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 4×10-2 mm (1.575×10-3 in.) when a tensile stress of 100 MPa (14500 psi) is applied?

Answer 1

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

$\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}$

Where:

ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)

σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi)

2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in)

Hence, the  maximum stress (σ) is:

$\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa$    

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!