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Chapter 9: Problem 79

Often, the properties of multiphase alloys may be approximated by the relationship $$ E(\text { alloy })=E_{\alpha} V_{\alpha}+E_{\beta} V_{\beta} $$ where \(E\) represents a specific property (modulus of elasticity, hardness, etc.), and \(V\) is the volume fraction. The subscripts \(\alpha\) and \(\beta\) denote the existing phases or microconstituents. Use this relationship to determine the approximate Brinell hardness of a \(99.75 \mathrm{wt} \% \mathrm{Fe}-0.25 \mathrm{wt} \%\) C alloy. Assume Brinell hardnesses of 80 and 280 for ferrite and pearlite, respectively, and that volume fractions may be approximated by mass fractions.

Short Answer

Expert verified
Answer: The approximate Brinell hardness of the alloy is 80.5.

Step by step solution

01

Write the given values#

To solve the problem, let's first write the given values: Weight percentage of \(\alpha\) (ferrite) in alloy: \(99.75\%\) Weight percentage of \(\beta\) (pearlite) in alloy: \(0.25\%\) Brinell hardness of \(\alpha\) (ferrite): \(E_\alpha = 80\) Brinell hardness of \(\beta\) (pearlite): \(E_\beta = 280\) Since volume fractions can be approximated by mass fractions in this exercise, we have \(V_\alpha = 99.75\%\) and \(V_\beta = 0.25\%\). Now, we can plug these values into the given relationship to find the Brinell hardness of the alloy, \(E(\text{alloy})\).
02

Apply the relationship and substitute the values#

We have the following relationship for the Brinell hardness of the alloy: $$ E(\text { alloy }) = E_{\alpha} V_{\alpha} + E_{\beta} V_{\beta} $$ Now, substitute the given values into the formula: $$ E(\text { alloy }) = 80 \times 0.9975 + 280 \times 0.0025 $$
03

Calculate the Brinell hardness of the alloy#

To find the Brinell hardness of the alloy, perform the calculations: $$ E(\text { alloy }) = 79.8 + 0.7 $$ $$ E(\text { alloy }) \approx 80.5 $$ So, the approximate Brinell hardness of the \(99.75\% \ \mathrm{Fe} - 0.25\% \ \mathrm{C}\) alloy is about 80.5.

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Most popular questions from this chapter

The room-temperature tensile strengths of pure copper and pure silver are 209 and \(125 \mathrm{MPa}\), respectively. (a) Make a schematic graph of the room-temperature tensile strength versus composition for all compositions between pure copper and pure silver. (Hint: You may want to consult Sections \(9.10\) and 9.11, as well as Equation \(9.24\) in Problem 9.79.) (b) On this same graph, schematically plot tensile strength versus composition at \(600^{\circ} \mathrm{C}\). (c) Explain the shapes of these two curves as well as any differences between them.

The mass fraction of eutectoid ferrite in an ironcarbon alloy is \(0.71\). On the basis of this information, is it possible to determine the composition of the alloy? If so, what is its composition? If this is not possible, explain why.

Construct the hypothetical phase diagram for metals \(\mathrm{A}\) and \(\mathrm{B}\) between room temperature \(\left(20^{\circ} \mathrm{C}\right)\) and \(700^{\circ} \mathrm{C}\), given the following information: \- The melting temperature of metal \(\mathrm{A}\) is \(480^{\circ} \mathrm{C}\). \- The maximum solubility of \(B\) in \(A\) is 4 wt \(\%\) B, which occurs at \(420^{\circ} \mathrm{C}\). \- The solubility of \(\mathrm{B}\) in \(\mathrm{A}\) at room temperature is 0 wt \(\%\) B. \- One eutectic occurs at \(420^{\circ} \mathrm{C}\) and \(18 \mathrm{wt} \%\) B-82 wt \(\%\) A. \- A second eutectic occurs at \(475^{\circ} \mathrm{C}\) and \(42 \mathrm{wt} \%\) B- \(58 \mathrm{wt} \% \mathrm{~A}\) \- The intermetallic compound AB exists at a composition of \(30 \mathrm{wt} \% \mathrm{~B}-70 \mathrm{wt} \% \mathrm{~A}\), and melts congruently at \(525^{\circ} \mathrm{C}\). \- The melting temperature of metal B is \(600^{\circ} \mathrm{C} .\) \- The maximum solubility of \(\mathrm{A}\) in \(\mathrm{B}\) is \(13 \mathrm{wt} \% \mathrm{~A}\), which occurs at \(475^{\circ} \mathrm{C}\). \- The solubility of \(\mathrm{A}\) in \(\mathrm{B}\) at room temperature is \(3 \mathrm{wt} \% \mathrm{~A}\)

For a 64 wt \% Zn-36 wt\% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: \(900^{\circ} \mathrm{C}\left(1650^{\circ} \mathrm{F}\right), 820^{\circ} \mathrm{C}\) \(\left(1510^{\circ} \mathrm{F}\right), 750^{\circ} \mathrm{C}\left(1380^{\circ} \mathrm{F}\right)\), and \(600^{\circ} \mathrm{C}\left(1100^{\circ} \mathrm{F}\right)\) Label all phases and indicate their approximate compositions.

Consider \(6.0 \mathrm{~kg}\) of austenite containing \(0.45 \mathrm{wt} \%\) \(\mathrm{C}\) and cooled to less than \(727^{\circ} \mathrm{C}\left(1341^{\circ} \mathrm{F}\right)\). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (d) Schematically sketch and label the resulting microstructure.
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