Lug Analysis beyond Bruhn / Niu

[andrimitum]

I am working through lug analysis again.... and need to analize a 6061 T6.

Where do I go to determine the Efficiency Factor for Tension, fig D1.12 Bruhn, fig 7.5.4 Niu. And Efficiency Factor for transverse loads, fig D1.15 Bruhn, Fig 7.5.9 Niu. As these tables do not seem to corealte to material strengths and nothing is close to 6061.

thanks

 

[edbgtr]

andrimitum

I have data for 6061-T4 sheet, -T6 bar, die forgings and extrusions, as well as -T62 plate, with associated graphs for the factors you seek. Provide me with material form you're using, grain direction in the lug axial direction and the W/D dimensions for a "classical" lug (as per Fig D1.12, Bruhn) and I'll post the factors you need.

Regards,

Ed

 

[andrimitum]

Thanks Ed,

6061 T6 Bar, grain direction L, W = 1", D = .25".

If you can provide the KTU results from Bruhn D1.15. I would appreciate that as well.
The Aav/Abr = 1.6

Thanks a lot,

 

[edbgtr]

andrimitum

Apologies for the delay. I suspect we are in very different time-zones and I needed to cross reference the method and data I have from one source with public domain data and other US data to ensure alignment with one another. Most lug curves, universally, are cut off at Aav/Abr = 1.4, so I extrapolated the curves to align with your given lug configuration. The data I have for 6061-T6 is based on aerospace practice and in-house testing. I could not find any US lug curve data for 6061 material.

Note that the D1.15 curve uses a common curve for the Ktu and Kty for all aluminium and steel alloys, so you could use curves 10 and 3 for your Ktu or Kty respectively. The data I have used suggests that the transverse allowable load should be the smaller of the values calculated using the Ktu or the Kty and then combine that value with the axial component as per the interaction MS equation (13) in Bruhn.

The use of the permanent set condition vs. the ultimate strength is often used where the “definition of failure” is concerned and I have encountered this in a number of joint analysis cases where high bearing loads are concerned. The fact that once the pin in the lug becomes a rattle fit, the lug, as a single load path component, becomes a very vulnerable item. If your lug is to be subjected to cyclic loading, a frozen-in bush improves its fatigue life considerably.

For the allowable axial load, the Kt for 6061-T6 bar is 0.84 for W/D = 4. For the area ratio of 1.6, the Ktu = 1.2 and the Kty = 2.1. My data has a common curve for all aluminium alloys only and individual curves for other metals. The allowables associated with this material are 245 and 295 MPa (35.5 and 42.8 ksi) for yield and ultimate respectively.

As an additional note on single load path lugs, don’t try to save weight on such a fitting, use the usual fitting factors and take special care with your allowables if the lug is welded onto other structure, you can lose the T6 condition very rapidly depending on the proximity of the weld.

Hope this helps.
Ed

 

[andrimitum]

Thanks a lot Ed,
I agree, we must be in different time zones. In spite of the delay I really do appreciate the help.

I had taken very conservative curves and determined a combined Margin of Safety of 1.67, with the numbers you provided my combined MS went up to 4.83 (including FF). So I am comfortable with my analysis.

This lug sees very small load in normal use, the load I am analyzing gets used once, in accordance with FAR 23.561 so the issue of permanent set vs ultimate is not a problem.

I admit I am undereducated when it comes to my job, I do get everything checked by someone qualified so the safety is not really an issue. I do however want to understand stress analysis and learn as much as possible so pardon me if I ask some questions that show my ignorance.

I am puzzled by the use of a common curve for Ktu for all aluminum and steel alloys (Fig D1.15). I did get that Kty is a common curve, no problem there but in all of my lug analyses so far my Ktu has fallen above curve 10 so I have used the actual curve instead of the cantilever curve. For some reason I misunderstood that if my actual material curve fell below the cantilever curve I was required to do a cantilever beam calculation, something I have not done yet. I expected that cantilever beam calculation to yield a number lower than curve 10 and more in line with curves 13 and 14. In other words I though that curve 10 was a cutoff line not an actual ultimate efficiency factor. I guess if I had done a calculation as a cantilever beam I would have realized my error but I had never had cause to. Can you confirm that I was wrong in my assumption?

One more thing, I can see that extrapolating Ktu off the chart to an Aav/Abr of 1.6 following line 10 would give me the Ktu of 1.2 that you provided. However I can not extrapolate the Kty line (curve 3) and come up with 2.1, I end up closer to 1.4 could you explain this to me?

Thanks a lot for your help,
Robert

 

[edbgtr]

Hi Robert

You’re welcome.

Doing this in my spare time over Christmas and New Year has been a bit problematic, so bear with me as I try to answer your questions in part.

As I am working with data generated on the other side of “The Pond” from you, there may be differences in approach and possibly conservatism in some areas and optimism in others wherever their testing and analysis has led them to reach different conclusions. The data that I’m using comes from a long-established and very experienced “Methods Group” that pools their information with others across this continent, very much like the US, but perhaps on a more formal level. Like you, I have been raised on US data and methods and when I find anomalies in others’ work I chase after them until I’m satisfied with their validity. However, in this case, I have not had the time to chase the lug analysis differences yet, but will get there eventually. I can only assume that the variance in the Kty line originates from different test data. I did notice the discrepancy between my data and that of Bruhn and, when in doubt, would use the more conservative one.

Lugs located in critical locations, and loaded up to the theoretical limit of their strength, are usually thoroughly tested for static strength and fatigue endurance before they go into service. From the description of your application, it sounds like you have a once-off crash case condition on a cabin interior structure or similar, so your approach is good.

I do know that the Methods Group (that generated the data I used) works very closely with a neighbouring university and related research centre that is world renowned for their testing of fastener joints, including single pin-jointed ones. In addition, my recent experience in determining fastener joint yield values from tests has been scattered, to say the least, and may account for the discrepancies between the D1.15 curve and the one I have. The local tendency is to use yield values and factor them up by 1.5, but also check the ultimate allowable value and then use the lesser of the two. This works particularly well when determining allowables for fastened joints.

Lugs work at their best when they only carry axial load and often designs try to align the structure to ensure that the lug works as close to this condition as possible. Recent sustaining work experience has shown that some aircraft designs go so far as to align separate axial lugs side by side with one another to carry what would have been the axial and transverse loads on a single lug. This allows for fail-safe DT design of the joint and also allows the lugs to work together optimally when intact.

In all the company manuals I have seen, only one actually produces guidelines for setting up the cantilever analysis of one side of the lug. From the quadratic shape of the curve I suspect that the Ftu curve has been based on a single cantilever beam analysis using fixed relationships between ultimate, yield and bearing values. The transverse load analysis is a complex mixture of the classical fastener “edge distance” problem (pin shear-out) combined with various stiffness variations of 2 parallel cantilevers joined by the lug “hoop” that provides the connection at the “free end” of the cantilevers. Obviously thin hoop lugs, i.e. small e/D values cannot provide the same coupling between the upper and lower cantilevers as would a lug with a thick hoop or high e/D value.

Ed

 

[andrimitum]

Thanks a lot Ed,

I certainly apprectiate you taking the time to answer my questions.
Thanks and have a happy new year.
Robert