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

Answer:

the critical flaw length is 10.06 mm

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