Crack Paths 2012

where 5=1—3,B+3,B2.

Hence, the normal contact stress acting on crack faces equals

0'.(h—h0)=—P=—qfl3/5~

(5)

Stresses 0', and a2, and displacements U(r,z) and W(r,z) of a plate in the

uncracked domain (re (l;R]) , according to the model [6], are as follows

N M ZG* 2

2

2 G’

o' = — ’ + ’ z +z —0.6h

— A h — ;

6 ( )

.

2h 1

3, (

)[q2

q2

1 + _ 6.=qi+Z[3Z—;j-q2; q1=5(q —q ), qz=(q +q ); z z3 1 + _

2 1—v* U(r,z)=u(r)—z @ —l—(l%—V*)Z—2 — ( i I ) % Z 3 ;

dr dr

3h

8 E h dr

2

‘I1 a0'q2 W r,z = wr + 2 a - — +-A — A w + — - B , ( ) (1 OZ E, 2 8Eh (Z) I Z

7 ()

I

4

where

A’: V ;

B(z)=6B2z2—B3Z—2;

G*= l

[£,—v”(3+v));

(1-v)

h

4(1-v) G

,

~

~

A'E'

e 0 = ; [ 4 £(7—,v)];— v w=w+1.582q2h/E,

B 2 = 1 + i , ;

20(1-v) G

2a0G

v"A'E'

h

d2w v dw

B =B _

;

a =0.5—v'-A';

M = ZO'dZ=—D + — ——eh2 ,

3

2

4 a 0 G

0

r

i r

(drz r dr) 0 ‘12

h

- du u

dw

N, = jo'rdz=2Eh(—+v—)+2A'hq, ,Q, = K ' — ’are the bending moment, normal

_h

dr r

dr

. . and shear forces, respectively; A = ——2 + —is a—differential operator; u is a 1+v dr r dr 1 d2 v d . . .

tangential displacement of the median surface of the uncracked domain of the plate.

The general solution of Eqs. (2) and (4) has the following form

4

W, =A,r2+i§,r21nl+c,1nl+1<,+

qiz'” ,(i=1,2)

R

64 ,

,

(8)

. * wi” = Ai” + Bi” In;— qizr2 /4K,', u, : Er+ L,r_1+ u, , . . I"

where integration constants A,,B,,C,,K,, 21;”, BE”, F}, L, are determined from the

boundary conditions, and are the particular solutions.

Fromthe displacement and bending momentboundedness conditions it follows, that four integration constants are equal to zero: B, = C, = Bil’ =L, = 0. The notations used

here should be extended to account values corresponding to the upper and lower parts of

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