**Problem: Isothermal transformation diagram for an iron–carbon alloy of eutectoid composition (0.76 wt% C).** 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.1. **Sketch the following treatment on the figure:** Rapidly cool to 575°C, hold for 20s, rapidly cool to 350°C, hold for 100s, then quench to room temperature. Specify microconstituents with percentage.2. **Sketch the following treatment on the figure:** Rapidly cool to 600°C, hold 4s, rapidly cool to 450°C, hold for 10s, and rapidly cool to room temperature. Specify microconstituents with percentage.3. **Sketch the following treatment on the figure:** Rapidly cool to 450°C, hold 10s, rapidly cool to room temperature, heat to 680°C, hold 30hrs, and cool to room temperature. Specify microconstituents with percentage.4. **Rank from the highest to the lowest hardness:** - (a) 0.2 wt% C with spheroidite - (b) 0.2 wt% C with fine pearlite - (c) 0.76 wt% C with fine pearlite - (d) 0.76 wt% C with tempered martensite**Diagram Explanation:**- The graph is an isothermal transformation diagram, showing the temperature in degrees Celsius on the vertical axis and the time in seconds on the horizontal axis.- The diagram includes several curves illustrating the transformation boundaries for different microstructures: austenite (A), pearlite (P), and bainite (B).- The blue curves represent the start and finish lines for pearlite and bainite transformations.- Red lines indicate the martensite start (M(start)) and transformation percentages from 50% (M(50%)) to 90% (M(90%)).- The gray dotted line marks the eutectoid temperature.- Labels within different zones indicate the phases present in each region, such as A (austenite), P (pearlite), and B (bainite), as well as mixtures like M + A (martensite and austenite).

Given:To find:Rapid cooled to 575oC Leaves 100% austenite. Holding for 20 seconds completely transforms the austenite to ferrite and Fe3C in the pearlite structure and the lower temperature makes a fine pearlite. The additional cooling and isothermal hold change nothing as there is no more austenite left to transform.100% fine pearlite is formed.50% pearlite, 25% bainite and 25% martensite is formed when the specimen is cooled rapidly to 600oC from 760oC and held for 4 s and then rapidly cooled to 450oC and held for 10 s and quenched to room temperature. 50% austentite and 50% Bainite.110 HB hardness is for 0.2 wt% C with spheroidite.135 HB hardness is for 0.2 wt% C with fine pearlite.265 HB hardness is for 0.76 wt% C with fine pearlite.690 HB hardness is for 0.76 wt% C with tempered martensite.

Engineering

Mechanical Engineering

Problem Isothermal transformation diagram for an iron-carbon alloy of eutectoid com- position (0.76 wt% C). 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. 1. Sketch the following treatment on the figure: Rapidly cool to 575°C, hold for 20s, rapidly cool to 350°C, hold for 100s, then quench to room temperature. Specify microcon- tituents with percentage. 2. Sketch the following treatment on the figure: rapidly cool to 600°C, hold 4s, rapidly cool to 450°C, hold for 10s, and rapidly cool to room temperature. Specify microcontituents with percentage. 3. Sketch the following treatment on the figure: rapidly cool to 450°C, hold 10s, rapidly cool to room temperature, heat to 680°C, hold 30hrs, and cool to room temperature. Specify microcontituents with percentage. Temperature (°C) 4. (a) 0.2 wt% C with spheroidite (b) 0.2 wt% C with fine pearlite (e) 0.76 wt% C with fine pearlite (d) 0.76 wt% C with tempered martensite 800 700 600 500 400 300 200 100 10-1 M(start) M(50%) M(90%) Rank from the highest to the lowest hardness: 10 M+A -Eutectoid temperature P 10² Time (s) B 50% 10³ 104 105

Problem Isothermal transformation diagram for an iron-carbon alloy of eutectoid com- position (0.76 wt% C). 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. 1. Sketch the following treatment on the figure: Rapidly cool to 575°C, hold for 20s, rapidly cool to 350°C, hold for 100s, then quench to room temperature. Specify microcon- tituents with percentage. 2. Sketch the following treatment on the figure: rapidly cool to 600°C, hold 4s, rapidly cool to 450°C, hold for 10s, and rapidly cool to room temperature. Specify microcontituents with percentage. 3. Sketch the following treatment on the figure: rapidly cool to 450°C, hold 10s, rapidly cool to room temperature, heat to 680°C, hold 30hrs, and cool to room temperature. Specify microcontituents with percentage. Temperature (°C) 4. (a) 0.2 wt% C with spheroidite (b) 0.2 wt% C with fine pearlite (e) 0.76 wt% C with fine pearlite (d) 0.76 wt% C with tempered martensite 800 700 600 500 400 300 200 100 10-1 M(start) M(50%) M(90%) Rank from the highest to the lowest hardness: 10 M+A -Eutectoid temperature P 10² Time (s) B 50% 10³ 104 105

Elements Of Electromagnetics

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Concept explainers

Strengthening of Alloys

Strengthening of the alloys is required as metals become weak due to various lattice defects, metal bonding phenomena. The strength of the metals is not enough for any mechanical work. The internal dislocations of any metals arise due to forming processes. Due to the movement of these dislocations, the metals are plastically deformed.

Question

**Problem: Isothermal transformation diagram for an iron–carbon alloy of eutectoid composition (0.76 wt% C).** 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.

1. **Sketch the following treatment on the figure:** Rapidly cool to 575°C, hold for 20s, rapidly cool to 350°C, hold for 100s, then quench to room temperature. Specify microconstituents with percentage.

2. **Sketch the following treatment on the figure:** Rapidly cool to 600°C, hold 4s, rapidly cool to 450°C, hold for 10s, and rapidly cool to room temperature. Specify microconstituents with percentage.

3. **Sketch the following treatment on the figure:** Rapidly cool to 450°C, hold 10s, rapidly cool to room temperature, heat to 680°C, hold 30hrs, and cool to room temperature. Specify microconstituents with percentage.

4. **Rank from the highest to the lowest hardness:**
- (a) 0.2 wt% C with spheroidite
- (b) 0.2 wt% C with fine pearlite
- (c) 0.76 wt% C with fine pearlite
- (d) 0.76 wt% C with tempered martensite

**Diagram Explanation:**

- The graph is an isothermal transformation diagram, showing the temperature in degrees Celsius on the vertical axis and the time in seconds on the horizontal axis.
- The diagram includes several curves illustrating the transformation boundaries for different microstructures: austenite (A), pearlite (P), and bainite (B).
- The blue curves represent the start and finish lines for pearlite and bainite transformations.
- Red lines indicate the martensite start (M(start)) and transformation percentages from 50% (M(50%)) to 90% (M(90%)).
- The gray dotted line marks the eutectoid temperature.
- Labels within different zones indicate the phases present in each region, such as A (austenite), P (pearlite), and B (bainite), as well as mixtures like M + A (martensite and austenite).

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