Cite one reason why ceramic materials are, in general, harder yet more brittle than metals.

Short Answer

Expert verified
Question: Explain one reason why ceramic materials tend to be harder yet more brittle than metals, considering their atomic structure and bonding. Answer: One reason why ceramic materials are harder yet more brittle than metals is due to their strong and directional ionic or covalent bonds. These bonds form a rigid and strongly bonded structure, which results in higher resistance to deformation as the individual atoms or ions find it more difficult to slide past each other, contributing to higher hardness. However, the same strong and rigid bonding structure also makes ceramics more brittle, as their bonds are less flexible and more prone to breaking under stress, limiting their ability to deform plastically before fracturing. In contrast, metals have non-directional and more flexible metallic bonds, allowing them to deform plastically, making them less brittle and more ductile compared to ceramic materials.

Step by step solution

01

Understand the properties of ceramic materials and metals

In general, ceramic materials are known for their hardness, wear resistance, and brittleness, while metals are known for their malleability, ductility, and toughness.
02

Explore the atomic structure and bonding in ceramic materials and metals

Ceramic materials are composed of ionic or covalent bonds, while metals primarily consist of metallic bonds. Ionic and covalent bonds are much stronger and more directional than metallic bonds, which leads to the differences in properties between ceramic materials and metals.
03

Explain one reason why ceramic materials are harder than metals

One reason why ceramic materials are generally harder than metals is their strong and directional bonding. The strong ionic or covalent bonds in ceramic materials result in a rigid and strongly bonded structure. This structure makes it more difficult for the individual atoms or ions to slide past each other, which contributes to a higher resistance to deformation and higher hardness compared to metals with non-directional metallic bonds.
04

Explain one reason why ceramic materials are more brittle than metals

The same strong and rigid bonding structure that gives ceramic materials their hardness also contributes to their brittleness. Since the ionic or covalent bonds are less flexible and more prone to breaking under stress, ceramic materials have limited ability to deform plastically before fracturing. On the other hand, the non-directional and more flexible metallic bonds allow metals to deform plastically, which makes them less brittle and more ductile compared to ceramic materials.

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

In your own words, briefly define the term stoichiometric.

If cupric oxide \((\mathrm{CuO})\) is exposed to reducing atmospheres at elevated temperatures, some of the \(\mathrm{Cu}^{2+}\) ions will become \(\mathrm{Cu}^{+}\). (a) Under these conditions, name one crystalline defect that you would expect to form in order to maintain charge neutrality. (b) How many \(\mathrm{Cu}^{+}\)ions are required for the creation of each defect? (c) How would you express the chemical formula for this nonstoichiometric material?

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius \(3.5 \mathrm{~mm}\) (0.14 in.); the specimen fractured at a load of \(950 \mathrm{~N}\left(215 \mathrm{lb}_{\mathrm{i}}\right)\) when the distance between the support points was \(50 \mathrm{~mm}\) (2.0 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of \(12 \mathrm{~mm}\) ( \(0.47\) in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is \(40 \mathrm{~mm}\) (1.6 in.)?

The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{cc} \hline\(\sigma_{f_{s}}(M P a)\) & \(P\) \\ \hline 100 & \(0.05\) \\ 50 & \(0.20\) \\ \hline \end{tabular} (a) Compute the flexural strength for a completely nonporous specimen of this material. (b) Compute the flexural strength for a \(0.10\) volume fraction porosity.

(a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of \(390 \mathrm{MPa}(56,600 \mathrm{psi})\). If the specimen radius is \(2.5 \mathrm{~mm}\) (0.10 in.) and the support point separation distance is 30 \(\mathrm{mm}\) (1.2 in.), predict whether you would expect the specimen to fracture when a load of \(620 \mathrm{~N}\left(140 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Justify your prediction. (b) Would you be \(100 \%\) certain of the prediction in part (a)? Why or why not?

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