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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10–4 mm (7.5 × 10–6 in.) and a crack length of 3.8 × 10–2 mm (1.5 × 10–3 in.) when a tensile stress of 140 MPa (20,000 psi) is applied?

Respuesta :

matthewadeyinka10

Answer:

Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)

where

σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa

α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m

ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m

Substituting these values into the formula, we can calculate the max stress as

 ====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)

σₐ  = 24.4MPa