Question

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 32 . It has been determined that fracture results at a stress of 219 MPa when the maximum (or critical) internal crack length is 2.68 mm. a) Determine the value of for this same component and alloy at a stress level of 284 MPa when the maximum internal crack length is 1.34 mm.

Answer 1

`(32)/(219x0.006488)`Answer:

The answer is 309MPa

Step-by-step explanation:

It first become necessary to solve for the parameter Y for the conditions under which the fracture occurred using below equation

Y = δ
`(kic)/(δ√(\pia) )`
`\pi a`

=
`(32)/(219√(\pi )(2.68x10^-3)/(2) )`

=
`(32)/(219√(0.00421) )`

=
`(32)/(219*0.06488)`

`(32)/(14.2097) =2.25`

`(Kic)/(y√(\pia ) )`
`\pi`a =
`\frac{32}{2.25\sqrt{\pi (1.34*10^-3)/(2) } }`

`(32)/(2.25√(\pi*0.00067 ) )`

=
`(32)/(2.25√(0.00211) )`

=
`(32)/(2.25 * 0.0459)` =
`(32)/(0.1034)`

= 309MPa