Applications and processing of metal alloys problem solutions



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CHAPTER 11
APPLICATIONS AND PROCESSING OF METAL ALLOYS
PROBLEM SOLUTIONS
Ferrous Alloys
11.1 (a) List the four classifications of steels. (b) For each, briefly describe the properties and typical applications.
Solution

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: hard and wear resistant; resistant to corrosion in a large variety of environments.

Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools.

11.2 (a) Cite three reasons why ferrous alloys are used so extensively. (b) Cite three characteristics of ferrous alloys that limit their utilization.


Solution

(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 a relatively high density.

(3) They have relatively low electrical conductivities.

11.3 What is the function of alloying elements in tool steels?


Solution

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.

11.4 Compute the volume percent of graphite VGr in a 3.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Solution

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




Conversion from weight fraction to volume fraction of graphite is possible using Equation 9.6a as


= 0.111 or 11.1 vol%

11.5 On the basis of microstructure, briefly explain why gray iron is brittle and weak in tension.


Solution

Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration.

11.6 Compare gray and malleable cast irons with respect to (a) composition and heat treatment, (b) microstructure, and (c) mechanical characteristics.
Solution

This question asks us to compare various aspects of gray and malleable cast irons.

(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 900C 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.

11.7 Compare white and nodular cast irons with respect to (a) composition and heat treatment, (b) microstructure, and (c) mechanical characteristics.


Solution

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 700C 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.

11.8 Is it possible to produce malleable cast iron in pieces having large cross-sectional dimensions? Why or why not?


Solution

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.



Nonferrous Alloys
11.9 What is the principal difference between wrought and cast alloys?
Solution

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.

11.10 Why must rivets of a 2017 aluminum alloy be refrigerated before they are used?
Solution

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.

11.11 What is the chief difference between heat-treatable and non-heat-treatable alloys?
Solution

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.

11.12 Give the distinctive features, limitations, and applications of the following alloy groups: titanium alloys, refractory metals, superalloys, and noble metals.
Solution

Titanium Alloys

Distinctive features: relatively low density, 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.

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.



Forming Operations
11.13 Cite advantages and disadvantages of hot working and cold working.
Solution

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.

11.14 (a) Cite advantages of forming metals by extrusion as opposed to rolling. (b) Cite some disadvantages.


Solution

(a) The advantages of extrusion as opposed to 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 a cross-section of an extruded piece.

Casting
11.15 List four situations in which casting is the preferred fabrication technique.
Solution

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.

11.16 Compare sand, die, investment, lost foam, and continuous casting techniques.
Solution

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 lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes.

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.



Miscellaneous Techniques
11.17 If it is assumed that, for steel alloys, the average cooling rate of the heat-affected zone in the vicinity of a weld is 10°C/s, compare the microstructures and associated properties that will result for 1080 (eutectoid) and 4340 alloys in their HAZs.
Solution

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 10C/s. Figure 10.27 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 10C/s (which is less than 35C/s) the resulting microstructure will be totally pearlite--probably a reasonably fine pearlite. On the other hand, in Figure 10.28 is shown the CCT diagram for a 4340 steel. From this diagram it may be noted that a cooling rate of 10C/s produces a totally martensitic structure. Pearlite is softer and more ductile than martensite, and, therefore, is most likely more desirable.

11.18 Describe one problem that might exist with a steel weld that was cooled very rapidly.
Solution

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.



Annealing Processes
11.19 In your own words describe the following heat treatment procedures for steels and, for each, the intended final microstructure: full annealing, normalizing, quenching, and tempering.
Solution

Full annealing--Heat to about 50C above the A3 line, Figure 11.10 (if the concentration of carbon is less than the eutectoid) or above the A1 line (if the concentration of carbon is greater than the eutectoid) until the alloy comes to equilibrium; then furnace cool to room temperature. The final microstructure is coarse pearlite.

Normalizing--Heat to at least 55C above the A3 line Figure 11.10 (if the concentration of carbon is less than the eutectoid) or above the Acm line (if the concentration of carbon is greater than the eutectoid) until the alloy completely transforms 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 650C, for the time necessary to achieve the desired hardness. The final microstructure is tempered martensite.

11.20 Cite three sources of internal residual stresses in metal components. What are two possible adverse consequences of these stresses?


Solution

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.

11.21 Give the approximate minimum temperature at which it is possible to austenitize each of the following iron–carbon alloys during a normalizing heat treatment: (a) 0.20 wt% C, (b) 0.76 wt% C, and (c) 0.95 wt% C.


Solution

(a) For 0.20 wt% C, heat to at least 905C (1660F) since the A3 temperature is 850C (1560F).

(b) For 0.76 wt% C, heat to at least 782C (1440F) since the A3 temperature is 727C (1340F).

(c) For 0.95 wt% C, heat to at least 840C (1545F) since the Acm temperature is 785C (1445F).

11.22 Give the approximate temperature at which it is desirable to heat each of the following iron–carbon alloys during a full anneal heat treatment: (a) 0.25 wt% C, (b) 0.45 wt% C, (c) 0.85 wt% C, and (d) 1.10 wt% C.
Solution

(a) For 0.25 wt% C, heat to about 880C (1510F) since the A3 temperature is 830C (1420F).

(b) For 0.45 wt% C, heat to about 830C (1525F) since the A3 temperature is 780C (1435F).

(c) For 0.85 wt% C, heat to about 777C (1430F) since the A1 temperature is 727C (1340F).

(d) For 1.10 wt% C, heat to about 777C (1430F) since the A1 temperature is 727C (1340F).

11.23 What is the purpose of a spheroidizing heat treatment? On what classes of alloys is it normally used?


Solution

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.


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