Principle Stress
Principle Stress
(OP)
I just need a quick sanity check to make sure I remember what I learned in my strengths of materials course. In order to determine if a part will fail under a static combined loading case, you first must calculate the principle stresses and apply a failure theory such as distortion-energy to determione the factor of safety against yielding (ductile material), correct? Thanks.





RE: Principle Stress
Yes thats one way to skin a cat.
regards desertfox
RE: Principle Stress
RE: Principle Stress
I'd say if you have calculated Maximum Principle Stress, then just apply this against the Ftu of the material. Also check the Max Shear against Fsu to ensure failure is not predicated by the shear mode.
jetmaker
RE: Principle Stress
I applied the two failure theories (distortion energy and max shear) using the principle stresses. I reason I bring this question up is because I have a case where I ahve two positive principle stresses. Applying the distortion energy theory give me a lower failure stress that the max principle stress (i.e. sigam1 = 120000 and sigma2 = 60000). I am having a hard time visualizing how the critical stress can be lower than one ot he principle stresses. However, using the max shear stress theory does indeed utilize the max principle stress since the theory states to use abs(sigma1) if sigma1 = sigma2. Am I making a simple mistake? Comments?
RE: Principle Stress
Your visualization troubles stem from the nature of "failure" in metals. They yield due to shear, so the max principle stress is meaningless for yielding. For your example (assuming σ3 = 0), then the equivalent stress is 104 000. If your material's yield stress is greater than 104 000, then your component will not yield.
Regards,
Cory
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RE: Principle Stress
Maxium principle stress is not meaningless, it just doesn't tell the whole story. As shown with Mohr's circle, maximum shear stress is dependent upon the principle stress values by definition. If you want to "see" the explanation check out this
http://www.efunda.com/formulae/solid_mechanics/fai...
It's also important to point out that if the material is not ductile, or the relative degree of ductility is unknown, then max principle stress (or minimum) may be the all important value.
RE: Principle Stress
My sentence you quoted is correct. Metals yield by shear. The maximum principle stress is irrelevant, however the square root of the sum of the squares of the differences of the principle stresses is important. Look at the von Mises (also known as the octahedral shear stress or distortion energy) criterion:
σe = {[(σ1-σ2)2+(σ2-σ3)2+(σ3-σ1)2]0.5}/√2
Assume a material's yield stress σy = 5
If σ1 = 5, σ2 = 0, & σ3 = 0, then σe = 5 and yielding occurs.
If σ1 = 10, σ2 = 9, & σ3 = 9, then σe = 1.2 and yielding does not occur.
Thus, maximum principle stress is not relevant to yielding, however the square root of the sum of the squares of the differences is relevant.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: Principle Stress
I agree with what you have presented mathematically. I was playing a bit of the devils advocate. However, I think it is fair to say that not all metals are ductile (cast iron jumps to mind). In the that case, it may be important to check the maximum normal stress criterion. I have found in my (limited) experience that it is best to check both for metals where ductility is in question, like with some steel alloys. This is completely the same as jetmaker suggested. I have no idea how to make the sigma's etc, so let me post a link.
http://www.efunda.com/formulae/solid_mechanics/fai...
Note the discussion of ductile vs. brittle. So a (completely) brittle material would yield in the examples you present. Certainly even a mildly brittle material would yield for the second example, right?
Also, to help a bit more with creigbm's visualization problem, I find Mohr's circle useful. It provides a graphical understanding of planar stress transformations from principle stress to any angle. And of course max shear occurs at 45 degrees between sigma 1 and sigma 2 (planar stress again). The link below is a neat little tool for Mohr's circle. If you are unfamiliar with it, it is important to keep in mind that angles are plotted as (2xactual) so you must divide the ange in half to make a physical understanding of the circle; e.g. the sigma2 is plotted at 180 degrees and max shear occurs on the plot at 90 degrees instead of 45.
http://www.aoe.vt.edu/~jing/MohrCircle.html
RE: Principle Stress
You seem to have a good understanding of this topic. My question to you is: how do you establish material failure at ultimate?
Regards,
jetmaker
RE: Principle Stress
"Failure" is a term that requires definition. Most people assume erroneously that it is equal to fracture. I suppose your question is how to establish fracture stress under multiaxial loading, since von Mises/distortion energy/octahedral shear stress is meaningful up to but not beyond yield. There is no easy answer. Physical testing is important. Many times one of the principle stresses dominates after yield. The Cockcroft-Latham criterion is one popular method for this type of analysis. The Cockcroft-Latham criterion was described in NEL Report Number 240, National Engineering Laboratory, Scotland, 1966. This criterion states that there is a critical (constant) value C* of the work done by the maximum (principle) tensile stress σt through the effective strain εe irrespective of deformation mode that causes fracture:
C* = ∫ ε σt dεe
Regards,
Cory
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RE: Principle Stress
a hollow tube being bent offers an interesting test for the above discussion.
RE: Principle Stress
The problem you run into starting from scratch is knowing what factor of safety to apply. If you don't know the accuracy of the failure criteria you are using, then you'll just wind up pulling a number out of the air anyway. You can get residual stresses due to the manufacturing processes used. You'll have some inaccuracies in your analysis. You may have local yielding under the initial load that does not constitute failure of the part.
Perhaps it would help to know, in a general way, what it is you're analyzing?
RE: Principle Stress
RE: Principle Stress
Some might say it's no big deal. However my view is that there is a principal at stake
RE: Principle Stress
RE: Principle Stress
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: Principle Stress
RE: Principle Stress