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fatigue calculations on shaft in torsion 1
- Thread starter gio1
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1
- #2
Whether to use von Mises, principal stress or some other criterion like Tresca depends on what material we are using, whether it is britle or ductile. I came across this problem some time ago and I seem to remember that for britle material like concrete, you use principal stress, and for ductile material, like steel you use von Mises. But must confess I do not remember it very well now.
Since your shaft is very likely to be a metal like steel, I think it will be ductile; and the appropriate stress should be von Mises.
I hope some others join in with their views.
Since your shaft is very likely to be a metal like steel, I think it will be ductile; and the appropriate stress should be von Mises.
I hope some others join in with their views.
Normm,
Von Mises criterion is only for strength assessment. For fatigue assessment, sigma 1 or sigma 3 to be used, regardless your material is brittle or ductile. Note that it is also important to determine the principal stress direction. Typically, fatigue crack initiation occurs in the direction perpendicular to the principal stress (sigma 1 or 3) direction. Cheers!
Yugabalan
Von Mises criterion is only for strength assessment. For fatigue assessment, sigma 1 or sigma 3 to be used, regardless your material is brittle or ductile. Note that it is also important to determine the principal stress direction. Typically, fatigue crack initiation occurs in the direction perpendicular to the principal stress (sigma 1 or 3) direction. Cheers!
Yugabalan
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- #5
personally i wouldn't use stress3-stress1 as a stress range since they occur in different directions and at the same time.
btw, isn't stress3 zero at the surface ? where stress1 and 2 are maximum, and where the fatigue failure will initiate ??
i'd look up Petersen (or similar) for shafts loaded in torsion and see what is the nominal stress.
btw, isn't stress3 zero at the surface ? where stress1 and 2 are maximum, and where the fatigue failure will initiate ??
i'd look up Petersen (or similar) for shafts loaded in torsion and see what is the nominal stress.
Stringmaker
Mechanical
Use principal stress as others have recommended. What does your stress cycle look like? Since torsion produces predominantly shear stress it makes sense that VonMises and Principal stresses don't match closely. In a state of tension/compression they should.
- Thread starter
- #12
The stress cycle is a simple +/- 5 degrees reversed twist (pure torsion) applied on a cantilever shaft.
Stresses are very easy to calculate and do not require FEA.
Using Tresca as the equivalent fatigue stress (as feajob suggests) is the same as using max/min principal stresses because for pure torsion Sigma1=MaxShear.
However this is not the most conservative, because Von Mises is definitely higher (for pure torsion:
VM=Sigma1 x sqrt(3)
and gives a hugely shorter fatigue life...
Of course VM is the most conservative, but which one is more correct?
Stresses are very easy to calculate and do not require FEA.
Using Tresca as the equivalent fatigue stress (as feajob suggests) is the same as using max/min principal stresses because for pure torsion Sigma1=MaxShear.
However this is not the most conservative, because Von Mises is definitely higher (for pure torsion:
VM=Sigma1 x sqrt(3)
and gives a hugely shorter fatigue life...
Of course VM is the most conservative, but which one is more correct?
the consensus would appear to be max. principal ... that's conservative enough given your loading, since the directions of the max. principal from +5deg and the min. principal for -5deg are so different.
if you wanted to fuss the matter, you could determine the stresses from mohr's circle and see which direction gives you the largest range (in axial stress)
if you wanted to fuss the matter, you could determine the stresses from mohr's circle and see which direction gives you the largest range (in axial stress)
According to Gio1, for pure torsion s3 and s1 is of the same magnitude. Why do we need to consider sigma2 in fatigue computation? The magnitude will not be as high as s1 and s3 in the first place.
I would say that a double amplitude stress range between s1 and s3 is valid in this case. I do realise that the stress direction is different between s1 and s3. But that is a secondary issue. The fact remains that the structure is responding to both s1 and s3 with same high magnitdude.
Now if the concern is with the stress direction, or more specifically crack initiation path, then my say is that the path would initiate in the resultant direction.
Yugabalan K
I would say that a double amplitude stress range between s1 and s3 is valid in this case. I do realise that the stress direction is different between s1 and s3. But that is a secondary issue. The fact remains that the structure is responding to both s1 and s3 with same high magnitdude.
Now if the concern is with the stress direction, or more specifically crack initiation path, then my say is that the path would initiate in the resultant direction.
Yugabalan K
I think...
(1) You have a material with a published fatigue strength (in bending).
(2) You correct this figure for surface finish, size of shaft, etc, etc.
(3) You multiply it by 0.577 to get endurance limit in shear.
(4) You divide this endurance limit in shear by your stress amplitude to get your safety factor (in your case, as there will be no mean stress as such, from what I understand of your description of the problem).
(5) If Tau(1)=-Tau(3) then your torsional stress amplitude is Tau(1) (this doesn't tally with what others are suggesting though).
All of the above assumes "full elasticity". Are there are stress concentrations on your shaft, for example?
Cheers.
(1) You have a material with a published fatigue strength (in bending).
(2) You correct this figure for surface finish, size of shaft, etc, etc.
(3) You multiply it by 0.577 to get endurance limit in shear.
(4) You divide this endurance limit in shear by your stress amplitude to get your safety factor (in your case, as there will be no mean stress as such, from what I understand of your description of the problem).
(5) If Tau(1)=-Tau(3) then your torsional stress amplitude is Tau(1) (this doesn't tally with what others are suggesting though).
All of the above assumes "full elasticity". Are there are stress concentrations on your shaft, for example?
Cheers.
- Thread starter
- #16
Correct me if I am wrong, but I think that sdra's approach is the same as using Von Mises as equivalent stress, as instead of multiplying the principal stress by sqrt(3) (to obtain Von Mises stress) he divides the endurance limit by the same number (1/sqrt(3)=0.577) so effectively the safety factor is the same. So sdra suggests Von Mises equivalent stress
Hi
I think there is a little bit of confusion with terminology here.
First let me apologise about the 0.577 figure: I wasn't trying to be misleading. What I was trying to suggest was that, given a fatigue STRENGTH in bending, von Mises theory allows a good estimate for fatigue STRENGTH in torsion. If you have fatigue data for your particular material in torsion, then this should be used instead. The von Mises theory correlation is very near though.
Like feajob and yugbalan have already stated, it is principal stresses that should be used for the calculation as, like you yourself have stated, Tau(1)=Sigma(1).
I think the terminology is getting confused between stress range and stress amplitude. yugbalan is talking about stress range, which does equal 2 x Sigma(1). Your stress amplitude, however, is simply Sigma(1) {or [[Sigma(1)--Sigma(3)]/2]. This is all tied up with whatever fatigue stress life assessment method you are to use but, as I mentioned earlier, for pure torsion with no mean stress, you should expect a fatigue failure once Tau(1) {or Sigma(1), or Sigma(Amplitude)} exceeds your fatigue STRENGTH in torsion.
Indeed, for steels at least, upto some limit on mean stress (which I assume to be material dependent), the mean stress itself has no effect on fatigue, for pure torsion only.
I think there is a little bit of confusion with terminology here.
First let me apologise about the 0.577 figure: I wasn't trying to be misleading. What I was trying to suggest was that, given a fatigue STRENGTH in bending, von Mises theory allows a good estimate for fatigue STRENGTH in torsion. If you have fatigue data for your particular material in torsion, then this should be used instead. The von Mises theory correlation is very near though.
Like feajob and yugbalan have already stated, it is principal stresses that should be used for the calculation as, like you yourself have stated, Tau(1)=Sigma(1).
I think the terminology is getting confused between stress range and stress amplitude. yugbalan is talking about stress range, which does equal 2 x Sigma(1). Your stress amplitude, however, is simply Sigma(1) {or [[Sigma(1)--Sigma(3)]/2]. This is all tied up with whatever fatigue stress life assessment method you are to use but, as I mentioned earlier, for pure torsion with no mean stress, you should expect a fatigue failure once Tau(1) {or Sigma(1), or Sigma(Amplitude)} exceeds your fatigue STRENGTH in torsion.
Indeed, for steels at least, upto some limit on mean stress (which I assume to be material dependent), the mean stress itself has no effect on fatigue, for pure torsion only.
gurmeet2003
Mechanical
sdra2,
If there was a mean stress, would there be a correction, for pure torsion? Some people recommend that for torsion mean stress correction should not be applied. I can see the logic in that because sum of principal stresses is zero.
sigma1+ sigma3=0
Gurmeet
If there was a mean stress, would there be a correction, for pure torsion? Some people recommend that for torsion mean stress correction should not be applied. I can see the logic in that because sum of principal stresses is zero.
sigma1+ sigma3=0
Gurmeet
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