11
APPLICATIONS CHAPTER AND PROCESSING OF METAL ALLOYS
PROBLEM SOLUTIONS
11.1 This question asks that we list four classifications of steels, and, for each, to describe properties and cite typical applications. Low-Carbon Steels Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans. Medium-Carbon Steels Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts. High-Carbon Steels Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades. High-Alloy Steels (Stainless and Tool) Properties: environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools. hard and wear resistant; resistant to corrosion in a large variety of
11.2 (a) Ferrous alloys are used extensively because: 1) Iron ores exist in abundant quantities. 2) Economical extraction, refining, and fabrication techniques are available. 3) The alloys may be tailored to have a wide range of properties. (b) Disadvantages of ferrous alloys are: 1) They are susceptible to corrosion. 2) They have relatively high densities. 3) They have relatively low electrical conductivities.
11.3 Ferritic and austenitic stainless steels are not heat treatable since "heat treatable" is taken to mean that martensite may be made to form with relative ease upon quenching austenite from an elevated temperature. 298
For ferritic stainless steels, austenite does not form upon heating, and, therefore, the austenite-to-martensite transformation is not possible. For austenitic stainless steels, the austenite phase field extends to such low temperatures that the martensitic transformation does not occur.
11.4 The alloying elements in tool steels (e.g., Cr, V, W, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.
11.5 We are asked to compute the volume percent graphite in a 3.5 wt% C cast iron. It first becomes necessary to compute mass fractions using the lever rule. From the iron-carbon phase diagram (Figure 11.2), the tie-line in the and graphite phase field extends from essentially 0 wt% C to 100 wt% C. Thus, for a 3.5 wt% C cast iron
CGr Co C Gr C 100 3.5 = 0.965 100 0
W =
=
WGr =
C o C CGr C
=
3.5 0 = 0.035 100 0
Conversion from weight fraction to volume fraction of graphite is possible using Equation (9.6a) as
WGr VGr = Gr W W Gr Gr
0.035 = 2.3 g / cm 3 0.965 0.035 3 7.9 g / cm 2.3 g / cm3
= 0.111 or 11.1 vol%
11.6 Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration.
11.7 This question asks us to compare various aspects of gray and malleable cast irons.
299
(a) With respect to composition and heat treatment: Gray iron--2.5 to 4.0 wt% C and 1.0 to 3.0 wt% Si. For most gray irons there is no heat treatment after solidification. Malleable iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and 900C for an extended time period. (b) With respect to microstructure: Gray iron--Graphite flakes are embedded in a ferrite or pearlite matrix. Malleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix. (c) With respect to mechanical characteristics: Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility.
11.8 Yes, it is possible to produce cast irons that consist of a martensite matrix in which graphite is embedded in either flake, nodule, or rosette form. For graphite flakes, gray cast iron is formed (as described in Section 11.2), which is then heated to a temperature at which the ferrite transforms to austenite; the austenite is then rapidly quenched, which transforms to martensite. For graphite nodules and rosettes, nodular and malleable cast irons are first formed (again as described in Section 11.2), which are then austenitized and rapidly quenched.
11.9 This question asks us to compare white and nodular cast irons. (a) With regard to composition and heat treatment: White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700C may be necessary to produce a ferritic matrix. (b) With regard to microstructure: White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix. (c) With respect to mechanical characteristics: White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility.
11.10 It is not possible to produce malleable iron in pieces having large cross-sectional dimensions. White cast iron is the precursor of malleable iron, and a rapid cooling rate is necessary for the formation of white iron, which may not be accomplished at interior regions of thick cross-sections.
300
11.11 The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot- or cold-worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting.
11.12 Both brasses and bronzes are copper-based alloys. For brasses, the principal alloying element is zinc, whereas the bronzes are alloyed with other elements such as tin, aluminum, silicon, or nickel.
11.13 Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven.
11.14 Strengthening of a 3003 aluminum alloy is accomplished by cold working. Welding a structure of a cold-worked 3003 alloy will cause it to experience recrystallization, and a resultant loss of strength.
11.15 The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.
11.16 This question asks us for the distinctive features, limitations, and applications of several alloy groups. Titanium Alloys Distinctive features: relatively low densities, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries. Refractory Metals Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes. Superalloys Distinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods. 301
Applications: aircraft turbines, nuclear reactors, and petrochemical equipment. Noble Metals Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples.
11.17 The advantages of cold working are: 1) A high quality surface finish. 2) The mechanical properties may be varied. 3) Close dimensional tolerances. The disadvantages of cold working are: 1) High deformation energy requirements. 2) Large deformations must be accomplished in steps, which may be expensive. 3) A loss of ductility. The advantages of hot working are: 1) Large deformations are possible, which may be repeated. 2) Deformation energy requirements are relatively low. The disadvantages of hot working are: 1) A poor surface finish. 2) A variety of mechanical properties is not possible.
11.18 (a) The advantages of extrusion over rolling are as follows: 1) Pieces having more complicated cross-sectional geometries may be formed. 2) Seamless tubing may be produced. (b) The disadvantages of extrusion over rolling are as follows: 1) Nonuniform deformation over the cross-section. 2) A variation in properties may result over the cross-section of an extruded piece.
11.19 Four situations in which casting is the preferred fabrication technique are: 1) For large pieces and/or complicated shapes. 2) When mechanical strength is not an important consideration. 3) For alloys having low ductilities. 4) When it is the most economical fabrication technique.
11.20 This question asks us to compare sand, die, investment, and continuous casting techniques. 302
For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast. For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast. For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low. For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and
mechanical properties are relatively uniform throughout the cross-section.
11.21 (a) Some of the advantages of powder metallurgy over casting are as follows: 1) It is used for alloys having high melting temperatures. 2) Better dimensional tolerances result. 3) Porosity may be introduced, the degree of which may be controlled (which is desirable in some applications such as self-lubricating bearings). (b) Some of the disadvantages of powder metallurgy over casting are as follows: 1) Production of the powder is expensive. 2) Heat treatment after compaction is necessary.
11.22 This question asks for the principal differences between welding, brazing, and soldering. For welding, there is melting of the pieces to be joined in the vicinity of the bond; a filler material may or may not be used. For brazing, a filler material is used which has a melting temperature in excess of about 425C (800F); the filler material is melted, whereas the pieces to be joined are not melted. For soldering, a filler material is used which has a melting temperature less than about 425C (800F); the filler material is melted, whereas the pieces to be joined are not.
11.23 This problem asks that we specify and compare the microstructures and mechanical properties in the heat-affected weld zones for 1080 and 4340 alloys assuming that the average cooling rate is 10C/s. Figure 10.18 shows the continuous cooling transformation diagram for an iron-carbon alloy of eutectoid composition (1080), and, in addition, cooling curves that delineate changes in microstructure. For a cooling rate of 10C/s (which is less than 35C/s) the resulting microstructure will be totally pearlite--probably a reasonably fine pearlite. On the other hand, in Figure 10.19 is 303
shown the CCT diagram for a 4340 steel. From this diagram it may be noted that a cooling rate of 10C/s produces a totally martensitic structure. Pearlite is softer and more ductile than martensite, and, therefore, is most likely more desirable.
11.24 If a steel weld is cooled very rapidly, martensite may form, which is very brittle. In some situations, cracks may form in the weld region as it cools. 11.25 Full annealing--Heat to between 15 and 40C above the A line (if the concentration of carbon is
3
less than the eutectoid) or above the A line (if the concentration of carbon is greater than the
1
eutectoid) until the alloy comes to equilibrium; then furnace cool to room temperature. The final microstructure is coarse pearlite. Normalizing--Heat to between 55 and 85C above the upper critical temperature until the specimen has fully transformed to austenite, then cool in air. The final microstructure is fine pearlite. Quenching--Heat to a temperature within the austenite phase region and allow the specimen to fully austenitize, then quench to room temperature in oil or water. The final microstructure is martensite. Tempering--Heat a quenched (martensitic) specimen, to a temperature between 450 and 650C, for the time necessary to achieve the desired hardness. martensite. The final microstructure is tempered
11.26
Three sources of residual stresses in metal components are plastic deformation processes,
nonuniform cooling of a piece that was cooled from an elevated temperature, and a phase transformation in which parent and product phases have different densities. Two adverse consequences of these stresses are distortion (or warpage) and fracture.
11.27 This question asks that we cite the temperature range over which it is desirable to austenitize several iron-carbon alloys during a normalizing heat treatment. (a) For 0.20 wt% C, heat to between 890 and 920C (1635 and 1690F) since the A temperature is
3
835C (1535F). (b) For 0.76 wt% C, heat to between 782 and 812C (1440 and 1494F) since the A temperature is
3
727C (1340F). (c) For 0.95 wt% C, heat to between 840 and 870C (1545 and 1600F) since A (1445F).
cm
is 785C
11.28
We are asked for the temperature range over which several iron-carbon alloys should be
austenitized during a full-anneal heat treatment.
304
(a) For 0.25 wt% C, heat to between 845 and 870C (1555 and 1600F) since the A temperature is
3
830C (1525F). (b) For 0.45 wt% C, heat to between 790 and 815C (1450 and 1500F) since the A temperature is
3
775C (1425F). (c) For 0.85 wt% C, heat to between 742 and 767C (1368 and 1413F) since the A temperature is
1
727C (1340F). (d) For 1.10 wt% C, heat to between 742 and 767C (1368 and 1413F) since the A temperature is
1
727C (1340F).
11.29 The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure. It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.
11.30
Hardness is a measure of a material's resistance to localized surface deformation, whereas
hardenability is a measure of the depth to which a ferrous alloy may be hardened by the formation of martensite. Hardenability is determined from hardness tests.
11.31 The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite.
11.32 A decrease of austenite grain size will decrease the hardenability. Pearlite normally nucleates at grain boundaries, and the smaller the grain size, the greater the grain boundary area, and, consequently, the easier it is for pearlite to form.
11.33 The three factors that influence the degree to which martensite is formed are as follows: 1) Alloying elements; adding alloying elements increases the extent to which martensite forms. 2) Specimen size and shape; the extent of martensite formation increases as the specimen cross-section decreases and as the degree of shape irregularity increases. 3) Quenching medium; the more severe the quench, the more martensite is formed. Water provides a more severe quench than does oil, which is followed by air. Agitating the medium also enhances the severity of quench.
11.34 The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity. 305
11.35 (a) This part of the problem calls for us to construct a radial hardness profile for a 50 mm (2 in.) diameter cylindrical specimen of an 8640 steel that has been quenched in moderately agitated oil. In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.13 and 11.16(b).
Radial Position Surface 3/4 R Midradius Center
Equivalent Distance, mm (in.) 7 (5/16) 11 (7/16) 14 (9/16) 16 (10/16)
HRC Hardness 54 50 45 44
The resulting profile is plotted here.
(b) The radial hardness profile for a 75 mm (3 in.) diameter specimen of a 5140 steel that has been quenched in moderately agitated oil is desired. The equivalent distances and hardnesses for the various radial positions, as determined using Figures 11.13 and 11.16(b), are tabulated below. Radial Position Surface 3/4 R Midradius Center Equivalent Distance, mm (in.) 13 (1/2) 19 (3/4) 22 (14/16) 25 (1) HRC Hardness 41 35 33 31
306
The resulting profile is plotted here.
(c) The radial hardness profile for a 90 mm (3-1/2 in.) diameter specimen of an 8630 steel that has been quenched in moderately agitated water is desired. The equivalent distances and hardnesses for the various radial positions, as determined using Figures 11.14 and 11.16(a) are tabulated below. Radial Position Surface 3/4 R Midradius Center The resulting profile is plotted here. Equivalent Distance, in. (mm) 3 (1/8) 10 (3/8) 17 (11/16) 20 (13/16) HRC Hardness 50 38 30 28
307
(d) The radial hardness profile for a 100 mm (4 in.) diameter specimen of a 8660 steel that has been quenched in moderately agitated water is desired. The equivalent distances and hardnesses for the various radial positions, as determined using Figures 11.14 and 11.16(a), are tabulated below.
Radial Position Surface 3/4 R Midradius Center
Equivalent Distance, in. (mm) 3 (1/8) 11 (7/16) 20 (13/16) 27 (1-1/16)
HRC Hardness 63 61 57 53
The resulting profile is plotted here.
11.36 We are asked to compare the effectiveness of quenching in moderately agitated water and oil by graphing, on a single plot, the hardness profiles for 75 mm (3 in.) diameter cylindrical specimens of an 8640 steel that had been quenched in both media. For moderately agitated water, the equivalent distances and hardnesses for the several radial positions [Figures 11.16(a) and 11.13] are tabulated below. Radial Position Surface 3/4 R Midradius Center Equivalent Distance, mm 3 9 13.5 16.5 HRC Hardness 56 53 47 43
308
While for moderately agitated oil, the equivalent distances and hardnesses for the several radial positions [Figures 11.16(b) and 11.13] are tabulated below.
Radial Position Surface 3/4 R Midradius Center
Equivalent Distance, mm 13 19 22 25
HRC Hardness 48 41 38 37
These data are plotted here.
11.37 This problem asks us to compare various aspects of precipitation hardening, and the quenching and tempering of steel. (a) With regard to the total heat treatment procedure, the steps for the hardening of steel are as follows: 1) Austenitize the above 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.
309
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.
11.38 For precipitation hardening, natural aging is allowing the precipitation process to occur at the ambient temperature; artificial aging is carried out at an elevated temperature.
Design Problems
11.D1 This problem calls for us to select, from a list, the best alloy for each of several applications and then to justify each choice.
310
(a) Gray cast iron would be the best choice for a milling machine base because it effectively absorbs vibrations and is inexpensive. (b) Stainless steel would be the best choice for the walls of a steam boiler because it is corrosion resistant to the steam and condensate. (c) Titanium alloys are the best choice for high-speed aircraft because they are light weight, strong, and easily fabricated. However, one drawback is their cost. (d) A tool steel would be the best choice for a drill bit because it is very hard and wear resistant, and, thus, will retain a sharp cutting edge. (e) For a cryogenic (low-temperature) container, an aluminum alloy would be the best choice; aluminum alloys have the FCC crystal structure, and therefore, are ductile down to very low temperatures. (f) As a pyrotechnic in flares and fireworks, magnesium is the best choice because it ignites easily and burns readily in air. (g) Platinum is the best choice for high-temperature furnace elements to be used in oxidizing atmospheres because it is very ductile, has a relatively high melting temperature, and is highly resistant to oxidation.
11.D2 (a) Important characteristics required of metal alloys that are used for coins are as follows: they must be hard, somewhat ductile, corrosion and oxidation resistant, and nontoxic. (b) Some of the metals and alloys that are used and have been used for coins are gold, silver, copper, nickel, copper-nickel alloys, and brass alloys.
11.D3 The first part of this question asks for a description of the shape memory phenomenon. A part having some shape and that is fabricated from a metal alloy that displays this phenomenon is plastically deformed. It can be made to return to its original shape by heating to an elevated
temperature. Thus, the material has a shape memory, or "remembers" its previous shape. Next we are asked to explain the mechanism for this phenomenon. A shape memory alloy is polymorphic (Section 3.6)--that is, it can exist having two crystal structures. One is body-centered cubic structure (termed an austenite phase) that exists at elevated temperatures; upon cooling, and at some temperature above the ambient, it transforms to a martensitic structure. Furthermore, this martensitic phase is highly twinned. Upon application of a stress to this low-temperature martensitic phase, plastic deformation is accomplished by the migration of twin boundaries to some preferred orientation. Once the stress is removed, the deformed shape will be retained at this temperature. When this deformed martensite is subsequently heated to above the phase transformation temperature, the alloy reverts back to the BCC phase, and assumes the original shape. procedure may then be repeated. 311 The
One material that exhibits this behavior is a nickel-titanium alloy. Furthermore, the desired "memory" shape may is established by forming the material above its phase transition temperature. Several applications for alloys displaying this effect are eyeglass frames, shrink-to-fit pipe couplers, tooth-straightening braces, collapsible antennas, greenhouse window openers, antiscald control valves on showers, women's foundations, and fire sprinkler valves.
11.D4 (a) Compositionally, the metallic glass materials are rather complex; several compositions are as follows: Fe80B20, Fe72Cr8P13C7, Fe67Co18B14Si, Pd77.5Cu6.0Si16.5, and 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 2) inasmuch as very rapid cooling rates are required, at least one 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, reinforcements for pressure vessels and tires, shields for electromagnetic interference, security tapes for library books. (e) Production techniques include centrifuge melt spinning, planar-flow casting, rapid pressure
application, arc melt spinning.
11.D5 This question provides us with a list of several metal alloys, and then asks us to pick those that may be strengthened by heat treatment, by cold work, or both. Those alloys that may be heat treated are either those noted as "heat treatable" (Tables 11.6 through 11.9), or as martensitic stainless steels (Table 11.4). Alloys that may be strengthened by cold working must not be
exceptionally brittle, and, furthermore, must have recrystallization temperatures above room temperature (which immediately eliminates lead).
Heat Treatable 6150 Steel C17200 Be-Cu 6061 Al
Cold Workable 6150 Steel C17200 Be-Cu 6061 Al 304 Stainless Steel R50500 Ti C51000 Phosphor Bronze AZ31B Mg
Both 6150 Steel C17200 Be-Cu 6061 Al
312
11.D6 This problem asks us to select from four alloys (brass, steel, titanium, and aluminum), the one that will support a 50,000 N (11,250 lb ) load without plastically deforming, and having the minimum f weight. From Equation (6.1), the cross-sectional area (A ) must necessarily carry the load (F) o without exceeding the yield strength ( ), as y
F
Ao =
y
Now, given the length l, the volume of material required (V) is just
lF V = l Ao =
y
Finally, the mass of the member (m) is
lF
y
m = V =
in which is the density. Using the values given for these alloys
m(brass) =
8.5 g/ cm3 (10 cm)(50,000 N) = 102 g 1 m 2 6 N / m2 415 x 10 102 cm
m(steel) =
7.9 g / cm3 (10 cm)(50, 000 N) = 46 g 1m 2 6 N/ m 2 860 x 10 102 cm
2.7 g / cm3 (10 cm)(50,000 N) = 43.5 g 1 m 2 310 x 106 N/ m2 2 10 cm
m(aluminum) =
m(titanium) =
4.5 g / cm3 (10 cm)(50, 000 N) = 40.9 g 1 m 2 550 x 106 N/ m 2 2 10 cm
313
Thus, titanium would have the minimum weight (or mass), followed by aluminum, steel, and brass.
11.D7 This question asks for us to decide whether or not it would be advisable to hot-work or cold-work several metals and alloys. Tin would almost always be hot-worked. Even deformation at room temperature would be considered hot-working inasmuch as its recrystallization temperature is below room temperature (Table 7.2). Tungsten is hard and strong at room temperature, has a high recrystallization temperature, and experiences oxidation at elevated temperatures. Cold-working is difficult because of its
strength, and hot-working is not practical because of oxidation problems. Most tungsten articles are fabricated by powder metallurgy, or by using cold-working followed by 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, magnesium alloys have relatively low recrystallization temperatures. A 4140 steel could be cold-worked in an over-tempered state which leaves it soft and relatively ductile, after which quenching and tempering heat treatments may be employed to strengthen and harden it. This steel would probably have a relatively high recrystallization temperature, and hot-working may cause oxidation.
11.D8 A one-inch diameter steel specimen is to be quenched in moderately agitated oil. We are to decide which of five different steels will have surface and center hardnesses of at least 55 and 50 HRC, respectively. In moderately agitated oil, the equivalent distances from the quenched end for a one-inch diameter bar for surface and center positions are 3 mm (1/8 in.) and 8 mm (11/32 in.), respectively [Figure 11.16(b)]. The hardnesses at these two positions for the alloys cited (as determined using Figure 11.13) are given below.
Alloy 1040 5140 4340 4140 8640
Surface Hardness (HRC) 50 55 57 56 56
Center Hardness (HRC) 30 47 57 54 52.5
314
Thus, alloys 4340, 4140, and 8640 will satisfy the criteria for both surface and center hardnesses.
11.D9 (a) This problem calls for us to decide which of 8660, 8640, 8630, and 8620 alloys may be fabricated into a cylindrical piece 57 mm (2-1/4 in.) in diameter which, when quenched in mildly agitated water, will produce a minimum hardness of 45 HRC throughout the entire piece. The center of the steel cylinder will cool the slowest and therefore will be the softest. In moderately agitated water the equivalent distance from the quenched end for a 57 mm diameter bar for the center position is about 11 mm (7/16 in.) [Figure 11.16(a)]. The hardnesses at this position for the alloys cited (Figure 11.14) are given below.
Alloy 8660 8640 8630 8620
Center Hardness (HRC) 61 49 36 25
Therefore, only 8660 and 8640 alloys will have a minimum of 45 HRC at the center, and therefore, throughout the entire cylinder. (b) This part of the problem asks us to do the same thing for moderately agitated oil. In moderately agitated oil the equivalent distance from the quenched end for a 57 mm diameter bar at the center position is about 17.5 mm (11.16 in.) [Figure 11.16(b)]. The hardnesses at this position for the alloys cited (Figure 11.14) are given below. Alloy 8660 8640 8630 8620 Center Hardness (HRC) 59 42 30 21
Therefore, only the 8660 alloy will have a minimum of 45 HRC at the center, and therefore, throughout the entire cylinder.
315
11.D10 A thirty-eight millimeter diameter cylindrical steel specimen is to be heat treated such that the microstructure throughout will be at least 80% martensite. We are to decide which of several alloys will satisfy this criterion if the quenching medium is moderately agitated (a) oil, and (b) water. (a) Since the cooling rate is lowest at the center, we want a minimum of 80% martensite at the center position. From Figure 11.16(b), the cooling rate is equal to an equivalent distance from the quenched end of 12 mm (1/2 in.). According to Figure 11.13, the hardness corresponding to 80% martensite for these alloys is 50 HRC. Thus, all we need do is to determine which of the alloys have a 50 HRC hardness at an equivalent distance from the quenched end of 12 mm (1/2 in.). At an equivalent distance of 12 mm (1/2 in.), the following hardnesses are determined from Figure 11.13 for the various alloys.
Alloy 4340 4140 8640 5140 1040
Hardness (HRC) 55 52 48 42 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 8 mm (5/16 in.) equivalent distance from the quenched end [Figure 11.16(a)]. At this position, the following hardnesses are determined from Figure 11.13 for the several alloys.
Alloy 4340 4140 8640 5140 1040
Hardness (HRC) 56 55 54 49 32
It is still necessary to have a hardness of 50 HRC or greater at the center; thus, alloys 4340, 4140, and 8640 qualify.
11.D11 A fifty-millimeter (two-inch) diameter cylindrical steel specimen is to be quenched in moderately agitated water. We are to decide which of eight different steels will have surface and center
hardnesses of at least 50 and 40 HRC, respectively. 316
In moderately agitated water, the equivalent distances from the quenched end for a 50 mm diameter bar for surface and center positions are 2 mm (1/16 in.) and 10 mm (3/8 in.), respectively [Figure 11.16(a)]. The hardnesses at these two positions for the alloys cited are given below.
Alloy 1040 5140 4340 4140 8620 8630 8640 8660
Surface Hardness (HRC) 53 57 57 57 43 52 57 64
Center Hardness (HRC) 28 46 56 54 28 38 52 62
Thus, alloys 5140, 4340, 4140, 8640, and 8660 will satisfy the criteria for both surface and center hardnesses.
11.D12 We are asked to determine the maximum diameter possible for a cylindrical piece of 4140 steel that is to be quenched in moderately agitated oil such that the microstructure will consist of at least 50% martensite throughout the entire piece. From Figure 11.13, the equivalent distance from the quenched end of a 4140 steel to give 50% martensite (or a 42.5 HRC hardness) is 26 mm (1-1/16 in.). Thus, the quenching rate at the center of the specimen should correspond to this equivalent distance. Using Figure 11.16(b), the center specimen curve takes on a value of 26 mm (1-1/16 in.) equivalent distance at a diameter of about 75 mm (3 in.).
11.D13 In this problem we are asked to describe a heat treatment that may be used on a 45 mm diameter steel shaft of a 1040 steel such that it will have a uniform tensile strength of at least 620 MPa across the entirety of its cross-section. First of all, if the steel is heat treated so as to produce martensite or tempered martensite, there will undoubtedly be a variation of tensile strength over the cross-section. Thus, let us determine if either martensite or tempered martensite is required to give this desired tensile strength, and, if not, what microstructure is necessary. From Equation (6.20a) a tensile strength of 620 MPa corresponds to a Brinell hardness of about 180. Upon consultation of Figure 10.21(a), we note that for an alloy of composition of 0.40 wt% C, in order to achieve a hardness of 180 HB, a microstructure of fine pearlite is required. One possible heat treatment that may be used to produce fine pearlite is a continuous cooling one. In
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Figure 10.29 is shown a continuous cooling transformation diagram for a 0.35 wt% C alloy, which would be very similar to that for a 1040 steel. According to this diagram, we would want to
austenitize the alloy at approximately 900C, and then continuously cool to room temperature such that the cooling curve would pass through the fine pearlite region of the diagram (per Figure 10.17). This would correspond to a cooling rate of approximately 3C/s (Figure 10.18), which would For this
probably result from air-cooling the shaft from the 900C austenitizing temperature.
relatively low cooling rate, the microstructure, and therefore the tensile strength, will be quite uniform across the specimen cross section.
11.D14 We are to determine, for a cylindrical piece of 8660 steel, the minimum allowable diameter possible in order yield a surface hardness of 58 HRC, when the quenching is carried out in moderately agitated oil. From Figure 11.14, the equivalent distance from the quenched end of an 8660 steel to give a hardness of 58 HRC is about 18 mm (3/4 in.). Thus, the quenching rate at the surface of the specimen should correspond to this equivalent distance. Using Figure 11.16(b), the surface
specimen curve takes on a value of 18 mm equivalent distance at a diameter of about 95 mm (3.75 in.).
11.D15 This problem is concerned with the precipitation-hardening of copper-rich Cu-Be alloys. It is necessary for us to use the Cu-Be phase diagram (Figure 11.26 ). (a) The range of compositions over which these alloys may be precipitation hardened is between approximately 0.2 wt% Be (the maximum solubility of Be in Cu at about 300C) 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 heat treatment must be carried out at a temperature within the phase region, after which, the specimen is quenched to room temperature. Finally, the precipitation heat treatment is conducted at a temperature within the + 2 phase region. For example, for a 1.5 wt% Be-98.5 wt% Cu alloy, the solution heat treating temperature must be between about 600C (1110F) and 900C (1650F), while the precipitation heat treatment would be below 600C (1110F), and probably above 300C (570F). Below 300C, diffusion rates are low, and heat treatment times would be relatively long. 11.D16 We are asked to specify a practical heat treatment for a 2014 aluminum alloy that will produce a minimum yield strength of 345 MPa (50,000 psi), and a minimum ductility of 12%EL. From Figure 11.25(a), the following heat treating temperatures and time ranges are possible to the give the required yield strength.
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Temperature (C) 260 204 149 121
Time Range (h) not possible 0.3-15 10-700 300-?
With regard to temperatures and times to give the desired ductility [Figure 11.25(b)]:
Temperature (C) 260 204 149 121
Time Range (h) <0.02, >10 <0.4, >350 <20 <1000
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 overlap of the two sets of time ranges. At 204C, the heat treating time would need to be about 0.4 h, which is practical. At 149C, the time range is between 10 and 20 h, which is a little on the long side. Finally, at 121C, the time range is unpractically long (300 to 1000 h).
11.D17 This problem inquires as to the possibility of producing a precipitation-hardened 2014 aluminum alloy having a minimum yield strength of 380 MPa (55,000 psi) and a ductility of at least 15%EL. In order to solve this problem it is necessary to consult Figures 11.25(a) and 11.25(b). Below are tabulated the times required at the various temperatures to achieve the stipulated yield strength.
Temperature (C) 260 204 149 121
Time Range (h) not possible 0.5-7 10-250 500-2500
With regard to temperatures and times to give the desired ductility:
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Temperature (C) 260 204 149 121
Time Range (h) <0.005 <0.13 <10 <450
Therefore, an alloy having this combination of yield strength and ductility is marginally possible. A heat treatment at 149C for 10 h would probably just achieve the stipulated ductility and yield strength.
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