Answer:
[tex]\mathbf{\sigma_{max} =638.308 \ MPa}[/tex]
Explanation:
For any large steel plate with an infinite length and width and a center through the crack under tension, the stress intensity factor can be expressed as:
[tex]K_I = \sigma \sqrt{\pi a}[/tex]
where;
a = 2.5 mm/2
a = 1.25 mm
a = 1.25 × 10⁻³ m
Therefore; the maximum stress in tension capacity can be computed as;
[tex]\sigma_{max} = \dfrac{K_{I_C}}{\sqrt{\pi a_c}}[/tex]
[tex]\sigma_{max} = \dfrac{40}{\sqrt{\pi \times 1.25 \times 10^{-3}}}[/tex]
[tex]\mathbf{\sigma_{max} =638.308 \ MPa}[/tex]
