Chapter 4 Introduction to Dislocations The discrepancy between the theoretical and observed yield stresses of crystals Dislocations The Burgers vector Vector notation for dislocations Dislocations in the face-centered cubic lattice Intrinsic and extrinsic stacking faults in face-centered cubic metals Extended dislocations in hexagonal metals Climb of edge dislocations Dislocation interactions The stress field of a screw dislocation The stress field of an edge dislocation The force on a dislocation The strain energy of a screw dislocation The strain energy of an edge dislocation 1
Discrepancy between Theoretical & Observed Yield Stresses Discrepancy between the theoretical and observed yield stresses of crystals - The stress-strain curve of a typical magnesium single crystal, oriented with the basal plane inclined at 45 to the stress axis and strained in tension, is shown in Fig. 4.1. At the low tensile stress of 0.7 MPa, the crystal yields plastically and then easily stretches out to a narrow ribbon which may be 4 or 5 times longer than the original crystal. - If one examines the surface of the deformed crystal, markings can be seen which run more or less continuously around the specimen in the form of ellipses (Fig. 4.2) Fig. 4.1 Tensile stress-strain curve for a Mg single crystal. Fig. 4.2 Slip 2lines on Mg crystal.
Discrepancy between Theoretical & Observed Yield Stresses (A) (B) - These markings (in Fig. 4.2) recognized as a series of fine steps that have formed on the surface. The nature of these steps is shown schematically in Fig. 4.3. Evidently, as a result of the applied force, the crystal has been sheared on a number of parallel planes. Crystallographic analyses of the markings, furthermore, show that these are basal (0002) planes and, therefore, the closest packed plane of the crystal. - When this type of deformation occurs, the crystal is said to have undergone slip, the visible markings on the surface are called slip lines, or slip traces, and the crystallographic plane on which the shear has occurred is called the slip plane ( 滑移面 ). - The shear stress at which plastic flow begins in a single crystal is amazingly small when compared to the theoretical shear strength of a perfect crystal. Fig. 4.3 (A) Magnified schematic view of slip lines (side view). (B) Magnified schematic view of slip lines (front view) 3
Discrepancy between Theoretical & Observed Yield Stresses - An estimate of the strength can be obtained in the following manner. Fig. 4.4A shows 2 adjacent planes of a hypothetic crystal. A shearing stress, acting as indicated by the vectors marked,ends to move the atoms of the upper plane to the left. Each atom of the upper plane rises to a maximum position (Fig. 4.4B) as it slides over its neighbor in the plane below. This maximum position represents a saddle point. A shear of one atomic distance requires that the atoms of the upper plane in Fig. 4.4A be brought to a position equivalent to that in Fig. 4.4B, after which they move on their own accord into the next equilibrium position, Fig. 4.4C. Since the separation of the 2 planes is of the order of 2 atomic radii, the shear strain at the saddle point is approximately equal to one half. where is shear strain. a 2a 1 2 (4.1) 4
Discrepancy between Theoretical & Observed Yield Stresses (A) (B) (c) Fig. 4.4 (A) Initial position of the atoms on a slip plane. (B) The saddle point for the shear of one plane of 5 atoms over another. (C) Final position of the atoms after shear by one atomic distance.
Discrepancy between Theoretical & Observed Yield Stresses - In a perfectly elastic crystal, the ratio of shear stress to shear strain is equal to the shear modulus: (4.2) where is shear strain, is shear stress, and is shear modulus. - Substituting the value ½ for the shear strain, and the value 17.2 GPa for, which is of the order of magnitude of the shear modulus for magnesium, we obtain for the stress at the saddle point, 1 2 17,200MPa 8.6 10 MPa - Real crystals deform at small fractions of their theoretical strengths (1/1000 to 1/100,000). 0.7MPa 5 8.1 10 3 8.6 10 MPa 3 6
Dislocations Dislocations ( 差排, 錯位 ) - The discrepancy between the computed and real yield stresses is because real crystals contain defects. - If the transmission foil has been prepared properly and contains a section of a slip plane, where it is examined in the microscope one may obtain a photograph of the type shown schematically in Fig. 4.5A. The lines (a-a and b-b) have been drawn on the figure to indicate the positions where the slip plane intersects the foil surfaces. - It should be noted that the drawing in Fig. 4.5A is a 2- dimensional projection of a 3-dimensional specimen. - Fig. 4.5 B demonstrates that the dark lines in the photograph run across the slip plane from the top to the bottom surfaces of the foil. - In a crystal which has undergone slip, lattice defects tend to accumulate along the slip planes. These defects are called dislocations. Fig. 4.5(A) 7
Dislocations - The points where dislocations intersect a specimen surface can often be made visible by etching the surface with a suitable etching solution. As a result, etch pits may form (Fig. 4.5C). (A) (B) (C) Fig. 4.5 (A) Schematic representation of an electron microscope photograph showing a section of a slip plane. (B) A 3-dimensional view of the same slip plane section. (C) Termination of dislocations can also be revealed by etch pits. 8
Dislocations - Fig. 4.6 shows a portion of a foil of an aluminum specimen with a grain containing a slip plane with dislocations. The specimen was polycrystalline. The dark region at the upper right-hand corner represents a second grain. - The best evidence now indicates that dislocations are boundaries on the slip planes where a shearing operation has ended. Fig. 4.6 An electron micrograph of a foil removed from an aluminum specimen. Note the dislocations lying along a slip plane, in agreement with Fig. 4.5. 9
Dislocations Edge dislocation ( 刃差排 ) - Fig. 4.7A represents a simple cubic crystal that is assumed to be subjected to shearing stresses,, on its upper and lower surfaces. The line SP represents a possible slip plane in the crystal. As a result of the applied shear stress, the righthand half of the crystal is displaced along SP so that the part above the slip plane is moved to the left with respect to the part below the slip plane. The amount of this shear is assumed to equal one interatomic spacing in a direction parallel to the slip plane. - As may be seen in Figs. 4.7B and C, this will leave an extra half-plane cd below the slip plane at the right and outside the crystal. It will also form an extra vertical halfplane ab above the slip plane and in the center of the crystal. - Fig. 4.7B clearly shows that the crystal is badly distorted where this half-plane terminates at the slip plane. It can also be deduced that this distortion decreases in intensity as one moves away from the edge of this half-plane. This is because at large distances from this lower edge of the extra plane, the atoms tend to be arranged as they would be in a perfect crystal. The distortion in the crystal is centered around the edge of the extra plane. This boundary of the additional plane is 10 called an edge dislocation.
(A) Dislocations (B) (C) Fig. 4.7 An edge dislocation. (A) A perfect crystal. (B) When the crystal is sheared one atomic distance over part of the distance S-P, an edge dislocation is formed. (C) 3 dimensional view of slip. 11
Dislocations - Fig. 4.8 represents a 3-dimensional sketch of the edge dislocations of Fig. 4.7. The figure clearly shows that the dislocation has the dimensions of a line. Another important fact is that the dislocation line marks the boundary between the sheared and unsheared parts of the slip plane. - In fact, a dislocation may be defined as a line that forms a boundary on a slip plane between a region that has slipped and one that has not. Fig. 4.8 This 3-dimensional view of a crystal containing an edge dislocation shows that the dislocation forms the boundary on the slip plane between a region that has been sheared and a region that has not been sheared. 12
Dislocations - As a result of the applied stress, atom c in Fig. 4.9A may move to the position marked c in Fig. 4.9B. The final result is that the crystal is sheared across the slip plane by one atomic distance, as shown in Fig. 4.9C. - Each step in the motion of the dislocation (Fig. 4.9) requires only a slight rearrangement of the atoms in the neighborhood of the extra plane. As a result, a very small force will move a dislocation. - In 1934, Orowan, Polyani, and Taylor presented papers which are said to have laid the foundation for the modern theory of slip due to dislocations. (A) (B) (C) Fig. 4.9 Three stages in the movement of an edge dislocation through a crystal. 13
Dislocations Screw dislocation ( 螺旋差排 ) - The movement of a single dislocation completely through a crystal produces a step on the surface, the depth of which is one atomic distance. Many hundreds or thousands of dislocations must move across a slip plane in order to produce a visible slip line. - Fig. 4.10A shows a screw dislocation, where each small cube can be considered to represent an atom. Fig. 4.10B represents the same crystal with the position of the dislocation line marked by the line DC. (A) (B) Fig. 4.10 Two representations of a screw dislocation. Notice that the planes in this dislocation spiral around the dislocation like a left-hand screw. 14
Dislocations - The designation screw for this lattice defect is derived from the fact that the lattice planes of the crystal spiral the dislocation line DC. This statement can be proved by starting an point x in Fig. 4.10A and then proceeding upward and around the crystal in the direction Fig. 4.10(A) of the arrows. One circuit of the crystal ends at point y; continued circuits will finally end at point z. - Fig. 4.10B plainly shows that a dislocation in a screw orientation also represents the boundary between a slipped and an unslipped area. Here the dislocation, centered along line DC, separates the slipped area ABCD from the remainder of the slip plane in back of the dislocation. Fig. 4.10(B) 15
Materials Science and Engineering Dislocations-Linear Defects Dislocation line could be seen as a line to separate the deformed and undeformed regions. Figure 4.4 (a) A screw dislocation within a crystal. (b) The screw dislocation in (a) as viewed from above. The dislocation line extends along line AB. Atom positions above 16 the slip plane are designated by open circles, those below by solid circles.
Dislocations Dislocations - The edge dislocation shown in Fig. 4.7B has an incomplete plane which lies above the slip plane. It is also possible to have the incomplete plane below the slip plane. The 2 cases are differentiated by calling the former a positive edge dislocation, and the latter a negative edge dislocation. Fig. 4.7(B) Symbols representing these 2 forms are and, respectively, where the horizontal line represents the slip plane and the vertical line the incomplete plane. - The screw dislocation shown in Fig. 4.10 has lattice planes that spiral the line DC like a left-hand screw. - Both forms of the edge and the screw dislocations, respectively, are shown in Fig. 4.12. Fig. 4.10(B) 17
Materials Science and Engineering Dislocations-Linear Defects Positive edge dislocation Negative edge dislocation 18
Dislocations (A) (B) (C) (D) Fig. 4.12 The ways that the 4 basic orientations of a dislocation move under the same applied stress: 19 (A) Positive edge, (B) Negative edge, (C) Left-hand screw, and (D) Right-hand screw.
Dislocations - Fig. 4.12 shows that a positive edge dislocation moves to the left when the upper half of the lattice is sheared to the left. On the other hand, a negative edge dislocation moves to the right, but produces the identical shear of the crystal. The right-hand screw moves forward and the left-hand screw moves to the rear, again producing the same shear of the lattice. - Dislocations cannot end inside a crystal. - In Fig. 4.13, the 2 dislocation segments a and b form a continuous path through the crystal from front to top surfaces. Fig. 4.13 Dislocations can vary in direction. This shaded extra plane forms a dislocation with edge components a and b. 20
Dislocations - It is also possible for all 4 edges of an incomplete plane to lie inside a crystal, forming a 4-sided closed edge dislocation at the boundaries of the plane. - Furthermore, a dislocation that is an edge in one orientation can change to a screw in another orientation, as is illustrated in Figs. 4.14 and 4.15. Fig. 4.14 A 2-component dislocation composed of an edge and a screw component. Fig. 4.15 Atomic configuration corresponding to the dislocation of Fig. 4.14 viewed from above. Opencircle atoms are above the slip plane, dot atoms are below the slip plane. 21
Dislocations - A dislocation does not need to be either pure screw or pure edge, but may have orientations intermediate to both. Fig. 4.16 shows a change in orientation from edge to screw, but here the change is not abrupt. - Fig. 4.17 consists of the 4 elementary types of dislocations. Sides a and c are positive and negative edge dislocations, respectively, while b and d are right- and left-hand screws, respectively. Fig. 4.17 A closed dislocation loop consisting of (a) positive edge, (b) right-hand screw, (c) negative edge, and (d) left-hand screw. Fig. 4.16 A dislocation that changes its orientation from a screw to an edge as viewed from above looking down on its slip plane. 22
Dislocations - A dislocation cannot inside a crystal. This is because a dislocation represents the boundary between a slipped and an unslipped area. - If the slipped area on the slip plane does not touch the specimen surface, as in Fig. 4.18, then its boundary is continuous and the dislocation has to be a closed loop. Fig. 4.18 A curved dislocation loop lying in a slip plane. 23
Burgers Vector The Burgers vector - In Fig. 4.17, although the dislocation varies in orientation in the slip plane ABCD, the variation in shear across the dislocation is everywhere the same, and the slip vector b is therefore a characteristic property of the dislocation. By definition, this vector is called the Burgers vector of the dislocation. - The Burgers vector of a dislocation is an important property of a dislocation because, if the Burgers vector and the orientation of the dislocation line are known, the dislocation is completely described. Fig. 4.17 24
Burgers Vector - In Fig. 4.19A a counterclockwise circuit of atom-to-atom steps in a perfect crystal closes, but when the same step-by-step circuit is made around a dislocation in an imperfect crystal (Fig. 4.19B), the end point of the circuit fails to coincide with the starting point. The vector b connecting the end point with the starting point is the Burgers vector of the dislocation. (A) (B) 25 Fig. 4.19 The Burgers circuit for an edge dislocation: (A) Perfect crystal and (B) crystal with dislocation.
Burgers Vector - The procedure can be used to find the Burgers vectors of any dislocation if the following rules are observed: 1. The circuit is traversed in the same manner as a rotating right-hand screw advancing in the positive direction of the dislocation. 2. The circuit must close in a perfect crystal and must go completely around the dislocation in the real crystal. 3. The vector that closes the circuit in the imperfect crystal (by connecting the end point to the starting point) is the Burgers vector. - The above convention involving a right-hand (RH) circuit around the dislocation line yields a Burgers vector pointing from the finish to the start (FS) of the circuit, and because the closure failure is measured in an imperfect crystal, it is called a local Burger vector or more completely a RHFS local Burgers vector. 26
Burgers Vector - Fig. 4.20 shows a Burgers circuit around a left-hand screw dislocation. In Fig. 4.20A, the circuit is indicated for the perfect crystal. Fig. 4.20B shows the same circuit transferred to a crystal containing a screw dislocation. (A) (B) Fig. 4.20 The Burgers circuit for a dislocation in a screw orientation. (A) Perfect crystal and (B) crystal with dislocation. 27
Burgers Vector Characteristics of both edge and screw dislocations - Edge dislocation 1. An edge dislocation lies perpendicular to its Burgers vector. 2. An edge dislocation moves (in its slip plane) in the direction of the Burgers vector (slip direction). Under a shear-stress sense a positive dislocation moves to the right, a negative one to the left. - Screw dislocation 1. A screw dislocation lies parallel to its Burgers vector. 2. A screw dislocation moves (in the slip plane) in a direction perpendicular to the Burgers vector (slip direction). 28
Materials Science and Engineering Dislocations-Linear Defects Edge dislocation Dislocation line Burgers vector Slip direction Screw dislocation Dislocation line Burgers vector Slip direction 29
Burgers Vector Slip plane ( 滑移面 ) - The slip plane is the plane containing both the Burgers vector and the dislocation. - The slip plane of an edge dislocation is thus uniquely defined because the Burgers vector and the dislocation are perpendicular. - The slip plane of a screw dislocation can be any plane containing the dislocation because the Burgers vector and dislocation have the same direction. - Edge dislocations are confined to move or glide in a unique plane, but screw dislocations can glide in any direction as long as they move parallel to their original orientation. 30
Vector Notation for Dislocations Vector notation for dislocations - In any crystal form, the distance between atoms in a close-packed direction corresponds to the smallest shear distance that will preserve the crystal structure during a slip movement. Dislocations with Burgers vectors equal to this shear are energetically the most favored in a given crystal structure. - With regard to the vector notation, the direction of a Burgers vector can be represented by the Miller indices of its direction, and the length of the vector can be expressed by a suitable numerical factor placed in front of the Miller indices. - In a simple cubic lattice, the distance between atoms in a close-packed direction equals the length of one edge of a unit cell. A dislocation with a Burgers vector in a simple cubic lattice is represented by [100]. 31
Vector Notation for Dislocations - In Fig. 4.21, the close-packed direction in a face-centered cubic lattice is a face diagonal, and the distance between atoms in this direction is equal to one-half the length of the face diagonal. As a result, a dislocation in a face-centered cubic lattice having a Burgers vector lying in the [101] direction should be written ½[101]. - In the body-centered cubic lattice, the close-packed direction is a cube diagonal, or a direction of the form <111>. The distance between atoms in these directions is onehalf the length of the diagonal, so that a dislocation having a Burgers vector parallel to [111] is written ½[111]. 001 011 101 111 000 010 Fig. 4.21 The spacing between atoms in the close-packed directions of the different cubic systems: face-centered cubic, body-centered cubic, and simple cubic. 100 110 32
Dislocations in the Face-Centered Cubic Lattice Dislocations in the face-centered cubic lattice - The primary slip plane in the face-centered cubic lattice is the octahedral plane {111}. Fig. 4.22 shows a plane of this type looking down on the extra plane of an edge dislocation. Notice that in the latter plane a zigzag row of atoms is missing. This corresponds to the missing plane of the edge dislocation. Fig. 4.22 A total dislocation (edge orientation) in a face-centered cubic lattice as viewed when looking down on the slip plane. 33
Dislocations in the Face-Centered Cubic Lattice - The vector b represents the Burgers vector of the dislocation, which is designed as ½[110] (Fig. 4.24). As is to be expected, this dislocation movement shears the upper half of the crystal (above the plane of the paper) one unit b to the right relative to the bottom half (below the plane of the paper). - In Fig. 4.22, the movement of a zigzag plane of atoms, such as aa, through the horizontal distance b would involve a very large lattice strain, because each white atom at the slip plane would be forced to climb over the dark atom below it and to its right. - What actually is believed to happen is that the indicated plane of atoms makes the move indicated by the vectors marked c in Fig. 4.23. This movement can occur with a much smaller strain of the lattice. A second movement of the same type, indicated by the vectors marked d, brings the atoms to the same final positions as the single displacement b of Fig. 4.22. 34 Fig. 4.22
Dislocations in the Face-Centered Cubic Lattice - The atom arrangement of Fig. 4.23 is particularly significant because it shows how a single-unit dislocation can break down into a pair of partial dislocations. - In Fig. 4.24, the Burgers vector of the total dislocation equals the distance B 1 B 2, while the Burgers vectors of the 2 partial dislocations c and d of Fig. 4.23 are the same as the distance B 1 C and CB 2. The Burgers vector of the total dislocation is ½[110]. The B 1 C lies in the [121] direction. Since B 1 C is just one-third of line mn, the Burgers vector for this partial dislocation, c, is 1/6[121]. In the same manner, the partial dislocation, d, can be represented as 1/6[211]. 001 101 [110] 011 1 1 /2 1 /2 1 /21 1 /2 100 010 110 Fig. 4.23 Partial dislocation in a face-centered cubic lattice. Fig. 4.24 The orientation relationship between the Burgers vectors of a 35 total dislocation and its partial dislocations.
Dislocations in the Face-Centered Cubic Lattice - The total face-centered cubic dislocation ½[110] is thus able to dissociate into 2 partial dislocations according to the relation 1 2 1 1 110 121 211 6 6 (4.4) - When a total dislocation breaks down into a pair of partials, the strain energy of the lattice is decreased. This results because the energy of a dislocation is proportional to the square of its Burgers vector and because the square of the Burgers vector of the total dislocation is more than twice as large as the square of the Burgers vector of a partial dislocation. 2 1 1 1 110 1 1 0 2 4 2 1 6 1 6 1 2 2 1 121 1 4 1 2 1 211 4 1 1 1 6 1 6 36 36 1 3 1 6 1 6 36
Dislocations in the Face-Centered Cubic Lattice - A total dislocation that has dissociated into a pair of separated partials like those in Fig. 4.25 is known as an extended dislocation ( 擴展差排 ). - If we assume that the dark-colored atoms occupy A positions in a stacking sequence and the white atoms at either end of the figure, B positions, then the white atoms between the 2 partial dislocations lie on C positions. In this region, the ABCABCABC stacking sequence of the face-centered cubic lattice suffers a discontinuity and becomes ABCA CABCA. The arrows indicate the discontinuity. - Discontinuity in the stacking order of the {111}, or closepacked planes, are called stacking faults ( 疊差 ). Fig. 4.25 An extended dislocation. 37
Dislocations in the Face-Centered Cubic Lattice - The stacking fault occurs on the slip plane (between the dark and white atoms) and is bounded at its end by what are known as Shockley partial dislocations. - In all cases, if a stacking fault terminates inside a crystal, its boundaries will form a partial dislocation. - The partial dislocations of stacking faults, in general, may be either of the Shockley type, with the Burgers vector of the dislocation lying in the plane of the fault, or of the Frank type, with the Burgers vector normal to the stacking fault. Fig. 4.25 38
Dislocations in the Face-Centered Cubic Lattice - The defect marked by A in Fig. 4.26 shows stacking faults bordered by 2 dislocations in the lower-left- and upper-right-hand side. - The defect marked by B shows a narrow region of stacking fault surrounded by Shockley partial dislocations. Fig. 4.26 An electron micrograph of a thin foil of lightly deformed Cu + 15.6 at.% Al alloy. The beam direction is close to [101] and g is indicated. 44,000. 39
Dislocations in the Face-Centered Cubic Lattice Stacking fault ( 疊差 ) - Since the atoms on either side of a stacking fault are not at the positions they would normally occupy in a perfect lattice, a stacking fault possesses a surface energy which, in general, is small compared with that of an ordinary grain boundary, but nevertheless finite. - The stacking-fault energy plays an important part in determining the size of an extended dislocation. The larger the separation between the partial dislocations, the smaller is the repulsive force between them. On the other hand, the total surface energy associated with the stacking fault increases with the distance between partial dislocations. The separation between the 2 partials thus represents an equilibrium between the repulsive energy of the dislocations and the surface energy of the fault. - Seeger and Schoeck have shown that the separation of the partial dislocations in an extended dislocation depends on a dimensionless parameter I c/gb 2, where I is the specific surface energy of the stacking fault, c is the separation between adjoining slip planes, G is the shear modulus in the slip plane, and b is the magnitude of the Burgers vector. 40
Dislocations in the Face-Centered Cubic Lattice In certain face-centered cubic metals typified by aluminum, this parameter is larger than 10-2 and the separation between dislocations is only of the order of a single atomic distance. These metals are said to have high stacking-fault energies. When the parameter is less than 10-2, a metal is said to have a low stacking-fault energy. Movement of an extended dislocation - If the moving dislocation meets obstacles, such as other dislocations, or even second-phase particles, the width of the stacking fault should vary. - Thermal vibrations may also cause the width of the stacking fault to vary locally along the dislocation, the vibration being a function of time. - An extend dislocation can be pictured as a pair of partial dislocations, separated by a finite distance, which move in consort through the crystal. 41
Intrinsic and Extrinsic Stacking Faults in FCC Metals Intrinsic and extrinsic stacking faults in face-centered cubic metals - The movement of a Shockley partial across the slip plane of a fcc metal has been shown to produce a stacking sequence ABCA CABC. A fault of this type is called an intrinsic stacking fault. An intrinsic stacking fault may also be developed in a fcc crystal by removing part of a close-packed plane, as shown in Fig. 4.27A. Its Burgers vector is equal to one-third of a total dislocation and therefore may be written 1/3<111>. Fig. 4.27A An intrinsic stacking fault can also be formed in a face-centered cubic crystal by removing 42 part of a close-packed plane.
Intrinsic and Extrinsic Stacking Faults in FCC Metals - The addition of a portion of an octahedral plane produces a different type of stacking sequence which is ABCA C BCABC. In this fault (Fig. 4.27B) a plane has been inserted that is not correctly stacked with respect to the planes on either side of the fault. This type of fault is called an extrinsic or double stacking fault. The Burgers vector for the extrinsic fault, shown in Fig. 4.27B, is 1/3<111>. An extrinsic stacking fault could be formed by the precipitation of interstitial atoms on an octahedral plane. Fig. 4.27B The addition of a portion of an extra close-packed plane to a face-centered cubic crystal produces an extrinsic stacking fault. 43
Extended Dislocations in Hexagonal Metals Extended dislocations in hexagonal metals - In the hexagonal system, the dissociation of a total dislocation into a pair of partials on the basal plane is expressed in the following fashion: 1 3 1 1 1210 0110 1100 3 (4.5) - For the face-centered cubic system, the dissociation of a total dislocation into a pair of partials is expressed in the following fashion: 1 2 3 1 1 110 121 211 6 6 (4.6) 44
Climb of Edge Dislocations Climb of edge dislocations - The slip plane of a dislocation is defined as the plane that contains both the dislocation and its Burgers vector. - Since the Burgers vector is parallel to a screw dislocation, any plane containing the dislocation is a possible slip plane (Fig. 4.28A). - The Burgers vector of an edge dislocation is perpendicular to the dislocation, and there is only one possible slip plane (Fig. 4.28B). (A) (B) Fig. 4.28 (A) Any plane containing the dislocation is a slip plane for a screw dislocation. (B) There is 45only one slip plane for an edge dislocation. It contains both the Burgers vector and the dislocation.
Climb of Edge Dislocations - A screw dislocation may move by slip or glide in any direction perpendicular to its self, but an edge dislocation can only glide in its single slip plane. - Fig. 4.29 demonstrates a positive climb of an edge dislocation and results in a decrease in size of the extra plane. - Negative climb corresponds to the opposite of the above in that the extra plane grows in size instead of shrinking. A mechanism for negative climb is illustrated in Fig. 4.30. (A) (B) (C) Fig. 4.29 Positive climb of an edge dislocation. 46
Climb of Edge Dislocations - Because we are removing material from inside the crystal as the extra plane itself grows smaller, the effect of positive climb on the crystal is to cause it to shrink in a direction parallel to the slip plane (perpendicular to the extra plane). Positive climb is therefore associated with a compressive strain and will be promoted by a compressive stress component perpendicular to the extra plane. - A tensile stress applied perpendicular to the extra plane of an edge dislocation promotes the growth of the plane and thus negative climb. - Slip occurs as the result of shear stress; climb as the result of a normal stress (tensile or compressive). (A) (B) Fig. 4.30 Negative climb of an edge dislocation. 47
Climb of Edge Dislocations - Both positive and negative climb require that vacancies move through the lattice, toward the dislocation in the first case and away form it in the second case. - If the concentration of vacancies and their jump rate is very low, then it is not expected that edge dislocation will climb. - Climb is a phenomenon that becomes increasingly important as the temperature rises. Slip, on the other hand, is only slightly influenced by temperature. 48
Dislocation Intersections Dislocation intersections - In Fig. 4.31 it is assumed that a dislocation has moved across the slip plane ABCD, thereby shearing the top half of the rectangular crystal relative to the bottom half by the length of its Burgers vector b. A second (vertical) dislocation, having a loop that intersects the slip plane at 2 points, is shown in Fig. 4.31. It is assumed that this loop is in the edge orientation where it intersects the slip plane. - The indicated displacement of the crystal (Fig. 4.31) also shears the top half of the vertical-dislocation loop relative to its bottom half by the amount of the Burgers vector b. The displacement lengthens the vertical-dislocation loop by an amount equal to the 2 horizontal steps (Fig. 4.31). Fig. 4.31 In the figure a dislocation is assumed to have moved across the horizontal plane ABCD and, in cutting through the vertical-dislocation loop, it forms a pair of jogs in the latter. 49
Dislocation Intersections This result is characteristic of the intersection of dislocations, for whenever a dislocation cuts another dislocation, both dislocations acquire steps of a size equal to the other s Burgers vector. - The first case, where the step lies in the slip plane of a dislocation, is called a kink (Fig. 4.32). The second case, where the step is normal to the slip plane of a dislocation, is called a jog (Fig. 4.33). - The kink in the edge dislocation has a screw orientation (Burgers vector parallel to line on), while the step in the screw dislocation has an edge orientation (Burgers vector normal to the line on) (Fig. 4.32). (A) Fig. 4.32 Dislocations with kinks that lie in the slip plane of the dislocations. (B) 50
Dislocation Intersections Both of these steps can easily be eliminated by moving line mn over to the position of the dashed line. This movement in both cases can occur by simple slip. Since the elimination of a step lowers the energy of the crystal by the amount of the strain energy associated with a step, it can be assumed that steps of this type may tend to disappear. - An edge and a screw dislocation, with steps normal to the primary slip plane, are shown in Fig. 4.33. This type of discontinuity is called a jog. (A) Fig. 4.33 Dislocations with jogs normal to their slip planes. (B) 51
Dislocation Intersections - The edge dislocation with a jog (Fig. 4.33A) is free to move on the stepped surface, for all 3 segments of the dislocation, mn, no, and op, are in a simple edge orientation with their respective Burgers vectors lying in the crystal planes that contain the dislocation segments. - The screw dislocation with a jog (Fig. 4.33B) represents quite a different case. Here the jog is an edge dislocation with an incomplete plane lying in the stepped surface. Here the jog (line no), which is an edge orientation, is not capable of gliding along the vertical surface because its Burgers vector is not in the surface of the step but is normal to it. The only way that the jog can move across the surface of the step is for it to move by dislocation climb. Fig. 4.33 52
Stress Field of a Screw Dislocation Stress field of a screw dislocation - The elastic strain of a screw dislocation is shown in Fig. 4.35. Consider the circular Burgers circuit shown in Fig. 4.35. Such a path results in an advance (parallel to the dislocation line) equal to the Burgers vector b. The strain in the lattice is the advance divided by the distance around the dislocation. b 2 r (4.7) where r is the radius of the Burgers circuit. This strain is accompanied by a corresponding state of stress in the crystal. Fig. 4.35 Shear strain associated with a screw dislocation. 53
Stress Field of a Screw Dislocation - Assuming the crystals to be homogeneous isotropic bodies, the elastic stress field surrounding a screw dislocation is written: b 2 r (4.8) where id the shear modulus of the material of the crystal. - The analysis of the stress close to the center of the dislocation is extremely difficult, and no completely satisfactory theory has yet been developed. 54