Crack Paths 2012

620 6 0

6 9

“ + 3 ’ = — ’ E ’ — o r a o r

11 ( )

@142

I"

, E where E'=E and Ot'=0t for plane stress, whereas E : l_v2 and or '= Ot(1+\/) for

plane strain.

Fromequation (6), we find

6 9 ( r , t ) : — q e H

(12)

6r 2rr7tr

The equilibrium equation of the thermo-mechanical problem is then

2 I i r 6 0 r + 3 a 0 r = qE0 te . I

(13)

6,42

6r 2Ir7tr

A nanalytical solution ofthe thermo-mechanicalproblem

Ranc et al. [6] have shownthat the solution of the differential equation (11) is

(WM z 2r2

z

2

821

I’

0.(r,t)=

+£2+G

(14)

E. I

It)»

4at

1”

with F and G two integration constants. Equation (7) allows one to find the

circumferential stress:

5,40

(15)

It is possible nowto use the mechanical boundary conditions in order to express the

two constants F and G. Because there is no thermal stress when r tends to infinity and

because of the mean stress relaxation in the reverse cyclic plastic zone, when t>0, the

boundary conditions are,

iimmm0,(r,t)=0 and o,(rR,t)=0

(16)

The first previous condition implies G=0. With equation (14), the second condition

allows one to express the constant, F, as follows, which is in fact a function depending

on time to respect the boundary conditions whatever the time t :

53