## Materials Science and Engineering: An Introduction

$σ_{fs} = 59.86 MPa$
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: $σ_{fs} = 59.86 MPa$