Question

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10^-4 mm (0.9843 × 10^-5 in.) and a crack length of 5 × 10^-2 mm (1.969 × 10^-3 in.) when a tensile stress of 130 MPa (18860 psi) is applied?

Answer 1

2600 MPa

Step-by-step explanation:

The formula to be used for the question is

σ(m) = 2 * σ(o) * [α/ρ(t)]^0.5, where

σ(m) = maximum stress

σ(o) = maximum applied tensile stress

α = length of surface crack

ρ(t) = radius of curvature of the crack

It's an easy one, as we have all the values given from the question, and all we do is plug them in directly

σ(m) = 2 * 130 * [(0.05/2)/0.00025]^0.5

σ(m) = 260 * [0.025/0.00025]^0.5

σ(m) = 260 * 100^0.5

σ(m) = 260 * 10

σ(m) = 2600 MPa