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Rasmus ÿstergaard forside 100%.indd - Solid Mechanics

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314 R.C. Østergaard, B.F. Sørensen<br />

Fig. A.1 Definition of<br />

coordinate axis y1, y2 and<br />

y3 for sections k = 1, k = 2<br />

and k = 3, respectively<br />

(Fig. 2) to modulus-argument form we get<br />

u1 + iu2 = |u|e iφ<br />

c1 + c2<br />

=<br />

2 √ 2π √ 1 + 4ɛ2 cosh πɛ Kriɛ+1/2e −iϕ ,<br />

(B.1)<br />

where ϕ = tan−1 <br />

2ɛ, |u| = u2 1 + u2 2 and φ is<br />

the phase angle of the openings given by<br />

φ = arctan u1<br />

.<br />

u2<br />

Now, by combining (B.1) with (7) we get the mode<br />

mixity<br />

ψ = φ − ɛ ln r/l + arctan 2ɛ.<br />

Furthermore, by combining (B.1) with (6) we can<br />

find the energy release rate from the crack face displacements<br />

via<br />

G = u2 2 + u2 1<br />

r<br />

(1 + 4ɛ2 )π<br />

. (B.2)<br />

2(c1 + c2)<br />

The crack surface displacement components (u1<br />

and u2) are found with the finite element method.<br />

However, when calculating G and ψ it turns out that<br />

these are not invariants of r. This is because the neartip<br />

stress and displacement fields given by (4) and (5)<br />

are only the first (but singular) term in a Taylor series<br />

(Rice 1988). The singular stress and displacement<br />

fields become more and more dominant as approaching<br />

the crack tip. Consequently, for r → 0 the singularity<br />

dominate. Based on this reasoning we suggest a CSDE<br />

method where the quantities G(r) and ψ(r) are extrapolated<br />

to r → 0.<br />

The FE analysis was carried out using the Ansys 6.0<br />

FE package. Figure B.1 shows a typical mesh used in<br />

the calculations. The elements were 8 noded isoparametric<br />

linear elastic elements. Typically, the quadratic<br />

fine mesh around the crack tip had a width of H/4 and<br />

contained 40 rings with decreasing spacing towards the<br />

crack tip. The distance between the inner rings was<br />

1/100 of the distance between the outer rings. The size<br />

of the inner elements was H/500. The elements outside<br />

123<br />

P1<br />

P2<br />

M1<br />

M2<br />

y1<br />

y2<br />

δ<br />

k = 1<br />

k = 2<br />

x2<br />

x1<br />

H<br />

h<br />

H<br />

k = 3<br />

the fine region had a rectangular shape and were slightly<br />

elongating towards the ends of the mesh.<br />

It is well known that in a linear elastic analysis of a<br />

sharp interface crack, the elements closest to the crack<br />

tip cannot resolve the displacement accurately. Therefore,<br />

a few nodes closest to the crack should not be<br />

used for the extrapolation. In our calculations, the 10<br />

inner node pairs were typically omitted from the calculations.<br />

The present results were typically obtained<br />

by using nodes in the range 1/25 < r/H < 1/10.<br />

The accuracy of the method was evaluated by checking<br />

the deviation between the exact G calculated from<br />

(16) and the G calculated with the CSDE method. For<br />

all the computations the deviation was less that 0.5%.<br />

C Estimating the error on the analytical determined<br />

energy release rate for short specimens<br />

In Sect. 4.4, curves are given that represent the uncracked<br />

length of the sandwich that result in an error<br />

of 5% on the analytical G. Here, equations are given<br />

for determining the combination of the loads P and M<br />

that result in the greatest error on G and calculate the<br />

error for a sandwich specimen.<br />

For any sandwich length L and crack length a, linearity<br />

dictates that the complex stress intensity factor<br />

can be written as<br />

K = K1(P, M) + iK2(P, M)<br />

= (k1HP+ k2M) + i(k3HP+ k4M), (C.1)<br />

where k1, k2, k3 and k4 are real constants that depend<br />

on α, β, η, (L − a)/H and a/H. These k’s can generally<br />

not be determined by analytical means. Therefore,<br />

a numerical method e.g the finite element method<br />

must be employed. In order to determine the four k’s<br />

by the finite element method, two different load situations<br />

must be solved for each set of α, β, η, (L − a)/H<br />

and a/H. First, using the load situation in Fig. 3c with<br />

P = 1/H, M = 0, k1 and k3 are given directly as<br />

the real part and the imaginary of the complex stress<br />

y3<br />

M3<br />

P3

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