A rectangular bar of 7075 aluminum is simply supported with

A rectangular bar of 7075 aluminum is simply supported with a centered point load and has dimensions 1=8 in, h= 1.25 in, f=0.5 in and an edge crack of depth a= 0.125 in. The load F= 1000 Ibf. Determine the FOS against fracture for the existing crack geometry. Determine the FOS against fracture if the bar is ground down, across its full length, such as to reduce the crack to o=0.06 in Determine the FOS against yield if the bar is ground down, across its full length, such as to remove the crack entirely.

Solution

Load F = 1000 lbf

Length = 8 in

Height = 1.25 int

Thickness t = 0.5 in

Ultimate tensile strength for 7075 material = 290 Mpa

Cross sectional Area = L*B

= 8 * (1.25-0.125)

= 9 in

Working stress = Load/Area

= 1000 / 9

= 111 lbf/in

Factor of Safety = Ultimate tensile strength / Working stress

= 290 / 111

= 2.61

(The design fail at twice the design load)

Cross sectional Area = L*B

= 8 * (1.25-0.06)

= 9.52 in

Working stress = Load/Area

= 1000 / 9.52

= 105 lbf/in

Factor of Safety = Ultimate tensile strength / Working stress

= 290 / 105

= 2.76

(The design fail at twice the design load)

Cross sectional Area = L*B

= 8 * 1.25

= 10 in

Working stress = Load/Area

= 1000 / 10

= 100 lbf/in

Factor of Safety = Ultimate tensile strength / Working stress

= 290 / 100

= 2.92

(The design fail at twice the design load)