Single and double shear joint questions

[koopas]

Good Monday all,

I was thinking about single and double shear fastened joints and came up with a few questions:

1. Mil-Hdnbk 5 and the SRM give joint allowables for material/fastener combinations for single and double shear loading conditions.

Essentially, the lower value of the fastener shear strength (Fsu_fastener * Pi * D^2 / 4) or the material bearing strength in the thinest sheet (Fbru * D * t) is listed as the critical value. A value that's listed that is other than the fastener shear strength represents a material bearing failure around the fastener.

My question is this: you will notice that for a given material thickness/material/fastener combination, the double-shear value is much higher than the single-shear value. For instance, assume the single shear joint strength of an AD6 rivet in 0.063" 2024-T6 sheet is 892 lb. (bearing critical). The double shear strength may be listed as 1,500 lb. The fastener shear strength of the AD6 is say 920 lb. (All these numbers are assumed!)

If you assume a failure in bearing, the bearing failure is only dependent upon the material's Fbru (ultimate bearing strength) and skin thickness. When testing a joint in single and double shear, both Fbru and the skin thicknesses (0.063&quot were the same.

So how can you yield a joint strength that is almost twice as high in double vs. single shear if the bearing failure is calculated as Fbru * t * fastener_diameter, where all three variables are constant in both single and double shear? In other words, how can you get a much higher joint strength in a double shear joint versus a single shear joint if both joints' bearing strengths are mathematically equal? (since bearing strength only depends on material Fbru, D, and t, and all three were constant in both single and double shear tests).

Would it be because in a double shear test, the joint clamp-up and preload is inducing an additional shear surface to transfer the load via friction, when compared to a single shear joint (where only one shear surface exists)? This additional shearing surface in a double shear joint "unloads" the fastener and adds load capability to the joint, beyond its single shear bearing strength.

2. How do you explain why the joint double shear strength exceeds the fastener shear strength rating? Using my (fake/illustrative) numbers above, the fastener shear strength is 920 lb. while the double shear strength is 1,500 lb. Is it because you're now dealing with two shear surfaces, thereby doubling the fastener shear strength to 920 lb. x 2 or 1,840 lb?

3. My second question: Is using Fbru * t * D acceptable to calculate joint strength in both single and double shear? What is your reasoning? I have been told that Fbru was obtained from double shear tests, thereby overestimating single shear joint strength.

4. Would a fuselage cutout repair involving both an internal and external doubler (net doubler thickness = one gage above nominal skin thickness) be considered as a double-shear joint?

Thanks for clarifying,
Alex

 

[meechu]

Dear koopas,
Single and double shear strengths of a rivetted joint is based on the following formulas.
For single shear, the lower of FSU*Pi*D^2/4 where D is the diameter of the rivet and FSU is the Ultimate shear strength of the rivet material and FBRU*D*t where FBRU is the Ultimate bearing allowable of the 'sheet metal'(not the fastener material)and t is the thickness of the sheet metal again.It is obvious that for double shear where the load transfer takes place from one source to two sheets on either side, the load is halved and the bearing stress is also halved and shearing surfaces become double and hence the critical condition may become the almost double shear strength of the fastener material.
Regarding the fourth question, the two doubler plates on either side of the parent skin, may share the source load depending on their relative thicknesses with the skin, and again transfer the load back to the skin with number of rivets connecting the doublers to the skin doing that job. These are general engineering methods and rivets in the immediate viscinity of the source load may take more loads than the rivets which are far off from the source of load.
I hope this answers your questions fully. Otherwise, please write to me at rms@cadestech.com.
meechu

 

[koopas]

Hi all,

Alright, I'll reduce all my ramblings to TWO questions, which I've rephrased:

Consider a single and double shear joint, joined by an AD6 rivet.

Fact 1: One of the sheets in the single shear joint is 0.063" 2024-T3. Doesn't matter what the other sheet is.

Fact 2: In the double shear joint, the middle sheet that's seeing double shear (i.e. the sheet that's sandwiched between the two outer sheets) is also 0.063" 2024-T3. Doesn't matter what the outer sheets are.

**QUESTION 1**

Now, find the bearing allowable of the 0.063" sheet in both cases (single & double shear), using either SRM Chapter 51 or MIL-HDBK 5 tables for single and double shear respectively. DON'T use the bearing allowable formula Pall = Fbru x D x t.

What results do you get? You should find out that the bearing allowable for the double shear joint of the AD6 rivet in 0.063" 2024-T6 sheet is much higher (almost twice as high) than the bearing allowable of the AD6 rivet in 0.063" 2024-T6 sheet for the single shear joint.

Question: Why? What causes the double shear bearing allowable to be higher? In both shear cases, the 0.063" sheet is "seeing" the same bearing area (D x t) against the rivet shank.



**QUESTION 2**


Let me now reverse the question: Repeat the exercise using the bearing allowable formula Pall = Fbru x D x t for both the single and double shear case. Indeed, Fbru, D, and t are all constant, so you end up with the same bearing allowable value for both types of joints.

Question: Why is the result different than the results obtained from *question 1* ?


Best regards,
Alex

 

[jetmaker]

Koopas,

Let me see if I can answer your question in a general manner.

1) a rivet in double shear has twice the shear area compared to single shear. This is simply because you have two shear load paths for the load to transfer from.

2) assume rivets have zero friction force due to clampup. Rivets are primary for shear load transfer, and only secondary tension loads (i.e. from small joint eccentricites).

3) for single shear joints, the sheet thickness is the thinest one... for double shear, it is the thickness of the middle sheet.

4) I actually expect a double shear lap joint to be stronger than a single shear lap joint. The reason being that you minimize joint eccentricity. This in turn minimizes fastener bending.. and limits the peaking factor on the hole. The peaking factor is an indication of the non-uniformity of the bearing load within the hole. A double shear joint has a more uniform bearing load distribution than does a single shear joint.

5) be careful that you are not looking at countersunk fasteners. If you are, the actual joint strength can not be calculated using the Fbr*t*d. Use the tables.

I know these are rambling answers, and do not specifically address your questions. However, you seem to be pretty quick to catch on... so if this info does not help... let me know... and we can narrow it down.

jetmaker

 

[koopas]

Jetmaker,

Thanks for your reply, and your optimism that I'd catch on quick.

I'll be frank, I'd really appreciate if you could tackle my previous two questions specifically. It appears that your point # 4 touches on the issue a bit. Is that your answer? So in my example, the 0.063" sheet in double shear has "reduced fastener bending which limits the peaking factor on the hole". You go on to write that "the peaking factor is an indication of the non-uniformity of the bearing load in the hole." Hmm...I am not sure I grasp that.

In layman's terms, does that mean that you have a greater material to fastener shank "contact" or bearing area for load transfer in the double shear joint due to reduced joint eccentricity?

Cheers!
Alex

 

[jetmaker]

Koopas,

Let's see what I can do for you on your questions then.

Q1) Why? What causes the double shear bearing allowable to be higher? In both shear cases, the 0.063" sheet is "seeing" the same bearing area (D x t) against the rivet shank.

A1) Contact pressure between the rivet and hole is not uniform in through-thickness. There is greater contact pressure towards the faying surface, which gradually reduces as you move away from the faying surface. For a single shear joint, the contact pressure at the faying surface is much higher than for a double shear joint. The increase in bearing pressure compared to a nominal value is refered to as a PEAKING FACTOR. Check out page 240, Fig. 7.7.27 in Michael Niu's "Airframe Structural Design" book for the graphical representation. The more non-uniform the contact pressure, the sooner the hole will begin to demonstrate failure in bearing. Please note... this is assuming that the joint fails in bearing well before the fastener fails. Single shear joints have a greater tendency to pop fastener heads off due to greater overturning loads.


Q2)Why is the result different than the results obtained from *question 1* when calculating using Fbru*t*d?

A2) Agreed... they should be equal IF the contact pressure distribution were the same. Once you check the Figure in Niu's book, it should become clear that you are missing a term in your bearing load calculation. The proper equation is Fbru*t*d*alpha, where alpha is the PEAKING FACTOR.

Hope this helps.

Jetmaker

 

[koopas]

Jetmaker,

Yes it does help. Thanks for the analysis. Some additional questions:

Q1. Why do you call the variable alpha the "peaking factor"? Alpha is only a hole condition factor (pp. 240). What do you define as the peaking factor? Is it the same as "peaking effect" defined on page 231 as "the large fastener loads developed at the ends of the doubler"?

With regards to the severity factor, I can see that sigma_max or "max. local stress on the considered element" as given in the equation on page 238 is dependent upon local stresses causes by load transfer AND bypass load.

Q2: In paragraph (1) of pp. 238, looking at the second bullet, what is meant by "detail design"?

Q3: Again looking at the same paragraph, third bullet, how does the severity factor (SF) take into account the "fastener load distribution to avoid the peaking effect"? Does Niu simply mean that the SF takes into account the loads at a particular fastener via the variables P and delta P found the sigma_max equation?

Q4: I don't see one of the bullets mentioning that the SF takes into account the joint type (single or double shear). Is that implied in bullet 4 (minimization of stress concentration caused by both local and bypass loads)?

Q5: What is meant by the "considered element" under paragraph (2) of pp. 238 ? Where is that ambiguous location? Is that the material surrounding the fastener?

Q6: Just so that there's no confusion, is the "load transfer" defined as the load transferred via bearing by the fastener into the other sheet whereas the bypass load is that load that remains in that sheet to be transferred by the other fasteners?

In addition, these local stresses are amplified with K factors (Ktg, Ktb) due to the geometry of the joint (fig. 7.7.25-26), as well as the factor theta.

The factor theta is the variable that takes into account whether the joint is in single or double shear. For a given t/d, the double shear joint sees a lower theta and hence, a lower max. stress (fig. 7.7.27)

Q7. Is that a correct description? Unfortunately, Niu doesn't dwell much into why a double shear joint yields a higher bearing allowable than a single shear joint. The only thing I can infer is what I've described above, based fig. 7.7.27. How does a lower max. local stress translate in increased bearing load transfer in a double shear joint? I am missing that connection.

Q8: What is that arrow "M" in fig. 7.7.24.a ? Some sort of induced moment in the hole?

I wish Niu could write like Flabel! Overall, it's an informative book but I find it hard to read. The explanations are not very intuitive. For instance, in example I on page 241, how does Niu obtain a Ktg of 3.0 from fig. 7.7.26? That would require the ratio r/c to equal zero or "r" to equal zero. How does "r" equal zero? The open hole still has a finite radius. Unless fig. 7.7.26 applies to holes filled with a solid fastener (i.e. r is the radius of a solid fastener...if none, use r=0 but again, that's not mentioned anywhere)

Thanks for any assistance.
Alex

 

[jetmaker]

koopas,

Q1) sorry for the confusion... I use alpha in my equation.. it is just an arbitrary symbol... to align with Nui, use theta. The "peaking factor" would be Sigma_peak/Sigma_nom in accordance with the image on Fig. 7.7.27. Again... this is mostly differences in terminology.

Q2) unable to help you on that one. I would guess that it has to do with the local design.. so is the hole near a fillet, or in a radius, etc....

Q3) the SF value will be different for each fastener in a loaded joint. For example, a 3 row lap will have a load distribution of 35%, 30%, 35%. However, the critical row has the highest bypass stress. So, the SF value would be higher for the critical row than for the third.

Q4) yep... if you check out Figure 7.7.24, you will see the use of theta in the equation.. which related to joint configuration.. ie. single/double shear.

Q5) considered element is just refering to the fastener that you are analyzing.

Q6) correct.

Q7) it works in the same way that increasing the hole diameter does. Essentially, you are lowering the bearing stress by making it more uniform. A good way of looking a this is to ask this question... which will yield first... a bar loaded in tension containing a small radius'd notch... or one that has the same depth notch, but a large radius? Answer this question from both a fatigue and ultimate standpoint.

Q8) The "M" is the bolt bending moment... notice it has a double arrow head.

In response to how you calculate a kt of 3 for an open hole from fig. 7.7.26, you are thinking the wrong way... it is not that r approaches zero, rather c approaches infinity. Niu's sketch is bad in that it implies a finite width not much larger than 3D. Also, a hole filled fastener has a different Kt than an open hole... but that's another story.

Have fun.

jetmaker

 

[koopas]

Good day Jetmaker,

Q1. Referring to Q7 and your last answer: (I'll paraphrase, please correct me if I am wrong). A single shear joint (with a higher theta) will increase the local stress in the element. This increased stress will raise the stress concentration, and consequently reduce the load carrying capability of the joint. Conversely, the double shear joint with its lower theta factor allows for a lower stress concentration and superior load carrying capability. Is this a proper characterization?

Q2. Referring to Q3, it can be seen from the sigma_max equation on pp. 238 that in order for the local stress to be maximized, both load transfer delta P and bypass load P must be at their maximum. Typically, in a lap joint, this would occur at the first fastener row in the thinnest sheet. Do you concur?

Q3. To sum this up, the worst fatigue life would occur at the fastener with the highest bearing and bypass load (typically first fastener row), a high d/w ratio (high Ktb), a high r/c ratio (high Ktg), and a high theta (single-shear joint). Is that correct?

Q4: How did you obtain Pall = Fbru * d * t * theta? I can't find the theta correction factor applied to the bearing allowable formula anywhere in Niu.

Thanks,
Alex

 

[jetmaker]

koopas,

Q1) yes.

Q2) yes... however, you must also consider csk fasteners. If your first row in not csk, but your second row is... you should check both rows because the csk detail may make the 2nd row critical.

Q3) yes. That is what I would expect. I can not think of any configuration where this would not hold true. If someone knows different, please share.

Q4) it is not in Niu, and the equation is in error. It should be Pall = Fbru * d * t / theta. This would more or less match the equation in Fig.7.7.24(a). How much I would rely on the values of theta presented in Niu is questionable. Generally, I use these charts for comparative purposes.. to see what kind of improvement I could expect if I changed from a single shear to double shear configuration.

Have fun.

jetmaker

 

[koopas]

Hi Jetmaker,

I'd like to resuscitate this thread. In particular, referring to your last reply's answer #4:

1. Did you deduct the Pall = Fbru * d * t / theta yourself or is it explicitly stated in a reference book so as to constitute "acceptable data" that I could use in a repair substantiation? Is there some other generally accepted method to quantify the difference in bearing strength between a single and double shear joint?

2. Does this mean that using a double shear joint with a lower theta will increase your joint allowable, thereby allowing you to use less fasteners? Do you always opt for a double-shear joint in say a skin repair? That said, when do you NOT choose to use a double-shear joint? (since the latter are "better&quot

3. For a skin repair to be a "true" double shear joint, do you need to have an internal as well as external doubler over the cutout? Same thicknesses for both doublers? What about an external doubler and external tripler? Does that configuration constitute a double-shear joint? (that last one is doubtful)

Thanks,
Alex

 

[Compositepro]

One important point to keep in mind when applying repair patches is that you do not want the patch to be much stiffer than the surrounding area. This would amplify the stresses on the patch. If you make the patch too stiff and strong it will cause a failure in the material arround the patch.

 

[jetmaker]

koopas,

Q1) The Pall equation is basically from Niu figure 7.7.24 or 7.7.27. In general, I do not use this function as the plate thickness to fastener diameter ratio is sufficiently small as to make the reduced strength from single shear not significant. It play a bigger role if you are working with thick plates and small diameter fasteners. This is rare in aircraft structures, and when it does occur, bolts/Hi-Loks with sufficient clamp-up are most often used. Another reason NOT to use Niu's theta values from figure 7.7.27 is that there is no source for the data, and therefore all information must be treated as suspect with regards to accuracy. It is useful for comparison purposes tho.

Q2) Switching to a double shear joint, assuming that the other plates are identical, does increase your allowable as the theta value decreases for any given t/d ratio (according to Niu). It is generally preferable to use a double shear joint over a single shear. There are many reasons for this: elimination of secondary bending, higher fastener capability, thinner doubler thicknesses, etc.... However, there are many reasons not too: accessibility, inspectability, cost, etc....

Q3) To be a "true" double shear joint, yes... you need a doubler on each side of the main part. Furthermore, both doublers need to be of the same thickness, size, material, etc.... Variations in those parameters will shift more load into one doubler than the other, thereby causing a slight load offset. The ability to quantify the exact amount is not trivial, and is often calculated using joint analysis.

Hope this all helps.

BTW... Compositepro makes an excellent statement... you want to try and maintain the same stiffness in your repair as the original structure. If you add too much stiffeness to your repair, it will draw load towards it, and may result in premature failure. So, something that was good for ultimate, and repair for ultimate, yet overly stiffens the structure, may draw additional load from elsewhere as to exceed the ultimate capability of the basic structure or the repair.

Best of luck.

jetmaker

 

[koopas]

Hello again Jetmaker,

Thanks for your continued tutelage and patience.

In your first answer (to Q1), you write "This is rare in aircraft structures, and when it does occur, bolts/Hi-Loks with sufficient clamp-up are most often used."

First off, let me ask if "bolts" mean threaded bolts or hi-lok-type bolts?

Second, I will assume that you're using bolts/Hi-Loks to minimize the severity factor SF discussed in eq. 7.7.7 (pp 240). Reducing SF improves fatigue life. Using double-shear joints vs. single-shear joints lowers theta and also reduces SF. But I digress...

I am reading in Niu fig. 7.4.1 that hi-loks have better clamp-up over say, rivets. How does clamp-up affect the SF equation? The hole filling factor beta is not really affected by using rivets or hi-loks (beta = 0.75). In fact, using threaded bolts (beta = 0.75 to 0.9) increases the SF. Essentially, I don't see clamp-up directly affecting the SF equation.

However, I do vaguely remember a discussion that I started regarding clamp-up. With higher clamp-up, you seem to experience improved shear load transfer via FRICTION between the metal sheets in the viscinity of the fastener hole, which "unloaded" or reduced the bearing stresses in the fastener and fastener hole (which would improve fatigue life). Is that correct?

Alex

 

[jetmaker]

koopas,

When I refer to a bolt, I am talking about structural bolts. These bolts have a smooth shank with a threaded portion at the bottom. An example per Boeing call-out would be a BACB30US4 or a BACB30MR4. These bolts have a wrenching head, and take a nut as opposed to an optional collar.

Ok... now for the reason a bolt would have a lower SF than a rivet: the head of the fastener. Bolts and tension Hi-Loks have a better ability to resist fastener overturning as the bolt is loaded in shear. The head/nut reacts a portion of the load thereby reducing the bolt bending and helps to force more of the shear load to be carried uniformly. The more clamp-up, the lower the SF would tend to be. Again, the quantative measure of this benefit is hard to provide, but it is conservative to ignore it, and know that it is there.

You have to be very careful here. The SF factor is a fatigue characteristic. Hi-Loks and rivets can have significantly different hole fill factors. In general, a rivet has a better hole fill over a transition fit Hi-Lok. The exception is when you have large diameter rivets, or large stack-up to diameter ratios.

There was a discussion about clamp-up improving fatigue capability due to the load transfer mechanism being through friction at the faying surfaces rather than through bearing/shear of the fastener. It is fairly accepted in industry that increased clamp-up improves fatigue performance. However, do not rely upon this improved performance unless the clamp-up is monitored during fastener installation. Failure to properly torque bolts often occurs in the field. Niu's equation for the SF does not include clamp-up effects, because I assume he is choosing to remain conservative because of the condition mentioned above.

Hope this is of use. Keep up with the good questions.

jetmaker

 

[koopas]

Greetings again jetmaker,

Let me try to paraphrase what you last wrote:

1. You're exploiting the fact that tension head fasteners with "big heads" (such as tension bolts and tension-head hiloks) resist fastener bending well, and you are counting on that to offset the higher theta factor found in high t/d stackups when using single shear joints. Thus, a tension-head fastener yields less bending and a lower SF. (btw, stackup refers to the total joint thickness, right?)

2. You shouldn't take into account higher friction from better clamp-up in your calculations (even though it's nice to know it's there)

3. Sorry but I missed what you were trying to say in your paragraph "You have to be very careful here. The SF factor is a fatigue characteristic. Hi-Loks and rivets can have significantly different hole fill factors. In general, a rivet has a better hole fill over a transition fit Hi-Lok. The exception is when you have large diameter rivets, or large stack-up to diameter ratios."

I was under the impression that hi-loks in aluminum structure were always installed in interference fit holes. Overall, I can imagine that a rivet would "fill" a hole better than a solid shank hi-lok. However, Niu does list the rivet and hi-lok at the same "beta" factor. Do you disagree with Niu in terms of the beta factor he assigned to hi-loks? (i.e. hi-lok's beta value is under-estimated). All else being equal, do you think that a rivet provides superior fatigue performance via a relatively lower beta and thus, lower SF?

Alex

 

[jetmaker]

koopas,

1) A tension headed fastener does resist bending better than a shear headed fastener, and significantly more than a rivet. Where I work, as part of our fatigue analysis there are different curves that we use if we have a bolt/Hi-Lok or a rivet. That is how we account for the better performance of a bolt in fatigue. Now keep in mind, Bolts/Hi-Loks are considerably stiffer than aluminum rivets, and I am sure that that is also playing a role, not just the head. And yes.. stack-up is the total joint thickness.

2) Correct, in general, friction effects are neglected on fatigue performance. If you are going to take advantage of the friction effect on fatigue, the assembly of the joint must be carefully inspected to ensure all fasteners are torqued to the right values, and that backing off of the torque during sustained operations is not likely.

3) Yes.. I disagree with Niu in that Hi-Loks and Rivets have the same hole factor. In general, for thin stack-up joints, rivets have a better hole fill factor than a Hi-Lok. I will try and use rivets for these application, but like everything, there is a trade-off which is looked at.

I often install Hi-Loks in transition fit holes if there is sufficiently low fatigue concern. As the fatigue concern grows, I switch to rivets, or install Hi-Loks in an interference fit. The degree of interference used determines the fatigue characteristic obtained.

hope this was helpful,

jetmaker

 

[koopas]

Jetmaker,

Yes, this was helpful.

Thank you.
Alex