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CENG151 Introduction to Materials Science and Selection Tutorial 1 14 th September, 2007.

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Presentation on theme: "CENG151 Introduction to Materials Science and Selection Tutorial 1 14 th September, 2007."— Presentation transcript:

1 CENG151 Introduction to Materials Science and Selection Tutorial 1 14 th September, 2007

2 Teaching Assistants Candy Lin  Rm 7250  Bryan Wei  Rm 7111 

3 Crystal In chemistry and mineralogy, a crystal is a solid in which the constituent atoms, molecules, or ions are packed in a regularly ordered, repeating pattern extending in all three spatial dimensions. Insulin crystals Synthetic bismuth crystal Gallium, a metal that easily forms large single crystals Quartz crystal

4 7 crystal system &14 Bravais lattice triclinic monoclinic orthorhombic hexagonal rhombohedral tetragonal cubic In all, there are 14 possible Bravais lattices that fill 3D space, and they classified by 7 kind of crystal system.

5 Comparing Different Atomic Arrangement Different atom arrangement can alter chemical and physical properties. Carbon is a very good example of this!

6 Carbon - Graphite Large parallel sheets of hexagonal rings.  The sheets are held together by weak Van der Waals forces.  Can be broken easily.  Because the layers of carbon rings can rub over each other, graphite is a good lubricant.

7 Carbon - Diamond (c) 2003 Brooks/Cole Publishing / Thomson Learning (a)Tetrahedron (b)The diamond cubic (DC) unit cell. Each carbon atom covalently bonded to four other carbons in a tetrahedral arrangement to form a tight tetrahedron lattice.  Very strong bonds  strongest natural material on earth. Diamond can be cleaved along its planes, but it cannot flake apart into layers because of this tetrahedral arrangement of carbons.

8 Carbon - Buckminsterfullerene The 1996 Nobel Prize in Chemistry has been awarded to three chemists (R. Smalley, R. Curl, H. Kroto) for discovery of fullerenes. Shaped like a soccer ball, with 20 hexagons and 12 pentagons. A total of 60 carbons in this molecules. Molecule is extremely stable and can withstand very high temperatures and pressures. Can react with other atoms and molecules.

9 Packing Factor (Packing Density) Packing – geometrical arrangement of these (atomic) spheres. Packing Factor/Density – proportion of space occupied by these solid spheres in a unit cell.

10 SC, BCC, and FCC structures when one atom is located at each lattice point. Determining the Relationship between Atomic Radius and Lattice Parameters (c) 2003 Brooks/Cole Publishing / Thomson Learning™ The relationships between the atomic radius and the Lattice parameter in cubic systems.

11 Atoms touch along the edge of the cube in SC structures. In FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this length—two radii from the face-centered atom and one radius from each corner, so: In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so

12 Packing Factor Packing Factor – the fraction of space occupied by atoms, assuming that atoms are hard spheres sized so that they touch their closest neighbor. Density (of a material) – can be calculated using properties of the crystal structure:

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14 Coordination Number = no. of touching neighbors

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19 HCP and FCC Greatest packing density obtainable (closest packed) with spheres of equal size is 74%. 1 atom surrounded by 12 others of equal size, all touching the reference sphere along the diameter. 2 related crystal structures obtain this packing density:  Hexagonal close packed (HCP)  Face centered cubic (FCC)

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21 Lattice Positions Coordinates of selected points in the unit cell. The number refers to the distance from the origin in terms of lattice parameters.

22 In Class Exercise Try to find A, B, C and D.

23 Answers A = 1, 0, 0 B = 0, 0, 1 C = ½, 1, 0 D = 1, 1, 0

24 Miller Indices for directions Miller Indices are shorthand notations used to describe directions. Follow these rules:  Using right-handed coordinate system, determine the coordinates of the two points that lie on the direction.  Substract the coordinates of the “tail” point from the coordinates of the “head” point. “head” – “tail”  Clear fractions and/or reduce the results obtained from the substraction to lowest integers.  Enclose the numbers in square brackets []. If negative sign is produced, put a bar over the number.

25 Direction A 1. Two points are 1, 0, 0, and 0, 0, , 0, 0, - 0, 0, 0 = 1, 0, 0 3. No fractions to clear or integers to reduce 4. [100] Lattice Directions Example: Determine the direction of A in the figure. (c) 2003 Brooks/Cole Publishing / Thomson Learning™

26 Direction B 1. Two points are 1, 1, 1, and 0, 0, , 1, 1, - 0, 0, 0 = 1, 1, 1 3. No fractions to clear or integers to reduce 4. [111] Lattice Directions Determine the direction of B in the figure. (c) 2003 Brooks/Cole Publishing / Thomson Learning™

27 Lattice Directions Direction C 1. Two points are 0, 0, 1, and ½, 1, , 0, 1, - ½, 1, 0 = -½, -1, 1 3. Fractions for clearing: 2(-½, -1, 1) = -1, -2, 2 (c) 2003 Brooks/Cole Publishing / Thomson Learning™ Determine the direction of C in the figure.

28 Exercise: Try to determine the directions for A, B, C, and D. A: 0,0,1 – 1,0,0 = -1,0,1 = [101] B: 1,0,1 – ½,1,0 = ½,-1,1 = [122] C: 1,0,0 – 0,¾,1 = 1,-¾,-1 = [434] D: 0,1,½ – 1,0,0 = -1,1,½ = [221] (c) 2003 Brooks/Cole Publishing / Thomson Learning

29 Miller Indices for planes Miller Indices are used to describe planes as well. Follow these rules:  Identify the points at which the plane intercepts the x, y, z coordinates. If the plane passes through the origin, the origin must be moved!  Take reciprocals of these intercepts.  Clear fractions and do NOT reduce to lowest integers.  Enclose the numbers in parentheses (). If negative sign is produced, put a bar over the number, like before.

30 Lattice Planes Intercepting points at: x= ∞, y=1, z= ∞ Take reciprocals: x=0, y=1, z=0 No fractions to clear, Enclose numbers in parentheses: (010) Lattice positions  no brackets Lattice directions  square brackets Lattice planes  parentheses Intercepting points at: x=-1, y= ∞, z= ∞ Take reciprocals: x=- 1, y=0, z=0 No fractions to clear, Enclose numbers in parentheses: (100)

31 Plane 1. x = 1, y = 1, z = ∞ 2.1/x = 1, 1/y = 1, 1 /z = 0 3. No fractions to clear 4. (110) Plane 1. X = ⅔, y = 1, z = ∞ 2. 1/x = 3/2, 1/y=1, 1/z=0 3. Clear fractions: 1/x = 3, 1/y = 2, 1/z = 0 4. (320)

32 Plane (origin in orange) 1. x = 1, y = -1, z = ∞ 2.1/x = 1, 1/y = -1, 1/z = 0 3. No fractions to clear 4. (110) Plane (origin in red) 1. x = -1, y = -1, z = ∞ 2.1/x = -1, 1/y = -1, 1/z = 0 3. No fractions to clear 4. (110) Plane (origin in blue) 1. x = 1, y = 1, z = ∞ 2.1/x = 1, 1/y = 1, 1/z = 0 3. No fractions to clear 4. (110) Plane (origin in green) 1. x = -1, y = 1, z = ∞ 2.1/x = -1, 1/y = 1, 1/z = 0 3. No fractions to clear 4. (110) y

33 Determine the Miller indices of planes A, B, and C in Figure. Exercise: Determining Miller Indices of Planes Plane B 1. The plane never intercepts the z axis, so x = 1, y = 2, and z =∞ 2.1/x = 1, 1/y =1/2, 1/z = 0 3. Clear fractions: 1/x = 2, 1/y = 1, 1/z = 0 4. (210) Plane A 1. x = 1, y = 1, z = 1 2.1/x = 1, 1/y = 1, 1/z = 1 3. No fractions to clear 4. (111)

34 (c) 2003 Brooks/Cole Publishing / Thomson Learning™ Plane C (origin in red) 1. We must move the origin, since the plane passes through 0, 0, 0. Let’s move the origin one lattice parameter in the y-direction. Then, x= ∞, y=-1, and z= ∞ 2. 1/x = 0, 1/y = -1, 1/z = 0 3. No fractions to clear. 4. (010) Exercise: Determining Miller Indices of Planes Determine the Miller indices of planes A, B, and C in Figure.

35 More Exercise! Try these! (c) 2003 Brooks/Cole Publishing / Thomson Learning Plane A (origin in red) 1. x = 1, y = -1, z = 1 2.1/x = 1, 1/y = -1, 1/z = 1 3. No fractions to clear 4. (111) Plane B (origin in orange) 1. x = ∞, y = ⅓, and z =∞ 2.1/x = 0, 1/y =3, 1/z = 0 3. No fractions to clear 4. (030) Plane C (origin in green) 1. x=1, y=∞, and z=-½ 2. 1/x = 1, 1/y = 0, 1/z = No fractions to clear. 4. (102), (102) Note: When origin in orange 

36 End of Tutorial 1 Thank you!


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