Von Mises /Maximum distortion theorem? 2

[geguy]

Hi,

Equivalent stress = (0.5^0.5))*((s1-s2)^2 +(s2-s3)^2 +(s3-s1)^2)^0.5

I am using the above formula to add principle stresses.

CASE 1
If I have a very long tube and want to find the stress in the wall.
I get hoop stress = 100 MPa. (Neglect radial stresses).
Therefore if my material yield stress is 90 MPa, tube will fail.

CASE2
Now I have a short tube with its ends capped off, so I have a hoop stress of 100MPa and a longitudinal/axial stress of say 50 MPa. Using Von Misses (s1=100, s2=50, s3-0) equation I get 86 MPa
Therefore if my material yield stress is 90 MPa, my tube will be OK.

Whats going on?

In case 1 the material only has 1 (2 include radial) stress acting on it, yet it will fail.
Where as in case 2 (where I would have thought it to be less safe) the 2 (3 inc radial) stresses combine to give a lower equivalent stress.

Now if I think about a rectangular section of the wall of each tube (& ignore radial stress)

CASE 1
The rectangle is being stretched around the circumference of the tube

CASE2
The rectangle is being stretched around the circumference of the tube and pulled axially

Is it the fact that there is a given amount of material that can deform, and (CASE2) by being pulled axially there is less material to deform circumferentially (therefore less strain & hoop stress)?

Or is there some other explanation for the lower equivalent stress in CASE 2?

Thank you

Guy

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http://www.mysite.com/images/happy.gif

 

[rb1957]

looking at your original post, i think you're talking about case 1 (not having longitudinal stresses).

anyways, i thought the point to the problem (oops, question) was that bi-axial tension relieves the uni-axial stresses (due to poisson effects) and so the bi-axial stress state may be safe even if the uni-axail component is unsafe (> yield).

note, too, the other conclusion; that tension/compression loading is more severe than uni-axial tension.

 

[corus]

desertfox,
If you look at this web site

http://www.efunda.com/formulae/solid_mechanics/failure_criteria/failure_criteria_ductile.cfm

then you'll see a definition of Von Mises yield criteria. There is clearly shows that the yield stress is greater than or equal to the function of principal stresses. It does not state that the yield stress is equal to that function as you quote. In your post of 16 July you clearly state that :

"If you use the formula and transpose to find the yield stress you will get the same results as the OP that yield stress for case 1 is 100mpa (failure) and that for case2 86Mpa (failure). The reason for the lower yield stress in case 2 is that a longitudinal stress as been introduced.
The OP seems to think that case2 is safe and case1 as failed.
What I am saying he has found the yield stress for both cases."

He hadn't found the yield stress for both cases and the yield stress wasn't lower in one case than the other. In your post of 19 July you then state that he has calculated the yield stress criteria with the known yield stress of that material. In that you are correct.

PS
Please don't use acronyms when when they're generally not known or needed. An OP is an Old Person in my book and reference to age should not be tolerated, anywhere. (If that's what you're referring to when using OP)

corus

 

[rb1957]

i thought OP meant "Original Poster", and i thought it was in pretty general usage o this site. i doubt that desertfox has any inside knowledge about geguy to refer to him (assumption) as an old person, and in context "original poster" is reasonable.

 

[prost]

Is there a problem with semantics here? The yield stress, defined most often as the stress when a uniaxial tension test causes a 0.2% strain (which of course never changes except for stochastic variations), is not normally the stress that causes yield in a structure with multiaxial states of stress. People use the phrase "Von Mises stress," or "equivalent stress" to suggest an engineering quantity that can be calculated in the more general case of multiaxial stress _and_ compared to the "yield stress" obtained in the uniaxial tension test--this comparison is the test for a particular Failure Theory, the Maximum Energy of Distortion. I've heard this also called 'the von Mises failure theory.' Recognize that although Max. Energy of Distortion is used quite often, that this theory is just one of many theories for estimating failure in structures with multiaxial stress states.

I would distinguish the 'yield stress' which doesn't change, from the 'stress of yield', which can change, depending on the structure. 'Stress of yield' in my definition is the _applied stress_ that causes the von Mises stress to go above the value of the 'yield stress' somewhere in the structure. In this usage, the 'stress of yield' could be the pressure in a pressurized container that causes the von Mises stress to go above the 'yield stress' someplace in the container.

 

[Cockroach]

Corus, this is the "effective stress" equation given in Advanced Mechanics of Materials, Cook & Young ISBN 0-13-396961-4, pg 41 as equation [2.6-12]

Se = (1/sqrt(2))[(Sx-Sy)^2 + (Sy-Sz)^2 + (Sz-Sx)^2
+ 6(Txy^2 + Tyz^2 + Tzx^2)]^0.5

We can clearly see the VonMises-Hencky model associated with the strain energy of distortion in the deviatoric state. Obviously there is a shear associated with the normal stress matrix, X .

Typically I have seen noted sites such as API, et al using the shear term as 6T^2 rather than 6(Txy^2 + Tyz^2 + Tzx^2).

But you are also correct in noting principle stresses have no shear component. Clearly the usage of principle stresses will reduce this second term to zero, hence the more conventional equation found in the literature, 2 S^2 = grad S for S=Sx i + Sy j + Sz k in the triaxial basis <x,y,z>.

I don't totally agree with everything said in the forum, I do support DesertFox position in the application and interpretation of the theory to the problem at hand. The dropping of longitudinal stresses from the equation thus reducing triaxial to biaxial state, I believe, is not valid. We are not talking of movable pistons in cylinders, I have treated the problem as a pressure vessel, ends closed and capped. In the case of infinitely long vessels such as endless pipelines, I think are not practical since you are blocking and retaining pressure over a length considerably shorter than infinite.

But I've enjoyed this discussion/thread, which is why this forum is of use to all of us as Engineers.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[desertfox]

Hi

Thanks rb1957 for your comments on the term OP you have hit the nail on the head.

Thanks also to cockroach for his comments.

Corus I have looked at the site and yes I agree that it states what you have said in your last post.
However if you look at this site then it shows a similar formula but with an = sign.

http://www.engineersedge.com/strength_of_materials.htm

I would point out that as far as I am concerned you can calculate Von Mises yield criteria without any reference to a particular material, therefore that would be a yield situation for a given set of circumstances, you would then look at the yield strength of a particular material to see whether it could be used or not.
The fact remains that if you put different figures into the righthand side of the equation it still equals stress yield for that set of circumstances.
the fact the the numbers on the righthand side of the equation doesn't arrive at a exact value of yield stress for a particular material is irrelevant.

 

[corus]

desertfox,
The formula on your site quotes the formula for Von Mises stress or design stress, and not the stress yield, as you say. If you read the text above the formula it states that yielding occurs when the Von Mises stress (as calculated on the formula) exceeds the yield stress in tension. If the material had failed and you had put stresses that had caused yield into the formula then all it would show is that yield had been exceeded. It doesn't equal yield for that set of circumstances. There is a big difference between equal (or not quite the exact value as you infer) and being greater than a value.

corus

 

[desertfox]

corus

I stated in the 19th of july post he compared his calculated yield stress with a known yield stress of a material and not that he had calculated with a known yield stress.

Further if we take the OP,s original figures 100Mpa and 86Mpa and we had materials with those values of yield stresses respectively, would you then use those materials
to manufacture components ?
I think not.
Therefore when you calculate Von Mises you are calculating a yield stress or elastic limit and you would not choose to use a material with a yield stress equal to that which you have calculated.

desertfox

 

[rb1957]

what a glorious bun fight !

"ding, ding, ... 2nds out, round n+1 ... "

doesn't vM define a failure criteria, such that if the equivalent stress caused by the applied local stresses, combined by the formula, equals the yield strength of the material then failure is assumed to occur.

in case 1, the vm "stress" is 100ksi > 90ksi yield strength of the material; not good.
in case 2, the vM "stress" (reduced by bi-axial tension) is 86ksi < 90 ksi yield strength; therefore ok.

btw, i disagree with cockroach in that the OP (ie Original Poster) defined the problem, and we just tried to create a physical situation that would account for the problem.

 

[corus]

No desertfox, he didn't calculate a yield stress. If you look at the web site you quoted it has the expression
2*(Stress(Von Mises))^2= ((s1-s2)^2 +(s2-s3)^2 +(s3-s1)^2
(or equivalent thereof) and not the expression you quoted of
2*(stress(yield))^2= ((s1-s2)^2 +(s2-s3)^2 +(s3-s1)^2. I think your text book must be wrong or you've misinterpreted it.

Your last sentence is almost correct but I'd alter it to say "Therefore when you calculate Von Mises stress you would not choose to use a material with a yield stress less than or equal to that which you have calculated."

rb1957 is quite correct when he/she says that Von Mises stress defines a failure criteria. Simply that.

To return to the Old Person's question, it is an interesting subject and one that I'm not altogether convinced of. Take for instance the case of tri-axial stress where hypothetically all the principal stresses are all equal to 10^6 MPa (for example) or a squillion psi for those still working in imperial units. Both Von Mises and Von Tresca would be equal to zero and so would satisfy the failure criteria of both tests. Can't be right surely?

corus

 

[tlee123]

What an interesting thread.

It works out that for a given principal stress s1 that the minimum possible von mises stress occurs when s2=s1/2, which just happens to be the condition you get with a cylinder with end caps under uniform pressure. This assumes bi-axial stress only (s3=0).

This (s2=s1/2) can be shown easily by using a spreadsheet or by:

1. Take the part of the von mises formula (s1-s2)^2 +(s2-s3)^2 +(s3-s1)^2.

2. Set s3=0.

3. Expand terms. 2*s1^2-2*s1*s2+2*s2^2.

4. Differentiate with respect to s2 (assume s1=constant) and set equal to 0.

5. Simplify to s2 = s1/2

For the physical explanation, I think rb1957 got it right a long time ago when he said, "isn't this a case of bi-axial stress, where the transverse tension stress is relieving the longitudinal stress (due to poisson effects) ..."

 

[desertfox]

Corus

I didn't state that you wouldn't use a material with a yield lower than that calculated as I thought that was obvious.

Schaum's Outline Series
Theory & problems of Strength of materials 2nd edition

page 349

Von Mises Theory:-

For an element subject to principal stresses [&#963;]1,[&#963;]2,[&#963;]3 this theory states that yielding begins
when:-


([&#963;]1-[&#963;]2)^2 + ([&#963;]2-[&#963;]3)^2 + ([&#963;]1-[&#963;]3)^2 = 2(([&#963;]yp)^2

where [&#963;]yp is the yield point of the material.

ref:- Mechanics Of Materials 2nd Edition By EJ Hearn Vol 1 page 404 (15.4)

ref:- Formula's for Stress & Strain 5th edition by Roark & Young page 24 quote:-

Theory of Constant Energy of Distortion,which states that elastic failure occurs when principle stresses [&#963;]1,[&#963;]2,[&#963;]3, satisfy the equation :- ([&#963;]1-[&#963;]2)^2 + ([&#963;]2-[&#963;]3)^2 + ([&#963;]1-[&#963;]3)^2 = 2(([&#963;]y)^2

where [&#963;]y is the yield point of the material.

Engineering Stress Analysis by D.N. Fenner page 54
quotes exactly the same formula with reference to [&#963;]y

I can't believe that all these references are wrong too.


desertfox

 

[prex]

This is my answer to the question raised by the OP:
-the failure criterion of von Mises (or of Guest or of Ros-Eichinger) is known to best represent experimental failure data for biaxial states of stress where the failure mechanism is the excessive deformation or yielding of plastic materials
-however the preceding statement, though I didn't find this in a short and fast literature search, is valid only for linear structures (beams, columns) and fails for plane structures (plates and shells)
-for the latter the Tresca criterion is used (this is the basis of the well known formula S=PD/2t for a circular shell) and is indeed enforced for stress analysis by most pressure vessels codes
-I don't think there is any Poisson effect in a pipe with and without a longitudinal stress (by the way the value of the Poisson's ratio has no influence on the stresses in this case): both pipes will fail at substantially the same pressure when the circumferential stress reaches the yield stress (though rupture will in fact occur at a somewhat higher pressure due to strain hardening)
-the case of uniform pressure that has been raised by Corus is treated by Div.2 by the additional criterion that the algebraic sum of the three primary principal stresses shall not exceed 4S_m (S_m being the allowable stress): one should recall that failure criteria such as those of von Mises or Tresca have been developed for biaxial states of stress, and are not necessarily applicable to general triaxial states

prex

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