Ups?
Yes, I was totally wrong about this. Von Mises stress can be QUITE HIGHER than principal stresses.
Why?
Principal stresses are a three dimensional representation of the stress state of a point, what Von Mises stress equation does is create an uni-axial equivalent stress based on these principal stresses. That "equivalent stress" can be higher than the principal stresses because you are transforming a 3 dimensional stress state into an UNIAXIAL EQUIVALENT stress state that you will use later to compare it with the yield stress and find the safety factor. This equivalent uniaxial stress has no physical components, and obviously has no direction. In your real part, you can't see it and in certain way, it doesn't exist.
I was wrong because I though that this "equivalent stress" was the result of an average stress of these principal stresses, which in such case can't be higher, but this is totally a misunderstanding about Von Mises stress theory and it's totally my fault. Patrick Asselman
Sorry for my mistake, I will edit my previous answer. Denis Jaunin
Now, I have questions,
Let's say that you graph the ellipse formed by the Von Mises stress equation. Can you graphically find the Von Mises Stress parameter? For example, when you graph a circle the equation is as follows: r^2= x^2 + y^2 and you know that r is the radius of the circle. The Von Mises stress equation for plane stresses forms an ellipse, but as with the circle, what does it mean the left side of the equation (Svm)? Is the major axis? The minor axis? Where is it?
And now we have just created this important issue:
Let us remember that Von Mises Stress theory is based on the deviatoric component of the stress state, so, according to Von Mises a piece will fail only if a deviatoric component exist, but what about hydrostatic stresses? I've been reading about this, and there is few information about it.
Norton (Strength of Materials Book) Says that rocks under the sea support such enormous pressure without fail (hydrostatic stresses), and when the same rock is tested under a uniaxial stress, it fails at lower values. So, what's happening here? How can you design something and predict failure for hydrostatic loads?
I know that there are low probabilities to design a piece that will support only hydrostatic pressure. But what about if that happens? You certainly cannot apply Von Mises stress theory in this case. It has no deviatoric component so it will not predict anything! What you should do in this case?
More important than this, what happens when principal stress has the same values? That's an hydrostatic case. And there are more probabilities to find this in real life.
And even more important than everything. Lets say that you have these principal stresses:
S1= 70
S2= 69.9
S3 = 70.1
It is quite similar to the hydrostatic case. And as we did with that case, we certainly cannot apply the von mises stress theory in this case.
Now you have the following situation:
S1= 70
S2= 69.8
S3= 70.2
As we did before, the deviatoric component is so tiny, that obviously Von Mises theory cannot be valid. Which lead us to this final question:
What do you use to decide that the deviatoric component is higher enough so Von Mises stress theory can be valid?