One slip system for the BCC crystal structure is \(\\{110\\}(111\rangle\). In a manner similar to Figure \(7.6 b\), sketch a \\{110\\}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different \(\langle 111\rangle\) slip directions within this plane.

Short Answer

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Question: Sketch a {110} plane for the BCC crystal structure and indicate two different <111> slip directions within this plane. Represent atom positions using circles. Answer: For the BCC crystal structure, a sketch of the {110} plane shows a triangle formed by cutting the cube along the front bottom edge to the back top edge and then to the front top edge. The atom positions within the {110} plane are distributed at two front vertices of the cube along the bottom edge, one at the center of the cube, and one at the back top vertex. Two different <111> slip directions within the plane can be indicated by arrows extending from the front bottom left corner towards the center of the cube, and from the front bottom right corner towards the center of the cube.

Step by step solution

01

BCC crystal structure

A BCC (Body-Centered Cubic) crystal structure consists of atoms at the corners of a cube and an additional atom at the center of the cube. Therefore, there are two atoms per unit cell in BCC structure. This structure is common in metals such as iron, chromium, and tungsten.
02

Identifying the {110} plane

The Miller indices {110} represent a family of parallel planes. These planes include (110), (011), (101), (1-10), (0-11), and (-101) planes, among others. These planes cut the crystal lattice along different directions, such as along perpendicular axes, diagonally along one axis and centered perpendicular along another axis. However, to simplify the sketch, we will consider just one vertex-sharing plane from the (110) family.
03

Sketching the {110} plane

To sketch the {110} plane for the BCC crystal structure, follow these steps: 1. Draw a cube representing the unit cell. 2. Sketch the atoms, where each vertex of the cube has an atom (8 in total), and an additional one placed at its center. 3. Show the plane, in this case (110), which cuts the cube along the front bottom edge to the back top edge and then to the front top edge forming a triangle.
04

Representing atom positions using circles

Now, on the sketch, represent atom positions within the {110} plane using circles. For the (110) plane, there will be four circles, representing four atoms positioned on two front vertices of the cube along the bottom edge, one at the center of the cube, and one at the back top vertex.
05

Identifying slip directions within the {110} plane

To identify the <111> slip directions within the {110} plane, we find the shortest vector direction within the plane that connects the closest neighboring atoms. In the BCC structure case, these would be the vectors connecting the atoms along diagonal directions within the plane. These represent the primary slip direction.
06

Indicating two different slip directions using arrows

On the sketch of the {110} plane, indicate two different <111> slip directions within the plane using arrows. One arrow would extend from the front bottom left corner towards the center of the cube, and the other arrow would extend from the front bottom right corner towards the center of the cube. These arrows illustrate two of the most likely slip directions in the <111> family along which dislocations could move.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Miller Indices
Miller indices are a mathematical notation system to define the orientation of planes in a crystal lattice. In simple terms, these are like addresses for planes within a crystal structure. The indices are denoted in parentheses, such as (hkl), where each letter represents an integer. These integers are inversely proportional to the intercepts that the plane makes with the crystal's axes.

For instance, the Miller indices {110} describe planes that intersect the x and y axes at one unit cell distance but never intersect the z axis. In a BCC crystal, such planes are vital in understanding slip, which is the movement of dislocations leading to plastic deformation. Properly visualizing these planes can be challenging, so simplifying the sketch to focus on specific examples, like the (110) plane, aids in grasping the concept without overwhelming details.
Slip Direction
Slip direction refers to the specific linear direction along which dislocations move within a crystal lattice. In a BCC structure, the <111> slip direction is significant because it represents the path of least resistance due to the close packing of atoms in this direction. When a force is applied, dislocations tend to move along the slip direction contributing to the plasticity of the material.

Visualizing the slip direction involves identifying the vector that connects the closest atoms within the plane. In our exercise, two different <111> slip directions were identified, emphasizing the three-dimensional complexity of the crystal structure and helping students conceptualize how materials deform on a microscopic level.
Crystal Lattice
The concept of a crystal lattice is foundational in understanding crystalline materials. It's an arrangement of points that defines the periodic structure in which atoms are positioned in a material. Imagine a three-dimensional grid where atoms sit at each grid point; that's what a crystal lattice is like.

A BCC lattice is one such arrangement where each lattice point has an atom at the corners and a central atom in the body of the cubic cell. This specific organization affects the material's properties, making it essential to visualize and understand how the atoms are arranged within a crystal structure. Such understanding provides the groundwork for comprehending more complex topics like slip systems and the mechanical properties of materials.
Unit Cell
A unit cell is the smallest structural unit that represents the symmetry and structure of a crystal lattice. Think of it as a building block from which the entire crystal can be constructed by translating it in three-dimensional space. For a BCC crystal, this cell is cubic with an atom at each corner and one atom at the very center of the cube.

The concept of a unit cell is crucial for simplifying the visualization and analysis of complex crystal structures. In our case, understanding that the BCC unit cell contains two atoms—eight shared corner atoms, each of which counts as one eighth, and the central atom counting as one whole—allows students to come to terms with the intricate but orderly world of crystallography.

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Most popular questions from this chapter

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: $$ \begin{array}{lcr} \hline & \begin{array}{c} \text { Circular } \\ \text { (diameter, } \mathbf{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \text { Original dimensions } & 15.2 & 125 \times 175 \\ \text { Deformed dimensions } & 11.4 & 75 \times 200 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the \([110]\) direction. If the critical resolved shear stress for this material is \(1.75 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [1\overline{110], [10\overline{1} ] } \text { and } [ 0 1 \overline { 1 } ] \text { directions. }

Briefly cite the differences between recovery and recrystallization processes.

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