Engineering Metallurgy

MM207Pratik Babhulkar(140100013)

Importance of Material Science

• We come across number of materials in our day to day life. • In various fields of engineering, one has to select material to be used.• One needs to know the material properties like strength, density and some special characteristics. • Thus it is important for us to learn the material science.

Approx Classification of Materials

Metals

Ceramics

Elastomers

Glasses

Polymers Hybrids

•Materials are broadly classified in these categories

•Hybrids are the materials obtained from composition of two or more types of materials so that we can incorporate best of these materials in this hybrid

•Comparison of some properties of these materials is shown in next slides

Density of various materials

Young’s Modulus comparison

Tensile Strength Comparison

Metals

• Materials in this group are composed of one or more metallic elements (e.g., iron,

▪ aluminum, copper, titanium, gold, and nickel), and often also nonmetallic elements

▪ (e.g., carbon, nitrogen, and oxygen) in relatively small amounts.

• These are the crystalline materials in which ions are indirectly connected through the field of free electrons surrounding them.

Ceramics

▪ Ceramic is an inorganic, non metallic solid with crystalline, partly crystalline or amorphous structure.

▪ Ceramics are compounds between metallic and non-metallic elements; they are most frequently oxides, nitrides, and carbides.

Properties

▪ High elastic moduli

▪ Brittle.

▪ Tensile (Fracture) strength is less than compressive (Crushing) strength.

Polymers

▪ Polymer is a large molecule composed of repeating structural units.

▪ Many of them are organic compounds that are chemically based on carbon, hydrogen, and other non-metallic elements

Properties

▪ Low elastic moduli

▪ Large elastic deflections

Elastomers

Elastomers are the material which have low elastic moduli, which show great extensibility and flexibility when it is stressed but return to original dimensions or almost so when deforming stress is removed.

They are a type of polymer. They are amorphous in nature.

Properties

▪ Young’s moduli are too low

Metallic foam(Advanced material)

o A metal foam is a cellular structure consisting of a solid metal, frequently aluminium, as well as a large volume fraction of gas-filled pores.

o The pores can be sealed (closed-cell foam), or they can form an interconnected network (open-cell foam).

Uses

▪ Can be used in aerospace applications, prosthetics, design, architecture.

Hierarchy of metallurgical length scales

AtomicStructur

e(0.1 nm-

1 nm)

Nano-

structur

e(1

nm-100 nm)

Micro-

structur

e(0.1 µm-100µm)

Macro-

structur

e(0.1 mm-10mm)

Dimensional Range-Scientific and engineering disciplines

Time and distance dimensional ranges

Stress vs Strain

•The graph shown aside shows stress strain behaviour of a typical material•The linear part upto A is the elastic region where the material obeys the hooke’s law. A is the proportional limit.•After A the non-linear deformation starts. i.e. Plastic deformation starts. The process just after elastic region is called yielding.•Then stress necessary to continue plastic deformation increases and eventually reaches its maximum called Ultimate stress.

A standard tensile specimen with circular cross section

All the deformation until ultimate stress is uniform throughout the tensile test specimen.At the ultimate stress a small neck (constriction) occurs.Further deformation gets confined to the neck area and eventually the sample fractures. The fracture strength corresponds to the stress during the fracture.

True stress-strain vs engineering stress-strain

oAs we reach the ultimate stress, the deformation is confined to neck area and thus the actual instantaneous area (A) is decreasing rapidly.oSo the stress associated with the force after considering this cross sectional area is called TRUE STRESS.oTrue Strain is instantaneous strain. oTrue stress and strain are defined in next page

Engineering stress is the stress by considering original cross sectional area of the specimen.S- Engineering stress E-engineering strainɛ- True strain σ-True stress

True stress=σ=S(1+E)

True strain=ɛ=ln(1+E)

S=P/A P→LoadA →Original area of cross section The upper one

shows the true stress and lower one is engineering stress

Bonding forces and Energies

▪ While two atoms are approaching, the net force between them is summation of attractive and repulsive forces.

▪ When the forces balance each other, net force is zero and atoms are in equilibrium at some specific inter-atomic distance.

▪ Thus the atomic bond is formed. The potential energy of this system is related to the forces as:

•The dependence of force and potential energy on inter-atomic distance is shown by this graph.•The dotted lines show individual attractive and repulsive energies and forces.•The bond is formed at minimum potential energy or zero net force.

Elastic deformation at Atomic scale

•On an atomic scale, macroscopic elastic strain is manifested as small changes in the inter-atomic spacing and the stretching of inter-atomic bonds.•The magnitude of the modulus of elasticity is a measure of the resistanceto separation of adjacent atoms, that is, the inter-atomic bonding forces.•Hence ,

Strength vs Density Specific strength=Strength/Weight

Metals

▪ Atomic Structure (key points)

1. Metals have loosely bound valence electrons.

2. They are easily available and referred as free electrons.

▪ Atomic Bonding in Solid

1. The type of bond in metals is metallic bond. In metals, valence electrons are not bound to any particular atom in the solid and are more or less free to drift throughout the entire metal.

2. Thus these electrons create a pool (electron cloud) of negative charge.

•The remaining non-valence electrons and atomic nuclei form what are called ion cores, which possess a net positive charge equal in magnitude to the total valence electron charge per atom

•The sea of electrons shields the positive core from repulsive forces.

•Again it also acts as glue by creating a strong attractive force with positive cores.

•Those makes the metal strong.

Crystal Structure of metals

•The atoms of metallic substances are closely positioned to neighbouring atoms in one of two common arrangements.

•The first arrangement is known as body-centred cubic. In this arrangement, each atom is positioned at the centre of eight others.

•The other is known as face-centred cubic. In this arrangement, each atom is positioned in the centre of six faces.

Bravaic Lattice

A crystalline material is one in which the atoms are situated in a repeating or periodic array over large atomic distances. There are following 7 broad ways in which atoms are found arranged in nature.

The 7 lattice systems are :

* Cubic * Hexagonal

* Rhombohedral * Triclinic

* Tetragonal * Orthorhombic

* Monoclinic

Cubic lattice

Simple cubic

Body centred

Face-centred

Effective no of atoms per unit cell=1

Packing factor=52.4%

Co-ordination no=6

Effective no of atoms per unit cell=2

Packing factor=68%

Co-ordination no=8

Effective no of atoms per unit cell=4

Packing factor=74%

Co-ordination no=12

Effective no of atoms per unit cell=6

Packing factor=74%

Co-ordination no=12

Hexagonal Triclinic Rhombohedral

Tetragonal (Simple)

Tetragonal –Body centred

Orthorhombic Crystal

Simple Base-centred Body centred Face centred

Monoclinic Crystal

Base centredSimple

Engineering Metallurgy

PPT 3-5MM207140100013Pratik Babhulkar

Content

▪ Polymorphic transformation

▪ Stacking sequence

▪ Effect of temperature on Young’s Modulus (Modulus of elasticity)

▪ Miller Indices

Polymorphic transformations

•δ-iron: When iron first solidifies at 1538 °C, it is in B.C.C. (body-centred cubic)•γ-iron: As it is cooled, it changes to F.C.C.  (912 ° C – 1394 °C)•α-iron: Then it becomes B.C.C. (non-magnetic) •α-iron: Becomes magnetic at 768 °C (Curie’s Temperature for iron)•Curie’s Temperature: Ferromagnetic atom becomes paramagnetic at this temperature•Lattice Parameter: Unit dimension of crystal structure

Temperature range(°C)

Name Lattice parameter

(Å)

Crystal Structur

e

Upto 768α-Iron(Ferr-

omagnetic)

2.86 BCC

768-912α-Iron(Para-

magnetic)2.86 BCC

912-1394 γ-Iron 3.65 FCC

1394-1538(Until it liqifies) δ-Iron 2.93 BCC

Stacking sequence(3D)

ABCABC......ABAB......

ABAB... Packing creates the hexagonal closed packing while ABCABC... Packing creates Face-centred cubic packing

Radius ratio Vs Possible co-ordination no Vs Possible geometry prediction

• Generally , Elastic modulus of a material decreases on increasing its temperature.

Young’s modulus dependence on temperature

Young’s modulus dependence on temperature for different materials

PointsDirections

Planes

Miller Indices

Indexing the Points

The position of any point located within a unit cell may be specified in terms of its coordinates as fractional multiples of the unit cell edge lengths (i.e., in terms of a, b,and c).Thus, the position of P is designatedusing coordinates q r s with values that are less than or equal to unity.Do not separate these coordinates by commas.The example of how the point co-ordinates are assigned is shown in next slide

The manner in which the q, r,and s coordinates at point P within theunit cell are determined is shown in this figureThe q coordinate(which is a fraction) corresponds to thedistance qa along the x axis, where a isthe unit cell edge length.

The respective r and s coordinates for the y and z axes are determined similarly. The co-ordinates of some special points are given in next slide

Point Number from above

figure

Fractional lengths Point co-ordinates

X axis Y-axis Z-axis

1 0 0 0 0 0 0

2 1 0 0 1 0 0

3 1 1 0 1 1 0

4 0 1 0 0 1 0

5 ½ ½ ½ ½ ½ ½

6 0 0 1 0 0 1

7 1 0 1 1 0 1

8 1 1 1 1 1 1

9 0 1 1 0 1 1

The table given below shows the fractional lengths and point co-ordinates of some special points in cubic crystal.

Crystallographic direction

A crystallographic direction is defined as a line between two points, or a vector.The following steps are used to determine the three directional indices:1. A vector of convenient length is positioned such

that it passes through the origin of the coordinate system. Any vector may be translated throughout the crystal lattice without alteration, if parallelism is maintained.

2. The length of the vector projection on each of the three axes is determined; these are measured in terms of the unit cell dimensions a, b, and c.

3. These three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values.

4. The three indices, not separated by commas, are enclosed in square brackets, thus: [u v w]. The u, v, and w integers correspond to the reduced projections along the x, y, and z axes, respectively.

5. negative indices are also possible, which are represented by a bar over the appropriate index.

6. e.g. [1 1 1] and [1 1 1] are directly opposite directions. This figure shows some common

directions like •edge of the unit cell [1 0 0]•Diagonal of the face [1 1 0]•Body diagonal of the unit cell [1 1 1]

Crystallographic Planes

Crystallographic planes are specified by three Miller indices as (hkl). The procedure used to determine the h, k, and l index numbers is as follows:1. If the plane passes through the selected origin,

either another parallel plane must be constructed within the unit cell by an appropriate translation, or a new origin must be established at the corner of another unit cell.

2. At this point the crystallographic plane either intersects or parallels each of the three axes; the length of the planar intercept for each axis is determined in terms of the lattice parameters a, b, and c.

3. The reciprocals of these numbers are taken. A plane that parallels an axis may be considered to have an infinite intercept

4. If necessary, these three numbers are changed to the set of smallest integers by multiplication or division by a common factor

5. Finally, the integer indices, not separated by commas, are enclosed within parentheses like (hkl).

6. negative side of the origin is indicated by a bar or minus sign positioned over the appropriate index

7. Reversing the digits (Sign) gives another parallel plane on opposite side and both these planes are equidistance from origin.

Some of the special planes are shown in next slide. Parallel planes have same miller indices as shown in next slide.

Some special planes

Surface of the unit cell(0 0 1)

Plane is parallel to X and Y axis

So intercepts are (∞,∞,1), hence the plane is

represented by (0 0 1)

(1 1 0)Plane is parallel to

Z axisintercepts are

(1,1,∞), hence the plane is

represented by (1 1 0)

(1 1 1)intercepts are

(1,1,1), hence the plane is

represented by (1 1 1)

Family of directionsFor some crystal, several non-parallel directions with different indices are actually equivalent.For example, in Cubic [1 0 0] , [0 1 0], [0 0 1] and negatives of these are equivalent.Such directions are grouped in a family.They are represented by enclosing in triangular brackets

<1 0 0>

For some crystal, several non-parallel planes with different indices are actually equivalent. The arrangement of atoms is same.For example, in Cubic [1 0 0] , [0 1 0], [0 0 1] and negatives of these are equivalent.Such directions are grouped in a family.They are represented by enclosing in curly brackets

{1 0 0}

Family of Planes

Linear density in crystal structure

Directional equivalency is related to linear density in the sense that, for a particular material, equivalent directions have identical linear densities. Linear density (LD) is defined as the number of atoms per unit length whose centres lie on the direction vector for a specific crystallographic direction.

Units of linear density are reciprocal length (e.g., nm-1 , m-1) An example is shown in next slide.

LD= Number of atoms centred on the direction vector Length of direction vector

Though here, there are three atoms centred on the line [1 1 0], atoms X and Y are half on the line. Thus, for linear density, we need to consider the linear image of the system.In other words, along this direction [1 1 0], X and Z are shared with one more unit cell respectively. So, there is ½ share of X and Z in this unit cell along this direction. Thus,

Planar density in crystal structure

Planes having the same planar density values are also equivalent. Planar density (PD) is taken as the number of atomsper unit area that are centred on a particular crystallographic plane

Units of planar density are nm-2 , m-2

An example is shown in next slide.

PD= Number of atoms centred on the PlaneArea of the plane

Here, we are finding planar density for FCC for the plane shown in figure.It is (1 1 0).There are 6 atoms centred on this plane. But, all the corner atoms lie in 3 neighbouring unit cells and are centred on (1 1 0) itself. Thus they are shared in 4 unit cells when we consider (1 1 0).Hence , no of atoms= 2*1/2+4*1/4=2 atoms

Now considering the dimensions in terms of R, length of that plane is 4R while the breadth is edge length of the crystal i.e. 2 sqrt(2) R.

Miller Bravais scheme for hexagonal structure

▪ In hexagonal crystal lattice some crystallographic equivalent directions will not have the same set of indices.

▪ This problem is solved by using this Co-ordinate scheme which involves 4 co-ordinate axes.

▪ The three a1, a2, and a3 axes are all contained within a single plane (called the basal plane) and are at 120 angles to one another. The z axis is perpendicular to this basal plane.

▪ Directional indices are shown by 4 digits like [u v t w] where u,v,t,w are the projections along a1, a2, a3 and z axis respectively.

Conversion from the three-index system to the four-index system

•The conversion from 3 axes co-ordinate system to 4 index system is done using these formula.•u’,v’ and w’ represent the three indices system while u,v,t,w represent four index system.•The u’, v’, w’ are given by conventional way and then u, v, w, t are found by these formula.

Reduced-scale coordinate axissystem for hexagonal

•A grid is constructed on the basal plane that consists of sets of lines parallel to each of the a1, a2, and a3 axes.•Z-axis is also trisected at points m and n.•This scale is referred as reduced scale co-ordinate system.•While using this system, we use a/3, c/3 instead of a and c. (Lattice parameters).

Planes in hexagonal crystal

•The plane is represented by four indices like (hkil).•These indices are determined in the similar way by finding the intercepts along a1, a2, z-axis.•And then the reciprocal is taken.•The index ‘i‘ is found by summation of h and k index as follows,

•These are represented by nearest integers.

Directions and Planes in Hexagonal crystal

Perfect crystal

•A perfect crystal has all the atoms situated at there designated place.•There are no vacancies or other crystallographic imperfections.•All the atoms are in their minimum potential energy state so that they are not dislocated.•An ideally perfect crystal is not possible in reality.

Slip in perfect lattice

▪ We can make reasonable estimate of shear stress required for slip in perfect lattice if we consider that the slip occurs by the translation of one plane of atoms over the another.

▪ Shear stress acts in slip plane along the slip direction.

▪ b is the distance between atoms in one slip plane, a is the distance between the two adjacent slip planes and x is the distance over which the plane has slipped.

▪ The stress is zero when two layers are co-incident and also when they move distance “b” relative to each other. It is zero in the midway between two atoms as well.

▪ An approximation of this is shown in next slides.

•The shear stress can be assumed to be sinusoidally realated to the displacement. So,

τm is the maximum shear stress and b is the period.•At small displacements, hooke’s law will apply ,

•For small small x/b,

•Combining these equations, the expression for the maximum shear stress at which slip should occur is ,

•Assuming b=a, shear strength of a perfect crystal is approximately ,

Relationship between shearing stress and displacement of the layer.

Imperfection in crystals

Point defects(Types)

Self interstitial

Interstitial impurity

Substitutional impurity Vacancy

Divacancy

Vacancy Defects

•The equilibrium number of vacancies for a given quantity of material depends on and increases with temperature according to equation,

•N is total number of atomic sites, Nv is total no of vacancies,Qa is the activation energy for the vacancy to be formed, k is Boltzman constant, T is absolute temperature.•As the temperature increases, energy of the atoms increases and they are no more in their minimum potential energy level and leave their locations.

•Vacancy is the simple defect where an atom is not present at the given place•Interstitial Impurity: In this case some foreign atom occupies the interstitial place.•Self interstitial: In this case the same atom dislocates and occupies the interstitial location.

Substitution impurity : In this case, some foreign atom occupies the lattice point.Divacancy: In this case, the two atoms leave their locations.

Shottky defect: In this effect, equal number of cations and anions are missing from their sites.Frenkel defect: In this effect, cation leaves the atomic site and occupies an interstitial space.

Diffusion in ionic crystals

There are four types:

Vc = no of cation vacancies VA =no of anion vacancies

1. Vc =VA , Schottky defect

2. No of interstitial Anions=VA , Anion Frenkel Pair

3. No of interstitial Cations= Vc , Cation Frenkel pair.

4. Interstitial Cations= Interstitial Anions , Anti- Schotky defect.

Line defects(Dislocations)

•A dislocation is a linear or one-dimensional defect around which some of the atoms are misaligned.•Following are the main types of dislocation

Dislocations

Edge dislocationScrew

dislocation

Mixed Dislocati

on

Edge dislocation

•An extra portion of a plane of atoms, or half-plane, the edge of which terminates within the crystal.This is termed an edge dislocation•It is a linear defect that centers on the line that is defined along the end of the extra half-plane of atoms.•This is sometimes termed the dislocation line

Sometimes theedge dislocation in diagram is represented by the symbol” “ which also indicatesthe position of the dislocation line. An edge dislocation may also be formed by anextra half-plane of atoms that is included in the bottom portion of the crystal; itsdesignation is “Т” .

Burgers Vector

The Burgers vector of a dislocation is a crystal vector specified by miller indices, that quantifies the difference between the distorted lattice around dislocation and the perfect lattice. Equivalently, the Burgers vector denotes the direction and magnitude of atomic displacement that occurs when dislocation moves. This image gives idea what the burger’s vector is.

This is how Burger’s vector is defined to give the dislocation

Classification of dislocation

Dislocations

Edge dislocationScrew

dislocation

Mixed Dislocatio

n

Edge dislocation:• extra half-plane of atoms inserted in a crystal structure•b the berger’s vector is perpendicular to dislocationLine

Screw dislocation:•spiral planar ramp resulting from shear deformation•b is parallel to dislocation line

Burger’s vector, b: is a measure of lattice distortion and ismeasured as a distance along the close packed directionsin the lattice

Relative plane motion in Edge Dislocation under application of Shear stress

The spiral stacking ofcrystal planes leads to the Burgers vectorbeing parallel to the dislocation line. This is Screw dislocation.

Mixed dislocationThis dislocation has bothedge and screw character with a singleBurgers vector consistent with the pure edgeand pure screw regions.

Roll of Dislocation

Deformation Processes

Slip Fracture

Fatigue

Creep

Structural

Incoherent twin

Grain Boundary

Semi-Coherent interfaces

Disc of vacancies

Kinetics

Diffusion

Crystal Growth

Further ways of creep mechanism are in next slide

Creep mechanism

Cross Slip

Dislocation Climb

Vacancy diffusion

Grain Boundary Sliding

Plastic Deformation

Slip

Twinning Phase transformation

Creep mechanism

Plastic deformation occurs by following ways

Grain Boundary sliding

What are grains and grain boundaries?

Upon completion of solidification, grains having irregular shapes haveFormed. This is how they would look. Crystallographic orientation of grains is different

The grain structure as it would appear under the microscope; dark lines are the grain boundaries.

There exists some atomic mismatch within the region where two grainsmeet; this area, called a grain boundary

This figure shows small and high angle of misorientation .Small angle of misorientation results due to edge dislocation as shown in next slide.This is also known as tilt boundary

Demonstration of how a tilt boundary having an angle of misorientation results from analignment of edge dislocations

A twin boundary is a special type of grain boundary across which there is a specificmirror lattice symmetry

Grain boundary sliding

Grain Boundary Sliding is one of the deformation mechanisms of materials which includes displacement of grains against each other The main mechanism of grain boundary sliding is the motion of dislocations by glide and climb.

Types of dislocation movement

Dislocation Movement

Glide

Climb

Dislocation occurs along the surface which contains its line and burgers vector. This is conservative motion.

non-conservative motion. Occurs when the dislocationmoves out of the glide surface, and thus normal to the Burgers vector

Climb vs Glide

+ve climb

-ve climb

Gliding along the plane having dislocation line as well as burgers vector

Slip

▪ Glide of many dislocations results in slip, which is the most common manifestation of plastic deformation in crystalline solids.

▪ This is single edge dislocation moving

▪ When large number of such dislocations move, it is slip as shown below

Motion of a plane

•For edge dislocation

• We have seen that there is a critical stress required to move a dislocation.• At the fundamental level the motion of a dislocation involves the

rearrangement of bonds requires application of shear stress on the slip plane.

• When ‘sufficient stress’ is applied the dislocation moves from one metastable energy minimum to another.

• Slip occurs along the plane with higher planar densities.• In above formula ,

b

w

PN eG

2

G → shear modulus of the crystal w → width of the dislocation b → |b|

Bubble raft model of dislocation

Bubble rafts demonstrate materials' microstructural and atomic length-scale behaviour by modelling the {111} plane of a close-packed crystal.These images give the demo of dislocations. Ref: https://youtu.be/-qcys9XuNe8

Questions

What are the indices for the directions indicatedby the two vectors in this sketch?

For direction 1:Projection along x=0Projection along y=(1/2)bProjection along z=c(a,b,c are edge lengths along x,y,z axes.)So, indices are 0 ½ 1After multiplying by 2, the direction has indices [0 1 2].

For direction 2: Projection along x=(1/2) aProjection along y=(1/2)bProjection along z= -c(a,b,c are edge lengths along x,y,z

axes.)So direction is: [2 1 2]

Q).What is grain boundary ? Explain what is twin grain boundary.Ans:

•When the solid is formed by solidification, grains of irregular shapes are formed.•These grains are oriented randomly in the solid. So, their crystallographic orientation might be different.•Hence, the region between such two grains has atomic mismatching.• Grain boundary is the region between two grains where atomic mismatch occurs.

•A twin boundary is a special type of grain boundary across which there is a specific mirror lattice symmetry.• Atoms on one side of the boundary are located at the mirror image position of other side and vice versa.