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The document details the fundamentals of crystal and amorphous structures within materials science, particularly focusing on unit cells, types of crystal systems, and atomic arrangements. It explains various unit cell configurations, including body-centered cubic (bcc), face-centered cubic (fcc), and hexagonal close-packed (hcp), alongside their atomic packing factors and coordination numbers. Additionally, it introduces Miller indices for identifying crystal planes and outlines the use of four indices in hexagonal unit cells.

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Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display CHAPTER 3 Crystal and Amorphous Structure in Materials Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 2 The Space Lattice and Unit Cells • Atoms, arranged in repetitive 3-Dimensional pattern, in long range order (LRO) give rise to crystal structure. • Properties of solids depends upon crystal structure and
bonding force. • An imaginary network of lines, with atoms at intersection of lines, representing the arrangement of atoms is called space lattice. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Building blocks 3 LEGO block Unit cell http://loyalkng.com/2010/03/21/lego-ironman-from-brickshelfs- legodreams-the-ironman-goes-lego/ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 4 Unit cell
• Unit cell is that block of atoms which repeats itself to form space lattice. • Materials arranged in short range order are called amorphous materials Unit Cell Space Lattice Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5 Crystal Systems and Bravais Lattice • Only seven different types of unit cells are necessary to create all point lattices. • According to Bravais (1811-1863) fourteen standard unit cells can describe all possible lattice networks. • The four basic types of unit cells are
Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6 Types of Unit Cells • Cubic Unit Cell • Tetragonal Simple Body Centered Face centered Simple Body Centered Figure 3.2 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
7 Types of Unit Cells (Cont..) • Orthorhombic ≠ b ≠ c • Rhombohedral Simple Base Centered Face Centered Body Centered Simple Figure 3.2 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 8 Types of Unit Cells (Cont..) • Hexagonal
• Monoclinic • Triclinic Simple Simple Simple Base Centered Figure 3.2 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 9 Principal Metallic Crystal Structures • 90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure.
• HCP is denser version of simple hexagonal crystal structure. BCC Structure FCC Structure HCP Structure Figure 3.3 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 10 Body Centered Cubic (BCC) Crystal Structure • Represented as one atom at each corner of cube and one at the center of cube. • Each atom has 8 nearest neighbors. • Therefore, coordination number is 8. • Examples :- Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission
required for reproduction or display 11 BCC Crystal Structure (Cont..) • Each unit cell has eight 1/8 atom at corners and 1 full atom at the center. • Therefore each unit cell has • Atoms contact each other at cube diagonal (8x1/8 ) + 1 = 2 atoms 3 4RTherefore, a = Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 12 Atomic Packing Factor of BCC Structure Atomic Packing Factor = Volume of atoms in unit cell Volume of unit cell
Vatoms = = 8.373R3 3 3 4 = 12.32 R3 Therefore APF = 8.723 R 3 12.32 R3 = 0.68 V unit cell = a3 =
3 4 .2 3R Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 13 Face Centered Cubic (FCC) Crystal Structure • FCC structure is represented as one atom each at the corner of cube and at the center of each cube face. • Coordination number for FCC structure is 12 • Atomic Packing Factor is 0.74 • Examples :- old (a = 0.408)
Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 14 FCC Crystal Structure (Cont..) • Each unit cell has eight 1/8 atom at corners and six ½ atoms at the center of six faces. • Therefore each unit cell has • Atoms contact each other across cubic face diagonal (8 x 1/8)+ (6 x ½) = 4 atoms 2 4RTherefore, lattice constant a = Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission
required for reproduction or display 15 Hexagonal Close-Packed Structure • The HCP structure is represented as an atom at each of 12 corners of a hexagonal prism, 2 atoms at top and bottom face and 3 atoms in between top and bottom face. • Atoms attain higher APF by attaining HCP structure than simple hexagonal structure. • The coordination number is 12, APF = 0.74. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 16 HCP Crystal Structure (Cont..) • Each atom has six 1/6 atoms at each of top and bottom layer, two half atoms at top and bottom layer and 3 full atoms at the middle layer. • Therefore each HCP unit cell has • Examples:-
• Ideal c/a ratio is 1.633. (2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display YouTube videos links • http://www.youtube.com/watch?v=Rm- i1c7zr6Q • http://www.youtube.com/watch?v=F4Du 4zI4GJ0 • http://www.youtube.com/watch?v=vYky UqUa6vU • http://www.youtube.com/watch?v=NYV SI83KiKU 17 http://www.youtube.com/watch?v=Rm-i1c7zr6Q http://www.youtube.com/watch?v=F4Du4zI4GJ0 http://www.youtube.com/watch?v=vYkyUqUa6vU http://www.youtube.com/watch?v=NYVSI83KiKU Foundations of Materials Science and Engineering, 5th Edn.
Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 18 Atom Positions in Cubic Unit Cells • Cartesian coordinate system is use to locate atoms. • In a cubic unit cell posite of positive directions. • Atom positions are located using unit distances along the axes. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 19 Directions in Cubic Unit Cells • In cubic crystals, Direction Indices are vector components of directions resolved along each axes, resolved to smallest
integers. • Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, converted to integers. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 20 Procedure to Find Direction Indices (0,0,0) (1,1/2,1) z Produce the direction vector till it emerges from surface of cubic cell Determine the coordinates of point of emergence and origin Subtract coordinates of point of Emergence by that of origin (1,1/2,1) - (0,0,0) = (1,1/2,1) Are all are
integers? Convert them to smallest possible integer by multiplying by an integer. 2 x (1,1/2,1) = (2,1,2) Are any of the direction vectors negative? Represent the indices in a square bracket without comas with a over negative index (Eg: [121]) Represent the indices in a square bracket without comas (Eg: [212] ) The direction indices are [212] x y YES NO YES NO
Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 21 Direction Indices - Example • Determine direction indices of the given vector. Origin coordinates are (3/4 , 0 , 1/4). Emergence coordinates are (1/4, 1/2, 1/2). Subtracting origin coordinates from emergence coordinates, (3/4 , 0 , 1/4) - (1/4, 1/2, 1/2) = (-1/2, 1/2, 1/4) Multiply by 4 to convert all fractions to integers 4 x (-1/2, 1/2, 1/4) = (-2, 2, 1) Therefore, the direction indices are [ 2 2 1 ] Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 22 Miller Indices
• Miller Indices are are used to refer to specific lattice planes of atoms. • They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell. z x y Miller Indices =(111) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 23 Miller Indices - Procedure Choose a plane that does not pass through origin Determine the x,y and z intercepts of the plane Find the reciprocals of the intercepts
Fractions? Clear fractions by multiplying by an integer to determine smallest set of whole numbers Enclose in parenthesis (hkl)where h,k,l are miller indicesof cubic crystal plane forx,y and z axes. Eg: (111) Place a ‘bar’ over the Negative indices Yes Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Miller Indices - Examples • Intercepts of the plane at x,y & z axes are 1, ∞ and ∞ • Taking reciprocals we get (1,0,0). • Miller indices are (100).
******************* • Intercepts are 1/3, 2/3 & 1. • taking reciprocals we get (3, 3/2, 1). • Multiplying by 2 to clear fractions, we get (6,3,2). • Miller indices are (632). x x y z (100) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Miller Indices - Examples • Plot the plane (101) Taking reciprocals of the indices we get (1 ∞ 1). The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1.
****************************** • Plot the plane (2 2 1) Taking reciprocals of the indices we get (1/2 1/2 1). The intercepts of the plane are x=1/2, y= 1/2 and z=1. Figure EP3.7 a Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 26 Miller Indices - Example • Plot the plane (110) The reciprocals are (1,-1, ∞) The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis) To show this plane in a single unit cell, the origin is moved along the positive direction of y axis by 1 unit. x y z(110)
Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 27 Miller Indices – Important Relationship • Direction indices of a direction perpendicular to a crystal plane are same as miller indices of the plane. • Example:- • Interplanar spacing between parallel closest planes with same miller indices is given by lkh d a hkl 222 ++ = Figure EP3.7b Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 28 Planes and Directions in Hexagonal Unit Cells • Four indices are used (hkil) called as Miller- Bravais indices. • Four axes are used (a1, a2, a3 and c). • Reciprocal of the intercepts that a crystal plane makes with the a1, a2, a3 and c axes give the h,k,I and l indices respectively. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 29 Hexagonal Unit Cell - Examples • Basal Planes:- Intercepts a1 = ∞ a2 = ∞ a3 = ∞ c = 1 (hkli) = (0001)
• Prism Planes :- For plane ABCD, Intercepts a1 = 1 a2 = ∞ a3 = -1 c = ∞ (hkli) = (1010) Figure 3.17 a&b Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 30 Directions in HCP Unit Cells • Indicated by 4 indices [uvtw]. • u,v,t and w are lattice vectors in a1, a2, a3 and c directions respectively. • Example:- For a1, a2, a3 directions, the direction indices are [ 2 1 1 0], [1 2 1 0] and [ 1 1 2 0] respectively.
Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 31 Comparison of FCC and HCP crystals • Both FCC and HCP are close packed and have APF 0.74. • FCC crystal is close packed in (111) plane while HCP is close packed in (0001) plane. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 32 Structural Difference between HCP and FCC Consider a layer of atoms (Plane ‘A’) Another layer (plane ‘B’) of atoms is placed in ‘a’ Void of plane ‘A’
Third layer of Atoms placed in ‘b’ Voids of plane ‘B’. (Identical to plane ‘A’.) HCP crystal. Third layer of Atoms placed in ‘a’ voids of plane ‘B’. Resulting In 3rd Plane C. FCC crystal. Plane A ‘a’ void ‘b’ void Plane A Plane B ‘a’ void ‘b’ void Plane A Plane B Plane A Plane A Plane B Plane C Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
33 Volume Density • Volume density of metal = • Example:- Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. vρ Mass/Unit cellVolume/Unit cell= a= 2 4R = 2 1278.04 nm× = 0.361 nm Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3 vρ FCC unit cell has 4 atoms. Mass of unit cell = m =
× − g Mg molatmos molgatoms 6 23 10 /107.4 )/54.63)(4( = 4.22 x 10-28 Mg 33329 28 98.898.8 107.4 1022.4
cm g m Mg m Mg V m == × × == − − Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 34
Planar Atomic Density • Planar atomic density= • Example:- In Iron (BCC, a=0.287), The (100) plane intersects center of 5 atoms (Four ¼ and 1 full atom). Area of 110 plane = ρ p= Equivalent number of atoms whose centers are intersected by selected area Selected area 222 aaa =× ρ p ( )2287.02 2 = 2 13 2 1072.12.17 mmnm
atoms × == Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 35 Linear Atomic Density • Linear atomic density = • Example:- For a FCC copper crystal (a=0.361), the [110] direction intersects 2 half diameters and 1 full diameter. Length of line = ρ l = Number of atomic diameters intersected by selected length of line in direction of interest Selected length of line mm
atoms nm atoms nm atoms 61092.392.3 361.02 2 × == × =ρ l nm361.02 × Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 36 Polymorphism or Allotropy • Metals exist in more than one crystalline form. This is
caller polymorphism or allotropy. • Temperature and pressure leads to change in crystalline forms. • Example:- Iron exists in both BCC and FCC form depending on the temperature. -2730C 9120C 13940C 15390C α Iron BCC γ Iron FCC δ Iron BCC Liquid Iron Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 37 Crystal Structure Analysis • Information about crystal structure are obtained using X- Rays.
• The X-rays used are about the same wavelength (0.05- 0.25 nm) as distance between crystal lattice planes. 35 KV (Eg: Molybdenum) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display X-Ray Spectrum of Molybdenum • X-Ray spectrum of Molybdenum is obtained when Molybdenum is used as target metal. • Kα and Kβ are characteristic of an element. • For Molybdenum Kα occurs at wave length of about 0.07nm. • Electrons of n=1 shell of target metal are knocked out by bombarding electrons. • Electrons of higher level drop down by releasing energy to replace lost electrons
48 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 39 X-Ray Diffraction • Crystal planes of target metal act as mirrors reflecting X- ray beam. • If rays leaving a set of planes are out of phase (as in case of arbitrary angle of incidence) no reinforced beam is produced. • If rays leaving are in phase, reinforced beams are produced. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
40 X-Ray Diffraction (Cont..) • For rays reflected from different planes to be in phase, the extra distance traveled by a ray should be a integral multiple of wave length λ . nλ = MP + PN (n = 1,2…) n is order of diffraction If dhkl is interplanar distance, Then MP = PN = dhkl.Sinθ Therefore, λ = 2 dhkl.Sinθ After A.G. Guy and J.J. Hren, “Elements of Physical Metallurgy,” 3d ed., Addison-Wesley, 1974, p.201.) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 41 Interpreting Diffraction Data • We know that 2
2222 2 222 222 4 2 2 a lkh Sin lkh aSin dSin lkh a d hkl ++ = ++ =
= ++ = λ θ θ λ θλSince Note that the wavelength λ and lattice constant a are the same For both incoming and outgoing radiation. Substituting for d, Therefore Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 42 Interpreting Diffraction Data (Cont..) • For planes ‘A’ and ‘B’ we get two equations 2
2222 2 2 2222 2 4 )( 4 )( a lkh Sin a lkh Sin BBB B AAA A ++ = ++
= λ θ λ θ (For plane ‘A’) (For plane ‘B’) Dividing each other, we get )( )( 222 222 2 2 BBB AAA B A lkh lkh Sin
Sin ++ ++ = θ θ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 43 X-Ray Diffraction Analysis • Powdered specimen is used for X-ray diffraction analysis as the random orientation facilitates different angle of incidence. • Radiation counter detects angle and intensity of diffracted beam. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 44 Diffraction Condition for Cubic Cells • For BCC structure, diffraction occurs only on planes whose miller indices when added together total to an even number. I.e. (h+k+l) = even Reflections present (h+k+l) = odd Reflections absent • For FCC structure, diffraction occurs only on planes whose miller indices are either all even or all odd. I.e. (h,k,l) all even Reflections present (h,k,l) all odd Reflections present (h,k,l) not all even or all odd Reflections absent. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 45 Interpreting Experimental Data • For BCC crystals, the first two sets of diffracting planes are {110} and {200} planes.
Therefore • For FCC crystals the first two sets of diffracting planes are {111} and {200} planes Therefore 5.0 )002( )011( 222 222 2 2 = ++ ++ = B A Sin Sin θ θ
75.0 )002( )111( 222 222 2 2 = ++ ++ = B A Sin Sin θ θ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 46 Crystal Structure of Unknown Metal Unknown metal Crystallographic Analysis FCC Crystal Structure BCC Crystal Structure 75.02 2 = B A Sin Sin θ θ 5.0
2 2 = B A Sin Sin θ θ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 47 Amorphous Materials • Random spatial positions of atoms • Polymers: Secondary bonds do not allow formation of parallel and tightly packed chains during solidification. micrystalline.
• Glass is a ceramic made up of SiO4 4- tetrahedron subunits – limited mobility. • Rapid cooling of metals (10 8 K/s) can give rise to amorphous structure (metallic glass). • Metallic glass has superior metallic properties. Definitions • Crystal structure: the manner in which atoms, ions and molecules are spatially arranged • Lattice: a three dimensional array of points coinciding with atom positions • Allotropy: a property of an element to coexist in more than one crystal structure • Amorphous materials: materials like glasses that have no long range order or crystal structure • Isotropic: having same properties in all directions • Anisotropic: property of materials that have different properties in different directions • Unit cell: subdivision of a crystal structure into small repeat entities • Basis or motif: a group of atoms associated with a lattice point • Coordination number: number of nearest neighbors of an atom in a crystal structure • Atomic packing fraction: fraction of plane, volume or direction that is actually occupied by atoms • Lattice parameter: the unit cell edge lengths and angle between them • Closed packed directions: directions in a crystal lattice along which atoms are in contact
Definitions • Cubic site: an interstitial site with coordination number of eight • Tetrahedral site: an interstitial site with coordination number of four • Octahedral site: an interstitial site with coordination number of six • Bravais lattice: fourteen possible lattice arrangements that are possible using lattice points in 3D space • Miller indices: a notation used to describe specific crystallographic planes and directions • Miller indices of directions: are vector components of the directions resolved along each of the coordinate axes • Miller indices of planes: are defined as reciprocals of fractional intercepts that the planes makes with coordinate axes • Metallic glass: metals with an amorphous atomic structure • Linear atomic density: the number of lattice points per unit length along a direction • Planar density: the number of atoms whose centers are intersected by the plane per unit area • Volume density: the ratio of mass per unit volume of a unit cell Ch03-revised-D2LCHAPTER �3The Space Lattice and Unit CellsBuilding blocksUnit cellCrystal Systems and Bravais LatticeTypes of Unit CellsTypes of Unit Cells (Cont..)Types of Unit Cells (Cont..)Principal Metallic Crystal Structures Body Centered Cubic (BCC) Crystal StructureBCC Crystal Structure (Cont..)Atomic Packing Factor of BCC Structure Face Centered Cubic (FCC) Crystal StructureFCC Crystal Structure
(Cont..)Hexagonal Close-Packed StructureHCP Crystal Structure (Cont..)YouTube videos linksAtom Positions in Cubic Unit CellsDirections in Cubic Unit CellsProcedure to Find Direction IndicesDirection Indices - ExampleMiller IndicesMiller Indices - ProcedureMiller Indices - ExamplesMiller Indices - ExamplesMiller Indices - ExampleMiller Indices – Important RelationshipPlanes and Directions in Hexagonal Unit Cells Hexagonal Unit Cell - ExamplesDirections in HCP Unit CellsComparison of FCC and HCP crystalsStructural Difference between HCP and FCCVolume DensityPlanar Atomic DensityLinear Atomic DensityPolymorphism or AllotropyCrystal Structure AnalysisX- Ray Spectrum of MolybdenumX-Ray DiffractionX-Ray Diffraction (Cont..)Interpreting Diffraction DataInterpreting Diffraction Data (Cont..)X-Ray Diffraction AnalysisDiffraction Condition for Cubic CellsInterpreting Experimental DataCrystal Structure of Unknown MetalAmorphous MaterialsDefinitions- ch-3DefinitionsDefinitions ENGR 1210 HW 3 1 Homework 3 on chapter 3; (Total points: 100) Reading assignments: Chapter 3 from textbook, notes and slides Student Name: _______________________________ MTSU ID: ____________________ Honor pledge: I _______________________________ acknowledge that I have neither received
nor given any unauthorized help/aid during this assignment. Part 1 Key Engineering Terms: fill in the blanks – 25 points 1. Crystal structure: The manner in which atoms, ions and molecules are _____________________________________ arranged. 2. Lattice: A three dimensional array of points coinciding with ________________________________ positions. 3. Unit cell: A ____________________ of lattice into small _________________________ entities still retaining overall characteristics of the entire lattice. 4. Basis or motif: a group of atoms associated with _____________________________. 5. Crystalline material: A crystalline material is one in which the atoms are situated in a _____________________________________ or periodic array over ________________________________ atomic distances. 6. Coordination number: number of ________________________________ neighbors of an atom in a crystal structure. 7. Lattice parameter: The unit cell
___________________________ and the _________________________ between the edges. 8. Closed packed direction: direction in a crystal lattice along which atoms are in _________________________________. 2 9. Bravais lattices: _______________________________ possible lattices arrangements using lattice points in 3 dimensional space. 10. Atomic packing fraction: a fraction of a plane, volume or direction that is _____________________________ occupied with atoms. 11. Allotropy or Polymorphism: a property of an element to coexist in more than ________________________________ crystal structures. 12. Amorphous Materials: Materials including glasses that have ______________________________ long range order or crystal structure. 13. Isotropic: Having _________________________________
properties in all directions. 14. Anisotropic: Having _________________________________ properties in different directions. 15. Miller Indices of Directions: Are _____________________________ components of the directions resolved along each of the coordinate axes. 16. Miller Indices of Planes: Are _____________________________ of fractional _________________________________ that plane makes with coordinate axes. 17. Linear density: It is defined as the number of lattice points or atomic diameters per unit ____________________________________ of the direction. 18. Planar density: The number of atoms whose ________________________________ are intersected by the plane per unit _____________________________. 19. X-ray diffraction: A technique to analyze the _____________________________ of materials using a beam of X-rays as source. 20. Metallic glass: Metals with ____________________________ crystal structure.
3 Part 2 Learning concepts/reflections/problems – 65 points 1. Sketch the unit cells of Simple Cubic, Body Centered Cubic and Face Centered Cubic lattices and determine and show calculations for following: A. Number of atoms per unit cell B. Equation/relation for radius and lattice parameter (unit cell edge length) C. Atomic Packing Fraction (APF) 15 points 2. Calculate the radius of a palladium (Pd) 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. Assume Avogadro’s number = 6.022 x 1023 atoms/mol 5 points 3. Determine the Miller Indices of directions A, B, C and D as shown in figure below: 10 points
4 4. Determine Miller Indices of planes as shown in figure below: 10 points 5. Determine Miller Bravais Indices [hkil] of directions as shown in figure below: 10 points 5 6. Determine Miller Bravais Indices (hkil) of planes as shown in figure below: 10 points 7. Define diffraction. Write an expression for Bragg’s law of diffraction and explain the terms of the equation. 5 points
Part 3 Materials design – 10 points 1. Why metallic crystal structure are dense? Explain. 5 points 2. Give 2 materials examples of each of the following crystal structures: a. Simple cubic (SC) b. Body Centered Cubuc (BCC) c. Face Centered Cubic (FCC) d. Hexagonal Closed Packed (HCP) 5 points 6 Bonus questions: 1. Explain why the properties of polycrystalline materials are most often isotropic? 5 points 2. Explain two differences between an optical microscope and a scanning electron microscope (SEM). 5 points 3. Calculate theoretical density of Chromium which has Body
Centered Cubic structure and atomic weight of 52 g/mol and atomic radius of 0.125 nm. Assume Avogadro’s number = 6.022 x 1023 atoms/mol 5 points Note: Please use extra sheets wherever necessary to answer the questions. All HW must be uploaded via D2L Dropbox system. Students are highly encouraged to refer textbook, notes and slides.

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Foundations of Materials Science and Engineering, 5th Edn. Sm.docx

  • 1. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display CHAPTER 3 Crystal and Amorphous Structure in Materials Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 2 The Space Lattice and Unit Cells • Atoms, arranged in repetitive 3-Dimensional pattern, in long range order (LRO) give rise to crystal structure. • Properties of solids depends upon crystal structure and
  • 2. bonding force. • An imaginary network of lines, with atoms at intersection of lines, representing the arrangement of atoms is called space lattice. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Building blocks 3 LEGO block Unit cell http://loyalkng.com/2010/03/21/lego-ironman-from-brickshelfs- legodreams-the-ironman-goes-lego/ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 4 Unit cell
  • 3. • Unit cell is that block of atoms which repeats itself to form space lattice. • Materials arranged in short range order are called amorphous materials Unit Cell Space Lattice Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5 Crystal Systems and Bravais Lattice • Only seven different types of unit cells are necessary to create all point lattices. • According to Bravais (1811-1863) fourteen standard unit cells can describe all possible lattice networks. • The four basic types of unit cells are
  • 4. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6 Types of Unit Cells • Cubic Unit Cell • Tetragonal Simple Body Centered Face centered Simple Body Centered Figure 3.2 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
  • 5. 7 Types of Unit Cells (Cont..) • Orthorhombic ≠ b ≠ c • Rhombohedral Simple Base Centered Face Centered Body Centered Simple Figure 3.2 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 8 Types of Unit Cells (Cont..) • Hexagonal
  • 6. • Monoclinic • Triclinic Simple Simple Simple Base Centered Figure 3.2 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 9 Principal Metallic Crystal Structures • 90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure.
  • 7. • HCP is denser version of simple hexagonal crystal structure. BCC Structure FCC Structure HCP Structure Figure 3.3 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 10 Body Centered Cubic (BCC) Crystal Structure • Represented as one atom at each corner of cube and one at the center of cube. • Each atom has 8 nearest neighbors. • Therefore, coordination number is 8. • Examples :- Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission
  • 8. required for reproduction or display 11 BCC Crystal Structure (Cont..) • Each unit cell has eight 1/8 atom at corners and 1 full atom at the center. • Therefore each unit cell has • Atoms contact each other at cube diagonal (8x1/8 ) + 1 = 2 atoms 3 4RTherefore, a = Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 12 Atomic Packing Factor of BCC Structure Atomic Packing Factor = Volume of atoms in unit cell Volume of unit cell
  • 9. Vatoms = = 8.373R3 3 3 4 = 12.32 R3 Therefore APF = 8.723 R 3 12.32 R3 = 0.68 V unit cell = a3 =
  • 10. 3 4 .2 3R Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 13 Face Centered Cubic (FCC) Crystal Structure • FCC structure is represented as one atom each at the corner of cube and at the center of each cube face. • Coordination number for FCC structure is 12 • Atomic Packing Factor is 0.74 • Examples :- old (a = 0.408)
  • 11. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 14 FCC Crystal Structure (Cont..) • Each unit cell has eight 1/8 atom at corners and six ½ atoms at the center of six faces. • Therefore each unit cell has • Atoms contact each other across cubic face diagonal (8 x 1/8)+ (6 x ½) = 4 atoms 2 4RTherefore, lattice constant a = Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission
  • 12. required for reproduction or display 15 Hexagonal Close-Packed Structure • The HCP structure is represented as an atom at each of 12 corners of a hexagonal prism, 2 atoms at top and bottom face and 3 atoms in between top and bottom face. • Atoms attain higher APF by attaining HCP structure than simple hexagonal structure. • The coordination number is 12, APF = 0.74. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 16 HCP Crystal Structure (Cont..) • Each atom has six 1/6 atoms at each of top and bottom layer, two half atoms at top and bottom layer and 3 full atoms at the middle layer. • Therefore each HCP unit cell has • Examples:-
  • 13. • Ideal c/a ratio is 1.633. (2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display YouTube videos links • http://www.youtube.com/watch?v=Rm- i1c7zr6Q • http://www.youtube.com/watch?v=F4Du 4zI4GJ0 • http://www.youtube.com/watch?v=vYky UqUa6vU • http://www.youtube.com/watch?v=NYV SI83KiKU 17 http://www.youtube.com/watch?v=Rm-i1c7zr6Q http://www.youtube.com/watch?v=F4Du4zI4GJ0 http://www.youtube.com/watch?v=vYkyUqUa6vU http://www.youtube.com/watch?v=NYVSI83KiKU Foundations of Materials Science and Engineering, 5th Edn.
  • 14. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 18 Atom Positions in Cubic Unit Cells • Cartesian coordinate system is use to locate atoms. • In a cubic unit cell posite of positive directions. • Atom positions are located using unit distances along the axes. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 19 Directions in Cubic Unit Cells • In cubic crystals, Direction Indices are vector components of directions resolved along each axes, resolved to smallest
  • 15. integers. • Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, converted to integers. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 20 Procedure to Find Direction Indices (0,0,0) (1,1/2,1) z Produce the direction vector till it emerges from surface of cubic cell Determine the coordinates of point of emergence and origin Subtract coordinates of point of Emergence by that of origin (1,1/2,1) - (0,0,0) = (1,1/2,1) Are all are
  • 16. integers? Convert them to smallest possible integer by multiplying by an integer. 2 x (1,1/2,1) = (2,1,2) Are any of the direction vectors negative? Represent the indices in a square bracket without comas with a over negative index (Eg: [121]) Represent the indices in a square bracket without comas (Eg: [212] ) The direction indices are [212] x y YES NO YES NO
  • 17. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 21 Direction Indices - Example • Determine direction indices of the given vector. Origin coordinates are (3/4 , 0 , 1/4). Emergence coordinates are (1/4, 1/2, 1/2). Subtracting origin coordinates from emergence coordinates, (3/4 , 0 , 1/4) - (1/4, 1/2, 1/2) = (-1/2, 1/2, 1/4) Multiply by 4 to convert all fractions to integers 4 x (-1/2, 1/2, 1/4) = (-2, 2, 1) Therefore, the direction indices are [ 2 2 1 ] Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 22 Miller Indices
  • 18. • Miller Indices are are used to refer to specific lattice planes of atoms. • They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell. z x y Miller Indices =(111) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 23 Miller Indices - Procedure Choose a plane that does not pass through origin Determine the x,y and z intercepts of the plane Find the reciprocals of the intercepts
  • 19. Fractions? Clear fractions by multiplying by an integer to determine smallest set of whole numbers Enclose in parenthesis (hkl)where h,k,l are miller indicesof cubic crystal plane forx,y and z axes. Eg: (111) Place a ‘bar’ over the Negative indices Yes Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Miller Indices - Examples • Intercepts of the plane at x,y & z axes are 1, ∞ and ∞ • Taking reciprocals we get (1,0,0). • Miller indices are (100).
  • 20. ******************* • Intercepts are 1/3, 2/3 & 1. • taking reciprocals we get (3, 3/2, 1). • Multiplying by 2 to clear fractions, we get (6,3,2). • Miller indices are (632). x x y z (100) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Miller Indices - Examples • Plot the plane (101) Taking reciprocals of the indices we get (1 ∞ 1). The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1.
  • 21. ****************************** • Plot the plane (2 2 1) Taking reciprocals of the indices we get (1/2 1/2 1). The intercepts of the plane are x=1/2, y= 1/2 and z=1. Figure EP3.7 a Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 26 Miller Indices - Example • Plot the plane (110) The reciprocals are (1,-1, ∞) The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis) To show this plane in a single unit cell, the origin is moved along the positive direction of y axis by 1 unit. x y z(110)
  • 22. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 27 Miller Indices – Important Relationship • Direction indices of a direction perpendicular to a crystal plane are same as miller indices of the plane. • Example:- • Interplanar spacing between parallel closest planes with same miller indices is given by lkh d a hkl 222 ++ = Figure EP3.7b Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
  • 23. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 28 Planes and Directions in Hexagonal Unit Cells • Four indices are used (hkil) called as Miller- Bravais indices. • Four axes are used (a1, a2, a3 and c). • Reciprocal of the intercepts that a crystal plane makes with the a1, a2, a3 and c axes give the h,k,I and l indices respectively. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 29 Hexagonal Unit Cell - Examples • Basal Planes:- Intercepts a1 = ∞ a2 = ∞ a3 = ∞ c = 1 (hkli) = (0001)
  • 24. • Prism Planes :- For plane ABCD, Intercepts a1 = 1 a2 = ∞ a3 = -1 c = ∞ (hkli) = (1010) Figure 3.17 a&b Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 30 Directions in HCP Unit Cells • Indicated by 4 indices [uvtw]. • u,v,t and w are lattice vectors in a1, a2, a3 and c directions respectively. • Example:- For a1, a2, a3 directions, the direction indices are [ 2 1 1 0], [1 2 1 0] and [ 1 1 2 0] respectively.
  • 25. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 31 Comparison of FCC and HCP crystals • Both FCC and HCP are close packed and have APF 0.74. • FCC crystal is close packed in (111) plane while HCP is close packed in (0001) plane. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 32 Structural Difference between HCP and FCC Consider a layer of atoms (Plane ‘A’) Another layer (plane ‘B’) of atoms is placed in ‘a’ Void of plane ‘A’
  • 26. Third layer of Atoms placed in ‘b’ Voids of plane ‘B’. (Identical to plane ‘A’.) HCP crystal. Third layer of Atoms placed in ‘a’ voids of plane ‘B’. Resulting In 3rd Plane C. FCC crystal. Plane A ‘a’ void ‘b’ void Plane A Plane B ‘a’ void ‘b’ void Plane A Plane B Plane A Plane A Plane B Plane C Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
  • 27. 33 Volume Density • Volume density of metal = • Example:- Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. vρ Mass/Unit cellVolume/Unit cell= a= 2 4R = 2 1278.04 nm× = 0.361 nm Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3 vρ FCC unit cell has 4 atoms. Mass of unit cell = m =
  • 28. × − g Mg molatmos molgatoms 6 23 10 /107.4 )/54.63)(4( = 4.22 x 10-28 Mg 33329 28 98.898.8 107.4 1022.4
  • 29. cm g m Mg m Mg V m == × × == − − Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 34
  • 30. Planar Atomic Density • Planar atomic density= • Example:- In Iron (BCC, a=0.287), The (100) plane intersects center of 5 atoms (Four ¼ and 1 full atom). Area of 110 plane = ρ p= Equivalent number of atoms whose centers are intersected by selected area Selected area 222 aaa =× ρ p ( )2287.02 2 = 2 13 2 1072.12.17 mmnm
  • 31. atoms × == Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 35 Linear Atomic Density • Linear atomic density = • Example:- For a FCC copper crystal (a=0.361), the [110] direction intersects 2 half diameters and 1 full diameter. Length of line = ρ l = Number of atomic diameters intersected by selected length of line in direction of interest Selected length of line mm
  • 32. atoms nm atoms nm atoms 61092.392.3 361.02 2 × == × =ρ l nm361.02 × Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 36 Polymorphism or Allotropy • Metals exist in more than one crystalline form. This is
  • 33. caller polymorphism or allotropy. • Temperature and pressure leads to change in crystalline forms. • Example:- Iron exists in both BCC and FCC form depending on the temperature. -2730C 9120C 13940C 15390C α Iron BCC γ Iron FCC δ Iron BCC Liquid Iron Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 37 Crystal Structure Analysis • Information about crystal structure are obtained using X- Rays.
  • 34. • The X-rays used are about the same wavelength (0.05- 0.25 nm) as distance between crystal lattice planes. 35 KV (Eg: Molybdenum) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display X-Ray Spectrum of Molybdenum • X-Ray spectrum of Molybdenum is obtained when Molybdenum is used as target metal. • Kα and Kβ are characteristic of an element. • For Molybdenum Kα occurs at wave length of about 0.07nm. • Electrons of n=1 shell of target metal are knocked out by bombarding electrons. • Electrons of higher level drop down by releasing energy to replace lost electrons
  • 35. 48 Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 39 X-Ray Diffraction • Crystal planes of target metal act as mirrors reflecting X- ray beam. • If rays leaving a set of planes are out of phase (as in case of arbitrary angle of incidence) no reinforced beam is produced. • If rays leaving are in phase, reinforced beams are produced. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
  • 36. 40 X-Ray Diffraction (Cont..) • For rays reflected from different planes to be in phase, the extra distance traveled by a ray should be a integral multiple of wave length λ . nλ = MP + PN (n = 1,2…) n is order of diffraction If dhkl is interplanar distance, Then MP = PN = dhkl.Sinθ Therefore, λ = 2 dhkl.Sinθ After A.G. Guy and J.J. Hren, “Elements of Physical Metallurgy,” 3d ed., Addison-Wesley, 1974, p.201.) Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 41 Interpreting Diffraction Data • We know that 2
  • 38. = ++ = λ θ θ λ θλSince Note that the wavelength λ and lattice constant a are the same For both incoming and outgoing radiation. Substituting for d, Therefore Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 42 Interpreting Diffraction Data (Cont..) • For planes ‘A’ and ‘B’ we get two equations 2
  • 40. = λ θ λ θ (For plane ‘A’) (For plane ‘B’) Dividing each other, we get )( )( 222 222 2 2 BBB AAA B A lkh lkh Sin
  • 41. Sin ++ ++ = θ θ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 43 X-Ray Diffraction Analysis • Powdered specimen is used for X-ray diffraction analysis as the random orientation facilitates different angle of incidence. • Radiation counter detects angle and intensity of diffracted beam. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
  • 42. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 44 Diffraction Condition for Cubic Cells • For BCC structure, diffraction occurs only on planes whose miller indices when added together total to an even number. I.e. (h+k+l) = even Reflections present (h+k+l) = odd Reflections absent • For FCC structure, diffraction occurs only on planes whose miller indices are either all even or all odd. I.e. (h,k,l) all even Reflections present (h,k,l) all odd Reflections present (h,k,l) not all even or all odd Reflections absent. Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 45 Interpreting Experimental Data • For BCC crystals, the first two sets of diffracting planes are {110} and {200} planes.
  • 43. Therefore • For FCC crystals the first two sets of diffracting planes are {111} and {200} planes Therefore 5.0 )002( )011( 222 222 2 2 = ++ ++ = B A Sin Sin θ θ
  • 44. 75.0 )002( )111( 222 222 2 2 = ++ ++ = B A Sin Sin θ θ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
  • 45. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 46 Crystal Structure of Unknown Metal Unknown metal Crystallographic Analysis FCC Crystal Structure BCC Crystal Structure 75.02 2 = B A Sin Sin θ θ 5.0
  • 46. 2 2 = B A Sin Sin θ θ Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 47 Amorphous Materials • Random spatial positions of atoms • Polymers: Secondary bonds do not allow formation of parallel and tightly packed chains during solidification. micrystalline.
  • 47. • Glass is a ceramic made up of SiO4 4- tetrahedron subunits – limited mobility. • Rapid cooling of metals (10 8 K/s) can give rise to amorphous structure (metallic glass). • Metallic glass has superior metallic properties. Definitions • Crystal structure: the manner in which atoms, ions and molecules are spatially arranged • Lattice: a three dimensional array of points coinciding with atom positions • Allotropy: a property of an element to coexist in more than one crystal structure • Amorphous materials: materials like glasses that have no long range order or crystal structure • Isotropic: having same properties in all directions • Anisotropic: property of materials that have different properties in different directions • Unit cell: subdivision of a crystal structure into small repeat entities • Basis or motif: a group of atoms associated with a lattice point • Coordination number: number of nearest neighbors of an atom in a crystal structure • Atomic packing fraction: fraction of plane, volume or direction that is actually occupied by atoms • Lattice parameter: the unit cell edge lengths and angle between them • Closed packed directions: directions in a crystal lattice along which atoms are in contact
  • 48. Definitions • Cubic site: an interstitial site with coordination number of eight • Tetrahedral site: an interstitial site with coordination number of four • Octahedral site: an interstitial site with coordination number of six • Bravais lattice: fourteen possible lattice arrangements that are possible using lattice points in 3D space • Miller indices: a notation used to describe specific crystallographic planes and directions • Miller indices of directions: are vector components of the directions resolved along each of the coordinate axes • Miller indices of planes: are defined as reciprocals of fractional intercepts that the planes makes with coordinate axes • Metallic glass: metals with an amorphous atomic structure • Linear atomic density: the number of lattice points per unit length along a direction • Planar density: the number of atoms whose centers are intersected by the plane per unit area • Volume density: the ratio of mass per unit volume of a unit cell Ch03-revised-D2LCHAPTER �3The Space Lattice and Unit CellsBuilding blocksUnit cellCrystal Systems and Bravais LatticeTypes of Unit CellsTypes of Unit Cells (Cont..)Types of Unit Cells (Cont..)Principal Metallic Crystal Structures Body Centered Cubic (BCC) Crystal StructureBCC Crystal Structure (Cont..)Atomic Packing Factor of BCC Structure Face Centered Cubic (FCC) Crystal StructureFCC Crystal Structure
  • 49. (Cont..)Hexagonal Close-Packed StructureHCP Crystal Structure (Cont..)YouTube videos linksAtom Positions in Cubic Unit CellsDirections in Cubic Unit CellsProcedure to Find Direction IndicesDirection Indices - ExampleMiller IndicesMiller Indices - ProcedureMiller Indices - ExamplesMiller Indices - ExamplesMiller Indices - ExampleMiller Indices – Important RelationshipPlanes and Directions in Hexagonal Unit Cells Hexagonal Unit Cell - ExamplesDirections in HCP Unit CellsComparison of FCC and HCP crystalsStructural Difference between HCP and FCCVolume DensityPlanar Atomic DensityLinear Atomic DensityPolymorphism or AllotropyCrystal Structure AnalysisX- Ray Spectrum of MolybdenumX-Ray DiffractionX-Ray Diffraction (Cont..)Interpreting Diffraction DataInterpreting Diffraction Data (Cont..)X-Ray Diffraction AnalysisDiffraction Condition for Cubic CellsInterpreting Experimental DataCrystal Structure of Unknown MetalAmorphous MaterialsDefinitions- ch-3DefinitionsDefinitions ENGR 1210 HW 3 1 Homework 3 on chapter 3; (Total points: 100) Reading assignments: Chapter 3 from textbook, notes and slides Student Name: _______________________________ MTSU ID: ____________________ Honor pledge: I _______________________________ acknowledge that I have neither received
  • 50. nor given any unauthorized help/aid during this assignment. Part 1 Key Engineering Terms: fill in the blanks – 25 points 1. Crystal structure: The manner in which atoms, ions and molecules are _____________________________________ arranged. 2. Lattice: A three dimensional array of points coinciding with ________________________________ positions. 3. Unit cell: A ____________________ of lattice into small _________________________ entities still retaining overall characteristics of the entire lattice. 4. Basis or motif: a group of atoms associated with _____________________________. 5. Crystalline material: A crystalline material is one in which the atoms are situated in a _____________________________________ or periodic array over ________________________________ atomic distances. 6. Coordination number: number of ________________________________ neighbors of an atom in a crystal structure. 7. Lattice parameter: The unit cell
  • 51. ___________________________ and the _________________________ between the edges. 8. Closed packed direction: direction in a crystal lattice along which atoms are in _________________________________. 2 9. Bravais lattices: _______________________________ possible lattices arrangements using lattice points in 3 dimensional space. 10. Atomic packing fraction: a fraction of a plane, volume or direction that is _____________________________ occupied with atoms. 11. Allotropy or Polymorphism: a property of an element to coexist in more than ________________________________ crystal structures. 12. Amorphous Materials: Materials including glasses that have ______________________________ long range order or crystal structure. 13. Isotropic: Having _________________________________
  • 52. properties in all directions. 14. Anisotropic: Having _________________________________ properties in different directions. 15. Miller Indices of Directions: Are _____________________________ components of the directions resolved along each of the coordinate axes. 16. Miller Indices of Planes: Are _____________________________ of fractional _________________________________ that plane makes with coordinate axes. 17. Linear density: It is defined as the number of lattice points or atomic diameters per unit ____________________________________ of the direction. 18. Planar density: The number of atoms whose ________________________________ are intersected by the plane per unit _____________________________. 19. X-ray diffraction: A technique to analyze the _____________________________ of materials using a beam of X-rays as source. 20. Metallic glass: Metals with ____________________________ crystal structure.
  • 53. 3 Part 2 Learning concepts/reflections/problems – 65 points 1. Sketch the unit cells of Simple Cubic, Body Centered Cubic and Face Centered Cubic lattices and determine and show calculations for following: A. Number of atoms per unit cell B. Equation/relation for radius and lattice parameter (unit cell edge length) C. Atomic Packing Fraction (APF) 15 points 2. Calculate the radius of a palladium (Pd) 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. Assume Avogadro’s number = 6.022 x 1023 atoms/mol 5 points 3. Determine the Miller Indices of directions A, B, C and D as shown in figure below: 10 points
  • 54. 4 4. Determine Miller Indices of planes as shown in figure below: 10 points 5. Determine Miller Bravais Indices [hkil] of directions as shown in figure below: 10 points 5 6. Determine Miller Bravais Indices (hkil) of planes as shown in figure below: 10 points 7. Define diffraction. Write an expression for Bragg’s law of diffraction and explain the terms of the equation. 5 points
  • 55. Part 3 Materials design – 10 points 1. Why metallic crystal structure are dense? Explain. 5 points 2. Give 2 materials examples of each of the following crystal structures: a. Simple cubic (SC) b. Body Centered Cubuc (BCC) c. Face Centered Cubic (FCC) d. Hexagonal Closed Packed (HCP) 5 points 6 Bonus questions: 1. Explain why the properties of polycrystalline materials are most often isotropic? 5 points 2. Explain two differences between an optical microscope and a scanning electron microscope (SEM). 5 points 3. Calculate theoretical density of Chromium which has Body
  • 56. Centered Cubic structure and atomic weight of 52 g/mol and atomic radius of 0.125 nm. Assume Avogadro’s number = 6.022 x 1023 atoms/mol 5 points Note: Please use extra sheets wherever necessary to answer the questions. All HW must be uploaded via D2L Dropbox system. Students are highly encouraged to refer textbook, notes and slides.