Differential Yield Stress and Von Mises Criterion

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Discussion Overview

The discussion revolves around the relationship between differential yield stress and the von Mises yield criterion, particularly in the context of axial tension and plane stress conditions. Participants explore how to derive axial yield stress from differential yield stress and the implications of these relationships in material behavior.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether the axial yield stress is equal to the differential yield stress or if it should be calculated as √3 × σd / 2, given that σ3 = 0 in axial tension.
  • Another participant references external sources to support the idea that the axial yield stress equals the differential yield stress, suggesting a division by √3 to derive the parameter k.
  • There is a repeated concern about the rationale for reporting differential stress, with a participant expressing confusion over its necessity when axial yield stress could be added to hydrostatic stress to determine maximum principal stress.
  • A participant notes that the use of differential stress may be intended to make the concept more relatable to those less familiar with the topic.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between axial yield stress and differential yield stress, with no consensus reached on the correct interpretation or application of these concepts.

Contextual Notes

There are unresolved assumptions regarding the definitions of yield stress in different contexts and the implications of using differential stress versus axial yield stress in calculations.

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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