The document provides solutions to problems involving interpreting an isothermal transformation diagram for an iron-carbon alloy of eutectoid composition. In problem 1, it is determined that for 50% completion of the austenite to pearlite reaction it will take 6 seconds, and for 100% completion it will take 9 seconds. The hardness of fully transformed pearlite is estimated to be around 205 Brinell. Problem 2 specifies the microconstituents and percentages that would result from various time-temperature treatments. Problem 3 involves sketching time-temperature paths on the diagram that would produce 100% fine pearlite and a microstructure with 50% coarse pearlite, 25% bainite, and 25% martensite

1. Problem 1
Suppose that a steel of eutectoid composition is cooled to 600ºC from 760ºC in less than 0.5 s
and held at this
temperature.
(a) How long will it take for the austenite to pearlite reaction to go to 50% completion? To
100% completion?
(b) Estimate the hardness of the alloy that has completely transformed to pearlite.
Solution:
a) You can see from the picture 50% line crosses about 6 sec and 100% line crosses
about 9 sec. It means that half of austenite changes its form to pearlite at 6 s, but full
austenite at 9 s.
b) It is said that eutectoid composition so weight of carbon is 0.76 % and it is less than
620 Celsius so it is fine pearlte. Then we can easily see from the Brinell hardness
graph that hardness is about 205.

2. Problem 2
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 time– temperature treatments. In each case assume that the specimen begins at
760ºC and that it has been held at this temperature long enough to have achieved a complete
and homogeneous austenitic structure.
(a) Cool rapidly to 700ºC, hold for 5000 s, then quench to room temperature.
(b) Reheat the specimen in part (a) to 700ºC for 20 h.
(c) Rapidly cool to 600ºC, hold for 6 s, rapidly cool to 400ºC, hold for 60 s, then quench to
room temperature.
(d) Cool rapidly to 400ºC, hold for 2 s, then quench to room temperature.
(e) Cool rapidly to 400ºC, hold for 20 s, then quench to room temperature.
(f) Cool rapidly to 400ºC, hold for 300 s, then quench to room temperature.
Solution:
a) 50% Pearlite, 50%Martensite

3. b) We have 50% pearlite and and 50% martensite. When reheat them to 700 Celsius then
pearlite will be Spheroidite and martensite will be Tempered Martensite
c) 50% Pearlite 25% Bainite 25%Martensite

4. d) 100 % Martensite

5. e) 30% Bainite 70%Martensite

6. f) 100% Bainite

7. Problem 3
Make a copy of the isothermal transformation diagram for an iron–carbon alloy of eutectoid
composition (Figure 10.22) and then sketch and label time–temperature paths on this diagram
to produce the following.
microstructures:
(a) 100% fine pearlite
(b) 50% coarse pearlite, 25% bainite, and 25% martensite
a)

8. b)
760
580
500

9. 760
650
450
180
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