Question

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 88 () and a yield strength of 710 MPa (51490 psi). The flaw size resolution limit of the flaw detection apparatus is 4 mm (0.1575 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.07, what is the critical flaw length

Answer 1

Answer:

Critical Flaw Length=17.08 mm

The Critical flaw Length > 4mm, It means it is detectable.

Explanation:

Given Data:

Fracture Toughness=$K_{tc}$=88MPa

Yield Strength=σ=710 MPa

Y=1.07

Solution:

Formula:

$Critical\ Length=\frac{1}{\pi } *(\frac{K_{tc}}{Y*\sigma} )^2$

Since yield Strength is half, Critical Length will be:

$Critical\ Length=\frac{1}{\pi } *(\frac{K_{tc}}{\frac{\sigma}{2} *Y} )^2\\Critical\ Length=\frac{1}{\pi } *(\frac{88MPa}{\frac{710MPa}{2} *1.07} )^2\\\\Critical\ Length=0.01708\ m$

Critical Flaw Length=17.08 mm

The Critical flaw Length > 4mm, It means it is detectable.