the tests proposed was to PUSH out the screw.

Of course, that will likely give totally wrong information.

Foor one thing, the screw i use will be uder tension, which means it can break, while the compression load does not allow that. There are other differences as well, such as the way bolt stretch affects it....compression may not act the same interacting with the female threads.