Differential Yield Stress and Von Mises Criterion

AI Thread Summary
The discussion centers on the relationship between differential yield stress (σd) and the von Mises yield criterion, particularly in the context of axial yield stress. It is established that when σ3 = 0, the axial yield stress can be equated to the differential yield stress. To derive the constant k for von Mises, it is noted that k equals 0.5 × (σ1 - σ3) or σd/√3. For the Tresca criterion, k is simply σd/2. The conversation raises questions about the utility of reporting differential stress, suggesting that it may be more accessible for those unfamiliar with the concepts, despite potential confusion regarding its application in calculating maximum principal stress.
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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