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 [tex] = 98.9 MPa \sqrt{m} [/tex]
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:
[tex] oc = \frac{K}{Y \sqrt{pi * a}} [/tex]
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
[tex] a= \frac{1}{pi} * [\frac{k}{y*a}]^2 [/tex]
Substituting figures in the expression above, we have:
[tex] = \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2[/tex]
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection
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.
