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Precipitation

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 866C).
(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 600C (1110F), and probably above 300C (570F). Below 300C, 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 260C so as to produce the desired set of properties— there is no overl
At 204C, the heat treating time would be between 0.02 and 0.15 h; times lying within this range are impractical
At 149C, the time would be between 3 and 10 h.
Finally, at 121C, 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 260C, the heat treating time would need to be less than 0.02 h (1.2 min), which is impractically short.
At 204C, the heat treatment would need to be less than 0.4 h (24 min), which is a little on the short side.
At 149C, the time range would be between 1 and 20 h.
Finally, at 121C, this property combination is possible for virtually all times less than about 1000 h.
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