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This document discusses potential sources of error in creep testing, such as temperature fluctuations, light variability, and humidity changes. It explains the importance of analyzing creep experiments by differentiating between primary, secondary, and tertiary compliance, utilizing master curves for predictions at various temperatures. Additionally, the fatigue testing section covers cyclic stress loading applications, highlighting the significance of the Wöhler curve in estimating the life expectancy of components under cyclic loading conditions. Safety factors for fatigue testing are emphasized.
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2 a) possible additional sources of error (creep test) • temperature not constantover time • light (opticalstraingages!) maybe not constant • humidity (sample affected!) not constant • loadordisplacementmay not beconst. • machine/instrumentcreeping • creepingofstraingage • etc. (day / night!)
2 b) Analyzing a creep experiment tertiary secondary primary
creepcompliance notethatthevaluesofearegiven in %, so youhavetodividethemby 100 to evaluatecreepcomplianceisonlyuseful in thesecondaryregion (markedby ) D is in therangeof 1/MPa
2 c) i. result: mastercurveat 120°C mastercurveat 120°C
2 c) ii. predictcompliance after 5 days mastercurveat 120°C
mastercurveat 120°C 5 d = 5*24*60 min = 7200 min
D(5 d) ≈ 1/psi mastercurveat 120°C 5 d = 5*24*60 min = 7200 min
2 c) iii. possiblepredictionrangeat 120°C mastercurveat 120°C
mastercurveat 120°C 40000 min = = 0,9 months = = 0,1 years
forcomparison: possiblepredictionrangeat 90°C mastercurveat 90°C 3e+9 min = = 70000 months = = 5700 years
3. Fatiguetesting a) threeexamplesofapplicationscyclic stress loading wingsofairplanes turbine blades rotor blades (nicht sehr schwer, wenn zwei davon schon in b) genannt sind)
b) Wöhler curveandestimationoflife time (aim: toshowthatcyclicloadedcomponents must havehugesafety factors, iffatiguetestingis not performed) i. N(40 MPa) ≈ 9000 ii. N(27 MPa) ≈ 300000 i. N(40 MPa) ≈ 9000 tlife = N/f = 9000 * 1 / 0,5 s = 18000 s = = 5 hours ii. N(27 MPa) ≈ 300000 tlife = N/f = 300000 * 1 / 1000 s = 300 s = = 5 minutes