a.
100% pearlite
From figure 2 cooling rate 35/°C is very slow hence pearlite will form.
B.
100% Martensite
At 600F for 10 seconds No transformation take place hence after quenching all Austenite gets converted to Martensite.
C.
50% Martensite+ 50% Austenite
At 175°C half of austenite gets converted to Martensite and remaining half gets stablizes.
D.
50% Bainite+ 50 % Martensite
At 350°C holding for 100 seconds half of austenite gets converted to bainite and after quenching remaining half gets converted to Martensite.
E.
100% Pearlite
At 600°C holding for 10 seconds all Austenite gets converted to Pearlite.
F.
100% Spherodite
Reheating will improve properties of steel and spherodite steel formation will take place.
Austenite Pearlite Question 3 (24 pts): We are interested in performing different heat treatments to an...
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...
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...
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,...
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...
(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...
this is the TTT curve Given a plain carbon steel which has 8 % primary (or pro eutectoid) carbide, determine both overall carbon content and ferrite content. a. Assume that the steel was at equilibrium just above the eutectoid temperature and then cooled immediately to 600°C and held for 5 seconds, then quenched. What microconstituents would there be in the structure and how much of each? (Assume the cooling behavior follows the TTT curve shown) 800 А Eutectoid temperature 1400...
800 1400 Eutectoid temperature 700 A 1200 600 1000 500 B 800 400 A 600 300 M(start 50% 200 400 M+A M(50%) M(90%) 100 200 C 10-1 102 103 1 10 104 105 Time (s) Temperature (C) Temperature (°F) Rapidly cool to 625°C, hold for 10 s, rapidly cool to 450°C, hold for 10s, then quench to room temperature. % Pearlite 100 % Bainite % Martensite 100 % Austenite 0 % Tempered Martensite 10 Enter answers as a whole number...
Describe the heat treatment scheume that would provide the following property changes to 1080 steel. Refer to TTT diagram below. a) 100% pearlite to a miture of 50% pearlite and 50% martensite b) Mixture of 75% pearlite and 25% martensite to 100% martensite 800 1400 Eutectoid temperature 700 1200 600 1000 500 800 400 600 300 M(start) M(50%) M(90%) 50% 400 200 M + A 100 200 0 10-1 102 Time (s) 10 103 10 10
Please show all work! Thanks! Isothermal transformation diagram is given below. 13.8-1 Using the isothermal transformation diagram for a carbon steel of eutectoid composition (obtained from the In specify the final microstructure (in terms of microstructure pres tituent present) for small sa ent and approximate percentages of each phase or con. uent present) for small samples allowed to equilibrate at 760°C (1400 F) following each of the cooling a. Cool rapidly to 700°C (1290°F), hold for 10 seconds, then quench...
HW Module 10 Due: March 31, 2019 11:59PM 1- Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition (Following Figure), specify the nature of the final microstructure (in terms of micro constituents 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 (1400°F) and that it has been held at this temperature long enough to have achieved a...