#### Answer

$ρ_{t} = 1.36 \times 10^{-6} mm = 1.36 nm$

#### Work Step by Step

Required:
Critical crack tip radius for a glass specimen that experiences tensile fracture at an applied stress of 70 MPa (10,000 psi), assuming a critical surface crack length of $10^{-2} mm$ and a theoretical fracture strength of E/10, where E is the modulus of elasticity.
Solution:
Using Equation 8.1:
$σ_{m} = 2σ_{0}(\frac{a}{ρ_{t}})^{1/2}$
Fracture will occur when $σ_{m}$ reaches the fracture strength of the material, which is E/10. E equals 120 GPa (from Table 12.5). Thus:
$\frac{E}{10} = 2σ_{0}(\frac{a}{ρ_{t}})^{1/2}$
Solving for $ρ_{t}$:
$ρ_{t} = \frac{400 (10^{-2} mm) (70 MPa)^{2}}{(120 \times 10^{3} MPa)^{2}} =1.36 \times 10^{-6} mm = 1.36 nm$