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!

 

[IBD]

Agreed. I will likely go through with testing this for my own curiosity - setting up the test to properly demonstrate this is another ordeal.

However my original question with this is and still is 'What is best conservative approximation and calculation that can be made in this sense?'

I have outlined the following:

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

Which appears to provide a reasonable and conservative measure when compared to rivet allowables etc. So is this a good measure? or is there better?

 

[stressebookllc]

Well, make sure you post the results here IBD... I am sure the boys (any girls I may have missed?) are all curious..

http://www.stressebook.com

Stressing Stresslessly!

 

[WKTaylor]

Hmmmmm...

Everyone appears to be describing a protruding head rivet. My gut tells me this could get messy if these were flush shear or tension head rivets.

Also... what rivet alloy exactly are you proposing: B, AD, D, KE [E]?

Also... what is Your exact pull-thru fail criteria? Bucked or manufactured head external circumference tears-out sheet and rivet is essentially intact; or head external circumference tears-out sheet and is highly deformed... but not separated?

Pure speculation: What about if the soft sheet metal deforms into dimples around each installed head... and the holes expand as the tension load increased, to the point that the head pulls thru the over-expanded hole?

Regards, Wil Taylor

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