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 ( ) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half the yield strength and the value of Y is 1.0, determine whether a critical flaw for this plate is subject to detection.

Answer 1

In the question, the plane-strain fracture toughness (K) is missing. The value is 98.9Mpa√m

Answer:

The critical flaw can be detected since A_c = 16.84mm is much more greater than the flaw detection Apparatus's resolution limit of 3mm.

Explanation:

Length of surface creek (A_c) is given as;

A_c = (1/π)(K/Yσ)

Where K is toughness

σ is Yield strength

Frim question, Y = 1

Thus,

A_c = (1/π)(98.9Mpa√m/(1 x 860/2))²

A_c = (1/π)(98.9²m/(430²) = 0.01684

= 16.84mm

The critical flaw can be detected since A_c = 16.84mm is much more greater than the flaw detection Apparatus's resolution limit of 3mm.

Answer 2

Complete 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 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection  

Explanation:

We are given:

Plane strain fracture toughness K $= 98.9 MPa \sqrt{m}$

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

$oc = \frac{K}{Y \sqrt{pi * a}}$

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

$a= \frac{1}{pi} * [\frac{k}{y*a}]^2$

Substituting figures in the expression above, we have:

$= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2$

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection