12.07.2015 Views

applied fracture mechanics

applied fracture mechanics

applied fracture mechanics

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

166 Applied Fracture Mechanics22 Will-be-set-by-IN-TECH2k 2n+ ⋅10x no01−22−4−2 −1 0 1 2Figure 7. The dependence of the shear stress intensity factor k + 2n on the coordinate x0 n for the case of theboundary of half-plane loaded with normal p(x 0 )=π/4 and frictional τ(x 0 )=−λp(x 0 ) stresses,y 0 n = −0.2, α n = 0, q 0 = 0: λ = 0.1 - curve marked with 1, λ = 0.2 - curve marked with 2 (after Kudishand Covitch [1]). Reprinted with permission from CRC Press.4. Stress intensity factors and directions of fatigue crack propagationThe process of fatigue failure is usually subdivided into three major stages: the nucleationperiod, the period of slow pre-critical fatigue crack growth, and the short period of fastunstable crack growth ending in material loosing its integrity. The durations of the firsttwo stages of fatigue failure depend on a number of parameters such as material properties,specific environment, stress state, temperature, etc. Usually, the nucleation period is short[1]. We are interested in the main part of the process of fatigue failure which is due to slowpre-critical crack growth. In these cases max (k + 1 , k− 1 ) < K f , where K f is the material <strong>fracture</strong>toughness.During the pre-critical fatigue crack growth cracks remain small. Therefore, they can bemodeled by small straight cuts in the material. For such small subsurface cracks it is sufficientto use the one-term asymptotic approximations k ± 1 = k 1 and k ± 2 = k 2 for the stress intensityfactors from (43) and (44) which in dimensional variables take the formk 1 = √ l[Y r + q 0 sin 2 α]θ[Y r + q 0 sin 2 α], k 2 = √ l[Y i − q02 sin 2α],Y =π1 ∫a H[p(t)D 0 (t)+τ(t)G 0 (t)]dt, τ = −λp,−a H{Y r , Y i } = {Re(Y), Im(Y)},[]D 0 (t) =2i −t−X 1 + 1t−X − e−2iα (X−X),(t−X)[ ]2G 0 (t) = 1 12 t−X + 1−e−2iαt−X− e−2iα (t−X), X = x + iy,(t−X) 2(45)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!