Fatigue Crack Paths 2003

The angular position stiffness matrix Kc(§2t) has one constant term (the m e a nstiffness)

and up to five harmonic components which are considered in following calculations.

This model is called Flex Model.

The stiffness matrix (square, symmetrical, 12><12 elements) is represented in “Eq. 1”:

_a c p q w m — a c — p q — w — m _ x1

e — q r i

0 —c f q s

—i 0 8y,

b — d k n — p — q — b — d — k — n y1

h j

0 — q s

d g —j 0 8,,

t 0 — w —i — k —j —t 0 z1

Xl

u — m 0 — n 0 0 — u 8,,

[Kc(§2t)

I

(1)

X2

a — c p — q w m x2

e q r

i

0 8y,

b d k n y2

h j

0 0,2

t

0 z2

_

u _ 8 Z 2

wherethe coefficients are defined as:

12J E (1+ 401i

12J E : x 3 (1 + )1c

6J E (1+ 401.

6J E (1+ )1c : x 2

(4+ <|))J E (1 + )1c

a : y

C : y 2

e :—

2 — J E f:( o, g:< 01. h:( +401. 2 — E 4

12J E p:_ ..3 q:_ . 26) 6J E

E

(1+ )1.

(1+ )1.

(1 + 401.

(1 + )1c

(1+ )1c

4 J E 2 — J E E A GJ

12E

(1+ <|>)lc

(1+ <|))lc

18

1t

GSlc

i, j, k, m, n, w, are cross coupling coefficients whichneed to be tuned, the parameter

(I) accounts for the shear effects, E and G respectively are the Young’s modulus and the

shear modulus, S is the cross section area. The different lengths [6, la and l, responsible

for the direct stiffness have also to be tuned by means of the 3D model.

As an example the ratio of “equivalent” length 10 to the diameter as a result from 3D

calculation, is represented in Fig. 4 (right) as a function of the ratio of the crack depth to

the diameter.

The3 DModel

The 3D non linear model allows obviously to calculate also deflections and strains (by

taking into account the breathing behaviour).