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Out of plane shear of a cracked rectangular orthotropic block
Engirurring Fmclwr Mechanics Printed in GreatBritain.
Vol. 29, No. 6, pp. 641446,
1988
0013-7Y44/88 s3.tlo+ .tn, Pergamon Press pk.
OUT OF PLANE SHEAR OF A CRACKED RECTANGULAR ORTHOTROPIC BLOCK G. MELROSE and S. DAVIDSON Old Dominion University, Department of Mathematics and Statistics, Norfolk, VA 23529-0077. U.S.A. Abstract-The problem under consideration is the out of plane shear of a cracked rectangular orthotropic block. The exact solution is obtained by stating the problem in terms of a triple trigonometric series relation, which in turn can be shown to be equivalent to a singular integral equation whose solution is known. For the case of constant shear, the solution simplifies greatly and numerical results are given,
1. INTRODUCTION WE CONSIDER the out of plane shear of a cracked
coordinates
rectangular
orthotropic
solid. In Cartesian
the solid is given by the inequalities
while the crack which runs the length of the solid in the z direction is given by O
2. FORMULATION
AND STATEMENT
It is well known that for an out-of-plane u, =
and for an orthotropic
OF THE PROBLEM
shear the only non zero displacement w(x, Y)
is (2.1)
body the non zero stresses are given by 8W
uxxz= Gn -; ax
aw uyz = G23 ay
where G13, GZ3 are the shear moduli of the orthotropic therefore be satisfied provided 2
(2.2)
material. The equilibrium equations will
2
p2$+q=0
where p2=$. ay
(2.3 23
By symmetry, we now see that the problem is reduced to solving the following mixed boundary value problem. 641
G. MELROSE and S. DAVIDSON
642
p.d.e. b.c.
p2w,,+
(O
wyy=O
(2.4)
O
(1) w,(O, y) = w,(rr, y) = 0
(0 < y < A7r)
r(x) (2) w,(x, h7r) = -
(0 < x < 7r)
G23
(0 < x < a) U (b < x < 7~)
(3) w(x, 0) = 0 w,(x, 0) = 0 It is easily verified
(a -=cx < b).
that by writing
4x9
1 y) = --cW(x,
y) + 44x7
y)l
(2.5)
G23
where _ B, sinh nPy Boy + “;, np cash nphaCoS nx
w~(x, Y) =;
(2.6)
T(X) cos nx dx then 4(x,
y) must satisfy
p.d.e.
P2bX + 4yv = 0
(0 < x <
0 < y < h7T).
7-r;
b.c. (1) 4x(0, y) = 4xx(~ y) = o
(2.7)
(2.8)
(O< y
(2) 4+, Ad = 0
(0 < x -=c7T)
(3) $4x, 0) = 0
(0 < x < a)
4+(x, 0) = -f(x)
(a
44x, 0) = 0
(b
where f(x) = %
(X, 0) = i BO + c B, sech(nphn) II=1
3. DETERMINATION
OF THE
(2.9)
cos nx.
PERTURBATION
4
Clearly,
(3.1)
satisfies requires
the p.d.e. and the first two boundary conditions. that the {A,}; must satisfy the triple series
G(x) = AC)+ f
Furthermore,
n-’ A, cos nx = 0
boundary
condition
(3)
(O
n=l
F(x)
=
2
n=l
G(x)=A,+
A, tanh n/3hrrcos
nx = f(x)
i n-’ A, cos nx = 0 fl=l
(a < x < b) (b
(3.2)
Out of plane shear of orthotropic block
643
The solution of these triple series is given in[3] and is outlined here. By choosing
@(t)
(3.3)
dt
and b
A,=-27T (1 p(t)sinntdt
(3.4)
I
then we find 4(x, o) = G(x) = ~[(b - x)(x -
a,]J’;p(t)
dt
(3.5)
a
where H(u) is the Heaviside step function. We also find that,
Mt) dt (O
F(x)
=
i
(3.6)
where M(x, t) = - 2 c tanh(npAr) II=1
= -kiog
cos nx sin nr
tn(F)
+ tn(ff)
tn(F)
_ tn(F)
(3.7)
and tn(u) = tn(u, k) is a Jacobian elliptic function. (See appendix for more details.) It is now clear that the triple series will be satisfied if p(t) is given by the integral equation tn
J
-1 b 7T (1 &)&log
K” 0 7T
/Kx\
+tn
K’ 0) /it\
dt=-f(x).
(a < x < 6)
(3.X)
‘“W -‘“W I
From[3] we find that
(3.9) and
F (x)
3x,0) = -F(x)
rA,(x) =
I
(a < x < 6)
-f(x) -
(O
Knc($)dc(F) m&(x)
F*(x)
(b < x < ?7)
(3.10)
G. MELROSE and S. DAVIDSON
644
where
A(x)f(x) tn(F)
dx (3.11)
tn2(!c)
c =
_ tn2(g
.
_2K;LF)nc(F)dc(F) dt A(x)f(x) t(F) dx ,.b
Ib
I
A(x) = ([ tn’(F)
(3.12)
A(0
a
n tn2(F)
- tn2(F)][
tn2($)
- tn2(F)
- tn2($)]}“’
(3.13)
and
A,(X) = ([ m2(e)
- tn2(F)][tn2(F)
- tn2(e)])1’2
(3.14)
with K; = K(k;);
k, = tn(F)/tn(E).
4. STRESS INTENSITY
FACTORS
The stress intensity factors k, and k+ are given by the formulae, k, = lim ~a,,(x,O) x-a-
(4.1)
kb = lim -a,,(~,
0).
(4.2)
x-b+
For this problem, using (2.2) and (2.5) these equations become k, = lim m$ .X-a-
(x, 0)
(4.3)
0).
(4.4)
kb = lim Jz(r-b)$(x. x-b+
On substituting
(3.10) into (4.3) and (4.4) we obtain
kc, =
F,(a)
I
7r[tn2(F)
and
Kdn
(
Ka
- tn2(T)isn($)
112
)
(4.5) en(F)
Out of plane shear of orthotropic
kb = - F,(b)
645
block
I
5. NUMERICAL
(4.6)
RESULTS
FOR CONSTANT
SHEAR
For the special case of a constant shear the problem reduces greatly. First we observe that wr(x, y) = ry and hence f(x) = T. This allows us to simplify the singular integral terms F,(a) and Fr(b). Indeed, we find that
and
F,(b) -= 7
-2K;tn Kdn
(
fi
7r Ka ( ‘TT)
)
b
(5.2)
Z(6, k;) dt
Z(P,k;)-
with sin p = sn(F)
k2 = dn(F)/dn($); and sn’(F)
-sn’($)
sn2($)
- sn’($)
sin2 6 =
and Z(0, k) is Jacobi’s Zeta function.
1.2-
1.0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Fig. 1. Stress intensity vs aspect ratio for various values of (As),
0.6
G. MELROSEand S. DAVIDSON
646 1.60-
-
k,/k,
---
k,/k,
c-r/4
1.60-
Fig. 2. Stress intensity vs aspect ratio for various values of (hp).
Figures 1 and 2 give the scaled stress intensities vs aspect ratio for various values of (A/3). Here k,, = & where d = (b - a)/2 and c denotes the midpoint of the crack. For c = ~r/2, by symmetry, we have k, = kb which is the special case investigated by Chang[2]. The results given here are in agreement with those obtained by Chang. We note how quickly k//c, degenerates to the results of an infinite strip with a central crack (A --, m) and the effect of orthotropy in this case is not very significant. For the off-center crack, however, the orthotropic property seems to have more bearing. APPENDIX The following notation, although standard, is listed here for completeness elliptic functions and integrals can be obtained from[4] and [5]. sn(y) = sn(y, k) en(y) - cn(y, k) My)
= My,
and clarity. An excellent introduction to
Jacobian elliptic functions with modulus k.
k)
tn(y)2w My)
complete elliptic integral of the 1st kind with modulus k.
K = K(k)
=~[l+2n~,e-*f12]2 k
2[
$ .-*,.+q
in terms of A, ratio of rectangular sides. modulus in terms of A.
n (1 K, = K(k,)
complete elliptic integral of the 1st kind with modulus k,.
K’ = K(k’)
associated complete elliptic integral of the 1st kind with modulus k’.
k’ = m
complementary
modulus.
REFERENCES [l] [2] [3] [4] [5]
0. L. Bowie and C. E. Freese, Int. J. Fracrwre Mech. 8, 49-57 (1972). S. S. Chang, Engng Fracrure Mech. 22, 253-261 (1985). G. Melrose, Doctoral Dissertation, Old Dominion University, Norfolk, Virginia (1984). F. Bowman, Introduction To EIIipric Functions with Applications. Dover, New York (1961). L. C. Woods, Theory of Subsonic Plane Flow. Cambridge University Press (1961). (Received I4 July 1987)
Vol. 29, No. 6, pp. 641446,
1988
0013-7Y44/88 s3.tlo+ .tn, Pergamon Press pk.
OUT OF PLANE SHEAR OF A CRACKED RECTANGULAR ORTHOTROPIC BLOCK G. MELROSE and S. DAVIDSON Old Dominion University, Department of Mathematics and Statistics, Norfolk, VA 23529-0077. U.S.A. Abstract-The problem under consideration is the out of plane shear of a cracked rectangular orthotropic block. The exact solution is obtained by stating the problem in terms of a triple trigonometric series relation, which in turn can be shown to be equivalent to a singular integral equation whose solution is known. For the case of constant shear, the solution simplifies greatly and numerical results are given,
1. INTRODUCTION WE CONSIDER the out of plane shear of a cracked
coordinates
rectangular
orthotropic
solid. In Cartesian
the solid is given by the inequalities
while the crack which runs the length of the solid in the z direction is given by O
2. FORMULATION
AND STATEMENT
It is well known that for an out-of-plane u, =
and for an orthotropic
OF THE PROBLEM
shear the only non zero displacement w(x, Y)
is (2.1)
body the non zero stresses are given by 8W
uxxz= Gn -; ax
aw uyz = G23 ay
where G13, GZ3 are the shear moduli of the orthotropic therefore be satisfied provided 2
(2.2)
material. The equilibrium equations will
2
p2$+q=0
where p2=$. ay
(2.3 23
By symmetry, we now see that the problem is reduced to solving the following mixed boundary value problem. 641
G. MELROSE and S. DAVIDSON
642
p.d.e. b.c.
p2w,,+
(O
wyy=O
(2.4)
O
(1) w,(O, y) = w,(rr, y) = 0
(0 < y < A7r)
r(x) (2) w,(x, h7r) = -
(0 < x < 7r)
G23
(0 < x < a) U (b < x < 7~)
(3) w(x, 0) = 0 w,(x, 0) = 0 It is easily verified
(a -=cx < b).
that by writing
4x9
1 y) = --cW(x,
y) + 44x7
y)l
(2.5)
G23
where _ B, sinh nPy Boy + “;, np cash nphaCoS nx
w~(x, Y) =;
(2.6)
T(X) cos nx dx then 4(x,
y) must satisfy
p.d.e.
P2bX + 4yv = 0
(0 < x <
0 < y < h7T).
7-r;
b.c. (1) 4x(0, y) = 4xx(~ y) = o
(2.7)
(2.8)
(O< y
(2) 4+, Ad = 0
(0 < x -=c7T)
(3) $4x, 0) = 0
(0 < x < a)
4+(x, 0) = -f(x)
(a
44x, 0) = 0
(b
where f(x) = %
(X, 0) = i BO + c B, sech(nphn) II=1
3. DETERMINATION
OF THE
(2.9)
cos nx.
PERTURBATION
4
Clearly,
(3.1)
satisfies requires
the p.d.e. and the first two boundary conditions. that the {A,}; must satisfy the triple series
G(x) = AC)+ f
Furthermore,
n-’ A, cos nx = 0
boundary
condition
(3)
(O
n=l
F(x)
=
2
n=l
G(x)=A,+
A, tanh n/3hrrcos
nx = f(x)
i n-’ A, cos nx = 0 fl=l
(a < x < b) (b
(3.2)
Out of plane shear of orthotropic block
643
The solution of these triple series is given in[3] and is outlined here. By choosing
@(t)
(3.3)
dt
and b
A,=-27T (1 p(t)sinntdt
(3.4)
I
then we find 4(x, o) = G(x) = ~[(b - x)(x -
a,]J’;p(t)
dt
(3.5)
a
where H(u) is the Heaviside step function. We also find that,
Mt) dt (O
F(x)
=
i
(3.6)
where M(x, t) = - 2 c tanh(npAr) II=1
= -kiog
cos nx sin nr
tn(F)
+ tn(ff)
tn(F)
_ tn(F)
(3.7)
and tn(u) = tn(u, k) is a Jacobian elliptic function. (See appendix for more details.) It is now clear that the triple series will be satisfied if p(t) is given by the integral equation tn
J
-1 b 7T (1 &)&log
K” 0 7T
/Kx\
+tn
K’ 0) /it\
dt=-f(x).
(a < x < 6)
(3.X)
‘“W -‘“W I
From[3] we find that
(3.9) and
F (x)
3x,0) = -F(x)
rA,(x) =
I
(a < x < 6)
-f(x) -
(O
Knc($)dc(F) m&(x)
F*(x)
(b < x < ?7)
(3.10)
G. MELROSE and S. DAVIDSON
644
where
A(x)f(x) tn(F)
dx (3.11)
tn2(!c)
c =
_ tn2(g
.
_2K;LF)nc(F)dc(F) dt A(x)f(x) t(F) dx ,.b
Ib
I
A(x) = ([ tn’(F)
(3.12)
A(0
a
n tn2(F)
- tn2(F)][
tn2($)
- tn2(F)
- tn2($)]}“’
(3.13)
and
A,(X) = ([ m2(e)
- tn2(F)][tn2(F)
- tn2(e)])1’2
(3.14)
with K; = K(k;);
k, = tn(F)/tn(E).
4. STRESS INTENSITY
FACTORS
The stress intensity factors k, and k+ are given by the formulae, k, = lim ~a,,(x,O) x-a-
(4.1)
kb = lim -a,,(~,
0).
(4.2)
x-b+
For this problem, using (2.2) and (2.5) these equations become k, = lim m$ .X-a-
(x, 0)
(4.3)
0).
(4.4)
kb = lim Jz(r-b)$(x. x-b+
On substituting
(3.10) into (4.3) and (4.4) we obtain
kc, =
F,(a)
I
7r[tn2(F)
and
Kdn
(
Ka
- tn2(T)isn($)
112
)
(4.5) en(F)
Out of plane shear of orthotropic
kb = - F,(b)
645
block
I
5. NUMERICAL
(4.6)
RESULTS
FOR CONSTANT
SHEAR
For the special case of a constant shear the problem reduces greatly. First we observe that wr(x, y) = ry and hence f(x) = T. This allows us to simplify the singular integral terms F,(a) and Fr(b). Indeed, we find that
and
F,(b) -= 7
-2K;tn Kdn
(
fi
7r Ka ( ‘TT)
)
b
(5.2)
Z(6, k;) dt
Z(P,k;)-
with sin p = sn(F)
k2 = dn(F)/dn($); and sn’(F)
-sn’($)
sn2($)
- sn’($)
sin2 6 =
and Z(0, k) is Jacobi’s Zeta function.
1.2-
1.0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Fig. 1. Stress intensity vs aspect ratio for various values of (As),
0.6
G. MELROSEand S. DAVIDSON
646 1.60-
-
k,/k,
---
k,/k,
c-r/4
1.60-
Fig. 2. Stress intensity vs aspect ratio for various values of (hp).
Figures 1 and 2 give the scaled stress intensities vs aspect ratio for various values of (A/3). Here k,, = & where d = (b - a)/2 and c denotes the midpoint of the crack. For c = ~r/2, by symmetry, we have k, = kb which is the special case investigated by Chang[2]. The results given here are in agreement with those obtained by Chang. We note how quickly k//c, degenerates to the results of an infinite strip with a central crack (A --, m) and the effect of orthotropy in this case is not very significant. For the off-center crack, however, the orthotropic property seems to have more bearing. APPENDIX The following notation, although standard, is listed here for completeness elliptic functions and integrals can be obtained from[4] and [5]. sn(y) = sn(y, k) en(y) - cn(y, k) My)
= My,
and clarity. An excellent introduction to
Jacobian elliptic functions with modulus k.
k)
tn(y)2w My)
complete elliptic integral of the 1st kind with modulus k.
K = K(k)
=~[l+2n~,e-*f12]2 k
2[
$ .-*,.+q
in terms of A, ratio of rectangular sides. modulus in terms of A.
n (1 K, = K(k,)
complete elliptic integral of the 1st kind with modulus k,.
K’ = K(k’)
associated complete elliptic integral of the 1st kind with modulus k’.
k’ = m
complementary
modulus.
REFERENCES [l] [2] [3] [4] [5]
0. L. Bowie and C. E. Freese, Int. J. Fracrwre Mech. 8, 49-57 (1972). S. S. Chang, Engng Fracrure Mech. 22, 253-261 (1985). G. Melrose, Doctoral Dissertation, Old Dominion University, Norfolk, Virginia (1984). F. Bowman, Introduction To EIIipric Functions with Applications. Dover, New York (1961). L. C. Woods, Theory of Subsonic Plane Flow. Cambridge University Press (1961). (Received I4 July 1987)