Respuesta :
Answer:
The answer is "[tex]\bold{87.3906 \ MPa \sqrt{m}}[/tex]".
Explanation:
Given value:
[tex]\sigma = 1400 \ MPa \ \ \ \ \ \ \ \ where \ \sigma = failure \ strength\\\\a = 1 \ mm = 1 \times 10^{-3} \ m \ \ \ \ \ \ \ \ \ \ where\ a = crack\ length\\\\b= 12.5 \ cm = 125 \ mm = 0.125 \ m\\\\[/tex]
[tex]\to \alpha = \frac{a}{b} =\frac{1}{125} = 8 \times 10^{-3}\\\\[/tex]
[tex]k_{b} = \frac{1.12 + \alpha (2.62 \alpha -1.59)}{1-0.7 \alpha}\\[/tex]
[tex]= \frac{1.12 + (8\times 10^{-3}(2.62(8\times 10^{-3}) -1.59))}{1-(0.7 \times 8\times 10^{-3})}\\\\= \frac{1.12 + (8\times 10^{-3}(0.02096 -1.59))}{1-(0.7 \times 8\times 10^{-3})}\\\\= \frac{1.12 + (8\times 10^{-3}(-1.56904))}{1-(0.0056)}\\\\= \frac{1.12 + (-0.01255232)}{0.9944}\\\\= \frac{-1.10744768}{0.9944}\\\\= -1.11368431\\\\[/tex]
[tex]k_{ic} = \sigma \sqrt{\pi a} \ y_b[/tex]
[tex]=1400 \times \sqrt{\pi \times 1 \times 10^{-3} } \times -1.11368431\\\\=1400 \times 0.00177200451 \times -1.11368431\\\\=87.3906 \ MPa \sqrt{m}[/tex]
