Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 75 MPa (68.25 ksi). If the plate is exposed to a tensile stress of 361 MPa (52360 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.03 for Y.

Answer 1

Answer:

Explanation:

From the given information:

Strain fracture toughness $K_k$= 75 MPa$\sqrt{m}$

Tensile stress $\sigma$ = 361 MPa

Value of Y = 1.03

Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:

$a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}$