Midterm02PracticeBinder

Midterm02PracticeBinder - Professor R Gronsky UNIVERSITY OF...

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Unformatted text preview: Professor R. Gronsky UNIVERSITY OF CALIFORNIA College of Engineering Department of Materials Science & Engineering Fall Semester, 2005 Engineering 45 Midterm 02 N ame : ( P le a s e p r i n t) 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 M id te r m 02 In i t i a l s: 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. 2. Defects in Solids (10 points) For this problem you must draw and label all requested features directly on the figures provided. (a) ☐ ☐ ☐ Point defects in solids include vacancies dislocations both (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) 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 (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. (d) Dislocations in solids ☐ participate in plastic deformation ☐ decrease the entropy of a material ☐ both (e) Fick’s First Law for diffusion flux ∂c Jx = −D ∂x expresses the fact that ☐ diffusion requires a negative diffusivity ☐ mass flows down a concentration gradient ☐ both p ag e 2 o f 7 E 45 M id te r m 02 In i t i a l s: 3. Phases and Phase Equilibria (10 points) Refer to the Cu-Zn binary phase diagram below (from ASM Metals Handbook, 8th 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 !’ 903° 4. Phases and Phase Equilibria (10 points) These questions refer to the same Cu-Zn binary phase diagram from Problem 3. at % Zn 10 20 30 40 50 60 70 80 90 L 835° (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. ! " 456° 468° 700° 598° $ 558° # % 424° & 0 Cu 10 20 30 40 50 60 70 80 90 Zn wt % Zn (a) (2 points) What is the maximum concentration of Zn that can be dissolved in α brass? (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. (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. (c) (2 points) What phase(s) is (are) in equilibrium when β brass (50:50 composition) is held at 500°C. (d) (2 points) Write the reaction that occurs on cooling through the 700°C peritectic isotherm. (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? p ag e 3 o f 7 E 45 M id te r m 02 In i t i a l s: 5. Kinetics (10 points) The following phase diagram is from the ASM Metals Handbook, 8th edition, Vol. 8, (1973). T (°C) 1600 1500 1538° 1495° # 1400 1394° 1300 1200 1100 1000 900 800 700 0.02 0.77 " 600 500 400 300 200 100 0 Fe 1 2 3 4 wt % C 5 6 Fe3C 910° 727° 6.67 Cementite ! 2.11 1148° 4.30 6. Kinetics (10 points) Mark ☒ the ballot box corresponding to the best answer. Two (+2) points for correct answers, -1 if wrong, 0 if blank. (a) The temperature at which a heavily cold worked Cu alloy recrystallizes ☐ decreases as the amount of cold work increases ☐ decreases as grain size increases ☐ both (b) The driving force for grain growth during annealing is ☐ reduction in strain energy ☐ reduction in surface energy ☐ both (c) When metallic alloys are quenched following a homogenization treatment in a single phase region of the phase diagram ☐ a supersaturated solution solution results ☐ the alloy is ready for aging ☐ both T (°C) 1100 1000 900 800 700 600 500 400 300 " 456° 468° 903° L 1227° 6.67 at % Zn 10 20 30 40 50 60 70 80 90 L 835° (a) (5 points) The following TTT diagram describes the isothermal kinetics of a 1020 steel. Show directly on the plot a thermal treatment that begins with 100% austenite, and ends with 100% bainite. T (°C) 1100 1000 900 800 700 600 500 400 300 200 100 0 0.1 Ms 1 sec 1 min 1 hour 1 day " 727°C Coarse pearlite " + Fe3C ! # 700° 598° $ 558° % 424° & 200 ! + Fe3C 100 !’ "+ !+ Fine pearlite Bainite Fe 3C 0 Cu 10 20 30 40 50 60 70 80 90 Zn wt % Zn (d) Brass can be precipitation hardened if it has the following composition ☐ 35 wt% Zn ☐ 70 wt% Zn ☐ both (e) ☐ ☐ ☐ Glass ceramics are aged to relieve internal stress in the glass precipitate a crystalline phase both 1 10 102 103 time (seconds) 104 105 (b) (5 points) Will a quench rate of 1000°C/sec convert a sample with 100% austenite to 100% martensite? Show why or why not on the same TTT diagram above. p ag e 4 o f 7 E 45 M id te r m 02 In i t i a l s: 7. Metallic Alloys (10 points) 8. Metallic Alloys (10 points) Temper H1 H2 T3 T6 T9 Strain-hardened Definition Strengthened by Weakened by Porosity (casting) Strain-hardened, partially annealed Solution-treated, cold-worked, naturally aged Solution-treated, artificially aged Solution-treated, artificially-aged, cold worked Cold working Alloying Phase Transformations Annealing Welding Phase Transformations 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, 9th 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. _____ 2. _____ 3. _____ 4. _____ 5. _____ 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. p ag e 5 o f 7 E 45 M id te r m 02 In i t i a l s: 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 Mg2+, 4 Al3+ and 8 O2-. 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. 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 Tg Ts T 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. p ag e 6 o f 7 E 45 M id te r m 02 In i t i a l s: Worksheet No points will be given or deducted for work shown here. Please enter answers in the spaces provided. DO NOT WRITE BELOW THIS LINE Pro ble m # 1 2 3 4 5 6 7 8 9 10 To t a l Po s s i ble Po i n t s 10 10 10 10 10 10 10 10 10 10 100 Yo u r S c o re p ag e 7 o f 7 Name: (Please print) UNIVERSITY OF CALIFORNIA College of Engineering Department of Materials Science & Engineering Professor R. Gronsky Fall Semester 2007 Engineering 45 Midterm 02 INSTRUCTIONS DO NOT OPEN until “START” is announced. 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 (disable all sounds) and iPods. NO DISRUPTION rule.......... Questions cause too much of a disturbance to classmates in the room, so there will be no questions during the exam. Instead, write any concerns or alternative interpretations in the margins adjacent to your answers. PROFESSIONAL protocol.... Engineers do not cheat on the job. There will be no cheating on this exam. E45 Midterm 02 Fall 2007 Initials:__________ 1. Defects in Solids (20 points) The following figure is an atomic-level representation of a few dozen unit cells in an elemental metal that has been subjected to substantial cold work. (a) The figure contains five (5) defects that appear as isolated vacancies, and five more than appear as Frenkel pairs. (5 points) Locate and accurately label (V for isolated vacancy, P for Frenkel pair) all of them. page 2 of 12 E45 Midterm 02 Fall 2007 Initials:__________ (b) (5 points) Locate and accurately label the extra half plane associated with an edge dislocation in this figure. (c) (5 points) Trace a Burgers circuit in the finish-to-start-right-hand (FSRH) convention and label the Burgers vector associated with this edge dislocation. (d) (5 points) Locate and accurately label the candidate slip plane on which this dislocation glides, and indicate its direction of motion in that slip plane. page 3 of 12 E45 Midterm 02 Fall 2007 Initials:__________ 2. Phases and Phase Equilibria (20 points) Refer to the following phase diagram from the Metals Handbook, 8th Edition, Volume 8, American Society for Metals, Metals Park, Ohio (1973), to support your answers. °C 1800 1700 1600 1500 1400 1300 1200 1100 1000 900 800 700 Mo 10 20 30 40 50 at. % Ni 60 70 80 90 L 1455° 1364° µ ! 35.7 49.6 1320° 53.2 60.7 $ 911° # " 876° 10 20 30 40 50 60 wt. % Ni 70 80 90 Ni (a) Ni3Mo is an ordered intermetallic compound with an orthorhombic structure at room temperature. What single-phase field does it occupy in the binary Mo-Ni system? (5 points) Answer: _____________ (b) A 20% Ni alloy is slowly cooled through 1364°C. Specify the reaction that occurs at the peritectic isotherm. (5 points) Answer: _____________ page 4 of 12 E45 Midterm 02 Fall 2007 Initials:__________ (c) The phase diagram shows that when a 50 wt.% alloy is solidified, it crosses a eutectic reaction isotherm at 1320°C and a peritectoid reaction isotherm at 911°C. Assume that the cooling rate is slow enough to sustain equilibrium conditions throughout and sketch the microstructure of the alloy when it is fully cooled at room temperature. °C 1800 1700 1600 1500 1400 1300 1200 1100 1000 900 800 700 Mo 10 20 30 40 50 at. % Ni 60 70 80 90 L 1455° 1364° µ ! 35.7 49.6 1320° 53.2 60.7 $ 911° # " 876° 10 20 30 40 50 60 wt. % Ni 70 80 90 Ni (10 points) Show sketch here ↓ page 5 of 12 E45 Midterm 02 Fall 2007 Initials:__________ 3. Kinetics (20 points) The following transformation curve for a steel of eutectoid composition was determined experimentally by R.A. Grange and J.M. Keifer in their classic paper "Transformation of Austenite on Continuous Cooling and Its Relation to Transformation at Constant Temperature," Trans. Am Soc. Metals, 29, 85 (1941). T (°C) 800 700 600 500 400 300 200 100 0 727°C Pearlite start Pearlite finish 1 sec c 140°/se 3.3°/sec 1 min 1 hour 1 day 0.1 1 10 102 103 104 105 time (seconds) (a) Using Grange & Keifer's data, specify a continuous cooling rate for a fully austenitic sample that would generate a microstructure of 100% martensite in a eutectoid steel. (5 points) Answer: _____________ (b) If a sample of a eutectoid steel was instantaneously quenched from the eutectoid isotherm to 550°C, how long could the sample remain at this temperature before a 100% martensite product was unachievable? (5 points) Answer: _____________ page 6 of 12 E45 Midterm 02 Fall 2007 Initials:__________ (c) The following microstructure (magnification 700X) was obtained from another sample of a eutectoid steel that was continuously cooled from the eutectoid isotherm. Using Grange & Keifer's data, specify a continuous cooling rate for a fully austenitic sample that would generate this microstructure. (5 points) Answer: _____________ (d) Explain, citing the competing effects of driving force and diffusion kinetics, why microstructures generated at higher temperatures are spatially "coarse," while those generated at lower temperatures are spatially "fine." (5 points) Answer: page 7 of 12 E45 Midterm 02 Fall 2007 Initials:__________ 4. Metallic Alloys (20 points) Binary Al-Mg alloys in the 5XXX series undergo substantial strengthening with good ductility as a result of cold work, in addition to excellent corrosion resistance and weldability. However, these alloys are also known as "non-heat-treatable" alloys because they do not show significant precipitation hardening at dilute (< 7%) alloy concentrations. °C 700 600 500 400 300 660° L 35.0 451° 14.9 35.5 462° 59.8 67.7 437° 87.3 649° ! µ " 200 100 Al "’ # 10 20 30 40 50 60 wt. % Mg 70 80 90 Mg (a) Why does "cold work" result in substantial strengthening of the 5XXX series aluminum alloys? What causes the strength increase? (5 points)? Answer: (b) Using a dilute 5 wt% Mg alloy for illustration, why does precipitation hardening not work? Explain why precipitates cannot be produced, or if they can (tell how), why they do not result in "hardening." (5 points)? Answer: page 8 of 12 E45 Midterm 02 Fall 2007 Initials:__________ (c) The Ni-Ti binary alloy system is very popular in the medical device industry due to a peculiar property known as "shape-memory" that is exhibited by the stoichiometric NiTi intermetallic compound with an ordered cubic structure. °C 1700 10 20 30 40 at. % Ti 50 60 70 80 90 1600 1500 1455° 1400 1300 1200 1100 (Ni) 1000 900 800 700 600 Ni 10 20 30 40 12.6 1382° 11.5 1306° 21.4 1672° L 1312°, 44.9% 1120° 33.8 38.1 # 985° 45.9 62.1 63.4 71.6 $ ("-Ti) 943° 87.6 883° " 640° (!-Ti) 50 60 wt. % Ti 70 80 90 Ti How would you produce a solid ingot of pure γ phase? Be specific, citing both composition(s) and heat treatments. (5 points) Answer: I. (d) The Food and Drug Administration (FDA) has approved NiTi for use as vascular stents, to be implanted in the blood vessels of human patients suffering from atherosclerosis. How do you rationalize this as a safe practice, based upon the information presented in the phase diagram above? (5 points) Answer: page 9 of 12 E45 Midterm 02 Fall 2007 Initials:__________ 5. Ceramics & Glasses (20 points) One of the more successful examples of structural ceramic materials is based upon the ternary oxide system Li2O—Al2O3— SiO2 and sold under the trade name Corningware™ (Corning). The product is shaped by conventional glass-forming routes, and its superior properties are due to precipitates of either β-spodumene with stoichiometry Li2O•Al2O3•4SiO2 and very small thermal expansion coefficient, or βeucryptite with stoichiometry Li2O•Al2O3•SiO2 and a negative thermal expansion coefficient. Glass is known as a vitreous (Latin vitrum= "glass") material and the process by which these products derive their superior properties is known as "devitrification." (a) Explain what is meant by devitrification, and what thermal-mechanical processing steps are included to generate Corningware™ products. Include in your answer a schematic temperature-time plot to illustrate the sequencing of all steps in the process. (10 points) Answer: page 10 of 12 E45 Midterm 02 Fall 2007 Initials:__________ (b) Sintering (Greek sintar = "ash") is another method used for the production of structural ceramics, including orthopedic implants, dental crowns, pacemaker electrodes, high temperature combustion feedthroughs, and (pictured) porous filters to remove particulates from high temperature liquid or gaseous streams. Explain how the sintering process can be controlled to generate this product. In your answer, pay attention to ceramic powder sizing, compaction, and firing (temperature and time) during synthesis. (10 points) Answer: page 11 of 12 E45 Midterm 02 Fall 2007 Initials:__________ Worksheet Problem 1 2 3 4 5 Total Possible 20 20 20 20 20 100 Score page 12 of 12 Name: (Please print) UNIVERSITY OF CALIFORNIA College of Engineering Department of Materials Science & Engineering Professor R. Gronsky Fall Semester 2008 Engineering 45 Midterm 02 INSTRUCTIONS DO NOT OPEN until “START” is announced. SEATING during exam .............Be sure that seat directly in front of you is occupied and the seats to your left / right are vacant (except for border seats) CLOSED BOOK format ...........Allowed: writing instruments / eraser / straightedge Not allowed: books / reference materials / calculators / PDAs / cell phones (disable all sounds) / other electronic devices / headphones / ear buds. NO DISRUPTION rule .............Courtesy demands that there be no questions during the exam. Write concerns or alternative interpretations in exam margins. PROFESSIONAL protocol .......Engineers do not cheat on the job. There will be no cheating on this exam. E45 Midterm 02 Fall 2008 ! Initials:__________ 1.!Phase Diagrams (20 points) Industrial applications of refractories for thermal protection include a large quantity of ceramics based upon the MgO-Al2O3 ceramic alloy system. Using the grid below, construct the magnesia–alumina binary alloy phase diagram from the following experimental observations. All compositions are expressed in mole percentage. Be sure to label all single phase fields. • Pure magnesia melts at 2800°C. • Pure alumina melts at 2054°C. • Magnesia can dissolve up to 2% alumina, the maximum solubility occurring at 2000°C. The name given to this solid solution of alumina in magnesia is "periclase." • Magnesia has no solubility in alumina. • "Spinel" is the name normally associated with the compound MgAl2O4, but it is also the name of a phase in the magnesia-alumina binary system with solubility ranging from 40% alumina (at 2000°C) to 84% alumina (at 1800°C). The spinel phase with a composition of 50% alumina melts congruently at 2200°C. • At 2200°C, an 82% magnesia alloy is partially solidified, consisting of equal parts periclase and liquid. • At 2000°C, a 68% magnesia alloy transforms on cooling from a liquid phase to two solid phases. One of the solid phases is periclase with 2% alumina content and the other is spinel with 40% alumina content. • At 2000°C, a 73% alumina alloy is half liquid and half spinel. • At 1800°C, a 95% alumina alloy solidifies on cooling into two solid phases, one a spinel of 84% alumina, and the other, pure alumina. • At 1000°C, a 75% magnesia alloy has equal parts periclase, with a composition of 100% magnesia, and spinel, with a composition 50% magnesia. • At 1000°C, a 77% alumina alloy has equal parts alumina phase and spinel phase, the latter with a composition of 54% alumina. ! page 2 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 3000 2500 T (°C) 2000 1500 1000 MgO 10 20 30 40 50 60 70 80 90 Mole % Al2O3 Al2O3 ! page 3 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 2.!Kinetics (20 points) a. A 1080 steel is austenitized at 800°C for 4 hours in preparation for an isothermal treatment to develop a uniform microstructure of 100% fine pearlite. Plot directly on the TTT curve below the kinetic path of a treatment that would achieve the desired result within one hour. 1600 1500 1538° 1495° L 1227° 1148° 2.11 910° 727° 6.67 Cementite T (°C) 1400 1394° 1300 1200 1100 1000 T (°C) 4.30 6.67 900 800 900 800 700 600 500 400 300 200 100 Ms = 215°C 1 sec 1 min 1 hour 1 day + +F eC 3 700 0.02 0.77 600 500 400 300 200 100 0 Fe 1 2 3 4 wt % C 5 727°C Coarse pearlite + Fe3C Fine pearlite Bainite 6 Fe3C 0 0.1 1 10 102 103 time (seconds) 104 105 b. Now describe your heat treatment in succinct terms as though you were giving instructions to your furnace shop manager, explaining how to achieve each temperature-time segment of your kinetic path traced above. ! page 4 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 2. c. A fully austenitized 1020 steel is held isothermally at 600°C for 30 minutes, removed from the furnace, and allowed to air cool to room temperature over the next several hours. Describe the resulting microstructure in quantitative terms, citing the relative amounts of all constituents, and comment on its overall strength and ductility for structural applications. 1600 1500 1538° 1495° L 1227° 1148° 2.11 910° 727° 6.67 Cementite 4.30 6.67 1400 1394° 1300 1200 1100 1000 T (°C) 900 800 900 800 700 600 727°C + Coarse pearlite 700 0.02 0.77 600 500 400 300 200 100 0 Fe 1 2 3 4 wt % C 5 T (°C) 500 400 300 200 100 0 Ms + + Fe3C +F e3 C Fine pearlite Bainite 1 sec 1 min 1 hour 1 day 6 Fe3C 0.1 1 10 102 103 time (seconds) 104 105 d. How would the microstructure of the same fully austenitized 1020 steel change if, instead of the above heat treatment, it was plunged directly into an iced brine quenching bath at 0°C? How would its strength and ductility differ from (c)? Explain. ! page 5 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 3.!Metallic Alloys (20 points) a. Plain carbon steels (AISI designation 10XX) are used for many load-bearing applications, such as this assortment of small rods, in a "quenched and tempered" state. Explain why this is a preferred heat treatment, citing the effects of processing on the microstructure and properties of products with relatively small section size. b. Sheet steel used for appliance housings is sometimes subjected to a processing treatment known as "strain hardening." Use your understanding of stress-strain behavior and the microstructural origins of hardening to explain this process. What makes it an attractive processing option? ! page 6 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 3. c. 800 10 20 30 40 50 718° 60 70 at. % Li 80 90 95 700 660.37° 600 500 T (°C) 600° 4.0 9.9 20.4 521° 25.0 34.0 L 46.8 400 300 200 100 0 Al 10 20 30 40 50 60 wt. % Li 70 80 (Al) 179° 180.5° (Li) 90 Li The aircraft industry has been interested in light weight Al-Li alloys for several years because they can be heat treated to high strength. Microstructural investigations reveal the origin of their strength to be a high density of ordered Al3Li precipitates. Refer to the equilibrium Al-Li phase diagram above (from ASM Metals Handbook, Volume 8, 1973, p. 261) and specify an alloy composition and heat treatment that will generate the favored microstructure. Be specific. ! page 7 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 3. d. If a metallic alloy can be strengthened by workhardening or age-hardening, why not both? Describe the microstructures resulting from (i) a workhardening treatment followed by age-hardening and (ii) an age-hardening treatment followed by work-hardening. Which sequence would you expect to yield a stronger alloy, and why? 600 500 MPa) Stress, 400 300 200 100 0.02 0.06 0.04 Strain, (mm/mm) 0.08 ! page 8 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 4.!Ceramics & Glasses (20 points) a. The poor fracture toughness of ceramics is reflected in low values of the critical stress intensity factor (KIC) defined by √ KI C = Y σ f π a where Y is a dimensionless geometrical factor ≈1, σf is the applied tensile stress that causes failure and a is the length of a surface crack, or half the length of an internal crack. Values of KIC for alumina can be two orders of magnitude lower than fullyannealed copper. Explain this difference, citing the role of dislocations. b. In his more detailed investigation of brittle fracture, aeronautical engineer Alan Griffith proposed that the sharpness of the crack, defined by the crack tip radius ρ establishes the value of the maximum stress σm at the tip of a crack of length a under a macroscopic tensile stress of σ, according to the relation ￿ a σm = ￿ 2σ ρ How, based upon this model, does the microcrack toughening mechanism work in ceramic materials? Be specific. ! page 9 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 4. c. Many ceramics are susceptible to thermal shock, complete or partial fracture resulting from a sudden change in temperature, usually cooling. Consider a crystalline fireclay refractory furnace lining that is suddenly subjected to a rush of cold air as the furnace door is open, leading to the following temperature gradient across its thickness. 600 500 400 300 200 100 2 furnace lining air T (°C) 4 x (in) 6 8 Using concepts of thermal conductivity and thermal expansion, relate this temperature gradient to a stress gradient that results in fracture. ! page 10 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 4. d. Would it matter if the furnace lining was instead made of a vitreous (non-crystalline) ceramic with a glass transition temperature Tg of 100°C? Explain. L / L0 T Tg Ts ! page 11 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 5.!Polymers (20 points) a. Polystyrene is used in many applications including CH2 H C laboratory Petri dishes, CD and DVD "jewel" H H H C H cases, picnic cutlery, smoke detector housings, C C C C and license plate frames. One popular version has the fifty-year-old Dow Chemical trade name C C H Styrofoam® used as building insulation panels. H H C The styrene monomer is shown here in its full H expansion on the left and in "shorthand" on the right, representing the phenol group by its hexagonal ring symbol. Is polystyrene a more likely candidate for synthesis by chain growth or step growth? Explain. b. Now sketch polystyrene in both its isotactic and syndiotactic conformations. ! page 12 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ 5. c. The stress-strain behavior 600 of polystyrene is compared here to 500 aluminum alloy 2024. Polystyrene fails at a 400 total elongation of 300 approximately 1%, at a Al alloy 2024 tensile strength of less 200 than 50 MPa. Explain the deformation 100 Polystyrene behavior of polystyrene, citing the microstructural 0.06 0.02 0.04 0.08 Strain, (mm/mm) mechanisms responsible for the shape of its σ–ε curve and obvious differences compared to the curve for the aluminum alloy. Stress, MPa) d. From your answer to parts a and b above, would you expect polystyrene to be a thermoplastic or a thermoset? Is it recyclable? Why do you suppose the city of Berkeley banned all foamed polystyrene coffee cups within its metropolitan limits? ! page 13 of 14 E45 Midterm 02 Fall 2008 ! Initials:__________ Worksheet Problem Possible Score 1 2 3 4 5 TOTAL 20 20 20 20 20 100 ! page 14 of 14 ...
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Midterm02PracticeBinder - Professor R Gronsky UNIVERSITY OF...

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