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 0.7 mm.

critical stress= 20.23 MPa

Explanation:

Since there was an internal crack, we will divide the length of the internal crack by 2

Length of internal crack, a = 0.7mm,

Half length = 0.7mm/2= 0.35mm  changing to meters becomes

0.35/ 1000= 0.35 x 10 ^-3m

The formulae for critical stress is calculated using

σC = (2Eγs /πa) ¹/₂

σC = critical stress=?

Given

E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²

γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m

a= Half Length of crack=0.35 x 10 ^-3m

σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂

=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂

=(4.0920 x10 ^14)¹/₂

σC=20.23 x10^6 N/m² = 20.23 MPa

answer:

put more,