Crack Length & Safety Factor: Understanding Acceptable Limits for Cyclic Loading

[richard9678]

I'm not getting something, I'm looking for some hints.Component classified as in infinite plate. Maximum working stress is 50% of yield strength. Has crack 10.5mm long. Cyclic loading - mode 1.

Is the crack acceptable?
At what length the growing crack exceeds the safety factor?

I just want hints as to how to answer the questions.

I can calculate the crack length at which the working load leads to a sigma critical point. That's greater than 10.5mm. Okay, so there no imminent danger of cleavage fracture. But, that is not to say the crack is acceptable.

There must some point where the crack length puts it within a margin of safety. But, I'm lost to understand this.

I'm guessing that 10.5mm is beyond some safety factor. That there is a length smaller than 10.5mm which is acceptable. Just don't know how to get to it. Need some key words or something to help me. Thanks.

P.S. Possibly talking about residual strength - not sure. Might be something else.

 

[richard9678]

If I say that the working load is doubled, then using my formula, I get a much smaller crack length where cleavage fracture would occur.

So, I get a_critical = 13mm when load is 287.5 MPa - the designed load. I get that down to 3mm if I say the load is doubled to 575 MPa. And the current crack length is 10.5mm.

I'd be saying 3mm is the safety factor because I can stand a load of 575 MPa - twice the working load (Incidentally equal to Yield Strength). That's a safety factor of two.

So, the crack is unacceptable, because it's longer than 3mm. And the crack exceeds the safety factor by 7.5mm .

Best I can do. I could be wrong.

 

[Nidum]

Sorry but your postings as written are practically gibberish .

Please try to explain your problem more clearly and tell us what understanding of strength of materials and stress analysis you have so that any answers can be pitched at the right level .

 

[richard9678]

We have a metal plate, which can be considered infinite, and in it is a crack of 10.5mm from the edge. At a certain critical stress the crack will suddenly open up, very fast, which is a cleavage fracture. The value of that critical stress is a function of the crack length (a). I believe the following formula will give you the critical stress, the stress at which cleavage fracture will commence: a_critical = [ k_IC / Y x σ ]².

Using that formula, I get a critical crack length of 43.29 mm, meaning the crack will become a cleavage fracture at the working load (287.5 MN m^-2), if it gets to 43.29 mm.

Assuming everything is okay so far. the next thing is to figure out what length of crack would represent a safety factor boundary. Clearly we must not let the crack be anywhere near 43.29mm. To come up with the crack length that should not be exceeded, I imagined that the load was twice the working load. And used the above formula. That gave a crack length of 27.07mm. So, my reasoning is, with a crack that is 27.07mm, the working load represents half the load that would produce a cleavage fracture. And that to my mind is a safety factor of 2.

I wonder if all this passes muster.

k_IC was 104 MN ^=3/2 and Y was 1.12.

P.S. Just ignore post # 2. Figures are wrong.