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## Von Mise stress higher than Max Principal Stress?

Hi all, I am currently working on an analysis for a mechanical I designed, but when I run the design study, it gives me results showing the von mise is higher than the max principal. How can that be? I feel I have constrained it properly, but just wondering. Has anyone had this issue?

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## Re: Von Mise stress higher than Max Principal Stress?

Hi,

You need good constraints to get good answers but the relationship between Von Mises and  Principal stresses has nothing to do with the constraints whether the constraints be right or wrong.

It is possible to have biaxial Principal stresses greater than uniaxial yield without yielding (according to Von Mises).

It is possible to have zero Von Mises stress when the principal stresses are equal (hydrostatic).

I can't see how it is possible for Von Mises ever to be greater than the Max Principal.

This is maths. Has someone got a simple proof or the required logic?

More practically, Use measures to get the values of principal stresses and Von Mises stress at a point of interest.

Work out the Von Mises by hand.

## Re: Von Mise stress higher than Max Principal Stress?

If, for example, you have a Square block with uniaxial tensile stress, then von Mises = max principal. If you add a compressive stress from the sides, then the von Mises will be greater than the max principal, which stays the same. Look at the formula for von Mises. If I remember correctly, you will see something like Svm=0.5*sqrt( (S1-S2)^2+(S2-S3)^2+(S3-S1)^2), So you see that if S2, S3 are negative(i.e. comressive) then the term (S1-S2) will be greater than S1.

Sxx, Syy, S1, S2 etc.  are directional quantities, that can be either positive(tensile) or negative(comressive).

The von Mises (or Tresca) stress is a scalar number, always positive, and has no direction. Like "temperature". It tells you if you are close to, or above material yielding. If your calculated Svm is greater than the yield limit, obtained in a uniaxial test, then the material will yield. The purpose of the vonMises/Tresca yield criteria, is to determine if there is yield in a multiaxial stress state. If you pull the material in one direction and simultaneously compresses it in the perpendicular direction, then it is easier to obtain plastic deformation, compared to if only pull the material. The von Mises takes this into acount.

## Re: Von Mise stress higher than Max Principal Stress?

Yes Von Mises is unsigned; always positive and no direction.

You're right. (but the 0.5 is inside the sqrt ).

A simpler example would be where all principal stresses are compressive (negative).

Von Mises stress is positive; therefore greater. QED

Svm=sqrt(0.5*[ (S1-S2)^2+(S2-S3)^2+(S3-S1)^2])

if s1=s2=s3 (regardless of sign) Svm=0

if s1=1, s2=-1, s3=-1 (your example) then Svm=2

if we state that 1=yield and set s1=1.15, s2=0.5, s3=0 then Svm=1 and s1>uniaxial limit without failing

## Re: Von Mise stress higher than Max Principal Stress?

Excellent question,

Let us review the Von Mises Equation:

$\sigma_v^2 = \tfrac{1}{2}[(\sigma_{11} - \sigma_{22})^2 + (\sigma_{22} - \sigma_{33})^2 + (\sigma_{33} - \sigma_{11})^2 + 6(\sigma_{23}^2 + \sigma_{31}^2 + \sigma_{12}^2)]$
This equation defines the yield surface as a circular cylinder (See Figure) whose yield curve, or intersection with the deviatoric plane, is a circle with radius $\sqrt{2}k$, or $\sqrt{ \tfrac{2}{3}} \sigma_y$. This implies that the yield condition is independent of hydrostatic stresses.
Basically, all possible combinations of principal stresses (compression and tension) will form an ellipse. Each point of that ellipse is called "The Von Mises Stress" for each particular combination of principal stresses. With only two principal stresses the ellipse is as follow:
So, notice that no matter the combination of stresses, the most critical situation that you can find is that a principal stress has the same exact value than the von mises stress$\sigma_v = \sigma_1\!$ which means that you are in the edge of the ellipse, and your factor of safety according to Von Mises theory is exactly 1.
Of course, in real experiments a principal stress can be a little higher than the Von Mises stress, take a look at this graph and notice the points of failure outside the ellipse.
Now, the problem that you have. Remember that FEM it is only an approach to solve a quite complicate ordinary differential equation. You will not ever get the exact result. Also, it is your responsibility to check if the code of the engine solver is working right. Notice this about Creo Simulate:
"If the interpolating polynomial for the spatial variation of the displacement field is linear within an element, then the strains and stress within are constant. Furthermore, the stresses are only continuous (smooth) within an element. At the border of the neighbor element, stresses may become discontinuous (jump). This difference is usually smoothed away in the post-processor by different averaging techniques. This difference can also be used for error estimation and convergence improvement during the solution process. Simulate uses superconverged stresses for this purpose, as described in a later module."
So, check the elements, check the convergence, and check the simulation again, maybe what you get is a little difference caused by number noise. Otherwise, you should contact PTC directly to inform about this.
The other thing that may be happening, is that your piece is in failure, so check that also, hahaha.
Have a nice day!

## Re: Von Mise stress higher than Max Principal Stress?

 Ruben Alejandro Villarreal Barrios wrote: Excellent question, The answer is no.

What was the question?

(You make it sound like "No, Von Mises stress can never be higher than Max Principal stress" which is incorrect, as Charles Simpson pointed out in his example)

## Re: Von Mise stress higher than Max Principal Stress?

Oh but Charles Simpson point was exactly my point,

Quoting him "It is possible to have zero Von Mises stress when the principal stresses are equal (hydrostatic).

I can't see how it is possible for Von Mises ever to be greater than the Max Principal."

You can get higher principal stresses with the hydrostatic principle, yes. BUT! That's another question.

I asked two different teachers about if hydrostatic tension could cause failure.

The first one told me "No, you could through a screw into the ocean and hydrostatic pressures will not cause failures on it no matter the depth"

The second told me "If you go directly into a black hole where hydrostatic tension are infinite, something will happen to you"

The second, also told me that the Von Mises cylinder is not infinite, there are some theories that limit its region. I believe what the second teacher told me.

In the example below that answer, quoting it "If, for example, you have a Square block with uniaxial tensile stress, then von Mises = max principal. If you add a compressive stress from the sides, then the von Mises will be greater than the max principal, which stays the same. Look at the formula for von Mises."

Is wrong, adding another stress (either in tension or compression) will change automatically the value of the Von Mises Stress.

You understood my point. "No, Von Mises stress can never be higher than Max Principal stress". And I understand your point, if all stresses remains negative, YES, Von Mises (because of the sign) will be greater. But remember, we are talking here about absolute values. Thanks to you I can correct this.

"No, Von Mises Stress absolute value can never be higher than Max Principal stress absolute value"

However I forgot the most important thing here, notice in the last diagram that most of the failures occur INSIDE the ellipse. So the question is, it is your safety factor enough? Should you use Treska instead?

## Re: Von Mise stress higher than Max Principal Stress?

I was referring to this bit:

 Svm=sqrt(0.5*[ (S1-S2)^2+(S2-S3)^2+(S3-S1)^2]) if s1=s2=s3 (regardless of sign) Svm=0 if s1=1, s2=-1, s3=-1 (your example) then Svm=2