Fracture Mechanics - Residual Strength/Crack Propagation

[bykncn]

Hello,

I was reading dta guidelines about small antenna installation on top of the fuselage from Patrick Safarian and ı came across a problem and ı couldn't solve it. Can you help me about this?

"If deltaP=8.9 psi, fuselage Radius=128", pitch(spacing)=1.2", hole diameter=0.188", t=0.062", Ka=130 Ksi*in^0.5, Ftu=62 ksi and Fty=42 ksi show that the critical crack length is 0.242"?

 

[dcascap]

Hi,

For the example, the Ka is very high for normal aerospace grade aluminum, therefore, probably is net tension failure due to yielding.
So you need to calculate how much stress you have in the section next to the cracks (assume width of a pitch so you don't have interaction with secondary cracks), and compare it to the yield strength. For net tension yielding the whole section yields, so no need to apply Kt (stress concentration, don't confuse with Ka=Kic ) because of stress redistribution.
This would be the procedure (if you don't know the critical crack length)

Fty=sigma*pitch/(pitch-diameter-2*crack_length)

sigma= hoop stress = P*r/t

then you get crack_length=0.243" (according to my excel, probably there is some round off error or so).
If your crack would be longer, then the section at the crack portion would yield, and therefore fail (if you treat it conservatively).

I hope it helps

Regards,
Diego

 

[rb1957]

along the lines of "African or European" ... single or double crack ?

when did Ka replace Kc ?

this looks like 2024T3 ... Ftu, Fty, Kc.

Another thought ... single crack or multiple site (like every hole) ? There is a canned solution of an infinite array of cracked holes (I got mine from Broek's book).
I notice "MSD" in your snippet. Note the check for net section yield.

 

[dcascap]

It's double crack according to the picture. No clue why they replace KIc with Ka, never heard of it.
But yeah it looks like 2024 t3 except the KIc which is around 40 ksi sqrt(in) if I'm not wrong

But I think it was done as an illustrative example for net tension yielding...

 

[rb1957]

K_Ic is low (like 40), Kc (plane stress) is upwards of 100, I've used 120, 130 is a little high

 

[rb1957]

I guess it'd help to look at the picture.

couldn't find it (someone must've stole it ... or I put it down somewhere) ...

Then, K = σ*(πaeff)0.5*(W’/πaeff*tan(πaeff/W’))^0.5 .... tan(pi*a/W) is beta for an array of cracks
where W’ = W-D-2a+2aeff
and aeff = β_bowie2*a (as the bowie solution for a doubly cracked hole determines an equivalent crack length accounting for the hole …
K = σ*(πaeff)^0.5 = β_bowie2*σ*(πa)^0.5) ...

this replaces the doubly cracked hole with an equivalent crack, and keeps the same physical distance between the adjacent cracked holes.
Net section yield would work with the real geometry (net area = hole pitch-diameter-2a)
but an "interesting" question ... would the plastic zones from the crack tips meet up first ?

 

[bykncn]

Fty=sigma*pitch/(pitch-diameter-2*crack_length)

Hello first of all thank you very much for all the answers but is there a reference where this formula is written? or did you reach it from a ratio-proportion calculation? In other words, do we reach your formula when Fty*Agross=sigma*Anet and then t's cancel each other out?

 

[dcascap]

Hi, you get it from the ratio, thickness is constant so it cancels out, and then you have the whole width/(width that carries load at the crack position), and of course crack and hole cannot carry tension load.

 

[dcascap]

K_Ic is low (like 40), Kc (plane stress) is upwards of 100, I've used 120, 130 is a little high

Good observation, you're right! Thanks for the correction.

@bykncn maybe download the DTA Handbook from afgrow webpage (I think it's for free), it explains a lot on DTA for metallic structures

 

[bykncn]

@bykncn maybe download the DTA Handbook from afgrow webpage (I think it's for free), it explains a lot on DTA for metallic structures

Hello, I had previously looked at the reference you gave, but even though I understand what you did I still don't know why would Fty*Agross is equal to sigma*Anet. Yes, the force equation was made to be equal to yield, but why is it equal to sigma*Anet? When finding yield stress, is it found with gross area, but I didn't understand why hoop stress should be found and equal to net area?

I couldn't find it in AFGrow, or is there a reference with a similar calculation?

 

[dcascap]

A very simple way of modeling a fuselage in an aircraft is to model it as a pressurized cylinder, then you have two principal in-plane stresses:
-Axial=PR/2t
-Hoop=PR/t
Check wikipedia for cylinder stresses. As hoop is higher I took hoop as a reference stress in a portion of the skin.
Then you make your typical diagram of the plate with a hole with the two corner cracks (see image). For this model you take the far field stress as the hoop stress due to the previous simplification.
Then the next question is well what is the width of this plate? Well you don't want interaction with the other holes neither cracks, then conservatively can say that 1 pitch is the width. In reality you can go further than 1 pitch. (That width is also shown in your picture so I can imagine that the person that made the course did that assumption).

Once the width is defined, you need to think about which failures could occur (for residual strength analysis) :
1. Net section yielding: The section next to the crack is not able to carry the load and yields completely. Then you can calculate at what crack length does the part yield for the given far field hoop stress.
2. LEFM: unstable crack propagation, at the load specified that crack would be unstable and propagate in the material. Calculate crack length using Ka and LEFM
Max crack length = min between 1 and 2
I only looked at 1 and it gave the result, and due to the numbers I got I'm pretty sure that 2nd is longer crack length.

 

[bykncn]

Thank you so much for explaining it in a simple way so that I can understand your way of thinking.