International Journal of Solids and Structures 38 (2001) 4759±4776
www.elsevier.com/locate/ijsolstr
The spherical-cap crack revisited
P.A. Martin *
Department of Mathematical and Computer Sciences, Colorado School of Mines, Golden, CO 80401-1887, USA
Received 15 May 2000; in revised form 25 July 2000; accepted 11 August 2000
Abstract
A crack in the shape of a spherical cap is subjected to a static loading. The exact solution for the crack-opening
displacement is obtained using a method based on dual series equations and Laplace transforms. For shallow caps, the
solution agrees with an asymptotic theory for perturbed penny-shaped cracks. Ó 2001 Elsevier Science Ltd. All rights
reserved.
1. Introduction
Let X be a loaded crack in a three-dimensional elastic solid. Assume that X is bounded and that the crack
edge oX is a simple, smooth, closed curve. The form of the stresses near oX is known (Leblond and Torlai,
1992), but the stress-intensity factors themselves can only be found by solving a boundary-value problem:
they cannot be obtained from a local analysis near the edge.
There is only one non-planar X for which the boundary-value problem can be solved exactly, and that is
a spherical cap. In fact, there are several Russian papers on this problem. One of the earliest is that of
Ziuzin and Mossakovskii (1970), but their analysis was subsequently criticised by Prokhorova and Solov'ev
(1976); both papers consider axisymmetric loadings, and use representations in terms of analytic functions
of a complex variable. A method for non-axisymmetric problems was developed more recently by Popov
(1992). He reduced the problems to some one-dimensional integral equations, whose desired solutions were
shown to have non-integrable end-point singularities. A method based on dual series equations was
sketched by Martynenko and Ulitko (1979). Their method is simple, in principle (we use it below), but it is,
nevertheless, complicated when detailed results are required. A striking feature of these papers is that they
do not contain mutual comparisons. Thus, it is dicult to know what the correct solution actually is, for
any particular loading!
Consequently, we decided to re-work the problem. The main purpose of the calculation is to obtain a
benchmark solution, so that other techniques (such as those based on solving a boundary integral equation
numerically) can be validated. We have also con®rmed the results for a shallow spherical cap by comparing
with an asymptotic theory for perturbed penny-shaped cracks (Martin, 2000).
*
Fax: +1-303-273-3875.
E-mail address: pamartin@mines.edu (P.A. Martin).
0020-7683/01/$ - see front matter Ó 2001 Elsevier Science Ltd. All rights reserved.
PII: S 0 0 2 0 - 7 6 8 3 ( 0 0 ) 0 0 3 1 0 - 3
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
The method used here is ®rst described in the context of a model problem for Laplace's equation. It
consists of patching separated solutions together in spherical polar coordinates, leading to some dual series
equations, which are then reduced to an integro-di€erential equation. We solve this equation using Laplace
transforms. We then generalise the method to the elasticity problem, which is much more complicated. It
leads to a pair of coupled integro-di€erential equations, which are essentially those obtained previously by
Martynenko and Ulitko (1979). We solve these using Laplace transforms. We also derive asymptotic approximations for the stress-intensity factors, valid for shallow spherical-cap cracks.
2. A model problem: potential ¯ow past a rigid spherical cap
Before embarking on the elasticity problem, it is instructive to consider a simpler problem for Laplace's
equation. The techniques used to solve this problem will generalise to the much more complicated crack
problem. Thus, consider a rigid spherical cap, given by
r ˆ c;
0 6 h < a;
0 6 / < 2p;
where r, h and / are spherical polar coordinates. We want to solve Laplace's equation, r2 u ˆ 0, outside the
cap, with
ou=or ˆ
U cos h on both sides of the cap;
2:1†
so that Uz ‡ u is the velocity potential for axisymmetric uniform ¯ow past the cap; u is required to vanish as
r ! 1. Separation of variables gives the representations
1
X
n
An …r=c† Pn …cos h† for 0 6 r < c and
u…r; h† ˆ Uc
nˆ0
u…r; h† ˆ Uc
1
X
nˆ0
Cn …r=c†
n 1
Pn …cos h†
for r > c;
where Pn is a Legendre polynomial and the dimensionless coecients An and Cn are to be found. Continuity
of ou=or across r ˆ c for 0 6 h 6 p gives nAn ˆ …n ‡ 1†Cn .
De®ne the discontinuity in u across r ˆ c by
‰u…h†Š ˆ lim u…r; h†
r!c
ˆ Uc
lim u…r; h†
r!c‡
1
X
2n ‡ 1† Un Pn …cos h†;
nˆ0
where Un ˆ An …n ‡ 1† 1 . Using Eq. (2.1) and the fact that ‰u…h†Š ˆ 0 for a 6 h 6 p gives
1
X
n…n ‡ 1† Un Pn …cos h† ˆ cos h; 0 6 h < a;
nˆ0
1
X
nˆ0
2n ‡ 1† Un Pn …cos h† ˆ 0;
a 6 h 6 p:
2:2†
2:3†
These form a pair of dual series equations for Un (Sneddon, 1966). To solve them, we use an integral
representation for Un ,
Z a
1
1
Un ˆ
u…t† sin n ‡
t dt;
2:4†
2n ‡ 1 0
2
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
where u…t† is to be found. This representation ensures that Eq. (2.3) is satis®ed for any u, due to the
discontinuous sum (A.4). It also gives
Z a
u…t†
p dt; 0 6 h < a:
‰u…h†Š ˆ Uc
2 cos h 2 cos t
h
From this, we can readily show that we must have u…0† ˆ 0, otherwise ‰u…0†Š would be unbounded.
Substitution of Eq. (2.4) in Eq. (2.2) gives
Z
1
X
n…n ‡ 1† a
1
u…t† sin n ‡
t dt Pn …cos h† ˆ cos h; 0 6 h < a;
2n ‡ 1 0
2
nˆ0
an equation for u…t†. To solve it, we would like to interchange the order of integration and summation, so
as to obtain an integral equation for u; however, the resulting sum is divergent, so that we must proceed
indirectly. An integration by parts in Eq. (2.4) gives
Z a
2k2n Un ˆ
u0 …t† cos kn t dt u…a† cos kn a;
2:5†
0
where kn ˆ n ‡ …1=2† and we have used u…0† ˆ 0. Then, we write Eq. (2.2) as
1
1
X
1X
2k2n Un Pn …cos h†
Un Pn …cos h† ˆ 2 cos h; 0 6 h < a:
2 nˆ0
nˆ0
Using Eqs. (2.4), (2.5), (A.3) and (A.5) gives
Z
1 a
0
u…s† ds ˆ 2 cos h;
Tt!h u …t†
4 t
0 6 h < a;
where we have de®ned the Abel operator T by
Z h
/…t† dt
p :
T / Tt!h f/…t†g ˆ
2 cos t 2 cos h
0
2:6†
2:7†
This operator can be inverted: if T / ˆ f , we can solve for / as (Porter and Stirling, 1990, Section 9.2)
Z s
2 d
f …h† sin h
1
1
p dh:
/…s† ˆ T f Th!s f f …h†g ˆ
2:8†
p ds 0
2 cos h 2 cos s
Applying T
u0 …t†
1
to Eq. (2.6) gives
Z
1 a
3
t ;
u…s† ds ˆ …4=p† cos
4 t
2
2:9†
which is to be solved subject to u…0† ˆ 0.
One way to solve Eq. (2.9) is to di€erentiate with respect to t, leading to a second-order ordinary differential equation for u with constant coecients. The general solution of this equation contains two arbitrary constants; these are to be determined by imposing u…0† ˆ 0 and by requiring that the solution of the
di€erential equation actually solves Eq. (2.9).
We shall use an alternative method, based on Laplace transforms. First, we write Eq. (2.9) in convolution form as
u0 …t† ‡
1
4
Z
t
0
1
u…s† ds ˆ M1
4
4=p† cos
3
t ;
2
2:10†
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
where M1 ˆ
Ra
u…t† dt is an unknown constant. Next, we de®ne
Z 1
u…t† e pt dt;
U…p† ˆ Lfug ˆ
0
2:11†
0
where p is the transform variable. Then, taking the Laplace transform of Eq. (2.10), we obtain
4=p†p2 =…p2 ‡ 94†;
p2 ‡ 14† U…p† ˆ 14M1
whence
U…p† ˆ
M1
1
‡
2p
4
p2
1
‡ 14
9
1
:
2
2p p ‡ 94
Inverting, using Lfsin btg ˆ b=…p2 ‡ b2 †, we obtain
u…t† ˆ 12M1 ‡ p 1 sin…12t† …3=p† sin…32t†:
To determine M1 , we substitute for u in the de®nition of M1 and evaluate the integral, giving
4=p† sin a sin…12a†
M1 ˆ
and then
u…t† ˆ …1=p† sec…12a† cos…32a† sin…12t†
3=p† sin…32t†:
This solution agrees with the well known solution of Collins (1959).
3. Elastic ®eld representations
For the crack problem, we start with representations for the elastic displacement u in terms of potentials.
From Lur'e (1964, Section 6.2), we have general solutions for three-dimensional elasticity, in spherical polar
coordinates …r; h; /†, where u ˆ …ur ; uh ; u/ † is independent of the azimuthal angle, /. Thus, the following
representations can be obtained.
Interior solution: displacements
2 ‡ 4m† ‡ Bnrn 1 gPn ;
ur ˆ fArn‡1 …n ‡ 1†…n
4m† ‡ Brn 1 gPn1 :
fArn‡1 …n ‡ 5
uh ˆ
Exterior solution: displacements
ur ˆ fCr n n…n ‡ 3
uh ˆ fCr n …n
4m†
4 ‡ 4m†
Dr
n 2
Dr
n 2
n
2
n ‡ 1†gPn ;
gPn1 :
Interior solution: stresses
2l† 1 srr ˆ fA…n ‡ 1†…n2
1
2l† srh ˆ
fA…n2 ‡ 2n
2m†rn ‡ Bn…n
1 ‡ 2m†rn ‡ B…n
1†rn 2 gPn ;
1†rn 2 gPn1 :
Exterior solution: stresses
1
fCn…n2 ‡ 3n
1
fC…n2
2l† srr ˆ
2l† srh ˆ
2m†r
2 ‡ 2m†r
n 1
n 1
D…n ‡ 1†…n ‡ 2†r
D…n ‡ 2†r
n 3
gPn1 :
n 3
gPn ;
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
4763
All these solutions are valid for n ˆ 0; 1; 2; . . .. In them, m is Poisson's ratio (Lur'e uses m ˆ 1=m), l is the
shear modulus, Pn ˆ Pn …cos h† and Pn1 ˆ Pn1 …cos h† ˆ Pn0 …cos h† sin h ˆ …d=dh†Pn …cos h†; note that P01 0.
Expressions for the other stress components are given on p. 330 of Lur'e's book.
We consider a crack in the shape of a spherical cap, given by r ˆ c, 0 6 h < a and 0 6 / < 2p. Using
superscripts …1† and …2† for the regions r < c and r > c, respectively, we have the following representations
for the displacements and stresses:
1 r n‡1
r n 1 X
ur…1† ˆ A0 …4m 2†r ‡ c
n ‡ 1†…n 2 ‡ 4m† ‡ Bn
n Pn ;
3:1†
An
c
c
nˆ1
1 c 2
c n
c n‡2
X
2†
ur ˆ cD0
Cn
‡c
n…n ‡ 3 4m† Dn
n ‡ 1† Pn ;
3:2†
r
r
r
nˆ1
1 r n‡1
r n 1 X
1†
uh ˆ c
An
n ‡ 5 4m† ‡ Bn
Pn1 ;
3:3†
c
c
nˆ1
1 c n
c n‡2 X
2†
uh ˆ c
n 4 ‡ 4m† Dn
3:4†
Cn
Pn1 ;
r
r
nˆ1
1
1†
ˆ
2l† srr
1
2†
2l† srr
2…1 ‡ m†A0
1 r n
r n 2
X
2
n ‡ 1†…n
n 2 2m† ‡ Bn
n…n 1† Pn ;
An
‡
c
c
nˆ1
1 c 3 X
c n‡1
c n‡3
2
ˆ 2D0
Cn
n…n ‡ 3n 2m† Dn
n ‡ 1†…n ‡ 2† Pn ;
r
r
r
nˆ1
1†
1 r n
X
n2 ‡ 2n
An
c
nˆ1
2†
1 X
2l† 1 srh ˆ
2l† 1 srh ˆ
Cn
nˆ1
c n‡1
r
n2
1 ‡ 2m† ‡ Bn
2 ‡ 2m†
Dn
r n
c
c n‡3
r
2
n
1† Pn1 ;
3:5†
3:6†
n ‡ 2† Pn1 :
These formulas were obtained using A ˆ c n An , B ˆ c2 n Bn , C ˆ cn‡1 Cn and D ˆ cn‡3 Dn , where the coecients An , Bn , Cn and Dn are dimensionless. These coecients are to be determined by applying the boundary
conditions on r ˆ c. The ®rst of these is that the stresses should be continuous across r ˆ c for all h, when
1 ‡ m†A0 ˆ D0 ;
An …n ‡ 1†…n2
An …n2 ‡ 2n
n
3:7†
2
2m† ‡ Bn n…n
1 ‡ 2m† ‡ Bn …n
Cn n…n2 ‡ 3n
1† ˆ
1† ˆ Cn …n2
2 ‡ 2m†
2m† ‡ Dn …n ‡ 1†…n ‡ 2†;
Dn …n ‡ 2†;
3:8†
3:9†
for n ˆ 1; 2; . . .. From these, we obtain
Dn Cn ˆ …n
1†f…n ‡ 1†…2n ‡ 3†An ‡ …2n ‡ 1†Bn g;
n ‡ 2†Dn Dn ˆ …2n ‡ 1†fn…n ‡ 2†…n2
where Dn ˆ 2fn2
when n ˆ 0.
n ‡ …2n ‡ 1†…1
1† ‡ 4
4m2 gAn ‡ n…n
3:10†
1†…n ‡ 2†…2n
1†Bn ;
3:11†
m†g. In particular, C1 ˆ 0. Note that Eq. (3.11) reduces to Eq. (3.7)
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
If the material in r < c is di€erent from that in r > c, one obtains an interface crack on r ˆ c. The
corresponding problem has been considered by Altenbach et al. (1995), using the representations given
above.
4. The crack-opening displacement
We de®ne the crack-opening displacement ‰uŠ by
‰ur …h†Š ˆ ur…1† …c; h†
ur…2† …c; h† ˆ c
1†
2†
‰uh …h†Š ˆ uh …c; h†
uh …c; h† ˆ c
1
X
2n ‡ 1† Un Pn ;
4:1†
nˆ0
1
X
nˆ1
2n ‡ 1† Vn Pn1 ;
4:2†
where the dimensionless coecients Un and Vn are to be determined. Comparing with the representations
(3.1)±(3.4), we ®nd that U0 ˆ …4m 2†A0 ‡ D0 ˆ 3…1 m†A0 ,
2n ‡ 1†Un ˆ An …n ‡ 1†…n
2n ‡ 1†Vn ˆ An …n ‡ 5
2 ‡ 4m† ‡ Bn n
Cn n…n ‡ 3
4m† ‡ Bn ‡ Cn …n
4 ‡ 4m†
Dn :
4m† ‡ Dn …n ‡ 1†;
Substituting for Cn and Dn from Eqs. (3.10) and (3.11), respectively, we ®nd that
n ‡ 2†Dn Un ˆ
m†f…2n ‡ 3†…n2
2…1
n ‡ 2†Dn Vn ˆ 2…1
2
m†f…2n ‡ 3†…n ‡ 3n
2 ‡ 2m†…n ‡ 1†An ‡ …2n
2m†An ‡ …2n
1†…n ‡ 2†nBn g;
1†…n ‡ 2†Bn g:
We can rewrite these, giving An and Bn in terms of Un and Vn :
2…1
2…1
m†…2n ‡ 3†An ˆ
2
n ‡ 2†…Un ‡ nVn †;
2m†Un ‡ …n ‡ 1†…n2
1†Bn ˆ …n ‡ 3n
m†…2n
2 ‡ 2m†Vn :
We can now state the problem to be solved: ®nd Un and Vn , so that
‰ur …h†Š ˆ 0
and
‰uh …h†Š ˆ 0 for a < h 6 p;
4:3†
and
srr …h† ˆ
lqr …h† and
srh …h† ˆ
lqh …h†
on r ˆ c
for 0 6 h < a;
4:4†
where qr and qh are given functions of h. In particular, for uniaxial tension at in®nity in the z-direction (so
that s1
zz ˆ p0 , say), we have (Lur'e, 1964, p. 343)
qr …h† ˆ …p0 =l† cos2 h
qh …h† ˆ
and
p0 =l† sin h cos h:
4:5†
Using Eq. (4.4) in Eqs. (3.5) and (3.6) gives
1
qr ˆ
2
1
qh ˆ
2
2…1 ‡ m†A0 ‡
1
X
nˆ1
An …n2 ‡ 2n
1
X
nˆ1
An …n ‡ 1†…n2
1 ‡ 2m† ‡ Bn …n
n
2
1† Pn1 :
Eliminating An and Bn in favour of Un and Vn gives
2m† ‡ Bn n…n
1† Pn ;
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
1
1
2
m†qr ˆ
1
1
2
m†qh ˆ
1
X
nˆ0
1
X
nˆ1
2‰w2n
1†ŠUn ‡ wn ‰wn ‡ 1 ‡ m…4wn
1 ‡ m…2wn
f‰wn ‡ 1 ‡ m…4wn
5†ŠUn ‡ ‰…wn ‡ 1†…2wn
5†ŠVn
3† ‡ 3mŠVn g
Pn
;
4wn 3
Pn1
4wn 3
4765
4:6†
4:7†
for 0 6 h < a, where wn ˆ n…n ‡ 1†. These are to be solved subject to Eq. (4.3), wherein ‰uŠ is de®ned by Eqs.
(4.1) and (4.2). Of particular interest are the stress-intensity factors. We de®ne these by
p p
p p
‰ur Š Kn 2a c…a h† and ‰uh Š Ks 2a c…a h† as h ! a ;
4:8†
where a is a length scale and Kn and Ks are dimensionless stress-intensity factors. It is convenient to take
a ˆ c sin a, for we may then consider the limiting case of a penny-shaped crack of radius a, obtained by
taking the limits c ! 1 and a ! 0 with a ®xed.
The de®nitions of the stress-intensity
factors in Eq. (4.8) are convenient, but not standard. For example,
p
e n = 2pq0 as q0 ! 0, where q0 is distance from the crack edge oX. Making
it is usual to suppose that srr K
use of the known general relations between ‰uŠ behind oX and the stress components ahead of oX (see, for
example, Rice (1989, p. 32)), we ®nd that
p
e n ˆ 1l paKn =…1 m†:
K
4:9†
2
This formula and the corresponding formula for Ks can be used to obtain expressions for the standard
stress-intensity factors from the results derived below.
5. Reduction to integro-di€erential equations
We introduce representations (2.4) for Un and
Z a
1
1
w…t† cos n ‡
t dt
Vn ˆ
4n…n ‡ 1† 0
2
5:1†
for Vn , where the functions u and w are to be found. Substituting Eq. (2.4) in Eq. (4.1), followed by
evaluation of the sum using Eq. (A.4), shows that ‰ur …h†Š ˆ 0 for h > a, as required, for any choice of u; we
also obtain
Z a
u…t†
p dt; 0 6 h < a:
‰ur …h†Š ˆ c
5:2†
2
cos
h 2 cos t
h
‰ur …0†Š will be bounded provided that t 1 u…t† is integrable near t ˆ 0, so that, in particular,
u…0† ˆ 0:
5:3†
Similarly, substituting Eq. (5.1) in Eq. (4.2), followed by use of Eq. (A.7) shows that ‰uh …h†Š ˆ 0 for h > a,
as required, provided that w satis®es
Z a
1
t dt ˆ 0:
5:4†
w…t† cos
2
0
We also obtain
‰uh …h†Š ˆ
c
2 sin h
Z
a
h
w…t† sin t
p dt;
2 cos h 2 cos t
0 6 h < a;
it turns out that ‰uh …0†Š ˆ 0, as expected from symmetry considerations.
5:5†
4766
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
The stress-intensity factors, Kn and Ks , can be expressed directly in terms of u and w, respectively. Thus,
from Eq. (5.2), we have
p
1
‰ur …h†Š c u…a† …sin a†
2 cos h 2 cos a as h ! a
with a similar approximation for ‰uh Š. Then, comparison with Eq. (4.8) gives
Kn ˆ
u…a†
sin a
and
Ks ˆ
w…a†
:
2 sin a
5:6†
To obtain u and w, we substitute Eqs. (2.4) and (5.1) in Eqs. (4.6) and (4.7), and evaluate the series using
results from the appendix. To do this, we note the following partial-fraction expansions:
2‰w2n 1 ‡ m…2wn 1†Š kn
d1
d3
1
1
‡
ˆ
‡
;
5:7†
2n ‡ 1†…4wn 3†
4 4kn 8 kn 1 kn ‡ 1
wn ‡ 1 ‡ m…4wn 5†
d3
1
1
ˆ d0 ‡
;
5:8†
4wn 3
2 k n 1 kn ‡ 1
wn ‡ 1 ‡ m…4wn 5† d0 kn
d2
3d3
1
1
ˆ
‡
‡
‡
;
5:9†
2n ‡ 1†…4wn 3†
2wn 8wn kn 16wn kn 1 kn ‡ 1
wn ‡ 1†…2wn 3† ‡ 3m k2n 3d3
1
1
ˆ
:
5:10†
4wn 3
kn 1 kn ‡ 1
2
4
In these, wn ˆ n…n ‡ 1†, kn ˆ n ‡ 1=2,
d0 ˆ 14…1 ‡ 4m†;
d1 ˆ 163 …5 ‡ 8m†;
d2 ˆ 163 …1
8m†
and
d3 ˆ 161 …7
8m†:
However, prior to substitution, we note that the ®rst terms on the right-hand sides of Eqs. (5.7) and
(5.10) will lead to divergent series. To overcome this, we ®rst integrate by parts in Eqs. (2.4) and (5.1). Thus,
making use of u…0† ˆ 0, we obtain Eq. (2.5) and then
Z
1
X
kn
1 a 0
1
u…a† f0 …a; h†
2n ‡ 1†Un
Pn …cos h† ˆ
u …t† f0 …t; h† dt
4
4
4
0
nˆ0
Z
1 h
u0 …t† dt
p ;
ˆ
4 0
2 cos t 2 cos h
where f0 is de®ned by Eq. (A.2) and we have used Eq. (A.3), noting that h < a. Similarly, we have
Z a
w0 …t† sin kn t dt;
4wn kn Vn ˆ w…a† sin kn a
0
whence
1
X
nˆ1
Vn
k2n 1
1
P …cos h† ˆ
8 sin h
2 n
Z
h
0
w0 …t† sin t dt
p ;
2 cos t 2 cos h
where we have used Eqs. (A.6) and (5.4). Use of these results, together with Eqs. (2.4), (5.1) and (5.7)±
(5.10), in Eqs. (4.6) and (4.7) gives
Z a
1
1
0
1 m†qr ˆ Tt!h fu …t†g ‡
5:11†
fu…t† S11 …t; h† ‡ w…t† S12 …t; h†gdt;
2
4
0
Z a
1
1
1 m†qh ˆ
Tt!h fw0 …t† sin tg ‡
5:12†
fu…t† S21 …t; h† ‡ w…t† S22 …t; h†gdt;
2
8 sin h
0
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
4767
for 0 6 h < a, where
1 X
d1
d3
1
1
S11 ˆ
‡
Pn …cos h† sin kn t;
4kn 8 kn 1 kn ‡ 1
nˆ0
1 X
d0 d3
1
1
S12 ˆ
‡
Pn …cos h† cos kn t;
4
8 kn 1 kn ‡ 1
nˆ0
1 X
d0 kn
d2
3d3
1
1
sin kn t
‡
;
S21 ˆ
‡
‡
Pn1
k
k
wn
2
8k
16
1
‡
1
n
n
n
nˆ1
1 3d3 X
1
1
cos kn t
:
Pn1 …cos h†
S22 ˆ
wn
16 nˆ1 kn 1 kn ‡ 1
The Abel operator T and its inverse are de®ned by Eqs. (2.7) and (2.8), respectively.
The sums Sij can be evaluated explicitly using results from the appendix. We obtain
1
S11 ˆ
4
S12 ˆ
Z
t
f0 …s; h† fd1
0
d0
f0 …t; h†
4
d3
4
sin h S21 ˆ 12d0 f0 …t; h† sin t
3d3
sin h S22 ˆ
8
Z
0
d3 cos…t
Z
t
0
f0 …s; h† sin…t
s† ds
d3 sin3 …12h† sin t ‡
1
8
Z
t
f01 …s; h†
1
h ;
2
d3
1
h ;
cos t sin
2
2
d3
s†g ds ‡ sin t sin
2
sin…t
s† ds
d3 sin
2
t
0
f01 …s; h† fd2 ‡ 3d3 cos…t
1
h
2
cos
1
t
2
sin
s†g ds;
1
h
2
cos t ;
where f01 is de®ned by Eq. (A.8). Then, changing the order of integration gives
Z a
d3
1
h
u…t† sin t dt;
‰d1 d3 cos…t s†Š u…s† ds ‡ sin
2
2
0
0
t
Z a
Z a
Z a
1
d3
1
h
sin
w…t† S12 …t; h† dt ˆ Tt!h d0 w…t† ‡ d3
w…s† sin…t s† ds
w…t† cos t dt;
4
2
2
0
t
0
Z a
Z a
1b
sin h
‰d2 ‡ 3d3 cos…t s†Š u…s† ds
u…t† S21 …t; h† dt ˆ T t!h
8
t
0
Z a
1
1
3
h
u…t† sin t dt;
‡ d0 Tt!h fu…t† sin tg d3 sin
2
2
0
Z a
Z a
Z a
3d3 b
1
3
h
w…t† S22 …t; h† dt ˆ
T t!h
sin h
w…t† cos t dt;
w…s† sin…t s† ds ‡ d3 sin
2
8
0
0
t
Z
a
1
u…t† S11 …t; h† dt ˆ Tt!h
4
Z
a
where we have used Eq. (5.4) to obtain the last formula and the operator Tb is de®ned by
Tb / Tbt!h f/…t†g ˆ
Z
0
h
p
/…t† 2 cos t 2 cos h dt:
4768
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
Using these results, we apply 4T 1 to Eq. (5.11), giving
Z a
Z a
0
u …s† ‡ d0 w…s† ‡ d3 sin s
fu…t† sin t w…t† cos tg dt ‡
fd1
0
s
Z a
‡ d3
w…t† sin…s t† dt ˆ Qr …s†
d3 cos…s
t†g u…t† dt
s
1 1
1
for 0 < s < a, where Qr …s† ˆ 2…1 m† Th!s
sin 2 h ˆ …1=2† sin s.
fqr …h†g and we have used Th!s
1
Next, we apply 8…sin s† T 1 sin h to Eq. (5.12). To do this, we ®rst calculate that
Z s
1 b
T t!h f/…t†g ˆ sin s
Th!s
/…t† dt;
and
0
ˆ sin s sin2 …12 s†, whence
Z s
Z a
Z a
0
w …s† 4d0 u…s† d2
u…t† t dt ‡ s
u…t† dt
3d3 cos s
fu…t† sin t
0
s
0
Z s
Z a
3d3
w…t† dt 3d3
fu…t† sin…s t† ‡ w…t† cos…s t†g dt ˆ Qh …s†;
1
Th!s
sin3 12 h
5:13†
3
4
0
s
w…t† cos tg dt
5:14†
1
1
for 0 < s < a, where Qh …s† ˆ 4…1 m†…sin s† Th!s
fqh …h† sin hg.
Eqs. (5.13) and (5.14) have been obtained by Martynenko and Ulitko (1979), although few details were
given; see their Eq. (2.4), noting that our w is their … 4w†. These authors considered all-round tension at
in®nity.
For uniaxial tension at in®nity, qr and qh are given by Eq. (4.5); elementary calculations, using Eq. (2.8),
then give
Qr …s† ˆ 13g…cos…12s† ‡ 2 cos…52s††
with g ˆ
4…1
and
Qh …s† ˆ 85g sin…52s†;
5:15†
m†p0 =…pl†.
6. Solution of the integro-di€erential equations
Martynenko and Ulitko (1979) solved Eqs. (5.13) and (5.14) (for all-round tension) by repeated differentiation, leading to a pair of coupled homogeneous ordinary di€erential equations with constant coecients, relating u, u00 , u…iv† , w0 , w000 and w…v† . The general solution of this system contains 17 arbitrary
constants, and these were then determined by substituting back in Eqs. (5.13) and (5.14), together with the
conditions (5.3) and (5.4).
Here, we solve Eqs. (5.13) and (5.14) using Laplace transforms. First, we rewrite the integro-di€erential
equations in convolution form as
Z s
u0 …s† ‡ d0 w…s†
6:1†
fk11 …s t† u…t† ‡ k12 …s t† w…t†gdt ˆ g1 …s†;
Z0 s
w0 …s† 4d0 u…s†
6:2†
fk21 …s t† u…t† ‡ k22 …s t† w…t†gdt ˆ g2 …s†;
0
where
k11 …s† ˆ d1
d3 cos s;
k12 …s† ˆ d3 sin s;
k21 …s† ˆ d2 s 3d3 sin s; k22 …s† ˆ 3d3 …1 cos s†;
g1 …s† ˆ Qr …s† ‡ d3 M2 cos s d1 M1 ; g2 …s† ˆ Qh …s† ‡ 3d3 M2 sin s ‡ sd2 M1 ;
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
M1 ˆ
Z
a
u…t† dt
and
0
M2 ˆ
Z
0
4769
a
fu…t† cos t ‡ w…t† sin tg dt:
6:3†
Note that the constants M1 and M2 are unknown, as they depend on the solutions u and w.
Next, we introduce U…p† ˆ Lfug and W…p† ˆ Lfwg, the Laplace transforms of u and w, respectively,
de®ned by Eq. (2.11); we use upper-case letters to denote the Laplace transforms of other functions. Then,
taking the Laplace transform of Eqs. (6.1) and (6.2) gives
p
K11 †U ‡ …d0 K12 †W ˆ G1 ;
4d0 ‡ K21 †U ‡ …p K22 †W ˆ G2 ‡ w0 ;
where w0 ˆ w…0† and we have used u…0† ˆ 0. Hence
DU ˆ …p K22 †G1 …d0 K12 †…G2 ‡ w0 †;
DW ˆ …4d0 ‡ K21 †G1 ‡ …p K11 †…G2 ‡ w0 †;
where D ˆ …p
K11 †…p
K11 …p† ˆ
d1
p
K21 …p† ˆ
K22 † ‡ …d0
6:4†
6:5†
K12 †…4d0 ‡ K21 †. We have
d3 p
d3
;
; K12 …p† ˆ 2
p2 ‡ 1
p ‡1
d2
3d3
3d3
; K22 …p† ˆ
;
2
2
p
p ‡1
p…p2 ‡ 1†
whence
D…p† ˆ
p2 ‡ 94† R…p2 †
;
p2 …p2 ‡ 1†
6:6†
where
R…x† ˆ x2
8m2 †x ‡ 169 :
1
3
2
p
Note that R…x† ˆ 0 when x ˆ 34 2m2 2imc with c ˆ 12 3 4m2 .
From Eqs. (6.4)±(6.6), we obtain
U…p† ˆ
W…p† ˆ
‰p2 …p2 ‡ 1†
4…d0 p2
d2 †…p2 ‡ 14†G1 ‡ ‰p4 ‡ …1 2d0 †p2
p2 ‡ 94† R…p2 †
using d3 ˆ d0 ‡ d2 ˆ d1
G1 …p† ˆ g
G2 …p† ˆ g
p2
3d3 ŠpG1 ‰d0 p2 d2 Šp2 …G2 ‡ w0 †
;
p2 ‡ 94† R…p2 †
d1 Šp…G2 ‡ w0 †
2d0 . Also, for uniaxial loading, Eq. (5.15) gives
p…p2 ‡ 94†
d3 M2 p2 d1 M1 …p2 ‡ 1†
;
‡
1
25
2
p…p2 ‡ 1†
‡ 4†…p ‡ 4 †
4
3d3 M2 p2 ‡ d2 M1 …p2 ‡ 1†
:
‡
25
p2 …p2 ‡ 1†
p2 ‡ 4
Thus, it is convenient to write
U ˆ gU1 ‡ U2
and
W ˆ gW1 ‡ W2 :
We have
U1 ˆ
;
p2 A1 …p2 †
p2 ‡ 14†…p2 ‡ 94†…p2 ‡ 254† R…p2 †
and
U2 ˆ
A2 …p2 †
;
p2 ‡ 94† R…p2 †
4770
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
where
A1 …p2 † ˆ p6 ‡ 14p4 …9
2
2
A2 …p † ˆ p …p
2
16m† ‡ 161 p2 …23
2
3d0 †d3 M2
p …d0 p
88m†
2
d2 †w0
1
177
64
fp4 d1
120m†;
‡ p2 …d1 ‡ d0 d2 †
81
gM1 :
64
Next, we split into partial fractions, as
p2 ‡ 34†D1 ‡ …p2
A1
B1
C1
‡
‡
‡
1
9
25
R…p2 †
p 2 ‡ 4 p2 ‡ 4 p 2 ‡ 4
U1 ˆ
p2 ‡ 34†D2 ‡ …p2
B2
‡
R…p2 †
p2 ‡ 94
U2 ˆ
3
†E2
4
3
†E1
4
;
;
where
1
;
16…1 ‡ m†
A1 ˆ
B1 ˆ
3
16…1
m…3 ‡ 5m 10m2 †
;
4…1 m2 †…7 5m†
D1 ˆ
m†
E1 ˆ
25
C1 ˆ
;
4…7
5m†
;
13 25m 6m2 ‡ 20m3
;
8…1 m2 †…7 5m†
3
f16d3 M2 4w0 …3 ‡ 8m†M1 g;
64…1 m†
m
D2 ˆ
f16d3 M2 ‡ 4…3 4m†w0 ‡ …13 24m†M1 g;
32…1 m†
1
16d3 …1 2m†M2 4…1 ‡ 6m 8m2 †w0 …51
E2 ˆ
64…1 m†
B2 ˆ
14m
48m2 †M1 :
Note that
lim U1 …p† ˆ 1 ˆ A1 ‡ B1 ‡ C1 ‡ D1 ‡ E1 :
6:7†
p!1
Inverting the Laplace transforms, we obtain
u…t† ˆ 2gA1 sin…12t† ‡ 23…gB1 ‡ B2 † sin…32t† ‡ 25gC1 sin…52t† ‡ m 1 …gD1 ‡ D2 † cosh ct sin mt
‡ c 1 …gE1 ‡ E2 † sinh ct cos mt;
using Lfcosh ct sin mtg ˆ m…p2 ‡ 34†=R…p2 † and Lfsinh ct cos mtg ˆ c…p2
Similarly, for W, we have
p B1 …p2 †
9
†…p2 ‡ 254† R…p2 †
4
and
W2 ˆ
B1 …p2 † ˆ …5 ‡ 4m†p4 ‡ 72p2 …1 ‡ 2m†
3
29
16
4
W1 ˆ
p2
‡
p B2 …p2 †
;
‡ 94† R…p2 †
p2
where
2
B2 …p † ˆ …1 ‡ m†…4p
3
1
16
2
3†d3 M2 ‡ fp ‡ …1
‡ m†f4…1 ‡ 8m†p
Splitting into partial fractions,
2
40m†;
2d0 †p2
3 ‡ 40mgM1 :
d1 gw0
3
†=R…p2 †.
4
6:8†
4771
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
W1 ˆ
W2 ˆ
p B3
p C3
p…p2
‡
‡
p2 ‡ 94 p2 ‡ 254
p B4
p…p2
‡
p2 ‡ 94
16
B,
9 1
where B3 ˆ
D3 ˆ
11
3…1
C3 ˆ
c2 ‡ m2 †D3 ‡ 2pE3
;
R…p2 †
c2 ‡ m2 †D4 ‡ 2pE4
;
R…p2 †
24
C,
25 1
13m
;
m†…7 5m†
B4 ˆ
E3 ˆ
16
B,
9 2
m…1 2m†…15 17m†
;
6…1 m†…7 5m†
1
f16d3 M2 ‡ 4…2 3m†w0 …3 ‡ 8m†M1 g;
12…1 m†
m
16md3 M2 ‡ 4…3 ‡ 2m 6m2 †w0 ‡ …48 3m
E4 ˆ
24…1 m†
D4 ˆ
56m2 †M1 :
Inverting the Laplace transforms, we obtain
w…t† ˆ …gB3 ‡ B4 † cos…32t† ‡ gC3 cos…52t†
‡ …gD3 ‡ D4 † cosh ct cos mt ‡ …cm† 1 …gE3 ‡ E4 † sinh ct sin mt;
6:9†
using Lfcosh ct cos mtg ˆ p…p2 c2 ‡ m2 †=R…p2 † and Lfsinh ct sin mtg ˆ 2cmp=R…p2 †. Note that, setting
t ˆ 0 in Eq. (6.9), we obtain w…0† ˆ w0 , as expected, since B3 ‡ C3 ‡ D3 ˆ 0 and B4 ‡ D4 ˆ w0 .
6.1. Determination of M1 , M2 and w0
At this stage, we have expressions for u…t† and w…t† involving the three constants w0 , M1 and M2 . To
determine these, we use the de®nitions of M1 and M2 , Eq. (6.3), and constraint (5.4). We start by rewriting
Eqs. (6.8) and (6.9) so as to display the three constants. Thus, we have
u…t† ˆ gu1 …t† ‡ …1
w…t† ˆ gw1 …t† ‡ …1
1
m† fw0 h0 …t† ‡ M1 h1 …t† ‡ M2 h2 …t†g;
6:10†
1
m† fw0 `0 …t† ‡ M1 `1 …t† ‡ M2 `2 …t†g;
6:11†
where
u1 …t† ˆ 2A1 sin…12t† ‡ 23B1 sin…32t† ‡ 25C1 sin…52t† ‡ m 1 D1 cosh ct sin mt ‡ c 1 E1 sinh ct cos mt;
w1 …t† ˆ B3 cos…32t† ‡ C3 cos…52t† ‡ D3 cosh ct cos mt ‡ …cm† 1 E3 sinh ct sin mt;
h0 …t† ˆ
h1 …t† ˆ
h2 …t† ˆ
`0 …t† ˆ
`1 …t† ˆ
`2 …t† ˆ
1
fsin…32t† ‡ …3 4m† cosh ct sin mt ‡ 12c 1 …1 ‡ 6m 8m2 † sinh ct cos mtg;
8
1
f…3 ‡ 8m† sin…32t† ‡ …13 24m† cosh ct sin mt ‡ 12c 1 …51 14m 48m2 † sinh ct
32
1
d fsin…32t† cosh ct sin mt ‡ 12c 1 …1 2m† sinh ct cos mtg;
2 2
1
fcos…32t† ‡ …2 3m† cosh ct cos mt 12c 1 …3 ‡ 2m 6m2 † sinh ct sin mtg;
3
1
f…3 ‡ 8m†…cos…32t† cosh ct cos mt† 12c 1 …48 3m 56m2 † sinh ct sin mtg;
12
4
d fcos…32t† cosh ct cos mt ‡ 12…m=c† sinh ct sin mtg:
3 2
cos mtg;
Then, Eqs. (5.4) and (6.3) give
3
X
jˆ1
Aij xj ˆ ci ;
i ˆ 1; 2; 3;
where x1 ˆ M1 , x2 ˆ M2 , x3 ˆ w0 ,
6:12†
4772
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
c1 ˆ …1
m†g
Z
A11 ˆ 1
m
Z
Z
A21 ˆ
Z
A23 ˆ
A31 ˆ
Z
u1 dt;
c2 ˆ …1
h1 dt;
A12 ˆ
m†g
Z
h1 cos t ‡ `1 sin t† dt;
Z
u1 cos t ‡ w1 sin t† dt;
A13 ˆ
h2 dt;
A22 ˆ 1
Z
m
Z
c3 ˆ
1
m†g
Z
w1 cos 12t dt;
h0 dt;
h2 cos t ‡ `2 sin t† dt;
h0 cos t ‡ `0 sin t† dt;
`1 cos
1
t dt;
2
A32 ˆ
Z
`2 cos
1
t dt;
2
A33 ˆ
Z
`0 cos
1
t dt
2
and all the integrals are over the range 0 6 t 6 a; they are all elementary. Eq. (6.12) is a system of three
simultaneous algebraic equations for M1 , M2 and w0 . Then, u and w are given by Eqs. (6.10) and (6.11),
respectively, the components of the crack-opening displacement ‰uŠ are given by Eqs. (5.2) and (5.5), and the
stress-intensity factors are given by Eq. (5.6).
Rather than giving the explicit solution of system (6.12) (which is straightforward but tedious), we obtain
an approximate asymptotic solution for a shallow spherical cap.
7. The shallow spherical-cap crack
Suppose that the cap is shallow, which means that c ! 1 and a ! 0 with a ˆ c sin a ®xed. Then, the
stress-intensity factors are given by Eq. (5.6) in which a is small. We can solve the 3 3 system (6.12) in this
limit. We have
u1 …t† ˆ t ‡ O…t3 †
~ 2 ‡ O…t4 †
w1 …t† ˆ wt
and
as t ! 0;
making use of Eq. (6.7) and B3 ‡ C3 ‡ D3 ˆ 0; the constant w~ is given by
w~ ˆ
9
B
8 3
25
C
8 3
‡ 12…c2
m2 †D3 ‡ E3 ˆ 12…5 ‡ 4m†:
There are similar approximations for hi …t† and `i …t†, and these lead to small-a approximations for Aij and ci .
Hence, we obtain
M1 ˆ M2 ˆ 12ga2 ‡ O…a4 †
as a ! 0. It follows that
u…a† ˆ ga ‡ O…a3 †
and
and
w0 ˆ
and
‡ O…a4 †
~ 2 ‡ O…a4 †
w…a† ˆ 23gwa
as a ! 0, whence
Kn ˆ g ‡ O…a2 †
1 ~ 2
gwa
3
Ks ˆ
1
g…5
6
‡ 4m†a ‡ O…a3 † as a ! 0
7:1†
with g ˆ 4…1 m†p0 =…pl†. The leading-order term corresponds to the well-known stress-intensity factor
for a penny-shaped crack opened by a constant pressure; (Eq. (4.9)). The ®rst-order correction is seen to
occur in the tangential component. We have derived this ®rst-order correction by an independent method,
giving some credence to both approaches. In that method, we combined a perturbation expansion with an
exact hypersingular boundary integral equation for ‰uŠ, the method being designed for cracks that are
perturbations of ¯at circular cracks, namely, wrinkled penny-shaped cracks (Martin, 2000).
P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
4773
Appendix A. The Mehler±Dirichlet integral, sums and variants
The standard Mehler±Dirichlet integral is (Whittaker and Watson, 1927, Section 15.231)
Z
cos…n ‡ 12†t
2 h
p
Pn …cos h† ˆ
dt;
p 0
2 cos t 2 cos h
valid for n ˆ 0; 1; 2; . . . , and 0 6 h 6 p. De®ne kn ˆ n ‡ 12 and
1=2
; 0 < t < h;
f0 …t; h† ˆ …2 cos t 2 cos h†
0;
h < t 6 p;
Rp
so that Pn …cos h† ˆ …2=p† 0 f0 …t; h† cos kn t dt. De®ne f0 …t; h† for p 6 t < 2p by f0 …t; h† ˆ f0 …2p
then expand the resulting extended function as a half-range Fourier cosine series. The result is
f0 …t; h† ˆ
1
X
nˆ0
Pn …cos h† cos kn t;
0 < t;
h 6 p;
t 6ˆ h:
A:1†
A:2†
t; h† and
A:3†
Such discontinuous sums are useful when solving dual series equations.
Replacing
t and h in Eq. (A.1) by …p t† and …p h†, respectively, we obtain Pn …cos h† ˆ
Rp
2=p† 0 f1 …t; h† sin kn t dt, where
0;
0 6 t < h;
f1 …t; h† ˆ
1=2
2 cos h 2 cos t†
; h < t 6 p:
Extending f1 using f1 …t; h† ˆ f1 …2p
f1 …t; h† ˆ
1
X
nˆ0
t; h† for p 6 t 6 2p and then expanding as a half-range sine series gives
Pn …cos h† sin kn t;
0 6 t;
h < p;
t 6ˆ h:
From Eqs. (A.3) and (A.4), we have
Z t
1
X
sin kn t
f0 …s; h† ds;
ˆ
Pn …cos h†
kn
0
nˆ0
1
X
nˆ0
Pn …cos h†
cos kn t
ˆ
kn
Z
t
p
f1 …s; h† ds:
We also have
f1 …t; h† cos t f0 …t; h† sin t ˆ
f0 …t; h† cos t f1 …t; h† sin t ˆ
1
X
nˆ0
1
X
nˆ0
Pn …cos h† sin…kn 1†t;
Pn …cos h† cos…kn 1†t;
whence
1
X
nˆ0
and
Pn …cos h†
sin…kn 1†t
ˆ
kn 1
Z
t
0
ff0 …s; h† cos s f1 …s; h† sin sg ds ˆ S ;
A:4†
A:5†
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
1
X
nˆ0
Thus,
1
X
nˆ0
cos…kn 1†t
Pn …cos h†
ˆ
kn 1
Pn …cos h†
Z
p
ff1 …s; h† cos s f0 …s; h† sin sg ds ˆ C :
t
1
X
sin kn t
cos kn t
ˆ S cos t C sin t;
ˆ C cos t S sin t:
Pn …cos h†
kn 1
kn 1
nˆ0
In particular,
1 X
1
1
‡
Pn …cos h† sin kn t ˆ …S‡ ‡ S † cos t …C‡ C † sin t
kn 1 kn ‡ 1
nˆ0
Z t
ˆ2
f0 …s; h† cos…t s† ds 4 sin t sin 12h;
0
1 X
nˆ0
1
kn
1
1
kn ‡ 1
Pn …cos h† cos kn t ˆ …C
ˆ
C‡ † cos t
2
Z
S‡ ‡ S † sin t
t
0
f0 …s; h† sin…t
s† ds
4 cos t sin 12h:
We need similar formulas involving Pn1 . Use of the recurrence relation
2n ‡ 1† Pn1 …cos h† sin h ˆ n…n ‡ 1†fPn 1 …cos h†
Pn‡1 …cos h†g
gives
2n ‡ 1 1
4
Pn …cos h† sin h ˆ
n…n ‡ 1†
p
ˆ
Z
4
p
p
f0 …t; h† sin t sin kn t dt
0
Z
p
0
f1 …t; h† sin t cos kn t dt
for n ˆ 1; 2; . . .. It follows that
1
X
kn 1
1
1
1
sin t
tan
h sin
t ‡
f0 …t; h†
Pn …cos h† sin kn t ˆ
2
2
2
sin h
wn
nˆ1
1
X
kn 1
1
1
1
sin t
h cos
t
f1 …t; h†
Pn …cos h† cos kn t ˆ cot
2
2
2
sin
h
w
n
nˆ1
A:6†
A:7†
for 0 < t; h < p, t 6ˆ h, where wn ˆ n…n ‡ 1†. Integrating Eq. (A.6) once gives
1
X
cos kn t
1
1
f 1 …t; h†
h cos
t ‡ 0
;
ˆ tan
Pn1 …cos h†
wn
2
2
sin h
nˆ1
where
f01 …t; h†
ˆ
2 cos t
0;
1=2
2 cos h†
;
0 6 t < h;
h 6 t 6 p:
Integrating again gives
1
X
nˆ1
Pn1 …cos h†
sin kn t
ˆ
kn w n
2 tan
1
h
2
sin
Z t
1
1
t ‡
f 1 …s; h† ds:
2
sin h 0 0
A:8†
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
Similarly, Eq. (A.7) gives
1
X
nˆ1
Pn1 …cos h†
sin kn t
ˆ cot
wn
1
h
2
sin
1
t
2
f11 …t; h†
;
sin h
where
f11 …t; h† ˆ
0;
2 cos h
2 cos t†
1=2
0 6 t < h;
h 6 t 6 p:
;
We also have
sin h
1
X
nˆ1
Pn1 …cos h†
sin…kn 1†t
ˆ …1 ‡ cos h† sin
wn
1
t
2
1
t
2
cos t …1 ‡ cos h† sin
cos t …1
cos h† cos
sin t
1
t
2
f11 …t; h† cos t f01 …t; h† sin t;
sin h
1
X
nˆ1
Pn1 …cos h†
cos…kn 1†t
ˆ
wn
1
cos h† cos
1
t
2
sin t
‡ f01 …t; h† cos t f11 …t; h† sin t;
whence
sin h
1
X
nˆ1
Pn1 …cos h†
sin…kn 1†t
ˆ
wn …kn 1†
Z
t
1
t
cos s sin s ds …1 cos h† sin
2
0
1
1
1
1
1
‡ sin
t
1 ‡ cos h† sin
t
sin
t
3
2
2
3
2
f01 …s; h†
f11 …s; h†
ˆ S1 ;
and
sin h
1
X
nˆ1
Pn1 …cos h†
cos…kn 1†t
ˆ
wn …kn 1†
Z
p
t
‡
f11 …s; h†
1
cos
3
1
t
2
cos s ˆ C1 :
Thus,
sin h
1
X
Pn1 …cos h†
sin h
1
X
sin kn t
ˆ S1 cos t C1 sin t;
wn …kn 1†
Pn1 …cos h†
cos kn t
ˆ C1 cos t S1 sin t:
wn …kn 1†
nˆ1
nˆ1
In particular,
1
f01 …s; h†
sin s ds ‡ …1 ‡ cos h†
cos
1
1
1
cos h† cos
t ‡ cos
t
2
3
2
1
t
2
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P.A. Martin / International Journal of Solids and Structures 38 (2001) 4759±4776
sin h
1 X
nˆ1
sin h
1 X
nˆ1
1
1
‡
1 kn ‡ 1
1
1
kn ‡ 1
kn
kn
1
Pn1 …cos h†
Pn1 …cos h†
sin kn t
ˆ S1‡ ‡ S1 cos t ‡ C1 C1‡ sin t
wn
Z t
ˆ2
f01 …s; h† cos…t s† ds
0
16
1
1
1
1
sin2
h
sin
t
sin
h sin t ;
‡
3
2
2
2
2
cos kn t
S1‡ ‡ S1 sin t
ˆ C1 C1‡ cos t
wn
Z t
ˆ 2
f01 …s; h† sin…t s† ds
0
16
1
1
1
sin2
h
cos
t
sin
h cos t :
‡
3
2
2
2
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