Question

A specimen of a 4340-steel alloy with a plane strain fracture toughness of 54.8 MPa sqrt(m) (50 ksi sqrt(in.)) is exposed to a stress of 1030 MPa (150,000 psi). Will this specimen experience fracture if the largest surface crack is 0.5 mm (0.02 in.) long

Answer 1

critical stress
`\sigma _c` = 1382.67 MPa

Step-by-step explanation:

given data

plane strain fracture toughness = 54.8 MP

length of surface creak = 0.5 mm

we take here

parameter Y = 1.0

solution

we apply critical stress formula that is

critical stress
`\sigma _c` =
`(K)/(Y√(\pi * a) )` .............................1

here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1

critical stress
`\sigma _c` =
`\frac{54.8 * 10^6}{1 \sqrt{\pi * 5 * 106{-4}}}`

solve it we get

critical stress = 1382.67 MPa

As exposed stress 1030 MPa is less than critical stress 1382 MPa

so that fracture will not be occur here