von mises stress vs maximum shear stress 1

[meher634]

hello group,
I have applied pure torque to the circular rod constrained at opposite end and examined the stresses, but i found that von mises stress so different from maximum shear stress,i got von mises as 75ksi whereas maximum shear as 55ksi. I am just wondering how the von mises stress is so different from maximum shear stress.There are no other forces acting on the circular rod. I believe the von mises stresses accounts for all the stresses.can any one describe this behavior?

 

[Drej]

Why do you sound as though you expect the von mises and max shear stresses to be the same? The shear only accounts for one component of load, whereas the vm stresses account for the bi/tri-axial load state (ie 2D or 3D), hence you would expect the vm stresses generally to be higher. Strictly speaking it depends on the load case and the contribution of each principal stress. Check each component of the principals and do the calcs below to convince yourself.

Maximum shear = (SIGMA1-SIGMA3)/2
Von Mises in 3D = root((1/2(SIGMA1-SIGMA2)^2+(SIGMA2-SIGMA3)^2+(SIGMA3-SIGMA1)^2))

-- drej --

 

[corus]

The shear stress should relate to half the stress intensity however this is calculated using the Von Tresca method (the maximum of the absloute value of the principal stress differences). Von Mises isn't a conservative way of calculating stress intensity though I believe it's commonly used for steel. Some design codes specify using the Von Tresca method, presumably because of its conservatism.

corus

 

[meher634]

Drej,
vm stresses account for all the stresses:compressive,bending and shear.but in case of pure torsion to circular rod there should be only pure shear stresses, so what accounts for the rest that vm stresses are higher than shear stresses.
thanks

 

[joekm]

"Von Mises Stress" is a bit of a misnomer. It is really a "strain energy density" measure. It just so happens that it's relatable to the ultimate tensile stress and, as such, is commonly referred to as "Von Mises stress". It's really just telling you the stability of the material and, for typical metals, does a pretty good job in comparison to test data.

Tresca, is sort of a simplified Von Mises criterion - sometimes referred to as "Tresca's Hexagon" due to it's shape. Where the Von Mises criterion plots as a 45 degree ellipse, the Tresca criterion would be a hexagon inscribed in that ellipse.

Other criterion like Tsai-Hill and Tsai-Wu are similar to Von Mises except that they manage materials with direction dependant properties. Because of that, it's easier to work the material properties into the equation so that the result is relatable to unity. Thus, we have Tsai-Hill and Tsai-Wu indexes.

 

[vonlueke]

meher634: Your numbers tell me you're looking at an element that is not in pure torsion. If you were, then maximum shear stress (tau_max) would equal the shear stress (e.g., tau_xy), and tau_max (and tau_xy) would equal 0.577 times von Mises stress (sigma_vm). Are you, by any chance, reading stresses adjacent to or anywhere near a nodal constraint or nodal applied load, instead of sufficiently far away from these stress concentrations that could skew your results?

 

[Drej]

Throughout your structure your vm contains components of both normal and shear stresses. If an element is subject to pure torsion only, the vm value will indicate this (because it will be dominated by this component). As vonlueke states, it may well be that you are comparing your vm with shear at a point having end effects - at a constraint say. If this is the case, your vm stress at the constraint will pick up on e.g. out of plane stresses, which will not give you a fair comparison.

Cheers,

-- drej --

 

[btrueblood]

Well, I tried to do this by sticking to the hand-calc formulas: from Young (Roarke's handbook), and a couple of other mechanics of materials references, the state of stress at the outer fiber is tau_xy, let's say that I calculate this to be 55 ksi. The book says that all other stresses are zero.

How do you then calculate Von Mises' stress?

What I did was use Mohr's circle, with (tau_xy)=55, and calculated the equivalent principle stresses sigma1=55, and sigma2=-55 (sigma3=0). Using these values, Sigma_VonMises would then equal 95.3 ksi. In pure shear, supposedly the VonMises (aka max. octahedral shear stress) and Tresca (aka max. shear stress) theories are supposed to agree; they are also supposed to agree at conditions of pure tensile/compressive loading. If this material had a tensile yield strength of 100 ksi, the max. shear stress theory would predict yielding, but the VonMises would not. Where is my error?

By the way, I know where the error is, but am looking all over the place for a good description of why, and how to properly calculate the VonMises stress for a shaft in torsion.

 

[vonlueke]

btrueblood: Maximum shear theory and von Mises theory should not be expected to agree; they agree at certain points, but not most points. Maximum shear theory is overly conservative relative to test data, whereas von Mises theory agrees reasonably well with experimental test data for ductile materials.

Where is your error? Aside from your impression that these two theories should agree, you didn't make one. Notice that the results for your example and my results (my first post) agree.

The formula for von Mises stress under a biaxial stress state is sigma_vm = sqrt(sigma1^2 - sigma1*sigma2 + sigma2^2).

 

[btrueblood]

Doh! I had to go back and use the Mohr's circle...

Somewhere I thought I'd been taught that the Tresca and Von Mises criteria agreed at conditions of uniaxial tensile load and at pure shear stress. Oops. They agree at uniaxial tensile and conditions of _zero_ shear.

Okay. Somewhere else I've seen people say that there are axial shear stresses in pure torsion (even with circular sections); I thought this was only true for non-circular cross sections?

 

[sgdon]

Tresca and von Mises agree at 6 points in a planar failure surface in terms of priniciple stresses. In pure torsion, tresca will predict 0.5fy and Mises 0.577fy.

 

[johnhors]

btrueblood and sgdon

There is no such thing as Principle Stress!
But you do have Principal Stresses.

 

[sgdon]

Thanks johnhors. Silly me!

 

[btrueblood]

Oops again. Will watch my language from now on!

 

[sasank]

Hallo! all

when we look at the stresses in a component, generally we ignore the stresses at the bolt hole locations. The reason for this is the way we model, the bolt hole(with RBE2's). My question is how do we judge the stresses around the bolt holes? In some cases we may not ignore them!!!

 

[johnhors]

Sasank

Simple don't use RBE2's and any other artifical, mathematical conveniences that don't relate to the real physical world. Unfortunately (for you!) this requires a FEA package that can handle contact analyses (eg. Abaqus, Lusas, Marc etcetera).

 

[SWComposites]

Sasank,

It also means that you have to explicitly model the hole and the bolt with enough detail to accurately determine the stresses around the hole, and have to get rid of the RBE2 elements and replace them with a reasonable appproximation of the joint flexibility. This will typically increase the size adn complexity of the model be an order of magnitude or more. The other alternative is to make a separate model of the joint region and apply loads/displacements from the global part model. Or just take the fastener loads from your current model and do a hand analysis of the local stresses, though I would be very cautious of loads that result from rigid elements such as RBE2s - they almost always result in models that are too stiff and artificially attract too much load. Joint load distributions from models using rigid elements to connect the joint memeber are almost always wrong - its actually much harder to correctly model a joint with rigid elements than it is with springs, beams, etc.

 

[BobM2]

Are you also constraining the bar axially? Under torsion the bar will want to shorten and you will also get tensile stresses.