Answer:
Explanation:
From the given information:
The first thing we need to do is to use the formula used in-plane strain fracture toughness to determine the geometry factor [tex]\gamma[/tex]
i.e
[tex]K_1 = \gamma \sigma_{applied} \sqrt{Ra_1}[/tex]
where;
a = crack length
[tex]\gamma =[/tex] geometry factor
In the first scenario, where;
Plain fracture toughness [tex]K_1 = 45 \ MPa \sqrt{m}[/tex]
[tex]\sigma _{applied} = 300 \ MPa \\ \\ a_1 = 6.95 \times 10^{-3} m[/tex]
radius(R) = 3.142
Then, replacing it into the above equation, we have:
[tex]45 MPa = \gamma (300 \ MPa ) \sqrt{3.142 \times 0.95 \times 10^{-3}} \\ \\ 45 \ MPa = 16.3902715 \gamma \\ \\ \gamma = \dfrac{45 \ MPa}{16.3902715} \\ \\ \gamma \ (geometry factor)= 2.745[/tex]
Now, since we've determined the geometry factor, it will be easier to estimate the max. allowable surface length.
∴
[tex]K_2 = \gamma \sigma_{applied} \sqrt{Ra _2}[/tex]
[tex]67.5= 2.745 \times 300 \times \sqrt{3.142 \times a _2}[/tex]
[tex]67.5= 1459.710372 \sqrt{ a _2} \\ \\ \dfrac{67.5}{ 1459.710372}= \sqrt{ a _2} \\ \\ 0.0462420 = \sqrt{ a _2} \\\\ a_2 = 0.0462420^2 \\ \\ a _ 2 = 0.002138 \\ \\ a_2 = 2.138\times 10^{-3} \ m \\ \\ \mathbf{a_2 = 2.138\ mm}[/tex]
