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Unformatted text preview: EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 Name: SOLUTION Final Exam — 180 minutes; 300 points; 10 questions; 12 pages; 30% of course
This exam is to be taken closed-book. You may use a calculator, pencils or pens, eraser, and a straight
edge. Handwritten notes on both sides of a single 8.5″×11″ sheet of paper are permitted. Please turn in
your note page with your test. No other form of stored information is permitted. Write all answers on
these pages; use the back if more space is needed.
Wiedemann-Franz constant, £ = π2kB2/(3qe2) = 2.445×10-8 W Ω K-2
Boltzmann’s constant, kB = 1.381×10-23 J K-1 = 8.620×10-5 eV K-1 gas constant, R: 8.314 J mol-1 K-1
Charge on an electron, qe = –1.602×10-19 C
Avogadro’s number, NA = 6.023×1023 mol-1
Faraday’s constant, Y = 96,500 C mol
1) The unit cell of a polytype of silicon carbide called 2H SiC is shown below. 2a/3 a) (4 points) What is the Bravais lattice of 2Η SiC? a/3 simple hexagonal (2 pts per adjective)
b) (3 points) What is the name of this structure?
hexagonal zinc sulfide, or wurtzite c c) (9 points) On the sketch at left, draw one
direction in each of the following families:
i) <1 1 2 0> short face diagonal in basal
plane (3 pts) 120
a a ii) <1 1 2 1> short body diagonal(3 pts)
iii) <1 1 0 1> long body diagonal (3 pts) d) (10 points) In the space at right, sketch
the (1 1 2 0) plane. Label the atoms
(Si or C) in your sketch and show
where they touch each other. C iv) See sketch at right. It doesn't
matter which element is assigned
to black or white, as long as they
are distinct as shown Si e) (4 points) How many atoms of Si and C are there per unit cell? Si: 2 C: 2 f) (12 points) In the space at the top of the next page, compute the density of 2H SiC, given that
the lattice parameters are a = 0.3076 nm and c = 0.5048 nm, and the atomic weights are 12.011
for C and 28.09 for Si.) density = (FW × Z)/(NA × Vcell) (2 pts) where FW is the formula weight, Z is the
Department of Materials Science and Engineering 1 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 number of formula units per unit cell, NA is Avogadro's number, and Vcell is the
volume per unit cell FW = 12.011 + 28.09 = 40.10 g/mol (1 pt). Z = 2 (from part e)
Vcell = base × height (2 pts); height = c (1 pt)
base = 3a 2
2 density = 40.10( g / mol ) × 2 × 10 21 (nm 3 / cm 3 )
= 3.219( g / cm 3 )
× 0.3076 2 ⎥ (nm 3 )
6.023 × 10 23 mol –1 × ⎢0.5048 ×
⎦ (2 pts + 1 pt units) 2) The graph below shows the electrical and thermal conductivity of several metals (elements and
alloys) at 20°C. The solid line represents the Wiedemann-Franz law.
Ag 400 Ag-3Cu Cu Ag-20Cu Thermal
(W K-1 m-1) 300 κ
= T CWFL
Brass (Cu-30Zn) Ni
Hg 100 0 0 10
6 -1 -1
Electrical conductivity, σ × 10 Ω m 70 a) (7 points) The conductivities of the two copper-silver alloys (Ag-3%Cu and Ag-20%Cu) are
lower than those of either silver or copper alone. Explain. Is this result primarily an effect of
carrier concentration, mobility, or both?
Ag atoms in Cu, or Cu atoms in Ag, act as scatterers of electrons (impurity
scattering) (2 pts), regardless of their own electrical properties in the pure state
(1 pts). The lower conductivity of the alloy is primarily due to reduced electronic
mobility (2 pts); copper and silver will have nearly the same number of free
electrons per atom (2 pts).
continued on next page Department of Materials Science and Engineering 2 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 2) b) (10 points) Consider bronze and brass, two alloys of copper. Bronze has lower conductivity than
brass, despite having a lower concentration of alloying element (5% Sn) than does brass (30%
Zn). Explain. (The atomic radii of Cu, Zn, and Sn are 0.128, 0.133, and 0.151 nm, respectively.) Impurities will cause strain in the lattice (2 pts).
rSn > rZn ≅ rCu ⇒ Sn causes more strain in copper than does Zn (3 pts).
The greater the strain, the greater the scattering of electrons (3 pts). ∴Atom for atom, Sn decreases the conductivities of Cu more than does Zn (2 pts).
Assume that the data shown are for annealed metals.
c) (5 points) How would plastic deformation of these materials affect their conductivity? Explain. Plastic deformation of metals increases the concentration of dislocations (2 pts).
Dislocations scatter electrons (1 pt), which would lower both the thermal and the
electrical conductivities (2 pts). d) (6 points) Would the Wiedemann-Franz law hold for plastically deformed metals? Explain.
Yes (2 pts). The W-F law presumes that both heat and current are carried by the
same carrier (2 pts). Plastic deformation (i.e. increased dislocation density) in
metals will affect electrical conductivity the same way as it affects thermal
conductivity, because both are primarily due to transport of electrons in metals
(2 pts). e) (6 points) Does the Wiedemann-Franz law hold for semiconductors and insulators? Explain.
No (2 pts). In semiconductors and insulators, which have low concentrations of
free electrons, most of the thermal conduction is by phonons, while electrical
conduction is by electrons and/or holes in semiconductors, and by electrons, holes,
or ions in insulators (3 pts). Thus the basis of the WFL law (that one carrier
carries both current and heat) doesn't apply (1 pt). Department of Materials Science and Engineering 3 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 3) One half of an electrochemical cell consists of a pure nickel electrode immersed in a solution of Ni2+
ions; the other half is a pure cobalt electrode in a solution containing Co2+ ions. A voltmeter is
connected between the two electrodes.
a) If the concentrations of the two solutions are both 1 M and the temperature is 25 °C,
i) (4 points) write the spontaneous overall reaction; Ni2+ + Co → Ni + Co2+
ii) (4 points) calculate the voltage generated; –0.250 V – (–0.277 V) = 0.027 V (3 pts +
1 pt units)
iii) (4 points) identify which metal is the cathode (Ni) and which metal is the anode (Co).
(2 pts each)
The standard electrode potentials for nickel and cobalt are:
Ni2+ + 2e– → Ni
Co2+ + 2e– → Co V° = –0.250 V
V° = –0.277 V The metal with the more negative standard electrode potential (in this case, Co)
will be oxidized, and will therefore be the anode. The other metal (in this case, Ni),
will be reduced and will therefore be the cathode (up to 4 pts partial credit). b) If the Ni2+ and Co2+ concentrations are 1.0×10–3 M and 0.5 M respectively at 25 °C,
i) (4 points) write the spontaneous overall reaction;
ii) (7 points) calculate the voltage generated.
Calculating the voltage will enable us to determine which direction is spontaneous
for the electrochemical reaction. Use the Nernst equation (2 pts):
RT ⎡ M 1 ⎤
² V = (V2 ° – V1 °) –
nF ⎡ M 2 ⎦
⎣ Note that, even though Co is more anodic than Ni in a standard cell, the
concentrations here will favor the corrosion of the Ni (3 pts partial credit). Let’s
guess that Ni will be oxidized (i.e., that it will be metal 1 and Co will be metal 2).
Making the appropriate substitutions (3 pts) gives: 8.314( J • mol –1 • K –1 ) × 298( K ) [0.001( M )]
² V = ⎡–0.277 – (–0.250 )⎤ –
= 0.053(V ) (2 pts)
2 × 96, 500(C • mol –1 )
[0.5( M )]
The positive voltage indicates that we guessed right. Therefore, the spontaneous
reaction is the reverse of that for part a), i.e. Ni + Co2+ → Ni2+ + Co (4 pts).
continued on next page
Department of Materials Science and Engineering 4 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 3) c) Starting with the conditions given in part b), state whether or not the voltage could be brought
closer to zero if the following changes were made. Briefly justify your answer:
i) (6 points) raising the temperature of the cell;
Raising the temperature of the cell will increase the magnitude of the second term
in the Nernst equation (2 pts). That will make the voltage more positive (2 pts), i.e.
the corrosion of Ni will be even more spontaneous, the cell will be farther from
equilibrium, and the voltage will be further from zero (2 pts). ii) (6 points) raising the concentration of the Ni2+ solution.
Raising the concentration of the nickel solution will alleviate the concentration
difference that is driving the nickel to corrode. It will decrease the magnitude of
the second term in the Nernst equation (2 pts). That will make the voltage less
positive (2 pts), i.e. the corrosion of Ni will be even less spontaneous, the cell will
be closer to equilibrium, and the voltage will be closer to zero (2 pts). 4) When the World Trade Center was attacked on 11 September 2001, the maximum temperature
reached in the building is believed to have been about 1000 °C.
a) (5 points) If the steels in the WTC were plain carbon steels, could they have melted (partially or
completely) under the conditions that are believed to have occurred on 9/11? Explain briefly.
In the whole range of
relevant to steels (01.2 wt% C), no liquid
forms at 1000 °C or
below (3 pts). In fact,
there is no liquid in
until the temperature
reaches about 1300 °C
(3 pts). Therefore it is
highly unlikely that the
steel in the WTC
melted on 9/11 (2 pts;
5 pts max). continued on next page Department of Materials Science and Engineering 5 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 4) b) (10 points) Estimate the thermal stress in steel constrained from expanding on heating from
25 °C to 1000 °C under each of two different approximations:
i) assuming that the room-temperature property values given below can be used throughout the
temperature range of interest;
elastic modulus, GPa
coefficient of linear
thermal expansion, °C–1
Poisson’s ratio 207
0.30 Use σ = Eα(T0 – Tf) (3 pts). With numbers:
207 (GPa) × 11.7×10–6 (°C–1) × (1000 – 25) (°C) = 2.36 GPa (a huge stress!) (3 pts +
1 pt units)
(Assuming that this is a case of uniaxial stress, Poisson’s ratio was not needed to
answer the problem. The factor of 1/(1–ν) from the planar film solution in the
overheads applies to biaxial stress. If you used it, no points were taken off.) ii) using an “average” elastic modulus over the temperature range that is 70% of the roomtemperature value.
Use the same formula, but with 0.70×207 GPa for the modulus, giving 1.6 GPa —
still huge. (2 pts + 1 pt units). c) (5 points) At 1000 °C, the strength of steels is approximately one-tenth of its value at room
temperature. Could the steel in the WTC have yielded or failed at 1000 °C under the stresses
computed in part b)? Explain briefly. For reference, the room-temperature yield strengths and
ultimate tensile strengths of two steels are given below. 1020, annealed
and tempered yield strength, tensile strength,
1760 With strengths reduced to 10% of their room temperature values (29.5 MPa yield
and 39.5 MPa uts for 1020, 162 MPa yield and 176 MPa uts for 4340), both of
these steels would have failed utterly (3 pts), as the stresses computed in part b)
are 10 to 100 times the strengths of the alloys (2 pts). d) Briefly discuss how the following factors may have contributed to the collapse of the WTC.
(Continue into the space at the top of the next page.)
i) (7 points) Oxidation of the steel at elevated temperatures
ii) (6 points) Temperature gradients in the concrete parts of the structure
Oxidation of metals proceeds much more quickly at elevated temperatures (3 pts),
with the enhanced kinetics outweighing the decreased thermodynamic driving
force relative to room-temperature oxidation. Furthermore, the oxide that forms
Department of Materials Science and Engineering 6 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 on carbon steels is unprotective (3 pts), so the kinetics will be linear in time and
therefore faster than the parabolic kinetics of a protective oxide. If the oxide
adheres to the metal, it will be brittle and more likely to crack under stress, with a
probable net weakening effect on the member (2 pts). In the more likely case that
the oxide spalls off, the member will become reduced in cross-sectional dimensions,
which also would effectively weaken it (2 pts; 7 points max).
(In the immediate vicinity of the burning fuel and other combustibles, there was
probably little oxygen available to oxidize metal; but the heat undoubtedly spread
in all directions, to spaces where there was plenty of air to oxidize hot metal.)
Under the non-uniform heating conditions and high temperatures that prevailed on
9/11, temperature gradients were probably large and ubiquitous (2 pts). Such
gradients would have caused large thermal stresses. Concrete is brittle like a
ceramic (2 pts), so thermal stresses that put concrete in tension would have caused
cracking (2 pts). This would occur, for example, in concrete surrounding a steel
reinforcing bar that is conducting heat from a hotter part of the fire (2 pts). The
expanding steel will put the concrete in tension at the interface between the two
materials, which could fracture the concrete (6 pts max). 5) (20 points) Pick one of the Materials in the News topics that was presented in class or posted to the
web site this semester or previously (not one that you handed in). Discuss how that topic illustrates
one or more of the legs of the “Materials Tetrahedron”: processing-performance, structureperformance, or properties-performance. Support your point with specific, quantitative information. Department of Materials Science and Engineering 7 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 6) Gaseous hydrogen at a constant pressure of 1.013 MPa (10.00 atm) is to flow within the inside of a
thin-walled cylindrical tube of nickel that has a radius of 0.100 m. The temperature of the tube is to
be 300 °C and the pressure of hydrogen outside of the tube will be maintained at 0.01013 MPa
a) (16 points) If the diffusion flux is to be no greater than 1.00×10–7 mol m–2 s–1, calculate the
minimum wall thickness ∆x. The concentration of hydrogen in the nickel, CH (in moles of
hydrogen per m3 of nickel) is a function of hydrogen pressure, PH2 (in MPa) and absolute
temperature T according to ⎛ 12.3kJ / mol ⎞
C H = 30.8 PH 2 exp ⎜ –
Furthermore, the diffusion coefficient DH for the diffusion of hydrogen in nickel depends on
temperature as ⎛ 39.56 kJ / mol ⎞
DH ⎡m 2 s –1 ⎤ = 4.76 × 10 –7 exp ⎜ –
Start with Fick’s first law, noting that the wall thickness ∆x is the denominator in
the concentration gradient, and rearranging to give ∆x as a function of DH, CH, and
the flux J:
Δx = − D ΔC
J Computing the diffusion coefficient from the given relationship:
39, 560( J / mol )
DH = 4.76 × 10 –7 exp ⎜ –
⎟ = 1.18 × 10 ( m • s ) (4 pts)
⎝ 8.314( J • mol • K )573( K ) ⎠ Computing the concentration difference across the wall from the given relationship
(4 pts): ² C H = 30.8 × ( 1.013 – ⎛
12, 300 J / mol
0.01013 exp ⎜ –
⎟ = 2.11( mol • m )
⎝ 8.314( J • mol • K ) × 573( K ) ⎠ ) Inserting values, with J = 1.00×10–7 mol m–2 s–1, gives (3 pts + 1 pt units)
Δx = − 1.18 × 10 –10 ( m 2 • s –1 ) × 2.11( mol • m –3 )
= 2.49 × 10 –3 ( m ) = 2.49( mm )
1.00 × 10 ( mol • m • s ) continued on next page
Department of Materials Science and Engineering 8 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 6) b) (5 points) For thin-walled cylindrical tubes that are pressurized, the circumferential stress σ
varies with the pressure difference ∆P across the wall, cylinder radius r, and wall thickness ∆x as σ= r² p
4² x Compute the circumferential stress to which the walls of this pressurized cylinder are subjected.
Inserting numbers into the given relationship: σ= 0.100( m ) × (1.013 – 0.01013)( MPa )
= 10.1( MPa ) (4 pts + 1 pt units)
4 × 2.49 × 10 –3 ( m ) c) (6 points) The yield strength of nickel at room temperature is 100 MPa and diminishes by
1.0 MPa for every 10 °C rise in temperature. With the wall thickness computed in part a) and the
stress computed in part b), will the cylinder experience yielding? Show your work. The yield strength at 300 °C will be 100 – 300 − 25
( MPa ) = 72.5( MPa ) (2 pts + 1 pt units)
10 This is much larger than the stress computed in part b) ⇒ no, the tube will not
yield (3 pts).
d) (13 points) If the cylinder will not experience yielding under these conditions, compute the
minimum thickness that would prevent yielding of the tube walls, and compute the
corresponding increase in the diffusive flux.
On the other hand, if the cylinder will experience yielding under these conditions, compute the
minimum thickness that will prevent yielding, and compute the corresponding decrease in
diffusive flux. The tube will not yield when the wall is 2.5 mm thick, so we want to see how thin it
must be to yield under the given conditions. Set the circumferential stress equal to
the yield strength at 300 °C (3 pts) and compute the wall thickness: ² x= r ² p 0.100( m ) × (1.013 – 0.01013)( MPa )
= 0.346 mm (3 pts + 1 pt units)
4 × 72.5( MPa ) (In practice, one would build in a safety factor of perhaps 100%, so a minimum wall
thickness of 0.70 mm would be selected.)
The increased flux at the minimum thickness is computed from Fick’s first law
(2 pts): J =− D ΔC 1.18 × 10 –10 ( m 2 • s –1 ) × 2.11( mol • m –3 )
= 7.20 × 10 –7 ( mol • m –2 • s –1 )
0.000346( m )
(3 pts + 1 pt units) That is, the wall is one-seventh as thick, so the flux is seven times greater. Department of Materials Science and Engineering 9 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 7) a) (16 points) In the space below right, sketch the volume-vs.-temperature relationship (6 pts) for
a liquid that can be cooled to the glassy state without crystallizing. On your sketch, indicate the
following properties or phenomena:
fast cooling i) the equilibrium melting point Tm of the
substance (2 pts)
ii) the volume-vs.-temperature
relationship for the substance when
fully crystallized (2 pts) Liquid Volume 睼m Glass iii) the glass transition range for rapid
cooling and for slow cooling (2 pts
iv) features on the sketch that are proporexpansion
tional to the volume thermal expansion
coefficients of the liquid and the glass
(1 pt each) Glass transition range,
Crystal Temperature Tm b) (16 points) Based on the stress-strain
curves of polymethylmethacrylate
(PMMA) shown at right, estimate its
glass transition temperature.
Qualitatively justify your answer on
the basis of the temperature
dependence of at least three of the
mechanical properties illustrated in
Variations in mechanical
properties that indicate a glass
transition are exhibited between
20 °C and 4 °C:
1) The strain to failure drops
from several percent to ~1%
2) Elastic modulus (as indicated by the initial slope of the stress-strain curve)
appears to double (4 pts)
3) Yield strength nearly doubles. (It’s not clear whether the curve at 4 °C is cut
off at the top of the graph.) (4 pts)
All of these changes are consistent with the onset of stiffer, stronger, and
brittler — i.e., glassy — behavior. We would estimate the glass transition
temperature to be around 4 °C (4 pts). (Callister lists Tg of 3 °C for PMMA.) c) (8 points) Compute the modulus of toughness of PMMA at 50 °C and at 20 °C.
Reasonable estimates for total area under the stress-strain curve are at 20 °C, 53
MPa × 0.04 × ½ = 1.1 MPa; and at 50 °C, 22 MPa × 0.25 = 5.5 MPa. Clearly, the
toughness decreases significantly as temperature drops toward Tg — another
hallmark of the glass transition. Department of Materials Science and Engineering 10 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering 10 May 2006 8) The figure at right shows the variation
of carrier concentration with
temperature in semiconductors. On the
sketch, indicate the regions that
illustrate the following phenomena:
a) (3 points) ionization of dopants
b) (3 points) exhaustion of donors or
saturation of acceptors
c) (3 points) intrinsic conductivity
e) (5 points) how the curve would look
if the dopant concentration were
lower lower dopant level ionization d) (5 points) how to compute the value
of the band gap energy of the
semiconductor 9) Foamed polystyrene (Styrofoam®) is an excellent example of a material that was designed to
achieve a desirable combination of properties: extremely low thermal conductivity, low density, and
reasonable structural rigidity. Discuss how each of the following aspects of its structure contributes
to the indicated properties.
a) (7 points) glassy structure ⇒ low thermal conductivity
The mean free path δph for scattering of phonons is on the order of a few atomic
distances in a noncrystalline solid (3 pts), essentially as small as it can be in a solid.
Because the phonon thermal conductivity varies linearly with δph (2 pts), the
phonon thermal conductivity of a glassy materials are the lowest of all solids
You were not asked to discuss it, but it is relevant that polystyrene is an electrical
insulator: this prevents thermal conduction by electrons from contributing
appreciably to its total thermal conductivity (up to 4 pts extra credit). b) (7 points) closed porosity ⇒ low thermal conductivity and low density
The thermal conductivity of a trapped gas is even lower than that of a glassy solid
(3 pts). By replacing a large fraction of the volume of the material with sealed gasfilled pores (the “bubbles” in the polymer foam are not strong enough to contain a
vacuum), the thermal conductivity of the material is made to be lower still (2 pts).
Because the gas has virtually zero density compared to that of the polymer, this
drastically reduces the overall density of the material (2 pts). continued on next page
Department of Materials Science and Engineering 11 of 12 Case Western Reserve University EMSE 201 — Introduction to Materials Science and Engineering Name: SOLUTION 10 May 2006 9) c) (7 points) glassy structure with Tg = 100 °C ⇒ structural rigidity at room temperature
At room temperature, polystyrene is below its glass transition temperature (2 pts).
This makes it relatively rigid, resilient, and brittle (2 pts). Without these
characteristics, the foamed polymer would be too soft to retain its shape under light
loading and too pliable to spring back after the load is removed (3 pts). (No one would
call styrofoam a “hard” material, but foams of e.g. low-density polyethylene, with a Tg
of –110 °C, are much softer than styrofoam. 10) The figure below shows hardness indentations in a steel part that has been case-hardened (i.e., given
a surface carburization).
a) (10 points) Name and describe the hardness test that was
used to make these indentations.
These indentations were made with a Knoop indenter
(3 pts), an elongated diamond pyramid (2 pts)
microindenter. In this test, the user can select the
load (2 pts) with which the indenter is pressed into
the specimen, as well as the microscopic region of the
specimen (1 pt) with a resolution of several
micrometers (impossible with Rockwell and Brinell
tests) (2 pts). The elongated indenter is designed for
use on thin features such as surface layers or fibers
(3 pts). The hardness is computed from the load and
from the long dimension of the indentation (3 pts, 10
pts max.). b) (6 points) Which part of the picture (top or bottom) shows the case-hardened part of the steel?
Justify your answer.
The top indentation is smaller (1 pt), which indicates that the material there is harder (assuming
the load was the same for both indentations) (1 pt). Carburization increases the carbon content
of the surface of a steel (1 pt), which will increase its hardness (1 pt). Therefore, one can
conclude that the top part of the micrograph shows the case-hardened region (2 pts). 1 /42 2 /34 3 /35 4 8 /19 9 /21 10 /16 Department of Materials Science and Engineering /33 5 /20 6
TOTAL: 12 of 12 /40 7 /40
/300 Case Western Reserve University ...
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