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Mechanics of Materials Lab. Lecture 17 Fatigue Mechanical Behavior of Materials Sec. 9.6-9.7 Jiangyu Li University of Washington. S-N Diagram. S-N Diagram. Endurance limit. Endurance Limit. For steel. Fatigue Failure Criteria. Effect of Mean Stress. Effect of Mean Stress.
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Mechanics of Materials Lab Lecture 17 Fatigue Mechanical Behavior of Materials Sec. 9.6-9.7 Jiangyu Li University of Washington Jiangyu Li, University of Washington
S-N Diagram Jiangyu Li, University of Washington
S-N Diagram Endurance limit Jiangyu Li, University of Washington
Endurance Limit For steel Jiangyu Li, University of Washington
Fatigue Failure Criteria Jiangyu Li, University of Washington
Effect of Mean Stress Jiangyu Li, University of Washington
Effect of Mean Stress Jiangyu Li, University of Washington
Fatigue Failure Criteria Multiply the stress By safety factor n Jiangyu Li, University of Washington
Example: Gerber Line AISI 1050 cold-drawn bar, withstand a fluctuating axial load varying from 0 to16 kip. Kf=1.85; Find Sa and Sm and the safety factor using Gerber relation Sut=100kpsi; Sy=84kpsi; Se’=0.504Sut kpsi Table 7-10 2 Change over 1 3 Jiangyu Li, University of Washington
Example: ASME Elliptic Table 7-11 2 3 1 Jiangyu Li, University of Washington
Torsional Fatigue Strength Jiangyu Li, University of Washington
Combination of Loads • Use Se from bending; • Apply appropriate Kf for each mode; • Multiply axial stress component by 1/kc • Find the principle stresses • Find von Mises effective alternative stress • Use the fatigue failure criteria to determine safety factor Jiangyu Li, University of Washington
Example A rotating AISI 1018 cold-drawn steel tube (42x4 mm) has a 6mm diameter hole drilled transversely through it. The shaft is subjected to a torque Fluctuating from 20Nm to 160Nm, and a stead bending moment 150Nm Estimate safety factor Sut=440 Mpa, Syt=370MPa, Se’=0.504x440 =222 Mpa, Se=166MPa Stress concentration factor Kt=2.366, from Table A-16, for bending, Kts=1.75 for torsion; from the Fig. 7-20, 7-21, notch sensitivity q is 0.78 for bending, and 0.96 for torsion. Thus Kf=2.07 for bending, and Kfs=1.72 for torsion Jiangyu Li, University of Washington
Example Tm=(20+160)/2=90Nm; Ta=(160-20)/2=70Nm Jiangyu Li, University of Washington
Assignment Mechanical Behavior of Materials 9.30, 9.33 Jiangyu Li, University of Washington