Question

An MgO component must not fail when a tensile stress of 13.5 MPa (1960 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of MgO is 1.0 J/m2 . Take 225 GPa as the modulus of elasticity

Answer 1

Answer:

The maximum allowable surface crack length is 7.86 x 10⁻⁴ m

Explanation:

Given;

tensile stress, σ = 13.5 Mpa = 13.5 x 10⁶ Pa

specific surface energy, γ = 1.0 J/m²

modulus of elasticity, E = 225 GPa = 225 x 10⁹ Pa

maximum allowable surface crack length, d = ?

$\sigma = (\frac{2E \gamma}{\pi d} )^{\frac{1}{2}}\\\\d = \frac{2E \gamma}{\pi \sigma ^2} = \frac{2*225*10^9 *1}{\pi (13.5 *10^6)^2} =\frac{450*10^9}{\pi*182.25*10^{12}} = 7.86 *10^{-4} \ m$

Therefore, the maximum allowable surface crack length is 0.000786 m if MgO component must not fail when a tensile stress of 13.5 MPa is applied.

Answer 2

If the surface energy is 1.0 J/m², the max. allowable surface length will be "7.86 × 10⁻⁴ m".

Elasticity and Stress

According to the question,

Tensile stress, σ = 13.5 MPa or,

                           = 13.5 × 10⁶ Pa

Specific surface energy, γ = 1.0 J/m²

Elasticity modulus, E = 225 GPa or,

                                    = 225 × 10⁹ Pa

As we know the relation,

→ σ = $(\frac{2E \gamma }{\pi d} )^{\frac{1}{2} }$

Now, max. length will be:

→ d = $\frac{2E \gamma }{\pi \sigma^2}$

By substituting the values, we get

     = $\frac{2\times 225\times 10^{9}\times 1}{\pi (13.5\times 10^6)^2}$

     = $\frac{450\times 10^9}{\pi\times 182.25\times 10^{12}}$

     = 7.86 × 10⁻⁴ m  

Thus the response above is appropriate.

Find out more information about stress here:

https://brainly.com/question/14468674