Answer:
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Explanation:
The given values are,
σ=1.65 MPa
γs=0.60 J/m2
E= 2.0 GPa
The maximum possible length is calculated as:
[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
