module 14 crystal defects in metals v - nptelnptel.ac.in/courses/113105023/lecture14.pdf · module...

NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || | | |

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Module 14

Crystal defects in metals V

Lecture 14

Crystal defects in metals V


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Keywords : Experimental evidence of the presence of dislocations, Structure of grain boundary & sub‐

grain boundary, Estimation of stacking fault energy, Grain boundary energy, Peierls stress

Introduction

We have seen so far how with the introduction of the concept of crystal defects it is possible to explain

why the strength of real crystal is much lower than its theoretical strength. The question that comes up

is there any direct evidence of the presence of dislocation in real crystals. We shall learn about it in this

module. We would also look at how to estimate the concentration of vacancies and the magnitude of

stacking fault energy. We shall also look at the structure of grain boundaries and derive a simple relation

between dislocation density and the strength of metals.

Etch pit technique

Microstructures of pure metal often show etch pits within the grains. They have specific geometric

shapes. What do these represent? We know that dislocations are line defects. They exist within crystals

as continuous loops, or end at grain boundaries at free surfaces. When we look at a microstructure we

see a section of the grains. This may intersect several line defects present within the grain. On etching

polished surface pits develop at such points of intersection. This is an electrochemical process. The rate

of dissolution of metal depends on the local state of the material. Slide 1 shows with help of sketches

the origin of etch pits within the grain. The shape of the etch pits depends on the nature of the

dislocation and the orientation of the plane intersecting it. There are special electrolytes for different

metals and alloys. These are first used to polish the sample. At an appropriate applied voltage, current

density and temperature of the bath the surface of the metal shines like a mirror. Subsequently the

voltage is changed to etch the specimen. The sketch on the right in slide 1 shows a typical

microstructure of a metal exhibiting etch pits. The number of pits per unit area is a measure of

dislocation density. (There are excellent micrographs in the book; Introduction to Dislocations by Derek

Hull, Pergamon Press, Oxford, p25‐46).

Rate of dissolution = f(local strain field, local solute concentration, plane geometry)

Etch pit technique

Transmission electron microscope

Slide 1: The sketch on the left shows intersection

of an edge dislocation with the top and the

bottom surfaces. The way the atoms are arranged

around it is different from that in regions away

from it. This acts as anode and undergoes

preferential dissolution during etching. The sketch

in the centre shows the intersection of a screw

dislocation with the top / bottom surface. Atoms

around it are arranged in the form of a helix. This

shows up as pits having different shapes.


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Etch pit techniques although provided indirect evidence of the presence of dislocations in crystals the

direct evidence came with the availability of transmission electron microscope (TEM). Thin foils of metal

suitable for examination under transmission mode in TEM can be made by electro‐polishing / thinning

techniques. Slide 2 illustrates the basic principle that makes dislocations visible under TEM. The wave

length of electron beam is of the order of atomic spacing of the crystal. They get diffracted if the Bragg

conditions are met.

Transmission electron microscope

TEM can also be used to find the orientation of the foil plane by recording its diffraction pattern. This is

done by blocking the transmitted beam and removing the projection lens. The pattern consists of an

array of points. This can be indexed to find the indices of the foil plane. TEM also has a facility to tilt the

specimen. This helps look at the microstructures on different planes. Using this you may notice that

under certain conditions of the tilt the dislocation disappears. From such extinction conditions it is

possible to find the slip plane and Burgers vector. TEM has been extensively used to reveal dislocation

structure and several other crystal defects like stacking faults, dislocation network and dislocation

boundaries. Interested reader may refer to books on transmission electron microscope & electron

diffraction. (There are excellent micrographs in the book; Introduction to Dislocations by Derek Hull,

Pergamon Press, Oxford, p25‐46).

Measurement of electrical resistivity:

Several physical properties are dependent on the concentration of defects in crystalline solids. For

example if there are vacancies in a crystal lattice there is a corresponding reduction in its effective cross

sectional area. The electrical resistance of metals which depends on its cross sectional area therefore

can give an estimate of the concentration of vacancies. The number of vacancies present in metals

follows Boltzmann statistics. If qv is the activation energy needed for the creation of a vacancy, the

number of vacancies (nv) at a given temperature (T) is given by where n is the total

number of sites in the lattice and k is the Boltzmann constant. This can be used to derive a simple

correlation between the change in resistivity and the concentration of vacancies. Let v denote an

increase in resistivity due to creation of one vacancy. The relation between (change in resistivity) due to creation of nv number of vacancies is given by:

Slide 2: The top sketch shows how the crystal planes

are arranged within a thin foil around a dislocation.

The red lines with arrow marks denote the paths of

the electron beam. The beam passing through the

region away from the dislocation reaches the image

plane directly. However those passing through the

region where there is a disturbance in the

arrangements of the atomic layers get scattered /

diffracted away. Thus the area beneath the

dislocation appears as a dark line.


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(1)

This equation can be used to design a simple experiment to estimate the activation energy for the

creation of vacancies. Take a piece of metal (Au) heat it to a high temperature say (600°C). The number

of vacancies at this temperature is much higher. Quench it to subzero temperature to retain the

concentration of vacancies and measure its resistivity as early as possible. At subzero temperatures the

mobility of vacancies is low. Therefore it may be assumed that it retains the same number of vacancies

as that at 600°C. Repeat this by heating the piece of Au sample to various other temperatures. Plot against 1/T (see slide 3). The slope of the plot gives an estimate of qv.

Experimental evidence of point defects

exp vv v v

qn n

kT

Heat a sample (Au) to T & quench &

measure at 0 deg C

Lo

g

1/T

Slope gives qv

TEM of quenched gold foil exhibits Frank dislocation

loops

There are excess vacancies in quenched samples. These are mobile too. Therefore left to it the vacancies

may diffuse out or accumulate at certain locations. A common phenomenon is the formation of a disc

shaped region (see Fig 1). This can be seen under TEM. A schematic representation of the same is shown

in slide 3. Figure 1 show the change in stacking sequence across the region where vacancies accumulate.

If the thin foil has such a region the electron beams passing through this get diffracted. Therefore it has

a different contrast in the image recorded under TEM. Figure 1 illustrates how the region gets encircled

by an edge dislocation. Its Burgers vector is 111 . It is a partial dislocation. Its slip plane is a cylindrical

surface. The size of the loop depends on the concentration of excess vacancies and stacking fault (SFE).

SFE (stacking fault energy) can be estimated from the size of the loop.

What is a grain boundary?

Slide 3: Illustrates an experimental scheme for

the estimation of activation energy for the

creation of a vacancy by annealing Au at different

temperatures followed by quenching. Resistivity

is a strong function of temperature. It should

therefore be measured at a fixed temperature

(say 0°C). Vacancies in lattice are mobile. Lower

the temperature lower is the mobility of vacancy.

Therefore Au samples after quenching must be

stored at as low a temperature as possible.

A

B

C

A

B

C

Fig 1: A sketch showing the change in stacking due to accumulation of

vacancies in one of the close packed planes. In FCC metals (111) planes

are arranged in the sequence ‘ABCABC’. If vacancies accumulate in a part

of the plane ‘C’ the sequence across this becomes ‘ABABC’. The region

over which there is a fault in stacking is encircled by an edge dislocation.


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We have so far looked at three types of crystal defects; vacancy – point defect, dislocation – a line

defect and stacking fault – a surface defect. However the most common defect that is visible even under

optical microscope is the grain boundary. This is the surface of contact between two neighboring grains.

Figure 2 shows a typical microstructure of a pure metal.

Pure metals are made of several randomly oriented grains or crystals. The surface which is shared by

two neighboring grains is the boundary between the two. Microstructure represents a 2 dimensional

section of such an arrangement. The grains that are visible represent crystals and the lines are the

boundaries. The atoms in respective grains on the two sides of a boundary are arranged in a periodic

fashion. However there may not be a definite relation between the two. This means that just across the

boundary there is a sudden change in the atomic array. Therefore a grain boundary may represent a

surface where the atoms may not be arranged in a periodic fashion. This may also be visualized as a

network of dislocations although it may be difficult to guess its structure. Nevertheless the nature of the

boundary will certainly be affected by the shapes of the neighboring grains and their orientations. Slide

4 gives a schematic arrangement of atoms in two neighboring grains.

Grain boundary

2 cos2GB s

Surface defects: boundary between two crystals having different orientations.

s s

GB

2

2

1.58

0.53

s

sGB

GbJm

Jm

Every free surface has energy. This can be estimated from heat of sublimation (evaporation) or the

strength of the metallic bond. In view of the total disarray in atomic arrangement in the boundary it may

be assumed that atoms from this zone would evaporate more easily than those from the top face. If a

metal is heated in vacuum more number of atoms would evaporate from the boundary than from the

top surface. This will result in the formation of a groove at the boundary (see slide 4 & note that the

angle is ). Therefore equating the forces acting along the boundary an expression relating grain boundary energy to the energy of the free surface can be obtained (see slide 4).

2 (2)

Fig 2: Shows a typical microstructure of a pure metal. The enclosed

areas are the 2 dimensional sections of grains. The lines represent

grain boundaries.

GrainGB

Slide 4: The sketch on the left shows how atoms

are arranged in two adjacent grains. Note the lack

of order in atomic arrangement at the boundary.

The sketch on the right shows a sectional view of

the way the crystal planes are arranged on either

sides of the boundary. The arrows indicate forces

acting along the boundary. s denotes energy of free surface & GB denotes energy of the grain boundary. is the angle of the groove formed due

to heat etching.


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If = 120°; GB = S. However grain boundary energy is expected to be less than that of the free surface. This is of the order of GB/3. This corresponds to 160°. Grains on the two sides of such boundaries are randomly oriented. There is no orientation relationship between the two. We already have come

across two boundaries (stacking fault & twin boundaries) where there are perfect lattice matching

between the adjacent grains. They are known as coherent boundaries. There is a definite orientation

relationship between the two parts of the crystal on either sides of the boundary.

Tilt boundary:

Every dislocation has its own stress field. When several of these are present in a crystal they would try

to arrange themselves in such a way that the total energy of the configuration is the lowest. One of

these arrangements leads to formation of a tilt boundary where the orientations of the two sides differ

by a small angle. Slide 5 explains the formation of an array of edge dislocations arranged one over the

other on a plane perpendicular to the plane of sketch.

Low angle / tilt boundary

X1

X2

b

h

Grain 1 Grain 2

The sketch on the left in slide 5 shows the most stable configuration of edge dislocations. If any of these

is displaced a little either along x1 or against x1 there will be restoring force acting on it. This would bring

them back to the most stable state. Such an array of dislocations represents a low angle boundary.

There is a small angle of miss‐orientation between the two parts of the crystal separated by an array of

dislocations. If the distance between two dislocations is h, the angle of miss‐orientation is given by = b/h. (Note that the angle denotes shear strain which is given by displacement over the distance.)

How do we estimate the energy of this boundary? Note that the boundary lies along the plane x2x3. It is

perpendicular to x1. If the length of the boundary along x3 is l, there is one dislocation of length l in an

area of lh. Therefore the elastic stored energy per unit area of the boundary is given by (El/lh), where E

is the energy per unit length of an edge dislocation. Since h = b/ the energy of the low angle boundary (ELAB) is given by equation 3, where r0 is the size of dislocation core, G is shear modulus, is the Poisson ratio and A is constant that depends on b & r0.

1 (3)

Slide 5: The sketch on the right shows the direction of

the forces acting on dislocations lying in various

locations on the plane normal to axis x3 due to the

stress field of a dislocation lying along axis x3. There is a

force acting along axis x2 on each of the dislocations

lying above the line x1. Therefore they could climb up.

The direction of forces on dislocations lying below the

line x1 is just the opposite. This would help these climb

down. Once they cross the dotted lines the forces of

attraction would bring them to occupy positions one

over the other.


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The energy of such a boundary is a strong function of the angle of miss‐orientation or the tilt. Figure 3

gives a schematic plot showing how the energy varies with . Equating dE/d to zero it is possible to estimate for which the energy of low angle boundary is the maximum. It comes out to be around 30°.

We have just looked at a boundary made of an array of edge dislocations. We could also think of a

boundary made of screw dislocation. This is known as the twist boundary. A more general low angle

boundary may be made of dislocations of mixed character.

Fig 4: Shows TEM images from 9CrMo steel (b) before creep test (b) after creep test. Note that the initial structure has high dislocation density within lath boundaries. As a result of creep exposure dislocation density decreases and carbide precipitates have grown. The inset in (b) gives a diffraction pattern from the precipitate. This helps identify this to be (CrFe)23C6. (J Baral et al to be published).

Coincidence site lattice:

We have looked at the possible structure of grain boundaries. One of ways is to visualize it to be an

array of dislocations. Tilt boundary is the simplest possible representation of grain boundary. The angle

of miss‐orientation is a measure of the energy of the boundary. There is another way we may look at the

boundary. The ways the atoms are arranged on the two sides of the boundary are different. Figure 5

shows a sketch of atomic arrangements in two adjacent grains of a pure metal.

ELAB

Fig 3: A schematic sketch showing how the energy of

low angle grain boundary would change with the angle

of miss‐orientation. Although the derivation has several

simplifying assumptions there is experimental evidence

to support such a trend.


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If the lattice structures of the two adjacent grains as shown in fig 5 are extended to superimpose one

over the other there will be a few points that would coincide. A set of such points constitutes a

coincident site lattice. The distance between two points in this is much greater than that in the original

lattice. The ratio of the volume of the coincident lattice to that of the parent lattice is known as sigma

(). This may be a convenient way of characterizing grain boundaries. Slide 6 illustrates how by

superimposition of lattice arrays one could evaluate the magnitude of .

Coincidence site lattice=(volume of CSL unit cell) / (volume of normal unit cell)

22 2

3

45

a a a

a

=1 for low angle boundary

=3 for twin boundary

=36.87

Boundaries with low has lower energy

The magnitude of for the case shown in slide 6 is 5. This represents the boundary between two grains where the lattice of one is rotated by 36.87 degree about its cube axis with respect to the other. The

slide 6 gives the magnitudes of for a twin boundary. Boundaries having lower values of have lower energy.

Stacking fault energy:

Stack fault is a surface defect. It denotes a type of boundary. Amongst all forms of boundaries this is the

one whose energy can be estimated easily. We have seen that stacking fault lies between a pair of

Shockley partials. Figure 6 is a schematic representation of a stacking fault between a pair of

dislocations.

Grain 1 Grain 2

Fig 5: The line denotes the boundary between two adjacent

grains of a pure metal. The grids are the planar lattices denoting

the way the atoms are arranged on the plane of the

microstructure. Note that there is a definite angular relation

between the two. Imagine that one of these is super imposed

over the other. There will be a few points which would coincide.

Slide 6: The sketch shows two sets of arrays one is

denoted by white dots and the other by red dots.

The distance between the two nearest points is a.

Note that there are a few sites where both the red

and white dots are coincident. This constitutes the

coincident lattice site. The arrows represent the

distance between the two nearest point. Its length

is a. Note the distance between the two nearest

points in coincident site lattice. This consists of

one arrow along x axis & two arrows along y axis.

This is equal to √5 .

2110

6121

6211

Fig 6: A schematic diagram showing how a prefect dislocation

splits into two Shockley partials separated by a region having a

stacking fault. The slip plane is (111) & the crystal structure is

FCC.


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The distance (d) between the two partials is inversely proportional to the stacking fault energy (). Stacking faults are visible under TEM. It exhibits stacking fault fringes. The relation between the two is

given by the following equation, where is the angle between the Burgers vectors of the two partials.

(4)

This method works well for metals having low stacking fault energy (example: Cu), where the separation

between the partials is large. In metals like Al whose SFE is high the separation between partials is less.

In such cases more a precise estimation can be obtained from the measurement of radius of curvatures

of complex dislocation networks. Cross slip is more prevalent in metals having high stacking fault energy.

The partials must join together to move over to the cross slip plane and then split again forming a

constriction. This is how a complex dislocation structure may develop with several nodes. A typical

example is shown in slide 7.

A

A AB

C D

Estimation of stacking fault energy in FCC crystal

R C DC

B

C DC

B

2T Gb

R R

Relation between dislocation density & plastic deformation (strain):

Plastic deformation occurs due to glide motion of dislocations. There are large numbers of dislocations

present in metals even in its annealed state. They can be seen under TEM or as etch pits within the

grains in an optical microscope. Density of dislocations is represented as the total length of dislocations

in a unit volume or as the number of intersections it makes on unit area of an intersecting plane. How do

we relate the movement of these to measurable macroscopic strain?

Slide 7: The sketch on the left gives the notations used

in Thompson Tetrahedron to denote Burgers vector.

The dislocation structure that develops as a result of

intersection of a dislocation with Burgers vector DC in

plane with one on plane with Burgers vector CB. Initially these are split into partials in the respective

planes. When they intersect they must join together to

form a constriction. Subsequently CB cross slips and

dissociate in plane resulting in the formation of a set

of partials separated by stacking fault. SFE can be

obtained from the radius of curvature as shown.

X1

X2

Fig 7: A sketch showing several dislocations in a crystal. If a dislocation

glides through a distance X1 it develops a displacement of magnitude b.

Let it move through a distance i. The total displacement is ib/X1. If this is divided by height of the crystal it gives the magnitude of shear strain due

the movement of the ith dislocation.


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Figure 7 presents a schematic diagram of a crystal having several dislocations. Let X1 be the average

length of the slip plane and X2 is the height of the crystal where the displacement is measured. On the

basis of the reason given in the caption of fig 7 shear strain i is given by:

(5)

Total strain ∑ ∑ (6)

Note that the summation extends over all the mobile dislocations; n is the total number of dislocations;

δis the average glide distance of dislocations and (=n/(X1X2)) is the dislocation density (number of

dislocation in unit area). Since the shear strain and tensile strains are directly related the relation

between plastic strain () and dislocation density can be given by:

(7)

Note that average contribution to tensile displacement is . Differentiating equation 7 with time we get

the frequently used relation between strain rate, dislocation density and dislocation velocity (v).

(8)

Relation between shear strength and dislocation density:

Dislocation density increases with plastic deformation. Often these are arranged in the form of a

network. Higher the dislocation density higher is the number of dislocation nodes and shorter is the

average dislocation link length. Figure 8 shows the sketch of a dislocation network.

Average link length can be easily related to dislocation density. If is the dislocation density is the

average distance between dislocations. This in turn is equal to the link length. Therefore the shear

strength of metal is given by

(9)

This shows that strength of metal is proportional to the square root of dislocation density.

An estimate of the stress required for dislocation glide:

Slip or glide is the most common mechanism of plastic deformation of metals. We know that metals are

made of crystals where atoms are arranged in a periodic fashion. The shear stress () required to displace atoms in a crystal by a small distance (x) on the slip plane along the glide direction is given by:

lFig 8: A part of a typical dislocation network. Average link length of

dislocation is l. Such links may act as Frank Read source.


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(10)

G is the shear modulus and b is the slip vector (distance between the two nearest atoms along the slip

direction). If x < b/ 4 there is a restoring force that would bring the atoms back to its initial position

when the stress is withdrawn. However if x exceeds b/4 all the atoms above the slip plane would move

to the next position of equilibrium. This would result in a permanent deformation. The stress needed for

this to happen is GNo metal can withstand such a high stress. This is why the concept of dislocation

was introduced. What is the magnitude that dislocation would need to glide on a slip plane?

A crude approximation comes from the way the atoms are arranged around an edge dislocation. The

part of the crystal above the glide plane has an extra plane of atoms. Whereas the part below has a

plane where there is no atom. Therefore the atoms above the glide plane are under compressive stress

and those beneath the plane are under tensile stress. When the dislocation moves the work done by the

top half is positive and that by the bottom half is negative. Thus the net work done to move the

dislocation is zero. So is the stress. This certainly does not explain why we need a certain amount of

stress for the deformation to take place.

One of the earliest and the simplest estimate came from Peierls & Nabarro. Slide 8 introduces a concept

of dislocation width which features in the expression derived by them.

Peierls stress: stress to move a dislocation in a periodic lattice

2sin

2

G x

b

-b/4<x<b/4

0 2expp

wG

b

x

w

Assuming a simple sinusoidal force displacement relation and the concept of dislocation width the

following expression for the shear strength (p) was derived by Peierls.

≅ (11)

If hard ball approximation works in describing material behavior, w is large. If atoms are soft w is less.

For most FCC metals w is of the order of 5b. This predicts extremely low shear stress needed to move

dislocations. As against this there are solids where w is much less. They have extremely high strength.

They are difficult to deform (Example Alumina).

Summary

Slide 8: The dotted lines denote perfect lattice. The points

of intersection are the locations of atoms. The firm

vertical lines denote planes of atoms in a lattice having an

edge dislocation. There is an extra half plane of atoms. In

the presence of this, the planes near the dislocation are

distorted. This persists over a distance w. This is called the

width of the dislocation. Let x be the distance between the dotted and the firm line as shown in the sketch. The

width is the length of the region over which x lies between ±b/4.


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In this module we looked at the experimental evidences for the presence of crystal defects. Etch pits in

metals often seen under optical microscope are due to the presence of dislocations. Examination of thin

metal foils under TEM shows direct evidence of the presence of various types of defects like, dislocation

& stacking faults. The resolution of TEM has improved significantly over the years. With the availability

of atomic level resolution it is now possible to reveal atomic arrangements around a dislocation. You

may refer to a recent publication by Chien‐Chun Chen et al in Nature VOL 496 (2013) 74‐77. Introduction

to Dislocations by Derek Hull also has images of atomic arrangements around an edge dislocation

observed by Field Ion (Atom Probe) Microscope.

We have also seen how the presence of vacancies can be indirectly derived from the measurement of

electrical resistivity. We also looked at the structure of grain boundary and tilt boundary. Rough

estimates of the energies of such boundaries could be estimated. Methods of estimating stacking fault

energy have been introduced. Apart from this a simple relation between the strain, the strain rate and

the strength of metals with dislocation density has been derived. Five modules have been devoted to

introduce the concept of dislocation and other crystal defects that determine the properties of metals.

You would appreciate the usefulness of this concept in subsequent modules particularly those related to

the strengthening of metals and alloys.

Exercise:

1. Estimate the distance between dislocations in a tilt boundary of alumunium if the

misorientation angle is 5⁰. Given lattice parameter of Al = 0.405nm. Crystal structure is fcc.

2. A more precise expression for low energy grain biundary is given by

where A is an constant. This is valid over the range 0 < < 10⁰. Find a reasonable estimate of

A. Given lattice parameter of Ni (fcc) = 0.35nm, G = 76MPa Poisson ratio = 0.3 (Hint: assume

dislocation core radius as 5b & the minimum distance between dislocation to be twice this. The

doslocation core energy )

3. Estimate the dislocation spacing and energy of a low angle boundary in copper crystal (fcc b =

0.25nm) if tilt angle = 1⁰. Given G = 48MPa & = 0.3

4. Use the expression given in problem 2 to find our the tilt angle (max) at which the enegry of a

low angle boundary is the maximum. Hence show that 1

5. Estimate the energy of the free surface of polycrystalline copper from its heat of sublimation.

Does this vary from grain to grain? Given Ls = 338 kJ/mole; a = 0.36 nm

Answer:


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1. Burgers vector of a dislocation in a tilt boundary = 110√

.

√0.29 The spacing

between the two dislocations is given by.

180 3.32

2. Burgers vector =√

.

√0.25 When the dislocations are 10b apart energy of the low

angle boundary = (since boundary consists of one dislocation of unit length at every distance

of h). h = 2r0 =10b. Thus where = b/10b = 0.1rad & A

=‐1.42

3. Since poisson ratio is same as in the previous problem 1.42

.

. 1.42 0.13 J/m2

4. Differentiating the expression for E: 0 Thus 1 On

substituting the magtitude of A from the previous problem 5.1⁰ &

(note A = 1+ln max) 1 1

5. Energy of the free surface depends on the way atoms are arranged. This varies from grain to

grain depending on their orientations. If Z is the cordination number, the number of bonds of

type AA in one mole of pure metal = where N0 is Avogrado number. If is the energy of

one bond, where Ls is heat of sublimation. The free surface has a set of broken

bonds. Energy of a broken bond is approximately /2. The number depends on the indices of the

top surface. If it were (111) there will be 3 broken bonds / atom (There are 6 bonds on the plane

3 beneath & 3 above it). Energy of free surface is therefore = 3/2 J/atom. If na is number of

atom / unit area surcae free energy The arrangements of atom in (111) plane is

shown below. On substituion in expression for √

√ 2.5 /

√2

60⁰

0.5

0.5√2

60

4

√3

module 14 crystal defects in metals v - nptelnptel.ac.in/courses/113105023/lecture14.pdf · module...

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