Differential Yield Stress and Von Mises Criterion

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SUMMARY

The discussion centers on the relationship between differential yield stress (σd) and the von Mises yield criterion. It is established that the axial yield stress is equal to the differential yield stress when σ3 = 0, and that k can be derived as k = σd / √3. For the Tresca criterion, k equals σd / 2. The use of differential stress is emphasized as it provides a more relatable metric for those unfamiliar with the complexities of stress analysis.

PREREQUISITES
  • Understanding of differential yield stress (σd)
  • Familiarity with the von Mises yield criterion
  • Knowledge of the Tresca yield criterion
  • Basic concepts of axial and hydrostatic stress
NEXT STEPS
  • Research the derivation of the von Mises yield criterion in detail
  • Study the implications of using differential yield stress in material science
  • Explore the differences between the von Mises and Tresca yield criteria
  • Investigate applications of yield criteria in engineering design
USEFUL FOR

Mechanical engineers, materials scientists, and students studying stress analysis and yield criteria in materials. This discussion is particularly beneficial for those looking to deepen their understanding of yield stress concepts and their applications in engineering.

1350-F
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If I have the differential yield stress σd=(σ13) for a material, what can I derive from that for use with the von mises criterion? Do I have k=0.5×(σ13) or σy in axial tension, where σ3 = 0? Essentially is my axial yield stress = σd or √3 × σd /2? (in plane stress)
 
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Chestermiller said:

Reading this confirmed my suspicions that the axial yield would equal the differential yield and from there you would divide by √3 to get k. For tresca k would equal σd/2. In that case why report it as a differential stress? I think that's what was throwing me off. Assuming we would know to add the axial yield to any hydrostatic stress to get our maximum principal stress why give σd?
 
1350-F said:
Reading this confirmed my suspicions that the axial yield would equal the differential yield and from there you would divide by √3 to get k. For tresca k would equal σd/2. In that case why report it as a differential stress? I think that's what was throwing me off. Assuming we would know to add the axial yield to any hydrostatic stress to get our maximum principal stress why give σd?
The reason differential stress is used is that people out-of-the-know can relate to it easily.

Chet
 

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