Question

A magnesium oxide component must not fail when a tensile stress of 14 MPa is applied. Determine the maximum allowable surface crack length if the surface energy of magnesium oxide is 1.0 J/m2. The modulus of elasticity of this material is 225 GPa.

Answer 1

`x=0.730*10^(-3)m`

Step-by-step explanation:

From the question we are told that:

Tensile stress
`\sigma_c = 14 MPa=>14*10^6`

Modulus of elasticity
`E=225GPa=>225*10^9`

Surface energy of MgO
`\gamma=1N/m`

Generally the equation for maximum allowable surface crack length is mathematically given by

`x=(2E \gamma)/(\pi\sigma_c^2)`

`x=(2(225*10^9)(1))/(3.142*(14*10^6)^2)`

`x=0.730*10^(-3)m`