[email protected] 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

[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

[email protected]

Engineering 45

Chp 2 & 3Chp 2 & 3ProblemsProblems



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



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



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



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



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



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



% 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



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)



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



WhiteBoard WorkWhiteBoard Work

Problem 3.47• Given Three Plane-Views, Determine Xtal

Structure ccgmacro /91.18Also:



All Done for TodayAll Done for Today

½ Cubic BixbiteXtal

e.g. Indium Oxide



All Done for TodayAll Done for Today





Documents

[email protected] 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