Question

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa m. It has been determined that fracture results at a stress of 365 MPa when the maximum internal crack length is 2.5 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm. 90 MPa

Answer 1

`\sigma_2=288.39\,MPa`

Step-by-step explanation:

Given:

Strain fracture toughness,
`K=40\,MPa.√(m)`

maximum internal crack length in first case,
`l_1=2.5* 10^(-3)m`

stress level in the first case,
`\sigma_2 =365\,MPa`

maximum internal crack length in second case,
`l_2=4.0* 10^(-3)m`

stress level in the second case,
`\sigma_2 =?`

So, now we require a factor given as:

`Y=\frac{K}{\sigma.\sqrt{(\pi.l)/(2) } }`

`Y=\frac{40}{365\sqrt{(\pi* 2.5* 10^(-3))/(2) } }`

`Y=1.75`

Now,

`\sigma_2=\frac{K}{Y.\sqrt{(\pi.l)/(2) } }`

`\sigma_2=\frac{40}{1.75\sqrt{(\pi* 4* 10^(-3))/(2) } }`

`\sigma_2=288.39\,MPa`