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.
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Principle Stress 1
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Jetmaker,
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?
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?
Creigbm,
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 [σ]3 = 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
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
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 [σ]3 = 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
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
I do not think it is right to say outright that, "They yield due to shear, so the max principle stress is meaningless for yielding."
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
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.
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
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.
YoungTurk,
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:
[σ]e = {[([σ]1-[σ]2)2+([σ]2-[σ]3)2+([σ]3-[σ]1)2]0.5}/[√]2
Assume a material's yield stress [σ]y = 5
If [σ]1 = 5, [σ]2 = 0, & [σ]3 = 0, then [σ]e = 5 and yielding occurs.
If [σ]1 = 10, [σ]2 = 9, & [σ]3 = 9, then [σ]e = 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.
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:
[σ]e = {[([σ]1-[σ]2)2+([σ]2-[σ]3)2+([σ]3-[σ]1)2]0.5}/[√]2
Assume a material's yield stress [σ]y = 5
If [σ]1 = 5, [σ]2 = 0, & [σ]3 = 0, then [σ]e = 5 and yielding occurs.
If [σ]1 = 10, [σ]2 = 9, & [σ]3 = 9, then [σ]e = 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.
Cory,
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.
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.
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.
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.
jetmaker,
"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 [σ]t through the effective strain [ε]e irrespective of deformation mode that causes fracture:
C* = [∫] [ε] [σ]t d[ε]e
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
"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 [σ]t through the effective strain [ε]e irrespective of deformation mode that causes fracture:
C* = [∫] [ε] [σ]t d[ε]e
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
You might note that in many cases, there is some sort of design code that specifies allowable stresses (usually just based on maximum tensile, compression, or shear 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?
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?
EnglishMuffin
Mechanical
I realize that this is not the grammar forum, but just for the record, it's not "principle stress" - it's "principal stress" ! "Principal" as used here is an adjective, although it can also be used as a noun - but "principle" can be used only as a noun or verb.
Thank you English Muffin. I had been waiting for several days, hoping for someone to point this out, yet every contributor who attempted the word compounded the mistake.
Some might say it's no big deal. However my view is that there is a principal at stake (-:
Some might say it's no big deal. However my view is that there is a principal at stake (-:
EnglishMuffin
Mechanical
Denial : No, you mean "there is a principle at stake"!
Ugghh... I had a feeling I was using the wrong version, but I just used what was already in the post. Also, my last sentence from my Sept. 22 post used information from a recent journal article in which "principle" was used rather than "principal". Not that I was correct, just mindlessly propagating an error. Ugghh...
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
EnglishMuffin
Mechanical
Sorry - my apologies - didn't get it ! I guess I'm losing my sense of humor.
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