Question

Problem 9.11 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 92 MPa m (klein.)) and a yield strength of 900 MPa (65270 psi). The flaw size resolution limit of the flaw detection apparatus is 3 mm (0.1181 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.15, what is the critical flaw length

Answer 1

the critical flaw length is 10.06 mm

Step-by-step explanation:

Given the data in the question;

plane strain fracture toughness
`K_{tc` = 92 Mpa√m

yield strength σ
`_y` = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length
`a_c` = 1/π(
`K_{tc` / Yσ )²

we substitute

`a_c` = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²

`a_c` = 1/π( 92 Mpa√m / (517.5 Mpa )²

`a_c` = 1/π( 0.177777 )²

`a_c` = 1/π( 0.03160466 )

`a_c` = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{
`a_c` = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection