Page 139 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C003.fm Page 119 Monday, May 23, 2005 5:42 PM

Elastic-Plastic Fracture Mechanics 119

or

K 2 2 Ω p
J = I + ∫ MdΩ (3.38)
E b′ 0 p
Conversely, the prior analysis on the double-edged cracked plate in tension could have been written
in terms of ∆ and ∆ . Recall, however, that the dimensional analysis was simplified in each case
c
nc
(Equation (3.30) and Equation (3.36)) by assuming a negligible dependence on a/b. This turns out to
be a reasonable assumption for plastic displacements in deeply notched DENT panels, but less so for
elastic displacements. Thus while elastic and plastic displacements due to the crack can be combined
to compute J in bending (Equation (3.37)), it is not advisable to do so for tensile loading. The relative
accuracy and the limitations of Equation (3.32) and Equation (3.37) are evaluated in Chapter 9.
In general, the J integral for a variety of configurations can be written in the following form:
η U
J = c (3.39)
Bb

where h is a dimensionless constant. Note that Equation (3.39) contains the actual thickness, while the
above derivations assumed a unit thickness for convenience. Equation (3.39) expresses J as the energy
absorbed, divided by the cross-sectional area, times a dimensionless constant. For a deeply cracked
plate in pure bending, h = 2. Equation (3.39) can be separated into elastic and plastic components:

η U η U
J = el c el() + p p
Bb Bb
K 2 η U (3.40)
= I + p p
′ E Bb

EXAMPLE 3.1

Determine the plastic h factor for the DENT configuration, assuming the load-plastic displacement
curve follows a power law:

P C =∆ N p

Solution: The plastic energy absorbed by the specimen is given by

∆
p
U p ∫ ∆ p ∆ p N d = ∆ p = C ∆ N+1 = Pp
0 N +1 N +1
Comparing Equation (3.32) and Equation (3.40) and solving for h p gives

∆
Pp  2 − 1 
N +1
η =  Pp  =− N
1
∆
p
N +1
For a nonhardening material, N = 0 and h p = 1.