Question

7.a) Describe the formation of pearlite, bainite, and martensite in the isothermal transformation. Mark each of it at the rig

0 0
Add a comment Improve this question Transcribed image text
Answer #1

(A)Temp TE Eutectic temp 700 0 pearlite Gool 2 bainite 900 3 200 matensite lot 10 103 105 time

Pearlite is formed when the austenite is cooled slowly just below eutectice temprature. Line 1 and 2 shows the course pearlite and fine pearlite.

Bainite is formed when austenite is cooled very fast below the neck and then slowly cooled. Curve 3 shows the formation of bainite.

Martensite is formed when austenite is rapidly cooled to room temprature. Curve 4 shows the formation of martensite.

(B)

Course pearlite is prepared by cooling the austenite just below the eutectic temprature as shown in curve 1. Fine pearlite is prepared by cooling the austenite much below the eutectic temprature near about 540 degree celcius as shown in curve 2. Fine pearlite is harder than course beacause finer the pearlite it will have less tendency to slip which will lead to more sttong and hard.

Add a comment
Know the answer?
Add Answer to:
7.a) Describe the formation of pearlite, bainite, and martensite in the isothermal transformation. Mark each of...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Q2. Below is shown the isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy. List...

    Q2. Below is shown the isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy. List the microconstituent(s) present for the heat treatment labeled on this diagram. It is not necessary to state the proportion(s) of the microconstituents. 6 options are given for each case. Pick the correct components and briefly explain your answer. (15 points) 900 1600 HIV 500 ZA+B (c) Temperature (°C) (6) A 6 Temperature (°F) Mstart) 300M(50%) ©) b) (90%) TTTT (b) (c) '(a) 103 (b)...

  • Make a copy of the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition provided...

    Make a copy of the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition provided overleaf and then sketch and label time–temperature paths on this diagram to produce the following microstructures: a) 25% martensite and 75% austenite b) 100% fine pearlite c) 25% pearlite, 25% bainite, 45% martensite, and 5% austenite. d) 100% tempered martensite. 800 A 1400 Eutectoid temperature 700 A 1200 A P 600 1000 500 800 AXTIS 400 A 600 300 M(start 50% 200 400 M+A...

  • a) Based on time-temperature-transformation (TTT) diagram in Figure 2, briefly explain and label the time-temperature paths...

    a) Based on time-temperature-transformation (TTT) diagram in Figure 2, briefly explain and label the time-temperature paths to produce the following microstructures (a), (b), (c). (d), 100% coarse pearlite 50% bainite 50% martensite 50% fine 100% martensite perlite and 50% martensite (10 marks) 800 727 700 Coarse pearlite 600 a + Fe,C Fine pearlite 500 400 300 200 100 1 see I min 1 hour I day 10 102 10) 10s Time, seconds Figure 2: Time-Temperature Transformation (TTT) diagram for eutectoid...

  • LOO Etter 700 A certain steel of eutectoid composition transforms according to the isothermal transformation curve...

    LOO Etter 700 A certain steel of eutectoid composition transforms according to the isothermal transformation curve shown below. a. Describe a thermal treatment that would result in a structure having 50% pearlite and 25% bainite, and 25% martensite. 1200 1000 Tenger! 500 TO 300 200 00 MASON MO 100 b. Assuming I hold for 105 seconds, what approximate temperature do I need to quench to and hold to get 100% i. Coarse pearlite 200 10 ii. Martensite iii. Bainite

  • 1. Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition, specify the nature...

    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...

  • X Incorrect. Using the Animated Figure 10.40, the isothermal transformation diagram for a 0.45 wt% steel...

    X Incorrect. Using the Animated Figure 10.40, the isothermal transformation diagram for a 0.45 wt% steel alloy, specify the nature of the final microstructure in terms of the microconstituents present) of a small specimen that has been subjected to the following temperature treatments. In each case assume that the specimen begins at 845 °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 250 degrees...

  • Using the isothermal transformation diagram for an alloy steel (type 4340) specify the nature of the final microstruct...

    Using the isothermal transformation diagram for an alloy steel (type 4340) specify the nature of the final microstructure (in terms of micro-constituents present and approximate percentages) 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 austenite structure. (a) (a) Rapidly cool to 400°C, hold for 10 seconds, and...

  • 900 1600 A A+C 404 • Chapter 10 Phase Transformations Figure 10.39 Isothermal transformation diagram for...

    900 1600 A A+C 404 • Chapter 10 Phase Transformations Figure 10.39 Isothermal transformation diagram for a 1.13 wt% Ciron-carbon alloy: A, austenite; B, bainite; C, proeutectoid ce 800 mentite; M, martensite; P.pearlite. [Adapted from H. Boyer (Editor), Atlas of Isother- mal Transformation and Cooling Transformation Diagrams, 1977. Reproduced by permission of ASM International, Materials Park, OH.) ? 1400 200 A+P 1200 600 1000 500 A+B Temperature (°C) 800 Temperature (°F) 400 A 300 600 50% 200 M(start) 400 M(50%)...

  • Austenite Pearlite Question 3 (24 pts): We are interested in performing different heat treatments to an...

    Austenite Pearlite Question 3 (24 pts): We are interested in performing different heat treatments to an iron-carbon alloy of eutectoid compostion. Use the provided Temperature-Time Transformation (TTT) and Continuous Cooling Transformation (CCT) diagrams to specify the nature of the final microstructure in terms of microconstituents present and approximate percentages) of a small specimen that has been subjected to the following heat treatments. In each case, assume that the specimen begins at 760°C and that it has been at this temperature...

  • True/False on Phase Transformation The figure below depicts four major phase transformations in steels (1.2.3. @)....

    True/False on Phase Transformation The figure below depicts four major phase transformations in steels (1.2.3. @). Determine whether the following statements are true or false. Each question is 3 pts. 1 11. 1 725 1/ U Coarse 10 Pearlite + P Fine Upper Temperature B Bainite Lower ME @ y + Martensite Martensite Time MA Question 16 (3 points) The cooling rate of 3 is faster than the cooling rate of 2. Question 16 (3 points) The cooling rate of...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT