An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 mpa m (36.4 ksi in.
). it has been determined that fracture results at a stress of 300 mpa (43,500 psi) when the maximum (or critical) internal crack length is 4.0 mm (0.16 in.). for this same component and alloy, will fracture occur at a stress level of 260 mpa (38,000 psi) when the maximum internal crack length is 6.0 mm (0.24 in.)? why or why not?
We will first determine using the given if an aircraft component will fracture with a given stress level (260 MPa), maximum internal crack length (6.0 mm) and fracture toughness (40 MPa m ), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. First, it is necessary to solve for the parameter Y, using Equation 8.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and 2 a = 4.0 mm). Therefore,
Y = K(Ic)/ sqrt(π a) = 40 MPa( m ) / (300 MPa) sqrt(( π ) ((4 × 10-3 m)/2)) = 1.68
We will now solve for the product Y σ π a for the other set of conditions, so as to ascertain whether or not this value is greater than the K(Ic) for the alloy. Thus,
Y sqrt(π a) = (1.68)(260 MPa) sqrt (( π )[(6 × 10^-3 m)/ 2]) = 42.4 MPa sqrt (m) (39 ksi in. )
Therefore, fracture will occur since this value ( 42.4 MPa sqrt(m)) is greater than the K(Ic) of the material, 40 MPa sqrt(m).
The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.
