Page 359 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C007.fm Page 339 Monday, May 23, 2005 5:54 PM

Fracture Toughness Testing of Metals 339

Research over the past 20 years into the fracture of ferritic steels in the ductile-brittle transition
region has led to two important conclusions:
1. Scatter in fracture toughness data in the transition region follows a characteristic statis-
tical distribution that is the same for all ferritic steels.
2. The shape of the fracture toughness vs. temperature curve in the transition range is
virtually identical for all ferritic steels. The only difference between steels is the absolute
position of this curve on the temperature axis.

The term Fracture Toughness Master Curve [39–41] was coined to describe these two characteristics
of steel. ASTM E 1921 outlines a fracture toughness test method that is based on the master curve concept.
Due to the idiosyncrasies of the micromechanism of cleavage fracture (Chapter 5), the fracture
toughness at a fixed temperature in the transition region follows a 3-parameter Weibull distribution with
a slope of 4:
 B K  K −  4 
1
F =− exp −   Jc min   (7.28)
 25 K  o K − min   

where
F = cumulative probability
B = specimen thickness in millimeters
K = fracture toughness (critical J converted to the equivalent critical K) in MPa m
Jc
K = threshold toughness
min
K = Weibull mean toughness, which corresponds to the 63rd percentile toughness for a 25-mm
o
(1 in.) thick specimen
ASTM E 1920 fixes K at 20 MPa m (18.2 ksi in .). Therefore, Equation (7.28) has only one
min
unspecified parameter K . A statistical distribution that contains only one parameter can be fit with
o
a relatively small sample size.
The thickness dependence in Equation (7.28) stems from statistical sampling effects. Cleavage
fracture occurs by a weakest-link mechanism, and toughness tends to decrease with increasing
5
crack front length. Toughness data for various thicknesses can be converted to equivalent values
for a 25-mm thick (i.e., standard 1T) specimen using the following relationship:
K K + K = − K ( min)   B   1 4 (7.29)
Jc T) min Jc B()  25
1
(
Once toughness values at a fixed temperature have been converted to 1T equivalent values
using Equation (7.29), the Weibull mean toughness K can be computed as follows:
o
 N K − K ( ) 4  1 4

K = ∑ Jc i () N min   + K min (7.30)
o
 i  =1  
where N is the number of valid tests in the data set. The median toughness for a 1T specimen at
a fixed temperature is related to the Weibull mean as follows:
K K + K = − K ( )[ln( )] 1 4 (7.31)
2
Jc(med ) min o min

5 When assessing a flaw in a structural component, B is the crack front length, which generally is not equal to the section
thickness.