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