Exam 8

EMSE 201 — Introduction to Materials Science & Engineering 5 May 2004

Name:

Department of Materials Science and Engineering 1 of 12 Case Western Reserve University

Final Exam — 180 minutes; 250 points; 6 questions; 12 pages; 25% of course grade

No calculators or formula sheets are allowed. Where numerical answers are requested, full credit will begiven for correctly setting up the calculation and for specifying the correct units.

1) The unit cell of a compound of molybdenum (Mo) and silicon (Si) has the following characteristics:

Mo: 0, 0, 01

2,

1

2,

1

2Si: 0, 0,

1

30, 0,

2

3

1

2,

1

2,

1

6

1

2,

1

2,

5

6

a = b = 0.3202 nm c = 0.7851 nm

α = β = γ = 90°

a) (6 points) In the space at right, sketch theunit cell. Clearly identify the atoms.

b) (4 points) How many atoms of each typeare in the unit cell?

Mo: 2 Si: 4

Si

Mo

c) (4 points) What is the Bravais lattice of this structure? body-centered tetragonal

d) (2 points) How many nearest-neighbor Si atoms does each Mo atom have? 10

(Hint: the Si atoms at 0, 0, 1

3 and

1

2,

1

2,

1

6 are almost exactly equidistant from the Mo atom at

1

2,

1

2,

1

2.)

e) (2 points) How many nearest-neighbor Mo atoms does each Si atom have? 5

Sketch the arrangement of atoms on the specified planes in this structure. Label the atoms in your sketches.

f) (5 points) (100)

Si

Mo

g) (7 points) (110)

Si

Mo

continued on next page



h) (8 points) Set up the calculation of the density of this material. The atomic weights of Mo and Si are95.94 and 28.09 g mol–1, respectively. Avogadro’s number is 6.023×1023 mol–1.

density = (mass of atoms per cell)/(volume of cell) (2 pts)

= (2AWMo + 4AWSi)/(ca2×NA) (4 pts)

= (2×95.94 + 4×28.09)[g mol–1]/(0.7851×.32022)[10–21cm3]×6.023×1023[mol–1]

(1 pt for inserting values; 1 pt for units)

(This comes out to 6.275 g cm–3.)

i) (10 points) A compound of tungsten (W) and silicon forms with the same stoichiometry and structureas the compound discussed above. Would you expect these two compounds to form a continuous solidsolution with each other? Evaluate each of the Hume-Rothery rules to justify your answer.

atomicradius, nm

electronegativity allowedvalences

Mo 0.136 2.16 6, 5, 4, 3, 2

W 0.137 2.36 6, 5, 4, 3, 2

Same structure (given): √ (1+1 pts)

∆r/r < 15%: √ (1+1 pts)

∆EN < 0.4: √ (1+1 pts)

similar valences: √ (1+1 pts)

Yes, we expect MoSi2 and WSi2 to form a continuous solid solution (2 pts) (and theydo).

Various properties of the Mo-Si compound arelisted at right.

j) (8 points) In what respect(s) does thiscompound behave like a ceramic? Whataspects of the crystal structure mightcontribute to this behavior?

Answer on next page.

Melting point: 2230 °C

Young’s modulus (25 °C): 430 GPa

Ductility (25 °C): 0.0%

Ductility rises rapidly with temperature above 1200°C

Electrical conductivity (25 °C): 2.8×106 Ω–1 m–1

Electrical conductivity (1700 °C): 2.5×105 Ω–1 m–1

continued (and with more space) on next page



The very high Young’s modulus (over twice that of steel, whose modulus is among thehighest for metals) is typical of a ceramic (3 pts).

The brittleness (zero ductility) at room temperature is also typical of a ceramic (3pts).

The alternating arrangement of Mo and Si atoms is ceramic-like (1 pt). The somewhatlong c axis suggest that this material would have longer Burger’s vectors than simplemetals (1 pt), which would lead to brittle behavior at room temperature (1 pt; 8 ptsmax).

j) (8 points) In what respect(s) does this compound behave like a metal? What does this suggest aboutthe bonding in this material? (The electronegativity of Si is 1.9 on Pauling’s scale, which ranges from0.7 to 4.0 among all of the elements.)

The high electrical conductivity is typical of a metal (2 pts), as is the slight decreasein this property with increasing temperature (2 pts).

The appearance of ductility at high temperature is reminiscent of the brittle-to-ductile transition of body-centered metals, but occurring here at much highertemperatures (1200 °C vs. 0-100 °C for plain steels) (2 pts).

The electrons are apparently delocalized as in metallic bonding of (1 pt), and thedecrease of conductivity with increasing temperature suggests that the material doesnot have a band gap (1 pt). The low electronegativities of the elements (1 pt) and thelow difference in electronegativities (1 pt; 8 pts max)) further suggest metallicbonding.

k) (6 points) Would you expect this material to have a thermal expansion coefficient closer to those ofmetals or to those of ceramics? Justify your answer.

The high modulus (1 pt) and the high melting point (1 pt) suggest that this compound hasa deep, narrow interatomic potential (2 pts), which would give it a low thermalexpansion coefficient (2 pts), i.e. more like that of ceramics (2 pts) than metals (6 ptsmax).



2) The hardness of a cylindrical rod of plain carbon steel (0.3 wt% C; 1.4" dia.) is plotted below as a functionof radial distance from the perimeter. Given only this information, comment on the plausibility of thefollowing possible explanations (PE1, PE2, PE3) for how this hardness profile was achieved. Justify youranswer. You may find the charts on page 6 useful.

(Callister, Fig. 8.38)

a) (10 points) PE1: The rod had beencarburized.

This would lead to a higherconcentration of carbon on thesurface than in the interior(3 pts) (see Ch. 5 in Callister),which would raise the hardness ofthe surface relative to that ofthe interior (3 pts), as seen in thegraph on the upper left of p. 6. Soqualitatively PE1 moves in theright direction.

However, according to the graph on the upper left of p. 6, simply raising the carboncontent on the surface of this plain carbon steel without changing what phases arepresent would bring the Brinell hardness values up only to around 300 (2 pts), whichaccording to the graphs on the right of p. 6 correspond to Rockwell C values ofaround 33 (2 pts). Also, it would take a very long time to diffuse carbon into steelto depths of 0.2” (2 pts), and the erf concentration profile wouldn’t level off nearthe surface as the hardness data do (2 pts). So quantitatively, PE1 is probably notthe correct explanation (10 pts max).

b) (10 points) PE2: The surface of the rod had been cold worked (e.g. by surface machining).

Cold-working the surface would lead to a higher concentration of dislocations on thesurface (3 pts), which would raise the hardness of the surface relative to that of theinterior (3 pts). So qualitatively PE2 also moves in the right direction.

However, according to the graphs on the bottom of p. 6 (assuming the curve for 1040steel is similar to that of this steel, which is probably an xx30 — 2 pts), cold workalone would raise the strength to around 130 ksi (2 pts), corresponding to RC hardnessaround 30 (2 pts), lower than those observed on the surface of the rod. Also, it isquestionable that the hardening effect of surface machining (such as cutting ofthreads) would penetrate to 0.2” (2 pts). So quantitatively, PE2 is probably not thecorrect explanation (10 pts max).




c) (10 points) PE3: The surface of the rod had been heated, then rapidly quenched.

If the quenching is fast enough, martensite will form (3 pts), more likely on thesurface, which will be cooled more rapidly than the interior (2 pts). Martensite canexhibit RC values between 50 and 60 for a steel of 0.3 wt% C (3 pts), matching thesurface hardness here. In contrast, the interior region, which would have been quenchedmore slowly and therefore could have formed microconstituents like bainite or finepearlite, would have RC values of 20 or below (2 pts), again consistent with theobserved hardness profile. Heat transfer is faster than mass transfer in a metal, so itis more likely that the hardening effect of this treatment could penetrate to 0.2” thancould a carburizing treatment (2 pts). PE3 is probably the correct explanation (10 ptsmax).

d) (10 points) Subsequent examination under the optical microscope shows that the rod’s surfaceconsists of martensite, whereas the rod’s interior consists of ferrite, pearlite, and possibly bainite.Which of the three possible explanations now seems most plausible? Justify your answer.

Martensite in steels is typically obtained only by quenching (2 pts), and bainite byfairly rapid cooling (2 pts) (or heating at temperatures below the nose of the TTTcurve — an observation worth 1 pt, though it’s not directly relevant here). Wheremartensite and bainite did not form, the microstructure of a steel like this onecontaining 0.3 wt% C would consist of (proeutectoid) ferrite (2 pts) and pearlite(2 pts). The microstructure therefore supports PE3 (4 pts) as the most likelyexplanation for the hardness profile (10 pts max).





3) A recent article in the Observer noted that the glaze on an ancient Chinese statue is cracked. A glaze is athin layer of glass on the surface of a thicker polycrystalline ceramic piece. During firing, the glaze meltsand flows, coating the ceramic evenly.

a) (16 points) Explain how thermal stresses that arise during heat treatment of the glazed piece couldhave caused the cracks. Give the relevant equation for thermal stress in the glaze; define all terms andgive units.

If the thermal expansion coefficients of the glaze ( g) and the ceramic ( c) aredifferent, they will contract at different rates during cooling (1 pt). Because the glazeis much thinner than the ceramic but bonded to it, the ceramic will impose that part ofits dimensional change that is parallel to the glaze-ceramic interface on the glaze(1 pt). As long as the glaze is fluid, it will flow to accommodate the dimensional change(1 pt), but as it becomes rigid during cooling, the imposed strain will lead to stress inthe glaze (1 pt). If ag<ac, the stress in the glaze will be compressive (1 pt); if g> c,the stress in the glaze will be tensile (1 pt). If the stress exceeds the glaze’s fracturestrength (either in compression or in tension), the glaze will crack (1 pt). (The glaze’sfracture strength is probably lower than the shear strength of the glaze-ceramicinterface, or else the glaze would debond from the ceramic rather than crack internally(1 pt).) (5 pts max.)

The relevant equation is: σg = Eg(αc–αg)∆T (3 pts), where:

σg is the stress in the glaze (Pa); Eg is the Young’s modulus of the glaze (Pa); αg and αcare the linear thermal expansion coefficients of the glaze and the ceramic (m m–1 K–1),respectively; and ∆T = T f – Ti is the difference between the final and initialtemperatures (in K) over which stress is generated (1 pt each definition, 1 pt eachunits, 8 pts total).

b) (5 points) When will most of the stress develop: when the glaze is above its glass transitiontemperature Tg, or below Tg? Explain.

Because the glaze can still flow above Tg (2 pts), little stress will develop until theglaze it is cooled below Tg, (2 pts) i.e. when it becomes rigid (2 pts; 5 pts max).

c) (12 points) Which of the following options (keeping everything else constant) would have reduced theprobability that the glaze would crack? (Check all that apply.) (3 pts per correct check/uncheck)

√ Choosing a glaze with a thermalexpansion coefficient closer to that ofthe ceramic.

Choosing a glaze with a higher Young’smodulus. Lower glaze modulus woulddecrease the likelihood of cracks.

√ Choosing a glaze with a lower Tg. Choosing a ceramic with a higher

Young’s modulus. No effect.



4) (40 points) The eight types of corrosion are: uniform attack, intergranular corrosion, galvanic corrosion,selective leaching, crevice corrosion, erosion-corrosion, pitting, and stress corrosion . Briefly explain thereasons for the following corrosion phenomena and identify which of the eight types of corrosion is chieflyresponsible for each.

a) A piece of magnesium, in electrical contact with steel pipe and buried underground with it, inhibits thecorrosion of the pipe.

When connected in this way, magnesium and steel form a galvanic couple (3 pts).Because magnesium is below iron in the standard emf series, magnesium will be theanode (3 pts). This is a case of exploiting galvanic corrosion (3 pts; 8 pts max) of themagnesium to protect the steel.

b) The steel in a heavily traveled bridge corrodes faster than that in a lightly traveled one.

The bridge will be repeatedly loaded and unloaded as traffic moves over it. Anycorrosion being experienced by the steel will be exacerbated by the stress (2 pts),because of the increased energy stored in elastic deformation (2 pts) and indislocations during plastic deformation (2 pts). This is a case of stress corrosion (3pts; 8 pts max).

c) Corrosion is observed to occur preferentially under steel washers on steel that has experiencedprolonged exposure to water vapor.

The moisture under the washers will become stagnant and depleted in oxygen (2 pts)because of its limited exposure to the surrounding atmosphere (1 pt). This sets upan oxygen concentration cell with the metal under the oxygen-depleted liquid beingthe anode (3 pts). This is a case of crevice corrosion (3 pts; 8 pts max).

continued (and with more space) on the next page



d) Exhaust gases that carry soot cause corrosion faster than soot-free exhaust gases.

Corrosion by gases is worsened by the additional kinetic energy of the gas if it isflowing (2 pts). If the gas carries solid particles like soot, there is additionalmechanical attack from the solid particles (2 pts), including the potential to wear awayany protective oxide that may be present (2 pts). This is a case of erosion corrosion (3pts; 8 pts max).

There could also be erosion at work here, if we consider abrasion of the exposedsurfaces by the passing traffic (2-4 pts).

e) Copper gutters on old buildings are covered with green copper salts.

This is a case of continuous deposits (2 pts) forming over surfaces that are exposed tounchanging, uniform environments such as acid rain and polluted air (2 pts each). This isa case of uniform attack (3 pts; 8 pts max).



5) a) (12 points) Sketch the mer structures for the following polymers:

i) polyethylene ii) polypropylene iii) polyvinyl chloride iv) polystyrene

–C–C–H

H

H

H–C–C–H

H

CH

H–C–C–

H

H

Cl

H

–C–C–H

H

H

Mechanical properties for these four polymers at room temperature are listed below.

tensilemodulus, GPa

tensilestrength, MPa

yield strength,MPa

elongation atbreak, %

polyethylene, high-density 1.1 22-31 26-33 10-1200

polypropylene 1.1-1.6 31-41 31-38 100-600

polyvinyl chloride 2.4-4.1 41-52 41-45 40-80

polystyrene 2.3-3.3 36-52 — 1.2-2.5

b) (12 points) The glass transition temperatures Tg for these polymers (in no particular order) are 100,–90, 87, and –18 °C. Considering the values of modulus, strength, and elongation, match each polymerwith its Tg . Justify your answers.

The farther a polymer is above its Tg, the lower its strength and the higher itselongation; vice versa for a polymer below its Tg.

HDPE: –90 °C (highest elongation, lowest strength and modulus) (3 pts)PP: –18 °C (high elongation, low strength and modulus) (3 pts)PVC: 87 °C (low elongation, high strength and modulus) (3 pts)PS: 100 °C (lowest elongation, high strength and modulus) (3 pts)

c) (12 points) Further support your assignment of Tg for these polymers on the basis of their structures.

Given the same degree of linearity/crosslinking (all of these are linear polymers —2 pts) and the same type of chain (here, straight -C-C-C-; 2 pts), simple merstructures usually lead to low Tg (2 pts) and bulkier side groups tend to raise Tg (2pts). All of this indicates that Tg should increases from HDPE to PP to PVC to PS(4 pts).


3



d) (10 points) Many consumer products (shampoos, lotions, medicines) come in containers of high-density polyethylene with caps of polypropylene. Justify these material selections on the basis of themechanical properties of these two polymers.

PP is more rigid (1 pt) and strong (1 pt) — appropriate for the part of the packagethat must grip the opening (1 pt) and seal tightly (1 pt). HDPE is more flexible (1 pt)and ductile (1 pt), making it easy both to squeeze (1 pt) and to fabricate into complexshapes (1 pt). With the Tg of both polymers below room temperature (1 pt), they willbe ductile enough to be tough (1 pt) and not brittle (1 pt).

6) (21 points) Pick one of the Materials in the News topics that was presented in class or posted to the website this semester or previously (not one that you handed in). Discuss how that topic illustrates one or moreof the legs of the “Materials Tetrahedron”: processing-performance, structure-performance, orproperties-performance.



Name: S O L U T I O N

1 /70 2 /40 3 /33 4 /40 5 /46 6 /21 TOTAL: /250

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