JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III Year METALLURGY FOR NON METALLURGISTS Previous Question Paper (February/March - 2022 - R16)

B.Tech | III Year | II Semester | R16 | February/March - 2022

Answer PaperPaper Code: 136FPMaximum Marks: 75Time: 3 Hours

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1

Classify plain carbon steels based on amount of carbon and mention their properties and applications.

15 marks

Answer

Plain carbon steels are classified primarily based on their carbon content, which significantly influences their microstructure, mechanical properties, and suitability for various industrial applications. The classification is as follows:

1. Low Carbon Steel (Mild Steel): Contains 0.05% to 0.30% carbon. These steels are characterized by high ductility, excellent weldability, and good formability, though they have relatively low strength and hardness. They are widely used in the construction of structural steel beams, automobile body panels, wire products, and general-purpose machinery components.

2. Medium Carbon Steel: Contains 0.30% to 0.60% carbon. This range provides a balance between strength and ductility. These steels can be heat-treated (quenched and tempered) to improve mechanical properties. They are commonly used for manufacturing gears, axles, shafts, connecting rods, railway wheels, and structural components requiring higher strength than mild steel.

3. High Carbon Steel: Contains 0.60% to 1.00% carbon. These steels possess high hardness and wear resistance but exhibit low ductility and are difficult to weld. They are typically used in applications requiring high strength and hardness, such as cutting tools, high-strength wires, springs, knives, and punches.

4. Ultra-High Carbon Steel: Contains 1.00% to 2.00% carbon. These materials are extremely hard and brittle. They are primarily used for specialized applications such as high-quality cutlery, files, and specialized cutting tools that require extreme wear resistance.

2

Explain in detail the two types of dislocations in crystals.

15 marks

Answer

Dislocations are linear defects in a crystal lattice that significantly influence the mechanical properties of materials, particularly their ability to undergo plastic deformation. The two primary types of dislocations are edge dislocations and screw dislocations.

1. Edge Dislocation: An edge dislocation can be visualized as the insertion of an extra half-plane of atoms into the crystal lattice. The dislocation line is the edge of this extra half-plane. Above the slip plane, the atoms are in compression, while below the slip plane, they are in tension. The Burgers vector, which defines the magnitude and direction of the lattice distortion, is perpendicular to the dislocation line. When a shear stress is applied, the dislocation moves through the lattice by the successive breaking and reforming of atomic bonds, allowing the crystal to deform plastically.

2. Screw Dislocation: A screw dislocation is formed by applying a shear stress to a crystal, causing one part of the crystal to shift relative to the other by one atomic spacing. This creates a helical ramp or spiral structure of atomic planes around the dislocation line. In a screw dislocation, the Burgers vector is parallel to the dislocation line. The lattice distortion is characterized by a twisting motion rather than a simple compression-tension field. The movement of a screw dislocation under stress also contributes to plastic deformation, but the mechanism involves the dislocation line moving perpendicular to the applied shear stress.

In real crystalline materials, dislocations rarely exist as pure edge or pure screw types; they typically exist as mixed dislocations, which possess both edge and screw components. The character of the dislocation changes continuously along the dislocation line, with the angle between the Burgers vector and the dislocation line varying between 0 and 90 degrees.

3

Explain the three reactions that occur in iron-iron carbide diagram.

15 marks

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415 marks
a

Explain annealing process.

7 marks

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b

Explain normalizing process.

8 marks

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5

Explain the microstructure, properties and applications of malleable irons.

15 marks

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6

Explain the differences between ductile iron and compacted graphite iron with respect to microstructure and properties.

15 marks

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7

Explain the properties and applications of Ti.

15 marks

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8

Explain the procedure and working principle of fatigue testing method.

15 marks

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About this paper

This page contains the question paper of METALLURGY FOR NON METALLURGISTS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech CE III Year II Semester, February/March - 2022, R16.

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