Crack Paths 2006

form of the integral equation 8 with discrete distributions of u,, t, and Au, along the

boundary parts TB and Ft}, respectively). Therefore, to determine the displacement

discontinuity, Au,(R(6)), along 17);} an additional integral equation for the crack surface

tractions (i.e.,

t,(R(g))Io',j(Rg))nj(R$)))

must be derived. Such equation can be

obtained by means of Hooke’s law:

011(1) : 2luk,k(r)6ij + 1“ lui.j(r)+ uj,i(r)l

(k=112 ) -

(9)

1-2v

By differentiating u,(r) in Eq. (8) with respect to the source point r and then

substituting the derivatives: uk,k(r), u,,j(r) and uj,,(r) into the expression 9, the

additional integral equation for 0,]. (r) is given by

a, (r) = [19,, (r, R’)tk(R')dF(R’)

- [s,,.(r,R')u,,(R')dr(R')

+

F 3

F 5 — 18.-.1.Rstullstrvrsst.

(1°)

Flc'l

where

2 Dkij : 1_V§VUkl,l(r)6if' + 1” [Uki,j(r)+ Ukj,i(r)i (1:112)’

(11a)

2 Skij : % T k i A fi +d1ul-]1i,jj(r)+Tlg,1(r)l '

(11b)

Multiplying both sides of Eq. 10 by nj (r) gives

10 = [ lD1.(r.R')r1.(R')dF(R')]

1.1)

- [ 1s.dr]1.1)

+

F e

F e

(12)

- [s,,(r, R',i*i)Au,(R'Ci—})dr(R'CH)

n, (r) .

Ft’

The boundary forms of the integral equations 8 and 12 define the problem to be solved.

However, the fundamental drawback of the boundary form of Eq. 12 is that this

equation contains the r'2 singularity that is difficult to handle in numerical calculations.