Question

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.3 mm.

Answer 1

Answer:

The critical stress required for the propagation of an initial crack              $\sigma_{c}$ =  21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942  

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ =  21.84 $\frac{N}{mm^{2} }$

⇒ $\sigma_{c}$ =  21.84 M pa

This is the critical stress required for the propagation of an initial crack.