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Chapter 4: Problem 26

Cite the relative Burgers vector-dislocation line orientations for edge, screw, and mixed dislocations.

Short Answer

Expert verified
Answer: For an edge dislocation, the Burgers vector is perpendicular to the dislocation line. For a screw dislocation, the Burgers vector is parallel to the dislocation line. For a mixed dislocation, the Burgers vector lies at an angle, having components along both the parallel and perpendicular directions to the dislocation line.

Step by step solution

01

Edge Dislocation

An edge dislocation is a defect in the crystal lattice where an extra half-plane of atoms is introduced at an angle to the lattice planes. The Burgers vector of an edge dislocation is perpendicular to the dislocation line. The orientation can be represented as follows: for an edge dislocation with a dislocation line along the z-axis, the Burgers vector (b) will lie along the x-axis (b = bx\hat{i}).
02

Screw Dislocation

A screw dislocation is a defect in the crystal lattice where atoms are displaced along a helical path around the dislocation line. The Burgers vector of a screw dislocation is parallel to the dislocation line. The orientation can be demonstrated as follows: for a screw dislocation with a dislocation line along the z-axis, the Burgers vector (b) will also lie along the z-axis (b = bz\hat{k}).
03

Mixed Dislocation

A mixed dislocation is a combination of edge and screw dislocations, exhibiting characteristics of both types. The Burgers vector of a mixed dislocation is neither parallel nor perpendicular to the dislocation line but lies at an angle. The orientation can be represented as follows: for a mixed dislocation with a dislocation line along the z-axis, the Burgers vector (b) will have components along both the x and z directions (b = bx\hat{i} + bz\hat{k}).

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Most popular questions from this chapter

For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? (Note: You may want to consult the solution to Problem \(3.54\) at the end of Chapter 3.)

(a) Briefly describe a twin and a twin, boundary. (b) Cite the difference between mechanical and annealing twins.

Some hypothetical alloy is composed of \(12.5\) \(\mathrm{wt} \%\) of metal \(\mathrm{A}\) and \(87.5 \mathrm{wt} \%\) of metal \(\mathrm{B}\). If the densities of metals \(\mathrm{A}\) and \(\mathrm{B}\) are \(4.27\) and \(6.35 \mathrm{~g} / \mathrm{cm}^{3}\), respectively, whereas their respective atomic weights are \(61.4\) and \(125.7 \mathrm{~g} / \mathrm{mol}\), determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body- centered cubic. Assume a unit cell edge length of \(0.395 \mathrm{~nm}\).

Determine the approximate density of a highleaded brass that has a composition of \(64.5\) \(\mathrm{wt} \% \mathrm{Cu}, 33.5 \mathrm{wt} \% \mathrm{Zn}\), and \(2 \mathrm{wt} \% \mathrm{~Pb}\).

For both FCC and BCC crystal structures, there are two different types of interstitial sites. In each case, one site is larger than the other and is normally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at \(0 \frac{1}{2} \frac{1}{4}\) positions \(-\) that is, lying on \(\\{100\\}\) faces and situated midway between two unit cell edges on this face and one- quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystal structures, compute the radius \(r\) of an impurity atom that will just fit into one of these sites in terms of the atomic radius \(R\) of the host atom.
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