Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Textbook Question
Chapter 9.13, Problem 64AAP
A 1.10% C hypereutectoid plain-carbon steel is slowly cooled from 900°C to a temperature just slightly below 727°C.
- a. Calculate the weight percent proeutectoid cementite present in the steel.
- b. Calculate the weight percent eutectoid cementite and the weight percent eutectoid ferrite present in the steel.
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Question-6
A 0.25% C hypoeutectoid plain-carbon steel is slowly cooled from 950°C to a temperature just
slightly below 723°C.
(a) Calculate the weight percent proeutectoid ferrite in the steel.
(b) Calculate the weight percent eutectoid ferrite and weight percent eutectoid cementite in the
steel.
The chemical composition of X38CrMoV5-3 steel is given in the table below.
b) This steel is austenitized at 1050 ° C and hardened by giving water in oil (completely martensite structure has been formed and has a hardness of 57 HRC) and tempered at 510 ° C to reduce its hardness (54 HRC) and gain toughness. In addition, it is desired to harden the surface by nitration. In order to achieve the desired surface hardness, liquid nitration should be done at 600 ° C, gas nitration at 510 ° C and plasma nitration at 450 ° C. Which nitration method do you prefer? Why is that?
A 200g hypoeutectoid plain carbon steel contains 5.1wt% eutectoid ferrite.
1) What is its average carbon content in weight?
2) What is the weight of total ferrite?
1600
8+7
Liquid
1400
L+ Fe;C
L+y
1200
1148°C
y austenite
1000
2.0
4.3
y + Fe,C
800 e
723°C
600
0.8
0.02
400
a + Fe,C
Fe;C
200
0% C
100%
Fe
6 6.67
Fe,C
3
4
Weight percent carbon
Temperature ("C)
Chapter 9 Solutions
Foundations of Materials Science and Engineering
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