maixmum principal stress greater than von mises stress

[meher634]

hello,
I ran my static analysis on curved beam subjected to bending loads and found that the maximum principal stress is greater than the vonimises stress at a particular node.Both the stresses below the yield st. Can any one explain what accounted for this differnce.
meher.

 

[nagi]

Hi Meher,

That is perfectly acceptable. If the 3 principal stresses are S1, S2 and S3, the von Mises stress

Sv = sqrt( (S1-S2)^2 + (S1-S3)^3 + (S2-S3)^2 )/sqrt(2).

So, the principal stresses can be greater than the von Mises stress. Consider, for example, S1=S2=S3=1000. The von Mises stress in that case is 0!!

Nagi

 

[bvi]

It is hard to concieve of how a beam, albeit curved, can have a high stress triaxiality (all three principal stresses in the same direction, and having relatively similar values). So, from that respect, you may want to double check your results. Of course, as nadi explained, you can have principal stresses higher than the von mises stress intensity when the principal stresses are all the same sign (tension or compression).

Note that high stress triaxiality is associated with low ductility behavior, and some codes (e.g. Section VIII, Div 2, Appendix 4) specify limits on the sum of the principal stresses.

 

[nagi]

You're right cb4 the principal stress can't be that much higher than the von Mises stress for a curved beam since it is unlikely that the 3 principal stresses in that case be all comparable in magnitude (and same sign).

Meher, what was the value of the maximum principal stress, the v.Mises stresses and the other 2 principal stresses? They should satisfy the equation I stated earlier.