MSE2034 (STALEY) Test #3 Review 4/2/06

Transcription

1 MSE2034 (STALEY) Test #3 Review 4/2/06 The third test in this course will be a take-home assignment handed out at the end of class Wednesday, April 5, and due by Noon on Friday, April 7. It will be open book and notes. The test will cover the following new material from Chapters 9, 10, 11, 14: - Fracture and Failure Mechanisms of Materials Textbook, Ch.9 (sections , , & ONLY) Course Notes Module#8 HW#6 - Alloy Phase Diagrams Textbook, Ch.10 (sections , & ONLY) Course Notes Module #9 HW#7 - Steel Transformation Diagrams & Thermal Processing Textbook, Ch.11 (sections , & ONLY) Textbook, Ch.14 (section 14.5 ONLY) Course Notes Module #10 (sections A-H ONLY) HW#8 This new material is covered in the text in the assigned sections we have discussed since Spring Break, and corresponding Notes modules. However, it will be assumed that you have a working knowledge of related topics covered earlier in the course (such as stressstrain behavior, hardness determination, diffusion, etc.). Some problems thus may require you to use information found outside these specific sections of the book. Shown on the following pages are a large number of old test questions on these topics and their full solutions. These constitute much more than one test s worth of material. You can also assume that the HW assignments we have completed (including the EC assignment due on Weds, 4/6) will be useful guides to relevant methods. IMPORTANT NOTE: Since these old problems assume an older version of the textbook, figure and equation #s will not match with what you find in your new edition. However, the same information remains available to you in the new text under new headings. Descriptions in each question should point you to relevant info in each case.

2 SAMPLE Q1 Figure 9.18 shows the temperature-composition phase diagram for Mg-Pb alloys. Consider two alloys of composition: (i) 30 wt% Pb and (ii) 70 wt% Pb. Compare the phases present, phase compositions, weight percentage of phases, and microstructures that you expect to see for these two different alloy compositions at the following series of temperatures during equilibrium cooling: (a) 600 C, (b) 400 C, (c) 200 C. Be sure to draw the microstructures you expect, showing qualitatively the relative abundance of each phase you expect to be present. (3) Solution: The problem calls for an analysis of two alloys (30% Pb - 70% Mg and 70% Mg-30% Pb) at three different temperatures. The correct interpretation of the phase diagram is as follows: Key: L (liquid) α (solid) Mg2Pb (solid) 30% Pb-70% Mg (a) 600 C: Phases present: α+l. Compositions: α: 11% Pb, L: 39% Pb. Wt. Fractions: 25 wt% α, 75 wt% L. 70% Pb-30% Mg (a) 600 C: Phase present: L. Compositions: L: 70% Pb. Wt. Fraction: 100 wt% L. (b) 400 C: Phase Present: α Composition: α: 30% Pb Wt. Fraction: 100% α. (b) 400 C: Phases present: α+mg2pb. Compositions: α: 32% Pb, Mg2Pb: 81% Pb. Wt. Fractions: 22 wt% α, 78 wt% Mg2Pb. (c) 200 C: Phases present: α+mg2pb. Compositions: α: 4% Pb, Mg2Pb: 81% Pb. Wt. Fractions: 66 wt% α, 34 wt% Mg2Pb. (c) 200 C: Phases present: α+mg2pb. Compositions: α: 4% Pb, Mg2Pb: 81% Pb. Wt. Fractions: 14 wt% α, 86 wt% Mg2Pb.

3 SAMPLE Q2 Refer to Figures and for TTT diagrams of isothermal and continuous cooling procedures for eutectoid 1040 steel. (a) Describe the microconstituents that you would expect to be produced by the following cooling procedures from 800 C to RT: (i) Rapidly cool to 450 C, hold 200 sec. Cool to 300 C, hold 200 sec. Quench to RT. (ii) Continuously cool at 10 C/sec to RT. Reheat to 700 C and hold 48 hours. (iii) Continuously cool at 150 C/sec to RT. Reheat to 600 C and hold 10 hours. (iv) Rapidly cool to 650 C, hold 100 sec. Cool to 575 C, hold 100 sec. (v) Continuously cool at 30 C/sec to RT. (b) Note that the stress-strain behavior for pure iron at RT is shown in Fig 6.14 of the text. Generate a qualitative stress-strain diagram that compares the RT behavior of the five samples you have produced in (a) to the behavior of pure iron, using what you know about the relative strengths and ductilities of different steel structures. Explain your reasons for the placement of the different samples. (4) Solution: (a) Each of the processes produces a structure with only one microconstituent. The expected structures are as follows: (i) 100% Bainite (ii) 100% Spheroidite. (iii) 100% Tempered Martensite. (iv) 100% Coarse Pearlite (v) 100% Medium Pearlite. (b) Using your knowledge of the hardness and ductility of the above microconstituents, and those of pure iron, you should have constructed a set of stress-strain curves qualitatively similar to the figure shown below. The lowest curve will be for pure iron. Next will be spheroidite - with its high ductility and low strength. Then, coarse pearlite, followed by the stronger but less ductile medium pearlite. The second highest curve is bainite, which should be both stronger and more ductile than pearlite The highest curve is for tempered martensite, which will have a greater ductility than totally-brittle martensite but which will still be limited in ductility. The exact placement of the curves will probably differ on your solution, but the trends should be the same. See sections 10.7 and 10.8 for a review of the mechanical properties of steels. Stress (Mpa) Iron Spheroidite Coarse Pearlite Medium Pearlite Bainite Tempered Martensite Strain (unitless)

4 SAMPLE Q3 Consider a simple plain carbon steel alloy sample that at room temperature - contains only pearlite plus some Fe 3 C. b) What is the maximum amount (mass percentage) of proeutectoid material that could exist in a steel sample with this microstructure? c) What mass percentage of the pearlite microconstituent will be cementite? d) Describe the properties this steel composition and microstructure would have at room temperature. Draw the microstructure, showing all phases and microconstituents in approximately correct proportions and distributions. e) Explain how you might be able to use subsequent thermal or mechanical processing steps to alter the mechanical properties of this steel, without either altering the composition or turning the solid back into austenite. Solution: a) For this hypereutectoid alloy ( pearlite + cementite), the maximum amount of proeutectoid will exist at the highest possible carbon concentration - which for steel is 2.14% carbon. To determine the exact percentage, we perform a lever rule calculation in the ( γ + Fe 3 C) phase field, just above the eutectoid temperature (728 C). For these conditions, the total composition is 2.14% C, the austenite has 0.76% C, and the cementite is 6.7% C. The weight percent of the cementite proeutectoid is determined as: wt% Fe 3 C [ proeutectoid] = ( ) / ( ) x 100% = 23 % b) Clearly, from (a), 77 wt% of the total mass will be pearlite, but this pearlite itself will consist of ferrite and cementite in alternating layers. To determine the exact percentage of cementite, we again perform a lever rule calculation - this time in the ( α + Fe 3 C) phase field just below the eutectoid temperature (726 C). This gives: wt% Fe 3 C [ pearlite] = ( ) / ( ) x 100% = 11 % Note that this percentage of cementite will always exist in any pearlite structure, independent of the total composition. c) This extremely high carbon steel will be very hard and brittle, with relatively low toughness, as a result of the large fraction of pure cementite present. The microstucture will consist of two phases - ferrite & cementite - and two microconstituents - proeutectoid and pearlite - with almost 1/4 of the total being pure proeuctectoid cementite and the remaining 3/4 being alternating lamellae of about 1/10 cementite and 9/10 ferrite. d) Without austenitizing or changes in composition, the mechanical properties could be changed by stress relief annealing, recrystallization & grain growth, spheroidizing, and - with difficulty - cold working.

5 SAMPLE Q4 Figures and show, respectively, the isothermal and continuous cooling TTT diagrams for 4340 steel, while Figure 11.5 shows the hardenability of this same steel versus a series of other ones. Use these diagrams, and your knowledge of steel processing and classification to answer the following: (a) What are the significant alloy additions in this steel, and why might they be used? What is the carbon content? (b) Circle the appropriate designation for this material out of the following choices: - Low Carbon, Medium Carbon, or High Carbon - Carbon Steel, Alloy Steel, or Stainless Steel - Hypoeutectoid, Eutectoid, or Hypereutectoid (c) Explain the level of hardness variation for the 4340 steel, as shown in Fig 11.5, in terms of specific microstructural changes that are expected to occur between the surface and interior of the test sample. (d) Propose both isothermal and continuous cooling procedures to obtain the following microstructures, if possible. Be quantitative, using times, temperatures, and cooling rates as appropriate. If obtaining any of these structures is impossible, just say so. (I) Ferrite and Martensite (II) Ferrite, Bainite, & Martensite (III) Pearlite & Martensite (IV) Pure Bainite (e) Being as specific as possible, what would you call the following thermal processes that might be used for this steel, and what is the basic purpose of each?: (I) Heat from room temperature to approximately 180 C, hold, then cool. (II) Heat from RT to ~800 C, hold, then cool at approximately C/sec. (III) Heat from RT to ~800 C, hold, then cool at approximately 20 C/sec. Solution: a) Additions of Ni, Cr, Mo, & Mn are utilized in this steel to increase hardenability (delaying bainite and pearlite transitions), to enhance corrosion resistance, and to strengthen the alloy. The carbon content is 0.4% - as determined from the second two digits in the AISI-SAE steel designation. b) This is a medium carbon, hypoeutectoid, alloy steel. c) The hardness variation observed in this figure results from a transition between pure martensite at the surface to an approximately 80% martensite/20% bainite mixture in the interior. d) Your answers may vary slightly from the following: I) Cool from austenite to 650 C, hold 3000 sec, quench. Continuous cooling impossible. II) Cool from austenite to 650 C, hold 3000 sec; cool to 350 C, hold 3000 sec, quench. OR continuously cool at approx 0.1 C/sec. III) Impossible to achieve. IV) Cool from austenite to 320 C, hold 10,000 sec, quench. e) I) Stress Relief Annealing: Remove residual stresses below recrystallization T. II) Normalizing: Produce uniform fine pearlite with reliable properties. III) Austenitize & Quench: Produce pure martensite, single phase BCT structure.

6 SAMPLE Q5 The phase diagram for alloys of magnesium (Mg) with lead (Pb) is given in Chapter 9 of the text. Using this diagram, identify the following characteristics: a) What are the invariant points in this alloy system (what type are they) and where do they occur on the diagram? The only invariant points are two eutectics, one at 67wt% Pb and 460 C and the other at 97wt% Pb and 250 C. b) What are the melting points of each pure solid phase? Mg melts at 640 C, Pb melts at 325 C, and Mg 2 Pb melts at 540 C. c) What are the maximum solubility limits of each solid phase and what is the solute in each case? The α phase has a maximum solubility of 42% for Pb as the solute. The β phase has a maximum solubility of less than 1% for Mg as the solute. Neither Mg nor Pb are significantly soluble in Mg 2 Pb. d) In the spaces provided below, sketch the microstructure you would expect to result from the cooling of a Mg-Pb alloy of the four indicated compositions (see below) from 700 C to room temperature. Identify each phase present. Sketches not provided here see text for examples. Phases present are as follows: 70Mg-30Pb: Majority α phase, with Mg 2 Pb precipitate inclusions within α grains. 50Mg-50Pb: About 70% primary α, containing Mg 2 Pb precipitate inclusions, and 30% eutectic mixture of α + Mg 2 Pb. The α component of the eutectic will contain Mg 2 Pb precipitate inclusions as well. 30Mg-70Pb: About 20% primary Mg 2 Pb, and 80% eutectic mixture of α + Mg 2 Pb. The α component of the eutectic will contain Mg 2 Pb precipitate inclusions as well. 10Mg-90Pb: About 40% primary Mg 2 Pb, and 60% eutectic mixture of β + Mg 2 Pb. For the 30Mg-70Pb alloy cooled from 700 C to RT, determine the following information: f) At what temperature will primary solid begin to form, and what will this solid be? Mg 2 Pb solid will begin to form at about 490 C. g) What will be the final weight fraction of primary solid? (70-67)/(82-67) x100% = 20% primary solid by lever rule just above eutectic T. h) What will be the weight fraction of eutectic material formed?

7 (82-70)/(82-67) x100% = 80% eutectic solid by lever rule just above eutectic T. i) What will be the total weight fraction of the α phase present at RT? (82-70)/(82-1) x100% = 15% total α by lever rule at 25 C e) What will be the weight fraction of α in the eutectic material at RT? The eutectic material will always be at a composition of 67% Pb. calculation at this composition and 25 C gives: A lever rule (82-67)/(82-1) x 100% = 19% eutectic α. f) What will be the total weight fraction of the β phase present at RT? There is no β phase present at this alloy composition. SAMPLE Q6 A piece of 4340 steel is to be processed in a series of steps to produce a useful part for structural application. At the outset of the process, it has an unknown microstructure. For purposes of this question, you may assume that the phase behavior of this alloy steel is accurately represented by Figure You may also need to refer to Figs 10.14, 10.19, 11.9, 11.13, and [63 points total 9 per section]. (a) Describe an appropriate Normalizing annealing process for converting this steel to a fine pearlite microstructure, citing specific temperatures and continuous cooling rates required. This steel contains 0.40% carbon, and Figs 9.21 and 11.9 show this composition to be hypoeutectoid. Thus, a structure composed completely of fine pearlite will be impossible to achieve, since some proeutectoid ferrite will always form. However, formation of fine pearlite (plus ferrite) can be achieved by the following steps: Heat to a temperature between 780 and 860 C (Fig 11.9) to convert to austenite, then cool continuously at a rate slightly less than C per second (Fig 10.19) to obtain first ferrite, then fine pearlite. A cooling rate on the order of C/sec would be reasonable. Note that this would be a somewhat lengthy process (taking several hours to accomplish), but that it would be complete by the time a low end T of about C was achieved (after which time, the material could be swiftly cooled to room T without significant change. (b) Could the same microstructure produced in (a) be achieved by an isothermal process? If so describe how, citing required times and temperatures. If not, explain why. Yes, the same combination of ferrite and fine pearlite could be achieved by a series of isothermal steps. We would need again to heat to form austenite (780-

8 860 C). Then, the most efficient conversion to fine pearlite would be cool to about 640 C [Fig 10.14] and hold for about 20,000 sec (or 5.5 hrs). (c) Estimate the weight percentage and composition of proeutectoid material that will form during the Normalizing process in (a). Show your work. Estimation of the amount of proeutectoid material requires a lever rule calculation using Fig 9.21, at a temperature just above the eutectoid transition [about 728 C] and net composition of 0.40%C. As the figure shows, here, the ferrite (proeutectoid) composition will be approximately 0.022%C, and the composition of the remaining austenite will be 0.76%C. Thus, the weight percentage of proeutectoid will be: Wt% Ferrite = ( )/( ) x100% = 49 wt% proeutectoid Note that, in reality, the alloy additions in this steel will be expected to alter both the composition and temperature of the eutectoid point (see Figs 9.31 and 9.32). However, you were instructed to ignore these effects in this problem. (d) Describe the simplest process possible for converting the fine pearlite microstructure achieved in (a) to spheroidite, in preparation for drawing the material into a cylinder of 60 mm diameter. Cite specific temperatures and times. Spheroidite formation occurs at temperatures slightly below the eutectoid temperature after hold times of hours. To be on the safe side, an appropriate estimate here would be to hold for 25 hours between 700 and 720 C, as shown on Fig (e) Once the material has been formed into a cylinder with 60 mm diameter, it is quenched in oil to form a hardened microstructure. Estimate the minimum hardness level that would be expected in this material after this process. Minimum hardness for a quenched structure will occur at the core of the cylinder. The hardness achieved in this case can be estimated using Figs and 11.16b. First, using Fig 11.16b, we find that the centerline of a 60 mm cylinder quenched in oil will be equivalent to the 19 mm point on a standard Jominy test sample. Then, Fig indicates that the hardness of a standard 4340 steel sample at this 19 mm position would be about 53 (±0.5) HRC. (f) Describe the microstructure(s) you would expect to exist in the cylinder after the process in (e). If possible, estimate weight percentages of each microconstituent. The quenching process will produce a surface microstructure composed predominantly of martensite (a surface hardness estimate similar to that in part (e) gives a value of 57 HRC, equivalent to essentially pure martensite). As we

9 progress further into the core of the sample, some of the martensite will be replaced by bainite, due to the comparatively slower rate of cooling. The hardness value from (e) suggests a mixture of about 87% martensite and 13% bainite at the centerline. These estimates can be obtained in a variety of ways: from the right hand side of Fig 11.13; from the cooling rates listed at the top of Fig 11.13, along with the continuous cooling diagram in Fig 10.19; or - with more difficulty - from the information in Figs 10.21, 10.22, 10.23, and (g) Would you expect any additional processing of the cylinder to be required after the process in (e) to make the material useful as a structural component? If so, describe it specifically. If not, describe the mechanical properties of the material obtained in (e). Beginning with the martensite-bainite structure from part (e), we would need to temper the component at C to convert the completely brittle martensite to the more robust tempered martensite structure.

10 SAMPLE Q7 Consider a certain metal alloy that undergoes a ductile-brittle transition near room temperature [RT]. When a sample of this material is heated, it reaches a maximum impact energy absorption value of 190 J; and when a similar sample is cooled it reaches a minimum impact energy absorption value of 10 J. Both these values are determined by a series of Charpy V-Notch tests. Assume that these maximum and minimum values represent (i) the combined ductile+brittle and (ii) totally brittle fracture energies, respectively. Thus, we can use them to estimate the relative contributions of surface energy and plastic deformation energy in the crack propagation process. (a) If the critical stress for the ductile state is determined to be 280 MPa for a sample containing 0.1 mm surface cracks, what do you expect the critical stress to be for the same sample in the brittle state? (b) What would be the maximum radius of crack curvature that would generate failure of the brittle sample in (a) at an applied stress level of 5 MPa? Solution (a) Critical stress is specified by Eqn 8.3 for the brittle state and Eqn 8.4 for the ductile state. However, the Young's Modulus [E], the surface energy, and the plastic deformation energy are not given, so direct calculations are not possible. You are given that the relative contributions can be determined from the impact test data. Thus, (190 J)/(10 J) = 19. Since the critical stress for a completely ductile sample with crack length 0.1 mm is 280 MPa, the corresponding critical stress for same sample in the brittle state can be determined as follows: σ[ductile]/ σ [brittle] = (280 MPa)/(x) = (19)1/2/(1)1/2 = 4.36 x = σ [brittle] = (280 MPa)/(4.36) = 64.2 MPa (b) At exactly the critical stress, the crack will begin to spontaneously propagate. Using Eqn 8.1a, the maximum radius of curvature required to amplify the applied stress to the critical level can be obtained: σ= σ0[1+2(a/ρt)1/2] 64.2 MPa = (5 MPa) [1+2([0.1 mm]/ ρt)1/2] ρt = (0.1 mm)/(5.92)2 = mm

11 SAMPLE Q8 You are involved in the construction of a steel framework structure in the Ethiopian desert near the equator. At this location, the daily temperature fluctuation is virtually constant year-round, going from 33 C at midday to -5 C at night. (a) Given that the thermal expansion coefficient of the steel is 12.0x10-6 C-1 and the Young's modulus is 207 GPa, calculate the stress generated inside the steel structure by this thermal cycling. (b) Using the S-N curve given below, estimate the fatigue life (in days) of this material given your result for (a). Note that the x-axis of the graph is given in Base-10 logarithms. (c) Suggest three different material property changes that would increase the fatigue life of the framework. Solution #13: (a) The stress magnitude generated by a thermal cycle can be calculated using Eqn All of the required quantities are given: m = \l E T m = (12.0x10-6 C-1)(207 GPa)(33 - (-5)) m = (12.0x10-6 C-1)(207 GPa)(38) m = GPa = 94.4 MPa (b) At 94.4 MPa, the S-N curve predicts failure of the sample at approximately log[n]=7 (any value between 6.9 and 7.1 is acceptable, given the precision of the figure). Thus, N can be estimated as being between 8,000,000

12 and 13,000,000 cycles (with a median value of 10,000,000 cycles). Since the temperature cycling occurs daily, the structure can be expected to resist failure by thermal fatigue for about 10,000,000 days. Note that this is about 27,000 years, so that guarding against thermal fatigue should not be a significant factor in the design of the structure. (c) If we did want to improve the structure's resistance to thermal fatigue by manipulating material properties, there are many options. The surface could be shot-peened or case-hardened (both of which will reduce the incidence of surface crack sites), or a material with a lower coefficient of thermal expansion or a lower modulus could be chosen (thus reducing the stress level directly). Since the graph appears to show a fatigue limit near 75 MPa, raising the limiting stress to above 94 MPa (by choosing a different type of steel) might also be an option; this would prevent any fatigue failure at the expected stress level. In addition, the structure could be redesigned to reduce stress concentration sites, although this is technically not a material property change. SAMPLE Q9 A high carbon steel sample with KIc of 22 MPa m and my of 750 MPa is being assessed for its susceptibility to cracking in tension. The steel is to be used in the form of support struts 2 cm thick and 15 cm wide. Either edge cracks or internal cracks (thru-thickness in both cases) could occur. Calculate the maximum crack size allowable in each case without brittle fracture occurring. SOLUTION #14: In both cases, we use the critical crack length formula, Eqn 9.9 The only variable not given is the geometric factor (Y), so we can obtain a general relationship: ac = (22)2 / (Y2 (750)2 k) = ( m)(y-2) Although the sample is of given finite dimensions, the very small crack lengths that we will obtain justify the assumption that in relation to cracks the plate is semi-infinite. Even if you did not make that assumption, an iterative process would have guided you to a value of (a/b) approaching zero [the semi-infinite plate condition] This allows us to use the values Y[edge]=1.12 and Y[internal] = From these, we can easily find: Length[edge] = ac = ( m)(1.12-2) = 0.21 mm Length[internal] = 2 ac = 2 ( m)(1.00-2) = 0.56 mm