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Fatigue evaluation for a repair
4

Fatigue evaluation for a repair

Fatigue evaluation for a repair

(OP)
I ran across some analysis that substantiates a repair for fatigue based off a damage ratio.  The repair damage ratio is calculated based on design ratio multiplied by the ratio of design inertia divided by repair inertia raised to the 5th power.

 design damage ratio = .45
 repair damage ratio = .45 * (I_design/I_repair)^5

Does anyone know the significance of raising to the 5th power?

Thanks

RE: Fatigue evaluation for a repair

5th power is a rule of thumb for fatigue damage ...
fatigue damage is proportional to stress^5
fatigue life is inversely proportional to stress^5

obviously you have a beam (in beanding) ...
and bending stress is inversely proportional to section modulus (not MoI, unless the depth of the beam is the same)
section modulus = MoI/c

I'd be surprised if you could change the MoI withouth changing "c", the extreme fiber distance from the NA.  so i'd change the repair damage ratio to ...

design damage ratio * (Zdesign/Zrepair)^5

if Zrepair > Zdesign, then repair fatigue damage < design, which is a good thing;
and i'd rather use 4th power, as being slightly conservative.

good luck

RE: Fatigue evaluation for a repair

(OP)
Thanks for the info RB.  I agree with the extreme fiber distance (although conservative to ignore in this case because the eval is for a blend).  

Do you know the derivation of the 5th power?  Is this related to the shape of a s-n curve?  

The repair was done years ago.  I'm just trying to understand the analysis.

RE: Fatigue evaluation for a repair

"long, long ago, in a galaxy far, far away ..."

i think it comes form the very early work into fatigue damage.  if you check through MIL-HDBK-5 their equations for s/n curves typically have exponential co-efficients close to 4

RE: Fatigue evaluation for a repair

3
Just to expand a little...

The for a given mean stress, the fatigue curve can be idealised as

(Cycles To Crack Initiation) = (Constant) * (Alternating Stress) ^ (Beta)

So if you have some factor on stress, you can find the factor on fatigue life as a function of the factor on stress by...

Equation 1:

(Cycles To Crack Initiation 1) = (Constant) * (Alternating Stress 1) ^ (Beta)

Equation 2:

(Cycles To Crack Initiation 2) = (Constant) * (Alternating Stress 2) ^ (Beta)

Equation 3:

(Cycles To Crack Initiation 2) = (Factor On Fatigue Life) * ( Cycles To Crack Initiation 1)

Equation 4:

(Alternating Stress 2) = (Factor on Stress) * (Alternating Stress 1)

Divide Equation 2 by Equation 1:

Cycles To Crack Initiation 2      (Constant) * (Alternating Stress 2) ^ (Beta)
-------------------------------- = ------------------------------------------------------
Cycles To Crack Initiation 1      (Constant) * (Alternating Stress 1) ^ (Beta)

Cycles To Crack Initiation 2   
-------------------------------- = ( (Alternating Stress 2) / (Alternating Stress 1) ) ^ (Beta)
Cycles To Crack Initiation 1   

Rearrange & substitute Equations 3 & 4 to get:

(Factor On Fatigue Life) = (Factor on Stress) ^ (Beta)

For 2024 aluminium, (Beta) is typically around -5.  For 7000-series aluminium alloys, (Beta) can be around -3.5.  You need to have good empirical data, measured in the range of stress levels that you are dealing with, for the appropriate temper, bolting conditions, etc, to substantiate your chosen value of (Beta).  Obviously, small changes in (Beta) can lead to large differences in (Factor On Fatigue Life), and thus it is good to establish your best estimate of Beta, and then use some conservative value.  Conservative values will be different depending on whether you are considering an increase in stress (and thus you want to over-predict the reduction in fatigue life) or decreases in stress (where you want to under-predict the increase in fatigue life).  In either case, you want to calculate a new fatigue life that is conservative.

Be wary also when selecting a value of (Beta) if you try to determine the change in stress from a known change in fatigue life.

Other messages in this thread have talked about a 5th power, rather than a -5th power.  It just depends on how you treat the change in stress, ie

(Factor on Fatigue Life) = (Factor on Stress) ^ -5 = ( 1 / (Factor on Stress) ) ^ 5

Similar logic leads to analogous rules-of-thumb for use in damage tolerance.

Hope this is useful.

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