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Materiales the science and engineering of materials - Solution Manual

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Solution of problems, design problems and also computer problems.

Solution of problems, design problems and also computer problems.

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  • 1. 8/25/2008 MSE 280: Introduction to Engineering Materials Metal and Ceramic Structures Reading: Chapter 3 Part 1 – – – – Defining ordered atoms in crystalline solids. Unit cells, unit vectors, Coordinates, directions, planes, close packing… Densities: Crystal density, Line density, Planar density Crystal symmetry and families Crystallography Part 2 – – – – Ionic crystals. Lattice energy. Silica & silicates. Carbon 1 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 ENERGY AND PACKING • Non dense, random packing Energy typical neighbor bond length typical neighbor bond energy • Dense, regular packing r Energy typical neighbor bond length b dl h r typical neighbor bond energy Dense, regular-packed structures tend to have lower energy. 2 © 2007, 2008 Moonsub Shim, University of Illinois From Callister 6e resource CD. MSE280 1
  • 2. 8/25/2008 Packing atoms together Crystalline materials... • atoms pack in periodic, 3D arrays • t i l of: -metals typical f -many ceramics -some polymers crystalline SiO2 Adapted from Fig. 3.18(a), Callister 6e. Noncrystalline materials... • atoms have no periodic packing • occurs for: -complex structures -rapid cooling "Amorphous" = Noncrystalline © 2007, 2008 Moonsub Shim, University of Illinois Si Oxygen noncrystalline SiO2 Adapted from Fig. 3.18(b), Callister 6e. 3 MSE280 Crystalline materials: Unit Cell Unit Cell: The basic structural unit of a crystal structure. Its geometry and atomic positions define the crystal structure. Note: more than one unit cell can be chosen for a given crystal but by convention/convenience the one with the highest symmetry is chosen. 4 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 2
  • 3. 8/25/2008 Unit cells and unit vectors v c v b Lattice parameters axial lengths: a, b, c interaxial angles: α, β, γ v v v unit vectors: a b c In general: a≠b≠c α≠β≠γ v a 5 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Unit Cells: Bravais Lattices 6 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 3
  • 4. 8/25/2008 (trigonal) 7 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Unit cell types Primitive Body-centered Face-centered End-centered 8 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 4
  • 5. 8/25/2008 Number of atoms in a unit cell Simple cubic 8 atoms but each atom is shared by 8 unit cells. → 1 atom per unit cell p 9 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Coordination Number Number of nearest neighbor atoms Simple cubic: coordination number = 6 10 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 5
  • 6. 8/25/2008 Atomic Packing Factor (APF) APF = Vol. of atoms in unit cell Vol. of unit cell Depends on: • Crystal structure. • How “close” packed the atoms are. p • In simple close-packed structures with hard sphere atoms, independent of atomic radius 11 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Close packing of atoms Consider atoms as hard spheres. 12 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 6
  • 7. 8/25/2008 Face-Centered Cubic (FCC) Atoms at the corners of the cube + Atoms at the center of each face a 4R a = ||unit vector|| R = atomic radius a 2 + a 2 = ( 4 R )2 a a = 2 2R 13 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Face-Centered Cubic (FCC) Adapted from Fig. 3.1(a), Callister 6e. 14 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 7
  • 8. 8/25/2008 Number of atoms in an FCC unit cell • Each corner atom contributes as 1/8 1/8. There are 8 corner atoms in an FCC unit cell. • Each face atom contributes as 1/2. There are 6 face atoms. 1 1 × 8( corner _ atoms ) + × 6( face _ atoms ) = 4 atoms / unit _ cell 2 8 © 2007, 2008 Moonsub Shim, University of Illinois 15 MSE280 Coordination number for FCC 2 2R 2R 6 9 2 2 2R 10 5 3 1 4 7 12 11 8 Total 12 nearest neighbor atoms Coordination number = 12 16 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 8
  • 9. 8/25/2008 Atomic packing factor (APF) for FCC 2 2R 2R 2 2R a 4R ⎛4 ⎞ Vatoms = 4⎜ πR 3 ⎟ ⎝3 ⎠ 3 3 Vunit _ cell = a = (2 2 R) = 16 2 R 3 a APF = Vatoms Vunit _ cell ⎛ 16 3 ⎞ ⎜ πR ⎟ 3 ⎠ = π = 0.74 Independent of R and a! =⎝ 3 16 2 R 3 2 17 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Summary for FCC ||unit vector|| = a = 2 2R 4 atoms/unit cell a 4R Coordination number = 12 APF = 0.74 18 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 9
  • 10. 8/25/2008 Body-Centered Cubic (BCC) 2a Atoms at the corners of the cube + Atom at the center of the cube a 4R ( ) a2 + 2 a 4R a= 3 2 = (4 R )2 19 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Body-Centered Cubic (BCC) From Callister 6e resource CD. 20 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 10
  • 11. 8/25/2008 Number of atoms in a BCC unit cell • Each corner atom contributes as 1/8 1/8. There are 8 corner atoms in an FCC unit cell. • The center atom contributes as 1. There is only 1 center atom. atom 1 × 8( corner _ atoms ) + 1 ×1( center _ atom ) = 2 atoms / unit _ cell 8 © 2007, 2008 Moonsub Shim, University of Illinois 21 MSE280 Coordination number for BCC 2 1 3 4 6 5 Total 8 nearest neighbor atoms Coordination number = 8 7 8 22 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 11
  • 12. 8/25/2008 Atomic packing factor (APF) for BCC ⎛4 ⎞ Vatoms = 2⎜ πR 3 ⎟ ⎝3 ⎠ 2a a a 2 + ( 2a ) 2 = (4 R ) 2 4R a= 3 64 R 3 4R Vunit _ cell = a 3 = ( )3 = 3 3 3 4R ⎛8 3⎞ ⎜ πR ⎟ Vatoms 3 ⎠ = 3π = 0.68 APF = =⎝ 3 64 R Vunit _ cell 8 3 3 23 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Summary for BCC 2a ||unit vector|| = a 4R a= 4R 3 2 atoms/unit cell Coordination number = 8 APF = 0.68 24 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 12
  • 13. 8/25/2008 Hexagonal Close-Packing (HCP) Find a, c, number of atoms/unit cell, coordination number, and APF for HW. 25 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Point Coordinates to define a point within a unit cell…. Essentially same as Cartesian coordinates except values of x, y, and z are expressed as fractions of the magnitude of unit vector(s) (and x, y, and z not necessarily orthogonal) orthogonal). pt. coord. z x (a) y (b) z (c) b A c y = point “A” = origin 0 0 0 B a 1 0 0 C 1 1 1 D 1/2 0 1/2 = point “B” x = point “C” = point “D” © 2007, 2008 Moonsub Shim, University of Illinois 26 MSE280 13
  • 14. 8/25/2008 Crystallographic Directions To define a vector: 1. Start at the origin. 2. 2 Determine length of vector projection in each of 3 axes in units (or fractions) of a, b, and c. 3. Multiply or divide by a common factor to reduce the lengths to the smallest integer values. 4. Enclose in square brackets: [u v w] where u, v, and w are integers. Along unit vectors: a b c Note: in any of the 3 directions there are both positive and negative directions. Negative directions are denoted with a “bar” over the number. e.g. [1 1 1] 27 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Example 1: Assign the coordinates to the following crystallographic direction Along x: 1 a z Along y: 1 b Along z: 1 c b a c y [1 1 1] x 28 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 14
  • 15. 8/25/2008 Example 2: draw [1 1 0] direction. 1 unit vector length along x -1 unit vector length along y g g z 0 unit vector length along z b a y c [1 1 0] x origin 29 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Note: for some crystal structures, different directions can be equivalent. e.g. For cubic crystals: [1 0 0], [ 1 0 0], [0 1 0], [0 1 0], [0 0 1], [0 0 1 ] are all equivalent Families of crystallographic directions e.g. <1 0 0> Angled brackets denote a family of crystallographic directions. 30 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 15
  • 16. 8/25/2008 Families and Symmetry • Which directions belong in the same family is determined by the crystal symmetry. • What is symmetry? • Whi h one i more symmetric? Which is ti ? c a a a a b • The more symmetry operations there are the more symmetric it is. 1 3 1 1 etc… and 2 3 2 1 equivalent 2 3 3 2 equivalent • Which is more symmetric, cubic or tetragonal? 31 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Families and Symmetry z Cubic symmetry z [010] Rotate 90o about z-axis [100] x Rotate 90o about y-axis y y y x z [001] Symmetry operation can generate all the directions within in a family. x © 2007, 2008 Moonsub Shim, University of Illinois y Similarly for other equivalent directions 32 MSE280 16
  • 17. 8/25/2008 Crystallographic directions for hexagonal crystals Consider vectors q and r…. If we keep the 3-coordinate system: v q = [1 0 0] v r = [1 1 0] z Different set of indices. However, these two vectors are equivalent by symmetry (i.e. via 60o rotation). y v r v q x Choose a 4-coordinate system! 33 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Crystallographic directions for hexagonal crystals In the Miller-Bravais (4 axes) coordinate system, We have: v q = [2 1 1 0] z a2 a3 v q 2 along a1 -1 along a2 a1 Although the length is different the direction is the same. -1 along a3 © 2007, 2008 Moonsub Shim, University of Illinois 34 MSE280 17
  • 18. 8/25/2008 Crystallographic directions for hexagonal crystals v Similarly for r, in the Miller-Bravais (4 axes) coordinate system, we now have: v r = [1 1 2 0] z a2 -2 along a3 v r a3 1 along a1 a1 1 along a2 Again, consistent direction. 35 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Crystallographic directions for hexagonal crystals Miller-Bravais (4 axes) coordinate system We now have: v q = [2 1 1 0] v r = [1 1 2 0] z a2 a3 v q v r a1 Indices are now consistent within the family, but where do these numbers come from? 36 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 18
  • 19. 8/25/2008 To transform 3 indices to 4 indices [u’ v’ w’] v [u v t w] u = 1/3 (2u’ – v’) ( ) v = 1/3 (2v’ – u’) t = - (u + v) w = w’ May need to factor to reduce u, v, t, and w to their smallest integers. e.g. convert q = [1 0 0] to 4-coord. system. u = 1/3 (2x1 – 0) = 2/3 2 x3 v = 1/3 (2x0 – 1) = -1/3 1/3 -1 1 x3 t = -(2/3 + (-1/3)) = -1/3 w=0 z x3 a2 [2 1 1 0] -1 x3 0 v q a3 a1 37 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 transforming 3 indices to 4 indices convert v r = [1 1 0] to 4-coord. system. u = 1/3 (2x1 – 1) = 1/3 v = 1/3 (2x1 – 1) = 1/3 t = -(1/3 + 1/3) = -2/3 x3 x3 x3 x3 w=0 1 1 [1 1 2 0] -2 z 0 a2 a3 © 2007, 2008 Moonsub Shim, University of Illinois v r a1 38 MSE280 19
  • 20. 8/25/2008 Crystallographic Planes To define a plane (i.e. to assign Miller Indices): 1. If the plane passes through the origin, either a) Construct a parallel plane translated or b) choose another origin at the corner of adjacent unit cell z b a y c x origin origin 2. Now, the plane either intersects or parallels each of the three axes axes. 3. Determine length in terms of lattice parameters a, b and c. Parallel to x-axis = ∞ length along x Parallel to y-axis = ∞ length along y Intercepts z-axis at z = c 39 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Crystallographic Planes 4. Take the reciprocal of the lengths e.g. Al Along x-axis: l i length = ∞ th Along y-axis: length = ∞ Along z-axis: length = 1 0 0 1 5. If necessary, factor to get the smallest integers. divide through by 3 e.g. e g if we had 0 0 3 0, 0, 0, 0, 0 0 1 6. Enclose indices in parentheses w/o commas In the example on previous slide, the Miller index would be (0 0 1) 40 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 20
  • 21. 8/25/2008 Example 1 Assign Miller indices 1. 1 Shift origin f i i from O t O’ to z z’ 2. Determine lengths along each axis Along x = ∞ b Intersects y at -1 a Intersects z at 1/3 O c/3 c y 3. Take the reciprocal: 0, -1, 3 O’ x 4. No need to factor. 5. (0 1 3) x’ 41 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Example 2 Draw the (0 1 1) plane. 1. 1 along x = 0 The plane runs parallel to x-axis 2. along y = -1 (in units of b) take reciprocal: -1 y-intercept at -1 3. along z = 1 (in units of c) take reciprocal: 1 z-intercept z intercept at 1 z |c| |b| y x 42 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 21
  • 22. 8/25/2008 Crystallographic planes in FCC z y (1 0 0) plane Look down this direction (perpendicular to the plane) x © 2007, 2008 Moonsub Shim, University of Illinois 43 MSE280 Crystallographic planes in FCC z Look down this direction (perpendicular to the plane) y x © 2007, 2008 Moonsub Shim, University of Illinois (1 1 1) plane 44 MSE280 22
  • 23. 8/25/2008 Note: similar to crystallographic directions, planes that are parallel to each other are equivalent 45 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Crystallographic Planes for Hexagonal Crystals Similar to crystallographic directions for hexagonal crystals, 4-coordinate system is used. i.e. instead of (h k l) for 3-coordinate systems, use (h k i l). For integers h, k, and l, same procedure as 3-coordinate systems is used and i = - (h+k) 46 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 23
  • 24. 8/25/2008 Crystallographic planes in HCP z Name this plane… a2 a3 (0 0 0 1) plane a1 Parallel to a1, a2 and a3 -> h = k = i = 0 Intersects at z = 1 © 2007, 2008 Moonsub Shim, University of Illinois 47 MSE280 Crystallographic planes in HCP z a2 +1 in a1 a3 -1 in a2 a1 h = 1, k = -1, i = -(1+-1) = 0, l = 0 © 2007, 2008 Moonsub Shim, University of Illinois (1 1 0 0) plane 48 MSE280 24
  • 25. 8/25/2008 z (1 1 1) plane of FCC y x z (0 0 0 1) plane of HCP SAME THING!* a2 a3 a1 © 2007, 2008 Moonsub Shim, University of Illinois 49 MSE280 Close Packing Note: whether the atoms of 2nd layer fills B-sites or C-sites does not matter. They are equivalent until the 3rd layer is added! 50 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 25
  • 26. 8/25/2008 FCC 51 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 HCP 52 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 26
  • 27. 8/25/2008 Difference between FCC and HCP Looking down (0001) plane FCC HCP ABCABC… ABABAB… Looking down (111) plane! © 2007, 2008 Moonsub Shim, University of Illinois 53 MSE280 Densities •C t l d Crystals density (i 3D d it (i.e. density) i units of mass it ) in it f 3). per volume (e.g. g/cm • Linear density: number of atoms per unit length (e.g. cm-1). • Planar density: number of atoms per unit area (e g (e.g. cm-2). 54 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 27
  • 28. 8/25/2008 Crystals density (ρ) Mass of unit cell ρ= M V M= nA NA Volume of unit cell n = number of atoms in unit cell A = atomic weight NA = Avogadro’s number ρ= nA N AV 55 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Density example Calculate density of copper given: R = 0.128 nm A = 63.5 g/mol FCC structure Recall for FCC, there are 4 atoms per unit cell. Express unit cell volume in terms of atomic radius R. V = a 3 = (2 2 R )3 = 16 2 R 3 Then we have: ρ= = M nA = V N A16 2 R 3 4(65.3 g / mol ) (6.022 x10 23 mol −1 )(16 2 )(1.28 x10 −8 cm)3 = 8.89 g/cm3 Compare to actual value of 8.94 g/cm3! © 2007, 2008 Moonsub Shim, University of Illinois 56 MSE280 28
  • 29. 8/25/2008 Linear Density (LD) LD = Number of atoms centered on a direction vector Length of the direction vector Example: calculate the linear density of an FCC crystal along [1 1 0]. Length = 4R Effectively 2 atoms along the [1 1 0] vector. LD = 2/4R = 1/2R 57 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 Planar Density (PD) PD = Number of atoms centered on a given plane Area of the plane Example: calculate the planar density on (1 1 0) plane of an FCC crystal. a = 2 2R 4R Count only the parts of the atoms within the plane: 2 atoms area = 2 2 R × 4 R PD = © 2007, 2008 Moonsub Shim, University of Illinois 2 1 = (4 R)(2 2 R) 4 2 R 2 58 MSE280 29
  • 30. 8/25/2008 Example problem (101) (011) 0.36 nm 0.32 nm 0.2 nm (110) 0.39 nm 0.3 nm 0.253 nm Given above information answer the following q g questions. A) What is the crystal structure? B) If atomic radius is 0.08 nm, what is APF? C) If atomic weight is 43 g/mol, calculate density. D) What is the linear density along [210]? E) What is the planar density of (210)? 59 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 How to determine crystal structure The two slit experiment Same idea with crystals - Light gets scattered off atoms… λ d θ Incoming light Constructive interference when nλ = 2d sin θ (Bragg’s Law) λ and d have to be comparable lengths. But since d (atomic spacing) is on the order of angstroms: x-ray diffraction 60 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 30
  • 31. 8/25/2008 Diffraction Experiment and Signal Scan versus 2x Angle: Polycrystalline Cu Diffraction Experiment Measure 2θ Diffraction collects data in “reciprocal space” since it is the equivalent of p p q a Fourier transform of “real space”, i.e., eikr, as k ~ 1/r. How can 2θ scans help us determine crystal structure type and distances between Miller Indexed planes (I.e. structural parameters)? MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 Crystal Structure and Planar Distances Bravais Lattice Constructive Interference Destructive Interference Reflections present Reflections absent BCC (h + k + l) = Even (h + k + l) = Odd FCC (h,k,l) All Odd or All Even (h,k,l) Not All Odd or All Even HCP Any other (h,k,l) h+2k=3n, l = Odd n= integer h, k, l are the Miller Indices of the planes of atoms that scatter! So they determine the important planes of atoms, or symmetry. Distances between Miller Indexed planes For cubic crystals: For hexagonal crystals: dhkl = a a (h 2 + hk + k 2 ) + l 2 ⎛ ⎞ ⎝ c⎠ 3 4 MSE 280: Introduction to Engineering Materials 2 ©D.D. Johnson 2004, 2006-08 31
  • 32. 8/25/2008 Allowed (hkl) in FCC and BCC for principal scattering (n=1) (h k l) h2+k2+l2 h+k+l h,k,l all even or odd? 100 1 1 No 110 2 2 No 111 3 3 Yes Self-Assessment: From what crystal structure is this? 200 4 2 Yes 210 5 3 No 211 6 4 No 220 8 4 Yes 221 9 5 No 300 9 3 No 310 10 4 No 311 11 5 Yes 222 h + k + l was even and gave the labels on graph above, so crystal is BCC. MSE 280: Introduction to Engineering Materials 12 6 Yes 320 13 5 No 321 14 6 No ©D.D. Johnson 2004, 2006-08 Example problem: X-ray crystallography • Given the information below, determine the crystal structure. Consider only FCC and BCC structures as possibilities. – Lattice parameter: a = 0.4997nm – Powder x-ray: λ = 0.1542 nm 2θ (o) 31 36 51.8 61.6 64.8 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 32
  • 33. 8/25/2008 Diffraction of Single 2D Crystal Grain Lattice of Atoms: Crystal Grain Diffraction Pattern in (h,k) plane Powder diffraction figures taken from Th. Proffe and R.B. Neder (2001) http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/pow_a.html MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 Diffraction of Multiple Single 2D Crystal Grains (powders) Multiple Crystal Grains: Diffraction Pattern 4 Polycrystals in (h,k) plane (4 grains) Diffraction Pattern in (h,k) plane (40 grains) Powder diffraction figures taken from Th. Proffe and R.B. Neder (2001) http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/pow_a.html MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 33
  • 34. 8/25/2008 Example from Graphite Single Crystal Grain Mutlple Grains: Polycrystal Powder diffraction figures taken from Th. Proffe and R.B. Neder (2001) http://www.uni-wuerzburg.de/mineralogie/crystal/teaching/pow_a.html MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 Concepts to remember • • • • • • • • • • Unit cell, unit vector, and lattice parameters. Bravais Lattices. Counting number of atoms for a given unit cell. Coordination number = number of nearest neighbor atoms. Atomic Packing Factor (APF) = Volume of atoms in a unit cell/Volume of unit cell. Close-packing FCC, BCC, FCC BCC HCP Crystallographic coordinates, directions, and planes. Densities X-ray crystallogrphy 68 © 2007, 2008 Moonsub Shim, University of Illinois MSE280 34