Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media
A problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media, is solved. The initial problem is reduced to an isotropic one by the change of variables. First of all, the problem for small inhomogeneity is considered. Its solution is obtained by the iteration m...
Збережено в:
| Дата: | 2002 |
|---|---|
| Автори: | , |
| Формат: | Стаття |
| Мова: | English |
| Опубліковано: |
Інститут гідромеханіки НАН України
2002
|
| Онлайн доступ: | http://dspace.nbuv.gov.ua/handle/123456789/931 |
| Теги: |
Додати тег
Немає тегів, Будьте першим, хто поставить тег для цього запису!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Цитувати: | Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media / A. G. Bagdoev, S. G. Sahakyan // Акуст. вісн. — 2002. — Т. 5, N 4. — С. 61-71. — Бібліогр.: 11 назв. — англ. |
Репозитарії
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
irk-123456789-931 |
|---|---|
| record_format |
dspace |
| spelling |
irk-123456789-9312008-10-15T18:37:28Z Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media Bagdoev, A.G. Sahakyan, S.G. A problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media, is solved. The initial problem is reduced to an isotropic one by the change of variables. First of all, the problem for small inhomogeneity is considered. Its solution is obtained by the iteration method and is expressed by quadratures from the solution of the homogeneous case. The stresses outside the crack and displacements on its faces are obtained. Besides, the solution for an arbitrary value of the inhomogeneity parameter is obtained. It is shown that its first order approximation coincides with the solution obtained by the method of small parameter. Решена задача о трещине, распространяющейся с произвольной скоростью в анизотропной неоднородной эластичной среде. Исходная задача сведена изотропной с помощью замены переменной. Прежде всего, рассмотрена задача для малой неоднородности. Ее решение получено итерационным методом и записано в квадратурах относительно решения для однородного случая. Найдены напряжения вне трещины и перемещения на ее границе. Кроме того, получено решение для произвольного значения параметра неоднородности. Показано, что аппроксимация первого порядка для него совпадает с решением, полученным методом малого параметра. Розв'язано задачу про тріщину, яка поширюється з довільною швидкістю в анізотропному неоднорідному пружному середовищі. Вихідну задачу зведено до ізотропної за допомогою заміни змінної. Насамперед, розглянуто задачу для малої неоднорідності. °ї розв'язок отримано ітераційним методом і записано в квадратурах відносно рішення для однорідного випадку. Знайдено напруження поза тріщиною й переміщення на її границі. Окрім того, отримано розв'язок для довільного значення параметра неоднорідності. Показано, що апроксимація першого порядку для нього збігається з рішенням, отриманим методом малого параметра. 2002 Article Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media / A. G. Bagdoev, S. G. Sahakyan // Акуст. вісн. — 2002. — Т. 5, N 4. — С. 61-71. — Бібліогр.: 11 назв. — англ. 1028-7507 http://dspace.nbuv.gov.ua/handle/123456789/931 539.1 en Інститут гідромеханіки НАН України |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| language |
English |
| description |
A problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media, is solved. The initial problem is reduced to an isotropic one by the change of variables. First of all, the problem for small inhomogeneity is considered. Its solution is obtained by the iteration method and is expressed by quadratures from the solution of the homogeneous case. The stresses outside the crack and displacements on its faces are obtained. Besides, the solution for an arbitrary value of the inhomogeneity parameter is obtained. It is shown that its first order approximation coincides with the solution obtained by the method of small parameter. |
| format |
Article |
| author |
Bagdoev, A.G. Sahakyan, S.G. |
| spellingShingle |
Bagdoev, A.G. Sahakyan, S.G. Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| author_facet |
Bagdoev, A.G. Sahakyan, S.G. |
| author_sort |
Bagdoev, A.G. |
| title |
Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| title_short |
Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| title_full |
Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| title_fullStr |
Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| title_full_unstemmed |
Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| title_sort |
antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media |
| publisher |
Інститут гідромеханіки НАН України |
| publishDate |
2002 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/931 |
| citation_txt |
Antiplane problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media / A. G. Bagdoev, S. G. Sahakyan // Акуст. вісн. — 2002. — Т. 5, N 4. — С. 61-71. — Бібліогр.: 11 назв. — англ. |
| work_keys_str_mv |
AT bagdoevag antiplaneproblemonacrackpropagatingwithanarbitraryspeedinanisotropicinhomogeneouselasticmedia AT sahakyansg antiplaneproblemonacrackpropagatingwithanarbitraryspeedinanisotropicinhomogeneouselasticmedia |
| first_indexed |
2025-07-02T04:32:34Z |
| last_indexed |
2025-07-02T04:32:34Z |
| _version_ |
1836508255696715776 |
| fulltext |
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
UDC 539.1
ANTIPLANE PROBLEM ON A CRACK,
PROPAGATING WITH AN ARBITRARY SPEED IN
ANISOTROPIC INHOMOGENEOUS ELASTIC MEDIA1
A. G. B AGD OE V∗, S. G. SA HAKY A N∗∗
∗Institute of Mechanics of NAS of Armenia, Yerevan
∗∗Yerevan Institute of Architecture and Construction, Armenia
Received 10.05.1999 � Revised 2.09.2002
A problem on a crack, propagating with an arbitrary speed in anisotropic inhomogeneous elastic media, is solved. The
initial problem is reduced to an isotropic one by the change of variables. First of all, the problem for small inhomogeneity
is considered. Its solution is obtained by the iteration method and is expressed by quadratures from the solution of the
homogeneous case. The stresses outside the crack and displacements on its faces are obtained. Besides, the solution for
an arbitrary value of the inhomogeneity parameter is obtained. It is shown that its first order approximation coincides
with the solution obtained by the method of small parameter.
Решена задача о трещине, распространяющейся с произвольной скоростью в анизотропной неоднородной эластичной
среде. Исходная задача сведена изотропной с помощью замены переменной. Прежде всего, рассмотрена задача
для малой неоднородности. Ее решение получено итерационным методом и записано в квадратурах относительно
решения для однородного случая. Найдены напряжения вне трещины и перемещения на ее границе. Кроме того,
получено решение для произвольного значения параметра неоднородности. Показано, что аппроксимация первого
порядка для него совпадает с решением, полученным методом малого параметра.
Розв’язано задачу про трiщину, яка поширюється з довiльною швидкiстю в анiзотропному неоднорiдному пружному
середовищi. Вихiдну задачу зведено до iзотропної за допомогою замiни змiнної. Насамперед, розглянуто задачу для
малої неоднорiдностi. Її розв’язок отримано iтерацiйним методом i записано в квадратурах вiдносно рiшення для
однорiдного випадку. Знайдено напруження поза трiщиною й перемiщення на її границi. Окрiм того, отримано
розв’язок для довiльного значення параметра неоднорiдностi. Показано, що апроксимацiя першого порядку для
нього збiгається з рiшенням, отриманим методом малого параметра.
INTRODUCTION
Propagation of crack with arbitrary velocity is
very important problem for application in seismology
and the engineering use of metallic details. Because
all practical materials, as a rule are inhomogeneous, it
is interesting to study the influence of inhomogeneity
on the stress intensity coefficient near crack’s edge
and condition of opening of the crack.
A solution of the plane problem for crack, mov-
ing with arbitrary speed in isotropic homogeneous
elastic medium, is obtained in [1] by the convolution
method. Both the antiplane and plane problems for
homogeneous isotropic medium are considered in [2].
For the first time a solution of the antiplane problem
for homogeneous isotropic medium was given in [3].
Wide range of questions related with propagati-
on of cracks was considered in [4]. The solutions,
presented [2, 3], are based on technique developed in
study on flow around wing in [5]. An antiplane ani-
sotropic problem on crack in homogeneous medium
is considered in [6]. In this paper a crack propagat-
ing with arbitrary speed in an inhomogeneous ani-
sotropic elastic medium for is discussed. In particular,
1Published in English as an exclusion.Translation is revised
by V. N.Oliynik
an antiplane problem is considered. Solutions for dis-
placements on crack are obtained under conditions
of small and arbitrary inhomogeneities. Besides that,
a solution for stresses out of crack is obtained for
small inhomogeneity. By the convolution method the
alternative symmetric problem on crack with given
displacements on its faces is solved.
1. STATEMENT OF PROBLEM
Assume that the crack occupies some region along
x-axis in plane (x, y). At that its edges x= l1(t) and
x= l2(t) move with arbitrary speeds. Under menti-
oned conditions one component of displacement of
medium (u along z-axis) and two components of
stresses (τxz , τyz) exist. The following relation is
valid [6] for the stresses:
τxz
ρ
= a2
1
∂u
∂x
+ a2
12
∂u
∂y
,
τyz
ρ
= a2
12
∂u
∂x
+ a2
2
∂u
∂y
,
(1)
where ρ is a density of the medium; a2
1, a
2
2, a
2
12 are
some constants.
In what follows we consider semiplane y>0 and
c© A. G. Bagdoev, S. G. Sahakyan, 2002 61
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
Fig. 1. Characteristics pattern for propagating crack
take that ρ=ρ(y). Equation of motion is written in
the form
∂τxz
∂x
+
∂τyz
∂y
= ρ
∂2u
∂t2
. (2)
Substituting Eq. (1) into Eq. (2) and introducing new
variables
x1 = x− a2
12
a2
2
y, y1 = y − a
a2
2
y, (3)
we obtain that
τyz
ρ
= a
∂u
∂y1
, a2 = a2
1a
2
2 − a4
12, (4)
∂2u
∂x2
1
+
∂2u
∂y2
1
+
1
ρ
∂ρ
∂y1
∂u
∂y1
=
a2
2
a2
∂2u
∂t2
. (5)
So, the initial problem is reduced to problem for
isotropic inhomogeneous medium. Assuming that
ρ(y)=ρ0e
ky1 and introducing function
u = Pe(−k/2)y1 (6)
we obtain from Eqs (4), (5) the following relations:
τyz
ρ
= a
(
∂P
∂y1
− k
2
P
)
e(−k/2)y1, (7)
∂2P
∂x2
1
+
∂2P
∂y2
1
− k2
4
P =
a2
2
a2
∂2P
∂t2
. (8)
To complement the statement we write the boundary
conditions in the form
τyz
ρ
= a
∂u
∂y1
=
τ ′
(
x, t
)
ρ
for y1 = 0, l1(t) < x < l2(t),
u = 0 for x > l2(t), x > l1(t).
(9)
Obtained boundary problem (8), (9) in princi-
ple can be solved by the convolution method [1].
However, complex quadratures are obtained in the
solution due to inhomogeneity. Therefore, at first
let us assume that the inhomogeneity is small and
in only the first order terms with respect to the
inhomogeneity parameter k are retained in the soluti-
on. In so doing Eq. (8) can be written as an equation
for homogeneous isotropic elastic medium
∂2P
∂x2
1
+
∂2P
∂y2
1
=
1
c2
∂2P
∂t2
, c2 =
a2
a2
1
(10)
with the boundary conditions at y=0, derived from
Eq. (9):
∂P
∂y1
− k
2
P =
τ ′(x, t)
ρa
l1(t) < x < l2(t),
P = 0, x > l2
(
t
)
, x < l1(t).
(11)
Now let us consider the solution for semiinfinite
crack, putting l1(t)=−∞. From the physical point
of view such situation means that the edges of the
crack do not affect one another. For the finite crack
the obtained solution is valid for the right hand side
region of the characteristic curve AB (fig. 1). It is
expedient to introduce the functions v=∂P/∂y1. At
that we can put the following conditions for y=0:
v(x, 0, t) = v(x, t),
P = P+ + P−, v = v+ + v−.
(12)
Here, index “+” corresponds to the functions being
equal to zero for x<l2(t), while index “−” – to functi-
ons being equal to zero for x>l2(t).
After this substitution conditions (11) yield
v− − k
2
P−=f
(
x, t
)
, f
(
x, t
)
=
τ ′
(
x, t
)
ρa
, P+ =0. (13)
The boundary conditions (11) and Eq. (8) were obtai-
ned for the case of exponential law ρ(y1)=ρ0e
ky1 . It
is interesting to consider more general class of functi-
ons ρ(y1), leading to mentioned equations. Let the
displacement take the form
u
(
x, y1, t
)
= F
(
y1
)
P
(
x, y1, t
)
.
Putting it into Eq. (5) we obtain that
2F ′
F
+
ρ′
ρ
= 0, F
(
0
)
= 1, 2F ′
0 =
ρ′0
ρ0
,
where primes denote differentiation with respect to
y1; ρ0(0)=ρ0; ρ′0(0)=ρ′0.
Originating from the above assumptions one can
obtain the following relations.
A. For the law
ρ = ρ0
(
1 +
ρ′0
2ρ0
y1
)2
,
ρ′0
ρ0
= k (14)
62 A. G. Bagdoev, S. G. Sahakyan
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
the equation
∂2P
∂x2
+
∂2P
∂y2
1
=
1
c2
∂2P
∂t2
(15)
is valid (which is analogous to Eq. (10)) along
with boundary conditions (11) for y1 =0. For
order of k the obtained problem is reduced to
exponential problem considered above.
B. For the law
ρ = ρ0
(
ch
k
2
y1 +
ρ′0
ρ0k
sh
k
2
y1
)2
(16)
the equation (8) for P is obtained, and boundary
condition yields
∂P
∂y1
− ρ′0
2ρ0
P =
τ ′
(
x, t
)
ρa
at y1 = 0. (17)
As it is seen, this problem also is reduced to the
above exponential law problem with the coeffici-
ent ρ′0/ρ0 replacing the coefficient k in (11). So,
in order of k both statements of the boundary
problems coincide. In particular, for ρ′0/(ρ0k)=1
Eq. (16) gives ρ=ρ0e
ky1 , i. e. the same as in the
previous case.
C. For the law
ρ
ρ0
=
(
cos
k
2
y1 +
ρ′0
ρ0k
sin
k
2
y1
)2
(18)
we obtain the equation
∂2P
∂x2
+
∂2P
∂y2
1
+
k2
4
P =
1
c2
∂2P
∂t2
(19)
and the boundary condition (17). So, in order of
k this problem coincides with the problem (10),
(11).
This is the reason for considering in further the
problem stated by equation (8) in the aggregate with
the boundary conditions (11), from which in order of
k all mentioned cases of distribution of ρ(y1) can be
obtained.
2. SOLUTION OF THE PROBLEM FOR
SMALL k
Solution of Eq. (10) for the boundary condition
∂P/∂y1 = v(x, t) at y1 = 0,
yields the Possio integral form [3 –5]
P
(
x, t
)
= − c
π
∫ ∫
v
(
x′, t′
)
dx′dt′√
c2
(
t− t′
)2 −
(
x− x′
)2 . (20)
From formula (20), using the condition P =0 for
x>l2(t) and introducing the characteristic coordi-
nates [3,5]
ξ′ = ct′ − x′, η′ = ct′ + x′,
ξ0 = ct− x, η0 = ct+ x,
(21)
one can write the solution for v+ in the form
v+ = − 1
π
1√
η0 − η2
(
ξ0
) ×
×
η2(ξ0)∫
−ξ0
v−
(
x′, t′
)
√
η0
(
ξ0
)
− η′
η0 − η′
dη′.
(22)
Note, that, as in [3 –5] it is accepted that l̇2(t)<c
(dot denotes differentiation on t), the following
equality becomes valid:
η2
(
ξ0
)
= l2
(
t2
)
+ ct2, ct2 − l2
(
t2
)
= ct− x. (23)
Here, (l2(t2), t2 is a point of intersection of the
characteristic ξ′=ξ0=const with the curve x′= l2(t′),
representing a motion law for the edge of the
crack (fig. 1).
Changing the integration variable from η′ to x′ and
taking into account Eqs (21) and (23) we obtain that
η0 − η′ = 2
(
x0 − x′
)
,
η2
(
ξ0
)
− η′ = 2
[
l
(
t2
)
− x′
]
,
η0 − η2
(
ξ0
)
= 2
[
x− l2
(
t2
)]
(24)
along ξ′=ξ0 [3]. Moreover, from Eq. (22) it follows
that
v+ = − 1
π
1√
x− l2
(
t2
)×
×
l2(t2)∫
x−ct
v−
(
x′, t+
x′
c
− x
c
)
√
l2
(
t2
)
− x′
x− x′
dx′.
(25)
The above expression coincides with the solution
from [3] in physical dimensional coordinates. In the
present case v− is unknown. From Eq. (13) we can
write that
v−
(
x, t
)
= f
(
x, t
)
+
k
2
P−
(
x, t
)
, (26)
where P− is also unknown.
It should be noted that integration in Eq. (20)
is carried out over the domains (s0+s1+s2) (see
A. G. Bagdoev, S. G. Sahakyan 63
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
fig. 1) [5]. At the same time, originating from the
boundary condition P =0 for x>l2(t), we obtain from
Eq. (20), written with respect to the characteristic
variables [2], that
η2(ξ
′)∫
−ξ′
τ1
(
ξ′, η′
)
√
η0 − η′
dη′ +
η0∫
η2(ξ′)
τ1
(
ξ′, η′
)
√
η0 − η′
dη′ = 0, (27)
where τ1(ξ
′, η′)=v(x′, t′); η2(ξ
′)= l2(t
′
2)+ct
′
2;
ξ′= l2(t
′
2)−ct′2. Therefore, it can be shown that the
integrals over s1 and s2 in Eq. (20) are cancelled. So,
only the integral over s0 remains [5] (filled domain
in the figure). Performing integration in Eq. (20), as
it was made in [2,4,5], one can obtain that
P− = − 1
2π
ξ0∫
ξa(η0)
dξ′√
ξ0 − ξ′
η0∫
−ξ′
v1
(
ξ′, η′
)
√
η0 − η′
dη′, (28)
where
v1
(
x′, t′
)
= f
(
x′, t′
)
+
k
2
Φ−
(
ξ′, η′
)
;
Φ−
(
ξ′, η′
)
= P−
(
x′, t′
)
.
(29)
Thus, for Φ−(ξ′, η′) one can obtain the integral
equation
f1
(
ξ′, η′
)
= f
(
x′, t′
)
,
Φ−
(
ξ0, η0
)
= − 1
2π
ξ0∫
ξa(η0)
dξ′√
ξ0 − ξ′
×
×
η0∫
−ξ′
f1
(
ξ′, η′
)
+
(
k/2
)
Φ−
(
ξ′, η′
)
√
η0 − η′
dη′,
(30)
where
ξa
(
η0
)
= ct3 − l2
(
t3
)
; ct3 + l2
(
t3
)
= ct+ x (31)
along the characteristic η′=η0. In homogeneous case
(k=0) the formula
Φ0
−
(
ξ0, η0
)
= − 1
2π
ξ0∫
ξa(η0)
dξ′√
ξ0 − ξ′
η0∫
−ξ′
f1
(
ξ′, η′
)
√
η0 − η′
dη′ (32)
is valid. Passing to the edge of the crack x≈ l2(t) and
accounting for narrowness of the domain of integrati-
on in ξ′-direction and Eq. (24), one can obtain the
following expression in physical coordinate x:
Φ0
−
(
ξ0, η0
)
= −
√
2
π
√
ξ0 − ξa
(
η0
)
×
×
x∫
x−ct
f
(
x′, t+
x′
c
− x
c
)
dx′√
x− x′
,
(33)
where due to Eq. (31)
ξ0 − ξa
(
η0
)
≈ 2
(
l2
(
t
)
− x
)
1 + l̇2
(
t
)
/c
. (34)
Finally, near the crack’s edge one can obtain for
k=0 (zero approximation) that
Φ0
− = P 0
− = − 1
π
√
l2
(
t
)
− x
1 + l̇2
(
t
)
/c
×
×
x∫
x−ct
f
(
x′, t+
x′
c
− x
c
)
dx′√
x− x′
.
(35)
Hence, from Eq. (30) in the first order of k a closed
solution can be obtained, where Φ−(ξ′, η′) in Eq. (30)
should be taken from Eq. (32). Also, and in determi-
nation of Φ0
−(ξ′, η′) the domain of integration should
be chosen between the characteristics η′′=η′; ξ′′=ξ′;
ξ′′=ξa(ξ′). Finally, we obtain that
Φ0
−
(
ξ′, η′
)
=− 1
2π
ξ′∫
ξa(η′)
dξ′′√
ξ′ − ξ′′
η′∫
−ξ′′
f1
(
ξ′′, η′′
)
√
η′ − η′′
dη′′, (36)
where
ξa
(
η′
)
= ct′3 − l2
(
t′3
)
, ct′3 + l2
(
t′3
)
= η′.
Formula (36) is valid for the points of integrati-
on (ξ′, η′) in Eq. (30), for which the characteri-
stic η′′=η′ intersects with the curve x′= l2(t′) (i. e.,
for η′>ηe = l2(0), see fig. 1). For the points (ξ′, η′),
in which it intersects with x′-axis within region
x′<l2(0) (i. e. for η′<ηe) one should integrate in (36)
within the limits
ξ1
(
η′
)
< ξ′′ < ξ′, ξ1
(
η′
)
=−η′.
So, formula (30) gives for arbitrary point (x, t) the
value of P−(x, t) equal to Φ−(ξ0, η0). Therefore, for
small k the value Φ0
−(ξ′, η′) should be be taken from
Eq. (36). For x≈ l2(t) we obtain
P− = − 2
π
√
l2
(
t
)
− x
1 + l̇2
(
t
)
/c
×
×
x∫
x−ct
{
f
(
x′, t+
x′
c
− x
c
)
+
+
k
2
Φ0
−
(
ξ0, ξ0 + 2x′
)} dx′√
x− x′
.
(37)
Eqs (30) and (37) give the displacement on crack
u=P−. It is evident that the inhomogeneity essenti-
ally effects the solution.
64 A. G. Bagdoev, S. G. Sahakyan
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
Simultaneously, at arbitrary (x, t) for x>l2(t) the
value v+ is given by Eqs (22), (25), where
v−
(
x′, t+
x′
c
− x
c
)
=
= f
(
x′, t+
x′
c
− x
c
)
+
k
2
Φ0
−
(
ξ0, η
′).
(38)
Due to Eq. (36), etc. the following relations hold true:
Φ0
−
(
ξ0, η
′)=− 1
2π
ξ0∫
ξa(η′)
dξ′′√
ξ0 − ξ′′
×
×
η′∫
−ξ′′
f1
(
ξ′′, η′′
)
√
η′ − η′′
dη′′, η′ > ηe,
Φ0
−
(
ξ0, η
′)=− 1
2π
ξ0∫
−η′
dξ′′√
ξ0 − ξ′′
×
×
η′∫
−ξ′′
f1
(
ξ′′, η′′
)
√
η′ − η′′
dη′′, η′ < ηe.
(39)
It should be mentioned that for x≈ l2(t) value
ξa(η′), in contrary to ξa(η0), is not close to ξ0 and
inhomogeneity significantly changes the stress v+
near the crack’s edge.
3. CONDITION OF CRACK’S PROPAGATI-
ON AND SOLUTION FOR A CONSTANT
TRACTION
For x≈ l2(t) (x>l2(t)), originating from (25) we
can write the following:
τyz = v+ =
K
π
1√
x− l2
(
t
) ,
K = −
√
1 − l̇2
(
t
)
c
×
×
l2(t)∫
x−ct
f
(
x′, t+
x′
c
−x
c
)
+
k
2
Φ0
−
(
ξ0, ξ0+2x′
)
√
l2
(
t′
)
− x′
dx′,
(40)
where Φ0
−(ξ0, ξ0+2x′) is given by (39). From (37) we
obtain that
u− = P− =
2K
π
√
l2
(
t
)
− x
1 − l̇22
(
t
)
/c2
,
x < l2(t).
(41)
The Irvin’s condition [2] gives
K2
π
ρa√
1 − l̇22(t)/c
2
= 2γ′, (42)
where γ′ is the surface energy of opening of the crack.
In the problem for a constant traction on the crack
f=const the Eq. (25) leads to the following expressi-
on:
v+ =− 1
π
1√
x−l2
(
t2
)
l2(t2)∫
x−ct
√
l2
(
t2
)
−x′
x−x′ dx′−
− k
2π
1√
x−l2(t2)
l2(t)∫
x−ct
Φ0
−
(
ξ0, η
′)
√
l2
(
t2
)
−x′
x−x′ dx′.
(43)
Due to Eq. (39) Φ0
−(ξ0, η
′) can be calculated in form
Φ0
−
(
ξ0, η
′)= 1
π
f
{
√
ξ0−ξa(η′)
√
η′+ξa
(
η′
)
+
+
(
ξ0+η′
)
arctg
√
ξ0 − ξa
(
η′
)
η′+ξa
(
η′
)
}
for η′ > ηe
(44)
and
Φ0
−
(
ξ0, η
′) = −f
2
(
ξ0 + η′
)
for η′ < ηe. (45)
Here,
η′=ct′+x′, ξ0=ct′−x′=ct−x,
ξa
(
η′
)
=ct′3−l2
(
t′3
)
, ct′3+l2
(
t′3
)
=ct′+x′.
(46)
Using Eq. (43) and passing to physical variables
we obtain for the stress distribution out from crack
A. G. Bagdoev, S. G. Sahakyan 65
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
a b
Fig. 2. Calculations after Eq. (48), giving the stresses out of the crack for constant velocity of the crack (a) l̇(t)=v
and constant traction f on the crack (b); x0 =x/ct, M =v0/c: M <1, k̃=kct, x′ =ctξ
that
v+ = − 2
π
f√
x− l2
(
t2
)
{√
l2
(
t2
)
− x+ ct−
−
√
x− l2
(
t2
)
arctg
√
l2
(
t2
)
− x+ ct
x− l2
(
t2
)
}
+
+
k
π2
f√
x− l2
(
t2
)
l2(t2)∫
α
{√
l2
(
t′3
)
− x′×
×
√
ct− x+ 2x′ − l2
(
t′3
)
+
(
ct− x+ x′
)
×
× arctg
√
l2
(
t′3
)
− x′
ct− x+ 2x′ − l2
(
t′3
)
}
×
×
√
l2
(
t2
)
− x′
x− x′
dx′ +
k
2π
f√
x− l2
(
t2
) ×
×
α∫
x−ct
(
ct− x+ x′
)
√
l2
(
t2
)
− x′
x− x′
dx′,
(47)
where α=(ηe+x−ct)/2.
As it is seen from Eq. (47), the stress out of
crack has the same singularity near the crack’s
edge x≈ l2(t), as in homogeneous case, but with the
stress intensity coefficient essentially depending from
inhomogeneity.
4. CASE OF ARBITRARY VALUE OF
INHOMOGENEITY
In the case of non-small k Eq. (5) can be solved in
the form of the Laplace and the Fourier transformati-
ons:
UL =
∞∫
0
e−st′u dt′;
Ul
(
x, y1, s
)
=
1
2π
∞∫
−∞
ULF e
−i(α1x+βy1)dα1.
(48)
Here ULF is the Fourier transform for UL(x, 0, s);
(1/ρ)∂ρ/∂y1 =k.
Substituting Eq. (48) into Eq. (5) we obtain
β = −i k
2
− i
√
k2
4
+ α2
1 +
s2
c2
. (49)
Introducing the function ψ=∂u/∂y1 we follow to the
relation between the integral transformants of ψ and
u:
ψLF = −iβULF ,
ULF = SLFψLF ,
SLF = −
(
k
2
+
√
k2
4
+ α2
1 +
s2
c2
)−1
.
(50)
The originals are written as
S
(
x, t
)
=
1
4π2i
σ+i∞∫
σ−i∞
estds
∞∫
−∞
e−iα1xSLF dα1. (51)
On the boundary y1 =0 the following relation
between u(x, t) and derivative (∂u/∂y1)|y1=0 =ψ(x, t)
is valid:
u
(
x, t
)
=
∫ ∫
ψ
(
x′, t′
)
S
(
x− x′, t− t′
)
dx′dt′, (52)
and the integration procedure should be carried out
over the complete domain s0+s1+s2 (see fig. 1). Now
66 A. G. Bagdoev, S. G. Sahakyan
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
for x<l2(t) it can be written that
u
(
x, t
)
=
=
1
2c
ξa(η0)∫
ξe
dξ′
η2(ξ
′)∫
−ξ′
f1
(
ξ′, η′
)
S1
(
ξ0−ξ′, η0−η′
)
dη′+
+
1
2c
ξa(η0)∫
ξe
dξ′
η0∫
η2(ξ′)
ψ1
(
ξ′, η′
)
S1
(
ξ0−ξ′, η0−η′
)
dη′+
+
1
2c
ξ0∫
ξa(η0)
dξ′
η0∫
−ξ′
f1
(
ξ′, η′
)
S1
(
ξ0−ξ′, η0−η′
)
dη′,
(53)
where f1(ξ
′, η′)=f(x′, t′) is given in s0, s2 behind
the crack’s edge; ψ1(ξ
′, η′)=ψ(x′, t′) is a value of
(∂u/∂y1)|y1=0 for x′>l2(t
′), i. e., out of the crack,
which in the order of k is given by Eq. (25) and
S1(ξ
′, η′)=S(x′, t′).
So, the solution for u in the domain x<l2(t) is
reduced to determination of the function S(x, t), gi-
ven by Eqs (50) and (51).
From the boundary condition u=0 for x>l2(t) and
Eq. (52), as in Eq. (8), for |ξ0|> |ξa(η0)| we have the
following:
ξ0∫
ξe
dξ′
η2(ξ
′)∫
−ξ′
f1
(
ξ′, η′
)
S1
(
ξ0−ξ′, η0−η′
)
dη′+
+
ξ0∫
ξe
dξ′
η0∫
η2(ξ′)
ψ1
(
ξ′, η′
)
S1
(
ξ0−ξ′, η0−η′
)
dη′ = 0.
(54)
However, in contrary to Eq. (27), in which
S1
(
ξ0 − ξ′, η0 − η′
)
= − c
π
√(
ξ0 − ξ′
)(
η0 − η′
) ,
as it will be shown later, for arbitrary k 6=0 the multi-
plies (ξ0−ξ′)−1/2 and (η0−η′)−1/2 are not separated.
Therefore, the integral over the domain s2, which
contains complex solution ψ1 given by Eq. (25), can
not be excluded. This fact distinguishes the problem
with k 6=0, from the problem for homogeneous medi-
um [2,5] (its solution is given by Eqs (32), (35)).
It should be noted that Eq. (53) is valid behind the
crack’s edge, i. e., for |ξ0|< |ξa(η0)|, while Eq. (54) –
ahead the crack’s edge, i. e., for |ξ0|> |ξa(η0)|. That
is why this last expression can not be used for si-
mplification of Eq. (53). However, later it will be
shown that for the points close to the edge (x≈ l2(t))
Eq. (54) still can be used for simplificationof Eq. (53).
To calculate S(x, t) from Eqs (50) and (51)
the complex s=pc
√
k2/4+α1 should be introduced
instead of s. Moreover, integration with respect to
s should be replaced by the Laplace integral with
respect to p, multiplied by c
√
k2/4+α1.
Note that the inverse Laplace transform from
g
(
p
)
= −
(
k
2
+
√
k2
4
+ α2
1p
)−1
is
f
(
t′
)
= −c exp
− k c t′
2
√
k2
4
+ α2
1
.
Additionally, in view of [8, Eq. (38) on page 123]
g
(√
p2 + 1
)
= f
(
t′
)
−
t′∫
0
J1
(
u
)
f
(
t′
2 − u2
)
du′;
t′ = c
√
k2
4
+ α2
1 t
the inverse Laplace transform for SLF gives the
following:
SF = −ce− kc
2
t+
+ c
c
√
k2
4
+α2
1
t∫
0
J1
(
u
)
exp
−
k
2
√
t′2 − u2
√
k2
4
+ α2
1
du.
(55)
After substitution u=v
√
k2/4+α2
1 we have
SF = −ce− kc
2
t + c
√
k2
4
+ α2
1 ×
×
ct∫
0
J1
(
v
√
k2
4
+ α1
)
exp
(
−k
2
√
c2t2 − v2
)
dv.
(56)
Accounting the well known property of the Bessel’s
functions J1(x)=−J ′
0(x), and integrating by parts,
one can obtain
SF = −cJ0
(
ct
√
k2
4
+ α2
1
)
+
+
kc
2
ct∫
0
J0
(
v
√
k2
4
+ α2
1
)
×
× v√
c2t2 − v2
exp
(
−k
2
√
c2t2 − v2
)
dv.
(57)
A. G. Bagdoev, S. G. Sahakyan 67
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
Again, replacing
√
c2t2−v2 =u gives the following
relation:
SF = −cJ0
(
ct
√
k2
4
+ α2
1
)
+
+
kc
2
ct∫
0
J0
(
√
c2t2 − v2
√
k2
4
+ α2
1
)
e−
k
2
udu.
(58)
The function SF is even on α1. This allows to
replace the exponential Fourier transform on the cosi-
ne transform. So, in view of [8, Eq. (35) on page 57],
one can obtain the final formula for the function in
physical domain:
S
(
x, t
)
= −cH
(
ct − x
)
π
×
×
{
cos
(
k/2
√
c2t2 − x2
)
√
c2t2 − x2
−
−k
2
√
c2t2−x2∫
0
cos
(
k/2
√
c2t2 − u2
)
√
c2t2 − x2 − u2
e−
k
2
udu
}
.
(59)
Moreover, the function S1(ξ0−ξ′, η0−η′) in
Eq. (53) is given by the expression:
S1
(
ξ0 − ξ′, η0 − η′
)
= − c
π
(
1
T
cos
kT
2
−
−k
2
T∫
0
cos
(
k/2
√
T 2 − u2
)
√
T 2 − u2
e−
k
2
udu
)
,
(60)
where
T =
√(
ξ0 − ξ′
)(
η0 − η′
)
. (61)
For small k the the first order approximation from
Eqs (60) and (61) gives that
S1
(
ξ0 − ξ′, η0 − η′
)
= − c
π
(
1
T
− kπ
4
)
. (62)
Let us put in relation (54) that
ψ1 = v0
+ + v1
+,
where v0
+ is the solution for homogeneous medium
(k=0) given by (22), in which v− is replaced for
f1(ξ0, η
′). For this addendum Eq. (27) is fulfilled.
Retaining in Eq. (54) terms of the k-th order we
obtain the following:
− kπ
4
ξ0∫
ξe
dξ′
η2(ξ
′)∫
−ξ′
f1dη
′−
− kπ
4
ξ0∫
ξe
dξ′
η0∫
η2(ξ′)
v0
+
(
ξ′, η′
)
dη′+
+
ξ0∫
ξe
dξ′
η0∫
η2(ξ′)
v1
+
(
ξ′, η′
)
T
dη′ = 0.
(63)
So, from Eqs (53) and (62) the solution precise to the
first order of k can be written:
u = − 1
2π
ξa(η0)∫
ξe
dξ′
η2(ξ
′)∫
−ξ′
f1
(
ξ′, η′
)
T
dη′+
+
k
8
ξa(η0)∫
ξe
dξ′
η2(ξ′)∫
−ξ′
f1
(
ξ′, η′
)
dη′−
− 1
2π
ξa(η0)∫
ξe
dξ′
η0∫
η2(ξ′)
(
v0
+ + v1
+
)( 1
T
− kπ
4
)
dη′+
+
1
2c
ξ0∫
ξa(η0)
dξ′
η0∫
−ξ′
f1S1dη
′.
(64)
It worth noting that the first addendum in right-hand
side of Eq. (64) and the addendum with v0
+, both
having the zeroth order of k, are cancelled (see (27)).
From Eq. (64), accounting for Eq. (63), we obtain for
the terms of order of k:
u ≈ k
8
ξa(η0)∫
ξ0
dξ′
η2(ξ′)∫
−ξ′
f1dη
′ +
k
8
ξa(η0)∫
ξ0
dξ′
η0∫
η2(ξ′)
v0
+dη
′−
− 1
2π
ξa(η0)∫
ξ0
dξ′
η0∫
η2(ξ′)
v1
+
T
dη′ +
1
2c
ξ0∫
ξa(η0)
dξ′
η0∫
−ξ′
f1S1dη
′.
(65)
Near the crack’s edge x≈ l2(t), ξ0≈ξa(η0) only
the terms of order
√
ξ0−ξa(η0) should be retained.
Therefore, the first two terms in the right-hand side
of Eq. (65) can be dropped out. This gives
u ≈ 1
2π
ξ0∫
ξa(η0)
dξ′
η0∫
η2(ξ′)
v1
+
(
ξ′, η′
)
T
dη′+
+
1
2c
ξ0∫
ξa(η0)
dξ′
η0∫
ξ′
f1S1dη
′.
(66)
68 A. G. Bagdoev, S. G. Sahakyan
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
Due to Eqs (22) and (26) we find, keeping the terms
of order of k, that
v1
+
(
ξ′, η′
)
= − k
2π
√
η′ − η2
(
ξ′
) ×
×
η2(ξ′)∫
−ξ′
Φ0
−
(
ξ′, η′′
)
√
η2
(
ξ′
)
− η′′
η′ − η′′
dη′′.
Substituting this relation into Eq. (66), interchangi-
ng the integration order on η′ and η′′ and, finally,
accounting that ξ′≈ξ0, η2(ξ
′)≈η0, we obtain that
near the crack’s edge
u=− k
4π
ξ0∫
ξa(η0)
dξ′√
ξ0−ξ′
η0∫
−ξ0
Φ0
−
(
ξ0, η
′) dη′√
η0−η′
−
− 1
2π
ξ0∫
ξa(η0)
dξ′√
ξ0−ξ′
η0∫
−ξ0
f1
(
ξ0, η
′)
√
η0−η′
dη′.
(67)
In the last term of Eq. (67) it is used that
ξ′≈ξ0. Moreover, in accordance with Eq. (61) the
second addendum in Eq. (66) is the same that for
homogeneous medium.
The same expression is obtained from the solution
of the first order of k given by Eq. (30). So values
of u=Φ−(ξ0, η0) obtained by two methods coincide
near the crack’s edge.
For arbitrary k the relations (53) and (54) are
valid. They keep true for different ξ0, but for (ξ0, η0),
taken near the crack’s edge, one can believe that
both (53) and (54) relations hold. This gives the
following:
u
(
x, t
)
= − 1
2π
ξ0∫
ξa(η0)
dξ′
η2(ξ
′)∫
−ξ′
f1S1dη
′−
− 1
2c
ξ0∫
ξa(η0)
dξ′
η0∫
η2(ξ′)
ψ1S1dη
′ +
1
2π
ξ0∫
ξa(η0)
dξ′
η0∫
−ξ′
f1S1dη
′.
Let us denote the value of v+ for k=0 as v0
+.
Then, accounting for equalities ψ1=v+, v+ =v0
++v1
+
and using the approximation S1(ξ0−ξ′, η0−η′) at
ξ0≈ξa(η0) (see Eq. (62) under T ≈0), we can drop
the small quantities of order of (ξ0−ξa(η0)). This
procedure leads to Eq. (66). Therefore, the soluti-
on for u(x, t) near the crack’s edge formally coincide
with the solution for the first order of k. Although,
it should be mentioned that v1
+(ξ′, η′) may differ
from the similar value for the case of the small
inhomogeneity.
Besides calculation of the displacement on the
crack, the function ψ(x, t)=(∂u/∂y1)|y1=0 =ψ+ out
of the crack should be found. Following the Eq. [1,4]
and using the expressions using denotions (12) one
can write the solution for u, ψ in the form
u−
(
x, t
)
= S− ∗ ∗
[(
S+ ∗ ∗ψ− − P ′
− ∗ ∗u+
)
×
×H
(
l2 − x
)]
;
(68)
ψ+
(
x, t
)
= −P+ ∗ ∗
[(
S+ ∗ ∗ψ− − P ′
− ∗ ∗u+
)
×
×H
(
x− l2
)]
,
(69)
where l2 = l2(t); H(x) is a step function; asterisks
denote a convolution on x and t; S±(x, t) and
P ′
±(x, t) are respectively the originals of the functi-
ons SLF± and P ′
LF±, which represent factorization
of the functions SLF and P ′
LF =1/SLF .
Finally, according to [7,9] we obtain the following
expressions:
SLF = SLF±SLF−;
SLF± = ± 1√
α1 ±
√
γ
×
× exp
1
2πi
∓∞∫
∓√
γ
ln
2
√
γ+α′2
1+k
2
√
γ+α′2
1−k
dα′
1
α′
1−α1
,
(70)
where γ=k2/4+s2/c2.
The next step is to find the originals S±(x, t)
according to Eq. (51). By the same way the originals
P ′
±(x, t) can be determined.
The obtained solution for ψ+(x, t) see Eqs (51),
(69), (70)) is very complex. Therefore, it is more sui-
table to use the small parameter solution (25) for the
first approximation.
As it is seen from the solutions (25), (30) and (30),
influence of the inhomogeneity on the solution as near
the crack’s edge, as along the whole x-axis is essential.
5. SOLUTION OF THE PROBLEM FOR DI-
SPLACEMENTS GIVEN ON THE CRACK’S
FACES
It should be noted that the problem considered
in §1 is the antisymmetric one, for which same
tractions τ−yz are given on the crack’s faces and
the displacements have opposite signs. Correspond-
ing symmetric problem implies the stresses having
opposite signs and the same displacements imposed
on the crack’s faces. General case of different stresses
and displacements on the crack’s faces can be
A. G. Bagdoev, S. G. Sahakyan 69
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
represented as a sum of these two solutions. So,
instead of Eq. (9) we can put the following:
y1 = 0; u− = Φ1
(
t, x
)
, x < l2
(
t
)
,
ψ+ = 0, x > l2
(
t
)
.
(71)
Because of
ψ+ =
∂u
∂y1
∣∣∣∣
y1
= v+ − k
2
u+
one can rewrite Eq. (71) in the form
y1 = 0; u = P− = Φ1
(
x, t
)
, v+ =
k
2
u+. (72)
In contrast to Eqs (68), (69), the convolution
method [1] can be applied to the functions P , v. This
gives
P+ = S̃+ ∗ ∗
[(
S̃− ∗ ∗ v+ − P̃+ ∗ ∗P−
)
×
×H
(
x− l2
)]
,
(73)
v− = −P̃− ∗ ∗
[(
S̃− ∗ ∗ v+ − P̃+ ∗ ∗P−
)
×
×H
(
l2 − x
)]
.
(74)
Taking that v=∂P/∂y1 one can obtain from
Eq. (10) that
PLF = S̃LF vLF ; S̃LF = −
(√
α2
1 +
s2
c2
)−1
, (75)
instead of Eq. (50).
Besides, Eq. (75) there is another condition:
P̃LF =1/S̃LF .
Note that for homogeneous medium the original
functions S̃(x, t) and P̃ (x, t), corresponding to S̃LF
and P̃LF , are found from Eq. (4), and their factori-
zation yields
S̃+
(
x, t
)
= −H
(
x
)
δ
(
t− x/c
)
√
π
√
x
;
P̃+
(
x, t
)
=
δ
(
t− x/c
)
2
√
π
H
(
x
)
x3/2
.
(76)
Since S̃LF+S̃LF− = S̃LF corresponds to the origi-
nal, given by Eq. (59) for k=0, the first addendum
in (73) is as follows:
S̃+ ∗ ∗
(
S̃− ∗ ∗ v+
)
= − c
π
∫ ∫
v+dx
′dt′√
T
=
= − ck
2π
∫ ∫
u+dx
′dt′√
T
.
(77)
Substituting Eq. (76) into Eq. (73) and carry-
ing out lengthy calculations one can obtain the dis-
placements out of the crack, corresponding to the first
order of k:
P+
(
t, x
)
=− ck
2π
∫ ∫
S0
P 1
+
(
t′, x′
)
√
T
dx′dt′+P 1
+
(
t, x
)
, (78)
where
P 1
+
(
t, x
)
=
√
x− l2
(
t0
)
π
√
c
t0∫
0
Φ1
[
τ, x− c
(
t− τ
)]
(
t− τ
)√
t0 − τ
dτ (79)
and l2(t0)−x+ct=ct0.
Formula (79) gives known behavior of the dis-
placement near the crack’s edge ∼
√
x−l2(t) for
hte boundary function Φ1(t, x), being equal to zero
for x= l2(t). Moreover, the integral in Eq. (79) is
convergent for x= l2(t), t0≈ t. The stresses on the
crack can be found from Eq. (74). However, more
simple way is to obtain V−(t, x)=∂P/∂y1−(k/2)P
accounting that V satisfies Eq. (10) and the boundary
conditions, due to Eq. (72), are of the order of k:
y1 = 0,
∂V
∂y1
= f
(
t, x
)
− k
2
V, x < l2
(
t
)
,
V = 0, x > l2
(
t
)
,
(80)
where
f
(
t, x
)
=
1
c2
∂2Φ1
∂t2
− ∂2Φ1
∂x2
.
Then from Eq. (20), written for V (t, x), one can
obtain that
V
(
t, x
)
=− c
π
∫∫
f
(
t′, x′
)
−
(
k/2
)
V
(
t′, x′
)
√
T
dx′dt′. (81)
Putting k=0 in Eq. (81) leads to homogeneous
solution V 0(t, x), which can be substituted in Eq. (81)
instead of V (t′, x′). Because of V =0 for x>l(t2) the
integrand in Eq. (81) is the same as that in Eq. (30).
Hence, after introducing of characteristic variables we
find that
V1
(
ξ0, η0
)
= − 1
2π
ξ0∫
ξa(η0)
dξ′√
ξ0 − ξ′
×
×
η0∫
−ξ′
f1
(
ξ′, η′
)
−
(
k/2
)
V 0
−
(
ξ′, η′
)
√
η0 − η′
dη′,
(82)
where f1(ξ
′, η′)=f(t′, x′). Following Eq. (39) for
η′>ηe we obtain
V 0
−
(
ξ′, η′
)
=− 1
2π
ξ′∫
ξa(η′)
dξ′′√
ξ′ − ξ′′
η′∫
−ξ′′
f1
(
ξ′′, η′′
)
√
η′ − η′′
dη′′. (83)
70 A. G. Bagdoev, S. G. Sahakyan
ISSN 1028 -7507 Акустичний вiсник. 2002. Том 5, N 4. С. 61 – 71
As Eq. (39) shows, for η′<ηe the integration with
respect to ξ′′ should be performed within the limits
(−η′, ξ′). Using the formula
(
1
c2
∂2
∂t2
− ∂2
∂x2
)
1√
T
= T−3/2
one can obtain
V−
(
ξ0, η0
)
=
= − 1
2π
ξ0∫
ξa(η0)
dξ′
(
ξ0 − ξ′
)3/2
η0∫
−ξ′
Φ1
(
ξ′, η′
)
(
η0 − η′
)3/2
dη′+
+
k
4π
ξ0∫
ξa(η0)
dξ′√
ξ0 − ξ′
η0∫
−ξ′
V 0
−
(
ξ′, η′
)
√
η0 − η′
dη′,
where only finite part of the integral with respect to
ξ′ is retained.
For the function Φ1(t
′, x′), which behaves at the
edge of the crack as
Φ1
(
t′, x′
)
= B
(
t′, x′
)[
l2
(
t′
)
− x′
)
,
the integral with respect to η′ is finite for x′= l2(t′),
so, usual singularity for stresses is observed:
V−
(
t′, x′
)
∼ 1√
l2
(
t
)
− x
.
CONCLUSION
The problem on the antiplane crack propagating
with arbitrary velocity in anisotropic inhomogeneous
elastic media is considered. The solution is obtai-
ned by method of integral equations developed in the
theory of wing. For wide class of inhomogeneities the
solution is obtained in closed analytical form. The
displacements on the crack’s faces and the stresses
out of the crack are obtained. It is shown that the
inhomogeneity essentially effects the solution.
1. Сарайкин В. А., Слепян Л. И. Плоская задача о
динамике трещины в упругом теле // Известия
АН СССР. МТТ.– 1979.– N 4.– С. 54–73.
2. Aki K. and Richards P. J. Quantitative seismology:
Theory and methods: in 2 volumes.– San Francisco:
W. H. Freeman & Co., 1980.
3. Костров Б. В. Неустановившаяся трещина про-
дольного сдвига // ПММ.– 1966.– 30, вып. 6.–
С. 1048–1050.
4. Поручиков В. Б. Методы динамической теории
упругости.– М.: Наука, 1986.– 328 с.
5. Красильщикова Е. А. Тонкое крыло в сжимаемом
потоке.– М.: Наука, 1978.– 223 с.
6. Багдоев А. Г., Мовсисян Л. А. Приближенное ре-
шение антиплоской анизотропной задачи о рас-
пространении трещины // Механика. Ереван. гос.
ун-т.– 1989.– 7.– С. 3–7.
7. Noble B. Methods based on the Wiener – Hopf techni-
que for the solution of partial defferential equations.–
London: Pergamon Press, 1958.– 246 p.
8. Bateman H., Erdélyi A. Tables of Integral
Transforms. Vol. 1.– New York: McGraw-Hill,
1954.– 343 p.
9. Багдоев А. Г., Мовсисян Л. А. Задачи о трещинах
в вязкоупругой среде.– Изв. АН Арм. ССР. Меха-
ника: 1987, N 40.– 3–10 с.
10. Гавриленко В. В. Плоская симметричная задача
погружения затупленного твердого тела в жид-
кость с учетом отрыва // Акуст. вiсн.– 1998.– 1,
N 3.– С. 24–29.
11. Кубенко В. Д. Проникание упругих оболочек в
сжимаемую жидкость.– К.: Наук. думка, 1981.–
160 с.
A. G. Bagdoev, S. G. Sahakyan 71
|