Compliances, Energy Release Rate, and Stress Intensity Factor for a System of Loads

To calculate arbitrary displacements from Kj expressions, it is convenient to establish a more general framework in which a system of independent forces Pi (г — 1, …, n) is assumed to act on a cracked elastic body. Let the displacement of the load-point г in the direction of Pi be U;. The displacements may be written as linear functions of the loads:

П Ф

Щг — 1, ..,,n (3.5.12)

where Cij(a) are the elements of the compliance matrix (which depend upon the crack length a). As а consequence of the reciprocity theorem, the compliance matrix is symmetric: Cij = Cj{.

The elastic and complementary energies are equal, and are given by U = U* = J^"=1 PjU;/2. Substituting (3.5.12), the complementary energy is obtained as

– П П

г=1 j~ 1

Подпись: G Подпись: l 26 Подпись: =i j=t Подпись: dCij{a) da Подпись: (3.5.14)

From this expression and (2.1.21), we obtain the following expression for the energy release rate:

For a general system of loads, this equation is the equivalent of (2.1.34).

For the stress intensity factor, we may apply the superposition principle and write

(3.5.15)

Compliances, Energy Release Rate, and Stress Intensity Factor for a System of Loads Compliances, Energy Release Rate, and Stress Intensity Factor for a System of Loads Подпись: (3.5.16)

where Kn —• (Pi/b/D)ki(a) is the stress intensity factor due to Pi alone. Now, according to Irwin’s equation, Q — Kj/E’, and so

The values of the. PjS are arbitrary. So for the equality to hold for any Pi, the coefficients of the products PiPj on both sides of the equation must be identical. We thus find that

rfC^(a) ^ = …,n) (3.5.17)

which is the generalization of (3.5.2) to any system of forces. This equation can be integrated in the same

way as before, to obtain

2 ra/D,

Cij[a) = Cijo + —■•; J ki(a)kj(a)da (i, j = 1, …, n) (3.5.18)

where CijQ is the component of the compliance matrix for the uncracked body (a — 0). This equation

provides the means of obtaining the displacements Uj at various points caused by only one force, say P (all the remaining forces being zero). The result is, obviously, Uj — CjP.

The foregoing equation can be recast in terms of the full expression of the stress intensity factor, by setting that к (a) — b/D Кц/Рі and thus

. 2b fa KjiKjj. ,, . .

Cij(a) — Gij0— J pp Ca (гі 3 — ■■■ї70 (3.5.19)

which does not require a particular form of expressing the stress intensity factor, and can be directly used when Pi are generalized forces rather than point loads.

Note, again, that in the foregoing expressions a stands for the total crack length, and the energy release

rates correspond to a single crack tip. For the center cracked panel, the two crack tips must be made

explicit, as previously done for the single force loading, Eq. (3.5.11). The general expression for multiple loading is:

2 fa^D г* л л -і

Су (a) = СУо + — J^ [fc+(a)fct (a) + (a)fc,- (a)J da i, j = 1, …, n (3.5.20)

where kf (a) and fcr (a) are the shape factors for the stress intensity factor created by load Pi at the right and at the left crack tips, respectively.