MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Problem Set 1
Due: Tuesday, January 24, 11:00 AM
1. Crystal Structure
a) Look up the crystal structure of the following elements at STP: sodium, magnesium,
aluminum, titanium, iron, nickel, copper, zirconium, silver, tungsten, gold, lead.
FCC – Aluminum, nickel, copper, silver, gold, lead.
BCC – Sodium, iron, tungsten.
HCP – Magnesium, titanium, zirconium.
b) In three-dimensional representations, draw: i) (0 0 1) ; ii) (1 1 0) ; and iii) (3 1 1)
planes in the unit cells of BCC crystals. On part (iii), also draw the [3 1 1] direction.
i)
iii)
ii)
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
c) i) Using Table 3.1 in Callister, calculate the theoretical density of pure α-titanium. In
your calculation, assume that the c/a ratio is such that the atoms are as densely packed as
possible.
(sidenote: As seen below, HCP structures can be represented by 1/3 of the hexagonalbased cell. This is because all of the information is captured in this cell containing 2
atoms, and it can be repeated to build the entire cell.)
First, find the c/a ratio representing the closest packing. Recognizing that these atoms
form a tetrahedron, and assuming the parameter “a” is minimized (i.e. a is twice the
atomic radius, R), we can relate c to a.
√
√
Use this to find the volume of the unit cell, recalling that a = 2R, where RTi is 0.1445nm
or 1.445 x 10-8 cm.
√
Area of Trapezoidal base =
Height of Cell =c= 1.633a
Volume= A*h= 1.4143*a3=11.3139*R3=3.4136*10-23 cm3
Using the formula presented in lecture:
ii) What is the density of titanium as measured in the literature? Include your source. To
what would you attribute the difference?
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
4.50 g/cm3. Found at http://www.matweb.com/
The c/a ratio is likely not the ideal ratio as calculated in part (i). Also, the models we use
assume hard spheres with defined radii, but bonding will distort these interatomic
distances and is often anisotropic. In real materials, the various types of bonding, which
can occur to differing degrees simultaneously, give rise to non-ideal behavior.
(note: the real c/a ratio in α-titanium is 1.59, 3% less than our “ideal” calculation.)
iii) Above 882˚C, titanium undergoes an allotropic transition to having a stable BCC
structure as opposed to the hexagonal α-titanium. Ignoring the possible effects of thermal
expansion, what should the volume of the unit cell, a,3 of BCC titanium be?
The [111] direction is the close packed direction in the BCC crystal system. If we assume
the atoms are touching, the body diagonal of the cube should be 4R, where R is the
atomic radius.
( √ )
√
2. Surfboards
Modern materials used in the fabrication of surfboards are polyurethane foam cores Encased in
fiberglass.
a) Up until the 1950s and the advent of fiberglass, surfboards were typically made of balsa
wood. In terms of the four interrelated components of materials science and engineering,
in what respect is fiberglass similar to wood, and what is/are the advantage(s) of
fiberglass over wood?
Fiberglass is similar to wood in terms of processing; they are both easy to shape. It
defers and is superior to balsa wood in structure, properties and performance; it is
stiffer, lighter and stronger, making boards more responsive, resilient and easy to
transport.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
b) Fiberglass has a more technical name, what is it? As this name implies, it is a composite
material. What two basic material types make up this composite, what are their respective
virtues/shortcomings, and how do they compliment each other?
Glass-fiber reinforced plastic, or GFRP. The two basic types are ceramics (glass) and
polymers (a plastic matrix). The glass fibers provide strength and stiffness both in
compression and tension along their long axis, but are prone to shear due to their large
aspect ratio. Embedding layers of these glass fibers within a plastic matrix complements
their strengths and compensates for their weaknesses.
c) Another modern material similar to fiberglass is carbon fiber, or carbon fiber-reinforced
polymers (CFRP). It would appear that it is superior to fiberglass for a number of
reasons, so why don’t we see more carbon fiber surfboards at campus point?
CFRP is indeed stiffer and stronger than GFRP, but considerably more expensive. In
terms of product engineering, cost is often the bottom line.
d) What are the different types of bonding exhibited by the two basic material types in
fiberglass? Of these two bonding types, which is primarily responsible for the strength of
fiberglass?
The plastic matrix is typically composed of nonmetallic elemental molecules, such as EP,
UP, etc. Hence, the predominant form of bonding is covalent. Since glass is a ceramic,
the primary bonding is ionic. The strength of fiberglass is a result of the glass fibers and
ionic bonding.
e) Featured below is an illustration of the polyurethane molecule present in the foam core
of most modern surfboards. What classification of material is the polyurethane foam
core? What sort of bonding does it predominantly exhibit? What purpose does it serve in
your surfboard?
Polyurethane is a polymer, as the name implies. The bonding is covalent, as evident by
the large number of bonds between nonmetallic and dissimilar atoms in the molecular
structure. It provides buoyancy and form to the surfboard.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
3. Graphene
Graphene is a genuine 2-Dimensional material consisting of a single sheet of graphite, and is
widely researched due to its array of extraordinary properties. The 2010 Nobel Prize in Physics
was awarded for the discovery of graphene.
a) Look up the lattice structure of graphene and provide a simple sketch. Give a definition
for the basis of a lattice, and sketch the basis for graphene’s lattice structure?
http://solidstate.physics.sunysb.edu/book/prob/img26.gif
b) Suppose you were to stack multiple sheets of graphene together to form a crystal, identify
i) the crystal structure of this material; ii) the bonding mechanism/force between different
sheets of graphene (hint: think about why graphite shatters so easily).
i). HCP
ii) Van der waals
c) Draw the crystal structure of graphite and diamond. Compare the structures with
graphene. How does the structure and bond length of graphite and diamond affect their
materials properties (aka strength, hardness etc)?
Graphite:
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
http://www.benbest.com/cryonics/lessons.html
Boiling Point ~ 4200 Celsius
Hardness ~ 7-11 Vickers or kg/mm2 or 1.5 Mohs
Strength (Compressive) ~ 20-200 MPa
Bond length ~ 1.421 A
Distance between layers ~ .335 A
Diamond Structure:
http://cst-www.nrl.navy.mil/lattice/struk/a4.html
Hardness ~ 10,000 Vickers or kg/mm2or 10 Mohs
Strength (Tensile) ~ 1.2 GPa
Strength (Compressive) ~ 110 GPa
Boiling Point ~ 4827 Celsius
Bond Length ~ 1.545 A
Graphene is a two dimensional material where diamond and graphite and three
dimensional. Graphene is one sheet of graphite.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Diamond has 4 bonds each sharing one electron. Graphite has 3 bonds all of which are
sharing 4 electrons and weak bonds between the layers. This is why graphite has a
smaller bond length than diamond. Diamond has a higher melting temperature, strength
and hardness than graphite because diamond has a tighter three dimensional structure
than graphite.
4. Bonding
Describe both which type(s) of bonding are present in the following materials and how that type
of bonding affects the properties of that material. Cite any additional resources used.
a) H2O in its liquid, solid and gaseous forms.
Liquid and solid: Polar covalent bonding between the O-H, but also secondary hydrogen
bonding to neighboring molecules. When water freezes, it has an open rigid structure due to
the 4 tetrahedral hydrogen bonds between each molecule and therefore less dense solid
phase of water. Liquid water has of average less hydrogen bonds per molecule.
Intermolecular hydrogen bonds are also responsible for the high boiling point of water
compared to other similar molecules (like H2S).
b) NH3 in liquid, solid and gaseous forms. How does the difference between NH3 and H2O
allow for life on earth?
Polar covalent bonding between the N-H, but also secondary hydrogen bonding to
neighboring molecules. Ammonia is similar to water in all forms but cannot form 4 hydrogen
bonds in its solid form which is why its solid phase does not form the tetrahedral structure
and is denser than its liquid phase (681 vs 817 kg/m3). If ice was not less dense life couldn’t
exist on this earth as we know it. Animals would not have been protected below the water in
the freeze thaw cycles etc.
c) Hg, at room temperature and below -39C
At RT, weak metallic Hg-Hg bonding, this allows it to be a liquid at room temperature.
Below -39C, solid Hg is formed and should show stronger metallic bonding.
c) CuO, at room temperature
~50% ionic, 50% covalent
electronegativity difference, Xcu=1.9, XO=3.5, (from Fig2.6)
(eq 2.10) %ionic character= {1 – exp[-0.25(XCu-XO)2]}*100 =47%
Properties: high melting temperature (1200C), insulating (or semiconducting) in its solid
form due to tightly bound, and thus localized electrons.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
5. Forces
a) When two atoms are brought to a small separation distance from an infinite separation,
what are the two types of forces which arise due to the interaction of the two atoms?
What condition between the two forces determines the equilibrium interatomic distance?
There is an attractive force, FA, and a repulsive force, FR, due to the interaction between the
two atoms. The attractive force arises due to phenomena specific to the type of bonding. The
repulsive force arises due to interaction of the positively charged atom core at small
separation distances where the valence shell of each atom begins to overlap. The
equilibrium interatomic distance, ro, is found at a separation resulting in a zero net force:
FN=FA+FR=0 at r=ro.
b)
The Lennard-Jones pair-potential function is used to approximate the Van der Waals
interactions in closed shell atoms such as Ar (Derivation presented in “Cohesion,”
Lennard-Jones JE (1931) Proc Phys Soc 43:461–482.) Plot the net energy between two
Ar atoms using the Lennard-Jones pair potential function given below. On the plot
indicate the equilibrium interatomic distance, ro, and the bonding energy, Eo. Solve for
the bonding energy and equilibrium distance analytically. How would the plot change for
a material with a greater melting temperature? When comparing the plot of net energy as
a function of interatomic spacing, list another physical property that can be determined
relative of one material to the other and describe how you might infer this information.
[( )
( ) ]
(Hint: plot for r from 2.5Å to 8Å)
Sketch:
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
The equilibrium separation can be solved by finding the location at the minimum
of the energy function. The magnitude of the bonding energy is found by
evaluating the energy function at the equilibrium spacing.
dV  r 
  6 2 12 
 24  7  13   0
dr r  r
ro 
 ro
o
13
o
7
o
r
r

2
12
6
ro 6  2 6
1
ro  2 6   3.822 Å
 6
 12 
1 1
Eo  V  r0   4  6  12   4       1.032  102 eV
4 
2 4
 2
A material with an increased melting temperature will have an increased bonding
energy, Eo, resulting in a deeper “well” at ro. When comparing two materials’
Energy vs. spacing plot we can also compare their linear coefficient of thermal
expansion which tends to be increased when a deeper well (increased bonding
energy) with increased curvature (decrease radius of curvature) is present.
c) For the case of ionic bonding: What is the relationship between the energy due to the
attractive force, EA, and the interatomic distance, r? What is the relationship between the
energy due to the repulsive force, ER, and the interatomic spacing? Which force drops
off quicker as the interatomic spacing is increased?
EA=-A/r
ER=B/rn where n is approximately 8
As the interatomic spacing increases, the repulsive energy drops off significantly quicker.
Since the exponent of r in the attractive energy is -1 and in the repulsive energy is it -8,
the energy due to the repulsive force will drop significantly faster than that of the
attractive force as the spacing is increased.
d) What is the percent ionic character of GaN?
From equation 2.10 and figure 2.7
% ionic character
(
(
)
)
(
)