## Materials Science and Engineering: An Introduction

$σ_{fs} = 43.82 MPa$
Given: Refer to the given Table. From Item 12.52a, $σ_{0} = 59.86 MPa$, n = 1.56 Required: Flexural strength for a 0.20 volume fraction porosity. Solution: Rearranging Equation 12.10: $σ_{fs} = σ_{0} e^{(-nP)} = (59.86 MPa) e^{[-(1.56)(.20)] }= 43.82 MPa$