Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 32

Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength: (a) \(0.25 \mathrm{wt} \% \mathrm{C}\) with spheroidite (b) \(0.25 \mathrm{wt} \% \mathrm{C}\) with coarse pearlite (c) \(0.60 \mathrm{wt} \% \mathrm{C}\) with fine pearlite (d) \(0.60 \mathrm{wt} \% \mathrm{C}\) with coarse pearlite Justify this ranking.

Short Answer

Expert verified
Question: Rank the following iron-carbon alloys and associated microstructures in terms of their tensile strength, from highest to lowest: (a) 0.25 wt % C with spheroidite, (b) 0.25 wt % C with coarse pearlite, (c) 0.60 wt % C with fine pearlite, and (d) 0.60 wt % C with coarse pearlite. Answer: The ranking from highest to lowest tensile strength is: (c), (d), (b), (a).

Step by step solution

01

Understand the microstructures

Spheroidite microstructure consists of cementite (Fe3C) particles in a ferrite (α-Fe) matrix. Coarse pearlite and fine pearlite microstructures consist of alternating layers of ferrite and cementite. The difference between coarse and fine pearlite is the spacing of these alternating layers, with fine pearlite having a smaller spacing.
02

Analyze the effect of carbon content

Higher carbon content typically leads to increased strength, but also increased brittleness. In the case of alloys with the same microstructure, a higher carbon content results in higher tensile strength.
03

Analyze the effect of different microstructures

Fine pearlite has a higher tensile strength than coarse pearlite because its better dispersion of cementite particles gives the material a higher strength. Spheroidite has a lower tensile strength because the dispersed cementite particles do not provide a strengthening mechanism as the layered structure in pearlite does.
04

Rank the alloys based on tensile strength

Based on our analysis, we can rank the iron-carbon alloys and associated microstructures as follows: 1. \(0.60 \mathrm{wt} \% \mathrm{C}\) with fine pearlite (c): highest tensile strength due to the higher carbon content and fine pearlite microstructure. 2. \(0.60 \mathrm{wt} \% \mathrm{C}\) with coarse pearlite (d): higher carbon content, but coarse pearlite microstructure decreases its tensile strength compared to fine pearlite. 3. \(0.25 \mathrm{wt} \% \mathrm{C}\) with coarse pearlite (b): lower carbon content with coarse pearlite microstructure results in lower tensile strength than (c) and (d). 4. \(0.25 \mathrm{wt} \% \mathrm{C}\) with spheroidite (a): lowest tensile strength due to the spheroidite microstructure and the lowest carbon content. So, the ranking from highest to lowest tensile strength is: (c), (d), (b), (a).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Figure \(10.40\) shows the continuous cooling transformation diagram for a \(1.13 \mathrm{wt} \%\) C iron-carbon alloy. Make a copy of this figure and then sketch and label continuous cooling curves to yield the following microstructures: (a) Fine pearlite and proeutectoid cementite (b) Martensite (c) Martensite and proeutectoid cementite (d) Coarse pearlite and proeutectoid cementite (e) Martensite, fine pearlite, and proeutectoid cementite

Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates: (a) \(10^{\circ} \mathrm{C} / \mathrm{s}\) (b) \(1^{\circ} \mathrm{C} / \mathrm{s}\) (c) \(0.1^{\circ} \mathrm{C} / \mathrm{s}\) (d) \(0.01^{\circ} \mathrm{C} / \mathrm{s}\)

In terms of heat treatment and the development of microstructure, what are two major limitations of the iron-iron carbide phase diagram?

Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition (Figure \(10.22)\), specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following timetemperature treatments. In each case assume that the specimen begins at \(760^{\circ} \mathrm{C}\left(1400^{\circ} \mathrm{F}\right)\) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Cool rapidly to \(700^{\circ} \mathrm{C}\left(1290^{\circ} \mathrm{F}\right)\), hold for \(10^{4} \mathrm{~s}\), then quench to room temperature. (b) Reheat the specimen in part (a) to \(700^{\circ} \mathrm{C}\) \(\left(1290^{\circ} \mathrm{F}\right)\) for \(20 \mathrm{~h}\). (c) Rapidly cool to \(600^{\circ} \mathrm{C}\left(1110^{\circ} \mathrm{F}\right)\), hold for \(4 \mathrm{~s}\), rapidly cool to \(450^{\circ} \mathrm{C}\left(840^{\circ} \mathrm{F}\right)\), hold for \(10 \mathrm{~s}\), then quench to room temperature. (d) Cool rapidly to \(400^{\circ} \mathrm{C}\left(750^{\circ} \mathrm{F}\right)\), hold for \(2 \mathrm{~s}\), then quench to room temperature. (e) Cool rapidly to \(400^{\circ} \mathrm{C}\left(750^{\circ} \mathrm{F}\right)\), hold for \(20 \mathrm{~s}\), then quench to room temperature. (f) Cool rapidly to \(400^{\circ} \mathrm{C}\left(750^{\circ} \mathrm{F}\right)\), hold for \(200 \mathrm{~s}\), then quench to room temperature. (g) Rapidly cool to \(575^{\circ} \mathrm{C}\left(1065^{\circ} \mathrm{F}\right)\), hold for \(20 \mathrm{~s}\), rapidly cool to \(350^{\circ} \mathrm{C}\left(660^{\circ} \mathrm{F}\right)\), hold for \(100 \mathrm{~s}\), then quench to room temperature. (h) Rapidly cool to \(250^{\circ} \mathrm{C}\left(480^{\circ} \mathrm{F}\right)\), hold for \(100 \mathrm{~s}\), then quench to room temperature in water. Reheat to \(315^{\circ} \mathrm{C}\left(600^{\circ} \mathrm{F}\right)\) for \(1 \mathrm{~h}\) and slowly cool to room temperature.

On the basis of diffusion considerations, explain why fine pearlite forms for the moderate cooling of austenite through the eutectoid temperature, whereas coarse pearlite is the product for relatively slow cooling rates.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Work Energy and Power

Read Explanation

Magnetism

Read Explanation

Kinematics Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free