Question

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10-4 mm (1.378 × 10-5 in.) and a crack length of 5.5 × 10-2 mm (2.165 × 10-3 in.) when a tensile stress of 220 MPa (31910 psi) is applied?

Answer 1

Answer:

The magnitude of the maximum stress that exists at the tip of the internal crack is 3900.183MPa (565703.38 psi).

Explanation:

The magnitude of the maximum stress that exists at the tip of an internal crack can be estimated using fracture mechanics. If a material that have internal crack is loaded with tensile stress, the crack tends to widen due to the tensile force created by the stress, and this leads to a concentration of stress near the tip of the crack, thereby expanding the crack length and finally leading to failure of the material.

Fracture mechanics involves the study of propagation of cracks in a material. It is used to predict the conditions at which failure of a material will likely occur.

A crack may appear in the surface of the material (surface crack) or at the interior of the material (internal crack), but the same formula can be used to estimate the magnitude of the maximum stress at the tip of the crack.

The formula is given as Πm=2Πo√(a/r)

Where,

Πm is the maximum stress

Πo is the tensile stress

r is the radius of curvature

a is the crack length.

The major difference between the surface crack and the internal crack is the value for their crack length.

Crack length can be understood as the length of a crack at which the crack becomes unstable under certain applied stress.

For surface crack, crack length=a

For internal crack, crack length=2a because internal cracks forms two surfaces of a complete sphere, while the surface crack forms one surface of a sphere.

Given in the question,

Radius of curvature, r=3.5×10^(-4) mm=0.00035mm

Crack length, a=5.5×10^(-2) mm

For internal crack, a=[(crack length)/2]={[5.5×10^(-2)]/2}=0.0275mm

Tensile stress, Πo=220Mpa

Maximum stress=Πm

Using,

Πm=2Πo√(a/r)

=(2×220)√(0.0275/0.00035)

=440√78.571

=3900.183MPa (565703.38 psi)