Rasmus ÿstergaard forside 100%.indd - Solid Mechanics

304 R.C. Østergaard, B.F. Sørensen
Fig. 2 Definition of crack
face opening components
3 Analysis
3.1 The reduced problem
We now turn towards the specific problem outlined in
Fig. 1. We wish to determine energy release rate, G,
and mode mixity, ψ, as function of the load, geometry
and stiffness parameters. The crack-tip is assumed to
be remote from the ends, i.e. L/(2H + h) ≫ 1 and
a/H ≫ 1, whereby we ensure that the stresses at the
ends are uniform (independent of x1). Then, G and
ψ are independent of a and L. Later, in Sect. 4.4, we
determine specific conditions under which G and ψ are
independent of a and L.
Following the procedure by Suo and Hutchinson
(1990) we transform the general load situation to a reduced
problem. First, by static moment and force equilibrium
requirements, it is clear that two of the six loads
are static determined. For instance, we can express P2
and M2 by P1, M1, P3 and M3 as follows
P2 = P1 − P3
and
3H
M2 = P1 − h − δ − P3 H +
2 h
− δ
2
+M1 − M3,
where δ is the distance from the bottom of the bilayer
beam to the neutral axis.
Next, we recognize that the crack tip stress field of
the original problem contains stress singularities in the
stress components σ22 and σ12. Then, we notice that the
stress field of the intact sandwich specimen (Fig. 3b)
only contains stress components parallel to the crack
(only σ11 = 0 under plane stress and only σ11 = 0 and
σ33 = 0 under plane strain). Therefore, we can superimpose
the stress field of the intact sandwich specimen
(Fig. 3b) to the stress field of the original problem without
altering the stress components σ22 and σ12. Thus,
the stress components σ22 and σ12 of the reduced problem
(Fig. 3c) are identical to that of the original problem
(Fig. 3a). It follows that the stress singularity at the
123
Δu2
Δu1
x2
r
θ
x1
#2
#1
crack tip of the original problem depends only on the
load parameters P and M of the reduced problem:
P = P1 −
H
0
σ11(x2)dx2
H
M = M1 − σ11(x2) x2 −
0
H
dx2
2
(8)
where the stress σ11(x2) is obtained from Hooke’s law
in material #2 σ11 = Ē2ɛ11,3, where ɛ11,3 denotes the
ɛ11 strain in the uncracked beam end. The strain distribution
in the ends of the specimen is given in Appendix
A. Calculating the integrals in (8) we get
P =−P1 + C1 P3 + C2M3/h
M =−M1 + C3M3, (9)
where C1, C2 and C3 are dimensionless constants that
depend on the stiffness parameter and the thickness
ratio
η = h/H. (10)
The C’s are given by
C1 = 1
A3η , C2 = 1
2I3
1 1
+
η η2
and C3 =
1
,
12I3η3 where A3 and I3 are given in Appendix A.
Equilibrium is obtained with a reacting moment (see
Fig. 3c):
M ∗
3H
= P + h − δ = Phχ + M,
2
where χ = 3
2η + 1 − and
= δ/h =
1 + 2η + η2
.
2η(1 + η)
3.2 Analysis of the reduced problem
The reduced problem is specified in terms of six parameters,
P, M, H, η, and β. It is possible, by the use of