Page 331 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

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Fracture Toughness Testing of Metals 311

Additional validity requirements include the restrictions on the fatigue load mentioned earlier, as
well as limits on the fatigue crack curvature. If the test meets all of the requirements of ASTM E
399, then K = K .
Ic
Q
Because the size requirements of ASTM E 399 are very stringent, it is very difficult and
sometimes impossible to measure a valid K in most structural materials, as Example 7.1 and
Ic
Example 7.2 illustrate. A material must either be relatively brittle or the test specimen must be
very large for linear elastic fracture mechanics to be valid. In low- and medium-strength structural
steels, valid K tests are normally possible only on the lower shelf of toughness; in the ductile-
Ic
brittle transition and the upper shelf, elastic-plastic parameters such as the J integral and CTOD
are required to characterize fracture.
Because of the strict validity requirements, the K test is of limited value to structural metals.
Ic
The toughness and thickness of most materials precludes a valid K result. If, however, a valid
Ic
K test can be measured on a given material, it is probably too brittle for most structural
Ic
applications.
When attempting to measure fracture toughness using ASTM E 399, one runs the risk of invalid
results due to the stringent size requirements. Once a result is declared invalid, E 399 offers no
recourse for deriving useful information from the test. A more recent ASTM standard, E 1820 [4],
provides an alterative test methodology that permits valid fracture toughness estimates from sup-
posedly invalid K tests. ASTM E 1820 is a generalized test method for fracture toughness mea-
surement that combines K, J, and CTOD parameters in a single standard. This standard provides
a single test method for all three parameters and then offers a choice of post-test analysis procedures
that pertain to a range of material behavior. If a test specimen exhibits too much plasticity to
compute a valid K , the fracture toughness of the material can be characterized by J or CTOD.
Ic
The calculation procedure and size requirements for K are essentially the same in E 399 and E
Ic
1820, but the latter relaxes specimen geometry requirements somewhat. For example, E 1820
permits side-grooved specimens in K tests. The standard BS 7448: Part 1 [10] is the British
Ic
equivalent of ASTM E 1820.

EXAMPLE 7.1

Consider a structural steel with σ YS = 350 MPa (51 ksi). Estimate the specimen dimensions required
for a valid K Ic test. Assume that this material is on the upper shelf of toughness, where typical K Ic values
for the initiation of microvoid coalescence in these materials are around 200 MPa m .

Solution: Inserting the yield strength and estimated toughness into Equation (7.2a) gives

 200 MPa m  2
,
.
32
Ba = 25   350 MPa   = . 0 816 m ( .1 in. )
Since a/W ≈ 0.5, W = 1.63 m (64.2 in.)! Thus a very large specimen would be required for a valid
K Ic test. Materials are seldom available in such thicknesses. Even if a sufficiently large section
thickness were fabricated, testing such a large specimen would not be practical; machining would
be prohibitively expensive, and a special testing machine with a high load capacity would be
needed.

EXAMPLE 7.2

Suppose that the material in Example 7.1 is fabricated in 25 mm (1 in.) thick plate. Estimate the largest
valid K Ic that can be measured on such a specimen.