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Signed von Mises Stress vs Principal Stress Range for Fracture/Fatigue

flash3780 (Mechanical) (OP)
7 Dec 11 13:14
I'm working with one of our customers to determine the maximum allowable flaw size in our hardware during inspection to meet life requirements.

FEA models (and hand calculations) produce complex stress states which must be boiled down to a stress range for fracture/fatigue calculations. I've always used a signed von Mises stress to compute the stress range to be used in fracture/fatigue calculations (i.e. the von Mises stress given the sign of the principal stress with the largest amplitude). However, our customer uses the principal stress range between the principal stresses of the largest amplitude at each load state to determine the stress range.

I've heard of both approaches before, but I wonder which turns out to be more accurate in metals. Specifically, I'm working with titanium, but I'd imagine that the failure mode for most ductile materials is similar.

I find that fracture calculations in real geometries are often a bit of a crapshoot because you're applying a 3d stress state to a 2d crack growth model. Still, I'd be interested to know which method yields the most accurate results. Any takers?

(If it matters, the particular case that I'm looking at is constant-amplitude, proportional loading: I don't imagine that it would make much of a difference when choosing to use a signed von Mises range or a principal stress range, though.)

Christopher K. Hubley
Mechanical Engineer
Sunpower Incorporated
Athens, Ohio

salmon2 (Materials)
7 Dec 11 14:20
A little bit unclear about your customer's approach. Does stress range equal to "sigma_1 - signma_3"? Both signma_1 and sigma_3 are from the same one load condition. If yes, then it is Tresca (maximum) yield model. Tresca is more conservative or strict than von Mises, generally.

corus (Mechanical)
7 Dec 11 17:30
Principal stress range is always used. Von Mises stress is for a different failure criteria. Look up FE-Safe on google for examples and tutorials.

flash3780 (Mechanical) (OP)
7 Dec 11 18:10
I apologize for being unclear above, the two criteria are as follows:
Signed von Mises (for each stress state):

Principal stress range (for each stress state):
SPP =IF(ABS(S1)>ABS(S3),S1,S3)

VM = von Mises stress
S1  = Minimum Principal Stress
S3  = Maximum Principal Stress
SVM = Signed von Mises Stress
SPP = Largest magnitude principal stress

From another thread on Eng-Tips (circa 2005), I found this:

Quote (feajob):

for -1<alpha<0  Signed Tresca is very conservative
                Signed von Mises is conservative
for alpha=0 (uniaxial) both are O.K.   

for 0<alpha<1 Signed Tresca is O.K.
              Signed von Mises is non-conservative

for alpha=1 (equibiaxial) both are O.K.

Ref. MSC.Fatigue
From a similar thread:

feajob's statement agrees with the information provided by MSC.Fatigue here.

It looks as though the appropriate method is dependent on the stress ratio. I'll have to take that into account. According to MSC.Fatigue, the signed von Mises is okay to use in all situations except when "0 < (SAMP/SMEAN) < 1", when a signed Tresca Stress should be used. However, they do suggest that the absolute maximum principal strain (or stress in the linear regime) may be more accurate than the signed von Mises stress for "-1 < (SAMP/SMEAN) < 0".

They also note that in cases of pure shear, a critical plane method should be used, where the stresses/strains are determined based on a stress cube rotation to determine stresses in the direction which would cause a crack to open/close.

Christopher K. Hubley
Mechanical Engineer
Sunpower Incorporated
Athens, Ohio

flash3780 (Mechanical) (OP)
7 Dec 11 19:06
Ack, egads! Pardon my mistake: Alpha is NOT the same as the A-Ratio.

The biaxiality ratio is defined as:
"alpha = SP1_surf/SP2_surf"

Where SP1 and SP2 are the principal stresses in the plane of the surface. So it's necessary to rotate your stress state such that it lies along the surface, and then compute the planar principal stresses in the area of interest.

I think that makes sense...  

Christopher K. Hubley
Mechanical Engineer
Sunpower Incorporated
Athens, Ohio

corus (Mechanical)
8 Dec 11 4:06
Your definition of range is incorrect as it applies to different load cases. The only time your method is correct if you're looking at the stress range from a zero stress state.
Looking at the MSC Fatigue section, it states:

Stage I is a period of nucleation and crystallographically orientated growth following immediately after initiation and is confined to shear planes. In this phase, both the shear stresses and strains and the normal stresses and strains are the moduli which control the rate of crack extension.
Stage II growth is growth which occurs on planes which are orientated perpendicular to the maximum principal stress range. In this phase, the magnitude of the maximum principal stresses and strains dominate the crack growth process.

The way I read this is that you have Stage I, crack initiation, and Stage II, crack growth, which is dependent upon the principal stress range. As you're looking for the critical flaw size and/or fatigue life, then you're looking at crack growth, and as such the principal stress range.

rb1957 (Aerospace)
8 Dec 11 6:52
i'd use max principal.  how significant is the thru-thickness term ?
flash3780 (Mechanical) (OP)
8 Dec 11 16:54
Tara, can you elaborate on where I'm going wrong on the stress range? The particular case that I'm analyzing is a beam with a bolt tying it down at one end (like a diving board). There's a boss at the bolt connection to prevent fretting, and I'm looking at the stress in the fillet in that boss.

If you could follow all of that, I've attached a comparison of the signed von Mises and the range in the max amplitude principal stresses for the two stress states (as you travel along the 0.5mm fillet).

All that said, I suppose that I was looking for a general approach, rather than a solution to this specific problem; in this case, I'm merely using the most conservative of the two, which happens to be the principal stress range. I'm curious, however, which is the most accurate when compared with reality. Basing the choice on the biaxiality ratio seems like a reasonable approach... though maybe a bit tedious if you don't have fatigue software to do all of the stress rotations along the surfaces for you (s'pose I could write something, but... meh).

rb1957, the through-thickness term is essentially zero at the surface in this particular case (clearly you can't have normal stress pointing out of a surface; subsurface cracks on the other hand give me headaches).

Thanks for the help. I still think crack growth is a bit of a voodoo science in all but very simple cases (pressure vessels, and pipes for example), but we always do our best to be as accurate as possible. It seems like both signed VM and the principal stress range are good unless you're working with principal surface stresses which are either both in tension or both in compression (0<alpha<1), in which case MSC Fatigue suggests using a signed Tresca stress to define the state. Pure shear seems to be another tricky state which requires selecting a critical plane, and equal equibiaxial stresses (alpha=1) apparently give nonconservative results when you use the absolute principal stress range.

Christopher K. Hubley
Mechanical Engineer
Sunpower Incorporated
Athens, Ohio

rb1957 (Aerospace)
8 Dec 11 17:27
if both principals have the same sign then the combined stresses (like vM) are less than the principals.  if thru thickness is negligible then you've got 2D stress state.  for both reasons, i'd go with max principal.
corus (Mechanical)
9 Dec 11 3:58
It is slightly more complicated than simply taking the principal stress range, in that if the stress is compressive then it shouldn't contribute to crack propogation. This doesn't apply at welds due to residual stresses and 'partly so' in plain material. If I recall, fatigue design code standards advise using 60% of the compressive stress to allow for any residual stresses in the plain material from manufacture unless the stress cycle is wholly compressive when you'd ignore them. In your case you appear to have only one load case where the principal stresses are wholly compressive, and it presumably cycles from zero stress to the stresses quoted in your graph. How does a crack in plain material propogate in those circumstances?
When the principal stress direction changes then take the difference of the stress components (sxx, sxy etc) and calculate the principal stress range from those stress differences.
All this is in the the design standard BS7608, which states that you use principal stresses. MSC Fatigue also says that, as I said earlier.

flash3780 (Mechanical) (OP)
10 Dec 11 23:39
rb1957: Good point. I think the table on the MSC Fatigue page states that both the absolute max principal and the signed von Mises are anti-conservative when the signs of the principal stresses are the same (0<alpha<1): They're suggesting to use the Tresca stress in that case. For grins I plotted out signed von Mises vs absolute max principal for various biaxiality ratios (attached). Clearly, signed von Mises is more conservative for -1<alpha<0. That said, MSC Fatigue suggests that the max principal is more accurate in that range.

So, using the absolute max principal range seems to be fine unless you're in pure shear or have principal stresses of the same sign, at which point a max shear criteria is most accurate.

corus: I see what you're saying about compressive stresses not contributing to crack growth. I suppose that we only generally consider mode 1 crack growth in analyses. That said, they certainly do have an effect on fatigue lives (according to S-N curves, etc.). Perhaps lower fatigue lives for reversing stresses are due higher stresses driving mode 2 and mode 3 crack propagation in the .

All that said, there's a guy at Lockheed who's plugging the stress values that I give him into NASGRO, and I want to make sure that I'm giving him values that make sense. I'm life-ing my parts with S-N curves, as well. From the sound of it, S-N curves and crack growth models are perhaps interested in two different stress ranges. Is that correct?

Since crack growth models are interested in crack orientation and a particular stress state, I suppose that things get complicated. In my mind, the two-dimensional models that I'm familiar with are hard to relate to complex geometries. Still, I've not worked with predicting crack growth much, so perhaps I have a lot to learn.

Thanks for the insight, it really is helpful.

Christopher K. Hubley
Mechanical Engineer
Sunpower Incorporated
Athens, Ohio

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