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.

Fracture Mechanics Based Models of Structural and Contact FatigueFracture Mechanics Based Models of Structural and Contact Fatigue 17 161which can be easily solved by classical methods [1, 14]. We will limit ourselves to determiningonly the first two terms of the expansions in (42) in the case of Coulomb’s friction law givenby (38) and (39). Without getting into the details of the solution process (which can be foundin [1]) for the stress intensity factors k ± 1n and k± 2n we obtain the following analytical formulas[1]k ± 1 = cr 0 ± 1 2 δ 0c r 1 + ... if cr 0 > 0, k± 1 = 0 if cr 0 < 0,k ± √ √1= 3δ09 c r 1 [±7 − 3θ(cr 1 )] 1±θ(c1 r)1±3θ(c1 r) + ... if cr 0 = 0 and cr 1= 0,k ± 2 = ci 0 ± 1 2 δ 0c r 1 + ...,(43)c j =π1 ∫ 1[p(x)D j (x)+τ(x)G j (x)]dx + δ j02 q0 (1 − e −2iα ), j = 1, 2,−1c r j = Re(c j), c i j = Im(c j),where the kernels are determined according to the formulas[[D 0 (x) =2i − 1 + 1 − e−2iα (z 0 −z 0 )], Gx−z 0 x−z 0 (x−z 0 ) 2 0 (x) = 1 12 x−z 0][+ 1−e−2iα − e−2iα (x−z 0 ), Dx−z 0 (x−z 0 ) 2 1 (x) = ie−iα 1 − e −2iα2(x−z 0 ) 2] [][(44)− 2e−2iα (z 0 −z 0 )+ ieiαx−z 0 2− 1 +e−2iα , G(x−z 0 ) 2 (x−z 0 ) 2 1 (x) = e−iα 12(x−z 0 ) 2] []−e −2iα − 2e−2iα (x−z 0 )+ eiα 1x−z 0 2+e−2iα ,(x−z 0 ) 2 (x−z 0 ) 2and θ(x) is the step function (θ(x) =−1 for x < 0 and θ(x) =1 for x ≥ 0).3.2. Comparison of analytical asymptotic and numerical solutions for smallsubsurface cracksLet us compare the asymptotically (k ± 1a and k± 2a ) and numerically (k± 1n and k± 2n ) obtainedsolutions of the problem for the case when y 0 = −0.4, δ 0 = 0.1, α = π/2, λ = 0.1, andq 0 = −0.005. The numerical method used for calculating k ± 1n and k± 2n is described in detail in[1]. Both the numerical and asymptotic solutions are represented in Fig. 3. It follows fromFig. 3 that the asymptotic and numerical solutions are almost identical except for the regionwhere the numerically obtained k + 1 (x0 ) is close to zero. The difference is mostly caused bythe fact that the used asymptotic solution involve only two terms, i.e., the accuracy of theseasymptotic solutions is O(δ0 2) for small δ 0. However, according to the two-term asymptoticsolutions the maximum values of k ± 1 differ from the numerical ones by no more than 1.4%.One can expect to get much higher precision if δ 0 < 0.1 and | y 0 | δ 0 .Therefore, formulas (43) and (44) provide sufficient precision for most possible applicationsand can be used to substitute for numerically obtained values of k ± 1 and k± 2 .

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

Saved successfully!

Ooh no, something went wrong!