Page 85 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C02.fm Page 65 Thursday, April 14, 2005 6:28 PM

Linear Elastic Fracture Mechanics 65

Thus, the total stress intensity at each crack tip resulting from the closure stresses is obtained by
replacing a with a + ρ in Equation (2.73) and summing the contribution from both crack tips:

σ a+ρ a x +  a + ρ − ρ x + 
K =− YS ∫  +  dx
closure ( a + ) π ρ  a a x +  a − ρ + ρ x + 

a + ρ a +ρ d x
=−2σ YS π ∫ a a + ( 2 − ρ) x 2 (2.75)

Solving this integral gives

a  a + ρ 
K =−2σ cos −1   (2.76)
ρ
closure YS π  a + 
The stress intensity from the remote tensile stress, K σ σπ( a = + ρ) , must balance with K closure .
Therefore,
a  πσ 
a + ρ = cos   σ YS   (2.77)
2
Note that ρ approaches infinity as σ σ → YS . Let us explore the strip-yield model further by
performing a Taylor series expansion on Equation (2.77):

a =− 1  πσ  2 + 1  πσ  4 − 1  πσ  6 +



 σ
 σ
 σ
a + ρ 1 22 YS   42 YS   62 YS   (2.78)
!
!
!
Neglecting all but the first two terms and solving for the plastic zone size gives
πσ a π  K  2
2
2
ρ = =  I  (2.79)
8 σ 8  σ 
YS 2 YS
for σ << σ . Note the similarity between Equation (2.79) and Equation (2.66); since 1/π = 0.318
YS
and π/8 = 0.392, the Irwin and strip-yield approaches predict similar plastic zone sizes.
One way to estimate the effective stress intensity with the strip-yield model is to set a equal
eff
to a + ρ:
 πσ 
K = σπ sec   (2.80)
a
eff
 σ2 YS 
However, Equation (2.80) tends to overestimate K ; the actual a is somewhat less than a + ρ
eff
eff
because the strip-yield zone is loaded to σ . Burdekin and Stone [26] obtained a more realistic
YS
estimate of K for the strip-yield model
eff
  πσ   12 /
K eff YS π a = σ  8 lnsec    (2.81)
2
  π 2  σ YS   
Refer to Appendix 3.1 for a derivation of Equation (2.81).