Journal of Applied Mathematics and Physics, 2014, 2, 225-234
Published Online April 2014 in SciRes. http://www.scirp.org/journal/jamp
http://dx.doi.org/10.4236/jamp.2014.25028
How to cite this paper: Aridi, M.R., et al. (2014) Mode Stresses for the Interaction between Straight and Curved Cracks
Problem in Plane Elasticity. Journal of Applied Mathematics and Physics, 2, 225-234.
http://dx.doi.org/10.4236/jamp.2014.25028
Mode Stresses for the Interaction between
Straight and Curved Cracks Problem in
Plane Elasticity
M. R. Aridi1, N. M. A. Nik Long1,2, Z. K. Eshkuvatov1, 2
1Institute for Mathematical Research, Universiti Putra Malaysia, Serdang, Selangor, Malaysia
2Department of Mathematics, Universiti Putra Malaysia, Serdang, Selangor, Malaysia
Email: radz7cr@gmail.com, n masr i@u pm.ed u. my, ezaini@upm.edu.my
Received January 2014
Abstract
In this paper, the complex variable function method is used to obtain the hypersingular integral
equations for the interaction between straight and curved cracks problem in plane elasticity. The
curved length coordinate method and suitable quadrature rule are used to solve the integrals for
the unknown function, which are later used to evaluate the stress intensity factor, SIF. Three types
of stress modes are presented for the numerical results.
Keywords
Hypersingular Integral Equation, Complex Variable Function Method, Stress Intensity Factor
1. Introduction
In order to predict the safety and stability of components, many researchers paid attention to the crack geometry.
Several types of integral equations have been developed to formulate the single crack problem such as singular
integral equation [1], Fredholm integral equation [2], hypersingular integral equation [3] and interaction integral
equation [4]. Formulation for the multiple cracks problem are also obtained in terms of singular integral equa-
tion [5], Fredholm integral equation [6] and hypersingular integral equation [7] for various set of cracks posi-
tions.
These integral equations can be solved numerically using the polynomial approximation of the dislocation
distribution. This approach was achieved by taken the crack opening displacement (COD) as the unknown and
the resultant forces as the right term in the equations [8].
In this paper, the formulation for the interaction between straight and curved cracks in term of hypersingular
integral equations is obtained using the complex potential method. Then, the cracks are mapped into a straight
line using the curved length coordinate method. In order to solve the equations numerically, the quadrature rules
are applied with M + 1 collocation points. The obtained unknown coefficients will later be used in calculating
the SIF. In this paper, we consider three types of loading modes: Mode I, Mode II and Mode III for the numeri-
cal computations.
M. R. Aridi et al.
226
2. Complex Variable Function Method
Let
( )
Φ'( )zz
φ
=
and
( )
Ψ'( )zz
ψ
=
be two complex potentials. Then the stress
(,,)
xyz
σσσ
, the resultant
function
(,)XY
and the displacement
(,)uv
are related to
( )
Φz
and
as [9]
( )
4ReΦ
xy
z
σσ
+=
(1)
() (
)
'
22
ΦΨ
y xxy
izz z
σσ σ

−+ =+

(2)
( )( )( )
'
fY iXzzzzc
φ φψ
=−+ =+++
(3)
()( )( )()
'
2G uivKzzzz
φ φψ
+=−+
(4)
whe r e G is shear modulus of elasticity,
3Kv= −
for plane strain,
(3) / (1)K vv=−+
for plane stress and
v
is
Poisson’s ratio and
zx iy= +
. The derivative in a specified direction (DISD) is defined as
{}( )( )() ()
()
,, Φ ΦΦ Ψ
dz ddz
Jz zYiXzzzzzNiT
dz dzdz

=−+=+++=+


(5)
whe r e J denotes the normal and tangential tractions along the segment
,
z zdt
+
. Note that the value of J not only
depends on the position of point z, but also on the direction of the segment
/dz dz
[3].
Let the complex potentials
( )
z
φ
and
()
z
ψ
be
( )( )
1,
2L
g t dt
ztz
φπ
=
( )( )()( )
( )
2
1
2
L LL
gtdt gtdtgtdt
ztztztz
t
ψπ

= +−

−−


∫ ∫∫
whe r e
L
denotes the crack configuration and substitute into (4). By letting
z
approaches
0
t
+
and
0
t
, which
are located on the upper and lower sides of the crack faces (see Figure 1), then using the Plemelj equations [3],
and rewriting
0
t
as t, the following result is obtained [3]
( )( )
( )
()( ) ()
21, G u tiv tikg ttL+=+
(6)
and
( )( )
( )
( )( )
()
()( )
( )
u tiv tu tiv tu tiv t
+−
+ =+−+
denotes the crack opening displacement (COD) for the curved crack. It is well known that the COD possesses
the following properties:
( )
( )
1/2
at the crack tip ,1,2
j
Bj
gtOt tBj

=−=


(7)
3. Hypersingular Integral Equation
The hypersingular integral equation for a curved crack problem is obtained by placing two point dislocations at
point
zt=
and
zt dt= +
. It is given by [3]
Figure 1. A crack configuration.
M. R. Aridi et al.
227
()
( )( )
()
( )
( )
()
( )
1 020000
2
0
11 1
.. ,,,
22
LL L
g t dt
hpKttgt dtKttgtdtN tiTttL
tt
ππ π
++=+∈
∫∫ ∫
(8)
whe r e
()( )
0
10 22
0
0
0
11
,,
()
dt dt
K ttdt dt
tt
tt
=+
() ()
( )
0
00
20 23
00
0
0
2
1
,()
tt
dt dt
dt dt
K ttdtdtdt dt
tt
tt

= +−


and
( )
gt
is the dislocation distribution along the curved. In Equ a tio n (8), the first integral with h.p. denotes the
hypersingular integral and should be defined in the sense of Hadamart finite part integral. It easy to see that for a
straight crack, Equation (8) is reduced to
( )
( )( )( )
0 00
2
0
1.. ,
L
g t dt
h pNtiTttL
tt
π
=+∈
(9)
and it can be solved analytically. Now consider the interaction between straight and curved cracks problem (see
Figure 2).
For the crack-1, if the point dislocation is placed at point
10
zt=
and
10
dz dt=
and
11
()gt
is the dislocation
doublet distribution along crac k-
1
, and the traction is applied on the
10
t
then the hypersingular integral equa-
tion for crack-
1
is
( )
( )()( )
1
11 111 1011 10
2
1 10
1..
L
g t dt
h pNtiTt
tt
π
= +
(10)
where
( )( )
11 1011 10
NtiT t+
denotes the traction influence on crack-1 caused by dislocation doublet distribution,
11
()gt
, on crack-1. The influence from the dislocation doublet distribution on crac k-2 gives
( )
( )( )
( )
( )
()
( )( )
22 2
22 212102 222 2102 2212101210
2
2 10
11 1
,,
22
LL L
g t dtKttgtdtKt tgt dtNtiTt
tt
ππ π
++=+
∫∫ ∫
(11)
whe r e
( )()
12 1012 10
NtiT t+
denotes the traction influence on crack-
1
caused by dislocation doublet distribution,
22
()gt
, on crack-2 and
() ( )
10 2
121022
10 2
2 10
2 10
11
,,
()
dt dt
Kttdt dt
tt
tt
=+
Figure 2. Straight and curved cracks in plane
elasticity with configurations on a real axis-s.
Cracks with length 2a (straight) and 2b (curved)
are known as crack-1 and crack-2 resp ectivel y.
M. R. Aridi et al.
228
()()
( )
( )
2 10
10 10
22
22 1023
21010 2
2 102 10
2
1
,.
tt
dt dt
dt dt
K ttdtdtdtdt
tt tt

= +−

−−

Note that since
2 10
0tt−≠
, all three integrals in (11) are regular and that
( )
11
gt
and
( )
22
gt
satisfy (7).
By superposition of the dislocation doublet distr ib ution,
( )
11
gt
along the crack-1 and dislocation doublet dis-
tribution,
( )
22
gt
along the crack-2, we obtained the following hypersingular integral equation for crack-1
( )
( )
( )
( )( )
( )
( )
( )
( )
( )()
12 2
1112 221210222 2102 22110110
22
1 102 10
1 11
.., ,
2
LL L
g t dtgtdt
hpKttg tK ttg tdtNtiTt
tt tt
π ππ

+++ =+

−−


∫∫ ∫
(12)
whe r e
( )( )( )( )()()
( )
1 101 1011 10121011 1012 10
NtiTt NtNtiTtiTt+= + ++
is the traction applied at point
10
t
of crack-1, which is derived from the boundary condition. The first integrals
represents the effect on crack-1 caused by the dislocation on the crack-1 itself, whereas the second three inte-
grals represent the effect of the dislocations on crack-2.
Similarly, the hypersingular integral equation for crack-2 is
( )
( )()
( )
( )
( )( )
( )( )()
2221
2 22111
1220 2222220 22 2220220
22
2201 20
111 1
.. ,,
22
LLL L
gtdtg t dt
hpKttg t dtK ttg tdtN tiTt
tt tt
πππ π
++ +=+
−−
∫∫∫ ∫
(13)
whe r e
( )( )()()( )()
( )
2 202 20212022 2021202220
NtiTt NtNtiTtiTt+= +++
is the traction applied at point
20
t
of crack-2 and
()( )
20 2
122022
20 2
2 20
2 20
11
,,
()
dt dt
Kttdt dt
tt
tt
=+
()( )
( )
( )
2 20
20 20
22
22 2023
22020 2
2 202 20
2
1
,.
tt
dt dt
dt dt
K ttdtdtdt dt
tt tt

= +−

−−

The first three integrals in (13) represent the effect on crack-2 caused by the dislocation on crack-2 itself, and
the fourth integral represents the effect of the dislocation on crack-1. Equations (12) and (13) are to be solved
for
()
11
gt
and
()
22
.gt
It is obvious that if the two cracks are far apart, the last three integrals in Equation (12)
and the fourth integral in Equ atio n (13) vanish. Then the solutions for Equations (12) and (13) approach the so-
lution for a single crack problem, and a closed form solution is available [10].
By mapping the two cracks configurations on a real axis s with an interval 2a and 2b respectively, the map-
ping functions
11
()ts
and
22
()ts
are expressed as
( )
( )
111
22
111 11
| ()
t ts
gtasH s
== −
(14)
( )
( )
222
22
22222
| ()
t ts
gtbsHs
=
= −
(15)
whe r e
1 111 112 1
()() ()
H sHsiHs= +
and
2 2212222
()() ()H sHsiHs
= +
.
In solving the integral equations, we used the following integration rules [11], for the hypersingular and regu-
lar integrals, respectively,
( )
( )( )
( )
( )
22
00
21
0
1,
M
a
jj
aj
a sGsdsWsG ssa
ss
π
+
=
= <
(16)
( )
( )
( )
( )
1
222 20
1
11,,
2
aM
jj
j
a
a sGsdsa sGssa
M
π
+
=
−=− <
+
(17)
M. R. Aridi et al.
229
whe r e G(s) is a given regular function,
M
,
0
a cos,1,2,3,,1,
2
jj
j
ssj M
M
π

=== …+

+

and
( )
( )
0
00
21,
2
Mn
j jn
n
s
W snVU
Ma
=

=−+

+
whe r e
()
1
sin sin.
22
n
j
nj
j
VMM
π
π
+


=

++


Here
( )
n
Ut
is a Chebyshev polynomial of the second kind, defined by
( )
( 1)
sin,cos .
sin
n
n
Ut t
θθ
θ
+

= =


1()Hs
and
2
()Hs
can be evaluated using
( )
11
0
,
M
nn
n
s
H scUsa
a
=

= ≤


(18)
and
( )
22
0
,
M
nn
n
s
H scUsb
b
=

= ≤


(19)
whe r e
( )
1
1 11
1
2,
2
Mn
nj
j
cVH s
M
+
=
=+
(20)
( )
1
2 22
1
2
2
Mn
nj
j
cVH s
M
+
=
=+
(21)
and
( )
11
Hs
and
( )
22
Hs
are defined from (14) and (15) respectively.
4. Numerical Results
The stress intensity factor (SIF) of inner and outer cracks can be calculated respectively
()( )
'
121
2 lim
jj
jAj
AA
Att
KKiKttg t
π
=−=−
(22)
and
()( )
'
12 2
2 lim
jj
jBj
BB
Btt
KKiKt tg t
π
=−=−
(23)
where
( )
'
1
gt
and
( )
'
2
gt
can be obtained by using Equations (12) and (13) simultaneously.
As the two cracks are far apart, the formulations in Equatio ns (12) and (13) become an equation for a straight
and a curved crack respectively. For the straight crack with length 2a, we obtain
K1.000 πa=
which is simi-
lar to the exact solution
Ka
π
=
. Whereas, for the curved crack, we compare the result of the curved crack
with the exact solution using the remote traction
1
xy
σσ
∞∞
= =
, given by [9]
( )( )
1/21/2
12
22
cos sin
22
,
1 sin1 sin
22
aa
aa
KK
aa
ππ
 
 
 
= =
 
++
 
 
(24)
M. R. Aridi et al.
230
The numerical results are tabulated in Table 1. It can be seen that maximum error is not more than 1.0% and
the results show very good agreement.
4.1. Example 1: Mode I
Consider the traction applied is
1xy
p
σσ
∞∞
= =
and the calculated results for SIF at the crack tips
121
,,AAB
and
2
B
are respectively expressed as
()
1
/,,1,2
jj
iA iA
KFb apaij
π
= =
( )
1
/,,1,2
jj
iB iB
KFb apaij
π
= =
Figure 3 shows the nondimensional SIF for a curved crack whe n
0.8,0.9,1.0
c
a=
. It is found that the real
Table 1. The SIF for single curved crack: A comparison be-
tween exact and numerical results.
σ
1
1A
K
1
1A
K
(exact)
1
2A
K
1
2A
K
(exact)
0 0.9972 0.9956 0.0435 0.0413
10 0.9887 0.9840 0.0 865 0.0777
20 0.9748 0.9677 0.1 283 0.1091
30 0.9560 0.9493 0.1 686 0.1584
40 0.9326 0.9308 0.2 068 0.2046
50 0.9053 0.9133 0.2 426 0.2420
60 0.8746 0.8966 0.2 758 0.2560
70 0.8413 0.8788 0.3 062 0.3058
80 0.8059 0.8240 0.3 338 0.3411
90 0.7690 0.7786 0.3 806 0.3798
Figure 3. Nondimensional SIF when θ is changing.
M. R. Aridi et al.
231
parts of SIF at the left crack tip
1
B
is equal to its right crack tip
( )
12
21 1BB
BF F=
, whereas the imaginary parts
give opposite sign
()
21
22BB
FF= −
. The effect of the distance between both cracks is also studied and the result
is shown in Fig ure 4. The nondimensional SIF for the straight crack is decreased at the same rate when
0
10
θ
<
whereas increased with different rates when
0
10
θ
>
and the SIF becomes higher as the two cracks are close
together.
4.2. Example 2: Mode II
The traction applied for this mode is
2xy
p
σσ
∞∞
= =
and the calculated results for SIF at the crack tips
121
,,AAB
and
2
B
are respectively expressed as
( )
2
/,,1,2
jj
iA iA
KFb apaij
π
= =
( )
2
/,,1,2
jj
iB iB
KFb apaij
π
= =
Figure 5 shows the nondimensional SIF for a curved crack when
0.8,0.9,1.0
c
a=
and the effect of the dis-
tance between both cracks is shown in Figure 6. The nondimensional SIF for the straight crack is decreased
with different rates at the considered domain. As the two cracks are close together, the nondimensional SIF at
the crack tip is higher.
4.3. Example 3: Mode III
Consider the traction applied is
xy
q
σσ
∞∞
= =
and the calculated results for SIF at the crack tips
121
,,AAB
and
2
B
are respectively expressed as
( )
/,,1,2
jj
iA iA
KFb aqaij
π
= =
()
/,,1,2
jj
iB iB
KFb aqaij
π
= =
Figure 7 shows the nondimensional SIF for a curved crack when
0.8,0.9,1.0
c
a=
. It is found that the imagi-
Figure 4. Nondimensional SIF when θ is changing.
M. R. Aridi et al.
232
Figure 5. Nondimensional SIF when θ is changing.
Figure 6. Nondimensional SIF when θ is changing.
nary parts of SIF at the left crack tip
1
B
is equal to its right crack tip
12
22 2
()
BB
BF F=
, whereas the real parts
give opposite sign
21
11
()
BB
FF= −
. The effect of the distance between both cracks is shown in Figure 8. For
0
15
θ
<
the nondimensional SIF is decreased at the same rate but later increased at the different rates.
5. Conclusion
In this paper, the different types of loading modes have been applied to the interaction between straight and
curved cracks problem in plane elasticity. As the result, we obtained different results of nondimensional SIF due
to the different loading modes. We also observed that the SIF increases as both cracks become closer. For a
M. R. Aridi et al.
233
Figure 7. Nondimensional SIF when θ is changing.
Figure 8. Nondimensional SIF when θ is changing.
symmetry shape crack problem, the real parts of SIF for the left side of the crack is equal to its right side while
the imaginary parts give the opposite sign.
Acknowledgements
The second author would like to thank Ministry of Science, Technology and Innovation (MOSTI), Malaysia for
the Science Fund, Vot No. 5450657.
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