## Materials Science and Engineering: An Introduction

$σ_{0} = 126.66 MPa$
Required: Using the data in Table 12.5, determine the flexural strength for nonporous MgO, assuming a value of 3.75 for n in Equation 12.10. Solution: From Table 12.5, $P =0.05, \ σ_{fs} = 105 MPa$, and, for nonporous material, $P = 0$ and $ln σ_{0} = ln σ_{fs}$ Using Equation 12.10 to compute the value of $n$: $ln σ_{fs} = ln σ_{0} - nP = ln (105 MPa) + (3.75)(0.05) = 4.8415$ $σ_{0} = e^{4.8415} = 126.66 MPa$