Question

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 30 MPa (m) 1/2 . It has been determined that fracture results at a stress of 120 MPa when the maximum internal crack length is 8.4 mm. For this same component and alloy,

Compute the stress level at which fracture will occur for a critical internal crack length of 5.6 mm?

Answer 1

Given ,

fracture toughness (K_lc) = 30MPam

maximum internal cracking length =8.4mm

another internal crack length = 5.6mm

since fracture occurs for the same component using the same alloy at 120MPa. First we find the parameter Y for the condition at which fracture occurs

we have the equation Y= K_lc/(x(a))

where Y - is a dimensionless parameter of function that depends on both crack and specimen size and geometrics as well as the manner of load application

a - half the length of internal crack

Y = 30MPam/ (120x(x(0.0084/2)))

= 2.17

Now we will compute the stress level at which fracture occurs

   = K_lc/(Yx(a))

= 30MPam/ (2.17x(x(0.0056/2)))

= 147.40 MPa

so the stress level at which failure occurs for a critical internal crack length of 5.6mm in 147.40MPa

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