Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech II MATERIAL SCIENCE AND METALLURGY Previous Question Paper (August/September - 2022 - R18)

B.Tech | II | I | R18 | August/September - 2022

Answer PaperPaper Code: 153BCMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

The axis of a cylindrical crystal of copper lies in the (1 1 1) plane. Show that, when a tensile stress σ is applied along the axis of the crystal, the resolved shear stress τ in any direction on this plane is zero.

15 marks

Answer

To determine the resolved shear stress τ on a specific slip system, we use Schmid's Law, which is given by the formula:

τ = σ cos(φ) cos(λ)

where σ is the applied tensile stress, φ is the angle between the tensile axis and the normal to the slip plane, and λ is the angle between the tensile axis and the slip direction.

The problem states that the axis of the cylindrical crystal lies within the (1 1 1) plane. Let the tensile axis be represented by a unit vector A, and the normal to the (1 1 1) plane be represented by a unit vector N. Since the axis lies within the (1 1 1) plane, it is perpendicular to the plane's normal vector. Therefore, the dot product of A and N is zero:

A · N = |A||N| cos(φ) = 0

Since A and N are unit vectors, this implies cos(φ) = 0, meaning φ = 90°. Substituting this into the Schmid's Law equation:

τ = σ cos(90°) cos(λ) = σ(0) cos(λ) = 0

Furthermore, for any slip direction S that lies within the (1 1 1) plane, the resolved shear stress is also governed by the orientation of the tensile axis relative to that direction. Because the tensile axis A is contained within the (1 1 1) plane, any slip direction S lying in that same plane must satisfy specific geometric constraints. If the slip direction S is perpendicular to the tensile axis A, then cos(λ) = 0, resulting in τ = 0. If the slip direction S is not perpendicular to A, the resolved shear stress is non-zero; however, the question specifies the resolved shear stress in any direction on this plane. Since the normal to the plane is perpendicular to the tensile axis, the plane contains the tensile axis itself, ensuring that for the slip system defined by the plane normal, the resolved shear stress is zero regardless of the applied load.

2

Describe how slip occurs in single crystals.

15 marks

Answer

Slip is the primary mechanism of plastic deformation in crystalline materials, occurring through the movement of dislocations along specific crystallographic planes and directions, known as the slip system. In single crystals, this process is governed by the resolved shear stress acting on these systems.

When an external tensile or compressive force is applied to a single crystal, the stress is resolved into a shear stress component acting on the slip plane in the slip direction. This is defined by Schmid's Law, which states that slip begins when the resolved shear stress reaches a critical value, known as the Critical Resolved Shear Stress (CRSS).

τ = σ cos(φ) cos(λ)

In this formula, τ is the resolved shear stress, σ is the applied tensile stress, φ is the angle between the normal to the slip plane and the tensile axis, and λ is the angle between the slip direction and the tensile axis. The term cos(φ) cos(λ) is referred to as the Schmid factor.

The mechanism of slip involves the following stages:

  • Dislocation Initiation: Under applied stress, existing dislocations within the crystal lattice begin to move. If the crystal is highly perfect, new dislocations may be generated via sources such as the Frank-Read source.
  • Dislocation Glide: Dislocations move through the crystal lattice along the close-packed planes where the atomic density is highest, as these planes require the least amount of energy for atomic displacement.
  • Slip Band Formation: As dislocations reach the surface of the crystal, they create small steps. When many dislocations move along the same or closely spaced parallel planes, these steps accumulate to form visible slip bands or slip lines on the surface of the crystal.
  • Crystal Rotation: As deformation continues, the slip plane tends to rotate to align itself with the direction of the applied tensile force, which changes the values of φ and λ and affects the ease of further slip.

In single crystals, slip is highly anisotropic, meaning it occurs preferentially in specific orientations where the Schmid factor is maximized. Once the CRSS is exceeded, the crystal undergoes permanent plastic deformation without fracture until the lattice becomes sufficiently distorted to impede further dislocation movement.

315 marks
a

Define an alloy. Discuss about Hume-Rothery rules.

7 marks

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b

Write down the eutectic, peritectic, eutectoid and peritectoid reactions with suitable examples.

8 marks

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415 marks
a

Explain the different stages of tempering.

8 marks

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b

What is spheroidising?

7 marks

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5

Discuss the phase transformation of austenite to pearlite under equilibrium cooling.

15 marks

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615 marks
a

Discuss about continuous cooling curves and their importance.

8 marks

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b

Explain nitriding process.

7 marks

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715 marks
a

What are stainless steels and tool steels?

8 marks

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b

What are the properties of brass and bronze? What are their uses?

7 marks

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8

What are the properties and uses of:

15 marks
a

Titanium and its alloys

7 marks

Answer

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b

Maraging steels and white cast iron.

8 marks

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About this paper

This page contains the question paper of MATERIAL SCIENCE AND METALLURGY from Jawaharlal Nehru Technological University Hyderabad for B.Tech MECT-MCT-MEM II I, August/September - 2022, R18.

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