Page 420 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C009.fm Page 400 Monday, May 23, 2005 3:58 PM

400 Fracture Mechanics: Fundamentals and Applications

panel is given by

ba  P  n+1
J = αε σ o W ha W n , )   P   (9.32)
(/
o
pl
1
o
where, in the case of the center-cracked panel, a is the half crack length and W is the half width.
This modification was made in order to reduce the sensitivity of h to the crack length/width ratio.
1
The elastic J is equal to G(a ), the energy release rate for an effective crack length, which is
eff
based on a modified Irwin plastic zone correction:
a
I
a =+ 1 1  n −  1   K   2 (9.33)
PP ) βπ
eff
1 + (/ o 2  n +  1  σ o 
where b = 2 for plane stress and b = 6 for plane strain conditions. Equation (9.33) is a first-order
correction, in which a is computed from the elastic K , rather than K ; thus iteration is not
eff
eff
I
necessary. This is an empirical adjustment that was applied in order to match the J values from
the estimation procedure with corresponding values from rigorous elastic-plastic finite element
analysis. This correction has a relatively small effect on computed J values.
The CTOD can be estimated from a computed J value as follows:
J
δ = d n σ o (9.34)
where d is a dimensionless constant that depends on flow properties [27]. Figure 3.18 shows plots
n
of d for both plane stress and plane strain. Equation (9.34) must be regarded as approximate in
n
the elastic-plastic and fully plastic regimes, because the J-CTOD relationship is geometry dependent
in large-scale yielding.

EXAMPLE 9.1

Consider a single-edge-notched tensile panel with W = 1 m, B = 25 mm, and a = 125 mm. Calculate
J vs. applied load assuming plane stress conditions. Neglect the plastic zone correction.
Given: s o = 414 MPa, n = 10, a = 1.0, E = 207,000 MPa, e o = s o /E = 0.002
Solution: From Table A9.13, the reference load for this configuration is given by`

P o . o b B= 1 072ησ

where

a
η = + 1   2 − a = 0 867 for = ab / 125/ = 875 0 143
.
.
  b
b
Solving for P o gives
P o = 8.42 MN
For a/W = 0.125 and n = 10, h 1 = 4.14 (from Table A9.13). Thus the fully plastic J is given by

0
1
J pl = ( . )( . 0 002 )(414 000kPa ) (.875 m )( .125 m ) (.414 )   P   11
1 0
,
. 10 m  .842 MN 
= × . 2 486 10 −8 P 11