Final answer:
To determine if a critical flaw is detectable, we use the Irwin–Griffith equation with provided material properties. If the computed critical flaw size exceeds the flaw detection limit of 4.0 mm, it is undetectable; if less, it is detectable.
Step-by-step explanation:
The question deals with determining if a critical flaw in a steel plate, under a given design stress, is detectable using the resolution limit of the flaw detection apparatus. Given a plane strain fracture toughness of 77.0 MPa√m, a yield strength of 1400 MPa, a design stress which is half of the yield strength, and a flaw size resolution limit of 4.0 mm, one can use the Irwin–Griffith equation to estimate the critical flaw size that would cause fracture at a given stress level.
The formula for determining the critical crack size (a_c) is given by:
a_c = √(√(K_IC / (σ * Y))^2)
Where K_IC is the fracture toughness, σ is the applied stress, and Y is a geometry factor which is given as 1.0 in this problem. The applied design stress (σ) is one half of the yield strength, so σ = 700 MPa. Substituting these values into the equation gives us the critical flaw size.
If the calculated critical flaw size is greater than the flaw detection apparatus's resolution limit of 4.0 mm, then the critical flaw would go undetected; otherwise, it would be detectable.