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.

The value of  the maximum length of a surface flaw that is possible without fracture a = 1.02 × mm

Explanation:

Given data

Tensile stress = 50

Specific surface energy = 0.5 = 0.5 ×

Modulus of elasticity E = 80 ×

The critical stress is given by = ----- (1)

In the limiting case =

⇒ = ------ (2)

Put all the values in above formula we get,

⇒ = 2 × 80 × × 0.5 × × ×

⇒ a = 1.02 × mm

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

found in the nucleus

has mass of one amu