Engineering 45
Chp 2 & 3
Problems
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Prob 3.4
 For HCP Xtal Show c/a =1.633
 Lattice Constants a & c
 Consider the TETRAHEDRON
formed by the Atoms between
Planes
c
 The Tetrahedron height, J
J
a
Engineering-45: Materials of Engineering
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 Note that c = 2J
 Thus need only find the height
of a Tetrahedron with
Edge Length, a
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
The HCP Tetrahedron
 Consider Tetrahedron
ABCE with SideLength a
• Find height J
a
 By Trig Length of
CD = a/3
• CH = aCos(30º) = a3/2
• CD =(2/3)CH = a/3
 Then by Pythagorus
2
 a 
a 
  J2
 3
 Thus
2
2
CD =a/3
 
Also CH  a 2  a 2
2
Engineering-45: Materials of Engineering
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2
2
a
2
a
J 2  a2 

3
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
The HCP Tetrahedron cont
 Thus c
a
c
J
Engineering-45: Materials of Engineering
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2
c  2J  2 a
3
 Then the c/a Ratio
c
8

 1.633
a
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Prob 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
nA
ρ
VC N A
 Recall From Lecture 3:
• n = 6 at/cell for HCP
 From Text inside Front
Cover Find

nA
VC 
ρN A
 Running the numbers
find
VC  1.058 10 22 cc / cell
• ATi = 47.88 g/mol
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Prob 3.11 – HCP Titanium cont
 b) if for Ti c/a = 1.58, then find a & c
Find This Volume and Mult by 6
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Prob 3.11 – HCP Titanium cont
 The Basal (or base) Triangle is equilateral
Area  0.5  base  hgt  0.5  a  a cos 30  Area  3a 2 4
 Thus the 1/6th Volume, and with c = 1.58a
VC 6  Area  c  c  3a 2 4  1.58a  3a 2 4
 Use VC from before
1.058 10
28

m3 cell 6  1.58  3a 3 4  a  0.295nm
 And c = 1.58a → c = 0.467 nm
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
P2.15 MATLAB m-File
% 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 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 = -1.436./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([.1 .4 -8 8])
Prob 2.15 plot by MATLAB
ENGR45 Problem 2.14
8
EA
ER
Etot
6
4
Energy (eV)
2
0
Min @ (r0 = 0.24 nm, E0 = -5.3 ev)
-2
-4
-6
-8
0.1
0.15
Engineering-45: Materials of Engineering
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0.2
0.25
r (InterAtom Spacing)
0.3
0.35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
0.4
Prob 2.17 – Applied Calculus
 Given Net Potential Energy by expression
C
r / 
E N    De
r

Find Optimum (minimum) Energy, E0, at Optimum
InterAtomic distance, r0, In terms of
a)
b)
D, , r0 → Need to Eliminate C
C, , r0 → Need to Eliminate D
 Plan: Take 1st Derivative and Set to Zero to find r0
• Do On WhiteBoard
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
WhiteBoard Work
 Problem 3.47
• Given Three Plane-Views, Determine Xtal
Structure
Also:  macro  18.91 g / cc
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
All Done for Today
½ Cubic
Bixbite
Xtal
 e.g. Indium Oxide
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
All Done for Today
Engineering-45: Materials of Engineering
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Engineering-45: Materials of Engineering
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt
Engineering-45: Materials of Engineering
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Prob_3-5_3-12_2-16_Lab.ppt