Answer:
The critical stress required for the propagation of an initial crack [tex]\sigma_{c}[/tex] = 21.84 M pa
Explanation:
Given data
Modulus of elasticity E = 225 × [tex]10^{9}[/tex] [tex]\frac{N}{m^{2} }[/tex]
Specific surface energy for magnesium oxide is [tex]\gamma_{s}[/tex] = 1 [tex]\frac{J}{m^{2} }[/tex]
Crack length (a) = 0.3 mm = 0.0003 m
Critical stress is given by [tex]\sigma_{c}^{2} }[/tex] = [tex]\frac{2 E \gamma}{\pi a}[/tex] -------- (1)
⇒ 2 E [tex]\gamma_{s}[/tex] = 2 × 225 × [tex]10^{9}[/tex] × 1 = 450 × [tex]10^{9}[/tex]
⇒ [tex]\pi[/tex] a = 3.14 × 0.0003 = 0.000942
⇒ Put these values in equation 1 we get
⇒ [tex]\sigma_{c}^{2} }[/tex] = [tex]\frac{450 }{0.000942} 10^{9}[/tex]
⇒ [tex]\sigma_{c}^{2} }[/tex] = 4.77 × [tex]10^{14}[/tex]
⇒ [tex]\sigma_{c}[/tex] = 2.184 × [tex]10^{7}[/tex] [tex]\frac{N}{m^{2} }[/tex]
⇒ [tex]\sigma_{c}[/tex] = 21.84 [tex]\frac{N}{mm^{2} }[/tex]
⇒ [tex]\sigma_{c}[/tex] = 21.84 M pa
This is the critical stress required for the propagation of an initial crack.
