Embed Size (px)
Citation preview
Chapter 02 Crystallography
JYOTHIS ACADEMYERAYIL KADAVU JN.,
KOTTAYAM-1
Phone +91 93 8888 7363
Email [email protected]
www.amieindia.in
for AMIE
What is this chapter about ?
• To make any object/article a material is to be used.• Eg Table – made of wood/metal• A car- made by using various metals• A chair- made of plastic etc..
• How do we decide the Application of a materials for making a product ?In order that a particular material to be qualified for making a product , it must have certain Properties.
• How do (from where) a material get its properties ??• Answer.....From its Structure (fashion in which atoms are
arranged)
Structure-Property-Application
• Application- is decided by property• Property- is decided by structure,
and • Structure = how atoms are arranged.
Thus we see
are interrelated
Arrangement of atoms
Atoms in solids may be
1.Regularly (Orderly) arranged
(We then say they have crystalline structure)
2.Randomly arranged
(They have amorphouse structure)
• atoms pack in periodic, 3D arrays• Ex. :
Crystalline materials...
-metals-many ceramics-some polymers
• atoms have no periodic packing•
Noncrystalline materials...
crystalline
noncrystalline
"Amorphous" = Noncrystalline
Crystalline Vs Non-crystalline
Most METALS (>99%) are CRYSTALLINE.
CERAMICS are CRYSTALLINE except GLASSES which are AMORPHOUS.
POLYMERS (plastics) tend to be: either AMORPHOUS or a mixture of CRYSTALLINE + AMORPHOUS (known as Semi-crystalline)
Ideal Crystal
• An ideal crystal is a periodic array of structural units, such as atoms or molecules.
• It can be constructed by the infinite repetition of these identical structural units in space.
• Structure can be described in terms of a lattice, with a group of atoms attached to each lattice point.
A crystal lattice is a 3-D stack of unit cells
Crystal lattice is an imaginative grid system in three dimensions in which every point (or node) has an environment that is identical to that of any other point or node.
Unit cell
• Actual size of UNIT CELLS is VERY VERY SMALL!!
• Iron unit cell length (0.287 x 10-9 m) (0.287 nm)• 1 mm length of iron crystal has 3.5 million unit cells
Unit cell
Group crystals depending on shape of Unit Cell.x, y and z are three axes of lattice separated by angles , and .A unit cell will have sides of length a, b and c. (Note: for the cubic system all sides equal so a = b = c)
A crystal’s unit cell dimensions are defined by six numbers, the lengths of the 3 axes, a, b, and c, and the three interaxial angles, , and .
Unit cell is the smallest unit of a crystal, which, if repeated, could generate the whole crystal.
Crystals are made of infinite number of unit cells
SOME DEFINITIONS …• Lattice: 3D array of
regularly spaced points• Crystalline material: atoms
situated in a repeating 3D periodic array over large atomic distances
• Amorphous material: material with no such order
• Unit cell: basic building block unit that repeats in space to create the crystal structure; it is usually a parallelepiped or prizm
CRYSTAL SYSTEMS
• Based on shape of unit cell ignoring actual atomic locations
• Unit cell = 3-dimensional unit that repeats in space• Unit cell geometry completely specified by a, b, c &
lattice parameters or lattice constants)• Seven possible combinations of a, b, c & ,
resulting in seven crystal systems
Crystal systems
• Crystals are grouped into seven crystal systems, according to characteristic symmetry of their unit cell.
METALLIC CRYSTAL STRUCTURESMost metals crystallize into one of three densely packed structures:
BODY CENTERED CUBIC - BCC
FACE CENTERED CUBIC - FCC
HEXAGONAL (CLOSE PACKED) - HCP
• Rare due to poor packing • Close-packed directions are cube edges.
• Coordination # = 6 (# nearest neighbors)
SIMPLE CUBIC STRUCTURE (SC)
Body Centered Cubic
• Atoms are arranged at the corners of the cube with another atom at the cube center.
Atomic Radius (BCC)
• Since atoms are assumed to touch along the cube diagonal in BCC, the lattice parameter is related to atomic radius through:
3
4Ra
Coordination number BCC
• Coordination number for BCC is 8. Each center atom is surrounded by the eight corner atoms.
• The lower coordination number also results in a slightly lower APF for BCC structures. BCC has an APF of 0.68, rather than 0.74 in FCC
• Coordination # = 8
• Close packed directions are cube diagonals.--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.
BODY CENTERED CUBIC STRUCTURE (BCC)
• Coordination # = 12
• Close packed directions are face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.
FACE CENTERED CUBIC STRUCTURE (FCC)
ATOMIC PACKING FACTOR
• Fill a box with hard spheres– Packing factor = total volume of spheres in box
/ volume of box– Question: what is the maximum packing factor
you can expect?
• In crystalline materials:– Atomic packing factor = total volume of atoms
in unit cell / volume of unit cell– (as unit cell repeats in space)
[email protected] 40• APF for a simple cubic structure = 0.52
APF = a3
4
3(0.5a)31
atoms
unit cellatom
volume
unit cellvolume
ATOMIC PACKING FACTOR
contains 8 x 1/8 = 1 atom/unit cell
Adapted from Fig. 3.19, Callister 6e.
Lattice constant
close-packed directions
a
R=0.5a
Atomic packing Factor
APF = Volume of atoms in unit cell*
Volume of unit cell
*assume hard spheres
APF for a simple cubic structure = 0.52
APF = a3
4
3(0.5a)31
atoms
unit cellatom
volume
unit cellvolume
close-packed directions
a
R=0.5a
contains 8 x 1/8 = 1 atom/unit cell
aR
• APF for a body-centered cubic structure = 0.68
Close-packed directions: length = 4R
= 3 a
Unit cell contains: 1 + 8 x 1/8 = 2 atoms/unit cell
ATOMIC PACKING FACTOR: BCC
APF = a3
4
3( 3a/4)32
atoms
unit cell atomvolume
unit cell
volume
BCC STRUCTUREAtoms at cube corners and one in cube centre.
Lattice Constant for BCC:3
4Ra
e.g. Fe (BCC) a = 0.287 nm
Two atoms in Unit Cell.(1 x 1 (centre)) + (8 x 1/8 (corners)) = 2
Each atom in BCC is surrounded by 8 others. COORDINATION number of 8.Packing is not as good as FCC; APF = 0.68
BCC metals include:Iron (RT), Chromium, Tungsten, Vanadium
APF -FCC
• Atomic Packing Factor: the ratio of atomic sphere volume to unit cell volume, assuming a hard sphere model.– FCC systems have an APF of 0.74, the
maximum packing for a system in which all spheres have equal diameter.
APF = a3
4
3( 2a/4)34
atoms
unit cell atomvolume
unit cell
volume
Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell
a
• APF for a body-centered cubic structure = 0.74
Close-packed directions: length = 4R
= 2 a
ATOMIC PACKING FACTOR: FCC
FACE CENTERED CUBIC (FCC)e.g. copper, aluminium, gold, silver, lead, nickel
Lattice constant (length of cube side in FCC) “a” for FCC structure:
22 2
48 R
RRa
where R = atomic radiusEach type of metal crystal structure has its own lattice constant.
(1/8 at each corner x 8) + (½ at each face x 6 ) = 4So 4 atoms per Unit Cell.
Each atom touches 12 others. Co-ordination number = 12.
• ABCABC... Stacking Sequence• 2D Projection
A sites
B sites
C sitesB B
B
BB
B BC C
CA
A
• FCC Unit CellA
BC
FCC STACKING SEQUENCE
• Coordination # = 12
• ABAB... Stacking Sequence
• APF = 0.74
• 3D Projection • 2D Projection
A sites
B sites
A sites Bottom layer
Middle layer
Top layer
Adapted from Fig. 3.3, Callister 6e.
HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)
HEXAGONAL CLOSE PACKED
Note: not simple hexagonal but HCPSimple Hex. very inefficient; HCP has extra plane of atoms in middle.
1/6 of atom at each corner.So (1/6) x 12 corners = 2 atomsand (½) x (top + bottom) = 1 atomand (3) internal = 3 atoms
Total = 6 atoms/cell
Because of Hexagonal arrangement (not cubic), have 2 lattice parameters“a” , and “c”
a = basal side = 2Rc = cell height
By geometry, for IDEAL HCP: 1.633 a
c
but this varies slightly for some HCP Metals.
HCP metals include: Magnesium, Zinc, Titanium, Zirconium, Cobalt.
atomic packing factor for HCP = 0.74 (same as FCC)Atoms are packed as tightly as possible.Each atom surrounded by 12 other atoms so co-ordination number = 12.
CRYSTAL DENSITY
The true density, , of material (free from defects) can be calculated knowing its crystal structure.
Acc
A
NV
nA
V
NnA
cellunit of volume
cellunit of mass
n = number of atoms in unit cellA = Atomic Weight of element (g/mol)Vc = volume of unit cellNav = Avogadro’s number (6.023 x 1023 atoms/mol)
e.g., copper, FCC 4 atoms/celln = 4Cu atoms have mass 63.5 g/mol
Vol. of cell = a3 , for FCC a =2R2Atomic radius of copper = 0.128 nm
36
2339-1089.8
10x 6.0230.128x10216
5.634
gmxx
x
NV
nA
Ac
Cu
= 8.89 Mgm-3 (or 8.89 gcm-3 or 8890 kgm-3)
POLYMORPHISM / ALLOTROPY
Some elements/compounds can exist in more than one crystal form. Usually requires change in temperature or pressure.Carbon: Diamond (high pressure) or Graphite (low).
Can be IMPORTANT as some crystal structures more dense (better packing, higher APF) than others, so a change in crystal structure can often result in volume change of material.
APFe.g. Iron 913oC FCC 0.74
911oC BCC 0.68
i.e. expands on cooling!
CRYSTALLOGRAPHIC POINTS, DIRECTIONS & PLANES
• In crystalline materials, often necessary to specify points, directions and planes within unit cell and in crystal lattice
• Three numbers (or indices) used to designate points, directions (lines) or planes, based on basic geometric notions
• The three indices are determined by placing the origin at one of the corners of the unit cell, and the coordinate axes along the unit cell edges
POINT COORDINATES• Any point within a unit cell specified as fractional
multiples of the unit cell edge lengths• Position P specified as q r s; convention:
coordinates not separated by commas or punctuation marks
EXAMPLE: POINT COORDINATES
• Specify point coordinates for all atom positions for a BCC unit cell– Answer: 0 0 0, 1 0 0, 1 1 0, 0 1 0, ½ ½ ½,
0 0 1, 1 0 1, 1 1 1, 0 1 1
CRYSTALLOGRAPHIC DIRECTIONS
• Defined as line between two points: a vector• Steps for finding the 3 indices denoting a direction
– Determine the point positions of a beginning point (X1 Y1 Z1) and a ending point (X2 Y2 Z2) for direction, in terms of unit cell edges
– Calculate difference between ending and starting point– Multiply the differences by a common constant to convert them
to the smallest possible integers u, v, w– The three indices are not separated by commas and are
enclosed in square brackets: [uvw]– If any of the indices is negative, a bar is placed in top of that
index
EXAMPLES: DIRECTIONS• Draw a [1,-1,0] direction within a cubic unit cell
• Determine the indices for this direction– Answer: [120]
Crystallographic Directions
• Cubic systems– directions are named based upon the
projection of a vector from the origin of the crystal to another point in the cell.
• Conventionally, a right hand Cartesian coordinate system is used.– The chosen origin is arbitrary, but is always
selected for the easiest solution to the problem.
Crystallographic Directions
• Points within the lattice are written in the form h,k,l, where the three indices correspond to the fraction of the lattice parameters in the x,y,z direction.
Miller Indices
• Procedure for writing directions in Miller Indices– Determine the coordinates of the two points
in the direction. (Simplified if one of the points is the origin).
– Subtract the coordinates of the second point from those of the first.
– Clear fractions to give lowest integer values for all coordinates
Miller Indices
– Indices are written in square brackets without commas (ex: [hkl])
– Negative values are written with a bar over the integer.
• Ex: if h<0 then the direction is•
][ klh
Miller Indices
• Crystallographic Planes– Identify the coordinate intercepts of the plane
• the coordinates at which the plane intercepts the x, y and z axes.
• If a plane is parallel to an axis, its intercept is taken as .
• If a plane passes through the origin, choose an equivalent plane, or move the origin
– Take the reciprocal of the intercepts
Miller Indices
– Clear fractions due to the reciprocal, but do not reduce to lowest integer values.
– Planes are written in parentheses, with bars over the negative indices.
• Ex: (hkl) or if h<0 then it becomes
• ex: plane A is parallel to x, and intercepts y and z at 1, and therefore is the (011). Plane B passes through the origin, so the origin is moved to O’, thereby making the plane the
)( klh
)121(
CRYSTALLOGRAPHIC DIRECTIONS
Line between two points or vector.Using 3 coordinate axes, x, y, and z.
• Position vector so that it passes through origin (parallel vectors can be translated).
• Length of vector projected onto the three axes (x, y and z) is determined in terms of unit cell dimensions (a, b and c).
• Multiply or divide by common factor to reduce to lowest common integers.
• Enclose in SQUARE brackets with no commas [uvw], and minus numbers given by bar over number; e.g.
]21[2 [111], ],211[
Parallel vectors have same indices. Changing sign of all indices gives opposite direction.
If directions are similar, (i.e., same atomic arrangements - for example, the edges of a BCC cube) they belong to a FAMILY of directions:
100 ]1[00 [001], ],01[0 [010], ],001[ ,]100[
i.e. with < > brackets can change order and sign of integers.e.g. cube internal diagonals <111>
cube face diagonals <110>
CRYSTAL PLANES
Planes specified by Miller Indices (hkl) (Reciprocal Lattice).
Any two planes parallel to each other are equivalent and have identical Miller indices
CRYSTALLOGRAPHIC PLANES• Crystallographic planes specified by 3
Miller indices as (hkl)• Procedure for determining h,k and l:
– If plane passes through origin, translate plane or choose new origin
– Determine intercepts of planes on each of the axes in terms of unit cell edge lengths (lattice parameters). Note: if plane has no intercept to an axis (i.e., it is parallel to that axis), intercept is infinity (½ ¼ ½)
– Determine reciprocal of the three intercepts (2 4 2)
– If necessary, multiply these three numbers by a common factor which converts all the reciprocals to small integers (1 2 1)
– The three indices are not separated by commas and are enclosed in curved brackets: (hkl) (121)
– If any of the indices is negative, a bar is placed in top of that index
1 /2
1 /2
1 /4
(1 2 1 )X
Y
Z
FCC & BCC CRYSTAL PLANES
• Consider (110) plane
• Atomic packing different in the two cases• Family of planes: all planes that are
crystallographically equivalent—that is having the same atomic packing, indicated as {hkl}– For example, {100} includes (100), (010), (001) planes– {110} includes (110), (101), (011), etc.
LINEAR ATOMIC DENSITIES
Tells us how well packed atoms are in a given direction. If LD = 1 then atoms are touching each other.
Llength, selected
Lcentres, atom ngintersecti line of length LDsity,Linear Den c
l
PLANAR DENSITIES Tells us how well packed atoms are on a given plane. Similar to linear densities but on a plane rather than just a line.
p
cA area, selected
A plane,by dintersecte atoms ofArea PDdensity,Planar
gives fraction of area covered by atoms.
e.g., BCC unit cell, (110) plane:
2 whole atoms on plane in unit cell.
for BCC) 3
4=a (where
22)2R( 2 R
aPD
So Ac = 2(R2) AD = a, DE = a2
And so Ap = a22
PACKING ON PLANES
FCC and HCP are both CLOSE-PACKED structures. APF = 0.74 (This is the maximum if all atoms are same size).
Atoms are packed in CLOSE-PACKED planesIn FCC, {111} are close packed planesIn HCP, (0001) is close packed
Both made of close packed planes, but different stacking sequence.FCC planes stack as ABCABCABCHCP planes stack as ABABABABAB
BCC is not close packed (APF = 0.68)most densely packed plane is {110}
• Single Crystals-Properties vary with direction: anisotropic.
-Example: the modulus of elasticity (E) in BCC iron:
• Polycrystals-Properties may/may not vary with direction.-If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa)-If grains are textured, anisotropic.
E (diagonal) = 273 GPa
E (edge) = 125 GPa
200 m
SINGLE VS POLYCRYSTALS
SINGLE CRYSTALSThis is when a piece of material is made up of one crystal; all the unit cells are aligned up in the same orientation.
POLYCRYSTALMany small crystals (grains) withdifferent orientations joined together. Most materials/metals are POLYCRYSTALLINE. Grain boundary - Regions where grains (crystals) meet.
ANISOTROPYMany properties depend on direction in crystal in which they are measured. E.g. Stiffness (rigidity) electrical conductivity, refraction. If property varies with direction - Anisotropic.If no variation with direction - Isotropic
Single crystals show this variation.
Polycrystalline materials are usually randomly oriented so effect is evened out to give average values in all directions.
So can measure peak and determine dhkl and then “a”.
Distance between similar planes in the cubic systems, e.g., (110) planes in adjacent unit cells:
222 lkh
adhkl
Angle Between Two Directions
Take two lattice vectors: t1 = U1a + V1b + W1c [U1 V1 W1]
and t2 = U2a + V2b + W2c [U2 V2 W2]
The definition of the dot product is
t1·t2 = |t1||t2|cos
so
21
21costt
tt
Gypsum
Monoclinica = 5.68Å, b = 15.18Å, c = 6.29Å, = 113.83°
What is Angle Between [1 0 0] and [0 2 1]? −
564.390435.14
0432.2300
435.140262.32
G
2
120100
A435.14
1
2
0
001tt
G
70.94
00.3168.5
435.14cos
tt
ttcos120001
1
120100
1201001
The Angle Calculation
Miller indices for direction are specified in the following manner:
Set up a vector of arbitrary length in the direction of interest.
Decompose the vector into its components along the principal axes.
Using an appropriate multiplier, convert the component values into the smallest possible whole number set.
[hkl] – square brackets are used to designate specific direction within the crystal.
<hkl> - triangular brackets designate an equivalent set of directions.
The separation between adjacent planes in a cubic crystal is given by:
The angle between planes is given by:
222 lkh
ad
22
22
22
21
21
21
212121coslkhlkh
llkkhh
HEXAGONAL CRYSTALS
Use a 4-axis system (Miller-Bravais).
a1, a2 and a3 axes in basal plane at 120 to each other and z axis in vertical direction. Directions given by [uvtw] or [a1 a2 a3 c]Can convert from three-index to four index system.
t=-(u+v)
1.3 Index System for Crystal Planes – Miler Indices
The orientation of a crystal plane is determined by three points in the plane that are not collinear to each other.It is more useful to specify the orientation of a plane by the following rules:
Find the intercepts of the axes in terms of lattice constants a1, a2 and a3.
Take a reciprocal of these numbers and then reduce to three integers having the same ratio. The result (hkl) is called the index of a plane.
Planes equivalent by summetry are denoted in curly brackets around the indices {hkl}.
To find Miller Indices of a plane: • If the plane passes through the selected origin, construct a parallel plan in the unit cell or select an origin in another unit cell.• Determine where plane intercepts axes. (if no intercept i.e.., plane is parallel to axis, then ) e.g., axis x y z
intercept a b c
• Take reciprocals of intercepts (assume reciprocal of is 0): 1/a 1/b 1/c
• Multiply or divide to clear fractions: (hkl) Miller indices of plane
FAMILY of planes, use {hkl}These planes are crystallographically similar (same atomic arrangements).e.g., for cube faces: {100}
}100{ )1(00 (001), ),01(0 (010), ),001( ,)100(
NOTE: In CUBIC system only, directions are perpendicular to planes with same indices.e.g., [111] direction is perpendicular to the (111) plane.
HEXAGONAL CRYSTALSFour-index system similar to directions; (hkil)i = - (h+K)
HexagonalClosePacked
HCP lattice is not a Bravais lattice, because orientation of the environmentOf a point varies from layer to layer along the c-axis.
Miller indices of lattice plane
• The indices of a crystal plane (h,k,l) are defined to be a set of integers with no common factors, inversely proportional to the intercepts of the crystal plane along the crystal axes:
Simple Crystal Structures
• There are several crystal structures of common interest: sodium chloride, cesium chloride, hexagonal close-packed, diamond and cubic zinc sulfide.
• Each of these structures have many different realizations.
ISSUES TO ADDRESS...
• How do atoms assemble into solid structures? • How does the density of a material depend on its structure?
• When do material properties vary with the sample orientation?
TOPIC 3: STRUCTURE OF SOLIDS
Based on Chapter 3 (Callister)
• Cubic unit cell is 3D repeat unit • Rare (only Po has this structure)• Close-packed directions (directions along which atoms touch each other) are cube edges.
• Coordination # = 6 (# nearest neighbors)
SIMPLE CUBIC (SC)
• Coordination # = 8
• Close packed directions are cube diagonals.--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.
BODY CENTERED CUBIC STRUCTURE (BCC)
• ABCABC... Stacking Sequence
• FCC Unit CellA
BC
FCC STACKING SEQUENCE
A sites
B sites
C sitesB B
B
BB
B BC C
CA
A
• 2D Projection
HCP STRUCTURE
Ideally, c/a = 1.633 for close packingHowever, in most metals, c/a ratio deviates from this value
• Coordination # = 12
• ABAB... Stacking Sequence
• APF = 0.74, for ideal c/a ratio of 1.633
• 3D Projection • 2D Projection
A sites
B sites
A sites Bottom layer
Middle layer
Top layer
Adapted from Fig. 3.3, Callister 6e.
HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)
Close packed crystals
A plane
B plane
C plane
A plane
…ABCABCABC… packing[Face Centered Cubic (FCC)]
…ABABAB… packing[Hexagonal Close Packing (HCP)]
COMPARISON OF CRYSTAL STRUCTURES
Crystal structure coordination # packing factor close packed directions
• Simple Cubic (SC) 6 0.52 cube edges
• Body Centered Cubic (BCC) 8 0.68 body diagonal
• Face Centered Cubic (FCC) 12 0.74 face diagonal
• Hexagonal Close Pack (HCP) 12 0.74 hexagonal side
THEORETICAL DENSITY, Density = mass/volume
mass = number of atoms per unit cell * mass of each atom
mass of each atom = atomic weight/avogadro’s number
n AVcNA
# atoms/unit cell Atomic weight (g/mol)
Volume/unit cell
(cm3/unit cell)Avogadro's number (6.023 x 1023 atoms/mol)
Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen
Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H
At. Weight (amu) 26.98 39.95 137.33 9.012 10.81 79.90 112.41 40.08 12.011 132.91 35.45 52.00 58.93 63.55 19.00 69.72 72.59 196.97 4.003 1.008
Atomic radius (nm) 0.143 ------ 0.217 0.114 ------ ------ 0.149 0.197 0.071 0.265 ------ 0.125 0.125 0.128 ------ 0.122 0.122 0.144 ------ ------
Density (g/cm3) 2.71 ------ 3.5 1.85 2.34 ------ 8.65 1.55 2.25 1.87 ------ 7.19 8.9 8.94 ------ 5.90 5.32 19.32 ------ ------
Crystal Structure FCC ------ BCC HCP Rhomb ------ HCP FCC Hex BCC ------ BCC HCP FCC ------ Ortho. Dia. cubic FCC ------ ------
Adapted fromTable, "Charac-teristics ofSelectedElements",inside frontcover,Callister 6e.
Characteristics of Selected Elements at 20C
n AVcNA
# atoms/unit cell Atomic weight (g/mol)
Volume/unit cell
(cm3/unit cell)Avogadro's number (6.023 x 1023 atoms/mol)
Example: CopperData from Table inside front cover of Callister (see previous slide):
• crystal structure = FCC: 4 atoms/unit cell• atomic weight = 63.55 g/mol (1 amu = 1 g/mol)• atomic radius R = 0.128 nm (1 nm = 10 cm)-7
Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3
Compare to actual: Cu = 8.94 g/cm3Result: theoretical Cu = 8.89 g/cm3
THEORETICAL DENSITY,
(g
/cm
3)
Graphite/ Ceramics/ Semicond
Metals/ Alloys
Composites/ fibersPolymers
1
2
20
30Based on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass,
Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers
in an epoxy matrix). 10
3 4 5
0.3 0.4 0.5
Magnesium
Aluminum
Steels
Titanium
Cu,Ni
Tin, Zinc
Silver, Mo
Tantalum Gold, W Platinum
Graphite Silicon
Glass -soda Concrete
Si nitride Diamond Al oxide
Zirconia
HDPE, PS PP, LDPE
PC
PTFE
PET PVC Silicone
Wood
AFRE *
CFRE *
GFRE*
Glass fibers
Carbon fibers
Aramid fibers
Why? Metals have... • close-packing (metallic bonding) • large atomic mass Ceramics have... • less dense packing (covalent bonding) • often lighter elements Polymers have... • poor packing (often amorphous) • lighter elements (C,H,O) Composites have... • intermediate values Data from Table B1, Callister 6e.
DENSITIES OF MATERIAL CLASSESmetals> ceramics> polymers
CRYSTAL STRUCTURES
• Plenty of crystal structures available at: http://cst-www.nrl.navy.mil/lattice/
• Polymorphism– Same compound occurring in more than one
crystal structure
• Allotropy– Polymorphism in elemental solids (e.g., carbon)
• Atoms may assemble into crystalline, noncrystalline (or amorphous) structures.
• We can predict the density of a material, provided we know the atomic weight, atomic radius, and crystal geometry (e.g., FCC, BCC, HCP).
• Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but properties are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains.
SUMMARY
Crystal Structures
• Types of crystal structures– Face centered cubic (FCC)– Body centered cubic (BCC)– Hexagonal close packed (HCP)
Face Centered Cubic (FCC)
• Atoms are arranged at the corners and center of each cube face of the cell.– Atoms are assumed to touch along face
diagonals
Face Centered Cubic (FCC)
• The lattice parameter, a, is related to the radius of the atom in the cell through:
• Coordination number: the number of nearest neighbors to any atom. For FCC systems, the coordination number is 12.
22Ra
Hexagonal Close Packed
• Cell of an HCP lattice is visualized as a top and bottom plane of 7 atoms, forming a regular hexagon around a central atom. In between these planes is a half-hexagon of 3 atoms.
Hexagonal Close Packed
• There are two lattice parameters in HCP, a and c, representing the basal and height parameters respectively. In the ideal case, the c/a ratio is 1.633, however, deviations do occur.
• Coordination number and APF for HCP are exactly the same as those for FCC: 12 and 0.74 respectively.– This is because they are both considered close
packed structures.
Close Packed Structures
• Even though FCC and HCP are close packed structures, they are quite different in the manner of stacking their close packed planes.– Close packed stacking in HCP takes place
along the c direction ( the (0001) plane). FCC close packed planes are along the (111).
– First plane is visualized as an atom surrounded by 6 nearest neighbors in both HCP and FCC.
Close Packed Structures
– The second plane in both HCP and FCC is situated in the “holes” above the first plane of atoms.
– Two possible placements for the third plane of atoms
• Third plane is placed directly above the first plane of atoms
– ABA stacking -- HCP structure
• Third plane is placed above the “holes” of the first plane not covered by the second plane
– ABC stacking -- FCC structure
• Some engineering applications require single crystals:
• Crystal properties reveal features of atomic structure.
(Courtesy P.M. Anderson)
--Ex: Certain crystal planes in quartz fracture more easily than others.
--diamond single crystals for abrasives
--turbine bladesFig. 8.30(c), Callister 6e.(Fig. 8.30(c) courtesyof Pratt and Whitney).(Courtesy Martin Deakins,
GE Superabrasives, Worthington, OH. Used with permission.)
Single Vs Polycrystals• Single crystal: when the periodic and repeated arrangement of atoms is perfect and extends throughout the entirety of the specimen
• Single Crystals-Properties vary with direction: anisotropic.
-Example: the modulus of elasticity (E) in BCC iron:
• Polycrystals
-Properties may/may not vary with direction.-If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa)-If grains are textured, anisotropic.
E (diagonal) = 273 GPa
E (edge) = 125 GPa
200 mm
Data from Table 3.3, Callister 6e.(Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)
Adapted from Fig. 4.12(b), Callister 6e.(Fig. 4.12(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].)
SINGLE VS POLYCRYSTALS
POLYCRYSTALLINE MATERIALS• “Nuclei” form during solidification, each of which grows into crystals
Miller indices
A Miller index is a series of coprime integers that are inversely proportional to the intercepts of the crystal face or crystallographic planes with the edges of the unit cell.
It describes the orientation of a plane in the 3-D lattice with respect to the axes.
The general form of the Miller index is (h, k, l) where h, k, and l are integers related to the unit cell along the a, b, c crystal axes.
Miller Indices
Rules for determining Miller Indices: 1. Determine the intercepts of the face along the crystallographic axes, in terms of unit cell dimensions.2. Take the reciprocals3. Clear fractions4. Reduce to lowest terms
An example of the (111) plane (h=1, k=1, l=1) is shown on the right.
Rules for determining Miller Indices: 1. Determine the intercepts of the face along the crystallographic axes, in terms of unit cell dimensions.2. Take the reciprocals3. Clear fractions4. Reduce to lowest terms
Another example:
Lattices
• In 1848, Auguste Bravais demonstrated that in a 3-dimensional system there are fourteen possible lattices
• A Bravais lattice is an infinite array of discrete points with identical environment
• seven crystal systems + four lattice centering types = 14 Bravais lattices
• Lattices are characterized by translation symmetry
Auguste Bravais (1811-1863)
LINEAR & PLANAR DENSITIES
• Linear density (LD) = number of atoms centered on a direction vector / length of direction vector– LD (110) = 2 atoms/(4R) =
1/(2R)• Planar density (PD) = number of
atoms centered on a plane / area of plane– PD (110) = 2 atoms / [(4R)
(2R2)] = 2 atoms / (8R22) = 1/(4R22)
• LD and PD are important considerations during deformation and “slip”; planes tend to slip or slide along planes with high PD along directions with high LD
Linear and Planar Atomic Densities
Linear Density:Directional equivalency is related to the atomic linear density in the sense that equivalent directions have identical linear densities.The direction vector is positioned so as to pass through atom centers.The fraction of line length intersected by these atoms is equal to the linear density.
Planar Density:Crystallographic planes that are equivalent have the same atomic planar density. The plane of interest is positioned so as to pass through atom centers.Planar density is the fraction of total crystallographic plane area that is occupied by atoms. Linear and planar densities are one- and two-dimensional analogs of the atomic packing factor.
Linear Density for BCC
Calculate the linear density for the following directions:
a. [100]
b. [110]
c. [111]
Planar Density for BCC
Calculate the planar density for the following BCC planes:
a. (100)
b. (110)
Crystalline and Non-Crystalline Materials
Single Crystal:
The periodic and repeated arrangements of atoms is perfect or extends throughout the entirety of the specimen without interruption.
All unit cells interlock in the same way and have the same orientation.
Single crystals exist in nature, but they may also produced artificially.
They are ordinarily difficult to grow, because the environment must be carefully controlled.
Several Single Crystals of Fluorite
Single crystals are needed for modern technologies today.
Electronic micro-chips uses single crystals of silicon and other semiconductors.
Polycrystalline Materials Composed of a collection of many small crystals or grains.
AnisotropyPhysical properties of single crystals of some substances depend on the crystallographic direction in which measurements are made.
This directionality of properties is termed anisotropy, and it is associated with the variance of atomic or ionic spacing with crystallographic direction.
Substances in which measured properties are independent of the direction of measurement are isotropic.
