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Rasmus ÿstergaard forside 100%.indd - Solid Mechanics

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Interface crack in sandwich specimen 305<br />

Fig. 3 By superimposing<br />

the stress field in (b) onthe<br />

stress field of the original<br />

problem (a) the stress field<br />

of a reduced problem (c) is<br />

obtained<br />

linearity and dimensionality arguments as those used<br />

by Thouless et al. (1987) on a homogeneous specimen,<br />

to express the dependency of most parameters<br />

analytically. First, we note that, in accordance with (6),<br />

the complex stress intensity factor must have the SI<br />

units Pa m 1 2 −iɛ . Furthermore, since the model is made<br />

within linear elasticity, the stress intensity factors can<br />

be written as a linear combination of P/H and M/H 2 ,<br />

<br />

P<br />

K = K1 + iK2 =<br />

H z1 + M<br />

<br />

z1 H<br />

H 2 1/2−iɛ , (11)<br />

where z1 and z2 are (presently unknown) non-dimensional<br />

complex numbers that depend on (or α), β<br />

and η, but not on P, M and H.<br />

It turns out to be convenient to write z1 and z2 in<br />

polar form<br />

z1 = eiω<br />

√ U and z2 = ie(ω+γ)<br />

√ V , (12)<br />

where ω and γ are phase angles and 1/ √ U and 1/ √ V<br />

are the magnitude of z1 and z2, respectively. Then, the<br />

complex stress intensity factor can be written as<br />

K = P eiω<br />

√ U H −1/2−iɛ + M ei(ω+γ)<br />

√ V H −3/2−iɛ , (13)<br />

where ω, γ , U and V are non-dimensional parameters<br />

that depend on , β and η but not on P and M.<br />

Three of the four unknown non-dimensional parameters<br />

can be determined analytically by analyzing three<br />

P1<br />

M1<br />

P2<br />

M2<br />

P3<br />

M3<br />

M3 M3<br />

M<br />

P<br />

P<br />

M∗ δ<br />

(a)<br />

(b)<br />

(c)<br />

different load combinations. First, G is calculated by<br />

the J-integral along the external boundaries of the specimen<br />

with P = 0 and M = 0 using the strain distributions<br />

given in Appendix A. The result is<br />

G = 1 P<br />

2Ē2<br />

2 <br />

1<br />

+<br />

h A1<br />

1<br />

+<br />

A2<br />

χ 2 <br />

, (14)<br />

I2<br />

where A1, A2 and I2 are constants given in Appendix<br />

A.<br />

With P = 0 and M = 0, K from (13) becomes<br />

P<br />

K1 + iK2 =<br />

H 1/2√U ei(ω−ɛ ln H) , (15)<br />

so that<br />

|K | 2 = K ¯K = P2<br />

HU<br />

Inserting K 2 into (6) givesG. Setting this equation<br />

for G equal to (14) givesUaccording to<br />

1 1<br />

= +<br />

U A2<br />

1<br />

+<br />

A1<br />

χ 2<br />

.<br />

I2<br />

Next, following the same procedure for a load combination<br />

with P = 0 and M = 0givesVas 1 1<br />

= +<br />

V I2<br />

1<br />

.<br />

I1<br />

Finally, the energy release rate for any combination of<br />

P and M can be written as<br />

G = c2<br />

<br />

P2 M2<br />

+<br />

16 hU h3 <br />

PM<br />

+ 2√ sin γ , (16)<br />

V UVh2 P3<br />

P3<br />

123

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