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An edge crack in a rectangular orthotropic sheet under arbitrary shear stress longitudinal
Enghcering l+acture Mechanics Printed in Great Britain.
Vol. 30, No. 2, pp. 219-226,
0013-7944188 $3.00+ @I 1988 Pertymon Press
1988
.OO
plc.
AN EDGE CRACK IN A RECTANGULAR ORTHOTROPIC SHEET UNDER ARBITRARY SHEAR STRESS LONGITUDINAL
S. S. CHANG Beijing Institute of Civil Engineering,Beijing, China Abstract-The general solution of the stress intensity factor of a rectangular orthotropic sheet with an edge crack under anti-plane shear is found by means of the Fourier transform and Fourier series in the present paper. It is of interest to note that the general solution of this problem is identical with the result of a central crack in a rectangular orthotropic sheet for mode III. The solutions of a strip containing an edge crack of mode III can easily be derived from the general solution in this study.
INTRODUCTION
THE STRESS intensity factors of an infinite plate and a strip with an edge crack have been discussed in great detail in a series of papers[l-51. Each of these studies is based on the assumption that the edge crack is located in an isotropic elastic plate. Considering many engineering structures, particularly plates with stiffeners or corrugations should be treated as orthotropic bodies, some solutions of a central crack in a rectangular orthotropic sheet have been given by several authors[l, 61. But the problem of a rectangular orthotropic sheet containing an edge crack has not been considered as yet. In this paper the general solution of an edge crack in a rectangular orthotropic sheet under arbitrary longitudinal shear stress will be investigated in three stages. Firstly, solving the basic differential equation established by [ 11, we find that the only non-zero displacement component can be expressed in terms of the Fourier cosine transform and Fourier sine series. Secondly, the stress intensity factor of this problem is found to be related to the result of a Fredholm integral equation of the second kind. Thirdly, it is verified that the solutions of an edge crack in an orthotropic strip for tearing mode can be derived from the general solution in this study. When the anti-plane shear stress on the crack face is constant, the numerical solutions of the problem are given in Table 1.
Table 1. The variations of Z&/T~‘/* with A and a : b : h[l] A
a:b:h
A
1.0
1.2
1.5
1 : 12 : 10 2 : 12 : 10 3 : 12 : 10 4: 12; 10 5: 12: 10 6 : 12 : 10
1.0032 1.0130 1.0295 1.0533 1.0847 1.1246
1.0030 1.0121 1.0279 1.0510 1.0827 1.1241
1.0029 1.0117 1.0272 1.0500 1.0814 1.1232
1: 00: 2: 00: 3 :00: 4: 00: 5: 00: 6: 00:
1.0021 1.0081 1.0178 1.0306 1.0457 1.0622
1.0014 1.0056 1.0125 1.0217 1.0330 1.0457
1.0009 1.0036 1.0081 1.0141 1.0217 1.0306
10 10 10 10 10 10
0.8
1.0
1:
Formula
(47)
1:8:10 2:8: 10 3:8: 10 4:8: 10 5:8:10 6:8: 10
1.0068 1.0284 1.0683 1.1323 1.2277 1.3678
1.0066 1.0272 1.0645 1.1231 1.2093 1.3313
1.0065 1.0265 1.0618 1.1153 1.1910 1.2945
(47)
(49)
1: lo:oo 2 : 10: 00 3 : 10 : 00 4: 10: 00 5 : 10: 00 6: 10: 00
1.0041 1.0170 1.0396 1.0740 1.1229 1.1897
1.0041 1.0170 1.0396 1.0740 1.1229 1.1897
1.0041 1.0170 1.0396 1.0740 1.1229 1.1897
(56)
Formula
a :b:h
Km/ TUI’*
KII,/ralIz 219
220
S. S. CHANG
FORMULATION Consider a rectangular orthotropic sheet (2 h X b) with an edge crack of length a as shown in Fig. 1. For this problem, it is very convenient to choose the y-axis coinciding with one of the edges of the plate and the x-axis along the line of the crack. The z-axis is perpendicular to xy-plane, but it is not drawn in Fig. 1. Let the edge crack be subjected to an arbitrary longitudinal shear stress forming a self-balanced system on its face. In this case, the mixed boundary conditions may be given as follows: +(x,
OSx
0) = -7(x);
(1)
t&(x, 0) = 0;
a
(2)
7&x,
Osxsb
(3)
h) = 0;
GAO, Y) = 0;
Os(y(s
h
(4)
T,,(b, y) = 0;
O
h
(5)
Considering the planes of elastic symmetry in an orthotropic written by [l, 71:
body, the Hook’s law can be
all,
a12, a13
u.s
~21,
a22,
a23
0;
a31,
a32,
a33
.I0,
I-
=
J
a4 -+?!!Z
ay
ax
a$
au,
St-+ay au,
;
(6)
%67xy
=
a44ryz
.
(7)
au,
ax+az For the anti-plane shear problem, it is well known that the only non-zero component u, is independent of z, i.e.
(8)
UZ= r&(x, y).
ty (b,h)
(o,h)
Pt(C,J --_A
--_)x
0 p&J J
(b,-h)
(O.-h)
Fig. 1.
displacement
Edge crack in a rectangular orthotropic
sheet
221
Then we may get EX= Ey = I?*= YXY = 0.
(9)
Putting (8) and (9) into (6) and (7) leads to a, = uy = uz = TXY = 0;
TX, =
C55Yxr =-
(10)
au,c55;
(11)
ax
7YZ
=
C44Yys
=
au,
-
(12)
c44,
ay
where c44 = lla44; c55 = 1/a55. Substituting equilibrium equation
the stress components
aTxz
x+-=0,
the displacement
component
aryz
a2u, ax2
+
(13)
ay
u,(x, y) is governed
c55
(11) and (12) into the
c44
by
a*u z=
0.
(14)
aY*
According to the foregoing analysis, we must find a displacement component z&, y) to satisfy all the boundary conditions (l)-(5) and the partial differential equation (14) for solving this problem. ANALYTICAL
SOLUTION
It is clear that the solution of the partial differential eq. (14) may be represented the Fourier cosine transform and Fourier sine series, i.e. Chc~(($)Y)cos(&)d~+
in terms of
B(n)chzsinz,
i
(15)
n=1.3,...
in which A* = c55/c44. The corresponding
Lk
y)
=
-
c55 [I
stress components
sin(tx) dt-
chc’$!&y)
O1 tW5)
can be derived from expressions (11) and (12): “z 3 ($)
I
shEsin
~1;
0
(17) ryr(x, y) = -
c44
SAA(Z) [Im
sh ZW - y) ch( (Ah)
0
d,!j-
cos(cfx)
"_~,,.(~)B(n)ch~cos~].
From (17) and (18) we see that the boundary conditions (3) and (4) are obviously satisfied. Inserting (18) and (15) into (1) and (2) leads to a pair of dual integral equations:
c 44
[lo-
&A(5)
WAN
cos(&)
d5-
“=i
, . .
(k$
B(n)
ch
E]
=
T(X);
0
6
x
-C
a
(19)
m I0
A(5) cos(~x) de = 0.
bax>a
(20)
222
S. S. CHANG
The unknown functions A(l) and B(n) in (19) are not independent. condition (5) we get
Considering
the boundary
Employing the basic theorem of the Fourier series, we have B(n) =
2
A(5) sin(@)d5.
(22)
With the help of (22), the pair of dual integral equations may be modified as the following form:
llA(5) cos(&jc) d5
.-L..
m
IOc2
sin(tb) d[
-
mb
hAsh%
A(()cos(&) dt = 0.
I
ch z
= 7(x);
OSx
(23)
b>x>a
(24)
The only unknown function A(t) may be determined by solving the pair of dual eqs (23) and (24). Then, all the boundary conditions of this problem are satisfied. Letting
(25) and considering
one of the dual equations (24) is automatically results [l]
satisfied. Substituting (25) into (23) and using the
(27)
2 Jo(s{) sin(tb) d.$‘= te-“*‘2M
cm
lo (&q,
> an integral equation of the Abel type is obtained:
x
(29)
Edge crack in a rectangular orthotropic
223
sheet
With aid of the Abel integral equation pair [l]
’ p(x)dx * Mf)d5 -o 4t2 o J
x2)
=
(30)
dh
and the known formulas (31)
and the identical equation
we can find that the auxiliary function (p(J) in (29) is governed by a Fredholm integral equation of the second kind:
J a
(P(5)+
0
’
c r(x)dx
dd[Fl(~, A, t, h) - &Cc A, 5, h, b)l dv = 7r 0 @-x2) J
(32)
where
J
FI = q om5IW5W
- llJo(h)Jo(50
(33)
d5;
-mbl2hA
n=l.3...
M2sh2
nmb
“(2)
(34)
“(3’
For tearing mode, the stress intensity factor may be determined
by means of the formula
Km = lim [2(x - a)11’2TYz(x,0). X40 According once, i.e.
to the left hand side of the eq. (29), the stress component
(35) TJX, 0) may be obtained at
(36) in which I2 and I3 stand for the second and third terms on the left hand side of eq. (29). Putting (36) into (35) and noticing I2 and I3 without a singularity at x = a, we get Km = lim [2(x -
a)]“2
[
-
&
loa
$xJ!yj].
(37)
X-Cl
Making use of integration by parts, the stress intensity factor of a rectangular with an edge crack under arbitrary anti-plane shear stress is obtained as
&II = cp(a)a”2, where
q(a)
orthotropic
sheet
(38)
is the solution of the Fredholm integral equation of the second kind (32) as 5 =
a.
224
S. S. CHANG
NUMERICAL
SOLUTION
AND DISCUSSION
From formula (38), it is clear that the determination of the stress intensity factor is closely related to the result of the auxiliary function cp(a) = c~(LJ(~=~.When the longitudinal shear stress on the edge crack face is specified, the solution of the Fredholm integral eq. (32) may be found by the iterative scheme
(P(5) = f. (Pm(C),
(39)
in which
PO(t)=:
cPm(O
r
o I
~(x)dx (40)
&2__2);
(41)
= joa (~m--1(7))(~2-Wd71.
Applying the series expansion Jo(h).ro(SJ)=1_~g+94+4~~+C4B_...;
(42)
~o(E&o(!!&~+~(~)2+n4+4~~+~4(ET)4+...;
(43)
th( (Ah) _ 1 = _ 2(e_2”hZ - e-4”@ + e-e*M) _ . . . ),
(44)
to the Fredholm integral eq. (32) and employing m I0
t”eMVdt=%,
(45) Y
and f
!l=l
f
(-on-'_~.
n2
12’
(-1)“_‘_7?r4* n4 720’ . ’ * ’
(46)
II=1
and assuming T(X) = T = constant; a/h < 1, we can get the solution of the auxiliary function cp(a) in series form:
( a)
2.4674 hh
e-(m+n)
m?rt)
sh-
2hh
n + l.O146(;yn
+ 2.2930(;)4n3)
&
“Lb(6.088)mn(fg4+ 0(&i). sh-
(47)
2M
Comparing these formulas (32), (38) and (47) in the present paper with those expressions (3.18),
Edge crack in a rectangular orthotropic
sheet
225
(3.24) and (4.8) in ref. [l], it is verified that general solution of the stress intensity factor of this problem is identical with the result of a rectangular orthotropic plate (2h x 26) with a central crack of length 2a under arbitrary longitudinal shear stress. According to the representation of the function q(a) in series form, the variations of Knr/ruYz with A and a : b : h can easily be given as shown in Table 1. The solution of an orthotropic strip containing an edge crack parallel to the strip edges, can be derived from the general solution (38) by putting b +a. In this case, the Fredholm integral equation will be simplified as
(48) because of the function Fz in (32) is vanished, and the function q(a) in series form is 2 q(a)
Considering
=
-0.0909
T
($+
o(&)];
(49)
h + ~0in eq. (32) and assuming
the solution of an orthotropic strip with an edge crack perpendicular to its edges may also be obtained from the general solution in this paper. But the Fredholm integral equation will become [ 11:
(51) in which the kernel function
(52) Using the known results
=_
2
(e-266/A
+ e-4WA
+ e--6WA
+ . . . );
(53)
Io(~)Io($)=~+3y(~)2+~4+41j1~+~4(~)4+ . ..)
(54)
and
(55) and substituting them into (51) we may find that the function q(a) = (p(l)lL_ in series form is independent of the elastic constants c44 and c55, i.e. q(a)=
r[ 1+0.4112
(;)2+0.3212
(;)4+O($)].
In other words, the stress intensity factor of an orthotropic EPPL 30:2-G
;+
1
strip containing
(56) an edge crack
226
S. S. CHANG
perpendicular to its edges is not related to the elastic constant. This conduction has been proved by Liebowitz[7] according to the problem of a central crack in an infinite anisotropic plate. The solution of an edge crack in an isotropic rectangular sheet may be also obtained by putting A = 1 or c44 = c55 into the genera1 solution in the present paper. The stress intensity factor is Ku1
=
cp(a)a”*,
(57)
the Fredholm integral equation of the second kind is [8]
(58) where
F4= FI(T, 1, 5, h); Fs = Hq, 1, 5, h, b),
(59)
and the function in series form is
+7
2.4674n ;+
1.0146 $ + 2.283n3$
+0 $
10
,
(60)
REFERENCES [l] S. S. Chang, The solution of a rectangular orthotropic sheet with a central crack under anti-plane shear. Engng Fmch~reMech. 22, 251 (1985). [2] G. C. Sih, Handbook of Stress ZnrcnsityFac?ors, p. 2.4.2-l (1973). [3] X. S. Zhang (S. S. Chang), Scienria Sinica (English series A) VXXVIII, 86 (1985). [4] I. N. Sneddon and S. C. Das, Znt. .Z. engng Sci. 9, 25 (1971). [5] G. C. Sih, .Z. Sot. Znd. appl. Math. 12, 403 (1964). [6] 0. L. Bowie and C. E. Freese, Znr.Z. Fracture Me&. 8,49 (1972). [7] H. Liebowitz, Fracture, An Advanced Treatise, Vol. 2, p. 108 (1968). [S] S. S. Chang, A central crack at the interface between two different media in a rectangular sheet under anti-plane shear. Engng Fracture Mech. 19, 709 (1984). (Receioed 1 Zune 1987)
Vol. 30, No. 2, pp. 219-226,
0013-7944188 $3.00+ @I 1988 Pertymon Press
1988
.OO
plc.
AN EDGE CRACK IN A RECTANGULAR ORTHOTROPIC SHEET UNDER ARBITRARY SHEAR STRESS LONGITUDINAL
S. S. CHANG Beijing Institute of Civil Engineering,Beijing, China Abstract-The general solution of the stress intensity factor of a rectangular orthotropic sheet with an edge crack under anti-plane shear is found by means of the Fourier transform and Fourier series in the present paper. It is of interest to note that the general solution of this problem is identical with the result of a central crack in a rectangular orthotropic sheet for mode III. The solutions of a strip containing an edge crack of mode III can easily be derived from the general solution in this study.
INTRODUCTION
THE STRESS intensity factors of an infinite plate and a strip with an edge crack have been discussed in great detail in a series of papers[l-51. Each of these studies is based on the assumption that the edge crack is located in an isotropic elastic plate. Considering many engineering structures, particularly plates with stiffeners or corrugations should be treated as orthotropic bodies, some solutions of a central crack in a rectangular orthotropic sheet have been given by several authors[l, 61. But the problem of a rectangular orthotropic sheet containing an edge crack has not been considered as yet. In this paper the general solution of an edge crack in a rectangular orthotropic sheet under arbitrary longitudinal shear stress will be investigated in three stages. Firstly, solving the basic differential equation established by [ 11, we find that the only non-zero displacement component can be expressed in terms of the Fourier cosine transform and Fourier sine series. Secondly, the stress intensity factor of this problem is found to be related to the result of a Fredholm integral equation of the second kind. Thirdly, it is verified that the solutions of an edge crack in an orthotropic strip for tearing mode can be derived from the general solution in this study. When the anti-plane shear stress on the crack face is constant, the numerical solutions of the problem are given in Table 1.
Table 1. The variations of Z&/T~‘/* with A and a : b : h[l] A
a:b:h
A
1.0
1.2
1.5
1 : 12 : 10 2 : 12 : 10 3 : 12 : 10 4: 12; 10 5: 12: 10 6 : 12 : 10
1.0032 1.0130 1.0295 1.0533 1.0847 1.1246
1.0030 1.0121 1.0279 1.0510 1.0827 1.1241
1.0029 1.0117 1.0272 1.0500 1.0814 1.1232
1: 00: 2: 00: 3 :00: 4: 00: 5: 00: 6: 00:
1.0021 1.0081 1.0178 1.0306 1.0457 1.0622
1.0014 1.0056 1.0125 1.0217 1.0330 1.0457
1.0009 1.0036 1.0081 1.0141 1.0217 1.0306
10 10 10 10 10 10
0.8
1.0
1:
Formula
(47)
1:8:10 2:8: 10 3:8: 10 4:8: 10 5:8:10 6:8: 10
1.0068 1.0284 1.0683 1.1323 1.2277 1.3678
1.0066 1.0272 1.0645 1.1231 1.2093 1.3313
1.0065 1.0265 1.0618 1.1153 1.1910 1.2945
(47)
(49)
1: lo:oo 2 : 10: 00 3 : 10 : 00 4: 10: 00 5 : 10: 00 6: 10: 00
1.0041 1.0170 1.0396 1.0740 1.1229 1.1897
1.0041 1.0170 1.0396 1.0740 1.1229 1.1897
1.0041 1.0170 1.0396 1.0740 1.1229 1.1897
(56)
Formula
a :b:h
Km/ TUI’*
KII,/ralIz 219
220
S. S. CHANG
FORMULATION Consider a rectangular orthotropic sheet (2 h X b) with an edge crack of length a as shown in Fig. 1. For this problem, it is very convenient to choose the y-axis coinciding with one of the edges of the plate and the x-axis along the line of the crack. The z-axis is perpendicular to xy-plane, but it is not drawn in Fig. 1. Let the edge crack be subjected to an arbitrary longitudinal shear stress forming a self-balanced system on its face. In this case, the mixed boundary conditions may be given as follows: +(x,
OSx
0) = -7(x);
(1)
t&(x, 0) = 0;
a
(2)
7&x,
Osxsb
(3)
h) = 0;
GAO, Y) = 0;
Os(y(s
h
(4)
T,,(b, y) = 0;
O
h
(5)
Considering the planes of elastic symmetry in an orthotropic written by [l, 71:
body, the Hook’s law can be
all,
a12, a13
u.s
~21,
a22,
a23
0;
a31,
a32,
a33
.I0,
I-
=
J
a4 -+?!!Z
ay
ax
a$
au,
St-+ay au,
;
(6)
%67xy
=
a44ryz
.
(7)
au,
ax+az For the anti-plane shear problem, it is well known that the only non-zero component u, is independent of z, i.e.
(8)
UZ= r&(x, y).
ty (b,h)
(o,h)
Pt(C,J --_A
--_)x
0 p&J J
(b,-h)
(O.-h)
Fig. 1.
displacement
Edge crack in a rectangular orthotropic
sheet
221
Then we may get EX= Ey = I?*= YXY = 0.
(9)
Putting (8) and (9) into (6) and (7) leads to a, = uy = uz = TXY = 0;
TX, =
C55Yxr =-
(10)
au,c55;
(11)
ax
7YZ
=
C44Yys
=
au,
-
(12)
c44,
ay
where c44 = lla44; c55 = 1/a55. Substituting equilibrium equation
the stress components
aTxz
x+-=0,
the displacement
component
aryz
a2u, ax2
+
(13)
ay
u,(x, y) is governed
c55
(11) and (12) into the
c44
by
a*u z=
0.
(14)
aY*
According to the foregoing analysis, we must find a displacement component z&, y) to satisfy all the boundary conditions (l)-(5) and the partial differential equation (14) for solving this problem. ANALYTICAL
SOLUTION
It is clear that the solution of the partial differential eq. (14) may be represented the Fourier cosine transform and Fourier sine series, i.e. Chc~(($)Y)cos(&)d~+
in terms of
B(n)chzsinz,
i
(15)
n=1.3,...
in which A* = c55/c44. The corresponding
Lk
y)
=
-
c55 [I
stress components
sin(tx) dt-
chc’$!&y)
O1 tW5)
can be derived from expressions (11) and (12): “z 3 ($)
I
shEsin
~1;
0
(17) ryr(x, y) = -
c44
SAA(Z) [Im
sh ZW - y) ch( (Ah)
0
d,!j-
cos(cfx)
"_~,,.(~)B(n)ch~cos~].
From (17) and (18) we see that the boundary conditions (3) and (4) are obviously satisfied. Inserting (18) and (15) into (1) and (2) leads to a pair of dual integral equations:
c 44
[lo-
&A(5)
WAN
cos(&)
d5-
“=i
, . .
(k$
B(n)
ch
E]
=
T(X);
0
6
x
-C
a
(19)
m I0
A(5) cos(~x) de = 0.
bax>a
(20)
222
S. S. CHANG
The unknown functions A(l) and B(n) in (19) are not independent. condition (5) we get
Considering
the boundary
Employing the basic theorem of the Fourier series, we have B(n) =
2
A(5) sin(@)d5.
(22)
With the help of (22), the pair of dual integral equations may be modified as the following form:
llA(5) cos(&jc) d5
.-L..
m
IOc2
sin(tb) d[
-
mb
hAsh%
A(()cos(&) dt = 0.
I
ch z
= 7(x);
OSx
(23)
b>x>a
(24)
The only unknown function A(t) may be determined by solving the pair of dual eqs (23) and (24). Then, all the boundary conditions of this problem are satisfied. Letting
(25) and considering
one of the dual equations (24) is automatically results [l]
satisfied. Substituting (25) into (23) and using the
(27)
2 Jo(s{) sin(tb) d.$‘= te-“*‘2M
cm
lo (&q,
> an integral equation of the Abel type is obtained:
x
(29)
Edge crack in a rectangular orthotropic
223
sheet
With aid of the Abel integral equation pair [l]
’ p(x)dx * Mf)d5 -o 4t2 o J
x2)
=
(30)
dh
and the known formulas (31)
and the identical equation
we can find that the auxiliary function (p(J) in (29) is governed by a Fredholm integral equation of the second kind:
J a
(P(5)+
0
’
c r(x)dx
dd[Fl(~, A, t, h) - &Cc A, 5, h, b)l dv = 7r 0 @-x2) J
(32)
where
J
FI = q om5IW5W
- llJo(h)Jo(50
(33)
d5;
-mbl2hA
n=l.3...
M2sh2
nmb
“(2)
(34)
“(3’
For tearing mode, the stress intensity factor may be determined
by means of the formula
Km = lim [2(x - a)11’2TYz(x,0). X40 According once, i.e.
to the left hand side of the eq. (29), the stress component
(35) TJX, 0) may be obtained at
(36) in which I2 and I3 stand for the second and third terms on the left hand side of eq. (29). Putting (36) into (35) and noticing I2 and I3 without a singularity at x = a, we get Km = lim [2(x -
a)]“2
[
-
&
loa
$xJ!yj].
(37)
X-Cl
Making use of integration by parts, the stress intensity factor of a rectangular with an edge crack under arbitrary anti-plane shear stress is obtained as
&II = cp(a)a”2, where
q(a)
orthotropic
sheet
(38)
is the solution of the Fredholm integral equation of the second kind (32) as 5 =
a.
224
S. S. CHANG
NUMERICAL
SOLUTION
AND DISCUSSION
From formula (38), it is clear that the determination of the stress intensity factor is closely related to the result of the auxiliary function cp(a) = c~(LJ(~=~.When the longitudinal shear stress on the edge crack face is specified, the solution of the Fredholm integral eq. (32) may be found by the iterative scheme
(P(5) = f. (Pm(C),
(39)
in which
PO(t)=:
cPm(O
r
o I
~(x)dx (40)
&2__2);
(41)
= joa (~m--1(7))(~2-Wd71.
Applying the series expansion Jo(h).ro(SJ)=1_~g+94+4~~+C4B_...;
(42)
~o(E&o(!!&~+~(~)2+n4+4~~+~4(ET)4+...;
(43)
th( (Ah) _ 1 = _ 2(e_2”hZ - e-4”@ + e-e*M) _ . . . ),
(44)
to the Fredholm integral eq. (32) and employing m I0
t”eMVdt=%,
(45) Y
and f
!l=l
f
(-on-'_~.
n2
12’
(-1)“_‘_7?r4* n4 720’ . ’ * ’
(46)
II=1
and assuming T(X) = T = constant; a/h < 1, we can get the solution of the auxiliary function cp(a) in series form:
( a)
2.4674 hh
e-(m+n)
m?rt)
sh-
2hh
n + l.O146(;yn
+ 2.2930(;)4n3)
&
“Lb(6.088)mn(fg4+ 0(&i). sh-
(47)
2M
Comparing these formulas (32), (38) and (47) in the present paper with those expressions (3.18),
Edge crack in a rectangular orthotropic
sheet
225
(3.24) and (4.8) in ref. [l], it is verified that general solution of the stress intensity factor of this problem is identical with the result of a rectangular orthotropic plate (2h x 26) with a central crack of length 2a under arbitrary longitudinal shear stress. According to the representation of the function q(a) in series form, the variations of Knr/ruYz with A and a : b : h can easily be given as shown in Table 1. The solution of an orthotropic strip containing an edge crack parallel to the strip edges, can be derived from the general solution (38) by putting b +a. In this case, the Fredholm integral equation will be simplified as
(48) because of the function Fz in (32) is vanished, and the function q(a) in series form is 2 q(a)
Considering
=
-0.0909
T
($+
o(&)];
(49)
h + ~0in eq. (32) and assuming
the solution of an orthotropic strip with an edge crack perpendicular to its edges may also be obtained from the general solution in this paper. But the Fredholm integral equation will become [ 11:
(51) in which the kernel function
(52) Using the known results
=_
2
(e-266/A
+ e-4WA
+ e--6WA
+ . . . );
(53)
Io(~)Io($)=~+3y(~)2+~4+41j1~+~4(~)4+ . ..)
(54)
and
(55) and substituting them into (51) we may find that the function q(a) = (p(l)lL_ in series form is independent of the elastic constants c44 and c55, i.e. q(a)=
r[ 1+0.4112
(;)2+0.3212
(;)4+O($)].
In other words, the stress intensity factor of an orthotropic EPPL 30:2-G
;+
1
strip containing
(56) an edge crack
226
S. S. CHANG
perpendicular to its edges is not related to the elastic constant. This conduction has been proved by Liebowitz[7] according to the problem of a central crack in an infinite anisotropic plate. The solution of an edge crack in an isotropic rectangular sheet may be also obtained by putting A = 1 or c44 = c55 into the genera1 solution in the present paper. The stress intensity factor is Ku1
=
cp(a)a”*,
(57)
the Fredholm integral equation of the second kind is [8]
(58) where
F4= FI(T, 1, 5, h); Fs = Hq, 1, 5, h, b),
(59)
and the function in series form is
+7
2.4674n ;+
1.0146 $ + 2.283n3$
+0 $
10
,
(60)
REFERENCES [l] S. S. Chang, The solution of a rectangular orthotropic sheet with a central crack under anti-plane shear. Engng Fmch~reMech. 22, 251 (1985). [2] G. C. Sih, Handbook of Stress ZnrcnsityFac?ors, p. 2.4.2-l (1973). [3] X. S. Zhang (S. S. Chang), Scienria Sinica (English series A) VXXVIII, 86 (1985). [4] I. N. Sneddon and S. C. Das, Znt. .Z. engng Sci. 9, 25 (1971). [5] G. C. Sih, .Z. Sot. Znd. appl. Math. 12, 403 (1964). [6] 0. L. Bowie and C. E. Freese, Znr.Z. Fracture Me&. 8,49 (1972). [7] H. Liebowitz, Fracture, An Advanced Treatise, Vol. 2, p. 108 (1968). [S] S. S. Chang, A central crack at the interface between two different media in a rectangular sheet under anti-plane shear. Engng Fracture Mech. 19, 709 (1984). (Receioed 1 Zune 1987)