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Von Mises /Maximum distortion theorem?
2

Von Mises /Maximum distortion theorem?

Von Mises /Maximum distortion theorem?

(OP)
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

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

RE: Von Mises /Maximum distortion theorem?

Hi geguy

Your maths are correct however what the answers are saying is that in case 1 the vessel will yield at 100Mpa and in case 2 yielding will begin at 86Mpa.
the fornula is:-

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

from this formula your finding the yield for your case and because of the axial stress your yield stress for failure is reduced to 86Mpa not that its ok.


regards
desertfox

RE: Von Mises /Maximum distortion theorem?

desertfox implies that the yield strength varies with the stress. Yield strength is a material property for unixial tension and doesn't vary for that material. You have to look at the definition of Von Mises stress and the criteria for failure which is that when the energy of distortion reaches the same energy to cause yielding in unixial tension then it fails. In your case your interpretation of the effect of axial deformation is correct. The Von Tresca criteria would give you the same result of a stress intensity of 100 MPa incidentally, and so would have failed in either case. Von Mises has been shown to be more accurate for ductile material such as carbon steel and is the common method for assessment of structures. Some design codes however use the Von Tresca test as it is more conservative.

corus

RE: Von Mises /Maximum distortion theorem?

Hi

Corus the definition I have here for Von Mises states:-

for an element subject to the principal stresses S1,S2,S3
this theory states yielding begins when:-
    2*(stress(yield))^2= ((s1-s2)^2 +(s2-s3)^2 +(s3-s1)^2.

As the OP as transposed the formula to find the yield stress
for 2 cases ie one with only internal pressure and the other
with internal pressure and capped ends which introduces a longitudinal stress.
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.

regards

desertfox

RE: Von Mises /Maximum distortion theorem?

Maybe I are missing something here. In the first example, you did not account for the longitudinal (axial) stress, while in the second example you did. Even if the tube is "really long", the longitudinal stress will still be one half the hoop stress--if you don't think this is the case, then how is pressure being maintained in the tube? Even a really long tube must be capped off at the ends, otherwise the pressurized gas escapes.

RE: Von Mises /Maximum distortion theorem?

As DesertFox correctly states, the Von Mises-Hencky Equation states stress as a gradient of components compromising the stress vector.  Hence, 2S^2 = [S] X [S] where S=element stress, [S]=stress vector=Sxi + Syj + Skz in the three dimensional basis <i,j,k>.  Clearly, resolving the cross product of the stress vector is by cyclic permutation with the basis itself, that is, (Sx-Sy)^2 + (Sy-Sz)^2 + (Sz-Sx)^2.

There is a component to shear stress, which in practice usually dropped from the model.  The octagonal stress term is 6{[T].[T]}, where [T] is the strain energy of distortion associated with the shear vector, expressed as six times the dot product with itself.  In this forum thus far, the shear term has been dropped.  I surmise that the shear term is more important for the rectangular than for the circular cross section case.

So properly the Von Mises-Hencky or Effective Stress Model is expressed as 2S^2 = {[S] X [S]} + 6{[T].[T]}.  Most of the literature will simply collapse the second term to 6T^2, understood to be the total shear stress content experienced by the material element.  You easily see that shear ADDS to the overall stress content by a magnitude of six.

I have not worked out the situation of a rectangular cross section in Thick Wall Pressure Vessel theory.  What I do know is that I would expect heavy stress concentrations in the four corners of that cross section, something absent in the circular profile.  I am therefore lead to believe that your computation for the rectangular cross section would represent a material element sufficiently displaced from the corners of the profile.

Returning to the circular profile, I strongly disagree that longitudinal stress can be dropped.  Are you not holding or blocking pressure?  Then the end caps of the vessel or pipeline naturally induce longitudinal stress.  For the circular case, the Von Mises-Hencky Equation for triaxial states of stress using Thick Wall Pressure Vessel Theory reduce to: S = sqrt(3) P [R^2/(R^2-1)] where R=D/d and D=outer diameter, d=inner diameter of the circular profile.  P=internal pressure of the vessel or pipeline.

This will save you many pages of mathematics.

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

RE: Von Mises /Maximum distortion theorem?

isn't this a case of bi-axial stress, where the transverse tension stress is relieving the longitudinal stress (due to poisson effects) ...

RE: Von Mises /Maximum distortion theorem?

prost is quite correct in saying that a long tube will still produce a longitudinal stress just as a short one will. I think geguy is taking a hypothetical example where you can have a pure hoop stress with no longitudinal stress, as with an interference fit say.

I still say to desertfox that a material has only one yield stress and have yet to find a material that can have two or more in a unixial tensile test.  

I'm not sure what cockroach is referring to as the expression for Von Mises stress involves the principal stresses, which by their definition have no shear component. As such the shear term hasn't been dropped but is included in the derivation of the principal stresses. Also, I think when geguy refers to a rectangular section of the wall, they are considering an element of the wall, as an approximate rectangle, ie. as delta-r, delta-theta,delta-z. They aren't considering a rectangular cross-section so there are no stress concentrations in the four corners.

corus

RE: Von Mises /Maximum distortion theorem?

Hi corus

I am not suggeseting that a material has two yield stresses I have merely worked out the yield stress criteria for the two cases given or are suggesting that the formula cannot be transposed for the two different cases?

regards

desertfox

RE: Von Mises /Maximum distortion theorem?

The criteria for failure is that the Von Mises stress > yield stress (90 MPa). In the first case the stresses fail the criteria as the Von Mises stress is above yield. In the second it's less than yield and so is OK. The yield stress doesn't change.

corus

RE: Von Mises /Maximum distortion theorem?

When Corus, Cockroach, and Prost say there must be a longitudinal stress for Case 1, this is not necessarily true.  For example, if the tube is capped at one end and there is a piston that seals on the inside diameter, there will be no axial stress.

RE: Von Mises /Maximum distortion theorem?

Ugh: seem like equilibrium is not obtained without long. stress.

RE: Von Mises /Maximum distortion theorem?

i thought tlee had the pressure vessel end load carried by a piston to some other foundation so that the tube carried hoop stress only

RE: Von Mises /Maximum distortion theorem?

Sorry, I meant to say ... uhhhhhh.
 Isn't equilibrium on the cap violated, tlee123? A freebody of the cap itself tells me that there has to be a force counteracting the pressure on the cap. What force is that? In a closed pressurized cylinder, that would be the longitudinal stress in the shell of the cylinder.

RE: Von Mises /Maximum distortion theorem?

prost, not if you tied the crankshaft to the head (cap) and did not tie it to the piston sleeve.  The internal forces would bypass the sleeve and go straight to the "cap" (head) and thereby leave only hoop stress in the sleeve.  Think about it.

I agree with rb1957 - this is a case of poisson effects working against each other instead of in tandem.  If you recalculated with hoop stress and a compressive axial load on your cylinder, you'll find that the stresses are worse, i.e. S1 = 100, S2 = -50, S3 = 0.  This makes sense because the hoop stress is stretching in one direction while the compressive stress is compressing in the other direction, exacerbating the stretch from the hoop stress due to poisson.

RE: Von Mises /Maximum distortion theorem?

Hi corus

I ask again do you agree that the formula can be transposed to find the yield stress for both cases yes or no?


regards

desertfox

RE: Von Mises /Maximum distortion theorem?

no

corus

RE: Von Mises /Maximum distortion theorem?

Hi corus


If you look at the formula below:-

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

and check in in any mechanics book it clearly shows that the yield stress is denoted on the left of the = sign.
In order for the original OP to get his values of stress he had to divide the righthand side by 2 and find the square root of the righthand side that is transposition in my book.
As the symbol on the left is yield stress I go back to my original statement that the OP then calculate the yield stress critera for both cases, then compared them with a yield stress of a known material.
If the formula cannot be transposed then what are the stresses the the OP quoted and where did they come from?


RE: Von Mises /Maximum distortion theorem?

desertfox has written the vM failure criteria.  some people say there is no such thing as a vM stress, since vM is a failure criteria.

like corus, we plug the stresses into the RHS and if that produces a number less than the LHS (which is clearly a material constant) then everything is ok.

no ?

RE: Von Mises /Maximum distortion theorem?

(OP)
Hi,

Thank you for all your input on this.

I would like to make a few things more clear.

When I refer to "a rectangular section of the wall", I am indeed thinking of a hypothetical arc length by axial length.

I believe the von misses equivalent stress is limited by poissons ratio (in that there is only a given amount of material to deform)which seems reasonable.

I believe there may be some miss-understanding between yield stress (the point at which the material deviates from the elastic limit) and equivalent stress (the stress calculated using von misses.

--------------------------

I think I got my case 2 wrong assuming very long or infinately long pipes do not have axial stresses or do not require end caps.  Though there is a pressure drop (for there to be a drop one end must be at a higher pressure than the other) in very long pipes so maybe they do not require capping off- what do you think?

--------------------------

Thanks

GEguy

RE: Von Mises /Maximum distortion theorem?

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.

RE: Von Mises /Maximum distortion theorem?

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

RE: Von Mises /Maximum distortion theorem?

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.

RE: Von Mises /Maximum distortion theorem?

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.

RE: Von Mises /Maximum distortion theorem?

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, [S] X [S].

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

RE: Von Mises /Maximum distortion theorem?

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.

RE: Von Mises /Maximum distortion theorem?

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

RE: Von Mises /Maximum distortion theorem?

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

RE: Von Mises /Maximum distortion theorem?

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.

RE: Von Mises /Maximum distortion theorem?

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

RE: Von Mises /Maximum distortion theorem?

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) ..."




RE: Von Mises /Maximum distortion theorem?

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 σ1,σ2,σ3 this theory states that yielding begins
when:-


        (σ1-σ2)^2 + (σ2-σ3)^2 + (σ1-σ3)^2 = 2((σyp)^2

where σyp 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 σ1,σ2,σ3, satisfy the equation :- (σ1-σ2)^2 + (σ2-σ3)^2 + (σ1-σ3)^2 = 2((σy)^2

where σy is the yield point of the material.

Engineering Stress Analysis by D.N. Fenner page 54
quotes exactly the same formula with reference to σy

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


desertfox

RE: Von Mises /Maximum distortion theorem?

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 4Sm (Sm 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

http://www.xcalcs.com
Online tools for structural design

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