Modeling the brittle–ductile transition in ferritic steels. Part II: analysis of scatter in fracture toughness

Abstract

A dislocation simulation model has been proposed to predict the brittle–ductile transition in ferritic steels in Part I. Here we extend the model to address the problem of inherent scatter in fracture toughness measurements. We carried out a series of Monte Carlo simulations using distributions of microcracks situated on the plane of a main macrocrack. Detailed statistical analysis of the simulation results showed the following: (a) fracture is initiated at one of the microcracks whose size is at the tail of the size distribution function, and (b) the inherent scatter arises from the distribution in the size of the critical microcrack that initiates the fracture and not from the variation of the location of the critical microcrack. Utilizing the weakest-link theory, Weibull analysis shows good agreement with the Weibull modulus values obtained from fracture toughness measurements.

Appendix

To estimate the three parameters in Eq. 2 we proceed as follows: the fracture toughness (K _F ) estimated numerically from our model is first ranked in an ascending order. The range of data is then divided into four equal classes and a histogram of probability density function (PDF) f is plotted. (The class value is represented by the mean of this range.) The density f of the class is the number of K _F in the ith class n _i divided by the product of the sample N and the class width w. The cumulative density function (CDF) is the area under the histogram up to the class,

$$ P_{{\rm i}} (K_{F} ) = \frac{{f_{i} w_{i} }} {2} + {\sum\limits_{j = 1}^{i - 1} {f_{j} w_{j} } } $$

(A1)

where w _i is the width of the class with rank i and probability density f _i . At first, we assume K _c  = 0; then from the plot of lnln(1/1 − P) versus ln(K _F /K ₀) (see e.g. Fig. A1) then, we can estimate K _c as follows.

Fig. A1

lnln(1/(1−P)) versus ln(K _F ) for determination of K _c for the case of friction stress τ _y  = 600 MPa, the points; note that P ₃−P ₂ = P ₂−P ₁

We first locate three points A, B, C such that they are equally spaced along the ordinate or lnln(1/(1−P)) axis, thus by construction, we have

$$ P_{2} - P_{1} = P_{3} - P_{2} = d $$

(A2)

or the displacements from origin are given as

$$ P = S_{P} \ln \ln (1/1 - P) $$

(A3)

$$ K = S_{K} \ln (K_{F} - K_{c} /K_{0} - K_{c} ). $$

(A4)

Taking the logarithm twice of Eq. 2 yields

$$ \ln {\left[ {\ln {\left( {\frac{1} {{1 - P}}} \right)}} \right]} = m\ln {\left( {K_{F} - K_{c} } \right)} - m\ln (K_{0} - K_{c} ) $$

(A5)

Now substituting the (A3) (A2), we obtain,

$$ K_{c} = K_{2} - \frac{{(K_{3} - K_{2} )(K_{2} - K_{1} )}} {{(K_{3} - K_{2} ) - (K_{2} - K_{1} )}} $$

(A6)

Now we compute a new set of data points using (K _F −K _c ) and replot lnln(1/(1−P)) (see Fig. A2). The curve becomes linear implying that the data is indeed Weibullian. The slope of this line gives the Weibull modulus and the characteristic value (K ₀) can be obtained from the value of ln(K _F −K _c ) corresponding to lnln(1/(1−P)).

Fig. A2

lnln(1/(1−P)) versus ln(K _F −K _c ) for a friction stress τ _y  = 600 MPa