Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 12 - Structures and Properties of Ceramics - Questions and Problems - Page 508: 12.52a


$σ_{fs} = 59.86 MPa$

Work Step by Step

Given: Refer to the given table. Required: Flexural strength for a completely nonporous specimen of this material. Solution: Using Equation 12.10 and substituting the given values in the Table, two expressions will be written: $ln σ_{fs} = ln σ_{0} - nP $ $ln(70 MPa) = ln(σ_{0}) - (0.10)n$ $ln(60 MPa) = ln(σ_{0}) - (0.15)n$ Solving for $σ_{0}$ and $n$, $σ_{0} = 59.86 MPa$ $n = 1.56$ Considering that for a nonporous material, P= 0 and from Equation 12.10, $σ_{0} = σ_{fs}$, it follows: $σ_{fs} = 59.86 MPa$
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