--- title: "Slow cooling of 0.4% C steel" subject: "Material Science" chapter: 6 key: "semester_3/material_science:slow-cooling-04-carbon-steel" source: https://www.bctengineer.com/q/semester_3/material_science/slow-cooling-04-carbon-steel site: BCT Engineering --- # Slow cooling of 0.4% C steel Material Science · Chapter 6: Iron – Iron Carbide diagram and Heat Treatment of Steels - Asked in: year not given, 6 marks - Video: [Steel phase diagram cooling and Lever Rule example problem](https://www.youtube.com/watch?v=fjqeEYhaXcI) — Taylor Sparks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/material_science/slow-cooling-04-carbon-steel - Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-6 ## Question Describe the microstructural changes during slow cooling of a 0.4 wt% carbon steel from 950 °C to room temperature. Using the lever rule with ferrite 0.022 wt% C, eutectoid 0.76 wt% C and cementite 6.67 wt% C, calculate the fractions of pro-eutectoid ferrite and pearlite, and also the total ferrite and cementite just below 727 °C. ## Short answer On slow cooling, pro-eutectoid ferrite forms first, then remaining austenite transforms to pearlite at 727°C. Lever rule gives 48.8% pro-eutectoid ferrite and 51.2% pearlite; total ferrite 94.3% and cementite 5.7%. ## Answer #### Cooling path 1. At 950 °C the structure is 100% austenite ($\gamma$) of 0.4% C. 2. On cooling to the A$_3$ line (about 790 °C), pro-eutectoid (primary) **ferrite** nucleates at austenite grain boundaries. The remaining austenite is enriched in carbon. 3. Between A$_3$ and 727 °C more ferrite forms; austenite composition follows the GOS line toward 0.76% C. 4. Just above 727 °C the austenite has 0.76% C. At 727 °C it transforms by the eutectoid reaction $\gamma \rightarrow \alpha + \text{Fe}_3\text{C}$ into **pearlite**. 5. Below 727 °C: pro-eutectoid ferrite + pearlite, little change on cooling to room temperature. ```text 950C: austenite -> 790C: ferrite at grain boundaries +-----------+ +-----------+ | gamma | |f gamma f | +-----------+ +-----------+ 727C and below: ferrite (white) + pearlite (dark) ``` #### Phase fractions just below 727 °C Pro-eutectoid ferrite (using tie line from 0.4 to 0.76 and 0.022): $$ W_{\alpha'} = \frac{0.76 - 0.40}{0.76 - 0.022} = \frac{0.36}{0.738} = 0.488 $$ Pearlite (formed from the remaining austenite): $$ W_{P} = \frac{0.40 - 0.022}{0.76 - 0.022} = \frac{0.378}{0.738} = 0.512 $$ Total ferrite and cementite (phases, including those inside pearlite): $$ W_{\alpha} = \frac{6.67 - 0.40}{6.67 - 0.022} = \frac{6.27}{6.648} = 0.943 $$ $$ W_{Fe_3C} = \frac{0.40 - 0.022}{6.67 - 0.022} = \frac{0.378}{6.648} = 0.057 $$ | Quantity | Fraction | |---|---| | Pro-eutectoid ferrite | 0.488 (48.8%) | | Pearlite | 0.512 (51.2%) | | Total ferrite | 0.943 | | Total cementite | 0.057 | **Answer: 48.8% pro-eutectoid ferrite and 51.2% pearlite; total phases 94.3% ferrite and 5.7% cementite.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On slowly cooling 0.4% C austenite below the A3 line (about 790 °C), what forms first? A. Pearlite throughout the grains B. Pro-eutectoid cementite at grain boundaries C. Pro-eutectoid ferrite at austenite grain boundaries D. Martensite Answer: C. The alloy is hypoeutectoid, so ferrite nucleates at grain boundaries and the remaining austenite becomes richer in carbon. (Key point: First phase to form) 2. Just above 727 °C, what is the carbon content of the remaining austenite, and what does it become? A. 0.4% C, becomes ferrite B. 0.022% C, becomes cementite C. 0.76% C, becomes pearlite D. 6.67% C, becomes ledeburite Answer: C. The austenite composition follows the GOS line to 0.76% C and then undergoes the eutectoid reaction to pearlite. (Key point: Composition of austenite at 727 °C) 3. Which expression gives the fraction of pro-eutectoid ferrite for 0.4% C steel just below 727 °C? A. (0.76 - 0.40)/(0.76 - 0.022) = 0.488 B. (0.40 - 0.022)/(0.76 - 0.022) = 0.512 C. (6.67 - 0.40)/(6.67 - 0.022) = 0.943 D. (0.40 - 0.022)/(6.67 - 0.022) = 0.057 Answer: A. The tie line runs from 0.022 to 0.76; the ferrite fraction is the opposite arm over the whole, (0.76 - 0.40)/0.738 = 0.488. (Key point: Lever rule for pro-eutectoid ferrite) 4. Why is the total ferrite (94.3%) larger than the pro-eutectoid ferrite (48.8%)? A. Cementite changes to ferrite on cooling B. Total ferrite also counts the ferrite lamellae inside pearlite C. The lever rule is wrong for phases D. Austenite is counted as ferrite Answer: B. Pearlite is itself ferrite plus cementite, so the phase fraction uses the tie line 0.022 to 6.67 and includes that ferrite. (Key point: Total phases vs constituents) ## Related in Material Science - [Slow cooling of 1.2% C steel](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Ahypereutectoid-steel-12-carbon) — Material Science, chapter 6 - [Lever rule in Cu-Ni alloy](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Alever-rule-copper-nickel-alloy) — Material Science, chapter 5 - [Cooling of 3% C cast iron](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Ahypoeutectic-cast-iron-3-carbon) — Material Science, chapter 6 ## Related in other subjects - [Ferrite and pearlite in high strength steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Aferrite-pearlite-high-strength-steel) — Civil Engineering Materials, chapter 8 - [Phases in 0.80% carbon steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Acarbon-percent-0-80-phases) — Civil Engineering Materials, chapter 8 - [Phases in 0.33% carbon steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Acarbon-percent-0-33-phases) — Civil Engineering Materials, chapter 8 - [Phases in 0.35% carbon steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Acarbon-percent-0-35-phases) — Civil Engineering Materials, chapter 8