ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 1 Bruce Mayer, PE Engineering-45: Materials of Engineering Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 45 Chp 2 & 3 Problems

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 2 Bruce Mayer, PE Engineering-45: Materials of Engineering Prob 3.4  For HCP Xtal Show c/a =1.633 a J c  Lattice Constants a & c  Consider the TETRAHEDRON formed by the Atoms between Planes  The Tetrahedron height, J  Note that c = 2J  Thus need only find the height of a Tetrahedron with Edge Length, a

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 3 Bruce Mayer, PE Engineering-45: Materials of Engineering The HCP Tetrahedron a  Thus CD =a/  3  Consider Tetrahedron ABCE with SideLength a Find height J  By Trig Length of CD = a/  3 CH = aCos(30º) = a  3/2 CD =(2/3)CH = a/  3  Then by Pythagorus

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 4 Bruce Mayer, PE Engineering-45: Materials of Engineering The HCP Tetrahedron cont  Then the c/a Ratio c J a  Thus c

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 5 Bruce Mayer, PE Engineering-45: Materials of Engineering Prob 3.11 – HCP Titanium  a) Given  = 4.51 g/cc, Find the Volume of the Unit Cell, V C  Use Eqn 3.5 for Theoretical Density  Recall From Lecture 3: n = 6 at/cell for HCP  From Text inside Front Cover Find A Ti = g/mol  Running the numbers find

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 6 Bruce Mayer, PE Engineering-45: Materials of Engineering Prob 3.11 – HCP Titanium cont Find This Volume and Mult by 6  b) if for Ti c/a = 1.58, then find a & c

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 7 Bruce Mayer, PE Engineering-45: Materials of Engineering Prob 3.11 – HCP Titanium cont  The Basal (or base) Triangle is equilateral  Thus the 1/6th Volume, and with c = 1.58a  Use V C from before  And c = 1.58a → c = nm

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 8 Bruce Mayer, PE Engineering-45: Materials of Engineering % Program E45_Prob_2_15.m: % Plot EA and ER vs r * Verify r0 numerically % Bruce Mayer, PE ENGR455 24Jan10 % % Calc r0 numerically using anonymous fcn for En % the eqn in text book is for r in nm % A = 1.436; B = 7.32e-6; En B/r^8 - A/r % % find En,min at r = r0 us fminbnd command r0 = fminbnd(En, 0, 1); disp('InterAtomic spacing for Min E, r0 in nm =') disp (r0) % % Calc En,min = En(r0) En_min = En(r0); disp('Min E, En_min in eV =') disp (En_min)% % Set Plotting Vector as 300 points r_plt = linspace (0.1,.4, 300); % in nm % % The Energy Functions EA = /r_plt; ER = 7.32e-6./r_plt.^8; Etot = EA + ER; % % Plot on Same Graph plot(r_plt,EA, r_plt,ER, r_plt, Etot), xlabel('r (InterAtom Spacing)'),... ylabel('Energy'), title('ENGR45 Problem 2.14'), grid,... legend('EA', 'ER', 'Etot'), axis([ ]) P2.15 MATLAB m-File

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 9 Bruce Mayer, PE Engineering-45: Materials of Engineering Prob 2.15 plot by MATLAB

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 10 Bruce Mayer, PE Engineering-45: Materials of Engineering Prob 2.17 – Applied Calculus  Given Net Potential Energy by expression  Find Optimum (minimum) Energy, E 0, at Optimum InterAtomic distance, r 0, In terms of a)D, , r 0 → Need to Eliminate C b)C, , r 0 → Need to Eliminate D  Plan: Take 1 st Derivative and Set to Zero to find r 0 Do On WhiteBoard

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 11 Bruce Mayer, PE Engineering-45: Materials of Engineering WhiteBoard Work  Problem 3.47 Given Three Plane-Views, Determine Xtal Structure Also:

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 12 Bruce Mayer, PE Engineering-45: Materials of Engineering All Done for Today ½ Cubic Bixbite Xtal  e.g. Indium Oxide

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 13 Bruce Mayer, PE Engineering-45: Materials of Engineering All Done for Today

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 14 Bruce Mayer, PE Engineering-45: Materials of Engineering

ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt 15 Bruce Mayer, PE Engineering-45: Materials of Engineering