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A
awibroe1-Visitor
1-Visitor
August 6, 2018
Solved

Fatigue and Crack Growth

  • August 6, 2018
  • 2 replies
  • 17288 views

Hi to all,

 

Looking for some help from anyone with experience in fatigue calculations.

 

I have the attached problems as part of an assignment but the supporting literature I have been given is not overly helpful. Essentially I think that I have either done some of the programming incorrectly or equally likely my units are wrong. As I suggest, the material I am working from is pretty light in explaining the units.

 

Any help appreciated.

 

Andy.

Best answer by Fred_Kohlhepp

Yes.  You're basically doing a weighted average of stresses and cycles.  Think of it as a percentage of fatigue "life."  Each of the stress levels and number of cycles uses a certain percentage of life.  (Miner's rule.)  Each year there are (in your problem) six stress levels and each level has so many cycles.  What would the equivalent stress be for one level for all the cycles in the year?

2 replies

A
awibroe1-VisitorAuthor
1-Visitor
August 6, 2018

PS I am using MCP3.

 

A
awibroe1-VisitorAuthor
1-Visitor
August 6, 2018

Think I have had a bit of time to reflect on this and think that the issue is the way I have set up the calculation. I think the cubing function in the Paris law formula i.e. (dK)^m which is (dK)^3 in this case is causing the units of dK to be cubed. This I don't think should be the case and it is just the value of dK which should be raised to the power of m? In which case I get a sensible crack growth figure but can anyone comment/ tell me if this is correct and if so what I should be doing differently in MC to get this to work properly?

 

Thanks

 

A

F
Fred_Kohlhepp23-Emerald I
23-Emerald I
August 6, 2018

You've got it almost right.

 

I'd send your sheet, but I'm working in 4!  See attached xps!

A
awibroe1-VisitorAuthor
1-Visitor
August 6, 2018

Fred,

 

Many thanks for that. I think I tend to agree with you on the units for C although it does not seem well documented...

 

The last part of my problem is to calculate the size of the crack for each year.

 

As opposed to going through manually calculating each da and adding this to the current a i.e. a is the original crack length, da is the amount it increases in year 1 and then you would sum these to get the length of crack at the start of year 2 (a2). 

 

Instead of doing this manually I am trying to think of a more sophisticated way of getting Mathcad to do this for me. I am thinking of a for loop in a programme but it is a bit mind boggling as in order to calculate da you need dK and to get dK you need da (as you need the current a for the previous year). Any thoughts on how to approach this cyclic argument?

 

Thanks,

 

Andy.

F
Fred_Kohlhepp23-Emerald I
23-Emerald I
August 6, 2018

K is a function of applied stress and a.

da/dN = C*K(a)

 

this is a differential equation in a as a function of n.  suggest odesolve or rkadapt.

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