The maximum stress that exists at the tip of the internal crack is 3,400 Mpa.
The given parameters;
- radius of the internal crack, r = 2.5 x 10⁻⁴ mm
- crack length, l = 2.5 x 10⁻²
- tensile stress, σ = 170 MPa = 170 x 10⁶ N/m²
The maximum stress that exists at the tip of the internal crack is calculated as follows;
[tex]\sigma _{max} = 2\sigma \times \sqrt{(\frac{l}{r} )} \\\\\sigma _{max} = 2 \times 170 \times 10^6 \times \sqrt{(\frac{2.5\times 10^{-2}}{2.5 \times 10^{-4}})} \\\\\sigma _{max} = 3.4 \times 10^{9} \ Pa\\\\\sigma _{max} = 3,400\ Mpa[/tex]
Thus, the maximum stress that exists at the tip of the internal crack is 3,400 Mpa.
Learn more here:https://brainly.com/question/19264500
