Engineering Assignment
Biowizard901
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Strength and Stiffness
Stress is applied to a material by loading it
Strain – a change of shape – is its response
Stiffness is the resistance to change of shape that is elastic – the material will return to its original shape when unloaded
Strength is the resistance to permanent distortion or total failure
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Material Properties
Stress and strain are not material properties – they
describe a stimulus and a response
Stiffness and strength are material properties which are
measured by the elastic modulus (E), elastic
limit (σy), and tensile strength (σts)
Stiffness, strength, and density are three material
properties central to mechanical design
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Density
Figure 4.1
Mass per unit volume – kg/m3 or lb/in3
Double-weighing method for calculating density
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Modes of Loading
(a) – axial tension
(b) – compression
(c) – axial tension on one
side and compression
on the opposite side
(d) – torsion
(e) – bi-axial tension or
compression
Figure 4.2
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Stress
Figure 4.3
(a)
Force applied normal to surface
Positive F indicates tension
Negative F indicates compression
(b)
Force applied parallel to surface
Shaded plane carries the shear stress
(c)
Equally applied tensile and compressive forces on all six sides of a cubic element
Hydrostatic pressure
1 N/m2 = 1 Pascal (Pa)
106 Pa = 1 MPa
1 lb/in2 = 1 psi
103 psi = 1 ksi
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Class Exercise
A cylindrical brick chimney is 50 feet tall. The bricks have a density of 1800 kg/m3. What is the axial compressive stress at the base? Does the shape and size of the cross-section matter? Will the bricks at the base support the pressure?
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Class Exercise: Solution
Force = weight of chimney = mass x acceleration due to gravity = (volume x density) x acceleration due to gravity = Ahg
Stress = -Force /Area = -hg = (-15.24 x 1800 x 10 = 269,000 Pa = -0.269 MPa (‘-’ indicates compression)
The stress is independent of the shape or size of the cross-section
Will the brick at base hold? Yes – compressive strength of bricks is 10 – 70 MPa
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Strain
Figure 4.3
(a)
Tensile stress lengthens the element causing a tensile strain (+)
Compressive stress shortens the element causing a compressive strain (-)
Strain is the ratio of two lengths
and is therefore
dimensionless
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Hooke’s Law: within elastic regime, strain is proportional to stress:
= E
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Stress-Strain Curves
Figure 4.4
Initial portion of curve is approximately
linear and is elastic – the material
returns to its original shape once the
stress is removed
Within the linear elastic region, strain is
proportional to stress
E: Young’s modulus
G: shear modulus
K: bulk modulus
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Stress-Strain Curve – Brittle Response
Entire response is elastic –
no plastic deformation
Yield strength not reached
before failure
Young’s modulus determined
by calculating the slope of
this region
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Ductile Response
Permanent deformation occurs at stresses beyond the yield strength – material will not return to its original shape past this point
Tensile strength is maximum
stress on the curve
Yield strength determined by
standard offset methods
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Poisson’s Ratio
Relates the Young’s modulus, shear modulus, and
bulk modulus to one another
Negative of the ratio of transverse strain
to axial strain in tensile loading
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Class Exercise
In reference to chimney example, given that Young’s modulus of the bricks is 25 GPa and Poisson’s ratio is 0.2, what is the axial and transverse strain at the bottom of the chimney?
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Class Exercise - Solution
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-> Lateral strain is positive, which means that the bricks at the base of the chimney would show a small expansion.
Stress-Free Strain
Figure 4.5
In certain situations, strain is not caused by stress; however, stresses can develop if the body suffering the strain is constrained
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Material-Property Charts: Modulus - Density
Figure 4.6
Identifies materials that
are both stiff and light
Critical for material selection
of stiffness-limited designs
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Modulus – Relative Cost
Figure 4.7
Identifies materials that are both stiff and cheap
Useful when the objective is minimizing cost
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X-axis = Relative Cost Per Unit Volume Cv,R = (Price * Density for a material)/(Price*Density for Low Carbon Steel = $0.7/kg x 7800 kg/m3 = $5460/m3)
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Anisotropy
The properties of most materials – glasses, ceramics, polymers and metals – do not depend on the direction in which they are measured across the material
Certain materials are considered anisotropic – meaning their properties are dependant upon which direction in the material they are being measured
Woods are stiffer along the grain than with it;
fiber composites are stronger and stiffer parallel
to the direction of the fibers than perpendicular
to them
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What Determines Density
Density is mostly dependant on atomic weight
Metals are dense because their atoms are heavy –
iron has an atomic weight of 56
Polymers have low densities because they are made of light
atoms – carbon has an atomic weight of 12 while hydrogen
has an atomic weight of 1
The size of atoms and the way in which they are packed (atomic arrangement) also influence density, but to a much lesser degree
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Levels of Atomic Arrangement
No order
Short range order
Glass: Short range order
Metals: Long range order
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Classification of Materials Based on Order
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Long-Range Order
Atoms form a regular, repetitive, gridlike pattern, or lattice.
Lattice is a collection of points, called lattice points
Surroundings of each point are identical
Lattice differs from material to material
Shape, size depending on the size of the atoms and the type of bonding between them
Crystal structure refers to size, shape and atomic arrangement within the lattice
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Characterizing Packing
How tightly and efficiently atoms are packed together
Coordination number
The number of atoms touching a particular atom
Atomic packing factor (APF)
The fraction of the space occupied by the atoms
Unit Cell
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Atomic Packing
Most materials are crystalline – have a regularly
repeating pattern of structural units
Atoms often behave as if they are
hard and spherical
Layer A represents the close-packed
layer – there is no way to pack the atoms
more closely than this
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Close Packing
First layer is A
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Close-Packed Spheres
ABCABC sequence
Face Centered Cubic
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Figure 4.8
Atomic structures are close-packed in three dimension
Close-packed hexagonal: ABABAB stacking sequence
Face-centered cubic: ABCABC stacking sequence
Packing fraction for CPH and FCC structures is 0.74 – meaning
spheres occupy 74% of all available space
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Non Close-Packed Structures
Body-centered cubic:
ABABAB packing sequence
Packing fraction = 0.68
Amorphous structure:
Packing fraction ≤ 0.64
Figure 4.9
Figure 4.10
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Unit Cell
Figure 4.11
Red lines define the cell edges while spheres represent
individual atoms
Shaded regions represent close or closest packed plane
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Smallest unit demonstrating the symmetry of arrangement of atoms in 3D – when unit cells are repeated in 3D one obtains a crystal of a material
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The Unit Cell
the smallest repeating unit in a crystal
repeats in x, y or z directions
14 known geometries
3 are cubic
Cubic: a = b = c; = = = 90o
Hexagonal: a = b c; = = 90o, = 120o
Tetragonal: a = b c; = = = 90o
Rhombohedral (trigonal): a = b = c; = = 90o
Orthorhombic: a b c; = = = 90o
Monoclinic: a b c; = = 90o
Triclinic: a b c; 90o
Crystal Lattice
Figure 4.12
Lattice points are the points at which cell edges meet
(a): hexagonal cell
(b): cubic cell
(c): cell with different length edges
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Metallic Crystals
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• tend to be densely packed.
• have several reasons for dense packing:
-Typically, only one element is present, so all atomic
radii are the same.
-Metallic bonding is not directional.
-Nearest neighbor distances tend to be small in
order to lower bond energy.
• have the simplest crystal structures.
We will look at three such structures...
METALLIC CRYSTALS
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The Unit Cell
the smallest repeating unit in a crystal
repeats in x, y or z directions
14 known geometries
3 are cubic
Cubic: a = b = c; = = = 90o
Hexagonal: a = b c; = = 90o, = 120o
Tetragonal: a = b c; = = = 90o
Rhombohedral (trigonal): a = b = c; = = 90o
Orthorhombic: a b c; = = = 90o
Monoclinic: a b c; = = 90o
Triclinic: a b c; 90o
Examples of Crystals
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Examples of Crystals
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Examples of Crystals
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Examples of Crystals
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Examples of Crystals
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Examples of Crystals
Lattice Parameters –> numbers needed to define the size and shape of the lattice
1 parameter for cubic
2 parameters for hexagonal
3 parameters for orthorhombic
Need correlation between atomic radius and lattice parameter
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Class Exercise: Find relationship between atomic radius r and lattice parameter a0 for SC, BCC, and FCC
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Solution
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APF: Simple Cubic
r
unit cell volume
space occupied by atoms
atomic packing factor
a0
APF: Class Exercise
Show that APF for BCC is 0.68
Show that APF for FCC is 0.74
Try: show that APF for HCP is 0.74
Coordination Numbers
Simple Cubic: 6
Body Centered Cubic: 8
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Show that co-ordination number for FCC and HCP is 12
Sharing of Atoms
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Simple Cubic
Unit Cell
Lattice
Points
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8 x 1/8 = 1 atom/unit cell
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Number of Atoms Per Unit Cell for Simple Cubic (SC) Lattice
1+ 8 x 1/8 = 2 atoms/unit cell
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Number of Atoms Per Unit Cell for Body Centered Cubic (BCC) Lattice
Number of Atoms Per Unit Cell for Face Centered Cubic (FCC) Lattice
6×1/2 + 8×1/8 = 4 atoms/unit cell
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Close Packing
First layer is A
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Close-Packed Spheres
ABCABC sequence
Face Centered Cubic
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Hexagonal Close-Packed (HCP)
Top
View
c0
a0
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HCP
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• Coordination # = 12
• ABAB... Stacking Sequence
• APF = 0.74
• 3D Projection
• 2D Projection
Adapted from Fig. 3.3,
Callister 6e.
HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)
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HCP vs. BCC
hcp
bcc
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Common Metallic Crystals
Structure | Lattice Parameters | Atoms/cell | Coor. No. | APF | Typical Metals |
SC | a0 = 2r | 1 | 6 | 0.52 | |
BCC | a0 = 4r/√3 | 2 | 8 | 0.68 | Fe, Ti, W, Mo, Nb, Ta, K, Na, V, Cr, Zr |
FCC | a0 = 4r/√2 | 4 | 12 | 0.74 | Fe, Cu, Al, Au, Ag, Pb, Ni, Pt |
HCP | a0 = 2r c0=1.633a0 | 6 | 12 | 0.74 | Ti, Mg, Zn, Be, Co, Zr, Cd |
Allotropy: (Greek -> allos: other + tropos: manner) two or more distinct physical forms in the same physical state; Carbon can be diamond, graphite, carbon nanotube, fullerene, hexagonal diamond, carbyne, amorphous carbon, graphene
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Example: Copper
Data from Table inside front cover of Callister (see next 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)
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THEORETICAL DENSITY, r
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Class Exercise
Determine the density of BCC Iron which has a radius of 0.1241 nm. Atomic weight = 55.8 g/mol
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Class Exercise
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Class Exercise (contd.)
The measured density of BCC Iron is 7.870 g/cm3.
The slight difference is a result of the defects in the lattice
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Adapted from
Table, "Charac-
teristics of
Selected
Elements",
inside front
cover,
Callister 6e.
Characteristics of Selected Elements at 20 oC
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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 CLASSES
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H2O
>
>
Ceramic Crystals
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Atomic Packing in Ceramics
Figure 4.13
(a): Hexagonal unit cell with a W-C atom pair associated
with each lattice point
(b): Cubic unit cell with a Si-C atom pair associated with each
lattice point
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• Bonding:
--Mostly ionic and some covalent.
--% ionic character increases with difference in
electronegativity.
Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by
Cornell University.
• Large vs small ionic bond character:
BONDING IN CERAMICS
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Ceramic Crystal Structures: Intro
Ceramics are composed of at least two elements, so the lattice sites are occupied by ions, not atoms as in metals.
Metallic elements (Na, Ca, Ti, V, W, Fe) – tend to lose electrons (i.e. they are electropositive), become smaller in size, are positively charged, get attracted to cathode, hence called cations.
Nonmetallic elements (C, N, O) – tend to accept electrons (i.e. they are electronegative), become larger in size, are negatively charged, get attracted to anodes, hence called anions.
Chapter 12; pp. 414 – 434, Callister 7th ed
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Atomic Packing in Glasses
Amorphous silica (SiO2) is the basis of most glasses. Rapid cooling after melting allows material to maintain amorphous structure while slow cooling leads to crystalline structure.
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Regular hexagonal arrangement of atoms in crystalline silica
Irregular arrangement of atoms in amorphous or “glassy” silica
5 sides
6 sides
6 sides
7 sides
Addition of Na2O gives soda glass (windows and bottles) while addition of B2O5 gives borosilicate glass (pyrex)
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Carbon: Diamond
Carbon bonds with 4 other carbon atoms; 100% covalent bonding; ZnS type; corner, face and all four tetrahedral positions occupied by carbon
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Carbon: Graphite
Three covalently bonded, co-planar carbon atoms hexagonally arranged, fourth bonding electron forms weak Van der Waals type bond between the layers
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Carbon: Fullerene
Hollow spherical cluster of 60 carbon atoms to form a discrete C60 molecule, 20 hexagons and 5 pentagons. Such molecules (buckyballs) form crystalline solids when arranged in FCC structure.
Pure fullerene is electrically insulating, when impurities are added, it can become highly conductive or semiconductive.
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Carbon: Nanotube
Single sheet of graphite rolled into a tube with tube ends capped by C60 fullerene hemispheres. Tube diameter is of the order nanometers (10-9 m) 100 nm or less.
Each tube consists of millions of atoms and high l/d ratio. Nanotubes are extremely strong, stiff and ductile, in fact it is the strongest known material (Tensile Strength: 50,000 – 200,000 MPa), elastic modulus 1000 GPa, low density, the ultimate fiber!
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Carbon: Graphene
One atom thick planar sheets of carbon atoms forming honeycomb (hexagonal) lattice – it can be wrapped into fullerenes, rolled into nanotubes and stacked into graphite!
Potential applications: extremely strong sheets (200x strength of steel), electrically conducting, transparent: applications in electrical and electronics industry e.g. semiconductors, organic light emitting diodes, touch screens, bio-devices (DNA sequencing, antibacterial packaging), solar cells
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Elements of Crystallography
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Crystallographic Directions
Vector passes through the origin
The length of vector projection is determined in terms of a, b, c
Reduced to the smallest integer
Reported in brackets omitting commas (e.g. [101]
Negative directions indicated by a bar over the index [001]
Family of directions in angle brackets: <100> = [100], [010], [001]
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Crystallographic Directions
[010]
[001]
[111]
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Crystallographic Directions
[010]
[001]
[111]
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Green: [101]
Yellow: [210]
Light Green: [221]
Purple: [211] (direction coming out through the center of the front face)
Brick Red: [043]
Line Density (LD)
LD: number of atoms per unit length whose centers lie on the direction vector for a given crystallographic direction e.g. take FCC [110] direction:
[110]
[110]
LD[110] = #atoms centered on vector/length of vector = 2/4r = 1/2r
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Crystallographic Planes
Plane does not pass through the origin
Determine the intercept to the three axes
Take reciprocals of the intercepts
Find the smallest integers
Enclose them in parentheses (e.g. (111))
Family of planes in curly brackets: {111} = (111), (111), (111), (111)
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Crystallographic Planes
Intercepts
3 2
Reciprocals
1/2 1/3 1/2
Plane(multiply by 6)
(323)
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X
Y
Z
Crystallographic Planes
(100)
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MATTER Project
Try Miller indices of a plane at:
http://www.doitpoms.ac.uk/tlplib/miller_indices/lattice_index.php
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Plane Density (PD)
PD: number of atoms per unit area that are centered on a particular crystallographic plane e.g. take FCC (110) plane:
(110)
(110)
a
Why study LD and PD?
Most materials are anisotropic, i.e. their properties differ in different crystallographic directions – mechanical properties, magnetic, electrical, thermal, crystal growth and so on
e.g. study data given in the table below
Materials may be processed such that majority of the crystals are oriented in a particular crystallographic direction – such materials are called textured materials
Modulus of Elasticity (GPa) | |||
Metal | [100] | [110] | [111] |
Aluminum | 63.7 | 72.6 | 76.1 |
Copper | 66.7 | 130.3 | 191.1 |
Iron | 125.0 | 210.5 | 272.7 |
Tungsten | 384.6 | 384.6 | 384.6 |
Most Densely Packed Lines and Directions in BCC and FCC
BCC: {110} planes , <111> directions
FCC: {111} planes, <110> directions
3D Microstructure
ND
RD
TD
Copper component: Plane – direction: {112}<111>
Rollett, 3cub13; E&M; T=0.5; Emin=0.55
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Microstructure Evolution
Times: 100, 1000,10,000 105, 106, 107
Initial cube fraction = 0.06; Max. cube fraction = 0.48
3cub19
4° spread in cube grain orientations
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Single Crystal vs. Polycrystals
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• Demonstrates "polymorphism"
The same atoms can have more than one crystal structure.
HEATING AND COOLING OF AN IRON WIRE
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Isotopes, Allotropes and Polymorphs
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Isotope: atoms have different number of neutrons e.g. 126C, 146C; (this is of course quite different from isotropy which means having uniform properties in all directions)
Allotropes: different forms/shapes of crystal structure an element e.g. C: diamond, graphene, fullerene, graphite;
Polymorphs: general term meaning exhibiting different crystalline shapes, especially when applied to compounds e.g. SiO2 - -quartz, -quartz, citrine, rose quartz, milky quartz, amethyst,…
Question: water, ice and steam are examples of which of these terms?
-quartz
Citrine
Rose Quartz
Milky Quartz
Amethyst
• 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 blades
Fig. 8.30(c), Callister 6e.
(Fig. 8.30(c) courtesy
of Pratt and Whitney).
(Courtesy Martin Deakins,
GE Superabrasives, Worthington, OH. Used with permission.)
CRYSTALS AS BUILDING BLOCKS
Applications of Single Crystals:
Silicon: Semiconductors, Microprocessors
Sapphire: Laser Generation
Fluorite (CaF2): Refracting Telescopes
Ni-based Superalloys: Turbine Blades
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In jet engine use, single-crystal turbine airfoils have proven to have as much as nine times more relative life in terms of creep strength and thermal fatigue resistance and over three times more relative life for corrosion resistance, when compared to equiaxed crystal counterparts. Modern high turbine inlet temperature jet engines with long life (that is, 25,000 hours of operation between overhauls) would not be possible without the use of single-crystal turbine airfoils. By eliminating grain boundaries, single-crystal airfoils have longer thermal and fatigue life, are more corrosion resistant, can be cast with thinner walls—meaning less material and less weight—and have a higher melting point temperature. These improvements all contribute to higher efficiencies.
The beige single crystal blades on this GE 9H turbine — the world's largest — are approximately 18 inches long and weigh more than 30 pounds apiece. Material: Ni-based superalloy. Centrifugal forces of order 20,000g, operating temperature 3000 oF, service life between overhaul 25,000 hours. Applications: Boeing 767, Airbus A310
TURBINE BLADE SINGLE CRYSTALS
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• Most engineering materials are polycrystals.
• Nb-Hf-W plate with an electron beam weld.
• Each "grain" is a single crystal.
• If crystals are randomly oriented,
overall component properties are not directional.
• Crystal sizes typically range from 1 nm to 2 cm
(i.e., from a few to millions of atomic layers).
Adapted from Fig. K, color inset pages of Callister 6e.
(Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang Albany)
1 mm
POLYCRYSTALS
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• 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.
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
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• Incoming X-rays diffract (= reflect/bend around objects/obstacles and interfere with each other) from crystal planes to form patterns. Examples of diffraction of visible light: halo around the moon, colors of CD, colors on oil drop spread on water, colors on spider web)
• Measurement of:
Critical angles, qc,
for X-rays provide
atomic spacing, d.
Adapted from Fig. 3.2W, Callister 6e.
Using X-RAYS To Confirm Crystal Structure
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Exercise
Compute interplanar spacing for the (111) set of planes in -iron, atomic radius of -iron 0.1241 nm
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Solution
-iron is BCC structure, so the lattice parameter is:
a = 4r/3 = 0.2866 nm;
Exercise
The metal rhodium has an FCC crystal structure. If the angle of diffraction (2) for (311) set of planes is 36.12o for first order reflection when monochromatic X-ray radiation of wavelength 0.0711 nm is used, compute (a) the interplanar spacing for this set of planes and (b) the atomic radius of rhodium atom
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Solution
Polymer Structures
102
Polymer Structures: Intro
Natural polymers: wood, rubber, cotton, silk, wool, leather.
Polymers: consists of ‘mers’ (parts) of hydrocarbon compounds
Radical: a group of atoms that retains its identity during a chemical reaction e.g. CH3 (methyl), C2H5 (ethyl), C6H5 (phenyl)
Carbon: four valence electrons, saturated => single covalent bond (CH4 methane), unsaturated => double (C2H4 ethylene) or triple covalent (C2H2 acetylene) bonds;
Hydrogen: one valence electron
Paraffin Compounds (not polymers) : CnH2n+2
Common Hydrocarbon Groups
R and R’ represent organic radicals: radicals are groups of atoms that remain as a single unit and maintain their identity during chemical reactions. Example of singly bonded hydrocarbon radicals: methyl (CH3), ethyl (C2H5) and phenyl (C6H5) groups
CH3OH
C2H6O
CH3COOH
CH2O
C6H5OH