PSI - Issue 18

Johannes Scheel et al. / Procedia Structural Integrity 18 (2019) 268–273

271

4

J. Scheel et al. / Structural Integrity Procedia 00 (2019) 000–000

σ

F

a = 36

12

100

6

a)

F

96

a = 18

200

F

12

a = 6

b)

c)

96

σ

20

Fig. 2. a) DCB , b) three-point-bending, c) tension bar specimen with F = 10 N, σ = 0 . 5 MPa, E = 210000 MPa, ν = 0 . 3.

problem at hand, where only a 1 and a 3 remain to be evaluated. As node B, according to Fig. 1, is the first node behind the crack tip node A, it is not chosen due to crack tip inaccuracy. Nodes C and D are reasonable choices and will be used exemplarily for the following analytical calculations. With

a 1 −

2 µ   

a 3  

3 L 2

, π = u D

2

u 2

3

√ L 3 a

√ La 1 −

2 µ

3 ,

1 + κ

1 + κ

3 L 2

3 L

u 2 ( L , π ) = u C

 , (9)

2 =

2 =

where u C / D 2

are the nodal displacements of the FE solution on the positive crack face, the parameters a 1 and a 3 can be

easily calculated. With Eqs. (7) and (8), the SIF is finally obtained as

D 2

D 2

8 µ 3 u C

8 µ 2 u C

( ∆ a − r ) 3 r

∆ a

4 √

4 √

1 ∆ a

1 ∆ a ∆ a

6 u

6 u

2 −

2 −

∆ a − r r

lim ∆ a → 0

dr −

dr .

(10)

K I =

lim ∆ a → 0

(1 + κ ) √ 2 π L

(1 + κ ) √ 2 π L 3

Without the limit the integrals are easily calculated as ∆ a 0 ∆ a − r r dr = π 2 ∆ a , ∆ a 0

( ∆ a − r ) 3 r

3 π

a 2 .

dr =

(11)

8 ∆

For this approach, the limit ∆ a → 0 is not taken, assuming ∆ a to be equal to one element length L . This assumption and the just calculated integrals lead to the final formulation of the SIF, exploiting two nodal displacements of the FE solution:

µ (1 + κ )

6   .

2 π L   3 u C

2 u D 2 √

(12)

K I =

2 −

For the verification, the models of Fig. 2 are investigated under mode I loading. Numerical investigations with the limit ∆ a → 0 proved to produce low quality results and are omitted here. Additionally, approaches using only the first eigenfunction are investigated. Here only node C or node D of Fig. 1 are employed, i.e.

2 µ (1 + κ )

4 µ (1 + κ )

2 π L

π 3 L

u C

u D

K I =

K I =

(13)

2 ,

2 ,