Page 392 - Mechanical Behavior of Materials

Section 8.9 Extensions of Fracture Mechanics Beyond Linear Elasticity 393

Example 8.7
Problem 8.48 concerns a test on a double-edge-cracked plate of 7075-T651 aluminum, for which
√
a = 5.7, b = 15.9, and t = 6.35 mm. A value of K Q = 37.3MPa m is calculated for the force
P Q = 50.3 kN, but the test for applicability of LEFM (Eq. 8.39) is not met. Calculate the fully
plastic force. If it is reasonable to do so, also apply the plastic zone adjustment to obtain a revised
value of K Q .
Solution Figure A.16(a) also applies to the geometry of Fig. 8.12(b), so that the fully plastic
load is
a

P o = 2btσ o 1 −
b
5.7mm

P o = 2(0.0159 m)(0.00635 m)(505 MPa) 1 −
15.9mm
P o = 0.0654 MN = 65.4kN Ans.
where the yield strength of 7075-T651 Al from Table 8.1 is used. Comparing P o with P Q gives

P Q 50.3kN
= = 0.77
P o 65.4kN
Since this ratio is less than 0.80, it is reasonable to apply the plastic zone adjustment.
From Fig. 8.12(b), the value α = a/b = 0.358 that applies is well within the range α ≤ 0.6,
where F ≈ 1.12. Hence, F can be taken as unchanged for a e and Eq. 8.43 applies. Thus,
we have
√
K Q 37.3MPa m √
= 40.5MPa m Ans.
K Qe = =
2 2

1 FS 1 1.12 × 249 MPa

1 − 1 −
2 σ o 2 505 MPa
where S = S g = P Q /2bt is used.
√
Comment This adjusted K is 40% above the value from Table 8.1 of K Ic = 29 MPa m. The
probable explanation for the difference is that K Qe includes an effect of increased toughness for
plane stress.

8.9.2 The J-Integral
An advanced approach to fracture based on the J-integral concept is capable of handling even
large amounts of yielding. In a formal mathematical sense, the J-integral is defined as the quantity
obtained from evaluating a particular line integral around a path enclosing the crack tip. The material
is assumed to be elastic—that is, to recover all strain on unloading—but the stress–strain curve
may be nonlinear. For our present purposes, it is sufficient to define J as the generalization of the