Dislocation crack tip shielding and the Paris exponent

Materials Science and Engineering A 468–470 (2007) 59–63

Dislocation crack tip shielding and the Paris exponent Johannes Weertman a,b,∗ a

Department of Materials Science and Engineering, Northwestern University, Evanston, IL 60208, USA b Department of Geological Sciences, Northwestern University, Evanston, IL 60208, USA Received 4 April 2006; received in revised form 29 July 2006; accepted 10 August 2006

Abstract In a previous paper we considered how the crack blunting dislocation emission mechanism likely leads to a Paris fatigue crack growth rate law exponent n = 2 if dislocation shielding is not important and showed how dislocation crack tip shielding can cause a reduction in the fatigue crack growth rate and showed qualitatively how it might lead to an increase in the Paris exponent n. In this paper semi-quantitative calculations are made to show how the Paris exponent can be increased by dislocation shielding. © 2007 Elsevier B.V. All rights reserved. Keywords: Paris equation; Paris exponent; Dislocation shielding

1. Introduction The magic lengths of fracture mechanics [1] are the plastic zone size (PSZ), the crack opening displacement (COD) (where the plastic zone boundary crosses the crack plane behind the crack tip) and the blunting radius ρ of a crack tip arising from dislocation emission. These lengths are illustrated in Figs. 1 and 2 and are equal to K2 PZS = α 2 , σy COD = β

ρ=η

K2 , σy G

K2 . G2

(1)

(2)

K = gKgc = Kgb .

∗ Correspondence address: Department of Materials Science and Engineering, Northwestern University, Evanston, IL 60208, USA. Tel.: +1 847 491 3197; fax: +1 847 467 6573. E-mail address: [email protected].

(4)

Here Kgc is the critical K value for cleavage fracture for a Griffith–Inglis crack in an elastic solid, Kgb the critical K value for dislocation emission, and g is a constant of order of magnitude 1. Dislocation emission occurs before cleavage failure when g is smaller than 1. We argued that if the applied K is much greater than Kgb many dislocations will leave the crack tip but no further dislocation emission occurs once the radius ρ of the blunted crack tip reaches the value

(3)

Here G and σy are the shear modulus and the yield √stress of the solid, K the applied stress intensity factor (K = πa where a is the crack half length) and the constants α and β are of order of magnitude 1 and the constant η is of order of magnitude of 1–10.

0921-5093/$ – see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.msea.2006.08.128

The radius ρ of Eq. (3) was estimated by us [1,2] from the theory of Rice and Thomson [3] and of Armstrong [4] of dislocation emission from a sharp crack tip. According to their theory a dislocation leaves the crack tip (into previously plastically undeformed material) when the applied stress intensity factor K reaches the level

ρ≈

bK2 bK2 K2 = 2 2 = η 2. 2 g Kgc G Kgb

(5)

Here b is the Burgers vector length of a dislocation. This equation is found on noting that in an elastic solid the magnitude of the stress field near a crack tip is proportional to K and noting too that the crack tip radius of an atomically sharp crack is approximately b. The non-traction stress √ at the surface of the blunted crack tip is, therefore, a factor (b/ρ) smaller than that of the atomically sharp crack. Because it is the stress at the walls of the ‘tips’ of elongated elliptical holes in stressed elastic solids is proportional √ to K/ ρ. Hence dislocations can be expected to be emitted from

60

J. Weertman / Materials Science and Engineering A 468–470 (2007) 59–63

gives the COD where σ is equal to an arbitrarily chosen value σy .) Since dislocations can enter the plastic zone from the crack faces as well as the tip there is no reason to expect COD ≈ ρ. 2. Paris equation

Fig. 1. Plastic zone size and crack opening displacement (schematic).

Fig. 2. Crack tip blunting by dislocation emission. Table 1 Values of terms used in plots G (GPa) (for Al) σy = G/800 (MPa) Kthreshold (MPa m1/2 ) Kstatic (MPa m1/2 ) η b (nm) (for Al) ν (for Al)

27 33.75 0.5 100 4 0.286 0.346

a blunted crack tip until ρ attains the value given by Eq. (5). In Eqs. (4) and (5)Kgc is equal to  4γG Kgc = , (6) 1 − ν where γ is the surface energy of the solid and ν is Poisson’s ratio. (Note that the surface energy γ ≈ (1/2)σt dt where σt ≈ (1/5)G is the theoretical strength of the solid and dt ≈ 2b is the separation distance of the crystal planes beyond which interatomic √ forces are negligible. Thus, γ ≈ (1/5)Gb and Kgc ≈ G b ≈ 1/2 MPa m1/2 for the values of G, b and ν listed in Table 1.) It should be noted that the COD of Fig. 2 is identical to the magnitude of the total Burgers vector of all the dislocations within the elastic–plastic boundary situated at σ = σy where σ is the stress magnitude (see Fig. 3). The total of the dislocation Burgers vectors includes those of the crack plane dislocation within the elastic–plastic boundary. (For an elastic solid Eq. (2)

Fig. 3. Dislocations within plastic zone (schematic).

In a previous (part I) paper [5] we considered the application of the magic lengths of fracture mechanics to the fatigue crack growth rate. The Paris equation [6,7] can be expressed as     da

K n da = , (7) dN dN th Kth where n is the Paris exponent, a the crack half width, N the number of cycles, K the applied cyclic stress intensity factor, Kth the threshold cyclic stress intensity factor at which a fatigue crack starts to grow, and c is the rate of growth of a fatigue crack at threshold. The term Kth reasonably be expected to be of the order of, or somewhat larger than, the critical stress inten√ sity factor Kgc or Kgb . Thus, Kth ≈ (1/2)G b. Fatigue crack growth terminates catastrophically in one cycle, of course, once

K is as large as the static stress intensity factor Kst for fracture in a metal. The value of (da/dN)th , the fatigue crack growth rate per cycle at K = Kth , can be expected to be approximately equal to b if a fatigue crack grows each cycle. This, of course, is the slowest possible fatigue growth rate for a crack which grows in each cycle. Were the fatigue crack growth in each cycle to be of the order of the plastic zone size (of a statically loaded crack) the Paris equation would be of order   da

K 2 . (8) = dN σy The upper curve of Fig. 4 shows this law. Were the crack growth in a cycle of the order of the crack opening displacement the Paris equation is ( K)2 da = . dN σy G

Fig. 4. Various Paris laws plotted for values of the terms listed in Table 1.

(9)

J. Weertman / Materials Science and Engineering A 468–470 (2007) 59–63

61

The second from the top curve of Fig. 4 is given by this equation. Eqs. (8) and (9) predict fatigue crack growth rates that are orders of magnitude faster than measured experimentally. If the growth per cycle is of the order of the blunting radius produced by dislocation emission from a crack tip into previously undeformed material the Paris equation becomes   da

K 2 =η . (10) dN G An equation essentially the same as Eq. (10) was proposed by McEvily and his co-workers [8]. It is   8 K 2 da = , (11) dN π E where E is Young’s modulus. They analyzed fatigue crack growth data, including that of Barsom et al. [9], that can be described reasonably well by their second power Paris equation. Fig. 5. Shear sliding fatigue crack growth model of Laird and Smith [10] and Neumann [11,12].

3. Paris exponents > 2 The measured values of the Paris exponent n usually are about n ≈ 3–4. Moreover, the fatigue crack growth rate usually is much smaller than that given by Eq. (10). Typical values are of order da

K4 = 2 , dN G K02

(12)

where K02 ≡ η−1 ( Kst /G)2 . The value of K0 is picked to make Eq. (10), the second power Paris equation curve, intersect the curve of Eq. (11) at K = Kstatic . Eq. (12) is plotted as the bottom curve of Fig. 4. The explanation of exponents that are appreciably larger than n = 2 presents a challenge because the natural lengths of fracture mechanics all lead to second power Paris equations. Consider an explanation based on the shear sliding model.

they become stuck in the solid after leaving the crack tip. It can be seen in Fig. 6 the dislocations self cancel leaving none in the solid. The same situation holds were the dislocations not stuck in the solid and returned into the tip. Fig. 6 represents the ideal situation of perfect dislocation annihilation during fatigue crack growth. Fig. 7 of the next section represents the more realistic situation in which dislocation debris consisting of dislocations of both signs is left in the wake of the growing crack. The blunting radius ρ give by Eq. (5) should give a reasonable estimate for the crack advance distance δa in each cycle in Figs. 5 and 6 provided the dislocations move far away from the crack tip.

4. Shear sliding model In the shear sliding, striation producing, model (or blunting and sharpening model) of fatigue crack growth of Laird and Smith [10] and Neumann [11,12] dislocations are emitted from the crack tip region on two or more sets of slip planes. One interpretation of this model is shown in schematic Fig. 5. In the forward cycle dislocations are emitted from the crack tip (a) on the slip plane shown in (b) and then on the slip planes (c). In the reverse cycle dislocations of opposite sign are emitted first as shown in (d) and then on another plane shown in (e). The crack faces in (e) come together again. The crack tip has advanced a distance δa. In each repeated cycle the crack advances another increment as indicated in (f). The crack advance distance δa in Fig. 5 presumably is of the order of that found from Fig. 2 for crack tip blunting under monotonic loadings. Hence the fatigue crack growth predicted for Fig. 5 model is Eq. (5). In Fig. 5 only the moving dislocations are shown. In Fig. 6 are shown all the dislocations of Fig. 5 under the assumption

Fig. 6. Illustrating cancellation of dislocations in Fig. 5 processes.

62

J. Weertman / Materials Science and Engineering A 468–470 (2007) 59–63

When the crack stops advancing the shielding crack tip intensity factor L produced  dislocation  by the plastic zone should equal L = −K + η−1 ρG2 ≈ −K + η−1 δaG2 (see Eq. (14)). Were δa given by Eq. (10) then L = 0 and there is no crack tip shielding (L = 0). If δa is proportional to K4 , Eq. (17) is in play and only the c2 term is important in Eq. (16). If the crack advances a distance δa without change of the dislocation field the shielding is reduced. The change in shielding δL can be estimated by noting that the shielding L from a dislocation of Burgers vector b at a distance r from a crack tip is of order

Fig. 7. Growing fatigue crack with wake dislocations (schematic).

5. Effect of dislocation shielding

bG L→ √ . 2 πr

In my earlier paper [5] I argued that Paris exponents greater than n = 2 might be explained through dislocation crack tip shielding. The argument presented is: the local (crack tip) stress intensity factor Kt is equal to

(Exact expressions are given in Appendix A.) The radial dependence of the dislocation density field B (for both dislocations of radial and azimuthal Burgers vectors) is for the mode I (or mode II) HRR crack tip stress–strain-rotation field [13]:

Kt = K + L,

(13)

where L is the shielding (or antishielding) stress intensity factor arising from the dislocations within the plastic zone. Eq. (5) for the crack blunting radius becomes ρ ≈ δa ≈ η

(K + L)2 . G2

(14)

When the crack tip is perfectly shielded L = −K. If the tip is perfectly shielded the fatigue crack does not grow. Dislocations cannot be emitted from the tip. For the imperfectly shielded tip L = −K + f (K).

(15)

Here f is an unknown (positive sign) function of K. Suppose f (K) has a Taylor series expansion f = c1 K +

c2 K 2 c3 K 3 + + ..., Kst Kst2

(16)

where ci are constants. If only the first term is important in this series (with 0 < c1 < 1) Eq. (14) leads to a second power Paris equation with a growth rate smaller than given by Eq. (10). If c1 = 0 and only the c2 term in Eq. (16) is important the fatigue crack growth equation (for the blunting/shear sliding mechanism) becomes the fourth power Paris equation   da

(K + L) 2 c2 K4 =η = η 22 2 . (17) dN G G Kst Whatever is the form of f (K) it is reasonable to expect that the fatigue crack growth rate is reduced and the Paris exponent is n ≥ 2. Consider next why the fatigue growth rate can be reduced and the Paris exponent might be increased through dislocation shielding effects. Fig. 7 shows a growing fatigue crack, which is advancing by the dislocation emission crack tip blunting mechanism, and dislocations in the plastic zone and plastic wake.

B=

const. σy (r/r0 )m/(1+m) , r0 G (r/r0 )2

(18)

(19)

where m is the exponent of the stress σ versus strain relationship σ = σconst. m for a work hardening solid and r0 is of the order of PZS. (HRR refers to the Hutchinson [14,15] and Rice and Rosengren [16] papers.) Eq. (19) is valid for the monotonically loaded crack. A similar density should exist for the fatigue crack. The dislocation density field (19) must produce perfect shielding. If√it did not the stress singularity of the HRR field would go as 1/ r instead of as 1/rm/(1+m) . Thus, setting Eq. (19) (when azimuthal terms are included) into Eq. (18) and integrating over the plastic zone must give L = −K. On dimensional grounds the change in L when √ the crack tip moves a distance δa should be proportional K δa/r0 . That is, if δa is proportional K2 then √ the constant c1 in Eq. (16) is proportional to δa/r0 . On the 4 other hand if δa is proportional K √ then2c1 = 0 and the constant c2 in Eq. (16) is proportional to δa/K . It is not possible to go further in developing the Paris equation in ductile metals without quantitative expressions for the dislocation density field around the tips of fatigue cracks. 6. Conclusions The blunting radius in the crack tip shear sliding model of Laird and Smith and of Neumann sets an upper limit to the rate of growth of fatigue cracks that advance with the formation of fatigue striations. Dislocation crack tip shielding should cause the fatigue crack growth rate to be reduced. Dislocation crack tip shielding also may lead to an increase of the value of the Paris law exponent. Acknowledgements It is a pleasure to be able participate in this symposium held to honor the outstanding achievements in the field of fatigue of Professor Art McEvily.

J. Weertman / Materials Science and Engineering A 468–470 (2007) 59–63

Appendix A Except for the mode III crack in a perfectly plastic solid the radial dependence of the dislocation density B is always more strongly singular than an inverse radial dependence from the crack tip for a crack loaded statically in a work hardened solid. For the mode III crack the radial dependence is 1/r (1+2m)/(1+m) . This dependence suggeststhat the crack tip shielding is infi√ r nite because the integral 0 (1/r (1+2m)/(1+m) )(1/ r)r dθ dr is infinite. Yet despite this strong dependence the crack tip is perfectly shielded (that is L = −K) in small scale yielding (see [1], p. 220). The reason it is finite is because of the azimuthal dependence of B. For the case of a mode I crack in a linear work hardening (m = 1) in fully plastic solid the dislocations produce no shielding whatsoever. Near the crack tip the dislocation densities are (see [1], p. 326): K 1 Br = −const. √ sin θ, r r 2

(A.1)

K 1 Bθ = const. √ cos θ. r r 2

(A.2)

These fields are more singular than an inverse one. The mode I shielding factors for dislocations of radial and azimuthal Burgers vector br , bθ are, near the crack tip (see [1], p. 170): L=−

3br G 1 √ sin θ cos θ, 2 2(1 − ν) 2πr

(A.3)

L=−

bθ G 1 √ cos θ cos θ. 2 (1 − ν) 2πr

(A.4)

Since  π 1 1 1 −3 sin θ sin θ cos θ + 2 cos θ cos2 θ dθ = 0, 2 2 2 −π

(A.5)

63

the dislocation density field near the crack (and elsewhere) produces no shielding. The HRR dislocation density field for the mode I crack must lead to a similar result. The net shielding in this case should be finite and equal to −K rather than zero. There are two fields for the HRR case. One has the strong radial dependence given above in Eq. (18). The other has the weaker dependence of the mode III crack in a work hardening solid mentioned above. (The azimuthal dependency can only be found numerically. A numerical check that infinite shielding or antishielding does not occur in the HRR field has yet to be carried out.) References [1] J. Weertman, Dislocation Based Fracture Mechanics, World Scientific, Singapore, 1996. [2] J. Weertman, in: T. Mura (Ed.), Mechanics of Fatigue, AMD, vol. 47, American Society of Mechanical Engineers, New York, 1981, pp. 11–19. [3] J.R. Rice, R. Thomson, Phil. Magn. 29 (1974) 73–96. [4] R. Armstrong, Mater. Sci. Eng. 1 (1966) 251–256. [5] J. Weertman, in: T.S. Srivatsan, W. Sobojeyo (Eds.), The Paul Paris Symposium, Minerals, Metals & Materials Society (TMS) of the AIME, Warrendale, PA, 1997, pp. 41–48. [6] P.C. Paris, M.P. Gomez, W.P. Anderson, Trend Eng. 13 (1961) 9–14. [7] P.C. Paris, F. Erdogan, J. Basic Eng. 85 (1963) 528–534. [8] R.J. Donahue, H.M. Clark, P. Atanmo, R. Kumble, A.J. McEvily, Int. J. Fract. Mech. 8 (1972) 209–219. [9] J.M. Barsom, E.J. Imhof, S.T. Rolfe, Eng. Fract. Mech. 2 (1971) 301– 317. [10] C. Laird, G.C. Smith, Phil. Magn. 7 (1962) 847–857. [11] P. Neumann, Acta Metall. 22 (1974) 1155–1165. [12] P. Neumann, Acta Metall. 22 (1974) 1167–1178. [13] J. Weertman, J.A. Hurtado, in: S.N.G. Chu, P.K. Liaw, R.J. Arsenault, K. Sadananda, K.S. Chan, W.W. Gerberich, C.C. Chau, T.M. Kung (Eds.), James C.M. Li Symposium, Minerals, Metals & Materials Society of the AIME, Warrendale, PA, 1995, pp. 79–85. [14] J.W. Hutchinson, J. Mech. Phys. Solids 16 (1968) 13–31. [15] J.W. Hutchinson, J. Mech. Phys. Solids 16 (1968) 337–347. [16] J.R. Rice, G.F. Rosengren, J. Mech. Phys. Solids 16 (1968) 1–12.