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If a steel cylindrical specimen is stressed nominally to 53 MPa, what stress level exists at the tip of an elliptical surface flaw, whose length is 5.8 microns and radius of curvature is 1,065 nanometers ? Answer Format: X (no decimal) Unit: MPa Aside: You should be asking yourself if this magnified stress is a big deal or not ... Note that the yield strength of a typical low-carbon ductile steel is around 250 MPa.

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Olajidey

Answer:

227.9MPa

Explanation:

Length of the flaws is given by

2b = 5.8microns

b = 2.9 × 10⁻⁶m

The relation between the radius of curvature and length and width of the elliptical flaw

[tex]r = \frac{a^2}{b}[/tex]

[tex]a = \sqrt{rb}[/tex]

Equation for stress at the tip of an elliptical surface flaw

[tex]\sigma _t = \sigma(1 + 2\frac{b}{a} )\\\\\sigma _t = \sigma(1 + 2\frac{b}{\sqrt{rb} })\\\\\sigma _t = \sigma(1 + 2\frac{\sqrt{b} }{\sqrt{r} })\\\\\sigma _t = \sigma(1 + 2\sqrt{\frac{b}{r} })[/tex]

[tex]\sigma _t = 53 \times 10^6 (1 + 2\sqrt{\frac{2.9\times10^-^6 }{1065 \times 10^-^9} } \\\\\sigma _t = 227.9 \times 10^6\\\\\sigma _t = 227.9MPa[/tex]

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