Strength is an important property of a material and is very sensitive to the micro­structure of the material. Therefore, understanding the ways in which strength of a material can be improved is of great technological significance. To be specific, we will discuss the strengthening mechanisms in terms of their effect on the yield stress of the material (defined as the stress at which gross plastic deformation begins). However, in order to accomplish that objective, we need to understand the science of the strengthening mechanisms. Dislocations are a common factor in almost all the important strengthening mechanisms we will discuss here. If dislo­cations are able to move in a crystal with relative ease, it means the material does

not intrinsically offer any resistance to the dislocation movement and thus would be less strong. But if the microstructure is laden with various obstacles, dislocation movement will be effectively impeded and this movement obstruction will be trans­lated into the increase of the strength of the materials. This principle is applicable to a wide array of materials with crystalline structures. The obstacle to dislocation movement could be dislocations themselves (strain hardening), grain boundaries (Hall-Petch or grain size strengthening), solute atoms (solid solution strengthen­ing), precipitates (precipitate strengthening), and dispersions of fine stable particles (dispersion strengthening). There are a few other strengthening mechanisms like texture strengthening, composite strengthening, and so on, which are not directly but indirectly related to dislocations. However, they will not be discussed here for brevity.

4.4.1

1) The frequency of cyclic loading has only a small effect on the fatigue strength.

2) The form of stress cycle such as square, triangular, or sinusoidal wave has no effect on the fatigue life.

3) The environment in which the component undergoes stress reversals has a marked effect on fatigue life. The fatigue life in vacuum could be some 10 times more than that in moist air. Also note the effect of corrosion fatigue.

4) The thickness of the test specimen has an effect on the fatigue properties. Thin­ner samples show a decrease in the crack growth rate.

5) Stress concentrators such as key ways with sharp corners promote crack initia­tion, thereby decreasing the number of cycles to failure.

6) Surface smoothness (or roughness) has strong effect on fatigue life and polished surfaces will have improved fatigue life.

The neutron energy spectrum is affected by various factors, including reactor type, position in the reactor, and immediate surroundings, such as adjacent fuel, control rods, and empty surroundings. The overall shape of the neutron spectrum is influenced by the specific type of reactor. For reactors using mod­erators, such as heavy water, light water, or graphite, Figure 1.5a depicts ideal­ized curves for a normalized flux of neutrons as a function of neutron energy.

The neutron fission spectrum calculated by Watt is also superimposed on the graph for comparison. Convenient techniques such as assuming monoener­getic neutron flux and the arbitrary selection of neutron flux cutoff level (>1 MeV) are mostly general approximations. Remember that most neutron fluxes cited at irradiation damage studies are expressed in terms of >1 MeV. Figure 1.5b depicts the two flux spectra obtained from the Advanced Test Reactor (ATR). One spectrum is without the use of cadmium shroud and another one is with the cadmium shroud (of ~ 1.14 mm thickness). It is clear that fast (hard) spectrum is achieved with the use of cadmium shroud (i. e., irradiation jig wrapped into cadmium foil) due to its absorption of thermal neutrons, but not fast ones. Dosimetric experiments followed by calculations can generate the flux spectrum for a specific position in the reactor.

1.8

The rock salt or sodium chloride (NaCl) lattice structure consists of two inter­mingling cation (Na+) and anion (Cl-) FCC sublattices, as illustrated in Figure 2.20. Alternatively, the rock salt structure can also be described as an FCC anion lattice in which all its octahedral interstitial sites are filled up by the cations. Here nearest neighbors to each ion are the six ions of the opposite charge. Coordination numbers of cations and anions are six each in this struc­ture similar to that for simple cubic structure. The number of octahedral sites in an FCC unit cell is the same as the number of atoms (in this case, anions). Therefore, the stoichiometry of this crystal type is MX (M = metal, X = non­metal). This means that valencies of both cations and anions in this crystal type need to be the same. Some examples of the rock salt structure are KCl, LiF, and KBr (monovalent ions); MgO, CoO, and MnO (divalent ions); and UC (tetravalent ions). Note that UC is a potential nuclear fuel. The Struckturbericht notation of a rock salt structure is B1. There are four ion pairs per unit cell similar to an FCC structure. The lattice constant of a NaCl-type crystal struc­ture is given by 2(R+ + R-), where R+ and R — are the cation and anion radii,

respectively. In rock salt structures, the cation to anion radius ratio can vary between 0.414 and 0.732.

We have already gained some basic idea about slip from the previous chapters. Slip is nothing but the movement of one crystal part over another causing plastic deforma­tion. Slip occurs only when the shear stress on the slip plane along the slip direction attains a critical value (known as critical resolved shear stress (CRSS)). Generally, the slip planes are the crystallographic planes with the highest atomic density (closest-packed planes (CPPs)) in that particular crystal structure, and the slip directions are the closest-packed directions (CPDs) in the respective crystal structures.1* A combination of slip plane and slip direction is called a slip system. Due to the slip, steps are formed on the prepolished surface of a material that has been plastically deformed. Due to the height variations in the different slip steps, they are observable on the sample surface as lines, and hence known as slip lines. Several slip lines banding together are

1) This is because CPPs are farthest apart and the atoms are closest along CPDs so that the force/stress required for slip to occur on CPPs along CPDs will be the lowest.

An Introduction to Nuclear Materials: Fundamentals and Applications, First Edition.

K. Linga Murty and Indrajit Charit.

© 2013 Wiley-VCH Verlag GmbH & Co. KGaA. Published 2013 by Wiley-VCH Verlag GmbH & Co. KGaA.

called slip bands. The slip bands can be seen by observing the prepolished surface of a deformed sample with an optical microscope or a scanning electron microscope. Note that slip is not just a surface phenomenon, the manifestation of slip can be tracked by observing the slip steps on the rightly conditioned surface. If the surface is later repol­ished, the slip lines will be removed as the slip steps showing the height variations will no longer be present. This is demonstrated in Figure 4.1a. For example, a micro­graph of copper with slip lines is shown in Figure 4.1b.

Let us first discuss an example from an FCC metal. As seen in Chapter 1, the close-packed planes in the FCC crystal are {111} with the close-packed directions being {110) (face diagonals). They are the slip planes and slip directions in FCC, respectively. Planes {111} are called octahedral planes as they form the faces of an octahedron inside the FCC crystal. There are eight (effective number) such octahe­dral planes per FCC unit cell. However, one plane is parallel to the other plane, thus leaving four independent slip planes. Now each such {111} plane contains three {110) directions (reverse directions are not taken into consideration as they are essentially the same directions), as shown in Figure 4.2. Thus, an FCC crystal would have 12 (4 x 3) slip systems. Generally, FCC metals show straighter slip lines as shown in Figure 4.1b. An easy way whether a crystallographic direction [uvw] is

indeed a slip direction on a slip plane (hkl) is to satisfy the relation h-u + k-v + l-w = 0 (dot product of the direction and the plane normal). For example, [110] resides on the slip plane (111) as (—1)(1) + (1)(1) + (0)(1) = 0.

A BCC crystal does not have a close-packed plane. Its closest-packed plane is {110}, closely followed by {112} and {123} due to their relatively high atomic den­sity. However, the BCC crystal has only one slip direction, the close-packed direc­tion {111). There are 48 possible slip systems in a BCC crystal. As none of the slip planes is close packed in BCC crystals, higher shearing stresses are required to cre­ate slip. So, there are multiple slip planes, but slip always occurs in a close-packed direction. As (screw) dislocations in BCC crystals can move from one slip plane to another, the slip lines produced have irregular wavy appearance. By observing the slip in the {111) direction more or less independent of the slip plane, Taylor coined the term pencil glide for describing slip in BCC crystals. Screw dislocations can cross-slip readily from one plane to another, thus forming such slip bands.

In the HCP metals, the close-packed plane is basal plane {0002}. The а-axes are the close-packed directions having Miller index, (1120), serving as the slip direc­tion. Slip along this direction regardless of the slip plane does not produce any strain parallel to c-axis. Only certain HCP metals like zinc, magnesium, cobalt, and cadmium show basal slip. There is only one type of close-packed plane and three slip directions in HCP crystal leading to the total number of available slip systems to be only three. That is why they exhibit limited ductility and extreme orientation dependence of properties. They all have one thing in common — their c/а ratios are close to the ideal (1.633). Interestingly, beryllium with a c/а ratio quite less than the ideal ratio primarily shows basal slip closely followed by prismatic slip. The stress required for pyramidal slip is much greater and can lead to fracture. Figure 4.3 shows the scenario in terms of stress-strain curves. On the other hand, alpha-Ti

and alpha-Zr (c/a ratio less than the ideal ratio) under normal conditions undergo prismatic slip ({1010}(1120)). But Poirier downplayed the effect of c/a ratio on the slip behavior of HCP metals and attributed the diverse slip behaviors to the anisot­ropy of the HCP crystals. Twinning can produce small strains in the HCP crystals even along the c-axis, but the main role of twinning in HCP crystals is to help orient unfavorable slip systems favorably for slip to take place.

Additional slip systems can be activated depending on test temperature. For example, {110} slip planes in aluminum start taking part in slip deformation at elevated temperatures even though the crystal structure remains FCC. Magnesium is known for its basal slip, but at a higher temperature (~225 °C), secondary slip systems involving {1011} pyramidal planes get activated.

■ Example 4.1

Zr alloys with low c/a ratio exhibit prism slip ({1010}(1120)). Show on a single HCP crystal a slip system and the Miller indices of the plane and direction chosen.

Solution

In the following figure, the slip plane (ABDF) is (1010) and the slip direc­tion AB is [І210] so that the slip system is (1010)[І210].

4.1.1

In the beginning of the discussion on the fine particle strengthening, we have dis­tinguished between the precipitation and dispersion strengthening. If the particles are fine, stable, and incoherent, dispersion strengthening is applicable. Dispersion strengthening is a significant strengthening mechanism utilized for high — temperature alloys. Orowan bypassing mechanism at lower temperature regime is applicable here also. One example of dispersion strengthening system is thoria — dispersed nickel (T-D Ni). This tends to have much greater high temperature defor­mation resistance compared to the nickel matrix itself.

■ Example 4.7

A Li {Li (bcc): m = 32 GPa; а = 3A v = 0.29; stacking fault energy (Г) = 1250 mJ m~2; TM = 181 °C} sample exhibited an yield strength of 10 MPa.

a) Estimate the dislocation density assuming that the strengthening is all due to dislocations (strain hardening)?

Solution

b) Estimate the strengthening if the dislocation density increases by a factor of 100 following cold working (or exposure to radiation).

Solution

t — aGbyffi) — (1)(32000)(2.598 x 10~10)/1.447 x 1014 — (8.314 x 10~6)(1.203 x 107).

So, t — 100 MPa.

4.5

Summary

This chapter introduced the dislocation concept in a greater detail. Various aspects of dislocation theory, including dislocation energy, line tension, velocity, and so on, are discussed. Also, dislocation reactions in various lattices are discussed. The chapter concludes with the introduction of various strengthening (hardening) mechanisms. This understanding will lead us to better understand the effect of increased dislocation density on the properties of the irradiated materials in the subsequent chapters.

Problems

4.1 We considered a cylindrical Ni single crystal (FCC) with a diameter of 0.1 mm pulled in tension with a stress of 1000 MPa (see Figure 4.4). The loading direc­tion is along [101], while the slip system is (111)[Ї10]. We calculated the resolved shear stress along the slip direction in the slip plane.

a) What is the resolved shear stress along the slip direction in the slip plane?

b) If tCRSS for Ni is 550 MPa, would the crystal deform?

An edge dislocation (AB) is situated perpendicular to the slip direction of the slip system and pinned at two points (A and B) separated by 1000 A

c) If the Burgers vector of the dislocation is along the slip direction, what will be its direction and magnitude?

d) Determine the radius of curvature of the dislocation in part (c) above due to the applied stress?

e) Compute the minimum applied load (normal to the specimen cross section) required for the pinned dislocation (AB) to operate as a Frank-Read source.

f) If the dislocation line in the figure is a screw dislocation, what will be its Burgers vector (magnitude and representation)?

4.2 By increasing the temperature, the concentration of vacancies can be increased. Will this also increase the density of dislocations (be quantitative)?

If not, why not and how may the dislocations be multiplied?

4.3 A prismatic loop has Burgers vector perpendicular to the plane of the loop and thus can glide in that plane under an applied shear stress. True or false?

4.4 In the figure, the dislocation AB lies in the plane (111) with b along [110] (per­pendicular to the line AB).

a) Is the dislocation AB an edge or a screw or a mixed type?

b) What is the line vector of this dislocation?

c) If AB were a screw dislocation, what will be its Burger’s vector?

4.5 Show that a perfect dislocation, (a/2)[110], in an FCC lattice splits into two Shockley partials with Burgers vectors of (a/6)(112) type. (Write down the equation.)

a) Are the Shockley partials glissile or sessile? Why?

b) A Shockley partial (a/6)[112] reacts with a Frank partial (a/3)[111] to yield a perfect dislocation. What is the Burger’s vector of the product dislocation?

c) Show that the reaction in problem 4.3 is valid.

(This reaction occurs in irradiated stainless steel when heated. Similar reaction could also occur in quenched Al-Mg alloy from 550 °Cto —20 °C fol­lowed by heating (see Figure 5.12 in Ref. [3]). These reactions lead to unfaulting offaulted loops.)

4.6 a) Evaluate the force (magnitude and direction) acting on an edge dislocation

(b = bi, l = k) due to an external stress oxx.

b) How this dislocation may move due to this force?

c) Evaluate the force on the dislocation due to hydrostatic pressure p and comment on how the dislocation may move due to this force.

4.7 Evaluate the hardening (tc) of a Li (BCC, G = 32GPa, a = 3A) alloy with a microstructure consisting of a dislocation density of 3 x 1013m—2 (a? = 0.5) and 3 at %solutes (ac = 0.005).

180 I 4 Dislocation Theory
Bibliography

1 Dieter, G. E. (1988) Mechanical Metallurgy, McGraw-Hill.

2 Hosford, W. F. (2010) Mechanical Behavior of Materials, Cambridge University Press.

3 Hull, D. and Bacon, D. J. (1984) Introduction to Dislocations, 3rd edn, Butterworth-Heinemann.

4 Gilman, J. J., Johnston, W. G., and Sears, G. W. (1958) Journal of Applied Physics, 29, 747-754.

5 Amelinckx, S. (1958) Acta Metallurgica, 6, 34

6 Gollapudi S. et al. (2010) "Creep Mechanisms in Ti-3Al-2.5V Tubing Deformed under Closed-end Internal Gas Pressurization,” Acta Materialia, 56 (2008) 2406-2419.

7 Ashby M. F. and Jones D. R.H. Jones, Engineering Materials 1 — An Introduction to their Properties and Applications, International Series on Materials Science and Technology, Pergamon Press (1980).

8 Malis, T. and Tangri, K. (1979) Acta Metallurgica 27, 25-32.

9 Weertman, J. andWeertman, J. R. (1992) Elementary dislocation theories, Oxford University Press, New York, USA.

10 Raghavan, V. (1995) Physical Metallurgy: Principles and Practice, Prentice-Hall, New Delhi, India.

11 Humphreys, F. J. and Hatherly, M. (2004) Recrystallization and related annealing phenomena, Pergamon, Oxford, UK.

12 Roesler, J., Harders, H., and Baeker, M. (2010) Mechanical behavior of engineering materials — metals, ceramics, polymers and composites, Springer, Berlin, Germany.

13 Argon, A.S. (2008) Strengthening mechanisms in crystal plasticity, Oxford University Press, New York, USA.

14 Barrett, C. R., Nix, W. D., andTetelman A. S., (1973) The Principles ofEngineering Materials, Prentice Hall, Englewood Cliffs, NJ, USA.

15 Schmid, E. (1935) Kristallplastizitatmit besonderer Berucksichtigung der Metalle (in German), Springer-Verlag, Berlin, Germany.

16 Nabarro, F. R.N. (1947) Dislocations in a simplecubic lattice, Proc. Phys. Soc., 59, 256.

17 Peierls, R. E. (1940) The size ofa dislocation, Proc. Phys. Soc., 52, 34.

18 Johnston, W. G. and Gilman, J. H. (1959) Dislocation velocities, dislocation densities and plastic flow in lithium fluoride crystals, J. Appl Phys., 30, 129.

19 Heidenreich, R. D. and Shockley, W.

(1948) Report on Strength of Solids, Physical Society, London, UK.

20 Frank, F. C. (1949) Sessile Dislocations, Proc. Phys. Soc., 62A, 202.

21 Lomer, W.M.(1951)Philosophical Magazine, 42, 1327.

22 Cottrell, A. H. (1952) Philosophical Magazine, 43, 645.

23 Cottrell, A. H. (1958) Transactions of Metallurgical Society (AIME), 212, 192.

24 Petch, N. J. (1953) Cleavage strength of polycrystals, J. Iron Steel Institute, 173, 25.

25 Hall, E. O. (1951) Deformation and ageing of mild steel, Proc. Phys. Soc., 64B, 742.

26 Orowan, E. (1947) Discussion on Internal Stresses, Institute ofMetals, London.

27 Von Mises (1928) Z. Angew. Math. Mech., 8, 161.

28 Stein D. F. and Low J. M. (1960) Mobility of edge dislocations in silicon iron crystals,

J. Appl. Phys., 31, 362.

Additional Reading

Meyers, M. and Chawla, K. (2009) Mechanical Behavior ofMaterials, 2nd edn, Cambridge University Press.

Reed-Hill, R. E. and Abbaschian, R. (1994) Physical Metallurgy Principles, 3rd edn, PWS Publishing, Boston.

Intergranular Corrosion Intergranular corrosion occurs when corrosion attack develops along the grain boundaries. One important example is the sensitization of 18-8 austenitic stainless steel (SS 304). The steel contains 18 wt% Cr and 8wt% Ni with a low carbon content. But the weldability characteristics of these stainless steels are poor because the carbon content is enough to form deleterious carbides at the grain boundary areas of the heat-affected zone upon welding as observed in the quasi-binary phase diagram (Figure 5.63). When the material is fusion welded, the area near the weld zone (i. e., the heat affected zone or HAZ) experiences a high temperature, but does not melt the metal. As the welded steel is cooled down, chro­mium carbide (Cr23C6) formation takes place at the grain boundaries, as illustrated in Figure 5.64a. Because these particles contain a high amount of chromium, they make the regions adjacent to the grain boundary lean in chromium (<12 wt%), as illustrated in Figure 5.64b. In order for the stainless steel to retain its corrosion — resistant (stainless) property, it must have at least 12 wt% Cr. Even though the grain interiors contain higher Cr contents, the areas surrounding grain boundaries in the HAZ become lean in Cr. This condition is known as sensitization. This type of sensitized steel becomes susceptible to intergranular corrosion. This is a serious problem for this type of stainless steels.

Sensitization problem can be solved in three different ways: (i) After welding, the plate is cooled rapidly across the temperature range where chromium carbide forms. If the chromium carbide formation can be avoided, there will be no problem of sensitization. However, quenching the plate may have other undue conse­quences of residual stresses or distortion. (ii) Special stainless steel grades with very low carbon content (~0.01 wt% C) have been developed such as in 304L stain­less steel grades. If there is not much carbon present, there will be less amount of chromium carbide formation. (iii) Stabilized grades of austenitic stainless steels (like 316, 324, etc.) have been developed. Strong carbide-forming elements like molybdenum, titanium, niobium, and so on are added to the steel composition and during welding, corresponding carbides (TiC etc.) are preferentially formed throughout the bulk of the material (not just the grain boundaries) with not much carbon left for chromium carbide formation.

Stress Corrosion Cracking Stress corrosion cracking involves cracking of a material under static load by the combined action of stress and a chemical environment. SCC is generally found in alloys, not in pure metals. SCC occurs only in a specific environment for a given alloy. The presence of a tensile component of stress is nec­essary for SCC to take place. In majority of cases, the crack path is intergranular; transgranular cracking is rare. Cracking occurs in two stages: crack initiation and crack propagation. It has been noted that titanium alloys are immune to crack initi­ation in a chloride environment, but a precracked material shows susceptibility to crack propagation. Season cracking and caustic embrittlement are two examples of SCC. We will discuss this topic in detail in Section 6.4.4.

The crack growth mechanism that is involved is thought to be by anodic dissolu­tion at the crack tip. However, in materials that form passive films, the crack tip does get exposed to the corrosion medium as the plastic deformation at the crack tip exposes fresh metal surface. Thus, an active-passive cell gets created between the

crack tip and the crack faces. Since only a small area at the crack tip gets exposed, a very high current density is generated and corrosion occurs. In a laboratory environ­ment, slow strain rate tensile testing in the chosen chemical environment is used to determine the SCC properties of a material. Another way to study the time-depen­dent fracture in a corrosive environment is to precrack a specimen and surround it in the corrosion environment and keep the specimen under constant load. The measured time to fracture is plotted as a function of mode-I fracture toughness (Kj) in Figure 5.65a. If the applied stress corresponds to the critical fracture toughness (KIc), fracture occurs without any delay time. When Kj < KIc, fracture occurs after a delay period implying that crack grows slowly. However, when the stress intensity factor is below a critical level (Kjscc), no fracture occurs even after very long periods. Figure 5.65b shows the measured crack growth rates as a function of Kj. Below Kjscc, the crack growth rate is essentially zero. Above Kjscc in region-I, the crack propagates at an accelerating rate till it reaches region-II in which crack moves at a constant speed independent of the stress intensity factor. In region-II, the crack propagation is plausibly controlled by electrochemical factors. Finally, crack moves again at an accelerated rate resulting into fracture. The energy balance equation used to develop Griffith’s equation needs to be modified as electrochemical energy is released due to anodic dissolution at the crack tip. The final equation for KIscc is given by where cp is the plastic work term at the crack tip, E is the Young’s modulus, n! is the number of electrons taking part in the anodic dissolution process, F is Faraday’s con­stant, q is the density of the alloy, d is the opening at the crack tip, M is the atomic weight, and V is the electrode potential. Following Eq. (5.91), KIscc decreases with an increase in yield strength (through decreasing yp) and with an increase in propensity toward corrosion (via electrode potential V).

Fabricability includes a host of characteristics such as formability, weldability, machinability, and so on. If fabricability issues are not dealt with during the first stage, it may cause problems at the later stages. In many cases, some parts of the nuclear power plant are to be built at the site (also called field fabrication) from smaller parts. If the materials do not have the requisite fabricability, it would not be possible to use the material no matter what fantastic properties it may have!

1.9.1.2 Dimensional Stability

The material should have adequate stability in properties. For example, many nuclear components would work at higher temperatures for extended period of time. So, the creep deformation (i. e., time-dependent plastic deformation) may cause dimensional stability problems.

One should also recognize that the microstructure of a material changes as a function of temperature, time, and stresses. So, the effects of these factors on microstructure and the consequent effects on the properties need to be taken into account carefully.

Surface defects are two-dimensional defects, also known as planar defects. There are a variety of surface defects such as grain boundaries, twin boundaries, stacking faults, interphase boundaries (coherent, semicoherent, and incoherent), and anti­phase boundaries (specifically in intermetallic compounds). Each of these planar defects has an important role to play affecting various properties of crystalline materials.

The distortion associated with a dislocation implies that it must have certain strain energy. The total strain energy (Etotal) of a dislocation is composed of two parts — elastic strain energy (Eel) and core energy (Ecore). The elastic strain energy is stored outside the core region, and hence can be estimated using the linear elastic theo­ries. We have discussed the shear stress and strain components associated with a screw dislocation in Eqs. (4.10a) and (4.10b). Due to the complete radial symmetry of the stress/strain field associated with the screw dislocation, we derive its elastic strain energy using a simple calculation. Let us consider a volume element shell of the cylinder with an arbitrary radius r, thickness dr, and length l, and thus the elas­tic strain energy is given by

„ 1 ri , ,1 Gbb J N Gb21

Eel — -1ezVez ■ (volumeof theelement) ——————- (l ■ 2pr ■ dr) ——— dr.

2 2 2pr 2pr 4pr

Note the above equation is applicable only for an isotropic continuous medium. The hollow cylinder of length l, inner radius r1, and outer radius r2 will have an elastic strain energy determined by the following integral:

Г2 Gb2ldr _ Gb2l ln /r2 ‘ri 4p r 4p r1

As noted earlier, the dislocation core can be 1b-5b radius, and in most cases can be taken as ~1 nm radius. So, r1 value can be assumed to be of 1 nm. Furthermore, dislocations tend to form networks where the long-range stress fields are superim­posed and may get canceled. Thus, the effective energy of the dislocation can get reduced. That is why if one takes the value of r2 as approximately half the average spacing between dislocations in a random arrangement, a realistic value could be 106 nm. Then the value of r2/r 1 becomes 106. So, Eq. (4.12) is reduced to

13 8

Eei = (Gb2l) . = aGb2l,

v ‘ 4 x 3.14

where a is about ~1.1.

So, the elastic strain energy of a screw dislocation per unit length is

Eel = aGb2, (4.13)

where a = ~1 for screw dislocation.

An analysis can be performed for deriving the energy of an edge dislocation incorporating its complex stress field into the derivation. A similar expression can be obtained with a constant, a = ~1/(1 — v) = ~1.5 (for v = 0.33). So, in simple terms, the edge dislocation has greater energy than that of a screw dislocation of similar length. The elastic strain energy ofa mixed dislocation can also be obtained, and is generally between the energies of a pure edge and a pure screw dislocation.

However, the value of the constant may undergo changes depending on the pre­cise values of r1 and r2, which are difficult to determine. However, whatever their values are, the constant a would still remain close to unity.

The estimation of the dislocation core energy is only approximate as linear elastic theories are no longer valid in this region. Approximate calculations show that the core energy is estimated to be on the order of 0.5 eV per atom plane threaded by the dislocation, whereas the elastic strain energy of a dislocation is 5-10 eV per atom plane. Despite variation in the core energy during the dislocation movement, the core energy still remains a small fraction of the elastic strain energy. Thus, the dis­locations possess large positive strain energy increasing the overall free energy of the crystal. But the system would like to attain a more stable state via lowering its free energy. That is why given the chance the dislocation density can be reduced through dislocation annihilation processes such as annealing. The same is not true for point defects (such as vacancies) as they attain different equilibrium concentra­tions at different temperatures. That is why dislocations are thermodynamically unstable defects, whereas vacancies are thermodynamically stable defects.