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1656_C005.fm Page 242 Monday, May 23, 2005 5:47 PM
242 Fracture Mechanics: Fundamentals and Applications
distribution by assuming that ρ is small and V is large, an assumption that is easily satisfied in the
2
present problem. The critical particle size depends on stress, as Equation (5.17), Equation (5.18),
and Equation (5.20) indicate. In cases where stress varies with position, such as near the tip of a
crack, ρ must also vary with position. In such instances, the failure probability must be integrated
over individual volume elements ahead of the crack tip:
F − ∫ ρ d =− exp V 1 (5.22)
V
Consider a uniformly stressed sample of volume V . A two-parameter Weibull distribution [41]
o
can be used to represent the statistical variation of the fracture stress:
σ m
1
F =− exp − f (5.23)
σ u
where m and σ are material constants. Comparing Equation (5.21) and Equation (5.23), we see that
u
f
ρ = 1 σ (5.24)
V o σ
u
When stress varies with position, the cumulative fracture probability can be expressed in terms of
an integral of maximum principal stress over volume, which can then be equated to an equivalent
fracture stress:
σ m σ m
F =− exp − 1 ∫ 1 dV =− exp − w (5.25)
1
1
V o V f σ u σ u
where
V = volume of the fracture process zone in the geometry of interest
f
V = reference volume
o
σ = Weibull stress [41]
w
The Weibull stress can be viewed as the equivalent fracture stress for uniform loading on a sample
with volume V . Solving for Weibull stress gives
o
1 1 m
σ w ∫ σ = 1 m dV (5.26)
V o V f
The fracture process zone is defined as the region where the stresses are sufficiently high for a
realistic probability of cleavage.
Let us now consider the special case of a cracked body subject to applied loading. Assuming ρ
depends only on the local stress field, and the crack-tip conditions are uniquely defined by K or J,
2 For a detailed discussion of the Poisson assumption, consult any textbook on probability and statistics.