Materiales the science and engineering of materials - Solution Manual

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MSE 280: Introduction to Engineering Materials

Reading: Chapter 3Part 1

Metal and Ceramic StructuresPart 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

MSE280© 2007, 2008 Moonsub Shim, University of Illinois1

Part 2– Ionic crystals.– Lattice energy.– Silica & silicates.– Carbon

ENERGY AND PACKING• Non dense, random packing Energy

typical neighbor bond length

• Dense, regular packing

r

bond length

typical neighbor bond energy

Energy

typical neighbor b d l h


Dense, regular-packed structures tend to have lower energy.

r

bond length

typical neighbor bond energy

From Callister 6e resource CD.

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Packing atoms together

• atoms pack in periodic, 3D arrayst i l f

Crystalline materials...

• typical of: -metals-many ceramics-some polymers

Noncrystalline materials...Si Oxygen

crystalline SiO2Adapted from Fig. 3.18(a),Callister 6e.


• atoms have no periodic packing• occurs for: -complex structures

-rapid cooling

noncrystalline SiO2"Amorphous" = NoncrystallineAdapted from Fig. 3.18(b),Callister 6e.

Crystalline materials: Unit Cell

Unit Cell: The basic structuralUnit 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.

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Unit cells and unit vectors

bv

cvLattice parameters

axial lengths: a, b, cinteraxial angles: α, β, γunit vectors:

In general: a ≠ b ≠ cα ≠ β ≠ γ

av bv

cv


av

Unit Cells: Bravais Lattices


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(trigonal)


Unit cell types

Primitive Face-centered

Body-centered End-centered


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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

Coordination NumberNumber of nearest neighbor atoms


Simple cubic: coordination number = 6

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Atomic Packing Factor (APF)

APFVol. of atoms in unit cell

Depends on: • Crystal structure.• How “close” packed the atoms are.

APF =Vol. of unit cell


p• In simple close-packed structures with hard sphere atoms, independent of atomic radius

Close packing of atomsConsider atoms as hard spheres.


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Face-Centered Cubic (FCC)

Atoms at the corners of the cube+

Atoms at the center of each face

a a = ||unit vector||4R

R = atomic radius


a Ra

)R(aa

22

4 222

=

=+

Face-Centered Cubic (FCC)


Adapted from Fig. 3.1(a),Callister 6e.

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Number of atoms in an FCC unit cell

• Each corner atom contributes as 1/8contributes as 1/8.There are 8 corner atoms in an FCC unit cell.

• Each face atom contributes as 1/2.There are 6 face atoms.


cell_unit/atoms)atoms_face()atoms_corner( 46218

81

=×+×

Coordination number for FCC

R22

1 4

32

8

7

6

5

12 11

1092R

R22

R22


8

Total 12 nearest neighbor atomsCoordination number = 12

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Atomic packing factor (APF) for FCC

2R

R22

R22

a 4R

⎟⎠⎞

⎜⎝⎛= 3

344 RVatoms π


a

⎠⎝ 3333

_ 216)22( RRaV cellunit ===

74.023216

316

3

3

_

==⎟⎠⎞

⎜⎝⎛

==π

π

R

R

VVAPF

cellunit

atoms Independent of R and a!

Summary for FCC

a 4R

||unit vector|| =

4 atoms/unit cell

Coordination number = 12

APF = 0.74

Ra 22=


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Body-Centered Cubic (BCC)

a2

Atoms at the corners of the cube+

Atom at the center of the cubea

a2

R4


( ) ( )

34

42 222

Ra

Raa

=

=+

Body-Centered Cubic (BCC)


From Callister 6e resource CD.

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Number of atoms in a BCC unit cell

• Each corner atom contributes as 1/8contributes as 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.

cell_unit/atoms)atom_center()atoms_corner( 211881

=×+×

Coordination number for BCC

32

Total 8 nearest neighbor atomsCoordination number = 8

4

7

32

6

1


85

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Atomic packing factor (APF) for BCC

⎟⎠⎞

⎜⎝⎛= 3

342 RVatoms πa2

)4()2( 222 Raa =+

3364)

34(

333

_RRaV cellunit ===

⎞⎛

a R43

4)4()2(

Ra

Raa

=

=+


68.083

336438

3

3

_

==⎟⎠⎞

⎜⎝⎛

==π

π

R

R

VVAPF

cellunit

atoms

Summary for BCC

2

||unit vector|| =

2 atoms/unit cell

Coordination number = 8

34Ra =

a

a2

R4


APF = 0.68

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Hexagonal Close-Packing (HCP)


Find a, c, number of atoms/unit cell, coordination number, and APF for HW.

Point Coordinatesto 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)and z not necessarily orthogonal).

ab

y

zpt. coord.

x (a) y (b) z (c)

A 0 0 0

B 1 0 0

C 1 1 1


c

x

y

= point “A” = origin

= point “B”

= point “C”

= point “D”

D 1/2 0 1/2

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Crystallographic DirectionsTo define a vector:

1. Start at the origin.2 Determine length of vector projection in each of 3 axes in units2. 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

Example 1: Assign the coordinates to the following crystallographic direction

Along x: 1 az

ab

c y

Along y: 1 bAlong z: 1 c

[1 1 1]


x

[1 1 1]

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Example 2: draw [1 0] direction. 1

1 unit vector length along x

-1 unit vector length along y

ab

c y

z

g g y

0 unit vector length along z


c

x

y

[1 0]1origin

Note: for some crystal structures, different directions can be equivalent.

e.g. For cubic crystals:

[1 0 0], [ 0 0], [0 1 0], [0 0], [0 0 1], [0 0 ] are all equivalent

111

Families of crystallographic directionse.g. <1 0 0>


Angled brackets denote a family of crystallographic directions.

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Families and Symmetry• Which directions belong in the same family is determined by the

crystal symmetry.• What is symmetry?

Whi h i t i ?• Which one is more symmetric?

• The more symmetry operations there are the more symmetric it is.

a

a

a a

b

c

1 31 1


• Which is more symmetric, cubic or tetragonal?

23 12

and

23 32equivalentequivalent

etc…

Families and Symmetryz

zCubic symmetry

x

y[100]

Rotate 90o about z-axis

x

y

[010]

z

[001]Rotate 90o about y-axis

MSE280© 2007, 2008 Moonsub Shim, University of Illinois32x

y

[001]y

Similarly for other equivalent directions

Symmetry operation can generate all the directions within in a family.

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Crystallographic directions for hexagonal crystals

Consider vectors q and r….If we keep the 3-coordinate system:

y

z = [1 0 0]= [1 1 0]

qv

rv

Different set of indices. However, these two vectors are equivalent by symmetry (i.e. via 60o rotation).


xqv

rv

Choose a 4-coordinate system!


In the Miller-Bravais (4 axes) coordinate system, We have:

a2

z= [2 0]qv 11


qv

a1

a3

Although the length is different the direction is the same.

2 along a1

-1 along a2 -1 along a3

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Similarly for , in the Miller-Bravais (4 axes) coordinate rvsystem, we now have:

a2

z= [1 1 0]rv 2

-2 along a


rva1

a3

1 along a1

1 along a2

-2 along a3

Again, consistent direction.


Miller-Bravais (4 axes) coordinate system

a2

z

We now have:

= [2 0]= [1 1 0]

qv

rv12

1


qvrv

a1

a3

Indices are now consistent within the family, but where do these numbers come from?

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To transform 3 indices to 4 indices[u’ v’ w’] [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 = [1 0 0] to 4-coord. system.

u = 1/3 (2x1 – 0) = 2/3 2

v = 1/3 (2x0 1) = 1/3 1

x 3

x 3

qv z


v = 1/3 (2x0 – 1) = -1/3 -1

t = -(2/3 + (-1/3)) = -1/3 -1

w = 0 0

x 3

x 3

[2 0]1 1 a2

qv a1

a3

transforming 3 indices to 4 indices

convert = [1 1 0] to 4-coord. system. rv

u = 1/3 (2x1 – 1) = 1/3 1

v = 1/3 (2x1 – 1) = 1/3 1

t = -(1/3 + 1/3) = -2/3 -2

w = 0 0

x 3

x 3

x 3

x 3

[1 1 0]

z

2


a2

a1

a3rv

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Crystallographic PlanesTo define a plane (i.e. to assign Miller Indices):

1. If the plane passes through the origin, eitherza) Construct a parallel plane translated

ab

c y

b) choose another origin at the corner of adjacent unit cell

or

2. Now, the plane either intersects or parallels each of the three axes


origin

origin

the three axes.3. Determine length in terms of lattice parameters a, b and c.

x

Parallel to x-axis = length along x ∞Parallel to y-axis = length along y ∞Intercepts z-axis at z = c

Crystallographic Planes4. Take the reciprocal of the lengths

Al i l th 0e.g. Along x-axis: length = 0∞Along y-axis: length = 0∞Along z-axis: length = 1 1

5. If necessary, factor to get the smallest integers.

e g if we had 0 0 3 0 0 1divide through by 3


e.g. if we had 0, 0, 3 0, 0, 16. Enclose indices in parentheses w/o commas

In the example on previous slide, the Miller index would be (0 0 1)

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Example 1Assign Miller indices

1 Shift i i f O t O’

ab

z

c/3O

1. Shift origin from O to O’

z’ 2. Determine lengths along each axis

Along x = ∞Intersects y at -1 Intersects z at 1/3


c

x

yc/3O

O’

x’

3. Take the reciprocal: 0, -1, 3

4. No need to factor.5. (0 3)1

Example 2Draw the (0 1) plane.1

1 along x = 0

y

z1. along x 0

The plane runs parallel to x-axis2. along y = -1 (in units of b)

take reciprocal: -1y-intercept at -1

3. along z = 1 (in units of c)take reciprocal: 1z intercept at 1

|b||c|


x

z-intercept at 1

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Crystallographic planes in FCC

z


y

(1 0 0) planeLook down this direction (perpendicular to the plane)

Crystallographic planes in FCC

zLook down this direction(perpendicular to the plane)


y

(1 1 1) plane

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Note: similar to crystallographic directions, planes that are parallel to each other are equivalent


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)

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Crystallographic planes in HCPz

Name this plane…

a2


a1

a3(0 0 0 1) plane

Parallel to a1, a2 and a3 -> h = k = i = 0Intersects at z = 1

Crystallographic planes in HCP

z

a2


a1

a3

(1 1 0 0) plane

+1 in a1

-1 in a2

h = 1, l = 0i = -(1+-1) = 0,k = -1,

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(1 1 1) plane of FCCz

x

y

z (0 0 0 1) plane of HCPSAME THING!*

MSE280© 2007, 2008 Moonsub Shim, University of Illinois49a1

a2

a3

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!

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FCC


HCP


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Difference between FCC and HCPLooking down (0001) plane


HCPFCC

ABCABC… ABABAB…

Looking down (111) plane!

Densities

C t l d it (i 3D d it ) i it f• Crystals density (i.e. 3D density) in units of mass per volume (e.g. g/cm3).

• Linear density: number of atoms per unit length (e.g. cm-1).

• Planar density: number of atoms per unit area (e g


• Planar density: number of atoms per unit area (e.g. cm-2).

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Crystals density (ρ)Mass of unit cell

VM

=ρ Volume of unit cell

ANnAM =

n = number of atoms in unit cellA = atomic weightNA = Avogadro’s number


VNnA

A

=ρ

Density exampleCalculate density of copper given: R = 0.128 nm

A = 63.5 g/molFCC structure

Recall for FCC, there are 4 atoms per unit cell.

Express unit cell volume in terms of atomic radius R.3aV = 3)22( R= 3216 R=

nAM==ρ

Then we have:


3216 RNV A

==ρ

38123 )1028.1)(216)(10022.6()/3.65(4

cmxmolxmolg

−−= = 8.89 g/cm3

Compare to actual value of 8.94 g/cm3!

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Linear Density (LD)LD =

Number of atoms centered on a direction vectorLength of the direction vector

Example: calculate the linear density of an FCC crystal along [1 1 0].

Length = 4R

Effectively 2 atoms along the


atoms along the [1 1 0] vector.

LD = 2/4R = 1/2R

Planar Density (PD)PD =

Number of atoms centered on a given planeArea of the plane

Example: calculate the planar density on (1 1 0) plane of an FCC crystal.

Ra 22=


R4Count only the parts of the atoms within the plane: 2 atoms

RRarea 422 ×= 2241

)22)(4(2

RRRPD ==

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Example problem(110)

nm 0

(101)0.

(011)

0.39 nm

0.2

.32 nm0.3 nm

36 nm

0.253 nm

Given above information answer the following questions.


g qA) 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)?

How to determine crystal structureThe two slit experiment

Same idea with crystals- Light gets scattered off atoms…

Incoming light

dθ

λ


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

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Diffraction Experiment and Signal

Diffraction Experiment Scan versus 2x Angle: Polycrystalline Cu

Measure 2θ

Diffraction collects data in “reciprocal space” since it is the equivalent of

MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08

How can 2θ scans help us determine crystal structure type and distances between Miller Indexed planes (I.e. structural parameters)?

p p qa Fourier transform of “real space”, i.e., eikr, as k ~ 1/r.

Crystal Structure and Planar Distances

h, k, l are the Miller Indices of the planes of atoms that scatter!

BravaisLattice

Constructive Interference

Destructive Interference

Reflections present

Reflections absent

BCC (h + k + l) = Even (h + k + l) = Odd

Distances between Miller Indexed planes

of atoms that scatter!

So they determine the important planes of atoms, or symmetry.

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 = Oddn= integer


For cubic crystals:

dhkl =a

4

3(h2 + hk + k2 ) + l2 a

c⎛⎝

⎞⎠

2

For hexagonal crystals:

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(h k l) h2+k2+l2 h+k+l h,k,l all even or odd?

100 1 1 No110 2 2 No111 3 3 Yes

Allowed (hkl) in FCC and BCC for principal scattering (n=1)

Self-Assessment:From what crystal structure is this?

111 3 3 Yes200 4 2 Yes210 5 3 No211 6 4 No220 8 4 Yes221 9 5 No300 9 3 No310 10 4 No


310 10 4 No311 11 5 Yes222 12 6 Yes320 13 5 No321 14 6 No

h + k + l was even and gave the labels on graph above, so crystal is BCC.

Example problem: X-ray crystallography

• Given the information below, determine the crystal structure. Consider only FCC and BCC structures asstructure. Consider only FCC and BCC structures as possibilities. – Lattice parameter: a = 0.4997nm– Powder x-ray: λ = 0.1542 nm

2θ (o)3136

MSE280© 2007, 2008 Moonsub Shim, University of Illinois

3651.861.664.8

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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

Diffraction of Multiple Single 2D Crystal Grains (powders)

Multiple Crystal Grains:4 Polycrystals

Diffraction Patternin (h,k) plane (4 grains)

Diffraction Patternin (h,k) plane (40 grains)




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Example from Graphite

Single Crystal Grain Mutlple Grains: Polycrystal




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 HCP


• FCC, BCC, HCP• Crystallographic coordinates, directions, and planes.• Densities• X-ray crystallogrphy

Design

Materiales the science and engineering of materials - Solution Manual