(Solved): = min = The 2-mm thick bar shown below is loaded axially with a fluctuating force, Fmax = 10 kN an ...

= min = The 2-mm thick bar shown below is loaded axially with a fluctuating force, Fmax = 10 kN and F, = 0 kN The load ratio, given as R = F min/Fmax = 0, captures the state of fluctuating loads in a single number. The bar material is a 1045 Cold-Drawn steel with the following material properties: Elasticity: E = 205 GPa, v = 0.29, p = 7850 kg/m3 Static strength: Sy = 530 MPa, Sut= 625 MPa Fatigue: o'=1225 MPa, b = -0.095 It is desired to drill a hole through the center of the 40-mm face of the bar to allow a cable to pass through it. A 4-mm hole is sufficient for the cable to pass through, but we also want to evaluate larger diameter holes to increase clearance and reduce weight. d = 12 mm 1045 Cold-Drawn Steel 40 mm 1 mm rad B A 40 mm 34 mm 10 KN ? 80 mm 140 mm