Crack Paths 2012

The figure 3a shows the evolution of the circumferential stress along a radial axis for

two different times. Near the reverse cyclic plastic zone (rRI4 pm) the circumferential

stress is negative (Figure 4) because the temperature is high and through the

circumferential direction, the material is under compression due to the thermal

expansion and the constraint effect. Further from this zone, the temperature is lower and

the circumferential stress becomes positive (tension) due to the confinementof the

material near the crack tip (Figure 4). For times tIl s and tI10 s the circumferential

stress in the edge of the reverse cyclic plastic zone is respectively equal to -0.056 M P a

and -0.065 M P afor a unit line heat source. It has to be pointed out that all the previous

stress values are small because they are computed for a unit heat source q (per unit

length of crack front), but the stresses are proportional to q.

0.002

0.00

— l = 1 0 s

"001

401

- l=1s

Ts‘

.-.

0. 0.000

m n. 0.02

E

I‘

g

5"‘,

0.1131

n

|

W

L?

'

g 0.0:

1’ 41m?

'

E

l“

m

‘.2

.

E 004

9 0,003

l

E

‘Z

‘

.'

E 0.05

U

l

g 0.004

1

v

3

o

' I

_\_ 6

.‘;

U 0005

1/

41.06

| I

‘ n ' m I.

I

I

l

l

|

l

' o v w|

I

l

l

‘

I

l

I

I

l

I

0,0)

0,01

0172

0.03

0 H 0.05

0

5

10

15

20

2b

50

35

40

Radius [rrl]

Radius [pm]

-3-

_b_

Figure 4: The circumferential stress distribution at various times for a unit heat source a =1 I/V.m_l and m I 4 pm; a) general view, b) enlargement near the reverse cyclic

plastic zone.

T H EE F F E COTFT H ET H E R MSATLR E S S EOS NT H ES T R E S IS N T E N S I T Y

F A C T OURN D ECRY C L ILCO A D I N G

N o wwithin the heterogeneous stress field due to the thermal stresses, if we consider the

previous theoretical case of an infinite plate with a semi-infinite crack along a radial line

from rR to +00, the associated stress intensity factor, Kljemp, due to the temperature

gradient can be determined from the Wedge force (Green's function) solution (see [10]

page 87) as:

K,,,,,,,,,(t)=\/%]°°

M a r .

(20)

’R \/r—rR

From equations (19) and (20), the stress intensity factor due to thermal stresses is

expressed by:

55