What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4 mm (1.181 × 10-5 in.) and a crack length of 2.5 × 10-2 mm (0.9843 × 10-3 in.) when a tensile stress of 170 MPa (24660 psi) is applied?

2195 MPa

Explanation:

Given that:

Radius of curvature () = 3 × 10⁻⁴ mm,

Crack length () =  2.5 × 10⁻² mm,

Therefore the length of surface crack (a) = crack length / 2 = 2.5 × 10⁻² mm / 2 = 1.25 × 10⁻² mm,

Tensile stress () = 170 MPa = 170 × 10⁶ Pa

The magnitude of the maximum stress that exists at the tip of an internal

crack () is given by the equation:

Therefore the maximum stress is 2195 MPa

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