Question

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5 × 10-4 mm (1.969 × 10-5 in.) and a crack length of 4.5 × 10-2 mm (1.772 × 10-3 in.) when a tensile stress of 190 MPa (27560 psi) is applied?

Answer 1

The question is a problem that requires the principles of fracture mechanics.

and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.

Step-by-step explanation:

Max Stress, σ = 2σ₀√(α/ρ)

where,

σ₀ = Tensile stress = 190MPa = 1.9x10⁸Pa

α = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m

ρ = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m

Max Stress, σ = 2 x 1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵

Max Stress, σ = 2 x 1.9x10⁸ x 6.708 Pa

Max Stress, σ = 2549MPa

Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa