Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 29 MPa (26.39 ksi). It has been determined that fracture results at a stress of 118 MPa (17110 psi) when the maximum internal crack length is 8.7 mm (0.3425 in.). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 5.9 mm (0.2323 in.).

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m−−√

(23.7 ksi in.−−√

). It has been determined that fracture results at a stress of 112 MPa (16,240 psi) when the maximum internal crack length is 8.6 mm (0.34 in.). 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 (0.24 in.).

Explanation:

As we know

σ = / y√π a

y = / σ√π a

y =

y = 2.00

now,

σ = / y√π a

   =

   = 150.65 MPa

So the stress level at which the fracture will occur  for  acritical internal crack length of 5.9 mm is 150.65 MPa.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

the right answer is 0.3 meters per second