International Journal of Engineering Science 41 (2003) 1767–1783 www.elsevier.com/locate/ijengsci
On semi-infinite crack problems in elastic–plastic bodies; uniqueness and examples E. Turska *, K. Wisniewski Polish Academy of Sciences, IFTR, Swietokryska 21, Warsaw 00-049, Poland Received 19 August 2002; accepted 1 November 2002
Abstract First, the conditions for uniqueness of the stress boundary value problem for infinite cracks in elastic– plastic bodies are discussed on the basis of the Laurent series representation of the plastic boundary in an elastic perfectly plastic body in anti-plane strain (mode III). Next, for two cases, exact closed-form solutions of the shape of the elastic–plastic boundary are found in terms of elementary functions. A crack under shear stress acting on its surfaces, and a crack under constant remote shear stress r1 13 ¼ p, are considered. In the first case, also the complete stress distribution is obtained, for the second the physical coordinates as functions of stresses are found. The new elastic–plastic solutions are compared with ones predicted by linear elastic fracture mechanics. Ó 2003 Elsevier Science Ltd. All rights reserved. Keywords: Mode III; Infinite cracks; Elastic–plastic boundary; Uniqueness; Eigen-solution
1. Introduction The most often used concept for modelling the crack-tip plastic zone in elastic–plastic materials is the small scale yielding (SSY) one. It assumes that the plastic zone at the crack tip is small compared to characteristic geometric dimensions of the body, and that the linear elastic stress field outside the plastic region is a K-dominant one, see e.g. Rice [8] and Cherepanov [2]. Formally, a real crack in a finite body is replaced by an infinite crack in an infinite body, and real boundary conditions are replaced by asymptotic ones in infinity, e.g. for mode III cracks, r23 ¼ OðKIII ð2prÞ1=2 Þ, r13 ¼ OðKIII ð2prÞ1=2 Þ as r ! 1, where r is is the distance from the crack *
Corresponding author. Tel.: +48-022-8278182. E-mail addresses:
[email protected] (E. Turska),
[email protected] (K. Wisniewski).
0020-7225/03/$ - see front matter Ó 2003 Elsevier Science Ltd. All rights reserved. doi:10.1016/S0020-7225(03)00065-X
1768
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
tip. In this case the elastic–plastic boundary is a circle, RðhÞ ¼ R0 cos h, with the center located on the extension of the crack line. In the present paper, we examine the influence of the SSY assumption that the crack is semiinfinite on the proper formulation of the boundary value problem (BVP). It is known that the elastic BVP does not obey the St.Venant principle [2] for the limit case of an infinite crack in a body infinite in all directions, and that additional conditions of correctness (stability and uniqueness) must be specified. In linear elastic fracture mechanics, such conditions can be: the attenuation of stresses in infinity and the finiteness of displacements at the crack tip. To discuss a proper formulation of the BVPs for elastic–plastic infinite cracks we present a Laurent series representation of the elastic stress distribution near the crack-tip plastic zone for an arbitrarily loaded body, which can also contain other cracks. In elastic–plastic problems, the series gives us an infinite set of stress solutions all attenuating in infinity. Because, in contrast to linear elastic fracture mechanics, the displacements are finite at the crack tip, we must use another condition to ensure uniqueness. This can be e.g. a condition of the asymptotic behavior of stresses in infinity, which should include not only the rate of decay but also the ratio. Thus the BV conditions should contain additional functions gij ðr; hÞ that are asymptotic to rij ðr; hÞ in infinity. Please note that, because of some geometrical or loading conditions of the body with cracks, the order in infinity may not be of the reciprocal square root. To define the elastic BVP analogous to the elastic–plastic one, we use the same boundary conditions, and the condition that displacements are finite at the crack tip. Then, the order and the constant of stresses in infinity must also be matched; this is illustrated on an example in Section 4.2. It should be noted that to obtain the SSY model as a limit case for some bounded body under finite loading, it is not enough, as some authors do, to let the stresses decay to zero. One must additionally know how fast they decay. In the paper, two solutions with different orders in infinity are given for an analogous elastic problem in Section 4.2. The solution by which they differ is, in literature, sometimes called an Ôeigen-solutionÕ, a class of solutions often exploited in composite mechanics. Next, two new results of the stress fields in elastic–plastic bodies with infinite cracks are derived in a strict analytical manner. Such exact mode III solutions still draw attention, e.g., Hurtado [6], because they give a good insight into into the nature of more complicated ones, and can be used to construct approximate methods of wider practical applicability. The closed-form solutions for the shape of the elastic–plastic boundary are found in terms of elementary functions and the aforementioned Laurent series is not used. The analyzed cases include: (a) a crack loaded on its surfaces, and (b) a crack in a body under remote shear load. The stress field is obtained by transforming the problem from the physical plane to the stress plane, in which an analytic (holomorphic) function with given singular boundary values is found. This problem is solved by using conformal mappings (also the Schwarz–Christoffel formula) and the Keldysh–Sedov technique. The constructions of the applied conformal mappings are omitted, since they can be easily verified. The results of the same BVPs for a purely elastic material and the elastic–plastic ones are compared. For this purpose the plastic boundary, in the case of the elastic body, is predicted by substituting the elastic stresses into the yield condition. This simplified procedure is quite often used in fracture mechanics. Even the shapes of such ÔpredictedÕ plastic zones are applied to such
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
1769
precision demanding problems as the evaluation of the crack initiation angle. The second considered example shows the shortcomings of such ÔpredictionsÕ. 2. Basic equations The Cartesian coordinate system ðx1 ; x2 ; x3 Þ, is taken with its origin at the crack tip, and the negative part of the 0x1 axis is aligned with the crack. Let us suppose that the body B is linear elastic––perfectly plastic and the load level is sufficiently low, so that the plastic zone originates only at the crack tip. In the case of mode III loading, the only non-vanishing displacement is the third component w of the displacement vector u ¼ ½0; 0; w parallel to the 0x3 axis. In the elastic region, according to the HookeÕs law, the components of stress for wðx1 ; x2 Þ ¼ ð1=lÞIm½f ðzÞ, where z ¼ x1 þ ix2 are r23 ðx1 ; x2 Þ þ ir13 ðx1 ; x2 Þ ¼ f 0 ðzÞ:
ð1Þ
In polar coordinates with the origin at the crack tip, for z ¼ rðcos h þ i sinhÞ, p < h 6 p. The following formulae hold, for the stresses rh3 ðr; hÞ þ irr3 ðr; hÞ ¼ f 0 ðzÞeih ;
ð2Þ
the strains eh3 ðr; hÞ þ ier3 ðr; hÞ ¼
1 0 f ðzÞeih 2l
and the displacements 1 o o 1 þi wðr; hÞ ¼ f 0 ðzÞeih : r oh or l
ð3Þ
ð4Þ
Let us now consider the plastic zone P at the crack tip. The yield condition has the following form r213 þ r223 ¼ k 2 ;
ð5Þ
pffiffiffi where k ¼ r0 =2 for the Mises condition or k ¼ r0 = 3 for the Tresca–St.Venant condition, for the tensile yield stress r0 . In the plastic zone, stresses are constant along radial lines h ¼ const, and equal to rh3 ¼ k, rr3 ¼ 0, i.e. r13 ðr; hÞ ¼ k sin h;
r23 ðr; hÞ ¼ k cos h;
ð6Þ
see e.g. Hult and McClintock [5], Cherepanov [2], or Broberg [1]. The above equation, in a concise form reads r23 ðr; hÞ þ ir13 ðr; hÞ ¼ keih :
ð7Þ
1770
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
All components of stresses are continuous on the boundary L of the elastic and plastic regions. Assume that L is represented by L ¼ fz : z ¼ RðhÞeih ; p=2 6 h 6 p=2g. Thus, from Eqs. (1) and (7) we obtain that the stress function f 0 ðzÞ is f 0 ðzÞjL ¼ keih ;
ð8Þ
In the whole plastic region rr3 ¼ 0 so for monotonic loading the displacement w is also constant along the radial lines, i.e. wðr; hÞ ¼ wðhÞ. Thus knowing RðhÞ, we can determine the strain and displacement in the plastic region ow k ¼ RðhÞ; oh l
or
k wðhÞ ¼ l
Z
h
RðaÞ da:
ð9Þ
0
Also for the known RðhÞ, the crack opening displacement (COD) is fully determined, k COD ¼ l
Z
p=2
RðaÞ da:
ð10Þ
p=2
Let us now change our point of view and instead of seeking the stresses r23 þ ir13 as an analytic function of z, see Eq. (1), we shall look for z as an analytic function of r23 þ ir13 . Equivalently, z is an analytic function of r23 ir13 , or, to simplify the following calculations, a function of dimensionless variables r23 r13 i ; k k
ð11Þ
i:e: z ¼ x1 ix2 ¼ F ðxÞ;
ð12Þ
def
x ¼ n þ ig¼
where F ðxÞ is some analytic function. For points z on the elastic–plastic boundary L, the stresses are x ¼ eih . Then the shape of the elastic -plastic boundary can be expressed in terms of the Ôinverse elastic stress functionÕ F ðxÞ as RðhÞ ¼ eih F ðeih Þ:
ð13Þ
3. Properties of stress distribution in elastic part of body Consider an arbitrary bounded or infinite body B, containing one or more cracks, under antiplane loading conditions. Let us limit our attention to one crack, situate the origin of the rectangular cartesian coordinates ðx1 ; x2 Þ at the crack tip, and align the negative part of x1 axis with the crack. For some constant q < 1 we find a curve D ¼ fðx1 ; x2 Þ : r213 ðx1 ; x2 Þ þ r223 ðx1 ; x2 Þ ¼ qk 2 g. Now, with no loss of relevance, instead of analyzing the whole body we can discuss a reduced
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
1771
problem of only the part bounded by curve D on which are prescribed boundary conditions imposed by the general problem, see Fig. 1. In the stress plane x ¼ n þ ig the unknown elastic–plastic boundary maps into a unit semicircle and the curve D into a semicircle of radius r, see Fig. 2. The solution of the general problem of the crack is equivalent to the determination of an analytical function F ðxÞ in a semi-annulus––the Ôinverse stress functionÕ. Thus, we shall seek the function satisfying the following conditions: 1. the function F ðxÞ is analytic in the right semi-annulus AR ¼ fx : q 6 kxk 6 1; arg x 2< p=2; p=2 >g; 2. the stresses are continuous on the elastic–plastic boundary, see Eq. (13), and the function RðhÞ must be real and positive, i.e. Im½xF ðxÞ ¼ 0;
ð14Þ
Re½xF ðxÞ > 0 for x ¼ eih ; p=2 < h < p=2;
ð15Þ
3. the crack is situated on the negative part of the x1 -axis Plastic zone
crack
D
x2
crack
A B E D
C
D F
x1
Fig. 1. Reduction of problem for an arbitrary body to a problem of a body bounded by curve D and containing one crack.
Fig. 2. Semi-annulus in the stress plane x, which is the image of the body bounded by curve D.
1772
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
ImF ðxÞ ¼ 0;
Re½F ðxÞ < 0 for n ¼ 0; g 2 ðq; 1Þ [ ð1; qÞ:
ð16Þ
The analytic continuation F1 of the function F into the left semi-annulus AL , AL ¼ fx : q 6 jxj 6 1; arg x 2< p; p=2 > [ < p=2; p >g, is of the form F1 ðn þ igÞ ¼
F ðn þ igÞ for x 2 AR ; F ðn þ igÞ for x 2 AL
ð17Þ
and can be represented in the annulus A ¼ AR [ Al by the Laurent series F1 ðxÞ ¼
1 X n¼1
1 X 1 an n þ bn xn ; x n¼0
ð18Þ
where an , bn 2 C. After some tedious algebraic manipulations, which consist of splitting the complex coefficients an , bn into real and imaginary parts, exploiting conditions Eqs. (14) and (16) and renumbering the components of Eq. (18) we obtain the Ôinverse stress functionÕ F1 ðxÞ ¼
1 1 X 1 1 2k1 2k þ x þ iA x ; A2k1 2k x k¼1 x2k1 x2k
ð19Þ
where An 2 R are new coefficients which are algebraic combinations of the real and imaginary parts of an and bn . This equation was obtained in Turska–Klebek [12]. Remark. A similar expansion to Eq. (19) is given in Broberg [1, Eq. 5.3.43]. Formula (19) is more general; for a specific case of cracks situated on the symmetry plane of the body A2k ¼ 0 we obtain the series Eq. 5.3.43 of Broberg [1]. From the above relationship, Eq. (19) inserted into Eq. (13), with the conditional equation (15) gives a general form of the elastic–plastic boundary RðhÞ ¼ 2
1 X
½A2k1 cosð2k 1Þh þ A2k sin 2kh;
k¼1
where An ¼
1 2pi
Z c0
F1 ðfÞ df and fnþ2
p=2 6 h 6 p=2
for some c0 ¼ q0 eia ;
q < q0 < 1;
p < a < p:
ð20Þ
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
1773
The coefficients An have to be determined for each particular boundary value problem. From Eq. (20) we see that multipliers of the arguments of the cosine and sine function are either odd or even, and hence not all curves can be elastic–plastic boundaries. Eq.(19) allows us to construct boundary value problems for an elastic–plastic boundary given a priori. The simplest (one term) form of RðhÞ is for q ! 0, an infinite body, see Fig. 2, and A1 6¼ 0, An ¼ 0, n ¼ 2; 3; . . .. This is the celebrated (Hult and McClintock) small scale yielding solution 2 =ð2pk 2 Þ. for an infinite crack RðhÞ ¼ 2A1 cos h and z ¼ A1 ð1=x2 þ 1Þ, when we take A1 ¼ KIII Let us consider some other simple examples of an elastic–plastic boundary for an infinite body. We shall assume the shape of the boundary and then find the Ôinverse stress functionÕ F1 by using Eq. (19). If we take 1. RðhÞ ¼ 2A1 cos h þ 2A2 sin 2h, A1 P 2A2 , p=2 6 h 6 p=2; then 1 1 F1 ðxÞ ¼ A1 þ 1 þ iA2 x : x2 x3
ð21Þ
It is an analytic function everywhere except x ¼ 0, and limx!0 F ðxÞ ¼ 1. To establish the inverse function of Eq. (21), i.e. the stress function equation (1), one must select one of its branches. The shape of the elastic–plastic boundary for A1 =A2 ¼ 2; 3; 5; 7 is shown in Fig. 3. In infinity, the order of the stress function kF11 ðzÞ ¼ r32 ir31 is as z ! 1 :
const ffiffi : r32 ir31 ! p 3 z
ð22Þ
A1 / A2 = 7 A1 / A2 = 5 A1 / A2 = 3 A1 / A2 = 2
10
x2 / A2
5
0
-5 0
5
10
15
x1 / A2 Fig. 3. Shape of the elastic–plastic boundary for RðhÞ ¼ 2A1 cos h þ 2A2 sin 2h, A1 =A2 ¼ 2; 3; 5; 7.
1774
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
This means that Eq. (21) gives us the stress distribution in an infinite body with an infinite crack under remote loading of type Eq. (22). 2. RðhÞ ¼ 2A1 cos h þ 2A2 cos 3h, A1 P 3A2 , p=2 6 h 6 p=2, then F1 ðxÞ ¼ A1
1 1 2 þ 1 þ A2 þx : x2 x4
ð23Þ
The order, in infinity, of the stress function kF11 ðzÞ ¼ r32 ir31 is as z ! 1 :
const ffiffi : r32 ir31 ! p 4 z
ð24Þ
The shapes of the elastic–plastic boundaries are shown in Fig. 4, for A1 =A2 ¼ 3; 5; 7; 9. Eq. (31) gives us the stress distribution in an infinite body with an infinite crack under remote loading of type Eq. (24). As we see, the stress BVP for an elastic–plastic infinite body with an infinite external crack can have many solutions vanishing in infinity; Eqs. (21) and (23) are two examples of such solutions. The existence of the plastic zone guarantees that the displacements at the crack tip are finite. Thus, to ensure uniqueness (or to be able to distinguish particular solutions) it is necessary to set the order of the stresses in infinity and the value of the constant connected with the order (e.g. A2 of Eq. (22), A2 of Eq. (24)). Of course, Eqs. (21) and (23) (and other infinite crack cases) can be interpreted as models for the solution of some particular problems for finite bodies. These par15
A1 / A2 = 9 A1 / A2 = 7 A1 / A2 = 5 A1 / A2 = 3
10
x2 / A2
5
0
-5
-10 0
5
10 x1 / A2
15
20
Fig. 4. Shape of the elastic–plastic boundary for RðhÞ ¼ 2A1 cos h þ 2A2 cos 3h, A1 =A2 ¼ 3; 5; 7; 9.
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
1775
ticular problems should impose the order and constant of the stresses in infinity, sometimes not of the inverse square root type. For elastic bodies with more than one semi-infinite crack (with two collinear cracks) the problem of uniqueness has been discussed by Stallybrass [9–11], Comninou and Barber [3], and Fabrikant et al. [4]. It was found that is not enough, as it is for infinite bodies without cracks, to solve the stress BVP with vanishing remote stress. To have a unique stress distribution one must additionally, either (a) impose some condition relevant to the considered physical problem, or (b) set the constants and the rate of decay of the stresses in infinity. This means that the boundary conditions should contain additional functions gij ðzÞ that are asymptotic to rij ðzÞ in infinity, i.e. limz!1 rij ðzÞ=gij ðzÞ ¼ 1. For elastic–plastic bodies even with one semi-infinite crack we come to the same conclusions, and as it is seen from Eqs. (15) and (19) that gij ðzÞ cannot be a completely arbitrary function. In Turska and Wisniewski [14], two parallel semi-infinite cracks are analyzed. In this case the elastic problem requires the rate of decay and one additional constant, whereas the elastic–plastic one requires the rate of decay and two additional constants. 4. Crack surfaces subjected to shear stress In the next sections three different cases of elastic–plastic problems are solved explicitly as independently posed BVPs. The expansion RðhÞ given by Eq. (20) will not be used. 4.1. Elastic–plastic case Let us consider an elastic–plastic body with an infinite crack x1 < 0; x2 ¼ 0, under the following shear loads applied to its surfaces (a ÔpressurizedÕ crack), r23 ðx1 ; 0Þ ¼
ka ; x1 a
ð25Þ
where a > 0 is an arbitrary constant. We assume that the order of the stress in infinity is 0ðz1=2 Þ and the constant is in accordance with Eq. (25). The plastic zone must enter the crack tip at such an angle h0 that the stress continuity condition, r23 ð0; 0Þ ¼ k ¼ k cos h0 , (Eqs. (6) and (25)) is satisfied. This means that h0 ¼ p, otherwise than in the case of an unloaded crack where h0 ¼ p=2. Transforming the problem into the stress plane and by adding a term to the Hult and McClintock solution [5] the analytical Ôinverse stress functionÕ F ðxÞ introduced in Eq. (12) is of form
1 F ðxÞ ¼ a þ1 x
2 ð26Þ
1776
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
and thus the stress function is pffiffiffi a 0 f ðzÞ ¼ r23 þ ir13 ¼ k pffiffi pffiffiffi : z a
ð27Þ
In the elastic part of the body the stresses are pffiffi pffiffiffi pffiffiffi r cosðh=2Þ a pffiffiffiffiffi r23 ¼ k a ; r 2 ra cosðh=2Þ þ a pffiffi pffiffiffi r sinðh=2Þ pffiffiffiffiffi r13 ¼ k a r 2 ra cosðh=2Þ þ a
ð28Þ ð29Þ
for which we easily verify the boundary condition equation (25). The elastic–plastic boundary, Eq. (13), is a cardioid, RðhÞ ¼ 2aðcos h þ 1Þ;
for z ¼ rih ;
p 6 h 6 p
ð30Þ
and is shown in Fig. 5. 4.2. Elastic case Now we shall solve the same BVP but for an elastic medium, i.e. an infinite crack loaded by boundary conditions equation (25) and stresses vanishing in infinity. 3 2
x2 / a
1 0 -1 -2 -3 -1
0
1
2
3
4
x1 / a Fig. 5. Cardioid shape of the elastic–plastic boundary RðhÞ ¼ 2aðcos h þ 1Þ.
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
1777
Fig. 6. Mapping of the physical plane z with crack onto a half-plane X.
pffiffi First, with the mapping X ¼ i z we transform the plane z ¼ x1 þ ix2 with the crack, onto the upper half-plane X ¼ y1 þ iy2 , see Fig. 6. Then, the real part of the unknown function gðXÞ ¼ f 0 ðXðzÞÞ for the y1 -axis reads Re½gðXÞ ¼
y12
ka : þa
From the Poisson integral we obtain the real part of gðXÞ in the entire half-plane X: Z þ1 1 y2 ds: Re½gðXÞ ¼ 2 2 a þ s y2 þ ðy1 sÞ2 1
ð31Þ
ð32Þ
Next, from the real part of gðXÞ the imaginary part of function gðXÞ is derived. Transforming the function gðXÞ from variable X to variable z we find a solution to the BVP a r023 þ ir013 ¼ k pffiffi pffiffi pffiffiffi : zð z þ aÞ
ð33Þ
This is a solution which satisfies stress boundary conditions typical for finite cracks, i.e. given stress on the crack surface and zero stress in infinity, but for which the order of attenuation of the stresses is not specified. As we see, the order of Eq. (33) in infinity is Oðz1 Þ. To match the asymptotic pffiffiffi pffiffi behavior and constants of the stresses in Eqs. (27) and (33), we add to Eq. (33) the term k a= z, i.e. pffiffiffi pffiffiffi k a a r23 þ ir13 ¼ pffiffi pffiffi pffiffiffi þ 1 : ð34Þ z zþ a Note that the added term is a solution to an elastic mode III crack problem and that it does not change the value of r23 on the crack faces. Thus Eq. (34) is the elastic solution of the same BVP as the elastic–plastic one, Eq. (25). The lines of constant shear stress intensity for Eq. (34), r223 þ r213 ¼ c2 k 2 and c ¼ 1:0; 0:8; 0:6; 0:4; 0:2, are shown in Fig. 7. Comparing the elastic–plastic and elastic case one should note that the curves in Figs. 5 and 7 have the same horizontal and vertical diameters, but are shifted on the x1 -axis by the distance between the tip of the crack and the isoline for c ¼ 1:0. The isolines for Eq. (33) are shown in Fig. 8, and they are of a completely different size and shape than these of Fig. 7. For another example of such a problem, see Turska-Klebek and Sokolowski [13]. We can see that it is hazardous to solve infinite problems without knowing the
1778
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
c = 0.2 c = 0.4 c = 0.6 c = 0.8 c = 1.0
40 30
x2 / a
20 10 0 -10 -20 -30 -30 -20 -10
0
10 20 x1 / a
30
40
Fig. 7. Lines of constant stress intensity r223 þ r213 ¼ c2 k 2 , c ¼ 1:0; 0:8; 0:6; 0:4; 0:2, for elastic body with crack under surface shear, with matched orders and constants in infinity, Eq. (34).
0.8 c = 1.0 c = 0.8 c = 0.6 c = 0.4 c = 0.2
0.6
x2 / a
0.4 0.2 0 -0.2 -0.4 -0.6
-0.4
-0.2
0 x1 / a
0.2
0.4
0.6
Fig. 8. Lines of constant stress intensity r223 þ r213 ¼ c2 k 2 , c ¼ 1:0; 0:8; 0:6; 0:4; 0:2, for elastic body with crack under surface shear, without matched orders and constants in infinity, Eq. (33).
order of the solution in infinity. This may become an essential problem of approximate methods for infinite cracks.
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
1779
5. Crack under shear stress in infinity 5.1. Elastic–plastic case Let us consider an infinite body with an infinite crack. The crack x1 < 0, x2 ¼ 0 is free from traction but in infinity the stress r1 13 ¼ p is applied. To complete the infinite boundary conditions we add that, when p ¼ 0 then as z ! 1 :
KIII r23 þ ir13 ! pffiffiffiffiffiffiffi : 2pz
ð35Þ
We shall seek a function F ðxÞ which maps the semi-circle onto the stress plane, see Fig. 9, onto the physical plane z. F ðxÞ must satisfy the following conditions 1. the crack is free from traction Im½F ðxÞ ¼ 0;
x 2 A0 D0 C 0 ;
2. the stresses in infinity are r1 13 ¼ p, and when p ¼ 0 we obtain the typical small scale yielding stress asymptote F ðip=kÞ ¼ 1; and p ¼ 0 ) lim x2 F ðxÞ ¼ x!1
2 KIII ; 2pk 2
ð36Þ
3. the elastic–plastic boundary is a curve given by RðhÞ ¼ eih F ðeih Þ, Im½xF ðxÞ ¼ 0;
x 2 A0 B0 C 0 ;
4. the crack tip is located at point ð0; 0Þ F ðiÞ ¼ F ðiÞ ¼ 0: 2
2
With the mapping X ¼ ði þ ixÞ =ð1 ixÞ we transform the semi-circle x, shown in Fig. 9, onto the half-plane X ¼ y1 þ iy2 , shown in Fig. 10. Then, for an auxiliary function G ðXÞ ¼ xðXÞF ðxðXÞÞ, the conditions Eq. (36) become η
ω
A' D'
z
x2 (1)
B'
ξ
D D
A C
B
x1
C'
Fig. 9. Mapping F ðxÞ of semicircle in the stress plane x onto physical plane z with crack.
1780
E. Turska, K. Wisniewski / International Journal of Engineering Science 41 (2003) 1767–1783
Ω
y2 -1
C"(∞)
B"
(1- p/k)/(1+ p/k)
A"
D"
y1 C"(∞)
Fig. 10. Image of semicircle x of Fig. 9 onto half-plane X for transformation X ¼ ði þ ixÞ2 =ð1 ixÞ2 .
1. Re½G ðXÞ ¼ 0; X 2 A00 D00 C 00 1 p=k 1 K2 G ðXÞ ¼ III2 2i 2. G ¼ 1, X ¼ D00 with p ¼ 0 ) limX!1 1 þ p=k X1 pk 3. Im½G ðXÞ ¼ 0;
X 2 A00 B00 C 00
ð37Þ
4. G ð0Þ ¼ G ð1Þ ¼ 0: This a mixed problem of the theory of holomorphic functions for the half-plane. As in Muskhelishvili [7], we obtain that function G ðXÞ is equal to pffiffiffiffi 2 2KIII i X G ðXÞ ¼ ð38Þ pk 2 X ðð1 p=kÞ=ð1 þ p=kÞÞ and yields 2 KIII 1 ð1 þ p=kÞ2 1 þ F ðxÞ ¼ : x2 1 þ p2 =k 2 þ ðip=kÞðx 1=xÞ 2pk 2
ð39Þ
For p ¼ 0 the function Eq. (39) takes the same form as in the small scale yielding case. The elastic–plastic boundary is as follows RðhÞ ¼
2 KIII ð1 þ p=kÞ2 cos h ; pk 2 1 þ p2 =k 2 ð2p=kÞ sin h
p p
