Crack Paths 2009

1 d K can be calculated as K : m — X (—a). .

.

Now,by assumin o" I o-

for

exp

exp 1

C

1

g

tensile

a : 0 t o a : a 1 ,

K e x p l: [:1 U t e n s l l e f (ax17)cbc

and

KQXP2 : it? Ut@"Sil@f(x’a2 )dx + I: 0-12f(x’ a2 )dx

(2)

Where o"12 denotes the stress acting on the section from a : al to a : a2. Similarly,

K e X P :3 Ioal U t e n s i l e f (ax3’) d x+ ‘Liz U l 2 f (ax3)’d x+ I ; 3 0 - 2 3 fa( 3x)’d x

Fromthese equations, the stress values can be evaluated as:

K

exp 1

Utensile :

I: f(x: a1 )dx

a1

K e x p_2 ‘[0 U t e n s i l e f (ax ’2) d x

U 1 2

—

I : f(x’a2)dx

K°XP3 —Ioa1Ulerlsilef(x7 a3 )dx _ I: 0'12f(x, a, )cbc

623 I

Iff(x, a3 )dx

2

Andby induction:

a1

1T3 alr+l

Kexpi _.l0 atensflef(x’ai)cbc_z|:.l_a O-k,k+1f(x:ai

18:1

‘’

U -

I

i-l,z

1:1 1f(x. aJdX

Wheref(x, a,.) is the weight function for a crack length of al..

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