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: 12.42


$ρ_{t} = 1.36 \times 10^{-6} mm = 1.36 nm$

Work Step by Step

Required: Critical crack tip radius for a glass specimen that experiences tensile fracture at an applied stress of 70 MPa (10,000 psi), assuming a critical surface crack length of $10^{-2} mm$ and a theoretical fracture strength of E/10, where E is the modulus of elasticity. Solution: Using Equation 8.1: $σ_{m} = 2σ_{0}(\frac{a}{ρ_{t}})^{1/2}$ Fracture will occur when $σ_{m}$ reaches the fracture strength of the material, which is E/10. E equals 120 GPa (from Table 12.5). Thus: $\frac{E}{10} = 2σ_{0}(\frac{a}{ρ_{t}})^{1/2}$ Solving for $ρ_{t}$: $ρ_{t} = \frac{400 (10^{-2} mm) (70 MPa)^{2}}{(120 \times 10^{3} MPa)^{2}} =1.36 \times 10^{-6} mm = 1.36 nm$
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