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Mechanics of Materials Lab
Lecture 17 Fatigue Mechanical Behavior of Materials Sec Jiangyu Li University of Washington 1 Jiangyu Li, University of Washington
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Jiangyu Li, University of Washington
S-N Diagram 2 Jiangyu Li, University of Washington
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Jiangyu Li, University of Washington
S-N Diagram Endurance limit 3 Jiangyu Li, University of Washington
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Jiangyu Li, University of Washington
Endurance Limit For steel 4 Jiangyu Li, University of Washington
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Fatigue Failure Criteria
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Jiangyu Li, University of Washington
Effect of Mean Stress 6 Jiangyu Li, University of Washington
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Jiangyu Li, University of Washington
Effect of Mean Stress 7 Jiangyu Li, University of Washington
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Fatigue Failure Criteria
Multiply the stress By safety factor n 8 Jiangyu Li, University of Washington
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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 9 Jiangyu Li, University of Washington
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Example: ASME Elliptic
Table 7-11 2 3 1 10 Jiangyu Li, University of Washington
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Torsional Fatigue Strength
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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 12 Jiangyu Li, University of Washington
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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 13 Jiangyu Li, University of Washington
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Jiangyu Li, University of Washington
Example Tm=(20+160)/2=90Nm; Ta=(160-20)/2=70Nm 14 Jiangyu Li, University of Washington
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Jiangyu Li, University of Washington
Assignment Mechanical Behavior of Materials 9.30, 9.33 15 Jiangyu Li, University of Washington
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