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Maximum Stress at the Tip of an Internal Crack

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 (10⁻⁵ in.) and a crack length of 2.5 × 10⁻² mm (10⁻³ in.) when a tensile stress of 170 MPa (25,000 psi) is applied?
A) 240 MPa
B) 300 MPa
C) 190 MPa
D) 520 MPa




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1 Answer

  1. To calculate the magnitude of the maximum stress at the tip of an internal crack, we can use the principle of stress concentration. The formula to estimate the maximum stress, (sigma_{text{max}}), at the crack tip is given by:

    [

    sigma_{text{max}} = sigma_{text{applied}} left(1 + 2 frac{a}{r}right)

    ]

    where:

    – (sigma_{text{applied}}) is the applied tensile stress,

    – (a) is the length of the crack,

    – (r) is the radius of curvature at the crack tip.

    Given:

    – (sigma_{text{applied}} = 170 text{ MPa}),

    – (a = 2.5 times 10^{-2} text{ mm} = 0.025 text{ mm}),

    – (r = 2.5 times 10^{-4} text{ mm} = 0.00025 text{ mm}).

    Now, substituting the values into the formula:

    1. Calculate ( frac{a}{r} ):

    [

    frac{a}{r} = frac{0.025}{0.00025} = 100

    ]

    2. Substitute into the stress concentration formula:

    [

    sigma_{text{max}} = 170 text{ MPa}

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