Answer:
193901.39793 Pa
Explanation:
a = Crack length
[tex]\gamma[/tex] = Geometrical factor
[tex]\sigma[/tex] = Applied stress = 112 MPa
Plane strain fracture toughness is given by
[tex]K_f=\gamma \sigma \sqrt{\pi a}\\\Rightarrow \gamma=\frac{K_f}{\sigma \sqrt{\pi a}}\\\Rightarrow \gamma=\frac{26}{112\times \sqrt{\pi 0.0086}}\\\Rightarrow \gamma=1.41231[/tex]
When a = 6 mm
[tex]K_f=\gamma \sigma \sqrt{\pi a}\\\Rightarrow K_f=1.41231 112\times 10^6\sqrt{\pi 0.006}\\\Rightarrow K_f=193901.39793\ Pa[/tex]
the stress level at which fracture will occur is 193901.39793 Pa
