Bueckner’s Expression for the Weight Function

Подпись: dw(x,a) _ сгдг dvx(a) 1 да E' да 1)
Подпись: JE2k(a)kG(a,x/D) Bueckner’s Expression for the Weight Function Подпись: (3.5.46)

Bucckner (1970) devised a procedure to obtain the weight function kc(a, x/D) from the solution for the stress intensity factor and crack opening profile for arbitrary loading. This is the method exploited in the book by Wu and Carlsson (1991). Here it suffices to exploit (3.5.42) for demonstrating the simplest version of Bueckner’s result. Differentiating that equation with respect to a (and keeping in inind that a = a/D), we get:

Подпись: kG{a, x/D) Подпись: E'[D du>(x, a) ~KJ(a) Ida Подпись: (3.5.47)

where in the last expression we substituted (2.3.11), i. e., K[ — а^/Ок(а). Solving for ka{a, x/D), we get

The arguments a and x have been made explicit for clarity. Note that, compared to the expressions in other texts, our analysis is limited to pure mode I (structures and loadings symmetric with respect to the crack plane), and so the half crack opening w/2 is equal to half the displacement of the upper face, which is the usual variable included in the weight function expressions. We use the crack opening rather than the displacement of one crack face because in the following chapters the crack opening is the essential variable.

Consider now the center-cracked panel. Since it has two crack tips, we cannot get both weight functions from a single loading. This is clear from (3.5.44) where two unknowns are present, namely kg(a, x/D) and k/.(a, x/D). In particular, from the solution for a symmetric loading for which kZ'(a) = k~(a) =

ks{a), we can find only the symmetric part of the weight function. Indeed, proceeding as before, we get k^{a/D, x/D) + kc(a/D, x/D) = (3.5.48)

where subscript s indicates that the loading corresponding to this solution must be symmetric. The symmetric part of fcg is ail that is needed to obtain further crack opening profiles for other symmetric loadings. To obtain the complete expression for the right and left weight functions, we also need to solve the antisymmetric case, for which k j (a) = — (a). If this solution is available, it is easy to find the

Подпись: k£(a/D,x/D) - ka(a/D,x/D) Подпись: E'bD dwa(x,a) P/ki(a/D) 2 da Подпись: E,'/D dwa(x,a) Подпись: (3.5.49)

antisymmetric part as

where subscript a refers to antisymmetric loading. Combining (3.5.49) and (3.5.48), one can easily obtain the expression for the weight functions corresponding to both crack tips.

Exercises

3.15 The stress intensity factor lor a center cracked strip of width D, with a crack of length 2a subjected to remote uniaxial stress, may be approximated by the Feddersen-Tada expression (3.1.4) within 0.1%. Write the equation for the additional compliance of the strip due to the crack. Take two terms of the series expansion of the integrand in powers of a/D, calculate the additional compliance, and estimate the values a/D for which this result is accurate within 2%.

3.16 For the panel in Exercise 3.4, find the volume of hydraulic fluid injected into the jack for given crack length 2a and pressure p, assuming the fluid to be incompressible. Hints: Define P — 2pbc and и = V/26с. Watch the integration limits. Integrate by parts twice.

3.17 Find the volume of a centrally located crack in a large panel subjected to equal and opposite normal forces at the crack center.

3.18 Find the crack opening profile of a centrally located crack in a large panel subjected to equal and opposite normal forces at the crack center. Note the logarithmic singularity at x ■— 0.