Spheroidite can be obtained via which of the following thermal treatment (starting from the austenite phase field at high temperatures)? Rapidly cool to 575°C, hold for 20 s, rapidly cool to 350°C, hold for 100 s, then quench to room temperature. Cool to 650°C for 15 hours. Rapidly cool to room temperature. Cool to 700°C and hold for 20 hours.

10.35 Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Martensite to spheroidite (b) Spheroidite to martensite (c) Bainite to pearlite (d) Pearlite to bainite (e) Spheroidite to pearlite (f) Pearlite to spheroidite (g) Tempered martensite to martensite (h) Bainite to spheroidite

Bainite, spheroidite, coarse pearlite, and fine pearlite are all composed of ferrite and cementite. Sketch these four microstructures and label the two phases. For a given carbon content, the mechanical behavior of these microstructures differs, even though they contain the same amounts of ferrite and cementite. Explain why.

What is the driving force for the formation of spheroidite? the net reduction of the carbon concentration the net reduction in ferrite-cementite phase boundary area the net reduction of the grain boundary area None of above What is the common feature in pearlite, bainite, and spheroidite microstructures? layered structures consist of ferrite and cementite phases Ο Ο Ο …

Question: 2. Problem 2 Rank the following iron-carbon alloys and associated microstructures from the hardest to the softest: a) 0.25 wt% C with coarse pearlite b) 0.80 wt% C with spheroidite c) 0.25 wt% C with spheroidite d) 0.80 wt% C with fine pearlite. e) Justify this ranking.

Engineering; Mechanical Engineering; Mechanical Engineering questions and answers; Rank the following iron–carbon alloys and associated microstructures from the highest to the lowest tensile strength: (a) 0.25 wt%C with spheroidite, (b) 0.25 wt%C with coarse pearlite, (c) 0.60 wt%C with fine pearlite, and (d) 0.60 wt%C with coarse pearlite.

Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Spheroidite to tempered martensite (b) Tempered martensite to pearlite (c) Bainite to martensite (d) Martensite to pearlite (e) Pearlite to tempered martensite (f) Tempered martensite to pearlite (g) Bainite to tempered …

2. Explain how spheroidite is formed; and how is it different than layered pearlite? 3. Explain the main difference between intergranular and transgranular failure in materials. 4. Sketch and explain the three different types for fracture profiles, and provide an example of a material for each fracture profile. 5.

Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.

The particle size of cementite in tempered martensite (< 1 micron) is smaller than it is in spheroidite (> 1 micron) Tempered martensite has been cold worked, so additional dislocations have been introduced The ferritic matrix in tempered martensite is strained and distorted, in spheroidite it is strain free Tempered martensite is characterized ...