useful materials notes

Useful Materials Notes

Chapter 1:

Types of Materials: Metals

o High toughnesso Conductiveo Crystallineo Metallic bond

Ceramics & Glasseso Chemically stableo High temperature resistanceo Brittleo Low toughnesso Hardo Crystalline or amorphouso Ionic/covalent bond

Polymerso Synthetico Lightweighto Elastico Low melting pointo Covalent and secondary

Compositeso Combinations of metals, ceramics & glasses, and polymerso Covalent or covalent/ionic

Semiconductorso Chemical purity determines electronic properties

Chapter 2:

Types of Bonding: Primary (strong, high melting points)

o Ionic Non-directional S-shells

o Covalent Directional P-shells and d-shells Low CN

o Metallic Non-directional S-shells High CN Closely packed Electron gas

Secondary (weak, low melting points)o Hydrogen

Attraction of permanent dipoleso Van der Waals

Attraction of temporary dipoles

Coulombic Attraction

FC=k q1q2r2

Repulsive Force

FR=λe−rρ

Where λ and ρ are experimentally determined constants

Net ForceF=FC+F R

Coordination Number and Radius Ratio

Chapter 3:

Crystalline/Amorphous: Crystalline = regular repeating pattern Amorphous = irregular

Unit Cell Smallest repeating structure

Crystalline Structures 7 crystal systems (axial systems) 14 crystal lattices (Bravais lattices) Mainly SC (Simple Cubic), BCC (Body-Centred Cubic), and FCC (Face-Centred Cubic)

SC BCC FCC HPCAtoms/Unit Cell 1 2 4 2

APF 0.52 0.68 0.74 0.74CN 6 8 12 12a 2R 4r/sqrt(3) 2sqrt(2)rd 2Sqrt(3)r 4r 2sqrt(6)r

Pattern AAAAAA AAAAAA ABCABC ABABAB

Density

ρ= nAV C N A

ρ is density (g/cm3)

n is atoms/unit cell (atoms)

A is atomic weight (g/mol)

VC is volume/unit cell (cm3)

NA is 6.02x1023 (atoms/mol)

Ionic Packing Factor APF but when there are two different ions

Indices (bar on top for negative) Lattice Points: [111] Family of Directions <111>

o [111], [1(bar), 1,1], [1,1(bar), 1],… Lattice Planes: (111)

o Miller indices are the reciprocals of the intercepts Miller-Bravais Indices: (hkil)

o i = -(h+k)o used for hexagonal systems

Family of Planes {1,1,1}

a ∙b=¿a∨¿b∨cosθ

Where theta is the angle between the vectors

Linear Density 1/r where r is the repeat distance between adjacent atoms

Bragg’s Lawn λ=2dsinθ

n is the diffraction order

λ is the wavelength (nm)

d is the spacing between adjacent crystal planes (nm)

2ϴ is the diffraction angle

For a cubic system:

d= a

√h2+k 2+l2

Diffraction Occurs Diffraction Does not OccurBCC h+k+l is even h+k+l is oddFCC h,k,l are all even or all odd h,k,l are a mix of even and oddHPC (h+2k) = 3n (where n is an

integer), and l is oddThe rest of the time

Chapter 4:

Types of Materials: 0-D: point defects

o Impurities Substitutionals

Different atom in a crystal Interstitials

Small atom that fits into the crystalo Vacancies

Hole in the crystal 1-D: linear defects

o Edge dislocations Extra half plane of atoms

o Screw dislocations Spiral stacking of crystal planes

o Mixed dislocations Edge dislocation on one edge, screw on another

2-D: planar defectso Grain boundaries

Atoms on boundaries (surface) are not in equilibrium Tilt boundary

Low angle Caused by edge dislocations

Twist boundary Low angle Caused by screw dislocations

Twin Boundary Two crystal structures are mirror images

Stacking Faults: Annealing Twins Fault in stacking of planes ie. ABCABCAB ABCABC FCC

Mechanical Twins Part of the crystal structure bends and then bends back

Deformation Elastic

o Reversible

o Bonds are stretched but not broken Plastic

o Permanento Caused by dislocations moving through the substance

Burger’s Vector (b) Vector required to close the loop of an edge dislocation Perpendicular to the edge dislocation Parallel to screw dislocation Consistent with each dislocation (edge and screw) for a mixed dislocation

Hume-Rothery rules for complete miscibilityIn order for complete solid solubility to occur between 2 solid metals:

1. Similar atomic radii (~15%)2. Same crystal structure3. Similar electronegativites (~10%)4. Same valence

Chapter 6 (1):

Engineering Stress

σ= PA0

σ is engineering stress (Pa)

P is load (N)

A0 is initial cross-sectional area (m2)

Engineering Strain

ε=l−l0l0

=∆ ll0

ε is engineering strain (unit-less)

l0 is initial gauge length (m)

l is final gauge length (m)

Relationship between stress and strain

ε= σE

Stress-Strain Graph Yield Strength

o Point between elastic and plastic phaseo At the top of the linear part of the stress-strain graph

Ultimate Tensile Strengtho Top of the stress –strain curve

E, Young Moduluso Slope of the plastic phase

Fracture Stresso Stress at which it breaks

Ductilityo the strain that corresponds to the intersection of a line with E as its slope traced back

from the fracture pointo % elongation or % decrease in area

Toughnesso Integral of the whole curve

Modulus of Resilienceo Integral of the elastic phase (energy it can take before being deformed)

Specific Strength Strength/Density

Strain hardening (a.k.a. Work Hardening) Second bump in stress-strain curve Pulled to that bump on graph then let return so that it will be stronger

Necking Middle of test specimen starts to become thinner and longer

True Stress

σ T=PA

σT is true stress (Pa)

P is load (N)

A is cross-sectional area (m2)

True Strain

ε T= ln (A0A

)

εT is true strain (unit-less)

A0 is initial cross-sectional area (m2)

A is cross-sectional area (m2)

Poisson’s Ratio Contraction perpendicular to the extension under tensile stress

v=−ε x

ε z

εx is negative strain (strain from change in diameter)

εz is positive strain (from young’s modulus)

Shear Stress

τ=PS

AS

PS is the load on the sample

AS is the area of the sample

Shear StrainShear stress produces an angular displacement α, with the shear strain γ

γ=tanα=∆ yz0

≈a a→0

Shear Modulus

G= τγ

Relationship between moduli E and GE=2G(1+v )

Three Point Bend Test (modulus of rupture)

MOR= 3 FL2bh2

F is fracture force (N)

L is the distance between the two bottom supports (m)

b is the width of the beam (m)

h is the thickness of the beam (m)

Three Point Bend Test (flexural modulus)

EFlex=L3m4bh3

m is slope of the tangent of the linear part of the load –deflection curve (which is F/γ, where γ is the vertical deflection from the original horizontal position) (N/m)

L is the distance between the two bottom supports (m)

b is the width of the beam (m)

h is the thickness of the beam (m)

Griffith Crack Model

σ=12σm√ ρ

c

σ is the stress applied (Pa)

σm is the maximum stress of the material (Pa)

ρ is the radius of the tip (m)

c is the length of the crack (m)

Chapter 6 (2):

Plastic Deformation: Bonds are broken and reformed Does not return to its original position Mechanisms

o Crystalline structure – Slip mechanism Atoms slide over each Dislocations increase slipping In perfect crystal high stress because all the atoms slide at once, in imperfect

crystal low stress because one atom at a time slideso Non-crystalline structure

Viscous flow mechanism

Critical Shear Stress G (Shear Modulus) is 10x larger than τC (critical resolved shear stress)

Resolved Shear Stress

τ=σ cos λcosϕ= FAcos λcos ϕ

τ is the resolved shear stress (Pa)

σ is the applied stress (Pa)

λ is the angle between the tensile axis and the slip direction (°)

ϕ is the angle between the normal of the slip plane and the tensile axis (°)

F is the tensile force (N)

A is the cross-sectional area of the wire normally (not the shear plane)

Schmid’s Law

τC=σC cos λ cosϕ=FAcos λcos ϕ

τC is the critical resolved shear stress (Pa) (when the plastic stage starts)

σC is the critical applied stress (Pa)

λ is the angle between the tensile axis and the slip direction (°)

ϕ is the angle between the normal of the slip plane and the tensile axis (°)

F is the tensile force (N)

A is the cross-sectional area of the wire normally (not the shear plane)

Methods to Increase Yield Strength Use Defects

o Dislocation movement starts in the plane with the highest resolved shear stresso Hinder dislocation movement = increased yield strengtho Dislocations have to change directions to move beyond defects

Smaller Grainso Grain boundaries impede dislocation movement

o σ y=σ 0+K √ 1d

σy is yield strength (Pa) σ0 is a constant for the metal (Pa) K is a constant of the metal d is average grain diameter

Cold workingo Causes work hardening

Solution Hardeningo Impurity atomso Interstitialso Larger Substitutionals

Dispersion Strengthening (Second Phase Strengthening)

Hardness Resistance to indentation Not a fundamental property (depends on many factors)

Brinell Hardness (ball)

BHN= 2P

πD(D−√(D 2−d2 ))

P is load (in kgf “kilogram-force”) ( 1kgf = 9.81N)

D is the maximum diameter of the ball (mm)

d is the diameter of indentation

Creep Plastic Deformation at high temperatures under constant stress Metals ceramics polymers

o Can be caused by vacancy climb

o ε̇=∆ε∆ t ε (with a dot) is creep ε is strain t is time

Exhibits Arrhenius behaviour (faster when hotter)

o dεdt

=Ce−QRT

s is strain (m) C is a constant Q is the activation energy (J/mol) R is the universal gas constant (8.314 J/mol) T is temperature (K)

Stress Relaxation Decreasing stress over time for constant strain polymers

Grain Boundary Sliding During Creep creates voids at an inclusion trapped at the grain boundary creates a void at the triple point of where three grains meet

Lowering Creep Rates1. Keep temperatures low2. Keep loads low (i.e. low σ)3. Use high melting point alloys

4. Add thermally stable particles that impede dislocation mobility5. Minimize grain boundary sliding by using columnar grains or single crystals

Larson-Miller Parameter

L=T (A+Blnt)1000

L is the Larson-Miller Parameter

T is temperature (K)

A and B are experimental constants

T is time (hours)

Chapter 7, 9 and 10:

Relative Thermal Expansion∆ ll0

=α∗∆T

l is length (m)

α is the coefficient of thermal expansion (1/K)


Thermal Conductivity (Fourier’s Law)

k=−( dQ

dt)

A (dTdx

)

K is thermal conductivity (W/K)

Q is the heat (J)


A is area (m^2)

x is distance (m)

Gibbs Phase RuleWhen pressure and temperature are variable:

F=C−P+2

When only pressure or temperature is variable:

F=C−P+1

F is degrees of freedom

C is number of components

P is the number of phases

Phase Diagrams

Eutectic - points down - liquid to solid

Peritectic - points up - liquid to solid

Eutectoid - points down - solid to solid

Peritectoid - points up - solid to solid

Tie Line and Lever Rule

Draw a line at a given temperature

The percent composition of each component is the fraction of the distance away over the whole length of the line

Cold Working

%CW=A0−A f

A0∗100%

A0 is the area before plastic deformation (m2)

Af is the area after plastic deformation (m2)

Chapter 8:

Fracture ToughnessK=Yσ √ πa

K is stress intensity factor (MPa sqrt(m))

Y is geometric factor

σ is stress (MPa)

a is depth of crack(m)

****internal then sub in half of the length given

KC is the critical stress intensity factor

Fracture Toughness of Brittle MaterialsFor a Vickers hardness indentation

K IC=α 0√ EH

∗P

d32


α0 is geometric factor (0.016???)

E is elastic modulus (N/m2)

H is Vickers hardness (N/m2)

P is load (N)

d is distance from tip of crack to centre

Paris’ LawdadN

=A(∆ K )m


N is number of load cycles

A is a constant

m is a constant

a is crack length

Weibull

f=1−e−( σ

σ 0)m

f is probability of failure

σ is stress applied

σ0 is reference stress

m is a constant

Chapter 13 and 15:

Resistivity

ρ=RAl

ρ is resistivity (Ωm)

R is resistance (Ω)

A is area (m2)

l is length (m)

Conductivity

σ=1ρ

σ is conductivity (1/ Ωm)

ρ is resistivity (Ωm)

R is resistance (Ω)

Conductivity and Resistivity Depend on:

1. Defects2. Temperature3. Atomic Arrangement (lattice type, degree of crystallinity)4. Electron structure and atomic bonding

Current Density (Flux)

J= IA

=Eσ=|e|nv

J is current density (A/m2)

I is current (A)

A is area (m2)

E is electric field strength (V/m)


e is the charge of an electron (C)

n is the number of electrons per unit volume (1/m3)

v is velocity (average) (m/s)

Temperature Dependence of Resistivityρ (∆T )=ρ ref (1+αref ∆T )

ρ resistivity (Ωm)

T is temperature (K or °C)

αref is the temperature coefficient of resistivity (TCR)

Effect of an Impurity on Resistivityρ=ρ0(1+βx )

ρ0 resistivity of pure metal (Ωm)

B is a material constant

x is the amount of impurity addition

Effect of Temperature on Carriers

σ=σ 0e−Eg

2kT


Eg is bandgap energy (J)

K is boltzmann’s constant (1.38 x 10-23 J/K)


Chapter 14:

Magnetism

1. Diamagnetica. μ slightly < 1

2. Paramagnetic a. Ferromagnetic “true” Ferromagnetic

i. μ significantly > 1b. Ferrimagnetic

i. μ slightly > 1c. Antiferromagnetic

Magnetic Field

B⃗=μ H⃗

B is induction (T)

μ is the magnetic permeability of the material

H is magnetic field (T)

M⃗=(μr−1) H⃗

M is magnetization (T) – Volume of magnetic dipoles

μr is the relative permeability (μr/ μ) where μ = 4π x 107 (H/m)

μr - 1 is the magnetic susceptibility

H is magnetic field (T)

General

ASTM Grain Size (G)n=2G−1

n is the number of grains at 100x per sq. in

G is average grain size

Count the number of full grains + half of the number of partial grains in the image to get n

Conversion between other magnifications

n100=(mag100

)2

∗nmag

Hall-Petch Equation

σ y=σ 0+k y

√d

σy is yield stress

σ0 is a material constant

ky is a material constant

d is the ASTM average grain diameter

useful materials notes

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