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 direction of dislocation line-motion is the same as the direction of applied shear stress. For screw dislocations, the direction of dislocation line-motion is perpendicular to the direction of applied shear stress. For mixed dislocations, the motion of the dislocation line can be decomposed into two parts, one parallel to the direction of applied shear stress (edge component) and one perpendicular to the direction of applied shear stress (screw component).

Step by step solution

01

Edge Dislocation

In an edge dislocation, the extra half plane of atoms is perpendicular to the dislocation line. When an applied shear stress is acting on the material, it will cause the dislocation line to move in the direction of the applied stress. Therefore, the direction of dislocation line-motion is the same as the direction of applied shear stress.
02

Screw Dislocation

In a screw dislocation, the extra half plane of atoms is parallel to the dislocation line. When applied shear stress is acting on the material, it will cause the dislocation line to move in the direction that is perpendicular to both the dislocation line and the applied shear stress. Thus, the direction of dislocation line-motion is perpendicular to the direction of applied shear stress.
03

Mixed Dislocation

A mixed dislocation consists of both edge and screw dislocation components. The motion of the dislocation line is influenced by both the edge and screw dislocation components. When applied shear stress is acting on the material, the motion of the dislocation line can be decomposed into two parts: one parallel to the direction of applied shear stress (edge dislocation component) and one perpendicular to the direction of applied shear stress (screw dislocation component). In summary, for each type of dislocation: 1. Edge dislocation: The direction of dislocation line-motion is the same as the direction of applied shear stress. 2. Screw dislocation: The direction of dislocation line-motion is perpendicular to the direction of applied shear stress. 3. Mixed dislocation: The motion of the dislocation line can be decomposed into two parts, one parallel to the direction of applied shear stress (edge component) and one perpendicular to the direction of applied shear stress (screw component).

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 Vaia!

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

An aluminum bar \(125 \mathrm{~mm}(5.0\) in.) long and having a square cross section \(16.5 \mathrm{~mm}(0.65 \mathrm{in} .)\) on an edge is pulled in tension with a load of 66,700 \(\mathrm{N}\left(15,000 \mathrm{lb}_{f}\right)\) and experiences an elongation of \(0.43 \mathrm{~mm}\left(1.7 \times 10^{-2}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(30.00\) and \(30.04 \mathrm{~mm}\), respectively, and its final length is \(105.20 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(65.5\) and \(25.4\) GPa, respectively.

Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of \(5.12 \mathrm{MPa}\), compute the critical resolved shear stress.

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{\circ}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 \(\left.\mathrm{Ib}_{i}\right)\). If the length of the rod is \(500 \mathrm{~mm}(20.0 \mathrm{in}\).), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015\) in.)?

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is \(0.5 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [101], [10\overline{1} ] \text { , and } [ 0 \overline { 1 1 } ] \text { } directions.

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free