Question

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 26 MPa√m . It has been determined that the fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm

Answer 1

193901.39793 Pa

Step-by-step explanation:

a = Crack length

`\gamma` = Geometrical factor

`\sigma` = Applied stress = 112 MPa

Plane strain fracture toughness is given by

`K_f=\gamma \sigma √(\pi a)\\\Rightarrow \gamma=(K_f)/(\sigma √(\pi a))\\\Rightarrow \gamma=(26)/(112* √(\pi 0.0086))\\\Rightarrow \gamma=1.41231`

When a = 6 mm

`K_f=\gamma \sigma √(\pi a)\\\Rightarrow K_f=1.41231 112* 10^6√(\pi 0.006)\\\Rightarrow K_f=193901.39793\ Pa`

the stress level at which fracture will occur is 193901.39793 Pa