Answer:
critical stress required is 18.92 MPa
Explanation:
given data
specific surface energy = 1.0 J/m²
modulus of elasticity = 225 GPa
internal crack of length = 0.8 mm
solution
we get here one half length of internal crack that is
2a = 0.8 mm
so a = 0.4 mm = 0.4 × [tex]10^{-3}[/tex] m
so we get here critical stress that is
[tex]\sigma _c = \sqrt{\frac{2E \gamma }{\pi a}}[/tex] ...............1
put here value we get
[tex]\sigma _c[/tex] = [tex]\sqrt{\frac{2\times 225\times 10^9 \times 1 }{\pi \times 0.4\times 10^{-3}}}[/tex]
[tex]\sigma _c[/tex] = 18923493.9151 N/m²
[tex]\sigma _c[/tex] = 18.92 MPa