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Linköpings Universitet, Hållfasthetslära, IEI/IKP Tore Dahlberg ...

Linköpings Universitet, Hållfasthetslära, IEI/IKP Tore Dahlberg ...

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Tekniska Högskolan i Linköping, <strong>IEI</strong>/<strong>IKP</strong><br />

<strong>Tore</strong> <strong>Dahlberg</strong><br />

TENTAMEN i Mekaniska svängningar och utmattning, 071016 kl 14-18<br />

EXAMINATION in Mechanical Vibrations and Fatigue<br />

DEL 2 - (Problemdel med hjälpmedel)<br />

7.<br />

En stor plåt, belastad med en en-axlig spänning,<br />

spänning/stress (MPa)<br />

utsätts för en belastningssekvens enligt figur.<br />

300<br />

Denna sekvens upprepas. Materialet har en<br />

Wöhlerkurva som ges av sambandet<br />

200<br />

σ a<br />

=−50 logN + 400 (MPa)<br />

100<br />

där σ a är spänningsamplituden. Bestäm<br />

förväntat antal sekvenser till utmattningsbrott.<br />

0<br />

Använd Palmgren-Miners delskadehypotes.<br />

tid/time<br />

Inverkan av spänningens medelvärde får<br />

försummas.<br />

English:<br />

7. A large plate, loaded in uni-axial tension, is subjected to a load sequence<br />

according to the figure. This sequence is repeated. The material has a Wöhler<br />

curve given by the equation<br />

σ a<br />

=−50 logN + 400 (MPa)<br />

where σ a is the stress amplitude. Determine the expected number of sequences<br />

to fatigue failure. Use the Palmgren-Miner damage accumulation rule. The<br />

influence of the stress mean value can be neglected.<br />

Lösning/Solution:<br />

Rain-flow count gives<br />

1 cycle from 0 to 300 MPa, giving σ a = 150 MPa,<br />

1 cycle from 50 to 300 MPa, giving σ a = 125 MPa,<br />

2 cycles between 50 and 250 MPa, giving σ a = 100 MPa, and<br />

1 cycle between 50 and 200 MPa, giving σ a = 75 MPa.<br />

These stress amplitudes give (from the Wöhler curve) N = 100 000, 316 228,<br />

1000000, and 3 162 278 cycles, respectively<br />

The Palmgren-Miner damage accumulation rule gives<br />

1<br />

D =<br />

100 000 + 1<br />

316 228 + 2<br />

1 000 000 + 1<br />

3 162 278 = 1<br />

64 600<br />

Thus, failure is expected after approximately 64 000 sequences (giving 320<br />

000 cycles).<br />

12

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