International Journal of Trend in Scientific Research and Development(IJTSRD)
ISSN:2456-6470 —www.ijtsrd.com—Volume -2—Issue-3
CRACK PROBLEMS CONCERNING BOUNDARIES OF CONVEX LENS LIKE
FORMS
DOO-SUNG LEE
Department of Mathematics
College of Education, Konkuk University
120 Neungdong-Ro, Kwangjin-Gu, Seoul, Korea
e-mail address: dslee@kumail.konkuk.ac.kr
Abstract
Another important problem involving a circumferential edge crack is that concerned with a
spherical cavity. Atsumi and Shindo[4] investigated the singular stress problem of a peripheral
edge crack around a spherical cavity under uniaxial tension field. In more recent years, Wan
et al.[7] obtained the solution for cracks emanating from surface semi-spherical cavity in finite
body using energy release rate theory. In previous studies concerning the spherical cavity with
the circumferential edge crack, the cavity was a
full spherical shape. In this present analysis, we
are concerned with a cavity of a spherical portion, rather than a full spherical cavity. More
briefly describing it, the cavity looks like a convex lens. Here, we employ the known methods of
previous investigators to derive a singular integral equation of the first kind which was solved
numerically, and obtained the s.i.f. for various
spherical portions. It is also shown that when
this spherical portion becomes a full sphere, the
present solution completely agrees with the already known solution.
The singular stress problem of a peripheral edge
crack around a cavity of spherical portion in an
infinite elastic medium when the crack is subjected to a known pressure is investigated. The
problem is solved by using integral transforms
and is reduced to the solution of a singular integral equation of the first kind. The solution
of this equation is obtained numerically by the
method due to Erdogan, Gupta , and Cook, and
the stress intensity factors are displayed graphically.
Also investigated in this paper is the pennyshaped crack situated symmetrically on the central plane of a convex lens shaped elastic material.
Key words: cavity of spherical portion/ peripheral edge crack/penny-shaped crack /SIF.
1.Introduction.
The problem of determining the distribution
of stress in an elastic medium containing a circumferential edge crack has been investigated by
several researchers including the present author.
Among these investigations, the notable ones
are Keer et al.[1,2],Atsumi and Shindo[3,4], and
Lee[5,6]. Keer et al.[1] considered a circumferential edge crack in an extended elastic medium
with a cylindrical cavity the analysis of which
provides immediate application to the study of
cracking of pipes and nozzles if the crack is small.
2.Formulation of problem and reduction to
singular integral equation. 0
We employ cylindrical coordinates (r, φ, z)
with the plane z = 0 coinciding the plane of peripheral edge crack. The spherical coordinates
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INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
(ρ, θ, φ) are connected with the cylindrical coordinates by
z = ρ cos θ,
σrz =
r = ρ sin θ.
∂2φ
∂2ψ
∂ψ
+z
− (1 − 2ν) ,
∂r∂z
∂r∂z
∂r
(2.8)
∂2ψ
∂ψ
∂2φ
σzz =
+
z
− 2(1 − ν) .
(2.9)
Spherical coordinates (ζ, ϑ, φ) whose origin is
2
2
∂z
∂z
∂z
located at z = −δ, r = 0, and is the center of
(1)
(2)
the upper spherical surface, are also used. The The functions φ and φ for the regions z > 0
cavity is symmetrical with respect to the plane and z < 0, respectively, are chosen as follows:
z = 0.
Z ∞
The crack occupies the region z = 0, 1 ≤
(1)
φ (r, z) = (2ν − 1)
ξ −1 A(ξ)J0 (ξr)e−ξz dξ
r ≤ γ.
So
the
radius
of
the
spherical
cavity
is
0
√
ζ0 = 1 + δ 2 .
∞
X
Pn (cos θ)
+
an
,
(2.10)
The boundary conditions are:
ρn+1
n=0
On the plane z = 0, we want the continuity of
the shear stress, and the normal displacement:
Z ∞
(2)
φ (r, z) = (2ν − 1)
ξ −1 A(ξ)J0 (ξr)eξz dξ
0
uz (r, 0+ ) − uz (r, 0− ) = 0,
γ ≤ r < ∞, (2.1)
−
∞
X
an (−1)n
n=0
Pn (cos θ)
.
ρn+1
(2.11)
Here the superscripts (1) and (2) are taken for
1 ≤ r < ∞.
(2.2) the region z > 0 and z < 0, respectively. The
(1)
(2)
And the crack is subjected to a known pressure functions ψ and ψ are chosen as follows:
Z ∞
p(r), i.e.,
ψ (1) (r, z) =
A(ξ)J0 (ξr)e−ξz dξ
+
0
σzz (r, 0 ) = −p(r),
1 ≤ r ≤ γ.
(2.3)
∞
X Pn (cos θ)
On the surface of the spherical cavity, stresses
+
bn
,
(2.12)
are zero:
ρn+1
n=0
Z ∞
(2)
ψ (r, z) = −
A(ξ)J0 (ξr)eξz dξ
σ (ζ , ϑ) = 0,
(2.4)
σrz (r, 0+ ) − σrz (r, 0− ) = 0,
ζζ
0
σζϑ (ζ0 , ϑ) = 0.
0
(2.5)
+
∞
X
bn (−1)n
Pn (cos θ)
.
ρn+1
(2.13)
We can make use of the axially symmetric son=0
lution of the equations of elastic equilibrium due Then we can immediately satisfy condition (2.2)
to Green and Zerna [8] which states that if ϕ(r, z) by this choice of functions (2.10)-(2.13).
and ψ(r, z) are axisymmetric solutions of Laplace
Now the condition (2.1) requires
equation, then the equations
Z ∞
∂ϕ
∂ψ
A(ξ)J0 (ξr)dξ = 0, r > γ.
(2.14)
2µur =
+z
,
(2.6)
0
∂r
∂r
0
∂ϕ
∂ψ
2µuz =
+z
− (3 − 4ν)ψ,
(2.7)
Equation (2.14) is automatically satisfied by
∂z
∂z
where µ is the modulus of rigidity and ν is Pois- setting
son’s ratio, provide a possible displacement field.
Z γ
The needed components of stress tensor are given
A(ξ) =
tg(t)J1 (ξt)dt.
(2.15)
by the equations
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Then from the boundary condition (2.3), we
obtain
Z ∞
∞
X
(2n + 1)2 P2n (0)
ξA(ξ)J0 (ξr)dξ −
a2n
r2n+3
0
1 ∂2φ
1 ∂φ
∂2ψ
− 2
+ cos ϑ
ζ ∂ζ∂ϑ ζ ∂ϑ
∂ζ∂ϑ
∂ψ
cos ϑ ∂ψ
n=0
+(1 − 2ν) sin ϑ
− 2(1 − ν)
. (3.2)
∂ζ
ζ ∂ϑ
∞
X
P0
(0)
To satisfy boundary conditions on the spherical
−2(1 − ν)
b2n+1 2n+1
= −p(r),
2n+3
surface, it is needed to represent φ, ψ in (2.10)r
n=0
(2.13) in terms of ζ, ϑ variables. To do so we uti1 ≤ r ≤ γ,
(2.16) lize the following formula whose validity is shown
where prime indicates the differentiation with re- in the Appendix 1. An expression useful for the
present analysis is the following
spect to the argument.
σζϑ =
By substituting (2.15) into (2.16), it reduces
∞
to
−
2
π
Z
γ
tg(t)R(r, t)dt −
1
∞
X
r−(2n+3) {P2n (0)
Pn (cos θ) X
=
ρn+1
µ
k=0
n+k
k
¶
n=0
0
×(2n + 1)2 a2n + αb2n+1 P2n+1
(0)} = −p(r),
1 ≤ r ≤ γ,
Thus
∞
X
(2.17)
n=0
where
µ ¶
1
r
R(r, t) = 2
, t > r,
E
2
r −t
t
µ ¶
µ ¶
r 1
1
t
t
=
− K
,
E
2
2
tr −t
r
rt
r
where
r > t.
Aj =
(2.18)
K and E in (2.18) are complete elliptic integrals Also
Z
of the first and the second kind, respectively and
α = 2(1 − ν).
The solution will be complete, if the conditions on the surface of the spherical cavity are
satisfied.
j
X
n=0
∞
n=0
k=0
µ
n+k
k
∞
X
Pj (cos ϑ)
j=0
ζ j+1
ξ
Z
−1
∞
×
0
¶
Aj ,
j!
an δ j−n .
(j − n)!n!
Z
∞
0
∞
Pn (cos θ) X X
an
=
an
ρn+1
Pn+k (cos ϑ) k
δ =
ζ n+k+1
×
Pn+k (cos ϑ) k
δ .
ζ n+k+1
(3.3)
−ξz
A(ξ)J0 (ξr)e
(3.4)
(3.5)
γ
dξ =
tg(t)
1
ξ −1 J0 (ξr)J1 (ξt)e−ξz dξdt.
(3.6)
If we make use of the formula in Whittaker and
Watson[9,pp.395-396]
3.Conditions on the surface of the spheri- Z π
cal cavity.
exp{−ξ(z+ix cos u+iy sin u)}du = 2πe−ξz J0 (ξr),
−π
Equation (2.17) gives one relation connecting 0
to the inner integral of (3.6), it can be written
unknown coefficients an and bn . The stress comas,
if we are using the shortened notation
ponents besides (2.8) and (2.9) which are needed
for the present analysis are given by the following
equations
β = z + ix cos u + iy sin u,
σζζ
∂2φ
∂2ψ
∂ψ
=
+
ζ
cos
ϑ
− 2(1 − ν) cos ϑ
2
2
∂ζ
∂ζ
∂ζ
sin ϑ ∂ψ
+2ν
,
ζ ∂ϑ
(3.1)
then
Z
0
=
1
2π
∞
Z
ξ −1 J0 (ξr)J1 (ξt)e−ξz dξ
π
−π
Z
0
∞
ξ −1 J1 (ξt)e−ξβ dξdu
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INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
Z γ
(2n)!
g(t)
2n−k
×
(−δ)
dt,
(3.11)
2n−1
(2n − k)!k!
1 t
and [(k+1)/2] is the greatest integer ≤ (k +1)/2.
π
t
p
du
β 2 + t2
−π β +
r
¾
Z π½
1
β
β2
It is also necessary to express ψ (1) in terms of
− 1 + 2 du
=−
2π −π t
t
spherical coordinates (ζ, ϑ). Now, as in (3.4)
µ
¶
¾
Z π½
∞
2n
1
n
1
β X (− 2 )n (−1) β
=−
−
du.
∞
∞
2π −π t
n!
t
X
Pn (cos θ) X Pj (cos ϑ)
n=0
=
Bj ,
(3.12)
b
n
(3.7)
ρn+1
ζ j+1
n=0
j=0
As |β| < |t|, the above series expansion is valid.
Next,
where
∞
X
2n
2n
0 2n
j!
β = (−δ+Z+ix cos u+iy sin u) = (−δ+β )
Bj =
bn δ j−n .
(j
−
n)!n!
¶
µ
2n
n=0
X
2n
=
(−δ)2n−k (β 0 )k ,
(3.8)
k
We first express following integral in (ρ, θ) cok=0
ordinates
where (r, φ, Z) is the cylindrical coordinates sysZ ∞
Z γ
tem centered at (r, z) = (0, −δ). We have
A(ξ)J0 (ξr)e−ξz dξ =
tg(t)dt
0
1
the following formula from Whittaker and WatZ ∞
son[9,p.392]
Z π
Z π
×
J1 (ξt)J0 (ξr)e−ξz dξ.
(3.13)
0 n
n
0
(β ) du =
(Z + ix cos u + iy sin u) du
The inner integral on the right-hand side of
−π
−π
n
= 2πζ Pn (cos ϑ).
(3.9) (3.13) is
Z
∂
1
1 ∞
If we utilize (3.8) and (3.9) in (3.7), it can be
J0 (ξr)e−ξz J0 (ξt)dξ =
−
written as
t 0
∂ξ
t
Z ∞
1
Z ∞
{rJ1 (ξr) + zJ0 (ξr)}e−ξz J0 (ξt)dξ.
−
−1
−ξz
t 0
ξ J0 (ξr)J1 (ξt)e dξ
0
(3.14)
∞
X
(− 21 )n (−1)n 1
Using the equation in Erdélyi et al.[10] 0
=
Z
n!
t2n
2 ∞ sin(ξx)dx
n=0
√
J
(ξt)
=
,
0
¶
2 − t2
2n µ
π
x
X
t
2n
(−δ)2n−k ζ k Pn (cos ϑ)
×
k
equation (3.14) is equal to
k=0
Z ∞
ζ
1
1
2
dx
+δ − P1 (cos ϑ).
(3.10)
√
+
t
t
tπ t
x2 − t2
(1)
Z ∞
Thus finally, from (3.4), (3.6) and (3.10), φ
can be written in terms of spherical coordinates
×=
{rJ1 (ξr) + zJ0 (ξr)}e−ξ(z+ix) dξ
0
(ζ, ϑ) as
Z
¸
·
−ix
1
1
2 ∞
1
∞
X
Ak
√
=p
= +
dx
k
(1)
2 − t2
2 + (z + ix)2
φ =
Pk (cos ϑ) k+1 + Φk ζ
t
tπ t
x
r
ζ
Z
k=0
1 12 ∞
1
dx
Z γ
Z γ
√
= −
=q
2
2
ρ
t
t
π
+δ
tg(t)dt − ζP1 (cos ϑ)
g(t)dt,
x −t
t
1 − 2 i cos θ + ( ρ i)2
=
1
2π
Z
1
1
where
Φk = −(α − 1)
∞
X
n=[(k+1)/2]
(− 12 )n (−1)n
n!
x
x
∞ µ ¶
( 1 )n
1 1 X ρ 2n+1
(−1)n P2n+1 (cos θ) 2 ,
= −
t
t
t
n!
n=0
(3.15)
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INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
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where we used the generating function
−Ak+1
∞
X
1
√
=
Pn (cos θ)xn .
1 − 2x cos θ + x2 n=0
+
To express (3.15) in the (ζ, ϑ) spherical coordinates, we use the following formula whose validity is shown in the Appendix 2.
2n+1
X
(k + 2)(k + 3)
ζ0k+4
Bk (k + 1){2 − α − (k + 1)(k + 4)}
2k + 1
ζ0k+2
−
Bk+2 (k + 2)(k + 3)(2α + k + 2)
2k + 5
ζ0k+4
k(k + 1)(k + 1 − 2α)
2k + 1
k=0
(k + 2){(k + 2)(k − 1) + α − 2}
k+1
= 0.
×(−δ)2n+1−k ζ k Pk (cos ϑ).
(3.16) −ζ0 Ψk+2
2k + 5
0
(3.19b)
If we use (3.16) in (3.15), then with (3.12), we
(1)
Thus
if
we
multiply
(3.18)
by
k
+
2
and
subtract
get ψ in spherical coordinates (ζ, ϑ)
(3.19b) from the resulting equation we find
µ
¶ Z γ
∞
X
Bk
ψ (1) =
Pk (cos ϑ) k+1 + Ψk ζ k +
g(t)dt,
ζ
1
·
k=0
ζ0k+2 (2k + 1)
(3.17) Bk = −
k(2k+3)ζ0k−1 Φk+1
α(2k
+
3)
+
2k(k
+
1)
where
¸
∞
X
(−1)n ( 12 )n
(2n + 1)!
(2k + 3)k(k + 1 − 2α) k+1
k−1
Ψk = −
+ζ0 Ψk
+ζ0 Ψk+2 k(k+2) .
n!
k!(2n + 1 − k)!
2k + 1
n=[k/2]
(3.20)
Z γ
g(t)
If we solve (3.20) for bi using the theorem
×(−δ)2n+1−k
dt.
2n+1
1 t
which is in the Appendix 3 and the relation,
ρ2n+1 P2n+1 (cos θ) =
(2n + 1)!
k!(2n + 1 − k)!
−k(k+1)ζ0k−1 Φk+1 −ζ0k−1 Ψk
Then if we substitute these values of φ(1) and
α−1
Φk+1 =
Ψk ,
ψ (1) given by (3.11) and (3.17) into (3.2), and
k+1
simplify the results by using the properties of
Legendre polynomials, from the condition that we find
on the spherical surface ζ = ζ0 , the shear stress
σζϑ = 0, we obtain the following equation
i
X
bi
i!(−1)i−k 1
=
[N1 (k)Ψk + N2 (k)Ψk+2 ],
2
k+3
Bk α − (k + 1)
δi
(i − k)!k! δ k
−Ak+1 k+4 + k+2
k=0
2k + 1
ζ0
ζ0
(3.21)
−
Bk+2 (k + 3)(2α + k + 2)
2k + 5
ζ0k+4
k(k + 1 − 2α)
2k + 1
k
+
2
+
α
−
(k
+
2)(k
+ 3)
−ζ0k+1 Ψk+2
= 0.
2k + 5
(3.18)
+kζ0k−1 Φk+1 + ζ0k−1 Ψk
Likewise, from the condition σζζ (ζ0 , ϑ) = 0,
we get following two equations,
where
µ
¶
ζ02k+1 (2k + 3)k
−α + k 2
N1 (k) = −
,
α(2k + 3) + 2k(k + 1)
k+1
N2 (k) = −
ζ02k+3 (2k + 1)k(k + 2)
.
α(2k + 3) + 2k(k + 1)
Equation (3.21) can be written as
Z
i
X
i!(−1)i−k δ i γ g(t)
bi = −
(i − k)!k! δ k 1 t
k=0
B1 2(2α + 1) 1
2
−A0 3 − 3
− Ψ1 (α − 4) = 0,
3
3
ζ0
ζ0
(3.19a)
×
1
X
µ
[Nn+1 (k)fk+2n
n=0
¶
1 δ
,
dt,
2 t
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(3.22)
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INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
where
fk (c, x) =
(−δ)1−k
k!
∞
X
(−1)` (c)` (2` + 1)!x2`
(2` + 1 − k)!`!
`=[k/2]
Thus
∞
∞
X
a0 X
(2i + 1)2 P2i (0)
a2i
= 3+
(Ã1 Ψ1 + Ã2 Ψ3 )
r2i+3
r
i=0
i=1
(2i + 1)2 (−1)i ( 21 )i
r2i+3 i!
∞
X (2i + 1)2 (−1)i ( 1 )i
2
+
r2i+3 i!
with (c)` = c(c + 1)(c + 2) · · · (c + ` − 1).
×δ 2i
Now from (3.19a,b) we have
Ak+1 = M1 (k)Ψk + M2 (k)Ψk+2 + M3 (k)Ψk+4 ,
i=1
A0 = Ã1 Ψ1 + Ã2 Ψ3 ,
where
×
0
2i−1
X
k=0
ζ02k+3 k(−α + k 2 )
M1 (k) = −
(k + 2)(k + 3)(2k + 1)
·
¸
(2k + 3)g̃(k)
×
+1 ,
h(k)
(2i)!(−δ)2i−k−1
F̃ (k).
(2i − 1 − k)!(k + 1)!
If we interchange the order of summation in the
last term of the above equation, and use
µ
¶
Z γ
1 δ
g(t)
dt,
Ψ1 = −
f1
,
t
2 t
1
µ
¶
Z γ
1 δ
g(t)
2k+5 ·
ζ
g̃(k)
g̃(k − 2)
f3
,
dt,
Ψ3 = −
M2 (k) = 0
−
k(k+1)+
t
2 t
1
k+3
h(k)
2k + 5
we finally obtain
Z γ
∞
¸
X
(2i + 1)2 P2i (0)
(k + 2)(2k + 7)(2α + k + 2){−α + (k + 2)2 }
a2i
tg(t)T1 (r, t)dt,
=−
,
+
r2i+3
(2k + 5)h(k + 2)
1
i=0
ζ 2k+7 (k + 2)(k + 4)(2α + k + 2)
M3 (k) = 0
,
h(k + 2)
½
¾
ζ03
5(2α + 1)(−α + 1)
Ã1 = −
α−4−
,
6
h(1)
3(2α + 1)ζ05
Ã2 =
,
h(1)
with
where
T1 (r, t) =
1
r 3 t2
µ
·X
¶ µ ¶
2
∞ X
1 δ
δ
Mm+1 (k)fk+2m ,
×
hk
2 t
r
k=0 m=0
¶
µ
¶¾½
µ ¶¾¸
½
µ
1 δ
δ
1 δ
,
+ Ã2 f3
,
1+j
,
+ Ã1 f1
2 t
2 t
r
with
µ ¶
δ
(−δ)−k−1
g̃(k) = 2 − α − (k + 1)(k + 4),
hk
=
r
(k + 1)!
h(k) = α(2k + 3) + 2k(k + 1).
∞
2
X (2i + 1) (−1)i ( 1 )i (2i)! µ δ ¶2i
In a similar way a0i s can be found from these
2
×
,
(2i − 1 − k)!i!
r
equations as follows:
i=[k/2+1]
a2i = (Ã1 Ψ1 + Ã2 Ψ3 )δ 2i
2i−1
X (2i)!(−δ)2i−k−1
+
F̃ (k),
(2i − 1 − k)!(k + 1)!
k=0
where
·
µ
¶
γ
g(t)
1 δ
F̃ (k) = −
M1 (k)fk
,
+ M2 (k)
t
2 t
1
µ
¶
µ
¶¸
1 δ
1 δ
×fk+2
,
+ M3 (k)fk+4
,
dt.
2 t
2 t
Z
µ ¶ X
µ ¶
∞
(2i + 1)2 (−1)i ( 21 )i δ 2i
δ
j
=
.
r
i!
r
i=1
Also,
∞
X
∞
b2i+1
i=0
0
X (−1)i ( 3 )i
P2i+1
(0)
2
=
−
r2i+3
i!r2i+3
i=0
×
2i+1
X
k=0
(2i + 1)!(−δ)2i−k+1
(2i + 1 − k)!k!
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7
¾
G(k − 1) (−1)k ( 12 )k 1
×
4k + 3
k!
t2k
1
n=0
(2k + 1)2 (2k + 2)(2k + 3)(2α + 2k + 1)
If we change the order of summation in the above
+
H(k + 1)
equation, we get
¸
k+1
Z γ
∞
(−1) ( 21 )k+1 1
0
X
P2i+1
(0)
×
,
tg(t)T2 (r, t)dt,
b2i+1 2i+3 = −
(k + 1)!
t2k+2
r
1
i=0
where 0
where
H(k) = α(4k + 1) + 4k(2k − 1),
¶
µ
∞ 1
1 XX
3 δ
G(k) = 2 − α − 2k(2k + 3),
T2 (r, t) = 2 3
,
Nn+1 (k)fk
t r
2 r
k=0 n=0
F (k) = −α + (2k − 1)2 .
¶
µ
¶
µ
¶
µ
∞ X
1
1 δ
X
1 δ
3 δ
×fk+2n ,
.
B
=
lim
,
f
,
N
(k)f
n+1
k+2n
k
2 t
δ→0
2 r
2 t
k=0 n=0
Thus finally the singular integral equation (2.17)
·
∞
X
(−1)k ( 23 )k
(2k + 1)(4k + 5)
becomes
=
−
k!r2k
2(k + 1)
k=1
Z
2 γ
F (k + 1) (−1)k ( 21 )k 1 (4k + 3)(2k + 1)(2k + 3)
tg(t){R(r, t) + S(r, t)}dt = p(r),
×
−
π 1
H(k + 1)
k!
H(k + 1)
t2k
½
¸
¾
1
k+1
(−1) ( 2 )k+1 1
1 ≤ r ≤ γ,
(3.23)
9
1
×
−5F
(1)+
+
.
(k + 1)!
t2k+2
2H(1)
t2
where
π
Thus
S(r, t) = − {T1 (r, t) + αT2 (r, t)}.
∞
2
X
(−1)k ( 21 )k (2k + 1)
A
+
αB
=
When δ = 0, we briefly show that this equation
k!r2k
k=1
completely agrees with what Atsumi and Shindo
·
½
obtained. Using
(−1)k (2k − 1)(2k + 1)
×
k(k + 1)F (k)
µ ¶
H(k)k(k + 1)
(−1)k ( 12 )k (2k + 1)2 1
δ
lim h2k−1
=
,
1
1
δ→0
r
k!
r2k
{(2k − 1)(4k + 3)2kG(k)
− 2
t
4(4k
+
3)
µ
¶
1
¾ 1
(−1)k−1 ( 2 )k−1 1
1 δ
( 2 )k−1 1
lim f2k−1
,
=
,
−H(k)G(k − 1)}
δ→0
2 t
(k − 1)!
t2k−2
(k − 1)! t2k−2
µ
¶
µ
¶
½
∞
2
XX
1 δ
δ
1
F (k + 1)(2k + 1)
Mm+1 (k)fk+2m
,
hk
A = lim
(2k + 3)(2k + 2) 2
+(−1)k
δ→0
2 t
r
H(k + 1)(2k + 2)
t
k=0 m=0
¾ 1
¸
½
¾
∞
4k + 5 ( 2 )k 1
α
9
X (−1)k ( 1 )k (2k + 1)2
+F (k+1)
−5F (1)
+
2
=
4k + 3 k! t2k
2H(1) t2
k!r2k
½
k=1
∞
X
·
½
¾
2(2k + 1)2
1
1
=
k(k + 1)F (k) − 2
1
2k − 1 (4k + 1)G(k)
×
+1
H(k)(k + 1)
t 4(4k + 3)
k=1
(2k + 1)(2k + 2) 4k − 1
H(k)
¾
Z
γ
µ
¶
1
g(t) X
1 δ
Nn+1 (k)fk+2n
,
dt.
t
2 t
(−1)k−1 ( 12 )k−1 1
(k − 1)!
t2k−2
½
G(k)
+(2k + 1) −
(2k − 1)2k
H(k)
(2k + 1)(4k + 5)(2α + 2k + 1)
F (k + 1)
−
H(k + 1)(4k + 3)
×F (k)
+
×{(2k − 1)(4k + 3)2kG(k) − H(k)G(k − 1)}
×
1
(rt)2k−2 r2
µ
( 21 )k
k!
¶2
½
¾
∞
X
F (k)
1
4k + 1
+
(2k + 1)(2k − 1) 2 + F (k)
H(k)
t
4k − 1
k=2
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µ
½
¾
(3.24) and (3.26) to obtain the following expres1
α
9
+
−5F
(1)
+
. sion:
(rt)2k−2 2H(1)
t2
¶1
·
¸
Z µ
Using
a 1 1+τ 2
a
¶
µ
¶
µ
G̃(τ ) (τ + 1) + 1 [R(s, τ )
1
1
1
π −1 1 − τ
2
Ã1 f1
, 0 + Ã2 f3
, 0 = Ã1 − Ã2 2 ,
2
2
2t
+S(s, τ )]dτ = 1, −1 < s < 1.
(4.1)
we finally obtain
The numerical solution technique is based on
·
∞
X
2
2
1+ν 1
2(2k + 1)2 the collocation scheme for the solution of singu− S(r, t) =
+
π
3 r 3 t2
H(k) (k + 1)r2
lar integral equations given by Erdogan, Gupta,
k=1
and Cook [11]. 0 This amounts to applying a
½
1
1
× k(k + 1)F (k) − 2
{(2k − 1)(4k + 3) Gaussian quadrature formula for approximating
t 4(4k + 3)
the integral of a function f (τ ) with weight func¾
½
1
tion [(1+τ )/(1−τ )] 2 on the interval [-1,1]. Thus,
×2kG(k) − H(k)G(k − 1)} + kF (k) (2k + 2)
letting n be the number of quadrature points,
¾¸
1
4k + 1
×(2k − 1) 2 + F (k)
¶1
Z 1µ
n
t
4k − 1
2π X
1+τ 2
µ
¶
f (τ )dτ +
(1 + τk )f (τk ),
(2k − 1)!! 2 1
2n + 1
−1 1 − τ
k=1
.
×
(2k)!!
(rt)2k r
(4.2)
The above equation completely agrees with the where
¶
µ
result by Atsumi and Shindo.
2k − 1
π, k = 1, . . . n.
(4.3)
τk = cos
2n + 1
A quantity of physical interest is the stress intensity factor which is given as
The solution of the integral equation is obp
K = lim
2(r − γ)σzz (r, 0).
tained
by choosing the collocation points:
r→γ +
µ
¶
2iπ
We choose p(r) = p0 and put γ − 1 = a. We let
si = cos
, i = 1, . . . , n,
(4.4)
2n + 1
a
a
r = (s + 1) + 1, t = (τ + 1) + 1, (3.24)
and solving the matrix system for G∗ (τk ) :
2
2
n
and in order to facilitate numerical analysis, asX
2n + 1
[R(sj , τk ) + S(sj , τk )]G∗ (τk ) =
,
sume g(t) to have the following form:
2a
k=1
1
1
g(t) = p0 (t − 1) 2 (γ − t)− 2 G̃(t).
(3.25)
j = 1, . . . , n,
(4.5)
With the aid of (3.24), g(τ ) can be rewritten as
where
µ
¶1
1+τ 2
G∗ (τk )
g(τ ) = p0 G̃(τ )
.
(3.26)
G̃(τk ) =
.
(4.6)
1−τ
(1 + τk )[ a2 (τk + 1) + 1]
×2k
( 12 )k
k!
¶2
The stress intensity factors K can therefore be
5.Numerical results and consideration.
expressed in terms of G̃(t) as
√
Numerical calculations have been carried out
K/p0 = 2aG̃(γ),
(3.27) for ν = 0.3. The values of normalized stress
√
intensity factor K/p0 a versus a are shown in
or in terms of the quantity actually calculated
√
√
Fig.1-3 for various values of δ.
K/p0 a = 2G̃(γ).
(3.28)
√
Fig.1 shows the variation of K/p0 a with re4.Numerical analysis.
spect to a when δ = 0.
In order to obtain numerical solution of (3.23),
This figure shows that as a increases, S.I.F.
substitutions are made by the application of decreases steadily.
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9
Fig.2 and 3 deal with the cases when δ = 0.3
and δ = 0.5, respectively. Here we omit units.
We can see that the trend is similar. Theoretically the infinite series involved converges when
δ < 1 by comparison test. However, because of
the overflow, computations could not be accomplished beyond the values δ > 0.5. And we found
that the variation of SIF is very small with respect to the variation of δ.
0
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6.Penny-shaped crack.
also used. The elastic body is symmetrical with
In this section we are concerned with a penny- respect to the plane z = 0.
shaped crack in a convex lens shaped elastic maThe crack occupies the region z = 0, 0 ≤
terial. The problem of determining the distri- r ≤ p
1. The radius of the spherical portion is
bution of stress in an elastic sphere containing ζ0 = γ 2 + δ 2 . The boundary conditions are:
a penny-shaped crack or the mixed boundary
On the plane z = 0, we want the continuity of
value problems concerning a spherical boundary the shear stress, and the normal displacement:
has been investigated by several researchers. Srivastava and Dwivedi [12] considered the problem of a penny-shaped crack in an elastic sphere,
uz (r, 0+ ) − uz (r, 0− ) = 0,
1 ≤ r ≤ γ, (7.1)
whereas Dhaliwal et al.[13] solved the problem of
a penny-shaped crack in a sphere embedded in an
infinite medium. On the other hand, Srivastav σ (r, 0+ )−σ (r, 0− ) = 0,
0 ≤ r ≤ γ. (7.2)
rz
rz
and Narain [14] investigated the mixed boundary
And the crack is subjected to a known pressure
value problem of torsion of a hemisphere.
p(r), i.e.,
7.Formulation of problem and reduction to
σzz (r, 0+ ) = −p(r),
0 ≤ r ≤ 1.
(7.3)
a Fredholm integral equation of the sec0
ond kind. We employ cylindrical coordinates On the surface of the spherical portion, stresses
(r, φ, z) with the plane z = 0 coinciding the plane are zero:
of the penny shaped crack. The center of the
σζζ (ζ0 , ϑ) = 0,
(7.4)
crack is located at (r,z)=(0,0). As before, spherσζϑ (ζ0 , ϑ) = 0.
(7.5)
ical coordinates (ρ, θ, φ) are connected with the
cylindrical coordinates by
We can make use of (2.6)-(2.9) also, for the
present case. The functions φ(1) and φ(2) for the
regions z > 0 and z < 0, respectively, are chosen
z = ρ cos θ,
r = ρ sin θ.
as follows:
Spherical coordinates (ζ, ϑ, φ) whose origin is at
z = −δ, r = 0, and is the center of the upper
spherical surface of the convex elastic body, is
Z
φ(1) (r, z) = (2ν − 1)
0
∞
ξ −1 A(ξ)J0 (ξr)e−ξz dξ
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11
+
∞
X
an ρn Pn (cos θ),
n=0
Z
φ(2) (r, z) = (2ν − 1)
∞
X
By substituting (7.11) into (7.12), it reduces
∞
2X
(−1)n {2nt2n−1 a2n
π
g(t) −
∞
0
−
(7.6) to
ξ −1 A(ξ)J0 (ξr)eξz dξ
an (−1)n ρn Pn (cos θ).
n=0
+αb2n+1 t2n+1 } = h(t),
where
(7.7)
2
h(t) = −
π
n=0
Z
t
0
0 ≤ r ≤ 1,
(7.13)
rp(r)
√
dr.
t2 − r 2
Here the superscripts (1) and (2) are taken for
The solution will be complete, if the condithe region z > 0 and z < 0, respectively. The
tions
on the surface of the spherical portion are
functions ψ (1) and ψ (2) are chosen as follows:
satisfied.
Z ∞
(1)
ψ (r, z) =
A(ξ)J0 (ξr)e−ξz dξ
0
8.Conditions on the surface of the sphere.
∞
X
Equation (7.13) gives one relation connecting
+
bn ρn Pn (cos θ),
(7.8) unknown coefficients a and b . The stress comn
n
n=0
ponents besides (2.8) and (2.9) which are needed
for the present analysis are given by (3.1) and
Z ∞
(2)
ξz
(3.2). To satisfy boundary conditions on the
ψ (r, z) = −
A(ξ)J0 (ξr)e dξ
spherical surface, it is needed to represent φ, ψ
0
in (7.6)-(7.9) in terms of ζ, ϑ variables. To do so
∞
X
n n
+
bn (−1) ρ Pn (cos θ).
(7.9) we utilize the following formula in the Appendix
2. An expression useful for the present analysis
n=0
is the following
Then we can immediately satisfy condition (7.2)
by these choice of functions (7.6)-(7.9).
¶
n µ
X
Now the condition (7.1) requires
n
n
Pk (cos ϑ)ζ k (−δ)n−k .
P
(cos
θ)ρ
=
n
Z ∞
k
k=0
A(ξ)J0 (ξr)dξ = 0, r > 1.
(7.10)
(8.1)
0
Equation (7.10) is automatically satisfied by
setting
Z 1
A(ξ) =
g(t) sin(ξt)dt, g(0) = 0.
(7.11)
Thus
¶
∞
∞
∞ µ
X
X
X
n
n
Pk (cos ϑ)
an Pn (cos θ)ρ =
an
k
n=0
0
n=0
×ζ k (−δ)n−k =
Then from the boundary condition (7.3), we obtain
Z ∞
∞
X
ξA(ξ)J0 (ξr)dξ −
a2n (2n)2 P2n (0)r2n−2
0
−2(1 − ν)
Aj =
∞
X
Also
∂
−
∂t
n=0
0 ≤ r ≤ 1,
(7.12)
where the prime indicates the differentiation
with respect to the argument. 0
(8.2)
where
n!
an (−δ)n−j .
(n − j)!j!
n=j
0
b2n+1 P2n+1
(0)r2n = −p(r),
Pj (cos ϑ)ζ j Aj ,
j=0
n=0
∞
X
∞
X
k=0
Z
=−
Z
∞
0
Z
1
g(t)
0
0
(8.3)
ξ −1 A(ξ)J0 (ξr)e−ξz dξ
∞
J0 (ξr) cos(ξt)e−ξz dξdt. (8.4)
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The inner integral of the above equation is
Z ∞
J0 (ξr) cos(ξt)e−ξz dξ
0
Z
∞
=<
0
The inner integral on the right-hand side of
(8.10) is
Z ∞
−=
J0 (ξr)e−ξ(z+it) dξ
0
J0 (ξr)e−ξ(z+it) dξ
= <p
=
1
n=0
r2 + (z + it)2
1
= <√
2
2
r + z + 2zit − t2
1
=< q
−it 2
ρ 1 − 2 cos θ( −it
ρ )+( ρ )
∞ µ ¶
1 X t 2n
=
(−1)n P2n (cos θ).
ρ
ρ
∞
X
P2n+1 (cos θ)
(−1)n t2n+1 .
ρ2n+2
(8.11)
To express (8.11) in the (ζ, ϑ) spherical coordinates, we use the following formula in the
Appendix 1.
∞
P2n+1 (cos θ) X (2n + 1 + k)!
=
ρ2n+1
k!(2n + 1)!
k=0
(8.5)
n=0
Thus finally, from (8.2), and using the formula in Appendix, φ(1) can be written in terms
of spherical coordinates (ζ, ϑ) as
·
¸
∞
X
Φk
(1)
k
Pk (cos ϑ) k+1 + Ak ζ ,
φ =
(8.6)
ζ
k=0
P2n+1+k (cos ϑ)
.
(8.12)
×δ k
ζ 2n+2+k
If we use (8.8) and (8.12), we get ψ (1) in spherical
coordinates (ζ, ϑ) as
µ
¶
∞
X
Ψk
(1)
k
Pk (cos ϑ) k+1 + Bk ζ
ψ =
+ B0 ,
ζ
k=0
(8.13)
where
[(k−1)/2]
where
[k/2]
Φk = −(α − 1)
X
n=0
Z
×(−1)
n
1
k!
δ k−2n
(2n)!(k − 2n)!
g(t)t2n+1
dt.
2n + 1
Ψk =
X
k!
δ k−2n−1
(2n + 1)!(k − 2n − 1)!
n=0
Z 1
n
×(−1)
g(t)t2n+1 dt.
(8.14)
0
(8.7) Then if we substitute these values of φ(1) and
0
ψ (1) given by (8.6) and (8.13) into (3.2), and
(1)
It is also necessary to express ψ in terms of simplify the results by using the properties of
spherical coordinates (ζ, ϑ). Now, as in (8.2)
Legendre polynomials, from the condition that
on the spherical surface ζ = ζ0 , σζϑ = 0, we
obtain the following equation
∞
∞
X
X
n
j
2
bn Pn (cos θ)ρ =
Pj (cos ϑ)ζ Bj , (8.8)
k−1
k+1 α − (k + 2)
A
kζ
−
B
ζ
k+1
k+2
0
0
n=0
j=0
2k + 5
where
k(2α − k − 1) k + 3
−Bk ζ0k−1
− k+4 Φk+1
∞
X
2k + 1
n!
ζ0
Bj =
bn (−δ)n−j .
(8.9)
j!(n − j)!
Ψk α − (k + 1)2
Ψk+2 (k + 3)(k + 2 + 2α)
n=j
+ k+2
− k+4
2k + 5
2k + 1
ζ0
ζ0
0
We first express following integral in (ρ, θ) co= 0.
(8.15)
ordinates
Z ∞
Z 1
Likewise, from the condition σζζ (ζ0 , ϑ) = 0,
A(ξ)J0 (ξr)e−ξz dξ =
g(t)dt
we
get the following equation,
0
0
Z ∞
×
sin(ξt)J0 (ξr)e−ξz dξ.
(8.10)
−Ak+1 k(k + 1)ζ0k−1
0
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13
+ 2){2 − α − (k − 1)(k + 2)}
2k + 5
k(k
+
1)(2α
− k − 1)
+Bk ζ0k−1
(2k + 1)
(k + 2)(k + 3)
−
Φk+1
ζ0k+4
(k + 2)(k + 3)(k + 2 + 2α)
−Ψk+2
ζ0k+4 (2k + 5)
(k + 1){(k + 4)(k + 1) + α − 2}
−Ψk
ζ0k+2 (2k + 1)
= 0.
(8.16)
Thus if we multiply (8.15) by k + 1 and add
(8.16), from the resulting equation we find
¸
G(k)
G(k + 2)
+(k + 4)(k + 5)
+
,
H(k + 2)
2k + 5
¸
·
G(k)
(2k + 7) − 1
N (k) =
H(k + 2)
(k
+Bk+2 ζ0k+1
(2k + 5)
Bk+2 = k+1
ζ0 {−α(2k + 3) + 2(k + 2)(k + 3)}
·
(k + 3)(2k + 3)
×
Φk+1
ζ0k+4
(2k + 3)(k + 3)(k + 2 + 2α)
+Ψk+2
ζ0k+4 (2k + 5)
¸
Ψk
+ k+2 (k + 1)(k + 3) .
(8.17)
ζ0
0
Using the relation
Φk+1 = −
α−1
Ψk+2 ,
k+2
Bk+2 is
Bk+2 =
(2k + 5)
k+1
ζ0 {−α(2k + 3) + 2(k + 2)(k + 3)}
×
(k + 5)I(k + 2)
,
(2k + 9)(k + 2)(k + 3)
with
H(k) = −α(2k + 3) + 2(k + 2)(k + 3),
G(k) = 2−α−(k+1)(k+4), I(k) = −α+(k+3)2 .
Therefore using the formula in Theorem B of Appendix 4, we have
A=
∞
X
(−1)n 2nt2n−1 a2n
n=1
=
∞
X
(−1)n 2nt2n−1
n=1
=
∞
X
k!Ak δ k
1
δ 2n (2n)!
(k − 2n)!
k=2n
∞ ·[k/2]
X
X
k=2 n=1
×Ak δ k =
µ ¶2n ¸
k!(−1)n
t
1
(2n − 1)!(k − 2n)! δ
t
∞
X
fk (t)Ak+3 =
∞
X
fk (t)
k=−1
k=−1
·
¸
Ψk
Ψk+2
Ψk+4
× 2k+3 L(k) + 2k+5 M (k) + 2k+7 N (k) .
ζ0
ζ0
ζ0
(8.19)
In the above equation Ψk = 0, if k ≤ 0 and
[(k+3)/2]
fk (t) = δ
k+3
X
(k + 3)!(−1)n
(2n − 1)!(k + 3 − 2n)!
n=1
·
(2k + 3)(k + 3)((k + 3)2 − α)
µ ¶2n
× Ψk+2
t
1
k+4
×
.
ζ0 (2k + 5)(k + 2)
δ
t
¸
Ψk
+ k+2 (k + 1)(k + 3) .
(8.18) If we substitute the values of Ψk into (8.19),it
ζ0
reduces to
Z 1
From (8.15) we have
A
=
g(u)Ω1 (t, u)du,
Ψk
Ψk+2
Ψk+4
0
Ak+3 = 2k+3 L(k) + 2k+5 M (k) + 2k+7 N (k),
ζ0
ζ0
ζ0
where
·
∞
where
X
hk (u)
Ω1 (t, u) =
fk (t) 2k+3 L(k)
(2α − k − 3)
,
ζ0
L(k) = (k + 1)(k + 3)
k=−1
H(k)
¸
·
hk+2 (u)
hk+4 (u)
1
(2k + 3)(2α − k − 3)I(k)
+ 2k+5 M (k) + 2k+7 N (k) .
M (k) =
(k + 3)
ζ0
ζ0
k+2
(2k + 5)H(k)
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INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
½
µ ¶2 ¾
δ
δ
= ζ 2 1 − 2 cos ϑ +
,
ζ
ζ
[(k−1)/2]
n
X
(−1) k!
∞
hk (u) = δ k
X
1
1
δn
(2n + 1)!(k − 2n − 1)!
= q
=
Pn (cos ϑ) n+1 ,
n=0
ρ
ζ
ζ 1 − 2 cos ϑ ζδ + ( ζδ )2 n=0
µ ¶2n+1
u
µ ¶
.
×
Pn (cos θ)
(−1)n ∂ n 1
δ
=
ρn+1
n! ∂z n ρ
Also, using Theorem B of Appendix 4, we get
∞
∞
∞
X (−1)n ∂ n µ Pk (cos ϑ) ¶
X
X
t2n+1
n
2n+1
n
δk
=
(−1) b2n+1 t
=
(−1)
n! ∂z n
ζ k+1
(2n + 1)!δ 2n+1
k=0
n=0
n=0
µ ¶
∞
X
∞
(−1)n ∂ n+k 1 (−1)k k
X
k!Bk δ k
=
δ
×
n! ∂z n+k ζ
k!
(k − 2n − 1)!
k=0
k=2n+1
¶
∞ µ
X
k−1
n + k Pn+k (cos ϑ) k
]
µ ¶2n+1 ¸
∞ ·[X
2
n
X
=
δ .
(−1) k!
t
k
ζ n+k+1
=
Bk δ k
k=0
(2n + 1)!(k − 2n − 1)! δ
In the above equation hk (u) = 0, if k ≤ 0 and
k=1
n=0
∞
X
=
Appendix 2. Proof of (3.16).
hk+2 (t)Bk+2
k=−1
·
2k + 5
Ψk
hk+2 (t) k+1
=
(k + 1)(k + 3)
k+2
ζ
H(k)
ζ
0
0
k=−1
¸
Ψk+2 (k + 3)(2k + 3)
+ k+4
I(k)
2k + 5
ζ0
Z 1
=
g(u)Ω2 (t, u)du,
∞
X
Z
n
2πρ Pn (cos θ) =
Z
π
=
∞
X
k=0
π
n!
(−δ)n−k
(n − k)!k!
(Z + ix cos u + iy sin u)k du
hk (u)
−π
(k + 1)
n
k+2
X
ζ0
n!
k=−1
(−δ)n−k ζ k Pk (ϑ).
=
2π
¸
(n − k)!k!
hk+2 (u) (k + 3)(2k + 3)
k=0
×(k + 3) +
I(k) .
k+4
2k + 5
ζ0
In the above equation hk (u) = 0, if k ≤ 0. Thus Appendix 3. Theorem A. If the equation
k
(7.13) reduces to the following Fredholm integral
X
k!
B
=
ai ,
equation of the second kind
k
(k − i)!i!
Z 1
i=0
0
g(t) +
g(u)K(t, u)du = h(t),
is solved for ai s, it will be written as
Ω2 (t, u) =
hk+2 (t)
2k + 5
n
X
=
×
·
−π
(−δ + Z + ix cos u + iy sin u)n du
Z
where
(z + ix cos u + iy sin u)n du
−π
0
0
π
ζ0k+1 H(k)
0
where
2
K(t, u) = − {Ω1 (t, u) + αΩ2 (t, u)}.
π
Appendix 1. Proof of (3.3). Since
2
2
2
ρ = (ζ cos ϑ − δ) + (ζ sin ϑ)
ai =
i
X
k=0
i!
Bk (−1)i−k .
(i − k)!k!
Proof. We prove it by mathematical induction.
Suppose it is true for i, we will show that it is
also true for i + 1. Suppose Bi+1 is given by
Bi+1 =
i+1
X
k=0
(i + 1)!
ak = ai+1
(i + 1 − k)!k!
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INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
15
+
i
X
k=0
Proof. Let
∞
X
(i + 1)!
ak .
(i + 1 − k)!k!
Thus
n=0
ai+1 = Bi+1 −
i
X
k=0
×
k
X
= Bi+1 −
i
X
then
(i + 1)!
(i + 1 − k)!k!
an (−δ)n n! = f (n) (0),
and
k!
Bj
(−1)k−j
j!(k − j)!
j=0
(i + 1)!
j!
Bj
j=0
i
X
k=j
f
(k)
(1) =
i
X
(−1)k−j
.
(i + 1 − k)!(k − j)!
=
(−1)k−j
i−j
X
m=0
f (x) =
(n)
m=0
=
(i + 1 − m − j)!m!
k=n
∞
X
Bk (−1)k−n (−δ)k k!
f (k) (1)
(−1)k−n =
.
(k − n)!
(k − n)!
k=n
Therefore
(−1)i−j+1
,
(i − j + 1)!
an =
∞
X
References
Then (A.1) is equal to
=
j=0
[1] L. M. Keer, V.K. Luk, and J. M. Freedman,
Circumferential edge crack in a cylindrical cavity, J.Appl.Mech. 44, 1977, pp.250-254
i
X
Bi (i + 1)!(−1)i+1−j
j!(i + 1 − j)!
i+1
X
Bi (i + 1)!(−1)i+1−j
j!(i + 1 − j)!
[2] L. M. Keer and H. A. Watts, Mixed boundary
value problems for a cylindrical shell, Int.J.Solids
Structures, 12, 1976, pp.723-729
.
¤
Appendix 4. Theorem B. If the equation
¶
∞ µ
X
n
Bk =
(−δ)n−k an ,
k
is solved for
n=k
0
an s, it will
∞ µ
X
an =
k=n
be written as
¶
k
δ k−n Bk .
n
k!
Bk δ k−n .
n!(k − n)!
¤
(−1)m (i − j + 1)!
= (1 − 1)i−j+1 = 0.
(i + 1 − m − j)!m!
j=0
(x − 1)k .
¯
∞
X
¯
f (k) (1) dn
k¯
(x−1) ¯
(0) =
n
k! dx
x=0
k=n
ai+1 = Bi+1 +
k!
k=0
∞
X
since
i−j+1
X
n!
= k!Bk (−δ)k .
(n − k)!
∞
X
f (k) (1)
k=0
an (−δ) n! = f
(−1)m
=−
an (−δ)n
From Taylor’s series
n
(i + 1 − k)!(k − j)!
k=j
∞
X
n=k
(A.1)
The inner summation in the above equation can
written as, by changing the variable k − j = m Thus
0
an (−δ)n xn = f (x),
[3] A. Atsumi and Y. Shindo, Axisymmetric singular stresses in a thick-walled spherical
shell with an internal edge crack, J.Appl.Mech.
50,1983, pp.37-42.
[4] A. Atsumi and Y. Shindo, Peripheral edge
crack around spherical cavity under uniaxial tension field, Int. J. Engng Sci. 21(3), 1983, pp.269278
[5] D.-S. Lee, Torsion of a long circular cylinder
having a spherical cavity with a peripheral edge
crack, Eur.J.Mech.A/Solids 21, 2002, pp.961-969
0@IJTSRD— Available Online@www.ijtsrd.com—Volume-2—Issue-3—Mar-Apr 2018 Page:996
16
INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
[6] D.-S. Lee, Tension of a long circular cylinder having a spherical cavity with a peripheral
edge crack , Int. J. Solids Structures, 40, 2003,
pp.2659-2671
[11] F. Erdogan, G.D. Gupta, and T.S. Cook,
Numerical solution of singular integral equation.
Methods of analysis and solutions of crack problems, Noordhoff International Publishing, 1973,
[7] H. Wan, Q. Wang, and X. Zhang, Closed pp.368-425
form solution of stress intensity factors for [12] K.N. Srivastava and J.P. Dwivedi, The efcracks emanating from surface semi-spherical fect of a penny-shaped crack on the distribution
cavity in finite body with energy release of stress in an elastic sphere, Int.J.Engng. sci.,
rate method, Appl.Math.Mech.Engl.Ed., 37(12), 9, 1971, pp.399-420,
2016, pp.1689-1706
[13] Ranjit S.Dhaliwal, Jon G. Rokne, and Brij
[8] A.E. Green and W. Zerna, Theoretical Elas- M. Singh, Penny-shaped crack in a sphere emticity. New York, Oxford Univ. Press, 1975.
bedded in an infinite medium, Int.J.Engng Sci.
[9] E.T. Whittaker and G.N. Watson, A Course 17, 1979, pp.259-269
of Modern Analysis, Cambridge Univ. Press, [14] R.P.Srivastav and Prem Narain, Some mixed
Cambridge, 1973. 0
boundary value problems for regions with spher[10] A. Erdélyi, W. Magnus, F. Oberhettinger, ical boundary, Journal of Math. Mech., 14, 1965,
and F.G. Tricomi, Tables of Integral Transforms pp.613-627
Vol.I, New York, McGraw-Hill, 1954
0@IJTSRD— Available Online@www.ijtsrd.com—Volume-2—Issue-3—Mar-Apr 2018 Page:997