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JNTU B.Tech III HEAT TREATMENT Previous Question Paper (October - 2020 - R16)

B.Tech | III | I | R16 | October - 2020

Answer PaperPaper Code: 135ATMaximum Marks: 75Time: 2 hours

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Main

1

With the help of a suitable diagram, show normalizing and hardening temperature ranges for plain carbon hypereutectoid steels. What is the criteria for the selection of these temperature ranges?

15 marks

Answer

For hypereutectoid steels, which contain more than 0.8% carbon, the heat treatment temperature ranges are defined relative to the A1 (lower critical) and Acm (upper critical) lines on the iron-iron carbide phase diagram.

Normalizing: The temperature range for normalizing hypereutectoid steel is typically 30°C to 50°C above the Acm line. Heating into this single-phase austenite region ensures the complete dissolution of proeutectoid cementite into the austenite matrix, which then transforms into a uniform, fine pearlitic structure upon air cooling.

Hardening: The hardening temperature range is generally 30°C to 50°C above the A1 line. Unlike normalizing, hardening is performed within the austenite plus cementite (α + Fe3C) region. This range is selected to transform the pearlite into austenite while leaving the excess proeutectoid cementite undissolved, which increases the wear resistance and hardness of the final martensitic structure.

Criteria for Selection: The selection of these ranges is based on the following factors:

  • Microstructural Control: Normalizing aims for a homogeneous grain structure, requiring full austenitization above Acm. Hardening aims for maximum hardness, where undissolved cementite particles act as hard phases to improve abrasion resistance.
  • Grain Growth Prevention: Temperatures are kept strictly near the critical lines (30-50°C above) to prevent excessive austenite grain growth, which would otherwise result in a brittle, coarse martensitic structure upon quenching.
  • Energy Efficiency: Operating as close to the critical lines as possible minimizes energy consumption and thermal stress during the heating and cooling cycles.
2

Explain why hardening by quenching is followed by tempering treatment? How do mechanical properties vary with tempering temperature?

15 marks

Answer

Hardening by quenching involves rapid cooling of steel from the austenitizing temperature to form martensite, a body-centered tetragonal structure. While martensite provides high hardness and strength, it is inherently brittle, possesses high internal stresses due to the volume expansion during transformation, and lacks sufficient ductility and toughness for structural applications. Tempering is essential to transform this metastable martensite into a more stable tempered martensite, which relieves internal stresses and restores ductility.

The tempering process involves reheating the quenched steel to a temperature below the lower critical temperature (A1). The variations in mechanical properties with increasing tempering temperature are as follows:

  • Hardness and Strength: These properties decrease progressively as the tempering temperature increases. The decomposition of martensite into ferrite and fine cementite particles reduces the lattice strain and dislocation density, leading to a loss in hardness.
  • Ductility and Toughness: These properties increase significantly with higher tempering temperatures. The relief of internal quenching stresses and the growth of carbide particles allow for greater plastic deformation, thereby improving the material's impact resistance and elongation.
  • Internal Stresses: Residual stresses induced during quenching are progressively relieved as the tempering temperature rises, which significantly improves the dimensional stability of the component.

In summary, tempering allows for a controlled trade-off between hardness and toughness, enabling the material to meet specific engineering requirements by selecting an appropriate tempering temperature.

315 marks
a

What are TTT curves? Explain the construction of TTT curves for 0.8% C steel.

7 marks

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b

Explain the phase transformations in 0.8% steel cooled from austenite to room temperature under different cooling rates.

8 marks

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415 marks
a

What is bainite? Distinguish between bainite and martensite.

7 marks

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b

Calculate the volume change associated with austenite to martensite transformation in one percent carbon steel.

8 marks

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515 marks
a

Explain the difference in microstructure, composition and properties of white and gray cast iron.

8 marks

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b

Discuss the effect of the amount of free carbon on the properties of gray cast iron.

7 marks

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615 marks
a

Distinguish between first stage of graphitization and second stage of graphitization?

7 marks

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b

Explain the effect of malleabilizing cycle on microstructure.

8 marks

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7

What is meant by thermomechanical annealing? Compare it with conventional annealing and spheroidizing.

15 marks

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815 marks
a

What is nitriding? Explain the conditions for nitriding of steels.

8 marks

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b

Discuss about flame hardening treatment.

7 marks

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This page contains the question paper of HEAT TREATMENT from JNTU for B.Tech MME-MMT III I, October - 2020, R16.

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