Crack Paths 2012

for a complex eigenvalue )1 this simplifies to

110 ~ 2 R e ( K X+)... I 2 ( R e K R —e IX m K I m X ) + . . .

For each pair X1, X 2 of power-lawsolutions there exists a pair of dual power-law

solutions

Yi : 1:_’\i\I/'(g0),

t‘: 1, 2

Case 1

(8)

Y1(.’I;) I 1TX W ,Y2(ac) I 1:_>‘\I/(g0)

Case 2

(9)

Y1(ac) I 1“_>‘\I/1(

1:)‘ ln 111/200), Y2(ac) I 1“_>‘\I/2(g0).

Case 3

(10)

By Clayperon’s theorem the potential energy can be represented as an integral over

the external boundary:

1 1 U(—Aa)I — — / - pU_Aa d5, hence A UI — / p -(U_Aa — 110) ds,

(11)

2 r

2 P

which meanswe can calculate the energy release if we knowthe difference U_Aa— 110

on the external boundary I‘. To calculate this as least asymptotically we use the

m e t h o dof matchedasymptoticexpansions [9, 10, 11]. For small Aa, near the

external boundary 1“, the solution U_Aa will not differ too muchfrom the solution

110, hence U_Aais approximatedby an outer expansion

u_A°(ac) ~ 110(ac) + a1Q1(9c) + agggfir) + . .. ,

>> Aa.

(12)

The functions Qj are so-called weighting functions (cf. [12],[13]),

g1 : Yj + 51'

here Zj are solutions to the problem (1), (3) for Aa I 0 and p I —0'(Yj), hence Qj

are singular at the crack tip (0, 0), moreover

g j ~ Y j + m j k X k + . . .|9,c|—>0.

(13)

Dependingon the fixing of X9‘, the dual power law solution Yj can always be nor

malized in such a way that

K j / p - Q j d rsesp,. K / p - Z C Z S ,

(14)

F

P

which plugged into (11) gives

2 A UI K1a1(Aa) + K2a2(Aa) + . . ., or

AU : Re(Ka1(Aa))+ ...,

(15)

respectively, observe that in Case 2 w e have a2 I a—1. To determinethe coefficients

a we use the inner decomposition of U_Aa. Passing to the stretched coordinates

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