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

Answer 2

The maximum length of a surface flaw that is possible without fracture is; 8.24 × 10⁻⁶ m

What is the maximum length of the surface flaw?

We are given;

Modulus of elasticity; E = 69 GPa = 69 * 10⁹ Pa

Specific surface energy; δ_s  = 0.3 J/m²

Tensile stress; σ_c = 40 MPa = 40 * 10⁶ Pa

Formula to find the stress is;

σ_c  = √(2Eδ_s/(πa))

where a is length of surface flaw

a = [(2E * δ_s)/(π * σ_c)]²

a = (2 * 69 * 10⁹ * 0.3)/(π * 40 * 10⁶)²

a = 8.24 × 10⁻⁶ m

Read more about brittle fracture at; https://brainly.com/question/19953629