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The method of analysis of cracks in three-dimensional transversely isotropic media: boundary integral equation approach
,
r
Engineering Analysis with Boundary Elements 21 (1988) 169-178
.~, ,4, v!;~il PII: S 0 9 5 5
ELSEVIER
-7997(98)00033
© 1998 Elsevier Science Ltd. All rights reserved Printed in Great Britain 0955-7997/98/$19.00+0.00
-7
Research Note The method of analysis of cracks in three-dimensional transversely isotropic media: boundary integral equation approach M. H. Zhao a'*, Y. P. Shen a, Y. J. Liu b & G. N. Liu b axi 'an Jiaotong University, Xi'an 710049, People's Republic of China bZhengzhou Research Institute of Mechanical Engineering, Zhengzhou 450052, People's Republic of China (Received 27 September 1996; accepted 24 February 1997)
The fundamental solutions are obtained for a unit-concentrated displacement discontinuity in a three-dimensional transversely isotropic medium. The method of solution involves reducing the problem to a system of hyper-singular integral equations by application of these fundamental solutions. The near crack border elastic displacement and stress are obtained. Stress intensity factors can be expressed in terms of the displacement discontinuity on the crack surface. Analogy is established between the boundary integral equations for arbitrarily shaped cracks in a transversely isotropic and elastic medium such that once the stress intensity factors in the transversely isotropic medium can be determined directly from that of the isotropic elastic medium. Results for the penny-shaped crack are obtained as an example. © 1998 Elsevier Science Ltd. All rights reserved
Key words: Fundamental solution, crack, transversely isotropic medium, displacement discontinuity, hyper-singular integral equation. antiplane, l 1 torsion,~ 2 plate-bending~ 3 and three-dimensional problems. 14-16 Some books on the subject came out in print] 7 In this study, efforts are focused on the DDBIEM of a three-dimensional transversely isotropic elastic solid.
I INTRODUCTION The researchers of problems with transversely isotropic properties have received attention since 1948. In this field the potential function methods 1-4 are well known and widely used. With these methods many important results for crack problems have been obtained 5-8. What has not been accomplished is the method of solution for an arbitrarily shaped plane crack. The boundary integral equation-boundary element method ( B I E M - B E M ) has many advantages over domain methods to be used for the problems of concentration and stress singularity. 9 The displacement discontinuity (DD) B I E M - B E M is more effective in fracture mechanics, which describes the basic character of a crack that the displacement across its surfaces is discontinuous. 1° Many applications have been made to solve two-dimensional, t°
2 PRELIMINARY
CONSIDERATION
In the absence of body forces, the basic equations can be outlined as follows: 5 o0j=0
(1)
g6=Cuklekl
(2)
CO= ½(Uj,i+Ui,j)
(3)
where i,j,k = 1,2,3. The quantities ag~ egj,ug are the components of the stress, strain, and displacement respectively, and Cgjk/ appear as the elastic stiffness. For a transversely isotropic elastic medium, there are five elastic constants. In what follows only the transversely isotropic material will be examined.
*Corresponding author. Zhengzhou Research Institute of Mechanical Engineering, Zhongyuan Road, Zhengzhou, Henan, 450052, People's Republic of China. Fax: (+86) 0371 7449148. 169
M. H. Zhao et al.
170 2.1 Potential functions ~-s
2.2 Displacements and stress
Let the cartesian coordinate system (x,y,z) be coincident with the isotropic plane of the medium. Eqns (1)-(3) can be cast into component forms. Eqn (3) can be substituted into eqn (2). Further substitution into eqn (1) yields the governing equations:
Let X(x,y,z) = @ 3(x,y,z) and define
32Ux 1 32Ux 32Ux Cll 7 X2 4" 7 (cll -- CI2) C~'2- 4" C44--~-Z-2-Z2
32uv
1
32u.
(4)
3@
3X
3@
3x - 3y ' = ~ +
3X
=0.
1 -c,2) (cll
( 02 orr=
o~
3X 2 + c44 _"Z ZSY_2= 0
~13"Y4.32X\ OX ~y2)
["
32
13
(6)
. 32@,
azz
, .aflO@A + co,, - c,2)
// 02 13 1 32 \ , , C13t ~rrZ-r24. 7574- 70--0-~)(@1 4. "i'2) --
3z 2
1 32@j ,
If eqn (7) has a non-trivial solution, the following condition must be satisfied:
(8)
32@3
02~j
1 02@3
Ozr (c444,c44kj)----c44" 3r3z r 303z (1
thus the quadratic in k is
02
1_32~ (,i,1 4- `i'2)
oro (¢,,-c,2)\;37 o r230) 4. ~(C11__C12)(02@3
(9)
Or2
4" C44(Cl3 4" C44) = 0 with the two roots denoted by kj and k2. Corresponding to the two real or conjugate kl and k2, there are two real or conjugate values of h expressed by hi and X2. Therefore eqn (7) becomes
32@i
(@' + @2)
+ c33k, 02@, (7)
c')2"i"i
ax---T-t- O-~y2--4.X,--OTy-z2= 0 ( i = 1,2)
1 3~\.
000 = ~f,2 ~ + ~,, ; ~ + c,, 7 ~ )
, . f02@ 32@'k 02@ I,c13 +c44 +c44k) t-~ffx2 + -~--y2) +c33k~-z2 =O
C44(C134"C44)k24" [(C134"C44)24"C24--C11C33]k
, 3 / 1 3@3\
+c,,k,v- c<,,-c,21 ( ;
(5)
/ 32@ 32 @"~ . . . . 02@ cll t 0-~5-4" 7 ) 4" [c44 4- tcl3 + c44)k j ~ = 0
C44 4" (Cl3 4" c44)k -c33k -~- )k Cll C13+ c44 + caak
502)
,
02"i'j
k O@
3x u: =
13
cl, ff-r4"Cl2r-O-r4"Ci2r2 OO2j (@i 4-@2)
+
c]2@i
(13)
The stresses are
Putting eqn (5) into eqn (4) gives 2
r a0
O@l , . 0@2 b/z = k I ---~-g-I-K2 ~-g
The displacements may be expressed in terms of the potential functions ~ and X u<-
1 3@3
0
Ur= ~('I'~+@2)
1 3 .@ 3@3 " 0 = 7 ~ ( I 4"@2) 4" 0~-
02btz
IZO2u~
(12)
It follows that the displacements may be expressed in terms of the three potential functions @ ~, @> @3. Referred to a cylindrical polar coordinate system, the results are
¢44t 0-~4" 3y 2) 4"C33 0322
4"(C'3 4"¢44) to~O-Z4" 3yOz)
02@3 . , 32@3
374- T 2+ ^3-~-? = 0
1 OZu., oeuv O2uv X(cll -- C12)_~-~.2 J 4"C11 a~-.2 4"c44 - 2" z ax oy oz
#/32#A. 32btz"~
(11)
such that
32@3
4, X(Cll 4"C12)7~:-,~ 4 " ( C l 3 4 " C 4 4 ) . . ' = 0 z oxoy " oxoz
1 32u~ 32U4" 7(CII 4" CI2) O~V 4" (C13 4" C44) Oyez = 0
2c44 X 3 - - CII -- CI2
10@3"] r ~r ]" (14)
a repeated index denoting summation. Eqns (10) and (12) in cylindrical polar coordinates become
02@i (10)
102@3 r 2 002
1 O@i
1 02@i 02@i
07 + ; Gr + -y-~ + - ~ g = 0 i= 1,2,3
(15)
The method of analysis of cracks in 3D transversely isotropic media where
171
boundary conditions along the place z ----0 are
zi = siz and si = - -
1
(16)
1
Uz(r, O,Z)[z=0 = ~5(r) arzlz=o = 0 OzOIz=o = 0
(21)
When v/r 2 + z2 ---*~, the finite conditions require that
3 FUNDAMENTAL SOLUTIONS FOR THE UNITCONCENTRATED DISPLACEMENT DISCONTINUITY Consider a penny-shaped crack of radius a with the upper surface S + and lower surface S- shown in Fig. 1. Displacement discontinuity ~u~] across the surfaces may be expressed as
(22)
Orr : 4700 : lTzz ~ lYzr ~ (TzO : {TrO -~- 0
All the components depend only on (r,z) and independent of 0 owing to axis-symmetry. Hence, eqn (15) simplifies to the form 02~i
10~ i
02~i
(23)
Or2 + r--~-r + --~-zi2= 0 ( i = 1,2) { ~ux~:Ux(X,Y, O+ ) -ux(x,y,O- ) ~Uy]=Uy(x,y,O+ ) -Uy(X,y,O - ) (x,y) E S
(17)
I[uz]]=uz(x,y,O + ) -Uz(X,y,O- ) The fundamental solutions corresponding to a unit-concentrated displacement discontinuity should satisfy the governing equations of transversely isotropic materials and the following conditions: lim
a--,0
and ~3 = 0. Thus, az0 is automatically equal to zero. Apply the zero-order Hankel transform to eqn (23), and observe eqn (22). A suitable solution is ~i(~,zi)=Ai(~)
(24)
e -~zl ( i = 1 , 2 )
~li(~,Zi) is the zero-order Hankel transform of
where
qI i(r, zi).
Substituting eqn (24) into the results obtained by zeroorder Hankel transform of eqn (13), the result is
f{~u~],~uy],~Uz]} d s = {1,0,0}
(18)
2
uz = - ZAj(~)kjsj e-~ZJ lira
a--,O
~{~ux],~Uy],~Uz]} d s = {0, 1,0}
(25)
j=l
(19)
The inverse Hankel transform can be applied to eqn (24) to yield
lim~s[~ux],[uy~,~Uz] } d s = { 0 , 0 ,
a---.O
(20)
11
xlZi(r , Zi) ~-
Satisfaction of eqn (18) and eqn (19) refer to conditions of skew-symmetry and eqn (20) to axis-symmetry. The fundamental solution with axis-symmetry will be derived.
f~Ai(~)~ e - ~ZiJo(~r ) d~
(26)
Substituting eqn (26) into eqn (14), azr is obtained as
(7zr =
(c44-~c44kj)Aj(~)sje
0 42
-~z)
Jl(~r)dr
3.1 Fundamental solutions satisfying eqn (20) (27) Owing to the symmetry, consider only the part z -> 0. The
The boundary conditions in eqn (21) are the transformed values and can be applied using eqns (25) and (27). This gives 1
Z
siAl + s2A2- 4r~ klslA1 + k2s2A2 --
(28)
1
47r~
which can be solved to obtain ml
Y
AI=-ff-A2=T
m2
(29)
The constants mi are ml m
Fig. 1. Circular crack.
1+k 2
4rsl(kl -k2)
m 2 -.~
l+kl
4rs2(kl -k2)
(30)
M. H. Zhao et al.
172 With the aid of eqn (29),
XI't i
is found from eqn (26) as
mi ql i(r, Z,) -- V / ~ -}- Z2
(31)
such that fl and f2 must satisfy
32J) . 1 32f1 . I 02fl
j+
r-g-r +
02f,
+ o-7 =0 (37)
Substituting eqn (31) into eqns (13) and (14), the stress and displacement fields are obtained:
where j = 1,2. To satisfy eqns (37) and (22), fl and f2 can be taken as 5
2
ur = - r jy= l
&2 l &2 1 0212 &2 5 7 + ;-g-r + + Oz- = °
)3,2
( r2q- "
3¢
Uz
2
mjkjzj
'=
(r 2 q-g2) 3/2
~c
fl (r, 0, zi) = 7 o : fo (Plm(() e - ~ZiJm((r) d(-sin mO
f2(r, O, z3) = Z
2
~P2m(~) e -
m=0 ~
Orr "-~- ~_. mj[c,l( 2 r 2 - Z2)--CI2(F 2 -]- Z2)
~Z~Jm(~r) d~-cos mO (38)
j=l
+ c,3kj(2zf- r2)]/(r 2 + 2 ) s/2 2 000= ~ . m j [ c l 2 ( 2 r 2 - z 2 ) - C l , ( r 2 q - z 2)
(32)
j=l
Substituting eqn (38) into eqn (35) and making use of eqn (13), the non-vanishing displacements become b~r= 2 7 0
siam0 Io
~2{[Jm-l(~r)-Jm+l(~r)]
+ c~3ki(2z~ - r2)]/(r 2 + z~) 5/2 × [H, e -~''' q-n2e-~Z2]Plm(~)-l-[Jm_l(~r )
(2Z2 __ r 2) °zz = j= Z I ajmj ( r2 q_z2)5/2 2
+ Jm+l (~r)]p2,,(~) e-~z~ }d~
2
az~ = 3r Z (c44 + c44kj) j: 1
Uo= ~m~_0c°sm0 f2~.2{[Jm-I((r)+ J,n+|(]~r)]
(sjzjmj)
(r2 q- Z2)5/2'
where
aj = - c13 q- c33kjs 2
(33)
+ [J,, _~ ((r) - Jm +, ((r)]P2m(() e - ~z3} d(. (39) The displacement conditions in eqn (34) may be enforced on eqn (39). It can be concluded that only the case of m ---- 1 needs to be retained while p In and P2m are all zero for m =~ 1. It follows that
3.2 Unit-concentrated tangential (y-direction) displacement discontinuity This fundamental solution satisfying eqn (19) will be obtained using the boundary conditions for z > 0:
f ~ 2 { [Jo(~r)- J2(F~r)]Hl2Pll(f;) + [Jo((r) + J2(~r)]p21 (()} d~ = 6(r)
ozlz= o = 0 1 1 u~ = 2-6(r) sin 0 Uo = ~6(r) cos 0
(40)
(34) J o ~2 { [d°(~ r) q- J2(~r)lHl2Pl '(~)
where 6(0 is the Dirac delta function. Based on eqn (15) and the expression of ozz in eqn (14), the potential functions satisfying the first condition in eqn (34) are given by
+ [J0(~r) -- d2(~r)]P21(~)} d~ = 6(r) where HI2 = H l q - H 2. From eqn (40), Pll(~) and P21(~) are obtained:
qll(r,O, Zl)=Hlfl(r,O, zl) q/t2(r, 0, Z2) = H2fl (r, 0, z2)
(35)
q/3(r, 0, Z3) =f2( r, 0, z3)
1
1
(41)
P21(~) -- 27rH12
where H l = 1 H e --
1 1
al
a2
(36)
Eqn (41) may be inserted into eqn (38). Further application of inverse Hankel transform renders
The method of analysis of cracks in 3D transversely isotropic media { ~i(r, O, zi) =
ql i( r, O, Zi) = ~ 1/4j [1 27rH12
H-'----L~I[1 - zi(r 2 -t- Z2) - 112] sin 0 27rH1z r (i = 1, 2)
xl13(r,O, 23)=
173
-- zi( r 2 4- d)-
112]cos o
i=1,2
1 1 1/2] ~-~r[1 --z3(r2 4- Z~) cosO
ql3(r'O'z3)=
21r r 1 [1 - z3(r 2 4- z~)- l/Z] sin 0
(42) The displacements follow from eqn (14):
Ur=_~__.~{j~=lHJ rl
(45) The displacement is obtained
Ur---
-- ~-~{j~ H~z[ I
ZJ(2r24-22)]
H,2L
4-(1
Z3 1 -t-(l-(rZ4-z])m)}-fisinO
u0= -uo= ~ -
V zel} o,o 2
rsm0 5-
27rHj2j~=~kjsjHj (r2 4-
(46)
z3(2rZ+zb] 1 z~ 1 } ;~ sin 0
r cos 0 2 ~-~l-@~=lk)SjHj(r 2"2"-312 ;~ -fzj) •
uz =
=~)-312
(r2 ~-~32),12.)} ~-~2cos 0 n,2_l -- _(r 2 +
-[I
23(2r2 4- z2)1 •
uz =
(43)
will[1
Zj(2r2+zf)]-(fi~]
The stress is The stresses can be obtained by putting eqn (42) into eqn (14). It is found that
azO =
3r sin 0 ~-- ,. . 2 , 2,-5/2 azz = ----;7, - 7. njajzjtr t zj ) 2rrtl2 j = 1 1
1
2
z0= - = [ . - = Y
3r cos 0 ~ Hjajzj(r 2 + z2 ) _ 512 °zz -= 27rH12 j = 1
2
Z ~ L/--/12j= 1
1
(c44 + c44kj)sjHj(r 2 + z~) - 312
+ c44s3(z ] _ 2r2)(r 2 + z32)- 5/2] sin 0
2 -- 3/Z
+zj)
Ozr ~
+ C44S3(Z~ _ 2r2)(r 2 + Z2) - 5/2] COS 0
1 ~,0 Z (c44 + c44kj)sjHj(z2 - 2r 2) 27r 12j= 1 2~ - 3/2]| × (r 2 -I- 2Zj)2~ - 5/2 -I- C44S3ttr2 4- Z3~
Ozr =
27r ~
~" (c44 4- c44kj)sjaj(z2 - 2 r 2 )
..I
12j= 1 X ( r 2 4- 222) - 5/2 4- c44s3(r 2 4- z 2) -
cos 0. (47)
3/21 sin
0.
-I
(44)
This completes the foundation for the problem of unitconcentrated displacement discontinuity fundamental solutions.
4 BOUNDARY INTEGRAL EQUATION 3.3 Unit-concentrated tangential (x-direction) displacement discontinuity The fundamental solution for a unit-concentrated displacement discontinuity in the x-direction can be obtained in the same way as that in the y-direction discussed earlier. The potential functions are
Let a crack of arbitrary shape in coincidence with the plane of isotropy that lies in the xy-domain as shown in Fig. 2. Denoted by S + and S- are, respectively, the upper surface with outer normal directional cosine {0,0, - 1 } and lower crack surface with {0,0,1 }. The reciprocal relations in the absence of body forces are given by
M. H. Zhao et al.
174
The foregoing constants are related to the material properties. The displacement discontinuities are expressed as
Y
{ ~u~] = ux(x,y, 0 + ) - ux(x, y, O- ) ~Uy~=Uy(X,y,O + )
Uy(X,y,O-) (x,y) E S
(53)
~uzU= uz(x, y, 0 + ) - uz(x, y, O- )
-- X
It should be noted that
r 2 = (x -- ~)2 + (y _ ~/)2
(54)
The kernel functions in eqn (51) have the same order of singularity as those in the displacement discontinuity boundary integral equations for the corresponding elasticity problem. Hence eqn (51) can also be regarded as the hypersingular integral equations.
Fig. 2. Planar crack. /
fs I"~Pi(1)Ui(2)'~) ds+ fJs- ~Pi(1)I'li(2)~)ds (48)
~- I s + {~Pi(2).(1), t~i ) ds-~- ~ s- /~Pi(2)ui(I)'~) ds where ul l), @) and ul2) are two sets of solutions for the body while p(l) = a(Om. Consider u}]), al l)' asthe real solution of the problem and ul ~), el 2) as the unit-concentrated displacement discontinuity fundamental solutions in Section 3. It can be easily shown that p12)l,+ = -plZ) l~-
(49)
The tractions on the crack surfaces satisfy the conditions
pll)ls+ = --pll)ls-
(50)
that are assumed throughout this work. Inserting these conditions into eqn (48) gives the boundary integral equations for a plane crack of arbitrary shape:
5 CRACK BORDER SINGULAR BEHAVIOUR
Choose an arbitrary point o on the crack border I" for analysing the singular behaviour. The border I" of the crack is smooth at point o. Without loss in generality the cartesian coordinate system oxyz is placed such that the yaxis and x-axis are tangent and normal to I' respectively, whereas the z-axis is normal to the crack plane S as shown in Fig. 3. Let e denote the radius of a small circle r~ contained in S. The hyper-singular integral in eqn (51) should be finite in E and can be written as
f rLzz I~Uz]~1 ds(~, r/) = Rz(x , y) I~ [(Lzr3 cos 2 0 + Lzo3 0sin ) ~2u x ]
q- (Lzr3 - tzo3)
1
sin 0.cosO~uy]]7 ds(~, 7) = R~(x, y)
[ fs+LzzllUzU~ds=pz(x,Y) IE [(Lzr3 -- Lz03) sin 0-cos O~ux]] Is +{ [Lzr3 cos 2 0 -t- Lzo3 sin20]~ux~
1
+ (Lzr3 sin2 0 + Lzo3 cos 20)~uy~] ~ ds(~, ~/) + [Lzr3 - Lzo3] sin 0 cos O~uy]} ~ ds((, ~7)=px(x, y) Is
= Ry(x, y). (55)
{[tzr3--gz03] sinOcosO~Ux~ z
ds + [Lzr3 sin 20+Lzo 3 cos 20]~uy]} -7=py(x,y). (51) where
2
LzzI = ~. aimi
Y
i=1
tzr3=~ -2,= 1[
1 ~-L
~-~-12(l+kj)+s3 c44
(52)
]
Lzo3 = w - I ,-r-- 2_ sjl4j(1 + kj) - 2s3 c44.
zTr Lrtl2j= 1
Fig. 3. Crack-tip analysis.
The method of analysis of cracks in 3D transversely isotropic media where R x(x, y ), R y(x, y ) and R z(x, y ) are all finite functions for
(x,y) E E.
Az(o)
cot
175
Otz.Lzz1 = 0
[2Lzr3 + Lzo3]Ax(o) cot rC~x= 0
5.I Discontinuity
[Lzr3 + 2Lzo3]Ay( o ) c o t
The displacement near point o can be obtained by superposing plane strain and the antiplane displacements in the oxz plane./s The displacement discontinuities at the neighbourhood of point o are given by 15
{ ~Ux]=Ax(o)~x [Uy] =Ay(o)~ ~Uz]=Az(o)~"~
(56)
'
f L~_d~ d T = x a z - l A z ( o ) f ~ z d~ f~-o~
rOty= 0
5.2 Singular behaviour The values of Ax, Ay and Az are not zero in general. Moreover, the coefficients in eqn (58) are certainly not zero. Hence, non-trivial solutions of eqn (58) exist only when the following conditions are satisfied: cot 7rczx= cot
where Ax, Ay and Az depend on the location of point o. The indices ax, ay and a z refer to the unknown solutions and their values are between [0,1]. Inserting eqn (56) into eqn (55), in the Hadamardprinciple value sense, 15 the hyper-singular integrals can be evaluated with accuracy when ~ is sufficiently small and x is infinitesimal, such as
(58)
71"O/y =
cot rcaz = 0
(59)
Recall that the values of ax, a r and a z are within the interval [0,1]. Eqn (59) requires that
ax = ay = a z = ½
(60)
This result reveals that the displacement near the crack tip behaves as r 1/2, which is the same as for an elastic solid. Making use of the unit displacements solutions in Section 3, the stresses near the crack tip outside of the crack are obtained:
azz(X, y, O)= - ~ [Lzzll[Uz]] ~ d~ d 7
d~
×
azx(X,y, O) =
[(1 -- ~)2 + (Y-- ~)2] 3/2
Ig [(Lzr3 cOs2 0 + Lzo3 sin2 O)~Ux] 1
+ (Lzr3 - Lz03) sin 0 cos O~uy[] -~ d~ d7
= -- 2x c~=- tAz(o)Tr cot 7r~z ~r~cos2Od~d7=
~r ~u~]](x-~)2r5 d~ d7
azy(X, y, O) =
I~ [(Lzr3 - Lz°3) sin 0 cos O~ux]
(57)
+ (Lzr 3 s i n 2 0 + Lzo3 c o s 20)[uy~]
4 a~
= - -~x - tAx(o)rr cot tax
1
~-~ d~ d 7
(61) sin 0 c o s 0
~E ~Ux~ r 3
ds(L 7) = 0
The results
f ~Ux]sin2 0 d ~ d 7= ~ ~ux](Y-- 7)2d ~ r.
r3
z
r5
cos 2
2a
= - -~x "-lAx(o)r cot rot x
r
t° (1 ~- 0 2 dt = - lr cot 7rc¢ dt
2
oc (t 2 + X2)3/2 = X2
f
~
dt
f [[Ux]]r 3
0
2
d~ d7 = gAx(°)Trlv/-O
sin 0 COS 0 ~ ~uy]] d~ d 7 = O
where the integrals
f~
~ ~ - d~ dT = Traz(o)/v/O
~ d7
4
- ~ (t 2 + x 2 ) 5/2 - - 3X4
have been used in deriving eqn (57). Inserting eqn (57) into eqn (55) and letting x ---* 0, then results in
(62)
r~
IlUx] sin2 0 7r d~ d7 = :~ax(o)lv/-O F. r 3 when applied to eqn (61) gives the stress near the crack tip at point (-p,y) as
azz(X, y, O) = -- [LzzlAz(o)]Tr/v ~ 1
Ozx(X,y, O) = - [ ~Lzr3 + ~Lzo3]Ax(o)Tr/V/-O
2 azy(X,y,O)~-[ILzr3+~Lzo3]ay(o)'tr[V/O
(63)
M. H. Zhao et al.
176
Table 1. Constants of materials Material
Elastic constraints ( × 10 3 N m--~)
Magnesium Zinc Barium titanate /3-Quartz
CII
C33
C44
C13
CI2
41.193 111.021 115.920 80.454
42.573 42.090 130.41 76.176
11.316 26.427 37.674 24.909
14.973 34.569 48.990 22.632
18.078 23.115 53.820 11.523
Stress singularity of the order of r -j/2 is obtained near the crack tip, which corresponds to that for the isotropic elastic material.
5.3 Intensity factors The stress intensity factors near the c r a c k tip are g i v e n by K I = lin~ V/~rcrzz(x,
y, O)
Consider a planar crack of arbitrary shape with the surfaces parallel to the isotropic plane of a three-dimensional infinite transversely isotropic medium. The crack surface is loaded by pressure p(x,y) such that boundary condition is
Oz(X,y, 0 + ) = - p
V/~Tr[LzzlAz(o)]
E
871.( i ~_ p2)
(65)
Since Ax, A t, and A z in eqn (65) are known, it can be further shown that
gll = -
K['-
2 1 V/~.Tr lim [~Lzr 3 -I- ~Lzo3] [[ux]]/~-'~k o o j vr
r-OL5
,5
)
E
lim V~p~wpB
(70)
E
(71)
8r(1 - .2)U,,A'"
while ~u.~ is obtained as
v/r
Table 2. Values of the constants related to the materials
SI $2 Lzzl(10 3) L~3(103)
Lzr3(lO3) P
(69)
It follows that the solution to eqn (68) is
Lzzl ~Uz]= -j
s
8(1 - v 2) p-0
(66)
2 -
~Md~d~l=_tz(x,y
It has the same form as eqn (68). In eqn (69), E and v are the Young's module and Poisson's ratio respectively. The pressure on the crack surface is tz(x,y). Now, let t z = - p, which corresponds to [[Wp] as given by eqn (69). The mode I stress intensity factor K p along the crack border is
!imo[Lzz,~uz]]/V/~
1
(68)
Listed in Table 1 are constraints for several transversely isotropic m a t e r i a l s ) They can be used to evaluate the values of those in Table 2. The displacement discontinuity boundary integral equation for isotropic solid takes the form
v/~rOzy(X, y, O)
Ki = - ~ / ~ . r
(67)
I~ [Lzz.[luz]]]~ d~ drl= p(x, y)
(64)
It follows from eqn (63) that K ~--- --
6.1 Symmetric problems
The boundary integral equation is given by
K,, = lira V~ro~(x, y, O) K I u = lira
6 M E T H O D OF ANALYSIS
Magnesium
Zinc
Barium titanate
0.69840 1.40844 - 1.37954 - 0.92624 - 3.60787 0.32936
1.08769 - 0.66410i 1.08769 + 0.66410i - 1.86112 - 4.80316 - 6.66638 0.10274
0.96472 - 0.11013i 0.96472 + 0.11013i - 4.26593 - 2.84293 - 10.64441 0.32329
13-Quartz 0.77277 1.32989 - 2.71520 - 3.74574 - 6.49832 0.16441
The method of analysis of cracks in 3D transversely isotropic media 1 E LzzI 87r(1 - v2)~vvp]~"
(72)
With the aid of eqn (66), the stress intensity factors near the crack border are obtained: K~ =KIp
(73)
Eqn (73) is independent of the shape, the dimension, the geometry of the crack and the distribution of the mechanical load p. Here we can conclude that the stress intensity factors of a planar crack of arbitrary shape in transversely isotropic media can be calculated by eqn (73) with the corresponding solutions of elasticity. 5 The same holds for the penny-shaped or elliptical cracks. 5
177
They are hyper-singular boundary integral equations in elasticity. 14,15 As before, the solution of a crack subjected to antisymmetrical loading in transverse media can also be obtained from that for an isotropic elastic material. Knowing that E
= 8(l-
ffrr
8(1 +
m li,-u
(77) Uy~
the stress intensity factors in transversely isotropic material become
6.2 Anti-symmetrical problems The boundary integral equations of the crack subjected to anti-symmetrical loading are
Kn=-
r--.oW~r ~ux] ~ r[[2Lzr3 - - ~ - -+- -Lz03] - J lim (78)
r,.P
+ (Lzr 3 -
[ Lzr3 + 2Lz03] lim ~ / ~ u v ]
^h'=-~'[
f s [(Lzr3 COS2 0 + Lzo3 0sin ) [2u x ] 1
"i r---0V r
(74)
~s [(Lzr3 -- Lz°3) sin 0 cos O[u~]
=py(x,y). where p, and py are the tractions on the crack surface in the x- and y-directions respectively. The equivalent isotropic elastic constants E and v can be identified with the transverse material constants as 3v
p 4r cos 0 Ku -- ~-(2--~-~v/Tra
(80)
(75)
E
Lzr3 - Lz°3 --
Solutions to many crack problems can be obtained in the same way. 5 Suppose that a penny-shaped crack is sheared in the direction of the x-axis, with r being the shear stress. The stress intensity factors KH and Km in elastic solid are: 5
K}IIP= _ 4(1 -r(_~._ ~ v ) (sin r 0)X~
1 -- 2_.______v
Lzr3 - Lzo3
(79) Kll I ~-~KPl
+ (Lzr3 sin 2 0 + Lzo3 COS20)~uy]] ~ ds(~, )1)
Lz__03_
-
The stress intensity factors are related to those in elasticity as
Lz03) sin 0 cos 0[uy]] ~ ds(~, 7)
= px(x, Y)
~
Substituting eqn (80) into eqn (79), the stress intensity factors can be obtained. The constant ~, is listed in Table 2.
8r(1 - v2)3v
Taking eqn (75) and eqn (74) into account, it is found that 7 CONCLUSION
f~{ [(1 - 2v)+ 3v cos 20]~ux]] + 3v sin 0 cos O~ur]} -~ ds(~, ~t) 87r(1 -- v2)-~ (x,Y)t-'x E ~ { [3v sin 0 cos O~u~]]
1 + [(1 - 2v) + 3 sin 20]~Uy]} -'~ ds(~, 7) 8r(1 - v2) --
E
-py (x, y)
(76)
The Hankel transform is used to derive the fundamental solutions for a unit-concentrated displacement discontinuity. The displacement discontinuity boundary integral equations are derived for an arbitrarily shaped crack in a transversely isotropic solid. They correspond to the hypersingular equations. Displacement and stress near the crack border are obtained. The familiar r - m singular behaviour is found for the stress in front of the crack tip. More specially, the stress intensity factors are shown to be obtainable from the corresponding solutions for an isotropic elastic material.
178
M. H. Zhao et al.
ACKNOWLEDGEMENTS 9. The work was supported by the National Nature Science Foundation of the People's Republic of China and the Technical Development Foundation of the Machinery Building Industry of the People's Republic of China.
REFERENCES 1. Elliott, H. A. Three-dimensional stress distribution in hexagonal aeolotropic crystals. Proceedings of the Cambridge Philosophical Society, 1948, 44, 522-533. 2. Hu, H. C. On the three-dimensional problems of the theory of elasticity of a transversely isotropic body. Acta Scientia Sinica, 1953, 2, 145-151. 3. Lodge, A. S. The transformation to isotropic form of the equilibrium for a class of anisotropic elastic solids. Quarterly Journal of Mechanics and Applied Mathematics, 1955, 8, 211-225. 4. Chen, W. T. On some problems in transversely isotropic elastic materials. Journal of Applied Mechanics, 1966, 33, 347-355. 5. Kassir, M. K. & Sih, G. C. Three-dimensional crack problems. Mechanics of Fracture, Vol. 2. Noordhoff, 1977. 6. Podil'chuk, Yu. N. and Tkachenko, V. F. Stress of a transversely isotropic media with a parabolic crack under a linearly shear load. Int. Appl. Mech., 1995, 31(5), 338-348. 7. Podil'chuk, Yu. N. Stress state of transversely isotropic body with elliptical crack in the presence of a uniform heat flux at its surface. Int. AppL Mech., 1995, 31(3), 182-189. 8. Fabrikant, V. I. Complete solution to the problem of an external circular crack in a transversely isotropic body sub-
10.
I1. 12.
13.
14.
15. 16.
17. 18.
jected to arbitrary loading. Int. J. Solid Structure, 1996, 32(2), 167-191. Brebbia C. A. & Walker, S. Boundary Element Techniques in Engineering. Butterworth, London, 1980. Crouch, S.L. Solution of plane elasticity problems by the displacement discontinuity method. Int. J. Num. Meth. Engng, 1976, 10, 301-343. Liu, N. and Altiero, N. J. An integral equation method to mode III crack problems. Engng Frac. Mech., 1992, 41(4), 587-597. Yulan, L., Zhiqing, M. and Renji, T. The solution of integral equations with strongly singular kernals applied to torsion of cracked circular cylinder. Appl. Math. Math. Mech., 1993, 14(10), 853-859. Minghao, Z., Changjun, C. and Yuanjie, L. Moment stress intensity factors for collinear and parallel cracks in Reissner's plate: boundary integral equation approach. Theoretical and Applied Fracture Mechanics, 1995, 22, 261-266. Minghao, Z., Yuanjie, L. and Changjun, C. Boundary integral equations and boundary element method for threedimensional fracture mechanics. Engineering Analysis with Boundary Elements, 1994, 13(4), 333-338. Renji, T. and Taiyan, Q. Method of hypersingular equations in three-dimensional fracture mechanics. Acta Mechanica Sinica, 1993, 25(6), 665-675. (in Chinese). Pihua, W. and Taihua, F. The discontinuity displacement method applied to three-dimensional coplanar crack problem for any boundary value condition. Engng Frac. Mech., 1994, 48(5), 691-702. Wen, P. H. Displacement Discontinuity Method, Topic in Engineering, Vol. 29. Computational Mechanics Publications, 1996. Parton V. Z. and Perlin P. I. Mathematical of the Theory of Elasticity. Moscow, 1984.
r
Engineering Analysis with Boundary Elements 21 (1988) 169-178
.~, ,4, v!;~il PII: S 0 9 5 5
ELSEVIER
-7997(98)00033
© 1998 Elsevier Science Ltd. All rights reserved Printed in Great Britain 0955-7997/98/$19.00+0.00
-7
Research Note The method of analysis of cracks in three-dimensional transversely isotropic media: boundary integral equation approach M. H. Zhao a'*, Y. P. Shen a, Y. J. Liu b & G. N. Liu b axi 'an Jiaotong University, Xi'an 710049, People's Republic of China bZhengzhou Research Institute of Mechanical Engineering, Zhengzhou 450052, People's Republic of China (Received 27 September 1996; accepted 24 February 1997)
The fundamental solutions are obtained for a unit-concentrated displacement discontinuity in a three-dimensional transversely isotropic medium. The method of solution involves reducing the problem to a system of hyper-singular integral equations by application of these fundamental solutions. The near crack border elastic displacement and stress are obtained. Stress intensity factors can be expressed in terms of the displacement discontinuity on the crack surface. Analogy is established between the boundary integral equations for arbitrarily shaped cracks in a transversely isotropic and elastic medium such that once the stress intensity factors in the transversely isotropic medium can be determined directly from that of the isotropic elastic medium. Results for the penny-shaped crack are obtained as an example. © 1998 Elsevier Science Ltd. All rights reserved
Key words: Fundamental solution, crack, transversely isotropic medium, displacement discontinuity, hyper-singular integral equation. antiplane, l 1 torsion,~ 2 plate-bending~ 3 and three-dimensional problems. 14-16 Some books on the subject came out in print] 7 In this study, efforts are focused on the DDBIEM of a three-dimensional transversely isotropic elastic solid.
I INTRODUCTION The researchers of problems with transversely isotropic properties have received attention since 1948. In this field the potential function methods 1-4 are well known and widely used. With these methods many important results for crack problems have been obtained 5-8. What has not been accomplished is the method of solution for an arbitrarily shaped plane crack. The boundary integral equation-boundary element method ( B I E M - B E M ) has many advantages over domain methods to be used for the problems of concentration and stress singularity. 9 The displacement discontinuity (DD) B I E M - B E M is more effective in fracture mechanics, which describes the basic character of a crack that the displacement across its surfaces is discontinuous. 1° Many applications have been made to solve two-dimensional, t°
2 PRELIMINARY
CONSIDERATION
In the absence of body forces, the basic equations can be outlined as follows: 5 o0j=0
(1)
g6=Cuklekl
(2)
CO= ½(Uj,i+Ui,j)
(3)
where i,j,k = 1,2,3. The quantities ag~ egj,ug are the components of the stress, strain, and displacement respectively, and Cgjk/ appear as the elastic stiffness. For a transversely isotropic elastic medium, there are five elastic constants. In what follows only the transversely isotropic material will be examined.
*Corresponding author. Zhengzhou Research Institute of Mechanical Engineering, Zhongyuan Road, Zhengzhou, Henan, 450052, People's Republic of China. Fax: (+86) 0371 7449148. 169
M. H. Zhao et al.
170 2.1 Potential functions ~-s
2.2 Displacements and stress
Let the cartesian coordinate system (x,y,z) be coincident with the isotropic plane of the medium. Eqns (1)-(3) can be cast into component forms. Eqn (3) can be substituted into eqn (2). Further substitution into eqn (1) yields the governing equations:
Let X(x,y,z) = @ 3(x,y,z) and define
32Ux 1 32Ux 32Ux Cll 7 X2 4" 7 (cll -- CI2) C~'2- 4" C44--~-Z-2-Z2
32uv
1
32u.
(4)
3@
3X
3@
3x - 3y ' = ~ +
3X
=0.
1 -c,2) (cll
( 02 orr=
o~
3X 2 + c44 _"Z ZSY_2= 0
~13"Y4.32X\ OX ~y2)
["
32
13
(6)
. 32@,
azz
, .aflO@A + co,, - c,2)
// 02 13 1 32 \ , , C13t ~rrZ-r24. 7574- 70--0-~)(@1 4. "i'2) --
3z 2
1 32@j ,
If eqn (7) has a non-trivial solution, the following condition must be satisfied:
(8)
32@3
02~j
1 02@3
Ozr (c444,c44kj)----c44" 3r3z r 303z (1
thus the quadratic in k is
02
1_32~ (,i,1 4- `i'2)
oro (¢,,-c,2)\;37 o r230) 4. ~(C11__C12)(02@3
(9)
Or2
4" C44(Cl3 4" C44) = 0 with the two roots denoted by kj and k2. Corresponding to the two real or conjugate kl and k2, there are two real or conjugate values of h expressed by hi and X2. Therefore eqn (7) becomes
32@i
(@' + @2)
+ c33k, 02@, (7)
c')2"i"i
ax---T-t- O-~y2--4.X,--OTy-z2= 0 ( i = 1,2)
1 3~\.
000 = ~f,2 ~ + ~,, ; ~ + c,, 7 ~ )
, . f02@ 32@'k 02@ I,c13 +c44 +c44k) t-~ffx2 + -~--y2) +c33k~-z2 =O
C44(C134"C44)k24" [(C134"C44)24"C24--C11C33]k
, 3 / 1 3@3\
+c,,k,v- c<,,-c,21 ( ;
(5)
/ 32@ 32 @"~ . . . . 02@ cll t 0-~5-4" 7 ) 4" [c44 4- tcl3 + c44)k j ~ = 0
C44 4" (Cl3 4" c44)k -c33k -~- )k Cll C13+ c44 + caak
502)
,
02"i'j
k O@
3x u: =
13
cl, ff-r4"Cl2r-O-r4"Ci2r2 OO2j (@i 4-@2)
+
c]2@i
(13)
The stresses are
Putting eqn (5) into eqn (4) gives 2
r a0
O@l , . 0@2 b/z = k I ---~-g-I-K2 ~-g
The displacements may be expressed in terms of the potential functions ~ and X u<-
1 3@3
0
Ur= ~('I'~+@2)
1 3 .@ 3@3 " 0 = 7 ~ ( I 4"@2) 4" 0~-
02btz
IZO2u~
(12)
It follows that the displacements may be expressed in terms of the three potential functions @ ~, @> @3. Referred to a cylindrical polar coordinate system, the results are
¢44t 0-~4" 3y 2) 4"C33 0322
4"(C'3 4"¢44) to~O-Z4" 3yOz)
02@3 . , 32@3
374- T 2+ ^3-~-? = 0
1 OZu., oeuv O2uv X(cll -- C12)_~-~.2 J 4"C11 a~-.2 4"c44 - 2" z ax oy oz
#/32#A. 32btz"~
(11)
such that
32@3
4, X(Cll 4"C12)7~:-,~ 4 " ( C l 3 4 " C 4 4 ) . . ' = 0 z oxoy " oxoz
1 32u~ 32U4" 7(CII 4" CI2) O~V 4" (C13 4" C44) Oyez = 0
2c44 X 3 - - CII -- CI2
10@3"] r ~r ]" (14)
a repeated index denoting summation. Eqns (10) and (12) in cylindrical polar coordinates become
02@i (10)
102@3 r 2 002
1 O@i
1 02@i 02@i
07 + ; Gr + -y-~ + - ~ g = 0 i= 1,2,3
(15)
The method of analysis of cracks in 3D transversely isotropic media where
171
boundary conditions along the place z ----0 are
zi = siz and si = - -
1
(16)
1
Uz(r, O,Z)[z=0 = ~5(r) arzlz=o = 0 OzOIz=o = 0
(21)
When v/r 2 + z2 ---*~, the finite conditions require that
3 FUNDAMENTAL SOLUTIONS FOR THE UNITCONCENTRATED DISPLACEMENT DISCONTINUITY Consider a penny-shaped crack of radius a with the upper surface S + and lower surface S- shown in Fig. 1. Displacement discontinuity ~u~] across the surfaces may be expressed as
(22)
Orr : 4700 : lTzz ~ lYzr ~ (TzO : {TrO -~- 0
All the components depend only on (r,z) and independent of 0 owing to axis-symmetry. Hence, eqn (15) simplifies to the form 02~i
10~ i
02~i
(23)
Or2 + r--~-r + --~-zi2= 0 ( i = 1,2) { ~ux~:Ux(X,Y, O+ ) -ux(x,y,O- ) ~Uy]=Uy(x,y,O+ ) -Uy(X,y,O - ) (x,y) E S
(17)
I[uz]]=uz(x,y,O + ) -Uz(X,y,O- ) The fundamental solutions corresponding to a unit-concentrated displacement discontinuity should satisfy the governing equations of transversely isotropic materials and the following conditions: lim
a--,0
and ~3 = 0. Thus, az0 is automatically equal to zero. Apply the zero-order Hankel transform to eqn (23), and observe eqn (22). A suitable solution is ~i(~,zi)=Ai(~)
(24)
e -~zl ( i = 1 , 2 )
~li(~,Zi) is the zero-order Hankel transform of
where
qI i(r, zi).
Substituting eqn (24) into the results obtained by zeroorder Hankel transform of eqn (13), the result is
f{~u~],~uy],~Uz]} d s = {1,0,0}
(18)
2
uz = - ZAj(~)kjsj e-~ZJ lira
a--,O
~{~ux],~Uy],~Uz]} d s = {0, 1,0}
(25)
j=l
(19)
The inverse Hankel transform can be applied to eqn (24) to yield
lim~s[~ux],[uy~,~Uz] } d s = { 0 , 0 ,
a---.O
(20)
11
xlZi(r , Zi) ~-
Satisfaction of eqn (18) and eqn (19) refer to conditions of skew-symmetry and eqn (20) to axis-symmetry. The fundamental solution with axis-symmetry will be derived.
f~Ai(~)~ e - ~ZiJo(~r ) d~
(26)
Substituting eqn (26) into eqn (14), azr is obtained as
(7zr =
(c44-~c44kj)Aj(~)sje
0 42
-~z)
Jl(~r)dr
3.1 Fundamental solutions satisfying eqn (20) (27) Owing to the symmetry, consider only the part z -> 0. The
The boundary conditions in eqn (21) are the transformed values and can be applied using eqns (25) and (27). This gives 1
Z
siAl + s2A2- 4r~ klslA1 + k2s2A2 --
(28)
1
47r~
which can be solved to obtain ml
Y
AI=-ff-A2=T
m2
(29)
The constants mi are ml m
Fig. 1. Circular crack.
1+k 2
4rsl(kl -k2)
m 2 -.~
l+kl
4rs2(kl -k2)
(30)
M. H. Zhao et al.
172 With the aid of eqn (29),
XI't i
is found from eqn (26) as
mi ql i(r, Z,) -- V / ~ -}- Z2
(31)
such that fl and f2 must satisfy
32J) . 1 32f1 . I 02fl
j+
r-g-r +
02f,
+ o-7 =0 (37)
Substituting eqn (31) into eqns (13) and (14), the stress and displacement fields are obtained:
where j = 1,2. To satisfy eqns (37) and (22), fl and f2 can be taken as 5
2
ur = - r jy= l
&2 l &2 1 0212 &2 5 7 + ;-g-r + + Oz- = °
)3,2
( r2q- "
3¢
Uz
2
mjkjzj
'=
(r 2 q-g2) 3/2
~c
fl (r, 0, zi) = 7 o : fo (Plm(() e - ~ZiJm((r) d(-sin mO
f2(r, O, z3) = Z
2
~P2m(~) e -
m=0 ~
Orr "-~- ~_. mj[c,l( 2 r 2 - Z2)--CI2(F 2 -]- Z2)
~Z~Jm(~r) d~-cos mO (38)
j=l
+ c,3kj(2zf- r2)]/(r 2 + 2 ) s/2 2 000= ~ . m j [ c l 2 ( 2 r 2 - z 2 ) - C l , ( r 2 q - z 2)
(32)
j=l
Substituting eqn (38) into eqn (35) and making use of eqn (13), the non-vanishing displacements become b~r= 2 7 0
siam0 Io
~2{[Jm-l(~r)-Jm+l(~r)]
+ c~3ki(2z~ - r2)]/(r 2 + z~) 5/2 × [H, e -~''' q-n2e-~Z2]Plm(~)-l-[Jm_l(~r )
(2Z2 __ r 2) °zz = j= Z I ajmj ( r2 q_z2)5/2 2
+ Jm+l (~r)]p2,,(~) e-~z~ }d~
2
az~ = 3r Z (c44 + c44kj) j: 1
Uo= ~m~_0c°sm0 f2~.2{[Jm-I((r)+ J,n+|(]~r)]
(sjzjmj)
(r2 q- Z2)5/2'
where
aj = - c13 q- c33kjs 2
(33)
+ [J,, _~ ((r) - Jm +, ((r)]P2m(() e - ~z3} d(. (39) The displacement conditions in eqn (34) may be enforced on eqn (39). It can be concluded that only the case of m ---- 1 needs to be retained while p In and P2m are all zero for m =~ 1. It follows that
3.2 Unit-concentrated tangential (y-direction) displacement discontinuity This fundamental solution satisfying eqn (19) will be obtained using the boundary conditions for z > 0:
f ~ 2 { [Jo(~r)- J2(F~r)]Hl2Pll(f;) + [Jo((r) + J2(~r)]p21 (()} d~ = 6(r)
ozlz= o = 0 1 1 u~ = 2-6(r) sin 0 Uo = ~6(r) cos 0
(40)
(34) J o ~2 { [d°(~ r) q- J2(~r)lHl2Pl '(~)
where 6(0 is the Dirac delta function. Based on eqn (15) and the expression of ozz in eqn (14), the potential functions satisfying the first condition in eqn (34) are given by
+ [J0(~r) -- d2(~r)]P21(~)} d~ = 6(r) where HI2 = H l q - H 2. From eqn (40), Pll(~) and P21(~) are obtained:
qll(r,O, Zl)=Hlfl(r,O, zl) q/t2(r, 0, Z2) = H2fl (r, 0, z2)
(35)
q/3(r, 0, Z3) =f2( r, 0, z3)
1
1
(41)
P21(~) -- 27rH12
where H l = 1 H e --
1 1
al
a2
(36)
Eqn (41) may be inserted into eqn (38). Further application of inverse Hankel transform renders
The method of analysis of cracks in 3D transversely isotropic media { ~i(r, O, zi) =
ql i( r, O, Zi) = ~ 1/4j [1 27rH12
H-'----L~I[1 - zi(r 2 -t- Z2) - 112] sin 0 27rH1z r (i = 1, 2)
xl13(r,O, 23)=
173
-- zi( r 2 4- d)-
112]cos o
i=1,2
1 1 1/2] ~-~r[1 --z3(r2 4- Z~) cosO
ql3(r'O'z3)=
21r r 1 [1 - z3(r 2 4- z~)- l/Z] sin 0
(42) The displacements follow from eqn (14):
Ur=_~__.~{j~=lHJ rl
(45) The displacement is obtained
Ur---
-- ~-~{j~ H~z[ I
ZJ(2r24-22)]
H,2L
4-(1
Z3 1 -t-(l-(rZ4-z])m)}-fisinO
u0= -uo= ~ -
V zel} o,o 2
rsm0 5-
27rHj2j~=~kjsjHj (r2 4-
(46)
z3(2rZ+zb] 1 z~ 1 } ;~ sin 0
r cos 0 2 ~-~l-@~=lk)SjHj(r 2"2"-312 ;~ -fzj) •
uz =
=~)-312
(r2 ~-~32),12.)} ~-~2cos 0 n,2_l -- _(r 2 +
-[I
23(2r2 4- z2)1 •
uz =
(43)
will[1
Zj(2r2+zf)]-(fi~]
The stress is The stresses can be obtained by putting eqn (42) into eqn (14). It is found that
azO =
3r sin 0 ~-- ,. . 2 , 2,-5/2 azz = ----;7, - 7. njajzjtr t zj ) 2rrtl2 j = 1 1
1
2
z0= - = [ . - = Y
3r cos 0 ~ Hjajzj(r 2 + z2 ) _ 512 °zz -= 27rH12 j = 1
2
Z ~ L/--/12j= 1
1
(c44 + c44kj)sjHj(r 2 + z~) - 312
+ c44s3(z ] _ 2r2)(r 2 + z32)- 5/2] sin 0
2 -- 3/Z
+zj)
Ozr ~
+ C44S3(Z~ _ 2r2)(r 2 + Z2) - 5/2] COS 0
1 ~,0 Z (c44 + c44kj)sjHj(z2 - 2r 2) 27r 12j= 1 2~ - 3/2]| × (r 2 -I- 2Zj)2~ - 5/2 -I- C44S3ttr2 4- Z3~
Ozr =
27r ~
~" (c44 4- c44kj)sjaj(z2 - 2 r 2 )
..I
12j= 1 X ( r 2 4- 222) - 5/2 4- c44s3(r 2 4- z 2) -
cos 0. (47)
3/21 sin
0.
-I
(44)
This completes the foundation for the problem of unitconcentrated displacement discontinuity fundamental solutions.
4 BOUNDARY INTEGRAL EQUATION 3.3 Unit-concentrated tangential (x-direction) displacement discontinuity The fundamental solution for a unit-concentrated displacement discontinuity in the x-direction can be obtained in the same way as that in the y-direction discussed earlier. The potential functions are
Let a crack of arbitrary shape in coincidence with the plane of isotropy that lies in the xy-domain as shown in Fig. 2. Denoted by S + and S- are, respectively, the upper surface with outer normal directional cosine {0,0, - 1 } and lower crack surface with {0,0,1 }. The reciprocal relations in the absence of body forces are given by
M. H. Zhao et al.
174
The foregoing constants are related to the material properties. The displacement discontinuities are expressed as
Y
{ ~u~] = ux(x,y, 0 + ) - ux(x, y, O- ) ~Uy~=Uy(X,y,O + )
Uy(X,y,O-) (x,y) E S
(53)
~uzU= uz(x, y, 0 + ) - uz(x, y, O- )
-- X
It should be noted that
r 2 = (x -- ~)2 + (y _ ~/)2
(54)
The kernel functions in eqn (51) have the same order of singularity as those in the displacement discontinuity boundary integral equations for the corresponding elasticity problem. Hence eqn (51) can also be regarded as the hypersingular integral equations.
Fig. 2. Planar crack. /
fs I"~Pi(1)Ui(2)'~) ds+ fJs- ~Pi(1)I'li(2)~)ds (48)
~- I s + {~Pi(2).(1), t~i ) ds-~- ~ s- /~Pi(2)ui(I)'~) ds where ul l), @) and ul2) are two sets of solutions for the body while p(l) = a(Om. Consider u}]), al l)' asthe real solution of the problem and ul ~), el 2) as the unit-concentrated displacement discontinuity fundamental solutions in Section 3. It can be easily shown that p12)l,+ = -plZ) l~-
(49)
The tractions on the crack surfaces satisfy the conditions
pll)ls+ = --pll)ls-
(50)
that are assumed throughout this work. Inserting these conditions into eqn (48) gives the boundary integral equations for a plane crack of arbitrary shape:
5 CRACK BORDER SINGULAR BEHAVIOUR
Choose an arbitrary point o on the crack border I" for analysing the singular behaviour. The border I" of the crack is smooth at point o. Without loss in generality the cartesian coordinate system oxyz is placed such that the yaxis and x-axis are tangent and normal to I' respectively, whereas the z-axis is normal to the crack plane S as shown in Fig. 3. Let e denote the radius of a small circle r~ contained in S. The hyper-singular integral in eqn (51) should be finite in E and can be written as
f rLzz I~Uz]~1 ds(~, r/) = Rz(x , y) I~ [(Lzr3 cos 2 0 + Lzo3 0sin ) ~2u x ]
q- (Lzr3 - tzo3)
1
sin 0.cosO~uy]]7 ds(~, 7) = R~(x, y)
[ fs+LzzllUzU~ds=pz(x,Y) IE [(Lzr3 -- Lz03) sin 0-cos O~ux]] Is +{ [Lzr3 cos 2 0 -t- Lzo3 sin20]~ux~
1
+ (Lzr3 sin2 0 + Lzo3 cos 20)~uy~] ~ ds(~, ~/) + [Lzr3 - Lzo3] sin 0 cos O~uy]} ~ ds((, ~7)=px(x, y) Is
= Ry(x, y). (55)
{[tzr3--gz03] sinOcosO~Ux~ z
ds + [Lzr3 sin 20+Lzo 3 cos 20]~uy]} -7=py(x,y). (51) where
2
LzzI = ~. aimi
Y
i=1
tzr3=~ -2,= 1[
1 ~-L
~-~-12(l+kj)+s3 c44
(52)
]
Lzo3 = w - I ,-r-- 2_ sjl4j(1 + kj) - 2s3 c44.
zTr Lrtl2j= 1
Fig. 3. Crack-tip analysis.
The method of analysis of cracks in 3D transversely isotropic media where R x(x, y ), R y(x, y ) and R z(x, y ) are all finite functions for
(x,y) E E.
Az(o)
cot
175
Otz.Lzz1 = 0
[2Lzr3 + Lzo3]Ax(o) cot rC~x= 0
5.I Discontinuity
[Lzr3 + 2Lzo3]Ay( o ) c o t
The displacement near point o can be obtained by superposing plane strain and the antiplane displacements in the oxz plane./s The displacement discontinuities at the neighbourhood of point o are given by 15
{ ~Ux]=Ax(o)~x [Uy] =Ay(o)~ ~Uz]=Az(o)~"~
(56)
'
f L~_d~ d T = x a z - l A z ( o ) f ~ z d~ f~-o~
rOty= 0
5.2 Singular behaviour The values of Ax, Ay and Az are not zero in general. Moreover, the coefficients in eqn (58) are certainly not zero. Hence, non-trivial solutions of eqn (58) exist only when the following conditions are satisfied: cot 7rczx= cot
where Ax, Ay and Az depend on the location of point o. The indices ax, ay and a z refer to the unknown solutions and their values are between [0,1]. Inserting eqn (56) into eqn (55), in the Hadamardprinciple value sense, 15 the hyper-singular integrals can be evaluated with accuracy when ~ is sufficiently small and x is infinitesimal, such as
(58)
71"O/y =
cot rcaz = 0
(59)
Recall that the values of ax, a r and a z are within the interval [0,1]. Eqn (59) requires that
ax = ay = a z = ½
(60)
This result reveals that the displacement near the crack tip behaves as r 1/2, which is the same as for an elastic solid. Making use of the unit displacements solutions in Section 3, the stresses near the crack tip outside of the crack are obtained:
azz(X, y, O)= - ~ [Lzzll[Uz]] ~ d~ d 7
d~
×
azx(X,y, O) =
[(1 -- ~)2 + (Y-- ~)2] 3/2
Ig [(Lzr3 cOs2 0 + Lzo3 sin2 O)~Ux] 1
+ (Lzr3 - Lz03) sin 0 cos O~uy[] -~ d~ d7
= -- 2x c~=- tAz(o)Tr cot 7r~z ~r~cos2Od~d7=
~r ~u~]](x-~)2r5 d~ d7
azy(X, y, O) =
I~ [(Lzr3 - Lz°3) sin 0 cos O~ux]
(57)
+ (Lzr 3 s i n 2 0 + Lzo3 c o s 20)[uy~]
4 a~
= - -~x - tAx(o)rr cot tax
1
~-~ d~ d 7
(61) sin 0 c o s 0
~E ~Ux~ r 3
ds(L 7) = 0
The results
f ~Ux]sin2 0 d ~ d 7= ~ ~ux](Y-- 7)2d ~ r.
r3
z
r5
cos 2
2a
= - -~x "-lAx(o)r cot rot x
r
t° (1 ~- 0 2 dt = - lr cot 7rc¢ dt
2
oc (t 2 + X2)3/2 = X2
f
~
dt
f [[Ux]]r 3
0
2
d~ d7 = gAx(°)Trlv/-O
sin 0 COS 0 ~ ~uy]] d~ d 7 = O
where the integrals
f~
~ ~ - d~ dT = Traz(o)/v/O
~ d7
4
- ~ (t 2 + x 2 ) 5/2 - - 3X4
have been used in deriving eqn (57). Inserting eqn (57) into eqn (55) and letting x ---* 0, then results in
(62)
r~
IlUx] sin2 0 7r d~ d7 = :~ax(o)lv/-O F. r 3 when applied to eqn (61) gives the stress near the crack tip at point (-p,y) as
azz(X, y, O) = -- [LzzlAz(o)]Tr/v ~ 1
Ozx(X,y, O) = - [ ~Lzr3 + ~Lzo3]Ax(o)Tr/V/-O
2 azy(X,y,O)~-[ILzr3+~Lzo3]ay(o)'tr[V/O
(63)
M. H. Zhao et al.
176
Table 1. Constants of materials Material
Elastic constraints ( × 10 3 N m--~)
Magnesium Zinc Barium titanate /3-Quartz
CII
C33
C44
C13
CI2
41.193 111.021 115.920 80.454
42.573 42.090 130.41 76.176
11.316 26.427 37.674 24.909
14.973 34.569 48.990 22.632
18.078 23.115 53.820 11.523
Stress singularity of the order of r -j/2 is obtained near the crack tip, which corresponds to that for the isotropic elastic material.
5.3 Intensity factors The stress intensity factors near the c r a c k tip are g i v e n by K I = lin~ V/~rcrzz(x,
y, O)
Consider a planar crack of arbitrary shape with the surfaces parallel to the isotropic plane of a three-dimensional infinite transversely isotropic medium. The crack surface is loaded by pressure p(x,y) such that boundary condition is
Oz(X,y, 0 + ) = - p
V/~Tr[LzzlAz(o)]
E
871.( i ~_ p2)
(65)
Since Ax, A t, and A z in eqn (65) are known, it can be further shown that
gll = -
K['-
2 1 V/~.Tr lim [~Lzr 3 -I- ~Lzo3] [[ux]]/~-'~k o o j vr
r-OL5
,5
)
E
lim V~p~wpB
(70)
E
(71)
8r(1 - .2)U,,A'"
while ~u.~ is obtained as
v/r
Table 2. Values of the constants related to the materials
SI $2 Lzzl(10 3) L~3(103)
Lzr3(lO3) P
(69)
It follows that the solution to eqn (68) is
Lzzl ~Uz]= -j
s
8(1 - v 2) p-0
(66)
2 -
~Md~d~l=_tz(x,y
It has the same form as eqn (68). In eqn (69), E and v are the Young's module and Poisson's ratio respectively. The pressure on the crack surface is tz(x,y). Now, let t z = - p, which corresponds to [[Wp] as given by eqn (69). The mode I stress intensity factor K p along the crack border is
!imo[Lzz,~uz]]/V/~
1
(68)
Listed in Table 1 are constraints for several transversely isotropic m a t e r i a l s ) They can be used to evaluate the values of those in Table 2. The displacement discontinuity boundary integral equation for isotropic solid takes the form
v/~rOzy(X, y, O)
Ki = - ~ / ~ . r
(67)
I~ [Lzz.[luz]]]~ d~ drl= p(x, y)
(64)
It follows from eqn (63) that K ~--- --
6.1 Symmetric problems
The boundary integral equation is given by
K,, = lira V~ro~(x, y, O) K I u = lira
6 M E T H O D OF ANALYSIS
Magnesium
Zinc
Barium titanate
0.69840 1.40844 - 1.37954 - 0.92624 - 3.60787 0.32936
1.08769 - 0.66410i 1.08769 + 0.66410i - 1.86112 - 4.80316 - 6.66638 0.10274
0.96472 - 0.11013i 0.96472 + 0.11013i - 4.26593 - 2.84293 - 10.64441 0.32329
13-Quartz 0.77277 1.32989 - 2.71520 - 3.74574 - 6.49832 0.16441
The method of analysis of cracks in 3D transversely isotropic media 1 E LzzI 87r(1 - v2)~vvp]~"
(72)
With the aid of eqn (66), the stress intensity factors near the crack border are obtained: K~ =KIp
(73)
Eqn (73) is independent of the shape, the dimension, the geometry of the crack and the distribution of the mechanical load p. Here we can conclude that the stress intensity factors of a planar crack of arbitrary shape in transversely isotropic media can be calculated by eqn (73) with the corresponding solutions of elasticity. 5 The same holds for the penny-shaped or elliptical cracks. 5
177
They are hyper-singular boundary integral equations in elasticity. 14,15 As before, the solution of a crack subjected to antisymmetrical loading in transverse media can also be obtained from that for an isotropic elastic material. Knowing that E
= 8(l-
ffrr
8(1 +
m li,-u
(77) Uy~
the stress intensity factors in transversely isotropic material become
6.2 Anti-symmetrical problems The boundary integral equations of the crack subjected to anti-symmetrical loading are
Kn=-
r--.oW~r ~ux] ~ r[[2Lzr3 - - ~ - -+- -Lz03] - J lim (78)
r,.P
+ (Lzr 3 -
[ Lzr3 + 2Lz03] lim ~ / ~ u v ]
^h'=-~'[
f s [(Lzr3 COS2 0 + Lzo3 0sin ) [2u x ] 1
"i r---0V r
(74)
~s [(Lzr3 -- Lz°3) sin 0 cos O[u~]
=py(x,y). where p, and py are the tractions on the crack surface in the x- and y-directions respectively. The equivalent isotropic elastic constants E and v can be identified with the transverse material constants as 3v
p 4r cos 0 Ku -- ~-(2--~-~v/Tra
(80)
(75)
E
Lzr3 - Lz°3 --
Solutions to many crack problems can be obtained in the same way. 5 Suppose that a penny-shaped crack is sheared in the direction of the x-axis, with r being the shear stress. The stress intensity factors KH and Km in elastic solid are: 5
K}IIP= _ 4(1 -r(_~._ ~ v ) (sin r 0)X~
1 -- 2_.______v
Lzr3 - Lzo3
(79) Kll I ~-~KPl
+ (Lzr3 sin 2 0 + Lzo3 COS20)~uy]] ~ ds(~, )1)
Lz__03_
-
The stress intensity factors are related to those in elasticity as
Lz03) sin 0 cos 0[uy]] ~ ds(~, 7)
= px(x, Y)
~
Substituting eqn (80) into eqn (79), the stress intensity factors can be obtained. The constant ~, is listed in Table 2.
8r(1 - v2)3v
Taking eqn (75) and eqn (74) into account, it is found that 7 CONCLUSION
f~{ [(1 - 2v)+ 3v cos 20]~ux]] + 3v sin 0 cos O~ur]} -~ ds(~, ~t) 87r(1 -- v2)-~ (x,Y)t-'x E ~ { [3v sin 0 cos O~u~]]
1 + [(1 - 2v) + 3 sin 20]~Uy]} -'~ ds(~, 7) 8r(1 - v2) --
E
-py (x, y)
(76)
The Hankel transform is used to derive the fundamental solutions for a unit-concentrated displacement discontinuity. The displacement discontinuity boundary integral equations are derived for an arbitrarily shaped crack in a transversely isotropic solid. They correspond to the hypersingular equations. Displacement and stress near the crack border are obtained. The familiar r - m singular behaviour is found for the stress in front of the crack tip. More specially, the stress intensity factors are shown to be obtainable from the corresponding solutions for an isotropic elastic material.
178
M. H. Zhao et al.
ACKNOWLEDGEMENTS 9. The work was supported by the National Nature Science Foundation of the People's Republic of China and the Technical Development Foundation of the Machinery Building Industry of the People's Republic of China.
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