Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 20

List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.

Short Answer

Expert verified
Answer: 1. In twinning, deformation occurs through an abrupt change and symmetrical arrangement of atoms across a defined plane, while in slip, gradual translations along specific planes take place. 2. Twinning typically occurs under high stress or strain rates and low temperatures, whereas slip occurs at low-stress conditions, high temperatures, and slow applied strain rates. 3. Twinning is more prevalent in brittle materials such as metals with hcp or bcc structures, while slip is more common in ductile materials with a large number of available slip systems. 4. Deformation by twinning results in a twin boundary, affecting material strength, ductility, and toughness, whereas deformation by slip maintains the lattice orientation and can cause surface roughness and reduced material strength.

Step by step solution

01

1: Mechanism of deformation by twinning

In the deformation by twinning, the crystal lattice structure undergoes a sudden change, leading to a symmetrical arrangement of atoms across a defined plane. The atoms on the twin plane change their positions with respect to one another in a coordinated manner, resulting in a new, mirrored lattice orientation.
02

2: Mechanism of deformation by slip

In the deformation by slip, the crystal lattice structure undergoes gradual and incremental translations along specific planes known as slip planes. The atoms move along these planes in a linear trajectory, which eventually causes the lattice to deform. In contrast to twinning, the atoms don't undergo any abrupt or coordinated rearrangement in slip deformation.
03

3: Conditions of occurrence for twinning

Deformation by twinning typically occurs under high stress or strain rates and low temperatures. It is more prevalent in brittle materials such as metals with a hexagonal close-packed (hcp) or body-centered cubic (bcc) structure and some minerals and polymers that have limited slip systems available.
04

4: Conditions of occurrence for slip

Deformation by slip is common in materials with a large number of available slip systems, such as face-centered cubic (fcc) metals and ductile materials. Slip deformation generally occurs at low-stress conditions, high temperatures, and slow applied strain rates, which allow the atoms to slide past each other without significant resistance.
05

5: Final result of deformation by twinning

Deformation by twinning usually results in a well-defined twin boundary, which is visible under an optical microscope or through transmission electron microscopy. This can have implications on the macroscopic properties of the material, such as enhancing its strength, ductility, and toughness.
06

6: Final result of deformation by slip

After deformation by slip, the material's lattice structure maintains its overall orientation, although it becomes distorted due to the translation of the atoms along the slip planes. The slip steps formed on the surface of the material during deformation can cause it to appear rough. The overall strength and ductility of the material may be reduced due to the dislocations generated during slip deformation.

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.

Start your free trial

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

Consider a hypothetical material that has a grain diameter of \(2.1 \times 10^{-2} \mathrm{~mm}\). After a heat treatment at \(600^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\), the grain diameter has increased to \(7.2 \times 10^{-2} \mathrm{~mm}\). Compute the grain diameter when a specimen of this same original material (i.e., \(d_{0}=2.1 \times 10^{-2} \mathrm{~mm}\) ) is heated for \(1.7 \mathrm{~h}\) at \(600^{\circ} \mathrm{C}\). Assume the \(n\) grain diameter exponent has a value of 2

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\mathrm{crs}}\) is a function of the dislocation density \(\rho_{D}\) as $$ \tau_{\text {crss }}=\tau_{0}+A \sqrt{\rho_{D}} $$ where \(\tau_{0}\) and \(A\) are constants. For copper, the critical resolved shear stress is \(0.69 \mathrm{MPa}\) (100 psi) at a dislocation density of \(10^{4} \mathrm{~mm}^{-2}\). If it is known that the value of \(\tau_{0}\) for copper is \(0.069 \mathrm{MPa}\) (10 psi), compute \(\tau_{\mathrm{crss}}\) at a dislocation density of \(10^{6} \mathrm{~mm}^{-2}\).

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{6}\right.\) 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 lb_{f } ). If the length of the rod is \(500 \mathrm{~mm}\) (20.0 in.), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015 \mathrm{in} .)\) ?

An undeformed specimen of some alloy has an average grain diameter of \(0.050 \mathrm{~mm}\). You are asked to reduce its average grain diameter to \(0.020 \mathrm{~mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

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 rectangular. Their original and deformed dimensions are as follows: \begin{tabular}{lcc} \hline & Circular \((\) diameter, \(\boldsymbol{m m})\) & Rectangular (mm) \\ \hline Original dimensions & \(18.0\) & \(20 \times 50\) \\ \hline Deformed dimensions & \multicolumn{1}{c}{\(15.9\)} & \(13.7 \times 55.1\) \\\ \hline \end{tabular} Which of these specimens will be the hardest after plastic deformation, and why?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Quantum Physics

Read Explanation

Space Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electricity

Read Explanation

Further Mechanics and Thermal Physics

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.

Sign-up for free