Rasmus ÿstergaard forside 100%.indd - Solid Mechanics

Interface crack in sandwich specimen 315
Fig. B.1 Mesh used in the
analysis
intensity factor, respectively. Analogously, k2 and k4
are determined with P = 0 and M = 1.
The imaginary and the real parts of K are
ℑ(K ) = G 16 cosh2 1/2
πɛ
sin (ψ − ɛ ln H)
c1 + c2
and
ℜ(K ) = G 16 cosh2 1/2
πɛ
cos (ψ − ɛ ln H),
c1 + c2
respectively. G and ψ are determined by the CSDE
method described in Appendix B.
In order to characterize the deviation on the energy
release rate between a long sandwich specimen where
the homogeneous strain state is present and one with
a more complex strain field (e.g. caused by a too short
uncracked beam length), the following error measure
is introduced
ξ =
|K |
2 num −|K |2ana |K | 2
. (C.2)
num
Here |K | num is the modulus of K determined numerically
by the CSDE method and |K | ana is the modulus
of K determined analytically according to the results
in Sect. 3.
Having determined the k’s numerically and introducing
¯λ = PH
M , it is found that
Knum = (k1 ¯λ + k2) + i(k3 ¯λ + k4) M
and the modulus of Knum squared can then be written
as
|K | 2 num =
(k1 ¯λ + k2) 2 + (k3 ¯λ + k4) 2
M 2 , (C.3)
which we will rewrite in the form
|K | 2 num = B ¯λ 2 + C ¯λ + D, (C.4)
P
where
B = k 2 3 + k2 1
(a)
(b)
C = 2(k3k4 + k1k2)
D = k 2 4 + k2 2 .
Using (16) and (6), an analytical expression based
on the strain fields given in Appendix A for the modulus
of the complex stress intensity factor is found
|K | 2 ana = (b¯λ 2 + c¯λ + d)M 2
(C.5)
b = q 1
H 3
η
U
c = q 1
H 3
2sinγ
√
UVη2 d = q 1
H 3
1
η 3 V ,
where q = c2 cosh2 (πɛ)
. c1+c2
For fixed α, β, η and a/H the error, ξ, depends on
(L − a) and ¯λ. In order to determine the error for a
certain length, the worst case load combination, ¯λc, is
determined. ¯λc is found by differentiating (C.2) and
= 0. The derivative of (C.2)is
setting ∂ξ
∂ ¯λ
∂ξ
∂ ¯λ =−(dB − bD) ¯λ 2 + (2 cB − 2 bC) ¯λ + cD − dC
B ¯λ 2 + C + Dλ 2 Equating this to zero gives two values of ¯λc
bC − cB ± 2 b
¯λc = 2C 2 − 2 bCcB + c2 B2 − dBcD+ d2 BC + bD2c − bDdC
.
dB − bD
Inserting ¯λc into (C.4) and (C.5) gives|K | 2 num and
|K | 2 ana , respectively. Inserting those values into (C.2)
gives ξ.
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