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Value of fracture toughness for a finite plate

  1. Aug 27, 2011 #1
    heeeeeelp please.
    I am new here and hope to find a solution and an easy to understand explanation to my problem:confused:. Thank you all :smile:

    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
  2. jcsd
  3. Aug 27, 2011 #2


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    Staff: Mentor

    How about writing some equations and showing some work.
  4. Aug 28, 2011 #3
    Can you attempt any part of the question?
  5. Aug 28, 2011 #4

    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.:confused:

    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
  6. Aug 28, 2011 #5
    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.

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