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[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt1
Bruce Mayer, PE Engineering-45: Materials of Engineering
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 45
Chp 2 & 3Chp 2 & 3ProblemsProblems
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt2
Bruce Mayer, PE Engineering-45: Materials of Engineering
Prob 3.4Prob 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
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt3
Bruce Mayer, PE Engineering-45: Materials of Engineering
The HCP TetrahedronThe HCP Tetrahedron
a
Thus
2
2
2
3 J
aa
CD =a/3
3
2
3
2222 aaaJ 222
2 Also aaCH
Consider Tetrahedron ABCE with SideLength a• Find height J
By Trig Length of CD = a/3• CH = aCos(30º) = a3/2
• CD =(2/3)CH = a/3
Then by Pythagorus
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt4
Bruce Mayer, PE Engineering-45: Materials of Engineering
The HCP Tetrahedron contThe HCP Tetrahedron cont
Then the c/a Ratio
aJ3
222 c
c
J
a
633.13
8
c a
Thus c
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt5
Bruce Mayer, PE Engineering-45: Materials of Engineering
Prob 3.11 – HCP TitaniumProb 3.11 – HCP Titanium a) Given = 4.51 g/cc, Find the Volume of the Unit
Cell, VC
Use Eqn 3.5 for Theoretical Density
AC
AC ρN
nAV
NV
nAρ
Recall From Lecture 3:• n = 6 at/cell for HCP
From Text inside Front Cover Find• ATi = 47.88 g/mol
Running the numbers find
cellccVC /10058.1 22
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt6
Bruce Mayer, PE Engineering-45: Materials of Engineering
Prob 3.11 – HCP Titanium contProb 3.11 – HCP Titanium cont
Find This Volume and Mult by 6
b) if for Ti c/a = 1.58, then find a & c
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt7
Bruce Mayer, PE Engineering-45: Materials of Engineering
Prob 3.11 – HCP Titanium contProb 3.11 – HCP Titanium cont The Basal (or base) Triangle is equilateral
4330cos5.05.0 Area 2aAreaaahgtbase Thus the 1/6th Volume, and with c = 1.58a
4358.1436 22 aaaccAreaVC Use VC from before
nmaacellm 295.04358.1610058.1 3328
And c = 1.58a → c = 0.467 nm
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt8
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 = @(r) B/r^8 - A/r%% find En,min at r = r0 us fminbnd commandr0 = 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 pointsr_plt = linspace (0.1, .4, 300); % in nm%% The Energy FunctionsEA = -1.436./r_plt;ER = 7.32e-6./r_plt.^8;Etot = EA + ER;%% Plot on Same Graphplot(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([.1 .4 -8 8])
P2.15 M
AT
LA
B m
-File
P2.15 M
AT
LA
B m
-File
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt9
Bruce Mayer, PE Engineering-45: Materials of Engineering
Prob 2.15 plot by MATLABProb 2.15 plot by MATLAB
0.1 0.15 0.2 0.25 0.3 0.35 0.4-8
-6
-4
-2
0
2
4
6
8ENGR45 Problem 2.14
r (InterAtom Spacing)
Ene
rgy
(eV
)
EA
ER
Etot
Min @ (r0 = 0.24 nm, E0 = -5.3 ev)
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt10
Bruce Mayer, PE Engineering-45: Materials of Engineering
Prob 2.17 – Applied CalculusProb 2.17 – Applied Calculus Given Net Potential Energy by expression
Find Optimum (minimum) Energy, E0, at Optimum InterAtomic distance, r0, In terms of
a) D, , r0 → Need to Eliminate C
b) C, , r0 → Need to Eliminate D
/rN De
r
CE
Plan: Take 1st Derivative and Set to Zero to find r0
• Do On WhiteBoard
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt11
Bruce Mayer, PE Engineering-45: Materials of Engineering
WhiteBoard WorkWhiteBoard Work
Problem 3.47• Given Three Plane-Views, Determine Xtal
Structure ccgmacro /91.18Also:
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt12
Bruce Mayer, PE Engineering-45: Materials of Engineering
All Done for TodayAll Done for Today
½ Cubic BixbiteXtal
e.g. Indium Oxide
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt13
Bruce Mayer, PE Engineering-45: Materials of Engineering
All Done for TodayAll Done for Today
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt14
Bruce Mayer, PE Engineering-45: Materials of Engineering
[email protected] • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt15
Bruce Mayer, PE Engineering-45: Materials of Engineering