Analysis of interfacial cracks emanating from a hole in a bi-material plate 469
Figure 2. Two equal cracks emanating from a circular hole in a bi-material rectangular plate.
3. Numerical investigations
Two cracking cases, namely two-equal-crack and single-crack, for a circular hole in a rectangular plate are investigated. The non-dimensional SIF values K
I
K
o
and K
II
K
o
are calculated and presented in the following examples, in which
K
o
= σ √
π ar
−12∓iε
, 17
where σ is the applied stress. 3.1. Two equal cracks
A case of two equal cracks emanating form a circular hole in a rectangular plate shown in figure 2 is investigated. In order to ensure the strain component ε
x
is continuous along the interface, the applied loading and material constants should satisfy the following relations Rice and Sih, 1965,
σ
x 2
= E
2
E
1
σ
x 1
+ ν
2
− E
2
E
1
ν
1
σ
y
18
470 J.S. Kuang, Y.H. Wang
Figure 3. K
I
K
o
for different R and ra ratios aw = 0.1, two equal cracks, biaxial loading.
for plane stress, and σ
x 2
= E
2
E
1
1 − ν
2 1
1 − ν
2 2
σ
x 1
+ ν
2
1 − ν
2
− E
2
E
1
ν
1
1 + ν
1
1 − ν
2 2
σ
y
19 for plane strain.
Two loading cases in the analysis will be considered. One is the biaxial loading, in which σ
x 1
, σ
x 2
and σ
y
satisfy Eq. 18 or 19. The other is the uniaxial loading, where σ
x 1
= σ
x 2
= 0, but the elastic material constants should satisfy a certain relation described in Eq. 18 or 19.
3.1.1. Biaxial loading The dimensions of the plate, hole and cracks shown in figure 2 are as follows:
h
1
= h
2
= 1, and w
1
= w
2
= 1, relative crack length: aw = 0.1 to 0.8,
ratio of hole radius to crack length: ra = 0.1, 0.25 and 0.5. The material properties are:
ν
1
= ν
2
= 0.3, E
1
= 1, and E
2
= 1, 3, 4, 10, 25, 100 and 1000. The applied loads σ
x 1
and σ
y
are taken as unit, and σ
x 2
is determined by Eq. 18 for plane stress case. Moreover, the parameters for computation are as follows:
Analysis of interfacial cracks emanating from a hole in a bi-material plate 471
Figure 4. K
II
K
o
for different R and ra ratios aw = 0.1, two equal cracks, biaxial loading.
number of summation terms: M = N = 15,
number of points on the hole boundary: N
h
= 90, number of points on each plate side:
N
1
= N
2
= 40. Numerical results are obtained and presented for various bi-material constant R ratios R = E
2
E
1
. The computed values of K
I
and K
II
for different R ratios are shown in figures 3 and 4, where the relative crack length aw = 0.1. Figure 3 shows that the values of K
I
K
o
decreases as R increases, but R seems to have negligible effect on the K
I
K
o
values when R 25. It is also shown in figure 3 that with the same R ratio the smaller the ratio ra, the larger the K
I
K
o
value. Moreover, when the hole is very small, say ra = 0.1, the present results trend to agree with those of the central crack problem ra = 0 Erdogan, 1965. From figure 4
it can be seen that the value of K
II
K
o
increases rapidly with the increase in R ratios when R 25. This indicates that the shear effect is highly sensitive to the difference between the elastic moduli of two materials.
Variations of K
I
and K
II
for different crack length aw are given in figures 5 and 6. From figure 5 it is seen that K
I
K
o
increases as the value of aw increases. Figure 6 shows that when aw increases there is little change in K
II
K
o
. Until the value of aw becomes larger than 0.6, K
II
K
o
increases fast. Figures 5 and 6 also show that size of the hole ra has little influence on the stress intensity factors.
472 J.S. Kuang, Y.H. Wang
Figure 5. K
I
K
o
for different aw and ra ratios R = 10, two equal cracks, biaxial loading.
3.1.2. Uniaxial loading The same plate in Subsection 3.1.1 is considered for this loading case. The loading conditions are: σ
x 1
= σ
x 2
= 0, and σ
y
= 1. In the analysis it is assumed that E
1
= 1 and ν
2
= 0.3. E
2
and ν
1
can be determined using Eq. 6 and the continuous condition of strain ε
x
along the interface of two materials gives: ν
1
E
1
= ν
2
E
2
. Assuming aw = 0.1, the computed results of K
I
and K
II
various ratios of R are presented in figures 7 and 8. It is seen that the curves in figure 7 are similar to those in figure 3. With the same value of R, the larger the
ratio ra, the smaller the value of K
I
K
o
in biaxial loading case figure 3; while the larger the ratio ra, the larger the value of K
I
K
o
in uniaxial loading case figure 7. This is because that there is no lateral tension in
Analysis of interfacial cracks emanating from a hole in a bi-material plate 473
Figure 6. K
II
K
o
for different aw and ra ratios R = 10, two equal cracks, biaxial loading.
Figure 7. K
I
K
o
for different R and ra ratios aw = 0.1, two equal cracks, uniaxial loading.
474 J.S. Kuang, Y.H. Wang
Figure 8. K
II
K
o
for different R and ra ratios aw = 0.1, two equal cracks, uniaxial loading.
the plate under uniaxial loading. Figure 8 shows the values of K
II
K
o
with different ratio ra for various R. It can be seen that the variation of K
II
K
o
is similar to that in biaxial loading case figure 4. Figures 9 and 10 show the computed results of K
I
and K
II
for different ratios of aw. From figure 9 it is seen that the K
I
K
o
curves are similar to those in the biaxial loading case figure 5, but the ratio ra has a more significant effect on the K
I
K
o
values. Figure 10 shows that K
II
K
o
increases as the ratio aw increases; while for the biaxial loading case figure 6 there is little change in the K
II
K
o
value when aw 0.6. 3.2. Single crack
The proposed method and the procedure for analysing the problem of two equal cracks emanating from a circular hole in a rectangular plate can be used for solving the single crack problem. In this case, the left or
right crack can be considered as a fictional crack. Let the crack tip be in the hole. For convenience, it coincides with the origin of the coordinates. Thus, taking b = 0 the left crack is considered as a fictional crack in the
formulae, the calculation procedure is the same as that in the two-equal-crack case.
Consider the problem of a crack which is emanated from a circular hole in a rectangular plate under uniaxial loading. The crack length is a. The dimensions and material properties are the same as those in the example of
Subsection 3.1.1. Assuming that the crack length aw = 0.1, and the hole radius ra = 0.1, 0.25, and 0.5. The stress intensity
factors are calculated for various values of R. The results of K
I
K
o
and K
II
K
o
are presented in figures 11 and 12. Similar to those in the two-equal-crack case, figure 11 shows that the value of K
I
K
o
decreases with the increase in the value of R; but when R 25, there is little influence on K
I
K
o
. However, the radius of the hole
Analysis of interfacial cracks emanating from a hole in a bi-material plate 475
Figure 9. K
I
K
o
for different aw and ra ratios R = 10, two equal cracks, uniaxial loading.
has a significant effect on the value of K
I
K
o
; while there is little effect in the two-equal-crack case figure 7. Figure 12 shows the results of K
I
K
o
for different R ratios. It can be seen that the variation of K
II
K
o
is similar to that in the two-crack shown in figure 8.
The variations of SIFs for different crack length aw are plotted in figures 13 and 14. From figure 13, it is seen that the K
I
K
o
curves have similar shapes to those in the problem of two cracks figure 9, but their increasing rates are lower than those in figure 9. Figure 14 shows that the value of K
II
K
o
increases as the crack length aw increases, and the variation profile is very similar to that in the two crack case presented in
figure 10.
476 J.S. Kuang, Y.H. Wang
Figure 10. K
II
K
o
for different aw and ra ratios R = 10, two equal cracks, uniaxial loading.
4. Conclusions
