Threaded hole strength in 6061 aluminum 3

[shorton2]

It's been a long time since I had to do anything like this and my old ME training is rusty. I need to determine the tensile strength for a threaded hole. As the hole is under the normal rule-of-thumb depths for full bolt strength, AND the material is weaker than the bolt, I figure the weak link will be the threaded hole. I need to determine how much load this can handle.

Parameters and constraints:
- Standard steel M3 socket screw
- Threaded hole into a 6061 aluminum part.
- I'd plan to try to get a formed thread but to be conservative consider a medium fit cut-tap thread.
- Thread depth is 4.5mm max, I'd say allow .5mm for chamfer, etc., and call it 4mm of thread engagement possible. I can't make the hole any deeper.

Can someone help me calculate the tensile strength of this connection?

 

[CoryPad]

thread725-35222

 

[shorton2]

Thanks CoryPad, I'll digest that thread and see if I can do the calc.

 

[ctopher]

When I design tooling, I always use Heli-coil inserts in the alum parts.

Chris, CSWA
SolidWorks 13
ctopher's home
SolidWorks Legion

 

[shorton2]

Thanks Chris. Without going into the details, I'm basically having to check someone else's design/part to be sure it meets my expectations. I've got some, but pretty limited input on the details. I'll be doing good to get form threads. So helicoils probably aern't gong to happen. My hope is the calculation will show their design sufficient for the task so I don't have to worry about it failing.

 

[shorton2]

CoryPad:

At the end of that thread there is a discussion about the "addition of 0.8P to m[sub]eff[/sub]". Can you elaborate on that, or show me the equation with that added? It's not clear to me how to incorporate. Or if I am supposed to.

A[sub]shear,nut[/sub] = (π d m[sub]eff[/sub])/P · [P/2 + (d - D[sub]2[/sub])/√3]

substitute m[sub]eff[/sub] with (m[sub]eff[/sub] + 0.8P)? As:

A[sub]shear,nut[/sub] = (π d (m[sub]eff[/sub]+0.8P))/P · [P/2 + (d - D[sub]2[/sub])/√3]


And that last part, is that divided by the square root of 3?

 

[Tmoose]

For charts of what UNBRAKO felt in 1996 were reasonable tapped hole thread strength to consider for a quality socket head capscrew into brass, some flavors of aluminum, cast iron and mild steel, look at page 66/68 here -

http://www.dalecompany.com/pdf/UNBRAKO.pdf

Note their comment on page 64/66 about using figgerin' and cypherin' to come up with tapped hole thread strength being "not entirely satisfactory". Thus what they chose to present in the technical section of their "engieering guide" were tables created using empirical data from tests done with non-digital threaded test pieces, real bolts, and torque wrenches of some type.

Many other sources acknowledge there are inherent inequalities in thread load distribution
( midpage here -

http://mece1.bjtu.edu.cn/otherweb/j...r_eng/jxnr_search/jxnr_content/d05z/d5z_8.htm

).
Combine that with mismatches in moduluses of elasticity and I would be hesitant to carry many significant figures in a calculation of thread strength involving more than a couple of pitches.

http://www.roymech.co.uk/Useful_Tables/Screws/Thread_Calcs.html

"Various studies on thread loading have established that the shear stress is not evenly distributed across the threads. The first thread withstanding the load is the highest stressed and the next one is much less stressed and so on... . If the thread materials were very hard and did not yield the first thread could be withstanding nearly all the load. However because of material yielding there is some distribution of the load. A study (see link 2 below) has established that for a typical grade 8 nut the percentage of the load taken by consecutive threads are about 34%, 23%, 16%,11%,9%, 7% .... "

 

[ctopher]

Also a quick Google search comes up with a lot of info:
Link

Link

Chris, CSWA
SolidWorks 13
ctopher's home
SolidWorks Legion

 

[shorton2]

Yes, Googled extensively before asking here. All of those links drive at the minimum threads to equal the strength of the same material fastener, where teh fastener will fail first. Or in one case a different material factor. One promising link in #1 above doesn't actually go where it says, it goes to a shear area calculation for internal and external threads. Partial, but I didn't know where to go from there. There are similar stress are forumlas in my Machinery's Handbook, but again, they work toward the minimum to match the screw strength.

Corey's post above is the most I've seen explicit to the question I've asked. How strong is a threaded hole of x depth. If I can just work out that detail from the update to that related published standard.

 

[shorton2]

re Unbrako's chart, and making the assumption that metric will be the same as their UNC? thread tests for #0-#10, taking the conservative case of 1xD for depth (3mm), I get a conservative 170,000psi for Alu (doesn't say what kind of Alu). Converting to inch I get a strength of a M3 hole 1xD in Alu to be 1,317 lbs. Maybe it's right, but it "feels" high to me. I am reading it right yes?

 

[shorton2]

Well, using Corey's formula,
UTS = 45000psi or 310MPa
C1 Dilation factor = 1 (not a nut, hole in relatively large area)
C3 = .897 (don't know why for sure)
Thread major diameter (of nut) 2.9mm
Thread pitch 0.5
Meff = 4mm

I get an A-shear area of 0.04291 in2 or 27.689 mm2

I got Fmax-nut = 1039lbs

That does not seem to make sense since I *think* a steel M3 screw can only take about 475lbs. All the rules of thumb say that hole needs to be 2xD to equal the strength. If I did the math right (I wish someone would check me), I've getting 777 lbs in Alu for only 3mm of thread depth.

I've attached an excel spreadsheet with the formula, if I have it right.

 

[Sparweb]

No

Shear through the pitch diameter of the female thread.
USS is about 20 ksi (138 MPa)
PDmin is 2.675 mm
Engagement is much less than 4.5mm if you count chamfers on both the screw and hole and incomplete threads.
Helix area is less than 50% but I can't quickly find the means to calculate the exact figure (don't know your tolerance anyway).

Shear area = pi * PDmin * (4.5mm - 1mm) * 50% = 14.7 mm^2

Force = 138 MPa * 14.7 mm^2 = 2.0 kN = 450 pounds

I doubt you could consistently get better than 300 pounds if you were to set up a dozen holes and pull-test them.
You're only as good as your tools. Hand tapped holes... No heli-coils... in 6061 of unknown heat-treat... No reason for confidence.
And check my 50% guess - it's probably worse than that.


STF

 

[shorton2]

Can you elaborate on Helix Area. I don't understand what that is.

Thanks!

 

[MikeHalloran]

I would sooner trust a cut thread than a formed thread in 6061. ... because the thread material has to yield in order to make a formed thread, and aluminum doesn't behave as nicely beyond the yield point as soft steel does.

If that leaves you with too little shear area or too little thread length to be comfortable, use a thread insert. I'd use Keenserts instead of Helicoils if I had the room.




Mike Halloran
Pembroke Pines, FL, USA

 

[tbuelna]

shorton2-

You're asking about a small diameter thread form, and tolerance stack-up can be an issue. If you assume the combined worst case for contact area, pitch errors, etc. the situation probably looks rather grim. One thing that is sometimes done to improve the fatigue properties of threads in aluminum substrates is to mill the threads with a slight continuous progression in the pitch. When the mating steel stud/bolt is threaded into the hole, this produces an axial compression preload force in the local aluminum substrate, which improves fatigue life of the aluminum part.

 

[TVP]

Ultimate shear strength for Al alloys is ~ 0.7*UTS, so for 6061-T6 it should be more like 215 MPa, not 138 MPa.

 

[KENAT]

http://www.boltplanet.com

May allow your to double check the answer you get, but as it's free doing the math yourself is probably a better idea\.

Posting guidelines faq731-376

http://eng-tips.com/market.cfm?

(probably not aimed specifically at you)
What is Engineering anyway: faq1088-1484​

 

[CoryPad]

shorton2,

To answer your two questions to me:

If you rearrange the equation to solve for meff, then add 0.8P to that value to account for non-uniform loading.

Yes, that is the square root of three.

 

[shorton2]

Thanks Corypad, so solve for Meff, THEN add 0.8P. I think I can back into the revised Meff eq from that, I think

 

[3DDave]

You might find it worthwhile to create a custom insert of harder material and thread that in.

Other than that try a pull test. It's a cheap test to perform, and won't have any calculation problems.