a) Using a TTT diagram of steel, describe the phase transformation conditions of ferrite + pearlite,...
a) Using a TTT diagram of steel, describe the phase transformation conditions of ferrite + pearlite, upper bainite, lower bainite, and martensite.
B) Using schematic diagrams, differentiate the formation mechanisms of pearlite, upper bainite, and lower bainite.
PLEASE ANSWER BOTH PARTS WITH SUITABALE DIAGRAMS AND EXPLANTIONS
Homework Answers
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a) Using a TTT diagram of steel, describe the phase
transformation conditions of ferrite + pearlite,...
Using the Animated Figure 10.40, the isothermal transformation diagram for a 0.45 wt%C steel alloy, specify...
Using the Animated Figure 10.40, the isothermal transformation diagram for a 0.45 wt%C 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 700 degrees C, hold...
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)...
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...
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...
Martensite is on the TTT diagram but is not on the phase diagram of steel. How...
Martensite is on the TTT diagram but is not on the phase diagram of steel. How do you explain the formation of martensite in steel?
7. A hypereutectoid steel contains 0.8% C. When slow cooled from austenite the following transformation products...
7. A hypereutectoid steel contains 0.8% C. When slow cooled from austenite the following transformation products result. a) ferrite and bainite b) ferrite and martensite c) ferrite and pearlite d) pearlite and cementite 8. Pearlite is a combination of a) cementite and iron carbide b) ferrite and cementite c) ferrite and bainite d) ferrite and martensite 9. The process of hardening steel consists of what two steps? a) melting the steel followed by quenching b) melting the steel followed by...
7.a) Describe the formation of pearlite, bainite, and martensite in the isothermal transformation. Mark each of...
7.a) Describe the formation of pearlite, bainite, and martensite in the isothermal transformation. Mark each of it at the right locations in the diagram given below for the Iron-Carbon alloy by copying the figure. Points: 5 800 T(°C) TE 600 1 400 - 200 - 10-1 10 10 105 b) Briefly explain how coarse and fine pearlites are prepared and why fine pearlite is harder and stronger than coarse pearlite. Points: 5
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...
phase dtransformation diagram and heat treatment. Can you answer parts a b and c please. need...
phase dtransformation diagram and heat
treatment.
Can you answer parts a b and c please. need it done in
the next 30 mins
1: Using the phase transformation diagram below, describe heat treatment procedures to produce the following from eutectoid steel at 750°C 800 1400 А Eutectoid temperature 700 A 1200 P 600 1000 N 500 -- Temperature (°C) Temperature (°F) @ B 800 400 A 600 300 50% 400 200 100 104 10-1 105 1 103 10 102 Time...
Instructions: For the following problems utilize the listed TTT Diagram from your Equation Bookle...
Instructions: For the following problems
utilize the listed TTT Diagram from your Equation Booklet. The
treatment begins with austenitizing temperature. RT = Room
Tempearture. Use 1 for 0% Transformed as an
answer!
--Given Values--
Time Temperature Transformation Diagram: = 1335 Steel
Heat Treatment: = Quench to 904 F hold for 12 seconds then quench
to 700F hold for 30 seconds then quench to RT then reheat to 1200 F
for 18 hrs
Problem 1 Part A:
For the diagram and...
Using the Animated Figure 10.40, the isothermal transformation diagram for a 0.45 wt%C 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 700 degrees C, hold...
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)...
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...
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...
7. A hypereutectoid steel contains 0.8% C. When slow cooled from austenite the following transformation products result. a) ferrite and bainite b) ferrite and martensite c) ferrite and pearlite d) pearlite and cementite 8. Pearlite is a combination of a) cementite and iron carbide b) ferrite and cementite c) ferrite and bainite d) ferrite and martensite 9. The process of hardening steel consists of what two steps? a) melting the steel followed by quenching b) melting the steel followed by...
7.a) Describe the formation of pearlite, bainite, and martensite in the isothermal transformation. Mark each of it at the right locations in the diagram given below for the Iron-Carbon alloy by copying the figure. Points: 5 800 T(°C) TE 600 1 400 - 200 - 10-1 10 10 105 b) Briefly explain how coarse and fine pearlites are prepared and why fine pearlite is harder and stronger than coarse pearlite. Points: 5
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...
phase dtransformation diagram and heat
treatment.
Can you answer parts a b and c please. need it done in
the next 30 mins
1: Using the phase transformation diagram below, describe heat treatment procedures to produce the following from eutectoid steel at 750°C 800 1400 А Eutectoid temperature 700 A 1200 P 600 1000 N 500 -- Temperature (°C) Temperature (°F) @ B 800 400 A 600 300 50% 400 200 100 104 10-1 105 1 103 10 102 Time...
Instructions: For the following problems
utilize the listed TTT Diagram from your Equation Booklet. The
treatment begins with austenitizing temperature. RT = Room
Tempearture. Use 1 for 0% Transformed as an
answer!
--Given Values--
Time Temperature Transformation Diagram: = 1335 Steel
Heat Treatment: = Quench to 904 F hold for 12 seconds then quench
to 700F hold for 30 seconds then quench to RT then reheat to 1200 F
for 18 hrs
Problem 1 Part A:
For the diagram and...
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