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Study Guide

by: Tristan Collette

Study Guide MTGN 202

Tristan Collette
Colorado School of Mines
GPA 3.0

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Chapters 1-5
Engineered Materials
Steven Thompson
Study Guide
materials, Bonding, Diffusion, steady, state, Vacancies
50 ?




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This 13 page Study Guide was uploaded by Tristan Collette on Tuesday September 27, 2016. The Study Guide belongs to MTGN 202 at Colorado School of Mines taught by Steven Thompson in Fall 2016. Since its upload, it has received 10 views. For similar materials see Engineered Materials in Metallurgical Engineering at Colorado School of Mines.


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Date Created: 09/27/16
Tristan Collette Sept. 24, 2016 Engineering Materials Exam 1: Study Guide 1. Classification of Materials a. Metals i. Good conductors of both electricity and heat due to their free electrons b. Polymers i. Plastic and rubber materials that have low densities that are ductile and  pliable c. Ceramics i. Stiff and strong materials between metallic and nonmetallic elements d. Semi­Conductors i. Electrical properties that are intermediate between those of electrical  conductors and insulators 2. Electron Configuration a. Atomic Number = # of protons b. Number of neutrons = Atomic mass ­ # of protons c. Number of electrons = # of protons unless it is a positive or negative isotope d. Atomic mass unit (amu) = 1/12 of atomic weight (based upon carbon 12) 3. Bohr Atomic Model and Energy Level Diagram (Hydrogen like atom) a. Electrons orbiting positively charged proton i. Charge of electrons and protons: 1.602x10^­19 J = 1 eV Tristan Collette Sept. 24, 2016 Engineering Materials ii. Mass of Proton: 1.67x10^­27 iii. Mass of Electron: 9.11x10^­31 iv. Electrons are permitted to only have specific values of energy; they must  take a quantum jump to achieve a higher energy state v. Adjacent energies are separated by finite energies b. Shells/Orbitals Value of L (quantum #) M_l Subshell # of # of Electrons n/shell (n­1) (­L…+L) Orbitals 1/K 0 0 1s 1 2 2/L 0 0 2s 1 2 1 ­1,0,1 2p 3 6 3/M 0 0 3s 1 2 1 ­1,0,1 3p 3 6 2 ­2,­1,0,1,2 3d 5 10 4/N 0 0 4s 1 2 1 ­1,0,1 4p 3 6 2 ­2,­1,0,1,2 4d 5 10 3 ­3,­2,­1,0,1,2,3 4f 7 14 *M_s (spin): +1/2,­1/2 to accommodate for spin up and spin down Tristan Collette Sept. 24, 2016 Engineering Materials *Orbital Shapes c. Electron energy diagram  4. Electronegativity and Bonding Types a. Van Arkel­Ketelaar triangle/bonding triangle Tristan Collette Sept. 24, 2016 Engineering Materials *Ionic Character: %IC= {1­exp[­(0.25)(Xa­Xb)^2]}*100 (Xa and Xb are electronegativites)  b. Ionic i. Occurs between metallic and nonmetallic elements  ii. Donation and receiving of electrons between elements iii. Always stable; non­directional (negative charge surrounding positive) c. Covalent i. Occurs within elements that have small differences in electronegativity ii. Sharing of electrons iii. Directional (concentrated ions on one side of an atom) d. Metallic i. Metals and their alloys (first two groups of periodic table) ii. Sea of electrons shield the positively charged ion cores iii. Non­directional character iv. Small difference in electronegativity and low average electronegativity e. Secondary Bonding i. Physical ii. Weak iii. Vander Waals iv. Dipoles 5. Bond Energies and Forces Tristan Collette Sept. 24, 2016 Engineering Materials a. r_o = equilibrium spacing b.  Bonding energy (∆E) present dE/dr = 0 c. Force = dE/dr  d. Attraction: A= (1/(4πε))(q1*e)(q2*e) i. q = charge on ion; e = 1.602(10^19) C; ε = 8.85(10^­12) ii. F_a = A/r^2 iii. E_a = ­A/r e. Repulsion:  Tristan Collette Sept. 24, 2016 Engineering Materials i. E_r = B/r^n 1. n~8 ii. F_r = (D/ρ)*e^(r/ ρ) 6. Crystal Structures: based upon Hard Sphere Model a. Equations:  i. Density = (n*A)/(V*N)   n= # atoms per unit cell; A= atomic mass; V=  volume per unit cell; N= Avogadro’s Number ii. Density = mass/volume iii. # of atoms per unit cell for FCC and BCC: N=N_i+N_f/2+N_c/8 N_i = # interior atoms; N_f = # face atoms; N_c = # corner atoms iv. Atomic Packing Factor: APF= (vol. atoms per unit cell)/(total unit cell  vol.) b. Face Centered Cubic i. a = 2R*sqrt(2) ii. N = 0+6/2+8/8 = 4 whole atoms per unit cell iii. APF = 0.74 c. Body Centered Cubic Tristan Collette Sept. 24, 2016 Engineering Materials i. a = (4R)/sqrt(3) ii. N = 1+0+8/8 = 2 whole atoms per unit cell iii. APF = 0.68 d. Hexagonal Close Packing i. N = 3+2/2+12/6 = 6 whole atoms per unit cell ii. APF = 0.74 e. Indices i. Miller 1. Planes; 0 to 1 units or infinite units (parallel to axis) away from  designated axis 2. Take reciprocal to turn into whole number Tristan Collette Sept. 24, 2016 Engineering Materials 3. If negative, a bar may be placed over the number instead of the  negative sign in front ii. Point coordinates (measurements in x, y, z direction) 7. Vacancy a. Definition: A point vacancy that increases the entropy of a crystal b. Equilibrium # of vacancies: N_v = N exp(­Q_v/k*T)  N is the total # of atomic  sites; Q_v is the energy required for the formation of a vacancy; k is Boltzmann’s  constant (1.38*10^­23 J/atom*K or 8.62*10^­5 eV/atom*K); T is absolute  temperature in Kelvin c. Importance:  i. Number of vacancies increase exponentially with an increase of  temperature ii. Vacancies generally move towards the surface or top layer d. Fraction of vacant lattice sites = N_v/N e. How big of an atom will fit in a vacant site? f. What is the difference between the “vacancy mechanism for diffusion” as  compared with the “interstitial mechanism for diffusion”? 8. Dislocations a. Types i. Burger’s Vector: describes distance/displacement on atomic level ii. Tangent vector is parallel to Burgers vector Tristan Collette Sept. 24, 2016 Engineering Materials iii. Screw iv. Edge b. Size of dislocation: c. Dislocation Core: i. Typical dislocation core radius size: ii. Typical dislocation length: 9. Grain Boundary a. Definition: An atomic mismatch within the region where two grains meet b.   Measure the length of the scale bar in millimeters using a ruler; Convert the  length into microns (1000 micrometers per millimeter); M = (measured scale  length) / (# appearing by the scale bar) c. Properties i. Grain size ii. Grain strength iii. Grain Shape/Direction d. Boundaries serve as high energy sites 10. Diffusion a. Mass diffusion: i. Steady State Diffusion: 1. Two solids achieving uniform composition through vacancies  within materials Tristan Collette Sept. 24, 2016 Engineering Materials 2. Increased temperature is used to increase vacancies and  frequencies within materials ii. Fick’s first law of diffusion 1. Diffusion flux: J = M/ At  M is mass; A is area; t is time 2. J = ­D(dC/dx) D = diffusion coefficient; dC/dx = concentration  gradient 3. dC/dx = ∆C/∆x = (C_a – C_b) / (x_a – x_b) iii. Diffusivity: 1. D = D_o*exp(­Q_d/R*T)  D_o = pre­exponential; Q_d =  activation energy for diffusion; R = gas constant (8.31 J/mol*K or  8.62x10^­5 eV/atom*K); T = absolute temperature (K) Z2 E n−13.6 2(eV) n F =F +F net attractirepulsion E= ∫dr dE F= dr E net attractirepulsion −A 1 E attraction,where A= ( 1 )(|e2 ) r 4πϵ o e=1.602x10 −19C +B E repulsion ,wheren 8 r 2 %IC= 1{exp − [ 0.25){XA−X B}]}100 N N N=N +i f+ c 2 8 volumeof atoms∈aunit cell APF= totalunit cellvolume V = πr 3 spher3 n A ρ= V c A 23 ¿ N A6.022x10 mole pointcoordinateindices=qr se.g., 01 2 ´ ´ crystallographicdirection=uvwe.g., [211 ] Miller indices=hkle.g., 011 ) N =N exp −Q v v ( kT k=1.38x10 −23 J =8.62x10 −5 eV atom K atomK J cal R=8.31 =1.987 moleK moleK length:1m=10 mm=10 μm=10 nm=10 Å=10 pm 10 12 7 18 ener gy:1J=0.239cal=10 ergs=6.24x10 eV mass:1kg=2.205lb m force:1N=10 dynes=0.2248lb f stress:6.896MPa=1ksi temperature:T (K)=273.15+T (oC ) ❑ J= M At dC Jx=−D dx dC ∆C C AC B = = dx ∆ x xA−xB −Q d D=D eop ( ) RT


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