/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 For each of edge, screw, and mix... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 4

For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.

Short Answer

Expert verified
Answer: For edge dislocations, the dislocation line motion is perpendicular to the applied shear stress direction. In the case of screw dislocations, the dislocation line motion is parallel to the applied shear stress direction. For mixed dislocations, the direction of dislocation line motion depends on the angle between the applied shear stress direction and the dislocation line, and it consists of components in both edge and screw dislocation directions.

Step by step solution

01

Define different types of dislocations

An edge dislocation is a type of defect in a crystal lattice where an extra half-plane of atoms is inserted. A screw dislocation, on the other hand, is a defect caused by shear forces that twist the lattice such that lattice planes spiral around the dislocation. Mixed dislocations possess characteristics of both edge and screw dislocations.
02

Explain the relationship for edge dislocations

For an edge dislocation, the direction of dislocation line motion is perpendicular to both the dislocation line and the applied shear stress direction. The motion of the dislocation line will cause the extra half-plane of atoms to advance through the lattice, resulting in the material being deformed.
03

Explain the relationship for screw dislocations

In the case of a screw dislocation, the dislocation line motion is parallel to the applied shear stress direction. The motion of the dislocation line occurs by the sliding of one part of the crystal lattice relative to the other, resulting in a twisting effect in the lattice structure.
04

Explain the relationship for mixed dislocations

For mixed dislocations, the motion of the dislocation line will have components along both the edge and screw dislocation directions. The direction of dislocation line motion will depend on the applied shear stress direction, and the angle between the applied shear stress direction and the dislocation line determines the proportion of edge and screw character in the mixed dislocation. In conclusion, the direction of dislocation line motion is influenced by the type of dislocation as well as the direction of the applied shear stress. For edge dislocations, the motion is perpendicular to the applied shear stress direction, for screw dislocations it is parallel, and for mixed dislocations, it is determined by the angle between the applied shear stress direction and the dislocation line.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a single crystal of silver oriented such that a tensile stress is applied along a \([001]\) direction. If slip occurs on a (111) plane and in a [ \(\overline{101}\) ] direction, and is initiated at an applied tensile stress of \(1.1 \mathrm{MPa}\) (160 psi), compute the critical resolved shear stress.

The lower yield point for an iron that has an average grain diameter of \(5 \times 10^{-2} \mathrm{~mm}\) is 135 MPa (19,500 psi). At a grain diameter of \(8 \times\) \(10^{-3} \mathrm{~mm}\), the yield point increases to \(260 \mathrm{MPa}\) \((37,500 \mathrm{psi})\). At what grain diameter will the lower yield point be \(205 \mathrm{MPa}(30,000 \mathrm{psi})\) ?

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 aluminum is oriented for a tensile test such that its slip plane normal makes an angle of \(28.1^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(62.4^{\circ}, 72.0^{\circ}\), and \(81.1^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(1.95 \mathrm{MPa}(280 \mathrm{psi})\), determine the critical resolved shear stress for aluminum.

(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.
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Physics of Motion

Read Explanation

Energy Physics

Read Explanation

Force

Read Explanation

Linear Momentum

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.