3/15/2014 - M5zn · 3/15/2014 5 9 Metallic Crystal Structures Three relatively simple crystal...

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This course gives :

1.An introduction to the structure of metals, alloys and ceramics

in the solid state.

2.Full description for (Lattice systems, Crystals and their

defects, Solid solution, Phase diagrams,…).

3.Short description of the fundamentals of metals and alloys

processing (Solidification, Deformation, Diffusion), followed by

reviewing some of commercial metallic alloys (Steels, Cast

irons, Magnetic, Aluminum and Magnesium alloys).

4.Attention is devoted to the relationship between the

microstructures, processing, the physical properties and

performance of some metals and alloys. 2

Aims of this Course

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SubjectNumber of

hours

1. Introduction to the Crystal Structure of Metals

and Alloys6 h

2. Defects in Crystals 3 h

3. Alloys and Phase Diagrams 6 h

4. Processing of Metals and Alloys (Solidification

structures and basics of deformation )3 h

5. Properties of metals and alloys (mechanical,

electrical, thermal, magnetic and optical)9 h

6. Example for any binary alloy systems 6 h

7. Introduction to Ceramics 6 h

Lectures List

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Chapter 1:

The Structure of Crystalline Solids

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Sources & References: Materials Science and Engineering , 7th edition, William D. Callister, Jr. , 2007.Alloy Physics, Wolfgang Pfeiler , 2007.The Structures of Metals and Alloys, WILLIAM HUME-ROTHERY , G.V. RAYNOR , 1962.Nanostructured metals and alloys, Sung H. Whang , 2011.

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1. Describe the difference, in atomic/molecular structure, betweencrystalline and non-crystalline (amorphous) materials.

2. Draw unit cells for simple cubic (SC), face-centered cubic (FCC),body-centered cubic (BCC), and hexagonal close-packed (HCP) crystalstructures.

3. Derive the relationships between unit cell edge length and atomicradius for SC, FCC, BCC and HCP crystal structures.

4. Compute, theoretically, the densities for metals having FCC and BCCcrystal structures given their unit cell dimensions.

5. By given three direction index integers, sketch the directioncorresponding to these indices within a unit cell.

6. Specify the Miller indices for a plane that has been drawn within a unitcell.

7. Distinguish between single crystals and polycrystalline materials.

8. Determine the crystal structure by using X-ray diffraction (XRD).

Objectives of Chapter: 1

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After studying this chapter you should be able to do the following:

Notes:•When describing crystalline structures, atoms (or ions) arethought of as being solid spheres having well-defineddiameters.

•In the present chapter, we will discuss and deals withseveral common metallic crystal structures.

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Introduction and Fundamental Concepts

A crystalline material is one in which the atoms arearranged in a highly ordered manner relative to each other.

Non-crystalline or amorphous materials are notcrystallized; or the atoms are randomly arranged.

Why we studying crystal structures?Some of the properties of crystalline solids depend on thecrystal structure of the material, the manner in which atoms,ions, or molecules are spatially arranged.

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In this particular case, Hard sphere model, all the atoms

are identical.

Sometimes the term lattice is used in the context of crystalstructures; in this sense “lattice” means a three-dimensional array of points coinciding with atom positions(or sphere centers).

The Hard Sphere Model

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Unit Cells

The atomic order in crystalline solids indicates that smallgroups of atoms form a repetitive pattern.

Thus, in describing crystal structures, it is oftenconvenient to subdivide the structure into small repeatentities called unit cells.

The unit cell is the basic structural unit or building blockof the crystal structure and defines the crystal structureby virtue of its geometry and the atom positions within.

Primitive unit Cell: a unit cell that contains only one latticepoint.

Non Primitive Cell: a large cell and contains more that onelattice point

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Metallic Crystal Structures

Three relatively simple crystal structures are found for mostof the common metals: face centered cubic, body-centeredcubic and hexagonal close-packed.

The crystal lattice: is a regular array of points in space orthree-dimensional array of points.

Bravais lattice: all lattice points are equivalent and all atomsin the crystal are of the same kind.

Non Bravais lattice: lattice points are not equivalent.

The atomic packing factor (APF) is defined as the fractionof solid sphere volume in a unit cell, or

The atomic radius (R): The distance between the atomcenter and the atom surface.

The coordination number is the number of nearest-neighbor or touching atoms.

Definitions

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The unit cell geometry is completely defined in terms ofsix parameters (lattice parameters of a crystal structure):the three edge lengths a, b, and c, and the three inter-axial angles α, β and γ .

On this basis there are seven different possiblecombinations of a, b, and c, and α, β and γ each of which

represents a distinct crystal system.

A unit cell with x, y, and z coordinateaxes, showing axial lengths (a, b, andc) and inter-axial angles α, β and γ .

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Crystal Systems

These seven crystal systems are cubic, tetragonal,hexagonal, orthorhombic, rhombohedral, monoclinic, andtriclinic.

Crystal Systems

The lattice parameter relationships and unit cell sketchesfor each are represented in next slides.

The cubic system has the greatest degree of symmetrywhere a = b = c and α = β = Ɣ = 90˚.

The triclinic system has the Least symmetry where a ≠ b≠ c and α ≠ β ≠ Ɣ.

From the discussion of metallic crystal structures, itshould be apparent that both FCC and BCC structuresbelong to the cubic crystal system, whereas HCP fallswithin hexagonal

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Number of PravaisLattices

3(SC, BCC, FCC)

1

2(P, BCC)

*P≡ Primitive ; SC ≡ simple cubic ; BCC ≡ body center cubic ; FCC ≡ face center cubic

Crystal systems and Pravais Lattices

Crystal Systems

13

1

4(P, C, BCC,

FCC)

2(P, C)

1

*C≡ base centered cubic14

Crystal systems and Pravais Lattices

Crystal Systems

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Simple Cubic (SC) Crystal Structure

The simple cubic unit cell is a cube (all sides of the samelength and all face perpendicular to each other) with anatom at each corner of the unit cell.

The coordination number for this structure is 6.Number of atoms per unit Cell is 1.The relation between atomic radius, R, and unit cell length,a, given by: a=2RAtomic packing factor (APF) for this structure is 0.52.

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The Face-Centered Cubic (FCC) Crystal Structure

The coordination number for this structure is 12.Number of atoms per unit Cell is 4.The relation between atomic radius, R, and unit cell length, a, given by:

Atomic packing factor (APF) for this structure is 0.74.

The crystal structure found for many metals has a unit cellof cubic geometry, with atoms located at each of thecorners and the centers of all the cube faces. It is aptlycalled the face-centered cubic (FCC) crystal structure.

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The Body-Centered Cubic (BCC) Crystal Structure

The coordination number for this structure is 8.

Number of atoms per unit Cell is 2.

The relation between atomic radius (R) and unit cell length(a) given by:


Another common metallic crystal structure also has a cubicunit cell with atoms located at all eight corners and asingle atom at the cube center. This is called a body-centered cubic (BCC) crystal structure.

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The coordination number for this structure is 12.

Number of atoms per unit Cell is 6.

The relation between atomic radius R and unit cell length given by: a=2R


The ratio between c and a is 1.63.

The Hexagonal Close-Packed (HCP) Crystal Structure

Not all metals have unit cells with cubic symmetry; thefinal common metallic crystal structure to be discussed hasa unit cell that is hexagonal.

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Example Problem: 1

Determination of FCC Unit Cell Volume

Calculate the volume of an FCC unit cell in terms of theatomic radius R.

From the right triangle on the face,

or, solving for a,

The FCC unit cell volume, VC, may be computed from:

Example Problem: 2

Computation of the Atomic Packing Factor for FCC

Show that the atomic packing factor for the FCC crystalstructure is 0.74.

Solution:

Both the total atom volume, VS, and unit cell volume, VC,may be calculated in terms of the atomic radius R.

and

Therefore, the atomic packing factor is:

The APF is defined as the fraction of solid sphere volume in a unit cell, or:

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Density Computations

A knowledge of the crystal structure of a metallic solidpermits computation of its theoretical density through therelationship

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Example Problem: 3

Theoretical Density Computation for Copper

Copper has an atomic radius of 0.128 nm, an FCC crystalstructure, and an atomic weight of 63.5 g/mol. Compute itstheoretical density and compare the answer with itsmeasured density.

The literature value for the density of copper is 8.94 g/cm3, which is in very close agreement with the foregoing result.

Solution:

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Materials of Importance

A tin is a common metal that experiences an allotropic change. White tin, havinga body-centered tetragonal crystal structure at RT, transforms, at 13.2˚C(55.8F), to gray tin, which has diamond cubic crystal structure; thistransformation is represented schematically as follows:

1. What is the difference between atomic structure andcrystal structure?

2. “One can not prepare a perfect crystal”- why?

3. If the atomic radius of lead is 0.175 nm, calculate thevolume of its unit cell in cubic meters.

4. Show for the body-centered cubic crystal structure thatthe unit cell edge length a and the atomic radius R isrelated through .

5. For the HCP crystal structure, show that the ideal ratio is1.633.

6. Show that the atomic packing factor for BCC is 0.68.

7. Show that the atomic packing factor for HCP is 0.74.

Problems on Ch1: The Structure of Crystalline Solids

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8. Molybdenum has a BCC crystal structure, an atomic

radius of 0.1363 nm, and an atomic weight of 95.94

g/mol. Compute and compare its theoretical density with

the experimental value found inside the front cover.

9. Calculate the radius of a palladium atom, given that Pd has

an FCC crystal structure, a density of 12.0 g/cm3, and an

atomic weight of 106.4 g/mol.

10. Calculate the radius of a tantalum atom, given that Ta

has a BCC crystal structure, a density of 16.6 g/cm3, and

an atomic weight of 180.9 g/mol. 25


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11. Some hypothetical metal has the simple cubic crystalstructure shown in Figure below. If its atomic weight is74.5 g/mol and the atomic radius is 0.145 nm, computeits density.

12. Titanium has an HCP crystal structure and a density of4.51 g/cm3.

(a) What is the volume of its unit cell in cubic meters?

(b) If the c/a ratio is 1.58, compute the values of c and a.

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13. Using atomic weight, crystal structure, and atomic radiusdata tabulated inside the front cover, compute thetheoretical densities of aluminum, nickel, magnesium,and tungsten, and then compare these values with themeasured densities listed in this same table. The c/aratio for magnesium is 1.624.

14. Niobium has an atomic radius of 0.1430 nm and a densityof 8.57 g/cm3. Determine whether it has an FCC or BCCcrystal structure.

15. The unit cell for uranium has orthorhombic symmetry,with a, b, and c lattice parameters of 0.286, 0.587, and0.495 nm, respectively. If its density, atomic weight, andatomic radius are 19.05 g/cm3, 238.03 g/mol, and 0.1385nm, respectively, compute the atomic packing factor.

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17. Indium has a tetragonal unit cell for which the a and clattice parameters are 0.459 and 0.495 nm, respectively.

(a) If the atomic packing factor and atomic radius are 0.693and 0.1625 nm, respectively, determine the number ofatoms in each unit cell.

(b) The atomic weight of indium is 114.82 g/mol; compute itstheoretical density. 28

16. Below are listed the atomic weight, density, and atomicradius for three hypothetical alloys.

For each determine whether its crystal structure is FCC,BCC, or simple cubic and then justify your determination. Asimple cubic unit cell is shown in Figure below.


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18. Beryllium has an HCP unit cell for which the ratio of thelattice parameters is 1.568. If the radius of the Be atomis 0.1143 nm,

(a)Determine the unit cell volume.

(b)Calculate the theoretical density of Be and compare itwith the literature value.

19. Magnesium has an HCP crystal structure, a c/a ratio of1.624, and a density of 1.74 g/cm3. Compute the atomicradius for Mg.

20. Cobalt has an HCP crystal structure, an atomic radius of 0.1253 nm, and a c/a ratio of 1.623. Compute the volume of the unit cell for Co.

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21. Below is a unit cell for a hypothetical metal.(a) To which crystal system does this unit cell belong?(b) What would this crystal structure be called?(c) Calculate the density of the material, given that its

atomic weight is 141 g/mol.

22. Sketch a unit cell for the face-centered & orthorhombiccrystal structure.

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Chapter 1:

The Structure of Crystalline Solids

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Sources & References:

Materials Science and Engineering , 7th edition, William D. Callister, Jr. , 2007.

Alloy Physics, Wolfgang Pfeiler , 2007.

The Structures of Metals and Alloys, WILLIAM HUME-ROTHERY , G.V. RAYNOR , 1962.

Nanostructured metals and alloys, Sung H. Whang , 2011.

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1. Determine and sketch crystallographic points, directions, and planes within a

unit cell.

2. Specify the Miller indices for a plane that has been drawn within a unit cell.

3. Define isotropy and anisotropy with respect to material properties.

4. Distinguish between single crystals, polycrystalline and amorphous materials.

5. Determine the crystal structure by using the diffraction pattern of a beam of

radiation incident on the crystal, e.g. X-ray diffraction (XRD).

Objectives of lecture II

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After studying this lecture you should be able to do the following:

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Point Coordinates

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).

To illustrate, consider the unit cell and the point P situated therein as

shown in Figure below.

We specify the position of P in terms of the

generalized coordinates q, r, and s where q

is some fractional length of a along the x

axis and similarly for r and s.

Note that, the values of coordinates q r s

are less than or equal to unity.

The coordinates q r s not separate by

commas or any other punctuation marks.

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Example Problem: 4

Location of Point Having Specified Coordinates

For the unit cell shown in the accompanying sketch (a), locate the

point having coordinates .2

11

4

1

(An

swer

)

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Example Problem: 5

Specification of Point Coordinates

Specify point coordinates for all atom positions for a BCC unit

cell.

(Answer)

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The following steps are utilized for determining of the three

directional indices:

1.We choose one lattice point on the line as an origin (0,0,0).

2.We choose the lattice vector , that join the origin with

another point on the line.

3.The crystal direction is now specified by the triplet [u v w] after

removing the common factor.

4.The three indices, not separated by commas, are enclosed in square

brackets, thus: [u v w].

5.The u, v, and w integers correspond to the reduced projections along

the x, y, and z axes, respectively.

cwbvauR

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Crystallographic Directions

A crystallographic direction is defined as a line or a vector between

two points.

The[100],[110], and [111] directions are

common ones; they are drawn in the unit cell

shown in Figure below.

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Crystallographic Directions

For each of the three axes, there will exist both positive and negative

coordinates and the negative one represented by a bar over the

appropriate index.

For example, the direction would have a component in the -y

direction.

]111[

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Example Problem: 6

Determination of Directional Indices

Determine the indices for the direction shown in the accompanying

figure below.

Answer:

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Example Problem: 7

Construction of Specified Crystallographic Direction

Draw a direction within a cubic unit cell. 011

There is no z component to

the vector, since the z

projection is zero.

There is no z component to

the vector, since the z

projection is zero.

For this direction , the projections along the x, y, and z axes are a, -

a and 0a, respectively.

This direction is defined by a vector passing from the origin to point P,

which is located by first moving along the x axis a units, and from this

position, parallel to the y axis units, as indicated in the figure.

011

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1.Let x, y and z are the intercepts of the plane with the axis along a, b

and c, respectively.

2.Define the triplet (x/a ; y/b ; z/c).

3.Invert to obtain the triplet (a/x ; b/y ; c/z).

4.Multiply by a common factor to obtain Miller indices (hkl).

5.The obtain Miller indices, not separated by commas, are enclosed

within parentheses, thus: (hkl).11

Crystallographic planes

Crystallographic planes are specified by three Miller indices as (hkl).

Any two planes parallel to each other are equivalent and have identical

indices.

The procedure employed in determination of the h, k, and l index

numbers is as follows:

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Example Problem: 8

Determination of Planar (Miller) Indices

What are the indices for the two planes

drawn in the sketch below?

Plane 2 left to you

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Example Problem: 9

Construction of Specified Crystallographic Plane

Construct a (011) plane within a cubic unit cell.

To solve this problem, carry out the procedure used in the preceding

example, slide 12, in reverse order.

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Spacing between planes of the same Miller indices (d-

spacing)

The formula used to calculate the d-spacing (inter-planar distance)

depends on the crystal structure.

2

2

2

2

2

2

1

c

l

b

k

a

hdhkl

For successive planes:

For cubic system a=b=c, or:

222 lkh

adhkl

Prove that, left to you

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Example Problem: 10

d-spacing calculation

Aluminum has FCC cubic structure of lattice constant a=4.04

angstrom, calculate d100, d111, d200 and d220?

222 lkh

adhkl

By using this formula:

Ad 04.4001

04.4

)0()0()1(

04.4

222100

Ad 34.2111

04.4

)1()1()1(

04.4

222111

Ad 02.2004

04.4

)0()0()2(

04.4

222200

Ad 43.1044

04.4

)0()2()2(

04.4

222220

Note that the atomic packing is different for each case. 16

Atomic Arrangements

The atomic arrangement for a crystallographic plane, which is often of

interest, depends on the crystal structure.

The (110) atomic planes for FCC crystal structure

The (110) atomic planes for BCC crystal structure

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Of course, the units of linear density are reciprocal length (e.g., nm-1, m-

1,… ). 17

Linear and planar densities

Linear density (LD): is defined as the number of atoms per unit length

whose centers lie on the direction vector for a specific crystallographic

direction; that is,

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Planar density (PD): is taken as the number of atoms per unit area that are

centered on a particular crystallographic plane:

The units for planar density are reciprocal area (e.g., m-2, nm-2, …).

Linear and planar densities

The corresponding parameter for crystallographic planes is planar

density, and planes having the same planar density values are also

equivalent.

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Crystalline and Non-crystalline Materials

Noncrystalline or amorphous solids lack a systematic and regular

arrangement of atoms over relatively large atomic distances.

Two-dimensional schematic diagrams for both structures of SiO2

A crystalline material is one in which the atoms are arranged in a highly

ordered manner relative to each other.

Metals normally form crystalline solids, but some ceramic materials are

crystalline, whereas others, the inorganic glasses, are amorphous.

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For a crystalline solid, when the periodic

and repeated arrangement of atoms is

perfect or extends throughout the

entirety of the specimen without

interruption, the result is a single crystal

They are ordinarily difficult to grow, because the environment must be

carefully controlled.

All unit cells interlock in the same way and have the same orientation.

Single crystals exist in nature, but they may also be produced

artificially.

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Most crystalline solids are composed of a collection of many small

crystals or grains; such materials are termed polycrystalline.

Initially, small crystals or nuclei form at various positions.

These have random crystallographic orientations, as indicated by the

square grids.

The small grains grow by the successive addition from the surrounding

liquid of atoms to the structure of each.

The extremities of adjacent grains impinge on one another as the

solidification process approaches completion.

There exists some atomic mismatch within the region where two grains

meet; this area, called a grain boundary.


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Schematic diagrams of the various stages in the solidification of a

polycrystalline material; the square grids depict unit cells.

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The physical properties of single crystals of some substances depend on

the crystallographic direction in which measurements are taken.

For example, the elastic modulus, the electrical conductivity, and the

index of refraction may have different values in the [100] and [111]

directions.

This directionality of properties is termed anisotropy, and it is

associated with the variance of atomic or ionic spacing with

crystallographic direction.

Substances in which measured properties are independent of the

direction of measurement are isotropic.

Anisotropy, and isotropic

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X-ray diffraction determination of crystal structures

X-Ray Diffraction and Bragg’s Law

For example, for crystal structures that have

cubic symmetry,

The diffraction of X-rays,

beam of neutrons and

beam of electrons can be

used to determine the

crystal structure of

crystals materials.

X-rays are EM waves whose wave length

around 1 angstrom which of the order of the

lattice constant.

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X-ray diffraction determination of crystal structures

Diffraction pattern for powdered lead.

Schematic diagram of an x-ray diffractometer; T x-

ray source, S specimen, C detector, and O the axis

around which the specimen and detector rotate.

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Inter-planar Spacing and Diffraction Angle Computations

Example Problem: 10

For BCC iron, compute (a) the inter-planar spacing, and (b) the

diffraction angle for the (220) set of planes. The lattice parameter for Fe

is 0.2866 nm. Also, assume that monochromatic radiation having a

wavelength of 0.1790 nm is used, and the order of reflection is 1.

nm 0.10138

0.2866

)0()2()2(

0.2866

222222

lkh

adhkl

1013.02

1790.0)1(

2)sin(

hkld

n Or, 2Ɵ= 124.26˚

Answer:

(a)

(b)

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23.List the point coordinates for all atoms that are associated with the

FCC unit cell.

24.Sketch a tetragonal unit cell, and within that cell indicate locations of

the and point coordinates.

25.Draw an orthorhombic unit cell, and within that cell a direction

. 112

26.Sketch a monoclinic unit cell, and within that cell a direction . 110

2

11

2

1

4

3

2

1

4

1

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27.What are the indices for the directions indicated by the two vectors in

the sketch below?

28.Within a cubic unit cell, sketch the following directions:

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29.Determine the indices for the directions shown in the following cubic

unit cell:

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30.Determine the indices for the directions shown in the following cubic

unit cell:

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31.(a) Draw an orthorhombic unit cell, and within that cell a (210) plane.

(b)Draw a monoclinic unit cell, and within that cell a (002) plane.

32.Sketch within a cubic unit cell the following planes:


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33.Determine the Miller indices for the planes shown in the following

unit cell:

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unit cell:

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unit cell:

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36.(a) Derive linear density expressions for FCC [100] and [111]

directions in terms of the atomic radius R.

(b) Compute and compare linear density values for these same two

directions for silver.

37.(a) Derive linear density expressions for BCC [110] and [111]

directions in terms of the atomic radius R.

(b) Compute and compare linear density values for these same two

directions for tungsten.

38.Explain why the properties of polycrystalline materials are most often

isotropic.

39.Determine the expected diffraction angle for the first-order reflection

from the (113) set of planes for FCC platinum when monochromatic

radiation of wavelength 0.1542 nm is used.

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40.The metal iridium has an FCC crystal structure. If the angle of

diffraction for the (220) set of planes occurs at 69.22° (first-order

reflection) when monochromatic x-radiation having a wavelength of

0.1542 nm is used, compute (a) the interplanar spacing for this set of

planes, and (b) the atomic radius for an iridium atom.

41.For which set of crystallographic planes will a first-order diffraction

peak occur at a diffraction angle of 46.21° for BCC iron when

monochromatic radiation having a wavelength of 0.0711 nm is used?

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42.Below figure shows an x-ray diffraction pattern for α-iron taken using

a diffractometer and monochromatic x-radiation having a wavelength

of 0.1542nm; each diffraction peak on the pattern has been indexed.

Compute the interplanar spacing for each set of planes indexed; also

determine the lattice parameter of Fe for each of the peaks.

Diffraction pattern for polycrystalline Cu.

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43.Figure below shows the first four peaks of the XRD pattern for Cu,

which has an FCC crystal structure; monochromatic x-radiation

having a wavelength of 0.1542 nm was used.

(a)Index (i.e., give h, k, and l indices) for each of these peaks.

(b)Determine the interplanar spacing for each of the peaks.

(c)For each peak, determine the atomic radius for Cu

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Chapter 2:

Imperfections in Solids

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1.Describe both vacancy and self-interstitial crystalline defects.

2.Calculate the equilibrium number of vacancies in a material at some

specified temperatures.

3.Name the two types of solid solutions, and provide a brief written

definition and/or schematic sketch of each.

4.Calculate the weight and atom percent for each element in a metal alloy

by using the masses and atomic weights of elements in a metal alloy.

5.For each of edge, screw, and mixed dislocations:

(a) Describe and make a drawing of the dislocation.

(b) Note the location of the dislocation line.

(c) Indicate the direction along which the dislocation line extends.

6.Describe the atomic structure within the vicinity of:

(a) A grain boundary.

(b) A twin boundary.

7. Determine the average grain size by using SEM images. 3


Learning Objectives

Classification of crystalline imperfections is frequently made according

to geometry or dimensionality of the defect.

There several different imperfections, including point defects (those

associated with one or two atomic positions), dislocations, the external

surface, the grain boundaries, precipitates, large voids and second-

phase particles.

4

Introduction:

An idealized crystal solid does not exist; there is no such thing in

nature.

The properties of some materials are profoundly influenced by the

presence of imperfections and often specific characteristics can be

performed by controlled amounts or numbers of particular defects.

It is important to have a knowledge about the types of imperfections

that exist and the roles they play in affecting the behavior of materials.

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The 3-dimensional (3D) defects change the crystal pattern over a finite

volume, e.g. precipitates, which are small volumes of different crystal

structure, and also large voids or inclusions of second-phase particles.5

It is useful to classify crystal lattice defects by their dimension into

four different types:

The 0-dimensional (0D) defects affect isolated sites in the crystal

structure, and are hence called point defects, an example is a solute or

impurity atom, which alters the crystal pattern at a single point.

The 1-dimensional (1D) defects are called dislocations, they are lines

along which the crystal pattern is broken.

The 2-dimensional (2D) defects are surfaces, such as the external

surface and the grain boundaries along which distinct crystallites are

joined together.

The various defects that are found in crystalline solids

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Point Defects

Intrinsic defects: vacancy and self-interstitial

All crystalline solids contain vacancies

and, in fact, it is not possible to create

such a material that is free of these

defects.

A self-interstitial is an atom from the

crystal that is crowded into an interstitial

site, a small void space that under

ordinary circumstances is not occupied.

An intrinsic defect is formed when an atom is missing from a position

that ought to be filled in the crystal, creating a vacancy, or when an

atom occupies an interstitial site.

A point defect disturbs the crystal pattern at an isolated site.

It is useful to distinguish intrinsic defects from extrinsic defects,

which are caused by solute or impurity atoms.

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7

Point Defects

Intrinsic defects: vacancy and self-interstitial

Self-interstitial are high-energy defects that are relatively uncommon

due to the small interstitial sites in most crystalline solids.

Frenkel defect is equivalent to missing atom that leaves its original site

and migrates to an interstitial position, which this position not normally

occupied by an atom.

Vacancies are present in a significant concentration in all crystalline

materials.

Schottky defect is equivalent to missing atom that leaves its original

site and migrates to surface of the crystal.

Fre

nk

el

defe

ct

Sch

ott

ky

defe

ct

8

Point Defects

Extrinsic defects

The extrinsic point defects are foreign atoms, which are called solutes if

they are intentionally added to the material to form solid solution, and

are called impurities if they are not.

A solid solution forms when, as the solute atoms are added to the host

material, the crystal structure is maintained, and no new structures are

formed.

Impurity point defects are found

in solid solutions, of which there

are two types: substitutional and

interstitial.

For the substitutional type, solute

or impurity atoms replace or

substitute for the host atoms.

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1.Atomic size factor: As the size (atomic radii) difference between two

elements increases, the solid solubility becomes more restricted. For

extensive solid solubility the difference in atomic radii of two elements

should be ±15 %.

2.Crystal structure: For complete solid solubility the crystal structures for

metals of both atom types must be the same, i.e. both atoms have FCC,

BCC or HCP structures.

3.Electronegativity: Electronegativity difference close to 0 gives

maximum solubility, the more electropositive one element and the more

electronegative the other increase the probability to form an inter-

metallic compound instead of a substitutional solid solution.

4.Valences: A metal will have more of a tendency to dissolve another

metal of higher valency than one of a lower valency. 9

Point Defects

Extrinsic defects

There are several features of the solute and solvent atoms that determine

the degree to which the former dissolves in the latter, as follows:

•Carbon, C, forms an interstitial solid solution when added to iron, Fe;

the maximum concentration of C is about 2%.

•The atomic radius of the C atom is much less than that for Fe: 0.071 nm

versus 0.124 nm. 10

Copper, Cu, forms a substitutional solid solution when added to Nickel,

Ni; these two elements are completely soluble in one another at all

proportions.

With regard to the aforementioned rules that govern degree of

solubility, the atomic radii for Cu and Ni are 0.128 and 0.125 nm,

respectively, both have the FCC crystal structure, and their

electronegativities are 1.9 and 1.8 the most common valences are for

Cu (although it sometimes can be ) and for Ni.

Point Defects

Extrinsic defects

An example of a substitutional solid solution

An example of a interstitial solid solution

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11

The necessity of the existence of vacancies is explained using

principles of thermodynamics; the presence of vacancies increases the

entropy (i.e., the randomness) of the crystal.

The equilibrium number of vacancies Nv for a given quantity of

material depends on and increases with temperature according to:

Temperature dependence of the equilibrium number of vacancies

In this expression, N is the total number of atomic sites, Qv is the

activation energy required for the formation of a vacancy, T is the

absolute temperature in kelvins, and k is the Boltzmann’s constant.

Point Defects

12

Example Problem: 1

Number of Vacancies Computation at a Specified Temperature

Calculate the equilibrium number of vacancies per cubic meter for

copper at 1000 ˚C. The energy for vacancy formation is 0.9 eV/atom;

the atomic weight and density (at 1000˚C) for copper are 63.5 g/mol

and 8.4 g/cm3, respectively.

Solution

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13

Specification of composition

It is often necessary to express the composition (or concentration) of

an alloy in terms of its constituent elements.

The two most common ways to specify composition are weight (or

mass) percent and atom percent.

The basis for weight percent (wt%) is the weight of a particular

element relative to the total alloy weight.

For an alloy that contains two different atoms denoted by 1 and 2, the

concentration of 1 in wt%, is defined as C1:

where m1 and m2 represent the weight (or mass) of elements 1 and 2,

respectively.

The concentration of 2 would be computed in an analogous manner, C2:

14

The basis for atom percent (at%) calculations is the number of

moles of an element in relation to the total moles of the elements in the

alloy.

The number of moles in some specified mass of an element 1, nm1, and

an element 2, nm2, may be computed as follows:

Concentration in terms of atom percent of element 1 and 2 in an binary

alloy, are defined by C‘1 and C‘2:

Specification of composition

Here m‘1, m‘2 and A1, A2 denote the mass (in grams) and atomic weight,

respectively, for element 1 and element 2.

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15

Composition Conversions

Sometimes it is necessary to convert from one composition scheme to

another e.g. from weight percent to atom percent.

We will now present equations for making these conversions in terms

of the two hypothetical elements 1 and 2.

Conversion of weight percent to atom percent (for a binary alloy):

Conversion of atom percent to weight percent (for a binary alloy):

Since we are considering only two elements, computations involving

the preceding equations are simplified when it is realized that:

16

In addition, it sometimes becomes necessary to convert concentration

from weight percent to mass of one component per unit volume of

material (i.e., from units of wt% to kg/m3); this latter composition

scheme is often used in diffusion computations.

Concentrations in terms of this basis will be denoted using a double

prime (i.e., C”1 and C”2 ), and the relevant equations are as follows:

Composition Conversions

Where ρ1, ρ2 are the densities for element 1 and 2 , respectively.

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17

Computation of average density for a binary alloy systems

Furthermore, on occasion we desire to determine the density and atomic

weight of a binary alloy given the composition in terms of either weight

percent or atom percent.

Computation of density (for a two element metal alloy):

Computation of atomic weight (for a two-element metal alloy):

18

Composition Conversion, From Weight Percent to Atom Percent

Example Problem: 2

Determine the composition, in atom percent, of an alloy that consists

of 97 wt% aluminum and 3 wt% copper.

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19

Dislocations- Linear defects

Dislocations are linear defects (1D); they are lines through the crystal

along which crystallographic registry is lost.

There are three kinds of this type: edge dislocation, screw dislocation

and mixed dislocation.

Ed

ge

dis

loca

tio

n

A screw dislocation, which may be thought of as being formed by a

shear stress that is applied to produce the distortion.

Scr

ew d

islo

cati

on

20

Plane defects

Plane defects are boundaries that have two dimensions and normally

separate regions of the materials that have different crystal structures

and/or crystallographic orientations.

These imperfections include external surfaces, grain boundaries, twin

boundaries, stacking faults, and phase boundaries.

A twin boundary: is a special type of grain boundary across which there

is a specific mirror lattice symmetry.

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Precipitates are small particles that are introduced into the matrix by

solid state reactions.

While precipitates are used for several purposes, their most common

purpose is to increase the strength of structural alloys by acting as

obstacles to the motion of dislocations.

Their efficiency in doing this depends on their size, their internal

properties, and their distribution through the lattice.

However, their role in the microstructure is to modify the behavior of

the matrix rather than to act as separate phases in their own right. 21

Bulk or volume defects

Volume defects in crystals are three-dimensional aggregates of atoms or

vacancies.

It is common to divide them into four classes in an imprecise

classification that is based on a combination of the size and effect of the

particle.

1. Precipitates: which are a fraction of a micron in size and decorate the

crystal;

Voids (or pores) are caused by gases that are trapped during

solidification or by vacancy condensation in the solid state.

Their principal effect is to decrease mechanical strength and promote

fracture at small loads. 22


2. Second phase particles : which vary in size from a fraction of a

micron to the normal grain size (10-100μm), but are intentionally

introduced into the microstructure;

Second phase particles are larger particles that behave as a second

phase as well as influencing the behavior of the primary phase.

They may be large precipitates, grains, or poly-granular particles

distributed through the microstructure.

In this case, we consider the average of the properties of the dispersant

phase and the parent.

3. Voids: which are holes in the solid formed by trapped gases or by the

accumulation of vacancies.

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23


Inclusions are foreign particles or large precipitate particles.

They are usually undesirable constituents in the microstructure.

They are also often harmful in microelectronic devices since they

disturb the geometry of the device by interfering in manufacturing, or

alter its electrical properties by introducing undesirable properties of

their own.

4. Inclusions: which vary in size from a few microns to macroscopic

dimensions, and are relatively large, undesirable particles that

entered the system as dirt or formed by precipitation;

Several applications of microstructural examinations are as follows:

Ensure that the associations between the properties and structure (and defects)

are properly understood.

Predict the properties of materials once these relationships have been

established.

Design alloys with new property combinations.

Determine whether or not a material has been correctly heat treated.

Ascertain the mode of mechanical fracture.24

Microscopic Examination

Grain size and shape are only two features of what is termed the

microstructure.

Optical, electron, and scanning probe microscopes are commonly used in

microscopy.

Microscopic examination is an extremely useful tool in the study and

characterization of materials.

In most materials the constituent grains are of microscopic dimensions, having

diameters that may be on the order of microns ,1µm=10-6m, and their details

must be investigated using some type of microscope.

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25

Microscopic Techniques

Optical Microscopy

With optical microscopy, the light microscope is used to study the

microstructure; optical and illumination systems are its basic elements.

Investigations of this type are often termed metallographic, since

metals were first examined using this technique.

The specimen surface must first be ground and polished to a smooth

and mirror like finish.

Photomicrograph of a poly-

crystalline brass specimen (×60).

The maximum magnification possible with an

optical microscope is approximately 2000

times (2000×).

26

Electron Microscopy


Some structural elements are too fine or small to permit observation

using optical microscopy.

An image of the structure under investigation is formed using beams of

electrons instead of light radiation.

High magnifications and resolving powers of these microscopes are

consequences of the short wavelengths of electron beams.

Both transmission and reflection beam modes of operation are possible

for electron microscopes.

Under such circumstances the electron microscope, which is capable of

much higher magnifications, may be employed.

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Magnifications ranging from 10 to in excess of 50,000 times are

possible, as are also very great depths of field.27

The image seen with a transmission electron microscope (TEM) is

formed by an electron beam that passes through the specimen.

Transmission Electron Microscopy


Scanning Electron Microscopy

Magnifications approaching 106× are possible with transmission

electron microscopy, which is frequently utilized in the study of

dislocations.

A more recent and extremely useful investigative tool is the scanning

electron microscope (SEM).

The image on the screen, which may be photographed, represents the

surface features of the specimen.

Some of the features that differentiate the SPM from other microscopic

techniques are as follows:

•Examination on the nanometer scale is possible in as much as

magnifications as high as are possible; much better resolutions are

attainable than with other microscopic techniques.

•Three-dimensional magnified images are generated that provide

topographical information about features of interest.

•Some SPMs may be operated in a variety of environments (e.g.,

vacuum, air, liquid); thus, a particular specimen may be examined in

its most suitable environment. 28

Recently, the field of microscopy has experienced a revolution with the

development of a new family of scanning probe microscopes.

This scanning probe microscope (SPM), of which there are several

varieties, differs from the optical and electron microscopes in that

neither light nor electrons is used to form an image.

Scanning Probe Microscopy


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29

Grain size determination

The grain size is often determined when the properties of a

polycrystalline material are under consideration.

Let n represent the grain size number, and N the average number of grains

per square inch at a magnification of 100×, these two parameters are

related to each other through the expression:

Grain size may be estimated by using an intercept method, described

as follows: Straight lines all the same length are drawn through several

photomicrographs that show the grain structure.

The grains intersected by each line segment are counted; the line length is

then divided by an average of the number of grains intersected, taken over all

the line segments.

The average grain diameter is found by dividing this result by the linear

magnification of the photomicrographs.

American Society for Testing and Materials (ASTM) method

30

Computations of ASTM Grain Size Number and Number of Grains Per Unit Area

Example Problem: 2

(a) Determine the ASTM

grain size number of a metal

specimen if 45 grains per

square inch are measured at a

magnification of 100×.

(b) For this same specimen,

how many grains per square

inch will there be at a

magnification of 80×.

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31

Problems on Ch2: Imperfections in Solids

1.Calculate the fraction of atom sites that are vacant for lead at its

melting temperature of 327°C (600 K). Assume an energy for vacancy

formation of 0.55 eV/atom.

2.Calculate the number of vacancies per cubic meter in iron at 850°C.

The energy for vacancy formation is 1.08 eV/atom. Furthermore, the

density and atomic weight for Fe are 7.65 g/cm3 and 55.85 g/mol,

respectively.

3.Calculate the activation energy for vacancy formation in aluminum,

given that the equilibrium number of vacancies at 500°C (773 K) is

7.57×1023 m-3. The atomic weight and density (at 500°C) for

aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3.

4.What is the composition, in atom percent, of an alloy that consists of

30 wt% Zn and 70 wt% Cu?


5.What is the composition, in weight percent, of an alloy that consists of

6 at% Pb and 94 at% Sn?

6.Calculate the composition, in weight percent, of an alloy that contains

218.0 kg titanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.

7.What is the composition, in atom percent, of an alloy that contains 98 g

tin and 65 g of lead?

8.What is the composition, in atom percent, of an alloy that contains 99.7

lbm copper, 102 lbm zinc, and 2.1 lbm lead?

10.What is the composition, in atom percent, of an alloy that consists of

97 wt% Fe and 3 wt% Si?

9.Convert the atom percent composition in Problem 8 to weight percent.

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33


11.Calculate the number of atoms per cubic meter in aluminum.

12.The concentration of carbon in an iron-carbon alloy is 0.15 wt%.

What is the concentration in kilograms of carbon per cubic meter of

alloy?

13.Determine the approximate density of a high-leaded brass that has a

composition of 64.5 wt% Cu, 33.5 wt% Zn, and 2.0 wt% Pb.

14.Calculate the unit cell edge length for an 85 wt% Fe-15 wt% V alloy.

All of the vanadium is in solid solution, and, at room temperature the

crystal structure for this alloy is BCC.

15.Some hypothetical alloy is composed of 12.5 wt% of metal A and

87.5 wt% of metal B. If the densities of metals A and B are 4.27 and

6.35 g/cm3, respectively, whereas their respective atomic weights are

61.4 and 125.7 g/mol, determine whether the crystal structure for this

alloy is simple cubic, face-centered cubic, or body-centered cubic.

Assume a unit cell edge length of 0.395 nm.

34


12.Gold forms a substitutional solid solution with silver. Compute the

number of gold atoms per cubic centimeter for a silver-gold alloy that

contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and

silver are 19.32 and 10.49 g/cm3, respectively.

13.Germanium forms a substitutional solid solution with silicon.

Compute the number of germanium atoms per cubic centimeter for a

germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si. The

densities of pure germanium and silicon are 5.32 and 2.33 g/cm3,

respectively.

14.Silver and palladium both have the FCC crystal structure, and Pd

forms a substitutional solid solution for all concentrations at room

temperature. Compute the unit cell edge length for a 75 wt% Ag–25

wt% Pd alloy. The room-temperature density of Pd is 12.02 g/cm3,

and its atomic weight and atomic radius are 106.4 g/mol and 0.138

nm, respectively.

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35


15.Niobium forms a substitutional solid solution with vanadium.

Compute the weight percent of niobium that must be added to

vanadium to yield an alloy that contains 1.55 × 1022 Nb atoms per

cubic centimeter. The densities of pure Nb and V are 8.57 and 6.10

g/cm3, respectively.

16.Cite the relative Burgers vector–dislocation line orientations for edge,

screw, and mixed dislocations.

17.(a) For a given material, would you expect the surface energy to be

greater than, the same as, or less than the grain boundary energy?

Why?

(b) The grain boundary energy of a small-angle grain boundary is less

than for a high-angle one. Why is this so?

18.(a) Briefly describe a twin and a twin boundary.

(b) Cite the difference between mechanical and annealing twins.

36


19.(a) Using the intercept method, determine the average grain size, in

millimeters, of the specimen whose microstructure is shown in figure

below; use at least seven straight-line segments.

(b) Estimate the ASTM grain size number for this material.

Photomicrograph of the surface of a polished

and etched polycrystalline specimen of an iron

chromium alloy in which the grain boundaries

appear dark. (100X)

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37


19.For an ASTM grain size of 8, approximately how many grains would

there be per square inch at

(a) a magnification of 100, and

(b) without any magnification?

20.Determine the ASTM grain size number if 25 grains per square inch

are measured at a magnification of 600.

21.Determine the ASTM grain size number if 20 grains per square inch

are measured at a magnification of 50.

22.Aluminum–lithium alloys have been developed by the aircraft

industry to reduce the weight and improve the performance of its

aircraft. A commercial aircraft skin material having a density of 2.55

g/cm3 is desired. Compute the concentration of Li (in wt%) that is

required.

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1

Prepared by:

Dr. Alaa Abd-Elnaiem

Chapter 3:

Phase Diagrams

2






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1. (a)Schematically sketch different types of phase diagrams (e.g.

simple isomorphous and eutectic phase diagrams).

(b) Label the various phase regions on phase diagrams .

(c) Label liquidus, solidus, and solvus lines.

2.For any binary phase diagram at equilibrium, determine:

(a) What phase(s) is (are) present.

(b) The composition(s) of the phase(s).

(c) The mass fraction(s) of the phase(s).

3.For any binary phase diagram, do the following:

(a)Locate the temperatures and compositions of all eutectic,

eutectoid, peritectic and congruent phase transformations.

(b)Write reactions for all these transformations for either heating or

cooling process.3


Learning Objectives

Through this chapter we will present and discuss the following topics:

1)Terminology associated with phase diagrams and phase

transformations.

2)Pressure-temperature phase diagrams for pure materials.

3)The interpretation of phase diagrams.

4)Some of the common and relatively simple binary phase diagrams,

including that for the iron-carbon system.

5)The development of equilibrium microstructures, upon cooling, for

several situations. 4

Introduction

The understanding of phase diagrams for alloy systems is extremely

important because there is a strong correlation between microstructure and

mechanical properties.

Phase diagrams provide valuable information about melting, casting,

crystallization and other phenomena.

The development of microstructure of an alloy is related to the

characteristics of its phase diagram.

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“A solid solution” consists of atoms of at least two different types; the

solute atoms occupy either substitutional or interstitial positions in the

solvent lattice. 5

Definitions and Basic Concepts

“Components” are pure metals and/or compounds of which an alloy is

composed, e.g. in a Cu-Zn alloy, the components are Cu and Zn.

“System” may refer to a specific body of material under consideration

(e.g., a ladle of molten steel) or it may relate to the series of possible

alloys consisting of the same components, but without regard to alloy

composition (e.g., the Co-Cu system).

“A phase” in a material, is a region that differ in its microstructure and/

or composition from another region.

“A phase diagrams” is a type of graph used to show the equilibrium

conditions between the thermodynamically-distinct phases; or to show

what phases are present in the material system at various temperature,

pressure and compositions.

6

Solubility Limit

For many alloy systems and at some specific temperature, there is a

maximum concentration of solute atoms that may dissolve in the

solvent to form a solid solution; this is called a solubility limit.

The addition of solute in excess of this solubility limit results in the

formation of another solid solution or compound that has a distinctly

different composition.

• This solubility limit of sugar in

water depends on the

temperature of the water.

• Change of temperature can

change number of phases.

• Change of composition can


Example to illustrate Solubility

limit, the solubility of sugar in

a sugar-water syrup.

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7

Phases

As indicated before, a phase may be defined as a homogeneous portion

of a system that has uniform physical and chemical characteristics.

Every pure material is considered to be a phase;

so also is every solid, liquid, and gaseous

solution.

A single-phase system is termed “homogeneous”.

Systems composed of two or more phases are

termed “mixtures” or “heterogeneous systems”.

Most metallic alloys, ceramic, polymeric and

composite systems are heterogeneous.H2O(solid, ice) in H2O(liquid) ⇒ 2 phases

Each phase characterized by:

•Homogeneous in crystal structure and atomic arrangement.

•Have a definite interface and able to be mechanically separated from its

surroundings.

•Have same chemical and physical properties throughout.

Al2CuMg

Al

When only a single phase or solid solution is present, the texture will be

uniform, except for grain boundaries that may be revealed.8

As mentioned before, the physical properties and, in particular the

mechanical behavior, of a material depend on the microstructure.

Microstructure is subject to direct microscopic observation, using optical

or electron microscopes.

In metal alloys, microstructure is characterized by the number of phases

present, their proportions and the manner in which they are distributed or

arranged.

The microstructure of an alloy depends on the alloying elements present,

their concentrations and the heat treatment of the alloy.

After appropriate polishing and etching, the different phases may be

distinguished by their appearance; for example, for a two-phase alloy, one

phase may appear light and the other phase dark.

Microstructure

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In many metallurgical and materials systems of interest, phase

equilibrium involves just solid phases.

9

Phase Equilibrium

A system is at equilibrium if its free energy is at a minimum under some

specified combination of temperature (T), pressure (P) and composition

(C).

In a macroscopic sense, this means that the characteristics of the system

do not change with time but persist indefinitely; that is, the system is

stable.

A change in T, P, and/or C for a system in equilibrium will result in an

increase in the free energy and in a possible spontaneous change to

another state whereby the free energy is lowered.

Phase equilibrium is reflected by a constancy with time in the phase

characteristics of a system.

A metastable state or microstructure may persist indefinitely,

experiencing only extremely slight and almost imperceptible changes as

time progresses.

Often, metastable structures are of more practical significance than

equilibrium ones.

For example, some steel and Al alloys rely for their strength on the

development of metastable microstructures during carefully designed

heat treatments. 10

It is often the case, especially in solid systems, that a state of

equilibrium is never completely achieved because the rate of approach

to equilibrium is extremely slow; such a system is said to be in a non-

equilibrium or metastable state.

Phase Equilibrium

In this regard the state of the system is reflected in the characteristics of

the microstructure, which necessarily include not only the phases

present and their compositions but the relative phase amounts and

their spatial arrangement or distribution.

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1.Cite three variables that determine the microstructure of an alloy.

Solution:

Three variables that determine the microstructure of an alloy are (1) the

alloying elements present, (2) the concentrations of these alloying

elements, and (3) the heat treatment of the alloy.

2. What thermodynamic condition must be met for a state of

equilibrium to exist?

Solution:

In order for a system to exist in a state of equilibrium the free energy

must be a minimum for some specified combination of temperature,

pressure, and composition.

Examples 1, 2

12

One Component Phase Diagram

Much of the information about the control of the phase structure of a

particular system is conveniently displayed in what is called a phase

diagram or an equilibrium diagram.

There are three externally controllable parameters that will affect phase

structure: temperature, pressure and composition.

Perhaps the simplest and easiest type of phase diagram to understand is

that for a one-component system, in which composition is held

constant, this means that pressure and temperature are the variables.

One-component phase diagram (or unary phase diagram) is

represented as a two-dimensional plot of pressure, y-axis, versus

temperature , x-axis.

Phase diagrams are constructed when various combinations of these

parameters are plotted against one another, see next slide.

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13

Cooling curves:

• Used to determine phase transition temperature.

• Record T of material versus t, as it cools from its molten state through

solidification and finally to room temperature (at a constant pressure).

BC: plateau or region of thermal arrest; in this region material is in the form of solid

and liquid phases.

CD: solidification is completed, T drops.

One Component Phase DiagramHow to construct phase diagrams?

14

Here it may be noted that regions for three different phases solid, liquid

and vapor are located on the plot.


Each of these phases will exist under equilibrium conditions over the

temperature-pressure ranges of its corresponding area.

Deposition

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15

Binary Phase Diagrams

Binary phase diagrams are maps that represent the relationships

between temperature and the compositions; and quantities of phases at

equilibrium, which influence the microstructure of an alloy.

Binary phase diagrams are helpful in predicting phase transformations

and the resulting microstructures, which may have equilibrium or non-

equilibrium character.

Binary phase diagram for

A-B system.

16

Binary Isomorphous Systems

Example for binary systems:

Temperature is plotted along x-axis and the y-axis represents the

composition of the alloy, in weight percent (bottom) and atom percent

(top) of nickel.

Three different phase regions

appear on the diagram, an α

region, a liquid (L) region, and

a two-phase α +L region.

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The liquid L is a homogeneous liquid solution composed of both copper

and nickel.

The α phase is a substitutional solid solution consisting of both Cu and

Ni atoms, and having an FCC crystal structure.

At temperatures below 1085 ˚C, copper and nickel are mutually soluble

in each other in the solid state for all compositions.

The Cu-Ni system is termed isomorphous because of this complete

liquid and solid solubility of the two components.

The line separating the L and α+L phase regions is termed the liquidus

line, the liquid phase is present at all temperatures and compositions

above this line.

The solidus line is located between the α and α+L regions, below which

only the solid phase exists.


18

3.Given here are the solidus and liquidus temperatures for the

germanium-silicon system for different compositions.

Construct the phase diagram for this system and label each region.

Solution:

The Ge-Si phase diagram is

constructed below:

Example 3

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19

Interpretation of Phase Diagrams

For a binary system of known composition and temperature that is at

equilibrium, at least three kinds of information are available: (1) the

phases that are present, (2) the compositions of these phases, and (3)

the percentages or fractions of the phases.

20


Phases Present

One just locates the temperature-composition point on the diagram and

notes the phase(s) with which the corresponding phase region is labeled.

On the other hand, a 35 wt% Ni–65 wt% Cu alloy at (point B) will

consist of both α and liquid phases at equilibrium.

For example, an alloy of composition 60 wt% Ni–40 wt% Cu at 1100

˚C would be located at point A in figure below; since this is within the

α region, only the single phase (α) will be present.

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1.A tie line is constructed across the two-phase region at the temperature of the

alloy.

2.The intersections of the tie line and the phase boundaries on either side are

noted.

3.Perpendiculars are dropped from these intersections to the horizontal

composition axis, from which the composition of each of the respective phases

is read. 21


Determination of Phase Compositions

The first step in the determination of phase compositions is to locate the

temperature-composition point on the phase diagram.

If only one phase is present, the composition of this phase is the same

as the overall composition of the alloy.

In all two-phase regions, one may imagine a series of horizontal lines,

one at every temperature; each of these is known as a tie line, or an

isotherm line and extend across the two-phase region and terminate at

the phase boundary lines on either side, then do the following steps:

22

Example 4

4.Consider the 35 wt% Ni–65 wt% Cu alloy at 1250 ˚C,

Determine the composition (in wt% Ni and Cu) for both the and

liquid phases at this point.

Solution:

By applied the previous

steps, one can obtain the

required composition as

shown in figure.

CL:31.5 wt% Ni–68.5 wt% Cu

Cα:42.5 wt% Ni–57.5 wt% Cu

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12

1.The tie line is constructed across the two-phase region at the temperature

of the alloy.

2.The overall alloy composition is located on the tie line.

3.The fraction of one phase is computed by taking the length of tie line from

the overall alloy composition to the phase boundary for the other phase,

and dividing by the total tie line length.

4.The fraction of the other phase is determined in the same manner.

5.If phase percentages are desired, each phase fraction is multiplied by 100.23


Determination of Phase Amounts

The relative amounts (as fraction or percentage) of the phases present at

equilibrium may also be computed with the aid of phase diagrams.

If one phase is present, the alloy is composed entirely of that phase; or

the phase fraction is 1.0 and the percentage is 100%.

If the composition and temperature position is located within a two-

phase region, we can determine the phase amounts by using lever rule:

24

Example 5

5.Consider again the 35 wt% Ni–65 wt% Cu alloy at 1250 ˚C,

Compute the fraction of each of the α and liquid (L) phases.

Solution:

Let the overall alloy composition C0 and mass fractions be represented

by WL and Wα for the respective phases, so:

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b) As observed from phase diagram the composition of α phase , Cα is 10

wt% Sn, and for β phase, Cβ is 98 wt% Sn. 25

6. For a 40 wt% Sn–60 wt% Pb alloy at 150˚C (300F),

(a) What phase(s) is (are) present?

(b) What is (are) the composition(s) of the phase(s)?

Example 6

Solution:

a) The present phases at indicated point are the solids α and β phases.

26

Example 7

7.For a 40 wt% Sn–60 wt% Pb alloy at 150˚C (300F),

Calculate the relative amount of each phase present in terms of mass

fraction.

Solution:

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27

8. For a 90 wt% Zn-10 wt% Cu alloy at 400C,

Calculate the compositions and relative amount of each phase

present in terms of mass fraction.

Example 8

W =C C0

C C

=97 90

97 87= 0.70

W =C0 C

C C

=90 87

97 87= 0.30

The compositions of the phases, are

Cɛ = 87 wt% Zn-13 wt% Cu

Cη = 97 wt% Zn-3 wt% Cu

In as much as the composition of the

alloy C0 = 90 wt% Zn, application of

the appropriate lever rule expressions

(for compositions in weight percent

zinc) leads to:

Solution:

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1

Prepared by:


Chapter 3:

Phase Diagrams

2






“Phase Diagrams are road maps”

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1.know, what a phase diagram (or equilibrium diagram) is, and how to

use it.

2.(a)Schematically sketch different types of phase diagrams (e.g. simple

isomorphous and eutectic phase diagrams).

(b) Label the various phase regions on phase diagrams .

(c) Label liquidus, solidus, and solvus lines.

2.For any binary phase diagram at equilibrium, determine:

(a) What phase(s) is (are) present.

(b) The composition(s) of the phase(s).

(c) The mass fraction(s) of the phase(s).

3.For any binary phase diagram, do the following:

(a)Locate the temperatures and compositions of all eutectic,



cooling process.3


Learning Objectives

Through this chapter we will present and discuss the following topics:

1)Terminology associated with phase diagrams and phase

transformations.

2)Pressure-temperature phase diagrams for pure materials.

3)The interpretation of phase diagrams.

4)Some of the common and relatively simple binary phase diagrams,

including that for the iron-carbon system.

5)The development of equilibrium microstructures, upon cooling, for

several situations. 4

Introduction

The understanding of phase diagrams for alloy systems is extremely

important because there is a strong correlation between microstructure and

mechanical properties.

Phase diagrams provide valuable information about melting, casting,

crystallization and other phenomena.

The development of microstructure of an alloy is related to the

characteristics of its phase diagram.

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“A solid solution” consists of atoms of at least two different types; the solute

atoms occupy either substitutional or interstitial positions in the solvent lattice.

5

Definitions and Basic Concepts

“Components” the chemical elements which make up alloys, e.g. in a Cu-

Zn alloy, the components are Cu and Zn.

“System” may refer to a specific body of material under consideration

(e.g., a ladle of molten steel) or it may relate to the series of possible alloys

consisting of the same components, but without regard to alloy

composition (e.g., the Co-Cu system, binary system, ternary system,…).

“A phase” in a material, is a region that differ in its microstructure and/ or

composition from another region.

“A phase diagrams” is a type of graph used to show the equilibrium

conditions between the thermodynamically-distinct phases; or to show

what phases are present in the material system at various temperature,

pressure and compositions.

A metallic alloy is a mixture of a metal with other metals or non-metals, e.g.

Brass: a mixture of copper (Cu) and zinc (Zn).

6

Solubility Limit

For many alloy systems and at some specific temperature, there is a

maximum concentration of solute atoms that may dissolve in the

solvent to form a solid solution; this is called a solubility limit.

The addition of solute in excess of this solubility limit results in the

formation of another solid solution or compound that has a distinctly

different composition.

• This solubility limit of sugar in

water depends on the

temperature of the water.

• Change of temperature can


• Change of composition can


Example to illustrate Solubility

limit, the solubility of sugar in

a sugar-water syrup.

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7

Phases

As indicated before, a phase may be defined as a homogeneous portion

of a system that has uniform physical and chemical characteristics.

Every pure material is considered to be a phase;

so also is every solid, liquid, and gaseous

solution.

A single-phase system is termed “homogeneous”.

Systems composed of two or more phases are

termed “mixtures” or “heterogeneous systems”.

Most metallic alloys, ceramic, polymeric and

composite systems are heterogeneous.H2O(solid, ice) in H2O(liquid) ⇒ 2 phases

Each phase characterized by:

•Homogeneous in crystal structure and atomic arrangement.

•Have a definite interface and able to be mechanically separated from its

surroundings.

•Have same chemical and physical properties throughout.

Al2CuMg

Al

When only a single phase or solid solution is present, the texture will be

uniform, except for grain boundaries that may be revealed.8

As mentioned before, the physical properties and, in particular the

mechanical behavior, of a material depend on the microstructure.

Microstructure is subject to direct microscopic observation, using optical

or electron microscopes.

In metal alloys, microstructure is characterized by the number of phases

present, their proportions and the manner in which they are distributed or

arranged.

The microstructure of an alloy depends on the alloying elements present,

their concentrations and the heat treatment of the alloy.

After appropriate polishing and etching, the different phases may be

distinguished by their appearance; for example, for a two-phase alloy, one

phase may appear light and the other phase dark.

Microstructure

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In many metallurgical and materials systems of interest, phase

equilibrium involves just solid phases.

9

Phase Equilibrium

A system is at equilibrium if its free energy is at a minimum under some

specified combination of temperature (T), pressure (P) and composition

(C).

In a macroscopic sense, this means that the characteristics of the system

do not change with time but persist indefinitely; that is, the system is

stable.

A change in T, P, and/or C for a system in equilibrium will result in an

increase in the free energy and in a possible spontaneous change to

another state whereby the free energy is lowered.

Phase equilibrium is reflected by a constancy with time in the phase

characteristics of a system.

A metastable state or microstructure may persist indefinitely,

experiencing only extremely slight and almost imperceptible changes as

time progresses.

Often, metastable structures are of more practical significance than

equilibrium ones.

For example, some steel and Al alloys rely for their strength on the

development of metastable microstructures during carefully designed

heat treatments. 10

It is often the case, especially in solid systems, that a state of

equilibrium is never completely achieved because the rate of approach

to equilibrium is extremely slow; such a system is said to be in a non-

equilibrium or metastable state.

Phase Equilibrium

In this regard the state of the system is reflected in the characteristics of

the microstructure, which necessarily include not only the phases

present and their compositions but the relative phase amounts and

their spatial arrangement or distribution.

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11

1.Cite three variables that determine the microstructure of an alloy.

Solution:

Three variables that determine the microstructure of an alloy are (1) the

alloying elements present, (2) the concentrations of these alloying

elements, and (3) the heat treatment of the alloy.

2. What thermodynamic condition must be met for a state of

equilibrium to exist?

Solution:

In order for a system to exist in a state of equilibrium the free energy

must be a minimum for some specified combination of temperature,

pressure, and composition.

Examples 1, 2

12


Much of the information about the control of the phase structure of a

particular system is conveniently displayed in what is called a phase

diagram or an equilibrium diagram.

There are three externally controllable parameters that will affect phase

structure: temperature, pressure and composition.

Perhaps the simplest and easiest type of phase diagram to understand is

that for a one-component system, in which composition is held

constant, this means that pressure and temperature are the variables.

One-component phase diagram (or unary phase diagram) is

represented as a two-dimensional plot of pressure, y-axis, versus

temperature , x-axis.

Phase diagrams are constructed when various combinations of these

parameters are plotted against one another, see next slide.

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13

Cooling curves:

• Used to determine phase transition temperature.

• Record T of material versus t, as it cools from its molten state through

solidification and finally to room temperature (at a constant pressure).

BC: plateau or region of thermal arrest; in this region material is in the form of solid

and liquid phases.

CD: solidification is completed, T drops.

One Component Phase DiagramHow to construct phase diagrams?

14

Here it may be noted that regions for three different phases solid, liquid

and vapor are located on the plot.


Each of these phases will exist under equilibrium conditions over the

temperature-pressure ranges of its corresponding area.

Deposition

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15

Binary Phase Diagrams

Binary phase diagrams are maps that represent the relationships

between temperature and the compositions; and quantities of phases at

equilibrium, which influence the microstructure of an alloy.

Binary phase diagrams are helpful in predicting phase transformations

and the resulting microstructures, which may have equilibrium or non-

equilibrium character.

Binary phase diagram for

A-B system.

16


Example for binary systems:

Temperature is plotted along x-axis and the y-axis represents the

composition of the alloy, in weight percent (bottom) and atom percent

(top) of nickel.

Three different phase regions

appear on the diagram, an α

region, a liquid (L) region, and

a two-phase α +L region.

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17

The liquid L is a homogeneous liquid solution composed of both copper

and nickel.

The α phase is a substitutional solid solution consisting of both Cu and

Ni atoms, and having an FCC crystal structure.

At temperatures below 1085 ˚C, copper and nickel are mutually soluble

in each other in the solid state for all compositions.

The Cu-Ni system is termed isomorphous because of this complete

liquid and solid solubility of the two components.

The line separating the L and α+L phase regions is termed the liquidus

line, the liquid phase is present at all temperatures and compositions

above this line.

The solidus line is located between the α and α+L regions, below which

only the solid phase exists.


18

3.Given here are the solidus and liquidus temperatures for the

germanium-silicon system for different compositions.

Construct the phase diagram for this system and label each region.

Solution:

The Ge-Si phase diagram is

constructed below:

Example 3

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19


For a binary system of known composition and temperature that is at

equilibrium, at least three kinds of information are available: (1) the

phases that are present, (2) the compositions of these phases, and (3)

the percentages or fractions of the phases.

20


Phases Present

One just locates the temperature-composition point on the diagram and

notes the phase(s) with which the corresponding phase region is labeled.

On the other hand, a 35 wt% Ni–65 wt% Cu alloy at (point B) will

consist of both α and liquid phases at equilibrium.

For example, an alloy of composition 60 wt% Ni–40 wt% Cu at 1100

˚C would be located at point A in figure below; since this is within the

α region, only the single phase (α) will be present.

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1.A tie line is constructed across the two-phase region at the temperature of the

alloy.

2.The intersections of the tie line and the phase boundaries on either side are

noted.

3.Perpendiculars are dropped from these intersections to the horizontal

composition axis, from which the composition of each of the respective phases

is read. 21


Determination of Phase Compositions

The first step in the determination of phase compositions is to locate the

temperature-composition point on the phase diagram.

If only one phase is present, the composition of this phase is the same

as the overall composition of the alloy.

In all two-phase regions, one may imagine a series of horizontal lines,

one at every temperature; each of these is known as a tie line, or an

isotherm line and extend across the two-phase region and terminate at

the phase boundary lines on either side, then do the following steps:

22

Example 4

4.Consider the 35 wt% Ni–65 wt% Cu alloy at 1250 ˚C,

Determine the composition (in wt% Ni and Cu) for both the and

liquid phases at this point.

Solution:

By applied the previous

steps, one can obtain the

required composition as

shown in figure.

CL:31.5 wt% Ni–68.5 wt% Cu

Cα:42.5 wt% Ni–57.5 wt% Cu

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1.The tie line is constructed across the two-phase region at the temperature

of the alloy.

2.The overall alloy composition is located on the tie line.

3.The fraction of one phase is computed by taking the length of tie line from

the overall alloy composition to the phase boundary for the other phase,

and dividing by the total tie line length.

4.The fraction of the other phase is determined in the same manner.

5.If phase percentages are desired, each phase fraction is multiplied by 100.23


Determination of Phase Amounts

The relative amounts (as fraction or percentage) of the phases present at

equilibrium may also be computed with the aid of phase diagrams.

If one phase is present, the alloy is composed entirely of that phase; or

the phase fraction is 1.0 and the percentage is 100%.

If the composition and temperature position is located within a two-

phase region, we can determine the phase amounts by using lever rule:

24

Example 5

5.Consider again the 35 wt% Ni–65 wt% Cu alloy at 1250 ˚C,

Compute the fraction of each of the α and liquid (L) phases.

Solution:

Let the overall alloy composition C0 and mass fractions be represented

by WL and Wα for the respective phases, so:

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b) As observed from phase diagram the composition of α phase , Cα is 10

wt% Sn, and for β phase, Cβ is 98 wt% Sn. 25

6. For a 40 wt% Sn–60 wt% Pb alloy at 150˚C (300F),

(a) What phase(s) is (are) present?

(b) What is (are) the composition(s) of the phase(s)?

Example 6

Solution:

a) The present phases at indicated point are the solids α and β phases.

26

Example 7

7.For a 40 wt% Sn–60 wt% Pb alloy at 150˚C (300F),

Calculate the relative amount of each phase present in terms of mass

fraction.

Solution:

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27

8. For a 90 wt% Zn-10 wt% Cu alloy at 400C,

Calculate the compositions and relative amount of each phase

present in terms of mass fraction.

Example 8

W =C C0

C C

=97 90

97 87= 0.70

W =C0 C

C C

=90 87

97 87= 0.30

The compositions of the phases, are

Cɛ = 87 wt% Zn-13 wt% Cu

Cη = 97 wt% Zn-3 wt% Cu

In as much as the composition of the

alloy C0 = 90 wt% Zn, application of

the appropriate lever rule expressions

(for compositions in weight percent

zinc) leads to:

Solution:

Chapter 3:

Phase Diagrams

28






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1. Distinguish between equilibrium cooling and non-equilibrium cooling.

2. Examine the development of microstructure that occurs for

isomorphous, eutectic, eutectoid, peritectic and congruent alloys

during solidification.

3. Determine present phases, composition and the amount of each phase

by using Lever rule at any temperature and certain composition.

4. Apply the Gibbs phase rule for any alloy system to determine the

number of degrees of freedom.

5. For any binary phase diagram, do the following:

(a)Determine the temperatures and compositions of all eutectic,



cooling process.

6. State and describe the phases present in Fe-Fe3C phase diagram. 29

After studying this lecture you should be able to do the following:

Learning Objectives

30

Introduction

In a binary alloy (in the solid state) the microstructure can take one of

four forms:

i. A single solid solution.

ii. Two separated, essentially pure components.

iii. Two separated solid solutions.

iv. A chemical compound, together with a solid solution.

The properties of an alloy do not depend only on concentration of the

phases but how they are arranged structurally at the microscopy level.

The microstructure is specified by the number of phases, their

proportions, and their arrangement in space.

The way to tell the microstructure: 1) Cut the material, 2) Polish it to

a mirror finish, 3) Etch it in a weak acid (components etch at a different

rate) and 4) Observe the surface under a microscope.

The microstructure is depends on: alloying elements, concentrations,

heat treatment (temperature, heating/cooling rate etc.).

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31

The compositions of phases can be controlled by temperature.

These changes of microstructure occur through the process of diffusion.

Because diffusion is time-dependent, much more time is required, at

each temperature for compositional adjustments.

Diffusion rates decrease with temperature.

In case of fast cooling rates, there is little time to enable equilibrium

cooling and this leads to a non-uniform distribution of the two elements

for isomorphous alloys..

“Segregation”: distribution of the two elements within

the grains is non-uniform, where in the centre of each

grain, the high melting element solidifies first and the

concentration of the low-melting element increases with

position from this region to the grain boundary.

Introduction

32

Microstructures in Isomorphous Alloys

Schematic representation of the development of

Microstructure during the solidification of a 35 Ni-65

Cu alloy.

Let us consider the Co-Ni system

of composition 35 wt% Ni-65 wt%

Cu as it is cooled from 1300 ˚C.

At 1300˚C, point a, the alloy is

completely liquid (of composition

35 wt% Ni-65 wt% Cu) and has the

microstructure represented by the

circle inset in the Figure.

At 1260˚C, point b, the first solid α

begins to form and Cα= 46 wt% Ni,

CL= 35 wt% Ni.

Case 1: Equilibrium Cooling

(Solidification under low cooling rates)

Isomorphous alloys: Complete solubility in both liquid and solid states.

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Subsequent cooling will produce no microstructural or compositional

alterations. 33

At 1250˚C, point c, the

compositions of the liquid and α

phases are CL(32 wt% Ni) and Cα

(43 wt% Ni), respectively.

The solidification process is

complete at 1220˚C, point d; the

composition of the solid is ~35

wt% Ni and liquid is ~24 wt% Ni.

Schematic representation of the development of Microstructure

during the solidification of a 35 Ni–65 Cu alloy.


The remaining liquid will solidifies

after solidus line, e.g. at point e; the

final product then is a

polycrystalline-phase solid solution

(s.s.) with composition 35 wt% Ni.

34


At 1300˚C, point a′, the alloy is

completely liquid (of composition

35 wt% Ni–65 wt % Cu) and has

the microstructure represented by

the circle inset in the Figure.

At 1260˚C, point b′, the first solid

α begins to form and Cα = 46 wt %

Ni, CL= 35 wt % Ni.

At 1250˚C, point c′, the

compositions of the liquid and α

phases are CL=29 wt% Ni and Cα

=42 wt% Ni, respectively.

Case 2: Non-equilibrium Cooling

(Solidification under fast cooling rates)


during the solidification of a 35 Ni-65 Cu alloy.

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35


The composition of the grains has

continuously changed with radial

position, from 46 wt% Ni at grain

centers to 40 wt% Ni at the outer

grain boundary (Average Cα~42 Ni).

Non-equilibrium solidification

finally reaches completion at

1205˚C, point e′.

At 1220˚C, point d′, there is still an

appreciable proportion of liquid

remaining, and the α phase that is

forming has a composition of 35

wt% Ni, (Average Cα~38 Ni).

The inset at ~1170˚C, point f′,

shows the microstructure of the

totally solid material.


during the solidification of a 35 Ni-65 Cu alloy.

36

Binary Eutectic Systems

The simplest kind of system with two solid phases is called a eutectic

system.

A eutectic system contains two solid phases at low temperature.

These solid phases may have different crystal structures, or the same

crystal structure with different lattice parameters.

Examples for binary eutectic systems:

•Pb(FCC) and Sn(tetragonal), solder systems.

•Cu(FCC) and Ag(FCC), high temperature solder.

•Fe (BCC) and C (graphite -hexagonal), cast irons.

•Al (FCC) and Si (diamond cubic), cast aluminum alloys.

Binary systems: systems which contain two components.

Eutectic means “easily melted in Greek” or by other words has a special

composition with a minimum melting temperature.

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37

Binary Eutectic Systems: Cu-Ag Eutectic System

Three single-phase regions

are found on the diagram:

α (s.s. rich in Cu), β (s.s.

rich in Ag) and liquid.

•Solvus line: BC, GH

•Solidus line: AB, FG, BEG

•Liquidus line: AE, EF

•Isotherm line: BEG

•Eutectic Point: E

Eutectic reaction:

Cu and Ag are both FCC, but their lattice parameters and atomic radii

are very different, so they have limited solubility in the solid state.

There are two solid stable phases α and β, and at high temperatures

there is a eutectic reaction where the solids α, β and the liquid coexist.

38

9. For a 40 wt% Sn–60 wt% Pb alloy at 150˚C (300F), (a) What phase(s)

is (are) present? (b) What is (are) the composition(s) of the phase(s)?,

(c) calculate the relative amount of each phase present in terms of

mass fraction.

Example 9

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39

a) The phases present:

α and β

b)The compositions of the

α and β phases are:

Cα ~ 10 wt% Sn

Cβ ~ 98 wt%Sn

Example 9

c) The relative amount of

each phase present in

terms of mass fraction:

Solution

40

Microstructures in Eutectic Alloys

Depending on composition, several different types of microstructures are

possible for the slow cooling of alloys belonging to binary eutectic systems.

1) 2)

α+L

L

αα+L

L

α

α+β

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41


This microstructure consists of

alternating layers of a Pb rich -

phase solid solution (dark layers),

and a Sn rich -phase solid solution

(light layers). 375×.

3)

No changes above eutectic temperature, TE.

At TE liquid transforms to α and β phases (eutectic reaction).

42

This microstructure is composed of a

primary Pb-rich phase (large dark

regions) within a lamellar eutectic structure

consisting of a Sn-rich phase (light layers)

and a Pb-rich phase (dark layers). 400×.


4)

As the temperature is lowered to below the eutectic, the liquid phase,

which is of the eutectic composition, will transform to the eutectic

structure (i.e., alternating α and β layers); insignificant changes will

occur with the phase that formed during cooling through the α+L region.

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43


It is possible to compute the relative amounts of both eutectic and

primary α, microconstituents.

Since the eutectic microconstituent always forms from the liquid having

the eutectic composition, this microconstituent may be assumed to have

a composition of 61.9 wt% Sn.

The fraction of the eutectic

microconstituent We is just

the same as the fraction of

liquid WL from which it

transforms, or:

44


The fraction of primary α, Wα′ is just the fraction of the phase that

existed prior to the eutectic transformation or:

The fractions of total α, Wα (both

eutectic and primary), and total of β,

Wβ are determined by use of the

lever rule and a tie line that extends

entirely across the α+β phase field.

&

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45

Eutectoid and Peritectic Reactions

Eutectoid (eutectic-like in Greek) reaction similar to eutectic reaction.

Invariant point (the eutectoid)three solid phases in equilibrium.

Upon cooling at eutectoid point, a solid phase transforms into two other

solid phases:

Part of Cu-Zn phase diagram

46

Eutectoid and Peritectic Reactions

The peritectic reaction is another invariant reaction involving three

phases at equilibrium.

In this reaction, upon heating, one solid phase transforms into a liquid

phase and another solid phase:

Peritectics are not as common as eutectics and eutectiods.

Part of Cu-Zn phase diagram

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47

Congruent Phase Transformations

Incongruent transformation, at

least one phase changes

composition (eutectic, eutectoid,

peritectic).

Phase transformations, may be, classified according to whether or not

there is any change in composition for the phases involved.

Those for which there are no compositional alterations are said to be

congruent transformations.

Part of Ni-Ti phase diagram

48

The Gibbs Phase Rule

The construction of phase diagrams as well as some of the principles

governing the conditions for phase equilibrium are dictated by laws of

thermodynamics, e.g. the Gibbs phase rule.

This rule represents a criterion for the number of phases that will

coexist within a system at equilibrium, and is expressed by the simple

equation: P+F=C+N

where F is the possible number of

degrees of freedom (e.g. temperature,

pressure and composition), P is the

number of phases present, C is the

number of chemical components and

N is the number of non-

compositional variables (e.g.,

temperature and pressure).

Apply Gibbs Phase rule for binary Pb-Sn phase diagram

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49

Example 10

For this system, the number of components, C, is 1, whereas N, the number of non-compositional

variables, is 2, temperature and pressure.

Thus, the phase rule now becomes, P + F = 1 + 2 = 3 or F = 3 – P where P is the number of

phases present at equilibrium.

At point A, three phases are present (i.e. ice I, ice III, and liquid) and P = 3; thus, the number of

degrees of freedom is zero since F=0.

Thus, point A is an invariant point (in this case a triple point), and we have no choice in the

selection of externally controllable variables in order to define the system.

Point B, C left to you

10.In attach figure is shown the P-T

phase diagram for H2O. Apply the

Gibbs phase rule at points A, B, and C;

that is, specify the number of degrees

of freedom at each of the points-that is,

the number of externally controllable

variables that need be specified to

completely define the system.

Solution:

50

The Iron-Carbon System

The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram

Steels: alloys of Iron (Fe) and Carbon (C).

Fe-C phase diagram is complex.

Will only consider the steel part of the diagram, up to around 7% C.

Eutectic:

4.3wt% C, 1147 C

L + Fe3C

Eutectoid:

0.76 wt%C, 727 C

+ Fe3C

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-ferrite, s.s. of C in BCC Fe•Stable form of iron at room temperature.

•The maximum solubility of C is 0.022 wt%.

•Transforms to FCC -austenite at 912 C.

-austenite, s.s. of C in FCC Fe• The maximum solubility of C is 2.14 wt %.

• Transforms to BCC d-ferrite at 1395 C .

• Is not stable below the eutectic temperature (727 C) unless cooled rapidly .

d-ferrite, s.s. of C in BCC Fe• The same structure as -ferrite.

• Stable only at high T, above 1394 C.

• Melts at 1538 C.

Fe3C (iron carbide or cementite)• This intermetallic compound is metastable, it remains as a compound

indefinitely at room temperature, but decomposes (very slowly, within several

years) into -Fe and C (graphite) at 650 - 700 C.

Fe-C liquid solution

51

The Iron–Carbon System

Phases in Fe–Fe3C Phase Diagram

-ferrite -austenite

52

Microstructures in Iron-Carbon Alloys

Assume slow cooling or equilibrium cooling

Microstructure depends on composition (carbon content) and heat

treatment.

Consider, for example, an alloy of eutectoid

composition (0.76 wt% C) as it is cooled from a

temperature within the ɤ phase region, say, 800˚C-

that is, beginning at point a and moving down.

Initially, the alloy is composed entirely of the

austenite phase having a composition of 0.76 wt% C.

The dark areas are

Fe3C layers, the

light phase is α-

ferrite

Pearlite, layered structure of two phases: α-ferrite

and cementite (Fe3C) will form at point b.

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53


Compositions to the left of eutectoid

(0.022-0.76wt%C) hypo-eutectoid

(less than eutectoid -Greek) alloys.

+ + Fe3C

Hypoeutectoid contains proeutectoid

ferrite formed above eutectoid

temperature and eutectoid perlite that

contains ferrite and cementite.

54


Compositions to right of eutectoid

(0.76- 2.14 wt % C) hypereutectoid

(more than eutectoid-Greek) alloys.

+ Fe3C + Fe3C

Hypereutectoid contains proeutectoid

cementite (formed above eutectoid

temperature) plus perlite that contains

eutectoid ferrite and cementite.

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55


The relative amounts of the proeutectoid and pearlite may be

determined in a manner similar by using the lever rule

Let us consider an alloy of composition C0′, Thus, the fraction of

pearlite, Wp may be determined according to:

The fraction of proeutectoid

α, Wα′is computed as follows:

56

Example 11

11.For a 99.65 wt% Fe–0.35 wt% C alloy at a temperature just below the

eutectoid, determine the following:

(a) The fractions of total ferrite, Wα , and cementite, WFe3C, phases.

(b) The fractions of the proeutectoid ferrite, WP, and pearlite, Wα′.

(c) The fraction of eutectoid ferrite, Wαe.

(a)(b)

(c)

Solution:

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57

Ternary Phase Diagrams

A ternary phase diagram has three components.

The three components are usually compositions of elements, but may include

temperature or pressure also.

This type of diagram is three-dimensional but is illustrated in two-dimensions

for ease of drawing and reading.

Instead of being a rectangular plot, it is a triangle.

Stainless steel (Fe-Ni-Cr) is a perfect example of a metal alloy that is

represented by a ternary phase diagram.

Stainless Steel Pipes

1. A 1.5 kg sample of a-brass contains 0.45 kg of Zn, and the rest is Cu.

The atomic weight of Cu is 63.5 and Zn is 65.4

The concentration of copper in a-brass, in wt%: WCu = …………..

The concentration of copper in the a-brass, in at.%: XCu = ……….

The concentration of zinc in the a-brass, in at.%: XZn = …………

2. A special brazing alloy contains 63 wt% gold (Au) and 37 wt% of nickel

(Ni) (which is written Au-37 wt% Ni).

The atomic weight of Au (197.0) is more than three times that of Ni

(58.7). At a glance, which of the two compositions, in at.%, is likely to

be the right one?

(a) XAu=0.34, XNi=0.66

(b) XAu =0.66, XNi=0.34

3. An alloy consists of XA at.% of A with an atomic weight of aA, and XB

at.% of B with an atomic weight of aB. Derive an equation for the

concentration of A in wt%. By symmetry, write down the equation for the

concentration of B in wt%.

Problems

58

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59

4. Heat pure copper. At 1083˚C it starts to melt. While it is melting, solid

and liquid copper co-exist. Using the definition of a phase, are one or two

phases present? …………………….

Why ? ……………………………...

5.Three components A, B and C of an alloy dissolve completely when liquid

but have no mutual solubility when solid. They do not form any chemical

compounds. How many phases and of what compositions, do you think

would appear in the solid state?

Number of phases …………

Compositions ……………..

Problems

60

6. Use the Pb-Sn diagram in figure below to answer the following questions

(6 to 9).

What are the values of the two state variables at point 1?

Composition: …………………….

Temperature: ……………………..

Pb–Sn phase diagram

Problems

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7. Mark the constitution points for Pb-70 wt% Sn and Pb–20 wt% Sn alloys

at 250˚C.

What does the alloy consist of in each case?

Pb–70wt%Sn: …………………

Pb–20wt%Sn: …………………

8. The alloy corresponding initially to the constitution point 1 is cooled very

slowly to room temperature.

At which temperatures do changes in the number or type of phases occur?

…………………………………………………

What phases are present at point 2? ……………………

What phases are present at point 3? ……………………

9. (Similarly, the alloy corresponding to the constitution point 4 is cooled to

room temperature. Identify the following:

Initial composition and temperature: …………..

Initial phase(s): ……………………………..

Temperature at which change of phase occurs: …………………

Final phase(s): …………………………61

Problems

62

10. Use the Pb-Sn diagram in figure below to answer the following

questions (10 to ).

The constitution point for a Sn-70 wt% Pb alloy at 250˚C lies in a two-

phase field. Construct a tie-line on the figure and read off the two phases

and their compositions:

Phases: ………………………..

Composition of phase 1: ………………..

Composition of phase 2: ……………………

Pb–Sn phase diagram, showing a tie-line.

Problems

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63

11. The alloy is slowly cooled to 200˚C. At this temperature, identify the:

Phases: ………………

Composition of phase 1: ……………..

Composition of phase 2: ……………

12. The alloy is cooled further to 1508C. At this temperature, identify the:

Phases: …………….

Composition of phase 1: ……………..

Composition of phase 2: ……………

Problems

64

13. Indicate with arrows on the figure the lines along which:

(1) The composition of phase 1 moves

(2) The composition of phase 2 moves

The overall composition of the alloy stays the same, of course. How do

you think the compositions of the phases change? ………………….

Problems

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Prepared by:


Chapter 4:On Mechanical Properties of Metals and Alloys

2






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1. Define stress, strain, and Poisson’s ratio.

2. State Hooke’s law and note the conditions under which it is valid.

3. In an stress-strain diagram, determine the modulus of elasticity, the yield

strength and the tensile strength, and estimate the percent elongation.

4. For the tensile deformation of a ductile cylindrical specimen, describe

changes in specimen profile to the point of fracture.

5. Compute ductility in terms of both percent elongation and percent reduction

of area for a material that is loaded in tension to fracture.

6. For a specimen being loaded in tension, given the applied load, the

instantaneous cross-sectional dimensions, as well as original and

instantaneous lengths, be able to compute true stress and true strain values.

7. Name the two most common hardness-testing techniques; note two

differences between them.

8. Name and briefly describe the two different micro-indentation hardness

testing techniques, and cite situations for which these techniques are

generally used. 3


Learning Objectives

The charts provide information on the melting point, tensile strength,

electrical conductivity, magnetic properties, and other properties of a

particular metal or alloy 4

Introduction

The characteristics of metals and alloys are explained by

studying:

I. Physical properties

II. Chemical properties

The various properties of metals and alloys were determined in the

laboratories of manufacturers and by various societies interested in

metallurgical development.

Charts presenting the properties of a particular metal or alloy are

available in many commercially published reference books and

scientific articles.

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Many materials are subjected to forces or loads.

The mechanical behavior of a material reflects the relationship between

its deformations to an applied force.

Important mechanical properties are strength, hardness, ductility,

stiffness, toughness, elasticity, plasticity, brittleness, and malleability.

These properties are described in terms of the types of force or stress

that the metal must hold out and how these are resisted.

It is possible for the load to be tensile, compressive, or shear, and its

magnitude may be constant with time, or it may fluctuate continuously.

There is a relationship between the microstructure of materials and their

mechanical properties. 5

Introduction

6

There are three principal ways in which a load may be applied: namely,

tension, compression , shear and torsion.

Concepts of Stress and Strain

Tension Compression Shear Torsion

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The tensile testing machine is designed to elongate the specimen at a

constant rate, and to continuously and simultaneously measure the

instantaneous applied load (with a load cell) and the resulting

elongations (using an extensometer). 7

Tension Tests

Tension is one of the most common mechanical stress-strain tests.

A standard tensile specimen is shown below:

in which l0 is the original length before any load is applied, and li is the

instantaneous length.

Sometimes the quantity li-l0 is denoted as ∆l and is the deformation

elongation or change in length at some instant, as referenced to the

original length.8

Tension Tests

The output of such a tensile test is recorded as force versus elongation.

These load-deformation characteristics are dependent on the specimen

size.

Stress σ is defined by the relationship:

in which F is the instantaneous load applied perpendicular to the

specimen cross section, in units of Newton (N), and A0 is the original

cross-sectional area before any load is applied (m2).

Strain ϵ is defined according to:

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9

Compression Tests

A compression test is conducted in a manner similar to the tensile test,

except that the force is compressive and the specimen contracts along

the direction of the stress.

By convention, a compressive force is taken to be negative, which

yields a negative stress.

Furthermore, since li is greater than l0, then compressive strains are

necessarily also negative.

10

Shear and Torsional Tests

Shear stress: = F / Ao

F is applied parallel to upper and lower faces each having area A0.

Shear strain: = tan ( 100 %)

is strain angle.

Torsion: like shear.

Load: applied torque, T.

Strain: angle of twist, .

ShearTorsion

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11

Stress-Strain Behavior

Elastic deformation

Reversible:

Stress removed Material returns to original

size

Plastic deformation

Irreversible:

Stress removed Material does not return to

original dimensions.

There are two types of deformation: elastic deformation and inelastic

(plastic) deformation.

12

Elastic deformation

This is known as Hooke’s law, and the constant of proportionality E

(GPa or psi) is the modulus of elasticity, or Young’s modulus.

For most metals that are stressed in tension and at relatively low levels,

stress and strain are proportional to each other through the relationship:

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Example 1:

A piece of copper originally 305 mm long is towed in tension with a

stress of 276 MPa. If the deformation is entirely elastic, what will be

the resultant elongation? (Hint: Elastic modulus for Cu is 110 GPa)

Solution:

A cylindrical specimen of a Ti alloy having an elastic modulus of 107

Gpa and an original diameter of 3.8 mm will experience only elastic

deformation when a tensile load of 2000 N is applied. Compute the

maximum length of the specimen before deformation if the maximum

allowable elongation is 0.42 mm.

14

A specimen of Al having a rectangular cross section 10 mm × 12.7 mm

is pulled in tension with 35,500 N force, producing only elastic

deformation. (Hint: Elastic modulus for Al is 69 × 109 N/m2)

Calculate the resulting strain.

Example 2, 3:

The cross-sectional area is (10 mm) × (12.7 mm) = 127 mm2 (= 1.27 ×

10-4 m2) then the strain yields :

Solution:

Left to you

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15

Poisson’s ratio

A parameter termed Poisson’s ratio ʋ is defined as the ratio of the

lateral and axial strains, or:

For isotropic materials, shear , G,

and elastic, E, modulus are related

to each other and to Poisson’s

ratio, ʋ, according to:

16

Example 4:

A tensile stress is to be applied along the long axis of a cylindrical

brass rod that has a diameter of 10 mm (0.4 in.).

Determine the magnitude of the load required to produce a 2.5×10-3 mm

(10-4 in.) change in diameter if the deformation is entirely elastic.

This deformation situation is represented in

the accompanying drawing.

Solution

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Strength is the property that enables a metal to resist deformation

under load

The ultimate strength is the maximum strain a material can hold out.

Tensile strength is a measurement of the resistance to being pulled

apart when placed in a tension load.

Fatigue strength is the ability of material to resist various kinds of

rapidly changing stresses and is expressed by the magnitude of

alternating stress for a specified number of cycles.

Impact strength is the ability of a metal to resist suddenly applied

loads and is measured in foot-pounds of force.17

Strength

Toughness is the property that enables a material to withstand shock

and to be deformed without rupturing.

Toughness may be considered as a combination of strength and

plasticity.

When a material has a load applied to it, the load causes the material to

deform.

Elasticity is the ability of a material to return to its original shape after

the load is removed.

Theoretically, the elastic limit of a material is the limit to which a

material can be loaded and still recover its original shape after the load

is removed.

Plasticity is the ability of a material to deform permanently without

breaking or rupturing.18

Toughness, Elasticity & Plasticity

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Brittleness is the opposite of the property of plasticity.

A brittle metal is one that breaks or shatters before it deforms.

White cast iron and glass are good examples of brittle material.

Generally, brittle metals are high in compressive strength but low in

tensile strength.

As an example, you would not choose cast iron for fabricating support

beams in a bridge.

Ductility is the property that enables a material to stretch, bend or

twist without cracking or breaking.

This property makes it possible for a material to be drawn out into a

thin wire.

In comparison, malleability is the property that enables a material to

deform by compressive forces without developing defects.

A malleable material is one that can be stamped, hammered, forged,

pressed, or rolled into thin sheets. 19

Brittleness, Ductility & Malleability

20

Plastic Deformation

For most metallic materials, elastic deformation persists only to strains

of about 0.005.

As the material is deformed beyond this point, the stress is no longer

proportional to strain or plastic deformation occurs.

Plastic deformation:

•stress not proportional to strain.

•deformation is not reversible.

•deformation occurs by breaking

and re-arrangement of atomic

bonds (crystalline materials by

motion of defects).

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Yield stress, y , usually more important than tensile strength.

Once yield stress has been passed, structure has deformed beyond

acceptable limits. 21

Tensile Strength

Tensile strength or

max. stress

Fracture Strength

If stress maintained specimen will break

22

Ductility

Ductility is a measure of the degree of plastic deformation that has

been sustained at fracture.

100l

llEL%

0

0f

100A

AARA%

0

f0

percent elongation

or

percent reduction in area

Effect of temperature on ductility

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23

Hardness

Hardness, is a measure of a material’s resistance to localized plastic

deformation.

Hardness tests are performed more frequently than any other

mechanical test for several reasons:

1.They are simple and inexpensive ordinarily no special specimen need

be prepared, and the testing apparatus is relatively inexpensive.

2.The test is nondestructive the specimen is neither fractured nor

excessively deformed; a small indentation is the only deformation.

3.Other mechanical properties often may be estimated from hardness

data, such as tensile strength.

24

Hardness

Rockwell Hardness Tester

The Rockwell tests constitute the most

common method used to measure hardness

because they are so simple to perform and

require no special skills.

With this system, a hardness number is

determined by the difference in depth of

penetration resulting from the application of an

initial minor load followed by a larger major

load; utilization of a minor load enhances test

accuracy.

For Rockwell, the minor load is 10 kg, whereas

major loads are 60, 100, and 150 kg.

Rockwell Hardness Tests

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In Brinell tests, as in Rockwell measurements,

a hard, spherical indenter is forced into the

surface of the metal to be tested.

The Brinell hardness number, HB, is a function

of both the magnitude

of the load and the diameter of the resulting

indentation.

Standard loads range between 500 and 3000 kg

in 500-kg increments; during a test, the load is

maintained constant for a specified time

(between 10 and 30 s).

25

Brinell Hardness Tests

Hardness-Testing Techniques

Brinell Hardness Tester

26

Knoop and Vickers Microindentation Hardness Tests

For each test a very small diamond indenter having pyramidal geometry

is forced into the surface of the specimen.

Applied loads are much smaller than for Rockwell and Brinell, ranging

between 1 and1000 g.

The resulting impression is observed under a microscope and

measured; this measurement is then converted into a hardness number.


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The following table show different hardness testing techniques and the

formula for hardness number for each one.

28

(a) A 10 mm diameter Brinell hardness indenter produced an indentation

1.62 mm in diameter in a steel alloy when a load of 500 kg was used.

Compute the HB of this material.

(b) What will be the diameter of an indentation to yield a hardness of

450 HB when a 500 kg load is used?

Example 5:

Solution: (a)

(b)

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