Fastener allowables and Shear-out formula 2

[koopas]

Hello all,

For those who work as airline/repair station structural engineers, how do you handle short edge distance (ED) problems? I am getting a little confused.

The SRM or MIL-HDBK-5 join allowable tables gives you the strength of fastener/material combinations. From what I've read, the value is the lowest of the following failures modes:

-fastener shear (shear failure of fastener, based on Fsu_fastener)
-bearing (local yielding of adjacent structure, based on Fbrg or Fbru)
-shear-out (shear failure of adjacent structure, based on Fsu_sheet)
-Tension (tension failure of adjacent structure, based on Ftu)

So for a given material and fastener combination, the joint allowable tells you the value at which SOMETHING will fail (fastener, sheet, whatever).

Now comes the question:

When you have a short ED condition, you probably use the empirical formula for shearout:

Pall = Fsu_material * t * (2ED + 0.766D) or
Pall = Fsu_material * 2t * (ED + 0.383D)

Then, you ensure that the Pall you just calculated (using the actual, shorter ED) is greater than the Pall from the SRM/MILHDBK 5 (based on 2ED). If it is, the short ED condition is OK.

How can Pall from that empirical shearout formula above be higher than the SRM joint allowables?

I thought that shearout of the sheet material was the predominant mode of failure of joints since Fsu_sheet is usually the lowest of Fbrg, Fbru, Ftu, and Fsu_fastener.

Thus, wouldn't most joint allowable table values be reflective of a sheet shearout? Therefore, how can the "quick" shearout calculation via "Pall = Fsu_material * 2t * (ED + 0.383D)" turn up an allowable load greater than all other modes of failures? I am really confused here.

How do YOU handle short ED with loads both PARALLEL and PERPENDICULAR to the edge???

Thanks for the insight.
Alex

 

[dan100]

Hi!
Try to find out what ED the handbook values are based on. I do not have the HDBK at hand now, but I think it might be 1.5D. Not sure though. The designed ED is usually 2D+ some company figure, maybe 2.5mm to account for misdrilled holes, oversized repair rivets etc. I recollect that a 1.5D ED has 80% strength of a 2D ED. How small is your ED in this case?
Hope this incoherent rambling is of some help.
Dan100

 

[ChrisSpedding]

Don't forget that there is an area to take into account here. If you have a very thin sheet, bearing area can become significantly less than shear out area, and the sheet fails in bearing.

 

[Tads]

Hi,

Have a look at Niu's famous textbook for a step by step approach. I've encountered the problem before, and I think I did the four calculations mentioned at the start of the thread and picked the worse case - like any repair.

Depending on your engineering system, Niu is pretty much an accepted approach. Boeing may differ though from memory, but effectively the same.

 

[KSshopper]

One thing to remember about the shear-out formula: It may be unconservative for e/D < 1.5. This can be appreciated if you envision the &quot;ligament&quot; as a beam, under a distributed load from the fastener. For e/D > 1.5, this beam is a shear critical beam, while a smaller e/D becomes bending critical (a picture here would be helpful, I know). So, how do you analyze a shy e/D less than 1.5 you ask? Treat it as an idealized lug and analyze according to Bruhn.

For e/D perpendicular to the load path, look at Petersons handbook for stress concentrations. They have a case for a hole near an edge. If you take the Kt's for various e/D's and normalize to a 2D configuration, you have a relative stress factor. To take it one step further, if you knew the fatigue life of the original part (maybe DT inspection, Service Bulletins, or fleet history), you could take the inverse of the relative stress to the fourth power (typical for aluminum)and get a relative life ratio to multiply your unmodified life by. Fun Stuff!

 

[koopas]

All,

Okay, I did an simple experiment with a sample problem:

I used 0.071&quot; thick 2024-T3 clad with a 3/16 Hilok and found its allowable joint load and mode of failure. I went through all the failure modes scenarios according to the Boeing training manuals (which are close to Niu's methods).

1. Net tension failure: Not applicable here since I assumed enough material area was present around the fastener hole to carry the load. In other words, I wasn't dealing with a lug here.

2. Fastener shear failure: for a standard 95 ksi hilok, allowable load stood at 2,690 lbs.

3. Bearing failure: Using F_bru * D * t, I obtained 1,344 lbs.

4. Shearout failure: Using Fsu of 39 ksi for 2024-T3 CLAD, I got 1,679 lbs.

The allowable load is the lowest of all four above. Bearing failure would be the mode of failure, occuring at 1,344 lbs.

The SRM gives a value of 1,420 lbs., conservatively 5% above the calculated bearing failure. The literature I am reading also confirms that most joint failures occur by bearing.

I can now see that the shearout failure would not be the primary mode of joint failure. Rather, a failure in bearing would occur first, with local yielding of the material around the fastener.

The ED that would render this joint prone to a shearout failure is at 1.75D, according to the shearout formula in my original post. In other words, below an ED of 1.75D, tearout would occur before a bearing failure.

Please correct me if I am wrong here.

Thanks.
Alex

 

[koopas]

ChrisSpedding,

You say that:

&quot;Don't forget that there is an area to take into account here. If you have a very thin sheet, bearing area can become significantly less than shear out area, and the sheet fails in bearing. &quot;

I agree. And if you lack material or area AROUND the fastener, a net tension failure (Ftu * (net area - fastener diameter)) could occur before the bearing failure. If you have insufficient edge distance, then a shearout failure would be the culprit.

Alex

 

[koopas]

Hello all,

I was wondering if you guys/gals agreed with my last two posts. I didn't get much feedback.

Thanks.
Alex

 

[KSshopper]

Yes!

But...

What if it's parrallel to the load path?

 

[philjc]

Check capability of a short ed hole (sed)! We usually calculate what the fast'r is good for in the apparent material your analyzing (ref. Fsu of fast'r in mil handbook, thickness and material type of the metal the sed hole is in, whether the fast'r is csk or not csk, the size of the fast'r). Then check your Tearout and Bearing in the apparent material which you have the sed hole in (using size of hole, Fsu and (Fbru<2xD), etc). Compare and take the lesser of the two allowables for tearout and bearing for the bad condition. Then take that and compare that to what your fast'r is capable of transfering into that material (fast'r allowable). Statically your good if the fast'r shear allowable is less than your allowable for the material failure mode. In thin metal hilocks will almost always show bearing failure unless you pull the loads. If your confident with load path and you know the primary reference stress is perpendicular to the short edge and just bypass stress is parallel to the sed hole just check do a net tension failure (lug) check.
My two cents worth anyway...

 

[koopas]

Philjc:

I am a little confused when you say &quot;Statically your good if the fast'r shear allowable is less than your allowable for the material failure mode.&quot;

When you say &quot;fast'r shear allowable&quot;, do you mean the &quot;joint allowable&quot; (combination of fast'r & material) that you find in chapter 8 of MIL-HNDK 5 or SRM 51-40-07?

Here's what I do and please correct me if I am wrong: First, I get the Joint Allowable from the SRM or MIL-HNDK 5 as described above. That gives me the allowable for the fastener/material combination in lbs. Then, I would calculate the allowable load for a condition of shearout using P_all = Fsu*2*t*(ED-0.383*D). (Fsu < 2D)

I then make sure that the shearout load is MORE than the joint allowable extracted from the SRM or MIL-HNDK 5. If this is the case, the short ED condition is OK for strength as it is higher than the joint allowable. In other words, the shearout value is higher than the predicted failure mode load. Do you concur? I think that's what you were trying to tell me but I'd like to confirm. The thing is that Fsu also goes down as ED goes down so the shearout formula (a function of Fsu) does overestimate a little with short ED, which tends to compromise the whole analysis. Any thoughts on that?

------------------------------------------------------------

On another issue, when a joint allowable could not be found in the SRM or MIL-HNDK 5, how do you usually proceed to find the joint allowable? I've been told to use the bearing strength formula P_all = Fbru*D*t. However, I have found that that value is severely overestimated at times, most likely because Fbrg is based on a double-shear condition.

As an example, take a D=5/32 monel rivet (49 ksi) in 0.063&quot; 301 1/4 hard CRES steel. The joint allowable per my liaison handbook gives me 732 lb. per fastener. Let's say that I need the allowable for the same rivet in 2014 aluminum. Fbru for 2014-T6 is 102 ksi. So per the bearing formula, I would get an allowable of Pall = 102,000*(5/32)*0.063 or 1,004 lb. per fastener.

Does this make any sense? It's telling me that the monel rivet in aluminum combination is stronger than the monel rivet in steel!!! How can that be?

What would be the proper way to get the monel rivet in aluminum allowable?? The bearing allowable method seems to have overestimated the value. So I then tried using the shearout formula to get an allowable estimate. I get Fsu for 2014-T6 is 40 ksi.

P_all = 41,000*2*0.063*(2*(5/32)-0.383*(5/32))= 1,305 lbs.!!

That's even worse! Perhaps 2014-T6 material is indeed stronger than 301 1/4 hard steel! What are your thoughts on this?

Thanks for replying.
Alex

PS: where do you find all your allowables from? The SRM and MIL-HDBK 5 only seems to have limited fastener/material combinations. Any other &quot;secret&quot; source? (aside from running the test yourself or asking Boeing)

 

[koopas]

Philjc:

So you're saying that my previous post dated April 9 only handled short ED conditions with the load being perpendicular to the part's edge?

Would you please re-explain what to do if the load is parallel to the edge? You lost me when you said &quot;ligament&quot; and &quot;Treat it as an idealized lug and analyze according to Bruhn.&quot; [I will look into ordering the book soon]


Thanks!
Alex

 

[koopas]

Oops, that very last post was directed to &quot;philcondit&quot;, not &quot;philjc&quot;.

Alex

 

[philjc]

Sorry, Yes it is the joint allowable you find in mil handbk. When finding joint allowables of the fast'r use the less of the bearing allowable and shearout allowable.

Fsu of the material will not very with short edge margin. What varies or gets smaller is your area around the hole making your shear out allowable for the material less. In the bearing calculations, Fbru varies with edge margin.

In your example with the monel rivets, cres 301 has higher allowables than 2014-t6. In both cases, the fast'r bearing allowable will be higher than the shear allowable in which the fast'r will shear first.

Hope this helps...

 

[koopas]

philjc,

I agree with your first two paragraphs.

Regarding the third paragraph, how did you know that cres 301 has higher joint allowables than 2014-t6 with monel rivets? Where did you get the monel rivet in 2014-t6 sheet data? I can't find that allowable in MIL-HDBK 5.

When you say &quot;has higher allowables&quot;, do you mean &quot;joint allowables&quot;?

I am confused because you then go on to state that &quot;the fast'r bearing allowable will be higher than the shear allowable in which the fast'r will shear first.&quot;

There is no such thing as fast'r bearing allowable. There is a material bearing allowable. If you were referring to that, I don't understand what you mean by why you say &quot;the fast'r will shear first&quot;. A shear failure is a material failure, not a fastener shear failure. Would you mind clarifying that last paragraph?

Just to reiterate, in my last post, I was wondering why my joint strength calculations for monel in 2014-t6, using the bearing and shearout formulas, yielded higher joint allowables (1,003 lb. and 1,305 lb., respectively) than the joint allowable value I read from the SRM for monel in 301 steel (732 lb.). It didn't make any sense that monel in 2014-t6 is stronger than monel in steel. Thus, I was asking why the bearing and shearout formulas yielded such unusually high allowables.

Thanks for clarifying.

Alex

 

[philjc]

koopas

monel example:
the statement should of said, the material bearing allowable will be higher than the fast'r shear allowable of the fast'r, making the joint shear critical. Cres 301 mechanical properties are higher.

For fasteners not found in MIL-HDBK-5, similarities can be drawn and correction factors are applied to the published data
fast'r in 2014 t6 063...joint allowable. (ref mil-5, table 8.1.2.1 (a) unit bearing strenth of sheet on rivets Fbru=100 ksi and table 3.2.1.0(c))
Fbru@e/d=1.7 conservative..Fbru=102.6ksi...Fsu=49ksi..monel rivet

bearing material --> 2014-T6 unit (100 ksi)
bearing material (Fbru) 102.6 100 1.026
Fastener material (Fsu) 49 49 1.000
single shear

Bearing Allowable 1002lb x 1.026 =1028#

Shear Allowable lb =Fsu*pixd2/4=998#

thanks for keeping me straight...