Precipitation Hardening 11.30 Compare precipitation hardening (Section 11.9) and the hardening of steel by quenching and tempering (S (a) The total heat treatment procedure (b) The microstructures that develop (c) How the mechanical properties change during the several heat treatment stages Solution (a) With regard to the total heat treatment procedure, the steps for the hardening of steel are as follows: (1) Austenitize above the upper critical temperature. (2) Quench to a relatively low temperature. (3) Temper at a temperature below the eutectoid. (4) Cool to room temperature. With regard to precipitation hardening, the steps are as follows: (1) Solution heat treat by heating into the solid solution phase region. (2) Quench to a relatively low temperature. (3) Precipitation harden by heating to a temperature that is within the solid two-phase region. (4) Cool to room temperature. (b) For the hardening of steel, the microstructures that form at the various heat treating stages in part (a) are: (1) Austenite (2) Martensite (3) Tempered martensite (4) Tempered martensite For precipitation hardening, the microstructures that form at the various heat treating stages in part (a) are: (1) Single phase (2) Single phase--supersaturated (3) Small plate-like particles of a new phase within a matrix of the original phase. (4) Same as (3) (c) For the hardening of steel, the mechanical characteristics for the various steps in part (a) are as follows: (1) Not important (2) The steel becomes hard and brittle upon quenching. (3) During tempering, the alloy softens slightly and becomes more ductile. (4) No significant changes upon cooling to or maintaining at room temperature. For precipitation hardening, the mechanical characteristics for the various steps in part (a) are as follows: (1) Not important (2) The alloy is relatively soft. (3) The alloy hardens with increasing time (initially), and becomes more brittle; it may soften with overaging. (4) The alloy may continue to harden or overage at room temperature. 11.31 What is the principal difference between natural and artificial aging processes? Solution For precipitation hardening, natural aging is allowing the precipitation process to occur at the ambient temperatu Design Problems Ferrous Alloys Nonferrous Alloys 11.D1 Below is a list of metals and alloys: Plain carbon steel Magnesium Brass Zinc Gray cast iron Tool steel Platinum Aluminum Stainless steel Tungsten Titanium alloy Select from this list the one metal or alloy that is best suited for each of the following applications, and cite at le (a) The block of an internal combustion engine (b) Condensing heat exchanger for steam (c) Jet engine turbofan blades (d) Drill bit (e) Cryogenic (i.e., very low temperature) container (f) As a pyrotechnic (i.e., in flares and fireworks) (g) High-temperature furnace elements to be used in oxidizing atmospheres Solution (a) Gray cast iron would be the best choice for an engine block because it is relatively easy to cast, is wear resist inexpensive. (b) Stainless steel would be the best choice for a heat exchanger to condense steam because it is corrosion resista (c) Titanium alloys are the best choice for high-speed aircraft jet engine turbofan blades because they are light w However, one drawback is their cost. (d) A tool steel would be the best choice for a drill bit because it is very hard retains its hardness at high tempera edge. (e) For a cryogenic (low-temperature) container, an aluminum alloy would be the best choice; aluminum alloys h low temperatures. (f) As a pyrotechnic in flares and fireworks, magnesium is the best choice because it ignites easily and burns rea (g) Platinum is the best choice for high-temperature furnace elements to be used in oxidizing atmospheres becau and is highly resistant to oxidation. 11.D2 A group of new materials are the metallic glasses (or amorphous metals). Write an essay about these mat of some of the common metallic glasses, (2) characteristics of these materials that make them technologically at and potential uses, and (5) at least one technique that is used to produce metallic glasses. Solution (a) Compositionally, the metallic glass materials are rather complex; several compositions are as follows: Fe80B2 Fe40Ni38Mo4B18. (b) These materials are exceptionally strong and tough, extremely corrosion resistant, and are easily magnetized. (c) Principal drawbacks for these materials are 1) complicated and exotic fabrication techniques are required; and dimension of the material must be small--i.e., they are normally produced in ribbon form. (d) Potential uses include transformer cores, magnetic amplifiers, heads for magnetic tape players, reinforcemen interference, security tapes for library books. (e) Production techniques include centrifuge melt spinning, planar-flow casting, rapid pressure application, arc m 11.D3 Of the following alloys, pick the one(s) that may be strengthened by heat treatment, cold work, or both: R C51000 phosphor bronze, lead, 6150 steel, 304 stainless steel, and C17200 beryllium copper. Solution This question provides us with a list of several metal alloys, and then asks us to pick those that may be strengthe may be heat treated are either those noted as "heat treatable" (Tables 11.6 through 11.9), or as martensitic stainle working must not be exceptionally brittle, and, furthermore, must have recrystallization temperatures above room fall within the three classifications are as follows: Heat Treatable Cold Workable Both 6150 steel 6150 steel 6150 steel C17200 Be-Cu C17200 Be-Cu C17200 Be-Cu 6061 Al 6061 Al 6061 Al 304 stainless steel R50500 Ti C51000 phosphor bronze AZ31B Mg 11.D4 A structural member 100 mm (4 in.) long must be able to support a load of 50,000 N (11,250 lbf) without brass, steel, aluminum, and titanium, rank them from least to greatest weight in accordance with these criteria. Alloy Yield Strength Density [MPa (ksi)] (g/cm3) Brass 415 (60) 8.5 Steel 860 (125) 7.9 Aluminum 310 (45) 2.7 Titanium 550 (80) 4.5 Solution This problem asks us to rank four alloys (brass, steel, titanium, and aluminum), from least to greatest weight for without experiencing plastic deformation. From Equation 6.1, the cross-sectional area (A0) must necessarily carr Now, given the length l, the volume of material required (V) is just Finally, the mass of the member (m) is Here is the density. Using the values given for these alloys Thus, titanium would have the minimum weight (or mass), followed by aluminum, steel, and brass. 11.D5 Discuss whether it would be advisable to hot work or cold work the following metals and alloys on the ba and degree of brittleness: tin, tungsten, aluminum alloys, magnesium alloys, and a 4140 steel. Solution Tin would almost always be hot-worked. Even deformation at room temperature would be considered hot-worki temperature (Table 7.2). Tungsten is hard and strong at room temperature, has a high recrystallization temperature, and experiences oxida its strength, and hot-working is not practical because of oxidation problems. Most tungsten articles are fabricated annealing cycles. Most aluminum alloys may be cold-worked since they are ductile and have relatively low yield strengths. Magnesium alloys are normally hot-worked inasmuch as they are quite brittle at room temperature. Also, magne A 4140 steel could be cold-worked in an over-tempered state which leaves it soft and relatively ductile, after wh strengthen and harden it. This steel would probably have a relatively high recrystallization temperature, and hot- Heat Treatment of Steels 11.D6 A cylindrical piece of steel 25 mm (1.0 in.) in diameter is to be quenched in moderately agitated oil. Surfa respectively. Which of the following alloys will satisfy these requirements: 1040, 5140, 4340, 4140, and 8640? J Solution In moderately agitated oil, the equivalent distances from the quenched end for a one-inch diameter bar for surfac respectively [Figure 11.17b]. The hardnesses at these two positions for the alloys cited (as determined using Figu Surface Center Alloy Hardness (HRC) Hardness (HRC) 1040 50 30 5140 56 49 4340 57 57 4140 57 55 8640 57 53 Thus, alloys 4340, 4140, and 8640 will satisfy the criteria for both surface and center hardnesses. 11.D7 A cylindrical piece of steel 75 mm (3 in.) in diameter is to be austenitized and quenched such that a minim piece. Of the alloys 8660, 8640, 8630, and 8620, which will qualify if the quenching medium is (a) moderately a choice(s). Solution (a) This problem calls for us to decide which of 8660, 8640, 8630, and 8620 alloys may be fabricated into a cylin mildly agitated water, will produce a minimum hardness of 40 HRC throughout the entire piece. The center of the steel cylinder will cool the slowest and therefore will be the softest. In moderately agitated wat diameter bar for the center position is about 17 mm (11/16 in.) [Figure 11.17a]. The hardnesses at this position f Center Alloy Hardness (HRC) 8660 58 8640 42 8630 30 8620 22 Therefore, only 8660 and 8640 alloys will have a minimum of 40 HRC at the center, and therefore, throughout th (b) This part of the problem asks us to do the same thing for moderately agitated oil. In moderately agitated oil th diameter bar at the center position is about 25.5 mm (1-1/32 in.) [Figure 11.17b]. The hardnesses at this position Center Alloy Hardness (HRC) 8660 53 8640 37 8630 26 8620 < 20 Therefore, only the 8660 alloy will have a minimum of 40 HRC at the center, and therefore, throughout the entir 11.D8 A cylindrical piece of steel 38 mm (1 in.) in diameter is to be austenitized and quenched such that a mic throughout the entire piece. Of the alloys 4340, 4140, 8640, 5140, and 1040, which will qualify if the quenching water? Justify your choice(s). Solution (a) Since the cooling rate is lowest at the center, we want a minimum of 80% martensite at the center position. F distance from the quenched end of 12 mm (1/2 in.). According to Figure 11.14, the hardness corresponding to 80 to determine which of the alloys have a 50 HRC hardness at an equivalent distance from the quenched end of 12 following hardnesses are determined from Figure 11.14 for the various alloys. Alloy Hardness (HRC) 4340 56 4140 53 8640 49 5140 43 1040 25 Thus, only alloys 4340 and 4140 will qualify. (b) For moderately agitated water, the cooling rate at the center of a 38 mm diameter specimen is 7 mm (5/16 in. this position, the following hardnesses are determined from Figure 11.14 for the several alloys. Alloy Hardness (HRC) 4340 57 4140 55 8640 54 5140 51 1040 33 It is still necessary to have a hardness of 50 HRC or greater at the center; thus, alloys 4340, 4140, 8640, and 514 11.D9 A cylindrical piece of steel 90 mm (3 in.) in diameter is to be quenched in moderately agitated water. S respectively. Which of the following alloys will satisfy these requirements: 1040, 5140, 4340, 4140, 8620, 8630, Solution A ninety-millimeter (three and one-half inch) diameter cylindrical steel specimen is to be quenched in moderatel will have surface and center hardnesses of at least 55 and 40 HRC, respectively. In moderately agitated water, the equivalent distances from the quenched end for a 90 mm diameter bar for surfa respectively [Figure 11.17a]. The hardnesses at these two positions for the alloys cited are given below. The hard (as determined from Figures 11.14 and 11.15). Surface Center Alloy Hardness (HRC) Hardness (HRC) 1040 50 < 20 5140 56 34 4340 57 53 4140 57 45 8620 42 < 20 8630 51 28 8640 56 38 8660 64 55 Thus, alloys 4340, 4140, and 8660 will satisfy the criteria for both surface hardness (minimum 55 HRC) and cen 11.D10 A cylindrical piece of 4140 steel is to be austenitized and quenched in moderately agitated oil. If the mic entire piece, what is the maximum allowable diameter? Justify your answer. Solution From Figure 11.14, the equivalent distance from the quenched end of a 4140 steel to give 50% martensite (or a 4 at the center of the specimen should correspond to this equivalent distance. Using Figure 11.17b, the center spec distance at a diameter of about 83 mm (3.3 in.). 11.D11 A cylindrical piece of 8640 steel is to be austenitized and quenched in moderately agitated oil. If the har the maximum allowable diameter? Justify your answer. Solution We are to determine, for a cylindrical piece of 8640 steel, the minimum allowable diameter possible in order yie out in moderately agitated oil. From Figure 11.15, the equivalent distance from the quenched end of an 8640 steel to give a hardness of 49 HRC surface of the specimen should correspond to this equivalent distance. Using Figure 11.17b, the surface specime diameter of about 75 mm (3 in.). 11.D12 Is it possible to temper an oil-quenched 4140 steel cylindrical shaft 100 mm (4 in.) in diameter so as to g minimum ductility of 21%EL? If so, specify a tempering temperature. If this is not possible, then explain why. Solution This problem asks if it is possible to temper an oil-quenched 4140 steel cylindrical shaft 100 mm (4 in.) in diame psi) and a minimum ductility of 21%EL. In order to solve this problem it is necessary to use Figures 11.20a and versus tempering temperature. For the 100 mm diameter line of Figure 11.20a, tempering temperatures less than MPa. Furthermore, from Figure 11.20c, for the 100 mm diameter line, tempering temperatures greater than abou it is not possible to temper this alloy to produce the stipulated minimum tensile strength and ductility. To meet th ductility minimum, T(tempering) > 585°C; thus, there is no overlap of these tempering temperature ranges. 11.D13 Is it possible to temper an oil-quenched 4140 steel cylindrical shaft 12.5 mm (0.5 in.) in diameter so as t minimum ductility of 16%EL? If so, specify a tempering temperature. If this is not possible, then explain why. Solution This problem asks if it is possible to temper an oil-quenched 4140 steel cylindrical shaft 12.5 mm (0.5 in.) in dia (145,000 psi) and a minimum ductility of 16%EL. In order to solve this problem it is necessary to use Figures 11 ductility versus tempering temperature. For the 12.5 mm diameter line of Figure 11.20b, tempering temperatures least 1000 MPa. Furthermore, from Figure 11.20c, for the 12.5 mm diameter line, tempering temperatures greate Hence, it is possible to temper this alloy to produce the stipulated minimum yield strength and ductility; the temp Precipitation Hardening 11.D14 Copper-rich copper–beryllium alloys are precipitation hardenable. After consulting the portion of the ph (a) Specify the range of compositions over which these alloys may be precipitation hardened. (b) Briefly describe the heat-treatment procedures (in terms of temperatures) that would be used to precipitation within the range given for part (a). Solution This problem is concerned with the precipitation-hardening of copper-rich Cu-Be alloys. It is necessary for us to below. (a) The range of compositions over which these alloys may be precipitation hardened is between approximately and 2.7 wt% Be (the maximum solubility of Be in Cu at 866C). (b) The heat treatment procedure, of course, will depend on the composition chosen. First of all, the solution hea region, after which, the specimen is quenched to room temperature. Finally, the precipitation heat treatment is co For example, for a 1.5 wt% Be-98.5 wt% Cu alloy, the solution heat treating temperature must be between about treatment would be below 600C (1110F), and probably above 300C (570F). Below 300C, diffusion rates ar 11.D15 A solution heat-treated 2014 aluminum alloy is to be precipitation hardened to have a minimum tensile s 15%EL. Specify a practical precipitation heat treatment in terms of temperature and time that would give these m Solution We are asked to specify a practical heat treatment for a 2014 aluminum alloy that will produce a minimum tensil 15%EL. From Figure 11.27a, the following heat treating temperatures and time ranges are possible to the give th Temperature (C) Time Range (h) 260 0.02-0.2 204 0.02-10 149 3-600 121 > 35-? With regard to temperatures and times to give the desired ductility [Figure 11.27b]: Temperature (C) Time Range (h) 260 < 0.01, > 40 204 < 0.15 149 < 10 121 < 500 From these tabulations, the following may be concluded: It is not possible to heat treat this alloy at 260C so as to produce the desired set of properties— there is no overl At 204C, the heat treating time would be between 0.02 and 0.15 h; times lying within this range are impractical At 149C, the time would be between 3 and 10 h. Finally, at 121C, the time range is 35 to about 500 h. 11.D16 Is it possible to produce a precipitation-hardened 2014 aluminum alloy having a minimum tensile streng so, specify the precipitation heat treatment. If it is not possible, explain why. Solution This problem inquires as to the possibility of producing a precipitation-hardened 2014 aluminum alloy having a of at least 12%EL. In order to solve this problem it is necessary to consult Figures 11.27a and 11.27b. Below are achieve the stipulated tensile strength. Temperature (C) Time Range (h) 260 < 0.5 204 < 15 149 1-1000 121 > 35-? With regard to temperatures and times to give the desired ductility: Temperature (C) Time Range (h) 260 < 0.02, > 10 204 < 0.4, > 350 149 < 20 121 < 1000 From these tabulations, the following may be concluded: At 260C, the heat treating time would need to be less than 0.02 h (1.2 min), which is impractically short. At 204C, the heat treatment would need to be less than 0.4 h (24 min), which is a little on the short side. At 149C, the time range would be between 1 and 20 h. Finally, at 121C, this property combination is possible for virtually all times less than about 1000 h. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted b without the permission of the copyright owner is unlawful. 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