Mechanics and Mechanical Engineering

Vol. 21, No. 2 (2017) 363–377

⃝
c Lodz University of Technology

Axisymmetric Torsion of an Internally Cracked Elastic Medium

by Two Embedded Rigid Discs

Fateh Madani
Belkacem Kebli
Department of Mechanical Engineering
Ecole Natioanle Polytechnique
El-Harrach, Algiers, Algeria
madanifateh1984@yahoo.com
belkacem.kebli@g.enp.edu.dz

Received (18 January 2017)

Revised (15 March 2017)
Accepted (7 May 2017)

The present work aims to investigate a penny–shaped crack problem in the interior of
a homogeneous elastic material at the symmetry plane, under an axisymmetric torsion by
two circular rigid discs symmetrically located in the elastic medium. The two discs rotate
with the same angle in the different direction about the axis passing through their centers.
The general solution of this problem is obtained by using the Hankel transforms method.
The corresponding doubly mixed boundary value problem associated with the rigid disc
and the penny–shaped is reduced to a system of dual integral equations, which are
transformed, to a Fredholm integral equations of the second kind. Using the quadrature
rule, the resulting system is converted to a system of infnite algebraic equations. The
variation in the displacement, stress and stress intensity factor are presented for some
particular cases of the problem.
Keywords: elastic medium, axisymmetric torsion, penny–shaped crack, dual integral
equations, Fredholm integral equations, stress intensity factor.

1. Introduction
The problems relating to the static and dynamic torsional of an elastic half–space
under the action of rigid discs are of great interest in mechanical Engineering,
civil engineering, geomechanics and applied mathematics. The problem is a mixed
boundary–value problem with the stress and the displacement boundary conditions
of the rigid disc and the crack planes. Many different studies conducted by many
different researchers who have studied the problem associated with the torsional
rotation of a rigid disc in bonded contact with the elastic medium. The distribution
of stress in the interior of a semi-infinite elastic medium is determined when a load
is applied to the surface by means of a rigid disc was first considered by Reissner
364 Madani, F. and Kebli, B.

and Sagoci [1]. The same problem was solved by Sneddon [2] by a different method.
He used the Hankel transforms method for reduction the problem to a pair of dual
integral equations. Collins [3] treated the torsional problem of an elastic half–
space by supposing the displacement at any point in the half–space to be due to
a distribution of wave sources over the part of the free surface in contact with the
disc.
The solution of the forced vibration problem of elastic layer of finite thickness
when the lower face is either stress free or rigidly clamped was given by Gladwell
[4] who reduced the mixed boundary value problem to a Fredholm integral equation
by Noble’s method [5]. Pak and Saphores [6] provided an analytical formulation for
the general torsional problem of a rigid disc embedded in an isotropic half–space.
The quadrature numerical was used for solving the obtained Fredholm integral
equation. Besides, Bacci and Bennati [7] employed the Hankel transforms method
and the power series method with the truncation of the second term to consider the
torsional of circular rigid disc adhered to the upper surface of an elastic layer fixed
to an undefonnable support.
More recently, Singh et al. [8] studied the torsional of a non–homogeneous,
isotropic, half–space by rotating a circular part of its boundary surface. The solution
of the corresponding triple integral equations was reduced to the solution of two
simultaneous integral equations. Cai and Zue [9] discussed the torsional vibration of
a rigid disc bonded to a poro–elastic multilayered. They used the Hankel transforms
and transferring matrix method. Yu [10] studied the forced torsional oscillations
inside the multilayered solid. The elastodynamic Green’s function of the center
of rotation and a point load method were used to solve the problem. Pal and
Mandal [11] considered the forced torsional oscillations of a transversely isotropic
elastic half space under the action of an inside rigid disc. The studied problem was
transformed to dual integral equations system, which was reduced to a fredholm
integral equation. A similar problem with the rocking rotation was solved later on
by Ahmadi and Eskandari [12]. They used an appropiate Green’s function to write
the mixed boundary–value problem posed as a dual integral equation.
The torsional of elastic layers with a penny shaped crack was considered by
some researchers. Sih and Chen [13] studied the problem of a penny-shaped crack
in layered composite under a uniform torsional stress. The displacement and stress
fields throughout the composite were obtained by solving a standard Fredholm in-
tegral equation of the second kind. Low [14] investigated a problem of the effects of
embedded flaws in the form of an inclusion or a crack in an elastic half space sub-
jected to torsional deformations. The corresponding Fredholm integral equations
were solved numerically by quadrature approach. The same method was used by
Dhawan [15] for solving the problem of a rigid disc attached to an elastic half–space
with an internal crack. By using Hankel and Laplace transforms and taking nu-
merical inversion of Laplace transform, Basu and Mandal [16] treated the torsional
load on a penny–shaped crack in an elastic layer sandwiched between two elastic
half–spaces.
In this paper, we investigate the problem of a penny–shaped crack in the in-
terior of a homogeneous elastic medium under an axisymmetric torsion applied to
two symmetrical rigid discs. With the aid of the Hankel integral transformation
method. The mixed boundary–value problem is written as a system dual of in-
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 365

tegral equations. The corresponding system of Fredholm integral equations was

approached by sets of linear equations. After getting the unknown coefficients of
this system we obtain numerical results and display curves according to certain
pertinent parameters.

2. Formulation of the problem

We consider the axisymmetric torsion of two rigid coaxial discs of a radius b sym-
metrically located at z = ±h planes in an infinite, isotropic and homogeneous elastic
medium, containing a penny–shaped crack at the symmetry plane z = 0. The faces
of the crack are supposed stress free while the dics rotate with an equal angle ω,
but of opposite sign, about the z axis passing through their centers as shown in
Fig. 1. Due to symmetry about the z = 0 plane, it is sufficient to consider the

Figure 1 Geometry and coordinate system

problem in the upper half–space only z ≥ 0. For the static axisymmetric torsion
of a homogeneous isotropic material and linear elastic behaviour, the only nonzero
displacement component is uθ (r, z). Then the only non-zero components stresses
are related to the displacement component by:

∂uθ ∂ ( uθ )
τθz = G , τθr = Gr (1)
∂z ∂r r
where G is the shear modulus of the material. The displacement satisfies the fol-
lowing differential equation:

∂ 2 uθ ∂uθ uθ ∂ 2 uθ
+ − + =0 (2)
∂r2 r∂r r2 ∂z 2
By means of the Hankel’s transformation integral and its inverse given by [19]:
∫ ∞
F (λ, z) = f (r, z)rJ1 (λr)dr
0
366 Madani, F. and Kebli, B.

and ∫ ∞
f (r, z) = F (λ, z)λJ1 (λr)dλ
0
where J1 is the Bessel function of the first kind of order one. The general solution
of the above differential equation is furnished for the regions I(0 ≤ z ≤ h) and
II(z ≥ h) as shown in fig.1 by
∫ ∞
[ ]
Ai (λ)e−λz + Bi (λ)eλz J1 (λr)dλ
(i)
uθ (r, z) = i = 1, 2 (3)
0

where Ai and Bi are unknown functions.

3. Boundary and continuity conditions

Let us assume the contact between the disc and the elastic half–space is perfectly
bonded all along their common interface. We consider the symmetry plane condi-
tion, the regularity conditions at infinity, the boundary and continuity conditions
at z = 0 and z = h, as shown in the following. The regularity conditions at infinity
are given by:
(i) (i)
lim uθ (r, z) = 0 lim τθz (r, z) = 0 (4)
r,z→∞ r,z→∞

The boundary conditions of the problem are:

(1)
τθz (r, 0+ ) = 0 r<a (5)
(1)
uθ (r, 0) = 0 r≥a (6)
uθ (r, h− )
(2) (1)
uθ (r, h+ ) = = ωr r≤b (7)
The continuity conditions of the problem in the planes z = h can be written as:

τθz (r, h+ ) − τθz (r, h− ) = 0

(2) (1)
r>b (8)
uθ (r, h− )
(2) (1)
uθ (r, h+ ) − =0 r>b (9)
With the regularity conditions at infinity given by Eq. (4), the expressions of
displacements and stresses in the two regions take the following forms:
∫ ∞
[ ]
A1 (λ)e−λz + B1 (λ)eλz J1 (λr)dλ
(1)
uθ (r, z) = (10)
0
∫ ∞
[ ]
λ −A1 (λ)e−λz + B1 (λ)eλz J1 (λr)dλ
(1)
τθz (r, z) = G (11)
∫ ∞0
[ ]
A2 (λ)e−λz J1 (λr)dλ
(2)
uθ (r, z) = (12)
0
∫ ∞
[ ]
λ A2 (λ)e−λz J1 (λr)dλ
(2)
τθz (r, z) = −G (13)
0

The unknown functions A1 (λ), B1 (λ) and A2 (λ) can be determined from the bound-
ary and continuity conditions. The boundary and continuity conditions Eq. (7) and
Eq. (9) show that:
uθ (r, h+ ) − uθ (r, h− ) = 0
(2) (1)
(14)
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 367

From the above condition, we obtain

A2 (λ) = A1 (λ) + B1 (λ)e2λh (15)
The mixed boundary conditions Eqs. (6–8) are converted to the following system
of dual integral equations for obtained the two unknowns A1 and B1 :
∫ ∞
λ[B1 (λ) − A1 (λ)]J1 (λr)dλ = 0 0≤r<a (16)
∫0 ∞
[A1 (λ) + B1 (λ)]J1 (λr)dλ = 0 r≥a (17)
∫0 ∞
[A1 (λ)e−λh + B1 (λ)(eλh )]J1 (λr)dλ = ωr 0≤r≤b (18)
∫ ∞
0

λB1 (λ)eλh J1 (λr)dλ = 0 r>b (19)

0

4. Reduction of the problem to a system of Fredholm integral equations

In reducing the system of dual equations to a system of Fredholm integral equations,
we begin by introducing the the auxiliary functions ϕ(t) and ψ(t) such that:
√ ∫ a√
A1 (λ) + B1 (λ) = λ tϕ(t)J 32 (λt)dt (20)
0
√ ∫ b √
B1 (λ)eλh = λ tψ(t)J 12 (λt)dt (21)
0
witch give us:
√ ∫ a √ √ ∫ b √
A1 (λ) = λ tϕ(t)J 32 (λt)dt − e−λh λ tψ(t)J 21 (λt)dt (22)
0 0

−λh
√ ∫ b √
B1 (λ) = e λ tψ(t)J 12 (λt)dt (23)
0

where ϕ(t) and ψ(t) are continuous unknown functions of t, defined over two inter-
vals 0 ≤ t < a and 0 ≤ t ≤ b respectively. This choice implies that the homogeneous
equations Eq. (17) and Eq. (19) are identically satisfied while the equations Eq.
(16) and Eq. (18) lead to the Fredholm’s integral equations.
Substituting A1 (λ) and B1 (λ) in the equations Eq. (16)and Eq. (18), we get:
∫ a√ ∫ ∞
3
tϕ(t)dt λ 2 f11 (λ)J 32 (λt)J1 (λr)dλ
0 0
(24)
∫ √ b ∫ ∞
3
+ tψ(t)dt λ 2 f12 (λ)J 12 (λt)J1 (λr)dλ = 0 r<a
∫ a 0√ ∫ ∞0 √
tϕ(t)dt λf21 (λ)J 32 (λt)J1 (λr)dλ
0 0
(25)
∫ b √ ∫ ∞ √
+ tψ(t)dt λf22 (λ)J 12 (λt)J1 (λr)dλ = ωr r<b
0 0
368 Madani, F. and Kebli, B.

where:
f11 (λ) = 1 f12 (λ) = −2e−λh f21 (λ) = e−λh f22 (λ) = 1 − e−2λh
Eq. (24) can be converted to the Abel integral equation by means of the relation:
1 d 2
λJ1 (λr) = [r J2 (λr)]
r2 dr
and the integral formula:
∫ ∞√ { √ 3
2 √t 2
r 2 −t2
t<r
λJ 32 (λt)J2 (λr)dλ = πr 2

0 0 t>r
we obtain Abel equation corresponding to equation Eq. (24):
√ ∫ r ∫ a√ ∫ ∞
2 t2 ϕ(t) 1
√ dt + r2 tϕ(t)dt λ 2 (f11 (λ) − 1)J 32 (λt)J2 (λr)dλ
π 0 r 2 − t2 0 0
(26)
∫ b√ ∫ ∞
1
+r2 tψ(t)dt λ 2 f12 (λ)J 12 (λt)J2 (λr)dλ = 0 r<a
0 0

Next, we invert the last equation by applying the Abel’s transform formula:
∫ r ∫
f (t) 2 d t rg(r)
√ dt = g(r) then f (t) = √ dr
0 r 2 − t2 π dt 0 t2 − r 2
to obtain:
√ ∫ ∫ ∫
2 d t
r3 a √ ∞
1
2
t ϕ(t) = √ [− δϕ(δ)dδ λ 2 (f11 (λ) − 1)J 32 (λδ)J2 (λr)dλ
π dt 0 t − r2
2
0 0
(27)
∫ b √ ∫ ∞
1
− δψ(δ)dδ λ 2 f12 (λ)J 12 (λδ)J2 (λr)dλ]dr r<a
0 0

For the left hand side of the above equation, the integral is further simplified by
using the following relationship:
√ ∫
2 d t r3 √ 5
√ J2 (λr)dr = λt 2 J 23 (λt)
π dt 0 t −r
2 2

we obtain the first Fredholm integral equation of second kind:

√ ∫ a√ √ ∫ b√
ϕ(t) + t δϕ(δ)K(t, δ)dδ + t δψ(δ)L(t, δ)dδ = 0, r<a (28)
0 0

where:
∫ ∞
K(t, δ) = λ(f11 (λ) − 1)J 23 (λt)J 23 (λδ)dλ
∫ 0
∞
L(t, δ) = λf12 (λ)J 32 (λt)J 21 (λδ)dλ
0
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 369

Following the same procedure as before to reduce Eq. (24) to the second Fredholm
integral equation. Using the formula:
∫ ∞√ √
2t √ 1
π r (r 2 −t2 ) t<r
λJ 12 λtJ1 (λr)dλ = (29)
0 0 t>r

we obtain the following Abel type equation:

√ ∫ ∫ b√ ∫ ∞√
1 2 r tψ(t)
√ dt + tψ(t)dt λ(f22 (λ) − 1)J 12 (λt)J1 (λr)dλ
r π 0 r 2 − t2 0 0
(30)
∫ a√ ∫ ∞√
+ tϕ(t)dt λf21 (λ)J 32 (λt)J1 (λr)dλ = ωr r<b
0 0

Now, we invert the above equation by applying the Abel’s transform formula to get:
√ ∫
2 d t r2
tψ(t) = √ ωr
π dt 0 t2 − r2
∫ b√ ∫ ∞√
− δψ(δ)dδ λ(f22 (λ) − 1)J 21 (λδ)J1 (λr)dλ (31)
∫0 a √ ∫0 ∞ √
− δϕ(δ)dδ λf21 (λ)J 32 (λδ)J1 (λr)dλ]dr r < b
0 0

Using the following relationship:

∫
d t r3
√ dr = 2t2
dt 0 (t2 − r2 )
(32)
√ ∫
2 d t 2
r J (λr) √
√ 1 dr = t λtJ 12 (λt)
π dt 0 (t2 − r2 )
we finally get the second Fredholm integral equation of second kind:
√ ∫ a√ √ ∫ b√ 4ω
ψ(t)+ t δϕ(δ)M (t, δ)dδ+ t δψ(δ)N (t, δ)dδ = √ t, 0 < t < b (33)
0 0 2π
with the kernel:
∫ ∞
M (t, δ) = λf21 (λ)J 12 (λt)J 23 (λδ)dλ (34)
∫ 0∞
N (t, δ) = λ(f22 (λ) − 1)J 12 (λt)J 21 (λδ)dλ (35)
0

The system (28) and (31) can be written in the dimensionless form as follows.
By putting: {
δ = as, 0 < δ < a; t = au 0 < t < a
δ = bs, 0 < δ < b; t = bu 0<t<b
370 Madani, F. and Kebli, B.
√
2π
Next, we multiply the above two equations of the system , respectively by 4aω ϕ(au)
√
2π
and 4bω ψ(bu) and using the following substitutions:

{ √
2π
√
2π
Φ(u) = 4aω ϕ(au) Ψ(u) = 4bω ψ(bu) (36)
c = b λ = xb H = hb
a

hence:
∫ ∫ 1
√ 1 √ 1 √ √
2
Φ(u) + c u sΦ(s)K(u, s)ds + √ u sΨ(s)L(u, s)ds = 0, u < 1 (37)
0 c 0
∫ 1 ∫ 1
5√ √ √ √
Ψ(u) + c 2 u sΦ(s)M (u, s)ds + u sΨ(s)N (u, s)ds = u, u < 1 (38)
0 0

where:

K(u, s) = 0
∫ ∞
L(u, s) = xf12 (x)J 23 (xcu)J 12 (xs)dx
∫ ∞
0

M (u, s) = xf21 (x)J 12 (xu)J 23 (xcs)dx

∫ ∞0

N (u, s) = x(f22 (x) − 1)J 12 (xu)J 12 (xs)dx

0

5. Numerical results and discussion

The quadrature rule is used in evaluating the Fredholm integral equations given by
Eq. (33) and Eq. (34). Let us divide the interval [0, 1] into N subintervals of length
1 2m−1 2n−1
N , so that u = um = 2N , s = un = 2N m, n = 1, 2, ..., N . For simplicity, we
adopt the following notations:

Φ(um ) = Φm , Ψ(um ) = Ψm
L(um , un ) = Lmn , M (um , un ) = Mmn , N (um , un ) = Nmn

we get the following system of algebraic equations for determination of the coeffi-
cients Φ and Ψ:

1 √ ∑N
√
Φm + √ um un Ψn Lmn = 0 m = 1, 2, ..., N (39)
N c n=1
5
c2 √ ∑N
√ 1√ ∑N
√
Ψm + um un Φn Mmn + um un Ψn Nmn = um (40)
N n=1
N n=1
m = 1, 2, ..., N
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 371

where:
Kmn = 0
∫ ∞
Lmn = xf12 (x)J 32 (xcum )J 12 (xun )dx
∫0 ∞
Mmn = xf21 (x)J 12 (xum )J 23 (xcun )dx
∫ 0∞
Nmn = x(f22 (x) − 1)J 12 (xum )J 12 (xun )dx
0

Next, we evaluate numerically the infinite integral L, M , and N by simpson rule.

After solving the above system, the unknown coeffcients can be obtained. Then we
get the numerical approximation of the unknown functions A1 , B1 and A2 given by
Eq. (22), Eq. (23) and Eq. (15):

4b2 ω √ ∑ √
N
um [c 2 Φm J 32 (xcum ) − e−xH Ψm J 12 (xum )]
5
A1 (x) = √ x (41)
N 2π m=1

4b2 ω −xH √ ∑ √
N
B1 (x) = √ e x um Ψm J 12 (xum ) (42)
N 2π m=1
and A2 (x) = A1 (x) + B1 (x)e2λh (43)

5.1. Stress intensity factors

The stress intensity factors at the edge of the crack and at the rim of disc are defined
respectively as:
√ (1)
a
KIII = limr→a+ 2π(r − a)τθz (r, z)|z=0 (44)
√ (1)
KIII = limr→b− 2π(b − r)τθz (r, z)|z=h
b
(45)
On the plane z = 0 for r ≥ a, the expression of stress is given by:
∫ ∞ ∫ a√
(1) 3
τθz (r, 0) = G λ 2 [− tϕ(t)J 23 (λt)dt
0 0
(46)
∫ b √
+2e−λh tψ(t)J 12 (λt)dt]J1 (λr)dλ
0

On the plane z = h, the expression of stress is given by:

∫ ∞ ∫ a√
λ 2 [−e−λh
(1) 3
τθz (r, h) = G tϕ(t)J 32 (λt)dt
0 0
(47)
∫ b √
+(1 + e−2λh ) tψ(t)J 12 (λt)dt]J1 (λr)dλ
0

Using the relation:

1 d
J1 (λR) = − J0 (λR)
λ dR
372 Madani, F. and Kebli, B.

we obtain:
∫ a √ ∫ ∞
(1) 1
τθz (r, 0) = G tϕ(t)dt λ 2 J 32 (λt)J0 (λr)dλ
0 0
(48)
∫ b √ ∫ ∞
λ 2 e−λh J 12 (λt)dtJ1 (λr)dλ
3
+2G tψ(t)dt
0 0
∫ b √ ∫ ∞ ∫ a √
(1) 1
τθz (r, h) = −G tψ(t)dt λ J 12 (λt)J0 (λr)dλ − G
2 tϕ(t)dt
0 0 0
(49)
∫ ∞ ∫ b √ ∫ ∞
e−λh λ 2 J1 (λr)dλ + G λ 2 e−2λh J 12 (λt)J1 (λr)dλ
3 3
tψ(t)dt
0 0 0

We use the following asymptotic behavior of the Bessel function of the first kind,
for large values of λ: √
2 π π
Jν (λ) ≃ cos(λ − ν − )
λπ 2 4
then we get:
√ √
2 2
J3/2 (λt) ≃ cos(λt − π) = − cos(λt)
λtπ λtπ
√ √
2 π 2
J1/2 (λt) ≃ cos(λt − ) = sin(λt)
λtπ 2 λtπ
Next, we use the following integral formula for the first infinite integral in the right
part of the Eq. (44) and Eq. (45) respectively:
∫ ∞ {
√ 1 r>t
cos(λt)J0 (λr)dλ = r 2 −t2
0 0 r<t
∫ ∞ {
0 r>t
sin(λt)J0 (λr)dλ = √ 1
0 t2 −r 2
r<t
we obtain: √ ∫ a
(1) 2 d ϕ(t)
τθz (r, 0) =− G √ dt + R1 (r) (50)
π dr 0 r 2 − t2
√ ∫ b
(2) 1 d ψ(t)
τθz (r, h) =− G √ dt + R2 (r) (51)
2π dr 0 t2 − r 2
where:
∫ b √ ∫ ∞
λ 2 e−λh J 21 (λt)dtJ1 (λr)dλ
3
R1 (r) = 2G tψ(t)dt
∫0 a √ ∫0 ∞
e−λh λ 2 J1 (λr)dλ
3
R2 (r) = −G tϕ(t)dt
0 0
∫ b √ ∫ ∞
λ 2 e−2λh J 12 (λt)J1 (λr)dλ
3
+G tψ(t)dt
0 0
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 373

Now integrating by parts, we get:

√ ∫ a
(1) G 2 aϕ(a) tϕ′ (t)
τθz (r, h1 ) = √ [ √ − √ dt] + R1 (r) (52)
π r r2 − a2 0 r r −t
2 2

∫
(1) G bψ(b) b
1 tψ ′ (t)
τθz (r, h2 ) =√ [ √ − √ dt] + R2 (r) (53)
2π r b2 − r2 r r t2 − r 2
The stress intensity factors at r = a and at r = b, may be calculated as:
√
√ G 2 aϕ(a)
KIII = lim+ 2π(r − a) √ ( √
a
) (54)
r→a π r r2 − a2
√ G bψ(b)
b
KIII = lim− 2π(b − r) √ ( √ ) (55)
r→b 2π r b2 − r2
By using the following transformations:

4aω 4bω r
ϕ(a) = √ ΦN ψ(b) = √ ΨN ρ=
2π 2π b

we obtain:
√
4Gω a
a
KIII = √ ΦN (56)
π
√
4Gω b
b
KIII = √ ΨN (57)
π

Fig. 2 shows the results of the effect of the normalized crack size a/b on the stress
a
intensity factor KIII defined by Eq. (56) for different disc locations H = 1; 0.75; 0.5
and 0.25. It is observed that the values of the stress intensity factor versus a/b
increase, attain its maximum values and then decrease to zero. The effect of the
distance between the crack and the disc H on the stress intensity factor is also shown
in Fig. 2. The increase of the height H induces the decrease of stress intensity factor
for all values of a/b.
b
Fig. 3 illustrates the variation of the normalized stress intensity factor KIII at the
edge of the rigid inclusion defined by Eq. (57) versus a/b for H = 1; 0.75; 0.5 and
0.25. Relatively, small variation for smaller values of a/b and considerable variation
for larger values of a/b are observed. Also, it can be seen that the interaction
between the crack and the rigid discs is greater when the discs are closer to the
b
crack. In addition to the interaction, the stress intensity factor KIII decreases as
the crack radius increases.

5.1.1. Displacement and stress fields

By substituting the Eqs. (37)–(39) in the expressions of the displacements and the
stresses Eqs. (10)–(132), we get the numerical results of displacements and stresses
for the two regions.
374 Madani, F. and Kebli, B.

2
H=1
1.8 H=0.75
H=0.5
1.6 H=0.25

1.4

1.2
KIaI I /ω.a

0.8

0.6

0.4

0.2

0
0 0.5 1 1.5 2 2.5 3 3.5 4
a/b

a with a/b
Figure 2 Variation of the normalized stress intensity factor at the edge of the crack KIII

3
H=1
H=0.75
2.8 H=0.5
H=0.25

2.6
KIbI I /ω.a

2.4

2.2

1.8
0 0.5 1 1.5 2 2.5 3 3.5 4
a/b

b with
Figure 3 Variation of the normalized stress intensity factor at the edge of the rigid disc KIII
a/b

(i)
The results for the variation of the normalized displacement uθ (ρ, ξ)/ωa and
stress τ (i) (ρ, ξ)/ωa with ρ = r/b are shown graphically in Fig. 4 to Fig. 5 for
the different values of the dimensionless axial distances ξ = z/a. For each region,
five different axial distances are selected as I(ξ = 0; H/4; H/2; 3H/4; H)
and I(ξ = H; 5H/4; 3H/2; 7H/4; 2H), with the particular values of the height
H = 1 and the dimensionless disc size c = 1.
The variation of the normalized displacements are shown in Fig. 4 and Fig. 5. We
notice that the displacements in the two regions increase at first, reach maximum
values at ρ = 1 then decrease out of the disc band with increasing ρ.
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 375

C= 1; H= 1
0.8
ζ=0
0.7 ζ= H/4
ζ= H/2
0.6 ζ= 3H/4
ζ= H
0.5
Uθ1 (ρ, ζ)/ω.a

0.4

0.3

0.2

0.1

0
0 0.5 1 1.5 2 2.5 3 3.5 4
ρ

Figure 4 Tangential displacement u1θ versus ρ for various ξ, 0 ≤ z ≤ h

C= 1; H= 1
0.8
ζ=H
0.7 ζ=5H/4
ζ=3H/2
0.6 ζ=7H/4
ζ=2H
0.5
Uθ2 (ρ, ζ)/ω.a

0.4

0.3

0.2

0.1

0
0 0.5 1 1.5 2 2.5 3 3.5 4
ρ

Figure 5 Tangential displacement u2θ versus ρ for various ξ, z ≥ h

The distribution of the shear stresses in the elastic medium is also discussed and
shown in Fig. 6 and Fig. 7. The stresses are initially rise, attain its maximum
values and with the increase in the value of ρ the stresses go on decreasing.

6. Conclusion
In this study, the axisymmetric torsion problem of two rigid discs symmetrically
located embedded in the interior of a homogeneous elastic material is analytically
addressed. The medium is weakened by a penny-shaped crack located parallel to
the discs at the symmetry plane. Using the Hankel integral transformation method,
376 Madani, F. and Kebli, B.

C= 1; H= 1
3
ζ=0
ζ= H/4
2.5 ζ= H/2
ζ= 3H/4
ζ= H
2
(ρ, ζ)/Gω.a

1.5
τθz
1

0.5

0
0 0.5 1 1.5 2 2.5 3 3.5 4
ρ

1 versus ρ for various ξ, 0 ≤ z ≤ h

Figure 6 Shear stress τθz

C= 1; H= 1
0.5
ζ=H
ζ=5H/4
0
ζ= 3H/2
ζ=7H/4
−0.5 ζ=2H
(ρ, ζ)/Gω.a

−1

−1.5
τθz
2

−2

−2.5

−3
0 0.5 1 1.5 2 2.5 3 3.5 4
ρ

2 versus ρ for various ξ, z ≥ h

Figure 7 Tangential displacement τθz

the doubly mixed boundary value problem is reduced to a system of dual integral
equations, which are transformed, to a Fredholm integral equations system of the
second kind. The presented graphs show the variation of the displacements, the
stresses and the stress intensity factor at the edge of the crack and at the rim of
the disc for some dimensionless parameters. The numerical results show that the
discontinuities around the crack and the inclusion cause a large increase in the
stresses which decay with distance from the disc-loaded. Furthermore, it can be
seen the dependence of the stress intensity factor on the crack size and the distance
between the crack and the disc.
 Axisymmetric Torsion of an Internally Cracked Elastic Medium ... 377

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