2.4 Antiplane Crack
Consider a semi-infinite crack as shown in Fig. 2.10 [4, 5]. References [4, 5] are the
same. One is a minor revision of the other with some change of the title. They were
both published because Professor M. Kachanov, Editor-in-Chief of International
Journal of Fracture at the time, forwarded both versions to the publisher separately
by mistake. He also said that he had made the same mistake before when contacted
by the author. We limit ourselves to the case of antiplane problems of crystals of
class (6mm). The material is doped into a p-type semiconductor. The crack surfaces
are traction free and unelectroded. The electric field inside the crack is neglected.
There are no free charges and normal current on the crack faces.
The governing equations are from Eqs. (1.42), (1.37), and (1.36) with n 0 ¼ 0 and
Δn ¼ 0:
∇
2 Δp
ð Þ ¼ k
2
Δp
ð Þ,
ð2:44Þ
Àε∇
2 φ ¼ q Δp
ð Þ,
ð2:45Þ
∇
2 u ¼ À
e
c
∇
2 φ,
ð2:46Þ
where
k
2
¼
p 0 μ
p
D
p
q
ε
, ε ¼ ε þ
e
2
c
:
ð2:47Þ
In polar coordinates, Eq. (2.44) takes the following form:
∂
2
∂r 2 þ
∂
r∂r
þ
1
r 2
∂
2
∂θ
2
Δp
ð Þ À k
2
Δp
ð Þ ¼ 0:
ð2:48Þ
We look for
Δp r, θ
ð Þ ¼ Δp r
ð Þ sin
θ
2
:
ð2:49Þ
Substituting Eq. (2.49) into Eq. (2.48), we obtain
x1
x2
r
Fig. 2.10 A semi-infinite
crack
24
2 Exact Solutions
Consider a semi-infinite crack as shown in Fig. 2.10 [4, 5]. References [4, 5] are the
same. One is a minor revision of the other with some change of the title. They were
both published because Professor M. Kachanov, Editor-in-Chief of International
Journal of Fracture at the time, forwarded both versions to the publisher separately
by mistake. He also said that he had made the same mistake before when contacted
by the author. We limit ourselves to the case of antiplane problems of crystals of
class (6mm). The material is doped into a p-type semiconductor. The crack surfaces
are traction free and unelectroded. The electric field inside the crack is neglected.
There are no free charges and normal current on the crack faces.
The governing equations are from Eqs. (1.42), (1.37), and (1.36) with n 0 ¼ 0 and
Δn ¼ 0:
∇
2 Δp
ð Þ ¼ k
2
Δp
ð Þ,
ð2:44Þ
Àε∇
2 φ ¼ q Δp
ð Þ,
ð2:45Þ
∇
2 u ¼ À
e
c
∇
2 φ,
ð2:46Þ
where
k
2
¼
p 0 μ
p
D
p
q
ε
, ε ¼ ε þ
e
2
c
:
ð2:47Þ
In polar coordinates, Eq. (2.44) takes the following form:
∂
2
∂r 2 þ
∂
r∂r
þ
1
r 2
∂
2
∂θ
2
Δp
ð Þ À k
2
Δp
ð Þ ¼ 0:
ð2:48Þ
We look for
Δp r, θ
ð Þ ¼ Δp r
ð Þ sin
θ
2
:
ð2:49Þ
Substituting Eq. (2.49) into Eq. (2.48), we obtain
x1
x2
r
Fig. 2.10 A semi-infinite
crack
24
2 Exact Solutions