2. The following figures show the TTT and phase diagrams of a hypoeutectoidic st
ID: 1867450 • Letter: 2
Question
2. The following figures show the TTT and phase diagrams of a hypoeutectoidic steel. There are two solid-solution treatment temperatures to be used: Point A-one above the As line (full solid-solution) and Point B one below the As line (in the aty two phase zone). Please describe the microstructure and predict strength of the steel under the following four cooling histories: (30 points) (Please do not read the temperature and time on the plot but focus on describing materials microstructure) (1) Cool down slowly from Point A to room temperature (the cooling history is not shown in the plot (2) Cooling history #2, indicated in the left figure. (3) Cooling history #3, indicated in the left figure. (4) Cooling history #4, indicated in the left figure. 900 C 800 700 600 500 727 C 50 (100 4002 300 Ms 200 10 Time 10 10 0.1Explanation / Answer
a) In this situation, hypoeutectoid steel is cooled from austenite to room temperature. Hence when austenite reaches the austenite+ferrite region, peoeutectoid ferrite phase is formed. On cooling down below eutectoid temperature, remaining austenite is transformed to pearlite. It is an alternate layers of ferrite and cementite. So final microstructure constituents are ferrite and cementite.
b) Here initially on cooling rapidly 100% martensite is formed. It is having a body centred tetragonal shape and is hardest transformation product in steel alloys. It is then tempered to reduce its hardness and increase toughness. So final microstructure constituents are ferrite and cementite in the tempered martensite.
c) It is brought to 500 degree celsius from austenite region and held for sufficient time for 100% formation of bainite.
d) It is at eutectoid temperature and rapidly cooled to room temperature. Hence fine pearlite is the final product
The following is the increasing order of strength
1<4<2<3
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