Crack Paths 2012

Table 1

Thick. Imp. stress Force equ. Force equ. Max. stress

C (exemple)

m K1 pour 0 . 5 m m K1 pour 0.5mm

d N

m m (MPa)

(N)

(kg)

no crack, M p a (dKin MPa.sqrt(mm))

(MPa.sqrt(mm)) (MPa.sqrt(m)) for da = 0.1 m m

1

1

56.57

5.77

5.9

3.80 E-1 1

3

3.82

0.12

4.73E+08

2

1

1 13.14

1 1.53

3

1

169.71

17.3

4

1

226.27

23.07

1

10

565.69

57.66

59

3.80 E 11

3

38.18

1.21

4.73E+05

1

20

1131.37 115.33

118

3.80E-11

3

76.36

2.41

5.91 E+O4

1

30

1697.06 172.99

177

3.80 E-1 1

3

114.54

3.62

1.75E+04

1

40

2262.74 230.66

236

3.80 E-1 1

3

152.72

4.83

7.39 E+03

1

50

2828.43 288.32

295

3.80E-11

3

190.9

6.04

3.78E+03

1

100

5656.85 576.64

590

3.80 E 11

3

381.8

12.07

4.73E+02

2

10

1131.37 115.33

59

3.80E-11

3

38.18

1.21

4.73E+05

2

2 0

2 2 6 2 . 7 4 2 3 0 . 6 6

1 1 8

3 8 0 5 1 1 3

76.36

2.41

5.91E + O 4

2

30

3394.1 1

345.98

177

3.80 E-1 1

3

1 14.54

3.62

1 .75E+04

2

40

4525.48 461.31

236

3.80 E-1 1

3

152.72

4.83

7.39 E+03

This table can be used to choose a device depending on the load to be applied. The

plastic area has to be monitored as it should remain “small” for the LEFM.

I'M

s

.

it

-mm'im

Y

o

|

2

a

4

6

[L11

—

Figure 8: Von Mises stress, with crack Figure 9: Zoomon notch

For a 2mm-deepcrack, the stress concentration is about 3 at 0.5mmfrom the tip and

about 5 at 0.1mmfrom the tip (see Figures 8 and 9). These distances are quite small

regarding to the specimen geometry, thus an imposed stress of about a third of the yield

stress would fulfill the condition for confinedplasticity.

Loadingdefinition forphase 2

Preliminary simulations were done. Unlike the Figure 5, in the present case, the static

load is vertical (1 M P a )and the alternate load is horizontal (between0 and 0.1, 0.5, 1, 2,

10, 10000 MPa).

Starting from a crack of 2 m min a 5 m mnotch, Figures 10 and 11 show the pathes of

the different configurations (about 2 0 m min total do). The top curved line corresponds

to a m a x i m u mload of 0.1 M P aand the right curve corresponds to a m a x i m u mload of

10000MPa.In the case wherethe m a x i m ulomad is 2, as the m e a nload ofthat case is 1

M P ahorizontally and vertically, there is a symmetrythat leads to the 45°-line.

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