Question

1. Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition, 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 760°C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure (a) Rapidly cool to 400°C, hold for 1000 s, then quench to room temperature. (b) Reheat the specimen in part (a) to 700°C (1290°F) for 20h. (c) Cool rapidly to 665°C (1230°F), hold for 1000 s, then quench to room temperature. (d) Rapidly cool to 350°C (660°F), hold for 150 s, then quench to room temperature. Figure 10.22 The complete isothermal transformation diagram 800 A 1400 Eutectoid temperature 700 for an iron-carbon alloy of eutectoid composition: A austenite; B, bainite; M, martensite; P. pearlite A 1200 P 600 1000 500 B 800 400 A 600 300 Mistart) 50% 200 400 M+A M50%) M190 % ) 100 200 102 103 104 1o5 10-1 10 Time (s) Temperature ("C) Temperature ("F)

Answer 1

Two main concepts here:

1. Quenching: On quenching or rapid cooling, atoms and molecules get fixed onto their positions as they don't get enough time to readjust to the decreasing temperature. Hence, whatever was the composition before quenching stays as it is immediately on cooling/quenching, unless time is given to readjust.

2. Tie lines: When the composition of a binary mixture falls in between curves of two different phases, the % of each phase can be calculated using tie lines. It will be explained for case b and d of the question.

Now, lets go over each scenario

Case a: Starting from 760C, the specimen is cooled rapidly to 400C, held for 1000s (10^3 on x-axis), and quenched to room temperature (RT = 25C). These steps are shown with the red arrow marks in the image below.

Figure 10.22 The complete isothermal transformation diagram 800 A 1400 - Eutectoid temperatu re 700 for an iron-carbon alloy of eutectoid composition: A austenite; B, bainite; M martensite; P pearlite 1200 P 600 1000 500 800 400 600 300 M(start) 50% 400 200 M+A M(50%) M190%) 100 |200 102 103 104 10- 10 105 Time (s) Temperature ("C) Temperature ("F)

Since on quenching composition does not change immediately, just after quenching the composition will be almost 100% B.

Case b: The above specimen is then heated to 700C for 20 h (20x60x60 s =72000s). This step is shown by the blue line

Figure 10.22 The complete isothermal transformation diagram 800 A 1400 - Eutectoid temperatu re 700 for an iron-carbon alloy of eutectoid composition: A austenite; B, bainite; M martensite; P pearlite 1200 P 600 1000 500 800 400 600 300 M(start) 50% 400 200 M+A M(50%) M190%) 100 |200 102 103 104 10- 10 105 Time (s) Temperature ("C) Temperature ("F)

Note: The actual time from beginning is actually 73000 s (1000s for step1 and 72000s for step 2). But since the time axis is logarithmic, 72000 and 73000 are approximately same.

Since the end point is in between the red (A) curve and green (P curve), the composition is a mixture of A and P. To get an approximate value of each component in the specimen at this point, use the concept of tie line

А Overall composition, C P A' х1 x2 Temperature

If the specimen has overall composition of C consisting of mixture of A and P, then the %A and %P can be calculated by

(a) drawing a horizontal line through the point C intersecting the curves A and P at A' and P' respectively

(b) %A = distance between CP'/ (total distance between A' and P') = x1/(x1+x2), and

%P = distance between CA'/ (total distance between A' and P') = x2/(x1+x2),

essentially, to get % of A, look at distance from the point C to the other curve (P) and divide by the total distance between A and P.

Following the above logic for the question in hand, the % of A and P appears 10% A and 90% P.

Case c: Starting from 760C, cool rapidly to 665C, hold for 1000s, and quench to RT

Following similar approach to Case a, this problem can be seen as follows. Hence, immediately on quenching the composition is going to be solely of P.

Figure 10.22 The complete isothermal transformation diagram 800 A 1400 - Eutectoid temperatu re 700 for an iron-carbon alloy of eutectoid composition: A austenite; B, bainite; M martensite; P pearlite 1200 P 600 1000 500 800 400 600 300 M(start) 400 200 M+A M(50%) M190%) 100 |200 102 103 104 10- 10 105 Time (s) Temperature ("C) Temperature ("F)

Case d: From 760C rapidly cool to 350C, hold for 150s and quench to RT. Follow the path diagram as explained previously to get to the end point, as shown below.

Figure 10.22 The complete isothermal transformation diagram for an iron-carbon 800 A 1400 - Eutectoid temperatu re 700 alloy of eutectoid composition: A austenite; B, bainite; M. martensite: P. pearlite 1200 P 600 1000 500 800 400 600 300 M(start) 200 50% 400 M+A M(50% M190%) 100 200 10- 103 102 1 10 104 105 Time (s) Temperature ("C) Temperature ("F)

Immediately on quenching, the composition will be that shown by the star mark, i.e., a mixture of A and B. Using the concept of tie line mentioned above, the A and B are approximately 50% each for this specimen.