Crack Paths 2009

{ w: K w w i l (_fK2 w é ‘ a ) I

I

1

I

6 : K66 1(f1_ K a v w ) :(K65 _ KswKwwlKws (f1 _ (KewKww1Kwo")f2)

As is stated above, the solution of the problem requires the knowledge of the crack

jumpdisplacement vector w(x) .

W h e na crack starts, it is reasonable to assumethat the strain state present in the

finite element is “condensed” in a highly-localised strained area concentrated in a very

narrow band (crack location), i.e. a discrete displacement jumpcan be assumed. Let us

consider the F Emeannodal displacement values across the crack, projected respectively

in direction normal (uc) and parallel (V6) to the crack direction:

w, :u,+v, :u,-i+v,-j, with u, : [Q-(8-i)]/n,,

v, : [Q-(8-j)]/n,

(9)

where n” is the total numberof element nodes, the matrix Q is the nodal discontinuity

matrix, 5 is the element nodal displacement vector.

The nodal discontinuity matrix Q is determined by observing which nodes of the

finite element are in one or in the other side of the crack. By referring to Fig. 1b:

node i

"ode /

nudek nudel 1 4 %

—1 0 +1 0 +1 0 —1 0 (—dofsalongx Q I

(10)

O —l 0 +1 0 +1 0 — 1 ( — d 0 f s a l 0 n g y

To physically represent the presence of the crack, the stress state in the finite element

must be modified in order to have exactly the stress a, and the stress I, transmitted

respectively in direction perpendicular and parallel to the crack faces. The stress state

can be elastically-corrected as follows:

6 :0—C:lVS(N-5,,’S)]:o—C:lVS(N-s,,-5,)]:o—C:(sn-B-8,,)

(11)

rel,u

where c is the effective stress tensor, and 6 M I (sn 6”) is a fictitious nodal

displacement vector assumed to be proportional, through the coefficient s, , to the nodal

displacement vector 6,. Analogously, by introducing a fictitious displacement vector

6 m I (ss 6,), which is assumed to be proportional (through the coefficient ss) to the

nodal displacement vector 6, obtained by considering the projection 6v : (6 - j) j of the

current nodal displacementvector 6 on the direction j, the stress state becomes:

0 :c—C:lVS(N-8M)]:c—C:lVs(N-ss -6V)]:0—C:(sS-B-6V)

(12)

rel ,v

The normal stress 0",," and shear stress rm can be evaluated as follows:

a.. I16.. -i)-i I lC1(B'-6.,.)-il-i,

I.. I (6.. -i)-j I lC1(B'-6..)-il-.i

(13>

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