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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[desertfox]

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

 

[corus]

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

 

[desertfox]

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

 

[prost]

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.

 

[Cockroach]

As DesertFox correctly states, the Von Mises-Hencky Equation states stress as a gradient of components compromising the stress vector. Hence, 2S^2 = X where S=element stress, =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 = { X } + 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

 

[rb1957]

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

 

[corus]

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

 

[desertfox]

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

 

[corus]

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

 

[tlee123]

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.

 

[prost]

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

 

[rb1957]

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

 

[prost]

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.

 

[swearingen]

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.

 

[desertfox]

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

 

[corus]

no

corus

 

[desertfox]

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?

 

[rb1957]

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 ?

 

[geguy]

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