Question

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in uniaxial tension. For this alloy Kic = 24.1 MPa m1/2 and =363 MPa. Assume Y = . The answer should be in mm with 3 decimals of accuracy.

Answer 1

Explanation:

Formula to determine the critical crack is as follows.

          $K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

  $\gamma$ = 1,     $K_{IC}$ = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   $\sigma_{f} = 570 \times \frac{3}{4}$

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

        = $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

       = $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

            Largest size = 2a

                                 = $2 \times 10.13 \times 10^{-4}$

                                 = $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$.