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The problem of a concentric penny-shaped crack of mode III in a nonhomogeneous finite cylinder
Engineering Fracrure Mechanics Vol. 44, No. 2, pp. 289-295, 1993 Printed in Great Britain.
0013-7944/93 $6.00 + 0.00 Pergamon Press Ltd.
THE PROBLEM OF A CONCENTRIC PENNY-SHAPED CRACK OF MODE III IN A NONHOMOGENEOUS FINITE CYLINDER LIANG BAN and X. S. SHANG Beijing Institute of Civil Engineering, P.R.C. Abstrat-By using the Hankel transform and the Fourier series, the stress intensity factor of a concentric penny-shaped crack in a finite long cylinder of a finite radius for mode III is determined. It is proved that the concentric penny-shaped cracks in an infinite long cylinder and in an infinite circular plate am the special cases of the problem considered here.
1. INTRODUCTION IN THE past, several solutions of an infinite elastic body with a central crack have been obtained [l-3]. Zhang has given a series of analytical results of the central crack, the edge crack and the concentric penny-shaped crack for mode III in a homogeneous and a nonhomogeneous finite body in recent years [4-6]. The problem of a nonhomogeneous finite cylinder with a concentric penny-shaped crack of mode III has not yet been treated. In this paper, by using the basic theorem of the Hankel transform and the Fourier series, the problem of a nonhomogeneous, finite, long cylinder of finite radius has been discussed. The result of the problem may be given by means of a Fredholm integral equation of the second kind. The numerical solution of the stress intensity factor is computed using the standard procedure of the Fredholm integral equation and is given when the shear stress is a constant t.
2. BASIC EQUATIONS We consider a concentric penny-shaped crack of radius a in a nonhomogeneous finite cylinder of diameter 2b and length 2h, subjected to an arbitrary torque shear stress r(r), and a self-equilibrium system on its crack face (Fig. 1). In discussing the problem of a cylinder with a concentric penny-shaped crack, it is convenient to use cylindrical polar coordinates, assuming the origin of the coordinates coincides with the center of the cylinder, the X and Y axes are in the plane of the penny-shaped crack and the Z axis is perpendicular to the XY-plane. The position of a point in the cracked cylinder may be described by the distance r in the radial direction, the angle 8 between r and OX, and the distance Z in the axial direction. For this problem the displacement components in the r and z directions vanish everywhere and the only nonzero displacement component in the 8 direction is a function of r and z, i.e.
u, = u, = 0
u, = U&, The corresponding component We@,2):
stress components rtJz(r9r)=11
2).
can be expressed au&+, 2) az
where the shear modulus ~1= harm and h is a constant, m Z 0. 289
in terms of the displacement
LIANG BAN and X. S. SHANG
290
T
2
I+A ,/-a \\
X
--.
=-
-/
8
c N
Y
r
2a
tide-- -----*.
1
i \
2b
-
Fig. 1.
Putting zd(r, z) and zor(r, z) into the following equilibrium equation: ~~,(r,z)+~~,(r,z)+fTd(r,z)=O, the partial differential equation about Ue(r, z) can be obtained:
($+$ 1Uo(r,z)+(l+m)d,
8
Wr,z)
( ) r
=
o
.
(1)
The boundary conditions may be given as z,(r,
0) = -z(r)
0< r < a
(2)
U&,0)=0
b >r >a
(3)
o,(r,h)=O
O
(4)
U,(b,z)=O
O
(5)
It is obvious that the key of this problem is to seek a displacement function Ue(r, z) that can satisfy all of these boundary equations (2)-(j) and the partial differential equation (1). 3. THE ANALYTICAL
SOLUTION
Solving the partial differential equation (1) we can get the displacement component in terms of the Hankel transform and Fourier series, i.e. h Ue(r, z) = r +‘*
I+mdtr)dt
+
WV f n = 1,3,...
l+m12(!k!k)sin(?$)]
U&, z)
(6)
in which Js stands for the Sth order Bessel function of the first kind while Is is the Sth order modified Bessel function of the first kind. The corresponding stress components are T&(r, z) = -rmi2
Co [S
t&(r, z) = - rm’*
0
tfA(O
ch5@ -dJ ch
&
2+m,2(tr) d5
291
Concentric penny-shaped crack in a finite cylinder
where the following recursion formula [7, p. 9701 has been employed:
Putting (7) and (6) into (2) and (3) respectively, we have
s
=0
mA(~)Jl+m~2(Tr)dS 0
(10)
bar >a.
For convenience the pair of dual integral equations (9) and (10) may be rewritten as m tA(t)(th s0
A(C)r (2+m’2)J2+m,2(&) d< f
4) E
@ - 1)J,+m,2(~r)r(2+m/2)d~
(> ?Inr
mN+m/*
T&
.=,,...
s
mA(t)Jl+m,2(Tr)dt
=0
b >r
r
(2+m/2) = r*r(r),
0 < r
>a.
(11)
(12)
0
The unknown functions A(<) and B(n) in eq. (11) are not independent of eqs (5) and (6) and on application of the Fourier sine series pair:
we have (13)
B(n) =
Considering eq. (13), the dual integral equations (11) and (12) become A(0
=r*t(r)
(*+ ““*)J2+,,,,*(5r) d5 +
O
m rA(<)(th
(14)
s
mA(t)J,+m,2(tr)d5=0
bar>u.
(15)
0
Letting
s~cp(r)J,.,,,+,,,(rr)i-“+““*dS
A(5>=t1’*
0
(16)
292
LIANG BAN and X. S. SHANG
and substituting (16) into (15) and using the result [8, p. 1001
s
2
‘12
i”
r(l+m/2)~-(3/2+m/2)(~2_r2)-l~2
a5”zJ,+m:2(5r)J,+m,2+,,2(Si)d5 = 71
0
0
Irl 5,
it is not difficult to see that eq. (15) is automatically satisfied, i.e. the boundary condition (3) has been satisfied. Inserting (16) in (14) and employing [8, p. 1001
OD 5"252+m,2(rr)J,+,2+,,z(5r)de s 0
""'
=
I
and identical equations j;v(W,l~!jdC
=J;p:r&v
.,;!$)J;:;
we can obtain
‘[ . 0-2 ‘“$ s 72
q(c)($
_ [2)-‘/2dc
0
considering the formula [7, p. 6941
s
d5
0
=(5)-"'42+m,2($+,+m,2~~)
where KS is the Sth order modified Bessel function of the second kind. Using the following formulas [4,5]
s
( J, + (lr ) *r c2+m’2) (p - ry 0
(
f0
,,,,2
Z‘id
-l/2
(c2 -
cc3”+m’2)z(3,2 + 4)
r2)l12
and with the help of the Abel integral equation pair d ‘C-cp(C)dl =f(r) ;i; s o (r2 - c2)1/2 vs)
2
=a
s
t f(r) dr o ([2_ r2)l/2'
(18)
Concentric penny-shaped crack in a finite cylinder
293
we can find that the auxiliary function cp(C) is governed by the following Fredholm equation of the second kind:
m, h) + F2;z(tt, L mlh, b)l
drt =
G (
)I
( r2r(r)dr 0 ([’ - r2)“2
integral
(20)
where c U+m)/2
0
401, L m,h) = i
s m
0
t .C .(th th -
l)J,,,+,,(~rl)J,,2+,,2(5~)
dt
(21)
The stress intensity factor is determined by the well known formula K,,, = lim[2(y - u)]‘/~.Z&J, 0). I-r(l
(23)
Here t&y, 0) is eq. (19) divided by r2, i.e. (24) in which T2 and T3 denote the second and third terms on the left-hand side of eq. (19) divided by r2. getting (24) into (23) and considering that T2 and TXdo not have a singularity at r = a, the following simple expression is found:
With the aid of integration by parts, the stress intensity factor is obtained: K,,, = ; 0
“2q(a),a?
(25)
Here p(a) is the solution of eq. (20) as I; = a. 4. NUMJIRICAL SOLUTION In accordance with expression (25) it is clear that the stress intensity factor is closely related to the solution of the second kind, Fredholm integral equation (20). Therefore, the numerical solution of K,,r may be computed by use of the standard procedure of the Fredholm integral equation. For this, eq. (20) needs to be changed to a Fredholm integral equation with a symmetrical kernel function. Replacing [, q and 5 by al, aq and r/a respectively in eqs (20), (21) and (22), and assuming that the stress on the crack face is specified as z(r) = T = constant; then, introducing another function Jl(t;) as follows: I/2 &al) = Ic $r[(mP+‘) *CC), (26) 08 we have the following Fredholm integral equation of the second kind with the symmetrical kernel function: $(C) +
s
o’@(tt)(F; + G)dg
= Cl-‘“”
(27)
where F; = (t#2
mW(@/a) s0
-
1142+m,2(trl)
- 4,2+m,2Wd5
(28)
LIANG BAN and X. S. SHANG
294
Table 1. K,,, /($a I/*) a
1 3 4 5 6
: b
12 12 12 12 12
: h
m=O
10 10 10 10 10
m =OS
1.0000 1.0002 1.0011 1.0033 1.0081
m = 1.0
1.0000 1.0003 1.0018 1.0026 1.0178
1.0000 1.0001 1.0003 1.0011 1.0035
Expression (26) yields, by letting c = 1, &@(I).
P(U) = cPM)l(=, = jj 0 Then expression (25) can be rewritten as
K,,, = f * z . a”2+(1). Here +(l) is the solution of eq. (27) at [ = 1. Several values of &,/(~‘/~/2) with m and a : b :h are given in Table 1. 5. DISCUSSION The solution of the problem of a concentric penny-shaped crack of mode III in a nonhomogeneous infinite radius circular plate may be derived from eq. (27). Letting b-co, then lim
b-m
For this case, eq. (27) must be rewritten as i,+(c) + 01+(tj)*F; dq = C(‘-m’2). f Here F; is provided as (28). The results of a concentric penny-shaped crack in a nonhomogeneous be derived from eq. (27). Letting h-co and
(30)
in!Xte cylinder may
we get
I,W +
’ #(q)F,drj
s0
= C(‘-m’2) >
(31)
in which
If we consider that both h and b approach infinity, and using the above limit results, the solution of an infinite plate containing a penny-shaped crack for mode III will be found immediately by eq. (27): (l/(C) = {” -m/Z). (32) K,,, is given as K,,, = ;~a”‘.
(33)
Concentric penny-shaped crack in a finite cylinder
295
REFERENCES [I] G. C. Sih and E. P. Chen, Mechanics of Fracture Cracks in Composite Materials (1981). [2] [3] [4] [5] [6] [7j [8]
M. Lowengrub and I. N. Sneddon, Int. J. Engng Sci. 3, 451-460 (1965). F. W. Smith, J. appl. Meek. 34, 947-952 (1967). S. S. Chang, Engng Fracture Me&. 22, 571 (1985). X. S. Zhang, J. Betjing Inst. Civil Engng Architecture 2, 2 (1987). X. S. Zhang and Liang Ran, Engng Fracture Mech. 34, 269-274 (1989). I. S. Gradshteyn and I. M. Ryzhik, Table ofZntegrds,Series and Products (1980). W. Magnus, F. Oberhettinger and R. P. Soni, Formulas and Theorems for The Special Functions of Mathematical Physics (1966). (Received 5 December 1991)
0013-7944/93 $6.00 + 0.00 Pergamon Press Ltd.
THE PROBLEM OF A CONCENTRIC PENNY-SHAPED CRACK OF MODE III IN A NONHOMOGENEOUS FINITE CYLINDER LIANG BAN and X. S. SHANG Beijing Institute of Civil Engineering, P.R.C. Abstrat-By using the Hankel transform and the Fourier series, the stress intensity factor of a concentric penny-shaped crack in a finite long cylinder of a finite radius for mode III is determined. It is proved that the concentric penny-shaped cracks in an infinite long cylinder and in an infinite circular plate am the special cases of the problem considered here.
1. INTRODUCTION IN THE past, several solutions of an infinite elastic body with a central crack have been obtained [l-3]. Zhang has given a series of analytical results of the central crack, the edge crack and the concentric penny-shaped crack for mode III in a homogeneous and a nonhomogeneous finite body in recent years [4-6]. The problem of a nonhomogeneous finite cylinder with a concentric penny-shaped crack of mode III has not yet been treated. In this paper, by using the basic theorem of the Hankel transform and the Fourier series, the problem of a nonhomogeneous, finite, long cylinder of finite radius has been discussed. The result of the problem may be given by means of a Fredholm integral equation of the second kind. The numerical solution of the stress intensity factor is computed using the standard procedure of the Fredholm integral equation and is given when the shear stress is a constant t.
2. BASIC EQUATIONS We consider a concentric penny-shaped crack of radius a in a nonhomogeneous finite cylinder of diameter 2b and length 2h, subjected to an arbitrary torque shear stress r(r), and a self-equilibrium system on its crack face (Fig. 1). In discussing the problem of a cylinder with a concentric penny-shaped crack, it is convenient to use cylindrical polar coordinates, assuming the origin of the coordinates coincides with the center of the cylinder, the X and Y axes are in the plane of the penny-shaped crack and the Z axis is perpendicular to the XY-plane. The position of a point in the cracked cylinder may be described by the distance r in the radial direction, the angle 8 between r and OX, and the distance Z in the axial direction. For this problem the displacement components in the r and z directions vanish everywhere and the only nonzero displacement component in the 8 direction is a function of r and z, i.e.
u, = u, = 0
u, = U&, The corresponding component We@,2):
stress components rtJz(r9r)=11
2).
can be expressed au&+, 2) az
where the shear modulus ~1= harm and h is a constant, m Z 0. 289
in terms of the displacement
LIANG BAN and X. S. SHANG
290
T
2
I+A ,/-a \\
X
--.
=-
-/
8
c N
Y
r
2a
tide-- -----*.
1
i \
2b
-
Fig. 1.
Putting zd(r, z) and zor(r, z) into the following equilibrium equation: ~~,(r,z)+~~,(r,z)+fTd(r,z)=O, the partial differential equation about Ue(r, z) can be obtained:
($+$ 1Uo(r,z)+(l+m)d,
8
Wr,z)
( ) r
=
o
.
(1)
The boundary conditions may be given as z,(r,
0) = -z(r)
0< r < a
(2)
U&,0)=0
b >r >a
(3)
o,(r,h)=O
O
(4)
U,(b,z)=O
O
(5)
It is obvious that the key of this problem is to seek a displacement function Ue(r, z) that can satisfy all of these boundary equations (2)-(j) and the partial differential equation (1). 3. THE ANALYTICAL
SOLUTION
Solving the partial differential equation (1) we can get the displacement component in terms of the Hankel transform and Fourier series, i.e. h Ue(r, z) = r +‘*
I+mdtr)dt
+
WV f n = 1,3,...
l+m12(!k!k)sin(?$)]
U&, z)
(6)
in which Js stands for the Sth order Bessel function of the first kind while Is is the Sth order modified Bessel function of the first kind. The corresponding stress components are T&(r, z) = -rmi2
Co [S
t&(r, z) = - rm’*
0
tfA(O
ch5@ -dJ ch
&
2+m,2(tr) d5
291
Concentric penny-shaped crack in a finite cylinder
where the following recursion formula [7, p. 9701 has been employed:
Putting (7) and (6) into (2) and (3) respectively, we have
s
=0
mA(~)Jl+m~2(Tr)dS 0
(10)
bar >a.
For convenience the pair of dual integral equations (9) and (10) may be rewritten as m tA(t)(th s0
A(C)r (2+m’2)J2+m,2(&) d< f
4) E
@ - 1)J,+m,2(~r)r(2+m/2)d~
(> ?Inr
mN+m/*
T&
.=,,...
s
mA(t)Jl+m,2(Tr)dt
=0
b >r
r
(2+m/2) = r*r(r),
0 < r
>a.
(11)
(12)
0
The unknown functions A(<) and B(n) in eq. (11) are not independent of eqs (5) and (6) and on application of the Fourier sine series pair:
we have (13)
B(n) =
Considering eq. (13), the dual integral equations (11) and (12) become A(0
=r*t(r)
(*+ ““*)J2+,,,,*(5r) d5 +
O
m rA(<)(th
(14)
s
mA(t)J,+m,2(tr)d5=0
bar>u.
(15)
0
Letting
s~cp(r)J,.,,,+,,,(rr)i-“+““*dS
A(5>=t1’*
0
(16)
292
LIANG BAN and X. S. SHANG
and substituting (16) into (15) and using the result [8, p. 1001
s
2
‘12
i”
r(l+m/2)~-(3/2+m/2)(~2_r2)-l~2
a5”zJ,+m:2(5r)J,+m,2+,,2(Si)d5 = 71
0
0
Irl
it is not difficult to see that eq. (15) is automatically satisfied, i.e. the boundary condition (3) has been satisfied. Inserting (16) in (14) and employing [8, p. 1001
OD 5"252+m,2(rr)J,+,2+,,z(5r)de s 0
""'
=
I
and identical equations j;v(W,l~!jdC
=J;p:r&v
.,;!$)J;:;
we can obtain
‘[ . 0-2 ‘“$ s 72
q(c)($
_ [2)-‘/2dc
0
considering the formula [7, p. 6941
s
d5
0
=(5)-"'42+m,2($+,+m,2~~)
where KS is the Sth order modified Bessel function of the second kind. Using the following formulas [4,5]
s
( J, + (lr ) *r c2+m’2) (p - ry 0
(
f0
,,,,2
Z‘id
-l/2
(c2 -
cc3”+m’2)z(3,2 + 4)
r2)l12
and with the help of the Abel integral equation pair d ‘C-cp(C)dl =f(r) ;i; s o (r2 - c2)1/2 vs)
2
=a
s
t f(r) dr o ([2_ r2)l/2'
(18)
Concentric penny-shaped crack in a finite cylinder
293
we can find that the auxiliary function cp(C) is governed by the following Fredholm equation of the second kind:
m, h) + F2;z(tt, L mlh, b)l
drt =
G (
)I
( r2r(r)dr 0 ([’ - r2)“2
integral
(20)
where c U+m)/2
0
401, L m,h) = i
s m
0
t .C .(th th -
l)J,,,+,,(~rl)J,,2+,,2(5~)
dt
(21)
The stress intensity factor is determined by the well known formula K,,, = lim[2(y - u)]‘/~.Z&J, 0). I-r(l
(23)
Here t&y, 0) is eq. (19) divided by r2, i.e. (24) in which T2 and T3 denote the second and third terms on the left-hand side of eq. (19) divided by r2. getting (24) into (23) and considering that T2 and TXdo not have a singularity at r = a, the following simple expression is found:
With the aid of integration by parts, the stress intensity factor is obtained: K,,, = ; 0
“2q(a),a?
(25)
Here p(a) is the solution of eq. (20) as I; = a. 4. NUMJIRICAL SOLUTION In accordance with expression (25) it is clear that the stress intensity factor is closely related to the solution of the second kind, Fredholm integral equation (20). Therefore, the numerical solution of K,,r may be computed by use of the standard procedure of the Fredholm integral equation. For this, eq. (20) needs to be changed to a Fredholm integral equation with a symmetrical kernel function. Replacing [, q and 5 by al, aq and r/a respectively in eqs (20), (21) and (22), and assuming that the stress on the crack face is specified as z(r) = T = constant; then, introducing another function Jl(t;) as follows: I/2 &al) = Ic $r[(mP+‘) *CC), (26) 08 we have the following Fredholm integral equation of the second kind with the symmetrical kernel function: $(C) +
s
o’@(tt)(F; + G)dg
= Cl-‘“”
(27)
where F; = (t#2
mW(@/a) s0
-
1142+m,2(trl)
- 4,2+m,2Wd5
(28)
LIANG BAN and X. S. SHANG
294
Table 1. K,,, /($a I/*) a
1 3 4 5 6
: b
12 12 12 12 12
: h
m=O
10 10 10 10 10
m =OS
1.0000 1.0002 1.0011 1.0033 1.0081
m = 1.0
1.0000 1.0003 1.0018 1.0026 1.0178
1.0000 1.0001 1.0003 1.0011 1.0035
Expression (26) yields, by letting c = 1, &@(I).
P(U) = cPM)l(=, = jj 0 Then expression (25) can be rewritten as
K,,, = f * z . a”2+(1). Here +(l) is the solution of eq. (27) at [ = 1. Several values of &,/(~‘/~/2) with m and a : b :h are given in Table 1. 5. DISCUSSION The solution of the problem of a concentric penny-shaped crack of mode III in a nonhomogeneous infinite radius circular plate may be derived from eq. (27). Letting b-co, then lim
b-m
For this case, eq. (27) must be rewritten as i,+(c) + 01+(tj)*F; dq = C(‘-m’2). f Here F; is provided as (28). The results of a concentric penny-shaped crack in a nonhomogeneous be derived from eq. (27). Letting h-co and
(30)
in!Xte cylinder may
we get
I,W +
’ #(q)F,drj
s0
= C(‘-m’2) >
(31)
in which
If we consider that both h and b approach infinity, and using the above limit results, the solution of an infinite plate containing a penny-shaped crack for mode III will be found immediately by eq. (27): (l/(C) = {” -m/Z). (32) K,,, is given as K,,, = ;~a”‘.
(33)
Concentric penny-shaped crack in a finite cylinder
295
REFERENCES [I] G. C. Sih and E. P. Chen, Mechanics of Fracture Cracks in Composite Materials (1981). [2] [3] [4] [5] [6] [7j [8]
M. Lowengrub and I. N. Sneddon, Int. J. Engng Sci. 3, 451-460 (1965). F. W. Smith, J. appl. Meek. 34, 947-952 (1967). S. S. Chang, Engng Fracture Me&. 22, 571 (1985). X. S. Zhang, J. Betjing Inst. Civil Engng Architecture 2, 2 (1987). X. S. Zhang and Liang Ran, Engng Fracture Mech. 34, 269-274 (1989). I. S. Gradshteyn and I. M. Ryzhik, Table ofZntegrds,Series and Products (1980). W. Magnus, F. Oberhettinger and R. P. Soni, Formulas and Theorems for The Special Functions of Mathematical Physics (1966). (Received 5 December 1991)