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 heat treatment listed above give the percentage of Pearlite:
Problem 1 Part B:
For the diagram and heat treatment listed above give the percentage of Upper Bainite:
Problem 1 Part C:
For the diagram and heat treatment listed above give the percentage of Lower Bainite:
Problem 1 Part D:
For the diagram and heat treatment listed above give the percentage of Quenched Martensite:
Problem 1 Part E:
For the diagram and heat treatment listed above give the percentage of Tempered Martensite:
Problem 1 Part F:
For the diagram and heat treatment listed above give the percentage of Pro-Eutectoid Ferrite:
Problem 1 Part E:
For the diagram and heat treatment listed above give the percentage of Spheroidite:
Homework Answers
Part A )% pearlite
part B 50% upper bainite
part C around 30% lower bainite
part D around 20% martensite
Part f around 58% proeutectoid ferrite
Part E around 4.8% of spherodite
Add Answer to:
Instructions: For the following problems utilize the listed TTT Diagram from your Equation Bookle...
PLEASE SHOW ALL STEPS AND LOGICAL REASONING. Figure 14: Carbon Steels: 1050 800 700 1 8...
PLEASE SHOW ALL STEPS AND LOGICAL REASONING.
Figure 14: Carbon Steels: 1050 800 700 1 8 23 26 32 36 42 48 52 I2 60 IO A F C 800 3oo 600 Mgo 20040O T DIAGRAM 0200 Estimated 62 IO io Io TIME SECONDS Time Temperature Transformation Diagram: 1050 Steel Heat Treatment: = Quench to 900 F hold for 3 seconds then quench to RT Problem 1 Part A For the diagram and heat treatment listed above give the percentage...
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...
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,...
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...
5. Using the attached ITT diagram for 1.13 wt% C steel, name the structures present and...
5. Using the attached ITT diagram for 1.13 wt% C steel, name the structures present and approximate quantities after the following heat treatments: a) heat to 900°C, quench to 700°C, hold for 10 s, quench to 0°C (the iron-iron carbide phase diagram will help) b) heat to 900°C, quench to 300°C, hold for 1000 s, quench to 0°C c) heat to 900°C, quench to 100°C d) heat to 800°C, quench to 0°C (the iron-iron carbide phase diagram will help) 404...
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%)...
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...
(1s) The TTT d austenitized. Determine the final amount (approximate percentages) of each microstructure, and show basic path (draw on charts). 6. agram for an iron-carbon alloy w, h 0.76%C steel is...
(1s) The TTT d austenitized. Determine the final amount (approximate percentages) of each microstructure, and show basic path (draw on charts). 6. agram for an iron-carbon alloy w, h 0.76%C steel is given. Assume that the specimen gans above 800°C for each case and that it has been held at this temperature long enough to be fully microstructure (such as bauxite, magnetite, bainite, pearlite, um, martensite, spheroidite, cementite). Must specify microstructure name (not a letter). 800 a) Sample 1- Decrease...
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...
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...
PLEASE SHOW ALL STEPS AND LOGICAL REASONING.
Figure 14: Carbon Steels: 1050 800 700 1 8 23 26 32 36 42 48 52 I2 60 IO A F C 800 3oo 600 Mgo 20040O T DIAGRAM 0200 Estimated 62 IO io Io TIME SECONDS Time Temperature Transformation Diagram: 1050 Steel Heat Treatment: = Quench to 900 F hold for 3 seconds then quench to RT Problem 1 Part A For the diagram and heat treatment listed above give the percentage...
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...
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,...
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...
5. Using the attached ITT diagram for 1.13 wt% C steel, name the structures present and approximate quantities after the following heat treatments: a) heat to 900°C, quench to 700°C, hold for 10 s, quench to 0°C (the iron-iron carbide phase diagram will help) b) heat to 900°C, quench to 300°C, hold for 1000 s, quench to 0°C c) heat to 900°C, quench to 100°C d) heat to 800°C, quench to 0°C (the iron-iron carbide phase diagram will help) 404...
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%)...
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...
(1s) The TTT d austenitized. Determine the final amount (approximate percentages) of each microstructure, and show basic path (draw on charts). 6. agram for an iron-carbon alloy w, h 0.76%C steel is given. Assume that the specimen gans above 800°C for each case and that it has been held at this temperature long enough to be fully microstructure (such as bauxite, magnetite, bainite, pearlite, um, martensite, spheroidite, cementite). Must specify microstructure name (not a letter). 800 a) Sample 1- Decrease...
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...
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...
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