Aluminum Sheet Metal Pull through calculation question

[IBD]

Hey Guys,

I am trying to simplify a Calculation to get a Margin of Safety Calculation for an Aluminum sheet metal pull 'Thru' Calc to ensure enough fasteners are being used and the sheet metal is thick enough etc... I am taking a shot at this here so any suggestions would be greatly appreciated.

IN order to take worst case scenario I am using AL 5052 as my worst case analysis:

AL 5052 - Tensile U - 28KSI - (I pulled this value from a few material tables) is there a better value to use from MMPDS somewhere??
thickness - 0.032 inch
No. of Fasteners - 4
Fwd G load - 9G
weight of attachment - 13.6lbs total

So taking Pfwd load due to 9Gload:

Pfwd = 13.61lbs * 1.5 (FS) * 9G = 183.735 lbf

Ptf = P total force reaction load per fastener assuming all loads react equally.

Ptf = Pfwd / 4 = 45.93 lbf


Sheet Metal Pull-Out Area (assuming #10 washer area)
Circ. = Pi * 0.438 inch (diameter) = 1.376 inch
A = C * thickness of sheet metal = 1.376 * 0.032 = 0.044 in^2

Pull-Out Allowable = 28KSI * 0.044in^2 = 1232.9 lbs --> Which implies that a force of 1233# is required to pull this fastener through the sheet metal at 0.032inch thk... Does this seem a bit high??

Now I would like to calculate a MS for this as well and I am wondering what exactly to use for this:
- do I use the same method and calculate a lbf applied per fastener?
- using the Area of the washer head?
- or the area of the thickness of the sheet metal again?

i.e.:

A of washer head = pi/4 * D^2 = 0.1507 in^2
A of sheet metal (as before) = Pi * Diam * thk = Pi * .438 * .032 = 0.044 in^2

Stress = Pthr / A = 1232.9lbs / 0.1507 in^2 = 304.8 lbf

OR

Stress = 1232.9lbs / 0.044 in^2 = 1043.94 lbf

There fore MS = (Load Allowable / (Fapp * FS)) - 1

1) MS = (1232.9 / (304.8 * 1.5)) - 1 = +1.7 MS
2) MS = (1232.9 / (1043.94 * 1.5)) - 1 = - 0.21 MS


Any comments would be appreciated!

 

[rb1957]

5052 ... yech !

9G .. crash case ? ... then already ultimate

(pi*D*t)*Ftu isn't the allowable i'd use. the fastener should have a tension allowable (yes, i know they're hard to find for rivets) the skin "pull thru" is more like Fsu*pi*Dh*t, shearing around the head (or tail or nut) of the fastener. Fsu is more like 16-20 ksi. MMPDS has a full set of strengths (table 3.5.1.0); it looks like you've picked the lowest Ftu you could find (for O condition, not something you'd use in the real world).

with an applied load of <50 lbs per, you could easily test if you couldn't say "passed by suspicion"



another day in paradise, or is paradise one day closer ?

 

[SAITAETGrad]

Fsu*pi*Dh*t doesn't match up with test results. The mechanics of course are different. Allowables do exist. See Bruhn D1. If you are resourceful, you should be able to find other resources.

 

[IBD]

hey - thanks for the reply.

Yes I agree the reason for using the 5052 for calcs is because the actual material is unknown other than it is Aluminum by inspection and 0.032" the calculations I have done only to meet an absolute worst case scenario and prove the method to pass by analysis.

As I do not have current access to the aircraft for physical proof test an analysis is preferred.

In regard to using rivet values this is actually a box that has thru screws go thru the aluminum sheet (wall) and into the box that is the mass. The concern is to prove the screw head side will not tear through the Aluminum sheet. SO I was trying to come up with a way to determine the force that would be required by using the diameter of the washer head being a No. 10 fastener. Assuming a NAS1149 No. 10 washer or panhead / washer head No. 10 screw.

In that sense does the method I have outlined make sense?

 

[rb1957]

you can always test with a pirce of Al sheet and a rivet and a small brkt (something to interface between the rivet and a load); ie you don't need to test on aircraft.

following on the suggestion, using Bruhn, table A (pg D1.28), rivet tension allowables are less than Fsu*pi*D*t < Ftu*pi*D*t, so 1) your method is unconservative, and
2) you can use Bruhn's table to see that the failure mode is more the sheet being pulled over the head rather than shear around the edge of the head.

But's it's all very messy, picking the weakness sort of Al you can find 'cause you don't know the actual material. This isn't for certification, is it ? this is more owner/operator/builder ??

if there's a worry about things, add a washer under the head, to distribute the load further into the sheet. do you have access to the far side (or are you using blind fasteners) ?

another day in paradise, or is paradise one day closer ?

 

[IBD]

Sorry - Yes of course I could get a piece of 5052 and do a simple pull test.

I would also just like to complete the analysis as I am determined to prove it on paper now... And yes just an operator builder and there is no concern for the case just curiosity to prove this analytically... As it would originally appear to be a simple calculation.

Would using rivet allowables as in Bruhn D1.28 Table A be a more conservative approach then?

 

[IBD]

Additionally the size of the rivet is much smaller than the surface area of the washer head or actual fastener that would actually be used so to say that the Rivet Allowable used in the closest gauge and diameter @ 0.032" thk and 3/16" diam. with allowable of 354 lbs - is actually an value which would be far less than the that of the actual fastener used as the fastener is a No. 10 for one. and there will be a 'washer' head included. So my original assessment was that using such a Rivet allowable would be FAR to conservative an approach... ?

So the question arose - just to calculate this allowable instead.. And in that sense what is the best approach?

 

[rb1957]

table A is for 2024 skins.

the simplest calc (Fsu*pi*Dh*t) is unconservative (wrt table A). the failure mechanism doesn't appear to be sheet shearing around the head, but rather the sheet pulling itself around the head.

a simple test would give the most real answer, as opposed to "waffling" about this is conservative and is sort of like what i've got and oh, i'll throw in another ultimate factor (just for the halibut) ... something else to remember is the affect of the sheet support in this ... if you had a large (flexible) sheet, it'd expect the failure to be lower than a well stiffened panel.

if you're showing 50 lbs load, most anything can handle that !



another day in paradise, or is paradise one day closer ?

 

[IBD]

Thanks for pointing that out.

wrt to (Fsu*pi*Dh*t) what exactly is 'Dh'? the Diameter of the head I assume? whether it be washer head or rivet head? and What value of Fsu would you be referring to that would be unconservative?

As for the rest - absolutely agreed! A test would provide the real answer - this is for arguments sake how to show it on paper... And my thought as well would that the failure mechanism would be the sheet pulling itself around the head - however for calculations how would you show this? My assumption was that you would show the fastener pulling through the sheet?

And yes there is no risk here it is less than a 20lbs load....

 

[rb1957]

yes, Dh is the daimeter of the material being sheared (around the head, washer, whatever). I thought you were using Ftu, which is unconservative compared to Fsu.

i don't know how to calculate the sheet hole slipping over the head ...

another day in paradise, or is paradise one day closer ?

 

[stressebookllc]

My thought is about Fty if the sheet pulling out from under the head is the expected failure mode.

If there is a way to transform the calculation into hoop stress and use Fty for your margin, then you might have something there.

http://www.stressebook.com

Stressing Stresslessly!

 

[SAITAETGrad]

Roarks has some solutions for annular plates in bending that seem a bit closer of a model but IIRC, not quite the same thing. Maybe some code based on Timoshenko Shells & Plates w/ some extra stuff from his elasticity book would yield results close to experimental...or you could apply some emperical constants until it did.. Or, you could do some non-linear FE. Sounds like a Masters' topic so it's probably already been done.

 

[IBD]

I was hoping it would be a bit simpler than that to be honest. I have also used the Bruhn rivet tension allowables as a comparison i.e.:

Table A Bruhn D1.28 indicates a 3/16inch rivet Ultimate Tensile is 354 lbs. And the applied load per fastener is roughly 50lbs so therefore:

Even with using a factor of safety of 4x:

MS = 354 / (50*4) - 1 = + 0.77

However it would be nice to be able to calculate this allowable value....

If I use the original assumption Maximum pull out force allowable becomes:

Pa = Fsu * Af = Fsu * Pi * D * t = 16000 * Pi * 0.438 * 0.032 = 704.8 lbs; which still sounds a little high.... ?

When compared to my applied load per fastener of 50lbs even with a factor of safety of 4x is a MS of +2.52...

 

[rb1957]

yes, but ... the failure isn't tearing of the skin around the head, it is more a plastic deformation of the skin; a much harder thing to calc, and probably testing is the easiest way.

i agree with you that the load is low, and since you're "guessing" other critical components of the calc (the type of Al, etc) then i think the more reasonable thing to do (i Hate doing meaningless sums just to show a +ve MS) is to say "low load". If you wanted to really work it, pack washers around the rivets so that it is clearly "passed by suspicion".

another day in paradise, or is paradise one day closer ?

 

[IBD]

Doing some more reading on pull out strengths how does this sound:

Pov = Pull over strength = which is the force required to pull the sheet over the head of the screw / washer.

t1 = thickness of sheet metal = 0.032 in
Dws = screw head / washer head diameter = 0.438in
Dh = hole diameter = 0.2.01 in

Ftu = tensile ultimate = 16000psi (for a conservative value using FSu from 5052 'O' condition MMPDS 3.5.1.0)

Pov = t1 * Ftu * (Dws - Dh) = 120.8 lbs

with a applied load of roughly 50lbs per fastener:

MS = +.61

 

[rb1957]

that'd be Fsu.

FWIW it looks like a gathering of numbers that seems to give a not unreasonable answer. It's hard to see the physical meaning of it.

another day in paradise, or is paradise one day closer ?

 

[IBD]

probably a good description of it...

what do you mean by 'That'd be Fsu'??

I was assuming using the Fsu value from MMPDS 3.5.10 would be a Conservative approach for the calculation?

 

[rb1957]

"Ftu = tensile ultimate = 16000psi" ... that'd be Fsu (not Ftu)

another day in paradise, or is paradise one day closer ?

 

[IBD]

Ah ok thanks!

 

[SWComposites]

The pull thru failure mode is complicated and cannot be accurately predicted in metals or composites using basic material properties (I have tried vs lots of test data). One can make conservative approximations or better yet, test it