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a) Based on time-temperature-transformation (TTT) diagram in Figure 2, briefly explain and label the time-temperature paths...
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...
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
8. For an alloy with 1.25 wt% Cat temperature just below eutectoid, determine the following: a....
8. For an alloy with 1.25 wt% Cat temperature just below eutectoid, determine the following: a. The fractions of total ferrite and cementite phases b. The fractions of pro-eutectoid cementite and pearlite C. The fraction of eutectoid cementite 9. For an eutectoid steel, sketch and label a time-temperature path that will produce the following: a. 50% Pearlite & 50% Bainite b. 100% Martensite C. 100% Coarse Pearlite
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)...
800 A 1400 Eutectoid temperature 700 A 1200 P 600 1000 500 A 800 400 A 600 300 M(start) 50% 400 200 M+ A M(50%) M(...
800 A 1400 Eutectoid temperature 700 A 1200 P 600 1000 500 A 800 400 A 600 300 M(start) 50% 400 200 M+ A M(50%) M(90%) 100 200 105 102 103 104 10-1 10 1 Time (s) Temperature (°C) Temperature (°F) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for the following heat treatment in the boxes below. Rapidly cool to 625°C, hold for 10s, rapidly cool to 450°C,...
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
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...
Round to the nearest 5%--no need for something like 43.2...that would become 45. If there's none of that phase form...
Round to the nearest 5%--no need for something like 43.2...that
would become 45. If there's none of that phase formed, enter a 0
(zero). If it looks close enough then it is
800 1400 Eutectoid temperature 700 1200 1000 500 800 400 600 300 M(start) 200 400 M + A M( 5096) M19096) 100 10 102 03 104105 Time (s) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for...
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...
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...
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...
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
8. For an alloy with 1.25 wt% Cat temperature just below eutectoid, determine the following: a. The fractions of total ferrite and cementite phases b. The fractions of pro-eutectoid cementite and pearlite C. The fraction of eutectoid cementite 9. For an eutectoid steel, sketch and label a time-temperature path that will produce the following: a. 50% Pearlite & 50% Bainite b. 100% Martensite C. 100% Coarse Pearlite
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)...
800 A 1400 Eutectoid temperature 700 A 1200 P 600 1000 500 A 800 400 A 600 300 M(start) 50% 400 200 M+ A M(50%) M(90%) 100 200 105 102 103 104 10-1 10 1 Time (s) Temperature (°C) Temperature (°F) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for the following heat treatment in the boxes below. Rapidly cool to 625°C, hold for 10s, rapidly cool to 450°C,...
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
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...
Round to the nearest 5%--no need for something like 43.2...that
would become 45. If there's none of that phase formed, enter a 0
(zero). If it looks close enough then it is
800 1400 Eutectoid temperature 700 1200 1000 500 800 400 600 300 M(start) 200 400 M + A M( 5096) M19096) 100 10 102 03 104105 Time (s) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for...
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...
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...
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