PSI - Issue 2_A

J. Hein et al. / Procedia Structural Integrity 2 (2016) 2462–2254

2248

J. Hein, M. Kuna / Structural Integrity Procedia 00 (2016) 000–000

3

with

A C c A C c

σ i j

∂ T ∂ x 1

∂ x 1

δ i j q d A , (2)

I 1 = I 2 = I 3 =

∂ C i jkl ∂ x 1

∂ C i jkl ∂ T

∂ T ∂ x 1 −

∂ ∆ T ( x i )

1 2

1 2

∂α ∂ T

∂α ∂ x 1

m i j

m

m i j

m kl

∆ T ( x i ) + α ( x i , T ( x i ))

kl +

+

U δ 1 k − σ ik u i , 1 q , k d A , Un 1 − t i u i , 1 q d s .

(3)

C + + C −

(4)

Hereby, m th i j is the mechanical strain tensor, C i jkl elasticity tensor, σ i j stress tensor, α thermal expansion coe ffi cient, U the strain energy density, t i = σ i j n j traction vector and q = q ( x ) is a function that varies smoothly within A C c q ( x ) = 0 on C gr 1 on C . (5) i j = i j −

x 2

C +

x 1

r

A C c

C C gr

a

A C

t i

C

Fig. 1: Integration paths around the crack tip (compare Kuna (2013)).

In case of a crack propagation in x 1 -direction the energy release rate is described by G = J = J 1 . Thereby, J 1 is the first component of the J -vector

r → 0 C r → 0 C

r → 0 C

r → 0 C

m i j , T ) n m − t i u i , m d s = lim

Un m − σ i j n j u i , m d s = lim

U δ m j − σ i j u i , m n j d s

J m = lim

U ( x i ,

Q m j n j d s , m = 1 , 2 , 3 .

(6)

= lim

When dealing with 3D problems, the crack tip becomes a crack front line. If the contour C is moved along the crack front by a small segment ∆ s , the cylindrical surface S C is created as sketched in Fig. 2. Assuming a virtual crack extension ∆ l ( s ) in the crack plane inside the small crack front segment ∆ s (Fig. 2) with a crack propagation vector normal to the crack front

√ v m v m = 1 , ∆ l ( s ) = ∆ l m ( s ) ∆ l m ( s ) = l ( s ) ∆ a ,

∆ l m ( s ) = l ( s ) ∆ a v m with

(7)

the energy that is set free at a certain point s , is given by J m ( s ) ∆ l m ( s ) = lim r → 0 C

Q m j n j d s l ( s ) ∆ a v m ( s ) .

(8)

So, the total released energy for a crack, which is virtually extended by the area ∆ A along the segment ∆ s , amounts to − ∆Π = ∆ s J m ( s ) ∆ l m ( s )d s = ∆ s lim r → 0 C Q m j n j d s ∆ l m ( s )d s = lim r → 0 S C Q m j n j ∆ l m d S = J ∆ A (9)