Answer :
To solve this problem we will apply the concepts related to the fracture toughness of material that relates the nominal stress, half of the internal through crack and the dimensional geometric constant. The mathematical equation contained in this equation is,
[tex]K_{IC} = Y\sigma_f \sqrt{\pi a}MPa \sqrt{m}[/tex]
Here,
[tex]K_{IC}[/tex] = Fracture toughness of the material
[tex]\sigma_f[/tex] = Applied nominal stress
a = Half of an internal through crack
Y = Dimensional geometric constant
Rearranging to find the crack length we have that
[tex]a = \frac{1}{\pi} \cdot (\frac{K_{IC}}{Y \cdot \sigma_f})^2 m[/tex]
Our values,
[tex]K_{IC} = 24.7Mpa \sqrt{m}[/tex]
[tex]\sigma = 395Mpa[/tex]
[tex]Y = \sqrt{\pi}[/tex]
Replacing with our values,
[tex]a = \frac{1}{\pi} \cdot (\frac{24.7}{\sqrt{\pi} \cdot 395})^2 m[/tex]
[tex]a = 0.000396187m[/tex]
[tex]a = 0.396187mm[/tex]
Calculate the critical crack length for a through crack in a thick plate,
[tex]2a = 2(0.396187mm)[/tex]
[tex]2a = 0.79237mm[/tex]
Therefore the critical crack length for a through crack in a thick plate 0.79237mm
The critical crack length is the crack that grows just after the fracture. The critical crack length of the given metal plate is 3961 m.
Critical crack length can be calculated by the formula:
[tex]a = \dfrac 1\pi \times (\dfrac {K_{IC}}{Y\sigma _f}} )^2m[/tex]
Where,
[tex]K_{IC}[/tex] = Fracture toughness of the material =[tex]24.7 Mpa \sqrt m[/tex]
[tex]\sigma_f[/tex]= Applied nominal stress = [tex]395 Mpa[/tex]
[tex]a[/tex] = Half of an internal through the crack = ?
[tex]Y[/tex] = Dimensional geometric constant = [tex]\sqrt \pi[/tex]
Put the values in the equation,
[tex]a = \dfrac 1\pi \times (\dfrac { 24.7 Mpa \sqrt m}{\sqrt \pi\times 395\rm \ Mpa}} )^2m\\\\a = 3961 \rm \ m[/tex]
Therefore, the critical crack length of the given metal plate is 3961 m.
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