Three-dimensional problem of a running crack

hr. J Engng Sci. Vol. 17, pp. 59-71 0 Permnon Press Ltd.. 1979. Printed in Great Britain

THREE-DIMENSIONAL PROBLEM OF A RUNNING CRACK S. ITOUt Department of Mechanical Engineering, Hachinohe Institute of Technology, Hachinohe 031, Japan Abstract-The problem of an uniformly propagating finite crack in an infinite medium is solved using the dynamic equations of elasticity in 3dimensions. Equal and opposite tractions are applied arbitrarily to the crack surfaces. The problem is reduced to the dual integral equations and solved with the aid of the series method. The numerical examples are presented on some graphs.

I. INTRODUCTION

a sufficiently large transverse tension is applied to the material, an infinitesmally small micro-crack appears and the crack will continue to propagate in both directions along its length. Such a problem has been first considered by Yoffe under the assumption that the crack propagates to the right only remaining of constant length[l]. A similar problem has been treated by Ang[2]. He solved a transient problem in which a half-plane crack extends at constant velocity. Craggs obtained a solution for a semi-infinite crack subjected to surface loads with points of application moving at the same speed as the crack[3]. Broberg treated the crack as suddenly opening from zero length and symmetrically growing with constant velocity [4]. Baker obtained a solution for a semi-infinite crack which suddenly appears and grows at constant velocity in a stretched elastic body[S]. In the recent paper, Yoffe’s investigation was extended to the problem of a moving crack in a strip of finite width by Sih and Chen[6]. The interaction of stress waves with a running crack has also been investigated for a semi-infinite crack and a finite one by Sih and his coauthors[7-91. The above problems are solved within the 2-dimensional theory of elasticity. In most instances, however, the crack is subjected to a state of stress that is triaxial in nature. In the present paper, Yoffe’s model is solved in 3-dimensional elasto-dynamic theory, namely, a finite crack is pressed by an arbitrarily distributed traction on its surfaces. The 2-dimensional Fourier transforms are used to solve the problem and a set of mixed boundary values are reduced to the dual integral equations. These equations are usually convertible to Fredholm equations of the second kind which may be solved numerically[lO, 111. The present problem, however, is 3-dimensional and we must use an uncomplicated method. To solve the problem, the present author expands the transformed surface displacement in a series of Jacobi polynomials and uses the Schmidt method. Numerical calculations are carried out and compared with those given by Yoffe to reveal the 3-dimensional effect. WHEN

2. FUNDAMENTAL

EQUATION

With respect to a fixed rectangular coordinate system (x*, y*, z*), the equations of motion for a homogeneous isotropic elastic solid are, in the absence of body forces

a=u*a=u*a=u* a2u* (A+‘&($+$ +az* aw*) =PgT’ +cL(p+“Z+az*2 ay

>

a20* a2v* a2v* (“+~)~(~+~+~)+~(~+~+~)=P~. a2w* a2w* a2w* a2w* (A+~)~(~+~+~)+~(~+~+~)=P~ tAssistant Professor.

(2.1)

S. ITOU

60

where u*, u*, w* are the displacement components, A and p are Lame constants and p is the density of the solid. Consider the problem of an infinite elastic solid which contains a through crack of length 2a along the x*-axis as shown in Fig. 1. It is assumed that the crack is opened at one end and closed at the other with constant speed U. For a constant velocity crack, it is convenient to introduce a Galilean transformation x =x* - Ut*, with

(x,

moving

2 = z*,

Y = y*,

t = t*,

(2.2)

y, z) being the translating coordinate system attached to the moving crack. In the coordinates, the equations of motion become independent of the time variable t

a au av aw (A+~)a*(~+a,+~)+~(~+~+~)=pLit~, a au au aw a2v (A+“)~(~+~+jy +cL > (

a2v

a20

2

-jg+g+;i;z

=pu2$ )

(2.3)

(~+I”)~(~+~+~)+~(~+~+~)=pu’~.

The stresses are expressed o,=(A+2&+A$+A$ o, = Ag+(A

+2p)$+A$

oZ=A$+A$+(A+2&

(2.4)

ryz=F($+g), ra=/L($+$), r_=/,L

au au ax ay

(

-+-.

>

The boundary conditions for the problem to be studied are

uyyl(2p) = - P(X,21,

for y = 0,

1x15 a,

121cm,

(2.5.1)

V=o,

for y = 0,

Ix\> a,

Jzl< 9

(2.52)

7,” = 7,, = 0,

for y = 0,

Ixl
Izl
(2.5.3)

Fig. 1. Geometry and coordinate system.

Three-dimensional

problem of a running crack

61

3. ANALYSIS

To find the solutions of the wave eqn (2.3) we introduce the 2-dimensional Fourier transforms f(5, Y, 5) =

j-1[_lf(x, Y, z) e’@‘+“’ dx dz, (3.1)

Using this theory, we can reduce eqn (2.3) in x, y and z to the following ordinary differential equations in y

1-$-(U2-M~)52-~2 Ili-i(a2-l)[$-=(au’-l)&p=O, -i~(cZ-l)~+(~2$-(l--M~)~2-~2]i-i1(.2-*)~=0, -&(a2 - l)E - il(a2 - l)e+ dy

(3.2)

-$-&‘-(l-M:)f2}~=0. 1

with cz2= (A + @)/CL,p2 = P//-Land M$= p2U2. Combining eqns (3.2) results in the equation

($-,:)($-.:)cr,5, G,)=O,

(3.3)

n 1= {( 1 - M&.Y2)~2+ [2)“2, n2 = ((1 - M:)t2 + s3”‘.

(3.4)

with

Because of the symmetry conditions in eqn (2.9, it is possible to consider the problem for the half-space, y Z 0, only. The solutions of eqn (3.3) appropriate to y L 0 will be the following forms ii = A, e-“ly f B, e-qy, v’= A2 e-“ly + B2 e-9y,

(3.5)

6 = A3 e-V + Bs e-qy t where Al, Al, As, B,, B2, B3 are independent of y. Substituting from eqn (3.5) into eqn (3.2) gives

A&A I

4

A 3= $

27

AZ, 82 = -h

([B, + {B3).

(3.6)

Substituting eqns (3.5) and (3.6) into eqn (2.4), we obtain the transformed stress components. For example, Zy, FXyand Tyzhave the forms Gy/(2~) = $-{(M:-

2)t2 - 253A2 e-“ly + i([B, + lB,) e-9y,

F,.,/(~F) = -itA

e-“ly - &B,

+ (d + t2)BJ evay,

iyr/(2p) = -i&42

e-w - $-{,gBI

+ (n: + 52)B~lem?‘.

(3.7)

From the boundary condition eqn (2.5.3), B1 and B3 are represented by A2 BI = -i&2AJ{( 1 - 0.5M$2 + 52), Bs = -i&Ad{( 1 - 0.5M;)[2 + [q.

(3.8)

Consequently, the transformed components of stress and displacement can be represented by

62

S. ITOU

the single unknown coefficients A*(&,5). It is convenient, however, to represent these components by the transformed displacement I?,,on y = 0. The relation between $ and A2 is given by substituting eqns (3.6) and (3.8) into eqn (3.5) A2

=

-2{(1-

O.SM;))z*+ &/(M&$2)Cc,.

(3.9)

As a result, we obtain &/(2~) = -Q{(l - o.5M+)[2+ 52x2 + MXl - 2/a2)}/(it4$rI) e-“ly - 2n2/Mk-*Yl, Gy/(2~) = $[2{(1 - 0.5M2,))52 + ~2}2/(M2&2n1) e+ly - 2(2*+ J2)n2/(M2&*)e-7, &/(2~) = -$[{(l - O.SM$* + 53(2c2 + M2d2(l - 2/a2)]/(M2&2nr)e-“~y-212n&14~2) e-9y]. iyz(2~) = 2i@,[{(l - O.SM2,))5* + ~2)/(M2&2)(e-“1y - e-“2y)], I=/ = -25@o[{(l - 0.5M2,)t2+ &/(M$$*n,) e-“ly - n2/(M&$‘)e-“2Y], iXY/(2p)= 2i@d{(l - O.SM&* + ~2}/(M%2)(e-“1y- e-“ZY)J, u’= - 2i&7a[{(l- 0.5M2,))5* + ~2)/(M~*nr) e-“ly - n2/(M2d2)e-ay], u’= -2Q{(l - 0.5#))5* + [2}/(M2&2)eCy - (t2 + 52)/(Mg2) e-n2y], 6 = -2i51%{(1- 0.5M2,))z2+ ~*}/(M~*nl) eCy - n2/(Mg2) e-%y]. (3.10) The remaining boundary condition eqns (2.5.1) and (2.5.2) reduce to the dual integral equations

4m.4) =

$

$K(& I) e-* d&= -p(x, l), for 1x1S a,

-@

dt = 0,

for [xl> a,

(3.11.1) (3.11.2)

with K([, 5) = 2[{(1- 0.5M+)t2 +
for 1x1S 1.0, for (xl> 1.0,

(3.13)

where a,(l) are the unknown coefficients to be determined and Pi”*-I’*)(x) is a Jacobi polynomial. Fourier transformation of eqn (3.13) is [12] $(& I) = Im &(0(x, 5) ep dx --m = 2V7r z, &lU)(- lY-1 Hf

J2n-&S

(3.14)

where r and J,, are Gamma and Bessel function, respectively. Substituting eqn (3.14) into eqn (3.11.1), we obtain for 1x1S 1.0 after integration with respect to x fi,

a.(b)2(-vff1)“?(2n (2n - 2)! -4, 1 mp1 J2n-1(5M5r0 sin(Zx)dS = - [ P(X,0 dx.

(3.15)

To evaluate the semi-infinite integrals in eqn (3.15) numerically, we must replace the integrands so as to decay rapidly when 5 becomes large. For this purpose, using the relation

I

omfJ2._,(f) sin (6x) dt = &

sin ((2n - 1) sin-’ (x)],

(3.16)

Three-dimensional problem of a running crack we

63

rewrite the integrals $L-I(Z)&& = I0

I) sin (6x) d5 m g(5)y sin (B) d5 + K(6, f)/(6(2n - 1)) sin (2n - 1) sin-’ (x)},

(3.17)

with (3.18) where S represents the infinite limit of 5, or

= 2{(1 - 0.5M+)2- V(1 - M&x2)(1 - M;)}/{MW( 1 - MW)}.

(3.19)

The function g(t) in eqn (3.17) behaves as t-2 for large 6, so that the semi-infinite integrals can easily be evaluated by Filcn’s method. Then eqn (3.15) can be solved for the coefficients a,(l) by the Schmidt method as follows[l3]. For brevity, we rewrite eqn (3.15) as

forI4 5 l.09

(3.20)

where E,(& x) and u([, x) are known functions and the coefficients a,,(5) are unknown and to be determined. A set of functions Q&, x) which satisfy the orthogonality condition

I

o’ Q&,

x)Q&

xl dx = Nnhm

Nn = 1’ Q&, x) dx,

(3.21)

can be constructed from the function, I?,,([, x), such that (3.22) where Min is the cofactor of the element dimof 4, which is defined as

D,,=

;

,

(j”,...

..I

.

.

.

.

din=

I’ 0

Ml, xPn(& x) dx.

(3.23)

j”,

Using eqns (3.20) and (3.22), we obtain (3.24) with UG x)Qj(t x) dx.

(3.25)

S.ITOU

64

Consequently, for instance, when we put 0.593 on the value of I in eqn (3.20), ~~(0.593) are given through the above procedure. 4.STRESS-INTENSITYFACTOR

Once the coefficients a,([) are known, the entire stress field is given. In fracture theory, however, the significant quantity to be calculated is the direct stress acting across a radius from the tip of the crack. We write x=a+rcosd,

(4.1)

y = r sin 8, and consider the stresses for small values of r. The required stress CQis given by f7e= rr, sin* S f q cos2 B - 2rXy sin 8 cos 8.

(4.2)

For small value of r, it is shown fm

Jo

J2,&()

e-‘+m2)Kcos (5x) d+-$

~“‘~~~_~~~

-..~

~~~+q~),

(-l)“+’ V(-cos B-t t/( 1 - m* sin’ 8)) f o(to) Jzn-t({) e-‘“-m*)yCsin (&) dt -+ -2t/r t/( 1 - m2 sin2 0)

(4.3)

Using eqns (3.10), (4.2) and (4.3) we obtain the stress-intensity factor Ki

= MCOS

e f ql)/{iW( 1- M2Tla*)s,}[(l- M$/2H2 + M%l - 2/a2)} sin* B -2(1- M$2)2 cos2 8]- 4(1- M$f2)Q(-cos 0 + q,)/(~~~) sin B cos B +2tl( 1 - ~~)~(cos B+ q~~~(M2~*~cos2 B - sin’ 8) + 4(1- M$f2) x u’(-cos e -t 42)/(M2~2) sin 0 cos 01

with 41~ q2

=

tr(l -

M$fa2 sin2 @), v’(1 - M$ sin2 e),

(4.5)

where the semi-infinite integral is also evaluated by Filon’s method. Since a,(l) are functions of MT and p(x, z), it is impossibIe to reduce the quantities in the bracket [ 1 in eqn (4.4) to such a simple expression given by Yoffe for the plane problem. RESULTS For a numerical calculation, the function p(x, z) is assumed as 5.NUMERICALEXAMPLEAND

P (XV 2) = F/(1 + @2)X

(5.1)

where P is a constant and b is the parameter which governs the dis~bution of the applied load along the z-direction. Poisson’s ratio v is taken as 0.25. To perform the Schmidt procedure, we adopt the first five terms of the infinite series in eqn (3.20). For a check of the accuracy, the m

values of “, ~~(~)~~(~,x) and - u(& x) are given in Table 1 in the cases of l= 0.0 and 0.2 for E b = 1.0 and Mr = 0.4. From this, it is clear that the accuracy of the Schmidt method is perfect. In Table 2, the values of the coefficients a,([) are given for b = 1.0 and MT = 0.4. In Figs. 2 and 3, the stress-intensity factors at 2 = 0.0 are plotted vs MT for b = 0.01, 1.0 and B = O”,27”, 36”,

Three-dimensional

65

problem of a running crack

4.5”,54”. From these, it is seen that there is also a critical velocity at which the crack tends to become curved as pointed out by Yoffe[l]. The values of the velocity are about MT = 0.63 and 0.65 for b = 0.01 and 1.0, respectively. When the crack branching occurs, the mathematics model upon which the present paper is based, i.e. a straight running crack, breaks down. Hence, numerical c~cu~ations are presented by limiting the value MT up to 0.6 hereinafter. In Figs. 4 and 5, the stress-intensity factors at z = 0.0 and 1.O are plotted vs 8 for MT = 0.01, 0.6 and b = 0.01, 0.1, 1.0. Figures 6 and 7 show the stress-intensity factors at B = 0” and 72” vs z for MT = O.Of,O.6and b = 0.01, 0.1, 1.O,The results for b = 0.01 are well coincident with those which are calculated using eqn (15) in Ref. [I].? I’Yoffe’s numerical results are somewhat rough in part. Equation (15) in Ref. [I] has a typographical error.

Table 1. The values of #t, a,,(g)E,(&x) and -u(&

X) for

6~l.O~d~*=O.4

0.0

0.9 1.0

-0.12565 x IO’ -0.15706x 10’ -0.18847 x IO’ -0.21988x IO’ -0.25129x 10’ -0.28270 x IO’ -0.31408 x 10’

0.1) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 I.0

-o.GOooOxIO0 -0.25718X too -0.51435 x 100 -0.77153 x Id -0.10287x 10’ -0.12859x 10’ -0.15431 x 10’ -0.18002x IO’ - 0.20574x 10’ -0.23146x IO’ -0.25715 x 10’

-o.OmOx loo -0.25718 x ld -0.51435 x loo -0.77153x loo -0.10287x IO’ -0.12859x IO’ -0.15431 x 10’ -0&?002x IO’ - 0.20574x 10’ - 0.23146x 10’ -0.25715 x 10’

Table 2. The values of o,(l)

UES Vol. II, No. I-E

o.OOoOo x loo

-0.12565x 10’ - 0.15706x 10’ -0.18847 x IO’ -0.21988 x IO’ -0.25129x IO’ -0.28270 x IO’ -0.3l408 x 10’

0.4 0.5 0.6 0.7 0.8

0.2

IO0

-0.31411 x lot - 0.62823 x IO“ - 0.94234x loo

0.2 0.3

0.0

o.ooooOx -0.31411 x - 0.62823 x - 0.94234 x

100 IO0 100

0.1

for 6 = 1.0and Mr = 0.4

c

Q’(5)

0.0

0.2 0.4

0.26963x IO’ 0.21259x 10’ 0.16121x 10’

- 0.64547x 10-7 0.36844x 10-2 0.107~lx 10-l

2:o 3.0 4.0

0.16uo3 * x 100 0.42856x 10-l 0.123$x 10-i

0.16629x lo-’ 0.72827x 10-z 0.27!13x 1O-2

10.0 12.0 14.0

0.13153x 10-4 0.15020x 10-j 0.17554x IO+

0.51584x 10-j 0.63126x 1O-6 0.77678x lo-’

Q2(5)

*..... ,*...* .*....

OS(S)

. . ... . .

0.37076x I@ - 0.29Ml x 10-4 -0.88798 x 10-4

. . ... . .

-025158x10-~

.,....

-0.16169x 1O-3 -0.83829 x lo-’

. . .*. . .

- 0 35647x 10-6 -0.44248 x IO” -0.508s4 x 10-E

. ...*. I

...... ......

66

S. ITOU

The crack surface displacement uO(x,z) is obtained by using eqns (3,l) and (3.13) uo(x, 2) =

J- f$, 27F

a.(3)P~fi?$*‘(x)(l -x2)“2e-i’Cz’d&

(5.2)

where the semi-infinite integral is evaluated by Filon’s method. In Figs. 8-10, the crack opening displacements at z = 0.0 are plotted vs x for b = 0.01, 0.1, 1.0 and MT = 0.01, 0.2, 0.4, 0.6. Figures 11 and 12 show the surface displacements at x = 0.0 and 0.8 vs z for MT = 0.01,0.6 and b = 0.01, 0.1, 1.0. In conclusion, we get the following information; (i) In contrast to the plane problem, the

Fig. 2. Stress-intensity factor at z = 0.0 for 6 = 0.01 vs MT.

b = 1.0

Fig. 3. Stress-intensity factor at z = 0.0 for b = 1.0 vs MT.

Three-dimensional

problem

of a running

crack

67

1

IO M,

= 0.01

-

I = 0.0

______

z=,.O

7 -c s N ‘;

0.5

I: %

0.0

1

1

0'

90"

45' 8

Fig. 4. Stress-intensity

factor 1

at I = 0.0 and 1.0 for

MT = 0.01 vs0.

I

I

IO-

&f,

1

r

= 0.6

-z=o.o _____--I

=

IO

b =O.Ol

00'

I

I

I

0'

Fig.5. Stress-intensity

I 45"

factor

at z = 0.0 and 1.0 for MT = 0.6 vs B.

90'

S.ITOU

68

b =O.Ol 0.1

2.0

Fig. 6. Stress-intensity factor at B’=0” and 72”for MT = 0.01 vs z.

b =O.Ol 0.1 0

=

4 ._

-

2

0.01

N

------v-w_

o.o-

___

1.0

0.0

I

Fig. 7. Stress-intensity factor at 0 = 0” and 72“for MT = 0.6 vs L.

2S

Three-dimensional

problem

r

r

of a running

69

crack

I

I

I

.Z=O.Q b = 0.01

/W,=O.6

00 00

1

I

I

I

I

I

1

IO

0.5 x Fig. 8. Crack

opening

at I = 0.0 for b = 0.01vs x.

displacement

I

1

I

I

1

I

z=o.o b=O.l f-

%

c < 3

0.5

0.0

x Fig. 9. Crack

I

opening

displacement

at z = 0.0 for b = 0.1 vs x.

1

I

c

M, = 0.6

Fig. 10. Crack

opening

displackment

at z = 0.0 for b = 1.0 vs x.

S. ITOU

70

M,

=QOl

-x=00 _______x=O8 IO

-

b =OOl 01

<

7

1.0

‘;p 0.5

-

-----------10

00

----____

---__

_----_-______

---___

---_____

----------_________

I

I

L

10

00

20

I

Fig.

I I.

Crack

opening

displacement

at x = 0.0 and 0.8 for MT = 0.01 vs I

1

I

1

I

M, = 0.6 -x=00 _-__-_

x =O.B

b=OOl 0.1

10 -

00

I

I

,

I

00

I

1

I

I

20

IO .?

Fig. 12. Crack

opening

displacement

at x = 0.0 and 0.8

for MT = 0.6 vs i.

stress-intensity factor at f3= 0"isdependent on the propagating speed and becomes smaller as the speed increases. (ii) Both of the stress-intensity factor and the crack surface displacement along the z-direction tend to decrease more feebly than the value of the applied pressure, 1.O/{1.O+ (bz)‘}, when z increases. (iii) The crack opening displacement increases monotonously as the crack speed is increased. (iv) Since the stress-intensity factor is represented by the separation of variables form of fI and z, the variation of Kias a function of 0 is determined for any values of 2. Acknowledgemenfs-The author wishes to express his hearty thanks to Prof. A. Atsumi, suggestions and to the unknown reviewer, Inr. 1. Engng S-i., for his helpful comments.

REFERENCES [I] E. H. YOFFE. Phil. Mag. 42,739 (1951). 121 D. D. ANG, 1. Math. Phys. 38, 246 (1959). [31 J. W. CRAGGS. J. Mech. Phys. Solids 8,66 (1960). [41 K. B. BROBERG, Arkiu Fiir Fysik 18, 159 (1960).

Tohoku

University,

for

his

Three-dimensionalproblem of a runningcrack

71

[S] B. R. BAKER, J. appl. Mech. 29, 449 (1962). [6] G. C. SIH and E. P. CHEN, Int. J. Engng Sci. 10,537 (1972). [7] G. C. SIH and I. F. LOEBER, J. appl. Mech. 37,324 (1970). [8] E. P. CHEN and G. C. SIH, Int. J. Solids Stractures 9,89l(l973). [9] E. P. CHEN and G. C. Sih, J. appl. Mech. 42,705 (1975). [IO] G. C. SIH and J. F. LOEBER, Q, appl. Math. 27, 193 (1969). [I I] A. K. MAL, ht. 1. EngngSci. 8,763 (1970). [12] A. ERDELYI (editor), Tables ofIntegralTransforms, Vol. I, p. 122. McGraw-Hill, New York (1954). [I9 P. M. MORSE and H. FESHBACH, Method of Theoretical Physics. Vol. 1, p. 926. McGraw-Hill, New York (1958).

(Received 20 February

1978)