---
title: "Phases in 0.80% carbon steel"
subject: "Civil Engineering Materials"
chapter: 8
key: "semester_3/civil_engineering_materials:carbon-percent-0-80-phases"
source: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/carbon-percent-0-80-phases
site: BCT Engineering
---

# Phases in 0.80% carbon steel

Civil Engineering Materials · Chapter 8: Metals and Alloys

- Asked in: 2070 Chaitra (old course), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/carbon-percent-0-80-phases
- Whole chapter: https://www.bctengineer.com/md/semester_3/civil_engineering_materials/chapter-8

## Question

In a steel, the percentage of carbon content is 0.80%, then find out the percentage of ferrite, cementite and pearlite.

## Short answer

Using the lever rule: pearlite fraction = C/0.80 = 100%. Cementite = C/6.67 = 0.80/6.67 ≈ 11.99%; ferrite = 100 − 11.99 = 88.01% (all lamellar inside pearlite). Thus: pearlite 100%, cementite 11.99%, ferrite 88.01%.

## Answer

Steel with 0.80% carbon is the **eutectoid steel**. Pearlite itself contains 0.80% C, so the steel is entirely pearlite.

**Data:** carbon in steel $C = 0.80\%$; carbon in pearlite $= 0.80\%$; carbon in cementite ($Fe_3C$) $= 6.67\%$. Carbon dissolved in ferrite is neglected (about 0.02%).

#### Pearlite

$$
\text{Pearlite} = \frac{0.80}{0.80} \times 100 = 100\%
$$

#### Cementite

$$
\text{Cementite} = \frac{0.80}{6.67} \times 100 = 11.99\%
$$

#### Ferrite

$$
\text{Ferrite} = 100 - 11.99 = 88.01\%
$$

Free (proeutectoid) ferrite is $100 - 100 = 0\%$, because all the ferrite and cementite are present together as lamellae in pearlite.

**Answer: Pearlite = 100%; cementite = 11.99%; ferrite = 88.01% (all of the ferrite and cementite lie inside the pearlite).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Steel with 0.80% carbon is entirely:
   A. Ferrite
   B. Cementite
   C. Pearlite
   D. Martensite
   Answer: C. Pearlite contains 0.80% C, so 0.80% C steel is 100% pearlite. (Key point: Eutectoid steel)

2. Cementite percentage in 0.80% C steel is:
   A. 0.80/0.80 x 100
   B. 0.80/6.67 x 100 = 11.99%
   C. 6.67/0.80 x 100
   D. 100 - 0.80
   Answer: B. Cementite = C/6.67 x 100 = 11.99%. (Key point: Cementite formula)

3. Ferrite percentage in this steel is:
   A. 11.99%
   B. 100%
   C. 0%
   D. 88.01%
   Answer: D. Ferrite = 100 - 11.99 = 88.01%, all inside the pearlite. (Key point: Ferrite)

4. Free (proeutectoid) ferrite in 0.80% C steel is:
   A. 88.01%
   B. 0%
   C. 11.99%
   D. 50%
   Answer: B. All ferrite lies in pearlite lamellae, so free ferrite is 100 - 100 = 0%. (Key point: Free ferrite)

## Related in Civil Engineering Materials

- [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 0.33% 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
- [Function of carbon in steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Acarbon-function-in-steel) — Civil Engineering Materials, chapter 8, asked 2×
- [Modulus of toughness of medium carbon steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Anumerical-modulus-toughness-415) — Civil Engineering Materials, chapter 1
- [Plain carbon steel versus alloy steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Aplain-carbon-vs-alloy-steel) — Civil Engineering Materials, chapter 8

## Related in other subjects

- [Phases and constituents of steel](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Aferrite-austenite-pearlite-martensite) — Material Science, chapter 6
- [Slow cooling of 0.4% C steel](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Aslow-cooling-04-carbon-steel) — Material Science, chapter 6
- [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
- [Silicon content: transformer versus rotating machine steel](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Asilicon-content-transformer-rotating) — Electric Machine Design, chapter 1