Use the grid below to draw the Al-Li phase diagram based upon the ...

Practice Midterm 02 Solutions 2000 

Question 1 (30 points) 

Use the grid below to draw the Al-Li phase diagram based upon the following 

experimental observations (all compositions in weight %). Label all phase fields. 

T(°C) 

800 

600 

400 

200 

Pure Al melts at 660°C; pure Li melts at 181°C. At 600°C, liquid of 10 %Li solidifies to 

form α phase of 4 %Li and δ phase of 20 %Li. Pure δ phase containing 20 %Li melts 

congruently at 718°C. At 521°C, γ phase containing 34 %Li melts to form δ phase of 24 

%Li and liquid phase of 47 %Li. At 620°C, a 30 % Li alloy has equal parts of liquid 

phase and δ phase, while at 260°C, a 60 %Li alloy has equal parts of liquid phase and γ 

phase. At 179°C, liquid containing 98 %Li solidifies to form γ phase (34 %Li) and β 

phase (99 %Li). At 0°C, Al can dissolve up to 3 % Li, while Li can dissolve no more than 

0.5 % Al; however. Also at 0°C, the solubility of Li in the δ phase ranges from 20 to 22 

%Li. 

α +L 

α 

α + δ 

L+δ 

δ 

δ +L 

γ 

Liquid 

δ + γ L+β 

β 

Al 20 40 60 80 

Li 

wt% Li 

γ +L 

γ + β 

page 1 of 8


T(°C) 

800 

0.02 

600 

α 

400 

(ferrite) 

200 


Illustrate the microstructural evolution resulting from cooling a hypereutectoid AISI 

1095 steel through the eutectoid decomposition isotherm. Use the templates at right, 

which show the prior austenite grains, and accurately draw the equilibrium 

microstructures at each of the temperatures identified on the right. Label each phase, 

and label all relevant microstructural constituents by their common metallurgical 

names. 

1600 

1538° 

1495° 

δ 

1400 

1394° 

1200 

γ 

(austenite) 

1000 

0.77 

γ + 

2.1 

1148° 

α + 

727° 

4.3 

Fe 1 2 3 4 5 6 7 

wt% C 

L 

γ + 

L+ 

6.7 

Fe3 

(cementite) 

Austenite 

1150°C 

100% Austenite 

Proeutectoid cementite 

at grain boundaries 

750°C 

92% austenite; 8% cementite 


Pearlite 

450°C 

86% ferrite; 14% 

cementite 


Pearlite 

25°C 

86% ferrite; 14% cementite 

page 2 of 8


Sketch and label the isothermal cooling curves for the same AISI 1095 steel that 

produce the following microstructures: Note that there are three TTT curves here. 

Label the first with a) and b), the second with c) and d), and the third with e) and f). 

a) 100% martensite 

b) 5% proeuctectoid cementite and 95% martensite 

T (°C) 

800 

700 

600 

500 

400 

300 

200 

100 

840° 

727° 

Ms 

M90 

1 sec 

γ + Fe3C 

α + Fe3C 

1 min 1 hour 

1 day 

0.1 1 10 102 103 104 105 0 

(a) (b) 

Time (sec) 


page 3 of 8

c) 25% coarse pearlite, 75% martensite 

d) 100% fine pearlite 

T (°C) 

800 

700 

600 

500 

400 

300 

200 

100 

840° 

727° 

Ms 

M90 

1 sec 

γ + Fe3C 

α + Fe3C 

1 min 1 hour 

1 day 

0.1 1 10 102 103 104 105 0 

Time (sec) 

(c) (d) 


page 4 of 8

e) 25% upper bainite and 75% tempered martensite 

f) 25% fine pearlite, 25% lower bainite, 50% martensite 

T (°C) 

800 

700 

600 

500 

400 

300 

200 

100 

840° 

727° 

Ms 

M90 

1 sec 

γ + Fe3C 

α + Fe3C 

1 min 1 hour 

1 day 

0.1 1 10 102 103 104 105 0 

Time (sec) 

(f) 

(e) 


page 5 of 8


Mark the checkbox next to the best answer (only one) in each case. 

a) Martensite is a metastable phase produced in ferrous alloys by 

holding at a specific temperature for a specific time 

heating to a specific temperature over a specific time 

☒ cooling to a specific temperature over a specific time 

b) Martensite can be tempered by 

heating to a specific temperature over a specific time 

cooling to a specific temperature over a specific time 

☒ holding at a specific temperature for a specific time 

c) The microstructure of tempered martensite consists of 

pearlite and bainite 

only bainite 

☒ neither pearlite nor bainite 

d) Martensite is strong because 

it is not an equilibrium phase 

☒ it presents many obstacles to dislocation motion 

it is a highly distorted arrangement of Fe and C atoms 

e) Ferrous alloys can be precipitation-hardened by 

holding in the austenite+ferrite two-phase field 

☒ aging to produce alloy carbides 

quenching to produce martensite 

f) Nonferrous metallic alloys can be precipitation-hardened by 

☒ solutionizing, quenching and aging 

solutionizing slow-cooling and quenching 

quenching, aging and quenching 

g) Ni-base superalloys are known for their superior 

☒ strength at elevated temperatures 

toughness at room temperature 

wear-resistant surface finish 


page 6 of 8

h) Al-base alloys are protected from oxidation by 

☒ a passivating oxide film 

a high density of second phase particles 

an ASTM standard requiring painted surfaces 

i) During annealing, the driving force for the recovery stage is 

☒ reduction in strain energy 

reduction in surface energy 

reduction in chemical potential 

j) During annealing, the driving force for the recrystallization stage is 

☒ reduction in strain energy 

reduction in surface energy 


k) During annealing, the driving force for the grain growth stage is 

reduction in strain energy 

☒ reduction in surface energy 


l) Ceramics are often formed into complex parts by sintering at 

one-forth of the melting temperature 

☒ one-third of the melting temperature 

two-thirds of the melting temperature 

m) Glass-ceramics are subjected to an aging treatment in order to 

provide stress-relief 

remove surface cracks 

☒ precipitate a crystalline phase 

n) The most significant attribute associated with a Griffith crack is 

its location at the surface 

its length 

☒ its tip radius 


page 7 of 8

o) Addition polymerization is a reaction among small organic molecules involving 

isotropic growth 

☒ chain growth 

step growth 

p) Condensation polymerization involves 

a rapid “chain reaction” 

☒ individual chemical reactions between pairs of reactive monomers 

electron transport in the condensate 

q) The primary role of the initiator in a polymerization reaction is to 

begin the assembly of mers into chains 

☒ convert double bonds into single bonds 

nucleate the condensation reaction 

r) Linear polymers are the result of bonding between 

☒ bifunctional mers 

polyfunctional mers 

linear mers 

s) A block co-polymer can also be described as 

a solid solution 

☒ a two phase alloy 

a cubic crystal structure 

t) A polymer is described as “viscoelastic” when it shows 

☒ slow recovery of strain 

viscous fracture 

negative modulus of elasticity 


page 8 of 8

UNIVERSITY OF CALIFORNIA 

College of Engineering 

Department of Materials Science and Engineering 

Professor R. Gronsky Fall Semester 2001 

ENGINEERING 45

E45•Midterm #2 

Question 1 The silver-mercury system forms the basis of an important biomaterial 

known as dental amalgam, which has been used for decades in restorative 

dental fillings. The term “amalgam” is frequently used to depict many 

different alloys of mercury, originating from the “amalgamation” process to 

refine ores of the precious metals, primarily silver and gold. The finely 

crushed ores in aqueous solution are mixed with mercury, allowing the 

mercury to alloy with the precious metals, forming an amalgam. When the 

amalgam is heated, the mercury is driven off as a vapor (collected and 

recycled of course), leaving behind pure precious metal “sponge.” 

1000 

961.93° 

800 

600 

T 

(°C) 

400 

Silver (Ag, Z = 47) is fcc and melts at 961.93°C, mercury (Hg, Z = 80) 

freezes at –38.862°C. The equilibrium β phase has AgHg stoichiometry and 

an hcp Bravais lattice. The equilibrium γ phase has Ag2Hg3 stoichiometry and 

a bcc Bravais Lattice. 

Use this information and the equilibrium phase diagram for the silver-mercury 

binary system due to M. Hanson (Constitution of Binary Alloys, 2 nd . Edition, 

McGraw –Hill, 1958) shown below to choose the best answers for 2 points 

each. 

(Ag) 

200 

127° 

β 

0 

γ 

-38.9° 

-38.862° 

(Hg) 

-200 

Ag 10 20 30 40 50 60 70 80 90 Hg 

Weight Percentage Hg 

2 of 10 

52.4 

L 

276° 

Hg Boiling Point 357°


a. The number of components in this binary system is 

one (1) 

two (2) 

three (3) 

b. The phase called (Ag) is actually 

pure silver 

a silver-mercury solid solution 

a metastable phase 

c. Based upon the Gibbs Phase Rule (P+F = C+2), the number of degrees of 

freedom available to pure Hg at 357°C in this system is 

zero (0) 

one (1) 

two (2) 

d. The number of degrees of freedom available to a 40 wt.% Ag alloy at 276°C is 

zero (0) 

one (1) 

two (2) 

e. The maximum solubility of mercury in silver is 

infinitesimally small 

undefined because mercury is a liquid at room temperature 

established by the highest temperature reaction isotherm 

f. Both β phase and γ phase are in equilibrium with liquid Hg at a 

monotectic 

eutectic 

peritectic 

g. When β phase of 60 wt% Hg melts, it 

disappears completely 

passes through a β + L two-phase equilibrium 

generates mercury vapor 

h. A dental hygienist “mixes” amalgam immediately before the dentist applies it so 

that it will “harden” as a complete filling. The hardening process is due to 

solid state diffusion 

solidification of liquid Hg 

dislocations created during mixing 

i. An amalgam of 20 wt % Hg in equilibrium will decompose when heated above 

276°C 

576 °C 

776°C 

j. Dental fillings made of silver amalgam are at risk of leaking mercury into the 

mouth if the equilibrium mercury concentration exceeds 

52.4% 

62.4% 

72.4% 

3 of 10


Question 2 Use the same phase diagram to now sketch the microstructures expected 

during equilibrium solidification of the following alloys. The lever rule 

construction is important here. Draw a vertical line on the phase diagram for 

the specified composition and show all tie lines used to establish your 

equilibrium microstructures. Remember that the high temperature 

microstructure influences the lower temperature microstructure. Clearly label 

all phases for full credit. 

T = 500°C 

85% (Ag) 

+ 15% L 

T = 25°C 

40% (Ag) 

+ 60% β 

T = 200°C 

50% L 

+ 50% β 

T = 25°C 

33% L 

+ 67% γ 

Each sketch is worth 5 points. 

a. Ag-55 wt% Hg 

L 

(Ag) 

b. Hg-20 wt% Ag 

β L 

γ L 

(Ag) 

1000 

961.93° 

800 

600 

T 

(°C) 

400 

200 

127° 

β 

0 

-38.862° 

-38.9° 

γ 

(Hg) 

-200 

Ag 10 20 30 40 50 60 70 80 90 Hg 


1000 

961.93° 

800 

600 

T 

(°C) 

400 

Note that in all cases here, there are no microstructural remnants from higher 

temperature reactions to influence the products. All are peritectic reactions, which 

require longer times to reach equilibrium, but time is not a factor here since we are 

working from an equilibrium phase diagram. 

4 of 10 

(Ag) 

(Ag) 

52.4 

52.4 

L 

276° 

Hg Boiling Point 357° 

200 

127° 

β 

0 

-38.862° 

-38.9° 

γ 

(Hg) 

-200 

Ag 10 20 30 40 50 60 70 80 90 Hg 


L 

276° 

Hg Boiling Point 357°


Question 3 The same TTT curve for a plain carbon steel of eutectoid composition is 

reproduced multiple times below. Draw directly on these plots the cooling 

curves corresponding to the specified treatments. Remember that TTT curves 

are appropriate for isothermal transformations only. The actual start and 

termination of the decomposition of austenite during continuous cooling 

occurs at slightly lower temperatures and longer times than shown here. 

Each curve is worth 5 points. 

a. Austempering, a commercial 

treatment used to avoid quench 

cracking. Show and label the 

curves corresponding to both 

the surface and the interior of a 

thick cross section steel beam. 

b. Martempering, a commercial 

treatment used to avoid quench 

cracking. Show and label the 

curves corresponding to both 

the surface and the interior of a 

thick cross section steel beam. 

Both (a) and (b) are the classic, traditional methods used widely throughout the 

engineering community to prevent quench-cracking in large cross-section steel 

components. 

5 of 10 

°C 

800 

700 

600 

500 

727° 

γ 

γ 

α + Fe 3 C 

400 

upper bainite 

Surface Interior 

300 

215° 

200 

Transform 

lower 

bainite 

100 

martensite 

sec min 

Quench 

hour day 

°C 

800 

700 

600 

500 

0.1 1 10 10 2 10 3 10 4 10 5 

727° 

γ 

γ 

time (sec) 

α + Fe 3 C 

coarse pearlite 

fine pearlite 

400 

upper Temper bainite 

Surface Interior 

300 

215° 

200 

martensite 

100 sec 

Quench 

min hour 

lower 

bainite 

day 

0.1 1 10 10 2 10 3 10 4 10 5 

time (sec) 


fine pearlite


c. An “interrupted quench” 

technique that would generate a 

final microstructure of 50% 

lower bainite and 50% fine 

pearlite. In this case the 

workpiece is a small diameter 

rod to be used as an axle in a 

motor-generator set. 

The only way this treatment can be an “interrupted quench” is to begin with the high 

temperature transformation (fine pearlite), holding isothermally until the reaction 

progresses half way between the start and finish lines. Note that this amounts to 

only a few seconds. The untransformed austenite is then lowered in temperature 

and held isothermally until it fully transforms to lower bainite as shown. 

d. An approximation. Show 

without compensation for 

continuous cooling the T-t 

trajectory for a sample that was 

left in the furnace during a 

power outage. The initial 

temperature was 800°C at 

midnight when the power failed. 

The cooling rate for the wellinsulated 

furnace was previously 

calibrated at 100°C / minute. 

The sample was recovered 8 

hours later at the start of the 

daylight shift. 

In question (d), note that a linear rate of 100°C / min is dramatically curved on the 

log scale used here. Although this is only an “approximation” because the cooling 

treatment is not isothermal, it is reasonable to expect that the final microstructure 

would be fully pearlitic, and rather coarse. 

6 of 10 

°C 

800 

700 

600 

500 

400 

300 

200 

100 

°C 

800 

700 

600 

500 

400 

300 

200 

100 

727° 

γ 

215° 

γ 

α + Fe 3 C 

martensite 

sec min hour day 

0.1 1 10 10 2 10 3 10 4 10 5 

727° 

γ 

215° 

50% 

γ 

time (sec) 


fine pearlite 

upper bainite 

lower 

remaining 50% 

bainite 

100°C/min 


α + Fe3C fine pearlite 

martensite 

sec min hour day 

0.1 1 10 10 2 10 3 10 4 10 5 

time (sec) 

upper bainite 

lower 

bainite


Question 4 Pure zirconia undergoes an allotropic transformation at 1000°C from a high 

temperature tetragonal structure to a low temperature monoclinic structure. 

A massive volume expansion upon cooling through the tetragonal-tomonoclinic 

transformation literally results in the self destruction of the crystal. 

Thermal cycling through the transformation temperature pulverizes the 

material into a fine powder. 

However, alloying zirconia with 25 mol % calcia yields a cubic phase that 

persists up to its melting point with no structural instabilities. This 

“stabilized” zirconia is a very popular structural ceramic for that reason. 

T 

(°C) 

2500 

2000 

Tetragonal 

ZrO2 SS 

1500 

1000 

Monoclinic 

ZrO2 SS 

500 

ZrO 2 

10 

Cubic 

ZrO 2 SS 

20 30 40 50 

Mole Percentage CaO 

a. Using this information, explain how “partially-stabilized” zirconia, which 

contains some cubic, some tetragonal, and some monoclinic phases, is 

prepared. Be specific, using the equilibrium phase diagram above and your 

understanding of phase transformation kinetics for 10 points. 

The phase diagram indicates that it is not possible to find all three phases (cubic, 

tetragonal and monoclinic) in equilibrium. So this is a problem involving kinetics. At 

CaO concentrations less than 15 mol%, monoclinic and cubic phases are in equilibrium 

at room temperature. To include a tetragonal phase at room temperature, all that is 

required is to heat above the eutectoid temperature (900°C) where tetragonal 

phase is in equilibrium, then “quench” to retain it kinetically at lower temperature. 

7 of 10 

L 

ZrCaO 3


b. For another 10 points, explain in detail how the partially-stabilized zirconia 

structure leads to “transformation toughening” of an otherwise brittle ceramic 

material. Comment specifically on the volume expansion associated with the 

tetragonal to monoclinic transformation, and its role in causing an increase in 

fracture toughness. Recall that fracture toughness refers to the performance 

of a material with pre-existing flaws, and focus your attention on the crack 

shown in the following sketch. 

Recalling from above that only the cubic 

and monoclinic phases are in equilibrium 

at room temperature, the metastable 

tetragonal phase becomes the key 

constituent in this problem. 

As the crack advances through the 

forest of tetragonal precipitates, it 

relaxes the constraints of the cubic 

monoclinic 

matrix confining the tetragonal 

particles. This relaxation induces the 

cubic matrix 

transformation of the metastable 

tetragonal phase to the equilibrium 

monoclinic phase in the vicinity of the 

crack tip. The resulting volume 

expansion (as described in the problem 

statement) exerts a closing force on 

the crack tip, retarding its progress 

through the ceramic. It is for this reason that the mechanism described here is 

known as “transformation toughening.” The transformation from the tetragonal to 

monoclinic phases gives partially stabilized zirconia (PSZ) the highest fracture 

tougnhness of all structural ceramics. 

8 of 10 

tetragonal


Question 5 Polymers or “many mers” may result from chain growth, also known as 

addition polymerization, which generates linear or branched polymers, or 

from step growth, also known as condensation polymerization, which 

generates network polymers. The mechanical properties of polymers are, like 

all materials, linked to their structure. 

(a) The styrene mer is shown schematically below. Show how polystyrene results 

from a polymerization reaction by sketching the final product containing at 

least five mers. Is polystyrene a (check all that apply) 

linear, 

branched, or 

network polymer? 

Showing the correct polymeric structure and checking the correct box(es) 

above earns 10 points. 

The signature feature of the styrene mer that marks it as a candidate for 

polymerization is its double carbon bond. Conversion of this bond into two single 

bonds functionalizes the mer. Its resulting bifunctional character makes it a 

candidate for chain growth into a linear polymer, just like polyethylene. A segment 

of the polymer backbone chain is shown below. 

H H 

C C 

H 

H C CH 2 

This particular configuration is the isotactic one, with all of the large side groups on 

the same side. The syndiotactic configuration shown below, and the atactic 

configuration (random) are also acceptable. 

H H 

C C 

H 

H H 

C C 

H 

H 

H 

C C 

H 

9 of 10 

H H 

C C 

H 

H H 

C C 

H 

H H 

C C 

H 

H 

H 

C C 

H 

H H 

C C 

H 

H H 

C C 

H


(b) Polymers, like metals and ceramics, can be described by a lattice and motif, 

but they are never fully crystalline. In fact, numerous polymers are 

completely amorphous. Those that possess some crystallinity have their 

polymeric chains arranged in a regular, parallel alignment, and these 

crystallites or “fringed micelles” are embedded in an amorphous matrix. 

A partially crystallized thermoplastic polymer shows “viscoelastic” mechanical 

behavior. Using the coordinate axes below, plot the elastic modulus as a 

function of temperature for a viscoelastic polymer. Note particularly the 

changes seen at the glass transition temperature and the melting 

temperature. On your plot, label the four distinct regions of behavior 

associated with viscoelasticity, that is, rubbery, viscous, leathery, and rigid 

(or elastic). A correctly labeled plot is worth 10 points. 

Modulus 

of 

Elasticity 

(log scale) 

T g 

Temperature 

In its traditional definition, the glass transition temperature marks the transition 

from rigid “crystal-like” to viscous “glass-like” mechanical behavior. Crystals deform 

by dislocation motion, glasses deform by the viscous flow. 

At the melting point, viscous flow is most obvious because the material cannot 

sustain its own weight against gravity. The elastic modulus drops to zero. 

10 of 10 

Rigid (or "Elastic") 

T m 

Leathery 

Rubbery 

Viscous

UNIVERSITY OF CALIFORNIA 

College of Engineering 

Department of Materials Science & Engineering 

Professor R. Gronsky Fall Semester, 2005 

Engineering 45 Midterm 02 

SOLUTIONS 

INSTRUCTIONS 

LATTICE seating .................. Please be seated with occupied seats to your front and back, vacant seats to your left and right. 

CLOSED BOOK format ...... All you need are writing instruments and a straightedge. Please store all books, reference materials, 

calculators, PDAs, cell phones (OFF), and iPods. 

NO DISRUPTION rule ......... Questions cause too much of a disturbance to others in the room. Instead of asking questions, 

write any concerns or alternative interpretations in your answers. 

PROFESSIONAL protocol ... Engineers do not cheat on the job and they certainly don’t cheat on exams. 

Do not open until “START” is announced.

E 45 Midterm 02 SOLUTIONS: 

1. Defects in Solids (10 points) 

Mark ☒ the ballot box corresponding to the best answer. 

Two (+2) points for correct answers, -1 if wrong, 0 if blank. 

(a) Point defects in solids include 

☒ vacancies 

☐ dislocations 

☐ both 

(b) The defect known as a “Frenkel pair” is shown in this 

sketch to have 

☐ body-centered symmetry 

☒ an extended strain field 

☐ both 

(c) Vacancies in solids 

☐ participate in diffusion 

☐ increase the entropy of a material 

☒ both 

(d) Dislocations in solids 

☒ participate in plastic deformation 

☐ decrease the entropy of a material 

☐ both 

(e) Fick’s First Law for diffusion flux 

Jx = −D ∂c 

∂x 

expresses the fact that 

☐ diffusion requires a negative diffusivity 

☒ mass flows down a concentration gradient 

☐ both 

page 2 of 6 

2. Defects in Solids (10 points) 

For this problem you must draw and label all requested 

features directly on the figures provided. 

(a) (2 points) Fill in the atoms comprising the extra-half 

plane and label with the conventional symbol (⊥) the edge 

of the extra half-plane that establishes the line direction 

vector of the edge dislocation shown below. 

(b) (3 points) On the same figure below, trace and label a 

Burgers circuit in finish-start-right-hand (FSRH) convention, 

and identify the resulting Burgers vector (b). 

b 1 

F 

3 

S 

2 

1 

3 

(c) (2 points) Fill in the atoms comprising the extra-half 

plane and label with the conventional symbol (⊥) the edge 

of the extra half-plane that establishes the line direction 

vector of the edge dislocation shown below. 

(d) (3 points) On the same figure, draw in and label the 

location of the slip plane on which this edge dislocation 

glides. 

2 

2 

Slip 

Plane 

3 

1 

1 

3 

2


3. Phases and Phase Equilibria (10 points) 

Refer to the Cu-Zn binary phase diagram below (from ASM 

Metals Handbook, 8 th edition, Vol. 8, (1973), p. 301) to 

answer the following questions. Recall that “brass” is the 

common name applied to this alloy. 

T (°C) 

1100 

1000 

900 

800 

700 

600 

500 

400 

300 

200 

100 

0 

Cu 

at % Zn 

10 20 30 40 50 60 70 80 90 

" 

903° 

(a) 

456° 

(b) 

! 

!’ 

(c) 

468° 

10 20 30 40 50 60 70 80 90 

wt % Zn 

# 

L 

835° 

(d) 

700° 

598° 

$ 

558° 

424° 

% & 

(a) (2 points) What is the maximum concentration of Zn 

that can be dissolved in α brass? 

Ans: 39 wt. % Zn 

(b) (2 points) Apply the phase rule (F = C-P+1) to calculate 

the number of degrees of freedom available to a 40 

wt% Zn alloy at room temperature. 

Ans: F = 2 - 2 + 1 = 1 

(c) (2 points) What phase(s) is (are) in equilibrium when 

β brass (50:50 composition) is held at 500°C. 

Ans: β + γ 

(d) (2 points) Write the reaction that occurs on cooling 

through the 700°C peritectic isotherm. 

Ans: γ + L ➔ δ 

(e) (2 points) Both the ε and η phases of brass have hexagonal 

crystal structures. What composition would yield 

equal weight fractions of the these two phases in equilibrium 

at 100°C? 

Ans: 92.5 wt % Zn 

(e) 

Zn 

page 3 of 6 

4. Phases and Phase Equilibria (10 points) 

These questions refer to the same Cu-Zn binary phase diagram 

from Problem 3. 

(a) (5 points) Sketch the microstructure resulting when a 

70 wt% Zn alloy with initially large γ grains at 550°C, as 

shown below, is cooled slowly to room temperature. Label 

all phases. 

Slow cooling will result in grain 

boundary precipitation of the ε phase 

(volume fraction ≈ 25%) within the γ 

grains (volume fraction ≈ 75%).. 

(b) (5 points) The eutectoid composition at 558°C is 74.1 

wt % Zn. Sketch the microstructure resulting when an alloy 

of this composition, showing a δ phase morphology at 

560°C as shown below, is cooled slowly to room temperature. 

Label all phases. 

! 

" 

! " 

At the eutectoid isotherm, all of the δ 

phase will decompose to the ε + γ in a 

lamellar constituent, in volume fractions 

of 40% ε and 60% γ.


7. Metallic Alloys (10 points) 

Temper Definition 

H1 Strain-hardened 

H2 Strain-hardened, partially annealed 

T3 Solution-treated, cold-worked, naturally 

aged 

T6 Solution-treated, artificially aged 

T9 Solution-treated, artificially-aged, cold 

worked 

The above table is a partial list of the “temper designations” 

for aluminum alloys specified by the Aluminum Association 

and published in the ASM Metals Handbook, 9 th 

edition, Volume 2 (1979). Recognizing that the specified 

treatments appearing in the “definition” column occur in 

sequence, and in the order given, rank the various temper 

treatments in order of highest strength to lowest strength. 

1. T9 

2. T3 

3. T6 

4. H1 

5. H2 

Reasoning: In general, the combination of work-hardening 

and precipitation hardening gives the highest overall 

strength. The T3 temper designation is the one given to 

aircraft rivets, as described in lecture. Strain hardening 

alone can be considerable, but if an alloy was designed to 

be precipitation-hardened (by choice of suitable alloying 

additions in appropriate amounts), it is expected that an 

aging treatment can also be designed to maximize strength 

without the damage of cold work. Those alloys for which 

the only available strengthening option is strain hardening 

would be treated to earn an H designation, with a judicious 

amount of cold-work to avoid non-uniform stress distributions 

that might lead to cracking and/or brittle fracture. 

Annealing of course softens a material, putting it at the 

bottom of the list. 

page 5 of 6 

8. Metallic Alloys (10 points) 

Strengthened by Weakened by 

Porosity (casting) 

Cold working Annealing 

Alloying Welding 

Phase Transformations Phase Transformations 

The above table lists some of the general effects of “processing” 

of metals and alloys on their mechanical strength. 

For example, an alloy used in the “as-cast” condition is 

generally weaker due to the porosity that occurs because of 

air entrapment during solidification from the liquid phase. 

Similarly, welding is listed as a cause of “weakening,” due 

to the fact that the local application of heat needed to weld 

alloys causes significant atomic transport, including some 

in the liquid phase. 

Explain why “phase transformations” is listed in both columns. 

Answer: The phase transformations in the “strengthen” 

column are those that cause precipitation hardening. Such 

transformations result from carefully controlled thermal 

treatments, beginning with homogenization in a single 

phase field, followed by rapid quenching to generate a supersaturated 

solid solution, finishing with an aging treatment 

to produce a fine dispersion of second phase particles 

that impede dislocation motion. 

The phase transformations in the “weaken” column 

are those that are not carefully controlled, such as 

slow cooling from the homogenization temperature, which 

usually results in a detrimental distribution of second phase 

particles occurring exclusively at grain boundaries. The 

grain interiors are therefore made much less resistant to 

dislocation motion because they have no precipitate particles 

to serve as obstacles, and many fewer solute atoms that 

also serve as impediments to dislocation motion (stronger 

metallic bonding). When such a material is deformed, dislocations 

move freely through the grains, sometimes with 

even lower strength than the solution solution would have 

shown.


9. Ceramics and Glasses (10 points) 

An important family of “magnetic” ceramics is based upon 

the spinel (MgAl2O4) structure. Spinel is formed from an 

fcc Bravais lattice and a basis of fourteen (14) ions per lattice 

point: 2 Mg 2+ , 4 Al 3+ and 8 O 2- . The magnesium ions 

are tetrahedrally coordinated by four oxygen ions, while 

the aluminum ions are octahedrally coordinated by six 

oxygen ions. Note that both the motif and the chemical 

formula preserve charge neutrality. 

However, the magnetic (known as “ferrimagnetic”) version 

of this structure is called an “inverse spinel,” in which the 

octahedral sites are occupied by the divalent ions and onehalf 

of the trivalent ions, while the remaining trivalent ions 

are in the tetrahedral sites. 

One example of the ferrimagnetic ceramics is magnetite, 

the naturally occurring “lodestone” that was used to make 

the original compass. It is also found in meteorites. This 

could be the reason why, in Fox's hit series The X-Files, 

FBI investigators observed that magnetite could disrupt 

alien life forms, often causing their death or destruction (?). 

Magnetite is an inverse spinel, yet its chemical formula is 

Fe3O4. Explain. 

Answer: The apparent anomaly here is charge balance. If 

you count up the number of negative charges on the eight 

(8) O 2- ions (= -16) and compare that number with the 

number of positive charges on six (6) Fe 2+ ions (= +12), 

you arrive at a conundrum. However, taking a clue from 

the opening paragraph above, a spinel consists of both divalent 

and trivalent cations in the correct ratio (2 of the 

divalent species and 4 of the trivalent species (= +16) to 

achieve charge neutrality. So, if the chemical formula of 

magnetite is correct, its charge distribution must be reconciled. 

And the way to do that is to give Fe a multivalent 

character. Assume that it can be either divalent or trivalent. 

The answer then is that there are must be two (2) Fe 2+ ions 

and four (4) Fe 3+ ions in this inverse spinel. You can think 

of the chemical formula in the same format as that for the 

original spinel, MgAl2O4, which would be expressed as 

FeFe2O4, but the more efficient form is Fe3O4. 

page 6 of 6 

10. Ceramics and Glasses (10 points) 

The following plot shows the thermal expansion behavior 

typical of a glass. Note that strain is plotted on the vertical 

axis, with temperature increasing to the right on the horizontal 

axis. At the softening temperature (Ts), the glass is 

unable to support its own weight and flows freely, invalidating 

the thermal expansion data. 

!L / L 0 

During the processing of safety glass for windows, a treatment 

is used to place the surface of the glass in residual 

compression, so that it will not be as susceptible to fine 

surface cracks. In one of these treatments, the glass is first 

equilibrated above its glass transition temperature (Tg). 

Next it is subjected to cold air blast on both sides, a “surface 

quench,” to form a rigid but thin “skin” on both surfaces. 

The skin is cool enough to remain below the glass 

transition temperature while the interior of the glass is still 

above Tg. The glass is then allowed to cool slowly and 

uniformly to room temperature. 

Explain how this treatment causes residual surface compression. 

Answer: The schematic at 

right shows how the stress 

distribution varies with each 

thermal treatment. The residual 

surface compression 

comes during the final stages 

of cooling, when the larger 

interior mass of glass contracts, 

pulling the surface skin 

into compression. 

T g 

T 0 

T g 

T 0 

T g 

T 0 

T g 

T f 

Temp 

T s 

T 

Compression 

0 

Tension 

Compression 

0 

Tension 

Compression 

0 

Tension 

Stress

Use the grid below to draw the Al-Li phase diagram based upon the ...

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