Question

Properties for all problems Sut = 80 ksi, 5= 60 ksi, Sf = 40 ksi, q=0.85 (notch sensitivity), steel E = 30e6 psi Remember we use Modified-Goodman for all calculations Problem 1 A 1-inch diameter rod that is 8 inches long has a bearing at one end and another bearing at 6 inches (treat bearings as simple supports). A vertical load is applied at the end of the overhanging rod and a torque is transmitted through the length of the rod. (Fmax = 100 lbf Tmax = 50 ft-lbf) Calculate the fatigue factor of safety for these cases. a) Force constant Fmax, Torque varies from Tmax to 0 b) Torque constant Tmax, Force varies from Fmax to o c) Torque constant Tmax, Force constant at Fmax but shaft is rotating so normal stress is fully reversed Problem 2 A3 inch hollow tube with 1/16 inch wall thickness is used to make a 4 foot long drive shaft. The maximum torque on the is 400 ft-lbf. The shaft is loaded cyclically from 0 to max torque. a) Calculate the fatigue and yield factor of safety b) Assume that there is a 0.3 in hole drilled in the shaft and the curve (B) below describes the stress concentration. Calculate the fatigue and yield factor of safety. Remember notch sensitivity. c) A force is placed in the center of the shaft that causes the drive shaft to deflect 0.002 inches. Calculate the fatigue and yield factor of safety. Assume the shaft is rotating so normal stress is fully reversed. (Assume drive shaft is simple supported at the ends) Table A-15 Charts of Theoretical Stress-Concentration Factors (Continued) 4.0 Figure A-15-10 Round shaft in torsion with transverse hole. 3.6

Answer 1

Given! for the material Sut = 80 KSF Sf = 40KSI diameter q ood" da din length = sin Imana lco Ibf, Toonz so lbf-fr maximum finan moment &in B m= fmcm x 2 4 DA 6 in TBY mo ICO x 2 = 16.66 lbf-in tay bending stress, of= 32m 169 psr 32x 16.66 TX13 16x so 254.65ps8 Shear stress, て 165 দনও IT+13 Force constant Frman, Torque varies Thon to o. Ta 352.8 PSI ob + V24C2 169 PSI meen stress, m = 01+02 २ 91.9 PSE 261PSE stress amplitude, oua 01-02 2 how from good man theory- + Tur of th 91.9 20) + 80x103 40x100 * factor ofuafety (J) = 130 Torque constant Tomca, force varies firmon too. - = 352.8 PSI 129.33 PSI 113 PSE meen stress, m = Ita stress amplitude, tar -62 240 PSI

from coupes b theary- om our T = os + + 240 113 80x103 T= 40x103 f factor of safety, f = 134 ☺ Fran but Torque constant Tman, force constant at shaft is rotating so normal stress is fully reversed- hince 352.8 PSI 352.8 pur 'maits nosu oraz = 0 २ Stress amplitude, ou こ 352.8 PSE I-02 2 from good idman theary. om द + EI 0} of f 0 352.3 80x103 40 x103 = 1/2 113.4 factor of sefety. If=