Introduction to Materials Science and Selection

Teaching Assistants Candy Lin Bryan Wei Rm 7250 Email: kecandy@ust.hk Bryan Wei Rm 7111 Email: kebw@ust.hk

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. Gallium, a metal that easily forms large single crystals Insulin crystals Synthetic bismuth crystal Quartz crystal

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

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

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.

Carbon - Diamond 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. (c) 2003 Brooks/Cole Publishing / Thomson Learning Tetrahedron The diamond cubic (DC) unit cell.

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.

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.

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

Atoms touch along the edge of the cube in SC structures. 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 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:

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:

Coordination Number = no. of touching neighbors

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)

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

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

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

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.

Lattice Directions Direction A 1. Two points are 1, 0, 0, and 0, 0, 0 Example: Determine the direction of A in the figure. Direction A 1. Two points are 1, 0, 0, and 0, 0, 0 2. 1, 0, 0, - 0, 0, 0 = 1, 0, 0 3. No fractions to clear or integers to reduce 4. [100] (c) 2003 Brooks/Cole Publishing / Thomson Learning™

Lattice Directions Direction B 1. Two points are 1, 1, 1, and 0, 0, 0 Determine the direction of B in the figure. (c) 2003 Brooks/Cole Publishing / Thomson Learning™ Direction B 1. Two points are 1, 1, 1, and 0, 0, 0 2. 1, 1, 1, - 0, 0, 0 = 1, 1, 1 3. No fractions to clear or integers to reduce 4. [111]

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

Exercise: Try to determine the directions for A, B, C, and D. (c) 2003 Brooks/Cole Publishing / Thomson Learning

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.

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

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)

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 red) 1. x = -1, y = -1, z = ∞ 2.1/x = -1, 1/y = -1, 1/z = 0 3. No fractions to clear 4. (110) y 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) 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)

Exercise: Determining Miller Indices of Planes Determine the Miller indices of planes A, B, and C in Figure. 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) 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)

Exercise: Determining Miller Indices of Planes Determine the Miller indices of planes A, B, and C in Figure. (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)

More Exercise! Try these! 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) (c) 2003 Brooks/Cole Publishing / Thomson Learning 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 = -2 3. No fractions to clear. 4. (102), (102) Note: When origin in orange 

End of Tutorial 1 Thank you! Introduction to Materials Science and Selection Introduction to Materials Science and Selection Thank you!

CENG151 Introduction to Materials Science and Selection Tutorial 7 16th November, 2007

Reviewing Phase Diagram Terms Phase - Any portion including the whole of a system, which is physically homogeneous within it and bounded by a surface so that it is mechanically separable from any other portions. Component – The distinct chemical substance from which the phase is formed. Phase Diagram - A diagram (map) showing phases present under equilibrium conditions and the phase compositions at each combination of temperature and overall composition. Gibbs Phase Rule - Used to determine the degrees of freedom (F), the number of independent variables available to the system: F = C – P + 1 (for pressure at constant 1 atm.)

Solubility terms Solubility - The amount of one material that will completely dissolve in a second material without creating a second phase. Unlimited solubility - When the amount of one material that will dissolve in a second material without creating a second phase is unlimited. Limited solubility - When only a maximum amount of a solute material can be dissolved in a solvent material.

In-Class Exercise: Revision Apply the Gibbs phase rule to a sketch of the MgO-Al2O3 phase diagram. (Figure 9-26) Find out for each phase compartment the degrees of freedom.

Solution 9.5 (F=2-2+1=1) F=2-1+1=2 F=2-2+1=1 F=1-2+1=0 F=1-2+1=0

Phase diagram for unlimited solubility The composition of the phases in a two-phase region of the phase diagram are determined by a tie line (horizontal line connecting the phase compositions) at the system temperature.

Eutectic Diagram with No Solid Solution Another binary system with components that are so dissimilar that their solubility in each other in nearly negligible. There exists a 2-phase region for pure solids (A+B). Because A and B cannot dissolve in each other! The eutectic temperature (eutektos greek for “easily melted”) is the temperature that eutectic composition is fully melted.

Eutectic diagram with limited solid solution Many binary systems are partially soluble  intermediate phase diagram of the previous two! α and  individual phases are single solid solution phases! While α+ is a 2-phase region for pure solids α and .

Eutectic Diagram with Limited Solid Solution Lead (Pb) – Tin (Sn) equilibrium phase diagram. (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. Solidification and microstructure of a Pb-2% Sn alloy. The alloy is a single-phase solid solution.

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. Solidification, precipitation, and microstructure of a Pb-10% Sn alloy. Some dispersion strengthening occurs as the β solid precipitates.

The Lever Rule Tells us the amount of each phase there are in the alloy  wt% C0 must be made up of appropriate amounts of α at composition Cα and of liquid at composition CLiq.

The Lever Rule Basically, the proportion of the phases present are given by the relative lengths of the tie line. So, the proportions of α and L present on the diagram are: So, the left side of the tie line gives the proportion of the liquid phase, and the right side of the tie line gives the proportion of the alpha phase!

Example: Application of Lever Rule Calculate the amounts of and L at 1250oC in the Cu-40% Ni alloy shown in the figure below. ©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. A tie line 1250°C in the copper-nickel system that is used to find the amount of each phase.

Example: Solution x = mass fraction of the alloy that is in α phase. Since we have only two phases, α and L. Thus, the mass fraction of nickel in liquid will be 1 - x. Total mass of nickel in 100 grams of the alloy = mass of nickel in liquid + mass of nickel in α So, 100  (% Ni in alloy = [(100)(1 – x)](% Ni in L) + (100)[x](% Ni in α ) x = (40-32)/(45-32) = 8/13 = 0.62 If we convert from mass fraction to mass percent, the alloy at 1250oC contains 62% α and 38% L. Note that the concentration of Ni in α phase (at 1250oC) is 45% and concentration of nickel in liquid phase (at 1250oC) is 32% (read from the diagram).

Example: Limited Solubility Consider a 50:50 Pb – Sn solder. For a temperature of 200˚C, determine (i) the phases present, (ii) their compositions, and (iii) their relative amounts (expressed in weight percent). Repeat part (a) for 100˚C.

Example: Solution Reading off the phase diagram at 50wt% and 200˚C, Phases present are  and liquid. The composition of  is ~18wt% Sn and of L is ~ 54wt% Sn. Using the Eqn 9.9 and 9.10, we have: Lever Rule

Example: Solution (cont.) (b) Similarly, at 100C, we obtain: (i) Phases are  and . (ii)  is ~5wt% Sn and  is ~ 99wt% Sn. (iii)

©2003 Brooks/Cole, a division of Thomson Learning, Inc ©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. The change in structure of a Cu-40% Ni alloy during nonequilibrium solidification. Insufficient time for diffusion in the solid produces a segregated structure.

In-Class Exercise: Limited Solubility Consider 1kg of an aluminum casting alloy with 10wt% silicon. Upon cooling, at what temperature would the first solid appear? What is the first solid phase and what is its composition? At what temperature will the alloy completely solidify? How much proeutectic phase will be found in the microstructure? How is the silicon distributed in the microstructure at 576C?

In-Class Exercise: Solution Follow microstructural development with the aid of the phase diagram: For this composition, the liquidus is at ~595C. It is solid solution  with the composition of ~1wt% Si. At the eutectic temperature, 577C. Practically all of the proeutectic  will have developed by 578C. Using the equation: (eqn9.9), we obtain:

In-Class Exercise: Solution (cont.) (e) At 576C, the overall microstructure is  + . The amounts of each are: But we found in (d) that 236g of the  is in the form of relatively large grains of proeutectic phase, giving:

In-Class Exercise: Solution (cont.) The silicon distribution is then given by multiplying its weight fraction in each microstructural region by the amount of that region: Si in proeutectic  = (0.016)(236g) = 3.8g Si in eutectic  = (0.016)(679g) = 10.9g Si in eutectic  = (1.000)(85g) = 85.0g Finally, note that the total mass of silicon in the three regions sums to 99.7g rather than 100.0g (=10wt% of the total alloy) due to round-off errors.

Question 9.10 (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. (c) (b) (a) Describe qualitatively the microstructural development during the slow cooling of a melt composed of the following compositions. (See Figure 9-16) 10 wt% Pb – 90 wt% Sn, 40 wt% Pb – 60 wt% Sn, 50 wt% Pb – 50 wt% Sn.

Solution 9.10 (a) The first solid to precipitate is solid solution, -Sn near 210˚C. At the eutectic temperature (183˚C), the remaining liquid solidifies leaving a two phase microstructure microstructure of solid solutions α-Pb and -Sn. (b) The first solid to precipitate is solid solution α-Pb near 188˚C. At the eutectic temperature (183˚C), the remaining liquid solidifies leaving a two phase microstructure of solid solution α-Pb and -Sn. (c) The first solid to precipitate is solid solution α-Pb near 210˚C. At the eutectic temperature (183˚C), the remaining liquid solidifies leaving a two-phase microstructure of solid solution α-Pb and -Sn.

Question 9.13 (a) 20 wt% Mg – 80 wt% Al, (b) 80 wt% Mg – 20 wt% Al. Describe qualitatively the microstructural development that will occur upon the slow cooling of a melt composed of the following compositions. (See Figure 9-28) (a) (b)

Solution 9.13 (a) The first solid to precipitate is α. At eutectic temperature (450˚C), the remaining liquid solidifies leaving a two-phase microstructure of solid solutions α and β. (b) The first solid to precipitate is . At the eutectic temperature (437˚C), the remaining liquid solidifies leaving a two-phase microstructure of solid solution  and .

Question 9.17 Calculate the amount of each phase present in 1kg of a 50 wt% Ni – 50 wt% Cu alloy at the following temperatures. (See Figure 9-9) 1400˚C 1300˚C 1200˚C

Solution 9.17 (c) In the single (α) (a) In the single (L) region: phase region: (a) In the single (L) region: (b) Two phases exist:

Question 9.18 (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. Calculate the amount of each phase present in 1 kg of a 50 wt% Pb – 50 wt% Sn solder alloy at the following temperatures. (See Figure 9-16) 300˚C 200˚C 100˚C 0˚C

Solution 9.18 (a) In the single (L) region: (b) Two phases exist (α-Pb + L):

Solution 9.18 (cont.) (c) Two phases exist (α-Pb + -Sn): (d) Two phases exist (α-Pb + α-Sn):

End of Tutorial 7 Thank You!