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

Step-by-step explanation:

From the given information:

Strain fracture toughness
`K_k`= 75 MPa
`√(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 = (1)/(\pi) ( (k_k)/(\sigma Y))^2 \\ \\ a_c = (1)/(\pi) \Big [ (75 * √(10^3))/(361 * 1.03 ) \Big]^2 \\ \\ a_c = (1)/(\pi) \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}`