Question

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elasticity is 69 GPa, determine the maximum length of a surface flaw that is possible without fracture.

Answer 1

Answer:

The maximum length of a surface flaw is 8.24 μm

8.24 μm

Explanation:

Given that:

The modulus of elasticity E = 69 GPa

The specific surface energy $\delta_s$ = 0.3 J/m²

The length of the surface flaw "a" = ??

From the theory of the brittle fracture;

$\sigma _c = \bigg ( \dfrac{2E \delta_s}{\pi a} \bigg )^{1/2}$

Making a the subject of the formula; we have:

$a = \bigg ( \dfrac{2 \times E \times \delta_s}{\pi \sigma _c ^2} \bigg )$

$a= \bigg ( \dfrac{2 \times 69*10^9 \times 0.3}{\pi (40*10^6)^2} \bigg )$

a = 8.24 × 10⁻⁶ m

a = 8.24 μm

Thus; the maximum length of a surface flaw is 8.24 μm