Crack Paths 2009

Wf : gcrfg: fl C G J r l % { O - f [ % , u c s i n 2 6 — % , u c s i n 4 0 — 2 , u 2 sin4 6J+

1 1— 2 o22[Z,acsin 40 —5,ucsin 20 — 2,a2ccos4 6 + 2 g,ancos6,/l(l + l,,.,)] 1

(15)

1 _ + o'lo'2 [ 5 p csin 46 _ H26sin2 6 + 1 2luns;inéltan g6,ll(l + 1.9)}

Based on Eqs. (8), (11), (13), and (15), S11 , S,2,S22 can be solved by comparing the

coefficients of the quadratic terms as

_ (1+k)(1+v) 1 S l 1 _T m b q l B l + C 1 ) (16a)

1 + k 1 + 1 s22 : % m o t e , +c2 +A2D, +A3 +A4C3)

(16b)

1 + k 1 + 1 s12 : % m o +124h—A4C4e+A5C5), (16c)

The coefficients of A,,A2,A3,A4,A5 related to the length of tensional crack I,

B1,B2,B3,B4,C,,C2,C3,C4,C5,D1related to the initial crack length 2c, and crack

orientation 0 weregiven in the reference[10].

For a body containing N sliding cracks, and if crack interactions are not

. A considered, the total non-elastic strain can be calculated as N ><[ 61], the total A52 . ,

strains induced from the applied compressive loads can be calculated by

1

k — 3

$1 : gli+gli :(k‘l'DG‘l‘V)

l€+1 U1 + NS11 S12 61

(17)

e,

e; +e2

4 E k — 3 1

0'2

S12 S22 o"2

k +1

Under dynamic uniaxial compressive loads, 52 : 0 ,

and the constitutive

relationship of the alumina is reduced to the following forms

8 : (k+1)(1+v)

k — 3

,

4E [0",+—k+1o'2]+N(SUo',+S,2o'2)

(18)

Under uniaxial strain conditions, lateral stress 02 : (1—v)/(1—2v)o',, and then the

constitutive relationship with respect to the 0-1 direction can be given as

g _ (1+v)(1—2v) 1

(1_V)E (0',+N><(A,B,+A,B3—A4C4+A5C5+C1+B4)O-1)

(19)

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