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Elastic green's function for an infinite half-space of a hexagonal continuum with its basal plane as surface
ht. J. Ehgng ki Vol. 17, pp. 681489 @ PcrgamonPress Ltd., 1979. Pnnted in Great Britain
ELASTIC GREEN’S FUNCTION FOR AN INFINITE HALF-SPACE OF A HEXAGONAL CONTINUUM WITH ITS BASAL PLANE AS SURFACE N. G. LEE Theoretical
Physics
Division,
AERE,
(Communicated
Harwell,
Didcot,
Englandt
by I. N. SNEDDON)
Abstract-The elastic Green’s function of a continuum is useful for investigating distortions of a substrate due to localised surface forces. A closed analytic solution for the Green’s function is found for the specific case of an infinite half-space of a hexagonal continuum with its basal plane as surface. The problem differs from that of an infinite whole space as boundary conditions require that there are no normal components of stress at the surface. It has been verified that the result satisfies these boundary conditions and that, in the appropriate limit, it coincides with Mindlin’s solution for an isotropic continuum. An expression for the stresses in an infinite half-space of a hexagonal continuum with its basal plane as surface is appended. This expression has also been shown to agree with its isotropic counterpart in the appropriate limit.
1. INTRODUCTION
situations localised surface forces distort the substrate on which they act. Examples include catalytic systems where metal particles reside on surfaces, as well as cases where individual atoms are adsorbed. The substrate distortion is important in determining the various interaction energies involved and the precise configuration of adsoibed atoms expected[l]. All the important energies can be calculated from the Green’s function for the substrate. Here a specific case is analysed, namely a substrate consisting of an infinite half-space of a hexagonal continuum, with the surface as the basal plane. The results should be useful for studies of surface phenomena on hexagonal substrates like graphite, and as an approximation for (111) surfaces of cubic crystals (see Appendix 5). The Green’s function Q, for which a closed analytic solution is obtained, is defined by IN
MANY
practical
u(x) = ax,x’)f(x’) where &,x’ are points within the half-space and where u(x) is the displacement at 8 due to a point force f(x’) acting at 3’. The boundary conditions require that there are no normal components of stress at the surface. Numerical checks show that the solution agrees with that for the isotropic half-space121 in the appropriate limit and that the boundary conditions are satisfied. The expression for the stresses in an infinite half-space of a hexagonal continuum with its basal plane as surface, presented in Appendix 3, is similarly shown to agree with its isotropic counterpart[2] in the appropriate limit. 2. CALCULATION
OF
c(x.8’)
We choose a right-handed Cartesian co-ordinate system x1, x2 and x3 in which the half-space occupies x3 B 0, and apply a point force at (0, 0, h), where h 2 0. The equations that define the Green’s function Q are these CijtiGp.jl+ &pS(X,)S(xMX3- h) = 0, Ci3dkp.l
= 0,
x3 =
We shall find Q by decomposing it in the form
fOn attachment
from the School
of Mathematics,
University 681
of Bath.
0.
x320
(la)
(lb)
N. G. LEE
682
where 9” is the infinite body Green’s function and $7 satisfies x350
cii&G&,i,= 0%
SMG&J = - Ci3tiG&,t*
(2a)
x3 = 0.
WI
2.1 Form of Cm A solution for G” has been determined by Kriiner[3], and corrected by Willis[4]. G” is symmetric and in different notation has the form
ATx2(m: +23;
EN;
-h&+BN;
c,x,z
~9’
cNx2z (12
DN
mN =
/
.\/(aNq’+z2)
q2=x:tx:, z=x:,-h
u2,
a3
AN
=
(k-
BN
= (c44-
CN
=
(cl.?
DN
=
cc44 -
6
=
(al
-
ud(a3
-
at)
E2
= (a2
-
a3Hal
-
a3
E3
=
-
u,)(u2
-
03)
(a3
C,i)(C33
-
Wk)
aNdc33-
a&44)
+ (C13 +
+ d(c44
-
Qd2
+ UN@13
+ cd2
&V%)
&‘h)h4
-
&&d
are the roots of the quadratic equation c,,c~u~ t (~2:~+ ~c~~c~-- c,,c~~)u +
c33c44
=
0.
The formula is only valid when al, a2and u3 are distinct. Appendix 4 gives some relationships involving the parameters defined in the paper. When 4 = 0, an expression for G” is obtained by expanding the mN and using relationships given in Appendix 4. For this case ke find <- to be diagonal and proportional to I$‘.
The term G’ can be found via its Fourier transform 6’. Fist, writing I’ be verified film eqn (2a) that a general representation for 4’ is $h,
kz, x3)
=
R(F)
8
VM
=I
e-iHxx
for q’@, h), it can
(3)
where the VM (M = 1,2,3) are matrices to be found. This equation is derived in Appendix 1. In eqn (3) the matrix Z?(k) is the adjoint of the matrix L(k) which has components J-&C) = Ci&k~. If the roots of the equation det L(kr, k2, k3) = 0
(4)
ElasticGreen’sfunction for a hexagonalcontinuum
which lie in the upper half of the complex k3 plane are K?(k,,
PM= tkrtk2rKt’tk,,kd),
683
kz) (M = 1,2,3), then
(M = 1,2,3).
The corresponding expressions for the roots in the lower half of the complex k3 plane are K M+3= p
= (k,, kt, e(k,,
(M= 1,2,3).
k,)),
There are exactly three such roots K?(k,, k2). This follows from eqn (4), which is a sextic with real coefficients. Using the restrictions on the elastic constants given by Nye[5], it can be shown that for real non-zero &,det L(k) > 0, and hence that eqn (4) has no real roots. The matrices VM can be found using eqn (2b). It is shown in Appendix 2 that 3 F(lppR(~)VM
=
-
;
’
p
F =l
3
F(lJN)R(&N)eiKy’ 8
-1
(5)
det L(KN) ak,
where the matrix F(k) has components &(&) = CiJk,k,. Equation (5) may be satisfied by taking *
3 iKyh
v”=-&
=,
lNCktl
kJ
(6)
e
&
independently of M, where
The derivation of eqn (5) given in Appendix 2 assumes that the roots KY (M = 1,2,3) are distinct. This is so for the general transversely isotropic medium and for the cubic approximation to one. For a wholly isotropic medium there are three equal roots. We shall not develop our method of solution for the isotropic case as the Green’s function is well known[2]. From eqns (3) for 6’ and (6) for VM we have -3 @kr,
k2,
~3)
=
2
3 =,
$3
=,
l?(p)tN(k,,
kJe_
iqxj+iKqh
.
(7)
If we define & as a 2~imensional vector (k,, kd, the Fourier ~ansform of eqn (7) gives
where AMN = (x, kj f xrkz i x3- hK?)ilkj. We now evaluate the integrals to obtain the desired expression for c’(x). It may be noted that the functions K?(kl,k2) are homogeneous of degree one: Ky(hk,, hk2) = hK?(k,, k2) if A > 0. Since the algebraic part of the integrand in eqn (8) is homogeneous of degree - 1, we have (9) where 7 is the unit vector k/(kl _ _ and where
N. G. LEE
684
Equation (9) must be rewritten to allow for cases when the imaginary part of (x3=zero
hK,N) is
Hence Q’(x) = -
j-$$
i
N?r)6N(rl,,
lim f
M=I N=I c+o+
l?l=l X17)1+X272+
712) dS
X,b;,
-
hbN - ic
where bu = K?(vI,
with 191= 1,
712)
(M = 1,2,3).
Note that b’M= -a; (M = 1,2,3), so our demands that KY (M = 1,2,3) be distinct, and the aM (M = 1,2,3) likewise distinct, are equivalent. With lg(= 1, all of the integrals have a denominator of the form xlql +x2q2+ c, where c is independent of Q, and numerators containing at most terms in 1, nl, n2, vi, vi, 0~7)~.Letting n1 = cos 0, ~2 = sin 8 and z = e”, the integrals may be expressed in the form $I+~ r(z) dz, where r is a rational function. The integrals are easily evaluated to give the final result for Q’
ix1x2 44
4
(SMN - thfN) UMN :
(ShfN - tMN)UhfN:
ipMN(-h
x2):
X~UMNTMN
(10)
;
In eqn (10) the following definitions are used
(9
PMN (XI, X2) = E
+ -$ [SM#MN(~I,
X2) + t~Nhf~(X2,
x;tt’m - x:&N %Nblr
x2)
=
+ TMNb:
nMN
TMN=$(l-E) UMN= 2T*N -
2&N + q2 nMN
hfN
= d/(&N
+ q2)
TMN = - i(XjbM
(ii)
rMN =
bLHdN
SMN = bLH,@N
+ b&MAN
tm
-
= b:B&N
A&N
UMN = b #MPN
+ b&MXN
UMN = b&M(@N
+ PN) + b&DMAN
WMN = b&MPN
(iii)
+ hbN)
5 = alus
+ ~‘MDMXN
- P(a
+
r)l
-
* 2 xl)
X111
Elastic Green’s function for a hexagonal continuum
685
4N = (rP - Sa)a, PN =
&bN
+ aaN)
PN
=
@UN
+ UfiN)
AN
=
d~(~N
XN
= d&N
+ YNYN)+ (a + (0
+ ?)a,]
+ r)PNl
3
d=
(iv)
i%
bkdiv;
d = a, P, y, 8, (T
@, = b&BN BN
=
&33DN
-
c,3cN
YN=AN-CN SN
=
bN[C,3HN
UN
=
bN(DN
-
c33cN1
CN)
HN = BN - aN&.
With the exception of bL = - a; (M = 1,2,3), all other parameters are given in Section 2.1. Appendix 4 summarises some relations among the parameters. When 4 = 0, an expression for Q’ is obtained by expanding the nMN.For this case Q is found to be diagonal, with its elements consisting of sums of terms each of which is the reciprocal of a linear factor of x3 and h.
3. DISCUSSION
The Green’s function for an infinite half-space of a hexagonal continuum with its basal plane as surface is given by the sum of G”, given in Section 2.1, and Q’ of eqn (10). Both of these expressions are complicated, and it-is likely that practical use of them will involve numerical methods. We have used numerical methods for two specific checks. First, it has been demonstrated that the surface boundary condition is indeed satisfied. Secondly, we have shown that our results reduce to Mindlin’s for an isotropic system. Finally we observe that G&, x’) = G&‘, x). Acknowledgements-1
am very grateful to Prof. J. R. Willis and Dr. A. M. Stoneham for their invaluable advice.
REFERENCES [I] A. M. STONEHAM, Solid State Communications 24.425 (1977). [21 R. D. MINDLIN, Physics 7, 195 (1936). [3] E. KRONER, Zeit. fiir Physik 136,402 (1953). [4] J. R. WILLIS, Q. I. Mech. Appl. Math. 18,419 (l%S). [5] J. F. NYE, Physical Properties of Crystals, p. 142. Clarendon Press, Oxford (1957). (Received
28 September
APPENDIX
1978)
1
With x32 0 throughout, we have from eqn (2a) Qk,G$.&) = 0. On Fourier transforming we have Gik&J!(k) = 0 and
(Al)
N. G. LEE
686 This can be written
L(k)G’(k) = 0 so that in particular L(&M)Q’(&M)= 0,
(M= 1.2,..., 6).
Thus (M=l,2,...,6)
Q(&“, = R(rrM) V”,
tA2)
where VM (M = i,2,. . ., 4) are arbitrary matrices. From eqn (Al) we have c,,&p.,,(kt, kz,x3) so
= 0
that
or, in operator form k2, x,) = 0
W&k,,
(A3)
where the operator P(k) has com~nents
aI[ (WI+ 43k + i&3&I
E
P>ttk)= ctjkl(Sit+ &P$ + 43 ’ a*,
I
Since the @KM) are functions of kl, kz only, we may try
where the FM are functions to be found. Thus, from eqns (A2), (A3) and (A4) we have R(fJ”)VMFM(k,,
kz,xJ = 0.
Solutions of this equation are obtained by setting p(W+(k,,
k,, x,) = UB”)FM(k,, kr, x3.
(M= 1,2(..., 6).
This condition is satisfied if
$ 3 FM = - [KY FM,
$
FM = -(Kr)‘F”,
(M=1,2,....6)
3
and hence if FM = e-rK%,,
(M=
I,...,@.
Hence from eqns (A2), (A4) and (AS) we have
Since @(k,, kz,xJ-+O as xl++-,
we take @‘(k,,kz,x3 = $
=I
R(p)
Vu e-+~j z
as a general solution.
APPENDIX Defining
2
(AS)
Elastic Green’s function taking I, PO and writing y(x)
for a hexagonal continuum
687
for B”(J, h), it follows from eqn (la) that
@(k)= &
L!&)l-' enjh
which, inverted with respect to k3, gives
c”(k,,
k2, x,) = $
_‘- [L(k)]-’ I m
e-“@3-h’dk+
If we demand that x3 C h, we can evaluate the integral using Cauchy’s Theorem by closing the contour in the upper half of the complex k, plane. Thus
(A6)
From eqn (2b) we have, for xl = 0 c,,wG&., = - c,&%.i. Hence
and Q(k,. kzr x@(k,,
kz. x3 = - Q(k,, kz, x@-(k,,
(A7)
kz, x,),
where the operator Q(kl, kz. x3) has components
Qtdhv kz. ~3)= c,tkl Substituting
-
ik& + 6&t&-
a . axII
from eqns (3) and (A6) in eqn (A7) we have
Ok, kz,1,)
3 R(@-‘) e-‘K~W”
R(p)VMCqx3 =2 Q(k),k2,xd &=,
a
,
,-(KN)
det
x, = 0
ak, _
which gives eqn (5)
’ 3 F(&N)R(&N)e'Ky* VM= 2 J
F(p)R(p)
a det
=I
L(KN)
ak, where the
maWiX
F(k)
has components
&(k)
= C,jkk.
APPENDIX
3
Using the co-ordinate system defined in Section 2 the stresses set up in the continuum force I(?‘) at x’, are given by the symmetric matrix function (T(X) where u,(x) = c&)9 @I = Ull. 61 = lt.
and the symmetric
a2 = u22. c2 = Cm
(i,j=l,2
03 = 033, c3 = cm
,...,
u4 = u23. c4 = 2c23.
matrix of strains C(F) is given by
Q(X) = 1 [u,.J(x) •t u&)1
= i IGek.,(SX’)t G,d&
&‘)lfk(f)~
Thus, redefining [ as one third of its original value (see eqn (to)), we have
at a point J due to the action of a
6) a5 = (Tn. es = 2Q3,
06 = u12
l6 =
2c,z
N. G. LEE
688
where the operator g is given by
and ,,(xi,12~=xl[-2~~(tm:~~~+X~~)
+& (
r
WI,
x2) =
[vMNbnrcl,-CII(r~N~S~N)l+2CS6(SOIN_fIWN) 3 nMN q4
1 &[QN('h: -7 IT4 (
X2%
&
-
_i. yNf2x:-
_ aN [2ANx:z2 + ENq4] M',
ruin
4*
(
4X:) + Z2(4X: - 12x:)/#]
(
-3
q2
4 )
nMN
x:l_ [yNh*+
IbfN +
-
BNdl
(
mN
)
m'N
)
tMv1
&N
b'M
)I
TMN ~(d--X:hN-hfbhf~
+r [6
&* &-s(xdh,N t
mN
&MN-h,WI
d s,fx, x2j=c44
~N~~ll~~~~l3cN)
-hfN)I
(
~(~Y[SUNX:+~HNX:I-LIUN.X~)+~M~MN I)1 n
kt1-&6lffN
- Cd,
N )
(
4
)
>I
When q =0, an expression for $ is obtained by expanding the mN and the n&rN,and using relationships given in Appendix 4. For this case we find that
and that S3. ST,St* consist of sums of terms each of which is the reciprocal squared of a linear factor of xj and h.
APPENDIX Some relationships concerning the parameters defined in the paper
4
{b,, bx,b3t = f- 6,s - 62, - 63) Ez=B3=CL=Dr=H,=0 $,
AN/&=
$,
CN/EN
=O
a*=n3=B,=6,=a,=cr,+y,=O e2=83=~2=~~=~,=AI~~,=B,+p,=o bfk, + s,,fE -
bf(r,, + .s,,)E, = 0,
bf(r,+f,,)E,-b:(r,,+r,,)E,=o, bf~g,E, t bfU,iE, = 0,
(i,j=1,2,3) fi,j=
1,2,3)
(i,j=1,2,3)
Elastic Green’s function for a hexagonal continuum b;w,,E, - bfw& = 0,
r,, = t,, = 0,
689
(i,j= 1,2,3)
(i = I, 2,3; j = 2,3)
s,, = IA,,= If,, = w,, = 0,
(i=
1,2,3)
r,, =o Pdh
x2) = PdXh x2) = 0,
(i =
2,3).
APPENDIX 5 With reference to the co-ordinate system defined in Section 2, let cir denote the matrix of elastic constants of a hexagonal medium with its surface a basal plane. Similarly let ci, denote the matrix of elastic constants of a cubic medium with its surface a (I 11) plane. These matrices have a similar form, differing only in that some elements which are zero in the cji matrix are proportional to the anisotropy factor n = - c’,, + c;* t 2c$ in the c:, matrix. Thus, by setting to zero those elements of the c:, matrix which are proportional to n we may apply our results to cubic crystals with (Ill) surfaces. Clearly this approximation is likely to have greater validity for cubic crystals with weak anisotropy.
LJF3Vol. 17.No. 6-C
ELASTIC GREEN’S FUNCTION FOR AN INFINITE HALF-SPACE OF A HEXAGONAL CONTINUUM WITH ITS BASAL PLANE AS SURFACE N. G. LEE Theoretical
Physics
Division,
AERE,
(Communicated
Harwell,
Didcot,
Englandt
by I. N. SNEDDON)
Abstract-The elastic Green’s function of a continuum is useful for investigating distortions of a substrate due to localised surface forces. A closed analytic solution for the Green’s function is found for the specific case of an infinite half-space of a hexagonal continuum with its basal plane as surface. The problem differs from that of an infinite whole space as boundary conditions require that there are no normal components of stress at the surface. It has been verified that the result satisfies these boundary conditions and that, in the appropriate limit, it coincides with Mindlin’s solution for an isotropic continuum. An expression for the stresses in an infinite half-space of a hexagonal continuum with its basal plane as surface is appended. This expression has also been shown to agree with its isotropic counterpart in the appropriate limit.
1. INTRODUCTION
situations localised surface forces distort the substrate on which they act. Examples include catalytic systems where metal particles reside on surfaces, as well as cases where individual atoms are adsorbed. The substrate distortion is important in determining the various interaction energies involved and the precise configuration of adsoibed atoms expected[l]. All the important energies can be calculated from the Green’s function for the substrate. Here a specific case is analysed, namely a substrate consisting of an infinite half-space of a hexagonal continuum, with the surface as the basal plane. The results should be useful for studies of surface phenomena on hexagonal substrates like graphite, and as an approximation for (111) surfaces of cubic crystals (see Appendix 5). The Green’s function Q, for which a closed analytic solution is obtained, is defined by IN
MANY
practical
u(x) = ax,x’)f(x’) where &,x’ are points within the half-space and where u(x) is the displacement at 8 due to a point force f(x’) acting at 3’. The boundary conditions require that there are no normal components of stress at the surface. Numerical checks show that the solution agrees with that for the isotropic half-space121 in the appropriate limit and that the boundary conditions are satisfied. The expression for the stresses in an infinite half-space of a hexagonal continuum with its basal plane as surface, presented in Appendix 3, is similarly shown to agree with its isotropic counterpart[2] in the appropriate limit. 2. CALCULATION
OF
c(x.8’)
We choose a right-handed Cartesian co-ordinate system x1, x2 and x3 in which the half-space occupies x3 B 0, and apply a point force at (0, 0, h), where h 2 0. The equations that define the Green’s function Q are these CijtiGp.jl+ &pS(X,)S(xMX3- h) = 0, Ci3dkp.l
= 0,
x3 =
We shall find Q by decomposing it in the form
fOn attachment
from the School
of Mathematics,
University 681
of Bath.
0.
x320
(la)
(lb)
N. G. LEE
682
where 9” is the infinite body Green’s function and $7 satisfies x350
cii&G&,i,= 0%
SMG&J = - Ci3tiG&,t*
(2a)
x3 = 0.
WI
2.1 Form of Cm A solution for G” has been determined by Kriiner[3], and corrected by Willis[4]. G” is symmetric and in different notation has the form
ATx2(m: +23;
EN;
-h&+BN;
c,x,z
~9’
cNx2z (12
DN
mN =
/
.\/(aNq’+z2)
q2=x:tx:, z=x:,-h
u2,
a3
AN
=
(k-
BN
= (c44-
CN
=
(cl.?
DN
=
cc44 -
6
=
(al
-
ud(a3
-
at)
E2
= (a2
-
a3Hal
-
a3
E3
=
-
u,)(u2
-
03)
(a3
C,i)(C33
-
Wk)
aNdc33-
a&44)
+ (C13 +
+ d(c44
-
Qd2
+ UN@13
+ cd2
&V%)
&‘h)h4
-
&&d
are the roots of the quadratic equation c,,c~u~ t (~2:~+ ~c~~c~-- c,,c~~)u +
c33c44
=
0.
The formula is only valid when al, a2and u3 are distinct. Appendix 4 gives some relationships involving the parameters defined in the paper. When 4 = 0, an expression for G” is obtained by expanding the mN and using relationships given in Appendix 4. For this case ke find <- to be diagonal and proportional to I$‘.
The term G’ can be found via its Fourier transform 6’. Fist, writing I’ be verified film eqn (2a) that a general representation for 4’ is $h,
kz, x3)
=
R(F)
8
VM
=I
e-iHxx
for q’@, h), it can
(3)
where the VM (M = 1,2,3) are matrices to be found. This equation is derived in Appendix 1. In eqn (3) the matrix Z?(k) is the adjoint of the matrix L(k) which has components J-&C) = Ci&k~. If the roots of the equation det L(kr, k2, k3) = 0
(4)
ElasticGreen’sfunction for a hexagonalcontinuum
which lie in the upper half of the complex k3 plane are K?(k,,
PM= tkrtk2rKt’tk,,kd),
683
kz) (M = 1,2,3), then
(M = 1,2,3).
The corresponding expressions for the roots in the lower half of the complex k3 plane are K M+3= p
= (k,, kt, e(k,,
(M= 1,2,3).
k,)),
There are exactly three such roots K?(k,, k2). This follows from eqn (4), which is a sextic with real coefficients. Using the restrictions on the elastic constants given by Nye[5], it can be shown that for real non-zero &,det L(k) > 0, and hence that eqn (4) has no real roots. The matrices VM can be found using eqn (2b). It is shown in Appendix 2 that 3 F(lppR(~)VM
=
-
;
’
p
F =l
3
F(lJN)R(&N)eiKy’ 8
-1
(5)
det L(KN) ak,
where the matrix F(k) has components &(&) = CiJk,k,. Equation (5) may be satisfied by taking *
3 iKyh
v”=-&
=,
lNCktl
kJ
(6)
e
&
independently of M, where
The derivation of eqn (5) given in Appendix 2 assumes that the roots KY (M = 1,2,3) are distinct. This is so for the general transversely isotropic medium and for the cubic approximation to one. For a wholly isotropic medium there are three equal roots. We shall not develop our method of solution for the isotropic case as the Green’s function is well known[2]. From eqns (3) for 6’ and (6) for VM we have -3 @kr,
k2,
~3)
=
2
3 =,
$3
=,
l?(p)tN(k,,
kJe_
iqxj+iKqh
.
(7)
If we define & as a 2~imensional vector (k,, kd, the Fourier ~ansform of eqn (7) gives
where AMN = (x, kj f xrkz i x3- hK?)ilkj. We now evaluate the integrals to obtain the desired expression for c’(x). It may be noted that the functions K?(kl,k2) are homogeneous of degree one: Ky(hk,, hk2) = hK?(k,, k2) if A > 0. Since the algebraic part of the integrand in eqn (8) is homogeneous of degree - 1, we have (9) where 7 is the unit vector k/(kl _ _ and where
N. G. LEE
684
Equation (9) must be rewritten to allow for cases when the imaginary part of (x3=zero
hK,N) is
Hence Q’(x) = -
j-$$
i
N?r)6N(rl,,
lim f
M=I N=I c+o+
l?l=l X17)1+X272+
712) dS
X,b;,
-
hbN - ic
where bu = K?(vI,
with 191= 1,
712)
(M = 1,2,3).
Note that b’M= -a; (M = 1,2,3), so our demands that KY (M = 1,2,3) be distinct, and the aM (M = 1,2,3) likewise distinct, are equivalent. With lg(= 1, all of the integrals have a denominator of the form xlql +x2q2+ c, where c is independent of Q, and numerators containing at most terms in 1, nl, n2, vi, vi, 0~7)~.Letting n1 = cos 0, ~2 = sin 8 and z = e”, the integrals may be expressed in the form $I+~ r(z) dz, where r is a rational function. The integrals are easily evaluated to give the final result for Q’
ix1x2 44
4
(SMN - thfN) UMN :
(ShfN - tMN)UhfN:
ipMN(-h
x2):
X~UMNTMN
(10)
;
In eqn (10) the following definitions are used
(9
PMN (XI, X2) = E
+ -$ [SM#MN(~I,
X2) + t~Nhf~(X2,
x;tt’m - x:&N %Nblr
x2)
=
+ TMNb:
nMN
TMN=$(l-E) UMN= 2T*N -
2&N + q2 nMN
hfN
= d/(&N
+ q2)
TMN = - i(XjbM
(ii)
rMN =
bLHdN
SMN = bLH,@N
+ b&MAN
tm
-
= b:B&N
A&N
UMN = b #MPN
+ b&MXN
UMN = b&M(@N
+ PN) + b&DMAN
WMN = b&MPN
(iii)
+ hbN)
5 = alus
+ ~‘MDMXN
- P(a
+
r)l
-
* 2 xl)
X111
Elastic Green’s function for a hexagonal continuum
685
4N = (rP - Sa)a, PN =
&bN
+ aaN)
PN
=
@UN
+ UfiN)
AN
=
d~(~N
XN
= d&N
+ YNYN)+ (a + (0
+ ?)a,]
+ r)PNl
3
d=
(iv)
i%
bkdiv;
d = a, P, y, 8, (T
@, = b&BN BN
=
&33DN
-
c,3cN
YN=AN-CN SN
=
bN[C,3HN
UN
=
bN(DN
-
c33cN1
CN)
HN = BN - aN&.
With the exception of bL = - a; (M = 1,2,3), all other parameters are given in Section 2.1. Appendix 4 summarises some relations among the parameters. When 4 = 0, an expression for Q’ is obtained by expanding the nMN.For this case Q is found to be diagonal, with its elements consisting of sums of terms each of which is the reciprocal of a linear factor of x3 and h.
3. DISCUSSION
The Green’s function for an infinite half-space of a hexagonal continuum with its basal plane as surface is given by the sum of G”, given in Section 2.1, and Q’ of eqn (10). Both of these expressions are complicated, and it-is likely that practical use of them will involve numerical methods. We have used numerical methods for two specific checks. First, it has been demonstrated that the surface boundary condition is indeed satisfied. Secondly, we have shown that our results reduce to Mindlin’s for an isotropic system. Finally we observe that G&, x’) = G&‘, x). Acknowledgements-1
am very grateful to Prof. J. R. Willis and Dr. A. M. Stoneham for their invaluable advice.
REFERENCES [I] A. M. STONEHAM, Solid State Communications 24.425 (1977). [21 R. D. MINDLIN, Physics 7, 195 (1936). [3] E. KRONER, Zeit. fiir Physik 136,402 (1953). [4] J. R. WILLIS, Q. I. Mech. Appl. Math. 18,419 (l%S). [5] J. F. NYE, Physical Properties of Crystals, p. 142. Clarendon Press, Oxford (1957). (Received
28 September
APPENDIX
1978)
1
With x32 0 throughout, we have from eqn (2a) Qk,G$.&) = 0. On Fourier transforming we have Gik&J!(k) = 0 and
(Al)
N. G. LEE
686 This can be written
L(k)G’(k) = 0 so that in particular L(&M)Q’(&M)= 0,
(M= 1.2,..., 6).
Thus (M=l,2,...,6)
Q(&“, = R(rrM) V”,
tA2)
where VM (M = i,2,. . ., 4) are arbitrary matrices. From eqn (Al) we have c,,&p.,,(kt, kz,x3) so
= 0
that
or, in operator form k2, x,) = 0
W&k,,
(A3)
where the operator P(k) has com~nents
aI[ (WI+ 43k + i&3&I
E
P>ttk)= ctjkl(Sit+ &P$ + 43 ’ a*,
I
Since the @KM) are functions of kl, kz only, we may try
where the FM are functions to be found. Thus, from eqns (A2), (A3) and (A4) we have R(fJ”)VMFM(k,,
kz,xJ = 0.
Solutions of this equation are obtained by setting p(W+(k,,
k,, x,) = UB”)FM(k,, kr, x3.
(M= 1,2(..., 6).
This condition is satisfied if
$ 3 FM = - [KY FM,
$
FM = -(Kr)‘F”,
(M=1,2,....6)
3
and hence if FM = e-rK%,,
(M=
I,...,@.
Hence from eqns (A2), (A4) and (AS) we have
Since @(k,, kz,xJ-+O as xl++-,
we take @‘(k,,kz,x3 = $
=I
R(p)
Vu e-+~j z
as a general solution.
APPENDIX Defining
2
(AS)
Elastic Green’s function taking I, PO and writing y(x)
for a hexagonal continuum
687
for B”(J, h), it follows from eqn (la) that
@(k)= &
L!&)l-' enjh
which, inverted with respect to k3, gives
c”(k,,
k2, x,) = $
_‘- [L(k)]-’ I m
e-“@3-h’dk+
If we demand that x3 C h, we can evaluate the integral using Cauchy’s Theorem by closing the contour in the upper half of the complex k, plane. Thus
(A6)
From eqn (2b) we have, for xl = 0 c,,wG&., = - c,&%.i. Hence
and Q(k,. kzr x@(k,,
kz. x3 = - Q(k,, kz, x@-(k,,
(A7)
kz, x,),
where the operator Q(kl, kz. x3) has components
Qtdhv kz. ~3)= c,tkl Substituting
-
ik& + 6&t&-
a . axII
from eqns (3) and (A6) in eqn (A7) we have
Ok, kz,1,)
3 R(@-‘) e-‘K~W”
R(p)VMCqx3 =2 Q(k),k2,xd &=,
a
,
,-(KN)
det
x, = 0
ak, _
which gives eqn (5)
’ 3 F(&N)R(&N)e'Ky* VM= 2 J
F(p)R(p)
a det
=I
L(KN)
ak, where the
maWiX
F(k)
has components
&(k)
= C,jkk.
APPENDIX
3
Using the co-ordinate system defined in Section 2 the stresses set up in the continuum force I(?‘) at x’, are given by the symmetric matrix function (T(X) where u,(x) = c&)9 @I = Ull. 61 = lt.
and the symmetric
a2 = u22. c2 = Cm
(i,j=l,2
03 = 033, c3 = cm
,...,
u4 = u23. c4 = 2c23.
matrix of strains C(F) is given by
Q(X) = 1 [u,.J(x) •t u&)1
= i IGek.,(SX’)t G,d&
&‘)lfk(f)~
Thus, redefining [ as one third of its original value (see eqn (to)), we have
at a point J due to the action of a
6) a5 = (Tn. es = 2Q3,
06 = u12
l6 =
2c,z
N. G. LEE
688
where the operator g is given by
and ,,(xi,12~=xl[-2~~(tm:~~~+X~~)
+& (
r
WI,
x2) =
[vMNbnrcl,-CII(r~N~S~N)l+2CS6(SOIN_fIWN) 3 nMN q4
1 &[QN('h: -7 IT4 (
X2%
&
-
_i. yNf2x:-
_ aN [2ANx:z2 + ENq4] M',
ruin
4*
(
4X:) + Z2(4X: - 12x:)/#]
(
-3
q2
4 )
nMN
x:l_ [yNh*+
IbfN +
-
BNdl
(
mN
)
m'N
)
tMv1
&N
b'M
)I
TMN ~(d--X:hN-hfbhf~
+r [6
&* &-s(xdh,N t
mN
&MN-h,WI
d s,fx, x2j=c44
~N~~ll~~~~l3cN)
-hfN)I
(
~(~Y[SUNX:+~HNX:I-LIUN.X~)+~M~MN I)1 n
kt1-&6lffN
- Cd,
N )
(
4
)
>I
When q =0, an expression for $ is obtained by expanding the mN and the n&rN,and using relationships given in Appendix 4. For this case we find that
and that S3. ST,St* consist of sums of terms each of which is the reciprocal squared of a linear factor of xj and h.
APPENDIX Some relationships concerning the parameters defined in the paper
4
{b,, bx,b3t = f- 6,s - 62, - 63) Ez=B3=CL=Dr=H,=0 $,
AN/&=
$,
CN/EN
=O
a*=n3=B,=6,=a,=cr,+y,=O e2=83=~2=~~=~,=AI~~,=B,+p,=o bfk, + s,,fE -
bf(r,, + .s,,)E, = 0,
bf(r,+f,,)E,-b:(r,,+r,,)E,=o, bf~g,E, t bfU,iE, = 0,
(i,j=1,2,3) fi,j=
1,2,3)
(i,j=1,2,3)
Elastic Green’s function for a hexagonal continuum b;w,,E, - bfw& = 0,
r,, = t,, = 0,
689
(i,j= 1,2,3)
(i = I, 2,3; j = 2,3)
s,, = IA,,= If,, = w,, = 0,
(i=
1,2,3)
r,, =o Pdh
x2) = PdXh x2) = 0,
(i =
2,3).
APPENDIX 5 With reference to the co-ordinate system defined in Section 2, let cir denote the matrix of elastic constants of a hexagonal medium with its surface a basal plane. Similarly let ci, denote the matrix of elastic constants of a cubic medium with its surface a (I 11) plane. These matrices have a similar form, differing only in that some elements which are zero in the cji matrix are proportional to the anisotropy factor n = - c’,, + c;* t 2c$ in the c:, matrix. Thus, by setting to zero those elements of the c:, matrix which are proportional to n we may apply our results to cubic crystals with (Ill) surfaces. Clearly this approximation is likely to have greater validity for cubic crystals with weak anisotropy.
LJF3Vol. 17.No. 6-C