Answer:
![\mathbf{\sigma_(max) =638.308 \ MPa}](https://img.qammunity.org/2021/formulas/engineering/college/tw46xoe58ym9ve2slnh6n31r1iu6jutyap.png)
Step-by-step 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:
![K_I = \sigma √(\pi a)](https://img.qammunity.org/2021/formulas/engineering/college/y3wqkrz20qbb9xc6ik7g15ppr3fze7bt3g.png)
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;
![\sigma_(max) = (K_(I_C))/(√(\pi a_c))](https://img.qammunity.org/2021/formulas/engineering/college/pl1e97wdsqa8otrl8i3js61wgqsgey7wax.png)
![\sigma_(max) = \frac{40}{\sqrt{\pi * 1.25 * 10^(-3)}}](https://img.qammunity.org/2021/formulas/engineering/college/ruimgyfn24o5h9ig3y4iqoddf8mtmar17s.png)
![\mathbf{\sigma_(max) =638.308 \ MPa}](https://img.qammunity.org/2021/formulas/engineering/college/tw46xoe58ym9ve2slnh6n31r1iu6jutyap.png)