Callister - An introduction - 8th edition

244 • Chapter 8 / Failure
EXAMPLE PROBLEM 8.1
Maximum Flaw Length Computation
A relatively large plate of a glass is subjected to a tensile stress of 40 MPa. If
the specific surface energy and modulus of elasticity for this glass are 0.3 J/m 2
and 69 GPa, respectively, determine the maximum length of a surface flaw
that is possible without fracture.
Solution
To solve this problem it is necessary to employ Equation 8.3. Rearrangement
of this expression such that a is the dependent variable, and realizing that
40 MPa, s 0.3 J/m 2 , and E 69 GPa, leads to
a 2Eg s
ps 2
122169 109 N/m 2 210.3 N/m2
p140 10 6 N/m 2 2 2
8.2 10 6 m 0.0082 mm 8.2 mm
Fracture toughness—
dependence on
critical stress for
crack propagation
and crack length
fracture toughness
Fracture Toughness
Furthermore, using fracture mechanical principles, an expression has been developed
that relates this critical stress for crack propagation ( c ) and crack length (a) as
K c Ys c 2pa
(8.4)
In this expression K c is the fracture toughness, a property that is a measure of a
material’s resistance to brittle fracture when a crack is present. Worth noting is that
K c has the unusual units of MPa1m or psi1in. (alternatively, ksi1in. ). Furthermore,
Y is a dimensionless parameter or function that depends on both crack and
specimen sizes and geometries as well as the manner of load application.
Relative to this Y parameter, for planar specimens containing cracks that are
much shorter than the specimen width, Y has a value of approximately unity. For
example, for a plate of infinite width having a through-thickness crack (Figure 8.9a),
Figure 8.9 Schematic
representations of (a) an
interior crack in a plate of
infinite width and (b) an
edge crack in a plate of
semi-infinite width.
2a
a
(a)
(b)