Question

A relatively large plate of a glass is subjected to a tensile stress of 50 MPa. If the specific surface energy and modulus of elasticity for this glass are 0.5 J/m2 and 80 GPa, respectively, determine the maximum length of a surface flaw that is possible without fracture.

Answer 1

Answer:

The value of  the maximum length of a surface flaw that is possible without fracture a = 1.02 × $10^{-11}$ mm

Explanation:

Given data

Tensile stress $\sigma$ = 50 $\frac{N}{mm^{2} }$

Specific surface energy $\gamma_{s}$ = 0.5 $\frac{J}{m^{2} }$ = 0.5 × $10^{-6}$ $\frac{J}{mm^{2} }$

Modulus of elasticity E = 80 × $10^{3}$ $\frac{N}{mm^{2} }$

The critical stress is given by $\sigma_{c}^{2}$ = $\frac{2 E \gamma_{s} }{\pi a}$ ----- (1)

In the limiting case $\sigma$ = $\sigma_{c}$

⇒ $\sigma^{2}$ = $\frac{2 E \gamma_{s} }{\pi a}$ ------ (2)

Put all the values in above formula we get,

⇒ $50^{2}$ = 2 × 80 × $10^{3}$ × 0.5 × $10^{-6}$ × $\frac{1}{3.14}$ × $\frac{1}{a}$

⇒ a = 1.02 × $10^{-11}$ mm

This is the value of  the maximum length of a surface flaw that is possible without fracture.