Value of fracture toughness for a finite plate

[mahkum]

Hiii
heeeeeelp please.
I am new here and hope to find a solution and an easy to understand explanation to my problem. Thank you all

a. A specimen corresponding to a finite plate of width W = 65 mm is made from a steel with a yield strength of 650 MN m^-2 and contains an edge crack 8.5 mm long. A fracture toughness test is carried out on this specimen, and a failure stress of 295 MN m^-2 is measured. Determine the value of fracture toughness (K IC), for this particular steel.

b. A steel similar to the one used during the fracture toughness test described in part (a) is used to manufacture a large component that can be considered to be equivalent to an infinite plate. If the maximum stress that the component can be subjected to is 30 % of the yield strength, determine:
i. wheather a crack that has grown by fatigue to 12 mm long is acceptable
ii. at what length the growing crack exceeds the safety factor

Thank you

 

[Astronuc]

How about writing some equations and showing some work.

 

[Studiot]

Can you attempt any part of the question?

 

[mahkum]

Hii,

Yes I did actually calculated the results for the whole question but I am not confident if it is correct. My gut feeling is that I am wrong.

For (a) I have used the equation KIC = Y*σ critical *(sqrt π*a)

Y= a factor for the geometry of the specimen, which I have calculated as follows:

Y= (1.12-0.23*(a/w)+10.55*(a/w)^2 -21.72* (a/w)^3 +30.39* (a/w)^4) =1.23


a = length of crack
w= width of the specimen

and according to these, the value of fracture toughness calculates:
K IC = 1.23*295 MN m^-2 *(sqrt π*0.0085) = 59.29 MN m^-3/2


For (b i): I know the maximum stress the component will be subjected to is 30% of yield = 195 MN m^-2
when I calculate the critical stress for the component with a 12 mm crack, I can compare and it comes out that it is acceptable:
σc = K IC/Y* (sqrt π*a) = 272.65 MN m-2




For (b ii): π*a critical = ( K IC/ Y*σ max)^2 = 0.07370

0.07370/π = 0.023m (this is the crack length which would exceed the safety factor)

What do you guys think ?
Thank you

 

[Studiot]

Well in principle the methods seems right, though I am not sure where you got your Y geometry correction - it does not look like an Irwin or Dugdale factor.

Nor is the specimen plate in part (a) like as BS 5447 test specimen.